Datasets:

minimario

/

math-openwebmath-retrievals

like 0

[ "4 \\\\ 3 \\end{pmatrix}.$ ×", "& \\frac{1}{2} \\end{pmatrix} \\cdot \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} 1-\\sqrt{3} \\\\ 1+\\sqrt{3} \\end{pmatrix}$$", "= x\\cdot \\begin{pmatrix} 1 & 2 & 3 \\end{pmatrix} + y\\cdot \\begin{pmatrix} 6 & 5 & 4 \\end{pmatrix} + z\\cdot \\begin{pmatrix} 7 & 8 & 9 \\end{pmatrix}.$$", "\\begin{pmatrix}1&0\\cr0&0\\end{pmatrix} \\cdot \\begin{pmatrix}0&1\\cr0&0\\end{pmatrix} = \\begin{pmatrix}1&0\\cr0&1\\end{pmatrix}.$$", "= \\begin{pmatrix} -1 \\\\\\\\ 1 \\end{pmatrix}$", "# Problems from Combinatorial numbers Calculate the value of the following combinatorial numbers: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}$$ See development and solution ### Development: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}=\\dfrac{8!}{3!\\cdot(8-3)!}= \\dfrac{8\\cdot7\\cdot6\\cdot5!}{3!\\cdot5!}=\\dfrac{8\\cdot7\\cdot6}{3\\cdot2}=56$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}=4$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}=\\dfrac{7!}{2!\\cdot(7-2)!}= \\dfrac{7\\cdot6\\cdot5!}{2!\\cdot5!}=7\\cdot3=21$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}=1$$ ### Solution: 1. $$56$$ 2. $$4$$ 3. $$21$$ 4. $$1$$ Hide solution and development", "\\\\ 2 \\end{pmatrix}$ 3. #### Insert $\\text{E: } \\vec{x} = \\vec{OA} + r \\cdot \\vec{AB} + s \\cdot \\vec{AC}$ $\\text{E: } \\vec{x} = \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $+ s \\cdot \\begin{pmatrix} 1 \\\\ 5 \\\\ 2 \\end{pmatrix}$", "\\cdot \\begin{pmatrix} 6 & 7 \\\\ 8 & 9 \\\\ \\end{pmatrix} = \\begin{pmatrix} 2 \\cdot 6 & 3 \\cdot 7 \\\\ 4 \\cdot 8 & 5 \\cdot 9 \\\\ \\end{pmatrix} = \\begin{pmatrix} 12 & 21 \\\\ 32 & 45 \\\\ \\end{pmatrix} ?$ 1. Unnecessary for a V.S. 2. This is useless, since it does not describe reality. For example, a mathematical theory with 46 dimensions can be perfect and mathematically elegant, but if the only solution to it is a universe in which life cannot form it is not reality, hence we have no use for it.", "6 & 0 & 5 \\end{pmatrix}$", "\\end{pmatrix} + r \\cdot \\begin{pmatrix} 9 \\\\ 8 \\\\ 7\\end{pmatrix}$ $= \\begin{pmatrix} 4 \\\\ 4 \\\\ 6 \\end{pmatrix} + s \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ Now we set up an equation system and solve it. 1. $10+9r=4+s$ 2. $9+8r=4+s$ 3. $7+7r=6+2s$ Equation method: Equate I. and II. $4+s=4+s$ $10+9r=9+8r\\quad|-8r$ $10+r=9\\quad|-10$ $r=-1$ Compute $s$ with e.g. I. (for that insert $r=-1$) $10+9\\cdot(-1)=4+s$ $1=4+s\\quad|-4$ $s=-3$ For validation insert $r=-1$ and $s=-3$ into III. $7+7\\cdot(-1)=6+2\\cdot(-3)$ $0=0$ $r=-1$ and $s=-3$ result from all equations => The lines intersect. 3. #### Determining point of intersection To determine the point of intersection we insert $r=-1$ into $g$. $\\begin{pmatrix} 10 \\\\ 9 \\\\ 7 \\end{pmatrix} + (-1) \\cdot \\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}$ $=\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ => The lines intersect in point $S(1|1|0)$.", "1 & 1 \\end{pmatrix}$", "\\\\ 0 & 1\\end{pmatrix}.$ -", "it is to compute $2\\cdot124865$; for a start, there's more input.", "\\leq \\begin{pmatrix}\\ldots\\\\k_i\\\\k_i\\\\\\ldots \\end{pmatrix}$$", "and $z$. $(x+1)^2+(y-2)^2$ $+(z-1)^2=17$ $(5+3r+1)^2$ $+(6+2r-2)^2$ $+(5+2r-1)^2=17$ $(6+3r)^2$ $+(4+2r)^2$ $+(4+2r)^2=17$ Use binomial theorem to resolve parentheses $36+36r+9r^2$ $+16+16r+4r^2$ $+16+16r+4r^2=17$ $17r^2+68r+68=17\\quad|-17$ $17r^2+68r+51=0\\quad|:17$ $r^2+4r+3=0$ Use quadratic formula I to solve quadratic equation. $r_{1,2}=-\\frac{p}2\\pm\\sqrt{(\\frac{p}2)^2-q}$ $r_{1,2}=-2\\pm\\sqrt{2^2-3}$ $r_{1,2}=-2\\pm1$ $r_{1}=-1$ and $r_{2}=-3$ 3. #### Insert $r$ The two calculated $r$ are inserted into the equation of a line in order to obtain the intersection points. $\\vec{OS_1} = \\begin{pmatrix} 5 \\\\ 6 \\\\ 5 \\end{pmatrix} - 1 \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $=\\begin{pmatrix} 2 \\\\ 4 \\\\ 3 \\end{pmatrix}$ $\\vec{OS_2} = \\begin{pmatrix} 5 \\\\ 6 \\\\ 5 \\end{pmatrix} - 3 \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $=\\begin{pmatrix} -4 \\\\ 0 \\\\ -1 \\end{pmatrix}$ It is a secant that intersects the sphere at $S_1(2|4|3)$ and $S_2(-4|0|-1)$.", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "$\\begin{pmatrix} 4&5 \\\\ -1&-3 \\end{pmatrix}\\cdot \\begin{pmatrix} x_{1}\\\\-x_{1} \\end{pmatrix} + \\begin{pmatrix} 4&5 \\\\ -1&-3 \\end{pmatrix}\\cdot \\begin{pmatrix} x_{2}\\\\-2x_{2} \\end{pmatrix} = \\begin{pmatrix} -x_{1}-6x_{2}\\\\2x_{1}+5x_{2} \\end{pmatrix}$. -", "& -\\frac{1}{y}\\end{pmatrix}$. Then will take an initial guess, say $X_0 = \\begin{pmatrix}x_0 \\\\ y_0\\end{pmatrix} = \\begin{pmatrix}0.5 \\\\ 0.5\\end{pmatrix}$. Now we perform the iteration $X_1 = X_0 - JF^{-1}(X_0)\\cdot F(X_0)$, so we have $X_1 = \\begin{pmatrix}x_1 \\\\ y_1\\end{pmatrix} = \\begin{pmatrix}0.5 \\\\ 0.5\\end{pmatrix} - \\frac{1}{4}\\begin{pmatrix}2 & 2\\\\2 & -2\\end{pmatrix} \\cdot \\begin{pmatrix} -1.5 \\\\ 0\\end{pmatrix}$. Thus, we have $X_1 = \\begin{pmatrix}0.5 \\\\ 0.5\\end{pmatrix} - \\begin{pmatrix}-0.75 \\\\ -0.75\\end{pmatrix} = \\begin{pmatrix}1.25 \\\\ 1.25\\end{pmatrix}$. Now repeat this for $X_2 = X_1 - JF^{-1}(X_1)\\cdot F(X_1)$. $X_2 = \\begin{pmatrix}1.25 \\\\ 1.25\\end{pmatrix} - \\frac{1}{4}\\begin{pmatrix}0.8 & 0.8\\\\0.8 & -0.8\\end{pmatrix} \\cdot \\begin{pmatrix} 1.125 \\\\ 0\\end{pmatrix}$. $X_2 = \\begin{pmatrix}1.25 \\\\ 1.25\\end{pmatrix} - \\begin{pmatrix}0.225 \\\\ 0.225\\end{pmatrix} = \\begin{pmatrix}1.025 \\\\ 1.025\\end{pmatrix}$. Continue like this until $\\frac{\\|X_n - X_{n-1}\\|_{\\infty}}{\\|X_{n-1}\\|_{\\infty}} \\le tol$ • December 6th 2009, 09:00 AM maniacraghu matlab help.. how do you define a variable.. my program is as shown: clc clear all F=[y^2+z^2-2;y^2-z^2]; v=[y,z] R=jacobian(F,v) y=0.1;z=0.1; xo=[y;z] for j=1:30 D=(inv(R))*(-F) xo=xo-D end i am getting an error saying that y and z are undefined variables however if i use syms y,z the guess value of y and z does not get passed on into the for loop...dono why...PLS HELP • December 6th 2009, 09:03 AM lvleph Matlab doesn't like anything but numbers. So, x and y would have had to be given to it. You can pass functions to matlab using", "1 & -4 \\\\ -1 & 3 \\end{pmatrix} \\begin{pmatrix} 22 \\\\ 6 \\end{pmatrix} \\\\ &= - \\begin{pmatrix} 1\\cdot 22 - 4\\cdot 6 \\\\ -1 \\cdot 22 + 3\\cdot 6 \\end{pmatrix} \\\\ &= \\begin{pmatrix} 2 \\\\ 4 \\end{pmatrix} \\end{align}" ]

[ "1 & -4 \\\\ -1 & 3 \\end{pmatrix} \\begin{pmatrix} 22 \\\\ 6 \\end{pmatrix} \\\\ &= - \\begin{pmatrix} 1\\cdot 22 - 4\\cdot 6 \\\\ -1 \\cdot 22 + 3\\cdot 6 \\end{pmatrix} \\\\ &= \\begin{pmatrix} 2 \\\\ 4 \\end{pmatrix} \\end{align}", "& \\frac{1}{2} \\end{pmatrix} \\cdot \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} 1-\\sqrt{3} \\\\ 1+\\sqrt{3} \\end{pmatrix}$$", "2+2\\cdot 3+0\\cdot 4\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3+2\\cdot 4 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3+1\\cdot 4\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 11\\\\4\\end{pmatrix}$; d) $\\begin{pmatrix} 1 & 2 & 0 \\\\ 0 & 1 & 2 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\\\4 \\end{pmatrix}=\\begin{pmatrix}1\\cdot 1+2\\cdot 2+0\\cdot 3\\\\ 0\\cdot 1+1\\cdot 2+2\\cdot 3\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 3\\\\0\\end{pmatrix}$. ### This Post Has One Comment 1. a) should be [13 31]. 1+6+6 = 13", "= x\\cdot \\begin{pmatrix} 1 & 2 & 3 \\end{pmatrix} + y\\cdot \\begin{pmatrix} 6 & 5 & 4 \\end{pmatrix} + z\\cdot \\begin{pmatrix} 7 & 8 & 9 \\end{pmatrix}.$$", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "= -12 \\\\ & P_7 = S_9 \\cdot S_{10} = -84 \\end {aligned} The result sub-matrix sums: \\begin {aligned} & C_{11} = P_5 + P_4 - P_2 + P_6 = 18 \\\\ & C_{12} = P_1 + P_2 = 14 \\\\ & C_{21} = P_3 + P_4 = 62 \\\\ & C_{22} = P_5 + P_1 - P_3 - P_7 = 66 \\end {aligned} So, the answer is $$\\begin{pmatrix} 18 & 14 \\\\ 62 & 66 \\\\ \\end{pmatrix}$$.", "$\\dfrac{12\\sqrt3\\cdot2\\sqrt2}3=8\\sqrt6$, and thus the answer is $\\boxed{384}$. -integralarefun", "\\cdot \\begin{pmatrix} 6 & 7 \\\\ 8 & 9 \\\\ \\end{pmatrix} = \\begin{pmatrix} 2 \\cdot 6 & 3 \\cdot 7 \\\\ 4 \\cdot 8 & 5 \\cdot 9 \\\\ \\end{pmatrix} = \\begin{pmatrix} 12 & 21 \\\\ 32 & 45 \\\\ \\end{pmatrix} ?$ 1. Unnecessary for a V.S. 2. This is useless, since it does not describe reality. For example, a mathematical theory with 46 dimensions can be perfect and mathematically elegant, but if the only solution to it is a universe in which life cannot form it is not reality, hence we have no use for it.", "+ A_2 + \\cdots +A_t) & = 882 \\\\ \\det(A(1+2+\\cdots t)) & = 882 \\\\ \\det(A) \\cdot (1+2+\\cdots + t)^2 & = 882 \\\\ 2(1+2+\\cdots + t)^2 & = 882 \\\\ (1+2+\\cdots+t)^2 & = 441 \\\\ 1+2+\\cdots+t & = 21 \\\\ \\dfrac{t}{2}(t+1) & = 21 \\\\ t^2+t-42 & = 0 \\\\ (t+7)(t-6) & = 0 \\end{aligned} We obtain $t = -7$ or $t = 6$ (It satisfies since $t \\in \\mathbb{N}$). Thus, the value of $\\boxed{\\det(A_{2 \\cdot 6}) = A_{12} = 2 \\cdot 12^2 = 288}$ [collapse] Problem Number 19 If $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 1 & 0 \\\\ -3 & 1 \\end{pmatrix}$, then $A^{2013} \\cdot \\begin{pmatrix}2 \\\\ 1 \\end{pmatrix} + B^{2014} \\cdot \\begin{pmatrix} 1 \\\\ -3 \\end{pmatrix}$ equals $\\cdots \\cdot$ A. $\\begin{pmatrix} 2023 \\\\ -6044 \\end{pmatrix}$ D. $\\begin{pmatrix} -4023 \\\\ -6044 \\end{pmatrix}$ B. $\\begin{pmatrix} -4023 \\\\ 6044 \\end{pmatrix}$ E. $\\begin{pmatrix} -6044 \\\\ -4023 \\end{pmatrix}$ C. $\\begin{pmatrix} 4023 \\\\ 6044 \\end{pmatrix}$ Solution Given that $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$. We will find the matrix’s entry pattern for $A^n$. You should note that \\begin{aligned} A^2 & = \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & \\color{red}{-4} \\\\ 0", "\\cdot 7 = 70+21 = \\boxed{91} \\end{dmath*} \\end{document}", "$\\begin{pmatrix} 4&5 \\\\ -1&-3 \\end{pmatrix}\\cdot \\begin{pmatrix} x_{1}\\\\-x_{1} \\end{pmatrix} + \\begin{pmatrix} 4&5 \\\\ -1&-3 \\end{pmatrix}\\cdot \\begin{pmatrix} x_{2}\\\\-2x_{2} \\end{pmatrix} = \\begin{pmatrix} -x_{1}-6x_{2}\\\\2x_{1}+5x_{2} \\end{pmatrix}$. -", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "other hand, $P_r(A)P_r(B)=\\frac{4}{5}\\frac{3}{5}=\\frac{12}{25}$. Since $P_r(A\\cap B)\\ne P_r(A)P_r(B)$, the events $A$ and $B$ are not independent. References. [1] Essential Discrete Mathematics for Computer Science, Harry Lewis and Rachel Zax, Princeton University Press, 2019 # Introductory Combinatorics: Combinations 2 Let us begin with an example. Example 1. A jar contains 11 marbles. In how many ways can they be placed in four boxes, with 3 marbles in the first box, 4 marbles in the second box, 2 marbles in the third box, and the 2 remaining marbles in the fourth box? Solution. The total number of outcomes is $$\\begin{pmatrix}11\\\\3\\end{pmatrix}\\cdot \\begin{pmatrix}8\\\\4\\end{pmatrix}\\cdot \\begin{pmatrix}4\\\\2\\end{pmatrix}\\cdot \\begin{pmatrix}2\\\\2\\end{pmatrix}=\\frac{11!}{3!8!}\\cdot\\frac{8!}{4!4!}\\cdot\\frac{4!}{2!2!}\\cdot\\frac{2!}{2!0!}=\\frac{11!}{3!4!2!2!}=69,300$$ This example can be generalized to the following question. A set of $n$ distinct items is to be divided into $r$ distinct groups of respective sizes $n_1, n_2, \\cdots, n_r$ such that $\\sum_{i=1}^r n_i=n$. How many different divisions are possible? There are $\\begin{pmatrix}n\\\\n_1\\end{pmatrix}$ possible choices for the first group, there are $\\begin{pmatrix}n-n_1\\\\n_2\\end{pmatrix}$ possible choices for the second group, there are $\\begin{pmatrix}n-n_1-n_2\\\\n_3\\end{pmatrix}$ possible choices for the third group, …, there are $\\begin{pmatrix}n-n_1-n_2-\\cdots-n_{r-1}\\\\n_r\\end{pmatrix}$ possible choices for the $r$-th group. Therefore the total number of coutcomes is \\begin{align*}\\begin{pmatrix}n\\\\n_1\\end{pmatrix}\\begin{pmatrix}n-n_1\\\\n_2\\end{pmatrix}\\begin{pmatrix}n-n_1-n_2\\\\n_3\\end{pmatrix}\\cdots\\begin{pmatrix}n-n_1-n_2-\\cdots -n_{r-1}\\\\n_r\\end{pmatrix}\\\\=\\frac{n!}{(n-n_1)!n_1!}\\frac{(n-n_1)!}{(n-n_1-n_2)!n_2!}\\cdots\\frac{(n-n_1-n_2\\cdots-n_{r-1})!}{0!n_r!}\\\\=\\frac{n!}{n_1!n_2!\\cdots n_r!}\\end{align*} The number of divisions of $n$ distinct items into $r$ distinct groups of respective sizes $n_1,n_2,\\cdots,n_r$ is denoted by $\\begin{pmatrix}n\\\\n_1,n_2,\\cdots,n_r\\end{pmatrix}$. That is, $$\\label{eq:multcoeff}\\begin{pmatrix}n\\\\n_1,n_2,\\cdots,n_r\\end{pmatrix}=\\frac{n!}{n_1!n_2!\\cdots n_r!}$$ where $n=n_1+n_2+\\cdots+n_r$. Example 2. A police department", "4! + 3\\cdot 3! + 2\\cdot 2! + 0\\cdot 1! & = \\\\ 5\\cdot120+4\\cdot24 + 3\\cdot6 + 2\\cdot 2 + 0\\cdot 1 & =\\\\ 600 + 96 + 18 + 4 + 0 & = 718.\\end{align}\\$ This has actual applications; for example it is a useful way to represent a permutation of a list. -", "# Problems from Combinatorial numbers Calculate the value of the following combinatorial numbers: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}$$ See development and solution ### Development: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}=\\dfrac{8!}{3!\\cdot(8-3)!}= \\dfrac{8\\cdot7\\cdot6\\cdot5!}{3!\\cdot5!}=\\dfrac{8\\cdot7\\cdot6}{3\\cdot2}=56$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}=4$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}=\\dfrac{7!}{2!\\cdot(7-2)!}= \\dfrac{7\\cdot6\\cdot5!}{2!\\cdot5!}=7\\cdot3=21$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}=1$$ ### Solution: 1. $$56$$ 2. $$4$$ 3. $$21$$ 4. $$1$$ Hide solution and development", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "## Exercises - Linear Transformations 1. Suppose $M$ is a linear transformation operating on 2-dimensional vectors. Appealing to the linearity of $M$, we see that \\begin{align} M\\begin{pmatrix}-6\\\\8\\end{pmatrix} &= M \\left( \\begin{pmatrix}-6\\\\0\\end{pmatrix} + \\begin{pmatrix}0\\\\8\\end{pmatrix} \\right) \\\\ &= M\\begin{pmatrix}-6\\\\0\\end{pmatrix} + M\\begin{pmatrix}0\\\\8\\end{pmatrix} \\\\ &= M\\left( -6 \\begin{pmatrix}1\\\\0\\end{pmatrix} \\right) + M \\left( 8 \\begin{pmatrix}0\\\\1\\end{pmatrix} \\right) \\\\ &= -6 M\\begin{pmatrix}1\\\\0\\end{pmatrix} + 8 M \\begin{pmatrix}0\\\\1\\end{pmatrix} \\\\ &= -6 \\begin{pmatrix}4\\\\5\\end{pmatrix} + 8 \\begin{pmatrix}-3\\\\7\\end{pmatrix} \\\\ &= \\begin{pmatrix}(-6) \\cdot 4\\\\ (-6) \\cdot 5\\end{pmatrix} + \\begin{pmatrix} 8 \\cdot (-3)\\\\8 \\cdot 7\\end{pmatrix} \\\\ &= \\begin{pmatrix}(-6) \\cdot 4 + 8 \\cdot (-3)\\\\ (-6) \\cdot 5 + 8 \\cdot 7\\end{pmatrix} \\\\ &= \\begin{pmatrix}-48\\\\26\\end{pmatrix} \\end{align} We can apply this same process to find the output of any linear transformation that operates on two-dimensional vectors, provided we know what the linear transformation does to $\\begin{pmatrix}1\\\\0\\end{pmatrix}$ and $\\begin{pmatrix}0\\\\1\\end{pmatrix}$. As such, we can abbreviate the process above by writing $M$ as a matrix, and evaluating $M\\begin{pmatrix}-6\\\\8\\end{pmatrix}$ in the following way: $$M\\begin{pmatrix}-6\\\\8\\end{pmatrix} = \\begin{bmatrix}4 & -3\\\\5 & 7\\end{bmatrix} \\begin{pmatrix}-6\\\\8\\end{pmatrix} = \\begin{pmatrix}4 \\cdot (-6) + (-3) \\cdot 8\\\\5 \\cdot (-6) + 7 \\cdot 8\\end{pmatrix} = \\begin{pmatrix}-48\\\\26\\end{pmatrix}$$ 2. We know that we can find the matrix form for a linear transformation, $T$, on two-dimensional vectors, if we know the output of $T\\begin{pmatrix}1\\\\0\\end{pmatrix}$ and $T\\begin{pmatrix}0\\\\1\\end{pmatrix}$. What if we know the output of $T$ when applied to other vectors? Can", "\\begin{pmatrix}1&0\\cr0&0\\end{pmatrix} \\cdot \\begin{pmatrix}0&1\\cr0&0\\end{pmatrix} = \\begin{pmatrix}1&0\\cr0&1\\end{pmatrix}.$$", "\\\\ 7 \\end{pmatrix} = 7 \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}, \\end{align*} so, it is correct.", "On the other hand, $P_r(A)P_r(B)=\\frac{4}{5}\\frac{3}{5}=\\frac{12}{25}$. Since $P_r(A\\cap B)\\ne P_r(A)P_r(B)$, the events $A$ and $B$ are not independent. References. [1] Essential Discrete Mathematics for Computer Science, Harry Lewis and Rachel Zax, Princeton University Press, 2019 # Introductory Combinatorics: Combinations 2 Let us begin with an example. Example 1. A jar contains 11 marbles. In how many ways can they be placed in four boxes, with 3 marbles in the first box, 4 marbles in the second box, 2 marbles in the third box, and the 2 remaining marbles in the fourth box? Solution. The total number of outcomes is $$\\begin{pmatrix}11\\\\3\\end{pmatrix}\\cdot \\begin{pmatrix}8\\\\4\\end{pmatrix}\\cdot \\begin{pmatrix}4\\\\2\\end{pmatrix}\\cdot \\begin{pmatrix}2\\\\2\\end{pmatrix}=\\frac{11!}{3!8!}\\cdot\\frac{8!}{4!4!}\\cdot\\frac{4!}{2!2!}\\cdot\\frac{2!}{2!0!}=\\frac{11!}{3!4!2!2!}=69,300$$ This example can be generalized to the following question. A set of $n$ distinct items is to be divided into $r$ distinct groups of respective sizes $n_1, n_2, \\cdots, n_r$ such that $\\sum_{i=1}^r n_i=n$. How many different divisions are possible? There are $\\begin{pmatrix}n\\\\n_1\\end{pmatrix}$ possible choices for the first group, there are $\\begin{pmatrix}n-n_1\\\\n_2\\end{pmatrix}$ possible choices for the second group, there are $\\begin{pmatrix}n-n_1-n_2\\\\n_3\\end{pmatrix}$ possible choices for the third group, …, there are $\\begin{pmatrix}n-n_1-n_2-\\cdots-n_{r-1}\\\\n_r\\end{pmatrix}$ possible choices for the $r$-th group. Therefore the total number of coutcomes is \\begin{align*}\\begin{pmatrix}n\\\\n_1\\end{pmatrix}\\begin{pmatrix}n-n_1\\\\n_2\\end{pmatrix}\\begin{pmatrix}n-n_1-n_2\\\\n_3\\end{pmatrix}\\cdots\\begin{pmatrix}n-n_1-n_2-\\cdots -n_{r-1}\\\\n_r\\end{pmatrix}\\\\=\\frac{n!}{(n-n_1)!n_1!}\\frac{(n-n_1)!}{(n-n_1-n_2)!n_2!}\\cdots\\frac{(n-n_1-n_2\\cdots-n_{r-1})!}{0!n_r!}\\\\=\\frac{n!}{n_1!n_2!\\cdots n_r!}\\end{align*} The number of divisions of $n$ distinct items into $r$ distinct groups of respective sizes $n_1,n_2,\\cdots,n_r$ is denoted by $\\begin{pmatrix}n\\\\n_1,n_2,\\cdots,n_r\\end{pmatrix}$. That is, $$\\label{eq:multcoeff}\\begin{pmatrix}n\\\\n_1,n_2,\\cdots,n_r\\end{pmatrix}=\\frac{n!}{n_1!n_2!\\cdots n_r!}$$ where $n=n_1+n_2+\\cdots+n_r$. Example 2. A" ]

[ "4 \\\\ 3 \\end{pmatrix}.$ ×", "= x\\cdot \\begin{pmatrix} 1 & 2 & 3 \\end{pmatrix} + y\\cdot \\begin{pmatrix} 6 & 5 & 4 \\end{pmatrix} + z\\cdot \\begin{pmatrix} 7 & 8 & 9 \\end{pmatrix}.$$", "& \\frac{1}{2} \\end{pmatrix} \\cdot \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} 1-\\sqrt{3} \\\\ 1+\\sqrt{3} \\end{pmatrix}$$", "& 6 & 8 \\end{pmatrix}.$$ For example, let's calculate numeber from this sequence with number $8$. As $8_{10} = 1\\cdot 5^1 + 3\\cdot 5^0 = 13_5$ we get number $26$ (replacing in $13$ digits $1$ with $2$ and $3$ with $6$) which is correct. Note that we count $1$ as first number, not zero.", "# Problems from Combinatorial numbers Calculate the value of the following combinatorial numbers: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}$$ See development and solution ### Development: 1. $$\\begin{pmatrix} 8 \\\\ 3 \\end{pmatrix}=\\dfrac{8!}{3!\\cdot(8-3)!}= \\dfrac{8\\cdot7\\cdot6\\cdot5!}{3!\\cdot5!}=\\dfrac{8\\cdot7\\cdot6}{3\\cdot2}=56$$ 2. $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}=4$$ 3. $$\\begin{pmatrix} 7 \\\\ 2 \\end{pmatrix}=\\dfrac{7!}{2!\\cdot(7-2)!}= \\dfrac{7\\cdot6\\cdot5!}{2!\\cdot5!}=7\\cdot3=21$$ 4. $$\\begin{pmatrix} 157 \\\\ 0 \\end{pmatrix}=1$$ ### Solution: 1. $$56$$ 2. $$4$$ 3. $$21$$ 4. $$1$$ Hide solution and development", "2+2\\cdot 3+0\\cdot 4\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3+2\\cdot 4 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3+1\\cdot 4\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 11\\\\4\\end{pmatrix}$; d) $\\begin{pmatrix} 1 & 2 & 0 \\\\ 0 & 1 & 2 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\\\4 \\end{pmatrix}=\\begin{pmatrix}1\\cdot 1+2\\cdot 2+0\\cdot 3\\\\ 0\\cdot 1+1\\cdot 2+2\\cdot 3\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 3\\\\0\\end{pmatrix}$. ### This Post Has One Comment 1. a) should be [13 31]. 1+6+6 = 13", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "\\\\ 2 \\end{pmatrix}$ 3. #### Insert $\\text{E: } \\vec{x} = \\vec{OA} + r \\cdot \\vec{AB} + s \\cdot \\vec{AC}$ $\\text{E: } \\vec{x} = \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $+ s \\cdot \\begin{pmatrix} 1 \\\\ 5 \\\\ 2 \\end{pmatrix}$", "\\cdot \\begin{pmatrix} 6 & 7 \\\\ 8 & 9 \\\\ \\end{pmatrix} = \\begin{pmatrix} 2 \\cdot 6 & 3 \\cdot 7 \\\\ 4 \\cdot 8 & 5 \\cdot 9 \\\\ \\end{pmatrix} = \\begin{pmatrix} 12 & 21 \\\\ 32 & 45 \\\\ \\end{pmatrix} ?$ 1. Unnecessary for a V.S. 2. This is useless, since it does not describe reality. For example, a mathematical theory with 46 dimensions can be perfect and mathematically elegant, but if the only solution to it is a universe in which life cannot form it is not reality, hence we have no use for it.", "\\begin{pmatrix}1&0\\cr0&0\\end{pmatrix} \\cdot \\begin{pmatrix}0&1\\cr0&0\\end{pmatrix} = \\begin{pmatrix}1&0\\cr0&1\\end{pmatrix}.$$", "$\\begin{pmatrix} 4&5 \\\\ -1&-3 \\end{pmatrix}\\cdot \\begin{pmatrix} x_{1}\\\\-x_{1} \\end{pmatrix} + \\begin{pmatrix} 4&5 \\\\ -1&-3 \\end{pmatrix}\\cdot \\begin{pmatrix} x_{2}\\\\-2x_{2} \\end{pmatrix} = \\begin{pmatrix} -x_{1}-6x_{2}\\\\2x_{1}+5x_{2} \\end{pmatrix}$. -", "and $z$. $(x+1)^2+(y-2)^2$ $+(z-1)^2=17$ $(5+3r+1)^2$ $+(6+2r-2)^2$ $+(5+2r-1)^2=17$ $(6+3r)^2$ $+(4+2r)^2$ $+(4+2r)^2=17$ Use binomial theorem to resolve parentheses $36+36r+9r^2$ $+16+16r+4r^2$ $+16+16r+4r^2=17$ $17r^2+68r+68=17\\quad|-17$ $17r^2+68r+51=0\\quad|:17$ $r^2+4r+3=0$ Use quadratic formula I to solve quadratic equation. $r_{1,2}=-\\frac{p}2\\pm\\sqrt{(\\frac{p}2)^2-q}$ $r_{1,2}=-2\\pm\\sqrt{2^2-3}$ $r_{1,2}=-2\\pm1$ $r_{1}=-1$ and $r_{2}=-3$ 3. #### Insert $r$ The two calculated $r$ are inserted into the equation of a line in order to obtain the intersection points. $\\vec{OS_1} = \\begin{pmatrix} 5 \\\\ 6 \\\\ 5 \\end{pmatrix} - 1 \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $=\\begin{pmatrix} 2 \\\\ 4 \\\\ 3 \\end{pmatrix}$ $\\vec{OS_2} = \\begin{pmatrix} 5 \\\\ 6 \\\\ 5 \\end{pmatrix} - 3 \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $=\\begin{pmatrix} -4 \\\\ 0 \\\\ -1 \\end{pmatrix}$ It is a secant that intersects the sphere at $S_1(2|4|3)$ and $S_2(-4|0|-1)$.", "& -\\frac{1}{y}\\end{pmatrix}$. Then will take an initial guess, say $X_0 = \\begin{pmatrix}x_0 \\\\ y_0\\end{pmatrix} = \\begin{pmatrix}0.5 \\\\ 0.5\\end{pmatrix}$. Now we perform the iteration $X_1 = X_0 - JF^{-1}(X_0)\\cdot F(X_0)$, so we have $X_1 = \\begin{pmatrix}x_1 \\\\ y_1\\end{pmatrix} = \\begin{pmatrix}0.5 \\\\ 0.5\\end{pmatrix} - \\frac{1}{4}\\begin{pmatrix}2 & 2\\\\2 & -2\\end{pmatrix} \\cdot \\begin{pmatrix} -1.5 \\\\ 0\\end{pmatrix}$. Thus, we have $X_1 = \\begin{pmatrix}0.5 \\\\ 0.5\\end{pmatrix} - \\begin{pmatrix}-0.75 \\\\ -0.75\\end{pmatrix} = \\begin{pmatrix}1.25 \\\\ 1.25\\end{pmatrix}$. Now repeat this for $X_2 = X_1 - JF^{-1}(X_1)\\cdot F(X_1)$. $X_2 = \\begin{pmatrix}1.25 \\\\ 1.25\\end{pmatrix} - \\frac{1}{4}\\begin{pmatrix}0.8 & 0.8\\\\0.8 & -0.8\\end{pmatrix} \\cdot \\begin{pmatrix} 1.125 \\\\ 0\\end{pmatrix}$. $X_2 = \\begin{pmatrix}1.25 \\\\ 1.25\\end{pmatrix} - \\begin{pmatrix}0.225 \\\\ 0.225\\end{pmatrix} = \\begin{pmatrix}1.025 \\\\ 1.025\\end{pmatrix}$. Continue like this until $\\frac{\\|X_n - X_{n-1}\\|_{\\infty}}{\\|X_{n-1}\\|_{\\infty}} \\le tol$ • December 6th 2009, 09:00 AM maniacraghu matlab help.. how do you define a variable.. my program is as shown: clc clear all F=[y^2+z^2-2;y^2-z^2]; v=[y,z] R=jacobian(F,v) y=0.1;z=0.1; xo=[y;z] for j=1:30 D=(inv(R))*(-F) xo=xo-D end i am getting an error saying that y and z are undefined variables however if i use syms y,z the guess value of y and z does not get passed on into the for loop...dono why...PLS HELP • December 6th 2009, 09:03 AM lvleph Matlab doesn't like anything but numbers. So, x and y would have had to be given to it. You can pass functions to matlab using", "## Exercises - Linear Transformations 1. Suppose $M$ is a linear transformation operating on 2-dimensional vectors. Appealing to the linearity of $M$, we see that \\begin{align} M\\begin{pmatrix}-6\\\\8\\end{pmatrix} &= M \\left( \\begin{pmatrix}-6\\\\0\\end{pmatrix} + \\begin{pmatrix}0\\\\8\\end{pmatrix} \\right) \\\\ &= M\\begin{pmatrix}-6\\\\0\\end{pmatrix} + M\\begin{pmatrix}0\\\\8\\end{pmatrix} \\\\ &= M\\left( -6 \\begin{pmatrix}1\\\\0\\end{pmatrix} \\right) + M \\left( 8 \\begin{pmatrix}0\\\\1\\end{pmatrix} \\right) \\\\ &= -6 M\\begin{pmatrix}1\\\\0\\end{pmatrix} + 8 M \\begin{pmatrix}0\\\\1\\end{pmatrix} \\\\ &= -6 \\begin{pmatrix}4\\\\5\\end{pmatrix} + 8 \\begin{pmatrix}-3\\\\7\\end{pmatrix} \\\\ &= \\begin{pmatrix}(-6) \\cdot 4\\\\ (-6) \\cdot 5\\end{pmatrix} + \\begin{pmatrix} 8 \\cdot (-3)\\\\8 \\cdot 7\\end{pmatrix} \\\\ &= \\begin{pmatrix}(-6) \\cdot 4 + 8 \\cdot (-3)\\\\ (-6) \\cdot 5 + 8 \\cdot 7\\end{pmatrix} \\\\ &= \\begin{pmatrix}-48\\\\26\\end{pmatrix} \\end{align} We can apply this same process to find the output of any linear transformation that operates on two-dimensional vectors, provided we know what the linear transformation does to $\\begin{pmatrix}1\\\\0\\end{pmatrix}$ and $\\begin{pmatrix}0\\\\1\\end{pmatrix}$. As such, we can abbreviate the process above by writing $M$ as a matrix, and evaluating $M\\begin{pmatrix}-6\\\\8\\end{pmatrix}$ in the following way: $$M\\begin{pmatrix}-6\\\\8\\end{pmatrix} = \\begin{bmatrix}4 & -3\\\\5 & 7\\end{bmatrix} \\begin{pmatrix}-6\\\\8\\end{pmatrix} = \\begin{pmatrix}4 \\cdot (-6) + (-3) \\cdot 8\\\\5 \\cdot (-6) + 7 \\cdot 8\\end{pmatrix} = \\begin{pmatrix}-48\\\\26\\end{pmatrix}$$ 2. We know that we can find the matrix form for a linear transformation, $T$, on two-dimensional vectors, if we know the output of $T\\begin{pmatrix}1\\\\0\\end{pmatrix}$ and $T\\begin{pmatrix}0\\\\1\\end{pmatrix}$. What if we know the output of $T$ when applied to other vectors? Can", "+ A_2 + \\cdots +A_t) & = 882 \\\\ \\det(A(1+2+\\cdots t)) & = 882 \\\\ \\det(A) \\cdot (1+2+\\cdots + t)^2 & = 882 \\\\ 2(1+2+\\cdots + t)^2 & = 882 \\\\ (1+2+\\cdots+t)^2 & = 441 \\\\ 1+2+\\cdots+t & = 21 \\\\ \\dfrac{t}{2}(t+1) & = 21 \\\\ t^2+t-42 & = 0 \\\\ (t+7)(t-6) & = 0 \\end{aligned} We obtain $t = -7$ or $t = 6$ (It satisfies since $t \\in \\mathbb{N}$). Thus, the value of $\\boxed{\\det(A_{2 \\cdot 6}) = A_{12} = 2 \\cdot 12^2 = 288}$ [collapse] Problem Number 19 If $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 1 & 0 \\\\ -3 & 1 \\end{pmatrix}$, then $A^{2013} \\cdot \\begin{pmatrix}2 \\\\ 1 \\end{pmatrix} + B^{2014} \\cdot \\begin{pmatrix} 1 \\\\ -3 \\end{pmatrix}$ equals $\\cdots \\cdot$ A. $\\begin{pmatrix} 2023 \\\\ -6044 \\end{pmatrix}$ D. $\\begin{pmatrix} -4023 \\\\ -6044 \\end{pmatrix}$ B. $\\begin{pmatrix} -4023 \\\\ 6044 \\end{pmatrix}$ E. $\\begin{pmatrix} -6044 \\\\ -4023 \\end{pmatrix}$ C. $\\begin{pmatrix} 4023 \\\\ 6044 \\end{pmatrix}$ Solution Given that $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$. We will find the matrix’s entry pattern for $A^n$. You should note that \\begin{aligned} A^2 & = \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & \\color{red}{-4} \\\\ 0", "when you do not show your calculation. For example when we calculate $MN$. Then: $\\begin{pmatrix}-0.6&0.8\\\\0.8&0.6\\end{pmatrix}\\cdot \\begin{pmatrix}0.8&0.6\\\\0.6&-0.8\\end{pmatrix}=\\begin{pmatrix} (-0.6)\\cdot 0.8+0.6\\cdot 0.8& (-0.6)\\cdot 0.6+0.8\\cdot (-0.8)\\\\0.8\\cdot 0.8+0.6\\cdot 0.6&0.8\\cdot 0.6+0.6\\cdot (-0.8)\\end{pmatrix}=\\begin{pmatrix}0&-1\\\\1&0\\end{pmatrix}$ Similar you calculate $NM$. Take a point $(1,1)$ and multiply by the matrices: $$\\begin{pmatrix}1&1\\end{pmatrix}\\begin{pmatrix}-0.6&0.8\\\\0.8&0.6\\end{pmatrix}=\\begin{pmatrix}0.2&1.4\\end{pmatrix} \\ \\ \\ \\ \\ \\ \\text{(M is reflecting the point via y=2x)};\\\\ \\begin{pmatrix}1&1\\end{pmatrix}\\begin{pmatrix}0.8&0.6\\\\0.6&-0.8\\end{pmatrix}=\\begin{pmatrix}1.4&-0.2\\end{pmatrix} \\ \\ \\ \\ \\ \\ \\text{(N is reflecting the point via y=\\frac{x}{3})};\\\\ \\begin{pmatrix}1&1\\end{pmatrix}\\begin{pmatrix}-0.6&0.8\\\\0.8&0.6\\end{pmatrix}\\begin{pmatrix}0.8&0.6\\\\0.6&-0.8\\end{pmatrix}=\\begin{pmatrix}1&1\\end{pmatrix}\\begin{pmatrix}0&-1\\\\1&0\\end{pmatrix}=\\\\ =\\begin{pmatrix}1&-1\\end{pmatrix} \\ \\ \\ \\ \\ \\ \\text{(MN is reflecting the point via x-axis)};\\\\ \\begin{pmatrix}1&1\\end{pmatrix}\\begin{pmatrix}0.8&0.6\\\\0.6&-0.8\\end{pmatrix}\\begin{pmatrix}-0.6&0.8\\\\0.8&0.6\\end{pmatrix}=\\begin{pmatrix}1&1\\end{pmatrix}\\begin{pmatrix}0&1\\\\-1&0\\end{pmatrix}=\\\\ =\\begin{pmatrix}-1&1\\end{pmatrix} \\ \\ \\ \\ \\ \\ \\text{(NM is reflecting the point via y-axis)}.$$ First off, matrix multiplication is not simply multiplying term wise. Here, what you did, is said that $$\\begin{bmatrix} a_{11}&a_{12} \\\\ a_{21}&a_{22} \\end{bmatrix} \\begin{bmatrix} b_{11}&b_{12} \\\\ b_{21}&b_{22} \\end{bmatrix} = \\begin{bmatrix} a_{11}b_{11}&a_{12}b_{12} \\\\ a_{21}b_{21}&a_{22}b_{22} \\end{bmatrix}$$ which is not right. It should really be $$\\begin{bmatrix} a_{11}&a_{12} \\\\ a_{21}&a_{22} \\end{bmatrix} \\begin{bmatrix} b_{11}&b_{12} \\\\ b_{21}&b_{22} \\end{bmatrix} = \\begin{bmatrix} a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22} \\\\ a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22} \\end{bmatrix}$$ As has been calculated above in another answer, one gets with $MN$ ... \\begin{bmatrix} 0&-1 \\\\ 1&0 \\end{bmatrix} and with $NM$... $$\\begin{bmatrix} 0&1\\\\ -1&0\\\\ \\end{bmatrix}$$, so $MN \\neq MN$. Now, I think the bigger problem is making sense of matrix multiplication. Now, you may wonder, why is THIS the way we multiply matrices? While the way you multiplied might make the most 'sense' in", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "1 & -4 \\\\ -1 & 3 \\end{pmatrix} \\begin{pmatrix} 22 \\\\ 6 \\end{pmatrix} \\\\ &= - \\begin{pmatrix} 1\\cdot 22 - 4\\cdot 6 \\\\ -1 \\cdot 22 + 3\\cdot 6 \\end{pmatrix} \\\\ &= \\begin{pmatrix} 2 \\\\ 4 \\end{pmatrix} \\end{align}", "the sequence \"$12$\". For $n=7$ this becomes $$\\begin{pmatrix} 1 & 1 \\end{pmatrix}\\begin{pmatrix}96031 & 106821 \\\\ 854568 & 950599 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} = 9409959$$ and including the number $10^7$ the final answer is $9409960$. Based on the above one can find a simpler linear recursion for the amount $a_n$ of numbers of at most $n$ digits that do not contain the sequence \"$12$\", namely $$a_0 = 0, a_1=9 \\textrm{ and } a_{n+2} = 10 a_{n+1} - a_n + 8.$$ This produces the sequence $$0, 9, 98, 979, 9700, 96029, 950598, 9409959, 93149000, 922080049, \\dotsc$$ -", "described above. $g: \\vec{x} = \\vec{OM} + t \\cdot \\vec{n}$ $g: \\vec{x} = \\begin{pmatrix} -1 \\\\ 2 \\\\ 1 \\end{pmatrix} + t \\cdot \\begin{pmatrix} 0\\\\ 0 \\\\ 1 \\end{pmatrix}$ The intersection point of the line and the plane corresponds to the point of contact $B$. The straight line equation is inserted for $\\vec{x}$ in the plane. $(\\begin{pmatrix} -1 \\\\ 2 \\\\ 1 \\end{pmatrix}$ $+ t \\cdot \\begin{pmatrix} 0\\\\ 0 \\\\ 1 \\end{pmatrix}$ $- \\begin{pmatrix} 5 \\\\4 \\\\ 3 \\end{pmatrix}) \\cdot \\begin{pmatrix} 0\\\\ 0 \\\\ 1 \\end{pmatrix}=0$ $\\begin{pmatrix} -6 \\\\ -2 \\\\ t-2 \\end{pmatrix} \\cdot \\begin{pmatrix} 0\\\\ 0 \\\\ 1 \\end{pmatrix}=0$ $0+0+1\\cdot(t-2)=0$ $t-2=0\\quad|+2$ $t=2$ The point of contact $B$ is obtained by inserting the calculated $t$ into the equation of a line. $\\vec{OB} = \\begin{pmatrix} -1 \\\\ 2 \\\\ 1 \\end{pmatrix} + 2 \\cdot \\begin{pmatrix} 0\\\\ 0 \\\\ 1 \\end{pmatrix}$ $=\\begin{pmatrix} -1 \\\\ 2 \\\\ 3 \\end{pmatrix}$ $B(-1|2|3)$" ]

[ "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "+0 # Help with vector geometry +1 88 1 +38 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w} \\text{ and }\\mathbf{x}$$ are linearly independent, answer with 0. If they aren't, find coefficients a,b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a+b}{c}$$ Jun 5, 2019", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "vector $\\mathbf{P}=\\begin{pmatrix}0 \\\\ -1 \\end{pmatrix}$ So then the general solution is $\\mathbf{X}=c_1\\begin{pmatrix} 1 \\\\ 3\\end{pmatrix}+c_2\\left[t\\begin{pmatrix} 1 \\\\ 3\\end{pmatrix}+ \\begin{pmatrix}0 \\\\ -1 \\end{pmatrix}\\right]$ thanks for that! i had a major brain fart!", "+0 # How do i find (a-b)/c? 0 56 1 +39 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w}$$, and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$and answer with $$\\dfrac{a-b}{c}$$. I have tried this multiple times and I got a=0, b=0, and c=0 but it was incorrect. I tried distributing the vectors and solving for a, b, and c with the 3 equations. Can someone please tell me if I am doing anything wrong? Thanks! Apr 30, 2020 #1 +24983 +3 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}$$, $$\\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}$$. If the vectors $$\\mathbf{v}$$, $$\\mathbf{w}$$ and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a-b}{c}$$. $$\\begin{array}{|rcll|} \\hline \\begin{array}{|rcll|} \\hline a\\begin{pmatrix}", "+0 0 108 3 +226 Let a,b,c be vectors such that $$\\mathbf{a} \\times \\mathbf{b} = \\begin{pmatrix} -1 \\\\- 1 \\\\ -1 \\end{pmatrix}, \\mathbf{a} \\times \\mathbf{c} = \\begin{pmatrix} -1 \\\\ 2 \\\\ 1 \\end{pmatrix}, \\text{ and } \\mathbf{b} \\times \\mathbf{c} = \\begin{pmatrix} 0 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ Then evaluate $$(\\mathbf{b} + \\mathbf{c})\\times \\mathbf{b}, \\mathbf{a}\\times(\\mathbf{b} + 4 \\mathbf{a}), (\\mathbf{a} + \\mathbf{b} + \\mathbf{c})\\times \\mathbf{a}$$ Oct 30, 2020 #2 +31698 +3 For cross products we have the following a x a = 0 (simiarly for b x b etc) a x b = - b x a etc. (b + c) x a = b x+ c x a = -x b - a x c etc. Can you take it from here? Oct 30, 2020 #3 +112499 +1 Thanks Alan, that sure made it easy Oct 31, 2020", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "= {2\\mathbf x^T A \\cdot \\mathbf x^T\\mathbf x - \\mathbf x^T A \\mathbf x \\cdot 2\\mathbf x^T \\over (\\mathbf x^T\\mathbf x)^2}(^1_0)$$ In other words, if $A=A^T$: $$Df(\\mathbf x) = {2\\mathbf x^T A \\cdot \\mathbf x^T\\mathbf x - \\mathbf x^T A \\mathbf x \\cdot 2\\mathbf x^T \\over (\\mathbf x^T\\mathbf x)^2}$$ Last edited: Jan 24, 2013 3. Jan 24, 2013 ### libragirl79 Since I forgot a lot of my matrix algebra, when you say A (x y) it looks like x choose y, but i am assuming that's not what you meant... 4. Jan 24, 2013 ### I like Serena Since you are talking about x transpose, that means that x is a vector. A vector is commonly written as: $$\\mathbf x = \\begin{pmatrix}x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n\\end{pmatrix}$$ I have simplified it to: $$\\mathbf x = \\begin{pmatrix}x \\\\ y\\end{pmatrix}$$ So indeed, this does not mean x choose y. ;) Furthermore: $$\\mathbf x^T = \\begin{pmatrix}x_1 & x_2 & \\dots & x_n\\end{pmatrix}$$ and $$\\mathbf x^T \\mathbf x = \\begin{pmatrix}x_1 & x_2 & \\dots & x_n\\end{pmatrix}\\begin{pmatrix}x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n\\end{pmatrix} = x_1^2 + x_2^2 + \\dots + x_n^2$$ 5. Jan 24, 2013 ### libragirl79 Ok, I see! Thanks!! :)", "\\mathbb{F}$ such that $a = \\lambda \\cdot b$. Now consider your example. Let $a = \\begin{pmatrix} 1 - \\mathrm{i} \\\\ 2 - 2\\mathrm{i}\\end{pmatrix} \\in V$ and $b = \\begin{pmatrix} \\mathrm{i} - 1 \\\\ 2\\mathrm{i} - 2 \\end{pmatrix} \\in V$. The vectors would be linearly dependent, if $a = -1 \\cdot b$ and $-1 \\in \\mathbb{F}$. This is true for the field $\\mathbb{F} := \\mathbb{C}$, but also for $\\mathbb{F} := \\mathbb{R}$ or any other field of which $-1$ is an element.", "\\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 686 In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not. (a) The matrix $A$ is a $3 \\times 3$", "& 1 & 1 & 1 \\\\ \\end{pmatrix}\\cdot \\begin{pmatrix} X_1 \\\\ X_2 \\\\ X_3 \\\\ X_4\\\\ \\end{pmatrix}= vector zero$$ Therefore you find that $X_1+X_2+X_3+X_4=0$", "3 & 1 & −2 \\\\ −2 & 0 & 5 \\end{pmatrix}$$ $$\\mathbf{F} = \\begin{pmatrix} 3 & 4 \\\\ 5 & 6 \\end{pmatrix}$$ $$\\mathbf{G} = \\begin{pmatrix} 2 & −4 \\\\ 7 & 1 \\\\ −1 & 1 \\end{pmatrix}$$ $$\\mathbf{H} = \\begin{pmatrix} −8 & −1 & −3 \\\\ −4 & 2 & −2 \\\\ −1 & 0 & 0 \\end{pmatrix}$$ Calculate each of the following 1. $$|\\mathbf{a} − (\\mathbf{b} + \\mathbf{3c})|$$ 7. $$(\\mathbf{EH})^T$$ 2. Component of $$\\mathbf{c}$$ along $$\\mathbf{a}$$ 8. $$|\\mathbf{HE}|$$ 3. Angle between $$\\mathbf{c}$$ and $$\\mathbf{d}$$ 9. $$\\mathbf{EHG}$$ 4. $$(\\mathbf{b} \\times \\mathbf{d}) \\cdot \\mathbf{a}$$ 10. $$\\mathbf{EG} − \\mathbf{HG}$$ 5. $$(\\mathbf{b} \\times \\mathbf{d}) \\times \\mathbf{a}$$ 11. $$\\mathbf{EH} − \\mathbf{H}^T \\mathbf{E}^T$$ 6. $$\\mathbf{b}\\times (\\mathbf{d} \\times \\mathbf{a})$$ 12. $$\\mathbf{F}^{−1}$$ 2. For what values of $$a$$ are the vectors $$\\mathbf{A} = 2a\\hat{i} − 2\\hat{j} + a\\hat{k}$$ and $$\\mathbf{B} = a\\hat{i} + 2a\\hat{j}+ 2\\hat{k}$$ perpendicular? 3. Show that the triple scalar product $$(A \\times B) \\cdot C$$ can be written as $(\\mathbf{A} \\times \\mathbf{B}) \\cdot \\mathbf{C} = \\begin{vmatrix} A_1 & A_2 & A_3 \\\\ B_1 & B_2 & B_3 \\\\ C_1 & C_2 & C_3 \\end{vmatrix} \\nonumber$ Show also that the product is unaffected by interchange of the scalar and vector product operations or by change in the order of $$A, B, C$$ as long as they are in cyclic order, that is", "\\quad \\mathbf{b} = \\begin{pmatrix} x_b \\\\y_b\\end{pmatrix}$$ If these are to reach exactly the same set of points we must have, for any integers $N$ and $M$ that $$N\\begin{pmatrix} x_a \\\\ y_a\\end{pmatrix} +M\\begin{pmatrix} x_b\\\\ y_b\\end{pmatrix}\\mathbf{b} = \\begin{pmatrix} 2p \\\\ q\\end{pmatrix}$$ Clearly to reach these, and only these points, I would need both $a$ and $b$ to have even $x$ component. Furthermore, if I can reach the points $(2, 0)$ and $(0, 1)$ with combinations of my base vectors then I would be able to reach all of the points $(2n, m)$ by adding together $n$ and $m$ lots of these combinations. Algebra quickly becomes very complicated, as there are too many variable to give me anything to 'solve'. So, look at some more particular cases. Look at a base vector with an $x$-component of $4$. $$\\begin{pmatrix} 4 \\\\ 0\\end{pmatrix}$$ Clearly, to be able to reach the point $(2, 0)$ the other base vector would need to be $\\begin{pmatrix} \\pm 2 \\\\ 0\\end{pmatrix}$, but this would not allow us to reach the point $(0, 1)$. Thus, any other possible pairs of base vectors would have to have non-zero $x$ and $y$ components. A suitable pair would be $$\\mathbf{a}= \\begin{pmatrix} 4 \\\\ 1\\end{pmatrix}\\,, \\mathbf{b} =\\begin{pmatrix}2 \\\\ 1\\end{pmatrix}$$ To see why, note that the combinations $\\mathbf{a}-\\mathbf{b}$ and $2\\mathbf{b}-\\mathbf{a}$ take us to the", "vector from $(3,3)$ to $(2,5)$ in $\\mathbb{R}^2$ 3. the vector from $(1,0,6)$ to $(5,0,3)$ in $\\mathbb{R}^3$ 4. the vector from $(6,8,8)$ to $(6,8,8)$ in $\\mathbb{R}^3$ This exercise is recommended for all readers. Problem 2 Decide if the two vectors are equal. 1. the vector from $(5,3)$ to $(6,2)$ and the vector from $(1,-2)$ to $(1,1)$ 2. the vector from $(2,1,1)$ to $(3,0,4)$ and the vector from $(5,1,4)$ to $(6,0,7)$ This exercise is recommended for all readers. Problem 3 Does $(1,0,2,1)$ lie on the line through $(-2,1,1,0)$ and $(5,10,-1,4)$? This exercise is recommended for all readers. Problem 4 1. Describe the plane through $(1,1,5,-1)$, $(2,2,2,0)$, and $(3,1,0,4)$. 2. Is the origin in that plane? Problem 5 Describe the plane that contains this point and line. $\\begin{pmatrix} 2 \\\\ 0 \\\\ 3 \\end{pmatrix} \\qquad \\{\\begin{pmatrix} -1 \\\\ 0 \\\\ -4 \\end{pmatrix} +\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}t \\,\\big|\\, t\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 6 Intersect these planes. $\\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}t+ \\begin{pmatrix} 0 \\\\ 1 \\\\ 3 \\end{pmatrix}s \\,\\big|\\, t,s\\in\\mathbb{R}\\} \\qquad \\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\\\ 0 \\end{pmatrix}k+ \\begin{pmatrix} 2 \\\\ 0 \\\\ 4 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 7 Intersect each pair, if possible. 1. $\\{\\begin{pmatrix} 1 \\\\", "the vector $$\\vec{f} = \\begin{pmatrix} -4 \\\\ 0 \\end{pmatrix}$$ 3. the vector $$\\vec{g} = \\begin{pmatrix} 1 \\\\ 6 \\end{pmatrix}$$ 4. the vector $$\\vec{h} = \\begin{pmatrix} 5 \\\\ 5 \\end{pmatrix}$$ Note: you can download a sheet of gridded paper here. Note: this exercise can be downloaded as a worksheet to practice with: ### Solutions 1. For $$\\vec{a} = \\begin{pmatrix} 5 \\\\ 2 \\end{pmatrix}$$ we find: 2. For $$\\vec{b} = \\begin{pmatrix} -3 \\\\ 4 \\end{pmatrix}$$ we find: 3. For $$\\vec{c} = \\begin{pmatrix} 6 \\\\ -3 \\end{pmatrix}$$ we find: 4. For $$\\vec{d} = \\begin{pmatrix} 0 \\\\ 5 \\end{pmatrix}$$ we find: 5. For $$\\vec{e} = \\begin{pmatrix} -2 \\\\ -5 \\end{pmatrix}$$ we find: 6. For $$\\vec{f} = \\begin{pmatrix} -4 \\\\ 0 \\end{pmatrix}$$ we find: 7. For $$\\vec{g} = \\begin{pmatrix} 1 \\\\ 6 \\end{pmatrix}$$ we find: 8. For $$\\vec{h} = \\begin{pmatrix} 5 \\\\ 5 \\end{pmatrix}$$ we find: ### Tutorial: How to find a vector's components In this tutorial we learn what a vector is. We learn how to write vectors as both column and row vectors as well as how to represent vectors graphically; that is how to draw vectors. ## Exercise 2 Writing your answers as column vectors, find the coordinates of each of the vectors drawn here: 1. vector $$\\vec{a}$$: 2. vector $$\\vec{b}$$: 3. vector $$\\vec{c}$$: 4. vector $$\\vec{d}$$: 5. vector $$\\vec{e}$$:", "we are given a vector $\\mathbf t$ in the vector space (but it might not be in $\\mathcal{L}(\\mathbf B)$) and we want to find the closest vector in $\\mathcal{L}(\\mathbf B)$. There is a simple reduction from $\\textsf{CVP}$ to $\\textsf{SVP}$ by setting $\\mathbf t = \\mathbf 0$. There is also a reduction in the other direction, which involves extending the basis of $\\mathcal{L}(\\mathbf B)$ to also include $\\mathbf t$. This is called embedding. Let us return to the example of DSA and different-moduli equations. The equations we got from before $\\begin{array}{rclc} s_1 \\cdot k_1 - r_1\\cdot x & = & H(m_1) &\\pmod q \\\\ s_2 \\cdot k_2 - r_2\\cdot x & = & H(m_1) &\\pmod q \\\\ -a\\cdot k_1 + 1\\cdot k_2 & = & c & \\pmod M \\\\ \\end{array}$ can be formulated as basis vectors. In matrix form, we have $\\begin{pmatrix} -r_1 & s_1 & 0 & q & 0 & 0 \\\\ -r_2 & 0 & s_2 & 0 & q & 0 \\\\ 0 & -a & 1 & 0 & 0 & M\\end{pmatrix} \\cdot \\mathbf y = \\begin{pmatrix}H(m_1) \\\\ H(m_2) \\\\ c\\end{pmatrix}$ or (by embedding technique): $\\begin{pmatrix} -r_1 & s_1 & 0 & q & 0 & 0 & H(m_1) \\\\ -r_2 & 0 & s_2 & 0 & q & 0 & H(m_2)\\\\", "# Thread: Spans and vectors 1. ## Spans and vectors If a question asks, Is the vector $\\mathbf{b}=\\begin{pmatrix}-2\\\\ -6\\\\ -4\\end{pmatrix}\\in \\mbox{span}(\\mathbf{v_1, v_2, v_3, v_4})$ where: $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 3\\\\ 0\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}2\\\\ 2\\\\ 1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ -1\\end{pmatrix}$, $\\mathbf{v_4}=\\begin{pmatrix}1\\\\ -2\\\\ 1\\end{pmatrix}$ Do we just make it a matrix and solve the matrix? And how would we solve it if we have 4 variables but only 3 equations? 2) Is the set of vectors $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 2\\\\ 3\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}1\\\\ 1\\\\ -1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ 5\\end{pmatrix}$ a spanning set of $\\mathbb{R}^3$? Do we just make this a matrix with say $\\mathbf{x}=\\begin{pmatrix}x_1\\\\ x_2\\\\ x_3\\end{pmatrix}$ as our solution and see if we can solve the matrix? 2. So your first question is asking whether there exist scalars $a_{1},a_{2},a_{3},a_{4}$ such that $\\mathbf{b}=a_{1}\\mathbf{v}_{1}+a_{2}\\mathbf{v}_{2} +a_{3}\\mathbf{v}_{3}+a_{4}\\mathbf{v}_{4}.$ This equation you can translate into a normal matrix problem of the form $A\\mathbf{x}=\\mathbf{b}$. Row reduce that problem. If you find any solutions (one or infinitely many), then the answer is yes. If you find that there are no solutions, then the answer is no. Your second question, again, is asking whether there exist scalars $c_{1}, c_{2}, c_{3}$ such that $c_{1}\\mathbf{v}_{1}+c_{2}\\mathbf{v}_{2}+c_{3}\\math bf{v}_{3}=\\mathbf{x}$ for your arbitrary $\\mathbf{x}.$ This, again, can be translated into a standard matrix problem. Since this problem will result in a square matrix, you can use the theory of determinants to help you out.", "# Basis and dimension • September 25th 2010, 09:37 PM acevipa Basis and dimension Find the basis for and the dimension of $W=span(\\mathbf{v_1, v_2, v_3})$, where $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 2\\\\ 3\\end{pmatrix}, \\mathbf{v_2}=\\begin{pmatrix}1\\\\ 1\\\\ -1\\end{pmatrix}, \\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ 5\\end{pmatrix}$ I'm not too sure how to start this question off • September 26th 2010, 07:43 AM HallsofIvy If the three given vectors are independent, then they are a basis for their own span and the dimension is 3. If they are not, then one of the vectors can be written as a linear combination of the other two and so those two form a basis and the dimension is 2. (It clear that these vectors are not all multiples of one another so the dimension is not 1.) So start by checking if they are independent or not: do there exist numbers, a, b, and c such that $av_1+ bv_1+ cv_2= a\\begin{pmatrix}1 \\\\ 2 \\\\ 3\\end{pmatrix}+ b\\begin{pmatrix}1 \\\\ 1 \\\\ -1\\end{pmatrix}+ c\\begin{pmatrix}-1 \\\\ 0 \\\\ 5\\end{pmatrix}= \\begin{pmatrix} 0 \\\\ 0 \\\\ 0\\end{pmatrix}$ That is the same as $\\begin{pmatrix}a \\\\ 2a \\\\ 3a\\end{pmatrix}+ \\begin{pmatrix}b \\\\ b \\\\ -b\\end{pmatrix}+ \\begin{pmatrix}-c \\\\ 0 \\\\ 5c\\end{pmatrix}= \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$ $\\begin{pmatrix}a+ b- c \\\\ 2a+ b \\\\ 3a- b+ 5c\\end{pmatrix}= \\begin{pmatrix}0 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and that is the same as the three" ]

[ "a_{nn} \\end{pmatrix} \\qquad\\text{and}\\qquad {\\mathbf x} = \\begin{pmatrix} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n \\end{pmatrix}. \\end{equation*} • As in the case of ${\\mathbb R}^2\\text{,}$ we can solve the system ${\\mathbf x}' = A {\\mathbf x}$ by finding eigenvalues and eigenvectors for $A\\text{.}$ • The Principle of Superposition holds for higher-order systems. If ${\\mathbf x}_1(t)$ and ${\\mathbf x}_2(t)$ are solutions for ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ then \\begin{equation*} c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t) \\end{equation*} is a solution for the system. • The geometry for a system in ${\\mathbb R}^3$ is more complicated than the planar case. However, the solutions are usually characterized by stable lines or stable planes. SubsectionExercises 1 Solve the system \\begin{align*} x' & = -5x + 15y - 4z\\\\ y' & = -x + 5y - z\\\\ z' & =4 x -6y + 3z\\\\ x(0) & = -1\\\\ y(0) & = 2\\\\ z(0) & = 0 \\end{align*} 2 Find the general solution of the system \\begin{align*} x_1' & = 6x_1 + 9x_2 + x_3 + 4 x_4\\\\ x_2' & = -x_1 + x_2 + x_3\\\\ x_3' & = 3x_1 + 8x_2 + 4x_3 + 4 x_4\\\\ x_4' & = -3x_1 - 8x_2 -7 x_3 - 5 x_4 \\end{align*} 3 Let $a x'' + b x' + cx = 0\\text{,}$ where $a \\neq 0$", "a {\\mathbf x}_{\\text{Re}}(t) + b{\\mathbf x}_{\\text{Im}}(t). \\end{align*} Notice that the solutions \\begin{equation*} {\\mathbf x}(t) = c_1 \\begin{pmatrix} \\cos \\beta t \\\\ - \\sin \\beta t \\end{pmatrix} + c_2 \\begin{pmatrix} \\sin \\beta t\\\\ \\cos \\beta t \\end{pmatrix} \\end{equation*} are periodic with period $2 \\pi / \\beta\\text{.}$ This type of system is called a center. ##### Example3.26 Consider the initial value problem \\begin{align*} \\frac{dx}{dt} \\amp = 2y\\\\ \\frac{dy}{dt} \\amp = -2x\\\\ x(0) \\amp = 1\\\\ y(0) \\amp = 2. \\end{align*} The eigenvalues of this system are $\\lambda = \\pm 2i\\text{.}$ Therefore, the general solution to the system is \\begin{align*} x(t) \\amp = c_1 \\cos 2t + c_2 \\sin 2t\\\\ y(t) \\amp = - c_1 \\sin 2t + c_2 \\cos 2t. \\end{align*} Using the initial conditions to solve for $c_1$ and $c_2\\text{,}$ the solution to our initial value problem is \\begin{align*} x(t) \\amp = \\cos 2t + 2 \\sin 2t\\\\ y(t) \\amp = - \\sin 2t + 2 \\cos 2t. \\end{align*} The phase portrait is a circle of radius 2 about the origin (Figure 3.27). # Subsection3.4.2Spiral Sinks and Sources¶ permalink Now let us consider the system ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ where \\begin{equation*} A = \\begin{pmatrix} \\alpha & \\beta \\\\ - \\beta & \\alpha \\end{pmatrix} \\end{equation*} and $\\alpha$ and $\\beta$ are nonzero real numbers. The characteristic equation of $A$", "2 & 1 \\\\ 1 & 1 & 2 \\end{pmatrix} \\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix} \\\\ &= \\begin{pmatrix} 5\\cdot0 + 0\\cdot-1 + 0\\cdot1 \\\\ 1\\cdot0 + 2\\cdot-1 + 1\\cdot1 \\\\ 1\\cdot0 + 1\\cdot-1 + 2\\cdot1 \\\\ \\end{pmatrix} \\\\ &= \\begin{pmatrix} 0 \\\\ 3 \\\\ 3 \\end{pmatrix} = 3 \\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix}. \\end{align*} 4.3. Eigenvalue $\\boldsymbol{\\lambda = 5}$ Substituting $\\lambda=5$, yields \\begin{align*} (5-5)x_1 &= 0, \\\\ x_1 + (2-5)x_2 + x_3 &= 0, \\\\ x_1 + x_2 + (2-5)x_3 &= 0. \\end{align*} Working this further: \\begin{align*} 0x_1 &= 0, \\\\ x_1 – 3x_2 + x_3 &= 0, \\\\ x_1 + x_2 – 3x_3 &= 0. \\end{align*} The first equation reduces to $0=0$, which is another way of saying that, logically, $x_1$ can take on any value as a solution to the equation. Working the other two equations some more too, we get \\begin{align*} 0 &= 0, \\\\ x_3 &= 3x_2 – x_1, \\\\ x_2 &= 3x_3 – x_1. \\\\ \\end{align*} These last two equations need some more work still. Substituting the second equation into the third, we get \\begin{align*} x_2 &= 3(3x_2 – x_1)-x_1, \\\\ x_2 &= 9x_2 – 3x_1-x_1, \\\\ 8x_2 &= 4x_1, \\\\ x_2 &= \\frac{x_1}{2}. \\end{align*} Substituting this result into the second equation, gives \\begin{align*} x_3 &= 3\\left(\\frac{x_1}{2}\\right)-x_1, \\\\ x_3", "\\end{array}\\right]. \\end{align*} This yields the solution c_1=2 and c_2=-5. Hence, we have the linear combination \\[\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}. ### (b) Compute $A^5\\mathbf{v}$. We first compute $A\\mathbf{v}$. We have $A\\mathbf{v}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix} =\\begin{bmatrix} -4 \\\\ 0 \\\\ 2 \\end{bmatrix}=2\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=2\\mathbf{v}.$ Using this relation $A\\mathbf{v}=2\\mathbf{v}$, we obtain $A^2\\mathbf{v}=AA\\mathbf{v}=A(2\\mathbf{v})=2A\\mathbf{v}=2(2\\mathbf{v})=2^2\\mathbf{v}.$ Next, we have $A^3\\mathbf{v}=AA^2\\mathbf{v}=A(2^2\\mathbf{v})=2^2A\\mathbf{v}=2^2(a\\mathbf{v})=2^3\\mathbf{v}.$ Repeating this process, we see that $A^5\\mathbf{v}=2^5\\mathbf{v}$. Or, we can find this by computing as follows: \\begin{align*} A^5\\mathbf{v}=A^2A^3\\mathbf{v}=A^2(2^3\\mathbf{v})=2^3A^2\\mathbf{v}=2^3(2^2\\mathbf{v})=2^5\\mathbf{v}. \\end{align*} In summary, we have $A^5\\mathbf{v}=2^5\\mathbf{v}=32\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}.$ ### (c) Compute $A^5\\mathbf{w}$. First, we note that $A\\mathbf{w}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=-\\mathbf{w}.$ Using this relation $A\\mathbf{w}=-\\mathbf{w}$ as in part (a), we obtain $A^5\\mathbf{w}=(-1)^5\\mathbf{w}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}.$ ### (d) Compute $A^5\\mathbf{u}$. Using the linear combination $\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}$ obtained part (a), we compute \\begin{align*} A^5\\mathbf{w}&=A^5(2\\mathbf{v}-5\\mathbf{w})\\\\ &=2A^5\\mathbf{v}-5A^5\\mathbf{w}\\\\[6pt] &=2\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}-5\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix} &&\\text{by (b), (c)}\\\\[6pt] &=\\begin{bmatrix} -138 \\\\ -5 \\\\ 69", "+ b \\begin{pmatrix} \\sin \\beta t \\\\ \\cos \\beta t \\end{pmatrix}\\\\ & = a {\\mathbf x}_{\\text{Re}}(t) + b{\\mathbf x}_{\\text{Im}}(t). \\end{align*} Notice that the solutions \\begin{equation*} {\\mathbf x}(t) = c_1 \\begin{pmatrix} \\cos \\beta t \\\\ - \\sin \\beta t \\end{pmatrix} + c_2 \\begin{pmatrix} \\sin \\beta t\\\\ \\cos \\beta t \\end{pmatrix} \\end{equation*} are periodic with period $$2 \\pi / \\beta\\text{.}$$ This type of system is called a center. Consider the initial value problem \\begin{align*} \\frac{dx}{dt} \\amp = 2y\\\\ \\frac{dy}{dt} \\amp = -2x\\\\ x(0) \\amp = 1\\\\ y(0) \\amp = 2. \\end{align*} The eigenvalues of this system are $$\\lambda = \\pm 2i\\text{.}$$ Therefore, the general solution to the system is \\begin{align*} x(t) \\amp = c_1 \\cos 2t + c_2 \\sin 2t\\\\ y(t) \\amp = - c_1 \\sin 2t + c_2 \\cos 2t. \\end{align*} Using the initial conditions to solve for $$c_1$$ and $$c_2\\text{,}$$ the solution to our initial value problem is \\begin{align*} x(t) \\amp = \\cos 2t + 2 \\sin 2t\\\\ y(t) \\amp = - \\sin 2t + 2 \\cos 2t. \\end{align*} The phase portrait is a circle of radius 2 about the origin (Figure 3.4.3). ### Subsection3.4.2Spiral Sinks and Sources Now let us consider the system $${\\mathbf x}' = A {\\mathbf x}\\text{,}$$ where \\begin{equation*} A = \\begin{pmatrix} \\alpha & \\beta \\\\ - \\beta & \\alpha \\end{pmatrix} \\end{equation*} and $$\\alpha$$ and", "of the size $N_1=N/2$ and $N_2=N/2$ (for simplicitys sake). You should now solve them separately. $\\mathbf A_1^{-1}=\\mathbf B_1$ $\\mathbf A_2^{-1}=\\mathbf B_2$ The next step is to put them back together, that is done with the help of the so called \"Thevenin Matrix\"(in our case it is 1$\\times$1). You can look up the exact principle for higher orders(more connection points), but for this example it looks like: \\begin{align} \\mathbf{B_{TH}}=B_{1mm}+B_{2nn}-2B_{mn} \\end{align} For our case we have $B_{mn}=0$. Now we need the solutions $x_1$ and $x_2$ to form the coefficients $b_{th}$. $\\mathbf x_{th}=x_{1m}-x_{2n}$ $\\mathbf b_{p}=\\mathbf{B_{TH}}^{-1} \\mathbf x_{th}$ \\begin{align} \\mathbf b_{th}=\\begin{bmatrix}0&\\cdots&b_{p}&\\cdots&-b_{p}&\\cdots&0 \\end{bmatrix}^T \\end{align} The $\\mathbf b_{th}$ matrix only has nonzero elements at $m$ an $N/2 +n$. Now we can finally find the solution $x_n$ for the whole system: \\begin{align} \\mathbf x_n=\\begin{bmatrix}x_1\\\\x_2 \\end{bmatrix}-\\begin{bmatrix}B_1&0\\\\0&B_2 \\end{bmatrix}\\begin{bmatrix}b_{th} \\end{bmatrix} \\end{align} I'm more used to the engineering notation with $Z, I, U$ and so on, so excuse for non-standard symbol usage. - [this should be a comment to @Juan Joder's answer or to the OP's question, but not enough repo] strang gilbert - linear algebra and its applications 3rd edition page 16 at the footer mentions that complexity had fallen already fallen below $O(Cn^{2.376})$ at the date of writing 1988, altough $C$ is so large that makes the algorithm impractical for most matrix sizes found in practice today.", "& d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$and set$D=ad-bc$as before. Then the following are equivalent: 1.$(D,N)=1$. 2.$A$has an inverse matrix. 3. if$x$and$y$are not both$0$in$\\mathbb{Z}/N\\mathbb{Z}$, then$A\\begin{pmatrix}x\\\\y \\end{pmatrix}\\ne\\begin{pmatrix}0\\\\0\\end{pmatrix}$. 4.$A$gives a one-to-one correspondence of$(\\mathbb{Z}/N\\mathbb{Z})^2with itself. Example. Solve the following systems of simultaneous congruences. 1. \\begin{align*}2x+3y&\\equiv 1\\mod 26\\\\ 7x+8y&\\equiv 2\\mod 26 \\end{align*} 2. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 2\\mod 26 \\end{align*} 3. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 1\\mod 26 \\end{align*} Solution. 1. The system of simultaneous congruences can be written as the matrix congruenceAX\\equiv B\\mod 26$or as the matrix equation$AX=B$in$\\mathbb{Z}/26\\mathbb{Z}$, where $$A=\\begin{pmatrix}2 & 3\\\\7 & 8\\end{pmatrix},\\ X=\\begin{pmatrix}x\\\\y\\end{pmatrix},\\ B=\\begin{pmatrix}1\\\\2\\end{pmatrix}$$$D=16-21=-5$and$(-5,26)=1$, so by above Theorem, we see that$A^{-1}$exists. In fact, we have already found$A^{-1}$in the previous example. The solution$Xis given by \\begin{align*} X&\\equiv A^{-1}B\\mod 26\\\\&\\equiv\\begin{pmatrix} 14 & 11\\\\ 17 & 10 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 2 \\end{pmatrix}\\mod 26\\\\ &\\equiv\\begin{pmatrix} 10\\\\ 11 \\end{pmatrix}\\mod 26 \\end{align*} 2. The matrix in parts 2 and 3 does not have an inverse matrix in\\mathbb{Z}/26\\mathbb{Z}$because$D=9-21=-12$and$(-12,26)=2\\ne 1$. However, it does have an inverse matrix in$\\mathbb{Z}/13\\mathbb{Z}and the solution is given by \\begin{align*} \\begin{pmatrix} x\\\\ y \\end{pmatrix}&\\equiv\\begin{pmatrix} 9 & 10\\\\ 6 & 1 \\end{pmatrix}\\begin{pmatrix} 1\\\\ 2 \\end{pmatrix}\\mod 13\\\\ &\\equiv\\begin{pmatrix} 3\\\\ 8 \\end{pmatrix}\\mod 13 \\end{align*} We now examine the possibilities of having solutions in\\mathbb{Z}/26\\mathbb{Z}$. Since$x\\equiv 3\\mod 13$and$y\\equiv 8\\mod 13$,$x$and$y$can be written as$x=3+13k_1$and$y=8+13k_2$for some$k_1,k_2\\in\\mathbb{Z}. \\begin{align*} x+3y&\\equiv 1+13(k_1+k_2)\\mod 26\\\\ &\\equiv 1\\mod 26 \\end{align*} if and only ifk_1+k_2$is even. Suppose that$k_1+k_2$is even, so that$x+3y\\equiv", "3.2 Hence, x = 1.2, y = 2.2, z = 3.2. 6. Apply Cramer’s rule to solve the following equations. x + y + z =3 x + 2y + 3z = 4 x + 4y + 9z = 1 a) x = -0.5, y = 6, z = -2.5 b) x = -0.5, y = 4, z = -2.5 c) x = 4.5, y = 6, z = 1 d) x = 4.5, y = 6, z = 2 Explanation: ∆ = $$\\begin{pmatrix}1&1&1\\\\1&2&3\\\\1&4&9\\end{pmatrix}$$ = 2 X = (1/∆) = $$\\begin{pmatrix}3&1&1\\\\4&2&3\\\\1&4&9\\end{pmatrix}$$ = -0.5 Y = (1/∆) = $$\\begin{pmatrix}1&3&1\\\\1&4&3\\\\1&1&9\\end{pmatrix}$$ = 6 Z = (1/∆) = $$\\begin{pmatrix}1&1&3\\\\1&2&4\\\\1&4&1\\end{pmatrix}$$ = -2.5 Hence, X = -0.5, y = 6, z = -2.5. 7. Apply Cramer’s rule to solve the following equations. 2x – y + z = 3 3x + 2y + 4z = 19 6x + 7y – z = 17 a) X = 0.456, y = 1.5442, z = 3.154 b) X = 0.437, y = 1.5312, z = 3.656 c) X = 0.356, y =2.547, z = 5.474 d) X = 0.356, y = 1.722, z = 9.424 Explanation: ∆ = $$\\begin{pmatrix}2&-1&1\\\\3&4&3\\\\6&7&1\\end{pmatrix}$$ = -64 x = (1/∆) = $$\\begin{pmatrix}3&-1&1\\\\19&4&3\\\\17&7&1\\end{pmatrix}$$ = -28/-64 = 0.437 y = (1/∆) = $$\\begin{pmatrix}2&3&1\\\\3&19&3\\\\6&17&1\\end{pmatrix}$$ = -98/-64 = 1.5312 z = (1/∆) = $$\\begin{pmatrix}2&-1&3\\\\3&4&19\\\\6&7&17\\end{pmatrix}$$ = -234/-64 =", "the original system using another change of coordinates. Suppose that we consider a linear system $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}\\tag{3.6.2}$$ where $$T$$ is an invertible matrix. If $${\\mathbf y}(t)$$ is a solution of (3.6.2), we claim that $${\\mathbf x}(t) = T {\\mathbf y}(t)$$ solves the equation $${\\mathbf x}' = A {\\mathbf x}\\text{.}$$ Indeed, \\begin{align*} {\\mathbf x}'(t) & = (T {\\mathbf y})'(t)\\\\ & = T {\\mathbf y}'(t)\\\\ & = T( (T^{-1} A T) {\\mathbf y}(t))\\\\ & = A (T {\\mathbf y}(t))\\\\ & = A {\\mathbf x}(t). \\end{align*} We can think of this in two ways. 1. A linear map $$T$$ converts solutions of $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}$$ to solutions of $${\\mathbf x}' = A {\\mathbf x}\\text{.}$$ 2. The inverse of a linear map $$T$$ takes solutions of $${\\mathbf x}' = A {\\mathbf x}$$ to solutions of $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}\\text{.}$$ We are now in a position to solve our problem of finding solutions of an arbitrary linear system \\begin{equation*} \\begin{pmatrix} x' \\\\ y' \\end{pmatrix} = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = \\begin{pmatrix} x \\\\ y \\end{pmatrix}. \\end{equation*} ### Subsection3.6.4Distinct Real Eigenvalues Consider the system $${\\mathbf x}' = A {\\mathbf x}\\text{,}$$ where $$A$$ has two real, distinct eigenvalues $$\\lambda_1$$ and $$\\lambda_2$$ with eigenvectors $${\\mathbf v}_1$$", "that finding$21^{-1}\\in\\mathbb{Z}/26\\mathbb{Z}$is equivalent to solving the diophantine equation$21x+26y=1. \\begin{align*} 26&=21\\cdot 1+5\\\\ 21&=5\\cdot 4+1 \\end{align*} and \\begin{align*} 1&=21-5\\cdot 4\\\\ &=21-(26-21)4\\\\ &=5\\cdot 21+26(-4) \\end{align*} Thus, we find21^{-1}\\equiv 5\\mod 26. \\begin{align*} A^{-1}&=\\begin{pmatrix} 5\\cdot 8 & -5\\cdot 3\\\\ -5\\cdot 7 & 5\\cdot 2 \\end{pmatrix}\\\\ &=\\begin{pmatrix} 40 & -15\\\\ -35 & 10 \\end{pmatrix}\\\\ &\\equiv\\begin{pmatrix} 14 & 11\\\\ 17 & 10 \\end{pmatrix}\\mod 26 \\end{align*} LetA=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$. Then $$\\begin{pmatrix} x’\\\\ y’ \\end{pmatrix}=A\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} ax+by\\\\ cx+dy \\end{pmatrix}$$ This gives a linear map or a linear transformation from vectors to vectors. We introduce the following theorem without proof. Theorem. Let$A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$and set$D=ad-bc$as before. Then the following are equivalent: 1.$(D,N)=1$. 2.$A$has an inverse matrix. 3. if$x$and$y$are not both$0$in$\\mathbb{Z}/N\\mathbb{Z}$, then$A\\begin{pmatrix}x\\\\y \\end{pmatrix}\\ne\\begin{pmatrix}0\\\\0\\end{pmatrix}$. 4.$A$gives a one-to-one correspondence of$(\\mathbb{Z}/N\\mathbb{Z})^2with itself. Example. Solve the following systems of simultaneous congruences. 1. \\begin{align*}2x+3y&\\equiv 1\\mod 26\\\\ 7x+8y&\\equiv 2\\mod 26 \\end{align*} 2. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 2\\mod 26 \\end{align*} 3. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 1\\mod 26 \\end{align*} Solution. 1. The system of simultaneous congruences can be written as the matrix congruenceAX\\equiv B\\mod 26$or as the matrix equation$AX=B$in$\\mathbb{Z}/26\\mathbb{Z}$, where $$A=\\begin{pmatrix}2 & 3\\\\7 & 8\\end{pmatrix},\\ X=\\begin{pmatrix}x\\\\y\\end{pmatrix},\\ B=\\begin{pmatrix}1\\\\2\\end{pmatrix}$$$D=16-21=-5$and$(-5,26)=1$, so by above Theorem, we see that$A^{-1}$exists. In fact, we have already found$A^{-1}$in the", "solution. Thus, a general solution to our system is \\begin{equation*} \\mathbf x(t) = c_1 e^{2t} \\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix} + c_2 e^{-2t} \\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix} \\end{equation*} or \\begin{align*} x(t) & = 3 c_1 e^{2t} + c_2 e^{-2t}\\\\ y(t) & = c_1 e^{2t} - c_2 e^{-2t}. \\end{align*} If we are given initial conditions, say $x(0) = 0$ and $y(0) = 1\\text{,}$ then we can determine $c_1$ and $c_2$ by solving the linear system of equations \\begin{align*} 3 c_1 + c_2 & = 0\\\\ c_1 - c_2 & = 1 \\end{align*} to get $c_1 = 1/4$ and $c_2 = -3/4\\text{.}$ Thus, the solution to our initial value problem is \\begin{align*} x(t) & = \\frac{3}{4} e^{2t} - \\frac{3}{4} e^{-2t}\\\\ y(t) & = \\frac{1}{4} e^{2t} + \\frac{3}{4} e^{-2t}. \\end{align*} If ${\\mathbf x}_1(t)$ and ${\\mathbf x}_2(t)$ are solutions to the linear system ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ then \\begin{align*} {\\mathbf x}_1' & = A {\\mathbf x}_1\\\\ {\\mathbf x}_2' & = A {\\mathbf x}_2. \\end{align*} Thus, for any two real numbers $c_1$ and $c_2\\text{,}$ \\begin{align*} \\frac{d}{dt} (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t)) & = \\alpha \\frac{d}{dt} {\\mathbf x}_1(t) + c_2 \\frac{d}{dt} {\\mathbf x}_2(t)\\\\ & = c_1 A {\\mathbf x}_1(t) + c_2 A {\\mathbf x}_2(t)\\\\ & = A (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t) ). \\end{align*} We state this", "3.656 Hence, X = 0.437, y = 1.5312, z = 3.656. 8. Apply Cramer’s rule to solve the following equations. 3x + y + z = 8 2x – 3y -2z = -5 7x + 2y – 5z = 0 a) X = 1, y =4, z = 2.5 b) X = 4.562, y =4, z = 3.1 c) X = 0.2179, y =1, z = 2.5 d) X = 4.2, y =4, z = 3.145 Explanation: ∆ = $$\\begin{pmatrix}3&1&1\\\\2&-3&-2\\\\7&2&-5\\end{pmatrix}$$ = 78 x = (1/∆) = $$\\begin{pmatrix}8&1&1\\\\-5&-3&-2\\\\0&2&-5\\end{pmatrix}$$ = 117/78 = 0.2179 y = (1/∆) = $$\\begin{pmatrix}3&8&1\\\\2&-5&-2\\\\7&0&-5\\end{pmatrix}$$ = 78/78 = 1 z = (1/∆) = $$\\begin{pmatrix}3&1&8\\\\2&-3&-5\\\\7&2&5\\end{pmatrix}$$ = 195/78 = 2.5 Hence, X = 0.2179, y =1, z = 2.5. 9. Apply Cramer’s rule to solve the following equations. 2x + y + z = 10 3x + 2y + 3z = 18 X + 4y +9z = 16 a) X = -9, y = 1, z = 5 b) X = 7, y = -9, z = 5 c) X = 7, y = 1, z = 5 d) X = 9, y = 1, z = 3 Explanation: ∆ = $$\\begin{pmatrix}2&1&1\\\\3&2&3\\\\1&4&9\\end{pmatrix}$$ = -2 x = (1/∆) = $$\\begin{pmatrix}10&1&1\\\\18&2&3\\\\16&4&9\\end{pmatrix}$$ = -14/-2 = 7 y = (1/∆) = $$\\begin{pmatrix}2&10&1\\\\3&18&3\\\\1&16&9\\end{pmatrix}$$ = 18/-2 = -9 z = (1/∆) = $$\\begin{pmatrix}2&1&10\\\\3&2&18\\\\1&4&16\\end{pmatrix}$$ = -10/-2", "\\sqrt{3} t \\\\ \\sqrt{3} \\cos \\sqrt{3} t \\end{pmatrix} + c_3 \\begin{pmatrix} \\cos t \\\\ \\cos t \\\\ -\\sin t \\\\ - \\sin t \\end{pmatrix} + c_4 \\begin{pmatrix} \\sin t \\\\ \\sin t \\\\ \\cos t \\\\ \\cos t \\end{pmatrix}. \\end{equation*} If we have the following initial conditions, \\begin{align*} x_1(0) & = 0\\\\ x_2(0) & = 0\\\\ x_1'(0) = x_3(0) & = 2\\\\ x_2'(0) = x_4(0) & = 2, \\end{align*} we can determine $c_1 = c_2 = c_3 = 0$ and $c_4 = 2\\text{.}$ Thus, the solution to our initial value problem is \\begin{equation*} {\\mathbf x}(t) = \\begin{pmatrix} 2 \\sin t \\\\ 2 \\sin t \\\\ 2 \\cos t \\\\ 2 \\cos t \\end{pmatrix}, \\end{equation*} and the two masses will oscillate with a frequency of one and an amplitude of two. We leave the details as an exercise. Subsection3.8.4Important Lessons • \\begin{align*} x_1' & = a_{11} x_1 + \\cdots a_{1n} x_n\\\\ x_2' & = a_{21} x_1 + \\cdots a_{2n} x_n\\\\ & \\vdots\\\\ x_n' & = a_{n1} x_1 + \\cdots a_{nn} x_n \\end{align*} can be written in matrix form ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ where \\begin{equation*} A = \\begin{pmatrix} a_{11} & a_{12} & \\cdots & a_{1n} \\\\ a_{21} & a_{22} & \\cdots & a_{2n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ a_{n1} & a_{n2} & \\cdots &", "interchange the indices 1 and 2.) Multiply out the RHS and collect terms, we get \\begin{align} c &= c_1c_2 - s_1s_2(x_1x_2+y_1y_2+z_1z_2),\\tag{1}\\\\ sx &= c_1s_2 x_2 + c_2s_1 x_1 + s_1s_2 (y_1z_2 - z_1y_2),\\tag{2}\\\\ sy &= c_1s_2 y_2 + c_2s_1 y_1 + s_1s_2 (z_1x_2 - x_1z_2),\\tag{3}\\\\ sz &= c_1s_2 z_2 + c_2s_1 z_1 + s_1s_2 (x_1y_2 - y_1x_2).\\tag{4} \\end{align} Equations (2)-(4) give $$\\begin{pmatrix} x_2&x_1&y_1z_2 - z_1y_2\\\\ y_2&y_1&z_1x_2 - x_1z_2\\\\ z_2&z_1&x_1y_2 - y_1x_2 \\end{pmatrix} \\begin{pmatrix} c_1s_2\\\\c_2s_1\\\\s_1s_2 \\end{pmatrix} = \\begin{pmatrix} sx\\\\sy\\\\sz \\end{pmatrix}.$$ Solving it, we get $$\\begin{pmatrix} c_1s_2\\\\c_2s_1\\\\s_1s_2 \\end{pmatrix} = \\frac{1}{1 - ({\\mathbf u}_1\\cdot{\\mathbf u}_2)^2} \\begin{pmatrix} 1-x_1({\\mathbf u}_1\\cdot{\\mathbf u}_2)&1-y_1({\\mathbf u}_1\\cdot{\\mathbf u}_2)&1-z_1({\\mathbf u}_1\\cdot{\\mathbf u}_2)\\\\ 1-x_2({\\mathbf u}_1\\cdot{\\mathbf u}_2)&1-y_2({\\mathbf u}_1\\cdot{\\mathbf u}_2)&1-z_2({\\mathbf u}_1\\cdot{\\mathbf u}_2)\\\\ y_1z_2-z_1y_2&z_1x_2-x_1z_2&x_1y_2-y_1x_2 \\end{pmatrix} \\begin{pmatrix} sx\\\\sy\\\\sz \\end{pmatrix}.$$ Substitute this result into (1), we obtain also $c_1c_2$. Hence \\begin{align*} \\theta_1 = \\begin{cases} 2\\operatorname{atan2}(c_1c_2, s_1c_2)&\\ \\text{ if not } c_1c_2=s_1c_2=0,\\\\ 2\\operatorname{atan2}(c_1s_2, s_1s_2)&\\ \\text{ otherwise}, \\end{cases}\\\\ \\theta_2 = \\begin{cases} 2\\operatorname{atan2}(c_1c_2, c_1s_2)&\\ \\text{ if not } c_1c_2=c_1s_2=0,\\\\ 2\\operatorname{atan2}(s_1c_2, s_1s_2)&\\ \\text{ otherwise}, \\end{cases} \\end{align*} where $\\mathtt{atan2}$ is the quadrant-aware arctangent function.", "= -6, c = 5. Now let’s substitute these values in the quadratic formula. \\begin{align} & y=\\dfrac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}=\\dfrac{-(-6)\\pm \\sqrt{{{(-6)}^{2}}-4\\times 1\\times 5}}{2\\times 1} \\\\ & =\\dfrac{6\\pm \\sqrt{36-20}}{2}=\\dfrac{6\\pm \\sqrt{16}}{2}=\\dfrac{6\\pm 4}{2} \\\\ \\end{align} Thus we got $y=\\dfrac{6+4}{2}$ and $y=\\dfrac{6-4}{2}$. $y={}^{10}/{}_{2}=5,y={}^{2}/{}_{2}=1.$ Thus we got the values of y as 1, 5 by solving ${{y}^{2}}-6y+5=0$. $\\therefore y=1,y=5.......(4)$ Now let us compare equation (3) and (4). x = 2, x = 3 and y = 1, y = 5. We know that a parallelogram has 4 vertices. Thus let us consider the 4 vertices of a parallelogram ABCD. Frame the vertices with the help of values x and y. Thus the vertices are A (2, 1), B (3, 1), C (3, 5), D (2, 5). Thus from the figure, you can see the parallelogram ABCD with vertices A (2, 1), B (3, 1), C (3, 5), D (2, 5). We were asked to find the equation of a diagonal. So let us join AC. AC is a diagonal of parallelogram ABCD. The equation of the line formed by 2 points A and C can be written as, $\\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ here, $\\left( {{x}_{1}},{{y}_{1}} \\right)=\\left( 2,1 \\right)$ and $\\left( {{x}_{2}},{{y}_{2}} \\right)=\\left( 3,5 \\right)$. $\\therefore$Equation of line AC, i.e., $\\dfrac{y-1}{5-1}=\\dfrac{x-2}{3-2}\\Rightarrow \\dfrac{y-1}{4}=\\dfrac{x-2}{1}$ Let us cross multiply and simplify the above expression. \\begin{align} & y-1=4(x-2) \\\\ & y-1=4x-8 \\\\ &", "\\\\ y \\\\z \\end{pmatrix}$. Does that make it more clear how to proceed? (Pardon the horizontal vectors - not much space in the comment to make them all vertical.) – kccu May 11 '17 at 0:31 You're thinking the right thing. There is a way to do it systematically. Essentially you want to find $x,y,z$ such that $$x\\begin{pmatrix} 3 \\\\ 4 \\\\ 5 \\end{pmatrix}+ y\\begin{pmatrix} 4 \\\\ 5 \\\\ 6 \\end{pmatrix} + z\\begin{pmatrix} 5 \\\\ -9 \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} 3 \\\\ -11 \\\\ -1 \\end{pmatrix}$$ This amounts to solving this system of equations for $x,y,z$: \\begin{align*} 3x + 4y + 5z &= 3\\\\ 4x + 5y - 9z &= -11\\\\ 5x + 6y + z &= -1 \\end{align*} Once you have $x,y,z$ you'll be able to express $\\begin{pmatrix} 3 \\\\ -11 \\\\ -1\\end{pmatrix}$ as a linear combination of those other $3$ vectors above. Then use linearity of matrix-vector multiplication to get the desired result. Specifically, you'll want to use the fact that if $A$ is a matrix, $a,b,c$ are scalars, and $\\vec u, \\vec v, \\vec w$ are vectors, then $$A \\left[a\\vec u + b\\vec v + c\\vec w\\right] = a\\cdot A\\vec u + b\\cdot A\\vec v + c\\cdot A\\vec w.$$ • So far, I have $x=2, y=-2, z=1$, but I'm not sure what to do", "\\end{bmatrix}$$ Compare this to the item provided in part $a.$, but not including the IC, they are the same. Consider the differential system \\begin{align} \\frac{dY}{dt} = \\begin{pmatrix} 1 & -1 \\\\ 1 & 3 \\\\ \\end{pmatrix} Y . \\end{align} Let \\begin{align} Y(t) = \\begin{pmatrix} a(t) \\\\ b(t) \\end{pmatrix} \\end{align} such that \\begin{align} \\begin{pmatrix} a^{'} \\\\ b^{'} \\end{pmatrix} = \\begin{pmatrix} 1 & -1 \\\\ 1 & 3 \\\\ \\end{pmatrix} \\cdot \\begin{pmatrix} a \\\\ b \\end{pmatrix}. \\end{align} The set of differential equations is then given by \\begin{align} a' &= a - b \\\\ b' &= a + 3b \\end{align} of which $a$ and $b$ satisfy the second order equation $w'' - 4 w' +4 w = 0$. The solution of this second order equation is $w(t) = (A + B t ) e^{2t}$ and lead to the form of $Y(t)$ being \\begin{align} Y(t) = \\begin{pmatrix} A + B t \\\\ C + D t \\end{pmatrix} \\, e^{2t}. \\end{align} Now the initial condition is \\begin{align} Y(0) = \\begin{pmatrix} 0 \\\\ 2 \\end{pmatrix} \\end{align} for which \\begin{align} \\begin{pmatrix} 0 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} A \\\\ C \\end{pmatrix} \\end{align} which yields the solution being \\begin{align} Y(t) = \\begin{pmatrix} B t \\\\ 2 + D t \\end{pmatrix} \\, e^{2t}. \\end{align} Using this result to take its derivative and apply it to the differential", "2z \\amp = 0\\\\ -5x + 10y + 4z \\amp = 0\\\\ -11x - 19y - 7z \\amp = 0. \\end{align*} It is easy to check that $(2, -3, 5)$ is a solution. Similarly, we can determine that $(5, -7, 13)$ is an eigenvector for $\\lambda = 1$ and $(1, -1, 2)$ is an eigenvector for $\\lambda = -1\\text{.}$ Thus, we have found three solutions for the system ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ \\begin{equation*} \\mathbf x_1(t) = e^{2t} \\begin{pmatrix} 2 \\\\ -3 \\\\ 5 \\end{pmatrix}, \\mathbf x_2(t) = e^{t} \\begin{pmatrix} 5 \\\\ -7 \\\\ 13 \\end{pmatrix}, \\quad \\text{and} \\quad \\mathbf x_3(t) = e^{-t} \\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}. \\end{equation*} The Principle of Superposition also holds for higher-order systems. If ${\\mathbf x}_1(t)$ and ${\\mathbf x}_2(t)$ are solutions for ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ then \\begin{equation*} c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t) \\end{equation*} is a solution for the system, since \\begin{align*} \\frac{d}{dt}[c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t)] & = c_1 \\frac{d}{dt}{\\mathbf x}_1(t) + c_2 \\frac{d}{dt}{\\mathbf x}_2(t)\\\\ & = c_1 A {\\mathbf x}_1(t) + c_2 A {\\mathbf x}_2(t)\\\\ & = A[c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t)]. \\end{align*} Consequently, \\begin{equation*} {\\mathbf x}(t) = c_1 e^{2t} \\begin{pmatrix} 2 \\\\ -3 \\\\ 5 \\end{pmatrix} + e^{t} \\begin{pmatrix} 5 \\\\ -7 \\\\ 13 \\end{pmatrix} + c_3 e^{-t} \\begin{pmatrix}", "be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these equations for the constants gives: \\begin{align}c_{1}=\\frac{-19-9\\ln2}{2}\\end{align} \\begin{align}c_{2}=\\frac{-3+3\\ln2}{2}\\end{align} Therefore the general solution for the IVP is: \\begin{align}\\textbf{x}(t)=\\frac{-19-9\\ln2}{2}e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+\\frac{-3+3\\ln2}{2}e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+18\\ln(e^t+1)e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}-3e^{-t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+6e^{-2t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}-6\\ln(e^t+1)e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix} \\end{align}", "3 & 1 & −2 \\\\ −2 & 0 & 5 \\end{pmatrix}$$ $$\\mathbf{F} = \\begin{pmatrix} 3 & 4 \\\\ 5 & 6 \\end{pmatrix}$$ $$\\mathbf{G} = \\begin{pmatrix} 2 & −4 \\\\ 7 & 1 \\\\ −1 & 1 \\end{pmatrix}$$ $$\\mathbf{H} = \\begin{pmatrix} −8 & −1 & −3 \\\\ −4 & 2 & −2 \\\\ −1 & 0 & 0 \\end{pmatrix}$$ Calculate each of the following 1. $$|\\mathbf{a} − (\\mathbf{b} + \\mathbf{3c})|$$ 7. $$(\\mathbf{EH})^T$$ 2. Component of $$\\mathbf{c}$$ along $$\\mathbf{a}$$ 8. $$|\\mathbf{HE}|$$ 3. Angle between $$\\mathbf{c}$$ and $$\\mathbf{d}$$ 9. $$\\mathbf{EHG}$$ 4. $$(\\mathbf{b} \\times \\mathbf{d}) \\cdot \\mathbf{a}$$ 10. $$\\mathbf{EG} − \\mathbf{HG}$$ 5. $$(\\mathbf{b} \\times \\mathbf{d}) \\times \\mathbf{a}$$ 11. $$\\mathbf{EH} − \\mathbf{H}^T \\mathbf{E}^T$$ 6. $$\\mathbf{b}\\times (\\mathbf{d} \\times \\mathbf{a})$$ 12. $$\\mathbf{F}^{−1}$$ 2. For what values of $$a$$ are the vectors $$\\mathbf{A} = 2a\\hat{i} − 2\\hat{j} + a\\hat{k}$$ and $$\\mathbf{B} = a\\hat{i} + 2a\\hat{j}+ 2\\hat{k}$$ perpendicular? 3. Show that the triple scalar product $$(A \\times B) \\cdot C$$ can be written as $(\\mathbf{A} \\times \\mathbf{B}) \\cdot \\mathbf{C} = \\begin{vmatrix} A_1 & A_2 & A_3 \\\\ B_1 & B_2 & B_3 \\\\ C_1 & C_2 & C_3 \\end{vmatrix} \\nonumber$ Show also that the product is unaffected by interchange of the scalar and vector product operations or by change in the order of $$A, B, C$$ as long as they are in cyclic order, that is" ]

[ "be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these equations for the constants gives: \\begin{align}c_{1}=\\frac{-19-9\\ln2}{2}\\end{align} \\begin{align}c_{2}=\\frac{-3+3\\ln2}{2}\\end{align} Therefore the general solution for the IVP is: \\begin{align}\\textbf{x}(t)=\\frac{-19-9\\ln2}{2}e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+\\frac{-3+3\\ln2}{2}e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+18\\ln(e^t+1)e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}-3e^{-t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+6e^{-2t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}-6\\ln(e^t+1)e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix} \\end{align}", "3 & 1 & −2 \\\\ −2 & 0 & 5 \\end{pmatrix}$$ $$\\mathbf{F} = \\begin{pmatrix} 3 & 4 \\\\ 5 & 6 \\end{pmatrix}$$ $$\\mathbf{G} = \\begin{pmatrix} 2 & −4 \\\\ 7 & 1 \\\\ −1 & 1 \\end{pmatrix}$$ $$\\mathbf{H} = \\begin{pmatrix} −8 & −1 & −3 \\\\ −4 & 2 & −2 \\\\ −1 & 0 & 0 \\end{pmatrix}$$ Calculate each of the following 1. $$|\\mathbf{a} − (\\mathbf{b} + \\mathbf{3c})|$$ 7. $$(\\mathbf{EH})^T$$ 2. Component of $$\\mathbf{c}$$ along $$\\mathbf{a}$$ 8. $$|\\mathbf{HE}|$$ 3. Angle between $$\\mathbf{c}$$ and $$\\mathbf{d}$$ 9. $$\\mathbf{EHG}$$ 4. $$(\\mathbf{b} \\times \\mathbf{d}) \\cdot \\mathbf{a}$$ 10. $$\\mathbf{EG} − \\mathbf{HG}$$ 5. $$(\\mathbf{b} \\times \\mathbf{d}) \\times \\mathbf{a}$$ 11. $$\\mathbf{EH} − \\mathbf{H}^T \\mathbf{E}^T$$ 6. $$\\mathbf{b}\\times (\\mathbf{d} \\times \\mathbf{a})$$ 12. $$\\mathbf{F}^{−1}$$ 2. For what values of $$a$$ are the vectors $$\\mathbf{A} = 2a\\hat{i} − 2\\hat{j} + a\\hat{k}$$ and $$\\mathbf{B} = a\\hat{i} + 2a\\hat{j}+ 2\\hat{k}$$ perpendicular? 3. Show that the triple scalar product $$(A \\times B) \\cdot C$$ can be written as $(\\mathbf{A} \\times \\mathbf{B}) \\cdot \\mathbf{C} = \\begin{vmatrix} A_1 & A_2 & A_3 \\\\ B_1 & B_2 & B_3 \\\\ C_1 & C_2 & C_3 \\end{vmatrix} \\nonumber$ Show also that the product is unaffected by interchange of the scalar and vector product operations or by change in the order of $$A, B, C$$ as long as they are in cyclic order, that is", "# Decide a number $a, b, c$ (expressed by $\\lambda$) so that $a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$ See the following vectors in $$\\mathbb{R}^3$$: $$\\mathbf{u}_1= \\begin{pmatrix} 1 \\\\ -1\\\\ 1 \\\\ 1 \\end{pmatrix}$$, $$\\mathbf{u}_2 = \\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\\\ 2 \\end{pmatrix}$$, $$\\mathbf{u}_3 = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\end{pmatrix}$$ and $$\\mathbf{v}_1= \\begin{pmatrix} 5 \\\\ -3\\\\ -4 \\\\ -1 \\end{pmatrix}$$, $$\\mathbf{v}_2 = \\begin{pmatrix} 6 \\\\ 4 \\\\ 1 \\\\ 8 \\end{pmatrix}$$, $$\\mathbf{v}_3 = \\begin{pmatrix} 7 \\\\ 2 \\\\ -1 \\\\ 6 \\end{pmatrix}$$ We are told that $$\\mathcal{B} = (\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3)$$ and $$\\mathcal{C} = (\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3)$$ both are bases for the same subspace $$\\mathcal{U}$$. Let $$\\mathcal{\\lambda}$$ be an unknown number og see the vector $$\\mathbf{x} = \\mathbf{v}_1 + \\lambda\\mathbf{v}_2 - \\mathbf{v}_3$$. Decide a number $$a, b, c$$ (expressed by $$\\lambda$$) so that $$a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$$ 1. Find some $$\\alpha_i$$ such that $$v_1=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3$$ 2. Find some $$\\beta_i$$ such that $$v_2=\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3$$ 3. Find some $$\\gamma_i$$ such that $$v_3=\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3$$ 4. Simplify the expression $$x=v_1+\\lambda v_2-v_3=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3+\\lambda(\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3)-(\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3)$$ • Thanks 5xum! When I reduce it, I get $\\begin{pmatrix} 8 \\\\ 3\\\\ -9 \\\\ 1 \\end{pmatrix} + a\\begin{pmatrix} 2 \\\\ -2\\\\ 1 \\\\ 1 \\end{pmatrix} + b\\begin{pmatrix} 3 \\\\ 6\\\\ 0 \\\\ 6 \\end{pmatrix} + c\\begin{pmatrix} 1 \\\\", "= {2\\mathbf x^T A \\cdot \\mathbf x^T\\mathbf x - \\mathbf x^T A \\mathbf x \\cdot 2\\mathbf x^T \\over (\\mathbf x^T\\mathbf x)^2}(^1_0)$$ In other words, if $A=A^T$: $$Df(\\mathbf x) = {2\\mathbf x^T A \\cdot \\mathbf x^T\\mathbf x - \\mathbf x^T A \\mathbf x \\cdot 2\\mathbf x^T \\over (\\mathbf x^T\\mathbf x)^2}$$ Last edited: Jan 24, 2013 3. Jan 24, 2013 ### libragirl79 Since I forgot a lot of my matrix algebra, when you say A (x y) it looks like x choose y, but i am assuming that's not what you meant... 4. Jan 24, 2013 ### I like Serena Since you are talking about x transpose, that means that x is a vector. A vector is commonly written as: $$\\mathbf x = \\begin{pmatrix}x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n\\end{pmatrix}$$ I have simplified it to: $$\\mathbf x = \\begin{pmatrix}x \\\\ y\\end{pmatrix}$$ So indeed, this does not mean x choose y. ;) Furthermore: $$\\mathbf x^T = \\begin{pmatrix}x_1 & x_2 & \\dots & x_n\\end{pmatrix}$$ and $$\\mathbf x^T \\mathbf x = \\begin{pmatrix}x_1 & x_2 & \\dots & x_n\\end{pmatrix}\\begin{pmatrix}x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n\\end{pmatrix} = x_1^2 + x_2^2 + \\dots + x_n^2$$ 5. Jan 24, 2013 ### libragirl79 Ok, I see! Thanks!! :)", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "$\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix},\\quad M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x-y=0\\},\\quad N=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, 2x+y=0\\}$ 3. $\\begin{pmatrix} 3 \\\\ 0 \\\\ 1 \\end{pmatrix},\\quad M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, x+y=0\\},\\quad N=\\{c\\cdot\\begin{pmatrix} 1 \\\\ 0 \\\\ 1 \\end{pmatrix}\\,\\big|\\, c\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 2 Find $M^\\perp$. 1. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x+y=0 \\}$ 2. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, -2x+3y=0 \\}$ 3. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x-y=0 \\}$ 4. $M=\\{\\vec{0}\\,\\}$ 5. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x=0 \\}$ 6. $M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, -x+3y+z=0 \\}$ 7. $M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, x=0\\text{ and } y+z=0\\}$ Problem 3 This subsection shows how to project orthogonally in two ways, the method of Example 3.2 and 3.3, and the method of Theorem 3.8. To compare them, consider the plane $P$ specified by $3x+2y-z=0$ in $\\mathbb{R}^3$. 1. Find a basis for $P$. 2. Find $P^\\perp$ and a basis for $P^\\perp$. 3. Represent this vector with respect to the concatenation of the two bases from the prior item. $\\vec{v}=\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ 4. Find the orthogonal projection of $\\vec{v}$ onto $P$ by keeping only the $P$ part from the prior item. 5. Check that against the result from applying Theorem 3.8. This exercise is recommended for all", "of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^{\\rm t} {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}. \\end{align*} ##### Example8.19 Suppose that the words to be encoded consist of all binary 3-tuples and that our encoding scheme is even-parity. To encode an arbitrary 3-tuple, we add a fourth bit to obtain an even number of 1s. Notice that an arbitrary $n$-tuple ${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^{\\rm t}$ has an even number of 1s exactly when $x_1 + x_2 + \\cdots + x_n = 0$; hence, a 4-tuple ${\\mathbf x} = (x_1, x_2, x_3, x_4)^{\\rm t}$ has an even number of 1s if $x_1+ x_2+ x_3+ x_4 = 0$, or \\begin{equation*}{\\mathbf x} \\cdot {\\mathbf 1} = {\\mathbf x}^{\\rm t} {\\mathbf 1} = \\begin{pmatrix} x_1 & x_2 & x_3 & x_4 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\\\ 1 \\end{pmatrix} = 0.\\end{equation*} This example leads us to hope that there is a connection between matrices and coding theory. Let ${\\mathbb M}_{m \\times n}({\\mathbb Z}_2)$ denote the set of all $m \\times n$ matrices with entries in ${\\mathbb Z}_2$. We do matrix operations", "solution. Thus, a general solution to our system is \\begin{equation*} \\mathbf x(t) = c_1 e^{2t} \\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix} + c_2 e^{-2t} \\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix} \\end{equation*} or \\begin{align*} x(t) & = 3 c_1 e^{2t} + c_2 e^{-2t}\\\\ y(t) & = c_1 e^{2t} - c_2 e^{-2t}. \\end{align*} If we are given initial conditions, say $x(0) = 0$ and $y(0) = 1\\text{,}$ then we can determine $c_1$ and $c_2$ by solving the linear system of equations \\begin{align*} 3 c_1 + c_2 & = 0\\\\ c_1 - c_2 & = 1 \\end{align*} to get $c_1 = 1/4$ and $c_2 = -3/4\\text{.}$ Thus, the solution to our initial value problem is \\begin{align*} x(t) & = \\frac{3}{4} e^{2t} - \\frac{3}{4} e^{-2t}\\\\ y(t) & = \\frac{1}{4} e^{2t} + \\frac{3}{4} e^{-2t}. \\end{align*} If ${\\mathbf x}_1(t)$ and ${\\mathbf x}_2(t)$ are solutions to the linear system ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ then \\begin{align*} {\\mathbf x}_1' & = A {\\mathbf x}_1\\\\ {\\mathbf x}_2' & = A {\\mathbf x}_2. \\end{align*} Thus, for any two real numbers $c_1$ and $c_2\\text{,}$ \\begin{align*} \\frac{d}{dt} (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t)) & = \\alpha \\frac{d}{dt} {\\mathbf x}_1(t) + c_2 \\frac{d}{dt} {\\mathbf x}_2(t)\\\\ & = c_1 A {\\mathbf x}_1(t) + c_2 A {\\mathbf x}_2(t)\\\\ & = A (c_1 {\\mathbf x}_1(t) + c_2 {\\mathbf x}_2(t) ). \\end{align*} We state this", "e_{21} & e_{22} & e_{23} \\\\ e_{31} & e_{32} & e_{33} \\end{pmatrix}$ the constraint can also be rewritten as $y'_1 y_1 e_{11} + y'_1 y_2 e_{12} + y'_1 e_{13} + y'_2 y_1 e_{21} + y'_2 y_2 e_{22} + y'_2 e_{23} + y_1 e_{31} + y_2 e_{32} + e_{33} = 0 \\,$ or $\\mathbf{e} \\cdot \\tilde{\\mathbf{y}} = 0$ where $\\tilde{\\mathbf{y}} = \\begin{pmatrix} y'_1 y_1 \\\\ y'_1 y_2 \\\\ y'_1 \\\\ y'_2 y_1 \\\\ y'_2 y_2 \\\\ y'_2 \\\\ y_1 \\\\ y_2 \\\\ 1 \\end{pmatrix}$ and $\\mathbf{e} = \\begin{pmatrix} e_{11} \\\\ e_{12} \\\\ e_{13} \\\\ e_{21} \\\\ e_{22} \\\\ e_{23} \\\\ e_{31} \\\\ e_{32} \\\\ e_{33} \\end{pmatrix}$ that is, $\\mathbf{e}$ represents the essential matrix in the form of a 9-dimensional vector and this vector must be orthogonal to the vector $\\tilde{\\mathbf{y}}$ which can be seen as a vector representation of the $3 \\times 3$ matrix $\\mathbf{y}' \\, \\mathbf{y}^{T}$. Each pair of corresponding image points produces a vector $\\tilde{\\mathbf{y}}$. Given a set of 3D points $\\mathbf{P}_k$ this corresponds to a set of vectors $\\tilde{\\mathbf{y}}_{k}$ and all of them must satisfy $\\mathbf{e} \\cdot \\tilde{\\mathbf{y}}_{k} = 0$ for the vector $\\mathbf{e}$. Given sufficiently many (at least eight) linearly independent vectors $\\tilde{\\mathbf{y}}_{k}$ it is possible to determine $\\mathbf{e}$ in a straightforward way. Collect all vectors $\\tilde{\\mathbf{y}}_{k}$ as the columns of a matrix $\\mathbf{Y}$ and", "that finding$21^{-1}\\in\\mathbb{Z}/26\\mathbb{Z}$is equivalent to solving the diophantine equation$21x+26y=1. \\begin{align*} 26&=21\\cdot 1+5\\\\ 21&=5\\cdot 4+1 \\end{align*} and \\begin{align*} 1&=21-5\\cdot 4\\\\ &=21-(26-21)4\\\\ &=5\\cdot 21+26(-4) \\end{align*} Thus, we find21^{-1}\\equiv 5\\mod 26. \\begin{align*} A^{-1}&=\\begin{pmatrix} 5\\cdot 8 & -5\\cdot 3\\\\ -5\\cdot 7 & 5\\cdot 2 \\end{pmatrix}\\\\ &=\\begin{pmatrix} 40 & -15\\\\ -35 & 10 \\end{pmatrix}\\\\ &\\equiv\\begin{pmatrix} 14 & 11\\\\ 17 & 10 \\end{pmatrix}\\mod 26 \\end{align*} LetA=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$. Then $$\\begin{pmatrix} x’\\\\ y’ \\end{pmatrix}=A\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} ax+by\\\\ cx+dy \\end{pmatrix}$$ This gives a linear map or a linear transformation from vectors to vectors. We introduce the following theorem without proof. Theorem. Let$A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$and set$D=ad-bc$as before. Then the following are equivalent: 1.$(D,N)=1$. 2.$A$has an inverse matrix. 3. if$x$and$y$are not both$0$in$\\mathbb{Z}/N\\mathbb{Z}$, then$A\\begin{pmatrix}x\\\\y \\end{pmatrix}\\ne\\begin{pmatrix}0\\\\0\\end{pmatrix}$. 4.$A$gives a one-to-one correspondence of$(\\mathbb{Z}/N\\mathbb{Z})^2with itself. Example. Solve the following systems of simultaneous congruences. 1. \\begin{align*}2x+3y&\\equiv 1\\mod 26\\\\ 7x+8y&\\equiv 2\\mod 26 \\end{align*} 2. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 2\\mod 26 \\end{align*} 3. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 1\\mod 26 \\end{align*} Solution. 1. The system of simultaneous congruences can be written as the matrix congruenceAX\\equiv B\\mod 26$or as the matrix equation$AX=B$in$\\mathbb{Z}/26\\mathbb{Z}$, where $$A=\\begin{pmatrix}2 & 3\\\\7 & 8\\end{pmatrix},\\ X=\\begin{pmatrix}x\\\\y\\end{pmatrix},\\ B=\\begin{pmatrix}1\\\\2\\end{pmatrix}$$$D=16-21=-5$and$(-5,26)=1$, so by above Theorem, we see that$A^{-1}$exists. In fact, we have already found$A^{-1}$in the", "and $z$. $(x+1)^2+(y-2)^2$ $+(z-1)^2=17$ $(5+3r+1)^2$ $+(6+2r-2)^2$ $+(5+2r-1)^2=17$ $(6+3r)^2$ $+(4+2r)^2$ $+(4+2r)^2=17$ Use binomial theorem to resolve parentheses $36+36r+9r^2$ $+16+16r+4r^2$ $+16+16r+4r^2=17$ $17r^2+68r+68=17\\quad|-17$ $17r^2+68r+51=0\\quad|:17$ $r^2+4r+3=0$ Use quadratic formula I to solve quadratic equation. $r_{1,2}=-\\frac{p}2\\pm\\sqrt{(\\frac{p}2)^2-q}$ $r_{1,2}=-2\\pm\\sqrt{2^2-3}$ $r_{1,2}=-2\\pm1$ $r_{1}=-1$ and $r_{2}=-3$ 3. #### Insert $r$ The two calculated $r$ are inserted into the equation of a line in order to obtain the intersection points. $\\vec{OS_1} = \\begin{pmatrix} 5 \\\\ 6 \\\\ 5 \\end{pmatrix} - 1 \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $=\\begin{pmatrix} 2 \\\\ 4 \\\\ 3 \\end{pmatrix}$ $\\vec{OS_2} = \\begin{pmatrix} 5 \\\\ 6 \\\\ 5 \\end{pmatrix} - 3 \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $=\\begin{pmatrix} -4 \\\\ 0 \\\\ -1 \\end{pmatrix}$ It is a secant that intersects the sphere at $S_1(2|4|3)$ and $S_2(-4|0|-1)$.", "\\\\ y\\end{pmatrix} \\cdot d\\mathbf r + \\hat {\\mathbf y} \\oint \\begin{pmatrix}z \\\\ 0 \\\\ -x\\end{pmatrix} \\cdot d\\mathbf r + \\hat {\\mathbf z} \\oint \\begin{pmatrix}-y \\\\ x \\\\ 0\\end{pmatrix} \\cdot d\\mathbf r$$ Applying the Kelvin-Stokes theorem we get: $$\\oint \\mathbf r \\times d\\mathbf r = \\hat {\\mathbf x} \\iint (\\nabla \\times \\begin{pmatrix}0 \\\\ -z \\\\ y\\end{pmatrix}) \\cdot d\\mathbf S + \\hat {\\mathbf y} \\iint (\\nabla \\times \\begin{pmatrix}z \\\\ 0 \\\\ -x\\end{pmatrix}) \\cdot d\\mathbf S + \\hat {\\mathbf z} \\iint (\\nabla \\times \\begin{pmatrix}-y \\\\ x \\\\ 0\\end{pmatrix}) \\cdot d\\mathbf S$$ $$\\oint \\mathbf r \\times d\\mathbf r = \\hat {\\mathbf x} \\iint 2 \\hat {\\mathbf x} \\cdot d\\mathbf S + \\hat {\\mathbf y} \\iint 2 \\hat {\\mathbf y} \\cdot d\\mathbf S + \\hat {\\mathbf z} \\iint 2 \\hat {\\mathbf z} \\cdot d\\mathbf S$$ $$\\oint \\mathbf r \\times d\\mathbf r = 2 \\iint d\\mathbf S = 2 \\iint dS \\hat {\\mathbf n}$$ However, this is a factor 2 different from the equation you proposed. Last edited: Aug 28, 2011 5. Aug 29, 2011 daudaudaudau Sorry, I was missing a factor of two in the equation i posted, so your proof is correct :) 6. Sep 1, 2011 Anthony Alternatively, for arbitrary fixed vector $\\mathbf{a} \\in \\mathbf{R}^3$ $$\\mathbf{a} \\cdot \\oint \\mathbf{r} \\wedge \\mathrm{d}\\mathbf{r} = \\oint \\mathbf{a} \\cdot (\\mathbf{r} \\wedge \\mathrm{d}\\mathbf{r}) = \\oint (\\mathbf{a}", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "we have $\\mathbf{i} + \\mathbf{j} + \\mathbf{k} + a_i \\mathbf{i} + a_j \\mathbf{j} + a_k \\mathbf{k} = 2\\mathbf{i} - 3\\mathbf{j} - \\mathbf{k}$ Comparing co-efficients, we have $\\begin{cases} 1 + a_i = 2 \\\\ 1 + a_j = -3 \\\\ 1 + a_k = -1 \\end{cases}$ Thus $$a_i = 1$$, $$a_j = -4$$, $$a_k = -2$$, and therefore $$\\mathbf{x} = \\mathbf{i} - 4 \\mathbf{j} - 2 \\mathbf{k}$$ Exercise 10 $$\\mathbf{0} = \\begin{pmatrix}0 \\\\ 0 \\\\ 0\\end{pmatrix}\\quad \\mathbf{i} = \\begin{pmatrix}1 \\\\ 0 \\\\ 0\\end{pmatrix} \\quad \\mathbf{j} = \\begin{pmatrix}0 \\\\ 1 \\\\ 0\\end{pmatrix} \\quad \\mathbf{k} = \\begin{pmatrix}0 \\\\ 0 \\\\ 1\\end{pmatrix}$$ ### Vector Equation of a Straight Line If $$A$$ and $$B$$ are any two distinct points on a straight line in space with position vectors $$\\mathbf{a}$$ and $$\\mathbf{b}$$ respectively, then the vector equation of a straight line is $\\mathbf{r} = (1 - s)\\mathbf{a} + s \\mathbf{b}$ where $$\\mathbf{r}$$ represents the position vector of any point on the line. In Cartesian coordinate form, with $\\mathbf{a} = a_1 \\mathbf{i} + a_2 \\mathbf{j} + a_3 \\mathbf{k}$ $\\mathbf{b} = b_1 \\mathbf{i} + b_2 \\mathbf{j} + b_3 \\mathbf{k}$ $\\mathbf{r} = x \\mathbf{i} + y \\mathbf{j} + z \\mathbf{k}$ the equation of the line is $x = a_1 + s (b_1 - a_1)\\quad,\\quad y = a_2 + s (b_2 - a_2)\\quad,\\quad z = a_3 + s", "+ 3 x_{22}\\\\ \\end{pmatrix} \\end{align*} So our equation is $$\\begin{pmatrix} 9 & -3 \\\\ 5 & -5 \\end{pmatrix} - \\begin{pmatrix} -9 x_{11} +8 x_{12} & -2x_{11}+ 3 x_{12}\\\\ -9 x_{21} + 8 x_{22} & -2 x_{21} + 3 x_{22}\\\\ \\end{pmatrix}= \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ \\end{pmatrix}$$ Now we write it as a system of equations: \\begin{align*} 1&=9- (-9 x_{11} +8x_{12})\\\\ 0&= -3-(-2x_{11} +3x_{12}) \\\\ 0&=5-(-9x_{21} +8 x_{21}) \\\\ 1&=-5 - (-2x_{21}+3x_{22}) \\\\ \\end{align*} If you need help solving this system tell me. - Thank you very much for your excellent answer! – Lukas Arvidsson Mar 15 at 9:27 You are welcome :) – Dominic Michaelis Mar 15 at 9:28 First of all rearrange to get X[-9 -2, 8 5] = [8 -3, 5 -6]. Then find the inverse of [-9 -2, 8 5], and multiply both sides of the equation by this inverse on the right, which will leave X = [8 -3, 5 -6][-9 -2, 8 5]^-1 as your solution. - Denote $B=\\begin{pmatrix}-9&-2\\\\8&5\\end{pmatrix}$. Then we have: $$\\begin{pmatrix}9&-3\\\\5&-5\\end{pmatrix}-E=XB$$ Observe that $\\det B=-45+16\\neq0$ and hence $B$ is invertible. Multiplying by $B^{-1}$ we have: $$X=\\left(\\begin{pmatrix}9&-3\\\\5&-5\\end{pmatrix}-E\\right)B^{-1}$$ Do you know how to find $B^{-1}$? - Thanks for your answer! I get the final result to be: $\\left[ \\begin{array}{cc} \\frac{-64}{29} & \\frac{-43}{29}\\\\ \\frac{-48}{29} & \\frac{-54}{29}\\end{array} \\right]$ Is that correct?", "why $x^2+y^2=(-y)^2+x^2$ (?!), which makes for a rather poor question, I am sorry to say. Note also that every sequence $(\\mathbf x_n)$ defined by $\\mathbf x_{n+1}=L(\\mathbf x_n)$ has period $4$, as an elementary computation shows. I know that to solve the exercise I need to show that F is constant along the orbits of L. That is, I need to show that $F \\text{ o } L(x) = L(x)$. No. That $F$ is an invariant of the dynamical system associated to $L$ means that, if $$\\dot{\\mathbf x}(t)=L(\\mathbf x(t)),$$ then the quantity $F(\\mathbf x(t))$ does not depend on $t$. This is usually done showing that $u'(t)=0$ for every $t$, where $$u(t)=F(\\mathbf x(t)),$$ which is implied by the condition that, for every $\\mathbf x$, $$\\nabla F(\\mathbf x)\\cdot L(\\mathbf x)=0.$$ In your case, $$\\nabla F(\\mathbf x)=2\\mathbf x,$$ hence one is left with $$\\nabla F(\\mathbf x)\\cdot L(\\mathbf x)=2\\mathbf x\\cdot L(\\mathbf x)=2\\begin{pmatrix}x\\\\y\\end{pmatrix}\\cdot\\begin{pmatrix}y\\\\-x\\end{pmatrix}=0.$$ Here is a rough sketch of some solutions $(\\mathbf x(t))$: $\\hspace2in$", "the original system using another change of coordinates. Suppose that we consider a linear system $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}\\tag{3.6.2}$$ where $$T$$ is an invertible matrix. If $${\\mathbf y}(t)$$ is a solution of (3.6.2), we claim that $${\\mathbf x}(t) = T {\\mathbf y}(t)$$ solves the equation $${\\mathbf x}' = A {\\mathbf x}\\text{.}$$ Indeed, \\begin{align*} {\\mathbf x}'(t) & = (T {\\mathbf y})'(t)\\\\ & = T {\\mathbf y}'(t)\\\\ & = T( (T^{-1} A T) {\\mathbf y}(t))\\\\ & = A (T {\\mathbf y}(t))\\\\ & = A {\\mathbf x}(t). \\end{align*} We can think of this in two ways. 1. A linear map $$T$$ converts solutions of $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}$$ to solutions of $${\\mathbf x}' = A {\\mathbf x}\\text{.}$$ 2. The inverse of a linear map $$T$$ takes solutions of $${\\mathbf x}' = A {\\mathbf x}$$ to solutions of $${\\mathbf y}' = (T^{-1} A T) {\\mathbf y}\\text{.}$$ We are now in a position to solve our problem of finding solutions of an arbitrary linear system \\begin{equation*} \\begin{pmatrix} x' \\\\ y' \\end{pmatrix} = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = \\begin{pmatrix} x \\\\ y \\end{pmatrix}. \\end{equation*} ### Subsection3.6.4Distinct Real Eigenvalues Consider the system $${\\mathbf x}' = A {\\mathbf x}\\text{,}$$ where $$A$$ has two real, distinct eigenvalues $$\\lambda_1$$ and $$\\lambda_2$$ with eigenvectors $${\\mathbf v}_1$$", "&\\equiv \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}^{-1} \\begin{pmatrix}13\\\\12\\end{pmatrix}\\\\ &\\equiv \\begin{bmatrix}2 & 7\\\\5 & 3\\end{bmatrix} \\begin{pmatrix}13\\\\12\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}2 \\cdot 13 + 7 \\cdot 12\\\\5 \\cdot 13 + 3 \\cdot 12\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}110\\\\101\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}6\\\\23\\end{pmatrix} \\end{align*} So $a=6$ and $b=23$. In a similar manner, we can solve for $c$ and $d$... \\begin{align*} \\begin{pmatrix}c\\\\d\\end{pmatrix} &\\equiv \\begin{bmatrix}25 & 11\\\\19 & 8\\end{bmatrix}^{-1} \\begin{pmatrix}20\\\\1\\end{pmatrix}\\\\ &\\equiv \\begin{bmatrix}2 & 7\\\\5 & 3\\end{bmatrix} \\begin{pmatrix}20\\\\1\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}2 \\cdot 20 + 7 \\cdot 1\\\\5 \\cdot 20 + 3 \\cdot 1\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}47\\\\103\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}21\\\\25\\end{pmatrix} \\end{align*} So $c=21$ and $d=25$. Now we have our (decrypting) matrix! $$D = \\begin{bmatrix}a & b\\\\c & d\\end{bmatrix} = \\begin{bmatrix}6 & 23\\\\21 & 25\\end{bmatrix}$$ Decrypting the rest of the message is now easy! We simply convert the letter blocks to vectors and apply $D$ to them, and out pops the vectors corresponding to the decrypted letter blocks! So, for example, consider the third letter block \"ZO\", which we suspect should decrypt to \"ER\". Does it? \\begin{align*} D(\\textrm{\"ZO\"}) &= \\begin{bmatrix}6 & 23\\\\21 & 25\\end{bmatrix} \\begin{pmatrix}25\\\\14\\end{pmatrix} \\\\ &= \\begin{pmatrix}6 \\cdot 25 + 23 \\cdot 14\\\\21 \\cdot 25 + 25 \\cdot 14\\end{pmatrix} \\\\ &= \\begin{pmatrix}472\\\\875\\end{pmatrix} \\\\ &\\equiv \\begin{pmatrix}4\\\\17\\end{pmatrix} \\pmod{26} \\end{align*} which, is the letter block \"ER\"! Looks like our guess about the word \"NUMBER\" was right!", "\\mathbf {B} ={\\begin{pmatrix}B_{11}&B_{12}&\\cdots &B_{1p}\\\\B_{21}&B_{22}&\\cdots &B_{2p}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\B_{m1}&B_{m2}&\\cdots &B_{mp}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\cdots &\\mathbf {b} _{p}\\end{pmatrix}}}$ where ${\\displaystyle {\\mathbf {a} }_{i}={\\begin{pmatrix}A_{i1}&A_{i2}&\\cdots &A_{im}\\end{pmatrix}}\\,,\\quad {\\mathbf {b} }_{i}={\\begin{pmatrix}B_{1i}\\\\B_{2i}\\\\\\vdots \\\\B_{mi}\\end{pmatrix}}}$ Then: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\dots &\\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{p})\\\\(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{p})\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{p})\\end{pmatrix}}}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: ${\\displaystyle {\\mathbf {AB} }={\\begin{pmatrix}{\\mathbf {A} }{\\mathbf {b} }_{1}&{\\mathbf {A} }{\\mathbf {b} }_{2}&\\dots &{\\mathbf {A} }{\\mathbf {b} }_{p}\\end{pmatrix}}={\\begin{pmatrix}{\\mathbf {a} }_{1}{\\mathbf {B} }\\\\{\\mathbf {a} }_{2}{\\mathbf {B} }\\\\\\vdots \\\\{\\mathbf {a} }_{n}{\\mathbf {B} }\\end{pmatrix}}}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1).{{ safesubst:#invoke:Unsubst||date=__DATE__ |B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }} Outer product The outer product (also known as the dyadic product or tensor product) of two", "\\end{array}\\right]. \\end{align*} This yields the solution c_1=2 and c_2=-5. Hence, we have the linear combination \\[\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}. ### (b) Compute $A^5\\mathbf{v}$. We first compute $A\\mathbf{v}$. We have $A\\mathbf{v}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix} =\\begin{bmatrix} -4 \\\\ 0 \\\\ 2 \\end{bmatrix}=2\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=2\\mathbf{v}.$ Using this relation $A\\mathbf{v}=2\\mathbf{v}$, we obtain $A^2\\mathbf{v}=AA\\mathbf{v}=A(2\\mathbf{v})=2A\\mathbf{v}=2(2\\mathbf{v})=2^2\\mathbf{v}.$ Next, we have $A^3\\mathbf{v}=AA^2\\mathbf{v}=A(2^2\\mathbf{v})=2^2A\\mathbf{v}=2^2(a\\mathbf{v})=2^3\\mathbf{v}.$ Repeating this process, we see that $A^5\\mathbf{v}=2^5\\mathbf{v}$. Or, we can find this by computing as follows: \\begin{align*} A^5\\mathbf{v}=A^2A^3\\mathbf{v}=A^2(2^3\\mathbf{v})=2^3A^2\\mathbf{v}=2^3(2^2\\mathbf{v})=2^5\\mathbf{v}. \\end{align*} In summary, we have $A^5\\mathbf{v}=2^5\\mathbf{v}=32\\begin{bmatrix} -2 \\\\ 0 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}.$ ### (c) Compute $A^5\\mathbf{w}$. First, we note that $A\\mathbf{w}= \\begin{bmatrix} -4 & -6 & -12 \\\\ -2 &-1 &-4 \\\\ 2 & 3 & 6 \\end{bmatrix} \\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=-\\mathbf{w}.$ Using this relation $A\\mathbf{w}=-\\mathbf{w}$ as in part (a), we obtain $A^5\\mathbf{w}=(-1)^5\\mathbf{w}=-\\begin{bmatrix} -2 \\\\ -1 \\\\ 1 \\end{bmatrix}=\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix}.$ ### (d) Compute $A^5\\mathbf{u}$. Using the linear combination $\\mathbf{u}=2\\mathbf{v}-5\\mathbf{w}$ obtained part (a), we compute \\begin{align*} A^5\\mathbf{w}&=A^5(2\\mathbf{v}-5\\mathbf{w})\\\\ &=2A^5\\mathbf{v}-5A^5\\mathbf{w}\\\\[6pt] &=2\\begin{bmatrix} -64 \\\\ 0 \\\\ 32 \\end{bmatrix}-5\\begin{bmatrix} 2 \\\\ 1 \\\\ -1 \\end{bmatrix} &&\\text{by (b), (c)}\\\\[6pt] &=\\begin{bmatrix} -138 \\\\ -5 \\\\ 69" ]

[ "#### Girls Olympic ###### back to index Among all pairs of real numbers $(x, y)$ such that $\\sin\\sin x=\\sin\\sin y$ with $-10\\pi \\le x, y \\le 10\\pi$. Oleg randomly selected a pair $(X, Y)$. Compute the probability that $X = Y$. Joel selected an acute angle $x$ (strictly between 0 and 90 degrees) and wrote the value of $\\sin x$, $\\cos x$, and $\\tan x$ on three different cards. Then he gave those cards to three students, Malvina, Paulina, and Georgian, one card to each, and asked them to figure out which trigonometric function (sin, cos, tan) produced their cards. Even after sharing the values on their cards with each other, only Malvian was able to surly identify which function produced the value on her card. Compute the sum of all possible values that Joel wrote on Marlvina's card.", "sequence $(a_n)$ has $a_1=\\sin x$, $a_2=\\cos x$, and $a_3= \\tan x$ for some real number $x$. For what value of $n$ does $a_n=1+\\cos x$? Let $S$ be the set of all point $(x,y)$ in the coordinate plane such that $0 \\le x \\le \\frac{\\pi}{2}$ and $0 \\le y \\le \\frac{\\pi}{2}$. What is the area of the subset of $S$ for which $\\sin^2x-\\sin x \\sin y + \\sin^2y \\le \\frac34?$ In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG + JH + CD$? Let integer $n\\ge 2$, prove $$\\sin{\\frac{\\pi}{n}}\\cdot\\sin{\\frac{2\\pi}{n}}\\cdots\\sin{\\frac{(n-1)\\pi}{n}}=\\frac{n}{2^{n-1}}$$ Joel selected an acute angle $x$ (strictly between 0 and 90 degrees) and wrote the value of $\\sin x$, $\\cos x$, and $\\tan x$ on three different cards. Then he gave those cards to three students, Malvina, Paulina, and Georgian, one card to each, and asked them to figure out which trigonometric function (sin, cos, tan) produced their cards. Even after sharing the values on their cards with each other, only Malvian was able to surly identify which function produced the value on her card. Compute the sum of all possible values that Joel wrote on Marlvina's card. Prove the following identities \\begin{align} \\sin (3\\alpha) &= 4\\cdot \\sin(60-\\alpha)\\cdot \\sin\\alpha\\cdot \\sin(60+\\alpha)\\\\ \\cos (3\\alpha) &= 4 \\cdot\\cos(60-\\alpha)\\cdot \\cos\\alpha\\cdot \\cos(60+\\alpha)\\\\ \\tan (3\\alpha) &= \\tan(60-\\alpha) \\cdot\\tan\\alpha \\cdot\\tan(60+\\alpha) \\end{align}", "### Trigonometry: Compound Angles Fluency exercise Below are some function cards, some value cards, and some target cards. You may wish to print them out and cut them up, but keep them in the three separate piles. You can choose a total of six cards, split between function cards and value cards in any way you choose (perhaps one or two function cards would be a good choice to start with). Choose your cards at random (pick from face-down cards, or get a friend to choose). Then choose a random target card. Your challenge is now to make the target by evaluating functions from your selected cards at values from your selected cards. You can use each card at most once (you don’t have to use them all). For example, if you have the function card $\\sin(\\square + \\square)$ and the value cards $\\dfrac{\\pi}{6}$ and $\\dfrac{\\pi}{3}$, then you could evaluate $\\sin\\left(\\frac{\\pi}{6} + \\frac{\\pi}{3}\\right),$ but you would not be allowed to evaluate $\\sin\\left(\\frac{\\pi}{6} + \\frac{\\pi}{6}\\right)$ (unless you happen to have two $\\dfrac{\\pi}{6}$ cards). #### Function $\\sin(\\square)$ #### Function $\\cos(\\square)$ #### Function $\\tan(\\square)$ #### Function $\\sin(\\square + \\square)$ #### Function $\\cos(\\square + \\square)$ #### Function $\\tan(\\square + \\square)$ #### Function $\\sin(\\frac{\\square}{2})$ #### Function $\\cos(\\frac{\\square}{2})$ #### Function $\\tan(\\frac{\\square}{2})$ #### Value $0$ #### Value $\\dfrac{\\pi}{6}$ #### Value $\\dfrac{\\pi}{4}$ #### Value $\\dfrac{\\pi}{3}$ ####", "article by Maureen Macinnis, a teacher in Nova Scotia. It seems like a terrific activity, with a lasting impact. She hands out pre-made cards, with the sine and cosine values for the sixteen \"famous\" angles (0°, 30°, 45°, and so on, all the way to 330°.) The sines are on orange cards, the cosines on green cards. Students are asked to pair up in sine-cosine pairs that correspond to one angle. For example, the student with an orange $\\frac{{\\sqrt 3 }}{2}$ needs to pair up with a green $\\frac{1}{2}$ or $-\\frac{1}{2}$. (That is already a good start!) At this point, the lesson continues outside, where the teacher has pre-drawn a large chalk unit circle on the ground, with x and y axes. Student pairs find their position on the circle. This would be a good time to review who is at what angle. But wait, there's more! The teacher has also drawn a pair of axes, with the y's from -1 to 1, and the x's marked off in degrees or radians. The next step is for the orange-card students to place themselves at the appropriate place on the graph, making a sine curve as the green-card students watch. And then they switch.", "sorted their cards. If possible, choose a group that created a group of 45 degree angles (cards B, D, F, and H) and another group of 270 degree angles (cards A, C, E, and G). Attend to the language that students use to describe their categories, giving them opportunities to describe the categories more precisely. Highlight the use of terms like central angle and ratio. Then, if students have not already done so, ask them to calculate the central angle and the ratio of arc length to radius for each of cards B, C, G, and H. If time is short, consider dividing the problems amongst the class so each student has only 1 problem to work. Invite them to share observations about their results. Be sure the following points come up in the discussion: • Cards B and H each represent 45 degree angles and have arc length to radius ratios equivalent to $$\\frac{\\pi}{4}$$ or $$0.25\\pi$$. • Cards C and G each represent 270 degree angles and have arc length to radius ratios equivalent to $$\\frac{3\\pi}{2}$$ or $$1.5\\pi$$. • The ratio appears to be the same for a particular central angle. Ask students, “Suppose I gave you another card that showed an arc length of $$30\\pi$$ units and a radius of 20 units. Without actually doing a", "draw a sketch of the unit circle and use the correct sign of cosine and sine in quadrant 4. This helps students evaluate, and improve upon, the written mathematical arguments of others, as they clarify how to find the values of sine and tangent given cosine and quadrant on the unit circle. Design Principle(s): Optimize output (for explanation); Maximize meta-awareness ## 6.3: Card Sort: Where's the Point? (15 minutes) ### Activity The goal of this matching activity is for students to practice making calculations with the Pythagorean Identity and using the structure of the unit circle (MP7). Students take turns matching cards. One set of cards shows the value of sine, cosine, or tangent of an unknown angle. The other set of cards shows a quadrant number on the unit circle. Students then select one of their possible matches and calculate the values of the two other trigonometric ratios. As students trade roles explaining their thinking and listening, they have opportunities to explain their reasoning and critique the reasoning of others (MP3). ### Launch Arrange students in groups of 2. Demonstrate how to lay out all the cards face up with those showing cosine, sine, or tangent on one side and those showing a quadrant number on the other. Tell students that they are going to take turns", "you know if the pair is or is not possible on the unit circle. Once a pair is identified, place the cards in front of you to use later. 2. For each pair that your partner draws, listen carefully to their explanation. If you disagree, discuss your thinking and work to reach an agreement. 2. Once you and your partner have identified 4 pairs each, pick 1 of your possible matches and then calculate the values of the two missing trigonometric ratios for that match. When you finish, trade calculations with your partner and check each other's work. If you disagree, discuss your thinking and work to reach an agreement. ### Student Facing #### Are you ready for more? Suppose $$\\tan(\\theta)=2$$ and the angle $$\\theta$$ is in quadrant 1. Find the values of $$\\cos(\\theta)$$ and $$\\sin(\\theta)$$. ### Anticipated Misconceptions As they work their calculations, students may be unsure when to take the positive or the negative expression when identifying the value of sine or cosine. Remind these students to sketch the situation determined by the two cards if they have not already done so, and to use that sketch to determine if the value they are calculating is positive or negative. For example, if they are calculating a cosine value in quadrant 4, it must be a positive", "Lesson 6 The Pythagorean Identity (Part 2) Let’s use the Pythagorean Identity. 6.1: Math Talk: Which Quadrant? For an angle $$\\theta$$ in the quadrant indicated, use mental estimation to identify the values of $$\\cos(\\theta)$$, $$\\sin(\\theta)$$, and $$\\tan(\\theta)$$ as either positive or negative. 6.2: Andre's Calculations Suppose that the angle $$\\theta$$, in radians, is in quadrant 4 of the unit circle. If $$\\cos(\\theta)=0.28$$, what are the values of $$\\sin(\\theta)$$ and $$\\tan(\\theta)$$? Andre uses the Pythagorean Identity and determines that the value of $$\\sin(\\theta)$$ is -0.96. Using the values of sine and cosine, he then calculates the value of tangent: \\begin{align} \\tan(\\theta)&=\\dfrac{\\sin(\\theta)}{\\cos(\\theta)} \\\\ &=\\frac{\\text-0.96}{0.28} \\\\ &\\approx \\text-3.43 \\end{align} Do you agree with Andre? Explain or show your reasoning. 6.3: Card Sort: Where's the Point? Your teacher will give you a set of cards that should be arranged face up with cards showing values for sine, cosine, and tangent on one side and cards showing quadrants on the other. 1. Take turns with your partner matching pairs of cards. Identify 2 pairs that are possible on the unit circle and 2 pairs that are not possible, in any order. 1. For each pair, explain to your partner how you know if the pair is or is not possible on the unit circle. Once a pair is identified, place the cards in", "variants in the Latin text, including one in the Gloria Prayers begin here at 5:15 a Songs start at $0 Songs start at$0 none The C library function double sin(double x) returns the sine of a radian angle x com/6679/print-all-trigonometric-ratios-c-program/If value of an angle is input through the keyboard, write a c program to print all its Tri In this video we look at the Math functions of SIN, COS and TAN and how we use them in C Programming Sine Series: Sine Series is a series which is used to find the value of Sin (x) x − This is the floating point value representing an angle expressed in radians Write a program in C to draw sine wave using graphics In this case, the two vectors are said to be orthogonal But the range is 0-PI This function answers the following question: Project ve Dies irae, dies illa! 2) Sanctus Satanas Sanctus Satanas, Sanctus Dominus Diabolus Sabaoth The Missal chants will cover the ordinary chants and dialogues of the Mass Also note the variants in the Latin text, including one in the Gloria Prayers begin here at 5:15 a Songs start at $0 Songs start at$0 14 hours ago · The tangent line approximation of f (x) for x near a is called the first degree", "# Given that sin a = 3/5 and a is in the interval (90,180), calculate tan a. hala718 | High School Teacher | (Level 1) Educator Emeritus Posted on sina = 3/5 We know that : sin^2 + cos^2 = 1 ==> cosa = sqrt(1-sin^2a) = sqrt(1-9/25) = sqrt(16/25) = 4/5 ==> cosa = 4/5 Since a is in the second quadrant , then sin>0 and cos < 0 ==> cosa = -4/5 ==> tana = sina/cosa = (3/5) / (-4/5)= -3/4 => tan(a) = -3/4 william1941 | College Teacher | (Level 3) Valedictorian Posted on For an angle x that lies in the interval 90 degrees to 180 degrees, sin x has positive values , cos x on the other hand in negative. We have sin a = 3 / 5. We also know that sin^a+ cos^a =1. Therefore we can get cos a = sqrt ( 1- sin^a) = sqrt [ 1- (3/5 )^2] = sqrt [ 1- ( 9/ 25)] = sqrt [ (25 -9)/ 25] =sqrt (16/ 25) = - 4/5 Therefore tan a = sin a / cos a = (3/5)/ (-4/5) = 3/ -4 = -3/4. The value of tan a is -3/4 neela | High School Teacher | (Level 3) Valedictorian Posted on sina = 3/5 To calculate for the value", "# Lesson 8EThe Amazing Inverse Trig Function RaceSolidify Understanding ## Learning Focus Define the inverse sine, inverse cosine, and inverse tangent functions. What are the formal definitions for the inverse sine, inverse cosine, and inverse tangent functions? Why is not always equal to ? ## Open Up the Math: Launch, Explore, Discuss To entertain themselves on weekends at the archeological dig, Javier has invented a game called “Find My Stake.” The game consists of drawing two cards, one from a deck of cards that Javier calls “The Angle Specification” cards, and the other from a deck Javier calls the “Location” cards. Based on these two clues, Veronica and Alyce race to locate the position of the stake. The friend who finds the correct location first, wins a prize. Alyce wonders why they need to have two clues. Veronica wonders if two clues will always be enough. With a partner, play Javier’s game a few times using the two decks of cards that will be provided by your teacher. One of you will draw an “Angle Specification Card.” The other will draw a “Location Card.” See if you can determine the exact location of the stake that is described by the two clues given on the cards. Note that “Angle Specification” cards do not state an angle directly. Rather,", "ago, when I first taught Precalculus at the college level, I was starting a section on trigonometry by reminding my students of the acronym SOHCAHTOA for keeping the trig functions straight: $\\sin \\theta = \\displaystyle \\frac{\\hbox{Opposite}}{\\hbox{Hypotenuse}}$, $\\cos \\theta = \\displaystyle \\frac{\\hbox{Adjacent}}{\\hbox{Hypotenuse}}$, $\\tan \\theta = \\displaystyle \\frac{\\hbox{Opposite}}{\\hbox{Adjacent}}$. At this point, one of my students volunteered that a previous math teacher had taught her an acrostic to keep these straight: Some Old Hippie Caught Another Hippie Tripping On Acid. Needless to say, I’ve been passing this pearl of wisdom on to my students ever since. My Favorite One-Liners: Part 69 In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. This story, that I’ll share with my Precalculus students, comes from Fall 1996, my first semester as a college professor. I was teaching a Precalculus class, and the topic was vectors. I forget the exact problem (believe me, I wish I could remember it), but I was going over the solution of a problem that required finding $\\tan^{-1}(7)$. I told the class that I had worked this out ahead of time, and that the approximate answer was $82^o$. Then I used that angle for whatever I needed it for and continued until obtaining the", "## Trigonometric Mnemonic At school we had a saying with which to remember the trigonometric forms: • $\\sin(\\vartheta) = O/H$ • $\\cos(\\vartheta) = A/H$, and • $\\tan(\\vartheta) = O/A$ in a right-angled triangle, where • $\\vartheta$ is one of the other angles • $O$ is the short side opposite the angle • $A$ is the short side adjacent to the angle, and • $H$ is the hypotenuse (the long side). The saying was Some Old Horses Can Always Hear Their Owners Approach. ### One Response to “Trigonometric Mnemonic” 1. Justin Says: Alphabetically Cosine, Sine, Tangent A Happy Old Heap Of Apples Cos (A/H) Sin (O/H) Tan (O/A) This site uses Akismet to reduce spam. Learn how your comment data is processed.", "Feynman. From Genius by James Gleick: \"He and his friends traded mathematical tidbits like baseball cards. If a boy named Morrie Jacobs told him that the cosine of 20 degrees multiplied by the cosine of 40 degrees multiplied by the cosine of 80 degrees equaled exactly one-eighth, he would remember that curiosity for the rest of his life, and he would remember that he was standing in Morrie's father's leather shop when he heard it.\" • Hint: Generalisation: $\\cos(60-x)\\cos(60+x)\\cos(x)=\\cos(3x)/4$. It gets even more interesting with tangents: $\\tan(60-x)\\tan(60+x)\\tan(x)=\\tan(3x)$ – chubakueno Sep 29 '13 at 18:51 • Also $\\sin(60-x)\\sin(60+x)\\sin(x)=\\sin(3x)/4$ (just multiple both your $\\cos$ and $\\tan$ identities). – user26486 Apr 3 '15 at 13:09 One of the most worth-mentioning identities may probably be Carl Friedrich Gauss's compution of $\\cos(\\frac{2\\pi}{17})$: \\begin{align} \\cos(\\frac{2\\pi}{17})&= \\frac1{16}[-1+\\sqrt{17} + \\sqrt{34-2\\sqrt{17}}+2\\sqrt{17+3\\sqrt{17}-\\sqrt{34-2\\sqrt{17}}-2\\sqrt{34+2\\sqrt{17}}}]\\\\ \\end{align} which has a significant role in regular heptadecagon construction. • As Gauss himself noted, no one before him had ever found a regular polygon with an odd number, $n$, of sides, with $n>5$, that could be constructed with only drawing compass and unmarked straight-edge. – DanielWainfleet Nov 18 '17 at 8:33 This is the most surprising result that I am the discoverer of. Consider the diophantine equation $$x(x+1)...(x+n-1) -y^n = k$$ where $x, y, n,$ and $k$ are integers, $x \\ge 1$, $y \\ge 1$,", "picture, this would be .” Bonita wrote the trigonometric ratio and asked, “So, how do we find angle ? Together, the girls talked about how this was like thinking backward: instead of knowing an angle and using their calculators to find a trigonometric ratio like they did while working on the height of the tree problem, they now know the trigonometric ratio and need to find an unknown angle value. Bonita notices the button on her calculator and wonders if this might work like an “inverse trigonometric ratio” button, undoing the ratio to produce the angle. She decides to try it out and gets the following output on her calculator: ### 3. How might this output convince Bonita that her assumption about the calculator was correct? ### 4. Use the trigonometric ratio you found for to find the value of . ### 5. Find all unknown values for the given right triangle: ### 6. Bonita and Andrea started talking about all of the ways to find unknown values in right triangles and decided to make a list. What do you think should be on their list? Be specific and precise in your description. For example, “trigonometric ratios” is not specific enough. You may use the following sentence frame to assist with writing each item in your list: When", "of cosine and the quadrant of the angle. This activity is meant to prepare students for the following card sort activity in which they will make similar calculations (MP1). Monitor for students who: • use a sketch of the unit circle to reason about the value of sine and tangent • write out the Pythagorean Identity calculations to check Andre’s statements that $$\\sin(\\theta)=\\text-0.96$$ ### Launch Provide access to calculators. Give students quiet work time and then time to share their work with a partner. Engagement: Develop Effort and Persistence. Encourage and support opportunities for peer interactions. Prior to the whole-class discussion, invite students to share their work with a partner. Display sentence frames to support student conversation such as: “First, Andre _____ because . . .”, “I noticed . . .”, “Why did Andre . . .?”, “I agree/disagree because . . . .” Supports accessibility for: Language; Social-emotional skills ### Student Facing Suppose that the angle $$\\theta$$, in radians, is in quadrant 4 of the unit circle. If $$\\cos(\\theta)=0.28$$, what are the values of $$\\sin(\\theta)$$ and $$\\tan(\\theta)$$? Andre uses the Pythagorean Identity and determines that the value of $$\\sin(\\theta)$$ is -0.96. Using the values of sine and cosine, he then calculates the value of tangent: \\begin{align} \\tan(\\theta)&=\\dfrac{\\sin(\\theta)}{\\cos(\\theta)} \\\\ &=\\frac{\\text-0.96}{0.28} \\\\ &\\approx \\text-3.43 \\end{align} Do you agree with", "$$\\frac14$$ the radius for cards B and H). ### Launch Arrange students in groups of 2. Distribute pre-cut slips. Tell them that in this activity, they will sort some cards into categories of their choosing. When they sort the cards, they should work with their partner to come up with categories. Give students 7–8 minutes of work time, then follow with a whole-class discussion. Be sure to leave enough time for students to calculate arc length to radius ratios in the synthesis. Conversing: MLR2 Collect and Display. As students sort the cards into categories, listen for and collect the language students use to categorize cards B, C, G, and H. Capture student language that suggests that the arc length and radii measurements are related to the central angle of 45 or 270 degrees. Write the students’ words and phrases on a visual display. As students review the visual display, ask students to revise and improve how ideas are communicated. For example, a statement such as, “The numbers divide to equal $$\\frac{\\pi}{4}$$” can be improved with the statement, “The central angle of 45 degrees has an arc length to radius ratio of $$\\frac{\\pi}{4}$$.” This will help students use the mathematical language necessary to precisely describe the relationship between central angles, arc lengths, and radii. Design Principle(s): Optimize output (for", "Emile van Gerwen 17 Aug '10 - 10:02 Instead of playing the game, I looked into the source code on the card flipping code because I think this is really neat. I think it can be much simpler: using the sine rule, the relation between previous card angle alpha and next card angle beta is, with card length l and distance between the two cards d: sin(alpha-beta) = (d * sin(alpha)) / l. There is no need to calculate the contact point. (little thing: I think your arcTan method is available as Math.Atan2) Neat game! -Emile. Sign In·View Thread·Permalink Re: My vote of 5 Christ Kennedy 17 Aug '10 - 12:33 thanks for your comment (and your vote!) yes the sine-rule could definitely be used here but I get : sin(alpha-beta)/d = sin(pi-alpha)/l since the alpha is the acute angle (in the article diagram) and the sine rule uses the obtuse angle (pi-alpha). as for the atan2(double x, double y), I've tried it in the past and the result was always from 0 to pi/2, but I'll have a look at it again! my code is perfect until i don't find a bug... Sign In·View Thread·Permalink Re: My vote of 5 Emile van Gerwen 17 Aug '10 - 21:19 Hi, Christ Kennedy wrote:but I get : sin(alpha-beta)/d =", "the value of a trigo function , ie $tanA=\\frac{1}{2}$ you can draw a triangle and find cos A and sin A as well ? If u do , well , the question tells us that A is an acute ( smaller than 90 ) , so we know that tan A is in the first quadrant and everything is positive here . Note that $(cos A)^2=cos^2A$", "café suppliers, or rent. A roommate happily watch this all over again just to see these wonderful characters again mum Mathilde. Reeling from Ray 's heartfelt declaration the night before, Gemma, and engineering come a! The fundamental trigonometric identities are derived from the six trigonometric ratios A+B ) -sin ( A-B (... { 1 } \\ ): Cantor you must be a registered user to use the IMDb rating plugin month. Realises that she 's run out of options to save the café functions... Deals with the characters straight away of this page fell in love, but then a family throws. On your own site leads was believable and felt real and the supporting cast were just as good geometry... 'S rating on your own site these functions also play an important in! Supporting cast were just as good is given below: Trigonometry MCQs cosA... Great streaming picks, above Gemma ’ s café adventure nobody expected following ’. For Sum and Difference of Two angles ( 1 ) sin ( )! Warsaw in 1935, established itself as a British Citizen during the 3rd century BC from applications of the functions! Their wedding aspiring British writer, activate a long-planned escape plan and disappear together for a specific ‘... Apartment to a head in front of friends and family a" ]

[ "sequence $(a_n)$ has $a_1=\\sin x$, $a_2=\\cos x$, and $a_3= \\tan x$ for some real number $x$. For what value of $n$ does $a_n=1+\\cos x$? Let $S$ be the set of all point $(x,y)$ in the coordinate plane such that $0 \\le x \\le \\frac{\\pi}{2}$ and $0 \\le y \\le \\frac{\\pi}{2}$. What is the area of the subset of $S$ for which $\\sin^2x-\\sin x \\sin y + \\sin^2y \\le \\frac34?$ In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG + JH + CD$? Let integer $n\\ge 2$, prove $$\\sin{\\frac{\\pi}{n}}\\cdot\\sin{\\frac{2\\pi}{n}}\\cdots\\sin{\\frac{(n-1)\\pi}{n}}=\\frac{n}{2^{n-1}}$$ Joel selected an acute angle $x$ (strictly between 0 and 90 degrees) and wrote the value of $\\sin x$, $\\cos x$, and $\\tan x$ on three different cards. Then he gave those cards to three students, Malvina, Paulina, and Georgian, one card to each, and asked them to figure out which trigonometric function (sin, cos, tan) produced their cards. Even after sharing the values on their cards with each other, only Malvian was able to surly identify which function produced the value on her card. Compute the sum of all possible values that Joel wrote on Marlvina's card. Prove the following identities \\begin{align} \\sin (3\\alpha) &= 4\\cdot \\sin(60-\\alpha)\\cdot \\sin\\alpha\\cdot \\sin(60+\\alpha)\\\\ \\cos (3\\alpha) &= 4 \\cdot\\cos(60-\\alpha)\\cdot \\cos\\alpha\\cdot \\cos(60+\\alpha)\\\\ \\tan (3\\alpha) &= \\tan(60-\\alpha) \\cdot\\tan\\alpha \\cdot\\tan(60+\\alpha) \\end{align}", "# Compute $\\lim\\limits_{x\\to 0} \\frac{\\sqrt{\\sin(\\tan(x))}}{\\sqrt{\\tan(\\sin(x))}}$ $$\\lim_{x\\to 0} \\frac{\\sqrt{\\sin(\\tan(x))}}{\\sqrt{\\tan(\\sin(x))}}= \\lim_{x\\to 0} \\frac{\\sqrt{\\frac{\\sin(\\tan(x))}{\\tan(x)}\\tan(x)}}{\\sqrt{\\frac{\\tan(\\sin(x))}{\\sin(x)}\\sin(x)}} \\tag{*}$$ We know $\\lim_{x\\to 0} \\sin(x)\\to x$ and $\\lim_{x\\to 0} \\tan(x)\\to x$; then: $$\\lim_{x\\to 0} \\frac{\\sqrt{\\frac{\\sin(\\tan(x))}{\\tan(x)}\\tan(x)}}{\\sqrt{\\frac{\\tan(\\sin(x))}{\\sin(x)}\\sin(x)}}=\\lim_{x\\to 0} \\frac{\\sqrt{\\tan(x)}}{\\sqrt{\\sin(x)}}=\\lim_{x\\to 0} \\sqrt{\\frac{\\sin(x)}{\\cos(x)\\sin(x)}}=\\lim_{x\\to 0} \\sqrt{\\frac{1}{\\cos(x)}}=1$$ Can I do that? Is what I did at (*) legal because of the continuity of the discussed functions? Thanks You're having the right idea, but the realization is not completely sound. The function you want to compute the limit of is defined for $0<x<\\pi/2$ (disregarding other intervals that are not of a concern here). The limit has no meaning for $x<0$, so somebody would consider it as the limit for $x\\to 0^+$. On the other hand, if the function is written as $$\\sqrt{\\frac{\\sin(\\tan x)}{\\tan(\\sin x)}}$$ it is defined in a whole punctured neighborhood of $0$. Now it's a matter of computing $$\\lim_{x\\to0}\\frac{\\sin(\\tan x)}{\\tan(\\sin x)}$$ which we'll take the square root of after computing it. Recall that $\\sin t=t+o(t)$ and $\\tan t=t+o(t)$, so $$\\lim_{x\\to0}\\frac{\\sin(\\tan x)}{\\tan(\\sin x)}= \\lim_{x\\to0}\\frac{\\tan x+o(\\tan x)}{\\sin x+o(\\sin x)}= \\lim_{x\\to0}\\frac{x+o(x)+o(x+o(x))}{x+o(x)+o(x+o(x))}= \\lim_{x\\to0}\\frac{x+o(x)}{x+o(x)}=1$$ It's not really different from your method, which is just rewriting the function as $$\\lim_{x\\to0}\\sqrt{\\frac{\\sin(\\tan x)}{\\tan x}\\frac{\\sin x}{\\tan(\\sin x)} \\frac{\\tan x}{\\sin x}}$$ where each fraction under the square root sign has limit $1$. • I appreciate your detailed answer, and I learnt from it few new things. but one thing I couldn't", "understand, and it's about the interval of definition. why isn't it relevant to look at $x<0$? why is the limit defined as for $lim_{x\\to 0^+}$? – Ami Gold Jun 8 '16 at 18:00 • @Ami For $-\\pi/2<x<0$, both $\\sin x$ and $\\tan x$ are negative and so also $\\sin(\\tan x)$ and $\\tan(\\sin x)$ are negative. Therefore $\\sqrt{\\sin(\\tan x)}$ and $\\sqrt{\\tan(\\sin x)}$ are undefined. – egreg Jun 8 '16 at 18:06 • now I feel stupid. Just noticed that.. thanks! – Ami Gold Jun 9 '16 at 6:46 Well, what you did is OK, but if you are not capable of explaining why it's OK, then you should do more work. This is mathematics afterall, so we are only interested in things you can prove, not hunches. I can have a hunch about a great deal of things, but if I am not sure, and cannot prove them, they may as well not exist. If I were you, I would write down $$\\lim_{x\\to 0} \\frac{\\sqrt{\\frac{\\sin(\\tan(x))}{\\tan(x)}\\tan(x)}}{\\sqrt{\\frac{\\tan(\\sin(x))}{\\sin(x)}\\sin(x)}}=\\lim_{x\\to 0} \\frac{\\sqrt{\\tan(x)}}{\\sqrt{\\sin(x)}}\\cdot\\sqrt{\\frac{\\sin(\\tan x)}{\\tan x}}\\cdot \\sqrt{\\frac{\\sin x}{\\tan(\\sin x)}}$$ Then decompose the limit into a product of three limits and show each one is equal to $1$. Using equivalents, $\\sin x\\sim x$ and $\\tan x\\sim x$ $$\\lim\\limits_{x\\to 0} \\frac{\\sqrt{\\sin(\\tan(x))}}{\\sqrt{\\tan(\\sin(x))}}=\\lim\\limits_{x\\to 0} \\frac{\\sqrt{\\tan(x)}}{\\sqrt{\\sin(x)}}=\\lim\\limits_{x\\to 0} \\frac{1}{\\sqrt{\\cos(x)}}=1$$ What you have done is correct without any issues (user egreg points a", "tan 2^\\circ . tan 3^\\circ . . . . tan 89^\\circ = 1$13) Answer (A) The given situation can be represented through the following figure. We know that$Tan 45^\\circ =$h/BC => 1 = h/BC => BC = h Now after walking for 20m, the angle changed to 60°. So we have Tan 60 = h/BD =>$ \\sqrt{3} = \\frac{h}{h – 20}$=>$20\\sqrt{3} = h(\\sqrt{3} – 1)$=> h =$\\frac{20\\sqrt{3}}{\\sqrt{3} – 1}$=> h =$ \\frac{60 + 20\\sqrt{3}}{2}$=> h =$ 30 + 10\\sqrt{3}$14) Answer (B) Given cos x =$\\dfrac{12}{13}$Squaring on both sides$cos^{2}x = \\dfrac{144}{169}$We know that$sin^2 x = 1-cos^2 x$⇒$sin^2 x = 1-$\\dfrac{144}{169}$ ⇒ $sin^2 x =$\\dfrac{25}{169}$⇒$sin x = $\\dfrac{5}{13}$ We know that tan x = $\\dfrac{sin x}{cos x}$ tan x $= \\dfrac(\\dfrac{5}{13})}{(\\dfrac{12}{13})} = \\dfrac{5}{12}$ $1/sinA – SinA/Cos^2A$ = $(cos^2A-sin^2A)/(sinA.cos^2A)$ = $2(Cos2A)/(sin2A.CosA)$ = $2Cot2A.secA$ So the answer is option A. We know that cos x = $\\sqrt{1-sin^{2}x}$ $= \\sqrt{1-\\dfrac{9}{25}}$ $= \\sqrt{\\dfrac{16}{25}} = \\dfrac{4}{5}$ tan x = \\dfrac{sin x}{cos x} = \\dfrac{(\\dfrac{3}{5})}{(\\dfrac{4}{5})} = \\dfrac{3}{4}$17) Answer (D) W.K.T$sec^2\\theta – tan^2\\theta = 1(sec\\theta-tan\\theta)(sec\\theta+tan\\theta) = 1(sec\\theta-tan\\theta)(1/3) = 1sec\\theta-tan\\theta = 3$So the answer is option D. 18) Answer (C) sin2θ=$\\frac{\\sqrt3}{2}$then 2θ = 60° ==> θ = 30° cosecθ = cosec30° = 2 So the answer is option C. 19) Answer (A) If sinθ = 12/13 cosθ = 5/13 and secθ = 13/5 secθ+cosθ = 13/5 + 5/13", "A = 1.5,$ find $\\sin A.$ 53. If $\\cot A + \\cosec A = 5,$ find $\\cos A.$ 54. If $3\\sec^4 A + 8 = 10\\sec^2A,$ find the value of $\\tan A.$ 55. If $\\tan^2A + \\sec A = 5,$ find $\\cos A.$ 56. If $\\tan A + \\cot A = 2,$ find $\\sin A.$ 57. If $\\sec^2A = 2 + 2\\tan A,$ find $\\tan A.$ 58. If $\\tan A = \\frac{2x(x + 1)}{2x + 1},$ find $\\sin A$ and $\\cos A.$ 59. If $3\\sin A + 5\\cos A = 5,$ show that $5\\sin A - 3\\cos A = \\pm 3$ 60. If $\\sec A + \\tan A = \\sec A - \\tan A$ prove that each side is $\\pm 1$ 61. If $\\frac{\\cos^4 A}{\\cos^2 B} + \\frac{\\sin^4 A}{\\sin^2 B} = 1,$ prove that 1. $\\sin^4A + \\sin^4B = 2\\sin^2A \\sin^2B$ 2. $\\frac{\\cos^4 B}{\\cos^2 A} + \\frac{\\sin^4 B}{\\sin^2 A} = 1,$ 62. If $\\cos A + \\sin A = \\sqrt{2}\\cos A,$ prove that $\\cos A - \\sin A = \\pm \\sqrt{2}\\sin A$ 63. If $a\\cos A - b\\sin A = c,$ prove that $a\\sin A + b\\cos A = \\sqrt{a^2 + b ^2 - c^2}$ 64. If $1 - \\sin A = 1 + \\sin A,$ then prove that value of each side is $\\pm \\cos A$ 65.", "# Math Help - Trig Limit 1. ## Trig Limit Find the value of $\\displaystyle\\lim_{x\\to{0}}\\frac{\\sqrt{1+tan\\;x}-\\sqrt{1+sin\\;x}}{x^3}$ Is there an alternate way of doing this question without using l'Hosital's Rule or this: $\\displaystyle\\lim_{x\\to{0}}\\frac{\\sqrt{1+tan\\;x}-\\sqrt{1+sin\\;x}}{x^3}\\;\\frac{\\sqrt{1+tan\\;x}+\\sqrt {1+sin\\;x}}{\\sqrt{1+tan\\;x}+\\sqrt{1+sin\\;x}}$ and then having to go through a long simplification before getting the answer.... 2. $\\displaystyle\\lim_{x\\to 0}\\frac{\\tan x-\\sin x}{x^3}\\cdot\\frac{1}{\\sqrt{1+\\tan x}+\\sqrt{1+\\sin x}}=\\frac{1}{2}\\lim_{x\\to 0}\\frac{\\sin x(1-\\cos x)}{x^3\\cos x}=$ $\\displaystyle=\\frac{1}{2}\\lim_{x\\to 0}\\frac{\\sin x}{x}\\cdot\\frac{2\\sin^2\\frac{x}{2}}{x^2}=\\frac{1}{ 4}$ 3. Originally Posted by red_dog $\\displaystyle\\lim_{x\\to 0}\\frac{\\tan x-\\sin x}{x^3}\\cdot\\frac{1}{\\sqrt{1+\\tan x}+\\sqrt{1+\\sin x}}=\\frac{1}{2}\\lim_{x\\to 0}\\frac{\\sin x(1-\\cos x)}{x^3\\cos x}=$ $\\displaystyle=\\frac{1}{2}\\lim_{x\\to 0}\\frac{\\sin x}{x}\\cdot\\frac{2\\sin^2\\frac{x}{2}}{x^2}=\\frac{1}{ 4}$ When you take the limit of $\\frac{2\\sin^2\\frac{x}{2}}{x^2}$ doesn't that equal 0 so then how do you get $\\frac{1}{4}$? 4. Originally Posted by polymerase When you take the limit of $\\frac{2\\sin^2\\frac{x}{2}}{x^2}$ doesn't that equal 0 so then how do you get $\\frac{1}{4}$? $\\lim_{x \\to 0} \\frac{2 sin^2 \\left ( \\frac{x}{2} \\right )}{x^2}$ $= 2 \\lim_{x \\to 0} \\left ( \\frac{sin \\left ( \\frac{x}{2} \\right )}{x} \\right ) ^2$ Now change the variable slightly: $y = \\frac{x}{2}$ Soooo.... $\\lim_{x \\to 0} \\frac{2 sin^2 \\left ( \\frac{x}{2} \\right )}{x^2} = 2 \\lim_{y \\to 0} \\left ( \\frac{sin(y)}{2y} \\right )^2$ $= \\frac{2}{4} \\lim_{y \\to 0} \\left ( \\frac{sin(y)}{y} \\right )^2$ $= \\frac{1}{2} (1)^2 = \\frac{1}{2}$ With the other 1/2 out in front of the expression we get red_dog's answer of 1/4. -Dan 5. Originally Posted by topsquark $\\lim_{x \\to 0} \\frac{2 sin^2 \\left", "a little help? – Ghost Apr 20 '13 at 17:30 @Ghost, So, $$\\cos^2(A+B)=\\frac{4-2\\sqrt3}8=\\frac{(\\sqrt3-1)^2}8\\implies \\cos(A+B)=\\pm\\frac{\\sqrt3-1}{2\\sqrt2}$$ – lab bhattacharjee Apr 20 '13 at 17:52 It is given that tanA=1, and tan B=$/sqrt3$. This implies that A can be 45 degrees and B can be 60 degrees.( Taking inverse, and using principal values) Also cosA.cosB-sinAsinB=cos(A B) So, cos(A B)=cos(60 45)=cos(105)=-sin(15)=-($\\sqrt3$-1)/(2$\\sqrt2$) - since $tan A =1$, then $A = 45^0$ and since $tan B = \\sqrt{3}$, then $B = 60^0$. By drawing triangles you can find that $sin 45^0 = 1/\\sqrt{2}$ and $cos 45^0 = 1/\\sqrt{2}$ and $sin 60^0 = \\sqrt{3}/2$ and $cos 60^0 = 1/2$. - Without some additional information, you quite simply can't. The closest you can come is using the identity $$\\cos^2\\theta=\\frac1{\\tan^2\\theta+1}$$ (which is derived from the Pythagorean identity, and holds wherever $\\tan\\theta$ is defined), from which you can determine that $$\\cos\\theta=\\pm\\sqrt{\\frac1{1+\\tan^2\\theta}}.$$ Now, if you have an additional assumption, such as that $\\theta$ is acute, then this becomes $$\\cos\\theta=\\sqrt{\\frac1{1+\\tan^2\\theta}},$$ at which point you can use this together with the angle sum identity for cosine to solve your problem -", "x} = 4$$ tells you that $1 - \\sin x = 4(1 + \\sin x)$, from which $\\sin x = -3/5.$ Now write the equation $$\\displaystyle \\frac 1{\\cos x} - \\tan x = 2$$ as $$\\displaystyle \\frac {1 - \\sin x}{\\cos x} = 2$$ to deduce that $\\cos x = 4/5$. Then you can calculate $$\\displaystyle \\frac 1{\\cos x} + \\tan x = \\frac {1 + \\sin x}{\\cos x} = \\frac12.$$ All right thanks a lot, but one question: How come the act of squaring the equation did, this time, not produce any extraneous information? I would, understandably so, actually expect that to occur. #### MarkFL Staff member I am having a sudden brain freeze: Where did that equality come from? We had $$\\displaystyle z = \\frac {1 + \\sin x}{\\cos x}$$ and not $$\\displaystyle z = \\frac{\\cos(x)}{\\sin(x)-1}$$, or am I just barking up the wrong tree?... We are originally given: $$\\displaystyle \\frac{1}{\\cos(x)}-\\tan(x)=2$$ $$\\displaystyle \\frac{1-\\sin(x)}{\\cos(x)}=2$$ Negating and inverting, we may write: $$\\displaystyle \\frac{\\cos(x)}{\\sin(x)-1}=-\\frac{1}{2}$$ Now, we may use this value for $$\\displaystyle z=-\\frac{1}{2}$$ to see if the equation results in an identity or not. #### sweatingbear ##### Member Thank you MarkFL! That showed me how to carefully check the algebra. @Opalg: Would really appreciate if you could share your thoughts on: Your equation $$\\displaystyle \\frac {1- \\sin x}{1 +", "The local lines of symmetry on the graph of $\\cos(x)$ are every $180^\\circ$ starting at $0^\\circ$. If there is a solution at $60^\\circ$, for example, there is also one at $360-60=300^\\circ$. Both of these solutions then repeat every $360^\\circ$. $y=\\sin(x):$ The local lines of symmetry on the graph of $\\sin(x)$ are every $180^\\circ$ starting at $90^\\circ$. If there is a solution at $30^\\circ$, for example, there is also one at $180-30=150^\\circ$. Both of these solutions then repeat every $360^\\circ$. $y=\\tan(x):$ The solutions on the tan graph are easier to find, the default value simply repeats every $180^\\circ$. If there is a solution at $45^\\circ$, for example, there is another at $-135^\\circ$, $225^\\circ$, etc. The tan graph is different to the sin and cos graphs. Firstly, tan can take a value between $-\\infty$ and $\\infty$ whereas sin and cos take values between -1 and 1. The graph of tan also has asymptotes as a result of dividing by $\\cos(x)=0$ in the expression $\\tan(x)=\\frac{\\sin(x)}{\\cos(x)}$. ## A note on 'CAST' It may be that the argument of the trigonometric function is a function of $x$. In these cases it is very important to extend the interval of solution values so that the full set of solutions are obtained. See Examples 2 and 4. Some students prefer to use ‘CAST’ – this", "gives us: $$\\displaystyle \\cos^2(x)=\\sin^2(x)-1$$ and since this is not an identity, we must discard $$\\displaystyle z=-\\frac{1}{2}$$ as a solution. #### Opalg ##### MHB Oldtimer Staff member Find the value of $$\\displaystyle z = \\frac 1{\\cos x} + \\tan x$$ given $$\\displaystyle \\frac 1{\\cos x} - \\tan x = 2$$. By squaring and using trigonometric identities on the given equation, one can arrive at $$\\displaystyle \\frac {1- \\sin x}{1 + \\sin x} = 4$$ Your equation $$\\displaystyle \\frac {1- \\sin x}{1 + \\sin x} = 4$$ tells you that $1 - \\sin x = 4(1 + \\sin x)$, from which $\\sin x = -3/5.$ Now write the equation $$\\displaystyle \\frac 1{\\cos x} - \\tan x = 2$$ as $$\\displaystyle \\frac {1 - \\sin x}{\\cos x} = 2$$ to deduce that $\\cos x = 4/5$. Then you can calculate $$\\displaystyle \\frac 1{\\cos x} + \\tan x = \\frac {1 + \\sin x}{\\cos x} = \\frac12.$$ #### sweatingbear ##### Member One way you could eliminate the extraneous solution is to use: $$\\displaystyle z=-\\frac{1}{2}=\\frac{\\cos(x)}{\\sin(x)-1}$$ I am having a sudden brain freeze: Where did that equality come from? We had $$\\displaystyle z = \\frac {1 + \\sin x}{\\cos x}$$ and not $$\\displaystyle z = \\frac{\\cos(x)}{\\sin(x)-1}$$, or am I just barking up the wrong tree? Your equation $$\\displaystyle \\frac {1- \\sin x}{1 + \\sin", "# Simplifying $\\sin(2\\tan^{-1} x)$ I've been working on this for a while. The answer in the book is $\\frac{2x}{x^2 + 1}$ Here's my workings: $\\sin(2\\tan^{-1} x)$ Let $\\alpha = \\tan^{-1}x \\Rightarrow \\tan \\alpha = x$ $\\sin(2\\alpha) = 2\\sin\\alpha\\cos\\alpha = 2\\tan\\alpha\\cos^2\\alpha = 2x\\cos^2\\alpha$ I'm not sure how to proceed to turn that $cos^2\\alpha$ into $\\frac{1}{x^2 + 1}$ • $\\sin(\\Arctan x)=\\frac{x}{\\sqrt{x^2+1}}$ and $\\cos(\\Arctan x)=\\frac{1}{\\sqrt{x^2+1}}$... – long tom Apr 24 '13 at 21:20 Hint: $\\cos^2(\\alpha) = \\dfrac{\\cos^2(\\alpha)}{\\cos^2(\\alpha) + \\sin^2(\\alpha)}$ • Your hint combined with Mike's helped me to see the solution. Maybe my algebra is lacking, but I had to invert that fraction to allow me to manipulate it, then invert again. – PeteUK Apr 24 '13 at 21:38 Making a right-triangle picture can be helpful too, either for getting to the result quickly or checking up on the analytical result. With $\\tan \\alpha = x = \\frac{x}{1}$, we would have $x$ for the leg opposite $\\alpha$ and $1$ for the adjacent leg; the hypotenuse is then $\\sqrt{x^2 + 1}$ . We wish to evaluate $\\sin 2\\alpha$ , which is thus $$2 \\sin \\alpha \\cos \\alpha = 2 \\cdot \\frac{x}{\\sqrt{x^2 + 1}} \\cdot \\frac{1}{\\sqrt{x^2 + 1}} = \\frac{2x}{x^2 + 1} .$$ • Your post says $\\alpha = \\tan x = \\tan(\\frac{x}{1})$, but should that be $\\tan \\alpha = x = \\frac{x}{1}$?", "36) $$f(x)=\\sin(4x), g(x)=\\sin(5x)\\cos x-\\cos(5x)\\sin x$$ 37) $$f(x)=\\sin(2x), g(x)=2 \\sin x \\cos x$$ They are the same. 38) $$f(\\theta )=\\cos(2\\theta ), g(\\theta )=\\cos^2\\theta -\\sin^2\\theta$$ 39) $$f(\\theta )=\\tan(2\\theta ), g(\\theta )=\\dfrac{\\tan \\theta }{1+\\tan^2\\theta }$$ They are different, try $$g(\\theta )=\\dfrac{2\\tan \\theta }{1-\\tan^2\\theta }$$ 40) $$f(x)=\\sin(3x)\\sin x, g(x)=\\sin^2(2x)\\cos^2x-\\cos^2(2x)\\sin2x$$ 41) $$f(x)=\\tan(-x), g(x)=\\dfrac{\\tan x-\\tan(2x)}{1-\\tan x \\tan(2x)}$$ They are different, try $$g(x)=\\dfrac{\\tan x-\\tan(2x)}{1+\\tan x \\tan(2x)}$$ ### Technology For the exercises 42-46, find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point. 42) $$\\sin (75^{\\circ})$$ 43) $$\\sin (195^{\\circ})$$ $$-\\dfrac{\\sqrt{3}-1}{2\\sqrt{2}}$$, or $$-0.2588$$ 44) $$\\cos (165^{\\circ})$$ 45) $$\\cos (345^{\\circ})$$ $$\\dfrac{1+\\sqrt{3}}{2\\sqrt{2}}$$, or $$-0.9659$$ 46) $$\\tan (-15^{\\circ})$$ ### Extensions For the exercises 47-51, prove the identities provided. 47) $$\\tan \\left ( x+\\dfrac{\\pi }{4} \\right )=\\dfrac{\\tan x+1}{1-\\tan x}$$ \\begin{align*} \\tan \\left ( x+\\dfrac{\\pi }{4} \\right ) &= \\\\ \\dfrac{\\tan x + \\tan\\left (\\tfrac{\\pi}{4} \\right )}{1-\\tan x \\tan\\left (\\tfrac{\\pi}{4} \\right )} &= \\\\ \\dfrac{\\tan x+1}{1-\\tan x(1)} &= \\dfrac{\\tan x+1}{1-\\tan x} \\end{align*} 48) $$\\dfrac{\\tan (a+b)}{\\tan (a-b)}=\\dfrac{\\sin a \\cos a + \\sin b \\cos b}{\\sin a \\cos a - \\sin b \\cos b}$$ 49) $$\\dfrac{\\cos (a+b)}{\\cos a \\cos b}=1-\\tan a \\tan b$$ \\begin{align*} \\dfrac{\\cos (a+b)}{\\cos a \\cos b} &= \\\\ \\dfrac{\\cos a \\cos b}{\\cos a \\cos b}- \\dfrac{\\sin a \\sin b}{\\cos a \\cos b} &= 1-\\tan a \\tan b \\end{align*} 50) $$\\cos(x+y)\\cos(x-y)=\\cos^2x-\\sin^2y$$ 51) $$\\dfrac{\\cos(x+h)-\\cos(x)}{h}=\\cos x\\dfrac{\\cos", "# Find the value of $\\frac{\\tan A}{\\tan B}$, given $\\frac{\\sin A}{\\sin B}=5$ If $\\displaystyle \\frac{\\sin A}{\\sin B}=5$, then find the value of $\\displaystyle \\frac{\\tan A}{\\tan B}$ Try using the Componendo and Dividendo formula: $$\\frac{\\sin A+\\sin B}{\\sin A-\\sin B}=\\frac{3}{2}$$ $$\\frac{\\tan(A+B)/2}{\\tan(A-B)/2}=\\frac{3}{2}$$ Can someone help me find: $$\\frac{\\tan A}{\\tan B}$$ Thanks • I'm not certain that there is one value of $\\frac{\\tan A}{\\tan B}$. What if $A = \\pi/2$ and $B = \\sin^{-1}(1/5)$? Then $\\frac{\\tan A}{\\tan B}$ is undefined. – Theo Bendit Mar 23 '18 at 1:10 This is impossible to answer for general $A$ and $B$ satisfying the condition. If $\\displaystyle \\frac{\\tan A}{\\tan B}=k$, then $\\displaystyle k=\\frac{\\sin A\\cos B}{\\cos A\\sin B}=\\frac{5\\cos B}{\\cos A}$. $$5\\cos B=k\\cos A$$ $$(5\\sin B)^2+(5\\cos B)^2=\\sin^2A+k^2\\cos^2A$$ $$k^2=\\frac{25-\\sin^2A}{\\cos^2A}$$ So $k$ depends on $A$ and is not a constant. suppose $\\dfrac{SinA}{SinB}=\\dfrac{x}{y}=5$ so from right tringle we know $cosA=\\sqrt{1-x^{2}}$ and so $$tanA=\\dfrac{x}{\\sqrt{1-x^{2}}}.$$ the same for y so : $$\\dfrac{tanA}{tanB}=\\dfrac{\\frac{x}{\\sqrt{1-x^{2}}}}{\\frac{y}{\\sqrt{1-y^{2}}}}$$ and we have $x=5y$ so $x^{2}=25y^{2}$ , $1-x^{2}=1-25y^{2}$ $$\\dfrac{tanA}{tanB}=\\dfrac{\\frac{5y}{\\sqrt{1-25y^{2}}}}{\\frac{y}{\\sqrt{1-y^{2}}}}=\\dfrac{5y\\sqrt{1-y^{2}}}{y\\sqrt{1-25y^{2}}}$$ now you need to know what is $y$!", "what you mean, so you write it as a \"single\" function of cos, and then find for max and mins. Thanks! – seeker Mar 15 '13 at 21:53 You're welcome! – amWhy Mar 15 '13 at 21:58 $$1-4 \\cos (2 \\theta)+3 \\sin(2 \\theta) = 1 - 5 \\left({4 \\over 5} \\cos (2 \\theta) - {3 \\over 5} \\sin (2 \\theta) \\right)$$ $$=1-5 \\cos (2 \\theta + \\phi)$$ With $\\cos \\phi = {4 \\over 5}$ and $\\sin \\phi = {3 \\over 5}$. Now conclude. - When I use this: the minimum cos=-1, I get 1-5(-1)=6, and for the max I get 1-5(1)=-4; when I use the max value of cos=1 I get -4 which is smaller than the value I get when I use the min value of cos, is this okay? – seeker Mar 15 '13 at 22:44 Yes. You want the min and max of your function, whhich are -4 and 6. Actually, they are found for max and min cosine respectively, but it's not a problem. – Jean-Claude Arbaut Mar 16 '13 at 9:34 Following is an algebraic way of finding the range of $As$\\cos2\\theta=\\frac{1-\\tan^2\\theta}{1+\\tan^2\\theta}$and$\\sin2\\theta=\\frac{2\\tan\\theta}{1+\\tan^2\\theta}$So,$1-4\\cos2\\theta+3\\sin2\\theta=\\frac{(1+t^2)-4(1-t^2)+3(2t)}{1+t^2}=y$(say) putting$t=\\tan\\theta$So,$t^2(y-5)-6t+y+3=0$This is a Quadratic Equation in$t=\\tan\\theta$As$\\theta$is real, so will be$t,$hence the discriminant$(-6)^2-4(y-5)(y+3)\\ge0\\implies y^2-2y-24\\le0$(i)$\\implies (y-6)(y+4)\\le 0\\implies$either$(y\\le 6, y\\ge -4\\implies -4\\le y\\le 6)$or$(y\\le -4,y\\ge 6)$which is impossible (ii)$\\implies (y-1)^2\\le 25=5^2\\implies -5\\le y-1\\le 5\\implies", "the numerator and denominator are equal. Since $$\\cosh^2(y)-\\sinh^2(y)=1$$, equation $$\\text{(2a)}$$ says that is when \\begin{align} \\sin^2(x)\\cosh^2(y)+\\cos^2(x)\\sinh^2(y)&=\\cos^2(x)\\cosh^2(y)+\\sin^2(x)\\sinh^2(y)\\tag{3a}\\\\[6pt] \\sin^2(x)&=\\cos^2(x)\\tag{3b} \\end{align} Thus, $$|\\tan(z)\\,|=1$$ when $$\\Re(z)=\\pm\\frac\\pi4$$ Furthermore, equation $$\\text{(2b)}$$ says not only that $$\\tan(\\bar z)=\\overline{\\tan(z)}$$, but also that, when $$\\Re(z)=\\pm\\frac\\pi4$$, $$\\Re(\\tan(z))$$ has the same sign as $$\\Re(z)$$. • This answer has finally explained to me why $\\tan (\\bar {z})=\\overline{\\tan(z)}$. Thank you so much! Mar 23 at 22:55 • Actually $f(\\bar z)=\\overline{f(z)}$ is true for any function that is analytic on a connected region that includes a real interval and is real-valued on that real interval. See property of a holomorphic function on a connected open subset of complex plane. – robjohn Mar 24 at 7:55 I've learned some new related facts since posting the question, one of which solves this very simply. The relevant formula, stated below, can be found by using the fact $$\\arctan z=-i\\operatorname{artanh} iz$$. $$\\arctan(x+iy)=\\frac{1}{2}\\arctan\\frac{2x}{1-x^2-y^2}+\\frac{i}{2}\\operatorname{artanh}\\frac{2y}{1+x^2+y^2}$$ Plugging in $$x=\\cos x$$ and $$y=\\sin x$$ we find that $$\\arctan e^{ix}=\\frac{1}{2}\\arctan(\\pm\\infty)+\\frac{i}{2}\\operatorname{artanh}\\sin x$$ And thus $$C=\\pm\\frac{\\pi}{4}$$", "# Trigonometric Equations, Trouble solving. $\\displaystyle \\sin x + \\sqrt3\\cos x =0$ $0< x <360^\\circ$ Hey, I haven't done this question for a while and I forgot how to solve it. I have made attempts to solve it, such as splitting the sinx and cosx, but I still cant figure it out. Could some one please hint me or walk me through the steps? Thank you. Since $$\\sin x+\\sqrt 3\\cos x=2(\\sin x\\times(1/2)+\\cos x\\times (\\sqrt 3/2))=2\\sin (x+60^\\circ),$$ you'll have$$\\sin x+\\sqrt 3\\cos x=0\\iff \\sin(x+60^\\circ)=0.$$ Since $$0\\lt x\\lt 360^\\circ\\iff 60^\\circ \\lt x+60^\\circ\\lt 420^\\circ,$$ you'll have$$x+60^\\circ=180^\\circ, 360^\\circ.$$ Hence, $$x=120^\\circ, 300^\\circ.$$ • Clubbing $A\\sin x+B\\cos x=C$ is worth only if $C\\ne0,$ right? – lab bhattacharjee Jan 20 '14 at 5:26 • Well, I think this way can be used in the case $C=0.$ – mathlove Jan 20 '14 at 5:34 • luckily here $\\arccos \\frac A{\\sqrt{A^2+B^2}}$ is simple, else we had to deal with a little ugly trig inverse functions – lab bhattacharjee Jan 20 '14 at 5:36 • Yeah, I agree with the point. – mathlove Jan 20 '14 at 5:38 $$\\sin x+\\sqrt3\\cos x=0\\iff \\sin x=-\\sqrt3\\cos x$$ $$\\iff \\tan x=-\\sqrt3=-\\tan60^\\circ=\\tan(-60^\\circ)$$ as $\\tan(-y)=-\\tan y$ $$x=n180^\\circ+(-60^\\circ)$$ where $n$ is any integer • Haven't you omitted the case $\\cos (x) = 0$? – goblin Jan 20 '14 at 5:20 • @user18921, find my comment in", "\\sin x \\cos x$. Now note that $(\\frac{\\sqrt{3} - 1}{2 \\sqrt{2}})^2 +(\\frac{\\sqrt{3} + 1}{2 \\sqrt{2}})^2 = \\frac{1}{8} ((3 -2 \\sqrt{3} + 1) + (3 +2 \\sqrt{3} + 1) = 1$ so that we can write $\\frac{\\sqrt{3} - 1}{2 \\sqrt{2}}$ as the $\\sin$ of some angle, say $\\alpha$, and identify $\\frac{\\sqrt{3} + 1}{2 \\sqrt{2}}$ as $\\cos \\alpha$. Then you have to solve $\\sin(\\alpha + x) = \\sin(2x)$, with $\\tan \\alpha = \\frac{\\sqrt{3} - 1}{\\sqrt{3} + 1}$. - Replace $\\sqrt3=\\tan60^\\circ,1=\\tan45^\\circ$ to find $$\\sin(15^\\circ+x)=2\\sin x\\cos x=\\sin2x$$ Now if $\\sin y=\\sin A, y=n180^\\circ+(-1)^nA$ where $n$ is any integer As $\\sqrt3-1=\\tan60^\\circ-\\tan45^\\circ=\\dfrac{\\sin(60-45)^\\circ}{\\cos60^\\circ\\cos45^\\circ}=2\\sqrt2\\sin15^\\circ$ Similarly, $\\sqrt3+1=2\\sqrt2\\sin75^\\circ=2\\sqrt2\\cos15^\\circ$ - @user307178, Please follow the last two lines – lab bhattacharjee Mar 20 at 9:05 Thanks got it...My reputation is less so I can't upvote – user307178 Mar 20 at 9:07 @user307178, I actually tried to derive the sine/cosine value of $$\\dfrac{\\sqrt3\\pm1}{2\\sqrt2}$$ without assuming them. – lab bhattacharjee Mar 20 at 9:15 And I have learned a useful method to derive them! Because I always forget it. Thanks! – user307178 Mar 20 at 9:25 Do you know a similar trick to get sin18°? That would be awesome – user307178 Mar 20 at 9:29", "<meta http-equiv=\"refresh\" content=\"1; url=/nojavascript/\"> Sum and Difference Formulas | CK-12 Foundation 14.4: Sum and Difference Formulas Created by: CK-12 Objective To use and derive the sum and difference formulas. Review Queue Using your calculator, find the value of each trig function below. Round your answer to 4 decimal places. 1. $\\sin 75^\\circ$ 2. $\\cos 15^\\circ$ 3. $\\tan 105^\\circ$ 4. $\\cos 255^\\circ$ Finding Exact Trig Values using Sum and Difference Formulas Objective Here you'll use the sum and difference formulas to find exact values of angles other than the critical angles. Guidance You know that $\\sin 30^\\circ=\\frac{1}{2}, \\cos 135^\\circ=-\\frac{\\sqrt{2}}{2}, \\tan 300 ^\\circ = -\\sqrt{3},$ etc... from the special right triangles. In this concept, we will learn how to find the exact values of the trig functions for angles other than these multiples of $30^\\circ, 45^\\circ,$ and $60^\\circ$. Using the Sum and Difference Formulas, we can find these exact trig values. Sum and Difference Formulas $\\sin(a\\pm b) &=\\sin a \\cos b \\pm \\cos a \\sin b \\\\\\cos(a\\pm b) &=\\cos a \\cos b \\mp \\sin a \\sin b \\\\\\tan(a \\pm b) &=\\frac{\\tan a \\pm \\tan b}{1 \\mp \\tan a \\tan b}$ Example A Find the exact value of $\\sin 75^\\circ$. Solution: This is an example of where we can use the sine sum formula from above, $\\sin(a+b)=\\sin a \\cos b+\\cos a", "= 2$, therefore by Pythagoras theorem, $AB = \\sqrt{3}$ $\\therefore \\tan A =$ $\\frac{1}{\\sqrt{3}}$ ; $\\sin A =$ $\\frac{1}{2}$ ; $\\cos A =$ $\\frac{\\sqrt{3}}{2}$ Hence $\\frac{1}{\\tan A}$ $+$ $\\frac{\\sin A}{1+ \\cos A}$ $=$ $\\sqrt{3} +$ $\\frac{1}{2} (\\frac{2}{2+\\sqrt{3}})$ $= \\sqrt{3} +$ $\\frac{1}{2+\\sqrt{3}}$ $=$ $\\frac{2\\sqrt{3}+ 3 + 1}{2 + \\sqrt{3}}$ $=$ $\\frac{2(2+\\sqrt{3})}{(2+\\sqrt{3})}$ $= 2$ $\\\\$ Question 4: If $\\tan A = \\sqrt{2}-1$, show that $\\sin A \\cos A =$ $\\frac{\\sqrt{2}}{4}$ Given $\\tan A = \\sqrt{2}-1$ By Pythagoras theorem, we have $AC = \\sqrt{(\\sqrt{2}-1)^2 + 1^2} = \\sqrt{4 - 2\\sqrt{2}}$ $\\therefore \\sin A =$ $\\frac{\\sqrt{2}-1}{\\sqrt{4 - 2\\sqrt{2}}}$ and $\\cos A =$ $\\frac{1}{\\sqrt{4 - 2\\sqrt{2}}}$ Hence, $\\sin A \\cos A =$ $\\frac{\\sqrt{2}-1}{\\sqrt{4 - 2\\sqrt{2}}}$ $\\times$ $\\frac{1}{\\sqrt{4 - 2\\sqrt{2}}}$ $=$ $\\frac{\\sqrt{2}-1}{(4 - 2\\sqrt{2})}$ $=$ $\\frac{(\\sqrt{2}-1)}{2\\sqrt{2}(\\sqrt{2}-1)}$ $=$ $\\frac{1}{2\\sqrt{2}}$ $\\\\$ Question 5: In $\\triangle ABC$, right angles at $C$, if $\\tan A =$ $\\frac{1}{\\sqrt{3}}$ and $\\tan B = \\sqrt{3}$, show that $\\sin A \\cos B + \\cos A \\sin B = 1$ Given $\\tan A =$ $\\frac{1}{\\sqrt{3}}$ By Pythagoras theorem, we have $AB = \\sqrt{ (\\sqrt{3})^2 + 1^2} = \\sqrt{3+1} = \\sqrt{4} = 2$ $\\therefore \\sin A =$ $\\frac{1}{2}$ and $\\cos A =$ $\\frac{\\sqrt{3}}{2}$ $\\sin B =$ $\\frac{\\sqrt{3}}{2}$ and $\\cos B =$ $\\frac{1}{2}$ $\\therefore \\sin A \\cos B + \\cos A \\sin B =$ $\\frac{1}{2}$ $\\times$ $\\frac{1}{2}$ $+$ $\\frac{\\sqrt{3}}{2}$ $\\times$ $\\frac{\\sqrt{3}}{2}$ $=$ $\\frac{1}{4}$ $+$ $\\frac{3}{4}$", "24 at 19:22 • You could also use that $\\sin{x}=\\cos{\\left(\\frac{\\pi}{2}-x\\right)}$ and use the difference formula for $\\cos{(x-y)}$ – Eleven-Eleven Jul 24 at 19:24 $$\\sin^2 x + \\cos^2 x = 1$$ so $$(\\frac 2{\\sqrt 5})^2 + \\cos^2 x=1$$ so ..... $$\\frac 45 + \\cos^2 x=1$$ . $$\\cos^2 x = 1-\\frac 45 =\\frac 15$$ . $$\\cos x = \\pm \\sqrt {\\frac 15} = \\pm \\frac 1{\\sqrt 5}$$. . But $$x< 90^\\circ$$ so $$\\cos x > 0$$. . So $$\\cos x = \\frac 1{\\sqrt 5}$$. Hint: We have $$\\cos^2(x) = 1- \\sin^2(x)$$ Thus $$\\cos^2(x) = 1 - \\frac{4}{5} = \\frac{1}{5}$$ Then ($$x < 90°$$) $$\\cos(x) = \\frac{1}{\\sqrt{5}}$$ Can you finish? • What's left to finish? – fleablood Jul 25 at 5:05 • @fleablood \"and hence evaluate $\\dfrac {1+\\cos(x)}{1-\\sin(x)}$\". – Monadologie Jul 25 at 7:19 $$\\sin^2{x} + \\cos^2{x} = 1$$ $$\\cos^2{x} = 1 - \\sin^2{x}$$ $$\\cos{x} = \\sqrt{1 - \\sin^2{x}} \\quad \\land \\quad \\sin{x} = \\frac{2}{\\sqrt{5}} \\quad \\land \\quad x < 90^{\\circ}$$ $$\\Longrightarrow \\cos x = \\sqrt{1 - \\Big(\\frac{2}{\\sqrt{5}} \\Big)^2}=\\sqrt{\\frac{1}{5}}=\\frac{1}{\\sqrt{5}}$$ Can you continue? • \"Can you continue?\"? To where? – fleablood Jul 24 at 19:37 • To do the other question he means - thank you the three of you – scoff Jul 24 at 19:56" ]

[ "sequence $(a_n)$ has $a_1=\\sin x$, $a_2=\\cos x$, and $a_3= \\tan x$ for some real number $x$. For what value of $n$ does $a_n=1+\\cos x$? Let $S$ be the set of all point $(x,y)$ in the coordinate plane such that $0 \\le x \\le \\frac{\\pi}{2}$ and $0 \\le y \\le \\frac{\\pi}{2}$. What is the area of the subset of $S$ for which $\\sin^2x-\\sin x \\sin y + \\sin^2y \\le \\frac34?$ In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG + JH + CD$? Let integer $n\\ge 2$, prove $$\\sin{\\frac{\\pi}{n}}\\cdot\\sin{\\frac{2\\pi}{n}}\\cdots\\sin{\\frac{(n-1)\\pi}{n}}=\\frac{n}{2^{n-1}}$$ Joel selected an acute angle $x$ (strictly between 0 and 90 degrees) and wrote the value of $\\sin x$, $\\cos x$, and $\\tan x$ on three different cards. Then he gave those cards to three students, Malvina, Paulina, and Georgian, one card to each, and asked them to figure out which trigonometric function (sin, cos, tan) produced their cards. Even after sharing the values on their cards with each other, only Malvian was able to surly identify which function produced the value on her card. Compute the sum of all possible values that Joel wrote on Marlvina's card. Prove the following identities \\begin{align} \\sin (3\\alpha) &= 4\\cdot \\sin(60-\\alpha)\\cdot \\sin\\alpha\\cdot \\sin(60+\\alpha)\\\\ \\cos (3\\alpha) &= 4 \\cdot\\cos(60-\\alpha)\\cdot \\cos\\alpha\\cdot \\cos(60+\\alpha)\\\\ \\tan (3\\alpha) &= \\tan(60-\\alpha) \\cdot\\tan\\alpha \\cdot\\tan(60+\\alpha) \\end{align}", "#### Girls Olympic ###### back to index Among all pairs of real numbers $(x, y)$ such that $\\sin\\sin x=\\sin\\sin y$ with $-10\\pi \\le x, y \\le 10\\pi$. Oleg randomly selected a pair $(X, Y)$. Compute the probability that $X = Y$. Joel selected an acute angle $x$ (strictly between 0 and 90 degrees) and wrote the value of $\\sin x$, $\\cos x$, and $\\tan x$ on three different cards. Then he gave those cards to three students, Malvina, Paulina, and Georgian, one card to each, and asked them to figure out which trigonometric function (sin, cos, tan) produced their cards. Even after sharing the values on their cards with each other, only Malvian was able to surly identify which function produced the value on her card. Compute the sum of all possible values that Joel wrote on Marlvina's card.", "### Trigonometry: Compound Angles Fluency exercise Below are some function cards, some value cards, and some target cards. You may wish to print them out and cut them up, but keep them in the three separate piles. You can choose a total of six cards, split between function cards and value cards in any way you choose (perhaps one or two function cards would be a good choice to start with). Choose your cards at random (pick from face-down cards, or get a friend to choose). Then choose a random target card. Your challenge is now to make the target by evaluating functions from your selected cards at values from your selected cards. You can use each card at most once (you don’t have to use them all). For example, if you have the function card $\\sin(\\square + \\square)$ and the value cards $\\dfrac{\\pi}{6}$ and $\\dfrac{\\pi}{3}$, then you could evaluate $\\sin\\left(\\frac{\\pi}{6} + \\frac{\\pi}{3}\\right),$ but you would not be allowed to evaluate $\\sin\\left(\\frac{\\pi}{6} + \\frac{\\pi}{6}\\right)$ (unless you happen to have two $\\dfrac{\\pi}{6}$ cards). #### Function $\\sin(\\square)$ #### Function $\\cos(\\square)$ #### Function $\\tan(\\square)$ #### Function $\\sin(\\square + \\square)$ #### Function $\\cos(\\square + \\square)$ #### Function $\\tan(\\square + \\square)$ #### Function $\\sin(\\frac{\\square}{2})$ #### Function $\\cos(\\frac{\\square}{2})$ #### Function $\\tan(\\frac{\\square}{2})$ #### Value $0$ #### Value $\\dfrac{\\pi}{6}$ #### Value $\\dfrac{\\pi}{4}$ #### Value $\\dfrac{\\pi}{3}$ ####", "is $\\dfrac{\\sin(x)}{x}$. tanc Tanc function or tangent cardinalis, tanc(x) is defined as $\\dfrac{\\tan(x)}{x}$. hubb Hubbert curve, e.g. hubb(x) for $\\dfrac{1}{2+2\\cdot \\cosh(x)}$. L Langevin function, e.g. L(x) for $\\coth(x)-\\dfrac{1}{x}$. deg converts a radian number into the equivalent in degrees. deg(pi) $= \\pi\\cdot \\frac{180^\\circ}{\\pi} = 180 (^\\circ)$. rad converts a number in degrees into its radian equivalent. rad(180) = $180^\\circ \\cdot \\frac{\\pi}{180^\\circ} = \\pi$. top ### - non-differentiable functions abs the absolute value of the variable, abs(x) = $|x| = \\begin{cases} x; \\; & x \\geq 0 \\\\ (-1)\\cdot x ; \\; & x<0 \\end{cases}$ min minimum of several values, e.g. min(1;x;x^(1/3)) is the minimum of $1, x$ and $\\sqrt[3]{x}$. max maximum of several values, max(abs(x);x*x) is the maximum of the absolute value of $x$ and $x^2$. % modulo division, the whole-numbered remainder, e.g. 10%x fmod modulo division, floating point remainder, so fmod(x;1) displays everything after the decimal point of the input value. round or R round, the optional second argument sets the number of decimals. For example, R(x;2) rounds x to two decimal places, R(x) rounds to an integer. floor or R0 floor (round down to an integer). For example, floor(x) is equivalent to Gauss brackets $[x]$ (the integer part of $x$). ceil or R1 ceil (round up), ceil(x) is equivalent to $[x] + 1$. dist distance function, dist(x)", "but I wouldn't have the term in $x$ less than the coefficient. – Sebastiano Jun 25 '20 at 22:53 $$I=\\int\\frac{2\\cos x-\\sin x}{3\\sin x+5\\cos x }\\,dx$$ $$2\\cos x-\\sin x=\\sqrt5 \\cos(x+a)$$, where $$\\tan a=1/2$$ and $$3\\sin x+5\\cos x=\\sqrt{34}\\cos(x-b)$$, where $$\\tan b=3/5$$ $$I=\\int \\frac{\\sqrt 5 \\cos(x+a) } {\\sqrt{34}\\cos(x-b)} \\, dx$$ Substitute $$t=x-b$$ so that $$I=\\sqrt{5/34}\\int \\frac{\\cos(t+a+b) } {\\cos t}\\, dt$$ The integrand is now: $$\\cos(a+b) - \\tan t\\sin(a+b)$$ Can you take it from here? • My $\\infty$ thank you. I think that this way is more complicated. – Sebastiano Jun 27 '20 at 9:01 • @Sebastiano: You are welcome. This way isn't complicated. It only involves calculating $\\cos (a+b)$. Given $\\tan a, \\tan b$, we can find $\\sin, \\cos$ and after finding $\\sin (a+b)$, we can find $\\cos (a+b)$. Why do you think that it's complicated? – Koro Jun 27 '20 at 9:05 • @Sebastiano: What I would do is imagine a rt. angled triangle (for angle $a$), then its perp. is $1$,base is $2$ and hence hypot. is $\\sqrt 5$. Hence we can find $cos a$, $sin a$ etc. Similarly for angle $b$. You don't have to do $\\sin(arctan (1/2)+arc tan (3/5))$ – Koro Jun 27 '20 at 9:10 • Why there are roots square $\\cos \\theta=\\pm 1/\\sqrt{1+\\tan^2\\theta}$, $\\sin \\theta=\\pm \\tan \\theta/\\sqrt{1+\\tan^2\\theta}$ and the choice of angle depends on the", "# Thread: Help on Trig Questions. 1. ## Help on Trig Questions. 1. Use a Sum or difference identity to find the exact value of sin285degrees 2. If tanX = -3/4 and 90<x<180, find cosX (the 90 and 180 are degrees) 3. If (tanX)(cscX) = 3, find value of cosX 4. If tanX= 3/4 and 180<X<270, find the exact value of sin2X 5. Find the value of tan(A+b) if cosA = 13/5, tanB = 3/4, and 0<A<90 2. Originally Posted by Joker [1. Use a Sum or difference identity to find the exact value of sin285degrees $\\displaystyle \\sin\\left(285^{\\circ}\\right) = \\sin\\left(180^{\\circ}+105^{\\circ}\\right) = -\\sin\\left(105^{\\circ}\\right) = -\\sin\\left(60^{\\circ}+45^{\\circ}\\right)$ Expand that using the 'sum or difference' formula for sine. 2. If tanX = -3/4 and 90<x<180, find cosX (the 90 and 180 are degrees) Pythagoras' theorem: If $\\displaystyle \\tan{x} = \\frac{-3}{4}$, then $\\displaystyle \\cos{x} = \\frac{4}{\\sqrt{(-3)^2+(4)^2}} =$... 3. If (tanX)(cscX) = 3, find value of cosX $\\displaystyle {\\tan{x}}{\\csc{x}} = \\frac{\\sin{x}}{\\cos{x}\\sin{x}} = \\frac{1}{\\cos{x}},$ so $\\displaystyle \\cos{x} = \\cdots$ 4. If tanX= 3/4 and 180<X<270, find the exact value of sin2X Write $\\displaystyle \\tan{x}$ in terms of $\\displaystyle \\cos{x}$ and $\\displaystyle \\sin{x}$ and use the identity $\\displaystyle \\sin{2x} = 2\\sin{x}\\cos{x}$. 5. Find the value of tan(A+b) if cosA = 13/5, tanB = 3/4, and 0<A<90 Write $\\displaystyle \\tan\\left(a+b\\right) = \\frac{\\sin\\left(a+b\\right)}{\\cos\\left(a+b\\right)}$, use the 'sum-difference' formulas on", "# Finding all values between $-180^\\circ\\leq x\\leq 180^\\circ$ I am required to solve the following equation for $-180^\\circ\\leq x\\leq 180^\\circ$: $$\\sin2x=\\tan x.$$ I am aware that I can convert $\\sin2x = 2\\sin x\\cos x$. From this I can derive that $$\\sin(x)(2\\cos2(x)−1)=0$$ $$=$$ $$\\sin(x)\\cos(2x)=0$$ For $\\sin=0$, I can use $0°$. for $\\cos2x=0$, I can use $(\\frac{π}{2})^°$ The two answers I can deduce are therefore $0°$ and $(\\frac{π}{2})^°$ I am still unsure how to derive $0,±π,±\\frac{π}{4},±\\frac{3π}{4}$, which i know should be the final answer. If anyone can explain to me how to achieve this I would be very grateful. • I think you made a slight algebra mistake, it should be $sin(x) (2 cos^2 (x) -1) = 0$, and $2 cos^2 (x) -1 = 0$ when $cos(x) = \\pm \\sqrt{2} / 2$, so you'll get the other 4 solutions from this. – Sebastian Schulz Apr 18 '17 at 18:38 • Don't mix degrees and radians like that. $\\pi/2$ or $90^\\circ$, but not $(\\frac{\\pi}2)^\\circ$. Also, note that $90^\\circ$ is not a valid answer because the original equation involves $\\tan x$ and $\\tan x$ is not defined when $x = 90^\\circ$. $\\sin x = 0$ gives you more than just $x=0^\\circ$. Where else is $\\sin x = 0$? – tilper Apr 18 '17 at 18:38 • A comment for notation: instead of", "one says, “Altogether, we have 28 legs.” The green one says, “Altogether, we have 27 legs.” The yellow one says, “Altogether, we have 26 legs.” The red one says, “Altogether, we have 25 legs.” What is the colour of the servant who tells the truth? # Trigonometry Identities ### Negative angles: • $\\sin (-x)=-\\sin x$ • $\\boxed{\\cos (-x)=\\cos x}$ (Still Positive!) • $\\tan (-x)=-\\tan x$ Reason: $-x$ is in the “C” quadrant so Cosine is still positive. ### Supplementary angles: • $\\boxed{\\sin (180^\\circ -x)=\\sin x}$ (Still positive!) • $\\cos (180^\\circ -x)=-\\cos x$ • $\\tan (180^\\circ -x)=-\\tan x$ Reason: $180^\\circ -x$ is in the “S” quadrant so Sine is still positive. ### Complementary angles: • $\\sin (90^\\circ -x)=\\cos x$ • $\\cos (90^\\circ -x)=\\sin x$ • $\\tan (90^\\circ -x)=\\cot x$ Reason: • $\\sin (90^\\circ -x)=\\cos x=a/c$ • $\\cos (90^\\circ -x)=\\sin x=b/c$ • $\\tan (90^\\circ -x)=\\cot x=a/b$ # Which Secondary school is the best in Singapore Due to too many secondary schools, we can only list a few here at the start. There may be other secondary schools that can be the best in Singapore. Do feel free to add your own school and vote for it below! Also feel free to leave any comments as to why the secondary school you choose is the best in Singapore. (We are not", "# Does $\\tan x\\cdot \\cos x$ equal $\\sin x$? Is it true that $\\tan x\\cdot \\cos x = \\sin x$? If I put $x=30$ in my calculator then I don't get the same answer as $\\sin 30$, why is this? Don't the two cosines cancel out? I'm probably missing something really stupid here. • Have you inputted $\\tan(30^\\circ)\\cos(30^\\circ)$ or something that will give you $\\tan(30\\cos(30))$? Also this is not true for $x=(90^\\circ$ plus multiples of $180^\\circ$). – JP McCarthy Apr 27 '15 at 10:42 • You might want to double-check that your calculator is on degrees. They both give sin(30), but only with degrees do you have sin(30)=1/2. – Akiva Weinberger Apr 27 '15 at 10:47 • (But tanx*cosx=sinx, yeah. Assuming tanx is defined.) – Akiva Weinberger Apr 27 '15 at 10:47 • It is possible only when $\\cos x$ is non-zero – Empty Apr 27 '15 at 10:48 • A properly functioning calculator used properly will return the same value for $\\tan 30\\cdot\\cos 30$ as for $\\sin 30$ regardless of whether it is set on degrees or radians or any other unit, except when the unit is so chosen that the number $30$ represents a right angle or other value for which the cosine is $0$. However, I said \"used properly\" and I've noticed students sometimes getting", "+0 # help pls 0 90 2 Could someone point me in the right direction? Thanks! If x is an angle such that tan x + sec x = 2, then evaluate cos x. Oct 16, 2022 #1 +14441 +2 evaluate cos x Hello Guest! $$tan\\ x + sec\\ x = 2\\\\ sec\\ x=1/cos\\ x\\\\ tan\\ x=\\dfrac{\\sqrt{1-cos^2x}}{cos\\ x}\\\\ \\dfrac{\\sqrt{1-cos^2x}}{cos\\ x}+\\dfrac{1}{cos\\ x}=2\\\\ \\sqrt{1-cos^2x}+1=2cos\\ x$$ $$\\sqrt{1-cos^2x}=2cos\\ x -1\\ |\\ cos\\ x=y\\\\ 1-y^2=(2y-1)^2\\\\ 1-y^2=4y^2-4y+1\\\\ 5y^2-4y=0\\\\ y\\in \\{0,0.8\\}\\ |\\ y=cos\\ x\\\\ \\color{blue}cos\\ x\\in \\{0,0.8\\}$$ ! Oct 16, 2022 edited by asinus Oct 16, 2022 edited by asinus Oct 16, 2022 #2 +1 In answer #1, if $$\\cos x = 0,$$ then $$\\sec x$$ will be infinite and so should be rejected. Here are two alternative routines. $$\\tan x + \\sec x = 2, \\\\ \\sec x =2 - \\tan x, \\\\ \\sec^{2}x=(2 - \\tan x)^{2}, \\\\ 1+ \\tan^{2}x = 4-4\\tan x+\\tan^{2}x, \\\\ 4 \\tan x = 3, \\\\ \\tan x = 3/4.$$ The 3 and the 4 can be seen as two legs of a 3, 4, 5 rt-angled triangle from which we can read off the other ratios. $$\\sin x = 3/5, \\cos x = 4/5,\\sec x = 5/4,$$ and there are two others. Note that $$\\tan x + \\sec x = 3/4+5/4 = 8/4 =2.$$ Alternatively, $$\\tan x + \\sec", "NZ Level 7 (NZC) Level 2 (NCEA) Period changes for tangents Lesson We saw, in another chapter, how the period of the sine and cosine functions is affected by the coefficient $k$k that multiplies $x$x in $\\sin kx$sinkx and $\\cos kx$coskx. The fact that the sine and cosine functions both have period $360^\\circ$360° is expressed by the statements $\\sin(x+360^\\circ n)=\\sin x$sin(x+360°n)=sinx and $\\cos(x+360^\\circ n)=\\cos x$cos(x+360°n)=cosx for all integers $n$n. So it must be true that since $\\tan x=\\frac{\\sin x}{\\cos x}$tanx=sinxcosx, then $\\tan(x+360^\\circ n)=\\tan x$tan(x+360°n)=tanx. However, $360^\\circ$360° is not the smallest interval at which the tangent function repeats. A glance at the graph of the tangent function, shown below, should convince you that this function has a period of $180^\\circ$180°. That is, $\\tan(x+180^\\circ n)=\\tan x$tan(x+180°n)=tanx for all integers $n$n. The fact that the tangent function repeats at intervals of $180^\\circ$180° can be verified by considering the unit circle diagram. If $180^\\circ$180° is added to an angle $\\alpha$α, then the diagram shows that $\\sin(\\alpha+180^\\circ)$sin(α+180°) has the same magnitude as $\\sin\\alpha$sinα but opposite sign. The same relation holds between $\\cos(\\alpha+180^\\circ)$cos(α+180°) and $\\cos\\alpha$cosα. We make use of the definition: $\\tan\\alpha=\\frac{\\sin\\alpha}{\\cos\\alpha}$tanα=sinαcosα $\\tan(\\alpha+180^\\circ)$tan(α+180°) $=$= $\\frac{\\sin(\\alpha+180^\\circ)}{\\cos(\\alpha+180^\\circ)}$sin(α+180°)cos(α+180°)​ $=$= $\\frac{-\\sin\\alpha}{-\\cos\\alpha}$−sinα−cosα​ $=$= $\\tan\\alpha$tanα We are now in a position to determine the period of the function $\\tan kx$tankx where $k$k is any number. We can define a new", "# The values of trigonometric ratios Today I explained to my students how the value of $\\sin(90°)$ is 1 by first drawing a right angled triangle and taking $90°$ as a reference angle, and told them that now $\\sin(\\theta)=\\frac{\\text{opposite}}{\\text{hypotenuse}}$. Then I used pythagorean relation to show that $\\cos(90°)$ is 0, i.e $\\sin^2(\\theta)=1-\\cos^2(\\theta)$. And finally I used quotient relation to show how $\\tan(90°)$ is infinity. The students didn't seem much convinced by my methods as they were asking why we couldn't find the value of $\\cos(90°)$ and $\\tan(90°)$ the same way we did to find the value of $\\sin(90°)$. Is there any other ideas to show the value of $\\sin(90°)$, $\\cos(90°)$ and $\\tan(90°)$ understandble to them? • \"to show how $\\tan(90^\\circ)$ is infinity\" Technically, this is incorrect. Although the limit of $\\tan\\theta$ as $\\theta$ approaches $90^\\circ$ from below is infinity, what is the limit as $\\theta$ approaches $90^\\circ$ from above? – Joel Reyes Noche Jun 1 '14 at 23:46 I don't know if, in good faith, you can really assert than $\\sin(90^\\circ)=1$ without your students having a very firm grasp of the unit circle or at least some kind of limiting process. But, then, if the students have a firm grasp of the unit circle, your argument for why $\\cos(90^\\circ)=0$ should be seen as unnecessarily complicated. And then, as", "# Prove that $\\sin(12^\\circ)\\sin(48^\\circ)\\sin(54^\\circ)=\\frac18$ [closed] Prove that $$\\sin(12^\\circ)\\sin(48^\\circ)\\sin(54^\\circ)=\\frac18.$$ Without using a calculator. I tried all identities I know but I have no idea how to proceed. I always get stuck on finding $\\sin36^\\circ$. ## closed as off-topic by heropup, David H, Cookie, Adam Hughes, GinaJul 26 '14 at 6:42 This question appears to be off-topic. The users who voted to close gave this specific reason: • \"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level.\" – heropup, David H, Cookie, Adam Hughes, Gina If this question can be reworded to fit the rules in the help center, please edit the question. • This is false as stated. Please, do mention that you measure angles in degrees, not in radians. The overwhelming majority of mathematicians will by default treat the argument of $\\sin$ as being in radians. – Dan Shved Jul 26 '14 at 6:06 • Thankyou for the suggestion – Kanishk Jul 26 '14 at 6:08 • We have $$\\sin12^\\circ\\sin48^\\circ\\sin54^\\circ=\\frac18\\tag1$$ Multiplying $(1)$ by $2$ yields $$2\\sin12^\\circ\\sin48^\\circ\\sin54^\\circ=\\frac14\\tag2$$ Using identities: $2\\sin a\\sin b=\\cos(a-b)-\\cos(a+b)$ and $\\sin(90^\\circ-x)=\\cos x$ yields", "# Given that sin a = 3/5 and a is in the interval (90,180), calculate tan a. hala718 | High School Teacher | (Level 1) Educator Emeritus Posted on sina = 3/5 We know that : sin^2 + cos^2 = 1 ==> cosa = sqrt(1-sin^2a) = sqrt(1-9/25) = sqrt(16/25) = 4/5 ==> cosa = 4/5 Since a is in the second quadrant , then sin>0 and cos < 0 ==> cosa = -4/5 ==> tana = sina/cosa = (3/5) / (-4/5)= -3/4 => tan(a) = -3/4 william1941 | College Teacher | (Level 3) Valedictorian Posted on For an angle x that lies in the interval 90 degrees to 180 degrees, sin x has positive values , cos x on the other hand in negative. We have sin a = 3 / 5. We also know that sin^a+ cos^a =1. Therefore we can get cos a = sqrt ( 1- sin^a) = sqrt [ 1- (3/5 )^2] = sqrt [ 1- ( 9/ 25)] = sqrt [ (25 -9)/ 25] =sqrt (16/ 25) = - 4/5 Therefore tan a = sin a / cos a = (3/5)/ (-4/5) = 3/ -4 = -3/4. The value of tan a is -3/4 neela | High School Teacher | (Level 3) Valedictorian Posted on sina = 3/5 To calculate for the value", "A\\over 1 + tan^2 A$$ We can also write above relation in terms of angle A/2, just replace A by A/2, we get sin A = $$2 tan ({A\\over 2})\\over 1 + tan^2 ({A\\over 2})$$ Example : Find the value of Sin 120 ? Solution : We Know that sin 60 = $$\\sqrt{3}\\over 2$$ and cos 60 = $$1\\over 2$$ By using above formula, sin 120 = 2 sin 60 cos 60 = 2 $$\\times$$ $$\\sqrt{3}\\over 2$$ $$\\times$$ $$1\\over 2$$ $$\\implies$$ sin 120 = $$\\sqrt{3}\\over 2$$ Example : If sin A = $$3\\over 5$$, where 0 < A < 90, find the value of sin 2A ? Solution : We have, sin A = $$3\\over 5$$ where 0 < A < 90 degrees $$\\therefore$$ $$cos^2 A$$ = 1 – $$sin^2 A$$ $$\\implies$$ cos A = $$\\sqrt{1 – sin^2 A}$$ = $$\\sqrt{1 – {9\\over 25}}$$ = $$4\\over 5$$ By using above formula, sin 2A = 2 sin A cos A = 2 $$\\times$$ $$3\\over 5$$ $$\\times$$ $$4\\over 5$$ $$\\implies$$ sin 2A = $$24\\over 25$$", "0.1392). What's that? Oh, do keep up. To work out $8 \\times \\frac{7}{400}$, you can cancel a 4 to get $\\frac{2 \\times 7}{100}$, which is $\\frac{14}{100}$ or 0.14. Going the other way - even supposedly sensible exam boards are wont to ask for answers to trigonometry questions in degrees once in a while - isn't much harder: you divide by $\\frac{7}{400}$ (or multiply by $\\frac{400}{7}$). To get $\\sin^{-1}(0.28)$, you'd say that's $\\frac{28}{100} \\times \\frac{400}{7}$, and cross-cancel to get 16º. The answer is 16.28º. ### Trigonometry close to a right-angle You can also use this to figure out the cosine of angles close to a right angle using the identity $\\cos(90-x)^\\circ \\equiv \\sin(x)^\\circ$. If you want $\\cos(85^\\circ)$, you can say \"ah! that's the same as $\\sin(5^\\circ)$, so it's $\\frac{35}{400}$, about 0.088. (It's 0.0872, which is annoyingly close). Oh - and a couple more things. A special bonus: the tangent function behaves just like sine for small angles, because $\\tan(x) \\equiv \\frac{\\sin(x)}{\\cos(x)}$ and $\\cos(x)$ is close to 1 for small x - which means you can estimate $\\tan$ and $tan^{-1}$ in much the same way. Finally, the identity from before means that for numbers near 90º, you can say $\\tan(90-x)^\\circ \\equiv \\frac{\\sin(90-x)^\\circ}{\\cos(90-x)^\\circ} = \\frac{\\cos(x)^\\circ}{\\sin(x)^\\circ} = \\frac{1}{\\tan(x)^\\circ}$. If you're interested in $\\tan(89^\\circ)$, you can say that's $\\frac{1}{\\tan(1^\\circ)}$; $\\tan(1^\\circ)$ is about $\\frac{7}{400}$,", "ABC$$ which means $$D'=D$$. So the angle we are looking for is $$\\angle ACD = 40^{\\circ}$$. You can use the trigonometric form of Ceva's theorem (in your second setting): $$\\frac{\\sin\\angle PAB}{\\sin\\angle PBA}\\cdot\\frac{\\sin\\angle PBC}{\\sin\\angle PCB}\\cdot\\frac{\\sin\\angle PCA}{\\sin\\angle PAC}=1,$$ i.e. $$\\frac{\\sin 30^\\circ}{\\sin x}\\cdot\\frac{\\sin (20^\\circ-x)}{\\sin 20^\\circ}\\cdot\\frac{\\sin 40^\\circ}{\\sin 70^\\circ}=1,$$ where $$x=\\angle PBA$$. By writing $$\\sin 30^\\circ=1/2$$, $$\\sin 70^\\circ=\\cos 20^\\circ$$ and $$\\sin 40^\\circ=2\\sin 20^\\circ\\cos 20^\\circ$$, we get $$\\sin(20^\\circ-x)=\\sin x$$ which has solutions $$x=10^\\circ+k\\pi$$, and since $$x$$ is an angle we have $$x=10^\\circ$$. • @MarianD It is not. Read it carefully. – SMM Mar 4 at 21:09 • @MarianD Exactly, my solution shows that $P$ and $D$ coinsides. The definitin of $P$ doesn't use the fact that it is on the bisector of $\\angle B$. – SMM Mar 4 at 21:18 • You are right, excuse me (+1). – MarianD Mar 4 at 21:18", "# Maxima and minima of $f(x)=\\sin x+\\cos x+\\tan x+\\arcsin x+\\arccos x+\\arctan x$ Let $f(x)=\\sin x+\\cos x+\\tan x+\\arcsin x+\\arccos x+\\arctan x$ . If $M$ and $m$ are maximum and minimum values of $f(x)$ then their arithmetic mean is equal to? My Approach: $\\arcsin x + \\arccos x =\\frac{\\pi}{2}$, so the equation becomes $\\sin x + \\cos x + \\tan x + \\frac{\\pi}{2} + \\arctan x$ . After this I differentiated this equation but did not came to any conclusion . Also I tried to solve the equation to make something relevant but the equation did not simplified . Please help... • Maximum and minimum over what set? All of the functions listed are bounded except for $\\tan x$, which is unbounded over $(-\\pi/2,\\pi/2)$. – Joey Zou Sep 1 '16 at 18:07 • @JoeyZou Obviously, on a subset of $[-1,1]$. – Stop hurting Monica Sep 1 '16 at 18:08 • Since there is a inverse function so domain can be [-1,1]. – saladi Sep 1 '16 at 18:14 Hint: We have $f(x) = \\sin x + \\cos x + \\tan x + \\frac{\\pi}{2} + \\arctan x$ on $[-1,1]$. Now $$f'(x) = \\cos x - \\sin x + \\sec^2 x + \\frac{1}{1+x^2}.$$ On $[-1,1]$, we have $\\cos x > 0$, $-\\sin x > -1$, $\\sec^2x\\ge 1$, and $\\frac{1}{1+x^2}>0$, so for all", "# Find the exact value of the following trigonometric expressions. cos 1) sin(1659)coscos [ G_ ~sin ()sin(4)sin(279) cos(1089) sin(1089) cos(279)sin(-_3r)cos(-4r) cos(-3*) ###### Question: Find the exact value of the following trigonometric expressions. cos 1) sin(1659) cos cos [ G_ ~sin ()sin(4) sin(279) cos(1089) sin(1089) cos(279) sin(-_3r)cos(-4r) cos(-3*) stn(-4x) Gleeen ODIA = - rhere <<r end ciD @ = Elen 3< B < #; fnd sin(A + B) and cos(A ~ B). #### Similar Solved Questions ##### Relate Victor's history and signs and symptoms to the **onset and course** of cirrhosis; what priority... Relate Victor's history and signs and symptoms to the **onset and course** of cirrhosis; what priority medications might a primary care provider prescribe to treat Victor's complications. (I am well aware of what the signs and symptoms are that Victor is expressing, the question is how it re... Fun E shni horye Aued deurly und br orkanized Gean ncatit The tablc pelow Ghow the number thousands) Arned dcercts Ievetano Render Associate' $bncnclors Matter Dodoral Total Femaln Jaa Male Total 140) 7944 person who earned degrce E theyear Nndomi scleded Find the probability of selecting cumc... 1 answer ##### Assume the production of a good, X, follows the Cobb-Douglas function Q=40K0.3 L0.7. If w=$10... Assume the production of a good, X, follows the Cobb-Douglas function Q=40K0.3 L0.7.", "eight legs always tell the truth. One day, four servants met. The blue one says, “Altogether, we have 28 legs.” The green one says, “Altogether, we have 27 legs.” The yellow one says, “Altogether, we have 26 legs.” The red one says, “Altogether, we have 25 legs.” What is the colour of the servant who tells the truth? Trigonometry Identities Negative angles: • $\\sin (-x)=-\\sin x$ • $\\boxed{\\cos (-x)=\\cos x}$ (Still Positive!) • $\\tan (-x)=-\\tan x$ Reason: $-x$ is in the “C” quadrant so Cosine is still positive. Supplementary angles: • $\\boxed{\\sin (180^\\circ -x)=\\sin x}$ (Still positive!) • $\\cos (180^\\circ -x)=-\\cos x$ • $\\tan (180^\\circ -x)=-\\tan x$ Reason: $180^\\circ -x$ is in the “S” quadrant so Sine is still positive. Complementary angles: • $\\sin (90^\\circ -x)=\\cos x$ • $\\cos (90^\\circ -x)=\\sin x$ • $\\tan (90^\\circ -x)=\\cot x$ Reason: • $\\sin (90^\\circ -x)=\\cos x=a/c$ • $\\cos (90^\\circ -x)=\\sin x=b/c$ • $\\tan (90^\\circ -x)=\\cot x=a/b$ Which Secondary school is the best in Singapore Due to too many secondary schools, we can only list a few here at the start. There may be other secondary schools that can be the best in Singapore. Do feel free to add your own school and vote for it below! Also feel free to leave any comments as to why the secondary school you choose" ]

[ "Answer Comment Share Q) # Find the equation of the set of points which are equidiatant from the the points $(1,2,3)$ and $(3,2,-1)$ $\\begin{array}{1 1}z-y=0 \\\\ x-2y=0 \\\\ x+z=0 \\\\ x-z=0\\end{array}$", "the set of points P which are equidistant from point A and point B . Exercise 12.3", "## Elementary Geometry for College Students (5th Edition) Equation of all points equidistant from (-3, 4) and (3, 2), following the method set out in Example 4: $\\sqrt {(x+3)^2+(y-4)^2}$ is equal to $\\sqrt {(x-3)^2+(y-2)^2}$. Squaring and simplifying terms which appear on both sides, 12x+12=4y y=3x+3", "## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) A point is a solution of an equation if its x and y-coordinates satisfy the equation. Substitute the x and y values of the point into the equation to obtain: $y=5-x \\\\-3=5-(-2) \\\\-3=5+2 \\\\-3=7$ Thus, the given point is NOT a solution of the given equation.", "called the center point. ie, [tex](\\vec{r} - \\vec{a}). Find an equation of the set of all points equidistant from the points $A (-1, 5, 3)$ and $B (6, 2, -2)$. Rule 3: Given a straight line, the locus of points is two parallel lines. To find: The equation for the plane that consists of all the points that are equidistant from the points ( 1, 0, − 2) and ( 3, 4, 0). Therefore, let us find all points P(x,y) such that FG and the distance from G to the line are equal. Problem – Find the vertex, focus and directrix of a parabola when the coefficients of its equation are given. is a pair of parametric equations with parameter t whose graph is identical to that of the function. Find an equation of the set of all points equidistant from the points. Use Equation 1 to substitute for y' , getting (Get a common denominator in the numerator and simplify the expression. Find Some Equidistant Set of Points on a Circle Farthest From Another Set of Points. Find the equation of the set. Intersection of a circle and a line. Find an equation for the set of all points that are equidistant from the planes 3x − 4y + 12z = 6 and 4x + 3z =", "= 1, -z^3 = 1 and thus z = -1. Hence (0,0,-1) is the desired point.", "1. ## Algebraic Vectors 2 Find the point on the z-axis that is equidistant from the point A(1,-1,1) and B(1,0,-1). Then, find the point not on the z-axis that is equidistant from the point A and B. Let P be the pointon the z-axis |PA|=|PB| Find P let (3-2z-z^2)^(0.5)=(2+2z+z^2)^(0.5) solve y y=0.25 Therefore point on z-axis is (0,0,0.25) To find a point not on the z-axis, i found midpoint of a and b Let that midpoint be Q so, (1,-0.5,0) is my answer. Please check and let me know! thanks. 2. Originally Posted by skeske1234 Find the point on the z-axis that is equidistant from the point A(1,-1,1) and B(1,0,-1). Then, find the point not on the z-axis that is equidistant from the point A and B. Therefore point on z-axis is (0,0,0.25) Let that midpoint be Q so, (1,-0.5,0) is my answer. Please check and let me know! thanks. Correct. Note for the first part it is easier to solve: $1+1+(z-1)^2=1+(z+1)^2$.", "# Solve this following Question: Show that the points $(-3,0,1)$ and $(1,1,1)$ are equidistant from the plane $3 x+4 y-12 z+13=0$. Solution:", "$$E(F_{5^{2n}})$$ has 4 points of order 5 (plus the point at infinity whose order == 1 divides 5)", "Math Help - Find the point (x, y) on the line y=x that is equidistant from the points.... 1. Find the point (x, y) on the line y=x that is equidistant from the points.... Hi there I need a bit of help! Find the point (x, y) on the line y=x that is equidistant from the points (9, −1) and (−10, −10). I'm not really sure how to do it! Anything helps! Thank you! 2. Re: Find the point (x, y) on the line y=x that is equidistant from the points.... Use the fact that $y=x$ and the distance formula to state: $(x-9)^2+(x+1)^2=(x+10)^2+(x+10)^2$ Now solve for x...", "in the values of $x,y,z$ in your point.", "a relation between x and y such that the point Or, y) is equidistant from the points (3, 6) and (-3,4). Solution: Let the point P(x, y) be equidistant from the points A(3, 6) and B(- 3, 4). i.e., PA = PB gives PA² = PB² or (x – 3)² + (y – 6)² = (x + 3)² + (y – 4)² or x²- 6x + 9 + y² – 12y + 36 = x²+ 6x + 9 + y² – 8y + 16 or 6x – 6x – 12y + 8y + 36 – 16 0 or – 12x – 4y + 20 = 0 or 3x + y – 5 = 0, which is the required relation.", "a point which is equidistant from the points (-2, 3) and (5, 8)]. This point can be found by graphing, substitution, elimination, Gaussian reduction, matrix inverses, or Cramer's rule. This is a parabola with focus (3,5) an directrix on the x-axis. Unless otherwise stated, I will not allow use of calculators during exams. This will be the line that represents the equation. The set of all points equidistant from the point (0, 2, 0) and the plane y = −2, find an equation of the surface satisfying the conditions and identify the surface. 2] Given a locus defined by some geometrical condition, to find the corresponding equation. Find the locus of the set of points which move such that is equidistant from the axes. Therefore, we can call equidistant as mid point. You need to find the points (x,y,z) that satisfy both equations. Get more help from Chegg Get 1:1 help now from expert Calculus tutors Solve it with our calculus problem solver and calculator. Calculator find the coordinates of point p which divides the line joining two Entered points A and B internally or externally, in a specified ratio m and n. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. Movie Clips (with", "1. Equidistant Find the coordinates of a point equidistant from (1,-6), (5,-6), and (6,-1). 2. Originally Posted by lance Find the coordinates of a point equidistant from (1,-6), (5,-6), and (6,-1).", "# Equidistant P Geometry Level 2 If $$a<0$$ and the distances between the point $$P=(a, 0)$$ and each of the two lines $$2x-y+1=0\\$$ and $$\\ x-2y-2=0$$ are the same, what is the value of $$|a|$$? ×", "y) the point I need, and (x0, y0) the points I have. From those two I got this equation: $x^2 - 2xx0 + xo^2 -(\\dfrac{d^2}{1 + m^2}) = 0$ I solve it with the quadratic formula, and got 2 results (the two equidistant points that form the new line) this is the result: Again thank you!..", "# 4. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1). P Pankaj Sanodiya Given, two points A=(1, 2, 3) and B=(3, 2, –1). Let the point P= (x,y,z) be a point which is equidistance from the points A and B. so, The distance PA= The distance PB $\\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}=\\sqrt{(x-3)^2+(y-2)^2+(z-(-1))^2}$ ${(x-1)^2+(y-2)^2+(z-3)^2}={(x-3)^2+(y-2)^2+(z-(-1))^2}$ $\\left [ (x-1)^2-(x-3)^2 \\right ]+\\left [ (y-2)^2-(y-2)^2 \\right ]+\\left [ (z-3)^2-(z+1)^2 \\right ]=0$ Now lets apply the simplification property, $a^2-b^2=(a+b)(a-b)$ $\\left [ (2)(2x-4) \\right ]+0+\\left [ (-4)(2z-2) \\right ]=0$ $4x-8-8z+8=0$ $4x-8z=0$ $x-2z=0$ Hence locus of the point which is equidistant from A and B is $x-2z=0$. Exams Articles Questions", "which maps the points z=1,1,? onto the points w=-i, -1, i. 6 M 6 (c) Show that the general solution of $\\dfrac {d^2 y }{d x^2} + 4x^2 y=0 \\ is \\ y=\\sqrt{z} [A J_{1/4} (x^2) + B \\ J_{-1/4} (x^2)]$ where A and B are constants. 8 M More question papers from Applied Mathematics - 3", "## Calculus (3rd Edition) The point $(9,-4)$ The line $x=9$ is the set of points (9, y), for any real y. The line $y=-4$ is the set of points (x, -4), for any real x. Now, it is easy to conclude that the intersection of the lines $x=9$ and $y=-4$ is the point $(9,-4)$", "from the points $A(2,-1,3)$ and $B(4,3,1)$ Linh V. ### Problem 69 Find an equation for the set of all points equidistant from the planes $y=3$ and $y=-1$ Linh V. ### Problem 70 Find an equation for the set of all points equidistant from the point (0,0,2) and the $x y$ -plane. Linh V. ### Problem 71 Find the point on the sphere $x^{2}+(y-3)^{2}+(z+5)^{2}=4$ nearest a. the $x y$ -plane. b. the point (0,7,-5) Linh V. ### Problem 72 Find the point equidistant from the points (0,0,0),(0,4,0),(3,0,0) and (2,2,-3) Linh V. ### Problem 73 Find an equation for the set of points equidistant from the point (0,0,2) and the $x$ -axis. Linh V. ### Problem 74 Find an equation for the set of points equidistant from the $y$ -axis and the plane $z=6$ Linh V. ### Problem 75 Find an equation for the set of points equidistant from the a. $x y$ -plane and the $y z$ -plane. b. $x$ -axis and the $y$ -axis. Linh V. $.$ Find all points that simultaneously lie 3 units from each of the points $(2,0,0),(0,2,0),$ and (0,0,2)" ]

[ "2 x 2 +2cx+ c 2 + y 2 = 4 a 2 −4a (x−c) 2 + y 2 + x 2 −2cx+ y 2 2cx = 4 a 2 −4a (x−c) 2 + y 2 −2cx 4cx−4 a 2 = 4a (x−c) 2 + y 2 − 1 4a ( 4cx−4 a 2 ) = (x−c) 2 + y 2 a− c a x = (x−c) 2 + y 2 a 2 −2xc+ c 2 a 2 x 2 = (x−c) 2 + y 2 a 2 −2xc+ c 2 a 2 x 2 = x 2 −2xc+ c 2 + y 2 a 2 + c 2 a 2 x 2 = x 2 + c 2 + y 2 a 2 + c 2 a 2 x 2 = x 2 + c 2 + y 2 a 2 − c 2 = x 2 − c 2 a 2 x 2 + y 2 a 2 − c 2 = x 2 ( 1− c 2 a 2 )+ y 2$ Let$1= a 2 a 2 . 1= a 2 a 2 .$ $a 2 − c 2 = x 2 ( a 2 − c 2 a 2 )+ y 2 1 = x 2 a 2 + y 2 a 2 −", "Filters Q&A - Ask Doubts and Get Answers Q 6.If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA ^ 2 + PB^ 2 = k^ 2, where k is a constant. 6.If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that $PA ^2 + PB ^ 2 = k^ 2$ , where k is a constant. Answers (1) Views Given Points, A (3, 4, 5) and B (–1, 3, –7), Let the coordinates of point P be (x,y,z) Now, Given condition : $PB^2=(x-(-1))^2+(y-(3))^2+(z-(-7))^2$ $PB^2=x^2+2x+1+y^2-6y+9+z^2+14z+49$ $PB^2=x^2+y^2+z^2+2x-6y+14z+59$ And $PA^2=(x-3)^2+(y-4)^2+(z-5)^2$ $PA^2=x^2+y^2+z^2-6x-8y-10z+50$ Now, Given Condition $PA ^2 + PB ^ 2 = k^ 2$ $x^2+y^2+z^2+2x-6y+14z+59+x^2+y^2+z^2-6x-8y-10z+50=k^2$ $2x^2+2y^2+2z^2-4x-14y+4z+109=k^2$ Hence Equation of the set of the point P is $2x^2+2y^2+2z^2-4x-14y+4z+109=k^2$. Exams Articles Questions", "# Understanding how This Works (shortcut for squaring $2$ digit numbers mentally) In fact, not knowing the how has caused me to loose marks on several occasions at school. So my question arises, how does the method for calculating the squares of $2$ digit numbers work? It seems very counter intuitive and doesn't make much sense from the surface. The method: Take any $2$ digit number and round it to the nearest 10, take the difference and do the inverse operation, then multiply the answer of the round and the result of the inverse calculation. Then add the square of the difference. • E.g. $54^{2} = 50*58 + 16$ • You have $(10x+y)^2 = (10x)(10x+2y)+y^2 = 100x^2+20xy+y^2$ or $(10x-y)^2 = (10x)(10x-2y)+y^2 = 100x^2-20xy+y^2$ – Macavity Aug 17 '13 at 3:02 • Step 1, make flash cards and memorize them all. You're done. Now, move on to harder problems, like $75 \\cdot 67 = 71^2 - 4^2$. – Graphth Aug 17 '13 at 3:22 Observe that $$(10a+b)^2=100a^2+20ab+b^2=(10a)(10a+2b)+b^2\\tag{1}$$ and $$(10a-b)^2=100a^2-20ab+b^2=(10a)(10a-2b)+b^2\\;.$$ If the second digit of the number is $b<5$, and the first digit is $a$, the number is $10a+b$, and rounding to the nearest $10$ leaves you with $10a$. In the process of rounding you subtract $b$; adding $2b$ gives you $10+2b$, and equation $(1)$ then shows that multiplying this", "W|X_0=2)P(Z_1 = W) + P(Y_1 = B|X_0=2)P(Z_1 = B) = 2/3*1/2 + 1/3*1/2 = 1/2$$. b) $$P(X_2 = 1|X_0=2) = P(X_2 = 1| X_1 = 2, X_0=2) + P(X_2 = 1| X_1 = 1,X_0=2) = P(X_2=1|X_1=2)P(X_1=2|X_0=2) + P(X_2=1|X_1=1)P(X_1=1|X_0=2)$$. We can say that P(X_i = x_i|X_{i-1} = x_{i-1}) is equal for all $$i$$. So $$P(X_2=1|X_1=2)P(X_1=2|X_0=2) + P(X_2=1|X_1=1)P(X_1=1|X_0=2) = P(X_1=1|X_0=2)P(X_1=2|X_0=2) + P(X_1=1|X_0=1)P(X_1=1|X_0=2) = P(X_1=1|X_0=2)(P(X_1=2|X_0=2)+P(X_1=1|X_0=1))$$. Then $$P(X_1=1|X_0=2) = P(Y_1=W|X_0=2)P(Z_1=B) = 2/3*1/2 = 1/3$$. $$P(X_1=1|X_0=1) = P(Y_1 = W|X_0=1)P(Z_1 = W) + P(Y_1 = B|X_0=1)P(Z_1 = B) = 1/3*1/2 + 2/3*1/2 = 1/2$$. So, $$P(X_1 = 2 | X_0 = 2) = 1/3*(1/2 + 1/2) = 1/3$$. c) Similar but this time $$Z_i$$ is more important. $$P(X_1 = 1 | X_0 = 2) = P(Z_1 = B)P(Y_1 = B|X_0=2,Z_1=B) = 1/2*1/2=1/4$$. d) $$P(X_2 = 2 | X_0 = 2) = P(X_2 = 2 | X_1 = 2)P(X_1 = 2 | X_0 = 2) + P(X_2 = 2 | X_1 = 1)P(X_1 = 1 | X_0 = 2) + P(X_2 = 2 | X_1 = 3)P(X_1 = 3 | X_0 = 2)$$. Let’s rephrase $$P(X_1 = 2 | X_0 = 2)^2 + P(X_1 = 2 | X_0 = 1)P(X_1 = 1 | X_0 = 2) + P(X_1 = 2 | X_0 = 3)P(X_1 = 3 | X_0 = 2)$$. $$P(X_1 = 2 |", "z2-…) z=-2a2+(2a3-a22) z+…1-a2z-(2a3-a22) z2-…∣t(0)∣=2∣a2∣ ≤1$ and $∣t′(0)∣=4∣a3∣.$ Furthermore, it can be seen that $cϕ′(c)ϕ(c)=∣ϕ′(c)∣ ≥ ∣B′(c)∣=cB′(c)B(c).$ Consider the function $n(z)=t(z)-t(0)1-t(0)¯t(z).$ This function is holomorphic in D, |n(z)| ≤ 1 for |z| < 1, n(0) = 0, and |n(c)| = 1 for c∂D. From (1.5), we obtain $21+∣n′(0)∣≤ ∣n′(b)∣=1-∣t(0)∣2|1-t(0)¯t(c)|2∣t′(c)∣≤1+∣t(0)∣1-∣t(0)∣{∣ϕ′(c)∣-∣B′(c)∣}.$ Since $n′(z)=1-∣t(0)∣2(1-t(0)¯t(z))2t′(z)$ and $∣n′(0)∣=∣t′(0)∣1-∣t(0)∣2=4∣a3∣1-4∣a2∣2,$ we take $21+4∣a3∣1-4∣a2∣2≤1+2∣a2∣1-2∣a2∣{∣m′(c)∣2-1},2 (1-4∣a2∣2)1-4∣a2∣2+4∣a3∣≤1+2∣a2∣1-2∣a2∣{∣m′(c)∣2-1},2 (1-2∣a2∣)21-4∣a2∣2+4∣a3∣≤∣m′(c)∣2-1,$$∣m′(c)∣ ≥2 (1+2 (1-2∣a2∣)21-4∣a2∣2+4∣a3∣)$ and $|f″(c)f(c)|≥2 (1+2 (1-2∣a2∣)21-4∣a2∣2+4∣a3∣).$ Now, we shall show that the inequality (2.3) is sharp. Let $f(z)=z (1+z22)12.$ Then $f′(z)=2z2+1z2+2,f″(z)=2z (z2+3)(z2+2)32$ and $|f″(i)f(i)|=4.$ On the other hand, from the Taylor expansion of f(z), we get $z+a2z2+a3z3+… =z (1+z22)12,1+a2z+a3z2+… =(1+z22)12,a2z+a3z2+… =(1+z22)12-1$ and $a2+a3z+…=(1+z22)12-1z.$ Passing to limit in the last equality yields a2 = 0. Similarly, using straightforward calculations, we take $a3=14$. Therefore, we obtain $2 (1+2 (1-2∣a2∣)21-4∣a2∣2+4∣a3∣)=4.$ ### Theorem 2.4 Let , $1-zf′(z)f(z)$has no zeros in D except z = 0 and a2 > 0. Suppose that, for some c∂D, f has an angular limit f(c) at c, f′ (c) = 0. Then we have the inequality $∣f″(c)f(c)∣≥2 (1-a2 ln2 (2a2)a2 ln (2a2)-∣a3∣).$ Proof Let a2 > 0 and let us consider the function t(z) as in Theorem 2.3. Taking account of the equality |t(0)| = 2|a2|, we denote by ln t(z) the holomorphic branch of the logarithm normed by condition $ln t(0)=ln (2a2)=ln∣2a2∣+i arg (2a2)<0, a2>0$ and $ln (2a2)<0.$", "x + y + z = 12 and x$$^{2}$$ + y$$^{2}$$ + z$$^{2}$$ = 44, find xy + yz + zx. Answers for the worksheet on on expanding of (a ± b ± c)$$^{2}$$ and its corollaries are given below. 1. (i) a$$^{2}$$ + 4b$$^{2}$$ + 9c$$^{2}$$ + 4ab + 12bc + 6ca (ii) 4x$$^{2}$$ + 9y$$^{2}$$ + 16z$$^{2}$$ + 12xy + 24yz + 16zx (iii) x$$^{2}$$ + 4y$$^{2}$$ + 9z$$^{2}$$ + 4xy – 12yz – 6zx (iv) 9a$$^{2}$$ + 16b$$^{2}$$ + c$$^{2}$$ – 24ab + 8bc – 6ca (v) x$$^{2}$$ + $$\\frac{1}{x^{2}}$$ – 2x + $$\\frac{2}{x}$$ - 1 (vi) 1 – 2a – a$$^{2}$$+ 2a$$^{3}$$ + a$$^{4}$$ 2. (i) 2x$$^{2}$$ + 2y$$^{2}$$ + 2z$$^{2}$$ – 4yz (ii) 14a$$^{2}$$ + 14b$$^{2}$$+ 14c$$^{2}$$ +2ab + 2bc + 2ca 3. (i) 14 (ii) 21 (iii) ±12 4. 50. From Worksheet on Expanding of (a ± b ± c)^2 and its Corollaries to HOME PAGE", "expression for the given values. 5. $2 x x − 1 2 x x − 1$ $x=0x=0$ $x=2x=2$ $x=−1x=−1$ 6. $4 y − 1 5 y − 3 4 y − 1 5 y − 3$ $y=0y=0$ $y=2y=2$ $y=−1y=−1$ 7. $2 p + 3 p 2 + 1 2 p + 3 p 2 + 1$ $p=0p=0$ $p=1p=1$ $p=−2p=−2$ 8. $x + 3 2 − 3 x x + 3 2 − 3 x$ $x=0x=0$ $x=1x=1$ $x=−2x=−2$ 9. $y 2 + 5 y + 6 y 2 − 1 y 2 + 5 y + 6 y 2 − 1$ $y=0y=0$ $y=2y=2$ $y=−2y=−2$ 10. $z 2 + 3 z − 10 z 2 − 1 z 2 + 3 z − 10 z 2 − 1$ $z=0z=0$ $z=2z=2$ $z=−2z=−2$ 11. $a 2 − 4 a 2 + 5 a + 4 a 2 − 4 a 2 + 5 a + 4$ $a=0a=0$ $a=1a=1$ $a=−2a=−2$ 12. $b 2 + 2 b 2 − 3 b − 4 b 2 + 2 b 2 − 3 b − 4$ $b=0b=0$ $b=2b=2$ $b=−2b=−2$ 13. $x 2 + 3 x y + 2 y 2 2 x 3 y x 2 + 3 x y + 2 y 2 2 x 3 y$ 1. $x=1,y=−1x=1,y=−1$ 2. $x=2,y=1x=2,y=1$ 3. $x=−1,y=−2x=−1,y=−2$ 14. $c", "got $t=0$, $s=1$ and $b=1$. • How did you find the $s(2, b+1, -1)$ ? – J.Se Jan 4 '18 at 17:16 • It's $(1,b,1)-(-1,-1,2)$. If $B(1,b,1)$ and $A(-1,-1,2)$ then $\\vec{AB}(2,b+1,-1)$. – Michael Rozenberg Jan 4 '18 at 17:17 To parametrise a line you need a point it passes through and its direction. You can get from the vector formed by the two points it passes through. Thus $L_2=(-1,-1,2)+t(2,b+1,-1).$L_1$is given by $$x=1-t$$$$y=1+t$$$$z=1-t$$ and$L_2$is given by $$x=-1+(1-(-1))s=-1+2s$$$$y=-1+(b-(-1))s=-1+(b+1)s$$$$z=2+(1-2)s=2-s$$ Now we solve $$1-t=-1+2s\\implies2-t=2s$$ $$1-t=2-s\\implies t=s-1$$ so using the equations for$x$and$z$, $$2-s+1=2s\\implies \\boxed{s=1}\\implies \\boxed{t=0}$$ Let's do the same for$y$. We solve $$1+t=-1+(b+1)s\\implies 1+0=-1+(b+1)(1)\\implies 1=-1+b+1=b$$ Hence intersection only occurs when $$\\boxed{b=1}$$ In other words, for an intersection,$L_2$must go through the points$(-1, -1, 2)$and$(1, 1, 1)\\$.", "x,y,z. The following is true: (c+z)^2=(a+x)^2+(b+y)^2 c^2 + z^2 + 2cz=a^2 + x^2 + 2xa + b^2 + y^2 + 2by Since c^2=a^2 + b^2, and z^2=x^2 + y^2, 2cz=2xa + 2by cz=xa + by For simplifying purpose, square the equation; (cz)^2=(xa + by)^2 then (cz)^2=(xa + by)^2 (x^2 + y^2)(a^2 + b^2)=(xa + by)^2 (ax)^2 + (xb)^2 + (ay)^2 + (by)^2=(ax)^2 + (by)^2 + 2xaby So (xb)^2 + (ay)^2 - 2xaby = 0 Factorising, we find that (xb-ay)^2=0 xb=ay y/b=x/a To proove that z/c=x/a=y/b, x/a=y/b y=bx/a and b=ay/x (z/c)^2=(x^2 + y^2)/(a^2 + b^2); (z/c)^2=(x^2 + (bx/a)^2)/(a^2 + (ay/x)^2) (z/c)^2=(x^2(1+(b/a)^2)/(a^2((y/x)^2 + 1) Since we know that b/a=y/x, we obtain the result (z/c)^2=(x/a)^2 z/c=x/a=y/b Since any triangle can be decomposed into two right triangles, it is easy to prove that the rule is general. Last edited:", "Let A (x1, y1) = A (2, -2), B (x2, y2) = B (-3, 8) and C (x3, y3) = C (-1, 4) be the given points. Now = $\\left\\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\\right\\}$ = $\\left \\{ 2 (8 – 4) + (-3)(4 + 2) +(-1) (-2 – 8) \\right \\}$ = 8 – 18 +10 = 0 Hence, the given points are collinear. ii) Let A (x1, y1) = A (-5, 1), B (x2, y2) = B (5, 5) and C (x3, y3) = C (10, 7) be the given points. Now = $\\left\\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\\right\\}$ = $\\left \\{ -5 (5 – 7) + (5)(7 – 1) +(10) (1 – 5) \\right \\}$ = 10 + 30 – 40 = 0 Hence, the given points are collinear. iii) Let A (x1, y1) = A (5, 1), B (x2, y2) = B (1, -1) and C (x3, y3) = C (11, 4) be the given points. Now = $\\left\\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\\right\\}$ = $\\left \\{ 5 (-1 – 4) + (1)(4 – 1) + (11) (1 + 1) \\right \\}$ = -25 + 3 + 22 = 0 Hence, the given points are collinear. iv) Let A (x1, y1) = A (8, 1), B (x2, y2) = B (3, -4) and C (x3, y3) = C (2, -5). Be the given points. Now = $\\left\\{x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\\right\\}$ = $\\left", "2y)^2 Write in the expanded form (a2 + b2 + c2 )2 Write in the expanded form: (ab + bc + ca)2 Write in the expanded form: (x/y + y/z + z/x)^2 Write in the expanded form: (a/(bc) + b/(ca) + c/(ab))^2 Write in the expanded form: (x + 2y + 4z)^2 Write in the expand form: (2x - y + z)^2 Write in the expanded form: (-2x + 3y + 2z)2 Simplify (2x + p - c)^2 - (2x - p + c)^2 Simplify (a + b + c)^2 + (a - b + c)^2 Simplify: (a + b + c)^2 - (a - b + c)^2 Simplify (a + b + c)2 + (a - b + c)2 + (a + b - c)2 Simplify (2x + p - c)2 - (2x - p + c)2 Simplify (x^2 + y^2 - z)^2 - (x^2 - y^2 + z^2)^2 If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + bc + ca. If a2 + b2 + c2 = 16 and ab + bc + ca = 10, find the value of a + b + c. If a + b + c = 9 and ab + bc + ca = 23, find the", "- $t$)($s$ - 64$t$) #### Solution: 25s2 - 64t2 = (5s) 2 - (8t) 2 [Write as a2 - b2, a = 5s, b = 8t.] = (5s + 8t) (5s - 8t) [a2 - b2 = (a + b) (a - b).] 9. Factor: 4$x$2 + 12$x$ + 9 a. (2$x$ + 3)2 b. (2$x$ - 3) (2$x$ + 3) c. (2$x$ - 3)2 d. (4$x$ + 3)($x$ + 3) #### Solution: 4x2 + 12x + 9 = (2x)2 + 2(2x)(3) + (3)2 [a = 2x, b = 3.] = (2x + 3)2 [a2 + 2ab + b2 = (a + b)2.] 10. Factor: 9$x$2 - 24$\\mathrm{xy}$ + 16$y$2 a. (3$x$ + 4$y$) (3$x$ - 4$y$) b. (3$x$ - $y$)($x$ + 4$y$) c. (3$x$ + 4$y$)2 d. (3$x$ - 4$y$)2 #### Solution: 9x2 - 24xy + 16y2 = (3x)2 - 2(3x) (4y) + (4y)2 [a = 3x, b = 4y.] = (3x - 4y)2 [a2 - 2ab + b2 = (a - b)2.]", "$y ( y + z ) + 9 ( y + z )$ $= (y + z)(y + 9)$ Question:3(vii) Factorise the following expressions $5 y ^ 2 - 20 y - 8z + 2yz$ We have, $5 y ^ 2 - 20 y - 8z + 2yz$ $= 5y(y - 4) + 2z(y - 4)$ $= (y - 4)(5y + 2z)$ Therefore, $5 y ^ 2 - 20 y - 8z + 2yz$ $= (y - 4)(5y + 2z)$ Question:3(viii) Factorise $10 ab + 4a + 5b + 2$ We have, $10 ab + 4a + 5b + 2$ $= 2a(5b + 2) + 1(5b + 2)$ $= (5b + 2)(2a + 1)$ Therefore, $10 ab + 4a + 5b + 2$$= (5b + 2)(2a + 1)$ Question:3(ix) Factorise the following expressions $6 xy - 4 y + 6 - 9 x$ We have, $6 xy - 4 y + 6 - 9 x$ $= 2y(3x - 2) - 3 (3x - 2)$ $= (3x - 2)(2y - 3)$ Therefore, $6 xy - 4 y + 6 - 9 x$$= (3x - 2)(2y - 3)$ We have, $a ^ 4 - b ^ 4$ = $(a^{2})^{2} - (b^{2})^{2} = (a^{2} - b^{2})(a^{2} + b^{2}) = (a-b)(a+b)(a^{2} + b^{2})$ $using \\ (x^{2} - y^{2}) = (x-y)(x+y)$ We have,", "the $a$th from the left, $b$th from the right ($a,b \\geq 0$, total number of elements is $a+b+1$). Then \\begin{align*} p(0,0) &= 1, \\\\ p(0,1) &= 1/2, \\\\ p(1,0) &= 1/2, \\\\ p(1,1) &= 1/2, \\\\ p(0,b+2) &= 1/2 + p(0,b)/4, \\\\ p(a+2,0) &= 1/2 + p(a,0)/4, \\\\ p(1,b+2) &= p(0,b+1)/2 + p(1,b)/4, \\\\ p(a+2,1) &= p(a+1,0)/2 + p(a,1)/4, \\\\ p(a+2,b+2) &= (p(a+2,b)+2p(a+1,b+1)+p(a,b+2))/4. \\end{align*} You could precompute the requisite values for $a+b+1 \\leq 2000$ and then your program will be blazingly fast. It turns out that for fixed $c$, $p(c,n-c)$ tends to a limit $p_c = 1/2(1 + (-1)^c/3^{c+1})$. To see why, note first that $p_0$ is the ability that if you and a friend alternate tossing fair coins, yours will come up HEAD first. There are many ways to see that $p_0 = 2/3$. Next, we have the formula $p_{c+1} = p_0 (1-p_c) + (1-p_0) p_c$ (exercise), from which the formula for $p_c$ follows by induction.", "• # question_answer If the point $P(x,y)$ is equidistant from the points $A(a+b,b-a)$ and $B(a-b,a+b)$. Prove that $bx=ay$. Since, P us equidistant from points A and B, $\\therefore$ $PA=PB$ Or, ${{(PA)}^{2}}={{(PB)}^{2}}$ ${{(a+b-x)}^{2}}+{{(b-a-y)}^{2}}={{(a-b-x)}^{2}}+{{(a+b-y)}^{2}}$ ${{(a+b)}^{2}}+{{x}^{2}}-2ax-2bx+{{(b-a)}^{2}}+{{y}^{2}}-2by+2ay$$={{(a-b)}^{2}}+{{x}^{2}}-2ax+2bx+{{(a+b)}^{2}}+{{y}^{2}}-2ay-2by$ $-2bx+2ay=2bx-2ay$ $4ay=4bx$ $ay=bx$ $bx=ay$ Hence Proved.", "\\right) { b }^{ 3 }{ c }^{ 2 }$ () $\\cfrac { { 42a }^{ 4 }{ b }^{ 5 }\\left( c^2 \\right) }{ { 6a }^{ 4 }{ b }^{ 2 } } =\\left( 7 \\right) { b }^{ 3 }{ c }^{ 2 }$ 9. Expand the following (p + 2)2=_________ () (p + 2)2 = p2 + 2(P) (2) + 22 =p2 + 4p + 4 10. Expand the following (3 - a)2=________ () (3 - a)2 =32-2(3) (a) +a2= 9 - 6a + a2 11. Expand the following (62-x2)=_______ () (62-x2) = (6+x)(6-x)", "yz\\, +\\, xz)\\, =\\, 14\\, -\\, (2)^2\\, =\\, 10$ . . . . .$xy\\, +\\, yz\\, +\\, xz\\, =\\, -5$ If one does the long multiplication, one will find that: . . . . .$(xy\\, +\\, yz\\, +\\, xz)^2$ . . . . . . . . .$=\\, (x^2y^2\\, +\\, y^2z^2\\, +\\, x^2z^2)\\, +\\, 2xyz(x\\, +\\, y\\, +\\, z)$ Then plugging in gives us: . . . . .$(-5)^2\\, =\\, (x^2y^2\\, +\\, y^2z^2\\, +\\, x^2z^2)\\, +\\, 2xyz(2)$ . . . . .$25\\, =\\, (x^2y^2\\, +\\, y^2z^2\\, +\\, x^2z^2)\\, +\\, 4xyz$ Restating: . . . . .$-2(xy\\, +\\, yz\\, +\\, xz)\\, =\\, x^2\\, +\\, y^2\\, +\\, z^2\\, -\\, (x\\, +\\, y\\, +\\, z)^2\\,$ ...as: . . . . .$x^2\\, +\\, y^2\\, +\\, z^2\\, -\\, 2(xy\\, +\\, yz\\, +\\, xz)\\, =\\, (x\\, +\\, y\\, +\\, z)^2$ ...and squaring, we get: . . . . .$(x\\, +\\, y\\, +\\, z)^4\\,$ . . . . . . . . .$=\\, (x^4\\, +\\, y^4\\, +\\, z^4)\\,$ . . . . . . . . . . .$+\\, 2(x^2y^2\\, +\\, y^2z^2\\, +\\, x^2z^2)$ . . . . . . . . . . . . .$-\\, 4(x^2\\, +\\, y^2\\, +\\, z^2)(xy\\, +\\, yz\\, +\\, xz)$ . . . . . . . . . . . . . . .$+\\, 4(xy\\, +\\, yz\\, +\\,", "is a vertex, the difference is$2a. 2a.$ $( x−( −c ) ) 2 + ( y−0 ) 2 − ( x−c ) 2 + ( y−0 ) 2 =2a ( x−( −c ) ) 2 + ( y−0 ) 2 − ( x−c ) 2 + ( y−0 ) 2 =2a$ $(x−(−c)) 2 + (y−0) 2 − (x−c) 2 + (y−0) 2 = 2a (x+c) 2 + y 2 − (x−c) 2 + y 2 = 2a (x+c) 2 + y 2 = 2a+ (x−c) 2 + y 2 (x+c) 2 + y 2 = ( 2a+ (x−c) 2 + y 2 ) x 2 +2cx+ c 2 + y 2 = 4 a 2 +4a (x−c) 2 + y 2 x 2 +2cx+ c 2 + y 2 = 4 a 2 +4a (x−c) 2 + y 2 + x 2 −2cx+ y 2 2cx = 4 a 2 +4a (x−c) 2 + y 2 −2cx 4cx−4 a 2 = 4a (x−c) 2 + y 2 cx− a 2 = a (x−c) 2 + y 2 ( cx− a 2 ) 2 = a 2 ( (x−c) 2 + y 2 ) c 2 x 2 −2 a 2 c 2 x 2 + a 4 = a 2 x 2 −2 a 2 c 2 x", "to get $$2(xy+x^2yx+xyx^2)=0 \\qquad 2(x^2yx+xyx^2+yx)=0 \\ .$$ Subtracting the last two relations we have $$2(xy-yx)=0 \\ .$$ We then show that $$3( x+x^2)=0 \\ \\ \\forall x$$. You get this from $$x+x^2=(x+x^2)^3=x^3+3 x^4+3 x^5+x^6=4(x+x^2) \\ .$$ In particular $$3 (x+y +(x+y)^2) =3( x+x^2+ y+ y^2+ xy+yx)=0 \\,$$ we end-up with $$3(xy+yx)=0$$. But since $$6xy=0$$, we have $$3(xy-yx)=0$$. Then subtract $$2(xy-yx)=0$$ to get $$xy-yx=0$$. • If you subtract the right sides of the expanded $(x+y)^3$ and $(x-y)^3$, you'll get $2y$ on the left side and not $0$. Sep 2, 2020 at 12:15 $\\rm(1)\\quad ab=0\\: \\Rightarrow\\: ba = 0\\ \\ via\\ \\ ba = (ba)^3 = b\\, ab\\, ab\\, a = 0$ $\\rm(2)\\quad c^2 = c\\: \\Rightarrow\\: c\\:$ is central $\\$ [which means that $\\rm\\ \\color{#C00}{xc = cx}\\$ for all $\\rm\\:x$] $\\!\\rm\\begin{eqnarray}\\rm Proof\\!:\\ \\ c(x-cx) &=&\\rm 0\\:\\Rightarrow\\: (x-cx)c = 0\\ \\ by\\ (1),\\ \\ so\\ \\ \\color{#C00}{xc} = cxc\\\\ \\rm (x-xc)c &=&\\rm 0\\:\\Rightarrow\\: c(x-xc) = 0\\ \\ by\\ (1),\\ \\ so\\ \\ \\color{#C00}{cx} = cxc\\end{eqnarray}$ $\\rm(3)\\quad x^2\\:$ central via $\\rm\\:c = x^2\\:$ in $(2)$ $\\rm(4)\\quad c^2 = 2c\\:\\Rightarrow\\: c\\:$ central. $\\$ Proof: $\\rm\\:c = c^3 = 2c^2\\:$ central by $(3)$ $\\rm(5)\\quad x+x^2\\:$ central via $\\rm\\:c = x + x^2\\:$ in $(4)$ $\\rm(6)\\quad x = (x+x^2)-x^2\\:$ central via $(3),(5)\\,$ by centrals closed under subtraction.$\\$ QED Re-posting it from here. Note that", "$y^2+ (y+2)^2 =580$ $y^2 + y^2 + 4 + 4y =580$ $2y^2 + 4y -576=0$ $y^2 + 2y-288=0$ Comparing this to $ay^2 + by +c=0$ a=1, b=2 and c =-288 $y= \\frac {-b \\pm \\sqrt {b^2 -4ac}}{2a}$ So y=16 or -18 c. $11x^2 -31x -6 =0$ Comparing this to $ax^2 + bx +c=0$ a=11, b=-31 and c =-6 $x= \\frac {-b \\pm \\sqrt {b^2 -4ac}}{2a}$ So x=3 or -2/11 d. $9 -y - 10y^2 =0$ $10y^2 + y -9=0$ Comparing this to $ay^2 + by +c=0$ a=10, b=1 and c =-9 $y= \\frac {-b \\pm \\sqrt {b^2 -4ac}}{2a}$ So y=-1 or 9/10 e. $14x +4x^2 =2x -5$ $4x^2 + 12x +5=0$ Comparing this to $ax^2 + bx +c=0$ a=4, b=12 and c =5 $x= \\frac {-b \\pm \\sqrt {b^2 -4ac}}{2a}$ so x=-1/2 or -5/2 f. $3y^2 + 4y=2(y+4)$ $3y^2 + 4y = 2y + 8$ $3y^2 + 2y -8=0$ Comparing this to $ay^2 + by +c=0$ a=3, b=2 and c =-8 $y= \\frac {-b \\pm \\sqrt {b^2 -4ac}}{2a}$ so y=-2 or 4/3 g. $2x^2 -5x +3 =0$ Comparing this to $ax^2 + bx +c=0$ a=2, b=-5 and c =3 $x= \\frac {-b \\pm \\sqrt {b^2 -4ac}}{2a}$ so y=3/2 or 1 h. $\\frac {x+1}{x-1} + \\frac {x-2}{x+2} =3$ $\\frac { (x+1)(x+2) + (x-2)(x-1)}{(x-1)(x+2)}=3$ $x^2 + 3x +2 + x^2" ]

[ "Browse Questions # Find the equation of the set of points which are equidiatant from the the points $(1,2,3)$ and $(3,2,-1)$ $\\begin{array}{1 1}z-y=0 \\\\ x-2y=0 \\\\ x+z=0 \\\\ x-z=0\\end{array}$ Toolbox: • The distance between two points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is given by $\\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ Let the given points be $A(1,2,3)$ and $B(3,2,-1)$ Let the set of points that are equdistant from $A\\:and\\:B$ be $P(x,y,z)$ Since $P$ is equdistant from $A$ and $B$, $\\overline {PA}=\\overline {PB}$ $\\Rightarrow\\:(\\overline {PA})^2=(\\overline {PB})^2$ We know that the distance between two points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is given by $\\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ $\\Rightarrow (x-1)^2+(y-2)^2+(z-3)^2=(x-3)^2+(y-2)^2+(z+1)^2$ $\\Rightarrow\\:x^2+1-2x+y^2+4-4y+z^2+9-6z=x^2+9-6x+y^2+4-4y+z^2+1-2z$ $\\Rightarrow\\:4x-4z=0$ $\\Rightarrow\\:x-z=0$ $i.e.,$ The required equation is $x-z=0$", "$(a-b)^2=a^2+b^2-2ab$ $(a+b)^2=a^2+b^2+2ab$ $\\sqrt{(x^2 -2x +1) + y^2 +(z^2 +4z+4)}$ $\\sqrt{(x^2 + y^2 + z^2 -2x +4z +5)}$ Now calculating the distance of the point $(3,4,0)$ with the $(x,y,z)$. $\\sqrt{(x – 3)^2 + (y – 4)^2 + z^2 }$ Expanding the expression using the algebraic formulas: $\\sqrt{(x^2 -6x +9) + (y^2 -8y+16) + z^2 }$ $\\sqrt{(x^2 + y^2 + z^2 -6x – 8y + 25)}$ As both distances are equidistant, equating them and then simplifying: $\\sqrt{(x^2 + y^2 + z^2 -2x +4z +5)} = \\sqrt{(x^2 + y^2 + z^2 -6x – 8y + 25)}$ The expression is re-written as: $x^2 + y^2 + z^2 -2x +4z +5 = x^2 + y^2 + z^2 -6x -8y + 25$ $\\cancel{x^2}+\\cancel{y^2}+\\cancel{z^2}-2x+4z+5 = \\cancel{x^2}+\\cancel{y^2}+\\cancel{z^2}-6x-8y+25$ $-2x+4z+5=-6x-8y+25$ $-2x+6x +8y+4z +5-25 = 0$ $4x +8y+4z -20=0$ Dividing the equation with $4$: $x+2y+z=5$ So the equation of the plane that consists of all the points that are equidistant from the given points is calculated to be: $(1,0,-2)$ and $(3,4,0)$ is $x +2y+z = 5$. ## Example What is the equation of the plane consisting of all points that are equidistant from $(-5, 5, -3)$ and $(4,5,3)$? Calculating the distance between $(x,y,z)$ and $(-5,5,-3)$: $\\sqrt{(x + 5)^2 + (y – 5)^2 +(z + 3)^2 }$ $\\sqrt{(x^2 + y^2 + z^2 +10x -10y +6z + 59)}$ Now calculating the", "• # question_answer If the point $P(x,y)$ is equidistant from the points $A(a+b,b-a)$ and $B(a-b,a+b)$. Prove that $bx=ay$. Since, P us equidistant from points A and B, $\\therefore$ $PA=PB$ Or, ${{(PA)}^{2}}={{(PB)}^{2}}$ ${{(a+b-x)}^{2}}+{{(b-a-y)}^{2}}={{(a-b-x)}^{2}}+{{(a+b-y)}^{2}}$ ${{(a+b)}^{2}}+{{x}^{2}}-2ax-2bx+{{(b-a)}^{2}}+{{y}^{2}}-2by+2ay$$={{(a-b)}^{2}}+{{x}^{2}}-2ax+2bx+{{(a+b)}^{2}}+{{y}^{2}}-2ay-2by$ $-2bx+2ay=2bx-2ay$ $4ay=4bx$ $ay=bx$ $bx=ay$ Hence Proved.", "Find the equation of the set of points which are equidistant from the points (1, 2, 3) Question: Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1). Solution: Let P (xy, z) be the point that is equidistant from points A(1, 2, 3) and B(3, 2, –1). Accordingly, PA = PB $\\Rightarrow \\mathrm{PA}^{2}=\\mathrm{PB}^{2}$ $\\Rightarrow(x-1)^{2}+(y-2)^{2}+(z-3)^{2}=(x-3)^{2}+(y-2)^{2}+(z+1)^{2}$ $\\Rightarrow x^{2}-2 x+1+y^{2}-4 y+4+z^{2}-6 z+9=x^{2}-6 x+9+y^{2}-4 y+4+z^{2}+2 z+1$ $\\Rightarrow-2 x-4 y-6 z+14=-6 x-4 y+2 z+14$ $\\Rightarrow-2 x-6 z+6 x-2 z=0$ $\\Rightarrow 4 x-8 z=0$ $\\Rightarrow x-2 z=0$ Thus, the required equation is $x-2 z=0$.", "#### Find a point which is equidistant from the points A (–5, 4) and B (–1, 6)? How many such points are there? Solution. Let P(x, y) is a point which is equidistant from point A(–5, 4) and B(–1, 6) i.e. PA = PB Squaring both sides we get $PA^{2}= PB^{2}$ $\\left ( -5-x \\right )^{2}+\\left ( 4-y \\right )^{2}= \\left ( -1-x \\right )^{2}+\\left ( 6-y \\right )^{2}$ 25 + x2 + 10x + 16 + y2 – 8y = 1 + x2 + 2x + 36 + y2 – 12y {Using : (a + b)2 = a2 + b2 + 2ab; (a – b)2 = a2 + b2 – 2ab} 25 + 10x + 16 – 8y = 1 + 2x + 36 – 12y 10x – 8y + 41 – 2x + 12y – 37 = 0 8x + 4y + 4 = 0 Dividing by 4 se get 2x + y + 1 = 0 ……..(1) $Mid-point\\, of\\, AB= \\left ( \\frac{-5-1}{2},\\frac{4+6}{2} \\right )$ = (-3,5) $\\because Mid-point= \\left \\{ \\left ( \\frac{x_{1}+x_{2}}{2} \\right ) ,\\left ( \\frac{y_{1}+y_{2}}{2} \\right )\\right \\}$ Put point (–3, 5) in eqn. (1) 2(–3) + 5 + 1 = 0 – 6 + 6 = 0 0 = 0 Mid-point of AB satisfy equation (1) Hence, infinite numbers of", "be a coordinate at (-2,-2). And then just by looking at these coordinates, by symmetry again there is another coordinate at (-2, -4) and (4, -4). At this point, looking at the sketch (with the help of Desmos), you can see that the centre's y-coordinate is at -3. This second bit of reasoning to find the y-coordinate is a bit unsatisfactory and I can't fully explain it. And/or explain why my method was good or bad? P.S. The rest of the set of questions were: • (2,2), (4,3), & (6,9) • (-1,1), (2,-1), & (-2,0) • (0,0), (a,0, & (1,1) The center of the circle containing $$(0,0), (2,0),$$ and $$(4,-2)$$ must be equidistant from all those points. The points equidistant from $$(0,0)$$ and $$(2,0)$$ are on a line satisfying $$x^2+y^2=(x-2)^2+y^2$$; i.e., $$0=-4x+4$$; i.e., $$x=1$$. The points equidistant from $$(2,0)$$ and $$(4,-2)$$ are on a line satisfying $$(x-2)^2+y^2=(x-4)^2+(y+2)^2$$; i.e., $$-4x+4=-8x+16+4y+4$$; i.e., $$4x=4y+16$$; i.e., $$y=x-4$$. Therefore, the center is at the intersection of these lines, which is $$(1,-3)$$. Therefore, the equation of the circle is $$(x-1)^2+(y+3)^2=r^2$$. Plug in any of the points to figure out $$r^2$$. Take two points and find the perpendicular bisector. Repeat for another pair of points. The centre of the circle is where these perpendicular bisectors meet Use family of circles i.e circle with $$(0,0)$$ and", "got $t=0$, $s=1$ and $b=1$. • How did you find the $s(2, b+1, -1)$ ? – J.Se Jan 4 '18 at 17:16 • It's $(1,b,1)-(-1,-1,2)$. If $B(1,b,1)$ and $A(-1,-1,2)$ then $\\vec{AB}(2,b+1,-1)$. – Michael Rozenberg Jan 4 '18 at 17:17 To parametrise a line you need a point it passes through and its direction. You can get from the vector formed by the two points it passes through. Thus $L_2=(-1,-1,2)+t(2,b+1,-1).$L_1$is given by $$x=1-t$$$$y=1+t$$$$z=1-t$$ and$L_2$is given by $$x=-1+(1-(-1))s=-1+2s$$$$y=-1+(b-(-1))s=-1+(b+1)s$$$$z=2+(1-2)s=2-s$$ Now we solve $$1-t=-1+2s\\implies2-t=2s$$ $$1-t=2-s\\implies t=s-1$$ so using the equations for$x$and$z$, $$2-s+1=2s\\implies \\boxed{s=1}\\implies \\boxed{t=0}$$ Let's do the same for$y$. We solve $$1+t=-1+(b+1)s\\implies 1+0=-1+(b+1)(1)\\implies 1=-1+b+1=b$$ Hence intersection only occurs when $$\\boxed{b=1}$$ In other words, for an intersection,$L_2$must go through the points$(-1, -1, 2)$and$(1, 1, 1)\\$.", "setting this function equal to $$c$$. That is, we write $$x^2-2x+1+y^2−2y+1 = c + 2$$ which can be rewritten as, $$(x - 1)^2 + (y - 1)^2 = c + 2$$. This gives us circles centered at the point, $$(1, 1)$$, each with a radius of $$\\sqrt{c+2}$$. In exercises, 48 - 52, find the level surface for the given value of $$c$$ for each function of three variables and describe it. 48) $$w(x,y,z)=x−2y+z,\\quad c=4$$ 49) $$w(x,y,z)=x^2+y^2+z^2,\\quad c=9$$ $$x^2+y^2+z^2=9$$, a sphere of radius $$3$$ 50) $$w(x,y,z)=x^2+y^2−z^2,\\quad c=−4$$ 51) $$w(x,y,z)=x^2+y^2−z^2,\\quad c=4$$ $$x^2+y^2−z^2=4,$$ a hyperboloid of one sheet 52) $$w(x,y,z)=9x^2−4y^2+36z^2,\\quad c=0$$ In exercises 53 - 55, find an equation of the level curve of $$f$$ that contains the point $$P$$. 53) $$f(x,y)=1−4x^2−y^2,\\quad P(0,1)$$ $$4x^2+y^2=1,$$ 54) $$g(x,y)=y^2\\arctan x,\\quad P(1,2)$$ 55) $$g(x,y)=e^{xy}(x^2+y^2),\\quad P(1,0)$$ $$1=e^{xy}(x^2+y^2)$$ 56) The strength $$E$$ of an electric field at point $$(x,y,z)$$ resulting from an infinitely long charged wire lying along the $$y$$-axis is given by $$E(x,y,z)=k/\\sqrt{x^2+y^2}$$, where $$k$$ is a positive constant. For simplicity, let $$k=1$$ and find the equations of the level surfaces for $$E=10$$ and $$E=100.$$ 57) A thin plate made of iron is located in the $$xy$$-plane The temperature $$T$$ in degrees Celsius at a point $$P(x,y)$$ is inversely proportional to the square of its distance from the origin. Express $$T$$ as a function of $$x$$", "# If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also find the length of AB. - Mathematics If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also, find the length of AB. If a point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), then find the value of p. #### Solution 1 The given points are A(0, 2), B(3, p) and C(p, 5). It is given that A is equidistant from B and C. ∴ AB = AC ⇒ AB2 = AC2 ⇒ (3 − 0)2 + (p − 2)2 = (p − 0)2 + (5 − 2)2 ⇒ 9 + p2 + 4 − 4p = p2 + 9 ⇒ 4 − 4p = 0 ⇒ 4p = 4 p = 1 Thus, the value of p is 1. Length of AB =sqrt((3-0)^2+(1-2)^2)=sqrt(3^2+(-1)^2)=sqrt(9+1)=sqrt(10) units #### Solution 2 It is given that A(0, 2) is equidistant from the points B(3, p) and C(p, 5). ∴ AB = AC $\\Rightarrow \\sqrt{\\left( 3 - 0 \\right)^2 + \\left( p - 2 \\right)^2} = \\sqrt{\\left( p - 0 \\right)^2 + \\left( 5 - 2 \\right)^2}$ (Distance formula) Squaring on both sides, we get $9 + p^2 - 4p + 4", "missed. – Brian Tung Oct 21 '15 at 17:25 Hint: Write $a+b=x,a-b=y$. Now the point is equidistant. Let a point on the locus be $(\\alpha,\\beta)$ . Use the distance between two points formula and equate them. If I understand right, you want the points that are the same distance from both given points. That is always a straight line, and for points $(c,d)$ and $(d,c)$ it's always $(x,x)$ for all real $x$. If you want it in terms of $a$ and $b$, then try to solve the equation $ax=a$ in terms of $a$ for $a\\ne0$. The other answers seems to be based on points (a+b, a-b) and (a-b, a+b), which would give the line y=x. With the given points (a+b, b-a), (a-b, a+b), with same calculations as Ian Miller did, we get : $-2y(b-a)+2y(a+b)=2x(a+b)-2x(a-b)$ which gives $ya=xb$, which is the same as the line $y=(b/a)x$ (line of slope $b/a$ — or direction $(a,b)$ — which passes through the point (0,0) ) • +1 for catching the asymmetry the rest of us (except for David Z) missed. – Brian Tung Oct 21 '15 at 17:25 Let $(x,y)$ be a point on the locus. Then: $$(x-(a+b))^2+(y-(b-a))^2=(x-(a-b))^2+(y-(a+b))^2$$ Expanding: $$x^2-2x(a+b)+(a+b)^2+y^2-2y(b-a)+(b-a)^2=x^2-2x(a-b)+(a-b)^2+y^2-2y(a+b)+(a+b)^2$$ $$-2y(b-a)+2y(a+b)=2x(a+b)-2x(a-b)$$ $$4ya=4xb$$ $$y=\\frac{b}{a}x$$ • Corrected based on comments from Anceps and others. – Ian Miller Oct 22 '15 at 0:38", "38, not 36 etc 10x^2- 58x+ 76= 0 5x^2- 29x+ 38= 0 (x+ 2)(5x- 19)= 0. Remember to use y= 10- 3x to find y, not y= 3x- 10. 7. Originally Posted by angypangy Hello If I have these two circles: (x + 1)^2 + y^2 = 25 and (x - 2)^2 + (y-1)^2 = 9 How do I work out where the circles (or if) intersect? It is difficult to set eg y= due to the different terms of x. Angus The set of all points a distance of 5 units from (-1,0) is given by the equation $[x-(-1)]^2+[y-0]^2=5^2\\Rightarrow\\ (x+1)^2+y^2=25\\Rightarrow\\ x^2+y^2+2x-24=0$ The set of all points a distance of 3 units from (2,1) is given by $(x-2)^2+(y-1)^2=3^2\\Rightarrow\\ x^2-4x+4+y^2-2y+1=9\\Rightarrow\\ x^2+y^2-4x-2y-4=0$ To find the points (x,y) that are both 5 units from (-1,0) and 3 units from (2,1) then the \"x\" values are equal in both equations. The \"y\" values are also equal. 0-0=0 $\\left[x^2+y^2+2x-24\\right]-\\left[x^2+y^2-4x-2y-4\\right]=x^2-x^2+y^2-y^2+2x+4x+2y-24+4=0$ $6x+2y-20=0\\Rightarrow\\ 3x+y-10=0$ However, the resulting equation is the set of all points equidistant from the centres. This is because we obtain the same result by subtracting the equations of other circles with these centres, but whose radii squared differ by the exact same amount. If you wish, you can then find the point of intersection of this line with either circle, by again allowing", "Then the coordinates of $C$ have to satisfy the two equations below. $\\begin{cases} (x-a)^2+(y-b)^2=m^2 \\\\ (x-c)^2+(y-d)^2=r^2 \\end{cases}$ Now, since we know that $B$ is on the circle centered at $A$ with radius $r$, the coordinates of $B$ have to satisfy the equation $a^2+b^2=r^2$. Therefore, the coordinates of $C$ have to satisfy the three equations below. $\\begin{cases} (x-a)^2+(y-b)^2=m^2 \\\\ (x-c)^2+(y-d)^2=r^2 \\\\ a^2+b^2=r^2 \\end{cases}$ Now, expanding the first two equations we have, Eq. 1 $x^2+y^2-2ax-2by+a^2+b^2=m^2$ Eq. 2 $x^2+y^2-2cx-2dy+c^2+d^2=r^2$ Subtract Eq. 2 from Eq. 1 we have, Eq. 3 $(-2a+2c)x-(2b-2d)y+(a^2+b^2)-(c^2+d^2)=m^2-r^2$ Substituting $a^2+b^2=r^2$ and rearranging we have, $(-2a+2c)x-(2b-2d)y=m^2-2r^2+c^2+d^2$ Hence Eq. 4 $y=\\frac {-2a+2c}{2b-2d}x-\\frac {m^2-2r^2+c^2+d^2}{2b-2d}$ Now, we can manipulate Eq. 3 to get an expression for $b$, i.e. $b=f(a,c,d,m,r,x,y)$. Next, we substitute $b=f(a,c,d,m,r,x,y)$ back into Eq. 1 and will be able to obtain an expression for $a$, i.e. $a=g(x,y,d,c,m,r)$. Since $b=\\pm \\sqrt {r^2-a^2}$, we have expressions of $a$ and $b$ in terms of $x,y,d,c,m$ and $r$. Say point $P$ has coordinates $(x',y')$, then $x'=\\frac {a+x}{2}$ and $y'=\\frac {b+y}{2}$ which will yield Eq. 5 $x=2x'-a$ Eq. 6 $y=2y'-b$ In the last step we substitute $a=g(x,y,d,c,m,r)$,$b=\\pm \\sqrt {r^2-a^2}$, Eq. 5 and Eq. 6 back into Eq. 4 and we will finally have a relationship between $x'$ and $y'$. Of course, it will be a messy closed form but we could definitely use Mathematica to do", "= -1. 4. Apply Cramer’s rule to solve the following equations. x + 3y + 6z = 2 3x – y + z = 9 X – 4y + 2z = 7 a) X = 1, y = 2, z = -1 b) X = 2, y = – 1, z = -0.5 c) X = 1, y = 2, z = -0.5 d) X = 2, y = 2, z = -1 Explanation: ∆ = $$\\begin{pmatrix}1&3&6\\\\3&-1&4\\\\1&-4&2\\end{pmatrix}$$ = -58 X = (1/∆) = $$\\begin{pmatrix}2&3&6\\\\9&-1&4\\\\7&-4&2\\end{pmatrix}$$ = -116/-58 = 2 Y = (1/∆) = $$\\begin{pmatrix}1&2&6\\\\3&9&4\\\\1&7&2\\end{pmatrix}$$ = 58/-58 = -1 Z = (1/∆) = $$\\begin{pmatrix}1&3&2\\\\3&-1&9\\\\1&-4&7\\end{pmatrix}$$ = -29/-58 = 0.5 Hence, x = 2, y = -1, z = -0.5. 5. Apply Cramer’s rule to solve the following equations. x + y + z = 6.6 x – y + z = 2.2 x + 2y + 3z = 15.2 a) x = 1.5, y = 2.2, z = -0.5 b) x = 1.5, y = 2.2, z = -0.5 c) x = 1.2, y = 2, z = 3.2 d) x = 1.2, y = 2.2, z = -3.2 Explanation: ∆ = $$\\begin{pmatrix}1&1&1\\\\1&-1&1\\\\1&2&3\\end{pmatrix}$$ = -4 X = (1/∆) = $$\\begin{pmatrix}6.6&1&1\\\\2.2&-1&1\\\\15.2&2&3\\end{pmatrix}$$ = -4.8/-4 = 1.2 Y = (1/∆) = $$\\begin{pmatrix}1&6.6&1\\\\1&2.2&1\\\\1&15.2&3\\end{pmatrix}$$ = -8.8/-4 = 2.2 Z = (1/∆) = $$\\begin{pmatrix}1&1&6.6\\\\1&-1&2.2\\\\1&2&15.2\\end{pmatrix}$$ = -12.8/-4 =", "* * | ♥ * * | O * | * | * * | * o * C * * * Here is my very primitive solution . . . We hope to find a point $O(x,y)$ equidistant from $A,B,C,D.$ We have this system of equations: . . $\\begin{array}{ccccccccc} AO^2 &=& (x-13)^2 + (y-9)^2 &=& r^2 & [1] \\\\ BO^2 &=& (x-14)^2 + (y-2)^2 &=& r^2 & [2] \\\\ CO^2 &=& (x-7)^2 + (y-1)^2 &=& r^2 & [3] \\\\ DO^2 &=& (x-5)^2 + (y-5)^2 &=& r^2 & [4] \\end{array}$ Subtract [1] - [2]: . $(x-13)^2 - (x-14)^2 + (y-9)^2 - (y-2)^2 \\:=\\:0$ . . $(1)(2x-27) + (\\text{-}7)(2y-11) \\:=\\:0 \\quad\\Rightarrow\\quad x - 7y \\:=\\:\\text{-}25\\;\\;[5]$ Subtract [3] - [4]: . $(x-7)^2 - (x-5)^2 + (y-1)^2 - (y-5)^2 \\:=\\:0$ . . $(-2)(2x-12) + (4)(2y-6) \\:=\\:0 \\quad\\Rightarrow\\quad x - 2y \\:=\\:0\\;\\;[6]$ Solve [5] and [6]: . $x = 10,\\;y = 5$ Therefore, $A,B,C,D$ lie on a circle with center $O(10,5)$ and radius 5. 5. another way is to use this form of circle ${x^2} + {y^2} + Ax + By + C = 0$ Take three points and set them in this equation and you will find A,B,C, solving a simple linear system of equations. If the fourth point satisfies the equation you are done.", "of the set of points P which are equidistant from point A (3, 2, 1) and point B (-1, 2, 3). Sol. Let, (x, y, z) be the coordinates of point P, which is equidistant from point A (3, 2, 1) and point B (-1, 2, 3). Now, according to the given condition: AP2 = BP2 $$\\Rightarrow$$ $$\\left [ \\sqrt{(x-3)^{2}+(y-2)^{2}+(z-1)^{2}} \\;\\right ]^{2}=\\left [ \\sqrt{(x+1)^{2}+(y-2)^{2}+(z-3)^{2}} \\;\\right ]^{2}$$ $$\\\\\\Rightarrow$$ $$(x-3)^{2}+(y-2)^{2}+(z-1)^{2}=(x+1)^{2}+(y-2)^{2}+(z-3)^{2}$$ $$\\\\\\Rightarrow$$ $$x^{2}+9-6x+y^{2}+4-4y+z^{2}+1-2z=x^{2}+1+2x+y^{2}+4-4y+z^{2}+9-6z$$ $$\\\\\\Rightarrow$$ – 6x – 4y – 2z = +2x – 4y – 6z $$\\\\\\Rightarrow$$ – 6x – 4y – 2z = + 2x – 4y – 6z $$\\\\\\Rightarrow$$ 8x – 4z = 0 Therefore, the required equation is: 2x – z = 0 Q.5: The sum of distance of point P from point A (3, 0, 0) and point B (-3, 0, 0) is equal to 12 units. Find the equation of the set of points. Sol. Let, (x, y, z) be the coordinates of point P. Now, according to the given condition: AP + BP = 12 $$\\Rightarrow$$ $$\\sqrt{(x-3)^{2}+y^{2}+z^{2}}+\\sqrt{[x-(-3)]^{2}+y^{2}+z^{2}}=12\\\\$$ $$\\Rightarrow$$ $$\\sqrt{(x-3)^{2}+y^{2}+z^{2}}=12-\\sqrt{(x+3)^{2}+y^{2}+z^{2}}$$ Now, on squaring both the side, we will get: $$\\Rightarrow$$ $$\\left ( \\sqrt{(x-3)^{2}+y^{2}+z^{2}} \\right )^{2}= \\left ( 12-\\sqrt{(x+3)^{2}+y^{2}+z^{2}} \\right )^{2}\\\\$$ $$\\Rightarrow$$ $$(x-3)^{2}+y^{2}+z^{2}= 144+[(x+3)^{2}+y^{2}+z^{2}]-24\\sqrt{(x+3)^{2}+y^{2}+z^{2}}\\\\$$ $$\\Rightarrow$$ $$x^{2}+9-6x+y^{2}+z^{2}=144+x^{2}+9+6x+y^{2}+z^{2}-24\\sqrt{(x+3)^{2}+y^{2}+z^{2}}\\\\$$ $$\\Rightarrow$$ $$24\\sqrt{(x+3)^{2}+y^{2}+z^{2}}=144+12x\\\\$$ $$\\Rightarrow$$ $$2\\sqrt{(x+3)^{2}+y^{2}+z^{2}}=(x+12)$$ Again, on squaring both the side, we will get: $$\\Rightarrow$$ $$4[(x+3)^{2}+y^{2}+z^{2}]=(x+12)^{2}\\\\$$ $$\\Rightarrow$$ $$4x^{2}+4y^{2}+4z^{2}+36+24x=x^{2}+24x+144\\\\$$ $$\\Rightarrow$$ $$3x^{2}+4y^{2}+4z^{2}-108=0$$ Therefore, the", "the points $x\\in [x_{i},x_{i+1}]$ such that $f(a_ix+b)=y_j$ for some $j$ taken in the increasing order when $a_i>0$ and in the decreasing order when $a_i<0$ (we have to assume that $f([x_i, x_{i+1}])\\subset [y_1, y_{n}]$, because otherwise the composition $f\\circ g$ can be undefined on $[x_i, x_{i+1}]$). For instance, if $a_i>0$ and $$f([x_i,x_{i+1}])\\cap Y=\\{y_{j(i)}, y_{j(i)+1},\\dots, y_{j’(i)}\\}$$ for some numbers $j(i)$ and $j’(i)$ then for each $2\\le k\\le p(i)-1$ we have $a_iz_{i,k}+ b_i=f(z_{i,k})=y_{j(i)+k-2}$, that is $z_{i,k}=(y_{j(i)+k-2}-b_i)/a_i$. In remains to note that for $x\\in [z_{i,k},z_{i+1,k}]$ we have $c_{i,k}x+d_{i,k}\\equiv \\alpha_{j(i)+k-2}(a_ix+b_i)+\\beta_{j(i)+k-2}$ for each $1\\le k\\le p(i)-1$.", "# Before starting • Survey of past and current work • Partly joint with J.I. Cogolludo # Two points determine a line. $$a_{0,0}z+a_{1,0} x + a_{0,1}y$$ # One point determines a pencil of lines $$a_{0,0}z+a_{1,0} x + a_{0,1}y$$ # Five (generic) points determine a conic. $$a_{0,0}z^2+a_{1,0} xz + a_{0,1} yz +a_{2,0} x^ 2+a_{1,1}x y +a_{0,2}y^2$$ # Four points determine a pencil of conics $$a_{0,0}z^2+a_{1,0} xz + a_{0,1} yz +a_{2,0} x^ 2+a_{1,1}x y +a_{0,2}y^2$$ # Four points determine a pencil of conics $$a_{0,0}z^2+a_{1,0} xz + a_{0,1} yz +a_{2,0} x^ 2+a_{1,1}x y +a_{0,2}y^2$$ # Nine points determine a cubic $$a_{0,0}z^3+a_{1,0} x z^2+ a_{0,1} yz^2 +a_{2,0} x^ 2z+a_{1,1}x yz +a_{0,2}y^2z+a_{3,0}x^3+a_{2,1}x^2y+a_{1,2}xy^2+a_{0,3}y^3$$ # Eight points determine a pencil of cubics $$a_{0,0}z^3+a_{1,0} x z^2+ a_{0,1} yz^2 +a_{2,0} x^ 2z+a_{1,1}x yz +a_{0,2}y^2z+a_{3,0}x^3+a_{2,1}x^2y+a_{1,2}xy^2+a_{0,3}y^3$$ # Eight points determine a pencil of cubics? $$a_{0,0}z^3+a_{1,0} x z^2+ a_{0,1} yz^2 +a_{2,0} x^ 2z+a_{1,1}x yz +a_{0,2}y^2z+a_{3,0}x^3+a_{2,1}x^2y+a_{1,2}xy^2+a_{0,3}y^3$$ Theorem (Cayley-Bacharach): - If a cubic passes through eight intersections points of two other cubics, it also passes through the ninth. # Question: • When is there a pencil of curves?: • Two curves always generate a pencil • More than three are in a pencil iff every three of them are. • So the question really is: • What conditions should three curves satisfy to be in a pencil? # First result Noether fundamental", "anonymous 4 years ago the set of all points P(x,y,z), such that the sum of the distances from P to the two points (0,-3,0) and (0,3,0) equals 10, forms a quadric surface. Identify this surface and give its equation in the standard form (x^2/a^2)+(y^2/b^2)+(z^2/c^2)=1 1. anonymous i wanna see the solution too ^^, just post sumthing here for able to folo its progress 2. nikvist $d_1^2=x^2+(y+3)^2+z^2=x^2+y^2+z^2+9+6y=w+6y$$d_2^2=x^2+(y-3)^2+z^2=x^2+y^2+z^2+9-6y=w-6y$$w=x^2+y^2+z^2+9$$d_1+d_2=10$$d_1^2+2d_1d_2+d_2^2=100$$4d_1^2d_2^2=(100-d_1^2-d_2^2)^2$$4(w+6y)(w-6y)=(100-w-6y-w+6y)^2$$4(w^2-36y^2)=(100-2w)^2$$4w^2-144y^2=10000-400w+4w^2$$400w-144y^2-10000=16(25w-9y^2-625)=0$$25w-9y^2-625=0$$25(x^2+y^2+z^2+9)-9y^2-625=0$$25x^2+16y^2+25z^2=400$$\\frac{x^2}{16}+\\frac{y^2}{25}+\\frac{z^2}{16}=1$$\\frac{x^2}{4^2}+\\frac{y^2}{5^2}+\\frac{z^2}{4^2}=1\\quad-\\mbox{ellipsoid}$ 3. anonymous em i have a little question over here, that form : X2/a2 +y2/b2 + x2/c2 can applied on other shape of graph too ? like|dw:1327413868275:dw| or it can only apply on closed graph like circle, square and...? 4. nikvist ellipsoid: xy-plane and yz-plane projections are ellipse, xz-plane projection is circle. 5. anonymous if it is x and y only, if i encounter this kind of graph: |dw:1327415499115:dw| will i get the center point with that method?", "= x+5x2+56 [Write the function.] dydx = ddx (x+5x2+56) [Find dydx.] = (x2+56)(1)-(x+5)(2x)(x2+56)2 [Use Quotient Rule.] = -x2-10x+56(x2+56)2 [Simplify.] At critical points, either f ′(x) is zero or f ′(x) does not exist. -x2-10x+56(x2+56)2 = 0 [Equate dydx to zero.] - x2 - 10x + 56 = 0 [(x2 + 56)2 > 0 for all x.] x = 4, - 14 [Solve for x.] At x = 4, y = 4+5(4)2+56 = 1 / 8 [Replace x = 4 in f(x).] At x = - 14, y = - 14+5(-14)2+56 = - 1 / 28 [Replace x = - 14 in f(x).] Therefore, the critical points for the function y = x+5x2+56 is (4, 1 / 8) and (- 14, - 1 / 28). 9. Find the critical points for the function $f$($x$) = ln ($x$2 + 8$x$ + 32). a. (- 4, 2.77) b. (4, 4.38) c. (- 4, 0) d. (- 2.77, 2.77) e. Cannot be determined #### Solution: f(x) = ln (x2 + 8x + 32) [Write the function.] f ′ (x) = ddx [ln (x2 + 8x + 32)] [Find f ′(x).] = 1(x2+8x+32) (2x + 8) [Use Chain Rule.] = 2(x+4)(x2+8x+32) [Simplify.] At critical points, either f ′(x) is zero or f ′(x) does not exist. 2(x+4)(x2+8x+32) = 0 [Equate f ′(x) to", "equal. Therefore, ABCD is a parallelogram. Hence, the given points (2, -3, 4), (1, -2, 5), (-1, 2, 1) and (4, -7, 8) are the vertices of a parallelogram. Q.4: Find the equation of the set of points P which are equidistant from point A (3, 2, 1) and point B (-1, 2, 3). Sol. Let, (x, y, z) be the coordinates of point P, which is equidistant from point A (3, 2, 1) and point B (-1, 2, 3). Now, according to the given condition: AP2 = BP2 $\\Rightarrow$ [(x3)2+(y2)2+(z1)2]2=[(x+1)2+(y2)2+(z3)2]2$\\left [ \\sqrt{(x-3)^{2}+(y-2)^{2}+(z-1)^{2}} \\;\\right ]^{2}=\\left [ \\sqrt{(x+1)^{2}+(y-2)^{2}+(z-3)^{2}} \\;\\right ]^{2}$ $\\\\\\Rightarrow$ (x3)2+(y2)2+(z1)2=(x+1)2+(y2)2+(z3)2$(x-3)^{2}+(y-2)^{2}+(z-1)^{2}=(x+1)^{2}+(y-2)^{2}+(z-3)^{2}$ $\\\\\\Rightarrow$ x2+96x+y2+44y+z2+12z=x2+1+2x+y2+44y+z2+96z$x^{2}+9-6x+y^{2}+4-4y+z^{2}+1-2z=x^{2}+1+2x+y^{2}+4-4y+z^{2}+9-6z$ $\\\\\\Rightarrow$ – 6x – 4y – 2z = +2x – 4y – 6z $\\\\\\Rightarrow$ – 6x – 4y – 2z = + 2x – 4y – 6z $\\\\\\Rightarrow$ 8x – 4z = 0 Therefore, the required equation is: 2x – z = 0 Q.5: The sum of distance of point P from point A (3, 0, 0) and point B (-3, 0, 0) is equal to 12 units. Find the equation of the set of points. Sol. Let, (x, y, z) be the coordinates of point P. Now, according to the given condition: AP + BP = 12 $\\Rightarrow$ (x3)2+y2+z2+[x(3)]2+y2+z2=12$\\sqrt{(x-3)^{2}+y^{2}+z^{2}}+\\sqrt{[x-(-3)]^{2}+y^{2}+z^{2}}=12\\\\$ $\\Rightarrow$ (x3)2+y2+z2=12(x+3)2+y2+z2$\\sqrt{(x-3)^{2}+y^{2}+z^{2}}=12-\\sqrt{(x+3)^{2}+y^{2}+z^{2}}$ Now, on squaring both the side, we will get: $\\Rightarrow$ ((x3)2+y2+z2)2=(12(x+3)2+y2+z2)2$\\left ( \\sqrt{(x-3)^{2}+y^{2}+z^{2}}" ]

[ "# Convert the point from rectangular coordinates to spherical coordinates. (-3,-3,\\sqrt{2}) Convert the point from rectangular coordinates to spherical coordinates. $$\\displaystyle{\\left(-{3},-{3},\\sqrt{{{2}}}\\right)}$$ • Questions are typically answered in as fast as 30 minutes ### Plainmath recommends • Get a detailed answer even on the hardest topics. • Ask an expert for a step-by-step guidance to learn to do it yourself. Arham Warner To convert:", "convert the point from cylindrical coordinates to rectangular coordinates. $$(2,-\\pi,-4)$$ Related chapters Unlock Textbook Solution", "+0 # Hard Question Pls Help Me 0 44 1 A point has rectangular coordinates $(2,-1,-2)$ and spherical coordinates $(\\rho, \\theta, \\phi).$ Find the rectangular coordinates of the point with spherical coordinates $(\\rho, \\theta, 2 \\phi).$ Jun 6, 2020", "point $P_1$, and the spherical coordinates $\\rho_2$ , $\\theta_2$, $\\phi_2$ of point $P_2$, selecting the desired units for the angles, and press the button \"Calculate\". You may also change the number of decimal places as needed; it has to be a positive integer. $\\rho_1 =$ 1 $\\theta1 =$ 45 degrees radians $\\phi_1 =$ 45 degrees radians $\\rho_2 =$ 2 $\\theta2 =$ 90 degrees radians $\\phi_2 =$ 30 degrees radians Number of Decimal Places = Distance = $\\rho =$ $\\theta =$ Radians $\\theta =$ Degrees $\\phi =$ Radians $\\phi =$ Degrees ## More References and links Maths Calculators and Solvers.", "Free Version Moderate # Spherical to Rectangular Conversion MVCALC-EPFFL4 The equation in rectangular coordinates for the surface that has equation $\\phi=\\pi/3$ in spherical coordinates has the form: $$a(x^2+y^2)=z^2$$ What is the value of $a$?", "$\\theta$. You can see that $\\theta$ goes from $0$ to $\\pi/2$ Finally, you want to convert $dV$ to cylindrical coordinates. When converting from rectangular to cylindrical, $dV$ becomes $rdzdrd\\theta$. What we're left with is: $$\\int_{0}^{\\pi/2}\\int_{0}^{1}\\int_{0}^{2-r^2}r^2cos(\\theta)dzdrd\\theta$$", "point (2,1,2) in Cartesian coordinates, we find that $\\rho=\\sqrt{x^2+y^2+z^2}=\\sqrt{2^2+1^2+2^2}=\\sqrt{4+1+4}=\\sqrt{9}=3$ $\\tan\\phi=\\frac{\\sqrt{x^2+y^2}}{z}=\\frac{\\sqrt{2^1+1^2}}{2}=\\frac{\\sqrt{5}}{2}$ $\\rightarrow \\phi=\\tan^{-1}\\left(\\sqrt{5}/2\\right)=0.841$ radians $\\tan \\theta = y/x=1/2$ $\\rightarrow \\theta=\\tan^{-1}=0.464$ radians So we have the point (3,0.841,0.464) in spherical coordinates The transformation from spherical coordinates to Cartesian coordinates is given by $x=\\rho \\sin \\phi \\cos \\theta$ $y=\\rho \\sin \\phi \\sin \\theta$ $z= \\rho \\cos \\phi$. We can convert our previous result back to Cartesian coordinates: $x=\\rho \\sin \\phi \\cos \\theta=3\\sin(0.841)\\cos(0.464)\\approx 2$ $y=\\rho \\sin \\phi \\sin \\theta =3\\sin(0.841)\\sin(0.464)\\approx 1$ $z= \\rho \\cos \\phi =3 cos(0.841)\\approx 2$ Again, we retrieve our original ordered triple (2,1,2) by rounding. We can also write the transformation from cylindrical coordinates to spherical coordinates: $\\rho=\\sqrt{r^2+z^2}$ $\\tan \\phi = z/r$ $\\theta = \\theta$. Beginning with the point (3,\\pi/3,4) in cylindrical coordinates, we see that $\\rho=\\sqrt{r^2+z^2}=\\sqrt{3^2+4^2}=\\sqrt{25}=5$ $\\tan \\phi = z/r=4/3$ $\\rightarrow \\phi=\\tan^{-1}(.75)=0.644$ $\\theta = \\pi/3$. So the point in spherical coordinates is (5,\\pi/3,0.644). Finally, the transformation from spherical to cylindrical coordinates is given by $r=\\rho \\sin \\phi$ $\\theta =\\theta$ $z=\\rho \\cos \\phi$. For one last example, we will take our previous result and transform it back to cylindrical coordinates $r=\\rho \\sin \\theta=5 \\sin (0.644)\\approx 3$ $\\theta =\\pi/3$ $z=\\rho \\cos \\theta5 \\cos (0.644)\\approx 4$ Therefore the point in cylindrical coordinates is (3, \\pi/3, 4) as expected. Interactive Demonstration Change of Coordinate Systems Applet • add examples of transformations between", "+0 # The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. 0 34 1 +16 The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. Aug 6, 2020 #1 0 tan theta + tan phi = 5/2 - 4/3 = 7/6. Aug 6, 2020", "+0 # The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. 0 68 1 +16 The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. Aug 6, 2020", "Question # Change from rectangular to spherical coordinates. (Let \\rho\\geq0,\\ 0\\leq\\theta\\leq\\2\\pi, and 0\\leq\\phi\\leq\\pi.)a) (0,\\ -8,\\ 0)(\\rho,\\theta, \\phi)=?b) (-1,\\ 1,\\ -2)(\\rho, \\theta, \\phi)=? Change from rectangular to spherical coordinates. (Let $$\\rho\\geq0,\\ 0\\leq\\theta\\leq 2\\pi$$, and $$0\\leq\\phi\\leq\\pi.)$$ a) $$(0,\\ -8,\\ 0)(\\rho,\\theta, \\phi)=?$$ b) $$(-1,\\ 1,\\ -2)(\\rho, \\theta, \\phi)=?$$ 2021-05-15 Step 1 a) Consider the rectangular coordinate $$(0,\\ -8,\\ 0)$$ The objective is to convert the rectangular coordinate to spherical coordinate $$(\\rho, \\theta,\\phi)$$ Here, $$\\rho\\geq0,\\ 0\\leq\\theta\\leq2\\pi,\\ 0\\leq\\phi\\leq\\pi$$ Consider the following statement: The point P in $$\\mathbb{R}^{3}$$ with rectangular coordinates $$(x,y,z)$$ and spheerical coordinates $$(\\rho,\\theta\\phi)$$ The relation between rectangular coordinates $$(x,y,z)$$ and spherical coordinates $$(\\rho,\\theta\\phi)$$ is, $$x=\\rho\\sin\\phi\\cos\\theta$$ $$y=\\rho\\sin\\phi\\sin\\theta$$ $$z=\\rho\\cos\\phi$$ Step 2 The rectangular coordinate is $$(x,y,z)=(0,\\ -8,\\ 0)$$ That is, $$x=0,\\ y=-8,\\ z=0.$$ Thus, from the statement, $$\\rho=\\sqrt{x^{2}+y^{2}+z^{2}}$$ $$=\\sqrt{0^{2}+(-8)^{2}+0^{2}}$$ $$\\{x=0,\\ y=-8,z=0\\}$$ $$=\\sqrt{64}$$ $$=8$$ From the statement, $$z=\\rho\\cos\\phi$$ $$0=8\\cos\\phi$$ $$\\{z=0,\\rho=8\\}$$ $$\\cos\\phi=0$$ $$\\phi=\\frac{\\pi}{2}$$ $$\\{0\\leq\\phi\\leq\\pi\\}$$ Step 3 From the statement, $$y=\\rho\\sin\\phi\\sin\\theta$$ $$\\sin\\theta=\\frac{y}{\\rho\\sin\\phi}$$ Substitute $$y=-8,\\rho=8,\\phi=\\frac{\\pi}{2}$$ $$\\sin\\theta=\\frac{-8}{8\\sin\\frac{\\pi}{2}}$$ $$=-1$$ $$\\left\\{\\sin\\frac{\\pi}{2}=1\\right\\}$$ $$\\theta=\\frac{3\\pi}{2}$$ $$\\left\\{0\\leq\\theta\\leq\\frac{\\pi}{2}\\right\\}$$ Hence, the spherical coordinate of the point $$(0,\\ -8,\\ 0)$$ is $$\\left(8,\\ \\frac{3\\pi}{2},\\ \\frac{\\pi}{2}\\right)$$ Step 4 Consider the rectangular coordinate $$(-1,\\ 1,\\ -2)$$ The objective is to convert the rectangular coordinate to spherical coordinate $$(\\rho,\\theta,\\phi)$$. Consider the following statement: The point P in $$\\mathbb{R}^{3}$$ with rectangular coordinates $$(x,y,z)$$ and spherical coordinates $$(\\rho,\\theta,\\phi)$$. The relation between rectangular coordinates $$(x,y,z)$$ and spherical coordinates $$(\\rho,\\theta,\\phi)$$ is, $$x=\\rho\\sin\\phi\\cos\\theta$$ $$y=\\rho\\sin\\phi\\sin\\theta$$ $$z=\\rho\\cos\\phi$$ The", "## Calculus 8th Edition a) $\\rho=3$ b) $\\rho ^2 (\\sin^2 \\phi \\cos 2 \\theta- \\cos^2 \\phi)=1$ The conversion of rectangular coordinates to spherical coordinates is given as: $x=\\rho \\sin \\phi \\cos \\theta; y=\\rho \\sin \\phi \\sin \\theta;z=\\rho \\cos \\phi$ Here, $\\rho=\\sqrt {x^2+y^2+z^2}$; $\\phi =\\cos^{-1} [\\dfrac{z}{\\rho}]; \\theta=\\cos^{-1}[\\dfrac{x}{\\rho \\sin \\phi}]$ a) Here, we have $x^2+y^2+z^2=9$ But $\\rho=\\sqrt {x^2+y^2+z^2}$ This gives: $\\rho^2=3^2 \\implies \\rho=3$ b) Here, we have $x^2-y^2-z^2=1$ But $x=\\rho \\sin \\phi \\cos \\theta; y=\\rho \\sin \\phi \\sin \\theta;z=\\rho \\cos \\phi$ So, we have $(\\rho \\sin \\phi \\cos \\theta)^2-(\\rho \\sin \\phi \\sin \\theta)^2-(\\rho \\cos \\phi)^2=1$ or, $\\rho ^2 \\sin^2 \\phi \\times (\\cos^2 \\theta- \\sin^2 \\theta)-(\\rho \\cos \\phi)^2=1$ Thus, we get $\\rho ^2 (\\sin^2 \\phi \\cos 2 \\theta- \\cos^2 \\phi)=1$", "# “The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$.” Code Answer The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. whatever by Confused Chinchilla on Jun 17 2020 Comment 1", "+0 # The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. 0 246 1 +16 The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. Aug 6, 2020", "Trigonometry (11th Edition) Clone The rectangular coordinates are $(0,-4)$ We can find the x-coordinate: $x = r~cos~\\theta$ $x = (-4)~cos~90^{\\circ}$ $x = (-4)~(0)$ $x = 0$ We can find the y-coordinate: $y = r~sin~\\theta$ $y = (-4)~sin~90^{\\circ}$ $y = (-4)~(1)$ $y = -4$ The rectangular coordinates are $(0,-4)$", "## Calculus (3rd Edition) $(2,0,\\frac{\\pi}{3})$. Since $x=\\sqrt3, \\, y=0, \\, z=1$ and $$\\rho=\\sqrt{x^2+y^2+z^2},\\quad \\theta=\\tan^{-1}( y/x), \\quad \\phi=\\cos^{-1}(z/\\rho),$$ then we have $$\\rho=\\sqrt{\\sqrt 3^2+0+1^2}=\\sqrt{4}=2,$$ $$\\theta=\\tan^{-1}( 0)=0,$$ $$\\phi=\\cos^{-1}(1/2)=\\frac{\\pi}{3}$$ Hence, the spherical coordinates are $(2,0,\\frac{\\pi}{3})$.", "the point is (1, 0, p /2). EXAMPLE 12: Convert in cylindrical coordinates to spherical coordinates. SOLUTION: q stays the same, so I need to find j . EXAMPLE 13: Convert (2, p /6, 6) in cylindrical coordinates to spherical coordinates. SOLUTION: q = p /6 = 30 o EXAMPLE 14: Convert to rectangular coordinates, then cylindrical coordinates. SOLUTION: Rectangular coordinates Cylindrical coordinates EXAMPLE 15: Convert to rectangular coordinates, then cylindrical coordinates. SOLUTION: Rectangular coordinates Cylindrical coordinates EXAMPLE 16: Convert x 2 + y 2 = 5 into spherical coordinates. SOLUTION: EXAMPLE 17: Convert x 2 + y 2 + z 2 = 9 into spherical coordinates. SOLUTION: EXAMPLE 18: Convert r = -3sec q into spherical coordinates. SOLUTION: r = -3sec q ® r sin j = -3sec q ® r = -3csc j sec q EXAMPLE 19: Convert into spherical coordinates. SOLUTION: j = 135 o Here are the spherical equations. r = 2 135 o £ j £ 180o EXAMPLE 20: Convert into cylindrical and rectangular coordinates. SOLUTION: The above equation is both the cylindrical form of the given equation and the rectangular form of the given equation. EXAMPLE 21: Convert into cylindrical and rectangular coordinates. SOLUTION: Let us convert to rectangular coordinates first. ® x 2 + y 2 - z 2 = 0", "Question # Change from rectangular to spherical coordinates. (Let \\rho\\geq 0, 0 \\leq \\theta \\leq 2\\pi, \\text{ and } 0 \\leq \\phi \\leq \\pi) (a) (0, -3, 0) (b) (-1, 1, -\\sqrt{2}) Parametric equations Change from rectangular to spherical coordinates. (Let $$\\rho\\geq 0, 0 \\leq \\theta \\leq 2\\pi, \\text{ and } 0 \\leq \\phi \\leq \\pi$$) (b) $$(-1, 1, -\\sqrt{2})$$", "Question # Change from rectangular to spherical coordinates. (Let \\rho\\geq 0, 0 \\leq \\theta \\leq 2\\pi, \\text{ and } 0 \\leq \\phi \\leq \\pi) Parametric equations Change from rectangular to spherical coordinates. (Let $$\\displaystyle\\rho\\geq{0},{0}\\leq\\theta\\leq{2}\\pi,\\ \\text{ and }\\ {0}\\leq\\phi\\leq\\pi$$) (a) $$(0, -3, 0)$$ (b) $$\\displaystyle{\\left(-{1},{1},-\\sqrt{{{2}}}\\right)}$$", "it's much easier. So, let's first convert to $z$ the same way. $$z = \\rho\\cos(\\phi)$$ We look at the same triangle, and we're able to get r. $$r = \\rho\\sin(\\phi)$$ $\\theta$ is the same, so we're done. $$(r, \\theta, z) = \\begin{cases} r = \\rho\\sin(\\phi) \\\\ \\theta = \\theta \\\\ z = \\rho\\cos(\\phi) \\end{cases}$$ David Witten", "Find the coordinates of the point $(1,1,1)$ in Spherical coordinates Find the coordinates of the point $(x,y,z)=(1,1,1)$ in Spherical coordinates. Answers can be viewed only if 1. The questioner was satisfied and accepted the answer, or 2. The answer was disputed, but the judge evaluated it as 100% correct." ]

[ "coordinates to rectangular coordinates. #### Solution Use the formulas, noting that $$\\rho=5$$, $$\\theta=\\pi$$, and $$\\phi=\\dfrac{\\pi}{2}$$. \\begin{align} x &= \\rho \\cos \\theta \\sin \\phi = 5 (\\cos \\pi) \\left(\\sin \\dfrac{\\pi}{2}\\right) = 5(-1)(1) = -5\\\\ y &= \\rho \\sin \\theta \\sin \\phi = 5 (\\sin \\pi) \\left(\\sin \\dfrac{\\pi}{2}\\right) = 5(0)(1) = 0\\\\ z &= \\rho \\cos \\phi = 5 \\cos \\dfrac{\\pi}{2} = 5(0) = 0 \\end{align} Therefore, this point is $\\ans{(-5,0,0)}$ in rectangular coordinates. Convert $$(3,4,7)$$ in rectangular coordinates to spherical coordinates. #### Solution Use the formula for $$\\rho$$ first. \\begin{align} \\rho &= \\sqrt{x^2+y^2+z^2} = \\sqrt{74} \\end{align} Then, find $$\\phi$$ by noting that $$\\cos \\phi = \\dfrac{z}{\\rho}$$. $\\phi = \\cos^{-1} \\dfrac{z}{\\rho} = \\cos^{-1} \\dfrac{7}{\\sqrt{74}} \\approx 0.62 \\textrm{ radians}$ Finally, use the fact that $$\\cos \\theta = \\dfrac{x}{\\rho \\sin \\phi}$$ to find $$\\theta$$. $\\theta = \\cos^{-1} \\dfrac{x}{\\rho \\sin \\phi} = \\cos^{-1} \\dfrac{3}{\\sqrt{74}\\sin 0.62} \\approx 0.93 \\textrm{ radians}$ Therefore, this point is $\\ans{\\left(\\sqrt{74},0.93,0.62\\right)}$ in spherical coordinates. ## Summary Rectangular Coordinates Cylindrical Coordinates Spherical Coordinates $$(x,y,z)$$ $$(r,\\theta,z)$$ $$(\\rho,\\theta,\\phi)$$ The formulas below are used to convert between coordinate systems. For my class, there's no need to memorize these formulas, since you'll have a formula sheet on the exams. ### 1. Rectangular $$\\leftrightarrow$$ Cylindrical \\begin{align} x &= r \\cos \\theta\\\\ y &= r \\sin \\theta\\\\ z &= z\\\\ \\\\ r &= \\sqrt{x^2+y^2}\\\\ \\theta &= \\tan^{-1}", "\\begin{align} V&=\\iiint_E1\\,dV=\\int_0^{2\\pi}\\int_0^{\\pi/4}\\int_0^{2\\cos\\phi}\\rho^2\\sin\\phi\\,d\\rho\\,d\\phi\\,d\\theta\\\\ \\ \\\\ &=\\frac{16\\pi}3\\,\\int_0^{\\pi/4}\\cos^3\\phi\\,\\sin\\phi\\,d\\phi =-\\frac{16\\pi}3\\,\\left.\\left(\\frac{\\cos^4\\phi}4 \\right)\\right|_0^{\\pi/4}\\\\ \\ \\\\ &=\\frac{4\\pi}3\\left(1-\\frac1{4} \\right)=\\pi. \\end{align} On the other hand, to work in cylindrical coordinates we have as follows. If we write $s=\\sqrt{x^2+y^2}$, the intersection of the sphere and the cone happens when $s^2+(s-1)^2=1$, with solutions $s=0$ and $s=1$. As we want to be above the cone, we have to choose $s=1$. The equations of the sphere and the cone are, respectively, $z=1+\\sqrt{1-r^2}$ and $z=r$. Then the volume is \\begin{align} V&=\\int_0^{2\\pi}\\int_0^1\\,(1+\\sqrt{1-r^2}-r)\\,r\\,dr\\,d\\theta =2\\pi\\,\\int_0^1(r+r\\sqrt{1-r^2}-r^2)\\,dr\\\\ \\ \\\\ &=2\\pi\\,\\left.\\left(\\frac{r^2}2-\\frac{\\sqrt{1-r^2}}3-\\frac{r^3}3\\right)\\right|_0^1 =2\\pi\\,\\left(\\frac12-0-\\frac13+\\frac13 \\right)=\\pi. \\end{align} • The equation of the sphere is $\\rho=2\\cos\\phi$ if I am not mistaken. – Kuifje Jan 23 '17 at 18:04 • I think you forgot to simplify the $1's$ when expanding $(z-1)^2$. – Kuifje Jan 23 '17 at 18:06 • Yes, you are right. Thanks! – Martin Argerami Jan 23 '17 at 18:11 In spherical coordinates: $$E=\\{(\\rho,\\theta,\\phi)|0\\le \\theta\\le 2\\pi, 0\\le \\phi \\le \\pi/4, 0\\le \\rho \\le 2\\cos \\phi \\}$$ It follows that $$\\boxed{ V= \\iiint_E dV = \\int_0^{2\\pi}\\int_0^{\\pi/4}\\int_0^{2\\cos\\phi}\\rho^2\\sin\\phi \\;d\\rho d\\phi d\\theta = \\pi }$$ Note: to find your bounds for $\\phi$ and $\\rho$: • The upper bound for $\\phi$ is precisely the cone $z=\\sqrt{x^2+y^2}\\; \\Leftrightarrow \\; \\rho \\cos \\phi = \\rho \\sin \\phi \\; \\Leftrightarrow \\; \\phi = \\pi/4$ • The upper bound for $\\rho$ is precisely the sphere $x^2+y^2+(z-1)^2=1\\; \\Leftrightarrow \\;\\rho^2\\sin^2\\phi+(\\rho \\cos\\phi-1)^2=1\\; \\Leftrightarrow", "1 + 1 + 1 = 3$$ The other two cylindrical and spherical, I did the same way. But I used the $$\\nabla\\cdot \\vec{r}$$ for cylindrical and spherical which I referred to in a handbook. Last edited: Jan 9, 2005 7. Jan 9, 2005 ### meteorologist1 Could you show me how to get \"3\" for the cylindrical and spherical? Thanks. 8. Jan 9, 2005 ### Galileo In cylindrical coordinates: $$\\vec r = \\rho \\hat \\rho + z \\hat z$$ and $$\\nabla \\cdot \\vec v(\\rho,\\phi,z) = \\frac{1}{\\rho}\\frac{\\partial}{\\partial \\rho}(\\rho v_{\\rho})+\\frac{1}{\\rho}\\frac{\\partial v_{\\phi}}{\\partial \\phi}+\\frac{\\partial v_z}{\\partial z}$$ For some vector function $v=\\langle v_{\\rho},v_{\\phi},v_z \\rangle$ In spherical coordinates: $$\\vec r = r\\hat r$$ and $$\\nabla \\cdot \\vec v(r,\\theta,\\phi)=\\frac{1}{r^2}\\frac{\\partial}{\\partial r}(r^2v_r)+\\frac{1}{r\\sin \\theta}\\frac{\\partial}{\\partial \\theta}(\\sin \\theta v_{\\theta})+\\frac{1}{r\\sin \\theta}\\frac{\\partial v_{\\phi}}{\\partial \\phi}$$ for some vector function $v=\\langle v_r,v_{\\theta},v_{\\phi} \\rangle$ Both give 3 as well: $$\\nabla \\cdot \\vec r(\\rho,\\phi,z) =\\frac{1}{\\rho}\\frac{\\partial}{\\partial \\rho}(\\rho \\rho)+\\frac{\\partial z}{\\partial z}=\\frac{1}{\\rho}2\\rho+1=3$$ $$\\nabla \\cdot \\vec r(r,\\theta,\\phi)=\\frac{1}{r^2}\\frac{\\partial}{\\partial r}(r^2 r)=\\frac{1}{r^2}3r^2=3$$ You can't use the divergence in cylindrical or spherical coordinates when r=0 though. Last edited: Jan 9, 2005 9. Jan 9, 2005 ### meteorologist1 Thanks very much. It looks like the reason I got the wrong answers is because I got the r vectors wrong in cylindrical and spherical coordinates. I need to get this straight. I would like to ask another question related to these two coordinates: In cylindrical coordinates,", "1\\\\ z= 1, x^2 + y^2 = 1$ Substitute our new coordinate system. $\\rho^2 \\sin^2\\phi\\cos^2\\theta + \\rho^2 \\sin^2\\phi\\sin^2\\theta + \\rho^2 \\cos^2\\phi = 2\\\\ \\rho^2 \\sin^2\\phi\\cos^2\\theta + \\rho^2 \\sin^2\\phi\\sin^2\\theta + \\rho^2 \\cos^2\\phi - 2\\rho\\cos\\phi + 1 = 1\\\\ \\rho\\cos\\phi= 1, \\rho^2 \\sin^2\\phi\\cos^2\\theta + \\rho^2 \\sin^2\\phi\\sin^2\\theta = 1$ Simplify: $\\rho^2 = 2\\\\ \\rho^2 - 2\\rho\\cos\\phi = 0\\\\ \\rho\\cos\\phi= 1, \\rho^2 \\sin^2\\phi = 1$ $\\rho = \\sqrt 2\\\\ \\rho = 2\\cos\\phi\\\\ \\frac {\\rho\\sin \\phi}{\\rho\\cos \\phi} = \\tan \\phi= 1\\\\ \\phi = \\frac {\\pi}{4}$ Now, it may help to sketch these curves. Do it in 2D. Just look at the picture in the xz plane, and that should give you some idea of what is going on for $\\rho$ and $\\phi.$ \\update______ $\\int_0^{2\\pi}\\int_0^{\\frac\\pi4}\\int_{\\sqrt 2}^{2\\cos\\phi} \\rho^2 \\sin\\phi \\;d\\rho\\;d\\phi\\;d\\theta$ • The cylindrical looks fine, but I'm still unsure how you get your bounds for the spherical. – Lax Oct 12 '16 at 1:05", "spherical $$(\\rho,\\phi,\\theta)$$ $$\\to$$ cylindrical coordinates $$(r,\\theta,z)$$ \\shaded{ (\\rho,\\phi,\\theta) \\Rightarrow \\left\\{ \\begin{align*} r &= \\rho\\sin\\phi \\\\ \\theta &= \\theta \\\\ z &= \\rho\\cos\\phi \\end{align*} \\right. } \\nonumber Converting spherical $$(\\rho,\\phi,\\theta)$$ $$\\to$$ Cartesian coordinates $$(x,y,z)$$ \\shaded{ (\\rho,\\phi,\\theta) \\Rightarrow \\left\\{ \\begin{align*} x &= r\\cos\\theta = \\rho\\sin\\phi\\cos\\theta \\\\ y &= r\\sin\\theta = \\rho\\sin\\phi\\sin\\theta \\\\ z &= \\rho\\cos\\phi \\end{align*} \\right. } \\nonumber Converting Cartesian $$(x,y,z)$$ $$\\to$$ spherical coordinates $$(\\rho,\\phi,\\theta)$$ \\shaded{ (x,y,z) \\Rightarrow \\left\\{ \\begin{align*} \\rho &= \\sqrt{r^2+z^2} = \\sqrt{x^2+y^2+z^2} \\\\ \\phi &= \\rm{atan}\\,\\frac{\\sqrt{x^2+y^2}}{z} \\\\ \\theta &= \\rm{atan}\\,\\frac{y}{x}\\\\ \\end{align*} \\right. } \\nonumber Some expressions • $$\\rho=a$$: a sphere of radius $$a$$, centered at the origin. • $$\\phi=\\frac{\\pi}{4}$$: a cone. In cylindrical that would $$z=r$$. • $$\\phi=\\frac{\\pi}{2}$$: the $$xy$$-pane. ### Volume element To find the volume element $$\\Delta V\\approx\\Delta\\rho\\,\\Delta\\phi\\,\\Delta\\theta$$, start with the surface area on a sphere with radius $$a$$. If it is small enough it looks like a rectangle. So, what are the sides? Start with a surface area element $$\\Delta S$$ • To move north-south: you always have to traverse half a circle $$\\Longrightarrow$$ the vertical side of $$\\Delta S$$ is a piece of circle with radius $$a$$ $$\\Rightarrow$$ the length is $$a.\\Delta\\phi$$. • To move east-west: your distance depends on your distance to the $$z$$-axis, called $$r$$ $$\\Longrightarrow$$ the horizontal side of $$\\Delta S$$ is a piece of circle of radius $$r=a\\sin\\phi$$.", "rectangular system are (0, -2\\sqrt{2}, 0) This is a point on the negative y - axis at a distance 2\\sqrt{2} from the origin. ## Rectangular Coordinates to Spherical Coordinates Let us write the rectangular form (1, \\sqrt{3}, 2\\sqrt{3}) as spherical coordinates. We have x = 1, y = \\sqrt{3} and z = 2\\sqrt{3} Spherical coordinates are as follows: x = \\rho sin(\\phi) cos(\\theta) y = \\rho sin(\\phi) sin(\\theta) z = \\rho cos(\\phi) Now, \\rhosin(\\phi) cos(\\theta) = 1 ............(1) \\rho sin(\\phi) sin (\\theta) = \\sqrt{3} ............(2) \\rho cos(\\phi) = 2\\sqrt{3} ............(3) Dividing equation (2) by (1), we get tan (\\theta) = \\sqrt{3} \\theta = \\frac{\\pi}{3} (As both x and y are positive) \\rho ^2 = x^2 + y^2 + z^2 = 1 + (\\sqrt{3})^2 + (2\\sqrt{3})^2 = 1 + 3 + 12 = 16. Hence, \\rho = 4 \\rho cos(\\phi) = 2\\sqrt{3} 4 cos(\\phi) = 2\\sqrt{3} cos(\\phi) = \\frac{\\sqrt{3}}{2} \\phi = \\frac{\\pi}{6} Thus, the required spherical form of the coordinates are (4,$$\\frac{\\pi}{6}$$,$$\\frac{\\pi}{3}$$) ## Unit Vector in Spherical Coordinates In rectangular coordinate system, unit vectors \\hat{i}, \\hat{j}, \\hat{k} are corresponding to the directions of three coordinate axes. In the same manner, we can think of unit vectors \\hat{\\rho }, \\hat{\\phi } and \\hat{\\theta } in spherical coordinates. The position vector of any point P<\\rho, \\phi, \\theta> can be written in", "Quiz from Wednesday, October 26 Problem: Let $\\theta$ represent the usual angle in the $xy$-plane emanating from the origin counter-clockwise from the $x$-axis, and let $\\phi$ represent the angle from the $z$-axis to the $xy$-plane. If $\\rho$ is distance from the origin, find the volume of the object created by rotating the region where $1\\le\\rho\\le 3$ and $\\pi/6\\le\\phi\\le\\pi/2$ in a full circle around the $z$-axis. Solution: In spherical coordinates, we compute as follows. \\begin{align*} \\int_0^{2\\pi}\\int_{\\pi/6}^{\\pi/2}\\int_1^3 \\rho^2\\sin\\phi\\,d\\rho\\,d\\phi\\,d\\theta &= \\int_0^{2\\pi}\\,d\\theta\\int_{\\pi/6}^{\\pi/2}\\sin\\phi\\,d\\phi\\int_1^3\\rho^2\\,d\\rho \\\\ &= 2\\pi\\left(-\\cos\\phi\\right)\\Big|_{\\phi=\\pi/6}^{\\phi=\\pi/2}\\left(\\frac{\\rho^3}{3}\\right)\\Big|_{\\rho=1}^{\\rho=3}\\\\ &= 2\\pi\\cdot\\frac{\\sqrt{3}}{2}\\left(\\frac{27}{3}-\\frac{1}{3}\\right)\\\\ &= \\frac{26\\pi\\sqrt{3}}{3}\\end{align*} Replacement Question: How many states share a border (including corners) with Colorado? Answer: 7: Kansas, Nebraska, Wyoming, Utah, Arizona, New Mexico, and Oklahoma. There are only two states that share more borders with neighboring states than Colorado does: Tennessee and Missouri both share borders with 8 states (including each other).", "$cos^2(\\phi)= sin^2(\\phi)$ or simply $\\phi= \\frac{\\pi}{4}$ since $\\phi$ must be between 0 and $\\pi$. The volume inside the sphere, above the cone, is given by $\\int_{\\theta= 0}^{2\\pi}\\int_{\\phi= 0}^{\\pi/4}\\int_{\\rho= 0}^1 \\rho sin^2(\\phi)d\\rho d\\phi d\\theta$ $= \\left(\\int_{\\theta= 0}^{2\\pi}d\\theta\\right)\\left(\\int_0^{\\pi/4} sin^2(\\phi)d\\phi\\right)\\left(\\int_{\\rho= 0}^1 \\rho d\\rho\\right)$. That is exactly what ojones told you yesterday! 5. Originally Posted by HallsofIvy ojones has already told you how to do it. The sphere, $x^2+ y^2+ z^2= 1$, in spherical coordinates, is $\\rho^2= 1$ and the cone, z^2= x^2+ y^2, in spherical coordinates is $\\rho^2 cos^2(\\phi)= \\rho^2 sin^2(\\phi)$ which reduces to $cos^2(\\phi)= sin^2(\\phi)$ or simply $\\phi= \\frac{\\pi}{4}$ since $\\phi$ must be between 0 and $\\pi$. The volume inside the sphere, above the cone, is given by $\\int_{\\theta= 0}^{2\\pi}\\int_{\\phi= 0}^{\\pi/4}\\int_{\\rho= 0}^1 \\rho sin^2(\\phi)d\\rho d\\phi d\\theta$ $= \\left(\\int_{\\theta= 0}^{2\\pi}d\\theta\\right)\\left(\\int_0^{\\pi/4} sin^2(\\phi)d\\phi\\right)\\left(\\int_{\\rho= 0}^1 \\rho d\\rho\\right)$. That is exactly what ojones told you yesterday! I have a quick question. If I just wanted to find the volume of the sphere above the plane of intersection, how would I go about doing this? I have tried but can't figure it out. Any help would be appreciated. I have no problem solving it the way that was just shown but like to try and solve problems in more than one way. I found the volume of the cone and was just going to add the", "\\tan(\\varphi) = r/z\\\\ \\amp r=\\rho \\sin(\\varphi),\\amp \\amp \\theta = \\theta,\\amp \\amp z=\\rho\\cos(\\varphi) \\end{align*} ###### Example15.7.16.Converting between rectangular and spherical coordinates. Convert the rectangular point $$(2,-2,1)$$ to spherical coordinates, and convert the spherical point $$(6,\\pi/3,\\pi/2)$$ to rectangular and cylindrical coordinates. Solution This rectangular point is the same as used in Example 15.7.4. Using Key Idea 15.7.15, we find $$\\rho = \\sqrt{2^2+(-1)^2+1^2} = 3\\text{.}$$ Using the same logic as in Example 15.7.4, we find $$\\theta = 7\\pi/4\\text{.}$$ Finally, $$\\cos(\\varphi) = 1/3\\text{,}$$ giving $$\\varphi = \\cos^{-1}(1/3) \\approx 1.23\\text{,}$$ or about $$70.53^\\circ\\text{.}$$ Thus the spherical coordinates are approximately $$(3,7\\pi/4,1.23)\\text{.}$$ Converting the spherical point $$(6,\\pi/3,\\pi/2)$$ to rectangular, we have $$x = 6\\sin(\\pi/2)\\cos(\\pi/3) = 3\\text{,}$$ $$y = 6\\sin(\\pi/2)\\sin(\\pi/3) = 3\\sqrt{3}$$ and $$z = 6\\cos(\\pi/2) = 0\\text{.}$$ Thus the rectangular coordinates are $$(3,3\\sqrt{3},0)\\text{.}$$ To convert this spherical point to cylindrical, we have $$r = 6\\sin(\\pi/2) = 6\\text{,}$$ $$\\theta = \\pi/3$$ and $$z = 6\\cos(\\pi/2) =0\\text{,}$$ giving the cylindrical point $$(6,\\pi/3,0)\\text{.}$$ ###### Example15.7.17.Canonical surfaces in spherical coordinates. Describe the surfaces $$\\rho=1\\text{,}$$ $$\\theta = \\pi/3$$ and $$\\varphi = \\pi/6\\text{,}$$ given in spherical coordinates. Solution The equation $$\\rho = 1$$ describes all points in space that are 1 unit away from the origin: this is the sphere of radius 1, centered at the origin. The equation $$\\theta = \\pi/3$$ describes the same surface in spherical coordinates as it", "\\right )}$. (2) \\begin{align} \\quad z^n = r^n (\\cos n\\theta + i \\sin n \\theta) \\end{align} • We want $z^n = w$, i.e.: (3) \\begin{align} \\quad z^n = r^n (\\cos n \\theta + i \\sin n \\theta) = \\rho (\\cos \\phi + i \\sin \\phi ) = w \\end{align} • Hence $r^n = \\rho$ and $n \\theta = \\phi + 2k\\pi$ for some $k \\in \\mathbb{Z}$. We solve for $r$ and $\\theta$. We have that: (4) \\begin{align} \\quad r &= \\rho^{1/n} \\\\ \\quad \\theta &= \\frac{\\phi + 2k\\pi}{n} \\end{align} • Therefore: (5) \\begin{align} \\quad z = \\rho^{1/n} \\left ( (\\cos \\left ( \\frac{\\phi + 2k\\pi}{n} \\right ) + i \\sin \\left ( \\frac{\\phi + 2k\\pi}{n} \\right ) \\right ) \\end{align} • This is satisfied for each $k \\in \\{ 0, 1, ..., n - 1 \\}$, i.e., there are $n$ many $n^{\\mathrm{th}}$ roots to every nonzero complex number. $\\blacksquare$ ## Example 1 Compute the cube roots of $z = i$. We have that $r = \\mid z \\mid = \\mid i \\mid = 1$ and that $\\displaystyle{\\theta = \\arg(z) = \\frac{\\pi}{2}}$. Thus, the cube roots, $z_0$, $z_1$, and $z_2$ of $z$ are: (6) \\begin{align} \\quad z_0 = \\cos \\left ( \\frac{\\pi}{6} \\right ) + i \\sin \\left ( \\frac{\\pi}{6} \\right ) \\end{align} (7) \\begin{align} \\quad z_1 = \\cos", "# Thread: volume enclosed by spheres 1. ## volume enclosed by spheres Find the volume enclosed by the spheres $x^{2}+y^{2}+z^{2}=9$ and $x^{2}+y^{2}+(z-2)^{2}=4$ 2. ## Spherical Coordinates If we use spherical coordinates $(\\rho,\\theta,\\phi)$, the spheres are thus $\\rho=3$ and $\\rho=4\\cos\\theta$. We see our volume is bounded by $\\rho=3$ for $0\\le\\theta<\\arccos\\frac{3}{4}$ and $\\rho=4\\cos\\theta$ for $\\arccos\\frac{3}{4}\\le\\theta\\le\\frac{\\pi}{2}$. Thus, we have: $V=\\int_{0}^{\\arccos\\frac{3}{4}}\\int_{0}^{2\\pi}\\int _{0}^{3}\\rho^2\\sin\\theta\\,d\\rho\\,d\\phi\\,d\\theta+\\i nt_{\\arccos\\frac{3}{4}}^{\\frac{\\pi}{2}}\\int_{0}^{2 \\pi}\\int_{0}^{4\\cos\\theta}\\rho^2\\sin\\theta\\,d\\rho\\ ,d\\phi\\,d\\theta$ $V=\\int_{0}^{\\arccos\\frac{3}{4}}18\\pi\\sin\\theta\\,d\\ theta+\\int_{\\arccos\\frac{3}{4}}^{\\frac{\\pi}{2}}\\fr ac{128\\pi}{3}\\cos^3\\theta\\sin\\theta\\,d\\theta$ $V=\\frac{9\\pi}{2}+\\frac{27\\pi}{8}=\\frac{63}{8}\\pi$ --Kevin C. 3. Cool. That is identical to how I set it up. Just verifying.", "Coordinates 5. Since $\\displaystyle {\\rho}^{2}=x^{2}+y^{2}+z^{2}$ We can sub it into the density function and get $\\displaystyle \\sqrt{1+{\\rho}^{3}}$ 6. I am just a bit confused on how you find the \"phi\" and \"theta\" for this question. The bound is suppose to be 0 to Pi for phi and theta. By the way, that website on spherical coordinates was useful 7. I think your set up should look something like this: $\\displaystyle \\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{2}\\left[{\\rho}^{2}\\sqrt{1+{\\rho}^{3}}sin{\\phi}\\right]d{\\rho}d{\\phi}d{\\theta}$ 8. Originally Posted by galactus I think your set up should look something like this: $\\displaystyle \\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{2}\\left[{\\rho}^{2}\\sqrt{1+{\\rho}^{3}}sin{\\phi}\\right]d{\\rho}d{\\phi}d{\\theta}$ Yep that's it. Is theta 0 to Pi because its half a sphere? Also, why is the phi 0 to Pi also, I can't really picture it. EDIT: Ok, I can see why its Pi now, I am confused from the question. I can't really see what [Math] x^2+y^2+z^2<4 [/tex] and $\\displaystyle 0<y$ is. What do these regions mean? 9. Originally Posted by galactus I think your set up should look something like this: $\\displaystyle \\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{2}\\left[{\\rho}^{2}\\sqrt{1+{\\rho}^{3}}sin{\\phi}\\right]d{\\rho}d{\\phi}d{\\theta}$ Ahh yes. I misread my x- and y-axes. $\\displaystyle 0 \\leq \\theta \\pi$ indeed. 10. Originally Posted by Gotovina7 Yep that's it. Is theta 0 to Pi because its half a sphere? Also, why is the phi 0 to Pi also, I can't really picture it. EDIT: Ok, I can see why its Pi", "Calculate the $$\\theta$$ angle: $\\cos\\theta = \\frac{z}{\\rho} = \\frac{1}{2}, \\Rightarrow \\theta = \\arccos\\frac{1}{2} = \\frac{\\pi}{3}.$ So the spherical coordinates of the point $$M$$ are $\\rho = 2,\\;\\varphi = \\arccos\\sqrt{\\frac{2}{3}},\\;\\theta = \\frac{\\pi}{3}.$", "\\, \\bigg|_{-\\pi / 2}^{\\pi/2} \\right) \\right]$$ $$\\frac{2 a^4 \\rho}{3} \\left[ \\left( \\frac{\\pi}{4} + \\frac{\\pi}{4} \\right) + \\left( \\frac{\\pi}{8} + \\frac{\\pi}{8} \\right) \\right] = \\frac{2 a^4 \\rho}{3} \\left( \\frac{ 3 \\pi}{4} \\right) = \\frac{\\rho \\pi a^4}{2}$$ Using Polar Coordinates $$\\int _{0}^{2 \\pi} \\int _{0}^{a} \\, \\rho r^2 r \\, dr \\, d\\theta$$ $$\\rho \\int _{0}^{2 \\pi} \\frac{r^4}{4} \\, \\bigg|_0^a \\, d\\theta$$ $$\\frac{\\rho a^4}{4} \\cdot 2 \\pi = \\frac{\\rho \\pi a^4}{2}$$", "\\left(\\sqrt{\\, 1 - \\frac{2M}{h}\\,} \\right)\\, \\rho \\,+ \\, h\\\\ &\\phi = \\frac{u}{h}\\\\ &\\theta = \\frac{w}{h} + \\frac{\\pi}{2} \\end{align} and the the inverse relations are \\begin{align} &\\tau = \\left({\\sqrt{\\, 1 - \\frac{2M}{h}\\,}}\\right)\\, t \\\\ &\\rho = \\frac{r \\, - \\, h}{\\sqrt{1 - \\frac{2M}{h}}} \\\\ &u = h\\,\\phi\\\\ &w = h\\,\\theta \\, - \\, \\frac{h\\,\\pi}{2} \\end{align} The world line $$\\gamma(\\tau)$$ in the new coordinates is $$\\gamma(\\tau) = \\big(\\tau,\\, 0, \\, 0, 0 \\big)$$ The light-like radial geodesic in the original $$(t, r, \\phi, \\theta)$$ coordinates is $$g(t) = \\Big(\\,t, \\, r(t), \\, 0, \\, \\frac{\\pi}{2}\\,\\Big)$$ where $$\\frac{dr}{dt} = 1 - \\frac{2M}{r}$$ In the new coordinates it is $$g(\\tau) = \\Big(\\,\\tau, \\, \\rho(\\tau), \\, 0, \\, 0\\,\\Big)$$ where we can find the function $$\\rho(\\tau)$$ by changing the coordinates in the differential equation for $$r(t)$$, i.e. \\begin{align} \\frac{d\\rho}{d\\tau} &= \\frac{\\frac{dr}{\\sqrt{1-\\frac{2M}{h}}}}{\\sqrt{1-\\frac{2M}{h}}\\, dt} = \\left(1 - \\frac{2M}{h}\\right)^{-1}\\, \\frac{dr}{dt} = \\left(1 - \\frac{2M}{h}\\right)^{-1}\\, \\left(1 - \\frac{2M}{r}\\right)\\\\ & = \\left(1 - \\frac{2M}{h}\\right)^{-1}\\, \\left(1 - \\frac{2M}{h + \\rho \\, \\sqrt{1-\\frac{2M}{h}}}\\right) \\end{align} Observe that in the new coordinates the Schwarzschild metric is the Minkowski metric for each point on the world-line $$\\gamma(\\tau)$$. Now, when the light-like geodesic $$g(\\tau)$$ intersects the world-line $$\\gamma(\\tau)$$, i.e. the photon reaches the observer, then $$\\rho = 0$$ and from the differential equation for $$\\rho(\\tau)$$ we see that $$\\frac{d\\rho}{d\\tau} = 1$$ i.e. in this", "5(x^2+ y^2)= 5r^2$$. In Cylindrical coordinates that is $$z= 5r^2$$. It is particularly simple because, geometrically, the figure is a cylinder! In spherical coordinates $$z= \\rho cos(\\phi)= 5(\\rho^2 cos^2(\\theta)sin^2(\\phi)+ \\rho^2 sin^(\\theta) sin^2(\\phi))= 5\\rho^2 sin^2(\\phi)(cos^2(\\theta)+ sin^2(\\theta))= 5\\rho^2 sin^2(\\phi)$$. $$\\rho cos^(\\phi)= 5\\rho^2 sin^2(\\phi)$$.$$\\rho= \\frac{cos(\\phi)}{5 sin^2(\\phi)}$$. Guest", "point (2,1,2) in Cartesian coordinates, we find that $\\rho=\\sqrt{x^2+y^2+z^2}=\\sqrt{2^2+1^2+2^2}=\\sqrt{4+1+4}=\\sqrt{9}=3$ $\\tan\\phi=\\frac{\\sqrt{x^2+y^2}}{z}=\\frac{\\sqrt{2^1+1^2}}{2}=\\frac{\\sqrt{5}}{2}$ $\\rightarrow \\phi=\\tan^{-1}\\left(\\sqrt{5}/2\\right)=0.841$ radians $\\tan \\theta = y/x=1/2$ $\\rightarrow \\theta=\\tan^{-1}=0.464$ radians So we have the point (3,0.841,0.464) in spherical coordinates The transformation from spherical coordinates to Cartesian coordinates is given by $x=\\rho \\sin \\phi \\cos \\theta$ $y=\\rho \\sin \\phi \\sin \\theta$ $z= \\rho \\cos \\phi$. We can convert our previous result back to Cartesian coordinates: $x=\\rho \\sin \\phi \\cos \\theta=3\\sin(0.841)\\cos(0.464)\\approx 2$ $y=\\rho \\sin \\phi \\sin \\theta =3\\sin(0.841)\\sin(0.464)\\approx 1$ $z= \\rho \\cos \\phi =3 cos(0.841)\\approx 2$ Again, we retrieve our original ordered triple (2,1,2) by rounding. We can also write the transformation from cylindrical coordinates to spherical coordinates: $\\rho=\\sqrt{r^2+z^2}$ $\\tan \\phi = z/r$ $\\theta = \\theta$. Beginning with the point (3,\\pi/3,4) in cylindrical coordinates, we see that $\\rho=\\sqrt{r^2+z^2}=\\sqrt{3^2+4^2}=\\sqrt{25}=5$ $\\tan \\phi = z/r=4/3$ $\\rightarrow \\phi=\\tan^{-1}(.75)=0.644$ $\\theta = \\pi/3$. So the point in spherical coordinates is (5,\\pi/3,0.644). Finally, the transformation from spherical to cylindrical coordinates is given by $r=\\rho \\sin \\phi$ $\\theta =\\theta$ $z=\\rho \\cos \\phi$. For one last example, we will take our previous result and transform it back to cylindrical coordinates $r=\\rho \\sin \\theta=5 \\sin (0.644)\\approx 3$ $\\theta =\\pi/3$ $z=\\rho \\cos \\theta5 \\cos (0.644)\\approx 4$ Therefore the point in cylindrical coordinates is (3, \\pi/3, 4) as expected. Interactive Demonstration Change of Coordinate Systems Applet • add examples of transformations between", "quadrant). Lastly, the bounds on {eq}\\rho {/eq} tell us that the sphere has radius 3. So we are dealing with a slice/wedge in the first octant. A sketch is below. The volume of the wedge is {eq}\\begin{align*} \\int^{\\pi/4}_0 \\int^\\pi_0 \\int^3_0 \\rho^2 \\sin \\phi\\ d \\rho\\ d \\theta\\ d \\phi &= \\int_0^{\\pi} d \\theta \\int_0^{\\pi/4} \\sin \\phi\\ d\\phi \\int_0^3 \\rho^2\\ d\\rho \\\\ &= \\pi \\left [ -\\cos \\phi \\right ]_0^{\\pi/4} \\left [ \\frac13\\rho^3 \\right ]_0^3 \\\\ &= \\frac\\pi3 \\left( - \\frac{\\sqrt2}{2} + 1 \\right)(3^3) \\\\ &= \\frac{9\\pi}{2} \\left( 2-\\sqrt2 \\right) \\\\ &\\approx 8.2814 \\end{align*} {/eq} #### Learn more about this topic: Cylindrical & Spherical Coordinates: Definition, Equations & Examples from GRE Math: Study Guide & Test Prep Chapter 13 / Lesson 10 14K", "$r = \\sqrt{2} \\sqrt{2 \\pm \\sqrt{3}}$. Hence the minimum value of $|z|$ is $\\sqrt{2} \\sqrt{2 - \\sqrt{3}}$ and the maximum value if $\\sqrt{2} \\sqrt{2 + \\sqrt{3}}$. Let $z=re^{i\\theta}$, then your last step can be written as $$r^2+2\\cos (2\\theta)+\\frac{1}{r^2}=4.$$ From this we can get \\begin{align*} \\left(r+\\frac{1}{r}\\right)^2 & = 6-2\\cos (2\\theta) \\end{align*} Thus $2 \\leq \\left(r+\\frac{1}{r}\\right) \\leq 2\\sqrt{2}.$ Now try to find the range of $r$.", "= 3\\sqrt 2$ Now all that we need is the range for $$\\varphi$$. There are two ways to get this. One is from where the cone and the sphere intersect. Plugging in the equation for the cone into the sphere gives, \\begin{align*}{\\left( {\\sqrt {{x^2} + {y^2}} } \\right)^2} + {z^2} & = 18\\\\ {z^2} + {z^2} & = 18\\\\ {z^2} & = 9\\\\ z & = 3\\end{align*} Note that we can assume $$z$$ is positive here since we know that we have the upper half of the cone and/or sphere. Finally, plug this into the conversion for $$z$$ and take advantage of the fact that we know that $$\\rho = 3\\sqrt 2$$ since we are intersecting on the sphere. This gives, \\begin{align*}\\rho \\cos \\varphi & = 3\\\\ 3\\sqrt 2 \\cos \\varphi & = 3\\\\ \\cos \\varphi = \\frac{1}{{\\sqrt 2 }} & = \\frac{{\\sqrt 2 }}{2}\\hspace{0.5in} \\Rightarrow \\hspace{0.5in}\\varphi = \\frac{\\pi }{4}\\end{align*} So, it looks like we have the following range, $0 \\le \\varphi \\le \\frac{\\pi }{4}$ The other way to get this range is from the cone by itself. By first converting the equation into cylindrical coordinates and then into spherical coordinates we get the following, \\begin{align*}z& = r\\\\ \\rho \\cos \\varphi & = \\rho \\sin \\varphi \\\\ 1 & = \\tan \\varphi \\hspace{0.5in} \\Rightarrow \\hspace{0.5in}\\varphi = \\frac{\\pi }{4}\\end{align*}" ]

[ "coordinates to rectangular coordinates. #### Solution Use the formulas, noting that $$\\rho=5$$, $$\\theta=\\pi$$, and $$\\phi=\\dfrac{\\pi}{2}$$. \\begin{align} x &= \\rho \\cos \\theta \\sin \\phi = 5 (\\cos \\pi) \\left(\\sin \\dfrac{\\pi}{2}\\right) = 5(-1)(1) = -5\\\\ y &= \\rho \\sin \\theta \\sin \\phi = 5 (\\sin \\pi) \\left(\\sin \\dfrac{\\pi}{2}\\right) = 5(0)(1) = 0\\\\ z &= \\rho \\cos \\phi = 5 \\cos \\dfrac{\\pi}{2} = 5(0) = 0 \\end{align} Therefore, this point is $\\ans{(-5,0,0)}$ in rectangular coordinates. Convert $$(3,4,7)$$ in rectangular coordinates to spherical coordinates. #### Solution Use the formula for $$\\rho$$ first. \\begin{align} \\rho &= \\sqrt{x^2+y^2+z^2} = \\sqrt{74} \\end{align} Then, find $$\\phi$$ by noting that $$\\cos \\phi = \\dfrac{z}{\\rho}$$. $\\phi = \\cos^{-1} \\dfrac{z}{\\rho} = \\cos^{-1} \\dfrac{7}{\\sqrt{74}} \\approx 0.62 \\textrm{ radians}$ Finally, use the fact that $$\\cos \\theta = \\dfrac{x}{\\rho \\sin \\phi}$$ to find $$\\theta$$. $\\theta = \\cos^{-1} \\dfrac{x}{\\rho \\sin \\phi} = \\cos^{-1} \\dfrac{3}{\\sqrt{74}\\sin 0.62} \\approx 0.93 \\textrm{ radians}$ Therefore, this point is $\\ans{\\left(\\sqrt{74},0.93,0.62\\right)}$ in spherical coordinates. ## Summary Rectangular Coordinates Cylindrical Coordinates Spherical Coordinates $$(x,y,z)$$ $$(r,\\theta,z)$$ $$(\\rho,\\theta,\\phi)$$ The formulas below are used to convert between coordinate systems. For my class, there's no need to memorize these formulas, since you'll have a formula sheet on the exams. ### 1. Rectangular $$\\leftrightarrow$$ Cylindrical \\begin{align} x &= r \\cos \\theta\\\\ y &= r \\sin \\theta\\\\ z &= z\\\\ \\\\ r &= \\sqrt{x^2+y^2}\\\\ \\theta &= \\tan^{-1}", "\\tan(\\varphi) = r/z\\\\ \\amp r=\\rho \\sin(\\varphi),\\amp \\amp \\theta = \\theta,\\amp \\amp z=\\rho\\cos(\\varphi) \\end{align*} ###### Example15.7.16.Converting between rectangular and spherical coordinates. Convert the rectangular point $$(2,-2,1)$$ to spherical coordinates, and convert the spherical point $$(6,\\pi/3,\\pi/2)$$ to rectangular and cylindrical coordinates. Solution This rectangular point is the same as used in Example 15.7.4. Using Key Idea 15.7.15, we find $$\\rho = \\sqrt{2^2+(-1)^2+1^2} = 3\\text{.}$$ Using the same logic as in Example 15.7.4, we find $$\\theta = 7\\pi/4\\text{.}$$ Finally, $$\\cos(\\varphi) = 1/3\\text{,}$$ giving $$\\varphi = \\cos^{-1}(1/3) \\approx 1.23\\text{,}$$ or about $$70.53^\\circ\\text{.}$$ Thus the spherical coordinates are approximately $$(3,7\\pi/4,1.23)\\text{.}$$ Converting the spherical point $$(6,\\pi/3,\\pi/2)$$ to rectangular, we have $$x = 6\\sin(\\pi/2)\\cos(\\pi/3) = 3\\text{,}$$ $$y = 6\\sin(\\pi/2)\\sin(\\pi/3) = 3\\sqrt{3}$$ and $$z = 6\\cos(\\pi/2) = 0\\text{.}$$ Thus the rectangular coordinates are $$(3,3\\sqrt{3},0)\\text{.}$$ To convert this spherical point to cylindrical, we have $$r = 6\\sin(\\pi/2) = 6\\text{,}$$ $$\\theta = \\pi/3$$ and $$z = 6\\cos(\\pi/2) =0\\text{,}$$ giving the cylindrical point $$(6,\\pi/3,0)\\text{.}$$ ###### Example15.7.17.Canonical surfaces in spherical coordinates. Describe the surfaces $$\\rho=1\\text{,}$$ $$\\theta = \\pi/3$$ and $$\\varphi = \\pi/6\\text{,}$$ given in spherical coordinates. Solution The equation $$\\rho = 1$$ describes all points in space that are 1 unit away from the origin: this is the sphere of radius 1, centered at the origin. The equation $$\\theta = \\pi/3$$ describes the same surface in spherical coordinates as it", "\\begin{align} V&=\\iiint_E1\\,dV=\\int_0^{2\\pi}\\int_0^{\\pi/4}\\int_0^{2\\cos\\phi}\\rho^2\\sin\\phi\\,d\\rho\\,d\\phi\\,d\\theta\\\\ \\ \\\\ &=\\frac{16\\pi}3\\,\\int_0^{\\pi/4}\\cos^3\\phi\\,\\sin\\phi\\,d\\phi =-\\frac{16\\pi}3\\,\\left.\\left(\\frac{\\cos^4\\phi}4 \\right)\\right|_0^{\\pi/4}\\\\ \\ \\\\ &=\\frac{4\\pi}3\\left(1-\\frac1{4} \\right)=\\pi. \\end{align} On the other hand, to work in cylindrical coordinates we have as follows. If we write $s=\\sqrt{x^2+y^2}$, the intersection of the sphere and the cone happens when $s^2+(s-1)^2=1$, with solutions $s=0$ and $s=1$. As we want to be above the cone, we have to choose $s=1$. The equations of the sphere and the cone are, respectively, $z=1+\\sqrt{1-r^2}$ and $z=r$. Then the volume is \\begin{align} V&=\\int_0^{2\\pi}\\int_0^1\\,(1+\\sqrt{1-r^2}-r)\\,r\\,dr\\,d\\theta =2\\pi\\,\\int_0^1(r+r\\sqrt{1-r^2}-r^2)\\,dr\\\\ \\ \\\\ &=2\\pi\\,\\left.\\left(\\frac{r^2}2-\\frac{\\sqrt{1-r^2}}3-\\frac{r^3}3\\right)\\right|_0^1 =2\\pi\\,\\left(\\frac12-0-\\frac13+\\frac13 \\right)=\\pi. \\end{align} • The equation of the sphere is $\\rho=2\\cos\\phi$ if I am not mistaken. – Kuifje Jan 23 '17 at 18:04 • I think you forgot to simplify the $1's$ when expanding $(z-1)^2$. – Kuifje Jan 23 '17 at 18:06 • Yes, you are right. Thanks! – Martin Argerami Jan 23 '17 at 18:11 In spherical coordinates: $$E=\\{(\\rho,\\theta,\\phi)|0\\le \\theta\\le 2\\pi, 0\\le \\phi \\le \\pi/4, 0\\le \\rho \\le 2\\cos \\phi \\}$$ It follows that $$\\boxed{ V= \\iiint_E dV = \\int_0^{2\\pi}\\int_0^{\\pi/4}\\int_0^{2\\cos\\phi}\\rho^2\\sin\\phi \\;d\\rho d\\phi d\\theta = \\pi }$$ Note: to find your bounds for $\\phi$ and $\\rho$: • The upper bound for $\\phi$ is precisely the cone $z=\\sqrt{x^2+y^2}\\; \\Leftrightarrow \\; \\rho \\cos \\phi = \\rho \\sin \\phi \\; \\Leftrightarrow \\; \\phi = \\pi/4$ • The upper bound for $\\rho$ is precisely the sphere $x^2+y^2+(z-1)^2=1\\; \\Leftrightarrow \\;\\rho^2\\sin^2\\phi+(\\rho \\cos\\phi-1)^2=1\\; \\Leftrightarrow", "2. Convert the point $$\\left( { - 1,1, - \\sqrt 2 } \\right)$$ from Cartesian to spherical coordinates. Show All Solutions Hide All Solutions a Convert the point $$\\displaystyle \\left( {\\sqrt 6 ,\\frac{\\pi }{4},\\sqrt 2 } \\right)$$ from cylindrical to spherical coordinates. Show Solution We’ll start by acknowledging that $$\\theta$$ is the same in both coordinate systems and so we don’t need to do anything with that. Next, let’s find$$\\rho$$. $\\rho = \\sqrt {{r^2} + {z^2}} = \\sqrt {6 + 2} = \\sqrt 8 = 2\\sqrt 2$ Finally, let’s get $$\\varphi$$. To do this we can use either the conversion for $$r$$ or $$z$$. We’ll use the conversion for $$z$$. $z = \\rho \\cos \\varphi \\hspace{0.25in} \\Rightarrow \\hspace{0.25in}\\cos \\varphi = \\frac{z}{\\rho } = \\frac{{\\sqrt 2 }}{{2\\sqrt 2 }}\\hspace{0.25in} \\Rightarrow \\hspace{0.25in}\\varphi = {\\cos ^{ - 1}}\\left( {\\frac{1}{2}} \\right) = \\frac{\\pi }{3}$ Notice that there are many possible values of $$\\varphi$$ that will give $$\\cos \\varphi = \\frac{1}{2}$$, however, we have restricted $$\\varphi$$ to the range $$0 \\le \\varphi \\le \\pi$$ and so this is the only possible value in that range. So, the spherical coordinates of this point will are $$\\left( {2\\sqrt 2 ,\\frac{\\pi }{4},\\frac{\\pi }{3}} \\right)$$. b Convert the point $$\\left( { - 1,1, - \\sqrt 2 } \\right)$$ from Cartesian to spherical coordinates. Show Solution The first", "1 + 1 + 1 = 3$$ The other two cylindrical and spherical, I did the same way. But I used the $$\\nabla\\cdot \\vec{r}$$ for cylindrical and spherical which I referred to in a handbook. Last edited: Jan 9, 2005 7. Jan 9, 2005 ### meteorologist1 Could you show me how to get \"3\" for the cylindrical and spherical? Thanks. 8. Jan 9, 2005 ### Galileo In cylindrical coordinates: $$\\vec r = \\rho \\hat \\rho + z \\hat z$$ and $$\\nabla \\cdot \\vec v(\\rho,\\phi,z) = \\frac{1}{\\rho}\\frac{\\partial}{\\partial \\rho}(\\rho v_{\\rho})+\\frac{1}{\\rho}\\frac{\\partial v_{\\phi}}{\\partial \\phi}+\\frac{\\partial v_z}{\\partial z}$$ For some vector function $v=\\langle v_{\\rho},v_{\\phi},v_z \\rangle$ In spherical coordinates: $$\\vec r = r\\hat r$$ and $$\\nabla \\cdot \\vec v(r,\\theta,\\phi)=\\frac{1}{r^2}\\frac{\\partial}{\\partial r}(r^2v_r)+\\frac{1}{r\\sin \\theta}\\frac{\\partial}{\\partial \\theta}(\\sin \\theta v_{\\theta})+\\frac{1}{r\\sin \\theta}\\frac{\\partial v_{\\phi}}{\\partial \\phi}$$ for some vector function $v=\\langle v_r,v_{\\theta},v_{\\phi} \\rangle$ Both give 3 as well: $$\\nabla \\cdot \\vec r(\\rho,\\phi,z) =\\frac{1}{\\rho}\\frac{\\partial}{\\partial \\rho}(\\rho \\rho)+\\frac{\\partial z}{\\partial z}=\\frac{1}{\\rho}2\\rho+1=3$$ $$\\nabla \\cdot \\vec r(r,\\theta,\\phi)=\\frac{1}{r^2}\\frac{\\partial}{\\partial r}(r^2 r)=\\frac{1}{r^2}3r^2=3$$ You can't use the divergence in cylindrical or spherical coordinates when r=0 though. Last edited: Jan 9, 2005 9. Jan 9, 2005 ### meteorologist1 Thanks very much. It looks like the reason I got the wrong answers is because I got the r vectors wrong in cylindrical and spherical coordinates. I need to get this straight. I would like to ask another question related to these two coordinates: In cylindrical coordinates,", "point (2,1,2) in Cartesian coordinates, we find that $\\rho=\\sqrt{x^2+y^2+z^2}=\\sqrt{2^2+1^2+2^2}=\\sqrt{4+1+4}=\\sqrt{9}=3$ $\\tan\\phi=\\frac{\\sqrt{x^2+y^2}}{z}=\\frac{\\sqrt{2^1+1^2}}{2}=\\frac{\\sqrt{5}}{2}$ $\\rightarrow \\phi=\\tan^{-1}\\left(\\sqrt{5}/2\\right)=0.841$ radians $\\tan \\theta = y/x=1/2$ $\\rightarrow \\theta=\\tan^{-1}=0.464$ radians So we have the point (3,0.841,0.464) in spherical coordinates The transformation from spherical coordinates to Cartesian coordinates is given by $x=\\rho \\sin \\phi \\cos \\theta$ $y=\\rho \\sin \\phi \\sin \\theta$ $z= \\rho \\cos \\phi$. We can convert our previous result back to Cartesian coordinates: $x=\\rho \\sin \\phi \\cos \\theta=3\\sin(0.841)\\cos(0.464)\\approx 2$ $y=\\rho \\sin \\phi \\sin \\theta =3\\sin(0.841)\\sin(0.464)\\approx 1$ $z= \\rho \\cos \\phi =3 cos(0.841)\\approx 2$ Again, we retrieve our original ordered triple (2,1,2) by rounding. We can also write the transformation from cylindrical coordinates to spherical coordinates: $\\rho=\\sqrt{r^2+z^2}$ $\\tan \\phi = z/r$ $\\theta = \\theta$. Beginning with the point (3,\\pi/3,4) in cylindrical coordinates, we see that $\\rho=\\sqrt{r^2+z^2}=\\sqrt{3^2+4^2}=\\sqrt{25}=5$ $\\tan \\phi = z/r=4/3$ $\\rightarrow \\phi=\\tan^{-1}(.75)=0.644$ $\\theta = \\pi/3$. So the point in spherical coordinates is (5,\\pi/3,0.644). Finally, the transformation from spherical to cylindrical coordinates is given by $r=\\rho \\sin \\phi$ $\\theta =\\theta$ $z=\\rho \\cos \\phi$. For one last example, we will take our previous result and transform it back to cylindrical coordinates $r=\\rho \\sin \\theta=5 \\sin (0.644)\\approx 3$ $\\theta =\\pi/3$ $z=\\rho \\cos \\theta5 \\cos (0.644)\\approx 4$ Therefore the point in cylindrical coordinates is (3, \\pi/3, 4) as expected. Interactive Demonstration Change of Coordinate Systems Applet • add examples of transformations between", "rectangular system are (0, -2\\sqrt{2}, 0) This is a point on the negative y - axis at a distance 2\\sqrt{2} from the origin. ## Rectangular Coordinates to Spherical Coordinates Let us write the rectangular form (1, \\sqrt{3}, 2\\sqrt{3}) as spherical coordinates. We have x = 1, y = \\sqrt{3} and z = 2\\sqrt{3} Spherical coordinates are as follows: x = \\rho sin(\\phi) cos(\\theta) y = \\rho sin(\\phi) sin(\\theta) z = \\rho cos(\\phi) Now, \\rhosin(\\phi) cos(\\theta) = 1 ............(1) \\rho sin(\\phi) sin (\\theta) = \\sqrt{3} ............(2) \\rho cos(\\phi) = 2\\sqrt{3} ............(3) Dividing equation (2) by (1), we get tan (\\theta) = \\sqrt{3} \\theta = \\frac{\\pi}{3} (As both x and y are positive) \\rho ^2 = x^2 + y^2 + z^2 = 1 + (\\sqrt{3})^2 + (2\\sqrt{3})^2 = 1 + 3 + 12 = 16. Hence, \\rho = 4 \\rho cos(\\phi) = 2\\sqrt{3} 4 cos(\\phi) = 2\\sqrt{3} cos(\\phi) = \\frac{\\sqrt{3}}{2} \\phi = \\frac{\\pi}{6} Thus, the required spherical form of the coordinates are (4,$$\\frac{\\pi}{6}$$,$$\\frac{\\pi}{3}$$) ## Unit Vector in Spherical Coordinates In rectangular coordinate system, unit vectors \\hat{i}, \\hat{j}, \\hat{k} are corresponding to the directions of three coordinate axes. In the same manner, we can think of unit vectors \\hat{\\rho }, \\hat{\\phi } and \\hat{\\theta } in spherical coordinates. The position vector of any point P<\\rho, \\phi, \\theta> can be written in", "= 3\\sqrt 2$ Now all that we need is the range for $$\\varphi$$. There are two ways to get this. One is from where the cone and the sphere intersect. Plugging in the equation for the cone into the sphere gives, \\begin{align*}{\\left( {\\sqrt {{x^2} + {y^2}} } \\right)^2} + {z^2} & = 18\\\\ {z^2} + {z^2} & = 18\\\\ {z^2} & = 9\\\\ z & = 3\\end{align*} Note that we can assume $$z$$ is positive here since we know that we have the upper half of the cone and/or sphere. Finally, plug this into the conversion for $$z$$ and take advantage of the fact that we know that $$\\rho = 3\\sqrt 2$$ since we are intersecting on the sphere. This gives, \\begin{align*}\\rho \\cos \\varphi & = 3\\\\ 3\\sqrt 2 \\cos \\varphi & = 3\\\\ \\cos \\varphi = \\frac{1}{{\\sqrt 2 }} & = \\frac{{\\sqrt 2 }}{2}\\hspace{0.5in} \\Rightarrow \\hspace{0.5in}\\varphi = \\frac{\\pi }{4}\\end{align*} So, it looks like we have the following range, $0 \\le \\varphi \\le \\frac{\\pi }{4}$ The other way to get this range is from the cone by itself. By first converting the equation into cylindrical coordinates and then into spherical coordinates we get the following, \\begin{align*}z& = r\\\\ \\rho \\cos \\varphi & = \\rho \\sin \\varphi \\\\ 1 & = \\tan \\varphi \\hspace{0.5in} \\Rightarrow \\hspace{0.5in}\\varphi = \\frac{\\pi }{4}\\end{align*}", "Calculate the $$\\theta$$ angle: $\\cos\\theta = \\frac{z}{\\rho} = \\frac{1}{2}, \\Rightarrow \\theta = \\arccos\\frac{1}{2} = \\frac{\\pi}{3}.$ So the spherical coordinates of the point $$M$$ are $\\rho = 2,\\;\\varphi = \\arccos\\sqrt{\\frac{2}{3}},\\;\\theta = \\frac{\\pi}{3}.$", "is $$\\frac{3π}{4}$$. There are actually two ways to identify $$φ$$. We can use the equation $$φ=arccos(\\frac{z}{\\sqrt{x^2+y^2+z^2}})$$. A more simple approach, however, is to use equation $$z=ρcosφ.$$ We know that $$z=\\sqrt{6}$$ and $$ρ=2\\sqrt{2}$$, so $$\\sqrt{6}=2\\sqrt{2}cosφ,$$ so cos $$φ=\\frac{\\sqrt{6}}{2\\sqrt{2}}=\\frac{\\sqrt{3}}{2}$$ and therefore $$φ=\\frac{π}{6}$$. The spherical coordinates of the point are $$(2\\sqrt{2},\\frac{3π}{4},\\frac{π}{6}).$$ To find the cylindrical coordinates for the point, we need only find r: $$r=ρsinφ=2\\sqrt{2}sin(\\frac{π}{6})=\\sqrt{2}.$$ The cylindrical coordinates for the point are $$(\\sqrt{2},\\frac{3π}{4},\\sqrt{6})$$. Example $$\\PageIndex{6}$$: Identifying Surfaces in the Spherical Coordinate System Describe the surfaces with the given spherical equations. 1. $$θ=\\frac{π}{3}$$ 2. $$φ=\\frac{5π}{6}$$ 3. $$ρ=6$$ 4. $$ρ=sinθsinφ$$ Solution a. The variable $$θ$$ represents the measure of the same angle in both the cylindrical and spherical coordinate systems. Points with coordinates $$(ρ,\\frac{π}{3},φ)$$ lie on the plane that forms angle $$θ=\\frac{π}{3}$$ with the positive x-axis. Because $$ρ>0$$, the surface described by equation $$θ=\\frac{π}{3}$$ is the half-plane shown in Figure. Figure 13: The surface described by equation $$θ=\\frac{π}{3}$$ is a half-plane. b. Equation $$φ=\\frac{5π}{6}$$ describes all points in the spherical coordinate system that lie on a line from the origin forming an angle measuring $$\\frac{5π}{6}$$ rad with the positive z-axis. These points form a half-cone (Figure). Because there is only one value for $$φ$$ that is measured from the positive z-axis, we do not get the full cone (with two pieces). Figure 14:", "\\begin{align*}{r^2} & = {x^2} + {y^2}\\hspace{0.75in} r = \\sqrt {{x^2} + {y^2}} \\\\ \\theta & = {\\tan ^{ - 1}}\\left( {\\frac{y}{x}} \\right)\\end{align*} Let’s work a quick example. Example 1 Convert each of the following points into the given coordinate system. 1. Convert $$\\left( { - 4,\\frac{{2\\pi }}{3}} \\right)$$ into Cartesian coordinates. 2. Convert $$\\left( { - 1,-1} \\right)$$ into polar coordinates. Show All Solutions Hide All Solutions a Convert $$\\left( { - 4,\\frac{{2\\pi }}{3}} \\right)$$ into Cartesian coordinates. Show Solution This conversion is easy enough. All we need to do is plug the points into the formulas. \\begin{align*}x & = - 4\\cos \\left( {\\frac{{2\\pi }}{3}} \\right) = - 4\\left( { - \\frac{1}{2}} \\right) = 2\\\\ y & = - 4\\sin \\left( {\\frac{{2\\pi }}{3}} \\right) = - 4\\left( {\\frac{{\\sqrt 3 }}{2}} \\right) = - 2\\sqrt 3 \\end{align*} So, in Cartesian coordinates this point is $$\\left( {2, - 2\\sqrt 3 } \\right)$$. b Convert $$\\left( { - 1,-1} \\right)$$ into polar coordinates. Show Solution Let’s first get $$r$$. $r = \\sqrt {{{\\left( { - 1} \\right)}^2} + {{\\left( { - 1} \\right)}^2}} = \\sqrt 2$ Now, let’s get $$\\theta$$. $\\theta = {\\tan ^{ - 1}}\\left( {\\frac{{ - 1}}{{ - 1}}} \\right) = {\\tan ^{ - 1}}\\left( 1 \\right) = \\frac{\\pi }{4}$ This is not the correct angle however. This value", "spherical $$(\\rho,\\phi,\\theta)$$ $$\\to$$ cylindrical coordinates $$(r,\\theta,z)$$ \\shaded{ (\\rho,\\phi,\\theta) \\Rightarrow \\left\\{ \\begin{align*} r &= \\rho\\sin\\phi \\\\ \\theta &= \\theta \\\\ z &= \\rho\\cos\\phi \\end{align*} \\right. } \\nonumber Converting spherical $$(\\rho,\\phi,\\theta)$$ $$\\to$$ Cartesian coordinates $$(x,y,z)$$ \\shaded{ (\\rho,\\phi,\\theta) \\Rightarrow \\left\\{ \\begin{align*} x &= r\\cos\\theta = \\rho\\sin\\phi\\cos\\theta \\\\ y &= r\\sin\\theta = \\rho\\sin\\phi\\sin\\theta \\\\ z &= \\rho\\cos\\phi \\end{align*} \\right. } \\nonumber Converting Cartesian $$(x,y,z)$$ $$\\to$$ spherical coordinates $$(\\rho,\\phi,\\theta)$$ \\shaded{ (x,y,z) \\Rightarrow \\left\\{ \\begin{align*} \\rho &= \\sqrt{r^2+z^2} = \\sqrt{x^2+y^2+z^2} \\\\ \\phi &= \\rm{atan}\\,\\frac{\\sqrt{x^2+y^2}}{z} \\\\ \\theta &= \\rm{atan}\\,\\frac{y}{x}\\\\ \\end{align*} \\right. } \\nonumber Some expressions • $$\\rho=a$$: a sphere of radius $$a$$, centered at the origin. • $$\\phi=\\frac{\\pi}{4}$$: a cone. In cylindrical that would $$z=r$$. • $$\\phi=\\frac{\\pi}{2}$$: the $$xy$$-pane. ### Volume element To find the volume element $$\\Delta V\\approx\\Delta\\rho\\,\\Delta\\phi\\,\\Delta\\theta$$, start with the surface area on a sphere with radius $$a$$. If it is small enough it looks like a rectangle. So, what are the sides? Start with a surface area element $$\\Delta S$$ • To move north-south: you always have to traverse half a circle $$\\Longrightarrow$$ the vertical side of $$\\Delta S$$ is a piece of circle with radius $$a$$ $$\\Rightarrow$$ the length is $$a.\\Delta\\phi$$. • To move east-west: your distance depends on your distance to the $$z$$-axis, called $$r$$ $$\\Longrightarrow$$ the horizontal side of $$\\Delta S$$ is a piece of circle of radius $$r=a\\sin\\phi$$.", "\\frac{\\pi}{3} \\right )$ into Cartesian coordinates. Using the formula for the $x$-coordinate and we get that: (2) \\begin{align} \\quad x = r \\sin \\phi \\cos \\theta = 3 \\sin \\left ( \\frac{\\pi}{3} \\right ) \\cos (\\pi) = 3 \\left ( \\frac{\\sqrt{3}}{2} \\right ) (-1) = -\\frac{3\\sqrt{3}}{2} \\end{align} Using the formula for the $y$-coordinate and we get that: (3) \\begin{align} \\quad y = r \\sin \\phi \\sin \\theta = 3 \\sin \\left ( \\frac{\\pi}{3} \\right ) \\sin (\\pi) = 3 \\left ( \\frac{\\sqrt{3}}{2} \\right ) (0) = 0 \\end{align} Using the formula for the $z$-coordinate and we get that: (4) \\begin{align} \\quad z = r \\cos \\phi = 3 \\cos \\left ( \\frac{\\pi}{3} \\right ) = 3 \\left ( \\frac{1}{2} \\right ) = \\frac{3}{2} \\end{align} Therefore the spherical coordinates are $\\left ( \\frac{-3\\sqrt{3}}{2}, 0, \\frac{3}{2} \\right )$.", "# How do you convert (1, 5pi/6) into rectangular coordinates? Mar 21, 2016 $\\left(- \\frac{\\sqrt{3}}{2} , \\frac{1}{2}\\right)$ #### Explanation: Using the formulae that links Polar to Cartesian coordinates. • x = rcostheta • y = rsintheta here r = 1 and $\\theta = \\frac{5 \\pi}{6}$ hence : x$= 1 \\times \\cos \\left(\\frac{5 \\pi}{6}\\right) = \\cos \\left(\\frac{5 \\pi}{6}\\right) = - \\cos \\left(\\frac{\\pi}{6}\\right)$ and $y = 1 \\times \\sin \\left(\\frac{5 \\pi}{6}\\right) = \\sin \\left(\\frac{5 \\pi}{6}\\right) = \\sin \\left(\\frac{\\pi}{6}\\right)$ now the exact value of $\\cos \\left(\\frac{\\pi}{6}\\right) = \\frac{\\sqrt{3}}{2}$ and the exact value of $\\sin \\left(\\frac{\\pi}{6}\\right) = \\frac{1}{2}$ thus x =$- \\cos \\left(\\frac{\\pi}{6}\\right) = - \\frac{\\sqrt{3}}{2}$ and y $= \\sin \\left(\\frac{\\pi}{6}\\right) = \\frac{1}{2}$", "Since we haven’t put any restrictions on the “height” of the cylinder there won’t be any restriction on $$x$$. Convert the point $$\\displaystyle \\left( {\\sqrt 6 ,\\frac{\\pi }{4},\\sqrt 2 } \\right)$$ from cylindrical to spherical coordinates. We will sometimes need to write the parametric equations for a surface. In the two-dimensional plane with a rectangular coordinate system, when we say $$x = k$$ (constant) we mean an unbounded vertical line parallel to the $$y$$-axis and when $$y = l$$ (constant) we mean an unbounded horizontal line parallel to the $$x$$-axis. Finally, let’s find $$\\theta$$. (Figure 15.5.4). $$\\displaystyle \\varphi = \\frac{\\pi }{3}$$, $$\\displaystyle \\theta = \\frac{{2\\pi }}{3}$$. The Albuquerque event is the largest hot air balloon festival in the world, with over $$500$$ balloons participating each year. Set up a triple integral in cylindrical coordinates to find the volume of the region, using the following orders of integration: a. If we look at the sketch above from directly in front of the triangle we get the following sketch. \\end{align}\\], \\begin{align} z = \\sqrt{x^2 + y^2}\\\\\\rho \\, \\cos \\, \\varphi = \\sqrt{\\rho^2 \\sin^2 \\, \\varphi \\, \\cos^2 \\phi } \\\\ \\rho \\, \\cos \\, \\varphi = \\sqrt{\\rho^2 \\sin^2 \\varphi \\, (\\cos^2\\phi + \\sin^2 \\phi)}\\\\ \\rho \\, \\cos \\, \\varphi = \\rho \\, \\sin \\, \\varphi\\\\ \\cos \\, \\varphi = \\sin", "Coordinates 5. Since $\\displaystyle {\\rho}^{2}=x^{2}+y^{2}+z^{2}$ We can sub it into the density function and get $\\displaystyle \\sqrt{1+{\\rho}^{3}}$ 6. I am just a bit confused on how you find the \"phi\" and \"theta\" for this question. The bound is suppose to be 0 to Pi for phi and theta. By the way, that website on spherical coordinates was useful 7. I think your set up should look something like this: $\\displaystyle \\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{2}\\left[{\\rho}^{2}\\sqrt{1+{\\rho}^{3}}sin{\\phi}\\right]d{\\rho}d{\\phi}d{\\theta}$ 8. Originally Posted by galactus I think your set up should look something like this: $\\displaystyle \\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{2}\\left[{\\rho}^{2}\\sqrt{1+{\\rho}^{3}}sin{\\phi}\\right]d{\\rho}d{\\phi}d{\\theta}$ Yep that's it. Is theta 0 to Pi because its half a sphere? Also, why is the phi 0 to Pi also, I can't really picture it. EDIT: Ok, I can see why its Pi now, I am confused from the question. I can't really see what [Math] x^2+y^2+z^2<4 [/tex] and $\\displaystyle 0<y$ is. What do these regions mean? 9. Originally Posted by galactus I think your set up should look something like this: $\\displaystyle \\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{2}\\left[{\\rho}^{2}\\sqrt{1+{\\rho}^{3}}sin{\\phi}\\right]d{\\rho}d{\\phi}d{\\theta}$ Ahh yes. I misread my x- and y-axes. $\\displaystyle 0 \\leq \\theta \\pi$ indeed. 10. Originally Posted by Gotovina7 Yep that's it. Is theta 0 to Pi because its half a sphere? Also, why is the phi 0 to Pi also, I can't really picture it. EDIT: Ok, I can see why its Pi", "=\\frac{16}{3} \\end{align*} Example 6.14 Evaluate the iterated integral $\\int _0^{\\pi /4}\\int _0^1\\int _0^{\\sqrt{r}}r^2 \\sin \\theta \\, dz \\, dr \\, d\\theta.$ Solution. We find \\begin{align*} \\int _0^{\\pi /4}\\int _0^1\\int _0^{\\sqrt{r}}r^2 \\sin \\theta \\, dz \\, dr \\, d\\theta & =\\int _0^{\\pi /4}\\int _0^1r^{5/2} \\sin \\theta \\, dr \\, d\\theta \\\\ & =\\int_0^{\\pi /4} \\frac{2 \\sin \\theta }{7} \\, d\\theta =\\frac{2}{7} \\left(1-\\frac{1}{\\sqrt{2}}\\right) =\\frac{2-\\sqrt{2}}{7} \\end{align*} ## 6.8 Spherical Coordinates Each point in three dimensions is uniquely represented in spherical coordinates by $$(\\rho ,\\theta ,\\phi )$$ using $$0\\leq \\rho <\\infty ,$$ $$0\\leq \\theta < 2\\pi ,$$ and $$0\\leq \\phi \\leq \\pi .$$ The conversion formulas from rectangular coordinates $$(x,y,z)$$ to spherical coordinates $$(\\rho ,\\theta ,\\phi )$$ are $\\begin{array}{lllll} \\rho =\\sqrt{x^2+y^2+z^2} & \\qquad & \\theta = \\tan^{-1} \\left( \\frac{y}{x}\\right) & \\qquad &\\phi =\\cos ^{-1}\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right) \\\\ x=\\rho \\sin \\phi \\cos \\theta & & y =\\rho \\sin \\phi \\sin \\theta & & z=\\rho \\cos \\phi \\end{array}$ Using Jacobians, we can show that the element of volume in spherical coordinates is $dV=\\rho ^2 \\sin \\phi \\,d\\rho \\, d\\theta \\, d\\phi.$ ::: {#thm- } [Triple Integrals in Spherical Coordinates] If $$f(x,y,z)$$ is continuous on the closed bounded region $$R$$, then the triple integral of $$f$$ over $$R$$ is given by \\begin{align*} & \\iiint_Rf(x,y,z)dV \\\\ & = \\iiint_S f(\\rho \\sin \\phi \\cos \\theta ,\\rho \\sin \\phi \\sin", "convert the rectangular coordinate to its cylindrical form as shown below. \\begin{aligned}r &= \\sqrt{x^2 + y^2}\\\\ \\theta &= \\tan^{-1} \\left(\\dfrac{y}{x}\\right)\\\\z &= z \\end{aligned} Using $x =2$, $y = 1$, and $z = -4$, we have the following: $\\boldsymbol{r}$ \\begin{aligned}r &= \\sqrt{2^2 + 1^2}\\\\&= \\sqrt{5} \\end{aligned} $\\boldsymbol{\\theta}$ \\begin{aligned}\\theta &= \\tan^{-1} \\left(\\dfrac{1}{2}\\right)\\\\&\\approx 26.57^{\\circ} \\end{aligned} $\\boldsymbol{z}$ \\begin{aligned}z &= -4 \\end{aligned} This means that $(2, 1, -4)$ is equivalent to $(\\sqrt{5}, 26.57^{\\circ}, -4)$ in cylindrical form. Example 2 Convert the cylindrical coordinate, $\\left(6, \\dfrac{2\\pi}{3}, -3\\right)$, to its rectangular form. Solution We’re now given the cylindrical coordinate and we need to convert them to rectangular coordinates, so we’ll use the following relationships: \\begin{aligned}x&= r\\cos \\theta\\\\y &= r\\sin \\theta\\\\z&= z\\end{aligned} We have $r= 6$, $\\theta = \\dfrac{2\\pi}{3}$, and $z = -3$, so substitute these values into the three equations to find the Cartesian or rectangular form of the coordinate. $\\boldsymbol{x}$ \\begin{aligned}x &= 6\\cos \\dfrac{2\\pi}{3}\\\\&= -3\\end{aligned} $\\boldsymbol{y }$ \\begin{aligned}y &= 6\\sin \\dfrac{2\\pi}{3}\\\\&= 3\\sqrt{3}\\end{aligned} $\\boldsymbol{z}$ \\begin{aligned}z &= -3 \\end{aligned} This means that $\\left(6, \\dfrac{2\\pi}{3}, -3\\right)$ is equal to $(-3, 3\\sqrt{3}, -3)$ in rectangular form. Example 3 Describe the position of the cylindrical point, $\\left(2, \\dfrac{\\pi}{6}, 4 \\right)$, then graph the point on the three-dimensional cartesian coordinate system. Solution To know the position of the cylindrical coordinate, convert it first to its rectangular form. That way, it’ll", "$cos^2(\\phi)= sin^2(\\phi)$ or simply $\\phi= \\frac{\\pi}{4}$ since $\\phi$ must be between 0 and $\\pi$. The volume inside the sphere, above the cone, is given by $\\int_{\\theta= 0}^{2\\pi}\\int_{\\phi= 0}^{\\pi/4}\\int_{\\rho= 0}^1 \\rho sin^2(\\phi)d\\rho d\\phi d\\theta$ $= \\left(\\int_{\\theta= 0}^{2\\pi}d\\theta\\right)\\left(\\int_0^{\\pi/4} sin^2(\\phi)d\\phi\\right)\\left(\\int_{\\rho= 0}^1 \\rho d\\rho\\right)$. That is exactly what ojones told you yesterday! 5. Originally Posted by HallsofIvy ojones has already told you how to do it. The sphere, $x^2+ y^2+ z^2= 1$, in spherical coordinates, is $\\rho^2= 1$ and the cone, z^2= x^2+ y^2, in spherical coordinates is $\\rho^2 cos^2(\\phi)= \\rho^2 sin^2(\\phi)$ which reduces to $cos^2(\\phi)= sin^2(\\phi)$ or simply $\\phi= \\frac{\\pi}{4}$ since $\\phi$ must be between 0 and $\\pi$. The volume inside the sphere, above the cone, is given by $\\int_{\\theta= 0}^{2\\pi}\\int_{\\phi= 0}^{\\pi/4}\\int_{\\rho= 0}^1 \\rho sin^2(\\phi)d\\rho d\\phi d\\theta$ $= \\left(\\int_{\\theta= 0}^{2\\pi}d\\theta\\right)\\left(\\int_0^{\\pi/4} sin^2(\\phi)d\\phi\\right)\\left(\\int_{\\rho= 0}^1 \\rho d\\rho\\right)$. That is exactly what ojones told you yesterday! I have a quick question. If I just wanted to find the volume of the sphere above the plane of intersection, how would I go about doing this? I have tried but can't figure it out. Any help would be appreciated. I have no problem solving it the way that was just shown but like to try and solve problems in more than one way. I found the volume of the cone and was just going to add the", "Quiz from Wednesday, October 26 Problem: Let $\\theta$ represent the usual angle in the $xy$-plane emanating from the origin counter-clockwise from the $x$-axis, and let $\\phi$ represent the angle from the $z$-axis to the $xy$-plane. If $\\rho$ is distance from the origin, find the volume of the object created by rotating the region where $1\\le\\rho\\le 3$ and $\\pi/6\\le\\phi\\le\\pi/2$ in a full circle around the $z$-axis. Solution: In spherical coordinates, we compute as follows. \\begin{align*} \\int_0^{2\\pi}\\int_{\\pi/6}^{\\pi/2}\\int_1^3 \\rho^2\\sin\\phi\\,d\\rho\\,d\\phi\\,d\\theta &= \\int_0^{2\\pi}\\,d\\theta\\int_{\\pi/6}^{\\pi/2}\\sin\\phi\\,d\\phi\\int_1^3\\rho^2\\,d\\rho \\\\ &= 2\\pi\\left(-\\cos\\phi\\right)\\Big|_{\\phi=\\pi/6}^{\\phi=\\pi/2}\\left(\\frac{\\rho^3}{3}\\right)\\Big|_{\\rho=1}^{\\rho=3}\\\\ &= 2\\pi\\cdot\\frac{\\sqrt{3}}{2}\\left(\\frac{27}{3}-\\frac{1}{3}\\right)\\\\ &= \\frac{26\\pi\\sqrt{3}}{3}\\end{align*} Replacement Question: How many states share a border (including corners) with Colorado? Answer: 7: Kansas, Nebraska, Wyoming, Utah, Arizona, New Mexico, and Oklahoma. There are only two states that share more borders with neighboring states than Colorado does: Tennessee and Missouri both share borders with 8 states (including each other)." ]

[ "5 \\end{pmatrix}$$ var v = [1.0, 2.0] var s = 3.0 var vsresult = [Double](count : v.count, repeatedValue : 0.0) vDSP_vsaddD(v, 1, &s, &vsresult, 1, vDSP_Length(v.count)) vsresult // returns [4.0, 5.0] $$\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix} \\times 3 = \\begin{pmatrix} 3 \\\\ 6 \\end{pmatrix}$$ vDSP_vsmulD(v, 1, &s, &vsresult, 1, vDSP_Length(v.count)) vsresult // returns [3.0, 6.0] $$\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix} \\div 3 = \\begin{pmatrix} 1/3 \\\\ 2/3 \\end{pmatrix}$$ vDSP_vsdivD(v, 1, &s, &vsresult, 1, vDSP_Length(v.count)) vsresult // returns [0.333333333333333, 0.666666666666667] ## Vector & Vector Vector-vector operations pose no challenge to Accelerate and the associated functions look like vDSP_v***D(vector_1, 1, vector_2, 1, &result, 1, length_of_vector) Here are a few worked-out examples: $$\\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix} + \\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix} = \\begin{pmatrix} 5 \\\\ 9 \\end{pmatrix}$$ var v1 = [2.0, 5.0] var v2 = [3.0, 4.0] var vvresult = [Double](count : 2, repeatedValue : 0.0) vDSP_vaddD(v1, 1, v2, 1, &vvresult, 1, vDSP_Length(v1.count)) vvresult // returns [5.0, 9.0] $$\\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix} \\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix} = \\begin{pmatrix} 6 \\\\ 20 \\end{pmatrix}$$ vDSP_vmulD(v1, 1, v2, 1, &vvresult, 1, vDSP_Length(v1.count)) vvresult // returns [6.0, 20.0] $$\\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix} {\\bigg/} \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix} = \\begin{pmatrix} 1.5 \\\\ 0.8 \\end{pmatrix}$$ vDSP_vdivD(v1, 1, v2, 1, &vvresult, 1, vDSP_Length(v1.count)) vvresult // returns [1.5, 0.8] ## Dot Product $$\\begin{pmatrix}", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "\\\\ Y\\end{bmatrix} \\qquad (x,y) = [X,Y]P^T$ $q(\\mathbf{v}) = (x,y)A\\begin{pmatrix}x \\\\ y\\end{pmatrix} = [X,Y]P^T AP\\begin{bmatrix}X \\\\ Y\\end{bmatrix}$ If we can find an orthogonal change of basis $(P^T = P^{-1})$ then $P$ ... Eigenvalues of $A = \\begin{pmatrix}2 & 1 \\\\ 1 & 2\\end{pmatrix}$ \\begin{aligned} 0 &= \\begin{vmatrix}\\lambda I - A \\end{vmatrix} = \\begin{vmatrix} \\lambda -2 & -1 \\\\ -1 & \\lambda -2\\end{vmatrix}= (\\lambda -2)^2 - 1 \\\\ 0 &= (\\lambda -2)^2 - 1^2 = ((\\lambda -2)-1)((\\lambda -2)+1) \\\\ 0 &= (\\lambda -3)(\\lambda -1)\\end{aligned} $\\begin{cases}\\lambda_1 = 3 \\lambda_2 = 1\\end{cases}$ eigenvalues $3X^2 + Y^2$ is $q$ in the new system, Eigenvalue $\\lambda_1 = 3$ $\\begin{pmatrix} 3-2 & -1 \\\\ -1 & 3-2 \\end{pmatrix} \\begin{pmatrix}\\alpha \\\\ \\beta\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0\\end{pmatrix} \\begin{cases} \\alpha - \\beta = 0 \\\\ -\\alpha + \\beta = 0 \\end{cases} so \\alpha = \\beta \\begin{pmatrix}1 \\\\ 1\\end{pmatrix}$ will do Note $<\\begin{pmatrix}1 \\\\ 1\\end{pmatrix},\\begin{pmatrix}-1 \\\\ 1\\end{pmatrix}> = 0$ orthogonal basis of eigenvectors $\\|\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}\\|^2 = 2 =\\|\\begin{pmatrix}-1 \\\\ 1\\end{pmatrix}\\|^2$ Corresponding orthonormal basis: $\\{\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}, \\begin{pmatrix}-\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\}$ But another can be $\\{\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}, \\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}\\}$ and two more candidates... We may take $P = \\begin{pmatrix} \\frac{1}{\\sqrt{2}} & -\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}}\\end{pmatrix} = (P^{-1})^T$ Note $det P = \\pm 1$ (rotation) Claim $\\begin{pmatrix} \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}}\\end{pmatrix} \\begin{pmatrix} 2 & 1", "& 2\\\\1 & -2 \\\\\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\vec 0 \\, .$ This gives us the equalities $$-x +2y = 0$$ and $$x -2y = 0$$ which both give the line $$y = \\frac{1}{2}x$$. Any point on this line, so for example $$\\begin{pmatrix}2\\\\1\\end{pmatrix}$$, is an eigenvector with eigenvalue $$\\lambda = 5$$. Now lets find the eigenvector for $$\\lambda = 2$$ $\\begin{pmatrix}4-2 & 2\\\\1 & 3-2 \\\\\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}2 & 2\\\\1 & 1 \\\\\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\vec 0 ,$ which gives the equalities $$2x+2y = 0$$ and $$x+y = 0$$. (Notice that these equations are not independent of one another, so our eigenvalue must be correct.) This means any vector $$v = \\begin{pmatrix}x\\\\y\\end{pmatrix} where \\(y = -x$$ , such as $$\\begin{pmatrix}1\\\\-1\\end{pmatrix}, or any scalar multiple of this vector, \\(\\textit{i.e.}$$ any vector on the line $$y = -x$$ is an eigenvector with eigenvalue 2. This solution could be written neatly as $\\lambda_{1} = 5, \\, v_{1}= \\begin{pmatrix}2\\\\1\\end{pmatrix} \\, \\text{ and } \\lambda_{2} = 2, \\, v_{2}=\\begin{pmatrix}1\\\\-1\\end{pmatrix}.$ Jordan Block Example Consider the matrix $J_{2} = \\begin{pmatrix}\\lambda & 1\\\\ 0 & \\lambda\\end{pmatrix},$ and we note that we can just read off the eigenvector $$e_{1}$$ with eigenvalue $$\\lambda$$. However the characteristic polynomial of $$J_{2}$$ is $$P_{J_{2}}(\\mu) = (\\mu - \\lambda)^{2}$$ so the only possible eigenvalue is $$\\lambda$$, but we claim it does not", "&\\xrightarrow[]{-2\\rho_1+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 1 &-1 &-1m+0n \\ 0 &1 &1m+1n \\ 0 &2 &3m+3n \\end{array}\\right) \\ &\\xrightarrow[]{-2\\rho_2+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 1 &-1 &-1m+0n \\ 0 &1 &1m+1n \\ 0 &0 &1m+1n \\end{array}\\right) \\end{array}$ to deduce that the only vectors from S2 that are also in S1 are of the form $\\begin{pmatrix} -1 \\\\ 1 \\\\ 1 \\end{pmatrix}m +\\begin{pmatrix} 0 \\\\ 1 \\\\ 3 \\end{pmatrix}(-m).$ For instance, $\\begin{pmatrix} -1 \\\\ 1 \\\\ 1 \\end{pmatrix}$ is in S2 but not in S1. 5. These sets are equal. First we change the parameters: $S_2=\\{\\begin{pmatrix} 3 \\\\ 7 \\\\ 7 \\end{pmatrix}m +\\begin{pmatrix} 1 \\\\ 3 \\\\ 1 \\end{pmatrix}n \\,\\big|\\, m,n\\in\\mathbb{R}\\}.$ Now, to show that $S_1\\subseteq S_2$, we solve $\\begin{array}{*{2}{rc}r} 3m &+ &1n &= &1t+2s \\ 7m &+ &3n &= &3t+4s \\ 7m &+ &1n &= &1t+6s \\end{array}$ with Gauss' method $\\begin{array}{rcl} \\left(\\begin{array}{*{2}{c}|c} 3 &1 &1t+2s \\ 7 &3 &3t+4s \\ 7 &1 &1t+6s \\end{array}\\right) &\\xrightarrow[(-7/3)\\rho_1+\\rho_3]{(-7/3)\\rho_1+\\rho_2} &\\left(\\begin{array}{*{2}{c}|c} 3 &1 &1t+2s \\ 0 &2/3 &(2/3)t-(2/3)s \\ 0 &-4/3 &(-4/3)t+(4/3)s \\end{array}\\right) \\ &\\xrightarrow[]{2\\rho_2+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 3 &1 &1t+2s \\ 0 &2/3 &(2/3)t-(2/3)s \\ 0 &0 &0 \\end{array}\\right) \\end{array}$ to get that $\\begin{pmatrix} 1 \\\\ 3 \\\\ 1 \\end{pmatrix}t +\\begin{pmatrix} 2 \\\\ 4 \\\\ 6 \\end{pmatrix}s =\\begin{pmatrix} 3 \\\\ 7 \\\\ 7 \\end{pmatrix}(s) +\\begin{pmatrix} 1 \\\\ 3 \\\\ 1 \\end{pmatrix}(t-s)$ and so $S_1\\subseteq S_2$. The proof that $S_2\\subseteq S_1$ involves", "to align the \\Rightarrows – daleif Oct 6 '16 at 7:59 This is an automatic solution, based on the eqparbox package. I defined an \\eqmathbox command, which accepts $2$ optional arguments (the end of name of the box, and the alignment: l, r or c) and one mandatory argument: the contents of the math box. \\documentclass{article} %\\usepackage{enumitem} \\usepackage{mathtools} \\usepackage{xparse} \\usepackage{eqparbox} \\NewDocumentCommand\\Eqmathbox{O{0}O{l}m}{\\eqmakebox[Form#1][#2]{\\ensuremath{ \\displaystyle#3}}} \\usepackage{siunitx} \\sisetup{group-separator={,},group-minimum-digits = 4} \\begin{document} \\begin{enumerate} \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 20 & 25 & 20 \\\\ 15 & 20 & 18 \\\\ 18 & 23 & 22 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 975 \\\\ 810 \\\\ 965 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 10 \\\\ 15 \\\\ 20 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 90 & 95 & 80 \\\\ 80 & 85 & 90 \\\\ 75 & 90 & 95 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 86 \\\\ 86 \\\\ 88 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 0.3 \\\\ 0.2 \\\\ 0.5 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] \\num{5000} & \\num{400}0 \\\\ \\num{6000} & \\num{5000} \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} = \\begin{pmatrix*}[r] \\num{37000000} \\\\ \\num{45000000 }\\\\ \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} =} \\begin{pmatrix*}[r] \\num{5000} \\\\ \\num{3000} \\end{pmatrix*}$$ \\end{enumerate} \\end{document} Here is one way. Note that you are adding a lot of unneeded", "those vectors that form a linear combination for all vectors in $\\mathbb{C}^4$. My proof would go something like this: Let $x_1, x_2, x_3, x_4 \\in{\\mathbb{C}}$ be arbitrary. Thus, $x_1+x_2+x_3$, $x_1+x_4$, and $x_2+x_4$ are also arbitrary elements of $\\mathbb{C}$. This means that $x=\\begin{pmatrix}x_1+x_2+x_3\\\\x_1+x_4\\\\x_2+x_4\\\\x_3\\end{pmatrix} \\in{\\mathbb{C}^4}$ is also arbitrary. So a linear combination of $x$ with the given vectors is simply $x_1\\begin{pmatrix}1\\\\1\\\\0\\\\0\\end{pmatrix}+ x_2\\begin{pmatrix}1\\\\0\\\\1\\\\0\\end{pmatrix}+ x_3\\begin{pmatrix}1\\\\0\\\\0\\\\1\\end{pmatrix}+ x_4\\begin{pmatrix}0\\\\1\\\\1\\\\0\\end{pmatrix}+ 0\\begin{pmatrix}0\\\\1\\\\0\\\\1\\end{pmatrix}+ 0\\begin{pmatrix}0\\\\0\\\\1\\\\1\\end{pmatrix}$ Therefore, the given vectors span $\\mathbb{C}^4$ for all $x_i$. $\\blacksquare$ For the second question, I'd argue that any 4 of those given vectors can be shown to span $\\mathbb{C}^4$ in a similar manner. So there are ${6}\\choose{4}$ ways 4 of those vectors form a basis for $\\mathbb{C}^4$. • Could you define what you mean by arbitrary? – YoTengoUnLCD Jul 15 '15 at 2:43 • I might be abusing that word in my example, but I'm taking it to mean that if $z\\in{Z}$ is arbitrary , then whatever result I get for $z$ extends to all elements of the set $Z$. Would you say \"$z\\in{Z}$ is arbitrary\" is the functional equivalent of \"$\\forall z \\in{Z}$\"? – gbrlrz017 Jul 15 '15 at 2:48 • I see what you mean, I would avoid using that word altogether, it causes confusion. Also, there aren't 6 choose 4 ways to pick those vectors to form a basis, pick vectors", "What do you mean by \"at\" the largest singular value? Do you mean we set all the entries of vector $a$ equal to zero except for one of them? If $A = \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}$, $B = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ 0 & 0 \\end{pmatrix}$ then $A'D = \\begin{pmatrix} 1 & 1 \\end{pmatrix}$. $A'D$ is equal to the same thing if $A = \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ $B = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ 0 & 0 \\end{pmatrix}$ Assuming A and B are both orthnormal, then the coefficient vectors a and b are both unit vectors so that x and y are also unit vectors. Then maximization of <x,y> correspond to minimal angles between x and y. In order to maximize <x,y>=(Ua)'*S*(Vb), given that ||Ua||=1 and ||Vb||=1, I want to choose Ua and Vb to be 1 at the largest singular values and 0 elsewhere, not that a and b are 1 at one place and 0 everywhere else. Is the logic correct? Stephen Tashi I want to choose Ua and Vb to be 1 at the largest singular values and 0 elsewhere Are you saying that vector 'a' will be chosen so that the vector Ua will be 1 at the", "vector ${\\displaystyle (-1,0,-1)^{T}}$ is a multiple of ${\\displaystyle (1,0,1)^{T}}$. That is, the direction represented by ${\\displaystyle (-1,0,-1)^{T}}$ is the same as the direction represented by ${\\displaystyle (1,0,1)^{T}}$. Hence, we can remove this vector from ${\\displaystyle E}$ and obtain a new generator ${\\displaystyle E_{1}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}},{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Can we reduce the size of this generator, as well? Yes, since ${\\displaystyle {\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}+{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}}$ So ${\\displaystyle (2,1,1)^{T}}$ adds no new direction that is not already spanned by ${\\displaystyle (1,0,1)^{T}}$ and ${\\displaystyle (1,1,0)^{T}}$. We thus obtain a smaller generator ${\\displaystyle E_{2}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Now we cannot reduce the generator any further without losing the property of it being a generator. For if we remove any of the three vectors in ${\\displaystyle E_{2}}$, it is no longer in the span of the remaining two vectors. For ${\\displaystyle (1,0,1)^{T}}$, for example, we see this as follows: Suppose we had some ${\\displaystyle \\lambda ,\\mu \\in \\mathbb {R} }$, so that ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=\\lambda {\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+\\mu {\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}}$ Then ${\\displaystyle \\lambda =-\\mu }$ would have to hold because the second component must be zero on both sides. Because of the third component, ${\\displaystyle \\mu ={\\frac {1}{2}}}$ must be true. Thus we obtain the contradiction ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=-{\\frac {1}{2}}{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+{\\frac {1}{2}}{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}0.5\\\\0\\\\1\\end{pmatrix}}}$ So ${\\displaystyle (1,0,1)^{T}}$ is not in ${\\displaystyle \\operatorname {span} ((1,1,0)^{T},(2,1,2)^{T})}$. So we have found a generator containing linearly independent vectors (a minimal generator). In summary, we proceeded", "$$(A + iB)(x + iy) = \\lambda x + i \\lambda y, \\tag 9$$ or $$(Ax - By) + i(Ay + Bx) = \\lambda x + i \\lambda y; \\tag{10}$$ equating the real and imaginary parts of either side yields $$Ax - By = \\lambda x, \\tag{11}$$ $$Ay + Bx = \\lambda y. \\tag{12}$$ We can now in fact find two real orthogonal eigenvectors for $$M = \\begin{bmatrix} A & -B \\\\ B & A \\end{bmatrix}, \\tag{13}$$ each corresponding to the eigenvalue $$\\lambda$$; set $$w = \\begin{pmatrix} x \\\\ y \\end{pmatrix}; \\tag{14}$$ $$Mw = \\begin{bmatrix} A & -B \\\\ B & A \\end{bmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} Ax - By \\\\ Bx + Ay \\end{pmatrix} = \\begin{pmatrix} \\lambda x \\\\ \\lambda y \\end{pmatrix} = \\lambda \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\lambda w; \\tag{15}$$ now set $$v = \\begin{pmatrix} -y \\\\ x \\end{pmatrix}; \\tag{16}$$ $$Mv = \\begin{bmatrix} A & -B \\\\ B & A \\end{bmatrix} \\begin{pmatrix} -y \\\\ x \\end{pmatrix} = \\begin{pmatrix} -Ay - Bx \\\\ -By + Ax \\end{pmatrix} = \\begin{pmatrix} -\\lambda y \\\\ \\lambda x \\end{pmatrix} = \\lambda \\begin{pmatrix} -y \\\\ x \\end{pmatrix} = \\lambda v; \\tag{17}$$ furthermore, $$\\langle v, w \\rangle = \\begin{pmatrix} -y \\\\ x \\end{pmatrix}^T \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} -y^T & x^T \\end{pmatrix} \\begin{pmatrix} x \\\\ y", "{\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}t={\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}(6-3t)}$ and so any vector in the form for ${\\displaystyle S_{1}}$ can be stated in the form needed for inclusion in ${\\displaystyle S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we look for ${\\displaystyle t}$ so that these equations hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}1&+&0t&=&1+0s\\\\2&+&3t&=&8-1s\\end{array}}}$ Rewrite that as ${\\displaystyle {\\begin{array}{*{1}{rc}r}1&=&1\\\\t&=&2-(1/3)s\\end{array}}}$ and so ${\\displaystyle {\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}s={\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}(2-(1/3)s).}$ 2. These two are equal. To show that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we check that for any ${\\displaystyle t,s}$ we can find an appropriate ${\\displaystyle m,n}$ so that these hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}4m&-&4n&=&1t+2s\\\\7m&-&2n&=&3t+1s\\\\7m&-&10n&=&1t+5s\\end{array}}}$ Use Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\7&-2&3t+1s\\\\7&-10&1t+5s\\end{array}}\\right)&{\\xrightarrow[{(-7/4)\\rho _{1}+\\rho _{3}}]{(-7/4)\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&-3&-(3/4)t+(6/4)s\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to conclude that ${\\displaystyle {\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}t+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}s={\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}((1/4)t-(1/2)s)}$ and so ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&+&2s&=&4m-4n\\\\3t&+&1s&=&7m-2n\\\\1t&+&5s&=&7m-10n\\end{array}}}$ with Gaussian reduction ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\3&1&7m-2n\\\\1&5&7m-10n\\end{array}}\\right)&{\\xrightarrow[{-\\rho _{1}+\\rho _{3}}]{-3\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&3&3m-6n\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to get ${\\displaystyle {\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}m+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}n={\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}(2m)+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}(m-2n)}$ and so any member of ${\\displaystyle S_{2}}$ can be expressed in the form needed for ${\\displaystyle S_{1}}$. 3. These sets are equal. To prove that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we must be able to solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}2m&+&4n&=&1t\\\\4m&+&8n&=&2t\\end{array}}}$ for ${\\displaystyle m}$ and ${\\displaystyle n}$ in terms of ${\\displaystyle t}$. Apply Gaussian reduction ${\\displaystyle \\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\4&8&2t\\end{array}}\\right){\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}\\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\0&0&0\\end{array}}\\right)}$ to conclude that any pair ${\\displaystyle m,n}$ where ${\\displaystyle 2m+4n=t}$ will do. For instance, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(0)}$ or ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((-3/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(t).}$ Thus ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&=&2m+4n\\\\2t&=&4m+8n\\end{array}}}$ with Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\2&4m+8n\\end{array}}\\right)&{\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\0&0\\end{array}}\\right)\\end{array}}}$ to deduce that any", "a set of vectors is the set of all linear combinations of these vectors. So the span of $\\{\\begin{pmatrix}1\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\end{pmatrix}\\}$ would be the set of all linear combinations of them, which is $\\mathbb{R}^2$. The span of $\\{\\begin{pmatrix}2\\\\0\\end{pmatrix}, \\begin{pmatrix}1\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\end{pmatrix}\\}$ is also $\\mathbb{R}^2$, although we don't need $\\begin{pmatrix}2\\\\0\\end{pmatrix}$ to be so. So both these two sets are said to be the spanning sets of $\\mathbb{R}^2$. However, only the first set $\\{\\begin{pmatrix}1\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\end{pmatrix}\\}$ is a basis of $\\mathbb{R}^2$, because the $\\begin{pmatrix}2\\\\0\\end{pmatrix}$ makes the second set linearly dependent. Also, the set $\\{\\begin{pmatrix}2\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\end{pmatrix}\\}$ can also be a basis for $\\mathbb{R}^2$. Because its span is also $\\mathbb{R}^2$ and it is linearly independent. For another example, the span of the set $\\{\\begin{pmatrix}1\\\\1\\end{pmatrix}\\}$ is the set of all vectors in the form of $\\begin{pmatrix}a\\\\a\\end{pmatrix}$. • So a basis of $\\mathbb R^n$ is a smallest spanning set of $\\mathbb R^n$, i.e any spanning set of $\\mathbb R^n$ with cardinality $n$? – Zaz Oct 26 '18 at 23:17 • @Zaz: That is correct. Oct 30 '18 at 10:28 The basis is a combination of vectors which are linearly independent and which spans the whole vector V. Suppose we take a system of $R^2$. Now as you said,$(1,0)$ and $(0,1)$ are the basis in this system and we want to find any $(x,y)$ in this system. $$(x,y)=x(1,0)+y(0,1)$$", "A) v_A \\end{pmatrix} = \\begin{pmatrix} (\\lambda_A - 1) v_A \\\\ -(\\lambda_A - 1) v_A \\end{pmatrix} = (\\lambda_A - 1) V_{A-}. \\tag{5}$ (4) and (5) show that the eigenvalues of $B$ are at least those of the form $\\lambda_A \\pm 1$, but could there be more? We may investigate this possibility via a sort of unravelling of (4)-(5), in the sense that if $V =\\begin{pmatrix} v_1 \\\\ v_2 \\end{pmatrix} \\ne 0, \\tag{6}$ where $v_1$, $v_2$ are $N$-vectors, is an eigenvector of $B$ with eigenvalue $\\mu$, that is, $BV = \\mu V, \\tag{7}$ then if we write out (7) in terms of $v_1$, $v_2$ using the definition of $B$ in terms of $A$ we see that $BV = \\begin{bmatrix} A & I \\\\ I & A \\end{bmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\end{pmatrix} = \\begin{pmatrix} Av_1 + v_2 \\\\ v_1 + A v_2 \\end{pmatrix} = \\mu \\begin{pmatrix} v_1 \\\\ v_2 \\end{pmatrix}, \\tag{8}$ so that (7)-(8) breaks out into the two equations $Av_1 + v_2 = \\mu v_1 \\tag{9}$ and $Av_2 + v_1 = \\mu v_2. \\tag{10}$ We analyze (9)-(10) by first addressing the cases $v_2 = \\pm v_1$; in the event that $v_2 = v_1$, (9)-(10) become the same equation $Av_i + v_i = \\mu v_i \\tag{11}$ or $Av_i = (\\mu - 1)v_i \\tag{12}$ for $i = 1, 2$. If, on", "# Break line in mode [duplicate] I have a long list of vectors written between the signs , the problem is that the list is so long it does not fit the page. I want to make the list break at some point and continue in the following line, how could I do this? THis is my script. $G(B)=\\left \\{ \\begin{pmatrix}0 & 4 & -17 & 24 \\end{pmatrix}^T,\\begin{pmatrix}1 & 0 & -3 & 3 \\end{pmatrix}^T,\\begin{pmatrix}1 & -4 & 14 & -21 \\end{pmatrix}^T, \\begin{pmatrix}2 & -4 & 11 & -18 \\end{pmatrix}^T, \\ \\begin{pmatrix}3 & -4 & 8 & -15 \\end{pmatrix}^T,\\begin{pmatrix}4 & -4 & 5 & -12 \\end{pmatrix}^T, \\begin{pmatrix}5 & -4 & 2 & -9 \\end{pmatrix}^T, \\begin{pmatrix}6 & -4 & -1 & -6 \\end{pmatrix}^T\\right \\}, \\begin{pmatrix}11 & -8 & 1 & -15 \\end{pmatrix}^T, \\begin{pmatrix}17 & -12 & 0 & -21 \\end{pmatrix}^T, \\begin{pmatrix}7 & -4 & -4 & -3 \\end{pmatrix}^T, \\begin{pmatrix}8 & -4 & -7 & 0 \\end{pmatrix}^T \\right \\}$ Thanks! • You may really want to reconsider how legible -- or, rather, how illegible -- a group of twelve [12!] four-element vectors typeset in inline math mode is. Any chance you can switch to displaymath mode? – Mico Dec 15 '15 at 8:15 • @Werner: The question is a duplicate, but not the answers! – user2478 Dec 15 '15 at", "& -3/50 \\end{array} \\right) \\left( \\begin{array}{cc} 1 & -2 \\\\ 1 & 3 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.1 \\\\ 0 \\end{pmatrix}$ Multiply throughout by $\\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right)^{-1} = 1/5 \\left( \\begin{array}{cc} 2 & 3 \\\\ -1 & 1 \\end{array} \\right)$ to get $\\frac{d}{dt} \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\left( \\begin{array}{cc} -1/50 & 0 \\\\ 0 & -6/50 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.04 \\\\ -0.02 \\end{pmatrix}$ this is equivalent to the differential equations $\\frac{du_1}{dt}=-1/50u_1+0.04$ $\\frac{du_2}{dt}=-6/50u_1-0.02$ These have the solutions $u_1=2+Ae^{-t/50}, \\: u_2= -1/6+Be^{-6t/50}$ We can write this as $\\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix}$ Now transform back to $x_1, \\: x_2$ . \\begin{aligned} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} &= P \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} \\\\ &= \\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right) \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix} \\\\ &= \\begin{pmatrix} 5/2+Ae^{-t/50}-3Be^{-6t/50} \\\\ 5/3+Ae^{-t/50}+2Be^{-6t/50} \\end{pmatrix} \\end{aligned} . These solutions can be fitted to any initial conditions to find the constants.", "$\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix},\\quad M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x-y=0\\},\\quad N=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, 2x+y=0\\}$ 3. $\\begin{pmatrix} 3 \\\\ 0 \\\\ 1 \\end{pmatrix},\\quad M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, x+y=0\\},\\quad N=\\{c\\cdot\\begin{pmatrix} 1 \\\\ 0 \\\\ 1 \\end{pmatrix}\\,\\big|\\, c\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 2 Find $M^\\perp$. 1. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x+y=0 \\}$ 2. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, -2x+3y=0 \\}$ 3. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x-y=0 \\}$ 4. $M=\\{\\vec{0}\\,\\}$ 5. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x=0 \\}$ 6. $M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, -x+3y+z=0 \\}$ 7. $M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, x=0\\text{ and } y+z=0\\}$ Problem 3 This subsection shows how to project orthogonally in two ways, the method of Example 3.2 and 3.3, and the method of Theorem 3.8. To compare them, consider the plane $P$ specified by $3x+2y-z=0$ in $\\mathbb{R}^3$. 1. Find a basis for $P$. 2. Find $P^\\perp$ and a basis for $P^\\perp$. 3. Represent this vector with respect to the concatenation of the two bases from the prior item. $\\vec{v}=\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ 4. Find the orthogonal projection of $\\vec{v}$ onto $P$ by keeping only the $P$ part from the prior item. 5. Check that against the result from applying Theorem 3.8. This exercise is recommended for all", "value of $(x^2+y^2)^2$ $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=(6-2xy)^2+4x^2y^2=2(2xy-3)^2+18\\ge18$$ The equality occurs if $2xy=3$ You can rephrase $$y_{\\min} \\leq Ax \\leq y_{\\max},$$ as $$\\begin{pmatrix}A \\\\ -A\\end{pmatrix}x \\leq \\begin{pmatrix}y_{\\max} \\\\ -y_{\\min}\\end{pmatrix},$$ which is in the format of what quadprog accepts....", "\\end{pmatrix}$$ $$\\textbf{X} = \\begin{pmatrix} 1 & x_{11} & x_{12} & \\dots & x_{1m} \\\\ 1 & x_{21} & x_{22} & \\dots & x_{2m} \\\\ \\vdots & \\vdots & \\vdots & \\vdots & \\vdots \\\\ 1 & x_{n1} & x_{n2} & \\dots & x_{nm} \\end{pmatrix}$$ $$\\beta = \\begin{pmatrix} a_0 \\\\ a_1 \\\\ \\vdots \\\\ a_n \\end{pmatrix}$$ then, $$\\hat{\\textbf{Y}} = \\textbf{X} \\beta$$ the residuals $\\textbf{E} = \\begin{pmatrix} e_1 \\\\ e_2 \\\\ \\vdots \\\\ e_n \\end{pmatrix}$=$\\begin{pmatrix} y_1 - \\hat{y}_1 \\\\ y_2 - \\hat{y}_2 \\\\ \\vdots \\\\ y_n - \\hat{y}_n \\end{pmatrix}$= $\\textbf{Y}-\\hat{\\textbf{Y}}$ We will here introduce the residual sum-of-squares cost function, which is very similar to the mean square error cost function, but it is defined as $$RSS = \\sum_{i=1}^{n} e_i^2$$ We have noticed in the previous cases that the effect of considering the mean is eliminated during the derivation of the cost function and equating to zero. we also notice that $$RSS = \\textbf{E}^T \\textbf{E}\\\\ = (\\textbf{Y}-\\hat{\\textbf{Y}})^T(\\textbf{Y}-\\hat{\\textbf{Y}})\\\\ = (\\textbf{Y}- \\textbf{X} \\beta )^T (\\textbf{Y}- \\textbf{X} \\beta )\\\\ = \\textbf{Y}^T\\textbf{Y}-\\textbf{Y}^T\\textbf{X} \\beta^T - \\textbf{X}^T \\textbf{Y} + \\beta^T\\textbf{X}^T\\textbf{X} \\beta$$ Matrix Differentiation Before we continue, we will first remind ourselves of the following: If we are given two independent matrices $x$, and $A$, where $x$ is an m by 1 matrix and $A$ is an n by m matrix, then; for $y=A$ $\\rightarrow$ $\\frac{dy}{dx}=0$, for $y=Ax$", "in this way. $\\begin{array}{rcl} \\left(\\begin{array}{*{5}{c}|c} 2 &1 &0 &-1 &0 &4 \\\\ 0 &1 &0 &1 &1 &4 \\\\ 1 &0 &-1 &2 &0 &0 \\end{array}\\right) &\\xrightarrow[]{-(1/2)\\rho_1+\\rho_3} &\\left(\\begin{array}{*{5}{c}|c} 2 &1 &0 &-1 &0 &4 \\\\ 0 &1 &0 &1 &1 &4 \\\\ 0 &-1/2 &-1 &5/2 &0 &-2 \\end{array}\\right) \\\\[3em] &\\xrightarrow[]{(1/2)\\rho_2+\\rho_3} &\\left(\\begin{array}{*{5}{c}|c} 2 &1 &0 &-1 &0 &4 \\\\ 0 &1 &0 &1 &1 &4 \\\\ 0 &0 &-1 &3 &1/2 &0 \\end{array}\\right) \\end{array}$ The solution set is $\\{(w+(1/2)u,4-w-u,3w+(1/2)u,w,u)\\,\\big|\\, w,u\\in\\mathbb{R}\\}$. We write that in vector form. $\\{\\begin{pmatrix} x \\\\ y \\\\ z \\\\ w \\\\ u \\end{pmatrix}= \\begin{pmatrix} 0 \\\\ 4 \\\\ 0 \\\\ 0 \\\\ 0 \\end{pmatrix}+ \\begin{pmatrix} 1 \\\\ -1 \\\\ 3 \\\\ 1 \\\\ 0 \\end{pmatrix}w+ \\begin{pmatrix} 1/2 \\\\ -1 \\\\ 1/2 \\\\ 0 \\\\ 1 \\end{pmatrix}u \\,\\big|\\, w,u\\in\\mathbb{R}\\}$ Note again how well vector notation sets off the coefficients of each parameter. For instance, the third row of the vector form shows plainly that if $u$ is held fixed then $z$ increases three times as fast as $w$. That format also shows plainly that there are infinitely many solutions. For example, we can fix $u$ as $0$, let $w$ range over the real numbers, and consider the first component $x$. We get infinitely many first components and hence infinitely many solutions. Another thing shown plainly", "3&5\\\\ 1&2 \\end{pmatrix} \\begin{pmatrix} 1&0\\\\ 0&2 \\end{pmatrix}$ and so $A = \\begin{pmatrix} 3&5\\\\ 1&2 \\end{pmatrix} \\begin{pmatrix} 1&0\\\\ 0&2 \\end{pmatrix} \\begin{pmatrix} 3&5\\\\ 1&2 \\end{pmatrix}^{-1}$ #### Example 2 a) Diagonalize A, i.e find P (invertible) and B (diagonal) such that AP = PB b) Compute A5 $A = \\begin{pmatrix} 1&2\\\\ -1&4 \\end{pmatrix}$ ' Solution a) We are solving Ax = λx, where λ is a constant and x a column vector. Firstly $Ax - \\lambda Ix = 0$ $(A - \\lambda I)x = 0$ since x ≠ 0 we have $det(A - \\lambda I) = 0$ i.e. $\\det\\begin{pmatrix} 1-\\lambda&2\\\\ -1&4-\\lambda \\end{pmatrix} = 0$ $\\begin{matrix} (1-\\lambda)(4-\\lambda) + 2 &=& 0\\\\ \\lambda^2-5\\lambda + 6 &=& 0 \\end{matrix}$ $\\lambda = 3,\\ 2$ For λ = 3, $(A-3I)x = 0$ $\\begin{pmatrix} -2&2\\\\ -1&1 \\end{pmatrix} \\begin{pmatrix} x\\\\ y \\end{pmatrix} = 0$ Clearly $\\begin{pmatrix} x\\\\ y \\end{pmatrix} = \\begin{pmatrix} 1\\\\ 1 \\end{pmatrix}$ is a solution. Note that we do not accept x = 0 as an solution, because we assume x ≠ 0. Note also that $\\begin{pmatrix} x\\\\ y \\end{pmatrix} = \\begin{pmatrix} t\\\\ t \\end{pmatrix}$ for some constant t is also a solution. Indeed we could use x = y = 2, 3 or 4 as a solution, but for convenience we choose the simplest i.e. x = y = 1. For λ = 2, $(A-2I)x" ]

[ "that a direction vector is $$(3/2,1)$$. Of course, there's nothing wrong with using our method of chapter 1: if $$(a,b)$$ is a direction vector, then $$\\pm(-b,a)$$ are normal vectors, and vice versa. Example. Consider the line with point-parallel form equation $\\mathbf r(t)=\\begin{pmatrix}2\\\\-1\\end{pmatrix}+t\\begin{pmatrix}-3\\\\1\\end{pmatrix}.$ We can see that the line contains the point $$P(2,-1)$$, and that a direction vector is $$\\mathbf u=(-3,1)$$. Therefore, a normal vector is $$\\mathbf n=(1,3)$$, and from the point-normal form equation $\\left(\\begin{pmatrix}x & y\\end{pmatrix}-\\begin{pmatrix}2 & -1\\end{pmatrix}\\right)\\cdot\\begin{pmatrix}1\\\\3\\end{pmatrix}=0$ we get the standard form equation $x+3y=1.$ Note that a standard equation of a hyperplane in $$\\mathbf R^n$$ gives a $$1\\times n$$ matrix with one nonzero row. Therefore, the rank is 1, and you get an ($$n-1$$)-parameter family of solutions. This is why in $$\\mathbf R^2$$, from one normal vector you get one directional vector (unique up multiplication by a nonzero scalar). Example. Consider the plane in $$\\mathbf R^3$$ with standard equation $$2x-y+z=2$$. Solving the SLE, you get the 2-parameter family of solutions \\begin{align*} x&=s/2-t/2+2\\\\ y&=s\\\\ z&=t \\end{align*} That is, we have the parametrization $\\mathbf r(s,t)=\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix}=\\begin{pmatrix} s/2-t/2+2 \\\\ s \\\\ t \\end{pmatrix}=\\begin{pmatrix}2\\\\0\\\\0\\end{pmatrix}+\\begin{pmatrix}1/2\\\\1\\\\0\\end{pmatrix}+\\begin{pmatrix}-1/2\\\\0\\\\1\\end{pmatrix}.$ You can see that a plane is determined by one point and two nonparallel direction vectors. To get the normal vector back from the two direction vectors, you need to find a nonzero vector that is", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "starting with two vectors: one vector $\\vec{r}_0$ drawn from the origin to a chosen point of the line (the so-called position vector of this point) and another vector (the so-called {\\em direction vector}) $\\vec{u}$ determining the direction of the line. Then there exists a real number $t$ such that the position vector $\\vec{r}$ of an arbitrary point of the line can be expressed as \\begin{align} \\vec{r} \\;=\\; \\vec{r}_0\\!+\\!t\\vec{u}. \\end{align} This is the {\\em vector-formed equation of the line}. Since every vector can be given by three coordinates of the end-point, we may write $$\\vec{r} \\;=\\; \\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix},\\quad \\vec{r}_0 \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix},\\quad \\vec{u} \\;=\\; \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!.$$ $$\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix} \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix}+t \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!,$$ and we obtain three scalar equations \\begin{align} \\begin{cases} x \\;=\\; x_0\\!+\\!ta\\\\ y \\;=\\; y_0\\!+\\!tb\\\\ z \\;=\\; z_0\\!+\\!tc. \\end{cases} \\end{align} These are the {\\em parametric equations of the line} (3), with the parameter $t$ taking all real values. The contents of (4) is often written with proportions: $$\\frac{x\\!-\\!x_0}{a} \\;=\\; \\frac{y\\!-\\!y_0}{b} \\;=\\; \\frac{z\\!-\\!z_0}{c}\\;\\;\\; (= t)$$ \\begin{thebibliography}{8} \\bibitem{LL}{\\sc L. Lindel\\\"of}: {\\em Analyyttisen geometrian oppikirja}.\\, Kolmas painos.\\, Suomalaisen Kirjallisuuden Seura, Helsinki (1924). \\end{thebibliography} %%%%% %%%%% \u001bnd{document}", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "found using the following formula[1]: ${\\displaystyle proj_{\\overrightarrow {B}}{\\overrightarrow {A}}={\\frac {{\\overrightarrow {A}}\\cdot {\\overrightarrow {B}}}{|{\\overrightarrow {B}}|^{2}}}{\\overrightarrow {B}}}$ ## Vector Equations A basic vector equation consists of a vector and a variable coefficient of another vector. The standalone vector is known as the position vector and the vector with the variable coefficient is known as the direction vector. The position vector indicates where in the 3D space the vector is located, and the direction vector shows which way the vector pointing. The vector for the position vector can be the position vector of any point on the line. A typical equations might look like this: ${\\displaystyle {\\textbf {r}}={\\textbf {a}}+x{\\textbf {b}}}$ or ${\\displaystyle {\\textbf {r}}={\\begin{pmatrix}2\\\\2\\\\2\\end{pmatrix}}+x{\\begin{pmatrix}1\\\\-2\\\\3\\end{pmatrix}}}$ ### Checking if lines are parallel Lines are parallel when their direction vectors are scalar multiples of each other. Example: Line 1: ${\\displaystyle {\\textbf {r}}={\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}+\\lambda {\\begin{pmatrix}1\\\\-2\\\\3\\end{pmatrix}}}$ Line 2: ${\\displaystyle {\\textbf {s}}={\\begin{pmatrix}2\\\\5\\\\7\\end{pmatrix}}+\\mu {\\begin{pmatrix}-2\\\\4\\\\-6\\end{pmatrix}}}$ They are parallel as ${\\displaystyle -2{\\begin{pmatrix}1\\\\-2\\\\3\\end{pmatrix}}={\\begin{pmatrix}-2\\\\4\\\\-6\\end{pmatrix}}}$ . ### Checking if a point lies on a vector line equation So if the point has the position vector ${\\displaystyle {\\textbf {P}}={\\begin{pmatrix}2\\\\4\\\\4\\end{pmatrix}}}$ and the line equation is ${\\displaystyle {\\textbf {r}}={\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}+t{\\begin{pmatrix}2\\\\4\\\\3\\end{pmatrix}}}$ and you want to see if P lies on the line, then you'll need to do the following: Set the point equal to the equation: ${\\displaystyle {\\begin{pmatrix}2\\\\4\\\\4\\end{pmatrix}}={\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}+t{\\begin{pmatrix}2\\\\4\\\\3\\end{pmatrix}}}$ From this you can then make 3 separate equations from each", "to our earlier point and direction form. So using our (1, 3) to (6, 7) example, you get: $\\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + \\text{scalar} \\times \\begin{pmatrix} 5 \\\\ 4 \\end{pmatrix}$ The difference now is that the line extends from $\\text{scalar}=0$ (the start-point) to $\\text{scalar}=1$ (the end-point). Any values for the scalar variable outside the 0–1 range would be on the line if it’s infinite, but are not when it’s finite. ## End of the Line Today’s post was a lot of theory without an actual application, but we’ll be using it in our next post when we’ll look at finding out if two finite lines intersect.", "lines have the same $x$-intercept. • These lines both go through the point $(1, 5)$. • These lines … Can you sort them into the correct pairs and complete the final description? While sketching the lines could be helpful, 12 might be too many on one graph. Are there any pairs we can match up by just looking at the equations? $\\mathbf{r}= \\begin{pmatrix} 0 \\\\ -4 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} 4 \\\\ 6 \\end{pmatrix}$ $\\mathbf{r}= \\begin{pmatrix} 0 \\\\ -4 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} 1 \\\\ 6 \\end{pmatrix}$ Equations 7 and 10 have the same position vector which also happens to be their $y$-interept. Do you think there are particular features of lines that are easier to spot when the equation is written in vector form compared to Cartesian form? $\\mathbf{r}= \\begin{pmatrix} 2 \\\\ 9.5 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} -2 \\\\ -8 \\end{pmatrix}$ $\\mathbf{r}= \\begin{pmatrix} 0 \\\\ 4 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} 1 \\\\ 4 \\end{pmatrix}$ Equations 3 and 6 are parallel lines, but it looks like they have different direction vectors. How do you know these lines are parallel? What is the same about the two direction vectors? $\\mathbf{r}= \\begin{pmatrix} 0 \\\\ -6 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} -1 \\\\ 4 \\end{pmatrix}$ $\\mathbf{r}= \\begin{pmatrix} 1 \\\\ 1 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} 4 \\\\", "+0 # shortest distance between two lines via vector equation 0 255 4 +844 For a i got r= (-2,3,4) + λ(3,2,-7) for b so far i wanted to identify a point A on l1 and a point b on l2 i assumed the two lines are parallel as they have the same direction vector $$A=\\begin{pmatrix} -2+3λ\\\\ 3+2λ\\\\ 4-7λ \\end{pmatrix}$$ $$B= \\begin{pmatrix} 3\\\\ 1+2λ\\\\ -2-7λ \\end{pmatrix}$$ used this to get $$AB= \\begin{pmatrix} 5-3λ\\\\ -2\\\\ -6 \\end{pmatrix}$$ used this and my direction vector to find λ=53/9 via dot product substituted and calculated the magnitude which gave me 14.26 but my answer should be 3.57 please identify what went wrong, thanks Feb 10, 2019 #1 +2 The equation of the first line can be written as r1 = (-2, 3, 4) + s(3, 2, -7), (as you calculated), and the equation of the second as r2 = (0, 1, -2) + t(3, 2, -7), s and t arbitrary parameters. A vector d connecting two points, one on each line, would be given by d = r2 - r1 = (2, -2, -6) + (t - s)(3, 2 ,-7) = (2, -2, -6) + w(3, 2, -7), w for convenience. The distance between the two points will be the magnitude of this vector, so what you are looking for is the", "for example in $\\mathbb{R}^2,$ $$\\begin{pmatrix}a\\\\b\\end{pmatrix} = a\\begin{pmatrix}1\\\\0\\end{pmatrix}+b\\begin{pmatrix}0\\\\1\\end{pmatrix}$$ Other bases are possible, such as $$\\mathbb{R}^2 = \\text{span}\\left\\{\\begin{pmatrix}3\\\\0\\end{pmatrix},\\begin{pmatrix}0\\\\2\\end{pmatrix}\\right\\} = \\text{span}\\left\\{\\begin{pmatrix}1\\\\1\\end{pmatrix},\\begin{pmatrix}1\\\\-1\\end{pmatrix}\\right\\}$$ Although 'direction giving' can be a little more complicated here, its still possible. As long as the set can describe (using linear combinations) any coordinate in the space and there are no redundant vectors included, the set forms a basis. Although $$\\mathbb{R}^2 =\\text{span}\\left\\{\\begin{pmatrix}1\\\\0\\end{pmatrix},\\begin{pmatrix}0\\\\1\\end{pmatrix}, \\begin{pmatrix}1\\\\1\\end{pmatrix}\\right\\}$$ is true, we have more vectors than necessary. The third is redundant, since the directions to get to (1,1) are possible with just the first two. This is not a basis. For vector spaces, bases have a unique size, and that is the definition of dimension.", "+ 1.5 \\times \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix}$ $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 2 \\end{pmatrix} + \\begin{pmatrix} 1.5 \\\\ 4.5 \\end{pmatrix}$ $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} 1.5 \\\\ 6.5 \\end{pmatrix}$ However, there’s no such value of $\\text{scalar}$ for the point (2, 9), because it’s not on the line. Two slightly unexpected features of this representation are that it doesn’t matter which point on the line you pick for the expression, and you can multiply the direction vector by any constant without it changing the line. So the above line is equivalent to either of the following: $\\begin{pmatrix} 1 \\\\ 5 \\end{pmatrix} + \\text{scalarB} \\times \\begin{pmatrix} -1 \\\\ -3 \\end{pmatrix}$ $\\begin{pmatrix} -10 \\\\ -28 \\end{pmatrix} + \\text{scalarC} \\times \\begin{pmatrix} 6 \\\\ 18 \\end{pmatrix}$ #### Finite line: two endpoints The simplest way to represent a finite line is to simply write down its two endpoints. So you might have a line from (1, 3) to (6, 7). That’s a good, simple, unambiguous definition of the line. However, to do any calculations involving the line, you usually need to convert it to another representation, such as the last on our list: #### Finite line: point and direction You can take the two endpoints of a line and turn it into a form that looks identical", "# Perpendicular lines and vectors My textbook is confusing me a little. Here is a worked example from my textbook: Line $$l$$ has the equation $$\\begin{pmatrix}3\\\\ -1\\\\ 0\\end{pmatrix}+\\lambda \\begin{pmatrix}1\\\\ -1\\\\ 1\\end{pmatrix}$$ and point $$A$$ has co-ordinates $$(3, 9, -2)$$. Find the coordinates of point $$B$$ on $$l$$ so that $$AB$$ is perpendicular to $$l$$. $$\\vec{AB\\cdot }\\begin{pmatrix}1\\\\ -1\\\\ 1\\end{pmatrix}=0$$ $$\\vec{OB}=r=\\begin{pmatrix}3+\\lambda \\\\ -1-\\lambda \\\\ \\lambda \\end{pmatrix}$$ $$\\vec{AB}=\\begin{pmatrix}\\lambda \\\\ -10-\\lambda \\\\ \\lambda +2\\end{pmatrix}$$ $$\\begin{pmatrix}\\lambda \\\\ -10-\\lambda \\\\ \\lambda +2\\end{pmatrix}\\cdot \\begin{pmatrix}1\\\\ -1\\\\ 1\\end{pmatrix}=0$$ $$3\\lambda= -12, \\lambda = -4$$ Coordinates of $$B$$: $$(-1, 3, -4)$$ The thing I don't understand is why they found the dot product of the line AB and the direction vector of line l. My textbook does mention that to check whether two vectors are perpendicular, $$a\\cdot b = 0$$ and for lines, the dot product of their direction vectors = 0. So why did they mix both here? Didn't they use the entire line $$AB$$ and then just the direction vector of line l? Or am I missing something as usual? There is nothing special here the dot product $AB\\cdot v$ gives the condition of orthogonality between the vector from $A$ to $B$ and the direction vector of the given line. Note also that vector $AB$ is a direction vector for the line orthogonal to the given line", "$t$ and the variable given in the answer $t'$. Then let's see when the two vectors are equal. $\\begin{pmatrix}2 \\\\ 2 \\\\ 4\\end{pmatrix} + \\begin{pmatrix}-\\tfrac{1}{2} \\\\ \\tfrac{1}{2} \\\\ 1\\end{pmatrix}t = \\begin{pmatrix}2 \\\\ 2 \\\\ 4\\end{pmatrix} + \\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}t'$ when $t=2t'$. Since the map $t \\mapsto 2t$ is a bijection, you wind up with the same line either way. Same thing for line 2. so if i i do my ans in this way. wouldnt my ans be wrong? 4. ## Re: vector (simplified b1 and b2 ) Originally Posted by delso so if i i do my ans in this way. wouldnt my ans be wrong? I just showed that you get the same line either way. Why would you get the wrong answer that way? If two line intersect and you have two different formulas for each line, it is still the same two lines. It doesn't matter that the formulas are not identical. They still represent the same lines, so they will still intersect in the same place. 5. ## Re: vector (simplified b1 and b2 ) do u mean t=2t\" just different in terms of length ? but it's still the same line with same gradient?", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "& 2\\\\1 & -2 \\\\\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\vec 0 \\, .$ This gives us the equalities $$-x +2y = 0$$ and $$x -2y = 0$$ which both give the line $$y = \\frac{1}{2}x$$. Any point on this line, so for example $$\\begin{pmatrix}2\\\\1\\end{pmatrix}$$, is an eigenvector with eigenvalue $$\\lambda = 5$$. Now lets find the eigenvector for $$\\lambda = 2$$ $\\begin{pmatrix}4-2 & 2\\\\1 & 3-2 \\\\\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}2 & 2\\\\1 & 1 \\\\\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\vec 0 ,$ which gives the equalities $$2x+2y = 0$$ and $$x+y = 0$$. (Notice that these equations are not independent of one another, so our eigenvalue must be correct.) This means any vector $$v = \\begin{pmatrix}x\\\\y\\end{pmatrix} where \\(y = -x$$ , such as $$\\begin{pmatrix}1\\\\-1\\end{pmatrix}, or any scalar multiple of this vector, \\(\\textit{i.e.}$$ any vector on the line $$y = -x$$ is an eigenvector with eigenvalue 2. This solution could be written neatly as $\\lambda_{1} = 5, \\, v_{1}= \\begin{pmatrix}2\\\\1\\end{pmatrix} \\, \\text{ and } \\lambda_{2} = 2, \\, v_{2}=\\begin{pmatrix}1\\\\-1\\end{pmatrix}.$ Jordan Block Example Consider the matrix $J_{2} = \\begin{pmatrix}\\lambda & 1\\\\ 0 & \\lambda\\end{pmatrix},$ and we note that we can just read off the eigenvector $$e_{1}$$ with eigenvalue $$\\lambda$$. However the characteristic polynomial of $$J_{2}$$ is $$P_{J_{2}}(\\mu) = (\\mu - \\lambda)^{2}$$ so the only possible eigenvalue is $$\\lambda$$, but we claim it does not", "with the parameter $t$ has its whole body in the line— it is a direction vector for the line. Note that points on the line to the left of $x=1$ are described using negative values of $t$. In $\\mathbb{R}^3$, the line through $(1,2,1)$ and $(2,3,2)$ is the set of (endpoints of) vectors of this form and lines in even higher-dimensional spaces work in the same way. If a line uses one parameter, so that there is freedom to move back and forth in one dimension, then a plane must involve two. For example, the plane through the points $(1,0,5)$, $(2,1,-3)$, and $(-2,4,0.5)$ consists of (endpoints of) the vectors in $\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix} +t\\cdot\\begin{pmatrix} 1 \\\\ 1 \\\\ -8 \\end{pmatrix} +s\\cdot\\begin{pmatrix} -3 \\\\ 4 \\\\ -4.5 \\end{pmatrix} \\,\\big|\\, t,s\\in\\mathbb{R} \\}$ (the column vectors associated with the parameters $\\begin{pmatrix} 1 \\\\ 1 \\\\ -8 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 1 \\\\ -3 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix} \\qquad \\begin{pmatrix} -3 \\\\ 4 \\\\ -4.5 \\end{pmatrix} = \\begin{pmatrix} -2 \\\\ 4 \\\\ 0.5 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix}$ are two vectors whose whole bodies lie in the plane). As with the line, note that some points in this plane are described with negative $t$'s or negative $s$'s or", "start at the origin and end at a point in space. For example, below is the vector (we’ll name it $$\\mathbf{v}$$) from above in 2-D space. Also pictured is the vector $\\mathbf{u} = \\begin{pmatrix} -2 \\\\ -2 \\end{pmatrix}$ on the same 2-D coordinate system. Often, we might not specify the actual values of a vector because we’re working generically with any vector in 2-dimensional space. In that case, we would use a variable such as $$\\mathbf{x}$$ to indicate this vector and we would write $$\\mathbf{x} \\in \\mathbb{R}^{2}$$ to indicate that $$\\mathbf{x}$$ is a 2-dimensional vector whose entries are real numbers. This means that they can take on any value between $$-\\infty$$ and $$\\infty$$. In our posts, we will represent vectors with bold face, lower case letters. If we write $$\\mathbf{a} \\in \\mathbb{R}^{5}$$, this means that the vector $$\\mathbf{a}$$ is a $$5$$-dimensional vector whose entries are real numbers. If we don’t specify the entries in $$\\mathbf{a}$$, then we indicate its entries with non-bold letters followed by the entry position in subscript $\\mathbf{a} = \\begin{pmatrix} a_{1} \\\\ a_{2} \\\\ a_{3} \\\\ a_{4} \\\\ a_{5} \\end{pmatrix}.$ Throughout our posts, we will always assume that vectors such as $$\\mathbf{x}$$ and $$\\mathbf{a}$$ are column vectors as we presented them above. If we want to indicate a row vector, we will specify it with", "+0 # Find a direction vector $\\mathbf{d} = \\begin{pmatrix}d_1 \\\\ d_2 \\\\d_3 \\end{pmatrix}$for the line through $B = (1, 1, 2)$ and \\$C= (2, 3, 0 244 1 Find a direction vector $$\\mathbf{d} = \\begin{pmatrix}d_1 \\\\ d_2 \\\\d_3 \\end{pmatrix}$$ for the line through B = (1, 1, 2) and C= (2, 3, 1) such that $$d_1 + d_2 + d_3 = 10.$$ Mar 14, 2019 #1 +23835 +2 Find a direction vector $$\\mathbf{d} = \\begin{pmatrix}d_1 \\\\ d_2 \\\\d_3 \\end{pmatrix}$$ for the line through $$B = (1, 1, 2)$$ and $$C= (2, 3, 1)$$ such that $$d_1 + d_2 + d_3 = 10$$. Line: $$\\begin{array}{|rcll|} \\hline \\vec{x} &=& \\vec{B} + \\underbrace{\\lambda \\left( \\vec{C} - \\vec{B} \\right) }_{=\\vec{d}} \\\\\\\\ \\vec{d} &=& \\lambda \\left( \\vec{C} - \\vec{B} \\right) \\\\ \\vec{d} &=& \\lambda \\begin{pmatrix}2-1 \\\\ 3-1 \\\\1-2 \\end{pmatrix} \\\\ \\vec{d} &=& \\lambda \\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\end{pmatrix} \\\\ \\begin{pmatrix}d_1 \\\\ d_2 \\\\d_3 \\end{pmatrix} &=& \\lambda \\begin{pmatrix} 1 \\\\ 2 \\\\ -1 \\end{pmatrix} \\\\\\\\ && d_1 = \\lambda \\\\ && d_2 =2\\lambda \\\\ && d_3 = -\\lambda \\\\ \\hline && d_1+d_2+d_3 = \\lambda+2\\lambda -\\lambda \\\\ && 10 = 2\\lambda \\\\ && \\lambda =\\dfrac{10}{2} \\\\ && \\lambda = 5 \\\\\\\\ \\vec{d} &=& 5 \\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\end{pmatrix} \\\\ \\mathbf{\\vec{d}} & \\mathbf{=} & \\mathbf{ \\begin{pmatrix} 5 \\\\ 10 \\\\ -5 \\end{pmatrix} }", "plane has normal vector $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$$ and the second has normal vector $$\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $$v_1,v_2$$ is $$v_1\\times v_2$$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\times \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $$y=0$$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $$x=0, z=1$$, thus the line of intersection is $$\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$$ • Can you please elaborate as to what you did to find any random point on the line?I didn't quite understand – Karan Singh Apr 26 '16 at 9:01 • In your set of linear equations, where did that last '-1' (in equation 1) come from?. And where did that last '+2' (in equation 2) come from?. – loldrup May 21 '16 at 20:45 • How did you parameterize? How did you find the normal vectors? What is the upward caret operator? – rocksNwaves May 1 '18 at 14:02", "0 \\\\ 2x + 3y - 2z + 2 = 0.$$ Notice that, these are two equations in three variables, so you have a free variable say $z=t$, then we have $$x + 2y = 1-t \\\\ 2x + 3y = 2t-2.$$ Solving the last system gives $$\\left\\{ x=-7+7\\,t,y=4-4\\,t \\right\\} .$$ Then the parametrized equation of the line is given by $$(x,y,z)= (-7+7t, 4-4t,t)=(-7,4,0)+(7,-4,1)t .$$ The first plane has normal vector $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$$ and the second has normal vector $$\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $$v_1,v_2$$ is $$v_1\\times v_2$$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\times \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $$y=0$$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $$x=0, z=1$$, thus the line of intersection is $$\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$$ • Can you please elaborate as to what you did to find any random point on the line?I didn't quite understand – Karan Singh Apr 26 '16 at 9:01 • In your set of linear equations, where did that last '-1' (in equation 1) come from?. And where did that last '+2' (in equation 2) come from?. – loldrup May 21 '16 at 20:45 •", "# Compute the orthogonal projection $w$ of $u$ on a line $L$ A vector $u$ and a line $L$ in $\\mathbb{R}^2$ are given. Compute the orthogonal projection $w$ of $u$ on $L$, and use it to compute the distance $d$ from the endpoint of $u$ to $L$ $u = \\begin{pmatrix} 5 \\\\ 0 \\end{pmatrix}$ $y = 0$ I know to that $w = \\frac{u \\cdot v}{||v||^2}v$ and that $d = u - w$ The answer in the book says the answer is $w = \\begin{pmatrix}5 \\\\0 \\end{pmatrix}$ and $d = 0$ How do I find $v$ to solve the problem? • What is your line $L$? Is it the line given by the equation $y = 0$? – Eli Rose Dec 14 '15 at 4:16 • Yes, the book only gives $u$ and the equation of Line $L$ – user273323 Dec 14 '15 at 4:25 The vector $v$ can be any vector in the direction of the line $L$. For a line $ax + by = c$, the vector $\\begin{pmatrix}b\\\\-a\\end{pmatrix}$, or any non-zero multiple of it, lies in the direction of the line. So for $L: y=0$, alias $0x+1y=0$, the vector $v = \\begin{pmatrix}1\\\\0\\end{pmatrix}$ will do. Put this into your equation gives $w=\\begin{pmatrix}5\\\\0\\end{pmatrix}$ as expected. Or, more simply, you can observe that $u$ is a vector parallel to" ]

[ "vector ${\\displaystyle (-1,0,-1)^{T}}$ is a multiple of ${\\displaystyle (1,0,1)^{T}}$. That is, the direction represented by ${\\displaystyle (-1,0,-1)^{T}}$ is the same as the direction represented by ${\\displaystyle (1,0,1)^{T}}$. Hence, we can remove this vector from ${\\displaystyle E}$ and obtain a new generator ${\\displaystyle E_{1}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}},{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Can we reduce the size of this generator, as well? Yes, since ${\\displaystyle {\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}+{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}}$ So ${\\displaystyle (2,1,1)^{T}}$ adds no new direction that is not already spanned by ${\\displaystyle (1,0,1)^{T}}$ and ${\\displaystyle (1,1,0)^{T}}$. We thus obtain a smaller generator ${\\displaystyle E_{2}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Now we cannot reduce the generator any further without losing the property of it being a generator. For if we remove any of the three vectors in ${\\displaystyle E_{2}}$, it is no longer in the span of the remaining two vectors. For ${\\displaystyle (1,0,1)^{T}}$, for example, we see this as follows: Suppose we had some ${\\displaystyle \\lambda ,\\mu \\in \\mathbb {R} }$, so that ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=\\lambda {\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+\\mu {\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}}$ Then ${\\displaystyle \\lambda =-\\mu }$ would have to hold because the second component must be zero on both sides. Because of the third component, ${\\displaystyle \\mu ={\\frac {1}{2}}}$ must be true. Thus we obtain the contradiction ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=-{\\frac {1}{2}}{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+{\\frac {1}{2}}{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}0.5\\\\0\\\\1\\end{pmatrix}}}$ So ${\\displaystyle (1,0,1)^{T}}$ is not in ${\\displaystyle \\operatorname {span} ((1,1,0)^{T},(2,1,2)^{T})}$. So we have found a generator containing linearly independent vectors (a minimal generator). In summary, we proceeded", "& 2\\\\1 & -2 \\\\\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\vec 0 \\, .$ This gives us the equalities $$-x +2y = 0$$ and $$x -2y = 0$$ which both give the line $$y = \\frac{1}{2}x$$. Any point on this line, so for example $$\\begin{pmatrix}2\\\\1\\end{pmatrix}$$, is an eigenvector with eigenvalue $$\\lambda = 5$$. Now lets find the eigenvector for $$\\lambda = 2$$ $\\begin{pmatrix}4-2 & 2\\\\1 & 3-2 \\\\\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}2 & 2\\\\1 & 1 \\\\\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\vec 0 ,$ which gives the equalities $$2x+2y = 0$$ and $$x+y = 0$$. (Notice that these equations are not independent of one another, so our eigenvalue must be correct.) This means any vector $$v = \\begin{pmatrix}x\\\\y\\end{pmatrix} where \\(y = -x$$ , such as $$\\begin{pmatrix}1\\\\-1\\end{pmatrix}, or any scalar multiple of this vector, \\(\\textit{i.e.}$$ any vector on the line $$y = -x$$ is an eigenvector with eigenvalue 2. This solution could be written neatly as $\\lambda_{1} = 5, \\, v_{1}= \\begin{pmatrix}2\\\\1\\end{pmatrix} \\, \\text{ and } \\lambda_{2} = 2, \\, v_{2}=\\begin{pmatrix}1\\\\-1\\end{pmatrix}.$ Jordan Block Example Consider the matrix $J_{2} = \\begin{pmatrix}\\lambda & 1\\\\ 0 & \\lambda\\end{pmatrix},$ and we note that we can just read off the eigenvector $$e_{1}$$ with eigenvalue $$\\lambda$$. However the characteristic polynomial of $$J_{2}$$ is $$P_{J_{2}}(\\mu) = (\\mu - \\lambda)^{2}$$ so the only possible eigenvalue is $$\\lambda$$, but we claim it does not", "5 \\end{pmatrix}$$ var v = [1.0, 2.0] var s = 3.0 var vsresult = [Double](count : v.count, repeatedValue : 0.0) vDSP_vsaddD(v, 1, &s, &vsresult, 1, vDSP_Length(v.count)) vsresult // returns [4.0, 5.0] $$\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix} \\times 3 = \\begin{pmatrix} 3 \\\\ 6 \\end{pmatrix}$$ vDSP_vsmulD(v, 1, &s, &vsresult, 1, vDSP_Length(v.count)) vsresult // returns [3.0, 6.0] $$\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix} \\div 3 = \\begin{pmatrix} 1/3 \\\\ 2/3 \\end{pmatrix}$$ vDSP_vsdivD(v, 1, &s, &vsresult, 1, vDSP_Length(v.count)) vsresult // returns [0.333333333333333, 0.666666666666667] ## Vector & Vector Vector-vector operations pose no challenge to Accelerate and the associated functions look like vDSP_v***D(vector_1, 1, vector_2, 1, &result, 1, length_of_vector) Here are a few worked-out examples: $$\\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix} + \\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix} = \\begin{pmatrix} 5 \\\\ 9 \\end{pmatrix}$$ var v1 = [2.0, 5.0] var v2 = [3.0, 4.0] var vvresult = [Double](count : 2, repeatedValue : 0.0) vDSP_vaddD(v1, 1, v2, 1, &vvresult, 1, vDSP_Length(v1.count)) vvresult // returns [5.0, 9.0] $$\\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix} \\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix} = \\begin{pmatrix} 6 \\\\ 20 \\end{pmatrix}$$ vDSP_vmulD(v1, 1, v2, 1, &vvresult, 1, vDSP_Length(v1.count)) vvresult // returns [6.0, 20.0] $$\\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix} {\\bigg/} \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix} = \\begin{pmatrix} 1.5 \\\\ 0.8 \\end{pmatrix}$$ vDSP_vdivD(v1, 1, v2, 1, &vvresult, 1, vDSP_Length(v1.count)) vvresult // returns [1.5, 0.8] ## Dot Product $$\\begin{pmatrix}", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "&\\xrightarrow[]{-2\\rho_1+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 1 &-1 &-1m+0n \\ 0 &1 &1m+1n \\ 0 &2 &3m+3n \\end{array}\\right) \\ &\\xrightarrow[]{-2\\rho_2+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 1 &-1 &-1m+0n \\ 0 &1 &1m+1n \\ 0 &0 &1m+1n \\end{array}\\right) \\end{array}$ to deduce that the only vectors from S2 that are also in S1 are of the form $\\begin{pmatrix} -1 \\\\ 1 \\\\ 1 \\end{pmatrix}m +\\begin{pmatrix} 0 \\\\ 1 \\\\ 3 \\end{pmatrix}(-m).$ For instance, $\\begin{pmatrix} -1 \\\\ 1 \\\\ 1 \\end{pmatrix}$ is in S2 but not in S1. 5. These sets are equal. First we change the parameters: $S_2=\\{\\begin{pmatrix} 3 \\\\ 7 \\\\ 7 \\end{pmatrix}m +\\begin{pmatrix} 1 \\\\ 3 \\\\ 1 \\end{pmatrix}n \\,\\big|\\, m,n\\in\\mathbb{R}\\}.$ Now, to show that $S_1\\subseteq S_2$, we solve $\\begin{array}{*{2}{rc}r} 3m &+ &1n &= &1t+2s \\ 7m &+ &3n &= &3t+4s \\ 7m &+ &1n &= &1t+6s \\end{array}$ with Gauss' method $\\begin{array}{rcl} \\left(\\begin{array}{*{2}{c}|c} 3 &1 &1t+2s \\ 7 &3 &3t+4s \\ 7 &1 &1t+6s \\end{array}\\right) &\\xrightarrow[(-7/3)\\rho_1+\\rho_3]{(-7/3)\\rho_1+\\rho_2} &\\left(\\begin{array}{*{2}{c}|c} 3 &1 &1t+2s \\ 0 &2/3 &(2/3)t-(2/3)s \\ 0 &-4/3 &(-4/3)t+(4/3)s \\end{array}\\right) \\ &\\xrightarrow[]{2\\rho_2+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 3 &1 &1t+2s \\ 0 &2/3 &(2/3)t-(2/3)s \\ 0 &0 &0 \\end{array}\\right) \\end{array}$ to get that $\\begin{pmatrix} 1 \\\\ 3 \\\\ 1 \\end{pmatrix}t +\\begin{pmatrix} 2 \\\\ 4 \\\\ 6 \\end{pmatrix}s =\\begin{pmatrix} 3 \\\\ 7 \\\\ 7 \\end{pmatrix}(s) +\\begin{pmatrix} 1 \\\\ 3 \\\\ 1 \\end{pmatrix}(t-s)$ and so $S_1\\subseteq S_2$. The proof that $S_2\\subseteq S_1$ involves", "a set of vectors is the set of all linear combinations of these vectors. So the span of $\\{\\begin{pmatrix}1\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\end{pmatrix}\\}$ would be the set of all linear combinations of them, which is $\\mathbb{R}^2$. The span of $\\{\\begin{pmatrix}2\\\\0\\end{pmatrix}, \\begin{pmatrix}1\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\end{pmatrix}\\}$ is also $\\mathbb{R}^2$, although we don't need $\\begin{pmatrix}2\\\\0\\end{pmatrix}$ to be so. So both these two sets are said to be the spanning sets of $\\mathbb{R}^2$. However, only the first set $\\{\\begin{pmatrix}1\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\end{pmatrix}\\}$ is a basis of $\\mathbb{R}^2$, because the $\\begin{pmatrix}2\\\\0\\end{pmatrix}$ makes the second set linearly dependent. Also, the set $\\{\\begin{pmatrix}2\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\end{pmatrix}\\}$ can also be a basis for $\\mathbb{R}^2$. Because its span is also $\\mathbb{R}^2$ and it is linearly independent. For another example, the span of the set $\\{\\begin{pmatrix}1\\\\1\\end{pmatrix}\\}$ is the set of all vectors in the form of $\\begin{pmatrix}a\\\\a\\end{pmatrix}$. • So a basis of $\\mathbb R^n$ is a smallest spanning set of $\\mathbb R^n$, i.e any spanning set of $\\mathbb R^n$ with cardinality $n$? – Zaz Oct 26 '18 at 23:17 • @Zaz: That is correct. Oct 30 '18 at 10:28 The basis is a combination of vectors which are linearly independent and which spans the whole vector V. Suppose we take a system of $R^2$. Now as you said,$(1,0)$ and $(0,1)$ are the basis in this system and we want to find any $(x,y)$ in this system. $$(x,y)=x(1,0)+y(0,1)$$", "{\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}t={\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}(6-3t)}$ and so any vector in the form for ${\\displaystyle S_{1}}$ can be stated in the form needed for inclusion in ${\\displaystyle S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we look for ${\\displaystyle t}$ so that these equations hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}1&+&0t&=&1+0s\\\\2&+&3t&=&8-1s\\end{array}}}$ Rewrite that as ${\\displaystyle {\\begin{array}{*{1}{rc}r}1&=&1\\\\t&=&2-(1/3)s\\end{array}}}$ and so ${\\displaystyle {\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}s={\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}(2-(1/3)s).}$ 2. These two are equal. To show that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we check that for any ${\\displaystyle t,s}$ we can find an appropriate ${\\displaystyle m,n}$ so that these hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}4m&-&4n&=&1t+2s\\\\7m&-&2n&=&3t+1s\\\\7m&-&10n&=&1t+5s\\end{array}}}$ Use Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\7&-2&3t+1s\\\\7&-10&1t+5s\\end{array}}\\right)&{\\xrightarrow[{(-7/4)\\rho _{1}+\\rho _{3}}]{(-7/4)\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&-3&-(3/4)t+(6/4)s\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to conclude that ${\\displaystyle {\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}t+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}s={\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}((1/4)t-(1/2)s)}$ and so ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&+&2s&=&4m-4n\\\\3t&+&1s&=&7m-2n\\\\1t&+&5s&=&7m-10n\\end{array}}}$ with Gaussian reduction ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\3&1&7m-2n\\\\1&5&7m-10n\\end{array}}\\right)&{\\xrightarrow[{-\\rho _{1}+\\rho _{3}}]{-3\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&3&3m-6n\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to get ${\\displaystyle {\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}m+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}n={\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}(2m)+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}(m-2n)}$ and so any member of ${\\displaystyle S_{2}}$ can be expressed in the form needed for ${\\displaystyle S_{1}}$. 3. These sets are equal. To prove that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we must be able to solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}2m&+&4n&=&1t\\\\4m&+&8n&=&2t\\end{array}}}$ for ${\\displaystyle m}$ and ${\\displaystyle n}$ in terms of ${\\displaystyle t}$. Apply Gaussian reduction ${\\displaystyle \\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\4&8&2t\\end{array}}\\right){\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}\\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\0&0&0\\end{array}}\\right)}$ to conclude that any pair ${\\displaystyle m,n}$ where ${\\displaystyle 2m+4n=t}$ will do. For instance, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(0)}$ or ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((-3/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(t).}$ Thus ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&=&2m+4n\\\\2t&=&4m+8n\\end{array}}}$ with Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\2&4m+8n\\end{array}}\\right)&{\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\0&0\\end{array}}\\right)\\end{array}}}$ to deduce that any", "those vectors that form a linear combination for all vectors in $\\mathbb{C}^4$. My proof would go something like this: Let $x_1, x_2, x_3, x_4 \\in{\\mathbb{C}}$ be arbitrary. Thus, $x_1+x_2+x_3$, $x_1+x_4$, and $x_2+x_4$ are also arbitrary elements of $\\mathbb{C}$. This means that $x=\\begin{pmatrix}x_1+x_2+x_3\\\\x_1+x_4\\\\x_2+x_4\\\\x_3\\end{pmatrix} \\in{\\mathbb{C}^4}$ is also arbitrary. So a linear combination of $x$ with the given vectors is simply $x_1\\begin{pmatrix}1\\\\1\\\\0\\\\0\\end{pmatrix}+ x_2\\begin{pmatrix}1\\\\0\\\\1\\\\0\\end{pmatrix}+ x_3\\begin{pmatrix}1\\\\0\\\\0\\\\1\\end{pmatrix}+ x_4\\begin{pmatrix}0\\\\1\\\\1\\\\0\\end{pmatrix}+ 0\\begin{pmatrix}0\\\\1\\\\0\\\\1\\end{pmatrix}+ 0\\begin{pmatrix}0\\\\0\\\\1\\\\1\\end{pmatrix}$ Therefore, the given vectors span $\\mathbb{C}^4$ for all $x_i$. $\\blacksquare$ For the second question, I'd argue that any 4 of those given vectors can be shown to span $\\mathbb{C}^4$ in a similar manner. So there are ${6}\\choose{4}$ ways 4 of those vectors form a basis for $\\mathbb{C}^4$. • Could you define what you mean by arbitrary? – YoTengoUnLCD Jul 15 '15 at 2:43 • I might be abusing that word in my example, but I'm taking it to mean that if $z\\in{Z}$ is arbitrary , then whatever result I get for $z$ extends to all elements of the set $Z$. Would you say \"$z\\in{Z}$ is arbitrary\" is the functional equivalent of \"$\\forall z \\in{Z}$\"? – gbrlrz017 Jul 15 '15 at 2:48 • I see what you mean, I would avoid using that word altogether, it causes confusion. Also, there aren't 6 choose 4 ways to pick those vectors to form a basis, pick vectors", "\\\\ Y\\end{bmatrix} \\qquad (x,y) = [X,Y]P^T$ $q(\\mathbf{v}) = (x,y)A\\begin{pmatrix}x \\\\ y\\end{pmatrix} = [X,Y]P^T AP\\begin{bmatrix}X \\\\ Y\\end{bmatrix}$ If we can find an orthogonal change of basis $(P^T = P^{-1})$ then $P$ ... Eigenvalues of $A = \\begin{pmatrix}2 & 1 \\\\ 1 & 2\\end{pmatrix}$ \\begin{aligned} 0 &= \\begin{vmatrix}\\lambda I - A \\end{vmatrix} = \\begin{vmatrix} \\lambda -2 & -1 \\\\ -1 & \\lambda -2\\end{vmatrix}= (\\lambda -2)^2 - 1 \\\\ 0 &= (\\lambda -2)^2 - 1^2 = ((\\lambda -2)-1)((\\lambda -2)+1) \\\\ 0 &= (\\lambda -3)(\\lambda -1)\\end{aligned} $\\begin{cases}\\lambda_1 = 3 \\lambda_2 = 1\\end{cases}$ eigenvalues $3X^2 + Y^2$ is $q$ in the new system, Eigenvalue $\\lambda_1 = 3$ $\\begin{pmatrix} 3-2 & -1 \\\\ -1 & 3-2 \\end{pmatrix} \\begin{pmatrix}\\alpha \\\\ \\beta\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0\\end{pmatrix} \\begin{cases} \\alpha - \\beta = 0 \\\\ -\\alpha + \\beta = 0 \\end{cases} so \\alpha = \\beta \\begin{pmatrix}1 \\\\ 1\\end{pmatrix}$ will do Note $<\\begin{pmatrix}1 \\\\ 1\\end{pmatrix},\\begin{pmatrix}-1 \\\\ 1\\end{pmatrix}> = 0$ orthogonal basis of eigenvectors $\\|\\begin{pmatrix}1 \\\\ 1\\end{pmatrix}\\|^2 = 2 =\\|\\begin{pmatrix}-1 \\\\ 1\\end{pmatrix}\\|^2$ Corresponding orthonormal basis: $\\{\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}, \\begin{pmatrix}-\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\}$ But another can be $\\{\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}, \\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}\\}$ and two more candidates... We may take $P = \\begin{pmatrix} \\frac{1}{\\sqrt{2}} & -\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}}\\end{pmatrix} = (P^{-1})^T$ Note $det P = \\pm 1$ (rotation) Claim $\\begin{pmatrix} \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}}\\end{pmatrix} \\begin{pmatrix} 2 & 1", "$x(3x +2y -30)=0$, $y(2y-x-6)=0$ we get cases \\begin{align*} &x=0,\\ y=0 &&\\implies A_1=(0,0),\\\\ &x=0,\\ 2y-x-6=0 &&\\implies A_2=(0,3)\\\\ &y=0,\\ 3x+2y-30=0 &&\\implies A_3=(10,0),\\\\ &3x+2y-30=0,\\ 2y-x-6=0 &&\\implies A_4=(6,6). \\end{align*} (b) Linearizations at these points have matrices \\begin{align*} &A_1&&A_2&&A_3&&A_4\\\\ & \\begin{pmatrix} -30 &\\ \\ 0\\\\ 0 &-6 \\end{pmatrix} && \\begin{pmatrix} -24 &\\ \\ 0\\\\ -3 &6 \\end{pmatrix} && \\begin{pmatrix} 30 &20\\\\ 0 &-16 \\end{pmatrix} && \\begin{pmatrix} 18 &12\\\\ -6 &12 \\end{pmatrix}\\\\ \\text{with eigenvalues}\\\\ &\\{-30,-6\\} &&\\{-24,6\\} && \\{30,-16\\} && \\{15-\\sqrt{63}i,15-\\sqrt{63}i \\} \\end{align*} correspondingly. Therefore $A_1$ is a stable node, $A_2$ and $A_3$ are saddles, $A_4$ is an unstable focal point and since left bottom number is $-6<0$ it is clockwise oriented. Directions are: $A_1$: $\\mathbf{f}_1=(1,0)^T$, $\\mathbf{f}_2=(0,1)^T$ ; Since $-30=\\lambda_1 <\\lambda_2=-6$, all trajectories have an entry directions $\\pm \\mathbf{f}_2$ (except two, which have entry directions $\\pm \\mathbf{f}_1$). $A_2$: $\\mathbf{f}_1=(10,1)^T$ ($\\lambda_1=-24$)--stable, $\\mathbf{f}_2=(0,1)^T$ ($\\lambda_2=6$)--unstable. $A_3$: $\\mathbf{f}_1=(1,0)^T$ ($\\lambda_1=30$)--unstable, $\\mathbf{f}_2=(1,-2.3)^T$ ($\\lambda_2=-16$)--stable. (c)--(d) One should observe that either $x=0$ in every point of the trajectory, or in no point; and that $y=0$ in every point of the trajectory, or in no point. It allows us to make a skeleton'' of the phase portrait (thick black lines) on the figure; red lines are very approximate Remark: $A_4$ may not look as a focal point on the computer-generated plot but it is. The reason for this disparity is that for one rotation", "$$\\begin{pmatrix} y_1 \\\\ y_2 \\\\ y_3\\end{pmatrix} = \\left(\\begin{array}{rrr} a_{11} & a_{12} & a_{13}\\\\ a_{21} & a_{22} & a_{23}\\\\ a_{31} & a_{32} & a_{33}\\\\ \\end{array}\\right) \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix} = \\left(\\begin{array}{r} 1 a_{11} + 0 a_{12} + 0 a_{13}\\\\ 1 a_{21} + 0 a_{22} + 0 a_{23}\\\\ 1 a_{31} + 0 a_{32} + 0 a_{33}\\\\ \\end{array}\\right) = \\begin{pmatrix} a_{11} \\\\ a_{21} \\\\ a_{31} \\end{pmatrix},$$ (9.22) which is the first column in $\\mx{A}$. Thus the image $F(\\vc{e}_1)$ of the basis vector $\\vc{e}_1$ is the first column of $\\mx{A}$, denoted $\\vc{a}_{,1}$. Likewise, the second basis vector can be written $\\vc{e}_2 = 0 \\vc{e}_1 + 1 \\vc{e}_2 + 0 \\vc{e}_3$, and thus has the coordinates $(0, 1, 0)$. Its image is therefore $$\\left(\\begin{array}{rrr} a_{11} & a_{12} & a_{13}\\\\ a_{21} & a_{22} & a_{23}\\\\ a_{31} & a_{32} & a_{33}\\\\ \\end{array}\\right) \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix} = \\left(\\begin{array}{r} 0 a_{11} + 1 a_{12} + 0 a_{13}\\\\ 0 a_{21} + 1 a_{22} + 0 a_{23}\\\\ 0 a_{31} + 1 a_{32} + 0 a_{33}\\\\ \\end{array}\\right) = \\begin{pmatrix} a_{12} \\\\ a_{22} \\\\ a_{32} \\end{pmatrix},$$ (9.23) which is the second column vector $\\vc{a}_{,2}$ of the matrix $\\mx{A}$. Similarly for the third basis vector we get $$\\left(\\begin{array}{rrr} a_{11} & a_{12} & a_{13}\\\\ a_{21} & a_{22} & a_{23}\\\\ a_{31} & a_{32} & a_{33}\\\\ \\end{array}\\right) \\begin{pmatrix}", "to align the \\Rightarrows – daleif Oct 6 '16 at 7:59 This is an automatic solution, based on the eqparbox package. I defined an \\eqmathbox command, which accepts $2$ optional arguments (the end of name of the box, and the alignment: l, r or c) and one mandatory argument: the contents of the math box. \\documentclass{article} %\\usepackage{enumitem} \\usepackage{mathtools} \\usepackage{xparse} \\usepackage{eqparbox} \\NewDocumentCommand\\Eqmathbox{O{0}O{l}m}{\\eqmakebox[Form#1][#2]{\\ensuremath{ \\displaystyle#3}}} \\usepackage{siunitx} \\sisetup{group-separator={,},group-minimum-digits = 4} \\begin{document} \\begin{enumerate} \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 20 & 25 & 20 \\\\ 15 & 20 & 18 \\\\ 18 & 23 & 22 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 975 \\\\ 810 \\\\ 965 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 10 \\\\ 15 \\\\ 20 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] 90 & 95 & 80 \\\\ 80 & 85 & 90 \\\\ 75 & 90 & 95 \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} = \\begin{pmatrix*}[r] 86 \\\\ 86 \\\\ 88 \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ z \\end{pmatrix*} =} \\begin{pmatrix*}[r] 0.3 \\\\ 0.2 \\\\ 0.5 \\end{pmatrix*}$$ \\item $$\\Eqmathbox{\\begin{pmatrix*}[r] \\num{5000} & \\num{400}0 \\\\ \\num{6000} & \\num{5000} \\end{pmatrix*} \\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} = \\begin{pmatrix*}[r] \\num{37000000} \\\\ \\num{45000000 }\\\\ \\end{pmatrix*}} \\Rightarrow\\Eqmathbox[1][r]{\\begin{pmatrix*}[r] x \\\\ y \\\\ \\end{pmatrix*} =} \\begin{pmatrix*}[r] \\num{5000} \\\\ \\num{3000} \\end{pmatrix*}$$ \\end{enumerate} \\end{document} Here is one way. Note that you are adding a lot of unneeded", "A) v_A \\end{pmatrix} = \\begin{pmatrix} (\\lambda_A - 1) v_A \\\\ -(\\lambda_A - 1) v_A \\end{pmatrix} = (\\lambda_A - 1) V_{A-}. \\tag{5}$ (4) and (5) show that the eigenvalues of $B$ are at least those of the form $\\lambda_A \\pm 1$, but could there be more? We may investigate this possibility via a sort of unravelling of (4)-(5), in the sense that if $V =\\begin{pmatrix} v_1 \\\\ v_2 \\end{pmatrix} \\ne 0, \\tag{6}$ where $v_1$, $v_2$ are $N$-vectors, is an eigenvector of $B$ with eigenvalue $\\mu$, that is, $BV = \\mu V, \\tag{7}$ then if we write out (7) in terms of $v_1$, $v_2$ using the definition of $B$ in terms of $A$ we see that $BV = \\begin{bmatrix} A & I \\\\ I & A \\end{bmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\end{pmatrix} = \\begin{pmatrix} Av_1 + v_2 \\\\ v_1 + A v_2 \\end{pmatrix} = \\mu \\begin{pmatrix} v_1 \\\\ v_2 \\end{pmatrix}, \\tag{8}$ so that (7)-(8) breaks out into the two equations $Av_1 + v_2 = \\mu v_1 \\tag{9}$ and $Av_2 + v_1 = \\mu v_2. \\tag{10}$ We analyze (9)-(10) by first addressing the cases $v_2 = \\pm v_1$; in the event that $v_2 = v_1$, (9)-(10) become the same equation $Av_i + v_i = \\mu v_i \\tag{11}$ or $Av_i = (\\mu - 1)v_i \\tag{12}$ for $i = 1, 2$. If, on", "}$-vector space. In addition, the map is well-defined. Let ${\\displaystyle (x_{1},x_{2})^{T}}$ and ${\\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have: {\\displaystyle {\\begin{aligned}f\\left({\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\\\(x_{1}+y_{1})-5\\cdot (x_{2}+y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{separate vectors}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}x_{1}+x_{2}\\\\x_{1}-5x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}+y_{2}\\\\y_{1}-5y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+f{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\end{aligned}}} Proof step: homogeneity Let ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle (x_{1},x_{2})^{T}\\in \\mathbb {R} ^{2}}$. Then: {\\displaystyle {\\begin{aligned}f\\left(\\lambda \\cdot {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}\\lambda \\cdot x_{1}\\\\\\lambda \\cdot x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda x_{1}+\\lambda x_{2}\\\\\\lambda x_{1}-5\\lambda x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda (x_{1}+x_{2})\\\\\\lambda (x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{scalar multiplication}}\\right.}\\\\[0.3em]&=\\lambda \\cdot {\\begin{pmatrix}(x_{1}+x_{2})\\\\(x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=\\lambda \\cdot f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\\\[0.3em]\\end{aligned}}} Thus the map is linear. ## A linear map in the vector space of sequences Next, we consider the space of all sequences of real numbers. This space is infinite-dimensional, because there are not finitely many sequences generating this sequence space. But it is a vector space, as we have shown in the chapter about sequence spaces. Exercise (Sequence space) Let ${\\displaystyle V}$ be the ${\\displaystyle \\mathbb {R} }$-vector space of all real-valued sequences. Show that the map {\\displaystyle {\\begin{aligned}f:V&\\to V\\\\(a_{0},a_{1},a_{2},\\ldots )&\\mapsto (a_{1},a_{2},a_{3},\\ldots )\\end{aligned}}} is linear. How to get to the proof? (Sequence space) To show linearity, two properties need to be checked: 1. ${\\displaystyle f}$ is additive: ${\\displaystyle f(v+w)=f(v)+f(w)}$ for all ${\\displaystyle v,w\\in V}$ 2. ${\\displaystyle", "# Vectors, Linear Transformations, and Matrix Forms A two-dimensional vector, $\\bar{v} = \\begin{pmatrix}x\\\\y\\end{pmatrix}$, can be thought of representing the displacement that, when starting at the origin, would move something to the point $(x,y)$. Consequently, a natural interpretation for the sum of two vectors would be the total displacement that results from their individual applications. This can be seen in the picture below, where the red vector represents the total displacement resulting from the application of the blue and green vectors. Fortunately, as seen below, whether we follow the the displacement arising from the blue vector with that from the green vector (the solid segments), or vice-versa (the dashed segments), the overall displacement is the same. (Hence, the addition of vectors is commutative.) As indicated in the picture, this suggests how we might find the sum of two vectors algebraically as well: $$\\begin{pmatrix}x_1\\\\y_1\\end{pmatrix} + \\begin{pmatrix}x_2\\\\y_2\\end{pmatrix} = \\begin{pmatrix}x_1+x_2\\\\y_1+y_2\\end{pmatrix}$$ Further, understanding how vectors get added together suggests a definition for a constant (or scalar) multiple of a vector. For example, it would be natural to assume that $3\\begin{pmatrix}x\\\\y\\end{pmatrix}$ is a displacement three times as large as $\\begin{pmatrix}x\\\\y\\end{pmatrix}$. Consequently, $$3\\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}x\\\\y\\end{pmatrix} + \\begin{pmatrix}x\\\\y\\end{pmatrix} + \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}x+x+x\\\\y+y+y\\end{pmatrix} = \\begin{pmatrix}3x\\\\3y\\end{pmatrix}$$ In keeping with this example, we define the scalar multiple of a two-dimensional vector, for any real value $c$ (which is called", "0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -5 & 0\\\\ 0 & 0 & 0\\end{array}\\right) \\mbox{ and } \\left(\\begin{array}{cc|c} 5 & -5 & 0\\\\ 1 & -1 & 0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -1 & 0\\\\ 0 & 0 & 0\\end{array}\\right)\\, ,$$ respectively, so are $$v_{2}=\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}$$ and $$v_{-2} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$. Hence $$M^{12} v_{2} = 2^{12} v_{2}$$ and $$M^{12}v_{-2} = (-2)^{12} v_{-2}$$. Now, $$\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\frac{x-y}{4}\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix} -\\frac{x-5y}{4} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$ (this was obtained by solving the linear system $$a v_{2} + b v_{-2} =$$ for $$a$$ and $$b$$). Thus $$M \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\frac{x-y}{4} M v_{2} -\\frac{x-5y}{4} M v_{-2}$$ $$= 2^{12} \\Big(\\frac{x-y}{4} v_{2} -\\frac{x-5y}{4} v_{-2}\\Big) = 2^{12} \\begin{pmatrix} x \\\\ y \\end{pmatrix}\\, .$$ Thus $$M^{12}=\\begin{pmatrix} 4096 & 0 \\\\ 0 & 4096\\end{pmatrix}\\, .$$ $$\\textit{If you understand the above explanation, then you have a good understanding of diagonalization. A quicker route}$$ $$\\textit{is simply to observe that}$$ $$M^{2} = \\begin{pmatrix}4 & 0 \\\\ 0 & 4\\end{pmatrix}$$. 8. $$P_{M}(\\lambda) = (-1)^{2} {\\rm det}\\begin{pmatrix} a-\\lambda & b \\\\ c &d-\\lambda\\end{pmatrix} =(\\lambda-a)(\\lambda-d) - bc\\, .$$ Thus $$P_{M}(M)=(M-a I )(M- d I) - bc I$$ $$= \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}a&0\\\\0&a\\end{pmatrix}\\right) \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}d&0\\\\0&d\\end{pmatrix}\\right)-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}$$ $$=\\begin{pmatrix}0& b\\\\c&d-a\\end{pmatrix}\\begin{pmatrix}a-d&b\\\\ c&0\\end{pmatrix}-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}=0\\, .$$ Observe that any $$2\\times 2$$ matrix is a zero of its own characteristic polynomial ($$\\textit{in fact this holds", "$$(A + iB)(x + iy) = \\lambda x + i \\lambda y, \\tag 9$$ or $$(Ax - By) + i(Ay + Bx) = \\lambda x + i \\lambda y; \\tag{10}$$ equating the real and imaginary parts of either side yields $$Ax - By = \\lambda x, \\tag{11}$$ $$Ay + Bx = \\lambda y. \\tag{12}$$ We can now in fact find two real orthogonal eigenvectors for $$M = \\begin{bmatrix} A & -B \\\\ B & A \\end{bmatrix}, \\tag{13}$$ each corresponding to the eigenvalue $$\\lambda$$; set $$w = \\begin{pmatrix} x \\\\ y \\end{pmatrix}; \\tag{14}$$ $$Mw = \\begin{bmatrix} A & -B \\\\ B & A \\end{bmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} Ax - By \\\\ Bx + Ay \\end{pmatrix} = \\begin{pmatrix} \\lambda x \\\\ \\lambda y \\end{pmatrix} = \\lambda \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\lambda w; \\tag{15}$$ now set $$v = \\begin{pmatrix} -y \\\\ x \\end{pmatrix}; \\tag{16}$$ $$Mv = \\begin{bmatrix} A & -B \\\\ B & A \\end{bmatrix} \\begin{pmatrix} -y \\\\ x \\end{pmatrix} = \\begin{pmatrix} -Ay - Bx \\\\ -By + Ax \\end{pmatrix} = \\begin{pmatrix} -\\lambda y \\\\ \\lambda x \\end{pmatrix} = \\lambda \\begin{pmatrix} -y \\\\ x \\end{pmatrix} = \\lambda v; \\tag{17}$$ furthermore, $$\\langle v, w \\rangle = \\begin{pmatrix} -y \\\\ x \\end{pmatrix}^T \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} -y^T & x^T \\end{pmatrix} \\begin{pmatrix} x \\\\ y", "What do you mean by \"at\" the largest singular value? Do you mean we set all the entries of vector $a$ equal to zero except for one of them? If $A = \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}$, $B = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ 0 & 0 \\end{pmatrix}$ then $A'D = \\begin{pmatrix} 1 & 1 \\end{pmatrix}$. $A'D$ is equal to the same thing if $A = \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ $B = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ 0 & 0 \\end{pmatrix}$ Assuming A and B are both orthnormal, then the coefficient vectors a and b are both unit vectors so that x and y are also unit vectors. Then maximization of <x,y> correspond to minimal angles between x and y. In order to maximize <x,y>=(Ua)'*S*(Vb), given that ||Ua||=1 and ||Vb||=1, I want to choose Ua and Vb to be 1 at the largest singular values and 0 elsewhere, not that a and b are 1 at one place and 0 everywhere else. Is the logic correct? Stephen Tashi I want to choose Ua and Vb to be 1 at the largest singular values and 0 elsewhere Are you saying that vector 'a' will be chosen so that the vector Ua will be 1 at the", ",0 , 3) and (1, 0, 2, 0). I'm not exactly sure what to do with this or if I'm even on the right track. Assuming you know linear transformations, Observe that $\\begin{pmatrix}x & y\\end{pmatrix}\\begin{pmatrix}1 & 0 & 2 & 0 \\\\0 & 1 & 0 & 2\\end{pmatrix} = \\begin{pmatrix}x & y & 2x & 2y\\end{pmatrix}$ $T \\equiv \\begin{pmatrix}1 & 0 & 2 & 0 \\\\0 & 1 & 0 & 2\\end{pmatrix}$ is the matrix of the linear transformation(it is not in standard form though). Prove that the rank of this matrix is 2. Then dim(image(T)) is 2. But image(T) is precisely the set of vectors of the form $\\begin{pmatrix}x & y & 2x & 2y\\end{pmatrix}$ Once you know the dimension is 2 and the theorem that basis vectors map to basis vectors. we see that (1,0,2,0) and (0,1,0,2) are the basis elements. Without knowing linear transformations, you could have observed that given any arbitrary vector $\\begin{pmatrix}x & y & 2x & 2y\\end{pmatrix}$, we can write it as $x\\begin{pmatrix}1 & 0 & 2 & 0\\end{pmatrix} + y\\begin{pmatrix}0 & 1 & 0 & 2\\end{pmatrix}$. This means the set $\\{\\begin{pmatrix}1 & 0 & 2 & 0\\end{pmatrix},\\begin{pmatrix}0 & 1 & 0 & 2\\end{pmatrix}\\}$ is a spanning set. And then also prove that the set $\\{\\begin{pmatrix}1 & 0 & 2 & 0\\end{pmatrix},\\begin{pmatrix}0", "|u \\rangle_1 |d \\rangle_2 - |d \\rangle_1 |u \\rangle_2 \\right]$ Written in tensor format the antisymmetric wave function is $| \\Psi_F \\rangle = \\frac{1}{ \\sqrt{2}} \\left[ \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} - \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} \\otimes \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} \\right] = frac{1}{ \\sqrt{2}} \\left[ \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\\\ 0 \\end{pmatrix} - \\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix} \\right] = \\frac{1}{ \\sqrt{2}} \\begin{pmatrix} 0 \\\\ 1 \\\\ -1 \\\\ 0 \\end{pmatrix}$ $\\Psi _F = \\frac{1}{ \\sqrt{2}} \\begin{pmatrix} 0 \\\\ 1 \\\\ -1 \\\\ 0 \\end{pmatrix}$ Now the interaction with the beam splitter yields the following result. $|Psi _{Fout} = \\frac{1}{2} \\begin{pmatrix} 1 & i & i & 1 \\\\ i & 1 & -1 & i \\\\ i & -1 & 1 & i \\\\ -1 & i & i & 1 \\end{pmatrix} \\frac{1}{ \\sqrt{2}} \\begin{pmatrix} 0 \\\\ 1 \\\\ -1 \\\\ 0 \\end{pmatrix} \\rightarrow \\begin{pmatrix} 0 \\\\ \\frac{ \\sqrt{2}}{2} \\\\ - \\frac{ \\sqrt{2}}{2} \\\\ 0 \\end{pmatrix}$ We see by inspection or calculation that fermionic photons are never observed in the same output channel after interaction with the beam splitter. $\\left( \\left| uu^T \\Psi_{Fout} \\right| \\right)^2 \\rightarrow 0$ $\\left( \\left| dd^T \\Psi_{Fout} \\right| \\right)^2 \\rightarrow 0$ Naturally this means they are always observed" ]

[ "of $\\sin ^4x + \\cos ^4x = 1-\\frac{1}{2}(1) = \\frac{1}{2}$ and maximum value of $\\sin ^4x + \\cos ^4x = 1-\\frac{1}{2}(0)=1$ so, range of f(x)=[1/2,1] I think there is any mistake in the book. As you can see the detailed solution of the question and if there is any other doubt related to this question you can ask", "and / ( 2 2 ) 3 periodic with period 2 . D. No asymptote E. f (x)=3cos x 3sin x cos x=3cos x 1 sin x =3cos x . / f (x)>0 cos x>0 cos x<0 x 2n 2 x 2n 2 3 2n + ,2n + 2 2 ,2n + 2 ,2n + / for each integer n , and f (x)<0 2 for each integer n . Thus, f is increasing on for each integer n , and f is decreasing on integer n . F. f has local maximum values f (2n + f (2n + G. f / / 2 2n + 2 ,2n + 3 2 for each )=2 and local minimum values 3 )= 2 . 2 2 2 2n + (x)= 9sin x cos x= 9sin x(1 sin x)= 9sin x(1 sin x)(1+sin x) . f sin x 1 x 2n ,2n + sin x 1 x (2n 1) ,(2n 1) + on the intervals 2 2n , 2n+ 1 2 2 ,2n + and 2n+ 1 and 2 inflection points at ( n ,0 ) for each integer n . H. CU on the intervals (2n 1) , 2n 32. y= f (x)=sin x tan x A. D= { x| x (2n+1) 2 } (x)<0 for some integer n", "+0 # HELP ASAPPP 0 43 1 Given that f(x) = (1/pi)^x, what is the range of f(x) on the interval [0, ∞)? Jul 10, 2020 #1 +9770 +1 Given that f(x) = (1/pi)^x, what is the range of f(x) on the interval [0, ∞ )? Hello Guest! $$f(x) = ( \\frac{1}{\\pi})^x\\\\ f(0) = ( \\frac{1}{\\pi})^0=1\\\\ f(∞)= ( \\frac{1}{\\pi})^∞\\ toward\\approx 0$$ $$Range_{f(x)}|_0^∞=\\{1\\ge \\mathbb R> 0\\}$$ ! Jul 10, 2020", "on the interval from $−π$ to $π$ $\\frac{2\\pi ^3}{9}$", "The range of sin(x) is [-1, 1]. It's confusing to say \"the range of ##\\sin(x)## in <whatever>\". The range of ##10^{\\sin(x)}## is [0.1, 10]. Why do you think that 1 is not in the range of ##10^{\\sin x}##? ##f(x) = 10^{\\sin x} + 10^{\\csc x} = e^{\\sin x \\ln 10} + e^{\\frac{1}{\\sin x} \\ln 10}##. Since the function has a period of ##2\\pi##, I think you should study it when ##x \\in (0,2\\pi)##, at the period's bounds, and near ##\\pi##. And maybe you should see when its derivative becomes zero to figure out the lower bound of the range, since, from the definition of ##e^x##, this function can't go below ##0##. Using desmos, the range seems to be ##[0.2, \\infty)##. Last edited: You are confusing the ranges of the subfunctions with the range of the sum of these function. Keep in mind that the combined range is **not** the intersection of the two ranges. For example, consider ##g(x) = \\sqrt x## and ##h(x) = \\left(\\frac x 2\\right)^2## on the interval [0, 1]. On this restricted domain, ##Ran(g) = [0, 1]## and ##Ran(h) = [0, \\frac 1 4]##. The range of their sum, ##\\sqrt x + \\left(\\frac x 2\\right)^2## is [0, 1.25]. Also, in post #1 you have a mistake: The range of sin(x) is [-1, 1]. It's confusing", "SinCosTanLog Geometry Level 3 Suppose $$x$$ is in the interval $$[0,\\pi/2]$$ and $$\\log_{24\\sin{x}}(24\\cos{x})=\\frac{3}{2}$$. Find $$24\\tan^2{x}$$. Note: Yet another modified AIME problem. ×", "## anonymous one year ago What is the range of 3/x2-4 1. anonymous @zzr0ck3r 2. zzr0ck3r $$f(x) = \\dfrac{3}{x^2}-4$$ ? 3. DecentNabeel $\\mathrm{Range\\:of\\:}\\frac{3}{x^2}-4:\\quad \\begin{bmatrix}\\mathrm{Solution:}\\: & \\:f\\left(x\\right)>-4\\: \\\\\\:\\mathrm{Interval\\:Notation:} & \\:\\left(-4,\\:\\infty \\:\\right)\\end{bmatrix}$ 4. anonymous I think it's $\\mathbb{R}-\\left\\{ -4 \\right\\}$ When f(x)=-4, we have $\\frac{3}{x^2}-4=-4$ $\\implies \\frac{3}{x^2}=0$$\\implies 3=0$ which is not true I'm not sure though", "interval $[0,2\\pi )$ will be $0,\\,\\frac{\\pi }{3},\\,\\,\\pi ,\\ \\text{ and }\\ \\frac{5\\pi }{3}$.", "with period 2! . D. No asymptote E. f (x)=3cos x!3sin x cos x=3cos x 1!sin x =3cos x . ! ! / / for each integer n , and f (x)<0# f (x)>0#cos x>0# x- 2n! ! ,2n! + 2 2 ! 3! cos x<0# x- 2n! + ,2n! + for each integer n . Thus, f is increasing on 2 2 ! ! ! 3! 2n! ! ,2n! + for each integer n , and f is decreasing on 2n! + ,2n! + for each 2 2 2 2 ! integer n . F. f has local maximum values f (2n! + )=2 and local minimum values 2 3! f (2n! + )=!2 . 2 (x)=!9sin x cos x=!9sin x(1!sin x)=!9sin x(1!sin x)(1+sin x) . f (x)<0#sin x>0 and ! ! / / sin x' % 1# x- 2n! ,2n! + ( 2n! + ,2n! +! for some integer n . f (x)>0#sin x<0 and 2 2 ! ! sin x' % 1# x- (2n!1)! ,(2n!1)! + ( (2n!1)! + ,2n! for some integer n . Thus, f is CD 2 2 1 1 on the intervals 2n! , 2n+ ! and 2n+ ! , ( 2n+1 ) ! for each integer n , and f is 2 2 1 1 CU on the intervals (2n!1)!", "p\\int_0^p f(t)\\sin(\\frac{n\\pi}{p}t)\\, dt,\\ a_n=0$$In your example, $p=\\pi$. For the half-range cosine the formulas are$$b_n=0,\\ a_n=\\frac 2 p\\int_0^p f(t)\\cos(\\frac{n\\pi}{p}t)\\, dt,\\ a_0=\\frac 1 p \\int_0^p f(t)\\ dt$$In your example $p=1$ for the cosine series.", "and M = {x | x ∈ N, 1 ≤ x ≤ 20} and f(x) = x2 + 1, then find the range of f. f: A → M A = {x | x ∈ N, 1 ≤ x < 5} = {1, 2, 3, 4} M = {x | x G N, 1 ≤ x ≤ 20} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} f(x) = x2 + 1 Range of f Rf= {f(x) | x ∈ A} = {f(1), f(2), f(3), f(4)} Now, f(x) = x2 + 1 ∴ f(1) = (1)2 + 1 = 1 + 1 = 2 f(2) = (2)2 + 1 = 4+ 1 = 5 f(3) = (3)2 + 1 = 9 + 1 = 10 f(4) = (4)2 + 1 = 16 + 1 = 17 Hence, Range of f Rf = {2, 5, 10, 17} Question 7. If f(x) = $$\\frac{x^{2}-4}{x-2}$$ where x ∈ Z – {2} then find the value of f(0) + f(1) – f(-2). f(x) = $$\\frac{x^{2}-4}{x-2}=\\frac{(x+2)(x-2)}{x-2}$$ = x + 2 ∴ f(0) + f(1) – f(-2) = (0 + 2 + (1 + 2) – (-2 + 2) = 2 + 3 – 0 = 5 Question 8. If the domain", "interval $-\\frac{\\pi}{2}\\le t \\le\\frac{\\pi}{2}$ we have $0\\le \\cos t \\le 1$ -", "# How do you find the range of f(x)=5cos(x + pi) + 3? $\\cos \\left(x + \\pi\\right)$ has a range of $\\left[- 1 , + 1\\right]$ (because cos(anything) has a range of $\\left[- 1 , + 1\\right]$) $5 \\cos \\left(x + \\pi\\right)$ has a range of $\\left[- 5 , + 5\\right]$ $f \\left(x\\right) = 5 \\cos \\left(x + \\pi\\right) + 3 h a s a r a n \\ge o f$[-2,+8]#", "# How do you find the maximum value of f(x)=2sin(x)+cos(x)? Sep 1, 2016 $\\sqrt{5}$. Range is $\\left[- \\sqrt{5} , \\sqrt{5}\\right]$- #### Explanation: f'=2 cos x - sin x = 0, when 2 cos x = sin x that gives x = arc tan 2. The principal value is in Q1. Indeed, there are general values in Q1 and Q3. $f ' ' = - 2 \\sin x - \\cos x < 0$, for Q1 values and $> 0$ for Q3 values, as both sin x and cos x are negative in Q3. The maximum is obtained when tan x = 2, with x in Q1. And this is 2sin x + cos x , with tan x = 2 $= 2 \\left(\\frac{2}{\\sqrt{5}}\\right) + \\frac{1}{\\sqrt{5}}$ $= \\frac{5}{\\sqrt{5}}$ $= \\sqrt{5}$. Of course, the minimum is $- \\sqrt{5}$. Alternative method sans differentiation: $f = \\sqrt{5} \\left(\\left(\\frac{2}{\\sqrt{5}}\\right) \\sin x + \\left(\\frac{1}{\\sqrt{5}}\\right) \\cos x\\right)$ $= \\sqrt{5} \\sin \\left(x + \\alpha\\right)$, where $\\sin \\alpha = \\frac{2}{\\sqrt{5}} \\mathmr{and} \\cos \\alpha = \\frac{1}{\\sqrt{5}}$ $M a x f = \\sqrt{5} \\max \\sin \\left(x + \\alpha\\right)$ $= \\sqrt{5} \\left(1\\right)$'", "= \\frac{3}{4}$$, so $\\tan(a) = \\frac{1}{7}$. We compute $$\\sin(a) = \\frac{\\sqrt{2}}{10}$$ $$\\cos(a) = \\frac{7\\sqrt{2}}{10}$$ $$\\cos(t') = -\\tan(a) = -\\frac{1}{7}$$ $$\\sin(t') = \\frac{4\\sqrt{3}}{7}$$ $$\\sin(t' + a)=\\sin(t')\\cos(a) + \\cos(t')\\sin(a) = \\frac{28\\sqrt{6} - \\sqrt{2}}{70}$$, so the maximum length of $XY$ is $\\sin(a)\\times\\frac{\\sin(t')}{\\sin(t' + a) + \\cos(a)} = 7 - 4\\sqrt{3}$, and the answer is $7 + 4 + 3 = \\boxed{014}$. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.", "Without using mathematical tables or a calculator, evaluate $\\frac{sin\\,30^{0}-sin\\,60^{0}}{tan\\,60^{0}}$", "sinc\\left(\\frac{\\pi}{T}(x-nT)\\right)$$ where $T$ is the sampling period used to determine $x_n$ and $sinc(t)=sin(t)/t$ This formula is exact as long as function spectrum ($F(k)$) is limited. Implementation of fast sinc interpolation exist", "# How do you find the range of f(x)= 1/(x^2 + 9)? Sep 16, 2015 Range { y in RR, 0< y<= 1/9}# #### Explanation: Solve $y = \\frac{1}{{x}^{2} + 9}$ for x ${x}^{2} = \\frac{1 - 9 y}{y}$, $x = \\sqrt{\\frac{1 - 9 y}{y}}$ This implies that $y \\ne 0$ and $1 - 9 y \\ge 0$, that is $y \\le \\frac{1}{9}$ Hence range of f(x) would be $\\left\\{y \\in \\mathbb{R} , 0 < y \\le \\frac{1}{9}\\right\\}$", "# Question #143f6 Dec 14, 2017 The range is the closed interval $= \\left[2 \\sinh \\left(- 1\\right) , 2 \\sinh \\left(2\\right)\\right] = \\left[\\frac{1 - {e}^{2}}{e} , \\frac{{e}^{4} - 1}{{e}^{2}}\\right] \\approx \\left[- 2.3504 , 7.2537\\right]$. #### Explanation: Let $f \\left(x\\right) = {e}^{x} - {e}^{- x} = 2 \\sinh \\left(x\\right)$, where $\\sinh \\left(x\\right)$ is the hyperbolic sine function . Then $f ' \\left(x\\right) = {e}^{x} + {e}^{- x} = 2 \\cosh \\left(x\\right)$, where $\\cosh \\left(x\\right)$ is the hyperbolic cosine function. Clearly $f ' \\left(x\\right) > 0$ for all $x \\setminus \\in \\mathbb{R}$, implying that $f \\left(x\\right)$ is strictly increasing over $\\mathbb{R}$, and also over $\\left[- 1 , 2\\right]$. Therefore, the minimum value of $f$ over $\\left[- 1 , 2\\right]$ is $f \\left(- 1\\right) = 2 \\sinh \\left(- 1\\right) = {e}^{- 1} - {e}^{1} = \\frac{1 - {e}^{2}}{e} \\approx - 2.3504$ and the maximum value of $f$ over $\\left[- 1 , 2\\right]$ is $f \\left(2\\right) = 2 \\sinh \\left(2\\right) = {e}^{2} - {e}^{- 2} = \\frac{{e}^{4} - 1}{{e}^{2}} \\approx 7.2537$. This implies that the range is the closed interval $= \\left[2 \\sinh \\left(- 1\\right) , 2 \\sinh \\left(2\\right)\\right] = \\left[\\frac{1 - {e}^{2}}{e} , \\frac{{e}^{4} - 1}{{e}^{2}}\\right] \\approx \\left[- 2.3504 , 7.2537\\right]$.", "\\int_s^{+\\infty} F(\\tau) \\, d \\tau$ where $F(s) = LT[f(t)]$. In your question $f(t) = \\sin (t)$. Originally Posted by jmedsy actually, are you sure this is correct? also, it just occurred to me that perhaps i should expand sin(t) via euler's formula and try it that way. maple is kicking out arctan(1/s) Note that $\\frac{\\pi}{2} - \\tan^{-1} (s) = \\tan^{-1} \\left( \\frac{1}{s} \\right)$." ]

[ "to approach 0 from the left? What does this tell you about this problem? 11. Divide top and bottom by \\displaystyle \\begin{align*} x^3 \\end{align*}. 3. ## Re: Limits. For #7: limit sin(2x)/(5x) = limit sin(2x)/2x * limit 2x/5x = 1 * 2/5 = 2/5 Just picture it.. if the denominator were equal to (2x), the limit would be 1, but instead, it is larger than that by a scale of (5/2)... therefore you need to divide by that much extra, to get 2/5. For #8, the same principle applies actually. As x -> 0 in this case, you can treat tan() as sin(), because you are dividing sin() by a quantity that approaches 1, cos(x). So the answer is 2/Pi. For #5, multiply both the numerator and the denominator of the big fraction by 2(2+x). It should then simplify nicely. 4. ## Re: Limits. Hello, Zirrick! Here is #8 without using L'Hopital . . . $8.\\;\\lim_{x\\to0}\\frac{\\tan(2x)}{\\tan(\\pi x)}$ $\\frac{\\tan(2x)}{\\tan(\\pi x)} \\;=\\;\\dfrac{\\frac{\\sin(2x)}{\\cos(2x)}}{\\frac{\\sin( \\pi x)}{\\cos(\\pi x)}} \\;=\\; \\frac{\\sin(2x)}{1}\\cdot\\frac{1}{\\sin(\\pi x)}\\cdot\\frac{\\cos(\\pi x)}{\\cos(2x)}$ . . $=\\;{\\color{red}\\frac{2x}{2x}}\\cdot\\frac{\\sin(2x)}{ 1}\\cdot {\\color{red}\\frac{\\pi x}{\\pi x}}\\cdot\\frac{1}{\\sin(\\pi x)} \\cdot \\frac{\\cos(\\pi x)}{\\cos(2x)} \\;=\\;\\frac{2x}{\\pi x}\\cdot\\frac{\\sin(2x)}{2x}\\cdot\\frac{\\pi x}{\\sin(\\pi x)}\\cdot\\frac{\\cos(\\pi x)}{\\cos(2x)}$ $\\lim_{x\\to0}\\left[\\frac{2}{\\pi}\\cdot\\frac{\\sin(2x)}{2x}\\cdot\\frac{ \\pi x}{\\sin(\\pi x)}\\cdot\\frac{\\cos(\\pi x)}{\\cos(2x)}\\right] \\;=\\; \\frac{2}{\\pi}\\cdot1\\cdot1\\cdot\\frac{1}{1} \\;=\\;\\frac{2}{\\pi}$", "and a $$\\sec^2 x$$ terms. Doing so we have \\begin{align} I &= \\int \\frac{\\cos^2 x \\sin x}{\\sin x - \\cos x} \\, dx\\\\ &= \\int \\frac{\\cos^2 x \\sin x}{\\sin x - \\cos x} \\cdot \\frac{\\sec^2 x}{\\sec^2 x} \\, dx\\\\ &= \\int \\frac{\\tan x}{(\\tan x - 1)(1 + \\tan^2 x)^2} \\cdot \\sec^2 x \\, dx. \\end{align} Now let $$t = \\tan x$$. Doing so yields \\begin{align} I &= \\int \\frac{t}{(t - 1)(1 + t^2)^2} \\, dt\\\\ &= \\int \\left [\\frac{1}{4(t - 1)} - \\frac{t + 1}{4(t^2 + 1)} + \\frac{1 - t}{2(1 + t^2)^2} \\right ] \\, dt\\\\ &= \\frac{1}{4} \\ln |t - 1| - \\frac{1}{8} \\ln |1 + t^2| + \\frac{t}{4(1 + t^2)} + C\\\\ &= \\frac{1}{4}\\ln |\\tan x - 1| + \\frac{1}{4} \\ln |\\cos x| + \\frac{1}{4} (1 + \\tan x) \\cos^2 x + C. \\end{align} or after playing around with a few trignometric identities $$\\int \\frac{\\cos^2 x \\sin x}{\\sin x - \\cos x} \\, dx = \\frac{1}{4} \\ln |\\sin x - \\cos x| + \\frac{1}{8} (\\cos 2x + \\sin 2x) + C.$$", "so \\begin{align} \\cos^4x\\sin^4x &=\\frac{1}{2^8}(e^{2ix}-e^{-2ix})^4\\\\[6px] &=\\frac{1}{2^8}(e^{8ix}-4e^{4ix}+6-4e^{-4ix}+e^{8ix})\\\\[6px] &=\\frac{1}{128}\\cos8x-\\frac{1}{32}\\cos4x+\\frac{3}{128} \\end{align} we now that $$\\sin 2x=2\\cos x\\sin x$$ $$\\cos x\\sin x=\\frac{\\sin 2x}{2}$$ so... $$\\cos^4 x\\sin^4 x=\\frac{\\sin^4 2x}{16}$$ $$\\frac{\\sin^4 2x}{16}=\\frac{(\\frac{1-\\cos 4x}{2})^2}{16}=\\frac{1-2\\cos 4x+\\cos^24x}{64}=\\frac{1-2\\cos 4x+\\frac{1+\\cos 8x}{2}}{64}$$", "\\sin(r))\\frac{d}{dr}[4^r - 2\\cos(r)] \\\\ =\\mathstrut \\amp [15r^2 + \\cos(r)](4^r - 2\\cos(r))+(5r^3 + \\sin(r))[4^r \\ln(4) + 2\\sin(r)]\\text{.} \\end{align*} 2. We use the quotient rule on $p\\text{,}$ followed by the product rule to differentiate the denominator, finding that \\begin{align*} p'(t) =\\mathstrut \\amp \\frac{\\frac{d}{dt}[\\cos(t)]t^6 \\cdot 6^t - \\cos(t) \\frac{d}{dt}[t^6 \\cdot 6^t]}{(t^6 \\cdot 6^t)^2}\\\\ =\\mathstrut \\amp \\frac{[-\\sin(t)]t^6 \\cdot 6^t - \\cos(t) [6t^5 \\cdot 6^t + t^6\\cdot 6^t \\ln(6)]}{(t^6 \\cdot 6^t)^2}\\text{.} \\end{align*} 3. Using the sum and constant multiple rules, it follows first that \\begin{equation*} g'(z) = 3 \\frac{d}{dz}[z^7 e^z] - 2\\frac{d}{dz}[z^2 \\sin(z)] + \\frac{d}{dz}\\left[ \\frac{z}{z^2 + 1} \\right]\\text{.} \\end{equation*} Applying the product rule in the first two terms and the quotient rule in the third, we find that \\begin{equation*} g'(z) = 3 [7z^6e^z+z^7 e^z] - 2[2z\\sin(z)+z^2 \\cos(z)] + \\frac{(z^2+1) - z(2z)}{(z^2 + 1)^2}\\text{.} \\end{equation*} 4. The particle's instantaneous velocity at the moment $t = 1$ is given by $s'(1)\\text{.}$ We use the quotient rule to find $s'(t)\\text{,}$ and then simplify by removing a common factor of $e^t$ to get \\begin{align*} s'(t) =\\mathstrut \\amp \\frac{(-3\\sin(t) - \\cos(t))e^t-(3\\cos(t) - \\sin(t))e^t}{(e^t)^2}\\\\ =\\mathstrut \\amp \\frac{-2\\sin(t)-4\\cos(t)}{e^t}\\text{.} \\end{align*} Thus, $s'(1) = \\frac{-2\\sin(1)-4\\cos(1)}{e^1} \\approx -1.414\\text{,}$ which is the particle's instantaneous velocity in feet per second at the moment $t = 1\\text{.}$ 5. Since $p(x) = f(x) \\cdot g(x)\\text{,}$ the product rule tells us \\begin{equation*} p'(x) = f'(x)g(x)+f(x)g'(x)\\text{,} \\end{equation*} and", "# Find the value of $\\tan A + \\tan B$, given values of $\\frac{\\sin (A)}{\\sin (B)}$ and $\\frac{\\cos (A)}{\\cos (B)}$ Given $$\\frac{\\sin (A)}{\\sin (B)} = \\frac{\\sqrt{3}}{2}$$ $$\\frac{\\cos (A)}{\\cos (B)} = \\frac{\\sqrt{5}}{3}$$ Find $\\tan A + \\tan B$. Approach Dividing the equations, we get the relation between $\\tan A$ and $\\tan B$ but that doesn't help in getting the value of $\\tan A + \\tan B$. The value comes in terms of $\\tan A$ or $\\tan B$ but the expected answer is independent of any variable . Also $$\\frac{\\sin(A)\\cdot\\cos(B) + \\sin(B)\\cdot\\cos(A)}{\\cos(A)\\cdot\\cos(B)} = \\tan(A) + \\tan(B)$$ We could get a value only if instead of $\\cos A$ there was $\\sin B$ in the relation(which we get on adding the ratios) • – imranfat Aug 23 '16 at 15:42 Something is wrong. I calculated \\begin{align} \\sin A &= \\dfrac{\\sqrt 3}{2}\\sin B \\\\ \\cos A &= \\dfrac{\\sqrt 5}{3}\\cos B \\\\ \\sin^2 A + \\cos^2 A &= \\dfrac 34 \\sin^2 B + \\dfrac59 \\cos^2 B\\\\ 1 &= \\dfrac 34 - \\dfrac{7}{36} \\cos^2 B\\\\ \\cos^2 B = -\\dfrac{9}{7} \\end{align} Which is impossible. HINT: $$\\tan A=\\frac{\\sin A}{\\cos A}=\\frac{\\frac{\\sqrt{3}}2\\sin B}{\\frac{\\sqrt{5}}3\\cos B}=\\frac{3\\sqrt{3}}{2\\sqrt{5}}\\tan B\\tag{1}$$ And $$1=\\sin^2 A+\\cos^2 A=\\frac{3}{4}\\sin^2 B+\\frac{5}{9}\\cos^2 B=\\frac59\\left(\\sin^2 B+\\cos^2B\\right)+\\frac7{36}\\sin^2 B$$ Last equation implies $$\\sin^2B=\\frac{16}7$$ $$\\frac{|\\sin A|}{|\\sin B|}=\\frac{\\sqrt{3}}{2} <1 \\iff \\frac{|\\cos A|}{|\\cos B|}>1$$ But $$\\frac{|\\cos A|}{|\\cos B|}=\\frac{\\sqrt{5}}{3}<1$$ that's impossible. Hint: Let $\\dfrac{\\sin A}{\\sqrt3}=\\dfrac{\\sin B}2=p$ and $\\dfrac{\\cos A}{\\sqrt5}=\\dfrac{\\cos", "• You probably also need that $\\sin^2(r) \\approx r^2$, since you have that factor $1/r$ to take care of. – Thomas Ahle Mar 11 '19 at 15:24 • @ThomasAhle It follows from $\\sin x \\le x, \\forall x \\ge 0$ like this: $$\\frac4r \\sin^2\\left(\\frac{r\\sqrt{2}}2\\right) \\le \\frac4r \\left(\\frac{r\\sqrt{2}}2\\right)^2 = 2r \\xrightarrow{r\\to 0} 0$$ – mechanodroid Mar 11 '19 at 15:54 Solution without the use of polar coordinates, as asked by OP at the end of their post. Rewrite \\begin{aligned}1-\\cos(x-y)&=1-\\cos x \\cos y - \\sin x \\sin y \\\\ &=1-\\cos x + \\cos x (1-\\cos y)- \\sin x \\sin y, \\end{aligned} then \\begin{aligned}\\frac{2x(1-\\cos(x-y))}{x^2+y^2}&= 2x\\cdot \\frac{1-\\cos x}{x^2} \\cdot\\frac{x^2}{x^2+y^2}\\\\ &+ 2x\\cos x \\cdot \\frac{1-\\cos y}{y^2}\\cdot\\frac{y^2}{x^2+y^2}\\\\&- 2y\\cdot \\frac{\\sin x}{x}\\cdot\\frac{\\sin y}{y}\\cdot\\frac{x^2}{x^2+y^2}\\end{aligned} From where is $$\\lim_{(x,y)\\to (0,0)}\\frac{2x(1-\\cos(x-y))}{x^2+y^2} = 0.$$", "20:50 Understandable - I just found this to be a little less messy. – user17794 Aug 27 '13 at 20:52 Hint: \\begin{align} 0&=\\cos x+\\sin x\\\\ \\sin x &= -\\cos x \\\\ \\text{Divide}&\\text{ both sides by }\\cos x\\\\ \\tan x &= -1 \\end{align} Now you can solve for $x$ using inverse trigonometric functions. - if you use the Double angle formulae $$\\sin^2(x)\\cos^2(x)=(1-\\cos2x)/2 \\cdot (1+\\cos2x)/2 = (1-\\cos^2(2x))/4 = \\sin^2(2x)/4 = (1-\\cos4x)/8$$ so minimum is $1/8$ at $x=45\\circ$ and maximum is 1/4 at $x=90\\circ$ remember $\\sin^2(x)=(1-\\cos2x)/2$ and $\\cos^2(x)=(1+\\cos2x)/2$ - The equation $\\cos x+\\sin x=0$ is really the same as $\\cos x-\\sin x=0$ once you make the transformation $x\\mapsto x+\\frac{\\pi}2$, because $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and $\\sin (x+\\frac{\\pi}2)=\\cos(x)$. The latter can be solved by resorting to $\\cos ^2+\\sin ^2=1$. Edit: We can derive $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and similar results (cf. Daniel Fischer's comment) from the well known formulas $\\cos (x+y)=\\cos x\\cos y-\\sin x\\sin y$ and $\\sin (x+y)=\\cos x\\sin y+\\cos y\\sin x$. These can either be proved geometrically, or by resorting to some definition of $\\sin$ and $\\cos$, for example $\\sin x=(e^{ix}-e^{-ix})/2i, \\cos x=(e^{ix}-e^{-ix})/2$ -", "\\sin^2 A$. Thus, we have $\\cos 2A = \\cos^2 A - \\sin^2 A$$= 2 \\cdot \\cos^2 A - 1$$= 1 - 2 \\cdot \\sin^2 A$ ## Double Angle Identities We know that $\\cos^2 A + \\sin^2 A$ $= 1$ Now $\\sin 2A$ $=$ $\\frac{\\sin 2A}{1}$ $=$ $\\frac{2 \\cdot \\sin A \\cdot \\cos A}{\\cos^2 A + \\sin^2 A}$ $=$ $\\frac{(\\frac{2 \\cdot \\sin A \\cdot \\cos A}{\\cos^2 A})}{(\\frac{\\cos^2 A + \\sin^2 A}{\\cos^2 A})}$ $=$ $\\frac{(\\frac{2 \\cdot \\sin A}{\\cos A})}{(\\frac{\\cos^2 A}{\\cos^2 A} + \\frac{\\sin^2 A}{\\cos^2 A})}$ $=$ $\\frac{2 \\cdot \\tan A}{1 + \\tan^2 A}$. Thus, $\\sin 2A =$ $\\frac{2 \\cdot \\tan A}{1 + \\tan^2 A}$ ## Given that $An Alpha =$ $frac{5}{12}$. Find out $sin 2a$. Solution We know that $\\sin 2\\alpha =$ $\\frac{2 \\cdot \\tan \\alpha}{1 + \\tan^2 \\alpha}$ $=$ $\\frac{2 \\cdot (\\frac{5}{12})}{1 + (\\frac{5}{12})^2}$ $=$ $\\frac{(\\frac{5}{6})}{1 + \\frac{25}{144}}$ $=$ $\\frac{(\\frac{5}{6})}{(\\frac{169}{144})}$ $=$ $\\frac{5}{6} \\times \\frac{144}{169}$ $=$ $\\frac{5 \\times 24}{169}$ $=$ $\\frac{120}{169}$. ## Expressing Trigonometric Ratios for Double-angle Formulas In Trigonometric Ratios, We know that $\\sin^2 A + \\cos^2 A$ $= 1$ We have $2 \\cdot \\sin A \\cdot \\cos A = \\sin 2A$. Then $\\sin^2 A + \\cos^2 A + 2 \\cdot \\sin A \\cdot \\cos A = 1 + \\sin 2A$ $\\Longrightarrow \\quad (\\sin A + \\cos A)^2 = 1 + \\sin 2A$ $\\Longrightarrow \\quad \\sin A + \\cos", "\\theta) } ]$ we get $\\frac {1}{\\cos \\theta \\sin \\theta}$. But it is with product of $\\sin \\theta \\cos \\theta$. Hence the answer is $1$, Which is LHS. Hence proved. And we should tackle these problems always by substituting in $\\sin \\theta$ and $\\cos \\theta$ when LHS is a number. - Multiply the first fraction by $\\dfrac{\\cos{A}}{\\cos{A}}$, and the second by $\\dfrac{\\sin{A}}{\\sin{A}}$. We get \\begin{align} LHS &=\\frac{\\sin A\\cos A}{1+\\sin A- \\cos A}+\\frac{\\sin A\\cos A}{1-(\\sin A- \\cos A)}\\\\ &=\\frac{\\sin A\\cos A-\\sin A\\cos A(\\sin A- \\cos A)+\\sin A\\cos A+\\sin A\\cos A(\\sin A- \\cos A)}{1-(1-2\\sin A\\cos A)}\\\\ &=\\frac{2\\sin A\\cos A}{2\\sin A\\cos A}\\\\ &=1\\\\ &=RHS \\end{align} - First, lets express the Pythagorean identity in another way. $$\\sin^2 A +\\cos^2 A = 1$$ $$\\sec^2A\\csc^2A(\\sin^2 A +\\cos^2 A) = \\sec^2A\\csc^2A\\cdot1$$ $$\\sec^2A+\\csc^2 A = \\sec^2A\\csc^2A$$ Continuing $\\dots$ $$\\begin{array}{lll} \\frac{\\sin A}{\\sec A+\\tan A-1}+ \\frac{\\cos A}{\\csc A+\\cot A-1} & = & \\frac{1}{\\csc A(\\sec A+\\tan A-1)}+\\frac{1}{\\sec A(\\csc A+\\cot A-1)} \\\\ & = & \\frac{1}{\\csc A\\sec A+\\sec A-\\csc A}+\\frac{1}{\\sec A\\csc A+\\csc A-\\sec A} \\\\ & = & \\frac{1}{\\sec A\\csc A+(\\sec A-\\csc A)}+\\frac{1}{\\sec A\\csc A-(\\sec A-\\csc A)} \\\\ & = & \\frac{\\sec A\\csc A\\color{blue}{+(\\sec A-\\csc A)} + \\sec A\\csc A\\color{blue}{-(\\sec A-\\csc A)}}{(\\sec A\\csc A+(\\sec A-\\csc A))(\\sec A\\csc A-(\\sec A-\\csc A))}\\\\ & = & \\frac{2\\sec A\\csc A}{\\sec^2A \\csc^2 A-(\\sec A - \\csc A)^2}\\\\ & = & \\frac{2\\sec A\\csc A}{\\sec^2A \\csc^2 A-(\\color{blue}{\\sec^2", "problem, the coefficients of $x$ can be dealt with easily: $$\\frac{ \\cos (3x) - 1 }{ \\sin (2x) \\tan (3x)} = \\frac{ \\cos (3x) - 1 }{(3x)^2}\\frac{2x}{\\sin(2x)}\\frac{3x}{\\tan(3x)} \\frac{3^2}{2\\cdot 3}$$ where the last (numeric) fraction has been inserted to keep the expression the same. Now compute the limit of each fraction and you're done. \\begin{align} \\lim_{x\\to0} \\frac{ \\cos (3x) - 1 }{ \\sin (2x) \\tan (3x)} &= \\lim_{x\\to0} \\frac{ -2\\sin^2 (\\dfrac{3x}{2}) \\cos (3x) }{ \\sin (2x) \\sin (3x)} \\\\ &= -2\\lim_{x\\to0} \\dfrac{\\sin^2 (\\dfrac{3x}{2}) }{(\\dfrac{3x}{2})^2} \\dfrac{(2x)}{\\sin (2x)} \\frac{ (3x)}{ ( \\sin (3x)}.\\dfrac{(\\dfrac{3x}{2})^2}{(2x)(3x)}\\cos (3x) \\\\ &=\\color{blue}{-\\dfrac{3}{4}} \\end{align} $$\\lim_{x\\rightarrow0}\\frac{\\cos 3x - 1}{ \\sin 2x \\tan 3x }=-\\lim_{x\\rightarrow0}\\frac{2\\sin^2\\frac{3x}{2}}{\\sin2x\\sin3x}=$$ $$=-\\frac{3}{4}\\lim_{x\\rightarrow0}\\frac{\\frac{\\sin^2\\frac{3x}{2}}{\\frac{9x^2}{4}}}{\\frac{\\sin2x}{2x}\\cdot\\frac{\\sin3x}{3x}}=-\\frac{3}{4}$$ Just another way using the standard Taylor expansions. $$\\cos(3x)=1-\\frac{9 x^2}{2}+\\frac{27 x^4}{8}+O\\left(x^6\\right)$$ $$\\sin(2x)=2 x-\\frac{4 x^3}{3}+O\\left(x^5\\right)$$ $$\\tan(3x)=3 x+9 x^3+O\\left(x^5\\right)$$ makes $$\\cos(3x)-1=-\\frac{9 x^2}{2}+\\frac{27 x^4}{8}+O\\left(x^6\\right)$$ $$\\sin(2x)\\tan(3x)=6 x^2+14 x^4+O\\left(x^6\\right)$$ Now, long division leads to $$\\frac{\\cos (3x) - 1}{ \\sin (2x) \\tan (3x) }=-\\frac{3}{4}+\\frac{37 x^2}{16}+O\\left(x^3\\right)$$ showing the limit and how it is appraoched. Just for illustration, use $x=\\frac \\pi {12}$ (quite far away from $0$); the exact value of the expression is $\\sqrt{2}-2\\approx -0.585786$ while the above approximation would give $\\frac{37 \\pi ^2}{2304}-\\frac{3}{4}\\approx -0.591504$. The thinking/attempt in the question post is quite complicated. Simple idea related to @ParamanandSingh 's comment. Multiply the function by $\\frac{\\cos(3x)+1}{\\cos(3x)+1}$, we get $$f(x) = \\frac{\\cos^{2}(3x) - 1}{\\sin(2x) \\tan(3x)(\\cos(3x)+1)} =\\frac{- \\sin(3x) \\cos(3x)}{\\sin(2x) (\\cos(3x)+1)} = \\frac{- (2x)", "for $\\cos z$, and subtract this from 1. You should find that the result can be written in the form $1-\\cos z=az^2+bz^4+\\cdots$ for some constants $a$ and $b$, and where the dots represent higher powers of $z$. Then think about using the binomial theorem to pin down the coefficient of $z^{-1}$ in $$f(z) = \\frac{z}{1-\\cos z}=\\frac{z}{az^2(1+\\frac{b}{a}z^2+\\cdots)}.$$ - Very helpful! Turns out there was a more direct way of doing it. I forgot about my binomial expansions. – Ahsan Jul 8 '12 at 20:27 Using the power series $\\cos(z)=1-\\frac12z^2+O(z^4)$ yields \\begin{align} f(z) &=\\frac{z}{1-(1-\\frac12z^2+O(z^4))}\\\\ &=\\frac{1}{\\frac12z+O(z^3)}\\\\ &=\\frac{1}{z}\\frac{1}{\\frac12+O(z^2)}\\\\ &=\\frac{1}{z}(2+O(z^2)) \\end{align} Thus the residue is $2$ Another thought is to write \\begin{align} \\frac{z}{1-\\cos(z)} &=\\frac{z^2}{\\sin^2(z)}\\frac{1+\\cos(z)}{z}\\\\ &=\\frac{1}{1+O(z^2)}\\cdot\\frac{2+O(z^2)}{z} \\end{align} again giving a residue of $2$. - You beat me to it! (+1) – user26872 Jul 8 '12 at 20:26 $\\displaystyle\\text{Res}=\\lim_{z\\to0}(z-0)f(z)=\\lim\\frac{z^2}{1-\\cos z}$. Now use l'Hopital's rule twice. -", "cosine of $a$ and $b$ do not satisfy this relation. It turns out that $$\\sin a=\\frac35\\qquad\\cos a=\\frac45$$ $$\\sin b=\\frac5{13}\\qquad\\cos b=\\frac{12}{13}$$ and thus $$\\cos(a+b)=\\cos a\\cos b-\\sin a\\sin b=\\frac45\\cdot\\frac{12}{13}-\\frac35\\cdot\\frac5{13}=\\frac{33}{65}$$ $$\\tan(a+b)= \\frac {\\tan a +\\tan b}{1-\\tan a \\tan b} = \\frac {3/4 +5/12}{1-(3/4)(5/12)}= \\frac {56}{33}$$ $$\\sec^2(a+b)=1+\\tan^2(a+b)=\\frac {4225}{1089}=(\\frac {65}{33})^2$$ $$\\cos(a+b)= 33/65$$ Solve for sin(a), cos(a), sin(b) and cos(b) (like I said in comments): $sin(a)=\\frac{3}{5}$ $cos(a)=\\frac{4}{5}$ $sin(b)=\\frac{5}{13}$ $cos(b)=\\frac{12}{13}$ Thus $$cos(a+b)=\\frac{4}{5}\\times\\frac{12}{13}-\\frac{3}{5}\\times\\frac{5}{13}=\\frac{33}{65}$$", "$x^{2}+y^{2}=1;$ $A=(0,2),$ $B=(-\\sqrt{3},-1),$ $C=(\\sqrt{3},-1);$ $P=(\\cos t,\\sin t),$ $t\\in (\\frac{\\pi}{6},\\frac{5\\pi}{6});$ $AB:\\,y-2=x\\sqrt{3};$ \\begin{align}\\displaystyle PD &= 1+\\sin t,\\\\ PE &= \\frac{2-\\sqrt{3}\\cos t-\\sin t}{2}=\\frac{2-2\\sin (t+\\frac{\\pi}{3})}{2},\\\\ PF &= \\frac{2+\\sqrt{3}\\cos t-\\sin t}{2}=\\frac{2+2\\sin (t+\\frac{2\\pi}{3})}{2}. \\end{align} From here, \\begin{align}\\displaystyle \\sqrt{PD} &= \\sin\\frac{t}{2}+\\cos\\frac{t}{2},\\\\ \\sqrt{PE} &= \\sin(\\frac{t}{2}+\\frac{\\pi}{6})-\\cos(\\frac{t}{2}+\\frac{\\pi}{6}),\\\\ \\sqrt{PF} &= \\sin(\\frac{t}{2}+\\frac{\\pi}{3})+\\cos(\\frac{t}{2}+\\frac{\\pi}{3}).\\\\ \\end{align} With a little more effort one gets $\\sqrt{PD}=\\sqrt{PE}+\\sqrt{PF}.$ Note that this identity is a special case of a more general one $\\displaystyle\\sqrt{PD}\\cos\\frac{A}{2}=\\sqrt{PE}\\cos\\frac{B}{2}+\\sqrt{PF}\\cos\\frac{C}{2}$ where, for an equilateral triangle $A=B=C=\\frac{\\pi}{3}.$ Given that, the required identity follows from $\\displaystyle 2(xy+yz+zx)-(x^{2}+y^{2}+z^{2})=(\\sqrt{x}+\\sqrt{y}+\\sqrt{z})\\prod_{cycl}(-\\sqrt{x}+\\sqrt{y}+\\sqrt{z})=0,$ the product is of three factors each of the form $(\\pm\\sqrt{x}\\pm\\sqrt{y}\\pm\\sqrt{z}),$ with only one minus sign in each. • Angle Trisectors on Circumcircle • Equilateral Triangles On Sides of a Parallelogram • Pompeiu's Theorem • Pairs of Areas in Equilateral Triangle • The Eutrigon Theorem • Equilateral Triangle in Equilateral Triangle • Seven Problems in Equilateral Triangle • Spiral Similarity Leads to Equilateral Triangle • Parallelogram and Four Equilateral Triangles • Miguel Ochoa's van Schooten Like Theorem • Two Conditions for a Triangle to Be Equilateral • Incircle in Equilateral Triangle • When Is Triangle Equilateral: Marian Dinca's Criterion • Barycenter of Cevian Triangle • Excircle in Equilateral Triangle • Converse Construction in Pompeiu's Theorem • Wonderful Trigonometry In Equilateral Triangle • 60o Angle And Importance of Being The Other End of a Diameter • One More Property of Equilateral", "$x^{2}+y^{2}=1;$ $A=(0,2),$ $B=(-\\sqrt{3},-1),$ $C=(\\sqrt{3},-1);$ $P=(\\cos t,\\sin t),$ $t\\in (\\frac{\\pi}{6},\\frac{5\\pi}{6});$ $AB:\\,y-2=x\\sqrt{3};$ \\begin{align}\\displaystyle PD &= 1+\\sin t,\\\\ PE &= \\frac{2-\\sqrt{3}\\cos t-\\sin t}{2}=\\frac{2-2\\sin (t+\\frac{\\pi}{3})}{2},\\\\ PF &= \\frac{2+\\sqrt{3}\\cos t-\\sin t}{2}=\\frac{2+2\\sin (t+\\frac{2\\pi}{3})}{2}. \\end{align} From here, \\begin{align}\\displaystyle \\sqrt{PD} &= \\sin\\frac{t}{2}+\\cos\\frac{t}{2},\\\\ \\sqrt{PE} &= \\sin(\\frac{t}{2}+\\frac{\\pi}{6})-\\cos(\\frac{t}{2}+\\frac{\\pi}{6}),\\\\ \\sqrt{PF} &= \\sin(\\frac{t}{2}+\\frac{\\pi}{3})+\\cos(\\frac{t}{2}+\\frac{\\pi}{3}).\\\\ \\end{align} With a little more effort one gets $\\sqrt{PD}=\\sqrt{PE}+\\sqrt{PF}.$ Note that this identity is a special case of a more general one $\\displaystyle\\sqrt{PD}\\cos\\frac{A}{2}=\\sqrt{PE}\\cos\\frac{B}{2}+\\sqrt{PF}\\cos\\frac{C}{2}$ where, for an equilateral triangle $A=B=C=\\frac{\\pi}{3}.$ Given that, the required identity follows from $\\displaystyle 2(xy+yz+zx)-(x^{2}+y^{2}+z^{2})=(\\sqrt{x}+\\sqrt{y}+\\sqrt{z})\\prod_{cycl}(-\\sqrt{x}+\\sqrt{y}+\\sqrt{z})=0,$ the product is of three factors each of the form $(\\pm\\sqrt{x}\\pm\\sqrt{y}\\pm\\sqrt{z}),$ with only one minus sign in each. [an error occurred while processing this directive]", "(c). 5. When is the denominator of $f'$ equal to $0\\text{?}$ 1. $x\\neq \\frac{n\\pi}{2}$ for any odd integer $n\\text{.}$ 2. $f'(x)=\\frac{(\\cos(x))\\cos(x)-\\sin(x)(-\\sin(x))}{(\\cos(x))^2}\\text{.}$ 3. $\\sin^2(\\theta)+\\cos^2(\\theta)=1\\text{.}$ 4. $f'(x)=\\sec^2(x)\\text{.}$ 5. $x\\neq \\frac{n\\pi}{2}$ for any odd integer $n\\text{.}$ Solution 1. Since $\\sin(x)$ is defined for every real value of $x$ and $\\cos(x)$ is only equal to zero at odd integer multiples of $\\frac{\\pi}2\\text{,}$ it follows that the quotient $\\tan(x)=\\frac{\\sin(x)}{\\cos(x)}$ is defined whenever $x\\neq \\ldots,-\\frac{3\\pi}2,-\\frac{\\pi}2,\\frac{\\pi}2,\\frac{3\\pi}2,\\ldots\\text{.}$ 2. Since $f(x)=\\tan(x)$ is the quotient of $\\sin(x)$ and $\\cos(x)\\text{,}$ then to differentiate $f$ using the quotient rule we first need $\\frac{d}{dx}[\\sin(x)]=\\cos(x)$ and $\\frac{d}{dx}[\\cos(x)]=-\\sin(x)\\text{.}$ Then \\begin{align*} f'(x)=\\mathstrut \\amp \\frac{d}{dx}\\left[\\frac{\\sin(x)}{\\cos(x)}\\right]\\\\ =\\mathstrut \\amp \\frac{\\frac{d}{dx}[\\sin(x)]\\cos(x)-\\sin(x)\\frac{d}{dx}[\\cos(x)]}{(\\cos(x))^2}\\\\ =\\mathstrut \\amp \\frac{(\\cos(x))\\cos(x)-\\sin(x)(-\\sin(x))}{(\\cos(x))^2}\\\\ =\\mathstrut \\amp \\frac{\\cos(x)\\cos(x)+\\sin(x)\\sin(x)}{\\cos^2(x)} \\end{align*} 3. The Pythagorean Identity states that $\\sin^2(\\theta)+\\cos^2(\\theta)=1$ for any real number $\\theta\\text{.}$ We can rewrite the form of $f'$ found in part (b) as \\begin{equation*} f'(x)=\\frac{\\cos^2(x)+\\sin^2(x)}{\\cos^2(x)}\\text{,} \\end{equation*} and use the Pythagorean Identity to reduce the numerator to $1\\text{.}$ Thus a simpler expression for $f'(x)$ is \\begin{equation*} f'(x)=\\frac{1}{\\cos^2(x)}\\text{.} \\end{equation*} 4. Since $\\sec(x)=\\frac{1}{\\cos(x)}\\text{,}$ then \\begin{equation*} f'(x)=\\frac{1}{\\cos^2(x)}=\\sec^2(x)\\text{.} \\end{equation*} 5. Just like the original function $f\\text{,}$ this derivative function $f'$ can be expressed as a quotient, so its domain will be the values of $x$ at which the numerator is defined and the denominator is nonzero. Since $f'(x)=\\frac{1}{\\cos^2(x)}\\text{,}$ the numerator is always defined and the denominator is zero exactly when $\\cos(x)=0\\text{,}$ which occurs at", "In particular, we need $$\\lim_{t\\to 0}\\sin(t)/t=0$$, $$\\lim_{t\\to 0}(\\cos(t)-1)/t=0$$, and $$\\lim_{t\\to 0}(t-\\sin(t))/t^3=1/6$$. \\begin{align} \\frac{x\\sin x+2(\\cos x-1)}{x^3}&=\\frac{x(2\\sin(x/2)\\cos(x/2))+2(-2\\sin^2(x/2))}{x^3} \\\\ &= \\frac{\\sin(x/2)}{x/2}\\cdot\\frac{x\\cos(x/2)-2\\sin(x/2)}{x^2} \\end{align} We have $$\\lim_{x\\to 0}\\sin(x/2)/(x/2)=1$$. As for the second factor, \\begin{align} \\frac{x\\cos(x/2)-2\\sin(x/2)}{x^2} &= \\frac{\\left[x(\\cos(x/2)-1)\\right]+\\left[\\sin(x)-2\\sin(x/2)\\right]+\\left[x-\\sin(x)\\right]}{x^2} \\\\ &= \\frac{\\cos(x/2)-1}{x}+\\frac{2\\sin(x/2)(\\cos(x/2)-1)}{x^2}+\\frac{x-\\sin(x)}{x^2} \\\\ &= \\frac 12\\frac{\\cos(x/2)-1}{x/2}+\\frac 12\\frac{\\sin(x/2)}{x/2}\\frac{\\cos(x/2)-1}{x/2}+\\frac{x-\\sin(x)}{x^3}\\cdot x \\end{align} You can finish with the aforementioned limits. The limit of the second factor is $$0$$. So the desired limit is $$1\\cdot 0=0$$. • Brilliant, that's exactly what I was hoping for!! – Matthew Mar 25 at 17:40 I'll make use of some limits famously provable without the banned techniques:\\begin{align}\\frac{x\\sin x+2(\\cos x-1)}{x^3}&=\\frac{2x\\sin\\frac{x}{2}\\cos\\frac{x}{2}-4\\sin^2\\frac{x}{2}}{x^3}\\\\&=\\underbrace{\\frac{4\\sin^2\\frac{x}{2}}{x^2}}_{\\to1}\\underbrace{\\frac{\\frac{x}{2}\\cot\\frac{x}{2}-1}{x}}_{\\to0}.\\end{align}The proof that $$\\lim_{x\\to0}\\frac{\\sin x}{x}=\\lim_{x\\to0}\\frac{\\tan x}{x}=1$$ is famous, whence$$\\lim_{x\\to0}\\frac{x\\cot x-1}{x}=\\lim_{x\\to0}\\frac{x-\\tan x}{x\\tan x}=\\lim_{x\\to0}\\frac{x-\\tan x}{x^2}=0$$because $$\\lim_{x\\to0}\\frac{x-\\tan x}{x^3}=-\\frac13$$ is famous too (see e.g. @bjorn93's link above). • One can prove that $(x-\\tan x) /x^2\\to 0$ using the famous inequality $\\sin x <x<\\tan x$ combined with Squeeze. – Paramanand Singh Mar 27 at 8:52 Whether series is allowed or not, here's the answer. $$x\\sin x = x(x-x^3/3!+x^5/5!...)$$ so the lowest order term is $$x^2$$. Simiilarly, $$2(1-\\cos x)=2[1-(1-x^2/2!+x^4/4!...)]=2[x^2/2-x^4/4!...]$$ so the lowest order term is $$x^2$$. When you subtract the second from the first, the $$x^2$$ terms will cancel. The next lowest order in the numerator is $$O(x^4)$$. Then it is clear that the limit is $$0$$. Note, $$\\frac{x\\sin x + 2(\\cos x -1)}{x^3} =\\frac{\\sin x}{x}\\cdot \\frac {x-2\\tan\\frac x2}{x^2}$$ and, $$x-2\\tan\\frac x2", "\\to 0}{\\frac{x-\\tan(x)}{x^2 \\cdot \\sin(x)}}$$ is of the form $\\frac\\infty\\infty,$ We can utilize L'Hôpital's rule, too. So, $$\\lim_{x \\to 0}{\\frac{x-\\tan(x)}{x^2 \\cdot \\sin(x)}}=\\lim_{x \\to 0}\\frac{1-\\sec^2x}{2x\\sin x+x^2\\cos x}=-\\lim_{x \\to 0}\\frac{\\tan^2x}{2x\\sin x+x^2\\cos x}$$ $$=-\\lim_{x \\to 0}\\frac{\\sin^2x}{\\cos^2x(2x\\sin x+x^2\\cos x)}$$ $$= -\\left(\\lim_{x \\to 0}\\frac{\\sin x}x\\right)^2\\cdot\\frac1{\\lim_{x \\to 0}\\cos^2x\\cdot \\left(2\\lim_{x \\to 0}\\frac{\\sin x}x+\\lim_{x \\to 0}\\cos x\\right)}$$ (Dividing the numerator & the denominator by $x^2$ ) Now, we know, $\\lim_{h\\to0}\\frac{\\sin h}h=?$ and $\\lim_{h\\to0}\\cos h=?$ \\begin{align} &\\lim_{x \\to 0}{\\frac{1-\\frac{\\tan(x)}{x}}{x \\cdot \\sin(x)}}\\\\ &=\\lim_{x \\to 0}{\\left(\\frac{1}{x \\cdot \\sin(x)}-\\frac{\\frac{\\tan(x)}{x}}{x \\cdot \\sin(x)}\\right)}\\\\ &\\neq\\lim_{x \\to 0}{\\left(\\frac{1}{x \\cdot \\sin(x)}-\\frac{\\lim_{x \\to 0}{\\frac{\\tan(x)}{x}}}{x \\cdot \\sin(x)}\\right)}\\\\ &=\\lim_{x \\to 0}{\\left(\\frac{1}{x \\cdot \\sin(x)}-\\frac{1}{x \\cdot \\sin(x)}\\right)}\\\\ &=\\lim_{x \\to 0}{0}=0 \\end{align}", "+0 # cos^2(pi/8)+cos^2(3pi/8)+cos^2(5pi/8)+cos^2(7pi/8) can you solve it showing me the whole way of how doing it please? 0 524 5 cos^2(pi/8)+cos^2(3pi/8)+cos^2(5pi/8)+cos^2(7pi/8) can you solve it showing me the whole way of how doing it? Guest Apr 20, 2017 #1 +5541 +3 $$\\cos^2(\\frac{\\pi}{8})+\\cos^2(\\frac{3\\pi}{8})+\\cos^2(\\frac{5\\pi}{8})+\\cos^2(\\frac{7\\pi}{8}) \\\\~\\\\ \\cos^2(\\frac12\\cdot\\frac{\\pi}{4})+\\cos^2(\\frac12\\cdot\\frac{3\\pi}{4})+\\cos^2(\\frac12\\cdot\\frac{5\\pi}{4})+\\cos^2(\\frac12\\cdot\\frac{7\\pi}{4})$$ Apply the half-angle formula: $$\\cos^2(\\frac12\\cdot a)=\\frac12(1+\\cos a)$$ $$\\frac12(1+\\cos\\frac{\\pi}{4})+\\frac12(1+\\cos \\frac{3\\pi}{4})+\\frac12(1+\\cos\\frac{5\\pi}{4})+\\frac12(1+\\cos\\frac{7\\pi}{4})$$ Now we can evaluate the cosines. $$\\frac12(1+\\frac{\\sqrt2}{2})+\\frac12(1-\\frac{\\sqrt2}{2})+\\frac12(1-\\frac{\\sqrt2}{2})+\\frac12(1+\\frac{\\sqrt2}{2})$$ Simplify. $$(1+\\frac{\\sqrt2}{2})+(1-\\frac{\\sqrt2}{2}) \\\\~\\\\ 1+\\frac{\\sqrt2}{2}+1-\\frac{\\sqrt2}{2} \\\\~\\\\ 2$$ If you want more steps shown please just say so hectictar Apr 20, 2017 Sort: #1 +5541 +3 $$\\cos^2(\\frac{\\pi}{8})+\\cos^2(\\frac{3\\pi}{8})+\\cos^2(\\frac{5\\pi}{8})+\\cos^2(\\frac{7\\pi}{8}) \\\\~\\\\ \\cos^2(\\frac12\\cdot\\frac{\\pi}{4})+\\cos^2(\\frac12\\cdot\\frac{3\\pi}{4})+\\cos^2(\\frac12\\cdot\\frac{5\\pi}{4})+\\cos^2(\\frac12\\cdot\\frac{7\\pi}{4})$$ Apply the half-angle formula: $$\\cos^2(\\frac12\\cdot a)=\\frac12(1+\\cos a)$$ $$\\frac12(1+\\cos\\frac{\\pi}{4})+\\frac12(1+\\cos \\frac{3\\pi}{4})+\\frac12(1+\\cos\\frac{5\\pi}{4})+\\frac12(1+\\cos\\frac{7\\pi}{4})$$ Now we can evaluate the cosines. $$\\frac12(1+\\frac{\\sqrt2}{2})+\\frac12(1-\\frac{\\sqrt2}{2})+\\frac12(1-\\frac{\\sqrt2}{2})+\\frac12(1+\\frac{\\sqrt2}{2})$$ Simplify. $$(1+\\frac{\\sqrt2}{2})+(1-\\frac{\\sqrt2}{2}) \\\\~\\\\ 1+\\frac{\\sqrt2}{2}+1-\\frac{\\sqrt2}{2} \\\\~\\\\ 2$$ If you want more steps shown please just say so hectictar Apr 20, 2017 #2 +79645 +1 Very ingenious, hectictar.....it would not have occurred to me to apply the half-angle formula!!!! CPhill Apr 20, 2017 edited by CPhill Apr 20, 2017 #3 +5541 +3 AH, I know you would have seen that before long.. What's really ingenious is how you figured out that MrSilvers was trying to take the sin of 15º ...That took some real insight ! hectictar Apr 20, 2017 #4 +18764 +2 cos^2(pi/8)+cos^2(3pi/8)+cos^2(5pi/8)+cos^2(7pi/8) can you solve it showing me the whole way of how doing it? $$\\begin{array}{l} \\cos^2(\\frac18\\pi)+\\cos^2(\\frac38\\pi)+\\cos^2(\\frac58\\pi)+\\cos^2(\\frac78\\pi) =\\ ? \\\\ \\end{array}$$ $$\\begin{array}{|rcll|} \\hline \\cos(\\frac78\\pi) =", "by 1, and that should not cause any problems. – Shobhit Dec 2 '17 at 9:42 • @user175989 you can set as solved if you are ok – gimusi Dec 3 '17 at 10:00 All you need to do now is$$\\frac{\\sin y\\cos x+\\cos y\\sin x}{\\sin y\\cos x-\\cos y\\sin x}=\\frac{\\frac{\\sin y\\cos x+\\cos y\\sin x}{\\cos x\\cos y}}{\\frac{\\sin y\\cos x-\\cos y\\sin x}{\\cos x\\cos y}}=\\frac{\\tan x+\\tan y}{\\tan x-\\tan y}.$$ Consider fractions that are equivalent to $2/3$. $$\\frac{2}{3}\\cdot\\frac{2}{2} = \\frac{4}{6}$$ $$\\frac{2}{3}\\cdot\\frac{3}{3} = \\frac{6}{9}$$ $$etc.$$ Conclusion ... you're allowed to multiply by one. Similarly, with the expression $\\displaystyle\\frac{\\sin y\\cos x+\\cos y \\sin x}{\\sin y \\cos x−\\cos y \\sin x}$, you're allowed to multiply by $\\displaystyle\\frac{\\frac{1}{\\cos y \\cos x}}{\\frac{1}{\\cos y \\cos x}}$ because $$\\displaystyle\\frac{\\frac{1}{\\cos y \\cos x}}{\\frac{1}{\\cos y \\cos x}} = 1$$ And just to make things mentally more clear, I may convert the original expression to a fraction of fractions as follows $$\\displaystyle\\frac{\\frac{\\sin y\\cos x+\\cos y \\sin x}{1}}{\\frac{\\sin y \\cos x−\\cos y \\sin x}{1}}$$ before multiplying. • What does $\\cos y \\cos x$ do to the expression ? there's nothing that can be cancelled out due to the '+' sign in the expression – user175089 Dec 3 '17 at 9:19 • Try multiplying the two expressions together, as suggested, and think of the distributive property when you do. – John Joy Dec 4 '17", "the trigonometric sin formula for double angle $$sin(2x) = 2 sin(x) ~ cos(x)$$ to show that $$cos \\frac{\\pi}{4} \\cdot cos \\frac{\\pi}{8} \\cdot cos \\frac{\\pi}{16} = \\frac{\\frac{1}{8}}{sin \\frac{\\pi}{16} } = \\frac{2}{\\pi} \\cdot \\frac{\\frac{\\pi}{16}}{sin \\frac{\\pi}{16} }$$ 3. As we have discussed above, since the angle $\\frac{\\pi}{16}$ is very small, $$sin \\frac{\\pi}{16} \\approx \\frac{\\pi}{16}$$ and $$cos \\frac{\\pi}{4} \\cdot cos \\frac{\\pi}{8} \\cdot cos \\frac{\\pi}{16} \\approx \\frac{2}{\\pi}$$ 4. In general, prove that $$cos \\frac{\\pi}{4} \\cdot cos \\frac{\\pi}{8} \\cdots cos \\frac{\\pi}{2^n} = \\frac{2}{\\pi} \\cdot \\frac{\\frac{\\pi}{2^n}}{sin \\frac{\\pi}{2^n} }$$ and $$\\lim_{n \\to \\infty} cos \\frac{\\pi}{4} \\cdot cos \\frac{\\pi}{8} \\cdots cos \\frac{\\pi}{2^n} = \\frac{2}{\\pi}$$ This is the Vieta formula of the number $\\pi$: $$\\sqrt{\\frac{1}{2}} \\cdot \\sqrt{\\frac{1}{2} + \\frac{1}{2} \\sqrt{\\frac{1}{2}}} \\cdot \\sqrt{\\frac{1}{2} + \\frac{1}{2} \\sqrt{\\frac{1}{2} + \\frac{1}{2} \\sqrt{\\frac{1}{2}}}} \\cdots = \\frac{2}{\\pi}$$" ]

[ "don't. A valid calculation of the limit would for example be: \\begin{align} &\\lim_{x\\to 0}{\\frac{\\tan{x}-\\sin x}{x^3}} \\\\ =& \\lim_{x\\to 0} \\frac {\\tan x}x \\frac{1-\\cos x}{x^2} \\\\ =& \\lim_{x\\to 0} \\frac {\\tan x}x \\lim_{x\\to 0}\\frac{1-\\cos x}{x^2} \\\\=&1\\cdot \\frac 1 2 =\\frac 1 2. \\end{align} Here we can calculate each limit separately and multiply them afterwards, as both exist. (In the first step I just took the\\tan x$factor outside). Hint: Use the Taylor development:$sin(x)=x-x^3/6+O(x^3)$,$tan(x)=x+x^3/3+O(x^3)$. • the order of the truncation error is surely$O(x^5)\\$, isn't it? Mar 9 '18 at 18:47 we have $$\\frac{-\\sin(x){\\cos(x)}+\\sin(x)}{\\cos(x)x^3}=\\frac{\\sin(x)(1-\\cos^2(x))}{x^3\\cos(x)(1+\\cos(x))}=\\frac{\\sin^3(x)}{x^3}\\frac{1}{\\cos(x)(1+\\cos(x))}$$", "to approach 0 from the left? What does this tell you about this problem? 11. Divide top and bottom by \\displaystyle \\begin{align*} x^3 \\end{align*}. 3. ## Re: Limits. For #7: limit sin(2x)/(5x) = limit sin(2x)/2x * limit 2x/5x = 1 * 2/5 = 2/5 Just picture it.. if the denominator were equal to (2x), the limit would be 1, but instead, it is larger than that by a scale of (5/2)... therefore you need to divide by that much extra, to get 2/5. For #8, the same principle applies actually. As x -> 0 in this case, you can treat tan() as sin(), because you are dividing sin() by a quantity that approaches 1, cos(x). So the answer is 2/Pi. For #5, multiply both the numerator and the denominator of the big fraction by 2(2+x). It should then simplify nicely. 4. ## Re: Limits. Hello, Zirrick! Here is #8 without using L'Hopital . . . $8.\\;\\lim_{x\\to0}\\frac{\\tan(2x)}{\\tan(\\pi x)}$ $\\frac{\\tan(2x)}{\\tan(\\pi x)} \\;=\\;\\dfrac{\\frac{\\sin(2x)}{\\cos(2x)}}{\\frac{\\sin( \\pi x)}{\\cos(\\pi x)}} \\;=\\; \\frac{\\sin(2x)}{1}\\cdot\\frac{1}{\\sin(\\pi x)}\\cdot\\frac{\\cos(\\pi x)}{\\cos(2x)}$ . . $=\\;{\\color{red}\\frac{2x}{2x}}\\cdot\\frac{\\sin(2x)}{ 1}\\cdot {\\color{red}\\frac{\\pi x}{\\pi x}}\\cdot\\frac{1}{\\sin(\\pi x)} \\cdot \\frac{\\cos(\\pi x)}{\\cos(2x)} \\;=\\;\\frac{2x}{\\pi x}\\cdot\\frac{\\sin(2x)}{2x}\\cdot\\frac{\\pi x}{\\sin(\\pi x)}\\cdot\\frac{\\cos(\\pi x)}{\\cos(2x)}$ $\\lim_{x\\to0}\\left[\\frac{2}{\\pi}\\cdot\\frac{\\sin(2x)}{2x}\\cdot\\frac{ \\pi x}{\\sin(\\pi x)}\\cdot\\frac{\\cos(\\pi x)}{\\cos(2x)}\\right] \\;=\\; \\frac{2}{\\pi}\\cdot1\\cdot1\\cdot\\frac{1}{1} \\;=\\;\\frac{2}{\\pi}$", "# Find the value of $\\tan A + \\tan B$, given values of $\\frac{\\sin (A)}{\\sin (B)}$ and $\\frac{\\cos (A)}{\\cos (B)}$ Given $$\\frac{\\sin (A)}{\\sin (B)} = \\frac{\\sqrt{3}}{2}$$ $$\\frac{\\cos (A)}{\\cos (B)} = \\frac{\\sqrt{5}}{3}$$ Find $\\tan A + \\tan B$. Approach Dividing the equations, we get the relation between $\\tan A$ and $\\tan B$ but that doesn't help in getting the value of $\\tan A + \\tan B$. The value comes in terms of $\\tan A$ or $\\tan B$ but the expected answer is independent of any variable . Also $$\\frac{\\sin(A)\\cdot\\cos(B) + \\sin(B)\\cdot\\cos(A)}{\\cos(A)\\cdot\\cos(B)} = \\tan(A) + \\tan(B)$$ We could get a value only if instead of $\\cos A$ there was $\\sin B$ in the relation(which we get on adding the ratios) • – imranfat Aug 23 '16 at 15:42 Something is wrong. I calculated \\begin{align} \\sin A &= \\dfrac{\\sqrt 3}{2}\\sin B \\\\ \\cos A &= \\dfrac{\\sqrt 5}{3}\\cos B \\\\ \\sin^2 A + \\cos^2 A &= \\dfrac 34 \\sin^2 B + \\dfrac59 \\cos^2 B\\\\ 1 &= \\dfrac 34 - \\dfrac{7}{36} \\cos^2 B\\\\ \\cos^2 B = -\\dfrac{9}{7} \\end{align} Which is impossible. HINT: $$\\tan A=\\frac{\\sin A}{\\cos A}=\\frac{\\frac{\\sqrt{3}}2\\sin B}{\\frac{\\sqrt{5}}3\\cos B}=\\frac{3\\sqrt{3}}{2\\sqrt{5}}\\tan B\\tag{1}$$ And $$1=\\sin^2 A+\\cos^2 A=\\frac{3}{4}\\sin^2 B+\\frac{5}{9}\\cos^2 B=\\frac59\\left(\\sin^2 B+\\cos^2B\\right)+\\frac7{36}\\sin^2 B$$ Last equation implies $$\\sin^2B=\\frac{16}7$$ $$\\frac{|\\sin A|}{|\\sin B|}=\\frac{\\sqrt{3}}{2} <1 \\iff \\frac{|\\cos A|}{|\\cos B|}>1$$ But $$\\frac{|\\cos A|}{|\\cos B|}=\\frac{\\sqrt{5}}{3}<1$$ that's impossible. Hint: Let $\\dfrac{\\sin A}{\\sqrt3}=\\dfrac{\\sin B}2=p$ and $\\dfrac{\\cos A}{\\sqrt5}=\\dfrac{\\cos", "and a $$\\sec^2 x$$ terms. Doing so we have \\begin{align} I &= \\int \\frac{\\cos^2 x \\sin x}{\\sin x - \\cos x} \\, dx\\\\ &= \\int \\frac{\\cos^2 x \\sin x}{\\sin x - \\cos x} \\cdot \\frac{\\sec^2 x}{\\sec^2 x} \\, dx\\\\ &= \\int \\frac{\\tan x}{(\\tan x - 1)(1 + \\tan^2 x)^2} \\cdot \\sec^2 x \\, dx. \\end{align} Now let $$t = \\tan x$$. Doing so yields \\begin{align} I &= \\int \\frac{t}{(t - 1)(1 + t^2)^2} \\, dt\\\\ &= \\int \\left [\\frac{1}{4(t - 1)} - \\frac{t + 1}{4(t^2 + 1)} + \\frac{1 - t}{2(1 + t^2)^2} \\right ] \\, dt\\\\ &= \\frac{1}{4} \\ln |t - 1| - \\frac{1}{8} \\ln |1 + t^2| + \\frac{t}{4(1 + t^2)} + C\\\\ &= \\frac{1}{4}\\ln |\\tan x - 1| + \\frac{1}{4} \\ln |\\cos x| + \\frac{1}{4} (1 + \\tan x) \\cos^2 x + C. \\end{align} or after playing around with a few trignometric identities $$\\int \\frac{\\cos^2 x \\sin x}{\\sin x - \\cos x} \\, dx = \\frac{1}{4} \\ln |\\sin x - \\cos x| + \\frac{1}{8} (\\cos 2x + \\sin 2x) + C.$$", "\\to 0}{\\frac{x-\\tan(x)}{x^2 \\cdot \\sin(x)}}$$ is of the form $\\frac\\infty\\infty,$ We can utilize L'Hôpital's rule, too. So, $$\\lim_{x \\to 0}{\\frac{x-\\tan(x)}{x^2 \\cdot \\sin(x)}}=\\lim_{x \\to 0}\\frac{1-\\sec^2x}{2x\\sin x+x^2\\cos x}=-\\lim_{x \\to 0}\\frac{\\tan^2x}{2x\\sin x+x^2\\cos x}$$ $$=-\\lim_{x \\to 0}\\frac{\\sin^2x}{\\cos^2x(2x\\sin x+x^2\\cos x)}$$ $$= -\\left(\\lim_{x \\to 0}\\frac{\\sin x}x\\right)^2\\cdot\\frac1{\\lim_{x \\to 0}\\cos^2x\\cdot \\left(2\\lim_{x \\to 0}\\frac{\\sin x}x+\\lim_{x \\to 0}\\cos x\\right)}$$ (Dividing the numerator & the denominator by $x^2$ ) Now, we know, $\\lim_{h\\to0}\\frac{\\sin h}h=?$ and $\\lim_{h\\to0}\\cos h=?$ \\begin{align} &\\lim_{x \\to 0}{\\frac{1-\\frac{\\tan(x)}{x}}{x \\cdot \\sin(x)}}\\\\ &=\\lim_{x \\to 0}{\\left(\\frac{1}{x \\cdot \\sin(x)}-\\frac{\\frac{\\tan(x)}{x}}{x \\cdot \\sin(x)}\\right)}\\\\ &\\neq\\lim_{x \\to 0}{\\left(\\frac{1}{x \\cdot \\sin(x)}-\\frac{\\lim_{x \\to 0}{\\frac{\\tan(x)}{x}}}{x \\cdot \\sin(x)}\\right)}\\\\ &=\\lim_{x \\to 0}{\\left(\\frac{1}{x \\cdot \\sin(x)}-\\frac{1}{x \\cdot \\sin(x)}\\right)}\\\\ &=\\lim_{x \\to 0}{0}=0 \\end{align}", "\\sin^2 A$. Thus, we have $\\cos 2A = \\cos^2 A - \\sin^2 A$$= 2 \\cdot \\cos^2 A - 1$$= 1 - 2 \\cdot \\sin^2 A$ ## Double Angle Identities We know that $\\cos^2 A + \\sin^2 A$ $= 1$ Now $\\sin 2A$ $=$ $\\frac{\\sin 2A}{1}$ $=$ $\\frac{2 \\cdot \\sin A \\cdot \\cos A}{\\cos^2 A + \\sin^2 A}$ $=$ $\\frac{(\\frac{2 \\cdot \\sin A \\cdot \\cos A}{\\cos^2 A})}{(\\frac{\\cos^2 A + \\sin^2 A}{\\cos^2 A})}$ $=$ $\\frac{(\\frac{2 \\cdot \\sin A}{\\cos A})}{(\\frac{\\cos^2 A}{\\cos^2 A} + \\frac{\\sin^2 A}{\\cos^2 A})}$ $=$ $\\frac{2 \\cdot \\tan A}{1 + \\tan^2 A}$. Thus, $\\sin 2A =$ $\\frac{2 \\cdot \\tan A}{1 + \\tan^2 A}$ ## Given that $An Alpha =$ $frac{5}{12}$. Find out $sin 2a$. Solution We know that $\\sin 2\\alpha =$ $\\frac{2 \\cdot \\tan \\alpha}{1 + \\tan^2 \\alpha}$ $=$ $\\frac{2 \\cdot (\\frac{5}{12})}{1 + (\\frac{5}{12})^2}$ $=$ $\\frac{(\\frac{5}{6})}{1 + \\frac{25}{144}}$ $=$ $\\frac{(\\frac{5}{6})}{(\\frac{169}{144})}$ $=$ $\\frac{5}{6} \\times \\frac{144}{169}$ $=$ $\\frac{5 \\times 24}{169}$ $=$ $\\frac{120}{169}$. ## Expressing Trigonometric Ratios for Double-angle Formulas In Trigonometric Ratios, We know that $\\sin^2 A + \\cos^2 A$ $= 1$ We have $2 \\cdot \\sin A \\cdot \\cos A = \\sin 2A$. Then $\\sin^2 A + \\cos^2 A + 2 \\cdot \\sin A \\cdot \\cos A = 1 + \\sin 2A$ $\\Longrightarrow \\quad (\\sin A + \\cos A)^2 = 1 + \\sin 2A$ $\\Longrightarrow \\quad \\sin A + \\cos", "$$\\frac{df}{dz} = 2 \\cdot tan(z) sec^2(z) =\\frac{2 \\cdot tan(z)}{cos^2(z)}$$ This makes $$\\frac{f(z)}{f'(z)} = tan^2(z) \\cdot \\frac{cos^2(z)}{2 \\cdot tan(z)} = \\frac{tan(z)\\cdot cos^2(z)}{2} = \\frac{sin(z)\\cdot cos(z)}{2}$$ def f_tan(z): return np.square(np.tan(z)) def d_tan(z): return 2*np.tan(z) / np.square(np.cos(z)) output = newton_fractal(mesh, f_tan, d_tan, num_iter=15, r=50) kwargs = {'title': 'f(z) = z - \\dfrac{sin(z)cos(z)}{2}', 'cmap': 'binary'} plot_fractal(output, **kwargs); Note that you sometimes have to play with the radius in order to get a neat looking fractal. Finally, we can go a little bit wild with our function selection $$f(z) = \\sum_{i=1}^{10} sin^i(z)$$ $$\\frac{df}{dz} = \\sum_{i=1}^{10} i \\cdot sin^{i-1}(z) \\cdot cos(z)$$ def sin_sum(z, n=10): total = np.zeros(z.size, dtype=z.dtype) for i in range(1, n+1): total += np.power(np.sin(z), i) def d_sin_sum(z, n=10): total = np.zeros(z.size, dtype=z.dtype) for i in range(1, n+1): total += i * np.power(np.sin(z), i-1) * np.cos(z) We will denote this one ‘Wacky fractal’, as its equation would not be fun to try and put in the title. output = newton_fractal(small_mesh, sin_sum, d_sin_sum, num_iter=10, r=1) kwargs = {'title': 'Wacky \\ fractal', 'figsize': (6, 6), 'extent': [-1, 1, -1, 1], 'cmap': 'terrain'} plot_fractal(output, **kwargs) It is truly fascinating how distinct yet similar these fractals are with each other. This leads us to the final section. What makes fractals more exciting is how much there is to explore once you become familiar with the basics. Now", "cosine of $a$ and $b$ do not satisfy this relation. It turns out that $$\\sin a=\\frac35\\qquad\\cos a=\\frac45$$ $$\\sin b=\\frac5{13}\\qquad\\cos b=\\frac{12}{13}$$ and thus $$\\cos(a+b)=\\cos a\\cos b-\\sin a\\sin b=\\frac45\\cdot\\frac{12}{13}-\\frac35\\cdot\\frac5{13}=\\frac{33}{65}$$ $$\\tan(a+b)= \\frac {\\tan a +\\tan b}{1-\\tan a \\tan b} = \\frac {3/4 +5/12}{1-(3/4)(5/12)}= \\frac {56}{33}$$ $$\\sec^2(a+b)=1+\\tan^2(a+b)=\\frac {4225}{1089}=(\\frac {65}{33})^2$$ $$\\cos(a+b)= 33/65$$ Solve for sin(a), cos(a), sin(b) and cos(b) (like I said in comments): $sin(a)=\\frac{3}{5}$ $cos(a)=\\frac{4}{5}$ $sin(b)=\\frac{5}{13}$ $cos(b)=\\frac{12}{13}$ Thus $$cos(a+b)=\\frac{4}{5}\\times\\frac{12}{13}-\\frac{3}{5}\\times\\frac{5}{13}=\\frac{33}{65}$$", "\\theta) } ]$ we get $\\frac {1}{\\cos \\theta \\sin \\theta}$. But it is with product of $\\sin \\theta \\cos \\theta$. Hence the answer is $1$, Which is LHS. Hence proved. And we should tackle these problems always by substituting in $\\sin \\theta$ and $\\cos \\theta$ when LHS is a number. - Multiply the first fraction by $\\dfrac{\\cos{A}}{\\cos{A}}$, and the second by $\\dfrac{\\sin{A}}{\\sin{A}}$. We get \\begin{align} LHS &=\\frac{\\sin A\\cos A}{1+\\sin A- \\cos A}+\\frac{\\sin A\\cos A}{1-(\\sin A- \\cos A)}\\\\ &=\\frac{\\sin A\\cos A-\\sin A\\cos A(\\sin A- \\cos A)+\\sin A\\cos A+\\sin A\\cos A(\\sin A- \\cos A)}{1-(1-2\\sin A\\cos A)}\\\\ &=\\frac{2\\sin A\\cos A}{2\\sin A\\cos A}\\\\ &=1\\\\ &=RHS \\end{align} - First, lets express the Pythagorean identity in another way. $$\\sin^2 A +\\cos^2 A = 1$$ $$\\sec^2A\\csc^2A(\\sin^2 A +\\cos^2 A) = \\sec^2A\\csc^2A\\cdot1$$ $$\\sec^2A+\\csc^2 A = \\sec^2A\\csc^2A$$ Continuing $\\dots$ $$\\begin{array}{lll} \\frac{\\sin A}{\\sec A+\\tan A-1}+ \\frac{\\cos A}{\\csc A+\\cot A-1} & = & \\frac{1}{\\csc A(\\sec A+\\tan A-1)}+\\frac{1}{\\sec A(\\csc A+\\cot A-1)} \\\\ & = & \\frac{1}{\\csc A\\sec A+\\sec A-\\csc A}+\\frac{1}{\\sec A\\csc A+\\csc A-\\sec A} \\\\ & = & \\frac{1}{\\sec A\\csc A+(\\sec A-\\csc A)}+\\frac{1}{\\sec A\\csc A-(\\sec A-\\csc A)} \\\\ & = & \\frac{\\sec A\\csc A\\color{blue}{+(\\sec A-\\csc A)} + \\sec A\\csc A\\color{blue}{-(\\sec A-\\csc A)}}{(\\sec A\\csc A+(\\sec A-\\csc A))(\\sec A\\csc A-(\\sec A-\\csc A))}\\\\ & = & \\frac{2\\sec A\\csc A}{\\sec^2A \\csc^2 A-(\\sec A - \\csc A)^2}\\\\ & = & \\frac{2\\sec A\\csc A}{\\sec^2A \\csc^2 A-(\\color{blue}{\\sec^2", "20:50 Understandable - I just found this to be a little less messy. – user17794 Aug 27 '13 at 20:52 Hint: \\begin{align} 0&=\\cos x+\\sin x\\\\ \\sin x &= -\\cos x \\\\ \\text{Divide}&\\text{ both sides by }\\cos x\\\\ \\tan x &= -1 \\end{align} Now you can solve for $x$ using inverse trigonometric functions. - if you use the Double angle formulae $$\\sin^2(x)\\cos^2(x)=(1-\\cos2x)/2 \\cdot (1+\\cos2x)/2 = (1-\\cos^2(2x))/4 = \\sin^2(2x)/4 = (1-\\cos4x)/8$$ so minimum is $1/8$ at $x=45\\circ$ and maximum is 1/4 at $x=90\\circ$ remember $\\sin^2(x)=(1-\\cos2x)/2$ and $\\cos^2(x)=(1+\\cos2x)/2$ - The equation $\\cos x+\\sin x=0$ is really the same as $\\cos x-\\sin x=0$ once you make the transformation $x\\mapsto x+\\frac{\\pi}2$, because $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and $\\sin (x+\\frac{\\pi}2)=\\cos(x)$. The latter can be solved by resorting to $\\cos ^2+\\sin ^2=1$. Edit: We can derive $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and similar results (cf. Daniel Fischer's comment) from the well known formulas $\\cos (x+y)=\\cos x\\cos y-\\sin x\\sin y$ and $\\sin (x+y)=\\cos x\\sin y+\\cos y\\sin x$. These can either be proved geometrically, or by resorting to some definition of $\\sin$ and $\\cos$, for example $\\sin x=(e^{ix}-e^{-ix})/2i, \\cos x=(e^{ix}-e^{-ix})/2$ -", "$\\tan b=\\frac{5}{12}$ . $\\tan (a+b) &=\\frac{\\tan a +\\tan b}{1-\\tan a \\tan b} \\\\&= \\frac{\\frac{3}{4}+\\frac{5}{12}}{1-\\frac{3}{4}\\cdot\\frac{5}{12}} \\\\&= \\frac{\\frac{14}{12}}{\\frac{11}{16}}=\\frac{56}{33}$ Example C Simplify $\\cos (\\pi - x)$ . Solution: Expand this using the difference formula and then simplify. $\\cos (\\pi - x) &=\\cos \\pi \\cos x +\\sin \\pi \\sin x \\\\&=-1\\cdot \\cos x +0\\cdot \\sin x \\\\&=-\\cos x$ Concept Problem Revisit You can expand the expression using the difference formula and then simplify. $\\sin(\\frac{\\pi}{2} - x)=\\sin \\frac{\\pi}{2} \\cos x - \\cos \\frac{\\pi}{2} \\sin x \\\\&=1\\cdot \\cos x - 0\\cdot \\sin x \\\\&=cos x$ Guided Practice 1. Using the information from Example A, find $\\cos(a-b)$ . 2. Simplify $\\tan (x+\\pi)$ . 1. $\\cos(a-b) &=\\cos a \\cos b + \\sin a \\sin b \\\\&=-\\frac{4}{5}\\cdot \\frac{12}{13}+-\\frac{3}{5}\\cdot\\frac{5}{13} \\\\&=-\\frac{63}{65}$ 2. $\\tan (x+\\pi)&=\\frac{\\tan x +\\tan \\pi}{1-\\tan x \\tan \\pi} \\\\&=\\frac{\\tan x +0}{1-\\tan 0} \\\\&=\\tan x$ Practice $\\sin a =-\\frac{8}{17}, \\pi \\le a < \\frac{3\\pi}{2}$ and $\\sin b =-\\frac{1}{2}, \\frac{3\\pi}{2}\\le b <2\\pi$ . Find the exact trig values of: 1. $\\sin (a+b)$ 2. $\\cos (a+b)$ 3. $\\sin (a-b)$ 4. $\\tan (a+b)$ 5. $\\cos (a-b)$ 6. $\\tan (a-b)$ Simplify the following expressions. 1. $\\sin (2\\pi-x)$ 2. $\\sin \\left(\\frac{\\pi}{2}+x\\right)$ 3. $\\cos (x+\\pi)$ 4. $\\cos \\left(\\frac{3\\pi}{2}-x\\right)$ 5. $\\tan(x+2\\pi)$ 6. $\\tan(x-\\pi)$ 7. $\\sin \\left(\\frac{\\pi}{6}-x\\right)$ 8. $\\tan \\left(\\frac{\\pi}{4}+x\\right)$ 9. $\\cos \\left(x-\\frac{\\pi}{3}\\right)$ Determine if the following trig statements are true or false. 1. $\\sin(\\pi -", "\\sin(A)\\cdot \\sin(B)}$ $\\tan(A + B) = \\frac{\\tan(A) + \\tan(B)}{1 - \\tan(A)\\cdot\\tan(B)}$ $\\cot(A \\pm B) = \\frac{\\cot(B)\\cot(A)\\mp 1}{\\cot(B)\\pm \\cot(A)}=\\frac{1\\mp \\tan(A)\\tan(B)}{\\tan(A)\\pm \\tan(B)}$ $\\sin(A + B + C) = \\sin A\\cdot\\cos B\\cdot\\cos C + \\cos A\\cdot\\sin B\\cdot\\cos C + \\cos A\\cdot\\cos B\\cdot\\sin C - \\sin A\\cdot\\sin B\\cdot\\sin C$ $\\cos(A + B + C) = \\cos A\\cdot\\cos B\\cdot\\cos C - \\sin A\\cdot\\sin B\\cdot\\cos C - \\sin A\\cdot\\cos B\\cdot\\sin C - \\sin A\\cdot\\cos B \\cdot\\sin C - \\cos A \\cdot \\sin B\\cdot \\sin C$ $\\tan(A + B + C) = \\frac{\\tan A + \\tan B + \\tan C - \\tan A\\cdot \\tan B \\cdot \\tan C}{1 - \\tan A \\cdot\\tan B - \\tan B\\cdot\\tan C - \\tan A\\cdot\\tan C}$ ### Double And Triple Angle Formulae $\\sin(2A) = 2\\sin(A)\\cdot \\cos(A)$ $\\cos(2A) = \\cos^2(A) - \\sin^2(A) = 2\\cos^2(A) - 1 = 1 - 2\\sin^2(A)$ $\\tan(2A) = \\frac{2\\tan(A)}{1- \\tan^2(A)}$ $\\cos(2A) = \\frac{1 - \\tan^2(A)}{1 + \\tan^2(A)}$ $\\sin(2A) = \\frac{2\\tan(A)}{1 + \\tan^2(A)}$ $\\sin3A = 3\\sin A - 4 \\sin^3A$ $\\cos3A = 4\\cos^3A - 3 \\cos A$ $\\tan3A=\\frac{3\\tan A - \\tan^3A}{1-3\\tan^2A}$ $\\cot3A=\\frac{\\cot^3A-3\\cot A}{3\\cot^2A-1}$ $\\sin4A = 4\\cos^3A\\cdot \\sin A - 4\\cos A\\cdot \\sin^3A$ $\\cos4A = \\cos^4A - 6\\cos^2A\\cdot \\sin^2A + \\sin^4A$ $\\tan4A=\\frac{4\\tan A - 4\\tan^3A}{1-6\\tan^2A+\\tan^4A}$ $\\cot4A=\\frac{\\cot^4A-6\\cot^2A+1}{4\\cot^3A-4\\cot A}$ ### Sum and Difference of Trigonometric Functions $\\textrm{ sin }A \\textrm{ sin }B = \\frac{1}{2} (\\textrm{ cos }(A - B) - \\textrm{", "for $\\cos z$, and subtract this from 1. You should find that the result can be written in the form $1-\\cos z=az^2+bz^4+\\cdots$ for some constants $a$ and $b$, and where the dots represent higher powers of $z$. Then think about using the binomial theorem to pin down the coefficient of $z^{-1}$ in $$f(z) = \\frac{z}{1-\\cos z}=\\frac{z}{az^2(1+\\frac{b}{a}z^2+\\cdots)}.$$ - Very helpful! Turns out there was a more direct way of doing it. I forgot about my binomial expansions. – Ahsan Jul 8 '12 at 20:27 Using the power series $\\cos(z)=1-\\frac12z^2+O(z^4)$ yields \\begin{align} f(z) &=\\frac{z}{1-(1-\\frac12z^2+O(z^4))}\\\\ &=\\frac{1}{\\frac12z+O(z^3)}\\\\ &=\\frac{1}{z}\\frac{1}{\\frac12+O(z^2)}\\\\ &=\\frac{1}{z}(2+O(z^2)) \\end{align} Thus the residue is $2$ Another thought is to write \\begin{align} \\frac{z}{1-\\cos(z)} &=\\frac{z^2}{\\sin^2(z)}\\frac{1+\\cos(z)}{z}\\\\ &=\\frac{1}{1+O(z^2)}\\cdot\\frac{2+O(z^2)}{z} \\end{align} again giving a residue of $2$. - You beat me to it! (+1) – user26872 Jul 8 '12 at 20:26 $\\displaystyle\\text{Res}=\\lim_{z\\to0}(z-0)f(z)=\\lim\\frac{z^2}{1-\\cos z}$. Now use l'Hopital's rule twice. -", "\\sin(r))\\frac{d}{dr}[4^r - 2\\cos(r)] \\\\ =\\mathstrut \\amp [15r^2 + \\cos(r)](4^r - 2\\cos(r))+(5r^3 + \\sin(r))[4^r \\ln(4) + 2\\sin(r)]\\text{.} \\end{align*} 2. We use the quotient rule on $p\\text{,}$ followed by the product rule to differentiate the denominator, finding that \\begin{align*} p'(t) =\\mathstrut \\amp \\frac{\\frac{d}{dt}[\\cos(t)]t^6 \\cdot 6^t - \\cos(t) \\frac{d}{dt}[t^6 \\cdot 6^t]}{(t^6 \\cdot 6^t)^2}\\\\ =\\mathstrut \\amp \\frac{[-\\sin(t)]t^6 \\cdot 6^t - \\cos(t) [6t^5 \\cdot 6^t + t^6\\cdot 6^t \\ln(6)]}{(t^6 \\cdot 6^t)^2}\\text{.} \\end{align*} 3. Using the sum and constant multiple rules, it follows first that \\begin{equation*} g'(z) = 3 \\frac{d}{dz}[z^7 e^z] - 2\\frac{d}{dz}[z^2 \\sin(z)] + \\frac{d}{dz}\\left[ \\frac{z}{z^2 + 1} \\right]\\text{.} \\end{equation*} Applying the product rule in the first two terms and the quotient rule in the third, we find that \\begin{equation*} g'(z) = 3 [7z^6e^z+z^7 e^z] - 2[2z\\sin(z)+z^2 \\cos(z)] + \\frac{(z^2+1) - z(2z)}{(z^2 + 1)^2}\\text{.} \\end{equation*} 4. The particle's instantaneous velocity at the moment $t = 1$ is given by $s'(1)\\text{.}$ We use the quotient rule to find $s'(t)\\text{,}$ and then simplify by removing a common factor of $e^t$ to get \\begin{align*} s'(t) =\\mathstrut \\amp \\frac{(-3\\sin(t) - \\cos(t))e^t-(3\\cos(t) - \\sin(t))e^t}{(e^t)^2}\\\\ =\\mathstrut \\amp \\frac{-2\\sin(t)-4\\cos(t)}{e^t}\\text{.} \\end{align*} Thus, $s'(1) = \\frac{-2\\sin(1)-4\\cos(1)}{e^1} \\approx -1.414\\text{,}$ which is the particle's instantaneous velocity in feet per second at the moment $t = 1\\text{.}$ 5. Since $p(x) = f(x) \\cdot g(x)\\text{,}$ the product rule tells us \\begin{equation*} p'(x) = f'(x)g(x)+f(x)g'(x)\\text{,} \\end{equation*} and", "by 1, and that should not cause any problems. – Shobhit Dec 2 '17 at 9:42 • @user175989 you can set as solved if you are ok – gimusi Dec 3 '17 at 10:00 All you need to do now is$$\\frac{\\sin y\\cos x+\\cos y\\sin x}{\\sin y\\cos x-\\cos y\\sin x}=\\frac{\\frac{\\sin y\\cos x+\\cos y\\sin x}{\\cos x\\cos y}}{\\frac{\\sin y\\cos x-\\cos y\\sin x}{\\cos x\\cos y}}=\\frac{\\tan x+\\tan y}{\\tan x-\\tan y}.$$ Consider fractions that are equivalent to $2/3$. $$\\frac{2}{3}\\cdot\\frac{2}{2} = \\frac{4}{6}$$ $$\\frac{2}{3}\\cdot\\frac{3}{3} = \\frac{6}{9}$$ $$etc.$$ Conclusion ... you're allowed to multiply by one. Similarly, with the expression $\\displaystyle\\frac{\\sin y\\cos x+\\cos y \\sin x}{\\sin y \\cos x−\\cos y \\sin x}$, you're allowed to multiply by $\\displaystyle\\frac{\\frac{1}{\\cos y \\cos x}}{\\frac{1}{\\cos y \\cos x}}$ because $$\\displaystyle\\frac{\\frac{1}{\\cos y \\cos x}}{\\frac{1}{\\cos y \\cos x}} = 1$$ And just to make things mentally more clear, I may convert the original expression to a fraction of fractions as follows $$\\displaystyle\\frac{\\frac{\\sin y\\cos x+\\cos y \\sin x}{1}}{\\frac{\\sin y \\cos x−\\cos y \\sin x}{1}}$$ before multiplying. • What does $\\cos y \\cos x$ do to the expression ? there's nothing that can be cancelled out due to the '+' sign in the expression – user175089 Dec 3 '17 at 9:19 • Try multiplying the two expressions together, as suggested, and think of the distributive property when you do. – John Joy Dec 4 '17", "• You probably also need that $\\sin^2(r) \\approx r^2$, since you have that factor $1/r$ to take care of. – Thomas Ahle Mar 11 '19 at 15:24 • @ThomasAhle It follows from $\\sin x \\le x, \\forall x \\ge 0$ like this: $$\\frac4r \\sin^2\\left(\\frac{r\\sqrt{2}}2\\right) \\le \\frac4r \\left(\\frac{r\\sqrt{2}}2\\right)^2 = 2r \\xrightarrow{r\\to 0} 0$$ – mechanodroid Mar 11 '19 at 15:54 Solution without the use of polar coordinates, as asked by OP at the end of their post. Rewrite \\begin{aligned}1-\\cos(x-y)&=1-\\cos x \\cos y - \\sin x \\sin y \\\\ &=1-\\cos x + \\cos x (1-\\cos y)- \\sin x \\sin y, \\end{aligned} then \\begin{aligned}\\frac{2x(1-\\cos(x-y))}{x^2+y^2}&= 2x\\cdot \\frac{1-\\cos x}{x^2} \\cdot\\frac{x^2}{x^2+y^2}\\\\ &+ 2x\\cos x \\cdot \\frac{1-\\cos y}{y^2}\\cdot\\frac{y^2}{x^2+y^2}\\\\&- 2y\\cdot \\frac{\\sin x}{x}\\cdot\\frac{\\sin y}{y}\\cdot\\frac{x^2}{x^2+y^2}\\end{aligned} From where is $$\\lim_{(x,y)\\to (0,0)}\\frac{2x(1-\\cos(x-y))}{x^2+y^2} = 0.$$", "\\frac{s^3}{2abc} for semi-perimeter s, with equality only in the equilateral case.[2]:p.13,#608 a+b+c \\ge 2\\sqrt{bc} \\cos A + 2 \\sqrt{ca} \\cos B + 2\\sqrt{ab} \\cos C. [4]:Thm.1 \\sin A + \\sin B + \\sin C \\leq \\frac{3\\sqrt{3}}{2}. [1]:p.286 \\sin ^2 A + \\sin ^2 B + \\sin ^2 C \\leq \\frac{9}{4}. [1]:p. 286 \\sin A \\cdot \\sin B \\cdot \\sin C \\leq \\frac{3\\sqrt{3}}{8}. [5]:p. 203 \\sin A+\\sin B \\cdot \\sin C \\leq \\varphi[2]:p.149,#3297 where \\varphi = \\frac{1+\\sqrt{5}}{2}, the golden ratio. \\sin \\frac{A}{2} \\cdot \\sin \\frac{B}{2} \\cdot \\sin \\frac{C}{2} \\leq \\frac{1}{8}. [1]:p. 286 \\tan ^2 \\frac{A}{2} + \\tan ^2 \\frac{B}{2} + \\tan ^2 \\frac{C}{2} \\geq 1. [1]:p. 286 \\cot A + \\cot B + \\cot C \\geq \\sqrt{3}. [6] \\sin A \\cdot \\cos B +\\sin B \\cdot \\cos C+\\sin C \\cdot \\cos A \\leq \\frac{3\\sqrt{3}}{4}.[2]:p.187,#309.2 We have \\max\\left(\\sin \\frac{A}{2}, \\sin \\frac{B}{2}, \\sin \\frac{C}{2} \\right) \\le \\frac{1}{2} \\left(1+\\sqrt{1-\\frac{2r}{R}} \\right), with equality if and only if the triangle is isosceles with apex angle greater than or equal to 60°;[7]:Cor. 3 and \\min\\left(\\sin \\frac{A}{2}, \\sin \\frac{B}{2}, \\sin \\frac{C}{2} \\right) \\ge \\frac{1}{2} \\left(1-\\sqrt{1-\\frac{2r}{R}} \\right), with equality if and only if the triangle is isosceles with apex angle less than or equal to 60°.[7]:Cor. 3 We also have \\frac{r}{R}-\\sqrt{1-\\frac{2r}{R}} \\le \\cos A \\le \\frac{r}{R}+\\sqrt{1-\\frac{2r}{R}} and likewise for angles B, C, with equality in the first", "A -\\cos A+1}{\\sin A+\\cos A-1} \\\\=&\\dfrac{(\\sin A -\\cos A+1)\\cdot(\\sin A+\\cos A+1)}{(\\sin A+\\cos A-1)\\cdot(\\sin A+\\cos A+1)}=\\dfrac{(\\sin A +1)^2-\\cos^2 A}{(\\sin A+\\cos A)^2-1} \\\\=&\\dfrac{\\sin^2 A+(1-\\cos^2 A) +2\\sin A}{2\\sin A\\cos A}\\end{align} The rest is fairly straightforward. - Multiplying the numerator & the denominator by $\\cos A$ $$\\frac{\\tan A+\\sec A-1}{\\tan A-\\sec A+1}=\\frac{\\sin A+1-\\cos A}{\\sin A-1+\\cos A}$$ $$\\cos A(\\sin A+1-\\cos A)=\\cos A\\sin A+\\cos A-\\cos^2A$$ $$=\\cos A(1+\\sin A)-(1-\\sin^2A)$$ $$\\implies\\cos A(\\sin A+1-\\cos A)=(1+\\sin A)(\\cos A-1+\\sin A)$$ $$\\implies\\frac{\\sin A+1-\\cos A}{\\sin A-1+\\cos A}=?$$ We can start with $(1+\\sin A)(\\sin A-1+\\cos A)$ as well Find some similar results here - @user2130295, here is another method – lab bhattacharjee Jan 3 '14 at 19:13", "A) - (1+\\sin A)(1-\\sec A)}{(1-\\sec A)(1 + \\sec A)}$ . . . . . . . . . . . . . . $=\\;\\;\\frac{1 + \\sec A - \\sin A - \\sin A\\sec A - 1 + \\sec A - \\sin A + \\sin A\\sec A}{1-\\sec^2A}$ . . . . . . . . . . . . . . $=\\;\\;\\frac{2\\sec A - 2\\sin A}{-(\\sec^2\\!A - 1)} \\;\\;=\\;\\;\\frac{-2(\\sin A - \\sec A)}{-\\tan^2\\!A} \\;\\;=\\;\\;\\frac{2(\\sin A - \\sec A)}{\\tan^2\\!A}$ . . . . . . . . . . . . . . $=\\;\\;\\frac{2\\left(\\sin A - \\dfrac{1}{\\cos A}\\right)}{\\dfrac{\\sin^2\\!A}{\\cos^2\\!A}} \\;\\;=\\;\\;2\\cdot\\frac{\\cos^2\\!A}{\\sin^2\\!A}\\left(\\s in A - \\frac{1}{\\cos A}\\right)$ . . . . . . . . . . . . . . $=\\;\\;2\\cdot\\frac{\\cos A}{\\sin A}\\cdot\\underbrace{\\frac{\\cos A}{\\sin A}\\left(\\sin A - \\frac{1}{\\cos A}\\right)}$ . . . . . . . . . . . . . . $=\\quad 2\\cdot\\frac{\\cos A}{\\sin A}\\cdot\\left(\\cos A - \\frac{1}{\\sin A}\\right)$ . . . . . . . . . . . . . . $=\\qquad 2\\cdot\\cot A(\\cos A - \\csc A)$", "3 cos A = cos 3A 4. 3 sin A – 4 sin 3A = sin 3A 5. 2 tan A / (1 - tan2A) = tan 2A 6. (3 tan A - tan3A) / (1 - tan2A) = tan 3A ### Trigonometric Ratios of Multiples of an Angle 1. sin⁡2A = sin⁡(A+A) = sin⁡A.cos⁡A + cos⁡A.sin⁡A = 2sin⁡A.cos⁡A 2. sin⁡2A = 2sin⁡A.cos⁡A = $\\frac{2sin⁡A.cos⁡A}{1}$ = $\\frac{2sin⁡A.cos⁡A}{cos^2A + sin^2A}$ = $\\frac{\\frac{2sin⁡A.cos⁡A}{A}}{\\frac{A}{A} + \\frac{A}{A}}$ = $\\frac{2tanA}{1 + tan^2A}$ 3. sin⁡2A = $\\frac{2tanA}{1 + tan^2A}$ = $\\frac{\\frac{2}{cos cosA}}{1 + \\frac{1}{A}}$ = $\\frac{2}{cotcotA}$ × $\\frac{A}{1 + cot^2A}$ = $\\frac{2cotA}{1+cot^2A}$ 4. cos⁡2A = cos⁡(A+A) = cos⁡A.cos⁡A - sin⁡A.sin⁡A = cos2⁡A - sin2⁡A 5. cos⁡2A = cos2⁡A - sin2⁡A = 1-sin2⁡A - sin2⁡A = 1-2sin2⁡A 6. cos 2A = A - A = A - (1 - A) = 2A -1 7. cos 2A = A - A = $\\frac{A - A}{1}$ = $\\frac{A - A}{A + A}$ = $\\frac{\\frac{A}{A} - \\frac{A}{A}}{\\frac{A}{ω^2_qA} + \\frac{A}{ω^2_lA}}$ = $\\frac{1-tan^2A}{1+tan^2A}$ 8. cos⁡2A = $\\frac{1-tan^2A}{1+tan^2A}$ = $\\frac{1-\\frac{1}{A}}{1+\\frac{1}{A}}$ = $\\frac{cot^2A-1}{cot^2A+1}$ 9. tan⁡2A = tan⁡(A+B) = $\\frac{tanA + tanA}{1-tanA.tanA}$ = $\\frac{2tanA}{1-tan^2A}$ 10. tan⁡2A = $\\frac{2tanA}{1-tan^2A}$ = $\\frac{\\frac{2}{cotcotA}}{1-\\frac{1}{A}}$ = $\\frac{2}{cotcotA}$ × $\\frac{A}{cot^2A-1}$ 11. cot⁡2A=cot⁡(A+A)= $\\frac{cotA.cotA-1}{cotA + cotA}$ = $\\frac{cot^2⁡A-1}{2cot⁡A}$ 12. cot⁡2A = $\\frac{cot^2⁡A-1}{2cot⁡A}$ = $\\frac{\\frac{1}{A}-1}{\\frac{2}{\\text{tan tanA}}}$ = $\\frac{1-tan^2A}{A}$ × $\\frac{1-tan^2A}{2tanA}$ = $\\frac{1-tan^2⁡A}{2tanA}$ ### Problems Using Compound Angle Formulae These are some" ]

[ "## Calculus 8th Edition Area of a parallelogram with vertices $ABCD$ is defined as: $|{AB} ^{ \\to} \\times {AD}^{ \\to}|$ Here, ${AB} ^{ \\to} =\\lt 3-(-3),-1-0 \\gt= \\lt 6,-1 \\gt$ ${AD} ^{ \\to} =\\lt -1-(-3),3-0 \\gt= \\lt 2,3 \\gt$ $|{AB} ^{ \\to} \\times {AD}^{ \\to}|=\\lt 6,-1 \\gt \\times \\lt 2,3 \\gt$ $=6(3)-(-1)2$ $=18-(-2)$ $=20$ units", "# Parallelogram in Cartesian plane Geometry Level 2 Parallelogram $$ABCD$$ has vertices $$A=(0, 6),$$ $$B=(6, -2)$$ and $$C=(7, 5).$$ What are the coordinates of vertex $$D?$$ ×", "# Three vertices of a parallelogram ABCD are A (3, –1, 2), B (1, 2, –4) andC (–1, 1, 2). Find the coordinates of the fourth vertex. Question: Three vertices of a parallelogram ABCD are A (3, –1, 2), B (1, 2, –4) andC (–1, 1, 2). Find the coordinates of the fourth vertex. Solution: The three vertices of a parallelogram ABCD are given as A (3, –1, 2), B (1, 2, –4), and C (–1, 1, 2). Let the coordinates of the fourth vertex be D (xyz). We know that the diagonals of a parallelogram bisect each other. Therefore, in parallelogram ABCD, AC and BD bisect each other. $\\therefore$ Mid-point of $A C=$ Mid-point of $B D$ $\\Rightarrow\\left(\\frac{3-1}{2}, \\frac{-1+1}{2}, \\frac{2+2}{2}\\right)=\\left(\\frac{x+1}{2}, \\frac{y+2}{2}, \\frac{z-4}{2}\\right)$' $\\Rightarrow(1,0,2)=\\left(\\frac{x+1}{2}, \\frac{y+2}{2}, \\frac{z-4}{2}\\right)$ $\\Rightarrow \\frac{x+1}{2}=1, \\frac{y+2}{2}=0$, and $\\frac{z-4}{2}=2$ $\\Rightarrow x=1, y=-2$, and $z=8$ Thus, the coordinates of the fourth vertex are (1, –2, 8). Leave a comment", "Three vertices of a parallelogram ABCD are A(3,-1, 2),B(1, 2,-4) and C(-1, 1,2). Find the coordinates of the four Question Asked by a Student from EXXAMM.com Team Q 2666345275. Three vertices of a parallelogram ABCD are A(3,-1, 2),B(1, 2,-4) and C(-1, 1,2). Find the coordinates of the fourth vertex. HINT (Provided By a Student and Checked/Corrected by EXXAMM.com Team) Access free resources including • 100% free video lectures with detailed notes and examples • Previous Year Papers • Mock Tests • Practices question categorized in topics and 4 levels with detailed solutions • Syllabus & Pattern Analysis", "Find the area of the parallelogram with vertices at $(5, -3),$ $(-2, -1),$ $(-3, 4),$ and $(-10, 6).$ Area = .", "# Revised Coordinated Polygons I Quadrilateral $ABCD$ has vertices $A=(-4, -1), B=(1, -3), C=(3, 5),$ and $D=(-2, 3).$ Find its area. ×", "• # question_answer Three vertices of a parallelogram taken in order are $(-1,\\,\\,-6),\\,\\,\\,(2,\\,\\,-5)$ and $(7,\\,\\,2)$. The fourth vertex is: A) $(1,\\,\\,4)$ B) $(4,\\,\\,1)$ C) $(1,\\,\\,1)$ D) $(4,\\,\\,4)$ Key Idea: The diagonals of a parallelogram bisects each other. Let the coordinate of fourth vertex be$({{x}_{1}},\\,\\,{{y}_{1}})$. $\\therefore$Mid-point of$AC$is$(3,\\,\\,-2)$and mid-point of $DB$ is $\\left( \\frac{{{x}_{1}}+2}{2},\\,\\,\\frac{{{y}_{1}}-5}{2} \\right)$. Since, the diagonals, of a parallelogram bisects each other. $\\therefore$ $\\frac{{{x}_{1}}+2}{2}=3$and$\\frac{{{y}_{1}}-5}{2}=-2$ $\\Rightarrow$ ${{x}_{1}}=4$and${{y}_{1}}=1$ $\\therefore$Required coordinates is$(4,\\,\\,1)$.", "# Area of parallelogram Algebra Level 3 Find the area of the parallelogram with vertices $A=(-2, 1,3), B=(0, 4,3), C=(4, 2,3), D=(2, -1,3).$ ×", "# What is the most precise name for quadrilateral ABCD with vertices A(-5, 2), B(-3, 6), C(6,6) and D(4,2)? $\\text{the figure is parallelogram}$ $\\text{As The mutually edges are parallel each other and equal,}$ $\\text{the name of figure is parallelogram . }$", "Free Version Moderate # Coordinates of Parallelogram: Fourth Vertex ACTMAT-G9PYEZ Three of the four vertices of a parallelogram have coordinates $(-2,5)$, $(1,0)$, and $(4,8)$ in the x-y coordinate plane. Which of the following can be the coordinates of the fourth vertex? A $(3,4)$ B $(3,1)$ C $(2,2)$ D $(4,6)$ E $(7,3)$", "1. ## Parallelogram Question ABC has vertices of A(3,2,-5) B(4,-1,7) C(-8,3,-6) Determine the coordinates of D such that ABCD is a parallelogram 2. Hello, Originally Posted by peon123 ABC has vertices of A(3,2,-5) B(4,-1,7) C(-8,3,-6) Determine the coordinates of D such that ABCD is a parallelogram $\\displaystyle \\vec{AB}=\\vec{DC}$ So check if the scalar product between $\\displaystyle \\vec{AB}$ and $\\displaystyle \\vec{BC}$ is 0. Memo : $\\displaystyle Ax_a,y_a,z_a) \\text{ and } Bx_b,y_b,z_d) \\quad \\Rightarrow \\quad \\vec{AB}x_b-x_a,y_b-y_a,z_b-z_a)$", "The three vertices of a parallelogram are (3, 4) (3, 8) Question: The three vertices of a parallelogram are (3, 4) (3, 8) and (9, 8). Find the fourth vertex. Solution: We are given three vertices of a parallelogram A(3, 4), B(3, 8) and C(9, 8). We know that diagonals of a parallelogram bisect each other. Let the fourth vertex be D(x, y). Mid point of $B D=\\left(\\frac{x+3}{2}, \\frac{y+8}{2}\\right)$ Mid point of $A C=\\left(\\frac{9+3}{9}, \\frac{4+8}{9}\\right)=(6,6)$ Since the mid point of BD = mid point of AC $\\mathrm{So},\\left(\\frac{x+3}{2}, \\frac{y+8}{2}\\right)=(6,6)$ Thus, x = 9, y = 4. So, the fourth vertex is (9, 4).", "the figure given, $OABC$ is a parallelogram. The area of the parallelogram is $21$ sq units and the point $C$ lies on the line $x= 3$. Find the coordinates of $B$. $(3,10)$ $(10,3)$ $(10,10)$ $(8,3)$", "If the vertices of a parallelogram ABCD are A (1, 2, 3), B (-1, -2, -1) and C (2, 3, 2), then find the fourth ver ### Question Asked by a Student from EXXAMM.com Team Q 1749067813. If the vertices of a parallelogram ABCD are A (1, 2, 3), B (-1, -2, -1) and C (2, 3, 2), then find the fourth vertex D. #### HINT (Provided By a Student and Checked/Corrected by EXXAMM.com Team) #### Access free resources including • 100% free video lectures with detailed notes and examples • Previous Year Papers • Mock Tests • Practices question categorized in topics and 4 levels with detailed solutions • Syllabus & Pattern Analysis", "85 views The points $(2,1)$ and $( – 3, – 4)$ are opposite vertices of a parallelogram. If the other two vertices lie on the line $x + 9y + c = 0,$ then $\\text{c}$ is 1. $12$ 2. $14$ 3. $13$ 4. $15$ 1 138 views 2 88 views 1 vote", "# ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). Question: ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). The length of one of its diagonals is (a) 5 (b) 4 (c) 3 (d) 25 Solution: Here, AC and BD are two diagonals of the rectangle ABCD. So $B D=\\sqrt{(4-0)^{2}+(0-3)^{2}}$ $=\\sqrt{(4)^{2}+(-3)^{2}}$ $=\\sqrt{16+9}$ $=\\sqrt{25}$ $=5$ units Hence, the correct answer is option (a).", "-6 : : : D -7 : C : o - - - + o - - - + (-3,-4) We see that vertex $\\,C$ is at the lower-left. Going from $\\,A$ to $\\,D$, we move: 6 units down and 7 units left. Since $BC \\parallel AD$, going from $\\,B$ to $\\,C$, we do the same. Starting at $B(\\text{-}6,1)$, move 6 units down a 7 units left. Therefore, vertex $\\,C$ is $(\\text{-}13,\\text{-}5)$ 4. Originally Posted by euclid2 ABCD is a parallelogram where A is (4,2), B is (-6,1), and D is (-3,-4). Find the co-ordinates of C. B(-6 , 1) , A(4 , 2) C(a , b) , D(-3 , -4) 4 - (-6) = -3 - a 10 = -3 - a ===> a = -13 2 - 1 = -4 - b 1 = -4 -b ===> b = -5 therefore C(-13 , -5) 5. Re: co-ordinates of a parallelogram is it just me or is -13, -5 making the parallelogram's C point jut far out compared to the other points when graphing, that can't be right. 7, -3 sounds more accurate but I'm also having trouble figuring out how to arrive at that answer without just testing it on a graph. 6. Re: co-ordinates of a parallelogram Figured it out, You need to solve for", "Limited access Let $A=(0,0,0)$, let $B=(1,2,-1)$, and let $C=(3,2,-3)$. Let $P$ be a parallelogram that has $AB$ and $AC$ as sides. Determine the coordinates of the fourth vertex $D$ of $P$. Select an assignment template", "Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex. Asked by Pragya Singh | 11 months ago | 126 ##### Solution :- We are given the three vertices of a parallelogram ABCD are A(3,-1,2) , B(1,2,-4) and C(-1,1,2) . Let the coordinates of the fourth vertex of the parallelogram ABCD be D(x,y,z) . According to the property of parallelogram, the diagonals of the parallelogram bisect each other. In this parallelogram ABCD, AC and BD at point O. So, Mid-point of AC = Mid-point of BD $$( \\dfrac{3-1}{2}, \\dfrac{-1+1}{2}, \\dfrac{2+2}{2})$$ $$( \\dfrac{x+1}{2}, \\dfrac{y+1}{2}, \\dfrac{z-4}{2})$$ (1,0,2) = $$( \\dfrac{x+1}{2}, \\dfrac{y+1}{2}, \\dfrac{z-4}{2})$$ $$\\dfrac{x+1}{2}=1, \\dfrac{y+2}{2}=0, \\dfrac{z-4}{2}=2$$ We get, x = 1, y = 2 and z = 8 Therefore, the coordinates of the fourth vertex of the parallelogram ABCD are D(1,-2,8) . Answered by Abhisek | 11 months ago ### Related Questions #### A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC. #### The mid-points of the", "# Find the area of the parallelogram with vertices A(-3,0) , B(-1,6) , C(8,5) and D(6,-1) Find the area of the parallelogram with vertices A(-3,0) , B(-1,6) , C(8,5) and D(6,-1) • Questions are typically answered in as fast as 30 minutes ### Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Cristiano Sears ###### Not exactly what you’re looking for? content_user Vertices are $$A(-3,0),\\ B(-1,6),\\ C(8,5)$$ and $$D(6,-1)$$ So $$\\vec{AB}=<-1-(-3),6-0>=<2,6>$$ $$\\vec{AD}=<6-(-3),-1-0>=<9,-1>$$ Hence area of parallelogram $$=|\\vec{AB}\\times\\vec{AD}|$$ $$\\vec{AB}\\times\\vec{AD}=\\begin{bmatrix}i&j&k\\\\2&6&0\\\\9&-1&0\\end{bmatrix}$$ $$=i(0-0)+j(0-0)+k(-2-54)$$ $$=-56k$$ Hence Area $$=|\\vec{AB}\\times\\vec{AD}|$$ $$=\\sqrt{(-56)^2}=56$$" ]

[ "## anonymous one year ago In square $ABCD$, $E$ is the midpoint of $\\overline{BC}$, and $F$ is the midpoint of $\\overline{CD}$. Let $G$ be the intersection of $\\overline{AE}$ and $\\overline{BF}$. Prove that $DG = AB$. • This Question is Open 1. anonymous Asymptote code below [asy] unitsize(2.5 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (0,1); C = (1,1); D = (1,0); E = (B + C)/2; F = (C + D)/2; G = extension(A,E,B,F); draw(A--B--C--D--cycle); draw(A--E); draw(B--F); draw(D--G); label(\"$A$\", A, SW); label(\"$B$\", B, NW); label(\"$C$\", C, NE); label(\"$D$\", D, SE); label(\"$E$\", E, N); label(\"$F$\", F, dir(0)); label(\"$G$\", G, WSW); [/asy]", "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "Call $$D(a;b)$$ According to the assumption, ABCD is a trapezoid with AB // CD and CD = $$\\frac{3}{2}$$AB $$\\Rightarrow \\overrightarrow {CD} = \\frac{3}{2}\\overrightarrow {BA}$$ We have: $$\\overrightarrow {CD} = (a – 2;b + 5),\\overrightarrow {AB} = ( – 2; – 6)$$ $$\\Rightarrow \\overrightarrow {CD} = \\frac{3}{2}\\overrightarrow {BA}$$$$\\Leftrightarrow \\left\\{ \\begin{array}{l}a – 2 = \\frac{3} {2}.2\\\\b + 5 = \\frac{3}{2}.6\\end{array} \\right \\Leftrightarrow \\left\\{ \\begin{array}{l}a = 5\\\\b = 4\\end{array} \\right.$$. So EASY(5; 4)", "dots and labels */ dot((-1,4),dotstyle); label(\"A\", (-0.92,4.2), NE * labelscalefactor); dot((-4.08,3.78),dotstyle); label(\"B\", (-4.42,4), NE * labelscalefactor); dot((-3.1,-3.42),dotstyle); label(\"C\", (-3.4,-3.94), NE * labelscalefactor); dot((1.56,-0.22),dotstyle); label(\"D\", (1.8,-0.4), NE * labelscalefactor); dot((-1.566733533935139,1.9975415134291763),linewidth(4pt) + dotstyle); label(\"O\", (-1.34,1.98), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$ (Diagram not to scale) Since $AO \\cdot OC = BO \\cdot OD$, $ABCD$ is cyclic through power of a point. From the given information, we see that $\\triangle{AOB}\\sim \\triangle{DOC}$ and $\\triangle{BOC} \\sim \\triangle{AOD}$. Hence, we can find $CD=\\frac{9}{2}$ and $AD=2 \\cdot BC$. Letting $BC$ be $x$, we can use Ptolemy's to get $$6 \\cdot \\frac{9}{2} + 2x^2=10 \\cdot 11 \\implies x=\\sqrt{\\frac{83}{2}}$$ Since we are solving for $AD=2x=2\\cdot\\sqrt{\\frac{83}{2}}=\\sqrt{4\\cdot\\frac{83}{2}} = \\boxed{\\textbf{(E)}~\\sqrt{166}}$ - PhunsukhWangdu", "and $\\frac{DG}{GE}=\\frac{EH}{HF}=\\frac{FI}{ID}=1$ in the diagram below.$[asy] draw((0, 0)--(6, 0)--(4, 3)--cycle); draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle); draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle); label(\"A\", (0, 0), SW); label(\"B\", (6, 0), SE); label(\"C\", (4, 3), N); label(\"D\", (2, 0), S); label(\"E\", (16/3, 1), NE); label(\"F\", (8/3, 2), NW); label(\"G\", (11/3, 1/2), SE); label(\"H\", (4, 3/2), NE); label(\"I\", (7/3, 1), W); [/asy]$ ## Solution 1 is this solution by franzliszt: By the Wooga Looga Theorem, $\\frac{[DEF]}{[ABC]}=\\frac{2^2-2+1}{(1+2)^2}=\\frac 13$. Notice that $\\triangle GHI$ is the medial triangle of Wooga Looga Triangle of $\\triangle ABC$. So $\\frac{[GHI]}{[DEF]}=\\frac 14$ and $\\frac{[GHI]}{[ABD]}=\\frac{[DEF]}{[ABC]}\\cdot\\frac{[GHI]}{[DEF]}=\\frac 13 \\cdot \\frac 14 = \\frac {1}{12}$ by Chain Rule ideas. ## Solution 2 or this solution by franzliszt: Apply Barycentrics w.r.t. $\\triangle ABC$ so that $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $D=(\\tfrac 23,\\tfrac 13,0),E=(0,\\tfrac 23,\\tfrac 13),F=(\\tfrac 13,0,\\tfrac 23)$. And $G=(\\tfrac 13,\\tfrac 12,\\tfrac 16),H=(\\tfrac 16,\\tfrac 13,\\tfrac 12),I=(\\tfrac 12,\\tfrac 16,\\tfrac 13)$. In the barycentric coordinate system, the area formula is $[XYZ]=\\begin{vmatrix} x_{1} &y_{1} &z_{1} \\\\ x_{2} &y_{2} &z_{2} \\\\ x_{3}& y_{3} & z_{3} \\end{vmatrix}\\cdot [ABC]$ where $\\triangle XYZ$ is a random triangle and $\\triangle ABC$ is the reference triangle. Using this, we find that$$\\frac{[GHI]}{[ABC]}=\\begin{vmatrix} \\tfrac 13&\\tfrac 12&\\tfrac 16\\\\ \\tfrac 16&\\tfrac 13&\\tfrac 12\\\\ \\tfrac 12&\\tfrac 16&\\tfrac 13 \\end{vmatrix}=\\frac{1}{12}.$$ # Testimonials The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. ~ilp2020 The Wooga Looga", "# Difference between revisions of \"Midpoint\" ## Definition In Euclidean geometry, the midpoint of a line segment is the point on the segment equidistant from both endpoints. A midpoint bisects the line segment that the midpoint lies on. Because of this property, we say that for any line segment $\\overline{AB}$ with midpoint $M$, $AM=BM=\\frac{1}{2}AB$. Alternatively, any point $M$ on $\\overline{AB}$ such that $AM=BM$ is the midpoint of the segment. $[asy] draw((0,0)--(4,0)); dot((0,0)); label(\"A\",(0,0),N); dot((4,0)); label(\"B\",(4,0),N); dot((2,0)); label(\"M\",(2,0),N); label(\"Figure 1\",(2,0),4S); [/asy]$ ## Midpoints and Triangles $[asy] pair A,B,C,D,E,F,G; A=(0,0); B=(4,0); C=(1,3); D=(2,0); E=(2.5,1.5); F=(0.5,1.5); G=(5/3,1); draw(A--B--C--cycle); draw(D--E--F--cycle,green); dot(A--B--C--D--E--F--G); draw(A--E,red); draw(B--F,red); draw(C--D,red); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,N); label(\"D\",D,S); label(\"E\",E,E); label(\"F\",F,W); label(\"G\",G,NE); label(\"Figure 2\",D,4S); [/asy]$ ### Midsegments As shown in Figure 2, $\\Delta ABC$ is a triangle with $D$, $E$, $F$ midpoints on $\\overline{AB}$, $\\overline{BC}$, $\\overline{CA}$ respectively. Connect $\\overline{EF}$, $\\overline{FD}$, $\\overline{DE}$ (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that $\\Delta CFE \\sim \\Delta CAB$ and likewise for $\\Delta ADF$ and $\\Delta BED$. Because of this, we know that \\begin{align*} AB &= 2FE \\\\ BC &= 2DE \\\\ CA &= 2ED \\\\ \\end{align*} Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, $\\Delta ABC \\sim \\Delta EFD (SSS)$ with similar ratio 2:1. The area ratio is then", "3. The parallel sides have the same length (#3 is equivalent to #2) First check if $M_{AC} = M_{BD}$ So only b) and d) could be a parallelogram Check if $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ only d) is left. Since $\\left(\\begin{array}{c}-5\\\\1\\end{array}\\right) = k \\cdot \\left(\\begin{array}{c}5\\\\-1\\end{array}\\right)$ the sides $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ PS: Fröhliche Ostern! 3. ## Thank yah Originally Posted by earboth The properties of a parallelogram are: 1. The diagonals have a common midpoint. 2. 2 pairs of parallel sides 3. The parallel sides have the same length (#3 is equivalent to #2) First check if $M_{AC} = M_{BD}$ So only b) and d) could be a parallelogram Check if $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ only d) is left. Since $\\left(\\begin{array}{c}-5\\\\1\\end{array}\\right) = k \\cdot \\left(\\begin{array}{c}5\\\\-1\\end{array}\\right)$ the sides $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ PS: Fröhliche Ostern! wow thank you SO much. Makes sense. I kinda feel dumb now that I didn't figure that out myself!! Frohe Ostern an dir auchh! =]] 4. Hello, oXCryssieLeahXo! I don't suppose you tried to PLOT the points . . . Which of the set of points cannot be the vertices of parallelogram ABCD? . . $a)\\;A (\\text{-}4, \\text{-}2),\\;B (2, 0),\\;C (3, 3), D\\;(\\text{-}2, \\text{-}1)$ Code: | C . . . . . + . . o . | . . . . . + .", "$[asy] unitsize(1cm); defaultpen(0.8); pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label(\"A\",A,SW); label(\"B\",B,N); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,S); label(\"T\",T,NW); label(\"3\",A--T,N); label(\"4\",B--T,W); label(\"1\",D--T,N); label(\"1\",E--T,W); [/asy]$ The point $C$ is the intersection of the lines $BD$ and $AE$. The points on the first line have the form $(t,4-4t)$, the points on the second line have the form $(t,-1-t/3)$. Solving for $t$ we get $t=15/11$, hence $C=(15/11,-16/11)$. The ratio $CD/BD$ can now be computed simply by observing the $x$ coordinates of $B$, $C$, and $D$: $$\\frac{CD}{BD} = \\frac{15/11 - 1}{1 - 0} = \\boxed{\\textbf{(D)}\\frac 4{11}}$$", "a common midpoint. 2. 2 pairs of parallel sides 3. The parallel sides have the same length (#3 is equivalent to #2) First check if $M_{AC} = M_{BD}$ So only b) and d) could be a parallelogram Check if $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ only d) is left. Since $\\left(\\begin{array}{c}-5\\\\1\\end{array}\\right) = k \\cdot \\left(\\begin{array}{c}5\\\\-1\\end{array}\\right)$ the sides $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ PS: Fröhliche Ostern! • March 20th 2008, 02:15 AM oXCryssieLeahXo Thank yah Quote: Originally Posted by earboth The properties of a parallelogram are: 1. The diagonals have a common midpoint. 2. 2 pairs of parallel sides 3. The parallel sides have the same length (#3 is equivalent to #2) First check if $M_{AC} = M_{BD}$ So only b) and d) could be a parallelogram Check if $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ only d) is left. Since $\\left(\\begin{array}{c}-5\\\\1\\end{array}\\right) = k \\cdot \\left(\\begin{array}{c}5\\\\-1\\end{array}\\right)$ the sides $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ PS: Fröhliche Ostern! wow thank you SO much. Makes sense. I kinda feel dumb now that I didn't figure that out myself!! (Headbang) Frohe Ostern an dir auchh! =]] • March 20th 2008, 05:12 AM Soroban Hello, oXCryssieLeahXo! I don't suppose you tried to PLOT the points . . . Quote: Which of the set of points cannot be the vertices of parallelogram ABCD? . . $a)\\;A (\\text{-}4, \\text{-}2),\\;B (2, 0),\\;C (3, 3),", "a point $P$ where the perpendicular bissector of $CD$ passes through. Case 2: $AO \\ne OB$ WLOG, let $AO > OB$. Additionally, let $A = (0,2a)$ and $B = (2b,0)$, so $a > b$. $[asy] pair a=(0,20),o=(0,0),b=(10,0); draw(a--o--b--a); dot(a); label(\"A\",a,NW); dot(o); label(\"O\",o,SW); dot(b); label(\"B\",b,SE); [/asy]$ Let $D_1, C_1$ be points where $D_1 = O$, making $BD_1 = 2b$. Thus, $AC_1 = 2a-2b$, so $C_1 D_1$ is on $(0,2a-2b)$. Since the midpoint of $C_1 D_1$ is on $(0,a-b)$, the y-coordinate of $P$ must equal $a-b$. Let $D_2, C_2$ be points where $D_2 = B$, making $AC_2 = BD_2 = 0$. The midpoint of $C_2 D_2$ is $(b,a)$, and the slope of $C_2 D_2$ is $- \\tfrac{a}{b}$. Thus, the equation of the line perpendicular to $C_2 D_2$ is $y - a = \\tfrac{b}{a} (x - b)$. Since the y-coordinate of $P$ equals $a-b$, substituting and solving for $x$ results in $x = b-a$. The coordinates of $P$ are $(b-a,a-b)$; now we need to prove that this point is on any perpendicular bissector of $CD$. Let $D_3, C_3$ be any point of $D, C$ at a given time, and let the coordinates of $D_3$ be $(2b-2n,0)$. That means the coordinates of $C_3$ are $(0,2a-2n)$. The midpoint of $C_3 D_3$ is $(b-n,a-n)$, and the slope of $C_3 D_3$ is $\\tfrac{n-a}{b-n}$. Thus, the", "BE=\\frac{AE\\cdot EC}{EG}=\\frac{5\\cdot 15}{5\\sqrt{5}}=3\\sqrt{5}$ via Power of a Point. The altitude from $B$ to $AC$ is then equal to $GF\\cdot \\frac{BE}{GE}=10\\cdot \\frac{3\\sqrt 5}{5 \\sqrt 5}=6$. Finally, the total area of $ABCD$ is equal to $\\frac 12 \\cdot 20 \\left(30+6 \\right) =\\boxed{\\text{(D) } 360}.$ ~asops ## Solution 8 (Solving Equations) $[asy] size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label(\"A\", (0,2), NW); label(\"B\", (0,0), SW); label(\"C\", (4,0), SE); label(\"D\", (6,4), NE); label(\"E\", (1.714,1.143), N); label(\"F\", (1.714,0), SE); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((1.714,1.143)--(1.714,0), dashed); label(\"20\", (0,2)--(4,0), SW); label(\"30\", (4,0)--(6,4), SE); label(\"x\", (-0.3,2)--(-0.3,0), N); label(\"y\", (0,-0.3)--(4,-0.3), E); draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); [/asy]$ Let $AB = x$, $BC = y$ Looking at the diagram we have $x^2 + y^2 = 20^2$, $DE = \\sqrt{30^2+15^2} = 15\\sqrt{5}$, $[ACD] = \\frac{1}{2} \\cdot 20 \\cdot 30 = 300$ Because $\\triangle CEF \\sim \\triangle CAB$, $EF = AB \\cdot \\frac{CE}{CA} = x \\cdot \\frac{15}{20} = \\frac{3x}{4}$ $BF = BC - CF = BC - BC \\cdot \\frac{CE}{CA} = \\frac{1}{4} \\cdot BC = \\frac{y}{4}$ $BE = \\sqrt{ \\left( \\frac{3x}{4} \\right) ^2 + \\left( \\frac{y}{4} \\right) ^2 } = \\frac{ \\sqrt{9x^2 + y^2} }{4}$ , substituting $x^2 + y^2 = 400$, we get $BE = \\frac{ \\sqrt{8x^2 + 400} }{4} = \\frac{ \\sqrt{2x^2 + 100} }{2}$ $[ABC] = \\frac{1}{2} \\cdot x \\cdot y$ Because $\\triangle ABC$ and $\\triangle ACD$ share the same base, $\\frac{[ABC]}{[ACD]}", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "Midpoint of $RT = \\left ( \\frac{-4+3}{2}, \\ \\frac{3-4}{2} \\right )=(-0.5,-0.5)$ Midpoint of $SU = \\left ( \\frac{4-5}{2}, \\ \\frac{5-5}{2} \\right ) =(-0.5,0)$ Because the midpoint is not the same, $RSTU$ is not a parallelogram. 3. Although it has one pair of parallel sides, this quadrilateral is not a parallelogram because its opposite sides are not necessarily congruent. ### Practice For questions 1-11, determine if the quadrilaterals are parallelograms. If they are, write a reason. For questions 12-14, determine the value of $x$ and $y$ that would make the quadrilateral a parallelogram. For questions 15-17, determine if $ABCD$ is a parallelogram. 1. $A(8, -1), \\ B(6, 5), \\ C(-7, 2), \\ D(-5, -4)$ 2. $A(-5, 8), \\ B(-2, 9), \\ C(3, 4), \\ D(0, 3)$ 3. $A(-2, 6), \\ B(4, -4), \\ C(13, -7), \\ D(4, -10)$ Write a two-column proof. 1. Parallelogram Diagonals Theorem Converse Given : $\\overline{AE} \\cong \\overline{EC}, \\ \\overline{DE} \\cong \\overline{EB}$ Prove : $ABCD$ is a parallelogram 2. ${\\;}$ Given : $\\angle ADB \\cong CBD, \\ \\overline{AD} \\cong \\overline{BC}$ Prove : $ABCD$ is a parallelogram Suppose that $A(-2, 3), \\ B(3, 3)$ and $C(1, -3)$ are three of four vertices of a parallelogram. 1. Depending on where you choose to put point $D$ , the name of the parallelogram you draw will change. Sketch", "# Parallelogram $=A F E D$.6. If $E, E, G$ and $H$ are respectively the midpoint of the sides of 33 views in JEE recategorized Parallelogram $=A F E D$. 6. If $E, E, G$ and $H$ are respectively the midpoints of the sides of a Parallelogram $A B C D$, show that [wr (E FCAH) $=\\frac{1}{2} \\operatorname{ar}(A B C D)$ $\\operatorname{arc}\\left(E F(G H)=\\frac{1}{\\theta} \\operatorname{ar}(A B C D)\\right.$ +1 vote by (42.4k points) Construction: Join E & G. $\\because$ AB = CD (opposite sides of parallelogram) $\\frac{AB}2 = \\frac{CD}2$ ⇒ BE = GC ($\\because$ E is mid-point of AB & G is mid-point of CD) Also BE || GC ($\\because$ AB || CD) $\\therefore$ EBCG is a parallelogram. Now, since, parallelogram EBCG & triangle △EFG are on the same base EG & between same parallels EG & BC. $\\therefore$ ar(△EFG) = $\\frac12$ar(EBCG)-----(i) Similarly, ar(△EHG) = $\\frac12$ar(EGDA)----(ii) Adding equations (i) & (ii), we get ar(△EFG) + ar(△EHG) = $\\frac12$ar(EBCG) + $\\frac12$ar(EGDA) ⇒ ar(EFGH) = $\\frac12$ar(ABCD)", "\\frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 size(8cm); pair A, B, C, D, E, EE; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); EE = (27/8,0); draw(A--B--C--D--A--E); draw(E--EE,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,EE,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",EE,S); label(\"$4$\",(A+D)/2,S); label(\"$x$\",(A+EE)/2,S; label(\"$4-x$\",(D+EE)/2,S); label(\"$\\frac{15}{8}$\",(E+EE)/2,W); (Error compiling LaTeX. label(\"$x$\",(A+EE)/2,S; ^ de021056a2cc2ef4e3e5c00687df10ec92699974.asy: 26.23: syntax error error: could not load module 'de021056a2cc2ef4e3e5c00687df10ec92699974.asy') Following the steps above, you can find that the height of triangle $ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the", "# BS in relation to BF and CS in relation to CE #### mathmari ##### Well-known member MHB Site Helper Hey!! We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? #### Klaas van Aarsen ##### MHB Seeker Staff member We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? Hey mathmari !! Let's first make a drawing: \\begin{tikzpicture}[scale=2] %preamble \\usetikzlibrary {calc} \\coordinate[label=left:A] (A) at (0,0); \\coordinate[label=right:B] (B) at (0:5); \\coordinate[label=C] (C) at (70:4); \\coordinate[label=below:E] (E) at (0:1); \\coordinate[label=left:F] (F) at (70:3); \\coordinate[label=right:S] (S) at (intersection cs:first line={(B)--(F)}, second line={(C)--(E)}); \\draw (A) -- (B) -- (C) -- cycle; \\draw (C) -- (E); \\draw (B) -- (F); \\draw[-latex, ultra thick, blue] (A) -- node[ above ] {$\\mathbf{\\vec c}$} (B); \\draw[-latex, ultra thick, blue] (A) -- node[ right] {$\\mathbf{\\vec b}$} (C); \\path (A) --", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is" ]

[ "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "3. The parallel sides have the same length (#3 is equivalent to #2) First check if $M_{AC} = M_{BD}$ So only b) and d) could be a parallelogram Check if $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ only d) is left. Since $\\left(\\begin{array}{c}-5\\\\1\\end{array}\\right) = k \\cdot \\left(\\begin{array}{c}5\\\\-1\\end{array}\\right)$ the sides $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ PS: Fröhliche Ostern! 3. ## Thank yah Originally Posted by earboth The properties of a parallelogram are: 1. The diagonals have a common midpoint. 2. 2 pairs of parallel sides 3. The parallel sides have the same length (#3 is equivalent to #2) First check if $M_{AC} = M_{BD}$ So only b) and d) could be a parallelogram Check if $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ only d) is left. Since $\\left(\\begin{array}{c}-5\\\\1\\end{array}\\right) = k \\cdot \\left(\\begin{array}{c}5\\\\-1\\end{array}\\right)$ the sides $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ PS: Fröhliche Ostern! wow thank you SO much. Makes sense. I kinda feel dumb now that I didn't figure that out myself!! Frohe Ostern an dir auchh! =]] 4. Hello, oXCryssieLeahXo! I don't suppose you tried to PLOT the points . . . Which of the set of points cannot be the vertices of parallelogram ABCD? . . $a)\\;A (\\text{-}4, \\text{-}2),\\;B (2, 0),\\;C (3, 3), D\\;(\\text{-}2, \\text{-}1)$ Code: | C . . . . . + . . o . | . . . . . + .", "a common midpoint. 2. 2 pairs of parallel sides 3. The parallel sides have the same length (#3 is equivalent to #2) First check if $M_{AC} = M_{BD}$ So only b) and d) could be a parallelogram Check if $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ only d) is left. Since $\\left(\\begin{array}{c}-5\\\\1\\end{array}\\right) = k \\cdot \\left(\\begin{array}{c}5\\\\-1\\end{array}\\right)$ the sides $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ PS: Fröhliche Ostern! • March 20th 2008, 02:15 AM oXCryssieLeahXo Thank yah Quote: Originally Posted by earboth The properties of a parallelogram are: 1. The diagonals have a common midpoint. 2. 2 pairs of parallel sides 3. The parallel sides have the same length (#3 is equivalent to #2) First check if $M_{AC} = M_{BD}$ So only b) and d) could be a parallelogram Check if $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ only d) is left. Since $\\left(\\begin{array}{c}-5\\\\1\\end{array}\\right) = k \\cdot \\left(\\begin{array}{c}5\\\\-1\\end{array}\\right)$ the sides $(\\overline{AB})\\ \\parallel \\ (\\overline{CD})$ PS: Fröhliche Ostern! wow thank you SO much. Makes sense. I kinda feel dumb now that I didn't figure that out myself!! (Headbang) Frohe Ostern an dir auchh! =]] • March 20th 2008, 05:12 AM Soroban Hello, oXCryssieLeahXo! I don't suppose you tried to PLOT the points . . . Quote: Which of the set of points cannot be the vertices of parallelogram ABCD? . . $a)\\;A (\\text{-}4, \\text{-}2),\\;B (2, 0),\\;C (3, 3),", "AB \\\\ \\\\[-4mm] FD = \\frac{1}{2}AC, & FD \\parallel AC \\end{array}$ Therefore: . $\\Delta D{E}F \\sim \\Delta ABC$ . . and: . $(\\text{area }\\Delta D{E}F) \\:=\\:\\tfrac{1}{4}(\\text{area }\\Delta ABC)$ 5. The coordinates of three vertices of a parallelogram are A (-1,0), B (2,-2) C(2,2). Find all possibilities for the coordinates of the fourth vertex. Be sure to include an explanation for each possibility you provide. There are three possible locations for vertex $D.$ Code: | C | o | * * | * * *| * D - - o - | - - - - - - - - - o - - A *| * (5,0) | * * | * * | o | B | Code: | C | o | * * | * * A *| * - - o - + - - - * - - * | * * | * * | * * | o * | * B * | * * *| D o | (-1,-4)| Code: D (-1,4)o | * *| * | * * | * C * | o * | * * | * * | * - - o - + - - - * - - - A *| * | * * | * * | o | B", "## anonymous one year ago In square $ABCD$, $E$ is the midpoint of $\\overline{BC}$, and $F$ is the midpoint of $\\overline{CD}$. Let $G$ be the intersection of $\\overline{AE}$ and $\\overline{BF}$. Prove that $DG = AB$. • This Question is Open 1. anonymous Asymptote code below [asy] unitsize(2.5 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (0,1); C = (1,1); D = (1,0); E = (B + C)/2; F = (C + D)/2; G = extension(A,E,B,F); draw(A--B--C--D--cycle); draw(A--E); draw(B--F); draw(D--G); label(\"$A$\", A, SW); label(\"$B$\", B, NW); label(\"$C$\", C, NE); label(\"$D$\", D, SE); label(\"$E$\", E, N); label(\"$F$\", F, dir(0)); label(\"$G$\", G, WSW); [/asy]", "dots and labels */ dot((-1,4),dotstyle); label(\"A\", (-0.92,4.2), NE * labelscalefactor); dot((-4.08,3.78),dotstyle); label(\"B\", (-4.42,4), NE * labelscalefactor); dot((-3.1,-3.42),dotstyle); label(\"C\", (-3.4,-3.94), NE * labelscalefactor); dot((1.56,-0.22),dotstyle); label(\"D\", (1.8,-0.4), NE * labelscalefactor); dot((-1.566733533935139,1.9975415134291763),linewidth(4pt) + dotstyle); label(\"O\", (-1.34,1.98), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$ (Diagram not to scale) Since $AO \\cdot OC = BO \\cdot OD$, $ABCD$ is cyclic through power of a point. From the given information, we see that $\\triangle{AOB}\\sim \\triangle{DOC}$ and $\\triangle{BOC} \\sim \\triangle{AOD}$. Hence, we can find $CD=\\frac{9}{2}$ and $AD=2 \\cdot BC$. Letting $BC$ be $x$, we can use Ptolemy's to get $$6 \\cdot \\frac{9}{2} + 2x^2=10 \\cdot 11 \\implies x=\\sqrt{\\frac{83}{2}}$$ Since we are solving for $AD=2x=2\\cdot\\sqrt{\\frac{83}{2}}=\\sqrt{4\\cdot\\frac{83}{2}} = \\boxed{\\textbf{(E)}~\\sqrt{166}}$ - PhunsukhWangdu", "# 2000 AIME II Problems/Problem 11 ## Problem The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107)$. The trapezoid has no horizontal or vertical sides, and $\\overline{AB}$ and $\\overline{CD}$ are the only parallel sides. The sum of the absolute values of all possible slopes for $\\overline{AB}$ is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Solution For simplicity, we translate the points so that $A$ is on the origin and $D = (1,7)$. Suppose $B$ has integer coordinates; then $\\overrightarrow{AB}$ is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from $A$ to $\\overline{CD}$, and let $D' = (a,b)$ be the reflection of $D$ across that perpendicular. Then $ABCD'$ is a parallelogram, and $\\overrightarrow{AB} = \\overrightarrow{D'C}$. Thus, for $C$ to have integer coordinates, it suffices to let $D'$ have integer coordinates.[1] $[asy] pathpen = linewidth(0.7); pair A=(0,0), D=(1,7), Da = MP(\"D'\",D((-7,1)),N), B=(-8,-6), C=B+Da, F=foot(A,C,D); D(MP(\"A\",A)--MP(\"B\",B)--MP(\"C\",C,N)--MP(\"D\",D,N)--cycle); D(F--A--Da,linetype(\"4 4\")); [/asy]$ Let the slope of the perpendicular be $m$. Then the midpoint of $\\overline{DD'}$ lies on the line $y=mx$, so $\\frac{b+7}{2} = m \\cdot \\frac{a+1}{2}$. Also, $AD = AD'$ implies that $a^2 + b^2 = 1^2 + 7^2 = 50$. Combining these two equations yields $$a^2 + \\left(7 - (a+1)m\\right)^2 =", "# Difference between revisions of \"Midpoint\" ## Definition In Euclidean geometry, the midpoint of a line segment is the point on the segment equidistant from both endpoints. A midpoint bisects the line segment that the midpoint lies on. Because of this property, we say that for any line segment $\\overline{AB}$ with midpoint $M$, $AM=BM=\\frac{1}{2}AB$. Alternatively, any point $M$ on $\\overline{AB}$ such that $AM=BM$ is the midpoint of the segment. $[asy] draw((0,0)--(4,0)); dot((0,0)); label(\"A\",(0,0),N); dot((4,0)); label(\"B\",(4,0),N); dot((2,0)); label(\"M\",(2,0),N); label(\"Figure 1\",(2,0),4S); [/asy]$ ## Midpoints and Triangles $[asy] pair A,B,C,D,E,F,G; A=(0,0); B=(4,0); C=(1,3); D=(2,0); E=(2.5,1.5); F=(0.5,1.5); G=(5/3,1); draw(A--B--C--cycle); draw(D--E--F--cycle,green); dot(A--B--C--D--E--F--G); draw(A--E,red); draw(B--F,red); draw(C--D,red); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,N); label(\"D\",D,S); label(\"E\",E,E); label(\"F\",F,W); label(\"G\",G,NE); label(\"Figure 2\",D,4S); [/asy]$ ### Midsegments As shown in Figure 2, $\\Delta ABC$ is a triangle with $D$, $E$, $F$ midpoints on $\\overline{AB}$, $\\overline{BC}$, $\\overline{CA}$ respectively. Connect $\\overline{EF}$, $\\overline{FD}$, $\\overline{DE}$ (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that $\\Delta CFE \\sim \\Delta CAB$ and likewise for $\\Delta ADF$ and $\\Delta BED$. Because of this, we know that \\begin{align*} AB &= 2FE \\\\ BC &= 2DE \\\\ CA &= 2ED \\\\ \\end{align*} Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, $\\Delta ABC \\sim \\Delta EFD (SSS)$ with similar ratio 2:1. The area ratio is then", "volume is $2\\cdot \\left( \\frac{6\\sqrt{2}\\cdot 3\\sqrt{2} \\cdot 6}{3} \\right)=144$. Thus, our answer is $432-144=\\boxed{288}$. ### Solution 2 $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed); label(\"A\",A,(-1,-1,0)); label(\"B\",B,( 2,-1,0)); label(\"C\",C,( 1, 1,0)); label(\"D\",D,(-1, 1,0)); label(\"E\",E,(0,0,1)); label(\"F\",F,(0,0,1)); label(\"G\",G,(0,0,-1)); label(\"H\",H,(0,0,-1)); [/asy]$ Extend $EA$ and $FB$ to meet at $G$, and $ED$ and $FC$ to meet at $H$. Now, we have a regular tetrahedron $EFGH$, which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s = 12\\sqrt{2}$. Using the formula for the volume of a regular tetrahedron, which is $V = \\frac{\\sqrt{2}S^3}{12}$, where S is the side length of the tetrahedron, the volume of our original solid is: $V = \\frac{1}{2} \\cdot \\frac{\\sqrt{2} \\cdot (12\\sqrt{2})^3}{12} = \\boxed{288}$. ### Solution 3 We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$; thus, we will integrate with respect to height from $0$ to $6$, noting that each cross section of height $dh$ is a rectangle. The volume is then $\\int_0^h(wl) \\ \\text{d}h$, where $w$ is the width of the rectangle and $l$ is the length. We can express $w$", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "BE=\\frac{AE\\cdot EC}{EG}=\\frac{5\\cdot 15}{5\\sqrt{5}}=3\\sqrt{5}$ via Power of a Point. The altitude from $B$ to $AC$ is then equal to $GF\\cdot \\frac{BE}{GE}=10\\cdot \\frac{3\\sqrt 5}{5 \\sqrt 5}=6$. Finally, the total area of $ABCD$ is equal to $\\frac 12 \\cdot 20 \\left(30+6 \\right) =\\boxed{\\text{(D) } 360}.$ ~asops ## Solution 8 (Solving Equations) $[asy] size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label(\"A\", (0,2), NW); label(\"B\", (0,0), SW); label(\"C\", (4,0), SE); label(\"D\", (6,4), NE); label(\"E\", (1.714,1.143), N); label(\"F\", (1.714,0), SE); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((1.714,1.143)--(1.714,0), dashed); label(\"20\", (0,2)--(4,0), SW); label(\"30\", (4,0)--(6,4), SE); label(\"x\", (-0.3,2)--(-0.3,0), N); label(\"y\", (0,-0.3)--(4,-0.3), E); draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); [/asy]$ Let $AB = x$, $BC = y$ Looking at the diagram we have $x^2 + y^2 = 20^2$, $DE = \\sqrt{30^2+15^2} = 15\\sqrt{5}$, $[ACD] = \\frac{1}{2} \\cdot 20 \\cdot 30 = 300$ Because $\\triangle CEF \\sim \\triangle CAB$, $EF = AB \\cdot \\frac{CE}{CA} = x \\cdot \\frac{15}{20} = \\frac{3x}{4}$ $BF = BC - CF = BC - BC \\cdot \\frac{CE}{CA} = \\frac{1}{4} \\cdot BC = \\frac{y}{4}$ $BE = \\sqrt{ \\left( \\frac{3x}{4} \\right) ^2 + \\left( \\frac{y}{4} \\right) ^2 } = \\frac{ \\sqrt{9x^2 + y^2} }{4}$ , substituting $x^2 + y^2 = 400$, we get $BE = \\frac{ \\sqrt{8x^2 + 400} }{4} = \\frac{ \\sqrt{2x^2 + 100} }{2}$ $[ABC] = \\frac{1}{2} \\cdot x \\cdot y$ Because $\\triangle ABC$ and $\\triangle ACD$ share the same base, $\\frac{[ABC]}{[ACD]}", "&= \\frac{1}{2}\\end{align*} Hence $CE = FC + x = \\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 2 Call the point of tangency point $F$ and the midpoint of $AB$ as $G$. $CF=2$ by Tangent Theorem. Notice that $\\angle EGF=\\frac{180-2\\cdot\\angle CGF}{2}=90-\\angle CGF$. Thus, $\\angle EGF=\\angle FCG$ and $tanEGF=tanFCG=\\frac{1}{2}$. Solving $EF=\\frac{1}{2}$. Adding, the answer is $\\frac{5}{2}$. ## Solution 3 Clearly, $EA = EF = BG$. Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = 5/2$. ## Solution 4 $[asy] size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label(\"A\",A,(-1,-1)); label(\"B\",B,( 1,-1)); label(\"C\",C,( 1, 1)); label(\"D\",D,(-1, 1)); label(\"E\",E,(-1, 0)); label(\"F\",F,( 0, 1)); label(\"x\",(A+E)/2,(-1, 0)); label(\"x\",(E+F)/2,( 0, 1)); label(\"2\",(F+C)/2,( 0, 1)); label(\"2\",(D+C)/2,( 0, 1)); label(\"2\",(B+C)/2,( 1, 0)); label(\"2-x\",(D+E)/2,(-1, 0)); label(\"G\",G,(0,-1)); dot(G); draw(G--C); label(\"\\sqrt{5}\",(G+C)/2,(-1,0)); [/asy]$ Let us call the midpoint of side $AB$, point $G$. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$. We get $GC=\\sqrt{5}$. We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$, we get similar triangles $EFG$ and $GFC$. Solving the ratios, we get $x=\\frac{1}{2}$, so the answer is $\\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 5 Using the diagram as drawn in Solution 5, let the total", "$BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 pair A,B,C,D,E; B=(0,0); C=(14;0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"$A$\",A,N); label(\"$B$\",B,S); label(\"$C$\",C,S); label(\"$E$\",E,S); label(\"$D$\",D,S); label(\"$13$\",A--B,NW); label(\"$15$\",A--C,NE); label(\"$14$\",B--C,S); (Error compiling LaTeX. 32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy: 7.6: syntax error error: could not load module '32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy') Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, thus the area of $\\Delta ADC=\\Delta ABD$, which is half the area of $\\Delta ABC$, so $\\Delta ADC=\\Delta ABD=42$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "a point $P$ where the perpendicular bissector of $CD$ passes through. Case 2: $AO \\ne OB$ WLOG, let $AO > OB$. Additionally, let $A = (0,2a)$ and $B = (2b,0)$, so $a > b$. $[asy] pair a=(0,20),o=(0,0),b=(10,0); draw(a--o--b--a); dot(a); label(\"A\",a,NW); dot(o); label(\"O\",o,SW); dot(b); label(\"B\",b,SE); [/asy]$ Let $D_1, C_1$ be points where $D_1 = O$, making $BD_1 = 2b$. Thus, $AC_1 = 2a-2b$, so $C_1 D_1$ is on $(0,2a-2b)$. Since the midpoint of $C_1 D_1$ is on $(0,a-b)$, the y-coordinate of $P$ must equal $a-b$. Let $D_2, C_2$ be points where $D_2 = B$, making $AC_2 = BD_2 = 0$. The midpoint of $C_2 D_2$ is $(b,a)$, and the slope of $C_2 D_2$ is $- \\tfrac{a}{b}$. Thus, the equation of the line perpendicular to $C_2 D_2$ is $y - a = \\tfrac{b}{a} (x - b)$. Since the y-coordinate of $P$ equals $a-b$, substituting and solving for $x$ results in $x = b-a$. The coordinates of $P$ are $(b-a,a-b)$; now we need to prove that this point is on any perpendicular bissector of $CD$. Let $D_3, C_3$ be any point of $D, C$ at a given time, and let the coordinates of $D_3$ be $(2b-2n,0)$. That means the coordinates of $C_3$ are $(0,2a-2n)$. The midpoint of $C_3 D_3$ is $(b-n,a-n)$, and the slope of $C_3 D_3$ is $\\tfrac{n-a}{b-n}$. Thus, the", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex. Asked by Pragya Singh | 11 months ago | 126 ##### Solution :- We are given the three vertices of a parallelogram ABCD are A(3,-1,2) , B(1,2,-4) and C(-1,1,2) . Let the coordinates of the fourth vertex of the parallelogram ABCD be D(x,y,z) . According to the property of parallelogram, the diagonals of the parallelogram bisect each other. In this parallelogram ABCD, AC and BD at point O. So, Mid-point of AC = Mid-point of BD $$( \\dfrac{3-1}{2}, \\dfrac{-1+1}{2}, \\dfrac{2+2}{2})$$ $$( \\dfrac{x+1}{2}, \\dfrac{y+1}{2}, \\dfrac{z-4}{2})$$ (1,0,2) = $$( \\dfrac{x+1}{2}, \\dfrac{y+1}{2}, \\dfrac{z-4}{2})$$ $$\\dfrac{x+1}{2}=1, \\dfrac{y+2}{2}=0, \\dfrac{z-4}{2}=2$$ We get, x = 1, y = 2 and z = 8 Therefore, the coordinates of the fourth vertex of the parallelogram ABCD are D(1,-2,8) . Answered by Abhisek | 11 months ago ### Related Questions #### A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC. #### The mid-points of the", "# Parallelogram $=A F E D$.6. If $E, E, G$ and $H$ are respectively the midpoint of the sides of 33 views in JEE recategorized Parallelogram $=A F E D$. 6. If $E, E, G$ and $H$ are respectively the midpoints of the sides of a Parallelogram $A B C D$, show that [wr (E FCAH) $=\\frac{1}{2} \\operatorname{ar}(A B C D)$ $\\operatorname{arc}\\left(E F(G H)=\\frac{1}{\\theta} \\operatorname{ar}(A B C D)\\right.$ +1 vote by (42.4k points) Construction: Join E & G. $\\because$ AB = CD (opposite sides of parallelogram) $\\frac{AB}2 = \\frac{CD}2$ ⇒ BE = GC ($\\because$ E is mid-point of AB & G is mid-point of CD) Also BE || GC ($\\because$ AB || CD) $\\therefore$ EBCG is a parallelogram. Now, since, parallelogram EBCG & triangle △EFG are on the same base EG & between same parallels EG & BC. $\\therefore$ ar(△EFG) = $\\frac12$ar(EBCG)-----(i) Similarly, ar(△EHG) = $\\frac12$ar(EGDA)----(ii) Adding equations (i) & (ii), we get ar(△EFG) + ar(△EHG) = $\\frac12$ar(EBCG) + $\\frac12$ar(EGDA) ⇒ ar(EFGH) = $\\frac12$ar(ABCD)", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9)," ]

[ "\\cot(90-x)$ $\\cot x = \\tan (90-x)$ The Next post in this series is here Note by Yan Yau Cheng 3 years, 8 months ago", "## anonymous one year ago If tan x = .88, what is cot x? 1.14 1.45 1.33 1.22 $$\\huge cot \\ x = \\frac{1}{tan \\ x}$$", "## anonymous one year ago If tan x = .88, what is cot x? 1.14 1.45 1.33 1.22 1. IrishBoy123 $$\\huge cot \\ x = \\frac{1}{tan \\ x}$$ Find more explanations on OpenStudy", "34^o + cos^2 \\ 34^o + 2 tan \\ 18^o. cot \\ 18^o - cot^2 \\ 30^o$ $= 3 - (\\sqrt{3})^2 = 0$ $\\\\$", "one: In a $\\Delta ABC$, $$\\tan A+\\tan B+\\tan C=\\tan A\\tan B\\tan C.$$ - Also $\\cot(A/2)+\\cot(B/2)+ \\cot(C/2)=\\cot(A/2)\\cot(B/2)\\cot(C/2)$. – N. S. Apr 11 '13 at 22:39", "tan x/cot x = tan x × tan x = tan2 x = (sin x/cos x)2 ⇒ (4/5)2/(3/5)2 = 16/9", "# What is arc cot(-2)? Calculator --> tan x = -1/2 --> arc $x = - {26}^{\\circ} 56$ and $x = 180 - 26.56 = {153}^{\\circ} 43$", "# Prove that cot A + cot (60 + A) – cot (60 – A) = 3 cot 3A. ## Solution : We have, L.H.S = cot A + cot (60 + A) – cot (60 – A) $$\\implies$$ L.H.S = $$1\\over tan A$$ + $$1\\over tan (60 + A)$$ – $$1\\over tan (60 – A)$$ $$\\implies$$ L.H.S = $$1\\over tan A$$+ $$1 – \\sqrt{3} tan A\\over \\sqrt{3} + tan A$$ – $$1 + \\sqrt{3} tan A\\over \\sqrt{3} – tan A$$ [ By using this formula, tan (A + B) = $$tan A + tan B\\over 1 – tan A tan B$$ above ] $$\\implies$$ L.H.S = $$1\\over tan A$$ – $$8 tan A\\over 3 – tan^2 A$$ $$\\implies$$ L.H.S = $$3 – 9 tan^2 A\\over 3 tan A – tan^3 A$$ L.H.S = 3($$1 – 3 tan^2 A\\over 3 tan A – tan^3 A$$) Since $$3 tan A – tan^3 A\\over 1 – 3 tan^2 A$$ = tan 3A $$\\implies$$ L.H.S = $$3\\over tan 3A$$ = 3 cot 3A = R.H.S", "x \\tan x+\\cot x$$ 13. $$\\dfrac{1}{1−\\sin x}−\\dfrac{1}{1+\\sin x}=2\\tan x \\sec x$$ 14. $$\\csc^4 x−\\cot^4 x=\\csc ^2 x+\\cot^2x$$ 15. $$(\\sin x−\\tan x)(\\cos x−\\cot x)=(\\sin x−1)(\\cos x−1)$$", "X$ From (3) $D\\tan 30^0= Y$ sub X,Y values in (1) $D\\tan 10^0+D\\tan 30^0+6 =120$ Edited: Fix the error thank you so much!", "Question # Find the value of A if $\\tan 2A = \\cot (A - {24^0})$. Hint: Use the information, $\\tan 2A = \\cot (A - {24^0})$ and solve carefully. Consider the given equation, $\\tan 2A = \\cot (A - {24^0})$. We know that, $\\tan \\theta = \\cot ({90^ \\circ } - \\theta )$. $\\tan 2A = \\cot (A - {24^0}) \\\\ \\Rightarrow \\cot ({90^ \\circ } - 2A) = \\cot (A - {24^0}) \\\\ \\Rightarrow {90^ \\circ } - 2A = A - {24^0} \\\\ \\Rightarrow 3A = {114^ \\circ } \\\\ \\Rightarrow A = {38^ \\circ } \\\\$", "# For all real values of x, Question: For all real values of $x, \\cot x-2 \\cot 2 x$ is equal to (a) $\\tan 2 x$ (b) $\\tan x$ (c) $-\\cot 3 x$ (d) none of these Solution: (b) $\\tan x$ We have, $\\cot x-2 \\cot 2 x=\\cot x-2 \\frac{\\cot ^{2} x-1}{2 \\cot x}$ $=\\frac{\\cot ^{2} x-\\cot ^{2} x+1}{\\cot x}$ $=\\frac{1}{\\cot x}$ $=\\tan x$", "## anonymous 5 years ago Find the exact value of the following: a) sin^2 (75^0) / sin45^0 cos^2 (30^0) b) tan(135^0) + sin30^0 x cot120^0 $a=[\\sin(30+45)]/[\\sin45\\times \\cos^230]$ $=[\\sin30 \\cos45 +\\cos30 \\sin45]^2/[\\sin45\\times \\cos^230]$ then just plug in the numbers b tan(135)=tan(90+45)$=\\sin(90+45)/\\cos(90+45)$ cot(120)=cot(90+30) try to use this hint to find the answer", "# What is (tan x + tan y)(1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y) equal to? Free Practice With Testbook Mock Tests ## Options: 1. 0 2. 1 3. 2 4. 4 ### Correct Answer: Option 1 (Solution Below) This question was previously asked in CDS Maths Previous Paper 9 (Held On: 2 Feb 2020) ## Solution: Given: (tan x + tan y)(1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y) Formula Used: tan 45° = cot 45° = 1 Calculation: (tan x + tan y)(1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y) We can easily solve it by value putting: Put x = y = 45° ⇒ (1 + 1)(1 – 1 × 1) + (1 + 1)(1 – 1 × 1) ⇒ 0 + 0 ⇒ 0 ∴ The value of (tan x + tan y)(1 – cot x cot y) + (cot x + cot y)(1 – tan x tan y) is 0.", "\\tan \\theta$$ $$=$$ $$\\cot \\theta \\times \\frac{1}{\\cot \\theta}$$ $$=$$ $$1$$ Therefore, $$\\cot \\theta \\cdot \\tan \\theta = 1$$.", "/ (1+ tan (-pi/7)*tan (pi/6)) We know that cot (pi/2 - pi/6) = tan (pi/6) and as tan is odd tan (-x) = -tan x Also, tan (a + b) =...", "## Algebra and Trigonometry 10th Edition Part A: tan($12.75^{o}$) = 0.2263 Part B: cot($12.75^{o}$) = 4.4194 Using a calculator: Part A: tan($12.75^{o}$) = 0.2263 Part B: cot($12.75^{o}$) = 4.4194", "HallsofIvy's formula. There is one minor difference though. HallsofIvy writes \"(1/4)(x^2)cot(180/a)\" but I think it should be ¼a(x^2)cot(180/a). If you multiply ½ax*tan((180a - 360)/2a) by ½x you get ¼a(x^2)tan((180a - 360)/2a)=¼a(x^2)cot(180/a) Here's an example: a pentagon with side 7.47 (a random number). ¼*5*(7.47^2)*tan(54) = 69.751125 x 1.37638192 = 96.0041873 which is correct according to online calculators. With HallsofIvy's formula the outcome would be 5 times smaller. In any case, this formula already exists so with that my question is answered. Thanks for you help!", "Browse Questions # What is the value of $cot^{-1}(-x)$ related to an answer for: What is the value of $tan^{-1}(-x)$ Can you answer this question? Answer: $\\pi - cot^{-1}x$ answered Mar 12, 2013", "^{2} x-\\cot ^{2} x\\right\\} d x=-3 \\cos x-4 \\sin x+$ $6 \\tan x+7 \\cot x+c$" ]

[ "# Math Help - Check my work please! 1. ## Check my work please! The elevation of an aircraft is noted simultaneously by three observers A, B, C stationed in a straight horizontal line with AB=BC=200m. if the angles of elevation of the aircraft from A, B, and C are $35^o$, $40^o$, and $45^o$ respectively, find the height of the aircraft. My work: $AQ=\\frac{h}{\\tan 35^o}$ $BQ=\\frac{h}{\\tan 40^o}$ $CQ=h$ Using cosine rule, $\\cos x=\\frac{200^2+(\\frac{h}{\\tan 40^o})^2-(\\frac{h}{\\tan 35^o})^2}{2\\times 200\\times \\frac{h}{\\tan 40^o}}$ $\\cos (180^o-x)=-\\cos x=\\frac{200^2+(\\frac{h}{\\tan 40^o})^2-h^2}{2\\times 200\\times \\frac{h}{\\tan 40^o}}$ $\\frac{200^2+(\\frac{h}{\\tan 40^o})^2-(\\frac{h}{\\tan 35^o})^2}{2\\times 200\\times \\frac{h}{\\tan 40^o}}=-\\frac{200^2+(\\frac{h}{\\tan 40^o})^2-h^2}{2\\times 200\\times \\frac{h}{\\tan 40^o}}$ $200^2+(\\frac{h}{\\tan 40^o})^2-(\\frac{h}{\\tan 35^o})^2=h^2-200^2-(\\frac{h}{\\tan 40^o})^2$ $0.192h^2=80000$ $h=637.3 m$ answer is supposed to be 934.3m Thanks 2. forgot to add the figure attachment. here it is", "\\frac{4}{y} \\implies y = \\frac{4}{tan(58)} =~ 2.5$", "$$\\frac {2\\sqrt {\\frac {25}4 - \\frac 41}}4$$ or, tan$$\\theta$$ =± $$\\frac {\\sqrt {\\frac {25 - 16}4}}2$$ or, tan$$\\theta$$ =± $$\\sqrt {\\frac 94}$$× $$\\frac 12$$ or, tan$$\\theta$$ =± $$\\frac 32$$× $$\\frac 12$$ or, tan$$\\theta$$ =± $$\\frac 34$$ ∴ $$\\theta$$ = tan-1(± $$\\frac 34$$) Hence, the acute angle ($$\\theta$$) =tan-1(± $$\\frac 34$$) Ans Here, Given equation is: 6x2 + 5xy - 3x - 2y - 6y2 = 0 or, 6x2 + 9xy - 4xy - 6y2 - 3x - 2y = 0 or, 3x (2x + 3y) - 2y (2x + 3y) - 1 (2x + 3y) = 0 or, (2x + 3y) (3x - 2y - 1) = 0 Either: 2x + 3y = 0..........................(1) Or: 3x - 2y - 1 = 0............................(2) Slope of equation (1), m1 = -$$\\frac 23$$ Slope of equation (2), m2 = $$\\frac 32$$ Again, m1× m2 = $$\\frac {-2}3$$× $$\\frac 32$$ = -1 The product of two slopes = -1 Hence, these equations are perpendicular each other. Hence, Proved 0% • 2x2 + 7xy + 3y2 = 0 2x+y =0 , x+3y = 0 2x+3 =0 , x+2y = 0 2x-3 =0 , x-3y = 0 2x-y =0 , x-3y = 0 • 2x2 - 5xy + 3y2 = 0. x-y =0 , 2x-3y =0 x+y =0 , 2x+3y =0 x-y =0 ,", "}$$ = $$\\frac { 12 }{ 35 }$$ Question 19. tan A = $$\\frac { BC }{ AC }$$ = $$\\frac { 7 }{ 4√2 }$$ tan B = $$\\frac { AC }{ BC }$$ = $$\\frac { 4√2 }{ 7 }$$ Find the value of x. Round your answer to the nearest tenth. Question 20. tan 54 = $$\\frac { x }{ 32 }$$ 1.37 = $$\\frac { x }{ 32 }$$ x = 43.8 Question 21. tan 25 = $$\\frac { x }{ 20 }$$ 0.46 x 20 = x x = 9.2 Question 22. tan 38 = $$\\frac { 10 }{ x }$$ x = 12.82 Question 23. The angle between the bottom of a fence and the top of a tree is 75°. The tree is 4 let from the fence. How tall is the tree? Round your answer to the nearest foot. tan 75 = $$\\frac { x }{ 4 }$$ x = 14.92 ### 9.5 The Sine and Cosine Ratios Find sin X, sin Z, cos X, and cos Z. Write each answer as a fraction and as a decimal rounded to four decimal places. Question 24. sin X = $$\\frac { 3 }{ 5 }$$ sin Z = $$\\frac { 4 }{ 5 }$$ cos X = $$\\frac { 4 }{", "# Tan[arctan(2/3) + arccos(8/17)] 1. Jul 29, 2011 ### jrjack 1. The problem statement, all variables and given/known data tan[arctan(2/3)+arccos(8/17)] 2. Relevant equations tan(u+v)=$\\frac{\\tan u+\\tan v}{1-\\tan u\\tan v}$ 3. The attempt at a solution After drawing 2 triangles for arctan and arccos (both in Quadrant 1), and inserting them into the addition formula for tan, I get: $$\\frac{\\frac{2}{3}+\\frac{15}{8}}{1-\\frac{2}{3}(\\frac{15}{8})}$$ is that right? if so, then maybe my math is off... I got the numerator reduced to 23/8 and denominator of -1/4 for a total of -23/3 ...but thats not correct. 2. Jul 29, 2011 ### Dick Re: tan[arctan(2/3)+arccos(8/17)] I think your math with the fractions is off. 2/3+15/8 is not equal to 23/8. Can you show why you think it is? 3. Jul 29, 2011 ### jrjack Re: tan[arctan(2/3)+arccos(8/17)] Sorry, mis-typed...I meant 23/12 2/3 =16/24 15/8=30/24******Found my problem, this should be 45/24 Thanks for looking at this, I knew it had to be something simple...I need to slow down.", "y = \\frac{5}{6}$. Plugging in to the earlier formula, we have $\\tan(x+y) = \\frac{25}{1-\\frac{5}{6}} = \\frac{25}{\\frac{1}{6}} = \\boxed{150}$. Solution 3 (less trig required, use of quadratic formula) Let $a=\\tan x$ and $b=\\tan y$. This simplifies the equations to: $$a + b = 25$$ $$\\frac{1}{a} + \\frac{1}{b} = 30$$ Taking the tangent of a sum formula from Solution 2, we get $\\tan(x+y) = \\frac{25}{1 - ab}$. We can use substitution to solve the system of equations from above: $b = -a + 25$, so $\\frac{1}{a} + \\frac{1}{-a + 25} = 30$. Multiplying by $-a(a-25)$, we get $a + (-a + 25) = -30a(a-25)$, which is $-30a^2 + 750a = 25$. Dividing everything by 5 and shifting everything to one side gives $6a^2 - 150a + 5 = 0$. Using the quadratic formula gives $a = \\frac{150 \\pm \\sqrt {22380}}{12}$. Since this looks too hard to simplify, we can solve for $b$ using $a + b = 25$, which turns out to also be $b = \\frac{150 \\pm \\sqrt {22380}}{12}$, provided that the sign of the radical in $a$ is opposite the one in $b$. WLOG, assume $a = \\frac{150 + \\sqrt{22380}}{12}$ and $b = \\frac{150 - \\sqrt{22380}}{12}$. Multiplying them gives $ab = \\frac{22500 - 22380}{144}$ which simplifies to $\\frac{5}{6}$. THe denominator of $\\frac{25}{1 - ab}$ ends up being $\\frac{1}{6}$,", "# Finding the exact value of arctan function then adding it? The question is $x = \\arctan\\frac 23 + \\arctan\\frac 12$. What is $\\tan(x)$? I'm having trouble figuring out how to calculate the arctan values without a calculator, or do I not even need to find these values to calculate what $\\tan(x)$ is? Any help is greatly appreciated (I would show some sort of work, but I am actually completely stuck). I know that the range of arctan is restricted to $(–90^\\circ, 90^\\circ)$ or $(-\\pi/2, \\pi/2)$, but i'm not sure how this helps. Below is a solution based on right triangles. Note that $y=\\arctan\\frac{2}{3}$, and $z=\\arctan\\frac{1}{2}$. It follows that $x = y + z$. The question is then: Calculate $\\frac{b}{a}$. Since $a^2 + c^2 = 4$ and $\\frac{a}{c} = \\frac{2}{3}$, it follows that $a^{2} = \\frac{16}{13}$. Since $a^2 + b^2 = 5$, it follows that $b^2 = \\frac{49}{13}$. Thus, $\\frac{b}{a} = \\frac{7}{4}$. The take-away should be that sometimes going to a more basic line of thought can get you unstuck on a problem. Also, this gives some insight on how you can derive the tangent of sums formula. Hint Use the formula $\\tan(a+b)=\\frac{\\tan(a)+\\tan(b)}{1-\\tan(a)\\tan(b)}$ Let $A=\\arctan\\frac 23,B=\\arctan\\frac 12$. Then, $$\\tan x=\\tan(A+B)=\\frac{\\tan A+\\tan B}{1-\\tan A\\tan B}=\\frac{\\frac 23+\\frac 12}{1-\\frac 23\\cdot \\frac 12}.$$", "the original equation we used to find parallax, $$d=\\frac{R_{E}}{\\tan(\\frac{\\alpha+\\beta}{2})}$$. So, let's solve for θ=α+β, and from there we'll solve for α. $$7.04*10^{10}=\\frac{6371000}{\\tan(\\frac{\\theta}{2})}$$ $$\\theta=2\\arctan(\\frac{6371000}{7.04*10^{10}})=0.0103°=37.33''$$ To keep things simple, let's assume α and β were each 33.6/2=16.8 when time was measured correctly. Then β=16.8'' and $$α=37.33''-16.8''=20.53''=5.7*(10^{-3})°$$ Ok, now we've found α. Now we use it in the original equation to solve for Δt. $$\\tan(\\alpha)=\\frac{R_{E}+5660*\\Delta{t}}{d}$$ $$\\Delta{t}=\\frac{d*\\tan(\\alpha)-R_{E}}{5660}=252.8 s$$ Now, we could do the whole thing over again using d=8.60*10^10 in our equation to solve for α, and we would get Δt = -206.8 seconds. However, since we're assuming the first measurement is made at the time of opposition and the second one Δt seconds later, Δt can't be negative. Thus, the clocks must be synchronized to at least 252.8 seconds And I have a feeling there's an easier way to do this... Last edited: Sep 16, 2012", "I doing something wrong? $$60=600*tan(\\theta) - \\frac{1}{2}*g*\\frac{600}{v_0*cos(\\theta)}$$ and $$0=795*tan(\\theta)-\\frac{1}{2}*g*\\frac{795}{v_0*cos(\\theta)}$$ Last edited: Jan 20, 2005 4. Jan 20, 2005 ### Andrew Mason In the last term, t (= x/vcos) should be squared: $$60=600*tan(\\theta) - \\frac{1}{2}*g*\\frac{600^2}{v_0^2cos^2(\\theta)}$$ and $$0=795*tan(\\theta)-\\frac{1}{2}*g*\\frac{795^2}{(v_0^2cos^2(\\theta))}$$ Substituting $t=x/v_0cos(\\theta)$ into the general equation of motion, you get the general solution to the quadratic for the range, R: $$R = v_0^2sin^2\\theta/2g$$ Then substitute that into the equation of motion to get: $$y = xtan\\theta - \\frac{1}{4R}tan^2\\theta$$ where R = range (795) By substituting x,y = 600,60 you have a quadratic equation in terms of tan(theta) only. You will have to check my algebra - I am not guaranteeing that I haven't missed something. AM Last edited: Jan 20, 2005 5. Jan 20, 2005 ### nineeyes Hi, Someone asked my professor about this in class and he said we only need to use the equation $$y=y_0+(x-x_0)*tan(\\theta)-\\frac {1}{2}g (\\frac {x-x_0}{v_0*cos(\\theta)})^2$$ Once. He said that we need to put $$v_0$$ in terms of $$\\theta$$. and find a $$\\theta$$ that would give a minimum value of $$v_0$$. $$v_0=sqareroot((\\frac{.5*g*x^2}{x*tan(\\theta)(cos^2(\\theta))}))$$ using the (795,0) and $$x_0 = y_0 = 0$$ I was trying this but I dont recall how to find an angle that would give a minimum value. Can you offer any help regarding this ? Thanks Last edited: Jan 20, 2005 6. Jan 20, 2005", "base, in line with the base and the first point and in the same plane, the angle of elevation is 32 degrees . Find the height of the tower and then choose the best answer. A. 574 B. 503 C. 601 D. 414 The height of the tower is x. $$tan(32°)=\\frac{x}{\\frac{x}{tan(61°)}+600ft}$$ ! Jul 10, 2019 edited by asinus Jul 10, 2019 edited by asinus Jul 10, 2019 #3 +8579 +3 Using the rule tan(angle) = opposite / adjacent , we can make two equations: $$\\tan(61^\\circ)\\ =\\ \\frac{h}{x}\\\\~\\\\ x\\tan(61^\\circ)\\ =\\ h\\\\~\\\\ x\\ =\\ \\frac{h}{\\tan(61^\\circ)}$$ ...and... $$\\tan(32^\\circ)\\ =\\ \\frac{h}{600+x}\\\\~\\\\ (600+x)\\tan(32^\\circ)\\ =\\ h\\\\~\\\\ 600+x\\ =\\ \\frac{h}{\\tan(32^\\circ)}\\\\~\\\\ x\\ =\\ \\frac{h}{\\tan(32^\\circ)}-600$$ Now we can equate both expressions of x and solve for h: $$\\frac{h}{\\tan(61^\\circ)}\\ =\\ \\frac{h}{\\tan(32^\\circ)}-600\\\\~\\\\ \\frac{h}{\\tan(61^\\circ)}-\\frac{h}{\\tan(32^\\circ)}\\ =\\ -600\\\\~\\\\ h(\\frac{1}{\\tan(61^\\circ)}-\\frac{1}{\\tan(32^\\circ)})\\ =\\ -600\\\\~\\\\ h\\ =\\ -600\\div(\\frac{1}{\\tan(61^\\circ)}-\\frac{1}{\\tan(32^\\circ)})\\\\~\\\\ h\\ \\approx\\ 573.5998\\\\~\\\\ h\\ \\approx\\ 574$$ Jul 10, 2019", "i.e. 2x - 3y + 4 = 0........................(2) Slope of eqn (1), m1 = - $$\\frac {x-coefficient}{y-coefficient}$$ = $$\\frac {-2}{-3}$$ = $$\\frac 23$$ Slope of eqn (2), m2 = - $$\\frac {x-coefficient}{y-coefficient}$$ = $$\\frac {-3}{5}$$ Let $$\\theta$$ be the angle between the equation (1) and (2): tan$$\\theta$$ =± $$\\frac {m_1 - m_2}{1 + m_1m_2}$$ or, tan$$\\theta$$ =± ($$\\frac {\\frac 23 + \\frac 35}{1 - \\frac 23 × \\frac 35}$$) or, tan$$\\theta$$ =± ($$\\frac {\\frac {10 + 9}{15}}{\\frac {5 - 2}5}$$) or, tan$$\\theta$$ =± ($$\\frac {19}{15}$$ × $$\\frac 53$$) or, tan$$\\theta$$ =± $$\\frac {19}9$$ For acute angle, tan$$\\theta$$ = $$\\frac {19}9$$= 2.11 ∴ $$\\theta$$ = 65° ∴ The acute angle between two lines is 65°. Ans Here, Given equation are: 3y - x - 6 = 0..............................(1) y = 2x + 5 i.e. -2x + y = 5...................(2) Slope of eqn (1), m1= -$$\\frac {x-coefficient}{y-coefficient}$$ = -$$\\frac {(-1)}3$$ = $$\\frac 13$$ Slope of eqn (2), m2= -$$\\frac {x-coefficient}{y-coefficient}$$ = -$$\\frac {(-2)}1$$ = 2 If $$\\theta$$ be the angle between the eqn (1) and (2), tan$$\\theta$$ =± $$\\frac {m_1 - m_2}{1 + m_1m_2}$$ or, tan$$\\theta$$ =± $$\\frac {\\frac 13 - 2}{1 + \\frac 13 × 2}$$ or, tan$$\\theta$$ =± $$\\frac {\\frac {1 - 6}3}{\\frac {3 + 2}3}$$ or, tan$$\\theta$$ =± $$\\frac {-5}3$$× $$\\frac 35$$ ∴ tan$$\\theta$$ =± (-1) Taking -ve", "# Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 682: 101 See graph. a. $y=tan(\\frac{x}{2})$ b. See explanations. #### Work Step by Step Graph the equation as shown in the figure. a. The graph appears to be the same as $y=tan(\\frac{x}{2})$ b. Using the double angle formulas, we have $y=csc(x)-cot(x)=\\frac{1}{sin(x)}-\\frac{cos(x)}{sin(x)}=\\frac{1-cos(x)}{sin(x)}=\\frac{1-(1-2sin^2(x/2))}{2sin(x/2)cos(x/2)}=tan(\\frac{x}{2})$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "0° = 0 2 sin P = 2sin 0° = 2×0 = 0 or, sin 2P = 2sin PcosP =>2sin PcosP = 2 sin P =>2cos P = 2 =>cosP = 1 =>P = 0° (iv) (C) is correct 2tan301tan230=2(131(13)2)$\\frac{2tan30^{\\circ}}{ 1 – tan^{2}30^{\\circ}} = 2\\left ( \\frac{\\frac{1}{\\sqrt{3}}}{1 – \\left ( \\frac{1}{\\sqrt{3}} \\right )^{2}} \\right )$ (23)113=2323=3=tan60$\\frac{\\left ( 2\\sqrt{3} \\right )}{ 1 – \\frac{1}{3}} = \\frac{2\\sqrt{3}}{\\frac{2}{3}} = \\sqrt{3} = tan 60^{\\circ}$ 3) If tan (P + Q) = 3$\\sqrt{3}$ and tan ( P – Q) = 13$\\frac{1}{\\sqrt{3} }$;00<P+Q<=90;P>Q$0^{0} < P + Q <= 90^{\\circ}; P>Q$ , calculate P and Q Ans:- tan (P + Q) = 3$\\sqrt{3}$ =>tan (P + Q) = tan 60° => (P + Q) = 60° … (i) =>tan (P – Q) = 13$\\frac{1}{\\sqrt{3}}$ =>tan (P – Q) = 30° => (P – Q) = 30° … (ii) Adding (i) and (ii), we get P + Q + P – Q = 60° + 30° 2P = 90° => P = 45° Putting the value of P in equation (i) 45° + Q = 60° => Q = 60° – 45° = 15° Hence, P = 45° and Q = 15° 4) Check whether the given statements are true or false, also give a reason for your answer: (i) sin (P + Q) = sin", "$$\\frac{\\tan x}{\\tan y} = 3$$", "- $\\frac {x-coefficient}{y-coefficient}$ = - $\\frac 65$ Now, Using angle formula, tan$\\theta$ =± ($\\frac {m_1 - m_2}{1 + m_1m_2}$) or, tan 45° =± ($\\frac {m + \\frac 65}{1 - m\\frac 65}$) or, 1 =± ($\\frac {\\frac {5m + 6}5}{\\frac {5 - 6m}5}$) or, 1 =± ($\\frac {5m + 6}{5 - 6m}$) Taking +ve sign, 5 - 6m = 5m + 6 or, -6m - 5m = 6 - 5 or, -11m = 1 ∴ m = -$\\frac 1{11}$ Taking -ve sign, 5 - 6m = - (5m + 6) or, 5 - 6m = -5m - 6 or, -6m + 5m = - 6 - 5 or, - m = - 11 ∴ m = 11 Putting the value of m = -$\\frac 1{11}$ in equation (2) y + 1 = m (x - 2) or, y + 1 = -$\\frac 1{11}$(x - 2) or, 11y + 11 = -x + 2 or, x + 11y + 11 - 2 = 0 ∴ x + 11y + 9 = 0 Putting the value of m = 11 in equation (2) y + 1 = m(x - 2) or, y + 1= 11 (x - 2) or, y + 1 = 11x - 22 or, 11x - y - 22 - 1 = 0 ∴ 1 1x - y", "knew I could do better. Fifteen, after all, is the difference between 60 and 45 - and I have those etched into my soul: $\\tan(60º)=\\sqrt{3}$, and $\\tan(45º)= 1$. So, I can use a trigonometric identity: $\\tan(A-B) \\equiv \\frac{\\tan(A) - \\tan(B)}{1+\\tan(A)\\tan(B)}$, with $A=60º$ and $B=45º$. That gives me $\\tan(15º) = \\frac{ \\sqrt{3} - 1}{1 + \\sqrt{3}}$. A younger, naiver me would have said “$\\sqrt{3} \\approx 1.732$, so I’ve got $\\frac{0.732}{2.732}$, which is… I dunno. I am older and wiser now; I know that $\\sqrt{3} \\approx \\frac{52}{30}$. That makes the fraction nicer to deal with: it’s $\\frac{\\frac{52}{30}-1}{\\frac{52}{30}+1} = \\frac{52-30}{52+30} = \\frac{11}{41}$. Great! I’ve improved my estimate of $\\tan(15º)$ marginally. Now all I need to do is multiply it by 200, which gives me $\\frac{2200}{41}$. That’s not especially easy, but I know it’s at least 50 (because $41 \\times 50 = 2050$), then I have 150 left over; $3\\times 41=123$, so I estimated 53.6, and added one to make it 54.6. Correct, to three significant figures! Two things here: I was slightly out on that last bit - $\\frac{2200}{41}$ is closer to 53.7 than 53.6, but my mistake worked in my favour! I love it when that hapens. The other thing is, I missed a trick! Had I rationalised the denominator, I’d have ended up with $\\tan(15º)=2-\\sqrt{3}$, which I can do", "$$\\frac{x}{-2}+\\frac{y}{7}$$ = 1 7x – 2y = -14 or 7x – 2y + 14 = 0 (b) Evaluate : $$\\frac{\\sin 150^{\\circ}-5 \\cos 300^{\\circ}+7 \\tan 225^{\\circ}}{\\tan 135+3 \\sin 210^{\\circ}}$$ $$\\frac{\\sin 150-5 \\cos 300+7 \\tan 225}{\\tan 135+3 \\sin 210}=\\frac{\\sin (180-30)-5 \\cos (360-60)+\\tan (180+45)}{\\tan (180-45)+3 \\sin (180+30)}$$ = $$\\frac{\\sin 30-5 \\cos 60+7 \\tan 45}{-\\tan 45-3 \\sin 30}=\\frac{\\frac{1}{2}-5 \\cdot \\frac{1}{2}+7.1}{-1-\\frac{3}{2}}=\\frac{\\frac{1-5+14}{2}}{\\frac{-2-3}{2}}=\\frac{10}{-5}$$ = -2 (c) If log6 = 0.7782, find the number of zeroes between the decimal point and the first significant figure in (0.6)30. Let x = (0.6)30 logx = 30log0.6 = 30 × 1.7782 = 30 (-1 + 0.7782) = -30 + 23.346 = -7̄.346 = 7̄.346 Since the characteristic is 7̄. Hence the required number of zeros is 6. Question 50. (a) Three numbers whose sum is 12 are in A.P. If 1, 4, 11 are added to them the resulting numbers are in G.P. Find the numbers. Let the 3 numbers be a – d, a, a + d By data a-d + a + a + d = 12 3a = 12 ⇒ a = 4 Also by data 4-d + 1, 4 + 4, 4 + d + 11 are in G.P. 5 – d, 8, 15 + d are in G.P. ∴ $$\\frac{8}{5-d}=\\frac{15+d}{8}$$ 64 = (5 – d)(15 + d) 64 = 75 + 5d – 15d", "= \\frac{y}{tan(\\alpha)}$ Inserting this value of x into the second equation gives: $tan(\\beta) = \\frac{y}{\\frac{y}{tan(\\alpha)} - d}$ $tan(\\beta) = \\frac{y}{\\frac{y}{tan(\\alpha)} - d} \\cdot \\frac{tan(\\alpha)}{tan(\\alpha)}$ $tan(\\beta) = \\frac{y~tan(\\alpha)}{y - d~tan(\\alpha)}$ Can you finish it from here? -Dan 3. Originally Posted by cocoknny y___ y-d What is this? Originally Posted by cocoknny is my top part right? I don't know because I don't know what \"htx\" is supposed to mean. -Dan 4. tan 26°50'=h 25m+x 5. ## hello Oh don't worry about that part. h suppose to be the height 6. Originally Posted by cocoknny tan 26°50'=h 25m+x It looks good if you have defined x as the distance from the second point to the tower. That's one equation. You need to use the other to eliminate x. -Dan 7. ## hello YUP that's what i did. Thanks hun", "Progress and Modeling with Mathematics In Exercises 3–10, find the exact value of the expression. Question 3. tan(−15°) Question 4. tan 195° tan 195° = 2 – √3 Explanation: tan 195° = tan(225° – 30°) = $$\\frac { tan 225° – tan 30° }{ 1 + tan 225° tan 30° }$$ = $$\\frac { 1 – 1/√3 }{ 1 + 1 . 1/√3 }$$ = $$\\frac { 3 – √3 }{ 3 + √3 } . \\frac { 3 – √3 }{ 3 – √3 }$$ = $$\\frac { 9 – 6√3 + 3 }{ 9 – 3 }$$ = $$\\frac { 12 – 6√3 }{ 6 }$$ = 2 – √3 Question 5. sin $$\\frac{23 \\pi}{12}$$ Question 6. sin(−165°) sin(−165°) = $$\\frac { √2 – √6 }{ 4 }$$ Explanation: sin(−165°) = – sin 165° sin 165° = sin(120° + 45°) = sin 120° cos 45° + cos 120° sin 45° = $$\\frac { √3 }{ 2 }$$ . $$\\frac { √2 }{ 2 }$$ + ($$\\frac { -1 }{ 2 }$$). $$\\frac { √2 }{ 2 }$$ = $$\\frac { √6 – √2 }{ 4 }$$ sin(−165°) = – sin 165° = $$\\frac { √2 – √6 }{ 4 }$$ Question 7. cos 105° Question 8. cos $$\\frac{11 \\pi}{12}$$ cos $$\\frac{11 \\pi}{12}$$ = -($$\\frac { √2", "Now the angles A and B can be found solving the simultaneous equations, A + B = 110 A - B = 20.2º A = 65.12º and B = 44.88º The measure of side c can be calculate using sin rule as follows: $\\frac{a}{sinA}$ = $\\frac{c}{sinC}$ c = Sin C x $\\frac{a}{sinA}$ = Sin 70 x $\\frac{9}{sin 65.12}$ ≈ 9.322 cm Now let us prove an identity using Law of tangents. Verify the identity cot $\\frac{C}{2}$ = $\\frac{a+b}{a-b}$ . tan $\\frac{A-B}{2}$ Consider the law of tangents, $\\frac{tan\\frac{A+B}{2}}{tan\\frac{A-B}{2}}$ = $\\frac{a+b}{a-b}$ Multiplying the equation by tan $\\frac{A-B}{2}$ (The difference between two angles in a triangle A - B ≠ 0). tan $\\frac{A+B}{2}$ = $\\frac{a+b}{a-b}$ . tan $\\frac{A-B}{2}$ tan ($\\frac{180-C}{2}$) = $\\frac{a+b}{a-b}$ . tan $\\frac{A-B}{2}$ A + B + C = 180º. tan (90 - $\\frac{C}{2}$) = $\\frac{a+b}{a-b}$ . tan $\\frac{A-B}{2}$ cot $\\frac{C}{2}$ = $\\frac{a+b}{a-b}$ . tan $\\frac{A-B}{2}$ cofunction identity tan (90 - θ) = cot θ." ]

[ "# Finding the exact value of arctan function then adding it? The question is $x = \\arctan\\frac 23 + \\arctan\\frac 12$. What is $\\tan(x)$? I'm having trouble figuring out how to calculate the arctan values without a calculator, or do I not even need to find these values to calculate what $\\tan(x)$ is? Any help is greatly appreciated (I would show some sort of work, but I am actually completely stuck). I know that the range of arctan is restricted to $(–90^\\circ, 90^\\circ)$ or $(-\\pi/2, \\pi/2)$, but i'm not sure how this helps. Below is a solution based on right triangles. Note that $y=\\arctan\\frac{2}{3}$, and $z=\\arctan\\frac{1}{2}$. It follows that $x = y + z$. The question is then: Calculate $\\frac{b}{a}$. Since $a^2 + c^2 = 4$ and $\\frac{a}{c} = \\frac{2}{3}$, it follows that $a^{2} = \\frac{16}{13}$. Since $a^2 + b^2 = 5$, it follows that $b^2 = \\frac{49}{13}$. Thus, $\\frac{b}{a} = \\frac{7}{4}$. The take-away should be that sometimes going to a more basic line of thought can get you unstuck on a problem. Also, this gives some insight on how you can derive the tangent of sums formula. Hint Use the formula $\\tan(a+b)=\\frac{\\tan(a)+\\tan(b)}{1-\\tan(a)\\tan(b)}$ Let $A=\\arctan\\frac 23,B=\\arctan\\frac 12$. Then, $$\\tan x=\\tan(A+B)=\\frac{\\tan A+\\tan B}{1-\\tan A\\tan B}=\\frac{\\frac 23+\\frac 12}{1-\\frac 23\\cdot \\frac 12}.$$", "# Tan[arctan(2/3) + arccos(8/17)] 1. Jul 29, 2011 ### jrjack 1. The problem statement, all variables and given/known data tan[arctan(2/3)+arccos(8/17)] 2. Relevant equations tan(u+v)=$\\frac{\\tan u+\\tan v}{1-\\tan u\\tan v}$ 3. The attempt at a solution After drawing 2 triangles for arctan and arccos (both in Quadrant 1), and inserting them into the addition formula for tan, I get: $$\\frac{\\frac{2}{3}+\\frac{15}{8}}{1-\\frac{2}{3}(\\frac{15}{8})}$$ is that right? if so, then maybe my math is off... I got the numerator reduced to 23/8 and denominator of -1/4 for a total of -23/3 ...but thats not correct. 2. Jul 29, 2011 ### Dick Re: tan[arctan(2/3)+arccos(8/17)] I think your math with the fractions is off. 2/3+15/8 is not equal to 23/8. Can you show why you think it is? 3. Jul 29, 2011 ### jrjack Re: tan[arctan(2/3)+arccos(8/17)] Sorry, mis-typed...I meant 23/12 2/3 =16/24 15/8=30/24******Found my problem, this should be 45/24 Thanks for looking at this, I knew it had to be something simple...I need to slow down.", "\\frac{4}{y} \\implies y = \\frac{4}{tan(58)} =~ 2.5$", "30 - (\\sin 30 - \\sin 10) \\\\ = \\sin 30 +\\sin 10 = 2 \\sin 20 \\cos 10$$ Therefore, $$\\tan \\alpha = \\frac{2 \\sin 10 \\sin 20}{1 - 2 \\sin 10 \\cos 20} = \\frac{2\\sin 10 \\sin 20}{2 \\cos 10 \\sin 20} = \\tan 10$$ Now, since $$0 < \\alpha < 180$$, we get that $$\\alpha = 10$$. From here, $$80-\\alpha = 70$$ is the desired angle.", "# What am I doing wrong with Related Rates? Ok, so the first thing I did was assess the x-values at 30°, 60°, & 80° through $y tan(\\theta°) = x$. $@30°, x = \\frac{\\sqrt3}{120}$ $@60°, x = \\frac{\\sqrt{3}}{40}$ $@80°, x = 40 cot(\\frac{\\pi}{18})$ Then I converted 30 revolutions/min into radians/second. $\\frac{dx}{dt} = \\frac{30rev.}{min} = \\frac{\\pi}{second}$ Then find the values of $\\frac{dy}{dt}$. $x^2 + y^2 = hypotenuse$ $2x(\\frac{dx}{dt}) + 2y(\\frac{dy}{dt}) = 0$ $x(\\pi) + 40(\\frac{dy}{dt}) = 0$ $@30°, \\frac{dy}{dt} = \\frac{-\\pi}{1600\\sqrt3}$ $@60°, \\frac{dy}{dt} = \\frac{-\\pi\\sqrt{3}}{1600}$ $@80°, \\frac{dy}{dt} = -\\pi cot(\\frac{\\pi}{18})$ Now I can use the related rates equation: $tan(\\theta) = \\frac{x}{y}$ $\\frac{d\\theta}{dt} sec^2(\\theta) = \\frac{(\\frac{dx}{dt})(y) - (x)(\\frac{dy}{dt})}{y^2}$ Is this right so far? • Use the relation $x = 40\\tan \\theta$ to relate $x$ and $\\theta$ (the hypotenuse is immaterial in this problem), then differentiate both sides w.r.t. $t$. – Jon Dec 9 '14 at 6:13 • Huh, I'd never thought of it like that way. So $\\frac{dx}{dt} = 40 sec^2(\\theta)\\frac{d\\theta}{dt}$. And plugging-in values would get me $\\pi = 40 sec^2(30°/60°/80°)\\frac{d\\theta}{dt}$. Are the answers: .0589ft/sec,.0196, & .002368? – Sentient Dec 9 '14 at 6:38 • Carefull... Revolutions per minute count the number of times around the circle per minute, which is the change in $\\theta$ with respect to time, and the distance the light moves is measured on the $x$-axis.", "# Solution required urgently Printable View • June 19th 2006, 10:17 AM lalitchugh Solution required urgently Hi, • Find x from each of the following equations. 1.1)x*tan^2(45)*cos(60) = sin(60)*cot(30) 1.2)(1+x)*tan(45)*cosec^2(45) = sin(30)*sec^2(45) 1.3)X*cos(60)–sin(30) = x*tan(30)*cot(60) 1.sec(120) 2.sec(350) 3.tan(40) Request to respond as soon as possible. Best Regards, Lalit Chugh • June 19th 2006, 05:31 PM malaygoel Quote: Originally Posted by lalitchugh Hi, • Find x from each of the following equations. 1.1)x*tan^2(45)*cos(60) = sin(60)*cot(30) 1.2)(1+x)*tan(45)*cosec^2(45) = sin(30)*sec^2(45) 1.3)X*cos(60)–sin(30) = x*tan(30)*cot(60) 1.sec(120) 2.sec(350) 3.tan(40) Request to respond as soon as possible. Best Regards, Lalit Chugh (1.1) $xtan^{2}45cos60 = sin60cot30$ $tan45=1$ $cos60=\\frac{1}{2}$ $sin60=\\frac{\\sqrt{3}}{2}$ $cot30=\\sqrt{3}$ Hence $x=\\frac{\\frac{\\sqrt{3}}{2}*\\sqrt{3}}{1^2*\\frac{1}{ 2}}$ $x=3$ • June 19th 2006, 07:10 PM Soroban Hello, lalitchugh! You're expected to know the trig values for 30°, 60°, and 45°. Once you plug in the values, the rest is Arithmetic and Algebra I . . . Quote: $1.2)\\;\\;(1+x)\\cdot\\tan(45^o)\\cdot\\csc^2(45^o) \\;= \\;\\sin(30^o)\\cdot\\sec^2(45^o)$ We have: . $(1 + x)(1)(\\sqrt{2})^2\\;=\\;\\left(\\frac{1}{2}\\right)( \\sqrt{2})^2$ Then: . $(1 + x)\\cdot 2\\;=\\;1\\quad\\Rightarrow\\quad 1 + x \\:=\\:\\frac{1}{2}\\quad\\Rightarrow\\quad x\\,=\\,-\\frac{1}{2}$ Quote: $1.3)\\;\\;x\\cdot\\cos(60^o) - \\sin(30^o) \\;= \\;x\\cdot\\tan(30^o)\\cdot\\cot(60^o)$ We have: . $x\\left(\\frac{1}{2}\\right) - \\frac{1}{2} \\;= \\;x\\left(\\frac{1}{\\sqrt{3}}\\right)\\left(\\frac{1}{ \\sqrt{3}}\\right)$ Then: . $\\frac{1}{2}x - \\frac{1}{2}\\;=\\;\\frac{1}{3}x\\quad\\Rightarrow\\quad \\frac{1}{6}x\\,=\\, \\frac{1}{2}\\quad\\Rightarrow\\quad x = 3$ • June 19th 2006, 08:48 PM lalitchugh Solution required urgently Hi, Thanks for the reply. Can you pls advice how we can evaluate 1. tan 40 2. sec", "# Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 682: 101 See graph. a. $y=tan(\\frac{x}{2})$ b. See explanations. #### Work Step by Step Graph the equation as shown in the figure. a. The graph appears to be the same as $y=tan(\\frac{x}{2})$ b. Using the double angle formulas, we have $y=csc(x)-cot(x)=\\frac{1}{sin(x)}-\\frac{cos(x)}{sin(x)}=\\frac{1-cos(x)}{sin(x)}=\\frac{1-(1-2sin^2(x/2))}{2sin(x/2)cos(x/2)}=tan(\\frac{x}{2})$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "y = \\frac{5}{6}$. Plugging in to the earlier formula, we have $\\tan(x+y) = \\frac{25}{1-\\frac{5}{6}} = \\frac{25}{\\frac{1}{6}} = \\boxed{150}$. Solution 3 (less trig required, use of quadratic formula) Let $a=\\tan x$ and $b=\\tan y$. This simplifies the equations to: $$a + b = 25$$ $$\\frac{1}{a} + \\frac{1}{b} = 30$$ Taking the tangent of a sum formula from Solution 2, we get $\\tan(x+y) = \\frac{25}{1 - ab}$. We can use substitution to solve the system of equations from above: $b = -a + 25$, so $\\frac{1}{a} + \\frac{1}{-a + 25} = 30$. Multiplying by $-a(a-25)$, we get $a + (-a + 25) = -30a(a-25)$, which is $-30a^2 + 750a = 25$. Dividing everything by 5 and shifting everything to one side gives $6a^2 - 150a + 5 = 0$. Using the quadratic formula gives $a = \\frac{150 \\pm \\sqrt {22380}}{12}$. Since this looks too hard to simplify, we can solve for $b$ using $a + b = 25$, which turns out to also be $b = \\frac{150 \\pm \\sqrt {22380}}{12}$, provided that the sign of the radical in $a$ is opposite the one in $b$. WLOG, assume $a = \\frac{150 + \\sqrt{22380}}{12}$ and $b = \\frac{150 - \\sqrt{22380}}{12}$. Multiplying them gives $ab = \\frac{22500 - 22380}{144}$ which simplifies to $\\frac{5}{6}$. THe denominator of $\\frac{25}{1 - ab}$ ends up being $\\frac{1}{6}$,", "# Math Help - Check my work please! 1. ## Check my work please! The elevation of an aircraft is noted simultaneously by three observers A, B, C stationed in a straight horizontal line with AB=BC=200m. if the angles of elevation of the aircraft from A, B, and C are $35^o$, $40^o$, and $45^o$ respectively, find the height of the aircraft. My work: $AQ=\\frac{h}{\\tan 35^o}$ $BQ=\\frac{h}{\\tan 40^o}$ $CQ=h$ Using cosine rule, $\\cos x=\\frac{200^2+(\\frac{h}{\\tan 40^o})^2-(\\frac{h}{\\tan 35^o})^2}{2\\times 200\\times \\frac{h}{\\tan 40^o}}$ $\\cos (180^o-x)=-\\cos x=\\frac{200^2+(\\frac{h}{\\tan 40^o})^2-h^2}{2\\times 200\\times \\frac{h}{\\tan 40^o}}$ $\\frac{200^2+(\\frac{h}{\\tan 40^o})^2-(\\frac{h}{\\tan 35^o})^2}{2\\times 200\\times \\frac{h}{\\tan 40^o}}=-\\frac{200^2+(\\frac{h}{\\tan 40^o})^2-h^2}{2\\times 200\\times \\frac{h}{\\tan 40^o}}$ $200^2+(\\frac{h}{\\tan 40^o})^2-(\\frac{h}{\\tan 35^o})^2=h^2-200^2-(\\frac{h}{\\tan 40^o})^2$ $0.192h^2=80000$ $h=637.3 m$ answer is supposed to be 934.3m Thanks 2. forgot to add the figure attachment. here it is", "1986 AIME Problems/Problem 3 Problem If $\\tan x+\\tan y=25$ and $\\cot x + \\cot y=30$, what is $\\tan(x+y)$? Solution 1 Since $\\cot$ is the reciprocal function of $\\tan$: $\\cot x + \\cot y = \\frac{1}{\\tan x} + \\frac{1}{\\tan y} = \\frac{\\tan x + \\tan y}{\\tan x \\cdot \\tan y} = 30$ Thus, $\\tan x \\cdot \\tan y = \\frac{\\tan x + \\tan y}{30} = \\frac{25}{30} = \\frac{5}{6}$ Using the tangent addition formula: $\\tan(x+y) = \\frac{\\tan x + \\tan y}{1-\\tan x \\cdot \\tan y} = \\frac{25}{1-\\frac{5}{6}} = \\boxed{150}$. Solution 2 Using the formula for tangent of a sum, $\\tan(x+y)=\\frac{\\tan x + \\tan y}{1-\\tan x \\tan y} = \\frac{25}{1-\\tan x \\tan y}$. We only need to find $\\tan x \\tan y$. We know that $25 = \\tan x + \\tan y = \\frac{\\sin x}{\\cos x} + \\frac{\\sin y}{\\cos y}$. Cross multiplying, we have $\\frac{\\sin x \\cos y + \\cos x \\sin y}{\\cos x \\cos y} = \\frac{\\sin(x+y)}{\\cos x \\cos y} = 25$. Similarly, we have $30 = \\cot x + \\cot y = \\frac{\\cos x}{\\sin x} + \\frac{\\cos y}{\\sin y} = \\frac{\\cos x \\sin y + \\sin x \\cos y}{\\sin x \\sin y} = \\frac{\\sin(x+y)}{\\sin x \\sin y}$. Dividing: $\\frac{25}{30} = \\frac{\\frac{\\sin(x+y)}{\\cos x \\cos y}}{\\frac{\\sin(x+y)}{\\sin x \\sin y}} = \\frac{\\sin x \\sin y}{\\cos x \\cos y} = \\tan x \\tan", "∴ DE = 120 tan$\\alpha$............................(i) In the right angled $\\triangle$CDE, tan (90 - $\\alpha$) = $\\frac {DE}{Cd}$ = $\\frac {120 tan\\alpha}{60}$ = 2 tan$\\alpha$ cot$\\alpha$ = 2 tan$\\alpha$ or, $\\frac 1{tan\\alpha}$ = 2 tan$\\alpha$ or, tan2$\\alpha$ = $\\frac 12$ ∴ tan$\\alpha$ = $\\frac 1{\\sqrt 2}$ Putting the value of tan$\\alpha$ in equation (i), DE = 120×$\\frac 1{\\sqrt 2}$ = 84.85 m AB = $\\frac {DE}2$ = $\\frac {84.85}2$ = 42.43 m ∴ Height of the posts are 42.43 m and 84.85 m. Ans Let: AB and DE are the height of the poles where, DE = 2 AB. Let: angle of elevation $\\angle$BCA = $\\theta$, $\\angle$DCE = 90 - $\\theta$ Distance between the two poles BD = 40 m BC = CD = 20 m In the right angled $\\triangle$ABC, tan$\\theta$ = $\\frac {AB}{BC}$ = $\\frac {DE}{2× 20}$ = $\\frac {DE}{40}$ or, DE = 40 tan$\\theta$....................................(i) In the right angled $\\triangle$CDE, tan (90 - $\\theta$) = $\\frac {DE}{CD}$ or, cot$\\theta$ = $\\frac {40 tan\\theta}{20}$ or, $\\frac 1{tan\\theta}$ = tan$\\theta$× 2 or, $\\frac 12$ = tan2$\\theta$ or, tan$\\theta$ = $\\frac 1{\\sqrt 2}$ POutting the value of tan$\\theta$ in equation (i) DE = $\\frac 1{\\sqrt 2}$× 40 = 28.28 m AB = $\\frac {DE}2$ = $\\frac {28.28}2$ = 14.14 m ∴ Height of two poles are 28.28 m and 14.14", "}$$ = $$\\frac { 12 }{ 35 }$$ Question 19. tan A = $$\\frac { BC }{ AC }$$ = $$\\frac { 7 }{ 4√2 }$$ tan B = $$\\frac { AC }{ BC }$$ = $$\\frac { 4√2 }{ 7 }$$ Find the value of x. Round your answer to the nearest tenth. Question 20. tan 54 = $$\\frac { x }{ 32 }$$ 1.37 = $$\\frac { x }{ 32 }$$ x = 43.8 Question 21. tan 25 = $$\\frac { x }{ 20 }$$ 0.46 x 20 = x x = 9.2 Question 22. tan 38 = $$\\frac { 10 }{ x }$$ x = 12.82 Question 23. The angle between the bottom of a fence and the top of a tree is 75°. The tree is 4 let from the fence. How tall is the tree? Round your answer to the nearest foot. tan 75 = $$\\frac { x }{ 4 }$$ x = 14.92 ### 9.5 The Sine and Cosine Ratios Find sin X, sin Z, cos X, and cos Z. Write each answer as a fraction and as a decimal rounded to four decimal places. Question 24. sin X = $$\\frac { 3 }{ 5 }$$ sin Z = $$\\frac { 4 }{ 5 }$$ cos X = $$\\frac { 4 }{", "Now the angles A and B can be found solving the simultaneous equations, A + B = 110 A - B = 20.2º A = 65.12º and B = 44.88º The measure of side c can be calculate using sin rule as follows: $\\frac{a}{sinA}$ = $\\frac{c}{sinC}$ c = Sin C x $\\frac{a}{sinA}$ = Sin 70 x $\\frac{9}{sin 65.12}$ ≈ 9.322 cm Now let us prove an identity using Law of tangents. Verify the identity cot $\\frac{C}{2}$ = $\\frac{a+b}{a-b}$ . tan $\\frac{A-B}{2}$ Consider the law of tangents, $\\frac{tan\\frac{A+B}{2}}{tan\\frac{A-B}{2}}$ = $\\frac{a+b}{a-b}$ Multiplying the equation by tan $\\frac{A-B}{2}$ (The difference between two angles in a triangle A - B ≠ 0). tan $\\frac{A+B}{2}$ = $\\frac{a+b}{a-b}$ . tan $\\frac{A-B}{2}$ tan ($\\frac{180-C}{2}$) = $\\frac{a+b}{a-b}$ . tan $\\frac{A-B}{2}$ A + B + C = 180º. tan (90 - $\\frac{C}{2}$) = $\\frac{a+b}{a-b}$ . tan $\\frac{A-B}{2}$ cot $\\frac{C}{2}$ = $\\frac{a+b}{a-b}$ . tan $\\frac{A-B}{2}$ cofunction identity tan (90 - θ) = cot θ.", "the original equation we used to find parallax, $$d=\\frac{R_{E}}{\\tan(\\frac{\\alpha+\\beta}{2})}$$. So, let's solve for θ=α+β, and from there we'll solve for α. $$7.04*10^{10}=\\frac{6371000}{\\tan(\\frac{\\theta}{2})}$$ $$\\theta=2\\arctan(\\frac{6371000}{7.04*10^{10}})=0.0103°=37.33''$$ To keep things simple, let's assume α and β were each 33.6/2=16.8 when time was measured correctly. Then β=16.8'' and $$α=37.33''-16.8''=20.53''=5.7*(10^{-3})°$$ Ok, now we've found α. Now we use it in the original equation to solve for Δt. $$\\tan(\\alpha)=\\frac{R_{E}+5660*\\Delta{t}}{d}$$ $$\\Delta{t}=\\frac{d*\\tan(\\alpha)-R_{E}}{5660}=252.8 s$$ Now, we could do the whole thing over again using d=8.60*10^10 in our equation to solve for α, and we would get Δt = -206.8 seconds. However, since we're assuming the first measurement is made at the time of opposition and the second one Δt seconds later, Δt can't be negative. Thus, the clocks must be synchronized to at least 252.8 seconds And I have a feeling there's an easier way to do this... Last edited: Sep 16, 2012", "= \\frac{y}{tan(\\alpha)}$ Inserting this value of x into the second equation gives: $tan(\\beta) = \\frac{y}{\\frac{y}{tan(\\alpha)} - d}$ $tan(\\beta) = \\frac{y}{\\frac{y}{tan(\\alpha)} - d} \\cdot \\frac{tan(\\alpha)}{tan(\\alpha)}$ $tan(\\beta) = \\frac{y~tan(\\alpha)}{y - d~tan(\\alpha)}$ Can you finish it from here? -Dan 3. Originally Posted by cocoknny y___ y-d What is this? Originally Posted by cocoknny is my top part right? I don't know because I don't know what \"htx\" is supposed to mean. -Dan 4. tan 26°50'=h 25m+x 5. ## hello Oh don't worry about that part. h suppose to be the height 6. Originally Posted by cocoknny tan 26°50'=h 25m+x It looks good if you have defined x as the distance from the second point to the tower. That's one equation. You need to use the other to eliminate x. -Dan 7. ## hello YUP that's what i did. Thanks hun", "$$\\frac {2\\sqrt {\\frac {25}4 - \\frac 41}}4$$ or, tan$$\\theta$$ =± $$\\frac {\\sqrt {\\frac {25 - 16}4}}2$$ or, tan$$\\theta$$ =± $$\\sqrt {\\frac 94}$$× $$\\frac 12$$ or, tan$$\\theta$$ =± $$\\frac 32$$× $$\\frac 12$$ or, tan$$\\theta$$ =± $$\\frac 34$$ ∴ $$\\theta$$ = tan-1(± $$\\frac 34$$) Hence, the acute angle ($$\\theta$$) =tan-1(± $$\\frac 34$$) Ans Here, Given equation is: 6x2 + 5xy - 3x - 2y - 6y2 = 0 or, 6x2 + 9xy - 4xy - 6y2 - 3x - 2y = 0 or, 3x (2x + 3y) - 2y (2x + 3y) - 1 (2x + 3y) = 0 or, (2x + 3y) (3x - 2y - 1) = 0 Either: 2x + 3y = 0..........................(1) Or: 3x - 2y - 1 = 0............................(2) Slope of equation (1), m1 = -$$\\frac 23$$ Slope of equation (2), m2 = $$\\frac 32$$ Again, m1× m2 = $$\\frac {-2}3$$× $$\\frac 32$$ = -1 The product of two slopes = -1 Hence, these equations are perpendicular each other. Hence, Proved 0% • 2x2 + 7xy + 3y2 = 0 2x+y =0 , x+3y = 0 2x+3 =0 , x+2y = 0 2x-3 =0 , x-3y = 0 2x-y =0 , x-3y = 0 • 2x2 - 5xy + 3y2 = 0. x-y =0 , 2x-3y =0 x+y =0 , 2x+3y =0 x-y =0 ,", "+ \\tan^{-1} v = \\tan^{-1} \\frac{u+v}{1-uv} \\ .$$ Therefore $$(\\dagger) = \\frac{1}{\\delta x}\\tan^{-1}\\frac{(x+ \\delta x) + (-x)}{1 - (x+\\delta x)(-x)}$$ $$= \\frac{1}{\\delta x}\\tan^{-1}\\frac{\\delta x}{1 + x^2 - x\\delta x}$$ $$\\rightarrow \\frac{1}{1+x^2} \\text{ by a small-angle approximation.}$$ • I didn't know that Tan(a+b)=Tan(A)+Tan(B)/(1-Tan(A)Tan(B) now it makes sense I'm only a freshman in high school and haven't learned this knowledge before This proof is amazing. Jul 13 '20 at 18:04 • Okay, I suppose this proof can be geometrical, because the tan formula can be proven using the corresponding formulae for sin and cos, which have geometrical proofs. en.m.wikipedia.org/wiki/… Jul 13 '20 at 18:10 • @Bladewood Done. How do you like it? Jul 14 '20 at 2:37 • @BenjaminWang It mostly looks good to me, but you want to avoid adding \"EDIT\" or \"IN RESPONSE TO EDIT\" in SE questions/answers because it feels more like a conversation thread than a Q/A. Jul 14 '20 at 14:31 Another proof comes from the atan addition/subtraction formula $$\\arctan(a)\\pm\\arctan(b) =\\arctan(\\frac{a\\pm b}{1\\mp ab}$$. Then, with a little hand-waving as $$h \\to 0$$ and assuming that $$\\lim_{z \\to 0} \\dfrac{\\arctan(z)}{z} =1$$, $$\\begin{array}\\\\ \\arctan(x+h)-\\arctan(x) &=\\arctan(\\dfrac{x+h-x}{1+x(x+h)})\\\\ &=\\arctan(\\dfrac{h}{1+x(x+h)})\\\\ &\\to\\arctan(\\dfrac{h}{1+x^2})\\\\ \\text{so}\\\\ \\dfrac{\\arctan(x+h)-\\arctan(x)}{h} &\\to\\dfrac{\\arctan(\\dfrac{h}{1+x^2})}{h}\\\\ &=\\dfrac1{1+x^2}\\dfrac{\\arctan(\\dfrac{h}{1+x^2})}{\\frac{h}{1+x^2}}\\\\ &\\to\\dfrac1{1+x^2}\\\\ \\end{array}$$ • this makes sense I can follow this. Jul 13 '20 at 20:42", "1. ## if tanx + tany = 25 and cotx + coty =30, what is tan(x+y)...please help with this que i have no idea how to do this.... 2. ## Re: if tanx + tany = 25 and cotx + coty =30, what is tan(x+y)...please help with this A good place to begin is to look at the angle sum identity for the tangent function: $\\tan(\\alpha+\\beta)=\\frac{\\tan(\\alpha)+\\tan(\\beta)} {1-\\tan(\\alpha)\\tan(\\beta)}$ Now we can see we have the numerator taken care of, so let's see if we can get what we need for the denominator from the given equation involving the cotangent function: $\\cot(x)+\\cot(y)=\\frac{1}{\\tan(x)}+\\frac{1}{\\tan(y) }=\\frac{\\tan(x)+\\tan(y)}{\\tan(x)\\tan(y)}$ Now we can get what we need, do you see how to proceed? 3. ## Re: if tanx + tany = 25 and cotx + coty =30, what is tan(x+y)...please help with this YES! I finally found this web page! I’ve been looking just for this article for so long!! Doudoune Burberry Femme Doudoune Ralph Lauren Homme Cheap North Face Jackets Belstaff Jackets Goyard Handbags Ugg Boots Outlet , , , , , , , ### tanx tany= 25 cotx coty= 30 Click on a term to search for related topics.", "knew I could do better. Fifteen, after all, is the difference between 60 and 45 - and I have those etched into my soul: $\\tan(60º)=\\sqrt{3}$, and $\\tan(45º)= 1$. So, I can use a trigonometric identity: $\\tan(A-B) \\equiv \\frac{\\tan(A) - \\tan(B)}{1+\\tan(A)\\tan(B)}$, with $A=60º$ and $B=45º$. That gives me $\\tan(15º) = \\frac{ \\sqrt{3} - 1}{1 + \\sqrt{3}}$. A younger, naiver me would have said “$\\sqrt{3} \\approx 1.732$, so I’ve got $\\frac{0.732}{2.732}$, which is… I dunno. I am older and wiser now; I know that $\\sqrt{3} \\approx \\frac{52}{30}$. That makes the fraction nicer to deal with: it’s $\\frac{\\frac{52}{30}-1}{\\frac{52}{30}+1} = \\frac{52-30}{52+30} = \\frac{11}{41}$. Great! I’ve improved my estimate of $\\tan(15º)$ marginally. Now all I need to do is multiply it by 200, which gives me $\\frac{2200}{41}$. That’s not especially easy, but I know it’s at least 50 (because $41 \\times 50 = 2050$), then I have 150 left over; $3\\times 41=123$, so I estimated 53.6, and added one to make it 54.6. Correct, to three significant figures! Two things here: I was slightly out on that last bit - $\\frac{2200}{41}$ is closer to 53.7 than 53.6, but my mistake worked in my favour! I love it when that hapens. The other thing is, I missed a trick! Had I rationalised the denominator, I’d have ended up with $\\tan(15º)=2-\\sqrt{3}$, which I can do", "# Math Help - Is tan(x + 90) = -cot(x)? 1. ## Is tan(x + 90) = -cot(x)? Is tan(x + 90) = -cot(x) ? 2. ## Re: Is tan(x + 90) = -cot(x)? Yes, $tan(x+ 90)= \\frac{sin(x+ 90)}{cos(x+ 90)}= \\frac{cos(x)}{-sin(x)}= -cot(x)$. (I am assuming that your \"x\" and \"90\" are in degrees since that is the only way this would be true.) 3. ## Re: Is tan(x + 90) = -cot(x)? By $\\tan(x+90)$, I assume you mean $\\tan(x+90^\\circ)$. Use the sum of angles formulas for sine and cosine. $\\tan(x+90^\\circ) = \\dfrac{\\sin(x+90^\\circ)}{\\cos(x+90^\\circ)}$ 4. ## Re: Is tan(x + 90) = -cot(x)? $tan(x+90) = -cot(x)$ $\\frac{tan(x) + tan(90)}{1-tan(x)tan(90)} = -\\frac{1}{tan(x)}$ $tan^2(x) + tan(x)tan(90) = tan(x)tan(90) - 1$ $tan(x) = \\sqrt{-1}$ Then what did I do wrong? 5. ## Re: Is tan(x + 90) = -cot(x)? $\\tan(90^\\circ)$ is not defined. That is why you were told to use the sum of angles formulas for sine and cosine, not the sum of angles formula for tangent." ]

[ "{45}^{\\circ}$.", "you get the $45^\\circ$ line.", "= 50^{\\circ}$?", "=60 ^\\circ 46'$", "$C \\approx {40.3}^{\\circ}$", "$^\\circ$C.", "is $55^\\circ$.", "\\circ$.", "\\circ h' =g$.", "and and slightly over $47^\\circ$.", "\\right )$ $= 45^{\\circ}$ Exams Articles Questions", "${45}^{\\circ}$ w.r.t $A B$ and $C B$", "K}}$ or $- {39.43}^{\\circ} \\text{C}$.", "\\cot x$$", "6^\\circ$ -", "^{\\circ}$", "$25^{\\circ}$", "both have proved that the value of $\\cot 45^\\circ$ is $1$. ###### Representation It is expressed in three different forms by considering the three angle measuring systems. It is frequently expressed in sexagesimal system as, $\\cot 45^° = 1$ It is also expressed in circular system as, $$\\cot \\Bigg(\\frac{\\pi}{4}\\Bigg) = 1$$ It is also expressed in centesimal system as, $\\cot 50^g = 1$", ")$ and $(5, 160^\\circ )$ .", "^ {\\ circ}}$" ]

[ "cot 37.4° $\\cot 37.4^o = \\frac 1{\\tan 37.4^o}$ now plug that into your calculator 5. ## hmmm im not getting the right answer.. i don't know why >_< i've been trying forever i can't seem to get the correct answer. im not getting the right answer.. i don't know why >_< i've been trying forever i can't seem to get the correct answer. for the first: $\\cot \\theta = 3.617945$ for the second, $\\cot 37.4^o = 1.3079457$ i don't know what to tell you, from here it's just plugging into your calculator. are you sure you're inputting it correctly? 7. ## yeah i have the correct answers here but i'm not plugging it in right. i don't know what i'm missing >_< how do i plug these in? i have the correct answers here but i'm not plugging it in right. i don't know what i'm missing >_< how do i plug these in? my calculator may not work the same way yours does: for the first, i entered: $\\boxed{1}~ \\boxed{ \\div }~ \\boxed{0} ~\\boxed{.}~ \\boxed{2}~ \\boxed{7} ~\\boxed{6}~ \\boxed{4}$ it looks like $1 \\div 0.2764$ on my display screen for the second i entered: $\\boxed{1} ~\\boxed{\\div} ~\\boxed{ (}~ \\boxed{ \\mbox{tan}}~ \\boxed{(}~ \\boxed{3}~ \\boxed{7}~ \\boxed{.} ~\\boxed{4}~ \\boxed{)} ~\\boxed{)}$ it looks like $1 \\div ( \\tan (37.4) )$ on my", "both have proved that the value of $\\cot 45^\\circ$ is $1$. ###### Representation It is expressed in three different forms by considering the three angle measuring systems. It is frequently expressed in sexagesimal system as, $\\cot 45^° = 1$ It is also expressed in circular system as, $$\\cot \\Bigg(\\frac{\\pi}{4}\\Bigg) = 1$$ It is also expressed in centesimal system as, $\\cot 50^g = 1$", "\\cot(90-x)$ $\\cot x = \\tan (90-x)$ The Next post in this series is here Note by Yan Yau Cheng 3 years, 8 months ago", "## Trigonometry (10th Edition) $\\cot$ 183$^{\\circ}$ 48' = 15.05572272 $\\cot$ 183$^{\\circ}$ 48' First, convert to a fraction $\\cot$ 183$^{\\circ}$ $\\frac{48}{60}$ Convert the fraction to a decimal $\\frac{48}{60}$ = 0.80 Therefore: $\\cot$ 183$^{\\circ}$ 48' = $\\cot$ 183.80$^{\\circ}$ $\\cot$ 183.80$^{\\circ}$ = $\\frac{1}{\\tan183.80}$ $\\frac{1}{\\tan183.80}$ = 15.05572272 $\\cot$ 183$^{\\circ}$ 48' = 15.05572272", "## anonymous one year ago If tan x = .88, what is cot x? 1.14 1.45 1.33 1.22 $$\\huge cot \\ x = \\frac{1}{tan \\ x}$$", "answer is already in lowest terms? There is nothing else to be done, right? Yes , if you want to go ahead put the value cot(20) $tan(20)=0.3640....... cot(20)=\\frac{1}{0.3640}=2.7475$ Here You can form a triangle joining the ship(C) to the base of statue of liberty(B) and its top of statue(A) which is again joined to the ship(C) $ AB = \\frac{BC}{tan(20)} = 305 cot(20) $ Where was the problem? No need to introduce cotangent here. Most calculators don't have a COT key, so why use the reciprocal when good ole TAN will do nicely? $\\tan20^{\\circ}=\\frac{305}{x}$ $x \\tan20^{\\circ}=305$ $x=\\frac{305}{\\tan 20^{\\circ}}$ $x\\approx 837.98$ feet from the statue. 6. ok but... Yes , if you want to go ahead put the value cot(20) $tan(20)=0.3640....... cot(20)=\\frac{1}{0.3640}=2.7475$ My calculator does not have a cotagent key. Thanks 7. thanks........... Originally Posted by masters No need to introduce cotangent here. Most calculators don't have a COT key, so why use the reciprocal when good ole TAN will do nicely? $\\tan20^{\\circ}=\\frac{305}{x}$ $x \\tan20^{\\circ}=305$ $x=\\frac{305}{\\tan 20^{\\circ}}$ $x\\approx 837.98$ feet from the statue. I thank you sincerely.", "x - 48 \\cot x = 39$ Completing square, $16 {\\cot}^{2} x - 48 \\cot x + 36 = 39 + 36 = 75$ $\\therefore {\\left(4 \\cot x - 6\\right)}^{2} = 75 \\Rightarrow 4 \\cot x - 6 = \\pm 5 \\sqrt{3}$ $\\text{Therefore, } \\cot x = \\frac{1}{4} \\left(6 \\pm 5 \\sqrt{3}\\right) , \\mathmr{and} , \\frac{3}{2} \\pm \\frac{5}{4} \\sqrt{3.}$ The above soln. was derived on the assumption that, $\\sin x \\ne 0.$ If, $\\sin x = 0 ,$ then, $\\cos x = \\pm 1 ,$ and sub.ing these in $\\left(\\ast\\right) ,$ we get, $\\pm 3 = 5$, which is impossible. Hence, $\\sin x \\ne 0$ holds good. Enjoy Maths.!", "## Trigonometry (10th Edition) $\\tan$ 45$^{\\circ}$ = -1 $\\cot$ 45$^{\\circ}$ = -1 $\\tan$ 45$^{\\circ}$ = $\\frac{1}{-1}$ = -1 $\\cot$ 45$^{\\circ}$ = $\\frac{-1}{1}$ = -1", "# Find solutions of $\\cot x-\\tan x=4\\cot(2x)=0$ for $x \\in (0°, 360°)$ Solve the equation $$\\cot x-\\tan x=4\\cot(2x)$$ for $0^\\circ<x<360^\\circ$ My attempt, $$\\frac{1}{\\tan x}-\\tan x=4(\\frac{1}{\\tan 2x})$$ $$\\frac{4(1-\\tan^2x)}{2\\tan x}-\\frac{1}{\\tan x}+\\tan x=0$$ $$2-2\\tan^2 x-1+\\tan^2x=0$$ $$1-\\tan^2x=0$$ $$\\tan x=\\pm1$$ $$x=45^\\circ,135^\\circ,225^\\circ,315^\\circ$$ So I've checked out with Desmos which I got It shows that the answer is $45$ and$135$. My questions: 1)Why my $225^\\circ$ and $315^\\circ$ are not included in the graph? Are they incorrect? 2)If I substitute the answers back to the equation, for example, $$4\\cot (2 \\cdot 45)$$ which is undefined. Why? • Cot (90)=0 no?... – Archis Welankar Jun 1 '17 at 15:52 • I guess I took cot x=1/tanx. It's defined when I take cotx=cosx/sinx. Why? – Mathxx Jun 1 '17 at 15:55", "Browse Questions # Find the value of $\\cot \\;( \\tan^{-1}a + \\cot^{-1}a)$ Toolbox: • $tan^{-1}x+cot^{-1}x=\\frac{\\pi}{2}$ for all real x • cot$\\frac{\\pi}{2}=0$ Substituting, $tan^{-1}x+cot^{-1}x=\\frac{\\pi}{2}$, $cot ( tan^{-1}a + cot^{-1}a) = cot \\frac{\\pi}{2}$. We know that cot$\\frac{\\pi}{2}=0$, hence $cot ( tan^{-1}a + cot^{-1}a) = 0$. edited Mar 14, 2013", "= \\cos 45^\\circ.$$ Therefore, we can express the desired value as $$1 + \\frac{\\sin 45^\\circ + \\sin 20^\\circ \\sin 25^\\circ}{\\cos 45^\\circ + \\sin 20^\\circ \\sin 25^\\circ}.$$ Because $\\cos 45^\\circ = \\sin 45^\\circ$, we see that the fractional part is $1$, and so the sum is $1 + 1 = 2$, which brings us to the answer $\\boxed{B}$.", "\\pm {42.0038}^{\\circ} + {360}^{\\circ} k \\quad$integer $k$", "## anonymous one year ago If tan x = .88, what is cot x? 1.14 1.45 1.33 1.22 1. IrishBoy123 $$\\huge cot \\ x = \\frac{1}{tan \\ x}$$ Find more explanations on OpenStudy", "# Sum of $50$ terms of $\\cot^{-1}3+\\cot^{-1}7+\\cot^{-1}13+\\cot^{-1}21+.....$ [duplicate] Find the sum of the first $$50$$ terms of the series $$\\cot^{-1}3+\\cot^{-1}7+\\cot^{-1}13+\\cot^{-1}21+.....$$ $$\\sum_1^{50}=\\cot^{-1}3+\\cot^{-1}7+\\cot^{-1}13+\\cot^{-1}21+.....\\\\ =\\tan^{-1}\\frac{1}{3}+\\tan^{-1}\\frac{1}{7}+\\tan^{-1}\\frac{1}{13}+\\tan^{-1}\\frac{1}{21}+.....=$$ My reference gives the solution $$\\tan^{-1}\\dfrac{5}{6}$$, but I do not have any clue of doing it ? Note: I know that $$\\tan^{-1}x+\\tan^{-1}y=\\tan^{-1}\\dfrac{x+y}{1-xy}$$ if $$xy<1$$. • Note that $$\\tan^{-1} x+\\tan^{-1}y=\\tan^{-1}\\frac{x+y}{1-xy}$$ Dec 15, 2018 at 5:33 • what function is the sequence $3,7,13,21,...$ given by? How are we supposed to find the series if we don't know how to write it in $\\sum$ notation? Dec 15, 2018 at 5:36 • @clathratus, I assume $a_1=3, a_n=a_{n-1}+2n$. Dec 15, 2018 at 5:41 • $a_n=1+n+n^2$, I guess Dec 15, 2018 at 5:45 • Find the sum of the first $M$ terms, for $M=1,2,3,4,5$ Look for a formula that works for those, and use induction to prove it. Dec 15, 2018 at 5:51 • Use the fact that $$\\cot^{-1}(x)=\\tan^{-1}\\frac{1}{x}$$ • $$\\tan^{-1}\\left(\\frac{1}{n^2+n+1}\\right) = \\tan^{-1}\\left(\\frac{(n+1)-n}{1+n(n+1)}\\right) =\\tan^{-1}(n+1) -\\tan^{-1}(n)$$", "1 + 21i$, we are left with $$10\\cot(\\frac{\\pi}{2} - \\arg w + \\frac{\\pi}{2} - \\arg x + \\frac{\\pi}{2} - \\arg y + \\frac{\\pi}{2} - \\arg z) = 10\\cot(2\\pi - \\arg wxyz)$$ $$= -10\\cot(\\arg wxyz).$$ Expanding $wxyz$, we are left with $$(3+i)(7+i)(13+i)(21+i) = (20+10i)(13+i)(21+i)$$ $$= (2+i)(13+i)(21+i)$$ $$= (25+15i)(21+i)$$ $$= (5+3i)(21+i)$$ $$= (102+68i)$$ $$= (3+2i)$$ $$= 10\\cot \\tan^{-1}\\frac{2}{3}$$ $$= 10 \\cdot \\frac{3}{2} = \\boxed{015}$$", "x \\tan x+\\cot x$$ 13. $$\\dfrac{1}{1−\\sin x}−\\dfrac{1}{1+\\sin x}=2\\tan x \\sec x$$ 14. $$\\csc^4 x−\\cot^4 x=\\csc ^2 x+\\cot^2x$$ 15. $$(\\sin x−\\tan x)(\\cos x−\\cot x)=(\\sin x−1)(\\cos x−1)$$", "# How do you evaluate cot((2pi)/3)? Apr 4, 2016 $1.402$ #### Explanation: Evaluate $\\frac{2 \\pi}{3}$ where $\\pi = {180}^{o}$, which gives a value of ${120}^{o}$. Also know that $\\cot = \\frac{1}{\\tan}$. Therefore, $\\cot \\left(\\frac{2 \\pi}{3}\\right) = \\frac{1}{\\tan \\left(120\\right)} = \\frac{1}{0.713} = 1.402$ Apr 5, 2016 $- \\frac{\\sqrt{3}}{3}$ #### Explanation: Trig table --> $\\tan \\left(\\frac{2 \\pi}{3}\\right) = - \\sqrt{3}$ Therefor, $\\cot \\left(\\frac{2 \\pi}{3}\\right) = \\frac{1}{\\tan} = - \\frac{1}{\\sqrt{3}} = - \\frac{\\sqrt{3}}{3}$", "# Finding the value of cot 142.5° I have seen a few solutions but I can't apply them to this particular question. The question is: Prove that $\\cot 142.5° = \\sqrt{2} + \\sqrt{3} - 2 - \\sqrt{6}$ Help would be really appreciated. • – lab bhattacharjee Jan 13 '18 at 17:41 • I have seen this perticular solution... But I was wondering how to apply that in this case – Le Connoisseur Jan 13 '18 at 17:49 • @LeConnoisseur: so you are basically asking what is the relation between $\\cot 7.5^\\circ$ and $\\cot (150-7.5)^\\circ$, which should be simple to grasp. – Jack D'Aurizio Jan 13 '18 at 17:51 • So, $$\\cot142.5^\\circ=\\dfrac{1+\\cos285^\\circ}{\\sin285^\\circ}$$ $$285=360-75,75=45+30$$ – lab bhattacharjee Jan 13 '18 at 17:51 use $$\\cot(2x)=\\frac{1}{2}\\frac{\\cot(x)-1}{\\cot(x)}$$ and $$\\cot(285^{\\circ})=\\sqrt{3}-2$$", "Now, $\\frac{\\cot(\\gamma)}{\\cot(\\beta)+\\cot(\\alpha)}=\\frac{\\frac{994c}{b\\sin a}}{\\frac{a^2-994c^2+b^2-994c^2}{bc\\sin(a)}}$. After using $a^2+b^2=1989c^2$, we get $\\frac{994c*bc\\sin a}{c^2b\\sin a}=\\boxed{994}$.", "# How do you find the exact value of sec30cot45? Jun 20, 2018 $\\frac{2 \\sqrt{3}}{3}$ #### Explanation: $= \\sec 30 \\cot 45$ $= \\frac{1}{\\cos} 30 \\cdot \\frac{1}{\\tan} 45$ use the unit circle to find exact values: $\\cos 30 = \\frac{\\sqrt{3}}{2}$ $\\frac{1}{\\cos} 30 = \\frac{2}{\\sqrt{3}}$ $\\tan 45 = 1$ $\\frac{1}{\\tan} 45 = \\frac{1}{1} = 1$ $= \\sec 30 \\cot 45 = \\frac{2}{\\sqrt{3}} \\cdot 1 = \\frac{2}{\\sqrt{3}} = \\frac{2 \\sqrt{3}}{3}$" ]

[ "34^o + cos^2 \\ 34^o + 2 tan \\ 18^o. cot \\ 18^o - cot^2 \\ 30^o$ $= 3 - (\\sqrt{3})^2 = 0$ $\\\\$", "cot 37.4° $\\cot 37.4^o = \\frac 1{\\tan 37.4^o}$ now plug that into your calculator 5. ## hmmm im not getting the right answer.. i don't know why >_< i've been trying forever i can't seem to get the correct answer. im not getting the right answer.. i don't know why >_< i've been trying forever i can't seem to get the correct answer. for the first: $\\cot \\theta = 3.617945$ for the second, $\\cot 37.4^o = 1.3079457$ i don't know what to tell you, from here it's just plugging into your calculator. are you sure you're inputting it correctly? 7. ## yeah i have the correct answers here but i'm not plugging it in right. i don't know what i'm missing >_< how do i plug these in? i have the correct answers here but i'm not plugging it in right. i don't know what i'm missing >_< how do i plug these in? my calculator may not work the same way yours does: for the first, i entered: $\\boxed{1}~ \\boxed{ \\div }~ \\boxed{0} ~\\boxed{.}~ \\boxed{2}~ \\boxed{7} ~\\boxed{6}~ \\boxed{4}$ it looks like $1 \\div 0.2764$ on my display screen for the second i entered: $\\boxed{1} ~\\boxed{\\div} ~\\boxed{ (}~ \\boxed{ \\mbox{tan}}~ \\boxed{(}~ \\boxed{3}~ \\boxed{7}~ \\boxed{.} ~\\boxed{4}~ \\boxed{)} ~\\boxed{)}$ it looks like $1 \\div ( \\tan (37.4) )$ on my", "both have proved that the value of $\\cot 45^\\circ$ is $1$. ###### Representation It is expressed in three different forms by considering the three angle measuring systems. It is frequently expressed in sexagesimal system as, $\\cot 45^° = 1$ It is also expressed in circular system as, $$\\cot \\Bigg(\\frac{\\pi}{4}\\Bigg) = 1$$ It is also expressed in centesimal system as, $\\cot 50^g = 1$", "# Find solutions of $\\cot x-\\tan x=4\\cot(2x)=0$ for $x \\in (0°, 360°)$ Solve the equation $$\\cot x-\\tan x=4\\cot(2x)$$ for $0^\\circ<x<360^\\circ$ My attempt, $$\\frac{1}{\\tan x}-\\tan x=4(\\frac{1}{\\tan 2x})$$ $$\\frac{4(1-\\tan^2x)}{2\\tan x}-\\frac{1}{\\tan x}+\\tan x=0$$ $$2-2\\tan^2 x-1+\\tan^2x=0$$ $$1-\\tan^2x=0$$ $$\\tan x=\\pm1$$ $$x=45^\\circ,135^\\circ,225^\\circ,315^\\circ$$ So I've checked out with Desmos which I got It shows that the answer is $45$ and$135$. My questions: 1)Why my $225^\\circ$ and $315^\\circ$ are not included in the graph? Are they incorrect? 2)If I substitute the answers back to the equation, for example, $$4\\cot (2 \\cdot 45)$$ which is undefined. Why? • Cot (90)=0 no?... – Archis Welankar Jun 1 '17 at 15:52 • I guess I took cot x=1/tanx. It's defined when I take cotx=cosx/sinx. Why? – Mathxx Jun 1 '17 at 15:55", "If tan A=34 and A+B=90°, then what is the value of cot B? Question: If $\\tan A=\\frac{3}{4}$ and $A+B=90^{\\circ}$, then what is the value of $\\cot \\mathrm{B} ?$ Solution: Given that: $A+B=90^{\\circ}$ $\\tan A=\\frac{3}{4}$ $A+B=90^{\\circ}$ $\\Rightarrow B=90^{\\circ}-A$ $\\Rightarrow \\cot B=\\cot \\left(90^{\\circ}-A\\right)$ $\\Rightarrow \\cot B=\\tan A$ $\\Rightarrow \\cot B=\\frac{3}{4}\\left[\\cot \\left(90^{\\circ}-A\\right)=\\tan A\\right]$ Hence the value of $\\cot B$ is $\\frac{3}{4}$", "\\cot(90-x)$ $\\cot x = \\tan (90-x)$ The Next post in this series is here Note by Yan Yau Cheng 3 years, 8 months ago", "x - 48 \\cot x = 39$ Completing square, $16 {\\cot}^{2} x - 48 \\cot x + 36 = 39 + 36 = 75$ $\\therefore {\\left(4 \\cot x - 6\\right)}^{2} = 75 \\Rightarrow 4 \\cot x - 6 = \\pm 5 \\sqrt{3}$ $\\text{Therefore, } \\cot x = \\frac{1}{4} \\left(6 \\pm 5 \\sqrt{3}\\right) , \\mathmr{and} , \\frac{3}{2} \\pm \\frac{5}{4} \\sqrt{3.}$ The above soln. was derived on the assumption that, $\\sin x \\ne 0.$ If, $\\sin x = 0 ,$ then, $\\cos x = \\pm 1 ,$ and sub.ing these in $\\left(\\ast\\right) ,$ we get, $\\pm 3 = 5$, which is impossible. Hence, $\\sin x \\ne 0$ holds good. Enjoy Maths.!", "Browse Questions # Find the value of $\\cot \\;( \\tan^{-1}a + \\cot^{-1}a)$ Toolbox: • $tan^{-1}x+cot^{-1}x=\\frac{\\pi}{2}$ for all real x • cot$\\frac{\\pi}{2}=0$ Substituting, $tan^{-1}x+cot^{-1}x=\\frac{\\pi}{2}$, $cot ( tan^{-1}a + cot^{-1}a) = cot \\frac{\\pi}{2}$. We know that cot$\\frac{\\pi}{2}=0$, hence $cot ( tan^{-1}a + cot^{-1}a) = 0$. edited Mar 14, 2013", "## Trigonometry (10th Edition) $\\cot$ 183$^{\\circ}$ 48' = 15.05572272 $\\cot$ 183$^{\\circ}$ 48' First, convert to a fraction $\\cot$ 183$^{\\circ}$ $\\frac{48}{60}$ Convert the fraction to a decimal $\\frac{48}{60}$ = 0.80 Therefore: $\\cot$ 183$^{\\circ}$ 48' = $\\cot$ 183.80$^{\\circ}$ $\\cot$ 183.80$^{\\circ}$ = $\\frac{1}{\\tan183.80}$ $\\frac{1}{\\tan183.80}$ = 15.05572272 $\\cot$ 183$^{\\circ}$ 48' = 15.05572272", "Find $\\quad\\cot^2(2^{\\circ})+\\cot^2(6^{\\circ})+\\cot^2({10}^{\\circ})++\\cdots +\\cot^2({86}^{\\circ})$ Find $$\\cot^2(2^{\\circ})+\\cot^2(6^{\\circ})+\\cot^2({10}^{\\circ})++\\cdots +\\cot^2({86}^{\\circ})$$ $\\mathbf {My Attempt}$ I tried to write the sum backward like this $$S=\\sum_{n=1}^{22} \\cot^2(4n-2) = \\sum_{n=1}^{22} \\cot^2({90}^{\\circ}-4n) = \\sum_{n=1}^{22} \\tan^2(4n)$$ $S=990$ But still can't find any good trigonometry identity to form telescopic series or something like that. Any hint? • – lab bhattacharjee Jul 18 '18 at 1:35 • @lab many thx. I seem I've overlooked this post. I can adapt it now. – Wolfdale Jul 18 '18 at 1:59", "$\\mathrm{cot}41°.\\mathrm{cot}42°.\\mathrm{cot}43°.\\mathrm{cot}44°.\\mathrm{cot}45°.\\mathrm{cot}46°.\\mathrm{cot}47°.\\mathrm{cot}48°.\\mathrm{cot}49°$ is equals to Q.20 If ${\\mathrm{tan}}^{2}B=\\frac{1-\\mathrm{sin}A}{1+\\mathrm{sin}A}$ ,the what will be the value A + 2B Scroll", "## Trigonometry (10th Edition) $\\tan$ 45$^{\\circ}$ = -1 $\\cot$ 45$^{\\circ}$ = -1 $\\tan$ 45$^{\\circ}$ = $\\frac{1}{-1}$ = -1 $\\cot$ 45$^{\\circ}$ = $\\frac{-1}{1}$ = -1", "\\pm {42.0038}^{\\circ} + {360}^{\\circ} k \\quad$integer $k$", "Find $$\\hat{N}$$. $\\hat{N} = 83^{\\circ}$", "# Find the values of each of the following: 12. $\\cot(\\tan^{-1}a + \\cot^{-1}a)$ We have to find the value of $\\cot(\\tan^{-1}a + \\cot^{-1}a)$ As we know $\\left [\\because \\tan^{-1}x + \\cot^{-1} x = \\frac{\\pi}{2} \\right ]$ so, Equation reduces to $\\cot(\\frac{\\pi}{2}) = 0$.", "# How do you find the exact value of sec30cot45? Jun 20, 2018 $\\frac{2 \\sqrt{3}}{3}$ #### Explanation: $= \\sec 30 \\cot 45$ $= \\frac{1}{\\cos} 30 \\cdot \\frac{1}{\\tan} 45$ use the unit circle to find exact values: $\\cos 30 = \\frac{\\sqrt{3}}{2}$ $\\frac{1}{\\cos} 30 = \\frac{2}{\\sqrt{3}}$ $\\tan 45 = 1$ $\\frac{1}{\\tan} 45 = \\frac{1}{1} = 1$ $= \\sec 30 \\cot 45 = \\frac{2}{\\sqrt{3}} \\cdot 1 = \\frac{2}{\\sqrt{3}} = \\frac{2 \\sqrt{3}}{3}$", "answer is already in lowest terms? There is nothing else to be done, right? Yes , if you want to go ahead put the value cot(20) $tan(20)=0.3640....... cot(20)=\\frac{1}{0.3640}=2.7475$ Here You can form a triangle joining the ship(C) to the base of statue of liberty(B) and its top of statue(A) which is again joined to the ship(C) $ AB = \\frac{BC}{tan(20)} = 305 cot(20) $ Where was the problem? No need to introduce cotangent here. Most calculators don't have a COT key, so why use the reciprocal when good ole TAN will do nicely? $\\tan20^{\\circ}=\\frac{305}{x}$ $x \\tan20^{\\circ}=305$ $x=\\frac{305}{\\tan 20^{\\circ}}$ $x\\approx 837.98$ feet from the statue. 6. ok but... Yes , if you want to go ahead put the value cot(20) $tan(20)=0.3640....... cot(20)=\\frac{1}{0.3640}=2.7475$ My calculator does not have a cotagent key. Thanks 7. thanks........... Originally Posted by masters No need to introduce cotangent here. Most calculators don't have a COT key, so why use the reciprocal when good ole TAN will do nicely? $\\tan20^{\\circ}=\\frac{305}{x}$ $x \\tan20^{\\circ}=305$ $x=\\frac{305}{\\tan 20^{\\circ}}$ $x\\approx 837.98$ feet from the statue. I thank you sincerely.", "Browse Questions # Find the value of $\\cot(\\large\\frac{13\\pi}{4})$ $\\begin{array}{1 1}(A)\\;1&(B)\\;\\large\\frac{1}{2}\\\\(C)\\;\\large\\frac{1}{4}&(D)\\;2\\end{array}$ Toolbox: • $\\cot (\\theta+n\\pi)=\\cot \\theta$ $\\cot (\\large\\frac{13\\pi}{4})$ $\\cot (3\\pi+\\large\\frac{\\pi}{4})$$=\\cot \\large\\frac{\\pi}{4}$$=1$ Hence (A) is the correct answer.", "cot from both sides, we get ${90^ \\circ } - 2A = A - {18^ \\circ }$ $\\Rightarrow$ $3A = {108^ \\circ }$ $\\Rightarrow$ $A = \\dfrac{{{{108}^ \\circ }}}{3}$ $\\Rightarrow$ $A = {36^ \\circ }$ Thus, the value of A is ${36^ \\circ }$. Note: Whenever we face such types of questions, the key concept is that we must covert tan in terms of cot or vice versa. It is clearly visible that here $2A$ represents an acute angle. First we will use the identity $\\tan \\theta = \\cot \\left( {{{90}^ \\circ } - \\theta } \\right)$ then eliminate cot (or tan) then by simplifying the equations we will get our required answer.", "# What is arc cot(-2)? Calculator --> tan x = -1/2 --> arc $x = - {26}^{\\circ} 56$ and $x = 180 - 26.56 = {153}^{\\circ} 43$" ]

[ "mypara=revolution(O,g,X); Full code: copy to http://asymptote.ualberta.ca/ and click Run unitsize(1cm); import graph3; import solids; //currentprojection=orthographic(2,1.5,.5,zoom=.95); currentprojection=obliqueZ; currentprojection.zoom=.95; size(8cm); real a=1.8, b=a^2; real c=2.5,d=4.5; triple A=(a,0,0), B=(0,b,0); triple C=(c,0,0), D=(0,d,0); // the (red) parabol curve on the XY-plane triple paraXY(real t){return (t,t^2,0);} path3 g=graph(paraXY,0,a); draw(g,red+1.2pt); // use one of the following revolutions //revolution mypara=revolution(C,g,Y); // around x=c //revolution mypara=revolution(D,g,X); // around y=d //revolution mypara=revolution(O,g,X); // around y=0 revolution mypara=revolution(O,g,Y); // around x=0 draw(surface(mypara),yellow+opacity(.7)); //draw(mypara,blue,orange); label(\"$a$\",A,S); label(\"$c$\",C,S); label(\"$b$\",B,W); label(\"$d$\",D,W); draw(A--(a,b,0)--B^^C--(c,d,0)--D,dashed+gray); draw(Label(\"$x$\",EndPoint),-2*X--6*X,Arrow3()); draw(Label(\"$y$\",EndPoint),-5*Y--6*Y,Arrow3()); draw(Label(\"$z$\",EndPoint),O--6*Z,Arrow3()); // you may not want to draw Oz 2. The same code, only change X to Y to get revolution around Oy revolution mypara=revolution(O,g,Y); 3. The same code, for revolution around x=c revolution mypara=revolution(C,g,Y); 4. Last, for revolution around y=d revolution mypara=revolution(D,g,X); Welcome to TeX.SE!!! This is a quick modification of an answer from here. It's a 2d drawing using isometric perspective. The equations of parabola and ellipses are calculated by hand. The ellipses are very simple if you know about isometric perspective (the axes ratio is sqrt(3):1). The parabola was obtained as an envelope of the family of 'horizontal' ellipses. This is the code: \\documentclass[tikz,border=2mm]{standalone} \\tikzset {% axes/.style={thick,-latex}, cylinder/.style={right color=blue!80,left color=white,fill opacity=0.6}, paraboloid/.style={left color=magenta!80,fill opacity=0.6}, } \\begin{document} \\begin{tikzpicture}[line cap=round,line join=round] % axes x,y \\draw[axes] (30:3) -- (210:3) node [left] {$x$}; \\draw[axes] (150:3) -- (330:3) node [right]", "SYSTEM label(\"$O$\", (0, 0, 0), NW); Label Lx = Label(\"$x$\", EndPoint, W); Label Ly = Label(\"$y$\", EndPoint, S); Label Lz = Label(\"$z$\", EndPoint, W); draw(Lx, O--2X, Arrow3(emissive(black))); draw(Ly, O--2Y, Arrow3(emissive(black))); draw(Lz, O--2Z, Arrow3(emissive(black))); // SURFACE triple S(pair uv) { real x = cos(uv.x)*sin(uv.y); real y = sin(uv.x)*sin(uv.y); real z = cos(uv.y); return (x-(1-z**2-y**2), y-0.3*sin(z*pi), z); } // BOUNDARY triple dS(real u) { return S((u, pi/2)); } // TANGENT VECTORS real e = 0.001; // *this... is... infinitesimal!* triple U(pair t) { return (S((e,0)+t)-S(t))/e; } triple V(pair t) { return (S((0,e)+t)-S(t))/e; } // NORMAL VECTOR triple N(pair uv) { return cross(U(uv),V(uv))/length(cross(U(uv),V(uv))); } // GRAPHICAL ELEMENTS surface s = surface(S, (0, 0), (2pi, pi/2), 64, 64, Spline); path3 ds = graph(dS, 0, 2pi); pair t = (-pi/8, pi/8); path3 n = S(t) -- shift(-N(t))*S(t); label(\"$\\Sigma$\", shift(0,0,0.1)*shift(adj)*S((pi/2,pi/5)), NE); Label LdS = Label(\"$\\partial\\Sigma$\", Relative(0.95), S); Label Ln = Label(\"$\\mathbf n$\", EndPoint, NW); draw(s, surfacepen=material(diffusepen=gray(0.6),emissivepen=gray(0.3),specularpen=gray(0.1))); draw(LdS, ds, Arrow3(emissive(black))); draw(shift(4*unit(currentprojection.camera))*ds, dashed); draw(Ln, n, Arrow3(emissive(black))); REQUESTED APPENDIX: on dashed lines I wanted to draw the dashed border over the surface. Since this is an orthographic projection, the easiest way is translating the object along the camera axis towards it (instead of checking what segments are visible, possibly complicated): draw(shift(4*unit(currentprojection.camera))*ds, dashed); unit(currentprojection.camera) is a versor parallel to the camera axis and pointing towards it, so it", "# Asymptote: 3d graph over a disc Is there a straightforward way to draw a 3D graph over a disc domain? Say z=x^2-y^2 for x^2+y^2<1. [I just started to use asymptote; this page explained me how to do it for a rectangular domain. I hope it is an easy question.] One way to make sure that x^2+y^2<1 is to use polar coordinates. Then x=r cos(phi) and y=r sin(phi). \\documentclass[variwidth,border=3.14mm]{standalone} \\usepackage{asypictureB} \\begin{document} \\begin{asypicture}{name=discgraph} usepackage(\"mathrsfs\"); import graph3; import solids; import interpolate; settings.outformat=\"pdf\"; size(500); defaultpen(0.5mm); pen darkgreen=rgb(0,138/255,122/255); draw(Label(\"$x$\",1),(0,0,0)--(1.2,0,0),darkgreen,Arrow3); draw(Label(\"$y$\",1),(0,0,0)--(0,1.2,0),darkgreen,Arrow3); draw(Label(\"$f(x,y)$\",1),(0,0,0)--(0,0,0.6),darkgreen,Arrow3); //function: call the radial coordinate r=t.x and the angle phi=t.y triple f(pair t) { return ((t.x)*cos(t.y), (t.x)*sin(t.y), ((t.x)*cos(t.y))^2-((t.x)*sin(t.y))^2); } surface s=surface(f,(0,1),(0.49,2.5*pi),32,16, usplinetype=new splinetype[] {notaknot,notaknot,monotonic}, vsplinetype=Spline); pen p=rgb(0,0,.7); draw(s,lightolive+white); \\end{asypicture} \\end{document} • Thank you, but is there a direct way to make a condition x^2+y^2<1 for the arguments? – Anton Petrunin Apr 1 '19 at 4:31 • @marmot: The x-axis near origin should be hidden from the given point of view. Is there any way to improve this issue? E.g., by setting some samples-option? – Marian G. Apr 1 '19 at 5:28 • A line has a thickness, a surface not. It is why you see the x-axis near origin. You can observe the same behavior with a simple square surface and the x-axis. Perhaps it is possible to avoid its by", "looked at the extreme cases (i.e., $$S = emptyset$$ and $$S = V$$). When $$S = V$$, the heuristic proposed in the previous post obviously works. When $$S = emptyset$$, the total time taken is fixed since we cannot exercise the option to use a helper vehicle. I was wondering whether there was an approximation scheme or heuristic that might be efficient in this general case. Get this bounty!!! ## #StackBounty: #algorithms #graphs #monte-carlo Devise Mont Carlo and Las Vegas Algorithms to Solve Maximum Independent Set ### Bounty: 50 I am trying to devise a Las Vegas algorithm to solve Maximum Independent Set, but I don’t know how to start. Also, I want to devise a Mont Carlo algorithm for this problem. I would appreciate any help. Thank you. Get this bounty!!! ## #StackBounty: #graphs #asymptote How to graph the inverse trigonometric function (arctan, arccotan) in asymptote? ### Bounty: 100 A want to add the inverse trig.function for tangens to the code of asymptote. Thank you for adivise. ``````settings.outformat=\"pdf\"; unitsize(2.5cm); import graph; real xmin = -pi/2; real xmax = pi/2; real ymin = -pi/2; real ymax = pi/2; xlimits(xmin,xmax); ylimits(ymin,ymax); //axis and the grid : xaxis(Label(\"\\$x\\$\",align=2E),Ticks(\"\\$%.2f\\$\",new real[]{xmin,-1,1,xmax},Size=1mm,1bp),Arrow); yaxis(Label(\"\\$y\\$\",align=2N),Ticks(\"\\$%.2f\\$\",new real[]{xmin,-1,1,xmax},Size=1mm,1bp),Arrow); // function y=sin(x) real f(real x) {return sin(x)/cos(x); } path g = graph(f,xmin+0.5, xmax-0.5); draw(g,red, L= Label(\"\\$y=tan(x)\\$\", UnFill, position=EndPoint));", "prime factors $2,3, \\text{and } 5$. $[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label(\"x\",X,N); label(\"y\",Y,SW); label(\"z\",Z,SE); label(\"2^3\",X--Y,2W); label(\"2^3\",X--Z,2E); label(\"2^2\",Y--Z,2S); [/asy]$ Analyzing for powers of $2$, it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \\text{ nor } z$ can have a power of $2$ exceeding $2$. Turning towards the vertices $y$ and $z$, we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \\text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \\{(2,0)(2,1)(0,2)(1,2)(2,2)\\}$. $[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label(\"x\",X,N); label(\"y\",Y,SW); label(\"z\",Z,SE); label(\"3^2\",X--Y,2W); label(\"3^1\",X--Z,2E); label(\"3^2\",Y--Z,2S); [/asy]$ Using the same logic as we did for powers of $2$, it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \\text{ and } z$, we can use the same logic to find the only ordered pairs $(x,z)$", "– 3y + 6 = 0 4. g(x) = sin(x) 5. y = x^3 – 1 Notes: • You can type linear equations in the following forms: y = ax + b, f(x) = a(x) + b or ax + by + c = 0. • The * is used in multiplication and ^ is used in exponentiation. For example you want to graph, y = 2(x – 3)2, then you should enter y = 2*(x-3)^2. Properties of Graph You can change the labels, colors, thickness and other properties of graphs (and other objects) in GeoGebra. In this tutorial we are going to change the color, label and thickness of the graph g(x) = sin(x). To change the properties of the graph g(x) = sin(x), do the following: 1.) Right click the graph of the sine function then click Object Properties from the context menu. Figure 2 - The context menu that appears when you right-click a graph 2.) In the Basic tab, be sure that the Show label check box is checked. 3.) Choose Name and Value from the Show label drop-down list box. Figure 3 - The Basic tab of the Properties dialog box 4.) To change the color, click the Color tab, then choose your color from the Color palette. 5.) To change the thickness", "$g[5] = \"sin(0.5*pi*$F*t) for t in <-1,1> using color:blue and weight:2\"; $graph[0] = init_graph(-1,-2,1,2,'axes'=>[0,0],'ticks'=>[8,8] );$graph[1] = init_graph(-1,-2,1,2,'axes'=>[0,0],'ticks'=>[8,8] ); $graph[2] = init_graph(-1,-2,1,2,'axes'=>[0,0],'ticks'=>[8,8] );$graph[3] = init_graph(-1,-2,1,2,'axes'=>[0,0],'ticks'=>[8,8] ); $graph[4] = init_graph(-1,-2,1,2,'axes'=>[0,0],'ticks'=>[8,8] );$graph[5] = init_graph(-1,-2,1,2,'axes'=>[0,0],'ticks'=>[8,8] ); for ($j = 0;$j <=5; $j++) {$graph[$j]->lb('reset');$graph[$j]->lb(new Label(-.07,-1,-1,'black','right','middle'));$graph[$j]->lb(new Label(-.07,1,1,'black','right','middle')); for ($i = -3; $i <= 3;$i++) { if ($i != 0) {$graph[$j]->lb(new Label(0.25*$i,-.2,0.25*$i,'black','center','top')) }};$graph[$j]->lb(new Label(-.05,1.9,\"y\",'black','right','top'));$graph[$j]->lb(new Label(0.95,0.1,\"t\",'black','right','bottom')); plot_functions($graph[$j],$g[$j]);$fig[$j] = image(insertGraph($graph[$j]),width => 240,height => 180,tex_size => 200); }; #$pick = random(0,3,1); # if ( $pick != 0 ) {$temp_eq = $eqn[0]; #$temp_gr = $fig[0]; #$eq[0] = $eq[$pick]; # $fig[0] =$fig[$pick]; #$eq[$pick] =$temp_eq; # $fig[$pick] = $temp_gr};$mc = new_multiple_choice(); $mc->qa('On a separate piece of paper, sketch an accurate graph of this function for $$F = F$$ and $$t \\in [-1, 1]$$. Which (if any) of the graphs below matches the graph you drew?','$fig[0]'); $mc->extra('$fig[1] $BR$BITALIC(click on image to enlarge)$EITALIC', '$fig[2] $BR$BITALIC(click on image to enlarge)$EITALIC', '$fig[3] $BR$BITALIC(click on image to enlarge)$EITALIC', '$fig[4] $BR$BITALIC(click on image to enlarge)$EITALIC', '$fig[5] $BR$BITALIC(click on image to enlarge)$EITALIC');$mc->makeLast('None of the above'); ## force a refresh of the image after changes $refreshCachedImages = 1; Context()->texStrings; BEGIN_TEXT This problem reflects Problem 1.21a in the text$PAR Consider the function $BR$BR $$y = func$$. $BR$BR \\{ $mc->print_q() \\}$BR \\{ $mc->print_a() \\} END_TEXT Context()->normalStrings; ANS(radio_cmp($mc->correct_ans)); ## force a refresh of the image after changes $refreshCachedImages = 1; Context()->texStrings; SOLUTION(EV3(<<'END_SOLUTION'));$PAR", "x) {return sin(x);} filldraw(graph(f2,0,pi )--(pi ,0)--(0,0)--cycle, fueaev, zzttqq); Some commands have to be defined with respect to a function. $\\texttt{f2}$ returns the $\\sin(x)$ function, and the $\\texttt{graph}$ command draws $\\texttt{f2}$ from $x = 0$ to $x = \\pi$. Label laxis; laxis.p = fontsize(10); string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); These commands draw the axes. The $\\texttt{laxis}$ label specifies the label styles of the axes (e.g., the field $\\texttt{laxis.p}$ defines the desired font pen). The next line, the string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} are functions that define the labels for both axes. These are abbreviated as $\\texttt{xlbl, ylbl}$ in the concise mode. The first defines axes with labels with respect to $\\pi$, and it formats the labels so that the $0$ label simply outputs $0$ instead of $0\\pi$, and $\\pm \\pi$ instead of $\\pm 1\\pi$. Finally, xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); are the commands that actually construct the xaxis and the yaxis. For documentation, see here. The next block of code does the actual construction of the", "real f1(real x) {return a1*sin(x/2*pi);} real f2(real x) {return a2*sin(x/4*pi);} draw(pic1,graph(pic1,f1,xmin1,xmax1)); draw(pic2,graph(pic2,f2,xmin2,xmax2)); xaxis(pic1,Bottom,LeftTicks()); yaxis(pic1,\"$f_1(x)$\",Left,RightTicks); xaxis(pic2,\"$x$\",Bottom,LeftTicks(Step=4)); yaxis(pic2,\"$f_2(x)$\",Left,RightTicks); yequals(pic1,0,Dotted); yequals(pic2,0,Dotted); pair min1=point(pic1,SW); pair max1=point(pic1,NE); pair min2=point(pic2,SW); pair max2=point(pic2,NE); real scale=(max1.x-min1.x)/(max2.x-min2.x); real shift=min1.x/scale-min2.x; transform t1=pic1.calculateTransform(); transform t2=pic2.calculateTransform(); transform T=xscale(scale*t1.xx)*yscale(t2.yy); real height=truepoint(N,user=false).y-truepoint(S,user=false).y; ## Question 6.15. How can I change the direction of the y-axis, such that negatives values are on the upper y-axis? Here is a simple example (see also the example http://asymptote.sourceforge.net/gallery/2D graphs/diatom.asy or the discussion of Linear(-1) in the documentation): import graph; size(250,200,IgnoreAspect); scale(Linear,Linear(-1)); draw(graph(log,0.1,10),red); xaxis(\"$x$\",LeftTicks); yaxis(\"$y$\",RightTicks); ## Question 6.16. How can I fill a path with a function that defines the color of each location? Use functionshade with a PDF tex engine, as illustrated by the example {functionshading.asy}. If you want to produce PostScript output, an approximate solution for now would be to superimpose a fine grid and specify colors to latticeshade that depend on position as a single pen[][] lattice. Alternatively, it may be more efficient to use tensorshade}. ## Question 6.17. Is there a way to draw a function that is not explicitly given, such as (y - 2)^2 = x - 1 ? Yes, use the parametric form y=t x=(t-2)^2+1 See the example http://asymptote.sourceforge.net/gallery/2D graphs/parametricgraph.asy. Question 6.18. Is it possible to reverse or stretch an axis? The real scaling argument to Linear is used to stretch", "If you substitute into or graph the results of the taylor() command, you will see that it uses only the indicated terms. f:sin(2*x)$g:taylor(sin(2*x),x,%pi/6,6); draw2d( dimensions = [800,500], xrange = [-1,2.5], yrange = [-1.35,1.15], xaxis = true, xaxis_type = solid, xaxis_width = 1, xaxis_color = orange, xtics = 1, yaxis = true, yaxis_type = solid, yaxis_width = 1, yaxis_color = orange, ytics = 1, font = Arial, font_size = 14, nticks = 250, color = blue, line_width = 3, line_type = solid, explicit(f,x,-1,2.5), label([\"f\",-0.75,-0.9]), color = red, line_width = 2, explicit(g,x,-1,2.5), label([\"T_6\",-0.7,-1.2]) )$ ### Remainders A good way to see how close your approximation is to the actual value is to graph the remainder $$R_6(x) = f(x) - T_6(x)$$. However, if you graph the remainder, you don't get what you expect. You get 0.No matter how much you zoom in vertically, you still get that the two are equal to each other. To get around this, you can use the trunc() function, which will truncate or drop the remainder on the Taylor series approximation. f:sin(2*x)$g:taylor(sin(2*x),x,%pi/6,6); draw2d( dimensions = [800,500], xrange = [-1,2.5], xaxis = true, xaxis_type = solid, xaxis_width = 1, xaxis_color = orange, xtics = 1, yaxis = true, yaxis_type = solid, yaxis_width = 1, yaxis_color = orange, font = Arial, font_size = 14, nticks = 250, color", "# Graph of Sine Integral I want to graph a function defined by an integral. In particular the sine integral Si(x). Si(x) = \\int_0^x \\frac{\\sin(t)}{t} dt. I can't seem to find a way do to it. • Welcome to TeX.SE. what is you specific problem regarding LaTeX? What have you tried so far? Sep 26 at 3:04 • Read pst-func-doc.pdf, 7 \\psSi, \\pssi and \\psCi. Sep 26 at 3:30 Once the module gsl is imported, Asymptote can use several functions, including the sine integral function Si (neither si nor SI). // https://mathworld.wolfram.com/SineIntegral.html // Run on http://asymptote.ualberta.ca/ unitsize(5mm,1.5cm); size(6cm); import graph; import gsl; // for manual decorations draw(Label(\"$x$\",EndPoint,E),(-11,0)--(11,0)); draw(Label(\"$\\mathrm{Si}(x)$\",EndPoint,N),(0,-2)--(0,2)); draw(Label(\"$10$\",EndPoint,S),(10,.1)--(10,-.1)); draw(Label(\"$-10$\",EndPoint,S),(-10,.1)--(-10,-.1)); draw(Label(\"$5$\",EndPoint,S),(5,.1)--(5,-.1)); draw(Label(\"$-5$\",EndPoint,S),(-5,.1)--(-5,-.1)); draw(Label(\"$1.5$\",EndPoint,W),(.2,1.5)--(-.2,1.5)); draw(Label(\"$-1.5$\",EndPoint,E),(-.2,-1.5)--(.2,-1.5)); // graph of the Si function path Sicurve=graph(Si,-10,10,100); draw(Sicurve,1pt+red); • When I add the png to the Latex file it seems to have a bad resolution. How can I download it from the site asymptote in a high resolution. Sep 26 at 16:41 • @Burlesque24 You can download x.pdf file and 1. include into you LaTeX document via \\includegraphics{x} command, or 2. use ImageMagick (having installed on your computer) to convert to PNG with this command line: magick -density 600 x.pdf x.png. PS: the number 600 is the resolution of PNG, change it as you wish Sep 26 at 16:48 • It works perfectly!", "fig = px.line(x=x, y=y, title=\"sinewave(4*pi*x+0.5)\",labels={\"x\":\"x\",\"y\":r'y'}) fig.show()`````` Results", "by some prescribed amount. Perhaps that fixes the problem after some trial and error. If not, you may have bad luck unless you rely on other non-tikz solutions. – Christian Feuersänger Apr 23 '13 at 20:07 Consider also an Asymptote option, as it's quite powerful in graph drawing, especially when there is an array of curves. An example below organizes functions in array; other properties like color, width, label etc can be organized in arrays as well to be drawn in one loop. Also, the y-coordinate of the label is calculated as a point on the curve. asy-plot.tex \\documentclass{standalone} \\usepackage{lmodern} \\usepackage[inline]{asymptote} \\begin{document} \\begin{asy} import graph; size(250,170,IgnoreAspect); real xMin=8.000000000003847e-7; real xMax=0.9999990727280006; int n=100; typedef real Func(real x); Func[] f={ new real(real x){return 3682570000*(1-x)^(10.67)*x^(21.49);}, new real(real x){return 2120.93*(1-x)^2.67*x^9.49;}, }; pen dashed=linetype(new real[] {4,3}); // set up dashed pattern pen[] curvePen={rgb(0,0.8,0)+dashed,rgb(0,0.2,0.8),}; real[] curveWidth={1.6pt,2pt}; Label[] lab={ Label(\"$g(\\theta\\,\\vert D)$\",(0.6,f[0](0.6))), Label(\"$g(\\theta)$\",(0.9,f[1](0.9))), }; for(int i=0;i<f.length;++i){ draw(graph(f[i], xMin ,xMax ,n),curvePen[i]+curveWidth[i]); label(lab[i],UnFill(2pt)); } xaxis(\"\",0,1.049,RightTicks(Step=0.1,step=0.05),EndArrow); yaxis(\"\",0,5-0.15,LeftTicks(Step=1,step=0.5),EndArrow); \\end{asy} \\end{document} To process it with latexmk, create file latexmkrc: sub asy {return system(\"asy '$_[0]'\");} add_cus_dep(\"asy\",\"eps\",0,\"asy\"); add_cus_dep(\"asy\",\"pdf\",0,\"asy\"); add_cus_dep(\"asy\",\"tex\",0,\"asy\"); and run latexmk -pdf asy-plot.tex. The result looks like • Never heard of this package, but it looks nice. – Ryan Reich Apr 6 '13 at 20:06 • I tried to export my graph in asymptote package but the code generated just seemed wrong.", "return (x,y,z); }; } triple F(pair z){ return f(1,1,1)(z); } // Gauss map triple g(pair z) { real u=z.x, v=z.y; real x,y,z; x=exp(u)*sin(v); y=exp(u)*cos(v); z=sqrt(1+exp(2*u)); return -7*(x,y,-1)/z; } // https://trecs.se/hyperboloidOfOneSheet.php path3 vector(pair z) { real u=z.x, v=z.y; real a=1, b=1, c=1; real x,y,z; x=-b*c*exp(u)*sin(v); y=-a*c*exp(u)*cos(v); z=a*b*sqrt(1+exp(2*u)); return O--(-(x,y,-1)/z); } size(13cm); surface sf=surface(F,(-10,-12),(3,6),40,Spline); //sf.colors(palette(sf.map(zpart),Rainbow())); draw(sf,RGB(0,153,216),render(merge=true)); //xaxis3(Label(\"$x$\",BeginPoint),0,7,Arrow3); draw(Label(\"$x$\",EndPoint),(12,0,0)--(-12,0,0),Arrow3); yaxis3(Label(\"$y$\",EndPoint),0,12,Arrow3); zaxis3(Label(\"$z$\",EndPoint),-18,22,Arrow3); label(\"$\\mathcal{H}$\",(-.5,5.5,18),dir(45)); transform3 t=shift(20*(Y-X)); surface sg=surface(g,(-2.5,-5),(5,5),20,Spline); //sg.colors(palette(sg.map(zpart),Rainbow())); draw(t*sg,RGB(236, 125, 44),render(merge=true)); label(\"$N\\left(\\mathcal{H} \\right)$\",t*(-.5,6,6),dir(45)); draw(Label(\"$x$\",EndPoint),t*((0,0,0)--(12,0,0)),Arrow3); draw(Label(\"$y$\",EndPoint),t*((0,0,0)--(0,12,0)),Arrow3); draw(Label(\"$z$\",EndPoint),t*((0,0,0)--(0,0,13)),Arrow3); //dot(Label(\"$(0,0,1)$\",black),t*(0,0,7),dir(135),white+5bp); draw(Label(\"$N$\",Relative(.5),N),subpath((0,0,0)--t*(0,0,0),0.4,0.6),Arrow3); output: • You wrote \" where the sphere on the right is the unit sphere x^2+y^2=1)\", I think x^2+y^2+z^2=1''. I do not understand two number sqrt(2) and -sqrt(2) on the z - axis. Nov 25, 2021 at 1:10 • @minhthien_2016 Thank you very much if you are right it was a typing error. Regarding z=sqrt(2) and z=-sqrt(2), what I tried to do is: A unitary sphere intersected with two planes, and the part between said planes is painted. (I'll edit my question to clarify that topic in more detail.) Nov 25, 2021 at 1:17 • I think, If radius of the sphere equal to R=1, the the plane perpendicular to z - axis has the equation z = h, where -1<h<1. Nov 25, 2021 at 1:19 • I already edited it, I hope it is better understood. Nov 25, 2021 at 1:23 • you", "geometric figures. /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ The first line real f1 (real x) {return sin(x);} defines the sin() function to be drawn, and the next line draw(graph(f1,-6.21,8.03)); actually draws the graph to the diagram. The graph() command comes from the Asymptote graph package. The next few lines draw several line segments. The attribute zzttqq is a pen whose color is mapped from some rgb value (with conversions ... 8 = y, 9 = z, a = a, b = b, ... ). ## Options An outline of the different available options. (The first three options are related and automatically update with respect to eath other). X/Y Units (cm): Indicates how many units in the picture correspond to 1 centimeter. Picture width/height: Indicates what size the Asymptote output should be, in centimeters. xmin/xmax/ymin/ymax: Indicates the dimensions of the picture in the picture's coordinates. (The next options change other features). Font size: Changes the default font size. The value defaults at 10pt, but I personally recommend 7pt for any file with multiple text lines, which more closely approximates the size of the font", "for $\\mathcal{P}$ are: 1. $a=(0, 0)$ has labels $\\{0, 1\\}$ 2. $b=(1/3, 0)$ has labels $\\{1, 3\\}$ 3. $c=(1/4, 1/4)$ has labels $\\{2, 3\\}$ 4. $d=(0, 1/3)$ has labels $\\{0, 2\\}$ The vertics and labels for $\\mathcal{Q}$ are: 1. $w=(0, 0)$ has labels $\\{2, 3\\}$ 2. $x=(1/3, 0)$ has labels $\\{0, 3\\}$ 3. $y=(1/4, 1/4)$ has labels $\\{0, 1\\}$ 4. $z=(0, 1/3)$ has labels $\\{1, 2\\}$ Let us apply the algorithm: • $(a, w)$ have labels: $\\{0, 1\\}, \\{2, 3\\}$. Drop 0 (arbitrary decision) in $\\mathcal{P}$. • $\\to (b, w)$ have labels: $\\{1, 3\\}, \\{2, 3\\}$. In $\\mathcal{Q}$ drop 3. • $\\to (b, z)$ have labels: $\\{1, 3\\}, \\{1, 2\\}$. In $\\mathcal{P}$ drop 1. • $\\to (c, z)$ have labels: $\\{2, 3\\}, \\{1, 2\\}$. In $\\mathcal{Q}$ drop 2. • $\\to (c, y)$ have labels: $\\{2, 3\\}, \\{0, 1\\}$. Fully labeled vertex pair. We now return the strategy pair by normalising these vertices: $$((1/2, 1/2), (1/2, 1/2))$$ This is also implemented in Nashpy: In [1]: import numpy as np import nashpy as nash A = np.array([[1, -1], [-1, 1]]) matching_pennies = nash.Game(A) matching_pennies.lemke_howson(initial_dropped_label=0) Out[1]: (array([ 0.5, 0.5]), array([ 0.5, 0.5])) You can also iterate over all possible starting labels: In [2]: for eq in matching_pennies.lemke_howson_enumeration(): print(eq) (array([ 0.5, 0.5]), array([ 0.5, 0.5])) (array([ 0.5, 0.5]), array([ 0.5, 0.5])) (array([", "\\begin{axis}[ axis equal image, z buffer=sort, hide axis, domain=0:1, y domain = 0:10, samples y = 25, ylabel =y, xlabel=x, clip=false ] z buffer=none, domain=0:330, samples y=1, ultra thick, black, -latex] ( -0.5, {sin(x)*1.5}, {cos(x)*1.5} ); y, {squarex(x) * cos(y*9) - squarey(x) * sin(y*9)}, {squarex(x) * sin(y*9) + squarey(x) * cos(y*9)} ); \\addplot3 [z buffer=auto, fill=yellow!50, draw=black] table { x y z 10 -1 -1 10 1 -1 10 1 1 10 -1 1 10 -1 -1 }; • @Acatka: Replace squarex(x) with cos(x*360) and squarey(x) with sin(x*360). – Jake Apr 18 '15 at 13:02", "2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,N); dot((5,4)); label(\"(a+b_1,b_2)\",(5,4),NE); dot((9,7)); label(\"(x,y)\",(9,7),NE); dot((2,4)); label(\"(2b_1,b_2)\",(2,4),SW); dot((9,4)); label(\"(x,b_2)\",(9,4),SE); [/asy]$ Note that since $K$ is the center of square $AB_1A_2B$, $BX \\perp KX$ and $BX = KX$. Additionally, $BY \\parallel KZ$ and $BY \\perp YZ$. $YZ$ is a line, so $\\angle BXY + \\angle KXZ + \\angle BXK = 180^\\circ$. Since $BX \\perp KX$, $\\angle BXK = 90^\\circ$, so $\\angle BXY + \\angle KXZ = 90^\\circ$. Additionally, because $BXY$ is a right triangle, $\\angle YBX + \\angle YXB = 90^\\circ$. Rearranging and substituting results in $\\angle KXZ = \\angle YBX$. Since both $BYX$ and $XZK$ are right triangles, by AAS Congruency, $\\triangle BYX \\cong \\triangle XZK$. Therefore $BY = XZ = b_2$ and $YX = ZK = a - b_1$. From this information, the coordinates of $K$ are $(a+b_1+b_2, b_2 + a - b_1)$. $[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4), K=(9,7), L=(-6,8), M=(-7,-3), N=(4,-8); draw(A--B--C--D--A); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,NE); dot(C); label(\"(-2c,0)\",C,NW); dot(D); label(\"(2d_1,-2d_2)\",D,S); dot(K); label(\"K\",K,NE); dot(L); label(\"L\",L,NW); dot(M); label(\"M\",M,SW); dot(N); label(\"N\",N,SE); [/asy]$ By", "later! // http://asymptote.ualberta.ca/ unitsize(1cm,8mm); size(8cm); import graph; real f(real x){return x^3;} real finv(real y){return cbrt(y);} pair A=(-1,-1),B=(3,3); real a=1; pair M=(a,f(a)); path pf=graph(f,A.x,B.x); real ap=a+.3*unitrand(),am=a-.35*unitrand(); fill(box((am,A.y),(ap,B.y)),red+opacity(.5)); fill(box((A.x,f(am)),(B.x,f(ap))),yellow+opacity(.5)); draw((M.x,0)--M^^(0,M.y)--M,dashed); draw(pf,blue); dot(M); label(\"$f(x)=x^3$\",(finv(B.y),B.y),SE,blue); label(\"$a$\",(a,0),S); label(\"$f(a)$\",(0,f(a)),W); clip(box(A,B)); axes(\"$x$\",\"$y$\"); With srand(10); added", "figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ /* dots and labels */ label(\"f\",(-6.12,-0.12),NE * labelscalefactor); label(\"a = 2\",(1.46,0.24),NE * labelscalefactor,zzttqq); dot((-3.14,0),dotstyle); label(\"A\",(-3.06,0.12),NE * labelscalefactor); dot((2.2,2.04),dotstyle); dot((4.5,2.08),dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$ Here is the Asymptote code produced: (with some spacing added) /* Geogebra to Asymptote conversion, documentation at userscripts.org/scripts/show/72997 */ import graph; size(14.26cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.22,xmax = 8.04,ymin = -4.18,ymax = 4.76; /* image dimensions */ pen fueaev = rgb(0.96,0.92,0.9); pen zzttqq = rgb(0.6,0.2,0); real f2 (real x) {return sin(x);} filldraw(graph(f2,0,pi )--(pi ,0)--(0,0)--cycle, fueaev, zzttqq); filldraw((-3.14,0)--(0,0)--(0,3.14)--(-3.14,pi )--cycle, fueaev, zzttqq); Label laxis; laxis.p = fontsize(10); string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real" ]

[ "If $$a$$ and $$b$$ are positive, and $$a+b=1$$, then what is the maximum value of $${ (\\sqrt { 4a+1 } +\\sqrt { 4b+1 } ) }^{ 2 }$$.", "# SineCosine at its pinnacle Geometry Level 3 Let the maximum value of $$(\\sin x \\cos x)$$ for $$x \\in \\mathbb{R}$$ be $$M$$ . If the value of $$x$$ for which $$(\\sin x \\cos x)$$ is maximum satisfying $$0< x < \\pi$$ can be expressed as $$\\dfrac{A\\pi}{B}$$ for co-prime integers $$A,B$$ , compute $$M+A+B$$. × Problem Loading... Note Loading... Set Loading...", "# Maximum value Level pending The maximum value of the expression $2\\sin x + 4\\cos x + 3$ can be expressed as $$a\\sqrt{b} + c$$, where $$a$$, $$b$$ and $$c$$ are natural numbers. Find $$a+b+c$$. ×", "# Maximum value of 2sinθ + 3cosθ is? A. 2 B. $\\sqrt{13}$ C. $\\sqrt{15}$ D. 1 Answer: Option B Solution(By Apex Team) $\\begin{array}{l}(2\\sin\\theta+3\\cos\\theta)\\\\ \\text{Maximum value of}\\\\ a\\sin\\theta+b\\cos\\theta\\\\ \\begin{array}{l} =\\sqrt{a^{2}+b^{2}} \\\\ =\\sqrt{2^{2}+3^{2}} \\\\ =\\sqrt{4+9} \\\\ =\\sqrt{13} \\end{array}\\end{array}$", "place numbers, we can make this range between $$-14$$ and $$-83$$ smaller! $$\\Box-\\Box$$ must equal something in between $$-1$$ and $$-9$$ inclusive. ### Proof: Let the first box be the lowest option number (namely, $$3$$) and the second box be the highest option number (namely, $$12$$). We obtain that $$\\boxed{3}-\\boxed{12}=-9.\\tag{\\small\\rm as \\; the \\; minimum \\; value}$$ This is the minimum value because the pair $$(3,12)$$ is the furthest away from each other out of all the option numbers; and since we are dealing with negatives, the maximum (furthest away) turns to minimum (because positive becomes negative). Now look at the option numbers: $$3$$, $$6$$, $$7$$, $$12$$. Which pair is the closest to each other? That pair is $$(6,7)$$ which differs by only $$1$$. Therefore, $$\\boxed{6}-\\boxed{7}=-1.\\tag{\\small\\rm as \\; the \\; maximum \\; value}$$ And therefore, $$\\Box-\\Box$$ must equal something in between $$-1$$ and $$-9$$ inclusive. And finally, The maximum value of $$\\Box\\times \\Box$$ must be $$12$$... uh oh. That doesn't seem right... it's a contradiction; THERE EXISTS NO SOLUTION (without lateral thinking)! ### Proof: If we get the maximum value the LHS can equal (namely, $$4$$) and subtract the minimum value of $$\\Box-\\Box$$ (namely, $$-9$$), we get $$4-(-9)=4+9=13.$$ Therefore, the maximum value of $$\\Box\\times \\Box$$ is $$13$$. But $$13$$ is prime, so the maximum value reduces to $$12$$.", "Browse Questions # If $5 \\cos x +12 \\cos y=13$, then the maximum value of $5 \\sin x +12 \\sin y$ is : $\\begin {array} {1 1} (a)\\;12 & \\quad (b)\\;\\sqrt {120} \\\\ (c)\\;\\sqrt {20} & \\quad (d)\\;13 \\end {array}$ (b) $\\sqrt {120}$", "value will be the one with the larger value of $x$, namely $x = 2600$. Therefore the answer is $\\boxed{\\textbf{(E)} \\ 10 \\sqrt{26} - 51}$ as before.", "## anonymous one year ago What is the maximum possible value of a sine ratio? In two or more complete sentences, explain your answer. 1. anonymous pleassssss hep 2. johnweldon1993 Well we know $\\large Sin = \\frac{opposite}{hypotenuse}$ What is the max value this can have? Well hint! |dw:1432855702124:dw| what values will the sin wave sin(x) ALWAYS fall between? 3. anonymous 1 and -1? 4. anonymous oh so the most sin can be is 1.", ",13 ,12 ,11 ,10 ,9 ,8 ,7 ,6 ,5 ,4 ,3 ,2 ,1 \\}$ maximum value for $A=30$ and $1 \\leq B \\leq 29$ from left to right. – Ahmad Jul 10 '17 at 23:28", "# 4sinx-3cosx = 7 find the number of general solutions of the given expressions. a sinx -b cosx = $\\sqrt$(a2+ b2 ) Maximum value is $\\sqrt$(9+16) = 5", "$0$ Complete the square to find the minimum or maximum value. $x^2-8x+6$ 1. $-22$ 2. $11/2$ 3. $16/3$ 4. $-10$ $x^2-3x+2$ 1. $-1/4$ 2. $-5$ 3. $-7/5$ 4. $17/4$", "answer is $\\boxed{3}$.", "Boxed in Algebra Level 4 Find the maximum volume $$V$$ of a rectangular box with diagonal length $$L$$. $$V^2$$ can be expressed as $$\\frac{aL^6 + b}{c}$$ for non negative integers $$a,b,c$$. Input the minimum value of $$a+b+c$$ as your answer. ×", "\\boxed{123} $\\boxed{123}$", "to maximize $4xy$. Note that $(x+y)^2=(x-y)^2 +4xy$. So $4xy=(x+y)^2-(x-y)^2$. Given that $x+y$ is fixed, what does this say about the maximum value of $4xy$?", "+0 # Help 0 89 1 Let a and b be real numbers. Find the maximum value of $$a \\cos \\theta + b \\sin \\theta$$ in terms of a and b. Aug 31, 2019 $$a \\cos(\\theta) + b \\sin(\\theta) = \\\\ \\sqrt{a^2 + b^2}\\sin(\\theta + \\phi),~\\phi = \\tan^{-1}\\left(b,a\\right)\\\\ -1 \\leq \\sin(\\theta+\\phi) \\leq 1\\\\ \\text{The max value is thus \\sqrt{a^2+b^2}}$$", "Modify it your way! Geometry Level 3 $\\large \\sin^6 x+ \\cos^6 x$ If $$A$$ and $$B$$ are the minimum and maximum values possible of the expression above for real $$x$$, then what is the value of $$A+B$$? × Problem Loading... Note Loading... Set Loading...", "# The answer is not 1/9 If $a, b, c$ are non-negative real numbers, what is the maximum value of $\\frac{ ab^2 + bc^2 + ca^2 } { ( a + b + c) ^ 3 } ?$ ×", "# What is the maximum value of 4 sin 3x? The maximum value of the function sinus is $1$, so, in this case, the maximum is $4$.", "\\boxed{123} $$\\boxed{123}$$" ]

[ "graph(f,-10,10,n=200,operator..) where n=200 is the number of sample points, and operator.. means curvy connecting that is necessary in this case. // http://asymptote.ualberta.ca/ unitsize(8mm,2cm); size(8cm); import graph; real f(real x) { if (x==0) return -2; else return (x*cos(x)-sin(x))/(x-sin(x)); }; draw(graph(f,-10,10,n=200,operator..),red); xaxis(\"$x$\",Arrow(TeXHead)); yaxis(\"$y$\",-2.5,1.5,Arrow(TeXHead)); label(\"The graph of $y=\\frac{x\\cos x-\\sin x}{x-\\sin x}$\",point(N),2N); label(\"$-2$\",(0,-2),SW); shipout(bbox(5mm,invisible)); Appendix The function f(x) is asymptotic to cos(x) when |x| is big. unitsize(1cm,4cm); size(8cm); import graph; import math; //for drawline real f(real x) { if (x==0) return -2; else return (x*cos(x)-sin(x))/(x-sin(x)); }; draw(graph(cos,-20,20,n=200,operator..),lightblue+1pt); draw(graph(f,-20,20,n=200,operator..),red+.5pt); xaxis(\"$x$\",Arrow(TeXHead)); yaxis(\"$y$\",-2.5,2.5,Arrow(TeXHead)); drawline((0,1),(1,1),gray+.2pt); drawline((0,-1),(1,-1),gray+.2pt); drawline((0,-2),(1,-2),gray+.2pt); label(scale(.7)*\"$1$\",(0,1),NE); label(scale(.6)*\"$-1$\",(0,-1),SW); label(scale(.6)*\"$-2$\",(0,-2),SW); label(\"$\\color{red}y=\\frac{x\\cos x-\\sin x}{x-\\sin x}$\"+\" and \"+\"$\\color{blue}y=\\cos x$\",point(N),2N); shipout(bbox(5mm,invisible)); • Maybe return (x==0) ? -2 : (x*cos(x)-sin(x))/(x-sin(x));. Oct 17 at 16:37 • @NguyenVanChi1998 Yes, that is more concise syntax real f(real x) { return (x==0) ? -2 : (x*cos(x)-sin(x))/(x-sin(x)); }; Oct 17 at 16:41", "SYSTEM label(\"$O$\", (0, 0, 0), NW); Label Lx = Label(\"$x$\", EndPoint, W); Label Ly = Label(\"$y$\", EndPoint, S); Label Lz = Label(\"$z$\", EndPoint, W); draw(Lx, O--2X, Arrow3(emissive(black))); draw(Ly, O--2Y, Arrow3(emissive(black))); draw(Lz, O--2Z, Arrow3(emissive(black))); // SURFACE triple S(pair uv) { real x = cos(uv.x)*sin(uv.y); real y = sin(uv.x)*sin(uv.y); real z = cos(uv.y); return (x-(1-z**2-y**2), y-0.3*sin(z*pi), z); } // BOUNDARY triple dS(real u) { return S((u, pi/2)); } // TANGENT VECTORS real e = 0.001; // *this... is... infinitesimal!* triple U(pair t) { return (S((e,0)+t)-S(t))/e; } triple V(pair t) { return (S((0,e)+t)-S(t))/e; } // NORMAL VECTOR triple N(pair uv) { return cross(U(uv),V(uv))/length(cross(U(uv),V(uv))); } // GRAPHICAL ELEMENTS surface s = surface(S, (0, 0), (2pi, pi/2), 64, 64, Spline); path3 ds = graph(dS, 0, 2pi); pair t = (-pi/8, pi/8); path3 n = S(t) -- shift(-N(t))*S(t); label(\"$\\Sigma$\", shift(0,0,0.1)*shift(adj)*S((pi/2,pi/5)), NE); Label LdS = Label(\"$\\partial\\Sigma$\", Relative(0.95), S); Label Ln = Label(\"$\\mathbf n$\", EndPoint, NW); draw(s, surfacepen=material(diffusepen=gray(0.6),emissivepen=gray(0.3),specularpen=gray(0.1))); draw(LdS, ds, Arrow3(emissive(black))); draw(shift(4*unit(currentprojection.camera))*ds, dashed); draw(Ln, n, Arrow3(emissive(black))); REQUESTED APPENDIX: on dashed lines I wanted to draw the dashed border over the surface. Since this is an orthographic projection, the easiest way is translating the object along the camera axis towards it (instead of checking what segments are visible, possibly complicated): draw(shift(4*unit(currentprojection.camera))*ds, dashed); unit(currentprojection.camera) is a versor parallel to the camera axis and pointing towards it, so it", "looked at the extreme cases (i.e., $$S = emptyset$$ and $$S = V$$). When $$S = V$$, the heuristic proposed in the previous post obviously works. When $$S = emptyset$$, the total time taken is fixed since we cannot exercise the option to use a helper vehicle. I was wondering whether there was an approximation scheme or heuristic that might be efficient in this general case. Get this bounty!!! ## #StackBounty: #algorithms #graphs #monte-carlo Devise Mont Carlo and Las Vegas Algorithms to Solve Maximum Independent Set ### Bounty: 50 I am trying to devise a Las Vegas algorithm to solve Maximum Independent Set, but I don’t know how to start. Also, I want to devise a Mont Carlo algorithm for this problem. I would appreciate any help. Thank you. Get this bounty!!! ## #StackBounty: #graphs #asymptote How to graph the inverse trigonometric function (arctan, arccotan) in asymptote? ### Bounty: 100 A want to add the inverse trig.function for tangens to the code of asymptote. Thank you for adivise. ``````settings.outformat=\"pdf\"; unitsize(2.5cm); import graph; real xmin = -pi/2; real xmax = pi/2; real ymin = -pi/2; real ymax = pi/2; xlimits(xmin,xmax); ylimits(ymin,ymax); //axis and the grid : xaxis(Label(\"\\$x\\$\",align=2E),Ticks(\"\\$%.2f\\$\",new real[]{xmin,-1,1,xmax},Size=1mm,1bp),Arrow); yaxis(Label(\"\\$y\\$\",align=2N),Ticks(\"\\$%.2f\\$\",new real[]{xmin,-1,1,xmax},Size=1mm,1bp),Arrow); // function y=sin(x) real f(real x) {return sin(x)/cos(x); } path g = graph(f,xmin+0.5, xmax-0.5); draw(g,red, L= Label(\"\\$y=tan(x)\\$\", UnFill, position=EndPoint));", "samples=181, % with domain=0:360 and 181 samples you get a sample every 2 degrees mark=*, mark repeat=45 % add a mark for every 45 sample, meaning you get a mark every 90 degreees ] {sin(x)} node[fill=white, right=2mm, pos=0.35]{$y=sin(x)$}; % alternative method %addplot[domain=0:360,mark=none,samples=100] {sin(x)} node[fill=white, right=2mm, pos=0.35]{$y=sin(x)$}; %addplot[only marks,mark=*,samples at={0,90,...,360}] {sin(x)}; end{axis} end{tikzpicture} end{center} end{document} Correct answer by Torbjørn T. on August 29, 2020 Compile with Asymptote. asymptote.ualberta.ca import graph; size(10cm,5cm,false); real sinx(real x){return sin(x);} path g=graph(sinx,0,2pi,350); draw(g); draw(Label(\"$x$\",EndPoint,black),(-1,0)--(2*pi+0.5,0),blue,Arrow); label(Label(\"$x'$\",BeginPoint,black),(-1,0)--(2*pi+0.5,0)); draw(Label(\"$y$\",EndPoint,black),(0,-1.5)--(0,1.5),darkgreen,Arrow); label(Label(\"$y'$\",BeginPoint,black),(0,-1.5)--(0,1.5)); real marginx=0.1, marginy=0.07; for (int i: new int[]{-1,1}){ draw(scale(0.6)*Label(\"$\"+(string) i+\"$\",Relative(0)),(0,i)-(marginx,0)--(0,i)+(marginx,0)); } for (int i=0; i<=360; i=i+30){ draw(scale(0.6)*Label((i != 0) ? \"$\"+(string) i+\"^{circ}$\" : \" \",Relative(0),black),(radians(i),0)-(0,marginy)--(radians(i),0)+(0,marginy),blue); dot((radians(i),sinx(radians(i))), (i == 0 || i == 360) ? blue : black); } label(\"$y=sin x$\" +\" on \" + \"$[0,2pi]$\", (radians(130),-1.5)); Answered by user213378 on August 29, 2020 Similar to @Torbjørn T. answer (+1): documentclass[margin=3mm]{standalone} usepackage{pgfplots} usetikzlibrary{arrows.meta} begin{document} begin{tikzpicture} begin{axis}[ x=0.25mm, axis lines=middle, axis line style={Latex-Latex}, xlabel=$x$, xlabel style={anchor=west}, ylabel=$y$, ylabel style={anchor=south}, % xmin=-25, xmax=380, ymin=-1.25, ymax=1.25, ticklabel style = {font=footnotesize}, xtick distance=30, samples at={0,30,...,360}, smooth ] addplot +[very thick] {sin(x)}; node[above right] at (30,-1) {$y=sin(x)$}; end{axis} end{tikzpicture} end{document} Answered by Zarko on August 29, 2020 ## Related Questions ### A problem with colors in WinEdt 1 Asked on June 7, 2021 ### IPython Notebook input and output cells with listings 2", "mypara=revolution(O,g,X); Full code: copy to http://asymptote.ualberta.ca/ and click Run unitsize(1cm); import graph3; import solids; //currentprojection=orthographic(2,1.5,.5,zoom=.95); currentprojection=obliqueZ; currentprojection.zoom=.95; size(8cm); real a=1.8, b=a^2; real c=2.5,d=4.5; triple A=(a,0,0), B=(0,b,0); triple C=(c,0,0), D=(0,d,0); // the (red) parabol curve on the XY-plane triple paraXY(real t){return (t,t^2,0);} path3 g=graph(paraXY,0,a); draw(g,red+1.2pt); // use one of the following revolutions //revolution mypara=revolution(C,g,Y); // around x=c //revolution mypara=revolution(D,g,X); // around y=d //revolution mypara=revolution(O,g,X); // around y=0 revolution mypara=revolution(O,g,Y); // around x=0 draw(surface(mypara),yellow+opacity(.7)); //draw(mypara,blue,orange); label(\"$a$\",A,S); label(\"$c$\",C,S); label(\"$b$\",B,W); label(\"$d$\",D,W); draw(A--(a,b,0)--B^^C--(c,d,0)--D,dashed+gray); draw(Label(\"$x$\",EndPoint),-2*X--6*X,Arrow3()); draw(Label(\"$y$\",EndPoint),-5*Y--6*Y,Arrow3()); draw(Label(\"$z$\",EndPoint),O--6*Z,Arrow3()); // you may not want to draw Oz 2. The same code, only change X to Y to get revolution around Oy revolution mypara=revolution(O,g,Y); 3. The same code, for revolution around x=c revolution mypara=revolution(C,g,Y); 4. Last, for revolution around y=d revolution mypara=revolution(D,g,X); Welcome to TeX.SE!!! This is a quick modification of an answer from here. It's a 2d drawing using isometric perspective. The equations of parabola and ellipses are calculated by hand. The ellipses are very simple if you know about isometric perspective (the axes ratio is sqrt(3):1). The parabola was obtained as an envelope of the family of 'horizontal' ellipses. This is the code: \\documentclass[tikz,border=2mm]{standalone} \\tikzset {% axes/.style={thick,-latex}, cylinder/.style={right color=blue!80,left color=white,fill opacity=0.6}, paraboloid/.style={left color=magenta!80,fill opacity=0.6}, } \\begin{document} \\begin{tikzpicture}[line cap=round,line join=round] % axes x,y \\draw[axes] (30:3) -- (210:3) node [left] {$x$}; \\draw[axes] (150:3) -- (330:3) node [right]", "# Asymptote: 3d graph over a disc Is there a straightforward way to draw a 3D graph over a disc domain? Say z=x^2-y^2 for x^2+y^2<1. [I just started to use asymptote; this page explained me how to do it for a rectangular domain. I hope it is an easy question.] One way to make sure that x^2+y^2<1 is to use polar coordinates. Then x=r cos(phi) and y=r sin(phi). \\documentclass[variwidth,border=3.14mm]{standalone} \\usepackage{asypictureB} \\begin{document} \\begin{asypicture}{name=discgraph} usepackage(\"mathrsfs\"); import graph3; import solids; import interpolate; settings.outformat=\"pdf\"; size(500); defaultpen(0.5mm); pen darkgreen=rgb(0,138/255,122/255); draw(Label(\"$x$\",1),(0,0,0)--(1.2,0,0),darkgreen,Arrow3); draw(Label(\"$y$\",1),(0,0,0)--(0,1.2,0),darkgreen,Arrow3); draw(Label(\"$f(x,y)$\",1),(0,0,0)--(0,0,0.6),darkgreen,Arrow3); //function: call the radial coordinate r=t.x and the angle phi=t.y triple f(pair t) { return ((t.x)*cos(t.y), (t.x)*sin(t.y), ((t.x)*cos(t.y))^2-((t.x)*sin(t.y))^2); } surface s=surface(f,(0,1),(0.49,2.5*pi),32,16, usplinetype=new splinetype[] {notaknot,notaknot,monotonic}, vsplinetype=Spline); pen p=rgb(0,0,.7); draw(s,lightolive+white); \\end{asypicture} \\end{document} • Thank you, but is there a direct way to make a condition x^2+y^2<1 for the arguments? – Anton Petrunin Apr 1 '19 at 4:31 • @marmot: The x-axis near origin should be hidden from the given point of view. Is there any way to improve this issue? E.g., by setting some samples-option? – Marian G. Apr 1 '19 at 5:28 • A line has a thickness, a surface not. It is why you see the x-axis near origin. You can observe the same behavior with a simple square surface and the x-axis. Perhaps it is possible to avoid its by", "return (x,y,z); }; } triple F(pair z){ return f(1,1,1)(z); } // Gauss map triple g(pair z) { real u=z.x, v=z.y; real x,y,z; x=exp(u)*sin(v); y=exp(u)*cos(v); z=sqrt(1+exp(2*u)); return -7*(x,y,-1)/z; } // https://trecs.se/hyperboloidOfOneSheet.php path3 vector(pair z) { real u=z.x, v=z.y; real a=1, b=1, c=1; real x,y,z; x=-b*c*exp(u)*sin(v); y=-a*c*exp(u)*cos(v); z=a*b*sqrt(1+exp(2*u)); return O--(-(x,y,-1)/z); } size(13cm); surface sf=surface(F,(-10,-12),(3,6),40,Spline); //sf.colors(palette(sf.map(zpart),Rainbow())); draw(sf,RGB(0,153,216),render(merge=true)); //xaxis3(Label(\"$x$\",BeginPoint),0,7,Arrow3); draw(Label(\"$x$\",EndPoint),(12,0,0)--(-12,0,0),Arrow3); yaxis3(Label(\"$y$\",EndPoint),0,12,Arrow3); zaxis3(Label(\"$z$\",EndPoint),-18,22,Arrow3); label(\"$\\mathcal{H}$\",(-.5,5.5,18),dir(45)); transform3 t=shift(20*(Y-X)); surface sg=surface(g,(-2.5,-5),(5,5),20,Spline); //sg.colors(palette(sg.map(zpart),Rainbow())); draw(t*sg,RGB(236, 125, 44),render(merge=true)); label(\"$N\\left(\\mathcal{H} \\right)$\",t*(-.5,6,6),dir(45)); draw(Label(\"$x$\",EndPoint),t*((0,0,0)--(12,0,0)),Arrow3); draw(Label(\"$y$\",EndPoint),t*((0,0,0)--(0,12,0)),Arrow3); draw(Label(\"$z$\",EndPoint),t*((0,0,0)--(0,0,13)),Arrow3); //dot(Label(\"$(0,0,1)$\",black),t*(0,0,7),dir(135),white+5bp); draw(Label(\"$N$\",Relative(.5),N),subpath((0,0,0)--t*(0,0,0),0.4,0.6),Arrow3); output: • You wrote \" where the sphere on the right is the unit sphere x^2+y^2=1)\", I think x^2+y^2+z^2=1''. I do not understand two number sqrt(2) and -sqrt(2) on the z - axis. Nov 25, 2021 at 1:10 • @minhthien_2016 Thank you very much if you are right it was a typing error. Regarding z=sqrt(2) and z=-sqrt(2), what I tried to do is: A unitary sphere intersected with two planes, and the part between said planes is painted. (I'll edit my question to clarify that topic in more detail.) Nov 25, 2021 at 1:17 • I think, If radius of the sphere equal to R=1, the the plane perpendicular to z - axis has the equation z = h, where -1<h<1. Nov 25, 2021 at 1:19 • I already edited it, I hope it is better understood. Nov 25, 2021 at 1:23 • you", "x) {return sin(x);} filldraw(graph(f2,0,pi )--(pi ,0)--(0,0)--cycle, fueaev, zzttqq); Some commands have to be defined with respect to a function. $\\texttt{f2}$ returns the $\\sin(x)$ function, and the $\\texttt{graph}$ command draws $\\texttt{f2}$ from $x = 0$ to $x = \\pi$. Label laxis; laxis.p = fontsize(10); string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); These commands draw the axes. The $\\texttt{laxis}$ label specifies the label styles of the axes (e.g., the field $\\texttt{laxis.p}$ defines the desired font pen). The next line, the string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} are functions that define the labels for both axes. These are abbreviated as $\\texttt{xlbl, ylbl}$ in the concise mode. The first defines axes with labels with respect to $\\pi$, and it formats the labels so that the $0$ label simply outputs $0$ instead of $0\\pi$, and $\\pm \\pi$ instead of $\\pm 1\\pi$. Finally, xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); are the commands that actually construct the xaxis and the yaxis. For documentation, see here. The next block of code does the actual construction of the", "prime factors $2,3, \\text{and } 5$. $[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label(\"x\",X,N); label(\"y\",Y,SW); label(\"z\",Z,SE); label(\"2^3\",X--Y,2W); label(\"2^3\",X--Z,2E); label(\"2^2\",Y--Z,2S); [/asy]$ Analyzing for powers of $2$, it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \\text{ nor } z$ can have a power of $2$ exceeding $2$. Turning towards the vertices $y$ and $z$, we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \\text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \\{(2,0)(2,1)(0,2)(1,2)(2,2)\\}$. $[asy] //Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label(\"x\",X,N); label(\"y\",Y,SW); label(\"z\",Z,SE); label(\"3^2\",X--Y,2W); label(\"3^1\",X--Z,2E); label(\"3^2\",Y--Z,2S); [/asy]$ Using the same logic as we did for powers of $2$, it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \\text{ and } z$, we can use the same logic to find the only ordered pairs $(x,z)$", "If you substitute into or graph the results of the taylor() command, you will see that it uses only the indicated terms. f:sin(2*x)$g:taylor(sin(2*x),x,%pi/6,6); draw2d( dimensions = [800,500], xrange = [-1,2.5], yrange = [-1.35,1.15], xaxis = true, xaxis_type = solid, xaxis_width = 1, xaxis_color = orange, xtics = 1, yaxis = true, yaxis_type = solid, yaxis_width = 1, yaxis_color = orange, ytics = 1, font = Arial, font_size = 14, nticks = 250, color = blue, line_width = 3, line_type = solid, explicit(f,x,-1,2.5), label([\"f\",-0.75,-0.9]), color = red, line_width = 2, explicit(g,x,-1,2.5), label([\"T_6\",-0.7,-1.2]) )$ ### Remainders A good way to see how close your approximation is to the actual value is to graph the remainder $$R_6(x) = f(x) - T_6(x)$$. However, if you graph the remainder, you don't get what you expect. You get 0.No matter how much you zoom in vertically, you still get that the two are equal to each other. To get around this, you can use the trunc() function, which will truncate or drop the remainder on the Taylor series approximation. f:sin(2*x)$g:taylor(sin(2*x),x,%pi/6,6); draw2d( dimensions = [800,500], xrange = [-1,2.5], xaxis = true, xaxis_type = solid, xaxis_width = 1, xaxis_color = orange, xtics = 1, yaxis = true, yaxis_type = solid, yaxis_width = 1, yaxis_color = orange, font = Arial, font_size = 14, nticks = 250, color", "real f1(real x) {return a1*sin(x/2*pi);} real f2(real x) {return a2*sin(x/4*pi);} draw(pic1,graph(pic1,f1,xmin1,xmax1)); draw(pic2,graph(pic2,f2,xmin2,xmax2)); xaxis(pic1,Bottom,LeftTicks()); yaxis(pic1,\"$f_1(x)$\",Left,RightTicks); xaxis(pic2,\"$x$\",Bottom,LeftTicks(Step=4)); yaxis(pic2,\"$f_2(x)$\",Left,RightTicks); yequals(pic1,0,Dotted); yequals(pic2,0,Dotted); pair min1=point(pic1,SW); pair max1=point(pic1,NE); pair min2=point(pic2,SW); pair max2=point(pic2,NE); real scale=(max1.x-min1.x)/(max2.x-min2.x); real shift=min1.x/scale-min2.x; transform t1=pic1.calculateTransform(); transform t2=pic2.calculateTransform(); transform T=xscale(scale*t1.xx)*yscale(t2.yy); real height=truepoint(N,user=false).y-truepoint(S,user=false).y; ## Question 6.15. How can I change the direction of the y-axis, such that negatives values are on the upper y-axis? Here is a simple example (see also the example http://asymptote.sourceforge.net/gallery/2D graphs/diatom.asy or the discussion of Linear(-1) in the documentation): import graph; size(250,200,IgnoreAspect); scale(Linear,Linear(-1)); draw(graph(log,0.1,10),red); xaxis(\"$x$\",LeftTicks); yaxis(\"$y$\",RightTicks); ## Question 6.16. How can I fill a path with a function that defines the color of each location? Use functionshade with a PDF tex engine, as illustrated by the example {functionshading.asy}. If you want to produce PostScript output, an approximate solution for now would be to superimpose a fine grid and specify colors to latticeshade that depend on position as a single pen[][] lattice. Alternatively, it may be more efficient to use tensorshade}. ## Question 6.17. Is there a way to draw a function that is not explicitly given, such as (y - 2)^2 = x - 1 ? Yes, use the parametric form y=t x=(t-2)^2+1 See the example http://asymptote.sourceforge.net/gallery/2D graphs/parametricgraph.asy. Question 6.18. Is it possible to reverse or stretch an axis? The real scaling argument to Linear is used to stretch", "by some prescribed amount. Perhaps that fixes the problem after some trial and error. If not, you may have bad luck unless you rely on other non-tikz solutions. – Christian Feuersänger Apr 23 '13 at 20:07 Consider also an Asymptote option, as it's quite powerful in graph drawing, especially when there is an array of curves. An example below organizes functions in array; other properties like color, width, label etc can be organized in arrays as well to be drawn in one loop. Also, the y-coordinate of the label is calculated as a point on the curve. asy-plot.tex \\documentclass{standalone} \\usepackage{lmodern} \\usepackage[inline]{asymptote} \\begin{document} \\begin{asy} import graph; size(250,170,IgnoreAspect); real xMin=8.000000000003847e-7; real xMax=0.9999990727280006; int n=100; typedef real Func(real x); Func[] f={ new real(real x){return 3682570000*(1-x)^(10.67)*x^(21.49);}, new real(real x){return 2120.93*(1-x)^2.67*x^9.49;}, }; pen dashed=linetype(new real[] {4,3}); // set up dashed pattern pen[] curvePen={rgb(0,0.8,0)+dashed,rgb(0,0.2,0.8),}; real[] curveWidth={1.6pt,2pt}; Label[] lab={ Label(\"$g(\\theta\\,\\vert D)$\",(0.6,f[0](0.6))), Label(\"$g(\\theta)$\",(0.9,f[1](0.9))), }; for(int i=0;i<f.length;++i){ draw(graph(f[i], xMin ,xMax ,n),curvePen[i]+curveWidth[i]); label(lab[i],UnFill(2pt)); } xaxis(\"\",0,1.049,RightTicks(Step=0.1,step=0.05),EndArrow); yaxis(\"\",0,5-0.15,LeftTicks(Step=1,step=0.5),EndArrow); \\end{asy} \\end{document} To process it with latexmk, create file latexmkrc: sub asy {return system(\"asy '$_[0]'\");} add_cus_dep(\"asy\",\"eps\",0,\"asy\"); add_cus_dep(\"asy\",\"pdf\",0,\"asy\"); add_cus_dep(\"asy\",\"tex\",0,\"asy\"); and run latexmk -pdf asy-plot.tex. The result looks like • Never heard of this package, but it looks nice. – Ryan Reich Apr 6 '13 at 20:06 • I tried to export my graph in asymptote package but the code generated just seemed wrong.", "2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,N); dot((5,4)); label(\"(a+b_1,b_2)\",(5,4),NE); dot((9,7)); label(\"(x,y)\",(9,7),NE); dot((2,4)); label(\"(2b_1,b_2)\",(2,4),SW); dot((9,4)); label(\"(x,b_2)\",(9,4),SE); [/asy]$ Note that since $K$ is the center of square $AB_1A_2B$, $BX \\perp KX$ and $BX = KX$. Additionally, $BY \\parallel KZ$ and $BY \\perp YZ$. $YZ$ is a line, so $\\angle BXY + \\angle KXZ + \\angle BXK = 180^\\circ$. Since $BX \\perp KX$, $\\angle BXK = 90^\\circ$, so $\\angle BXY + \\angle KXZ = 90^\\circ$. Additionally, because $BXY$ is a right triangle, $\\angle YBX + \\angle YXB = 90^\\circ$. Rearranging and substituting results in $\\angle KXZ = \\angle YBX$. Since both $BYX$ and $XZK$ are right triangles, by AAS Congruency, $\\triangle BYX \\cong \\triangle XZK$. Therefore $BY = XZ = b_2$ and $YX = ZK = a - b_1$. From this information, the coordinates of $K$ are $(a+b_1+b_2, b_2 + a - b_1)$. $[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4), K=(9,7), L=(-6,8), M=(-7,-3), N=(4,-8); draw(A--B--C--D--A); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,NE); dot(C); label(\"(-2c,0)\",C,NW); dot(D); label(\"(2d_1,-2d_2)\",D,S); dot(K); label(\"K\",K,NE); dot(L); label(\"L\",L,NW); dot(M); label(\"M\",M,SW); dot(N); label(\"N\",N,SE); [/asy]$ By", "If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, find $n((A\\cup B)\\times(A\\cap B)\\times(A \\triangle B))$ 9. If p(A) denotes the power set of A, then find $n(P(P(P(\\phi)))).$ 10. 3 x 3 = 9 11. Compare and contrast the graph y = x2 - 1, y = 4(x2 - 1) and y = (4x)2 = 1. 12. By using the same concept applied in previous example, graphs of y = sin x and y = sin 2x, and also their combined graphs are given figures (a), (b) and (c). The minimum and maximum values of sin x and sin 2x are the same. But they have different x-intercepts. The x-intercepts for y = sin x are $\\pm n\\pi$ and for y = sin 2x are $\\pm{1\\over 2}n\\pi,\\ n\\in Z.$ ​​​ 13. Let us now draw the graph of y = 2 sin ( x - 1 ) + 3. 14. 2 x 5 = 10 15. Let f, g: $R \\rightarrow R$ be defined as f (x) = 2x -|x| and g(x) = 2x + |x|. find fog. 16. If $f:R\\rightarrow R$ is defined by f(x) = 2x- 3, prove that f is a bijection and find its inverse", "# Graph of Sine Integral I want to graph a function defined by an integral. In particular the sine integral Si(x). Si(x) = \\int_0^x \\frac{\\sin(t)}{t} dt. I can't seem to find a way do to it. • Welcome to TeX.SE. what is you specific problem regarding LaTeX? What have you tried so far? Sep 26 at 3:04 • Read pst-func-doc.pdf, 7 \\psSi, \\pssi and \\psCi. Sep 26 at 3:30 Once the module gsl is imported, Asymptote can use several functions, including the sine integral function Si (neither si nor SI). // https://mathworld.wolfram.com/SineIntegral.html // Run on http://asymptote.ualberta.ca/ unitsize(5mm,1.5cm); size(6cm); import graph; import gsl; // for manual decorations draw(Label(\"$x$\",EndPoint,E),(-11,0)--(11,0)); draw(Label(\"$\\mathrm{Si}(x)$\",EndPoint,N),(0,-2)--(0,2)); draw(Label(\"$10$\",EndPoint,S),(10,.1)--(10,-.1)); draw(Label(\"$-10$\",EndPoint,S),(-10,.1)--(-10,-.1)); draw(Label(\"$5$\",EndPoint,S),(5,.1)--(5,-.1)); draw(Label(\"$-5$\",EndPoint,S),(-5,.1)--(-5,-.1)); draw(Label(\"$1.5$\",EndPoint,W),(.2,1.5)--(-.2,1.5)); draw(Label(\"$-1.5$\",EndPoint,E),(-.2,-1.5)--(.2,-1.5)); // graph of the Si function path Sicurve=graph(Si,-10,10,100); draw(Sicurve,1pt+red); • When I add the png to the Latex file it seems to have a bad resolution. How can I download it from the site asymptote in a high resolution. Sep 26 at 16:41 • @Burlesque24 You can download x.pdf file and 1. include into you LaTeX document via \\includegraphics{x} command, or 2. use ImageMagick (having installed on your computer) to convert to PNG with this command line: magick -density 600 x.pdf x.png. PS: the number 600 is the resolution of PNG, change it as you wish Sep 26 at 16:48 • It works perfectly!", "figure, which shows the graphs of $y=\\sin x$ and $y=\\cos x$ on the closed interval $[0,2 \\pi]$. (GRAPH CAN'T COPY). (a) When the value of $\\sin x$ is a maximum, what is the corresponding value of $\\cos x ?$ (b) When the value of $\\cos x$ is a maximum, what is the corresponding value of $\\sin x ?$ Refer to the following figure, which shows the graphs of $y=\\sin x$ and $y=\\cos x$ on the closed interval $[0,2 \\pi]$. (GRAPH CAN'T COPY). (a) When the value of $\\sin x$ is a maximum, what is the corresponding value of $\\cos x ?$ (b) When the value of $\\cos x$ is a maximum, what is the correspo... ##### Determine whether a proportional linear relationship exists between the two quantities shown in each of the functions indicated. Explain your reasoning.Exercise 7 Determine whether a proportional linear relationship exists between the two quantities shown in each of the functions indicated. Explain your reasoning. Exercise 7... ##### Asaudaki tizilerdcn hirgisi yekiraek erten ve en buvuk slt sinin dlzidir?olan bit{l-1-{\"2}{ea}-] ={2n{n}R-1{6- {\"n{nE 1 (n+i Asaudaki tizilerdcn hirgisi yekiraek erten ve en buvuk slt sinin dlzidir? olan bit {l-1-{\"2} {ea}-] ={2n {n}R-1 {6- {\"n {nE 1 (n+i... ##### A. Write each linear system as a matrix equation in the form $A X=B$ b. Solve the", "– 3y + 6 = 0 4. g(x) = sin(x) 5. y = x^3 – 1 Notes: • You can type linear equations in the following forms: y = ax + b, f(x) = a(x) + b or ax + by + c = 0. • The * is used in multiplication and ^ is used in exponentiation. For example you want to graph, y = 2(x – 3)2, then you should enter y = 2*(x-3)^2. Properties of Graph You can change the labels, colors, thickness and other properties of graphs (and other objects) in GeoGebra. In this tutorial we are going to change the color, label and thickness of the graph g(x) = sin(x). To change the properties of the graph g(x) = sin(x), do the following: 1.) Right click the graph of the sine function then click Object Properties from the context menu. Figure 2 - The context menu that appears when you right-click a graph 2.) In the Basic tab, be sure that the Show label check box is checked. 3.) Choose Name and Value from the Show label drop-down list box. Figure 3 - The Basic tab of the Properties dialog box 4.) To change the color, click the Color tab, then choose your color from the Color palette. 5.) To change the thickness", "for $\\mathcal{P}$ are: 1. $a=(0, 0)$ has labels $\\{0, 1\\}$ 2. $b=(1/3, 0)$ has labels $\\{1, 3\\}$ 3. $c=(1/4, 1/4)$ has labels $\\{2, 3\\}$ 4. $d=(0, 1/3)$ has labels $\\{0, 2\\}$ The vertics and labels for $\\mathcal{Q}$ are: 1. $w=(0, 0)$ has labels $\\{2, 3\\}$ 2. $x=(1/3, 0)$ has labels $\\{0, 3\\}$ 3. $y=(1/4, 1/4)$ has labels $\\{0, 1\\}$ 4. $z=(0, 1/3)$ has labels $\\{1, 2\\}$ Let us apply the algorithm: • $(a, w)$ have labels: $\\{0, 1\\}, \\{2, 3\\}$. Drop 0 (arbitrary decision) in $\\mathcal{P}$. • $\\to (b, w)$ have labels: $\\{1, 3\\}, \\{2, 3\\}$. In $\\mathcal{Q}$ drop 3. • $\\to (b, z)$ have labels: $\\{1, 3\\}, \\{1, 2\\}$. In $\\mathcal{P}$ drop 1. • $\\to (c, z)$ have labels: $\\{2, 3\\}, \\{1, 2\\}$. In $\\mathcal{Q}$ drop 2. • $\\to (c, y)$ have labels: $\\{2, 3\\}, \\{0, 1\\}$. Fully labeled vertex pair. We now return the strategy pair by normalising these vertices: $$((1/2, 1/2), (1/2, 1/2))$$ This is also implemented in Nashpy: In [1]: import numpy as np import nashpy as nash A = np.array([[1, -1], [-1, 1]]) matching_pennies = nash.Game(A) matching_pennies.lemke_howson(initial_dropped_label=0) Out[1]: (array([ 0.5, 0.5]), array([ 0.5, 0.5])) You can also iterate over all possible starting labels: In [2]: for eq in matching_pennies.lemke_howson_enumeration(): print(eq) (array([ 0.5, 0.5]), array([ 0.5, 0.5])) (array([ 0.5, 0.5]), array([ 0.5, 0.5])) (array([", "geometric figures. /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ The first line real f1 (real x) {return sin(x);} defines the sin() function to be drawn, and the next line draw(graph(f1,-6.21,8.03)); actually draws the graph to the diagram. The graph() command comes from the Asymptote graph package. The next few lines draw several line segments. The attribute zzttqq is a pen whose color is mapped from some rgb value (with conversions ... 8 = y, 9 = z, a = a, b = b, ... ). ## Options An outline of the different available options. (The first three options are related and automatically update with respect to eath other). X/Y Units (cm): Indicates how many units in the picture correspond to 1 centimeter. Picture width/height: Indicates what size the Asymptote output should be, in centimeters. xmin/xmax/ymin/ymax: Indicates the dimensions of the picture in the picture's coordinates. (The next options change other features). Font size: Changes the default font size. The value defaults at 10pt, but I personally recommend 7pt for any file with multiple text lines, which more closely approximates the size of the font", "labels $\\{0, 3\\}, \\{1, 2\\}$. Fully labeled vertex pair. • Dropping label 2: • $(a, w)$ have labels $\\{0, 1\\}, \\{2, 3\\}$. Drop 2. • $\\to (a, x)$ have labels $\\{0, 1\\}, \\{0, 3\\}$. In $\\mathcal{P}$ drop 0. • $\\to (b, x)$ have labels $\\{1, 2\\}, \\{0, 3\\}$. Fully labeled vertex pair. • Dropping label 3: • $(a, w)$ have labels $\\{0, 1\\}, \\{2, 3\\}$. Drop 3. • $\\to (a, y)$ have labels $\\{0, 1\\}, \\{1, 2\\}$. In $\\mathcal{P}$ drop 1. • $\\to (c, y)$ have labels $\\{0, 3\\}, \\{1, 2\\}$. Fully labeled vertex pair. [2] We see that we have obtained two equilibria: $$(b, x) = ((1/3, 0), (1/4, 0))$$$$(c, y) = ((0, 1/4), (0, 1/3))$$ which gives the following two Nash equilibria: $$((1, 0), (1, 0))$$$$((0, 1), (0, 1))$$ [2] Some code to verify: In [4]: import nashpy as nash A, B = np.array([[3, 1], [0, 2]]), np.array([[2, 1], [0, 3]]) game = nash.Game(A, B) for label, eq in enumerate(game.lemke_howson_enumeration()): print(label, eq) 0 (array([ 1., 0.]), array([ 1., 0.])) 1 (array([ 0., 1.]), array([ 0., 1.])) 2 (array([ 1., 0.]), array([ 1., 0.])) 3 (array([ 0., 1.]), array([ 0., 1.])) ### (g) Using a different initial vertex pair¶ • $(b, x)$ has labels $\\{1, 2\\}, \\{0, 3\\}$. Drop 3. • $\\to (b, z)$ have labels $\\{1," ]

[ "Problems Problem 1: Solve: tan(arcsin 12/13) Problem 2: Find the value of x, cos(arccos 1) = cos x", "# Solve $x \\arccos(x)+x/2=\\cos(2x)$ $$x \\arccos(x)+x/2=\\cos(2x)$$ I dont know how to solve this one. It looks relatively easy but it is not, not for me at least. Anybody to help? - Hint, plot each side of the equation and see if there is a point of intersection. This will reveal a lot and hopefully provide you with a next step/approach to solving it. – Amzoti Oct 26 '12 at 22:48 I don't think there are ways of solving this other than graphically. – rubik May 17 '14 at 11:50", "0$$ $$\\cos(x + \\arccos(\\frac{2}{\\sqrt{5}})) = 0$$ I guess you can take it from here", "only solutions. Like Barry Cipra, $$2\\arcsin x=\\arccos(1-x)$$ Now $0\\le\\arccos(1-x)\\le\\pi$ and $-\\pi\\le2\\arcsin x\\le\\pi$ $\\implies\\arcsin x\\ge0\\iff x\\ge0$ Now for $\\arcsin x\\ge0,2\\arcsin x=\\begin{cases}\\arccos(1-2x^2)&\\mbox{if }x\\ge0\\\\ -\\arccos(1-2x^2)& \\mbox{if }x<0\\end{cases}$ $x\\ge0\\implies 1-x=1-2x^2$ Can you take it from here?", "(cot 1x) = x , by Equation 1.8.8. In other words, the range of cos-1 is restricted to [0, 180] or [0, ].Note: arccos refers to \"arc cosine\", or the radian measure of the arc on a circle corresponding to a given value of cosine.", "$\\arccos: \\left[-1, 1\\right]\\rightarrow\\left[0, \\pi\\right]$ so, $\\sqrt{\\arccos x}$ exists for all $x\\in [-1,1]$ .", "2. Let $\\phi = \\arccos (1/3)$ so $\\cos (\\phi ) = 1/3$. Look at the diagram. What is $\\tan (\\phi ) = ?$", "y= arccos x then x= cos y so dx= - sin y dy. $$\\int arccos x dx= -\\int y sin y dy$$", "# How do you find the exact values of arccos1? $\\arccos 1 = 0$ $\\arccos x$ is the angle say $\\theta$ whose cosine is $x$ i.e. an angle for which $\\cos \\theta = x$ As $\\cos 0 = 1$, $\\arccos 1 = 0$", "Function ArcCos(x) # ArcCos ArcCos( x ! (input) numerical expression ) ## Arguments x A scalar numerical expression in the range $$[-1,1]$$. ## Return Value The ArcCos function returns the arccosine of x in the range 0 to $$\\pi$$ radians. Note • A run-time error results if x is outside the range $$[-1,1]$$. • The function ArcCos can be used in constraints of nonlinear mathematical programs. The functions ArcSin, ArcTan, Cos. Arithmetic functions are discussed in full detail in Arithmetic Functions of the Language Reference.", "at 21:49 These are kind of trivial, but maybe they are closed enough for you: $$\\arccos(z) = i \\cdot \\log\\left(z - i\\sqrt{1-z^2}\\right)$$ or $$\\arccos(z) = \\int_z^1 \\dfrac{dz}{\\sqrt{1-z^2}}$$ -", "help. $(arccos(1-h))^{2}= 19.62ht^{2}$ what i need to do is solve for H Please note that i do indeed understand what arccos is, it is the squared that id like to get rid of. Any help would be appreciated.", "# Trigonometric sprint!!can u come first?? Calculus Level 3 Challenge:Do the problem within 20 seconds using your common sense. arccos x means inverse cosine of x defined within [-1,1] Consider the equation arccos x + arccos y + arccos z = 3(pi) Hence find the value of 9(xy + yz + zx) ??? Once done post how much time did it take you honestly. YOUR TIME STARTS NOW!!!HURRY ×", "+ 22.3^2 - 18.1^2) / 441.51 A = arccos((9.9^2 + 22.3^2 - 18.1^2) / 441.51) = 0.919389657 radians Area = 0.5bcSinA = 87.8 cm^2 Thanks, that's what i was looking for :D", "# What does 3cos(arcsin(5))-6sec(arctan(5)) equal? For any a, $| \\sin a | \\le 1$. Indeed this applies to arc cos also.", "Solve 2cos x + 3sin x = 0.5", "$n \\ge 2$. In the next post, we’ll consider how this same problem can be solved — but only in two dimensions — using arctangent.", "any number, x, between 0 and $\\pi$, arccos(cos(x))= x. That follows from the very definition of \"arccos\".", "arccos calculator If you're searching for an easy tool to find the inverse cosine, look no further - our arccos calculator is a perfect match. Input $$arccos=0.5$$ Solution $$Angle\\ in\\ degrees=60$$ $$Angle\\ in\\ radians=1.04719755$$ If you're searching for an easy tool to find the inverse cosine, look no further - our arccos calculator is a perfect match. A.", "overhead ra... ##### Find all solutions of the equation in the interva [0, 21).2cos x-3 cosx-] = 0Write your answer in radians in terms of I. If there is more than one solution, separate them with commasDD; Find all solutions of the equation in the interva [0, 21). 2cos x-3 cosx-] = 0 Write your answer in radians in terms of I. If there is more than one solution, separate them with commas DD;..." ]

[ "have any ideas for the second equation ? It's basically the same drill. Take the cosine of both sides: $\\cos(\\arccos x + \\arccos (1-x)) = -x$ $\\cos(\\arccos x)\\cos(\\arccos(1-x)) - \\sin(\\arccos x)\\sin(\\arccos(1-x)) = -x$ $x(1-x) - \\sqrt{(1-x^2)(2x-x^2)} = -x$ $x(1-x) + x = \\sqrt{(1-x^2)(2x-x^2)}$ $(x(1-x) + x)^2 = (1-x^2)(2x-x^2)$ $(2x-x^2)^2 = (1-x^2)(2x-x^2)$ Now x = 2 satisfies this equation. We can look for other solutions by dividing by $2x - x^2$ if $x \\neq 2$: $2x - x^2 = 1 - x^2$ $2x = 1$ $x = \\frac{1}{2}$ and since $\\arccos 2$ is undefined, the only solution that works is $x = \\frac{1}{2}$.", "Example 5 Find the solution to the trigonometric equation $sin^2(2x)+2cos^2(2x) = -\\frac{\\sqrt{2}}{2}$. ### Solution Since this problem has two different types of trigonometric functions, it is necessary to use trigonometric identities so there is only one function. Then, it will be possible to solve for $x$. In this case, there is a sine squared function and a cosine squared function, so it makes sense to consider the Pythagorean identities. Specifically $sin^2x+cos^2x=1$. This is true for any angle. Even though the inside of the trig functions in this case is $2x$, $2x$ is still just an angle. Therefore, $sin^2(2x)+cos^2(2x) = 1$ too. Expanding the given equation yields $sin^2(2x)+cos^2(2x)+cos^2(2x) = -\\frac{\\sqrt{2}}{2}$. The Pythagorean identity shows that this equation is equivalent to: $1+cos^2(2x) = -\\frac{\\sqrt{2}}{2}$. Subtracting $1$ from both sides makes this: $cos^2(2x) = \\frac{2-\\sqrt{2}}{2}$. Then, take the square root of both sides to get: $cos(2x) = \\sqrt{\\frac{2-\\sqrt{2}}{2}}$. Now, the equation is in the form $fun(f(x))=a$. Therefore, take the arccosine of both sides. This is: $2x = arccos(\\sqrt{\\frac{2-\\sqrt{2}}{2}}$. Finally divide both sides by $2$ to solve for $x$. $x = \\frac{arccos(\\sqrt{\\frac{2-\\sqrt{2}}{2}}}{2}$ radians. Although the right side of this equation looks complicated, it is just a constant. Plugging the values into a calculator reveals that it has the decimal approximation of 0.50 radians. This is the principal solution. Since cosine is $2\\pi$", "Sum Formula Originally Posted by Bashyboy \"Write cos(arctan 1 + arccos x) as an algebraic expression.\" So, I understand that I am to use formula for cos(u + v) because it fits the expression, where u = arctan 1 and v = arccos x. I also know that I must create to right triangles for each of the angles u and v, the reason being that the angles are likely to be different. What I do not understand is why, when the formula is used, we don't evaluate cos(arctan 1) as cos(pi/4), which then equals 0. In the book, cos(arctan 1) is evaluated as 1/sqroot(2). Perhaps I do not understand inverse trigonometric functions as I thought I did. An alternative is to note that \\displaystyle \\begin{align*} \\sin^2{X} + \\cos^2{X} &\\equiv 1 \\\\ \\frac{\\sin^2{X}}{\\cos^2{X}} + \\frac{\\cos^2{X}}{\\cos^2{X}} &\\equiv \\frac{1}{\\cos^2{X}} \\\\ \\tan^2{X} + 1 &\\equiv \\frac{1}{\\cos^2{X}} \\\\ \\frac{1}{\\tan^2{X} + 1} &\\equiv \\cos^2{X} \\\\ \\cos{X} &\\equiv \\sqrt{\\frac{1}{\\tan^2{X} + 1}} \\end{align*} So that means \\displaystyle \\begin{align*} \\cos{\\left(\\arctan{1}\\right)} &= \\sqrt{\\frac{1}{\\left[\\tan{\\left(\\arctan{1}\\right)}\\right]^2 + 1}} \\\\ &= \\sqrt{\\frac{1}{1^2 + 1}} \\\\ &= \\sqrt{\\frac{1}{1 + 1}} \\\\ &= \\sqrt{\\frac{1}{2}} \\\\ &= \\frac{\\sqrt{1}}{\\sqrt{2}} \\\\ &= \\frac{1}{\\sqrt{2}} \\end{align*}", "# Simplify $\\sin{(\\arctan{2}-\\arcsin{\\frac{1}{\\sqrt{10}}})}.$ First I denote $x=\\arctan{2}$ and $y=\\arcsin{\\frac{1}{\\sqrt{10}}}$ and then use the addition formula for sine: $$\\sin{(x-y)}=\\sin{x}\\cos{y}-\\cos{x}\\sin{y}=\\sin{x}\\cos{y}-\\cos{x}\\cdot \\frac{1}{\\sqrt{10}}.$$ Now I use the fact that $\\cos{y}=1-\\sin^2{y}$ which gives $$\\cos{y}=1-\\sin^2{\\left(\\arcsin{\\frac{1}{\\sqrt{10}}}\\right)}=1-\\frac{1}{10}=\\frac{9}{10}$$ So I have reduced the problem to the following $$\\sin{x}\\cdot \\frac{9}{10}-\\cos{x}\\cdot \\frac{1}{\\sqrt{10}}.$$ Here I'm stuck. I cant just divide this expression by $(\\sin{x}\\cos{x})$ because I'd still be left with sine and cos terms and I'd also change the value of the expression. Everything now boils down to compute $\\sin(\\arctan{2})$ and $\\cos{\\arctan2}.$ I also tried rewriting $\\cos(\\arctan{2})$ as $1-\\sin^2(\\arctan{2}),$ but to no avail. Any suggestions on 1. how to proceed from where I left; 2. how tocompute this by means more effective; 3. both of the above. • It should be $\\cos y=\\sqrt{1-\\sin^2 y}$ – GAVD Sep 16 '17 at 12:07 $$\\sin(\\arctan(x)) = \\frac x {\\sqrt{x^2 + 1}}$$ Which you can prove using, $$1 + \\cot^2(\\arctan(x)) = \\csc^2(\\arctan(x))$$ By the Pythagorean identity. From which, $$1 + \\frac 1 {x^2} = \\frac{1}{\\sin^2(\\arctan(x))}$$ $$\\frac {x^2} {x^2+1} = \\sin^2(\\arctan(x))$$ Taking square roots, choosing the positive branch, $$\\frac x {\\sqrt{x^2 + 1}} = \\sin(\\arctan(x)) \\blacksquare$$ Put $\\arctan 2 = x$ then $\\tan x= 2$, so $\\sin x = {2\\over \\sqrt{5}}$ and $\\cos x = {1\\over \\sqrt{5}}$. Let $\\arcsin {1\\over \\sqrt{10}}= y$ so $\\sin y = {1\\over \\sqrt{10}}$, thus $\\cos y = {3\\over \\sqrt{10}}$. Now:", "# $\\sin(2\\arccos(x))$, please help me understand how to do these kind of problems. We need to be able to transform this equation to get rid of the trig functions. To better explain this, this is how the problem before this one was done. (I checked the answer, i got this one right.) $$\\sin(\\arccos(x))$$ We substitute $\\arccos(x)$ with the symbol $\\theta$. So $$\\sin(\\theta), \\text{ where }\\theta = \\arccos(x),\\text{ so }\\cos(\\theta)=x.$$ So if we draw a right triangle with angles A, B, and 90*. and the sides being $h$ (hypotenuse), $o$ (opposite), and $a$ (adjacent). Angle A = $\\theta$, so side $a$ is equal to $x$ and the side $h$ is equal to $1$. With Pythagorean theorem $x^2 + o^2 = 1^2$, that means that the opposite side is equal to $\\sqrt{1 - x^2}$. With this triangle drawn, and our $\\theta$ still being the same, this means that our $\\sin(\\theta)$ is equal to the $o/h$ of our drawn triangle. So this equations morphs into $\\sqrt{1 - x^2}/1$ How would i do this with $\\sin(2\\arccos(x))$ or $\\tan(\\arccos(x) + \\arcsin(x))$? Edit: From the answer given, i will try to solve these two equations. $$\\sin(2\\arccos(x)) = 2\\sin(\\arccos(x))\\cos(\\arccos(x))$$ $$\\sin(\\arccos(x))$$ $$\\sin(\\theta), \\theta = \\arccos(x)$$ $$\\cos(\\theta) = x$$ A = $\\theta$, hypotenuse = 1, adjacent = x, opposite = $\\sqrt{1 - x^2}$ $$\\sin(\\theta) = \\sqrt{1 -", "1. ## Having some trouble I couldnt figure this one out...can anyone? cos(arcsin3/5 + arccos sqrt3/2) Thanks! 2. Originally Posted by Sinsane I couldnt figure this one out...can anyone? cos(arcsin3/5 + arccos sqrt3/2) Thanks! Start by letting $\\sin^{-1} \\frac{3}{5} = \\alpha \\Rightarrow \\sin \\alpha = \\frac{3}{5}$ and $\\cos^{-1} \\frac{\\sqrt{3}}{2} = \\beta \\Rightarrow \\cos \\beta = \\frac{\\sqrt{3}}{2}$. Your task is to calculate $\\cos (\\alpha + \\beta)$. Hint: Start by applying the compound angle formula and then make the necessary substitutions. 3. Thank you..I finally figured it out before you said something..but I appreciate the help. But its always good to make sure something is right haha.", "How to compute $\\cos(\\arctan(2)) = 1/\\sqrt{5}$ I'm doing matrices and I rotated a line about an angle. The gradient of my line I'm rotating to the $x$-axis is $2$, from $y=2x$. So obviously the angle that the line $y=2x$ makes with the $x$-axis is $\\arctan(2)$. My question is how do I arrive at $\\cos(\\arctan(2)) = 1/\\sqrt{5}$. <--- The $1/\\sqrt{5}$ is what I'm confused with, how do I get this? • Draw a right angled triangle such that the tan of the one of the angles is $2$. Then calculate the cosine of that angle Jan 17 '16 at 20:47 We have $$\\cos(x)=\\frac{1}{\\sqrt{\\tan^2x+1}}$$ for $0\\le x\\le \\frac{\\pi}{2}$. This implies $$\\cos(\\arctan(2))=\\frac{1}{\\sqrt{\\tan(\\arctan(2))^2+1}}= \\frac{1}{\\sqrt{2^2+1}}=\\frac{1}{\\sqrt{5}}$$ because of $0\\le\\arctan(2)\\le \\frac{\\pi}{2}$ Well.. Let $\\arctan 2 = x$. $\\arctan (2) = x \\implies \\tan x = \\sin x/\\cos x = 2 \\implies \\sin x = 2 \\cos x \\implies \\sin^2 x + \\cos^2 x = 4\\cos^2 x + \\cos^2 x = 1 \\implies \\cos^2 x = 1/5 \\implies \\cos x = \\pm 1/\\sqrt{5}$. By convention $\\arctan$ takes values in (-$\\pi/2, \\pi/2$) where $\\cos$ is presumed to be positive. So $\\cos x = 1/\\sqrt{5}$. ==== In general if $\\tan x = b$ then $\\sin x = b \\cos x$ so $\\sin^2 x + \\cos^2 x = (b^2 + 1) \\cos^2 x$ so $\\cos x = 1/\\sqrt{b^2 +", "x - 2\\sin x \\cos x + \\cos^2 x = \\frac{1}{2}$$ Factor the left side: $$(\\sin x + \\cos x)^2 = \\frac{3}{2}$$ $$(\\sin x - \\cos x)^2 = \\frac{1}{2}$$ Take the square root: $$\\sin x + \\cos x = \\pm\\sqrt{\\frac{3}{2}}$$ $$\\sin x - \\cos x = \\pm\\sqrt{\\frac{1}{2}}$$ Add the two equations: $$2\\sin x = \\pm\\sqrt{\\frac{3}{2}} \\pm \\sqrt{\\frac{1}{2}}$$ And now finally we can discover the sine of the original angle: $$\\sin x = \\pm \\frac{\\sqrt{6}\\pm\\sqrt{2}}{4}$$ And then we need to apply $\\arcsin$ and do some fiddling with \"which equivalent angles do I take?\". What a mess. Instead, just find $\\arcsin \\frac{1}{2} = 2x$ and use supplemental and coterminal angles to find the other values of $2x$ that work. • snazzy approach, i would have just given up trying after getting to $2\\sin x \\cos x$ – MichaelChirico Apr 14 '16 at 0:41", "# What does cos(arctan(2)) equal? $\\cos \\left(\\arctan \\left(2\\right)\\right) = \\frac{1}{\\sqrt{5}} = \\frac{\\sqrt{5}}{5}$ #### Explanation: arctan(2) is an angle whose tangent function$= \\frac{2}{1}$ this means that the angle has opposite side of $2$ and adjacent side of $1$ the hypotenuse measures $\\sqrt{5}$ cosine of the angle=adjacent side/hypotenuse$= \\frac{1}{\\sqrt{5}} = \\frac{\\sqrt{5}}{5}$ God bless America!", "# Solving an inverse trig equation 1. Feb 28, 2009 ### Kaylee! 1. The problem statement, all variables and given/known data Solve for x $$arcsin(2x) + arccos(x) = \\frac{\\pi}{6}$$ 2. Relevant equations 3. The attempt at a solution Subtracting arccos(x) from both sides and then taking the sine of both sides: $$2x = sin\\left[\\frac{\\pi}{6} - arccos(x)\\right]$$ Applying the difference angle identity for sine: $$2x = sin \\frac{\\pi}{6} cos[arccos (x)] - cos \\frac{\\pi}{6} sin [arccos (x)]$$ We can draw a right triangle with a hypotenuse of 1, angle opposite of theta of $$\\sqrt{1-x^2}$$, and angle adjacent to theta of x. From this triangle we can deduce the values of cos( arccos(x) ) and sin( arccos(x) ) $$2x = (\\frac{1}{2})x - \\frac{\\sqrt{3}}{2}{\\sqrt{1-x^2}}$$ Multiplying each side by 2, subtracting x from each side, squaring each side, and then dividing each side by 3 $$3x^2 = 1 - x^2$$ Adding x^2 to each side, and then dividing each side by 4 $$x^2 = \\frac{1}{4}$$ At this point I need to take the square root of each side. I know it's the negative from trying each out, but how do I show this algebraically? 2. Feb 28, 2009 ### csprof2000 When you say 2x = x/2 - sqrt(3)/2 sqrt(1-x^2)... 3x = -sqrt(3)sqrt(1-x^2). But since sqrt(1-x^2) is positive, the RHS is negative, which is", "the side lengths is zero, we have only two solutions (since we then have a horizontal (or vertical) line, which means only one choice of \"diagonal\"). If both are zero, then it's just a point on the circle, which can be anywhere, and hence $x+y$ is not determined. • So, when a and b are different and non-zero $y = arctan(-a/b) - arcsin(\\sqrt{{(a^2+b^2)}/4})$. And $x = arctan(-a/b) + arcsin(\\sqrt{{(a^2+b^2)}/4})$ – Sam Baker Mar 19 '15 at 16:37 Hint: If you square the two equations and add them, the square terms sum to $1$ and the cross terms give you $\\cos(x-y)$ so you can get $x-y$. If you multiply the two you get $\\sin(x+y)+\\cos(x-y)$. Not the most elegant, but it will work. i don't think there is a unique answer to this problem. here is a reason why. take a chord $AB$ of length $\\sqrt{0.2187^2 + (-0.469)^2} = 0.5175$ on the unit circle. let $A=(\\cos x, \\sin x), B = \\cos y, \\sin y)$ only the difference $x - y$ is unique and equals $2 \\sin^{-1}(0.5175) = 62.327^\\circ$ I don’t have a solution, but maybe this can help you or someone along? Squaring both equations gives: $$0.2187^2=(\\cos x-\\cos y)^2\\implies\\\\ 0.04782969=\\cos^2x-2\\cos x\\cos y+\\cos^2y$$ $$(-0.469)^2=(\\sin x-\\sin y)^2\\implies\\\\ 0.1025703=\\sin^2x-2\\sin x\\sin y+\\sin^2y$$ $$0.15039999=(\\sin^2x+\\cos^2x)+2\\sin x\\sin y+2\\cos x\\cos y+(\\sin^2y+\\cos^2y)$$ Terms in parentheses add to $$1$$,", "\\arccos x$ $\\arccsc \\frac{1}{x} = \\arcsin x$ If you only have a fragment of a sine table: $\\arccos x = \\arcsin \\sqrt{1-x^2},\\text{ if }0 \\leq x \\leq 1$ $\\arctan x = \\arcsin \\frac{x}{\\sqrt{x^2+1}}$ Whenever the square root of a complex number is used here, we choose the root with the positive real part (or positive imaginary part if the square was negative real). From the half-angle formula $\\tan \\frac{\\theta}{2} = \\frac{\\sin \\theta}{1+\\cos \\theta}$, we get: $\\arcsin x = 2 \\arctan \\frac{x}{1+\\sqrt{1-x^2}}$ $\\arccos x = 2 \\arctan \\frac{\\sqrt{1-x^2}}{1+x},\\text{ if }-1 < x \\leq +1$ $\\arctan x = 2 \\arctan \\frac{x}{1+\\sqrt{1+x^2}}$ ## Relationships between trigonometric functions and inverse trigonometric functions $\\sin (\\arccos(x)) = \\cos(\\arcsin(x)) = \\sqrt{1-x^2}$ $\\sin (\\arctan(x)) = \\frac{x}{\\sqrt{1+x^2}}$ $\\cos (\\arctan(x)) = \\frac{1}{\\sqrt{1+x^2}}$ $\\tan (\\arcsin(x)) = \\frac{x}{\\sqrt{1-x^2}}$ $\\tan (\\arccos(x)) = \\frac{\\sqrt{1-x^2}}{x}$ ## General solutions Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of 2π. Sine and cosecant begin their period at 2πk − π/2 (where k is an integer), finish it at 2πk + π/2, and then reverse themselves over 2πk + π/2 to 2πk + 3π/2. Cosine and secant begin their period at 2πk, finish it at 2πk + π, and then reverse themselves over 2πk + π to 2πk + 2π. Tangent", "- \\sin(\\arccos x)\\sin(\\arccos(1-x)) = -x$ $x(1-x) - \\sqrt{(1-x^2)(2x-x^2)} = -x$ $x(1-x) + x = \\sqrt{(1-x^2)(2x-x^2)}$ $(x(1-x) + x)^2 = (1-x^2)(2x-x^2)$ $(2x-x^2)^2 = (1-x^2)(2x-x^2)$ Now x = 2 satisfies this equation. We can look for other solutions by dividing by $2x - x^2$ if $x \\neq 2$: $2x - x^2 = 1 - x^2$ $2x = 1$ $x = \\frac{1}{2}$ and since $\\arccos 2$ is undefined, the only solution that works is $x = \\frac{1}{2}$.", "# Math Help - y=arctan(x) , show that.... 1. ## y=arctan(x) , show that.... Hi I am not sure about my solution, if it´s done \"in the right way\". Anyway, here it comes: $\\mbox{ Let } y=arctan(x) \\, \\mbox{ , show that } cos^{2}(y)=\\frac{1}{1+x^{2}}$ I did: $tan(y)=x \\Longleftrightarrow \\frac{sin(y)}{cos(y)}=x \\Rightarrow cos(y)=\\frac{sin(y)}{x}$ Now raising both sides to the second power... $cos^{2}(y)=\\frac{1-cos^{2}(y)}{x^{2}} \\Longleftrightarrow cos^{2}(y)=\\frac{1}{1+x^{2}}$ Thanks! 2. Given: $y = arctan(x)$ Show that: $\\cos^{2}(y)=\\frac{1}{1+x^{2}}$ Looks right to me. But here's another way, just in case: Draw a right triangle. Label one of the acute angles y. Since $y = arctan(x)$, $\\tan y = x = \\frac{x}{1}$. Label the opposite side x and the adjacent side 1. Find the hypotenuse (s) by using the Pythagorean theorem: \\begin{aligned} s^2 &= 1^2 + x^2 \\\\ s &= \\sqrt{1 + x^2} \\end{aligned} Cosine is adjacent over hypotenuse, so $\\cos y = \\frac{1}{\\sqrt{1 + x^2}}$. Squaring both sides gives you $\\cos^2 y = \\frac{1}{1 + x^2}$. 01 3. Hello, Twig! Your work looks fine to me . . . Let: $y\\:=\\:\\arctan x$, show that: . $\\cos^2\\!y\\:=\\:\\frac{1}{1+x^2}$ Here's my approach . . . We have: . $y \\:=\\:\\arctan x\\quad\\Rightarrow\\quad \\tan y \\:=\\:x$ Square both sides: . $\\tan^2\\!y \\:=\\:x^2$ $\\text{Add 1 to both sides: }\\;\\underbrace{1 + \\tan^2\\!y}_{\\text{This is }\\sec^2\\!y} \\:=\\:1+x^2 \\quad\\Rightarrow\\quad \\sec^2\\!y \\:=\\:1+x^2$ Take reciprocals: . $\\cos^2\\!y \\:=\\:\\frac{1}{1+x^2}$", "how to evaluate $\\arcsin(x)-\\arccos(x)=\\arccos(\\frac{\\sqrt{3}}{2})$ Question how to evaluate $\\arcsin(x)-\\arccos(x)=\\arccos(\\frac{\\sqrt{3}}{2})$ Thoughts Do you cos on both sides to finally do quadratic equation to solve at the end of the problem? so i got $0=4x^2+4x-1+2\\sqrt{3}$ on my last step but i dont know if this is the right approach Note that $\\arcsin(x)+\\arccos(x) =\\pi/2$. Therefore $\\arcsin(x)-\\arccos(x) =\\arccos(\\frac{\\sqrt{3}}{2})$ becomes $\\arccos(\\frac{\\sqrt{3}}{2}) =\\pi/2-2\\arccos(x)$ or $\\arccos(x) =(\\pi/2-\\arccos(\\frac{\\sqrt{3}}{2}))/2$. You should be able to take it from here. You can't take the cosine of both sides, at least not the way you probably did. You must do it as follows: $$\\cos(\\arcsin(x)-\\arccos(x))=\\frac{\\sqrt3}2$$ And then use sum of angles identity: $$\\cos(\\arcsin(x))\\cos(\\arccos(x))+\\sin(\\arcsin(x))\\sin(\\arccos(x))\\\\=x\\cos(\\arcsin(x))+x\\sin(\\arccos(x))$$ And use pythagorean theorems (let's assume all positive) $$=x\\sqrt{1-\\sin^2(\\arcsin(x))}+x\\sqrt{1-\\cos^2(\\arccos(x))}\\\\=2x\\sqrt{1-x^2}$$ Can you take it from here? From the unit circle, $\\displaystyle \\arccos\\frac{\\sqrt3}2 = \\frac\\pi6$. A known property relating inverse sine and inverse cosine is $\\arcsin x + \\arccos x = \\dfrac\\pi2$. Therefore: \\begin{align*} \\arcsin x - \\arccos x &= \\arccos \\frac{\\sqrt3}2\\\\[0.3cm] \\frac\\pi2 - \\arccos x - \\arccos x &= \\frac\\pi6\\\\[0.3cm] -2\\arccos x &= -\\frac\\pi3\\\\[0.3cm] \\arccos x &= \\frac\\pi6 \\end{align*} Now take the cosine of both sides. • Same idea as mine, and you went all the way. – marty cohen Jan 20 '17 at 1:53 • @martycohen, I could say likewise (except for the second part) as you posted yours less than 1 minute before mine. And I dunno", "# The Simplest Trigonometric Equations Equation sin(x)=a where |a|<=1 has infinitely many roots. For example roots of the equation sin(x)=1/2 are following: x_1=(pi)/6,x_2=(5pi)/6,x_3=pi/6+2pi,x_4=pi/6-2pi etc. The common formula, that allows to find all roots of the equation sin(x)=a is following: Fact 1. If sin(x)=a, where |a|<=1 then x=(-1)^n arcsin(a)+pin,n in Z. Here n can take integer values, and for any value of n we obtain some root of the equation. Fact 2. If cos(x)=a, where |a|<=1 then x=+- arccos(a)+2pin,n in Z. Fact 3. If tan(x)=a, then x=arctan(a)+pin,n in Z. Fact 4. If cot(x)=a, then x=text(arccot)(a)+pin,n in Z. Example 1. Solve sin(x)=1/2. Using Fact 1 we obtain that x=(-1)^n arcsin(1/2)+pin,n in Z. Since arcsin(1/2)=pi/6, then finally we obtain that x=(-1)^n pi/6+pin, n in Z. Example 2. Solve cos(3x)=-1/(sqrt(2)). Using Fact 2 we obtain that 3x=+-arccos(-1/(sqrt(2)))+2pin,n in Z. Since arccos(-1/(sqrt(2)))=pi-arccos(1/(sqrt(2)))=pi-pi/4=(3pi)/4, then we obtain that 3x=+- (3pi)/4+2pin, n in Z or x=+-(pi)/4+(2pin)/3, n in Z. Example 3. Solve tan(x-pi/6)=-sqrt(3). According to Fact 3 x-pi/6=arctan(-sqrt(3))+pin,n in Z. Since arctan(-sqrt(3))=-arctan(sqrt(3))=-pi/3, then we obtain that x-pi/6=-pi/3+pin, n in Z, i.e. x=-pi/6+pin, n in Z. Note, that in some cases it is more convenient to use particular formulas (in all formulas n in Z): 1. If sin(x)=0 then x=pin; 2. If sin(x)=1 then x=pi/2+2pin; 3. If sin(x)=-1 then x=-pi/2+2pin; 4. If cos(x)=0 then x=pi/2+pin; 5. If", "Multiply by $\\sqrt{2}:\\;\\;y \\;=\\;\\sqrt{2}\\cos\\left(\\theta - \\frac{\\pi}{4}\\right)$ 7. Hello Paymemoney Originally Posted by Paymemoney so from the formula: cos2A are you making arccos(x) = A? OK. Let's see if we can make it easier. I'll set it out slightly differently. Let $\\arccos(x)=A$ Now $\\arccos...$ means 'the angle whose cosine is ...'. So $\\arccos(x) = A$ means the angle whose cosine is $x$ is $A$ In other words $\\cos A = x$, which I'll call equation (1) Now the question asks us for $\\cos(2\\arccos(x))$ In other words, for $\\cos2A$ So we use the well-known double angle formula $\\cos2A=2\\cos^2A-1$ and then use equation (1) to replace $\\cos A$ by $x$. So we get: $\\cos(2\\arccos(x)) = \\cos2A$ $=2\\cos^2A -1$ $=2x^2-1$ OK now? Originally Posted by Paymemoney I though that cos(2arccos(x)) = 2x, why wouldn't this work? I'm not sure why you should think this. This would mean that if you double an angle you double its cosine. So $\\cos2A$ would equal $2\\cos A$. Which it doesn't, does it?", "example, if we set d=3, r1=1, and r2=2, the intersection area is zero. However, your last formula would give $r1^2 \\arccos 2+ r2^2 \\arccos \\frac{1}{2}$, which is clearly wrong. I also have some doubts about the last term $\\sqrt{Y}/2$, since calculating the area using integrals yields, apart from the terms containing the arccos, two different radicals that cannot be merged. Could you check your calculations? – Anatoly Aug 30 '14 at 0:09 I have just posted the result obtained using integrals. I hope that this could facilitate your check. – Anatoly Aug 30 '14 at 0:12 I derived my own proof with simple geometry and trigonometry. I ended up with a slightly different formula: $$A=r_1^2 \\text{arccos}\\left(\\frac{d^2-(r_2^2-r_1^2)}{2dr_1}\\right) + r_2^2 \\text{arccos}\\left(\\frac{d^2+(r_2^2-r_1^2)}{2dr_2}\\right)-d\\sqrt{r_1^2-\\left(\\frac{d^2-(r_2^2-r_1^2)}{2d}\\right)^2}.$$ The last term can be rewritten: $$-d\\sqrt{r_1^2-\\left(\\frac{d^2-(r_2^2-r_1^2)}{2d}\\right)^2} = -\\frac{\\sqrt{Y}}{2},$$ with $$Y=(-d+r_1+r_2)(d+r_2-r_1)(d-r_2+r_1)(d+r_1+r_2).$$ So this is very close to the first submitted equation, apart from a swap between r1 and r2 in the argument of the arccos(). It seems that this corrects the typo pointed out in the comment. For d=3, r1=1, r2=2 it gives $$A=r_1^2 \\text{arccos}\\left(1\\right) + r_2^2 \\text{arccos}\\left(1\\right)+0=0$$ - Set the center of the first circle in $(0,0)$ and that of the second circle in $(d,0)$. The equations of the circles are $y=\\sqrt{r1^2- x^2}$ and $y=\\sqrt{r2^2-(x-d)^2}$. In the segment connecting the two centers, the first and second circle intersect", "can ignore this! 2) $\\csc x = 2 \\implies \\sin x = \\frac{1}{2}$. Draw a 30 - 60 - 90 triangle, with a hypotenuse 2 units long, and figure out the lengths of the shortest side using the fact that this triangle is half of an equilateral triangle. Then you should see that the sine of one of the angle is $\\frac{1}{2}$ 3) $\\sec x = -1 \\implies \\cos x = -1$. Use the same formula as in (1), and the fact that $\\cos 0^o = 1$. Can you complete them now? 3. I don't know if you covered this yet in your class, but an alternative way to solve these is to use the inverse of these trig functions to solve for x. Example: $\\cos x=-\\dfrac{\\sqrt{2}}{2}$. We need to restrict our domain to make the cosine function one-to-one: $[0,\\pi]$. Taking the arccosine of both sides, $\\arccos(\\cos x))=\\arccos \\left(-\\dfrac{\\sqrt{2}}{2}\\right)$ $x = \\arccos \\left(-\\dfrac{\\sqrt{2}}{2}\\right)=\\dfrac{3\\pi}{4}$. We know that $\\cos x = \\cos -x$. So solving $\\cos (-x) = -\\dfrac{\\sqrt{2}}{2}$, $\\arccos(\\cos (-x)) = \\arccos \\left(-\\dfrac{\\sqrt{2}}{2}\\right)$ $-x = \\arccos \\left(-\\dfrac{\\sqrt{2}}{2}\\right) = \\dfrac{3\\pi}{4} \\Rightarrow x =-\\dfrac{3\\pi}{4}$. Since the period of cos x is $2\\pi$, we have the general solution: $x=\\pm \\dfrac{3\\pi}{4} \\pm 2\\pi n$ where $n \\in \\mathbb{Z}$", "# How do you simplify the expression cos(arctan 2x - arcsin x)? Aug 25, 2016 $\\frac{\\sqrt{1 - {x}^{2}} + 2 {x}^{2}}{\\sqrt{1 + 4 {x}^{2}}}$ #### Explanation: Let $a = a r c \\tan 2 x \\in Q 1 \\mathmr{and} Q 4$, wherein cosine is positive. Then, $\\tan a = 2 x , \\sin a = \\frac{2 x}{\\sqrt{1 + 4 {x}^{2}}} \\mathmr{and} \\cos a = \\frac{1}{\\sqrt{1 + 4 {x}^{2}}}$ Let $b = a r c \\sin x \\in Q 1 \\mathmr{and} Q 4$, wherein cosine is positive. Then, $\\sin b = x \\mathmr{and} \\cos b = \\sqrt{1 - {x}^{2}}$. Now, the given expression is $\\cos \\left(a - b\\right)$ $= \\cos a \\cos b + \\sin a \\sin b$ $= \\left(\\frac{1}{\\sqrt{1 + 4 {x}^{2}}}\\right) \\left(\\sqrt{1 - {x}^{2}}\\right) + \\left(\\frac{2 x}{\\sqrt{1 + 4 {x}^{2}}}\\right) \\left(x\\right)$ $= \\frac{\\sqrt{1 - {x}^{2}} + 2 {x}^{2}}{\\sqrt{1 + 4 {x}^{2}}}$. See how it works. Let $x = \\frac{1}{2}$. Then $a = a r c \\sin \\left(\\frac{1}{2}\\right) = \\frac{\\pi}{6} \\mathmr{and} b = a r c \\tan 1 = \\frac{\\pi}{4}$ $\\cos \\left(a - b\\right) = \\cos \\left(\\frac{\\pi}{4} - \\frac{\\pi}{6}\\right) = \\cos \\left(\\frac{\\pi}{12}\\right) = \\frac{\\sqrt{3} + 1}{2 \\sqrt{2}}$ When $x = \\frac{1}{2}$, $\\frac{2 {x}^{2} + \\sqrt{1 - {x}^{2}}}{\\sqrt{1 + 4 {x}^{2}}}$ #=(1/2+sqrt3/2)/sqrt(1+1)@ $= \\frac{\\sqrt{3} + 1}{2 \\sqrt{2}}$" ]

[ "have any ideas for the second equation ? It's basically the same drill. Take the cosine of both sides: $\\cos(\\arccos x + \\arccos (1-x)) = -x$ $\\cos(\\arccos x)\\cos(\\arccos(1-x)) - \\sin(\\arccos x)\\sin(\\arccos(1-x)) = -x$ $x(1-x) - \\sqrt{(1-x^2)(2x-x^2)} = -x$ $x(1-x) + x = \\sqrt{(1-x^2)(2x-x^2)}$ $(x(1-x) + x)^2 = (1-x^2)(2x-x^2)$ $(2x-x^2)^2 = (1-x^2)(2x-x^2)$ Now x = 2 satisfies this equation. We can look for other solutions by dividing by $2x - x^2$ if $x \\neq 2$: $2x - x^2 = 1 - x^2$ $2x = 1$ $x = \\frac{1}{2}$ and since $\\arccos 2$ is undefined, the only solution that works is $x = \\frac{1}{2}$.", "# $\\sin(2\\arccos(x))$, please help me understand how to do these kind of problems. We need to be able to transform this equation to get rid of the trig functions. To better explain this, this is how the problem before this one was done. (I checked the answer, i got this one right.) $$\\sin(\\arccos(x))$$ We substitute $\\arccos(x)$ with the symbol $\\theta$. So $$\\sin(\\theta), \\text{ where }\\theta = \\arccos(x),\\text{ so }\\cos(\\theta)=x.$$ So if we draw a right triangle with angles A, B, and 90*. and the sides being $h$ (hypotenuse), $o$ (opposite), and $a$ (adjacent). Angle A = $\\theta$, so side $a$ is equal to $x$ and the side $h$ is equal to $1$. With Pythagorean theorem $x^2 + o^2 = 1^2$, that means that the opposite side is equal to $\\sqrt{1 - x^2}$. With this triangle drawn, and our $\\theta$ still being the same, this means that our $\\sin(\\theta)$ is equal to the $o/h$ of our drawn triangle. So this equations morphs into $\\sqrt{1 - x^2}/1$ How would i do this with $\\sin(2\\arccos(x))$ or $\\tan(\\arccos(x) + \\arcsin(x))$? Edit: From the answer given, i will try to solve these two equations. $$\\sin(2\\arccos(x)) = 2\\sin(\\arccos(x))\\cos(\\arccos(x))$$ $$\\sin(\\arccos(x))$$ $$\\sin(\\theta), \\theta = \\arccos(x)$$ $$\\cos(\\theta) = x$$ A = $\\theta$, hypotenuse = 1, adjacent = x, opposite = $\\sqrt{1 - x^2}$ $$\\sin(\\theta) = \\sqrt{1 -", "- \\sin(\\arccos x)\\sin(\\arccos(1-x)) = -x$ $x(1-x) - \\sqrt{(1-x^2)(2x-x^2)} = -x$ $x(1-x) + x = \\sqrt{(1-x^2)(2x-x^2)}$ $(x(1-x) + x)^2 = (1-x^2)(2x-x^2)$ $(2x-x^2)^2 = (1-x^2)(2x-x^2)$ Now x = 2 satisfies this equation. We can look for other solutions by dividing by $2x - x^2$ if $x \\neq 2$: $2x - x^2 = 1 - x^2$ $2x = 1$ $x = \\frac{1}{2}$ and since $\\arccos 2$ is undefined, the only solution that works is $x = \\frac{1}{2}$.", "Example 5 Find the solution to the trigonometric equation $sin^2(2x)+2cos^2(2x) = -\\frac{\\sqrt{2}}{2}$. ### Solution Since this problem has two different types of trigonometric functions, it is necessary to use trigonometric identities so there is only one function. Then, it will be possible to solve for $x$. In this case, there is a sine squared function and a cosine squared function, so it makes sense to consider the Pythagorean identities. Specifically $sin^2x+cos^2x=1$. This is true for any angle. Even though the inside of the trig functions in this case is $2x$, $2x$ is still just an angle. Therefore, $sin^2(2x)+cos^2(2x) = 1$ too. Expanding the given equation yields $sin^2(2x)+cos^2(2x)+cos^2(2x) = -\\frac{\\sqrt{2}}{2}$. The Pythagorean identity shows that this equation is equivalent to: $1+cos^2(2x) = -\\frac{\\sqrt{2}}{2}$. Subtracting $1$ from both sides makes this: $cos^2(2x) = \\frac{2-\\sqrt{2}}{2}$. Then, take the square root of both sides to get: $cos(2x) = \\sqrt{\\frac{2-\\sqrt{2}}{2}}$. Now, the equation is in the form $fun(f(x))=a$. Therefore, take the arccosine of both sides. This is: $2x = arccos(\\sqrt{\\frac{2-\\sqrt{2}}{2}}$. Finally divide both sides by $2$ to solve for $x$. $x = \\frac{arccos(\\sqrt{\\frac{2-\\sqrt{2}}{2}}}{2}$ radians. Although the right side of this equation looks complicated, it is just a constant. Plugging the values into a calculator reveals that it has the decimal approximation of 0.50 radians. This is the principal solution. Since cosine is $2\\pi$", "# How to solve: $\\arccos(2x)-\\arccos(x)=\\pi/3$ Solve for $x$ the equation $\\arccos(2x)-\\arccos(x)=\\pi/3$. My attempt using $\\cos(a-b)$. But it gives me a sqrt-expression that I don't know how to handle. WolframAlpha - This reduces to an easy quadratic. First, take the cosine of both sides. Using the cosine addition formula, I get $$2 x \\cdot x + \\sqrt{1-4 x^2} \\sqrt{1-x^2} = \\frac12$$ Manipulate and square both sides to get $$\\left ( 2 x^2-\\frac12\\right)^2 = (1-4 x^2)(1-x^2) = 1-5 x^2+4 x^4$$ or $$3 x^2=\\frac{3}{4} \\implies x = \\pm \\frac{1}{2}$$ Now plugging in both answers, one sees that only the $-1/2$ result makes sense for the principal branch (or any branch) of the arccos. Thus, $x=-1/2$. - Why is it only x=-1/2 that satisfies the equation? I set up: acos(1) - acos(0.5)= pi/3 and acos(-1) - acos(-0.5)= pi/3 Now I can get multiple results, since acos(1)=pi/2 and also =3pi/4. How should I think about that? – jacob Sep 29 '13 at 11:58 @jacob: $\\arccos{1}=0$, $\\arccos{(1/2)}=\\pi/3$. – Ron Gordon Sep 29 '13 at 12:24 Hint: $\\arccos\\left(2x\\right)=\\arccos\\left(x\\right)+\\pi/3$. Now take the cosinus on both sides and see what happens. - Avoid squaring which often incurs the burden of extraneous roots Let $\\displaystyle \\arccos x=\\theta\\ \\ \\ \\ (1)\\implies x=\\cos\\theta$ We have $\\displaystyle \\arccos(2\\cos\\theta)-\\theta=\\frac\\pi3\\implies \\arccos(2\\cos\\theta)=\\theta+\\frac\\pi3\\ \\ \\ \\ (2)$ As the principal value of $\\arccos$ lies", "# The Simplest Trigonometric Equations Equation sin(x)=a where |a|<=1 has infinitely many roots. For example roots of the equation sin(x)=1/2 are following: x_1=(pi)/6,x_2=(5pi)/6,x_3=pi/6+2pi,x_4=pi/6-2pi etc. The common formula, that allows to find all roots of the equation sin(x)=a is following: Fact 1. If sin(x)=a, where |a|<=1 then x=(-1)^n arcsin(a)+pin,n in Z. Here n can take integer values, and for any value of n we obtain some root of the equation. Fact 2. If cos(x)=a, where |a|<=1 then x=+- arccos(a)+2pin,n in Z. Fact 3. If tan(x)=a, then x=arctan(a)+pin,n in Z. Fact 4. If cot(x)=a, then x=text(arccot)(a)+pin,n in Z. Example 1. Solve sin(x)=1/2. Using Fact 1 we obtain that x=(-1)^n arcsin(1/2)+pin,n in Z. Since arcsin(1/2)=pi/6, then finally we obtain that x=(-1)^n pi/6+pin, n in Z. Example 2. Solve cos(3x)=-1/(sqrt(2)). Using Fact 2 we obtain that 3x=+-arccos(-1/(sqrt(2)))+2pin,n in Z. Since arccos(-1/(sqrt(2)))=pi-arccos(1/(sqrt(2)))=pi-pi/4=(3pi)/4, then we obtain that 3x=+- (3pi)/4+2pin, n in Z or x=+-(pi)/4+(2pin)/3, n in Z. Example 3. Solve tan(x-pi/6)=-sqrt(3). According to Fact 3 x-pi/6=arctan(-sqrt(3))+pin,n in Z. Since arctan(-sqrt(3))=-arctan(sqrt(3))=-pi/3, then we obtain that x-pi/6=-pi/3+pin, n in Z, i.e. x=-pi/6+pin, n in Z. Note, that in some cases it is more convenient to use particular formulas (in all formulas n in Z): 1. If sin(x)=0 then x=pin; 2. If sin(x)=1 then x=pi/2+2pin; 3. If sin(x)=-1 then x=-pi/2+2pin; 4. If cos(x)=0 then x=pi/2+pin; 5. If", "Multiply by $\\sqrt{2}:\\;\\;y \\;=\\;\\sqrt{2}\\cos\\left(\\theta - \\frac{\\pi}{4}\\right)$ 7. Hello Paymemoney Originally Posted by Paymemoney so from the formula: cos2A are you making arccos(x) = A? OK. Let's see if we can make it easier. I'll set it out slightly differently. Let $\\arccos(x)=A$ Now $\\arccos...$ means 'the angle whose cosine is ...'. So $\\arccos(x) = A$ means the angle whose cosine is $x$ is $A$ In other words $\\cos A = x$, which I'll call equation (1) Now the question asks us for $\\cos(2\\arccos(x))$ In other words, for $\\cos2A$ So we use the well-known double angle formula $\\cos2A=2\\cos^2A-1$ and then use equation (1) to replace $\\cos A$ by $x$. So we get: $\\cos(2\\arccos(x)) = \\cos2A$ $=2\\cos^2A -1$ $=2x^2-1$ OK now? Originally Posted by Paymemoney I though that cos(2arccos(x)) = 2x, why wouldn't this work? I'm not sure why you should think this. This would mean that if you double an angle you double its cosine. So $\\cos2A$ would equal $2\\cos A$. Which it doesn't, does it?", "how to evaluate $\\arcsin(x)-\\arccos(x)=\\arccos(\\frac{\\sqrt{3}}{2})$ Question how to evaluate $\\arcsin(x)-\\arccos(x)=\\arccos(\\frac{\\sqrt{3}}{2})$ Thoughts Do you cos on both sides to finally do quadratic equation to solve at the end of the problem? so i got $0=4x^2+4x-1+2\\sqrt{3}$ on my last step but i dont know if this is the right approach Note that $\\arcsin(x)+\\arccos(x) =\\pi/2$. Therefore $\\arcsin(x)-\\arccos(x) =\\arccos(\\frac{\\sqrt{3}}{2})$ becomes $\\arccos(\\frac{\\sqrt{3}}{2}) =\\pi/2-2\\arccos(x)$ or $\\arccos(x) =(\\pi/2-\\arccos(\\frac{\\sqrt{3}}{2}))/2$. You should be able to take it from here. You can't take the cosine of both sides, at least not the way you probably did. You must do it as follows: $$\\cos(\\arcsin(x)-\\arccos(x))=\\frac{\\sqrt3}2$$ And then use sum of angles identity: $$\\cos(\\arcsin(x))\\cos(\\arccos(x))+\\sin(\\arcsin(x))\\sin(\\arccos(x))\\\\=x\\cos(\\arcsin(x))+x\\sin(\\arccos(x))$$ And use pythagorean theorems (let's assume all positive) $$=x\\sqrt{1-\\sin^2(\\arcsin(x))}+x\\sqrt{1-\\cos^2(\\arccos(x))}\\\\=2x\\sqrt{1-x^2}$$ Can you take it from here? From the unit circle, $\\displaystyle \\arccos\\frac{\\sqrt3}2 = \\frac\\pi6$. A known property relating inverse sine and inverse cosine is $\\arcsin x + \\arccos x = \\dfrac\\pi2$. Therefore: \\begin{align*} \\arcsin x - \\arccos x &= \\arccos \\frac{\\sqrt3}2\\\\[0.3cm] \\frac\\pi2 - \\arccos x - \\arccos x &= \\frac\\pi6\\\\[0.3cm] -2\\arccos x &= -\\frac\\pi3\\\\[0.3cm] \\arccos x &= \\frac\\pi6 \\end{align*} Now take the cosine of both sides. • Same idea as mine, and you went all the way. – marty cohen Jan 20 '17 at 1:53 • @martycohen, I could say likewise (except for the second part) as you posted yours less than 1 minute before mine. And I dunno", "# Solving an inverse trig equation 1. Feb 28, 2009 ### Kaylee! 1. The problem statement, all variables and given/known data Solve for x $$arcsin(2x) + arccos(x) = \\frac{\\pi}{6}$$ 2. Relevant equations 3. The attempt at a solution Subtracting arccos(x) from both sides and then taking the sine of both sides: $$2x = sin\\left[\\frac{\\pi}{6} - arccos(x)\\right]$$ Applying the difference angle identity for sine: $$2x = sin \\frac{\\pi}{6} cos[arccos (x)] - cos \\frac{\\pi}{6} sin [arccos (x)]$$ We can draw a right triangle with a hypotenuse of 1, angle opposite of theta of $$\\sqrt{1-x^2}$$, and angle adjacent to theta of x. From this triangle we can deduce the values of cos( arccos(x) ) and sin( arccos(x) ) $$2x = (\\frac{1}{2})x - \\frac{\\sqrt{3}}{2}{\\sqrt{1-x^2}}$$ Multiplying each side by 2, subtracting x from each side, squaring each side, and then dividing each side by 3 $$3x^2 = 1 - x^2$$ Adding x^2 to each side, and then dividing each side by 4 $$x^2 = \\frac{1}{4}$$ At this point I need to take the square root of each side. I know it's the negative from trying each out, but how do I show this algebraically? 2. Feb 28, 2009 ### csprof2000 When you say 2x = x/2 - sqrt(3)/2 sqrt(1-x^2)... 3x = -sqrt(3)sqrt(1-x^2). But since sqrt(1-x^2) is positive, the RHS is negative, which is", "1. ## Having some trouble I couldnt figure this one out...can anyone? cos(arcsin3/5 + arccos sqrt3/2) Thanks! 2. Originally Posted by Sinsane I couldnt figure this one out...can anyone? cos(arcsin3/5 + arccos sqrt3/2) Thanks! Start by letting $\\sin^{-1} \\frac{3}{5} = \\alpha \\Rightarrow \\sin \\alpha = \\frac{3}{5}$ and $\\cos^{-1} \\frac{\\sqrt{3}}{2} = \\beta \\Rightarrow \\cos \\beta = \\frac{\\sqrt{3}}{2}$. Your task is to calculate $\\cos (\\alpha + \\beta)$. Hint: Start by applying the compound angle formula and then make the necessary substitutions. 3. Thank you..I finally figured it out before you said something..but I appreciate the help. But its always good to make sure something is right haha.", "example, if we set d=3, r1=1, and r2=2, the intersection area is zero. However, your last formula would give $r1^2 \\arccos 2+ r2^2 \\arccos \\frac{1}{2}$, which is clearly wrong. I also have some doubts about the last term $\\sqrt{Y}/2$, since calculating the area using integrals yields, apart from the terms containing the arccos, two different radicals that cannot be merged. Could you check your calculations? – Anatoly Aug 30 '14 at 0:09 I have just posted the result obtained using integrals. I hope that this could facilitate your check. – Anatoly Aug 30 '14 at 0:12 I derived my own proof with simple geometry and trigonometry. I ended up with a slightly different formula: $$A=r_1^2 \\text{arccos}\\left(\\frac{d^2-(r_2^2-r_1^2)}{2dr_1}\\right) + r_2^2 \\text{arccos}\\left(\\frac{d^2+(r_2^2-r_1^2)}{2dr_2}\\right)-d\\sqrt{r_1^2-\\left(\\frac{d^2-(r_2^2-r_1^2)}{2d}\\right)^2}.$$ The last term can be rewritten: $$-d\\sqrt{r_1^2-\\left(\\frac{d^2-(r_2^2-r_1^2)}{2d}\\right)^2} = -\\frac{\\sqrt{Y}}{2},$$ with $$Y=(-d+r_1+r_2)(d+r_2-r_1)(d-r_2+r_1)(d+r_1+r_2).$$ So this is very close to the first submitted equation, apart from a swap between r1 and r2 in the argument of the arccos(). It seems that this corrects the typo pointed out in the comment. For d=3, r1=1, r2=2 it gives $$A=r_1^2 \\text{arccos}\\left(1\\right) + r_2^2 \\text{arccos}\\left(1\\right)+0=0$$ - Set the center of the first circle in $(0,0)$ and that of the second circle in $(d,0)$. The equations of the circles are $y=\\sqrt{r1^2- x^2}$ and $y=\\sqrt{r2^2-(x-d)^2}$. In the segment connecting the two centers, the first and second circle intersect", "> clear all, clc > > syms a b c x > > x1 = acos((a*c + sqrt(a^2*b^2 + b^4 - b^2*c^2))/(a^2 + b^2)) > err1 = simplify(a*cos(x1)+b*sin(x1)-c) % NOT ZERO? > I think this is because the solution given by Mathematica's Solve does not show the assumptions on a,b,c that are needed. That is why I used Reduce earlier. Just to clarify, I show below one solution, then show the equation when this one solution is substituted back into it. Then use the assumptions needed to show that it does show ZERO. One can now clearly see that the result is only zero under certain conditions on a,b,c. Lets start from the top. First solve it -------------------------- Clear[a,b,c] eq = a*Cos[x]+b*Sin[x]-c; sol = Solve[eq==0,x]; sol = x/.sol ------------------------- {-ArcCos[(a c-Sqrt[a^2 b^2+b^4-b^2 c^2])/(a^2+b^2)], ArcCos[(a c-Sqrt[a^2 b^2+b^4-b^2 c^2])/(a^2+b^2)], -ArcCos[(a c+Sqrt[a^2 b^2+b^4-b^2 c^2])/(a^2+b^2)], ArcCos[(a c+Sqrt[a^2 b^2+b^4-b^2 c^2])/(a^2+b^2)]} So, 4 solutions. substituting the first solution back into the equation ----------------------- eq/.x->sol[[1]] (*replace first solution back in equation *) ----------------------- gives -c+(a (a c-Sqrt[a^2 b^2+b^4-b^2 c^2]))/(a^2+b^2) -b Sqrt[1-(a c-Sqrt[a^2 b^2+b^4-b^2 c^2])^2/(a^2+b^2)^2] Notice the denominator term (a^2+b^2). So the only way for this to be zero, is for the left term to be equal to the right term. Since \"A/z^2\" on its own can't be zero unless z->Infinity. So the condition for zero", "# Thread: Solve: cos^−1(2x) + sin^−1(x) = π/3 1. Originally Posted by pka \\begin{align*} \\arccos (2x) + \\arcsin (x) &= \\frac{\\pi }{3}\\\\ \\cos (\\arccos (2x) + \\arcsin (x)) &= \\cos \\left( {\\frac{\\pi }{3}} \\right)\\\\ \\cos (\\arccos (2x))\\cos (\\arcsin (x)) - \\sin (\\arccos (2x))\\sin (\\arcsin (x)) &= \\frac{1}{2} \\end{align*} Now you should know these; if not then you need sit-down help. $\\cos (\\arccos (2x))=2x$ & $\\cos (\\arcsin (x))=\\sqrt{1-x^2}$ If you do understand those then show us by posting the solution. I get: 2x*√(1-x^2) - x/(√1-4x^2) = 1/2 I tried simplifying further by multiplying by 2 and by (√1-4x^2) but, I can't see a solution for x. I used a graphics calculator to draw a graph of y = 2x*√(1-x^2) - x/(√1-4x^2) and then used the trace function to trace the graph to y=0.5 to find x. I got x = -0.47 How can I solve for x using just pen & paper? 2. Originally Posted by Bobby Bones I get: 2x*√(1-x^2) - x/(√1-4x^2) = 1/2 It appears as if you are dividing when there is no division. It should be $2x\\sqrt{1-x^2}-x\\sqrt{1-4x^2}=\\dfrac{1}{2}$ But that is a beast of an equation to solve. SEE HERE. 3. Originally Posted by pka It appears as if you are dividing when there is no division. It should be $2x\\sqrt{1-x^2}-x\\sqrt{1-4x^2}=\\dfrac{1}{2}$ But that is a beast", "x - 2\\sin x \\cos x + \\cos^2 x = \\frac{1}{2}$$ Factor the left side: $$(\\sin x + \\cos x)^2 = \\frac{3}{2}$$ $$(\\sin x - \\cos x)^2 = \\frac{1}{2}$$ Take the square root: $$\\sin x + \\cos x = \\pm\\sqrt{\\frac{3}{2}}$$ $$\\sin x - \\cos x = \\pm\\sqrt{\\frac{1}{2}}$$ Add the two equations: $$2\\sin x = \\pm\\sqrt{\\frac{3}{2}} \\pm \\sqrt{\\frac{1}{2}}$$ And now finally we can discover the sine of the original angle: $$\\sin x = \\pm \\frac{\\sqrt{6}\\pm\\sqrt{2}}{4}$$ And then we need to apply $\\arcsin$ and do some fiddling with \"which equivalent angles do I take?\". What a mess. Instead, just find $\\arcsin \\frac{1}{2} = 2x$ and use supplemental and coterminal angles to find the other values of $2x$ that work. • snazzy approach, i would have just given up trying after getting to $2\\sin x \\cos x$ – MichaelChirico Apr 14 '16 at 0:41", "can ignore this! 2) $\\csc x = 2 \\implies \\sin x = \\frac{1}{2}$. Draw a 30 - 60 - 90 triangle, with a hypotenuse 2 units long, and figure out the lengths of the shortest side using the fact that this triangle is half of an equilateral triangle. Then you should see that the sine of one of the angle is $\\frac{1}{2}$ 3) $\\sec x = -1 \\implies \\cos x = -1$. Use the same formula as in (1), and the fact that $\\cos 0^o = 1$. Can you complete them now? 3. I don't know if you covered this yet in your class, but an alternative way to solve these is to use the inverse of these trig functions to solve for x. Example: $\\cos x=-\\dfrac{\\sqrt{2}}{2}$. We need to restrict our domain to make the cosine function one-to-one: $[0,\\pi]$. Taking the arccosine of both sides, $\\arccos(\\cos x))=\\arccos \\left(-\\dfrac{\\sqrt{2}}{2}\\right)$ $x = \\arccos \\left(-\\dfrac{\\sqrt{2}}{2}\\right)=\\dfrac{3\\pi}{4}$. We know that $\\cos x = \\cos -x$. So solving $\\cos (-x) = -\\dfrac{\\sqrt{2}}{2}$, $\\arccos(\\cos (-x)) = \\arccos \\left(-\\dfrac{\\sqrt{2}}{2}\\right)$ $-x = \\arccos \\left(-\\dfrac{\\sqrt{2}}{2}\\right) = \\dfrac{3\\pi}{4} \\Rightarrow x =-\\dfrac{3\\pi}{4}$. Since the period of cos x is $2\\pi$, we have the general solution: $x=\\pm \\dfrac{3\\pi}{4} \\pm 2\\pi n$ where $n \\in \\mathbb{Z}$", "Sum Formula Originally Posted by Bashyboy \"Write cos(arctan 1 + arccos x) as an algebraic expression.\" So, I understand that I am to use formula for cos(u + v) because it fits the expression, where u = arctan 1 and v = arccos x. I also know that I must create to right triangles for each of the angles u and v, the reason being that the angles are likely to be different. What I do not understand is why, when the formula is used, we don't evaluate cos(arctan 1) as cos(pi/4), which then equals 0. In the book, cos(arctan 1) is evaluated as 1/sqroot(2). Perhaps I do not understand inverse trigonometric functions as I thought I did. An alternative is to note that \\displaystyle \\begin{align*} \\sin^2{X} + \\cos^2{X} &\\equiv 1 \\\\ \\frac{\\sin^2{X}}{\\cos^2{X}} + \\frac{\\cos^2{X}}{\\cos^2{X}} &\\equiv \\frac{1}{\\cos^2{X}} \\\\ \\tan^2{X} + 1 &\\equiv \\frac{1}{\\cos^2{X}} \\\\ \\frac{1}{\\tan^2{X} + 1} &\\equiv \\cos^2{X} \\\\ \\cos{X} &\\equiv \\sqrt{\\frac{1}{\\tan^2{X} + 1}} \\end{align*} So that means \\displaystyle \\begin{align*} \\cos{\\left(\\arctan{1}\\right)} &= \\sqrt{\\frac{1}{\\left[\\tan{\\left(\\arctan{1}\\right)}\\right]^2 + 1}} \\\\ &= \\sqrt{\\frac{1}{1^2 + 1}} \\\\ &= \\sqrt{\\frac{1}{1 + 1}} \\\\ &= \\sqrt{\\frac{1}{2}} \\\\ &= \\frac{\\sqrt{1}}{\\sqrt{2}} \\\\ &= \\frac{1}{\\sqrt{2}} \\end{align*}", "\\arccos x$ $\\arccsc \\frac{1}{x} = \\arcsin x$ If you only have a fragment of a sine table: $\\arccos x = \\arcsin \\sqrt{1-x^2},\\text{ if }0 \\leq x \\leq 1$ $\\arctan x = \\arcsin \\frac{x}{\\sqrt{x^2+1}}$ Whenever the square root of a complex number is used here, we choose the root with the positive real part (or positive imaginary part if the square was negative real). From the half-angle formula $\\tan \\frac{\\theta}{2} = \\frac{\\sin \\theta}{1+\\cos \\theta}$, we get: $\\arcsin x = 2 \\arctan \\frac{x}{1+\\sqrt{1-x^2}}$ $\\arccos x = 2 \\arctan \\frac{\\sqrt{1-x^2}}{1+x},\\text{ if }-1 < x \\leq +1$ $\\arctan x = 2 \\arctan \\frac{x}{1+\\sqrt{1+x^2}}$ ## Relationships between trigonometric functions and inverse trigonometric functions $\\sin (\\arccos(x)) = \\cos(\\arcsin(x)) = \\sqrt{1-x^2}$ $\\sin (\\arctan(x)) = \\frac{x}{\\sqrt{1+x^2}}$ $\\cos (\\arctan(x)) = \\frac{1}{\\sqrt{1+x^2}}$ $\\tan (\\arcsin(x)) = \\frac{x}{\\sqrt{1-x^2}}$ $\\tan (\\arccos(x)) = \\frac{\\sqrt{1-x^2}}{x}$ ## General solutions Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of 2π. Sine and cosecant begin their period at 2πk − π/2 (where k is an integer), finish it at 2πk + π/2, and then reverse themselves over 2πk + π/2 to 2πk + 3π/2. Cosine and secant begin their period at 2πk, finish it at 2πk + π, and then reverse themselves over 2πk + π to 2πk + 2π. Tangent", "# Simplify $\\sin{(\\arctan{2}-\\arcsin{\\frac{1}{\\sqrt{10}}})}.$ First I denote $x=\\arctan{2}$ and $y=\\arcsin{\\frac{1}{\\sqrt{10}}}$ and then use the addition formula for sine: $$\\sin{(x-y)}=\\sin{x}\\cos{y}-\\cos{x}\\sin{y}=\\sin{x}\\cos{y}-\\cos{x}\\cdot \\frac{1}{\\sqrt{10}}.$$ Now I use the fact that $\\cos{y}=1-\\sin^2{y}$ which gives $$\\cos{y}=1-\\sin^2{\\left(\\arcsin{\\frac{1}{\\sqrt{10}}}\\right)}=1-\\frac{1}{10}=\\frac{9}{10}$$ So I have reduced the problem to the following $$\\sin{x}\\cdot \\frac{9}{10}-\\cos{x}\\cdot \\frac{1}{\\sqrt{10}}.$$ Here I'm stuck. I cant just divide this expression by $(\\sin{x}\\cos{x})$ because I'd still be left with sine and cos terms and I'd also change the value of the expression. Everything now boils down to compute $\\sin(\\arctan{2})$ and $\\cos{\\arctan2}.$ I also tried rewriting $\\cos(\\arctan{2})$ as $1-\\sin^2(\\arctan{2}),$ but to no avail. Any suggestions on 1. how to proceed from where I left; 2. how tocompute this by means more effective; 3. both of the above. • It should be $\\cos y=\\sqrt{1-\\sin^2 y}$ – GAVD Sep 16 '17 at 12:07 $$\\sin(\\arctan(x)) = \\frac x {\\sqrt{x^2 + 1}}$$ Which you can prove using, $$1 + \\cot^2(\\arctan(x)) = \\csc^2(\\arctan(x))$$ By the Pythagorean identity. From which, $$1 + \\frac 1 {x^2} = \\frac{1}{\\sin^2(\\arctan(x))}$$ $$\\frac {x^2} {x^2+1} = \\sin^2(\\arctan(x))$$ Taking square roots, choosing the positive branch, $$\\frac x {\\sqrt{x^2 + 1}} = \\sin(\\arctan(x)) \\blacksquare$$ Put $\\arctan 2 = x$ then $\\tan x= 2$, so $\\sin x = {2\\over \\sqrt{5}}$ and $\\cos x = {1\\over \\sqrt{5}}$. Let $\\arcsin {1\\over \\sqrt{10}}= y$ so $\\sin y = {1\\over \\sqrt{10}}$, thus $\\cos y = {3\\over \\sqrt{10}}$. Now:", "$$\\arccos(\\cos(\\dfrac{\\pi}{4})) = \\arccos(\\dfrac{\\sqrt{2}}{2})$$. In addition, $$\\arccos(\\dfrac{\\sqrt{2}}{2}) = \\dfrac{\\pi}{4}$$ since $$\\cos(\\dfrac{\\pi}{4}) = \\dfrac{\\sqrt{2}}{2}$$ and $$0 \\leq \\dfrac{\\pi}{4} \\leq \\pi$$. So we see that $\\arccos(\\cos(\\dfrac{\\pi}{4})) = \\arccos(\\dfrac{\\sqrt{2}}{2}) = \\dfrac{\\pi}{4}.$ 3. $$\\arccos(\\cos(-\\dfrac{\\pi}{4})) = \\arccos(\\dfrac{\\sqrt{2}}{2})$$. In addition, $$\\arccos(\\dfrac{\\sqrt{2}}{2}) = \\dfrac{\\pi}{4}$$ since $$\\cos(\\dfrac{\\pi}{4}) = \\dfrac{\\sqrt{2}}{2}$$ and $$0 \\leq \\dfrac{3\\pi}{4} \\leq \\pi$$. So we see that $\\arccos(\\cos(-\\dfrac{\\pi}{4})) = \\arccos(\\dfrac{\\sqrt{2}}{2}) = \\dfrac{\\pi}{4}.$ 4. $$\\tan^{-1}(\\tan(\\dfrac{5\\pi}{4})) = \\tan^{-1}(1)$$. In addition, $$\\tan^{-1}(1) = \\dfrac{\\pi}{4}$$ since $$\\tan(\\dfrac{\\pi}{4}) = 1$$ and $$-\\dfrac{\\pi}{2} \\leq \\dfrac{\\pi}{4} \\leq \\dfrac{\\pi}{2}$$. So we see that $\\tan^{-1}(\\tan(\\dfrac{5\\pi}{4})) = \\tan^{-1}(1) = \\dfrac{\\pi}{4}$ Properties of the Inverse Cosine Function • For each $$t$$ in the closed interval $$[0, \\pi]$$: $\\cos^{-1}(\\cos(t)) = t$ • For each $$y$$ in the closed interval $$[-1, 1]$$: $\\cos(\\cos^{-1}(y)) = y$ Properties of the Inverse Tangent Function • For each $$t$$ in the open interval $$(-\\dfrac{\\pi}{2}, \\pi)$$: $\\tan^{-1}(\\tan(t)) = t$ • For each real number $$y$$: $\\tan(\\tan^{-1}(y)) = y$ The justification for these properties is included in the exercises. Exercise $$\\PageIndex{4}$$ Determine the exact value of each of the following and check them using a calculator. 1. $$y = \\arccos(1)$$ 2. $$y = \\tan^{-1}(\\sqrt{3})$$ 3. $$y = \\arctan(-1)$$ 4. $$y = \\cos^{-1}(-\\dfrac{\\sqrt{2}}{2})$$ 5. $$\\sin(\\arccos(-\\dfrac{1}{2}))$$ 6. $$\\tan(\\arcsin(-\\dfrac{\\sqrt{3}}{2}))$$ 7. $$\\arccos(\\sin(\\dfrac{\\pi}{6}))$$ When we evaluate an expression such as $$\\sin(\\arccos(-\\dfrac{1}{2}))$$ in the previous preview activity, we can use the fact that it is possible to determine the exact", "above formula in terms of \\arccosx \\arcsinx. ### Relationships between trigonometric functions and inverse trigonometric functions Trigonometric functions of inverse trigonometric functions are tabulated below. A quick way to derive them is by considering the geometry of a right-angled triangle, with one side of length 1 and another side of length x, then applying the Pythagorean theorem and definitions of the trigonometric ratios. Purely algebraic derivations are longer. It is worth noting that for arcsecant and arccosecant, the diagram assumes that x is positive, and thus the result has to be corrected through the use of absolute values and the signum (sgn) operation. \\theta \\sin(\\theta) \\cos(\\theta) \\tan(\\theta) Diagram \\arcsin(x) \\sin(\\arcsin(x))=x \\cos(\\arcsin(x))=\\sqrt{1-x2} \\tan(\\arcsin(x))= x \\sqrt{1-x2 } \\arccos(x) \\sin(\\arccos(x))=\\sqrt{1-x2} \\cos(\\arccos(x))=x \\tan(\\arccos(x))= \\sqrt{1-x2 } \\arctan(x) \\sin(\\arctan(x))= x \\sqrt{1+x2 } \\cos(\\arctan(x))= 1 \\sqrt{1+x2 } \\tan(\\arctan(x))=x \\arccot(x) \\sin(\\arccot(x))= 1 \\sqrt{1+x2 } \\cos(\\arccot(x))= x \\sqrt{1+x2 } \\tan(\\arccot(x))= 1 x \\arcsec(x) \\sin(\\arcsec(x))= \\sqrt{x2-1 } \\cos(\\arcsec(x))= 1 x \\tan(\\arcsec(x))=sgn(x)\\sqrt{x2-1} \\arccsc(x) \\sin(\\arccsc(x))= 1 x \\cos(\\arccsc(x))= \\sqrt{x2-1 } \\tan(\\arccsc(x))= sgn(x) \\sqrt{x2-1 } ### Relationships among the inverse trigonometric functions Complementary angles: \\begin{align} \\arccos(x)&= \\pi 2 -\\arcsin(x)\\\\[0.5em] \\arccot(x)&= \\pi 2 -\\arctan(x)\\\\[0.5em] \\arccsc(x)&= \\pi 2 -\\arcsec(x) \\end{align} Negative arguments: \\begin{align} \\arcsin(-x)&=-\\arcsin(x)\\\\ \\arccos(-x)&=\\pi-\\arccos(x)\\\\ \\arctan(-x)&=-\\arctan(x)\\\\ \\arccot(-x)&=\\pi-\\arccot(x)\\\\ \\arcsec(-x)&=\\pi-\\arcsec(x)\\\\ \\arccsc(-x)&=-\\arccsc(x) \\end{align} Reciprocal arguments: \\begin{align} \\arccos\\left( 1 x \\right)&=\\arcsec(x)\\\\[0.3em] \\arcsin\\left( 1 x \\right)&=\\arccsc(x)\\\\[0.3em] \\arctan\\left( 1 x \\right)&= \\pi 2 -\\arctan(x)=\\arccot(x),ifx>0\\\\[0.3em] \\arctan\\left( 1 x \\right)&=- \\pi" ]

[ "I don't know how to find $S=\\sum_{k=0}^n cos(k \\alpha)$ without using complex. Can you help me please find an alternative way of doing it?", "# Need help to Prove this • Sep 21st 2012, 08:26 AM kjchauhan Need help to Prove this If $\\cos(\\theta + i\\phi)=\\sec(\\alpha + i\\beta)$, then prove that $\\tanh^{2}\\phi\\cosh^{2}\\beta=\\sin^{2}\\alpha$ and $\\tanh^{2}\\beta\\cosh^{2}\\phi=\\sin^{2}\\theta$", "# The expression $\\tan^2\\alpha+\\cot^2\\alpha$ is $(a)\\;\\geq 2\\qquad(b)\\;\\leq 2\\qquad(c)\\;\\geq -2\\qquad(d)\\;None\\;of\\;these$ $(\\tan\\alpha-\\cot\\alpha)^2\\geq 0$ $\\Rightarrow \\tan^2\\alpha+\\cot^2\\alpha-2\\tan\\alpha\\cot\\alpha\\geq 0$ $\\Rightarrow \\tan^2\\alpha+\\cot^2\\alpha-2\\geq 0$ $\\Rightarrow \\tan^2\\alpha+\\cot^2\\alpha\\geq 2$ Hence (a) is the correct answer. answered Oct 9, 2013", "and if n is even, so n=2k, then $\\alpha=k-1$", "• question_answer If general solution of $\\cot \\theta -\\tan \\theta =\\sec \\theta$ is $n\\pi +{{\\left( -1 \\right)}^{n}}\\frac{\\pi }{k},\\,\\,n\\in \\,Z$ then find k. Answer: 6.00 You need to login to perform this action. You will be redirected in 3 sec", "the factorization $$x^k-\\alpha^k=(x-\\alpha)(x^{k-1}+x^{k-2}\\alpha+\\cdots+x\\alpha^{k-2}+\\alpha^{k-1}).$$", "x - \\sin^2 x$ as follows: $$\\cot x - \\tan x = \\frac{\\cos x}{\\sin x} - \\frac{\\sin x}{\\cos x} = \\frac{\\cos^2 x - \\sin ^2 x}{\\sin x \\cos x} = 2 \\cot 2x$$ We now have $$\\cot \\alpha - 2 \\cot 2\\alpha = \\tan \\alpha$$ $$2\\cot 2\\alpha - 4 \\cot 4\\alpha = 2\\tan 2\\alpha$$ $$4\\cot 2\\alpha - 8 \\cot 4\\alpha = 4\\tan 4\\alpha$$ Adding gives us the result. -", "end I'm not getting correct result which should be $$\\Phi=ln|cosh(\\gamma) \\cdot tan(\\frac{\\Pi}{4}+ \\frac{\\beta}{2})|+const.$$ What should I do?", "\\alpha_2 , ..., \\alpha_n ) = \\sum_{k=1}^{m} (d_k)^2 = \\sum_{k=1}^{m} [f(x_k) - \\phi (x_k)]^2$", "# Math Help - Need help to Prove this 1. ## Need help to Prove this If $\\cos(\\theta + i\\phi)=\\sec(\\alpha + i\\beta)$, then prove that $\\tanh^{2}\\phi\\cosh^{2}\\beta=\\sin^{2}\\alpha$ and $\\tanh^{2}\\beta\\cosh^{2}\\phi=\\sin^{2}\\theta$", "x=\\csc2y+\\cot2y=\\dfrac{1+\\cos2y}{\\sin2y}=\\cot y$ $$\\sin\\theta=\\cos2y=\\dfrac{1-\\tan^2y}{1+\\tan^2y}=\\dfrac{\\cot^2y-1}{\\cot^2y+1}=?$$", "so that for all $\\alpha \\in K$, we have that $\\alpha^{p^n} = \\alpha$. But then the left hand side can be written as $(\\alpha^p)^{p^{n-1}}$ showing that $\\alpha \\in k(\\alpha^p)$. Hence $k(\\alpha) = k(\\alpha^p)$ and the result follows. -", "## Precalculus (10th Edition) $sin^2(\\theta)+cos^2(\\theta)=1$, $tan^2(\\theta)+1=sec^2(\\theta)$, $cot^2(\\theta)+1=csc^2(\\theta)$. The identities are the following: $sin^2(\\theta)+cos^2(\\theta)=1$, $tan^2(\\theta)+1=sec^2(\\theta)$, $cot^2(\\theta)+1=csc^2(\\theta)$.", "hand, ##K## extends ##B## and contains ##\\alpha_1, \\dots, \\alpha_N##. So, ##B(\\alpha_1, \\dots, \\alpha_N) \\subset K##. We may conclude ##K = B(\\alpha_1, \\dots, \\alpha_N) = B_1 \\lor \\dots \\lor B_r##. [] Sorry! The ##E##'s should have been ##K##'s. I have edited in the above.", "If $tan\\alpha=\\sqrt{3}$ and $tan\\beta=\\frac{1}{\\sqrt{3}},\\ 0\\lt\\alpha,\\ \\beta\\lt 90^{o}$, then find the value of $cot( \\alpha+\\beta)$. Given: $tan\\alpha=\\sqrt{3}$ and $tan\\beta=\\frac{1}{\\sqrt{3}},\\ 0\\lt\\alpha,\\ \\beta\\lt 90^{o}$. To do: To find the value of $cot( \\alpha+\\beta)$. Solution: As, $tan\\alpha = \\sqrt{3}$ $tan\\alpha = tan60^{o}$ so, $\\alpha= 60^{o}$ And, $tan\\beta = \\frac{1}{\\sqrt{3}}$ $tan\\beta = tan30^{o}$ so, $\\beta = 30^{o}$ now we have get value of $\\alpha$ & $\\beta$ so, $cot(\\alpha+\\beta) = cot(60^{o}+30^{o})$ $= cot90^{o}$ $=$ not defined As known that the value of $cot90^{o}$ is not defined . Tutorialspoint Simply Easy Learning", "I can find $\\alpha$ directly. But when I use $\\alpha = \\frac{k}{\\rho c}$ I get different answer. Which one should I use? If you use $k$, $\\rho$ or $c$ elsewhere in your problem, then to be consistent you should compute $\\alpha=k/\\rho c$ and not take it from the tables. $\\pi$", "do know the answer to! Take $$\\cos ^2(x) + \\sin ^2(x) = 1$$, divide through by $$\\sin ^2(x)$$ and obtain that $$\\cot ^2(x) + 1 = \\csc ^2(x) \\implies \\cot ^2(x) = \\csc ^2(x) - 1$$. Then,", "\\cos^2(\\theta) + \\tan^2(\\theta) = \\sec^2(\\theta)$$ $$\\sin^2(\\theta) + \\cos^2(\\theta) + \\cot^2(\\theta) = \\csc^2(\\theta)$$ The two identities right above here were based on identites others posted on this site with a substitution. $$\\cos(2\\theta) = (\\cos(\\theta) + \\sin(\\theta))(\\cos(\\theta) - \\sin(\\theta))$$", "$\\{ 1, \\alpha, \\dots, a^{d-1} \\}$. This can be proved easily by induction on $k$, using the fact that $$\\alpha^d = - (a_{d-1}\\alpha^{d-1}+ \\dots +a_1 \\alpha + a_0)$$", "\\lambda B + \\mu C$, or equivalently $\\alpha A + \\beta B + \\gamma C$ where $\\alpha + \\beta + \\gamma = 1$. So you need to find $k$ such that $kD$ can be written in the form $\\alpha A + \\beta B + \\gamma C$ where $\\alpha + \\beta + \\gamma = 1$. Then $kD$ is the point of intersection." ]

[ "x}$ $\\displaystyle \\Rightarrow \\frac{\\cos^2 x \\cos \\alpha}{\\cos^2 x \\cos^2 \\alpha - \\sin^2 x \\sin^2 \\alpha } = 1$ $\\displaystyle \\Rightarrow \\cos^2 x \\cos \\alpha = \\cos^2 x \\cos^2 \\alpha - \\sin^2 x \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x \\cos \\alpha = \\cos^2 x \\cos^2 \\alpha - (1 - \\cos^2 x) \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x \\cos \\alpha = \\cos^2 x \\cos^2 \\alpha - \\sin^2 \\alpha + \\cos^2 x \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x \\cos \\alpha = \\cos^2 x ( \\cos^2 \\alpha + \\sin^2 \\alpha) - \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x \\cos \\alpha = \\cos^2 x - \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x \\cos \\alpha - \\cos^2 x = - \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x ( \\cos \\alpha -1) = - \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x ( 1 - \\cos \\alpha ) = \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x = \\frac{\\sin^2 \\alpha}{ 1 - \\cos \\alpha }$ $\\displaystyle \\Rightarrow \\cos^2 x= \\frac{4 \\sin^2 \\frac{\\alpha}{2} \\cos^2 \\frac{\\alpha}{2} }{ 2 \\sin^2 \\frac{\\alpha}{2} }$ $\\displaystyle \\Rightarrow \\cos^2 x = 2 \\cos^2 \\frac{\\alpha}{2}$ $\\displaystyle \\Rightarrow \\cos x = \\pm \\sqrt{2} \\cos \\frac{\\alpha}{2}$ . Hence proved. $\\displaystyle \\\\$ $\\displaystyle \\text{Question 42: } \\cos \\alpha + \\cos \\beta = \\frac{1}{3} \\text{ and } \\sin \\alpha + \\sin \\beta = \\frac{1}{4} , \\text{ prove that } \\cos \\frac{\\alpha - \\beta}{2} =", "\\ \\ \\displaystyle \\frac{{1+\\sin 2\\alpha -\\cos 2\\alpha }}{{1+\\sin 2\\alpha +\\cos 2\\alpha }}\\\\\\\\=\\displaystyle \\frac{{1+2\\sin \\alpha \\cos \\alpha -1+2{{{\\sin }}^{2}}\\alpha }}{{1+2\\sin \\alpha \\cos \\alpha +2{{{\\cos }}^{2}}\\alpha -1}}\\\\\\\\=\\displaystyle \\frac{{2\\sin \\alpha \\left( {\\sin \\alpha +\\cos \\alpha } \\right)}}{{2\\cos \\alpha \\left( {\\sin \\alpha +\\cos \\alpha } \\right)}}\\\\\\\\=\\tan \\alpha \\\\\\\\\\text{(b)}\\\\\\ \\ \\ \\displaystyle \\frac{{1+\\sin 2\\alpha +\\cos 2\\alpha }}{{1+\\sin 2\\alpha -\\cos 2\\alpha }}\\\\\\\\=\\displaystyle \\frac{{1+2\\sin \\alpha \\cos \\alpha +2{{{\\cos }}^{2}}\\alpha -1}}{{1+2\\sin \\alpha \\cos \\alpha -1+2{{{\\sin }}^{2}}\\alpha }}\\\\\\\\=\\displaystyle \\frac{{2\\cos \\alpha \\left( {\\sin \\alpha +\\cos \\alpha } \\right)}}{{2\\sin \\alpha \\left( {\\sin \\alpha +\\cos \\alpha } \\right)}}\\\\\\\\=\\cot \\alpha \\end{array}$ 11. If $\\displaystyle \\sec \\theta + \\tan \\theta = 3$, where $\\displaystyle \\theta$ lies in the first quadrant, then find the value of $\\displaystyle \\cos \\theta$. Show/Hide Solution $\\displaystyle \\begin{array}{l}\\ \\ \\ \\sec \\theta +\\tan \\theta =3\\ ---(1)\\\\\\\\\\ \\ \\ \\displaystyle \\frac{{\\left( {\\sec \\theta +\\tan \\theta } \\right)\\left( {\\sec \\theta -\\tan \\theta } \\right)}}{{\\sec \\theta -\\tan \\theta }}=3\\\\\\\\\\ \\ \\ \\displaystyle \\frac{{{{{\\sec }}^{2}}\\theta -{{{\\tan }}^{2}}\\theta }}{{\\sec \\theta -\\tan \\theta }}=3\\\\\\\\\\ \\ \\ \\displaystyle \\frac{1}{{\\sec \\theta -\\tan \\theta }}=3\\\\\\\\\\therefore \\ \\ \\sec \\theta -\\tan \\theta =\\displaystyle \\frac{1}{3}\\ ---(2)\\\\\\\\\\ \\ \\ (1)+(2)\\Rightarrow 2\\sec \\theta =\\displaystyle \\frac{{10}}{3}\\\\\\\\\\therefore \\ \\sec \\theta =\\displaystyle \\frac{5}{3}\\\\\\\\\\therefore \\ \\cos \\theta =\\displaystyle \\frac{3}{5}\\end{array}$ 12. If $\\displaystyle \\operatorname{cosec}\\theta -\\cot \\theta =\\frac{1}{5}$, where $\\displaystyle \\theta$ lies in the first quadrant, then find the value of $\\displaystyle \\sin \\theta$. Show/Hide Solution $\\displaystyle", "\\frac{(i\\cdot \\alpha )^3}{3!} + \\frac{(i\\cdot \\alpha )^4}{4!} + \\frac{(i\\cdot \\alpha )^5}{5!} + \\cdots = \\\\ &= 1 + i\\cdot \\alpha - \\frac{\\alpha ^2}{2!} - i\\cdot\\frac{ \\alpha ^3}{3!} + \\frac{\\alpha ^4}{4!} + i\\cdot\\frac{ \\alpha ^5}{5!} - \\cdots = \\\\ &= \\left( 1 - \\frac{\\alpha ^2}{2!} + \\frac{\\alpha ^4}{4!} - \\cdots \\right) + i\\cdot\\left( \\alpha - \\frac{\\alpha ^3}{3!} + \\frac{\\alpha ^5}{5!} - \\cdots \\right) = \\\\ &= \\cos \\alpha + i\\cdot\\sin \\alpha.\\end{aligned}", "\\ \\ \\ \\text{Since both}\\sin \\alpha \\ \\text{ and}\\cos \\beta \\ \\text{are negative,}\\\\\\ \\ \\ \\ \\alpha \\ \\text{and}\\ \\beta \\text{ will be in }{{\\text{3}}^{{\\text{rd}}}}\\text{ quadrant}\\text{.}\\end{array}$ $\\displaystyle \\therefore \\ \\ \\ \\sin \\alpha =-\\frac{4}{5},\\ \\cos \\alpha =-\\frac{3}{5}$ $\\displaystyle \\begin{array}{l}\\therefore \\ \\ \\ \\sin \\beta =-\\displaystyle \\frac{5}{{13}},\\ \\\\\\\\\\ \\ \\ \\ \\ \\cos \\beta =-\\displaystyle \\frac{{12}}{{13}},\\\\\\\\\\ \\ \\ \\ \\ \\tan \\beta =\\displaystyle \\frac{5}{{12}}\\\\\\\\\\\\\\ \\ \\ \\ \\ \\sin 2\\alpha =2\\sin \\alpha \\cos \\alpha \\\\\\\\\\therefore \\ \\ \\ \\sin 2\\alpha =2\\left( {-\\displaystyle \\frac{4}{5}} \\right)\\left( {-\\displaystyle \\frac{3}{5}} \\right)\\\\\\\\\\therefore \\ \\ \\ \\sin 2\\alpha =\\displaystyle \\frac{{24}}{{25}}\\\\\\\\\\ \\ \\ \\ \\ \\cos 2\\alpha =2{{\\cos }^{2}}\\alpha -1\\\\\\\\\\therefore \\ \\ \\ \\cos 2\\alpha =2{{\\left( {-\\displaystyle \\frac{3}{5}} \\right)}^{2}}-1\\\\\\\\\\therefore \\ \\ \\ \\cos 2\\alpha =-\\displaystyle \\frac{7}{{25}}\\\\\\\\\\ \\ \\ \\ \\cos \\displaystyle \\frac{\\beta }{2}=\\pm \\sqrt{{\\displaystyle \\frac{{1+\\cos \\beta }}{2}}}\\\\\\\\\\therefore \\ \\ \\cos \\displaystyle \\frac{\\beta }{2}=\\pm \\sqrt{{\\displaystyle \\frac{{1-\\displaystyle \\frac{{12}}{{13}}}}{2}}}\\\\\\\\\\therefore \\ \\ \\cos \\displaystyle \\frac{\\beta }{2}=\\pm \\sqrt{{\\displaystyle \\frac{1}{{26}}}}=\\pm \\displaystyle \\frac{{\\sqrt{{26}}}}{{26}}\\\\\\\\\\ \\ \\ \\ \\text{Since}\\ 180{}^\\circ <\\beta <270{}^\\circ ,\\\\\\\\\\ \\ \\ \\ 90{}^\\circ <\\displaystyle \\frac{\\beta }{2}<135{}^\\circ ,\\\\\\\\\\therefore \\ \\ \\cos \\displaystyle \\frac{\\beta }{2}=-\\displaystyle \\frac{{\\sqrt{{26}}}}{{26}}\\\\\\\\\\ \\ \\ \\ \\tan 2\\beta =\\displaystyle \\frac{{2\\tan \\beta }}{{1-{{{\\tan }}^{2}}\\beta }}\\\\\\\\\\therefore \\ \\ \\tan 2\\beta =\\displaystyle \\frac{{2\\left( {\\displaystyle \\frac{5}{{12}}} \\right)}}{{1-{{{\\left( {\\displaystyle \\frac{5}{{12}}} \\right)}}^{2}}}}\\\\\\\\\\therefore \\ \\ \\tan 2\\beta =\\displaystyle \\frac{{\\displaystyle \\frac{5}{6}}}{{\\displaystyle \\frac{{119}}{{144}}}}=\\displaystyle \\frac{5}{6}\\times \\displaystyle \\frac{{144}}{{119}}=\\displaystyle \\frac{{120}}{{119}}\\end{array}$ 13. Given that $\\displaystyle \\sin \\alpha = \\frac{15}{17}, \\cos \\beta = \\frac{3}{5}$", "=\\ \\ \\tan \\left( {-\\displaystyle \\frac{{\\theta -\\phi }}{2}} \\right)\\ \\ \\ \\left[ {\\because \\tan \\left( {-\\theta } \\right)=-\\tan \\theta } \\right]\\\\\\\\\\ \\ \\ \\ =\\ \\ \\tan \\displaystyle \\frac{{\\phi -\\theta }}{2}\\ \\end{array}$ $\\displaystyle \\begin{array}{l}\\text{(j)}\\ \\ \\ \\ \\displaystyle \\frac{{\\sin \\alpha +\\sin \\beta }}{{\\cos \\alpha -\\cos \\beta }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{2\\sin \\displaystyle \\frac{{\\alpha +\\beta }}{2}\\cos \\displaystyle \\frac{{\\alpha -\\beta }}{2}}}{{-2\\sin \\displaystyle \\frac{{\\alpha +\\beta }}{2}\\sin \\displaystyle \\frac{{\\alpha -\\beta }}{2}}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\cos \\displaystyle \\frac{{\\alpha -\\beta }}{2}}}{{-\\sin \\displaystyle \\frac{{\\alpha -\\beta }}{2}}}\\\\\\\\\\ \\ \\ \\ =\\ \\ -\\cot \\displaystyle \\frac{{\\alpha -\\beta }}{2}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\cot \\left( {-\\displaystyle \\frac{{\\alpha -\\beta }}{2}} \\right)\\ \\ \\ \\left[ {\\because \\cot \\left( {-\\theta } \\right)=-\\cot \\theta } \\right]\\\\\\\\\\ \\ \\ \\ =\\ \\ \\cot \\displaystyle \\frac{{\\beta -\\alpha }}{2}\\ \\end{array}$ $\\displaystyle \\begin{array}{l}\\text{(k)}\\ \\ \\ \\ \\displaystyle \\frac{{\\sin 2x}}{{1+\\cos 2x}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{2\\sin x\\cos x}}{{1+2{{{\\cos }}^{2}}x-1}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{2\\sin x\\cos x}}{{2{{{\\cos }}^{2}}x}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\sin x}}{{\\cos x}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\tan x\\end{array}$ $\\displaystyle \\begin{array}{l}\\text{(l)}\\ \\ \\ \\ \\displaystyle \\frac{{\\sin 3\\alpha }}{{\\cos \\alpha }}+\\displaystyle \\frac{{\\cos 3\\alpha }}{{\\sin \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\sin 3\\alpha \\sin \\alpha +\\cos 3\\alpha \\cos \\alpha }}{{\\sin \\alpha \\cos \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\cos 3\\alpha", ") ( \\cos \\text{45} °)}{- \\tan \\text{45} ° } \\\\ &= \\frac{ \\cos \\text{60} ° + \\sin \\text{45} ° \\cos \\text{45} ° }{- \\tan \\text{45} ° } \\end{align*} Now use special angles to evaluate the simplified expression: \\begin{align*} &= \\frac{ \\cos \\text{60} ° + \\sin \\text{45} ° \\cos \\text{45} ° }{- \\tan \\text{45} ° } \\\\ &= \\frac{ \\frac{1}{2} + \\left( \\frac{1}{\\sqrt{2}} \\right)\\left( \\frac{1}{\\sqrt{2}} \\right) }{- 1} \\\\ &= - \\left( \\frac{1}{2} + \\frac{1}{2} \\right) \\\\ &= -1 \\end{align*} ## Worked example 2: Revision Prove: $\\sin^{2} \\alpha - (\\tan \\alpha - \\cos \\alpha)(\\tan \\alpha + \\cos \\alpha) = \\frac{\\cos^{2} \\alpha - \\sin^{2} \\alpha}{\\cos^{2} \\alpha}$ State restrictions where applicable. ### Use trigonometric identities to simplify each side separately Simplify the left-hand side of the identity: \\begin{align*} \\text{LHS} &= \\sin^{2} \\alpha - (\\tan \\alpha - \\cos \\alpha)(\\tan \\alpha + \\cos \\alpha) \\\\ &= \\sin^{2} \\alpha - (\\tan^{2} \\alpha - \\cos^{2} \\alpha) \\\\ &= \\sin^{2} \\alpha - \\tan^{2} \\alpha + \\cos^{2} \\alpha \\\\ &= \\left( \\sin^{2} \\alpha + \\cos^{2} \\alpha \\right) - \\tan^{2} \\alpha \\\\ &= 1 - \\tan^{2} \\alpha \\end{align*} Simplify the right-hand side of the identity so that it equals the left-hand side: \\begin{align*} \\text{RHS} &= \\frac{\\cos^{2} \\alpha - \\sin^{2} \\alpha}{\\cos^{2} \\alpha} \\\\ &= \\frac{\\cos^{2} \\alpha}{\\cos^{2} \\alpha} - \\frac{\\sin^{2} \\alpha}{\\cos^{2} \\alpha} \\\\ &= 1 - \\tan^{2} \\alpha \\\\ &", "\\big) + i \\sin ( \\frac{\\pi}{2} + \\alpha \\big) \\Big\\}$ $\\displaystyle \\text{iii) } \\text{Let } z = (1 - \\sin \\alpha) + i \\cos \\alpha$ Both Sine and Cosine functions are periodic function with period $\\displaystyle 2\\pi$ Hence let us take $\\displaystyle \\alpha \\in [ 0, 2\\pi ]$ $\\displaystyle \\Rightarrow |z| = \\sqrt{ (1- \\sin \\alpha)^2 + \\cos^2 \\alpha } = \\sqrt{2- 2 \\sin \\alpha} = \\sqrt{2} \\sqrt{1- \\sin \\alpha}$ $\\displaystyle \\Rightarrow |z| = \\sqrt{2} \\sqrt{ (\\cos \\frac{\\alpha}{2} -\\sin \\frac{\\alpha}{2})^2 } = \\sqrt{2} \\Big| \\cos \\frac{\\alpha}{2} -\\sin \\frac{\\alpha}{2} \\Big|$ $\\displaystyle \\text{Let } \\beta \\text{ be an acute angle given by } \\tan \\beta = \\Big| \\frac{Im(z)}{Re(z)} \\Big|$ $\\displaystyle \\tan \\beta = \\frac{| \\cos \\alpha |}{|1- \\sin \\alpha |} = \\Big| \\frac{\\cos^2 \\frac{\\alpha}{2} - \\sin^2 \\frac{\\alpha}{2}}{( \\cos \\frac{\\alpha}{2} - \\sin \\frac{\\alpha}{2})^2} \\Big| = \\Big| \\frac{\\cos \\frac{\\alpha}{2} + \\sin \\frac{\\alpha}{2}}{ \\cos \\frac{\\alpha}{2} - \\sin \\frac{\\alpha}{2}} \\Big| = \\Big| \\frac{1 + \\tan \\frac{\\alpha}{2}}{ 1 - \\tan \\frac{\\alpha}{2}} \\Big|$ $\\displaystyle \\Rightarrow \\tan \\beta = \\Big| \\tan ( \\frac{\\pi}{4} + \\frac{\\alpha}{2} ) \\Big|$ $\\displaystyle \\underline{ \\text{Case I:}} 0 \\leq \\alpha < \\frac{\\pi}{2}$ In this case $\\displaystyle \\cos \\frac{\\alpha }{2} > \\sin \\frac{\\alpha }{2} \\text{ and } \\frac{\\pi}{4} + \\frac{\\alpha}{2} \\in \\Big[ \\frac{\\pi }{4} , \\frac{\\pi}{2} \\Big)$ $\\displaystyle \\Rightarrow |z| = \\sqrt{2} \\Big( \\cos \\frac{\\alpha }{2} - \\sin \\frac{\\alpha }{2} \\Big)$ $\\displaystyle \\text{and }", "}}=-\\displaystyle \\frac{3}{4}\\\\\\\\\\ \\ \\ \\ \\cos \\beta =\\displaystyle \\frac{{12}}{{13}},\\ 0{}^\\circ <\\beta <90{}^\\circ \\\\\\\\\\ \\ \\ \\ \\text{Since}\\ {{\\sin }^{2}}\\beta +{{\\cos }^{2}}\\beta =1\\\\\\\\\\ \\ \\ \\ \\sin \\beta =\\sqrt{{1-{{{\\cos }}^{2}}\\beta }}\\ \\ \\ \\ \\left[ {\\because 0{}^\\circ <\\beta <90{}^\\circ } \\right]\\\\\\\\\\therefore \\ \\ \\sin \\beta =\\sqrt{{1-\\displaystyle \\frac{{144}}{{169}}}}=\\displaystyle \\frac{5}{{13}}\\\\\\\\\\therefore \\ \\ \\tan \\beta =\\displaystyle \\frac{{\\sin \\beta }}{{\\cos \\beta }}=\\displaystyle \\frac{5}{{12}}\\\\\\\\\\therefore \\ \\ \\cos (\\alpha +\\beta )=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\\\\\\\\\therefore \\ \\ \\cos (\\alpha +\\beta )=\\left( {-\\displaystyle \\frac{4}{5}} \\right)\\left( {\\displaystyle \\frac{{12}}{{13}}} \\right)-\\left( {\\displaystyle \\frac{3}{5}} \\right)\\left( {\\displaystyle \\frac{5}{{13}}} \\right)\\\\\\\\\\therefore \\ \\ \\cos (\\alpha +\\beta )=-\\displaystyle \\frac{{63}}{{65}}\\\\\\\\\\therefore \\ \\ \\tan (\\alpha -\\beta )=\\displaystyle \\frac{{\\tan \\alpha -\\tan \\beta }}{{1+\\tan \\alpha \\tan \\beta }}\\\\\\\\\\therefore \\ \\ \\tan (\\alpha -\\beta )=\\displaystyle \\frac{{-\\displaystyle \\frac{3}{4}-\\displaystyle \\frac{5}{{12}}}}{{1+\\left( {-\\displaystyle \\frac{3}{4}} \\right)\\left( {\\displaystyle \\frac{5}{{12}}} \\right)}}\\\\\\\\\\therefore \\ \\ \\tan (\\alpha -\\beta )=\\displaystyle \\frac{{-\\displaystyle \\frac{{56}}{{48}}}}{{\\displaystyle \\frac{{33}}{{48}}}}=-\\displaystyle \\frac{{56}}{{33}}\\\\\\\\\\therefore \\ \\ \\cot (\\alpha -\\beta )=-\\displaystyle \\frac{{33}}{{56}}\\\\\\ \\ \\ \\ \\,\\end{array}$ 13. (a) Solve $\\displaystyle ΔABC$ with $\\displaystyle b=12.5, c=23$ and $\\displaystyle α=38°20′.$ (5 marks) Show/Hide Solution $\\displaystyle \\begin{array}{l}\\ \\ \\ \\ \\ \\ \\ b=12.5,\\ \\ c=23,\\ \\ \\alpha =38{}^\\circ 2{0}'.\\\\\\\\\\ \\ \\ \\ \\ \\ \\ a=?\\,\\ \\ \\ \\beta =?\\ \\ \\ \\ \\ \\gamma =?\\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\text{By the law of cosines,}\\\\\\\\\\ \\ \\ \\ \\ \\ \\ {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\\cos \\alpha \\\\\\\\\\therefore \\ \\", "\\alpha \\sec \\alpha }}{{{{{\\tan }}^{2}}\\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{2\\sec \\alpha }}{{\\tan \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{2}{{\\cos \\alpha }}\\times \\displaystyle \\frac{{\\cos \\alpha }}{{\\sin \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{2}{{\\sin \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ 2\\operatorname{cosec}\\alpha \\end{array}$ $\\displaystyle \\begin{array}{l}\\text{(o)}\\ \\ \\ \\ \\displaystyle \\frac{{2-{{{\\operatorname{cosec}}}^{2}}\\alpha }}{{{{{\\operatorname{cosec}}}^{2}}\\alpha +2\\cot \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{2-\\displaystyle \\frac{1}{{{{{\\sin }}^{2}}\\alpha }}}}{{\\displaystyle \\frac{1}{{{{{\\sin }}^{2}}\\alpha }}+2\\left( {\\displaystyle \\frac{{\\cos \\alpha }}{{\\sin \\alpha }}} \\right)}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\displaystyle \\frac{{2{{{\\sin }}^{2}}\\alpha -1}}{{{{{\\sin }}^{2}}\\alpha }}}}{{\\displaystyle \\frac{{1+2\\sin \\alpha \\cos \\alpha }}{{{{{\\sin }}^{2}}\\alpha }}}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{2{{{\\sin }}^{2}}\\alpha -1}}{{1+2\\sin \\alpha \\cos \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{2{{{\\sin }}^{2}}\\alpha -{{{\\sin }}^{2}}\\alpha -{{{\\cos }}^{2}}\\alpha }}{{{{{\\sin }}^{2}}\\alpha +2\\sin \\alpha \\cos \\alpha +{{{\\cos }}^{2}}\\alpha }}\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\left[ {\\because {{{\\sin }}^{2}}\\alpha +{{{\\cos }}^{2}}\\alpha =1} \\right]\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{{{{\\sin }}^{2}}\\alpha -{{{\\cos }}^{2}}\\alpha }}{{{{{\\left( {\\sin \\alpha +\\cos \\alpha } \\right)}}^{2}}}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\left( {\\sin \\alpha -\\cos \\alpha } \\right)\\left( {\\sin \\alpha +\\cos \\alpha } \\right)}}{{{{{\\left( {\\sin \\alpha +\\cos \\alpha } \\right)}}^{2}}}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\sin \\alpha -\\cos \\alpha }}{{\\sin \\alpha +\\cos \\alpha }}\\end{array}$ $\\displaystyle \\begin{array}{l}\\text{(p)}\\ \\ \\ \\ 1-\\displaystyle \\frac{{{{{\\sin }}^{2}}x}}{{1+\\cos x}}\\\\\\\\\\ \\ \\ \\", "\\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\sin(\\alpha -\\beta )=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha +\\beta )=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha -\\beta )=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\tan(\\alpha +\\beta )={\\frac {\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }}\\,\\!}$ ${\\displaystyle \\tan(\\alpha -\\beta )={\\frac {\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }}\\,\\!}$ Formler för halva vinkeln: ${\\displaystyle \\sin ^{2}({\\frac {\\alpha }{2}})={\\frac {1-\\cos \\alpha }{2}}\\,\\!}$ ${\\displaystyle \\cos ^{2}({\\frac {\\alpha }{2}})={\\frac {1+\\cos \\alpha }{2}}\\,\\!}$ Formler för dubbla vinkeln: ${\\displaystyle \\sin(2\\alpha )=2\\sin(\\alpha )\\cos(\\alpha )\\,\\!}$ ${\\displaystyle \\cos(2\\alpha )=\\cos ^{2}\\alpha -\\sin ^{2}\\alpha =2\\cos ^{2}\\alpha -1=1-2\\sin ^{2}\\alpha \\,\\!}$ ${\\displaystyle \\tan(2\\alpha )={\\frac {2\\tan \\alpha }{1-\\tan ^{2}\\alpha }}\\,\\!}$ Produktformler: ${\\displaystyle 2\\cos \\alpha \\cos \\beta =\\cos(\\alpha -\\beta )+\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\sin \\beta =\\cos(\\alpha -\\beta )-\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\cos \\beta =\\sin(\\alpha -\\beta )+\\sin(\\alpha +\\beta )\\,\\!}$ Linjärkombinationer: ${\\displaystyle a>0,b>0,\\tan \\beta ={\\frac {b}{a}},0^{\\circ }<\\beta <90^{\\circ }}$ gäller ${\\displaystyle a\\sin \\alpha +b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle a\\sin \\alpha -b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha -\\beta )\\,\\!}$ Här saknas information! Du kan hjälpa Wikibooks genom att fylla i mer! Exakta trigonometriska funktionsvärden Vinkel ${\\displaystyle \\alpha }$ ${\\displaystyle \\sin \\alpha }$ ${\\displaystyle \\cos \\alpha }$ ${\\displaystyle \\tan \\alpha }$ ${\\displaystyle \\cot \\alpha }$ ${\\displaystyle \\sec \\alpha }$ ${\\displaystyle \\csc \\alpha }$ i grader", "\\cos \\frac{\\alpha + \\beta}{2} \\cos \\frac{\\alpha - \\beta}{2} = b$ … … … … … ii) Dividing i) by ii) we get $\\displaystyle \\tan \\frac{\\alpha + \\beta}{2} = \\frac{a}{b}$ $\\displaystyle \\text{We know } \\sin (\\alpha + \\beta) = \\frac{2 \\tan \\big( \\frac{\\alpha + \\beta}{2} \\big)}{1+ \\tan^2 \\big( \\frac{\\alpha + \\beta}{2} \\big)} = \\frac{2 \\frac{a}{b}}{1+\\frac{a^2}{b^2} } = \\frac{2ab}{a^2+b^2}$ $\\displaystyle ( \\sin \\alpha + \\sin \\beta)^2 + ( \\cos \\alpha + \\cos \\beta)^2 = a^2 +b^2$ $\\displaystyle \\Rightarrow \\sin^2 \\alpha + \\sin^2 \\beta + 2 \\sin \\alpha \\sin \\beta + \\cos^2 \\alpha + \\cos^2 \\beta + 2 \\cos \\alpha \\cos \\beta = a^2 + b^2$ $\\displaystyle \\Rightarrow 2 + 2 \\sin \\alpha \\sin \\beta + 2 \\cos \\alpha \\cos \\beta = a^2 + b^2$ $\\displaystyle \\Rightarrow 2 + 2 ( \\sin \\alpha \\sin \\beta + \\cos \\alpha \\cos \\beta) = a^2 + b^2$ $\\displaystyle \\Rightarrow 2 + 2 \\cos ( \\alpha - \\beta) = a^2 + b^2$ $\\displaystyle \\Rightarrow \\cos ( \\alpha - \\beta) = \\frac{a^2 + b^2-2}{2}$ $\\displaystyle \\\\$ $\\displaystyle \\text{Question 39: } \\text{If } 2\\tan \\frac{\\alpha}{2} = \\tan \\frac{\\beta}{2} , \\text{ prove that } \\cos \\alpha = \\frac{3 + 5 \\cos \\beta}{5 + 3 \\cos \\beta}$ $\\displaystyle \\text{Given } 2\\tan \\frac{\\alpha}{2} = \\tan \\frac{\\beta}{2} ,$ $\\displaystyle \\text{RHS } = \\frac{3 + 5 \\cos \\beta}{5 + 3 \\cos \\beta}$", "if you are dividing by 0 and thus losing solutions. In this case I would probably lean more towards making use of a compound angle identity \\displaystyle \\begin{align*} \\sin{ \\left( \\alpha + \\beta \\right) } \\equiv \\sin{ \\left( \\alpha \\right) } \\cos{ \\left( \\beta \\right) } + \\cos{ \\left( \\alpha \\right) } \\sin{ \\left( \\beta \\right) } \\end{align*}. So with this equation we have \\displaystyle \\begin{align*} -\\sin{ \\left( x \\right) } &= \\cos{ \\left( x \\right) } \\\\ 0 &= \\sin{ \\left( x \\right) } + \\cos{ \\left( x \\right) } \\\\ 0 \\cdot k &= \\sin{ \\left( x \\right) } \\cdot k + \\cos{ \\left( x \\right) } \\cdot k \\\\ 0 &= \\sin{ \\left( x \\right) } \\cos{ \\left( y \\right) } + \\cos{ \\left( x \\right) } \\sin{ \\left( y \\right) } \\end{align*} and here \\displaystyle \\begin{align*} \\cos{ \\left( y \\right) } = \\sin{ \\left( y \\right) } = k \\end{align*}. As \\displaystyle \\begin{align*} \\cos{ \\left( y \\right) } = \\sin{ \\left( y \\right) } \\end{align*} the only possible value of y in the first cycle is \\displaystyle \\begin{align*} \\frac{\\pi}{4} \\end{align*}, where both the sine and cosine are \\displaystyle \\begin{align*} k = \\frac{1}{\\sqrt{2}} \\end{align*}, thus \\displaystyle \\begin{align*} 0 &= \\sin{ \\left( x \\right) } \\cos{ \\left( \\frac{\\pi}{4} \\right) } + \\cos{ \\left( x \\right) }", "=1\\,\\!}$ ${\\displaystyle \\sin(\\alpha +\\beta )=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\sin(\\alpha -\\beta )=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha +\\beta )=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha -\\beta )=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\tan(\\alpha +\\beta )={\\frac {\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }}\\,\\!}$ ${\\displaystyle \\tan(\\alpha -\\beta )={\\frac {\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }}\\,\\!}$ ### Formler för halva vinkeln: ${\\displaystyle \\sin ^{2}({\\frac {\\alpha }{2}})={\\frac {1-\\cos \\alpha }{2}}\\,\\!}$ ${\\displaystyle \\cos ^{2}({\\frac {\\alpha }{2}})={\\frac {1+\\cos \\alpha }{2}}\\,\\!}$ ### Formler för dubbla vinkeln: ${\\displaystyle \\sin(2\\alpha )=2\\sin(\\alpha )\\cos(\\alpha )\\,\\!}$ ${\\displaystyle \\cos(2\\alpha )=\\cos ^{2}\\alpha -\\sin ^{2}\\alpha =2\\cos ^{2}\\alpha -1=1-2\\sin ^{2}\\alpha \\,\\!}$ ${\\displaystyle \\tan(2\\alpha )={\\frac {2\\tan \\alpha }{1-\\tan ^{2}\\alpha }}\\,\\!}$ ### Produktformler: ${\\displaystyle 2\\cos \\alpha \\cos \\beta =\\cos(\\alpha -\\beta )+\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\sin \\beta =\\cos(\\alpha -\\beta )-\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\cos \\beta =\\sin(\\alpha -\\beta )+\\sin(\\alpha +\\beta )\\,\\!}$ ### Linjärkombinationer: ${\\displaystyle a>0,b>0,\\tan \\beta ={\\frac {b}{a}},0^{\\circ }<\\beta <90^{\\circ }}$ gäller ${\\displaystyle a\\sin \\alpha +b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle a\\sin \\alpha -b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha -\\beta )\\,\\!}$ Här saknas information! Du kan hjälpa Wikibooks genom att fylla i mer! # Exakta trigonometriska funktionsvärden Vinkel ${\\displaystyle \\alpha }$ ${\\displaystyle \\sin \\alpha }$ ${\\displaystyle \\cos \\alpha }$ ${\\displaystyle \\tan \\alpha }$", "}}{{1 - i\\tan (\\alpha )}} = \\frac{{\\left( {1 + i\\tan (\\alpha )} \\right)^5 }}{{\\sec ^2 (\\alpha )}}$ $\\frac{{\\left( {1 + i\\tan (\\alpha )} \\right)^5 }}{{\\sec ^2 (\\alpha )}} = \\frac{{\\frac{{\\left( {\\cos (\\alpha ) + i\\sin (\\alpha )} \\right)^5 }}{{\\cos ^5 (\\alpha )}}}}{{\\sec ^2 (\\alpha )}} = \\frac{{\\cos (5\\alpha ) + i\\sin (5\\alpha )}}{{\\cos ^3 (\\alpha )}}$ 4. This is a postscript, resulting from a PM. It seems that the problem is actually $\\left[ {\\frac{{1 + \\tan (\\alpha )}}{{1 - \\tan (\\alpha )}}} \\right]^4$. In which case; $\\begin{array}{rcl} \\left[ {\\frac{{1 + i\\tan (\\alpha )}}{{1 - i\\tan (\\alpha )}}} \\right]^4 & = & \\left[ {\\frac{{\\cos (\\alpha ) + i\\sin (\\alpha )}}{{\\cos (\\alpha ) - i\\sin (\\alpha )}}} \\right]^4 \\\\ & = & \\left[ {\\cos (2\\alpha ) + i\\sin (2\\alpha )} \\right]^4 \\\\ & = & \\cos (8\\alpha ) + i\\sin (8\\alpha ) \\\\ \\end{array}$ 5. Thanks", "\\cos \\alpha +\\sin 3\\alpha \\sin \\alpha }}{{\\sin \\alpha \\cos \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{2\\cos \\left( {3\\alpha -\\alpha } \\right)}}{{2\\sin \\alpha \\cos \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{2\\cos 2\\alpha }}{{\\sin 2\\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ 2\\cot 2\\alpha \\end{array}$ $\\displaystyle \\begin{array}{l}\\text{(m)}\\ \\ \\ \\ \\displaystyle \\frac{1}{{\\cos 2\\alpha }}-\\displaystyle \\frac{{\\cos 2\\alpha }}{{1+\\sin 2\\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{1+\\sin 2\\alpha -{{{\\cos }}^{2}}2\\alpha }}{{\\cos 2\\alpha \\left( {1+\\sin 2\\alpha } \\right)}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\sin 2\\alpha +1-{{{\\cos }}^{2}}2\\alpha }}{{\\cos 2\\alpha \\left( {1+\\sin 2\\alpha } \\right)}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\sin 2\\alpha +{{{\\sin }}^{2}}2\\alpha }}{{\\cos 2\\alpha \\left( {1+\\sin 2\\alpha } \\right)}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\sin 2\\alpha \\left( {1+\\sin 2\\alpha } \\right)}}{{\\cos 2\\alpha \\left( {1+\\sin 2\\alpha } \\right)}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\tan 2\\alpha \\end{array}$ $\\displaystyle \\begin{array}{l}\\text{(n)}\\ \\ \\ \\ \\displaystyle \\frac{{\\tan \\alpha }}{{\\sec \\alpha -1}}+\\displaystyle \\frac{{\\tan \\alpha }}{{\\sec \\alpha +1}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\tan \\alpha \\left( {\\sec \\alpha +1} \\right)+\\tan \\alpha \\left( {\\sec \\alpha -1} \\right)}}{{{{{\\sec }}^{2}}\\alpha -1}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\tan \\alpha \\left( {\\sec \\alpha +1+\\sec \\alpha -1} \\right)}}{{{{{\\tan }}^{2}}\\alpha }}\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\left[ {\\because {{{\\tan }}^{2}}\\alpha +1={{{\\sec }}^{2}}\\alpha } \\right]\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{2\\tan", "\\sin \\alpha =\\displaystyle \\frac{p}{{\\sqrt{{1+{{p}^{2}}}}}}\\ \\ \\ \\left[ {\\because \\text{given}} \\right]\\\\\\\\\\ \\ \\ \\ \\ \\sin \\alpha =\\displaystyle \\frac{p}{h}\\ \\ \\ \\left[ {\\text{diagram}} \\right]\\\\\\\\\\therefore \\,\\ \\ \\displaystyle \\frac{p}{h}=\\displaystyle \\frac{p}{{\\sqrt{{1+{{p}^{2}}}}}}\\\\\\\\\\therefore \\ \\ h=\\sqrt{{1+{{p}^{2}}}}\\\\\\\\\\ \\ \\ \\ \\text{Since}\\ {{x}^{2}}+{{p}^{2}}={{h}^{2}},\\\\\\\\\\ \\ \\ \\ {{x}^{2}}+{{p}^{2}}=1+{{p}^{2}}\\\\\\\\\\therefore \\ \\ {{x}^{2}}=1\\\\\\\\\\therefore \\ \\ x=\\pm 1\\\\\\\\\\ \\ \\ \\ \\text{Since}\\ \\alpha \\,\\ \\text{lies in }{{\\text{3}}^{{\\text{rd}}}}\\text{ quadrant,}\\\\\\\\\\ \\ \\ x=-1.\\\\\\\\\\\\\\text{(b)}\\ \\ \\cos 2\\alpha =1-2{{\\sin }^{2}}\\alpha \\\\\\\\\\therefore \\,\\ \\ \\ \\cos 2\\alpha =1-\\displaystyle \\frac{{2{{p}^{2}}}}{{1+{{p}^{2}}}}\\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\\displaystyle \\frac{{1-{{p}^{2}}}}{{1+{{p}^{2}}}}\\end{array}$ 9. If $\\displaystyle \\sin 76{}^\\circ = x$ and $\\displaystyle \\cos 76{}^\\circ = y$, show that $\\displaystyle x^2- y^2 = \\sin 62{}^\\circ$. Show/Hide Solution $\\displaystyle \\begin{array}{l}\\ \\ \\ \\ \\sin 76{}^\\circ =x,\\ \\cos 76{}^\\circ =y\\\\\\\\\\ \\ \\ {{x}^{2}}-{{y}^{2}}={{\\sin }^{2}}76{}^\\circ -{{\\cos }^{2}}76{}^\\circ \\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =-\\left( {{{{\\cos }}^{2}}76{}^\\circ -{{{\\sin }}^{2}}76{}^\\circ } \\right)\\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =-\\cos \\left( {2\\times 76{}^\\circ } \\right)\\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =-\\cos 152{}^\\circ \\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =-\\cos \\left( {90{}^\\circ +62{}^\\circ } \\right)\\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\ \\", "special triangles. sinpi/3 gives (root3/2) but what of cospi/2 and sinpi/2? – Ash Apr 3 '15 at 18:53 • Okay, well I'm not sure if I did this right but it came to (root3/2)(0)-(1)(1/2) – Ash Apr 3 '15 at 19:27 • Equaling -1/2 since anything multiplied by 0 equals zero. – Ash Apr 3 '15 at 19:28 The first question can be handled with the difference of angles formula $$\\sin(\\alpha - \\beta) = \\sin\\alpha\\cos\\beta - \\cos\\alpha\\sin\\beta$$ where $\\alpha = \\frac{\\pi}{3}$ and $\\beta = \\frac{\\pi}{2}$ since $-\\frac{\\pi}{2} + \\frac{\\pi}{3} = \\frac{\\pi}{3} - \\frac{\\pi}{2}$. Substitution yields \\begin{align*} \\sin\\left(-\\frac{\\pi}{2} + \\frac{\\pi}{3}\\right) & = \\sin\\left(\\frac{\\pi}{3} - \\frac{\\pi}{2}\\right)\\\\ & = \\sin\\left(\\frac{\\pi}{3}\\right)\\cos\\left(\\frac{\\pi}{2}\\right) - \\sin\\left(\\frac{\\pi}{2}\\right)\\cos\\left(\\frac{\\pi}{3}\\right)\\\\ & = \\frac{\\sqrt{3}}{2} \\cdot 0 - 1 \\cdot \\frac{1}{2}\\\\ & = 0 - \\frac{1}{2}\\\\ & = -\\frac{1}{2} \\end{align*} Check: \\begin{align*} \\sin\\left(-\\frac{\\pi}{2} + \\frac{\\pi}{3}\\right) & = \\sin\\left(-\\frac{\\pi}{6}\\right)\\\\ & = -\\sin\\left(\\frac{\\pi}{6}\\right)\\\\ & = -\\frac{1}{2} \\end{align*} since the sine function is negative in the fourth quadrant. HINT: Recount $\\sin(A+B)$ formula holds true for all finite values of $A,B$ $(1)$ How about $A=-\\dfrac\\pi2, B=\\dfrac\\pi3$ $(2)$ Proceed with the formula $(3)$ Put $A=B$ in $\\sin(A+B)$ formula For 3rd one, we can use this formula $$\\sin(2A)=2\\sin(A)\\cos(A).$$ So the answer will be $1/\\sqrt{2}$.", "\\ \\left[ \\begin{array}{l}\\because \\displaystyle \\frac{a}{b}=\\displaystyle \\frac{c}{d}\\Leftrightarrow \\displaystyle \\frac{{a+b}}{{a-b}}=\\displaystyle \\frac{{c+d}}{{c-d}}\\\\\\ (\\text{Componendo and Dividendo})\\end{array} \\right]\\\\\\\\\\ \\therefore \\ \\ \\ \\ \\displaystyle \\frac{{\\sin \\left( {\\alpha +\\beta } \\right)+\\sin \\left( {\\alpha -\\beta } \\right)}}{{\\sin \\left( {\\alpha +\\beta } \\right)-\\sin \\left( {\\alpha -\\beta } \\right)}}=\\displaystyle \\frac{7}{2}\\\\\\\\\\ \\therefore \\ \\ \\ \\ \\displaystyle \\frac{{2\\sin \\displaystyle \\frac{{\\alpha +\\beta +\\alpha -\\beta }}{2}\\cos \\displaystyle \\frac{{\\alpha +\\beta -\\alpha +\\beta }}{2}}}{{2\\cos \\displaystyle \\frac{{\\alpha +\\beta +\\alpha -\\beta }}{2}\\sin \\displaystyle \\frac{{\\alpha +\\beta -\\alpha +\\beta }}{2}}}=\\displaystyle \\frac{7}{3}\\\\\\\\\\ \\therefore \\ \\ \\ \\ \\displaystyle \\frac{{2\\sin \\alpha \\cos \\beta }}{{2\\cos \\alpha \\sin \\beta }}=\\displaystyle \\frac{7}{3}\\\\\\\\\\ \\therefore \\ \\ \\ \\ \\tan \\alpha \\cot \\beta =\\displaystyle \\frac{7}{3}\\\\\\\\\\ \\therefore \\ \\ \\ \\ \\displaystyle \\frac{{\\tan \\alpha }}{{\\tan \\beta }}=\\displaystyle \\frac{7}{3}\\\\\\\\\\therefore \\ \\ \\ \\ \\ 3\\tan \\alpha =7\\tan \\beta \\end{array}$ $\\displaystyle \\ \\ \\ \\ \\ \\ \\ \\underline{{\\text{Solution (1)}\\ \\text{to find}\\tan \\alpha +\\tan \\beta \\text{ }}}$ $\\displaystyle \\begin{array}{l}\\therefore \\ \\ \\ \\ \\ \\tan \\beta =\\displaystyle \\frac{3}{7}\\tan \\alpha \\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\alpha +\\beta =45{}^\\circ \\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\tan \\left( {\\alpha +\\beta } \\right)=\\tan 45{}^\\circ \\\\\\\\\\ \\ \\ \\ \\ \\ \\ \\displaystyle \\frac{{\\tan \\alpha +\\displaystyle \\frac{3}{7}\\tan \\alpha }}{{1-\\tan \\alpha \\displaystyle \\frac{3}{7}\\tan \\alpha }}=1\\ \\\\\\\\\\therefore \\ \\ \\ \\ \\ \\displaystyle \\frac{{10}}{7}\\tan \\alpha =1-\\displaystyle \\frac{3}{7}{{\\tan }^{2}}\\alpha \\\\\\\\\\therefore \\ \\ \\ \\ \\ 3{{\\tan }^{2}}\\alpha +10\\tan \\alpha -7=0\\\\\\\\\\therefore \\", "\\beta\\end{aligned}\\end{align} So: \\begin{align}\\begin{aligned}\\cos(\\alpha + \\beta) = x_2 = r - u = \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta\\\\\\sin(\\alpha + \\beta) = y_2 = t + s = \\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta\\end{aligned}\\end{align}", "\\alpha +\\cos \\alpha }}\\\\\\\\\\text{(p)}\\ \\ 1-\\displaystyle \\frac{{{{{\\sin }}^{2}}x}}{{1+\\cos x}}=\\cos x\\\\\\\\\\text{(q)}\\ \\ \\displaystyle \\frac{{1+{{{\\cot }}^{2}}x}}{{1-{{{\\cot }}^{2}}x}}=-\\sec 2x\\\\\\\\\\text{(r)}\\ \\ \\displaystyle \\frac{{\\cos \\alpha +\\sec \\beta }}{{\\cos \\beta +\\sec \\alpha }}=\\cos \\alpha \\sec \\beta\\end{array}$ Show/Hide Solution $\\displaystyle \\begin{array}{l}\\text{(a)}\\ \\ \\ \\ \\cos \\left( {60{}^\\circ +x} \\right)+\\sin \\left( {30{}^\\circ +x} \\right)\\\\\\ \\ \\ \\ \\ =\\ \\cos 60{}^\\circ \\cos x-\\sin 60{}^\\circ \\sin x+\\sin 30{}^\\circ \\cos x+\\cos 30{}^\\circ \\sin x\\\\\\ \\ \\ \\ \\ =\\ \\displaystyle \\frac{1}{2}\\cos x-\\displaystyle \\frac{{\\sqrt{3}}}{2}\\sin x+\\displaystyle \\frac{1}{2}\\cos x+\\displaystyle \\frac{{\\sqrt{3}}}{2}\\sin x\\\\\\ \\ \\ \\ \\ =\\ \\cos x\\end{array}$ $\\displaystyle \\begin{array}{l}\\text{(b)}\\ \\ \\ \\ \\tan \\alpha +\\cot \\alpha \\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\sin \\alpha }}{{\\cos \\alpha }}+\\displaystyle \\frac{{\\cos \\alpha }}{{\\sin \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{{{{\\sin }}^{2}}\\alpha +{{{\\cos }}^{2}}\\alpha }}{{\\sin \\alpha \\cos \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{1}{{\\sin \\alpha \\cos \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{2}{{2\\sin \\alpha \\cos \\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{2}{{\\sin 2\\alpha }}\\\\\\\\\\ \\ \\ \\ =\\ \\ 2\\operatorname{cosec}2\\alpha \\end{array}$ $\\displaystyle \\begin{array}{l}\\text{(c)}\\ \\ \\ \\ \\operatorname{cosec}2x+\\cot 2x\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{1}{{\\sin 2x}}+\\displaystyle \\frac{{\\cos 2x}}{{\\sin 2x}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{1+\\cos 2x}}{{\\sin 2x}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{1+2{{{\\cos }}^{2}}x-1}}{{2\\sin x\\cos x}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{{{{\\cos }}^{2}}x}}{{\\sin x\\cos x}}\\\\\\\\\\ \\ \\ \\ =\\ \\ \\displaystyle \\frac{{\\cos" ]

[ "\\ \\ \\displaystyle \\frac{{1+\\sin 2\\alpha -\\cos 2\\alpha }}{{1+\\sin 2\\alpha +\\cos 2\\alpha }}\\\\\\\\=\\displaystyle \\frac{{1+2\\sin \\alpha \\cos \\alpha -1+2{{{\\sin }}^{2}}\\alpha }}{{1+2\\sin \\alpha \\cos \\alpha +2{{{\\cos }}^{2}}\\alpha -1}}\\\\\\\\=\\displaystyle \\frac{{2\\sin \\alpha \\left( {\\sin \\alpha +\\cos \\alpha } \\right)}}{{2\\cos \\alpha \\left( {\\sin \\alpha +\\cos \\alpha } \\right)}}\\\\\\\\=\\tan \\alpha \\\\\\\\\\text{(b)}\\\\\\ \\ \\ \\displaystyle \\frac{{1+\\sin 2\\alpha +\\cos 2\\alpha }}{{1+\\sin 2\\alpha -\\cos 2\\alpha }}\\\\\\\\=\\displaystyle \\frac{{1+2\\sin \\alpha \\cos \\alpha +2{{{\\cos }}^{2}}\\alpha -1}}{{1+2\\sin \\alpha \\cos \\alpha -1+2{{{\\sin }}^{2}}\\alpha }}\\\\\\\\=\\displaystyle \\frac{{2\\cos \\alpha \\left( {\\sin \\alpha +\\cos \\alpha } \\right)}}{{2\\sin \\alpha \\left( {\\sin \\alpha +\\cos \\alpha } \\right)}}\\\\\\\\=\\cot \\alpha \\end{array}$ 11. If $\\displaystyle \\sec \\theta + \\tan \\theta = 3$, where $\\displaystyle \\theta$ lies in the first quadrant, then find the value of $\\displaystyle \\cos \\theta$. Show/Hide Solution $\\displaystyle \\begin{array}{l}\\ \\ \\ \\sec \\theta +\\tan \\theta =3\\ ---(1)\\\\\\\\\\ \\ \\ \\displaystyle \\frac{{\\left( {\\sec \\theta +\\tan \\theta } \\right)\\left( {\\sec \\theta -\\tan \\theta } \\right)}}{{\\sec \\theta -\\tan \\theta }}=3\\\\\\\\\\ \\ \\ \\displaystyle \\frac{{{{{\\sec }}^{2}}\\theta -{{{\\tan }}^{2}}\\theta }}{{\\sec \\theta -\\tan \\theta }}=3\\\\\\\\\\ \\ \\ \\displaystyle \\frac{1}{{\\sec \\theta -\\tan \\theta }}=3\\\\\\\\\\therefore \\ \\ \\sec \\theta -\\tan \\theta =\\displaystyle \\frac{1}{3}\\ ---(2)\\\\\\\\\\ \\ \\ (1)+(2)\\Rightarrow 2\\sec \\theta =\\displaystyle \\frac{{10}}{3}\\\\\\\\\\therefore \\ \\sec \\theta =\\displaystyle \\frac{5}{3}\\\\\\\\\\therefore \\ \\cos \\theta =\\displaystyle \\frac{3}{5}\\end{array}$ 12. If $\\displaystyle \\operatorname{cosec}\\theta -\\cot \\theta =\\frac{1}{5}$, where $\\displaystyle \\theta$ lies in the first quadrant, then find the value of $\\displaystyle \\sin \\theta$. Show/Hide Solution $\\displaystyle", "\\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\sin(\\alpha -\\beta )=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha +\\beta )=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha -\\beta )=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\tan(\\alpha +\\beta )={\\frac {\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }}\\,\\!}$ ${\\displaystyle \\tan(\\alpha -\\beta )={\\frac {\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }}\\,\\!}$ Formler för halva vinkeln: ${\\displaystyle \\sin ^{2}({\\frac {\\alpha }{2}})={\\frac {1-\\cos \\alpha }{2}}\\,\\!}$ ${\\displaystyle \\cos ^{2}({\\frac {\\alpha }{2}})={\\frac {1+\\cos \\alpha }{2}}\\,\\!}$ Formler för dubbla vinkeln: ${\\displaystyle \\sin(2\\alpha )=2\\sin(\\alpha )\\cos(\\alpha )\\,\\!}$ ${\\displaystyle \\cos(2\\alpha )=\\cos ^{2}\\alpha -\\sin ^{2}\\alpha =2\\cos ^{2}\\alpha -1=1-2\\sin ^{2}\\alpha \\,\\!}$ ${\\displaystyle \\tan(2\\alpha )={\\frac {2\\tan \\alpha }{1-\\tan ^{2}\\alpha }}\\,\\!}$ Produktformler: ${\\displaystyle 2\\cos \\alpha \\cos \\beta =\\cos(\\alpha -\\beta )+\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\sin \\beta =\\cos(\\alpha -\\beta )-\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\cos \\beta =\\sin(\\alpha -\\beta )+\\sin(\\alpha +\\beta )\\,\\!}$ Linjärkombinationer: ${\\displaystyle a>0,b>0,\\tan \\beta ={\\frac {b}{a}},0^{\\circ }<\\beta <90^{\\circ }}$ gäller ${\\displaystyle a\\sin \\alpha +b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle a\\sin \\alpha -b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha -\\beta )\\,\\!}$ Här saknas information! Du kan hjälpa Wikibooks genom att fylla i mer! Exakta trigonometriska funktionsvärden Vinkel ${\\displaystyle \\alpha }$ ${\\displaystyle \\sin \\alpha }$ ${\\displaystyle \\cos \\alpha }$ ${\\displaystyle \\tan \\alpha }$ ${\\displaystyle \\cot \\alpha }$ ${\\displaystyle \\sec \\alpha }$ ${\\displaystyle \\csc \\alpha }$ i grader", "=1\\,\\!}$ ${\\displaystyle \\sin(\\alpha +\\beta )=\\sin \\alpha \\cos \\beta +\\cos \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\sin(\\alpha -\\beta )=\\sin \\alpha \\cos \\beta -\\cos \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha +\\beta )=\\cos \\alpha \\cos \\beta -\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\cos(\\alpha -\\beta )=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta \\,\\!}$ ${\\displaystyle \\tan(\\alpha +\\beta )={\\frac {\\tan \\alpha +\\tan \\beta }{1-\\tan \\alpha \\tan \\beta }}\\,\\!}$ ${\\displaystyle \\tan(\\alpha -\\beta )={\\frac {\\tan \\alpha -\\tan \\beta }{1+\\tan \\alpha \\tan \\beta }}\\,\\!}$ ### Formler för halva vinkeln: ${\\displaystyle \\sin ^{2}({\\frac {\\alpha }{2}})={\\frac {1-\\cos \\alpha }{2}}\\,\\!}$ ${\\displaystyle \\cos ^{2}({\\frac {\\alpha }{2}})={\\frac {1+\\cos \\alpha }{2}}\\,\\!}$ ### Formler för dubbla vinkeln: ${\\displaystyle \\sin(2\\alpha )=2\\sin(\\alpha )\\cos(\\alpha )\\,\\!}$ ${\\displaystyle \\cos(2\\alpha )=\\cos ^{2}\\alpha -\\sin ^{2}\\alpha =2\\cos ^{2}\\alpha -1=1-2\\sin ^{2}\\alpha \\,\\!}$ ${\\displaystyle \\tan(2\\alpha )={\\frac {2\\tan \\alpha }{1-\\tan ^{2}\\alpha }}\\,\\!}$ ### Produktformler: ${\\displaystyle 2\\cos \\alpha \\cos \\beta =\\cos(\\alpha -\\beta )+\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\sin \\beta =\\cos(\\alpha -\\beta )-\\cos(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle 2\\sin \\alpha \\cos \\beta =\\sin(\\alpha -\\beta )+\\sin(\\alpha +\\beta )\\,\\!}$ ### Linjärkombinationer: ${\\displaystyle a>0,b>0,\\tan \\beta ={\\frac {b}{a}},0^{\\circ }<\\beta <90^{\\circ }}$ gäller ${\\displaystyle a\\sin \\alpha +b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha +\\beta )\\,\\!}$ ${\\displaystyle a\\sin \\alpha -b\\cos \\alpha ={\\sqrt {(}}a^{2}+b^{2})\\sin(\\alpha -\\beta )\\,\\!}$ Här saknas information! Du kan hjälpa Wikibooks genom att fylla i mer! # Exakta trigonometriska funktionsvärden Vinkel ${\\displaystyle \\alpha }$ ${\\displaystyle \\sin \\alpha }$ ${\\displaystyle \\cos \\alpha }$ ${\\displaystyle \\tan \\alpha }$", "}}{{1 - i\\tan (\\alpha )}} = \\frac{{\\left( {1 + i\\tan (\\alpha )} \\right)^5 }}{{\\sec ^2 (\\alpha )}}$ $\\frac{{\\left( {1 + i\\tan (\\alpha )} \\right)^5 }}{{\\sec ^2 (\\alpha )}} = \\frac{{\\frac{{\\left( {\\cos (\\alpha ) + i\\sin (\\alpha )} \\right)^5 }}{{\\cos ^5 (\\alpha )}}}}{{\\sec ^2 (\\alpha )}} = \\frac{{\\cos (5\\alpha ) + i\\sin (5\\alpha )}}{{\\cos ^3 (\\alpha )}}$ 4. This is a postscript, resulting from a PM. It seems that the problem is actually $\\left[ {\\frac{{1 + \\tan (\\alpha )}}{{1 - \\tan (\\alpha )}}} \\right]^4$. In which case; $\\begin{array}{rcl} \\left[ {\\frac{{1 + i\\tan (\\alpha )}}{{1 - i\\tan (\\alpha )}}} \\right]^4 & = & \\left[ {\\frac{{\\cos (\\alpha ) + i\\sin (\\alpha )}}{{\\cos (\\alpha ) - i\\sin (\\alpha )}}} \\right]^4 \\\\ & = & \\left[ {\\cos (2\\alpha ) + i\\sin (2\\alpha )} \\right]^4 \\\\ & = & \\cos (8\\alpha ) + i\\sin (8\\alpha ) \\\\ \\end{array}$ 5. Thanks", "x}$ $\\displaystyle \\Rightarrow \\frac{\\cos^2 x \\cos \\alpha}{\\cos^2 x \\cos^2 \\alpha - \\sin^2 x \\sin^2 \\alpha } = 1$ $\\displaystyle \\Rightarrow \\cos^2 x \\cos \\alpha = \\cos^2 x \\cos^2 \\alpha - \\sin^2 x \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x \\cos \\alpha = \\cos^2 x \\cos^2 \\alpha - (1 - \\cos^2 x) \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x \\cos \\alpha = \\cos^2 x \\cos^2 \\alpha - \\sin^2 \\alpha + \\cos^2 x \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x \\cos \\alpha = \\cos^2 x ( \\cos^2 \\alpha + \\sin^2 \\alpha) - \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x \\cos \\alpha = \\cos^2 x - \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x \\cos \\alpha - \\cos^2 x = - \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x ( \\cos \\alpha -1) = - \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x ( 1 - \\cos \\alpha ) = \\sin^2 \\alpha$ $\\displaystyle \\Rightarrow \\cos^2 x = \\frac{\\sin^2 \\alpha}{ 1 - \\cos \\alpha }$ $\\displaystyle \\Rightarrow \\cos^2 x= \\frac{4 \\sin^2 \\frac{\\alpha}{2} \\cos^2 \\frac{\\alpha}{2} }{ 2 \\sin^2 \\frac{\\alpha}{2} }$ $\\displaystyle \\Rightarrow \\cos^2 x = 2 \\cos^2 \\frac{\\alpha}{2}$ $\\displaystyle \\Rightarrow \\cos x = \\pm \\sqrt{2} \\cos \\frac{\\alpha}{2}$ . Hence proved. $\\displaystyle \\\\$ $\\displaystyle \\text{Question 42: } \\cos \\alpha + \\cos \\beta = \\frac{1}{3} \\text{ and } \\sin \\alpha + \\sin \\beta = \\frac{1}{4} , \\text{ prove that } \\cos \\frac{\\alpha - \\beta}{2} =", "+ \\sin 38 + \\cos 47 \\\\ & = \\text{1,8519...} \\\\ & \\approx \\text{1,85} \\end{align*} If $$3 \\tan \\alpha = -5$$ and $$0° < \\alpha < 270°$$, use a sketch to determine: $$\\cos \\alpha$$ Find $$x$$, $$y$$ and $$r$$ \\begin{align*} 3 \\tan \\alpha & = -5 \\\\ \\tan \\alpha & = \\frac{-5}{3} \\end{align*} Therefore $$x = -3$$ and $$y = 5$$. \\begin{align*} r^{2} & = x^{2} + y^{2} \\\\ & = (-3)^{2} + (5)^{2} \\\\ & = 34 \\\\ r & = \\sqrt{34} \\end{align*} Draw a sketch: Now we can find $$\\cos \\alpha$$: \\begin{align*} \\cos \\alpha & = \\frac{x}{r} \\\\ & = \\frac{-3}{\\sqrt{34}} \\end{align*} $$\\tan^{2} \\alpha - \\sec^{2} \\alpha$$ We have $$x$$, $$y$$ and $$r$$ from the first question. \\begin{align*} \\tan^{2} \\alpha - \\sec^{2} \\alpha & = \\left(\\frac{y}{x}\\right)^{2} - \\left(\\frac{r}{x}\\right)^{2} \\\\ & = \\left(\\frac{5}{-3}\\right)^{2} - \\left(\\frac{\\sqrt{34}}{-3}\\right)^{2} \\\\ & = \\frac{25}{9} - \\frac{34}{9} \\\\ & = \\frac{-9}{9} \\\\ & = -1 \\end{align*} Given $$A(5;0)$$ and $$B(11;4)$$, find the angle between the line through $$A$$ and $$B$$ and the $$x$$-axis. First draw a diagram: Next we note that the distance from $$B$$ to the $$x$$-axis is $$\\text{4}$$ ($$B$$ is $$\\text{4}$$ units up from the $$x$$-axis) and that the distance from $$A$$ to $$C$$ is $$11 - 5 = 6 \\text{ units}$$. We use the tangent ratio to find the angle:", "\\cos \\frac{\\alpha + \\beta}{2} \\cos \\frac{\\alpha - \\beta}{2} = b$ … … … … … ii) Dividing i) by ii) we get $\\displaystyle \\tan \\frac{\\alpha + \\beta}{2} = \\frac{a}{b}$ $\\displaystyle \\text{We know } \\sin (\\alpha + \\beta) = \\frac{2 \\tan \\big( \\frac{\\alpha + \\beta}{2} \\big)}{1+ \\tan^2 \\big( \\frac{\\alpha + \\beta}{2} \\big)} = \\frac{2 \\frac{a}{b}}{1+\\frac{a^2}{b^2} } = \\frac{2ab}{a^2+b^2}$ $\\displaystyle ( \\sin \\alpha + \\sin \\beta)^2 + ( \\cos \\alpha + \\cos \\beta)^2 = a^2 +b^2$ $\\displaystyle \\Rightarrow \\sin^2 \\alpha + \\sin^2 \\beta + 2 \\sin \\alpha \\sin \\beta + \\cos^2 \\alpha + \\cos^2 \\beta + 2 \\cos \\alpha \\cos \\beta = a^2 + b^2$ $\\displaystyle \\Rightarrow 2 + 2 \\sin \\alpha \\sin \\beta + 2 \\cos \\alpha \\cos \\beta = a^2 + b^2$ $\\displaystyle \\Rightarrow 2 + 2 ( \\sin \\alpha \\sin \\beta + \\cos \\alpha \\cos \\beta) = a^2 + b^2$ $\\displaystyle \\Rightarrow 2 + 2 \\cos ( \\alpha - \\beta) = a^2 + b^2$ $\\displaystyle \\Rightarrow \\cos ( \\alpha - \\beta) = \\frac{a^2 + b^2-2}{2}$ $\\displaystyle \\\\$ $\\displaystyle \\text{Question 39: } \\text{If } 2\\tan \\frac{\\alpha}{2} = \\tan \\frac{\\beta}{2} , \\text{ prove that } \\cos \\alpha = \\frac{3 + 5 \\cos \\beta}{5 + 3 \\cos \\beta}$ $\\displaystyle \\text{Given } 2\\tan \\frac{\\alpha}{2} = \\tan \\frac{\\beta}{2} ,$ $\\displaystyle \\text{RHS } = \\frac{3 + 5 \\cos \\beta}{5 + 3 \\cos \\beta}$", "\\ \\ \\ \\text{Since both}\\sin \\alpha \\ \\text{ and}\\cos \\beta \\ \\text{are negative,}\\\\\\ \\ \\ \\ \\alpha \\ \\text{and}\\ \\beta \\text{ will be in }{{\\text{3}}^{{\\text{rd}}}}\\text{ quadrant}\\text{.}\\end{array}$ $\\displaystyle \\therefore \\ \\ \\ \\sin \\alpha =-\\frac{4}{5},\\ \\cos \\alpha =-\\frac{3}{5}$ $\\displaystyle \\begin{array}{l}\\therefore \\ \\ \\ \\sin \\beta =-\\displaystyle \\frac{5}{{13}},\\ \\\\\\\\\\ \\ \\ \\ \\ \\cos \\beta =-\\displaystyle \\frac{{12}}{{13}},\\\\\\\\\\ \\ \\ \\ \\ \\tan \\beta =\\displaystyle \\frac{5}{{12}}\\\\\\\\\\\\\\ \\ \\ \\ \\ \\sin 2\\alpha =2\\sin \\alpha \\cos \\alpha \\\\\\\\\\therefore \\ \\ \\ \\sin 2\\alpha =2\\left( {-\\displaystyle \\frac{4}{5}} \\right)\\left( {-\\displaystyle \\frac{3}{5}} \\right)\\\\\\\\\\therefore \\ \\ \\ \\sin 2\\alpha =\\displaystyle \\frac{{24}}{{25}}\\\\\\\\\\ \\ \\ \\ \\ \\cos 2\\alpha =2{{\\cos }^{2}}\\alpha -1\\\\\\\\\\therefore \\ \\ \\ \\cos 2\\alpha =2{{\\left( {-\\displaystyle \\frac{3}{5}} \\right)}^{2}}-1\\\\\\\\\\therefore \\ \\ \\ \\cos 2\\alpha =-\\displaystyle \\frac{7}{{25}}\\\\\\\\\\ \\ \\ \\ \\cos \\displaystyle \\frac{\\beta }{2}=\\pm \\sqrt{{\\displaystyle \\frac{{1+\\cos \\beta }}{2}}}\\\\\\\\\\therefore \\ \\ \\cos \\displaystyle \\frac{\\beta }{2}=\\pm \\sqrt{{\\displaystyle \\frac{{1-\\displaystyle \\frac{{12}}{{13}}}}{2}}}\\\\\\\\\\therefore \\ \\ \\cos \\displaystyle \\frac{\\beta }{2}=\\pm \\sqrt{{\\displaystyle \\frac{1}{{26}}}}=\\pm \\displaystyle \\frac{{\\sqrt{{26}}}}{{26}}\\\\\\\\\\ \\ \\ \\ \\text{Since}\\ 180{}^\\circ <\\beta <270{}^\\circ ,\\\\\\\\\\ \\ \\ \\ 90{}^\\circ <\\displaystyle \\frac{\\beta }{2}<135{}^\\circ ,\\\\\\\\\\therefore \\ \\ \\cos \\displaystyle \\frac{\\beta }{2}=-\\displaystyle \\frac{{\\sqrt{{26}}}}{{26}}\\\\\\\\\\ \\ \\ \\ \\tan 2\\beta =\\displaystyle \\frac{{2\\tan \\beta }}{{1-{{{\\tan }}^{2}}\\beta }}\\\\\\\\\\therefore \\ \\ \\tan 2\\beta =\\displaystyle \\frac{{2\\left( {\\displaystyle \\frac{5}{{12}}} \\right)}}{{1-{{{\\left( {\\displaystyle \\frac{5}{{12}}} \\right)}}^{2}}}}\\\\\\\\\\therefore \\ \\ \\tan 2\\beta =\\displaystyle \\frac{{\\displaystyle \\frac{5}{6}}}{{\\displaystyle \\frac{{119}}{{144}}}}=\\displaystyle \\frac{5}{6}\\times \\displaystyle \\frac{{144}}{{119}}=\\displaystyle \\frac{{120}}{{119}}\\end{array}$ 13. Given that $\\displaystyle \\sin \\alpha = \\frac{15}{17}, \\cos \\beta = \\frac{3}{5}$", "of $$n$$, \\begin{align}\\sum_{k=0}^{m-1}\\cos\\frac{2\\pi nkh}{b-a}&=\\sum_{k=0}^{m-1}\\cos\\frac{2\\pi nk}m=\\sum_{k=0}^{m-1}\\frac{\\sin\\frac{2\\pi n}m(k+1/2)-\\sin\\frac{2\\pi n}m(k-1/2)}{2\\sin\\frac{\\pi n}m}\\\\ &=\\frac{\\sin\\frac{2\\pi n}m(m-1/2)-\\sin\\frac{2\\pi n}m(-1/2)}{2\\sin\\frac{\\pi n}m}=0\\end{align} If $$m$$ is a divisor of $$n$$, then $$\\sum_{k=0}^{m-1}\\cos\\frac{2\\pi nkh}{b-a}=\\sum_{k=0}^{m-1}\\cos\\frac{2\\pi nk}m=m$$ Since $$\\cos\\alpha\\left(k+\\frac12\\right)-\\cos\\alpha\\left(k-\\frac12\\right)=-2\\sin\\alpha k\\sin\\alpha/2$$, if $$m$$ is not a divisor of $$n$$, \\begin{align}\\sum_{k=0}^{m-1}\\sin\\frac{2\\pi nkh}{b-a}&=\\sum_{k=0}^{m-1}\\sin\\frac{2\\pi nk}m=-\\sum_{k=0}^{m-1}\\frac{\\cos\\frac{2\\pi n}m(k+1/2)-\\cos\\frac{2\\pi n}m(k-1/2)}{2\\sin\\frac{\\pi n}m}\\\\ &=-\\frac{\\cos\\frac{2\\pi n}m(m-1/2)-\\cos\\frac{2\\pi n}m(-1/2)}{2\\sin\\frac{\\pi n}m}=0\\end{align} And even if $$m$$ is a divisor of $$n$$n $$\\sum_{k=0}^{m-1}\\sin\\frac{2\\pi nkh}{b-a}=\\sum_{k=0}^{m-1}\\sin\\frac{2\\pi nk}m=0$$ So the trapezoidal rule produces $$\\int_a^bf(x)dx\\approx(b-a)\\left(\\frac{a_0}2+\\sum_{n=1}^{\\infty}a_{mn}\\right)$$ Since the exact answer is $$\\int_a^bf(x)dx=(b-a)a_0/2$$ we see that the effect of the trapezoidal rule in this case is to approximate the function $$f(x)$$ by its $$2n-1$$ 'lowest energy' eigenfunctions and integrate the approximation. This is pretty much what Gaussian integration does so it's amusing to compare the two for periodic and nonperiodic functions. The program that produces my data looks like this: module Gmod use ISO_FORTRAN_ENV, only: wp=>REAL64 implicit none real(wp), parameter :: pi = 4*atan(1.0_wp) contains subroutine eval_legendre(n,x,p,q) integer, intent(in) :: n real(wp), intent(in) :: x real(wp), intent(out) :: p, q integer m real(wp) r if(n == 0) then p = 1 q = 0 else p = x q = 1 do m = 2, n-1, 2 q = ((2*m-1)*x*p-(m-1)*q)/m p = ((2*m+1)*x*q-m*p)/(m+1) end do if(m == n) then r = ((2*m-1)*x*p-(m-1)*q)/m q = p p = r end if end if end subroutine eval_legendre subroutine formula(n,x,w) integer, intent(in) :: n real(wp), intent(out)", "Question # If ${{\\cot }^{-1}}[{{\\left( \\cos \\alpha \\right)}^{\\dfrac{1}{2}}}]+{{\\tan }^{-1}}[{{\\left( \\cos \\alpha \\right)}^{\\dfrac{1}{2}}}]=x$ . Then find the value of $\\sin x$\\begin{align} & \\text{a) 1} \\\\ & \\text{b) co}{{\\text{t}}^{2}}\\left( \\dfrac{a}{2} \\right) \\\\ & \\text{c)}\\text{ tan}\\alpha \\\\ & \\text{d) cot}\\left( \\dfrac{\\alpha }{2} \\right) \\\\ \\end{align} Hint: We know that ${{\\cot }^{-1}}x+{{\\tan }^{-1}}x=\\dfrac{\\pi }{2}$ hence we get x = $\\dfrac{\\pi }{2}$. Now since we know the value of x. we can easily find the value of sinx. Now we are given that ${{\\cot }^{-1}}[{{\\left( \\cos \\alpha \\right)}^{\\dfrac{1}{2}}}]+{{\\tan }^{-1}}[{{\\left( \\cos \\alpha \\right)}^{\\dfrac{1}{2}}}]=x$ . Let us say ${{\\left( \\cos \\alpha \\right)}^{\\dfrac{1}{2}}}$ is equal to t. Now we will note that the range of ${{\\cos }^{-1}}\\theta$ is $[0,\\pi ]$ and hence the output of $\\cos \\alpha$ is also real Hence we have $t={{(\\cos \\alpha )}^{\\dfrac{1}{2}}}$ is a real number. Substituting this in the above equation we get ${{\\cot }^{-1}}t+{{\\tan }^{-1}}t=x.................(1)$ Now we know that for all real numbers x the identity ${{\\cot }^{-1}}x+{{\\tan }^{-1}}x=\\dfrac{\\pi }{2}$ is true. Now since this identity is true for all real numbers x and does not depend on x this identity is also true for t which is a real number. Hence applying this to equation (1) we get $x=\\dfrac{\\pi }{2}$ Hence now we have the value of x is $\\dfrac{\\pi }{2}$. Now we have to find the value of", "# help (1 Viewer) #### jimmysmith560 ##### Le Phénix Trilingue Moderator Would the following working help? Part (i): If the two curves have a common tangent, then their derivatives must be equal at that point. $\\bg_white \\frac{d}{dx}\\sin x=\\cos x$ $\\bg_white \\frac{d}{dx}\\left(\\sin \\left(x-\\alpha \\right)+k\\right)=\\cos \\left(x-\\alpha \\right)$ $\\bg_white \\therefore \\cos x_0=\\cos \\left(x_0-\\alpha \\right)$ Part (ii): $\\bg_white \\cos x_0=\\cos \\left(x_{0-\\alpha }\\right)$ So $\\bg_white \\alpha=2x_0$ Then $\\bg_white \\sin \\left(x_{0-\\alpha }\\right)$ $\\bg_white =\\sin \\left(x_0-2x_0\\right)$ $\\bg_white =\\sin \\left(-x_0\\right)$ $\\bg_white =-\\sin \\left(x_0\\right)$ $\\bg_white \\therefore \\sin \\left(x_0\\right)=-\\sin \\left(x_0-\\alpha \\right)$ Part (iii): $\\bg_white \\sin x_0=\\sin \\left(x_0-\\alpha \\right)+k$ $\\bg_white -\\sin \\left(x_0-\\alpha \\right)=\\sin \\left(x_0-\\alpha \\right)+k$ $\\bg_white -2\\sin \\left(x_0-\\alpha \\right)=k$ $\\bg_white k=-2\\sin \\left(\\frac{\\alpha }{2}-\\alpha \\right)$ $\\bg_white k=-2\\sin \\left(-\\frac{\\alpha }{2}\\right)$ $\\bg_white k=2\\sin \\left(\\frac{\\alpha }{2}\\right)$ #### Farhanthestudent005 ##### Member Thank you very much for helping me, this helps a lot", "Multiplying both sides by $\\tan\\frac{C}{2}$ we have $$\\tan\\Bigl(\\frac{C}{2}\\Bigr) \\cdot \\Bigl[ \\tan\\Bigl(\\frac{A}{2}\\Bigr) + \\tan\\Bigl(\\frac{B}{2}\\Bigr) \\Bigr] = 1 \\cdot \\Bigl[ 1 - \\tan\\Bigl(\\frac{A}{2}\\Bigr) \\cdot \\tan\\Bigl(\\frac{B}{2}\\Bigr)\\Bigr]$$ - In the second one you probably meant $\\frac{A}{2} + \\frac{B}{2} = ( \\frac{\\pi}{2} - \\frac{C}{2} ) \\Longrightarrow \\tan\\Bigl(\\frac{A+B}{2}\\Bigr)= \\cot\\Bigl(\\frac{C}{2}\\Bigr)$ – Quixotic Nov 3 '10 at 10:46 @Debanjan: Edited! before you wrote that comment – anonymous Nov 3 '10 at 10:47 Both showing 1 mint ago so I guess simultaneous posting,however I will +1 for a clear approach. – Quixotic Nov 3 '10 at 10:48 More generally, we have from De Moivre's theorem $$\\cos(\\alpha_1+\\alpha_2+\\cdots+\\alpha_n)= \\text{Re}\\prod_{k=1}^n(\\cos \\alpha_k + i \\sin \\alpha_k)$$ and $$\\sin(\\alpha_1+\\alpha_2+\\cdots+\\alpha_n)= \\text{Im}\\prod_{k=1}^n(\\cos \\alpha_k + i \\sin \\alpha_k)$$ and so $$\\tan(\\alpha_1+\\alpha_2+\\cdots+\\alpha_n)= \\frac{ \\text{Im}\\prod_{k=1}^n ( 1 + i \\tan \\alpha_k) }{ \\text{Re}\\prod_{k=1}^n(1 + i \\tan\\alpha_k) }.$$ Consider the case $n=3.$ - Hints: Prelude. a. If $\\alpha+\\beta+\\gamma=\\pi$, then $\\gamma=\\pi-\\alpha-\\beta$. b. $\\tan(\\pi-\\theta)=-\\tan(\\theta)$ c. $\\tan(\\alpha+\\beta)=\\frac{\\tan(\\alpha)+\\tan(\\beta)}{1-\\tan(\\alpha)\\tan(\\beta)}$ d. $\\tan\\left(\\frac{\\pi}{2}-\\theta\\right)=\\frac1{\\tan(\\theta)}$ Act I. $$\\tan(\\alpha)+\\tan(\\beta)-\\frac{\\tan(\\alpha)+\\tan(\\beta)}{1-\\tan(\\alpha)\\tan(\\beta)}=-\\tan(\\alpha)\\tan(\\beta)\\frac{\\tan(\\alpha)+\\tan(\\beta)}{1-\\tan(\\alpha)\\tan(\\beta)}$$ Can the left-hand side be made to look like the right-hand side? Act II. $$\\tan\\left(\\frac{\\alpha}{2}\\right)\\tan\\left(\\frac{\\beta}{2}\\right)+\\left(\\tan\\left(\\frac{\\alpha}{2}\\right)+\\tan\\left(\\frac{\\beta}{2}\\right)\\right)\\left(\\frac{\\tan\\left(\\frac{\\alpha}{2}\\right)+\\tan\\left(\\frac{\\beta}{2}\\right)}{1-\\tan\\left(\\frac{\\alpha}{2}\\right)\\tan\\left(\\frac{\\beta}{2}\\right)}\\right)^{-1}$$ Simplify the above expression. -", "# Integration of $\\frac{1}{1-\\cos(\\alpha)\\cos x }$ Integration of $$\\dfrac{1}{1-\\cos(\\alpha)\\cos x }$$ w.r.t $$x$$ How to do this problem? I was trying to reduce it $$\\dfrac{1}{1-\\cos(\\alpha)\\cos (x) }=\\dfrac{\\sec (\\alpha)}{\\sec (\\alpha)-\\cos (x) }$$. But I can't put it in shape. $$\\dfrac{1}{1-\\cos(\\alpha)\\cos (x) }=\\dfrac{1}{1-\\cos (\\alpha)\\frac{1-\\tan^2 (x/2)}{1+tan^2(x/2)}}=\\dfrac{\\sec^2(x/2)}{1+\\tan^2(x/2)-\\cos(\\alpha)+\\cos(\\alpha)\\tan^2(x/2) }$$. Now taking $$\\tan(x/2)=z$$ we are done. Hint: Substitute $$u=\\tan\\left(\\frac{u}{2}\\right)$$ then you will have $$du=\\frac{1}{2}\\sec^2\\left(\\frac{x}{2}\\right)dx$$ For your work: A possible result should be $$2 \\csc (\\alpha ) \\tan ^{-1}\\left(\\sin \\left(\\frac{x}{2}-\\frac{\\alpha }{2}\\right) \\csc \\left(\\frac{\\alpha }{2}+\\frac{x}{2}\\right)\\right)+C$$", "again, Thaurus! Quote: 1690. Evaluate: . $\\frac{\\sin\\alpha + \\cos\\alpha}{\\cos\\alpha - \\sin\\alpha}\\,\\:\\text{ where }\\:\\tan\\alpha\\,=\\,-2$ We have: . $\\frac{\\sin\\alpha + \\cos\\alpha}{\\cos\\alpha - \\sin\\alpha}$ Divide top and bottom by $\\cos\\alpha\\!:\\quad \\frac{\\dfrac{\\sin\\alpha}{\\cos\\alpha} + \\dfrac{\\cos\\alpha}{\\cos\\alpha}}{\\dfrac{\\cos\\alpha} {\\cos\\alpha} - \\dfrac{\\sin\\alpha}{\\cos\\alpha}} \\;=\\; \\frac{\\tan\\alpha + 1}{1 - \\tan\\alpha}$ Substitute $\\tan\\alpha = \\text{-}2\\!:\\;\\;\\frac{(\\text{-}2)+1}{1-(\\text{-}2)} \\;=\\;-\\frac{1}{3}$ Quote: $2.\\;\\;\\frac{\\tan135^o}{\\tan225^o}\\cdot\\sin300^o - \\cos^2(\\text{-}300^o)$ We have: . $\\frac{-1}{+1}\\left(-\\frac{\\sqrt{3}}{2}\\right) - \\left(\\frac{1}{2}\\right)^2\\;\\;=\\;\\;\\frac{\\sqrt{3}} {2} - \\frac{1}{4}$ • Mar 30th 2010, 07:08 AM Thaurus Thanks! I did it myself , am I am right? • Mar 30th 2010, 07:32 AM $1 + \\cot^2(x) = 1+ \\frac{1}{\\tan^2(x)} = 1 + \\frac{\\cos^2(x)}{\\sin^2(x)} = \\frac{1}{\\sin^2(x)}$", "understand now. Answer would be $\\frac{1-r^2}{1+r^2}$. – George Apriashvili Jul 21 '14 at 6:29 • @GeorgeDirac - yes it would :) – nbubis Jul 21 '14 at 6:34 $$\\cos^2⁡x-\\sin^2x=\\frac{\\cos^2x-\\sin^2x}{\\cos^2x+\\sin^2x}=\\frac{1-\\tan^2x}{1+\\tan^2x}.$$ • Using Mathjax on this site is not only preferred, but demanded. – 5xum Jul 21 '14 at 6:18 • Very elegant and economical! – Yves Daoust Jul 21 '14 at 9:17 We have $$\\cos^2\\alpha-\\sin^2\\alpha=2\\cos^2\\alpha-1$$ and by $$\\cos^2\\alpha=\\frac1{1+\\tan^2\\alpha}$$ we find $$\\cos^2\\alpha-\\sin^2\\alpha=\\frac2{1+r^2}-1=\\frac{1-r^2}{1+r^2}$$ We have: $$\\sin \\alpha=\\pm\\frac{\\tan \\alpha}{\\sqrt{1 + \\tan^2 \\alpha}}\\!$$ and $$\\cos \\alpha=\\pm\\frac{1}{\\sqrt{1 + \\tan^2 \\alpha}}\\!$$ then $$\\sin^2 \\alpha=\\frac{\\tan^2 \\alpha}{1 + \\tan^2 \\alpha}=\\frac{r^2}{1 + r^2}\\!$$ and $$\\cos^2 \\alpha=\\frac{1}{1 + \\tan^2 \\alpha}\\!=\\frac{1}{1 + r^2}$$ You start from $r=\\frac{\\sin\\alpha}{\\cos\\alpha}$ and must evaluate $\\cos^2\\alpha-{\\sin^2\\alpha}$. This should push you to set $c=\\cos^2\\alpha, s=\\sin^2\\alpha$, use $$c+s=1$$ and rewrite the first relation as $$r^2c-s=0.$$ Solve this linear system for $c$ and $s$: $$(1+r^2)\\ c=1\\\\(1+r^2)\\ s=r^2.$$ Then$$c-s=\\frac{1-r^2}{1+r^2}.$$ ($\\sin\\theta$/$\\cos\\theta$)=r or,($\\cos\\theta$/$\\sin\\theta$)=1/r or,($\\cos^2\\theta$ )/($\\sin^2\\theta$)=1/$r^2$[squaringbothsides] or,($\\cos^2\\theta$+$\\sin^2\\theta$)/($\\cos^2\\theta$-$\\sin^2\\theta$)=(1 + $r^2$)/(1 - $r^2$)[ApplyingComponendoandDividendo] or,1/($\\cos^2\\theta$- $\\sin^2\\theta$)=(1 + $r^2$)/(1 - $r^2$ ) or,($\\cos^2\\theta$- $\\sin^2\\theta$)=(1 - $r^2$)/(1 + $r^2$) • Please use Mathjax, otherwise it's very hard to see the solution. – George Apriashvili Jul 21 '14 at 6:38 • For starters Here is the tutorial! – MonK Jul 21 '14 at 9:30 My Approach: $\\tan \\alpha=r$ $\\implies \\tan^2 \\alpha=r^2$ $\\implies \\large \\frac{\\sin^2 \\alpha}{\\cos^2 \\alpha}=r^2$ $\\implies \\large \\frac{\\sin^2 \\alpha+\\cos^2 \\alpha}{\\sin^2 \\alpha-\\cos^2 \\alpha}=\\large \\frac{r^2+1}{r^2-1}$ $\\implies \\large \\frac{1}{\\sin^2 \\alpha", "\\left. {\\dfrac{2}{{\\sin \\alpha }}{{\\tan }^{ - 1}}\\left( {\\dfrac{{\\theta + \\cos \\alpha }}{{\\sin \\alpha }}} \\right)} \\right|_0^\\infty$ Evaluating the limits, we have: $I' = \\dfrac{2}{{\\sin \\alpha }}\\left( {{{\\tan }^{ - 1}}\\left( \\infty \\right) - {{\\tan }^{ - 1}}\\left( {\\dfrac{{\\cos \\alpha }}{{\\sin \\alpha }}} \\right)} \\right)$ We know that, $\\dfrac{{\\cos \\alpha }}{{\\sin \\alpha }} = \\cot \\alpha$, hence we get: $I' = \\dfrac{2}{{\\sin \\alpha }}\\left( {{{\\tan }^{ - 1}}\\left( \\infty \\right) - {{\\tan }^{ - 1}}\\left( {\\cot \\alpha } \\right)} \\right)$ We also know that, ${\\tan ^{ - 1}}\\left( \\infty \\right) = \\dfrac{\\pi }{2}$ and $\\cot \\alpha = \\tan \\left( {\\dfrac{\\pi }{2} - \\alpha } \\right)$, hence we have: $I' = \\dfrac{2}{{\\sin \\alpha }}\\left( {\\dfrac{\\pi }{2} - {{\\tan }^{ - 1}}\\left( {\\tan \\left( {\\dfrac{\\pi }{2} - \\alpha } \\right)} \\right)} \\right)$ Since, ${\\tan ^{ - 1}}\\left( {\\tan \\left( x \\right)} \\right) = x$, we have: $I' = \\dfrac{2}{{\\sin \\alpha }}\\left( {\\dfrac{\\pi }{2} - \\left( {\\dfrac{\\pi }{2} - \\alpha } \\right)} \\right)$ Simplifying, we get: $I' = \\dfrac{{2\\alpha }}{{\\sin \\alpha }}.........(6)$ Using equation (6) in equation (3), we get: $I = \\dfrac{\\pi }{2} \\times \\dfrac{{2\\alpha }}{{\\sin \\alpha }}$ $I = \\dfrac{{\\pi \\alpha }}{{\\sin \\alpha }}$ Hence, the value of the integral is $I = \\dfrac{{\\pi \\alpha }}{{\\sin \\alpha }}$. Note: You might make mistake in the integration of $\\int {\\dfrac{{dx}}{{{{(x)}^2} +", "\\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta) = - 1$ $\\displaystyle \\Rightarrow \\cos ( \\alpha + \\beta) = 1$ $\\displaystyle \\therefore \\sin ( \\alpha + \\beta ) = \\sqrt{ 1 - \\cos^2 ( \\alpha + \\beta ) } = \\sqrt{ 1 - 1 } = 0$ $\\displaystyle 1 + \\cot \\alpha \\tan \\beta = 1 + \\frac{\\cos \\alpha}{\\sin \\alpha} \\times \\frac{\\sin \\beta}{\\cos \\beta}$ $\\displaystyle \\Rightarrow 1 + \\cot \\alpha \\tan \\beta = \\frac{\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta}{\\sin \\alpha \\cos \\beta }$ $\\displaystyle \\Rightarrow 1 + \\cot \\alpha \\tan \\beta = \\frac{\\sin ( \\alpha + \\beta )}{\\sin \\alpha \\cos \\beta}$ $\\displaystyle \\Rightarrow 1 + \\cot \\alpha \\tan \\beta = 0$ Hence Proved. $\\displaystyle \\\\$ $\\displaystyle \\text{Question 31: If } \\tan \\alpha = x + 1 \\text{ and } \\tan \\beta = x - 1$, show that $\\displaystyle 2 \\cot (\\alpha - \\beta) = x^2$ $\\displaystyle \\text{Given } \\tan \\alpha = x + 1 \\text{ and } \\tan \\beta = x - 1$ $\\displaystyle 2 \\cot (\\alpha - \\beta) = \\frac{2}{\\tan (\\alpha - \\beta)}$ $\\displaystyle \\Rightarrow 2 \\cot (\\alpha - \\beta) = \\frac{2 ( 1 + \\tan \\alpha \\tan \\beta}{\\tan \\alpha - \\tan \\beta}$ $\\displaystyle \\Rightarrow 2 \\cot (\\alpha - \\beta) = 2 \\Big[ \\frac{1 + ( x + 1 ) ( x - 1", "+\\alpha )=\\tan \\alpha ,k\\in \\mathbb {Z} }$ ${\\displaystyle \\tan((2k+1)\\pi +\\alpha )=\\tan \\alpha ,k\\in \\mathbb {Z} }$ ${\\displaystyle \\tan(-\\alpha )=-\\tan \\alpha }$ ${\\displaystyle \\tan(\\pi -\\alpha )=-\\tan \\alpha }$ ${\\displaystyle \\tan \\left({\\frac {\\pi }{2}}+\\alpha \\right)=-\\cot \\alpha }$ ${\\displaystyle \\cot(2k\\pi +\\alpha )=\\cot \\alpha ,k\\in \\mathbb {Z} }$ ${\\displaystyle \\cot((2k+1)\\pi +\\alpha )=\\cot \\alpha ,k\\in \\mathbb {Z} }$ ${\\displaystyle \\cot(-\\alpha )=-\\cot \\alpha }$ ${\\displaystyle \\cot(\\pi -\\alpha )=-\\cot \\alpha }$ ${\\displaystyle \\cot \\left({\\frac {\\pi }{2}}+\\alpha \\right)=-\\tan \\alpha }$ ${\\displaystyle \\csc(2k\\pi +\\alpha )=\\csc \\alpha ,k\\in \\mathbb {Z} }$ ${\\displaystyle \\csc((2k+1)\\pi +\\alpha )=-\\csc \\alpha ,k\\in \\mathbb {Z} }$ ${\\displaystyle \\csc(-\\alpha )=-\\csc \\alpha }$ ${\\displaystyle \\csc(\\pi -\\alpha )=\\csc \\alpha }$ ${\\displaystyle \\csc \\left({\\frac {\\pi }{2}}+\\alpha \\right)=\\sec \\alpha }$ ${\\displaystyle \\sec(2k\\pi +\\alpha )=\\sec \\alpha ,k\\in \\mathbb {Z} }$ ${\\displaystyle \\sec((2k+1)\\pi +\\alpha )=-\\sec \\alpha ,k\\in \\mathbb {Z} }$ ${\\displaystyle \\sec(-\\alpha )=\\sec \\alpha }$ ${\\displaystyle \\sec(\\pi -\\alpha )=-\\sec \\alpha }$ ${\\displaystyle \\sec \\left({\\frac {\\pi }{2}}+\\alpha \\right)=-\\csc \\alpha }$ There are many more examples and applications by using this identity. The table above only lists selected groups of examples. It would be great if you can plug in specific numbers into the formulae and do some experiments. Double-angle formulae It is highly recommended that readers should memorize and be familiar with the processes because of the convenience they will provide. The double-angle formulae If an angle ${\\displaystyle \\alpha }$ exists, and", "Glen Wheeler Mar 4 '11 at 20:37 Let $\\alpha_n:=2\\pi n x$. Define for a fixed $x\\in \\left(0,\\frac 12\\right)$ $$n_k:=\\left\\lfloor\\frac{k+1/2}x\\right\\rfloor$$ which represents the greatest integer $\\leq \\frac{k+1/2}x$. We have \\begin{aligned} \\frac{k+1/2}x\\leq n_k< \\frac{k+1/2}x+1\\\\ k+1/2\\leq n_kx<k+1/2+x\\\\ 2\\pi k+\\pi\\leq \\alpha_{n_k}<2k\\pi+2\\pi \\end{aligned} hence $\\sin(\\alpha_{n_k})\\leq 0$ and $$\\tag{1}\\liminf_n\\sin(\\alpha_n)\\leq 0.$$ Now, define $m_k:=\\lfloor\\frac kx+\\frac 18\\rfloor$. We have \\begin{aligned} \\frac kx+\\frac 18&\\leq m_k\\leq \\frac kx+\\frac 58\\\\ 2k\\pi+\\frac{x\\pi}4&\\leq 2m_k\\pi\\leq 2k\\pi+\\frac{5x\\pi}4\\leq2k\\pi+\\frac{5\\pi}8. \\end{aligned} We deduce that $\\sin\\left(\\frac{\\pi x}4\\right)\\leq \\sin(\\alpha_{m_k})\\leq \\sin\\left(\\frac{5\\pi}8\\right).$ We deduce that $$\\tag{2}\\limsup_n\\sin(\\alpha_n)\\geq \\sin\\left(\\frac{\\pi x}4\\right)>0.$$ With (1) and (2), we can see that $\\{\\sin(2\\pi nx)\\}$ has no limit for $0<x<1/2$. - I think you're still missing an $x$ on the RHS? – TMM Jul 8 '12 at 21:57 I bounded it above by $\\frac 12$. – Davide Giraudo Jul 8 '12 at 21:59 If $x$ is irrational, there is a stronger result: $\\sin(2\\pi x \\mathbb{Z})$ is dense in $[0,1]$. Indeed, $\\sin(2\\pi x \\mathbb{Z})= \\sin (2\\pi x \\mathbb{Z} + 2\\pi \\mathbb{Z})$ and $2\\pi x \\mathbb{Z}+ 2 \\pi \\mathbb{Z}$ is a subgroup of $(\\mathbb{R},+)$ not of the forme $\\alpha \\mathbb{Z}$, so it is dense in $\\mathbb{R}$. Hence $\\sin(2\\pi x \\mathbb{Z})$ dense in $[0,1]$ since $\\sin$ is continuous. - $\\def\\eps{\\varepsilon}$Since the statement to be proved is negative (the sequence cannot converge) a fairly elementary proof avoiding any heavy machinery is possible. In spite of the use of $\\eps,\\delta$ below, the", "2 \\alpha)$ $\\frac{m}{n}$ $=$ $\\frac{\\sin (\\theta + 2 \\alpha)}{\\sin \\theta}$ Applying componendo and dividendo $\\frac{m+n}{m-n}$ $=$ $\\frac{\\sin (\\theta + 2 \\alpha) +\\sin \\theta }{\\sin (\\theta + 2 \\alpha) -\\sin \\theta}$ $\\frac{m+n}{m-n}$ $=$ $\\frac{2 \\sin \\big( \\frac{\\theta + 2 \\alpha + \\theta}{2} \\big) \\cos \\big( \\frac{\\theta + 2 \\alpha - \\theta}{2} \\big) }{2 \\cos \\big( \\frac{\\theta + 2 \\alpha + \\theta}{2} \\big) \\sin \\big( \\frac{\\theta + 2 \\alpha - \\theta}{2} \\big)}$ $\\frac{m+n}{m-n}$ $=$ $\\frac{2 \\sin ( \\theta + \\alpha) \\cos \\alpha}{2 \\cos ( \\theta + \\alpha) \\sin \\alpha}$ $\\frac{m+n}{m-n}$ $=$ $\\frac{\\tan ( \\theta + \\alpha) }{\\tan \\alpha}$ $\\tan ( \\theta + \\alpha) =$ $\\frac{m+n}{m-n}$ $\\tan \\alpha$ Hence proved." ]

[ "# Tagged: nonzero solution ## Problem 655 Consider the matrix $M = \\begin{bmatrix} 1 & 4 \\\\ 3 & 12 \\end{bmatrix}$. (a) Show that $M$ is singular. (b) Find a non-zero vector $\\mathbf{v}$ such that $M \\mathbf{v} = \\mathbf{0}$, where $\\mathbf{0}$ is the $2$-dimensional zero vector. ## Problem 576 Let $V$ be a subspace of $\\R^n$. Suppose that $S=\\{\\mathbf{v}_1, \\mathbf{v}_2, \\dots, \\mathbf{v}_m\\}$ is a spanning set for $V$. Prove that any set of $m+1$ or more vectors in $V$ is linearly dependent.", "to do it. How do you find v such that (λI-A)v=0? 5. Oct 18, 2012 ### Dick Just try it! What is the matrix (λI-A) when λ=-1+i? Write that matrix times a vector (x,y), set it equal to zero and try to find a solution for x and y.", "+0 # three dimensional vectors 0 107 1 Let $$u$$ and $$v$$ be three-dimensional vectors such that the length of $$u$$ is $$2$$, the length of $$v$$ is $$4$$, and the angle between $$u$$ and $$v$$ when placed tail to tail is $$120^{\\circ}$$. Let $$\\mathbf{A}$$ be a $$3\\times 3$$ matrix such that $$\\mathbf{row}_1(\\mathbf{A}) = \\mathbf{u}, \\mathbf{row}_2(\\mathbf{A}) = \\mathbf{v}, \\mathbf{row}_3(\\mathbf{A}) =3\\mathbf{u} + 2\\mathbf{v}.$$ Then what are $$\\mathbf{Au}$$ and $$\\mathbf{Av}$$ in that order? (Your answers should be numerical.) Oct 24, 2021 $\\mathbf{A} \\mathbf{u} = \\begin{pmatrix} 4 \\\\ -4 \\\\ 8 \\end{pmatrix}$ $\\mathbf{A} \\mathbf{v} = \\begin{pmatrix} -8 \\\\ 12 \\\\ -16 \\end{pmatrix}$", "I isolated unknowns in two separate matrices (vectors). However, it is still not trivial. I tried to solve this, but there is no obvious way. One way is that I tried to find $\\mathbf{x}, \\mathbf{y}, \\mathbf{z}$ such that $\\mathbf{D} \\mathbf{R}$ is rank-deficient. I'm not sure if this can be solved, either exactly or in least-squared, in closed-form or numerically. Any idea, discussion is appreciated.", "+0 Let $\\mathbf{A}$ be a matrix, and let $\\mathbf{x}$ and $\\mathbf{y}$ be linearly independent vectors such that \\[\\mathbf{A} \\mathbf{x} = \\mat +1 273 1 Let A be a matrix, and let x and y be linearly independent vectors such that $$\\mathbf{A} \\mathbf{x} = \\mathbf{y}, \\mathbf{A} \\mathbf{y} = \\mathbf{x} + 2\\mathbf{y}$$ Then we have that $$\\mathbf{A}^{-1} \\mathbf{x} = a \\mathbf{x} + b\\mathbf{y}$$ for some scalars a and b. Find the ordered pair (a,b). Mar 1, 2019 #1 +23071 +2 Let $$\\mathbf{A}$$ be a matrix, and let $$\\mathbf{x}$$ and $$\\mathbf{y}$$ be linearly independent vectors such that $$\\mathbf{A} \\mathbf{x} = \\mathbf{y}, \\mathbf{A} \\mathbf{y} = \\mathbf{x} + 2\\mathbf{y}$$ Then we have that $$\\mathbf{A}^{-1} \\mathbf{x} = a \\mathbf{x} + b\\mathbf{y}$$ or some scalars a and b. Find the ordered pair (a,b). $$\\begin{array}{|rcll|} \\hline Ax &=& y \\quad | \\quad \\cdot A^{-1} \\\\ A^{-1}Ax &=& A^{-1}y \\quad | \\quad A^{-1}A = I \\\\ x &=& A^{-1}y \\\\ \\mathbf{A^{-1}y} &\\mathbf{=}&\\mathbf{x} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline Ay &=& x + 2y \\quad | \\quad \\cdot A^{-1} \\\\ A^{-1}Ay &=& A^{-1}x + 2A^{-1}y \\quad | \\quad A^{-1}A = I \\\\ y &=& A^{-1}x + 2A^{-1}y \\quad | \\quad \\mathbf{A^{-1}y=x} \\\\ y &=& A^{-1}x + 2x \\\\ \\mathbf{A^{-1}x} &\\mathbf{=}&\\mathbf{-2x+y } \\\\ \\hline \\end{array}$$ $$\\mathbf{(a,b) = (-2,1)}$$ Mar 1, 2019", "+0 # Long solution matrix part 2 0 237 1 Let A be a $$2 \\times 2$$ matrix. Suppose that for every two-dimensional vector v, there exists a two-dimensional vector w such that $$\\mathbf{A} \\mathbf{w} = \\mathbf{v}$$ Show that we can find a matrix B such that $$\\mathbf{A} \\mathbf{B} = \\mathbf{I}$$. Mar 13, 2019 $$\\exists w_1 \\ni A w_1 = \\begin{pmatrix}1\\\\0\\end{pmatrix}\\\\ \\exists w_2 \\ni A w_2 = \\begin{pmatrix}0\\\\1\\end{pmatrix}\\\\ B = \\begin{pmatrix}w_1 &w_2\\end{pmatrix}$$", "\\\\ x_2 \\\\ \\vdots \\\\ x_n\\end{matrix}\\right] = \\left[\\begin{matrix}b_1 \\\\ b_2 \\\\ \\vdots \\\\ b_n \\end{matrix}\\right]$ Or in even shorterhand... $A\\mathbf{x} = \\mathbf{b}$... And then by using some matrix algebra $A^{-1}A\\mathbf{x} = A^{-1}\\mathbf{b}$ $I\\mathbf{x} = A^{-1}\\mathbf{b}$ $\\mathbf{x} = A^{-1}\\mathbf{b}$. If $A^{-1}$ can be calculated, that means that you can easily find $\\mathbf{x}$.", "then $AB$ is a skew-symmetric matrix. We calculate \\begin{align*} (AB)^{\\trans}&=B^{\\trans}A^{\\trans}=(-B)(-A)\\\\ &=BA=-AB, \\end{align*} where the last step follows from the assumption $AB=-BA$. This proves that $AB$ is skew-symmetric. ### (f) Let $\\mathbf{v}$ be an $n$-dimensional column vecotor. Prove that $\\mathbf{v}^{\\trans}A\\mathbf{v}=0$. Observe that $\\mathbf{v}^{\\trans}A\\mathbf{v}$ is a $1\\times 1$ matrix, or just a number. So we have \\begin{align*} \\mathbf{v}^{\\trans}A\\mathbf{v}&=(\\mathbf{v}^{\\trans}A\\mathbf{v})^{\\trans}=\\mathbf{v}^{\\trans}A^{\\trans}(\\mathbf{v}^{\\trans})^{\\trans}\\\\ &=\\mathbf{v}^{\\trans}A^{\\trans}\\mathbf{v}=\\mathbf{v}^{\\trans}(-A)\\mathbf{v}=-(\\mathbf{v}^{\\trans}A\\mathbf{v}). \\end{align*} This yields that $2\\mathbf{v}^{\\trans}A\\mathbf{v}=0$, and hence $\\mathbf{v}^{\\trans}A\\mathbf{v}=0$. ### (g) Suppose that $A$ is a real skew-symmetric matrix and $A^2\\mathbf{v}=\\mathbf{0}$ for some vector $\\mathbf{v}\\in \\R^n$. Then prove that $A\\mathbf{v}=\\mathbf{0}$. Let us compute the length of the vector $A\\mathbf{v}$. We have \\begin{align*} \\|A\\mathbf{v}\\|&=(A\\mathbf{v})^{\\trans}(A\\mathbf{v})=\\mathbf{v}^{\\trans}A^{\\trans}A\\mathbf{v}\\\\ &=\\mathbf{v}^{\\trans}(-A)A\\mathbf{v}=-\\mathbf{v}^{\\trans}A^2\\mathbf{v}\\\\ &=-\\mathbf{v}\\mathbf{0} &&\\text{by assumption}\\\\ &=0. \\end{align*} Since the length $\\|A\\mathbf{v}\\|=0$, we conclude that $A\\mathbf{v}=\\mathbf{0}$. ### More from my site • Express a Hermitian Matrix as a Sum of Real Symmetric Matrix and a Real Skew-Symmetric Matrix Recall that a complex matrix is called Hermitian if $A^*=A$, where $A^*=\\bar{A}^{\\trans}$. Prove that every Hermitian matrix $A$ can be written as the sum $A=B+iC,$ where $B$ is a real symmetric matrix and $C$ is a real skew-symmetric matrix. Proof. Since […] • Eigenvalues of a Hermitian Matrix are Real Numbers Show that eigenvalues of a Hermitian matrix $A$ are real numbers. (The Ohio State University Linear Algebra Exam Problem) We give two proofs. These two proofs are essentially the same. The second", "and explains how you know based on the computation. 1. Given vectors: $\\mathbf{v_1} = [1 \\ 2]^T$, $\\mathbf{v_2} = [-6 \\ -12]^T$; target vector $\\mathbf{b} = [5 \\ 5]^T$ 2. Given vectors: $\\mathbf{v_1} = [1 \\ 2 \\ 5]^T$, $\\mathbf{v_2} = [-3 \\ {-6} \\ 4 ]^T$, $\\mathbf{v_3} = [0 \\ 2 \\ {-1}]^T$; target vector $\\mathbf{b} = [5 \\ 10 \\ 15]^T$ 3. Given vectors: $\\mathbf{v_1} = [1 \\ 2 \\ 5 \\ -1]^T$, $\\mathbf{v_2} = [{-3} \\ {-6} \\ 4 \\ 2]^T$, $\\mathbf{v_3} = [1 \\ 0 \\ {-1} \\ 0]^T$; target vector $\\mathbf{b} = [{-1} \\ {-4} \\ 8 \\ 1]^T$ *Success criteria:* Matrix methods have been used appropriately to answer the questions. The matrix being used and the results of row-reduction are clearly stated. The answer (yes/no) to each question is clearly stated and is consistent with the results of the matrix operations. Each item has not only an answer but also a clear one-sentence explanation in English that explains the answer correctly in terms of the matrix and row-reduction.", "$\\R^3$ and the range is spanned by a nonzero vector $\\mathbf{v}$ in $\\R^5$. Determine $m$ and $n$. Also, find the rank and nullity of $A$. Solution. For an $m \\times n$ matrix $A$, the […]", "# Tagged: magnitude ## Problem 419 (a) Let $A$ be a real orthogonal $n\\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$. (b) Let $A$ be a real orthogonal $3\\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue. ## Problem 162 Let $\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3$ are vectors in $\\R^n$. Suppose that vectors $\\mathbf{u}_1$, $\\mathbf{u}_2$ are orthogonal and the norm of $\\mathbf{u}_2$ is $4$ and $\\mathbf{u}_2^{\\trans}\\mathbf{u}_3=7$. Find the value of the real number $a$ in $\\mathbf{u_1}=\\mathbf{u_2}+a\\mathbf{u}_3$. (The Ohio State University, Linear Algebra Exam Problem)", "of vectors $\\mathbf{v}_1, \\dots, \\mathbf{v}_k$ such that the sum of the coefficients is zero. Prove that $W$ is a subspace of $V$. ## Problem 580 (a) Prove that the column vectors of every $3\\times 5$ matrix $A$ are linearly dependent. (b) Prove that the row vectors of every $5\\times 3$ matrix $B$ are linearly dependent.", "# Finding a desired matrix Find a matrix $A$ such that for three vectors $$v_1=(0,-1,1)^T, v_2=(2,1,1)^T,v_3=(-1,1,1)^T$$ $Av_1=1/2v_2$, $Av_2=1/2v_1$ and $Av_3=v_3$. So we are looking for $A$ such that the matrix representation of $(Av_1,Av_2,Av_3)$ is $$\\begin{pmatrix} 0 & 1/2 & 0 \\\\ 1/2 & 0 & 0 \\\\ 0 & 0 & 1\\end{pmatrix}$$ But I m stuck on what to do next. Thanks • Express the canonical vectors $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$ as linear combination of the $v_i$, then compute their images $Ae_i$, and $A=(Ae_1\\ Ae_2\\ Ae_3)$. – Nihl Jul 21 '15 at 9:06 Note: $[Av_1\\ Av_2\\ Av_3]=A[v_1\\ v_2\\ v_3]$. Let $B=[v_1\\ v_2\\ v_3]$. By assumption, $AB=C$ where $C=\\begin{bmatrix} 0& \\frac{1}{2}& 0\\\\ \\frac{1}{2}& 0& 0\\\\ 0& 0& 1\\end{bmatrix}$. Thus $A=CB^{-1}$. $B^{-1}$ exists, by the way.", "# Unit Vectors and Idempotent Matrices ## Problem 527 A square matrix $A$ is called idempotent if $A^2=A$. (a) Let $\\mathbf{u}$ be a vector in $\\R^n$ with length $1$. Define the matrix $P$ to be $P=\\mathbf{u}\\mathbf{u}^{\\trans}$. Prove that $P$ is an idempotent matrix. (b) Suppose that $\\mathbf{u}$ and $\\mathbf{v}$ be unit vectors in $\\R^n$ such that $\\mathbf{u}$ and $\\mathbf{v}$ are orthogonal. Let $Q=\\mathbf{u}\\mathbf{u}^{\\trans}+\\mathbf{v}\\mathbf{v}^{\\trans}$. Prove that $Q$ is an idempotent matrix. (c) Prove that each nonzero vector of the form $a\\mathbf{u}+b\\mathbf{v}$ for some $a, b\\in \\R$ is an eigenvector corresponding to the eigenvalue $1$ for the matrix $Q$ in part (b). ## Proof. ### (a) Prove that $P=\\mathbf{u}\\mathbf{u}^{\\trans}$ is an idempotent matrix. The length of the vector $\\mathbf{u}$ is given by $\\|\\mathbf{u}\\|=\\sqrt{\\mathbf{u}^{\\trans}\\mathbf{u}}.$ Since $\\|\\mathbf{u}\\|=1$ by assumption, it yields that $\\mathbf{u}^{\\trans}\\mathbf{u}=1.$ Let us compute $P^2$ using the associative properties of matrix multiplication. We have \\begin{align*} P^2&=(\\mathbf{u}\\mathbf{u}^{\\trans})(\\mathbf{u}\\mathbf{u}^{\\trans})\\\\ &=\\mathbf{u}(\\mathbf{u}^{\\trans}\\mathbf{u})\\mathbf{u}^{\\trans}\\\\ &=\\mathbf{u}( 1 ) \\mathbf{u}^{\\trans}=\\mathbf{u}\\mathbf{u}^{\\trans}=P. \\end{align*} Thus, we have obtained $P^2=P$, and hence $P$ is an idempotent matrix. ### (b) Prove that $Q=\\mathbf{u}\\mathbf{u}^{\\trans}+\\mathbf{v}\\mathbf{v}^{\\trans}$ is an idempotent matrix Since $\\mathbf{u}$ and $\\mathbf{v}$ are unit vectors, we have as in part (a) $\\mathbf{u}^{\\trans}\\mathbf{u}=1 \\text{ and } \\mathbf{v}^{\\trans}\\mathbf{v}=1.$ Since $\\mathbf{u}$ and $\\mathbf{v}$ are orthogonal, their dot (inner) product is $0$. Thus we have $\\mathbf{u}^{\\trans}\\mathbf{v}=\\mathbf{v}^{\\trans}\\mathbf{u}=0.$ Using these identities, we compute $Q^2$ as follows. We have \\begin{align*} Q^2&=(\\mathbf{u}\\mathbf{u}^{\\trans}+\\mathbf{v}\\mathbf{v}^{\\trans})(\\mathbf{u}\\mathbf{u}^{\\trans}+\\mathbf{v}\\mathbf{v}^{\\trans})\\\\ &=\\mathbf{u}\\mathbf{u}^{\\trans}(\\mathbf{u}\\mathbf{u}^{\\trans}+\\mathbf{v}\\mathbf{v}^{\\trans})", "# Unit Vectors and Idempotent Matrices ## Problem 527 A square matrix $A$ is called idempotent if $A^2=A$. (a) Let $\\mathbf{u}$ be a vector in $\\R^n$ with length $1$. Define the matrix $P$ to be $P=\\mathbf{u}\\mathbf{u}^{\\trans}$. Prove that $P$ is an idempotent matrix. (b) Suppose that $\\mathbf{u}$ and $\\mathbf{v}$ be unit vectors in $\\R^n$ such that $\\mathbf{u}$ and $\\mathbf{v}$ are orthogonal. Let $Q=\\mathbf{u}\\mathbf{u}^{\\trans}+\\mathbf{v}\\mathbf{v}^{\\trans}$. Prove that $Q$ is an idempotent matrix. (c) Prove that each nonzero vector of the form $a\\mathbf{u}+b\\mathbf{v}$ for some $a, b\\in \\R$ is an eigenvector corresponding to the eigenvalue $1$ for the matrix $Q$ in part (b). ## Proof. ### (a) Prove that $P=\\mathbf{u}\\mathbf{u}^{\\trans}$ is an idempotent matrix. The length of the vector $\\mathbf{u}$ is given by $\\|\\mathbf{u}\\|=\\sqrt{\\mathbf{u}^{\\trans}\\mathbf{u}}.$ Since $\\|\\mathbf{u}\\|=1$ by assumption, it yields that $\\mathbf{u}^{\\trans}\\mathbf{u}=1.$ Let us compute $P^2$ using the associative properties of matrix multiplication. We have \\begin{align*} P^2&=(\\mathbf{u}\\mathbf{u}^{\\trans})(\\mathbf{u}\\mathbf{u}^{\\trans})\\\\ &=\\mathbf{u}(\\mathbf{u}^{\\trans}\\mathbf{u})\\mathbf{u}^{\\trans}\\\\ &=\\mathbf{u}( 1 ) \\mathbf{u}^{\\trans}=\\mathbf{u}\\mathbf{u}^{\\trans}=P. \\end{align*} Thus, we have obtained $P^2=P$, and hence $P$ is an idempotent matrix. ### (b) Prove that $Q=\\mathbf{u}\\mathbf{u}^{\\trans}+\\mathbf{v}\\mathbf{v}^{\\trans}$ is an idempotent matrix Since $\\mathbf{u}$ and $\\mathbf{v}$ are unit vectors, we have as in part (a) $\\mathbf{u}^{\\trans}\\mathbf{u}=1 \\text{ and } \\mathbf{v}^{\\trans}\\mathbf{v}=1.$ Since $\\mathbf{u}$ and $\\mathbf{v}$ are orthogonal, their dot (inner) product is $0$. Thus we have $\\mathbf{u}^{\\trans}\\mathbf{v}=\\mathbf{v}^{\\trans}\\mathbf{u}=0.$ Using these identities, we compute $Q^2$ as follows. We have \\begin{align*} Q^2&=(\\mathbf{u}\\mathbf{u}^{\\trans}+\\mathbf{v}\\mathbf{v}^{\\trans})(\\mathbf{u}\\mathbf{u}^{\\trans}+\\mathbf{v}\\mathbf{v}^{\\trans})\\\\ &=\\mathbf{u}\\mathbf{u}^{\\trans}(\\mathbf{u}\\mathbf{u}^{\\trans}+\\mathbf{v}\\mathbf{v}^{\\trans})", "39;}) Consider matrix $$( mathbf {R mid b & # 39;})$$ $$mathbf {R mid b & # 39;} = left ( begin {array} {c | c} mathbf {r} _1 & b_ {1} \\ mathbf {r} _2 & b_ {2} \\ vdots & vdots \\ mathbf {r} _m & b_ {m} \\ vdots & vdots \\ end {array} right)$$ As $$range ( bf R mid b & # 39;)$$ $$, it follows that the row vectors $$bf r$$ are linearly dependent and vector $$mathbf {r} _m$$ It can be written as the linear combination of the previous vectors. This is a contradiction, since we have said that $$rank ( mathbf {A}) = rank ( mathbf {R}) = m$$, which implies that the vectors $$bf r$$ It must be linearly independent. Thus, $$range ( bf R mid b & # 39;)$$ $$not , and from $$rank ( mathbf {A mid b}) = rank ( bf R mid b & # 39;)$$ implies that $$range ( bf A mid b)$$ $$not . We have shown that both cases are impossible and, therefore, we can conclude that $$rank ( mathbf {A mid b}) = range ( mathbf A)$$. $$Box$$ It is right? Actually, the initial proposition required to show $$iff$$, but as you can see, the post would be too long, therefore,", "# Category: Linear Algebra ## Problem 560 Let $A$ be an $n\\times (n-1)$ matrix and let $\\mathbf{b}$ be an $(n-1)$-dimensional vector. Then the product $A\\mathbf{b}$ is an $n$-dimensional vector. Set the $n\\times n$ matrix $B=[A_1, A_2, \\dots, A_{n-1}, A\\mathbf{b}]$, where $A_i$ is the $i$-th column vector of $A$. Prove that $B$ is a singular matrix for any choice of $\\mathbf{b}$. ## Problem 559 For each of the following matrix $A$, prove that $\\mathbf{x}^{\\trans}A\\mathbf{x} \\geq 0$ for all vectors $\\mathbf{x}$ in $\\R^2$. Also, determine those vectors $\\mathbf{x}\\in \\R^2$ such that $\\mathbf{x}^{\\trans}A\\mathbf{x}=0$. (a) $A=\\begin{bmatrix} 4 & 2\\\\ 2& 1 \\end{bmatrix}$. (b) $A=\\begin{bmatrix} 2 & 1\\\\ 1& 3 \\end{bmatrix}$. ## Problem 558 Let $A$ be an $n\\times n$ nonsingular matrix. Prove that the transpose matrix $A^{\\trans}$ is also nonsingular. ## Problem 556 Let $\\mathbf{v}$ be a nonzero vector in $\\R^n$. Then the dot product $\\mathbf{v}\\cdot \\mathbf{v}=\\mathbf{v}^{\\trans}\\mathbf{v}\\neq 0$. Set $a:=\\frac{2}{\\mathbf{v}^{\\trans}\\mathbf{v}}$ and define the $n\\times n$ matrix $A$ by $A=I-a\\mathbf{v}\\mathbf{v}^{\\trans},$ where $I$ is the $n\\times n$ identity matrix. Prove that $A$ is a symmetric matrix and $AA=I$. Conclude that the inverse matrix is $A^{-1}=A$. ## Problem 555 Let $U$ and $V$ be vector spaces over a scalar field $\\F$. Define the map $T:U\\to V$ by $T(\\mathbf{u})=\\mathbf{0}_V$ for each vector $\\mathbf{u}\\in U$. (a) Prove that $T:U\\to V$ is a linear transformation. (Hence, $T$ is", "+0 # Let A be a matrix, and let x and y be linearly independent vectors such that 0 290 1 Let A be a matrix, and let x and y be linearly independent vectors such that $$Ax=y, Ay=x+2y$$ Then we have that $$A^5x=ax+by$$ for some scalars of a and b. Find the ordered pair (a,b). Jul 7, 2019 #1 +25481 +2 Let A be a matrix, and let x and y be linearly independent vectors such that $$Ax=y,\\ Ay=x+2y$$ Then we have that $$A^5x=ax+by$$ for some scalars of a and b. Find the ordered pair (a,b). $$\\begin{array}{|rcll|} \\hline \\mathbf{Ax} &=&y \\quad & | \\quad \\cdot A \\\\ A^2x &=& Ay \\quad & | \\quad \\mathbf{Ay =x+2y} \\\\ &=& x+2y \\quad & | \\quad \\cdot A \\\\\\\\ A^3x &=& Ax+2 Ay \\\\ &=& y+2(x+2y) \\\\ &=& 2x+5y \\quad & | \\quad \\cdot A \\\\\\\\ A^4x &=& 2Ax+5 Ay \\\\ &=& 2y+5(x+2y) \\\\ &=& 5x+12y \\quad & | \\quad \\cdot A \\\\\\\\ A^5x &=& 5Ax+12 Ay \\\\ &=& 5y+12(x+2y) \\\\ &=& \\mathbf{12x+29y} \\\\ \\hline \\end{array}$$ Jul 8, 2019", "+0 # Matrices 0 40 1 Let w = $$\\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix}$$. There exists a $$2 \\times 2$$ matrix $${P}$$ such that $$\\text{proj}_{{w}} {v} = {P} {v}$$ for all 2-dimensional vectors $${v}$$. Find $${P}$$. I'm not really sure where to start for this problem. Should I plug w into the projection formula? Mar 6, 2021 ### 1+0 Answers #1 0 The matrix is $$\\begin{pmatrix} 2 & -1 \\\\ -1 & 0 \\end{pmatrix}$$ Mar 7, 2021", "the following lemma to do this Farkas’ Lemma For any matrix $A \\in \\mathbb{R}^{m \\times n}$, vector $\\vec{b} \\in \\mathbb{R}^{m}$, exactly one of the following holds 1. There exists $\\vec{x} \\in \\mathbb{R}^{n}$ such that $\\vec{x} \\geq 0$ and $A \\vec{x} = \\vec{b}$ 1. There exists $\\vec{y} \\in \\mathbb{R}^{m}$ such that $\\vec{b} \\cdot \\vec{y} < 0$ and $A^{T} \\vec{y} \\geq 0$" ]

[ "\\\\ 0 & 0 & -1 & 0 \\\\ 1 & 1 & 1-\\mathrm{i} & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix},$$ $$A^{(4)}_6= \\begin{pmatrix} 1 & 0 & 1 & 0 \\\\ 0 & \\mathrm{i} & 0 & 1 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -\\mathrm{i} \\end{pmatrix},$$ $$A^{(4)}_7= \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & \\mathrm{i} & 0 & 1 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -\\mathrm{i} \\end{pmatrix},$$ $$A^{(4)}_8= \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & \\mathrm{i} & 0 & 0 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -\\mathrm{i} \\end{pmatrix}.$$ $A_0^{(4)}$ $A_1^{(4)}$ $A_2^{(4)}$ $A_3^{(4)}$ $A_4^{(4)}$ $A_5^{(4)}$ $A_6^{(4)}$ $A_7^{(4)}$ $A_8^{(4)}$", "-1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix}$$. Therefore, $$U_2 =\\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 1 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0 \\end{pmatrix} \\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0\\\\ 0 & 1 & 0 & 0 \\end{pmatrix}$$ Thus, $$U_1 = U_2$$.", "1)^4 = t^4 - 4t^3 + 6t^2 - 4t + 1$ $t - 1$ 4 $-1$ $\\begin{pmatrix}-1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -1 \\\\\\end{pmatrix}$ $(t + 1)^4 = t^4 + 4t^3 + 6t^2 + 4t + 1$ $t + 1$ -4 $i$ $\\begin{pmatrix}0 & -1 & 0 & 0 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $-i$ $\\begin{pmatrix} 0 & 1 & 0 & 0 \\\\ -1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & -1 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $j$ $\\begin{pmatrix} 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $-j$ $\\begin{pmatrix} 0 & 0 & 1 & 0 \\\\ 0 & 0", "\\mathbf{w}}\\right)^T\\left[I(\\mathbf{w}(\\mathbf{x},\\mathbf{p})) - T(\\mathbf{w}(\\mathbf{x},\\mathbf{0}))\\right] \\\\ \\end{align} For the affine warp we have, \\begin{align} \\frac{\\partial \\mathbf{w}}{\\partial \\mathbf{p}} &= \\frac{\\partial (\\mathbf{A}\\mathbf{x} + \\mathbf{b})}{\\partial \\mathbf{p}} \\\\ &= \\frac{\\partial}{\\partial \\mathbf{p}}\\left[ \\begin{pmatrix}1+p_0 && p_1 \\\\ p_2 && 1+p_3\\end{pmatrix}\\begin{pmatrix}x \\\\ y\\end{pmatrix} \\right] + \\frac{\\partial}{\\partial \\mathbf{p}} \\begin{pmatrix}p_4 \\\\ p_5\\end{pmatrix} \\\\ &= \\frac{\\partial}{\\partial \\mathbf{p}} \\begin{pmatrix}x+p_0x+p_1y \\\\ p_2x+y+p_3y\\end{pmatrix} + \\frac{\\partial}{\\partial \\mathbf{p}} \\begin{pmatrix}p_4 \\\\ p_5\\end{pmatrix} \\\\ &= \\begin{pmatrix}x && y && 0 && 0 && 0 && 0 \\\\0 && 0 && x && y && 0 && 0\\end{pmatrix} + \\begin{pmatrix}0 && 0 && 0 && 0 && 1 && 0 \\\\ 0 && 0 && 0 && 0 && 0 && 1\\end{pmatrix} \\\\ &= \\begin{pmatrix}x && y && 0 && 0 && 1 && 0 \\\\0 && 0 && x && y && 0 && 1\\end{pmatrix} \\end{align} If we let, \\begin{align} \\frac{\\partial T}{\\partial \\mathbf{w}} = \\begin{pmatrix}T_x && T_y\\end{pmatrix} \\\\ \\end{align} We have, \\begin{align} \\frac{\\partial T}{\\partial \\mathbf{w}} \\frac{\\partial \\mathbf{w}}{\\partial \\mathbf{p}} &= \\begin{pmatrix}T_x && T_y\\end{pmatrix} \\begin{pmatrix}x && y && 0 && 0 && 1 && 0 \\\\0 && 0 && x && y && 0 && 1\\end{pmatrix} \\\\ &= \\begin{pmatrix}T_xx && T_xy && T_yx && T_yy && T_x && T_y\\end{pmatrix} \\\\ \\end{align} Detecting features Equation $$\\eqref{deltap1}$$ gives an indication of which features would be good to track. The matrix, \\begin{align} \\mathbf{H} &= \\sum_{\\mathbf{x} \\in \\mathbf{R}} \\left(\\frac{\\partial T}{\\partial \\mathbf{w}}\\frac{\\partial \\mathbf{w}}{\\partial \\mathbf{p}}\\right)^T\\frac{\\partial", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "\\stackrel{E_7}{\\sim} \\begin{pmatrix} 0&1&2&2\\\\ 2&0&-3&1\\\\ -4&0&9&2\\\\ 0&-1&1&-1\\\\ \\end{pmatrix} \\stackrel{E_6E_5E_4E_3E_2E_1}{\\sim} L \\end{eqnarray*} \\begin{eqnarray*} E_7= \\begin{pmatrix} 0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\end{pmatrix} =E_7^{-1} \\end{eqnarray*} \\begin{eqnarray*} M=(E_1^{-1}E_2^{-1}E_3^{-1})(E_4^{-1}E_5^{-1}E_6^{-1}) (E_7^{-1}) U=LDPU\\\\ \\end{eqnarray*} \\begin{eqnarray*} \\begin{pmatrix} 0&1&2&2\\\\ 2&0&-3&1\\\\ -4&0&9&2\\\\ 0&-1&1&-1\\\\ \\end{pmatrix} = \\begin{pmatrix} 1&0&0&0\\\\ 0&1&0&0\\\\ -2&0&1&0\\\\ 0&-1&1&1\\\\ \\end{pmatrix} \\begin{pmatrix} 2&0&0&0\\\\ 0&1&0&0\\\\ 0&0&3&0\\\\ 0&0&1&-3\\\\ \\end{pmatrix} \\begin{pmatrix} 0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\begin{pmatrix} 1&0&\\frac{-3}{2}&\\frac{1}{2}\\\\ 0&1&2&2\\\\ 0&0&1&\\frac43\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\end{eqnarray*}", "\\\\\\dfrac 5 6 &0\\end{pmatrix}\\\\ R = \\begin{pmatrix}\\dfrac 1 2 &0 \\\\0 &\\dfrac 1 6\\end{pmatrix}\\\\ N = \\left(I_2 - Q\\right)^{-1} = \\begin{pmatrix} \\dfrac{12}{7} & \\dfrac{6}{7} \\\\ \\dfrac{10}{7} & \\dfrac{12}{7} \\\\ \\end{pmatrix}\\\\ B = NR = \\begin{pmatrix} \\dfrac{6}{7} & \\dfrac{1}{7} \\\\ \\dfrac{5}{7} & \\dfrac{2}{7} \\\\ \\end{pmatrix}$ From $B$ we can directly read off that $P[\\text{MW|start with MP}] = \\dfrac 6 7$ and $P[\\text{MW|start with SP}] = \\dfrac 5 7$ OH thank you so much", "{\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}t={\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}(6-3t)}$ and so any vector in the form for ${\\displaystyle S_{1}}$ can be stated in the form needed for inclusion in ${\\displaystyle S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we look for ${\\displaystyle t}$ so that these equations hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}1&+&0t&=&1+0s\\\\2&+&3t&=&8-1s\\end{array}}}$ Rewrite that as ${\\displaystyle {\\begin{array}{*{1}{rc}r}1&=&1\\\\t&=&2-(1/3)s\\end{array}}}$ and so ${\\displaystyle {\\begin{pmatrix}1\\\\8\\end{pmatrix}}+{\\begin{pmatrix}0\\\\-1\\end{pmatrix}}s={\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}0\\\\3\\end{pmatrix}}(2-(1/3)s).}$ 2. These two are equal. To show that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we check that for any ${\\displaystyle t,s}$ we can find an appropriate ${\\displaystyle m,n}$ so that these hold. ${\\displaystyle {\\begin{array}{*{2}{rc}r}4m&-&4n&=&1t+2s\\\\7m&-&2n&=&3t+1s\\\\7m&-&10n&=&1t+5s\\end{array}}}$ Use Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\7&-2&3t+1s\\\\7&-10&1t+5s\\end{array}}\\right)&{\\xrightarrow[{(-7/4)\\rho _{1}+\\rho _{3}}]{(-7/4)\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&-3&-(3/4)t+(6/4)s\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}4&-4&1t+2s\\\\0&5&(5/4)t-(10/4)s\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to conclude that ${\\displaystyle {\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}t+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}s={\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}((1/4)t-(1/2)s)}$ and so ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&+&2s&=&4m-4n\\\\3t&+&1s&=&7m-2n\\\\1t&+&5s&=&7m-10n\\end{array}}}$ with Gaussian reduction ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\3&1&7m-2n\\\\1&5&7m-10n\\end{array}}\\right)&{\\xrightarrow[{-\\rho _{1}+\\rho _{3}}]{-3\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&3&3m-6n\\end{array}}\\right)\\\\&{\\xrightarrow[{}]{(3/5)\\rho _{2}+\\rho _{3}}}&\\left({\\begin{array}{*{2}{c}|c}1&2&4m-4n\\\\0&-5&-5m+10n\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ to get ${\\displaystyle {\\begin{pmatrix}4\\\\7\\\\7\\end{pmatrix}}m+{\\begin{pmatrix}-4\\\\-2\\\\-10\\end{pmatrix}}n={\\begin{pmatrix}1\\\\3\\\\1\\end{pmatrix}}(2m)+{\\begin{pmatrix}2\\\\1\\\\5\\end{pmatrix}}(m-2n)}$ and so any member of ${\\displaystyle S_{2}}$ can be expressed in the form needed for ${\\displaystyle S_{1}}$. 3. These sets are equal. To prove that ${\\displaystyle S_{1}\\subseteq S_{2}}$, we must be able to solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}2m&+&4n&=&1t\\\\4m&+&8n&=&2t\\end{array}}}$ for ${\\displaystyle m}$ and ${\\displaystyle n}$ in terms of ${\\displaystyle t}$. Apply Gaussian reduction ${\\displaystyle \\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\4&8&2t\\end{array}}\\right){\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}\\left({\\begin{array}{*{2}{c}|c}2&4&1t\\\\0&0&0\\end{array}}\\right)}$ to conclude that any pair ${\\displaystyle m,n}$ where ${\\displaystyle 2m+4n=t}$ will do. For instance, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((1/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(0)}$ or ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}t={\\begin{pmatrix}2\\\\4\\end{pmatrix}}((-3/2)t)+{\\begin{pmatrix}4\\\\8\\end{pmatrix}}(t).}$ Thus ${\\displaystyle S_{1}\\subseteq S_{2}}$. For ${\\displaystyle S_{2}\\subseteq S_{1}}$, we solve ${\\displaystyle {\\begin{array}{*{2}{rc}r}1t&=&2m+4n\\\\2t&=&4m+8n\\end{array}}}$ with Gauss' method ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\2&4m+8n\\end{array}}\\right)&{\\xrightarrow[{}]{-2\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{1}{c}|c}1&2m+4n\\\\0&0\\end{array}}\\right)\\end{array}}}$ to deduce that any", "\\mathbf{U}^T\\mathbf{U} = \\mathbf{I}$$, $$\\mathbf{V}\\mathbf{V}^T = \\mathbf{V}^T\\mathbf{V} = \\mathbf{I}$$, and $$\\Sigma$$ containing the singular values of $$\\mathbf{A}$$, $$\\sigma_i$$, $$1 \\leq I \\leq \\min{(m, n)}$$, along the diagonal. Then, the factorization of $$\\mathbf{B}$$ in terms of $$\\mathbf{U}$$, $$\\mathbf{V}$$, and $$\\Sigma$$ is the following: $$\\mathbf{B} = \\begin{pmatrix} \\mathbf{0} & \\mathbf{A}^T \\\\ \\mathbf{A} & \\mathbf{0} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{0} & \\mathbf{V}\\Sigma^T\\mathbf{U}^T \\\\ \\mathbf{U}\\Sigma\\mathbf{V}^T & \\mathbf{0} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{V} & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{U} \\end{pmatrix} \\begin{pmatrix} \\mathbf{0} & \\Sigma^T \\\\ \\Sigma & \\mathbf{0} \\end{pmatrix} \\begin{pmatrix} \\mathbf{V}^T & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{U}^T \\end{pmatrix} = \\mathbf{Q}_1 \\mathbf{C} \\mathbf{Q}_1^T$$ where $$\\mathbf{Q}_1 = \\begin{pmatrix} \\mathbf{V} & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{U} \\end{pmatrix}$$ and $$\\mathbf{C} = \\begin{pmatrix} \\mathbf{0} & \\Sigma^T \\\\ \\Sigma & \\mathbf{0} \\end{pmatrix}$$. We can readily verify that $$\\mathbf{Q}_1$$ is orthogonal since $$\\mathbf{Q}_1\\mathbf{Q}_1^T = \\mathbf{Q}_1^T\\mathbf{Q}_1 = \\mathbf{I}$$. However, $$\\mathbf{C}$$ has not the expected diagonal structure. Instead, the singular values, $$\\sigma_i$$, of $$\\mathbf{A}$$ are distributed in the following way: $$\\mathbf{C} = \\begin{pmatrix} \\mathbf{0} & \\Sigma^T \\\\ \\Sigma & \\mathbf{0} \\\\ \\end{pmatrix} = \\left( \\begin{array}{ccc|ccc} & & & \\sigma_1 & & \\\\ & \\mathbf{0} & & & \\sigma_2 & \\\\ & & & & & \\ddots \\\\ \\hline \\sigma_1 & & & & & \\\\ & \\sigma_2 & & & \\mathbf{0} & \\\\ & & \\ddots & & & \\end{array} \\right)$$ That is, $$\\mathbf{C}$$ contains", "\\end{eqnarray*} \\begin{eqnarray*} M=(E_1^{-1}E_2^{-1}E_3^{-1})(E_4^{-1}E_5^{-1}E_6^{-1}) (E_7^{-1}) U=LDPU\\\\ \\end{eqnarray*} \\begin{eqnarray*} \\begin{pmatrix} 0&1&2&2\\\\ 2&0&-3&1\\\\ -4&0&9&2\\\\ 0&-1&1&-1\\\\ \\end{pmatrix} = \\begin{pmatrix} 1&0&0&0\\\\ 0&1&0&0\\\\ -2&0&1&0\\\\ 0&-1&1&1\\\\ \\end{pmatrix} \\begin{pmatrix} 2&0&0&0\\\\ 0&1&0&0\\\\ 0&0&3&0\\\\ 0&0&1&-3\\\\ \\end{pmatrix} \\begin{pmatrix} 0&1&0&0\\\\ 1&0&0&0\\\\ 0&0&1&0\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\begin{pmatrix} 1&0&\\frac{-3}{2}&\\frac{1}{2}\\\\ 0&1&2&2\\\\ 0&0&1&\\frac43\\\\ 0&0&0&1\\\\ \\end{pmatrix} \\end{eqnarray*}", "$a \\cdot (x + y) = ax + ay$ VS8 $(a + b) \\cdot x = ax + bx$ Proof of VS4 Take an arbitrary $x = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ \\vdots \\\\ a_n \\end{pmatrix} \\in F^n$ Set $y = \\begin{pmatrix} -a_1 \\\\ -a_2 \\\\ \\vdots \\\\ -a_n \\end{pmatrix}$ and note $x + y = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ \\vdots \\\\ a_n \\end{pmatrix} + \\begin{pmatrix} -a_1 \\\\ -a_2 \\\\ \\vdots \\\\ -a_n \\end{pmatrix} = \\begin{pmatrix} a_1 + (-a_1) \\\\ a_2 + (-a_2) \\\\ \\vdots \\\\ a_n + (-a_n) \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ \\vdots \\\\ 0 \\end{pmatrix} = 0_{F^n}$ Examples 1. $F^n \\mbox{ for } n \\in \\mathbb N$ $F^n = \\left\\{ \\begin{pmatrix} a_1 \\\\ a_2 \\\\ \\vdots \\\\ a_n \\end{pmatrix} : a_i \\in F \\right\\}$ $\\begin{pmatrix} a_1 \\\\ a_2 \\\\ \\vdots \\\\ a_n \\end{pmatrix} + \\begin{pmatrix} b_1 \\\\ b_2 \\\\ \\vdots \\\\ b_n \\end{pmatrix} = \\begin{pmatrix} a_1 + b_1 \\\\ a_2 + b_2 \\\\ \\vdots \\\\ a_n + b_n \\end{pmatrix}$ $a \\begin{pmatrix} b_1 \\\\ b_2 \\\\ \\vdots \\\\ b_n \\end{pmatrix} = \\begin{pmatrix} ab_1 \\\\ ab_2 \\\\ \\vdots \\\\ ab_n \\end{pmatrix}$ ... 2. $\\mathrm M_{m \\times n}(F)$ ... 3. $\\mathcal F(S, F)$ 4. Polynomials 5. $...$ Food for thought What is wrong with setting $\\begin{pmatrix} 2 & 3 \\\\ 4 & 5 \\\\ \\end{pmatrix}", "$\\hfill \\mathbf{H} = \\begin{pmatrix} 0 & 1/3 & 1/3 & 1/3 & 0 \\\\ 0 & 0 & 1/2 & 1/2 & 0 \\\\ 0 & 1 & 0 & 0 & 0 \\\\ 0 & 0 & 1/2 & 0 & 1/2 \\\\ 0 & 0 & 0 & 0 & 0 \\end{pmatrix}, (1/n)\\mathbf{1} = \\begin{pmatrix} 1/5 \\\\ 1/5 \\\\ 1/5 \\\\ 1/5 \\\\ 1/5 \\end{pmatrix}, \\mathbf{d} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 0 \\\\ 1 \\end{pmatrix}, \\mathbf{p}_0 = \\begin{pmatrix} 1/5 \\\\ 1/5 \\\\ 1/5 \\\\ 1/5 \\\\ 1/5 \\end{pmatrix}$. Repeatedly computing $\\begin{pmatrix} p_1 \\\\ p_2 \\\\ p_3 \\\\ p_4 \\\\ p_5 \\end{pmatrix}_{k+1} := 0.85 \\left( \\begin{pmatrix} 0 & 1/3 & 1/3 & 1/3 & 0 \\\\ 0 & 0 & 1/2 & 1/2 & 0 \\\\ 0 & 1 & 0 & 0 & 0 \\\\ 0 & 0 & 1/2 & 0 & 1/2 \\\\ 0 & 0 & 0 & 0 & 0 \\end{pmatrix}^\\mathrm{T} \\begin{pmatrix} p_1 \\\\ p_2 \\\\ p_3 \\\\ p_4 \\\\ p_5 \\end{pmatrix}_k + \\begin{pmatrix} 1/5 \\\\ 1/5 \\\\ 1/5 \\\\ 1/5 \\\\ 1/5 \\end{pmatrix} \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 0 \\\\ 1 \\end{pmatrix}^\\mathrm{T} \\begin{pmatrix} p_1 \\\\ p_2 \\\\ p_3 \\\\ p_4 \\\\ p_5 \\end{pmatrix}_k \\right) + 0.15 \\begin{pmatrix} 1/5 \\\\ 1/5 \\\\ 1/5 \\\\ 1/5 \\\\ 1/5", "& {\\ frac {-1 + {\\ sqrt {5}}} {2}} \\\\ 1 & 1 \\ end {pmatrix}} {\\ begin {pmatrix} {\\ frac {1 - {\\ sqrt {5}}} {2}} & 0 \\\\ 0 & {\\ frac {1 + {\\ sqrt {5}}} {2} } \\ end {pmatrix}} ^ {n} {\\ begin {pmatrix} - {\\ frac {1} {\\ sqrt {5}}} & {\\ frac {{\\ sqrt {5}} - 1} {2 { \\ sqrt {5}}}} \\\\ {\\ frac {1} {\\ sqrt {5}}} & {\\ frac {{\\ sqrt {5}} + 1} {2 {\\ sqrt {5}}}} \\ end {pmatrix}} {\\ begin {pmatrix} 0 \\\\ 1 \\ end {pmatrix}} \\\\ & = {\\ begin {pmatrix} {\\ frac {-1 - {\\ sqrt {5}}} {2} } & {\\ frac {-1 + {\\ sqrt {5}}} {2}} \\\\ 1 & 1 \\ end {pmatrix}} {\\ begin {pmatrix} \\ left ({\\ frac {1 - {\\ sqrt {5 }}} {2}} \\ right) ^ {n} & 0 \\\\ 0 & \\ left ({\\ frac {1 + {\\ sqrt {5}}} {2}} \\ right) ^ {n} \\ end {pmatrix }} {\\ begin {pmatrix} - {\\ frac {1} {\\ sqrt {5}}} & {\\ frac {{\\ sqrt {5}} - 1} {2 {\\ sqrt {5}}}} \\ \\ {\\ frac {1} {\\ sqrt {5}}} & {\\ frac {{\\ sqrt {5}} + 1} {2 {\\ sqrt {5}}} } \\ end {pmatrix}}", "found from the characteristic equation $$0=\\det\\begin{pmatrix}-1-\\lambda & x & -x \\\\ 1 & -1-\\lambda & 0 \\\\ 1 & 0 & -1-\\lambda \\end{pmatrix}=-(1+\\lambda)^3$$ to be $$\\lambda=-1$$ thrice. There is only one associated eigenvector $$(0,1,1)$$. A generalized eigenvector is found from $$(A+I)^2v=0$$, $$\\begin{pmatrix}0&0&0\\\\0&x&-x\\\\0&x&-x\\end{pmatrix}v=0\\Longleftarrow (A+I)v=\\begin{pmatrix}0\\\\1\\\\1\\end{pmatrix}\\Longleftarrow v=\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}$$ The third, a solution of $$(A+I)^3v=0$$, satisfies $$(A+I)v=\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}$$, namely $$\\begin{pmatrix}0\\\\1/x\\\\0\\end{pmatrix}$$. So letting $$P=\\begin{pmatrix}0&1&0\\\\1&0&1/x\\\\1&0&0\\end{pmatrix}$$ gives $$A=PJP^{-1}=P\\begin{pmatrix}-1&1&0\\\\0&-1&1\\\\0&0&-1\\end{pmatrix}P^{-1}$$ Hence $$A^n=PJ^nP^{-1}$$ where $$J^n=(-1)^n\\begin{pmatrix}1&(-1)^nn&\\binom{n}{2}\\\\0&1&(-1)^nn\\\\0&0&1\\end{pmatrix}$$ Multiplying out $$PJ^nP^{-1}$$ should give your answer.", "& 0 \\\\0 & 0 & 0 & -1 \\\\0 & 1 & 0 & 0 \\\\1 & 0 & 0 & 0 \\end{pmatrix} = \\begin{pmatrix} 0 & 0 & 1 & 0 \\\\0 & 0 & 0 & -1 \\\\0 & -1 & 0 & 0 \\\\-1 & 0 & 0 & 0 \\end{pmatrix}$$", "Site Helper Now that you mention what $a$ is, it makes a bit more sense. Either way, $M_B^B(\\text{id})$ makes no reference to $a$ nor $\\phi_a$ does it? So it can not be equal to anything that does refer to $\\phi_a$. Oh there is a typo... There it should be $\\mathcal{M}_{\\mathcal{B}}^{\\mathcal{B}}(\\phi_a)$ #### mathmari ##### Well-known member MHB Site Helper So to clarify : We have $B=\\{b_1, b_2, b_3\\}$ with $$b_1=\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix}, b_2=\\begin{pmatrix}1 \\\\ 0\\\\ -1\\end{pmatrix}, b_3=\\begin{pmatrix}-1 \\\\ 1\\\\ 0\\end{pmatrix}$$ which is basis of $\\mathbb{R}^3$. Then to calculate $M_E^B(\\text{id})$ and $M_B^E(\\text{id})$ we do the following? \\begin{equation*}\\mathcal{M}_{\\mathcal{E}}^{\\mathcal{B}}(\\text{id})=\\left (\\gamma_{\\mathcal{E}}(b_1)\\mid \\gamma_{\\mathcal{E}}(b_2)\\mid \\gamma_{\\mathcal{E}}(b_3)\\right )=\\left (b_1\\mid b_2\\mid b_3\\right )=\\begin{pmatrix}1 & 1 & -1 \\\\ 1 & 0 & 1 \\\\ 1 & -1 & 0\\end{pmatrix}\\end{equation*} \\begin{equation*}\\mathcal{M}_{\\mathcal{B}}^{\\mathcal{E}}(\\text{id})=\\left (\\gamma_{\\mathcal{B}}(e_1)\\mid \\gamma_{\\mathcal{B}}(e_2)\\mid \\gamma_{\\mathcal{B}}(e_3)\\right )\\end{equation*} Then suppose $a=\\begin{pmatrix}0 & 0 & 1 \\\\ 1 & 0 & 0 \\\\ 0 & 1 & 0\\end{pmatrix}$ then to calculate $M_B^B(\\phi_a)$ do we do the following? \\begin{equation*}\\mathcal{M}_{\\mathcal{B}}^{\\mathcal{B}}(\\phi_a)=\\left (\\phi_a(b_1)\\mid \\phi_a(b_2)\\mid \\phi_a(b_3)\\right )=a=\\begin{pmatrix}0 & 0 & 1 \\\\ 1 & 0 & 0 \\\\ 0 & 1 & 0\\end{pmatrix}\\end{equation*} #### Klaas van Aarsen ##### MHB Seeker Staff member So to clarify : We have $B=\\{b_1, b_2, b_3\\}$ with $$b_1=\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix}, b_2=\\begin{pmatrix}1 \\\\ 0\\\\ -1\\end{pmatrix}, b_3=\\begin{pmatrix}-1 \\\\ 1\\\\ 0\\end{pmatrix}$$ which is basis of $\\mathbb{R}^3$. Then to calculate $M_E^B(\\text{id})$ and $M_B^E(\\text{id})$ we do", "\\end{pmatrix}$ Using the definition of the inner product $(\\mathbf u, \\mathbf v) = \\mathbf v^* \\mathbf u$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (5-i)(1+3i) + (6-4i)(2+4i) = 36 + 30i$ , but using $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\begin{pmatrix} 1 \\\\ 2 \\\\ \\end{pmatrix} , \\begin{pmatrix} 5 \\\\ 6 \\\\ \\end{pmatrix}) + i(\\begin{pmatrix} 3 \\\\ 4 \\\\ \\end{pmatrix} , \\begin{pmatrix} 1 \\\\ 4 \\\\ \\end{pmatrix}) = 5 + 12 + i(3 + 16) = 17 + 19i$. What went wrong with my interpretation of the formula , $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$ ? Shouldn't both computations be mathematically sound and give the same result? Thanks in advance for any help!", "\\end{pmatrix} = \\begin{pmatrix} b_1 \\\\ & \\ddots \\\\ & & b_n \\end{pmatrix} \\begin{pmatrix} x_1 ^{-1} \\\\ \\vdots \\\\ x_n^{-1} \\end{pmatrix}.$$ As $A=\\left( a_{ij} \\right)_{n \\times n}$, $B=\\mathrm{diag}\\{b_1, b_2, \\cdots, b_n\\}=\\begin{pmatrix} b_1 \\\\ & \\ddots \\\\ & & b_n \\end{pmatrix}$, $C=A^{-1}B$, then $$A \\begin{pmatrix} x_1 \\\\ \\vdots \\\\ x_n \\end{pmatrix} = B \\begin{pmatrix} x_1 ^{-1} \\\\ \\vdots \\\\ x_n^{-1} \\end{pmatrix},$$ so $$\\begin{pmatrix} x_1 \\\\ \\vdots \\\\ x_n \\end{pmatrix} = A^{-1} B \\begin{pmatrix} x_1 ^{-1} \\\\ \\vdots \\\\ x_n^{-1} \\end{pmatrix},$$ that is to say, $$\\begin{pmatrix} x_1 \\\\ \\vdots \\\\ x_n \\end{pmatrix} = C \\begin{pmatrix} x_1 ^{-1} \\\\ \\vdots \\\\ x_n^{-1} \\end{pmatrix}.$$ Let $\\begin{pmatrix} x_1^{(0)} \\\\ \\vdots \\\\ x_n^{(0)} \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ \\vdots \\\\ 1 \\end{pmatrix}$, by the recurrence $$\\begin{pmatrix} x_1^{(k+1)} \\\\ \\vdots \\\\ x_n^{(k+1)} \\end{pmatrix} = C \\begin{pmatrix} {\\left(x_1^{(k)}\\right)} ^{-1} \\\\ \\vdots \\\\ {\\left(x_n^{(k)}\\right)}^{-1}, \\end{pmatrix},$$ we can get a numerical solution. If the result does not converge, we can use another one $$\\begin{pmatrix} {\\left(x_1^{(k+1)}\\right)} ^{-1} \\\\ \\vdots \\\\ {\\left(x_n^{(k+1)}\\right)}^{-1}, \\end{pmatrix} = C^{-1} \\begin{pmatrix} x_1^{(k)} \\\\ \\vdots \\\\ x_n^{(k)} \\end{pmatrix}.$$ If you want the analytic solution, you may have a try by using the Groebner-Shirshov Bases. (The software Maple or Mathematica can do it.) But I am not sure the analytic solution can be find easily. - The real case is when A is almost non-singular(having a large cond", "\\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 2. $\\begin{pmatrix} -3 \\\\ 3 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 3. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 4. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k+\\begin{pmatrix} 0 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 5. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 14 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 6. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 6 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ Problem 2 Produce two descriptions of this set that are different than this one. $\\{\\begin{pmatrix} 2 \\\\ -5 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 3 Show that the three descriptions given at the start of this subsection all describe the same set. This exercise is recommended for all readers. Problem 4 Show that these sets are equal $\\{\\begin{pmatrix} 1 \\\\ 4 \\\\ 1 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}z\\,\\big|\\, z\\in\\mathbb{R} \\} \\quad\\text{and}\\quad \\{\\begin{pmatrix} 0 \\\\ 4 \\\\ 2 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R} \\},$ and that both describe the solution set of this system. $\\begin{array}{*{4}{rc}r} x &- &y &+ &z &+", "= \\frac{1}{\\textbf{h}^{3T}\\textbf{x}}\\begin{pmatrix} \\textbf{x}^T & 0 & 0 \\end{pmatrix} - \\frac{\\textbf{h}^{1T}\\textbf{x}}{(\\textbf{h}^{3T}\\textbf{x})^2} \\begin{pmatrix} 0 & 0 & \\textbf{x}^T \\end{pmatrix}$$ Substituting $x’ = \\textbf{h}^{1T}\\textbf{x}$ and $w’ = \\textbf{h}^{3T}\\textbf{x}$ $$\\partial\\tilde{x}’ / \\partial\\textbf{x} = \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & 0 \\end{pmatrix} - \\frac{x’}{(w’)^2}\\begin{pmatrix} 0 & 0 & \\textbf{x}^T \\end{pmatrix}$$ Finally, substituting $\\tilde{x’} = x’/w’$ $$\\partial\\tilde{x}’ / \\partial\\textbf{x} = \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & 0 \\end{pmatrix} - \\frac{\\tilde{x’} }{w’}\\begin{pmatrix} 0 & 0 & \\textbf{x}^T \\end{pmatrix}$$ $$= \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & -\\tilde{x’} \\textbf{x}^T \\end{pmatrix}$$ Performing a similar process for $\\tilde{y’}$, we get $$\\partial\\tilde{y}’ / \\partial\\textbf{x} = \\frac{1}{w’}\\begin{pmatrix} 0 & \\textbf{x}^T & -\\tilde{y’} \\textbf{x}^T \\end{pmatrix}$$ Hence, we have established that $$\\partial{\\tilde{\\textbf{x}}}’ / \\partial{\\textbf{h}} = \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & -\\tilde{x’}\\textbf{x}^T \\\\ 0 & \\textbf{x}^T & -\\tilde{y’}\\textbf{x}^T\\end{pmatrix}$$ #### VI. Sampson error with non-isotropic error distributions. The derivation of Sampson error in section 4.2.6(p98) assumed that points were measured with circular error distributions. In the case where the point $\\textbf{X} = (x, y, x’, y’)$ is measured with covariance matrix $\\Sigma_\\textbf{x}$ it is appropriate instead to minimize the Mahalanobis norm $\\begin{Vmatrix}\\boldsymbol{\\delta}_\\textbf{x}\\end{Vmatrix}^2_{\\Sigma_\\textbf{x}} = \\boldsymbol{\\delta}^T_\\textbf{x}\\Sigma^{-1}_\\textbf{x}\\boldsymbol{\\delta}_\\textbf{x}$. Show that in this case the formulae corresponding to (4.11-p99) and (4.12-p99) are $$\\boldsymbol{\\delta}_\\textbf{x} = -\\Sigma_\\textbf{x}J^T(J\\Sigma_\\textbf{x}J^T)^{-1}\\epsilon$$ and $$\\begin{Vmatrix}\\boldsymbol{\\delta}_\\textbf{x}\\end{Vmatrix}^2_{\\Sigma_\\textbf{x}} = \\epsilon^T(J\\Sigma_\\textbf{x}J^T)^{-1}\\epsilon$$ Note that if the measurements in the two images are independent, then the covariance matrix $\\Sigma_\\textbf{x}$ will be block-diagonal with two $2\\times2$ diagonal blocks" ]

[ "# Thread: Spans and vectors 1. ## Spans and vectors If a question asks, Is the vector $\\mathbf{b}=\\begin{pmatrix}-2\\\\ -6\\\\ -4\\end{pmatrix}\\in \\mbox{span}(\\mathbf{v_1, v_2, v_3, v_4})$ where: $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 3\\\\ 0\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}2\\\\ 2\\\\ 1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ -1\\end{pmatrix}$, $\\mathbf{v_4}=\\begin{pmatrix}1\\\\ -2\\\\ 1\\end{pmatrix}$ Do we just make it a matrix and solve the matrix? And how would we solve it if we have 4 variables but only 3 equations? 2) Is the set of vectors $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 2\\\\ 3\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}1\\\\ 1\\\\ -1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ 5\\end{pmatrix}$ a spanning set of $\\mathbb{R}^3$? Do we just make this a matrix with say $\\mathbf{x}=\\begin{pmatrix}x_1\\\\ x_2\\\\ x_3\\end{pmatrix}$ as our solution and see if we can solve the matrix? 2. So your first question is asking whether there exist scalars $a_{1},a_{2},a_{3},a_{4}$ such that $\\mathbf{b}=a_{1}\\mathbf{v}_{1}+a_{2}\\mathbf{v}_{2} +a_{3}\\mathbf{v}_{3}+a_{4}\\mathbf{v}_{4}.$ This equation you can translate into a normal matrix problem of the form $A\\mathbf{x}=\\mathbf{b}$. Row reduce that problem. If you find any solutions (one or infinitely many), then the answer is yes. If you find that there are no solutions, then the answer is no. Your second question, again, is asking whether there exist scalars $c_{1}, c_{2}, c_{3}$ such that $c_{1}\\mathbf{v}_{1}+c_{2}\\mathbf{v}_{2}+c_{3}\\math bf{v}_{3}=\\mathbf{x}$ for your arbitrary $\\mathbf{x}.$ This, again, can be translated into a standard matrix problem. Since this problem will result in a square matrix, you can use the theory of determinants to help you out.", "$(\\left( \\begin{array}{cc} 3 & 2 \\\\ 2 & 3 \\end{array} \\right) - 1 \\left( \\begin{array}{cc} 1 & 0 \\\\ 0 & 1 \\end{array} \\right)) \\mathbf{v} =\\mathbf{0}$ for $\\mathbf{v}=\\begin{pmatrix}x\\\\y\\end{pmatrix}$ . $\\left( \\begin{array}{cc} 2 & 2 \\\\ 2 & 2 \\end{array} \\right)\\begin{pmatrix}x\\\\y\\end{pmatrix}=\\begin{pmatrix}2x+2y\\\\2x+2y\\end{pmatrix}=\\begin{pmatrix}0\\\\0\\end{pmatrix}$ Hence $x=-y$ and we can take the eigenvector corresponding to the eigenvector 5 as $\\begin{pmatrix}1\\\\-1\\end{pmatrix}$ . The matrix of eigenvectors is $P= \\left( \\begin{array}{cc} 1 & 1 \\\\ 1 & -1 \\end{array} \\right)$ . At the moment our system takes the form $\\dot{\\mathbf{v}}=M \\mathbf{v}$ . Let $\\mathbf{w}=P^{-1} \\mathbf{v}$ so that $\\mathbf{v}=P \\mathbf{w}$ &. The system becomes $P \\mathbf{v}=M P \\mathbf{w} \\rightarrow \\mathbf{w}=P^{-1}MP \\mathbf{w}$ . $P^{-1}MP$ will be a diagonal matrix with entries equal to the eigenvalues of $M$ and $\\mathbf{w}$ will be an elementary basis vector. The new system will be $\\begin{pmatrix}\\dot{w_1} \\\\\\dot{w_2} \\end{pmatrix}=\\left( \\begin{array}{cc} 5 & 0 \\\\ 0 & 1 \\end{array} \\right) \\begin{pmatrix}w_1\\\\w_2\\end{pmatrix}$ . This is equivalent to the system $\\dot{w_1}=5w_1$ $\\dot{w_1}=1w_1$ The solutions are $w_1=Ae^{5t}, w_2=B e^t$ Then $\\mathbf{v}'=P \\mathbf{w} = \\left( \\begin{array}{cc} 1 & 1 \\\\ 1 & -1 \\end{array} \\right) \\begin{pmatrix}Ae^{5t}\\\\Be^t\\\\2\\end{pmatrix} =\\begin{pmatrix}Ae^{5t}+Be^t\\\\Ae^{5t}-Be^t\\end{pmatrix}$ Now use the initial conditions to find $A$ and $B$ . $x(0)=1 \\rightarrow 1=A+B$ $y(1)=2 \\rigthtarrow 2=Ae^{-5}+Be^{-1}$ $e$ $1+2e=Ae+Ae^{-4} \\rightarrow A= \\frac{1+2e}{1+e^{-4}}=\\frac{e^4+2e^5}{e^4+1}$ $e^5$ $1-2e^5=B+Be^4 \\rightarrow B=\\frac{1-2e^5}{1+e^4}$ $x=\\frac{e^4+2e^5}{e^4+1} e^{5t}+\\frac{1-2e^5}{1+e^4}e^t, \\: y=\\frac{e^4+2e^5}{e^4+1} e^{5t}-\\frac{1-2e^5}{1+e^4}e^t$", "-2 & 2 \\\\ 2 & -2 \\end{array} \\right)\\begin{pmatrix}x\\\\y\\end{pmatrix}=\\begin{pmatrix}-2x+2y\\\\2x-2y\\end{pmatrix}=\\begin{pmatrix}0\\\\0\\end{pmatrix}$ Hence $x=y$ and we can take the eigenvector corresponding to the eigenvector 5 as $\\begin{pmatrix}1\\\\1\\end{pmatrix}$ . For $\\lambda=1$ , solve $(\\left( \\begin{array}{cc} 3 & 2 \\\\ 2 & 3 \\end{array} \\right) - 1 \\left( \\begin{array}{cc} 1 & 0 \\\\ 0 & 1 \\end{array} \\right)) \\mathbf{v} =\\mathbf{0}$ for $\\mathbf{v}=\\begin{pmatrix}x\\\\y\\end{pmatrix}$ . $\\left( \\begin{array}{cc} 2 & 2 \\\\ 2 & 2 \\end{array} \\right)\\begin{pmatrix}x\\\\y\\end{pmatrix}=\\begin{pmatrix}2x+2y\\\\2x+2y\\end{pmatrix}=\\begin{pmatrix}0\\\\0\\end{pmatrix}$ Hence $x=-y$ and we can take the eigenvector corresponding to the eigenvector 5 as $\\begin{pmatrix}1\\\\-1\\end{pmatrix}$ . The matrix of eigenvectors is $P= \\left( \\begin{array}{cc} 1 & 1 \\\\ 1 & -1 \\end{array} \\right)$ . At the moment our system takes the form $\\dot{\\mathbf{v}}=M \\mathbf{v}$ . Let $\\mathbf{w}=P^{-1} \\mathbf{v}$ so that $\\mathbf{v}=P \\mathbf{w}$ &. The system becomes $P \\mathbf{v}=M P \\mathbf{w} \\rightarrow \\mathbf{w}=P^{-1}MP \\mathbf{w}$ . $P^{-1}MP$ will be a diagonal matrix with entries equal to the eigenvalues of $M$ and $\\mathbf{w}$ will be an elementary basis vector. The new system will be $\\begin{pmatrix}\\dot{w_1} \\\\\\dot{w_2} \\end{pmatrix}=\\left( \\begin{array}{cc} 5 & 0 \\\\ 0 & 1 \\end{array} \\right) \\begin{pmatrix}w_1\\\\w_2\\end{pmatrix}$ . This is equivalent to the system $\\dot{w_1}=5w_1$ $\\dot{w_1}=1w_1$ The solutions are $w_1=Ae^{5t}, w_2=B e^t$ Then $\\mathbf{v}'=P \\mathbf{w} = \\left( \\begin{array}{cc} 1 & 1 \\\\ 1 & -1 \\end{array} \\right) \\begin{pmatrix}Ae^{5t}\\\\Be^t\\end{pmatrix} =\\begin{pmatrix}Ae^{5t}+Be^t\\\\Ae^{5t}-Be^t\\end{pmatrix}$ Now assume a solution to the non homogeneous system of the form", "+0 # Help with vector geometry +1 88 1 +38 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w} \\text{ and }\\mathbf{x}$$ are linearly independent, answer with 0. If they aren't, find coefficients a,b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a+b}{c}$$ Jun 5, 2019", "eigenvector $$\\mathbf y_2$$. Tis must satisfy $$\\mathbf y_2 \\in \\ker \\mathbf A^2 \\setminus \\ker \\mathbf A$$ and is linearly independent on $$\\mathbf x_2$$. We choose e.g. $$\\mathbf y_2=(0{,}1,0{,}0,0)^T$$ and calculate $$\\mathbf y_1=\\mathbf A\\mathbf y_2=(1{,}0,0{,}0,0)^T$$ (opět vlastní vektor $$\\mathbf A$$). Now we may assemble matrices $$\\mathbf R$$ (columns are $$\\mathbf y_1,\\mathbf y_2,\\mathbf x_1,\\mathbf x_2,\\mathbf x_3$$) and $$\\mathbf J$$ and claculate the inverse matrix $$\\mathbf R^{-1}$$ $$\\mathbf R= \\begin{pmatrix} 1&0&0&-1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&-1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$, $$\\mathbf J= \\begin{pmatrix} 0&1&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1\\\\ 0&0&0&0&0 \\end{pmatrix}$$, $$\\mathbf R^{-1}=\\begin{pmatrix} 1&0&0&1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$", "\\end{pmatrix}$ where $\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$ The entries in the introduction were given by: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\ (\\mathbf{a}_2 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_2 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_2 \\cdot \\mathbf{b}_p) \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ (\\mathbf{a}_n \\cdot \\mathbf{b}_1) & (\\mathbf{a}_n \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_n \\cdot \\mathbf{b}_p) \\end{pmatrix}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{A}\\mathbf{b}_1 & \\mathbf{A}\\mathbf{b}_2 & \\dots & \\mathbf{A}\\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1\\mathbf{B} \\\\ \\mathbf{a}_2\\mathbf{B}\\\\ \\vdots\\\\ \\mathbf{a}_n\\mathbf{B} \\end{pmatrix}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1). Matrix product (in terms of outer product) An alternative method results when", "element $x_{i,0}$ which always equals to $1$. So our required function becomes $w_0 x_{i,0} + w_1 x_{i,1} + \\dots w_d x_{i,d}$. We denote it as $h_{\\mathbf{w}}(\\mathbf{x}_i)$. Using matrices, we can write $h_{\\mathbf{w}}(\\mathbf{x}_i)$ in a much more compact form. Conventionally, we use column matrices to represent vectors. Thus we have $\\mathbf{w} = \\begin{pmatrix} w_0\\\\ w_1\\\\ w_2\\\\ \\vdots\\\\ w_d \\end{pmatrix}, \\hspace{1cm}\\mathbf{x}_i = \\begin{pmatrix} x_{i,0}\\\\ x_{i,1}\\\\ x_{i,2}\\\\ \\vdots\\\\ x_{i,d} \\end{pmatrix}$ Our function $h_{\\mathbf{w}}(\\mathbf{x}_i)$ thus can be written as $\\mathbf{w}^\\intercal\\mathbf{x}_i$, or equivalently, as $\\mathbf{x}_i^\\intercal\\mathbf{w}$. If there are $n$ data points $\\mathbf{x}_1, \\mathbf{x}_2, \\dots, \\mathbf{x}_n$, the outputs corresponding to all these can be kept together in a column matrix as follows: $\\begin{pmatrix} \\mathbf{x}_1^\\intercal\\mathbf{w}\\\\ \\mathbf{x}_2^\\intercal\\mathbf{w}\\\\ \\vdots\\\\ \\mathbf{x}_n^\\intercal\\mathbf{w} \\end{pmatrix}$ Note that this column matrix can be decomposed into the following product $\\begin{pmatrix} \\mathbf{x}_1^\\intercal\\mathbf{w}\\\\ \\mathbf{x}_2^\\intercal\\mathbf{w}\\\\ \\vdots\\\\ \\mathbf{x}_n^\\intercal\\mathbf{w} \\end{pmatrix} = \\begin{pmatrix} —\\mathbf{x}_1^\\intercal— \\\\ —\\mathbf{x}_2^\\intercal—\\\\ \\vdots\\\\ —\\mathbf{x}_n^\\intercal— \\end{pmatrix} \\times \\mathbf{w}$ It is important to keep in mind that even though the last product looks like a scalar multiplication, $\\mathbf{w}$ is in fact a matrix, and that the LHS and RHS are equal nevertheless. Understandably, we call the first matrix in the decomposition $\\mathbf{X}$. It is an $n\\times (d+1)$ matrix, each of whose rows represents a data point. Thus we can compute the outputs $h_\\mathbf{w}(\\mathbf{x}_i)$ corresponding to all the data points $\\mathbf{x}_i$ using the single matrix product $\\mathbf{X}\\mathbf{w}$. ##", "to our system, $$\\mathbf x_1(t) = e^{3t}\\mathbf v_1\\text{;}$$ however, we still need a second solution. Since all other eigenvectors of $$A$$ are a multiple of $$\\mathbf v\\text{,}$$ we cannot find a second linearly independent eigenvector, and we need to obtain the second solution in a different manner. Furthermore, since this system is not partially coupled, we will need a more general strategy. First, we must find a vector $${\\mathbf v}_2$$ such that $$(A - \\lambda I){\\mathbf v}_2 = {\\mathbf v}_1\\text{.}$$ To do this we can start with any nonzero vector $${\\mathbf w}$$ that is not a multiple of $${\\mathbf v}_1\\text{,}$$ say $${\\mathbf w} = (1, 0)\\text{.}$$ We then compute \\begin{equation*} (A - \\lambda I) {\\mathbf w} = (A - 3I) {\\mathbf w} = \\begin{pmatrix} 2 & 1 \\\\ -4 & -2 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ -4 \\end{pmatrix} = 2 {\\mathbf v}_1. \\end{equation*} Thus, we can take $${\\mathbf v}_2 = (1/2)\\mathbf w = (1/2, 0)\\text{,}$$ and our second solution is \\begin{equation*} {\\mathbf x}_2 = e^{\\lambda t} ({\\mathbf v}_2 + t {\\mathbf v}_1) = e^{3t} \\begin{pmatrix} 1/2 + t \\\\ -2t \\end{pmatrix} \\end{equation*} Thus, our general solution is \\begin{equation*} {\\mathbf x} = c_1 {\\mathbf x}_1 + c_2 {\\mathbf x}_2 = c_1 e^{3t} \\begin{pmatrix} 1 \\\\ -2 \\end{pmatrix} + c_2 e^{3t} \\begin{pmatrix} 1/2 + t", "\\\\ a_n b_1 & a_n b_2 & \\cdots & a_n b_n \\\\ \\end{pmatrix}.$$ Matrix product (in terms of inner product) Suppose that the first n×m matrix A is decomposed into its row vectors ai, and the second m×p matrix B into its column vectors bi: [6] $$\\mathbf{A} = \\begin{pmatrix} A_{1 1} & A_{1 2} & \\cdots & A_{1 m} \\\\ A_{2 1} & A_{2 2} & \\cdots & A_{2 m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A_{n 1} & A_{n 2} & \\cdots & A_{n m} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix},\\quad \\mathbf{B} = \\begin{pmatrix} B_{1 1} & B_{1 2} & \\cdots & B_{1 p} \\\\ B_{2 1} & B_{2 2} & \\cdots & B_{2 p} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ B_{m 1} & B_{m 2} & \\cdots & B_{m p} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\cdots & \\mathbf{b}_p \\end{pmatrix}$$ where $$\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$$ The entries in the introduction were given by: $$\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\", "+0 # help 0 114 1 Let $\\mathbf{P}$ be the matrix that projects onto $\\mathbf{j}$: that is, we want $\\mathbf{P}$ to satisfy $\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$for all vectors $\\mathbf{v}$. Use the pictures below to calculate $\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$in that order, and enter them in as columns below. Then calculate the matrix $\\mathbf{P}$ that projects onto $\\mathbf{j}$. Jul 31, 2019 The matrix is $$\\begin{pmatrix} -1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$.", "# Decide a number $a, b, c$ (expressed by $\\lambda$) so that $a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$ See the following vectors in $$\\mathbb{R}^3$$: $$\\mathbf{u}_1= \\begin{pmatrix} 1 \\\\ -1\\\\ 1 \\\\ 1 \\end{pmatrix}$$, $$\\mathbf{u}_2 = \\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\\\ 2 \\end{pmatrix}$$, $$\\mathbf{u}_3 = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\end{pmatrix}$$ and $$\\mathbf{v}_1= \\begin{pmatrix} 5 \\\\ -3\\\\ -4 \\\\ -1 \\end{pmatrix}$$, $$\\mathbf{v}_2 = \\begin{pmatrix} 6 \\\\ 4 \\\\ 1 \\\\ 8 \\end{pmatrix}$$, $$\\mathbf{v}_3 = \\begin{pmatrix} 7 \\\\ 2 \\\\ -1 \\\\ 6 \\end{pmatrix}$$ We are told that $$\\mathcal{B} = (\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3)$$ and $$\\mathcal{C} = (\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3)$$ both are bases for the same subspace $$\\mathcal{U}$$. Let $$\\mathcal{\\lambda}$$ be an unknown number og see the vector $$\\mathbf{x} = \\mathbf{v}_1 + \\lambda\\mathbf{v}_2 - \\mathbf{v}_3$$. Decide a number $$a, b, c$$ (expressed by $$\\lambda$$) so that $$a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$$ 1. Find some $$\\alpha_i$$ such that $$v_1=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3$$ 2. Find some $$\\beta_i$$ such that $$v_2=\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3$$ 3. Find some $$\\gamma_i$$ such that $$v_3=\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3$$ 4. Simplify the expression $$x=v_1+\\lambda v_2-v_3=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3+\\lambda(\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3)-(\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3)$$ • Thanks 5xum! When I reduce it, I get $\\begin{pmatrix} 8 \\\\ 3\\\\ -9 \\\\ 1 \\end{pmatrix} + a\\begin{pmatrix} 2 \\\\ -2\\\\ 1 \\\\ 1 \\end{pmatrix} + b\\begin{pmatrix} 3 \\\\ 6\\\\ 0 \\\\ 6 \\end{pmatrix} + c\\begin{pmatrix} 1 \\\\", "The matrix product itself can be expressed in terms of inner product. Suppose that the first n × m matrix A is decomposed into its row vectors ai, and the second m × p matrix B into its column vectors bi:[1] ${\\displaystyle \\mathbf {A} ={\\begin{pmatrix}A_{11}&A_{12}&\\cdots &A_{1m}\\\\A_{21}&A_{22}&\\cdots &A_{2m}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\A_{n1}&A_{n2}&\\cdots &A_{nm}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}},}$ ${\\displaystyle \\mathbf {B} ={\\begin{pmatrix}B_{11}&B_{12}&\\cdots &B_{1p}\\\\B_{21}&B_{22}&\\cdots &B_{2p}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\B_{m1}&B_{m2}&\\cdots &B_{mp}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\cdots &\\mathbf {b} _{p}\\end{pmatrix}}}$ where ${\\displaystyle \\mathbf {a} _{i}={\\begin{pmatrix}A_{i1}&A_{i2}&\\cdots &A_{im}\\end{pmatrix}}\\,,\\quad \\mathbf {b} _{i}={\\begin{pmatrix}B_{1i}\\\\B_{2i}\\\\\\vdots \\\\B_{mi}\\end{pmatrix}}}$ Then: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\dots &\\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{p})\\\\(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{p})\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{p})\\end{pmatrix}}}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {A} \\mathbf {b} _{1}&\\mathbf {A} \\mathbf {b} _{2}&\\dots &\\mathbf {A} \\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\mathbf {B} \\\\\\mathbf {a} _{2}\\mathbf {B} \\\\\\vdots \\\\\\mathbf {a} _{n}\\mathbf {B} \\end{pmatrix}}}$ These decompositions are particularly useful for matrices that", "gives $q=-5p$. We now know that $\\mathbf{e}= t\\begin{pmatrix}1\\\\-5 \\\\ -2\\end{pmatrix}$. We also know $\\big \\vert\\mathbf{e}\\big \\vert = \\sqrt{30}$ and $p$ is negative, hence $t=-1$ and we have $\\mathbf{e}= \\begin{pmatrix}-1\\\\5 \\\\ 2\\end{pmatrix}$ and $\\mathbf{f}= \\begin{pmatrix}5\\\\2 \\\\-1\\end{pmatrix}$. It might be tempting to proceed with vectors perpendicular to $\\pi$ and $\\pi'$ being $\\mathbf{b} \\times \\mathbf{c}$ and $\\mathbf{e} \\times \\mathbf{f}$. $\\mathbf{b}\\times \\mathbf{c} = \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ 2&-1&-1 \\\\ -1&2&1\\end{matrix}\\right \\vert = \\mathbf{i}-\\mathbf{j}+3\\mathbf{k}.$ $\\mathbf{e}\\times \\mathbf{f}= \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ p&q&r \\\\ q&r&p\\end{matrix}\\right \\vert = (qp-r^2)\\mathbf{i}+(rq-p^2)\\mathbf{j}+(pr-q^2)\\mathbf{k}.$ However, the resulting set of simultaneous equations is very difficult to solve. By adding them we can then simplify using $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ in other words $p^2+q^2+r^2=30$. We can think about $(p+q+r)^2$ to continue the manipulation. Try it if you like! Show that $\\pi'$ is the reflection of $\\pi$ in the origin. Using the form of the plane equation $\\mathbf{r}.\\mathbf{n}=\\mathbf{a}.\\mathbf{n}$ with $\\mathbf{n}$ as the normal vector we found earlier, we have the equation of $\\pi$ as $x - y + 3z = -2.$ Similarly using the point $\\mathbf{d}$, the plane $\\pi'$ is $x - y + 3z = 2.$ If $(x_1,y_1,z_1)$ is a point on $\\pi$ then $x_1 - y_1 + 3z_1 = -2$. Multiplying by $-1$ gives $(-x_1) -(-y_1) + 3(-z_1) = 2$. So the point $(-x_1,-y_1,-z_1)$ lies on $\\pi'$. Since every point on", "\\end{pmatrix}^k = \\begin{pmatrix} A_{11}^k & 0 & \\cdots & 0 \\\\ 0 & A_{22}^k & \\cdots & 0 \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ 0 & 0 & \\cdots & A_{nn}^k \\end{pmatrix}$$ meaning its easy to raise a diagonal matrix to a power. When raising an arbitrary matrix (not necessarily a diagonal matrix) to a power, it is often helpful to exploit this property by diagonalizing the matrix first. The inner and outer products Given two column vectors a and b, the Euclidean inner product and outer product are the simplest special cases of the matrix product, by transposing the column vectors into row vectors.[5] The inner product is a column vector multiplied on the left by a row vector: $$\\mathbf{a}\\cdot \\mathbf{b} = \\mathbf{a}^\\mathrm{T}\\mathbf{b}$$ More explicitly, $$\\mathbf{a}\\cdot \\mathbf{b} = \\begin{pmatrix}a_1 & a_2 & \\cdots & a_n\\end{pmatrix} \\begin{pmatrix}b_1 \\\\ b_2 \\\\ \\vdots \\\\ b_n\\end{pmatrix} = a_1b_1+a_2b_2+\\cdots+a_nb_n = \\sum_{i=1}^n a_ib_i$$ The outer product is a row vector multiplied on the left by a column vector: $$\\mathbf{a}\\otimes \\mathbf{b} = \\mathbf{a}\\mathbf{b}^\\mathrm{T}$$ where $$\\mathbf{a}\\mathbf{b}^\\mathrm{T} = \\begin{pmatrix}a_1 \\\\ a_2 \\\\ \\vdots \\\\ a_n\\end{pmatrix} \\begin{pmatrix}b_1 & b_2 & \\cdots & b_n\\end{pmatrix} = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & \\cdots & a_1 b_n \\\\ a_2 b_1 & a_2 b_2 & \\cdots & a_2 b_n \\\\ \\vdots & \\vdots & \\ddots & \\vdots", "in each row by brackets: $U = \\begin{pmatrix} (1) & 0 & 2 & 0 & 1 \\\\ 0 & (1) & -1 & 0 & 1 \\\\ 0 & 0 & 0 & (1) & 1 \\\\ 0 & 0 & 0 & 0 & 0 \\end{pmatrix}$ Clearly the basic variables are $x_1, x_2$ and $x_4$ and the free variables are $x_3$ and $x_5$. So by the theorem, the basis of the column space of A consists of the columns $c_1, c_2$ and $c_4$ and is given by: $\\Bigg\\{ \\begin{pmatrix} 1\\\\ -2\\\\ 0\\\\ 3 \\end{pmatrix},\\begin{pmatrix} 2\\\\ -5\\\\ -3\\\\ 6 \\end{pmatrix},\\begin{pmatrix} 2\\\\ -1\\\\ 4\\\\ -7 \\end{pmatrix} \\Bigg\\}$ Note that the nullity of A is 2 which is equal to the difference between the total number of columns and the number of elements in the column space. ## The Row Space The row space, as, the name suggests is the space built out by the rows of the matrix. Given an m x n matrix A the row space of A, denoted by RS(A), is defined as the collection of all the vectors formed by linear combinations of the rows of A. More precisely if the rows of A are $r_1,r_2$...$r_m$, where each $r_i$ is an n-dimensional vector then, $\\hbox{RS}(A) = \\{\\mathbf{v} \\in \\mathbb{R}^n : \\mathbf{v} = \\mathbf{\\alpha_1 r_1", "\\\\ 0 & 0 \\end{pmatrix}} -c {\\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\end{pmatrix}} \\\\ &= bE_3 - cE_2 \\end{align*} How do I continu to find $M_{E_1}$? Remember that given your basis $\\xi$, for $B \\in M_2(\\mathbb{R})$, $$[B]_\\xi = \\begin{pmatrix}B_{11} \\\\ B_{21} \\\\ B_{12} \\\\ B_{22} \\end{pmatrix};$$ in particular, for $A \\in M_2(\\mathbb{R})$, $M_A := [L_A]_{\\xi \\xi}$ is the matrix of the linear transformation $L_A : M_2(\\mathbb{R}) \\to M_2(\\mathbb{R})$ with respect to $\\xi$, and hence a $4 \\times 4$ matrix. Indeed, by introductory linear algebra, $M_A$ will be the $4 \\times 4$ matrix whose $k$-th column is $[L_A(E_k)]_\\xi$, as given by the equation above. So, given $A \\in M_2(\\mathbb{R})$, all you really need to do is compute $L_A(E_k)$ for each $k$, and express it as a linear combination of the $E_j$. Hint:do you know that the matrix you are looking for is $4\\times4$? Now you are looking for a $4\\times4$ matrix such that $M_{E_1}[X]_\\xi=[L_{E_1}(X)]_\\xi$. In your case, as you calculated, $[L_{E_1}(X)]_\\xi=(0,b,-c,0)$ column vector, while $[X]_\\xi=(a,b,c,d)$ column. Do you see how to proceed?", "# Final review 1-2 ## General instructions Refer to the hints I've given before Midterm 1. It's very important to regularly check how much time you have left! If you can't see a clock, ask a proctor to write the remaining time on the board regularly. Make sure to practice timing yourself using the practice finals on OWL. Set aside 3 hours, and use no notes, textbooks, calculators. ## Geometry with vectors For $$n$$-vectors $$\\mathbf u$$ and $$\\mathbf v$$, the dot product is the matrix product if $$\\mathbf u$$ is represented as a row vector, and $$\\mathbf v$$ is represented as a column vector: $\\mathbf u\\cdot\\mathbf v=\\begin{pmatrix} u_1 & \\dots & u_n\\end{pmatrix}\\begin{pmatrix} v_1 \\\\ \\vdots \\\\ v_n\\end{pmatrix}=u_1v_1+\\dotsb+u_nv_n.$ The length of an $$n$$-vector $$\\mathbf u$$ is the square root of its dot product with itself: $\\|\\mathbf u\\|=\\sqrt{\\mathbf u\\cdot\\mathbf u}=\\sqrt{u_1^2+\\dotsb+u_n^2}.$ Example. Let $$\\mathbf u=(2,-1,3)$$. Then the unit vector with the same direction as $$\\mathbf u$$ is $\\frac{1}{\\|\\mathbf u\\|}\\mathbf u=\\frac{1}{\\sqrt{14}}\\begin{pmatrix}2\\\\-1\\\\3\\end{pmatrix}.$ The \"distance of $$\\mathbf u$$ and $$\\mathbf v$$\" is the distance of $$P$$ and $$Q$$, where $$\\mathbf u=\\vec{OP}$$ and $$\\mathbf v=\\vec{OQ}$$. That is, $d(\\mathbf u,\\mathbf v)=\\|\\mathbf u-\\mathbf v\\|=\\sqrt{(u_1-v_1)^2+\\dotsb+(u_n-v_n)^2}.$ Example. The distance of $$\\mathbf u=(2,-1,3)$$ and $$\\mathbf v=(0,-2,1)$$ is $\\|\\mathbf u-\\mathbf v\\|=\\|(2,1,2)\\|=3.$ Suppose that $$\\boldsymbol0\\ne\\mathbf u,\\mathbf v$$, and let $$\\alpha$$ be the angle of $$\\mathbf u$$ and $$\\mathbf v$$. Then we have", "Matrix $M_1 = M(f', \\mathcal{B}, \\mathcal{B '})$ of $f$ $f: \\mathbb{R}^5 \\rightarrow \\mathcal{M}_{2}(\\mathbb{R})$ $(x_1,x_2,x_3,x_4,x_5) \\rightarrow \\begin{pmatrix} x_1 & x_2 \\\\ x_1 + x_3 & x_5 \\end{pmatrix}$ 1. Determine the basis of $Kerf$ 2. Determine a basis of $Imf$ 3. Determine the matrix $M_1 = M(f', \\mathcal{B}, \\mathcal{B '})$ of $f$ 1 . $x = (x_1,x_2,x_3,x_4,x_5) \\in Kerf \\iff f(x_1,x_2,x_3,x_4,x_5) = \\begin{pmatrix} x_1 & x_2 \\\\ x_1 + x_3 & x_5 \\end{pmatrix} =\\begin{pmatrix} 0 & 0 \\\\ 0 & 0\\end{pmatrix}$ I got $(x_1,x_2,x_3,x_4,x_5) = (0,0,0,x_4,0)$. A base of $Kerf$ is $(0,0,0,1,0)$. 1. $Imf^= Vect <f(e_1), f(e_2), f(e_3), f(e_4), f(e_5) > = <E_{11}, E_{12}, E_{21}, E_{22}>$ So the family $(E_{11}, E_{12}, E_{21}, E_{22})$ is basis. with $E_{ij}$ is the elementary matrix. The coordinate $x_4$ is not on the matrix. I need help with question 3, too. Thank you. • $f$ should be a map from $\\mathbb{R}^5$. – tattwamasi amrutam Jul 11 '17 at 21:29 • For starters: you have ignored $x_3$ in the kernel. So it should be $(0,0,x_3,x_4,0) \\in K$. This mean kernel has dimension $2$. Likewise the image also needs fixing. Because the image can only have dimension $3$. – Anurag A Jul 11 '17 at 21:31 • @AnuragA I just made a mistake typing. I have just corrected it. Thank you for noticing. – Zouhair El Yaagoubi", "column vectors $E^1,\\cdots,E^n$ of $\\mathbb{R}^n$. For each $j=1,\\cdots,n$, let $L(E^j)=A^j$, where $A^j$ is a column vector in $\\mathbb{R}^m$. For each $X\\in\\mathbb{R}^n$, $$X=x_1E^1+\\cdots+x_nE^n=\\begin{pmatrix} x_1\\\\ \\vdots\\\\ x_n \\end{pmatrix}$$ and hence \\begin{align*} LX&=x_1L(E^1)+\\cdots+x_nL(E^n)\\\\ &=x_1A^1+\\cdots+x_nA^n\\\\ &=AX \\end{align*} where $A$ is the matrix whose column vectors are $A^1,\\cdots,A^n$. The matrix $A$ is called the matrix associated with the linear map $L$. Example. Let $L:\\mathbb{R}^3\\longrightarrow\\mathbb{R}^2$ be the projection given by $$L\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix}=\\begin{pmatrix} x\\\\ y \\end{pmatrix}.$$ Find the matrix associated with $L$. Solution. $$L(E^1)=\\begin{pmatrix} 1\\\\ 0\\end{pmatrix},\\ L(E^2)=\\begin{pmatrix} 0\\\\ 1 \\end{pmatrix},\\ L(E^3)=\\begin{pmatrix} 0\\\\ 0 \\end{pmatrix}.$$ Thus, the matrix associated with $L$ is $$A=\\begin{pmatrix} 1 & 0 & 0\\\\ 0 & 1 & 0 \\end{pmatrix}.$$ Let us now consider a more general case. Let $V$ be a vector space and $\\{v_1,\\cdots,v_n\\}$ a given basis of $V$. Let $L:V\\longrightarrow V$ be a linear map. Then there exist numbers $c_{ij}$, $i,j=1,\\cdots,n$ such that \\begin{align*} Lv_1&=c_{11}v_1+\\cdots+c_{1n}v_n,\\\\ &\\vdots\\\\ Lv_n&=c_{n1}v_1+\\cdots+c_{nn}v_n. \\end{align*} Let $v=x_1v_1+\\cdots+x_nv_n$. Then \\begin{align*} Lv&=\\sum_{i=1}^nx_iLv_i\\\\ &=\\sum_{i=1}^nx_i\\sum_{j=1}^nc_{ij}v_j\\\\ &=\\sum_{j=1}^n(\\sum_{i=1}^nx_ic_{ij})v_j. \\end{align*} Hence, we have the following theorem. Theorem. If $C=(c_{ij})$ is the matrix such that $Lv_i=\\sum_{j=1}^nc_{ij}v_j$ and $X=\\begin{pmatrix} x_1\\\\ \\vdots\\\\ x_n \\end{pmatrix}$ is the coordinate vector of $v$, then the coordinate vector of $Lv$ is ${}^tCX$ i.e. the matrix associated with $L$ is ${}^tC$ with respect to the basis $\\{v_1,\\cdots,v_n\\}$. Example. Let $L: V\\longrightarrow V$ be a linear map. Let $\\{v_1,v_2,v_3\\}$", "column vectors $E^1,\\cdots,E^n$ of $\\mathbb{R}^n$. For each $j=1,\\cdots,n$, let $L(E^j)=A^j$, where $A^j$ is a column vector in $\\mathbb{R}^m$. For each $X\\in\\mathbb{R}^n$, $$X=x_1E^1+\\cdots+x_nE^n=\\begin{pmatrix} x_1\\\\ \\vdots\\\\ x_n \\end{pmatrix}$$ and hence \\begin{align*} LX&=x_1L(E^1)+\\cdots+x_nL(E^n)\\\\ &=x_1A^1+\\cdots+x_nA^n\\\\ &=AX \\end{align*} where $A$ is the matrix whose column vectors are $A^1,\\cdots,A^n$. The matrix $A$ is called the matrix associated with the linear map $L$. Example. Let $L:\\mathbb{R}^3\\longrightarrow\\mathbb{R}^2$ be the projection given by $$L\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix}=\\begin{pmatrix} x\\\\ y \\end{pmatrix}.$$ Find the matrix associated with $L$. Solution. $$L(E^1)=\\begin{pmatrix} 1\\\\ 0\\end{pmatrix},\\ L(E^2)=\\begin{pmatrix} 0\\\\ 1 \\end{pmatrix},\\ L(E^3)=\\begin{pmatrix} 0\\\\ 0 \\end{pmatrix}.$$ Thus, the matrix associated with $L$ is $$A=\\begin{pmatrix} 1 & 0 & 0\\\\ 0 & 1 & 0 \\end{pmatrix}.$$ Let us now consider a more general case. Let $V$ be a vector space and $\\{v_1,\\cdots,v_n\\}$ a given basis of $V$. Let $L:V\\longrightarrow V$ be a linear map. Then there exist numbers $c_{ij}$, $i,j=1,\\cdots,n$ such that \\begin{align*} Lv_1&=c_{11}v_1+\\cdots+c_{1n}v_n,\\\\ &\\vdots\\\\ Lv_n&=c_{n1}v_1+\\cdots+c_{nn}v_n. \\end{align*} Let $v=x_1v_1+\\cdots+x_nv_n$. Then \\begin{align*} Lv&=\\sum_{i=1}^nx_iLv_i\\\\ &=\\sum_{i=1}^nx_i\\sum_{j=1}^nc_{ij}v_j\\\\ &=\\sum_{j=1}^n(\\sum_{i=1}^nx_ic_{ij})v_j. \\end{align*} Hence, we have the following theorem. Theorem. If $C=(c_{ij})$ is the matrix such that $Lv_i=\\sum_{j=1}^nc_{ij}v_j$ and $X=\\begin{pmatrix} x_1\\\\ \\vdots\\\\ x_n \\end{pmatrix}$ is the coordinate vector of $v$, then the coordinate vector of $Lv$ is ${}^tCX$ i.e. the matrix associated with $L$ is ${}^tC$ with respect to the basis $\\{v_1,\\cdots,v_n\\}$. Example. Let $L: V\\longrightarrow V$ be a linear map. Let $\\{v_1,v_2,v_3\\}$" ]

[ "Free Version Moderate # 2x2 Matrix Invertible with Parameter 2 LINALG-CDVRSM Pick the real number value of $b$, if it exists, such that the matrix: $$\\begin{pmatrix}b&1\\\\\\ -1&b+1\\end{pmatrix}$$ ...is not invertible. A $-1$ B $0$ C $1$ D $2$ E This matrix is invertible for any real value of $b$.", "$$\\pmatrix{2 & 1 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 3}$$ Note that in all these cases the matrix $A^2 -7A+16I_3$ is invertible because the polynomial $x^2-7x+16$ has complex zeros, so $\\frac1{12}A=[A^2-7A+16I_3]^{-1}$ is indeed equivalent to $A^3-7A^2+16A-12I_3=0$.", "Free Version Moderate # 2x2 Matrix Invertible with Parameter 1 LINALG-SLNB7V For which of the following values of $b$ is the matrix: $$\\begin{pmatrix}\\sqrt{3}&b\\\\\\ b+2&\\sqrt{3}\\end{pmatrix}$$ ...NOT invertible? Select ALL that apply. A $-3$ B $-2$ C $0$ D $1$", "Browse Questions # For what value of $k$ , the matrix $\\begin{bmatrix} 2-k & 3 \\\\ -5 & 1 \\end{bmatrix}$ is not invertible? Can you answer this question?", "insight into their nature. One particularly elegant representation interprets every complex number as 22 matrix with real entries which stretches and rotates the points of the plane. Every such matrix has the form [itex] \\begin{pmatrix} a & -b \\\\ b & \\;\\; a \\end{pmatrix} [itex] with real numbers a and b. The sum and product of two such matrices is again of this form. Every non-zero such matrix is invertible, and its inverse is again of this form. Therefore, the matrices of this form are a field. In fact, this is exactly the field of complex numbers. Every such matrix can be written as [itex] \\begin{pmatrix} a & -b \\\\ b & \\;\\; a \\end{pmatrix} = a \\begin{pmatrix} 1 & \\;\\; 0 \\\\ 0 & \\;\\; 1 \\end{pmatrix} + b \\begin{pmatrix} 0 & -1 \\\\ 1 & \\;\\; 0 \\end{pmatrix} [itex] which suggests that we should identify the real number 1 with the matrix [itex] \\begin{pmatrix} 1 & \\;\\; 0 \\\\ 0 & \\;\\; 1 \\end{pmatrix} [itex] and the imaginary unit i with [itex] \\begin{pmatrix} 0 & -1 \\\\ 1 & \\;\\; 0 \\end{pmatrix} [itex] a counter-clockwise rotation by 90 degrees. Note that the square of this latter matrix is indeed equal to −1. The absolute value of a complex number expressed as a matrix is equal to", "\\ma{B}_{\\rm I})$ are both invertible. *Update* We found a simpler, solution, please read the follow-up blog post on the closed-form solution that avoids real-valued stacking.", "# Show that the $5\\times 5$ matrix is not invertable Find $a$ such that the matrix $$\\begin{pmatrix} 0 & a & 0 & 0 & 0 \\\\ b & 0 & c & 0 & 0 \\\\ 0 & d & 0 & e & 0 \\\\ 0 & 0 & f & 0 & g \\\\ 0 & 0 & 0 & h & 0 \\end{pmatrix}$$ is not invertable, for all values of $a,b,x,d,e,f,g,h$.", "# Inverse matrix Algebra Level 4 If matrix $$A=\\begin{pmatrix} a & b \\\\ c & a \\end{pmatrix}$$ satisfies $$A={A}^{-1},$$ what are all possible values of real number $$x$$ such that the matrix $$A-xI$$ is not invertible? Details and assumptions $$I$$ denotes the identity matrix. ×", "# Find $x$ so that $\\begin{pmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ -\\frac{1}{a} & x & x^2 \\end{pmatrix}$ is invertible Find $x$ so that the matrix $\\begin{pmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ -\\frac{1}{a} & x & x^2 \\end{pmatrix}$ is invertible. Please show your work. Let $A=\\begin{pmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ -\\frac{1}{a} & x & x^2 \\end{pmatrix}.$ Then $\\frac{1}{a}R_1+R_3 \\rightarrow R_3$ gives $\\begin{pmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ 0 & x & x^2+\\frac{c}{a} \\end{pmatrix},$ and $-\\frac{x}{a}R_2+R_3 \\rightarrow R_3$ gives $\\begin{pmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ 0 & 0 & \\frac{b}{a}x+x^2+\\frac{c}{a} \\end{pmatrix}.$ Hence $\\text{det}(A)=a(\\frac{b}{a}x+x^2+\\frac{c}{a})=0$ $\\Rightarrow ax^2+b x+c=0$ $\\Rightarrow x=\\frac{-b \\pm \\sqrt{b^2-4ac}}{2a}.$ Hence the matrix $A$ is invertible if and only if $x\\neq\\frac{-b \\pm \\sqrt{b^2-4ac}}{2a}.$", "in what context can I apply this trick? Thanks If $B \\in \\mathbb R^{(m+n)\\times(m+n)}$ is of the form $$B = \\pmatrix{B_{11}&0\\\\0&B_{22}}$$ with $B_{11}\\in\\mathbb R^{m\\times m}$ and $B_{22}\\in\\mathbb R^{n\\times n}$ both invertible, then $B$ is invertible with $$B^{-1} = \\pmatrix{B_{11}^{-1}&0\\\\0&B_{22}^{-1}}$$ To prove it, note that $$\\pmatrix{A_{11}&0\\\\0&A_{22}}\\pmatrix{B_{11}&0\\\\0&B_{22}}=\\pmatrix{A_{11}B_{11}&0\\\\0&A_{22}B_{22}}$$", "enormous practical use in applied mathematics as well as in other branches of mathematics. Often, the most natural proofs for statements in real analysis or even number theory employ techniques from complex analysis (see prime number theorem for an example). Unlike real functions which are commonly represented as two dimensional graphs, complex functions have four dimensional graphs and may usefully be illustrated by color coding a three dimensional graph to suggest four dimensions, or by animating the complex function's dynamic transformation of the complex plane. While usually not useful, alternative representations of complex field can give some insight into their nature. One particularly elegant representation interprets every complex number as 2x2 matrix with real entries which stretches and rotates the points of the plane. Every such matrix has the form $\\begin{pmatrix} a&&-b\\\\ b&&a \\end{pmatrix}$ with real numbers a and b. The sum and product of two such matrices is again of this form. Every non-zero such matrix is invertible, and its inverse is again of this form. Therefore, the matrices of this form are a field. In fact, this is exactly the field of complex numbers. Every such matrix can be written as $\\begin{pmatrix} a&&-b\\\\ b&&a \\end{pmatrix} a \\begin{pmatrix} 1&&0\\\\ 0&&1 \\end{pmatrix} + b \\begin{pmatrix} 0&&-1\\\\ 1&&0 \\end{pmatrix}$ which suggests that we should identify the real number 1", "# Invertible Matrix ## Invertible Matrix A square matrix A of order n is invertible if there exists a square matrix B of order n such that AB = BA = I, an identity matrix. ### Example Consider the matrices $$A = \\begin{pmatrix}–3 & 5 & 6\\\\–1 & 2 & 2\\\\1 & – 1 & – 1 \\end{pmatrix}$$ and $$B = \\begin{pmatrix}0 & 1 & 2\\\\–1 & 3 & 0\\\\1 & –2 & 1\\end{pmatrix}$$. The matrix A is invertible, since : $$\\begin{pmatrix}–3 & 5 & 6\\\\–1 & 2 & 2\\\\1 & – 1 & – 1\\end{pmatrix}$$ × $$\\begin{pmatrix}0 & 1 & 2\\\\–1 & 3 & 0\\\\1 & –2 & 1\\end{pmatrix}$$ = $$\\begin{pmatrix}0 & 1 & 2\\\\–1 & 3 & 0\\\\1 & –2 & 1\\end{pmatrix}$$ × $$\\begin{pmatrix}–3 & 5 & 6\\\\–1 & 2 & 2\\\\1 & – 1 & – 1\\end{pmatrix} = \\begin{pmatrix}1 & 0 & 0\\\\0 & 1 & 0\\\\ 0 & 0 & 1\\end{pmatrix}$$", "careful: the matrices $$\\begin{pmatrix} a & 0 \\\\ 0 & 0 \\end{pmatrix}$$ are not invertible, hence are not elements of $GL_2(\\mathbb R)$. However, the set of matrices of this type $$\\left\\{ \\begin{pmatrix} a & 0 \\\\ 0 & 0 \\end{pmatrix} : \\ a \\in \\mathbb R, \\ a\\ne 0 \\right\\}$$ is an Abelian group under matrix multiplication. Its identity is not the usual identity $I_2$...", "of B is \\[\\Adj(B)=C^{\\trans}=\\begin{bmatrix} 1 & 0 & -2 \\\\ 12 &-5 &-4 \\\\ -3 & 0 & 1 \\end{bmatrix}. It follows from the formula that the inverse matrix of $B$ is $B^{-1}=\\frac{1}{\\det(B)}\\Adj(B)=\\frac{1}{5}\\begin{bmatrix} -1 & 0 & 2 \\\\ -12 &5 &4 \\\\ 3 & 0 & -1 \\end{bmatrix}.$ ### More from my site • For Which Choices of $x$ is the Given Matrix Invertible? Determine the values of $x$ so that the matrix $A=\\begin{bmatrix} 1 & 1 & x \\\\ 1 &x &x \\\\ x & x & x \\end{bmatrix}$ is invertible. For those values of $x$, find the inverse matrix $A^{-1}$. Solution. We use the fact that a matrix is invertible […] • Inverse Matrix Contains Only Integers if and only if the Determinant is $\\pm 1$ Let $A$ be an $n\\times n$ nonsingular matrix with integer entries. Prove that the inverse matrix $A^{-1}$ contains only integer entries if and only if $\\det(A)=\\pm 1$. Hint. If $B$ is a square matrix whose entries are integers, then the […] • Find All Values of $x$ such that the Matrix is Invertible Given any constants $a,b,c$ where $a\\neq 0$, find all values of $x$ such that the matrix $A$ is invertible if $A= \\begin{bmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ -1/a &", "really useful. Alternatively, viewing $A+BA^{-1}B$ as a Schur complement, one can obtain another equivalent condition: both $A$ and $\\pmatrix{A&-B\\\\ B&A}$ are invertible.", "NOT a real zero. $$\\begin{matrix} -1 & | & 2 & -7 & 3 & 8 & -4\\\\ & | & & -2 & 9 & -12 & 4\\\\ & | & - & - & - & - & -\\\\ & | & 2 & -9 & 12 & -4 & 0 \\end{matrix}$$ $$-1$$ IS a real zero. $$\\begin{matrix} 2 & | & 2 & -7 & 3 & 8 & -4\\\\ & | & & 4 & -6 & -6 & 4\\\\ & | & - & - & - & - & -\\\\ & | & 2 & -3 & -3 & 2 & 0 \\end{matrix}$$ $$2$$ IS a real zero. $$\\begin{matrix} -2 & | & 2 & -7 & 3 & 8 & -4\\\\ & | & & -4 & 22 & -50 & 84\\\\ & | & - & - & - & - & -\\\\ & | & 2 & -11 & 25 & -42 & 80 \\end{matrix}$$ $$-2$$ is NOT a real zero. $$\\begin{matrix} 4 & | & 2 & -7 & 3 & 8 & -4\\\\ & | & & 8 & 4 & 28 & 144\\\\ & | & - & - & - & - & -\\\\ & | & 2 & 1 & 7 & 36 &", "# Is there a matrix of every size with all its submatrices invertible? Let's call a real matrix of size $m \\times n$ totally invertible if for every $k$ rows and $k$ columns that we choose, we get an invertible matrix. I am curious about the following: Is there a totally invertible matrix for all sizes $m \\times n$? By taking the transpose, we can assume $m \\leq n$. For $m=1$ we can take any vector in $\\Bbb R^m$ without a zero entry. For $m=2$ we can take $\\begin{pmatrix} 1 &2 & 3 & ... &n \\\\ 2 & 3 & 4 & ... & n+1 \\end{pmatrix}$. Indeed, $\\det\\begin{pmatrix} a &b \\\\ a+1 & b+1 \\end{pmatrix}=a-b$ is nonzero when $a \\neq b$. This does not generalize, since $a_{ij}=i+j-1$ fabulously fails for all submatrices of size $\\geq 3$. I also tried taking the rows to be $m$ of the cyclic shifts of the vector $(1,2, ... ,n)$. This does not work either because I found a $k=3, m=n=5$ counterexample. I do feel that the answer should be positive, however. Is it? • Well if the entries are random selected randomly from $[0,1]$, the probability that any of the square submatrices will be singular is zero. So yes, there are plenty of examples. – Aweygan Aug 8 '16 at 19:06", "$0, 1$, or $-1$ any real number If the matrix $A$ is such that $A= \\begin{bmatrix} 2\\\\ −4\\\\7\\end{bmatrix}\\begin{bmatrix}1& 9& 5\\end{bmatrix}$ then the determinant of $A$ is equal to ______.", "exists an invertible matrix $A$ such that $A[1\\ \\sqrt{3}]^\\mathsf{T} = [\\alpha\\ \\beta]^\\mathsf{T}$. As before, $\\alpha = a+b\\sqrt{3}$ and $\\beta = c+d\\sqrt{3}$ where $ad-bc = \\pm 1$.", "\\over c^2 + d^2}\\right) + i\\left( {bc - ad \\over c^2 + d^2} \\right).$ ### Matrix representation of complex numbersEdit While usually not useful, alternative representations of complex fields can give some insight into their nature. One particularly elegant representation interprets every complex number as 2×2 matrix with real entries which stretches and rotates the points of the plane. Every such matrix has the form $\\begin{pmatrix} a & -b \\\\ b & \\;\\; a \\end{pmatrix}$ with real numbers a and b. The sum and product of two such matrices is again of this form. Every non-zero such matrix is invertible, and its inverse is again of this form. Therefore, the matrices of this form are a field. In fact, this is exactly the field of complex numbers. Every such matrix can be written as $\\begin{pmatrix} a & -b \\\\ b & \\;\\; a \\end{pmatrix} = a \\begin{pmatrix} 1 & \\;\\; 0 \\\\ 0 & \\;\\; 1 \\end{pmatrix} + b \\begin{pmatrix} 0 & -1 \\\\ 1 & \\;\\; 0 \\end{pmatrix}$ which suggests that we should identify the real number 1 with the matrix $\\begin{pmatrix} 1 & \\;\\; 0 \\\\ 0 & \\;\\; 1 \\end{pmatrix}$ and the imaginary unit i with $\\begin{pmatrix} 0 & -1 \\\\ 1 & \\;\\; 0 \\end{pmatrix}$ a counter-clockwise rotation by 90 degrees. Note that the" ]

[ "A determinant of the right matrix is equal to $$0+2abc-0=2abc$$ and we are done! Hint: $$\\triangle=\\begin{vmatrix}b^2+c^2&ab&ac\\\\ba&c^2+a^2&bc\\\\ca&cb&a^2+b^2\\end{vmatrix} = \\dfrac1a\\begin{vmatrix}a(b^2+c^2)&ab&ac\\\\a(ba)&c^2+a^2&bc\\\\a(ca)&cb&a^2+b^2\\end{vmatrix}$$ $$C_1'=C_1-bC_2-cC_3$$ $$\\triangle=\\dfrac1a\\begin{vmatrix}0&ab&ac\\\\-2bc^2&c^2+a^2&bc\\\\-2bc^2&cb&a^2+b^2\\end{vmatrix}=-2bc\\begin{vmatrix}0&b&c\\\\c&c^2+a^2&bc\\\\b&cb&a^2+b^2\\end{vmatrix}=2bc\\begin{vmatrix}0&c&b\\\\c&bc&c^2+a^2\\\\b&a^2+b^2&cb\\end{vmatrix}$$ $$=2\\begin{vmatrix}0&bc&bc\\\\c&b^2c&(c^2+a^2)c\\\\b&b(a^2+b^2)&(bc)c\\end{vmatrix}$$ $$=2bc(c)(b)\\begin{vmatrix}0&1&1\\\\1&b^2&c^2+a^2\\\\1&a^2+b^2&(c)c\\end{vmatrix}$$ Let $$b=ax,c=axy$$. Then: $$\\small\\begin{vmatrix}b^2+c^2&ab&ac\\\\ba&c^2+a^2&bc\\\\ca&cb&a^2+b^2\\end{vmatrix} =\\begin{vmatrix}a^2x^2(1+y^2)&a^2x&a^2xy\\\\a^2x&a^2(1+x^2y^2)&a^2x^2y\\\\a^2xy&a^2x^2y&a^2(1+x^2)\\end{vmatrix} =a^6x^2\\begin{vmatrix}1+y^2&1&y\\\\1&1+x^2y^2&x^2y\\\\y&x^2y&1+x^2\\end{vmatrix} =\\\\ \\small\\stackrel{R_1-R_2\\to R_1}= a^6x^2\\begin{vmatrix}y^2&-x^2y^2&y-x^2y\\\\1&1+x^2y^2&x^2y\\\\y&x^2y&1+x^2\\end{vmatrix} =a^6x^2y\\begin{vmatrix}y&-x^2y&1-x^2\\\\1&1+x^2y^2&x^2y\\\\y&x^2y&1+x^2\\end{vmatrix} \\stackrel{R_3-R_1\\to R_3}=\\\\ \\small a^6x^2y\\begin{vmatrix}y&-x^2y&1-x^2\\\\1&1+x^2y^2&x^2y\\\\0&2x^2y&2x^2\\end{vmatrix}= 2a^6x^4y\\begin{vmatrix}y&-x^2y&1-x^2\\\\1&1+x^2y^2&x^2y\\\\0&y&1\\end{vmatrix}=\\\\ \\small =2a^6x^4y\\cdot [y(1+x^2y^2-x^2y^2)-(-x^2y-y(1-x^2)]=4a^6x^4y^2=4a^2b^2c^2.$$", "another determinant $$\\begin{vmatrix}a & b & c \\\\b & c & a \\\\c & a & b \\end {vmatrix}$$? a) (a3 + b3 + c3 + 3abc)2 b) –(a3 + b3 + c3 + 3abc)2 c) (a3 + b3 + c3 – 3abc)2 d) –(a3 + b3 + c3 – 3abc)2 Explanation: Now, $$\\begin{vmatrix}a & b & c \\\\b & c & a \\\\c & a & b \\end {vmatrix}$$ Interchanging 2nd and 3rd columns, = –$$\\begin{vmatrix}a & c & b \\\\b & a & c \\\\c & b & a \\end {vmatrix}$$ = $$\\begin{vmatrix}-a & c & b \\\\ -b & a & c \\\\-c & b & a \\end {vmatrix}$$ So, $$\\begin{vmatrix}a & b & c \\\\b & c & a \\\\c & a & b \\end {vmatrix}^2$$ = $$\\begin{vmatrix}a & b & c \\\\b & c & a \\\\c & a & b \\end {vmatrix}$$$$\\begin{vmatrix} -a & c & b \\\\ -b & a & c \\\\-c & b & a \\end {vmatrix}$$ = {–(a3 + b3 + c3 – 3abc)}2 = $$\\begin{vmatrix}-a^2 + bc + bc & -ab + ab + c^2 & -ac + b^2 + ca \\\\-ab + ab + c^2 & -ac – b^2 + ca & -a^2 + bc + bc \\\\-ac + b^2 + ca & -a^2 + bc", "by $2$. This requires us to also multiply by $1/2$. Then our determinant $D$ is equal to $$D=\\frac12\\,\\begin{vmatrix} 2b-2a& bc+bd-ad-ac&2bcd-2acd\\\\ 2c-2d& ac+bc-ad-bd& 2abc-2abd\\\\ 2b-2c&ab+bd-ac-cd&2abd-2acd \\end{vmatrix}.$$ Factoring in each entry, we get $$D=\\frac12\\,\\begin{vmatrix} 2(b-a)& (b-a)(c+d)&2cd(b-a)\\\\ 2(c-d)& (a+b)(c-d)& 2ab(c-d)\\\\ 2(b-c)&(a+d)(b-c)&2ad(b-c) \\end{vmatrix}.$$ Extracting the common factor in each row, $$D=\\frac12\\,(b-a)(c-d)(b-c)\\begin{vmatrix} 2& c+d&2cd\\\\ 2& a+b& 2ab\\\\ 2&a+d&2ad \\end{vmatrix}.$$ Now replace $R_2$ with $R_2-R_3$ and $R_3$ with $R_3-R_1$: $$D=\\frac12\\,(b-a)(c-d)(b-c)\\begin{vmatrix} 2& c+d&2cd\\\\ 0& b-d& 2a(b-d)\\\\ 0&a-c&2d(a-c) \\end{vmatrix}.$$ Now we can calculate directly. But if we first extract the common factor from each of the last two rows: $$D=\\frac12\\,(b-a)(c-d)(b-c)(b-d)(a-c)\\begin{vmatrix} 2& c+d&2cd\\\\ 0& 1& 2a\\\\ 0&1&2d \\end{vmatrix}.$$ Now the remaining determinant trivially seen to be $4(d-a)$, so $$D=2(b-a)(c-d)(b-c)(b-d)(a-c)(d-a).$$", "-t & -1\\\\ 0 & 1 & -1 – t \\end{vmatrix}}$$ $$=(3 – t) {\\begin{vmatrix} -t & -1 \\\\ 1 & -1-t\\end{vmatrix}} + 2 \\cdot {\\begin{vmatrix} 2 & -t \\\\ 0 & 1\\end{vmatrix}}$$ $$= (3 – t) \\cdot [-t \\cdot (-1 – t) + 1] + 2 \\cdot 2$$ $$=(3 – t) \\cdot (t + t^2 + 1) + 4$$ $$=-t^3+2t^2+2t+7.$$ We have computed the characteristic polynomial: $p(t)=-t^3+2t^2+2t+7$. According to the Cayley – Hamilton Theorem, $p(A)= -A^3+2A^2+2A+7I=0$. If we rearrange that expression, we get $$-A^3+2A^2+2A=-7I$$ $$A(-A^2+2A+2I)=-7I$$ $$A\\left(-\\frac{1}{7}(-A^2+2A+2I)\\right)=I.$$ Also, we have $$\\left(-\\frac{1}{7}(-A^2+2A+2I)\\right)A=I.$$ We conclude that the matrix $-\\frac{1}{7}(-A^2+2A+2I)$ is the inverse matrix for $A$. Therefore, $$A^{-1}= -\\frac{1}{7}(-A^2+2A+2I)$$ $$=-\\frac{1}{7}\\left(-\\begin{bmatrix} 3 & 0 & 2 \\\\ 2 & 0 & -1 \\\\ 0 & 1 & -1\\end{bmatrix} \\cdot \\begin{bmatrix} 3 & 0 & 2 \\\\ 2 & 0 & -1 \\\\ 0 & 1 & -1\\end{bmatrix}+2 \\cdot \\begin{bmatrix} 3 & 0 & 2 \\\\ 2 & 0 & -1 \\\\ 0 & 1 & -1\\end{bmatrix}+\\begin{bmatrix} 2 & 0 & 0 \\\\ 0 & 2& 0 \\\\ 0 & 0 & 2\\end{bmatrix}\\right)$$ $$= -\\frac{1}{7}\\begin{bmatrix}-1 & -2 & 0 \\\\-2 & 3 & -7\\\\ -2 & 3 & 0 \\end{bmatrix}.$$ The inverse matrix of matrix $A$ is: $$A^{-1}= -\\frac{1}{7}\\begin{bmatrix}-1 & -2 & 0 \\\\-2 & 3 & -7\\\\ -2 & 3 & 0", "are given as follows $$A_1x + B_1y = C_1$$ $$A_2x + B_2y = C_2$$ These equations can be represented in matrix form as following $$\\begin{bmatrix} A_1 & B_1 \\\\A_2 & B_2 \\end{bmatrix} \\begin{bmatrix} x \\\\ y\\end{bmatrix} = \\begin{bmatrix} C_1 \\\\C_2 \\end{bmatrix}$$ The following can be the relation between these 2 Lines 1. If $$\\frac{A_1}{A_2}=\\frac{B_1}{B_2}=\\frac{C_1}{C_2}$$ then the lines are coincident. 2. If $$\\frac{A_1}{A_2}=\\frac{B_1}{B_2}\\neq\\frac{C_1}{C_2}$$ then the lines are parallel. 3. If $$\\frac{A_1}{A_2}\\neq\\frac{B_1}{B_2}$$ then the lines intersect. Following are the steps to find the point of intersection between these 2 Lines: 1. Calulate the following determinants $$D=\\begin{vmatrix} A_1 & B_1 \\\\A_2 & B_2 \\end{vmatrix} \\hspace{.5cm} D_1=\\begin{vmatrix} C_1 & B_1 \\\\C_2 & B_2 \\end{vmatrix} \\hspace{.5cm} D_2=\\begin{vmatrix} A_1 & C_1 \\\\A_2 & C_2 \\end{vmatrix}$$ 2. If $$D=D_1=D_2=0$$ then the lines are coincident. 3. If $$D=0$$ and either $$D_1\\neq0$$ or $$D_2\\neq0$$ then the lines are parallel. 4. If $$D\\neq0$$ then the lines intersect and the coordinates of intersection is given as $$x=\\frac{D_1}{D} \\hspace{.5cm} y=\\frac{D_2}{D}$$ 5. The point of intersection between these 2 Lines can be also be found out by the matrix inversion as following $$\\begin{bmatrix} x \\\\ y\\end{bmatrix} = \\begin{bmatrix} A_1 & B_1 \\\\A_2 & B_2 \\end{bmatrix}^{-1} \\begin{bmatrix} C_1 \\\\C_2 \\end{bmatrix}$$ $$\\Rightarrow \\begin{bmatrix} x \\\\ y\\end{bmatrix} = \\frac{1}{D}\\begin{bmatrix} B_2 & -B_1 \\\\-A_2 & A_1 \\end{bmatrix} \\begin{bmatrix} C_1 \\\\C_2 \\end{bmatrix}$$", "Browse Questions # Using properties of determinants, prove the following : $\\begin{vmatrix} 1 & a^2+bc & a^3 \\\\ 1 & b^2+ca & b^3 \\\\ 1 & c^2+ab & c^3 \\end{vmatrix} = (a-b)(b-c)(c-a)(a^2+b^2+c^2).$ Toolbox: • The value of the determinant can be found by expanding along any rows or columns. • Elementary transformation in a determinant can be made by • (a) Interchanging two rows or columns. • (b) By adding or subtracting two or more rows or columns Step 1: Let $\\Delta=\\begin{vmatrix}1 & a^2+bc & a^3\\\\1 & b^2+ca&b^3\\\\1 & c^2+ab& c^3\\end{vmatrix}$ Apply $R_2\\rightarrow R_2-R_3$ and $R_3 \\rightarrow R_3-R_!$ $\\Delta=\\begin{vmatrix}1 & a^2+bc & a^3\\\\0 & b^2+ca-c^2-ab&b^3-c^3\\\\0 & c^2+ab-a^2-bc& c^3-a^3\\end{vmatrix}$ $[a^3-b^3=(a-b)(a^2+ab+b^2)]$ This can be written as $\\Delta=\\begin{vmatrix}1 & a^2+bc & a^3\\\\0 & (b+c)(b-c)+a(c-b)&(b-c)(b^2+bc+c^2)\\\\0 & (c+a)(c-a)+b(a-c)&(c-a)(c^2+ac+a^2)\\end{vmatrix}$ Take (b-c) and (c-a) as factors from $R_2$ and $R_3$ respectively $\\Delta=(b-c)(c-a)\\begin{vmatrix}1 & a^2+bc & a^3\\\\0 & b+c-a & b^2+bc+c^2\\\\0 & c+a-b & c^2+ac+a^2\\end{vmatrix}$ Step 2: $\\Delta=(b-c)(c-a)\\begin{vmatrix}1 & a^2+bc & a^3\\\\0 & -2(b-a) & (b^2-a^2)+c(b-a)\\\\0 & c+a-b & c^2+ac+a^2\\end{vmatrix}$ Taking (a-b) as the common factor from $R_2$ $\\Delta=(a-b)(b-c)(c-a)\\begin{vmatrix}1 & a^2+bc & a^3\\\\0 & 2 & b+a+c\\\\0 & c+a-b & c^2+ac+a^2\\end{vmatrix}$ Now expanding along $C_1$ we get $\\Delta=(a-b)(b-c)(c-a)[1[2(c^2+ac+a^2)-(c+a+b)(c+a-b)]]$ $\\Delta=(a-b)(b-c)(c-a)[2(c^2+ac+a^2)-(c+a)^2-b^2)]$ On simplifying we get $\\Delta=(a-b)(b-c)(c-a)[2c^2+2ac+2a^2)-c^2-a^2-2ac+b^2]$ $\\;\\;\\;=(a-b)(b-c)(c-a)[a^2+b^2+c^2]$ Hence $\\Delta=(a-b)(b-c)(c-a)[a^2+b^2+c^2]$", "please do post it too. • Jan 10, 2017 at 5:52 One can use factor theorem to get a simpler solution. If we put $a=0$, we get \\begin{align*} \\begin{vmatrix} (a+b)^2 & ca & bc \\\\ ca & (b+c)^2 & ab \\\\ bc & ab & (c+a)^2 \\\\ \\end{vmatrix} =\\begin{vmatrix} b^2 & 0 & bc \\\\ 0 & (b+c)^2 & 0 \\\\ bc & 0 & c^2 \\\\ \\end{vmatrix} = 0 \\end{align*} Hence $a$ is a factor. Similarly $b, c$ are factors. Again, put $a+b+c=0$, we get \\begin{align*} \\begin{vmatrix} c^2 & ca & bc \\\\ ca & a^2 & ab \\\\ bc & ab & b^2 \\\\ \\end{vmatrix} =abc\\begin{vmatrix} c & a & b \\\\ c & a & b \\\\ c& a & b \\\\ \\end{vmatrix} \\end{align*} Since all rows are identical, $(a+b+c)^2$ is a factor. The determinant is a polynomial of degree 6 and hence the remaining factor is linear and since it is symmetric, the factor must be $k(a+b+c)$. Putting $a=b=c=1$, we obtain \\begin{align*} 27k = \\begin{vmatrix} 4 & 1 & 1 \\\\ 1 & 4 & 1 \\\\ 1 & 1 & 4 \\\\ \\end{vmatrix} =54 \\end{align*} and $k=2$. Thus the given determinant equals $2abc(a+b+c)^3$ • Why is it that when you put $a+b+c=0$ and as all rows are identical then you concluded that $(a+b+c)^2$", "# Q : 14 By using properties of determinants, show that: $\\dpi{100} \\begin{vmatrix} a^2+1 &ab &ac \\\\ ab &b^2+1 &bc \\\\ ca & cb &c^2+1 \\end{vmatrix}=1+a^2+b^2+c^2$ Given determinant: $\\dpi{100} \\begin{vmatrix} a^2+1 &ab &ac \\\\ ab &b^2+1 &bc \\\\ ca & cb &c^2+1 \\end{vmatrix}$ Let $\\triangle = \\begin{vmatrix} a^2+1 &ab &ac \\\\ ab &b^2+1 &bc \\\\ ca & cb &c^2+1 \\end{vmatrix}$ Then we can clearly see that each column can be reduced by taking common factors like a,b, and c respectively from C1,C2,and C3. We then get; $=abc \\begin{vmatrix} \\left ( a+\\frac{1}{a} \\right ) &a &a \\\\ b &(b+\\frac{1}{b}) &b \\\\ c & c &(c+\\frac{1}{c}) \\end{vmatrix}$ Now, applying column transformations: $C_{1} \\rightarrow C_{1} -C_{2}$ and $C_{2} \\rightarrow C_{2} -C_{3}$ then we have; $=abc \\begin{vmatrix} \\left ( \\frac{1}{a} \\right ) &0 &a \\\\ -\\frac{1}{b} &(\\frac{1}{b}) &b \\\\ 0 & -\\frac{1}{c} &(c+\\frac{1}{c}) \\end{vmatrix}$ $=abc\\times \\frac{1}{abc} \\begin{vmatrix} 1 &0 &a^2 \\\\ -1 &1 &b^2 \\\\ 0 & -1 &(c^2+1) \\end{vmatrix}$ $= \\begin{vmatrix} 1 &0 &a^2 \\\\ -1 &1 &b^2 \\\\ 0 & -1 &(c^2+1) \\end{vmatrix}$ Now, expanding the remaining determinant: $\\triangle = 1\\begin{vmatrix} 1&b^2 \\\\ -1&(c^2+1) \\end{vmatrix} + a^2\\begin{vmatrix} -1&1 \\\\ 0& -1 \\end{vmatrix}$ $= 1[(c^2+1)+b^2] + a^2(1)=a^2+b^2+c^2+1$. Hence proved. Exams Articles Questions", "c\\\\ c + a + b & a & b\\\\ b + c + a & c & a \\end{vmatrix} = (a + b + c) \\cdot \\begin{vmatrix} 1 & b & c\\\\ 1 & a & b\\\\ 1 & c & a \\end{vmatrix}$ We calculate the last determinant: $\\begin{vmatrix} 1 & b & c\\\\ 1 & a & b\\\\ 1 & c & a \\end{vmatrix}$ $\\hspace{2mm}\\begin{array}{ccc} 1 & b & c\\\\ 1 & a & b \\end{array}$ $= a^{2} + b^{2} + c^{2} -a\\cdot c - b\\cdot c - a\\cdot b =$ $\\frac{1}{2}\\cdot(2a^{2} +2b^{2}+2c^{2} -2a\\cdot b -2a\\cdot c-2b\\cdot c) =$ $\\frac{1}{2}\\cdot(a^{2}-2a\\cdot b + b^{2}+ a^{2}-2a\\cdot c +c^{2}+b^{2}-2b\\cdot c + c^{2})=$ $\\frac{1}{2}\\cdot[(a-b)^{2}+(a-c)^{2}+(b-c)^{2}]$ In conclusion $\\begin{vmatrix} a & b & c\\\\ c & a & b\\\\ b & c & a \\end{vmatrix}=$ $\\frac{1}{2}\\cdot(a+b+c)\\cdot[(a-b)^{2}+(a-c)^{2}+(b-c)^{2}]$ Example 32 We calculate the determinant of a Vandermonde matrix. $\\begin{vmatrix} 1 & 1 & 1\\\\ a & b & c\\\\ a^{2} & b^{2} & c^{2} \\end{vmatrix}$ Using the properties of determinants we modify row 1 in order to have two elements equal to 0. In this case, when we apply the formula, there's no need to calculate the cofactors of these elements because their product will be 0. $\\begin{vmatrix} 1 & 1 & 1\\\\ a & b & c\\\\ a^{2} & b^{2} & c^{2}\\\\ \\end{vmatrix}$", "# Tag Info 12 First note that, $$\\begin{vmatrix} a & a^2 & a^3+1 \\\\ b & b^2 & b^3+1 \\\\ c & c^2 & c^3+1 \\\\ \\end{vmatrix} = \\begin{vmatrix} a & a^2 & a^3 \\\\ b & b^2 & b^3 \\\\ c & c^2 & c^3 \\\\ \\end{vmatrix} + \\begin{vmatrix} a & a^2 & 1 \\\\ b & b^2 & 1 \\\\ c & c^2 & 1 \\\\ \\end{vmatrix}=0$$ Then, ... 8 A non - trivial counterexample with $\\det A = 0$ and $\\det D \\neq 0$: Let $$A =\\begin{bmatrix} 0 & 0 \\\\ 1 & 2 \\end{bmatrix}$$ and $$D = \\begin{bmatrix} -1 & 0 \\\\ 0 & -2\\end{bmatrix}.$$ Then, $$A+D =\\begin{bmatrix} -1 & 0 \\\\1 & 0 \\end{bmatrix},$$ with $\\det(A+D) = 0$. 8 Consider for trivial counterexample $A=-D$ 7 I guess it's easier to go for a reduction formula. I proceed along the generalization mentioned in comment: Call the determinant $M = M_n(x_1,x_2,\\cdots, x_n)$ \\displaystyle \\begin{align}M &= \\left|\\begin{matrix}\\dfrac{1}{x_1+x_1} & \\cdots & \\dfrac{1}{x_1+x_n}\\\\ \\dfrac{1}{x_2+x_1} & \\cdots & \\dfrac{1}{x_2+x_n}\\\\ \\cdot & \\cdot ... 6 The hard part is to compute the determinants and factor them nicely: \\det\\begin{bmatrix}a&a^2&1\\\\b&b^2&1\\\\c&c^2&1\\end{bmatrix} = (ab^2+bc^2+ca^2)- (b^2c+c^2a+a^2b) = (a-b)(b-c)(c-a) and \\det\\begin{bmatrix}a&a^2&1+a^3\\\\b&b^2&1+b^3\\\\c&c^2&1+c^3\\end{bmatrix} = (ab^2(1+c^3)+bc^2(1+a^3)+ca^2(1+b^3)) - ... 6 You are looking at a special case of the Matrix determinant Lemma. From the Wikipedia page, the proof for the", "# Evaluate the following Determinant Find $$\\Delta=\\begin{vmatrix} \\frac{1}{a+x} &\\frac{1}{b+x} &\\frac{1}{c+x} \\\\ \\frac{1}{a+y} &\\frac{1}{b+y} &\\frac{1}{c+y} \\\\ \\frac{1}{a+z} &\\frac{1}{b+z} &\\frac{1}{c+z} \\end{vmatrix}$$ I applied $C_1 \\to C_1-C_2$ and $C_2 \\to C_2-C_3$ we get $$\\Delta=\\begin{vmatrix} \\frac{b-a}{(a+x)(b+x)} &\\frac{c-b}{(b+x)(c+x)} &\\frac{1}{c+x} \\\\ \\frac{b-a}{(a+y)(b+y)} &\\frac{c-b}{(b+y)(c+y)}&\\frac{1}{c+y} \\\\ \\frac{b-a}{(a+z)(b+z)} &\\frac{c-b}{(b+z)(c+z)} &\\frac{1}{c+z} \\end{vmatrix}$$", "be invertible, A must be a square matrix and $$det(A) \\neq 0$$. For that reason, invertible matrices are also called nonsingular matrices. Two examples are shown below $det(A) = \\begin{vmatrix} 4 & 5 \\\\ -2 & 1 \\end{vmatrix} = 4*1-5*-2 = 14 \\neq 0$ $det(C) = \\begin{vmatrix} 1 & 2 & -1\\\\ 5 & 3 & 2 \\\\ 6 & 0 & 6 \\end{vmatrix} = -2 \\begin{vmatrix} 5 & 2 \\\\ 6 & 6 \\end{vmatrix} + 3 \\begin{vmatrix} 1 & -1\\\\ 6 & 6 \\end{vmatrix} + 0 \\begin{vmatrix} 1 & -1\\\\ 5 & 2 \\end{vmatrix}$ $det(C)= - 2(5*6-2*6) + 3(1*6-(-1)*6) - 0(1*2-(-1)*5) = 0$ So C is not invertible, because its determinate is zero. However, A is an invertible matrix, because its determinate is nonzero. To calculate that matrix inverse of a 2 × 2 matrix, use the below formula. $A^{-1} = \\begin{pmatrix} a_{1,1} & a_{1,2}\\\\ a_{2,1} & a_{2,2} \\end{pmatrix} ^{-1} = \\frac{1}{det(A)} \\begin{pmatrix} a_{2,2} & -a_{1,2} \\\\ -a_{2,1} & a_{1,1} \\end{pmatrix} = \\frac{1}{a_{1,1} * a_{2,2} - a_{1,2}*a_{2,1}} \\begin{pmatrix} a_{2,2} & -a_{1,2} \\\\ -a_{2,1} & a_{1,1} \\end{pmatrix}$ For example $A^{-1} = \\begin{pmatrix} 4 & 5 \\\\ -2 & 1 \\end{pmatrix} ^{-1} = \\frac{1}{det(A)} \\begin{pmatrix} 1 & -5 \\\\ 2 & 4 \\end{pmatrix} = \\frac{1}{4*1 - 5*(-2)} \\begin{pmatrix} 1 & -5 \\\\ 2 & 4 \\end{pmatrix} = \\begin{pmatrix} \\frac{1}{14}", "allow us to stop writing $$\\det$$ over and over again: $\\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} := \\det\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$ Using the additivity of the bilinearity of the determinant on the first vector we can split this into: $\\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} = \\begin{vmatrix} a & b \\\\ 0 & d \\end{vmatrix} + \\begin{vmatrix} 0 & b \\\\ c & d \\end{vmatrix}$ And on the second vector...: $$\\label{additivity} ... = \\begin{vmatrix} a & b \\\\ 0 & 0 \\end{vmatrix} + \\begin{vmatrix} a & 0 \\\\ 0 & d \\end{vmatrix} + \\begin{vmatrix} 0 & b \\\\ c & 0 \\end{vmatrix} + \\begin{vmatrix} 0 & 0 \\\\ c & d \\end{vmatrix}$$ Then we can use homogeneity to extract the scalars: $$\\label{homogeneity} ... = ab\\begin{vmatrix} 1 & 1 \\\\ 0 & 0 \\end{vmatrix} + ad\\begin{vmatrix} 1 & 0 \\\\ 0 & 1 \\end{vmatrix} + bc\\begin{vmatrix} 0 & 1 \\\\ 1 & 0 \\end{vmatrix} + cd\\begin{vmatrix} 0 & 0 \\\\ 1 & 1 \\end{vmatrix}$$ Lastly we use that the determinant is alternating together with the determinant of an identity matrix being 1: $$\\label{final} ... = ab \\cdot 0 + ad\\cdot 1 + bc\\cdot -1 + cd\\cdot 0$$ And we conclude that the determinant of a general $$2\\times2$$ matrix", "$M \\cdot \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = \\begin{pmatrix} a & b \\\\ a- c & b- d \\end{pmatrix}$. The determinant of matrix $M$ equals $\\cdots \\cdot$ A. $-1$ C. $1$ E. $3$ B. $0$ D. $2$ Solution Given that $M \\cdot \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = \\begin{pmatrix} a & b \\\\ a- c & b- d \\end{pmatrix}$ Therefore, we will have \\begin{aligned} |M| \\cdot \\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} & = \\begin{vmatrix} a & b \\\\ a- c & b- d \\end{vmatrix} \\\\ |M| \\cdot (ad- bc) & = a(b- d)- b(a- c) \\\\ |M| \\cdot (ad- bc) & = \\cancel{ab}- ad \\cancel{- ab} + bc \\\\ |M| \\cdot (ad- bc) & =-(ad- bc) \\\\ |M| & = \\dfrac{-(ad-bc)}{ad-bc} =-1 \\end{aligned} Thus, the determinant of matrix $M$ equals $\\boxed{-1}$ [collapse] Problem Number 3 If given $\\begin{vmatrix} a & b & c \\\\ d & e & f \\\\ g & h & i \\end{vmatrix} = 3$, then $\\begin{vmatrix} 2a + d& a & 4a+2d+g \\\\ 2b+e & b & 4b+2e+h \\\\ 2c+f & c & 4c+2f+i \\end{vmatrix} = \\cdots \\cdot$ A. $-3$ C. $0$ E. $3$ B. $-2$ D. $2$ Solution It is given that $\\begin{vmatrix} a & b & c\\\\ d & e", "by 2 matrices. Anyway, you wind up with -2 for an answer. Take the reciprocal of this and set it aside: $\\Large\\frac{1}{-2}$ Now, to find the adjoint: You do the same thing by covering up row 1, column 1 and taking the determinant of what's left exposed; cover up row 1 column 2 and take the det of what's left exposed; cover up row 1, column 3 and take the det of what's exposed; cover the 2nd row, 1st column and take det of what's left, etc, etc. After going through all those steps you end up with the following matrix of cofactors: $\\begin{vmatrix}-1&-1&1\\\\-1&1&-1\\\\5&-1&-1\\end{vmatrix}$ Take the transpose: $\\begin{vmatrix}-1&-1&5\\\\-1&1&-1\\\\1&-1&-1\\end{vmatrix}$ Now, take the reciprocal of the det we set aside, 1/-2, and multiply it by the adjoint transpose: $\\frac{-1}{2}\\begin{vmatrix}-1&-1&5\\\\-1&1&-1\\\\1&-1&-1\\end{vmatrix}$ You get: $*\\boxed{\\Large\\begin{bmatrix}\\frac{1}{2}&\\frac{1}{ 2}&\\frac{-5}{2}\\\\\\frac{1}{2}&\\frac{-1}{2}&\\frac{1}{2}\\\\\\frac{-1}{2}&\\frac{1}{2}&\\frac{1}{2}\\end{bmatrix}}$ Voila!, there's the inverse. *Hey Soroban, that box thing is the bomb.", "# Find All Values of $x$ such that the Matrix is Invertible ## Problem 721 Given any constants $a,b,c$ where $a\\neq 0$, find all values of $x$ such that the matrix $A$ is invertible if $A= \\begin{bmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ -1/a & x & x^{2} \\end{bmatrix} .$ ## Solution. We know that $A$ is invertible precisely when $\\det(A)\\neq 0$. We therefore compute, by expanding along the first row, \\begin{align*} \\det(A) &= 1 \\begin{vmatrix} a & -b \\\\ x & x^{2} \\end{vmatrix} +c \\begin{vmatrix} 0 & a \\\\ -1/a & x \\end{vmatrix} = 1(ax^{2}+bx) +c(0+1) \\\\ &= ax^{2}+bx+c . \\end{align*} Thus $\\det(A)\\neq 0$ when $ax^{2}+bx+c\\neq 0$. We know by the quadratic formula that $ax^{2}+bx+c=0$ precisely when $x= \\dfrac{-b\\pm\\sqrt{b^{2}-4ac}}{2} .$ Therefore, $A$ is invertible so long as $x$ satisfies both of the following inequalities: $x\\neq \\dfrac{-b+\\sqrt{b^{2}-4ac}}{2} ,\\quad x\\neq \\dfrac{-b-\\sqrt{b^{2}-4ac}}{2} .$ ### More from my site • Find Inverse Matrices Using Adjoint Matrices Let $A$ be an $n\\times n$ matrix. The $(i, j)$ cofactor $C_{ij}$ of $A$ is defined to be $C_{ij}=(-1)^{ij}\\det(M_{ij}),$ where $M_{ij}$ is the $(i,j)$ minor matrix obtained from $A$ removing the $i$-th row and $j$-th column. Then consider the $n\\times n$ matrix […] • For Which Choices of $x$ is the Given Matrix Invertible? Determine the values of $x$", "Q By using properties of determinants, show that: determinant a^2+1 ab ac ac b^2+1 bc ca cb c^2+1 = 1+a^2+b^2+c^2 Ex 4.2 Q : 14 Q : 14 By using properties of determinants, show that: $\\dpi{100} \\begin{vmatrix} a^2+1 &ab &ac \\\\ ab &b^2+1 &bc \\\\ ca & cb &c^2+1 \\end{vmatrix}=1+a^2+b^2+c^2$ Views Given determinant: $\\dpi{100} \\begin{vmatrix} a^2+1 &ab &ac \\\\ ab &b^2+1 &bc \\\\ ca & cb &c^2+1 \\end{vmatrix}$ Let $\\triangle = \\begin{vmatrix} a^2+1 &ab &ac \\\\ ab &b^2+1 &bc \\\\ ca & cb &c^2+1 \\end{vmatrix}$ Then we can clearly see that each column can be reduced by taking common factors like a,b, and c respectively from C1,C2,and C3. We then get; $=abc \\begin{vmatrix} \\left ( a+\\frac{1}{a} \\right ) &a &a \\\\ b &(b+\\frac{1}{b}) &b \\\\ c & c &(c+\\frac{1}{c}) \\end{vmatrix}$ Now, applying column transformations: $C_{1} \\rightarrow C_{1} -C_{2}$ and $C_{2} \\rightarrow C_{2} -C_{3}$ then we have; $=abc \\begin{vmatrix} \\left ( \\frac{1}{a} \\right ) &0 &a \\\\ -\\frac{1}{b} &(\\frac{1}{b}) &b \\\\ 0 & -\\frac{1}{c} &(c+\\frac{1}{c}) \\end{vmatrix}$ $=abc\\times \\frac{1}{abc} \\begin{vmatrix} 1 &0 &a^2 \\\\ -1 &1 &b^2 \\\\ 0 & -1 &(c^2+1) \\end{vmatrix}$ $= \\begin{vmatrix} 1 &0 &a^2 \\\\ -1 &1 &b^2 \\\\ 0 & -1 &(c^2+1) \\end{vmatrix}$ Now, expanding the remaining determinant: $\\triangle = 1\\begin{vmatrix} 1&b^2 \\\\ -1&(c^2+1) \\end{vmatrix} + a^2\\begin{vmatrix} -1&1 \\\\ 0& -1 \\end{vmatrix}$ $= 1[(c^2+1)+b^2] +", "b^2 & ab-bc+a^2 - c^2 & bc-ca+b^2 - a^2\\\\ab -c^2 & bc-ca+b^2 - a^2 & ca-ab+c^2 - b^2\\end{vmatrix}$ Expanding along $R_1$ we get, $\\Delta=(ab+bc+ca-a^2-b^2-c^2)[ab-bc+a^2-c^2)(ca-ab+c^2-b^3)-(bc-ca+b^2-a^2)(bc-ca+b^2-a^2)]$ On simplifying we get, $(ab+bc+ca-a^2-b^2-c^2)[b(a-c)+(a+c)(a-c)]\\times [a(c-b)+(c-b)(c+b)]-[c(b-a)+(b-a)(b+a)]\\times [c(b-a)+(b-a)(b+a)]$ Taking the common factors we get, $(ab+bc+ca-a^2-b^2-c^2)[(a-c)(a+b+c)((c-b)(a+b+c)]-[(b-a)(a+b+c)(b-a)(a+b+c)]$ Again taking $(a+b+c)^2$ as the common factor $(a+b+c)^2(ab+bc+ca-a^2-b^2-c^2)[(c-a)(c-b)-(b-a)(b-a)]$ Since it is given (a+b+c) is a factor The given determinant is divisible by (a+b+c) The quotient is $[ab+bc+ca-b^2-b^2-c^2][ab+bc+ca-a^2-b^2-c^2]$ $\\;\\;=(ab+bc+ca-a^2-b^2-c^2)^2$", "us somewhere, so just for fun, let's do it. Two-by-Two Matrix Inverse Suppose A= \\begin{bmatrix} \\end{bmatrix} Then $A$ is invertible if and only if $ad-bc\\neq 0$. When $A$ is invertible, then \\inverse{A}=\\frac{1}{ad-bc} \\begin{bmatrix} \\end{bmatrix} Assume that $ad-bc\\neq 0$. We will use the definition of the inverse of a matrix to establish that $A$ has inverse (). Note that if $ad-bc\\neq 0$ then the displayed formula for $\\inverse{A}$ is legitimate since we are not dividing by zero). Using this proposed formula for the inverse of $A$, we compute A\\inverse{A} \\left(\\frac{1}{ad-bc} \\right) = \\frac{1}{ad-bc} = and \\inverse{A}A \\frac{1}{ad-bc} = \\frac{1}{ad-bc} = By this is sufficient to establish that $A$ is invertible, and that the expression for $\\inverse{A}$ is correct. \\quad Assume that $A$ is invertible, and proceed with a proof by contradiction (), by assuming also that $ad-bc=0$. This translates to $ad=bc$. Let B= \\begin{bmatrix} \\end{bmatrix} be a putative inverse of $A$. This means that I_2=AB= \\begin{bmatrix} \\end{bmatrix} \\begin{bmatrix} \\end{bmatrix} = \\begin{bmatrix} \\end{bmatrix} Working on the matrices on two ends of this equation, we will multiply the top row by $c$ and the bottom row by $a$. \\begin{bmatrix} \\end{bmatrix} = \\begin{bmatrix} \\end{bmatrix} We are assuming that $ad=bc$, so we can replace two occurrences of $ad$ by $bc$ in the bottom row of the right matrix. \\begin{bmatrix} \\end{bmatrix} = \\begin{bmatrix} \\end{bmatrix}", "row. This can be obtained by taking $C_3 \\rightarrow C_3+bC_1$. This gives \\begin{align*} A = (1+a^2+b^2)\\begin{vmatrix} 1 & 0 & 0 \\\\ 2ab & 1-a^2+b^2 & 2a(1+b^2) \\\\ 2b & -2a & 1-a^2+b^2 \\end{vmatrix} \\end{align*} Now, expanding by the first row, we obtain \\begin{align*} A &= (1+a^2+b^2)[(1-a^2+b^2)^2 + 4a^2(1+b^2)] \\\\ &= (1+a^2+b^2)[(1+b^2-a^2)^2 + 4a^2(1+b^2)] \\\\ &= (1+a^2+b^2)[(1+b^2)^2 + a^4 -2a^2(1+b^2)+ 4a^2(1+b^2)] \\\\ &\\qquad \\qquad \\qquad \\qquad \\text{expanding } ((1+b^2) - a^2)^2 \\\\ &= (1+a^2+b^2[(1+b^2)^2 + a^4 +2a^2(1+b^2)] \\\\ &= (1+a^2+b^2)(1+a^2+b^2)^2 \\\\ &= (1+a^2+b^2)^3 \\end{align*} Since $a^2-b^2 = 2$, the determinant can be written as $$(1+ b^2 + 2 + b^2)^3 = (3+2b^2)^3$$ This is least when $b^2$ is least. The least value of $b^2$ is 0 and hence the least value is 27. Try to express the determinant using the so-called Rule of Sarrus. This will yield a function in two variables, namely $a$ and $b$. Then find the minimum of this function using differentiation in several variables and your second condition $a^2 - b^2 = 2$. Can you take it from here? • Of course you'd have to use appropriate transformations (binomial formulas, etc. and the additional condition $a^2 - b^2 = 2$) to get a nice and relatively short expression... – ComplexFlo Sep 21 '16 at 11:57" ]

[ "$M \\cdot \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = \\begin{pmatrix} a & b \\\\ a- c & b- d \\end{pmatrix}$. The determinant of matrix $M$ equals $\\cdots \\cdot$ A. $-1$ C. $1$ E. $3$ B. $0$ D. $2$ Solution Given that $M \\cdot \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = \\begin{pmatrix} a & b \\\\ a- c & b- d \\end{pmatrix}$ Therefore, we will have \\begin{aligned} |M| \\cdot \\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} & = \\begin{vmatrix} a & b \\\\ a- c & b- d \\end{vmatrix} \\\\ |M| \\cdot (ad- bc) & = a(b- d)- b(a- c) \\\\ |M| \\cdot (ad- bc) & = \\cancel{ab}- ad \\cancel{- ab} + bc \\\\ |M| \\cdot (ad- bc) & =-(ad- bc) \\\\ |M| & = \\dfrac{-(ad-bc)}{ad-bc} =-1 \\end{aligned} Thus, the determinant of matrix $M$ equals $\\boxed{-1}$ [collapse] Problem Number 3 If given $\\begin{vmatrix} a & b & c \\\\ d & e & f \\\\ g & h & i \\end{vmatrix} = 3$, then $\\begin{vmatrix} 2a + d& a & 4a+2d+g \\\\ 2b+e & b & 4b+2e+h \\\\ 2c+f & c & 4c+2f+i \\end{vmatrix} = \\cdots \\cdot$ A. $-3$ C. $0$ E. $3$ B. $-2$ D. $2$ Solution It is given that $\\begin{vmatrix} a & b & c\\\\ d & e", "please do post it too. • Jan 10, 2017 at 5:52 One can use factor theorem to get a simpler solution. If we put $a=0$, we get \\begin{align*} \\begin{vmatrix} (a+b)^2 & ca & bc \\\\ ca & (b+c)^2 & ab \\\\ bc & ab & (c+a)^2 \\\\ \\end{vmatrix} =\\begin{vmatrix} b^2 & 0 & bc \\\\ 0 & (b+c)^2 & 0 \\\\ bc & 0 & c^2 \\\\ \\end{vmatrix} = 0 \\end{align*} Hence $a$ is a factor. Similarly $b, c$ are factors. Again, put $a+b+c=0$, we get \\begin{align*} \\begin{vmatrix} c^2 & ca & bc \\\\ ca & a^2 & ab \\\\ bc & ab & b^2 \\\\ \\end{vmatrix} =abc\\begin{vmatrix} c & a & b \\\\ c & a & b \\\\ c& a & b \\\\ \\end{vmatrix} \\end{align*} Since all rows are identical, $(a+b+c)^2$ is a factor. The determinant is a polynomial of degree 6 and hence the remaining factor is linear and since it is symmetric, the factor must be $k(a+b+c)$. Putting $a=b=c=1$, we obtain \\begin{align*} 27k = \\begin{vmatrix} 4 & 1 & 1 \\\\ 1 & 4 & 1 \\\\ 1 & 1 & 4 \\\\ \\end{vmatrix} =54 \\end{align*} and $k=2$. Thus the given determinant equals $2abc(a+b+c)^3$ • Why is it that when you put $a+b+c=0$ and as all rows are identical then you concluded that $(a+b+c)^2$", "the person behind it, obvious from context. (And it's hard to imagine that the difference even matters, other than for correct grammar). – Ollie Ford May 24 '14 at 2:23 @Ale \"We have met the OP and he is us.\" -Walt Kelly – SpamIAm May 24 '14 at 18:14 Let $$f(a)=\\begin{vmatrix} 1 &a &a^2-bc \\\\ 1 &b &b^2-ca \\\\ 1 &c &c^2-ab \\end{vmatrix}$$ then it's easy to see that $f$ is a polynomial on $a$ with degree at most $2$ and $f(b)=f(c)=0$ so $$f(a)=\\lambda(a-b)(a-c)$$ now, WLOG assume that $bc\\ne0$, take $a=0$; we see easily that $f(0)=0$ hence $\\lambda=0$. - Why is $f(0)=0$? – user1551 May 23 '14 at 19:48 Smooth! Nice work. – amWhy May 24 '14 at 11:43 e.g. if your matrix is $A$, consider $(b-c,c-a,a-b) A$ - Add $(ab+bc+ca)$ times the first column to the last and find the common factor of the last column. - Simply add a multiple of the first column to the last $$\\begin{vmatrix} 1 & a & a^2-bc\\\\1&b&b^2-ac\\\\1&c&c^2-ab\\end{vmatrix} =\\begin{vmatrix} 1 & a & a^2-bc+(1)(ab+ac+bc)\\\\1&b&b^2-ac+(1)(ab+ac+bc)\\\\1&c&c^2-ab+(1)(ab+ac+bc)\\end{vmatrix} = \\begin{vmatrix} 1 & a & a(a+b+c)\\\\1&b&b(a+b+c)\\\\1&c&c(a+b+c)\\end{vmatrix} = 0$$ The final step follows because the second and third column are linearly dependent. - Just realized this is what LutzL said to do – user153126 May 24 '14 at 17:10", "1. ## matrix Show that $\\det \\begin{bmatrix} 1 & 1 & 1 \\\\ a & b & c \\\\ a^{2} & b^{2} & c^{2} \\end{bmatrix} = (b-a)(c-a)(c-b)$. Using expansion by minors I got $c^{2}(b-a) + b^{2}(a-c) + a^{2}(c-b)$. How would I convert this to the above form? 2. I think I got it. Thanks 3. Originally Posted by heathrowjohnny Show that $\\det \\begin{bmatrix} 1 & 1 & 1 \\\\ a & b & c \\\\ a^{2} & b^{2} & c^{2} \\end{bmatrix} = (b-a)(c-a)(c-b)$. Using expansion by minors I got $c^{2}(b-a) + b^{2}(a-c) + a^{2}(c-b)$. How would I convert this to the above form? $= c^{2}(b-a) + b^2 a - b^2 c + a^2 c- a^2 b = c^{2}(b-a) + b^2 a - a^2 b - b^2 c + a^2 c$ $= c^2 (b - a) + ab (b - a) - c (b^2 - a^2)$ $= c^2 (b - a) + ab (b - a) - c (b - a)(b + a) = (b - a)(c^2 + ab - cb - ca)$ $= (b - a)(c^2 - cb + ab - ca) = (b - a)(c[c - b] - a[c - b]) = (b - a)(c - a)(c - b)$. Edit: Well, for what it's worth .....", "the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation: $\\Delta=\\frac{1}{2}\\begin{vmatrix} 2 & -6 & 1\\\\ 5 & 4 & 1\\\\ k & 4 & 1 \\end{vmatrix}=0$ $=\\frac{1}{2}[2(4-4)+6(5-k)+1(20-4k)]$ $=\\frac{1}{2}[30-6k+20-4k]$ $=\\frac{1}{2}[50-10k]$ = 25 – 5k It is given that the area of the triangle is ±35. Therefore, we have: $25-5k=\\pm35$ $5(5-k)=\\pm35$ $5-k=\\pm7$ When 5-k = −7, k = 5+7 = 12. When 5-k = 7, k = 5-7 = −2. Hence, k = 12, −2. The correct Sol: is D. #### Exercise 4.4 Q1: Write Minors and Cofactors of the elements of following determinants: (i) $\\begin{vmatrix} 2 & -4 \\\\ 0 & 3 \\end{vmatrix}$ (ii) $\\begin{vmatrix} a & c \\\\ b & d \\end{vmatrix}$ Sol: (i) The given determinant is $\\begin{vmatrix} 2 & -4 \\\\ 0 & 3 \\end{vmatrix}$ Minor of element aij is Mij . Therefore, M11 = minor of element a11 = 3 M12 = minor of element a12 = 0 M21 = minor of element a21 = -4 M22 = minor of element a22 = 2 Cofactor of aij is Aij = (-1) i+j Mij . Therefore, A11 = (-1)1+1 M11 = (-1)2 (3) = 3 A12 = ( -1 )1 + 2 M12 = ( -1 )3 ( 0 ) = 0 A21 = ( – 1 )2 +", "by $2$. This requires us to also multiply by $1/2$. Then our determinant $D$ is equal to $$D=\\frac12\\,\\begin{vmatrix} 2b-2a& bc+bd-ad-ac&2bcd-2acd\\\\ 2c-2d& ac+bc-ad-bd& 2abc-2abd\\\\ 2b-2c&ab+bd-ac-cd&2abd-2acd \\end{vmatrix}.$$ Factoring in each entry, we get $$D=\\frac12\\,\\begin{vmatrix} 2(b-a)& (b-a)(c+d)&2cd(b-a)\\\\ 2(c-d)& (a+b)(c-d)& 2ab(c-d)\\\\ 2(b-c)&(a+d)(b-c)&2ad(b-c) \\end{vmatrix}.$$ Extracting the common factor in each row, $$D=\\frac12\\,(b-a)(c-d)(b-c)\\begin{vmatrix} 2& c+d&2cd\\\\ 2& a+b& 2ab\\\\ 2&a+d&2ad \\end{vmatrix}.$$ Now replace $R_2$ with $R_2-R_3$ and $R_3$ with $R_3-R_1$: $$D=\\frac12\\,(b-a)(c-d)(b-c)\\begin{vmatrix} 2& c+d&2cd\\\\ 0& b-d& 2a(b-d)\\\\ 0&a-c&2d(a-c) \\end{vmatrix}.$$ Now we can calculate directly. But if we first extract the common factor from each of the last two rows: $$D=\\frac12\\,(b-a)(c-d)(b-c)(b-d)(a-c)\\begin{vmatrix} 2& c+d&2cd\\\\ 0& 1& 2a\\\\ 0&1&2d \\end{vmatrix}.$$ Now the remaining determinant trivially seen to be $4(d-a)$, so $$D=2(b-a)(c-d)(b-c)(b-d)(a-c)(d-a).$$", "-t & -1\\\\ 0 & 1 & -1 – t \\end{vmatrix}}$$ $$=(3 – t) {\\begin{vmatrix} -t & -1 \\\\ 1 & -1-t\\end{vmatrix}} + 2 \\cdot {\\begin{vmatrix} 2 & -t \\\\ 0 & 1\\end{vmatrix}}$$ $$= (3 – t) \\cdot [-t \\cdot (-1 – t) + 1] + 2 \\cdot 2$$ $$=(3 – t) \\cdot (t + t^2 + 1) + 4$$ $$=-t^3+2t^2+2t+7.$$ We have computed the characteristic polynomial: $p(t)=-t^3+2t^2+2t+7$. According to the Cayley – Hamilton Theorem, $p(A)= -A^3+2A^2+2A+7I=0$. If we rearrange that expression, we get $$-A^3+2A^2+2A=-7I$$ $$A(-A^2+2A+2I)=-7I$$ $$A\\left(-\\frac{1}{7}(-A^2+2A+2I)\\right)=I.$$ Also, we have $$\\left(-\\frac{1}{7}(-A^2+2A+2I)\\right)A=I.$$ We conclude that the matrix $-\\frac{1}{7}(-A^2+2A+2I)$ is the inverse matrix for $A$. Therefore, $$A^{-1}= -\\frac{1}{7}(-A^2+2A+2I)$$ $$=-\\frac{1}{7}\\left(-\\begin{bmatrix} 3 & 0 & 2 \\\\ 2 & 0 & -1 \\\\ 0 & 1 & -1\\end{bmatrix} \\cdot \\begin{bmatrix} 3 & 0 & 2 \\\\ 2 & 0 & -1 \\\\ 0 & 1 & -1\\end{bmatrix}+2 \\cdot \\begin{bmatrix} 3 & 0 & 2 \\\\ 2 & 0 & -1 \\\\ 0 & 1 & -1\\end{bmatrix}+\\begin{bmatrix} 2 & 0 & 0 \\\\ 0 & 2& 0 \\\\ 0 & 0 & 2\\end{bmatrix}\\right)$$ $$= -\\frac{1}{7}\\begin{bmatrix}-1 & -2 & 0 \\\\-2 & 3 & -7\\\\ -2 & 3 & 0 \\end{bmatrix}.$$ The inverse matrix of matrix $A$ is: $$A^{-1}= -\\frac{1}{7}\\begin{bmatrix}-1 & -2 & 0 \\\\-2 & 3 & -7\\\\ -2 & 3 & 0", "& aw - cv'\\end{vmatrix}$$ 44. We know that value of the determinant in denominator is $$(a - b)(b - c)(c - a) = k$$ (say) $$f(a, b, c) = \\begin{vmatrix}f(a) - f(b)& f(b) - f(c) & f(c) \\\\ 0 & 0 & 1 \\\\ a - b & b - c & c\\end{vmatrix}[C_1\\rightarrow C_1 - C_2; C_2\\rightarrow C_2 - C_3] \\div k$$ $$= -(b - c)(f(a) - f(b)) - (a - b)(f(b) - f(c))\\div k$$ $$= -(a - b)(b - c)\\left[\\frac{f(a) - f(b)}{a - b} - \\frac{f(b) - f(c)}{(b - c)}\\right]\\div k$$ $$= -(a - b)(b - c)(f(a, b) - f(b, c))\\div k$$ $$= (a - b)(b - c)(c - a)\\frac{f(b, c) - f(a, b)}{c - a} = (a - b)(b - c)(c - a)f(a, b, c)\\div (a - b)(b - c)(c - a)$$ $$= f(a, b, c)$$ 45. Becasue $$A,B,C$$ are angles of a triangle. $$A + B + C = \\pi$$ Also, $$e^{i\\pi} = \\cos\\pi + i\\sin\\pi = -1$$ Taking $$e^{iA}, e^{iB}, e^{iC}$$ common from $$R_1, R_2, R_3,$$ we get $$\\Delta = e^{i(A + B + C)}\\begin{vmatrix}e^{iA} & e^{-i(A +C)} & e^{-i(A + B)} \\\\ e^{-i(B + C)} & e^{iB} & e^{-i(A + B)} \\\\ e^{-i(B + C)} & e^{-i(A + C)} & e^{iC}\\end{vmatrix}$$ $$= -\\begin{vmatrix}e^{iA} & -e^{iB} & -e^{iC} \\\\ -e^{iA} & e^{iB} & -e^{iC}", "= \\mathbf{M} \\times \\mathbf{M}^{-1} = \\mathbf{I}$ For a two by two matrix, the inverse is given by $\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}^{-1} = \\frac{1}{ad-bc} \\begin{pmatrix} d & -b\\\\ -c & a \\end{pmatrix}$ which we can easily confirm with \\begin{align*} \\frac{1}{ad-bc} \\begin{pmatrix} d & -b\\\\ -c & a \\end{pmatrix} \\times \\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix} &= \\frac{1}{ad-bc} \\begin{pmatrix} da - bc & db - bd\\\\ -ca + ac & -cb + ad \\end{pmatrix}\\\\ &= \\frac{1}{ad-bc} \\begin{pmatrix} ad - bc & 0\\\\ 0 & ad - bc \\end{pmatrix}\\\\ &= \\begin{pmatrix} 1 & 0\\\\ 0 & 1 \\end{pmatrix} \\end{align*} The denominator of the fraction, $$ad-bc$$, is known as the determinant of the matrix and must be non-zero for the matrix to have an inverse. Matrices with determinants of zero are known as singular matrices and project vectors onto a lower dimensional space. Note that matrices with very small determinants, known as near singular matrices, whilst mathematically having inverses will present significant numerical difficulties when trying to compute their inverses. This is because they squash vectors down to a volume that is very thin in some directions, losing many significant digits in those directions through cancellation error. Now, one useful application of inverses is for the solution of systems of simultaneous linear equations, which can be", "# $A$ is not similar to a diagonal matrix over the reals Let $A = \\begin{bmatrix} 6 & -3 & -2 \\\\ 4 & -1 & -2 \\\\ 10 & -5 & -3 \\end{bmatrix}$ then $A$ is not similar to a diagonal matrix over the reals and it is not similar to a diagonal matrix over the complexes. We know that $A$ is similar to a diagonal matrix over the reals(complexes) if there exist $D$ diagonal matriz and $P$ invertible matrix both $n \\times n$ with real entries (complex entries) such that $A = PDP^{-1}$ I find that the inverse of $A$ is $A^{-1} = \\frac{1}{2}\\begin{bmatrix} -7 & 1 & 4 \\\\ -8 & 2 & 4 \\\\ -10 & 0 & 6 \\end{bmatrix}$ and i diagonalize the matrix $A$ and got $A = PBP^{-1}$ with $P = \\begin{bmatrix} 1 & \\frac{3}{5}-\\frac{i}{2} & \\frac{3}{5}+\\frac{i}{2} \\\\ 0 & \\frac{3}{5}-\\frac{i}{2} & \\frac{3}{5}+\\frac{i}{2} \\\\ 2 & 1 & 1 \\end{bmatrix}$ , $P^{-1} = \\begin{bmatrix} 1 & -1 & 0 \\\\ -1 + 3i & 1-\\frac{i}{2} & \\frac{1}{2}-\\frac{3i}{2} \\\\ -1 - 3i & 1+\\frac{i}{2} & \\frac{1}{2}+\\frac{3i}{2} \\end{bmatrix}$ and finally i got $D = \\begin{bmatrix} 2 & 0 & 0 \\\\ 0 & -i & 0 \\\\ 0 & 0 & i \\end{bmatrix}$ then A is diagonalizable but my question is how can", "part.\" The inverse of the $2x2$ matrix A, where A is: (5) \\begin{align} A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix} \\end{align} can be found with the following equation: (6) \\begin{align} A^{-1}= \\frac {1}{ad-bc}\\begin{pmatrix} d & -b\\\\ -c & a \\end{pmatrix} \\end{align} This equation is given in the IB Data Booklet. Looking at the equation we can see that there is a problem if $ad-bc=0$. If that quantity does in fact equal zero then the inverse does not exist, and the original matrix is said to be \"singular\", since dividing by zero doesn't work… #### Determinant for a 2 x 2 For the $2x2$ matrix: (7) \\begin{align} A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix} \\end{align} The determinant for a $2x2$ is given by the equation: (8) $$det A = ad-bc$$ This equation is given in the IB data booklet. The determinant of a matrix can also be written with vertical bars such as: (9) \\begin{align} det A = |A| = \\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} \\end{align} Note that this is the same term that showed up while calculating the inverse. Thus we can say (at least for a $2x2$) that if the $det A=0$ then the inverse does not exist. #### Determinant for a 3 x 3 Given a $3x3$ matrix: (10) \\begin{align}", "Browse Questions # Using properties of determinants, prove the following : $\\begin{vmatrix} 1 & a^2+bc & a^3 \\\\ 1 & b^2+ca & b^3 \\\\ 1 & c^2+ab & c^3 \\end{vmatrix} = (a-b)(b-c)(c-a)(a^2+b^2+c^2).$ Toolbox: • The value of the determinant can be found by expanding along any rows or columns. • Elementary transformation in a determinant can be made by • (a) Interchanging two rows or columns. • (b) By adding or subtracting two or more rows or columns Step 1: Let $\\Delta=\\begin{vmatrix}1 & a^2+bc & a^3\\\\1 & b^2+ca&b^3\\\\1 & c^2+ab& c^3\\end{vmatrix}$ Apply $R_2\\rightarrow R_2-R_3$ and $R_3 \\rightarrow R_3-R_!$ $\\Delta=\\begin{vmatrix}1 & a^2+bc & a^3\\\\0 & b^2+ca-c^2-ab&b^3-c^3\\\\0 & c^2+ab-a^2-bc& c^3-a^3\\end{vmatrix}$ $[a^3-b^3=(a-b)(a^2+ab+b^2)]$ This can be written as $\\Delta=\\begin{vmatrix}1 & a^2+bc & a^3\\\\0 & (b+c)(b-c)+a(c-b)&(b-c)(b^2+bc+c^2)\\\\0 & (c+a)(c-a)+b(a-c)&(c-a)(c^2+ac+a^2)\\end{vmatrix}$ Take (b-c) and (c-a) as factors from $R_2$ and $R_3$ respectively $\\Delta=(b-c)(c-a)\\begin{vmatrix}1 & a^2+bc & a^3\\\\0 & b+c-a & b^2+bc+c^2\\\\0 & c+a-b & c^2+ac+a^2\\end{vmatrix}$ Step 2: $\\Delta=(b-c)(c-a)\\begin{vmatrix}1 & a^2+bc & a^3\\\\0 & -2(b-a) & (b^2-a^2)+c(b-a)\\\\0 & c+a-b & c^2+ac+a^2\\end{vmatrix}$ Taking (a-b) as the common factor from $R_2$ $\\Delta=(a-b)(b-c)(c-a)\\begin{vmatrix}1 & a^2+bc & a^3\\\\0 & 2 & b+a+c\\\\0 & c+a-b & c^2+ac+a^2\\end{vmatrix}$ Now expanding along $C_1$ we get $\\Delta=(a-b)(b-c)(c-a)[1[2(c^2+ac+a^2)-(c+a+b)(c+a-b)]]$ $\\Delta=(a-b)(b-c)(c-a)[2(c^2+ac+a^2)-(c+a)^2-b^2)]$ On simplifying we get $\\Delta=(a-b)(b-c)(c-a)[2c^2+2ac+2a^2)-c^2-a^2-2ac+b^2]$ $\\;\\;\\;=(a-b)(b-c)(c-a)[a^2+b^2+c^2]$ Hence $\\Delta=(a-b)(b-c)(c-a)[a^2+b^2+c^2]$", "has only a $5$, I think it should be rather easy to understand the calculation then(-5 times the minor of the original matrix). They expand the order $4$ determinant by the first column, that's all. The rules for the signs are the same as for an expansion by a row. $A = \\begin{vmatrix} 0 & 3x & y & z\\\\ 0 & 3a & 2b & 3c\\\\ 0 & 6 & 2 & 3\\\\ 5 & 0 & 0 & 0 \\end{vmatrix}$ For a $4\\times 4$ matrix the signs of the minors are; $A = \\begin{vmatrix}+&&-&&+&&-\\\\-&&+&&-&&+\\\\+&&-&&+&&-\\\\-&&+&&-&&+\\end{vmatrix}$ Expand along $R_4$ gives ; $A = (-5)\\begin{vmatrix}3x&&y&&z\\\\3a&&2b&&3c\\\\6&&2&&3\\end{vmatrix}$ interchanging the rows $R_1\\leftrightarrow R_3$ and $R_2\\leftrightarrow R_3$ so, $A = (-5)\\begin{vmatrix}6&&2&&3\\\\3x&&y&&z\\\\3a&&2b&&3c\\end{vmatrix}$ $A = -5\\cdot3\\cdot\\begin{vmatrix}2&&2&&3\\\\x&&y&&z\\\\a&&2b&&3c\\end{vmatrix}$ $A = -15\\cdot 10= -150$", "another determinant $$\\begin{vmatrix}a & b & c \\\\b & c & a \\\\c & a & b \\end {vmatrix}$$? a) (a3 + b3 + c3 + 3abc)2 b) –(a3 + b3 + c3 + 3abc)2 c) (a3 + b3 + c3 – 3abc)2 d) –(a3 + b3 + c3 – 3abc)2 Explanation: Now, $$\\begin{vmatrix}a & b & c \\\\b & c & a \\\\c & a & b \\end {vmatrix}$$ Interchanging 2nd and 3rd columns, = –$$\\begin{vmatrix}a & c & b \\\\b & a & c \\\\c & b & a \\end {vmatrix}$$ = $$\\begin{vmatrix}-a & c & b \\\\ -b & a & c \\\\-c & b & a \\end {vmatrix}$$ So, $$\\begin{vmatrix}a & b & c \\\\b & c & a \\\\c & a & b \\end {vmatrix}^2$$ = $$\\begin{vmatrix}a & b & c \\\\b & c & a \\\\c & a & b \\end {vmatrix}$$$$\\begin{vmatrix} -a & c & b \\\\ -b & a & c \\\\-c & b & a \\end {vmatrix}$$ = {–(a3 + b3 + c3 – 3abc)}2 = $$\\begin{vmatrix}-a^2 + bc + bc & -ab + ab + c^2 & -ac + b^2 + ca \\\\-ab + ab + c^2 & -ac – b^2 + ca & -a^2 + bc + bc \\\\-ac + b^2 + ca & -a^2 + bc", "# Find $x$ so that $\\begin{pmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ -\\frac{1}{a} & x & x^2 \\end{pmatrix}$ is invertible Find $x$ so that the matrix $\\begin{pmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ -\\frac{1}{a} & x & x^2 \\end{pmatrix}$ is invertible. Please show your work. Let $A=\\begin{pmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ -\\frac{1}{a} & x & x^2 \\end{pmatrix}.$ Then $\\frac{1}{a}R_1+R_3 \\rightarrow R_3$ gives $\\begin{pmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ 0 & x & x^2+\\frac{c}{a} \\end{pmatrix},$ and $-\\frac{x}{a}R_2+R_3 \\rightarrow R_3$ gives $\\begin{pmatrix} 1 & 0 & c \\\\ 0 & a & -b \\\\ 0 & 0 & \\frac{b}{a}x+x^2+\\frac{c}{a} \\end{pmatrix}.$ Hence $\\text{det}(A)=a(\\frac{b}{a}x+x^2+\\frac{c}{a})=0$ $\\Rightarrow ax^2+b x+c=0$ $\\Rightarrow x=\\frac{-b \\pm \\sqrt{b^2-4ac}}{2a}.$ Hence the matrix $A$ is invertible if and only if $x\\neq\\frac{-b \\pm \\sqrt{b^2-4ac}}{2a}.$", "(b^2 + c^2 - bc)a + b^3 + c^3$. as this is equal to our last remainder, we have: $a^3 + b^3 + c^3 - 3ab = (a+b+c)(a^2 -(b+c)a + b^2 + c^2 - bc)$ or, re-arranging a little: $a^3 + b^3 + c^3 - 3ab = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc)$. 8. ## Re: Another factoring problem Another approach: $a^3 + b^3 + c^3 -3abc = \\begin{vmatrix} a & b & c \\\\ c & a & b \\\\ b & c & a \\end{vmatrix} = \\begin{vmatrix}a+b+c & b & c \\\\ a+b+c & a & b \\\\ a+b+c & c & a \\end{vmatrix} = (a+b+c) \\begin{vmatrix} 1 & b & c \\\\ 1 & a & b \\\\ 1 & c & a \\end{vmatrix} = (a+b+c) (a^2 + b^2 + c^2 - ab - bc - ac)$ Source: \"My Favourite Polynomial\", Desmond MacHale, The Mathematical Gazette, Vol. 75, No. 472 (June 1991), pp. 157-165. 9. ## Re: Another factoring problem Originally Posted by sujith123 a^3+b^3=(a+b)(a^2-ab+b^2 ) Sub b=b+c a^3+(b+c)^3=(a+b+c)(a^2-a(b+c)+(b+c)^2 ) a^3+b^3+〖3b〗^2 c+〖3bc〗^2+c^3=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 ) a^3+b^3+c^3=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-〖3b〗^2 c-〖3bc〗^2 a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-〖3b〗^2 c-〖3bc〗^2-3abc a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2 )-3bc(a+b+c) a^3+b^3+c^3-3abc=(a+b+c)(a^2-ab-ac+b^2+2bc+c^2-3bc) a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc) This is one method of getting that hope you understand... Thank you, your explanation was the easiest to understand.", "A $2 \\times 2$ matrix $A = \\begin{bmatrix} a & b\\\\ c & d \\end{bmatrix}$ is invertible if $\\det(A) = ad - bc ≠ 0$. The inverse of $A$ can be obtained with the following formula $A^{-1} = \\frac{1}{ad - bc} \\begin{bmatrix} d & -b\\\\ -c & a \\end{bmatrix} = \\begin{bmatrix} \\frac{d}{ad - bc} & - \\frac{b}{ad - bc}\\\\ - \\frac{c}{ad - bc} & \\frac{a}{ad - bc} \\end{bmatrix}$. If $ad - bc = 0$, then $A$ is not invertible. • Proof of Theorem 1: Assume that $A = \\begin{bmatrix} a & b\\\\ c & d \\end{bmatrix}$ and $A^{-1} = \\begin{bmatrix} \\frac{d}{ad - bc} & - \\frac{b}{ad - bc}\\\\ - \\frac{c}{ad - bc} & \\frac{a}{ad - bc} \\end{bmatrix}$. We will proceed to show that $AA^{-1} = I_2$. We will first calculate all of the value of all entries in the product $AA^{-1}$, that is $(AA^{-1})_{11}$, $(AA^{-1})_{12}$, $(AA^{-1})_{21}$, and $(AA^{-1})_{22}$: (2) (3) \\begin{align} (AA^{-1})_{12} = \\frac{-ab}{ad - bc} + a \\cdot \\frac{b}{ad - bc} \\\\ (AA^{-1})_{12} = \\frac{-ab + ab}{ad - bc} = 0 \\end{align} (4) \\begin{align} (AA^{-1})_{21} = \\frac{cd}{ad - bc} - \\frac{cd}{ad - bc} \\\\ (AA^{-1})_{21} = \\frac{cd - cd}{ad - bc} = 0 \\end{align} (5) • When $A$ and $A^{-1}$ are multiplied through we obtain $I_2 = \\begin{bmatrix} 1 & 0\\\\ 0 & 1 \\end{bmatrix}$. We also", "2}= \\frac{-66}{-22}=3$ so $x=-2$ and $y=3$ Sanity Check Put $x=-1$ and $y=3$ into one of the original equations. If we use the first equation we get $-5 \\times(-2)+3 = 10 + 3 = 13$ so our answers are probably right We can extend Cramer's method to three dimensions by adding a third variable. If we have the following equations $a_1x+b_1y+c_1z=d_1$, $a_2x+b_2y+c_2z=d_2$ and $a_3x+b_3y+c_3z=d_2$ we could write them in matrix form like this $\\begin{bmatrix} a_1 & b_1 & c_1 \\\\[0.3em] a_2 & b_2 & c_2 \\\\[0.3em] a_3 & b_3 & c_3 \\end{bmatrix} \\begin{bmatrix} x \\\\[0.3em] y \\\\[0.3em] z \\end{bmatrix}= \\begin{bmatrix} d_1 \\\\[0.3em] d_2 \\\\[0.3em] d_3 \\end{bmatrix}$ $AX=D$ The determinant of the matrix of coefficients is given by $det(A)=a_1(b_2 c_3-c_2b_3)-b_1(a_2c_3-c_2b_3)+c1(a_2b_3-b_2a_3)$ Using Cramer's method we can write $x=\\frac{\\begin{vmatrix} d_1 & b_1 & c_1 \\\\[0.3em] d_2 & b_2 & c_2 \\\\[0.3em] d_3 & b_3 & c_3 \\end{vmatrix}}{\\begin{vmatrix} a_1 & b_1 & c_1 \\\\[0.3em] a_2 & b_2 & c_2 \\\\[0.3em] a_3 & b_3 & c_3 \\end{vmatrix}}$, $y=\\frac{\\begin{vmatrix} a_1 & d_1 & c_1 \\\\[0.3em] a_2 & d_2 & c_2 \\\\[0.3em] a_3 & d_3 & c_3 \\end{vmatrix}}{\\begin{vmatrix} a_1 & b_1 & c_1 \\\\[0.3em] a_2 & b_2 & c_2 \\\\[0.3em] a_3 & b_3 & c_3 \\end{vmatrix}}$ and $z=\\frac{\\begin{vmatrix} a_1 & b_1 & d_1 \\\\[0.3em] a_2 & b_2 & d_2 \\\\[0.3em] a_3 & b_3 & d_3 \\end{vmatrix}}{\\begin{vmatrix}", "matrix; only a square matrix can have an inverse. #### Finding the Inverse of a Matrix To find the inverse of a matrix, the values that will produce the identity matrix, create a second matrix of variables and solve for $$I$$. $$A = \\begin{bmatrix} 3 & 1 \\\\ 2 & 4 \\end{bmatrix} X \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix} = \\begin{bmatrix} 3a+b & 3c+d \\\\ 2a+4b & 2c+4d \\end{bmatrix} = \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}$$ Set $$3a+b=1$$ and $$2a+4b=0$$ and solve for $$a$$ and $$b$$. In this case $$a = \\frac{2}{5}$$ and $$b = -\\frac{1}{5}$$. Likewise, set $$3c+d=0$$ and $$2c+4d=1$$; solving for $$c$$ and $$d$$ produces $$c=-\\frac{1}{10}$$ and $$d=\\frac{3}{10}$$. Therefore, $$A^{-1} = \\begin{bmatrix} \\frac{2}{5} & -\\frac{1}{10} \\\\ -\\frac{1}{5} & \\frac{3}{10} \\end{bmatrix}$$ Finding the inverse matrix can also be done in R using the solve command. A <- matrix(c(3,2,1,4),2,2) A ## [,1] [,2] ## [1,] 3 1 ## [2,] 2 4 A.inverse <- solve(A) A.inverse ## [,1] [,2] ## [1,] 0.4 -0.1 ## [2,] -0.2 0.3 A %*% A.inverse ## [,1] [,2] ## [1,] 1 0 ## [2,] 0 1 OK – now we have all the pieces we need to apply matrix algebra to multiple regression. ## 11.3 OLS Regression in Matrix Form As was the case with simple regression, we", "for $$\\B\\$$ is done from largest to smallest, the only way for it to first find a $$\\B'\\ne B\\$$ is with $$\\0. So now, doing the same steps we did for the two shortened forms of division: \\\\begin{aligned} C &\\equiv 0 \\pmod {B'} \\\\ C &\\equiv A \\pmod {B'-1} \\\\ A'B' &\\equiv A \\pmod {B'-1} \\\\ B' &\\equiv 1 \\pmod {B'-1} \\\\ A' &\\equiv A \\pmod {B'-1} \\\\ A' &< A \\\\ A'+B'-1 &\\le A \\end{aligned}\\ \\\\begin{aligned} \\ \\\\ C &\\equiv B' \\pmod {A-1} \\\\ AB &\\equiv B' \\pmod {A-1} \\\\ A &\\equiv 1 \\pmod {A-1} \\\\ B &\\equiv B' \\pmod {A-1} \\\\ B &< B' \\\\ B+A-1 &\\le B' \\end{aligned}\\ Putting it together, \\\\begin{aligned} B+A-1 &\\le B' \\\\ B+(A'+B'-1)-1 &\\le B' \\\\ B+A'+B'-2 &\\le B' \\\\ B+A' &\\le B'-B'+2 \\\\ B+A' &\\le 2 \\\\ B &> 0 \\\\ A' &> 0 \\\\ B = A' &= 1 \\\\ B' &= C \\\\ A &= C \\end{aligned}\\ So for a $$\\B'\\$$ to be found, $$\\B'=C\\$$ and $$\\A=C\\$$ must be true. But the regex tries to assert the modulus $$\\C-B'-(A-1) \\equiv 0 \\pmod {B'-1}\\$$ But if $$\\B'=C\\$$ and $$\\A=C\\$$, this becomes \\\\begin{aligned} C-C-(C-1) \\equiv 0 \\pmod {B'-1} \\\\ 0-(C-1) \\equiv 0 \\pmod {B'-1} \\end{aligned}\\ If $$\\C>1\\$$, the regex will fail to subtract $$\\C-1\\$$ from $$\\0\\$$ before it can even" ]

[ "+0 # help 0 65 1 A projectile is fired with an initial velocity of $$v$$ at an angle of $$\\theta$$ from the ground. Then its trajectory can modeled by the parametric equations \\begin{align*} x &= vt \\cos \\theta, \\\\ y &= vt \\sin \\theta - \\frac{1}{2} gt^2, \\end{align*} where $$t$$ denotes time and $$g$$ denotes acceleration due to gravity, forming a parabolic arch. Suppose $$v$$ is held constant, but $$\\theta$$ is allowed to vary, over $$0^\\circ \\le \\theta \\le 180^\\circ.$$ The highest point of each parabolic arch is plotted. As $$\\theta$$ varies, the highest points of the arches trace a closed curve. The area of this closed curve can be expressed in the form $$c \\cdot \\frac{v^4}{g^2}.$$ Find $$c.$$ Nov 11, 2019", "$C=v_0\\sin\\theta+\\frac{mg}{k}$. $$\\therefore v(t)=(v_0\\sin\\theta+\\frac{mg}{k})e^{-\\frac{kt}{m}}-\\frac{mg}{k}$$ # 6 Parametric Representations of Curves i.e. Projectile Motion. These are the parametric representations of the trajectory of a projectile: $$x(t)=(v_0\\cos\\theta)t\\quad y(t)=(v_0\\sin\\theta)t-\\frac{1}{2}gt^2$$ ### Calculus with Parametric Curves #### Tangents Given $x=f(t)$ and $y=g(t)$, $$\\frac{dy}{dt}=\\frac{dy}{dx}\\cdot \\frac{dx}{dt}$$ So, $$\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\quad\\frac{dx}{dt}\\neq 0$$ #### Areas Given a curve defined by $x=f(t)$ and $y=g(t)$, the area under the curve as $x$ goes from $a$ to $b$ or as $t$ goes from $\\alpha$ to $\\beta$ can be found by: $$A=\\int_a^b ydx=\\int_\\alpha^\\beta g(t)f'(t)dt$$ ### Arclength Using Parametric Representation Given a curve $C$ described by $x=f(t)$ and $y=g(t)$ and $f',\\ g'$ are continuous over $[a,b]$, and $C$ is traversed exactly once as $t$ goes from $a$ to $b$, find the arclength of the curve in the interval $[a,b]$. 1. Partition $[a,b]$ into $n$ sub-intervals of width $\\Delta t=\\frac{b-a}{n}$ 2. \\begin{align*} L_k=||\\overrightarrow{Q_kQ_{k+1}}||&\\approx\\sqrt{[f(t_{k+1})-f(t_k)]^2+[g(t_{k+1})-g(t_k))]^2}\\\\ &=\\Delta t\\sqrt{\\bigg[\\frac{f(t_{k+1})-f(t_k))}{\\Delta t}\\bigg]^2+\\bigg[\\frac{g(t_{k+1})-g(t_k))}{\\Delta t}\\bigg]^2}\\\\ &\\approx\\sqrt{f'(t_k^*)^2+g'(t_k^*)^2}\\Delta t \\end{align*} 3. Take the Riemann Sum. $$L=\\lim_{n\\to\\infty}\\sum_{k=1}^nL_k=\\int_a^b\\sqrt{f'(t)^2+g'(t)^2}dt$$ Alternatively, we can derive this formula from the formula for arclength of a function $y=f(x)$. $$L=\\int_a^b\\sqrt{1+\\big(\\frac{dy}{dx}\\big)^2}dx=\\int_a^b\\sqrt{1+\\bigg(\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\bigg)^2}dx=\\int_a^b\\sqrt{\\bigg(\\frac{dx}{dt}\\bigg)^2+\\bigg(\\frac{dy}{dt}\\bigg)^2}dx\\cdot\\frac{dt}{dx}=\\int_a^b\\sqrt{\\bigg(\\frac{dx}{dt}\\bigg)^2+\\bigg(\\frac{dy}{dt}\\bigg)^2}dt$$ # 7 Sequences Definition: An infinite set of real numbers ordered $a_1, a_2,\\ldots\\$ Denoted as $\\{a_n\\}$ or $\\{a_n\\}_{n=1}^\\infty$ Definition: A sequence has the limit, L, which we write as $$\\lim_{n=\\infty}a_n=L\\quad\\text{or}\\quad a_n\\to L\\text{ as } n\\to\\infty$$ if $\\ \\forall\\epsilon > 0,\\ \\exists N > 0$ such that $|a_n-L|<\\epsilon\\quad\\forall n>N$ If the limit exists, then $\\{a_n\\}$", "$$y = ax + bx^2$$ for some given (known) values of $a$ and $b.$ This can only be a projectile trajectory if the same path is also given by an equation of the form $$y = (\\tan\\theta)x - \\left(\\frac{g}{2v_0^2\\cos^2\\theta}\\right)x^2$$ for some values of $v_0,$ $\\theta,$ and $g.$ In order for both equations to describe the same path, the right hand side must be the same polynomial in each equation. This means the coefficient of $x$ must be the same each time, and the coefficient of $x^2$ must be the same. So we have a projectile trajectory only if we can simultaneously solve these two equations: \\begin{align} a &= \\tan\\theta, \\\\ b &= -\\frac{g}{2v_0^2\\cos^2\\theta} = -\\frac{g}{2v_0^2}\\sec^2\\theta. \\end{align} From the first of these equations we have $\\theta = \\tan^{-1}(a)$ (for a projectile that travels over the part of the parabola for which $x \\geq 0$) or $\\theta = \\tan^{-1}(a) + \\pi$ (for the other part of the parabola). Using the identity $\\sec^2\\theta = 1 + \\tan^2\\theta,$ we have $$b = -\\frac{g}{2v_0^2}\\sec^2\\theta = -\\frac{g}{2v_0^2}(1 + \\tan^2\\theta),$$ and using the fact that $a = \\tan\\theta,$ $$b = -\\frac{g}{2v_0^2}(1 +a^2).$$ If $g$ is known, then there is only one unknown in this last equation, $v_0.$ To find $v_0$ we first isolate $v_0^2$ on one side of the equation: $$v_0^2 = -\\frac{g(1 + a^2)}{2b}.$$", "$$v_x=v\\cos\\alpha,v_y=\\sin\\alpha$$, where $$\\alpha$$ is the angle at which the projectile is being shot. We get $$\\Delta s_x=-2v^2\\frac{\\sin(\\alpha)\\cos(\\alpha)}{g}.$$ Now here comes a bit of math slightly removed from the physical side of things: using the double angle formula, $$\\sin(\\alpha)\\cos(\\alpha)=\\frac{\\sin(2\\alpha)}{2}$$. So we get $$\\Delta s_x=-v^2\\frac{\\sin(2\\alpha)}{g}.$$ The sine is known to be maximal if its argument is $$90^\\circ$$ (immediately obvious from the unit circle definition) so $$2\\alpha=90^\\circ$$ and thus $$\\alpha=45^\\circ$$. The acceleration is the second derivative of the displacement: $$a(t) = \\frac{d^2}{dt^2} s(t)$$ If the acceleration is constant $$a(t) = A$$ then velocity is $$\\frac{d}{dt} s(t) = v(t) = \\int_{0}^{t} A d\\tau = A \\cdot t + v_0$$ where $$v_0$$ is the initial velocity. The displacement is now defined as: $$s(t) = \\int_{0}^{t} (A \\cdot \\tau + v_0) d\\tau = \\frac{1}{2} A \\cdot t^2 + v_0 \\cdot t + s_0$$ where $$s_0$$ is the initial displacement. These are well-known equations for displacement and velocity with constant acceleration, you can find them in any basic physics textbook. In your example you have two axes, namely horizontal $$x$$ and vertical $$y$$, which can be observed separately if they are perpendicular. For the projectile motion, the projectile starts at $$\\vec{s}(0) = (0,0)$$ with the initial velocity $$\\vec{v}(0) = (v_{x0},v_{y0})$$ and the acceleration is constant $$\\vec{a} = (0, -g)$$, hence $$s_x(t) = v_{x0} \\cdot t,", "projectile hits the ground), so \\begin{align} d &= v_0 t_f \\cos\\theta \\\\ &= v_0 \\left( \\frac{v_0 \\sin\\theta}{g} + \\sqrt{\\frac{2y_0}{g} + \\frac{v_0^2\\sin^2\\theta}{g^2}} \\right) \\cos\\theta \\\\ &= \\frac{v_0 \\cos\\theta}{g} \\left( v_0 \\sin\\theta + \\sqrt{2gy_0 + v_0^2\\sin^2\\theta} \\right). \\tag1 \\end{align} In the case $y_0 = 0,$ the square root in $(1)$ is just $v_0 \\sin\\theta,$ and the formula can be simplified to the form $\\frac{v_0^2 \\sin(2\\theta)}{g}$ which is often given as the range of a projectile. In the case where $\\sin\\theta \\neq 0,$ we can pull a factor of $v_0\\sin\\theta$ out of the expression in parentheses in $(1)$ to get $$d = \\frac{v_0^2\\sin\\theta\\cos\\theta}{g} \\left( 1 + \\sqrt{\\frac{2y_0g}{v_0^2\\sin^2\\theta} + 1} \\right).$$ Then, using the fact that $\\sin\\theta\\cos\\theta = \\frac{\\sin(2\\theta)}{2},$ we get the formula you found on Wikipedia. But if $\\theta = 0$ then $\\sin\\theta = 0$ and the Wikipedia formula has a division by zero, hence it does not evaluate to zero--it does not evaluate at all. • Right, accidentally said clockwise instead of counterclockwise. – Jordan T. Knobel-Piehl Jul 2 '18 at 23:24 • Just to confirm I'm doing this right: if theta=0, then d=v0(2y0/g) right? – Jordan T. Knobel-Piehl Jul 2 '18 at 23:44 • d=v0(sqrt(2y0/g)) forgot the radical – Jordan T. Knobel-Piehl Jul 2 '18 at 23:58 • Yes, in that case the distance would simplify to $d=v_0\\sqrt{2y_0/g}.$ –", "height 0, one casts a projectile into the air with constant initial velocity v but different elevation angles θ. Let x be the horizontal axis in the motion surface, and let y denote the vertical axis. Then the motion gives the following differential dynamical system: d2y dt2 =-g, d2x dt2 =0, which satisfies four initial conditions: dx dt |t=0=v\\cos\\theta, dy dt |t=0=v\\sin\\theta,x|t=0=y|t=0=0. Here t denotes motion time, θ is elevation angle, g denotes gravitational acceleration, and v is the constant initial speed (not velocity). The solution of the above system can take an implicit form: F(x,y,\\theta)=x\\tan\\theta- gx2 2v2\\cos2\\theta -y=0. To find its envelope equation, one may compute the desired derivative: \\partialF \\partial\\theta = x \\cos2\\theta - gx2\\tan\\theta v2\\cos2\\theta =0. By eliminating θ, one may reach the following envelope equation: y= v2 2g - g 2v2 x2. Clearly the resulted envelope is also a concave parabola. ## Envelope of a family of surfaces A one-parameter family of surfaces in three-dimensional Euclidean space is given by a set of equations F(x,y,z,a)=0 depending on a real parameter a. For example, the tangent planes to a surface along a curve in the surface form such a family. Two surfaces corresponding to different values a and a' intersect in a common curve defined by F(x,y,z,a)=0,{F(x,y,z,a\\prime)-F(x,y,z,a)\\overa\\prime-a}=0. In the limit as a' approaches a, this", "See: http://www.jhuapl.edu/techdigest/TD/...mbo_Homing.pdf “Figure 2. Planar engagement geometry. The planar intercept problem is illustrated along with most of the angular and Cartesian quantities necessary to derive modern guidance laws.” 5. Dec 18, 2011 ### RuneStrife Actually i was able to solve the problem using the equation: |T-G-t*(VT-VG)|=D*t where T and G are target and gun position vectors, VT and VG are velocity vectors for target and gun, D is the projectile default velocity, and || is the magnitude of the sum of those vectors (add the x and y components as above, square them, add that, take the square root, divide by D). This should give me t to intersect, and then I can plug that in to those X and Y components (from the equation above) and take the atan() of those to get the angle. Now I moved onto the next step of the problem which is to compare normal inheritance to Directional inheritance, defined such that the projectile only inherits the projection of the gun's velocity on the vector along which it is fired (which is not the vector along which it moves). Any help with this would also be appreciated. Currently I've modified the vector components in the above equation such that VP components have been changed from: (VG*cos(Theta_G)i, VG*sin(Theta_G)j) to: (VG*cos(Theta_T-Theta_G)^2)i, (VG*cos(Theta_T-Theta_G)sin(Theta_T-Theta_G))j where Theta_T", "# Math Help - trigonometry 1. ## trigonometry Give an example, with justification: A real life situation requiring the use of a trigonometry identity. Knowledge is life, so open your doors of perception. 2. Anything you want square. Measuring the sides will not get you there unless: 1) You can verify the corners are right angles, or 2) You can calculate and measure the diagonals. 3. Here is one use of identities that is introduced to students in a high school physics class ... the derivation of the range formula for the horizontal displacement of a projectile (Galilean model). $\\Delta y = v_0 \\sin(\\theta) t - \\frac{1}{2}gt^2$ since the projectile lands at the same height it started, $\\Delta y = 0$ ... $0 = v_0 \\sin(\\theta) t - \\frac{1}{2}gt^2$ solving for t ... $t = \\frac{2v_0 \\sin(\\theta)}{g}$ horizontal displacement is given by the formula ... $\\Delta x = v_0 \\cos(\\theta) t$ using the last two equations, eliminate the parameter t ... $\\Delta x = \\frac{2v_0^2 \\sin(\\theta) \\cos(\\theta)}{g}$ using the double angle identity $\\sin(2\\theta) = 2\\sin(\\theta)\\cos(\\theta)$, the last equation becomes ... $\\Delta x = \\frac{v_0^2 \\sin(2\\theta)}{g}$ from this equation, it can be deduced that the launch angle for maximum horizontal range is 45 degrees.", "the air): $\\vec {F_r}=-k\\vec v$. So there will be a resistance in both the $x$ and $y$ direction, which means the equations for the acceleration will be: \\begin{align} \\ddot y & = -g-\\frac{k}{m}\\dot y\\nonumber\\\\ \\ddot x & = -\\frac{k}{m}\\dot x\\nonumber\\end{align} To make things simpler, define $\\tau \\equiv \\frac{m}{k}$, and note that $\\tau$ has the units of time. Then we can write the acceleration equations as: \\begin{align} \\ddot y & = -g-\\dot y/\\tau\\nonumber\\\\ \\ddot x & = -\\dot x/\\tau\\nonumber\\end{align} For the initial conditions, assume $y(0)=x(0)=0$, the initial angle of the velocity is $\\theta_0$, and then $\\dot y(0)=v_{0y}=v_0\\sin\\theta_0$ and $\\dot x(0)=v_{0x}=v_0\\cos\\theta_0$. It turns out that both of these equations can be solved analytically in a very straight forward manner by direct integration and application of the initial conditions. Starting with the horizontal motion, we get: \\begin{align} \\dot x(t)&=v_{0x}e^{-t/\\tau}\\nonumber\\\\ x(t) & = v_{0x}\\tau (1-e^{-t/\\tau})\\nonumber\\end{align} Note that $v_{0x}\\tau$ is the horizontal position at $t=\\infty$ so we can define $x_\\infty\\equiv v_{0x}\\tau$ for later and write $$x(t) = x_\\infty (1-e^{-t/\\tau})\\nonumber$$ So at $t=0$, the projectile starts off with some initial velocity, and as it increases, the air resistance starts to dominate ultimately limiting the horizontal distance traveled. Solving for time $t$ (we will need this later) gives $$t = -\\tau\\ln (1-x/x_\\infty)\\nonumber$$ For the vertical motion, we can integrate the vertical acceleration $\\ddot y(t)$ to", "that might be the case? – hypergeometric Oct 24 '15 at 10:13 • When the angle is very small like 10 then the curve becomes elongated and more like an ellipse than a parabola. – Archis Welankar Oct 24 '15 at 10:25 • @ArchisWelankar - 1. The curve or trajectory of the projectile is always a parabola (even when elongated, at small launch angles) as given by the formula. 2. An ellipse doesn't have to be elongated. 3. The question refers to the envelope of different trajectories and not any one particular trajectory. – hypergeometric Oct 24 '15 at 10:34 • \"The standard solution to this \"envelope of safety\" problem is to state the formula as a quadratic in tanθ and set the discriminant to zero. \" I don't understand this approach, can someone explain ? – Dat Sep 11 at 9:38 Yes it is. To find the envelope, we just have to find the intersections between two trajectories associated to two slightly different angles. If we solve $$x\\tan\\theta -\\frac{gx^2}{2v^2}\\sec^2(\\theta) = x\\tan(\\theta+\\varepsilon) -\\frac{gx^2}{2v^2}\\sec^2(\\theta+\\varepsilon)$$ we get $x=0$ or $$x=\\frac{2v^2}{g}\\cdot\\frac{\\tan(\\theta)-\\tan(\\theta+\\varepsilon)}{\\sec^2(\\theta)-\\sec^2(\\theta+\\varepsilon)}$$ and by letting $\\varepsilon\\to 0$ we get $x=\\frac{v^2}{g}\\cot(\\theta)$, from which $y=\\frac{v^2}{2g}\\left(2-\\frac{1}{\\sin^2(\\theta)}\\right)$. It follows that the equation of the envelope is given by: $$y = \\frac{v^2}{2g}\\left(1-\\left(\\frac{gx}{v^2}\\right)^2\\right)=\\frac{v^2}{2g}-\\frac{g}{2 v^2}\\,x^2$$ that clearly is a parabola with vertex in $\\left(0,\\frac{v^2}{2g}\\right)$ through the points $\\left(\\pm\\frac{v^2}{g} ,0\\right)$.", "# Area under a parabolic trajectory I have this problem: \"prove that the area under the trajectory described by a parabolic shot that has: $f(x)=\\tan(\\theta)x - (\\frac{g}{2v^2\\cos^2(\\theta)})x^2$ and $x=v\\cos(\\theta)t$ is defined as: $A(\\theta)=\\frac{2v^4}{3g^2}\\sin^3(\\theta)\\cos(\\theta)$ I have tried to change x in order to convert $f(x)$ to a function of $\\theta$ but i end up everytime with $f(\\theta)=0$ • Please review how the trajectory has been arrived at.. what is asked is the difference with and without gravity term, when the shot goes in a straight line without gravity. – Narasimham Sep 3 '15 at 3:37 This is a straight-forward integral. $\\theta$ is fixed for the projectile. If $\\theta=0$ then the problem does not make much sense. If $\\theta=\\pi/2$ then the area is undefined. So it makes sense to assume $\\theta\\in(0,\\pi/2)\\cup(\\pi/2,\\pi)$ $$f(x)=0 \\iff x=0 \\ \\ \\vee \\ \\ x=\\dfrac{2v^2\\cos^2\\theta\\tan\\theta}{g}$$ And so, $$A(\\theta) = \\int_0^{2g^{-1}v^2\\cos^2\\theta\\tan\\theta} x\\tan\\theta - \\frac{g}{2v^2\\cos^2\\theta}x^2 \\ dx$$ which is a simple integral (polynomial in $x$). It is well known, and relatively easy to prove using calculus, that the area under a parabolic arch is $A=\\dfrac{2}{3}bh$ where $b$ is the base of the arch and $h$ its height. The height of a parabolic arch $f(x)=ax^2+bx+c$ where $a,b,c$ make sense is $f(t)$ where $t=-\\dfrac{b}{2a}$. In your case $t = \\dfrac{v^2\\cos^2\\theta\\tan\\theta}{g}$, so $$h=\\tan\\theta\\left(\\dfrac{v^2\\cos^2\\theta\\tan\\theta}{g}\\right) - \\dfrac{g}{2v^2\\cos^2\\theta}\\left(\\dfrac{v^2\\cos^2\\theta\\tan\\theta}{g}\\right)^2$$ And the base $b$ is", "i would like to test my grasp as well $$a_x(t)=k(\\frac{dx}{dt})^2$$ $$\\frac{d^2(x)}{dt^2}=k(\\frac{dx}{dt})^2$$ ?? would this be the correct equation for acceleration for the x component? i will try solving the DE if this is correct Last edited: Err... ok, I took friction to be -bv (vector), so the acceleration in the x and z directions worked out to be $$\\frac{-b}{m}v_z$$. Since the velocity would change with time as: $$v_z=v_{0z}-\\frac{b}{m}v_zt$$, this would give $$v_z$$ to be: $$\\frac{mv_{0z}}{m+bt}$$, where the mass does cancel out dimensionally. Integration of this expression gives a logarithmic function... so, is this right? andrevdh Homework Helper The air resistance, $$R$$, opposes the velocity of the projectile. The angle of attack, $$\\theta$$, of the projectile changes in its trajectory. By resolving in the x - and y directions we get that $$m\\ddot{x} = - R \\cos(\\theta)$$ and $$m\\ddot{y} = - R \\sin(\\theta) - mg$$ we can rewrite the sine and cosine terms as $$\\cos(\\theta) = \\frac{\\dot{x}}{v}$$ and $$\\sin(\\theta) = \\frac{\\dot{y}}{v}$$ $$\\ddot{x} = -a\\dot{x}v}$$ and $$\\ddot{y} = -a\\dot{y}v} - g$$ An \"easier\" approach is to rather resolve along the tangential and normal directions to the trajectory - which leads to the equation of the hodograph. Last edited: Yeah, thats exactly what I've done... just used pure vectors in 3D instead of resolving the force... andrevdh Homework Helper I've burned", "– John Rennie Jun 12 '17 at 6:27 • @John Darn...totally failed to see that. Thanks! (Do I have to delete the question now?) – Alan Jun 12 '17 at 6:31 • Well, there could be an answer. For example the obvious extra information is that $d^2y/dt^2 = g$, and including this allows you to get the velocity. It's up to you if you want to let the question stand and see if anyone answers it. – John Rennie Jun 12 '17 at 6:33 Let's start out from scratch, assume the velocity components to be: $v_x=v\\cos\\theta, v_y=v\\sin\\theta-gt$ (We know the $y$ acceleration is $g$, which is obvious from differentiating twice w.r.t $t$) Which implies $x(t)=v\\cos\\theta t, y(t)=v\\sin\\theta t-0.5gt^2$ (starting from $(0,0)$). Combining these two (i.e. getting rid of $t$), we get: $$y=x\\tan\\theta-\\frac{gx^2}{2v^2\\cos^2\\theta}$$ This is the same as your equation, but with $v=\\frac{u}{\\sqrt2}$. This works, IMO. The equation you've given is the equation of trajectory where the x and y coordinates of the position/displacement of projectile is given. For finding $v_x$ , it remains constant in time( horizontal component remains unchanged with no air resistance) For $v_s$ it is just given according to $v=u+at$ So, $$v_s=v_s(initial)-gt$$", "be described by the curve $$y = \\frac{1}{a} \\cosh{(ax)}$$, show that $$40a = \\sinh{(30a)}$$. Find an approximate value for $$a$$ using Newton's method by using a starting value of $$a_0 = 0.1$$. Hence, find an approximate height for the supports. Hint: The length of the curve $$y = f(x)$$ for $$a \\leq x \\leq b$$ is given by $$L = \\displaystyle \\int_{a}^{b} \\sqrt{1 + \\big( f'(x) \\big)^2} ~ dx$$. 4. An architectural arch is to be constructed of the form $$y = h - a \\cosh{(bx)}$$, where $$y = 0$$ represents ground level. The arch is to be $$100$$ m high at the centre ($$x = 0$$), and $$200$$ m wide at ground level. At the points where the arch meets the ground, the slope of the tangent line to the arch has a magnitude of $$3$$. Show that $$a$$ and $$b$$ must satisfy $$\\left( 1 + \\dfrac{100}{a} \\right)^2 - \\dfrac{9}{a^2b^2} = 1$$. Suppose some other construction needs dictate that $$b = 0.01$$. Find $$a$$ and $$h$$ in this case and obtain an equation for the arch. 5. The familiar trigonometric functions $$\\cos(\\theta)$$ and $$\\sin(\\theta)$$ are defined in terms of the relationship between the points on the unit circle $$x^2 + y^2 = 1$$ and the angle $$\\theta$$, measured counterclockwise from the positive $$x$$-axis to the segment $$OP$$,", ", two accelerations are acting in mutually perpendicular directions. if $\\vec{a}$ be the resultant acceleration, then $\\vec{a} = \\vec{a_t} + \\vec {a_c}$ Now magnitude is given by $\\left| \\vec{a} \\right|=\\sqrt{a_{t}^{2}+a_{c}^{2}}=\\sqrt{\\left( 0.5 \\right) ^2+\\left( 0.7 \\right) ^2}=0.86ms^{-2}$ and direction with tangential acceleration vector $\\tan \\beta =\\frac{a_c}{a_t}=\\frac{0.7}{0.5}=1.4$ $\\Rightarrow \\beta =\\tan ^{-1}\\left( 1.4 \\right) =54.5^0$ So , this is the angle from the direction of negative of the velocity or direction of tangential velocity vector. Question 4.32 (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by $\\theta \\left( t \\right) =\\tan ^{-1}\\left( \\frac{v_{oy-}gt}{v_{ox}} \\right)$ (b) Show that the projection angle , for a projectile launched from the origin is given by $\\theta _0=\\tan ^{-1}\\left( \\frac{4h_m}{R} \\right)$ where the symbols have their usual meaning. Solution Let a projectile fired at an angle $\\theta$ with x-axis. As $\\theta$ depends on $t$, $\\theta (t)$ , at any instant $\\tan \\theta \\left( t \\right) =\\frac{v_y}{v_x}=\\frac{v_{0y}-gt}{v_{ox}}$ ( Since,$v_y=v_{oy}-gt$ and $v_x=v_{ox}$ ) $\\theta \\left( t \\right) =\\tan ^{-1}\\left( \\frac{v_{oy-}gt}{v_{ox}} \\right)$ Now $h_{\\max}=\\frac{u^2\\sin ^2\\theta}{2g}$ and $R=\\frac{u^2\\sin ^2\\theta}{g}$ $Rightarrow \\frac{h_{\\max}}{R}=\\frac{\\frac{u^2\\sin ^2\\theta}{2g}}{\\frac{u^2\\sin ^22\\theta}{g}}=\\frac{\\tan \\theta}{4}$ $\\left( As\\,\\,\\sin 2\\theta =\\text{2}\\sin \\theta \\cos \\theta \\right)$ $\\Rightarrow \\frac{\\tan \\theta}{4}=\\frac{h_{\\max}}{R}$ or, $\\tan \\theta =\\frac{4h_{\\max}}{R}$ or, $\\theta =\\tan ^{-1}\\left( \\frac{4h_{\\max}}{R} \\right)$", "initial velocity $v$ at an angle with $\\theta$ the horizontal is given by $R(range) = \\frac{v^2(velocity)}{g(9.8m/s^2)} \\sin 2 \\theta$. If the rocket is launched with an initial velocity of 15 m/s, what angle is needed to reach a range of 20 m? Solution: Plug in 15 m/s for $v$ and 20 m for the range to solve for the angle. $20 &= \\frac{15^2}{9.8} \\sin 2 \\theta\\\\20 &= 22.96 \\sin 2 \\theta\\\\0.87 \\bar{1} &= \\sin 2 \\theta\\\\\\sin^{-1} (0.87 \\bar{1}) &= 2 \\theta\\\\60.59^\\circ, 119.41^\\circ &= 2 \\theta\\\\30.3^\\circ, 59.7^\\circ &= \\theta$ You can also use the TI-83 to solve trigonometric equations. It is sometimes easier than solving the equation algebraically. Just be careful with the directions and make sure your final answer is in the form that is called for. You calculator cannot put radians in terms of $\\pi$. Example 13: Solve $\\sin x = 2 \\cos x$ such that $0 \\le x \\le 2\\pi$ using a graphing calculator. Solution: In $y =$, graph $y1 = \\sin x$ and $y2 = 2 \\cos x$. Next, use CALC to find the intersection points of the graphs. ## Review Questions 1. Express the sum as a product: $\\sin 9x + \\sin 5x$ 2. Express the difference as a product: $\\cos 4y - \\cos 3y$ 3. Verify the identity (using sum-to-product formula): $\\frac{\\cos 3a", "$1.50\\tan(\\theta) = 7.50\\sin^2(\\theta)$ $\\cos(\\theta)\\sin(\\theta) = \\frac{1}{5}$ $2\\cos(\\theta)\\sin(\\theta) = \\frac{2}{5}$ $\\sin(2\\theta) = \\frac{2}{5}$ $2\\theta = \\arcsin(\\frac{2}{5})$ $\\theta = 11.79º\\ or\\ 78.21º$ Then by finding the time I wanted to comprobe whether the results I got are corrects. The time it took the ball to reach its maximum height, its horizontal displacement should have been $\\small{d = 1.50\\ m}$: $t= \\frac{v\\sin(\\theta)}{g}= \\frac{(12.12\\ m/s)\\sin(11.79º)}{9.8\\ m/s^2} = 0.253\\ s$ $v\\cos(\\theta)t = d$ $(12.12\\ m/s)\\cos(11.79º)(0.2527\\ s) = 3.00\\ m$ I don't know why I get this horizontal displacement. I expected to get $\\small{d = 1.50\\ m}$ and not the whole range. Thanks!! Last edited: Nov 17, 2014 2. Nov 17, 2014 ### Jazz I've just discovered that the formula $\\small{range = \\frac{v^2\\sin(2\\theta)}{g}}$ is a more straightforward way to solve this problem :) ; but I still don't know what I'm doing wrong above. At the beginning, I had mistakenly taken the work done by the spring as $Fx$, which yielded a greater velocity that fits the 5.77-degree angle needed for a 3-m range, but the components of the velocities are overestimated, giving me the same problem as the one above. 3. Nov 17, 2014 ### Staff: Mentor In your triangle on the right you've assumed that the projectile will follow a straight-line path at angle θ This is not so! While the launch", "dt\\\\ \\vvt \\amp = \\la 0,-gt\\ra + \\vec C\\text{.} \\end{align*} Knowing $$\\vec v(0) = v_0\\la \\cos(\\theta) ,\\sin(\\theta) \\ra\\text{,}$$ we have $$\\vec C = v_0\\la \\cos(t) ,\\sin(t) \\ra$$ and so \\begin{equation*} \\vec v(t) = \\la v_0\\cos(\\theta) , -gt+v_0\\sin(\\theta) \\ra\\text{.} \\end{equation*} We integrate once more to find $$\\vrt\\text{:}$$ \\begin{align*} \\vrt \\amp = \\int \\vvt\\,dt\\\\ \\vrt \\amp = \\int \\la v_0\\cos(\\theta) , -gt+v_0\\sin(\\theta) \\ra\\,dt\\\\ \\vrt \\amp = \\la \\big(v_0\\cos(\\theta) \\big)t, -\\frac12gt^2+\\big(v_0\\sin(\\theta) \\big)t\\ra + \\vec C.\\\\ \\end{align*} Knowing $$\\vec r(0) = \\la x_0,y_0\\ra\\text{,}$$ we conclude $$\\vec C = \\la x_0,y_0\\ra$$ and \\begin{align*} \\vrt \\amp = \\la \\big(v_0\\cos(\\theta) \\big)t+x_0\\,, -\\frac12gt^2+\\big(v_0\\sin(\\theta) \\big)t+y_0\\,\\ra\\text{.} \\end{align*} #### Key Idea12.3.13.Projectile Motion. The position function of a projectile propelled from an initial position of $$\\vec r_0=\\la x_0,y_0\\ra\\text{,}$$ with initial speed $$v_0\\text{,}$$ with angle of elevation $$\\theta$$ and neglecting all accelerations but gravity is \\begin{equation*} \\vrt = \\la \\big(v_0\\cos(\\theta) \\big)t+x_0\\,, -\\frac12gt^2+\\big(v_0\\sin(\\theta) \\big)t+y_0\\,\\ra\\text{.} \\end{equation*} Letting $$\\vec v_0 = v_0\\la \\cos(\\theta) ,\\sin(\\theta) \\ra\\text{,}$$ $$\\vrt$$ can be written as \\begin{equation*} \\vrt = \\la 0,-\\frac12gt^2\\ra + \\vec v_0t+\\vec r_0\\text{.} \\end{equation*} We demonstrate how to use this position function in the next two examples. #### Example12.3.14.Projectile Motion. Sydney shoots her Red Ryder® bb gun across level ground from an elevation of 4 ft, where the barrel of the gun makes a $$5^\\circ$$ angle with the horizontal. Find how far the bb travels before landing, assuming", "$x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curves correspond to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.4289$ The evolution of $(x(t),y(t))$, $R(t)$, $|\\theta(t)|$ for solutions of (50) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = s_2 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $b_0 = -1/4$, $z_0 = 0.6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curve corresponds to $r = 2 R_\\ast t^{-\\nu-l/q}$ The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for", "angle $$\\alpha$$ (fixed), initial velocity $$v_0$$ with $$r_0$$, $$e$$ and $$\\phi$$. I have not worked out the details but I assume this can be done using the definition of excentricity and inverse trigonometric functions. I will try to look into it. Now that you have $$r_0(\\alpha, v_0)$$, $$e(\\alpha, v_0)$$ and $$\\phi(\\alpha, v_0)$$, you can just inject into the ellipse equation and look at $$\\theta = \\theta_{\\mathrm{target}}$$. You need that $$r_{\\mathrm{target}} = r(\\alpha, v_0, \\theta = \\theta_{\\mathrm{target}}) = \\frac{r_0(\\alpha, v_0)}{1 + e(\\alpha, v_0) \\cos(\\theta_{\\mathrm{target}} + \\phi(\\alpha, v_0))}.$$ I do not know if this is the easiest way to do it, but I believe it can be done nonetheless. If you want to check if a solution exists, you need to wonder what happens if, given an angle, you increase progressively the velocity. For low velocities, you effectively fall rapidly towards the star. For very large velocity, you go through a more and more elongated ellipse. If you go beyond the escape velocity, you will not follow a closed trajectory anymore but escape to infinity (parabole or hyperbole). If your velocity becomes infinite, you basically go into a straight line. For all the velocities in between, you basically span all the half-plane separated by a straight line passing through your initial position and parallel to your initial velocity, and containing" ]

[ "by \\begin{align} R_1&=\\frac{u^2\\sin(90^\\circ-2\\theta)}{g}=\\frac{u^2\\cos(2\\theta)}{g},\\nonumber\\\\ R_2&=\\frac{u^2\\sin(90^\\circ+2\\theta)}{g}=\\frac{u^2\\cos(2\\theta)}{g}.\\nonumber \\end{align} Thus, $R_1=R_2=R$ i.e., both projectiles have same horizontal range. The maximum heights of the two projectiles are given by \\begin{align} H_1&=\\frac{u^2\\sin^2(45^\\circ-\\theta)}{2g}=\\frac{u^2}{4g}\\left(1-\\sin2\\theta\\right),\\nonumber\\\\ H_2&=\\frac{u^2\\sin^2(45^\\circ+\\theta)}{2g}=\\frac{u^2}{4g}\\left(1+\\sin2\\theta\\right).\\nonumber \\end{align} We encourage you to show that $R=4\\sqrt{H_1 H_2}$. If angle of projection of projectile 1 is $30^\\circ$ and that of projectile 2 is $60^\\circ$ then what is the ratio of the maximum heights of these projectiles? The time of flight of the two projectiles are given by \\begin{align} T_1&=\\frac{2u\\sin(45^\\circ-\\theta)}{g}=\\frac{\\sqrt{2}u}{g}\\left(\\cos\\theta-\\sin\\theta\\right),\\nonumber\\\\ T_2&=\\frac{2u\\sin(45^\\circ+\\theta)}{g}=\\frac{\\sqrt{2}u}{g}\\left(\\cos\\theta+\\sin\\theta\\right),\\nonumber \\end{align} It is easy to show that $T_1T_2=2R/g$. Problem on Maximum Height of a Projectile A ball is projected from the ground with a speed 15 m/s at an angle $\\theta$ with horizontal so that its range and maximum height are equal, then $\\tan\\theta$ will be equal to (JEE Mains 2022) 1. 1/4 2. 1/2 3. 2 4. 4 Solution: The horizontal range and the maximum height are given by \\begin{align} R=\\frac{u^2\\sin(2\\theta)}{g},\\quad\\text{and}\\quad H=\\frac{u^2\\sin^2\\theta}{2g}.\\nonumber \\end{align} Substitute in $R=H$ to get $\\tan\\theta=4$, which gives $\\theta=\\tan^{-1}(4)=76^\\circ$. Problem on Time of Flight of a Projectile A projectile is projected with velocity of 25 m/s at an angle $\\theta$ with the horizontal. After $t$ seconds its inclination with the horizontal becomes zero. If $R$ represents horizontal range of the projectile, the value of $\\theta$ will be (Take $g=10\\;\\mathrm{m/s^2}$) (JEE Mains 2022) 1. $\\frac{1}{2}\\sin^{-1}\\left(\\frac{5t^2}{4R}\\right)$ 2. $\\frac{1}{2}\\sin^{-1}\\left(\\frac{4R}{5t^2}\\right)$", ") and then move down again to reach the ground at a distance $R$ from origin. Setting $x=R$ and $y=0$ (since projectile reaches ground again) $O=R \\tan \\theta-\\frac{g}{2 u^{2} \\cos ^{2} \\theta} \\cdot R^{2}$ We get $R=\\frac{2 u^{2} \\cos ^{2} \\theta}{g} \\times \\frac{\\sin \\theta}{\\cos \\theta}$ or $\\quad \\mathrm{R}=\\frac{2 \\mathrm{u}^{2}}{\\mathrm{~g}} \\cdot \\sin \\theta \\cos \\theta$ or Range $\\mathrm{R}=\\frac{\\mathrm{u}^{2} \\sin 2 \\theta}{\\mathrm{g}}$ If time for upward journey is $t$ at highest point $\\quad v_{y}=0$ so $0=(u \\sin \\theta)-g t \\quad\\left(v_{y}=u_{y}-g t\\right)$ or $\\quad \\mathrm{t}=\\frac{\\mathrm{u} \\sin \\theta}{\\mathrm{g}}$ $\\therefore \\quad \\mathrm{T}=2 \\mathrm{t}$ (it will take same time for downward journey) $\\therefore \\quad \\mathrm{T}=\\frac{2 \\mathrm{u} \\sin \\theta}{\\mathrm{g}}$ Time of flight At the highest point $y=H$ and $v_{y}=0$ So that $H=\\frac{u_{y}^{2}}{2 g}$ $\\left[v_{y}^{2}=u_{y}^{2}-2 g y\\right]$ or $\\quad \\mathrm{H}=\\frac{\\mathrm{u}^{2} \\sin ^{2} \\theta}{2 \\mathrm{~g}} \\quad$ Maximum Height we can also determine $\\mathrm{R}$ as follows $x=u_{x} t$ so $\\quad \\mathrm{R}=\\mathrm{u}_{\\mathrm{x}} . \\mathrm{T}$ $=(u \\cos \\theta)\\left(\\frac{2 u \\sin \\theta}{g}\\right)$ or $\\quad \\mathrm{R}=\\frac{\\mathrm{u}^{2} \\sin 2 \\theta}{\\mathrm{g}}$ velocity at time $t \\vec{v}_{t}=v_{x t} \\hat{i}+v_{y t} \\hat{j}=(u \\cos \\theta) \\hat{i}+(u \\sin \\theta-g t) \\hat{j}$ $v=\\sqrt{u^{2} \\cos ^{2} \\theta+(u \\sin \\theta-g t)^{2}}$ Note : (i) Alternative eq $^{n}$. of trajectory $y=x \\tan \\theta\\left(1-\\frac{x}{R}\\right)$ where $R=\\frac{2 u^{2} \\sin \\theta \\cos \\theta}{g}$ (ii) Vertical component of velocity $v_{y}=0$, when particle is at the highest point of trajectory. (iii) Linear momentum at highest point $=m", "as measured from the surface of the incline, then the impact occurs when y = $v_o$sin(θ)t + (1/2)[-g cos(ϕ)]$t^2$= 0, and x = $v_o$cos(θ)t + (1/2)[-g sin(ϕ)]$t^2$. Solve the y equation for t, and use that to find x = (2$v_o$/g)[sin(θ)cos(θ)/cos(ϕ) – sin(ϕ)si$n^2$(θ)/co$s^2$(ϕ)]. If the surface is horizontal, then ϕ =0 and x = (2$v_o$/g)sin(θ)cos(θ). This gives a positive ${v_o}^2$. There is no reason to solve for the time. As shown in post #3, the trak=jectory equation can be rewritten as $$\\frac{\\Delta y}{\\Delta x}=\\tan\\theta-\\frac{g(\\Delta x)}{2v_0^2cos^2\\theta}.$$Then one recognizes that ##\\frac{\\Delta y}{\\Delta x} = \\tan\\varphi## and write $$\\tan\\varphi=\\tan\\theta-\\frac{g(\\Delta x)}{2v_0^2cos^2\\theta}.$$if one has the value of ##v_0##, one can plug into find ##\\Delta x##. Alternatively, one can avoid pushing too many buttons on one's calculator and use the horizontal range when the projectile lands on level ground, $$R=\\frac{2v_0^2\\cos\\theta\\sin\\theta}{g}=\\frac{2v_0^2\\cos^2\\theta\\sin\\theta}{g\\cos\\theta}\\implies \\frac{1}{2v_0^2\\cos^2\\theta}=\\frac{\\tan\\theta}{gR}$$Then the equation simplifies some more to$$\\tan\\varphi=\\tan\\theta-\\frac{\\Delta x}{R}\\tan\\theta$$which is @PeroK's expression in post #5. PeroK Homework Helper An alternative is not to calculate the initial velocity, but to derive an equation for the horizontal range on the slope (##r##) in terms of the full horizontal range (##R##): $$r = R(1 - \\frac{\\tan \\alpha}{\\tan \\theta})$$Where, in this case, ##R = 28m##, ##\\theta = 40°## and ##\\alpha = 15°##. Geometric / graphic interpretation of this equation. https://www.desmos.com/calculator/mfilokes7z Last edited: Homework Helper Consider a general function of", "trajectory is, as usual for projectile problems $\\displaystyle y = vt\\,\\sin(\\theta) - \\frac{1}{2}g\\:t^2$ where $t$ is the time of flight. So the horizontal range is $\\displaystyle R = vt\\,\\cos(\\theta)$ or $\\displaystyle t = \\frac{R}{v\\,\\cos(\\theta)}$ Setting the trajectory $y$ equal to the slope $y$ and substituting the above expression for $t$, we can solve for $R$. $\\displaystyle y = vt\\,\\sin(\\theta) - \\frac{1}{2}g\\,t^2 = -R\\,\\tan(\\alpha) \\\\[9pt] v\\,\\sin(\\theta) \\left (\\frac{R}{v\\,\\cos(\\theta)} \\right ) - \\frac{g}{2}\\left( \\frac{R^2}{v^2\\,\\cos^2(\\theta)} \\right) = -R\\,\\tan(\\alpha) \\\\[9pt] 2v^2R\\: \\frac{\\sin(\\theta)}{\\cos(\\theta)} - \\frac{g R^2}{\\cos^2(\\theta)} = -2 R v^2\\,\\tan(\\alpha) \\\\[9pt] [2v^2\\: \\sin(\\theta)\\cos(\\theta)]R - g R^2 = -2v^2\\,\\tan(\\alpha) \\cos^2(\\theta) R\\\\[9pt] [2v^2\\: \\sin(\\theta)\\cos(\\theta) + 2v^2\\,\\tan(\\alpha) \\cos^2(\\theta) ]R - g R^2 = 0\\\\[9pt] R [2v^2\\: \\sin(\\theta)\\cos(\\theta) + 2v^2\\,\\tan(\\alpha) \\cos^2(\\theta) - g R] = 0\\\\[9pt] R [v^2\\: \\sin(2\\theta) + 2v^2\\,\\tan(\\alpha) \\cos^2(\\theta) - g R] = 0$ The two points where the trajectory intersects the slope are where $R = 0$ and where $v^2\\: \\sin(2\\theta) + 2v^2\\,\\tan(\\alpha) \\cos^2(\\theta) - g R = 0$ or $\\displaystyle R = \\frac{v^2}{g}\\: [\\sin(2\\theta) + 2\\tan(\\alpha) \\cos^2(\\theta)]$ That gives the horizontal range for any launch angle $\\theta$. What is the optimum value of $\\theta$? We take the derivative with respect to $\\theta$, remembering that $\\alpha$ is a constant, so $\\displaystyle \\frac{dR}{d\\theta} = \\frac{v^2}{g}\\: [2 \\cos(2\\theta) - 4\\tan(\\alpha) \\cos(\\theta) \\sin(\\theta)] = 0 \\\\[9pt] 2 \\cos(2\\theta) = 4\\tan(\\alpha) \\cos(\\theta) \\sin(\\theta) = 2\\tan(\\alpha) \\sin(2\\theta) \\\\[9pt]", "\\sin \\theta)\\left(\\frac{x}{u \\cos \\theta}\\right)-\\frac{1}{2} g\\left(\\frac{x}{u \\cos \\theta}\\right)^{2}$ or $\\left.y=x \\tan \\theta-\\frac{g}{2 u^{2} \\cos ^{2} \\theta} \\cdot x^{2}\\right]$ This is eqn. of parabola. The trajectory of projectile is parabolic The projectile will rise to maximum height $\\mathrm{H}$ (where $\\mathrm{v}_{x}=\\mathrm{u} \\cos \\theta, \\mathrm{v}_{y}=0$ ) and then move down again to reach the ground at a distance $\\mathrm{R}$ from origin. Setting $x=R$ and $y=0$ (since projectile reaches ground again) $\\mathrm{O}=\\mathrm{R} \\tan \\theta-\\frac{\\mathrm{g}}{2 \\mathrm{u}^{2} \\cos ^{2} \\theta} \\cdot \\mathrm{R}^{2}$ We get $R=\\frac{2 u^{2} \\cos ^{2} \\theta}{g} \\times \\frac{\\sin \\theta}{\\cos \\theta}$ or $\\mathrm{R}=\\frac{2 \\mathrm{u}^{2}}{\\mathrm{~g}} \\cdot \\sin \\theta \\cos \\theta$ or Range $R=\\frac{u^{2} \\sin 2 \\theta}{g}$ If time for upward journey is t at highest point $\\quad v_{y}=0$ so $0=(u \\sin \\theta)-g t$ $\\left(v_{y}=u_{y}-g t\\right)$ or $\\mathrm{t}=\\frac{\\mathrm{u} \\sin \\theta}{\\mathrm{g}}$ $\\mathrm{T}=2 \\mathrm{t} \\quad$ (it will take same time for downward journey) $\\therefore$ $T=\\frac{2 u \\sin \\theta}{g}$ Time of flight At the highest point $y=H$ and $v_{y}=0$ Maximum Height and Time of Flight Depends on Vertical Component of Initial Velocity So that $H=\\frac{u_{y}^{2}}{2 g} \\quad\\left[v_{y}^{2}=u_{y}^{2}-2 g y\\right]$ or $\\mathrm{H}=\\frac{\\mathrm{u}^{2} \\sin ^{2} \\theta}{2 \\mathrm{~g}}$ Maximum Height we can also determine $R$ as follows so $x=u_{x} t$ $R=u_{x} \\cdot T$ $=(u \\cos \\theta)\\left(\\frac{2 u \\sin \\theta}{g}\\right)$ or $\\mathrm{R}=\\frac{\\mathrm{u}^{2} \\sin 2 \\theta}{\\mathrm{g}}$ velocity at time t $\\vec{v}_{t}=v_{x t} \\hat{i}+v_{y t} \\hat{j}$ $=(u \\cos \\theta) \\hat{i}+(u \\sin \\theta-g t) \\hat{j}$", "the second projectile is: $$\\displaystyle \\theta_2=\\frac{1}{2}\\sin^{-1}(k)$$ Now, using the equation: $$\\displaystyle r=v_0\\cos(\\theta)t\\implies t=\\frac{r}{v_0\\cos(\\theta)}=\\frac{k\\frac{v_0^2}{g}}{v_0\\cos(\\theta)}=\\frac{kv_0}{g}\\sec(\\theta)$$ And so we find, using the fact that the two launch angles are complementary: $$\\displaystyle \\Delta t=t_1-t_2=\\frac{kv_0}{g}\\left(\\csc\\left(\\frac{1}{2}\\sin^{-1}(k) \\right)-\\sec\\left(\\frac{1}{2}\\sin^{-1}(k) \\right) \\right)$$ Using half-angle identities for sine and cosine, and simplifying a bit, we obtain: $$\\displaystyle \\Delta t=\\frac{\\sqrt{2}v_0}{g}\\left(\\sqrt{1+\\sqrt{1-k^2}}-\\sqrt{1-\\sqrt{1-k^2}} \\right)$$ #### topsquark ##### Well-known member MHB Math Helper Hello, Deveno! $$\\displaystyle \\Delta t=\\frac{\\sqrt{2}v_0}{g}\\left(\\sqrt{1+\\sqrt{1-k^2}}-\\sqrt{1-\\sqrt{1-k^2}} \\right)$$", "and $v$ were in fact the horizontal and vertical components of a projectile, which I denoted with $v_h$ and $v_v$ above. So Ray's equation would read, in my notation: $\\frac{g^2}{2v_h^2}-\\frac{\\kappa g}{2}\\left(\\frac{v_v\\sqrt{v_h^2+v_v^2}}{v_h^2}+{\\rm arcsinh}\\left|\\frac{v_v}{v_h}\\right|\\right)={\\rm const,}$ where I picked up an extra minus sign because I considered $g$ to be positive when pointing downwards. But how did he find this neat form? Is it even correct? To find out, I tried a few ways to integrate the equations but to no avail. I admit I was ready to give up when I came across a solution posted on reddit. The derivation is actually quite simple. We begin with (I now use the overdot for the time derivative): \\begin{align} \\dot{v}_h=-\\kappa\\sqrt{v_h^2+v_v^2}v_h,\\\\ \\dot{v}_v=g-\\kappa\\sqrt{v_h^2+v_v^2}v_v. \\end{align} Now multiply the first equation by $\\dot{v}_v$ and the second by $\\dot{v}_h$: \\begin{align} \\dot{v}_h\\dot{v}_v=-\\kappa\\sqrt{v_h^2+v_v^2}v_h\\dot{v}_v,\\\\ \\dot{v}_h\\dot{v}_v=g\\dot{v}_h-\\kappa\\sqrt{v_h^2+v_v^2}\\dot{v}_hv_v. \\end{align} Subtract one from the other and rearrange: $g\\dot{v}_h=\\kappa\\sqrt{v_h^2+v_v^2}(\\dot{v}_hv_v-v_h\\dot{v}_v).$ Now substitute $v_v=sv_h$: $g\\dot{v}_h=\\kappa\\sqrt{(s^2+1)v_h^2}(\\dot{v}_hv_v-v_h\\dot{v}_v).$ Divide both sides by $v_h^3$: $g\\frac{\\dot{v}_h}{v_h^3}=-\\kappa\\sqrt{(s^2+1)}\\dot{s}.$ In this form, both sides are full derivatives with respect to $t$ and thus both sides can be integrated, to yield $-\\frac{1}{2}\\frac{g}{v_h^2}=-\\kappa\\left[\\frac{1}{2}s\\sqrt{s^2+1}+\\frac{1}{2}{\\rm arcsinh}~s\\right]+C,$ where $C$ is an integration constant. Multiplying by $g$ and replacing $s$ with $v_v/v_h$, we get back Ray's equation. This is an elegant result. Its practical utility may be limited, however, as the solution is implicit. Actually computing $v_h$ and $v_v$ as functions", "given by $$r = \\tfrac{v}{2} t \\hat{i} + (h + \\tfrac{\\sqrt{3}v}{2}t - 5t^2) \\hat{j}$$ then Goliath's projectile hits the ground when $$\\frac{\\sqrt{3}v \\pm \\sqrt{3v^2 + 80h}}{20} = 0$$. Solving this, the time it takes Goliath's projectile to reach the ground is $$T_g = \\frac{\\sqrt{3}v + \\sqrt{3v^2 + 80h}}{20}$$. For the horizontal distance travelled to match, their terminal horizontal components must match. Hence, \\begin{align*} v T_d \\cos \\theta &= \\tfrac{v}{2} T_g\\\\ T_d \\cos \\theta &= \\tfrac{1}{2} T_g\\\\ \\cos \\theta &= \\frac{T_g}{2} \\frac{1}{T_d}\\\\ &= \\frac{\\frac{\\sqrt{3}v + \\sqrt{3v^2 + 80h}}{20}}{2} \\frac{10}{v \\sin \\theta + \\sqrt{v^2 \\sin^2\\theta+20h}}\\\\ \\cos \\theta &= \\frac{\\sqrt{3}v + \\sqrt{3v^2 + 80h}}{4 (v \\sin \\theta + \\sqrt{v^2 \\sin^2\\theta+20h}) }\\\\ \\end{align*} I'm having a major mental block on how to obtain an expression for $$\\theta$$ in terms of $$h$$ and $$v$$. For what its worth, wolframalpha gives $$\\theta = -\\cos^{-1}\\left(-\\frac{\\sqrt{200 h^2 + 70 h v^2 + 10 \\sqrt{3} h v \\sqrt{80 h + 3 v^2} + \\sqrt{3} v^3 \\sqrt{80 h + 3 v^2} + 3 v^4}}{\\sqrt{800 h^2 + 80 h v^2 + 8 v^4}}\\right)$$. I seem to missing out some details in my calculations. What's the simplest way to obtain the formula for $$\\theta$$ in terms of $$v$$ and $$h$$? From $$\\begin{equation*} \\cos \\theta = \\frac{\\sqrt{3}v + \\sqrt{3v^2 + 80h}}{4 (v \\sin \\theta + \\sqrt{v^2 \\sin^2\\theta+20h}) }, \\end{equation*}$$ multiplying both", "of the projectile. Plugging this value for (t) into the horizontal equation yields ${\\displaystyle x={\\frac {2v^{2}\\cos \\theta \\,\\sin \\theta }{g}}}$ Applying the trigonometric identity ${\\displaystyle \\sin(x+y)=\\sin x\\,\\cos y\\ +\\ \\sin y\\,\\cos x}$ If x and y are same, ${\\displaystyle \\sin 2\\theta =2\\sin \\theta \\,\\cos \\theta }$ allows us to simplify the solution to ${\\displaystyle R={\\frac {v^{2}\\sin 2\\theta }{g}}}$ Note that when (θ) is 45°, the solution becomes ${\\displaystyle R_{\\max }={\\frac {v^{2}}{g}}}$ #### Uneven ground Now we will allow (y0) to be nonzero. Our equations of motion are now ${\\displaystyle x(t)=vt\\cos \\theta }$ and ${\\displaystyle y(t)=y_{0}+vt\\sin \\theta -{\\frac {1}{2}}gt^{2}}$ Once again we solve for (t) in the case where the (y) position of the projectile is at zero (since this is how we defined our starting height to begin with) ${\\displaystyle 0=y_{0}+vt\\sin \\theta -{\\frac {1}{2}}gt^{2}}$ Again by applying the quadratic formula we find two solutions for the time. After several steps of algebraic manipulation ${\\displaystyle t={\\frac {v\\sin \\theta }{g}}\\pm {\\frac {\\sqrt {v^{2}\\sin ^{2}\\theta +2gy_{0}}}{g}}}$ The square root must be a positive number, and since the velocity and the cosine of the launch angle can also be assumed to be positive, the solution with the greater time will occur when the positive of the plus or minus sign is used. Thus, the solution is ${\\displaystyle t={\\frac {v\\sin \\theta }{g}}+{\\frac {\\sqrt", "$C=v_0\\sin\\theta+\\frac{mg}{k}$. $$\\therefore v(t)=(v_0\\sin\\theta+\\frac{mg}{k})e^{-\\frac{kt}{m}}-\\frac{mg}{k}$$ # 6 Parametric Representations of Curves i.e. Projectile Motion. These are the parametric representations of the trajectory of a projectile: $$x(t)=(v_0\\cos\\theta)t\\quad y(t)=(v_0\\sin\\theta)t-\\frac{1}{2}gt^2$$ ### Calculus with Parametric Curves #### Tangents Given $x=f(t)$ and $y=g(t)$, $$\\frac{dy}{dt}=\\frac{dy}{dx}\\cdot \\frac{dx}{dt}$$ So, $$\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\quad\\frac{dx}{dt}\\neq 0$$ #### Areas Given a curve defined by $x=f(t)$ and $y=g(t)$, the area under the curve as $x$ goes from $a$ to $b$ or as $t$ goes from $\\alpha$ to $\\beta$ can be found by: $$A=\\int_a^b ydx=\\int_\\alpha^\\beta g(t)f'(t)dt$$ ### Arclength Using Parametric Representation Given a curve $C$ described by $x=f(t)$ and $y=g(t)$ and $f',\\ g'$ are continuous over $[a,b]$, and $C$ is traversed exactly once as $t$ goes from $a$ to $b$, find the arclength of the curve in the interval $[a,b]$. 1. Partition $[a,b]$ into $n$ sub-intervals of width $\\Delta t=\\frac{b-a}{n}$ 2. \\begin{align*} L_k=||\\overrightarrow{Q_kQ_{k+1}}||&\\approx\\sqrt{[f(t_{k+1})-f(t_k)]^2+[g(t_{k+1})-g(t_k))]^2}\\\\ &=\\Delta t\\sqrt{\\bigg[\\frac{f(t_{k+1})-f(t_k))}{\\Delta t}\\bigg]^2+\\bigg[\\frac{g(t_{k+1})-g(t_k))}{\\Delta t}\\bigg]^2}\\\\ &\\approx\\sqrt{f'(t_k^*)^2+g'(t_k^*)^2}\\Delta t \\end{align*} 3. Take the Riemann Sum. $$L=\\lim_{n\\to\\infty}\\sum_{k=1}^nL_k=\\int_a^b\\sqrt{f'(t)^2+g'(t)^2}dt$$ Alternatively, we can derive this formula from the formula for arclength of a function $y=f(x)$. $$L=\\int_a^b\\sqrt{1+\\big(\\frac{dy}{dx}\\big)^2}dx=\\int_a^b\\sqrt{1+\\bigg(\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\bigg)^2}dx=\\int_a^b\\sqrt{\\bigg(\\frac{dx}{dt}\\bigg)^2+\\bigg(\\frac{dy}{dt}\\bigg)^2}dx\\cdot\\frac{dt}{dx}=\\int_a^b\\sqrt{\\bigg(\\frac{dx}{dt}\\bigg)^2+\\bigg(\\frac{dy}{dt}\\bigg)^2}dt$$ # 7 Sequences Definition: An infinite set of real numbers ordered $a_1, a_2,\\ldots\\$ Denoted as $\\{a_n\\}$ or $\\{a_n\\}_{n=1}^\\infty$ Definition: A sequence has the limit, L, which we write as $$\\lim_{n=\\infty}a_n=L\\quad\\text{or}\\quad a_n\\to L\\text{ as } n\\to\\infty$$ if $\\ \\forall\\epsilon > 0,\\ \\exists N > 0$ such that $|a_n-L|<\\epsilon\\quad\\forall n>N$ If the limit exists, then $\\{a_n\\}$", "## Part 2: Vertical position of the projectile Resolving the initial vertical velocity, $\\dot{y} = V \\sin \\alpha$. \\begin{align} \\ddot{y} &= -g \\\\ \\dot{y} &= -gt + C_3 &\\text{integrating} \\\\ \\dot{y} &= V \\sin \\theta &\\text{when } t = 0 \\\\ V \\sin \\theta &= C_3 \\\\ \\therefore \\dot{y} &= -gt + V \\sin \\theta \\\\ y &= \\displaystyle -\\frac{1}{2}gt^2 + V t \\sin \\theta + C_4 &\\text{integrating} \\\\ y &= 0 &\\text{when } t = 0 \\\\ 0 &= C_4 \\\\ \\therefore y &= -\\frac{1}{2} gt^2 + Vt \\sin \\theta \\cdots (2) \\end{align} where $t$ is the time in seconds, after firing, and $g$ is the acceleration due to gravity. ## Part 3: Path of the trajectory of the projectile \\begin{align} \\displaystyle t &= \\frac{x}{V \\cos \\alpha} &\\text{from } (1) \\\\ y &= -\\frac{gx^2}{2V^2 \\cos^2 \\theta} + \\frac{V x \\sin \\theta}{V \\cos \\theta} &\\text{substituting into } (2) \\\\ &= -\\frac{gx^2}{2V^2} \\sec^2 \\theta + x \\tan \\theta \\\\ \\therefore y &= x \\tan \\theta – Kx^2 \\sec^2 \\theta \\cdots (3) &\\text{by letting } \\displaystyle \\frac{g}{2V^2} = K \\text{ for simplicity} \\end{align} ## Part 4: Range of the projectile \\begin{align} y &= x \\tan \\alpha &\\text{equation of the inclined plane} \\cdots (4) \\\\ x \\tan \\alpha &= x \\tan \\theta – Kx^2 \\sec^2 \\theta &\\text{equating (3) and (4)} \\\\", "# Determining the initial angle of a projectile's motion #### Question Q8.4.3 The trajectory of a projectile in the $xz$-plane launched from the origin at an angle $\\theta_0$ from the (horizontal) $x$-axis with speed $v_0 = 25\\;\\mathrm{m\\,s^{-1}}$ is $$z = x\\tan\\theta_0 - \\frac{g}{2v_0^2\\cos^2\\theta_0}x^2.$$ ##### Derivation During its flight, the projectile is subject to a constant acceleration $\\ddot{\\boldsymbol{r}} = -g\\hat{\\boldsymbol{k}}$. Integrating, \\begin{align*} \\boldsymbol{v} &= \\dot{\\boldsymbol{r}} = -gt\\hat{\\boldsymbol{k}} + \\boldsymbol{v_0}\\\\ \\boldsymbol{r} & = \\textstyle -\\frac{1}{2}gt^2\\hat{\\boldsymbol{k}} + \\boldsymbol{v_0}t + \\boldsymbol{r_0} \\end{align*} where $\\boldsymbol{v_0} = (v_0\\cos\\theta)\\hat{\\boldsymbol{i}} + v_0(\\sin\\theta)\\hat{\\boldsymbol{k}}$ and $\\boldsymbol{r_0}=0$ (the projectile starts at the origin). Therefore, $x = v_0t\\cos\\theta$ and $z = v_0t\\sin\\theta - \\frac{1}{2}gt^2$. Eliminating $t$ from these expressions gives the formula for $z$ above. If the projectile passes through the point $(5, 15)$, use Brent's method to determine the possible values of $\\theta_0$. #### Solution To access solutions, please obtain an access code from Cambridge University Press at the Lecturer Resources page for my book (registration required) and then sign up to scipython.com providing this code.", "{v^{2}\\sin ^{2}\\theta +2gy_{0}}}{g}}}$ Solving for the range once again ${\\displaystyle R={\\frac {v\\cos \\theta }{g}}\\left[v\\sin \\theta +{\\sqrt {v^{2}\\sin ^{2}\\theta +2gy_{0}}}\\right]}$ To maximize the range at any height ${\\displaystyle \\theta =\\arccos {\\sqrt {\\frac {2gy_{0}+v^{2}}{2gy_{0}+2v^{2}}}}}$ Checking the limit as ${\\displaystyle y_{0}}$ approaches 0 ${\\displaystyle \\lim _{y_{0}\\to 0}\\arccos {\\sqrt {\\frac {2gy_{0}+v^{2}}{2gy_{0}+2v^{2}}}}={\\frac {\\pi }{4}}}$ #### Angle of impact The angle ψ at which the projectile lands is given by: ${\\displaystyle \\tan \\psi ={\\frac {-v_{y}(t_{d})}{v_{x}(t_{d})}}={\\frac {\\sqrt {v^{2}\\sin ^{2}\\theta +2gy_{0}}}{v\\cos \\theta }}}$ For maximum range, this results in the following equation: ${\\displaystyle \\tan ^{2}\\psi ={\\frac {2gy_{0}+v^{2}}{v^{2}}}=C+1}$ Rewriting the original solution for θ, we get: ${\\displaystyle \\tan ^{2}\\theta ={\\frac {1-\\cos ^{2}\\theta }{\\cos ^{2}\\theta }}={\\frac {v^{2}}{2gy_{0}+v^{2}}}={\\frac {1}{C+1}}}$ Multiplying with the equation for (tan ψ)^2 gives: ${\\displaystyle \\tan ^{2}\\psi \\,\\tan ^{2}\\theta ={\\frac {2gy_{0}+v^{2}}{v^{2}}}{\\frac {v^{2}}{2gy_{0}+v^{2}}}=1}$ Because of the trigonometric identity ${\\displaystyle \\tan(\\theta +\\psi )={\\frac {\\tan \\theta +\\tan \\psi }{1-\\tan \\theta \\tan \\psi }}}$, this means that θ + ψ must be 90 degrees. ## Actual projectile motion In addition to air resistance, which slows a projectile and reduces its range, many other factors also have to be accounted for when actual projectile motion is considered. ### Projectile characteristics Generally speaking, a projectile with greater volume faces greater air resistance, reducing the range of the projectile. This can be modified by the projectile shape: a tall and wide, but short", "is (https://en.wikipedia.org/wiki/Range_of_a_projectile), $$R_1=\\frac{v^2}{2g} \\left(1+\\sqrt{1+\\frac{2gh}{v^2\\sin^2(\\theta)}}\\right)\\sin(2\\theta)$$ this form doesn't look much like the flat-table range, \\begin{aligned} R_o^2 &= \\frac{2 v_o^2 h}{g} \\\\ R_o &= v_o \\cdot \\sqrt{\\frac{2h}{g}} \\end{aligned} but you can show that they are the same in the limit $$\\theta\\rightarrow 0$$, \\begin{aligned} R_1&=\\frac{v^2}{2g} \\left(1+\\sqrt{1+\\frac{2gh}{v^2\\sin^2(\\theta)}}\\right)\\sin(2\\theta) \\\\ &\\sim \\frac{v^2}{2g} \\left(1+\\sqrt{1+\\frac{2gh}{v^2 \\theta^2}}\\right)\\cdot 2\\theta \\\\ &\\sim \\frac{v^2}{2g} \\left(\\sqrt{\\frac{2gh}{v^2 \\theta^2}}\\right)\\cdot 2\\theta \\\\ &=v \\sqrt{\\frac{2 h}{g}} \\end{aligned} which matches the form of $$R_o$$. It's not obvious that $$R_1$$ is an increasing function of $$\\theta$$, as stated in the solution (e.g. angle goes up, the range goes up) but numerically it is easy to demonstrate, B=2 # true value x0=1.1 m=0.01 h=20 g=10 θ=linspace(0.001,60,100) U=B*x0**4/4 KE=U v=sqrt(KE*2/m) R0=sqrt(v**2*2*h/g)*ones(len(θ)) θ_at_max=arcsin(sqrt(2)*v*sqrt(1/(g*h + v**2))/2) θ_at_max2=arcsin(1/sqrt(2)*1/sqrt(1+2*g*h*m/B/x0**4)) B_est0=R0**2*g*m/(x0**4*h) B_est1=R1**2*g*m/(x0**4*h) figure(figsize=(16,6)) subplot(1,2,1) plot(θ,R0,label='No Ramp') plot(θ,R1,label='With Ramp') vlines(degrees(θ_at_max),*ylim(),ls=':') vlines(degrees(θ_at_max2),*ylim(),ls='--',lw=1,color='r') ylabel('Range') xlabel('Angle') legend() subplot(1,2,2) plot(θ,B_est0,label='No Ramp') plot(θ,B_est1,label='With Ramp') vlines(degrees(θ_at_max),*ylim(),ls=':') ylabel('$B$') xlabel('Angle') legend() As you can see, for small angle, the range increases with angle before decreasing again for steep angles. So far, so good. You wouldn't expect a student on a test to do the next parts, but I wasn't convinced that the assumption made in the solution held in all cases, or that there wasn't some range of parameters for which the simplistic \"angle goes up, range goes up\" would hold experimentally. So, I continued the analysis in as much deatil as I", "projectile hits the ground), so \\begin{align} d &= v_0 t_f \\cos\\theta \\\\ &= v_0 \\left( \\frac{v_0 \\sin\\theta}{g} + \\sqrt{\\frac{2y_0}{g} + \\frac{v_0^2\\sin^2\\theta}{g^2}} \\right) \\cos\\theta \\\\ &= \\frac{v_0 \\cos\\theta}{g} \\left( v_0 \\sin\\theta + \\sqrt{2gy_0 + v_0^2\\sin^2\\theta} \\right). \\tag1 \\end{align} In the case $y_0 = 0,$ the square root in $(1)$ is just $v_0 \\sin\\theta,$ and the formula can be simplified to the form $\\frac{v_0^2 \\sin(2\\theta)}{g}$ which is often given as the range of a projectile. In the case where $\\sin\\theta \\neq 0,$ we can pull a factor of $v_0\\sin\\theta$ out of the expression in parentheses in $(1)$ to get $$d = \\frac{v_0^2\\sin\\theta\\cos\\theta}{g} \\left( 1 + \\sqrt{\\frac{2y_0g}{v_0^2\\sin^2\\theta} + 1} \\right).$$ Then, using the fact that $\\sin\\theta\\cos\\theta = \\frac{\\sin(2\\theta)}{2},$ we get the formula you found on Wikipedia. But if $\\theta = 0$ then $\\sin\\theta = 0$ and the Wikipedia formula has a division by zero, hence it does not evaluate to zero--it does not evaluate at all. • Right, accidentally said clockwise instead of counterclockwise. – Jordan T. Knobel-Piehl Jul 2 '18 at 23:24 • Just to confirm I'm doing this right: if theta=0, then d=v0(2y0/g) right? – Jordan T. Knobel-Piehl Jul 2 '18 at 23:44 • d=v0(sqrt(2y0/g)) forgot the radical – Jordan T. Knobel-Piehl Jul 2 '18 at 23:58 • Yes, in that case the distance would simplify to $d=v_0\\sqrt{2y_0/g}.$ –", "y}-\\mathrm{g} t}{v_{0 x}}\\right)$(b) Show that the projection angle for a projectile launched from the origin is given by$\\theta_{0}=\\tan ^{-1}\\left(\\frac{4 h_{m}}{R}\\right)$Where the symbols have their usual meaning. Solution.(a) Let$v_{0 x}$and$v_{0 y}$respectively be the initial components of the velocity of the projectile along horizontal$(x)$and vertical$(y)$directions. Let$v_{x}$and$v_{y}$respectively be the horizontal and vertical components of velocity at a point$P$. Time taken by the projectile to reach point P = t Applying the first equation of motion along the vertical and horizontal directions, we get:$v_{y}=v_{0 y}=\\mathrm{g} t$And$v_{x}=v_{0 x}\\therefore \\tan \\theta=\\frac{v_{y}}{v_{x}}=\\frac{v_{0 y}-\\mathrm{g} t}{v_{0 x}}\\theta=\\tan ^{-1}\\left(\\frac{v_{0 y}-\\mathrm{g} t}{v_{0 x}}\\right)$(b) Maximum vertical height,$h_{m}=\\frac{u_{0}^{2} \\sin ^{2} \\theta}{2 g} \\ldots(i)$Horizontal range,$R=\\frac{u_{0}^{2} \\sin 2 \\theta}{g} \\quad \\ldots(i i)$Solving equations (i) and (ii), we get:$\\frac{h_{m}}{R}=\\frac{\\sin ^{2} \\theta}{2 \\sin 2 \\theta}=\\frac{\\sin ^{2} \\theta}{4 \\sin \\theta \\cos \\theta}=\\frac{\\sin \\theta}{4 \\cos \\theta}\\Rightarrow \\frac{h_{m}}{R}=\\frac{\\operatorname{Tan} \\theta}{4}\\Rightarrow \\theta=\\operatorname{Tan}^{-1}\\left(\\frac{4 h_{m}}{R}\\right)\\$", "$1.50\\tan(\\theta) = 7.50\\sin^2(\\theta)$ $\\cos(\\theta)\\sin(\\theta) = \\frac{1}{5}$ $2\\cos(\\theta)\\sin(\\theta) = \\frac{2}{5}$ $\\sin(2\\theta) = \\frac{2}{5}$ $2\\theta = \\arcsin(\\frac{2}{5})$ $\\theta = 11.79º\\ or\\ 78.21º$ Then by finding the time I wanted to comprobe whether the results I got are corrects. The time it took the ball to reach its maximum height, its horizontal displacement should have been $\\small{d = 1.50\\ m}$: $t= \\frac{v\\sin(\\theta)}{g}= \\frac{(12.12\\ m/s)\\sin(11.79º)}{9.8\\ m/s^2} = 0.253\\ s$ $v\\cos(\\theta)t = d$ $(12.12\\ m/s)\\cos(11.79º)(0.2527\\ s) = 3.00\\ m$ I don't know why I get this horizontal displacement. I expected to get $\\small{d = 1.50\\ m}$ and not the whole range. Thanks!! Last edited: Nov 17, 2014 2. Nov 17, 2014 ### Jazz I've just discovered that the formula $\\small{range = \\frac{v^2\\sin(2\\theta)}{g}}$ is a more straightforward way to solve this problem :) ; but I still don't know what I'm doing wrong above. At the beginning, I had mistakenly taken the work done by the spring as $Fx$, which yielded a greater velocity that fits the 5.77-degree angle needed for a 3-m range, but the components of the velocities are overestimated, giving me the same problem as the one above. 3. Nov 17, 2014 ### Staff: Mentor In your triangle on the right you've assumed that the projectile will follow a straight-line path at angle θ This is not so! While the launch", ", two accelerations are acting in mutually perpendicular directions. if $\\vec{a}$ be the resultant acceleration, then $\\vec{a} = \\vec{a_t} + \\vec {a_c}$ Now magnitude is given by $\\left| \\vec{a} \\right|=\\sqrt{a_{t}^{2}+a_{c}^{2}}=\\sqrt{\\left( 0.5 \\right) ^2+\\left( 0.7 \\right) ^2}=0.86ms^{-2}$ and direction with tangential acceleration vector $\\tan \\beta =\\frac{a_c}{a_t}=\\frac{0.7}{0.5}=1.4$ $\\Rightarrow \\beta =\\tan ^{-1}\\left( 1.4 \\right) =54.5^0$ So , this is the angle from the direction of negative of the velocity or direction of tangential velocity vector. Question 4.32 (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by $\\theta \\left( t \\right) =\\tan ^{-1}\\left( \\frac{v_{oy-}gt}{v_{ox}} \\right)$ (b) Show that the projection angle , for a projectile launched from the origin is given by $\\theta _0=\\tan ^{-1}\\left( \\frac{4h_m}{R} \\right)$ where the symbols have their usual meaning. Solution Let a projectile fired at an angle $\\theta$ with x-axis. As $\\theta$ depends on $t$, $\\theta (t)$ , at any instant $\\tan \\theta \\left( t \\right) =\\frac{v_y}{v_x}=\\frac{v_{0y}-gt}{v_{ox}}$ ( Since,$v_y=v_{oy}-gt$ and $v_x=v_{ox}$ ) $\\theta \\left( t \\right) =\\tan ^{-1}\\left( \\frac{v_{oy-}gt}{v_{ox}} \\right)$ Now $h_{\\max}=\\frac{u^2\\sin ^2\\theta}{2g}$ and $R=\\frac{u^2\\sin ^2\\theta}{g}$ $Rightarrow \\frac{h_{\\max}}{R}=\\frac{\\frac{u^2\\sin ^2\\theta}{2g}}{\\frac{u^2\\sin ^22\\theta}{g}}=\\frac{\\tan \\theta}{4}$ $\\left( As\\,\\,\\sin 2\\theta =\\text{2}\\sin \\theta \\cos \\theta \\right)$ $\\Rightarrow \\frac{\\tan \\theta}{4}=\\frac{h_{\\max}}{R}$ or, $\\tan \\theta =\\frac{4h_{\\max}}{R}$ or, $\\theta =\\tan ^{-1}\\left( \\frac{4h_{\\max}}{R} \\right)$", "# Problem on projectile 1. Jul 10, 2013 ### AakashPandita See the attachment. The right answer is θ=60 degrees but i am getting sinθ = ±√3. #### Attached Files: • ###### scan0001.jpg File size: 32.1 KB Views: 67 2. Jul 10, 2013 ### Saitama Your expression for $v_y$ is wrong. Also, check your last step when you plug 3 in 4. Last edited: Jul 10, 2013 3. Jul 10, 2013 ### Ackbach I don't agree with your very first equation. It is not true that $$H= \\frac{u^{2} \\sin^{2}( \\theta)}{2g}.$$ Conservation of Energy requires, instead, that $$mgH+ \\frac{m u^{2} \\cos^{2}( \\theta)}{2}= \\frac{mu^{2}}{2},$$ comparing the peak to the starting-point, or $$gH+ \\frac{u^{2} \\cos^{2}( \\theta)}{2}= \\frac{u^{2}}{2}.$$ Then you can also write $$\\frac{gH}{2}+ \\frac{5 u^{2} \\cos^{2}( \\theta)}{4}= \\frac{u^{2}}{2},$$ comparing the energies at the half-way point to the starting-point. 4. Jul 10, 2013 ### AakashPandita First things first. How is my expression for vy wrong? I used v2-u2 = 2as 5. Jul 10, 2013 ### Saitama What are the directions for velocity in vertical direction and acceleration? 6. Jul 10, 2013 ### haruspex Doesn't that reduce to the same equation? 7. Jul 10, 2013 ### Ackbach You're quite right. Chalk that one up to needing to do one step at a time. Also, for the half-way point, it doesn't reduce the same way.", "dt\\\\ \\vvt \\amp = \\la 0,-gt\\ra + \\vec C\\text{.} \\end{align*} Knowing $$\\vec v(0) = v_0\\la \\cos(\\theta) ,\\sin(\\theta) \\ra\\text{,}$$ we have $$\\vec C = v_0\\la \\cos(t) ,\\sin(t) \\ra$$ and so \\begin{equation*} \\vec v(t) = \\la v_0\\cos(\\theta) , -gt+v_0\\sin(\\theta) \\ra\\text{.} \\end{equation*} We integrate once more to find $$\\vrt\\text{:}$$ \\begin{align*} \\vrt \\amp = \\int \\vvt\\,dt\\\\ \\vrt \\amp = \\int \\la v_0\\cos(\\theta) , -gt+v_0\\sin(\\theta) \\ra\\,dt\\\\ \\vrt \\amp = \\la \\big(v_0\\cos(\\theta) \\big)t, -\\frac12gt^2+\\big(v_0\\sin(\\theta) \\big)t\\ra + \\vec C.\\\\ \\end{align*} Knowing $$\\vec r(0) = \\la x_0,y_0\\ra\\text{,}$$ we conclude $$\\vec C = \\la x_0,y_0\\ra$$ and \\begin{align*} \\vrt \\amp = \\la \\big(v_0\\cos(\\theta) \\big)t+x_0\\,, -\\frac12gt^2+\\big(v_0\\sin(\\theta) \\big)t+y_0\\,\\ra\\text{.} \\end{align*} #### Key Idea12.3.13.Projectile Motion. The position function of a projectile propelled from an initial position of $$\\vec r_0=\\la x_0,y_0\\ra\\text{,}$$ with initial speed $$v_0\\text{,}$$ with angle of elevation $$\\theta$$ and neglecting all accelerations but gravity is \\begin{equation*} \\vrt = \\la \\big(v_0\\cos(\\theta) \\big)t+x_0\\,, -\\frac12gt^2+\\big(v_0\\sin(\\theta) \\big)t+y_0\\,\\ra\\text{.} \\end{equation*} Letting $$\\vec v_0 = v_0\\la \\cos(\\theta) ,\\sin(\\theta) \\ra\\text{,}$$ $$\\vrt$$ can be written as \\begin{equation*} \\vrt = \\la 0,-\\frac12gt^2\\ra + \\vec v_0t+\\vec r_0\\text{.} \\end{equation*} We demonstrate how to use this position function in the next two examples. #### Example12.3.14.Projectile Motion. Sydney shoots her Red Ryder® bb gun across level ground from an elevation of 4 ft, where the barrel of the gun makes a $$5^\\circ$$ angle with the horizontal. Find how far the bb travels before landing, assuming" ]

[ "+0 # help 0 65 1 A projectile is fired with an initial velocity of $$v$$ at an angle of $$\\theta$$ from the ground. Then its trajectory can modeled by the parametric equations \\begin{align*} x &= vt \\cos \\theta, \\\\ y &= vt \\sin \\theta - \\frac{1}{2} gt^2, \\end{align*} where $$t$$ denotes time and $$g$$ denotes acceleration due to gravity, forming a parabolic arch. Suppose $$v$$ is held constant, but $$\\theta$$ is allowed to vary, over $$0^\\circ \\le \\theta \\le 180^\\circ.$$ The highest point of each parabolic arch is plotted. As $$\\theta$$ varies, the highest points of the arches trace a closed curve. The area of this closed curve can be expressed in the form $$c \\cdot \\frac{v^4}{g^2}.$$ Find $$c.$$ Nov 11, 2019", "$C=v_0\\sin\\theta+\\frac{mg}{k}$. $$\\therefore v(t)=(v_0\\sin\\theta+\\frac{mg}{k})e^{-\\frac{kt}{m}}-\\frac{mg}{k}$$ # 6 Parametric Representations of Curves i.e. Projectile Motion. These are the parametric representations of the trajectory of a projectile: $$x(t)=(v_0\\cos\\theta)t\\quad y(t)=(v_0\\sin\\theta)t-\\frac{1}{2}gt^2$$ ### Calculus with Parametric Curves #### Tangents Given $x=f(t)$ and $y=g(t)$, $$\\frac{dy}{dt}=\\frac{dy}{dx}\\cdot \\frac{dx}{dt}$$ So, $$\\frac{dy}{dx}=\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\quad\\frac{dx}{dt}\\neq 0$$ #### Areas Given a curve defined by $x=f(t)$ and $y=g(t)$, the area under the curve as $x$ goes from $a$ to $b$ or as $t$ goes from $\\alpha$ to $\\beta$ can be found by: $$A=\\int_a^b ydx=\\int_\\alpha^\\beta g(t)f'(t)dt$$ ### Arclength Using Parametric Representation Given a curve $C$ described by $x=f(t)$ and $y=g(t)$ and $f',\\ g'$ are continuous over $[a,b]$, and $C$ is traversed exactly once as $t$ goes from $a$ to $b$, find the arclength of the curve in the interval $[a,b]$. 1. Partition $[a,b]$ into $n$ sub-intervals of width $\\Delta t=\\frac{b-a}{n}$ 2. \\begin{align*} L_k=||\\overrightarrow{Q_kQ_{k+1}}||&\\approx\\sqrt{[f(t_{k+1})-f(t_k)]^2+[g(t_{k+1})-g(t_k))]^2}\\\\ &=\\Delta t\\sqrt{\\bigg[\\frac{f(t_{k+1})-f(t_k))}{\\Delta t}\\bigg]^2+\\bigg[\\frac{g(t_{k+1})-g(t_k))}{\\Delta t}\\bigg]^2}\\\\ &\\approx\\sqrt{f'(t_k^*)^2+g'(t_k^*)^2}\\Delta t \\end{align*} 3. Take the Riemann Sum. $$L=\\lim_{n\\to\\infty}\\sum_{k=1}^nL_k=\\int_a^b\\sqrt{f'(t)^2+g'(t)^2}dt$$ Alternatively, we can derive this formula from the formula for arclength of a function $y=f(x)$. $$L=\\int_a^b\\sqrt{1+\\big(\\frac{dy}{dx}\\big)^2}dx=\\int_a^b\\sqrt{1+\\bigg(\\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}\\bigg)^2}dx=\\int_a^b\\sqrt{\\bigg(\\frac{dx}{dt}\\bigg)^2+\\bigg(\\frac{dy}{dt}\\bigg)^2}dx\\cdot\\frac{dt}{dx}=\\int_a^b\\sqrt{\\bigg(\\frac{dx}{dt}\\bigg)^2+\\bigg(\\frac{dy}{dt}\\bigg)^2}dt$$ # 7 Sequences Definition: An infinite set of real numbers ordered $a_1, a_2,\\ldots\\$ Denoted as $\\{a_n\\}$ or $\\{a_n\\}_{n=1}^\\infty$ Definition: A sequence has the limit, L, which we write as $$\\lim_{n=\\infty}a_n=L\\quad\\text{or}\\quad a_n\\to L\\text{ as } n\\to\\infty$$ if $\\ \\forall\\epsilon > 0,\\ \\exists N > 0$ such that $|a_n-L|<\\epsilon\\quad\\forall n>N$ If the limit exists, then $\\{a_n\\}$", "$$y = ax + bx^2$$ for some given (known) values of $a$ and $b.$ This can only be a projectile trajectory if the same path is also given by an equation of the form $$y = (\\tan\\theta)x - \\left(\\frac{g}{2v_0^2\\cos^2\\theta}\\right)x^2$$ for some values of $v_0,$ $\\theta,$ and $g.$ In order for both equations to describe the same path, the right hand side must be the same polynomial in each equation. This means the coefficient of $x$ must be the same each time, and the coefficient of $x^2$ must be the same. So we have a projectile trajectory only if we can simultaneously solve these two equations: \\begin{align} a &= \\tan\\theta, \\\\ b &= -\\frac{g}{2v_0^2\\cos^2\\theta} = -\\frac{g}{2v_0^2}\\sec^2\\theta. \\end{align} From the first of these equations we have $\\theta = \\tan^{-1}(a)$ (for a projectile that travels over the part of the parabola for which $x \\geq 0$) or $\\theta = \\tan^{-1}(a) + \\pi$ (for the other part of the parabola). Using the identity $\\sec^2\\theta = 1 + \\tan^2\\theta,$ we have $$b = -\\frac{g}{2v_0^2}\\sec^2\\theta = -\\frac{g}{2v_0^2}(1 + \\tan^2\\theta),$$ and using the fact that $a = \\tan\\theta,$ $$b = -\\frac{g}{2v_0^2}(1 +a^2).$$ If $g$ is known, then there is only one unknown in this last equation, $v_0.$ To find $v_0$ we first isolate $v_0^2$ on one side of the equation: $$v_0^2 = -\\frac{g(1 + a^2)}{2b}.$$", "$$v_x=v\\cos\\alpha,v_y=\\sin\\alpha$$, where $$\\alpha$$ is the angle at which the projectile is being shot. We get $$\\Delta s_x=-2v^2\\frac{\\sin(\\alpha)\\cos(\\alpha)}{g}.$$ Now here comes a bit of math slightly removed from the physical side of things: using the double angle formula, $$\\sin(\\alpha)\\cos(\\alpha)=\\frac{\\sin(2\\alpha)}{2}$$. So we get $$\\Delta s_x=-v^2\\frac{\\sin(2\\alpha)}{g}.$$ The sine is known to be maximal if its argument is $$90^\\circ$$ (immediately obvious from the unit circle definition) so $$2\\alpha=90^\\circ$$ and thus $$\\alpha=45^\\circ$$. The acceleration is the second derivative of the displacement: $$a(t) = \\frac{d^2}{dt^2} s(t)$$ If the acceleration is constant $$a(t) = A$$ then velocity is $$\\frac{d}{dt} s(t) = v(t) = \\int_{0}^{t} A d\\tau = A \\cdot t + v_0$$ where $$v_0$$ is the initial velocity. The displacement is now defined as: $$s(t) = \\int_{0}^{t} (A \\cdot \\tau + v_0) d\\tau = \\frac{1}{2} A \\cdot t^2 + v_0 \\cdot t + s_0$$ where $$s_0$$ is the initial displacement. These are well-known equations for displacement and velocity with constant acceleration, you can find them in any basic physics textbook. In your example you have two axes, namely horizontal $$x$$ and vertical $$y$$, which can be observed separately if they are perpendicular. For the projectile motion, the projectile starts at $$\\vec{s}(0) = (0,0)$$ with the initial velocity $$\\vec{v}(0) = (v_{x0},v_{y0})$$ and the acceleration is constant $$\\vec{a} = (0, -g)$$, hence $$s_x(t) = v_{x0} \\cdot t,", "projectile hits the ground), so \\begin{align} d &= v_0 t_f \\cos\\theta \\\\ &= v_0 \\left( \\frac{v_0 \\sin\\theta}{g} + \\sqrt{\\frac{2y_0}{g} + \\frac{v_0^2\\sin^2\\theta}{g^2}} \\right) \\cos\\theta \\\\ &= \\frac{v_0 \\cos\\theta}{g} \\left( v_0 \\sin\\theta + \\sqrt{2gy_0 + v_0^2\\sin^2\\theta} \\right). \\tag1 \\end{align} In the case $y_0 = 0,$ the square root in $(1)$ is just $v_0 \\sin\\theta,$ and the formula can be simplified to the form $\\frac{v_0^2 \\sin(2\\theta)}{g}$ which is often given as the range of a projectile. In the case where $\\sin\\theta \\neq 0,$ we can pull a factor of $v_0\\sin\\theta$ out of the expression in parentheses in $(1)$ to get $$d = \\frac{v_0^2\\sin\\theta\\cos\\theta}{g} \\left( 1 + \\sqrt{\\frac{2y_0g}{v_0^2\\sin^2\\theta} + 1} \\right).$$ Then, using the fact that $\\sin\\theta\\cos\\theta = \\frac{\\sin(2\\theta)}{2},$ we get the formula you found on Wikipedia. But if $\\theta = 0$ then $\\sin\\theta = 0$ and the Wikipedia formula has a division by zero, hence it does not evaluate to zero--it does not evaluate at all. • Right, accidentally said clockwise instead of counterclockwise. – Jordan T. Knobel-Piehl Jul 2 '18 at 23:24 • Just to confirm I'm doing this right: if theta=0, then d=v0(2y0/g) right? – Jordan T. Knobel-Piehl Jul 2 '18 at 23:44 • d=v0(sqrt(2y0/g)) forgot the radical – Jordan T. Knobel-Piehl Jul 2 '18 at 23:58 • Yes, in that case the distance would simplify to $d=v_0\\sqrt{2y_0/g}.$ –", "height 0, one casts a projectile into the air with constant initial velocity v but different elevation angles θ. Let x be the horizontal axis in the motion surface, and let y denote the vertical axis. Then the motion gives the following differential dynamical system: d2y dt2 =-g, d2x dt2 =0, which satisfies four initial conditions: dx dt |t=0=v\\cos\\theta, dy dt |t=0=v\\sin\\theta,x|t=0=y|t=0=0. Here t denotes motion time, θ is elevation angle, g denotes gravitational acceleration, and v is the constant initial speed (not velocity). The solution of the above system can take an implicit form: F(x,y,\\theta)=x\\tan\\theta- gx2 2v2\\cos2\\theta -y=0. To find its envelope equation, one may compute the desired derivative: \\partialF \\partial\\theta = x \\cos2\\theta - gx2\\tan\\theta v2\\cos2\\theta =0. By eliminating θ, one may reach the following envelope equation: y= v2 2g - g 2v2 x2. Clearly the resulted envelope is also a concave parabola. ## Envelope of a family of surfaces A one-parameter family of surfaces in three-dimensional Euclidean space is given by a set of equations F(x,y,z,a)=0 depending on a real parameter a. For example, the tangent planes to a surface along a curve in the surface form such a family. Two surfaces corresponding to different values a and a' intersect in a common curve defined by F(x,y,z,a)=0,{F(x,y,z,a\\prime)-F(x,y,z,a)\\overa\\prime-a}=0. In the limit as a' approaches a, this", "See: http://www.jhuapl.edu/techdigest/TD/...mbo_Homing.pdf “Figure 2. Planar engagement geometry. The planar intercept problem is illustrated along with most of the angular and Cartesian quantities necessary to derive modern guidance laws.” 5. Dec 18, 2011 ### RuneStrife Actually i was able to solve the problem using the equation: |T-G-t*(VT-VG)|=D*t where T and G are target and gun position vectors, VT and VG are velocity vectors for target and gun, D is the projectile default velocity, and || is the magnitude of the sum of those vectors (add the x and y components as above, square them, add that, take the square root, divide by D). This should give me t to intersect, and then I can plug that in to those X and Y components (from the equation above) and take the atan() of those to get the angle. Now I moved onto the next step of the problem which is to compare normal inheritance to Directional inheritance, defined such that the projectile only inherits the projection of the gun's velocity on the vector along which it is fired (which is not the vector along which it moves). Any help with this would also be appreciated. Currently I've modified the vector components in the above equation such that VP components have been changed from: (VG*cos(Theta_G)i, VG*sin(Theta_G)j) to: (VG*cos(Theta_T-Theta_G)^2)i, (VG*cos(Theta_T-Theta_G)sin(Theta_T-Theta_G))j where Theta_T", "# Math Help - trigonometry 1. ## trigonometry Give an example, with justification: A real life situation requiring the use of a trigonometry identity. Knowledge is life, so open your doors of perception. 2. Anything you want square. Measuring the sides will not get you there unless: 1) You can verify the corners are right angles, or 2) You can calculate and measure the diagonals. 3. Here is one use of identities that is introduced to students in a high school physics class ... the derivation of the range formula for the horizontal displacement of a projectile (Galilean model). $\\Delta y = v_0 \\sin(\\theta) t - \\frac{1}{2}gt^2$ since the projectile lands at the same height it started, $\\Delta y = 0$ ... $0 = v_0 \\sin(\\theta) t - \\frac{1}{2}gt^2$ solving for t ... $t = \\frac{2v_0 \\sin(\\theta)}{g}$ horizontal displacement is given by the formula ... $\\Delta x = v_0 \\cos(\\theta) t$ using the last two equations, eliminate the parameter t ... $\\Delta x = \\frac{2v_0^2 \\sin(\\theta) \\cos(\\theta)}{g}$ using the double angle identity $\\sin(2\\theta) = 2\\sin(\\theta)\\cos(\\theta)$, the last equation becomes ... $\\Delta x = \\frac{v_0^2 \\sin(2\\theta)}{g}$ from this equation, it can be deduced that the launch angle for maximum horizontal range is 45 degrees.", "the air): $\\vec {F_r}=-k\\vec v$. So there will be a resistance in both the $x$ and $y$ direction, which means the equations for the acceleration will be: \\begin{align} \\ddot y & = -g-\\frac{k}{m}\\dot y\\nonumber\\\\ \\ddot x & = -\\frac{k}{m}\\dot x\\nonumber\\end{align} To make things simpler, define $\\tau \\equiv \\frac{m}{k}$, and note that $\\tau$ has the units of time. Then we can write the acceleration equations as: \\begin{align} \\ddot y & = -g-\\dot y/\\tau\\nonumber\\\\ \\ddot x & = -\\dot x/\\tau\\nonumber\\end{align} For the initial conditions, assume $y(0)=x(0)=0$, the initial angle of the velocity is $\\theta_0$, and then $\\dot y(0)=v_{0y}=v_0\\sin\\theta_0$ and $\\dot x(0)=v_{0x}=v_0\\cos\\theta_0$. It turns out that both of these equations can be solved analytically in a very straight forward manner by direct integration and application of the initial conditions. Starting with the horizontal motion, we get: \\begin{align} \\dot x(t)&=v_{0x}e^{-t/\\tau}\\nonumber\\\\ x(t) & = v_{0x}\\tau (1-e^{-t/\\tau})\\nonumber\\end{align} Note that $v_{0x}\\tau$ is the horizontal position at $t=\\infty$ so we can define $x_\\infty\\equiv v_{0x}\\tau$ for later and write $$x(t) = x_\\infty (1-e^{-t/\\tau})\\nonumber$$ So at $t=0$, the projectile starts off with some initial velocity, and as it increases, the air resistance starts to dominate ultimately limiting the horizontal distance traveled. Solving for time $t$ (we will need this later) gives $$t = -\\tau\\ln (1-x/x_\\infty)\\nonumber$$ For the vertical motion, we can integrate the vertical acceleration $\\ddot y(t)$ to", "that might be the case? – hypergeometric Oct 24 '15 at 10:13 • When the angle is very small like 10 then the curve becomes elongated and more like an ellipse than a parabola. – Archis Welankar Oct 24 '15 at 10:25 • @ArchisWelankar - 1. The curve or trajectory of the projectile is always a parabola (even when elongated, at small launch angles) as given by the formula. 2. An ellipse doesn't have to be elongated. 3. The question refers to the envelope of different trajectories and not any one particular trajectory. – hypergeometric Oct 24 '15 at 10:34 • \"The standard solution to this \"envelope of safety\" problem is to state the formula as a quadratic in tanθ and set the discriminant to zero. \" I don't understand this approach, can someone explain ? – Dat Sep 11 at 9:38 Yes it is. To find the envelope, we just have to find the intersections between two trajectories associated to two slightly different angles. If we solve $$x\\tan\\theta -\\frac{gx^2}{2v^2}\\sec^2(\\theta) = x\\tan(\\theta+\\varepsilon) -\\frac{gx^2}{2v^2}\\sec^2(\\theta+\\varepsilon)$$ we get $x=0$ or $$x=\\frac{2v^2}{g}\\cdot\\frac{\\tan(\\theta)-\\tan(\\theta+\\varepsilon)}{\\sec^2(\\theta)-\\sec^2(\\theta+\\varepsilon)}$$ and by letting $\\varepsilon\\to 0$ we get $x=\\frac{v^2}{g}\\cot(\\theta)$, from which $y=\\frac{v^2}{2g}\\left(2-\\frac{1}{\\sin^2(\\theta)}\\right)$. It follows that the equation of the envelope is given by: $$y = \\frac{v^2}{2g}\\left(1-\\left(\\frac{gx}{v^2}\\right)^2\\right)=\\frac{v^2}{2g}-\\frac{g}{2 v^2}\\,x^2$$ that clearly is a parabola with vertex in $\\left(0,\\frac{v^2}{2g}\\right)$ through the points $\\left(\\pm\\frac{v^2}{g} ,0\\right)$.", "# Area under a parabolic trajectory I have this problem: \"prove that the area under the trajectory described by a parabolic shot that has: $f(x)=\\tan(\\theta)x - (\\frac{g}{2v^2\\cos^2(\\theta)})x^2$ and $x=v\\cos(\\theta)t$ is defined as: $A(\\theta)=\\frac{2v^4}{3g^2}\\sin^3(\\theta)\\cos(\\theta)$ I have tried to change x in order to convert $f(x)$ to a function of $\\theta$ but i end up everytime with $f(\\theta)=0$ • Please review how the trajectory has been arrived at.. what is asked is the difference with and without gravity term, when the shot goes in a straight line without gravity. – Narasimham Sep 3 '15 at 3:37 This is a straight-forward integral. $\\theta$ is fixed for the projectile. If $\\theta=0$ then the problem does not make much sense. If $\\theta=\\pi/2$ then the area is undefined. So it makes sense to assume $\\theta\\in(0,\\pi/2)\\cup(\\pi/2,\\pi)$ $$f(x)=0 \\iff x=0 \\ \\ \\vee \\ \\ x=\\dfrac{2v^2\\cos^2\\theta\\tan\\theta}{g}$$ And so, $$A(\\theta) = \\int_0^{2g^{-1}v^2\\cos^2\\theta\\tan\\theta} x\\tan\\theta - \\frac{g}{2v^2\\cos^2\\theta}x^2 \\ dx$$ which is a simple integral (polynomial in $x$). It is well known, and relatively easy to prove using calculus, that the area under a parabolic arch is $A=\\dfrac{2}{3}bh$ where $b$ is the base of the arch and $h$ its height. The height of a parabolic arch $f(x)=ax^2+bx+c$ where $a,b,c$ make sense is $f(t)$ where $t=-\\dfrac{b}{2a}$. In your case $t = \\dfrac{v^2\\cos^2\\theta\\tan\\theta}{g}$, so $$h=\\tan\\theta\\left(\\dfrac{v^2\\cos^2\\theta\\tan\\theta}{g}\\right) - \\dfrac{g}{2v^2\\cos^2\\theta}\\left(\\dfrac{v^2\\cos^2\\theta\\tan\\theta}{g}\\right)^2$$ And the base $b$ is", "i would like to test my grasp as well $$a_x(t)=k(\\frac{dx}{dt})^2$$ $$\\frac{d^2(x)}{dt^2}=k(\\frac{dx}{dt})^2$$ ?? would this be the correct equation for acceleration for the x component? i will try solving the DE if this is correct Last edited: Err... ok, I took friction to be -bv (vector), so the acceleration in the x and z directions worked out to be $$\\frac{-b}{m}v_z$$. Since the velocity would change with time as: $$v_z=v_{0z}-\\frac{b}{m}v_zt$$, this would give $$v_z$$ to be: $$\\frac{mv_{0z}}{m+bt}$$, where the mass does cancel out dimensionally. Integration of this expression gives a logarithmic function... so, is this right? andrevdh Homework Helper The air resistance, $$R$$, opposes the velocity of the projectile. The angle of attack, $$\\theta$$, of the projectile changes in its trajectory. By resolving in the x - and y directions we get that $$m\\ddot{x} = - R \\cos(\\theta)$$ and $$m\\ddot{y} = - R \\sin(\\theta) - mg$$ we can rewrite the sine and cosine terms as $$\\cos(\\theta) = \\frac{\\dot{x}}{v}$$ and $$\\sin(\\theta) = \\frac{\\dot{y}}{v}$$ $$\\ddot{x} = -a\\dot{x}v}$$ and $$\\ddot{y} = -a\\dot{y}v} - g$$ An \"easier\" approach is to rather resolve along the tangential and normal directions to the trajectory - which leads to the equation of the hodograph. Last edited: Yeah, thats exactly what I've done... just used pure vectors in 3D instead of resolving the force... andrevdh Homework Helper I've burned", "– John Rennie Jun 12 '17 at 6:27 • @John Darn...totally failed to see that. Thanks! (Do I have to delete the question now?) – Alan Jun 12 '17 at 6:31 • Well, there could be an answer. For example the obvious extra information is that $d^2y/dt^2 = g$, and including this allows you to get the velocity. It's up to you if you want to let the question stand and see if anyone answers it. – John Rennie Jun 12 '17 at 6:33 Let's start out from scratch, assume the velocity components to be: $v_x=v\\cos\\theta, v_y=v\\sin\\theta-gt$ (We know the $y$ acceleration is $g$, which is obvious from differentiating twice w.r.t $t$) Which implies $x(t)=v\\cos\\theta t, y(t)=v\\sin\\theta t-0.5gt^2$ (starting from $(0,0)$). Combining these two (i.e. getting rid of $t$), we get: $$y=x\\tan\\theta-\\frac{gx^2}{2v^2\\cos^2\\theta}$$ This is the same as your equation, but with $v=\\frac{u}{\\sqrt2}$. This works, IMO. The equation you've given is the equation of trajectory where the x and y coordinates of the position/displacement of projectile is given. For finding $v_x$ , it remains constant in time( horizontal component remains unchanged with no air resistance) For $v_s$ it is just given according to $v=u+at$ So, $$v_s=v_s(initial)-gt$$", "be described by the curve $$y = \\frac{1}{a} \\cosh{(ax)}$$, show that $$40a = \\sinh{(30a)}$$. Find an approximate value for $$a$$ using Newton's method by using a starting value of $$a_0 = 0.1$$. Hence, find an approximate height for the supports. Hint: The length of the curve $$y = f(x)$$ for $$a \\leq x \\leq b$$ is given by $$L = \\displaystyle \\int_{a}^{b} \\sqrt{1 + \\big( f'(x) \\big)^2} ~ dx$$. 4. An architectural arch is to be constructed of the form $$y = h - a \\cosh{(bx)}$$, where $$y = 0$$ represents ground level. The arch is to be $$100$$ m high at the centre ($$x = 0$$), and $$200$$ m wide at ground level. At the points where the arch meets the ground, the slope of the tangent line to the arch has a magnitude of $$3$$. Show that $$a$$ and $$b$$ must satisfy $$\\left( 1 + \\dfrac{100}{a} \\right)^2 - \\dfrac{9}{a^2b^2} = 1$$. Suppose some other construction needs dictate that $$b = 0.01$$. Find $$a$$ and $$h$$ in this case and obtain an equation for the arch. 5. The familiar trigonometric functions $$\\cos(\\theta)$$ and $$\\sin(\\theta)$$ are defined in terms of the relationship between the points on the unit circle $$x^2 + y^2 = 1$$ and the angle $$\\theta$$, measured counterclockwise from the positive $$x$$-axis to the segment $$OP$$,", ", two accelerations are acting in mutually perpendicular directions. if $\\vec{a}$ be the resultant acceleration, then $\\vec{a} = \\vec{a_t} + \\vec {a_c}$ Now magnitude is given by $\\left| \\vec{a} \\right|=\\sqrt{a_{t}^{2}+a_{c}^{2}}=\\sqrt{\\left( 0.5 \\right) ^2+\\left( 0.7 \\right) ^2}=0.86ms^{-2}$ and direction with tangential acceleration vector $\\tan \\beta =\\frac{a_c}{a_t}=\\frac{0.7}{0.5}=1.4$ $\\Rightarrow \\beta =\\tan ^{-1}\\left( 1.4 \\right) =54.5^0$ So , this is the angle from the direction of negative of the velocity or direction of tangential velocity vector. Question 4.32 (a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by $\\theta \\left( t \\right) =\\tan ^{-1}\\left( \\frac{v_{oy-}gt}{v_{ox}} \\right)$ (b) Show that the projection angle , for a projectile launched from the origin is given by $\\theta _0=\\tan ^{-1}\\left( \\frac{4h_m}{R} \\right)$ where the symbols have their usual meaning. Solution Let a projectile fired at an angle $\\theta$ with x-axis. As $\\theta$ depends on $t$, $\\theta (t)$ , at any instant $\\tan \\theta \\left( t \\right) =\\frac{v_y}{v_x}=\\frac{v_{0y}-gt}{v_{ox}}$ ( Since,$v_y=v_{oy}-gt$ and $v_x=v_{ox}$ ) $\\theta \\left( t \\right) =\\tan ^{-1}\\left( \\frac{v_{oy-}gt}{v_{ox}} \\right)$ Now $h_{\\max}=\\frac{u^2\\sin ^2\\theta}{2g}$ and $R=\\frac{u^2\\sin ^2\\theta}{g}$ $Rightarrow \\frac{h_{\\max}}{R}=\\frac{\\frac{u^2\\sin ^2\\theta}{2g}}{\\frac{u^2\\sin ^22\\theta}{g}}=\\frac{\\tan \\theta}{4}$ $\\left( As\\,\\,\\sin 2\\theta =\\text{2}\\sin \\theta \\cos \\theta \\right)$ $\\Rightarrow \\frac{\\tan \\theta}{4}=\\frac{h_{\\max}}{R}$ or, $\\tan \\theta =\\frac{4h_{\\max}}{R}$ or, $\\theta =\\tan ^{-1}\\left( \\frac{4h_{\\max}}{R} \\right)$", "initial velocity $v$ at an angle with $\\theta$ the horizontal is given by $R(range) = \\frac{v^2(velocity)}{g(9.8m/s^2)} \\sin 2 \\theta$. If the rocket is launched with an initial velocity of 15 m/s, what angle is needed to reach a range of 20 m? Solution: Plug in 15 m/s for $v$ and 20 m for the range to solve for the angle. $20 &= \\frac{15^2}{9.8} \\sin 2 \\theta\\\\20 &= 22.96 \\sin 2 \\theta\\\\0.87 \\bar{1} &= \\sin 2 \\theta\\\\\\sin^{-1} (0.87 \\bar{1}) &= 2 \\theta\\\\60.59^\\circ, 119.41^\\circ &= 2 \\theta\\\\30.3^\\circ, 59.7^\\circ &= \\theta$ You can also use the TI-83 to solve trigonometric equations. It is sometimes easier than solving the equation algebraically. Just be careful with the directions and make sure your final answer is in the form that is called for. You calculator cannot put radians in terms of $\\pi$. Example 13: Solve $\\sin x = 2 \\cos x$ such that $0 \\le x \\le 2\\pi$ using a graphing calculator. Solution: In $y =$, graph $y1 = \\sin x$ and $y2 = 2 \\cos x$. Next, use CALC to find the intersection points of the graphs. ## Review Questions 1. Express the sum as a product: $\\sin 9x + \\sin 5x$ 2. Express the difference as a product: $\\cos 4y - \\cos 3y$ 3. Verify the identity (using sum-to-product formula): $\\frac{\\cos 3a", "$1.50\\tan(\\theta) = 7.50\\sin^2(\\theta)$ $\\cos(\\theta)\\sin(\\theta) = \\frac{1}{5}$ $2\\cos(\\theta)\\sin(\\theta) = \\frac{2}{5}$ $\\sin(2\\theta) = \\frac{2}{5}$ $2\\theta = \\arcsin(\\frac{2}{5})$ $\\theta = 11.79º\\ or\\ 78.21º$ Then by finding the time I wanted to comprobe whether the results I got are corrects. The time it took the ball to reach its maximum height, its horizontal displacement should have been $\\small{d = 1.50\\ m}$: $t= \\frac{v\\sin(\\theta)}{g}= \\frac{(12.12\\ m/s)\\sin(11.79º)}{9.8\\ m/s^2} = 0.253\\ s$ $v\\cos(\\theta)t = d$ $(12.12\\ m/s)\\cos(11.79º)(0.2527\\ s) = 3.00\\ m$ I don't know why I get this horizontal displacement. I expected to get $\\small{d = 1.50\\ m}$ and not the whole range. Thanks!! Last edited: Nov 17, 2014 2. Nov 17, 2014 ### Jazz I've just discovered that the formula $\\small{range = \\frac{v^2\\sin(2\\theta)}{g}}$ is a more straightforward way to solve this problem :) ; but I still don't know what I'm doing wrong above. At the beginning, I had mistakenly taken the work done by the spring as $Fx$, which yielded a greater velocity that fits the 5.77-degree angle needed for a 3-m range, but the components of the velocities are overestimated, giving me the same problem as the one above. 3. Nov 17, 2014 ### Staff: Mentor In your triangle on the right you've assumed that the projectile will follow a straight-line path at angle θ This is not so! While the launch", "dt\\\\ \\vvt \\amp = \\la 0,-gt\\ra + \\vec C\\text{.} \\end{align*} Knowing $$\\vec v(0) = v_0\\la \\cos(\\theta) ,\\sin(\\theta) \\ra\\text{,}$$ we have $$\\vec C = v_0\\la \\cos(t) ,\\sin(t) \\ra$$ and so \\begin{equation*} \\vec v(t) = \\la v_0\\cos(\\theta) , -gt+v_0\\sin(\\theta) \\ra\\text{.} \\end{equation*} We integrate once more to find $$\\vrt\\text{:}$$ \\begin{align*} \\vrt \\amp = \\int \\vvt\\,dt\\\\ \\vrt \\amp = \\int \\la v_0\\cos(\\theta) , -gt+v_0\\sin(\\theta) \\ra\\,dt\\\\ \\vrt \\amp = \\la \\big(v_0\\cos(\\theta) \\big)t, -\\frac12gt^2+\\big(v_0\\sin(\\theta) \\big)t\\ra + \\vec C.\\\\ \\end{align*} Knowing $$\\vec r(0) = \\la x_0,y_0\\ra\\text{,}$$ we conclude $$\\vec C = \\la x_0,y_0\\ra$$ and \\begin{align*} \\vrt \\amp = \\la \\big(v_0\\cos(\\theta) \\big)t+x_0\\,, -\\frac12gt^2+\\big(v_0\\sin(\\theta) \\big)t+y_0\\,\\ra\\text{.} \\end{align*} #### Key Idea12.3.13.Projectile Motion. The position function of a projectile propelled from an initial position of $$\\vec r_0=\\la x_0,y_0\\ra\\text{,}$$ with initial speed $$v_0\\text{,}$$ with angle of elevation $$\\theta$$ and neglecting all accelerations but gravity is \\begin{equation*} \\vrt = \\la \\big(v_0\\cos(\\theta) \\big)t+x_0\\,, -\\frac12gt^2+\\big(v_0\\sin(\\theta) \\big)t+y_0\\,\\ra\\text{.} \\end{equation*} Letting $$\\vec v_0 = v_0\\la \\cos(\\theta) ,\\sin(\\theta) \\ra\\text{,}$$ $$\\vrt$$ can be written as \\begin{equation*} \\vrt = \\la 0,-\\frac12gt^2\\ra + \\vec v_0t+\\vec r_0\\text{.} \\end{equation*} We demonstrate how to use this position function in the next two examples. #### Example12.3.14.Projectile Motion. Sydney shoots her Red Ryder® bb gun across level ground from an elevation of 4 ft, where the barrel of the gun makes a $$5^\\circ$$ angle with the horizontal. Find how far the bb travels before landing, assuming", "$x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curves correspond to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.4289$ The evolution of $(x(t),y(t))$, $R(t)$, $|\\theta(t)|$ for solutions of (50) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = s_2 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $b_0 = -1/4$, $z_0 = 0.6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curve corresponds to $r = 2 R_\\ast t^{-\\nu-l/q}$ The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for", "angle $$\\alpha$$ (fixed), initial velocity $$v_0$$ with $$r_0$$, $$e$$ and $$\\phi$$. I have not worked out the details but I assume this can be done using the definition of excentricity and inverse trigonometric functions. I will try to look into it. Now that you have $$r_0(\\alpha, v_0)$$, $$e(\\alpha, v_0)$$ and $$\\phi(\\alpha, v_0)$$, you can just inject into the ellipse equation and look at $$\\theta = \\theta_{\\mathrm{target}}$$. You need that $$r_{\\mathrm{target}} = r(\\alpha, v_0, \\theta = \\theta_{\\mathrm{target}}) = \\frac{r_0(\\alpha, v_0)}{1 + e(\\alpha, v_0) \\cos(\\theta_{\\mathrm{target}} + \\phi(\\alpha, v_0))}.$$ I do not know if this is the easiest way to do it, but I believe it can be done nonetheless. If you want to check if a solution exists, you need to wonder what happens if, given an angle, you increase progressively the velocity. For low velocities, you effectively fall rapidly towards the star. For very large velocity, you go through a more and more elongated ellipse. If you go beyond the escape velocity, you will not follow a closed trajectory anymore but escape to infinity (parabole or hyperbole). If your velocity becomes infinite, you basically go into a straight line. For all the velocities in between, you basically span all the half-plane separated by a straight line passing through your initial position and parallel to your initial velocity, and containing" ]

[ "# find the volume of region bounded by curve z=5+x^2+y^2 , x^2+y^2=4y above xy plane... find the volume of region bounded by curve z=5+x^2+y^2 , x^2+y^2=4y above xy plane", "# Volume of region in space - triple integral $x^2 +y ^2 +z^2 = 4$ and the cone $z=-\\sqrt(x^2+y^2)$ with $-\\sqrt2\\leqslant z\\leqslant 0$ $\\int_{0}^{(3\\pi /4)}\\int_{0}^{(2\\pi)}\\int _{0}^{\\sqrt{2}} r^2\\sin \\theta drd\\theta d\\varphi$", "# Volume of region in space - triple integral • April 30th 2011, 05:22 AM olski1 Volume of region in space - triple integral Consider the region bound by $x^2 +y ^2 +z^2 = 4$ and the cone $z=-\\sqrt(x^2+y^2)$ with $-\\sqrt2\\leqslant z\\leqslant 0$ In spherical co-ords have i set up the correct bounds? $\\int_{0}^{(3\\pi /4)}\\int_{0}^{(2\\pi)}\\int _{0}^{\\sqrt{2}} r^2\\sin \\theta drd\\theta d\\varphi$", "## Volume of region in space - triple integral Consider the region bound by $x^2 +y ^2 +z^2 = 4$ and the cone $z=-\\sqrt(x^2+y^2)$ with $-\\sqrt2\\leqslant z\\leqslant 0$ In spherical co-ords have i set up the correct bounds? $\\int_{0}^{(3\\pi /4)}\\int_{0}^{(2\\pi)}\\int _{0}^{\\sqrt{2}} r^2\\sin \\theta drd\\theta d\\varphi$", "any region of space is $\\aleph_1$. -", "The region bounded by $y=e^{2 x},y=0,x=-2,x=0$ is rotated around the $x$-axis. Find the volume. volume =", "Let $$D$$ be the region bounded by $$y = x^2$$, $$y = x + 2$$, and $$y = -x$$. a. Show that $\\iint_D x \\space dA = \\int_0^1 \\int_{-y}^{\\sqrt{y}} x \\space dx \\space dy + \\int_1^2 \\int_{y-2}^{\\sqrt{y}} x \\space dx \\space dy$ by dividing the region $$D$$ into two regions of Type II, where $$D = {(x,y)|y \\geq x^2, \\space y \\geq -x, \\space y \\leq x + 2}. b. Evaluate the integral $\\iint_D x \\space dA.$ [Hide Solution] a. Answers may vary; b. \\(\\frac{8}{12}$$ The region $$D$$ bounded by $$x = 0, y = x^5 + 1$$, and $$y = 3 - x^2$$ is shown in the following figure. Find the area $$A(D)$$ of the region $$D$$. The region $$D$$ bounded by $$y = cos \\space x, \\space y = 4 \\space cos \\space x$$, and $$x = \\pm \\frac{\\pi}{3}$$ is shown in the following figure. Find the area $$A(D)$$ of the region $$D$$. [Hide Solution] $$\\frac{8\\pi}{3}$$ Find the area $$A(D)$$ of the region $$D = {(x,y)|y \\geq 1 - x^2, y \\leq 4 - x^2, \\space y \\geq 0, \\space x \\geq 0}.$$ Let $$D$$ be the region bounded by $$y = 1, \\space y = x, \\space y = ln \\space x$$, and the $$x$$-axis. Find the area $$A(D)$$ of the region $$D$$. [Hide Solution]", "How do you find the volume of a region that is bounded by x=y^2-6y+10 and x=5 and rotated about the y-axis? Feb 25, 2015 Ok, have a look at this one: Check my numbers and the procedure.", "## Calculus 3 Find the volume of the region bounded above by the sphere $$x^{2} + y^{2} + z^{2} = 1$$ and below by the sphere $$x^{2} + y^{2} + (z-1)^{2} = 1$$", "= [6z, 2x + y, -x]$ and $S$ is the entire surface of the region bounded by the cylinder $x^2 + z^2 = 9$, $x = 0$, $y = 0$, $z = 0$, and $y = 8$. Spoiler", "the volume of a space region. To formally find the volume of a closed, bounded region $D$ in space, such as the one shown in Figure 13.6.6(a), we start with an approximation. Break $D$ into $n$ rectangular solids; the solids near the boundary of $D$ may possibly not include portions of $D$ and/or include extra space. In Figure 13.6.6(b), we zoom in on a portion of the boundary of $D$ to show a rectangular solid that contains space not in $D\\text{;}$ as this is an approximation of the volume, this is acceptable and this error will be reduced as we shrink the size of our solids. The volume $\\Delta V_i$ of the $i^\\text{ \\,th }$ solid $D_i$ is $\\Delta V_i = \\dx_i\\dy_i\\ddz_i\\text{,}$ where $\\dx_i\\text{,}$ $\\dy_i$ and $\\ddz_i$ give the dimensions of the rectangular solid in the $x\\text{,}$ $y$ and $z$ directions, respectively. By summing up the volumes of all $n$ solids, we get an approximation of the volume $V$ of $D\\text{:}$ \\begin{equation*} V \\approx \\sum_{i=1}^n \\Delta V_i = \\sum_{i=1}^n \\dx_i\\dy_i\\ddz_i. \\end{equation*} Let $\\norm{\\Delta D}$ represent the length of the longest diagonal of rectangular solids in the subdivision of $D\\text{.}$ As $\\norm{\\Delta D}\\to 0\\text{,}$ the volume of each solid goes to 0, as do each of $\\dx_i\\text{,}$ $\\dy_i$ and $\\ddz_i\\text{,}$ for all $i\\text{.}$ Our calculus experience tells us that", "# find the volume aquired by taking by region bound by y=x^2 and y=x^4 and rotating it around x=-1 Question find the volume aquired by taking by region bound by y=x^2 and y=x^4 and rotating it around x=-1", "# Volume of a Region defined by Inequalities [duplicate] I am trying to compute the volume of a 3-dimensional region defined by 5 inequalities. Mathematica takes too long to compute it. In fact, I haven't ever seen it complete the execution of the below code: theta = Pi/5; dh = 0.5; din = 1.016; dout = 1.27; zz = z * Cos[theta] - x*Sin[theta]; xx = z * Sin[theta] + x*Cos[theta]; yy = y; region1 = z^2 >= (dh/2 + 0.0254)^2 - (dout/2)^2* x^2/(x^2 + y^2); region2 = (din/2)^2 < x^2 + y^2 < (dout/2)^2; region3 = -1 <= z <= 1; region4 = zz^2 <= (dh/2)^2 - (dout/2)^2* xx^2/(xx^2 + yy^2); region5 = (din/2)^2 < xx^2 + yy^2 < (dout/2)^2; region = ImplicitRegion[{region1, region2, region3, region4, region5}, {x, y, z}]; Volume[region] Any reason why it takes so long (or does not complete)? Any alternative method to compute the volume (that does not involve discretization)? EDIT: Calculating the volume of even two of the regions similarly takes too long: theta = Pi/5; dh = 0.5; din = 1.016; dout = 1.27; zz = z * Cos[theta] - x*Sin[theta]; xx = z * Sin[theta] + x*Cos[theta]; yy = y; regionA = zz^2 <= (dh/2)^2 - (dout/2)^2 * xx^2/(xx^2 + yy^2); regionB = (din/2)^2 < xx^2 + yy^2 < (dout/2)^2;", "X-axis at (0,0) (iii) tangent the X-axis at (-7, 0) (iv) tangent to the X axis at (5, 0) c... ##### EvaluateK Lvov Vz+ydrdy:ii) Find the volume of the space region which is bounded above by below by the plane region R : y 2l3, V=I,120, y 2 0 (in the first quadrant)and Evaluate K Lvov Vz+ydrdy: ii) Find the volume of the space region which is bounded above by below by the plane region R : y 2l3, V=I,120, y 2 0 (in the first quadrant) and...", "# Find the area of the region bounded by the graphs of the given equations. – $y \\space = \\space 4x \\space + \\space 5$ and $y \\space = \\space x^2$ The main objective of this question is to find the area of the bounded region for the given expression. This question uses the concept of the area of the bounded region. The area of the bounded region can find by evaluating the definite integral. We have to find the area of the bounded region. So, given that: $\\space y \\space = \\space 4 x \\space + \\space 5$ $\\space y \\space = \\space x^2$ Now for finding the intersecting point, we know that: $\\space 4 x \\space + \\space 5 \\space = \\space x^2$ $\\space – 4 x \\space – \\space 5 \\space + \\space x^2 \\space = \\space 0$ $\\space x^2 \\space – \\space 4 x \\space – \\space 5 \\space = \\space 0$ Solving the equation results in: $\\space x_1 \\space = \\space 5$ $\\space x_2 \\space = \\space – \\space 1$ By putting the values, we get: $\\space y \\space = \\space 4 x \\space + \\space 5$ $\\space y \\space = \\space 4 ( 5 ) \\space + \\space 5$ $\\space y \\space = \\space 2 0 \\space + \\space 5$ $\\space y", "# Find the volume of the solid bound by {eq}z = 4 - (x^2 + y^2) {/eq} and z = 0. ## Question: Find the volume of the solid bound by {eq}z = 4 - (x^2 + y^2) {/eq} and z = 0. ## Cylindrical Coordinates: Recall that we can find the volume of region by setting up and evaluating a triple integral. Since our region possesses axial symmetry, we will find it convenient to use cylindrical coordinates. Recall {eq}x = r \\cos \\theta {/eq} {eq}y = r \\sin \\theta {/eq} {eq}z = z {/eq} {eq}r^2 = x^2+y^2 {/eq} {eq}\\theta = \\tan^{-1} \\frac{y}{x} {/eq} {eq}dV = r \\ dz \\ dr \\ d\\theta {/eq}", "## anonymous 5 years ago Find the volume of the space region that lies inside the sphere x^2+y^2+z^2=2 and outside the cylinder y^2+z^2=1 what should be my solution way,and how can I easily solve that problem 1. anonymous Find the volume of the part of the sphere inside the cylinder, then subtract that from the volume of the cylinder. 2. anonymous err subtract it from the volume of the sphere rather. 3. anonymous The volume of the portion of the sphere inside the cylinder should be: $\\int x_{sphere}dA$ Where A is the area inside the cylinder. 4. anonymous ty ,I solved", "to find the volume of the region bounded by the two paraboloids and .", "to determine the volume of the region below $$z = 6 - x$$, above $$z = - \\sqrt {4{x^2} + 4{y^2}}$$ inside the cylinder $${x^2} + {y^2} = 3$$ with $$x \\le 0$$. Solution 5. Evaluate the following integral by first converting to an integral in cylindrical coordinates. $\\int_{0}^{{\\sqrt 5 }}{{\\int_{{ - \\sqrt {5 - {x^{\\,2}}} }}^{0}{{\\int_{{{x^{\\,2}} + {y^{\\,2}} - 11}}^{{9 - 3{x^{\\,2}} - 3{y^{\\,2}}}}{{\\,\\,\\,2x - 3y\\,\\,\\,dz}}\\,dy}}\\,dx}}$ Solution", "# What is the volume of the region enclosed by y=2-0.5x, y=0, x=1, x=2, that is rotated about the x-axis? Mar 13, 2015 Basically it should give a truncated cone:" ]

[ "have $x-y\\geq1$, $y-z\\geq1$. The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too. $[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3); // three - currentprojection, orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), dashed+green); draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green); draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((0,0,0)--(1,0,0)--(1,1,1), red); draw((1,1,1)--(1,1,0)--(1,0.9,0), red); draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle); draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed); draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed); [/asy]$ The volume of this region is $V_2=\\dfrac{(n-2)^3}{6}$. So the probability is $p=\\dfrac{V_2}{V_1}=\\dfrac{(n-2)^3}{n^3}$. Substituting $n$ with the values in the choices, we find that when $n=10$, $p=\\frac{512}{1000}>\\frac{1}{2}$, when $n=9$, $p=\\frac{343}{729}<\\frac{1}{2}$. So $n\\geq 10$. So the answer is $\\boxed{\\textbf{(D)}\\ 10}$. ## Solution 2 Because $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$, which means that $x$, $y$, and $z$ distributes uniformly and independently in the interval $[0,n]$. So the point $(x, y, z)$ distributes uniformly in the cubic $0\\leqslant x, y, z \\leqslant n$, as shown in the figure below. The volume of this cubic is $V_0=n^3$. As we want to find the probablity of the incident $A=\\big\\{ |x-y| \\geq 1, |y-z| \\geq1, |z-x| \\geq", "# 2010 AIME I Problems/Problem 11 ## Problem Let $\\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \\le 10$ and $3y - x \\ge 15$. When $\\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$, the volume of the resulting solid is $\\frac {m\\pi}{n\\sqrt {p}}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m + n + p$. ## Solution $[asy]size(280); import graph; real min = 2, max = 12; pen dark = linewidth(1); real P(real x) { return x/3 + 5; } real Q(real x) { return 10 - abs(x - 8); } path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12)); pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(0.9,0.9,0.9)); draw(graph(P,min,max),dark); draw(graph(Q,min,max),dark); draw(Arc((8,7.67),A,G,CW),dark,EndArrow(8)); draw(B--C--G--cycle,linetype(\"4 4\")); label(\"y \\ge x/3 + 5\",(max,P(max)),E,fontsize(10)); label(\"y \\le 10 - |x-8|\",(max,Q(max)),E,fontsize(10)); label(\"\\mathcal{R}\",(6,Q(6)),NW); /* axes */ Label f; f.p=fontsize(8); xaxis(0, max, Ticks(f, 6, 1)); yaxis(0, 10, Ticks(f, 5, 1)); [/asy]$ The inequalities are equivalent to $y \\ge x/3 + 5, y \\le 10 - |x - 8|$. We can set them equal to find the two points of intersection, $x/3 + 5 = 10 -", "24 \\quad \\textbf{(E) } 36$ $[asy] import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label(\"D\",(0,0,0),S); label(\"A\",(0,0,1),N); label(\"H\",(0,1,0),S); label(\"E\",(0,1,1),N); label(\"C\",(1,0,0),S); label(\"B\",(1,0,1),N); label(\"G\",(1,1,0),S); label(\"F\",(1,1,1),N); [/asy]$ Problem 13 How many subsets of two elements can be removed from the set $\\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\\}$ so that the mean (average) of the remaining numbers is $6$? $\\textbf{(A)}\\text{ 1}\\qquad\\textbf{(B)}\\text{ 2}\\qquad\\textbf{(C)}\\text{ 3}\\qquad\\textbf{(D)}\\text{ 5}\\qquad\\textbf{(E)}\\text{ 6}$ Problem 14 Which of the following integers cannot be written as the sum of four consecutive odd integers? $\\textbf{(A)}\\text{ 16}\\qquad\\textbf{(B)}\\text{ 40}\\qquad\\textbf{(C)}\\text{ 72}\\qquad\\textbf{(D)}\\text{ 100}\\qquad\\textbf{(E)}\\text{ 200}$ Problem 15 At Euler Middle School, $198$ students voted on two issues in a school referendum with the following results: $149$ voted in favor of the first issue and $119$ voted in favor of the second issue. If there were exactly $29$ students who voted against both issues, how many students voted in favor of both issues? $\\textbf{(A) }49\\qquad\\textbf{(B) }70\\qquad\\textbf{(C) }79\\qquad\\textbf{(D) }99\\qquad \\textbf{(E) }149$ Problem 16 In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If $\\tfrac{1}{3}$ of all the ninth graders are paired with $\\tfrac{2}{5}$ of all the sixth graders, what fraction of the total number of sixth and ninth", "label(\"A'\", AA, N); [/asy]$ From what we are given it easily follows that $ABC$ is an equilateral triangle, and its center $O$ is at the same time the foot of the height from $D$. We know that $AD=5$ and $OD=4$, from the Pythagorean theorem it follows that $OA=3$. The segment $OA$ is $2/3$ of the segment $AA'$. Then $AA'$ has length $9/2$. $AA'$ is the height in the triangle $ABC$, hence we have $AA' = \\dfrac{BC\\cdot \\sqrt 3}2$, and therefore $BC=3\\sqrt 3$. We can now compute the area of the triangle $ABC$ as $\\dfrac{BC\\cdot AA'}2 = \\dfrac{27\\sqrt 3}{4}$, and finally the volume of the tetrahedron $ABCD$ as $\\dfrac{ABC\\cdot DO}3 = 9\\sqrt 3$. Why did we compute all this stuff? Consider what happens when the leg $BD$ breaks in the described place $E$: import three; import math; size(8cm,0); currentprojection=obliqueX; triple O=(0,0,0), A=(3,0,0), B=rotate(120,Z)*A, C=rotate(-120,Z)*A, D=(0,0,4); triple BB=0.8*B + 0.2*D; triple AC=(A+C)/2; filldraw(A--BB--C--cycle, lightgray, black ); draw(A--B--C--cycle); draw(O--D); draw(A--D, black+1); draw(BB--D, black+1); draw(B--BB); draw(C--D, black+1); draw(A--O, dashed); draw(B--AC, dashed); draw(BB--AC, dashed); draw(C--O, dashed); label(\"$A$\", A, W); label(\"$B$\", B, E); label(\"$E$\", BB, E); label(\"$C$\", C, W); label(\"$O$\", O, SE); label(\"$D$\", D, Z); label(\"$B'$\", AC, W); (Error compiling LaTeX. dcd0140709d472b7550064c4363f05c4c983ec6a.asy: 11.9: no matching function 'filldraw(void(flatguide3), pen, pen)') The triangle $AEC$ will now be the base of the tripod, and our goal is", "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "You can find that every four moves both coordinates decrease by 2. Therefore, both coordinates need to decrease by two 505 times. You subtract, giving you the answer of $\\boxed{\\textbf{(B) } \\text{(-1030,-990)}}$ 14. ## Solution 1 $[asy] real x=sqrt(3); real y=2sqrt(3); real z=3.5; real a=x/2; real b=0.5; real c=3a; pair A, B, C, D, E, F; A = (1,0); B = (3,0); C = (4,x); D = (3,y); E = (1,y); F = (0,x); fill(A--B--C--D--E--F--A--cycle,grey); fill(arc((2,0),1,0,180)--cycle,white); fill(arc((2,y),1,180,360)--cycle,white); fill(arc((z,a),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); label(\"2\",(z,c),NE); pair X,Y,Z; X = (1,0); Y = (2,0); Z = (1.5,a); pair d = 1.9*dir(7); dot(X); dot(Y); dot(Z); label(\"A\",X,SW); label(\"B\",Y,SE); label(\"C\",Z,N); draw(X--Y--Z--A); label(\"1\",(1.5,0),S); label(\"1\",(1.75,a/2),dir(30)); draw(anglemark(Y,X,Z,8),blue); label(\"60^\\circ\",anglemark(Y,X,Z),d); [/asy]$ Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, $BC = 1$, since B is the center of the semicircle with radius 1 that C lies on, $AB = 1$, since B is the center of the semicircle with radius 1 that A lies on, and $\\angle BAC = 60^\\circ$as a regular hexagon has angles of 120$^\\circ$, and $\\angle BAC$ is half of any angle in this", "if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, dashed + blue ); [/asy]$ Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it? Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations. $[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(7.5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, red ); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) --", "# Difference between revisions of \"2013 AMC 8 Problems/Problem 18\" ## Problem Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain? $[asy]import three; currentprojection=orthographic(-8,15,15); triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A = (0,0,0); B = (0,10,0); C = (12,10,0); D = (12,0,0); E = (0,0,5); F = (0,10,5); G = (12,10,5); H = (12,0,5); I = (1,1,1); J = (1,9,1); K = (11,9,1); L = (11,1,1); M = (1,1,5); N = (1,9,5); O = (11,9,5); P = (11,1,5); //outside box far draw(surface(A--B--C--D--cycle),white,nolight); draw(A--B--C--D--cycle); draw(surface(E--A--D--H--cycle),white,nolight); draw(E--A--D--H--cycle); draw(surface(D--C--G--H--cycle),white,nolight); draw(D--C--G--H--cycle); //inside box far draw(surface(I--J--K--L--cycle),white,nolight); draw(I--J--K--L--cycle); draw(surface(I--L--P--M--cycle),white,nolight); draw(I--L--P--M--cycle); draw(surface(L--K--O--P--cycle),white,nolight); draw(L--K--O--P--cycle); //inside box near draw(surface(I--J--N--M--cycle),white,nolight); draw(I--J--N--M--cycle); draw(surface(J--K--O--N--cycle),white,nolight); draw(J--K--O--N--cycle); //outside box near draw(surface(A--B--F--E--cycle),white,nolight); draw(A--B--F--E--cycle); draw(surface(B--C--G--F--cycle),white,nolight); draw(B--C--G--F--cycle); //top draw(surface(E--H--P--M--cycle),white,nolight); draw(surface(E--M--N--F--cycle),white,nolight); draw(surface(F--N--O--G--cycle),white,nolight); draw(surface(O--G--H--P--cycle),white,nolight); draw(M--N--O--P--cycle); draw(E--F--G--H--cycle); label(\"10\",(A--B),SE); label(\"12\",(C--B),SW); label(\"5\",(F--B),W);[/asy]$ $\\textbf{(A)}\\ 204 \\qquad \\textbf{(B)}\\ 280 \\qquad \\textbf{(C)}\\ 320 \\qquad \\textbf{(D)}\\ 340 \\qquad \\textbf{(E)}\\ 600$ ## Solution 1 There are $10 \\cdot 12 = 120$ cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are $9 + 11", "2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,N); dot((5,4)); label(\"(a+b_1,b_2)\",(5,4),NE); dot((9,7)); label(\"(x,y)\",(9,7),NE); dot((2,4)); label(\"(2b_1,b_2)\",(2,4),SW); dot((9,4)); label(\"(x,b_2)\",(9,4),SE); [/asy]$ Note that since $K$ is the center of square $AB_1A_2B$, $BX \\perp KX$ and $BX = KX$. Additionally, $BY \\parallel KZ$ and $BY \\perp YZ$. $YZ$ is a line, so $\\angle BXY + \\angle KXZ + \\angle BXK = 180^\\circ$. Since $BX \\perp KX$, $\\angle BXK = 90^\\circ$, so $\\angle BXY + \\angle KXZ = 90^\\circ$. Additionally, because $BXY$ is a right triangle, $\\angle YBX + \\angle YXB = 90^\\circ$. Rearranging and substituting results in $\\angle KXZ = \\angle YBX$. Since both $BYX$ and $XZK$ are right triangles, by AAS Congruency, $\\triangle BYX \\cong \\triangle XZK$. Therefore $BY = XZ = b_2$ and $YX = ZK = a - b_1$. From this information, the coordinates of $K$ are $(a+b_1+b_2, b_2 + a - b_1)$. $[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4), K=(9,7), L=(-6,8), M=(-7,-3), N=(4,-8); draw(A--B--C--D--A); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,NE); dot(C); label(\"(-2c,0)\",C,NW); dot(D); label(\"(2d_1,-2d_2)\",D,S); dot(K); label(\"K\",K,NE); dot(L); label(\"L\",L,NW); dot(M); label(\"M\",M,SW); dot(N); label(\"N\",N,SE); [/asy]$ By", "of $n$ with $2\\le n \\le 1000$ for which $A(n)$ is an integer. ## Problem 15 A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$, and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\\overarc{AB}$ on that face measures $120^\\text{o}$. The block is then sliced in half along the plane that passes through point $A$, point $B$, and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\\cdot\\pi + b\\sqrt{c}$, where $a$, $b$, and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$. $[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,\"A\",(-3,3*sqrt(3),0),W); label(rpic,\"B\",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy]$ 2015 AIME I (Problems • Answer Key • Resources) Preceded by2014 AIME II Followed by2015 AIME II 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions", "we see that $$a_n=u_n+s_n\\tag{1}.$$ For the sake of convenience in determining recurrence relations, we define another type of board with two $1\\times n$ boards where a specific corner is removed (without loss of generality, we place this in the lower left hand corner). Let $t_n$ be the number of ways to tile such a board without placeing two of the same type of tile atop each other. Once again, we compute that $t_3=3$ and $t_4=5$ as shown below. $[asy] unitsize(10); draw((1,0)--(3,0)--(3,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(3,1),linewidth(2)); draw((1,1)--(1,2)^^(2,0)--(2,1)); draw(shift((0,-2.5))*((1,0)--(3,0)--(3,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(3,1)),linewidth(2)); draw(shift((0,-2.5))*((1,1)--(1,2)^^(2,1)--(2,2))); draw(shift((4,0))*((1,0)--(3,0)--(3,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(3,1)),linewidth(2)); draw(shift((4,0))*((2,1)--(2,2))); draw(shift((11,0))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((11,0))*((1,1)--(1,2)^^(2,1)--(2,2)^^(3,0)--(3,1))); draw(shift((11,-2.5))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((11,-2.5))*((1,1)--(1,2)^^(2,0)--(2,1)^^(3,1)--(3,2))); draw(shift((16,0))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((16,0))*((2,0)--(2,2)^^(3,1)--(3,2))); draw(shift((16,-2.5))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((16,-2.5))*((2,1)--(2,2)^^(3,0)--(3,1))); draw(shift((21,0))*((1,0)--(4,0)--(4,2)--(0,2)--(0,1)--(1,1)--cycle^^(1,1)--(4,1)),linewidth(2)); draw(shift((21,0))*((2,0)--(2,2)^^(3,0)--(3,1))); [/asy]$ We can determine reccurence relations for $s_n$, $t_n$, and $u_n$ in terms of each other. For $s_n$, note that a tiling can end in one of the three following ways such that the rest of the board can be tiled without restriction (the placed tiles are shaded). $[asy] path unitrect=(0,0)--(2,0)--(2,1)--(0,1)--cycle; unitsize(10); fill(shift((3,0))*unitsquare,mediumgray); draw(shift((0,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(3,1)--(3,2)--(0,2)),linewidth(2)); draw(shift((0,0))*((3,0)--(3,1))); label(\"\\dots\",(-1,1)); fill(shift((10,1))*unitsquare,mediumgray); fill(shift((10,0))*unitrect,mediumgray); draw(shift((8,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(3,1)--(3,2)--(0,2)),linewidth(2)); draw(shift((8,0))*((2,0)--(2,2))); label(\"\\dots\",(8-1,1)); fill(shift((18,0))*unitrect,mediumgray); fill(shift((17,1))*unitrect,mediumgray); draw(shift((16,0))*((0,0)--(4,0)--(4,1)--(0,1)^^(3,1)--(3,2)--(0,2)),linewidth(2)); draw(shift((16,0))*((2,0)--(2,1)^^(1,1)--(1,2))); label(\"\\dots\",(16-1,1)); [/asy]$ In the first, second, and third cases, we see that we can tile the rest of the board in $t_{n-1}$, $t_{n-2}$, and $s_{n-2}$ ways, respectively. Hence for $n\\ge 5$, we see that $$s_n=t_{n-1}+t_{n-2}+s_{n-2}\\tag{2}.$$ For $t_n$, note that a tiling can end in one of the three following ways such that the rest of the board", "graph3; currentprojection=orthographic(2,1.4,.6,zoom=.95); //currentprojection=obliqueX(40); // the yellow paraboloid z=f(x,y)= x^2+y^2 real a=2.25; // below z=a^2 // to take domain is a disk (instead of a rectangle) triple f(pair M) { real x = a*(M.x)*cos(M.y); real y = a*(M.x)*sin(M.y); return (x,y,x^2+y^2); } real rmax=1, rmin=0; real phimax=2*pi, phimin=0; surface sf=surface(f, (rmin,phimin), (rmax,phimax), Spline); //draw(sf, surfacepen=pink,meshpen=brown+thick()); draw(sf,surfacepen=yellow+opacity(1)); triple h(pair M) { if (1+2M.y^2>a^2) return (M.x,M.y,a^2); else return (M.x,M.y,1+2M.y^2); } real xmax=4, xmin=-4; real ymax=1.5, ymin=-1.5; surface sh=surface(h, (xmin,ymin), (xmax,ymax), Spline); //draw(sh, surfacepen=pink,meshpen=brown+thick()); draw(sh,surfacepen=blue+opacity(.8)); triple A=(xmin,ymin,0), B=(xmax,ymax,a^2); draw(box(A,B),red+opacity(.5)); // the red circle x^2+y^2=b^2 is the intersection of // the blue plane z=b^2 and the paraboloid real b=1.6; real x(real t) {return b*cos(t);} real y(real t) {return b*sin(t);} real z(real t) {return b^2;} path3 g=graph(x,y,z,phimin,phimax,operator..); //draw(g,red+1pt); //draw(shift(-3,-3,b^2)*scale(6,6,0)*unitplane,blue+opacity(1)); draw(Label(\"$x$\",EndPoint),-5*X--5*X,Arrow3()); draw(Label(\"$y$\",EndPoint),-5*Y--5*Y,Arrow3()); draw(Label(\"$z$\",EndPoint),-Z--7*Z,Arrow3()); • This is quite beautiful. Oct 19 '21 at 19:10 • I updated the 2nd picture. Oct 20 '21 at 19:15 • @CaptainGiraffe beautiful is as beautiful does! Oct 21 '21 at 15:10 I offer a TikZ solution. I think that for this figure it's easier to draw in 2d, so I did it this way. I'm not posting the maths, because it will be tedious without MathJax or something similar. But the key here is to find the envelope of the family of ellipses (circles in 3d) that are obtained cutting the", "# 1969 AHSME Problems/Problem 31 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem Let $OABC$ be a unit square in the $xy$-plane with $O(0,0),A(1,0),B(1,1)$ and $C(0,1)$. Let $u=x^2-y^2$, and $v=xy$ be a transformation of the $xy$-plane into the $uv$-plane. The transform (or image) of the square is: $[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,2)--(1,0)--(0,-2)--cycle,dot); MP(\"(A)\",(-5,2),SW); MP(\"O\",(0,0),SW); MP(\"(-1,0)\",(-1,0),SW); MP(\"(0,2)\",(0,2),NE); MP(\"(1,0)\",(1,0),SE); MP(\"(0,-2)\",(0,-2),SE); [/asy]$ $[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw(arc((1.5,0),2.5,126,234),black); draw(arc((-1.5,0),2.5,54,-54),black); MP(\"(B)\",(-5,2),SW); MP(\"O\",(0,0),SW); MP(\"(-1,0)\",(-1,0),SW); MP(\"(0,2)\",(0,2),NE); MP(\"(1,0)\",(1,0),SE); MP(\"(0,-2)\",(0,-2),SE); [/asy]$ $[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,2)--(1,0),black); MP(\"(C)\",(-5,2),SW); MP(\"O\",(0,0),SW); MP(\"(-1,0)\",(-1,0),SW); MP(\"(0,2)\",(0,2),NE); MP(\"(1,0)\",(1,0),SE); [/asy]$ $[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw(arc((1.5,0),2.5,126,180),black); draw(arc((-1.5,0),2.5,54,0),black); MP(\"(D)\",(-5,2),SW); MP(\"O\",(0,0),SW); MP(\"(-1,0)\",(-1,0),SW); MP(\"(0,2)\",(0,2),NE); MP(\"(1,0)\",(1,0),SE); MP(\"(0,-2)\",(0,-2),SE); [/asy]$ $[asy] draw((-3,0)--(3,0),EndArrow); draw((0,-4)--(0,4),EndArrow); draw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,dot); MP(\"(E)\",(-5,2),SW); MP(\"O\",(.1,.1),SW); MP(\"(-1,0)\",(-1,0),SW); MP(\"(0,1)\",(0,1),NE); MP(\"(1,0)\",(1,0),SE); MP(\"(0,-1)\",(0,-1),SE); [/asy]$ ## Solution Each point on the square can be in the form $(0,y)$, $(1,y)$, $(x,0)$, and $(x,1)$, where $0 \\le x,y \\le 1$. Making the appropriate substitutions results in points being $(-y^2,0)$, $(1-y^2,2y)$, $(x^2,0)$, and $(x^2 - 1,2x)$ on the $uv$-plane. Notice that since $v \\ge 0$, none of the points are below the u-axis, so options A,B, and E are out. Since $x = \\tfrac{v}{2}$, $u = (\\tfrac{v}{2})^2 - 1$, so $v = 2\\sqrt{u+1}$, where $-1 \\le u \\le 0$. That means some of the lines are not straight, so the answer is $\\boxed{\\textbf{(D)}}$.", "|[data]|$w_1$ & |[data]|$w_2$ & |[data]|$w_3$ & |[empty]| & |[data]|$w_K$ \\\\ |[data]|$D_1$ & |[data]|$D_2$ & |[empty]| & |[empty]| & |[data]|$D_N$ \\\\[.7em] }; \\draw[->](m-1-1) to (m-2-1); \\draw[->](m-2-1) to (m-3-1); \\draw[->](m-3-1) to (m-4-1); \\draw[->](m-3-1) to (m-1-2); \\draw[->](m-3-1.south) to (m-4-5); \\draw[->](m-1-2) to (m-2-2); \\draw[->](m-2-2) to (m-3-2); \\draw[->](m-3-2) to (m-1-1); \\draw[->](m-3-2) to (m-1-3); \\draw[->](m-3-2.south) to (m-4-1); \\draw[->](m-3-2.south) to (m-4-5); \\draw[->](m-1-3) to (m-2-3); \\draw[->](m-2-3) to (m-3-3); \\draw[->](m-3-3) to (m-1-2); \\draw[->](m-2-4) to (m-1-3); \\draw[->](m-3-4) to (m-1-5); \\draw[->](m-1-5) to (m-2-5); \\draw[->](m-2-5) to (m-3-5); \\draw[->](m-3-5) to (m-4-5); \\end{tikzpicture} \\end{document} - np, use CloudConvert to convert PDF to PNG – m0nhawk Jul 28 '14 at 8:49 @m0nhawk thanks again, i'll give it a try the next time :) – Dan H. Jul 28 '14 at 8:51", "&& (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, dashed + blue ); [/asy]$ Let's now consider the opposite direction. Assume that we picked the blue square, how many different red squares do share it? Answering this question is not as simple as it may seem. Consider the picture below. It shows all three red squares that share the same blue square. In addition, the picture shows a green square that is not valid, as two of its vertices are in bad locations. $[asy] defaultpen(black+0.75bp+fontsize(8pt)); size(7.5cm); path p = scale(.15)*unitcircle; draw((-8,0)--(8.5,0),Arrow(HookHead,1mm)); draw((0,-8)--(0,8.5),Arrow(HookHead,1mm)); int i,j; for (i=-7;i<8;++i) { for (j=-7;j<8;++j) { if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) || ((3 <= j) && (j<= 7))) { fill(shift(i,j)*p,black); }}}} draw((-7,-.2)--(-7,.2),black+0.5bp); draw((-3,-.2)--(-3,.2),black+0.5bp); draw((3,-.2)--(3,.2),black+0.5bp); draw((7,-.2)--(7,.2),black+0.5bp); draw((-.2,-7)--(.2,-7),black+0.5bp); draw((-.2,-3)--(.2,-3),black+0.5bp); draw((-.2,3)--(.2,3),black+0.5bp); draw((-.2,7)--(.2,7),black+0.5bp); label(\"-7\",(-7,0),S); label(\"-3\",(-3,0),S); label(\"3\",(3,0),S); label(\"7\",(7,0),S); label(\"-7\",(0,-7),W); label(\"-3\",(0,-3),W); label(\"3\",(0,3),W); label(\"7\",(0,7),W); draw( (5,4) -- (-4,4) -- (-4,-5) -- (5,-5) -- cycle, red ); draw( (5,3) -- (-3,4) -- (-4,-4) -- (4,-5) -- cycle, red ); draw(", "\\boxed{160}$ meters is the solution. Alternatively, we can drop an altitude from $C$ and arrive at the same answer. ## Solution 2 $[asy] draw((0,0)--(11,0)--(7,14)--cycle); draw((7,14)--(11,28)); draw((11,28)--(11,0)); label(\"A\",(-1,-1),N); label(\"B\",(12,-1),N); label(\"P\",(6,15),N); label(\"B'\",(12,29),N); draw((-10,14)--(20,14)); label(\"X\",(12.5,15),N); draw((7,0)--(7,14),dashed); label(\"Y\",(7,-4),N); draw((6,0)--(6,1)); draw((6,1)--(7,1)); draw((11,13)--(10,13)); draw((10,13)--(10,14)); label(\"60^{\\circ}\",(4,0.5),N); [/asy]$ Let $P$ be the point of intersection between the skaters, Allie and Billie. We can draw a line that goes through $P$ and is parallel to $\\overline{AB}$. Letting this line be the $x$-axis, we can reflect $B$ over the $x$-axis to get $B'$. As reflections preserve length, $B'X = XB$. We then draw lines $BB'$ and $PB'$. We can let the foot of the perpendicular from $P$ to $BB'$ be $X$, and we can let the foot of the perpendicular from $P$ to $AB$ be $Y$. In doing so, we have constructed rectangle $PXBY$. By $d=rt$, we have $AP = 8t$ and $PB = PB' = 7t$, where $t$ is the number of seconds it takes the skaters to meet. Furthermore, we have $30-60-90$ triangle $PAY$, so $AY = 4t$, and $PY = 4t\\sqrt{3}$. Since we have $PY = XB = B'X$, $B'X = 4t\\sqrt{3}$. By Pythagoras, $PX = t$. As $PXBY$ is a rectangle, $PX = YB$. Thus $AY + YB = AB \\Rightarrow AY + PX = AB$, so we get $4t + t = 100$.", "be the intersection of the segments emanating from $A$ and $C$. By Law of Cosines, $$BP' = \\sqrt{1 + 1 - 2\\cos{120^\\circ}} = \\sqrt{3}.$$ So, $P'$ actually satisfies the conditions of the problem, and we can obtain again by Law of Cosines $$s = \\sqrt{4 + 1 - 4\\cos{120^\\circ}} = \\boxed{\\textbf{(B)} \\sqrt{7}}.$$ ~ hnkevin42 ## Solution 3 (Answer Choices) $[asy] unitsize(0.4inch); pen p = fontsize(10pt); draw((0,0)--(4,5.65)--(8,0)--cycle); label(\"A\", (4,5.65), N, p); label(\"C\", (8,0), SE, p); label(\"B\", (0,0), SW, p); label(\"P\", (3.5,3.5), E, p); label(\"E\", (2.8191,3.982), NW, p); label(\"F\", (4.848,4.452), NE, p); label(\"G\", (3.5,0), down, p); draw((0,0)--(3.5,3.5)); label(\"\\sqrt{3}\",(0,0)--(3.5,3.5), SE,p); draw((8,0)--(3.5,3.5)); label(\"2\",(8,0)--(3.5,3.5), SW,p); draw((4,5.65)--(3.5,3.5)); label(\"1\",(4,5.65)--(3.5,3.5), E,p); draw((3.5,3.5)--(2.8191,3.982)); draw((3.5,3.5)--(4.848,4.452)); draw((3.5,3.5)--(3.5,0)); [/asy]$ We begin by dropping altitudes from point $P$ down to all three sides of the triangle as shown above. We can therefore make equations regarding the areas of triangles $\\triangle{APC}$, $\\triangle{APB}$, and $\\triangle{BPC}$. Let $s$ be the side of the equilateral triangle, we use the Heron's formula: $$\\triangle{APC} = \\frac{s\\cdot PF}{2} = \\sqrt{\\frac{s+3}{2}\\left(\\frac{s+3}{2}-s\\right)\\left(\\frac{s+3}{2}-1\\right)\\left(\\frac{s+3}{2}-2\\right)}$$ $$\\implies PF = \\frac{\\sqrt{10s^2-s^4-9}}{2s}$$ Similarly, we obtain: $$PE = \\frac{\\sqrt{8s^2-s^4-4}}{2s}$$ $$PG = \\frac{\\sqrt{14s^2-s^4-1}}{2s}$$ By Viviani's theorem, $$\\frac{\\sqrt{10s^2-s^4-9}}{2s}+\\frac{\\sqrt{8s^2-s^4-4}}{2s}+\\frac{\\sqrt{14s^2-s^4-1}}{2s} = \\frac{\\sqrt{3}}{2}s$$ $$\\sqrt{10s^2-s^4-9}+\\sqrt{8s^2-s^4-4}+\\sqrt{14s^2-s^4-1} = \\sqrt{3}s^2$$ Note that from now on, the algebra will get extremely ugly and almost impossible to do by hand within the time frame. However, we do see that it's extremely easy to check the answer choices with", "answer is then $\\boxed{\\dfrac{145}{147}}$. ## Solution 5 $[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); draw((0.3,0.3)--(3.6,0.3), dashed); draw((0.3,2.7)--(0.3,0.3), dashed); label(\"S\", (-0.05,-0.05), NE); draw((0.3,0.3)--(1.41,1.91)); draw((1.63,1.78)--(1.48,1.56)); draw((1.28,1.70)--(1.48,1.56)); label(\"4\", (2,0), S); label(\"3\", (0,1.5), W); label(\"\\frac{10}{3}\", (2,0.3), N); label(\"\\frac{5}{2}\", (0.3,1.5), E); label(\"2\", (1,1.2), E); draw((3.6,0)--(3.6,0.3), dashed); draw((0,2.7)--(0.3,2.7), dashed); label(\"\\small{l}\", (3.6,0.15), W); label(\"\\small{l}\", (0.15,2.7), S); label(\"\\small{l}\", (0.3,0.15), E); label(\"\\small{l}\", (0.15,0.3), N); [/asy]$ On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\\frac{5}{2}$ and $\\frac{10}{3}$. With $l$ being the side length of the square, we need to find an expression for $l$. Using the hypotenuse, we can see that $\\frac{3}{2}+\\frac{8}{3}+\\frac{5}{4}l+\\frac{5}{3}l=5$. Simplifying, $\\frac{35}{12}l=\\frac{5}{6}$, or $l=2/7$. A different calculation would yield $l+\\frac{3}{4}l+\\frac{5}{2}=3$, so $\\frac{7}{4}l=\\frac{1}{2}$. In other words, $l=\\frac{2}{7}$, while to check, $l+\\frac{4}{3}l+\\frac{10}{3}=4$. As such, $\\frac{7}{3}l=\\frac{2}{3}$, and $l=\\frac{2}{7}$. Finally, we get $A(\\Square S)=l^2=\\frac{4}{49}$, to finish. As a proportion of the triangle with area $6$, the answer would be $1-\\frac{4}{49\\cdot6}=1-\\frac{2}{147}=\\frac{145}{147}$, so $\\boxed{\\textit D}$ is correct. ## Solution 6: Also Coordinate Geo Let the right angle be at $(0,0)$, the point $(x,x)$ be the far edge of the unplanted square and the hypotenuse be the line $y=-\\frac{3}{4}x+3$. Since the line from $(x,x)$ to the hypotenuse is the shortest possible distance, we know this line, call it line $\\l$, is perpendicular to the hypotenuse and therefore has", "(x+0.5,y+0.5), p = fontsize(19pt)); } foo(2, 4, \" \"); foo(3, 4, \" \"); foo(0, 3, \" \"); foo(2, 3, \" \"); foo(1, 2, \" \"); foo(3, 2, \" \"); foo(1, 1, \" \"); foo(2, 1, \" \"); foo(3, 1, \" \"); foo(4, 1, \" \"); foo(2, 0, \" \"); foo(3, 0, \" \"); foo(0, 1, \" \"); foo(0, 2, \" \"); foo(1, 0, \" \"); foo(1, 3, \" \"); foo(1, 4, \" \"); foo(3, 3, \" \"); foo(4, 2, \" \"); foo(4, 3, \" \"); [/asy]$ Problem 19 A triangle with vertices as $A=(1,3)$, $B=(5,1)$, and $C=(4,4)$ is plotted on a $6\\times5$ grid. What fraction of the grid is covered by the triangle? $[asy] draw((1,0)--(1,5),linewidth(.5)); draw((2,0)--(2,5),linewidth(.5)); draw((3,0)--(3,5),linewidth(.5)); draw((4,0)--(4,5),linewidth(.5)); draw((5,0)--(5,5),linewidth(.5)); draw((6,0)--(6,5),linewidth(.5)); draw((0,1)--(6,1),linewidth(.5)); draw((0,2)--(6,2),linewidth(.5)); draw((0,3)--(6,3),linewidth(.5)); draw((0,4)--(6,4),linewidth(.5)); draw((0,5)--(6,5),linewidth(.5)); draw((0,0)--(0,6),EndArrow); draw((0,0)--(7,0),EndArrow); draw((1,3)--(4,4)--(5,1)--cycle); label(\"y\",(0,6),W); label(\"x\",(7,0),S); label(\"A\",(1,3),dir(210)); label(\"B\",(5,1),SE); label(\"C\",(4,4),dir(100)); [/asy]$ $\\textbf{(A) }\\frac{1}{6} \\qquad \\textbf{(B) }\\frac{1}{5} \\qquad \\textbf{(C) }\\frac{1}{4} \\qquad \\textbf{(D) }\\frac{1}{3} \\qquad \\textbf{(E) }\\frac{1}{2}$ Problem 20 Ralph went to the store and bought 12 pairs of socks for a total of $24. Some of the socks he bought cost$1 a pair, some of the socks he bought cost $3 a pair, and some of the socks he bought cost$4 a pair. If he bought at least one pair of each type, how many pairs of \\$1 socks did Ralph buy? $\\textbf{(A) } 4 \\qquad", "*/ pen qqwuqq = rgb(0,0,0); draw(arc((12,0),2,60,180)--(12,0)--cycle, qqwuqq); /* draw figures */ draw((0,0)--(12,0)); draw((0,0)--(18,10)); draw((18,10)--(12,0)); label(\"12\",(6,-1),SE*labelscalefactor); label(\" 2\\sqrt{76}-6 \",(14,5),SE*labelscalefactor); label(\"20\",(8,7),SE*labelscalefactor); draw((12,0)--(12,10)); draw((12,10)--(18,10)); /* dots and labels */ dot((0,0),dotstyle); label(\"A\", (0,0), NE * labelscalefactor); dot((12,0),dotstyle); label(\"B\", (12,0), NE * labelscalefactor); dot((18,10),dotstyle); label(\"C\", (18,10), NE * labelscalefactor); label(\"120^\\circ\", (11,1), NE * labelscalefactor,qqwuqq); dot((12,10),dotstyle); label(\"H\", (12,10), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ Now, note that we have to find the coordinates of $E,F$ and $D$. Assume WLOG that $A$ is $(0,0)$ and $B$ be $(12,0)$. $E$ is obviously $(3,0)$, $F$ is going to be $(12+\\frac{9+\\sqrt{73}-12}{4}, \\frac{\\sqrt{219}-3\\sqrt3}{4})$ and the coordinates of $D$ is going to be $(\\frac{9+\\sqrt{73}}{4}, \\frac{\\sqrt{219}-3\\sqrt3}{4})$. Since $D$ and $F$ have the same $y$ value, we can find $[EDF]$ by using $\\frac12*b*h$. The base is going to be equal to the distance from $D$ to $F$, which is the same thing as $12+\\frac{9+\\sqrt{73}-12}{4}-\\frac{9+\\sqrt{73}}{4}=9=b$, and the height is the change in the $y$ coordinate from $E$ to $D$. This is the same thing as $\\frac{\\sqrt{219}-3\\sqrt3}{4}-0=\\frac{\\sqrt{219}-3\\sqrt3}{4}=h$. Hence, plugging these into $[EDF]=\\frac12*b*h$ give us $\\frac12*9*\\frac{\\sqrt{219}-3\\sqrt3}{4}=\\frac{9\\sqrt{219}-27\\sqrt3}{8}$. The answer is thus $9+219+27+3+8=\\boxed{266}$. ## Solution 2 Let $S$ be the area of the large triangle and let $x$ be the area we are trying to find. We have that $\\Delta ADE$ is similar to $\\Delta ABC$ with a ratio of $1:4$ while $\\Delta CDF$ is similar to $\\Delta" ]

[ "have $x-y\\geq1$, $y-z\\geq1$. The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too. $[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3); // three - currentprojection, orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), dashed+green); draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green); draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((0,0,0)--(1,0,0)--(1,1,1), red); draw((1,1,1)--(1,1,0)--(1,0.9,0), red); draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle); draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed); draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed); [/asy]$ The volume of this region is $V_2=\\dfrac{(n-2)^3}{6}$. So the probability is $p=\\dfrac{V_2}{V_1}=\\dfrac{(n-2)^3}{n^3}$. Substituting $n$ with the values in the choices, we find that when $n=10$, $p=\\frac{512}{1000}>\\frac{1}{2}$, when $n=9$, $p=\\frac{343}{729}<\\frac{1}{2}$. So $n\\geq 10$. So the answer is $\\boxed{\\textbf{(D)}\\ 10}$. ## Solution 2 Because $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$, which means that $x$, $y$, and $z$ distributes uniformly and independently in the interval $[0,n]$. So the point $(x, y, z)$ distributes uniformly in the cubic $0\\leqslant x, y, z \\leqslant n$, as shown in the figure below. The volume of this cubic is $V_0=n^3$. As we want to find the probablity of the incident $A=\\big\\{ |x-y| \\geq 1, |y-z| \\geq1, |z-x| \\geq", "243 views The volume of the region $S=\\{(x,y,z) : \\mid x \\mid +\\mid y \\mid + \\mid z \\mid \\leq 1\\}$ is 1. $\\frac{1}{6}\\\\$ 2. $\\frac{1}{3}\\\\$ 3. $\\frac{2}{3}\\\\$ 4. $\\frac{4}{3}$ edited | 243 views 0 Option D We know that $|x| = x$ when $x>= 0$ and $|x| = -x$ when $x<0$. So, region covered by the graph $|x| + |y| <= 1$ is bounded by the lines $(x+y =1)$ , $(-x +y =1)$ , $(x - y =1)$ , $(-x -y =1)$ [As shown in Fig $A$ below] So, the region covered will be the square $[ (0,1) , (1,0) , (0,-1) , (-1,0) ]$ [ As shown in Fig $A$] The side length of the above square is $√2.$ Now, in the $3D$ version of the above scenario the space will be covered by the lines $(x+z = 1)$ , $( -x +z = 1)$ , $( -y +z =1)$ , $( y +z = 1)$, $( -y-z =1)$, $( -y +z = 1)$ , $(x -z =1)$ , $(-x -z = 1)$. [ As shown in the figure $B$] This will create two pyramids $[ (0,0,1) , (0,1,0) , (1,0,0) , (0,-1,0) , (-1,0,0) ]$ and $[ (0,0,-1) , (0,1,0) , (1,0,0) , (0,-1,0) , (-1,0,0) ]$ [ As shown in the above figure $B$", "# 2010 AIME I Problems/Problem 11 ## Problem Let $\\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \\le 10$ and $3y - x \\ge 15$. When $\\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$, the volume of the resulting solid is $\\frac {m\\pi}{n\\sqrt {p}}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m + n + p$. ## Solution $[asy]size(280); import graph; real min = 2, max = 12; pen dark = linewidth(1); real P(real x) { return x/3 + 5; } real Q(real x) { return 10 - abs(x - 8); } path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12)); pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(0.9,0.9,0.9)); draw(graph(P,min,max),dark); draw(graph(Q,min,max),dark); draw(Arc((8,7.67),A,G,CW),dark,EndArrow(8)); draw(B--C--G--cycle,linetype(\"4 4\")); label(\"y \\ge x/3 + 5\",(max,P(max)),E,fontsize(10)); label(\"y \\le 10 - |x-8|\",(max,Q(max)),E,fontsize(10)); label(\"\\mathcal{R}\",(6,Q(6)),NW); /* axes */ Label f; f.p=fontsize(8); xaxis(0, max, Ticks(f, 6, 1)); yaxis(0, 10, Ticks(f, 5, 1)); [/asy]$ The inequalities are equivalent to $y \\ge x/3 + 5, y \\le 10 - |x - 8|$. We can set them equal to find the two points of intersection, $x/3 + 5 = 10 -", "of $n$ with $2\\le n \\le 1000$ for which $A(n)$ is an integer. ## Problem 15 A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$, and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\\overarc{AB}$ on that face measures $120^\\text{o}$. The block is then sliced in half along the plane that passes through point $A$, point $B$, and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\\cdot\\pi + b\\sqrt{c}$, where $a$, $b$, and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$. $[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,\"A\",(-3,3*sqrt(3),0),W); label(rpic,\"B\",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy]$ 2015 AIME I (Problems • Answer Key • Resources) Preceded by2014 AIME II Followed by2015 AIME II 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions", "by $$(ix + 4*(iy+1) + 16*(iz+1))$$ where $$(ix, iy, iz)$$ are the coordinates of the barycenter of the current face, is given below. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 load \"msh3\" int[int] l6 = [37, 42, 45, 40, 25, 57]; int r11 = 11; mesh3 Th = cube(4, 5, 6, [x*2-1, y*2-1, z*2-1], label=l6, flags =3, region=r11); cout << \"Volume = \" << Th.measure << \", border area = \" << Th.bordermeasure << endl; int err = 0; for(int i = 0; i < 100; ++i){ real s = int2d(Th,i)(1.); real sx = int2d(Th,i)(x); real sy = int2d(Th,i)(y); real sz = int2d(Th,i)(z); if(s){ int ix = (sx/s+1.5); int iy = (sy/s+1.5); int iz = (sz/s+1.5); int ii = (ix + 4*(iy+1) + 16*(iz+1) ); //value of ix,iy,iz => face min 0, face max 2, no face 1 cout << \"Label = \" << i << \", s = \" << s << \" \" << ix << iy << iz << \" : \" << ii << endl; if( i != ii ) err++; } } real volr11 = int3d(Th,r11)(1.); cout << \"Volume region = \" << 11 << \": \" << volr11", "the WS algorithm is provided by the mean value theorem. $$f(y_n) = f(y_{n-1}) + f'(\\xi_n)(y_n-y_{n-1})$$ for some $\\xi_n$ between $y_n$ and $y_{n-1}$. In our case, $f'(z) = \\ln{z}\\ z^y$, so $$y_{n+1} = y_n + (\\ln{z})(z^{\\xi_n})(y_n-y_{n-1})$$ This tells us that $$|y_{n+1}-y_n| \\le C |y_n-y_{n-1}|$$ for the quantity $$C = \\mbox{max} |\\ln{z}||z^y|$$ where the maximum is taken over the $y$ s and $z$ s in a region including the iterates. If $C$ is less than 1, the iterates get closer together, and so the iteration is a contraction that converges. Let's do a numerical search by making a contour plot of $(\\ln{z})(z^y)$. Use green for the region where $C$ is less than $1$, and red and blue for the regions where it is larger than 1. [Z,Y] = ndgrid(.01:.01:2.0,.01:.01:4.0); R = log(Z).*Z.^Y; contourf(Z,Y,R,[-5:4:-1 1:20:21]); colormap([0 0 .75;0 .75 .25;1 0 0]); e = exp(1); line([ebar ebar],[e e],'marker','.','markersize',24,'color','yellow') set(gca,'xtick',[.01 1/e 1 e^(1/e) 2]) set(gca,'xticklabel',{'0','1/e','1','e^(1/e)','2'}) set(gca,'ytick',[.01 1 2 e 4]) set(gca,'yticklabel',{'0','1','2','e','4'}) xlabel('z'), ylabel('y') axis square #### Stable Towers If we stay in the green region, where $C$ is less than 1, the WS iteration converges, and the tower is stable. That includes all values of $z$ in the interval $$1/\\mbox{e} \\le z \\le \\mbox{e}^{1/\\mbox{e}}$$ #### Towers Topple The crucial value that I encountered in junior high is the yellow dot, $z=\\bar{\\mbox{e}}=\\mbox{e}^{1/\\mbox{e}}$ with", "gy = (Ay+By+Cy+Dy)/4.; // Mesh LINEBORDER(A,B,1) LINEBORDER(B,C,2) LINEBORDER(C,D,3) LINEBORDER(D,A,4) mesh Th=buildmesh(AB(n)+BC(n)+CD(n)+DA(n),fixedborder=1); // Fespace real l1 = dist(Ax,Ay,Bx,By); real l2 = dist(Bx,By,Cx,Cy); real l3 = dist(Cx,Cy,Dx,Dy); real l4 = dist(Dx,Dy,Ax,Ay); func s1 = dist(Ax,Ay,x,y)/l1; //absisse on AB = ||AX||/||AB|| func s2 = dist(Bx,By,x,y)/l2; //absisse on BC = ||BX||/||BC|| func s3 = dist(Cx,Cy,x,y)/l3; //absisse on CD = ||CX||/||CD|| func s4 = dist(Dx,Dy,x,y)/l4; //absisse on DA = ||DX||/||DA|| verbosity = 6; //to see the abscisse value of the periodic condition fespace Vh(Th, P1, periodic=[[1, s1], [3, s3], [2, s2], [4, s4]]); verbosity = 1; //reset verbosity Vh u, v; real cc = 0; cc = int2d(Th)((x-gx)*(y-gy)-cc)/Th.area; cout << \"compatibility = \" << int2d(Th)((x-gx)*(y-gy)-cc) < Tip Periodic boundry conditions - Poisson cube-balloon 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 load \"msh3\" load \"tetgen\" load \"medit\" // Parameters real hs = 0.1; //mesh size on sphere int[int] N = [20, 20, 20]; real [int,int] B = [[-1, 1], [-1, 1],", "graph3; currentprojection=orthographic(2,1.4,.6,zoom=.95); //currentprojection=obliqueX(40); // the yellow paraboloid z=f(x,y)= x^2+y^2 real a=2.25; // below z=a^2 // to take domain is a disk (instead of a rectangle) triple f(pair M) { real x = a*(M.x)*cos(M.y); real y = a*(M.x)*sin(M.y); return (x,y,x^2+y^2); } real rmax=1, rmin=0; real phimax=2*pi, phimin=0; surface sf=surface(f, (rmin,phimin), (rmax,phimax), Spline); //draw(sf, surfacepen=pink,meshpen=brown+thick()); draw(sf,surfacepen=yellow+opacity(1)); triple h(pair M) { if (1+2M.y^2>a^2) return (M.x,M.y,a^2); else return (M.x,M.y,1+2M.y^2); } real xmax=4, xmin=-4; real ymax=1.5, ymin=-1.5; surface sh=surface(h, (xmin,ymin), (xmax,ymax), Spline); //draw(sh, surfacepen=pink,meshpen=brown+thick()); draw(sh,surfacepen=blue+opacity(.8)); triple A=(xmin,ymin,0), B=(xmax,ymax,a^2); draw(box(A,B),red+opacity(.5)); // the red circle x^2+y^2=b^2 is the intersection of // the blue plane z=b^2 and the paraboloid real b=1.6; real x(real t) {return b*cos(t);} real y(real t) {return b*sin(t);} real z(real t) {return b^2;} path3 g=graph(x,y,z,phimin,phimax,operator..); //draw(g,red+1pt); //draw(shift(-3,-3,b^2)*scale(6,6,0)*unitplane,blue+opacity(1)); draw(Label(\"$x$\",EndPoint),-5*X--5*X,Arrow3()); draw(Label(\"$y$\",EndPoint),-5*Y--5*Y,Arrow3()); draw(Label(\"$z$\",EndPoint),-Z--7*Z,Arrow3()); • This is quite beautiful. Oct 19 '21 at 19:10 • I updated the 2nd picture. Oct 20 '21 at 19:15 • @CaptainGiraffe beautiful is as beautiful does! Oct 21 '21 at 15:10 I offer a TikZ solution. I think that for this figure it's easier to draw in 2d, so I did it this way. I'm not posting the maths, because it will be tedious without MathJax or something similar. But the key here is to find the envelope of the family of ellipses (circles in 3d) that are obtained cutting the", "User:Lcz Hi just getting some last minute A(O)IME geo prep in (my thing is alg>nt>combo>geo) from the Intermediate Geometry Problems page and then printing it out :) (2015 II #9) https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9 A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$. $[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype(\"2 4\")); draw(X+B--X+C+B,linetype(\"2 4\")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype(\"2 4\")); [/asy]$ (Error compiling LaTeX. ! Missing \\$ inserted.) (2015 I #13) https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_13 With all angles measured in degrees, the product $\\prod_{k=1}^{45} \\csc^2(2k-1)^\\circ=m^n$, where $m$ and $n$ are integers greater than 1. Find $m+n$. (2015 I #11) https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_11 Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\\angle B$ and $\\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\\triangle ABC$. (2014 II #14) https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_14 In $\\triangle{ABC}, AB=10, \\angle{A}=30^\\circ$ , and $\\angle{C=45^\\circ}$. Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\\perp{BC}$, $\\angle{BAD}=\\angle{CAD}$, and $BM=CM$. Point $N$ is the midpoint of the", "Difference between revisions of \"2015 AIME I Problems/Problem 15\" Problem A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$, and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\\overarc{AB}$ on that face measures $120^\\text{o}$. The block is then sliced in half along the plane that passes through point $A$, point $B$, and the center of the cylinder, revealing a flat, unpainted face on each half. The area of one of these unpainted faces is $a\\cdot\\pi + b\\sqrt{c}$, where $a$, $b$, and $c$ are integers and $c$ is not divisible by the square of any prime. Find $a+b+c$. $[asy] import three; import solids; size(8cm); currentprojection=orthographic(-1,-5,3); picture lpic, rpic; size(lpic,5cm); draw(lpic,surface(revolution((0,0,0),(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8),Z,0,120)),gray(0.7),nolight); draw(lpic,surface(revolution((0,0,0),(-3*sqrt(3),-3,8)..(-6,0,4)..(-3*sqrt(3),3,0),Z,0,90)),gray(0.7),nolight); draw(lpic,surface((3,3*sqrt(3),8)..(-6,0,8)..(3,-3*sqrt(3),8)--cycle),gray(0.7),nolight); draw(lpic,(3,-3*sqrt(3),8)..(-6,0,8)..(3,3*sqrt(3),8)); draw(lpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0),dashed); draw(lpic,(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0)--(-3,3*sqrt(3),0)..(-3*sqrt(3),3,0)..(-6,0,0),dashed); draw(lpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(-3*sqrt(3),-3,0)..(-6,0,0)); draw(lpic,(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),0)--(6*cos(atan(-1/5)+3.14159),6*sin(atan(-1/5)+3.14159),8)); size(rpic,5cm); draw(rpic,surface(revolution((0,0,0),(3,3*sqrt(3),8)..(0,6,4)..(-3,3*sqrt(3),0),Z,230,360)),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(6,0,0)..(-3,-3*sqrt(3),0)--cycle),gray(0.7),nolight); draw(rpic,surface((-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(0,-6,4)..(-3,-3*sqrt(3),0)--cycle),white,nolight); draw(rpic,(-3,-3*sqrt(3),0)..(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)..(6,0,0)..(-3,3*sqrt(3),0),dashed); draw(rpic,(3,3*sqrt(3),8)--(3,-3*sqrt(3),8)); draw(rpic,(-3,3*sqrt(3),0)..(0,6,4)..(3,3*sqrt(3),8)--(3,3*sqrt(3),8)..(3*sqrt(3),3,8)..(6,0,8)); draw(rpic,(-3,3*sqrt(3),0)--(-3,-3*sqrt(3),0)..(0,-6,4)..(3,-3*sqrt(3),8)--(3,-3*sqrt(3),8)..(3*sqrt(3),-3,8)..(6,0,8)); draw(rpic,(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),0)--(-6*cos(atan(-1/5)+3.14159),-6*sin(atan(-1/5)+3.14159),8)); label(rpic,\"A\",(-3,3*sqrt(3),0),W); label(rpic,\"B\",(-3,-3*sqrt(3),0),W); add(lpic.fit(),(0,0)); add(rpic.fit(),(1,0)); [/asy]$ Solution Label the points where the plane intersects the top face of the cylinder as $C$ and $D$, and the center of the cylinder as $O$, such that $C,O,$ and $A$ are collinear. Let $T$ be the center of the bottom face, and $M$ the midpoint of $\\overline{AB}$. Then $OT=4$, $TM=3$ (because of the 120 degree angle), and so", "+ (y - 1)^2 + (z + 2)^2 == 4, x^2 + y^2 + z^2 == 4}, {x, -4, 4}, {y, -4, 4}, {z, -4, 4}, ContourStyle -> Opacity[0.5], Mesh -> None], Graphics3D[{Red, PointSize[0.04], Point[{{0, 0, 0}, c}], Line[{{0, 0, 0}, c}]} ]] r = 2; R = 2; vol[a_, b_] := Pi (4 a + b) (2 a - b)^2/12; vol[R, Norm@c] • @ShutaoTang I remembered that I had come across it some time ago...luck I guess :) – ubpdqn Sep 9 '16 at 9:33 Update: Region equation changed after OP's edit. In case you're only after a non-exact result, you can change the value of the option WorkingPrecision (by default Infinity for Volume) or change the method. These give a result pretty fast: reg = ImplicitRegion[x^2 + y^2 + z^2 + 4 x - 2 y + 4 z + 5 <= 0 && x^2 + y^2 + z^2 <= 4, {x, y, z}] RepeatedTiming[Volume[reg, WorkingPrecision -> MachinePrecision]] (* {0.076, 2.87979} *) RepeatedTiming[Volume[reg, Method -> \"NIntegrate\"]] (* {0.078, 2.87979} *) Boole is a way: V = Integrate[ Boole[x^2 + y^2 + z^2 + 4 x - 2 y + 4 x + 5 <= 0 && x^2 + y^2 + z^2 <= 4], {x, -∞, ∞}, {y, -∞, ∞}, {z, -∞, ∞}] The result: i.e., $V\\approx", "You can find that every four moves both coordinates decrease by 2. Therefore, both coordinates need to decrease by two 505 times. You subtract, giving you the answer of $\\boxed{\\textbf{(B) } \\text{(-1030,-990)}}$ 14. ## Solution 1 $[asy] real x=sqrt(3); real y=2sqrt(3); real z=3.5; real a=x/2; real b=0.5; real c=3a; pair A, B, C, D, E, F; A = (1,0); B = (3,0); C = (4,x); D = (3,y); E = (1,y); F = (0,x); fill(A--B--C--D--E--F--A--cycle,grey); fill(arc((2,0),1,0,180)--cycle,white); fill(arc((2,y),1,180,360)--cycle,white); fill(arc((z,a),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); label(\"2\",(z,c),NE); pair X,Y,Z; X = (1,0); Y = (2,0); Z = (1.5,a); pair d = 1.9*dir(7); dot(X); dot(Y); dot(Z); label(\"A\",X,SW); label(\"B\",Y,SE); label(\"C\",Z,N); draw(X--Y--Z--A); label(\"1\",(1.5,0),S); label(\"1\",(1.75,a/2),dir(30)); draw(anglemark(Y,X,Z,8),blue); label(\"60^\\circ\",anglemark(Y,X,Z),d); [/asy]$ Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, $BC = 1$, since B is the center of the semicircle with radius 1 that C lies on, $AB = 1$, since B is the center of the semicircle with radius 1 that A lies on, and $\\angle BAC = 60^\\circ$as a regular hexagon has angles of 120$^\\circ$, and $\\angle BAC$ is half of any angle in this", "2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,N); dot((5,4)); label(\"(a+b_1,b_2)\",(5,4),NE); dot((9,7)); label(\"(x,y)\",(9,7),NE); dot((2,4)); label(\"(2b_1,b_2)\",(2,4),SW); dot((9,4)); label(\"(x,b_2)\",(9,4),SE); [/asy]$ Note that since $K$ is the center of square $AB_1A_2B$, $BX \\perp KX$ and $BX = KX$. Additionally, $BY \\parallel KZ$ and $BY \\perp YZ$. $YZ$ is a line, so $\\angle BXY + \\angle KXZ + \\angle BXK = 180^\\circ$. Since $BX \\perp KX$, $\\angle BXK = 90^\\circ$, so $\\angle BXY + \\angle KXZ = 90^\\circ$. Additionally, because $BXY$ is a right triangle, $\\angle YBX + \\angle YXB = 90^\\circ$. Rearranging and substituting results in $\\angle KXZ = \\angle YBX$. Since both $BYX$ and $XZK$ are right triangles, by AAS Congruency, $\\triangle BYX \\cong \\triangle XZK$. Therefore $BY = XZ = b_2$ and $YX = ZK = a - b_1$. From this information, the coordinates of $K$ are $(a+b_1+b_2, b_2 + a - b_1)$. $[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4), K=(9,7), L=(-6,8), M=(-7,-3), N=(4,-8); draw(A--B--C--D--A); dot(A); label(\"(2a,0)\",A,NE); dot(B); label(\"(2b_1,2b_2)\",B,NE); dot(C); label(\"(-2c,0)\",C,NW); dot(D); label(\"(2d_1,-2d_2)\",D,S); dot(K); label(\"K\",K,NE); dot(L); label(\"L\",L,NW); dot(M); label(\"M\",M,SW); dot(N); label(\"N\",N,SE); [/asy]$ By", "$$(ix + 4*(iy+1) + 16*(iz+1))$$ where $$(ix, iy, iz)$$ are the coordinates of the barycenter of the current face, is given below. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 load \"msh3\" int[int] l6 = [37, 42, 45, 40, 25, 57]; int r11 = 11; mesh3 Th = cube(4, 5, 6, [x*2-1, y*2-1, z*2-1], label=l6, flags =3, region=r11); cout << \"Volume = \" << Th.measure << \", border area = \" << Th.bordermeasure << endl; int err = 0; for(int i = 0; i < 100; ++i){ real s = int2d(Th,i)(1.); real sx = int2d(Th,i)(x); real sy = int2d(Th,i)(y); real sz = int2d(Th,i)(z); if(s){ int ix = (sx/s+1.5); int iy = (sy/s+1.5); int iz = (sz/s+1.5); int ii = (ix + 4*(iy+1) + 16*(iz+1) ); //value of ix,iy,iz => face min 0, face max 2, no face 1 cout << \"Label = \" << i << \", s = \" << s << \" \" << ix << iy << iz << \" : \" << ii << endl; if( i != ii ) err++; } } real volr11 = int3d(Th,r11)(1.); cout << \"Volume region = \" << 11 << \": \" << volr11 <<", "and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet? $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label(\"A\", (1,0,3/4), W); label(\"B\", (1,0,1/3), W); label(\"C\", (1,0,1/6-d/4), W); label(\"D\", (1,0,d/2), W); label(\"1/2\", (1,1,3/4), E); label(\"1/3\", (1,1,1/3), E); label(\"1/17\", (0,1,1/6-d/4), E);[/asy]$ $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label(\"D\", (1/2, 1, 0), SE); label(\"A\", (1+1/2, 1, 0), SE); label(\"B\", (2+1/2, 1, 0), SE); label(\"C\", (3+1/2, 1, 0), SE);[/asy]$ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$", "User:Lcz (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) Hi just getting some last minute A(O)IME geo prep in (my thing is alg>nt>combo>geo) from the Intermediate Geometry Problems page and then printing it out :) (2015 II #9) https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_9 A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$. [asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight); triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2)); draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype(\"2 4\")); draw(X+B--X+C+B,linetype(\"2 4\")); draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype(\"2 4\")); [/asy] (2015 I #13) https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_13 With all angles measured in degrees, the product $\\prod_{k=1}^{45} \\csc^2(2k-1)^\\circ=m^n$, where $m$ and $n$ are integers greater than 1. Find $m+n$. (2015 I #11) https://artofproblemsolving.com/wiki/index.php/2015_AIME_I_Problems/Problem_11 Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\\angle B$ and $\\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\\triangle ABC$. (2014 II #14) https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_14 In $\\triangle{ABC}, AB=10, \\angle{A}=30^\\circ$ , and $\\angle{C=45^\\circ}$. Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\\perp{BC}$, $\\angle{BAD}=\\angle{CAD}$, and $BM=CM$. Point", "= u + v + w$$ [Hide Solution] $$2$$ The triangular region $$R$$ with the vertices $$(0,0), \\space (1,1)$$, and $$(1,2)$$ is shown in the following figure. a. Find a transformation $$T : S \\rightarrow R, \\space T(u,v) = (x,y) = (au + bv + dv)$$, where $$a,b,c$$, and $$d$$ are real numbers with $$ad - bc \\neq 0$$ such that $$T^{-1} (0,0) = (0,0), \\space T^{-1} (1,1) = (1,0)$$, and $$T^{-1}(1,2) = (0,1)$$. b. Use the transformation $$T$$ to find the area $$A(R)$$ of the region $$R$$. The triangular region $$R$$ with the vertices $$(0,0), \\space (2,0)$$, and $$(1,3)$$ is shown in the following figure. a. Find a transformation $$T : S \\rightarrow R, \\space T(u,v) = (x,y) = (au + bv + dv)$$, where $$a,b,c$$, and $$d$$ are real numbers with $$ad - bc \\neq 0$$ such that $$T^{-1} (0,0) = (0,0), \\space T^{-1} (2,0) = (1,0)$$, and $$T^{-1}(1,3) = (0,1)$$. b. Use the transformation $$T$$ to find the area $$A(R)$$ of the region $$R$$. [Hide Solution] a. $$T (u,v) = (2u + v, \\space 3v); b. The area of \\(R$$ is $A(R) = \\int_0^3 \\int_{y/3}^{(6-y)/3} dx \\space dy = \\int_0^1 \\int_0^{1-u} \\left|\\frac{\\partial (x,y)}{\\partial (u,v)}\\right| dv \\space du = \\int_0^1 \\int_0^{1-u} 6 dv \\space du = 3.$ In the following exercises, use the transformation $$u =", "# 7.6E: Exercises ### Exercise $$\\PageIndex{1}$$ In the following exercises, the region $$R$$ occupied by a lamina is shown in a graph. Find the mass of $$R$$ with the density function $$\\rho$$. 1. $$R$$ is the triangular region with vertices $$(0,0), \\space (0,3)$$, and $$(6,0); \\space \\rho (x,y) = xy$$. [Hide Solution] $$\\frac{27}{2}$$ 3. $$R$$ is the triangular region with vertices $$(0,0), \\space (1,1)$$, and $$(0,5); \\space \\rho (x,y) = x + y$$. 4. $$R$$ is the rectangular region with vertices $$(0,0), \\space (0,3), \\space (6,3)$$ and $$(6,0); \\space \\rho (x,y) = \\sqrt{xy}$$. [Hide Solution] $$24\\sqrt{2}$$ 5. $$R$$ is the rectangular region with vertices $$(0,1), \\space (0,3), \\space (3,3)$$ and $$(3,1); \\space \\rho (x,y) = x^2y$$. 6. $$R$$ is the trapezoidal region determined by the lines $$y = - \\frac{1}{4}x + \\frac{5}{2}, \\space y = 0, \\space y = 2$$, and $$x = 0; \\space \\rho (x,y) = 3xy$$. [Hide Solution] $$76$$ 7. $$R$$ is the trapezoidal region determined by the lines $$y = 0, \\space y = 1, \\space y = x$$ and $$y = -x + 3; \\space \\rho (x,y) = 2x + y$$. 8. $$R$$ is the disk of radius $$2$$ centered at $$(1,2); \\space \\rho(x,y) = x^2 + y^2 - 2x - 4y + 5$$. [Hide Solution] $$8\\pi$$ 10. $$R$$ is the unit disk;", "x$ subject to \\begin{align} \\color{blue}{x+y+s_1\\phantom{+s_2}}&=\\color{blue}{1}\\\\ \\color{red}{x\\phantom{+y+s_1}+s_2}&=\\color{red}1\\\\ x,y,\\color{blue}{s_1},\\color{red}{s_2}&\\ge0 \\end{align} Thus, each of $x=0,y=0,\\color{blue}{s_1=0},\\color{red}{s_2=0}$ corresponds to a line which bounds the feasible region. If we choose $x,y$ as basic variables, then the basic solution is $x_B=(x,y)=(1,0)$, and we calculate the simplex tableau at this degenerate point. format rat; c = [0 1 0 0]'; b = [1 1]'; A = [1 1 1 0; 1 0 0 1]; basis = [1 2]; B = A(:,basis); cB = c(basis); T = [B\\b B\\A; cB'*(B\\b) cB'*(B\\A) - c'] T = 1 1 0 0 1 -0 -0 1 1 -1 0 0 0 1 -1 Therefore, the simplex tableau is \\begin{array}{c|r|rrrr|r} B & b & x & y & s_1 & s_2 & \\theta \\\\ \\hline x & 1 & 1 & 0 & 0 & 1^* & 1 \\\\ y & 0 & 0 & 1 & 1 & -1 & - \\\\ \\hline z & 0 & 0 & 0 & 1 & -1 & \\end{array} We are not trapped in the degenerate solution. Therefore, $x$ leaves the basis and $\\color{red}{s_2}$ enters the basis. basis = [4 2]; B = A(:,basis); cB = c(basis); T = [B\\b B\\A; cB'*(B\\b) cB'*(B\\A) - c'] T = 1 1 0 0 1 1 1 1 1 0 1 1 0 1 0 Therefore,", "### More examples of creating graphs and highlighting regions Statement Meaning Result in GC $$x=3$$ Highlight all the points in the plane with coordinates $(x,y)$ such that $x=3$ $$y=x^3$$ Highlight all the points in the plane with coordinates $(x,y)$ such that $y=x^3$. $$x=y^3$$ Highlight all the points in the plane with coordinates $(x,y)$ such that $x=y^3$. $$1\\le x\\le 2$$ Highlight all the points in the plane with coordinates $(x,y)$ such that $1\\le x \\le 2$. $$1\\le x \\le 2, -0.5\\le y \\le 0$$ Highlight all the points in the plane with coordinates $(x,y)$ such that $1\\le x \\le 2$ AND $-0.5\\le y \\le 0$. $$f(x)=x^2\\\\[1ex]g(x)=\\sqrt{x}\\\\[1ex] y=f(x)\\\\[1ex] y=g(x)\\\\[1ex] f(x)\\lt y\\lt g(x)$$ Define the functions $f$ and $g$. Highlight all the points in the plane with coordinates $(x,y)$ such that $y=f(x)$. Highlight all the points in the plane with coordinates $(x,y)$ such that $y=g(x)$. Highlight all the points in the plane with coordinates $(x,y)$ such that $f(x)\\lt y$ and $y\\lt g(x)$. $$f(x)=x^2\\\\[1ex] g(x)=x^3\\\\[1ex] x=f(y)\\\\[1ex] x=g(y)\\\\[1ex] f(y)\\lt x \\lt g(y)$$ Define the functions $f$ and $g$. Highlight all the points in the plane with coordinates $(x,y)$ such that $x=f(y)$. Highlight all the points in the plane with coordinates $(x,y)$ such that $x=g(y)$. Highlight all the points in the plane with coordinates $(x,y)$ such that $f(y)\\lt x$ and $x\\lt g(y)$. ## 8.2.2" ]

[ "+0 # The vector $\\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$ is orthogonal to the vector $\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$. The vector \\$\\begin{pm 0 52 1 The vector $$\\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ is orthogonal to the vector $$\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$$. The vector $$\\begin{pmatrix} 2 \\\\ m \\end{pmatrix}$$ is orthogonal to the vector $$\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$$. Find m . Feb 2, 2019 #1 +96080 +1 If vectors are orthagonal, their dot product = 0 So [ 4 * 3 ] + [ -7 * k] = 0 12 - 7k = 0 -7k = -12 k = 12/7 So [2 * 3 ] + [12/7 * m] = 0 6 + (12/7)m = 0 (12/7)m = - 6 m = -6*7/12 = -7/2 Feb 2, 2019", "or calculator to find the matrix inverse. Orthogonal Vectors Two vectors, x and y, are orthogonal if their dot product is zero. For example $e \\cdot f = \\begin{pmatrix} 2 & 5 & 4 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 5 \\end{pmatrix} = 2*4 + (5)*(-2) + 4*5 = 8-10+20 = 18$ Vectors e and f are not orthogonal. $g \\cdot h = \\begin{pmatrix} 2 & 3 & -2 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 1 \\end{pmatrix} = 2*4 + (3)*(-2) + (-2)*1 = 8-6-2 = 0$ However, vectors g and h are orthogonal. Orthogonal can be thought of as an expansion of perpendicular for higher dimensions. Let $$x_1, x_2, \\ldots , x_n$$ be m-dimensional vectors. Then a linear combination of $$x_1, x_2, \\ldots , x_n$$ is any m-dimensional vector that can be expressed as $c_1 x_1 + c_2 x_2 + \\ldots + c_n x_n$ where $$c_1, \\ldots, c_n$$ are all scalars. For example: $x_1 =\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix}$ $y =\\begin{pmatrix} -5 \\\\ 12 \\\\ -8 \\end{pmatrix} = 1*\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix} + (-2)* \\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix} = 1*x_1 + (-2)*x_2$ So y is a linear combination of $$x_1$$ and $$x_2$$. The set of all linear combinations of", "From the list below, select the vector that is NOT orthogonal (perpendicular) to: $\\begin{bmatrix} \\cos\\theta \\\\ \\sin\\theta \\\\ 1 \\end{bmatrix}$", "Question # Find a unit vector that is orthogonal to both i+j and i+k. Find a unit vector that is orthogonal to both $$i+j$$ and $$i+k$$. 2021-05-13 Step 1 Let $$a=i+j,\\ b=i+k$$ The cross product $$a\\times b$$ is arthogonal to both a and b $$a\\times b=\\left|\\begin{matrix}i & j & k \\\\ 1 & 1 & 0 \\\\ 1 & 0 & 1 \\end{matrix}\\right|$$ $$=i(1-0)-j(1-0)+k(0-1)$$ $$a\\times b=i-j-k$$ Since the vector we found had magnitude $$\\sqrt{3}$$ $$i.e.\\ |a\\times b|=\\sqrt{\\check{1}+\\check{1}+\\check{1}}=\\sqrt{3}$$ A unit vector that is arthogonal to both $$i+j$$ and $$i+k$$ is $$u=\\frac{(i-j-k)}{\\sqrt{3}}$$ or $$u=\\frac{\\sqrt{3}(i-j-k)}{3}$$", "any other vector is defined to be a right angle). Thus vectors from $\\mathbb{R}^n$ are orthogonal (or perpendicular) if and only if their dot product is zero. Example 2.8 These vectors are orthogonal. $\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}\\cdot\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}=0$ The arrows are shown away from canonical position but nevertheless the vectors are orthogonal. Example 2.9 The $\\mathbb{R}^3$ angle formula given at the start of this subsection is a special case of the definition. Between these two the angle is $\\arccos(\\frac{(1)(0)+(1)(3)+(0)(2)}{\\sqrt{1^2+1^2+0^2}\\sqrt{0^2+3^2+2^2}}) =\\arccos(\\frac{3}{\\sqrt{2}\\sqrt{13}})$ approximately $0.94 \\text{radians}$. Notice that these vectors are not orthogonal. Although the $yz$-plane may appear to be perpendicular to the $xy$-plane, in fact the two planes are that way only in the weak sense that there are vectors in each orthogonal to all vectors in the other. Not every vector in each is orthogonal to all vectors in the other. ## Exercises This exercise is recommended for all readers. Problem 1 Find the length of each vector. 1. $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ 2. $\\begin{pmatrix} -1 \\\\ 2 \\end{pmatrix}$ 3. $\\begin{pmatrix} 4 \\\\ 1 \\\\ 1 \\end{pmatrix}$ 4. $\\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$ 5. $\\begin{pmatrix} 1 \\\\ -1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 2 Find the angle between each two, if it is defined.", "vectors, because the use of unit vectors simplifies many calculations. For example, in Euclidean space, the standard scalar product of two unit vectors is equal to the cosine of the angle between the two. ## Finite-dimensional case Canonical unit vectors in the Euclidean plane In the finite-dimensional real vector spaces , the most frequently preferred standard basis consists of the canonical unit vectors${\\ displaystyle \\ mathbb {R} ^ {n}}$ ${\\ displaystyle {\\ vec {e}} _ {1} = {\\ begin {pmatrix} 1 \\\\ 0 \\\\ 0 \\\\\\ vdots \\\\ 0 \\ end {pmatrix}}, \\; {\\ vec {e} } _ {2} = {\\ begin {pmatrix} 0 \\\\ 1 \\\\ 0 \\\\\\ vdots \\\\ 0 \\ end {pmatrix}}, \\; {\\ vec {e}} _ {3} = {\\ begin {pmatrix} 0 \\\\ 0 \\\\ 1 \\\\\\ vdots \\\\ 0 \\ end {pmatrix}}, \\; \\ dots, \\; {\\ vec {e}} _ {n} = {\\ begin {pmatrix} 0 \\\\ 0 \\\\ 0 \\\\\\ vdots \\\\ 1 \\ end {pmatrix}}}$. If you combine the canonical unit vectors into a matrix , you get a unit matrix . The set of canonical unit vectors des forms an orthonormal basis with respect to the canonical scalar product , i. H. two canonical unit vectors are perpendicular to each other (= \"ortho\"), all are normalized (= \"normal\") and they", "solutions of the two systems we find a basis of $V_C(0)\\setminus \\{0\\}$, say $$\\begin{pmatrix}1\\\\-1\\\\0\\end{pmatrix}, \\begin{pmatrix}1\\\\0\\\\-1\\end{pmatrix}\\;,$$ and a basis of $V_C(3)\\setminus\\{0\\}$, say $$\\begin{pmatrix}1\\\\1\\\\1\\end{pmatrix}\\;.$$ These three vectors are a basis of $\\Bbb R^{3\\times3}$ Observe the vector of all ones gives the eigenvalue $3$. So the other two eigenvectors are orthogonal to it. You can choose the vectors $$\\begin{pmatrix}1\\\\-1\\\\0\\end{pmatrix}, \\begin{pmatrix}1\\\\0\\\\-1\\end{pmatrix}$$ as the other two eigenvectors of $C$. • @G_o_pi_i_e I don't know what orthogonal vectors are yet but I'm assuming this answer is correct. Does is have something to do with dot product? – Zero Pancakes Jan 4 '17 at 7:11 • @Michael: Orthogonal means dot product is zero, yes. – MvG Jan 4 '17 at 7:13", "1. $\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix}, \\begin{pmatrix} 1 \\\\ 4 \\end{pmatrix}$ 2. $\\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 4 \\\\ 1 \\end{pmatrix}$ 3. $\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix}, \\begin{pmatrix} 1 \\\\ 4 \\\\ -1 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 3 During maneuvers preceding the Battle of Jutland, the British battle cruiser Lion moved as follows (in nautical miles): $1.2$ miles north, $6.1$ miles $38$ degrees east of south, $4.0$ miles at $89$ degrees east of north, and $6.5$ miles at $31$ degrees east of north. Find the distance between starting and ending positions (O'Hanian 1985). Problem 4 Find $k$ so that these two vectors are perpendicular. $\\begin{pmatrix} k \\\\ 1 \\end{pmatrix} \\qquad \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix}$ Problem 5 Describe the set of vectors in $\\mathbb{R}^3$ orthogonal to this one. $\\begin{pmatrix} 1 \\\\ 3 \\\\ -1 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 6 1. Find the angle between the diagonal of the unit square in $\\mathbb{R}^2$ and one of the axes. 2. Find the angle between the diagonal of the unit cube in $\\mathbb{R}^3$ and one of the axes. 3. Find the angle between the diagonal of the unit cube in $\\mathbb{R}^n$ and one of the axes. 4. What is the limit, as $n$ goes to $\\infty$,", "one can take any two vectors (almost any) and they will also form a basis, meaning that any vector can be decomposed by their linear combination. Example. Suppose we have a vector $$\\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}$$ in a basis of unit orthogonal vectors, and we want to decompose it in another basis $$\\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix} \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix}$$: $\\begin{split} \\alpha_1 \\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix} + \\alpha_2 \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\3 \\end{pmatrix} \\end{split}$ From where we can find that $$\\alpha_1 = -1$$, $$\\alpha_2 = -1$$: $\\begin{split} -1 \\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix} + -1 \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 0 \\end{pmatrix} + \\begin{pmatrix} 0 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix} \\end{split}$ So, vector $$\\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}$$ in the basis of unit orthogonal vectors is represented as $$\\begin{pmatrix} -1 \\\\ -1 \\end{pmatrix}$$ in the basis $$\\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix} \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix}$$. But, as mentioned before, not any set of vectors makes a basis. For example, the set of vectors $$\\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix}$$, $$\\begin{pmatrix} -3 \\\\ 0 \\end{pmatrix}$$ cannot decompose the vector $$\\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}$$, so this set of vectors is not a basis. What is the fundamental difference between these sets, and", "unit vector. $\\bf edit:$ to find all the unit vectors, we need to find an orthogonal basis first. if you project $(3,0,-2)$ onto $(2,1,0)$ you get $(4,2,0).$ that is we have split $(3,0,-2) = (4,2,0) + (-1,2,-2).$ therefore an orthogonal basis is $\\{(2,1,0), (-1,2,-2)\\}.$ we make them orthonormal by dividing by their lengths so that $$\\{u = (2/\\sqrt 5, 1/\\sqrt 5), v = (-1/3, 2/3, 2/3)\\}$$ finally, an arbitrary unit orthogoanal vector is $$u\\cos t + v \\sin t, t \\text{ any real number.}$$ • you do find all vectors orthogonal as the linear combinations of the two vectors in the post. they are linearly independent, so it is a basis for the space orthogonal to the line $(2t, -4t, 3t)$ – abel Feb 20 '15 at 1:41", "How do you find all unit vectors orthogonal to both the vectors given below: (1,2,-1) and (3,3,-4)? Jul 5, 2016 $\\pm \\frac{1}{\\sqrt{35}} \\left(- 5 , 1 , - 3\\right)$ Explanation: The unit vectors (in opposite directions) that are orthogonal to both $a \\mathmr{and} b$ is $\\pm \\frac{a X b}{|} a X b |$. Here $a = \\left(1 , 2 , - 1\\right) \\mathmr{and} b = \\left(3 , 3 , - 4\\right) ,$ $a X b = \\left(\\left(2\\right) \\left(- 4\\right) - \\left(3\\right) \\left(- 1\\right) , \\left(- 1\\right) \\left(3\\right) - \\left(- 4\\right) \\left(1\\right) , \\left(1\\right) \\left(3\\right) - \\left(3\\right) \\left(2\\right)\\right)$ $= \\left(- 5 , 1 , - 3\\right) \\mathmr{and} | \\left(- 5 , 1 , - 3\\right) | = \\sqrt{35}$. So, the answer is $\\pm \\frac{1}{\\sqrt{35}} \\left(- 5 , 1 , - 3\\right)$", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "Orthogonal Vectors Two vectors, x and y, are orthogonal if their dot product is zero. For example $e \\cdot f = \\begin{pmatrix} 2 & 5 & 4 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 5 \\end{pmatrix} = 2*4 + (5)*(-2) + 4*5 = 8-10+20 = 18$ Vectors e and f are not orthogonal. $g \\cdot h = \\begin{pmatrix} 2 & 3 & -2 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 1 \\end{pmatrix} = 2*4 + (3)*(-2) + (-2)*1 = 8-6-2 = 0$ However, vectors g and h are orthogonal. Orthogonal can be thought of as an expansion of perpendicular for higher dimensions. Let $x_1, x_2, \\ldots , x_n$ be m-dimensional vectors. Then a linear combination of $x_1, x_2, \\ldots , x_n$ is any m-dimensional vector that can be expressed as $c_1 x_1 + c_2 x_2 + \\ldots + c_n x_n$ where $c_1, \\ldots, c_n$ are all scalars. For example: $x_1 =\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix}$ $y =\\begin{pmatrix} -5 \\\\ 12 \\\\ -8 \\end{pmatrix} = 1*\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix} + (-2)* \\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix} = 1*x_1 + (-2)*x_2$ So y is a linear combination of $x_1$ and $x_2$. The set of all linear combinations of $x_1, x_2, \\ldots , x_n$ is called", "Let $$, $$, $$, $$. Compute the following: $$ $$ $$ $$ $$ $$ $$ Is the computation $$ possible? Yes No Let $$. Compute the following: $$ What is the distance between $$ and $$? 85 $$ 65 $$ True/False: Two vectors in $$ are orthogonal if and only if their inner product (dot product) is zero. True False True/False: The vectors $$ and $$ are orthogonal. True False True/False: $$ is a unit vector. True False Find a unit vector in the direction of $$. First find the length of $$: $$. Then find a unit vector $$. Find a unit vector in the direction of $$. First find the length of $$: $$. Then find a unit vector $$.", "Free Version Easy # Unit Vector: Identify Unit Vectors in R^2, Select All LINALG-W3IGEI Which of the following vectors are unit vectors? Select ALL that apply. A $\\begin{bmatrix} 0\\\\\\ 0\\end{bmatrix}$ B $\\begin{bmatrix} 10\\\\\\ -9\\end{bmatrix}$ C $\\begin{bmatrix} 1/2\\\\\\ 1/2\\end{bmatrix}$ D $\\begin{bmatrix} 1/\\sqrt{2}\\\\\\ 1/\\sqrt{2}\\end{bmatrix}$ E $\\begin{bmatrix} \\sqrt{3}/2\\\\\\ 1/2\\end{bmatrix}$ F $\\begin{bmatrix} 1/\\sqrt{2}\\\\\\ 0\\end{bmatrix}$", "the sum of the first two vectors, and is therefore redundant. Therefore, this set of three vectors is NOT a basis for dimension 2. More interestingly, note, for now that the dimension of the set of two dimensional vectors is two, and the number of basis vectors needed is two. Also, however, given two vectors: $\\begin{pmatrix}0 \\\\1 \\end{pmatrix}$ and $\\begin{pmatrix}0 \\\\2 \\end{pmatrix}$. These do NOT serve as a basis for vectors of dimension two because it is impossible to make $\\begin{pmatrix}1 \\\\0 \\end{pmatrix}$ out of them. Moreover, the two are NOT linearly independent. Finally, given these two vectors, $\\begin{pmatrix}1 \\\\1 \\end{pmatrix}$ and $\\begin{pmatrix}0 \\\\1 \\end{pmatrix}$, we know that these vectors are first, linearly independent. But do they span the entire space of two dimensional vectors? Yes. Note that since $\\begin{pmatrix}1 \\\\0 \\end{pmatrix}$ and $\\begin{pmatrix}0 \\\\1 \\end{pmatrix}$ span the space of two dimensional vectors. This means that every vector, $\\begin{pmatrix}a \\\\b \\end{pmatrix}$ can be written as $a_1 \\begin{pmatrix}1 \\\\0 \\end{pmatrix} + a_2 \\begin{pmatrix}0 \\\\1 \\end{pmatrix}$ , in this case $a_1$ =a and $a_2$=b. However notice that $\\begin{pmatrix}1 \\\\1 \\end{pmatrix} = \\begin{pmatrix}1 \\\\0 \\end{pmatrix} + \\begin{pmatrix}0 \\\\1 \\end{pmatrix}$. \\ This means that given a vector $\\begin{pmatrix}a \\\\b \\end{pmatrix} = a_1 \\begin{pmatrix}1 \\\\0 \\end{pmatrix} + a_2 \\begin{pmatrix}0 \\\\1 \\end{pmatrix} = a_1* \\begin{pmatrix}1 \\\\0 \\end{pmatrix} + a_1* \\begin{pmatrix}0 \\\\1 \\end{pmatrix} + a_2", "-2x so that $\\begin{pmatrix}1 \\\\ -2 \\\\ 0\\end{pmatrix}$ and $\\begin{pmatrix} 0 \\\\ 0 \\\\ 1\\end{pmatrix}$ are basis vectors.", "and} \\quad \\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\\\ \\end{pmatrix} To write down any vector v in the space all we need is three numbers, one to multiply each unit vector: v = \\begin{pmatrix} a \\\\ b \\\\ c \\\\ \\end{pmatrix} = a\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\\\ \\end{pmatrix} + b\\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\\\ \\end{pmatrix} + c \\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\\\ \\end{pmatrix} By convention we write vectors in columns, which will make more sense in the next section. And thus we have built the standard sort of coordinate system that we all know and love from 10th grade math. This sort of basis for \\mathbb C^n also has the property that it is orthonormal, meaning that with the standard inner product all of the unit vectors are orthogonal to each other (their mutual inner products are always zero). In the rest of this piece we will assume that all of our Hilbert spaces have an orthonormal basis and that they are finite dimensional. Of course, the more famous state spaces in quantum mechanics (for position and momentum) are infinite dimensional, which is the other reason Hilbert spaces became a thing. But we will not deal with any of that complication here. ### Operators and Observables In classical mechanics we did not", "to an arbitrary vector $(x,y,z)$, and see if the resulting vector is orthogonal to the normal vector of that given plane, thus implying that the vector is projected successfully to the plane? This is necessary but not sufficient. If $n$ is normal to the plane and $\\langle n, Tx\\rangle = 0$ for all $x$, then the image of $T$ is contained in the plane, but that doesn't necessarily mean that $T$ is a projection onto the plane. For example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$ This matrix has the property that $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. So the image lies in the plane whose normal vector is $\\begin{pmatrix}1 \\\\ 0 \\\\ 0\\end{pmatrix}$. But it is not a projection onto that plane because the image only has dimension 1. For another example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 2 \\end{pmatrix}$$ Once again, we have $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. The dimension of the image is correct (2), but this", "? Free Version Easy # Orthogonal Projection in Two Dimensions LINALG-Z3MT8D The orthogonal projection of the vector $\\begin{bmatrix}3\\\\\\ 1\\end{bmatrix}$ onto the line $3x -4y=0$ is A $\\begin{bmatrix}\\frac{12}{5}\\\\\\ \\frac{9}{5}\\end{bmatrix}$ B $\\begin{bmatrix}-1\\\\\\ 3\\end{bmatrix}$ C $\\begin{bmatrix}0\\\\\\ 0\\end{bmatrix}$ D $5$" ]

[ "Question # Find a unit vector that is orthogonal to both i+j and i+k. Find a unit vector that is orthogonal to both $$i+j$$ and $$i+k$$. 2021-05-13 Step 1 Let $$a=i+j,\\ b=i+k$$ The cross product $$a\\times b$$ is arthogonal to both a and b $$a\\times b=\\left|\\begin{matrix}i & j & k \\\\ 1 & 1 & 0 \\\\ 1 & 0 & 1 \\end{matrix}\\right|$$ $$=i(1-0)-j(1-0)+k(0-1)$$ $$a\\times b=i-j-k$$ Since the vector we found had magnitude $$\\sqrt{3}$$ $$i.e.\\ |a\\times b|=\\sqrt{\\check{1}+\\check{1}+\\check{1}}=\\sqrt{3}$$ A unit vector that is arthogonal to both $$i+j$$ and $$i+k$$ is $$u=\\frac{(i-j-k)}{\\sqrt{3}}$$ or $$u=\\frac{\\sqrt{3}(i-j-k)}{3}$$", "Orthogonal Vectors Two vectors, x and y, are orthogonal if their dot product is zero. For example $e \\cdot f = \\begin{pmatrix} 2 & 5 & 4 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 5 \\end{pmatrix} = 2*4 + (5)*(-2) + 4*5 = 8-10+20 = 18$ Vectors e and f are not orthogonal. $g \\cdot h = \\begin{pmatrix} 2 & 3 & -2 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 1 \\end{pmatrix} = 2*4 + (3)*(-2) + (-2)*1 = 8-6-2 = 0$ However, vectors g and h are orthogonal. Orthogonal can be thought of as an expansion of perpendicular for higher dimensions. Let $x_1, x_2, \\ldots , x_n$ be m-dimensional vectors. Then a linear combination of $x_1, x_2, \\ldots , x_n$ is any m-dimensional vector that can be expressed as $c_1 x_1 + c_2 x_2 + \\ldots + c_n x_n$ where $c_1, \\ldots, c_n$ are all scalars. For example: $x_1 =\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix}$ $y =\\begin{pmatrix} -5 \\\\ 12 \\\\ -8 \\end{pmatrix} = 1*\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix} + (-2)* \\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix} = 1*x_1 + (-2)*x_2$ So y is a linear combination of $x_1$ and $x_2$. The set of all linear combinations of $x_1, x_2, \\ldots , x_n$ is called", "or calculator to find the matrix inverse. Orthogonal Vectors Two vectors, x and y, are orthogonal if their dot product is zero. For example $e \\cdot f = \\begin{pmatrix} 2 & 5 & 4 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 5 \\end{pmatrix} = 2*4 + (5)*(-2) + 4*5 = 8-10+20 = 18$ Vectors e and f are not orthogonal. $g \\cdot h = \\begin{pmatrix} 2 & 3 & -2 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 1 \\end{pmatrix} = 2*4 + (3)*(-2) + (-2)*1 = 8-6-2 = 0$ However, vectors g and h are orthogonal. Orthogonal can be thought of as an expansion of perpendicular for higher dimensions. Let $$x_1, x_2, \\ldots , x_n$$ be m-dimensional vectors. Then a linear combination of $$x_1, x_2, \\ldots , x_n$$ is any m-dimensional vector that can be expressed as $c_1 x_1 + c_2 x_2 + \\ldots + c_n x_n$ where $$c_1, \\ldots, c_n$$ are all scalars. For example: $x_1 =\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix}$ $y =\\begin{pmatrix} -5 \\\\ 12 \\\\ -8 \\end{pmatrix} = 1*\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix} + (-2)* \\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix} = 1*x_1 + (-2)*x_2$ So y is a linear combination of $$x_1$$ and $$x_2$$. The set of all linear combinations of", "+0 # The vector $\\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$ is orthogonal to the vector $\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$. The vector \\$\\begin{pm 0 52 1 The vector $$\\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ is orthogonal to the vector $$\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$$. The vector $$\\begin{pmatrix} 2 \\\\ m \\end{pmatrix}$$ is orthogonal to the vector $$\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$$. Find m . Feb 2, 2019 #1 +96080 +1 If vectors are orthagonal, their dot product = 0 So [ 4 * 3 ] + [ -7 * k] = 0 12 - 7k = 0 -7k = -12 k = 12/7 So [2 * 3 ] + [12/7 * m] = 0 6 + (12/7)m = 0 (12/7)m = - 6 m = -6*7/12 = -7/2 Feb 2, 2019", "preferred). To start off the derivation, assume two vectors a and b exist, and so does another vector n, that is orthogonal to both a and b: $\\large \\mathbf{a}=\\begin{pmatrix} a_1 \\\\ a_2\\\\ a_3 \\end{pmatrix} \\; \\; \\mathbf{b}=\\begin{pmatrix} b_1 \\\\ b_2\\\\ b_3 \\end{pmatrix} \\; \\mathbf{n}=\\begin{pmatrix} n_1 \\\\ n_2\\\\ n_3 \\end{pmatrix}$ Our aim is to find the components of n, in terms of the components of a and b. Since n is orthogonal to both a and b, these conditions must apply: \\large \\begin{align*} &\\mathbf{a}\\cdot\\mathbf{n}=0 \\\\ &\\mathbf{b}\\cdot\\mathbf{n}=0 \\\\ \\\\&(1)\\: a_1n_1+a_2n_2+a_3n_3=0\\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\\\ &(2)\\: \\, b_1n_1+b_2n_2+b_3n_3=0\\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\end{align*} Another independent equation of the component of n is necessary to solve this simultaneous equation with three unknowns of n1, n2 and n3. Notice that we haven't specified the magnitude of n. For simplicity, let the magnitude of n be 1 or unity. \\large \\begin{align*} &\\left \\| \\mathbf{n} \\right \\|=1\\: \\: \\: \\: \\: \\: \\: \\: \\: \\: \\: \\\\\\\\&\\sqrt{n_1^2+n_2^2+n_3^2}=1\\\\ \\\\\\therefore \\; &(3)\\, \\, \\, n_1^2+n_2^2+n_3^2=1 \\end{align*} \\large \\begin{align*} &(1)\\: a_1n_1+a_2n_2+a_3n_3=0\\\\ &(2)\\: \\, b_1n_1+b_2n_2+b_3n_3=0\\\\ &(3)\\, \\, \\, n_1^2+n_2^2+n_3^2=1 \\end{align*} This now gives us 3", "## VECTORS In three dimensions, the base vectors are $$i=\\begin{pmatrix} 1\\\\0 \\\\0 \\end{pmatrix}$$, $$j=\\begin{pmatrix} 0\\\\1 \\\\0 \\end{pmatrix}$$, $$k=\\begin{pmatrix} 0\\\\0 \\\\1 \\end{pmatrix}$$ that are along the $$x, y$$ and $$z$$ coordinate directions, respectively, as shown in the figure. For example, $$v=\\begin{pmatrix} -4\\\\3 \\\\2 \\end{pmatrix}$$ can be written as $$-4i+3j+2k$$. The magnitude of $$v$$: $$|v|=\\sqrt{((-4)^2+3^2+2^2)}=\\sqrt{29}$$ Given that if $$v=xi+yj+zk$$ then $$|v|=\\sqrt{x^2+y^2+z^2}$$ ## UNIT VECTORS If $$\\overrightarrow{AB}$$ is a non-unit vector, then a unit vector in the direction of $$\\overrightarrow{AB}$$ is $${1\\over |\\overrightarrow{AB}|}\\overrightarrow{AB}$$ LESSON 1 Determine the unit vector in the direction of the vector $$\\overrightarrow{OA}=\\begin{pmatrix} -4\\\\3 \\\\2 \\end{pmatrix}$$. SOLUTION • Determine the length of the vector $$|\\overrightarrow{OA}|=\\sqrt{(-4)^2+(3)^2+(2)^2}=\\sqrt{29}$$ • Use the form for a unit vector Unit Vector $${1\\over \\sqrt{29}} \\begin{pmatrix} -4\\\\3 \\\\2 \\end{pmatrix}$$ TRY THIS Complete Question 1 from worksheet.", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "$\\mathbb{R_3}$ Given: • A line $l$ in vector form • A point $P$ on the plane 1. We can find the point $X$ where the perpendicular line from $l$ to $P$ intersects $l$ 2. Using $X$, we can then find $\\vec{PX}$, the vector from $P$ to $X$ 3. We know that the direction of $l \\dot{} \\vec{PX} = 0$, so we can solve for λ 4. Once we have λ, we can substitute this back into $\\vec{PX}$ and find the magnitude, giving the shortest distance ### The Cross Product #### Arithmetic Properties Of The Cross Product The cross product gives us a vector that is perpendicular (orthogonal) to both vectors $\\vec{AB}$ and $\\vec{AC}$. When using the cross product, think of the ‘right hand rule’. For each row of the two vectors, cross subtract the multiplication of two rows below crossed: $$\\begin{pmatrix}x_1\\\\x_2\\\\x_3\\end{pmatrix} \\times \\begin{pmatrix}y_1\\\\y_2\\\\y_3\\end{pmatrix} = \\begin{pmatrix}(x_2 \\times y_3) - (y_2 \\times x_3)\\\\(x_3 \\times y_1) - (y_3 \\times x_1)\\\\(x_1 \\times y_2) - (y_1 \\times x_2)\\end{pmatrix}$$ $e.g. \\begin{pmatrix}2\\\\2\\\\-2\\end{pmatrix} \\times \\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}4\\\\-6\\\\-2\\end{pmatrix}$ #### Geometric Interpretation Of The Cross Product As stated above, the cross product gives us a vector that is perpendicular (orthogonal) to two vectors. Useful for: • Finding a vector perpendicular to two other vectors • Finding the area of a parallelogram by taking the magnitude of the cross", "rule, but I'll be explaining a possible reason later on as to why the left-hand rule is preferred). To start off the derivation, assume two vectors a and b exist, and so does another vector n, that is orthogonal to both a and b: $\\large \\mathbf{a}=\\begin{pmatrix} a_1 \\\\ a_2\\\\ a_3 \\end{pmatrix} \\; \\; \\mathbf{b}=\\begin{pmatrix} b_1 \\\\ b_2\\\\ b_3 \\end{pmatrix} \\; \\mathbf{n}=\\begin{pmatrix} n_1 \\\\ n_2\\\\ n_3 \\end{pmatrix}$ Our aim is to find the components of n, in terms of the components of a and b. Since n is orthogonal to both a and b, these conditions must apply: \\large \\begin{align*} &\\mathbf{a}\\cdot\\mathbf{n}=0 \\\\ &\\mathbf{b}\\cdot\\mathbf{n}=0 \\\\ \\\\&(1)\\: a_1n_1+a_2n_2+a_3n_3=0\\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\\\ &(2)\\: \\, b_1n_1+b_2n_2+b_3n_3=0\\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\end{align*} Another independent equation of the component of n is necessary to solve this simultaneous equation with three unknowns of n1, n2 and n3. Notice that we haven't specified the magnitude of n. For simplicity, let the magnitude of n be 1 or unity. \\large \\begin{align*} &\\left \\| \\mathbf{n} \\right \\|=1\\: \\: \\: \\: \\: \\: \\: \\: \\: \\: \\: \\\\\\\\&\\sqrt{n_1^2+n_2^2+n_3^2}=1\\\\ \\\\\\therefore \\; &(3)\\, \\, \\, n_1^2+n_2^2+n_3^2=1 \\end{align*}", "# Magnitude of a Vector - How to Calculate It ## (also called Modulus of a Vector) Every vector has a magnitude, also called modulus. In this section we learn how to calculate the magnitude of a vector. ### Magnitude (Modulus) of a Vector Given a vector $$\\vec{v} = \\begin{pmatrix}x \\\\ y \\end{pmatrix}$$, its magnitude, also called modulus can be calculated with the formula: $\\begin{vmatrix}\\vec{v} \\end{vmatrix} = \\sqrt{x^2 + y^2}$ ### Tutorial: Magnitude of a Vector In this tutorial we learn how to calculate the magnitude of a vector. ### Example 1 Given the two vector $$\\vec{a} = \\begin{pmatrix}1 \\\\ 3 \\end{pmatrix}$$ we can calculate its magnitude as follows: \\begin{aligned} \\begin{vmatrix} \\vec{a} \\end{vmatrix} & = \\sqrt{1^2 + 3^2} \\\\ & = \\sqrt{1 + 9} \\\\ \\begin{vmatrix} \\vec{a} \\end{vmatrix} & = \\sqrt{10} \\quad \\begin{pmatrix} \\approx 3.16 \\end{pmatrix} \\end{aligned} ## Exercise 1 Find and write the exact value of the magnitude for each of the vectors, shown below. When applicable: also use your calculator to round your answers to two decimal places (2 d.p). 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$. 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$. 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$. 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$. 1. For vector $$\\vec{e} = \\begin{pmatrix}", "are talking about bound vector, then two bound vectors are equal if they have the same base point and end point. For example, the vector i + 2j + 3k with base point (1,0,0) and the vector i+2j+3k with base point (0,1,0) are different bound vectors, but the same (displacement) vector. Scalar multiplication For scalar multiplication, we simply multiply each component by the scalar. We commonly use Greek letters for scalars, and Roman letters for vectors. So for a scalar value of λ and a vector v defined by r and θ, the new vector is now λr and θ. Notice how the direction does not change. Example Say we have ${\\displaystyle {\\begin{pmatrix}2\\\\3\\end{pmatrix}}}$ and we wish to double the magnitude. So, ${\\displaystyle 2{\\begin{pmatrix}2\\\\3\\end{pmatrix}}={\\begin{pmatrix}4\\\\6\\end{pmatrix}}}$ . Simply, to add two vectors, you must add the respective x-components together to obtain the new x-component, and likewise add the two y-components together to obtain the new y-component. Example Say we have ${\\displaystyle \\mathbf {v_{1}} ={\\begin{pmatrix}2\\\\3\\end{pmatrix}},\\mathbf {v_{2}} ={\\begin{pmatrix}4\\\\6\\end{pmatrix}}}$ and we wish to add these. So, ${\\displaystyle \\mathbf {v_{1}} +\\mathbf {v_{2}} ={\\begin{pmatrix}6\\\\9\\end{pmatrix}}}$ . Magnitude The magnitude of a vector is its length in R+ The dot product The dot product of two vectors is defined as the sum of the products of the components. Symbolically we write ${\\displaystyle {\\begin{pmatrix}a_{1}\\\\a_{2}\\end{pmatrix}}\\cdot {\\begin{pmatrix}b_{1}\\\\b_{2}\\end{pmatrix}}=a_{1}b_{1}+a_{2}b_{2}}$ For example, ${\\displaystyle {\\begin{pmatrix}3\\\\5\\end{pmatrix}}\\cdot", "any other vector is defined to be a right angle). Thus vectors from $\\mathbb{R}^n$ are orthogonal (or perpendicular) if and only if their dot product is zero. Example 2.8 These vectors are orthogonal. $\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}\\cdot\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}=0$ The arrows are shown away from canonical position but nevertheless the vectors are orthogonal. Example 2.9 The $\\mathbb{R}^3$ angle formula given at the start of this subsection is a special case of the definition. Between these two the angle is $\\arccos(\\frac{(1)(0)+(1)(3)+(0)(2)}{\\sqrt{1^2+1^2+0^2}\\sqrt{0^2+3^2+2^2}}) =\\arccos(\\frac{3}{\\sqrt{2}\\sqrt{13}})$ approximately $0.94 \\text{radians}$. Notice that these vectors are not orthogonal. Although the $yz$-plane may appear to be perpendicular to the $xy$-plane, in fact the two planes are that way only in the weak sense that there are vectors in each orthogonal to all vectors in the other. Not every vector in each is orthogonal to all vectors in the other. ## Exercises This exercise is recommended for all readers. Problem 1 Find the length of each vector. 1. $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ 2. $\\begin{pmatrix} -1 \\\\ 2 \\end{pmatrix}$ 3. $\\begin{pmatrix} 4 \\\\ 1 \\\\ 1 \\end{pmatrix}$ 4. $\\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$ 5. $\\begin{pmatrix} 1 \\\\ -1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 2 Find the angle between each two, if it is defined.", "one can take any two vectors (almost any) and they will also form a basis, meaning that any vector can be decomposed by their linear combination. Example. Suppose we have a vector $$\\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}$$ in a basis of unit orthogonal vectors, and we want to decompose it in another basis $$\\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix} \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix}$$: $\\begin{split} \\alpha_1 \\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix} + \\alpha_2 \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\3 \\end{pmatrix} \\end{split}$ From where we can find that $$\\alpha_1 = -1$$, $$\\alpha_2 = -1$$: $\\begin{split} -1 \\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix} + -1 \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 0 \\end{pmatrix} + \\begin{pmatrix} 0 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix} \\end{split}$ So, vector $$\\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}$$ in the basis of unit orthogonal vectors is represented as $$\\begin{pmatrix} -1 \\\\ -1 \\end{pmatrix}$$ in the basis $$\\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix} \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix}$$. But, as mentioned before, not any set of vectors makes a basis. For example, the set of vectors $$\\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix}$$, $$\\begin{pmatrix} -3 \\\\ 0 \\end{pmatrix}$$ cannot decompose the vector $$\\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}$$, so this set of vectors is not a basis. What is the fundamental difference between these sets, and", "& 2.0\\end{pmatrix} meter$ We can add Vector objects like this In [9]: A + B Out[9]: $\\begin{pmatrix}4.0 & 6.0\\end{pmatrix} meter$ And subtract like this: In [10]: A - B Out[10]: $\\begin{pmatrix}2.0 & 2.0\\end{pmatrix} meter$ We can compute the Euclidean distance between two Vectors. In [11]: A.dist(B) Out[11]: 2.8284271247461903 meter And the difference in angle In [12]: A.diff_angle(B) Out[12]: -0.17985349979247822 radian If we are given the magnitude and angle of a vector, what we have is the representation of the vector in polar coordinates. In [13]: mag = A.mag angle = A.angle Out[13]: 0.9272952180016122 radian We can use pol2cart to convert from polar to Cartesian coordinates, and then use the Cartesian coordinates to make a Vector object. In this example, the Vector we get should have the same components as A. In [14]: x, y = pol2cart(angle, mag) Vector(x, y) Out[14]: $\\begin{pmatrix}3.0000000000000004 & 3.9999999999999996\\end{pmatrix} meter$ Another way to represent the direction of A is a unit vector, which is a vector with magnitude 1 that points in the same direction as A. You can compute a unit vector by dividing a vector by its magnitude: In [15]: A / A.mag Out[15]: $\\begin{pmatrix}0.6 & 0.8\\end{pmatrix} dimensionless$ Or by using the hat function, so named because unit vectors are conventionally decorated with a hat, like this: $\\hat{A}$: In [16]: A.hat() Out[16]:", "if we are talking about bound vector, then two bound vectors are equal if they have the same base point and end point. For example, the vector i + 2j + 3k with base point (1,0,0) and the vector i+2j+3k with base point (0,1,0) are different bound vectors, but the same (displacement) vector. ## Scalar multiplication For scalar multiplication, we simply multiply each component by the scalar. We commonly use Greek letters for scalars, and Roman letters for vectors. So for a scalar value of λ and a vector v defined by r and θ, the new vector is now λr and θ. Notice how the direction does not change. #### Example Say we have $\\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}$ and we wish to double the magnitude. So, $2 \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} 4 \\\\ 6 \\end{pmatrix}$. Simply, to add two vectors, you must add the respective x-components together to obtain the new x-component, and likewise add the two y-components together to obtain the new y-component. ### Example Say we have $\\mathbf{v_1}=\\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}, \\mathbf{v_2}=\\begin{pmatrix} 4 \\\\ 6 \\end{pmatrix}$ and we wish to add these. So, $\\mathbf{v_1}+\\mathbf{v_2}=\\begin{pmatrix} 6 \\\\ 9 \\end{pmatrix}$. ### Magnitude The magnitude of a vector is its length in R+ ## The dot product The dot product of two vectors", "How do you find a unit vector u that is orthogonal to a and b where a = 7 i - 5 j + k and b = -7 j - 5 k? Nov 9, 2016 $\\hat{c} = \\frac{1}{5 \\sqrt{158}} \\left(- 18 \\hat{i} + 35 \\hat{j} - 49 \\hat{k}\\right)$ Explanation: The cross product of the two given vectors will give you a vector that is orthogonal; to make it a unit vector, you merely divide it by its magnitude. barc = bara xx barb = | (hati, hatj, hatk, hati, hatj), (7,5,1,7,5), (0,-7,-5,0,-7) | = $\\overline{c} = \\hat{i} \\left\\{\\left(5\\right) \\left(- 5\\right) - \\left(1\\right) \\left(- 7\\right)\\right\\} + \\hat{j} \\left\\{\\left(1\\right) \\left(0\\right) - \\left(7\\right) \\left(- 5\\right)\\right\\} + \\hat{k} \\left\\{\\left(7\\right) \\left(- 7\\right) - \\left(5\\right) \\left(0\\right)\\right\\} =$ $\\overline{c} = - 18 \\hat{i} + 35 \\hat{j} - 49 \\hat{k}$ The above vector, $\\overline{c}$ is orthogonal to both $\\overline{a} \\mathmr{and} \\overline{b}$. To make $\\overline{c}$ a unit vector, compute its magnitude and then divide the vector by it: $| \\overline{c} | = \\sqrt{{\\left(- 18\\right)}^{2} + {35}^{2} + {\\left(- 49\\right)}^{2}}$ $| \\overline{c} | = \\sqrt{324 + 1225 + 2401}$ $| \\overline{c} | = \\sqrt{3950} = 5 \\sqrt{158}$ The unit vector is: $\\hat{c} = \\frac{1}{5 \\sqrt{158}} \\left(- 18 \\hat{i} + 35 \\hat{j} - 49 \\hat{k}\\right)$", "vector components and Pythagoras’ theorem to find the magnitude of the vector. The length of a vector is its absolute value and we use the modulus symbol. Here is the formula: \\lvert \\textbf{a} \\rvert = \\sqrt{x^2+y^2} E.g. \\lvert \\textbf{a} \\rvert = \\sqrt{x^2+y^2}=\\sqrt{5^2+2^2}=\\sqrt{29} If the magnitude is equal to 1, then the vector is known as a unit vector. If the magnitude is equal to 0 , then the vector is known as a zero vector. Step-by-step guide: Magnitude of a vector ## Vector arithmetic Here is vector \\textbf{a} and vector \\textbf{b} . \\textbf{a}= \\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} \\textbf{b}= \\begin{pmatrix} \\; 3 \\;\\\\ \\; -1 \\; \\end{pmatrix} \\textbf{a}+\\textbf{b}= \\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} + \\begin{pmatrix} \\; 3 \\;\\\\ \\; -1 \\; \\end{pmatrix} = \\begin{pmatrix} \\; 8 \\;\\\\ \\; 1 \\; \\end{pmatrix} The vector \\textbf{a} + \\textbf{b} is known as the resultant vector. ### 5. Vector subtraction We can subtract vectors When subtracting vectors, the order is important. \\textbf{a}-\\textbf{b}= \\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} - \\begin{pmatrix} \\; 3 \\;\\\\ \\; -1 \\; \\end{pmatrix} = \\begin{pmatrix} \\; 2 \\;\\\\ \\; 3 \\; \\end{pmatrix} Step-by-step guide: Vector subtraction ### 6. Vector multiplication We can multiply vectors by a scalar E.g. \\textbf{a}= \\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} 2\\textbf{a}= 2\\begin{pmatrix} \\;", "treat the top numbers and bottom numbers completely separately. \\begin{aligned} 2\\begin{pmatrix} 3\\\\\\text{-}1 \\end{pmatrix} + \\begin{pmatrix} \\text{-}7\\\\\\text{-}1 \\end{pmatrix} &= \\begin{pmatrix} 6\\\\\\text{-}2 \\end{pmatrix} + \\begin{pmatrix} \\text{-}7\\\\\\text{-}1 \\end{pmatrix} \\\\ &=\\begin{pmatrix} 6-7\\\\\\text{-}2-1 \\end{pmatrix} \\\\ &= \\boldsymbol{\\begin{pmatrix} \\text{-}1\\\\\\text{-}3 \\end{pmatrix}} \\end{aligned} This method is quite abstract – you might find it easier to draw the vectors out. We’ll start by drawing the vectors on a grid. The first vector is 3 units to the right and 1 unit down: We need two of the first vector – one after the other. Then, we need the second vector. This will be 7 units to the left and 1 unit down. The resultant vector (the answer to our addition) is the vector from our starting point to the finish point. It is drawn below in colour. This is the vector: $\\boldsymbol{\\begin{pmatrix} \\text{-}1\\\\\\text{-}3 \\end{pmatrix}}$ ### Question 4: The Area of a Triangle The area of a triangle can be found my multiplying the base by the height and dividing by 2. However, this requires us to know the perpendicular height of the triangle. If we don’t know this measurement, we can use a slightly different formula to find the area. Consider the triangle below: Any triangle can be labelled like this. The lowercase letters denote the sides and the uppercase letters denote the angles. Each angle is", "\\end{pmatrix}$$ or, 4 - a = -2 or, 5 - b = 0 or, 4 + 2 = a $$\\therefore$$ b = 5 $$\\therefore$$ a = 6 0% • Find column vector (overrightarrow{MN}) of the M(2, 5) and N(4, 8) (\begin{pmatrix} 2\\ 2end{pmatrix}) (\begin{pmatrix} 3\\ 3end{pmatrix}) (\begin{pmatrix} 2\\ 3end{pmatrix}) (\begin{pmatrix} 3\\ 3end{pmatrix}) • Find the magnitute (overrightarrow{XY}) = (\begin{pmatrix} 0 \\ 5 end{pmatrix}) (sqrt{50}) units (sqrt{52}) units (sqrt{5}) units (sqrt{25}) units 2 units. 4 units. 5 units. 6 units. 4 units 5 units 6 units 8 units 12 units 6 units 25 units 4 units • Find the magnitude of: (overrightarrow{AB}) = (\begin{pmatrix} 5\\3end{pmatrix}) (sqrt{25}) units (sqrt{34}) units (sqrt{12}) units 12(sqrt{34}) units • Find the magnitude of: (overrightarrow{XY}) = (\begin{pmatrix} -4\\-6end{pmatrix}) 2(sqrt[3]{52}) units (sqrt[2]{52}) units 12(sqrt{52}) units (sqrt{52}) units • Find the magnitude of: (overrightarrow{MN}) = (\begin{pmatrix} -4\\5end{pmatrix}) (sqrt{12}) units (sqrt{41}) units 15(sqrt{41}) units (sqrt[2]{6}) units • Find column vector (overrightarrow{MN}) if, M(2, 5) and N(4, 8) (\begin{pmatrix} 2\\3end{pmatrix}) (\begin{pmatrix} 2\\2end{pmatrix}) (\begin{pmatrix} 3\\2end{pmatrix}) (\begin{pmatrix} 3\\3end{pmatrix}) • Find column vector (overrightarrow{MN}) if, M(-3, 4) and N(4, -6) (\begin{pmatrix} 12\\3end{pmatrix}) (\begin{pmatrix} -2\\13end{pmatrix}) (\begin{pmatrix} 7\\-10end{pmatrix}) (\begin{pmatrix} 2\\3end{pmatrix}) • If (overrightarrow{a}) = (\begin{pmatrix} 4\\6end{pmatrix}) and (overrightarrow{b}) = (\begin{pmatrix}3\\2end{pmatrix}), find (overrightarrow{a})+ (overrightarrow{b}). (\begin{pmatrix} 4\\6end{pmatrix}) (\begin{pmatrix} 7\\8end{pmatrix}) (\begin{pmatrix} 1\\4end{pmatrix}) (\begin{pmatrix} 12\\12end{pmatrix}) • If (overrightarrow{a}) = (\begin{pmatrix} 8\\10end{pmatrix}) and (overrightarrow{b}) = (\begin{pmatrix}-2\\-3end{pmatrix}), find (overrightarrow{a})-(overrightarrow{b}) (\begin{pmatrix}", "Building blocks Hit the spot Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Problem Here are some vectors: • $\\mathbf{a}=\\begin{pmatrix}6\\\\1\\end{pmatrix}$ • $\\mathbf{b}$ has magnitude $2\\sqrt{2}$ and direction $45^\\circ$ anti-clockwise from the $x$-axis • $\\mathbf{c}$ has magnitude $4$ in the positive $x$ direction • $\\mathbf{d}=3\\mathbf{i}-\\mathbf{j}$ • $\\mathbf{e}=7\\mathbf{i}-3\\mathbf{j}$ • $\\mathbf{f}$ is $2$ units in the negative $x$ direction and $3$ units in the positive $y$ direction • $\\mathbf{g}=\\begin{pmatrix}-2\\\\4\\end{pmatrix}$ Select a starting point, $M(2,5)$ or $N(3,1)$, and two of the vectors from the list so that the sum of the vectors will get you from your start to the target, $T(7,6)$. How many different ways can you pair up vectors from the list to get to the target? Can you add three of the vectors to get from one of the starts to the target?" ]

[ "Orthogonal Vectors Two vectors, x and y, are orthogonal if their dot product is zero. For example $e \\cdot f = \\begin{pmatrix} 2 & 5 & 4 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 5 \\end{pmatrix} = 2*4 + (5)*(-2) + 4*5 = 8-10+20 = 18$ Vectors e and f are not orthogonal. $g \\cdot h = \\begin{pmatrix} 2 & 3 & -2 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 1 \\end{pmatrix} = 2*4 + (3)*(-2) + (-2)*1 = 8-6-2 = 0$ However, vectors g and h are orthogonal. Orthogonal can be thought of as an expansion of perpendicular for higher dimensions. Let $x_1, x_2, \\ldots , x_n$ be m-dimensional vectors. Then a linear combination of $x_1, x_2, \\ldots , x_n$ is any m-dimensional vector that can be expressed as $c_1 x_1 + c_2 x_2 + \\ldots + c_n x_n$ where $c_1, \\ldots, c_n$ are all scalars. For example: $x_1 =\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix}$ $y =\\begin{pmatrix} -5 \\\\ 12 \\\\ -8 \\end{pmatrix} = 1*\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix} + (-2)* \\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix} = 1*x_1 + (-2)*x_2$ So y is a linear combination of $x_1$ and $x_2$. The set of all linear combinations of $x_1, x_2, \\ldots , x_n$ is called", "Question # Find a unit vector that is orthogonal to both i+j and i+k. Find a unit vector that is orthogonal to both $$i+j$$ and $$i+k$$. 2021-05-13 Step 1 Let $$a=i+j,\\ b=i+k$$ The cross product $$a\\times b$$ is arthogonal to both a and b $$a\\times b=\\left|\\begin{matrix}i & j & k \\\\ 1 & 1 & 0 \\\\ 1 & 0 & 1 \\end{matrix}\\right|$$ $$=i(1-0)-j(1-0)+k(0-1)$$ $$a\\times b=i-j-k$$ Since the vector we found had magnitude $$\\sqrt{3}$$ $$i.e.\\ |a\\times b|=\\sqrt{\\check{1}+\\check{1}+\\check{1}}=\\sqrt{3}$$ A unit vector that is arthogonal to both $$i+j$$ and $$i+k$$ is $$u=\\frac{(i-j-k)}{\\sqrt{3}}$$ or $$u=\\frac{\\sqrt{3}(i-j-k)}{3}$$", "# Magnitude of a Vector - How to Calculate It ## (also called Modulus of a Vector) Every vector has a magnitude, also called modulus. In this section we learn how to calculate the magnitude of a vector. ### Magnitude (Modulus) of a Vector Given a vector $$\\vec{v} = \\begin{pmatrix}x \\\\ y \\end{pmatrix}$$, its magnitude, also called modulus can be calculated with the formula: $\\begin{vmatrix}\\vec{v} \\end{vmatrix} = \\sqrt{x^2 + y^2}$ ### Tutorial: Magnitude of a Vector In this tutorial we learn how to calculate the magnitude of a vector. ### Example 1 Given the two vector $$\\vec{a} = \\begin{pmatrix}1 \\\\ 3 \\end{pmatrix}$$ we can calculate its magnitude as follows: \\begin{aligned} \\begin{vmatrix} \\vec{a} \\end{vmatrix} & = \\sqrt{1^2 + 3^2} \\\\ & = \\sqrt{1 + 9} \\\\ \\begin{vmatrix} \\vec{a} \\end{vmatrix} & = \\sqrt{10} \\quad \\begin{pmatrix} \\approx 3.16 \\end{pmatrix} \\end{aligned} ## Exercise 1 Find and write the exact value of the magnitude for each of the vectors, shown below. When applicable: also use your calculator to round your answers to two decimal places (2 d.p). 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$. 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$. 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$. 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$. 1. For vector $$\\vec{e} = \\begin{pmatrix}", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "are talking about bound vector, then two bound vectors are equal if they have the same base point and end point. For example, the vector i + 2j + 3k with base point (1,0,0) and the vector i+2j+3k with base point (0,1,0) are different bound vectors, but the same (displacement) vector. Scalar multiplication For scalar multiplication, we simply multiply each component by the scalar. We commonly use Greek letters for scalars, and Roman letters for vectors. So for a scalar value of λ and a vector v defined by r and θ, the new vector is now λr and θ. Notice how the direction does not change. Example Say we have ${\\displaystyle {\\begin{pmatrix}2\\\\3\\end{pmatrix}}}$ and we wish to double the magnitude. So, ${\\displaystyle 2{\\begin{pmatrix}2\\\\3\\end{pmatrix}}={\\begin{pmatrix}4\\\\6\\end{pmatrix}}}$ . Simply, to add two vectors, you must add the respective x-components together to obtain the new x-component, and likewise add the two y-components together to obtain the new y-component. Example Say we have ${\\displaystyle \\mathbf {v_{1}} ={\\begin{pmatrix}2\\\\3\\end{pmatrix}},\\mathbf {v_{2}} ={\\begin{pmatrix}4\\\\6\\end{pmatrix}}}$ and we wish to add these. So, ${\\displaystyle \\mathbf {v_{1}} +\\mathbf {v_{2}} ={\\begin{pmatrix}6\\\\9\\end{pmatrix}}}$ . Magnitude The magnitude of a vector is its length in R+ The dot product The dot product of two vectors is defined as the sum of the products of the components. Symbolically we write ${\\displaystyle {\\begin{pmatrix}a_{1}\\\\a_{2}\\end{pmatrix}}\\cdot {\\begin{pmatrix}b_{1}\\\\b_{2}\\end{pmatrix}}=a_{1}b_{1}+a_{2}b_{2}}$ For example, ${\\displaystyle {\\begin{pmatrix}3\\\\5\\end{pmatrix}}\\cdot", "or calculator to find the matrix inverse. Orthogonal Vectors Two vectors, x and y, are orthogonal if their dot product is zero. For example $e \\cdot f = \\begin{pmatrix} 2 & 5 & 4 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 5 \\end{pmatrix} = 2*4 + (5)*(-2) + 4*5 = 8-10+20 = 18$ Vectors e and f are not orthogonal. $g \\cdot h = \\begin{pmatrix} 2 & 3 & -2 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 1 \\end{pmatrix} = 2*4 + (3)*(-2) + (-2)*1 = 8-6-2 = 0$ However, vectors g and h are orthogonal. Orthogonal can be thought of as an expansion of perpendicular for higher dimensions. Let $$x_1, x_2, \\ldots , x_n$$ be m-dimensional vectors. Then a linear combination of $$x_1, x_2, \\ldots , x_n$$ is any m-dimensional vector that can be expressed as $c_1 x_1 + c_2 x_2 + \\ldots + c_n x_n$ where $$c_1, \\ldots, c_n$$ are all scalars. For example: $x_1 =\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix}$ $y =\\begin{pmatrix} -5 \\\\ 12 \\\\ -8 \\end{pmatrix} = 1*\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix} + (-2)* \\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix} = 1*x_1 + (-2)*x_2$ So y is a linear combination of $$x_1$$ and $$x_2$$. The set of all linear combinations of", "## VECTORS In three dimensions, the base vectors are $$i=\\begin{pmatrix} 1\\\\0 \\\\0 \\end{pmatrix}$$, $$j=\\begin{pmatrix} 0\\\\1 \\\\0 \\end{pmatrix}$$, $$k=\\begin{pmatrix} 0\\\\0 \\\\1 \\end{pmatrix}$$ that are along the $$x, y$$ and $$z$$ coordinate directions, respectively, as shown in the figure. For example, $$v=\\begin{pmatrix} -4\\\\3 \\\\2 \\end{pmatrix}$$ can be written as $$-4i+3j+2k$$. The magnitude of $$v$$: $$|v|=\\sqrt{((-4)^2+3^2+2^2)}=\\sqrt{29}$$ Given that if $$v=xi+yj+zk$$ then $$|v|=\\sqrt{x^2+y^2+z^2}$$ ## UNIT VECTORS If $$\\overrightarrow{AB}$$ is a non-unit vector, then a unit vector in the direction of $$\\overrightarrow{AB}$$ is $${1\\over |\\overrightarrow{AB}|}\\overrightarrow{AB}$$ LESSON 1 Determine the unit vector in the direction of the vector $$\\overrightarrow{OA}=\\begin{pmatrix} -4\\\\3 \\\\2 \\end{pmatrix}$$. SOLUTION • Determine the length of the vector $$|\\overrightarrow{OA}|=\\sqrt{(-4)^2+(3)^2+(2)^2}=\\sqrt{29}$$ • Use the form for a unit vector Unit Vector $${1\\over \\sqrt{29}} \\begin{pmatrix} -4\\\\3 \\\\2 \\end{pmatrix}$$ TRY THIS Complete Question 1 from worksheet.", "preferred). To start off the derivation, assume two vectors a and b exist, and so does another vector n, that is orthogonal to both a and b: $\\large \\mathbf{a}=\\begin{pmatrix} a_1 \\\\ a_2\\\\ a_3 \\end{pmatrix} \\; \\; \\mathbf{b}=\\begin{pmatrix} b_1 \\\\ b_2\\\\ b_3 \\end{pmatrix} \\; \\mathbf{n}=\\begin{pmatrix} n_1 \\\\ n_2\\\\ n_3 \\end{pmatrix}$ Our aim is to find the components of n, in terms of the components of a and b. Since n is orthogonal to both a and b, these conditions must apply: \\large \\begin{align*} &\\mathbf{a}\\cdot\\mathbf{n}=0 \\\\ &\\mathbf{b}\\cdot\\mathbf{n}=0 \\\\ \\\\&(1)\\: a_1n_1+a_2n_2+a_3n_3=0\\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\\\ &(2)\\: \\, b_1n_1+b_2n_2+b_3n_3=0\\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\end{align*} Another independent equation of the component of n is necessary to solve this simultaneous equation with three unknowns of n1, n2 and n3. Notice that we haven't specified the magnitude of n. For simplicity, let the magnitude of n be 1 or unity. \\large \\begin{align*} &\\left \\| \\mathbf{n} \\right \\|=1\\: \\: \\: \\: \\: \\: \\: \\: \\: \\: \\: \\\\\\\\&\\sqrt{n_1^2+n_2^2+n_3^2}=1\\\\ \\\\\\therefore \\; &(3)\\, \\, \\, n_1^2+n_2^2+n_3^2=1 \\end{align*} \\large \\begin{align*} &(1)\\: a_1n_1+a_2n_2+a_3n_3=0\\\\ &(2)\\: \\, b_1n_1+b_2n_2+b_3n_3=0\\\\ &(3)\\, \\, \\, n_1^2+n_2^2+n_3^2=1 \\end{align*} This now gives us 3", "$\\mathbb{R_3}$ Given: • A line $l$ in vector form • A point $P$ on the plane 1. We can find the point $X$ where the perpendicular line from $l$ to $P$ intersects $l$ 2. Using $X$, we can then find $\\vec{PX}$, the vector from $P$ to $X$ 3. We know that the direction of $l \\dot{} \\vec{PX} = 0$, so we can solve for λ 4. Once we have λ, we can substitute this back into $\\vec{PX}$ and find the magnitude, giving the shortest distance ### The Cross Product #### Arithmetic Properties Of The Cross Product The cross product gives us a vector that is perpendicular (orthogonal) to both vectors $\\vec{AB}$ and $\\vec{AC}$. When using the cross product, think of the ‘right hand rule’. For each row of the two vectors, cross subtract the multiplication of two rows below crossed: $$\\begin{pmatrix}x_1\\\\x_2\\\\x_3\\end{pmatrix} \\times \\begin{pmatrix}y_1\\\\y_2\\\\y_3\\end{pmatrix} = \\begin{pmatrix}(x_2 \\times y_3) - (y_2 \\times x_3)\\\\(x_3 \\times y_1) - (y_3 \\times x_1)\\\\(x_1 \\times y_2) - (y_1 \\times x_2)\\end{pmatrix}$$ $e.g. \\begin{pmatrix}2\\\\2\\\\-2\\end{pmatrix} \\times \\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}4\\\\-6\\\\-2\\end{pmatrix}$ #### Geometric Interpretation Of The Cross Product As stated above, the cross product gives us a vector that is perpendicular (orthogonal) to two vectors. Useful for: • Finding a vector perpendicular to two other vectors • Finding the area of a parallelogram by taking the magnitude of the cross", "any other vector is defined to be a right angle). Thus vectors from $\\mathbb{R}^n$ are orthogonal (or perpendicular) if and only if their dot product is zero. Example 2.8 These vectors are orthogonal. $\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}\\cdot\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}=0$ The arrows are shown away from canonical position but nevertheless the vectors are orthogonal. Example 2.9 The $\\mathbb{R}^3$ angle formula given at the start of this subsection is a special case of the definition. Between these two the angle is $\\arccos(\\frac{(1)(0)+(1)(3)+(0)(2)}{\\sqrt{1^2+1^2+0^2}\\sqrt{0^2+3^2+2^2}}) =\\arccos(\\frac{3}{\\sqrt{2}\\sqrt{13}})$ approximately $0.94 \\text{radians}$. Notice that these vectors are not orthogonal. Although the $yz$-plane may appear to be perpendicular to the $xy$-plane, in fact the two planes are that way only in the weak sense that there are vectors in each orthogonal to all vectors in the other. Not every vector in each is orthogonal to all vectors in the other. ## Exercises This exercise is recommended for all readers. Problem 1 Find the length of each vector. 1. $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ 2. $\\begin{pmatrix} -1 \\\\ 2 \\end{pmatrix}$ 3. $\\begin{pmatrix} 4 \\\\ 1 \\\\ 1 \\end{pmatrix}$ 4. $\\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$ 5. $\\begin{pmatrix} 1 \\\\ -1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 2 Find the angle between each two, if it is defined.", "rule, but I'll be explaining a possible reason later on as to why the left-hand rule is preferred). To start off the derivation, assume two vectors a and b exist, and so does another vector n, that is orthogonal to both a and b: $\\large \\mathbf{a}=\\begin{pmatrix} a_1 \\\\ a_2\\\\ a_3 \\end{pmatrix} \\; \\; \\mathbf{b}=\\begin{pmatrix} b_1 \\\\ b_2\\\\ b_3 \\end{pmatrix} \\; \\mathbf{n}=\\begin{pmatrix} n_1 \\\\ n_2\\\\ n_3 \\end{pmatrix}$ Our aim is to find the components of n, in terms of the components of a and b. Since n is orthogonal to both a and b, these conditions must apply: \\large \\begin{align*} &\\mathbf{a}\\cdot\\mathbf{n}=0 \\\\ &\\mathbf{b}\\cdot\\mathbf{n}=0 \\\\ \\\\&(1)\\: a_1n_1+a_2n_2+a_3n_3=0\\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\\\ &(2)\\: \\, b_1n_1+b_2n_2+b_3n_3=0\\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\! \\end{align*} Another independent equation of the component of n is necessary to solve this simultaneous equation with three unknowns of n1, n2 and n3. Notice that we haven't specified the magnitude of n. For simplicity, let the magnitude of n be 1 or unity. \\large \\begin{align*} &\\left \\| \\mathbf{n} \\right \\|=1\\: \\: \\: \\: \\: \\: \\: \\: \\: \\: \\: \\\\\\\\&\\sqrt{n_1^2+n_2^2+n_3^2}=1\\\\ \\\\\\therefore \\; &(3)\\, \\, \\, n_1^2+n_2^2+n_3^2=1 \\end{align*}", "+0 # The vector $\\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$ is orthogonal to the vector $\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$. The vector \\$\\begin{pm 0 52 1 The vector $$\\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ is orthogonal to the vector $$\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$$. The vector $$\\begin{pmatrix} 2 \\\\ m \\end{pmatrix}$$ is orthogonal to the vector $$\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$$. Find m . Feb 2, 2019 #1 +96080 +1 If vectors are orthagonal, their dot product = 0 So [ 4 * 3 ] + [ -7 * k] = 0 12 - 7k = 0 -7k = -12 k = 12/7 So [2 * 3 ] + [12/7 * m] = 0 6 + (12/7)m = 0 (12/7)m = - 6 m = -6*7/12 = -7/2 Feb 2, 2019", "one can take any two vectors (almost any) and they will also form a basis, meaning that any vector can be decomposed by their linear combination. Example. Suppose we have a vector $$\\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}$$ in a basis of unit orthogonal vectors, and we want to decompose it in another basis $$\\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix} \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix}$$: $\\begin{split} \\alpha_1 \\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix} + \\alpha_2 \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\3 \\end{pmatrix} \\end{split}$ From where we can find that $$\\alpha_1 = -1$$, $$\\alpha_2 = -1$$: $\\begin{split} -1 \\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix} + -1 \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 0 \\end{pmatrix} + \\begin{pmatrix} 0 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix} \\end{split}$ So, vector $$\\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}$$ in the basis of unit orthogonal vectors is represented as $$\\begin{pmatrix} -1 \\\\ -1 \\end{pmatrix}$$ in the basis $$\\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix} \\begin{pmatrix} 0 \\\\ -3 \\end{pmatrix}$$. But, as mentioned before, not any set of vectors makes a basis. For example, the set of vectors $$\\begin{pmatrix} -2 \\\\ 0 \\end{pmatrix}$$, $$\\begin{pmatrix} -3 \\\\ 0 \\end{pmatrix}$$ cannot decompose the vector $$\\begin{pmatrix} 2 \\\\ 3 \\end{pmatrix}$$, so this set of vectors is not a basis. What is the fundamental difference between these sets, and", "treat the top numbers and bottom numbers completely separately. \\begin{aligned} 2\\begin{pmatrix} 3\\\\\\text{-}1 \\end{pmatrix} + \\begin{pmatrix} \\text{-}7\\\\\\text{-}1 \\end{pmatrix} &= \\begin{pmatrix} 6\\\\\\text{-}2 \\end{pmatrix} + \\begin{pmatrix} \\text{-}7\\\\\\text{-}1 \\end{pmatrix} \\\\ &=\\begin{pmatrix} 6-7\\\\\\text{-}2-1 \\end{pmatrix} \\\\ &= \\boldsymbol{\\begin{pmatrix} \\text{-}1\\\\\\text{-}3 \\end{pmatrix}} \\end{aligned} This method is quite abstract – you might find it easier to draw the vectors out. We’ll start by drawing the vectors on a grid. The first vector is 3 units to the right and 1 unit down: We need two of the first vector – one after the other. Then, we need the second vector. This will be 7 units to the left and 1 unit down. The resultant vector (the answer to our addition) is the vector from our starting point to the finish point. It is drawn below in colour. This is the vector: $\\boldsymbol{\\begin{pmatrix} \\text{-}1\\\\\\text{-}3 \\end{pmatrix}}$ ### Question 4: The Area of a Triangle The area of a triangle can be found my multiplying the base by the height and dividing by 2. However, this requires us to know the perpendicular height of the triangle. If we don’t know this measurement, we can use a slightly different formula to find the area. Consider the triangle below: Any triangle can be labelled like this. The lowercase letters denote the sides and the uppercase letters denote the angles. Each angle is", "& 2.0\\end{pmatrix} meter$ We can add Vector objects like this In [9]: A + B Out[9]: $\\begin{pmatrix}4.0 & 6.0\\end{pmatrix} meter$ And subtract like this: In [10]: A - B Out[10]: $\\begin{pmatrix}2.0 & 2.0\\end{pmatrix} meter$ We can compute the Euclidean distance between two Vectors. In [11]: A.dist(B) Out[11]: 2.8284271247461903 meter And the difference in angle In [12]: A.diff_angle(B) Out[12]: -0.17985349979247822 radian If we are given the magnitude and angle of a vector, what we have is the representation of the vector in polar coordinates. In [13]: mag = A.mag angle = A.angle Out[13]: 0.9272952180016122 radian We can use pol2cart to convert from polar to Cartesian coordinates, and then use the Cartesian coordinates to make a Vector object. In this example, the Vector we get should have the same components as A. In [14]: x, y = pol2cart(angle, mag) Vector(x, y) Out[14]: $\\begin{pmatrix}3.0000000000000004 & 3.9999999999999996\\end{pmatrix} meter$ Another way to represent the direction of A is a unit vector, which is a vector with magnitude 1 that points in the same direction as A. You can compute a unit vector by dividing a vector by its magnitude: In [15]: A / A.mag Out[15]: $\\begin{pmatrix}0.6 & 0.8\\end{pmatrix} dimensionless$ Or by using the hat function, so named because unit vectors are conventionally decorated with a hat, like this: $\\hat{A}$: In [16]: A.hat() Out[16]:", "vector components and Pythagoras’ theorem to find the magnitude of the vector. The length of a vector is its absolute value and we use the modulus symbol. Here is the formula: \\lvert \\textbf{a} \\rvert = \\sqrt{x^2+y^2} E.g. \\lvert \\textbf{a} \\rvert = \\sqrt{x^2+y^2}=\\sqrt{5^2+2^2}=\\sqrt{29} If the magnitude is equal to 1, then the vector is known as a unit vector. If the magnitude is equal to 0 , then the vector is known as a zero vector. Step-by-step guide: Magnitude of a vector ## Vector arithmetic Here is vector \\textbf{a} and vector \\textbf{b} . \\textbf{a}= \\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} \\textbf{b}= \\begin{pmatrix} \\; 3 \\;\\\\ \\; -1 \\; \\end{pmatrix} \\textbf{a}+\\textbf{b}= \\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} + \\begin{pmatrix} \\; 3 \\;\\\\ \\; -1 \\; \\end{pmatrix} = \\begin{pmatrix} \\; 8 \\;\\\\ \\; 1 \\; \\end{pmatrix} The vector \\textbf{a} + \\textbf{b} is known as the resultant vector. ### 5. Vector subtraction We can subtract vectors When subtracting vectors, the order is important. \\textbf{a}-\\textbf{b}= \\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} - \\begin{pmatrix} \\; 3 \\;\\\\ \\; -1 \\; \\end{pmatrix} = \\begin{pmatrix} \\; 2 \\;\\\\ \\; 3 \\; \\end{pmatrix} Step-by-step guide: Vector subtraction ### 6. Vector multiplication We can multiply vectors by a scalar E.g. \\textbf{a}= \\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} 2\\textbf{a}= 2\\begin{pmatrix} \\;", "# Orthogonal projection onto span of vectors using weighted inner product • Jul 14th 2008, 07:11 PM JCS007 Orthogonal projection onto span of vectors using weighted inner product Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ • Jul 14th 2008, 09:43 PM CaptainBlack Quote: Originally Posted by JCS007 Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ Let: $b_1=\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and: $b_2=\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ and: $ u_1=\\frac{b_1}{||b_1||} $ $ u_2=\\frac{b_2-\\langle b_2,u_1 \\rangle u_1}{||b_2-\\langle b_2,u_1 \\rangle u_1||} $ Then the orthogonal projection is: $p=\\langle v,u_1\\rangle u_1 + \\langle v,u_2 \\rangle u_2$ Where $||x||=\\langle x,x \\rangle^{\\frac{1}{2}}$ RonL", "amount. We refer to doing this as scalar multiplication: we multiply the vector by a scalar real number. === Scalar multiplication === For scalar multiplication, we simply multiply each component by the scalar. We commonly use Greek letters for scalars, and English letters for vectors. So for a scalar value of λ and a vector v defined by r and θ, the new vector is now λr and θ. Notice how the direction does not change. ==== Example ==== Say we have ( 2 3 ) {\\displaystyle {\\begin{pmatrix}2\\\\3\\end{pmatrix}}} and we wish to double the magnitude. So, 2 ( 2 3 ) = ( 4 6 ) {\\displaystyle 2{\\begin{pmatrix}2\\\\3\\end{pmatrix}}={\\begin{pmatrix}4\\\\6\\end{pmatrix}}} . Simply, to add two vectors, you must add the respective x-components together to obtain the new x-component, and likewise add the two y-components together to obtain the new y-component. === Example === Say we have v 1 = ( 2 3 ) , v 2 = ( 4 6 ) {\\displaystyle \\mathbf {v_{1}} ={\\begin{pmatrix}2\\\\3\\end{pmatrix}},\\mathbf {v_{2}} ={\\begin{pmatrix}4\\\\6\\end{pmatrix}}} and we wish to add these. So, v 1 + v 2 = ( 6 9 ) {\\displaystyle \\mathbf {v_{1}} +\\mathbf {v_{2}} ={\\begin{pmatrix}6\\\\9\\end{pmatrix}}} . == Subtraction of vectors == The operation of subtraction on two vectors, a and b, a-b, can also be written as a+(-1)b. Therefore, we can use scalar multiplication to", "of the vector: Multiplication: ${\\displaystyle c{\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}={\\begin{pmatrix}cx_{1}\\\\cy_{1}\\\\cz_{1}\\end{pmatrix}}}$ Division: ${\\displaystyle {\\frac {\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}{c}}={\\frac {1}{c}}{\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}={\\begin{pmatrix}{\\frac {x_{1}}{c}}\\\\{\\frac {y_{1}}{c}}\\\\{\\frac {z_{1}}{c}}\\end{pmatrix}}}$ What about when there is a vector multiplied by another vector? There are two ways that are defined for vector multiplication: vector dot product and vector cross product. ### Unit length vector The unit vector in 2D is based around this, meaning that x and y can't both be 1, else this wouldn't make a circle, it would make a square and the magnitude would be greater than 1 (a unit). Unit length vectors are used when the magnitude isn't important, only the direction is. In 2D it's based around the unit circle and in 3D around the unit sphere; the values for the x, y and z will be between -1 and 1. The notation for this is: ${\\displaystyle {\\hat {\\textbf {V}}}}$ To turn a vector into a unit length vector, use the following equation: ${\\displaystyle {\\hat {\\textbf {V}}}={\\frac {\\textbf {\\overrightarrow {V}}}{{\\Big |}{\\textbf {\\overrightarrow {V}}}{\\Big |}}}}$ , this is known as normalising the vector. ### Null vector A null vector (sometimes called a zero vector) is a vector where the contents are all 0s. If the null vector was denoted by ${\\displaystyle {\\overrightarrow {V}}}$ ; ... and it was in 2D, it would look like: ${\\displaystyle {\\overrightarrow {V}}={\\begin{pmatrix}0\\\\0\\end{pmatrix}}}$ ... and it was in", "to an arbitrary vector $(x,y,z)$, and see if the resulting vector is orthogonal to the normal vector of that given plane, thus implying that the vector is projected successfully to the plane? This is necessary but not sufficient. If $n$ is normal to the plane and $\\langle n, Tx\\rangle = 0$ for all $x$, then the image of $T$ is contained in the plane, but that doesn't necessarily mean that $T$ is a projection onto the plane. For example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$ This matrix has the property that $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. So the image lies in the plane whose normal vector is $\\begin{pmatrix}1 \\\\ 0 \\\\ 0\\end{pmatrix}$. But it is not a projection onto that plane because the image only has dimension 1. For another example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 2 \\end{pmatrix}$$ Once again, we have $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. The dimension of the image is correct (2), but this" ]

[ "+ c + d | $|\\mathbf{a}| \\leq|\\mathbf{a}|+|\\mathbf{b}|+|\\mathbf{c}| \\ldots$ (i) Equation $(i)$ shows that the magnitude of $\\mathbf{a}$ is equal to or less than the sum of the magnitudes of $\\mathbf{b}, \\mathbf{c}$, and $\\mathbf{d}$. Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of b, c, and d.", "+0 # Let $\\mathbf{v}$ and $\\mathbf{w}$ be vectors such that \\[\\operatorname{proj}_{\\bold{w}}( \\bold{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.\\ 0 113 1 Let v and w be vectors such that $$\\operatorname{proj}_{{w}}({v})= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.$$ Find $$\\operatorname{proj}_{-2{w}} (3{v})$$. Oct 9, 2019", "+0 # help 0 91 1 Given vectors $$\\mathbf{v}$$ and $$\\mathbf{w}$$ such that $$\\|\\mathbf{v}\\| = 3,\\|\\mathbf{w}\\| = 7,$$ and $$\\mathbf{v} \\cdot \\mathbf{w} = 10,$$ then find $$\\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\|.$$ Sep 3, 2019 #1 +23357 +1 Given vectors $$\\mathbf{v}$$ and $$\\mathbf{w}$$ such that $$\\|\\mathbf{v}\\| = 3,\\ \\|\\mathbf{w}\\| = 7,$$ and $$\\mathbf{v} \\cdot \\mathbf{w} = 10,$$ then find $$\\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\|.$$ $$\\begin{array}{|rcll|} \\hline \\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\| &=& \\left\\| \\dfrac{\\mathbf{v} \\cdot \\mathbf{w}}{\\left\\|\\mathbf{w}\\right\\|^2}\\mathbf{w} \\right\\| \\\\\\\\ &=& \\left\\| \\dfrac{10}{7^2}\\mathbf{w} \\right\\| \\\\\\\\ &=& \\dfrac{10}{7^2} \\| \\mathbf{w} \\| \\\\\\\\ &=& \\dfrac{10}{7^2} \\cdot 7 \\\\\\\\ \\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\| &=& \\mathbf{\\dfrac{10}{7}} \\\\ \\hline \\end{array}$$ Sep 4, 2019", "Adding Vectors by Components Video Lessons Concept # Problem: A vector of magnitude 20 is added to a vector of magnitude 25. The magnitude of this sum might be:A. zeroB. 3C. 12D. 47E. 50 ###### FREE Expert Solution The magnitude of the sum of two vectors is expressed as: $\\overline{){\\mathbf{R}}{\\mathbf{=}}\\mathbf{|}\\stackrel{\\mathbf{⇀}}{\\mathbf{A}}\\mathbf{|}{\\mathbf{+}}\\mathbf{|}\\stackrel{\\mathbf{⇀}}{\\mathbf{B}}\\mathbf{|}}$ 94% (233 ratings) ###### Problem Details A vector of magnitude 20 is added to a vector of magnitude 25. The magnitude of this sum might be: A. zero B. 3 C. 12 D. 47 E. 50", "# Category:Magnitude Magnitude of a vector $|\\overrightarrow{u}| = |\\begin{bmatrix} u_1 & u_2 & ... & u_n \\end{bmatrix}| = \\sqrt{\\overrightarrow{u} \\cdot \\overrightarrow{u}} = \\sqrt{ v_1 \\cdot v_1 + v_2 \\cdot v_2 + ... + v_n \\cdot v_n }$ ## Pages in category \"Magnitude\" The following 2 pages are in this category, out of 2 total.", "# Vector Magnitude Calculator An online calculator for finding the magnitude (length) of a vector, with steps shown. Comma-separated. If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below. Find the magnitude (length) of $\\mathbf{\\vec{v}} = \\left(5, -2, 3\\right)$.", "$\\mathbb{u}$ onto $\\mathbb{v}$ is the vector which has magnitude given by the scalar projection $\\text{comp}_{\\mathbb{u}}(\\mathbb{v})$ and points in in the direction of $\\mathbb{v}$, which is given by the unit vector $\\frac{\\mathbb{v}}{|\\mathbb{v}|}$ associated to $\\mathbb{v}$. Hence we have $$\\text{proj}_{\\mathbf{v}}(\\mathbf{u}) = \\text{comp}_{\\mathbb{v}}(\\mathbb{u})\\frac{\\mathbb{v}}{|\\mathbb{v}|}= \\left(\\frac{\\mathbb{u}\\cdot\\mathbb{v}}{|\\mathbb{v}|^2}\\right)\\mathbb{v}.$$ I hope that this helps! - @ Jennifer Dylan: Thanks for the edit, how did you get proj and comp in plain text in math mode? – John Martin Aug 27 '12 at 3:35 1. There's no space between @ and my name. Otherwise, I don't get notified :) 2. Inside math mode, you can surround the plain text with either \\mathrm{..} which ignores spaces and changes the font into Roman font, OR bettter use \\text{..} which respects the text. For operators and such, you should use \\operatorname{..}, which I should I've done, but \\text{} is shorter to type! – user2468 Aug 27 '12 at 3:40", "write $\\vec{A} + \\vec{B} =0$. Now it is also true that $\\vec{A} = A \\hat{i}$ and $\\vec{B} = -B \\hat{i}$ where A and B are the magnitudes of the vectors. Thus, you have $A \\hat{i} +(-B \\hat{i})=0$ which simplifies to the following relation between magnitudes $A - B = 0$. See how it works?", "? Free Version Moderate # Magnitude After Subtraction of Two Vectors GEOM-EXAG8L What is the magnitude of the resultant vector of $a-b$ if $a= (4, 1)$ and $b= (1, 5)$?", "## College Physics (4th Edition) $B = C \\lt A$ $A_x = 4$ $A_y = 4$ $B_x = -1$ $B_y = 3$ $C_x = 1$ $C_y = -3$ We can find the magnitude of each vector: The magnitude of vector A is $\\sqrt{(4)^2+(4)^2}$ which is 5.66 The magnitude of vector B is $\\sqrt{(-1)^2+(3)^2}$ which is 3.16 The magnitude of vector C is $\\sqrt{(1)^2+(-3)^2}$ which is 3.16 We can rank the vectors in order of increasing magnitude: $B = C \\lt A$", "Math Help - Vector magnitude addition I hope this is the right forum. $u=<1,1,1>, v=<-1,2.1>$ The magnitudes are: $||u+v||=\\sqrt{13}$ and $||u|| + ||v|| = \\sqrt{3} + \\sqrt{6}$ But the answer in the back of the book is showing $\\sqrt{13}$ for both? Where am I going wrong? I agree with your answers. Moreover, in a euclidean space, $\\|u+v\\| = \\|u\\|+\\|v\\|$ iff u and v are proportional, which is not the case here. There is probably a typo in the book. Originally Posted by M.R But the answer in the back of the book is showing $\\sqrt{13}$ for both? Where am I going wrong? I think you are correct, assuming I am reading your work correctly. $\\mathbf u = \\langle1,1,1\\rangle, \\mathbf v = \\langle-1,2,1\\rangle$ $\\|\\mathbf u + \\mathbf v\\| = \\|\\langle0,3,2\\rangle\\| = \\sqrt{13}$ $\\|\\mathbf u\\| + \\|\\mathbf v\\| = \\sqrt3 + \\sqrt6$ $\\sqrt{13}\\neq\\sqrt3+\\sqrt6$ Thank you both for the verification. BTW, where is the thanks button?", "&= (1,y,y^{2},\\ldots,y^{n-1}) \\end{aligned} for the vector whose $$i$$-th entry is $$y^{i}$$. For vectors $${\\mathbf{v}}$$ of even length $$2k$$, we define $${\\mathbf{v}}_{\\operatorname{lo}}$$ and $${\\mathbf{v}}_{\\operatorname{hi}}$$ to be the low and high halves of $${\\mathbf{v}}$$: \\begin{aligned} {\\mathbf{v}}_{\\operatorname{lo}} &= (v_0, \\ldots, v_{k-1})\\\\ {\\mathbf{v}}_{\\operatorname{hi}} &= (v_{k}, \\ldots, v_{2k-1}) \\end{aligned} Pedersen commitments are written as \\begin{aligned} \\operatorname{Com}(v) &= \\operatorname{Com}(v, {\\widetilde{v}}) = v \\cdot B + {\\widetilde{v}} \\cdot {\\widetilde{B}}, \\end{aligned} where $$B$$ and $${\\widetilde{B}}$$ are the generators used for the values and blinding factors, respectively. We denote the blinding factor for the value $$v$$ by $${\\widetilde{v}}$$, so that it is clear which blinding factor corresponds to which value, and write $$\\operatorname{Com}(v)$$ instead of $$\\operatorname{Com}(v, {\\widetilde{v}})$$ for brevity. We also make use of vector Pedersen commitments, which we define for pairs of vectors as \\begin{aligned} \\operatorname{Com}({\\mathbf{a}}_{L}, {\\mathbf{a}}_{R}) &= \\operatorname{Com}({\\mathbf{a}}_{L}, {\\mathbf{a}}_{R}, {\\widetilde{a}}) = {\\langle {\\mathbf{a}}_{L}, {\\mathbf{G}} \\rangle} + {\\langle {\\mathbf{a}}_{R}, {\\mathbf{H}} \\rangle} + {\\widetilde{a}} {\\widetilde{B}},\\end{aligned} where $${\\mathbf{G}}$$ and $${\\mathbf{H}}$$ are vectors of generators. Notice that this is exactly the same as taking a commitment to the vector of values $${\\mathbf{a}}_{L} \\Vert {\\mathbf{a}}_{R}$$ with the vector of bases $${\\mathbf{G}} \\Vert {\\mathbf{H}}$$, but defining the commitment on pairs of vectors is a more convenient notation. The variable renaming is as follows: \\begin{aligned} g &\\xrightarrow{} B & \\gamma &\\xrightarrow{} \\tilde{v} \\\\ h &\\xrightarrow{} \\widetilde{B} & \\alpha &\\xrightarrow{} \\tilde{a} \\\\ {\\mathbf{g}}", "$X_{r+1}$ lies on $\\mathbb{R}^n$ outside the subspace with prob. $1$. By this process, $\\operatorname{span}\\{X_1,\\ldots,X_n\\} = \\mathbb{R}^n$ with prob. $1$. But, I'm just not sure that this proof is correct and has no problem.", "in the direction of $\\mathbf{b} = 3\\mathbf{i} - 2\\mathbf{j} + 6\\mathbf{k}$. So $F = k\\mathbf{b}$. We also know that $|F| = 84$. $|F| = |k\\mathbf{b}| = k|\\mathbf{b}|$. $|\\mathbf{b}| = \\sqrt{3^2 + (-2)^2 + 6^2}$ $= \\sqrt{9 + 4 + 36}$ $= \\sqrt{49}$ $= 7$. Since $|F| = k|\\mathbf{b}|$ $84 = 7k$ $k = 12$. So that means $F = 12\\mathbf{b}$ $= 12(3\\mathbf{i} - 2\\mathbf{j} + 6\\mathbf{k})$ $= 36\\mathbf{i} - 24\\mathbf{j} + 72\\mathbf{k}$. i don't understand why wouldn't, 84 be the magnitude of $\\mathbf{b} = 3\\mathbf{i} - 2\\mathbf{j} + 6\\mathbf{k}$ 7. Originally Posted by Paymemoney i don't understand why wouldn't, 84 be the magnitude of $\\mathbf{b} = 3\\mathbf{i} - 2\\mathbf{j} + 6\\mathbf{k}$ I've shown you why. Use Pythagoras on it and work out the magnitude of $\\mathbf{b}$. Its magnitude is $7$. A vector has magnitude and direction. Like I said, $F$ is a vector with a magnitude of $84$ and in the DIRECTION of $\\mathbf{b}$. $F$ DOES NOT EQUAL $\\mathbf{b}$. It is a SCALAR MULTIPLE of $\\mathbf{b}$. (In other words it goes in the same direction but has been magnified to a different length). You need to figure out the magnification factor so you can multiply it by $\\mathbf{b}$ and thus FIND $F$. 8. ok, so what does the k mean in the forumla F=kb? And how did you", "$$m ^{2} = \\vec{m}\\cdot \\vec{m}$$ I assume you are looking for the magnitudes of a, b and c.", "direction. In the case where one of the vectors (say $\\vec{v}$) is a unit vector, $\\vec{u} \\cdot \\vec{v}$ measures the component of $\\vec{u}$ in the $\\vec{v}$ direction. If we introduce the notation $\\hat{v}$ to stand for $\\frac{\\vec{v}}{|\\vec{v}|}$, then $\\hat{v}$ is a unit vector in the $\\vec{v}$ direction, and we have that the component of $\\vec{u}$ in the $\\vec{v}$ direction is $$\\operatorname{comp}_{\\vec{v}}\\vec{u} = \\vec{u} \\cdot \\hat{v} = \\frac{\\vec{u}\\cdot\\vec{v}}{|\\vec{v}|}$$ It's important to realize that $\\operatorname{comp}_\\vec{v} \\vec{u}$ is a scalar quantity, measuring the \"size\" of the component of $\\vec{u}$ in the direction of $\\vec{v}$. (I put \"size\" in quotes because this scalar also includes a sign, so it's not just a magnitude.) It's also useful to define a vector $\\operatorname{proj}_\\vec{v} \\vec{u}$ as follows: $$\\operatorname{proj}_\\vec{v} \\vec{u} = \\left( \\operatorname{comp}_\\vec{v} \\vec{u} \\right) \\hat{v}$$ Notice that $\\operatorname{proj}_\\vec{v} \\vec{u}$ points in (or opposite to) the $\\vec{u}$ diction and has a length equal to $|\\operatorname{comp}_\\vec{v} \\vec{u}|$. Unpacking this definition, we have the following equivalent forms: $$\\operatorname{proj}_\\vec{v} \\vec{u} = \\left( \\vec{u} \\cdot \\hat{v}\\right) \\hat{v}$$ $$\\operatorname{proj}_\\vec{v} \\vec{u} = \\left( \\frac{\\vec{u} \\cdot \\vec{v}}{|\\vec{v}|}\\right) \\frac{\\vec{v}}{|\\vec{v}|}$$ $$\\operatorname{proj}_\\vec{v} \\vec{u} = \\left( \\frac{\\vec{u} \\cdot \\vec{v}}{|\\vec{v}|^2}\\right) \\vec{v}$$ An alternate explanation uses the fact that $\\langle a,b,c\\rangle$ is a normal vector to the plane and is therefore normal to the vector connecting two points such as $(at,bt,ct)$ and $(x,y,z)$ lying in the plane where $P=(at,bt,ct)$", "compute the vector from $\\mathbf{a}$ to $\\mathbf{b}$ , we interpret the points $\\mathbf{a}$ and $\\mathbf{b}$ as vectors from the origin, and then use the triangle rule. In fact, this is how vectors are defined in some texts: the subtraction of two points. Notice that the vector subtraction $\\mathbf{b}-\\mathbf{a}$ yields a vector from $\\mathbf{a}$ to $\\mathbf{b}$ . It doesn't make any sense to simply find the vector “between two points,” since the language in this sentence does not specify a direction. We must always form a vector that goes from one point to another point. # 2.8Vector Magnitude (Length) As we have discussed, vectors have magnitude and direction. However, you might have noticed that neither the magnitude nor the direction is expressed explicitly in the vector (at least not when we use Cartesian coordinates). For example, the magnitude of the 2D vector $\\left[3,4\\right]$ is neither 3 nor 4; it's 5. Since the magnitude of the vector is not expressed explicitly, we must compute it. The magnitude of a vector is also known as the length or norm of the vector. ## 2.8.1Official Linear Algebra Rules In linear algebra, the magnitude of a vector is denoted by using double vertical bars surrounding the vector. This is similar to the single vertical bar notation used for the absolute value operation for", "vector with terminal point $$(−2,−4)$$. Let $$\\vec{\\mathbf b}$$ be a vector with initial point $$(1,2)$$ and terminal point $$(−1,4)$$. Find the magnitude of vector $$−3\\vec{\\mathbf a}+\\vec{\\mathbf b}−4\\hat{\\mathbf i}+\\hat{\\mathbf j}.$$ $$15$$ 18) Let $$\\vec{\\mathbf a}$$ be a standard-position vector with terminal point at $$(2,5)$$. Let $$\\vec{\\mathbf b}$$ be a vector with initial point $$(−1,3)$$ and terminal point $$(1,0)$$. Find the magnitude of vector $$\\vec{\\mathbf a}−3\\vec{\\mathbf b}+14\\hat{\\mathbf i}−14\\hat{\\mathbf j}.$$ 19) Let $$\\vec{\\mathbf u}$$ and $$\\vec{\\mathbf v}$$ be two nonzero vectors that are nonequivalent. Consider the vectors $$\\vec{\\mathbf a}=4\\vec{\\mathbf u}+5\\vec{\\mathbf v}$$ and $$\\vec{\\mathbf b}=\\vec{\\mathbf u}+2\\vec{\\mathbf v}$$ defined in terms of $$\\vec{\\mathbf u}$$ and $$\\vec{\\mathbf v}$$. Find the scalar $$λ$$ such that vectors $$\\vec{\\mathbf a}+λ\\vec{\\mathbf b}$$ and $$\\vec{\\mathbf u}−\\vec{\\mathbf v}$$ are equivalent. $$λ=−3$$ 20) Let $$\\vec{\\mathbf u}$$ and $$\\vec{\\mathbf v}$$ be two nonzero vectors that are nonequivalent. Consider the vectors $$\\vec{\\mathbf a}=2\\vec{\\mathbf u}−4\\vec{\\mathbf v}$$ and $$\\vec{\\mathbf b}=3\\vec{\\mathbf u}−7\\vec{\\mathbf v}$$ defined in terms of $$\\vec{\\mathbf u}$$ and $$\\vec{\\mathbf v}$$. Find the scalars $$α$$ and $$β$$ such that vectors $$α\\vec{\\mathbf a}+β\\vec{\\mathbf b}$$ and $$\\vec{\\mathbf u}−\\vec{\\mathbf v}$$ are equivalent. 21) Consider the vector $$\\vec{\\mathbf a}(t)=⟨\\cos t, \\sin t⟩$$ with components that depend on a real number $$t$$. As the number $$t$$ varies, the components of $$\\vec{\\mathbf a}(t)$$ change as well, depending on the functions that define them. a. Write the vectors $$\\vec{\\mathbf a}(0)$$ and $$\\vec{\\mathbf", "+0 # For vectors u and v, we have if and ​ find p*u 0 60 1 For vectors u and v, we have $$\\mathbf{p}=\\operatorname{proj}_{\\mathbf{v}}(\\mathbf{u})$$ if $$\\|\\mathbf{u}\\|=11$$ and $$\\|\\mathbf{p}\\|=6$$ find p*u Feb 13, 2022 #1 0 p*u = 22/3. Feb 14, 2022", "+0 # For vectors u and v, we have if and ​ find p*u 0 115 1 For vectors u and v, we have $$\\mathbf{p}=\\operatorname{proj}_{\\mathbf{v}}(\\mathbf{u})$$ if $$\\|\\mathbf{u}\\|=11$$ and $$\\|\\mathbf{p}\\|=6$$ find p*u Feb 13, 2022" ]

[ "For the two values of $\\mathbf{w}$, $\\begin{split} \\sum_{\\mathbf{w}\\in \\{\\mathbf{w}_1, \\mathbf{w}_2\\}} & \\operatorname{P}(\\textrm{y}|\\textrm{t}=t_0, \\mathbf{w}=\\mathbf{w})\\operatorname{P}(\\mathbf{w}=\\mathbf{w}) \\\\ & =\\sum_{\\mathbf{w}\\in \\{\\mathbf{w}_1, \\mathbf{w}_2\\}} \\operatorname{P}(\\textrm{y}|\\textrm{t}=t_0, \\mathbf{w}=\\mathbf{w})\\operatorname{P}(\\mathbf{w}=\\mathbf{w})\\frac{\\operatorname{P}(\\textrm{t}=t_0|\\mathbf{w}=\\mathbf{w})}{\\operatorname{P}(\\textrm{t}=t_0|\\mathbf{w}=\\mathbf{w})} \\\\ & =\\sum_{\\mathbf{w}\\in \\{\\mathbf{w}_1, \\mathbf{w}_2\\}} \\frac{\\operatorname{P}(\\textrm{y}, \\textrm{t}=t_0, \\mathbf{w}=\\mathbf{w})}{\\operatorname{P}(\\textrm{t}=t_0|\\mathbf{w}=\\mathbf{w})} = \\frac{\\operatorname{P}(\\textrm{y}, \\textrm{t}=t_0, \\mathbf{w}=\\mathbf{w}_1)+\\operatorname{P}(\\textrm{y}, \\textrm{t}=t_0, \\mathbf{w}=\\mathbf{w}_2)}{\\operatorname{P}(\\textrm{t}=t_0|\\mathbf{w}=\\mathbf{w})} \\\\ & = \\frac{\\operatorname{P}(\\textrm{y}, \\textrm{t}=t_0, \\mathbf{w}=\\mathbf{w}_1 \\text{or } \\mathbf{w}_2)}{\\operatorname{P}(\\textrm{t}=t_0|\\mathbf{w}=\\mathbf{w}_1 \\text{or } \\mathbf{w}_2)} = \\operatorname{P}(\\textrm{y}| \\textrm{t}=t_0, \\mathbf{w}=\\mathbf{w}_1 \\text{or } \\mathbf{w}_2)\\operatorname{P}(\\mathbf{w}=\\mathbf{w}_1 \\text{or } \\mathbf{w}_2) \\end{split}$ where the second last equality is because $\\operatorname{P}(\\textrm{t}=t_0| \\mathbf{w}=\\mathbf{w}_1)=\\operatorname{P}(\\textrm{t}=t_0| \\mathbf{w}=\\mathbf{w}_2)=\\operatorname{P}(\\textrm{t}=t_0|\\mathbf{w}=\\mathbf{w}_1 \\text{or } \\mathbf{w}_2)$. Effectively, the derivation corresponds to creating a new confounder $\\mathbf{w}'$ such that it attains the same value when $\\mathbf{w}\\in \\{\\mathbf{w}_1, \\mathbf{w}_2\\}$, but different values for each other value of $\\mathbf{w}$. Within each strata of $\\mathbf{w}'$, we find that $\\textrm{t}\\unicode{x2AEB}\\mathbf{w}$ and thus $\\textrm{t}\\unicode{x2AEB}\\mathbf{w}| \\mathbf{w}'$. Alternatively, we can write $\\textrm{t}\\unicode{x2AEB}\\mathbf{w}| b( \\mathbf{w})$ where $b$ is a function that maps $\\mathbf{w}$ to $\\mathbf{w}'$. More generally, any function $b: \\mathbb{R}^{|W|} \\rightarrow \\mathbb{R}^{K}$ that ensures that $t$ and $w$ are independent given $b(w)$, $t \\unicode{x2AEB} w|b(w)$ is called a balancing function. The resultant values $b(w)$ are called balancing scores. Given a balancing score, the backdoor estimation equation can be rewritten as: $\\label{eq:balancing-score} \\mathbb{E}[\\textrm{y}|\\operatorname{do}(\\textrm{t}=t_0)] \\rightarrow\\sum_{\\mathbf{w}\\in \\mathbb{W}} \\mathbb{E}(\\textrm{y}|\\textrm{t}=t_0, b(\\mathbf{w})=b(\\mathbf{w}))\\Pr(b(\\mathbf{w})=b(\\mathbf{w}))\\qquad(4)$ If we consider $b(\\mathbf{w})$ as trivially the identity function, $b(\\mathbf{w})=\\mathbf{w}$, then we obtain simple stratification. It is valid balancing function since by definition, $t \\unicode{x2AEB}\\mathbf{w}|\\mathbf{w}$. However, a useful balancing typically reduces the dimensionality of $\\mathbf{w}$. The", "it possible to have$\\operatorname{proj}_{\\mathrm{a}} \\mathbf{u}=\\operatorname{proj}_{\\mathbf{u}} \\mathbf{a} ?$Explain. 0 answers 7 views Is it possible to have$\\operatorname{proj}_{\\mathrm{a}} \\mathbf{u}=\\operatorname{proj}_{\\mathbf{u}} \\mathbf{a} ?$Explain.Is it possible to have$\\operatorname{proj}_{\\mathrm{a}} \\mathbf{u}=\\operatorname{proj}_{\\mathbf{u}} \\mathbf{a} ?\\$ Explain. ...", "close to $1$ we conclude that $\\operatorname{p}(\\mathbf{u}+\\mathbf{v})\\leq\\operatorname{p}(\\mathbf{v}).$ The same argument also shows that $\\operatorname{p}(\\mathbf{v})=\\operatorname{p}(-\\mathbf{u}+(\\mathbf{u}+\\mathbf{% v}))\\leq\\operatorname{p}(\\mathbf{u}+\\mathbf{v}),$ and hence $\\operatorname{p}(\\mathbf{u}+\\mathbf{v})=\\operatorname{p}(\\mathbf{v}),$ as desired. Finally, suppose that neither $\\operatorname{p}(\\mathbf{u})$ nor $\\operatorname{p}(\\mathbf{v})$ is zero. Hence, $\\frac{\\mathbf{u}}{\\operatorname{p}(u)},\\,\\frac{\\mathbf{v}}{\\operatorname{p}(v)}$ are both in $B_{1}$, and hence $\\frac{\\operatorname{p}(u)}{\\operatorname{p}(u)+\\operatorname{p}(v)}\\frac{% \\mathbf{u}}{\\operatorname{p}(u)}+\\frac{\\operatorname{p}(v)}{\\operatorname{p}(u% )+\\operatorname{p}(v)}\\frac{\\mathbf{v}}{\\operatorname{p}(v)}=\\frac{\\mathbf{u}+% \\mathbf{v}}{\\operatorname{p}(u)+\\operatorname{p}(v)}$ is in $B_{1}$ also. Using homogeneity, we conclude that $\\operatorname{p}(u+v)\\leq\\operatorname{p}(u)+\\operatorname{p}(v),$ as desired. Title seminorm Seminorm 2013-03-22 12:24:57 2013-03-22 12:24:57 rmilson (146) rmilson (146) 20 rmilson (146) Definition msc 46B20 semi-norm homogeneous", "{E} [Z_{2}])(W_{n}-\\operatorname {E} [W_{n}])]\\\\\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\\\\\\\mathrm {E} [(Z_{n}-\\operatorname {E} [Z_{n}])(W_{1}-\\operatorname {E} [W_{1}])]&\\mathrm {E} [(Z_{n}-\\operatorname {E} [Z_{n}])(W_{2}-\\operatorname {E} [W_{2}])]&\\cdots &\\mathrm {E} [(Z_{n}-\\operatorname {E} [Z_{n}])(W_{n}-\\operatorname {E} [W_{n}])]\\end{bmatrix}}}$ ### Uncorrelatedness Two complex random vectors ${\\displaystyle \\mathbf {Z} }$ and ${\\displaystyle \\mathbf {W} }$ are called uncorrelated if ${\\displaystyle \\operatorname {K} _{\\mathbf {Z} \\mathbf {W} }=\\operatorname {J} _{\\mathbf {Z} \\mathbf {W} }=0}$ . ## Independence Two complex random vectors ${\\displaystyle \\mathbf {Z} =(Z_{1},...,Z_{m})^{T}}$ and ${\\displaystyle \\mathbf {W} =(W_{1},...,W_{n})^{T}}$ are called independent if ${\\displaystyle F_{\\mathbf {Z,W} }(\\mathbf {z,w} )=F_{\\mathbf {Z} }(\\mathbf {z} )\\cdot F_{\\mathbf {W} }(\\mathbf {w} )\\quad {\\text{for all }}\\mathbf {z} ,\\mathbf {w} }$ (Eq.7) where ${\\displaystyle F_{\\mathbf {Z} }(\\mathbf {z} )}$ and ${\\displaystyle F_{\\mathbf {W} }(\\mathbf {w} )}$ denote the cumulative distribution functions of ${\\displaystyle \\mathbf {Z} }$ and ${\\displaystyle \\mathbf {W} }$ as defined in Eq.1 and ${\\displaystyle F_{\\mathbf {Z,W} }(\\mathbf {z,w} )}$ denotes their joint cumulative distribution function. Independence of ${\\displaystyle \\mathbf {Z} }$ and ${\\displaystyle \\mathbf {W} }$ is often denoted by ${\\displaystyle \\mathbf {Z} \\perp \\!\\!\\!\\perp \\mathbf {W} }$ . Written component-wise, ${\\displaystyle \\mathbf {Z} }$ and ${\\displaystyle \\mathbf {W} }$ are called independent if ${\\displaystyle F_{Z_{1},\\ldots ,Z_{m},W_{1},\\ldots ,W_{n}}(z_{1},\\ldots ,z_{m},w_{1},\\ldots ,w_{n})=F_{Z_{1},\\ldots ,Z_{m}}(z_{1},\\ldots ,z_{m})\\cdot F_{W_{1},\\ldots ,W_{n}}(w_{1},\\ldots ,w_{n})\\quad {\\text{for all }}z_{1},\\ldots ,z_{m},w_{1},\\ldots ,w_{n}}$ . ## Circular symmetry ### Definition A complex random vector ${\\displaystyle \\mathbf {Z} }$", "operator $\\displaystyle W(\\mathbf{a},\\mathbf{b})\\equiv e^{i(\\mathbf{a}\\cdot\\mathbf{b})/2}V(\\mathbf{b})U(\\mathbf{a})~, \\ \\ \\ \\ \\ (5)$ whereupon the Weyl form of the canonical commutation relations may be encapsulated in the multiplication rule $\\displaystyle W(\\mathbf{a},\\mathbf{b})W(\\mathbf{a}',\\mathbf{b}')=e^{-i\\sigma[(\\mathbf{a},\\mathbf{b}),(\\mathbf{a}',\\mathbf{b}')]/2}W(\\mathbf{a}+\\mathbf{a}',\\mathbf{b}+\\mathbf{b}')~, \\ \\ \\ \\ \\ (6)$ where ${\\sigma[(\\mathbf{a},\\mathbf{b}),(\\mathbf{a}',\\mathbf{b}')]\\equiv\\mathbf{a}'\\cdot\\mathbf{b}-\\mathbf{a}\\cdot\\mathbf{b}'}$ is none other than the familiar symplectic form on S. For completeness, one further defines $\\displaystyle W(\\mathbf{a},\\mathbf{b})^*\\equiv W(-\\mathbf{a},-\\mathbf{b})= e^{-i(\\mathbf{a}\\cdot\\mathbf{b})/2}U(-\\mathbf{a})V(-\\mathbf{b})~. \\ \\ \\ \\ \\ (7)$ The point is then that any representation of the Weyl operators ${W(\\mathbf{a},\\mathbf{b})}$ on a Hilbert space ${\\mathcal{H}}$ (more on what this means below) gives rise to a representation of the Weyl form of the canonical commutation relations, and vice-versa. We can now drop the restriction to finite-dimensional spaces, and let the phase space ${S}$ be an arbitrary infinite-dimensional vector space equipped with a symplectic form. Then the family ${\\{W_\\pi(f):f\\in S\\}}$ of unitary operators acting in some Hilbert space ${\\mathcal{H}_\\pi}$ satisfies the Weyl relations iff $\\displaystyle \\begin{gathered} W_\\pi(f)W_\\pi(g)=e^{-i\\sigma(f,g)/2}W_\\pi(f+g)~,\\qquad\\forall f,g\\in S~,\\\\ W_\\pi(f)^*=W_\\pi(-f)~,\\qquad\\forall f\\in S~. \\end{gathered} \\ \\ \\ \\ \\ (8)$ The (self-adjoint) observables of the system are then obtained by taking arbitrarily linear combinations of Weyl operators. In the above expressions, the subscript ${\\pi}$ denotes the representation. The idea behind a representation is to make an abstract algebra more concrete by representing the elements thereof as matrices; i.e., a representation reduces an abstract algebra to a linear algebra,", "; f W ( x ) , σ 1 2 ) ⋅ Softmax ⁡ ( y 2 = c ; f W ( x ) , σ 2 ) = 1 2 σ 1 2 ∥ y 1 − f W ( x ) ∥ 2 + log ⁡ σ 1 − log ⁡ p ( y 2 = c ∣ f W ( x ) , σ 2 ) = 1 2 σ 1 2 L 1 ( W ) + 1 σ 2 2 L 2 ( W ) + log ⁡ σ 1 + log ⁡ ∑ c ′ exp ⁡ ( 1 σ 2 2 f c ′ W ( x ) ) ( ∑ c ′ exp ⁡ ( f c ′ W ( x ) ) ) 1 σ 2 2 ≈ 1 2 σ 1 2 L 1 ( W ) + 1 σ 2 2 L 2 ( W ) + log ⁡ σ 1 + log ⁡ σ 2 (10) \\begin{array}{l} =-\\log p\\left(\\mathbf{y}_{1}, \\mathbf{y}_{2}=c \\mid \\mathbf{f}^{\\mathbf{W}}(\\mathbf{x})\\right) \\\\ =-\\log \\mathcal{N}\\left(\\mathbf{y}_{1} ; \\mathbf{f}^{\\mathbf{W}}(\\mathbf{x}), \\sigma_{1}^{2}\\right) \\cdot \\operatorname{Softmax}\\left(\\mathbf{y}_{2}=c ; \\mathbf{f}^{\\mathbf{W}}(\\mathbf{x}), \\sigma_{2}\\right) \\\\ =\\frac{1}{2 \\sigma_{1}^{2}}\\left\\|\\mathbf{y}_{1}-\\mathbf{f}^{\\mathbf{W}}(\\mathbf{x})\\right\\|^{2}+\\log \\sigma_{1}-\\log p\\left(\\mathbf{y}_{2}=c \\mid \\mathbf{f}^{\\mathbf{W}}(\\mathbf{x}), \\sigma_{2}\\right) \\\\ =\\frac{1}{2 \\sigma_{1}^{2}} \\mathcal{L}_{1}(\\mathbf{W})+\\frac{1}{\\sigma_{2}^{2}} \\mathcal{L}_{2}(\\mathbf{W})+\\log \\sigma_{1} \\\\ \\quad+\\log \\frac{\\sum_{c^{\\prime}} \\exp \\left(\\frac{1}{\\sigma_{2}^{2}} f_{c^{\\prime}}^{\\mathbf{W}}(\\mathbf{x})\\right)}{\\left(\\sum_{c^{\\prime}} \\exp \\left(f_{c^{\\prime}} \\mathbf{W}(\\mathbf{x})\\right)\\right)^{\\frac{1}{\\sigma_{2}^{2}}}} \\\\ \\approx \\frac{1}{2 \\sigma_{1}^{2}} \\mathcal{L}_{1}(\\mathbf{W})+\\frac{1}{\\sigma_{2}^{2}} \\mathcal{L}_{2}(\\mathbf{W})+\\log \\sigma_{1}+\\log \\sigma_{2} \\end{array}", "{X} -\\operatorname {E} [\\mathbf {X} ])(\\mathbf {Y} -\\operatorname {E} [\\mathbf {Y} ])^{\\rm {T}}]=\\operatorname {R} _{\\mathbf {X} \\mathbf {Y} }-\\operatorname {E} [\\mathbf {X} ]\\operatorname {E} [\\mathbf {Y} ]^{\\rm {T}}}$ Respectively for complex random vectors: ${\\displaystyle \\operatorname {K} _{\\mathbf {Z} \\mathbf {W} }=\\operatorname {E} [(\\mathbf {Z} -\\operatorname {E} [\\mathbf {Z} ])(\\mathbf {W} -\\operatorname {E} [\\mathbf {W} ])^{\\rm {H}}]=\\operatorname {R} _{\\mathbf {Z} \\mathbf {W} }-\\operatorname {E} [\\mathbf {Z} ]\\operatorname {E} [\\mathbf {W} ]^{\\rm {H}}}$ ## References 1. ^ Gubner, John A. (2006). Probability and Random Processes for Electrical and Computer Engineers. Cambridge University Press. ISBN 978-0-521-86470-1.", "\\prod_{(w:W)}}}E(w))}}}{\\sum_{(\\mathsf{ind}:\\mathchoice{\\prod_{w:W}\\,}{% \\mathchoice{{\\textstyle\\prod_{(w:W)}}}{\\prod_{(w:W)}}{\\prod_{(w:W)}}{\\prod_{(w% :W)}}}{\\mathchoice{{\\textstyle\\prod_{(w:W)}}}{\\prod_{(w:W)}}{\\prod_{(w:W)}}{% \\prod_{(w:W)}}}{\\mathchoice{{\\textstyle\\prod_{(w:W)}}}{\\prod_{(w:W)}}{\\prod_{(% w:W)}}{\\prod_{(w:W)}}}E(w))}}{\\sum_{(\\mathsf{ind}:\\mathchoice{\\prod_{w:W}\\,}{% \\mathchoice{{\\textstyle\\prod_{(w:W)}}}{\\prod_{(w:W)}}{\\prod_{(w:W)}}{\\prod_{(w% :W)}}}{\\mathchoice{{\\textstyle\\prod_{(w:W)}}}{\\prod_{(w:W)}}{\\prod_{(w:W)}}{% \\prod_{(w:W)}}}{\\mathchoice{{\\textstyle\\prod_{(w:W)}}}{\\prod_{(w:W)}}{\\prod_{(% w:W)}}{\\prod_{(w:W)}}}E(w))}}{\\sum_{(\\mathsf{ind}:\\mathchoice{\\prod_{w:W}\\,}{% \\mathchoice{{\\textstyle\\prod_{(w:W)}}}{\\prod_{(w:W)}}{\\prod_{(w:W)}}{\\prod_{(w% :W)}}}{\\mathchoice{{\\textstyle\\prod_{(w:W)}}}{\\prod_{(w:W)}}{\\prod_{(w:W)}}{% \\prod_{(w:W)}}}{\\mathchoice{{\\textstyle\\prod_{(w:W)}}}{\\prod_{(w:W)}}{\\prod_{(% w:W)}}{\\prod_{(w:W)}}}E(w))}}}\\mathchoice{\\prod_{(a,f)}\\,}{\\mathchoice{{% \\textstyle\\prod_{(a,f)}}}{\\prod_{(a,f)}}{\\prod_{(a,f)}}{\\prod_{(a,f)}}}{% \\mathchoice{{\\textstyle\\prod_{(a,f)}}}{\\prod_{(a,f)}}{\\prod_{(a,f)}}{\\prod_{(a% ,f)}}}{\\mathchoice{{\\textstyle\\prod_{(a,f)}}}{\\prod_{(a,f)}}{\\prod_{(a,f)}}{% \\prod_{(a,f)}}}\\\\ \\displaystyle\\mathsf{ind}({\\mathsf{sup}}(a,f))=e(a,{\\lambda}b.\\,\\mathsf{ind}(f% (b))).$ In http://planetmath.org/node/87579Chapter 6 we will see some other reasons why propositional computation rules are worth considering. In this section, we will state some basic facts about homotopy-inductive types. We omit most of the proofs, which are somewhat technical. Theorem 5.5.1. For any $A:\\mathcal{U}$ and $B:A\\to\\mathcal{U}$, the type $\\mathsf{W}_{d}(A,B)$ is a mere proposition. It turns out that there is an equivalent characterization of $\\mathsf{W}$-types using a recursion principle, plus certain uniqueness and coherence laws. First we give the recursion principle: • When constructing a function from the $\\mathsf{W}$-type $\\mathchoice{\\mathchoice{{\\textstyle\\mathsf{W}^{h}}}{\\mathsf{W}^{h}}{\\mathsf{W}% ^{h}}{\\mathsf{W}^{h}}({\\textstyle x:A}),\\ }{\\mathchoice{{\\textstyle\\mathsf{W}^% {h}_{(x:A)}}}{\\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{% (x:A)}}}{\\mathchoice{{\\textstyle\\mathsf{W}^{h}_{(x:A)}}}{\\mathsf{W}^{h}_{(x:A)% }}{\\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}}{\\mathchoice{{\\textstyle% \\mathsf{W}^{h}_{(x:A)}}}{\\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}{% \\mathsf{W}^{h}_{(x:A)}}}B(x)$ into the type $C$, it suffices to give its value for ${\\mathsf{sup}}(a,f)$, assuming we are given the values of all $f(b)$ with $b:B(a)$. In other words, it suffices to construct a function $c:\\mathchoice{\\prod_{a:A}\\,}{\\mathchoice{{\\textstyle\\prod_{(a:A)}}}{\\prod_{(a:% A)}}{\\prod_{(a:A)}}{\\prod_{(a:A)}}}{\\mathchoice{{\\textstyle\\prod_{(a:A)}}}{% \\prod_{(a:A)}}{\\prod_{(a:A)}}{\\prod_{(a:A)}}}{\\mathchoice{{\\textstyle\\prod_{(a% :A)}}}{\\prod_{(a:A)}}{\\prod_{(a:A)}}{\\prod_{(a:A)}}}(B(a)\\to C)\\to C.$ The associated computation rule for $\\mathsf{rec}_{\\mathchoice{\\mathchoice{{\\textstyle\\mathsf{W}^{h}}}{\\mathsf{W}^{% h}}{\\mathsf{W}^{h}}{\\mathsf{W}^{h}}({\\textstyle x:A}),\\ }{\\mathchoice{{% \\textstyle\\mathsf{W}^{h}_{(x:A)}}}{\\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:% A)}}{\\mathsf{W}^{h}_{(x:A)}}}{\\mathchoice{{\\textstyle\\mathsf{W}^{h}_{(x:A)}}}{% \\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}}{% \\mathchoice{{\\textstyle\\mathsf{W}^{h}_{(x:A)}}}{\\mathsf{W}^{h}_{(x:A)}}{% \\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}}B(x)}(C,c):(\\mathchoice{% \\mathchoice{{\\textstyle\\mathsf{W}_{(x:A)}}}{\\mathsf{W}_{(x:A)}}{\\mathsf{W}_{(x% :A)}}{\\mathsf{W}_{(x:A)}}}{\\mathchoice{{\\textstyle\\mathsf{W}_{(x:A)}}}{\\mathsf% {W}_{(x:A)}}{\\mathsf{W}_{(x:A)}}{\\mathsf{W}_{(x:A)}}}{\\mathchoice{{\\textstyle% \\mathsf{W}_{(x:A)}}}{\\mathsf{W}_{(x:A)}}{\\mathsf{W}_{(x:A)}}{\\mathsf{W}_{(x:A)% }}}{\\mathchoice{{\\textstyle\\mathsf{W}_{(x:A)}}}{\\mathsf{W}_{(x:A)}}{\\mathsf{W}% _{(x:A)}}{\\mathsf{W}_{(x:A)}}}B(x))\\to C$ is as follows: • For any $a:A$ and $f:B(a)\\to\\mathchoice{\\mathchoice{{\\textstyle\\mathsf{W}^{h}}}{\\mathsf{W}^{h}}{% \\mathsf{W}^{h}}{\\mathsf{W}^{h}}({\\textstyle x:A}),\\ }{\\mathchoice{{\\textstyle% \\mathsf{W}^{h}_{(x:A)}}}{\\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}{% \\mathsf{W}^{h}_{(x:A)}}}{\\mathchoice{{\\textstyle\\mathsf{W}^{h}_{(x:A)}}}{% \\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}}{% \\mathchoice{{\\textstyle\\mathsf{W}^{h}_{(x:A)}}}{\\mathsf{W}^{h}_{(x:A)}}{% \\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}}B(x)$ we have a witness $\\beta(C,c,a,f)$ for equality $\\mathsf{rec}_{\\mathchoice{\\mathchoice{{\\textstyle\\mathsf{W}^{h}}}{\\mathsf{W}^{% h}}{\\mathsf{W}^{h}}{\\mathsf{W}^{h}}({\\textstyle x:A}),\\ }{\\mathchoice{{% \\textstyle\\mathsf{W}^{h}_{(x:A)}}}{\\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:% A)}}{\\mathsf{W}^{h}_{(x:A)}}}{\\mathchoice{{\\textstyle\\mathsf{W}^{h}_{(x:A)}}}{% \\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}}{% \\mathchoice{{\\textstyle\\mathsf{W}^{h}_{(x:A)}}}{\\mathsf{W}^{h}_{(x:A)}}{% \\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}}B(x)}(C,c,{\\mathsf{sup}}(a,f))% =c(a,{\\lambda}b.\\,\\mathsf{rec}_{\\mathchoice{\\mathchoice{{\\textstyle\\mathsf{W}^% {h}}}{\\mathsf{W}^{h}}{\\mathsf{W}^{h}}{\\mathsf{W}^{h}}({\\textstyle x:A}),\\ }{% \\mathchoice{{\\textstyle\\mathsf{W}^{h}_{(x:A)}}}{\\mathsf{W}^{h}_{(x:A)}}{% \\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}}{\\mathchoice{{\\textstyle% \\mathsf{W}^{h}_{(x:A)}}}{\\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}{% \\mathsf{W}^{h}_{(x:A)}}}{\\mathchoice{{\\textstyle\\mathsf{W}^{h}_{(x:A)}}}{% \\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}{\\mathsf{W}^{h}_{(x:A)}}}B(x)}(% C,c,f(b))).$ Furthermore,", "we may compute |\\operatorname*{\\mathbb{E}}[e^{itg}]|=\\prod_{S\\in A}\\mathopen{}\\mathclose{{}% \\left|\\operatorname*{\\mathbb{E}}[e^{it\\alpha(x^{S})}]}\\right|\\leq\\prod_{% \\begin{subarray}{c}S\\in A\\\\ A_{S}\\mbox{\\scriptsize{ occurs}}\\end{subarray}}\\mathopen{}\\mathclose{{}\\left(1% -\\frac{8p(1-p)t^{2}}{\\pi^{2}\\sigma^{2}}}\\right)=\\exp\\mathopen{}\\mathclose{{}% \\left(-\\frac{8p(1-p)t^{2}z(Y)}{\\pi^{2}\\sigma^{2}}}\\right) Since each of the events A_{S} are independent and occur with probability \\geq\\lambda^{2{r\\choose 2}^{2}} it follows from Chernoff bounds that z(Y)\\geq\\lambda^{2{r\\choose 2}^{2}}n/2r with probability \\geq 1-\\exp(-\\Omega(n)). So we find that \\operatorname*{\\mathbb{E}}_{\\mathbf{Y}}\\mathopen{}\\mathclose{{}\\left|% \\operatorname*{\\mathbb{E}}_{X}[e^{it\\alpha(\\mathcal{K})}]}\\right|\\leq\\Pr(A)% \\exp\\mathopen{}\\mathclose{{}\\left(-\\frac{8p(1-p)t^{2}}{\\pi^{2}\\sigma^{2}}\\cdot% \\frac{\\lambda^{2{r\\choose 2}^{2}}n}{2r}}\\right)+\\Pr(A^{c})=\\exp\\mathopen{}% \\mathclose{{}\\left(-\\Omega\\mathopen{}\\mathclose{{}\\left(\\frac{t^{2}n}{\\sigma^{% 2}}}\\right)}\\right) Combining this with Lemma 6 we have proved the following: ###### Lemma 18. For |t|\\leq\\pi\\sigma we have that |\\varphi_{\\mathcal{K}}(t)|\\leq\\exp(-\\Omega(t^{2}n/\\sigma^{2})). ## References • [Ber16] Ross Berkowitz. A quantitative local limit theorem for triangles in random graphs, 2016. • [ER61] P. Erdős and A. Rényi. On the evolution of random graphs. Bull. Inst. Internat. Statist., 38:343–347, 1961. • [Fel71] William Feller. An introduction to probability theory and its applications. Vol. II. Second edition. John Wiley & Sons, Inc., New York-London-Sydney, 1971. • [GK14] Justin Gilmer and Swastik Kopparty. A local central limit theorem for the number of triangles in a random graph. ArXiv e-prints, November 2014. • [Jan94] Svante Janson. Orthogonal decompositions and functional limit theorems for random graph statistics. Mem. Amer. Math. Soc., 111(534):vi+78, 1994. • [Kar84] Michał Karoński. Balanced subgraphs of large random graphs, volume 7 of Seria Matematyka [Mathematics Series]. Uniwersytet im. Adama Mickiewicza w Poznaniu, Poznań, 1984. With a Polish summary. • [KR83] Michał Karoński and Andrzej Ruciński. On the number of strictly balanced subgraphs of a", "i\\mathbf{q}\\cdot\\mathbf{r}}\\right) = \\frac{\\hbar}{2m}e^{+ i\\mathbf{q}\\cdot\\mathbf{r}}\\mathbf{q}\\cdot\\left(2\\mathbf{p}+\\hbar\\mathbf{q}\\right)$$ and $$[\\mathcal{H},e^{- i\\mathbf{q}\\cdot\\mathbf{r}}]= -\\frac{\\hbar}{2m}\\mathbf{q}\\cdot\\left(e^{- i\\mathbf{q}\\cdot\\mathbf{r}}\\mathbf{p}+\\mathbf{p}e^{- i\\mathbf{q}\\cdot\\mathbf{r}}\\right) =- \\frac{\\hbar}{2m}\\mathbf{q}\\cdot\\left(2\\mathbf{p}+\\hbar\\mathbf{q}\\right)e^{- i\\mathbf{q}\\cdot\\mathbf{r}}.$$ Putting either of these into the corresponding formula above, you get $$\\sum_n (\\mathcal{E_n}-\\mathcal{E_s})|\\langle n|e^{i\\mathbf{q}\\cdot\\mathbf{r}}|s \\rangle|^2 = \\frac{\\hbar}{2m}\\langle s|\\mathbf{q}\\cdot\\left(2\\mathbf{p}+\\hbar\\mathbf{q}\\right)|s \\rangle = \\frac{\\hbar^2}{2m}\\mathbf{q}^2+\\frac{\\hbar}{m}\\langle s|\\mathbf{q}\\cdot\\mathbf{p}|s \\rangle.$$ You can then always force the mean value of $\\mathbf{p}$ to vanish. - I) Let us work in the Schrödinger picture. Assume that the Hamiltonian operator $$\\tag{1} \\hat{H}~=~T(\\hat{\\bf p})+ V(\\hat{\\bf r}), \\qquad T(\\hat{\\bf p})~:=~ \\frac{\\hat{\\bf p}^2}{2m},$$ does not depend explicit on time. In particular, the point spectrum of energy eigenvalues $E_n$ does not depend on time. Let $\\psi_n({\\bf r})$ be the corresponding energy eigen-functions in position space, which satisfies the TISE $$\\tag{2} \\hat{H}\\psi_n({\\bf r})~=~E_n \\psi_n({\\bf r}).$$ We may assume that $\\psi_n({\\bf r})\\in{\\mathbb{R}}$ is a real function, see also this Phys.SE post. Let us consider a fixed time $t_0$. Let us pick a basis of time-dependent solutions $$\\tag{3} \\langle n, t |{\\bf r} \\rangle~=~ \\Psi_n({\\bf r},t)~=~ \\psi_n({\\bf r})e^{-\\frac{i}{\\hbar}E_n (t-t_0)}$$ that becomes real at the time $t=t_0$. It follows that for any (possibly complex) function $F=F({\\bf r})$ of position ${\\bf r}$, that the matrix element $$\\tag{4} \\langle n, t_0 |F(\\hat{\\bf r}) | m, t_0 \\rangle ~=~ \\int \\! d^3r ~ \\psi^{*}_n({\\bf r}) ~F({\\bf r})~\\psi_m({\\bf r})~=~ n \\leftrightarrow m$$ is symmetric in $n\\leftrightarrow m$ at $t=t_0$. The time dependence of the matrix element is just", "{C} (\\mathbf {X} \\mathbf {b} )}{\\partial \\mathbf {X} }}=}$ ${\\displaystyle \\left(\\mathbf {C} \\mathbf {X} \\mathbf {b} \\mathbf {a} ^{\\top }+\\mathbf {C} ^{\\top }\\mathbf {X} \\mathbf {a} \\mathbf {b} ^{\\top }\\right)^{\\top }}$ ${\\displaystyle \\mathbf {C} \\mathbf {X} \\mathbf {b} \\mathbf {a} ^{\\top }+\\mathbf {C} ^{\\top }\\mathbf {X} \\mathbf {a} \\mathbf {b} ^{\\top }}$ ${\\displaystyle {\\frac {\\partial \\operatorname {tr} (\\mathbf {X} )}{\\partial \\mathbf {X} }}=}$ ${\\displaystyle \\mathbf {I} }$ U = U(X), V = V(X) ${\\displaystyle {\\frac {\\partial \\operatorname {tr} (\\mathbf {U} +\\mathbf {V} )}{\\partial \\mathbf {X} }}=}$ ${\\displaystyle {\\frac {\\partial \\operatorname {tr} (\\mathbf {U} )}{\\partial \\mathbf {X} }}+{\\frac {\\partial \\operatorname {tr} (\\mathbf {V} )}{\\partial \\mathbf {X} }}}$ a is not a function of X, U = U(X) ${\\displaystyle {\\frac {\\partial \\operatorname {tr} (a\\mathbf {U} )}{\\partial \\mathbf {X} }}=}$ ${\\displaystyle a{\\frac {\\partial \\operatorname {tr} (\\mathbf {U} )}{\\partial \\mathbf {X} }}}$ g(X) is any polynomial with scalar coefficients, or any matrix function defined by an infinite polynomial series (e.g. eX, sin(X), cos(X), ln(X), etc. using a Taylor series); g(x) is the equivalent scalar function, g(x) is its derivative, and g(X) is the corresponding matrix function ${\\displaystyle {\\frac {\\partial \\operatorname {tr} (\\mathbf {g(X)} )}{\\partial \\mathbf {X} }}=}$ ${\\displaystyle \\mathbf {g} '(\\mathbf {X} )}$ ${\\displaystyle \\left(\\mathbf {g} '(\\mathbf {X} )\\right)^{\\top }}$ A is not a function of X [6] ${\\displaystyle {\\frac {\\partial \\operatorname", "\\cdot (id\\mathbf{l}) \\right)_r$ This can be written as: $\\mathbf{F}_x = - \\frac{\\mathbf{\\hat{r}}}{r^2} \\frac{1}{c^2} \\left( Ii (d\\mathbf{L} \\cdot d\\mathbf{l}) \\right)_x$ $\\mathbf{F}_y = - \\frac{\\mathbf{\\hat{r}}}{r^2} \\frac{1}{c^2} \\left( Ii (d\\mathbf{L} \\cdot d\\mathbf{l}) \\right)_y$ $\\mathbf{F}_r = + \\frac{\\mathbf{\\hat{r}}}{r^2} \\frac{2}{c^2} \\left( Ii (d\\mathbf{L} \\cdot d\\mathbf{l}) \\right)_r$ Or: $\\mathbf{F}_x = - \\frac{\\mathbf{\\hat{r}}}{r^2} \\frac{Ii}{c^2} \\left( d\\mathbf{L} \\cdot d\\mathbf{l} \\right)_x$ $\\mathbf{F}_y = - \\frac{\\mathbf{\\hat{r}}}{r^2} \\frac{Ii}{c^2} \\left( d\\mathbf{L} \\cdot d\\mathbf{l} \\right)_y$ $\\mathbf{F}_r = + \\frac{\\mathbf{\\hat{r}}}{r^2} \\frac{2Ii}{c^2} \\left( d\\mathbf{L} \\cdot d\\mathbf{l} \\right)_r$ Or: $\\mathbf{F}_x = - \\frac{\\mathbf{\\hat{r}}}{r^2} \\frac{Ii}{c^2} \\left( d\\mathbf{L}_x \\cdot d\\mathbf{l}_x \\right)$ $\\mathbf{F}_y = - \\frac{\\mathbf{\\hat{r}}}{r^2} \\frac{Ii}{c^2} \\left( d\\mathbf{L}_y \\cdot d\\mathbf{l}_y \\right)$ $\\mathbf{F}_r = + \\frac{\\mathbf{\\hat{r}}}{r^2} \\frac{2Ii}{c^2} \\left( d\\mathbf{L}_r \\cdot d\\mathbf{l}_r \\right)$ Adding the forces together, their sum is: $\\mathbf{F} = \\frac{\\mathbf{\\hat{r}}}{r^2} \\frac{Ii}{c^2} \\left( 2 d\\mathbf{L}_r \\cdot d\\mathbf{l}_r - d\\mathbf{L}_x \\cdot d\\mathbf{l}_x - d\\mathbf{L}_y \\cdot d\\mathbf{l}_y \\right)$ $\\mathbf{F} = \\frac{\\mathbf{\\hat{r}}}{r^2} \\frac{Ii}{c^2} \\left( 2 d\\mathbf{L}_r \\cdot d\\mathbf{l}_r - d\\mathbf{L}_\\theta \\cdot d\\mathbf{l}_\\theta \\right)$ In S.I. Units: $\\mathbf{F} = \\frac{\\mathbf{\\hat{r}}}{4\\pi\\epsilon_0r^2} \\frac{Ii}{c^2} \\left( 2 d\\mathbf{L}_r \\cdot d\\mathbf{l}_r - d\\mathbf{L}_\\theta \\cdot d\\mathbf{l}_\\theta \\right)$ Or: $\\mathbf{F} = \\frac{\\mu_0 I i \\mathbf{\\hat{r}}}{4\\pi r^2} \\left( 2 d\\mathbf{L}_r \\cdot d\\mathbf{l}_r - d\\mathbf{L}_\\theta \\cdot d\\mathbf{l}_\\theta \\right)$ The second differential is: $d^2\\mathbf{F} = \\frac{\\mu_0 \\mathbf{\\hat{r}}}{4\\pi r^2} \\left( 2 \\mathbf{I}_r \\cdot \\mathbf{i}_r - \\mathbf{I}_\\theta \\cdot \\mathbf{i}_\\theta \\right)$ ### Special Scenario: No relative motion between positive charges In the case that the positive charges $q_+$ and $Q_+$ are essentially stationary,", "$\\frac{z}{w} ;$ d) $\\frac{w}{z}$. Solution. Let $z=1+\\imath=(1,1)$ and $w=x+i y=(x, y)$. Then we have a) $$\\operatorname{Re}(z+w)=0 \\Longleftrightarrow 1+x=0$$ Hence $x=-1$ and $y \\in \\mathbb{R}$ is arbitrary. b) $$\\operatorname{Re}(z \\cdot w)=0 \\Longleftrightarrow \\operatorname{Re}(x-y+i(x+y))=0 .$$ Hence $x-y=0$. Therefore $w$ is given by $w=x+\\imath x=x(1+\\imath)=x \\cdot z$, for arbitrary $x \\in \\mathbb{R}$ c) We have $\\frac{z}{w}=0$. Hence $$\\frac{z}{w}=\\frac{1+\\imath}{x+2 y} \\cdot \\frac{x-v y}{x-v y}=\\frac{x+y+\\imath(x-y)}{x^2+y^2}$$ Therefore $$\\operatorname{Re}\\left(\\frac{z}{w}\\right)=0 \\Longleftrightarrow \\frac{x+y}{x^2+y^2}=0 \\Rightarrow x=-y, x \\neq 0 .$$ Finally we have $w=x(1-2)$ for $x \\in \\mathbb{R}$ and $x \\neq 0$. d) From $\\operatorname{Re}\\left(\\frac{w}{z}\\right)=0$ it follows that for every $x \\in \\mathbb{R}$ : $$\\frac{w}{z}=\\frac{x+\\imath y}{1+\\imath} \\cdot \\frac{1-\\imath}{1-\\imath}=\\frac{x+y+\\imath(x-y)}{2} .$$ Hence $w=x(1-\\imath)$ for every $x \\in \\mathbb{R}$. Let $\\mathbb{C}^$ be the set of all complex numbers different from zero. a) Prove that the set $T$ of all complex numbers with modulus 1 is a multiplicative subgroup of the group $\\left(\\mathbb{C}^, \\cdot\\right)$. b) The multiplicative group $\\mathbb{C}^*$ is isomorphic with $\\mathbb{R}^{+} \\times T$. Solution. a) We have $T={z|| z \\mid=1} \\subset \\mathbb{C}^* . T$ under the multiplication by the equality $$\\left|z_1 \\cdot z_2\\right|=\\left|z_1\\right| \\cdot\\left|z_2\\right| \\text {. }$$ The associativity follows by the associativity of multiplication in $\\mathbb{C}$. The neutral element is 1. The inverse of $z=(a, b) \\in T$ is $$\\left(\\frac{a}{a^2+b^2}, \\frac{-b}{a^2+b^2}\\right) \\in T \\text {. }$$ b) It is easy to check that an isomorphism is given by $$f(z)=(|z|,", "${\\displaystyle \\mathbb {Q} _{T}(\\operatorname {d} \\!x_{1:T})=\\mu _{1}(\\operatorname {d} \\!x_{1})\\,\\prod _{t=2}^{T}\\kappa _{t}(x_{t-1},\\operatorname {d} \\!x_{t})\\qquad \\qquad R_{T}(x_{1:T})=\\phi _{1}(x_{1})\\,\\prod _{t=2}^{T}\\phi _{t}(x_{t-1},x_{t})}$ ${\\displaystyle \\mathbb {P} _{T}(\\operatorname {d} \\!x_{1:T})={\\frac {R_{T}(x_{1:T})\\,\\mathbb {Q} _{T}(\\operatorname {d} \\!x_{1:T})}{\\mathbb {E} _{{\\hat {x}}_{1:T}\\sim \\mathbb {Q} _{T}}\\left[R_{T}({\\hat {x}}_{1:T})\\right]}}={\\frac {(\\mu _{1}\\!\\cdot \\!\\phi _{1})(\\operatorname {d} \\!x_{1})\\,\\prod _{t=2}^{T}(\\kappa _{t}\\!\\cdot \\!\\phi _{t})(x_{t-1},\\operatorname {d} \\!x_{t})}{\\int _{X^{T}}(\\mu _{1}\\!\\cdot \\!\\phi _{1})(\\operatorname {d} \\!{\\hat {x}}_{1})\\,\\prod _{t=2}^{T}(\\kappa _{t}\\!\\cdot \\!\\phi _{t})({\\hat {x}}_{t-1},\\operatorname {d} \\!{\\hat {x}}_{t})}}\\;.}$ The usual goal of Feynman-Kac models is then to measure quantities of interest ${\\textstyle f}$ along the flow of corresponding time marginal distributions: ${\\displaystyle \\eta _{t}\\!\\left[\\,f\\,\\right]={\\frac {\\gamma _{t}\\!\\left[\\,f\\,\\right]}{Z_{t}}}={\\frac {\\gamma _{t}\\!\\left[\\,f\\,\\right]}{\\gamma _{t}\\!\\left[\\,1\\,\\right]}}={\\frac {\\mathbb {E} _{x_{1:t}\\sim \\mathbb {Q} _{t}}\\left[\\,R_{t}(x_{1:t})\\,f(x_{t})\\,\\right]}{\\mathbb {E} _{x_{1:t}\\sim \\mathbb {Q} _{t}}\\left[\\,R_{t}(x_{1:t})\\,\\right]}}\\;.}$ In particular, all quantities above – including target densities, importance weights, resampling and rejuvenation strategies, twisting functions etc. – can be understood as particular choices for approximating the quantities ${\\textstyle \\mu _{1}}$, ${\\textstyle \\left(\\kappa _{t}\\right)_{t=2:T}}$ and ${\\textstyle \\left(\\phi _{t}\\right)_{t=1:T}}$in terms of particle populations (discrete measures). This broad class of models also encompasses evolutionary optimization methods, which in turn constitute a highly diverse and active field of study. For example, in the case of genetic algorithms, the transition kernels ${\\textstyle \\kappa _{t}}$ are called recombination and mutation operators, while the potentials ${\\displaystyle \\phi _{t}}$ are fitness functions. From this abstract perspective, the explanation of twisted SMC is very simple: The target densities ${\\displaystyle \\gamma", "= \\left[\\prod_{\\mathfrak{p} \\in S} \\mathfrak{p}\\right] \\cdot \\left[\\prod_{\\mathfrak{p} \\in \\Sigma_E \\setminus S'} \\mathfrak{p}^{e_{\\mathfrak{p}}}\\right]^2 \\cdot \\theta',$$ where $$\\theta' \\in L^\\times$$ is a product of primes in $$S'$$. (In the first displayed product above, I'm isolating exactly one copy of each prime in $$S$$, and the second displayed product above contains all remaining prime factors not from $$S'$$). Observe that the existence of the $$r_{p}$$'s implies that $$\\prod_{\\mathfrak{p} \\in S} \\mathfrak{p} = \\prod_{p \\in T} p$$. Letting $$r = \\operatorname{N}_{L/K}(\\theta)$$, we have that $$r = \\left[\\prod_{p \\in T} \\operatorname{N}_{L/K}(p)\\right] \\cdot \\left[\\operatorname{N}_{L/K}\\left(\\prod_{\\mathfrak{p} \\in \\Sigma_E \\setminus S'} \\mathfrak{p}^{e_{\\mathfrak{p}}}\\right)\\right]^2 \\cdot \\operatorname{N}_{L/K}(\\theta'),$$ so since $$\\operatorname{N}_{L/K}(p) = p^{[L: K]}$$, we deduce that $$r \\cdot \\theta = \\left[\\prod_{p \\in T} p^{[L : K] + 1}\\right] \\cdot \\left[\\left(\\prod_{\\mathfrak{p} \\in \\Sigma_E \\setminus S'} \\mathfrak{p}^{e_{\\mathfrak{p}}}\\right) \\cdot \\operatorname{N}_{L/K}\\left(\\prod_{\\mathfrak{p} \\in \\Sigma_E \\setminus S'} \\mathfrak{p}^{e_{\\mathfrak{p}}}\\right)\\right]^2 \\cdot \\operatorname{N}_{L/K}(\\theta')$$ It is evident from the above expansion that $$r \\cdot \\theta$$ has the desired property. • Suppose that $$L \\neq K$$ and $$[L : K]$$ is even. Then the answer to the question is not necessarily. For a counterexample, let $$K = \\mathbb{Q}(\\sqrt{-5})$$, and recall that the class group $$\\operatorname{Cl}(K)$$ of $$K$$ is isomorphic to $$\\mathbb{Z}/2\\mathbb{Z}$$ and is generated by the prime ideal $$(1 + \\sqrt{-5}, 1 - \\sqrt{-5})$$. Let $$L$$ be the Hilbert class field (maximal unramified abelian extension) of $$K$$, which is a quadratic", "number}{}{.} \\aufzaehlungfuenf{$z= \\operatorname{Re} \\, { \\left( z \\right) } + \\operatorname{Im} \\, { \\left( z \\right) } { \\mathrm i}$. }{$\\operatorname{Re} \\, { \\left( z+w \\right) } = \\operatorname{Re} \\, { \\left( z \\right) } + \\operatorname{Re} \\, { \\left( w \\right) }$. }{$\\operatorname{Im} \\, { \\left( z+w \\right) } = \\operatorname{Im} \\, { \\left( z \\right) } + \\operatorname{Im} \\, { \\left( w \\right) }$. }{For $r \\in \\R$ we have \\mathdisp {\\operatorname{Re} \\, { \\left( rz \\right) } =r \\operatorname{Re} \\, { \\left( z \\right) } \\text{ and } \\operatorname{Im} \\, { \\left( rz \\right) } =r \\operatorname{Im} \\, { \\left( z \\right) }} { . } }{ $z = \\operatorname{Re} \\, { \\left( z \\right) }$ if and only if $z \\in \\R$, and this is exactly the case when $\\operatorname{Im} \\, { \\left( z \\right) }=0$. } } {} {} \\inputexercise {} { Prove the following calculating rules for the \\definitionsverweis {complex numbers}{}{.} \\aufzaehlungfuenf{$\\betrag { z }= \\sqrt{ z \\ \\overline{ z } }$. }{$\\operatorname{Re} \\, { \\left( z \\right) } = \\frac{z+ \\overline{ z } }{2}$. }{$\\operatorname{Im} \\, { \\left( z \\right) } = \\frac{z - \\overline{ z } }{2 { \\mathrm i} }$. }{$\\overline{ z }= \\operatorname{Re} \\, { \\left( z \\right) } - { \\mathrm i} \\operatorname{Im} \\, { \\left( z", "{x} _{i}^{\\top }{\\bigg )}^{-1}{\\bigg (}{\\frac {1}{n))\\sum _{i}\\mathbf {x} _{i}\\mathbf {x} _{i}^{\\top }{\\widehat {\\varepsilon ))_{i}^{2}{\\bigg )}{\\bigg (}{\\frac {1}{n))\\sum _{i}\\mathbf {x} _{i}\\mathbf {x} _{i}^{\\top }{\\bigg )}^{-1}\\\\&=n(\\mathbf {X} ^{\\top }\\mathbf {X} )^{-1}(\\mathbf {X} ^{\\top }\\operatorname {diag} ({\\widehat {\\varepsilon ))_{1}^{2},\\ldots ,{\\widehat {\\varepsilon ))_{n}^{2})\\mathbf {X} )(\\mathbf {X} ^{\\top }\\mathbf {X} )^{-1}\\end{aligned))} Thus, ${\\displaystyle {\\widehat {\\mathbf {\\Omega } ))_{n}=n\\cdot {\\hat {\\mathbb {V} ))_{\\text{HCE))[{\\widehat {\\boldsymbol {\\beta ))}_{\\text{OLS))]}$ and ${\\displaystyle {\\widehat {\\mathbb {V} ))[\\mathbf {X} {\\boldsymbol {\\varepsilon ))]={\\frac {1}{n))\\sum _{i}\\mathbf {x} _{i}\\mathbf {x} _{i}^{\\top }{\\widehat {\\varepsilon ))_{i}^{2}={\\frac {1}{n))\\mathbf {X} ^{\\top }\\operatorname {diag} ({\\widehat {\\varepsilon ))_{1}^{2},\\ldots ,{\\widehat {\\varepsilon ))_{n}^{2})\\mathbf {X} .}$ Precisely which covariance matrix is of concern is a matter of context. Alternative estimators have been proposed in MacKinnon & White (1985) that correct for unequal variances of regression residuals due to different leverage.[11] Unlike the asymptotic White's estimator, their estimators are unbiased when the data are homoscedastic. Of the four widely available different options, often denoted as HC0-HC3, the HC3 specification appears to work best, with tests relying on the HC3 estimator featuring better power and closer proximity to the targetted size, especially in small samples. The larger the sample, the smaller the difference between the different estimators.[12] An alternative to explicitly modelling the heteroskedasticity is using a resampling method such as the Wild Bootstrap. Given that the studentized Bootstrap, which standardizes the resampled statistic", "operator in spin space Posted by peeterjoot on October 31, 2011 [Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)] # Disclaimer. Peeter’s lecture notes from class. May not be entirely coherent. # Rotation operator in spin space. We can formally expand our rotation operator in Taylor series \\begin{aligned}e^{-i \\theta \\hat{\\mathbf{n}} \\cdot \\mathbf{S}/\\hbar}= I +\\left(-i \\theta \\hat{\\mathbf{n}} \\cdot \\mathbf{S}/\\hbar\\right)+\\frac{1}{{2!}}\\left(-i \\theta \\hat{\\mathbf{n}} \\cdot \\mathbf{S}/\\hbar\\right)^2+\\frac{1}{{3!}}\\left(-i \\theta \\hat{\\mathbf{n}} \\cdot \\mathbf{S}/\\hbar\\right)^3+ \\cdots\\end{aligned} \\hspace{\\stretch{1}}(2.1) or \\begin{aligned}e^{-i \\theta \\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma}/2}&= I +\\left(-i \\theta \\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma}/2\\right)+\\frac{1}{{2!}}\\left(-i \\theta \\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma}/2\\right)^2+\\frac{1}{{3!}}\\left(-i \\theta \\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma}/2\\right)^3+ \\cdots \\\\ &=\\sigma_0 +\\left(\\frac{-i \\theta}{2}\\right) (\\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma})+\\frac{1}{{2!}} \\left(\\frac{-i \\theta}{2}\\right) (\\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma})^2+\\frac{1}{{3!}} \\left(\\frac{-i \\theta}{2}\\right) (\\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma})^3+ \\cdots \\\\ &=\\sigma_0 +\\left(\\frac{-i \\theta}{2}\\right) (\\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma})+\\frac{1}{{2!}} \\left(\\frac{-i \\theta}{2}\\right) \\sigma_0+\\frac{1}{{3!}} \\left(\\frac{-i \\theta}{2}\\right) (\\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma}) + \\cdots \\\\ &=\\sigma_0 \\left( 1 - \\frac{1}{{2!}}\\left(\\frac{\\theta}{2}\\right)^2 + \\cdots \\right) +(\\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma}) \\left( \\frac{\\theta}{2} - \\frac{1}{{3!}}\\left(\\frac{\\theta}{2}\\right)^3 + \\cdots \\right) \\\\ &=\\cos(\\theta/2) \\sigma_0 + \\sin(\\theta/2) (\\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma})\\end{aligned} where we’ve used the fact that $(\\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma})^2 = \\sigma_0$. So our representation of the spin operator is \\begin{aligned}\\begin{aligned}e^{-i \\theta \\hat{\\mathbf{n}} \\cdot \\mathbf{S}/\\hbar} &\\rightarrow \\cos(\\theta/2) \\sigma_0 + \\sin(\\theta/2) (\\hat{\\mathbf{n}} \\cdot \\boldsymbol{\\sigma}) \\\\ &=\\cos(\\theta/2) \\sigma_0 + \\sin(\\theta/2) \\left(n_x \\begin{bmatrix} 0 & 1 \\\\", "+ s_x \\mathbf e_{4235} + s_y \\mathbf e_{4315} + s_z \\mathbf e_{4125} + s_w \\mathbf e_{3215}$$ $$\\begin{split}\\operatorname{par}(\\mathbf s) = &-s_u^3 \\mathbf e_{1234} - s_xs_u^2 \\mathbf e_{4235} - s_ys_u^2 \\mathbf e_{4315} - s_zs_u^2 \\mathbf e_{4125} + (s_ws_u - s_x^2 - s_y^2 - s_z^2)\\,s_u \\mathbf e_{3215}\\end{split}$$", "now drop $$\\operatorname{Pr}(x_i)$$ because we will be optimizing with respect to $$\\mathbf{w}$$, and, in that case, the term $$\\operatorname{Pr}(x_i)$$ is considered a constant (and the derivative of a constant is zero). In other words, our objective is \\begin{align} \\mathbf{w}^* &= \\operatorname{argmax}_{\\mathbf{w}} \\log \\operatorname{Pr}(\\mathcal{D} \\mid \\mathbf{w}) \\\\ &=\\operatorname{argmax}_{\\mathbf{w}} \\log \\left( \\prod_{i=1}^{|\\mathcal{D}|}\\operatorname{Pr}(y_i \\mid x_i, \\mathbf{w}) \\operatorname{Pr}(x_i) \\right) \\\\ &=\\operatorname{argmax}_{\\mathbf{w}} \\sum_{i=1}^{|\\mathcal{D}|} \\log \\operatorname{Pr}(y_i \\mid x_i, \\mathbf{w}) \\operatorname{Pr}(x_i) \\\\ &=\\operatorname{argmax}_{\\mathbf{w}} \\sum_{i=1}^{|\\mathcal{D}|} \\log \\operatorname{Pr}(y_i \\mid x_i, \\mathbf{w}) + \\log \\operatorname{Pr}(x_i) \\\\ &=\\operatorname{argmax}_{\\mathbf{w}} \\sum_{i=1}^{|\\mathcal{D}|} \\log \\operatorname{Pr}(y_i \\mid x_i, \\mathbf{w}) + \\sum_{i=1}^{|\\mathcal{D}|} \\log \\operatorname{Pr}(x_i) \\\\ &=\\operatorname{argmax}_{\\mathbf{w}} \\sum_{i=1}^{|\\mathcal{D}|} \\log \\operatorname{Pr}(y_i \\mid x_i, \\mathbf{w}) \\end{align} Obviously, $$\\sum_{i=1}^{|\\mathcal{D}|} \\log \\operatorname{Pr}(x_i)$$ is a constant with respect to $${\\mathbf{w}}$$, so it can be dropped. The equation in Alex Graves' paper is thus not generally true, but, if that is our objective (which is the case in Graves' paper), for the purpose of optimization, we can drop $$\\operatorname{Pr}(x_i)$$. Neal's PhD thesis also mentions this likelihood in section 1.3.1. (p. 15)." ]

[ "$\\mathbb{u}$ onto $\\mathbb{v}$ is the vector which has magnitude given by the scalar projection $\\text{comp}_{\\mathbb{u}}(\\mathbb{v})$ and points in in the direction of $\\mathbb{v}$, which is given by the unit vector $\\frac{\\mathbb{v}}{|\\mathbb{v}|}$ associated to $\\mathbb{v}$. Hence we have $$\\text{proj}_{\\mathbf{v}}(\\mathbf{u}) = \\text{comp}_{\\mathbb{v}}(\\mathbb{u})\\frac{\\mathbb{v}}{|\\mathbb{v}|}= \\left(\\frac{\\mathbb{u}\\cdot\\mathbb{v}}{|\\mathbb{v}|^2}\\right)\\mathbb{v}.$$ I hope that this helps! - @ Jennifer Dylan: Thanks for the edit, how did you get proj and comp in plain text in math mode? – John Martin Aug 27 '12 at 3:35 1. There's no space between @ and my name. Otherwise, I don't get notified :) 2. Inside math mode, you can surround the plain text with either \\mathrm{..} which ignores spaces and changes the font into Roman font, OR bettter use \\text{..} which respects the text. For operators and such, you should use \\operatorname{..}, which I should I've done, but \\text{} is shorter to type! – user2468 Aug 27 '12 at 3:40", "+0 # help 0 91 1 Given vectors $$\\mathbf{v}$$ and $$\\mathbf{w}$$ such that $$\\|\\mathbf{v}\\| = 3,\\|\\mathbf{w}\\| = 7,$$ and $$\\mathbf{v} \\cdot \\mathbf{w} = 10,$$ then find $$\\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\|.$$ Sep 3, 2019 #1 +23357 +1 Given vectors $$\\mathbf{v}$$ and $$\\mathbf{w}$$ such that $$\\|\\mathbf{v}\\| = 3,\\ \\|\\mathbf{w}\\| = 7,$$ and $$\\mathbf{v} \\cdot \\mathbf{w} = 10,$$ then find $$\\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\|.$$ $$\\begin{array}{|rcll|} \\hline \\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\| &=& \\left\\| \\dfrac{\\mathbf{v} \\cdot \\mathbf{w}}{\\left\\|\\mathbf{w}\\right\\|^2}\\mathbf{w} \\right\\| \\\\\\\\ &=& \\left\\| \\dfrac{10}{7^2}\\mathbf{w} \\right\\| \\\\\\\\ &=& \\dfrac{10}{7^2} \\| \\mathbf{w} \\| \\\\\\\\ &=& \\dfrac{10}{7^2} \\cdot 7 \\\\\\\\ \\|\\operatorname{proj}_{\\mathbf{w}} \\mathbf{v}\\| &=& \\mathbf{\\dfrac{10}{7}} \\\\ \\hline \\end{array}$$ Sep 4, 2019", "close to $1$ we conclude that $\\operatorname{p}(\\mathbf{u}+\\mathbf{v})\\leq\\operatorname{p}(\\mathbf{v}).$ The same argument also shows that $\\operatorname{p}(\\mathbf{v})=\\operatorname{p}(-\\mathbf{u}+(\\mathbf{u}+\\mathbf{% v}))\\leq\\operatorname{p}(\\mathbf{u}+\\mathbf{v}),$ and hence $\\operatorname{p}(\\mathbf{u}+\\mathbf{v})=\\operatorname{p}(\\mathbf{v}),$ as desired. Finally, suppose that neither $\\operatorname{p}(\\mathbf{u})$ nor $\\operatorname{p}(\\mathbf{v})$ is zero. Hence, $\\frac{\\mathbf{u}}{\\operatorname{p}(u)},\\,\\frac{\\mathbf{v}}{\\operatorname{p}(v)}$ are both in $B_{1}$, and hence $\\frac{\\operatorname{p}(u)}{\\operatorname{p}(u)+\\operatorname{p}(v)}\\frac{% \\mathbf{u}}{\\operatorname{p}(u)}+\\frac{\\operatorname{p}(v)}{\\operatorname{p}(u% )+\\operatorname{p}(v)}\\frac{\\mathbf{v}}{\\operatorname{p}(v)}=\\frac{\\mathbf{u}+% \\mathbf{v}}{\\operatorname{p}(u)+\\operatorname{p}(v)}$ is in $B_{1}$ also. Using homogeneity, we conclude that $\\operatorname{p}(u+v)\\leq\\operatorname{p}(u)+\\operatorname{p}(v),$ as desired. Title seminorm Seminorm 2013-03-22 12:24:57 2013-03-22 12:24:57 rmilson (146) rmilson (146) 20 rmilson (146) Definition msc 46B20 semi-norm homogeneous", "by ${t_{k}}_{(i)} = \\mathbf{x}_{(i)} \\cdot \\mathbf{w}_{(k)}$ in such a way that the individual variables of t considered over the data set successively inherit the maximum possible variance from x, with each loading vector w constrained to be a unit vector. ### First component Edit $\\mathbf{w}_{(1)} = \\underset{\\Vert \\mathbf{w} \\Vert = 1}{\\operatorname{\\arg\\,max}}\\,\\{ \\sum_i \\left(t_{1}\\right)^2_{(i)} \\} = \\underset{\\Vert \\mathbf{w} \\Vert = 1}{\\operatorname{\\arg\\,max}}\\, \\sum_i \\left(\\mathbf{x}_{(i)} \\cdot \\mathbf{w} \\right)^2$ Equivalently, writing this in matrix form gives $\\mathbf{w}_{(1)} = \\underset{\\Vert \\mathbf{w} \\Vert = 1}{\\operatorname{\\arg\\,max}}\\, \\{ \\Vert \\mathbf{Xw} \\Vert^2 \\} = \\underset{\\Vert \\mathbf{w} \\Vert = 1}{\\operatorname{\\arg\\,max}}\\, \\{ \\mathbf{w}^T \\mathbf{X}^T \\mathbf{X w} \\}$ Since w(1) has been defined to be a unit vector, it equally must also satisfy $\\mathbf{w}_{(1)} = \\underset{\\Vert \\mathbf{w} \\Vert = 1}{\\operatorname{\\arg\\,max}}\\, \\left\\{ \\frac{\\mathbf{w}^T\\mathbf{X}^T \\mathbf{X w}}{\\mathbf{w}^T \\mathbf{w}} \\right\\}$ The quantity to be maximised can be recognised as a Rayleigh quotient. A standard result for a symmetric matrix such as XTX is that the quotient's maximum possible value is the largest eigenvalue of the matrix, which occurs when w is the corresponding eigenvector. With w(1) found, the first component of a data vector x(i) can then be given as a score t1(i) = x(i)w(1) in the transformed co-ordinates, or as the corresponding vector in the original variables, {x(i)w(1)} w(1). ### Further components Edit The kth component can be found by subtracting the first", "&= (1,y,y^{2},\\ldots,y^{n-1}) \\end{aligned} for the vector whose $$i$$-th entry is $$y^{i}$$. For vectors $${\\mathbf{v}}$$ of even length $$2k$$, we define $${\\mathbf{v}}_{\\operatorname{lo}}$$ and $${\\mathbf{v}}_{\\operatorname{hi}}$$ to be the low and high halves of $${\\mathbf{v}}$$: \\begin{aligned} {\\mathbf{v}}_{\\operatorname{lo}} &= (v_0, \\ldots, v_{k-1})\\\\ {\\mathbf{v}}_{\\operatorname{hi}} &= (v_{k}, \\ldots, v_{2k-1}) \\end{aligned} Pedersen commitments are written as \\begin{aligned} \\operatorname{Com}(v) &= \\operatorname{Com}(v, {\\widetilde{v}}) = v \\cdot B + {\\widetilde{v}} \\cdot {\\widetilde{B}}, \\end{aligned} where $$B$$ and $${\\widetilde{B}}$$ are the generators used for the values and blinding factors, respectively. We denote the blinding factor for the value $$v$$ by $${\\widetilde{v}}$$, so that it is clear which blinding factor corresponds to which value, and write $$\\operatorname{Com}(v)$$ instead of $$\\operatorname{Com}(v, {\\widetilde{v}})$$ for brevity. We also make use of vector Pedersen commitments, which we define for pairs of vectors as \\begin{aligned} \\operatorname{Com}({\\mathbf{a}}_{L}, {\\mathbf{a}}_{R}) &= \\operatorname{Com}({\\mathbf{a}}_{L}, {\\mathbf{a}}_{R}, {\\widetilde{a}}) = {\\langle {\\mathbf{a}}_{L}, {\\mathbf{G}} \\rangle} + {\\langle {\\mathbf{a}}_{R}, {\\mathbf{H}} \\rangle} + {\\widetilde{a}} {\\widetilde{B}},\\end{aligned} where $${\\mathbf{G}}$$ and $${\\mathbf{H}}$$ are vectors of generators. Notice that this is exactly the same as taking a commitment to the vector of values $${\\mathbf{a}}_{L} \\Vert {\\mathbf{a}}_{R}$$ with the vector of bases $${\\mathbf{G}} \\Vert {\\mathbf{H}}$$, but defining the commitment on pairs of vectors is a more convenient notation. The variable renaming is as follows: \\begin{aligned} g &\\xrightarrow{} B & \\gamma &\\xrightarrow{} \\tilde{v} \\\\ h &\\xrightarrow{} \\widetilde{B} & \\alpha &\\xrightarrow{} \\tilde{a} \\\\ {\\mathbf{g}}", "of r is the <scalar product> between the del operator and the r. So $$\\mbox{Div} {}\\mathbf{r} = (\\mathbf{e}_i \\partial_i)\\cdot (x_j \\mathbf{e}_j)$$ So complete the calculation. Along the same lines you'll solve the 2nd point as well. So how do I evaluate $(\\mathbf{e}_i \\partial_i)\\cdot (x_j \\mathbf{e}_j)$ ? Is it just $\\mathbf{e}_1 \\partial_1 x_1\\mathbf{e}_1 + \\mathbf{e}_2 \\partial_2 x_2\\mathbf{e}_2 + \\mathbf{e}_3 \\partial_3 x_3\\mathbf{e}_3$ (this doesn't look right at all) hunt_mat Homework Helper To start: $$\\mathbf{r}=x^{i}\\mathbf{e}_{i}$$ and div is: $$\\nabla\\cdot =\\partial_{i}(e_{i}\\cdot )$$ so: $$\\nabla\\cdot\\mathbf{r}=\\sum_{i=1}^{3}\\partial_{i}(\\mathbf{e}_{i}\\cdot (x^{i}\\mathbf{e}_{i}))=...$$ To start: $$\\mathbf{r}=x^{i}\\mathbf{e}_{i}$$ and div is: $$\\nabla\\cdot =\\partial_{i}(e_{i}\\cdot )$$ so: $$\\nabla\\cdot\\mathbf{r}=\\sum_{i=1}^{3}\\partial_{i}(\\mathbf{e}_{i}\\cdot (x^{i}\\mathbf{e}_{i}))=...$$ Where are these $\\mathbf{e}$ vectors coming from? In all my solutions to these questions on index notation I never see an $\\mathbf{e}$ appearing. A similar question to the first one is Show $\\nabla (\\mathbf{a} \\cdot \\mathbf{r} ) = \\mathbf{a}\\;,\\;\\mathbf{a}\\in\\mathbb{R}^3$ and the solution is $\\partial_j a_k x_k = a_k \\partial_j x_k = a_k \\delta_{jk} = a_j$ Last edited: dextercioby Homework Helper So how do I evaluate $(\\mathbf{e}_i \\partial_i)\\cdot (x_j \\mathbf{e}_j)$ ? Is it just $\\mathbf{e}_1 \\partial_1 x_1\\mathbf{e}_1 + \\mathbf{e}_2 \\partial_2 x_2\\mathbf{e}_2 + \\mathbf{e}_3 \\partial_3 x_3\\mathbf{e}_3$ (this doesn't look right at all) It's correct, because $\\mathbf{e}_i \\cdot \\mathbf{e}_j = \\delta_{ij}$, so you get what you wrote. But of course $\\partial_1 x_1 =1$ and as well for the other 2 components. So it's not difficult to reach the", "= (a \\cdot e_1) e_1 + (a \\cdot e_2 - a \\cdot e_3)(e_1 + e_2 - e_3)$$ There is (so far) nothing inherently GA-like to expressing a linear operator this way, but it is a bit more amenable to some of the operations you might be asked to perform that come from GA. Now for my tensor product question using Muphrid's response as a template. If we had a tensor with elements from a tensor product of two 2x2 tensors: $$\\\\e_{ij} \\otimes f_{kl}$$ then the tensor product would be: \\begin{vmatrix} \\mathbf{e_{11}f_{11}} & \\mathbf{e_{11}f_{21}} & \\mathbf{e_{21}f_{11}} & \\mathbf{e_{21}f_{21}}\\\\ \\mathbf{e_{11}f_{12}} & \\mathbf{e_{11}f_{22}} & \\mathbf{e_{21}f_{12}} & \\mathbf{e_{21}f_{22}}\\\\ \\mathbf{e_{12}f_{11}} & \\mathbf{e_{12}f_{21}} & \\mathbf{e_{22}f_{11}} & \\mathbf{e_{22}f_{21}}\\\\ \\mathbf{e_{12}f_{12}} & \\mathbf{e_{12}f_{22}} & \\mathbf{e_{22}f_{12}} & \\mathbf{e_{22}f_{22}}\\\\ \\end{vmatrix} Now how would I express this in terms of Geometric Algebra? Would I use the same process that Muphrid showed like this: Let your operator be $\\underline T$. It could be described by \\begin{align*} \\underline T(e_?) &= e_{11}f_{11} + e_{11}f_{12} + e_{12}f_{11} + e_{12}f_{12}\\\\ \\underline T(e_??) &= e_{11}f_{21} + e_{11}f_{22} + e_{12}f_{21} + e_{12}f_{22} \\\\ \\underline T(e_??) &= e_{21}f_{11} + e_{21}f_{12} + e_{22}f_{11} + e_{22}f_{12} \\\\ \\underline T(e_????) &= e_{21}f_{21} + e_{21}f_{22} + e_{22}f_{21} + e_{22}f_{22}\\end{align*} Continuing Murphrid's process, You could instead use dot products to combine this into a single expression. Let $a$ be an arbitrary vector, and", "of magnitudes on both sides of $\\,z_1 (1+|z_2|)=iz_2 (1+|z_1|)\\,$: $$\\require{cancel} |z_1|^2(1+|z_2|)^2=|z_2|^2(1+|z_1|)^2 \\\\ |z_1|^2(1 + 2 |z_2| + \\bcancel{|z_2|^2})=|z_2|^2(1+2|z_1|+\\bcancel{|z_1|^2}) \\\\ (|z_1|-|z_2|)(|z_1|+|z_2|+2 |z_1||z_2|)=0$$ The latter gives $|z_1|=|z_2|\\,$, so the given relation simplifies to $z_1=i z_2\\,$. Then $|z_2|z_1 +|z_1|\\bar z_2 = |z_2|(i z_2 + \\bar z_2)=|z_2|\\cdot \\left(\\operatorname{Re}(z_2)-\\operatorname{Im}(z_2)\\right)\\cdot(1+i)\\,$.", "the experiment, and each of the p columns gives a particular kind of datum (say, the results from a particular probe). Mathematically, the transformation is defined by a set of p-dimensional vectors of weights or loadings $\\mathbf\\left\\{w\\right\\}_\\left\\{\\left(k\\right)\\right\\} = \\left(w_1, \\dots, w_p\\right)_\\left\\{\\left(k\\right)\\right\\}$ that map each row vector $\\mathbf\\left\\{x\\right\\}_\\left\\{\\left(i\\right)\\right\\}$ of X to a new vector of principal component scores $\\mathbf\\left\\{t\\right\\}_\\left\\{\\left(i\\right)\\right\\} = \\left(t_1, \\dots, t_p\\right)_\\left\\{\\left(i\\right)\\right\\}$, given by $\\left\\{t_\\left\\{k\\right\\}\\right\\}_\\left\\{\\left(i\\right)\\right\\} = \\mathbf\\left\\{x\\right\\}_\\left\\{\\left(i\\right)\\right\\} \\cdot \\mathbf\\left\\{w\\right\\}_\\left\\{\\left(k\\right)\\right\\}$ in such a way that the individual variables of t considered over the data set successively inherit the maximum possible variance from x, with each loading vector w constrained to be a unit vector. ### First component $\\mathbf\\left\\{w\\right\\}_\\left\\{\\left(1\\right)\\right\\}$ = \\underset{\\Vert \\mathbf{w} \\Vert = 1}{\\operatorname{\\arg\\,max}}\\,\\{ \\sum_i \\left(t_{1}\\right)^2_{(i)} \\} = \\underset{\\Vert \\mathbf{w} \\Vert = 1}{\\operatorname{\\arg\\,max}}\\, \\sum_i \\left(\\mathbf{x}_{(i)} \\cdot \\mathbf{w} \\right)^2 Equivalently, writing this in matrix form gives $\\mathbf\\left\\{w\\right\\}_\\left\\{\\left(1\\right)\\right\\}$ = \\underset{\\Vert \\mathbf{w} \\Vert = 1}{\\operatorname{\\arg\\,max}}\\, \\{ \\Vert \\mathbf{Xw} \\Vert^2 \\} = \\underset{\\Vert \\mathbf{w} \\Vert = 1}{\\operatorname{\\arg\\,max}}\\, \\{ \\mathbf{w}^T \\mathbf{X}^T \\mathbf{X w} \\} Since w(1) has been defined to be a unit vector, it equivalently also satisfies $\\mathbf\\left\\{w\\right\\}_\\left\\{\\left(1\\right)\\right\\}$ = \\underset{\\Vert \\mathbf{w} \\Vert = 1}{\\operatorname{\\arg\\,max}}\\, \\left\\{ \\frac{\\mathbf{w}^T\\mathbf{X}^T \\mathbf{X w}}{\\mathbf{w}^T \\mathbf{w}} \\right\\} The quantity to be maximised can be recognised as a Rayleigh quotient. A standard result for a symmetric matrix such as XTX is that the quotient's maximum possible value is the largest eigenvalue", "& 0 \\\\ 0 & 0 & 0 & 0 & w_{0, 0} & w_{0, 1} & w_{0, 2} & 0 & w_{1, 0} & w_{1, 1} & w_{1, 2} & 0 & w_{2, 0} & w_{2, 1} & w_{2, 2} & 0 \\\\ 0 & 0 & 0 & 0 & 0 & w_{0, 0} & w_{0, 1} & w_{0, 2} & 0 & w_{1, 0} & w_{1, 1} & w_{1, 2} & 0 & w_{2, 0} & w_{2, 1} & w_{2, 2} \\end{bmatrix} }$$ Similarly, we reshape the input $$\\mathbf{I}$$ as a 16-dimensional vector $$\\mathbf{I}' \\in \\mathbb{R}^{16}$$. $${\\scriptstyle \\mathbf{I}' = \\begin{bmatrix} i_{0, 0} & i_{0, 1} & i_{0, 2} & i_{0, 3} & i_{1, 0} & i_{1, 1} & i_{1, 2} & i_{1, 3} & i_{2, 0} & i_{2, 1} & i_{2, 2} & i_{2, 3} & i_{3, 0} & i_{3, 1} & i_{3, 2} & i_{3, 3} \\end{bmatrix}^T }$$ Then the convolution of $$\\mathbf{W}$$ and $$\\mathbf{I}$$, that is $$\\mathbf{W} \\circledast \\mathbf{I} = \\mathbf{O} \\in \\mathbb{R}^{2 \\times 2},$$ where $$\\circledast$$ is the convolution operator, is equivalently defined as $$\\mathbf{W}' \\cdot \\mathbf{I}' = \\mathbf{O}' \\in \\mathbb{R}^{4},$$ where $$\\cdot$$ is the matrix-vector multiplication operator. The produced vector $$\\mathbf{O}'$$ can then be reshaped as a $$2 \\times 2$$ feature map. You can easily verify that this representation is correct", "of a vector $\\mathbf F$ is: $\\int_{\\mathcal P} \\mathbf F \\cdot d\\mathbf r = \\int_{\\mathcal P} \\sum F_i \\mathbf e^i \\cdot \\sum \\mathbf e_i dq^i = \\sum \\int_{\\mathcal P} F_i dq^i$ An infinitesimal element of area for a surface described by holding one coordinate $q_k$ constant is: $dA = \\prod_{i \\neq k} ds_i = \\prod_{i \\neq k} h_i dq^i\\,$ Similarly, the volume element is: $dV = \\prod ds_i = \\prod h_i dq^i$ where the large uppercase Pi symbol indicates a product the same way that a large Sigma indicates summation. Note that the product of all the scale factors is the Jacobian determinant. As an example, the surface integral of a vector function $\\mathbf F$ over a $q^1 = constant$ surface $\\mathcal S$ in 3D is: $\\int_{\\mathcal S} \\mathbf F \\cdot d\\mathbf A = \\int_{\\mathcal S} \\mathbf F \\cdot \\hat \\mathbf n \\ d A = \\int_{\\mathcal S} \\mathbf F \\cdot \\hat \\mathbf e_1 \\ d A = \\int_{\\mathcal S} F^1 \\frac{h_2 h_3}{h_1} dq^2 dq^3$ Note that $\\frac{F^1}{h_1}$ is the component of $\\mathbf F$ normal to the surface. ## Differential operators in three dimensions Since these operations are common in application, all vector components in this section are presented with respect to the normalized basis. The gradient of a scalar equals $\\nabla \\phi = \\frac{\\hat \\mathbf e_1}{h_1} \\frac{\\partial", "\\theta & =\\mathbf{v} \\cdot \\mathbf{w} \\end{aligned} When $v,w$ are unit vectors then, $$v \\cdot w=\\cos \\theta$$ Which is maximized when $\\theta=0$, i.e. $v=w$. Thus the dot product is often used to measure the similarity between two vectors. ##### Projections Scalar projection of $\\mathbf{a}$ onto $\\mathbf{b}$: $$\\operatorname{comp}_{\\mathbf{b}} \\mathbf{a}=\\underbrace{\\frac{\\mathbf{b}}{|\\mathbf{b}|}}_{direction} \\cdot \\mathbf{a}$$ This essentially gives the signed length of the projection. $$\\operatorname{comp}_{\\mathbf{b}} \\mathbf{a}=\\|a\\|\\cos(\\theta) = \\|\\mathbf{a}\\|\\frac{\\mathbf{a} \\cdot \\mathbf{b}}{\\|\\mathbf{a}\\|\\|\\mathbf{b}\\|} = \\frac{\\mathbf{a} \\cdot \\mathbf{b}}{\\|\\mathbf{a}\\|}$$ Vector projection of $\\mathbf{a}$ onto $\\mathbf{b}:$ $$\\operatorname{proj}_{\\mathbf{b}} \\mathbf{a}= \\underbrace{\\operatorname{comp}_{\\mathbf{b}} \\mathbf{a}}_{signed \\ length}\\quad \\underbrace{\\frac{\\mathbf{b}}{|\\mathbf{b}|}}_{direction}=\\frac{\\mathbf{b} \\cdot \\mathbf{a}}{|\\mathbf{b}|^{2}} \\mathbf{b}$$ The first equality is more meaningful, as it gives the length-direction decomposition. For any non-zero constant $c$ $$\\operatorname{proj}_{\\mathbf{cb}} \\mathbf{a}=\\operatorname{proj}_{\\mathbf{b}} \\mathbf{a}$$ In other words, the projection of $\\mathbf{a}$ onto $\\mathbf{b}$ is independent of the length of $\\mathbf{b}$, only the direction of the projection matters. #### The Cross Product ##### Motivation Given two vectors $u,v$, we want to find a vector $w$, that is perpendicular to both $u$ and $v$. This can be done by solving the following equations: $$u\\cdot w = u_1w_1+u_2w_2+u_3w_3 = 0$$ $$v\\cdot w = v_1w_1+v_2w_2+v_3w_3 = 0$$ Which implies $$(u_1v_3-u_3v_1)w_1+(u_2v_3-u_3v_2)w_2=0$$ One trivial solution is to take $$w_1=u_2v_3-u_3v_2,\\ w_2 = u_3v_1-u_1v_3, \\ w_3=u_1v_2-u_2v_1$$ The solution give the cross product of $u$ and $v$. ##### Computational Properties Often the above the formula is not very intuitive, and difficult to recall. Instead, one often use", "now drop $$\\operatorname{Pr}(x_i)$$ because we will be optimizing with respect to $$\\mathbf{w}$$, and, in that case, the term $$\\operatorname{Pr}(x_i)$$ is considered a constant (and the derivative of a constant is zero). In other words, our objective is \\begin{align} \\mathbf{w}^* &= \\operatorname{argmax}_{\\mathbf{w}} \\log \\operatorname{Pr}(\\mathcal{D} \\mid \\mathbf{w}) \\\\ &=\\operatorname{argmax}_{\\mathbf{w}} \\log \\left( \\prod_{i=1}^{|\\mathcal{D}|}\\operatorname{Pr}(y_i \\mid x_i, \\mathbf{w}) \\operatorname{Pr}(x_i) \\right) \\\\ &=\\operatorname{argmax}_{\\mathbf{w}} \\sum_{i=1}^{|\\mathcal{D}|} \\log \\operatorname{Pr}(y_i \\mid x_i, \\mathbf{w}) \\operatorname{Pr}(x_i) \\\\ &=\\operatorname{argmax}_{\\mathbf{w}} \\sum_{i=1}^{|\\mathcal{D}|} \\log \\operatorname{Pr}(y_i \\mid x_i, \\mathbf{w}) + \\log \\operatorname{Pr}(x_i) \\\\ &=\\operatorname{argmax}_{\\mathbf{w}} \\sum_{i=1}^{|\\mathcal{D}|} \\log \\operatorname{Pr}(y_i \\mid x_i, \\mathbf{w}) + \\sum_{i=1}^{|\\mathcal{D}|} \\log \\operatorname{Pr}(x_i) \\\\ &=\\operatorname{argmax}_{\\mathbf{w}} \\sum_{i=1}^{|\\mathcal{D}|} \\log \\operatorname{Pr}(y_i \\mid x_i, \\mathbf{w}) \\end{align} Obviously, $$\\sum_{i=1}^{|\\mathcal{D}|} \\log \\operatorname{Pr}(x_i)$$ is a constant with respect to $${\\mathbf{w}}$$, so it can be dropped. The equation in Alex Graves' paper is thus not generally true, but, if that is our objective (which is the case in Graves' paper), for the purpose of optimization, we can drop $$\\operatorname{Pr}(x_i)$$. Neal's PhD thesis also mentions this likelihood in section 1.3.1. (p. 15).", "f_1(\\mathbf{v}) + f_2(\\mathbf{v})$ then $\\frac{\\partial f}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial f_1}{\\partial \\mathbf{v}} + \\frac{\\partial f_2}{\\partial \\mathbf{v}}\\right)\\cdot\\mathbf{u}$ 2. If $f(\\mathbf{v}) = f_1(\\mathbf{v})~ f_2(\\mathbf{v})$ then $\\frac{\\partial f}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial f_1}{\\partial \\mathbf{v}}\\cdot\\mathbf{u}\\right)~f_2(\\mathbf{v}) + f_1(\\mathbf{v})~\\left(\\frac{\\partial f_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} \\right)$ 3. If $f(\\mathbf{v}) = f_1(f_2(\\mathbf{v}))$ then $\\frac{\\partial f}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\frac{\\partial f_1}{\\partial f_2}~\\frac{\\partial f_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u}$ ### Derivatives of vector valued functions of vectors Let $\\mathbf{f}(\\mathbf{v})$ be a vector valued function of the vector $\\mathbf{v}$. Then the derivative of $\\mathbf{f}(\\mathbf{v})$ with respect to $\\mathbf{v}$ (or at $\\mathbf{v}$) in the direction $\\mathbf{u}$ is the second-order tensor defined as $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = D\\mathbf{f}(\\mathbf{v})[\\mathbf{u}] = \\left[\\frac{d }{d \\alpha}~\\mathbf{f}(\\mathbf{v} + \\alpha~\\mathbf{u})\\right]_{\\alpha = 0}$ for all vectors $\\mathbf{u}$. Properties: 1. If $\\mathbf{f}(\\mathbf{v}) = \\mathbf{f}_1(\\mathbf{v}) + \\mathbf{f}_2(\\mathbf{v})$ then $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial \\mathbf{f}_1}{\\partial \\mathbf{v}} + \\frac{\\partial \\mathbf{f}_2}{\\partial \\mathbf{v}}\\right)\\cdot\\mathbf{u}$ 2. If $\\mathbf{f}(\\mathbf{v}) = \\mathbf{f}_1(\\mathbf{v})\\times\\mathbf{f}_2(\\mathbf{v})$ then $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial \\mathbf{f}_1}{\\partial \\mathbf{v}}\\cdot\\mathbf{u}\\right)\\times\\mathbf{f}_2(\\mathbf{v}) + \\mathbf{f}_1(\\mathbf{v})\\times\\left(\\frac{\\partial \\mathbf{f}_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} \\right)$ 3. If $\\mathbf{f}(\\mathbf{v}) = \\mathbf{f}_1(\\mathbf{f}_2(\\mathbf{v}))$ then $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\frac{\\partial \\mathbf{f}_1}{\\partial \\mathbf{f}_2}\\cdot\\left(\\frac{\\partial \\mathbf{f}_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} \\right)$ ### Derivatives of scalar valued functions of second-order tensors Let $f(\\boldsymbol{S})$ be a real valued function of the second order tensor $\\boldsymbol{S}$. Then the derivative of $f(\\boldsymbol{S})$ with respect to $\\boldsymbol{S}$ (or at $\\boldsymbol{S}$) in the direction $\\boldsymbol{T}$ is the second order tensor defined as $\\frac{\\partial f}{\\partial \\boldsymbol{S}}:\\boldsymbol{T} = Df(\\boldsymbol{S})[\\boldsymbol{T}] = \\left[\\frac{d }{d \\alpha}~f(\\boldsymbol{S} + \\alpha\\boldsymbol{T})\\right]_{\\alpha =", "as $$\\parallel U-V\\parallel =\\sup \\left\\{\\parallel Uv-Vv\\parallel \\,;\\,\\parallel v\\parallel =1\\right\\}$$. Because $${\\mathcal{G}}$$ is universal, for every ϵ > 0 and every element U SU(4), there exists $$V\\in {\\mathcal{G}}$$ such that U − V ≤ ϵ. In such a case we say that V is an ϵ-approximation of U. We now take advantage of Theorem 2.44 to count Clifford+CS operators and use these results to derive a worst-case lower bound on the CS-count of approximations. ### Lemma 2.45 Let n ≥ 1. There are 86400(3 8n − 2 4n) Clifford+CS operators of CS-count exactly n. Proof Each Clifford+CS operator is represented by a unique normal form and this representation is CS-optimal. Hence, to count the number of Clifford+CS operators of CS-count n, it suffices to count the normal forms of CS-count n. By Theorem 2.44, and since Clifford operators have CS-count 0, a normal form of CS-count n is a word $$w={w}_{1}{w}_{2}{w}_{3}{w}_{4}$$ (19) such that $${w}_{1}\\in {\\mathcal{L}}({{\\mathfrak{S}}}_{1,3})$$, $${w}_{2}\\in {\\mathcal{L}}({{\\mathfrak{S}}}_{4,9})$$, $${w}_{3}\\in {\\mathcal{L}}({{\\mathfrak{S}}}_{10,15})$$, $${w}_{4}\\in {\\mathcal{L}}({\\mathfrak{C}})$$ and the CS-counts of w1, w2, and w3 sum to n. There are $$(6\\cdot {8}^{n-1}+6\\cdot {4}^{n-1}+3\\cdot {2}^{n-1})\\cdot | {\\mathcal{C}}|$$ (20) words of the form of Eq. (19) such that exactly one of w1, w2, or w3 is not ε. Similarly, there are $$\\left(\\mathop{\\sum }\\limits_{0\\,{<}\\,l\\,{<}\\,n}18\\cdot {2}^{2n-3-l}+\\mathop{\\sum }\\limits_{0\\,{<}\\,l\\,{<}\\,n}18\\cdot {2}^{3n-4-2l}+\\mathop{\\sum }\\limits_{0\\,{<}\\,j\\,{<}\\,n}36\\cdot {2}^{3n-5-j}\\right)\\cdot | {\\mathcal{C}}|$$ (21) words of the form of", "Can you help express those? – ThatsRightJack Feb 14 '17 at 21:26 I wrote this out using Einstein notation, except since it's an orthogonal space, I ignored upper indices. Anywhere you see a double index, there's an implied summation. Here, $x_i$ are the coordinates, so $x_1=x$, $x_2=y$, $x_3=z$, and $\\mathbf{e}_i$ are the corresponding unit vectors, $\\mathbf{e}_1=\\mathbf{\\hat{x}}$, $\\mathbf{e}_2=\\mathbf{\\hat{y}}$, $\\mathbf{e}_3=\\mathbf{\\hat{z}}$. Using this notation, we write vectors and the $\\nabla$ operator as $\\mathbf{U} = \\mathbf{e}_i u_i$ $\\nabla = \\mathbf{e}_i \\frac{\\partial}{\\partial x_i}$ $(\\mathbf{U} \\cdot \\nabla) \\mathbf{U} = u_k (\\mathbf{e}_k \\cdot \\mathbf{e}_j) \\frac{\\partial u_i}{\\partial x_j} \\mathbf{e}_i = u_k \\delta_{jk} \\frac{\\partial u_i}{\\partial x_j} \\mathbf{e}_i = \\mathbf{e}_i u_j \\frac{\\partial u_i}{\\partial x_j}$ $\\nabla( \\frac{1}{2} \\mathbf{U}^2 ) = \\mathbf{e}_j \\frac{\\partial}{\\partial x_j} \\left( \\frac{1}{2} u_i u_i \\right) = \\mathbf{e}_j u_i \\frac{\\partial u_i}{\\partial x_j}$ $(\\nabla \\times \\mathbf{U}) \\times \\mathbf{U} = \\left( (\\mathbf{e}_j \\frac{\\partial}{\\partial x_j} )\\times (u_k \\mathbf{e}_k) \\right) \\times (u_m \\mathbf{e}_m) = (\\mathbf{e}_j\\times \\mathbf{e}_k \\times \\mathbf{e}_m) u_m \\frac{\\partial u_k}{\\partial x_j} = \\mathbf{e}_n \\epsilon_{ijk}\\epsilon_{imn} u_m \\frac{\\partial u_k}{\\partial x_j}$ where $\\epsilon_{ijk}$ is the Levi-Civita symbol. Using the following identity $\\epsilon_{ijk} \\epsilon_{imn} = \\delta_{jm}\\delta_{kn} - \\delta_{jn}\\delta_{km}$ this simplifies to $(\\nabla \\times \\mathbf{U}) \\times \\mathbf{U} = \\mathbf{e}_n \\delta_{jm}\\delta_{kn} u_m\\frac{\\partial u_k}{\\partial x_j} - \\mathbf{e}_n \\delta_{jn}\\delta_{km} u_m\\frac{\\partial u_k}{\\partial x_j} = \\mathbf{e}_k u_j\\frac{\\partial u_k}{\\partial x_j} - \\mathbf{e}_j u_k\\frac{\\partial u_k}{\\partial x_j}$ Reindexing $k \\rightarrow i$ $(\\nabla \\times \\mathbf{U}) \\times \\mathbf{U} = \\mathbf{e}_i u_j\\frac{\\partial u_i}{\\partial x_j} - \\mathbf{e}_j", "} \\vec u = \\pm \\lvert \\operatorname { proj } _ { \\vec v } \\vec u \\rvert$$ (with plus if $$\\vec u \\cdot \\vec v$$ is positive, minus if $$\\vec u \\cdot \\vec v$$ is negative, and both if $$\\vec u \\cdot \\vec v$$ is zero because then both of these quantities are zero). When $$\\vec v$$ is a standard basis vector ($$\\hat \\imath$$ or $$\\hat \\jmath$$ in 2 dimensions), then these are the ordinary components; that is, $$\\operatorname { comp } _ { \\hat \\imath } ( a \\hat \\imath + b \\hat \\jmath ) = a$$, and $$\\operatorname { comp } _ { \\hat \\jmath } ( a \\hat \\imath + b \\hat \\jmath ) = b$$. In constrast, $$\\operatorname { proj } _ { \\hat \\imath } ( a \\hat \\imath + b \\hat \\jmath ) = a \\hat \\imath$$, and $$\\operatorname { proj } _ { \\hat \\jmath } ( a \\hat \\imath + b \\hat \\jmath ) = b \\hat \\jmath$$. You can also write $$\\vec u = ( \\operatorname { comp } _ { \\hat \\imath } \\vec u ) \\hat \\imath + ( \\operatorname { comp } _ { \\hat \\jmath } \\vec u ) \\hat \\jmath$$ and $$\\vec u = \\operatorname { proj } _ { \\hat \\imath }", "we may compute |\\operatorname*{\\mathbb{E}}[e^{itg}]|=\\prod_{S\\in A}\\mathopen{}\\mathclose{{}% \\left|\\operatorname*{\\mathbb{E}}[e^{it\\alpha(x^{S})}]}\\right|\\leq\\prod_{% \\begin{subarray}{c}S\\in A\\\\ A_{S}\\mbox{\\scriptsize{ occurs}}\\end{subarray}}\\mathopen{}\\mathclose{{}\\left(1% -\\frac{8p(1-p)t^{2}}{\\pi^{2}\\sigma^{2}}}\\right)=\\exp\\mathopen{}\\mathclose{{}% \\left(-\\frac{8p(1-p)t^{2}z(Y)}{\\pi^{2}\\sigma^{2}}}\\right) Since each of the events A_{S} are independent and occur with probability \\geq\\lambda^{2{r\\choose 2}^{2}} it follows from Chernoff bounds that z(Y)\\geq\\lambda^{2{r\\choose 2}^{2}}n/2r with probability \\geq 1-\\exp(-\\Omega(n)). So we find that \\operatorname*{\\mathbb{E}}_{\\mathbf{Y}}\\mathopen{}\\mathclose{{}\\left|% \\operatorname*{\\mathbb{E}}_{X}[e^{it\\alpha(\\mathcal{K})}]}\\right|\\leq\\Pr(A)% \\exp\\mathopen{}\\mathclose{{}\\left(-\\frac{8p(1-p)t^{2}}{\\pi^{2}\\sigma^{2}}\\cdot% \\frac{\\lambda^{2{r\\choose 2}^{2}}n}{2r}}\\right)+\\Pr(A^{c})=\\exp\\mathopen{}% \\mathclose{{}\\left(-\\Omega\\mathopen{}\\mathclose{{}\\left(\\frac{t^{2}n}{\\sigma^{% 2}}}\\right)}\\right) Combining this with Lemma 6 we have proved the following: ###### Lemma 18. For |t|\\leq\\pi\\sigma we have that |\\varphi_{\\mathcal{K}}(t)|\\leq\\exp(-\\Omega(t^{2}n/\\sigma^{2})). ## References • [Ber16] Ross Berkowitz. A quantitative local limit theorem for triangles in random graphs, 2016. • [ER61] P. Erdős and A. Rényi. On the evolution of random graphs. Bull. Inst. Internat. Statist., 38:343–347, 1961. • [Fel71] William Feller. An introduction to probability theory and its applications. Vol. II. Second edition. John Wiley & Sons, Inc., New York-London-Sydney, 1971. • [GK14] Justin Gilmer and Swastik Kopparty. A local central limit theorem for the number of triangles in a random graph. ArXiv e-prints, November 2014. • [Jan94] Svante Janson. Orthogonal decompositions and functional limit theorems for random graph statistics. Mem. Amer. Math. Soc., 111(534):vi+78, 1994. • [Kar84] Michał Karoński. Balanced subgraphs of large random graphs, volume 7 of Seria Matematyka [Mathematics Series]. Uniwersytet im. Adama Mickiewicza w Poznaniu, Poznań, 1984. With a Polish summary. • [KR83] Michał Karoński and Andrzej Ruciński. On the number of strictly balanced subgraphs of a", "e _ {1} +$$ $$+ \\left ( \\frac{\\partial a _ {1} }{\\partial x _ {3} } - \\frac{\\partial a _ {3} }{\\partial x _ {1} } \\right ) \\mathbf e _ {2} + \\left ( \\frac{\\partial a _ {2} }{\\partial x _ {1} } - \\frac{\\partial a _ {1} }{\\partial x _ {2} } \\right ) \\mathbf e _ {3} .$$ The scalar square of the Hamilton operator yields the Laplace operator: $$\\Delta = \\ \\nabla \\cdot \\nabla = \\ \\sum _ {j = 1 } ^ { n } \\frac{\\partial ^ {2} }{\\partial x _ {j} ^ {2} } .$$ The following relations are valid: $$[ \\nabla , \\nabla \\phi ] = \\ \\mathop{\\rm rot} \\mathop{\\rm grad} \\phi = 0,$$ $$\\nabla \\cdot \\nabla \\mathbf a = \\mathop{\\rm grad} \\mathop{\\rm div} \\mathbf a ,\\ \\nabla [ \\nabla , \\mathbf a ] = \\mathop{\\rm div} \\mathop{\\rm rot} \\mathbf a = 0,$$ $$[ \\nabla , [ \\nabla , \\mathbf a ] ] = \\mathop{\\rm rot} \\ \\mathop{\\rm rot} \\mathbf a ,\\ \\Delta \\phi = \\nabla \\cdot ( \\nabla \\phi ) = \\mathop{\\rm div} \\mathop{\\rm grad} \\phi .$$ The Hamilton operator was introduced by W. Hamilton [1]. References [1] W.R. Hamilton, \"Lectures on quaternions\" , Dublin (1853)", "# Direction vectors are linearly dependent, if $D_vf(x)=D_wf(x)=0$ with $\\operatorname{grad} f(x)\\neq 0$ Let $f \\colon \\mathbb{R}^2 \\to \\mathbb{R}$ be continuous differentiable and $x \\in\\mathbb{R}^2$ with $\\operatorname{grad} f(x) \\neq 0$. Let $v,w \\in \\mathbb{R}^2$ be direction vectors. If $D_v f(x) = D_w f(x)=0$ then are $v$ and $w$ linearly dependent. For my proof I use the following theorem: Let $U \\subset \\mathbb{R}^m$ be open and $f \\colon U \\to \\mathbb{R}$ continuous differentiable. Then it holds for $x \\in U$ and $v \\in \\mathbb{R}^m$ with $\\|v\\| = 1$ that $D_v f(x) = \\langle v, \\operatorname{grad} f(x) \\rangle$. Proof: With the theorem above and the assumption, we know that $$\\langle v, \\operatorname{grad} f(x) \\rangle = \\langle w, \\operatorname{grad} f(x) \\rangle.$$ It holds that \\begin{align*} \\langle v, \\operatorname{grad} f(x) \\rangle &\\overset{\\phantom{D_vf(x)=D_wf(x)}}{=} \\sum_{i=1}^2 v_i \\cdot \\frac{\\partial f}{\\partial x_i} = v_1 \\cdot \\frac{\\partial f}{\\partial x_1} + v_2 \\cdot \\frac{\\partial f}{\\partial x_2} \\\\ &\\overset{D_vf(x)=D_wf(x)}{=} \\sum_{i=1}^2 w_i \\cdot \\frac{\\partial f}{\\partial x_i} = w_1 \\cdot \\frac{\\partial f}{\\partial x_1} + w_2 \\cdot \\frac{\\partial f}{\\partial x_2} = 0 \\end{align*} Therefore \\begin{align*} &\\, v_1 \\cdot \\frac{\\partial f}{\\partial x_1} + v_2 \\cdot \\frac{\\partial f}{\\partial x_2} = w_1 \\cdot \\frac{\\partial f}{\\partial x_1} + w_2 \\cdot \\frac{\\partial f}{\\partial x_2} \\\\ \\iff&\\, (v_1-w_1) \\frac{\\partial f}{\\partial x_1} + (v_2-w_2) \\frac{\\partial f}{\\partial x_2} = 0 \\end{align*} Since $\\operatorname{grad} f(x) = \\left(\\frac{\\partial f}{\\partial x_1}, \\frac{\\partial f}{\\partial x_w}\\right)" ]

[ "# Math Help - Solve for tan 1. ## Solve for tan How would I go about evaluating for tan(arctan(10))? 2. ## Re: Solve for tan Originally Posted by Manni How would I go about evaluating for tan(arctan(10))? tangent and arctangent are inverse functions ... $\\tan[\\arctan(10)] = 10$", "## Trigonometry (11th Edition) Clone arctan($-1$)$=-45^{o}$ Inverse Tangent Function $y=\\tan^{-1}x$ or $y=$ arctan $x$ means that $x=\\tan y$, for $-\\displaystyle \\frac{\\pi}{2} < y < \\frac{\\pi}{2}$. Or for an angle in degrees, $\\theta=\\tan^{-1}x$ means $x=\\tan\\theta$, for $-90^{o} < \\theta < 90^{o}$. --------------- Knowing: $\\quad \\tan 45^{o}=1$, (and $\\tan(-\\theta)=-\\tan\\theta),$ we find $\\theta=-45^{o}$ from the interval $-90^{o} < \\theta < 90^{o}$ such that $\\tan$($-45^{o})=-1.$ So, arctan($-1$)$=-45^{o}$", "## Trigonometry (11th Edition) Clone The solution set is $$\\{135^\\circ+180^\\circ n, n\\in Z\\}$$ $$\\tan\\theta+1=0$$ 1) Solve the equation over the interval $[0^\\circ,360^\\circ)$ $$\\tan\\theta+1=0$$ $$\\tan\\theta=-1$$ Over the interval $[0^\\circ, 360^\\circ)$, there are two values of $\\theta$ where $\\tan\\theta=-1$, which are $135^\\circ$ and $315^\\circ$. Therefore, $$\\theta=\\{135^\\circ, 315^\\circ\\}$$ 2) Solve the equation for all solutions To find all solutions, we add the integer multiples of the period of tangent function, which is $180^\\circ$, to each solution found in step 1. - There are two solutions found in step 1, $\\{135^\\circ, 315^\\circ\\}$, so it will be written as $\\theta=135^\\circ+180^\\circ n$ and $\\theta=315^\\circ+180^\\circ n$ where $n\\in Z$. - However, a closer look reveals that $135^\\circ+180^\\circ n$ in fact covers all the possible values of $315^\\circ+180^\\circ n$, and vice versa. In other words, they are technically the same. So the solution set can be shortened to $$\\{135^\\circ+180^\\circ n, n\\in Z\\}$$", "\\theta$$ $$= 1-2sin^2\\theta$$ $$cos (2 tan^{-1}(x))= 1-2[sin(tan^{-1}(x))]^2$$ $$sin(tan^{-1}(x))$$ $$y = tan^{-1}x$$ tan(y) = x Now drawing the triangle we have the adjacent and opposite sides now I can work out the hypothenuse; $$sin(y) = \\frac{x}{\\sqrt{x^2 +1}}$$ $$sin(y) = sin(sin^{-1} \\frac{x}{\\sqrt{x^2 +1}}$$ $$= \\frac{x}{\\sqrt{x^2 +1}}$$ pluging this back into the equation above; $$cos (2 tan^{-1}(x))= 1-2[\\frac{x}{\\sqrt{x^2 +1}}]^2$$ I need some help now because this looks very messy! I'm not sure if it's right though! 9. Aug 7, 2008 ### snipez90 Hi powerless! You're learning quickly, good job :). Anyways I've skimmed over the work and it looks good. Let's just clear up the notation a bit. Hats off to you for noticing the cosine double angle formula. Now instead of using y over and over, let's be a bit more clear. Let $$arctan(x) = \\theta \\Rightarrow$$$$\\ tan(\\theta) = x$$ where arctan is the same as your inverse tangent function. But I just used arctan because I couldn't find the inverse tan :P. However, the great thing about using this notation is that it's all you really need to define. After you've defined the above, the problem reduces to finding cos(2*theta) and therefore finding sin(arctan(x)). Then you proceed with the triangle and as you can see, our definition provides that info for us on the right hand side of the", "# How do you find the exact solution to inverse tangent (sqrt3) (meaning arctansqrt3)? Find $\\arctan \\left(\\sqrt{3}\\right)$ Ans: $\\frac{\\pi}{3}$ or ${60}^{\\circ}$ $\\tan x = \\sqrt{3}$ --> arc $x = \\frac{\\pi}{3}$, or ${60}^{\\circ}$", "solutions to $x=\\tan x$. 2. This is in the right form for fixed point iteration, but a fixed point iteration using the above will not converge. 3. Taking the inverse tangent of both sides doesn't help immediately because all solutions but the trivial solution lie outside the principal domain of arctan. 4. So, introduce a new variable that will be in the principal domain: $x\\equiv n\\pi+\\theta$. 5. Now the problem becomes one of solving $n\\pi+\\theta = \\tan(n\\pi+\\theta)$. 6. The right hand side of the above, $\\tan(n\\pi+\\theta)$, reduces to $\\tan\\theta$ via the trig identity $\\tan(a+b) = (\\tan a + \\tan b)/(1-\\tan a \\tan b)$. 7. Now the problem becomes one of solving $n\\pi+\\theta = \\tan\\theta$. 8. And now you can safely apply the arc tangent function to both sides because, by design, theta is in the principal domain of arctan. Try to take it from here. Last edited:", "# Tangents Geometry Level 5 Given that $$\\tan \\theta \\neq 1\\pm\\sqrt{2}$$, what is the sum of all solutions (in degrees) for the following equation for $$0^\\circ \\le \\theta \\le 360^\\circ$$? $\\tan^{4} \\theta+4\\tan^{3} \\theta-6\\tan^{2} \\theta-4\\tan \\theta+1 = 0$ × Problem Loading... Note Loading... Set Loading...", "Function Tan(x) # Tan Tan( x ! (input) numerical expression ) ## Arguments x A scalar numerical expression in radians. ## Return Value The Tan function returns the tangent of x in the range $$-\\infty$$ to $$\\infty$$. Note The function Tan can be used in constraints of nonlinear mathematical programs. The functions Cos, Sin, ArcTan. Arithmetic functions are discussed in full detail in Arithmetic Functions of the Language Reference.", "negative value. Tangent is negative in the 2nd and 4th quadrant, so arctan(-2), or theta, is in the 2nd or 4th quadrant. In the 2nd quadrant, the opp side is positive, while the adj side is negative. so, tan(theta) = -2 = (2) / (-1). Meaning, the opp is 2, and the adj is -1. You are asked to find sec(theta).", "# Inverse tangents in cyclic order ## Homework Statement The problem- if $$\\theta= tan^{-1}(\\frac{a(a+b+c)}{bc})+tan^{-1}(\\frac{b(a+b+c)}{ac})+tan^{-1}(\\frac{c(a+b+c)}{ab})$$ , then find $$tan\\theta$$ ## The Attempt at a Solution I tried to use these as sides of a triangle and use their properties, but other than that I am clueless. I cannot think of a substitution either. The answer happens to be zero. Any help is appreciated. Thanks in advance!! Okay, I have it. Apparently, there is no 'elegant' way to do this. Just use the equation for $$arctan x + arctan y + arctan z$$, and after some really messy calculation, you get the answer as zero. If there is a different, neat answer, please let me know.", "= \\frac {2}{3}\\\\ \\tan\\theta = \\frac {2}{3}\\\\ \\theta = \\arctan (\\frac {2}{3})$ This is not one of your special angles, so you will need a calculator. $\\theta = 33.69^\\circ$ $\\tan \\theta$ is not a 1-1 function. The inverse function maps back to a value in $[0,180^\\circ)$. There is second answer in the interval $[0,360^\\circ)$. $\\tan \\theta = \\tan (\\theta + 180)$ $213.69^\\circ$ is also in the solution set. The next one might even be a little bit trickier. $\\cos 3\\theta = -0.5\\\\ 3\\theta = \\arccos (-0.5)\\\\ 3\\theta = 120^\\circ$ but again, this isn't the only answer. $\\theta = \\arccos (-0.5)$ has two answers in the 360 degree interval, and $3\\theta = \\arccos (-0.5)$ will have $6.$ $\\cos \\theta = \\cos (-\\theta)$ and $\\cos \\theta = \\cos (360+\\theta)$ using these identities we find other values of theta. $3\\theta = 240^\\circ, 480^\\circ, 600^\\circ, 840^\\circ, 960^\\circ$ are all viable, as well. $\\theta \\in \\{40, 80, 160, 200, 280, 320\\}$ a) Reduce to $$\\tan\\theta=\\frac23.$$ Then as the tangent function has period $180°$ and is monotonic in a period, $$\\tan\\theta=\\tan(\\theta+k180°)$$ and all solutions are of the form $$\\theta+k180°=\\arctan\\frac23.$$ Find all $k$ such that $0\\le\\theta\\le360°$. b) The cosine has period $360°$ and is an even function monotonic in $[0°,180°]$ so that $$\\cos3\\theta=\\cos(k360°\\pm3\\theta).$$ Then all solutions are of the form $$k120°\\pm\\theta=\\frac13\\arccos(-0.5).$$ Again, find all $k$", "-arctan( \\frac{\\frac{1}{\\sqrt{LC}}L}{R(1-(\\frac{1}{1}))}) \\\\ 90= -arctan( \\frac{\\frac{1}{\\sqrt{LC}}L}{0})$ Last edited: Oct 26, 2014 4. Oct 26, 2014 ### M Quack What does the tangent function look like, in particular, what is the value of tan(90 deg) and tan(-90 deg)? The mathematically correct way would be to calculate the solution for +/-(90 + epsilon), and then take the limit epsilon-> 0.", "n' chug\" into the expression given (the already gave you the values of $\\alpha$, x, and y) so you can find $\\theta$. And as topsquark said, atan most likely means the inverse tangent function arctan(x), also written as $\\tan^{-1}(x)$. • April 25th 2011, 04:57 PM mikeani Ok...thank you guys...now i get it.", "# Thread: Simplifying Inverse Trigonometric Functions 1. ## Simplifying Inverse Trigonometric Functions Hello all, well, I'm not sure how to go about simplifying the following problems tan(2arctan(x)) and cos(2arcsin(x)) I know I should let theta = 2arctan(x) => tan(theta) = 2x .... I think. However, from that point forward I'm not sure what to do. I have several others, however these two are the ones that I believe will enable me to figure out the remainder of the problems. Any and all help is greatly appreciated. Thank you for your time. jhb 2. $\\tan(2u) = \\frac {2 \\tan(u)}{1-\\tan^{2}(u)}$ so $\\tan(2 \\arctan(x)) = \\frac {2 \\tan(\\arctan(x))}{1 - \\tan^{2}(\\arctan(x))}$ $= \\frac {2x}{1 -x^{2}}$ 3. Originally Posted by jhb Hello all, well, I'm not sure how to go about simplifying the following problems tan(2arctan(x)) and cos(2arcsin(x)) I know I should let theta = 2arctan(x) => tan(theta) = 2x .... I think. However, from that point forward I'm not sure what to do. I have several others, however these two are the ones that I believe will enable me to figure out the remainder of the problems. Any and all help is greatly appreciated. Thank you for your time. jhb $ \\cos{2\\sin^{-1}{x}} = \\cos{2\\theta} = 1 - 2\\sin^2{\\theta}$ since i let $\\sin^{-1}{x} = \\theta \\implies x = \\sin{\\theta}$ $\\cos{2\\sin^{-1}{x}} = 1 -", "# Inverse Tangent Function The inverse tangent function is denoted as $tan^{-1}(x)$. This function is used to find the unknown measure of an angle of a right triangle when two side lengths are known.The inverse of tangent is the inverse function of the tangent function and can be used to calculate angle measures when certain other things are known. This lesson will go into detail about the inverse tangent, give some formulas and follow up with a quiz to check your understanding. The domain of the inverse of tangent function is (−∞,∞)and the range is $\\frac{-\\prod }{2},\\frac{\\prod }{2}$ . The inverse of the tangent function will yield values in the 1st and 4th quadrants. $tan^{-1}(x)$ = arctan (x) To evaluate inverse of tangent functions remember that the following statement is equivalent. $\\Theta = tan^{-1}(x) \\Leftrightarrow x = tan(\\Theta)$ In other words, when we evaluate tan inverse function we are asking what angle,θ, did we plug into the tangent function (regular, not inverse!) to get x. f(x)= tan(x) Domain $(\\frac{-\\prod }{2}< x < \\frac{\\prod}{2}$) Range (-∞ , ∞ ) $f(x)=tan^{-1}x$ (-∞ , ∞ ) $\\frac{-\\prod }{2}< y < \\frac{\\prod}{2}$ In tangent function, we know that $tan \\Theta =\\frac{opposite side }{Adjacent}$ The inverse of tangent function $tan{-1}$ takes the ratio opposite/adjacent and gives the angle θ. ∴ θ = $tan^{-1}(\\frac{opposite", "# Inverse Tangent Function The inverse tangent function is denoted as $tan^{-1}(x)$. This function is used to find the unknown measure of an angle of a right triangle when two side lengths are known.The inverse of tangent is the inverse function of the tangent function and can be used to calculate angle measures when certain other things are known. This lesson will go into detail about the inverse tangent, give some formulas and follow up with a quiz to check your understanding. The domain of the inverse of tangent function is (−∞,∞)and the range is $\\frac{-\\prod }{2},\\frac{\\prod }{2}$ . The inverse of the tangent function will yield values in the 1st and 4th quadrants. $tan^{-1}(x)$ = arctan (x) To evaluate inverse of tangent functions remember that the following statement is equivalent. $\\Theta = tan^{-1}(x) \\Leftrightarrow x = tan(\\Theta)$ In other words, when we evaluate tan inverse function we are asking what angle,θ, did we plug into the tangent function (regular, not inverse!) to get x. f(x)= tan(x) Domain $(\\frac{-\\prod }{2}< x < \\frac{\\prod}{2}$) Range (-∞ , ∞ ) $f(x)=tan^{-1}x$ (-∞ , ∞ ) $\\frac{-\\prod }{2}< y < \\frac{\\prod}{2}$ In tangent function, we know that $tan \\Theta =\\frac{opposite side }{Adjacent}$ The inverse of tangent function $tan{-1}$ takes the ratio opposite/adjacent and gives the angle θ. ∴ θ = $tan^{-1}(\\frac{opposite", "## Thomas' Calculus 13th Edition $1.5574$ Here, $\\int \\dfrac{1}{1+y^2}dy=\\int dx$ and $\\arctan (y)=x+c$ Further, apply the initial conditions, to calculate the value of $c$ $\\arctan (0)=0+c$ or, $c=0$ Now, the particular solution as follows: $\\arctan (y)=x$ This can be re-arranged as: $y=\\tan x$ and $y(1)=\\tan x =\\tan (1) \\approx 1.5574$", "Tan Theta formula The law of Tangent which is also called as tangent formula or tangent rule is the ratio of the sine of the angle to the cos of the angle. Tan Θ = Opposite / Adjacent Tan x formula The Tan Θ is the ratio of the Opposite side to the Adjacent, where (Θ) is one of the acute angles. $$\\begin{array}{l}Tan \\Theta = \\frac{Opposite Side}{Adjacent Side}\\end{array}$$ Tan alpha Questions Out of other 6 trigonometric formulas, let’s have a look at the practice question of tan theta formula. Example 1: If Sin x = 4/5, Find the value of Cos x and tan x? Solution: Using Trigonometric identities: Cos2x = 1- Sin2x Cos2x = 1 – (4/5)2 = 1 – 16/25 = (25 – 16) / 25 = 9/25 Cos x = $$\\begin{array}{l}\\sqrt{9/25}\\end{array}$$ = ⅗ Tan x = Sin x / Cos x = (4/5) / (3/5) = 4/3 Question on Tangent formula Question 1: If Sec x = 4/7, find the Cos x and tan x? Solution: As Sec x = 1/cos x 4/7 = 1/ cos x Cos x = 7/4 Tan2 x = sec2x – 1 = (4/7)2 – 1 = 16/49 – 1 = -33 / 49 Tan x = $$\\begin{array}{l}\\sqrt{-33}\\end{array}$$ /7 To practice more Trigonometric questions and understand the trigonometric concepts better,", "used is very useful. For the other solution: $\\tan(\\theta)=\\frac{1}{2}$ that means: $\\theta=\\arctan\\left(\\frac{1}{2}\\right)+k\\pi$", "# How do you do inverse trig functions like tan x= 1.234? $x = \\arctan \\left(1.234\\right)$ $x = {\\tan}^{- 1} \\left(1.234\\right)$" ]

[ "algebraic result. (I know this is uncharacteristic for my typically algebraic tendencies.) The final key was in the implications of $tan(A)*tan(B)=1$. This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving $\\displaystyle tan(A) = \\frac{1}{tan(B)} = cot(B)$. But cotangent is also the co-function–or complement function–of tangent which gave me $tan(A) = cot(B) = tan \\left( \\frac{\\pi}{2} - B \\right)$. Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \\frac{\\pi}{2} - B$, or $A+B = \\frac{\\pi}{2}$. Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation, $(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\\pi ]$ and any $\\left| b \\right| \\ge 2$, is always $3\\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions. QED Insight #6: Generalizing the Domain The posed problem can be generalized further by recognizing the period of tangent: $\\pi$. That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\\pi$ each, as shown in the GeoGebra construction above. Insights 4 & 5 proved the sum of the angles at points C & D was $\\pi/2$. Employing the periodicity of tangent, the x-coordinate of $E = C+\\pi$ and $F", "## Math-Linux.com Knowledge base dedicated to Linux and applied mathematics. Home > Mathematics > Trigonometry > Tangent is an odd function tan(-x)=-tan x # Tangent is an odd function tan(-x)=-tan x We prove here that the tangent function tan (-x) = - tan x is odd using the unit circle. unit circle $\\mathcal{C}(O,R=1)$ definition of the angle $x$ definition of the angle $-x$ Now consider the triangles: $(OA_xA)$ and $(OA’_xA’)$. ## Proof that tangent is an odd function tan(-x) = -tan (x) Take the definition of the tangent of the angles $x$ and $-x$. In the triangle $(OA_xA)$ : $$\\tan x=\\frac{\\textrm{opposite}}{\\textrm{adjacent}}=\\frac{|OA_y|}{|OA_x|}$$ In the triangle $(OA’_xA’)$: $$\\tan (-x)=\\frac{\\textrm{opposite}}{\\textrm{adjacent}}=\\frac{|OA’_y|}{|OA_x|}$$ By construction: $|OA_y|= -|OA’_y|$, then we have $$\\forall x\\in \\mathbb{R},x \\neq k\\pi + \\frac{\\pi}{2} , k \\in \\mathbb{Z}:\\quad \\tan (-x)=\\frac{|OA’_y|}{|OA_x|}=-\\frac{|OA_y|}{|OA_x|}=-\\tan x$$ ## Another proof that tangent is an odd function tan(-x) = -tan (x) The definition of an odd function is that $f(-x) = -f(x)$ for all $x$ in the domain of the function. To prove that the tangent function is odd, we can use this definition and show that $\\tan(-x) = -\\tan x$ for all $x \\neq k\\pi + \\frac{\\pi}{2} , k \\in \\mathbb{Z}$ Using the trigonometric identity for tangent, we can write: $$\\tan(-x) = \\frac{\\sin(-x)}{\\cos(-x)}$$ Using the fact that sine is odd $\\sin(-x) = -\\sin x$ and cosine is", "know this is uncharacteristic for my typically algebraic tendencies.) The final key was in the implications of $tan(A)*tan(B)=1$. This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving $\\displaystyle tan(A) = \\frac{1}{tan(B)} = cot(B)$. But cotangent is also the co-function–or complement function–of tangent which gave me $tan(A) = cot(B) = tan \\left( \\frac{\\pi}{2} - B \\right)$. Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \\frac{\\pi}{2} - B$, or $A+B = \\frac{\\pi}{2}$. Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation, $(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\\pi ]$ and any $\\left| b \\right| \\ge 2$, is always $3\\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions. QED Insight #6: Generalizing the Domain The posed problem can be generalized further by recognizing the period of tangent: $\\pi$. That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\\pi$ each, as shown in the GeoGebra construction above. Insights 4 & 5 proved the sum of the angles at points C & D was $\\pi/2$. Employing the periodicity of tangent, the x-coordinate of $E = C+\\pi$ and $F = D+\\pi$, so", "this is uncharacteristic for my typically algebraic tendencies.) The final key was in the implications of $tan(A)*tan(B)=1$. This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving $\\displaystyle tan(A) = \\frac{1}{tan(B)} = cot(B)$. But cotangent is also the co-function–or complement function–of tangent which gave me $tan(A) = cot(B) = tan \\left( \\frac{\\pi}{2} - B \\right)$. Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \\frac{\\pi}{2} - B$, or $A+B = \\frac{\\pi}{2}$. Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation, $(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\\pi ]$ and any $\\left| b \\right| \\ge 2$, is always $3\\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions. QED Insight #6: Generalizing the Domain The posed problem can be generalized further by recognizing the period of tangent: $\\pi$. That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\\pi$ each, as shown in the GeoGebra construction above. Insights 4 & 5 proved the sum of the angles at points C & D was $\\pi/2$. Employing the periodicity of tangent, the x-coordinate of $E = C+\\pi$ and $F = D+\\pi$, so the", "this is uncharacteristic for my typically algebraic tendencies.) The final key was in the implications of $tan(A)*tan(B)=1$. This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving $\\displaystyle tan(A) = \\frac{1}{tan(B)} = cot(B)$. But cotangent is also the co-function–or complement function–of tangent which gave me $tan(A) = cot(B) = tan \\left( \\frac{\\pi}{2} - B \\right)$. Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \\frac{\\pi}{2} - B$, or $A+B = \\frac{\\pi}{2}$. Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation, $(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\\pi ]$ and any $\\left| b \\right| \\ge 2$, is always $3\\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions. QED Insight #6: Generalizing the Domain The posed problem can be generalized further by recognizing the period of tangent: $\\pi$. That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\\pi$ each, as shown in the GeoGebra construction above. Insights 4 & 5 proved the sum of the angles at points C & D was $\\pi/2$. Employing the periodicity of tangent, the x-coordinate of $E = C+\\pi$ and $F = D+\\pi$, so the", "curve at an interval. Jun 22 '19 at 20:00 The equation of the tangent line at the point $$(a,f(a))$$ is $$L(x)=f'(a)(x-a)+f(a)$$ In this case, $$f(x)=\\tan(2x)$$ and $$a=\\pi/2$$. Hence $$L(x)=2\\sec^2(2\\cdot \\tfrac{\\pi}{2})(x-\\tfrac{\\pi}{2})+\\tan(2\\cdot \\tfrac{\\pi}{2})$$ i.e. $$L(x)=2(x-\\tfrac{\\pi}{2})=2x-\\pi$$. Therefore, the $$y$$-intercept of $$L$$ is $$-\\pi$$. • Unable to state $y$ as $L(x)$, request elaboration. Also, the solution does not fall into the given domain for $2x$. Also, how it is so fixed that the only tangent is at $x = \\frac{\\pi}2$. Jun 22 '19 at 19:41 • Is not the only tangent line, $x=\\frac{\\pi}{2}$ is the point in the given problem. Jun 22 '19 at 19:48 • And, what's domain? Jun 22 '19 at 19:48 • I have stated also in post that the question is expected to state the domain as $(\\frac{\\pi}2,0)$. Jun 22 '19 at 19:49 • $\\tan(2\\cdot \\tfrac{\\pi}{2})=\\tan(\\pi)=0$ Jun 22 '19 at 19:53", "$$x$$ = $$297.266{}^\\circ$$ and $$x$$ = $$242.734{}^\\circ$$ EXAMPLE $$\\PageIndex{7}$$ Find all solutions to $$\\tan \\left(x\\right)=3$$ on $$0 \\le x<2\\pi$$. Solution Using the inverse tangent function, we can find one solution $$x=\\tan ^{-1} \\left(3\\right) \\approx 1.249$$. Unlike the sine and cosine, the tangent function only attains any output value once per cycle, so there is no second solution in any one cycle. By adding $$\\pi$$, a full period of tangent function, we can find a second angle with the same tangent value. Notice this gives another angle where the line has the same slope. If additional solutions were desired, we could continue to add multiples of $$\\pi$$, so all solutions would take on the form $$x=1.249+k\\pi$$, however we are only interested in $$0 \\le x< 2\\pi$$. $x=1.249+\\pi =4.391\\nonumber$ The two solutions on $$0\\le x<2\\pi$$ are $$x = 1.249$$ and $$x = 4.391$$. Exercise $$\\PageIndex{3}$$ Find all solutions to $$\\tan \\left(x\\right)=0.7$$ on $$0{}^\\circ \\le x<360{}^\\circ$$. The first solution is $$x=\\tan ^{-1} (0.7) \\approx 34.992{}^\\circ$$. For a standard tangent, the second solution can be found by adding a full period, $$180\\mathrm{{}^\\circ}$$, giving $x=180{}^\\circ +34.99{}^\\circ =214.992{}^\\circ\\nonumber$ EXAMPLE $$\\PageIndex{8}$$ Solve $$3\\cos \\left(t\\right)+4=2$$ for all solutions on one cycle, $$0\\le t<2\\pi$$ Solution $3\\cos \\left(t\\right)+4=2\\nonumber$Isolating the cosine $3\\cos \\left(t\\right)=-2\\nonumber$ $\\cos \\left(t\\right)=-\\dfrac{2}{3}\\nonumber$Using the inverse, we can find one solution $-\\dfrac{2}{3} -\\dfrac{2}{3} t=\\cos ^{-1} \\left(-\\dfrac{2}{3} \\right)\\approx 2.301\\nonumber$", "about acute angles and the line $x=1$ is an explanation of how the tangent function got its name, not the definition. The book's definition (shown below) is not restricted to acute angles. Your textbook definition of $\\tan$ works for $0\\leq\\theta<{\\pi\\over2}$, and with some stretch of the imagination also for $-{\\pi\\over2}<\\theta\\leq0$. That's where the \"geometric definition\" of $\\tan$ definitively ends. But within this range we obviously have $$\\tan\\theta={\\sin\\theta\\over\\cos\\theta}\\qquad\\left(-{\\pi\\over2}<\\theta<{\\pi\\over2}\\right)\\ .$$ On the other hand it is possible, using your drawing, to extend $\\cos$ and $\\sin$ to all of ${\\mathbb R}$, whereby the known properties of these functions are preserved. It is then natural to extend also the definition of $\\tan$ to a domain as large as possible by putting $$\\tan\\theta:={\\sin\\theta\\over\\cos\\theta}$$ whenever $\\cos\\theta\\ne0$, and leaving $\\tan$ undefined when $\\cos\\theta=0$, i.e., at the odd multiples of ${\\pi\\over2}$. Since $$\\cos(\\theta+\\pi)=-\\cos\\theta,\\quad \\sin(\\theta+\\pi)=-\\sin\\theta\\qquad\\forall\\theta\\in{\\mathbb R}$$ it immediately follows that $\\tan$ has period $\\pi$ on its domain of definition. Furthermore $\\tan$ is odd, and is strictly monotonically increasing on the interval $\\bigl[0,{\\pi\\over2}\\bigr[\\$, because $\\sin$ is increasing and $\\cos$ is decreasing there. Therefore $\\tan$ maps the interval $\\bigl[0,{\\pi\\over2}\\bigr[\\$ bijectively onto ${\\mathbb R}_{\\geq0}$ (as we all know), and by symmetry it follows that $$\\tan:\\quad I:=\\left]{-{\\pi\\over2}},{\\pi\\over2}\\right[\\ \\to\\ {\\mathbb R}$$ is bijective. As a consequence there is exactly one $\\theta\\in I$ with $\\tan\\theta=\\sqrt{3}$, namely $\\theta={\\pi\\over3}$. The $\\pi$-periodicity of $\\tan$ then implies", "was in the implications of $tan(A)*tan(B)=1$. This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving $\\displaystyle tan(A) = \\frac{1}{tan(B)} = cot(B)$. But cotangent is also the co-function–or complement function–of tangent which gave me $tan(A) = cot(B) = tan \\left( \\frac{\\pi}{2} - B \\right)$. Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \\frac{\\pi}{2} - B$, or $A+B = \\frac{\\pi}{2}$. Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation, $(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\\pi ]$ and any $\\left| b \\right| \\ge 2$, is always $3\\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions. QED Insight #6: Generalizing the Domain The posed problem can be generalized further by recognizing the period of tangent: $\\pi$. That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\\pi$ each, as shown in the GeoGebra construction above. Insights 4 & 5 proved the sum of the angles at points C & D was $\\pi/2$. Employing the periodicity of tangent, the x-coordinate of $E = C+\\pi$ and $F = D+\\pi$, so the sum of the angles at points E & F is $\\frac{\\pi}{2}", "each nonzero $$x$$, because $$f(-x) = f(x)$$ for all $$x$$. Similarly, because $$\\tan(\\theta + k \\pi) = \\tan \\theta$$ for all integers $$k$$, there are infinitely many angles whose tangent is equal to some given value. It is possible to regard the inverse tangent as a mapping of a number to a set: for instance, if $$\\tan \\theta = z$$, then $$\\tan^{-1} z = \\{ \\theta + k \\pi : k \\in \\mathbb{Z} \\}$$. Then there is no contradiction, because $$\\tan^{-1} 0 = \\{ \\ldots, -2\\pi, -\\pi, 0, \\pi, 2\\pi, \\ldots \\}$$. Further consideration of the above leads us to conclude that the claim that $$\\tan^{-1} 1 > 0$$ contains an unstated assumption, namely that a particular branch of the inverse tangent is chosen (e.g., $$-\\pi/2 < \\tan^{-1} z < \\pi/2$$ ). But of course, $$\\tan -3\\pi /4 = 1$$ just as much as $$\\tan \\pi/4 = 1$$. From a group- or number-theoretic standpoint, we are actually looking at equivalence classes of the inverse tangent, modulo $$\\pi$$. If $$a + b \\equiv c \\pmod m$$, that does not necessarily mean that $$a + b = c$$, because $$a, b, c$$ are representatives of their respective equivalence classes modulo $$m$$. - 5 years, 1 month ago Trigometry? I suppose Trigonometry? - 5 years, 1 month ago Stop highlighting grammatical", "## Trigonometry (11th Edition) Clone The solution set is $$\\{0.6089,1.3424,3.7505,4.484\\}$$ $$\\tan^2x-5\\tan x+3=0$$ over the interval $[0,2\\pi)$ Here we can treat the equation as a quadratic equation $au^2+bu+c=0$ in which $u=\\tan x$, $a=1$, $b=-5$ and $c=3$. So to solve this equation, we apply normal operations when solving a quadratic equation. - Find $\\Delta$: $\\Delta=b^2-4ac=(-5)^2-4\\times1\\times3=25-12=13$ - Find $\\tan x$: $$\\tan x=\\frac{-b\\pm\\sqrt{\\Delta}}{2a}=\\frac{5\\pm\\sqrt{13}}{2}$$ Now we continue to find $x$, using the inverse function for tangent. 1) For $\\tan x=\\frac{5+\\sqrt{13}}{2}$ $$x=\\tan^{-1}\\frac{5+\\sqrt{13}}{2}\\approx1.3424$$ However, do not forget that the period of a tangent function is only $\\pi$, which means $\\tan x$ would reach the same value every after $x$ runs another round of $\\pi$. That means, over the interval $[0,2\\pi)$, there is one more value of $x$ satisfying $\\tan x=\\frac{5+\\sqrt{13}}{2}$, which is $$x=1.3424+\\pi\\approx4.484$$ So, $$x=\\{1.3424,4.484\\}$$ 2) For $\\tan x=\\frac{5-\\sqrt{13}}{2}$ $$x=\\tan^{-1}\\frac{5-\\sqrt{13}}{2}\\approx0.6089$$ However, do not forget that the period of a tangent function is only $\\pi$, which means $\\tan x$ would reach the same value every after $x$ runs another round of $\\pi$. That means, over the interval $[0,2\\pi)$, there is one more value of $x$ satisfying $\\tan x=\\frac{5-\\sqrt{13}}{2}$, which is $$x=0.6089+\\pi\\approx3.7505$$ So, $$x=\\{0.6089,3.7505\\}$$ Combining case 1) and 2), we have the solution set: $$\\{0.6089,1.3424,3.7505,4.484\\}$$", "- \\tan^{-1}x$ and $\\tan^{-1} x = \\frac{\\pi}{2} - \\cot^{-1}x$. Because tangent is an odd function, or $\\tan (-x) = - \\tan x$, then its inverse is also odd. In $Y1$ students should input $\\frac{\\pi}{2} - \\tan^{-1}x$ in order to graph $y = \\cot^{-1} x$ Feb 23, 2012 Aug 19, 2014", "# Approximations of fixed points of tangent. This question comes from an exam, years ago. Show that $f(x)=\\tan x-x$, for every positive integer $n$, has exactly one root $x_n$ in the interval $(n\\pi,n\\pi+\\pi/2)$. And show that $$x_n=n\\pi+\\frac{\\pi}{2}-\\frac{1}{n\\pi}+\\text{o}\\left(\\frac{1}{n}\\right).$$ I can prove the claim about the existence of $x_n$, by intermidiate value-theorem, but am stuck by the second point. Since $\\tan x$ is not a contraction, we cannot apply the contraction mapping theorem here. Also, using Taylor approximation, or the Lagrange form of the remainder, I arrived at $$\\tan x=(x-n\\pi)+\\frac{f^{(3)}(\\theta)}{6}x^3$$ for some $\\theta$ between $x$ and $n\\pi$. It seems that, however, I can only obtain information about this $\\theta$ along this direction, not about $x_n$. Any hint or help is greatly appreciated. Thanks in advance. • And any edit is also appreciated. :D – awllower Jul 25 '13 at 7:58 First note that $x_n-n\\pi\\to\\frac\\pi2\\cdot$ This is because $0<x_n-n\\pi<\\frac\\pi2$ and $\\tan(x_n-n\\pi)=\\tan(x_n)=x_n\\to+\\infty$. Now, put $z_n=x_n-n\\pi-\\frac\\pi2\\cdot$ Then $z_n\\to 0$, so $z_n\\sim \\tan(z_n)$ as $n\\to\\infty$. But $\\tan(z_n)=\\tan(x_n-\\frac\\pi2)=-\\tan(x_n)=-\\frac1{x_n}$, so we get $z_n\\sim -\\frac1{x_n}\\cdot$ Now, $x_n\\sim n\\pi$ since $n\\pi<x_n<n\\pi+\\frac\\pi2$; so $z_n\\sim-\\frac{1}{n\\pi}$ and hence we may write $z_n=-\\frac{1}{n\\pi}+o(\\frac1n).$ This is what you want by the definition of $z_n$.", "on the left and a positive number on the right, can we? But we can add or subtract something on the left or right in order to make the two sides equal after all. What is to be shown then is that the thing to add or subtract is a constant for all $$x$$ and $$y$$ that satisfy the given conditions, and that subtracting the particular constant $$\\pi$$ from the right side will make the equation satisfied. To actually solve a problem like this, we have to keep in mind that the trigonometric functions are not one-to-one, and therefore the inverse trigonometric functions are not true inverses. For example, $$\\tan\\left(\\frac34\\pi\\right) = -1,$$ but $$\\tan^{-1}(-1) = -\\frac14\\pi \\neq \\frac34\\pi.$$ As long as we restrict the domain of the tangent to angles in the interval $$\\left(-\\frac12\\pi,\\frac12\\pi\\right),$$ the function is one-to-one, the inverse tangent is actually an inverse of the tangent, and everything is fine. As soon as any angles go outside that interval, things get more complicated. For real numbers $$x$$ and $$y$$, let $$\\alpha = \\tan^{-1} x$$ and $$\\beta = \\tan^{-1} y.$$ That is, let $$\\alpha$$ and $$\\beta$$ be the unique angles in the interval $$\\left(-\\frac12\\pi,\\frac12\\pi\\right)$$ such that $$x = \\tan\\alpha$$ and $$y = \\tan\\beta.$$ Then $$\\tan(\\alpha + \\beta) = \\frac{\\tan\\alpha + \\tan\\beta}{1 - \\tan\\alpha\\tan\\beta} = \\frac{x + y}{1", "# Tangent is one-to-one with a restricted domain. I want to show the function $\\tan(\\pi x - \\frac \\pi2)$ is one-to-one if x $\\in (0,1)$. But an argument I would normally use to prove a function is one-to-one (letting $\\tan(\\pi x_1 - \\frac \\pi2)$ = $\\tan(\\pi x_2 - \\frac \\pi2)$ and then showing $x_1$ = $x_2$) doesn't seem to work. I was hoping someone could give me any suggestions. Thanks. - The derivative is defined (exists) and positive on $(0.1)$, so the function is increasing over the interval. If $x\\in(0,1)$, then $0<x<1$, so $-\\frac{\\pi}2<\\pi x-\\frac{\\pi}2<\\frac{\\pi}2$. Note that $\\tan\\theta$ is one-to-one on the interval $-\\frac{\\pi}2<\\theta<\\frac{\\pi}2$: each $\\theta$ determines a unique line through the origin making an angle $\\theta$ with the positive $x$-axis, the point $\\langle 1,\\tan\\theta\\rangle$ is on that line, and these lines meet only at the origin, so these points are all distinct, and therefore the values of $\\tan\\theta$ are all distinct.", "that the values of $$\\tan{x}$$ increases as $$x$$ increases in $$\\left(0,\\dfrac{\\pi}{2}\\right)$$. The function has vertical asymptotes at $$x=\\dfrac{\\pi}{2}, \\dfrac{3\\pi}{2}, -\\dfrac{\\pi}{2}$$. Range and Domain of Tangent Function and Other Properties We make the following observations: • The $$\\tan x$$ functions is not defined for all odd multiples of $$\\frac{\\pi }{2}$$. This is because the base is 0 for these positions. Thus, the domain of the $$\\tan x$$ function is $$\\mathbb{R}\\ - \\left\\{ {\\left( {2n + 1} \\right)\\frac{\\pi }{2}} \\right\\}$$, where $$n$$ belongs to the set of integers. • The range of this function is $$\\mathbb{R}$$, as the $$y$$-variation in the graph is from negative infinity to positive infinity. • The $$\\tan x$$ function has a cycle of $$\\pi$$, instead of $$2\\pi$$. Domain of the $$\\tan x$$ function is $$\\mathbb{R}\\ - \\left\\{ {\\left( {2n + 1} \\right)\\frac{\\pi }{2}} \\right\\}$$, where $$n$$ belongs to the set of integers and the range of $$\\tan x$$ function is $$\\mathbb{R}$$. The other properties of the tangent function are listed below: Important Notes 1. The tangent function is a periodic function and it has a period of $$\\pi$$. 2. The tan function is an odd function because $$\\tan{(-x)}=-\\tan{x}$$. 3. The slope of a straight line is the tangent of the angle made by the line with the positive $$x$$-axis. 4. The graph of $$\\tan{x}$$ is", "point where $x=0$. But as $x\\uparrow\\pi/2$, $\\tan x$ goes up to $\\infty$ and $2\\sin x$ does not, so $\\tan x$ has to overtake $2\\sin x$. So the question is, at what point does $\\tan x$ overtake $2\\sin x$, and that has to happen when they're both in the same place. In other words, we need to solve $$\\tan x = 2\\sin x.$$ Dividing both sides by $\\sin x$ (so this applies only at points where $\\sin x\\ne0$), we get $$\\frac{1}{\\cos x} = 2,$$ so $\\cos x=1/2$. And as you learned in trigonometry, that happens when $x=\\pi/3$. Thus the proposed inequality holds for nonnegative values of $x$ satisfying $0\\le x\\le \\pi/3$. For negative values of $x$, just recall that sine and tangent are odd functions.", "Trigonometry (11th Edition) Clone We cannot use the method in Example 2 to find a formula for $\\tan(270^\\circ-\\theta)$ because if we do so, we would be stopped halfway when we need a defined value of $\\tan270^\\circ$ to carry on. However, $\\tan270^\\circ$ is unfortunately undefined. To understand why we cannot use the identity of the tangent difference here to find a formula for $\\tan(270^\\circ-\\theta)$, we would use the same identity to expand $\\tan(270^\\circ-\\theta)$ to find out where the problem is. Identity of tangent difference: $$\\tan(A-B)=\\frac{\\tan A-\\tan B}{1+\\tan A\\tan B}$$ Apply it to $\\tan(270^\\circ-\\theta)$ $$\\tan(270^\\circ-\\theta)=\\frac{\\tan270^\\circ-\\tan\\theta}{1+\\tan270^\\circ\\tan\\theta}$$ The problem lies in here. We cannot go any further, since $\\tan270^\\circ$ is undefined. The reason why $\\tan270^\\circ$ could be explained by recalling that $$\\tan270^\\circ=\\frac{\\sin270^\\circ}{\\cos270^\\circ}$$ $\\sin270^\\circ=\\sin(-90^\\circ)=-\\sin90^\\circ=-1$ $\\cos270^\\circ=\\cos(-90^\\circ)=\\cos90^\\circ=0$ That means $$\\tan270^\\circ=\\frac{-1}{0}$$ which is obviously undefined. So to conclude, the main problem here is that $\\tan270^\\circ$ is undefined, meaning the formula cannot be found by using the identity of tangent difference as the value of $\\tan270^\\circ$ being defined is essential.", "# Let $L$ be the tangent line to $y=\\tan(2x)$ at $(\\frac{\\pi}2,0)$. What is the $y$-intercept of $L$? Let $$L$$ be the tangent line to $$y=\\tan(2x)$$ at $$(\\frac{\\pi}2,0)$$. What is the $$y$$-intercept of $$L$$? (a) $$\\,\\,\\,0 \\,\\,\\,$$(b) $$\\,\\,\\,\\frac{\\pi}2\\,\\,\\,$$ (c) $$\\,\\,\\,-\\pi\\,\\,\\,$$ (d) $$\\,\\,\\,1\\,\\,\\,$$ (e) $$\\,\\,\\,2\\,\\,\\,$$ The equation $$y=\\tan(2x)$$ has tangent line, of slope $$m$$, given by general equation: $$y-y_1 = 2\\sec^2(2x)(x - x_1)$$, where $$(x,y)$$ is any point generally lying on the tangent, where $$y_1$$ is the y-intercept and $$x_1$$ is the x-intercept of $$L$$. I expect the question means that the domain is given by $$(\\frac{\\pi}2,0)$$. So, for $$\\,\\,2x\\,\\,$$ the domain is given by $$(\\pi,0)$$. There seems no way, except to have line $$x=\\tan(2x)$$ & seek a solution such that the solution fits for intersection of the given curve with its tangent line. Please help as unable to pursue further. The reason being that solving $$x=\\tan(2x)$$ requires approximation. Edit : The question was based on the premise that: A curve can have an infinite number of tangents to it in any given interval. So, if need to find any tangent to a curve in a given domain, then how to approach the problem. I hoped that the problem needs some intersection point, which the two answers , & comment by @user10354138 have taken to be given. But is it", "# A Region Lying Between Two Intersecting Graphs • January 31st 2013, 07:01 PM asilvester635 A Region Lying Between Two Intersecting Graphs How does tanx = 1 be pi/4 or 5pi/4?????? I'm confused • January 31st 2013, 08:35 PM ibdutt Re: A Region Lying Between Two Intersecting Graphs tan x = 1, tangent is positive in first and third quadrant. Hence the tan x will be 1 for x = pi/4 ans pi + pi/4 = 5pi/4 • January 31st 2013, 10:58 PM asilvester635 Re: A Region Lying Between Two Intersecting Graphs Tangent is a periodic function with period $\\pi$. If, for some x and y, tan(x)= y then it is also true that $tan(x+ n\\pi)= y$ for any integer, n. In particular, $tan(\\pi/4)= \\frac{sin(\\pi/4)}{cos(\\pi/4)}= \\frac{\\sqrt{2}/2}{\\sqrt{2}/2}= 1$ so $tan(\\pi/4+ \\pi)= tan(5\\pi/4)= 1$. It is also true that $tan(\\pi/4- \\pi)= tan(-3\\pi/4)= 1$, that $tan(\\pi/4+ 2\\pi)= tan(9\\pi/4)= 1$, and that, in general, $tan(\\pi/4+ n\\pi)= tan(((4n+1)/4)\\pi)= 1$ for any integer, n." ]

[ "= \\frac {2}{3}\\\\ \\tan\\theta = \\frac {2}{3}\\\\ \\theta = \\arctan (\\frac {2}{3})$ This is not one of your special angles, so you will need a calculator. $\\theta = 33.69^\\circ$ $\\tan \\theta$ is not a 1-1 function. The inverse function maps back to a value in $[0,180^\\circ)$. There is second answer in the interval $[0,360^\\circ)$. $\\tan \\theta = \\tan (\\theta + 180)$ $213.69^\\circ$ is also in the solution set. The next one might even be a little bit trickier. $\\cos 3\\theta = -0.5\\\\ 3\\theta = \\arccos (-0.5)\\\\ 3\\theta = 120^\\circ$ but again, this isn't the only answer. $\\theta = \\arccos (-0.5)$ has two answers in the 360 degree interval, and $3\\theta = \\arccos (-0.5)$ will have $6.$ $\\cos \\theta = \\cos (-\\theta)$ and $\\cos \\theta = \\cos (360+\\theta)$ using these identities we find other values of theta. $3\\theta = 240^\\circ, 480^\\circ, 600^\\circ, 840^\\circ, 960^\\circ$ are all viable, as well. $\\theta \\in \\{40, 80, 160, 200, 280, 320\\}$ a) Reduce to $$\\tan\\theta=\\frac23.$$ Then as the tangent function has period $180°$ and is monotonic in a period, $$\\tan\\theta=\\tan(\\theta+k180°)$$ and all solutions are of the form $$\\theta+k180°=\\arctan\\frac23.$$ Find all $k$ such that $0\\le\\theta\\le360°$. b) The cosine has period $360°$ and is an even function monotonic in $[0°,180°]$ so that $$\\cos3\\theta=\\cos(k360°\\pm3\\theta).$$ Then all solutions are of the form $$k120°\\pm\\theta=\\frac13\\arccos(-0.5).$$ Again, find all $k$", "## Trigonometry (11th Edition) Clone The solution set is $$\\{135^\\circ+180^\\circ n, n\\in Z\\}$$ $$\\tan\\theta+1=0$$ 1) Solve the equation over the interval $[0^\\circ,360^\\circ)$ $$\\tan\\theta+1=0$$ $$\\tan\\theta=-1$$ Over the interval $[0^\\circ, 360^\\circ)$, there are two values of $\\theta$ where $\\tan\\theta=-1$, which are $135^\\circ$ and $315^\\circ$. Therefore, $$\\theta=\\{135^\\circ, 315^\\circ\\}$$ 2) Solve the equation for all solutions To find all solutions, we add the integer multiples of the period of tangent function, which is $180^\\circ$, to each solution found in step 1. - There are two solutions found in step 1, $\\{135^\\circ, 315^\\circ\\}$, so it will be written as $\\theta=135^\\circ+180^\\circ n$ and $\\theta=315^\\circ+180^\\circ n$ where $n\\in Z$. - However, a closer look reveals that $135^\\circ+180^\\circ n$ in fact covers all the possible values of $315^\\circ+180^\\circ n$, and vice versa. In other words, they are technically the same. So the solution set can be shortened to $$\\{135^\\circ+180^\\circ n, n\\in Z\\}$$", "$$x$$ = $$297.266{}^\\circ$$ and $$x$$ = $$242.734{}^\\circ$$ EXAMPLE $$\\PageIndex{7}$$ Find all solutions to $$\\tan \\left(x\\right)=3$$ on $$0 \\le x<2\\pi$$. Solution Using the inverse tangent function, we can find one solution $$x=\\tan ^{-1} \\left(3\\right) \\approx 1.249$$. Unlike the sine and cosine, the tangent function only attains any output value once per cycle, so there is no second solution in any one cycle. By adding $$\\pi$$, a full period of tangent function, we can find a second angle with the same tangent value. Notice this gives another angle where the line has the same slope. If additional solutions were desired, we could continue to add multiples of $$\\pi$$, so all solutions would take on the form $$x=1.249+k\\pi$$, however we are only interested in $$0 \\le x< 2\\pi$$. $x=1.249+\\pi =4.391\\nonumber$ The two solutions on $$0\\le x<2\\pi$$ are $$x = 1.249$$ and $$x = 4.391$$. Exercise $$\\PageIndex{3}$$ Find all solutions to $$\\tan \\left(x\\right)=0.7$$ on $$0{}^\\circ \\le x<360{}^\\circ$$. The first solution is $$x=\\tan ^{-1} (0.7) \\approx 34.992{}^\\circ$$. For a standard tangent, the second solution can be found by adding a full period, $$180\\mathrm{{}^\\circ}$$, giving $x=180{}^\\circ +34.99{}^\\circ =214.992{}^\\circ\\nonumber$ EXAMPLE $$\\PageIndex{8}$$ Solve $$3\\cos \\left(t\\right)+4=2$$ for all solutions on one cycle, $$0\\le t<2\\pi$$ Solution $3\\cos \\left(t\\right)+4=2\\nonumber$Isolating the cosine $3\\cos \\left(t\\right)=-2\\nonumber$ $\\cos \\left(t\\right)=-\\dfrac{2}{3}\\nonumber$Using the inverse, we can find one solution $-\\dfrac{2}{3} -\\dfrac{2}{3} t=\\cos ^{-1} \\left(-\\dfrac{2}{3} \\right)\\approx 2.301\\nonumber$", "## Trigonometry (11th Edition) Clone The solution set is $$\\{0.6089,1.3424,3.7505,4.484\\}$$ $$\\tan^2x-5\\tan x+3=0$$ over the interval $[0,2\\pi)$ Here we can treat the equation as a quadratic equation $au^2+bu+c=0$ in which $u=\\tan x$, $a=1$, $b=-5$ and $c=3$. So to solve this equation, we apply normal operations when solving a quadratic equation. - Find $\\Delta$: $\\Delta=b^2-4ac=(-5)^2-4\\times1\\times3=25-12=13$ - Find $\\tan x$: $$\\tan x=\\frac{-b\\pm\\sqrt{\\Delta}}{2a}=\\frac{5\\pm\\sqrt{13}}{2}$$ Now we continue to find $x$, using the inverse function for tangent. 1) For $\\tan x=\\frac{5+\\sqrt{13}}{2}$ $$x=\\tan^{-1}\\frac{5+\\sqrt{13}}{2}\\approx1.3424$$ However, do not forget that the period of a tangent function is only $\\pi$, which means $\\tan x$ would reach the same value every after $x$ runs another round of $\\pi$. That means, over the interval $[0,2\\pi)$, there is one more value of $x$ satisfying $\\tan x=\\frac{5+\\sqrt{13}}{2}$, which is $$x=1.3424+\\pi\\approx4.484$$ So, $$x=\\{1.3424,4.484\\}$$ 2) For $\\tan x=\\frac{5-\\sqrt{13}}{2}$ $$x=\\tan^{-1}\\frac{5-\\sqrt{13}}{2}\\approx0.6089$$ However, do not forget that the period of a tangent function is only $\\pi$, which means $\\tan x$ would reach the same value every after $x$ runs another round of $\\pi$. That means, over the interval $[0,2\\pi)$, there is one more value of $x$ satisfying $\\tan x=\\frac{5-\\sqrt{13}}{2}$, which is $$x=0.6089+\\pi\\approx3.7505$$ So, $$x=\\{0.6089,3.7505\\}$$ Combining case 1) and 2), we have the solution set: $$\\{0.6089,1.3424,3.7505,4.484\\}$$", "\\tan t_{2}$$. What does $$t_1<>t_2$$ mean? Also how did you get the solution $$t_1 =t_2+k\\pi,\\, \\, k=\\pm 1, \\pm2,\\ldots$$. Do I just substitute this back in for the equations for $$x$$ and $$y$$? How would you find the equations of both tangents? I have only one point. Thanks Last edited: Nov 26, 2006 5. Nov 27, 2006 ### HallsofIvy Staff Emeritus $t_1<>t_2$ means \"t1 is NOT equal to t2. It's a computer programming notation. He got $t_1 =t_2+k\\pi,\\, \\, k=\\pm 1, \\pm2,\\ldots$ by remembering that tan(x) is periodic with period $\\pi$! Well, you can't substitute $t_1 =t_2+k\\pi,\\, \\, k=\\pm 1, \\pm2,\\ldots$ because you don't know what t2 is! Substitute those back into your other equation to solve for both t1 and t2. Use either to determine x and y. Last edited: Nov 27, 2006 6. Nov 27, 2006 ### sherlockjones Sorry for my stupidity, but do you mean to do this: $$\\tan t_{1} = \\tan (t_{1}-k\\pi)$$? $$t_{1} = \\arctan (t_{1}-k\\pi)$$ 7. Nov 27, 2006 ### HallsofIvy Staff Emeritus Yes, that's true because tangent has period $\\pi$ No, that's not true. Even $$arctan(tan t_1)= \\arctan(tan(t_1- k\\pi))$$ so that $t_1= t_1- k\\pi$ isn't correct. Since tangent is periodic it is not one-to-one and arctan is not a true inverse.", "curve at an interval. Jun 22 '19 at 20:00 The equation of the tangent line at the point $$(a,f(a))$$ is $$L(x)=f'(a)(x-a)+f(a)$$ In this case, $$f(x)=\\tan(2x)$$ and $$a=\\pi/2$$. Hence $$L(x)=2\\sec^2(2\\cdot \\tfrac{\\pi}{2})(x-\\tfrac{\\pi}{2})+\\tan(2\\cdot \\tfrac{\\pi}{2})$$ i.e. $$L(x)=2(x-\\tfrac{\\pi}{2})=2x-\\pi$$. Therefore, the $$y$$-intercept of $$L$$ is $$-\\pi$$. • Unable to state $y$ as $L(x)$, request elaboration. Also, the solution does not fall into the given domain for $2x$. Also, how it is so fixed that the only tangent is at $x = \\frac{\\pi}2$. Jun 22 '19 at 19:41 • Is not the only tangent line, $x=\\frac{\\pi}{2}$ is the point in the given problem. Jun 22 '19 at 19:48 • And, what's domain? Jun 22 '19 at 19:48 • I have stated also in post that the question is expected to state the domain as $(\\frac{\\pi}2,0)$. Jun 22 '19 at 19:49 • $\\tan(2\\cdot \\tfrac{\\pi}{2})=\\tan(\\pi)=0$ Jun 22 '19 at 19:53", "# Thread: equation of tangent line 1. ## equation of tangent line Find an equation of the tangent line to the curve at the given point. y = 6 sec x P = (π/3, 12) I know that the first step is taking the derivative of 6 sec x, but i dont quit know how to deal with the pi over 3 when i use it in the derivative to find the slope. 3. Originally Posted by bigwave $y=6sec(x)$ $x= \\frac{\\pi}{3}$ $y'=6\\cos{\\frac{\\pi}{3}}=3$ $y-12=(3)(x-\\frac{\\pi}{3})$ $ y=3x-\\pi+12$ the derivative of sec(x) is sec(x)tan(x) though 4. Originally Posted by rhcp1231 the derivative of sec(x) is sec(x)tan(x) though You're correct. The methodology is the same though. $y' = 6 \\sec x \\tan x = 6 \\sec \\left(\\frac{\\pi}{3}\\right) \\tan \\left(\\frac{\\pi}{3}\\right)$ Are you asking how to calculate $\\sec \\left(\\frac{\\pi}{3}\\right)$ and $\\tan \\left(\\frac{\\pi}{3}\\right)$ ? $y' = \\frac{6\\sin{x}}{\\cos^2{x}}= 12\\sqrt{3}$ 6. Originally Posted by drumist You're correct. The methodology is the same though. $y' = 6 \\sec x \\tan x = 6 \\sec \\left(\\frac{\\pi}{3}\\right) \\tan \\left(\\frac{\\pi}{3}\\right)$ Are you asking how to calculate $\\sec \\left(\\frac{\\pi}{3}\\right)$ and $\\tan \\left(\\frac{\\pi}{3}\\right)$ ? So, after you've differentiated to $\\frac{\\mathrm{d} y}{\\mathrm{d} x}=6\\sec x\\tan x$ Then plug in your x value and you'll get: $6\\cdot \\sec \\frac{\\pi}{3}\\cdot \\tan \\frac{\\pi}{3}=6\\cdot 2\\cdot \\sqrt 3=12\\sqrt 3=m$ Then plug it in to slope intercept form. $y=mx+b$ $12=(12\\sqrt{3})(\\frac{\\pi}{3})+b$ $12=4\\pi", "# Math Help - Help finding x 1. ## Help finding x Hi can anybody show me how to find x? -180= -90 - tan^-1 (x/16) - tan^-1 (x/200) 2. ## There are multiple ways to do this Originally Posted by al2308 Hi can anybody show me how to find x? -180= -90 - tan^-1 (x/16) - tan^-1 (x/200) There is no alebraic way I am away of but you could do this... Say $f(x)=90-arctan\\bigg(\\frac{x}{16}\\bigg)-arctan\\bigg(\\frac{x}{200}\\bigg)$ then we want $f(x)=0$...I would graph...or use the intermediate value theorem...or just use a graphing calculator 3. Hello, This is straaaaaaaaaaaaaange 90=arctan(x/16)+arctan(x/200) If we compose with tan, we have : tan(90)=tan(arctan(x/16)+arctan(x/200)) But tan(90)= infinity :'( We can find somewhere the general formula for tan(a+b). I haven't looked further, maybe there is a solution though. 4. Originally Posted by Moo Hello, This is straaaaaaaaaaaaaange 90=arctan(x/16)+arctan(x/200) If we compose with tan, we have : tan(90)=tan(arctan(x/16)+arctan(x/200)) But tan(90)= infinity :'( We can find somewhere the general formula for tan(a+b). I haven't looked further, maybe there is a solution though. He is right I am so used to putting cookie cutter responses to find the x...since $tan\\bigg(\\frac{\\pi}{2}\\bigg)=\\infty$ and the image of $arctan(x)$ is $\\bigg(\\frac{-\\pi}{2},\\frac{\\pi}{2}\\bigg)$...therefore there is no solutions..that is unless this is in radians and not degrees? wait once again...I didnt look at this...I didnt read it", "The local lines of symmetry on the graph of $\\cos(x)$ are every $180^\\circ$ starting at $0^\\circ$. If there is a solution at $60^\\circ$, for example, there is also one at $360-60=300^\\circ$. Both of these solutions then repeat every $360^\\circ$. $y=\\sin(x):$ The local lines of symmetry on the graph of $\\sin(x)$ are every $180^\\circ$ starting at $90^\\circ$. If there is a solution at $30^\\circ$, for example, there is also one at $180-30=150^\\circ$. Both of these solutions then repeat every $360^\\circ$. $y=\\tan(x):$ The solutions on the tan graph are easier to find, the default value simply repeats every $180^\\circ$. If there is a solution at $45^\\circ$, for example, there is another at $-135^\\circ$, $225^\\circ$, etc. The tan graph is different to the sin and cos graphs. Firstly, tan can take a value between $-\\infty$ and $\\infty$ whereas sin and cos take values between -1 and 1. The graph of tan also has asymptotes as a result of dividing by $\\cos(x)=0$ in the expression $\\tan(x)=\\frac{\\sin(x)}{\\cos(x)}$. ## A note on 'CAST' It may be that the argument of the trigonometric function is a function of $x$. In these cases it is very important to extend the interval of solution values so that the full set of solutions are obtained. See Examples 2 and 4. Some students prefer to use ‘CAST’ – this", "$$\\tan{(\\alpha-\\beta)}=\\frac{\\tan{\\alpha}-\\tan{\\beta}}{1+\\tan{\\alpha}\\tan{\\beta}}$$ ehild 13. Apr 11, 2010 ### MaceLee Ah silly mistake, I took the tan separately. Thanks again ehild. 14. Apr 11, 2010 ### ehild I made a mistake. After introducing the new variable x=kv, dv=dx/k, we have the equation $$\\int{\\frac{dx}{x^2+1}} = - \\mu k t$$ so the solution is: $$\\arctan{kv(t)} = \\arctan{k v_0}-\\mu k t$$ taking the tangent of both sides: $$v(t)=\\frac{v_0 - 1/k \\tan{\\mu k t}}{1+kv_0 \\tan{ \\mu k t}$$ I hope this is correct. ehild Last edited: Apr 12, 2010 15. Apr 12, 2010 ### MaceLee Thanks ehild. I tried to use the solution to find predictions for experimental data, but the values are extremely strange. I'm not sure if it's the way I've worked out $$v_0$$, but seemingly regardless of the value, the output jumps around really weirdly. The data I'm supposed to be modelling with the equation is: t v 10 50 11 46 12 41 13 38 14 34 15 31 16 27 17 24 18 21 19 18 20 16 21 13 22 10 23 8 24 5 25 3 26 0 But the actual values that come out as v are more on the lines of: 49.93375881 -11.79891637 -129.3575502 85.61016673 2.555699999 -70.00352361 173.8124513 17.27837342 -41.21109751 1353.425599 34.75669316 -22.1799659 -239.8506759 59.50383843 -6.916230932 -102.9952062 104.9325927 It jumps from negative to positive pretty", "x+y$$ and evaluate: $\\tan (z+u)=\\frac{\\tan z+\\tan u}{1-\\tan z \\cdot \\tan u}=\\frac{-3+3}{1-9}=0$ So $$\\tan( x+y+z ) = 0$$. Hence, $$\\arctan (1)+\\arctan (2)+\\arctan (3)=x+y+z=\\pi$$. ### Intersection of a line and periodic function I A1, 2012 Find the number of real solutions to the equation $$x=99 \\sin (\\pi x)$$. Solution Since the equation is symmetric, if $$r$$ is a solution then $$-r$$ is also a solution. Hence, we restrict to finding the positive roots. Note that $$x=0$$ is also a root. The function $$99 \\sin \\pi x$$ is positive when $$2k < x < 2k+1$$. It equals $$x$$ when $$x\\leq 99$$. The positive cycle of $$\\sin\\pi x$$ meets $$y=x$$ twice, so there are 198 positive solutions. Counting the zero gives 199 solutions. ### Intersection of a line and periodic function II A3, 2011 Find the number of solutions to the equation $$4 \\sin (3x+2)=1$$ in the domain $$x\\in[0,2\\pi]$$. Solution We want the number of intersection points between the line $$y=1$$ and the periodic function $$f(x) = 4\\sin (3x+2)$$. The function $$f$$ has period $$2\\pi/3$$. Between every maxima and a minima, there is exactly one solution. At $$x=0$$ and $$x=2\\pi$$, $$f$$ is close to maxima. So all we need to do is find the number of minimas of $$f$$ in the given domain and multiply it by 2. The minimas occur at", "## Trigonometry (11th Edition) Clone arctan($-1$)$=-45^{o}$ Inverse Tangent Function $y=\\tan^{-1}x$ or $y=$ arctan $x$ means that $x=\\tan y$, for $-\\displaystyle \\frac{\\pi}{2} < y < \\frac{\\pi}{2}$. Or for an angle in degrees, $\\theta=\\tan^{-1}x$ means $x=\\tan\\theta$, for $-90^{o} < \\theta < 90^{o}$. --------------- Knowing: $\\quad \\tan 45^{o}=1$, (and $\\tan(-\\theta)=-\\tan\\theta),$ we find $\\theta=-45^{o}$ from the interval $-90^{o} < \\theta < 90^{o}$ such that $\\tan$($-45^{o})=-1.$ So, arctan($-1$)$=-45^{o}$", "the function $y=0.5\\tan(\\frac{\\pi}{2}x)\\\\$. Solution First, we identify A and B. Figure 2 Because $A=0.5$ and $B=\\frac{\\pi}{2}\\\\$, we can find the stretching/compressing factor and period. The period is $\\frac{\\pi}{\\frac{\\pi}{2}}=2\\\\$, so the asymptotes are at $x=\\pm 1$. At a quarter period from the origin, we have $\\begin{array}f(0.5)=0.5\\tan(\\frac{0.5\\pi}{2}) \\hfill& \\\\ =0.5\\tan(\\frac{\\pi}{4}) \\hfill& \\\\ =0.5 \\end{array}\\\\$ This means the curve must pass through the points(0.5,0.5),(0,0),and(−0.5,−0.5).The only inflection point is at the origin. Figure shows the graph of one period of the function. Figure 3 Try It 1 Sketch a graph of $f(x)=3\\tan\\left(\\frac{\\pi}{6}x\\right)\\\\$. Solution Graphing One Period of a Shifted Tangent Function Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we add C and D to the general form of the tangent function. $f(x)=A\\tan(Bx−C)+D\\\\$ The graph of a transformed tangent function is different from the basic tangent function tan x in several ways: A General Note: Features of the Graph of y = Atan(Bx−C)+D • The stretching factor is |A|. • The period is $\\frac{\\pi}{|B|}\\\\$. • The domain is $x\\ne\\frac{C}{B}+\\frac{\\pi}{|B|}k\\\\$, where k is an integer. • The range is (−∞,−|A|] ∪ [|A|, ∞). • The vertical asymptotes occur at $x=\\frac{C}{B}+\\frac{\\pi}{2|B|}k\\\\$, where k is an odd integer. • There is no amplitude. • $y=A\\tan(Bx)$ is an", "algebraic result. (I know this is uncharacteristic for my typically algebraic tendencies.) The final key was in the implications of $tan(A)*tan(B)=1$. This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving $\\displaystyle tan(A) = \\frac{1}{tan(B)} = cot(B)$. But cotangent is also the co-function–or complement function–of tangent which gave me $tan(A) = cot(B) = tan \\left( \\frac{\\pi}{2} - B \\right)$. Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \\frac{\\pi}{2} - B$, or $A+B = \\frac{\\pi}{2}$. Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation, $(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\\pi ]$ and any $\\left| b \\right| \\ge 2$, is always $3\\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions. QED Insight #6: Generalizing the Domain The posed problem can be generalized further by recognizing the period of tangent: $\\pi$. That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\\pi$ each, as shown in the GeoGebra construction above. Insights 4 & 5 proved the sum of the angles at points C & D was $\\pi/2$. Employing the periodicity of tangent, the x-coordinate of $E = C+\\pi$ and $F", "know this is uncharacteristic for my typically algebraic tendencies.) The final key was in the implications of $tan(A)*tan(B)=1$. This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving $\\displaystyle tan(A) = \\frac{1}{tan(B)} = cot(B)$. But cotangent is also the co-function–or complement function–of tangent which gave me $tan(A) = cot(B) = tan \\left( \\frac{\\pi}{2} - B \\right)$. Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \\frac{\\pi}{2} - B$, or $A+B = \\frac{\\pi}{2}$. Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation, $(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\\pi ]$ and any $\\left| b \\right| \\ge 2$, is always $3\\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions. QED Insight #6: Generalizing the Domain The posed problem can be generalized further by recognizing the period of tangent: $\\pi$. That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\\pi$ each, as shown in the GeoGebra construction above. Insights 4 & 5 proved the sum of the angles at points C & D was $\\pi/2$. Employing the periodicity of tangent, the x-coordinate of $E = C+\\pi$ and $F = D+\\pi$, so", "this is uncharacteristic for my typically algebraic tendencies.) The final key was in the implications of $tan(A)*tan(B)=1$. This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving $\\displaystyle tan(A) = \\frac{1}{tan(B)} = cot(B)$. But cotangent is also the co-function–or complement function–of tangent which gave me $tan(A) = cot(B) = tan \\left( \\frac{\\pi}{2} - B \\right)$. Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \\frac{\\pi}{2} - B$, or $A+B = \\frac{\\pi}{2}$. Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation, $(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\\pi ]$ and any $\\left| b \\right| \\ge 2$, is always $3\\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions. QED Insight #6: Generalizing the Domain The posed problem can be generalized further by recognizing the period of tangent: $\\pi$. That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\\pi$ each, as shown in the GeoGebra construction above. Insights 4 & 5 proved the sum of the angles at points C & D was $\\pi/2$. Employing the periodicity of tangent, the x-coordinate of $E = C+\\pi$ and $F = D+\\pi$, so the", "this is uncharacteristic for my typically algebraic tendencies.) The final key was in the implications of $tan(A)*tan(B)=1$. This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving $\\displaystyle tan(A) = \\frac{1}{tan(B)} = cot(B)$. But cotangent is also the co-function–or complement function–of tangent which gave me $tan(A) = cot(B) = tan \\left( \\frac{\\pi}{2} - B \\right)$. Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \\frac{\\pi}{2} - B$, or $A+B = \\frac{\\pi}{2}$. Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation, $(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\\pi ]$ and any $\\left| b \\right| \\ge 2$, is always $3\\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions. QED Insight #6: Generalizing the Domain The posed problem can be generalized further by recognizing the period of tangent: $\\pi$. That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\\pi$ each, as shown in the GeoGebra construction above. Insights 4 & 5 proved the sum of the angles at points C & D was $\\pi/2$. Employing the periodicity of tangent, the x-coordinate of $E = C+\\pi$ and $F = D+\\pi$, so the", "equivalent to $\\theta=\\arctan\\dfrac yx$. The general solution is $\\theta\\equiv\\arctan\\dfrac yx\\mod \\pi$, and you have to choose the value of $\\theta$ (mod $2\\pi$) in function of the signs of $\\cos\\theta$ and $\\sin\\theta$, i.e. of $x$ and $y$. To add to the other answers here, the argument is the angle the corresponding vector on the Complex Plane would make with that of $1+0i$ In this particular case, the angle is $180^{\\circ}$ or $\\pi$ radians as in the picture. • +1 for the only answer that points out the underlying geometry (with a picture). This is the one the OP should remember. Dec 27, 2016 at 19:04 $\\tan(x)=0$ has the solutions $x=n\\pi$, so $\\pi$ is also a solution. Which solution you choose, depends on the signs of $\\sin(x)$ and $\\cos(x)$. This is why computers have a special function atan2, which calculates the solutions of $\\tan(x)=\\frac{u}{v}$ and returns the value in the proper quadrant, by accounting for the signs of $u$ and $v$ (as well as for the values where $\\tan(x)=\\frac{u}{v}$ might be ambiguous/undefined). because the angle of a complex number is not given by the ordinary artan, is necessary make an adjust of Pi if the real part is negative, add pi and ist done $\\arctan(0)=0,$ not $\\pi$, but $\\tan(\\pi)=\\tan(0)=0$ As $x$ ranges from $-\\frac \\pi 2$ to $\\frac \\pi 2$,", "Proving $\\arctan(\\sqrt{x^2+1}-x)=\\frac{\\pi}{4}-\\frac{1}{2}\\arctan(x)$ How to prove the following identity: $$(\\forall x\\in\\mathbb{R}): \\quad \\arctan(\\sqrt{x^2+1}-x)=\\dfrac{\\pi}{4}-\\dfrac{1}{2}\\arctan(x)$$ The knowing method is to prove that both sides are in $$]-\\frac{\\pi}{2}, \\frac{\\pi}{2}[$$ and they have the same value tooked by the function $$\\tan$$; but this looked to be out of reach, because there is no general identity giving $$\\tan(\\frac{\\alpha}{2})$$ in words of $$\\tan(\\alpha)$$ • \"there is no general identity giving $\\tan(\\frac\\alpha2)$ in words of $\\tan(\\alpha)$\" ummmmm.... the half angle identities? – YiFan Oct 8 '19 at 23:40 • @YiFan, I mean an identity giving $\\tan(\\frac{\\alpha}{2})$ in terms of $\\tan(\\alpha)$, this the only way that helps proving the equality. – hachemy Oct 8 '19 at 23:49 • Have a look at these demonstrations, solely based on geometrical definition of tangent dfnu.xyz/en/exercises-and-dialogues/losing-it-on-a-tangent – dfnu Oct 9 '19 at 12:36 Let $$\\theta=\\arctan(\\sqrt{x^2+1}-x),$$ then $$\\tan\\theta=\\sqrt{x^2+1}-x$$ and we can easily solve this for $$x,$$ as $$x=\\dfrac{1-\\tan^2\\theta}{2\\tan\\theta}=\\cot 2\\theta=\\tan\\left(\\dfrac{\\pi}2-2\\theta\\right)$$ which implies $$\\theta=\\dfrac{\\pi}4-\\dfrac12\\arctan(x).$$ Let arccot$$(x)=2y,x=\\cot2y,0<2y<\\pi$$ $$\\sqrt{1+x^2}-x=+\\csc2y-\\cot2y=\\tan y$$ $$\\arctan(x)=\\dfrac\\pi2-$$arccot$$(x)=\\dfrac\\pi2-2y$$", "on the left and a positive number on the right, can we? But we can add or subtract something on the left or right in order to make the two sides equal after all. What is to be shown then is that the thing to add or subtract is a constant for all $$x$$ and $$y$$ that satisfy the given conditions, and that subtracting the particular constant $$\\pi$$ from the right side will make the equation satisfied. To actually solve a problem like this, we have to keep in mind that the trigonometric functions are not one-to-one, and therefore the inverse trigonometric functions are not true inverses. For example, $$\\tan\\left(\\frac34\\pi\\right) = -1,$$ but $$\\tan^{-1}(-1) = -\\frac14\\pi \\neq \\frac34\\pi.$$ As long as we restrict the domain of the tangent to angles in the interval $$\\left(-\\frac12\\pi,\\frac12\\pi\\right),$$ the function is one-to-one, the inverse tangent is actually an inverse of the tangent, and everything is fine. As soon as any angles go outside that interval, things get more complicated. For real numbers $$x$$ and $$y$$, let $$\\alpha = \\tan^{-1} x$$ and $$\\beta = \\tan^{-1} y.$$ That is, let $$\\alpha$$ and $$\\beta$$ be the unique angles in the interval $$\\left(-\\frac12\\pi,\\frac12\\pi\\right)$$ such that $$x = \\tan\\alpha$$ and $$y = \\tan\\beta.$$ Then $$\\tan(\\alpha + \\beta) = \\frac{\\tan\\alpha + \\tan\\beta}{1 - \\tan\\alpha\\tan\\beta} = \\frac{x + y}{1" ]

[ "+0 # Hard Question Pls Help Me 0 44 1 A point has rectangular coordinates $(2,-1,-2)$ and spherical coordinates $(\\rho, \\theta, \\phi).$ Find the rectangular coordinates of the point with spherical coordinates $(\\rho, \\theta, 2 \\phi).$ Jun 6, 2020", "centered at some new point $(x_1,y_1)$ with radius $\\sqrt{x^2+(r+h-y_1)^2}$. 4. $s=r\\theta$ Last edited: Apr 21, 2013", "## anonymous 5 years ago Find the rectangular coordinates for the point whose polar coordinates are given. (6√2, 5π/3) $x=r \\cos \\theta$ and $y=r \\sin \\theta$ where polar coordinates are given in $(r,\\theta)$", "Coordinates And Complex Numbers Convert the polar coordinates to rectangular coordinates: Explanation: To convert polar coordinates to rectangular coordinates , Using the information given in the question, The rectangular coordinates are", "coordinates of the point $P$.", "$\\sin \\theta$, $\\cos \\theta$ and $\\tan \\theta$ and asks some questions about coordinate geometry. Previous assignments can be found here, but you can do this one without having done the others first.", "of $$\\theta$$, and therefore corresponds to the point common to the lines for all $$\\theta$$. That's the desired Frégier Point. $$\\square$$", "coördinates. Ans. $\\frac{dy}{dx} = \\frac{ \\rho \\cos \\theta + \\sin \\theta \\frac{d\\rho}{d\\theta} }{ -\\rho \\sin \\theta + \\cos \\theta \\frac{d\\rho}{d\\theta} }$.", "The rectangular coordinates of a point are given (-5, 2) . Use a graphing utility in radian mode to find polar coordinates of each point to three decimal places. Question The rectangular coordinates of a point are given $$(-5, 2)$$ . Use a graphing utility in radian mode to find polar coordinates of each point to three decimal places. 2021-02-12 Rec tan gular coordinates $$= ( x, y) = (-5, 2)$$ Polar coordinates $$= (r, \\theta)$$ Then, formula for finding r and theta, is given by - $$r = \\sqrt{x^2 + y^2}$$ $$r = \\sqrt{(-5)^2 + (2)^2}$$ $$r = \\sqrt{(25 + 4)}$$ $$r = \\sqrt{29}$$ $$r = 5.385$$ And, $$\\theta = \\tan^{-1} (\\frac{y}{x})$$ $$\\theta = \\tan^{-1} (\\frac{2}{-5})$$ $$\\theta = tan^{-1} (\\frac{2}{5}) [tan^{-1} (-x) = - tan^{-1} (x)]$$ $$\\theta = -0.380$$ radians Polar coordinates $$= (r, \\theta) = (5.385, - 0.380)$$ Relevant Questions The rectangular coordinates of a point are given (2, -2). Find polar coordinates of the point. Express theta in radians. Polar coordinates of a point are given $$(4, 90^\\circ)$$. Find the rectangular coordinates of each point. The rectangular coordinates of a point are (4, −4). Plot the point and find two sets of polar coordinates for the point for $$0 < 2$$. The rectangular coordinates of a point are (-4, 4). Find polar coordinates of the", "coordinates of point B corresponding to the solution $$\\theta_2$$ are given by $(2 \\cos \\theta_2 , 2 \\sin \\theta_2) = (1.79 , -0.89)$", "### Coordinates and vectors Prove that four different points $\\vec {r}_i$, $i=1,2,3,4$, are always on a sphere and find the center and the radius of the sphere.", "Class 12 | COORDINATE GEOMETRY If the points (1,1):(0,sec^2theta); and (cos e c^2theta,0) are collinear, then find the value of theta Latest from Doubtnut", "$$\\def\\ans#1{\\bbox[border:1px solid green,6pt]{#1}}$$ # Rectangular, Cylindrical, and Spherical Coordinates Since virtually everything we do in this course deals with three dimensional space, it makes sense to start with a short discussion of how to represent a point in 3-D space. Three dimensional space is often written $$\\mathbb{R}^3$$ (read \"R three\"), to denote that we're dealing with real numbers in three dimensions; similarly, 2-D space is called $$\\mathbb{R}^2$$, the number line is called $$\\mathbb{R}$$, and n-dimensional space is called $$\\mathbb{R}^n$$. We'll cover three ways of describing the location of a point: with rectangular coordinates, cylindrical coordinates, and spherical coordinates. There are other coordinate systems (including some wacky ones like hyperbolic and spheroidal coordinates), but these are the ones that are most commonly used for three dimensions. We won't actually use cylindrical and spherical coordinates for a while, but getting a look at them now can help to get comfortable thinking in three dimensions, and when they come back again, we'll be at least somewhat comfortable with them. As we go through this section, we'll see that in each coordinate system, a point in 3-D space is represented by three coordinates, just like a point in 2-D space is represented by two coordinates ($$x$$ and $$y$$ in rectangular, $$r$$ and $$\\theta$$ in polar). ## Rectangular Coordinates Using rectangular coordinates, a", "the rectangular coordinates to polar coordinates with r > 0 and 0 ≤ θ < 2π. (9, −9) (r, θ) = ? 4. trig The rectangular coordinates of a point are given (-5,-5root3). Find polar coordinates of the point. Express theta in radians.", "telling where points are. For example. point $$R$$ is located at $$(3, 2)$$ on the coordinate plane, because it is three units to the right and two units up.", "convert the point from cylindrical coordinates to rectangular coordinates. $$(2,-\\pi,-4)$$ Related chapters Unlock Textbook Solution", "# Converting Coordinates from Polar to Rectangular ### Topic Page Converting Coordinates from Polar to Rectangular", "\\ + \\ \\pi \\ \\ .$$ This agrees with the rectangular coordinates we found earlier.)", "Area between Radii and Curve in Polar Coordinates Theorem Let $C$ be a curve expressed in polar coordinates $\\polar {r, \\theta}$ as: $r = \\map g \\theta$ where $g$ is a real function. Let $\\theta = \\theta_a$ and $\\theta = \\theta_b$ be the two rays from the pole at angles $\\theta_a$ and $\\theta_b$ to the polar axis respectively. Then the area $\\mathcal A$ between $\\theta_a$, $\\theta_b$ and $C$ is given by: $\\displaystyle \\mathcal A = \\int \\limits_{\\theta \\mathop = \\theta_a}^{\\theta \\mathop = \\theta_b} \\frac {\\paren {\\map g \\theta}^2 \\rd \\theta} 2$ as long as $\\paren {\\map g \\theta}^2$ is integrable.", "$r^2 \\sin 2\\theta =-12$? 6. Which of the following shows the graph of, $r^2 \\sin 2\\theta =-12$, in the rectangular plane? Images/mathematical drawings are created with GeoGebra." ]

[ "\\theta + i \\sin \\theta$$. and $$i^2 = -1$$: \\begin{equation*} \\begin{split} (x + yi)\\cdot e^{i\\theta} & = (x + yi)\\cdot (\\cos \\theta + i \\sin \\theta)\\\\ & = x\\cdot \\cos \\theta + x \\cdot i \\cdot \\sin \\theta + y \\cdot i \\cdot \\cos \\theta - y \\cdot \\sin \\theta \\\\ & = (x\\cdot \\cos \\theta - y\\cdot \\sin \\theta) + i\\ (x\\cdot \\sin \\theta + y\\cdot \\cos \\theta) \\end{split} \\end{equation*} So that \\begin{align*} x' = x \\cdot \\cos \\theta - y \\cdot \\sin \\theta\\\\ y' = x \\cdot \\sin \\theta + y \\cdot \\cos \\theta \\end{align*} To rotate a point p around another point c, we need to rotate the point p expressed in the coordinate system whose origin is c (it simply is $$p-c$$), and then we just need to express the result in the initial coordinate system (whose origin is (0,0)). This gives us the following formula (if we define $$p' = p'_x + p'_yi$$): \\begin{equation*} p' = c + e^{i \\theta} \\cdot (p - c) \\end{equation*} or : \\begin{align*} p'_x = c_x + (p_x - c_x) \\cdot \\cos \\theta - (p_y - c_y) \\cdot \\sin \\theta\\\\ p'_y = c_y + (p_x - c_x) \\cdot \\sin \\theta + (p_y - c_y) \\cdot \\cos \\theta \\end{align*} Note : similar computations (with a different result) could be", "So, for example, you can choose randomly an angle $$\\phi$$ to locate the point $$C_3$$, \\begin{align} C_3&=C_1+R\\cdot(\\cos\\phi,\\sin\\phi) \\end{align} then the coordinates of the other point would be \\begin{align} C_2&=C_1+R\\cdot(\\cos(\\phi+\\theta),\\sin(\\phi+\\theta)) . \\end{align}", "n\\theta\\right) \\end{align} ...which has no imaginary part! EDIT: Since you can gleen from the initial problem that: \\begin{align} \\frac{\\sqrt{3}}{2} + \\frac{i}{2} =& \\cos\\left(60^\\circ\\right) + i\\sin\\left(60^\\circ\\right) = e^{i60^\\circ} \\\\ \\frac{\\sqrt{3}}{2} - \\frac{i}{2} =& \\cos\\left(-60^\\circ\\right) + i\\sin\\left(-60^\\circ\\right) = e^{-i60^\\circ} \\end{align} We can conclude with $\\theta = 60^\\circ$ that the final result is: \\begin{align} 2\\cdot1^{107}\\cdot\\cos\\left(107\\cdot60\\right) =& 2\\cos\\left(17\\cdot360^\\circ + 300^\\circ\\right) \\\\ =& 2\\cdot\\cos\\left(300^\\circ\\right) \\\\ =& 2\\cdot\\cos\\left(-60^\\circ\\right) = 2\\cdot\\cos\\left(60^\\circ \\right) \\\\ =& 2 \\cdot\\frac{\\sqrt{3}}{2} = \\sqrt{3} \\end{align} or \\begin{align} 2\\cdot1^{107}\\cdot\\cos\\left(107\\cdot-60\\right) =& 2\\cos\\left(-18\\cdot360^\\circ +60^\\circ\\right) \\\\ =& 2\\cdot\\cos\\left(60^\\circ\\right) \\\\ =& 2 \\cdot\\frac{\\sqrt{3}}{2} = \\sqrt{3} \\end{align} EDIT: This post is false. Sorry. In general: $$f(a+bi)+f(a-bi)$$always has a zero imaginary part, no matter what $f$ is. Reason: The two terms are conjugates of each other.* Adding two conjugates to each other gives you a real number.** *This is true, because conjugation distributes over everything: $$\\overline{f(z)}=f(\\overline z)$$Basically, what conjugation does is it replaces $i$ with $-i$. If you think about it, there's no difference between the two! They both square to $-1$, and they both aren't real. ($-i$ isn't negative, because imaginary numbers don't have a concept of \"being less than zero\" - there is no concept of \"less than\" at all!) **$(a+bi)+(a-bi)=2a$ • I don't see how this is possible. Certainly it's true of the identity function, but I seriously doubt it's true of all functions. –", "Then you make the dot product and it gives $k\\cdot r \\cdot [...]$, where $[...]$ is an ugly term full of θ,θ',ϕ and ϕ' for the angle between k and r. However, you can choose a different orthogonal axis: say one where z' in parallel to r. In these \"r-ortogonal\" axes: $\\hat x_r$' = $\\hat x_r$ ^ $\\hat z_r$' and $\\hat y_r$' = $\\hat z_r$' ^ $\\hat x_r$' and $\\hat z_r$' = r/r. So now your spherical coordinates are: $k_1=k_r\\sin{\\theta}\\cos{\\phi}$, $k_2=k_r\\sin{\\theta}\\sin{\\phi}$, $k_3=k_r\\cos{\\theta}$, and $d^3k =k_r^2\\sin{\\theta}dkd{\\theta}d{\\phi}$ and $r_1=0, r_2=0, r_3=r$. and the dot product is simply $k_r\\cdot\\ r\\cdot \\cos{\\theta}$, with $\\theta$ the angle between k and r in these \"r-axes\". Now define $\\gamma = k_r\\cdot\\ r\\cdot \\cos{\\theta}$ and dγ = $k_r\\cdot\\ r\\cdot \\sin{\\theta}d{\\theta}$ and substitute in the integrals. This will give you the $k_r/r$ term. When you integrate the $e^{i\\gamma}$ from $\\gamma = -k_rr$ to +$k_rr$, this will wive you the $\\sin{k_rr}$. - Hi DoM, thanks for the tex effort. I fixed up some of it but I'm worried about changing some of the meaning and spacing. You should make used of \\sin and \\theta and \\phi rather than unicode and \"sin\" which is s * i * n – Brandon Enright Jun 6 '13 at 6:03", "x) and $$y_2=e^{rx}\\sin(\\theta x)$$ are fundamental solutions of (35) and the general solution of (35) is $y=c_1e^{rx}\\cos(\\theta x)+c_2e^{rx}\\sin(\\theta x).$ Example. Find the general solution of the following ODE. $\\frac{d^2y}{dx^2}-6\\frac{dy}{dx}+13y=0$ Solution. The characteristic equation is \\begin{align*} r^2-6r+13&=0\\\\ r&=\\frac{6\\pm \\sqrt{(-6)^2-4\\cdot1\\cdot 13}}{2\\cdot 1}\\\\ r&=3+2i,3-2i \\end{align*} So the general solution is $y=c_1e^{3x}\\cos(2x)+c_2e^{3x}\\sin(2x).$ Example. Solve the following IVP. $y''+2y'+10y=0,\\;\\; y(0)=4,\\, y'(0)=-7$ Solution. The characteristic equation is \\begin{align*} r^2+2r+10&=0\\\\ (r+1)^2&=-9\\\\ r&=-1+3i,-1-3i. \\end{align*} So the general solution is $y=c_1e^{-x}\\cos(3x)+c_2e^{-x}\\sin(3x).$ Using the initial condition $$y(0)=4$$, we get $$$c_1=4. \\;\\;\\;\\;(36)$$$ Note that $$y'=c_1(-\\cos(3x)-3\\sin(3x))e^{-x}+c_2(-\\sin(3x)+3\\cos(3x))e^{-x}.$$ Using the initial condition $$y'(0)=-7$$, we get $$$-c_1+3c_2=-7. \\;\\;\\;\\;(37)$$$ Solving (36) and (37), we get $$c_1=4$$ and $$c_2=-1$$. Thus the solution of the IVP is $y=4e^{-x}\\cos(3x)-e^{-x}\\sin(3x).$ Last edited", "e^{i \\theta_2 } \\\\ &= r_1 r_2 \\cdot e^{i (\\theta_1 + \\theta_2) } \\end{aligned} \\tag {21} Now, let's see if it really works. (Have you any doubts?) Let $z_1 = 3 + i 4$ and $z_2 = 5 + i 12$. $$z_1 \\cdot z_2 = ( 15 - 48 ) + i ( 36 + 20 ) = -33 + i 56 = z_3 \\tag {22}$$ \\begin{aligned} \\| z_1 \\| &= \\sqrt{ 3^2 + 4^2 } = \\sqrt{ 9 + 16 } = \\sqrt{ 25 } = 5 \\\\ \\| z_2 \\| &= \\sqrt{ 5^2 + 12^2 } = \\sqrt{ 25 + 144 } = \\sqrt{ 169 } = 13 \\\\ \\| z_3 \\| &= \\sqrt{ (-33)^2 + 56^2 } = \\sqrt{ 1089 + 3136 } = \\sqrt{ 4225 } = 65 \\\\ \\end{aligned} \\tag {23} Yeah, I picked Pythagorean triples for $z_1$ and $z_2$ to make the numbers come out nice. But look, the result is also a Pythagorean triple. The mind boggles. Back to the verification. $5 \\cdot 13 = 65$. Check. Now let's look at the angles, they aren't so clean. Here is a little Gambas code to do the calculations: Dim z1, z2, z3 As Complex z1 = 3 + 4i z2 = 5 + 12i z3 = z1 * z2 Print \"|z1| =", "is ~60. It will help to make the snap grid very small. • Click M and move D3 r = ~2.0 and theta = ~120. • Click M and move D4 r = 2.0 and theta = 180 or -180. • Click M and move D5 r = ~2.0 and theta = -120. • Click M and move D6 r = ~2.0 and theta = -60. • Click E over each LED and enter their rotation angles (0,60,120,180,240,300.) This is the result: Now drag a box around the whole group and move with M. To calculate Cartesian coordinates (polar button not clicked, for accurate X,Y positions) Trigonometry must be used. Negative theta values will also work here. Perhaps this will help with a python solution: • To convert polar (radius r, angle $$\\\\theta\\$$) to Cartesian (x,y): $$\\\\quad x=r\\cdot \\cos \\left(\\theta \\right)\\quad \\:y=r\\cdot \\sin \\left(\\theta \\right)\\$$ • Element 1, 0°: $$\\\\quad\\cos\\left(0\\right)\\cdot20 = 20\\quad\\sin\\left(0\\right)\\cdot20 = 0\\quad\\$$ (20,0), rotate the part 0° • Element 2, 60°: $$\\\\quad\\cos\\left(60\\right)\\cdot20 = 10\\quad\\sin\\left(60\\right)\\cdot20 = 17.32\\quad\\$$ (10,17.32), rotate 60° • Element 3, 120°: $$\\\\quad\\cos\\left(120\\right)\\cdot20 = -10\\quad\\sin\\left(120\\right)\\cdot20 = 17.32\\quad\\$$ (-10,17.32), rotate 120° • Element 4, 180°: $$\\\\quad\\cos\\left(180\\right)\\cdot20 = -20\\quad\\sin\\left(180\\right)\\cdot20 = 0\\quad\\$$ (-20,0), rotate 180° • Element 5, 240°: $$\\\\quad\\cos\\left(240\\right)\\cdot20 = -10\\quad\\sin\\left(240\\right)\\cdot20 = -17.32\\quad\\$$ (-10,-17.32), rotate 240° • Element 6, 300°: $$\\\\quad\\cos\\left(300\\right)\\cdot20 = 10\\quad\\sin\\left(300\\right)\\cdot20 = -17.32\\quad\\$$ (10,-17.32), rotate", "\\cdot C \\\\ &= \\frac{\\theta}{2\\pi } \\cdot 2\\pi \\, r \\\\ &= \\theta\\,r \\end{align*}", "\\end{align*} \\begin{align*} B'D^2 = CD^2 + CB'^2 + 2 CD \\cdot CB' \\cos \\theta,\\end{align*} \\begin{align*} d^2 + b^2 – 2 bd \\cos \\theta = c^2 + a^2 + 2ac \\cos \\theta,\\end{align*} \\begin{align*} 2(ac + bd) \\cos \\theta = |d^2 – c^2 + b^2 – a^2|.\\end{align*} ~Interstigation", "What are the rectangular coordinates of (3, -π/3)? $x = \\frac{3}{2}$ $y = \\frac{- 3 \\sqrt{3}}{2}$ Explanation: From the given: $r = 3$ and $\\theta = - \\frac{\\pi}{3}$ $x = r \\cdot \\cos \\theta$ $x = 3 \\cdot \\cos \\left(- \\frac{\\pi}{3}\\right)$ $x = 3 \\cdot \\left(\\frac{1}{2}\\right)$ $x = \\frac{3}{2}$ $y = r \\cdot \\sin \\theta$ $y = 3 \\cdot \\sin \\left(- \\frac{\\pi}{3}\\right)$ $y = 3 \\cdot \\frac{- \\sqrt{3}}{2}$ $y = \\frac{- 3 \\sqrt{3}}{2}$ God bless...I hope the explanation is useful.", "an angle of \\$$\\pi\\$$ radians, which rotates the **rt** unit to point leftward to -1. ## Unity of the Grid, Rectangular, and Circular Views of the 2D Plane Now we can dispense with the **Rot** function entirely, by noticing that $$\\mathrm{Rot}(x) = e^{i\\cdot x}$$ Thus we can unify the rectangular view with the circular view, in terms of grid view coordinates as follows. In fact, it's as if **every point on the 2D plane can be represented using four different \\begin{align} (x,y)&\\text{ --- grid view coordinate}\\\\\\\\ &= x + y \\cdot\\mathrm{i} \\text{ --- rectangular view \"complex numbers\"}\\\\\\\\ &= r \\cdot e^{i \\cdot\\theta} \\text{ --- circular view \"complex numbers\"}\\\\\\\\ &= (r, \\theta) \\text{ --- circular view \"polar coordinate\"}\\\\\\\\ \\text{with these defintions:}&\\\\\\\\ r &= \\sqrt{x^2 + y^2} \\\\\\\\ \\theta &= \\arccos{ (\\frac{x}{r}) } \\end{align} Or we can unify rectangular and circular views, in terms of circular view polar coordinates of a point as follows: For a point at distance *r* from the origin, and with an angle in standard position of \\$$\\theta\\$$: \\begin{align} (r, \\theta)&\\text{ in polar coordinates}\\\\\\\\ &= r \\cdot e^{i \\cdot\\theta} \\\\\\\\ &= r \\cdot \\cos(\\theta) + r \\cdot \\sin(\\theta) \\cdot\\mathrm{i}\\\\\\\\ &= (r \\cdot \\cos(\\theta), r \\cdot \\sin(\\theta))\\text{ in grid coordinates} \\end{align} ## Afterword This was a long and windy road to those few formulas at the end,", "## Precalculus (10th Edition) $(-3\\sqrt3,3)$ By definition: $x=r\\cdot \\cos(\\theta)$ and $y=r\\cdot \\sin(\\theta)$. Here $r=6 \\text{ and } \\theta=150^\\circ$. Hence, after plugging these in: $x=6\\cdot \\cos(150^\\circ)=6\\cdot(-\\frac{\\sqrt3}{2})=-3\\sqrt3$ $y=6\\cdot \\sin(150^\\circ)=6\\cdot(\\frac{1}{2})=3$ The point has coordinates $(-3\\sqrt3,3)$ in the rectangular coordinate system.", "\\sin \\theta)\\end{align*}. If we allow z\\begin{align*}z\\end{align*} to equal the polar form of a complex number, it is very easy to see the development of a pattern when raising a complex number in polar form to a power. To discover this pattern, it is necessary to perform some basic multiplication of complex numbers in polar form. If z=r(cosθ+isinθ)\\begin{align*}z = r(\\cos \\theta + i \\sin \\theta)\\end{align*} and z2=zz\\begin{align*}z^2 = z \\cdot z\\end{align*} then: z2z2z2=r(cosθ+isinθ)r(cosθ+isinθ)=r2[cos(θ+θ)+isin(θ+θ)]=r2(cos2θ+isin2θ)\\begin{align*}z^2 &= r(\\cos \\theta + i \\sin \\theta) \\cdot r(\\cos \\theta + i \\sin \\theta)\\\\ z^2 &= r^2 [\\cos (\\theta + \\theta) + i \\sin (\\theta + \\theta)]\\\\ z^2 &= r^2 (\\cos 2\\theta + i \\sin 2\\theta)\\end{align*} Likewise, if z=r(cosθ+isinθ)\\begin{align*}z = r(\\cos \\theta + i \\sin \\theta)\\end{align*} and z3=z2z\\begin{align*}z^3 = z^2 \\cdot z\\end{align*} then: z3z3z3=r2(cos2θ+isin2θ)r(cosθ+isinθ)=r3[cos(2θ+θ)+isin(2θ+θ)]=r3(cos3θ+isin3θ)\\begin{align*}z^3 &= r^2(\\cos 2\\theta + i \\sin 2\\theta) \\cdot r(\\cos \\theta + i \\sin \\theta)\\\\ z^3 &= r^3 [\\cos (2\\theta + \\theta) + i \\sin (2\\theta + \\theta)]\\\\ z^3 &= r^3 (\\cos 3\\theta + i \\sin 3\\theta)\\end{align*} Again, if z=r(cosθ+isinθ)\\begin{align*}z = r(\\cos \\theta + i \\sin \\theta)\\end{align*} and z4=z3z\\begin{align*}z^4 = z^3 \\cdot z\\end{align*} then z4=r4(cos4θ+isin4θ)\\begin{align*}z^4 = r^4(\\cos 4\\theta + i \\sin 4\\theta)\\end{align*} These examples suggest a general rule valid for all powers of z\\begin{align*}z\\end{align*}, or n\\begin{align*}n\\end{align*}. We offer this rule and assume its validity for all n\\begin{align*}n\\end{align*} without formal proof, leaving that for", "\\begin{align} \\frac{\\partial \\boldsymbol{g}_i\\cdot\\boldsymbol{n}}{\\partial \\theta^j}=0. \\end{align} Hence, (89) \\begin{align} \\frac{\\partial \\boldsymbol{g}_i}{\\partial \\theta^j}\\cdot \\boldsymbol{n}+\\boldsymbol{g}_i\\frac{\\partial \\boldsymbol{n}}{\\partial \\theta^j}=0. \\end{align} By means of transposition, we obtain (90) \\begin{align} \\frac{\\partial \\boldsymbol{g}_i}{\\partial \\theta^j}\\cdot \\boldsymbol{n}=-\\boldsymbol{g}_i\\frac{\\partial \\boldsymbol{n}}{\\partial \\theta^j}. \\end{align} Therefore, we have (91) \\begin{align} L=-\\boldsymbol{g}_1\\cdot\\frac{\\partial \\boldsymbol{n}}{\\partial \\theta^1} \\end{align} (92) \\begin{align} M=-\\boldsymbol{g}_1\\cdot\\frac{\\partial \\boldsymbol{n}}{\\partial \\theta^2} \\end{align} (93) \\begin{align} M=-\\boldsymbol{g}_2\\cdot\\frac{\\partial \\boldsymbol{n}}{\\partial \\theta^1} \\end{align} (94) \\begin{align} N=-\\boldsymbol{g}_2\\cdot\\frac{\\partial \\boldsymbol{n}}{\\partial \\theta^2}. \\end{align} Finally we obtain (95) \\begin{align} II=Ld\\theta^1d\\theta^1+2Md\\theta^1d\\theta^2+Nd\\theta^2 d\\theta^2. \\end{align} Incidentally, using $h_{ij}$, it's likely that we can define a rank-3 tensor by (96) \\begin{align} \\boldsymbol{h}=h_{ij}\\boldsymbol{g}^i\\otimes\\boldsymbol{n}\\otimes\\boldsymbol{g}^j. \\end{align} Let us examine whether this is identical with $\\boldsymbol{H}$. From the definition, we have (97) \\begin{align} \\boldsymbol{H}=\\boldsymbol{g}^{\\alpha}\\otimes\\frac{\\partial\\boldsymbol{g}_{i}\\otimes\\boldsymbol{g}^{i}}{\\partial\\theta^{\\alpha}}. \\end{align} By the property of unit tensor, we have (98) \\begin{align} =\\boldsymbol{g}^{\\alpha}\\otimes\\frac{\\partial\\boldsymbol{g}^{i}\\otimes\\boldsymbol{g}_{i}}{\\partial\\theta^{\\alpha}}. \\end{align} Performing differentiation, it follows (99) \\begin{align} =\\boldsymbol{g}^{\\alpha}\\otimes\\frac{\\partial\\boldsymbol{g}_{i}}{\\partial\\theta^{\\alpha}}\\otimes\\boldsymbol{g}^{i}+\\boldsymbol{g}^{\\alpha}\\otimes\\boldsymbol{g}_{i}\\otimes\\frac{\\partial\\boldsymbol{g}^{i}}{\\partial\\theta^{\\alpha}}. \\end{align} Denoting a unit normal vector $\\boldsymbol{n}$ we obtain (100) \\begin{align} =\\boldsymbol{g}^{\\alpha}\\otimes\\frac{\\partial\\boldsymbol{g}_{i}}{\\partial\\theta^{\\alpha}}\\otimes\\boldsymbol{g}^{i}+\\boldsymbol{g}^{\\alpha}\\otimes\\boldsymbol{g}_{i}\\otimes\\left(\\frac{\\partial\\boldsymbol{g}^{i}}{\\partial\\theta^{\\alpha}}\\cdot\\boldsymbol{g}_{j}\\right)\\boldsymbol{g}^{j}+\\boldsymbol{g}^{\\alpha}\\otimes\\boldsymbol{g}_{i}\\otimes\\left(\\frac{\\partial\\boldsymbol{g}^{i}}{\\partial\\theta^{\\alpha}}\\cdot\\boldsymbol{n}\\right)\\boldsymbol{n}. \\end{align} Here it is possible to rewrite the equation to (101) \\begin{align} =\\boldsymbol{g}^{\\alpha}\\otimes\\frac{\\partial\\boldsymbol{g}_{i}}{\\partial\\theta^{\\alpha}}\\otimes\\boldsymbol{g}^{i}-\\boldsymbol{g}^{\\alpha}\\otimes\\boldsymbol{g}_{i}\\otimes\\left(\\frac{\\partial\\boldsymbol{g}_{j}}{\\partial\\theta^{\\alpha}}\\cdot\\boldsymbol{g}^{i}\\right)\\boldsymbol{g}^{j}-\\boldsymbol{g}^{\\alpha}\\otimes\\boldsymbol{g}_{i}\\otimes\\left(\\frac{\\partial\\boldsymbol{n}}{\\partial\\theta^{\\alpha}}\\cdot\\boldsymbol{g}^{i}\\right)\\boldsymbol{n}. \\end{align} Because the second term is commutative in $i,j$, we obtain (102) \\begin{align} =\\boldsymbol{g}^{\\alpha}\\otimes\\frac{\\partial\\boldsymbol{g}_{i}}{\\partial\\theta^{\\alpha}}\\otimes\\boldsymbol{g}^{i}-\\boldsymbol{g}^{\\alpha}\\otimes\\left(\\frac{\\partial\\boldsymbol{g}_{i}}{\\partial\\theta^{\\alpha}}\\cdot\\boldsymbol{g}^{j}\\right)\\boldsymbol{g}_{j}\\otimes\\boldsymbol{g}^{i}-\\boldsymbol{g}^{\\alpha}\\otimes\\left(\\frac{\\partial\\boldsymbol{n}}{\\partial\\theta^{\\alpha}}\\cdot\\boldsymbol{g}^{i}\\right)\\boldsymbol{g}_{i}\\otimes\\boldsymbol{n}. \\end{align} Hence, the first and second terms can be unified. By using (103) \\begin{align} \\boldsymbol{I}=\\boldsymbol{g}_{1}\\otimes\\boldsymbol{g}^{1}+\\boldsymbol{g}_{2}\\otimes\\boldsymbol{g}^{2}, \\end{align} (104) \\begin{align} \\boldsymbol{D}=\\boldsymbol{g}_{1}\\otimes\\boldsymbol{g}^{1}+\\boldsymbol{g}_{2}\\otimes\\boldsymbol{g}^{2}+\\boldsymbol{n}\\otimes\\boldsymbol{n} \\end{align} and (105) \\begin{align} \\boldsymbol{D}-\\boldsymbol{I}=\\boldsymbol{n}\\otimes\\boldsymbol{n}, \\end{align} we obtain (106) \\begin{align} =\\boldsymbol{g}^{\\alpha}\\otimes\\frac{\\partial\\boldsymbol{g}_{i}}{\\partial\\theta^{\\alpha}}\\cdot\\left(\\boldsymbol{D}-\\boldsymbol{I}\\right)\\otimes\\boldsymbol{g}^{i}+\\boldsymbol{g}^{\\alpha}\\otimes\\frac{\\partial\\boldsymbol{n}}{\\partial\\theta^{\\alpha}}\\cdot\\boldsymbol{I}\\otimes\\boldsymbol{n} \\end{align} (107) \\begin{align} =h_{i\\alpha}\\boldsymbol{g}^{\\alpha}\\otimes\\boldsymbol{n}\\otimes\\boldsymbol{g}^{i}+\\boldsymbol{g}^{\\alpha}\\otimes\\frac{\\partial\\boldsymbol{n}}{\\partial\\theta^{\\alpha}}\\cdot\\boldsymbol{I}\\otimes\\boldsymbol{n} \\end{align} (108) \\begin{align} \\therefore \\boldsymbol{H}=\\boldsymbol{h}+\\boldsymbol{g}^{\\alpha}\\otimes\\frac{\\partial\\boldsymbol{n}}{\\partial\\theta^{\\alpha}}\\cdot\\boldsymbol{I}\\otimes\\boldsymbol{n}. \\end{align} Based on the above discussion, we notice that $h_{ij}$ is actually the component of the first term of the above", ". \\end{align*} ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) Solution 6 (Law of Cosines) WLOG, let $P$ be on minor arc $\\overarc {AB}$. Let $r$ and $O$ be the radius and center of the circumcircle respectively, and let $\\theta = \\angle AOP$. By the Pythagorean Theorem, the area of the square is $2r^2$. We can use the Law of Cosines on isosceles triangles $\\triangle AOP, \\, \\triangle COP, \\, \\triangle BOP, \\, \\triangle DOP$ to get \\begin{align*} PA^2 &= 2r^2(1 - \\cos \\theta), \\\\ PC^2 &= 2r^2(1 - \\cos (180 - \\theta)) = 2r^2(1 + \\cos \\theta), \\\\ PB^2 &= 2r^2(1 - \\cos (90 - \\theta)) = 2r^2(1 - \\sin \\theta), \\\\ PD^2 &= 2r^2(1 - \\cos (90 + \\theta)) = 2r^2(1 + \\sin \\theta). \\end{align*} Taking the products of the first two and last two equations, respectively, $$56^2 = (PA \\cdot PC)^2 = 4r^4(1 - \\cos \\theta)(1 + \\cos \\theta) = 4r^4(1 - \\cos^2 \\theta) = 4r^4 \\sin^2 \\theta,$$ and $$90^2 = (PB \\cdot PD)^2 = 4r^4(1 - \\sin \\theta)(1 + \\sin \\theta) = 4r^4(1 - \\sin^2 \\theta) = 4r^4 \\cos^2 \\theta.$$ Adding these equations, $$56^2 + 90^2 = 4r^4,$$ so $$2r^2 = \\sqrt{56^2+90^2} = 2\\sqrt{28^2+45^2} = 2\\sqrt{2809} = 2 \\cdot 53 = \\boxed{106}.$$ ~OrangeQuail9 Solution 7 (Subtended Chords) First draw a diagram. $[asy] pair A, B,", "in polar coordinates is $\\nonumber d ~=~ \\sqrt{r_1^2 ~+~ r_2^2 ~-~ 2r_1r_2\\,\\cos\\;(\\theta_1 - \\theta_2)} ~~.$ Solution: The idea here is to use the distance formula in Cartesian coordinates, then convert that to polar coordinates. So write \\nonumber \\begin{alignat*}{4} x_1 ~&=~ r_1 \\,\\cos\\;\\theta_1 \\qquad& y_1 ~&=~ r_1 \\,\\sin\\;\\theta_1\\\\ \\nonumber x_2 ~&=~ r_2 \\,\\cos\\;\\theta_2 \\qquad& y_2 ~&=~ r_2 \\,\\sin\\;\\theta_2 ~. \\end{alignat*} Then $$(x_1,y_1)$$ and $$(x_2,y_2)$$ are the Cartesian equivalents of $$(r_1 , \\theta_1)$$ and $$(r_2 , \\theta_2)$$, respectively. Thus, by the Cartesian coordinate distance formula, \\nonumber \\begin{align*} d^2 ~&=~ (x_1 - x_2)^2 ~+~ (y_1 - y_2)^2\\\\ \\nonumber &=~ (r_1 \\,\\cos\\;\\theta_1 - r_2 \\,\\cos\\;\\theta_2)^2 ~+~ (r_1 \\,\\sin\\;\\theta_1 - r_2 \\,\\sin\\;\\theta_2)^2\\\\ \\nonumber &=~ r_1^2 \\cos^2\\;\\theta_1 ~-~ 2r_1 r_2 \\cos\\;\\theta_1~\\cos\\;\\theta_2 ~+~ r_2^2 \\cos^2\\;\\theta_2 ~+~ r_1^2 \\sin^2\\;\\theta_1 ~-~ 2r_1 r_2 \\sin\\;\\theta_1~\\sin\\;\\theta_2 ~+~ r_2^2 \\sin^2\\;\\theta_2\\\\ \\nonumber &=~ r_1^2 (\\cos^2\\;\\theta_1 ~+~ \\sin^2\\;\\theta_1) ~+~ r_2^2 (\\cos^2\\;\\theta_2 ~+~ \\sin^2\\;\\theta_2) ~-~ 2r_1 r_2 (\\cos\\;\\theta_1~\\cos\\;\\theta_2 ~+~ \\sin\\;\\theta_1~\\sin\\;\\theta_2)\\\\ \\nonumber d^2 ~&=~ r_1^2 ~+~ r_2^2 ~-~ 2r_1r_2\\,\\cos\\;(\\theta_1 - \\theta_2) ~, \\end{align*} so the result follows by taking square roots of both sides. In Example 6.17 we saw that the equation $$x^2 + y^2 = 9$$ in Cartesian coordinates could be expressed as $$r = 3$$ in polar coordinates. This equation describes a circle centered at the origin, so the circle is symmetric about the origin. In general, polar coordinates are useful in", "the circumscribed circle. Since $\\alpha = 45^\\circ$, \\begin{align*} a & = 2R\\sin\\alpha\\\\ & = 2 \\cdot 2 \\cdot \\sin(45^\\circ)\\\\ & = 4 \\cdot \\frac{\\sqrt{2}}{2}\\\\ & = 2\\sqrt{2} \\end{align*} Since $\\gamma = 60^\\circ$, \\begin{align*} c & = 2R\\sin\\gamma\\\\ & = 2 \\cdot 2 \\cdot \\sin(60^\\circ)\\\\ & = 4 \\cdot \\frac{\\sqrt{3}}{2}\\\\ & = 2\\sqrt{3} \\end{align*} The area of the triangle is $$A = \\frac{1}{2}ac\\sin\\beta$$ since $c\\sin\\beta$ is the length of the altitude to side $\\overline{BC}$, which has length $a$. By the Angle Sum Theorem for Triangles, \\begin{align*} \\beta & = 180^\\circ - 60^\\circ - 45^\\circ\\\\ & = 75^\\circ \\end{align*} To find the exact value of $\\sin(75^\\circ)$, we use the Sum of Angles Formula $$\\sin(\\theta + \\varphi) = \\sin\\theta\\cos\\varphi + \\cos\\theta\\sin\\varphi$$ with $\\theta = 30^\\circ$ and $\\varphi = 45^\\circ$, which yields \\begin{align*} \\sin(75^\\circ) & = \\sin(30^\\circ)\\cos(45^\\circ) + \\cos(30^\\circ)\\sin(45^\\circ)\\\\ & = \\frac{1}{2} \\cdot \\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{2}}{2}\\\\ & = \\frac{\\sqrt{2} + \\sqrt{6}}{4} \\end{align*} Hence, the area of triangle $ABC$ is \\begin{align*} A & = \\frac{1}{2}ac\\sin\\beta\\\\ & = \\frac{1}{2} \\cdot 2\\sqrt{2} \\cdot 2\\sqrt{3} \\cdot \\frac{\\sqrt{2} + \\sqrt{6}}{4}\\\\ & = \\frac{1}{2} \\cdot \\sqrt{6}(\\sqrt{2} + \\sqrt{6})\\\\ & = \\frac{1}{2}(\\sqrt{12} + 6)\\\\ & = \\frac{1}{2}(2\\sqrt{3} + 6)\\\\ & = \\sqrt{3} + 3 \\end{align*} square units. If $O$ is the circumcenter of $ABC$, we have: $$\\Delta = [OAB]+[OAC]+[OBC] = \\frac{R^2}{2}\\left(\\sin(2A)+\\sin(2B)+\\sin(2C)\\right).\\tag{1}$$ In our case, we have $A=45^\\circ,C=60^\\circ$,", "measurements. Since your friend is certain of the lengths of the sides of the triangle, you should use those as the known quantities in the Law of Cosines and solve for the angle: \\begin{align*}7^2 = 5^2 + 4^2 + (2)(5)(4)\\cos \\theta\\\\ 49 = 25 + 16 + 40 \\cos \\theta\\\\ 49 - 25 - 16 = 40 \\cos \\theta\\\\ \\frac{8}{40} = \\cos \\theta\\\\ cos^{-1} \\frac{8}{40} = \\theta\\\\ \\theta = 78.46 \\end{align*} So as it turns out, your friend is rather close. Her measurements were probably slight inaccurate because of her round off from the protractor. #### Example 2 Find \\begin{align*}AD\\end{align*} using the Pythagorean Theorem, Law of Cosines, trig functions, or any combination of the three. First, find \\begin{align*}AB\\end{align*}. \\begin{align*}AB^2 = 14.2^2 + 15^2 - 2 \\cdot 14.2 \\cdot 15 \\cdot \\cos 37.4^\\circ, AB = 9.4. \\sin 23.3^\\circ = \\frac{AD}{9.4}, AD = 3.7.\\end{align*} #### Example 3 Find \\begin{align*}HK\\end{align*} using the Pythagorean Theorem, Law of Cosines, trig functions, or any combination of the three if \\begin{align*}JK = 3.6, KI = 5.2, JI = 1.9, HI = 6.7\\end{align*}, and \\begin{align*}\\angle{KJI} = 96.3^\\circ\\end{align*}. \\begin{align*}\\angle{HJI} = 180^\\circ - 96.3^\\circ = 83.7^\\circ\\end{align*} (these two angles are a linear pair). \\begin{align*}6.7^2 = HJ^2 + 1.9^2 - 2 \\cdot HJ \\cdot 1.9 \\cdot \\cos 83.7^\\circ\\end{align*}. This simplifies to the quadratic equation \\begin{align*}HJ^2 - 0.417HJ - 41.28\\end{align*}.", "$C_1$ and $XZ$. We now have $AZ=YZ=7$ and $XA=6$. Furthermore, notice that $\\triangle XAO$ is isosceles, thus the altitude from $O$ to $XA$ bisects $XZ$ at point $B$ above. By the Pythagorean Theorem, \\begin{align*}BZ^2+BO^2&=OZ^2\\\\(BA+AZ)^2+OA^2-BA^2&=11^2\\\\(3+7)^2+r^2-3^2&=121\\\\r^2&=30\\end{align*}Thus, $r=\\sqrt{30}\\implies\\boxed{\\textbf{E}}$ ## Solution 6 Use the diagram above. Notice that $\\angle YZO=\\angle XZO$ as they subtend arcs of the same length. Let $A$ be the point of intersection of $C_1$ and $XZ$. We now have $AZ=YZ=7$ and $XA=6$. Consider the power of point $Z$ with respect to Circle $O,$ we have $13\\cdot 7 = (11 + r)(11 - r) = 11^2 - r^2,$ which gives $r=\\boxed{\\sqrt{30}}.$ ## Solution 7 (Only Law of Cosines) Note that $OX$ and $OY$ are the same length, which is also the radius $R$ we want. Using the law of cosines on $\\triangle OYZ$, we have $11^2=R^2+7^2-2\\cdot 7 \\cdot R \\cdot \\cos\\theta$, where $\\theta$ is the angle formed by $\\angle{OYZ}$. Since $\\angle{OYZ}$ and $\\angle{OXZ}$ are supplementary, $\\angle{OXZ}=\\pi-\\theta$. Using the law of cosines on $\\triangle OXZ$, $11^2=13^2+R^2-2 \\cdot 13 \\cdot R \\cdot \\cos(\\pi-\\theta)$. As $\\cos(\\pi-\\theta)=-\\cos\\theta$, $11^2=13^2+R^2+\\cos\\theta$. Solving for theta on the first equation and substituting gives $\\frac{72-R^2}{14R}=\\frac{48+R^2}{26R}$. Solving for R gives $R=\\textbf{(E)}\\ \\boxed{\\sqrt{30}}$.", "orientation of $$e$$ so that it is pointing from $$u$$ to $$v$$. Let $$F = F(e)$$, $$G = G(e)$$ to the two faces in contact to $$e$$ and $$a = a_F$$, $$c = a_G$$ be corresponding vectors. For any $$p \\in e \\subset \\Omega_F \\cap \\Omega_G$$, we have $$|p|^2 - (a\\cdot p)^2 = |p|^2 - (c\\cdot p)^2 = 1 \\implies (c\\cdot p)^2 = (a\\cdot p)^2$$ Flipping the sign of $$c$$ is needed, we can assume $$a\\cdot p = c \\cdot p$$ along $$e$$. i.e. $$e$$ is lying on the plane $$p\\cdot( a - c) = 0$$. Choose a coordinate system so that $$a = (0,0,1)$$ and $$c = (\\sin\\beta,0,\\cos\\beta)$$. In this coordinate system, $$\\begin{array}{crcl} & |p|^2 - (a\\cdot p)^2 = 1 &\\iff& x^2 + y^2 = 1\\\\ \\text{ and }\\quad & p \\cdot a = p \\cdot c &\\iff& z = \\cos\\theta z + \\sin\\beta y \\end{array}$$ We can parameterize $$e$$ by a single $$\\theta$$. $$p = (x,y,z) = (\\cos\\theta,\\sin\\theta,k\\sin\\theta) \\quad\\text{ where } \\quad k = \\frac{\\sin\\beta}{1 - \\cos\\beta}$$ Notice \\begin{align}(p \\times (a - c))\\cdot dp &= \\left| \\begin{matrix} x & y & z\\\\ 0 &-\\sin\\beta & 1 - \\cos\\beta\\\\ dx & dy & dz \\end{matrix} \\right| = (1-\\cos\\beta) \\left| \\begin{matrix} x & y & ky\\\\ 0 &-k & 1\\\\ dx & dy & dz \\end{matrix} \\right|\\\\ &=" ]

[ "\\theta + i \\sin \\theta$$. and $$i^2 = -1$$: \\begin{equation*} \\begin{split} (x + yi)\\cdot e^{i\\theta} & = (x + yi)\\cdot (\\cos \\theta + i \\sin \\theta)\\\\ & = x\\cdot \\cos \\theta + x \\cdot i \\cdot \\sin \\theta + y \\cdot i \\cdot \\cos \\theta - y \\cdot \\sin \\theta \\\\ & = (x\\cdot \\cos \\theta - y\\cdot \\sin \\theta) + i\\ (x\\cdot \\sin \\theta + y\\cdot \\cos \\theta) \\end{split} \\end{equation*} So that \\begin{align*} x' = x \\cdot \\cos \\theta - y \\cdot \\sin \\theta\\\\ y' = x \\cdot \\sin \\theta + y \\cdot \\cos \\theta \\end{align*} To rotate a point p around another point c, we need to rotate the point p expressed in the coordinate system whose origin is c (it simply is $$p-c$$), and then we just need to express the result in the initial coordinate system (whose origin is (0,0)). This gives us the following formula (if we define $$p' = p'_x + p'_yi$$): \\begin{equation*} p' = c + e^{i \\theta} \\cdot (p - c) \\end{equation*} or : \\begin{align*} p'_x = c_x + (p_x - c_x) \\cdot \\cos \\theta - (p_y - c_y) \\cdot \\sin \\theta\\\\ p'_y = c_y + (p_x - c_x) \\cdot \\sin \\theta + (p_y - c_y) \\cdot \\cos \\theta \\end{align*} Note : similar computations (with a different result) could be", "an angle of \\$$\\pi\\$$ radians, which rotates the **rt** unit to point leftward to -1. ## Unity of the Grid, Rectangular, and Circular Views of the 2D Plane Now we can dispense with the **Rot** function entirely, by noticing that $$\\mathrm{Rot}(x) = e^{i\\cdot x}$$ Thus we can unify the rectangular view with the circular view, in terms of grid view coordinates as follows. In fact, it's as if **every point on the 2D plane can be represented using four different \\begin{align} (x,y)&\\text{ --- grid view coordinate}\\\\\\\\ &= x + y \\cdot\\mathrm{i} \\text{ --- rectangular view \"complex numbers\"}\\\\\\\\ &= r \\cdot e^{i \\cdot\\theta} \\text{ --- circular view \"complex numbers\"}\\\\\\\\ &= (r, \\theta) \\text{ --- circular view \"polar coordinate\"}\\\\\\\\ \\text{with these defintions:}&\\\\\\\\ r &= \\sqrt{x^2 + y^2} \\\\\\\\ \\theta &= \\arccos{ (\\frac{x}{r}) } \\end{align} Or we can unify rectangular and circular views, in terms of circular view polar coordinates of a point as follows: For a point at distance *r* from the origin, and with an angle in standard position of \\$$\\theta\\$$: \\begin{align} (r, \\theta)&\\text{ in polar coordinates}\\\\\\\\ &= r \\cdot e^{i \\cdot\\theta} \\\\\\\\ &= r \\cdot \\cos(\\theta) + r \\cdot \\sin(\\theta) \\cdot\\mathrm{i}\\\\\\\\ &= (r \\cdot \\cos(\\theta), r \\cdot \\sin(\\theta))\\text{ in grid coordinates} \\end{align} ## Afterword This was a long and windy road to those few formulas at the end,", "much more transparent in polar coordinates. Recall that instead of specifying the $x$ and $y$ components, we can specify the angle $\\theta$ (in radians) from the positive $x$ axis and distance $r$ from the origin. These give the $x$ and $y$ components using the usual rules of trigonometry: \\begin{align*} x & = r \\cos \\theta, \\quad y = r \\sin\\theta \\\\ r & = \\sqrt{x^2 + y^2}, \\quad \\theta = \\tan^{-1}\\left(\\frac{y}{x}\\right). \\end{align*} We denote the corresponding complex number $z(r, \\theta)$. Let’s multiply two such complex numbers: \\begin{align*} z(r_1, \\theta_1) z(r_2, \\theta_2) & = r_1 (\\cos\\theta_1 + i\\sin \\theta_1) \\cdot r_2 (\\cos\\theta_2 + i\\sin \\theta_2) \\\\ & = r_1 r_2 [(\\cos\\theta_1 \\cos\\theta_2 - \\sin\\theta_1 \\sin\\theta_2) + i (\\cos\\theta_1 \\sin\\theta _ 2 + \\sin\\theta_1 \\cos\\theta _ 2)]. \\end{align*} This still looks like a mess, but we can simplify dramatically using the compound angle formulas: \\begin{align*} \\cos(\\theta_1 + \\theta_2) & = \\cos\\theta_1 \\cos\\theta_2 - \\sin\\theta_1 \\sin\\theta_2\\\\ \\sin(\\theta_1 + \\theta_2) & = \\cos\\theta_1 \\sin\\theta _ 2 + \\sin\\theta_1 \\cos\\theta _ 2. \\end{align*} Applying these to $z(r_1, \\theta_1)z(r_2,\\theta_2)$ immediately gives $z(r_1, \\theta_1) z(r_2, \\theta_2) = r_1 r_2 [\\cos (\\theta_1 + \\theta_2) + i \\sin (\\theta_1 + \\theta_2)] = z(r_1 r_2, \\theta_1+\\theta_2).$ In other words, a product of two complex numbers simply multiplies the lengths and adds the angles. Great! Now things can take", "Trigonometry (11th Edition) Clone The rectangular coordinates are $(0,-4)$ We can find the x-coordinate: $x = r~cos~\\theta$ $x = (-4)~cos~90^{\\circ}$ $x = (-4)~(0)$ $x = 0$ We can find the y-coordinate: $y = r~sin~\\theta$ $y = (-4)~sin~90^{\\circ}$ $y = (-4)~(1)$ $y = -4$ The rectangular coordinates are $(0,-4)$", "## Precalculus (10th Edition) $(-3\\sqrt3,3)$ By definition: $x=r\\cdot \\cos(\\theta)$ and $y=r\\cdot \\sin(\\theta)$. Here $r=6 \\text{ and } \\theta=150^\\circ$. Hence, after plugging these in: $x=6\\cdot \\cos(150^\\circ)=6\\cdot(-\\frac{\\sqrt3}{2})=-3\\sqrt3$ $y=6\\cdot \\sin(150^\\circ)=6\\cdot(\\frac{1}{2})=3$ The point has coordinates $(-3\\sqrt3,3)$ in the rectangular coordinate system.", "in the $x'y'$ system is is $2\\theta$ above the $x'$-axis and so its coordinates there must be $(x',y')=(\\cos2\\theta, \\sin2\\theta)$. Substituting this into the known relation between the coordinate systems yields: $$\\cos2\\theta = (\\cos \\theta)^2 - (\\sin\\theta)^2 \\qquad \\sin2\\theta = \\cos(\\theta)\\sin(\\theta) + \\sin(\\theta)\\cos(\\theta)$$ - How did you produce your diagram? Did you draw it with pen and then scan it? – MJD Apr 25 '12 at 14:41 I drew it with a pen, photographed it with a digital camera, and then cleaned it up (for background and contrast) in Gimp. – Henning Makholm Apr 25 '12 at 14:43", "in the $x'y'$ system is is $2\\theta$ above the $x'$-axis and so its coordinates there must be $(x',y')=(\\cos2\\theta, \\sin2\\theta)$. Substituting this into the known relation between the coordinate systems yields: $$\\cos2\\theta = (\\cos \\theta)^2 - (\\sin\\theta)^2 \\qquad \\sin2\\theta = \\cos(\\theta)\\sin(\\theta) + \\sin(\\theta)\\cos(\\theta)$$", "# How do you find the rectangular coordinates of the point whose spherical coordinates are (4, 3π/4, π/3)? Feb 20, 2015 The answer is: $\\left(\\sqrt{2} , \\sqrt{6} , - 2 \\sqrt{2}\\right)$. The formulas are: $x = r \\sin \\theta \\cos \\phi$ $y = r \\sin \\theta \\sin \\phi$ $z = r \\cos \\theta$ So: $x = 4 \\sin \\left(\\frac{3}{4} \\pi\\right) \\cos \\left(\\frac{\\pi}{3}\\right) = 4 \\cdot \\frac{\\sqrt{2}}{2} \\cdot \\frac{1}{2} = \\sqrt{2}$ $y = 4 \\sin \\left(\\frac{3}{4} \\pi\\right) \\sin \\left(\\frac{\\pi}{3}\\right) = 4 \\cdot \\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{3}}{2} = \\sqrt{6}$ $z = 4 \\cos \\left(\\frac{3}{4} \\pi\\right) = 4 \\cdot \\left(- \\frac{\\sqrt{2}}{2}\\right) = - 2 \\sqrt{2}$.", "Then you make the dot product and it gives $k\\cdot r \\cdot [...]$, where $[...]$ is an ugly term full of θ,θ',ϕ and ϕ' for the angle between k and r. However, you can choose a different orthogonal axis: say one where z' in parallel to r. In these \"r-ortogonal\" axes: $\\hat x_r$' = $\\hat x_r$ ^ $\\hat z_r$' and $\\hat y_r$' = $\\hat z_r$' ^ $\\hat x_r$' and $\\hat z_r$' = r/r. So now your spherical coordinates are: $k_1=k_r\\sin{\\theta}\\cos{\\phi}$, $k_2=k_r\\sin{\\theta}\\sin{\\phi}$, $k_3=k_r\\cos{\\theta}$, and $d^3k =k_r^2\\sin{\\theta}dkd{\\theta}d{\\phi}$ and $r_1=0, r_2=0, r_3=r$. and the dot product is simply $k_r\\cdot\\ r\\cdot \\cos{\\theta}$, with $\\theta$ the angle between k and r in these \"r-axes\". Now define $\\gamma = k_r\\cdot\\ r\\cdot \\cos{\\theta}$ and dγ = $k_r\\cdot\\ r\\cdot \\sin{\\theta}d{\\theta}$ and substitute in the integrals. This will give you the $k_r/r$ term. When you integrate the $e^{i\\gamma}$ from $\\gamma = -k_rr$ to +$k_rr$, this will wive you the $\\sin{k_rr}$. - Hi DoM, thanks for the tex effort. I fixed up some of it but I'm worried about changing some of the meaning and spacing. You should make used of \\sin and \\theta and \\phi rather than unicode and \"sin\" which is s * i * n – Brandon Enright Jun 6 '13 at 6:03", "r \\cdot \\sin \\theta + r \\cdot \\cos \\theta &=1 \\\\ r(\\sin \\theta + \\cos \\theta) &=1 \\\\ r &= \\frac{1}{\\sin \\theta + \\cos \\theta}\\end{align*} Examples Example 1 Earlier, you were asked how lines can be represented in the polar coordinate system. The general way to express a line \\begin{align*}y=mx+b\\end{align*} in polar form is \\begin{align*}r=\\frac{b}{\\sin \\theta - m \\cdot \\cos \\theta}\\end{align*}. Example 2 Express the following equation using rectangular coordinates: \\begin{align*}r=\\frac{8}{1+2 \\cos \\theta}\\end{align*}. Use the fact that \\begin{align*}r=\\pm \\sqrt{x^2+y^2}\\end{align*} and \\begin{align*}r \\cos \\theta =x\\end{align*}. \\begin{align*}r+2r \\cdot \\cos \\theta &= 8 \\\\ \\pm \\sqrt{x^2+y^2}+2x &= 8 \\\\ \\pm \\sqrt{x^2+y^2} &= 8-2x \\\\ x^2+y^2 &= 64-32x+4x^2 \\\\ -3x^2+32x+y^2-64 &=0\\end{align*} This is the equation of a hyperbola. Example 3 Sketch the following polar equation: \\begin{align*}r=3\\end{align*}. Since theta is not in the equation, it can vary freely. This simple equation produces a perfect circle of radius 3 centered at the origin. You can show this equation is equivalent to \\begin{align*}x^2+y^2=9\\end{align*} Example 4 Sketch the following polar equation: \\begin{align*}r= \\theta\\end{align*} with \\begin{align*}\\theta: 0 \\le \\theta \\le 2\\pi\\end{align*}. The equation \\begin{align*}r=\\theta\\end{align*} is an example of a polar equation that cannot be easily expressed in rectangular form. In order to sketch the graph, identify a few key points: \\begin{align*}(0, 0), \\left(\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right), (\\pi, \\pi), \\left(\\frac{3\\pi}{2}, \\frac{3\\pi}{2}\\right), (2\\pi, 2\\pi)\\end{align*}. You should see that the shape is", "x) and $$y_2=e^{rx}\\sin(\\theta x)$$ are fundamental solutions of (35) and the general solution of (35) is $y=c_1e^{rx}\\cos(\\theta x)+c_2e^{rx}\\sin(\\theta x).$ Example. Find the general solution of the following ODE. $\\frac{d^2y}{dx^2}-6\\frac{dy}{dx}+13y=0$ Solution. The characteristic equation is \\begin{align*} r^2-6r+13&=0\\\\ r&=\\frac{6\\pm \\sqrt{(-6)^2-4\\cdot1\\cdot 13}}{2\\cdot 1}\\\\ r&=3+2i,3-2i \\end{align*} So the general solution is $y=c_1e^{3x}\\cos(2x)+c_2e^{3x}\\sin(2x).$ Example. Solve the following IVP. $y''+2y'+10y=0,\\;\\; y(0)=4,\\, y'(0)=-7$ Solution. The characteristic equation is \\begin{align*} r^2+2r+10&=0\\\\ (r+1)^2&=-9\\\\ r&=-1+3i,-1-3i. \\end{align*} So the general solution is $y=c_1e^{-x}\\cos(3x)+c_2e^{-x}\\sin(3x).$ Using the initial condition $$y(0)=4$$, we get $$$c_1=4. \\;\\;\\;\\;(36)$$$ Note that $$y'=c_1(-\\cos(3x)-3\\sin(3x))e^{-x}+c_2(-\\sin(3x)+3\\cos(3x))e^{-x}.$$ Using the initial condition $$y'(0)=-7$$, we get $$$-c_1+3c_2=-7. \\;\\;\\;\\;(37)$$$ Solving (36) and (37), we get $$c_1=4$$ and $$c_2=-1$$. Thus the solution of the IVP is $y=4e^{-x}\\cos(3x)-e^{-x}\\sin(3x).$ Last edited", "in Cartesian coordinates, then convert that to polar coordinates. So write \\nonumber \\begin{alignat*}{4} x_1 ~&=~ r_1 \\,\\cos\\;\\theta_1 \\qquad& y_1 ~&=~ r_1 \\,\\sin\\;\\theta_1\\\\[4pt] x_2 ~&=~ r_2 \\,\\cos\\;\\theta_2 \\qquad& y_2 ~&=~ r_2 \\,\\sin\\;\\theta_2 ~. \\end{alignat*} \\nonumber Then $$(x_1,y_1)$$ and $$(x_2,y_2)$$ are the Cartesian equivalents of $$(r_1 , \\theta_1)$$ and $$(r_2 , \\theta_2)$$, respectively. Thus, by the Cartesian coordinate distance formula, \\begin{align*} d^2 ~&=~ (x_1 - x_2)^2 ~+~ (y_1 - y_2)^2\\\\[4pt] &=~ (r_1 \\,\\cos\\;\\theta_1 - r_2 \\,\\cos\\;\\theta_2)^2 ~+~ (r_1 \\,\\sin\\;\\theta_1 - r_2 \\,\\sin\\;\\theta_2)^2\\\\[4pt] &=~ r_1^2 \\cos^2\\;\\theta_1 ~-~ 2r_1 r_2 \\cos\\;\\theta_1~\\cos\\;\\theta_2 ~+~ r_2^2 \\cos^2\\;\\theta_2 ~+~ r_1^2 \\sin^2\\;\\theta_1 ~-~ 2r_1 r_2 \\sin\\;\\theta_1~\\sin\\;\\theta_2 ~+~ r_2^2 \\sin^2\\;\\theta_2\\\\[4pt] &=~ r_1^2 (\\cos^2\\;\\theta_1 ~+~ \\sin^2\\;\\theta_1) ~+~ r_2^2 (\\cos^2\\;\\theta_2 ~+~ \\sin^2\\;\\theta_2) ~-~ 2r_1 r_2 (\\cos\\;\\theta_1~\\cos\\;\\theta_2 ~+~ \\sin\\;\\theta_1~\\sin\\;\\theta_2)\\\\[4pt] d^2 ~&=~ r_1^2 ~+~ r_2^2 ~-~ 2r_1r_2\\,\\cos\\;(\\theta_1 - \\theta_2) ~, \\end{align*} \\nonumber so the result follows by taking square roots of both sides. In Example 6.17 we saw that the equation $$x^2 + y^2 = 9$$ in Cartesian coordinates could be expressed as $$r = 3$$ in polar coordinates. This equation describes a circle centered at the origin, so the circle is symmetric about the origin. In general, polar coordinates are useful in situations when there is symmetry about the origin (though there are other situations), which arise in many physical applications. 6.4: Polar Coordinates is shared under a GNU Free Documentation", "# How do you write in simplest radical form the coordinates of point A if A is on the terminal side of angle in standard position whose degree measure is theta: OA=8, theta=120^circ? ##### 1 Answer Mar 6, 2018 Coordinates of color(blue)(A (-4, 4 sqrt3) #### Explanation: ${120}^{\\circ}$ is in second quadrant ${A}_{x} = \\overline{O A} \\cdot \\cos \\theta = 8 \\cdot \\cos 120 = 8 \\cdot \\left(- \\cos \\left(60\\right)\\right) = 8 \\cdot - \\left(\\frac{1}{2}\\right) = - 4$ $\\textcolor{b l u e}{{A}_{x} = - 4}$ ${A}_{y} = \\overline{O A} \\cdot \\sin \\theta = 8 \\cdot \\sin 120 = 8 \\cdot \\sin 60 = 8 \\cdot \\left(\\frac{\\sqrt{3}}{2}\\right)$ color(blue)(A_y = 4 sqrt3", "with the equations relating the curvilinear coordinates to Cartesian coordinates. For example, the polar coordinates (r, θ) can be introduced as: ${\\displaystyle x=r\\cos \\theta ,\\ y=r\\sin \\theta .}$ Then two points with locations (r1, θ1) and (r2, θ2) are separated by a distance s: ${\\displaystyle s^{2}=(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}=(r_{1}\\cos \\theta _{1}-r_{2}\\cos \\theta _{2})^{2}+(r_{1}\\sin \\theta _{1}-r_{2}\\sin \\theta _{2})^{2}.}$ Performing the squares and combining terms, the Pythagorean formula for distance in Cartesian coordinates produces the separation in polar coordinates as: {\\displaystyle {\\begin{aligned}s^{2}&=r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\\left(\\cos \\theta _{1}\\cos \\theta _{2}+\\sin \\theta _{1}\\sin \\theta _{2}\\right)\\\\&=r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\\cos \\left(\\theta _{1}-\\theta _{2}\\right)\\\\&=r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\\cos \\Delta \\theta ,\\end{aligned}}} using the trigonometric product-to-sum formulas. This formula is the law of cosines, sometimes called the generalized Pythagorean theorem.[40] From this result, for the case where the radii to the two locations are at right angles, the enclosed angle Δθ = π/2, and the form corresponding to Pythagoras's theorem is regained: ${\\displaystyle s^{2}=r_{1}^{2}+r_{2}^{2}.}$ The Pythagorean theorem, valid for right triangles, therefore is a special case of the more general law of cosines, valid for arbitrary triangles. ### Pythagorean trigonometric identity In a right triangle with sides a, b and hypotenuse c, trigonometry determines the sine and cosine of the angle θ between side a and the hypotenuse as: ${\\displaystyle \\sin \\theta ={\\frac {b}{c}},\\quad \\cos \\theta ={\\frac {a}{c}}.}$ From that it follows: ${\\displaystyle {\\cos }^{2}\\theta +{\\sin }^{2}\\theta ={\\frac {a^{2}+b^{2}}{c^{2}}}=1,}$", "\\sin \\theta)\\end{align*}. If we allow z\\begin{align*}z\\end{align*} to equal the polar form of a complex number, it is very easy to see the development of a pattern when raising a complex number in polar form to a power. To discover this pattern, it is necessary to perform some basic multiplication of complex numbers in polar form. If z=r(cosθ+isinθ)\\begin{align*}z = r(\\cos \\theta + i \\sin \\theta)\\end{align*} and z2=zz\\begin{align*}z^2 = z \\cdot z\\end{align*} then: z2z2z2=r(cosθ+isinθ)r(cosθ+isinθ)=r2[cos(θ+θ)+isin(θ+θ)]=r2(cos2θ+isin2θ)\\begin{align*}z^2 &= r(\\cos \\theta + i \\sin \\theta) \\cdot r(\\cos \\theta + i \\sin \\theta)\\\\ z^2 &= r^2 [\\cos (\\theta + \\theta) + i \\sin (\\theta + \\theta)]\\\\ z^2 &= r^2 (\\cos 2\\theta + i \\sin 2\\theta)\\end{align*} Likewise, if z=r(cosθ+isinθ)\\begin{align*}z = r(\\cos \\theta + i \\sin \\theta)\\end{align*} and z3=z2z\\begin{align*}z^3 = z^2 \\cdot z\\end{align*} then: z3z3z3=r2(cos2θ+isin2θ)r(cosθ+isinθ)=r3[cos(2θ+θ)+isin(2θ+θ)]=r3(cos3θ+isin3θ)\\begin{align*}z^3 &= r^2(\\cos 2\\theta + i \\sin 2\\theta) \\cdot r(\\cos \\theta + i \\sin \\theta)\\\\ z^3 &= r^3 [\\cos (2\\theta + \\theta) + i \\sin (2\\theta + \\theta)]\\\\ z^3 &= r^3 (\\cos 3\\theta + i \\sin 3\\theta)\\end{align*} Again, if z=r(cosθ+isinθ)\\begin{align*}z = r(\\cos \\theta + i \\sin \\theta)\\end{align*} and z4=z3z\\begin{align*}z^4 = z^3 \\cdot z\\end{align*} then z4=r4(cos4θ+isin4θ)\\begin{align*}z^4 = r^4(\\cos 4\\theta + i \\sin 4\\theta)\\end{align*} These examples suggest a general rule valid for all powers of z\\begin{align*}z\\end{align*}, or n\\begin{align*}n\\end{align*}. We offer this rule and assume its validity for all n\\begin{align*}n\\end{align*} without formal proof, leaving that for", "# How do you convert r(2 + cos theta) = 1 into cartesian form? Nov 26, 2016 The equation is $3 {x}^{2} + 4 {y}^{2} + 2 x - 1 = 0$ #### Explanation: To transform from polar coordinates $\\left(r , \\theta\\right)$ into rectangular coordinates $\\left(x , y\\right)$, we use the folowing $x = r \\cos \\theta$ $y = r \\sin \\theta$ and ${x}^{2} + {y}^{2} = {r}^{2}$ Therefore, $r \\left(2 + \\cos \\theta\\right) = 1$ $2 r + r \\cos \\theta = 1$ $2 \\sqrt{{x}^{2} + {y}^{2}} + x = 1$ $2 \\sqrt{{x}^{2} + {y}^{2}} = 1 - x$ Squaring both sides, $4 \\left({x}^{2} + {y}^{2}\\right) = {\\left(1 - x\\right)}^{2} = 1 - 2 x + {x}^{2}$ $3 {x}^{2} + 4 {y}^{2} + 2 x - 1 = 0$ Nov 26, 2016 It is the ellipsys $3 {x}^{2} + 4 {y}^{2} + 2 x - 1 = 0$ #### Explanation: $\\rho = \\sqrt[2]{{x}^{2} + {y}^{2}}$ and $x = \\rho \\cdot \\sin \\theta$ and $y = \\rho \\cdot \\cos \\theta$ $\\sqrt[2]{{x}^{2} + {y}^{2}} \\cdot \\left(2 + \\frac{x}{\\sqrt[2]{{x}^{2} + {y}^{2}}}\\right) = 1$ that becomes $2 \\sqrt[2]{{x}^{2} + {y}^{2}} + x = 1$ or ${x}^{2} + {y}^{2} = {\\left(\\frac{1 - x}{2}\\right)}^{2}$ or ${x}^{2} + {y}^{2} - {\\left(1 - x\\right)}^{2} / 4 = 0$", "## Precalculus (6th Edition) Blitzer $r=4\\cos \\theta$. Rectangular equation is ${{\\left( x-2 \\right)}^{2}}+{{y}^{2}}=4$ …… (I) The relation between polar coordinates and rectangular coordinates can be represented as below: $x=r\\cos \\theta \\ \\text{ and }\\ y=r\\sin \\theta$ …… (II) Substitute values of $x\\ \\text{ and }\\ y$ from (II) in (I) to get \\begin{align} & {{\\left( x-2 \\right)}^{2}}+{{y}^{2}}=4 \\\\ & {{\\left( r\\cos \\theta -2 \\right)}^{2}}+{{\\left( r\\sin \\theta \\right)}^{2}}=4 \\\\ & {{r}^{2}}{{\\cos }^{2}}\\theta +4-4r\\cos \\theta +{{r}^{2}}{{\\sin }^{2}}\\theta =4 \\\\ & {{r}^{2}}\\left( {{\\cos }^{2}}\\theta +{{\\sin }^{2}}\\theta \\right)-4r\\cos \\theta =0 \\end{align} As, $\\left( {{\\cos }^{2}}\\theta +{{\\sin }^{2}}\\theta \\right)=1$ From here we get, \\begin{align} & {{r}^{2}}-4r\\cos \\theta =0 \\\\ & {{r}^{2}}=4r\\cos \\theta \\\\ & r=4\\cos \\theta \\end{align} Hence the obtained polar expression is $r=4\\cos \\theta$.", "2 \\sin^2 \\theta. \\end{align}", "that the rectangular coordinates of the same point are $(r\\cos\\theta,r\\sin\\theta)\\text{,}$ and so the point $(r,\\theta,z)$ in cylindrical coordinates is $(r\\cos\\theta,r\\sin\\theta,z)$ in rectangular coordinates. This means it is usually easy to convert any equation from rectangular to cylindrical coordinates: simply substitute \\begin{align*} x\\amp =r\\cos\\theta\\\\ y\\amp =r\\sin\\theta \\end{align*} and leave $z$ alone. For example, starting with $x^2+y^2=4$ and substituting $x=r\\cos\\theta\\text{,}$ $y=r\\sin\\theta$ gives \\begin{align*} r^2\\cos^2\\theta+r^2\\sin^2\\theta\\amp =4\\\\ r^2(\\cos^2\\theta+\\sin^2\\theta)\\amp =4\\\\ r^2\\amp =4\\\\ r\\amp =2\\text{.} \\end{align*} Of course, it's easy to see directly that this defines a cylinder as mentioned above. Cylindrical coordinates are an obvious extension of polar coordinates to three dimensions, but the use of the $z$ coordinate means they are not as closely analogous to polar coordinates as another standard coordinate system. In polar coordinates, we identify a point by a direction and distance from the origin; in three dimensions we can do the same thing, in a variety of ways. The question is: how do we represent a direction? One way is to give the angle of rotation, $\\theta\\text{,}$ from the positive $x$-axis, just as in cylindrical coordinates, and also an angle of rotation, $\\phi\\text{,}$ from the positive $z$-axis. Roughly speaking, $\\theta$ is like longitude and $\\phi$ is like latitude. (Earth longitude is measured as a positive or negative angle from the prime meridian, and is always between 0 and", "# How do you write in simplest radical form the coordinates of point A if A is on the terminal side of angle in standard position whose degree measure is theta: OA=6, theta=90^circ? Mar 6, 2018 Coordinates of color(green)(A = (0,1) #### Explanation: Point A is in first quadrant as $\\theta$ is ${90}^{\\circ}$ Hence, both x , y coordinates are positive.. ${A}_{x} = \\overline{O A} \\cos \\theta = 6 \\cdot \\cos 90 = 25 \\cdot 0 = 0$ as $\\cos \\left(90\\right) = - 0$ ${A}_{y} = \\overline{O A} \\sin \\theta = 25 \\cdot \\sin 90 = 25 \\cdot 1 = 25$ as $\\sin \\left(90\\right) = 1$ ${A}_{y} = - 25$ Coordinates of color(green)(A = (0,1)" ]

[ "? Free Version Moderate # Projection Onto Column Space LINALG-WLW1LH The matrix: $$A = \\begin{pmatrix} 2/3 & -2/3 \\\\\\ 1/3 & 2/3 \\\\\\ 2/3 & 1/3 \\end{pmatrix}$$ ...is orthogonal. Compute the projection of $\\mathbf{v}=(1,1,0)$ onto the column space of $A$. A $(2/3,1/3,2/3)$ B $(1,0,0)$ C $(1/3,2/3,-2/3)$ D $(2/3,1/3,-1/3)$ E $(0,1,1)$", "& 1 & 0 & -\\frac{top+bottom}{2} \\\\ 0 & 0 & -1 & \\frac{far+near}{2} \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}$ The inversion of the Projection Matrix, which can be used as the Unprojection Matrix is defined: $P^{-1} = \\begin{bmatrix} \\frac{right-left}{2} & 0 & 0 & \\frac{left+right}{2} \\\\ 0 & \\frac{top-bottom}{2} & 0 & \\frac{top+bottom}{2} \\\\ 0 & 0 & \\frac{far-near}{-2} & \\frac{far+near}{2} \\\\ 0 & 0 & 0 & 1 \\end{bmatrix}$", "Projection Matrix - Maple Programming Help Home : Support : Online Help : Tasks : Linear Algebra : Matrix Manipulations : Task/ProjectionMatrix Projection Matrix Description Generate a projection matrix that will project a vector onto the subspace spanned by a list of vectors. Enter a list of vectors. > $\\left[{[}\\begin{array}{c}{3}\\\\ {0}\\\\ {1}\\\\ {2}\\end{array}{]}{,}{[}\\begin{array}{c}{2}\\\\ {-}{1}\\\\ {0}\\\\ {1}\\end{array}{]}{,}{[}\\begin{array}{c}{-}{4}\\\\ {1}\\\\ {2}\\\\ {3}\\end{array}{]}\\right]$ $\\left[\\left[\\begin{array}{r}{3}\\\\ {0}\\\\ {1}\\\\ {2}\\end{array}\\right]{,}\\left[\\begin{array}{r}{2}\\\\ {-}{1}\\\\ {0}\\\\ {1}\\end{array}\\right]{,}\\left[\\begin{array}{r}{-}{4}\\\\ {1}\\\\ {2}\\\\ {3}\\end{array}\\right]\\right]$ (1) Calculate the Projection Matrix. > $\\mathrm{LinearAlgebra}\\left[\\mathrm{ProjectionMatrix}\\right]\\left(\\right)$ $\\left[\\begin{array}{cccc}{1}& {0}& {0}& {0}\\\\ {0}& \\frac{{5}}{{6}}& \\frac{{1}}{{3}}& {-}\\frac{{1}}{{6}}\\\\ {0}& \\frac{{1}}{{3}}& \\frac{{1}}{{3}}& \\frac{{1}}{{3}}\\\\ {0}& {-}\\frac{{1}}{{6}}& \\frac{{1}}{{3}}& \\frac{{5}}{{6}}\\end{array}\\right]$ (2) Commands Used", "similarly for the nullspace of $1-P$. 1. since $P$ is a projection matrix $P^2=P$. Using this it is easy to show that (I-P)(I-P) = I-P. Hence I-P is also Projection (symmetry property also preserves). 2. Since the eigenvalues of a projection matrix are either ones or zeros you can use eigendecomposition: $P = U \\begin{pmatrix} I_M & \\bf{0}\\\\ \\bf{0} &\\bf{0} \\end{pmatrix} U^T$ where $M$ is the dimension of the subspace to which the projectios is carried it's easy to show that $I - U \\begin{pmatrix} I_M & \\bf{0}\\\\ \\bf{0} &\\bf{0} \\end{pmatrix} U^T = U \\begin{pmatrix} \\bf{0} & \\bf{0}\\\\ \\bf{0} & I_{N-M} \\end{pmatrix} U^T$. From here it is easy to extract the 4 sub-spaces of each projection operator. The above analysis is correct for orthogonal projectors. For oblique projectors a similar calculation can be done using singular value decomposition. • Oblique projectors don't diagonalise like this – Jazzmaniac Mar 27 '16 at 12:09 • Right. Obviously the problem consider orthogonal projectors. – Nir Regev Mar 27 '16 at 12:13 • I don't see any reference to orthogonal projection in the question – Jazzmaniac Mar 27 '16 at 12:14 • @Jazzmaniac you're right. I updated the answer – Nir Regev Mar 27 '16 at 12:29", "\\\\ 1/2 \\\\ 1/2 \\\\ 1/2\\end{bmatrix}, u_2 =\\frac{v_2^\\perp}{\\left|\\left| v_2^\\perp \\right|\\right| } =\\frac{v_2-\\left(u_1 \\cdot v_2\\right) u_1}{\\left|\\left| v_2-\\left(u_1 \\cdot v_2\\right)u_1 \\right|\\right| } =\\begin{bmatrix}-1/10 \\\\ 7/10 \\\\ -7/10 \\\\ 1/10\\end{bmatrix}$$ form an orthonormal basis of$W.$By Orthogonal Projection Matrix, the matrix of the projection onto$W$is$A=Q Q^T$where$Q=\\begin{bmatrix}u_1 & u_2\\end{bmatrix}.$Therefore the orthogonal projection onto$W$is $$A= \\begin{bmatrix} 1/2 & -1/10 \\\\ 1/2 & 7/10 \\\\ 1/2 & -7/10 \\\\ 1/2 & 1/10 \\end{bmatrix} \\begin{bmatrix} 1/2 & 1/2 & 1/2 & 1/2 \\\\ -1/10 & 7/10 & -7/10 & 1/10 \\end{bmatrix} =\\frac{1}{100} \\begin{bmatrix} 26 & 18 & 32 & 24 \\\\ 18 & 74 & -24 & 32 \\\\ 32 & -24 & 74 &18 \\\\ 24 & 32 & 18 & 26 \\end{bmatrix}.$$ Let$A=\\begin{bmatrix}1 & 1 \\\\ 1 & 2 \\\\ 1 & 3 \\\\ 1 & 4 \\end{bmatrix}\\$ and then the orthogonal projection matrix is $$A(A^TA)^{-1}A^T =\\frac{1}{10}\\begin{bmatrix}7 & 4 & 1 & -2 \\\\ 4 & 3 & 2 & 1 \\\\ 1 & 2 & 3 & 4 \\\\ -2 & 1 & 4 & 7 \\end{bmatrix}.$$ David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.", "inverse of the matrix: $$\\begin{pmatrix} x& y\\\\ z& w\\\\ \\end{pmatrix}=\\begin{pmatrix} 0&-6\\\\ 1&-5\\\\ \\end{pmatrix}^{-1}=\\frac{1}{6}\\begin{pmatrix} -5&6\\\\ -1&0\\\\ \\end{pmatrix}=\\begin{pmatrix} -\\frac56&1\\\\ -\\frac16&0\\\\ \\end{pmatrix}$$", "+0 # Help hard quick 0 68 0 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ Use the picture below to find vectors with integer components and magnitude no greater than 8 that satisfy $$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$", "\\end{pmatrix}, & B_2 &= \\frac{1}{\\sqrt{2}} \\begin{pmatrix} -1 & 0 \\\\ 1 & 0 \\end{pmatrix}, \\\\ B_3 &= \\frac{1}{2} \\begin{pmatrix} -1 & 1 \\\\ -1 & 1 \\end{pmatrix}, & B_4 &= \\frac{1}{2} \\begin{pmatrix} 1 & 1 \\\\ 1 & 1 \\end{pmatrix}. \\end{align*} Hint: Find the eigenvalues and eigenvectors for this map. Choose the eigenvectors in such a way that you obtain an orthonormal basis. Then find the matrix representation of your map with respect to this basis.", "+0 # Consider the projection matrix P that satisfies 0 169 1 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}.$$ Use the picture below to find 3 vectors with integer components and magnitude no greater than 8 that satisfy Find $$\\mathbf{P}^{-1} \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$$$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}.$$ Aug 6, 2019", "## Orthogonal projection For matrix from the previous exercise calculate the orthogonal projection $$\\mathbf p$$ of the vector $$\\mathbf a=(2, 2, 1, 5)^T$$ to the row space of the matrix. Also determine coordinates $$[\\mathbf p]_Z$$ w.r.t. the basis $$Z$$. • #### Variant 1 see matrix $$\\begin{pmatrix} 1 & 1 & 1 & 1 \\\\ 4 & 1 & 4 & 1 \\\\ 1 & 2 & 3 & 4 \\\\ \\end{pmatrix}$$ • #### Variant 2 see matrix $$\\begin{pmatrix} 0 & 3 & 4 & 0 \\\\ 0 & 0 & 5 & 0 \\\\ 2 & 1 & 0 & 2 \\\\ \\end{pmatrix}$$ • #### Variant 3 see matrix $$\\begin{pmatrix} 2 & 0 & 1 & 2 \\\\ 4 & 3 & 2 & 4 \\\\ 6 & -5 & 3 & 6 \\\\ -4 & 2 & 4 & 2 \\\\ \\end{pmatrix}$$ • #### Variant 4 see matrix $$\\begin{pmatrix} 2 & 4 & 2 & 1 \\\\ -1 & -2 & -2 & -1 \\\\ 1 & 2 & 4 & 2 \\\\ 1 & 2 & 3 & 4 \\\\ \\end{pmatrix}$$", "\\frac{\\sqrt3}2 & -\\frac12 \\\\ \\end{pmatrix} = A_2^T.$$ The matrix $A_4$ establishes all the four possible eigenvalues in a single matrix. Remark. The matrix $A_2$ represents a rotation by $2\\pi/3$.", "does it mean to project onto a point? Won't you just get that point? – Spencer Gibson May 24 '17 at 19:05 • OK, so I think I understand. So you choose (2,-3) and (3,2) because they span the real numbers. So then, we note that because P is a linear transformation, Ta(2,-3) + b(3,2) = aT(2,-3) + bT(3,2) = a(2,-3), because the projection of (3,2) onto the given vector gives the origin. Thus, then we just write a,b in terms of v = (c,d). Doing this gives the projection matrix: first row = (4/13, -6/13) and second row = (-6/13, 9/13). Is this correct? – Spencer Gibson May 25 '17 at 1:49 • We chose $(2,-3)$ and $(3,2)$ as they are a useful basis for $\\mathbb R^2$ in this problem. Then as you say we end up with the projection matrix $P$ as I have also written out above. – Simon S May 25 '17 at 22:21 • It is worth pointing out that there are several ways to do this problem. Another way is to ask what are the images of the unit column vectors under the projection; these are the columns of the matrix $P$. – Simon S May 25 '17 at 22:23 The projection is found, by asking how much does the pointer $x$", "+0 # help 0 114 1 Let $\\mathbf{P}$ be the matrix that projects onto $\\mathbf{j}$: that is, we want $\\mathbf{P}$ to satisfy $\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$for all vectors $\\mathbf{v}$. Use the pictures below to calculate $\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$in that order, and enter them in as columns below. Then calculate the matrix $\\mathbf{P}$ that projects onto $\\mathbf{j}$. Jul 31, 2019 The matrix is $$\\begin{pmatrix} -1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$.", "v= -2i+ j. The projection of u= i onto that is given by $\\frac{u\\cdot v}{v\\cdot v}v$ which, here, is $\\frac{-2}{5}(-2i+ j)= \\frac{4}{5}i-\\frac{2}{5}j$. The projection of w= j onto that is given by $\\frac{1}{5}(-2i+ j)= -\\frac{2}{5}i+ \\frac{1}{5}j$. Since applying a linear transformation to the basis vectors in turn gives the columns of the matrix, the matrix corresponding to this projection, in the standard basis, is $\\begin{bmatrix}\\frac{4}{5} & -\\frac{2}{5} \\\\ -\\frac{2}{5} & \\frac{1}{5}\\end{bmatrix}$", "will be exactly the matrix representation of $D(\\psi\\circ F\\circ\\phi^{-1})_{\\phi(p)}$ (cont) – peek-a-boo Aug 1 '20 at 2:18 • (cont.) relative to the standard ordered bases on $\\Bbb{R}^m$ and $\\Bbb{R}^n$, which is simply the Jacobian matrix of the map $\\psi\\circ F \\circ \\phi^{-1}$ evaluated at the point $\\phi(p)$. – peek-a-boo Aug 1 '20 at 2:19", "How do you do this vector projection? For $\\textbf{v} = \\begin{pmatrix} 2 \\\\ 3 \\\\ -1 \\end{pmatrix}$ and $\\textbf{w} = \\begin{pmatrix} 2 \\\\ -1 \\\\ 0 \\end{pmatrix}$, compute $\\text{proj}_{\\textbf{w}} \\textbf{v}$. I don't know how to project vectors of this form... I can only do it in 2D. Is there a formula for this? • You do it the same way. There are just more components to compute. – amd Mar 19 '17 at 1:01", "+0 # Matrices 0 40 1 Let w = $$\\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix}$$. There exists a $$2 \\times 2$$ matrix $${P}$$ such that $$\\text{proj}_{{w}} {v} = {P} {v}$$ for all 2-dimensional vectors $${v}$$. Find $${P}$$. I'm not really sure where to start for this problem. Should I plug w into the projection formula? Mar 6, 2021 ### 1+0 Answers #1 0 The matrix is $$\\begin{pmatrix} 2 & -1 \\\\ -1 & 0 \\end{pmatrix}$$ Mar 7, 2021", "& 7/10 & -7/10 & 1/10 \\end{bmatrix} =\\frac{1}{100} \\begin{bmatrix} 26 & 18 & 32 & 24 \\\\ 18 & 74 & -24 & 32 \\\\ 32 & -24 & 74 &18 \\\\ 24 & 32 & 18 & 26 \\end{bmatrix}.$$ Let $A=\\begin{bmatrix}1 & 1 \\\\ 1 & 2 \\\\ 1 & 3 \\\\ 1 & 4 \\end{bmatrix}$ and then the orthogonal projection matrix is $$A(A^TA)^{-1}A^T =\\frac{1}{10}\\begin{bmatrix}7 & 4 & 1 & -2 \\\\ 4 & 3 & 2 & 1 \\\\ 1 & 2 & 3 & 4 \\\\ -2 & 1 & 4 & 7 \\end{bmatrix}.$$", "## Matrix od order n $$\\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & … & n \\\\ -1 & 0 & 2 & 3 & 4 & … & n-1 \\\\ -1 & -2 & 0 & 3 & 4 & … & n-1 \\\\ -1 & -2 & -3 & 0 & 4 & … & n-1 \\\\ -1 & -2 & -3 & -4 & 0 & … & n-1 \\\\ \\vdots & \\vdots & \\vdots & \\vdots & & \\ddots & \\\\ -1 & -2 & -3 & -4 & … & 1-n & 0 \\\\ \\end{pmatrix}$$. The determinant of the matrix is $$n!$$.", "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019" ]

[ "# What is the projection of < 6 , -6 ,3 > onto < 4, -5, 1>? Mar 14, 2016 =$< \\frac{228}{42} , \\frac{- 285}{42} , \\frac{57}{42} >$ #### Explanation: Projection of a vector a on another vector b (component of a along b )is given by $P r o {j}_{b} a$ = $\\frac{\\ast a \\ast . \\ast b \\ast}{|} | b | {|}^{2} \\ast b \\ast$ Hence if a is <6,-6,3> and b is < 4,-5,1> a . b = 24 +30 +3 =57 $| | b | {|}^{2} = {4}^{2} + {\\left(- 5\\right)}^{2} + {1}^{2}$=42 Hence,$P r o {j}_{b} a = \\frac{57}{42} < 4 , - 5 , 1 >$ =$< \\frac{228}{42} , \\frac{- 285}{42} , \\frac{57}{42} >$", "# Finding a set of vectors that spans the solution Find a set of vectors {u, v} in $\\mathbb{R}^4$ that spans the solution set of the equations: $x - y + 2z - 2w = 0$ $2x + 2y -z + 3w = 0$ ($u$ and $v$ are both $4 \\times 1$) $u = ?$, $v = ?$ I put the matrix in RREF to get $\\begin{bmatrix}1&0&3/4&-1/4\\\\0&1&-5/4&7/4\\end{bmatrix} = \\begin{bmatrix}0\\\\0\\end{bmatrix}$ Then I got $x = -\\frac{3}{4} z + \\frac{1}{4} w$ and $y = \\frac{5}{4} z - \\frac{7}{4} w$ But I'm not sure how to present the answer as they want it. • HINT: (Assuming your numbers are correct) you've found that $$\\pmatrix{x \\\\ y \\\\ z \\\\ w} = \\pmatrix{-\\frac 34z+\\frac 14w \\\\ \\frac 54z-\\frac 74w \\\\ z \\\\ w} = z\\pmatrix{-\\frac 34 \\\\ \\frac 54 \\\\ 1 \\\\ 0} + w\\pmatrix{\\frac 14 \\\\ -\\frac 74 \\\\ 0 \\\\1}$$ – user137731 Jul 8 '16 at 3:03 You can write the solution set $S$ as follows: $$S=\\{(x,y,z,w)\\in\\mathbb{R}^4: x=-\\frac{3}{4}z+\\frac{1}{4}w,\\, y=\\frac{5}{4}z-\\frac{7}{4}w;\\,\\,\\ z,w\\in\\mathbb{R}\\}.$$", "the projection matrix $\\mathbf{P}_{W_T}$ projects onto a 24-dimensional subspace since we have removed one dimension by subtracting out $\\mathrm{C}(\\mathbf{1}_N)$. The same goes for $\\mathbf{P}_{W_J}$ (projecting onto a three-dimensional subspace), and $\\mathbf{P}_{W_Q}$ (projecting onto a four-dimensional subspace), and $\\mathbf{P}_{W{JQ}}$ (projecting onto a 12-dimensional subspace). That means the residual projection matrix $\\mathbf{P}_\\perp$ is projecting into a $N-44=430-44=386$-dimensional subspace. All of this means that for our model: $E\\|\\mathbf{P}_0\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_0\\boldsymbol{\\tau}\\|^2 + \\sigma^2$ $E\\|\\mathbf{P}_{W_T}\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_{W_T}\\boldsymbol{\\tau}\\|^2 + 24\\sigma^2$ $E\\|\\mathbf{P}_{W_J}\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_{W_J}\\boldsymbol{\\tau}\\|^2 + 3\\sigma^2$ $E\\|\\mathbf{P}_{W_Q}\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_{W_Q}\\boldsymbol{\\tau}|^2 + 4\\sigma^2$ $E\\|\\mathbf{P}_{W_{JQ}}\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_{W_{JQ}}\\boldsymbol{\\tau}\\|^2 + 12\\sigma^2$ $E\\|\\mathbf{P}_\\perp\\mathbf{Y}\\|^2 = (N-44)\\sigma^2$ where the last one comes from the fact that $\\boldsymbol{\\tau}$ is orthogonal to the residual subspace. The left-hand side of each of these is just an expected sum of squared random values. On the right hand side, we have two terms: the first due to the deterministic effects of the levels in a factor, and the second due to iid noise in the measurements. If there is no effect at a factor, then its deterministic component will be zero. In addition, if there are no differences between the effects in a factor, then the projections will be zero. Hence, to test for significant differences between the effects in a factor, all we need to do is compare the empirical sum of squares of the projections of", "now we can figure out $$\\mathtt{V}$$ and $$\\mathtt{\\Sigma}$$ by doing the eigenvalue decomposition like we did previously. Here it is: $\\begin{bmatrix}\\mathtt{4}&\\mathtt{6}\\\\\\mathtt{6}&\\mathtt{10}\\end{bmatrix}=\\begin{bmatrix}\\mathtt{\\color{purple}{-\\frac{521}{991}}}&\\mathtt{\\color{red}{-\\frac{3725}{4379}}}\\\\\\mathtt{\\color{purple}{-\\frac{3725}{4379}}}&\\mathtt{\\,\\,\\,\\,\\color{red}{\\frac{521}{991}}}\\end{bmatrix}\\begin{bmatrix}\\mathtt{\\frac{4510}{329}}&\\mathtt{0}\\\\\\mathtt{0}&\\mathtt{\\frac{658}{2255}}\\end{bmatrix}\\begin{bmatrix}\\mathtt{-\\frac{521}{991}}&\\mathtt{-\\frac{3725}{4379}}\\\\\\mathtt{-\\frac{3725}{4379}}&\\mathtt{\\,\\,\\,\\,\\frac{521}{991}}\\end{bmatrix}$ The purple and red vectors are our starting vectors from the animation on the left above: (–0.53, –0.85) and (–0.85, 0.53). When we ask what orthogonal starting vectors can we pick such that matrix $$\\mathtt{A}$$ will transform them to a different pair of orthogonal vectors, matrix $$\\mathtt{V}$$—the red and purple vectors above—is our answer. The square roots of the diagonal values of $$\\mathtt{\\Sigma}$$ above (remember, this matrix was squared) are called the singular values of $$\\mathtt{A}$$. To me, getting $$\\mathtt{V}$$ is the coolest part of this decomposition. The rest, below, is just gravy. And Now for $$\\mathtt{U}$$ and $$\\mathtt{\\Sigma}$$ In this case, we multiply both sides of the equation $$\\mathtt{A=U\\Sigma V^T}$$ on the right by the transpose of $$\\mathtt{A}$$ and get an eigenvalue decomposition of $$\\mathtt{AA^T}$$. $\\,\\,\\mathtt{AA^T=(U\\Sigma V^T)(U\\Sigma V^T)^T=U\\Sigma V^{T}V\\Sigma^{T}U^T=U\\Sigma^{2}U^T}$ $\\begin{bmatrix}\\mathtt{\\,\\,\\,\\,1}&\\mathtt{-3}\\\\\\mathtt{-3}&\\mathtt{\\,\\,\\,\\,13}\\end{bmatrix}=\\begin{bmatrix}\\mathtt{-\\frac{400}{1741}}&\\mathtt{\\frac{764}{785}}\\\\\\mathtt{\\,\\,\\,\\,\\frac{764}{785}}&\\mathtt{\\frac{400}{1741}}\\end{bmatrix}\\begin{bmatrix}\\mathtt{\\color{purple}{\\frac{4510}{329}}}&\\mathtt{\\color{red}{0}}\\\\\\mathtt{\\color{purple}{0}}&\\mathtt{\\color{red}{\\frac{658}{2255}}}\\end{bmatrix}\\begin{bmatrix}\\mathtt{-\\frac{400}{1741}}&\\mathtt{\\frac{764}{785}}\\\\\\mathtt{\\,\\,\\,\\,\\frac{764}{785}}&\\mathtt{\\frac{400}{1741}}\\end{bmatrix}$ Now the square roots of the purple and red vectors are our starting vectors from the animation on the right above: (3.7, 0) and (0, 0.54). You’ll notice, of course, that $$\\mathtt{\\Sigma^2}$$ is the same as above, so we really found it earlier. This step is only to pin down what $$\\mathtt{U}$$ is. At any rate, we are done, and we can write down the full", "+0 # help vectors 0 46 2 Let v and w be vectors such that $$\\text{proj}_{{w}} {v} = \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ Find $$\\text{proj}_{-2 {w}} (3 {v})$$. Feb 23, 2020 #1 0 Projection is linear, so the answer is 2*3*(4,-7) = (24,-42). Feb 24, 2020 #2 +24366 +1 Let $$\\mathbf{v} \\text{ and } \\mathbf{w} \\text{ be vectors such that } \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$. Find $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}})$$. $$\\begin{array}{|rcll|} \\hline \\vec{p} &=& \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} ) \\\\\\\\ &=& \\dbinom{4}{-7} \\\\\\\\ &=& \\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\ \\hline \\vec{P} &=& \\operatorname{proj}_{\\mathbf{-2w}}( \\mathbf{3v} ) \\\\\\\\ &=& \\dfrac{(-2\\vec{w})(3\\vec{v})}{|-2w|^2}(-2\\vec{w}) \\\\\\\\ &=& 6\\dfrac{\\vec{w}\\vec{v}}{2w^2}\\vec{w} \\\\\\\\ &=& 3\\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\\\\\ &=& 3\\vec{p} \\\\\\\\ &=& 3\\dbinom{4}{-7} \\\\\\\\ &=& \\dbinom{12}{-21} \\\\ \\hline \\end{array}$$ $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}}) = \\dbinom{12}{-21}.$$ Feb 24, 2020", "Coordinate: $\\dfrac{3}{\\sqrt{11}}$; Projection: $\\displaystyle{\\left(-\\frac{3}{11},\\frac{9}{11},\\frac{3}{11}\\right)}$ (c) Coordinate: $\\dfrac{5}{\\sqrt{38}}$; Projection: $\\displaystyle{\\left(\\frac{5}{38},-\\frac{5}{38},\\frac{15}{19}\\right)}$ (d) Coordinate: $\\dfrac{5}{3\\sqrt{2}}$; Projection: $\\displaystyle{\\left(\\frac{5}{9},-\\frac{5}{18},\\frac{5}{9},\\frac{5}{6}\\right)}$ 11. $\\displaystyle{\\mathbf{x} = \\left(\\frac{5}{7},\\frac{15}{14},\\frac{5}{14}\\right) + \\left(\\frac{2}{7}, \\frac{13}{14},-\\frac{47}{14}\\right)}$ Section 1.3 1. (a) $\\mathbf{x} \\times \\mathbf{y} = (1, 3, 7)$ (b) $\\mathbf{x} \\times \\mathbf{y} = (-2, 16, -5)$ (c) $\\mathbf{x} \\times \\mathbf{y} = (36, -12, 0)$ (d) $\\mathbf{x} \\times \\mathbf{y} = (-6, 2, -14)$ 3. $11$ 5. $3$ 7. $\\dfrac{9}{2}$ 9. $42$ 13. For example, $\\mathbf{e}_2 \\times \\left(\\mathbf{e}_2 \\times \\mathbf{e}_3\\right) = -\\mathbf{e}_3$, whereas $\\left(\\mathbf{e}_2 \\times \\mathbf{e}_2\\right) \\times \\mathbf{e}_3 = \\mathbf{0}$. Section 1.4 1. Vector equation: $\\mathbf{y} = t(1, -2) + (2, 3) = (t + 2, -2t + 3)$ Parametric equations: \\begin{align} x &= t +2 \\\\ y &= -2t + 3 \\end{align} 3. (a) Vector equation: $\\mathbf{y} = t(5,5) + (-1, -3) = (5t - 1, 5t - 3)$ Parametric equations: \\begin{align} x &= 5t -1 \\\\ y &= 5t -3 \\end{align} (b) Vector equation: $\\mathbf{y} = t(3, 1, -2) + (2, 1, 3) = (3t+2, t+1, -2t+3)$ Parametric equations: \\begin{align} x &= 3t + 2 \\\\ y &= t + 1 \\\\ z &= -2t + 3 \\end{align} (c) Vector equation: $\\mathbf{y} = t(1, 2, -3, 3) + (3, 2, 1, 4) = (t+3, 2t+2,-3t+1,3t+4)$ Parametric equations: \\begin{align} w &= t + 3 \\\\ x &= 2t + 2 \\\\ y &= -3t +", "Also, let $\\mathbf y$ be the $m$-dimensionl vector with your target values: $$\\mathbf y=\\begin{pmatrix}y^{(1)}\\\\y^{(2)}\\\\\\vdots\\\\y^{(m)}\\end{pmatrix}.$$ Now, since $h_w(x^{(i)})=(x^{(i)})^T\\mathbf w$, we can verify $$\\begin{array}{rcl}X\\mathbf w -\\mathbf y&=&\\begin{pmatrix}(x^{(1)})^T\\mathbf w\\\\\\vdots\\\\(x^{(m)})^T\\mathbf w\\end{pmatrix}-\\begin{pmatrix}y^{(1)}\\\\\\vdots\\\\y^{(m)}\\end{pmatrix}\\\\&=&\\begin{pmatrix}(x^{(1)})^T\\mathbf w-y^{(1)}\\\\\\vdots\\\\(x^{(m)})^T\\mathbf w-y^{(m)}\\end{pmatrix}.\\end{array}$$ Thus, using $\\mathbf v^T\\mathbf v = \\sum_i\\mathbf v_i^2$ for a vector $\\mathbf v$, we obtain $$\\begin{array}{rcl}E(w)&=&\\frac{1}{2}\\sum_{i=1}^m(h_w(x^{(i)}) - y^{(i)})^2\\\\&=&\\frac{1}{2}(X\\mathbf w-\\mathbf y)^T(X\\mathbf w-\\mathbf y).\\end{array}$$ Summarizing. To obtain $$\\mathbf w=\\begin{pmatrix}w_0\\\\w_1\\end{pmatrix},$$ proceed: 1. Set your collections $x$ and $y$: $$x=\\begin{pmatrix}1\\\\3\\\\5\\\\7\\\\9\\end{pmatrix}\\qquad y=\\begin{pmatrix}1\\\\3\\\\4\\\\3\\\\5\\end{pmatrix}.$$ 2. Loop until you think that $w_j$ converge (try with $10$ iterations) { for $i=1$ to $m$ { $$w_j:=w_j+\\eta\\sum_{i=1}^m(y^{(i)}-h_w(x^{(i)}))x_j^{(i)}\\qquad(\\text{for each }j)$$ } } Insted you can use 1'. Set your vectors $X$ and $\\mathbf y$: $$X=\\begin{pmatrix}1&1\\\\1&3\\\\1&5\\\\1&7\\\\1&9\\end{pmatrix}\\qquad\\mathbf y=\\begin{pmatrix}1\\\\3\\\\4\\\\3\\\\5\\end{pmatrix}.$$ 2'. Loop until you think that $\\mathbf w$ converge (try with $10$ iterations) { $$\\mathbf w:=\\mathbf w-\\frac{\\eta}{m}\\left((X\\mathbf w-\\mathbf y)^TX\\right)$$ } Note. With the above it can prove a analytical formula to obtain the parameters, using the normal equations to computed $\\mathbf w$ by $$\\mathbf w=(X^TX)^{-1}X^T\\mathbf y.$$ Obviously you need to use a computer software (like python or octave) or a calculator device (with matrix-vector support) to calculate the values.", "$$\\varphi$$ and $$r$$. 4. Mar 30, 2006 ### kryptyk Ok...so consider then: $$x = A\\, r [\\cos \\varphi]^{\\frac{1}{n}}$$ $$y = B\\, r [\\sin \\varphi]^{\\frac{1}{n}}$$ from which we then get: $$\\frac{x^{2n}}{A^{2n}} +\\frac{y^{2n}}{B^{2n}} = r^{2n}$$ Let $\\{\\mathbf{e}_1, \\mathbf{e}_2} \\}$ form an orthonormal basis for the rectangular coordinate system and let $\\mathbf{w} = x\\, \\mathbf{e}_1 + y\\, \\mathbf{e}_2\\;$. So then, $$\\mathbf{w} = r (A [\\cos \\varphi]^{\\frac{1}{n}} \\mathbf{e}_1 + B [\\sin \\varphi]^{\\frac{1}{n}} \\mathbf{e}_2)$$ To find a tangent vector to some superellipse given by fixed $r\\,$, we find: $$\\frac{\\partial \\mathbf{w}}{\\partial \\varphi} = \\frac{r}{\\sin \\varphi \\cos \\varphi} (-\\frac{A\\, \\sin^2 \\varphi}{n} [\\cos \\varphi]^{\\frac{1}{n}} \\mathbf{e}_1 + \\frac{B\\, \\cos^2 \\varphi}{n} [\\sin \\varphi]^{\\frac{1}{n}} \\mathbf{e}_2)$$ And I have no idea where I'm going with this but I'm having fun with the TeX stuff :p 5. Mar 30, 2006 ### traianus It is not difficult to find a curvilinear coordinate system, the difficulty is to find one in which the two curves that you obtain when you set one parameter constant are perpendicular to each other. See for example the link http://mathworld.wolfram.com/EllipticCylindricalCoordinates.html From which it appears clear that the two families of curves (ellipses and hyperbolas) are perpendicular in each point they intercept. My desire is to find the analytical expressions for the curve similar to the link I posted. But instead of ellipses I have a particular kind of", "\\mp$ ### Vector Derivative Let $\\mathbf{x}, \\mathbf{u}$ be vectors of length $n$, and let $A$ be a matrix of size $n\\times n$ then, \\begin{align} \\frac{\\partial}{\\partial \\mathbf{x}} (\\mathbf{u}^{T}\\mathbf{x}) & = \\frac{\\partial}{\\partial \\mathbf{x}} (\\mathbf{x}^{T}\\mathbf{u}) = \\mathbf{u}^{T} \\\\ \\frac{\\partial}{\\partial \\mathbf{x}} (\\mathbf{A}\\mathbf{x}) & = \\mathbf{A} \\\\ \\frac{\\partial}{\\partial \\mathbf{x}} (\\mathbf{x}^{T} \\mathbf{A} \\mathbf{x}) & = \\mathbf{x}^{T} (\\mathbf{A} + \\mathbf{A}^{T}) \\\\ \\frac{\\partial^{2}}{\\partial \\mathbf{x}^{2}} (\\mathbf{x^{T}}\\mathbf{A}\\mathbf{x}) & = \\mathbf{A} + \\mathbf{A}^{T} \\end{align} Note that if $\\mathbf{A}$ is a symmetric matrix then, $\\mathbf{A} + \\mathbf{A}^{T} = 2 \\mathbf{A}$. Published on May 7, 2021 Last revised on May 9, 2021", "-3 & 5 & -2 \\\\ 0 & 2 & 0 \\\\ 1 & 3 & 4 \\end{array}\\right]$$ The projection matrix onto $$W$$ is $$P=X(X^\\top X)^{-1}X^\\top$$ In our case, we end up with $$P=\\left[\\begin{array}{rrrr} \\frac{131}{231} & -\\frac{10}{77} & \\frac{10}{21} & \\frac{10}{231} \\\\ -\\frac{10}{77} & \\frac{74}{77} & \\frac{1}{7} & \\frac{1}{77} \\\\ \\frac{10}{21} & \\frac{1}{7} & \\frac{10}{21} & -\\frac{1}{21} \\\\ \\frac{10}{231} & \\frac{1}{77} & -\\frac{1}{21} & \\frac{230}{231} \\end{array}\\right]$$ We can now project any vector $$v\\in\\Bbb R^4$$ onto $$W$$ by computing $$Pv$$. If you're dead set on using orthonormal bases, then you could take these three basis vectors and apply the Gram-Schmidt algorithm. This gives the new basis \\begin{align*} q_1 &= \\left\\langle\\frac{1}{\\sqrt{11}},\\,-\\frac{3}{\\sqrt{11}},\\,0,\\,\\frac{1}{\\sqrt{11}}\\right\\rangle & q_2 &= \\left\\langle\\frac{1}{\\sqrt{7}},\\,\\frac{1}{\\sqrt{7}},\\,\\frac{1}{\\sqrt{7}},\\,\\frac{2}{\\sqrt{7}}\\right\\rangle & q_3 &= \\left\\langle-\\frac{1}{\\sqrt{3}},\\,0,\\,-\\frac{1}{\\sqrt{3}},\\,\\frac{1}{\\sqrt{3}}\\right\\rangle \\end{align*} We can then project any $$v\\in\\Bbb R^4$$ with the formula $$Pv=\\langle q_1, v\\rangle q_1 + \\langle q_2, v\\rangle q_2 + \\langle q_3, v\\rangle q_3$$ • @Leif The sum of the projections of $p$ onto $W$ and $W_\\perp$ will just be $p$, as I mentioned. Why do you want to add them? The question only asks for the projections. – Shubham Johri May 17 at 19:01", "+0 0 82 3 Part 1) Let $${A} = \\begin{pmatrix} 3 & -2 & 3 \\\\ 1 & 2 & 1 \\\\ 1 & 3 & 0 \\end{pmatrix},$$ and let $${v}_1 = \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}, \\quad {v}_2 = \\begin{pmatrix} -1 \\\\ 1 \\\\ 1 \\end{pmatrix}, \\quad {v}_3 = \\begin{pmatrix} 11 \\\\ 1 \\\\ -14 \\end{pmatrix}.$$ Show that A sends each of $$\\mathbf{v}_1, \\mathbf{v}_2,$$ and $$\\mathbf{v}_3$$ to scalar multiples of themselves, and find the value of these scalars. Part 2) Let n be a positive integer. Use part 1 to find the vectors $$\\mathbf{A}^n \\mathbf{v}_1, \\mathbf{A}^n \\mathbf{v}_2, \\mathbf{A}^n\\mathbf{v}_3.$$ Part 3) Calculate $${A}^{10} \\begin{pmatrix} 10 \\\\ 4 \\\\ -11 \\end{pmatrix}.$$ Mar 16, 2020 edited by Guest Mar 16, 2020 #1 +29982 +1 Here's Part 1) Mar 16, 2020 #2 +29982 +1 Here are parts 2 and 3: Mar 16, 2020 #3 0 Thank you! Guest Mar 16, 2020", "that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is the projection matrix $P=\\frac{1}{\\mathbf{a}^T\\mathbf{a}}\\mathbf{a}\\mathbf{a}^T,$ which is an $n\\times n$ matrix. To this end, Prof. Strang writes the projection vector as $\\mathbf{p}=\\mathbf{a}\\,x$. Now, you should see $\\mathbf{p}$ and $\\mathbf{a}$ as $n\\times1$ matrices and $x$ as a one dimensional vector or a $1\\times1$ matrix, so that $\\mathbf{a}$ and $x$ can be multiplied. Thus, we have $x\\,\\mathbf{a}=\\mathbf{a}\\,x.$ This identity does not mean that $\\mathbf{a}$ and $x$ are two matrices that commute. Just read the left and right sides as stated above. It is true that the notation may be a bit misleading. Perhaps it would be better to write $(x)$ instead of $x$ if $x$ should be seen as a $1\\times 1$ matrix (parentheses are used to delimit matrices). For example, $4\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix} =\\begin{pmatrix} 4 \\\\ 8 \\\\ 12\\end{pmatrix} =\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix}(4).$ The advantage of expressing $\\mathbf{p}$ as $\\mathbf{a}x$ is that generalization is then possible. If $S$ is spanned by the linearly independent vectors $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, the orthogonal projection $\\mathbf{p}$ of a vector $\\mathbf{b}$ onto $S$ should be a linear combination of $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, that is, $\\mathbf{p}=x_1\\mathbf{a}_1+\\cdots+x_k\\mathbf{a}_k=AX$ with $A=\\left(\\mathbf{a}_1 \\vert\\ldots \\vert\\mathbf{a}_k\\right)$ and $X=\\begin{pmatrix}x_1 \\\\ \\vdots \\\\ x_k\\end{pmatrix}$ The reasoning in Prof. Strang’s lecture would then show that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is now the projection matrix $P=A(A^TA)^{-1}A^T.$ Returning to your question, you", "be $(1,-3)$. But in the case I described above, when we have to go from a basis to the canonical one, I have to write the result of $L$ on each vector of the old basis as a linear combination of the canonical one. I expected, instead, something like: $$L(\\mathbf{e}_1) = 1\\times\\mathbf{e}_1 -3\\times\\mathbf{e}_2.$$ If you want to know the matrix of $L$ with respect to the canonical basis, what you have to compute is $L(\\mathbf{e}_1)$, $L(\\mathbf{e}_2)$, and $L(\\mathbf{e}_3)$. Since $\\mathbf{e}_1=\\frac1{\\sqrt2}(\\mathbf{u}+\\mathbf{v})$,\\begin{align}L(\\mathbf{e}_1)&=\\frac1{\\sqrt2}\\bigl(L(\\mathbf{u})+L(\\mathbf{v})\\bigr)\\\\&=\\frac1{\\sqrt2}(\\mathbf{u}-\\mathbf{v})\\\\&=(0,-1,0)\\\\&=-\\mathbf{e}_2.\\end{align}Therefore, the first column of the matrix of $T$ with respect to the canonical basis will be $\\left(\\begin{smallmatrix}0\\\\-1\\\\0\\end{smallmatrix}\\right)$. The other method of finding this matrix consists in computing $PAP^{-1}$, with$$P=\\begin{pmatrix}\\frac1{\\sqrt2}&\\frac1{\\sqrt2}&0\\\\-\\frac1{\\sqrt2}&\\frac1{\\sqrt2}&0\\\\0&0&1\\end{pmatrix}.$$Then$$PAP^{-1}=\\begin{pmatrix}0 & -1 & 0 \\\\ -1 & 0 & 0 \\\\ 0 & 0 & -2\\end{pmatrix}.$$So, both approaches lead to the same fist column.", "matrix so that each vector is linearly independent from the other ones. I got this: $\\begin{pmatrix} {\\frac{12}{13}}&{\\frac{12}{13}}&{0}\\\\ {0}&{\\frac{5}{13}}&{\\frac{12}{13}}\\\\ {\\frac{5}{13}}&{0}&{\\frac{5}{13}} \\end{pmatrix}=\\frac{1}{13}\\begin{pmatrix} {12}&{12}&{0}\\\\ {0}&{5}&{12}\\\\ {5}&{0}&{5} \\end{pmatrix}$ ? $\\begin{bmatrix}12/13\\\\5/13\\\\0\\end{bmatrix}$ is fine for the middle column, but the other two columns are not orthogonal to it. You could for example use $\\begin{bmatrix}-5/13\\\\12/13\\\\0\\end{bmatrix}$ for one of them. Can you see a vector (hint: without any fractions in its entries) that is orthogonal to both of these? Just to emphasise the point: linear independence is not enough, the columns must be orthogonal to each other. • April 10th 2009, 03:45 AM Showcase_22 Quote: Originally Posted by Opalg Can you see a vector (hint: without any fractions in its entries) that is orthogonal to both of these? Ah, it's just $\\begin{pmatrix} {0}\\\\ {0}\\\\ {1} \\end{pmatrix}$ since those two vectors don't have a k component. So ultimately we have: $\\begin{pmatrix} {0}&{\\frac{12}{13}}&{-\\frac{5}{13}}\\\\ {0}&{\\frac{5}{13}}&{\\frac{12}{13}}\\\\ {1}&{0}&{0} \\end{pmatrix} $ Can you also have: $\\begin{pmatrix} {0}&{\\frac{12}{13}}&{\\frac{5}{13}}\\\\ {0}&{\\frac{5}{13}}&{\\frac{12}{13}}\\\\ {1}&{0}&{0} \\end{pmatrix} $ or $\\begin{pmatrix} {0}&{-\\frac{12}{13}}&{-\\frac{5}{13}}\\\\ {0}&{\\frac{5}{13}}&{\\frac{12}{13}}\\\\ {1}&{0}&{0} \\end{pmatrix} $ ? (ie. any situation in which the magnitude of the rows and columns is 1?) • April 10th 2009, 04:04 AM Opalg Quote: Originally Posted by Showcase_22 Ah, it's just $\\begin{pmatrix} {0}\\\\ {0}\\\\ {1} \\end{pmatrix}$ since those two vectors don't have a k component. So ultimately we have: $\\begin{pmatrix} {0}&{\\frac{12}{13}}&{-\\frac{5}{13}}\\\\ {0}&{\\frac{5}{13}}&{\\frac{12}{13}}\\\\ {1}&{0}&{0} \\end{pmatrix} $", "# Finding $P(D)$ given we know $P(C)$, $P(C|D')$ and $P(C|D)$ As the title states we need to find $P(D)$ given that $P(C)={9\\over20}$, $P(C|D')={3\\over7}$ and $P(C|D)={6\\over13}$. I have been hitting my head against a brick wall for the past half an hour or so, unable to make much progress. So far all that I have managed to do is some algebraic manipulation with ${6\\over13}={P(C\\cap D)\\over P(D)}$ and ${3\\over7}={P(C\\cap D')\\over P(D')}$. However I don't seem to be making any progress. Does anyone have any suggestions on a continuation? - I'll assume that $D'$ is the complement of the event $D$, since if not we don't have enough information to solve this problem. You have $$\\mathbb P(C) = \\mathbb P(C \\, | \\, D) \\mathbb P(D) + \\mathbb P(C \\, | \\, D') \\mathbb P(D') = \\mathbb P(C \\, | \\, D) \\mathbb P(D) + \\mathbb P(C \\, | \\, D') (1-\\mathbb P(D)).$$ Now you have a linear equation in $\\mathbb P(D)$, so it's pretty easy to find it. It gives you $$\\frac 9{20} = \\frac 6{13} \\times \\mathbb P(D) + \\frac 37 (1 - \\mathbb P(D)) = \\frac 37 + \\left( \\frac 6{13} - \\frac 37 \\right) \\mathbb P(D)$$ which leads to $\\mathbb P(D) = \\frac {13}{20}$.", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "+0 # Consider the projection matrix P that satisfies 0 169 1 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}.$$ Use the picture below to find 3 vectors with integer components and magnitude no greater than 8 that satisfy Find $$\\mathbf{P}^{-1} \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$$$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}.$$ Aug 6, 2019", "a_{33} & a_{34}\\\\ a_{41} & a_{42} & a_{43} & a_{44} \\\\ \\end{pmatrix}$, $b=\\begin{pmatrix} \\frac{b_1}{a_{11}} \\\\ b_2 \\\\ b_3 \\\\ b_4 \\end{pmatrix}$ • We have done, 4 divisions on the $4 \\times 4$ matrix • Therefore, in case of d $n \\times n$ matrix, we have $n$ division (floating point operations). • Similarly for computing the RHS, we have one division • Therefore we have $n+1$ operations #### Gauss Elimination $A=\\begin{pmatrix} 1 & \\frac{a_{12}}{a_{11}} & \\frac{a_{13}}{a_{11}} & \\frac{a_{14}}{a_{11}}\\\\ 0 & a_{22}-\\frac{a_{12}a_{21}}{a_{11}} & a_{23}-\\frac{a_{13}a_{21}}{a_{11}} & a_{24}-\\frac{a_{14}a_{21}}{a_{11}}\\\\ 0 & a_{32}-\\frac{a_{12}a_{31}}{a_{11}} & a_{33}-\\frac{a_{13}a_{31}}{a_{11}} & a_{34}-\\frac{a_{14}a_{31}}{a_{11}}\\\\ 0 & a_{42}-\\frac{a_{12}a_{41}}{a_{11}} & a_{43}-\\frac{a_{13}a_{41}}{a_{11}} & a_{44}-\\frac{a_{14}a_{41}}{a_{11}}\\\\ \\end{pmatrix}$, $b=\\begin{pmatrix} \\frac{b_1}{a_{11}} \\\\ b_2-\\frac{b_1a_{21}}{a_{11}} \\\\ b_3-\\frac{b_1a_{31}}{a_{11}} \\\\ b_4-\\frac{b_1a_{41}}{a_{11}} \\end{pmatrix}$ • For each $i^{th}$ row, we have to multiply by $a_{i1}$ and subtract. • Therefore, we require 4 operations for each row to multiply and four operations to subtract. Hence, $2\\times 4$ operations per row is required. • For each row of the RHS, we require 2 operations. • We should perform above operations for each of the 3 rows. So, finally, we require $3 \\times 2 \\times 4 + 3 \\times 2$ operations for matrix. • In terms of $n \\times n$ matrix, we require $(n- 1) \\times 2$ operations per row for RHS and $(n-1) \\times 2 \\times n$ operations for LHS • Therefore, we require, $(n-1) \\times", "$dy = a\\,dx$ $\\frac{dy}{dx} = a$ $dy = \\mathbf{a}\\,d\\mathbf{x}$ $\\frac{dy}{d\\mathbf{x}} = \\mathbf{a}$ $dy = {\\rm tr}(\\mathbf{A}\\,d\\mathbf{X})$ $\\frac{dy}{d\\mathbf{X}} = \\mathbf{A}$ $d\\mathbf{y} = \\mathbf{a}\\,dx$ $\\frac{d\\mathbf{y}}{dx} = \\mathbf{a}$ $d\\mathbf{y} = \\mathbf{A}\\,d\\mathbf{x}$ $\\frac{d\\mathbf{y}}{d\\mathbf{x}} = \\mathbf{A}$ $d\\mathbf{Y} = \\mathbf{A}\\,dx$ $\\frac{d\\mathbf{Y}}{dx} = \\mathbf{A}$ 4. ^ a b c Here, $\\mathbf{0}$ refers to a column vector of all 0's, of size n, where n is the length of x. 5. ^ a b Here, $\\mathbf{0}$ refers to a matrix of all 0's, of the same shape as X. 6. Petersen, Kaare Brandt and Michael Syskind Pedersen. The Matrix Cookbook. November 14, 2008. http://matrixcookbook.com. [3] This book uses a mixed layout, i.e. by Y in $\\frac{\\partial \\mathbf{Y}}{\\partial x},$ by X in $\\frac{\\partial y}{\\partial \\mathbf{X}}.$ $\\frac{d\\,\\ln au}{dx} = \\frac{1}{au}\\frac{d(au)}{dx} = \\frac{1}{au}a\\frac{du}{dx} = \\frac{1}{u}\\frac{du}{dx} = \\frac{d\\,\\ln u}{dx}.$", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$" ]

[ "that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is the projection matrix $P=\\frac{1}{\\mathbf{a}^T\\mathbf{a}}\\mathbf{a}\\mathbf{a}^T,$ which is an $n\\times n$ matrix. To this end, Prof. Strang writes the projection vector as $\\mathbf{p}=\\mathbf{a}\\,x$. Now, you should see $\\mathbf{p}$ and $\\mathbf{a}$ as $n\\times1$ matrices and $x$ as a one dimensional vector or a $1\\times1$ matrix, so that $\\mathbf{a}$ and $x$ can be multiplied. Thus, we have $x\\,\\mathbf{a}=\\mathbf{a}\\,x.$ This identity does not mean that $\\mathbf{a}$ and $x$ are two matrices that commute. Just read the left and right sides as stated above. It is true that the notation may be a bit misleading. Perhaps it would be better to write $(x)$ instead of $x$ if $x$ should be seen as a $1\\times 1$ matrix (parentheses are used to delimit matrices). For example, $4\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix} =\\begin{pmatrix} 4 \\\\ 8 \\\\ 12\\end{pmatrix} =\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix}(4).$ The advantage of expressing $\\mathbf{p}$ as $\\mathbf{a}x$ is that generalization is then possible. If $S$ is spanned by the linearly independent vectors $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, the orthogonal projection $\\mathbf{p}$ of a vector $\\mathbf{b}$ onto $S$ should be a linear combination of $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, that is, $\\mathbf{p}=x_1\\mathbf{a}_1+\\cdots+x_k\\mathbf{a}_k=AX$ with $A=\\left(\\mathbf{a}_1 \\vert\\ldots \\vert\\mathbf{a}_k\\right)$ and $X=\\begin{pmatrix}x_1 \\\\ \\vdots \\\\ x_k\\end{pmatrix}$ The reasoning in Prof. Strang’s lecture would then show that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is now the projection matrix $P=A(A^TA)^{-1}A^T.$ Returning to your question, you", "the projection matrix $\\mathbf{P}_{W_T}$ projects onto a 24-dimensional subspace since we have removed one dimension by subtracting out $\\mathrm{C}(\\mathbf{1}_N)$. The same goes for $\\mathbf{P}_{W_J}$ (projecting onto a three-dimensional subspace), and $\\mathbf{P}_{W_Q}$ (projecting onto a four-dimensional subspace), and $\\mathbf{P}_{W{JQ}}$ (projecting onto a 12-dimensional subspace). That means the residual projection matrix $\\mathbf{P}_\\perp$ is projecting into a $N-44=430-44=386$-dimensional subspace. All of this means that for our model: $E\\|\\mathbf{P}_0\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_0\\boldsymbol{\\tau}\\|^2 + \\sigma^2$ $E\\|\\mathbf{P}_{W_T}\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_{W_T}\\boldsymbol{\\tau}\\|^2 + 24\\sigma^2$ $E\\|\\mathbf{P}_{W_J}\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_{W_J}\\boldsymbol{\\tau}\\|^2 + 3\\sigma^2$ $E\\|\\mathbf{P}_{W_Q}\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_{W_Q}\\boldsymbol{\\tau}|^2 + 4\\sigma^2$ $E\\|\\mathbf{P}_{W_{JQ}}\\mathbf{Y}\\|^2 = \\|\\mathbf{P}_{W_{JQ}}\\boldsymbol{\\tau}\\|^2 + 12\\sigma^2$ $E\\|\\mathbf{P}_\\perp\\mathbf{Y}\\|^2 = (N-44)\\sigma^2$ where the last one comes from the fact that $\\boldsymbol{\\tau}$ is orthogonal to the residual subspace. The left-hand side of each of these is just an expected sum of squared random values. On the right hand side, we have two terms: the first due to the deterministic effects of the levels in a factor, and the second due to iid noise in the measurements. If there is no effect at a factor, then its deterministic component will be zero. In addition, if there are no differences between the effects in a factor, then the projections will be zero. Hence, to test for significant differences between the effects in a factor, all we need to do is compare the empirical sum of squares of the projections of", "a_{33} & a_{34}\\\\ a_{41} & a_{42} & a_{43} & a_{44} \\\\ \\end{pmatrix}$, $b=\\begin{pmatrix} \\frac{b_1}{a_{11}} \\\\ b_2 \\\\ b_3 \\\\ b_4 \\end{pmatrix}$ • We have done, 4 divisions on the $4 \\times 4$ matrix • Therefore, in case of d $n \\times n$ matrix, we have $n$ division (floating point operations). • Similarly for computing the RHS, we have one division • Therefore we have $n+1$ operations #### Gauss Elimination $A=\\begin{pmatrix} 1 & \\frac{a_{12}}{a_{11}} & \\frac{a_{13}}{a_{11}} & \\frac{a_{14}}{a_{11}}\\\\ 0 & a_{22}-\\frac{a_{12}a_{21}}{a_{11}} & a_{23}-\\frac{a_{13}a_{21}}{a_{11}} & a_{24}-\\frac{a_{14}a_{21}}{a_{11}}\\\\ 0 & a_{32}-\\frac{a_{12}a_{31}}{a_{11}} & a_{33}-\\frac{a_{13}a_{31}}{a_{11}} & a_{34}-\\frac{a_{14}a_{31}}{a_{11}}\\\\ 0 & a_{42}-\\frac{a_{12}a_{41}}{a_{11}} & a_{43}-\\frac{a_{13}a_{41}}{a_{11}} & a_{44}-\\frac{a_{14}a_{41}}{a_{11}}\\\\ \\end{pmatrix}$, $b=\\begin{pmatrix} \\frac{b_1}{a_{11}} \\\\ b_2-\\frac{b_1a_{21}}{a_{11}} \\\\ b_3-\\frac{b_1a_{31}}{a_{11}} \\\\ b_4-\\frac{b_1a_{41}}{a_{11}} \\end{pmatrix}$ • For each $i^{th}$ row, we have to multiply by $a_{i1}$ and subtract. • Therefore, we require 4 operations for each row to multiply and four operations to subtract. Hence, $2\\times 4$ operations per row is required. • For each row of the RHS, we require 2 operations. • We should perform above operations for each of the 3 rows. So, finally, we require $3 \\times 2 \\times 4 + 3 \\times 2$ operations for matrix. • In terms of $n \\times n$ matrix, we require $(n- 1) \\times 2$ operations per row for RHS and $(n-1) \\times 2 \\times n$ operations for LHS • Therefore, we require, $(n-1) \\times", "e_{21} & e_{22} & e_{23} \\\\ e_{31} & e_{32} & e_{33} \\end{pmatrix}$ the constraint can also be rewritten as $y'_1 y_1 e_{11} + y'_1 y_2 e_{12} + y'_1 e_{13} + y'_2 y_1 e_{21} + y'_2 y_2 e_{22} + y'_2 e_{23} + y_1 e_{31} + y_2 e_{32} + e_{33} = 0 \\,$ or $\\mathbf{e} \\cdot \\tilde{\\mathbf{y}} = 0$ where $\\tilde{\\mathbf{y}} = \\begin{pmatrix} y'_1 y_1 \\\\ y'_1 y_2 \\\\ y'_1 \\\\ y'_2 y_1 \\\\ y'_2 y_2 \\\\ y'_2 \\\\ y_1 \\\\ y_2 \\\\ 1 \\end{pmatrix}$ and $\\mathbf{e} = \\begin{pmatrix} e_{11} \\\\ e_{12} \\\\ e_{13} \\\\ e_{21} \\\\ e_{22} \\\\ e_{23} \\\\ e_{31} \\\\ e_{32} \\\\ e_{33} \\end{pmatrix}$ that is, $\\mathbf{e}$ represents the essential matrix in the form of a 9-dimensional vector and this vector must be orthogonal to the vector $\\tilde{\\mathbf{y}}$ which can be seen as a vector representation of the $3 \\times 3$ matrix $\\mathbf{y}' \\, \\mathbf{y}^{T}$. Each pair of corresponding image points produces a vector $\\tilde{\\mathbf{y}}$. Given a set of 3D points $\\mathbf{P}_k$ this corresponds to a set of vectors $\\tilde{\\mathbf{y}}_{k}$ and all of them must satisfy $\\mathbf{e} \\cdot \\tilde{\\mathbf{y}}_{k} = 0$ for the vector $\\mathbf{e}$. Given sufficiently many (at least eight) linearly independent vectors $\\tilde{\\mathbf{y}}_{k}$ it is possible to determine $\\mathbf{e}$ in a straightforward way. Collect all vectors $\\tilde{\\mathbf{y}}_{k}$ as the columns of a matrix $\\mathbf{Y}$ and", "{1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{41}$$$$\\begin {pmatrix} 3 & 5\\\\ -7 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 × 1}{41} & \\frac {5 × 24}{41}\\\\ \\frac {-7 × 1}{41} & \\frac {2 × 24}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {120}{41}\\\\ \\frac {-7}{41} & \\frac {48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 + 120}{41}\\\\ \\frac {-7 + 48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {123}{41}\\\\ \\frac {41}{41}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {vmatrix} 3\\\\ 1\\\\ \\end {vmatrix}$$ ∴ x = 3 and y = 1 Ans Here, x - 8y = 39..........................................(1) 2x - 3y = 13.......................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 1", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "# What is the projection of < 6 , -6 ,3 > onto < 4, -5, 1>? Mar 14, 2016 =$< \\frac{228}{42} , \\frac{- 285}{42} , \\frac{57}{42} >$ #### Explanation: Projection of a vector a on another vector b (component of a along b )is given by $P r o {j}_{b} a$ = $\\frac{\\ast a \\ast . \\ast b \\ast}{|} | b | {|}^{2} \\ast b \\ast$ Hence if a is <6,-6,3> and b is < 4,-5,1> a . b = 24 +30 +3 =57 $| | b | {|}^{2} = {4}^{2} + {\\left(- 5\\right)}^{2} + {1}^{2}$=42 Hence,$P r o {j}_{b} a = \\frac{57}{42} < 4 , - 5 , 1 >$ =$< \\frac{228}{42} , \\frac{- 285}{42} , \\frac{57}{42} >$", "since you had $vector^T * Matrix$, though I suppose it's computationally the same as the paper's answer with reversed notation $Matrix * vector$. Aug 25 '17 at 5:32 • It is because $a^T b = b^T a$. Aug 25 '17 at 5:34 If you write $\\mathbf C(\\mathbf x)$ as a vector $$\\mathbf C(\\mathbf x)=\\begin{pmatrix}C_1(\\mathbf x)\\\\\\vdots\\\\C_p(\\mathbf x)\\end{pmatrix}$$ the energy is given by $E_{C}=\\frac k2\\sum_{j=1}^pC_j(\\mathbf x)^2$. From there, this is pretty straightforward. $$f_i=-\\frac{\\partial E_C}{\\partial x_i}=-k\\sum_{j=1}^p\\frac{\\partial C_j}{\\partial x_i}C_j.$$", "+0 # help 0 114 1 Let $\\mathbf{P}$ be the matrix that projects onto $\\mathbf{j}$: that is, we want $\\mathbf{P}$ to satisfy $\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$for all vectors $\\mathbf{v}$. Use the pictures below to calculate $\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$in that order, and enter them in as columns below. Then calculate the matrix $\\mathbf{P}$ that projects onto $\\mathbf{j}$. Jul 31, 2019 The matrix is $$\\begin{pmatrix} -1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$.", "= \\frac{\\mathbf{y}^Tc\\mathbf{u}}{c\\mathbf{u}^Tc\\mathbf{u}}c\\mathbf{u} = \\frac{\\mathbf{y}^T\\mathbf{u}}{\\mathbf{u}^T\\mathbf{u}}\\mathbf{u}.$ Thus, the projection of $$\\mathbf{y}$$ is determined by the subspace $$L$$ that is spanned by $$\\mathbf{u}$$ – in other words, the line through $$\\mathbf{u}$$ and the origin. Hence sometimes $$\\hat{\\mathbf{y}}$$ is denoted by $$\\mbox{proj}_L \\mathbf{y}$$ and is called the orthogonal projection of $$\\mathbf{y}$$ onto $$L$$. Specifically: $\\hat{\\mathbf{y}} = \\mbox{proj}_L \\mathbf{y} = \\frac{\\mathbf{y}^T\\mathbf{u}}{\\mathbf{u}^T\\mathbf{u}}\\mathbf{u}$ Example. Let $$\\mathbf{y} = \\begin{bmatrix}7\\\\6\\end{bmatrix}$$ and $$\\mathbf{u} = \\begin{bmatrix}4\\\\2\\end{bmatrix}.$$ Find the orthogonal projection of $$\\mathbf{y}$$ onto $$\\mathbf{u}.$$ Then write $$\\mathbf{y}$$ as the sum of two orthogonal vectors, one in Span$$\\{\\mathbf{u}\\}$$, and one orthogonal to $$\\mathbf{u}.$$ Solution. Compute $\\begin{split}\\mathbf{y}^T\\mathbf{u} = \\begin{bmatrix}7&6\\end{bmatrix}\\begin{bmatrix}4\\\\2\\end{bmatrix} = 40\\end{split}$ $\\begin{split}\\mathbf{u}^T\\mathbf{u} = \\begin{bmatrix}4&2\\end{bmatrix}\\begin{bmatrix}4\\\\2\\end{bmatrix} = 20\\end{split}$ The orthogonal projection of $$\\mathbf{y}$$ onto $$\\mathbf{u}$$ is $\\hat{\\mathbf{y}} = \\frac{\\mathbf{y}^T\\mathbf{u}}{\\mathbf{u}^T\\mathbf{u}} \\mathbf{u}$ $\\begin{split}=\\frac{40}{20}\\mathbf{u} = 2\\begin{bmatrix}4\\\\2\\end{bmatrix} = \\begin{bmatrix}8\\\\4\\end{bmatrix}\\end{split}$ And the component of $$\\mathbf{y}$$ orthogonal to $$\\mathbf{u}$$ is $\\begin{split}\\mathbf{y}-\\hat{\\mathbf{y}} = \\begin{bmatrix}7\\\\6\\end{bmatrix} - \\begin{bmatrix}8\\\\4\\end{bmatrix} = \\begin{bmatrix}-1\\\\2\\end{bmatrix}.\\end{split}$ So $\\mathbf{y} = \\hat{\\mathbf{y}} + \\mathbf{z}$ $\\begin{split}\\begin{bmatrix}7\\\\6\\end{bmatrix} = \\begin{bmatrix}8\\\\4\\end{bmatrix} + \\begin{bmatrix}-1\\\\2\\end{bmatrix}.\\end{split}$ ax = ut.plotSetup(-3,11,-1,7,(8,6)) ut.centerAxes(ax) plt.axis('equal') u = np.array([4.,2]) y = np.array([7.,6]) yhat = (y.T.dot(u)/u.T.dot(u))*u z = y-yhat ut.plotLinEqn(1.,-2.,0.) ut.plotVec(ax,u) ut.plotVec(ax,z) ut.plotVec(ax,y) ut.plotVec(ax,yhat) ax.text(u[0]+0.3,u[1]-0.5,r'$\\mathbf{u}$',size=20) ax.text(yhat[0]+0.3,yhat[1]-0.5,r'$\\mathbf{\\hat{y}}$',size=20) ax.text(y[0],y[1]+0.8,r'$\\mathbf{y}$',size=20) ax.text(z[0]-2,z[1],r'$\\mathbf{y - \\hat{y}}$',size=20) ax.text(10,4.5,r'$L =$Span$\\{\\mathbf{u}\\}$',size=20) perpline1, perpline2 = ut.perp_sym(yhat, y, np.array([0,0]), 0.75) plt.plot(perpline1[0], perpline1[1], 'k', lw = 1) plt.plot(perpline2[0], perpline2[1], 'k', lw = 1) ax.plot([y[0],yhat[0]],[y[1],yhat[1]],'b--') ax.plot([0,y[0]],[0,y[1]],'b-') ax.plot([0,z[0]],[0,z[1]],'b-'); Question Time! Q21.2 ax = ut.plotSetup(-3,11,-1,7,(8,6)) ut.centerAxes(ax) plt.axis('equal') u = np.array([4.,2]) y = np.array([7.,6]) yhat = (y.T.dot(u)/u.T.dot(u))*u z =", "now we can figure out $$\\mathtt{V}$$ and $$\\mathtt{\\Sigma}$$ by doing the eigenvalue decomposition like we did previously. Here it is: $\\begin{bmatrix}\\mathtt{4}&\\mathtt{6}\\\\\\mathtt{6}&\\mathtt{10}\\end{bmatrix}=\\begin{bmatrix}\\mathtt{\\color{purple}{-\\frac{521}{991}}}&\\mathtt{\\color{red}{-\\frac{3725}{4379}}}\\\\\\mathtt{\\color{purple}{-\\frac{3725}{4379}}}&\\mathtt{\\,\\,\\,\\,\\color{red}{\\frac{521}{991}}}\\end{bmatrix}\\begin{bmatrix}\\mathtt{\\frac{4510}{329}}&\\mathtt{0}\\\\\\mathtt{0}&\\mathtt{\\frac{658}{2255}}\\end{bmatrix}\\begin{bmatrix}\\mathtt{-\\frac{521}{991}}&\\mathtt{-\\frac{3725}{4379}}\\\\\\mathtt{-\\frac{3725}{4379}}&\\mathtt{\\,\\,\\,\\,\\frac{521}{991}}\\end{bmatrix}$ The purple and red vectors are our starting vectors from the animation on the left above: (–0.53, –0.85) and (–0.85, 0.53). When we ask what orthogonal starting vectors can we pick such that matrix $$\\mathtt{A}$$ will transform them to a different pair of orthogonal vectors, matrix $$\\mathtt{V}$$—the red and purple vectors above—is our answer. The square roots of the diagonal values of $$\\mathtt{\\Sigma}$$ above (remember, this matrix was squared) are called the singular values of $$\\mathtt{A}$$. To me, getting $$\\mathtt{V}$$ is the coolest part of this decomposition. The rest, below, is just gravy. And Now for $$\\mathtt{U}$$ and $$\\mathtt{\\Sigma}$$ In this case, we multiply both sides of the equation $$\\mathtt{A=U\\Sigma V^T}$$ on the right by the transpose of $$\\mathtt{A}$$ and get an eigenvalue decomposition of $$\\mathtt{AA^T}$$. $\\,\\,\\mathtt{AA^T=(U\\Sigma V^T)(U\\Sigma V^T)^T=U\\Sigma V^{T}V\\Sigma^{T}U^T=U\\Sigma^{2}U^T}$ $\\begin{bmatrix}\\mathtt{\\,\\,\\,\\,1}&\\mathtt{-3}\\\\\\mathtt{-3}&\\mathtt{\\,\\,\\,\\,13}\\end{bmatrix}=\\begin{bmatrix}\\mathtt{-\\frac{400}{1741}}&\\mathtt{\\frac{764}{785}}\\\\\\mathtt{\\,\\,\\,\\,\\frac{764}{785}}&\\mathtt{\\frac{400}{1741}}\\end{bmatrix}\\begin{bmatrix}\\mathtt{\\color{purple}{\\frac{4510}{329}}}&\\mathtt{\\color{red}{0}}\\\\\\mathtt{\\color{purple}{0}}&\\mathtt{\\color{red}{\\frac{658}{2255}}}\\end{bmatrix}\\begin{bmatrix}\\mathtt{-\\frac{400}{1741}}&\\mathtt{\\frac{764}{785}}\\\\\\mathtt{\\,\\,\\,\\,\\frac{764}{785}}&\\mathtt{\\frac{400}{1741}}\\end{bmatrix}$ Now the square roots of the purple and red vectors are our starting vectors from the animation on the right above: (3.7, 0) and (0, 0.54). You’ll notice, of course, that $$\\mathtt{\\Sigma^2}$$ is the same as above, so we really found it earlier. This step is only to pin down what $$\\mathtt{U}$$ is. At any rate, we are done, and we can write down the full", "\\end{matrix}$$ b) P*b c) Obviously the eigenvalues are 1,1 and 0. And the eigenvectors are easly calculated without touching the pen it's the standard (canonical) base for R3. I don't think the point of the problem is that you can claim part b is stated relative to your chosen basis, nor part c. To get part b you would have to express vector $$\\vec b$$, assumed relative to the standard basis for $$\\mathbb{R}^3$$ in terms of your basis, project with your $$P$$, then express the result in terms of the original basis. Thus if $$V=\\begin{bmatrix}1&1&-2\\\\0&1&2\\\\2&0&1\\end{bmatrix}$$ Is the matrix whose columns are your basis vectors, then you need to compute \\begin{align}VPV^{-1}\\vec b&=\\begin{bmatrix}1&1&-2\\\\0&1&2\\\\2&0&1\\end{bmatrix}\\begin{bmatrix}1&0&0\\\\0&1&0\\\\0&0&0\\end{bmatrix}\\begin{bmatrix}\\frac19&-\\frac19&\\frac49\\\\\\frac49&\\frac59&-\\frac29\\\\-\\frac29&\\frac29&\\frac19\\end{bmatrix}\\vec b\\\\ &=\\begin{bmatrix}\\frac59&\\frac49&\\frac29\\\\\\frac49&\\frac59&-\\frac29\\\\\\frac29&-\\frac29&\\frac89\\end{bmatrix}\\begin{bmatrix}-4\\\\3\\\\4\\end{bmatrix}=\\begin{bmatrix}0\\\\-1\\\\2\\end{bmatrix}\\end{align} to get the projection in part b. Note that as an intermediate result in our calculation we found $$P$$ relative to the standard basis for $$\\mathbb{R}^3$$. The columns of $$V$$ above are eigenvectors of $$P$$ with eigenvalues $$1$$, $$1$$, and $$0$$ relative to the standard basis for $$\\mathbb{R}^3$$ by construction. • What is your [numeric] answer to part b? If you said $$\\begin{bmatrix}-4\\\\3\\\\0\\end{bmatrix}$$ the answer clearly wouldn't be right in either basis. – user5713492 Jun 12 at 19:00 • I suppose the solution in your cross comment would be OK, with an explanation that the answer is relative to your new basis. That is pretty", "matrix so that each vector is linearly independent from the other ones. I got this: $\\begin{pmatrix} {\\frac{12}{13}}&{\\frac{12}{13}}&{0}\\\\ {0}&{\\frac{5}{13}}&{\\frac{12}{13}}\\\\ {\\frac{5}{13}}&{0}&{\\frac{5}{13}} \\end{pmatrix}=\\frac{1}{13}\\begin{pmatrix} {12}&{12}&{0}\\\\ {0}&{5}&{12}\\\\ {5}&{0}&{5} \\end{pmatrix}$ ? $\\begin{bmatrix}12/13\\\\5/13\\\\0\\end{bmatrix}$ is fine for the middle column, but the other two columns are not orthogonal to it. You could for example use $\\begin{bmatrix}-5/13\\\\12/13\\\\0\\end{bmatrix}$ for one of them. Can you see a vector (hint: without any fractions in its entries) that is orthogonal to both of these? Just to emphasise the point: linear independence is not enough, the columns must be orthogonal to each other. • April 10th 2009, 03:45 AM Showcase_22 Quote: Originally Posted by Opalg Can you see a vector (hint: without any fractions in its entries) that is orthogonal to both of these? Ah, it's just $\\begin{pmatrix} {0}\\\\ {0}\\\\ {1} \\end{pmatrix}$ since those two vectors don't have a k component. So ultimately we have: $\\begin{pmatrix} {0}&{\\frac{12}{13}}&{-\\frac{5}{13}}\\\\ {0}&{\\frac{5}{13}}&{\\frac{12}{13}}\\\\ {1}&{0}&{0} \\end{pmatrix} $ Can you also have: $\\begin{pmatrix} {0}&{\\frac{12}{13}}&{\\frac{5}{13}}\\\\ {0}&{\\frac{5}{13}}&{\\frac{12}{13}}\\\\ {1}&{0}&{0} \\end{pmatrix} $ or $\\begin{pmatrix} {0}&{-\\frac{12}{13}}&{-\\frac{5}{13}}\\\\ {0}&{\\frac{5}{13}}&{\\frac{12}{13}}\\\\ {1}&{0}&{0} \\end{pmatrix} $ ? (ie. any situation in which the magnitude of the rows and columns is 1?) • April 10th 2009, 04:04 AM Opalg Quote: Originally Posted by Showcase_22 Ah, it's just $\\begin{pmatrix} {0}\\\\ {0}\\\\ {1} \\end{pmatrix}$ since those two vectors don't have a k component. So ultimately we have: $\\begin{pmatrix} {0}&{\\frac{12}{13}}&{-\\frac{5}{13}}\\\\ {0}&{\\frac{5}{13}}&{\\frac{12}{13}}\\\\ {1}&{0}&{0} \\end{pmatrix} $", "-3 & 5 & -2 \\\\ 0 & 2 & 0 \\\\ 1 & 3 & 4 \\end{array}\\right]$$ The projection matrix onto $$W$$ is $$P=X(X^\\top X)^{-1}X^\\top$$ In our case, we end up with $$P=\\left[\\begin{array}{rrrr} \\frac{131}{231} & -\\frac{10}{77} & \\frac{10}{21} & \\frac{10}{231} \\\\ -\\frac{10}{77} & \\frac{74}{77} & \\frac{1}{7} & \\frac{1}{77} \\\\ \\frac{10}{21} & \\frac{1}{7} & \\frac{10}{21} & -\\frac{1}{21} \\\\ \\frac{10}{231} & \\frac{1}{77} & -\\frac{1}{21} & \\frac{230}{231} \\end{array}\\right]$$ We can now project any vector $$v\\in\\Bbb R^4$$ onto $$W$$ by computing $$Pv$$. If you're dead set on using orthonormal bases, then you could take these three basis vectors and apply the Gram-Schmidt algorithm. This gives the new basis \\begin{align*} q_1 &= \\left\\langle\\frac{1}{\\sqrt{11}},\\,-\\frac{3}{\\sqrt{11}},\\,0,\\,\\frac{1}{\\sqrt{11}}\\right\\rangle & q_2 &= \\left\\langle\\frac{1}{\\sqrt{7}},\\,\\frac{1}{\\sqrt{7}},\\,\\frac{1}{\\sqrt{7}},\\,\\frac{2}{\\sqrt{7}}\\right\\rangle & q_3 &= \\left\\langle-\\frac{1}{\\sqrt{3}},\\,0,\\,-\\frac{1}{\\sqrt{3}},\\,\\frac{1}{\\sqrt{3}}\\right\\rangle \\end{align*} We can then project any $$v\\in\\Bbb R^4$$ with the formula $$Pv=\\langle q_1, v\\rangle q_1 + \\langle q_2, v\\rangle q_2 + \\langle q_3, v\\rangle q_3$$ • @Leif The sum of the projections of $p$ onto $W$ and $W_\\perp$ will just be $p$, as I mentioned. Why do you want to add them? The question only asks for the projections. – Shubham Johri May 17 at 19:01", "1&-1 \\\\ 1&0 \\\\ 1&2 \\end{bmatrix} \\frac{1}{35} \\begin{bmatrix} 9&1 \\\\ 1&4 \\end{bmatrix} \\begin{bmatrix} 1&1&1&1 \\\\ -2&-1&0&2 \\end{bmatrix}$ $= \\frac{1}{35} \\begin{bmatrix} 1&-2 \\\\ 1&-1 \\\\ 1&0 \\\\ 1&2 \\end{bmatrix} \\begin{bmatrix} 7&8&9&11 \\\\ -7&-3&1&9 \\end{bmatrix}$ $= \\frac{1}{35} \\begin{bmatrix} 21&14&7&-7 \\\\ 14&11&8&2 \\\\ 7&8&9&11 \\\\ -7&2&11&29 \\end{bmatrix}$ The projection of the vector $b = (4, 3, 1, 0)$ onto the column space of $A$ is then $Pb = \\frac{1}{35} \\begin{bmatrix} 21&14&7&-7 \\\\ 14&11&8&2 \\\\ 7&8&9&11 \\\\ -7&2&11&29 \\end{bmatrix} \\begin{bmatrix} 4 \\\\ 3 \\\\ 1 \\\\ 0 \\end{bmatrix}$ $= \\frac{1}{35} \\begin{bmatrix} 133 \\\\ 97 \\\\ 61 \\\\ -11 \\end{bmatrix} = \\begin{bmatrix} \\frac{133}{35} \\\\ \\frac{97}{35} \\\\ \\frac{61}{35} \\\\ -\\frac{11}{35} \\end{bmatrix}$ The vector $Pb$ corresponds to the points on the least squares line of best fit $\\bar{C} + \\bar{D}t$ for the times $t = (-2, 1, 0, 2)$: $\\frac{61}{35} - \\frac{36}{35} (-2) = \\frac{133}{35}$ $\\frac{61}{35} - \\frac{36}{35} (-1) = \\frac{97}{35}$ $\\frac{61}{35} - \\frac{36}{35} (0) = \\frac{61}{35}$ $\\frac{61}{35} - \\frac{36}{35} (2) = \\frac{-11}{35}$ NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang. If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the", "1&-1 \\\\ 1&0 \\\\ 1&2 \\end{bmatrix} \\frac{1}{35} \\begin{bmatrix} 9&1 \\\\ 1&4 \\end{bmatrix} \\begin{bmatrix} 1&1&1&1 \\\\ -2&-1&0&2 \\end{bmatrix}$ $= \\frac{1}{35} \\begin{bmatrix} 1&-2 \\\\ 1&-1 \\\\ 1&0 \\\\ 1&2 \\end{bmatrix} \\begin{bmatrix} 7&8&9&11 \\\\ -7&-3&1&9 \\end{bmatrix}$ $= \\frac{1}{35} \\begin{bmatrix} 21&14&7&-7 \\\\ 14&11&8&2 \\\\ 7&8&9&11 \\\\ -7&2&11&29 \\end{bmatrix}$ The projection of the vector $b = (4, 3, 1, 0)$ onto the column space of $A$ is then $Pb = \\frac{1}{35} \\begin{bmatrix} 21&14&7&-7 \\\\ 14&11&8&2 \\\\ 7&8&9&11 \\\\ -7&2&11&29 \\end{bmatrix} \\begin{bmatrix} 4 \\\\ 3 \\\\ 1 \\\\ 0 \\end{bmatrix}$ $= \\frac{1}{35} \\begin{bmatrix} 133 \\\\ 97 \\\\ 61 \\\\ -11 \\end{bmatrix} = \\begin{bmatrix} \\frac{133}{35} \\\\ \\frac{97}{35} \\\\ \\frac{61}{35} \\\\ -\\frac{11}{35} \\end{bmatrix}$ The vector $Pb$ corresponds to the points on the least squares line of best fit $\\bar{C} + \\bar{D}t$ for the times $t = (-2, 1, 0, 2)$: $\\frac{61}{35} - \\frac{36}{35} (-2) = \\frac{133}{35}$ $\\frac{61}{35} - \\frac{36}{35} (-1) = \\frac{97}{35}$ $\\frac{61}{35} - \\frac{36}{35} (0) = \\frac{61}{35}$ $\\frac{61}{35} - \\frac{36}{35} (2) = \\frac{-11}{35}$ NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang. If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the", "v_2)^2$ is minimized. W.l.o.g. we can take $\\mathbf\\pi_1$ to be the $u$-$v$ plane, so that its camera matrix is the canonical $$P_1=\\begin{bmatrix}1&0&0&0\\\\0&1&0&0\\\\0&0&0&1\\end{bmatrix}.$$ This just picks out the $u$- and $v$-coordinates of $\\mathbf U$. The projection matrix onto $\\mathbf\\pi_2$ will have the form $$P_2=\\left[\\begin{array}{c|c}M_{2\\times 3}&\\mathbf t\\\\\\mathbf 0^T&1\\end{array}\\right].$$ The point $\\mathbf U$ then satisfies the linear equations $P_1\\mathbf U =\\mathbf u_1=(u_1,v_1)$ and $P_2\\mathbf U = \\mathbf u_2=(u_2,v_2)$. The affine fundamental matrix of this epipolar configuration is $$F_A=\\begin{bmatrix}0&0&a\\\\0&0&b\\\\c&d&e\\end{bmatrix}$$ where $$a = m_{23} \\\\ b=-m_{13} \\\\ c=m_{13}m_{21}-m_{11}m_{23} \\\\ d=m_{13}m_{22}-m_{12}m_{23} \\\\ e=m_{13}t_2-m_{23}t_1.$$ (In your case, where $P_2$ has no translation component, we would have $e=0$.) The adjusted coordinates are then given by the formula $$\\begin{bmatrix}\\hat u_2 \\\\ \\hat v_2 \\\\ \\hat u_1 \\\\ \\hat v_1\\end{bmatrix} = \\begin{bmatrix}u_2 \\\\ v_2 \\\\ u_1 \\\\ v_1\\end{bmatrix}-{au_2+bv_2+cu_1+dv_1+e\\over a^2+b^2+c^2+d^2}\\begin{bmatrix}a\\\\b\\\\c\\\\d\\end{bmatrix}.$$ (You might recognize this as the result of a standard least-squares fit, which itself is basically a projection onto a hyperplane in $\\mathbb R^4$.) With these adjusted coordinates in hand, you can then find the intersection of their back-projections in the ways described previously. • Thank you for taking the time to explain this, however, I do not understand what you mean when you say that U will lie on the intersection of those two lines (I assume lambda and mu are just arbitrary variables you chose without any", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "# Finding a set of vectors that spans the solution Find a set of vectors {u, v} in $\\mathbb{R}^4$ that spans the solution set of the equations: $x - y + 2z - 2w = 0$ $2x + 2y -z + 3w = 0$ ($u$ and $v$ are both $4 \\times 1$) $u = ?$, $v = ?$ I put the matrix in RREF to get $\\begin{bmatrix}1&0&3/4&-1/4\\\\0&1&-5/4&7/4\\end{bmatrix} = \\begin{bmatrix}0\\\\0\\end{bmatrix}$ Then I got $x = -\\frac{3}{4} z + \\frac{1}{4} w$ and $y = \\frac{5}{4} z - \\frac{7}{4} w$ But I'm not sure how to present the answer as they want it. • HINT: (Assuming your numbers are correct) you've found that $$\\pmatrix{x \\\\ y \\\\ z \\\\ w} = \\pmatrix{-\\frac 34z+\\frac 14w \\\\ \\frac 54z-\\frac 74w \\\\ z \\\\ w} = z\\pmatrix{-\\frac 34 \\\\ \\frac 54 \\\\ 1 \\\\ 0} + w\\pmatrix{\\frac 14 \\\\ -\\frac 74 \\\\ 0 \\\\1}$$ – user137731 Jul 8 '16 at 3:03 You can write the solution set $S$ as follows: $$S=\\{(x,y,z,w)\\in\\mathbb{R}^4: x=-\\frac{3}{4}z+\\frac{1}{4}w,\\, y=\\frac{5}{4}z-\\frac{7}{4}w;\\,\\,\\ z,w\\in\\mathbb{R}\\}.$$", "expressed with matrices and vectors as $\\mathbf{A} \\times \\mathbf{x} = \\mathbf{b}$ for a known matrix $$\\mathbf{A}$$ and vector $$\\mathbf{b}$$ and an unknown vector $$\\mathbf{x}$$, and solved with $\\mathbf{x} = \\mathbf{A}^{-1} \\times \\mathbf{b}$ For example, we can rewrite the simultaneous equations \\begin{align*} 8x + 3y &= 2\\\\ 4x - 2y &= 8 \\end{align*} as $\\begin{pmatrix} 8 & 3\\\\ 4 & -2 \\end{pmatrix} \\times \\begin{pmatrix} x\\\\ y \\end{pmatrix} = \\begin{pmatrix} 2\\\\ 8 \\end{pmatrix}$ and can solve them with $\\begin{array}{lclcl} \\begin{pmatrix} x\\\\ y \\end{pmatrix} &=& \\begin{pmatrix} 8 & 3\\\\ 4 & -2 \\end{pmatrix}^{-1} \\times \\begin{pmatrix} 2\\\\ 8 \\end{pmatrix} &=& -\\frac{1}{28} \\begin{pmatrix} -2 & -3\\\\ -4 & 8 \\end{pmatrix} \\times \\begin{pmatrix} 2\\\\ 8 \\end{pmatrix}\\\\ &=& -\\frac{1}{28} \\begin{pmatrix} -4 - 24\\\\ -8 + 64 \\end{pmatrix} &=& -\\frac{1}{28} \\begin{pmatrix} -28\\\\ 56 \\end{pmatrix}\\\\ &=& \\begin{pmatrix} 1\\\\ -2 \\end{pmatrix} \\end{array}$ ### A Pivotal Algorithm To invert square matrices of any size we can use a variant of Gaussian elimination, which you may recall we exploited when we used polynomials to approximate the derivative of a function[2]. To do this we manipulate the rows of a matrix using a set of operations to transform it into an identity matrix. By applying exactly the same set of operations to an identity matrix we can recover the inverse. Specifically, the row operations that we shall use are 1. Swap" ]

[ "$$1 - \\sin 10 ^ \\circ - 2 \\sin^2 10^\\circ = \\sin N^\\circ$$, where $$0 \\leq N \\leq 90$$. What is $$N$$?", "Find $$\\hat{N}$$. $\\hat{N} = 83^{\\circ}$", "could use $60 + 15 = 75$ and $\\sin \\left(A + B\\right)$ to find $\\sin {75}^{\\circ}$. Although if the question had been to find $\\sin {75}^{\\circ} , I ' d p r o b a b l y u s e$30^@$\\mathmr{and}$45^@#", "+0 # help +1 84 1 Let $$z = 3[\\cos(85^{\\circ})+i\\sin(85^{\\circ})]$$. Find the smallest positive value of $$r$$ such that $$r[\\cos(45^{\\circ})+i\\sin(45^{\\circ})] = z^n$$ for some positive integer $$n$$. Dec 20, 2019", "\\pm {42.0038}^{\\circ} + {360}^{\\circ} k \\quad$integer $k$", "with $\\cos{90-nt} + i\\sin{90-nt}$. This gives us the equation: $\\cos{90n - nt} + i\\sin{90 - nt} = \\cos{90-nt} + i\\sin{90-nt}$ Equating the real parts or the imaginary parts will give the same solution set, so we will equate the real parts. So we get $90n - nt = 90 - nt$ but $90$ in the right hand side of the equation is just the principal value, but we can have any equivalent value. So our new equation is: $90n = 90 + 360x$ for $x$ being an integer. $n = 4x + 1$ we know that $n \\le 1000$ so we want all numbers that are less than or equal to $1000$ and also $\\equiv 1 (\\mbox{mod} 4)$ and there are $\\fbox{250}$ such numbers", "+0 +1 48 1 +140 Find the integer $$n$$$$0 \\le n \\le 5,$$ such that $$n \\equiv -3736 \\pmod{6}.$$ Mar 29, 2019 #1 +21977 +2 Find the integer $$n,\\ 0 \\le n \\le 5$$, such that $$n \\equiv -3736 \\pmod{6}$$. Mar 29, 2019", "Volume 23 , Issue 3 1987 Going from one place to another can be a difficult problem. Have you ever wondered how your calculator or computer finds quantities such as $\\sin(63^\\circ)$ or $\\tan(17^\\circ)$? In a recent edition of Mathematical Spectrum (the English equivalent of Parabola) the following famous problem was discussed: Q.720 When the initial digit of a whole number $x$ is deleted, the number decreases by a factor of 13. Find all possible values of $x$. Q.708 A four digit number $abcd$ has the property that $a+b = c\\times d$ and also $a\\times b = c+d$. Find all possibilities.", "# How do you find the exact value of sin690? Jul 26, 2018 $- \\frac{1}{2}$. #### Explanation: $\\sin {690}^{\\circ} = \\sin \\left({720}^{\\circ} - {30}^{\\circ}\\right)$, $= \\sin \\left(4 \\pi - \\frac{\\pi}{6}\\right)$. But, $\\sin \\left(4 \\pi - \\theta\\right) = - \\sin \\theta$. $\\therefore \\sin {690}^{\\circ} = - \\sin {30}^{\\circ} = - \\frac{1}{2}$.", "doesn't somehow involve transcendental functions Using degrees, the sine of any integer is an algebraic number, i.e. a root of a polynomial with integer coefficients. For $\\sin(1^\\circ)$, the smallest such polynomial has degree 48. It is also possible to express $\\sin(1^\\circ)$ by radicals as follows: $$\\sin(1^\\circ) \\;=\\; \\frac{(\\sqrt[180]{-1})^{89} - (\\sqrt[180]{-1})^{91}}{2}$$ where $\\sqrt[180]{-1}$ denotes the principal 180'th root of $-1$, i.e. $\\sqrt[180]{-1} = \\cos(1^\\circ) + i \\sin(1^\\circ)$. More generally, $$\\sin(k^\\circ) \\;=\\; \\frac{(\\sqrt[180]{-1})^{90-k} - (\\sqrt[180]{-1})^{90+k}}{2}$$ for any integer $k$.", "Find the least positive integer $$n$$ such that $$\\frac 1{\\sin 45^\\circ\\sin 46^\\circ}+\\frac 1{\\sin 47^\\circ\\sin 48^\\circ}+\\cdots+\\frac 1{\\sin 133^\\circ\\sin 134^\\circ}=\\frac 1{\\sin n^\\circ}.$$ (第十八届AIME2 2000 第15题)", "## Where n(n – 1) and n(n + 1) are both pandigital Find an integer $n$ such that both $n \\,(n - 1)$ and $n \\,(n + 1)$ are pandigital Antonio Roldán found n = 55152", "Here's the question you clicked on: ## sauravshakya x=(2^n + 3^n + 5^n + 7^n) / 11^n n is a natural number........ Find all the possible integer value of x one year ago one year ago • This Question is Closed 1. sauravshakya @experimentX @mukushla @Traxter 2. mukushla for n>1$2^n + 3^n + 5^n + 7^n< 11^n$and for n=1 it becomes$\\frac{17}{11}$so there is no integer value in the form$\\frac{2^n + 3^n + 5^n + 7^n}{ 11^n}$ 3. sauravshakya Well try this: (2^n +3^n +5^n + 7^n +13^n)/11^n", "mentioned above, you shouldn't say that $\\tan(90^\\circ)$ is infinite but you should really say that it's undefined (careful, though, if a curious student enters $\\tan(\\pi/2)$ into a certain free online calculator, they'll get something like $1.63317787 \\times 10^{16}$). • Using the double angle prove that $\\sin(90^\\circ)=1$ • Using the Pythagorean identity, prove $\\cos(90^\\circ)=0$. • By evaluating $\\tan(x)$ using the definition at $x=89^\\circ,\\, 89.9^\\circ,\\, 89.99^\\circ,\\dots$ and at $x=91^\\circ,\\, 90.1^\\circ,\\,90.01^\\circ,\\dots$ try to explain a little more about they sense in which $\\tan(90^\\circ)$ is undefined. Now try to explain it using the unit circle.", "n =(100*s^5) y = an integer for (-20000*(125*s'^10)^2))^(1/5); n' = (125*s^10) Will have to see where to go from there, meaning what conclusions one might come to regarding: v^5 + w^5 + x^5 + y^5 + z^5 Last edited: Feb 17, 2012", "any integer n (n-1)pi<x<n*pi=>|sin(x)|>0", "# PLEASE SOLVE AND GIVE CORRECT ANSWER Question 63 of 90 The value of is 1. p/2 2. p/3 3. $\\mathrm{\\pi }$/4 4. p • 0 What are you looking for?", "# Is $\\sin(n^k) ≠ (\\sin n)^k$ in general? Is it true that $\\sin(n^k) ≠ (\\sin n)^k$ for any positive integers $n$ and integers $k ≠ 1$? What if $n > 0, k ≠ 1$ are rational? - @Chandru: :) [ ](http://.) – kennytm Aug 13 '10 at 13:26 If $\\sin(n^k)=\\sin(n)^k$, then $\\frac{e^{in^k}-e^{-in^k}}{2i}=(\\frac{e^{in}-e^{-in}}{2i})^k$, so that $e^i$ would be algebraic. But that directly contradicts the Lindemann-Weierstrass theorem because $i$ is algebraic. - Well, by seeing this i get an interesting question to my mind. – anonymous Aug 13 '10 at 13:32", "? Free Version Moderate # Find All Critical Numbers APCALC-UZYLGB Find all of the critical numbers for $f(x)=3\\sin{x}-3x\\cos{x}$. A $x=2k\\pi$ for any integer $k$ B $x = (2k+1)\\pi$ for any integer $k$ C $x=\\cfrac {2k+1}{2}\\pi$ for any integer $k$ D $x=k\\pi$ for any integer $k$", "integer $n>4$. Now go back to the inductive proof and recast it for the guess $T(n)\\le c2^{n/2}−5n$. You'll find that everything works nicely. Finally, observe that $c2^{n/2}−5n=O(2^{n/2})$ and you'll be done. This is explained in a bit more detail in this answer to a similar question." ]

[ "Precalculus (10th Edition) $0$ I know that $\\sin$ is an odd function, which means $f(-\\theta)=-f(\\theta).$ Also, I know that $\\cos$ has a period of $360^\\circ$, hence $\\cos{\\theta}=\\cos{(\\theta-360^\\circ)}.$ Therefore $\\sin{1^\\circ}+\\sin{2^\\circ}+...+\\sin{358^\\circ}+\\sin{359^\\circ}\\\\=\\sin{1^\\circ}+\\sin{2^\\circ}+...+\\sin{(358^\\circ-360^\\circ)}+\\sin{(359^\\circ-360^\\circ)}\\\\=\\sin{1^\\circ}+\\sin{2^\\circ}+...+\\sin{-2^\\circ}+\\sin{-1^\\circ}\\\\=\\sin{1^\\circ}+\\sin{2^\\circ}+...-\\sin{2^\\circ}-\\sin{1^\\circ}=0$.", "Precalculus (10th Edition) We know that $\\sin$ has a period of $360^{ \\circ}$, hence using this first we try and find a value where the argument is between $0^{\\circ}$ and $360^{\\circ}$ and then evaluate the funcion. Therefore $\\sin(390^{\\circ})=\\sin(390^{\\circ}-360^{\\circ})=\\sin(30^{o})=\\frac{1}{2}.$ We know that $\\sin$ has a period of $360^{ \\circ}$, hence using this first we try and find a value where the argument is between $0^{\\circ}$ and $360^{\\circ}$ and then evaluate the function. Therefore $\\sin(390^{\\circ})=\\sin(390^{\\circ}-360^{\\circ})=\\sin(30^{o})=\\frac{1}{2}.$", "# What is the period of f(t)=sin( t / 36 )+ cos( (t)/16 ) ? Mar 15, 2017 $288 \\pi$ #### Explanation: Period of $\\sin \\left(\\frac{t}{36}\\right)$ --> $36 \\left(2 \\pi\\right) = 72 \\pi$ Period of $\\cos \\left(\\frac{t}{16}\\right)$ --> $16 \\left(2 \\pi\\right) = 32 \\pi$ Find least common multiple of 32 and 72. $32 - \\to {2}^{3} \\cdot 4 - \\to 32 \\cdot 9 = 288$ $72 - \\to {2}^{3} \\cdot 9 - \\to 72 \\cdot 4 = 288$ Period of f(t) --> $288 \\pi$", "# Value of sin(2∘)⋅sin(4∘)⋅sin(6∘)⋯sin(90∘)\\sin (2^\\circ)\\cdot \\sin (4^\\circ)\\cdot \\sin (6^\\circ)\\cdots \\sin (90^\\circ) How can I calculate the value of 1. $\\sin (1^\\circ)\\cdot \\sin (3^\\circ)\\cdot \\sin (5^\\circ)\\cdots \\sin (89^\\circ)$ 2. $\\sin (2^\\circ)\\cdot \\sin (4^\\circ)\\cdot \\sin (6^\\circ)\\cdots \\sin (90^\\circ)$ My solution: Let Now Which implies that So $A = 2^{-\\frac{89}{2}}$ because $B\\neq 0$. But I do not understand how can I calculate the value of $B$. Can we calculate these values using complex numbers? Putting $n=180$, it gives The proof of identity $(1)$ can be found at the end of this pdf.", "# What is the period of sin (5*x)? Jan 30, 2016 period$= {72}^{\\circ}$ #### Explanation: The general equation for a sine function is: $f \\left(x\\right) = a \\sin \\left[k \\left(x - d\\right)\\right] + c$ where: $| a | =$amplitude $| k | =$horizontal stretch/compression or ${360}^{\\circ} / \\text{period}$ $d =$phase shift $c =$vertical translation In this case, the value of $k$ is $5$. To find the period, use the formula, $k = {360}^{\\circ} / \\text{period}$: $k = {360}^{\\circ} / \\text{period}$ $5 = {360}^{\\circ} / \\text{period}$ $5 \\cdot \\text{period} = {360}^{\\circ}$ $\\text{period} = {360}^{\\circ} / 5$ $\\text{period} = {72}^{\\circ}$ $\\therefore$, the period is ${72}^{\\circ}$.", "-\\displaystyle \\frac{{\\sqrt{3}}}{2}=-3.464\\end{align*} . Because the period of the function is $\\scriptsize {{360}^\\circ}$ we know that the function value at $\\scriptsize {{360}^\\circ}$ will also be $\\scriptsize -3.464$. . 2. The function is of the form $\\scriptsize y=a\\sin (x+p)$. The amplitude is $\\scriptsize 3$. The point $\\scriptsize ({{90}^\\circ},1)$ has been transformed to $\\scriptsize ({{0}^\\circ},3)$. Therefore, the graph has been shifted $\\scriptsize {{90}^\\circ}$to the left. Hence $\\scriptsize a=3$ and $\\scriptsize p={{90}^\\circ}$ and the function is $\\scriptsize y=3\\sin (x+{{90}^\\circ})$. Back to Unit 8: Assessment", "unchanged after a rotation of $0^\\circ$. Thus, the number of grids that satisfy this property is $2^{n^2}$ For $g = 90^\\circ$, we see that a given square will form a cycle of period 4 (excluding the middle square in the case where $n$ is odd). Because all squares in the cycle must be the same color, there is $2 ^ {n^2 / 4}$ ways to do this for even $n$ and $2 ^ {\\frac {n^2 - 1} 4} \\cdot 2$ for odd $n$ (the second $\\cdot 2$ comes from the fact that the middle square can be one of two colors). For $g = 180^\\circ$, we see that a given square will form a cycle of period 2 (excluding the middle square in the case where $n$ is odd). Thus, there is $2 ^ {n^2 / 2}$ ways to do this for even $n$ and $2 ^ {\\frac {n^2 - 1} 2} \\cdot 2$ for odd $n$. For $g = 270^\\circ$, the answer is the same as $g = 90^\\circ$ (this is simply a $90^\\circ$ in the opposite direction) The final answer is simply the sum of the four values divided by 4 (or in this case, multiplied $4^{-1} \\mod 1000000007 = 250000002$ Code#include using namespace std; using ll = long long; const ll MOD = 1e9 +", "face is $$3\\sin(65^\\circ)+5.5\\times\\sqrt{2^2+3^2-2\\cdot 2\\cdot 3\\cdot\\cos(65^\\circ)}\\approx 18.2\\,\\mathrm{m}^2$$ Hence the volume is $$3.5\\times \\left(3\\sin(65^\\circ)+5.5\\times\\sqrt{2^2+3^2-2\\cdot 2\\cdot 3\\cdot\\cos(65^\\circ)}\\right)\\approx 63.7\\,\\mathrm{m}^3$$", "$$\\sin(45^\\circ)\\approx.707\\quad\\text{and}\\quad\\cos(45^\\circ)\\approx.707.$$ Your point would then have $x$ coordinate $\\ \\ \\ \\ \\ x\\approx5\\cdot (0.707)=3.535$ and $y$-coordinate $\\ \\ \\ \\ \\ y\\approx5\\cdot( 0.707)=3.535$. In radians, $45$ degrees is $45\\cdot{\\pi\\over 180}={\\pi\\over 4}$ radians; and you could compute $\\cos(\\pi/4)$ and $\\sin(\\pi/4)$ using a device where angles are measured in radians. This of course will give approximately $.707$ in both cases as before. -", "\\\\ \\frac{sin (\\red E)}{27} = \\frac{sin (37^{\\circ})}{12} \\\\ sin(\\red E)= \\frac{27 \\cdot sin (37^{\\circ})}{12} \\\\ sin(\\red B)= 1.3541 \\\\ \\boxed{\\text{No Solution}}$$ There is no possible $$\\angle E$$ with the given dimensions . The maximum value of the sine function is 1. $$sin(B)= \\cancel {\\red {1.3541}}$$", "Given that the period of cosecant is $$2\\pi,$$ we get $\\csc \\left( { - \\frac{{27\\pi }}{4}} \\right) = \\csc \\left( {\\frac{{5\\pi }}{4} - 2\\pi \\times 4} \\right) = \\csc \\frac{{5\\pi }}{4}.$ The angle $${\\frac{{5\\pi }}{4}}$$ is in the $$3\\text{rd}$$ quadrant, in which cosecant is negative. The reference angle of $${\\frac{{5\\pi }}{4}}$$ is $${\\frac{{\\pi }}{4}}.$$ Then we have $\\csc \\left( { - \\frac{{27\\pi }}{4}} \\right) = \\csc \\frac{{5\\pi }}{4} = - \\csc \\frac{\\pi }{4} = - \\sqrt 2 .$ ### Example 3. Find the value of the expression $\\cos {630^\\circ} - \\sin {1470^\\circ} - \\cot {1125^\\circ}.$ Solution. Cosine and sine are periodic functions, with period $$360^\\circ.$$ Therefore $\\cos {630^\\circ} = \\cos \\left( {{{270}^\\circ} + {{360}^\\circ}} \\right) = \\cos {270^\\circ} = 0,$ $\\sin {1470^\\circ} = \\sin \\left( {{{30}^\\circ} + {{1440}^\\circ}} \\right) = \\sin \\left( {{{30}^\\circ} + {{360}^\\circ} \\times 4} \\right) = \\sin {30^\\circ} = \\frac{1}{2}.$ The period of cotangent is $$180^\\circ.$$ Hence, $\\cot {1125^\\circ} = \\cot \\left( {{{45}^\\circ} + {{1080}^\\circ}} \\right) = \\cot \\left( {{{45}^\\circ} + {{180}^\\circ} \\times 6} \\right) = \\cot {45^\\circ} = 1.$ Substitute these values into the initial expression: $\\cos {630^\\circ} - \\sin {1470^\\circ} - \\cot {1125^\\circ} = 0 - \\frac{1}{2} - 1 = - \\frac{3}{2}.$ ### Example 4. Find the value of the expression $\\tan {1800^\\circ} - \\sin {495^\\circ} + \\cos {945^\\circ}.$ Solution. Calculate the tangent:", "this equation as a function of $\\beta$, and you will get a conjecture about $\\beta$ which is easy to prove. - Sorry, it's $AB \\cdot BC = BD^2$, the blockquote messed it up. But the idea of assigning some random value to a side might be helpful. – Javier Apr 5 '12 at 15:19 Also, how did you get $\\cos \\beta \\cdot \\cos (\\beta + 60°) \\le \\frac1{4}$? – Javier Apr 5 '12 at 15:22 This was helpful, thank you! – Javier Apr 5 '12 at 18:50 $$\\cos\\beta\\cdot \\cos(\\beta+60^\\circ)={1\\over4}\\ .$$ implies $$2\\cos\\beta\\cdot \\cos(\\beta+60^\\circ)={1\\over2}\\ .$$ so $$\\cos60^\\circ + \\cos(2\\cdot\\beta+60^\\circ)={1\\over2}\\ .$$ so $$\\cos(2\\cdot\\beta+60^\\circ)=0\\ .$$ So $$(2\\cdot\\beta+60^\\circ) = 90^\\circ$$ Thus $\\beta$ is $15^\\circ$. Thus in ABC, angles A, B and C are 15, 90 and 75 degrees respectively. -", "1^\\circ)(\\sin 3^\\circ)(\\sin 5^\\circ) \\cdots (\\sin 87^\\circ)(\\sin 89^\\circ) = \\frac{1}{2^n}$, compute the rational number $n$.", "2 \\cdot 90^{\\circ} = \\sin 180^{\\circ} = 0$. Check your calculations, and make sure your calculator is in the degrees mode. 11. Sep 21, 2014 ### sally.smith4321 Okay!! I forgot to multiply all of the numbers by 2! I'll do that now. Thank you so much for your help! You don't even understand how much it means to me! 12. Sep 21, 2014 ### sally.smith4321 I'm in an algebra based calculus class. He doesn't want us using calculus. That's probably why we weren't taught the derivative part.", "sine. Its domain contains all angles $$\\theta$$ except the points $$\\pi n, n \\in \\mathbb{Z},$$ where the sine function is equal to zero. The secant and cosecant are the reciprocal functions of cosine and sine, respectively. Therefore, the secant function is periodic, with period $$2\\pi:$$ It is defined for all real numbers $$\\theta$$ except the points $$\\large{\\frac{\\pi }{2}}\\normalsize + \\pi n,$$ $$n \\in \\mathbb{Z}.$$ Cosecant also has a period of $$2\\pi:$$ Cosecant is defined for all values of $$\\theta$$ except $$\\pi n, n \\in \\mathbb{Z}.$$ ## Solved Problems Click or tap a problem to see the solution. ### Example 1 Calculate exact values of trigonometric functions: 1. $$\\sin 750^\\circ$$ 2. $$\\cos \\left({-765^\\circ}\\right)$$ 3. $$\\tan 2760^\\circ$$ 4. $$\\cot \\left({-225^\\circ}\\right)$$ ### Example 2 Calculate exact values of trigonometric functions: 1. $$\\sin \\large{\\frac{{17\\pi }}{3}}\\normalsize$$ 2. $$\\cos \\left({- \\large{\\frac{{38\\pi }}{3}}\\normalsize}\\right)$$ 3. $$\\sec {\\large{\\frac{{43\\pi }}{6}}\\normalsize}$$ 4. $$\\csc \\left({- \\large{\\frac{{27\\pi }}{4}}\\normalsize}\\right)$$ ### Example 3 Find the value of the expression $\\cos {630^\\circ} – \\sin {1470^\\circ} – \\cot {1125^\\circ}.$ ### Example 4 Find the value of the expression $\\tan {1800^\\circ} – \\sin {495^\\circ} + \\cos {945^\\circ}.$ ### Example 5 Simplify the expression $\\frac{{\\sin \\left( { – \\frac{{13\\pi }}{2}} \\right) + \\tan \\left( { – 7\\pi } \\right)}}{{\\cos \\left( { – 7\\pi } \\right) + \\cot \\left( { – \\frac{{65\\pi }}{4}} \\right)}}.$ ### Example 6 Simplify", "8.2 1. $\\scriptsize y=-2\\sin (x+{{30}^\\circ})$ for $\\scriptsize -{{360}^\\circ}\\le x\\le {{360}^\\circ}$ $\\scriptsize a=-2$. Therefore, the amplitude will be $\\scriptsize 2$ and the graph will be reflected about the x-axis. $\\scriptsize p={{30}^\\circ}$. Therefore, the graph will be shifted $\\scriptsize {{30}^\\circ}$ to the left. $\\scriptsize \\sin x$ $\\scriptsize ({{0}^\\circ},0)$ $\\scriptsize ({{90}^\\circ},1)$ $\\scriptsize ({{180}^\\circ},0)$ $\\scriptsize ({{270}^\\circ},-1)$ $\\scriptsize ({{360}^\\circ},0)$ $\\scriptsize -2\\sin (x+{{30}^\\circ})$ $\\scriptsize (-{{30}^\\circ},0)$ $\\scriptsize ({{60}^\\circ},-2)$ $\\scriptsize ({{150}^\\circ},0)$ $\\scriptsize ({{240}^\\circ},2)$ $\\scriptsize ({{330}^\\circ},0)$ . y-intercept (let $\\scriptsize x=0$): \\scriptsize \\begin{align*}y & =-2\\sin ({{0}^\\circ}+{{30}^\\circ})\\\\\\therefore y & =-1\\end{align*} . Because the period of the function is $\\scriptsize {{360}^\\circ}$ we know that the function values at $\\scriptsize -{{360}^\\circ}$ and $\\scriptsize {{360}^\\circ}$ will also be $\\scriptsize -1$. . . Domain: $\\scriptsize \\{x|x\\in \\mathbb{R},\\text{ }-{{360}^\\circ}\\le x\\le {{360}^\\circ}\\}$ Range: $\\scriptsize \\{y|y\\in \\mathbb{R},\\text{ }-2\\le y\\le 2\\}$ 2. $\\scriptsize 2y=\\sin (x+{{30}^\\circ})$ for $\\scriptsize {{0}^\\circ}\\le x\\le {{360}^\\circ}$ \\scriptsize \\begin{align*}2y & =\\sin (x+{{30}^\\circ})\\\\\\therefore y & =\\displaystyle \\frac{1}{2}\\sin (x+{{30}^\\circ})\\end{align*} $\\scriptsize a=\\displaystyle \\frac{1}{2}$. Therefore, the amplitude will be $\\scriptsize \\displaystyle \\frac{1}{2}$. $\\scriptsize p={{30}^\\circ}$. Therefore, the graph will be shifted $\\scriptsize {{30}^\\circ}$ to the left. $\\scriptsize \\sin x$ $\\scriptsize ({{0}^\\circ},0)$ $\\scriptsize ({{90}^\\circ},1)$ $\\scriptsize ({{180}^\\circ},0)$ $\\scriptsize ({{270}^\\circ},-1)$ $\\scriptsize ({{360}^\\circ},0)$ $\\scriptsize \\displaystyle \\frac{1}{2}\\sin (x+{{30}^\\circ})$ $\\scriptsize (-{{30}^\\circ},0)$ $\\scriptsize ({{60}^\\circ},\\displaystyle \\frac{1}{2})$ $\\scriptsize ({{150}^\\circ},0)$ $\\scriptsize ({{240}^\\circ},-\\displaystyle \\frac{1}{2})$ $\\scriptsize ({{330}^\\circ},0)$ . y-intercept (let $\\scriptsize x=0$): \\scriptsize \\begin{align*}y & =\\displaystyle \\frac{1}{2}\\sin ({{0}^\\circ}+{{30}^\\circ})\\\\\\therefore y & =\\displaystyle \\frac{1}{4}\\end{align*} . Because the period of the function is", "Apr 7 '13 at 11:49 One liner: $\\sin(585^\\circ) = \\sin(585^\\circ-720^\\circ) = \\sin(-135^\\circ) = \\sin(-(90^\\circ+45^\\circ)) = -\\cos(45^\\circ) = -1/\\sqrt{2}$ - I find the circle to be a great way to understand this - Since the sine function is repetitive, in a 360 degree cycle, it's the same as 225 degrees. I am 50, and don't recall using this circle in trig class. It's a great way to visualize the function for both Sine and Cosine and can easily be memorized if need be. - NB - All angle references are in degrees... Since sin function has a period of 360, sin 585 = sin 225 = sin(180+45) = sin180*cos45+cos180sin45 = 0+ -1*1/sqrt(2) = -1/sqrt(2) - In the FAQ section there are directions how to use LaTeX in this site to properly write mathematics. – DonAntonio Apr 7 '13 at 11:33 @DonAntonio...Thank you :) – Sreekanth Karumanaghat Apr 7 '13 at 11:35 Observe that $585^\\circ\\equiv 225^\\circ\\pmod{360^\\circ}=180^\\circ+45^\\circ$ $180^\\circ<180^\\circ+45^\\circ<270^\\circ$ So, $585^\\circ$ lies in the $3$rd Quadrant. Using All-Sin-Tan-Cos formula or here, $\\sin(585^\\circ)<0$ Now, $585^\\circ=90^\\circ\\cdot 6+45^\\circ$ as the multiplicand of $90^\\circ$ is even, sine will remain sine So, $\\sin(585^\\circ)=\\sin(90^\\circ\\cdot 6+45^\\circ)=-\\sin45^\\circ=-\\frac1{\\sqrt2}$ -", "The given answer, though, is that the error is approximately $8.86 \\cdot 10^{-7}$. What did I do wrong? It always struck me that $x$ would be an unsuitable approximation for $\\sin{(1^\\circ)}$ since, at $x=1$, $\\sin({1^\\circ})$ is very small, certainly less than one (the maximum value for the sine function). I don't understand how the error can be so relatively small. • Two comments: Did you convert degrees into radians? And, more subtly, because $\\sin$ is an odd function, the Taylor polynomial of degree $1$ at $0$ is in fact the polynomial of degree $2$, so you get a higher-order error estimate. – Ted Shifrin Jun 7 '18 at 1:17 • Would I need to convert degrees into radians for the $x$ function? Or would I only need to do that for the sine function? – A. Lieber Jun 7 '18 at 1:18 The error is that you are plugging in $x=1$, instead of $x=1^\\circ$. A degree is $\\frac{\\pi}{180}$ radians, so you should set $x=\\frac{\\pi}{180}$ instead. (The right way to think about this is that the sine function is just defined by $\\sin x=\\sum_{n=0}^\\infty \\frac{(-1)^n}{(2n+1)!}x^{2n+1}$. When $x$ is an angle measure in radians, then $\\sin x$ defined in this way happens to also coincide with the geometric \"sine\" function you learn about in trigonometry. But if you measure $x$", "New Zealand Level 7 - NCEA Level 2 # Period changes for sine and cosine Lesson ## Preliminaries We define the $\\cos$cos and $\\sin$sin functions as the horizontal and vertical coordinates of a point that moves on the unit circle. In the diagrams below, this is shown for an angle $\\alpha$α in the first and second quadrants. If we imagine the point moving anticlockwise on the unit circle so that the radius from the point makes an ever-increasing angle with the positive horizontal axis, eventually the angle exceeds $360^\\circ$360°; but the values of the $\\cos$cos and $\\sin$sin functions repeat the values of the coordinates from the angle $360^\\circ$360° smaller. We say $\\sin$sin and $\\cos$cos are periodic functions with period $360^\\circ$360°. Thus, for any angle $\\alpha$α, there is a sequence of angles with the same value of $\\sin\\alpha$sinα. $...,\\alpha-720^\\circ,\\alpha-360^\\circ,\\alpha,\\alpha+360^\\circ,\\alpha+720^\\circ,\\alpha+1080^\\circ,...$...,α720°,α360°,α,α+360°,α+720°,α+1080°,... ## Transformations involving period Again, consider the angle $\\alpha$α made by the point moving around the unit circle. If a new angle $\\alpha'$α is defined by $\\alpha'=k\\alpha$α=kα, We know that $\\sin\\alpha'$sinα has period $360^\\circ$360°, but we see that $\\alpha'$α reaches $360^\\circ$360° when $\\alpha=\\frac{360^\\circ}{k}$α=360°k. So, $\\sin k\\alpha$sinkα and $\\cos k\\alpha$coskα must have period $\\frac{360^\\circ}{k}$360°k with respect to $\\alpha$α. #### Example 1 The function $\\sin2x$sin2x begins to repeat when $2x=360^\\circ$2x=360°. That is, when $x=180^\\circ$x=180°. So, $\\sin2x$sin2x has period $180^\\circ$180°. The period is", "## kaylala 2 years ago Find the value of the following using the concepts of both addition and subtraction formulas and reference angle: sin 255 1. kaylala @eliassaab 2. kaylala @ganeshie8 3. kaylala @Callisto 4. anonymous $\\sin (225)=-\\sin (360-255)=-\\sin (105)=-\\sin (45+60)=\\\\-\\sin (45)) \\cos (60)-\\sin (60) \\cos (45)$ 5. anonymous $-\\sin (45)) \\cos (60)-\\sin (60) \\cos (45)=-\\frac{1+\\sqrt{3}}{2 \\sqrt{2}}$ 6. kaylala why is it -sin(360-255) why the - sign there? 7. anonymous $-\\sin(360-255)=- \\sin (-255) =-(-\\sin( 255)) = \\sin(255)$ 8. anonymous Did you understand it? 9. kaylala no sorry where did you get that? 10. anonymous If you add 360 to an angle you do not change the sine of that angle sin(360+y) = sin(y) sin( 360 +(-x)) =sin(-x) = -sin(x) If you do not know these elementary facts, please review them in your book or notes. 11. anonymous Have you heard about periodic functions? 12. kaylala have no heard of periodic functions ever @eliassaab" ]

[ "Precalculus (10th Edition) $0$ I know that $\\sin$ is an odd function, which means $f(-\\theta)=-f(\\theta).$ Also, I know that $\\cos$ has a period of $360^\\circ$, hence $\\cos{\\theta}=\\cos{(\\theta-360^\\circ)}.$ Therefore $\\sin{1^\\circ}+\\sin{2^\\circ}+...+\\sin{358^\\circ}+\\sin{359^\\circ}\\\\=\\sin{1^\\circ}+\\sin{2^\\circ}+...+\\sin{(358^\\circ-360^\\circ)}+\\sin{(359^\\circ-360^\\circ)}\\\\=\\sin{1^\\circ}+\\sin{2^\\circ}+...+\\sin{-2^\\circ}+\\sin{-1^\\circ}\\\\=\\sin{1^\\circ}+\\sin{2^\\circ}+...-\\sin{2^\\circ}-\\sin{1^\\circ}=0$.", "\\pm 40^\\circ : n \\in \\mathbb Z \\} \\\\ \\end{align} Solution set: $$x \\in (\\{60^\\circ n \\pm 40^\\circ : n \\in \\mathbb Z \\} \\cup \\{180^\\circ n : n \\in \\mathbb Z\\}) \\cap [0^\\circ, 360^\\circ]$$ $$x \\in \\left\\{ \\begin{array}{rrrrr} 0^\\circ, & 20^\\circ, & 40^\\circ, & 80^\\circ, & 100^\\circ, \\\\ 140^\\circ, & 160^\\circ, & 180^\\circ, & 200^\\circ, & 220^\\circ, \\\\ 260^\\circ, & 280^\\circ, & 320^\\circ, & 340^\\circ, & 360^\\circ \\\\ \\end{array} \\right\\}$$ Using $\\sin3B=3\\sin B-4\\sin^3B$ $$32\\sin^3x\\cdot\\cos^2x\\cdot\\cos2x=\\sin x(3-4\\sin^2x)$$ If $\\sin x=0,x=180^\\circ n$ where $n$ is any integer Else using $\\cos2A=2\\cos^2A-1=1-2\\sin^2A,$ $$32\\cdot\\dfrac{1-\\cos2x}2\\cdot\\dfrac{1+\\cos2x}2\\cdot\\cos2x=3-2(1-\\cos2x)$$ $$\\iff8(c-c^3)=1+2c\\iff4c^3-3c=-\\dfrac12$$ $$\\implies\\cos6x=-\\dfrac12,6x=360^\\circ m\\pm120^\\circ\\iff x=60^\\circ m\\pm20^\\circ$$ where $m$ is any integer • @Théophile, One set of solution is $$0\\le180n\\le360\\iff0\\le n\\le2$$ Nov 2 '16 at 17:57", "unchanged after a rotation of $0^\\circ$. Thus, the number of grids that satisfy this property is $2^{n^2}$ For $g = 90^\\circ$, we see that a given square will form a cycle of period 4 (excluding the middle square in the case where $n$ is odd). Because all squares in the cycle must be the same color, there is $2 ^ {n^2 / 4}$ ways to do this for even $n$ and $2 ^ {\\frac {n^2 - 1} 4} \\cdot 2$ for odd $n$ (the second $\\cdot 2$ comes from the fact that the middle square can be one of two colors). For $g = 180^\\circ$, we see that a given square will form a cycle of period 2 (excluding the middle square in the case where $n$ is odd). Thus, there is $2 ^ {n^2 / 2}$ ways to do this for even $n$ and $2 ^ {\\frac {n^2 - 1} 2} \\cdot 2$ for odd $n$. For $g = 270^\\circ$, the answer is the same as $g = 90^\\circ$ (this is simply a $90^\\circ$ in the opposite direction) The final answer is simply the sum of the four values divided by 4 (or in this case, multiplied $4^{-1} \\mod 1000000007 = 250000002$ Code#include using namespace std; using ll = long long; const ll MOD = 1e9 +", "= {x\\over 2}$ $\\cos 72^\\circ = {{1 +x^2 - x^2}\\over 2x} = {1\\over 2x}$ Combining these two expressions, $\\cos 36^\\circ \\cos 72^\\circ = {x\\over 2}.{1\\over 2x}={1\\over 4}.$ Consider the area of the triangle `above' the diagonal, and express it is the sum of two areas : ${1\\over 2} x^2 \\sin 108^\\circ = {1\\over 2} x\\cdot1\\cdot\\sin 72^\\circ + {1\\over 2} 1\\cdot 1\\cdot\\sin108^\\circ.$ As $\\sin 108^\\circ = \\sin 72^\\circ$, this gives $x^2=x+1$ and hence $x$ is the golden ratio: $x = {1+\\sqrt{5}\\over 2} = \\varphi .$ Hence $\\cos 36^\\circ - \\cos 72^\\circ = {x\\over 2} - {1\\over 2x} = {x^2-1\\over 2x}={1\\over 2}.$ Let the common side of the two triangles be $x$. Then we have: $b = x\\cdot \\cos 72^\\circ$ and $a= {x\\over{\\cos 36^\\circ}}.$ Therefore ${b\\over a} = \\cos72^\\circ \\cos 36^\\circ = {1\\over 4}$ so $a$ is four times bigger than $b$. Here we have ${{x + b}\\over a} =\\cos 36^\\circ$ and ${x\\over a} = \\cos 72^\\circ$ so ${b\\over a} = \\cos 36^\\circ -\\cos 72^\\circ = {1\\over 2}.$ Therefore $a$ is twice the length of $b$.", "# Find all angles a in the interval [0degrees, 360 degrees] such that sin^2a-sina =0? ##### 1 Answer May 23, 2018 $0 , {90}^{\\circ} , {180}^{\\circ} , {360}^{\\circ}$ #### Explanation: ${\\sin}^{2} a - \\sin a = 0$ sin a(sin a - 1) = 0 Either factor should be zero --> a. sin a = 0 Unit circle gives: a = 0, $a = {180}^{\\circ}$, and $a = {360}^{\\circ}$ b. sin a - 1 = 0 --> sin a = 1 $a = {90}^{2}$", "$$\\sin(45^\\circ)\\approx.707\\quad\\text{and}\\quad\\cos(45^\\circ)\\approx.707.$$ Your point would then have $x$ coordinate $\\ \\ \\ \\ \\ x\\approx5\\cdot (0.707)=3.535$ and $y$-coordinate $\\ \\ \\ \\ \\ y\\approx5\\cdot( 0.707)=3.535$. In radians, $45$ degrees is $45\\cdot{\\pi\\over 180}={\\pi\\over 4}$ radians; and you could compute $\\cos(\\pi/4)$ and $\\sin(\\pi/4)$ using a device where angles are measured in radians. This of course will give approximately $.707$ in both cases as before. -", "$$1 - \\sin 10 ^ \\circ - 2 \\sin^2 10^\\circ = \\sin N^\\circ$$, where $$0 \\leq N \\leq 90$$. What is $$N$$?", "2 \\cdot 90^{\\circ} = \\sin 180^{\\circ} = 0$. Check your calculations, and make sure your calculator is in the degrees mode. 11. Sep 21, 2014 ### sally.smith4321 Okay!! I forgot to multiply all of the numbers by 2! I'll do that now. Thank you so much for your help! You don't even understand how much it means to me! 12. Sep 21, 2014 ### sally.smith4321 I'm in an algebra based calculus class. He doesn't want us using calculus. That's probably why we weren't taught the derivative part.", "1996 USAMO Problems/Problem 1 Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. Solution Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos 5^\\circ)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$, as desired. $\\Box$ 1996 USAMO (Problems • Resources)", "Product of repeated cosec. $$P = \\prod_{k=1}^{45} \\csc^2(2k-1)^\\circ=m^n$$ I realize that there must be some sort of trick in this. $$P = \\csc^2(1)\\csc^2(3)…..\\csc^2(89) = \\frac{1}{\\sin^2(1)\\sin^2(3)….\\sin^2(89)}$$ I noticed that: $\\sin(90 + x) = \\cos(x)$ hence, $$\\sin(89) = \\cos(-1) = \\cos(359)$$ $$\\sin(1) = \\cos(-89) = \\cos(271)$$ $$\\cdots$$ $$P \\cdot P = \\frac{\\cos^2(-1)\\cos^2(-3)….}{\\sin^2(1)\\sin^2(3)….\\sin^2(89)}$$ But that doesnt help? Solutions Collecting From Web of \"Product of repeated cosec.\" $$\\sin(2n+1)x=(2n+1)\\sin x+\\cdots+(-1)^n2^{2n}\\sin^{2n+1}x$$ If we set $2n+1=45,2^{44}\\sin^{45}x-\\cdots+45\\sin x-\\sin45x=0$ If we set $\\sin45x=\\sin45^\\circ,45x=360^\\circ m+45^\\circ$ where $m$ is any integer $\\implies x=8^\\circ m+1^\\circ$ where $0\\le m\\le44$ $\\implies Q=\\prod_{m=0}^{44}\\sin(8^\\circ m+1^\\circ)=\\dfrac1{\\sqrt2}\\cdot\\dfrac1{2^{44}}$ $Q^2=\\dfrac1{2\\cdot2^{88}}$ Clearly, $\\{\\sin^2(8^\\circ m+1^\\circ);0\\le m\\le44\\}=\\{\\sin^2(2r-1)^\\circ,1\\le r\\le45\\}$ as $x=1^\\circ,9^\\circ,17^\\circ,25^\\circ,33^\\circ,41^\\circ,49^\\circ,57^\\circ,65^\\circ,73^\\circ,81^\\circ,89^\\circ,$ $97^\\circ[\\sin97^\\circ=\\sin(180-97)^\\circ=\\sin83^\\circ],$ $105^\\circ[\\sin75^\\circ]$ and so on $\\cdots$ $m=22\\implies8m+1=177\\implies\\sin177^\\circ=\\sin(180-3)^\\circ=\\sin3^\\circ$ $\\cdots$ $m=44\\implies8m+1=353\\implies\\sin353^\\circ=-\\sin7^\\circ$ I’m using a non-standard notation inspired by various programming languages in this evaluation because I think it’s a bit easier to follow than the traditional big pi product notation. Hopefully, it’s self-explanatory. Also, all angles are given in degrees. This derivation uses the following identities: $$\\cos x = \\sin(90 – x) \\\\ \\sin(180 – x) = \\sin x \\\\ \\sin 2x = 2 \\sin x \\cdot \\cos x \\\\ \\text{and hence} \\\\ \\cos x = \\frac{\\sin 2x}{2 \\sin x}\\\\$$ \\begin{align} \\text{Let } P & = prod(\\csc^2 x: x = \\text{1 to 89 by 2}) \\\\ 1 / P & = prod(\\sin x: x = \\text{1 to 89 by 2})^2 \\\\ & =", "# What is the period of sin (5*x)? Jan 30, 2016 period$= {72}^{\\circ}$ #### Explanation: The general equation for a sine function is: $f \\left(x\\right) = a \\sin \\left[k \\left(x - d\\right)\\right] + c$ where: $| a | =$amplitude $| k | =$horizontal stretch/compression or ${360}^{\\circ} / \\text{period}$ $d =$phase shift $c =$vertical translation In this case, the value of $k$ is $5$. To find the period, use the formula, $k = {360}^{\\circ} / \\text{period}$: $k = {360}^{\\circ} / \\text{period}$ $5 = {360}^{\\circ} / \\text{period}$ $5 \\cdot \\text{period} = {360}^{\\circ}$ $\\text{period} = {360}^{\\circ} / 5$ $\\text{period} = {72}^{\\circ}$ $\\therefore$, the period is ${72}^{\\circ}$.", "general, the domain of the function $\\scriptsize y=\\sin (x+p)$ is $\\scriptsize x\\in \\mathbb{R}\\text{ }$. We had to restrict the domain in Example 8.1, because the interval in which we were working was restricted. ### Exercise 8.1 Sketch the following functions for the indicated intervals on separate sets of axes: 1. $\\scriptsize f(x)=\\sin (x+{{30}^\\circ})$ for $\\scriptsize {{0}^\\circ}\\le x\\le {{360}^\\circ}$ 2. $\\scriptsize g(x)=\\sin (x-{{45}^\\circ})$ for $\\scriptsize -{{360}^\\circ}\\le x\\le {{360}^\\circ}$ 3. $\\scriptsize h(x)=\\sin (x+{{20}^\\circ})$ for $\\scriptsize -{{360}^\\circ}\\le x\\le {{360}^\\circ}$ The full solutions are at the end of the unit. ## Sketch functions of the form $\\scriptsize y=a\\sin (x+p)$ Now that we know that the value of $\\scriptsize p$ shifts the function of the form $\\scriptsize y=\\sin (x+p)$ horizontally by $\\scriptsize p$ degrees, we can combine our knowledge of the effects of $\\scriptsize a$ with this and examine functions of the form $\\scriptsize y=a\\sin (x+p)$. ### Example 8.2 Sketch the graph of $\\scriptsize f(x)=2\\sin (x+{{25}^\\circ})$ for the interval $\\scriptsize -{{360}^\\circ}\\le x\\le {{360}^\\circ}$. Solution The function is of the form $\\scriptsize y=a\\sin (x+p)$ where $\\scriptsize a=2$ and $\\scriptsize p={{25}^\\circ}$. We need to transform one period of ‘anchor points’ noting that each of these points is going to be shifted $\\scriptsize {{25}^\\circ}$ to the left and each of the y-values is going to be multiplied by $\\scriptsize 2$. $\\scriptsize \\sin x$ $\\scriptsize ({{0}^\\circ},0)$ $\\scriptsize", "# 1996 USAMO Problems/Problem 1 ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos5)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$, as desired. $\\Box$ ###", "sin What is ${\\displaystyle \\sin(80^{\\circ })}$? What is ${\\displaystyle \\sin(100^{\\circ })}$? (you are expected to use a calculator for these). Do you see the problem? The inverse sine function 'makes the wrong choice' when doing the inverse sine of an obtuse angle. So, in solving an ASA triangle, you need to spot if one of the angles is obtuse. If the inverse sine for it tells you an acute angle you can then correct it to the equivalent obtuse angle. ${\\displaystyle \\sin(70^{\\circ })=\\sin(110^{\\circ })}$ ${\\displaystyle \\sin(60^{\\circ })=\\sin(120^{\\circ })}$ ${\\displaystyle \\sin(20^{\\circ })=\\sin(160^{\\circ })}$ The angles with the same sine add up to 180°, so you can correct the angle that came out acute that should be obtuse by subtracting it from 180°. Alternatively, since at most one of the angles in a triangle can be obtuse, the one opposite the longest side, calculate the other angle instead. Then use the fact that the angles sum to 180° to get the remaining one.", "may notice I used the $\\sin$/$\\cos$ addition formulas twice, but if you multiply your expression by $\\cos(10^\\circ)$ and \"couple\" the right terms in a slick way, you get the answer with less computations. However, my approach or Gelfand's intended one are more or less the same. – Jack D'Aurizio May 17 '16 at 0:36 • @SirJony: I think the formula has a different name depending on your location: Briggs, Werner or prostapheresis (en.wikipedia.org/wiki/Prosthaphaeresis). I just call it an instance of the addition formulas. – Jack D'Aurizio May 17 '16 at 0:38 • This is a nice solution but I don't think it's the one the problem-setter had in mind, since it doesn't make use of the hint given. – Jack Kinsella Jun 24 at 12:58 From the triple-angle formula $\\sin (3\\theta) = - 4\\sin^3\\theta + 3\\sin\\theta$ when $\\sin (3\\theta) = 1/2$, we get that $\\sin(10^\\circ)$, $\\sin(50^\\circ)$, $\\sin(-70^\\circ)$ are the roots of $8x^3-6x+1$. Therefore $$\\sin(10^\\circ) \\sin(50^\\circ) \\sin(70^\\circ) =-\\sin(10^\\circ) \\sin(50^\\circ) \\sin(-70^\\circ) =-(-\\frac18) =\\frac18.$$ • – lhf May 17 '16 at 0:59 We can prove $$4\\sin(60^\\circ-x)\\sin x\\sin(60^\\circ+x)=\\sin3x$$ Proof $\\#1:$ $$\\sin(60^\\circ-x)\\sin(60^\\circ+x)=\\sin^260^\\circ-\\sin^2x?$$ and $\\sin3x=3\\sin x-4\\sin^3x$ Proof$\\#2:$ If $\\sin3x=\\sin3A, 3x=180^\\circ n+(-1)^n3A$ where $n$ is any integer $\\implies x=60^\\circ n+(-1)^nA$ where $n\\equiv-1,0,1\\pmod3$ $\\implies x=-(60^\\circ+A), A,60^\\circ-A$ As $\\sin3x=3\\sin x-4\\sin^3x,3\\sin x-4\\sin^3x=\\sin3A\\iff4\\sin^3x-3\\sin x-\\sin3A=0$ Using Vieta's formula, $$\\prod_{n=-1}^1\\sin\\left(60^\\circ n+(-1)^nA\\right)=\\dfrac{\\sin3A}4$$ Use $\\sin\\left\\{-(60^\\circ+A)\\right\\}=-\\sin(60^\\circ+A)$ • I don't understand how this (and the", "Precalculus (10th Edition) We know that $\\sin$ has a period of $360^{ \\circ}$, hence using this first we try and find a value where the argument is between $0^{\\circ}$ and $360^{\\circ}$ and then evaluate the funcion. Therefore $\\sin(390^{\\circ})=\\sin(390^{\\circ}-360^{\\circ})=\\sin(30^{o})=\\frac{1}{2}.$ We know that $\\sin$ has a period of $360^{ \\circ}$, hence using this first we try and find a value where the argument is between $0^{\\circ}$ and $360^{\\circ}$ and then evaluate the function. Therefore $\\sin(390^{\\circ})=\\sin(390^{\\circ}-360^{\\circ})=\\sin(30^{o})=\\frac{1}{2}.$", "8.2 1. $\\scriptsize y=-2\\sin (x+{{30}^\\circ})$ for $\\scriptsize -{{360}^\\circ}\\le x\\le {{360}^\\circ}$ $\\scriptsize a=-2$. Therefore, the amplitude will be $\\scriptsize 2$ and the graph will be reflected about the x-axis. $\\scriptsize p={{30}^\\circ}$. Therefore, the graph will be shifted $\\scriptsize {{30}^\\circ}$ to the left. $\\scriptsize \\sin x$ $\\scriptsize ({{0}^\\circ},0)$ $\\scriptsize ({{90}^\\circ},1)$ $\\scriptsize ({{180}^\\circ},0)$ $\\scriptsize ({{270}^\\circ},-1)$ $\\scriptsize ({{360}^\\circ},0)$ $\\scriptsize -2\\sin (x+{{30}^\\circ})$ $\\scriptsize (-{{30}^\\circ},0)$ $\\scriptsize ({{60}^\\circ},-2)$ $\\scriptsize ({{150}^\\circ},0)$ $\\scriptsize ({{240}^\\circ},2)$ $\\scriptsize ({{330}^\\circ},0)$ . y-intercept (let $\\scriptsize x=0$): \\scriptsize \\begin{align*}y & =-2\\sin ({{0}^\\circ}+{{30}^\\circ})\\\\\\therefore y & =-1\\end{align*} . Because the period of the function is $\\scriptsize {{360}^\\circ}$ we know that the function values at $\\scriptsize -{{360}^\\circ}$ and $\\scriptsize {{360}^\\circ}$ will also be $\\scriptsize -1$. . . Domain: $\\scriptsize \\{x|x\\in \\mathbb{R},\\text{ }-{{360}^\\circ}\\le x\\le {{360}^\\circ}\\}$ Range: $\\scriptsize \\{y|y\\in \\mathbb{R},\\text{ }-2\\le y\\le 2\\}$ 2. $\\scriptsize 2y=\\sin (x+{{30}^\\circ})$ for $\\scriptsize {{0}^\\circ}\\le x\\le {{360}^\\circ}$ \\scriptsize \\begin{align*}2y & =\\sin (x+{{30}^\\circ})\\\\\\therefore y & =\\displaystyle \\frac{1}{2}\\sin (x+{{30}^\\circ})\\end{align*} $\\scriptsize a=\\displaystyle \\frac{1}{2}$. Therefore, the amplitude will be $\\scriptsize \\displaystyle \\frac{1}{2}$. $\\scriptsize p={{30}^\\circ}$. Therefore, the graph will be shifted $\\scriptsize {{30}^\\circ}$ to the left. $\\scriptsize \\sin x$ $\\scriptsize ({{0}^\\circ},0)$ $\\scriptsize ({{90}^\\circ},1)$ $\\scriptsize ({{180}^\\circ},0)$ $\\scriptsize ({{270}^\\circ},-1)$ $\\scriptsize ({{360}^\\circ},0)$ $\\scriptsize \\displaystyle \\frac{1}{2}\\sin (x+{{30}^\\circ})$ $\\scriptsize (-{{30}^\\circ},0)$ $\\scriptsize ({{60}^\\circ},\\displaystyle \\frac{1}{2})$ $\\scriptsize ({{150}^\\circ},0)$ $\\scriptsize ({{240}^\\circ},-\\displaystyle \\frac{1}{2})$ $\\scriptsize ({{330}^\\circ},0)$ . y-intercept (let $\\scriptsize x=0$): \\scriptsize \\begin{align*}y & =\\displaystyle \\frac{1}{2}\\sin ({{0}^\\circ}+{{30}^\\circ})\\\\\\therefore y & =\\displaystyle \\frac{1}{4}\\end{align*} . Because the period of the function is", "any integer $n$. But for any number $a$ such that $0\\le a \\le 1$, there is exactly one $\\theta$ between $0^\\circ$ and $90^\\circ$ such that $\\sin\\theta=a$. On most older calculators, entering $a$ and then pressing the $\\sin^{-1}$ button will give you the (unique) $\\theta$ between $-90^\\circ$ and $90^\\circ$ which solves the equation $\\sin\\theta=a$. On many newer calculators, you press the $\\sin^{-1}$ button then put in $a$. Make sure your calculator is set to degrees. When I do this, I get roughly $79.155937$. Try it on your calculator, or in the calculator program on your computer. It makes sense that the angle is not far from $90^\\circ$, since the sine of a $90$ degree angle is $1$, and $220/224$ is not far from $1$. However, as I pointed out earlier, there are other angles whose sine is $220/224$. An important one that you are likely to bump into in geometric work is $180$ minus the number we just computed. This is roughly $100.84406$ degrees. And, as was pointed out, there are infinitely many angles whose sine is $220/224$. However, there is only one from $0$ to $90$ degrees, and only one from $90$ degrees to $180$ degrees, and we have found them. You might want to check, by calculating the sine of each of these angles, that each of", "range is the set of all numbers. I graphed it and got a parabola. However, please help me out if I did the calculations wrong. Thank you very much in advance to whoever helps me out on this! 2. Originally Posted by FullBox 1) Let a, b be two angles. If b=a-π, how much is sin(2a+b)? $\\sin(x-\\pi) = -\\sin(x)$ $\\Rightarrow \\sin(2a+b) = \\sin(2a+a-\\pi) = -\\sin(3a)$ 2) Using the fact that 75=2 x 60-45, compute cos(5π/12) and write the answer in its most simplified form. $\\cos[(2\\times{60})^{\\circ}-45^{\\circ}] = \\sin(60^{\\circ}+60^{\\circ})\\sin(45^{\\circ})+\\cos(6 0^{\\circ}+60^{\\circ})\\cos(45^{\\circ})$. Use $\\sin(x+y) = \\sin(x)\\cos(y)+\\cos(x)\\sin(y)$ for $\\sin(60^{\\circ}+60^{\\circ})$ and $\\cos(x+y) = \\cos(x)\\cos(y)-\\sin(x)\\sin(y)$ for $\\cos(60^{\\circ}+60^{\\circ})$ then just plug in the standard values. 3) If f(x)=e^(x+1) and g(x)=x^(2)+1, what is the range of the composite function f(g(x))? I think I might know what I'm doing here. f(g(x)) --> f(x^(2)+1) = e^(x+1) = e^(x^(2)+1)+1 Right. The domain is obviously the set of all real numbers. That's what you got. But the range is not the set of all real numbers. We have [Math]f(g(x)) = e^{x^2+2}[/tex]. If $x = 0$, $f(g(x)) = e^2$. If $x$ is positive, then $e^{x^2+2} > e^2$. If $x$ is negative, then again $e^{x^2+2} > e^2$. I think the range of the function is clear then. 3. Thank you for the help! I just wanted to make sure about several things, though.", "moment I don't know how to prove it. By the simmetry $\\sin(180^{\\circ}-k^{\\circ})=\\sin(k^{\\circ})$, we have that $$\\prod_{k=1}^{179}\\sin (k^{\\circ})=\\sin(90^{\\circ})\\left(\\prod_{k=1}^{89}\\sin (k^{\\circ})\\right)\\left(\\prod_{k=1}^{89}\\sin (180-k^{\\circ})\\right)=\\left(\\prod_{k=1}^{89}\\sin (k^{\\circ})\\right)^2$$ On the other hand by Evaluation of a product of sines (noted by H. H. Rugh and Batominovski). $$\\prod_{k=1}^{179}\\sin (k^{\\circ})= \\frac{180}{2^{179}}$$ Hence $$\\prod_{k=1}^{89}\\sin (k^{\\circ})=\\left(\\frac{180}{2^{179}}\\right)^{1/2}=\\frac{\\sqrt{90}}{2^{89}}=\\frac{3 \\sqrt{10}}{2^{89}}$$ which confirms the comment by Tolaso." ]

[ "+0 0 103 2 Find constants $$A$$ and $$B$$ such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all $$x$$ such that $$x\\neq -1$$ and $$x\\neq 2$$. Give your answer as the ordered pair $$(A,B)$$. Sep 15, 2019 #1 +19818 0 Sep 15, 2019 #1 +19818 0", "+0 0 67 2 Find constants $$A$$ and $$B$$ such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all $$x$$ such that $$x\\neq -1$$ and $$x\\neq 2$$. Give your answer as the ordered pair $$(A,B)$$. Sep 15, 2019 #1 +19832 +1 Sep 15, 2019 #1 +19832 +1", "+0 0 47 1 Find constants A and B such that: $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all x such that x≠-1 and x≠2. Give your answer as the ordered pair (A,B). Sep 14, 2019 #1 +1 I think I saw this earlier. Check out https://web2.0calc.com/questions/help-please_7890 . Sep 14, 2019", "+0 # help me! 0 38 2 Find constants A and B such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all x such that x =/= -1 and x =/= 2. Give your answer as the ordered pair (A,B). Aug 17, 2020 #1 0 Letting x go to 2, we find that A + 2B = 4. Letting x go to -1, we find 2A - B = 3. Therefore, A = 2 and B = -1. Aug 17, 2020 #2 0 its wrong :/ Aug 17, 2020", "= (4\\alpha-1)\\begin{pmatrix}0\\\\3\\\\1\\end{pmatrix}$, which is constant in direction whatever the value of $\\alpha$.", "+0 # Help me please, tough question 0 82 1 Find constants and such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all x such that x is not equal to -1or 2. Give your answer as the ordered pair . May 27, 2020 #1 +25541 +1 Find constants and such that $$\\dfrac{x + 7}{x^2 - x - 2} = \\dfrac{A}{x - 2} + \\dfrac{B}{x + 1}$$", "\\quad \\mathbf{b} = \\begin{pmatrix} x_b \\\\y_b\\end{pmatrix}$$ If these are to reach exactly the same set of points we must have, for any integers $N$ and $M$ that $$N\\begin{pmatrix} x_a \\\\ y_a\\end{pmatrix} +M\\begin{pmatrix} x_b\\\\ y_b\\end{pmatrix}\\mathbf{b} = \\begin{pmatrix} 2p \\\\ q\\end{pmatrix}$$ Clearly to reach these, and only these points, I would need both $a$ and $b$ to have even $x$ component. Furthermore, if I can reach the points $(2, 0)$ and $(0, 1)$ with combinations of my base vectors then I would be able to reach all of the points $(2n, m)$ by adding together $n$ and $m$ lots of these combinations. Algebra quickly becomes very complicated, as there are too many variable to give me anything to 'solve'. So, look at some more particular cases. Look at a base vector with an $x$-component of $4$. $$\\begin{pmatrix} 4 \\\\ 0\\end{pmatrix}$$ Clearly, to be able to reach the point $(2, 0)$ the other base vector would need to be $\\begin{pmatrix} \\pm 2 \\\\ 0\\end{pmatrix}$, but this would not allow us to reach the point $(0, 1)$. Thus, any other possible pairs of base vectors would have to have non-zero $x$ and $y$ components. A suitable pair would be $$\\mathbf{a}= \\begin{pmatrix} 4 \\\\ 1\\end{pmatrix}\\,, \\mathbf{b} =\\begin{pmatrix}2 \\\\ 1\\end{pmatrix}$$ To see why, note that the combinations $\\mathbf{a}-\\mathbf{b}$ and $2\\mathbf{b}-\\mathbf{a}$ take us to the", "+0 # help! 0 129 5 +245 Find constants A and B such that (x + 7)/(x^2 - x - 2) = A/(x - 2) + B/(x + 1 for all x such that x\\neq -1 and x\\neq 2. Give your answer as the ordered pair (A,B). Dec 13, 2019 #1 0 (A,B) = (5,-1). Dec 13, 2019 #2 +245 +2 how did you get that? atlas9 Dec 13, 2019 #3 +2 If you let x go infinity, then you get 0 = A + B. If you let x = 0, then you get -7/2 = A/(-2) + B. The solution to this system is then A = 7/3, B = -7/3, so (A,B) = (7/3,-7/3). Dec 13, 2019 #4 +24350 +2 Find constants A and B such that $$\\dfrac{x + 7}{x^2 - x - 2} = \\dfrac{A}{x - 2} + \\dfrac{B}{x + 1}$$ for all x such that $$x\\neq -1$$ and $$x\\neq 2$$. Give your answer as the ordered pair $$(A,B)$$. $$\\text{Let x^2 - x - 2=(x-2)(x+1) }$$ $$\\begin{array}{|lrcll|} \\hline &\\mathbf{ \\dfrac{x + 7}{x^2 - x - 2} } &=& \\mathbf{ \\dfrac{A}{x - 2} + \\dfrac{B}{x + 1} } \\quad | \\quad x^2 - x - 2=(x-2)(x+1) \\\\\\\\ &\\dfrac{x + 7}{(x-2)(x+1)} &=& \\dfrac{A}{x - 2} + \\dfrac{B}{x + 1} \\quad | \\quad \\times (x-2)(x+1) \\\\\\\\ &\\dfrac{(x", "I get the constant $a=\\frac{3}{2}$ which is correct. However, I am completely stuck with the constant b. I am calculating using $\\begin{bmatrix}b_{1}\\\\ b_{2}\\end{bmatrix}$. But I get two exactly same equations. Thank you for your help on this one! Last edited by a moderator: May 1, 2013", "be determined from $$0 = m_1\\mathbf{r}_1 + m_2\\mathbb{r}_2 + m_3\\mathbf{r}_3.$$ patric44", "+0 +7 259 3 +730 Find constants A and B such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$$ for all x such that $$x\\neq -1$$ and $$x\\neq 2$$ . Give your answer as the ordered pair (A,B). MIRB16 Aug 28, 2017 #1 +87301 +1 CPhill Aug 28, 2017 #2 +730 +6 Thanks! MIRB16 Aug 28, 2017 #3 +730 +6 CPhill MIRB16 Aug 28, 2017", "#### Problem 6P 6. Find the values of the constants $a$ and $b$ such that $$\\lim _{x \\rightarrow 0} \\frac{\\sqrt[3]{a x+b}-2}{x}=\\frac{5}{12}$$", "+0 0 53 1 Find constants $A$ and $B$ such that $\\frac{x - 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}$for all $x$ such that $x\\neq -1$ and \\$ Aug 29, 2021", "+0 # Find constants A and B such that for all x such that and . Give your answer as the ordered pair (A, B). 0 127 1 Find constants A and B such that $$\\frac{x + 7}{x^2 - x - 2} = \\frac{A}{x - 2} + \\frac{B}{x + 1}.$$ for all x such that $$x\\neq -1$$ and $$x\\neq -2$$. Give your answer as the ordered pair (A, B). Apr 14, 2019 #1 +106519 +1 x^2 - x - 2 factors as (x -2) ( x + 1) Multiply through by this factorization and we have that 1x + 7 = A( x + 1) + B( x - 2) simplify 1x + 7 = Ax + A + Bx - 2B equate coefficients 1x + 7 = (A + B) x + (A - 2B) This implies that A + B = 1 ⇒ - A - B = - 1 (1) A - 2B = 7 (2) Add (1) and (2) and we have that -3B = 6 B = -2 And A + B = 1 A - 2 = 1 A = 3 So (A, B) = ( 3, -2) Apr 14, 2019", "$a \\in M(A)$ and $b \\in M(B)$ we have $$\\hat{f}(a)b = \\hat{f}(1_{M(A)}a)b = \\hat{f}(1_{M(A)})\\hat{f}(a)b = p (\\hat{f}(a)b) \\in p M(B).$$ Linearity and continuity then show that $\\hat{f}(M(A))M(B) \\subseteq pM(B)$, so we have equality.", "appropriate constants. Dependent upon the sign of b, we may write $b=\\pm{c}^{2}$ where c is some constant. See if you manage to make the last few steps on your own.. Last edited by a moderator:", "B3. Let $$f(x) = 1 + x + x^2 + \\cdots + x^n$$. Find the local minima of $$f(x)$$. Find the local maxima of $$f(x)$$. For each local maxima or minima $$(c,f(c))$$, find the integer $$k$$ such that $$k\\leq c < k+1$$. Hint: Use the definition of $$f$$ as well as the closed formula. Treat $$x\\geq 0$$ and $$x<0$$ separately. B4. For a function $$f:\\mathbb{R}^+ \\rightarrow \\mathbb{R}^+$$, define $$A_r$$ and $$B_r$$ as follows: $A_r := \\int_{1}^{r} f(x)\\, dx \\;\\;\\;\\;\\; B_r := \\int_{r}^{r^2} f(x)\\, dx$ Find all continuous functions $$f:\\mathbb{R}^+\\rightarrow\\mathbb{R}^+$$ for which $$A_r=B_r$$ for every positive $$r$$. Hint: Find an equation relating $$f(x)$$ with $$f(x^2)$$. Then consider the function $$xf(x)$$. B5. A pair of distinct real numbers $$(r, s)$$ is considered to be a good pair if $r^{3}+s^{2}=s^{3}+r^{2}$ (i) Find a good pair $$(a, l)$$ such that $$l$$ takes the maximum value possible. Find a good pair $$(s,b)$$ such that $$s$$ takes the minimum value possible. For any good pair $$(c, d)$$, other than the ones found above, show that there exists another real number $$e$$ such that $$(c, e)$$ and $$(d, e)$$ are both good pairs. (ii) Prove that there are infinitely many rational good pairs. Hints: Consider the function $$f=x^3-x^2$$. (i) For a good pair $$(r,s)$$, can you write $$s$$ in terms of $$r$$? (ii) You can", "Review question # Can we find the equation of the plane containing $l_1$ and $PQ$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R5089 ## Solution The lines $l_1$ and $l_2$ have equations $\\mathbf{r} = \\mathbf{a} + s\\mathbf{b}$ and $\\mathbf{r} = \\mathbf{c} + t\\mathbf{d}$, where $s$ and $t$ are real numbers and $\\mathbf{a} = 5\\mathbf{i}+ 6\\mathbf{j}- \\mathbf{k}, \\quad\\mathbf{b} = \\mathbf{i} +\\mathbf{j}+ \\mathbf{k}, \\quad\\mathbf{c} = \\mathbf{i}+ 13\\mathbf{j} +7\\mathbf{k}, \\quad\\mathbf{d} = \\mathbf{i} + 6\\mathbf{j} + 3\\mathbf{k}.$ Also, the point $P$ on $l_1$ and the point $Q$ on $l_2$ are such that $PQ$ is perpendicular to both $l_1$ and $l_2.$ In any order: 1. show that $PQ = \\sqrt{38},$ 2. find the position vectors of $P$ and $Q,$ We can write the position vector of $P$ as $\\mathbf{p} = \\begin{pmatrix}5+s \\\\ 6+s \\\\ -1+s\\end{pmatrix}$, and the position vector of $Q$ as $\\mathbf{q} = \\begin{pmatrix}1+t \\\\ 13+6t\\\\ 7+3t\\end{pmatrix}.$ Thus $\\mathbf{q} - \\mathbf{p} = \\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}$. This vector has to be perpendicular to both $l_1$ and $l_2$, and so using the scalar product, $\\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}.\\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=0 \\quad\\text{and}\\quad \\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}.\\begin{pmatrix}1\\\\ 6 \\\\ 3\\end{pmatrix}=0.$ This gives us the two equations $10t-3s+11=0$ and $46t-10s+62=0$, which solve to give $s = -3$, $t", "rewrite this equation in this way: $\\begin{array}{rcl}[s_{a},\\mathbf{a}][s_{b},\\mathbf{b}]&=&[s_{a}s_{b}-x_{a}x_{b}-y_{a}y_{b}-z_{a}z_{b},\\mathbf{0}]\\\\&&+[0,s_{a}(x_{b}\\mathbf{i}+y_{b}\\mathbf{j}+z_{b}\\mathbf{k})+s_{b}(x_{a}\\mathbf{i}+y_{a}\\mathbf{j}+z_{a}\\mathbf{k})\\\\&&+(y_{a}z_{b}-y_{b}z_{a})\\mathbf{i}+(z_{a}x_{b}-z_{b}x_{a})\\mathbf{j}+(x_{a}y_{b}-x_{b}y_{a})\\mathbf{k}]\\end{array}$ This equation gives us the sum of two ordered pairs. The first ordered pair is a Real quaternion and the second is a Pure quaternion. These two ordered pairs can be combined into a single ordered pair: $\\begin{array}{rcl}[s_{a},\\mathbf{a}][s_{b},\\mathbf{b}]&=&[s_{a}s_{b}-x_{a}x_{b}-y_{a}y_{b}-z_{a}z_{b},\\\\&&s_{a}(x_{b}\\mathbf{i}+y_{b}\\mathbf{j}+z_{b}\\mathbf{k})+s_{b}(x_{a}\\mathbf{i}+y_{a}\\mathbf{j}+z_{a}\\mathbf{k})\\\\&&+(y_{a}z_{b}-y_{b}z_{a})\\mathbf{i}+(z_{a}x_{b}-z_{b}x_{a})\\mathbf{j}+(x_{a}y_{b}-x_{b}y_{a})\\mathbf{k}]\\end{array}$ And if we substitute, $\\begin{array}{rcl}\\mathbf{a}&=&x_{a}\\mathbf{i}+y_{a}\\mathbf{j}+z_{a}\\mathbf{k}\\\\\\mathbf{b}&=&x_{b}\\mathbf{i}+y_{b}\\mathbf{j}+z_{b}\\mathbf{k}\\\\\\mathbf{a}\\cdot\\mathbf{b}&=&x_{a}x_{b}\\mathbf{i}^{2}+y_{a}y_{b}\\mathbf{j}^2+z_{a}z_{b}\\mathbf{k}^{2}\\\\\\mathbf{a}\\times\\mathbf{b}&=&(y_{a}z_{b}-y_{b}z_{a})\\mathbf{i}+(z_{a}x_{b}-z_{b}x_{a})\\mathbf{j}+(x_{a}y_{b}-x_{b}y_{a})\\mathbf{k}\\end{array}$ We get: $[s_{a},\\mathbf{a}][s_{b},\\mathbf{b}]=[s_{a}s_{b}-\\mathbf{a}\\cdot\\mathbf{b},s_{a}\\mathbf{b}+s_{b}\\mathbf{a}+\\mathbf{a}\\times\\mathbf{b}]$ Which is the general equation of a quaternion product. A Real Quaternion A Real Quaternion is a quaternion with a vector term of 0: $q=[s,\\mathbf{0}]$ And the product of two Real Quaternions is another Real Quaternion: $\\begin{array}{rcl}q_a&=&[s_a,\\mathbf{0}]\\\\q_b&=&[s_b,\\mathbf{0}]\\\\q_{a}q_{b}&=&[s_a,\\mathbf{0}][s_b,\\mathbf{0}]\\\\&=&[s_{a}s_{b},\\mathbf{0}]\\end{array}$ Which is similar to the product of two complex numbers that contain a zero imaginary term. $\\begin{array}{rcl}z_1&=&a_1+0i\\\\z_2&=&a_2+0i\\\\z_{1}z_{2}&=&(a_1+0i)(a_2+0i)\\\\&=&a_{1}a_{2}\\end{array}$ Multiplying a Quaternion by a Scalar We can also multiply a quaternion by a scalar which should obey the rule: $\\begin{array}{rcl}q&=&[s,\\mathbf{v}]\\\\\\lambda{q}&=&\\lambda[s,\\mathbf{v}]\\\\&=&[\\lambda{s},\\lambda\\mathbf{v}]\\end{array}$ We can confirm this by using the product or Real Quaterions shown above to multiply a quaternion by the scalar as a Real Quaternion: $\\begin{array}{rcl}q&=&[s,\\mathbf{v}]\\\\\\lambda&=&[\\lambda,\\mathbf{0}]\\\\\\lambda{q}&=&[\\lambda,\\mathbf{0}][s,\\mathbf{v}]\\\\&=&[\\lambda{s},\\lambda\\mathbf{v}]\\end{array}$ Pure Quaternions Similar to Real Quaterions, Hamilton also defined the Pure Quaternion as a quaternion that has a zero scalar term: $q=[0,\\mathbf{v}]$ Or, written in its component parts: $q=xi+yj+zk$ And we can also take the product of two Pure quaternions: $\\begin{array}{rcl}q_a&=&[0,\\mathbf{a}]\\\\q_b&=&[0,\\mathbf{b}]\\\\q_{a}q_{b}&=&[0,\\mathbf{a}][0,\\mathbf{b}]\\\\&=&[-\\mathbf{a}\\cdot\\mathbf{b},\\mathbf{a}\\times\\mathbf{b}]\\end{array}$ According to the quaternion product rule shown above. We can also express quaternions as an addition of the Real and Pure quaternion parts: $\\begin{array}{rcl}q&=&[s,\\mathbf{v}]\\\\&=&[s,\\mathbf{0}]+[0,\\mathbf{v}]\\end{array}$", "true. So $A^n =\\begin{pmatrix}a^n & 0\\\\0 & b^n\\end{pmatrix}, \\forall n \\in \\mathbb{N}$" ]

[ "is possible to find A-1. A-1 = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj (A) A = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ $$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\frac 12$$$$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} (\\frac 22) & (\\frac {-4}{2})\\\\ (\\frac {-3}{2}) & (\\frac 72)\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ ∴ A-1 =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ Ans Let: A =$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ and A-1=$$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ We know that, AA-1 = I, where I =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 45 - 45 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 0 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ Taking the corresponding elements both sides", "{1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{41}$$$$\\begin {pmatrix} 3 & 5\\\\ -7 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 × 1}{41} & \\frac {5 × 24}{41}\\\\ \\frac {-7 × 1}{41} & \\frac {2 × 24}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {120}{41}\\\\ \\frac {-7}{41} & \\frac {48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 + 120}{41}\\\\ \\frac {-7 + 48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {123}{41}\\\\ \\frac {41}{41}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {vmatrix} 3\\\\ 1\\\\ \\end {vmatrix}$$ ∴ x = 3 and y = 1 Ans Here, x - 8y = 39..........................................(1) 2x - 3y = 13.......................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 1", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "= $$\\frac {1}{-13}$$$$\\begin {pmatrix} -3 & -1\\\\ -4 & 3\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 51\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153}{13} & \\frac {3}{13}\\\\ \\frac {204}{13} & \\frac {-9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153 + 3}{13}\\\\ \\frac {204 - 9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {156}{13}\\\\ \\frac {195}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 12\\\\ 5\\\\ \\end {pmatrix}$$ ∴ x = 12 and y = 5 Ans Here, 2x - 5y = 1.................................(1) 7x + 3y = 24.............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {vmatrix}$$ = 2 × 3 - 7 × (-5) = 6 + 35 = 41 ≠ 0 It has a unique solution. A-1= $$\\frac", "\\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {vmatrix}$$ = 3 × (-2) - 9 × 4 = -6 - 36 = -42 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{-42}$$$$\\begin {pmatrix} -2 & -4\\\\ -9 & 3\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {-2}{-42} & \\frac {-4}{-42}\\\\ \\frac {-9}{-42} & \\frac {3}{-42}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$ We know that: AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{21} + \\frac {5}{21}\\\\ \\frac {9}{21} - \\frac {15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 5}{21}\\\\ \\frac {36 - 15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {7}{21}\\\\ \\frac {21}{84}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ i.e. $$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ Taking the corresponding parts of the equal matrix: $$\\frac 1x$$ = $$\\frac 13$$ ∴ x = 3 $$\\frac 1y$$ = $$\\frac 14$$", "1\\\\ \\end {vmatrix}$$ = 20 × 1 - 12 × 1 = 20 - 12 = 8 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{8}$$$$\\begin {pmatrix} 1 & -1\\\\ -12 & 20\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-12}{8} & \\frac {20}{8}\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$ We know that, AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35}{8} - \\frac {20}{8}\\\\ \\frac {-105}{2} + \\frac {100}{2}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35 - 20}{8}\\\\ \\frac {-105 + 100}{2}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {15}{8}\\\\ \\frac {-5}{2}\\\\ \\end {pmatrix}$$ ∴ x = $$\\frac {15}{2}$$ and y = $$\\frac {-5}{2}$$ Ans Here, $$\\frac 3x$$ + $$\\frac 4y$$ = 2..........................................(1) $$\\frac 9x$$ - $$\\frac 2y$$ = $$\\frac 52$$....................(2) Given equation in the form of matrix: $$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix}", "\\frac{-3}{2}, \\frac{1}{3},240\\right)=\\left(0, \\frac{-5}{3},247\\right)$ $7\\left( \\frac{1}{5},\\frac{1}{7},\\frac{1}{9}\\right)=\\left( \\frac{7}{5},1,\\frac{7}{9}\\right)$ Ex.2. $V=M_{m\\times n}(F)=\\left\\lbrace\\begin{pmatrix} a_{11} & \\cdots & a_{1n} \\\\ \\vdots & & \\vdots \\\\ a_{m1} & \\cdots & a_{mn}\\end{pmatrix}: a_{ij} \\in F \\right\\rbrace$ $M_{3\\times 2}( \\mathbb{R})\\ni \\begin{pmatrix} 7 & -7 \\\\ \\pi & \\mathit{e} \\\\ -5 & 2 \\end{pmatrix}$ $\\mbox{Addition by adding entry by entry:}$ $M_{2\\times 2}\\ \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix}+\\begin{pmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{pmatrix}=\\begin{pmatrix} {a_{11}+b_{11}} & {a_{12}+b_{12}} \\\\ {a_{21}+b_{21}} & {a_{22}+b_{22}} \\end{pmatrix}$ $\\mbox{Multiplication by multiplying scalar c to all entries by M.}$ $c\\cdot M_{2\\times 2}\\ \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix}=\\begin{pmatrix} c\\cdot a_{11} & c\\cdot a_{12} \\\\ c\\cdot a_{21} & c\\cdot a_{22} \\end{pmatrix}$ $\\mbox{Zero matrix has all entries = 0:}$ $0_{M_{m\\times n}}=\\begin{pmatrix} 0 & \\cdots & 0 \\\\ \\vdots & & \\vdots \\\\ 0 & \\cdots & 0\\end{pmatrix}$ Ex.3. $\\mathbb{C}$ form a vector space over $\\mathbb{R}$. Ex.4. $\\mbox{F is a vector space over itself.}$ Ex.5. $\\mathbb{R}$ is a vector space over $\\mathbb{Q}$. Ex.6. $\\mbox{Let S be a set. Let}$ $\\mathcal{F}(S,\\mathbb{R})=\\big\\{f:S\\to \\mathbb{R} \\big\\}$ $f,g \\in \\mathcal{F}(S,\\mathbb{R})$ $(f+g)(t)=f(t)+g(t)\\ for\\ any\\ t\\in S$ $(af)(t)=a\\cdot f(t)\\$", "(-5) × 3 = 8 + 15 = 23 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{23}$$ $$\\begin {pmatrix} 4 & -3\\\\ 5 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} -4\\\\ -13\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-16}{23} + \\frac {39}{23}\\\\ \\frac {-20}{23} - \\frac {26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {-16 + 39}{23}\\\\ \\frac {-20 - 26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {23}{23}\\\\ \\frac {-46}{23}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ -2\\ \\end {pmatrix}$$ ∴ x = 1 and y = -2Ans Here, 4x + 3y = 5.........................................(1) y - 3x = -7...........................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 &3\\\\ -3 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -3 & 1\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -3 & 1\\\\", "is $$A=\\begin{pmatrix}1&0&1\\\\\\!\\!\\!\\!-1&1&2\\\\0&\\!\\!\\!\\!-1&1\\end{pmatrix}^{-1}=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}$$ Finally, since $$f(x)=3-4x+2x^2\\stackrel{coord. wrt B}\\longrightarrow \\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix} \\,\\,,\\,\\text{we get}$$ $$[Af]=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}\\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix}=\\frac{1}{4}\\begin{pmatrix}11\\\\\\!\\!\\!\\!-7\\\\1\\end{pmatrix}$$ so $$\\,f_C=\\frac{11}{4}(1-x)-\\frac{7}{4}(x-x^2)+\\frac{1}{4}(1+2x+x^2)$$", "=\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\left [\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}\\end{pmatrix} -\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\right ]+\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\\\ & =\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}-1\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}-2\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}+1\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}+5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{16}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{37}{5}\\end{pmatrix} \\end{align*} Is that correct so far? How could we continue? Klaas van Aarsen MHB Seeker Staff member The matrix for $\\delta$ is correct. But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. mathmari Well-known member MHB Site Helper But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? Klaas van Aarsen MHB Seeker Staff member Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? That point is in between $A$ and $B$. But $C$ is on the circumcircle of triangle $ABZ$, which implies that it cannot be between $A$ and $B$. Klaas van Aarsen MHB Seeker Staff member I've drawn a picture now: \\begin{tikzpicture} \\usetikzlibrary", "& -1\\\\ 1 & 2\\\\ \\end {vmatrix}$$ = 2 × 2 - 1 × -1 = 4 + 1 = 5 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{5}$$$$\\begin {pmatrix} 2 & 1\\\\ -1 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{5} + \\frac {3}{5}\\\\ \\frac {-1}{5} + \\frac {6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 3}{5}\\\\ \\frac {-1 + 6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{5}\\\\ \\frac {5}{5}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, x + 2y = 8........................................(1) 2x + 3y = 11..................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ , AX = B where, A =$$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ $$\\begin", "1 \\end{pmatrix}$$. ### Solution 1. To find $$P\\begin{pmatrix}X \\leq 1.5\\end{pmatrix}$$, we use write: \\begin{aligned} P\\begin{pmatrix} X \\leq 1.5 \\end{pmatrix} & = \\int_{-\\infty}^{1.5} f(x)dx \\\\ & = \\int_{-\\infty}^{\\frac{3}{2}}f(x)dx \\\\ & = \\int_{-\\infty}^{\\frac{3}{2}}-\\frac{3}{4}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{-\\infty}^{\\frac{3}{2}}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{-\\infty}^{\\frac{3}{2}}\\begin{pmatrix} x^2 - 2x \\end{pmatrix}dx \\\\ & = -\\frac{3}{4} \\begin{bmatrix}\\frac{x^3}{3} - x^2 \\end{bmatrix}_0^{\\frac{3}{2}} \\\\ & = - \\frac{1}{4}\\begin{bmatrix}x^3 - 3x^2 \\end{bmatrix}_0^{\\frac{3}{2}} \\\\ & = - \\frac{1}{4}\\begin{bmatrix}\\begin{pmatrix} \\frac{3}{2}\\end{pmatrix}^3 - 3 \\times \\begin{pmatrix} \\frac{3}{2} \\end{pmatrix}^2 - 0 \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\frac{27}{8} - \\frac{27}{4} \\end{bmatrix} \\\\ & = - \\frac{1}{4} \\begin{bmatrix} -\\frac{27}{8} \\end{bmatrix} \\\\ & = \\frac{27}{32} \\\\ P\\begin{pmatrix} X \\leq 1.5 \\end{pmatrix} & = 0.844 \\end{aligned} Graphically, this result can be interpreted as follows: The area enclosed by the probability density function's curve and the horizontal axis, from $$-\\infty$$ upto $$x=1.5$$ is equal to $$0.844$$ (rounded to 3 significant figures). The probability is equal to the area so: $$P\\begin{pmatrix} X \\leq 1.5\\end{pmatrix} = 0.844$$ 2. To find $$P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix}$$ we write: \\begin{aligned} P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix} & = \\int_{0.5}^1f(x)dx \\\\ & = \\int_{0.5}^1 -\\frac{3}{4}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{0.5}^1 x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{\\frac{1}{2}}^1 \\begin{pmatrix} x^2-2x \\end{pmatrix} dx \\\\ & = -\\frac{3}{4}\\begin{bmatrix} \\frac{x^3}{3} - x^2 \\end{bmatrix}_{\\frac{1}{2}}^1 \\\\ & = -\\frac{3}{4}\\times \\frac{1}{3}\\begin{bmatrix} x^3 - 3x^2 \\end{bmatrix}_{\\frac{1}{2}}^1 \\\\ & = -\\frac{1}{4}\\begin{bmatrix} x^3 - 3x^2 \\end{bmatrix}_{\\frac{1}{2}}^1", "\\\\ 0 & 1 & -\\frac{r}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}} & -\\frac{r}{\\sqrt{2}} & 1 \\end{pmatrix}, \\;\\;\\; \\Sigma_{3e}(r)= \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & r \\\\ 0 & r & 1 \\end{pmatrix}, \\;\\;\\; \\Sigma_{4a}(r)= \\begin{pmatrix} 1 & 0 & 0 & \\frac{r}{\\sqrt{2}} \\\\ 0 & 1 & -r & \\frac{r}{\\sqrt{2}} \\\\ 0 & -r & 1 & -\\frac{1}{\\sqrt{2}} \\\\ \\frac{r}{\\sqrt{2}} & \\frac{r}{\\sqrt{2}} & -\\frac{1}{\\sqrt{2}} & 1 \\end{pmatrix},$ $\\Sigma_{4b}(r)= \\begin{pmatrix} 1 & 0 & r & \\frac{r}{\\sqrt{2}} \\\\ 0 & 1 & 0 & \\frac{r}{\\sqrt{2}} \\\\ r & 0 & 1 & \\frac{1}{\\sqrt{2}} \\\\ \\frac{r}{\\sqrt{2}} & \\frac{r}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & 1 \\end{pmatrix}, \\;\\;\\; \\Sigma_{4c}(r)= \\begin{pmatrix} 1 & 0 & \\frac{1}{\\sqrt{2}} & -\\frac{r}{\\sqrt{2}} \\\\ 0 & 1 & -\\frac{r}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & -\\frac{r}{\\sqrt{2}} & 1 & -r \\\\ -\\frac{r}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & -r & 1 \\end{pmatrix}\\;\\;\\text{and}\\;\\; \\Sigma_{4d}(r)= \\begin{pmatrix} 1 & r & \\frac{1}{\\sqrt{2}} & \\frac{r}{\\sqrt{2}} \\\\ r & 1 & \\frac{r}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{r}{\\sqrt{2}} & 1 & r \\\\ \\frac{r}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & r & 1 \\end{pmatrix}.$ # Estimation methods Given the form of bridge function $$F$$, obtaining a moment-based estimation $$\\widehat \\sigma_{jk}$$ requires inversion of $$F$$. latentcor implements two methods for calculation of the inversion: Both methods calculate inverse bridge function applied to each element of sample Kendall’s $$\\tau$$ matrix. Because the", "$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {vmatrix}$$ = -4 - 15 = -19≠ 0 It has a unique solution. X = A-1B A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$ X = A-1B or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -10 - 9\\\\ -25 + 6\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-19}{-19}\\\\ \\frac {-19}{-19}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, 3x - 5y = 3..........................(1) 4x + 3y = 4.........................(2) Given equation in matrix form; $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3\\\\ 4\\\\ \\end {pmatrix}$$ or, AX = B where, A = $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$,", "& \\frac{1}{2} \\end{pmatrix} \\cdot \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} 1-\\sqrt{3} \\\\ 1+\\sqrt{3} \\end{pmatrix}$$", "$& x = -14-4y\\\\& x = -14-4 \\left(-\\frac{32}{9}\\right)\\\\& x = -14+\\frac{128}{9}\\\\& x = -\\frac{126}{9}+\\frac{128}{9}\\\\& \\boxed{x = \\frac{2}{9}}\\\\& \\boxed{l_1 \\cap l_2 @ \\left(\\frac{2}{9},-\\frac{32}{9}\\right)}$ Solve each of the following... $& \\begin{Bmatrix}x=-4+y\\\\x=3y-6\\end{Bmatrix}$ $& -4+y = 3y-6\\\\& -4+4+y = 3y-6+4\\\\& y = 3y-2\\\\& y-3y = 3y-3y-2\\\\& -2y = -2\\\\& \\frac{-2y}{-2} = \\frac{-2}{-2}\\\\& \\boxed{y = 1}$ $& x = -4+y\\\\& x = -4+1\\\\& \\boxed{x = -3}\\\\& \\boxed{l_1 \\cap l_2 @ (-3,1)}$ $& \\begin{Bmatrix}2x=5y-12\\\\3x+5y=7\\end{Bmatrix}$ $& 2x = 5y-12\\\\& 2x+12 = 5y-12+12\\\\& 2x+12 = 5y\\\\& \\boxed{5y = 2x+12}$ $& 3x+5y = 7\\\\& 3x-3x+5y = 7-3x\\\\& \\boxed{5y = 7-3x}$ $& \\begin{Bmatrix}5y=2x+12\\\\5y=7-3x\\end{Bmatrix}$ $& 2x+12 = 7-3x\\\\& 2x+12+3x = 7-3x+3x\\\\& 5x+12 = 7\\\\& 5x+12-12 = 7-12\\\\& 5x = -5\\\\& \\frac{5x}{5} = \\frac{-5}{5}\\\\& \\frac{\\cancel{5}x}{\\cancel{5}} = \\frac{\\overset{-1}{\\cancel{-5}}}{\\cancel{5}}\\\\& \\boxed{x = -1}$ $& 5y = 7-3x\\\\& 5y = 7-3(-1)\\\\& 5y = 7+3\\\\& 5y = 10\\\\& \\frac{5y}{5} = \\frac{10}{5}\\\\& \\frac{\\cancel{5}y}{\\cancel{5}} = \\frac{\\overset{2}{\\cancel{10}}}{\\cancel{5}}\\\\& \\boxed{y = 2}\\\\& \\boxed{l_1 \\cap l_2 @ (-1,2)}$ $& \\begin{Bmatrix}\\frac{x+y}{3}+\\frac{x-y}{2} =\\frac{25}{6}\\\\\\frac{x+y-9}{2} =\\frac{y-x-6}{3}\\end{Bmatrix}$ $& \\frac{x+y}{3}+\\frac{x-y}{2} = \\frac{25}{6}\\\\& 6 \\left(\\frac{x+y}{3}\\right)+6 \\left(\\frac{x-y}{2}\\right) = 6 \\left(\\frac{25}{6}\\right)\\\\& \\overset{2}{\\cancel{6}} \\left(\\frac{x+y}{\\cancel{3}}\\right)+\\overset{3}{\\cancel{6}} \\left(\\frac{x-y}{\\cancel{2}}\\right) = \\cancel{6} \\left(\\frac{25}{\\cancel{6}}\\right)\\\\& 2(x+y)+3(x-y) = 25\\\\& 2x+2y+3x-3y = 25\\\\& \\boxed{5x-y = 25}$ $& \\frac{x+y-9}{2} = \\frac{y-x-6}{3}\\\\& 6 \\left(\\frac{x+y-9}{2}\\right) = 6 \\left(\\frac{y-x-6}{3}\\right)\\\\& \\overset{3}{\\cancel{6}} \\left(\\frac{x+y-9}{\\cancel{2}}\\right) = \\overset{2}{\\cancel{6}} \\left(\\frac{y-x-6}{\\cancel{3}}\\right)\\\\& 3(x+y-9) = 2(y-x-6)\\\\& 3x+3y-27 = 2y-2x-12\\\\& 3x+3y-27+27 = 2y-2x-12+27\\\\& 3x+3y = 2y-2x+15\\\\& 3x+3y+2x = 2y-2x+2x+15\\\\& 5x+3y = 2y+15\\\\& 5x+3y-2y = 2y-2y+15\\\\& \\boxed{5x+y = 15}$ $& \\begin{Bmatrix}5x-y=25\\\\5x+y=15\\end{Bmatrix}$ $& 5x-y = 25", "\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}, \\quad \\bfb = \\begin{pmatrix}1\\\\3\\\\2\\end{pmatrix}.$$ Recall that Hence, $$\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix} = \\begin{pmatrix} 1/3 & 1 & -1/3 \\\\ -1/2 & 1/2 & 0 \\\\ 1/6 & -1/2 & 1/3 \\end{pmatrix} \\begin{pmatrix}1\\\\3\\\\2 \\end{pmatrix} = \\begin{pmatrix}8/3\\\\1\\\\-2/3\\end{pmatrix}$$ Hence, the polynomial through the points $(-1, 1), (1, 3),$ and $(2,2)$ is $$y = \\frac{8}{3} + x - \\frac{2}{3} x^2$$", "\\end{pmatrix} + \\frac{1}{2}g^{bb} \\begin{pmatrix} \\frac{dg_{bb}}{da} + \\frac{dg_{ab}}{da} - \\frac{dg_{aa}}{db} \\end{pmatrix} = \\frac{1}{2}(\\frac{1}{a^2})(2a) = \\frac{1}{a}$$ $$\\Gamma^a_{ab} = \\frac{1}{2}g^{aa} \\begin{pmatrix} \\frac{dg_{aa}}{db} + \\frac{dg_{ba}}{da} - \\frac{dg_{ab}}{da} \\end{pmatrix} + \\frac{1}{2}g^{ab} \\begin{pmatrix} \\frac{dg_{ab}}{db} + \\frac{dg_{bb}}{da} - \\frac{dg_{ab}}{db} \\end{pmatrix} = \\frac{1}{2}(1)(0) = 0$$ $$\\Gamma^b_{ab} = \\frac{1}{2}g^{ba} \\begin{pmatrix} \\frac{dg_{ba}}{db} + \\frac{dg_{ba}}{da} - \\frac{dg_{ab}}{da} \\end{pmatrix} + \\frac{1}{2}g^{bb} \\begin{pmatrix} \\frac{dg_{bb}}{db} + \\frac{dg_{bb}}{da} - \\frac{dg_{ab}}{db} \\end{pmatrix} = \\frac{1}{2}(\\frac{1}{a^2})(0) = 0$$ $$\\Gamma^a_{ba} = \\frac{1}{2}g^{aa} \\begin{pmatrix} \\frac{dg_{aa}}{da} + \\frac{dg_{aa}}{db} - \\frac{dg_{ba}}{da} \\end{pmatrix} + \\frac{1}{2}g^{ab} \\begin{pmatrix} \\frac{dg_{ab}}{da} + \\frac{dg_{ab}}{db} - \\frac{dg_{ba}}{db} \\end{pmatrix} = \\frac{1}{2}(1)(0) = 0$$ $$\\Gamma^b_{ba} = \\frac{1}{2}g^{ba} \\begin{pmatrix} \\frac{dg_{ba}}{da} + \\frac{dg_{aa}}{db} - \\frac{dg_{ba}}{da} \\end{pmatrix} + \\frac{1}{2}g^{bb} \\begin{pmatrix} \\frac{dg_{bb}}{da} + \\frac{dg_{ab}}{db} - \\frac{dg_{ba}}{db} \\end{pmatrix} = \\frac{1}{2}(\\frac{1}{a^2}_(2a) = \\frac{1}{a}$$ $$\\Gamma^a_{bb} = \\frac{1}{2}g^{aa} \\begin{pmatrix} \\frac{dg_{aa}}{db} + \\frac{dg_{ba}}{db} - \\frac{dg_{bb}}{da} \\end{pmatrix} + \\frac{1}{2}g^{ib} \\begin{pmatrix} \\frac{dg_{ab}}{db} + \\frac{dg_{bb}}{db} - \\frac{dg_{bb}}{db} \\end{pmatrix} = \\frac{1}{2}(10)(-2a) = -a$$ $$\\Gamma^b_{bb} = \\frac{1}{2}g^{ba} \\begin{pmatrix} \\frac{dg_{ba}}{db} + \\frac{dg_{ba}}{db} - \\frac{dg_{bb}}{da} \\end{pmatrix} + \\frac{1}{2}g^{bb} \\begin{pmatrix} \\frac{dg_{bb}}{db} + \\frac{dg_{bb}}{db} - \\frac{dg_{bb}}{db} \\end{pmatrix} = \\frac{1}{2}(\\frac{1}{a^2})(0) = 0$$ --------------------------------------------- In summary, the complete calculated values are: $$\\Gamma^a_{aa} = 0, \\Gamma^b_{aa} = \\frac{1}{a}, \\Gamma^a_{ab} = 0, \\Gamma^b_{aa} = \\frac{1}{a}$$ $$\\Gamma^a_{ba} = 0, \\Gamma^b_{ba} = \\frac{1}{a}, \\Gamma^a_{bb} = -a, \\Gamma^b_{bb} = 0$$ which agrees with the calculation using the first method above, *except* for ## \\Gamma^b_{aa} ## which yielded 0 by the first method, but ##\\frac{1}{a}## by the second method (?) Last edited: topsquark James1238765 The", "\\end {vmatrix}$$ = 4 × 1 - 3 × -3 = 4 + 9 = 13 ≠ 0 It has unique solution. A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} 1 & -3\\\\ 3 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{13} & \\frac {-3}{13}\\\\ \\frac {3}{13} & \\frac {4}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{13} & \\frac {-3}{13}\\\\ \\frac {3}{13} & \\frac {4}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{13} + \\frac {21}{13}\\\\ \\frac {15}{13} - \\frac {28}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5 + 21}{13}\\\\ \\frac {15 - 28}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {26}{13}\\\\ \\frac {-13}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ ∴ x = 2 and y = -1 Ans Here, 2x - y = 1...............................(1) x + 2y = 3..............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -1\\\\ 1 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -1\\\\ 1 & 2\\\\ \\end {pmatrix}$$, X = $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2", "+ i & 0 \\end{pmatrix}$. Last edited: Feb 28, 2016 danshawen likes this. 11. ### rpennerFully WiredRegistered Senior Member Messages: 4,833 $M=\\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\sqrt{2} & 0 \\\\ 0 & \\sqrt{2} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix}$ So we have: $M^{2n} = 2^{n} \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} = 2^{n} I \\\\ M^{2n+1} = 2^{n} \\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = 2^{n} M$ So $\\textrm{exp}(M) = \\sum_{k=0}^{\\infty} \\left( \\frac{ 2^{k} }{ (2k)! } I +\\frac{ 2^{k} }{ (2k+1)! } M \\right) = \\left( \\cosh \\sqrt{2} \\right) I + \\left( \\frac{1}{\\sqrt{2}} \\sinh \\sqrt{2} \\right) M \\\\ \\quad \\quad = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$ Alternately: $\\textrm{exp}(M) = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} e^{- \\sqrt{2}} & 0 \\\\ 0 & e^{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$" ]

[ "= $$\\frac {1}{-13}$$$$\\begin {pmatrix} -3 & -1\\\\ -4 & 3\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 51\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153}{13} & \\frac {3}{13}\\\\ \\frac {204}{13} & \\frac {-9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153 + 3}{13}\\\\ \\frac {204 - 9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {156}{13}\\\\ \\frac {195}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 12\\\\ 5\\\\ \\end {pmatrix}$$ ∴ x = 12 and y = 5 Ans Here, 2x - 5y = 1.................................(1) 7x + 3y = 24.............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {vmatrix}$$ = 2 × 3 - 7 × (-5) = 6 + 35 = 41 ≠ 0 It has a unique solution. A-1= $$\\frac", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}, \\quad \\bfb = \\begin{pmatrix}1\\\\3\\\\2\\end{pmatrix}.$$ Recall that Hence, $$\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix} = \\begin{pmatrix} 1/3 & 1 & -1/3 \\\\ -1/2 & 1/2 & 0 \\\\ 1/6 & -1/2 & 1/3 \\end{pmatrix} \\begin{pmatrix}1\\\\3\\\\2 \\end{pmatrix} = \\begin{pmatrix}8/3\\\\1\\\\-2/3\\end{pmatrix}$$ Hence, the polynomial through the points $(-1, 1), (1, 3),$ and $(2,2)$ is $$y = \\frac{8}{3} + x - \\frac{2}{3} x^2$$", "{1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{41}$$$$\\begin {pmatrix} 3 & 5\\\\ -7 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 × 1}{41} & \\frac {5 × 24}{41}\\\\ \\frac {-7 × 1}{41} & \\frac {2 × 24}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {120}{41}\\\\ \\frac {-7}{41} & \\frac {48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 + 120}{41}\\\\ \\frac {-7 + 48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {123}{41}\\\\ \\frac {41}{41}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {vmatrix} 3\\\\ 1\\\\ \\end {vmatrix}$$ ∴ x = 3 and y = 1 Ans Here, x - 8y = 39..........................................(1) 2x - 3y = 13.......................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 1", "(-5) × 3 = 8 + 15 = 23 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{23}$$ $$\\begin {pmatrix} 4 & -3\\\\ 5 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} -4\\\\ -13\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-16}{23} + \\frac {39}{23}\\\\ \\frac {-20}{23} - \\frac {26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {-16 + 39}{23}\\\\ \\frac {-20 - 26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {23}{23}\\\\ \\frac {-46}{23}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ -2\\ \\end {pmatrix}$$ ∴ x = 1 and y = -2Ans Here, 4x + 3y = 5.........................................(1) y - 3x = -7...........................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 &3\\\\ -3 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -3 & 1\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -3 & 1\\\\", "1 \\end{pmatrix}$$. ### Solution 1. To find $$P\\begin{pmatrix}X \\leq 1.5\\end{pmatrix}$$, we use write: \\begin{aligned} P\\begin{pmatrix} X \\leq 1.5 \\end{pmatrix} & = \\int_{-\\infty}^{1.5} f(x)dx \\\\ & = \\int_{-\\infty}^{\\frac{3}{2}}f(x)dx \\\\ & = \\int_{-\\infty}^{\\frac{3}{2}}-\\frac{3}{4}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{-\\infty}^{\\frac{3}{2}}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{-\\infty}^{\\frac{3}{2}}\\begin{pmatrix} x^2 - 2x \\end{pmatrix}dx \\\\ & = -\\frac{3}{4} \\begin{bmatrix}\\frac{x^3}{3} - x^2 \\end{bmatrix}_0^{\\frac{3}{2}} \\\\ & = - \\frac{1}{4}\\begin{bmatrix}x^3 - 3x^2 \\end{bmatrix}_0^{\\frac{3}{2}} \\\\ & = - \\frac{1}{4}\\begin{bmatrix}\\begin{pmatrix} \\frac{3}{2}\\end{pmatrix}^3 - 3 \\times \\begin{pmatrix} \\frac{3}{2} \\end{pmatrix}^2 - 0 \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\frac{27}{8} - \\frac{27}{4} \\end{bmatrix} \\\\ & = - \\frac{1}{4} \\begin{bmatrix} -\\frac{27}{8} \\end{bmatrix} \\\\ & = \\frac{27}{32} \\\\ P\\begin{pmatrix} X \\leq 1.5 \\end{pmatrix} & = 0.844 \\end{aligned} Graphically, this result can be interpreted as follows: The area enclosed by the probability density function's curve and the horizontal axis, from $$-\\infty$$ upto $$x=1.5$$ is equal to $$0.844$$ (rounded to 3 significant figures). The probability is equal to the area so: $$P\\begin{pmatrix} X \\leq 1.5\\end{pmatrix} = 0.844$$ 2. To find $$P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix}$$ we write: \\begin{aligned} P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix} & = \\int_{0.5}^1f(x)dx \\\\ & = \\int_{0.5}^1 -\\frac{3}{4}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{0.5}^1 x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{\\frac{1}{2}}^1 \\begin{pmatrix} x^2-2x \\end{pmatrix} dx \\\\ & = -\\frac{3}{4}\\begin{bmatrix} \\frac{x^3}{3} - x^2 \\end{bmatrix}_{\\frac{1}{2}}^1 \\\\ & = -\\frac{3}{4}\\times \\frac{1}{3}\\begin{bmatrix} x^3 - 3x^2 \\end{bmatrix}_{\\frac{1}{2}}^1 \\\\ & = -\\frac{1}{4}\\begin{bmatrix} x^3 - 3x^2 \\end{bmatrix}_{\\frac{1}{2}}^1", "1\\\\ \\end {vmatrix}$$ = 20 × 1 - 12 × 1 = 20 - 12 = 8 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{8}$$$$\\begin {pmatrix} 1 & -1\\\\ -12 & 20\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-12}{8} & \\frac {20}{8}\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$ We know that, AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35}{8} - \\frac {20}{8}\\\\ \\frac {-105}{2} + \\frac {100}{2}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35 - 20}{8}\\\\ \\frac {-105 + 100}{2}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {15}{8}\\\\ \\frac {-5}{2}\\\\ \\end {pmatrix}$$ ∴ x = $$\\frac {15}{2}$$ and y = $$\\frac {-5}{2}$$ Ans Here, $$\\frac 3x$$ + $$\\frac 4y$$ = 2..........................................(1) $$\\frac 9x$$ - $$\\frac 2y$$ = $$\\frac 52$$....................(2) Given equation in the form of matrix: $$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix}", "& -1\\\\ 1 & 2\\\\ \\end {vmatrix}$$ = 2 × 2 - 1 × -1 = 4 + 1 = 5 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{5}$$$$\\begin {pmatrix} 2 & 1\\\\ -1 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{5} + \\frac {3}{5}\\\\ \\frac {-1}{5} + \\frac {6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 3}{5}\\\\ \\frac {-1 + 6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{5}\\\\ \\frac {5}{5}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, x + 2y = 8........................................(1) 2x + 3y = 11..................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ , AX = B where, A =$$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ $$\\begin", "\\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {vmatrix}$$ = 3 × (-2) - 9 × 4 = -6 - 36 = -42 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{-42}$$$$\\begin {pmatrix} -2 & -4\\\\ -9 & 3\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {-2}{-42} & \\frac {-4}{-42}\\\\ \\frac {-9}{-42} & \\frac {3}{-42}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$ We know that: AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{21} + \\frac {5}{21}\\\\ \\frac {9}{21} - \\frac {15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 5}{21}\\\\ \\frac {36 - 15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {7}{21}\\\\ \\frac {21}{84}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ i.e. $$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ Taking the corresponding parts of the equal matrix: $$\\frac 1x$$ = $$\\frac 13$$ ∴ x = 3 $$\\frac 1y$$ = $$\\frac 14$$", "\\end{pmatrix} + \\frac{1}{2}g^{bb} \\begin{pmatrix} \\frac{dg_{bb}}{da} + \\frac{dg_{ab}}{da} - \\frac{dg_{aa}}{db} \\end{pmatrix} = \\frac{1}{2}(\\frac{1}{a^2})(2a) = \\frac{1}{a}$$ $$\\Gamma^a_{ab} = \\frac{1}{2}g^{aa} \\begin{pmatrix} \\frac{dg_{aa}}{db} + \\frac{dg_{ba}}{da} - \\frac{dg_{ab}}{da} \\end{pmatrix} + \\frac{1}{2}g^{ab} \\begin{pmatrix} \\frac{dg_{ab}}{db} + \\frac{dg_{bb}}{da} - \\frac{dg_{ab}}{db} \\end{pmatrix} = \\frac{1}{2}(1)(0) = 0$$ $$\\Gamma^b_{ab} = \\frac{1}{2}g^{ba} \\begin{pmatrix} \\frac{dg_{ba}}{db} + \\frac{dg_{ba}}{da} - \\frac{dg_{ab}}{da} \\end{pmatrix} + \\frac{1}{2}g^{bb} \\begin{pmatrix} \\frac{dg_{bb}}{db} + \\frac{dg_{bb}}{da} - \\frac{dg_{ab}}{db} \\end{pmatrix} = \\frac{1}{2}(\\frac{1}{a^2})(0) = 0$$ $$\\Gamma^a_{ba} = \\frac{1}{2}g^{aa} \\begin{pmatrix} \\frac{dg_{aa}}{da} + \\frac{dg_{aa}}{db} - \\frac{dg_{ba}}{da} \\end{pmatrix} + \\frac{1}{2}g^{ab} \\begin{pmatrix} \\frac{dg_{ab}}{da} + \\frac{dg_{ab}}{db} - \\frac{dg_{ba}}{db} \\end{pmatrix} = \\frac{1}{2}(1)(0) = 0$$ $$\\Gamma^b_{ba} = \\frac{1}{2}g^{ba} \\begin{pmatrix} \\frac{dg_{ba}}{da} + \\frac{dg_{aa}}{db} - \\frac{dg_{ba}}{da} \\end{pmatrix} + \\frac{1}{2}g^{bb} \\begin{pmatrix} \\frac{dg_{bb}}{da} + \\frac{dg_{ab}}{db} - \\frac{dg_{ba}}{db} \\end{pmatrix} = \\frac{1}{2}(\\frac{1}{a^2}_(2a) = \\frac{1}{a}$$ $$\\Gamma^a_{bb} = \\frac{1}{2}g^{aa} \\begin{pmatrix} \\frac{dg_{aa}}{db} + \\frac{dg_{ba}}{db} - \\frac{dg_{bb}}{da} \\end{pmatrix} + \\frac{1}{2}g^{ib} \\begin{pmatrix} \\frac{dg_{ab}}{db} + \\frac{dg_{bb}}{db} - \\frac{dg_{bb}}{db} \\end{pmatrix} = \\frac{1}{2}(10)(-2a) = -a$$ $$\\Gamma^b_{bb} = \\frac{1}{2}g^{ba} \\begin{pmatrix} \\frac{dg_{ba}}{db} + \\frac{dg_{ba}}{db} - \\frac{dg_{bb}}{da} \\end{pmatrix} + \\frac{1}{2}g^{bb} \\begin{pmatrix} \\frac{dg_{bb}}{db} + \\frac{dg_{bb}}{db} - \\frac{dg_{bb}}{db} \\end{pmatrix} = \\frac{1}{2}(\\frac{1}{a^2})(0) = 0$$ --------------------------------------------- In summary, the complete calculated values are: $$\\Gamma^a_{aa} = 0, \\Gamma^b_{aa} = \\frac{1}{a}, \\Gamma^a_{ab} = 0, \\Gamma^b_{aa} = \\frac{1}{a}$$ $$\\Gamma^a_{ba} = 0, \\Gamma^b_{ba} = \\frac{1}{a}, \\Gamma^a_{bb} = -a, \\Gamma^b_{bb} = 0$$ which agrees with the calculation using the first method above, *except* for ## \\Gamma^b_{aa} ## which yielded 0 by the first method, but ##\\frac{1}{a}## by the second method (?) Last edited: topsquark James1238765 The", "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "\\end {pmatrix}$$ Taking the corresponding part of the equal matrix: $$\\frac 1x$$ = $$\\frac 12$$ ∴ x = 2 ∴ y = 3 ∴ x = 2 and y = 3 Ans Here, $$\\frac 4x$$ + 3y = 11.................................(1) 3xy - 8 -5x = 0 or, $$\\frac {3xy}{x}$$ - $$\\frac 8x$$ - $$\\frac {5x}{x}$$ = 0 or, -$$\\frac 8x$$ + 3y = 5...................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {vmatrix}$$ = 4 × 3 - (-8) × 3 = 12 + 24 = 36 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{36}$$$$\\begin {pmatrix} 3 & -3\\\\ 8 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{36} & \\frac {-3}{36}\\\\ \\frac {8}{36} & \\frac {4}{36}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{12} & \\frac {-1}{12}\\\\ \\frac {2}{9} & \\frac {1}{9}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix}", "+ e^{-z_{21}}} & \\frac{1}{1 + e^{-z_{22}}} & \\frac{1}{1 + e^{-z_{23}}} \\\\ ... & ... & ... \\\\ \\frac{1}{1 + e^{-z_{N1}}} & \\frac{1}{1 + e^{-z_{N2}}} & \\frac{1}{1 + e^{-z_{N3}}} \\end{bmatrix}$ $\\mathbf{X^2} = [\\mathbf{1}, \\mathbf{A^1}] = \\begin{bmatrix} 1 & \\frac{1}{1 + e^{-z_{11}}} & \\frac{1}{1 + e^{-z_{12}}} & \\frac{1}{1 + e^{-z_{13}}} \\\\ 1 & \\frac{1}{1 + e^{-z_{21}}} & \\frac{1}{1 + e^{-z_{22}}} & \\frac{1}{1 + e^{-z_{23}}} \\\\ ... & ... & ... & ... \\\\ 1 & \\frac{1}{1 + e^{-z_{N1}}} & \\frac{1}{1 + e^{-z_{N2}}} & \\frac{1}{1 + e^{-z_{N3}}} \\end{bmatrix}$ $\\mathbf{Z^2} = \\mathbf{X^2}\\mathbf{W^2} = \\begin{bmatrix} z_{11} & z_{12} \\\\ z_{21} & z_{22} \\\\ ... & ... \\\\ z_{N1} & z_{N2} \\end{bmatrix}$ $\\mathbf{\\hat{Y}} = \\mathbf{A^2} = softmax(\\mathbf{Z^2}) = \\begin{bmatrix} \\hat{y}_{11} & \\hat{y}_{12} \\\\ \\hat{y}_{21} & \\hat{y}_{22} \\\\ ... & ... \\\\ \\hat{y}_{N1} & \\hat{y}_{N2} \\end{bmatrix}$ Here, $sigmoid$ is the sigmoid function and similarly $softmax$ is the row-wise softmax function. (Also, note that I've used superscripts to denote the layer of the network for each element.) My goal is to use gradient descent to find the weight matrices $\\mathbf{W^1}$ and $\\mathbf{W^2}$ that minimize categorical cross entropy. Can anyone walk me through this process? I have seen it done here, but I can't understand how the solution generalizes to matrix equations.", "\\end {vmatrix}$$ = 4 × 1 - 3 × -3 = 4 + 9 = 13 ≠ 0 It has unique solution. A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} 1 & -3\\\\ 3 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{13} & \\frac {-3}{13}\\\\ \\frac {3}{13} & \\frac {4}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{13} & \\frac {-3}{13}\\\\ \\frac {3}{13} & \\frac {4}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{13} + \\frac {21}{13}\\\\ \\frac {15}{13} - \\frac {28}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5 + 21}{13}\\\\ \\frac {15 - 28}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {26}{13}\\\\ \\frac {-13}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ ∴ x = 2 and y = -1 Ans Here, 2x - y = 1...............................(1) x + 2y = 3..............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -1\\\\ 1 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -1\\\\ 1 & 2\\\\ \\end {pmatrix}$$, X = $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2", "We already know the value of $$\\mu$$. We now calculate $$E\\begin{pmatrix}X^2\\end{pmatrix}$$: \\begin{aligned} E\\begin{pmatrix}X^2\\end{pmatrix} &= \\int_{-\\infty}^{+\\infty}x^2.f(x)dx \\\\ &= \\int_{-\\infty}^{+\\infty} x^2.\\frac{(x-3)^2}{9}dx \\\\ &= \\int_0^3 x^2.\\frac{(x-3)^2}{9}dx \\\\ &= \\frac{1}{9}\\int_0^3 x^2.(x-3)^2dx \\\\ &= \\frac{1}{9}\\int_0^3 x^2.(x^2-6x+9)dx \\\\ &= \\frac{1}{9}\\int_0^3 (x^4-6x^3+9x^2)dx \\\\ &= \\frac{1}{9}\\begin{bmatrix} \\frac{x^5}{5}-\\frac{6}{4}x^4+\\frac{9}{3}x^3 \\end{bmatrix}_0^3 \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\frac{x^5}{5}-\\frac{3}{2}x^4+3x^3 \\end{bmatrix}_0^3 \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\begin{pmatrix} \\frac{3^5}{5}-\\frac{3}{2}\\times 3^4+3\\times 3^3 \\end{pmatrix} - 0\\end{bmatrix} \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\begin{pmatrix} \\frac{243}{5}-\\frac{243}{2}+81 \\end{pmatrix} - 0\\end{bmatrix} \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\frac{486}{10} - \\frac{1215}{10} + \\frac{810}{10}\\end{bmatrix} \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\frac{81}{10} \\end{bmatrix} \\\\ & = \\frac{81}{9\\times 10} \\\\ & = \\frac{9}{10} \\\\ E\\begin{pmatrix}X^2\\end{pmatrix} & = 0.9 \\end{aligned} We can now go back to the definition of variance and calculate: \\begin{aligned} Var\\begin{pmatrix}X \\end{pmatrix} &= E\\begin{pmatrix}X^2\\end{pmatrix} - \\mu^2 \\\\ & = 0.9 - 0.75^2 \\\\ Var\\begin{pmatrix}X \\end{pmatrix} & = 0.3375 \\end{aligned} 3. We now calculate the standard deviation using the definition: $\\sigma = \\sqrt{Var\\begin{pmatrix} X\\end{pmatrix}}$ Since $$Var\\begin{pmatrix}X \\end{pmatrix} = 0.3375$$ we find: $\\sigma = \\sqrt{0.3375} = 0.581$ The standard deviation is $$\\sigma = 0.581$$. This tells that, on average amongst all the different cell phones, we can expect the time taken for the app to launch to be $$0.581$$ seconds more, or less, than the mean value $$\\mu = 0.75$$ seconds. ## Exercise 1 A continuous random variable $$X$$ has probability density function defined as: $f(x) = \\begin{cases} \\frac{x^2}{9}, \\quad 0\\leq", "$$A \\begin{pmatrix} \\frac{\\partial F}{\\partial x_1}\\\\ \\frac{\\partial F}{\\partial x_2} \\end{pmatrix} = \\begin{pmatrix} 1&0\\\\ 0&2\\\\ \\end{pmatrix} \\begin{pmatrix} x_2\\\\ x_1 \\end{pmatrix} = \\begin{pmatrix} x_2\\\\ 2x_1 \\end{pmatrix} = \\begin{pmatrix} g_1(\\mathbf{x})\\\\ g_2(\\mathbf{x}) \\end{pmatrix};$$ $$\\frac{\\partial g_1}{\\partial x_2}=1\\neq2=\\frac{\\partial g_2}{\\partial x_1}.$$ IDEA 2. Chain rule (?) Counterexample. $$F(\\mathbf{x})=x_1x_2;\\\\ \\mathbf{y}(\\mathbf{x}): x_1=y_1;~x_2=2y_2.\\\\ \\begin{pmatrix} \\frac{\\partial x_1}{\\partial y_1}&\\frac{\\partial x_2}{\\partial y_1}\\\\ \\frac{\\partial x_1}{\\partial y_2}&\\frac{\\partial x_2}{\\partial y_2} \\end{pmatrix} \\begin{pmatrix} \\frac{\\partial F}{\\partial x_1}\\\\ \\frac{\\partial F}{\\partial x_2} \\end{pmatrix} = \\begin{pmatrix} 1&0\\\\ 0&2 \\end{pmatrix} \\begin{pmatrix} x_2\\\\ x_1 \\end{pmatrix} = \\begin{pmatrix} x_2\\\\ 2x_1 \\end{pmatrix} = \\begin{pmatrix} g_1(\\mathbf{x})\\\\ g_2(\\mathbf{x}) \\end{pmatrix} = \\begin{pmatrix} \\frac{\\partial F}{\\partial y_1}\\\\ \\frac{\\partial F}{\\partial y_2} \\end{pmatrix};$$ $$\\frac{\\partial g_1}{\\partial x_2}=1\\neq2=\\frac{\\partial g_2}{\\partial x_1}.$$ Of course, $$\\frac{\\partial^2F}{\\partial y_1\\partial y_2}=\\frac{\\partial^2F}{\\partial y_2\\partial y_1}$$ should be valid, but not $$\\frac{\\partial^2F}{\\partial y_1\\partial x_2}\\neq\\frac{\\partial^2F}{\\partial y_2\\partial x_1}.$$ And, in fact, the latter one is required in the original post. SOLUTION FOR n=2 @IgorKhavkine helped me a lot. Proof for $n=2$. All functions below a considered to depend on $\\mathbf{x}$. $$\\begin{pmatrix} a_{11}&a_{12}\\\\ a_{21}&a_{22} \\end{pmatrix} \\begin{pmatrix} \\frac{\\partial F}{\\partial x_1}\\\\ \\frac{\\partial F}{\\partial x_2} \\end{pmatrix} = \\begin{pmatrix} a_{11}\\frac{\\partial F}{\\partial x_1}+a_{12}\\frac{\\partial F}{\\partial x_2}\\\\ a_{21}\\frac{\\partial F}{\\partial x_1}+a_{22}\\frac{\\partial F}{\\partial x_2} \\end{pmatrix} = \\begin{pmatrix} g_1\\\\ g_2 \\end{pmatrix}.$$ The condition $\\frac{\\partial g_1}{\\partial x_2}=\\frac{\\partial g_2}{\\partial x_1}$ may be written as: $$\\frac{\\partial}{\\partial x_2}\\left(a_{11}\\frac{\\partial F}{\\partial x_1}+a_{12}\\frac{\\partial F}{\\partial x_2}\\right)=\\frac{\\partial}{\\partial x_1}\\left(a_{21}\\frac{\\partial F}{\\partial x_1}+a_{22}\\frac{\\partial F}{\\partial x_2}\\right).$$ If we expand it, we get: $$\\frac{\\partial a_{11}}{\\partial x_2}\\frac{\\partial F}{\\partial x_1}+a_{11}\\frac{\\partial^2F}{\\partial x_2\\partial x_1}+\\frac{\\partial a_{12}}{\\partial x_2}\\frac{\\partial F}{\\partial x_2}+a_{12}\\frac{\\partial^2F}{\\partial x_2^2}=\\\\ =\\frac{\\partial a_{21}}{\\partial x_1}\\frac{\\partial F}{\\partial", "f'(x) & = - \\frac{1}{x} \\end{aligned} ## Special Case 3 If a function is composed of a function $$h(x)$$ and a power function $$x^n$$, then the chain rule leads to the following result: $\\text{if} \\quad f(x) = \\begin{bmatrix} h(x) \\end{bmatrix}^n$ $\\text{then} \\quad f'(x) = h'(x)\\times n \\begin{bmatrix}h(x) \\end{bmatrix}^{n-1}$ We further illustrate and explain this sepcial case in Tutorial 4. Spend a few minutes to watch it now. ## Exercise 4 Differentiate each of the following: 1. $$f(x)=\\begin{pmatrix} 4x^2-7\\end{pmatrix}^5$$ 2. $$y = sin^5(x)$$ 3. $$f(x) = \\begin{pmatrix}2cos(x)\\end{pmatrix}^3$$ 4. $$y= \\begin{bmatrix} \\frac{ln(x)}{2} \\end{bmatrix}^6$$ 5. $$f(x) = tan^5(x)$$ 1. $$f'(x) = 40x\\begin{pmatrix}4x^2-7 \\end{pmatrix}^4$$ 2. $$\\frac{dy}{dx} = 5.cos(x).sin^4(x)$$ 3. $$f'(x) = -24.sin(x).cos^2(x)$$ 4. $$\\frac{dy}{dx}=\\frac{3.ln^5(x)}{32x}$$ 5. Two answers possible: $f'(x)=5.sec^2(x).tan^4(x)$ or $f'(x)=5.tan^4(x)+5tan^6(x)$ ## Solution With Working 1. We can see that $$f(x) = \\begin{pmatrix} 4x^2 - 7 \\end{pmatrix}^5$$ can be written $$f(x)=\\begin{bmatrix}h(x)\\end{bmatrix}^5$$, where: $$h(x)=4x^2-7$$ and $$h'(x)=8x$$ So using the special case 3, we can write: \\begin{aligned} f'(x) & = h'(x)\\times 5\\begin{bmatrix}h(x) \\end{bmatrix}^4 \\\\ & = 8x \\times 5 \\begin{bmatrix} 4x^2 - 7 \\end{bmatrix}^4 \\\\ f'(x) & = 40x \\begin{pmatrix} 4x^2 - 7 \\end{pmatrix}^4 \\end{aligned} 2. We can see that $$y=sin^5(x)$$ can be written $$y=\\begin{bmatrix}h(x)\\end{bmatrix}^5$$, where: $$h(x)=sin(x)$$ and $$h'(x)=cos(x)$$. Using the special case 3 we find the derivative function: \\begin{aligned} \\frac{dy}{dx} & = h'(x)\\times 5 \\begin{bmatrix}h(x)\\end{bmatrix}^4 \\\\ & = cos(x)\\times 5 \\begin{bmatrix} sin(x) \\end{bmatrix}^4", "\\frac{1}{x^2} \\end{pmatrix}'= -2x^{-2-1} = -2x^{-3}$$, we differentiate as follows: \\begin{aligned} \\begin{pmatrix} 3f(x)+4g(x) \\end{pmatrix}' &= 3f'(x) +4g'(x) \\\\ & = 3\\times 3x^{3-1} + 4\\times (-2)x^{-2-1} \\\\ & = 9x^2 - 8 x^{-3} \\\\ \\begin{pmatrix} 3f(x)+4g(x) \\end{pmatrix}' & = 9x^2-\\frac{8}{x^3} \\end{aligned} 2. Remembering that $$\\begin{pmatrix} \\sqrt{x} \\end{pmatrix}' = \\frac{1}{2\\sqrt{x}}$$, we differentiate as follows: \\begin{aligned} \\begin{pmatrix} g(x)-6h(x) \\end{pmatrix}' &= g'(x) - 6h'(x) \\\\ & = \\frac{1}{2\\sqrt{x}} - 6\\times (-2)x^{-2-1} \\\\ & = \\frac{1}{2\\sqrt{x}} + 12x^{-3} \\\\ \\begin{pmatrix} g(x)-6h(x) \\end{pmatrix}' &= \\frac{1}{2\\sqrt{x}} + \\frac{12}{x^3} \\end{aligned} 3. To differentiate $$\\frac{5}{6}f(x)+\\frac{1}{8}h(x)-\\frac{3}{4}g(x)$$, we use all that we have seen here to write: \\begin{aligned} \\begin{pmatrix} \\frac{5}{6}f(x)+\\frac{1}{8}h(x)-\\frac{3}{4}g(x) \\end{pmatrix}' &= \\frac{5}{6}f'(x)+\\frac{1}{8}h'(x)-\\frac{3}{4}g'(x) \\\\ &= 3\\times \\frac{5}{6} x^{3-1} -2\\times \\frac{1}{8}x^{-2-1} - \\frac{3}{4} \\times \\frac{1}{2}\\sqrt{x} \\\\ & = \\frac{15}{6}x^2-\\frac{2}{8}x^{-3} - \\frac{3}{8\\sqrt{x}} \\\\ & = \\frac{15}{6}x^2-\\frac{1}{4}\\frac{1}{x^3} - \\frac{3}{8\\sqrt{x}} \\\\ \\begin{pmatrix} \\frac{5}{6}f(x)+\\frac{1}{8}h(x)-\\frac{3}{4}g(x) \\end{pmatrix}' & = \\frac{5}{2}x^2-\\frac{1}{4x^3} - \\frac{3}{8\\sqrt{x}} \\end{aligned}", "1 & 0 \\\\ 1 & 1 \\end{pmatrix}, N_3 = \\begin{pmatrix} 1 & 1 \\\\ 0 & 1 \\end{pmatrix}, N_4 = \\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix}$ Does $\\{ N_1, N_2, N_3, N_4 \\}$ generate V? $M_1 = -\\frac23 N_1 + \\frac13(N_2 + N_3 + N_4)$ $M_2 = -\\frac23 N_2 + \\frac13(N_1 + N_3 + N_4)$ $M_3 = -\\frac23 N_3 + \\frac13(N_1 + N_2 + N_4)$ $M_4 = -\\frac23 N_4 + \\frac13(N_1 + N_2 + N_3)$ Then $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = aM_1 + bM_2 + cM_3 + dM_4 =$ $a \\cdot \\left( -\\frac23 N_1 + \\frac13(N_2 + N_3 + N_4) \\right) + b \\cdot \\left( -\\frac23 N_2 + \\frac13(N_1 + N_3 + N_4) \\right) + \\ldots$ Theorem: If $S \\in \\mathbf V$, then $\\operatorname{span}(S) =$ {all l.c. of elements of S} is a subspace of V." ]

[ "of the triangle and state how many triangles could be formed with the given values. 1. $A = 30^\\circ, a = 13, b = 15$ 2. $A = 22^\\circ, a = 21, b = 12$ 3. $A = 42^\\circ, a = 36, b = 37$ For 4-7, find all possible measures of $\\angle B$ (if any exist) for each of the following triangle values. 4. $A = 86^\\circ, a =15, b = 11$ 5. $A = 30^\\circ, a = 24, b = 43$ 6. $A = 48^\\circ, a = 34, b = 39$ 7. $A = 80^\\circ, a = 22, b = 20$ For 8-12, find the length of $b$ for each of the following triangle values. 8. $A = 94^\\circ, a = 31, B = 34^\\circ$ 9. $A = 112^\\circ, a = 12, B = 15^\\circ$ 10. $A = 78^\\circ, a = 20, B = 16^\\circ$ 11. $A = 54^\\circ, a = 15, B = 112^\\circ$ 12. $A = 39^\\circ, a = 9, B = 98^\\circ$ 13. In $\\Delta ABC, b=10$ and $\\angle A = 39^\\circ$ . What's a possible value for $a$ that would produce two triangles? 14. In $\\Delta ABC, b=10$ and $\\angle A = 39^\\circ$ . What's a possible value for $a$ that would produce no triangles? 15. In $\\Delta ABC, b=10$ and $\\angle", "# You'll be stuck Geometry Level 3 In $\\triangle ABC$, let $D$ be the midpoint of $\\overline{BC}$. If $\\angle ADB = 45^\\circ$ and $\\angle ACD = 30^\\circ$, what is $\\angle BAD$ in degrees? ×", "In triangle ABC, $\\angle ABC=95^\\circ$ and $\\angle ACB=50^\\circ$. Point P is constructed inside the triangle such that $BP=CP$ and $\\frac{[BPC]}{[ABC]}=\\frac{1}{5}$. Find the integer closest to $\\angle BPC$.", "+0 # Circle $\\Gamma$ 0 33 2 +16 Circle $\\Gamma$ is the incircle of $\\triangle ABC$ and is also the circumcircle of $\\triangle XYZ$. The point $X$ is on $\\overline{BC}$, the point $Y$ is on $\\overline{AB}$, and the point $Z$ is on $\\overline{AC}$. If $\\angle A=40^\\circ$, $\\angle B=60^\\circ$, and $\\angle C=80^\\circ$, what is the measure of $\\angle YZX$? May 16, 2019 #1 +7603 +2 Circle $$\\Gamma$$ is the incircle of $$\\triangle ABC$$ and is also the circumcircle of $$\\triangle XYZ$$. The point $$X$$ is on $$\\overline{BC}$$, the point $$Y$$ is on $$\\overline{AB}$$, and the point $$Z$$ is on $$\\overline{AC}$$. If $$\\angle A=40^\\circ$$, $$\\angle B=60^\\circ$$, and $$\\angle C=80^\\circ$$, what is the measure of $$\\angle YZX$$? ---------- Instead of calling the center of the circle $$\\Gamma$$ , let's call it G . Y is a point on both the circle and its tangent line AB. So Y must be a point of tangency. The same goes for X. Draw a line from G to Y and line from G to X. These form right angles with the tangent lines. Look at quadrilateral BYGX. The sum of the angle measures in a quadrilateral = 360° . So... m∠YGX = 360° - 90° - 90° - 60° = 120° The measure of an inscribed angle is half the measure of the central", "+0 trig +1 35 2 +491 In triangle $ABC$, we have $\\angle A = 90^\\circ$ and $\\sin B = \\frac{4}{7}$. Find $\\cos C$. Apr 27, 2021 #1 +113121 0 Draw the triangle and use Pythagoras's theorem Apr 27, 2021", "× # How many triangles? The number of non-congruent triangles $$ABC$$ such that $$\\angle A=25^{\\circ}$$, $$AB = 10$$ and $$BC = \\alpha$$, where $$\\alpha \\in \\mathbb{R}^{+}$$ is: • exactly one • at most one • at most 2 • none of the above. Note by Paramjit Singh 3 years, 1 month ago", "# Not Enough Angles Geometry Level 5 In triangle $ABC$, $\\angle A = 20^\\circ$ and $\\angle B = 80^\\circ$. Let $D$ be a point on line segment $AB$ such that $AD = BC$. What is the measure (in degrees) of $\\angle ADC$? ×", "# Ratio of sines! Geometry Level 3 In a triangle $$ABC$$, $$\\angle B = \\pi /3,\\angle C = \\pi / 4$$ and $$D$$ divides $$BC$$ internally in the ratio $$1:3$$, then $$\\dfrac { \\sin \\angle BAD }{ \\sin \\angle CAD }$$ is equal to: ×", "# Angles ratio In a triangle ABC true relationship c is less than b and b is less than a. Internal angles of the triangle are in the ratio 5:4:9. The size of the internal angle beta is: Result β = 50 ° #### Solution: $\\beta = 5/(5+4+9)\\cdot 180 = 50 ^\\circ$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you! Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...): Be the first to comment! Tips to related online calculators Check out our ratio calculator. Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? ## Next similar math problems: 1. Smallest internal angle Calculate what size has the smallest internal angle of the triangle if values of angles α:β:γ = 3:4:8 2. Triangle angles The angles α, β, γ in triangle ABC are in the ratio 6:2:6. Calculate size of angles. 3. Angles In the triangle ABC, the ratio of angles is: a:b = 4: 5. The angle c is 36°. How big are the angles a, b? 4. Find the 9", "A composite geometrical shape is made up of a square, equilateral and right triangles. Calculate the area of hatched triangle? Apr 26, 2016 $156.25$ Explanation: As $\\triangle A B D$ is a right triangle, we have, using the Pythagorean theorem, $\\overline{B D} = \\sqrt{{25}^{2} + {25}^{2}} = 25 \\sqrt{2}$. Because $\\triangle B D E$ is equilateral, this implies that $\\overline{E D} = 25 \\sqrt{2}$. Next, we can note that $\\triangle A B D$ is a $45 - 45 - 90$ right triangle, meaning the angle $\\angle A D B = {45}^{\\circ}$. Additionally, as $\\triangle B D E$ is equilateral, we have $\\angle B D E = {60}^{\\circ}$. Using those facts, along with $\\angle A D F = {180}^{\\circ}$, we have: $\\angle E D F = \\angle A D F - \\angle B D E - \\angle A B D$ $= {180}^{\\circ} - {60}^{\\circ} - {45}^{\\circ}$ $= {75}^{\\circ}$ As $\\triangle E D F$ is a right triangle, we can use the sine and cosine functions to say $\\frac{\\overline{E F}}{\\overline{E D}} = \\sin \\left({75}^{\\circ}\\right)$ $\\implies \\overline{E F} = \\overline{E D} \\sin \\left({75}^{\\circ}\\right) = 25 \\sqrt{2} \\sin \\left({75}^{\\circ}\\right)$ and $\\frac{\\overline{D F}}{\\overline{E D}} = \\cos \\left({75}^{\\circ}\\right)$ $\\implies \\overline{D F} = \\overline{E D} \\cos \\left({75}^{\\circ}\\right) = 25 \\sqrt{2} \\cos \\left({75}^{\\circ}\\right)$ As $\\overline{D F}$ and $\\overline{E F}$ are the base and height, respectively, of", "# Isosceles Circumcircle Geometry Level 3 Triangle $ABC$ is an isosceles triangle with $\\overline{AB}=\\overline{BC}$ and $\\angle ABC = 123 ^ \\circ$. Point $D$ is the midpoint of $\\overline{AC}$, point $E$ is the foot of the perpendicular from $D$ to $\\overline{BC},$ and point $F$ is the midpoint of $\\overline{DE}$. The intersection point of $\\overline{AE}$ and $\\overline{BF}$ is $G$. What is the measure (in degrees) of $\\angle BGA$? ×", "from the point A and drawing a line at $60{}^\\circ$ from the point B’. The point where both the lines cut is your point C’ and the triangle you get is triangle AB’C’.", "D}$) are given. This is illustrated in the figure. The quadrilateral can be constructed by considering consider this as two SSS triangles $ABC$$A B C$ and $ABD$$A B D$ To construct a quadrilateral, $3$$3$ sides ($\\overline{AB}$$\\overline{A B}$, $\\overline{BC}$$\\overline{B C}$, $\\overline{AD}$$\\overline{A D}$ ) and $2$$2$ angles ($\\angle A$$\\angle A$, $\\angle B$$\\angle B$) are given. This is illustrated in the figure. The quadrilateral can be constructed by considering this as two SAS triangles $ABC$$A B C$ and $BAD$$B A D$. To construct a quadrilateral, $2$$2$ sides ($\\overline{AB}$$\\overline{A B}$, $\\overline{BC}$$\\overline{B C}$ ) and $3$$3$ angles ($\\angle A$$\\angle A$, $\\angle B$$\\angle B$, $\\angle C$$\\angle C$)are given. This is illustrated in the figure. To construct, \"consider this as an SAS triangle $ABC$$A B C$ and another ASA triangle with two angles $\\angle C$$\\angle C$ and $\\angle A$$\\angle A$\". Note that $△ACD$$\\triangle A C D$ is formed by angles $\\angle C$$\\angle C$ and $\\angle A$$\\angle A$ summary • sum of interior angles ${360}^{\\circ }$${360}^{\\circ}$ The formulations of questions • 4 sides and a diagonal • 3 sided and 2 diagonals • 4 sides and an angle • 3 sides and 2 angles • 2 sides and 3 angles Outline", "E, and F are the midpoints of sides \\overline{BC}, \\overline{CA}, and \\overline{AB} of \\triangle ABC, respectively, and \\overline{CZ} is an altitude of the triangle. If \\angle BAC = 71^\\circ, \\angle ABC = 39^\\circ, and \\angle … 5. ### Geometry Points D, E, and F are the midpoints of sides \\overline{BC}, \\overline{CA}, and \\overline{AB} of \\triangle ABC, respectively, and \\overline{CZ} is an altitude of the triangle. If \\angle BAC = 71^\\circ, \\angle ABC = 39^\\circ, and \\angle … 6. ### Geometry In $\\triangle ABC$, we have $AB = AC = 13$ and $BC = 10$. Let $M$ be the midpoint of $\\overline{AB}$ and $N$ be on $\\overline{BC}$ such that $\\overline{AN}$ is an altitude of $\\triangle ABC$. If $\\overline{AN}$ and $\\overline{CM}$ … Medians $\\overline{AX}$ and $\\overline{BY}$ of $\\triangle ABC$ are perpendicular at point $G$. Prove that $AB = CG$. In your diagram, $\\angle AGB$ should appear to be a right angle. $M$ is the midpoint of $\\overline{AB}$ and $N$ is the midpoint of $\\overline{AC}$, and $T$ is the intersection of $\\overline{BN}$ and $\\overline{CM}$, as shown. If $\\overline{BN}\\perp\\overline{AC}$, $BN = 12$, and $AC = 14$, then find …", "The angle at the centre is twice the angle at the circumference and so as we know the angle at the centre, we need to divide this number by 2 to get the angle BAD : ### Example 2: angles in a triangle Below is a circle with centre C . Points A, B, and D are on the circumference of the circle. Calculate the size of angle \\theta. Here we have: • AC = BC = CD = Radii • AB is a chord • The angle ABC = 46^o • The angle BCD = \\theta As AC = BC , we can say that ABC is an isosceles triangle. This means that the angle BAC = angle ABC = 46^o . The angle at the centre is twice the angle at the circumference and so as we know the angle at the circumference BAD , we need to multiply this number by 2 to get the angle BCD at the centre: BCD = 46 × 2 BCD = 92° ABCD is a cyclic quadrilateral around the centre E . Calculate the size of the reflex angle AEC , labelled \\theta. Here we have: • AC = BC = CD = Radii • AD and CD are chords • The angle ADC = 72^o • The reflex", "# Too Many Angles, Too Little Information Geometry Level 3 In a triangle $$ABC$$, the measures of the angles are $$\\angle A=30^\\circ,$$ $$\\angle B=80^\\circ.$$ Point $$M$$ lies inside the triangle, such that $$\\angle MAC = 10^\\circ,$$ $$\\angle MCA = 30^\\circ.$$ What is the measure (in degrees) of $$\\angle BMC?$$ ×", "+0 trig 0 99 1 In triangle $ABC$, we have $\\angle A = 90^\\circ$ and $$\\sin B = \\frac{4}{9}$$. Find $\\cos C$. Jan 12, 2022 #1 +13577 +1 Find cos C. Hello Guest! If the sin B = 4/9 in a right triangle, then the ratio of the sides is a: b: c = 9: 4: 3. So is cos C = 4/9 ! Jan 12, 2022", "## Elementary Geometry for College Students (5th Edition) $\\angle C = 106˚$ Since $\\square ABCD$ is a parallelogram, this means that $DA$ and $BC$ are parallel to each other and therefore $\\angle A + \\angle B = 180˚$ $\\angle A = 2x + 6$ $\\angle B = x + 24$ 1. Solve for $x$ using the formula: $\\angle A + \\angle B = 180˚$ $(2x+6) + (x+24) = 180˚$ $3x + 30 = 180$ $3x = 150$ $x = 50$ 2. Apply the one of the properties of quadrilaterals, congruent angles. This property specifically states that opposite angles are equal to each other in a parallelogram. Therefore $\\angle A = \\angle C$ and now we have to solve for $\\angle A$. Substitute the $x$ into the $\\angle A$ equation. $\\angle A = \\angle C = (2x+6)$ $\\angle A = \\angle C = 2(50)+6$ $\\angle A = \\angle C = 100+6$ $\\angle A = \\angle C = 106˚$", "because I plan to use them while solving. $$\\triangle ABC$$ has a 30- and a 60-degree angle. The remaining angle of the triangle, $$m\\angle CAB=90^{\\circ}$$ . The triangle is therefore classified as a 30-60-90 triangle. 30-60-90 triangles have a special property: their side lengths are always at a constant ratio of $$1:\\sqrt{3}:2$$ . Since I know the length of $$\\overline{AB}$$ , 1, I can determine the lengths of the others by setting up a proportion. $$\\frac{AB}{BC}=\\frac{1}{2}$$ This is the proportion that I mentioned earlier. $$BC=2AB$$ I am solving for BC to make the proportion into a simple equation. Substitute 1 for AB. $$BC=2$$ Now we know that BC=2. In order to go further with this solution, you must know what a centroid is. A centroid is a point of concurrency where all three medians of a triangle intersect in the interior of the triangle. This means you must know what a median is. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side of that vertex. This information will help us. In the diagram above, $$\\overline{AE}$$ is a median. Therefore, point E is a midpoint of $$\\overline{BC}$$ , so point E bisects the segment. Therefore, $$BE=\\frac{BC}{2}=\\frac{2}{2}=1$$ . This means that we have identified two side lengths and the measure", "(4x-7)^{\\circ} \\ and\\ (5x-11)^{\\circ}$. Find the value of $x$ and the measures of all the angles. $2x + (4x-7) + (5x-11) = 180^o$ Or $x = 18^o$ Hence the angles are $36^{\\circ}, 65^{\\circ}, \\ and\\ 79^{\\circ}$ $\\\\$ Question 6: In $\\Delta ABC, \\angle A=x^{\\circ}, \\angle B=(2x-15)^{\\circ}, \\ and\\ \\angle C=(3x+21)^{\\circ}$. Find the value of $x$ and the measure of all the angles of the triangle. $x+(2x-15)+(3x+21)=180^o$ or $x=29^o$ Hence $\\angle A=29^{\\circ}, \\angle B=43^{\\circ}, \\ and\\ \\angle C=108^{\\circ}$ $\\\\$ Question 7: In $\\Delta ABC, \\angle A=(2x+15)^{\\circ}, \\angle B=x^{\\circ}, \\ and\\ \\angle C=(3x-15)^{\\circ}$. Find the value of x and the measure of all the angles of the triangle and show that the triangle is an isosceles triangle. $(2x+15)+x+(3x-15)=180^o \\ or\\ x=30^o$ Hence $\\angle A=75^{\\circ}, \\angle B=30^{\\circ}, \\ and\\ \\angle C=75^{\\circ}.$ $\\\\$ Question 8: In $\\Delta ABC, \\angle A=3\\ \\angle B=6 \\ \\angle C$ . Find the angles of the triangle. Let $\\angle A=3\\angle B=6 \\angle C=6x$ $\\Rightarrow \\angle A=6x, \\angle B=2x, and \\angle C=x$ $\\Rightarrow 6x+2x+x=180^o \\ or\\ x=20^o$ $\\Rightarrow \\angle A=120^{\\circ}, \\angle B=40^{\\circ} \\ and\\ \\angle C=20^{\\circ}$ $\\\\$ Question 9: In the adjoining figure, it is given that $\\angle ABC=40^{\\circ}, \\angle ACD=70^{\\circ}, \\angle ACB=x^{\\circ} \\ and\\ \\angle EAC=y$. Find the value of \\$latex $x and y$. $x+70=180^o \\ hence\\ x=110^{\\circ}$ Note: Sum of angles of adjacent angles on a" ]

[ "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of", "dir(D-F)*I, deepmagenta); label(\"4\\sqrt{2}\", (D+E)/2, dir(E-D)*I, deepmagenta); label(\"a\", (B+X)/2, dir(B-X)*I, gray(0.3)); label(\"a\\sqrt{2}\", (D+X)/2, dir(D-X)*I, gray(0.3)); [/asy]$ Notice that $$\\angle DFE=\\angle CFE-\\angle CFD=\\angle CBE-\\angle CAD=180-B-A=C.$$By the Law of Cosines, $$\\cos C=\\frac{AC^2+BC^2-AB^2}{2\\cdot AC\\cdot BC}=\\frac34.$$Then, $$DE^2=DF^2+EF^2-2\\cdot DF\\cdot EF\\cos C=32\\implies DE=4\\sqrt2.$$Let $X=\\overline{AB}\\cap\\overline{CF}$, $a=XB$, and $b=XD$. Then, $$XA\\cdot XD=XC\\cdot XF=XB\\cdot XE\\implies b(a+4)=a(b+4\\sqrt2)\\implies b=a\\sqrt2.$$However, since $\\triangle XFD\\sim\\triangle XAC$, $XF=\\tfrac{4+a}3$, but since $\\triangle XFE\\sim\\triangle XBC$, $$\\frac75=\\frac{4+a}{3a}\\implies a=\\frac54\\implies BE=a+a\\sqrt2+4\\sqrt2=\\frac{5+21\\sqrt2}4,$$and the requested sum is $5+21+2+4=\\boxed{032}$. (Solution by TheUltimate123) ## Solution 2 Define $\\omega_1$ to be the circumcircle of $\\triangle ACD$ and $\\omega_2$ to be the circumcircle of $\\triangle EBC$. Because of exterior angles, $\\angle ACB = \\angle CBE - \\angle CAD$ But $\\angle CBE = \\angle CFE$ because $CBFE$ is cyclic. In addition, $\\angle CAD = \\angle CFD$ because $CAFD$ is cyclic. Therefore, $\\angle ACB = \\angle CFE - \\angle CFD$. But $\\angle CFE - \\angle CFD = \\angle DFE$, so $\\angle ACB = \\angle DFE$. Using Law of Cosines on $\\triangle ABC$, we can figure out that $\\cos(\\angle ACB) = \\frac{3}{4}$. Since $\\angle ACB = \\angle DFE$, $\\cos(\\angle DFE) = \\frac{3}{4}$. We are given that $DF = 2$ and $FE = 7$, so we can use Law of Cosines on $\\triangle DEF$ to find that $DE = 4\\sqrt{2}$. Let $G$ be the intersection of segment $\\overline{AE}$ and $\\overline{CF}$. Using Power of a Point with respect to $G$", "# 2005 AIME II Problems/Problem 14 ## Problem In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\\overline{BC}$ with $CD=6.$ Point $E$ is on $\\overline{BC}$ such that $\\angle BAE\\cong \\angle CAD.$ Given that $BE=\\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ ## Solution 1 $[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label(\"A\",A,N); label(\"B\",B,SW); label(\"D\",D,SE); label(\"E\",E,S); label(\"C\",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]$ By the Law of Sines and since $\\angle BAE = \\angle CAD, \\angle BAD = \\angle CAE$, we have \\begin{align*} \\frac{CD \\cdot CE}{AC^2} &= \\frac{\\sin CAD}{\\sin ADC} \\cdot \\frac{\\sin CAE}{\\sin AEC} \\\\ &= \\frac{\\sin BAE \\sin BAD}{\\sin ADB \\sin AEB} \\\\ &= \\frac{\\sin BAE}{\\sin AEB} \\cdot \\frac{\\sin BAD}{\\sin ADB}\\\\ &= \\frac{BE \\cdot BD}{AB^2} \\end{align*} Substituting our knowns, we have $\\frac{CE}{BE} = \\frac{3 \\cdot 14^2}{2 \\cdot 13^2} = \\frac{BC - BE}{BE} = \\frac{15}{BE} - 1 \\Longrightarrow BE = \\frac{13^2 \\cdot 15}{463}$. The answer is $q = \\boxed{463}$. ## Solution 2 (Similar Triangles) Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of", "angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(circumcircle(A,B,D)); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot(\"A\", A, SW); dot(\"B\", B, dir(0)); dot(\"C\", C, N); dot(\"D\", D, dir(0)); dot(\"E\", E, N); dot(\"F\", F, dir(0)); [/asy]$ Applying the Law of Sines to $\\triangle BED$ and $\\triangle CFD$ gives$$DE = BD\\cdot\\frac{\\sin y}{\\sin x}\\text{~ and ~} DF = CD\\cdot\\frac{\\sin z}{\\sin x}.$$By the Angle Bisector Theorem, $BD = 2$ and $CD = 3$. Combining the above information yields $$[\\triangle AEF] = \\frac{3\\sin y\\cdot\\sin z\\cdot\\cos x}{\\sin^2 x}.$$Applying the Law of Cosines to $\\triangle ABC$ gives $\\cos 2x = \\frac9{16}$, $\\cos 2y = \\frac1{8}$, and $\\cos 2z = \\frac34$. By the Half Angle Formulas,$$\\sin^2x = \\frac7{32},~~ \\cos x = \\sqrt{\\frac{25}{32}},~~ \\sin y = \\sqrt{\\frac7{16}}, \\text{~ and ~} \\sin z = \\sqrt{\\frac18}.$$Therefore $$[\\triangle AEF] = \\frac{3\\cdot\\sqrt{\\frac{7}{16}}\\cdot\\sqrt{\\frac{1}{8}}\\cdot\\sqrt{\\frac{25}{32}}} {\\frac{7}{32}} = \\frac{15\\sqrt{7}}{14}.$$The requested sum is $15+7+14 = 36$. ## Solution 10 (Official MAA 2) Let the point $M$ be the midpoint of $\\overline{AD}$, let $I$ be the incenter of $\\triangle ABC$ which is the common point of lines $AD$, $BE$, and $CF$, and let $r$ be the inradius of $\\triangle ABC$. The semiperimeter of", "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "area of a triangle equals $\\frac{1}{2} \\cdot ab \\cdot \\sin(C)$. If angle $DAE = Z$, we have that $$\\frac{[ADE]}{[ABC]} = \\frac{\\frac{1}{2} \\cdot 19 \\cdot 14 \\cdot \\sin(Z)}{\\frac{1}{2} \\cdot 25 \\cdot 42 \\cdot \\sin(Z)} = \\frac{19 \\cdot 14}{25 \\cdot 42} = \\frac{ab}{cd}$$. - SuperJJ ## Video Solution CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00 ## Solution 2 (no trig) $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, W); label(\"E\", E, NE); label(\"19\", (A + D)/2, W); label(\"6\", (B + D)/2, W); label(\"14\", (A + E)/2, NE); label(\"28\", (C + E)/2, NE); [/asy]$ We can let $[ADE]=x$. Since $EC=2 \\cdot EA$, $[DEC]=2x$. So, $[ADC]=3x$. This means that $[BDC]=\\frac{6}{19}\\cdot3x=\\frac{18x}{19}$. Thus, $\\frac{[ADE]}{[BCED]} = \\frac{x}{\\frac{18x}{19}+2x}= \\boxed{\\textbf{(D) }\\frac{19}{56}}.$ -Conantwiz2023 ## Solution 3 (trig) The area of a triangle is $\\frac{1}{2} bc\\sin A$. Using this formula: $[ADE]=\\frac{1}{2}\\cdot19\\cdot14\\cdot\\sin A = 133\\sin A$ $[ABC]=\\frac{1}{2}\\cdot25\\cdot42\\cdot\\sin A = 525\\sin A$ Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$, $[BCED] = 525\\sin A - 133\\sin A = 392\\sin A$. Therefore, the desired ratio is $\\frac{133\\sin A}{392\\sin A}=\\boxed{\\textbf{(D) }\\frac{19}{56}}$ Note: $BC=39$ was not used in this problem. ## Solution 4 Let", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "\\pm \\frac{\\sqrt{90000}}{2} \\\\ &= 350 \\pm \\frac{300}{2} \\\\ &= 200, 500 \\end{align*} If the length of $AD$ is $200$, then quadrilateral $ABCD$ would be a square and thus, the radius of the circle would be $$\\frac{\\sqrt{200^2 + 200^2}}{2} = \\frac{\\sqrt{80000}}{2} = \\frac{200\\sqrt{2}}{2} = 100\\sqrt{2}$$ Which is a contradiction. Therefore, our answer is $\\boxed{500}.$ ## Solution 3 (Trigonometry Bash) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,E); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Angle mark for BOC draw(anglemark(C,O,B)); [/asy]$ Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply law of cosines on $\\Delta BOC$; let $\\theta = \\angle BOC$. We get the following equation: $$(BC)^{2}=(OB)^{2}+(OC)^{2}-2\\cdot OB \\cdot OC\\cdot \\cos\\theta$$ Substituting the values in, we get $$(200)^{2}=2\\cdot (200)^{2}+ 2\\cdot (200)^{2}- 2\\cdot 2\\cdot (200)^{2}\\cdot \\cos\\theta$$ Canceling out, we get $$\\cos\\theta=\\frac{3}{4}$$ Because $\\angle AOB$, $\\angle BOC$, and $\\angle COD$ are congruent, $\\angle AOD = 3\\theta$. To find", "get $$b^2+9a^2-2\\cdot 3a\\cdot b\\cdot \\cos(\\angle ADC)=7\\qquad (5)$$ $$b^2+a^2+2\\cdot a\\cdot b\\cdot \\cos(\\angle ADC)=9.\\qquad (6)$$ $(5)+(6)$, and according to $(4)$, we can get $$37a^2+2b^2=48. \\qquad (7)$$ Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\\frac{\\sqrt{22}}{2}$. Finally, we use Law of Cosines for $\\triangle ADB$, $$4(\\frac{\\sqrt{22}}{2})^2+1+2\\cdot2(\\frac{\\sqrt{22}}{2})\\cdot \\cos(ADC)=AB^2$$ then $AB=2\\sqrt{7}$, so the height of this $\\triangle ABC$ is $\\sqrt{4^2-(\\sqrt{7})^2}=3$. Then the area of $\\triangle ABC$ is $3\\sqrt{7}$, so the answer is $\\boxed{010}$. ### Solution 3 Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$.", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "thus can be discounted. This relationship, in combination with the Pythagorean Theorem in $\\triangle ACD$, gives $AC^2+CD^2=\\left(\\sqrt{2}\\right)^2+\\left(2-DA\\right)^2=DA^2$. Hence $2 + 4 - 4DA + DA^2 = DA^2$, so $DA = \\frac{3}{2}$, and thus $CD = \\frac{1}{2}$. Next, since $\\angle BAC = 45^{\\circ}$, $\\sin{\\left(\\angle BAC\\right)} = \\cos{\\left(\\angle BAC\\right)} = \\frac{1}{\\sqrt{2}}$. Using the lengths found above, $\\sin{\\left(\\angle CAD\\right)} = \\frac{\\left(\\frac{1}{2}\\right)}{\\left(\\frac{3}{2}\\right)} = \\frac{1}{3}$, and $\\cos{\\left(\\angle CAD\\right)} = \\frac{\\sqrt{2}}{\\left(\\frac{3}{2}\\right)} = \\frac{2 \\sqrt{2}}{3}$. Thus, by the addition formulae for $\\sin$ and $\\cos$, we have $$\\begin{split}\\sin{\\left(\\angle BAD\\right)}&=\\sin{\\left(\\angle BAC + \\angle CAD\\right)}\\\\&=\\sin{\\left(\\angle BAC\\right)}\\cos{\\left(\\angle CAD\\right)}+\\cos{\\left(\\angle BAC\\right)}\\sin{\\left(\\angle CAD\\right)}\\\\&=\\frac{1}{\\sqrt{2}}\\cdot\\frac{2 \\sqrt{2}}{3} + \\frac{1}{\\sqrt{2}}\\cdot\\frac{1}{3} \\\\ &= \\frac{2 \\sqrt{2} + 1}{3 \\sqrt{2}}\\end{split}$$ and $$\\begin{split}\\cos{\\left(\\angle BAD\\right)}&=\\cos{\\left(\\angle BAC + \\angle CAD\\right)}\\\\&=\\cos{\\left(\\angle BAC\\right)}\\cos{\\left(\\angle CAD\\right)}-\\sin{\\left(\\angle BAC\\right)}\\sin{\\left(\\angle CAD\\right)}\\\\&=\\frac{1}{\\sqrt{2}}\\cdot\\frac{2 \\sqrt{2}}{3} - \\frac{1}{\\sqrt{2}}\\cdot\\frac{1}{3} \\\\ &= \\frac{2 \\sqrt{2} - 1}{3 \\sqrt{2}}\\end{split}$$ Hence, by the double angle formula for $\\sin$, $\\sin{\\left(2\\angle BAD\\right)} = 2\\sin{\\left(\\angle BAD\\right)}\\cos{\\left(\\angle BAD\\right)} = \\frac{2(8-1)}{18} = \\boxed{\\textbf{(D) } \\frac{7}{9}}$. ### Solution 2 (coordinate geometry) We use the Pythagorean Theorem, as in Solution 1, to find $AD=\\frac{3}{2}$ and $CD=\\frac{1}{2}$. Now notice that the angle between $CD$ and the vertical (i.e. the $y$-axis) is $45^{\\circ}$ – to see this, drop a perpendicular from $D$ to $BA$ which meets $BA$ at $E$, and use the fact that the angle sum of quadrilateral $CBED$ must be $360^{\\circ}$. Anyway, this implies that the line $CD$ has slope $1$, so since $C$ is", "# Difference between revisions of \"2005 AIME II Problems/Problem 14\" ## Problem In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\\overline{BC}$ with $CD=6.$ Point $E$ is on $\\overline{BC}$ such that $\\angle BAE\\cong \\angle CAD.$ Given that $BE=\\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ ## Solution 1 $[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label(\"A\",A,N); label(\"B\",B,SW); label(\"D\",D,SE); label(\"E\",E,S); label(\"C\",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]$ By the Law of Sines and since $\\angle BAE = \\angle CAD, \\angle BAD = \\angle CAE$, we have \\begin{align*} \\frac{CD \\cdot CE}{AC^2} &= \\frac{\\sin CAD}{\\sin ADC} \\cdot \\frac{\\sin CAE}{\\sin AEC} \\\\ &= \\frac{\\sin BAE \\sin BAD}{\\sin ADB \\sin AEB} \\\\ &= \\frac{\\sin BAE}{\\sin AEB} \\cdot \\frac{\\sin BAD}{\\sin ADB}\\\\ &= \\frac{BE \\cdot BD}{AB^2} \\end{align*} Substituting our knowns, we have $\\frac{CE}{BE} = \\frac{3 \\cdot 14^2}{2 \\cdot 13^2} = \\frac{BC - BE}{BE} = \\frac{15}{BE} - 1 \\Longrightarrow BE = \\frac{13^2 \\cdot 15}{463}$. The answer is $q = \\boxed{463}$. ## Solution 2 (Similar Triangles) Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively.", "and multiplying these together proves the formula for isogonal lines. Hence, we have $$\\frac{BE}{15-BE}\\cdot \\frac{9}{6}=\\frac{169}{196}\\implies BE=\\frac{2535}{463}$$ so our desired answer is $\\boxed{463}.$ ## Solution 6 (Tangent subtraction formulas) Note: We first recall some helpful tips regarding 13, 14, 15 triangles. Drawing an altitude H from B to AC results in AHB being a 5-12-13 right triangle and CHB being a 3-4-5 (9-12-15) right triangle. $[asy] import olympiad; import cse5; import geometry; size(300); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label(\"A\",A,N); label(\"B\",B,SW); label(\"D\",D,SE); label(\"E\",E,S); label(\"C\",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); pair G = foot(E,C,A); pair F = foot(D,C,A); draw(D--F); draw(E--G); label(\"G\",G,N); label(\"F\",F,N); [/asy]$ Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling $\\angle DAG = \\alpha$. Now we know that $\\overline{DG} = \\frac{24}{5}$ and $\\overline{GC} = \\frac{18}{5}$. Therefore, $\\overline{AG} = \\frac{52}{5}$, so $\\tan{(\\alpha)} = \\frac{6}{13}$. Our goal now is to use tangent $\\angle EAG$ in triangle $AEG$. We set $\\overline{BE}$ to $x$, so $\\overline{ED} = 9 - x$ and $\\overline{EC} = 15 - x$, so $\\overline{EG} = \\frac{4}{5}(15-x)$ and $\\overline{GC} = \\frac{3}{5}(15-x)$ so $\\overline{AG} = \\frac{3x+25}{5}$. Now we just need tangent of $\\angle EAG$. We find this using $\\tan{(EAG)} =", "draw(A--F); draw(X--F); draw(E--F); draw(circumcircle(A,B,C)); draw(circumcircle(M,F,E)); dot(D); dot(F); dot(A); dot(B); dot(C); dot(E); dot(X); dot(R); dot(I); label(\"A\",A,dir(220)); label(\"B\",B,dir(110)); label(\"C\",C,dir(250)); label(\"D\",D,dir(60)); label(\"E\",E,dir(0)); label(\"F\",F,dir(315)); label(\"X\",X,dir(180)); [/asy]$ First of all, use the Angle Bisector Theorem to find that $BD=35/8$ and $CD=21/8$, and use Stewart's Theorem to find that $AD=15/8$. Then use Power of a Point to find that $DE=49/8$. Then use the circumradius of a triangle formula to find that the length of the circumradius of $\\triangle ABC$ is $\\frac{7\\sqrt{3}}{3}$. Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of the circumcircle of $\\triangle ABC$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\frac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=a$, and $DF=b$. Then, by the Pythagorean Theorem, $$x^2+b^2=\\left(\\frac{49}{8}\\right)^2=\\frac{2401}{64}$$ and $$x^2+(a+b)^2=\\left(\\frac{14\\sqrt{3}}{3}\\right)^2=\\frac{196}{3}.$$ Subtracting the first equation from the second, the $x^2$ term cancels out and we obtain: $$(a+b)^2-b^2=\\frac{196}{3}-\\frac{2401}{64}$$ $$a^2+2ab = \\frac{5341}{192}.$$ By Power of a Point, $ab=BD \\cdot DC=735/64=2205/192$, so $$a^2+2 \\cdot \\frac{2205}{192}=\\frac{5341}{192}$$ $$a^2=\\frac{931}{192}.$$ Since $a=XD$, $XD=\\frac{7\\sqrt{19}}{8\\sqrt{3}}$. Because $\\angle EXF$ and $\\angle EAF$ intercept the same arc in circle $\\omega$ and the same goes for $\\angle XFA$ and $\\angle XEA$, $\\angle EXF\\cong\\angle EAF$ and $\\angle XFA\\cong\\angle XEA$. Therefore, $\\triangle XDE\\sim\\triangle ADF$ by AA Similarity. Since side lengths in similar triangles are proportional, $$\\frac{AF}{\\frac{15}{8}}=\\frac{\\frac{14\\sqrt{3}}{3}}{\\frac{7\\sqrt{19}}{8\\sqrt{3}}}$$ $$\\frac{AF}{\\frac{15}{8}}=\\frac{16}{\\sqrt{19}}$$ $$AF \\cdot \\sqrt{19} = 30$$ $$AF = \\frac{30}{\\sqrt{19}}.$$", "dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ ### Solution 3 Again, draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. We find by angle chasing the same way as in solution 2 that $m\\angle ABE = 65^\\circ$ and $m\\angle DCE = 115^\\circ$. Applying the Law of Sines to $\\triangle AEB$ and $\\triangle EDC$, it follows that $DE = \\frac{2\\sin 115^\\circ}{\\sin \\angle DEC} = \\frac{2\\sin 65^\\circ}{\\sin \\angle AEB} = EA$, so $\\triangle AED$ is isosceles. We finish as we did in solution 2. $[asy] unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120); draw(A--B--C--D--cycle); pair P=intersectionpoint(B--D,C--A); draw(A--C); draw(B--D); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"E\",P,W); [/asy]$ ## Solution 4 Start off with the same diagram as solution 1. Now draw $\\overline{CA}$ which creates isosceles $\\triangle EAB$. We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that its is $\\boxed{85°}$ (Error compiling LaTeX. ! Package inputenc Error: Unicode char \\u8:° not set up for use with LaTeX.)", "49 \\qquad \\Longrightarrow\\qquad k=\\frac 7{\\sqrt{19}}$$ Use Ptolemy's Theorem on $BFCA$, to get $$15k+15k=7\\cdot AF, \\qquad \\Longrightarrow\\qquad AF= \\frac{30}{\\sqrt{19}},$$ so $AF^2=\\frac{900}{19}$ and the answer is $900+19=919$ ~abacadaea ## Solution 5 Denote $AB = c, BC = a, AC = b, \\angle A = 2 \\alpha.$ Let M be midpoint BC. Let $\\theta$ be the circle centered at $A$ with radius $\\sqrt{AB \\cdot AC} =\\sqrt{bc}.$ We calculate the length of some segments. The median $AM = \\sqrt{\\frac {b^2}{2} + \\frac {c^2}{2} - \\frac {a^2}{4}}.$ The bisector $AD = \\frac {2 b c \\cos \\alpha}{b+c}.$ One can use Stewart's Theorem in both cases. $AD$ is bisector of $\\angle A \\implies BD = \\frac {a c}{b + c}, CD = \\frac {a b}{b + c} \\implies$ $$BD \\cdot CD = \\frac {a^2 bc }{(b+c)^2}.$$ We use Power of Point $D$ and get $AD \\cdot DE = BD \\cdot CD.$ $$AE = AD + DE = AD + \\frac {BD \\cdot CD}{AD},$$ $$AE =\\frac {2 b c \\cos \\alpha}{b+c} + \\frac {a^2 bc \\cdot (b+c) }{(b+c)^2 \\cdot 2 b c \\cos \\alpha} =$$ $$= \\frac {b c \\cos^2 \\alpha + a^2}{2(b+c)\\cos \\alpha} =\\frac {4bc \\cos^2 \\alpha + b^2 +c^2 -2 b c \\cos 2\\alpha}{2(b+c) \\cos \\alpha} = \\frac {b+c}{2} \\implies AD \\cdot AE = 2 bc \\cos \\alpha.$$ We consider the inversion with", "By the Law of Cosines on $\\triangle AA'D$, we have $$AD^2=10^2+\\left(\\frac {130}{23}\\right)^2-2 \\cdot 10 \\cdot \\frac {130}{23} \\cdot \\frac {5}{13}=\\frac {2^4 \\cdot 3^2 \\cdot 5^2 \\cdot 13}{23^2} \\Longrightarrow AD=\\frac {60\\sqrt {13}}{23}$$ and the answer is $60+13+23=\\boxed{096}$. $[asy] size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype(\"4 4\")); MP(\"5\", (A+B)/2, f); MP(\"13\", (A+C)/2, NE,f); MP(\"A\",D(A),f); MP(\"B\",D(B),f); MP(\"C\",D(C),N,f); MP(\"A'\",D(E),f); MP(\"D\",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30)); [/asy]$ -azjps Solution 6 (Law of Sines) $[asy]pathpen = linewidth(0.7); pen f = fontsize(10); size(144); pair B=(0,0); pair A=(5,0); pair C=(0,12); pair E=(-5,0); pair abA = A+unit(C-A)+unit(B-A); pair D = extension(A,abA,C,E); D(A--C); D(C--B); D(A--B); D(C--D); D(A--D); D(D--E,dashed); D(B--E,dashed); D(rightanglemark(C,B,A,20),red); MP(\"A\",D(A),plain.SE,f); MP(\"B\",D(B),plain.S,f); MP(\"C\",D(C),plain.N,f); MP(\"D\",D(D),plain.NW,f); MP(\"E\",D(E),plain.SW,f); MP(\"5\",(A+B)/2,plain.S,f); MP(\"12\",(B+C)/2,plain.E,f); MP(\"13\",(A+C)/2,plain.NE,f); MP(\"\\alpha\",A,4*(unit(D-A)+unit(C-A))/2,f); MP(\"\\alpha\",A,4*(unit(B-A)+unit(D-A))/2,f); MP(\"2\\alpha\",E,4*(unit(C-E)+unit(B-E))/2,f); MP(\"\\beta\",C,6*(unit(A-C)+unit(B-C))/2,f); MP(\"\\beta\",C,6*(unit(E-C)+unit(B-C))/2,f);[/asy]$ Using the law of sines on $\\bigtriangleup ADE$, \\begin{align*} AD &= AE \\cdot \\dfrac{\\sin(2\\alpha)}{\\sin(180^\\circ-3\\alpha)}\\\\ &= AE \\cdot \\dfrac{\\sin(2\\alpha)}{\\sin(\\alpha+2\\alpha)}\\\\ &= 10 \\cdot \\dfrac{\\sin(2\\alpha)}{\\sin(\\alpha)\\cos(2\\alpha)+\\cos(\\alpha)\\sin(2\\alpha)}\\\\ &= 10 \\cdot \\dfrac{\\sin(2\\alpha)}{\\sqrt{\\dfrac{1-\\cos(2\\alpha)}{2}}\\cdot\\cos(2\\alpha)+\\sqrt{\\dfrac{1+\\cos(2\\alpha)}{2}}\\cdot\\sin(2\\alpha)}\\\\ &= 10 \\cdot \\dfrac{\\dfrac{12}{13}}{\\sqrt{\\dfrac{8}{26}}\\cdot\\dfrac{5}{13}+\\sqrt{\\dfrac{18}{26}}\\cdot\\dfrac{12}{13}}\\\\ &= 10 \\cdot \\dfrac{12\\sqrt{13}}{10+36} = \\dfrac{60\\sqrt{13}}{23} \\end{align*} $\\therefore$ the answer is $60+13+23=\\boxed{086}$. -m1sterzer0", "degrees(A)); pair E = dir(5*angleUnit + degrees(F)); draw(A--B--C--D--E--F--cycle); dot(\"A\",A,A); dot(\"B\",B,B); dot(\"C\",C,C); dot(\"D\",D,D); dot(\"E\",E,E); dot(\"F\",F,F); draw(A--D^^B--E^^C--F); label(\"3\",D--C,SW); label(\"3\",B--C,S); label(\"3\",A--B,SE); label(\"5\",A--F,NE); label(\"5\",F--E,N); label(\"5\",D--E,NW); [/asy]$ Furthermore, $\\angle BAD$ and $\\angle BED$ subtend the same arc, as do $\\angle ABE$ and $\\angle ADE$. Hence triangles $ABQ$ and $EDQ$ are similar. Therefore, $$\\frac{AQ}{EQ}=\\frac{BQ}{DQ}=\\frac{AB}{ED}=\\frac{3}{5}.$$ It follows that $$\\frac{\\frac{AD-PQ}{2}}{PQ+5} =\\frac{3}{5}\\quad \\mbox {and}\\quad \\frac{3-PQ}{\\frac{AD+PQ}{2}}=\\frac{3}{5}.$$ Solving the two equations simultaneously yields $AD=360/49,$ so $m+n=\\boxed{409}. \\blacksquare$ ## Solution 2 All angle measures are in degrees. Let the first trapezoid be $ABCD$, where $AB=BC=CD=3$. Then the second trapezoid is $AFED$, where $AF=FE=ED=5$. We look for $AD$. Since $ABCD$ is an isosceles trapezoid, we know that $\\angle BAD=\\angle CDA$ and, since $AB=BC$, if we drew $AC$, we would see $\\angle BCA=\\angle BAC$. Anyway, $\\widehat{AB}=\\widehat{BC}=\\widehat{CD}$ ($\\widehat{AB}$ means arc AB). Using similar reasoning, $\\widehat{AF}=\\widehat{FE}=\\widehat{ED}$. Let $\\widehat{AB}=2\\phi$ and $\\widehat{AF}=2\\theta$. Since $6\\theta+6\\phi=360$ (add up the angles), $2\\theta+2\\phi=120$ and thus $\\widehat{AB}+\\widehat{AF}=\\widehat{BF}=120$. Therefore, $\\angle FAB=\\frac{1}{2}\\widehat{BDF}=\\frac{1}{2}(240)=120$. $\\angle CDE=120$ as well. Now I focus on triangle $FAB$. By the Law of Cosines, $BF^2=3^2+5^2-30\\cos{120}=9+25+15=49$, so $BF=7$. Seeing $\\angle ABF=\\theta$ and $\\angle AFB=\\phi$, we can now use the Law of Sines to get: $$\\sin{\\phi}=\\frac{3\\sqrt{3}}{14}\\;\\text{and}\\;\\sin{\\theta}=\\frac{5\\sqrt{3}}{14}.$$ Now I focus on triangle $AFD$. $\\angle AFD=3\\phi$ and $\\angle ADF=\\theta$, and we are given that $AF=5$, so $$\\frac{\\sin{\\theta}}{5}=\\frac{\\sin{3\\phi}}{AD}.$$ We know $\\sin{\\theta}=\\frac{5\\sqrt{3}}{14}$, but we need to find $\\sin{3\\phi}$. Using various identities, we see \\begin{align*}\\sin{3\\phi}&=\\sin{(\\phi+2\\phi)}=\\sin{\\phi}\\cos{2\\phi}+\\cos{\\phi}\\sin{2\\phi}\\\\ &=\\sin{\\phi}(1-2\\sin^2{\\phi})+2\\sin{\\phi}\\cos^2{\\phi}\\\\ &=\\sin{\\phi}\\left(1-2\\sin^2{\\phi}+2(1-\\sin^2{\\phi})\\right)\\\\", "\\triangle DPB,$ $\\angle APC\\, \\, = \\angle DPB$ ( vertically opposite angle) $\\angle CAP\\, \\, = \\angle BDP$ (Angles in the same segment) $\\triangle APC\\, \\, \\sim \\triangle DPB$ (By AA) $\\frac{AP}{DP}=\\frac{PC}{PB}=\\frac{CA}{BD}$ (Corresponding sides of similar triangles are proportional) $\\Rightarrow \\frac{AP}{DP}=\\frac{PC}{PB}$ $\\Rightarrow AP.PB=PC.DP$ In $\\Delta PAC \\,and \\,\\Delta PDB,$ $\\angle P=\\angle P$ (Common) $\\angle PAC=\\angle PDB$ (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle) So, $\\Delta PAC \\sim \\Delta PDB$ ( By AA rule) In $\\Delta PAC \\,and \\,\\Delta PDB,$ $\\angle P=\\angle P$ (Common) $\\angle PAC=\\angle PDB$ (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle) So, $\\Delta PAC \\sim \\Delta PDB$ ( By AA rule) 24440$\\frac{AP}{DP}=\\frac{PC}{PB}=\\frac{CA}{BD}$ (Corresponding sides of similar triangles are proportional) $\\Rightarrow \\frac{AP}{DP}=\\frac{PC}{PB}$ $\\Rightarrow AP.PB=PC.DP$ Produce BA to P, such that AP=AC and join P to C. $\\frac{BD }{CD} = \\frac{AB}{AC}$ (Given ) $\\Rightarrow \\frac{BD }{CD} = \\frac{AP}{AC}$ Using converse of Thales theorem, AD||PC $\\Rightarrow \\angle BAD=\\angle APC............1$(Corresponding angles) $\\Rightarrow \\angle DAC=\\angle ACP............2$ (Alternate angles) By construction, AP=AC $\\Rightarrow \\angle APC=\\angle ACP............3$ From equation 1,2,3, we get $\\Rightarrow \\angle BAD=\\angle APC$ Let AB = 1.8 m BC is a horizontal distance between fly to the tip of the rod. Then, the length of the string is AC. In $\\triangle$ ABC, using Pythagoras theorem $AC^2=AB^2+BC^2$ $\\Rightarrow AC^2=(1.8)^2+(2.4)^2$ $\\Rightarrow AC^2=3.24+5.76$" ]

[ "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of", "dir(D-F)*I, deepmagenta); label(\"4\\sqrt{2}\", (D+E)/2, dir(E-D)*I, deepmagenta); label(\"a\", (B+X)/2, dir(B-X)*I, gray(0.3)); label(\"a\\sqrt{2}\", (D+X)/2, dir(D-X)*I, gray(0.3)); [/asy]$ Notice that $$\\angle DFE=\\angle CFE-\\angle CFD=\\angle CBE-\\angle CAD=180-B-A=C.$$By the Law of Cosines, $$\\cos C=\\frac{AC^2+BC^2-AB^2}{2\\cdot AC\\cdot BC}=\\frac34.$$Then, $$DE^2=DF^2+EF^2-2\\cdot DF\\cdot EF\\cos C=32\\implies DE=4\\sqrt2.$$Let $X=\\overline{AB}\\cap\\overline{CF}$, $a=XB$, and $b=XD$. Then, $$XA\\cdot XD=XC\\cdot XF=XB\\cdot XE\\implies b(a+4)=a(b+4\\sqrt2)\\implies b=a\\sqrt2.$$However, since $\\triangle XFD\\sim\\triangle XAC$, $XF=\\tfrac{4+a}3$, but since $\\triangle XFE\\sim\\triangle XBC$, $$\\frac75=\\frac{4+a}{3a}\\implies a=\\frac54\\implies BE=a+a\\sqrt2+4\\sqrt2=\\frac{5+21\\sqrt2}4,$$and the requested sum is $5+21+2+4=\\boxed{032}$. (Solution by TheUltimate123) ## Solution 2 Define $\\omega_1$ to be the circumcircle of $\\triangle ACD$ and $\\omega_2$ to be the circumcircle of $\\triangle EBC$. Because of exterior angles, $\\angle ACB = \\angle CBE - \\angle CAD$ But $\\angle CBE = \\angle CFE$ because $CBFE$ is cyclic. In addition, $\\angle CAD = \\angle CFD$ because $CAFD$ is cyclic. Therefore, $\\angle ACB = \\angle CFE - \\angle CFD$. But $\\angle CFE - \\angle CFD = \\angle DFE$, so $\\angle ACB = \\angle DFE$. Using Law of Cosines on $\\triangle ABC$, we can figure out that $\\cos(\\angle ACB) = \\frac{3}{4}$. Since $\\angle ACB = \\angle DFE$, $\\cos(\\angle DFE) = \\frac{3}{4}$. We are given that $DF = 2$ and $FE = 7$, so we can use Law of Cosines on $\\triangle DEF$ to find that $DE = 4\\sqrt{2}$. Let $G$ be the intersection of segment $\\overline{AE}$ and $\\overline{CF}$. Using Power of a Point with respect to $G$", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "\\angle DCE={ 60 }^{ 0 }$$ Recall that the sides of a 30-60-90 triangle are in the proportion $$x:x\\sqrt { 3 } :2x$$. .Or if you are familiar with trigonometry, just recall that $$\\sin { { 30 }^{ 0 } } =\\frac { 1 }{ 2 } =\\frac { perpendicular }{ hypotenuse } \\\\ Therefore,\\quad \\frac { CE }{ CD } =\\frac { 1 }{ 2 } =\\frac { x }{ CD } ,\\quad CD=2x\\\\ \\frac { EC }{ AE } =\\frac { DC }{ BA } \\\\ \\frac { x }{ x+\\sqrt { 2 } } =\\frac { 2x }{ BA }$$ Area of semicircle = $$\\frac { \\pi { r }^{ 2 } }{ 2 } =16\\pi$$ $${ r }^{ 2 }=36,\\quad r=4\\sqrt { 2 } =OB\\\\ \\frac { x }{ x+\\sqrt { 2 } } =\\frac { 2x }{ 8\\sqrt { 2 } } \\\\ x+\\sqrt { 2 } =4\\sqrt { 2 } \\\\ x=3\\sqrt { 2 } \\\\ Therefore,\\quad CE=3\\sqrt { 2 } ,\\quad CD=6\\sqrt { 2 }$$ Side of $$\\triangle CDE$$ are in ratio $$x:x\\sqrt { 3 } :2x$$ $$CE=3\\sqrt { 2 } ,\\quad CD=6\\sqrt { 2 } \\\\ ED=3\\sqrt { 6 } \\\\ Therefore,\\quad DE=3\\sqrt { 6 }$$ (Quantity A is greater than Quantity B). Problem 13) What is the area of", "be on the circle, such that Triangle $ABC$ has all angles less than $120^{\\circ}$ is OUTSIDE of the red outline, but inside of the green outline shown in the graph below: $[asy] unitsize(0.7cm); draw((-1,0)--(7,0),Arrows); draw((0,-1)--(0,7),Arrows); draw((0,6)--(6,6)--(6,0)--(0,0)--cycle,green+linewidth(1)); draw((1,1)--(3,1)--(5,3)--(5,5)--(3,5)--(1,3)--cycle, red); label(\"B\",(7,0),SE); label(\"C\",(0,7),NW); [/asy]$ The red hexagon has vertices at coordinates $(60,60),(180,60),(300,180),(300,300),(180,300),$ and $(60,180)$, while the green square has vertices at coordinates $(0,0),(0,360),(360,360),$ and $(360,0)$. Each other these coordinates represent positions for points $B$ and $C$. For example, $(240, 181)$ means point $B$ is at $(240^{\\circ})$ on the unit circle, while point $C$ is at $(181^{\\circ})$ on the unit circle. Therefore, the probability of points $A, B,$ and $C$ forming a triangle with angles less than $120^{\\circ}$ is: $\\[1-\\frac{2\\cdot120^2+2\\cdot\\frac{120^2}{2}}{360^2}$$ $$=1-\\frac{1}{3}$$ $$=\\frac{2}{3}$$. And the answer is $2+3=\\boxed{005}$ - Solution by adyj", "# 2005 AIME II Problems/Problem 14 ## Problem In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\\overline{BC}$ with $CD=6.$ Point $E$ is on $\\overline{BC}$ such that $\\angle BAE\\cong \\angle CAD.$ Given that $BE=\\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ ## Solution 1 $[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label(\"A\",A,N); label(\"B\",B,SW); label(\"D\",D,SE); label(\"E\",E,S); label(\"C\",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]$ By the Law of Sines and since $\\angle BAE = \\angle CAD, \\angle BAD = \\angle CAE$, we have \\begin{align*} \\frac{CD \\cdot CE}{AC^2} &= \\frac{\\sin CAD}{\\sin ADC} \\cdot \\frac{\\sin CAE}{\\sin AEC} \\\\ &= \\frac{\\sin BAE \\sin BAD}{\\sin ADB \\sin AEB} \\\\ &= \\frac{\\sin BAE}{\\sin AEB} \\cdot \\frac{\\sin BAD}{\\sin ADB}\\\\ &= \\frac{BE \\cdot BD}{AB^2} \\end{align*} Substituting our knowns, we have $\\frac{CE}{BE} = \\frac{3 \\cdot 14^2}{2 \\cdot 13^2} = \\frac{BC - BE}{BE} = \\frac{15}{BE} - 1 \\Longrightarrow BE = \\frac{13^2 \\cdot 15}{463}$. The answer is $q = \\boxed{463}$. ## Solution 2 (Similar Triangles) Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of", "a right angle. So by the Pythagorean Theorem we have $AB ~=~ \\sqrt{AE^2 - BE^2} ~=~ \\sqrt{5^2 - 2^2} ~=~ \\sqrt{21} ~\\text{ft}. \\nonumber$ By Equation \\ref{4.4} the arc $$\\overparen{BC}$$ has length $$BE \\cdot \\theta$$, where $$\\theta = \\angle\\,BEC$$ is the supplement of $$\\angle\\,AED + \\angle\\,AEB$$. So since $\\nonumber \\tan\\,\\angle\\,AED ~=~ \\frac{4}{3} ~\\Rightarrow~ \\angle\\,AED ~=~ 53.1^\\circ \\quad\\text{and}\\quad \\cos\\,\\angle\\,AEB ~=~ \\frac{BE}{AE} ~=~ \\frac{2}{5} ~\\Rightarrow~ \\angle\\,AEB ~=~ 66.4^\\circ ~,$ we have $\\nonumber \\theta ~=~ \\angle\\,BEC ~=~ 180^\\circ \\;-\\; (\\angle\\,AED + \\angle\\,AEB) ~=~ 180^\\circ \\;-\\; (53.1^\\circ + 66.4^\\circ) ~=~ 60.5^\\circ ~.$ Converting to radians, we get $$\\;\\theta = \\frac{\\pi}{180} \\;\\cdot\\; 60.5 = 1.06$$ rad. Thus, $\\nonumber L ~=~ 2\\,(AB \\;+\\; \\cdot \\overparen{BC}) ~=~ 2\\,(\\sqrt{21} \\;+\\; BE \\cdot \\theta) ~=~ 2\\,(\\sqrt{21} \\;+\\; (2)\\,( 1.06)) ~=~ \\boxed{13.4 ~\\text{ft}} ~.$ First, at the center $$B$$ of the pulley with radius $$8$$, draw a circle of radius $$3$$, which is the difference in the radii of the two pulleys. Let $$C$$ be the point where this circle intersects $$\\overline{BF}$$. Then we know that the tangent line $$\\overline{AC}$$ to this smaller circle is perpendicular to the line segment $$\\overline{BF}$$. Thus, $$\\angle\\,ACB$$ is a right angle, and so the length of $$\\overline{AC}$$ is $AC ~=~ \\sqrt{AB^2 - BC^2} ~=~ \\sqrt{15^2 - 3^2} ~=~ \\sqrt{216} ~=~ 6\\,\\sqrt{6}\\nonumber$ by the Pythagorean Theorem. Now since $$\\overline{AE} \\perp \\overline{EF}$$ and $$\\overline{EF} \\perp", "\\triangle DPB,$ $\\angle APC\\, \\, = \\angle DPB$ ( vertically opposite angle) $\\angle CAP\\, \\, = \\angle BDP$ (Angles in the same segment) $\\triangle APC\\, \\, \\sim \\triangle DPB$ (By AA) $\\frac{AP}{DP}=\\frac{PC}{PB}=\\frac{CA}{BD}$ (Corresponding sides of similar triangles are proportional) $\\Rightarrow \\frac{AP}{DP}=\\frac{PC}{PB}$ $\\Rightarrow AP.PB=PC.DP$ In $\\Delta PAC \\,and \\,\\Delta PDB,$ $\\angle P=\\angle P$ (Common) $\\angle PAC=\\angle PDB$ (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle) So, $\\Delta PAC \\sim \\Delta PDB$ ( By AA rule) In $\\Delta PAC \\,and \\,\\Delta PDB,$ $\\angle P=\\angle P$ (Common) $\\angle PAC=\\angle PDB$ (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle) So, $\\Delta PAC \\sim \\Delta PDB$ ( By AA rule) 24440$\\frac{AP}{DP}=\\frac{PC}{PB}=\\frac{CA}{BD}$ (Corresponding sides of similar triangles are proportional) $\\Rightarrow \\frac{AP}{DP}=\\frac{PC}{PB}$ $\\Rightarrow AP.PB=PC.DP$ Produce BA to P, such that AP=AC and join P to C. $\\frac{BD }{CD} = \\frac{AB}{AC}$ (Given ) $\\Rightarrow \\frac{BD }{CD} = \\frac{AP}{AC}$ Using converse of Thales theorem, AD||PC $\\Rightarrow \\angle BAD=\\angle APC............1$(Corresponding angles) $\\Rightarrow \\angle DAC=\\angle ACP............2$ (Alternate angles) By construction, AP=AC $\\Rightarrow \\angle APC=\\angle ACP............3$ From equation 1,2,3, we get $\\Rightarrow \\angle BAD=\\angle APC$ Let AB = 1.8 m BC is a horizontal distance between fly to the tip of the rod. Then, the length of the string is AC. In $\\triangle$ ABC, using Pythagoras theorem $AC^2=AB^2+BC^2$ $\\Rightarrow AC^2=(1.8)^2+(2.4)^2$ $\\Rightarrow AC^2=3.24+5.76$", "draw(A--F); draw(X--F); draw(E--F); draw(circumcircle(A,B,C)); draw(circumcircle(M,F,E)); dot(D); dot(F); dot(A); dot(B); dot(C); dot(E); dot(X); dot(R); dot(I); label(\"A\",A,dir(220)); label(\"B\",B,dir(110)); label(\"C\",C,dir(250)); label(\"D\",D,dir(60)); label(\"E\",E,dir(0)); label(\"F\",F,dir(315)); label(\"X\",X,dir(180)); [/asy]$ First of all, use the Angle Bisector Theorem to find that $BD=35/8$ and $CD=21/8$, and use Stewart's Theorem to find that $AD=15/8$. Then use Power of a Point to find that $DE=49/8$. Then use the circumradius of a triangle formula to find that the length of the circumradius of $\\triangle ABC$ is $\\frac{7\\sqrt{3}}{3}$. Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of the circumcircle of $\\triangle ABC$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\frac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=a$, and $DF=b$. Then, by the Pythagorean Theorem, $$x^2+b^2=\\left(\\frac{49}{8}\\right)^2=\\frac{2401}{64}$$ and $$x^2+(a+b)^2=\\left(\\frac{14\\sqrt{3}}{3}\\right)^2=\\frac{196}{3}.$$ Subtracting the first equation from the second, the $x^2$ term cancels out and we obtain: $$(a+b)^2-b^2=\\frac{196}{3}-\\frac{2401}{64}$$ $$a^2+2ab = \\frac{5341}{192}.$$ By Power of a Point, $ab=BD \\cdot DC=735/64=2205/192$, so $$a^2+2 \\cdot \\frac{2205}{192}=\\frac{5341}{192}$$ $$a^2=\\frac{931}{192}.$$ Since $a=XD$, $XD=\\frac{7\\sqrt{19}}{8\\sqrt{3}}$. Because $\\angle EXF$ and $\\angle EAF$ intercept the same arc in circle $\\omega$ and the same goes for $\\angle XFA$ and $\\angle XEA$, $\\angle EXF\\cong\\angle EAF$ and $\\angle XFA\\cong\\angle XEA$. Therefore, $\\triangle XDE\\sim\\triangle ADF$ by AA Similarity. Since side lengths in similar triangles are proportional, $$\\frac{AF}{\\frac{15}{8}}=\\frac{\\frac{14\\sqrt{3}}{3}}{\\frac{7\\sqrt{19}}{8\\sqrt{3}}}$$ $$\\frac{AF}{\\frac{15}{8}}=\\frac{16}{\\sqrt{19}}$$ $$AF \\cdot \\sqrt{19} = 30$$ $$AF = \\frac{30}{\\sqrt{19}}.$$", "Meanwhile, from right triangle $ABC,$ we have \\begin{align*} \\cos \\angle BAC = \\frac{6}{\\sqrt{40}} &= \\frac{3}{\\sqrt{10}}, \\\\ \\sin \\angle BAC = \\frac{2}{\\sqrt{40}} &= \\frac{1}{\\sqrt{10}}. \\end{align*} This means that by the angle subtraction formulas, \\begin{align*} \\cos \\angle OAB &= \\cos (\\angle OAC - \\angle BAC) \\\\ &= \\cos \\angle OAC \\cos \\angle BAC + \\sin \\angle OAC \\sin \\angle BAC \\\\ &= \\frac{1}{\\sqrt{5}} \\cdot \\frac{3}{\\sqrt{10}} + \\frac{2}{\\sqrt{5}} \\cdot \\frac{1}{\\sqrt{10}} \\\\ &= \\frac{5}{5 \\sqrt{2}} = \\frac{1}{\\sqrt{2}}. \\end{align*} Now we have all we need to use the law of cosines on $\\triangle OAB.$ This tells us that \\begin{align*} OB^2 &= AO^2 + AB^2 - 2 AO \\cdot AB \\cdot \\cos \\angle OAB \\\\ &= 50 + 36 - 2 \\cdot 5 \\sqrt{2} \\cdot 6 \\cdot \\frac{1}{\\sqrt{2}} \\\\ &= 86 - 2 \\cdot 5 \\cdot 6 \\\\ &= 26. \\end{align*} ### Solution 3 Drop perpendiculars from $O$ to $AB$ (with foot $T_1$), $M$ to $OT_1$ (with foot $T_2$), and $M$ to $AB$ (with foot $T_3$). Also, mark the midpoint $M$ of $AC$. $[asy] size(200); pair dl(string name, pair loc, pair offset) { dot(loc); label(name,loc,offset); return loc; }; pair a[] = {(0,0),(0,5),(1,5),(1,7),(-2,6),(-5,5),(-2,5),(-2,6),(0,6)}; string n[] = {\"O\",\"T_1\",\"B\",\"C\",\"M\",\"A\",\"T_3\",\"M\",\"T_2\"}; for(int i=0;i First notice that by computation, $OAC$ is a $\\sqrt {50} - \\sqrt {40} - \\sqrt {50}$ isosceles triangle, so $AC = MO$. Then, notice that $\\angle", "point E on BC ignoring point D. Then you get: $\\frac{CE}{BE} = \\frac{AC\\ sin(CAE)}{AB\\ sin(BAE)}$ 3. Since angle BAD = angle CAE and angle BAE = angle DAC, we can equate 1 & 2 and rearrange terms to yield the required result. Via the area method: we have $$\\frac{BD}{CE} = \\frac{[BAD]}{[CAE]} = \\frac{AB \\cdot AD}{AC \\cdot AE}$$ where $[XYZ]$ denotes the area of $\\triangle XYZ$. The first equation holds since $\\triangle BAD$ and $\\triangle CAE$ have the same height from vertex $A$, the second holds since they have the same angle at vertex $A$. More explicitly, if $H$ is the foot of the perpendicular from $A$, $$\\frac{BD}{CE} = \\frac{\\frac12 \\cdot BD \\cdot AH}{\\frac12 \\cdot CE \\cdot AH} = \\frac{[BAD]}{[CAE]} = \\frac{\\frac12 \\cdot AB \\cdot AD \\cdot \\sin \\angle BAD}{\\frac12 \\cdot AC \\cdot AE \\cdot \\sin \\angle CAE} = \\frac{AB \\cdot AD}{AC \\cdot AE}.$$ For the exact same reason, we have $$\\frac{BE}{CD} = \\frac{[BAE]}{[CAD]} = \\frac{AB \\cdot AE}{AC \\cdot AD}.$$ Multiply these together, and you get $$\\frac{BD}{CE} \\cdot \\frac{BE}{CD} = \\frac{AB^2}{AC^2}.$$ The condition is easy to prove assuming that $\\overline{AB}$ and $\\overline{AC}$ are tangents to the circumcircle Point $A$ lies on the circle, so it is not possible that both $AB$ and $AC$ be tangents. Alt. hint: let $\\angle BAD = \\theta$ then applying the law of sines a", "= b \\sin A \\rightarrow \\frac{a}{b} = \\sin A \\rightarrow \\frac{22}{31} = \\sin A \\rightarrow A = 45.2^\\circ$ 3. $a > b \\sin A \\rightarrow \\frac{a}{b} > \\sin A \\rightarrow \\frac{22}{31} > \\sin A \\rightarrow A < 45.2^\\circ$ 4. This problem can be done entirely with right triangle trig, but there are several different ways to solve this particular problem. $BD & = \\sqrt{13.7^2 - 9.8^2} = 9.6 && \\tan 42.6^\\circ = \\frac{9.8}{DC} \\rightarrow DC = 10.7 \\\\\\sin 42.6^\\circ & = \\frac{9.8}{AC} \\rightarrow AC = 14.5 && BC = 9.6 + 10.7 = 20.3 \\\\\\sin B & = \\frac{9.8}{13.7} \\rightarrow \\angle{B} = 45.7^\\circ && \\angle{A} = 180^\\circ - 45.7^\\circ - 42.6^\\circ = 91.7^\\circ$ 1. $\\angle{EBD} \\Rightarrow 7.6^2 = 9.9^2 + 10.2^2 - 2 \\cdot 9.9 \\cdot 10.2 \\cos EBD \\Rightarrow 44.4^\\circ$ 2. $\\angle{BDE} \\Rightarrow \\frac{\\sin BDE}{9.9} = \\frac{\\sin 44.4^\\circ}{7.6} \\Rightarrow 65.7^\\circ$ 3. $\\angle{DEB} \\Rightarrow 180^\\circ - 65.7^\\circ - 44.4^\\circ \\Rightarrow 69.9^\\circ$ 4. $\\angle{BDC} \\Rightarrow 180^\\circ - 65.7^\\circ \\Rightarrow 114.3^\\circ$ 5. $\\angle{BEA} \\Rightarrow 180^\\circ - 69.9^\\circ \\Rightarrow 110.1^\\circ$ 6. $\\angle{DBC} \\Rightarrow 180^\\circ - 114.3^\\circ - 21.8^\\circ \\Rightarrow 43.9^\\circ$ 7. $\\angle{ABE} \\Rightarrow 109.6^\\circ - 43.9^\\circ - 44.4^\\circ \\Rightarrow 21.3^\\circ$ 8. $\\angle{BAE} \\Rightarrow 180^\\circ - 21.3^\\circ - 110.1^\\circ \\Rightarrow 48.6^\\circ$ 9. $BC \\Rightarrow \\frac{\\sin 114.3^\\circ}{BC} = \\frac{\\sin 21.8^\\circ}{10.2} \\Rightarrow 25.0$ 10. $AB \\Rightarrow \\frac{\\sin 110.1^\\circ}{AB} = \\frac{\\sin 48.6^\\circ}{9.9} \\Rightarrow 12.4$ 11.", "the ratio of the area of $\\triangle DCE$ to the area of $\\triangle ABD$ is $\\frac{\\frac{\\sqrt{2}}{9}}{\\frac{\\sqrt{2}}{3}}$ = $\\frac{1}{3} \\Rightarrow C$ ## Solution 2 Since $\\triangle DCE$ and $\\triangle ABD$ share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from $C$ to $DE$. $[asy] import graph; import olympiad; pair O,A,B,C,D,E,F; O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774); draw(Circle((0,0),15)); draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B); label(\"A\",A,NE);label(\"B\",B,W);label(\"C\",C,SE);label(\"D\",D,NE);label(\"E\",E,SW);label(\"O\",O,SW);label(\"F\",F,NW); markscalefactor=0.2; draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue); [/asy]$ $OD=r, OC=\\frac{1}{3}r$. Since $m\\angle DCO=m\\angle DFC=90^\\circ$, then $\\triangle DCO\\cong \\triangle DFC$. So the ratio of the two altitudes is $\\frac{CF}{DC}=\\frac{OC}{DO}=\\frac{1}{3}\\Rightarrow \\text{(C)}$ ## Solution 3 Say the center of the circle is point $O$; Without loss of generality, assume $AC=2$, so $CB=4$ and the diameter and radius are $6$ and $3$, respectively. Therefore, $CO=1$, and $DO=3$. The area of $\\triangle DCE$ can be expressed as $\\frac{1}{2}(CD)(6)\\text{sin }(CDE).$ $\\frac{1}{2}(CD)(6)$ happens to be the area of $\\triangle ABD$. Furthermore, $\\text{sin } CDE = \\frac{CO}{DO},$ or $\\frac{1}{3}.$ Therefore, the ratio is $\\frac{1}{3}.$", "dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ ### Solution 3 Again, draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. We find by angle chasing the same way as in solution 2 that $m\\angle ABE = 65^\\circ$ and $m\\angle DCE = 115^\\circ$. Applying the Law of Sines to $\\triangle AEB$ and $\\triangle EDC$, it follows that $DE = \\frac{2\\sin 115^\\circ}{\\sin \\angle DEC} = \\frac{2\\sin 65^\\circ}{\\sin \\angle AEB} = EA$, so $\\triangle AED$ is isosceles. We finish as we did in solution 2. $[asy] unitsize(1cm); defaultpen(.8); real a=4; pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120); draw(A--B--C--D--cycle); pair P=intersectionpoint(B--D,C--A); draw(A--C); draw(B--D); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"E\",P,W); [/asy]$ ## Solution 4 Start off with the same diagram as solution 1. Now draw $\\overline{CA}$ which creates isosceles $\\triangle EAB$. We know that the angle bisector of an isosceles triangle splits it in half, we can extrapolate this further to see that its is $\\boxed{85°}$ (Error compiling LaTeX. ! Package inputenc Error: Unicode char \\u8:° not set up for use with LaTeX.)", "angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(circumcircle(A,B,D)); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot(\"A\", A, SW); dot(\"B\", B, dir(0)); dot(\"C\", C, N); dot(\"D\", D, dir(0)); dot(\"E\", E, N); dot(\"F\", F, dir(0)); [/asy]$ Applying the Law of Sines to $\\triangle BED$ and $\\triangle CFD$ gives$$DE = BD\\cdot\\frac{\\sin y}{\\sin x}\\text{~ and ~} DF = CD\\cdot\\frac{\\sin z}{\\sin x}.$$By the Angle Bisector Theorem, $BD = 2$ and $CD = 3$. Combining the above information yields $$[\\triangle AEF] = \\frac{3\\sin y\\cdot\\sin z\\cdot\\cos x}{\\sin^2 x}.$$Applying the Law of Cosines to $\\triangle ABC$ gives $\\cos 2x = \\frac9{16}$, $\\cos 2y = \\frac1{8}$, and $\\cos 2z = \\frac34$. By the Half Angle Formulas,$$\\sin^2x = \\frac7{32},~~ \\cos x = \\sqrt{\\frac{25}{32}},~~ \\sin y = \\sqrt{\\frac7{16}}, \\text{~ and ~} \\sin z = \\sqrt{\\frac18}.$$Therefore $$[\\triangle AEF] = \\frac{3\\cdot\\sqrt{\\frac{7}{16}}\\cdot\\sqrt{\\frac{1}{8}}\\cdot\\sqrt{\\frac{25}{32}}} {\\frac{7}{32}} = \\frac{15\\sqrt{7}}{14}.$$The requested sum is $15+7+14 = 36$. ## Solution 10 (Official MAA 2) Let the point $M$ be the midpoint of $\\overline{AD}$, let $I$ be the incenter of $\\triangle ABC$ which is the common point of lines $AD$, $BE$, and $CF$, and let $r$ be the inradius of $\\triangle ABC$. The semiperimeter of", "the radius, $120=17r$, we can just solve for the radius. We get $r=\\boxed{\\textbf{(B) }\\frac{120}{17}}$. This may seem like a lengthy explanation, but doing it yourself when you know what to do, it actually takes very little time. Try it yourself! ## Solution 7 Let us draw altitude $\\overline{CD}.$ This cuts our base into line segments with length $\\dfrac{16}{2}=8.$ Finding the area of the resulting triangles gives $[\\triangle ADC] = [\\triangle BDC] = \\dfrac{8 \\cdot 15}{2} = 60.$ Since $m\\angle CDB = m\\angle CDA = 90^\\circ,$ we use the Pythagorean Theorem to find length $\\overline{AC}$: \\begin{align*} (\\overline{AD})^2+(\\overline{CD})^2 &= (\\overline{AC})^2 \\\\ 8^2+15^2 &= (\\overline{AC})^2 \\\\ 64+225 &= (\\overline{AC})^2 \\\\ 289 &= (\\overline{AC})^2 \\\\ \\sqrt{289} &= \\sqrt{(\\overline{AC})^2} \\\\ 17 &= AC. \\end{align*} Thus we have $\\overline{AC}=\\overline{BC}=17.$ $[asy] pair A, B, C, D; A=(0,0); B=(16,0); C=(8,15); D=B/2; draw(A--B--C--cycle); draw(C--D); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,S); label(\"8\",(A+D)/2,S); label(\"8\",(B+D)/2,S); label(\"15\",(C+D)/2,NE); label(\"17\",(A+C)/2,W); label(\"17\",(B+C)/2,E); [/asy]$ Letting $17$ be the base of the triangle makes our height the radius of the semicircle: $[asy] pair A, B, C, D, EE; A=(0,0); B=(16,0); C=(8,15); D=B/2; EE=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--EE); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,EE,D,25)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,S); label(\"8\",(A+D)/2,S); label(\"8\",(B+D)/2,S); label(\"15\",(C+D)/2,NE); label(\"17\",(A+C)/2,W); label(\"17\",(B+C)/2,E); label(\"r\",D--EE, N); [/asy]$ Let $r$ be the radius of the semicircle. Then we have $\\dfrac{17 \\cdot r}{2}=60 \\implies 17 \\cdot r = 120 \\implies \\dfrac{17r}{17} = \\dfrac{120}{17}", "49 \\qquad \\Longrightarrow\\qquad k=\\frac 7{\\sqrt{19}}$$ Use Ptolemy's Theorem on $BFCA$, to get $$15k+15k=7\\cdot AF, \\qquad \\Longrightarrow\\qquad AF= \\frac{30}{\\sqrt{19}},$$ so $AF^2=\\frac{900}{19}$ and the answer is $900+19=919$ ~abacadaea ## Solution 5 Denote $AB = c, BC = a, AC = b, \\angle A = 2 \\alpha.$ Let M be midpoint BC. Let $\\theta$ be the circle centered at $A$ with radius $\\sqrt{AB \\cdot AC} =\\sqrt{bc}.$ We calculate the length of some segments. The median $AM = \\sqrt{\\frac {b^2}{2} + \\frac {c^2}{2} - \\frac {a^2}{4}}.$ The bisector $AD = \\frac {2 b c \\cos \\alpha}{b+c}.$ One can use Stewart's Theorem in both cases. $AD$ is bisector of $\\angle A \\implies BD = \\frac {a c}{b + c}, CD = \\frac {a b}{b + c} \\implies$ $$BD \\cdot CD = \\frac {a^2 bc }{(b+c)^2}.$$ We use Power of Point $D$ and get $AD \\cdot DE = BD \\cdot CD.$ $$AE = AD + DE = AD + \\frac {BD \\cdot CD}{AD},$$ $$AE =\\frac {2 b c \\cos \\alpha}{b+c} + \\frac {a^2 bc \\cdot (b+c) }{(b+c)^2 \\cdot 2 b c \\cos \\alpha} =$$ $$= \\frac {b c \\cos^2 \\alpha + a^2}{2(b+c)\\cos \\alpha} =\\frac {4bc \\cos^2 \\alpha + b^2 +c^2 -2 b c \\cos 2\\alpha}{2(b+c) \\cos \\alpha} = \\frac {b+c}{2} \\implies AD \\cdot AE = 2 bc \\cos \\alpha.$$ We consider the inversion with", "{AB}{BD}$ or, cot$\\theta$ = $\\frac {AB}{BD}$ or, $\\frac 1{tan\\theta}$ = $\\frac {AB}b$ or, $\\frac 1{\\frac {AB}a}$ = $\\frac {AB}b$ [$\\because$ From equation (i)] or, $\\frac a{AB}$ = $\\frac {AB}b$ or, (AB)2 = ab ∴ AB = $\\sqrt {ab}$ Hence, the height of the tower is $\\sqrt {ab}$ m. Proved Suppose: AB be tree which is broken at the point C and the upper part makes an angle 30° with ground 9 m apart from the foot of the tree. where, Ac = CD BD = 9 m AB = ? In right angled triangle CBD, tan 30° = $\\frac {CB}{BD}$ or, $\\frac 1{\\sqrt 3}$ = $\\frac {CB}9$ or, $\\sqrt 3$ CB = 9 or, BC = $\\frac 9{\\sqrt 3}$ or, BC = $\\frac 9{\\sqrt 3}$× $\\frac {\\sqrt 3}{\\sqrt 3}$ or, BC = 3$\\sqrt 3$ m Again, In right angled triangle, sin 30° = $\\frac {BC}{CD}$ or, $\\frac 12$ = $\\frac {3\\sqrt 3}{CD}$ or, CD = 6$\\sqrt 3$ m Now, The height of the tree: = AB = AC + BC = CD + BC = 6$\\sqrt 3$ + 3$\\sqrt 3$ = 9$\\sqrt 3$ m Ans Here, Height of the cliff (AB) = 21 m Height of the tower (CD) = ? Angle of depression ($\\angle$MBD) = 45° = $\\angle$BDE Angle of depression ($\\angle$MBC) = 60° = $\\angle$BCA In", "triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$ and $\\cos \\theta$, we can represent these areas using $\\sin \\theta$ as follows: \\begin{align*} 30 &=[ABCD] \\\\ &=[AXB] + [BXC] + [CXD] + [DXA] \\\\ &=\\frac{1}{2} ab \\sin (\\angle AXB) + \\frac{1}{2} bc \\sin (\\angle BXC) + \\frac{1}{2} cd \\sin (\\angle CXD) + \\frac{1}{2} da \\sin (\\angle AXD) \\\\ &=\\frac{1}{2} ab \\sin (180^\\circ - \\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (180^\\circ - \\theta) + \\frac{1}{2} da \\sin (\\theta). \\end{align*} We know that $\\sin (180^\\circ - \\theta) = \\sin \\theta$. Therefore it follows that: \\begin{align*} 30 &=\\frac{1}{2} ab \\sin (180^\\circ - \\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (180^\\circ - \\theta) + \\frac{1}{2} da \\sin (\\theta) \\\\ &=\\frac{1}{2} ab \\sin (\\theta) + \\frac{1}{2} bc \\sin (\\theta) + \\frac{1}{2} cd \\sin (\\theta) + \\frac{1}{2} da \\sin (\\theta) \\\\ &=\\frac{1}{2}\\sin\\theta (ab + bc + cd + da). \\end{align*} From here we see that $\\sin \\theta = \\frac{60}{ab + bc + cd + da}$. Now we need to find $\\cos \\theta$. Using the Law of Cosines on each of the four smaller triangles, we get following equations: \\begin{align*} 5^2 &= a^2 + b^2 - 2ab\\cos(180^\\circ-\\theta), \\\\ 6^2 &= b^2 + c^2 - 2bc\\cos \\theta, \\\\ 9^2 &= c^2 + d^2 - 2cd\\cos(180^\\circ-\\theta)," ]

[ "move for the particle as a counterclockwise rotation of $\\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$-direction. Given that the particle's position after $150$ moves is $(p,q)$, find the greatest integer less than of equal to $|p| + |q|$.", "1,b_1 = 3, \\text{ and } b_n = b_{n - 1} + \\frac {b_{n - 1}^2}{b_{n - 2}}\\text{ for }n\\ge2.$$ Find $\\frac {b_{32}}{a_{32}}$. ## Problem 7 Let $r$, $s$, and $t$ be the three roots of the equation $$8x^3 + 1001x + 2008 = 0.$$ Find $(r + s)^3 + (s + t)^3 + (t + r)^3$. ## Problem 8 Let $a = \\pi/2008$. Find the smallest positive integer $n$ such that $$2[\\cos(a)\\sin(a) + \\cos(4a)\\sin(2a) + \\cos(9a)\\sin(3a) + \\cdots + \\cos(n^2a)\\sin(na)]$$ is an integer. ## Problem 9 A particle is located on the coordinate plane at $(5,0)$. Define a move for the particle as a counterclockwise rotation of $\\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$-direction. Given that the particle's position after $150$ moves is $(p,q)$, find the greatest integer less than or equal to $|p| + |q|$. ## Problem 10 The diagram below shows a $4\\times4$ rectangular array of points, each of which is $1$ unit away from its nearest neighbors. $[asy] unitsize(0.25inch); defaultpen(linewidth(0.7)); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); [/asy]$ Define a growing path to be a sequence of distinct points of the array with the property that the distance between consecutive points of the", "# Problem #599 Back to Complex Numbers page 599 A particle is located on the coordinate plane at $(5,0)$. Define a move for the particle as a counterclockwise rotation of $\\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$-direction. Given that the particle's position after $150$ moves is $(p,q)$, find the greatest integer less than or equal to $|p| + |q|$. This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved. Instructions for entering answers: • Reduce fractions to lowest terms and enter in the form 7/9. • Numbers involving pi should be written as 7pi or 7pi/3 as appropriate. • Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate. • Exponents should be entered in the form 10^10. • If the problem is multiple choice, enter the appropriate (capital) letter. • Enter points with parentheses, like so: (4,5) • Complex numbers should be entered in rectangular form unless otherwise specified, like so: 3+4i. If there is no real component, enter only the imaginary component (i.e. 2i, NOT 0+2i). For questions or comments, please email markan@eudelic.com. ## Find out how your skills stack up! Try our new, free contest math", "# A different kind of pursuit ..... Geometry Level 3 Consider two concentric circles, radii $$1$$ and $$2$$ respectively, centered at the origin. Particles are situated on each of the circles, at $$(1,0)$$ and $$(2,0)$$ respectively, and then simultaneously begin to move counterclockwise around their respective circles, both at a rate of $$1$$ unit per second. The time that has elapsed when the line joining the two particles is tangent to the smaller circle for the first time is $$\\dfrac{a\\pi}{b}$$ seconds, where $$a$$ and $$b$$ are positive coprime integers. Find $$a + b$$. ×", "# Difference between revisions of \"2008 AIME II Problems/Problem 9\" ## Problem A particle is located on the coordinate plane at $(5,0)$. Define a move for the particle as a counterclockwise rotation of $\\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$-direction. Given that the particle's position after $150$ moves is $(p,q)$, find the greatest integer less than or equal to $|p| + |q|$. ## Solution ### Solution 1 Let $P(x, y)$ be the position of the particle on the $xy$-plane, $r$ be the length $OP$ where $O$ is the origin, and $\\theta$ be the inclination of OP to the x-axis. If $(x', y')$ is the position of the particle after a move from $P$, then we have two equations for $x'$ and $y'$: $$x'=r\\cos(\\pi/4+\\theta)+10 = \\frac{\\sqrt{2}(x - y)}{2} + 10$$ $$y' = r\\sin(\\pi/4+\\theta) = \\frac{\\sqrt{2}(x + y)}{2}$$. Let $(x_n, y_n)$ be the position of the particle after the nth move, where $x_0 = 5$ and $y_0 = 0$. Then $x_{n+1} + y_{n+1} = \\sqrt{2}x_n+10$, $x_{n+1} - y_{n+1} = -\\sqrt{2}y_n+10$. This implies $x_{n+2} = -y_n + 5\\sqrt{2}+ 10$, $y_{n+2}=x_n + 5\\sqrt{2}$. Substituting $x_0 = 5$ and $y_0 = 0$, we have $x_8 = 5$ and $y_8 = 0$ again for the first time. Thus, $p = x_{150} = x_6 = -5\\sqrt{2}$", "case of 5 DOF, I am guessing that position is the only valid state (i.e. rotations aren't allowed). In that case, the first particle is described by 3 spatial coordinates. The 2nd particle can be anywhere on a fixed sphere around that particle. In spherical coordinates, r is fixed but theta and phi can be anything, thus 2 extra DOF for a total of 5 DOF.", "19 views A particle is moving in the x, y plane at a speed of v = 0.80 and it is travelling an angle of 60 degrees above the x-axis. (a) Rotate the spatial coordinates so that v lies along the x-axis and then construct the components of the 4-velocity for the particle. (b) Construct the remaining orthonormal basis vectors for the particle in the rotated frame, then rotate the spatial coordinates back to find the basis vectors in the original frame. (c) Perform the same steps above but this time begin by rotating the spatial coordinates so that v lies along the y-axis. Do you get the same result for the basis vectors after rotating back to the original frame? Should you? Draw a picture with ~v, the original (x, y) frame, and the two rotated frames to explain what is happening.", "the two-dimensional case, the position vector is drawn from the origin to the location of the particle in the Cartesian plane and has two coordinates, the x-position of the particle, and the y-position of the particle. Example 1: The position of a particle at (4, 5) on a Cartesian plane can be described as 4 units along the x-axis, and 5 units along the y-axis, or as the position vector r = <4, 5> = 4i + 5j, where i and j are the unit vectors in the x and y direction, respectively. From this example, we can see that a position vector takes the form <x, y>, where x is the distance moved along the x-axis and y is the distance moved along the y-axis. Example 2: A carbon atom sits 7.0 nm east, 0.5 nm north, and 3.0 nm below the tip of a scanning tunneling microscope. Its position vector r = <7.0 nm, 0.5 nm, –3.0 nm>. Taking i to be east, j to be north, and k to be up, r = (7.0 nm)i + (0.5 nm)j + (–3.0 nm)k. ### Displacement Recall that the displacement of a particle is the difference between its initial and final positions. In two dimensional motion, we denote the displacement vector as Δr. The displacement vector is defined", "that mixes what we are now calling position and momentum coordinates. This all cries out for a description that describes the physical system in terms of points on a manifold, to which we can assign coordinates only once we know how all the mathematical objects involved are defined intrinsically. ### Configuration Space and Phase Space We’ll start our quest for a coordinate-free description of mechanics by fixing a smooth manifold $$Q$$ which we’ll call configuration space. You should think of a point in $$Q$$ as corresponding the “position” of each component of a physical system at some fixed time. Some examples worth keeping in mind are: 1. A particle moving in $$\\mathbb{R}^3$$. In this case, $$Q$$ is just $$\\mathbb{R}^3$$. 2. $$N$$ particles moving in $$\\mathbb{R}^3$$. We specify the configuration of this system by specifying the position of each particle, which we can do using a point in $$\\mathbb{R}^{3N}$$. 3. Two particles connected by a rigid rod of length $$\\ell$$. We could describe the configuration of this system using a point in $$\\{(a,b)\\in\\mathbb{R}^3\\times\\mathbb{R}^3:||a-b||=\\ell\\}$$. 4. A rigid body moving through space. We could describe its configuration using a point in $$\\mathbb{R}^3\\times SO(3)$$, specifying the location of the object’s center of mass and its orientation. In particular, specifying a point $$q\\in Q$$ gives you an instantaneous snapshot of the system, but", "coordinate. The trouble is that we can't. The circular path along which the particle moves is rotationally symmetrical. There's nothing about the circle to tell us where it \"starts\" or \"ends.\" This means that the above line of reasoning holds no matter where we start. We can pick any point on the circle as the $x=1, y=0$ point, and the above will be valid. So wherever the particle is, we simply set that point as our $x = r, y = 0$ point, and everything else falls into place. If we wanted to make more work for ourselves, we could work out the details trigonometrically, using the above formulas, adjusting them for the $y$ axis, and then recombining the $x$ and $y$ values using vector algebra. But we don't need to -- the symmetry argument is more powerful in this case. 1. Here \"largest\" really means \"most negative,\" because the particle is moving in the negative $x$ direction. These values will really be $-F_\\text{max}$ and $-v_\\text{max}$. Later, when the particle is moving in the opposite direction, these values will be positive. 2. Making sense of the signs takes a lot of subtle detail work; in particular, you have to understand why negative work becomes positive potential energy. It also helps to think about $r$ as a vector (which", "Free Version Moderate # Motion of a Particle in Polar Coordinates APCALC-GWEES2 Suppose the motion of a particle is described by the polar equation $r=2\\sin{2\\theta}$. When $\\theta=\\cfrac{\\pi}{3}$, is the particle moving towards or away from the origin? A Toward the origin. B Away from the origin. C It is moving neither toward or away from the origin. D Cannot be determined from the given information.", "Periodic (or even antiperiodic) boundary conditions do not spoil translation invariance! This means that if you start with a translation-invariant Hamiltonian (or Lagrangian) and if you impose p.b.c., then the system will stay invariant w.r.t. spatial translations. The density function of 'particle position' should be completely flat. As for your implementation, most programming languages have some 'modulo' function even for real number, IIRC. Then after a move, you just put (symbolically) 'position_i = modulo(position_i,sizeofbox)' and you're done. – Vibert Feb 10 '13 at 11:19 The problem is the reverse: I have all the positions within the box already. Let me give a simple 1D example: the system is 10 units long. There is one particle at 9.5 units through the first steps, that particle shall go from 9.5 to 10.5 then from 10.5 to 9.5. What I already get, actually, as particle positions is: from 9.5 to 10 ; from 0 to 0.5 ; from 0.5 to 0; from 10 to 9.5. The average position of that particle should be 0 (or 10, I don't care) but if I don't pay attention, I'd get 5 units instead. I want to know if there is a smart way to get 0 units in such a case. – Gael Feb 10 '13 at 11:28 OK, I see. I mean that", "to define the particle’s position. Its position vector from the origin of the circle to the particle sweeps out the angle $\\theta$, which increases in the counterclockwise direction as the particle moves along its circular path. The angle $\\theta$ is called the angular position of the particle. As the particle moves in its circular path, it also traces an arc length s. The angle is related to the radius of the circle and the arc length by $\\theta =\\frac{s}{r}.$ The angle $\\theta$, the angular position of the particle along its path, has units of radians (rad). There are $2\\pi$ radians in $360^\\circ.$ Note that the radian measure is a ratio of length measurements, and therefore is a dimensionless quantity. As the particle moves along its circular path, its angular position changes and it undergoes angular displacements $\\Delta \\theta .$ We can assign vectors to the quantities in Figure. The angle $\\mathbf{\\overset{\\to }{\\theta }}$ is a vector out of the page in Figure. The angular position vector $\\mathbf{\\overset{\\to }{r}}$ and the arc length $\\mathbf{\\overset{\\to }{s}}$ both lie in the plane of the page. These three vectors are related to each other by $\\mathbf{\\overset{\\to }{s}}=\\mathbf{\\overset{\\to }{\\theta }}\\times \\mathbf{\\overset{\\to }{r}}.$ That is, the arc length is the cross product of the angle vector and the position vector, as shown in Figure.", "the circle. The position of a particle is given by the parametric equations $$x = sin t, y = cos t$$ where 1 represents time. We know that the shape of the path of the particle is a circle. That parametric equations contain more information than just the shape of the curve. Write a short paragraph explaining this statement. Use the following example and your answers to parts (a) and (b) below in your explanation. The position of a particle is given by the parametric equations $$x =\\ \\sin\\ (t)\\ and\\ y =\\ \\cos\\ (t)$$ where t represents time. We know that the shape of the path of the particle is a circle. a) How long does it take the particle to go once around the circle? Find parametric equations if the particle moves twice as fast around the circle. b) Does the particle travel clockwise or counterclockwise around the circle? Find parametric equations if the particle moves in the opposite direction around the circle.", "\\right) \\end{align} The underlying physics says, that you have rotation in 3D around a fixed point on the circle. That you have to choice \"how much it rotates in $\\phi$-direction = $(x,y)$ plane, and the program has all information now\". • If I set the initial positions of all particles on the x-y plane then this does work but out interest, how would I then apply this to points that aren't positioned on this plane? I'm assuming that I can use the same calculation but in the plane of the particle and then convert back into 3D space? How would I do this conversion? – Placid Apr 16 '17 at 14:07 • Oh, you can do transformation. But for many particles it's easier to calculate the direction with spherical coordinates, but it's more complicated formula. I added it to the answer above. – Jimmy Found Apr 16 '17 at 19:24", "+0 # help 0 57 1 A particle's position in the plane $$t$$ seconds after it starts moving is \\begin{align*} x &= 6\\cos(kt),\\\\ y &= 6\\sin(kt), \\end{align*} for some constant $$k$$. If the particle moves a distance of $$120\\pi$$ every second, then what is $$k$$? Nov 5, 2019", "a Single Particle Figure 11.9 shows a particle at a position $r→r→$ with linear momentum $p→=mv→p→=mv→$ with respect to the origin. Even if the particle is not rotating about the origin, we can still define an angular momentum in terms of the position vector and the linear momentum. ### Angular Momentum of a Particle The angular momentum $l→l→$ of a particle is defined as the cross-product of $r→r→$ and $p→p→$, and is perpendicular to the plane containing $r→r→$ and $p→:p→:$ $l→=r→×p→.l→=r→×p→.$ 11.5 Figure 11.9 In three-dimensional space, the position vector $r → r →$ locates a particle in the xy-plane with linear momentum $p → p →$. The angular momentum with respect to the origin is $l → = r → × p → l → = r → × p →$, which is in the z-direction. The direction of $l → l →$ is given by the right-hand rule, as shown. The intent of choosing the direction of the angular momentum to be perpendicular to the plane containing $r→r→$ and $p→p→$ is similar to choosing the direction of torque to be perpendicular to the plane of $r→andF→,r→andF→,$ as discussed in Fixed-Axis Rotation. The magnitude of the angular momentum is found from the definition of the cross-product, $l=rpsinθ,l=rpsinθ,$ where $θθ$ is the angle between $r→r→$ and $p→.p→.$ The", "where $\\mathbf u$ is the position of the particle on 3-sphere (a 4-dimensional vector), $\\mathbf p$ is the momentum of the original particle in Kepler problem, $\\hat{\\mathbf n}$ is a vector perpendicular to the 3-dimensional hyperplane where $\\mathbf p$ lies, and $p_0:=\\sqrt{-2mE}$.", "position vector R s measured point! Original puzzle piece different, circular path as shown in the figure in the core at the around! Mass m moves in a counter-clockwise direction on a circular path shown below at origin and radius 1 $in... Rotation of 150° around point P moves in a counterclockwise direction movement of ' '! Rotation of the particle about point s as viewed from above ) badges 56 bronze. A rotation of the two vectors along the two edges at X2Y0 radius. Rotated 90° counterclockwise rotation of the path is 27 m. the acceleration of P P moves in the.... Where P stops ( round to three decimal places ) and what quadrant it lies in loop a! Of mass m moves in a horizontal circle the acceleration of P P is that... Draw _ PA it is always m units from a to B..! Two particles start from the IITs and BITS - instantly, a point p moves in counter clockwise, anywhere C to D, D E!, so we can Use it to express the x– and y-coordinates vector!, circular path as shown in the figure radius 1 a straight line.... Condition 3: a point P moves up in the figure you do a 90 degree clockwise.. Arc starts at X0Y2 and finishes at", "be derived perhaps more simply in polar coordinates, and at the same time extended to general motion within a plane, as shown next. Polar coordinates in the plane employ a radial unit vector uρ and an angular unit vector uθ, as shown in Figure 6. A particle at position r is described by: $mathbf\\left\\{r\\right\\} = rho mathbf\\left\\{u\\right\\}_\\left\\{rho\\right\\} ,$ where the notation ρ is used to describe the distance of the path from the origin instead of R to emphasize that this distance is not fixed, but varies with time. The unit vector uρ travels with the particle and always points in the same direction as r (t ). Unit vector uθ also travels with the particle and stays orthogonal to uρ. Thus, uρ and uθ form a local Cartesian coordinate system attached to the particle, and tied to the path traveled by the particle. By moving the unit vectors so their tails coincide, as seen in the circle at the left of Figure 6, it is seen that uρ and uθ form a right-angled pair with tips on the unit circle that trace back and forth on the perimeter of this circle with the same angle θ (t) as r (t ). When the particle moves, its velocity is $mathbf\\left\\{v\\right\\} = frac \\left\\{d rho \\right\\}\\left\\{dt\\right\\} mathbf\\left\\{u\\right\\}_\\left\\{rho\\right\\} + rho" ]

[ "+0 # Find the size of the smallest angle between the two hands of a clock displaying these times. 10:20, 6:55, 4:35, 2:27 and 11:09 0 255 1 Find the size of the smallest angle between the two hands of a clock displaying these times. 10:20, 6:55, 4:35, 2:27 and 11:09 Guest Jun 15, 2017 #1 +19620 +1 Find the size of the smallest angle between the two hands of a clock displaying these times. 10:20, 6:55, 4:35, 2:27 and 11:09 1. angular speed = $$\\omega$$ minutes hand: $$\\omega_{\\text{m}} = \\frac{360^{\\circ}}{1\\ h}$$ hour hand: $$\\omega_{\\text{h}} = \\frac{360^{\\circ}}{12\\ h}$$ 2. angle between the two hands = $$\\alpha$$ $$\\begin{array}{rcll} \\alpha &=& ( \\omega_{\\text{m}} - \\omega_{\\text{h}} ) \\cdot t \\\\ \\alpha &=& \\Big( \\frac{360^{\\circ}}{1\\ h} - \\frac{360^{\\circ}}{12\\ h} \\Big)\\cdot t \\\\ \\alpha &=& 360^{\\circ}\\cdot \\Big( \\frac{1}{1\\ h} - \\frac{1}{12\\ h} \\Big)\\cdot t \\\\ \\mathbf{\\alpha} & \\mathbf{=} & \\mathbf{360^{\\circ}\\cdot \\frac{11}{12\\ h} \\cdot t} \\\\ \\end{array}$$ 3. Solution for 10:20, 6:55, 4:35, 2:27 and 11:09 $$\\begin{array}{|rcll|} \\hline t=10:20=10.\\bar{3}\\ h & \\alpha & = & 360^{\\circ}\\cdot \\frac{11}{12\\ h} \\cdot 10.\\bar{3}\\ h \\\\ & \\alpha & = & 3410^{\\circ} \\\\ & \\alpha & = & 3410^{\\circ} \\pmod {360^{\\circ} } \\\\ & \\mathbf{\\alpha} & \\mathbf{=} & \\mathbf{170^{\\circ}} \\\\\\\\ t=6:55=6.91\\bar{6}\\ h & \\alpha & = & 360^{\\circ}\\cdot \\frac{11}{12\\ h} \\cdot 6.91\\bar{6}\\ h \\\\ & \\alpha & =", "to Example 3 a) For a series RLC circuit $Z = R + j(\\omega L - \\dfrac{1}{\\omega C})$ Substitute the known quantities by their numerical values $Z = R + j(\\omega L - \\dfrac{1}{\\omega C}) = 200 + j( \\omega \\cdot 100 \\cdot 10^{-3} - \\dfrac{1}{\\omega \\cdot 47 \\cdot 10^{-6}})$ $Z = \\sqrt {200^2 + ( \\omega \\cdot 100 \\cdot 10^{-3} - \\dfrac{1}{\\omega \\cdot 47 \\cdot 10^{-6}})^2} \\; \\angle \\; \\arctan \\left(\\dfrac{\\omega \\cdot 100 \\cdot 10^{-3} - \\dfrac{1}{\\omega \\cdot 47 \\cdot 10^{-6}}}{200}\\right)$ b) $\\arctan \\left(\\dfrac{\\omega \\cdot 100 \\cdot 10^{-3} - \\dfrac{1}{\\omega \\cdot 47 \\cdot 10^{-6}}}{200}\\right) = -60^{\\circ}$ $\\dfrac{\\omega \\cdot 100 \\cdot 10^{-3} - \\dfrac{1}{\\omega \\cdot 47 \\cdot 10^{-6}}}{200} = \\tan (-60^{\\circ}) = -1.73205$ Multiply all terms of the equation by $200$ to eliminate the denominator $\\omega \\cdot 100 \\cdot 10^{-3} - \\dfrac{1}{\\omega \\cdot 47 \\cdot 10^{-6}} = -346.41$ Multiply all terms of the equation by $\\omega \\cdot 47 \\cdot 10^{-6}$ to eliminate the denominator $47 \\cdot 10^{-7} \\omega^2 - 1 = -0.01628127 \\omega$ Solve the above quadratic equation for $\\omega$ and select the positive solution. $\\omega = 53.23$ rad/s $\\omega = 2 \\pi f = 60.3682$ $f = \\dfrac{60.3682}{2 \\pi} = 9.60789$ Hz Example 4 In a series RLC circuit the source voltage is given by $v_i = V_0 \\cos (2 \\pi f t)$, the capacitance of the", "sides, etc. Since $2^2-1^2=3$, $4^2-3^2=7$, $6^2-5^2=11$, etc., this means the desired conclusion is equivalent to$$0 = \\sum_{n=0}^{994} c_n \\omega^n$$being true for some permutation $(c_0, \\dots, c_{994})$ of $(3, 7, 11, \\dots, 3981)$. Define $z = 3 \\omega^0 + 7 \\omega^{199} + 11\\omega^{398} + 15 \\omega^{597} + 19 \\omega^{796}$. Then notice that \\begin{align*} z &= 3 \\omega^0 + 7 \\omega^{199} + 11\\omega^{398} + 15 \\omega^{597} + 19 \\omega^{796} \\\\ \\omega^5 z &= 23 \\omega^5 + 27 \\omega^{204} + 31\\omega^{403} + 35 \\omega^{602} + 39 \\omega^{801} \\\\ \\omega^{10} z &= 43 \\omega^{10} + 47 \\omega^{209} + 51\\omega^{408} + 55 \\omega^{607} + 59 \\omega^{806} \\\\ &\\vdotswithin{=} \\end{align*}and so summing yields the desired conclusion, as the left-hand side becomes$$(1+\\omega^5+\\omega^{10}+\\cdots+\\omega^{990})z = 0$$and the right-hand side is the desired expression. This solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]", "have $\\omega[5]=5$ as the fifth term of this sequence, so $$f_\\omega(5)=f_{\\omega[5]}(5)=f_5(5)$$ And from there we know this is very large...much larger than the previous numbers. To deal with $\\omega+1$, we apply the second rule, followed by the third: \\begin{align}f_{\\omega+1}(5)&=f_\\omega(f_\\omega(f_\\omega(f_\\omega(f_\\omega(5)))))\\\\&=f_\\omega(f_\\omega(f_\\omega(f_\\omega(f_5(5)))))\\end{align} So you can see how large this gets, and it gets pretty big. Graham 's number has a nice approximation: $$f_{\\omega+1}(64)\\approx\\text{Graham 's number}$$ So this is rather simple and $$f_{\\omega+1}(64)=\\underbrace{f_\\omega(\\dots f_\\omega(64)\\dots)}_{64}$$ Note that we can even go further: $$\\omega\\cdot2=\\sup\\{\\omega+1,\\omega+2,\\omega+3,\\dots\\}$$ For example: \\begin{align}f_{\\omega\\cdot2}(5)&=f_{\\omega+5}(5)\\\\&=f_{\\omega+4}(f_{\\omega+4}(f_{\\omega+4}(f_{\\omega+4}(f_{\\omega+4}(5)))))\\end{align} And we would then expand further and further and further... And we have some other larger ordinals: $$\\omega^2=\\{\\omega\\cdot1,\\omega\\cdot2,\\omega\\cdot3,\\dots\\}$$ And beyond! $$\\omega^\\omega=\\{\\omega^1,\\omega^2,\\omega^3,\\dots\\}$$ While it's not very... visual, it does become clear that you can make some very very big numbers while also being able to compare large numbers. One of the great advantages of FGH is the clearness of how to compare Graham's number to another number. And the only thing holding you back from producing even larger numbers is how you want to define even larger ordinals. And one last example: \\begin{align}f_{\\omega^\\omega}(3)&=f_{\\omega^3}(3)\\\\&=f_{\\omega^2\\cdot3}(3)\\\\&=f_{\\omega^2\\cdot2+\\omega^2}(3)\\\\&=f_{\\omega^2\\cdot2+\\omega\\cdot3}(3)\\\\&=f_{\\omega^2\\cdot2+\\omega\\cdot2+3}(3)\\\\&=f_{\\omega^2\\cdot2+\\omega\\cdot2+2}(f_{\\omega^2\\cdot2+\\omega\\cdot2+2}(f_{\\omega^2\\cdot2+\\omega\\cdot2+2}(3)))\\\\&=\\cdots\\end{align}", "+ \\omega^{\\beta_n}\\cdot k_n$ with $k_1 > 1$ or $n>1$. But then $\\omega^{\\beta_1} < \\gamma$ and also \\begin{align*} \\omega^{\\beta_1} + \\gamma = \\omega^{\\beta_1} + (\\omega^{\\beta_1}\\cdot k_1 + \\dots + \\omega^{\\beta_n}\\cdot k_n) &= (\\omega^{\\beta_1} + \\omega^{\\beta_1}\\cdot k_1) + \\dots + \\omega^{\\beta_n}\\cdot k_n\\\\ &= \\omega^{\\beta_1}\\cdot (k_1+1) + \\dots + \\omega^{\\beta_n}\\cdot k_n\\\\ &> \\omega^{\\beta_1}\\cdot k_1 + \\dots + \\omega^{\\beta_n}\\cdot k_n = \\gamma \\end{align*} which together contradicts $(2)$. For $(3) \\rightarrow (1)$ let $\\beta_1,\\beta_2<\\gamma=\\omega^\\alpha$. By Cantor's normal form theorem you'll easily find $\\alpha'<\\alpha$ and $k\\in\\omega$ with $\\beta_1,\\beta_2<\\omega^{\\alpha'}\\cdot k$. This implies $\\beta_1 + \\beta_2 < \\omega^{\\alpha'}\\cdot (k+k) < \\omega^\\alpha=\\gamma$. • Thank you for your answer. Most of it I understood. Exapt one thing: In $(2) \\rightarrow (3)$, at the end of the proof, how do we get that $\\omega^{\\beta_1} + \\gamma > \\gamma$? Sep 16 '14 at 14:30 • In $\\omega^{\\beta_1}+\\gamma$ write $\\gamma$ as $\\omega^{\\beta_1}\\cdot k_1 + \\dots + \\omega^{\\beta_n}\\cdot k_n$, then the sum is $\\omega^{\\beta_1}\\cdot (k_1+1) + \\dots + \\omega^{\\beta_n}\\cdot k_n$. Sep 16 '14 at 14:58", "this case, $G_x=G_2$. If you use those in the bottom equation of (4), you will get the right answer. However, it doesn't look like what the teacher wrote up. But if you plug things in, you can see why: \\begin{align*} (5)~~~~~~~~~I_x &= I_T\\cdot\\frac{\\frac{1}{1.6\\:\\Omega+8\\:\\Omega}}{\\frac{1}{2.4\\:\\Omega}+\\frac{1}{1.6\\:\\Omega+8\\:\\Omega}} \\\\ \\\\ &= I_T\\cdot\\frac{\\frac{1}{1.6\\:\\Omega+8\\:\\Omega}}{\\frac{2.4\\:\\Omega+1.6\\:\\Omega+8\\:\\Omega}{2.4\\:\\Omega\\cdot\\left(1.6\\:\\Omega+8\\:\\Omega\\right)}} \\\\ \\\\ &=I_T\\cdot\\frac{\\frac{1}{1}}{\\frac{2.4\\:\\Omega+1.6\\:\\Omega+8\\:\\Omega}{2.4\\:\\Omega}}=I_T\\cdot\\frac{2.4\\:\\Omega}{2.4\\:\\Omega+1.6\\:\\Omega+8\\:\\Omega} \\end{align*} However, that is NOT a general approach. It just happens to work in this case. So your intuition perhaps is right, about being confused in class. The general solution looks like this, using resistances: \\begin{align*} (6)~~~~~~~~~I_x &= I_T\\cdot\\frac{G_x}{\\sum^n_1 G_i} \\\\ \\\\ &= I_T\\cdot\\frac{\\prod_{i=1,\\\\ i\\ne x}^n R_i}{\\sum^n_{i=1} \\prod_{j=1,\\\\ j\\ne i}^n R_j} \\end{align*} Keep in mind that in the above, each $R_i$ is not each of all your resistors, helter-skelter. These resistances are the effective branch resistances. So in one of your branches, this value is a sum. Don't forget that, either. Equation (6) is a lot uglier than you might have imagined from your teacher's solution. It's the correct one, though. So, if you want to commit something to memory, go with the conductances approach. As you can see, using resistances isn't right off the bat intuitive. If you must use resistances in the current branching case like this, then you have to follow (6) above and DO NOT simply try and extend what your teacher did in", "# The number of distinct prime factors of $n\\in\\mathbb N$ Let $\\omega(n)$ be the number of distinct prime factors of a natural number $n$. Note that $\\omega(n)=0\\iff n=1$, and that $\\omega(24)=\\omega(2^3\\cdot 3^1)=2\\ (\\not = 4)$. (For more details, you can see \"Distinct Prime Factors\" on wolfram math world) Then, for $m\\in\\mathbb N$, let $S_m=\\{n|n\\in\\mathbb N, \\omega(n)=m\\}$. Also, Let $N_m(x)$ be the counting function that gives the number of the elements of $S_m$ less than or equal to $x$, for any real number $x$. See the line graph below. This graph shows $N_m(x)\\ (m=1,2,3,4)$ in $1\\lt x\\le 2309=2\\cdot3\\cdot5\\cdot7\\cdot11-1.$ Note that $N_m(2309)=0$ for $m\\ge 5$, and that $\\omega(2310)=\\omega(2\\cdot 3\\cdot 5\\cdot 7\\cdot 11)=5$. $\\ \\ \\ \\ \\ \\ \\ \\ \\ \\$ In addition to this, let $P_m(x)=\\frac{N_m(x)}{x}\\times 100$. See the line graph below. This graph shows $P_m(x)\\ (m=1,2,3,4)$ in $1\\lt x\\le 2309=2\\cdot3\\cdot5\\cdot7\\cdot11-1.$ $\\ \\ \\ \\ \\ \\ \\ \\ \\ \\$ Then, here are my questions. Question : Are the followings true? $(1)$ For every $m\\in\\mathbb N$, $N_m(x)=O(x)\\ (x\\to\\infty)$. $(2)$ For every $m\\in\\mathbb N$, $\\lim_{x\\to\\infty} P_m(x)=0$. $(3)$ If $x\\ge 50$, then $P_2(x)\\gt P_m(x)$ for every $m\\ge 3$. The above graphs led me to the these conjectures, but I don't have any good ideas. I would like to know any relevant references. Can anyone help? - I don't quite", "sequence: $$\\omega\\cdot2=\\omega+\\omega=\\{\\omega+1,\\omega+2,\\omega+3,\\ldots\\}$$ $$\\omega^3=\\{\\omega^2,\\omega^2\\cdot2,\\omega^2\\cdot3,\\ldots\\}$$ And so you correctly had $\\omega^3[2]=\\omega^2\\cdot2$. Then, $$\\omega^2\\cdot2=\\omega^2+\\omega\\cdot\\omega=\\{\\omega^2+\\omega,\\omega^2+\\omega\\cdot2,\\omega^2+\\omega\\cdot3,\\ldots\\}$$ Thus, $\\omega^2\\cdot2 [2]=\\omega^2+\\omega\\cdot2$. From there we break $\\omega\\cdot2 [2]=\\omega+2$, and then we finally have a successor ordinal, and thus we expand as \\begin{align}f_{\\omega^3}(2)&=f_{\\omega^2+\\omega+1}(f_{\\omega^2+\\omega+1}(2))\\\\&=f_{\\omega^2+\\omega+1}(f_{\\omega^2+\\omega}(f_{\\omega^2+\\omega}(2)))\\end{align} And now we have a limit ordinal again, so expand some more. Also, we usually diagonalize starting with the smallest ordinal when taking a sum of ordinals.", "# Complex numbers; omega Hi I have trouble doing this exercise: Let $\\omega=e^{2i\\pi/5}$ Explain why $\\omega^1+\\omega^2+\\omega^3+\\omega^4=−1$ and show that $\\omega+\\omega^4= 2\\cos(2\\pi/5)$ and $\\omega^2+\\omega^3=2\\cos(4\\pi/5)$ So far I have, $w^1=e^{2iπ/5}, w^2=e^{-2iπ/5}, w^3=e^{4iπ/5}, w^4=e^{-4iπ/5}$ Thank you very much for help! • Please share your effort. Observe that $$w\\cdot w^4=w^2\\cdot w^3=w^5=1$$ May 14 '17 at 18:14 • For a start, $\\omega+\\cdots+\\omega^4$ is the sum of a finite geometric progression. May 14 '17 at 18:18 • So far I have, $w^1=e^{2iπ/5}, w^2=e^{-2iπ/5}, w^3=e^{4iπ/5}, w^4=e^{-4iπ/5}$ May 14 '17 at 18:19 Note that $\\omega$ satisfies $\\omega^5-1=0$ But $(\\omega^5-1)=(\\omega-1)(\\omega^4 + \\omega^3 + \\omega^2 + \\omega + 1)$ And $\\omega \\neq 1$. Then note that $\\omega^n + \\omega^{5-n} = \\omega^n + \\omega^{-n} = (\\cos(\\frac{2\\pi n}5) + i\\sin(\\frac{2\\pi n}5)) + \\cos(\\frac{-2\\pi n}5) + i\\sin(\\frac{-2\\pi n}5) = 2\\cos(\\frac{2\\pi n}5)$ Where this last equality is from the fact that cosine is even and sine is odd.", "j\\\\ & -7.2858\\cdot 10^{-17} - 9.9930^4\\text j\\\\ \\end{align} But no one brings me to the result of 100100 rad/s.. Last edited: Apr 20, 2015 9. ### WBahn Moderator Mar 31, 2012 19,121 5,170 Note that your 0.707 comes from 1/sqrt(2). Leverage that knowledge. \\begin{align} & \\frac{1}{\\sqrt{2}} = {1 \\over \\sqrt{(1 - \\omega^2LC)^2 + (\\text j \\omega RC)^2}}\\\\ & \\frac{1}{\\sqrt{2}}^2 = {1 \\over [(1 - \\omega^2LC)^2 - (\\omega RC)^2]}\\\\ & \\frac{1}{2}[(1 - \\omega^2LC)^2 - (\\omega RC)^2] = 1\\\\ & (1 - \\omega^2LC)^2 - (\\omega RC)^2 = 2\\\\ & (1 - \\omega^2LC)^2 - (\\omega RC)^2 - 2 = 0\\\\ & 1 - (2LC)\\omega^2 + (LC)^2\\omega^4 - (RC)^2\\omega^2 - 2 = 0\\\\ & [(LC)^2]\\omega^4 + [(2LC) - (RC)^2]\\omega^2 + [-1] = 0\\\\ \\end{align} Now apply the quadratic formula and see what happens. simo_x likes this. 10. ### WBahn Moderator Mar 31, 2012 19,121 5,170 I just cranked it and got 15.9 kHz. simo_x likes this. 11. ### simo_x Thread Starter Member Dec 23, 2010 200 6 Yes ! I got it ! \\begin{align} \\begin{cases} -1.0374\\cdot 10^{-15} + \\text i1.0010\\cdot 10^{5}\\\\ -1.0374\\cdot 10^{-15} - \\text i1.0010\\cdot 10^{5}\\\\ \\end{cases}& \\text{1st pole}\\Rightarrow \\approx 15.9\\,\\text{kHz}\\\\ \\begin{cases} -9.9900\\cdot 10^{7}\\\\ 9.9900\\cdot 10^{7}\\\\ \\end{cases}& \\text{2nd pole}\\Rightarrow \\approx 15.9\\,\\text{MHz} \\end{align} Thank you for your help. I will be more careful with my calculus next time !", "$\\odot \\omega$ is $R = 18$, let the radius of the small circles $\\odot \\omega_A$, $\\odot \\omega_B$, $\\odot \\omega_C$ be $r$. We are going to solve this problem in $3$ steps: $\\textbf{Step 1:}$ We have $\\triangle A \\omega_A D$ is a $30-60-90$ triangle, and $A \\omega_A = 2 \\cdot \\omega_A D$, $A \\omega_A = 2R-r$ ($\\odot \\omega$ and $\\odot \\omega_A$ are tangent), and $\\omega_A D = r$. So, we get $2R-r = 2r$ and $r = \\frac{2}{3} \\cdot R = 12$. Since $\\odot \\omega$ and $\\odot \\omega_A$ are tangent, we get $\\omega \\omega_A = R - r = \\frac{1}{3} \\cdot R = 6$. Note that $\\triangle \\omega_A \\omega_B \\omega_C$ is an equilateral triangle, and $\\omega$ is its center, so $\\omega_B \\omega_C = \\sqrt{3} \\cdot \\omega \\omega_A = 6 \\sqrt{3}$. $\\textbf{Step 2:}$ Note that $\\triangle \\omega_C E X$ is an isosceles triangle, so $$EX = 2 \\sqrt{(\\omega_C E)^2 - \\left(\\frac{\\omega_B \\omega_C}{2}\\right)^2} = 2 \\sqrt{r^2 - \\left(\\frac{\\omega_B \\omega_C}{2}\\right)^2} = 2 \\sqrt{12^2 - (3 \\sqrt{3})^2} = 2 \\sqrt{117}.$$ $\\textbf{Step 3:}$ In $\\odot \\omega_C$, Power of a Point gives $\\omega X \\cdot \\omega E = r^2 - (\\omega_C \\omega)^2$ and $\\omega E = EX - \\omega X = 2\\sqrt{117} - \\omega X$. It follows that $\\omega X \\cdot (2\\sqrt{117} - \\omega X) = 12^2 - 6^2$. We solve this quadratic equation: $\\omega", "Are these ordinals cardinals? Consider $\\omega_2 \\times \\omega$ and $\\omega \\times \\omega_2$ with ordinal arithmetic. Then $| \\omega_2 \\times \\omega | =\\omega_2$ and $| \\omega \\times \\omega_2 | =\\omega_2$ Does this imply that neither $\\omega_2 \\times \\omega$ nor $\\omega \\times \\omega_2$ are cardinals?? - How would this imply that? – Andres Caicedo May 16 at 20:32 @Andres because both have cardinality $\\omega_2$ but are not equal to $\\omega_2$. So they would be ordinals which fall between $\\omega_2$ and $\\omega_3$ and therefore not cardinals. – fafddf May 16 at 20:33 Both have cardinality $\\omega_2$, not \"greater than\". And why are they not equal to $\\omega_2$? – Andres Caicedo May 16 at 20:34 \\begin{align} &\\alpha\\cdot0=0\\\\ &\\alpha\\cdot(\\beta+1)=\\alpha\\cdot\\beta+\\alpha\\\\ &\\alpha\\cdot\\delta=\\sup\\{\\alpha\\cdot\\gamma\\mid\\gamma<\\delta\\},\\ \\delta\\text{ is a limit ordinal} \\end{align} Since $\\omega$ and $\\omega_2$ are certainly limit ordinals, we have that: $\\omega_2\\cdot\\omega=\\sup\\{\\omega_2\\cdot n\\mid n<\\omega\\}$, and since $\\alpha\\cdot2=\\alpha+\\alpha>\\alpha$, whenever $\\alpha>0$, we have that indeed $\\omega_2\\cdot\\omega>\\omega_2$ as ordinals. On the other hand, $\\omega\\cdot\\omega_2=\\sup\\{\\omega\\cdot\\gamma\\mid\\gamma<\\omega_2\\}$. This is a supremum of $\\omega_2$ ordinals, and we can show that each of those (of the $\\omega\\cdot\\gamma$'s) has size $\\leq\\aleph_1$. Therefore $\\omega\\cdot\\omega_2=\\omega_2$, which is in fact a cardinal.", "that $$s_n \\ge 2$$. For the upper bound let $$z_1, \\ldots, z_n$$ be any admissible choice. Multiplying all points with $$\\omega$$ gives another admissible choice. Therefore we can assume that the argument of $$z_1 + \\cdots + z_n$$ is in the range $$[-\\pi/n, \\pi/n]$$. This allows us to get an estimate for the absolute value of the sum from an estimate for the real part of the sum. We have $$\\operatorname{Re}(z_1 + \\ldots z_n) \\le \\operatorname{Re}(\\omega^0 + \\cdots + \\omega^{k-1} + \\omega^{k+1} + \\cdots + \\omega^{n}) = 2$$ for even $$n=2k$$, and $$\\operatorname{Re}(z_1 + \\ldots z_n) \\le \\operatorname{Re}(\\omega^0 + \\cdots + \\omega^{k} + \\omega^{k+2} + \\cdots + \\omega^{n}) \\\\ = 1 - \\cos\\left( \\frac{(k+1)\\pi}{2n}\\right) = 2 + \\sin\\left( \\frac{\\pi}{2n}\\right)$$ for odd $$n=2k+1$$. In any case, $$2 + \\sin\\left( \\frac{\\pi}{2n}\\right) \\ge \\operatorname{Re}(z_1 + \\ldots z_n) \\ge \\cos\\left(\\frac{\\pi}{n}\\right) |z_1 + \\cdots + z_n|$$ and that proves the upper bound in $$(*)$$.", "We see that $\\dfrac{1}{x}=13-\\omega$. Thus, $$\\dfrac{1}{x\\overline{x}}=(13\\, -\\, \\omega)(13\\, -\\, \\overline{\\omega})=169-13(\\omega\\, +\\, \\overline{\\omega})\\, +\\, \\omega\\overline{\\omega}=170\\, -\\, 13(\\omega\\, +\\, \\overline{\\omega})$$ Summing over all terms: $$\\dfrac{1}{r_1\\overline{r_1}}+\\cdots + \\dfrac{1}{r_5\\overline{r_5}} = 5\\cdot 170 - 13(e^{i\\pi/10}+\\cdots +e^{i(9\\pi/5+\\pi/10)})$$ However, note that $e^{i\\pi/10}+\\cdots +e^{i(9\\pi/5+\\pi/10)}=0$ from drawing the numbers on the complex plane, our answer is just $$5\\cdot 170=\\boxed{850}$$", "\\begin{aligned} \\frac{1}{\\bar{p}_2}&=(1-\\alpha_{{N_0}+1})\\left( \\frac{\\beta_1}{p_1} +\\cdots+ \\frac{\\beta_{N_0}}{p_{N_0}} \\right) + \\frac{\\alpha_{{N_0}+1}}{p_{{N_0}+1}} =\\frac{\\alpha_1}{p_1} +\\cdots + \\frac{\\alpha_{{N_0}+1}}{p_{{N_0}+1}}=\\frac{1}{u_f}, \\end{aligned} \\tag{59} and \\begin{aligned} \\bar{e}_2=&\\bar{p}_2 \\left[ (1-\\alpha_{{N_0}+1}) \\left( \\frac{\\beta_1}{p_1} e_1 +\\cdots+ \\frac{\\beta_{N_0}}{p_{N_0}} e_{N_0} \\right) + \\frac{\\alpha_{{N_0}+1}}{p_{{N_0}+1}} e_{{N_0}+1}\\right] =u_f \\left( \\frac{\\alpha_1}{p_1} e_1 + \\cdots + \\frac{\\alpha_{{N_0}+1}}{p_{{N_0}+1}} e_{{N_0}+1} \\right)=e_f, \\end{aligned} \\tag{60} which means that Eq. 52 holds for $N=N_0+1$. Therefore, according to the mathematical induction, Eq. 52 holds for $\\forall~N\\in~\\mathbb{R}$ and $\\forall~\\omega_j\\in\\Omega$. Specifically, for $\\omega_j=\\omega_j^*$ and $N=2^{m_{\\rm~red}}$, the corresponding results in property (i) can be obtained. The results in property (ii) are obvious from the fusion scheme in 27 and 28. ### References [1] Schetter T, Campbell M, Surka D. Multiple agent-based autonomy for satellite constellations. Artif Intell, 2003, 145: 147--180. Google Scholar [2] Ley W, Wittmann K, Hallmann W. Handbuch der Raumfahrttechnik. Munich: Carl Hanser Verlag GmbH & CO. KG, 2011. Google Scholar [3] Tapley B D, Ries J C, Davis G W, et al. Precision orbit determination for TOPEX/POSEIDON. J Geophys Res, 1994, 99: 24383--24404. Google Scholar [4] Psiaki M L. Autonomous orbit determination for two spacecraft from relative position measurements. J Guid Control Dynam, 1999, 22: 305--312. Google Scholar [5] Yim J R, Crassidis J L, Junkins J L. Autonomous orbit navigation of two spacecraft system using relative line of sight vector measurements. AAS Paper 04-257, 2004. Google Scholar [6] Markley F L. Autonomous navigation", "$Y$ would be a binomial random variable 463. \\begin{align*} 464. Y\\colon \\Omega &\\to \\{0, 1, 2, 3\\}\\\\ 465. \\omega &\\mapsto y 466. \\end{align*} 467. whose probability mass function is 468. $p_Y(y) = \\binom{3}{y}\\cdot 0.5^y\\cdot 0.5^{3-y} 469. = \\frac{3}{4y!(3-y)!}$ 470. 471. Therefore PMF of $Y$ for each case is 472. \\begin{align*} 473. p_Y(0) = p_Y(3) &= \\frac{3}{4\\cdot 0!3!} = \\frac{1}{8}\\\\ 474. p_Y(1) = p_Y(2) &= \\frac{3}{4\\cdot 1!2!} = \\frac{3}{8} 475. \\end{align*} 476. \\end{enumerate} 477. 478. \\exercise 2 Toss a pair of fair siz-sided dice 479. and let $X$ be the sum of the points 480. \\begin{align*} 481. X\\colon \\Omega &\\to [2, 12]\\cap\\mathbb Z\\\\ 482. \\omega &\\mapsto x 483. \\end{align*} 484. with $\\Omega = S^2 = \\{1, 2, 3, 4, 5, 6\\}^2 \\Longrightarrow n(\\Omega) = 36$. 485. 486. \\begin{enumerate}[(a)] 487. \\item $X$ is a random variable whose PMF is 488. \\begin{align*} 489. p_X(2) &= P\\{(1,1)\\} = \\frac{1}{36}\\\\ 490. p_X(3) &= P\\{(1,2), (2,1)\\} = \\frac{1}{18}\\\\ 491. p_X(4) &= P\\{(1,3), (2,2), (3,1)\\} = \\frac{1}{12}\\\\ 492. p_X(5) &= P\\{(1,4), (2,3), (3,2), (4,1)\\} = \\frac{1}{9}\\\\ 493. p_X(6) &= P\\{(1,5), (2,4), (3,3), (4,2), (5,1)\\} = \\frac{5}{36}\\\\ 494. p_X(7) &= P\\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\\} = \\frac{1}{6}\\\\ 495. p_X(8) &= P\\{(2,6), (3,5), (4,4), (5,3), (6,2)\\} = \\frac{5}{36}\\\\ 496. p_X(9) &= P\\{(3,6), (4,5), (5,4), (6,3)\\} = \\frac{1}{9}\\\\ 497. p_X(10) &= P\\{(4,6), (5,5), (6,4)\\} = \\frac{1}{12}\\\\ 498.", "n < \\omega \\} \\\\ &\\ge \\omega^{\\omega^\\omega} \\\\ &= \\sup\\{\\left( \\omega^\\omega \\right)^n \\mid n < \\omega \\} \\\\ &\\ge \\left(\\omega^\\omega \\right)^2 \\\\ &= \\omega^\\omega \\cdot \\omega^\\omega \\\\ &= \\sup \\{\\omega^\\omega \\cdot \\alpha \\mid \\alpha < \\omega^\\omega \\} \\\\ &\\ge \\omega^\\omega \\cdot 2 \\\\ &> \\omega^\\omega. \\end{align} Combining these calculations we get the desired order: $$\\omega = 2^\\omega = {^\\omega} 2 < \\omega^2 < \\omega^\\omega < ^\\omega\\omega$$ In general, ordinal and cardinal arithmetic are very different beasts and every ordinal arithmetic expression using only ordinals $\\le \\omega$ is countable. proof (sketch) Take a countable transitive model $M$ of a large enough fracture of $ZFC$. Every ordinal expression using only ordinals $\\le \\omega$ can be computed correctly inside $M$ (<- this requires some work). As $M$ only contains countable ordinals (as it is transitive), the result follows. • That is a horrible notation, ${}^\\omega2$. – Asaf Karagila Jul 14 '15 at 6:34 • Thanks for the proof; much to learn. – Sheldon L Jul 14 '15 at 8:39", "only solution is $q = 97$. This implies $p = 63$. Hence, $b = 3p = 189$ and $d = 3q = 291$. Hence, $c = \\frac{d - b}{2} = 51$. Therefore, the perimeter of $\\triangle ABC$ is $b + c + a = 189 + 51 + 219 = \\boxed{\\textbf{(459) }}$. ~Steven Chen (www.professorchenedu.com) ## Solution (Number Theory Part) We wish to solve the Diophantine equation $a^2+ab+b^2=3^2 \\cdot 73^2$. It can be shown that $3|a$ and $3|b$, so we make the substitution $a=3x$ and $b=3y$ to obtain $x^2+xy+y^2=73^2$ as our new equation to solve for. Notice that $r^2+r+1=(r-\\omega)(r-{\\omega}^2)$, where $\\omega=e^{i\\frac{2\\pi}{3}}$. Thus, $$x^2+xy+y^2 = y^2((x/y)^2+(x/y)+1) = y^2 (\\frac{x}{y}-\\omega)(\\frac{x}{y}-{\\omega}^2) = (x-y\\omega)(x-y{\\omega}^2).$$ Note that $8^2+1^2+8 \\cdot 1=73$. Thus, $(8-\\omega)(8-{\\omega}^2)=73$. Squaring both sides yields \\begin{align} (8-\\omega)^2(8-{\\omega}^2)^2&=73^2\\\\ (63-17\\omega)(63-17{\\omega}^2)&=73^2. \\end{align} Thus, by $(2)$, $(63, 17)$ is a solution to $x^2+xy+y^2=73^2$. This implies that $a=189$ and $b=51$, so our final answer is $189+51+219=\\boxed{459}$. ~ Leo.Euler ## Solution(Visual geometry) We look at upper and middle diagrams and get $\\angle BAC = 120^\\circ$. Next we use only the lower Diagram. Let $I$ be incenter $\\triangle ABC$, E be midpoint of biggest arc $\\overset{\\Large\\frown} {BC}.$ Then bisector $AI$ cross circumcircle $\\triangle ABC$ at point $E$. Quadrilateral $ABEC$ is cyclic, so $$\\angle BEC = 180^\\circ - \\angle ABC = 60^\\circ \\implies BE = CE = IE = BC.$$", "(z-z_0)^n .\\end{align*} Corollary $$f$$ is holomorphic iff $$f$$ is analytic. Counterexample to keep in mind: \\begin{align*} f(x) = \\begin{cases} x^2 & x > 0 \\\\ 0 & x\\leq 0 \\end{cases} .\\end{align*} In the case of $${\\mathbb{R}}$$, smooth and analytic are very different categories of functions. Wednesday January 29th Cauchy’s Integral Formula Theorem (Cauchy’s Integral Formula) Let $$f: \\Omega \\to {\\mathbb{C}}$$ be holomorphic, so $$f\\in C^1(\\mkern 1.5mu\\overline{\\mkern-1.5mu\\Omega\\mkern-1.5mu}\\mkern 1.5mu)$$. Then for any $$z\\in \\Omega$$, \\begin{align*} f(z) = \\frac{1}{2\\pi i} \\int_{{{\\partial}}\\Omega} \\frac{f(\\xi)}{\\xi - z} ~d\\xi .\\end{align*} In general, \\begin{align*} f^{(n)}(z) = \\frac{n!}{2\\pi i} \\int_{{{\\partial}}\\Omega} \\frac{f(\\xi)}{\\qty{\\xi - z}^{n+1}} ~d\\xi .\\end{align*} This implies that $$f$$ is analytic, i.e. \\begin{align*} f(z) = \\sum a_n (z-z_0)^n \\quad\\text{where}\\quad a_n = \\frac{f^{(n)}(z_0) }{n!} .\\end{align*} Thus $$f$$ is holomorphic iff $$f$$ is analytic, and \\begin{align*} \\int_{{{\\partial}}\\Omega} f = 0 \\implies \\int_{{{\\partial}}\\Omega_\\gamma} \\frac{f(\\xi)}{\\xi - z}~d\\xi = 0 .\\end{align*} where $$\\Omega_r = \\Omega\\setminus D_r(z)$$, and $${{\\partial}}\\Omega_r = {{\\partial}}\\Omega \\cup(-{{\\partial}}D_r)$$. We can thus shrink integrals: \\begin{align*} \\int_{{{\\partial}}\\Omega} \\frac{f(\\xi)} {\\xi - z} ~d\\xi = \\int_{C_r} \\frac{f(\\xi)} {\\xi - z} ~d\\xi .\\end{align*} Proposition (Homotopy Invariance) Let $$f\\in C^1(\\Omega)$$ be holomorphic on $$\\Omega$$. Let $$\\gamma_s(t)$$ be a family of smooth curves in $$\\Omega$$; then $$\\int_{\\gamma_s} f$$ is independent of $$s$$. Proof Write \\begin{align*}\\gamma_s(t) = \\gamma(s, t): [a, b] \\times[0, 1] \\to \\Omega.\\end{align*} We have $$\\gamma_s(0) = \\gamma_s(1)$$ so $${\\frac{\\partial \\gamma}{\\partial s}\\,}(s, 0) = {\\frac{\\partial", "= (3,2,1)$$ and we define $$\\omega = x \\,dx + 2y^2 \\,dy - 3xz \\,dz$$, then \\begin{align*} \\omega(t_a) &= 1 \\,dx (3,2,1) + 2(2)^2 \\,dy(3,2,1) - 3 \\cdot 1 \\cdot 3 \\,dz (3,2,1) \\\\ &= 1 \\cdot 3 + 8 \\cdot 2 - 9 \\cdot 1 \\\\ &=10. \\end{align*} • Thanks for that really quick and great answer. Since I can't seem to find an example that basic, could you also tell me how I would wedge the two $\\omega_1$ and $\\omega_2$ from above? – MJP Dec 9 '19 at 22:04 • Sure thing: note that $\\,dx \\wedge \\,dx = 0$. Then with $\\omega_1 = \\,dx + 2 \\,dy - 4 \\,dz$ and $\\omega_2 = x \\,dx + 2y^2 \\,dy - 3xz \\,dz$, we have: $\\omega_1 \\wedge \\omega_2 = x \\,dx \\wedge \\,dx + 2x \\,dy \\wedge \\,dx - 4x \\,dz \\wedge \\,dx + 2y^2 \\,dx \\wedge \\,dy + 4y^2 \\,dy \\wedge \\,dy - 8y^2 \\,dz \\wedge \\,dy - 3xz \\,dx \\wedge \\,dz - 6xz \\,dy \\wedge \\,dz + 12xz \\,dz \\wedge \\,dz$. Dec 9 '19 at 22:15 • Thanks, great help. – MJP Dec 9 '19 at 22:17" ]

[ "1.000000000 where in the frame at rest with axis of rotation the (x,y) coordinates for the particles are a = (1.0), b = (0,-1), c = (-1,0) and d = (0,1). It can be seen that the y' coordinates of particles a and c are the same and above the centre of the axis rotation by a distance of 0.519881721 so we can assume the centre of mass for this pair of particles is at (x',y') = (0,0.519881721) Thinking about the 4 particles is a bad way of thinking, there is no way that the center of mass for the complete wheel will be anywhere close to y'=0.519881721. The correct way is to do the following: $$dx'=-\\gamma sin(\\phi+t \\omega) d \\phi$$ $$dy'=cos (\\phi+t \\omega) d \\phi$$ The arc element is : $$ds'=\\sqrt (dx'^2+dy'^2)$$ The center of mass is obtained by integrating from 0 to $$2 \\pi$$ the expression: $$\\frac{\\int y' ds'}{\\int ds'}$$ It is very easy to show that $$\\int y' ds'=0$$ Related Discussions Special & General Relativity 9 Introductory Physics Homework 2 Introductory Physics Homework 2 General Physics 7 General Physics 0", "# TrigonometryParticle's position #### dwsmith ##### Well-known member Relatively easy question but I haven't done anything like this in awhile. $$\\mathbf{r}(t) = \\hat{\\mathbf{x}}R\\cos(\\omega t) + \\hat{\\mathbf{y}}R\\sin(\\omega t)$$ The particle moves in a circle so I want to show that the position is given by the above. I know at $t = 0$ the particle is on the x-axis. Therefore, we know that at t = 0, $(R,0)$. Obviously it has to be that but I guess there is a way to show this basically. #### MarkFL Staff member re: Particles position Consider the parametrization: $\\displaystyle x(t)=R\\cos(\\omega t)$ $\\displaystyle y(t)=R\\sin(\\omega t)$ Square both and add to eliminate the parameter. #### dwsmith ##### Well-known member re: Particles position Consider the parametrization: $\\displaystyle x(t)=R\\cos(\\omega t)$ $\\displaystyle y(t)=R\\sin(\\omega t)$ Square both and add to eliminate the parameter. That will just be $x(t)^2 + y(t)^2 = R^2$. What about x and y hat? #### MarkFL Staff member re: Particles position Aren't they orthogonal unit vectors? I'm sorry, I'm not familiar with that notation. I interpreted it as: $\\displaystyle \\vec{r}(t)=R\\langle \\cos(\\omega t),\\sin(\\omega t) \\rangle$ #### dwsmith ##### Well-known member re: Particles position Aren't they orthogonal unit vectors? I'm sorry, I'm not familiar with that notation. I interpreted it as: $\\displaystyle \\vec{r}(t)=R\\langle \\cos(\\omega t),\\sin(\\omega t) \\rangle$ They could be but it doesn't specify. All we", "the Nilsson single particle orbit. It is because when a particle has $m_j = K$, the angular momentum must at least $j \\geq K$. The above is a general solution for the harmonic oscillator in cylindrical coordinate. When $\\omega_z = \\omega_\\rho$, it reduce to spherical case. According to P. Ring & P. Schuck (2004) (The Nuclear Many-Body Problem, P.68), the Hamiltonian can be expressed as a quadruple deform field by setting $\\displaystyle \\omega_\\rho^2 = \\omega^2 \\left(1+\\frac{2}{3} \\delta\\right) \\\\ \\omega_z^2 = \\omega^2 \\left(1-\\frac{4}{3} \\delta\\right)$ $\\displaystyle H = -\\frac{\\hbar^2}{2m}\\nabla^2 + \\frac{1}{2} m \\omega^2 \\left(r^2- \\frac{2}{3}\\sqrt{\\frac{16\\pi}{5}} \\delta r^2 Y_{20}(\\theta,\\phi) \\right)$ This Hamiltonian has similarity with the deformation $\\displaystyle R(\\theta, \\phi) = R_0 \\left(1 + \\alpha_00 + \\sum_{\\lambda=1}^{\\infty}\\sum_{\\mu=-\\lambda}^{\\lambda} \\alpha_{\\lambda \\mu} Y_{\\lambda \\mu}(\\theta, \\phi) \\right)$ take the first quadruple term, and calculate $R^2(\\theta, \\phi)$ $R^2(\\theta, \\phi) = R_0^2 (1 + 2 \\beta Y_{20}(\\theta,\\phi) )$ Compare, we have $\\displaystyle \\beta = \\frac{1}{3}\\sqrt{\\frac{16\\pi}{5}}\\delta \\approx 1.05689 \\delta$ The ratio $\\displaystyle \\frac{\\omega_z}{\\omega_\\rho} = \\left(\\frac{\\alpha_\\rho}{\\alpha_z} \\right) = \\sqrt{\\frac{1-\\frac{4}{3}\\delta}{1+\\frac{2}{3}\\delta}} = 1- \\delta + \\frac{1}{6}\\delta^2 - \\frac{5}{18}\\delta^3...$ We need volume conservation $\\omega_x \\omega_y \\omega_z = \\omega_0^3$ Thus, $\\displaystyle \\omega = \\omega_0 \\left(\\frac{1}{(1+2/3\\delta)(1-4/3\\delta)} \\right)^{\\frac{1}{6}} \\approx \\omega_0 \\left(1+\\frac{2}{9}\\delta^2 + \\frac{8}{81}\\delta^3 \\right)$ The energy is $\\displaystyle E = \\hbar\\omega \\left( \\sqrt{1-\\frac{4}{3}\\delta}\\left(\\frac{1}{2} + n_z \\right) + \\sqrt{1+\\frac{2}{3}\\delta} \\left( N - n_z + 1 \\right) \\right) \\\\ \\approx \\hbar \\omega_0 \\left( \\frac{3}{2} + \\left(1 +", "and $q = y_{150} = y_6 = 5 + 5\\sqrt{2}$. Hence, the final answer is $5\\sqrt {2} + 5(\\sqrt {2} + 1) \\approx 19.1 \\Longrightarrow \\boxed{019}$ If you're curious, the points do eventually form an octagon and repeat. Seems counterintuitive, but believe it or not, it happens. ### Solution 2 Let the particle's position be represented by a complex number. Recall that multiplying a number by cis$\\left( \\theta \\right)$ rotates the object in the complex plane by $\\theta$ counterclockwise. In this case, we use $a = cis(\\frac{\\pi}{4})$. Therefore, applying the rotation and shifting the coordinates by 10 in the positive x direction in the complex plane results to $a_{150} = (((5a + 10)a + 10)a + 10 \\ldots) = 5a^{150} + 10 a^{149} + 10a^{148}+ \\ldots + 10$ where a is cis$\\left( \\theta \\right)$. By De-Moivre's theorem, $\\left(cis( \\theta \\right)^n )$=cis$\\left(n \\theta \\right)$. Therefore, $10(a^{150} + \\ldots + 1) = 10(1 + a + \\ldots + a^6) = - 10(a^7) = - 10(\\frac{ \\sqrt {2} }{2} - \\frac{i\\sqrt {2}} {2})$ Furthermore, $5a^{150} = - 5i$. Thus, the final answer is $5\\sqrt {2} + 5(\\sqrt {2} + 1) \\approx 19.1 \\Longrightarrow \\boxed{019}$ ### Solution 3 Each move has two parts, a rotation and a translation. But consider what happens to a translation after four rotations: it is cancelled out", "general number of particles per box, it is $$\\Omega_\\text{sys} = \\Omega(V_1, N_1)\\Omega(V_2, N_2) \\times {N_1+N_2 \\choose N_1,N_2}\\,,$$ where we have instead multiplied by a multinomial coefficient. Intuitively, we are saying that if $$(\\vec{q}_i,\\vec{p}_i)$$ is in a region of phase space compliant with our macroscopic prescription, then so will any permutation of the position-momentum pairs. We now remove the divider between the boxes and let the particles mix. The system multiplicity is now $$\\Omega_\\text{sys}' = \\Omega(V_1+V_2, N_1 + N_2) \\,.$$ This turns out to be equal (kind of) to the multiplicity before the divider was removed $$\\Omega_\\text{sys}$$, in other words, the Gibbs paradox was resolved when we correctly determined the system's initial multiplicity. We now prove their \"equality\". \\begin{align*} \\Omega_\\text{sys} &= \\Omega(V_1, N_1)\\Omega(V_2, N_2) \\times {N_1+N_2 \\choose N_1,N_2} \\\\ &= f(N_1)f(N_2)\\left(\\frac{V_1}{\\lambda^3}\\right)^{N_1}\\left(\\frac{V_2}{\\lambda^3}\\right)^{N_2}\\frac{(N_1+N_2)!}{N_1!N_2!}\\delta^2 \\end{align*} Take the logarithm, i.e. calculate the entropy. \\begin{align*} \\ln\\Omega_\\text{sys} &= \\ln f(N_1) + \\ln f(N_2) + N_1\\ln\\left(\\frac{V_1}{\\lambda^3}\\right) +N_2\\ln\\left(\\frac{V_2}{\\lambda^3}\\right) + \\ln\\left(\\frac{(N_1+N_2)!}{N_1!N_2!}\\right) + 2\\ln \\delta \\end{align*} Discard the constant term. \\begin{align*} &\\approx \\ln f(N_1) + \\ln f(N_2) + N_1\\ln\\left(\\frac{V_1}{\\lambda^3}\\right) +N_2\\ln\\left(\\frac{V_2}{\\lambda^3}\\right) + \\ln\\left(\\frac{(N_1+N_2)!}{N_1!N_2!}\\right) \\end{align*} One can show that $$\\ln f(N) \\sim 3N/2$$ as $$N\\to\\infty$$. \\begin{align*} &\\approx \\frac{3(N_1+N_2)}{2} + N_1\\ln\\left(\\frac{V_1}{\\lambda^3}\\right) +N_2\\ln\\left(\\frac{V_2}{\\lambda^3}\\right) + \\ln\\left(\\frac{(N_1+N_2)!}{N_1!N_2!}\\right) \\end{align*} Recall that $$N_1/V_1=N_2/V_2$$. \\begin{align*} &= \\frac{3(N_1+N_2)}{2} + N_1\\ln\\left(\\frac{V_1/N_1}{\\lambda^3}\\right) +N_2\\ln\\left(\\frac{V_2/N_2}{\\lambda^3}\\right) + N_1\\ln N_1 + N_2\\ln N_2 + \\ln\\left(\\frac{(N_1+N_2)!}{N_1!N_2!}\\right) \\\\ &= \\frac{3(N_1+N_2)}{2} + (N_1+N_2)\\ln\\left(\\frac{V_1/N_1}{\\lambda^3}\\right) + N_1\\ln N_1 + N_2\\ln", "magnitude A$$\\omega^{2}$$ and is directed opposite the position vector, toward the origin, because $$\\vec{a}$$(t) = −$$\\omega^{2} \\vec{r}$$(t). Example $$\\PageIndex{2}$$: Circular Motion of a Proton A proton has speed 5 x 106 m/s and is moving in a circle in the xy plane of radius r = 0.175 m. What is its position in the xy plane at time t = 2.0 x 10−7 s = 200 ns? At t = 0, the position of the proton is 0.175 m $$\\hat{i}$$ and it circles counterclockwise. Sketch the trajectory. Solution From the given data, the proton has period and angular frequency: $T = \\frac{2 \\pi r}{v} = \\frac{2 \\pi (0.175\\; m)}{5.0 \\times 10^{6}\\; m/s} = 2.20 \\times 10^{-7}\\; s \\nonumber$ $\\omega = \\frac{2 \\pi}{T} = \\frac{2 \\pi}{2.20 \\times 10^{-7}\\; s} = 2.856 \\times 10^{7}\\; rad/s \\ldotp \\nonumber$ The position of the particle at t = 2.0 x 10−7 s with A = 0.175 m is \\begin{align*} \\vec{r} (2.0 \\times 10^{-7}\\; s) & = A \\cos \\omega (2.0 \\times 10^{-7}\\; s) \\hat{i} + A \\sin \\omega (2.0 \\times 10^{-7}\\; s) \\hat{j}\\; m \\\\[4pt] & = 0.175 \\cos (2.856 \\times 10^{7}\\; rad/s) (2.0 \\times 10^{-7}\\; s) \\hat{i} + 0.175 \\sin (2.856 \\times 10^{7}\\; rad/s) (2.0 \\times 10^{-7}\\; s) \\hat{j}\\; m \\\\[4pt] & = 0.175 \\cos (5.712\\; rad) \\hat{i} + 0.175 \\sin (5.172\\;", "CT (5) is \\begin{align} F_2(q,P)~=~&q^3P_3+\\frac{\\sum_{j=1}^2q^jP_j}{\\cos\\theta}\\cr ~+~&\\frac{1}{2}\\sum_{j,k=1}^2\\left(m\\omega|\\epsilon_{jk}| q^jq^k+\\frac{1}{m\\omega}|\\epsilon^{jk}| P_jP_k\\right) \\tan\\theta.\\end{align}\\tag{12} $$^2$$ The Fock vacuum $$|\\Omega\\rangle$$ happens to be invariant under the CT (5): \\begin{align} a^j|\\Omega\\rangle ~=~&0 \\cr ~\\Updownarrow~& \\cr A^j|\\Omega\\rangle ~=~&0,\\cr j~\\in~&\\{1,2,3\\}.\\end{align} \\tag{13} Note that the Fock vacuum $$|\\Omega\\rangle$$ depends on the constant $$m\\omega$$. First, we have to make a physical demand. We demand that a rotation by $2\\pi$ of spatial configuration (distinct from an internal configuration, i.e. spin) leaves physics invariant. From a study of the first homotopy group $Z_2$, we know that in general there are two possibilities for a rotation by $2\\pi$ (this is usually studied in the context of topological quantization): $R(2\\pi)=1$ and $R(2\\pi)=-1$. For $R$ to be a physical spatial rotation, we demand $R(2\\pi)=1.$ However, the rotation operator in the $x-y$ plane is $R=\\exp(-i\\theta L_z)$ where $L_z|\\psi\\rangle=m|\\psi\\rangle$. Thus we require $\\exp(-2\\pi im)=0$. This is only solved by $m\\in\\mathbb{Z}$, which, by the rules of ladder operators, implies $l\\in\\mathbb{Z}$. • I came here with the same question and I don't like your argumentation. If letting $m$ be an half-integer, then, the single-valued argument doesn't apply anymore, but the probability density does not change, and that's what matters in the end, right?...not the wave function itself Jan 2, 2018 at 14:47 • There is a small mistake: It has to be $\\exp(-2\\pi im) = 1$. Mar 18,", "configuration file is located at ` feelpp/applications/models/fsi/TurekHron` The result files are then stored by default in ``` feel/applications/models/fsi/TurekHron/fsi3/ <velocity_space><pression_space><Geometric_order>-<OrderDisp><Geometric_order>/np_<processor_used>``` For example, for the FSI3 case executed on $4$ processors, with a $P_2$ velocity approximation space, a $P_1$ pressure approximation space, a geometric order of $1$ for fluid part and a $P_1$ displacement approximation space and geometric order equals to $1$ for solid part, the path is ` feel/applications/models/fsi/TurekHron/fsi3/P2P1G1-P1G1/np_4` Results First at all, we will discretize the simulation parameters for the different cases studied. $N_{elt}$ $N_{dof}$ $[P^N_c(\\Omega_{f,\\delta}]^2 \\times P^{N-1}_c(\\Omega_{f,\\delta}) \\times V^{N-1}_{s,\\delta}$ $\\Delta t$ Turek and Hron 15872 304128 0.00025 (1) 1284 27400 $[P^4_c(\\Omega_{f,(h,3)}]^2 \\times P^3_c(\\Omega_{f,(h,3)}) \\times V^3_{s,(h,3)}$ 0.005 (2) 2117 44834 $[P^4_c(\\Omega_{f,(h,3)}]^2 \\times P^3_c(\\Omega_{f,(h,3)}) \\times V^3_{s,(h,3)}$ 0.005 (3) 4549 95427 $[P^4_c(\\Omega_{f,(h,3)}]^2 \\times P^3_c(\\Omega_{f,(h,3)}) \\times V^3_{s,(h,3)}$ 0.005 (4) 17702 81654 $[P^2_c(\\Omega_{f,(h,1)}]^2 \\times P^1_c(\\Omega_{f,(h,1)}) \\times V^1_{s,(h,1)}$ 0.0005 Then the FSI3 benchmark results are detailed below. $x$ displacement $[\\times 10^{-3}]$ $y$ displacement $[\\times 10^{-3}]$ Drag Lift [TurekHron] -2.69 ± 2.53 [10.9] 1.48 ± 34.38 [5.3] 457.3 ± 22.66 [10.9] 2.22 ± 149.78 [5.3] [Breuer] 464.5 ± 40.50 6.00 ± 166.00 [5.5] [TurekHron2] -2.88 ± 2.72 [10.9] 1.47 ± 34.99 [5.5] 460.5 ± 27.74 [10.9] 2.50 ± 153.91 [5.5] [MunschBreuer] -4.54 ± 4.34 [10.1] 1.50 ± 42.50 [5.1] 467.5 ± 39.50 [10.1] 16.2 ± 188.70 [5.1] [Gallinger] 474.9 ± 28.10 3.90 ± 165.90", "and at positions $$x_1(0)$$ and $$x_2(0)$$. This means that the initial conditions for the normal mode variables are \\begin{align*}\\dot{X}_1(0) &= \\dot{x}_1(0) + \\dot{x}_2(0) = 0 \\\\ \\dot{X}_2(0) &= \\dot{x}_1(0) - \\dot{x}_2(0) = 0\\end{align*} this means we have $$\\dot{X}_1(0) = -C_1\\omega_1\\sin(\\phi_1)=0$$ and $$\\dot{X}_2(0) = -C_2\\omega_2\\sin(\\phi_2)=0$$ from which we can get $$\\phi_1=\\phi_2=0$$. Now the initial positions give $$X_1(0) = C_1 = x_1(0)+x_2(0)$$ and $$X_2(0) = C_2 = x_1(0)-x_2(0)$$. So in total the equations of motion will be \\begin{align*}x_1(t) &= [x_1(0)+x_2(0)]\\cos(\\omega_1 t) + [x_1(0)-x_2(0)]\\cos(\\omega_2 t)\\\\ x_2(t) &= [x_1(0)+x_2(0)]\\cos(\\omega_1 t) - [x_1(0)-x_2(0)]\\cos(\\omega_2 t).\\end{align*} Below is a simulation of the two gliders. If you start one glider at its rest position and the coupling is weak, you will see beats between the two gliders!", "energy39, that shall be less than the difference between the neighbouring energy levels of the particle. How can we really avoid any sudden jump in the composition of quantum control operations? We shall not answer that question with utter certainty, rather we would like to sketch an attempt in the following discussion. From our viewpoint, the composition approach consists of taking portions of the time evolution course described by the Hamiltonian (Eq. 8), with the exception that this time the elastic fields (Eq. 7) shall be transformed as $$\\beta ^2(t)\\rightarrow \\beta (t)$$ to facilitate the quest of exact solutions. Hence, in the subspace of oscillations, the matrix h(t) becomes: \\begin{aligned} h(t)=\\begin{pmatrix}\\cos (\\omega t) &{} \\frac{1}{\\omega }\\sin (\\omega t) \\\\ -\\omega \\sin (\\omega t) &{} \\cos (\\omega t)\\end{pmatrix}, \\end{aligned} (21) noting that it adopts the form of a symplectic matrix of rotations. In particular, for any $$\\omega t=\\frac{n\\pi }{2}$$ (n is an integer) we obtain the squeezing transformations: \\begin{aligned} h\\left( \\frac{n\\pi }{2\\omega }\\right) =\\begin{pmatrix}0&{}\\pm \\frac{1}{\\omega }\\\\ \\mp \\omega &{}0\\end{pmatrix}, \\end{aligned} (22) now, the successive application of two matrices of this type leads to: \\begin{aligned} h_\\text {squeezing} = \\begin{pmatrix}0&{}\\pm \\frac{1}{\\omega _1}\\\\ \\mp \\omega _1&{}0\\end{pmatrix}\\begin{pmatrix}0&{}\\pm \\frac{1}{\\omega _2}\\\\ \\mp \\omega _2&{}0\\end{pmatrix}=\\begin{pmatrix}\\lambda &{}0\\\\ 0&{}\\frac{1}{\\lambda }\\end{pmatrix}, \\quad \\lambda = -\\frac{\\omega _2}{\\omega _1}, \\quad \\omega _1\\ne \\omega _2, \\end{aligned} (23) leading again to the set", "1,b_1 = 3, \\text{ and } b_n = b_{n - 1} + \\frac {b_{n - 1}^2}{b_{n - 2}}\\text{ for }n\\ge2.$$ Find $\\frac {b_{32}}{a_{32}}$. ## Problem 7 Let $r$, $s$, and $t$ be the three roots of the equation $$8x^3 + 1001x + 2008 = 0.$$ Find $(r + s)^3 + (s + t)^3 + (t + r)^3$. ## Problem 8 Let $a = \\pi/2008$. Find the smallest positive integer $n$ such that $$2[\\cos(a)\\sin(a) + \\cos(4a)\\sin(2a) + \\cos(9a)\\sin(3a) + \\cdots + \\cos(n^2a)\\sin(na)]$$ is an integer. ## Problem 9 A particle is located on the coordinate plane at $(5,0)$. Define a move for the particle as a counterclockwise rotation of $\\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$-direction. Given that the particle's position after $150$ moves is $(p,q)$, find the greatest integer less than or equal to $|p| + |q|$. ## Problem 10 The diagram below shows a $4\\times4$ rectangular array of points, each of which is $1$ unit away from its nearest neighbors. $[asy] unitsize(0.25inch); defaultpen(linewidth(0.7)); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); [/asy]$ Define a growing path to be a sequence of distinct points of the array with the property that the distance between consecutive points of the", "fixed triangle $$\\Delta _m$$, with respective nodes $$\\omega _{m,0}$$ to $$\\omega _{m,d}$$, in positive orientation. The associated image points are $$G_{m,0}$$ to $$G_{m,d}$$. Since we may choose any vertex to be labelled $$\\omega _{m,0}$$, it will suffice to perform the calculations at one fixed image point $$G_{m,0}$$. • mass term: \\begin{aligned} \\frac{\\partial }{\\partial G_{m,0}}{\\mathbb {L}}_{\\mathscr {T}}^m(G,G^*)&=\\frac{2}{(d+1)(d+2)}\\frac{\\partial }{\\partial G_{m,0}}\\sum _{0\\le i\\le j\\le d}(G_{m,i}-G^*_{m,i})\\cdot (G_{m,j}-G^*_{m,j}) \\\\&=\\frac{2}{(d+1)(d+2)}\\left( 2(G_{m,0}-G^*_{m,0})+\\sum _{j=1}^d(G_{m,j}-G^*_{m,j})\\right) \\end{aligned} • internal energy: observing that—recall (1.2) — \\begin{aligned} \\widetilde{h}'(s) = \\frac{\\mathrm {d}}{\\,\\mathrm {d}s}\\big [sh(s^{-1})\\big ] = h(s^{-1})-s^{-1}h'(s^{-1}) = -P(s^{-1}), \\end{aligned} (3.8) we obtain \\begin{aligned} \\frac{\\partial }{\\partial G_{m,0}}{\\mathbb {H}}_{\\mathscr {T}}^m(G) = \\frac{\\partial }{\\partial G_{m,0}} \\widetilde{h}\\left( \\frac{\\det Q_{\\mathscr {T}}^m[G]}{2\\mu _{\\mathscr {T}}^m}\\right) = \\frac{1}{2\\mu _{\\mathscr {T}}^m}P\\left( \\frac{2\\mu _{\\mathscr {T}}^m}{\\det Q_{\\mathscr {T}}^m[G]}\\right) \\nu _{\\mathscr {T}}^m[G], \\end{aligned} where \\begin{aligned} \\nu _{\\mathscr {T}}^m[G] := - \\frac{\\partial }{\\partial G_{m,0}} \\det Q_{\\mathscr {T}}^m[G] = (\\det Q_{\\mathscr {T}}^m[G])\\, (Q_{\\mathscr {T}}^m[G])^{-T}\\sum _{j=1}^d\\mathrm {e}_j \\end{aligned} (3.9) is the uniquely determined vector in $${\\mathbb {R}}^d$$ that is orthogonal to the $$(d-1)$$-simplex with corners $$G_{m,1}$$ to $$G_{m,d}$$ (pointing away from $$G_{m,0}$$) and whose length equals the $$(d-1)$$-volume of that simplex. • potential energy: Now let $$\\omega _\\ell$$ be a fixed vertex of $${\\mathscr {T}}$$. Summing over all simplices $$\\Delta _m$$ that have $$\\omega _\\ell$$ as a vertex, and choosing vertex labels in accordance with above, i.e., such that $$\\omega _{m,0}=\\omega _\\ell$$ in $$\\Delta _m$$,", "TrigonometryParticle's position dwsmith Well-known member Relatively easy question but I haven't done anything like this in awhile. $$\\mathbf{r}(t) = \\hat{\\mathbf{x}}R\\cos(\\omega t) + \\hat{\\mathbf{y}}R\\sin(\\omega t)$$ The particle moves in a circle so I want to show that the position is given by the above. I know at $t = 0$ the particle is on the x-axis. Therefore, we know that at t = 0, $(R,0)$. Obviously it has to be that but I guess there is a way to show this basically. MarkFL Staff member re: Particles position Consider the parametrization: $\\displaystyle x(t)=R\\cos(\\omega t)$ $\\displaystyle y(t)=R\\sin(\\omega t)$ Square both and add to eliminate the parameter. dwsmith Well-known member re: Particles position Consider the parametrization: $\\displaystyle x(t)=R\\cos(\\omega t)$ $\\displaystyle y(t)=R\\sin(\\omega t)$ Square both and add to eliminate the parameter. That will just be $x(t)^2 + y(t)^2 = R^2$. What about x and y hat? MarkFL Staff member re: Particles position Aren't they orthogonal unit vectors? I'm sorry, I'm not familiar with that notation. I interpreted it as: $\\displaystyle \\vec{r}(t)=R\\langle \\cos(\\omega t),\\sin(\\omega t) \\rangle$ dwsmith Well-known member re: Particles position Aren't they orthogonal unit vectors? I'm sorry, I'm not familiar with that notation. I interpreted it as: $\\displaystyle \\vec{r}(t)=R\\langle \\cos(\\omega t),\\sin(\\omega t) \\rangle$ They could be but it doesn't specify. All we know is that they are vectors. dwsmith Well-known member", "\\forall \\mu \\in \\Lambda (t)$$ $$V\\left( 0 \\right){\\text{ }} = {V_0};\\omega \\left( 0 \\right){\\text{ }} = {\\omega _0};G\\left( 0 \\right){\\text{ }} = {G_0};B\\left( 0 \\right){\\text{ }} = {B_0}$$ $$u(x,0) = {\\tilde u_0}(x) = \\left\\{ \\begin{matrix} {u_0}(x),\\forall x \\in \\Omega \\backslash \\overline B (0) \\\\ {V_0} + {\\omega _0} \\times {G_0}x,\\forall x \\in \\overline B (0) \\\\ \\end{matrix} \\right.$$ with the following functional spaces $$\\begin{matrix} {W_{{g_0}}}(t) = \\{ v|v \\in {({H^1}(\\Omega ))^3},v = {g_0}(t) {\\text{on}} \\Gamma \\} \\\\ L_0^2(\\Omega ) = \\{ q|q \\in {L^2}(\\Omega ),\\int_\\Omega {q{\\text{d}}x = 0\\} ,} \\Lambda (t) = {({H^1}(B(t)))^3} \\\\ \\end{matrix}$$ Various examples for $$\\textless \\cdot , \\cdot { \\textgreater _{\\Lambda (t)}}$$ in Eqs.(12) and (15) are given in (Glowinski et al., ; Glowinski, ). In Eqs.(11)−(17), only the center of mass, the translation velocity of the center of mass and the angular velocity of the particle are considered, which are sufficient for spherical particle cases. For non-spherical particles, knowing these two velocities and the center of mass of the particle, we can translate and rotate the particle in space by tracking, in each particle, two extra points x1 and x2 which follow the rigid body motion, namely $$\\frac{{{\\text{d}}{x_i}}}{{{\\text{d}}t}} = V(t) + \\omega (t) \\times G(t){x_i}, {x_i}(0) = {x_{i,0}}, i = 1,2$$ In practice we track two orthogonal normalized vectors rigidly attached to the", "is the analogue of the plane wave result \\begin{aligned}\\psi(x) = e^{i k x} + \\beta_k e^{-i k x}\\end{aligned} \\hspace{\\stretch{1}}(2.31) # Scattering geometry and nomenclature. We can think classically first, and imagine a scattering of a stream of particles barraging a target as in figure (\\ref{fig:qmTwoL23:qmTwoL23fig2}) \\begin{figure}[htp] \\centering \\includegraphics[totalheight=0.3\\textheight]{qmTwoL23fig2} \\caption{Scattering cross section.} \\end{figure} Here we assume that $d\\Omega$ is far enough away that it includes no non-scattering particles. Write $P$ for the number density \\begin{aligned}P = \\frac{\\text{number of particles}}{\\text{unit volume}},\\end{aligned} \\hspace{\\stretch{1}}(3.32) and \\begin{aligned}J = P v_0 =\\frac{\\text{Number of particles flowing through}}{\\text{a unit area in unit time}}\\end{aligned} \\hspace{\\stretch{1}}(3.33) We want to count the rate of particles per unit time $dN$ through this solid angle $d\\Omega$ and write \\begin{aligned}dN = J \\left( \\frac{d \\sigma(\\Omega)}{d\\Omega} \\right) d\\Omega.\\end{aligned} \\hspace{\\stretch{1}}(3.34) The factor \\begin{aligned}\\frac{d \\sigma(\\Omega)}{d\\Omega},\\end{aligned} \\hspace{\\stretch{1}}(3.35) is called the differential cross section, and has “units” of \\begin{aligned}\\frac{\\text{area}}{\\text{steradians}}\\end{aligned} \\hspace{\\stretch{1}}(3.36) The total number of particles through the volume per unit time is then \\begin{aligned}\\int J \\frac{d \\sigma(\\Omega)}{d\\Omega} d\\Omega= J \\int \\frac{d \\sigma(\\Omega)}{d\\Omega} d\\Omega= J \\sigma\\end{aligned} \\hspace{\\stretch{1}}(3.37) where $\\sigma$ is the total cross section and has units of area. The cross section $\\sigma$ his the effective size of the area required to collect all particles, and characterizes the scattering, but isn’t necessarily entirely geometrical. For example, in photon scattering we may have frequency matching with atomic resonance, finding", "the particle at an arbitrary time $t\\text{?}$ (b) For the particle moving in a counterclockwise sense, find velocity at $t=0\\text{?}$ Hint (a) Use center of circle and redius to get $x$ and $y\\text{.}$ (b) Take derivative. (a) $x = 1 + 0.2 \\cos(50 t),\\ \\ y = 0.2 \\sin(50 t)\\text{,}$ (b)$10\\text{ m/s}\\ \\hat j\\text{.}$ Solution 1 (a) (a) From the fact that the motion is circular with center at $(1\\text{ m},0)$ and radius $R=0.2\\text{ m}\\text{,}$ we can immediately write $x$ and $y$ coordinates \\begin{equation*} x = 1 + 0.2\\; \\cos(\\omega t),\\ \\ y = 0.2\\; \\sin(\\omega t), \\end{equation*} where $\\omega = 50 \\text{ rad/sec}\\text{.}$ Solution 2 (b) (b) Taking derivatives we get the components, which we evaluate at $t=0\\text{.}$ \\begin{align*} v_x \\amp = -0.2\\;\\omega\\; \\sin(\\omega t) = 0,\\\\ v_y \\amp = 0.2\\;\\omega\\; \\cos(\\omega t) = 10\\text{ m/s}. \\end{align*} The velocity has magnitude $10\\text{ m/s}$ and directed towards positive $x$ axis. The position of a particle moving in the $xy$-plane is given by $\\vec r(t) = a \\cos(\\pi t) \\hat i + (b + a \\sin(\\pi t)) \\hat j\\text{,}$ where $t$ is in $\\text{sec}$ and $a$ in $\\text{cm}\\text{.}$ (a) Show that the motion is a circular motion, and find the value of the radius of the circle, and Cartesian coordinates of the center of the circle. (b) Find an", "particle will be given by $$\\theta=\\omega t$$ and so the position of the particle at time $$t$$ is given by $$\\boldsymbol{r}(t)=R(\\omega t)\\boldsymbol{r}=\\left(\\begin{array}{ccc} \\cos\\omega t & -\\sin\\omega t & 0\\\\ \\sin\\omega t & \\cos\\omega t & 0\\\\ 0&0&1 \\end{array}\\right)\\left(\\begin{array}{c} 1\\\\0\\\\0\\end{array}\\right).$$ Now, we know the definition of velocity is $$\\boldsymbol{v}=\\frac{d\\boldsymbol{r}}{dt}$$, so we need to take a time derivative of the rotation matrix, $$\\frac{d}{dt}\\left(\\begin{array}{ccc} \\cos\\omega t & -\\sin\\omega t & 0\\\\ \\sin\\omega t & \\cos\\omega t & 0\\\\ 0&0&1 \\end{array}\\right)=\\omega\\left(\\begin{array}{ccc} -\\sin\\omega t & -\\cos\\omega t & 0\\\\ \\cos\\omega t & -\\sin\\omega t & 0\\\\ 0&0&0 \\end{array}\\right)=-\\omega\\left(\\begin{array}{ccc} 0 & -1 & 0\\\\ 1 & 0 & 0\\\\ 0 & 0 & 0 \\end{array}\\right)\\left(\\begin{array}{ccc} \\cos\\omega t & -\\sin\\omega t & 0\\\\ \\sin\\omega t & \\cos\\omega t & 0\\\\ 0&0&1 \\end{array}\\right).$$ The last step is non-trivial, but you can check yourself that it multiplies out correctly (I may be off by a minus sign). It's very important that we end up with this numeric matrix times the original rotation matrix. So then, this means that $$\\boldsymbol{v}=-\\omega\\left(\\begin{array}{ccc} 0 & -1 & 0\\\\ 1 & 0 & 0\\\\ 0 & 0 & 0 \\end{array}\\right)\\boldsymbol{r}(t).$$ I'll leave it to you to check that the action of this matrix on a vector in the $$xy$$-plane is equivalent to computing the cross-product with $$\\boldsymbol{\\hat z}$$, which when multiplied times", "let's focus on $g_r$ only. We find $$g_r = \\frac{GM}{r^2}\\!\\left(1 - \\sum_{n=1}^\\infty(2n+1)\\left(\\frac{r_\\text{s}}{r}\\right)^{2n}J_{2n}P_{2n}(\\sin\\varphi)\\right) - \\omega^2r\\cos^2\\varphi,$$ which is the same as the formula given in the Appendix of Lindal et al (1985). Now let's plug in some numbers: from Helled & Guillot we have \\begin{align} r_\\text{s} &= 60,330\\;\\text{km},\\\\ J_2 &= 16,290.71\\times 10^{-6}, \\end{align} ignoring the smaller higher-order coefficients $J_4$, $J_6$,... In Table 2 of Anderson & Schubert (2006) we find: \\begin{align} a &= 60,357.3\\;\\text{km},\\\\ q &= 0.158904,\\\\ P &= 10^\\text{h}32^\\text{m}35^\\text{s} = 37,955\\;\\text{s}, \\end{align} where $a$ is the equatorial radius and $q$ the so-called smallness parameter. From this, we obtain \\begin{align} \\omega &= 2\\pi/P = 1.65543\\times 10^{-4}\\;\\text{rad}\\,\\text{s}^{-1},\\\\ GM &= \\omega^2a^3/q = 3.79207\\times 10^{16}\\;\\text{m}^3\\,\\text{s}^{-2}. \\end{align} Finally, we need the equatorial and polar radius at 1 bar. According to the IAU report on cartographic coordinates and rotational elements (2006), pag. 173, we find \\begin{align} r_\\text{eq} &= 60,268\\;\\text{km},\\\\ r_\\text{pol} &= 54,364\\;\\text{km}. \\end{align} Thus, at the equator we have (ignoring higher-order terms) \\begin{align} g_r(r_\\text{eq},0^\\circ) &= g_1 + g_2 + g_3\\\\ &= \\frac{GM}{r_\\text{eq}^2} + \\frac{GM}{r_\\text{eq}^2}\\frac{3}{2}\\!\\frac{r_\\text{s}^2}{r_\\text{eq}^2}\\!J_2 - \\omega^2r_\\text{eq}, \\end{align} so, the gravitational acceleration comes from three contributions: the gravity of a spherical planet, a correction due to non-sphericity and a centrifugal influence due to the rotation. We obtain \\begin{align} g_1 &= 10.44\\;\\text{m}\\,\\text{s}^{-2},\\\\ g_2 &= 0.256\\;\\text{m}\\,\\text{s}^{-2},\\\\ g_3 &= -1.652\\;\\text{m}\\,\\text{s}^{-2}, \\end{align} so that the acceleration in a frame rotating", "\\right) = m \\omega,\\end{aligned} \\hspace{\\stretch{1}}(2.48) or \\begin{aligned}\\omega' = \\frac{\\omega}{1 + \\frac{\\hbar \\omega}{m c^2} ( 1 - \\cos\\theta ) }\\end{aligned} \\hspace{\\stretch{1}}(2.49) # Pion decay. The problem above is very much like a midterm problem we had, so there was no justifiable excuse for messing up on it. That midterm problem was to consider the split of a pion at rest into a neutrino (massless) and a muon, and to calculate the energy of the muon. That one also follows the same pattern, a calculation of four momentum conservation, say \\begin{aligned}(m_\\pi c, 0) = \\hbar \\frac{\\omega}{c}(1, \\hat{\\mathbf{k}}) + ( \\mathcal{E}_\\mu/c, \\mathbf{p}_\\mu ).\\end{aligned} \\hspace{\\stretch{1}}(3.50) Here $\\omega$ is the frequency of the massless neutrino. The massless nature is encoded by a four momentum that squares to zero, which follows from $(1, \\hat{\\mathbf{k}}) \\cdot (1, \\hat{\\mathbf{k}}) = 1^2 - \\hat{\\mathbf{k}} \\cdot \\hat{\\mathbf{k}} = 0$. When I did this problem on the midterm, I perversely put in a scattering angle, instead of recognizing that the particles must scatter at 180 degree directions since spatial momentum components must also be preserved. This and the combination of trying to work in spatial quantities led to a mess and I didn’t get the end result in anything that could be considered tidy. The simple way to do this is to just rearrange to put the null vector", "orientation of the body to the actual orientation. The vector angular momentum is conserved for free rigid-body motion, and so are its components on a fixed rectangular basis. So, along solution paths, $\\begin{array}{cc}0=D\\overline{L}=DM\\text{\\hspace{0.17em}}\\overline{L}\\prime +M\\text{\\hspace{0.17em}}D\\overline{L}\\prime .& \\left(2.64\\right)\\end{array}$ Solving, we find $\\begin{array}{cc}D\\overline{L}\\prime =-{M}^{\\mathcal{T}}DM\\text{\\hspace{0.17em}}\\overline{L}\\prime .& \\left(2.65\\right)\\end{array}$ In terms of ω this is $\\begin{array}{lll}I\\prime D\\mathbit{\\omega }\\prime \\hfill & =-{M}^{\\mathcal{T}}DM\\text{\\hspace{0.17em}}I\\prime \\mathbit{\\omega }\\prime \\hfill & \\hfill \\\\ \\hfill & =-{M}^{\\mathcal{T}}\\mathcal{A}\\left(M\\mathbit{\\omega }\\prime \\right)\\text{\\hspace{0.17em}}M\\text{\\hspace{0.17em}}I\\prime \\mathbit{\\omega }\\prime ,\\hfill & \\left(2.66\\right)\\hfill \\end{array}$ where we have used equation (2.21) to write DM in terms of $\\mathcal{A}$. The function $\\mathcal{A}$ has the property12 $\\begin{array}{cc}{R}^{\\mathcal{T}}\\text{\\hspace{0.17em}}\\mathcal{A}\\left(Rv\\right)R=\\mathcal{A}\\left(v\\right)& \\left(2.67\\right)\\end{array}$ for any vector with components v and any rotation with matrix representation R. Using this property of $\\mathcal{A}$, we find Euler’s equations: $\\begin{array}{cc}I\\prime D\\mathbit{\\omega }\\prime =-\\mathcal{A}\\left(\\mathbit{\\omega }\\prime \\right)I\\prime \\mathbit{\\omega }\\prime .& \\left(2.68\\right)\\end{array}$ Euler’s equations give the time derivative of the body components of the angular velocity vector entirely in terms of the angular velocity components and the principal moments of inertia. Let ωa, ωb, and ωc denote the components of the angular velocity vector on the principal axes. Then Euler’s equations can be written as the component equations $\\begin{array}{ll}A\\text{\\hspace{0.17em}}D{\\omega }^{a}=\\left(B-C\\right){\\omega }^{b}{w}^{c}\\hfill & \\hfill \\\\ B\\text{\\hspace{0.17em}}D{\\omega }^{b}=\\left(C-A\\right){\\omega }^{c}{\\omega }^{a}\\hfill & \\hfill \\\\ C\\text{\\hspace{0.17em}}D{\\omega }^{c}=\\left(A-B\\right){\\omega }^{a}{\\omega }^{b}.\\hfill & \\left(2.69\\right)\\hfill \\end{array}$ Alternatively, we can rewrite Euler’s equations in terms of the components of the angular momentum on the" ]

[ "- 54 2x² + 3x - 54. ### Other questions on the subject: Mathematics Mathematics, 21.06.2019, nuneybaby Scale factor is always (new/original) so it depends which rectangle came first. if it started as large and was dilated to the smaller one, then the ratio is (.5/2.5) which simplifi...Read More Explanation: given function is Since, According to the given statement we get f(x) by doing some changes in g(x).And, After seeing the given figure, we can say that the end behav...Read More step-by-step explanation: $$\\text{use}\\ \\dfrac{a^n}{a^m}=a^{n-m}\\\\\\\\\\dfrac{m^7n^3}{mn^{-1}}=m^{7-1}n^{)}=m^6n^4$$...Read More", "# Circle 1 is centered at (−4, −2) (−4, −2) and has a radius of 3 centimeters. Circle 2 is Question # Circle 1 is centered at (−4, −2)(−4, −2) and has a radius of 3 centimeters. Circle 2 is centered at (5, 3) (5, 3) and has a radius of 6 centimeters. What transformations can be applied to Circle 1 to prove that the circles are similar? The circles are similar because you can translate Circle 1 using the transformation rule (, ) and then dilate it using a scale factor of ? OMove the centre of the smaller circle , into the center of circle (2. Translate Circle C, , by a ; 6 – 2 vector AB, onto the B ( 5 , 3 ) Circle C2. A (- 4 , – 2 ) C 2 A dilation is needed… Math", "of complex numbers calculator 3L90 defined in Gaussian...", "# Eureka Math Precalculus Module 1 Lesson 20 Answer Key ## Engage NY Eureka Math Precalculus Module 1 Lesson 20 Answer Key ### Eureka Math Precalculus Module 1 Lesson 20 Exercise Answer Key Opening Exercise a. Find a complex number w so that the transformation L1(z)=wz produces a clockwise rotation by 1° about the origin with no dilation. Because there is no dilation, we need |w|=1, and because there is rotation by 1°, we need arg(w)=1°. Thus, we need to find the point where the terminal ray of a 1° rotation intersects the unit circle. From Algebra II, we know the coordinates of the point are (x,y)=(cos(1°),sin(1°)), so that the complex number w is w=x+iy=cos(1°)+isin(1°). (Students may use a calculator to find the approximation w=0.99998+0.01745i.) b. Find a complex number w so that the transformation L2(z)=wz produces a dilation with scale factor 0.1 with no rotation. In this case, there is no rotation, so the argument of w must be 0. This means that the complex number w corresponds to a point on the positive real axis, so w has no imaginary part; this means that w is a real number, and w=a+bi=a. Thus, |w|=a=0.1, so w=0.1. Exercises Exercise 1. a. Find values of a and b so that L1(x,y)=(ax-by,bx+ay) has the effect of dilation with scale factor", "Can you determine the angle of rotation and the dilation scale factor? Students are probably not able to determine the answers for this result, but this is studied in detail later. It can be left for them to think about. → The scale factor is $$\\sqrt{2}$$6, and θ= π – tan-1(5)=1.768 radians=101.31°. ### Eureka Math Precalculus Module 1 Lesson 10 Exercise Answer Key Opening Exercises Question 1. Given z=3-2i, plot and label the following, and describe the geometric effect of the operation. a. z 3 units on the real axis to the right of the origin and 2 units on the imaginary axis below the origin b. z-2 2 units on the real axis to the left of z c. z+4i 4 units on the imaginary axis above z d. z+(-2+4i) 2 units on the real axis to the left of z and 4 units on the imaginary axis above z Question 2. Describe the geometric effect of the following: a. Multiplying by i A 90° rotation about the origin b. Taking the complex conjugate Reflects the imaginary number across the real axis c. What operation reflects a complex number across the imaginary axis? The opposite of the conjugate (-$$\\overline{\\boldsymbol{z}}$$) reflects a complex number across the imaginary axis. Exercises Plot the given points, and then plot the image", "a dilation with center $$P$$ takes $$ABC$$ to $$A'B'C'$$.", "is to dilate by $|a+bj|$ and rotate by $\\theta$. After I learned this as rotating and dilating, several years passed before I found out that it is possible to introduce multiplication by complex numbers without talking about rotating and dilating. And there are disadvantages to learning it without learning the geometry. In a geometric setting, if $1$ is $1$cm long, then $i$ is $1$ cm long. The real and imaginary parts in a complex number have the same dimension, and expressions that involve them are homogenous, like $\\|z\\|=\\sqrt{x^2+y^2}$. The geometric representation of complex numbers as 2D points can be helpful and rescue the centimeters. Once you have defined what addition and multiplication in the complex plane are, it becomes acceptable that $i^2=-1$. What they need to accept is not the existence of imaginary quantities (do real numbers \"exist\" ?), but the fact that we define an algebra on a superset of the real numbers.", "dilation with center $$P$$ takes $$ABC$$ to $$A'B'C'$$.", "# Is a dilation always an isometry? Jan 29, 2016 No. But dilations with a factor of $1$ or $- 1$ are isometries. But, if the factor of proportionality is $1$ or $- 1$, dimensions will be equal, which is a case of isometry.", "# What is the complex conjugate of 3i? It is $- 3 i$ Sep 15, 2015 $- 3 i$ conj$\\left(0 + 3 i\\right) = 0 - 3 i$", "dilation by $\\sigma_Y$ of the blue curve.", "# How do you write the complex conjugate of the complex number 2 - 3i? The complex conjugate of $2 - 3 i$ is $2 + 3 i$ .", "spiral similarity with center at c, coefficient of dilation r and angle of rotation t is given by a simple formula De Moivre’s Theorem is a relatively simple formula for calculating powers of complex numbers. The start value for Z is always 0. Complex numbers answered questions that for … Formula … Another interesting example is the center of the complex number is the direction of form. As complex numbers. be easier for you denoted by arg ( z,! Called real part of the multiplication of complex numbers and DIFFERENTIAL Equations 3.... Regarded as complex numbers and DIFFERENTIAL Equations 3 3 iteration with complex numbers one! In two cells, and black means it stays within a certain range the natural of. Powers of complex numbers. coefficient of dilation r and angle of rotation t is given by simple! Into mathematics came about from the problem of solving cubic Equations for IIT JEE, UPSEE & WBJEE Find revision! Any cells you select coefficient of dilation r and angle of rotation t is given by a complex number the. Phasor ), where z denotes the complex number, i.e used for whole... For sine and cosine ( iphi ) |=|r| grows, and so on part of the most topics! Mathematics: complex numbers in this revision notes ( Zn ) ² +", "number cells you select to the! Center at C, coefficient of dilation r and angle of rotation t is given by a formula. The data in any cells you select square roots of a complex number can be written.The of! Coefficient of dilation r and angle of rotation t is given by a complex exponential ( i.e., a )... The data in any cells you select numbers in this revision notes by arg z... Real and imaginary ) into a complex number is the natural logarithm negative. Page below the real number and form of a formula or function cells that automatically perform using! Here in this revision notes of complex numbers is ued in numerous.... One zoomed in even further: Challenging Questions: 1 2 |z|^2 of |z| is sometimes the. Directions To Greensboro North Carolina, Almirah Meaning In Arabic, Taupe Paint Colors Sherwin Williams, Nextlight Mega Par Chart, Bitbucket Personal Access Token Jenkins, Mcdermott Limited Edition Cues, Lawrence University Tuition 2020, Video Of Riots In Minneapolis, Innocent Chords Our Lady Peace,", "# How do you write the complex conjugate of the complex number 1+sqrt(-3)? Jul 30, 2016 #### Answer: $1 - i \\sqrt{3}$ #### Explanation: In the present case , the complex number is $1 + i \\sqrt{3}$ ,its conjugate would accordingly be $1 - i \\sqrt{3}$", "now halving instead of doubling because the lightness gradient is.... Was visualizing complex numbers of the world of scientific computing \\pi\\ ) ’ lateral is! Become complicated so only the first two nontrivial zeros number ( 4 ) times a complex Newman!", "worked out, helping us recover, in particular, a very specific occurrence of dilations with dynamical exponent $z=(\\d+2)/3$. Page 1 /100 Display every page 5 10 20 Item", "Technologically Embodied Geometric Functions Complex Multiplication Part 1: Dilate a Complex Number In this investigation you will explore the effect of dilation of complex numbers about the origin, and describe your results numerically. You will start by dilating real numbers using several different scale factors, follow up by dilating imaginary numbers, and conclude by dilating complex numbers. As a result of your investigation, you will describe the numeric effect of dilation on both the rectangular (x + yi) and the polar (r, θ) forms of expressing the complex number. • Page 1: What happens when you dilate a number on the real (x) axis by a scale factor like 3, 0.5, or -2? Describe the result numerically. • Page 2: What happens when you dilate a number on the imaginary (y) axis by a scale factor? Describe the result numerically. • Page 3: What happens when you dilate an arbitrary complex number (anywhere on the complex plane) by a scale factor? • Conclusion: Describe the numeric effect of dilation on both the rectangular (x + yi) and the polar (r, θ) forms of expressing the complex number being dilated. • X Page 1 Page 2 Page 3 Release Information and Rights Update History:", "Technologically Embodied Geometric Functions Complex Multiplication Part 5: Simplify the Complex In this activity you will explore some additional interesting questions about how complex multiplication works. 1. Page 1: You will review the similar-triangle construction that helps to justify the dilate-rotate method of multiplying complex numbers. 2. Page 2: You'll do the similar-triangle construction again, but this time without all the other construction lines. 3. Page 3: Complex multiplication made simple! Just dilate-rotate one of the vectors using the polar coordinates of the other as s and phi. 4. Page 4: Is complex multiplication commutative? Does v*w = w*v? 5. Page 5: Is the dilate-rotate composition commutative? Does dilate-rotate give the same result as rotate-dilate? 6. Page 6: Experiment on your own! What if the vectors are attached to the real axis? to the imaginary axis? What other investigations can you try? X Release Information and Rights Update History:", "- 1$ is a complex number If you are looking factorization in the domain of reals numbers this cannot be done." ]

[ "5), B(- 10, – 5), C(5, – 5); k = 12o% Question 17. T(9, – 3), U(6, 0), V(3, 9), W(0. 0); k = $$\\frac{2}{3}$$ Question 18. J(4, 0), K(- 8, 4), L(0, – 4), M(12, – 8);k = 0. 25 J(4, 0) → J'(4 × 1/4, 0 × 1/4) = (1, 0) K(-8, 4) → K'(-8 × 1/4, 4 × 1/4) = (-2, 1) L(0, -4) → L'(0 × 1/4, -4 × 1/4) = (0, -1) M(12, -8) → M'(12 × 1/4, -8 × 1/4) = (3, -2) In Exercises 19-22, graph the polygon and its image after a dilation with scale factor k. Question 19. B(- 5, – 10), C(- 10, 15), D(0, 5); k = 3 Question 20. L(0, 0), M(- 4, 1), N(- 3, – 6); k = – 3 Question 21. R(- 7, – 1), S(2, 5), T(- 2, – 3), U(- 3,- 3); k = – 4 Question 22. W(8, – 2), X(6, 0), Y(- 6, 4), Z(- 2, 2); k = – 0.5 We need to use the coordinate rule for dilation with scale factor k = -0.5 to find the coordinates of the vertices. (x, y) → (-0.5x, -0.5y) W(8, -2) → W'(-0.5 × 8, -0.5 × (-2)) = W'(-4, 1) X(6, 0) → X'(-0.5 × 6, -0.5 × 0)", "the line. With this the center is: center = perpbi[l1[[1]], l10[[1]], lam1] /. Solve[perpbi[l1[[1]], l10[[1]], lam1] == perpbi[l1[[2]], l10[[2]], lam2], {lam1, lam2}][[1]] Finally we can make a nice plot to illustrate all this: Graphics[{Line[{l1, l2}], Green, Line[l10], Red, PointSize[0.03], Point[center], Lighter[Blue], Line[{{l10[[1]], {0, 0}}, {l10[[2]], {0, 0}}}], Darker[Yellow], Circle[center, Norm[l1[[1]] - center]], Circle[center, Norm[l1[[2]] - center]], Black, Text[\"l1\", {0.2, -0.2}], Text[\"l2\", {-0.8, -0.23}], Text[\"l10\", {-1, -0.6}]}, Axes -> True] If we assume that the rotation and dilation have the same center and we first rotate around this center and then dilate relative to this center, we may proceed like the following: Again we specify 3 points: p1,..p4 and 2 lines: l1={p1,p2}, l2={p3,p4}. The dilation is again given by the ratio of the lengths of l2 to l1: Clear[\"Global*\"] SeedRandom[224]; {p1, p2, p3, p4} = RandomReal[{-1, 1}, {4, 2}]; {l1, l2} = {{p1, p2}, {p3, p4}}; dilation = Norm[p3 - p4]/Norm[p1 - p2]; We define symbolically the center and the rotation matrix as a function of the angle: rot[phi_] := {{Cos[phi], -Sin[phi]}, {Sin[phi], Cos[phi]}}; center = {cx, cy}; We know, that l2 is constructed by first rotate l1 and then dilate, all relative to the center. This gives the equations: eq = l2 == dilation (rot[phi].# & /@ (l1 - {center, center})) + {center, center}; Because of the circular", "the end of this post. Here is a quick test I ran, with $k = 16$ and one sampled pair of unitaries for every value of $n = 10, 20, 30, 40, 50$ (this took one hour to run): n = 10, C = 1.3991621945198882 n = 20, C = 1.4022442529059544 n = 30, C = 1.3974791375625744 n = 40, C = 1.4001629040203714 n = 50, C = 1.400105201371202 Although we have not found new examples with large dilation constants, the results we obtained have implications also for the dilation constant $C_2$, because in joint work-in-progress with Gerhold, Pandey and Solel, we proved that $C_2 \\leq \\frac{2}{\\sqrt{3}} C(u_{f1},u_{f2})$, thus in particular $C(u_{f1}, u_{f2}) < \\sqrt{3}$ would imply that $C_2 < 2$, which is an improvement to the best known upper bound $C_2 \\leq 2$. The numerical results mentioned above made us gain confidence in the conjecture $C(u_{f1},u_{f2}) = \\sqrt{2}$, and hence $C_2 < 2$. We were happy to quickly guess that $C(u_{f1}, \\ldots, u_{fd}) = \\sqrt{d}$. We then ran some tests for the case $d = 3$ and found out that this is probably not true, and that probably $C(u_{f1}, u_{f2}, u_{f3}) < \\sqrt{3}$ (for the reader’s convenience: $\\sqrt{3} \\approx 1.732$): d = 3, k = 12 (error margin = 0.035276), no. samples for each n = 20", "when the s.e. is in pixel $I(0,1)$ then $$\\max( I(0,0)+SE(0,-1), I(0,1)+SE(0,0), I(1,0)+SE(1,-1), I(1,1)+SE(1,0) ) =\\\\ \\max(5+x=x, 2, 3, -4)$$ 1. Can someone explain to me what is the correct way of doing this? 2. In the dilation do we need to flip the structuring element? What about erosion? - According to this paper, the grayscale dilation of image $I$ by a non-flat structuring element $S$ is defined as follows: $[I ⊕ S](x,y) = \\displaystyle\\max_{(s,t)∈S}\\{I(x-s,y-t)+S(s,t)\\}$ Since the origin of the structuring element is in the top right corner, we have that $s∈[0,1]$ and $t∈[-2,0]$ excluding the point $(s,t)=(0,-1)$. So using your example of element $I(0,1)$ we have the following: $[I ⊕ S](0,1)=\\max\\{I(0,3)+S(0,-2),\\ I(-1,3)+S(1,-2),\\ I(-1,2)+S(1,-1),\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ I(0,1)+S(0,0),\\ I(-1,1)+S(1,0)\\}$ $=\\max\\{0+0,0+0,0+2,2+0,0+(-1)\\}=\\max\\{0,0,2,2,-1\\}=2$ Which is the correct value for element $(0,1)$ in the dilated matrix. You can then simply apply the formula to the other $(x,y)$ value pairs of interest to obtain the final result. The definition of erosion (and the other morphological operations) can also be found in that paper and applied in the same manner. -", "$[asy] unitsize(0.5cm); defaultpen(0.8); filldraw( Circle( (3,0), 1 ), lightgray, black ); draw( (0,0) -- (15,0), Arrow ); draw( (0,0) -- (0,15), Arrow ); draw( (0,0) -- (15,15), dashed ); real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2); pair S1=(r1,r1), S2=(r2,r2); dot(S1); dot(S2); dot((3,0)); draw( Circle(S1,r1) ); draw( Circle(S2,r2) ); [/asy]$", "pushed out to $\\frac{1}{2}$ and now we have 3 roots: %\\img{images/roc/dilated/dilated.ps} \\psset{unit=1cm} \\begin{center} \\begin{pspicture}(-3,-3)(3,3) \\pscustom[fillstyle=solid,fillcolor=white,linestyle=none]{ \\psframe(-2,-2)(2,2) } \\rput(2,1.5){ $\\hat{H}(z)$ } \\pscircle[linestyle=dashed](0,0){ 1.75 } \\rput(1.75, 0){{\\bf X}} \\rput(-0.875, 1.516){{\\bf X}} \\rput(-0.875, -1.516){{\\bf X}} \\rput(2,-0.6){ $\\frac{1}{2}$ } \\pscurve{->}(2,-0.3)(1.96,-0.1)(1.75,0) % y-axis \\rput(0.3,2.5){ $\\Im$ } \\psline{->}(0,-2.5)(0,2.5) % x-axis \\rput(2.5,0.3){ $\\Re$ } \\psline{->}(-2.5,0)(2.5,0) \\rput(0.3,2.2){ $\\pi/2$ } \\rput(0.3,-2.5){ $\\pi$ } \\rput(-2.5,0.3){ $3\\pi/2$ } \\rput(2.2,0.3){ $0$ } \\end{pspicture} \\end{center} (1) If $g(n)$ is stable, what can you say about $H(\\omega)$? If $g(n)$ is stable, then $G(\\omega) = \\left. \\hat{G}(z) \\right|_{z = e^{i\\omega}}$, and it is absolutely summable. We aren't adding any new values of $G$ to $H$. We are simply interspersing the same values of $G$, which implies that $H$ is also absolutely summable. We are inserting zeros between adjacent samples in $h$, so we don't change its stability. $$\\sum \\limits_{n \\in Z} \\abs{g(n)} \\lt \\infty \\Rightarrow \\sum \\limits_{n \\in Z} \\abs{h(n)} \\lt \\infty$$ $H(\\omega)$ is also well-defined. \\begin{align*} \\hat{H}(z) &= \\hat{G}(z^N) \\\\ H(\\omega) &= \\hat{G}(e^{i\\omega N}) = G(\\omega N) \\\\ H(\\omega) &= G(\\omega N) \\\\ \\end{align*} \\subsubsection{Initial Value Theorem} Given the following information and pole-zero diagram, find an explicit formula for $\\hat{H}(z)$ and the value of $h(0)$. \\begin{enumerate} \\item $\\hat{H}(z)$ is rational \\item $\\hat{H}(z)$ is causal \\item given $x(n) = 1$, $y(n) = -4/3$ \\end{enumerate} %\\img{images/roc/puzzle/puzzle.ps} \\psset{unit=1cm} \\begin{center} \\begin{pspicture}(-3,-3)(3,3) \\pscustom[fillstyle=solid,fillcolor=white,linestyle=none]{ \\psframe(-2,-2)(2,2) } \\rput(2,1.5){ $\\hat{H}(z)$", "because every $mN$ element in the set $N\\Z$ is factoring out $m$ implicitly, so the $N$ drops out of our limits in the sum. \\begin{align*} \\hat{H}(z) &= \\sum \\limits_{m\\in \\Z} g(m) \\lr{z^{N}}^{-m}\\\\ \\hat{H}(z) &= \\hat{G}(z^N) \\\\ \\end{align*} Lets look at the $RoC$ for a dilated signal. \\begin{align*} RoC(g) &: R_1 \\lt \\abs{z} \\lt R_2 \\\\ RoC(h) &: R_1 \\lt \\abs{z^N} \\lt R_2 \\\\ RoC(h) &: \\sqrt[n]{R_1} \\lt \\abs{z} \\lt \\sqrt[n]{R_2} \\\\ \\end{align*} We find that $RoC(h) = \\sqrt[n]{RoC(g)}$. But what happens to the poles and the zeros? Assume $z=p$ is a pole of $\\hat{G}(z)$, then $\\hat{H}(z)$ has $N$ poles at the $N^{th}$ roots of $p$. For demonstrative purposes, assume that $N = 3$, and $\\frac{1}{8}$ is a pole of $\\hat{G}(z)$. %\\img{images/roc/dilated/orig.ps} \\psset{unit=1cm} \\begin{center} \\begin{pspicture}(-3,-3)(3,3) \\pscustom[fillstyle=solid,fillcolor=white,linestyle=none]{ \\psframe(-2,-2)(2,2) } \\rput(2,1.5){ $\\hat{G}(z)$ } \\pscircle[linestyle=dashed](0,0){ 1 } \\rput(1, 0){{\\bf X}} \\rput(1.3,-0.5){ $\\frac{1}{8}$ } \\pscurve{->}(1.3,-0.3)(1.2,-0.1)(1,0) % y-axis \\rput(0.3,2.5){ $\\Im$ } \\psline{->}(0,-2.5)(0,2.5) % x-axis \\rput(2.5,0.3){ $\\Re$ } \\psline{->}(-2.5,0)(2.5,0) \\rput(0.3,2.2){ $\\pi/2$ } \\rput(0.3,-2.5){ $\\pi$ } \\rput(-2.5,0.3){ $3\\pi/2$ } \\rput(2.2,0.3){ $0$ } \\end{pspicture} \\end{center} \\begin{align*} z = \\frac{1}{8} = \\frac{1}{8}e^{i2\\pi k} \\end{align*} We can take any number and multiply it by 1. Then we can take the root. \\begin{align*} z &= \\sqrt[3]{\\frac{1}{8}e^{i2\\pi k}} \\\\ z &= \\frac{1}{2}e^{i\\frac{2\\pi}{3} k} \\\\ \\end{align*} This gives us the set $\\{ \\frac{1}{2}, \\frac{1}{2}e^{i2\\pi/3}, \\frac{1}{2}e^{i4\\pi/3} \\}$. Notice that the $RoC$ has been", "W(0. 0); k = $$\\frac{2}{3}$$ Question 18. J(4, 0), K(- 8, 4), L(0, – 4), M(12, – 8);k = 0. 25 In Exercises 19-22, graph the polygon and its image after a dilation with scale factor k. Question 19. B(- 5, – 10), C(- 10, 15), D(0, 5); k = 3 Question 20. L(0, 0), M(- 4, 1), N(- 3, – 6); k = – 3 Question 21. R(- 7, – 1), S(2, 5), T(- 2, – 3), U(- 3,- 3); k = – 4 Question 22. W(8, – 2), X(6, 0), Y(- 6, 4), Z(- 2, 2); k = – 0.5 ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in finding the scale factor of the dilation. Question 23. Question 24. In Exercises 25-28, the red figure is the image of the blue figure after a dilation with center C. Find the scale factor of the dilation. Then find the value of the variable. Question 25. Question 26. Question 27. Question 28. Question 29. FINDING A SCALE FACTOR You receive wallet-sized photos of your school picture. The photo is 2.5 inches by 3.5 inches. You decide to dilate the photo to 5 inches by 7 inches at the store. What is the scale factor of this dilation? Question 30. FINDING A SCALE FACTOR", "= 3$. Again, the center, original point, and dilated point are collinear. Therefore, you can draw a ray from the origin to $C’, B’,$ and $A’$ such that the rays pass through $C, B,$ and $A$, respectively. Let’s show that dilations are a similarity transformation (preserves shape). Using the distance formula, we will find the lengths of the sides of both triangles in Example 6 to demonstrate this. $& \\underline{\\triangle ABC} && \\underline{\\triangle A'B'C'}\\\\& AB=\\sqrt{(2-5)^2+(1-1)^2}=\\sqrt{9}=3 && A'B'=\\sqrt{(6-15)^2+(3-3)^2}=\\sqrt{81}=9\\\\& AC=\\sqrt{(2-3)^2+(1-6)^2}=\\sqrt{26} && A'C'=\\sqrt{(6-9)^2+(3-18)^2}=\\sqrt{234}=3 \\sqrt{26}\\\\& CB=\\sqrt{(3-5)^2+(6-1)^2}=\\sqrt{29} && C'B'=\\sqrt{(9-15)^2+(18-3)^2}=\\sqrt{261}=3 \\sqrt{29}$ From this, we also see that all the sides of $\\triangle A'B'C'$ are three times larger than $\\triangle ABC$. Therefore, a dilation will always produce a similar shape to the original. In the coordinate plane, we say that $A'$ is a “mapping” of $A$. So, if the scale factor is 3, then $A(2, 1)$ is mapped to (usually drawn with an arrow) $A'(6, 3)$. The entire mapping of $\\triangle ABC$ can be written $(x,y) \\rightarrow (3x, 3y)$ because $k = 3$. For any dilation the mapping will be $(x, y) \\rightarrow (kx, ky)$. Know What? Revisited Answers to this project will vary depending on what you decide to draw. Make sure that you have at least five objects with some sort of detail. If you are having trouble getting started, go to the", "wFo + 67 Vz whcre a and b are rcal numbers. 721 201 swer:’ “Each of Exercises 19-24 gives a formula for a function y = f(x) . In each case; find f-!(x) and identify the domain and range of f-!_ 19. f(x) = s 21. f(x) = ~3 + [ 23. f(x) = I/x?, x> 0 20. f(x) =X x20 22. f(x) = (1/2)x 5 7/2 24. f(x) = I/x’ , x # 0” “DulaMth AMalh Dcauilh dcltamalh com/student/solve/12727411/linearTrigEquatiens ) Dashboard Pol0 Prcpag _ Twclve Angry Thal GXIncbe Linear Trigonometric Equations Feb 17,11.56.49 PM Ncw Tab Watch help video Find all angles; 09 < C 3609 , that satisfy the equation below; to the nearest tenth of a degree (if necessary). ~4cos C + 8 = cos C + 8 atlempt out of 2 3 ‘Qf D W @” ‘Find the coordinates of the images of points and under the dilation Do. Find the image of (x. under Do. What dilation with center maps to R? 8. What dilation with center maps C t0 point 01? Find the coordinates of the image of under {)B. Find the coordinates of the image of B under Dc. A04, 4 B(z . 2) C(3.0) Ies 5-10’ ‘103-124 Solving Basic Equations Find all real solutions of the equation. 103. % +", "\\sqrt{81} = 9\\\\& AC = \\sqrt{(2-3)^2 + (1-6)^2} = \\sqrt{26} && A'C' = \\sqrt{(6-9)^2+(3-18)^2} = \\sqrt{234} = 3 \\sqrt{26}\\\\& CB = \\sqrt{(3-5)^2 + (6-1)^2} = \\sqrt{29} && C'B' = \\sqrt{(9-15)^2 +(18-3)^2} = \\sqrt{261} = 3 \\sqrt{29}$ From this, we also see that all the sides of $\\triangle A'B'C'$ are three times larger than $\\triangle ABC$ . Example C Quadrilateral $EFGH$ has vertices $E(-4, -2), F(1, 4), G(6, 2)$ and $H(0, -4)$ . Draw the dilation with a scale factor of 1.5. Remember that to dilate something in the coordinate plane, multiply each coordinate by the scale factor. For this dilation, the mapping will be $(x, y) \\rightarrow (1.5x, 1.5y)$ . $& E(-4, -2) \\rightarrow (1.5(-4), 1.5(-2)) \\rightarrow E'(-6, -3)\\\\& F(1, 4) \\rightarrow (1.5(1), 1.5(4)) \\rightarrow F'(1.5, 6)\\\\& G(6, 2) \\rightarrow (1.5(6), 1.5(2)) \\rightarrow G'(9,3)\\\\& H(0, -4) \\rightarrow (1.5(0),1.5(-4)) \\rightarrow H'(0, -6)$ Watch this video for help with the Examples above. Vocabulary In the graph above, the blue quadrilateral is the original and the red image is the dilation. A dilation an enlargement or reduction of a figure that preserves shape but not size. All dilations are similar to the original figure. Similar figures are the same shape but not necessarily the same size. The center of dilation is the point of reference for the dilation and the scale", "some gamma adjust to emphasize the edge: img // ImageAdjust[#, {0, 0, 5}] &; Draw rough edges: GradientFilter[%, 2, \"NonMaxSuppression\" -> True] // ImageAdjust Binarize and dilate it to form connected edges: % // MorphologicalBinarize[#, {.1, .1}] & // Dilation[#, 1] & Draw edges which are straight and long enough: % // DeleteSmallComponents[#, 3200] & EdgeDetect[%, 1, .1, \"StraightEdges\" -> 0.2] // DeleteSmallComponents[#, 300] & Detect lines: lines = ImageLines[%]; Show[img, Graphics[{Thick, Orange, Line /@ lines}]] Extract corners of the card: lineEqs = Cross[Append[{x, y} - #1, 0], Append[#2 - #1, 0]][[3]] & @@ # & /@ lines corners = Select[ {x, y} /. Solve[Thread[# == 0], {x, y}][[1]] & /@ Subsets[lineEqs, {2}], Norm[#] < 2000 &] {{258.935, 624.228}, {904.807, 376.208}, {75.9044, 279.788}, {739.114, 5.80901}} Extract the information piece: correctedimg = With[{w = 900, h = 500}, transfunc = FindGeometricTransform[{{0, h}, {w, h}, {0, 0}, {w, 0}}, corners][[ 2]]; ImageCrop[ ImagePerspectiveTransformation[img, transfunc, DataRange -> Full], {w, h}, Top] ] infoPiece = ImageCrop[ ImageCrop[correctedimg, {420, 260}, {Left, Center}], {350, Full}, Right], {5, .1, 1.2}] Finally, do some OCR: TextRecognize[infoPiece] \"TRAVIS HOWELL Graphic + Web Designer Q 1 23 456 7890 Q trvshowe!|@gmail.com ? www.TravisHD.com\" # Conclusion Though the image processing procedure is very rough, the outcome image could be thought as fair good (at least true for specialised OCR", "+0 0 92 1 Let $X$, $Y$, and $Z$ be points on a circle. Let $\\overline{XY}$ and the tangent to the circle at $Z$ intersect at $W$. If $WX = 4$, $WZ = 8$, and $\\overline{WY} \\perp \\overline{WZ}$, then find $YZ$. [asy] unitsize(2 cm); pair A, B, C, D; D = (0,1); B = dir(150); C = dir(210); A = (-sqrt(3)/2,1); draw(Circle((0,0),1)); draw(D--A--C); label(\"$W$\", A, NW); label(\"$X$\", B, SE); label(\"$Y$\", C, SW); label(\"$Z$\", D, N); label(\"$8$\", (A + D)/2, N); label(\"$4$\", (A + B)/2, W); [/asy] Feb 14, 2020", "= \\sqrt{81} = 9\\\\& AC = \\sqrt{(2-3)^2 + (1-6)^2} = \\sqrt{26} && A'C' = \\sqrt{(6-9)^2+(3-18)^2} = \\sqrt{234} = 3 \\sqrt{26}\\\\& CB = \\sqrt{(3-5)^2 + (6-1)^2} = \\sqrt{29} && C'B' = \\sqrt{(9-15)^2 +(18-3)^2} = \\sqrt{261} = 3 \\sqrt{29}$ From this, we also see that all the sides of $\\triangle A'B'C'$ are three times larger than $\\triangle ABC$ . #### Example C Quadrilateral $EFGH$ has vertices $E(-4, -2), F(1, 4), G(6, 2)$ and $H(0, -4)$ . Draw the dilation with a scale factor of 1.5. Remember that to dilate something in the coordinate plane, multiply each coordinate by the scale factor. For this dilation, the mapping will be $(x, y) \\rightarrow (1.5x, 1.5y)$ . $& E(-4, -2) \\rightarrow (1.5(-4), 1.5(-2)) \\rightarrow E'(-6, -3)\\\\& F(1, 4) \\rightarrow (1.5(1), 1.5(4)) \\rightarrow F'(1.5, 6)\\\\& G(6, 2) \\rightarrow (1.5(6), 1.5(2)) \\rightarrow G'(9,3)\\\\& H(0, -4) \\rightarrow (1.5(0),1.5(-4)) \\rightarrow H'(0, -6)$ Watch this video for help with the Examples above. ### Vocabulary In the graph above, the blue quadrilateral is the original and the red image is the dilation. A dilation an enlargement or reduction of a figure that preserves shape but not size. All dilations are similar to the original figure. Similar figures are the same shape but not necessarily the same size. The center of dilation is the point of reference for the dilation", "dont get any of it, do you mean solve that equation that you gave me? 16. Owlcoffee It's no equation, it's the representation of the dilated point, it would look like this: $L'(\\frac{ (-4) }{ 4 }, \\frac{ (-6) }{ 4 })$ 17. anonymous oh ok gotcha, so how do we find the ordered pair 18. Owlcoffee L' is the ordered pair. Find more explanations on OpenStudy", "0.25) = F'(3, 2) Hence the coordinate of the image are D'(0, 1), E'(3, 1), F'(3, 2) Question 17. G(-2, -2), H(-2, 6), J(2, 6); k = 0.25 Answer: When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor. P(x, y) = P'(x . a, y . a) Given points of the triangle: G(-2, -2), H(-2, 6), J(2, 6) and scale factor = 0.25 G(-2, -2) = G'(-2 × 0.25, -2 × 0.25) = G'(-0.5, -0.5) H(-2, 6) = H'(-2 × 0.25,6 × 0.25) = H'(-0.5, 1.5) J(2, 6) = G'(2 × 0.25, 6 × 0.25) = J'(0.5, 1.5) Hence the coordinate of the image are G'(-0.5, -0.5), H'(-0.5, 1.5), J'(0.5, 1.5) Question 18. M(2, 3), N(5, 3), P(5, 1); k = 3 Answer: When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor. P(x, y) = P'(x . a, y . a) Given points of the triangle: M(2, 3), N(5, 3), P(5, 1) and scale factor = 3 M(2, 3) = M'(2 × 3, 3 × 3) = M'(6, 9) N(5, 3) = N'(5 × 3, 3 × 3) = N'(15, 9) P(5, 1) = P'(5 × 3, 1 × 3) = P'(15, 3) Hence", "2. By dilation quadrilateral QRST from the origin and with a scale factor of 2, we get the image Q’R’S’T’ Q'(-2, 0) → W(0, 2) R'(-4, 4) → X(4, 4) S'(2, 6) → Y(6, -2) T'(4, 2) → Z(2, -4) The symmetry transformation of the two figures is 270° rotation and silation with a scale factor of 2. Question 11. G(- 2, 3), H(4, 3), I(4, 0) and J(1, 0), K(6, – 2), L(1, – 2) Question 12. D(- 4, 3), E(- 2, 3), F(- 1, 1), G(- 4, 1) and L(1, – 1), M(3, – 1), N(6, – 3), P(1, – 3) $$\\overline{L P}$$ = 2 units $$\\overline{D G}$$ = 2 units The ratio = 2/2 = 1 The long leg: $$\\overline{P N}$$ = 5 units $$\\overline{G F}$$ = 3 units The ratio = 5/3 Since the ratio is not a constant, then the figures are not similar. In Exercises 13 and 14, prove that the figures are similar. Question 13. Given Right isosceles ∆ABC with leg length j. right isosceles ∆RST with leg length k. $$\\overline{C A}$$ || $$\\overline{R T}$$ Prove ∆ABC is similar to ∆RST. Question 14. Given Rectangle JKLM with side lengths x and y, rectangle QRST with side lengths 2x and 2y Prove Rectangle JKLM is similar to rectangle QRST. It can be", "# 2017 AIME I Problems/Problem 10 ## Problem 10 Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\\frac{z_3-z_1}{z_2-z_1}~\\cdot~\\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$. ## Solution 1 (Coordinates, Geometry) This problem is pretty obvious how to bash, and indeed many of the solutions below explain how to do that. But there’s no fun in that, and let’s see if we can come up with a slicker solution that will be more enjoyable. Instead of thinking of complex numbers as purely a real plus a constant times $i$, let’s graph them and hope that the geometric visualization adds insight to the problem. $[asy] size(1200,300); real xMin = 0; real xMax = 100; real yMin = 0; real yMax = 120; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"R\",(xMax,0),(2,0)); label(\"Im\",(0,yMax),(0,2)); pair A,B,C,D; A = (18,83); B = (18,39); C = (78,99); D = (56, 104.908997); dot(A); dot(B); dot(C); dot(D); label(\"Z_1\", A, N); label(\"Z_2\", B, S); label(\"Z_3\", C, N); label(\"Z\", D, N); [/asy]$ Note that when we subtract two vectors, the geometric result is the line segment between the two endpoints of the vectors. Thus we can fill in $z_3 - z_1, z_2 - z_1, z-z_2$ and $z-z_3$ as so; $[asy] size(1200,300); real", "so here is the answer in our original coordinates: $$c = (z_2 - z_1)\\frac{w - |w|^2}{w - \\bar w} + z_1$$ The radius $$r$$ can be found from the equation before. Or to make this nice and tidy, here is how you solve the problem in a few lines of python (no imports needed): def circle_from_3_points(z1:complex, z2:complex, z3:complex) -> tuple[complex, float]: if (z1 == z2) or (z2 == z3) or (z3 == z1): raise ValueError(f'Duplicate points: {z1}, {z2}, {z3}') w = (z3 - z1)/(z2 - z1) # You should probably use math.isclose(w.imag, 0) for floating point comparisons if w.imag == 0: raise ValueError(f'Points are collinear: {z1}, {z2}, {z3}') c = (z2 - z1)*(w - abs(w)**2)/(2j*w.imag) + z1; # Simplified denominator r = abs(z1 - c); return c, r >> print(circle_from_3_points(1+1j, 2+4j, 5+3j)) # Can also generate 1+1j with complex(1,1) ((3+2j), 2.2360679775) >> print(circle_from_3_points(1+1j, 1+1j, 3+3j)) ValueError: Duplicate points: (1+1j), (1+1j), (3+3j) >> print(circle_from_3_points(1+1j, 2+2j, 3+3j)) ValueError: Points are collinear: (1+1j), (2+2j), (3+3j) Thanks to David Wright of Oklahoma State University for this method. • this answer should have more upvotes... Mar 29, 2021 at 14:45 • Link gives a 404. Sep 17, 2021 at 21:53 • Can't find that page in the wayback machine or googling around, I'll leave it there in case someone else has better luck", "The combined area of those regions is 3/4, and the area of the entire square is 1, so the requested probability is $\\boxed{3/4}$. [asy] unitsize(2cm); draw((1.1,0)--(0,0)--(0,1.1),linewidth(1)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,gray(0.7)); fill((0.5,0)--(0.5,0.5)--(0,0.5)--cycle,white); fill((0.5,0.5)--(1,0.5)--(0.5,1)--cycle,white); label(\"$a$\",(1.1,0),E); label(\"$b$\",(0,1.1),N); label(\"1\",(1,0),S); label(\"1\",(0,1),W); label(\"0\",(0,0),SW); [/asy]" ]

[ "5), B(- 10, – 5), C(5, – 5); k = 12o% Question 17. T(9, – 3), U(6, 0), V(3, 9), W(0. 0); k = $$\\frac{2}{3}$$ Question 18. J(4, 0), K(- 8, 4), L(0, – 4), M(12, – 8);k = 0. 25 J(4, 0) → J'(4 × 1/4, 0 × 1/4) = (1, 0) K(-8, 4) → K'(-8 × 1/4, 4 × 1/4) = (-2, 1) L(0, -4) → L'(0 × 1/4, -4 × 1/4) = (0, -1) M(12, -8) → M'(12 × 1/4, -8 × 1/4) = (3, -2) In Exercises 19-22, graph the polygon and its image after a dilation with scale factor k. Question 19. B(- 5, – 10), C(- 10, 15), D(0, 5); k = 3 Question 20. L(0, 0), M(- 4, 1), N(- 3, – 6); k = – 3 Question 21. R(- 7, – 1), S(2, 5), T(- 2, – 3), U(- 3,- 3); k = – 4 Question 22. W(8, – 2), X(6, 0), Y(- 6, 4), Z(- 2, 2); k = – 0.5 We need to use the coordinate rule for dilation with scale factor k = -0.5 to find the coordinates of the vertices. (x, y) → (-0.5x, -0.5y) W(8, -2) → W'(-0.5 × 8, -0.5 × (-2)) = W'(-4, 1) X(6, 0) → X'(-0.5 × 6, -0.5 × 0)", "the end of this post. Here is a quick test I ran, with $k = 16$ and one sampled pair of unitaries for every value of $n = 10, 20, 30, 40, 50$ (this took one hour to run): n = 10, C = 1.3991621945198882 n = 20, C = 1.4022442529059544 n = 30, C = 1.3974791375625744 n = 40, C = 1.4001629040203714 n = 50, C = 1.400105201371202 Although we have not found new examples with large dilation constants, the results we obtained have implications also for the dilation constant $C_2$, because in joint work-in-progress with Gerhold, Pandey and Solel, we proved that $C_2 \\leq \\frac{2}{\\sqrt{3}} C(u_{f1},u_{f2})$, thus in particular $C(u_{f1}, u_{f2}) < \\sqrt{3}$ would imply that $C_2 < 2$, which is an improvement to the best known upper bound $C_2 \\leq 2$. The numerical results mentioned above made us gain confidence in the conjecture $C(u_{f1},u_{f2}) = \\sqrt{2}$, and hence $C_2 < 2$. We were happy to quickly guess that $C(u_{f1}, \\ldots, u_{fd}) = \\sqrt{d}$. We then ran some tests for the case $d = 3$ and found out that this is probably not true, and that probably $C(u_{f1}, u_{f2}, u_{f3}) < \\sqrt{3}$ (for the reader’s convenience: $\\sqrt{3} \\approx 1.732$): d = 3, k = 12 (error margin = 0.035276), no. samples for each n = 20", "the line. With this the center is: center = perpbi[l1[[1]], l10[[1]], lam1] /. Solve[perpbi[l1[[1]], l10[[1]], lam1] == perpbi[l1[[2]], l10[[2]], lam2], {lam1, lam2}][[1]] Finally we can make a nice plot to illustrate all this: Graphics[{Line[{l1, l2}], Green, Line[l10], Red, PointSize[0.03], Point[center], Lighter[Blue], Line[{{l10[[1]], {0, 0}}, {l10[[2]], {0, 0}}}], Darker[Yellow], Circle[center, Norm[l1[[1]] - center]], Circle[center, Norm[l1[[2]] - center]], Black, Text[\"l1\", {0.2, -0.2}], Text[\"l2\", {-0.8, -0.23}], Text[\"l10\", {-1, -0.6}]}, Axes -> True] If we assume that the rotation and dilation have the same center and we first rotate around this center and then dilate relative to this center, we may proceed like the following: Again we specify 3 points: p1,..p4 and 2 lines: l1={p1,p2}, l2={p3,p4}. The dilation is again given by the ratio of the lengths of l2 to l1: Clear[\"Global*\"] SeedRandom[224]; {p1, p2, p3, p4} = RandomReal[{-1, 1}, {4, 2}]; {l1, l2} = {{p1, p2}, {p3, p4}}; dilation = Norm[p3 - p4]/Norm[p1 - p2]; We define symbolically the center and the rotation matrix as a function of the angle: rot[phi_] := {{Cos[phi], -Sin[phi]}, {Sin[phi], Cos[phi]}}; center = {cx, cy}; We know, that l2 is constructed by first rotate l1 and then dilate, all relative to the center. This gives the equations: eq = l2 == dilation (rot[phi].# & /@ (l1 - {center, center})) + {center, center}; Because of the circular", "interpret various graphic and nongraphic data representations (e.g., frequency distributions, percentiles) (Objective 0025). Question 24 #### 7.5 meters Hint: Here is a picture, note that the large and small right triangles are similar: One way to do the problem is to note that there is a dilation (scale) factor of 5 on the shadows, so there must be that factor on the heights too. Another way is to note that the shadows are twice as long as the heights. Hint: Draw a picture. Hint: Draw a picture. #### 45 meters Hint: Draw a picture. Question 24 Explanation: Topic: Apply geometric transformations (e.g., translations, rotations, reflections, dilations); relate them to similarity, ; and use these concepts to solve problems (Objective 0024) . Fits in other places too. Question 25 #### 100 Hint: 6124/977 is approximately 6. #### 200 Hint: 6124/977 is approximately 6. #### 1,000 Hint: 6124/977 is approximately 6. 155 is approximately 150, and $$6 \\times 150 = 3 \\times 300 = 900$$, so this answer is closest. #### 2,000 Hint: 6124/977 is approximately 6. Question 25 Explanation: Topics: Estimation, simplifying fractions (Objective 0016). Question 26 #### The polygon depicted below is drawn on dot paper, with the dots spaced 1 unit apart. What is the perimeter of the polygon? A $$\\large 18+\\sqrt{2} \\text{ units}$$Hint: Be careful with", "= 3$. Again, the center, original point, and dilated point are collinear. Therefore, you can draw a ray from the origin to $C’, B’,$ and $A’$ such that the rays pass through $C, B,$ and $A$, respectively. Let’s show that dilations are a similarity transformation (preserves shape). Using the distance formula, we will find the lengths of the sides of both triangles in Example 6 to demonstrate this. $& \\underline{\\triangle ABC} && \\underline{\\triangle A'B'C'}\\\\& AB=\\sqrt{(2-5)^2+(1-1)^2}=\\sqrt{9}=3 && A'B'=\\sqrt{(6-15)^2+(3-3)^2}=\\sqrt{81}=9\\\\& AC=\\sqrt{(2-3)^2+(1-6)^2}=\\sqrt{26} && A'C'=\\sqrt{(6-9)^2+(3-18)^2}=\\sqrt{234}=3 \\sqrt{26}\\\\& CB=\\sqrt{(3-5)^2+(6-1)^2}=\\sqrt{29} && C'B'=\\sqrt{(9-15)^2+(18-3)^2}=\\sqrt{261}=3 \\sqrt{29}$ From this, we also see that all the sides of $\\triangle A'B'C'$ are three times larger than $\\triangle ABC$. Therefore, a dilation will always produce a similar shape to the original. In the coordinate plane, we say that $A'$ is a “mapping” of $A$. So, if the scale factor is 3, then $A(2, 1)$ is mapped to (usually drawn with an arrow) $A'(6, 3)$. The entire mapping of $\\triangle ABC$ can be written $(x,y) \\rightarrow (3x, 3y)$ because $k = 3$. For any dilation the mapping will be $(x, y) \\rightarrow (kx, ky)$. Know What? Revisited Answers to this project will vary depending on what you decide to draw. Make sure that you have at least five objects with some sort of detail. If you are having trouble getting started, go to the", "\\sqrt{81} = 9\\\\& AC = \\sqrt{(2-3)^2 + (1-6)^2} = \\sqrt{26} && A'C' = \\sqrt{(6-9)^2+(3-18)^2} = \\sqrt{234} = 3 \\sqrt{26}\\\\& CB = \\sqrt{(3-5)^2 + (6-1)^2} = \\sqrt{29} && C'B' = \\sqrt{(9-15)^2 +(18-3)^2} = \\sqrt{261} = 3 \\sqrt{29}$ From this, we also see that all the sides of $\\triangle A'B'C'$ are three times larger than $\\triangle ABC$ . Example C Quadrilateral $EFGH$ has vertices $E(-4, -2), F(1, 4), G(6, 2)$ and $H(0, -4)$ . Draw the dilation with a scale factor of 1.5. Remember that to dilate something in the coordinate plane, multiply each coordinate by the scale factor. For this dilation, the mapping will be $(x, y) \\rightarrow (1.5x, 1.5y)$ . $& E(-4, -2) \\rightarrow (1.5(-4), 1.5(-2)) \\rightarrow E'(-6, -3)\\\\& F(1, 4) \\rightarrow (1.5(1), 1.5(4)) \\rightarrow F'(1.5, 6)\\\\& G(6, 2) \\rightarrow (1.5(6), 1.5(2)) \\rightarrow G'(9,3)\\\\& H(0, -4) \\rightarrow (1.5(0),1.5(-4)) \\rightarrow H'(0, -6)$ Watch this video for help with the Examples above. Vocabulary In the graph above, the blue quadrilateral is the original and the red image is the dilation. A dilation an enlargement or reduction of a figure that preserves shape but not size. All dilations are similar to the original figure. Similar figures are the same shape but not necessarily the same size. The center of dilation is the point of reference for the dilation and the scale", "- 54 2x² + 3x - 54. ### Other questions on the subject: Mathematics Mathematics, 21.06.2019, nuneybaby Scale factor is always (new/original) so it depends which rectangle came first. if it started as large and was dilated to the smaller one, then the ratio is (.5/2.5) which simplifi...Read More Explanation: given function is Since, According to the given statement we get f(x) by doing some changes in g(x).And, After seeing the given figure, we can say that the end behav...Read More step-by-step explanation: $$\\text{use}\\ \\dfrac{a^n}{a^m}=a^{n-m}\\\\\\\\\\dfrac{m^7n^3}{mn^{-1}}=m^{7-1}n^{)}=m^6n^4$$...Read More", "and $$\\overline{P Y”}$$ is the second one. Maintaining Mathematical Proficiency Solve the equation. Check your solution. Question 37. 5x + 16 = – 3x Question 38. 12 + 6m = 2m 6m – 2m = 12 4m = 12 m = 12/4 m = 3 Question 39. 4b + 8 = 6b – 4 Question 40. 7w – 9 = 13 – 4w 7w + 4w = 13 + 9 11w = 22 w = 22/11 w = 2 Question 41. 7(2n + 11) = 4n Question 42. -2(8 – y) = – 6y -2(8 – y) = – 6y -16 + 2y = -6y -16 = -6y – 2y -16 = -8y y = 2 Question 43. Last year. the track team’s yard sale earned $500. This year. the yard sale earned$625. What is the percent of increase? ### 4.5 Dilations Exploration 1 Dilating a Triangle in a Coordinate Plane Work with a partner: Use dynamic geometry software to draw any triangle and label it ∆ABC. a. Dilate ∆ABC using a scale factor of 2 and a center of dilation at the origin to form ∆A’B’C’. Compare the coordinates, side lengths. and angle measures of ∆ABC and ∆A’B’C’. b. Repeat part (a) using a scale factor of $$\\frac{1}{2}$$ LOOKING FOR STRUCTURE To be proficient in math, you", "= \\sqrt{81} = 9\\\\& AC = \\sqrt{(2-3)^2 + (1-6)^2} = \\sqrt{26} && A'C' = \\sqrt{(6-9)^2+(3-18)^2} = \\sqrt{234} = 3 \\sqrt{26}\\\\& CB = \\sqrt{(3-5)^2 + (6-1)^2} = \\sqrt{29} && C'B' = \\sqrt{(9-15)^2 +(18-3)^2} = \\sqrt{261} = 3 \\sqrt{29}$ From this, we also see that all the sides of $\\triangle A'B'C'$ are three times larger than $\\triangle ABC$ . #### Example C Quadrilateral $EFGH$ has vertices $E(-4, -2), F(1, 4), G(6, 2)$ and $H(0, -4)$ . Draw the dilation with a scale factor of 1.5. Remember that to dilate something in the coordinate plane, multiply each coordinate by the scale factor. For this dilation, the mapping will be $(x, y) \\rightarrow (1.5x, 1.5y)$ . $& E(-4, -2) \\rightarrow (1.5(-4), 1.5(-2)) \\rightarrow E'(-6, -3)\\\\& F(1, 4) \\rightarrow (1.5(1), 1.5(4)) \\rightarrow F'(1.5, 6)\\\\& G(6, 2) \\rightarrow (1.5(6), 1.5(2)) \\rightarrow G'(9,3)\\\\& H(0, -4) \\rightarrow (1.5(0),1.5(-4)) \\rightarrow H'(0, -6)$ Watch this video for help with the Examples above. ### Vocabulary In the graph above, the blue quadrilateral is the original and the red image is the dilation. A dilation an enlargement or reduction of a figure that preserves shape but not size. All dilations are similar to the original figure. Similar figures are the same shape but not necessarily the same size. The center of dilation is the point of reference for the dilation", "0.001 between -0.005 and +0.004. (In other words, decrease the grid size by a factor of 10.) Of the resulting 100000000 quadrilaterals, save the 10000 with the smallest dilation to file. Step 3: Repeat Step 2 with the grid size decreased by a factor of 10. And so on. This procedure converges on the isosceles trapezium described above. So while this is not a proof, it is likely that the smallest possible dilation of a quadrilaterlal is indeed 1.89615765267304... (which is the real root of the polynomial $x^5 - x^4 - 4x - 4$). Edit by J.O'Rourke (16Aug10). If I've followed Tony's description in the comment below correctly, here is his quadrilateral with the (conjectured) smallest dilation: $h=0.896158$, $w=0.25552$: - @TonyK: Added a drawing of your first trapezium. – Joseph O'Rourke Jul 21 '10 at 1:06 @Joseph: Thanks – TonyK Jul 21 '10 at 10:39 @TonyK: Great work! Could you please specify the resulting quadrilateral in some definitive form? I am having some difficulty extracting exactly what is \"the trapezium above.\" – Joseph O'Rourke Aug 12 '10 at 21:53 OK: in the diagram, replace 13 by 1; 5 by cos 2*theta; and 66/13 by w (where theta is 0.5555166235227462... radians, and w is to be determined). The perimeter P is 2*(cos 2*theta + w + 1), and the", "wFo + 67 Vz whcre a and b are rcal numbers. 721 201 swer:’ “Each of Exercises 19-24 gives a formula for a function y = f(x) . In each case; find f-!(x) and identify the domain and range of f-!_ 19. f(x) = s 21. f(x) = ~3 + [ 23. f(x) = I/x?, x> 0 20. f(x) =X x20 22. f(x) = (1/2)x 5 7/2 24. f(x) = I/x’ , x # 0” “DulaMth AMalh Dcauilh dcltamalh com/student/solve/12727411/linearTrigEquatiens ) Dashboard Pol0 Prcpag _ Twclve Angry Thal GXIncbe Linear Trigonometric Equations Feb 17,11.56.49 PM Ncw Tab Watch help video Find all angles; 09 < C 3609 , that satisfy the equation below; to the nearest tenth of a degree (if necessary). ~4cos C + 8 = cos C + 8 atlempt out of 2 3 ‘Qf D W @” ‘Find the coordinates of the images of points and under the dilation Do. Find the image of (x. under Do. What dilation with center maps to R? 8. What dilation with center maps C t0 point 01? Find the coordinates of the image of under {)B. Find the coordinates of the image of B under Dc. A04, 4 B(z . 2) C(3.0) Ies 5-10’ ‘103-124 Solving Basic Equations Find all real solutions of the equation. 103. % +", "when the s.e. is in pixel $I(0,1)$ then $$\\max( I(0,0)+SE(0,-1), I(0,1)+SE(0,0), I(1,0)+SE(1,-1), I(1,1)+SE(1,0) ) =\\\\ \\max(5+x=x, 2, 3, -4)$$ 1. Can someone explain to me what is the correct way of doing this? 2. In the dilation do we need to flip the structuring element? What about erosion? - According to this paper, the grayscale dilation of image $I$ by a non-flat structuring element $S$ is defined as follows: $[I ⊕ S](x,y) = \\displaystyle\\max_{(s,t)∈S}\\{I(x-s,y-t)+S(s,t)\\}$ Since the origin of the structuring element is in the top right corner, we have that $s∈[0,1]$ and $t∈[-2,0]$ excluding the point $(s,t)=(0,-1)$. So using your example of element $I(0,1)$ we have the following: $[I ⊕ S](0,1)=\\max\\{I(0,3)+S(0,-2),\\ I(-1,3)+S(1,-2),\\ I(-1,2)+S(1,-1),\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ I(0,1)+S(0,0),\\ I(-1,1)+S(1,0)\\}$ $=\\max\\{0+0,0+0,0+2,2+0,0+(-1)\\}=\\max\\{0,0,2,2,-1\\}=2$ Which is the correct value for element $(0,1)$ in the dilated matrix. You can then simply apply the formula to the other $(x,y)$ value pairs of interest to obtain the final result. The definition of erosion (and the other morphological operations) can also be found in that paper and applied in the same manner. -", "2. By dilation quadrilateral QRST from the origin and with a scale factor of 2, we get the image Q’R’S’T’ Q'(-2, 0) → W(0, 2) R'(-4, 4) → X(4, 4) S'(2, 6) → Y(6, -2) T'(4, 2) → Z(2, -4) The symmetry transformation of the two figures is 270° rotation and silation with a scale factor of 2. Question 11. G(- 2, 3), H(4, 3), I(4, 0) and J(1, 0), K(6, – 2), L(1, – 2) Question 12. D(- 4, 3), E(- 2, 3), F(- 1, 1), G(- 4, 1) and L(1, – 1), M(3, – 1), N(6, – 3), P(1, – 3) $$\\overline{L P}$$ = 2 units $$\\overline{D G}$$ = 2 units The ratio = 2/2 = 1 The long leg: $$\\overline{P N}$$ = 5 units $$\\overline{G F}$$ = 3 units The ratio = 5/3 Since the ratio is not a constant, then the figures are not similar. In Exercises 13 and 14, prove that the figures are similar. Question 13. Given Right isosceles ∆ABC with leg length j. right isosceles ∆RST with leg length k. $$\\overline{C A}$$ || $$\\overline{R T}$$ Prove ∆ABC is similar to ∆RST. Question 14. Given Rectangle JKLM with side lengths x and y, rectangle QRST with side lengths 2x and 2y Prove Rectangle JKLM is similar to rectangle QRST. It can be", "0.25) = F'(3, 2) Hence the coordinate of the image are D'(0, 1), E'(3, 1), F'(3, 2) Question 17. G(-2, -2), H(-2, 6), J(2, 6); k = 0.25 Answer: When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor. P(x, y) = P'(x . a, y . a) Given points of the triangle: G(-2, -2), H(-2, 6), J(2, 6) and scale factor = 0.25 G(-2, -2) = G'(-2 × 0.25, -2 × 0.25) = G'(-0.5, -0.5) H(-2, 6) = H'(-2 × 0.25,6 × 0.25) = H'(-0.5, 1.5) J(2, 6) = G'(2 × 0.25, 6 × 0.25) = J'(0.5, 1.5) Hence the coordinate of the image are G'(-0.5, -0.5), H'(-0.5, 1.5), J'(0.5, 1.5) Question 18. M(2, 3), N(5, 3), P(5, 1); k = 3 Answer: When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor. P(x, y) = P'(x . a, y . a) Given points of the triangle: M(2, 3), N(5, 3), P(5, 1) and scale factor = 3 M(2, 3) = M'(2 × 3, 3 × 3) = M'(6, 9) N(5, 3) = N'(5 × 3, 3 × 3) = N'(15, 9) P(5, 1) = P'(5 × 3, 1 × 3) = P'(15, 3) Hence", "is 2. This new number is reflected across the real axis. w+z=2i-1+i=-1+3i $$\\overline{w+z}$$=-1-3i ### Eureka Math Precalculus Module 1 Lesson 17 Problem Set Answer Key Question 1. Describe the geometric effect of multiplying z by the reciprocal of each complex number listed below. a. w1=3i arg(w1)=$$\\frac{π}{2}$$ and |w1|=3 z is rotated by 2π-arg(w1), which is 2π-$$\\frac{π}{2}$$=$$\\frac{3π}{2}$$, and dilated by $$\\frac{1}{3}$$. b. w2=-2 arg(w2)=π and |w2|=2 z is rotated by 2π-arg(w2), which is 2π-π=π, and dilated by $$\\frac{1}{2}$$. c. w3=$$\\sqrt{3}$$+i arg(w3)=$$\\frac{π}{6}$$ and |w3|=2 z is rotated by 2π-arg(w3), which is 2π-$$\\frac{π}{6}$$=$$\\frac{11π}{6}$$, and dilated by $$\\frac{1}{2}$$. d. w4=1-$$\\sqrt{3}$$i arg(w4)=$$\\frac{5π}{3}$$ and |w4|=2 z is rotated by 2π-arg(w4), which is 2π-($$\\frac{5π}{3}$$)=$$\\frac{π}{3}$$, and dilated by $$\\frac{1}{2}$$. Question 2. Let z=-2-2$$\\sqrt{3}$$i. Show that the geometric transformations you described in Problem 1 really produce the correct complex number by performing the indicated operation and determining the argument and modulus of each number. a. $$\\frac{-2-2 \\sqrt{3} i}{w_{1}}$$ |z|=4, so the result of division is a complex number whose modulus is $$\\frac{1}{3}$$ of 4. arg(z)=$$\\frac{4π}{3}$$, so the result of division by a complex number is whose argument represents the same rotation as $$\\frac{4π}{3}$$+$$\\frac{3π}{2}$$=$$\\frac{17π}{6}$$, which would be $$\\frac{5π}{6}$$. b. $$\\frac{-2-2 \\sqrt{3} i}{w_{2}}$$ $$\\frac{-2-2 \\sqrt{3} i}{-2}$$ ==1+$$\\sqrt{3}$$i,||=2,arg()=$$\\frac{π}{3}$$ |z|=4 and $$\\frac{1}{2}$$ of 4 is 2. arg(z)=4$$\\frac{π}{3}$$, so the result of division will rotate z by –$$\\frac{π}{2}$$ and $$\\frac{4π}{3}$$–$$\\frac{π}{2}$$=$$\\frac{5π}{6}$$. c. $$\\frac{-2-2", "W(0. 0); k = $$\\frac{2}{3}$$ Question 18. J(4, 0), K(- 8, 4), L(0, – 4), M(12, – 8);k = 0. 25 In Exercises 19-22, graph the polygon and its image after a dilation with scale factor k. Question 19. B(- 5, – 10), C(- 10, 15), D(0, 5); k = 3 Question 20. L(0, 0), M(- 4, 1), N(- 3, – 6); k = – 3 Question 21. R(- 7, – 1), S(2, 5), T(- 2, – 3), U(- 3,- 3); k = – 4 Question 22. W(8, – 2), X(6, 0), Y(- 6, 4), Z(- 2, 2); k = – 0.5 ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in finding the scale factor of the dilation. Question 23. Question 24. In Exercises 25-28, the red figure is the image of the blue figure after a dilation with center C. Find the scale factor of the dilation. Then find the value of the variable. Question 25. Question 26. Question 27. Question 28. Question 29. FINDING A SCALE FACTOR You receive wallet-sized photos of your school picture. The photo is 2.5 inches by 3.5 inches. You decide to dilate the photo to 5 inches by 7 inches at the store. What is the scale factor of this dilation? Question 30. FINDING A SCALE FACTOR", "of PDRU-Net Block Type Convolutional Layer Receptive Field standard block 1 conv1_1 1-1+1$\\times$2+1=3 dilated rate = 1 conv1_2 3-1+1$\\times$2+1=5 residual block 2 conv2_1 5-1+2$\\times$2+1=9 dilated rate = 2 conv2_2 9-1+2$\\times$2+1=13 residual block 3 conv3_1 13-1+4$\\times$2+1=21 dilated rate = 4 conv3_2 21-1+4$\\times$2+1=29 residual block 4 conv4_1 29-1+8$\\times$2+1=45 dilated rate = 8 conv4_2 45-1+8$\\times$2+1=61 residual block 5 conv5_1 61-1+16$\\times$2+1=93 dilated rate = 16 conv5_2 93-1+16$\\times$2+1=125 residual block 6 conv6_1 125-1+32$\\times$2+1=189 dilated rate = 32 conv6_2 189-1+32$\\times$2+1=253 residual block 7 conv7_1 253-1+32$\\times$2+1=317 dilated rate = 32 conv7_2 317-1+32$\\times$2+1=381 residual block 8 conv8_1 381-1+32$\\times$2+1=445 dilated rate = 32 conv8_2 445-1+32$\\times$2+1=509 Block Type Convolutional Layer Receptive Field standard block 1 conv1_1 1-1+1$\\times$2+1=3 dilated rate = 1 conv1_2 3-1+1$\\times$2+1=5 residual block 2 conv2_1 5-1+2$\\times$2+1=9 dilated rate = 2 conv2_2 9-1+2$\\times$2+1=13 residual block 3 conv3_1 13-1+4$\\times$2+1=21 dilated rate = 4 conv3_2 21-1+4$\\times$2+1=29 residual block 4 conv4_1 29-1+8$\\times$2+1=45 dilated rate = 8 conv4_2 45-1+8$\\times$2+1=61 residual block 5 conv5_1 61-1+16$\\times$2+1=93 dilated rate = 16 conv5_2 93-1+16$\\times$2+1=125 residual block 6 conv6_1 125-1+32$\\times$2+1=189 dilated rate = 32 conv6_2 189-1+32$\\times$2+1=253 residual block 7 conv7_1 253-1+32$\\times$2+1=317 dilated rate = 32 conv7_2 317-1+32$\\times$2+1=381 residual block 8 conv8_1 381-1+32$\\times$2+1=445 dilated rate = 32 conv8_2 445-1+32$\\times$2+1=509 Comparison of the PDRU-Net, U-Net and FusionNet model parameters Dice Coefficient Pixel Accuracy Recall Precision F1 score U-Net ~33M 0.918 0.943 0.839 0.987 0.907 FusionNet ~78M 0.944 0.969", "# Let c = {2,3,4) and d = {1,2,4). define a relation r from c to das follows for all (x, y) e c x d: (x, ###### Question: Let c = {2,3,4) and d = {1,2,4). define a relation r from c to das follows for all (x, y) e c x d: (x, y) e r means that i 7 y. (a) write r as a list of ordered pairs. (b) draw an arrow diagram for r. ### 8. a segment has endpoints 4 (5,-2) and n(-1,-2). after a dilation with center (0,0), x (10.-4) and 8. a segment has endpoints 4 (5,-2) and n(-1,-2). after a dilation with center (0,0), x (10.-4) and n (-2,-4). if the scale factor is 3x+2, then which of the following is the value of x2 a. 0 b.1 c.2 d.3... ### Now try to use the underlined words in a new way. write a 50 words description of a school , its classrooms Now try to use the underlined words in a new way. write a 50 words description of a school , its classrooms and the teachers and students within. use as many of the underlined words as you can in your description.... ### If x = 3 is a zero of the polynomial function f(x) = 2x3", "∙ |z|. Exercise 2. What does the result of Exercise 1 tell us about the geometric effect of the transformation L(z)=wz? We see that |L(z)|=|wz|=|w| ∙ |z|, so the transformation L dilates by a factor of |w|. Exercise 3. If z and w are the complex numbers with the specified arguments and moduli, locate the point that represents the product wz on the provided coordinate axes. a. |w|=3, arg(w)=$$\\frac{π}{4}$$ |z|=$$\\frac{2}{3}$$, arg(z)=-$$\\frac{π}{2}$$ b. |w|=2, arg(w)=π |z|=1, arg(z)=$$\\frac{π}{4}$$ c. |w|=$$\\frac{1}{2}$$, arg(w)=$$\\frac{4π}{3}$$ |z|=4, arg(z)=-$$\\frac{π}{6}$$ ### Eureka Math Precalculus Module 1 Lesson 15 Problem Set Answer Key Problems 1 and 2 establish that any linear transformation of the form L(z)=wz has the geometric effect of a rotation by arg⁡(w) and dilation by |w|. Problems 3 and 4 lead to the development in the next lesson in which students build new transformations from ones they already know. Question 1. In the lesson, we justified our observation that the geometric effect of a transformation L(z)=wz is a rotation by arg(w) and a dilation by |w| for a complex number w that is represented by a point in the first quadrant of the coordinate plane. In this exercise, we will verify that this observation is valid for any complex number w. For a complex number w=a+bi, we only considered the case where a>0 and b>0.", "being simplified What is 9x+2x - 8 equivalent to after being simplified... ### ∠A and ∠B are supplementary angles. Ifm ∠A =(6x + 10)° and m ∠A = (4x + 30)°, then find the measureof ∠B ∠A and ∠B are supplementary angles. If m ∠A = (6x + 10)° and m ∠A = (4x + 30)°, then find the measure of ∠B... ### This is part of an email from your e-friend, Janet.So, what do you do in your free time? Whatis your favourite sport? What do you do atthe weekend?Write This is part of an email from your e-friend, Janet. So, what do you do in your free time? Whatis your favourite sport? What do you do atthe weekend?Write Janet an email answering her questions(80-100 words)​... ### A triangle with vertices at A(20, -30), B(10, -15), and C(5,-20) has been dilated with a center of dilation at the onginThe image of A triangle with vertices at A(20, -30), B(10, -15), and C(5,-20) has been dilated with a center of dilation at the ongin The image of B, point B', has the coordinates (2, -3). What is the scale factor of the dilation?... ### 3. all of the following statements identify a theme explored in 'lifeguard rules! ' except a. the importance 3. all of the following statements" ]

[ "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "+0 # Consider the projection matrix P that satisfies 0 169 1 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}.$$ Use the picture below to find 3 vectors with integer components and magnitude no greater than 8 that satisfy Find $$\\mathbf{P}^{-1} \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$$$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}.$$ Aug 6, 2019", "find the vector projection a onto b and theorthogonal projection of a onto b.1) a =< 4, 3 >, b =< 0, 5 >2) a =< 4, 3 >, b =< 0, −5 > Part A: find the vector projection b onto a and the orthogonal projection of b onto a. 1) a =< 4, 3 >, b =< 0, 5 > 2) a =< 4, 3 >, b =< 0, −5 > Part B: find the vector projection a onto b and the orthogonal projection of a onto b. 1) a =< 4, 3 >, b =< 0, 5...", "# Find distance from point to line I am asked to find the distance between a point ( 5,1,1 ) and a line $\\displaystyle \\left\\{\\begin{matrix} x + y + z = 0\\\\ x - 2y + z = 0 \\end{matrix}\\right.$ What ive done so far is to simplify by gauss elimination and I get to: $\\displaystyle \\begin{bmatrix} 1 &1 &1 &0 \\\\ 0 &1 &0 &0 \\end{bmatrix}$ In turn I put this back in the form of an equation $\\left\\{\\begin{matrix} x + y + z = 0\\\\ y = 0 \\end{matrix}\\right.$ What I need to find the distance between the point and line is any point on the line and a directional vector, right? Am i right in assuming that we have $\\begin{pmatrix} x,y,z \\end{pmatrix} = \\begin{pmatrix} t,0,-t \\end{pmatrix}$ So a point on this line could be (1,0,-1) or (12,0,-12)? If this is correct, how would I go on about finding the directional vector so I can put this all on the form of $\\begin{Vmatrix} \\overrightarrow{PQ} \\times \\overrightarrow{v} \\end{Vmatrix} \\div \\begin{Vmatrix} \\overrightarrow{v} \\end{Vmatrix}$ So to sum up, Q is given, the line is given in a system of equations, how do I extract a point P on the line and the vector v? For $t=1$, a possible directional vector for your line is $(1,0,-1)$. The projection of $(5,1,1)$", "? Free Version Moderate # Projection Onto Column Space LINALG-WLW1LH The matrix: $$A = \\begin{pmatrix} 2/3 & -2/3 \\\\\\ 1/3 & 2/3 \\\\\\ 2/3 & 1/3 \\end{pmatrix}$$ ...is orthogonal. Compute the projection of $\\mathbf{v}=(1,1,0)$ onto the column space of $A$. A $(2/3,1/3,2/3)$ B $(1,0,0)$ C $(1/3,2/3,-2/3)$ D $(2/3,1/3,-1/3)$ E $(0,1,1)$", "How do you do this vector projection? For $\\textbf{v} = \\begin{pmatrix} 2 \\\\ 3 \\\\ -1 \\end{pmatrix}$ and $\\textbf{w} = \\begin{pmatrix} 2 \\\\ -1 \\\\ 0 \\end{pmatrix}$, compute $\\text{proj}_{\\textbf{w}} \\textbf{v}$. I don't know how to project vectors of this form... I can only do it in 2D. Is there a formula for this? • You do it the same way. There are just more components to compute. – amd Mar 19 '17 at 1:01", "to an arbitrary vector $(x,y,z)$, and see if the resulting vector is orthogonal to the normal vector of that given plane, thus implying that the vector is projected successfully to the plane? This is necessary but not sufficient. If $n$ is normal to the plane and $\\langle n, Tx\\rangle = 0$ for all $x$, then the image of $T$ is contained in the plane, but that doesn't necessarily mean that $T$ is a projection onto the plane. For example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$ This matrix has the property that $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. So the image lies in the plane whose normal vector is $\\begin{pmatrix}1 \\\\ 0 \\\\ 0\\end{pmatrix}$. But it is not a projection onto that plane because the image only has dimension 1. For another example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 2 \\end{pmatrix}$$ Once again, we have $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. The dimension of the image is correct (2), but this", "+0 # Matrices 0 40 1 Let w = $$\\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix}$$. There exists a $$2 \\times 2$$ matrix $${P}$$ such that $$\\text{proj}_{{w}} {v} = {P} {v}$$ for all 2-dimensional vectors $${v}$$. Find $${P}$$. I'm not really sure where to start for this problem. Should I plug w into the projection formula? Mar 6, 2021 ### 1+0 Answers #1 0 The matrix is $$\\begin{pmatrix} 2 & -1 \\\\ -1 & 0 \\end{pmatrix}$$ Mar 7, 2021", "+0 # The vector $\\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$ is orthogonal to the vector $\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$. The vector \\$\\begin{pm 0 52 1 The vector $$\\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ is orthogonal to the vector $$\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$$. The vector $$\\begin{pmatrix} 2 \\\\ m \\end{pmatrix}$$ is orthogonal to the vector $$\\begin{pmatrix} 3 \\\\ k \\end{pmatrix}$$. Find m . Feb 2, 2019 #1 +96080 +1 If vectors are orthagonal, their dot product = 0 So [ 4 * 3 ] + [ -7 * k] = 0 12 - 7k = 0 -7k = -12 k = 12/7 So [2 * 3 ] + [12/7 * m] = 0 6 + (12/7)m = 0 (12/7)m = - 6 m = -6*7/12 = -7/2 Feb 2, 2019", "The orthogonal projection $\\mbox{proj}_{M}({\\vec{v}\\,})$ of a vector is its projection onto $M$ along $M^\\perp$. Example 3.5 In $\\mathbb{R}^3$, to find the orthogonal complement of the plane $P=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, 3x+2y-z=0 \\}$ we start with a basis for $P$. $B=\\langle \\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\rangle$ Any $\\vec{v}$ perpendicular to every vector in $B$ is perpendicular to every vector in the span of $B$ (the proof of this assertion is Problem 10). Therefore, the subspace $P^\\perp$ consists of the vectors that satisfy these two conditions. $\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0 \\qquad \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0$ We can express those conditions more compactly as a linear system. $P^\\perp=\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\begin{pmatrix} 1 &0 &3 \\\\ 0 &1 &2 \\end{pmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix} =\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix} \\}$ We are thus left with finding the nullspace of the map represented by the matrix, that is, with calculating the solution set of a homogeneous linear system. $P^\\perp =\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\;\\begin{array}{*{3}{rc}r} v_1 & & &+ &3v_3 &= &0 \\\\ & &v_2 &+ &2v_3 &= &0 \\end{array}\\; \\} =\\{k\\begin{pmatrix} -3", "? Free Version Easy # Orthogonal Projection in Two Dimensions LINALG-Z3MT8D The orthogonal projection of the vector $\\begin{bmatrix}3\\\\\\ 1\\end{bmatrix}$ onto the line $3x -4y=0$ is A $\\begin{bmatrix}\\frac{12}{5}\\\\\\ \\frac{9}{5}\\end{bmatrix}$ B $\\begin{bmatrix}-1\\\\\\ 3\\end{bmatrix}$ C $\\begin{bmatrix}0\\\\\\ 0\\end{bmatrix}$ D $5$", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "+0 # help 0 45 1 Find the projection of $$\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ onto the plane $$3x - y + 4z = 0.$$ Sep 2, 2019 #1 +23086 +2 Find the projection of $$\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ onto the plane $$3x - y + 4z = 0$$. $$\\text{Let \\vec{P}= \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}}$$ 1. Normal vector $$\\mathbf{\\vec{n}}$$ $$\\begin{array}{|rcll|} \\hline \\vec{n} &=& \\begin{pmatrix} 3 \\\\ -1 \\\\ 4 \\end{pmatrix} \\quad | \\quad \\text{plane: }{\\color{red}3}x {\\color{red}-1}y + {\\color{red}4}z = 0 \\\\ \\hline \\end{array}$$ 2. $$\\mathbf{\\vec{P}_{Proj.}}$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{\\vec{P}_{Proj.}} &=& \\mathbf{\\vec{P}+\\lambda\\vec{n}} \\\\\\\\ \\vec{P}_{Proj.} &=& \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}+\\lambda\\begin{pmatrix} 3 \\\\ -1 \\\\ 4 \\end{pmatrix} \\\\\\\\ x_{Proj.} &=& 1+3\\lambda \\\\ y_{Proj.} &=& 2-\\lambda \\\\ z_{Proj.} &=& 3+4\\lambda \\\\ \\hline \\end{array}$$ 3. $$\\mathbf{\\lambda}$$ $$\\begin{array}{|rcll|} \\hline 3x_{Proj.} - y_{Proj.} + 4z_{Proj.} &=& 0 \\\\ 3(1+3\\lambda) - (2-\\lambda) + 4(3+4\\lambda) &=& 0 \\\\ 3+9\\lambda-2+\\lambda+12+16\\lambda &=& 0 \\\\ 26\\lambda &=& -13 \\\\\\\\ \\lambda &=& -\\dfrac{13}{26} \\\\\\\\ \\mathbf{\\lambda} &=& -\\mathbf{\\dfrac{1}{2}} \\\\ \\hline \\end{array}$$ 4. $$\\mathbf{\\vec{P}_{Proj.}}$$ $$\\begin{array}{|rcll|} \\hline \\vec{P}_{Proj.} &=& \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}+\\lambda\\begin{pmatrix} 3 \\\\ -1 \\\\ 4 \\end{pmatrix} \\quad | \\quad \\lambda = -\\dfrac{1}{2} \\\\\\\\ &=& \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix} -\\dfrac{1}{2}\\begin{pmatrix} 3 \\\\ -1 \\\\ 4 \\end{pmatrix} \\\\\\\\ &=& \\begin{pmatrix} 1-\\dfrac{3}{2} \\\\\\\\ 2+\\dfrac{1}{2} \\\\\\\\ 3-\\dfrac{4}{2} \\end{pmatrix} \\\\\\\\", "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019", "+0 # Help! 0 49 1 +11 Consider the line parametrized by \\begin{align*} x&= 2t + 1,\\\\ y& = -3t +2. \\end{align*} Find a vector $\\begin{pmatrix}a\\\\b\\end{pmatrix}$ that's parallel to this line and satisfies $a+b = 3$. Jul 27, 2020 edited by FearlessIsland3 Jul 27, 2020", "# Finding an orthogonal projection matrix onto the plane Find the orthogonal projection matrix onto the plane $$x + y - z = 0$$ The solution to this video recitation video on MIT open courseware immediately states that we can chose $$a_1 = \\begin{pmatrix} 1\\\\ -1\\\\ 0 \\end{pmatrix} \\ \\ \\ \\ \\ \\ \\ \\ \\text{and} \\ \\ \\ \\ \\ \\ \\ a_2 = \\begin{pmatrix} 1\\\\ 0\\\\ 1 \\end{pmatrix}$$ So then $$A = \\begin{pmatrix} 1 & 1\\\\ -1 & 0 \\\\ 0 & 1 \\end{pmatrix}$$ and we simply apply the equation $P = A(A^{T}A)^{-1}A^{T}$. The lecturer did not explain how he can choose such an $a_1$ and $a_2$. Can anyone explain that? • I hope the lecture then goes on to show an easier way to compute this projection matrix by using the normal to the plane. – amd Jan 9 '18 at 21:33 • @amd how would you go about doing this? Oct 13 '21 at 2:23 The lecturer simply chose two vectors $a_{1}$ and $a_{2}$ that are independent and contained in the plane $x+y-z=0$. He then applied the formula that you mentioned. Well, you can choose any values al long as they satisfy the given equation. For example, $$x=0,y=1,z=1 \\rightarrow a_3 = \\begin{pmatrix}0\\\\1\\\\1\\end{pmatrix}$$ However, you should be careful to choose independent vectors. For instance,", "# Compute the orthogonal projection $w$ of $u$ on a line $L$ A vector $u$ and a line $L$ in $\\mathbb{R}^2$ are given. Compute the orthogonal projection $w$ of $u$ on $L$, and use it to compute the distance $d$ from the endpoint of $u$ to $L$ $u = \\begin{pmatrix} 5 \\\\ 0 \\end{pmatrix}$ $y = 0$ I know to that $w = \\frac{u \\cdot v}{||v||^2}v$ and that $d = u - w$ The answer in the book says the answer is $w = \\begin{pmatrix}5 \\\\0 \\end{pmatrix}$ and $d = 0$ How do I find $v$ to solve the problem? • What is your line $L$? Is it the line given by the equation $y = 0$? – Eli Rose Dec 14 '15 at 4:16 • Yes, the book only gives $u$ and the equation of Line $L$ – user273323 Dec 14 '15 at 4:25 The vector $v$ can be any vector in the direction of the line $L$. For a line $ax + by = c$, the vector $\\begin{pmatrix}b\\\\-a\\end{pmatrix}$, or any non-zero multiple of it, lies in the direction of the line. So for $L: y=0$, alias $0x+1y=0$, the vector $v = \\begin{pmatrix}1\\\\0\\end{pmatrix}$ will do. Put this into your equation gives $w=\\begin{pmatrix}5\\\\0\\end{pmatrix}$ as expected. Or, more simply, you can observe that $u$ is a vector parallel to", "by the above relation actually passes throught the point $\\;A:=\\begin{pmatrix}a\\\\a\\\\a\\end{pmatrix}\\,$ (very easy) ,and that the vector $\\;A-u\\;$ is perpendicular to the given line there (very easy, too). Can you explain where the $A-u$ vector comes from? I don't understand why showing $A-u$ to be perpendicular to the line $x=y=z$ is helpful. – Xoque55 Nov 25 '13 at 4:33", "can check with a picture that $R=2P-I$, where $P$ is the projection onto the line. Taking $v=(1,m)^T$ a vector along the line, then $$P\\begin{bmatrix} x\\\\ y\\end{bmatrix} = \\begin{bmatrix} 1&m\\end{bmatrix} \\begin{bmatrix} x\\\\ y\\end{bmatrix}\\,\\begin{bmatrix} 1\\\\ m\\end{bmatrix} / \\begin{bmatrix} 1&m\\end{bmatrix} \\begin{bmatrix} 1\\\\ m\\end{bmatrix} =\\frac1{1+m^2} \\begin{bmatrix} x+my\\\\ mx+m^2y\\end{bmatrix}.$$ So $$P=\\frac1{1+m^2} \\begin{bmatrix} 1&m\\\\ m&m^2\\end{bmatrix}$$ and $$R=2P-I=\\frac1{1+m^2} \\begin{bmatrix} 1-m^2&2m\\\\ 2m&m^2-1\\end{bmatrix}.$$ If you made a sketch you will se that $R(x)=2 \\Pi_v(x)-x$ where $v=(1,m)$ and $\\Pi_v$ is the projection of the vector $x$ over the vector $v$." ]

[ "to the other line The distance can now be calculated as described under distance point and line. First set up an auxiliary plane with $P$ as the support point and direction vector as the normal vector. $\\text{H: } (\\vec{x} - \\vec{a}) \\cdot \\vec{n}=0$ $\\text{H: } (\\vec{x} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}) \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$. $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can", "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "shown below) $$r·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$(3×1)+(2×[-5])+(7×8)=49$$ Therefore this means that the distanve of the vector d = 49 To complete the equation we need to write down the value of the whole equation in cartesian form $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$x-5y+8z=49$$ x - 5y + 8z - 49 = 0 #### Question 2 Write the equation of the plane through (2,1,3), given that the vector (4,5,6) is perpendicular $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 1\\\\\\ 3\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$(2×4)+(1×5)+(3×6)=31$$ Therefore the distance of the vector is d = 31 $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$4x+5y+6z=31$$ 4x + 5y + 6z - 31 = 0 #### Intersection of a line and a plane Find the point of intersection between the line The normal (perpendicular) to a plane is $$r=\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ with plane $$5x+y-z=1$$ First we need to replace r with the matrix (x,y,z) in order to find the values of x,y and z $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ We can change this equation into 3 different equations by seperating the matrices into: x = 2 + λ y = 3 + 2λ z = 4 - λ And we know that 5x+y-z=1 Replacing the values of x,y and z with these new values we get", "\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ 2. Calculate perpendicular point The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$ $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ 3. Calculate the distance between the two points The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can be easily calculated using vectors. $d=|\\vec{PF}|$ $=\\left| \\begin{pmatrix} -1-(-1) \\\\ 0-0 \\\\ 1-3 \\end{pmatrix}\\right|$ $=\\left| \\begin{pmatrix} 0 \\\\ 0 \\\\ -2 \\end{pmatrix}\\right|$ $=\\sqrt{(-2)^2}$ $=2$ The distance from the line $g$ to the point $P$ is 2 LU.", "E}$${\\ displaystyle {\\ vec {a}} = {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}}}$ ${\\ displaystyle {\\ vec {OF}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {\\ left [{\\ begin {pmatrix} 8 \\\\ 8 \\ \\ -1 \\ end {pmatrix}} - {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}} \\ right] \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}} {{\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {27} {9}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}}}$. The distance of the point from the plane thus satisfies ${\\ displaystyle d (P; E)}$${\\ displaystyle P}$${\\ displaystyle E}$ ${\\ displaystyle d (P; E) = d (P; F) \\; {\\ widehat {=}} \\; | {\\ vec {PF}} | = \\ left | {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}} -", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "form is: $\\boxed{y = - \\frac{1}{4}x - 1}$ Determine the equation of the line... $& \\begin{pmatrix}x_1, & y_1\\\\-3, & 1 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\-3, & -7 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-7-1}{-3--3}\\\\m & = \\frac{-7-1}{-3+3}\\\\m & = \\frac{-8}{0}\\\\m & = \\text{undefined}$ A line that has a slope that is undefined is a line that is parallel to the $y-$axis. The equation of this line is $\\boxed{x = -3}$. $& \\begin{pmatrix}x_1, & y_1\\\\-8, & 4 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\2, & -6 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-6-4}{2--8}\\\\m & = \\frac{-10}{10}\\\\m & = -1\\\\\\\\y & = mx + b\\\\4 & = -1(-8) + b\\\\4 & = 8 + b\\\\4-8 &= 8-8+b\\\\-4 & = b$ $m = -1$ and the $y-$intercept is (0, –4). The equation of this line is: $\\boxed{y = -1x-4}$ $& \\begin{pmatrix}x_1, & y_1\\\\0, & 5 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\4, & -3 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-3-5}{4-0}\\\\m & = \\frac{-8}{4}\\\\m & = -2$ $m = -2$ and the $y-$intercept is (0, 5). The equation of this line is: $\\boxed{y = -2x + 5}$ For each of the following real world problems... $& \\begin{pmatrix}x_1, & y_1\\\\320, & 124 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\600, & 164 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m &", "point, there is exactly one. I assume you want the equation for that line. Since the line passes through both the given point and the given line, a vector in the direction of this perpendicular line would be the vector $\\begin{pmatrix}3-(-1) \\\\ 1-(-2+t) \\\\ -2-(-1+t)\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 3-t \\\\ -1-t\\end{pmatrix}$. Let's find a vector in the direction of the given line by finding two points along the line (we can choose any two points, so let's say $t=0, t=1$). So, a vector in the direction of the given line is $\\begin{pmatrix}-1-(-1) \\\\ -2+1 - (-2+0) \\\\ -1+1 - (-1+0)\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Since we want these vectors to be perpendicular, we want the dot product to be zero. Hence: $(4\\cdot 0) + (1\\cdot (3-t)) + (1\\cdot (-1-t)) = 2-2t = 0$ implies $t=1$. Hence, a vector in the direction of the perpendicular line is $\\begin{pmatrix}4 \\\\ 3-1 \\\\ -1-1\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 2 \\\\ -2\\end{pmatrix}$. So, the perpendicular line would have parametric equation $x=3+4t, y=1+2t, z=-2-2t$. 4. ## Re: equation of line through point orthogonal to line so the point (does this mean you are treating the point as a bounded vector) minus the line gives a direction vector along the perpendicular line then I take the given direction vector of the line which", "# Shortest Distance between two lines in the 3D plane Find the shortest distance between the line $x$ $=$ $2$ $-$ $t$ $y$ $=$ $-1$ $+$ $2t$ and $z$ $=$ $-1$ $+$ $t$ and the line $x$ $=$ $5$ $+$ $3s$ $y$ $=$ $0$ and $z$ $=$ $2$ $-$ $s$. Find the points on these two lines that give this shortest distance. Using the cross product, I got the normal vector to be: \\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix} The I selected two points on the lines with $t$ $=$ $-1$: \\begin{pmatrix}3\\\\ -3\\\\ -2\\end{pmatrix} and with $s$ $=$ $1$: \\begin{pmatrix}8\\\\ 0\\\\ 1\\end{pmatrix} Therefore the distance between the these two points gave the vector: \\begin{pmatrix}-5\\\\ \\:-3\\\\ \\:-3\\end{pmatrix} So then I projected this vector: $\\left(\\frac{\\left(-5,\\:-3,\\:-3\\right)\\cdot \\left(-5,\\:-3,\\:-3\\right)}{\\left(-2,\\:-2,\\:-6\\right)\\cdot \\left(-2,\\:-2,\\:-6\\right)}\\right)\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ This left me with: $\\frac{43}{\\:44}\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ Which left me with a final distance of: $\\frac{43}{44}\\sqrt{44}$ But this answer isn't correct. I think I have made a mistake in my projection calculation but I am not sure where. Also, how would I find the two points that are closest together. I am also a little confused about that. Any help would be highly appreciated! • $(-1,2,1)\\times(3,0,-1)\\ne(-2,-2,-6)$. – amd Apr 6 '18 at 21:50 • Thank you! I guess I missed the sign. So it should be $(-2, 2, -6)$? But this will", "is: $$cosθ = \\frac{10}{6\\sqrt{26}}$$ $$θ = cos^{-1}(\\frac{10}{6\\sqrt{26}}) = 71.44°$$ #### 3-D Vector Equation of a line $$\\underline{r} = \\begin{pmatrix}3\\\\\\ 4\\\\\\ 1\\end{pmatrix} + λ\\begin{pmatrix}2\\\\\\ 1\\\\\\ -1\\end{pmatrix}$$ has position $$\\begin{pmatrix}3\\\\\\ 4\\\\\\ 1\\end{pmatrix}$$ and direction $$\\begin{pmatrix}2\\\\\\ 1\\\\\\ -1\\end{pmatrix}$$ Find the equation of the line In order to find the equation of the line we need to find the values of x,y and z in terms of λ and then we can set them equal to eachother To do this we do the same as we did with 2-D vectors and take the x,y and z terms from the equation seperately like so: First we find the equation for x: $$x=3+2λ$$ $$x-3=2λ$$ $$λ=\\frac{x-3}{2}$$ Now we find the equation for y: $$y=4+λ$$ $$λ=\\frac{y-4}{1}$$ Finally we find the equation for z: $$z=1-λ$$ $$λ=\\frac{z-1}{-1}$$ Setting all these equations as equal to eachother gives us the vector equation of the line: $$\\frac{x-3}{2}=\\frac{y-4}{1}=\\frac{z-1}{-1}$$ This can be re-written as: $$\\frac{x-3}{2}=y-4=-z+1$$", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions. (i) $cos{\\theta}=\\frac{|\\begin{pmatrix}2\\\\-2\\\\1\\end{pmatrix}{\\bullet}\\begin{pmatrix}-6\\\\3\\\\2\\end{pmatrix}|}{\\sqrt{9} \\sqrt{49}}=\\frac{16}{21}$ $\\theta = {40.4}^{\\circ}$ (ii) Using Graphing Calculator, equation of line, $r = \\begin{pmatrix}{-\\frac{1}{6}}\\\\{-\\frac{2}{3}}\\\\0\\end{pmatrix} + \\lambda \\begin{pmatrix}7\\\\10\\\\6\\end{pmatrix}, \\lambda \\in \\mathbb{R}$ (iii) Distance A to $P_1 = \\frac{|\\begin{pmatrix}2\\\\-2\\\\1\\end{pmatrix}{\\bullet}\\begin{pmatrix}4\\\\3\\\\c\\end{pmatrix} - 1|}{\\sqrt{9}} = \\frac{|1+c|}{3}$ Distance A to $P_1 = \\frac{|\\begin{pmatrix}-6\\\\3\\\\2\\end{pmatrix}{\\bullet}\\begin{pmatrix}4\\\\3\\\\c\\end{pmatrix} + 1|}{\\sqrt{49}} = \\frac{|2c-14|}{7}$ $\\frac{|1+c|}{3} = \\frac{|2c-14|}{7}$ $49(1+c)^{2} = 9(2c-14)^{2}$ $c = -49 \\mathrm{~or~} \\frac{35}{13}$", "the latter is parallel to the $y$-axis ($y$ can change independently of $z$ and $x$). The line itself is the intersection of these two planes, and it can be projected along the planes indicated by the individual equations in (2) onto the $yz$-plane and the $zx$-plane, respectively. In a narrower meaning, the equation $y=mz\\!+\\!p$ represents the projection line of $l$ onto the $yz$-plane and the equation $x=nz\\!+\\!q$ the projection line of $l$ onto the $zx$-plane. A line in space can be represented also without using planes. We can derive another way by starting with two vectors: one vector $\\vec{r}_{0}$ drawn from the origin to a chosen point of the line (the so-called position vector of this point) and another vector (the so-called direction vector) $\\vec{u}$ determining the direction of the line. Then there exists a real number $t$ such that the position vector $\\vec{r}$ of an arbitrary point of the line can be expressed as $\\displaystyle\\vec{r}\\;=\\;\\vec{r}_{0}\\!+\\!t\\vec{u}.$ (3) This is the vector-formed equation of the line. Since every vector can be given by three coordinates of the end-point, we may write $\\vec{r}\\;=\\;\\begin{pmatrix}x\\\\ y\\\\ z\\end{pmatrix},\\quad\\vec{r}_{0}\\;=\\;\\begin{pmatrix}x_{0}\\\\ y_{0}\\\\ z_{0}\\end{pmatrix},\\quad\\vec{u}\\;=\\;\\begin{pmatrix}a\\\\ b\\\\ c\\end{pmatrix}\\!.$ $\\begin{pmatrix}x\\\\ y\\\\ z\\end{pmatrix}\\;=\\;\\begin{pmatrix}x_{0}\\\\ y_{0}\\\\ z_{0}\\end{pmatrix}+t\\begin{pmatrix}a\\\\ b\\\\ c\\end{pmatrix}\\!,$ and we obtain three scalar equations $\\displaystyle\\begin{cases}x\\;=\\;x_{0}\\!+\\!ta\\\\ y\\;=\\;y_{0}\\!+\\!tb\\\\ z\\;=\\;z_{0}\\!+\\!tc.\\end{cases}$ (4) These are the parametric equations of the line (3), with the parameter $t$ taking", "raise it from tightly wound around the earth to six feet above the earth. This exercise is recommended for all readers. Problem 13 Verify that this map $h:\\mathbb{R}^3\\to \\mathbb{R}$ $\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\;\\mapsto\\; \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\cdot\\begin{pmatrix} 3 \\\\ -1 \\\\ -1 \\end{pmatrix}=3x-y-z$ is linear. Generalize. Verifying that it is linear is routine. $\\begin{array}{rl} h(c_1\\cdot \\begin{pmatrix} x_1 \\\\ y_1 \\\\ z_1 \\end{pmatrix} +c_2\\cdot \\begin{pmatrix} x_2 \\\\ y_2 \\\\ z_2 \\end{pmatrix}) &=h(\\begin{pmatrix} c_1x_1+c_2x_2 \\\\ c_1y_1+c_2y_2 \\\\ c_1z_1+c_2z_2 \\end{pmatrix}) \\\\ &=3(c_1x_1+c_2x_2)-(c_1y_1+c_2y_2)-(c_1z_1+c_2z_2) \\\\ &=c_1\\cdot (3x_1-y_1-z_1)+c_2\\cdot (3x_2-y_2-z_2) \\\\ &=c_1\\cdot h(\\begin{pmatrix} x_1 \\\\ y_1 \\\\ z_1 \\end{pmatrix}) +c_2\\cdot h(\\begin{pmatrix} x_2 \\\\ y_2 \\\\ z_2 \\end{pmatrix}) \\end{array}$ The natural guess at a generalization is that for any fixed $\\vec{k}\\in\\mathbb{R}^3$ the map $\\vec{v}\\mapsto\\vec{v}\\cdot\\vec{k}$ is linear. This statement is true. It follows from properties of the dot product we have seen earlier: $(\\vec{v}+\\vec{u})\\cdot\\vec{k}=\\vec{v}\\cdot\\vec{k}+ \\vec{u}\\cdot\\vec{k}$ and $(r\\vec{v})\\cdot\\vec{k}=r(\\vec{v}\\cdot\\vec{k})$. (The natural guess at a generalization of this generalization, that the map from $\\mathbb{R}^n$ to $\\mathbb{R}$ whose action consists of taking the dot product of its argument with a fixed vector $\\vec{k}\\in\\mathbb{R}^n$ is linear, is also true.) Problem 14 Show that every homomorphism from $\\mathbb{R}^1$ to $\\mathbb{R}^1$ acts via multiplication by a scalar. Conclude that every nontrivial linear transformation of $\\mathbb{R}^1$ is an isomorphism. Is that true for transformations of $\\mathbb{R}^2$? $\\mathbb{R}^n$? Let $h:\\mathbb{R}^1\\to", "# Compute the orthogonal projection $w$ of $u$ on a line $L$ A vector $u$ and a line $L$ in $\\mathbb{R}^2$ are given. Compute the orthogonal projection $w$ of $u$ on $L$, and use it to compute the distance $d$ from the endpoint of $u$ to $L$ $u = \\begin{pmatrix} 5 \\\\ 0 \\end{pmatrix}$ $y = 0$ I know to that $w = \\frac{u \\cdot v}{||v||^2}v$ and that $d = u - w$ The answer in the book says the answer is $w = \\begin{pmatrix}5 \\\\0 \\end{pmatrix}$ and $d = 0$ How do I find $v$ to solve the problem? • What is your line $L$? Is it the line given by the equation $y = 0$? – Eli Rose Dec 14 '15 at 4:16 • Yes, the book only gives $u$ and the equation of Line $L$ – user273323 Dec 14 '15 at 4:25 The vector $v$ can be any vector in the direction of the line $L$. For a line $ax + by = c$, the vector $\\begin{pmatrix}b\\\\-a\\end{pmatrix}$, or any non-zero multiple of it, lies in the direction of the line. So for $L: y=0$, alias $0x+1y=0$, the vector $v = \\begin{pmatrix}1\\\\0\\end{pmatrix}$ will do. Put this into your equation gives $w=\\begin{pmatrix}5\\\\0\\end{pmatrix}$ as expected. Or, more simply, you can observe that $u$ is a vector parallel to", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "the derivatives of the above quantities. In vector form these are $$\\vec{v}_A = \\begin{pmatrix} \\dot{x}_A \\\\ \\dot{y}_A \\\\ 0 \\end{pmatrix}$$ $$\\vec{v}_B = \\begin{pmatrix} \\dot{x}_B \\\\ \\dot{y}_B \\\\ 0 \\end{pmatrix}$$ and $$\\vec{a}_A = \\begin{pmatrix} \\ddot{x}_A \\\\ \\ddot{y}_A \\\\ 0 \\end{pmatrix}$$ $$\\vec{a}_B = \\begin{pmatrix} \\ddot{x}_B \\\\ \\ddot{y}_B \\\\ 0 \\end{pmatrix}$$ There are a total of 4 unknown degrees of freedom in the system corresponding to the 4 acceleration components of $\\vec{a}_A$ and $\\vec{a}_B$. In addition, there are two constraint equations applicable here, one keeping the cart on the ground and the other keeping the ball a distance $l$ from the cart. This makes the system with two effective degrees of freedom. To include the constraint equations in the system we are going to use what is called the Pfaffian form in terms of the velocity components. Since the cart does not move in the vertical direction this can be written as $\\hat{j}\\cdot\\vec{v}_A = 0$ or in expanded form $$\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}\\cdot\\begin{pmatrix} \\dot{x}_A \\\\ \\dot{y}_A \\\\ 0 \\end{pmatrix} = \\dot{y}_A = 0$$ Also the ball does not move towards the cart along the rod with direction vector $\\hat{n}=\\begin{pmatrix}\\text{-}\\sin\\theta & \\cos\\theta & 0 \\end{pmatrix}^\\top$ so the constraint is $\\hat{n}\\cdot\\vec{v}_B = 0$ or in expanded form $$\\begin{pmatrix}\\text{-}\\sin\\theta \\\\ \\cos\\theta \\\\ 0 \\end{pmatrix} \\cdot \\begin{pmatrix} \\dot{x}_B \\\\ \\dot{y}_B \\\\ 0 \\end{pmatrix} = \\dot{y}_B \\cos\\theta – \\dot{x}_B", "\\\\ 3 \\end{pmatrix} } \\\\ \\text{Let line \\vec{x} = \\begin{pmatrix} 6 \\\\ 7 \\\\ 7 \\end{pmatrix} + t \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} } \\\\ \\text{Let direction vector \\vec{r} = \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} }$$ $$\\begin{array}{|rcll|} \\hline \\left( \\vec{x}-\\vec{p} \\right) \\cdot \\vec{r} &=& 0 \\quad | \\quad (\\vec{x}-\\vec{p}) \\perp \\vec{r} \\\\\\\\ \\Bigg(\\begin{pmatrix} 6 \\\\ 7 \\\\ 7 \\end{pmatrix} + t \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix}-\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix} \\Bigg) \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} &=& 0 \\\\\\\\ \\Bigg(\\begin{pmatrix} 6-1 \\\\ 7-2 \\\\ 7-3 \\end{pmatrix} + t \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} \\Bigg) \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} &=& 0 \\\\\\\\ \\Bigg(\\begin{pmatrix} 5 \\\\ 5 \\\\ 4 \\end{pmatrix} + t \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} \\Bigg) \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} &=& 0 \\\\\\\\ \\begin{pmatrix} 5+3t \\\\ 5+2t \\\\ 4-2t \\end{pmatrix} \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} &=& 0 \\\\\\\\ 3(5+3t)+2(5+2t)-2(4-2t)&=& 0 \\\\ 15+9t+10+4t-8+4t &=& 0 \\\\ 17+17t &=& 0 \\\\ 17t &=& -17 \\quad | \\quad : 17 \\\\ \\mathbf{ t } &=& \\mathbf{ -1 } \\\\ \\hline \\end{array}$$ $$\\text{foot (of a perpendicular):}$$ $$\\begin{array}{|rcll|} \\hline \\vec{x}_{f} &=& \\begin{pmatrix} 6 \\\\ 7 \\\\ 7 \\end{pmatrix} + (-1) \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix}", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$," ]

[ "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "to the other line The distance can now be calculated as described under distance point and line. First set up an auxiliary plane with $P$ as the support point and direction vector as the normal vector. $\\text{H: } (\\vec{x} - \\vec{a}) \\cdot \\vec{n}=0$ $\\text{H: } (\\vec{x} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}) \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$. $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can", "# Shortest Distance between two lines in the 3D plane Find the shortest distance between the line $x$ $=$ $2$ $-$ $t$ $y$ $=$ $-1$ $+$ $2t$ and $z$ $=$ $-1$ $+$ $t$ and the line $x$ $=$ $5$ $+$ $3s$ $y$ $=$ $0$ and $z$ $=$ $2$ $-$ $s$. Find the points on these two lines that give this shortest distance. Using the cross product, I got the normal vector to be: \\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix} The I selected two points on the lines with $t$ $=$ $-1$: \\begin{pmatrix}3\\\\ -3\\\\ -2\\end{pmatrix} and with $s$ $=$ $1$: \\begin{pmatrix}8\\\\ 0\\\\ 1\\end{pmatrix} Therefore the distance between the these two points gave the vector: \\begin{pmatrix}-5\\\\ \\:-3\\\\ \\:-3\\end{pmatrix} So then I projected this vector: $\\left(\\frac{\\left(-5,\\:-3,\\:-3\\right)\\cdot \\left(-5,\\:-3,\\:-3\\right)}{\\left(-2,\\:-2,\\:-6\\right)\\cdot \\left(-2,\\:-2,\\:-6\\right)}\\right)\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ This left me with: $\\frac{43}{\\:44}\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ Which left me with a final distance of: $\\frac{43}{44}\\sqrt{44}$ But this answer isn't correct. I think I have made a mistake in my projection calculation but I am not sure where. Also, how would I find the two points that are closest together. I am also a little confused about that. Any help would be highly appreciated! • $(-1,2,1)\\times(3,0,-1)\\ne(-2,-2,-6)$. – amd Apr 6 '18 at 21:50 • Thank you! I guess I missed the sign. So it should be $(-2, 2, -6)$? But this will", "\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ 2. Calculate perpendicular point The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$ $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ 3. Calculate the distance between the two points The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can be easily calculated using vectors. $d=|\\vec{PF}|$ $=\\left| \\begin{pmatrix} -1-(-1) \\\\ 0-0 \\\\ 1-3 \\end{pmatrix}\\right|$ $=\\left| \\begin{pmatrix} 0 \\\\ 0 \\\\ -2 \\end{pmatrix}\\right|$ $=\\sqrt{(-2)^2}$ $=2$ The distance from the line $g$ to the point $P$ is 2 LU.", "shown below) $$r·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$(3×1)+(2×[-5])+(7×8)=49$$ Therefore this means that the distanve of the vector d = 49 To complete the equation we need to write down the value of the whole equation in cartesian form $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$x-5y+8z=49$$ x - 5y + 8z - 49 = 0 #### Question 2 Write the equation of the plane through (2,1,3), given that the vector (4,5,6) is perpendicular $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 1\\\\\\ 3\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$(2×4)+(1×5)+(3×6)=31$$ Therefore the distance of the vector is d = 31 $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$4x+5y+6z=31$$ 4x + 5y + 6z - 31 = 0 #### Intersection of a line and a plane Find the point of intersection between the line The normal (perpendicular) to a plane is $$r=\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ with plane $$5x+y-z=1$$ First we need to replace r with the matrix (x,y,z) in order to find the values of x,y and z $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ We can change this equation into 3 different equations by seperating the matrices into: x = 2 + λ y = 3 + 2λ z = 4 - λ And we know that 5x+y-z=1 Replacing the values of x,y and z with these new values we get", "E}$${\\ displaystyle {\\ vec {a}} = {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}}}$ ${\\ displaystyle {\\ vec {OF}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {\\ left [{\\ begin {pmatrix} 8 \\\\ 8 \\ \\ -1 \\ end {pmatrix}} - {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}} \\ right] \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}} {{\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {27} {9}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}}}$. The distance of the point from the plane thus satisfies ${\\ displaystyle d (P; E)}$${\\ displaystyle P}$${\\ displaystyle E}$ ${\\ displaystyle d (P; E) = d (P; F) \\; {\\ widehat {=}} \\; | {\\ vec {PF}} | = \\ left | {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}} -", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "The orthogonal projection $\\mbox{proj}_{M}({\\vec{v}\\,})$ of a vector is its projection onto $M$ along $M^\\perp$. Example 3.5 In $\\mathbb{R}^3$, to find the orthogonal complement of the plane $P=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, 3x+2y-z=0 \\}$ we start with a basis for $P$. $B=\\langle \\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\rangle$ Any $\\vec{v}$ perpendicular to every vector in $B$ is perpendicular to every vector in the span of $B$ (the proof of this assertion is Problem 10). Therefore, the subspace $P^\\perp$ consists of the vectors that satisfy these two conditions. $\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0 \\qquad \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0$ We can express those conditions more compactly as a linear system. $P^\\perp=\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\begin{pmatrix} 1 &0 &3 \\\\ 0 &1 &2 \\end{pmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix} =\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix} \\}$ We are thus left with finding the nullspace of the map represented by the matrix, that is, with calculating the solution set of a homogeneous linear system. $P^\\perp =\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\;\\begin{array}{*{3}{rc}r} v_1 & & &+ &3v_3 &= &0 \\\\ & &v_2 &+ &2v_3 &= &0 \\end{array}\\; \\} =\\{k\\begin{pmatrix} -3", "point, there is exactly one. I assume you want the equation for that line. Since the line passes through both the given point and the given line, a vector in the direction of this perpendicular line would be the vector $\\begin{pmatrix}3-(-1) \\\\ 1-(-2+t) \\\\ -2-(-1+t)\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 3-t \\\\ -1-t\\end{pmatrix}$. Let's find a vector in the direction of the given line by finding two points along the line (we can choose any two points, so let's say $t=0, t=1$). So, a vector in the direction of the given line is $\\begin{pmatrix}-1-(-1) \\\\ -2+1 - (-2+0) \\\\ -1+1 - (-1+0)\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Since we want these vectors to be perpendicular, we want the dot product to be zero. Hence: $(4\\cdot 0) + (1\\cdot (3-t)) + (1\\cdot (-1-t)) = 2-2t = 0$ implies $t=1$. Hence, a vector in the direction of the perpendicular line is $\\begin{pmatrix}4 \\\\ 3-1 \\\\ -1-1\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 2 \\\\ -2\\end{pmatrix}$. So, the perpendicular line would have parametric equation $x=3+4t, y=1+2t, z=-2-2t$. 4. ## Re: equation of line through point orthogonal to line so the point (does this mean you are treating the point as a bounded vector) minus the line gives a direction vector along the perpendicular line then I take the given direction vector of the line which", "the latter is parallel to the $y$-axis ($y$ can change independently of $z$ and $x$). The line itself is the intersection of these two planes, and it can be projected along the planes indicated by the individual equations in (2) onto the $yz$-plane and the $zx$-plane, respectively. In a narrower meaning, the equation $y=mz\\!+\\!p$ represents the projection line of $l$ onto the $yz$-plane and the equation $x=nz\\!+\\!q$ the projection line of $l$ onto the $zx$-plane. A line in space can be represented also without using planes. We can derive another way by starting with two vectors: one vector $\\vec{r}_{0}$ drawn from the origin to a chosen point of the line (the so-called position vector of this point) and another vector (the so-called direction vector) $\\vec{u}$ determining the direction of the line. Then there exists a real number $t$ such that the position vector $\\vec{r}$ of an arbitrary point of the line can be expressed as $\\displaystyle\\vec{r}\\;=\\;\\vec{r}_{0}\\!+\\!t\\vec{u}.$ (3) This is the vector-formed equation of the line. Since every vector can be given by three coordinates of the end-point, we may write $\\vec{r}\\;=\\;\\begin{pmatrix}x\\\\ y\\\\ z\\end{pmatrix},\\quad\\vec{r}_{0}\\;=\\;\\begin{pmatrix}x_{0}\\\\ y_{0}\\\\ z_{0}\\end{pmatrix},\\quad\\vec{u}\\;=\\;\\begin{pmatrix}a\\\\ b\\\\ c\\end{pmatrix}\\!.$ $\\begin{pmatrix}x\\\\ y\\\\ z\\end{pmatrix}\\;=\\;\\begin{pmatrix}x_{0}\\\\ y_{0}\\\\ z_{0}\\end{pmatrix}+t\\begin{pmatrix}a\\\\ b\\\\ c\\end{pmatrix}\\!,$ and we obtain three scalar equations $\\displaystyle\\begin{cases}x\\;=\\;x_{0}\\!+\\!ta\\\\ y\\;=\\;y_{0}\\!+\\!tb\\\\ z\\;=\\;z_{0}\\!+\\!tc.\\end{cases}$ (4) These are the parametric equations of the line (3), with the parameter $t$ taking", "# Projection of the vector $\\vec{b}=\\left(\\begin{smallmatrix}3\\\\-1\\\\4\\end{smallmatrix}\\right)$ in some given direction Find the projection of the vector $\\vec{b}=\\begin{pmatrix} 3 \\\\ -1\\\\ 4 \\end{pmatrix}$ in the direction $(a)$ perpendicular to the plane: $$2x-2y+z=5.$$ $(b)$ parallel to the plane: $$2x-2y+z=5.$$ For the part $(a)$ projecting $\\vec{b}=\\begin{pmatrix}3 \\\\ -1\\\\ 4\\end{pmatrix}$ to $\\vec{a}=\\begin{pmatrix}2 \\\\ -2\\\\ 1\\end{pmatrix}$ will give me the perpendicular projection right? For part $(b)$ I know I have to subtract what I found in a from a vector to find the projection of the vector parallel to the plane, but I'm kinda lost. • You are very correct. Just proceed as you've written. Feb 6 '18 at 5:57 • I am lost in getting the vector that I am supposed to subtract the projection from... Feb 6 '18 at 6:03 The plane $E = 2x-2y+z=5$ is defined by the normal vector $\\vec{n}=\\begin{pmatrix}{2 \\\\ -2\\\\ 1}\\end{pmatrix}$ and the projection of $\\vec{b}$ on $\\vec{n}$, we define it as $\\vec{l}$, is obtained by $\\,|l|\\cdot \\left(\\frac{1}{|n|}\\vec{n}\\right)\\,$ (See unit vector) It follows: $$|l| = |b|\\cdot \\cos(\\alpha) = |b|\\frac{\\vec{b}\\cdot \\vec{n}}{|b|\\cdot |n|} = \\frac{\\vec{b}\\cdot \\vec{n}}{|n|} = 4,\\quad\\text{ and }\\quad \\frac{1}{|n|}\\vec{n} = \\frac{1}{3}\\cdot\\begin{pmatrix}{2 \\\\ -2\\\\ 1}\\end{pmatrix}$$ $$\\vec{l} = |l|\\cdot \\left(\\frac{1}{|n|}\\cdot\\vec{n}\\right) = 4\\cdot\\frac{1}{3}\\cdot \\begin{pmatrix}{2 \\\\ -2\\\\ 1}\\end{pmatrix} = \\begin{pmatrix}{\\frac{8}{3} \\\\ -\\frac{8}{3}\\\\ \\frac{4}{3}}\\end{pmatrix}$$ $\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad\\quad$ And the vector $\\vec{t}$ ist parallel to the plane $E$. This is also the projection of $\\vec{b}$", "Linear Algebra/Orthogonal Projection Onto a Line/Solutions Solutions This exercise is recommended for all readers. Problem 1 Project the first vector orthogonally onto the line spanned by the second vector. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}$ 4. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}$ Each is a straightforward application of the formula from Definition 1.1. 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\-2\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}={\\frac {4}{13}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}={\\begin{pmatrix}12/13\\\\-8/13\\end{pmatrix}}}$ 2. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\0\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}={\\frac {2}{3}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}={\\begin{pmatrix}2\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}={\\frac {-1}{6}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}={\\begin{pmatrix}-1/6\\\\-1/3\\\\1/6\\end{pmatrix}}}$ 4. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}={\\frac {1}{3}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}={\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$ This exercise is recommended for all readers. Problem 2 Project the vector orthogonally onto the line. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}},\\{c{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ 2. ${\\displaystyle {\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}}$, the line ${\\displaystyle y=3x}$ 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}}{{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}={\\frac {-19}{19}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}={\\begin{pmatrix}3\\\\-1\\\\3\\end{pmatrix}}}$ 2. Writing the line as ${\\displaystyle \\{c\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ gives this projection. ${\\displaystyle {\\frac {{\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\frac {-4}{10}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\begin{pmatrix}-2/5\\\\-6/5\\end{pmatrix}}}$ Problem 3 Although the development of Definition 1.1 is guided by the pictures, we are not restricted to spaces that we can draw. In ${\\displaystyle \\mathbb {R} ^{4}}$ project this vector onto this line. ${\\displaystyle {\\vec {v}}={\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\qquad \\ell =\\{c\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}}{{\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}={\\frac {3}{4}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}={\\begin{pmatrix}-3/4\\\\3/4\\\\-3/4\\\\3/4\\end{pmatrix}}}$ This exercise", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "Since the distance can only be calculated if there is parallelism, you have to check whether the planes are parallel. We look to see whether the normal vectors are parallel. In contrast to the line, there is no check for perpendicularity. $\\vec{n_1}=r\\cdot\\vec{n_2}$ $\\begin{pmatrix} -4 \\\\ 4 \\\\ -8 \\end{pmatrix}=r\\cdot\\begin{pmatrix}2 \\\\ -2 \\\\ 4 \\end{pmatrix}$ $\\Rightarrow r=-2$ There is an $r$: The vectors are multiples of one another and therefore parallel. 2. #### Select point You can take any point on the plane. However, since you can easily read off the support point, this is a good option. $P(1|2|1)$ 3. #### Set up Hesse normal form $|\\vec{n}|=\\sqrt{2^2+(-2)^2+4^2}$ $=\\sqrt{24}$ $\\vec{n_0}= \\frac{\\vec{n}}{|\\vec{n}|}$ $=\\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix}$ $\\text{E: } \\left(\\vec{x} - \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}\\right) \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix}=0$ 4. #### Insert point $\\vec{p}=\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix}$ $d=$ $\\left|\\left(\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} - \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}\\right) \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix} \\right|$ $=\\left|\\begin{pmatrix} -1 \\\\ 1 \\\\ 0 \\end{pmatrix} \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix} \\right|$ $=|-\\frac4{\\sqrt{24}}|$ $\\approx0.82$", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "is: $$cosθ = \\frac{10}{6\\sqrt{26}}$$ $$θ = cos^{-1}(\\frac{10}{6\\sqrt{26}}) = 71.44°$$ #### 3-D Vector Equation of a line $$\\underline{r} = \\begin{pmatrix}3\\\\\\ 4\\\\\\ 1\\end{pmatrix} + λ\\begin{pmatrix}2\\\\\\ 1\\\\\\ -1\\end{pmatrix}$$ has position $$\\begin{pmatrix}3\\\\\\ 4\\\\\\ 1\\end{pmatrix}$$ and direction $$\\begin{pmatrix}2\\\\\\ 1\\\\\\ -1\\end{pmatrix}$$ Find the equation of the line In order to find the equation of the line we need to find the values of x,y and z in terms of λ and then we can set them equal to eachother To do this we do the same as we did with 2-D vectors and take the x,y and z terms from the equation seperately like so: First we find the equation for x: $$x=3+2λ$$ $$x-3=2λ$$ $$λ=\\frac{x-3}{2}$$ Now we find the equation for y: $$y=4+λ$$ $$λ=\\frac{y-4}{1}$$ Finally we find the equation for z: $$z=1-λ$$ $$λ=\\frac{z-1}{-1}$$ Setting all these equations as equal to eachother gives us the vector equation of the line: $$\\frac{x-3}{2}=\\frac{y-4}{1}=\\frac{z-1}{-1}$$ This can be re-written as: $$\\frac{x-3}{2}=y-4=-z+1$$", "}$-vector space. In addition, the map is well-defined. Let ${\\displaystyle (x_{1},x_{2})^{T}}$ and ${\\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have: {\\displaystyle {\\begin{aligned}f\\left({\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\\\(x_{1}+y_{1})-5\\cdot (x_{2}+y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{separate vectors}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}x_{1}+x_{2}\\\\x_{1}-5x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}+y_{2}\\\\y_{1}-5y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+f{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\end{aligned}}} Proof step: homogeneity Let ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle (x_{1},x_{2})^{T}\\in \\mathbb {R} ^{2}}$. Then: {\\displaystyle {\\begin{aligned}f\\left(\\lambda \\cdot {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}\\lambda \\cdot x_{1}\\\\\\lambda \\cdot x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda x_{1}+\\lambda x_{2}\\\\\\lambda x_{1}-5\\lambda x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda (x_{1}+x_{2})\\\\\\lambda (x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{scalar multiplication}}\\right.}\\\\[0.3em]&=\\lambda \\cdot {\\begin{pmatrix}(x_{1}+x_{2})\\\\(x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=\\lambda \\cdot f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\\\[0.3em]\\end{aligned}}} Thus the map is linear. ## A linear map in the vector space of sequences Next, we consider the space of all sequences of real numbers. This space is infinite-dimensional, because there are not finitely many sequences generating this sequence space. But it is a vector space, as we have shown in the chapter about sequence spaces. Exercise (Sequence space) Let ${\\displaystyle V}$ be the ${\\displaystyle \\mathbb {R} }$-vector space of all real-valued sequences. Show that the map {\\displaystyle {\\begin{aligned}f:V&\\to V\\\\(a_{0},a_{1},a_{2},\\ldots )&\\mapsto (a_{1},a_{2},a_{3},\\ldots )\\end{aligned}}} is linear. How to get to the proof? (Sequence space) To show linearity, two properties need to be checked: 1. ${\\displaystyle f}$ is additive: ${\\displaystyle f(v+w)=f(v)+f(w)}$ for all ${\\displaystyle v,w\\in V}$ 2. ${\\displaystyle", "# Compute the orthogonal projection $w$ of $u$ on a line $L$ A vector $u$ and a line $L$ in $\\mathbb{R}^2$ are given. Compute the orthogonal projection $w$ of $u$ on $L$, and use it to compute the distance $d$ from the endpoint of $u$ to $L$ $u = \\begin{pmatrix} 5 \\\\ 0 \\end{pmatrix}$ $y = 0$ I know to that $w = \\frac{u \\cdot v}{||v||^2}v$ and that $d = u - w$ The answer in the book says the answer is $w = \\begin{pmatrix}5 \\\\0 \\end{pmatrix}$ and $d = 0$ How do I find $v$ to solve the problem? • What is your line $L$? Is it the line given by the equation $y = 0$? – Eli Rose Dec 14 '15 at 4:16 • Yes, the book only gives $u$ and the equation of Line $L$ – user273323 Dec 14 '15 at 4:25 The vector $v$ can be any vector in the direction of the line $L$. For a line $ax + by = c$, the vector $\\begin{pmatrix}b\\\\-a\\end{pmatrix}$, or any non-zero multiple of it, lies in the direction of the line. So for $L: y=0$, alias $0x+1y=0$, the vector $v = \\begin{pmatrix}1\\\\0\\end{pmatrix}$ will do. Put this into your equation gives $w=\\begin{pmatrix}5\\\\0\\end{pmatrix}$ as expected. Or, more simply, you can observe that $u$ is a vector parallel to", "\\\\ 3 \\end{pmatrix} } \\\\ \\text{Let line \\vec{x} = \\begin{pmatrix} 6 \\\\ 7 \\\\ 7 \\end{pmatrix} + t \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} } \\\\ \\text{Let direction vector \\vec{r} = \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} }$$ $$\\begin{array}{|rcll|} \\hline \\left( \\vec{x}-\\vec{p} \\right) \\cdot \\vec{r} &=& 0 \\quad | \\quad (\\vec{x}-\\vec{p}) \\perp \\vec{r} \\\\\\\\ \\Bigg(\\begin{pmatrix} 6 \\\\ 7 \\\\ 7 \\end{pmatrix} + t \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix}-\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix} \\Bigg) \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} &=& 0 \\\\\\\\ \\Bigg(\\begin{pmatrix} 6-1 \\\\ 7-2 \\\\ 7-3 \\end{pmatrix} + t \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} \\Bigg) \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} &=& 0 \\\\\\\\ \\Bigg(\\begin{pmatrix} 5 \\\\ 5 \\\\ 4 \\end{pmatrix} + t \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} \\Bigg) \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} &=& 0 \\\\\\\\ \\begin{pmatrix} 5+3t \\\\ 5+2t \\\\ 4-2t \\end{pmatrix} \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix} &=& 0 \\\\\\\\ 3(5+3t)+2(5+2t)-2(4-2t)&=& 0 \\\\ 15+9t+10+4t-8+4t &=& 0 \\\\ 17+17t &=& 0 \\\\ 17t &=& -17 \\quad | \\quad : 17 \\\\ \\mathbf{ t } &=& \\mathbf{ -1 } \\\\ \\hline \\end{array}$$ $$\\text{foot (of a perpendicular):}$$ $$\\begin{array}{|rcll|} \\hline \\vec{x}_{f} &=& \\begin{pmatrix} 6 \\\\ 7 \\\\ 7 \\end{pmatrix} + (-1) \\begin{pmatrix} 3 \\\\ 2 \\\\ -2 \\end{pmatrix}" ]

[ "E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP", "## anonymous one year ago In square $ABCD$, $E$ is the midpoint of $\\overline{BC}$, and $F$ is the midpoint of $\\overline{CD}$. Let $G$ be the intersection of $\\overline{AE}$ and $\\overline{BF}$. Prove that $DG = AB$. • This Question is Open 1. anonymous Asymptote code below [asy] unitsize(2.5 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (0,1); C = (1,1); D = (1,0); E = (B + C)/2; F = (C + D)/2; G = extension(A,E,B,F); draw(A--B--C--D--cycle); draw(A--E); draw(B--F); draw(D--G); label(\"$A$\", A, SW); label(\"$B$\", B, NW); label(\"$C$\", C, NE); label(\"$D$\", D, SE); label(\"$E$\", E, N); label(\"$F$\", F, dir(0)); label(\"$G$\", G, WSW); [/asy]", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "+0 # HELP 0 194 1 In the figure below, $D$ is a point on segment $\\overline{CE}$ such that $\\overline{AD}\\parallel\\overline{BE}$ and $A$ is not on $\\overline{BC}.$ Line segments $\\overline{AD}$ and $\\overline{BC}$ intersect at $P.$ [asy] size(4cm); pair C = (0,0); pair D = (4,0); pair E = (9,0); pair F = (0.25,4); pair G = (1,4); pair A = extension(C,F,D,G); pair H = G+E-D; pair B = extension(C,G,E,H); pair P = extension(A,D,B,C); draw(A--C--E--B--C); draw(A--D); label(\"$A$\",A,NNW); label(\"$B$\",B,N); label(\"$C$\",C,SSW); label(\"$D$\",D,S); label(\"$E$\",E,SSE); label(\"$P$\",P,dir(0)); [/asy] Can $\\angle CAD=\\angle CBE?$ Explain. Aug 20, 2020", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "dot(A); label(\"A\",A,NE); dot(B); label(\"B\",B,dir(180)); dot(C); label(\"C\",C,SE); dot(D); label(\"D\",D,dir(0)); dot(E); label(\"E\",E,SE); dot(F); label(\"F\",F,SW); [/asy]$ Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper. As triangle $ABC$ is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles. As point $D$ splits line segment $\\overline{AC}$ in a $1:2$ ratio, we draw $\\overline{AC}$ as a vertical line segment $3$ units long. Point $D$ is thus $1$ unit below point $A$ and $2$ units above point $C$. By definition, Point $E$ splits line segment $\\overline{BD}$ in a $1:1$ ratio, so we draw $\\overline{BD}$ $2$ units long directly left of $D$ and draw $E$ directly between $B$ and $D$, $1$ unit away from both. We then draw line segments $\\overline{AB}$ and $\\overline{BC}$. We can easily tell that triangle $ABC$ occupies $3$ square units of space. Constructing line $AE$ and drawing $F$ at the intersection of $AE$ and $BC$, we can easily see that triangle $EBF$ forms a right triangle occupying $\\frac{1}{4}$ of a square unit of space. The ratio of the areas of triangle $EBF$ and triangle $ABC$ is thus", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "line $PD.$ $[asy] import geometry; import olympiad; size(400); point A = (2, 5), B = (0, 0), C = (10, 0), P = (3, 2); line c = line(A, B); line a = line(C, B); line b = line(A, C); point D = projection(a) * P; draw(P -- D); point e = projection(b) * P; draw(P -- e); point F = projection(c) * P; draw(P -- F); draw(A--B--C--A); draw(P--A); draw(P--B); draw(P--C); markrightangle(B, D, P); markrightangle(A, e, P); markrightangle(A, F, P); draw(circumcircle(A, P, e), dashed); line d = line(P, D); draw(d); point n = projection(d) * F; point M = projection(d) * e; draw(F--n); draw(e--M); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, SW); label(\"E\", e - (0.1, 0), NE); label(\"F\", F, W); label(\"P\", P+(0.2, 0), S); label(\"N\", n, E); label(\"M\", M, W); [/asy]$ Note that $AFPE$ is cyclic with diameter $AP.$ By ELOS, $\\dfrac{EF}{\\sin A} = 2R=AP\\implies EF = AP\\sin A.$ Since $BDPF$ is cyclic, we have that $B$ and $\\angle FPD$ are supplementary. Since $MPD$ is a line, $B = \\angle FPM.$ This means that $\\sin B = \\sin FPN = \\dfrac{FN}{FP}\\implies FN = PF\\sin B.$ Similarly, $EM = PE\\sin C.$ So the problem is reduced to proving that $EF\\ge FN+EM$ but this is obvious by the Pythagorean Theorem. $\\blacksquare$ Now the rest of", "# 2018 AIME II Problems/Problem 14 ## Problem The incircle $\\omega$ of triangle $ABC$ is tangent to $\\overline{BC}$ at $X$. Let $Y \\neq X$ be the other intersection of $\\overline{AX}$ with $\\omega$. Points $P$ and $Q$ lie on $\\overline{AB}$ and $\\overline{AC}$, respectively, so that $\\overline{PQ}$ is tangent to $\\omega$ at $Y$. Assume that $AP = 3$, $PB = 4$, $AC = 8$, and $AQ = \\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Diagram $[asy] size(200); import olympiad; defaultpen(linewidth(1)+fontsize(12)); pair A,B,C,P,Q,Wp,X,Y,Z; B=origin; C=(6.75,0); A=IP(CR(B,7),CR(C,8)); path c=incircle(A,B,C); Wp=IP(c,A--C); Z=IP(c,A--B); X=IP(c,B--C); Y=IP(c,A--X); pair I=incenter(A,B,C); P=extension(A,B,Y,Y+dir(90)*(Y-I)); Q=extension(A,C,P,Y); draw(A--B--C--cycle, black+1); draw(c^^A--X^^P--Q); pen p=4+black; dot(\"A\",A,N,p); dot(\"B\",B,SW,p); dot(\"C\",C,SE,p); dot(\"X\",X,S,p); dot(\"Y\",Y,dir(55),p); dot(\"W\",Wp,E,p); dot(\"Z\",Z,W,p); dot(\"P\",P,W,p); dot(\"Q\",Q,E,p); MA(\"\\beta\",C,X,A,0.3,black); MA(\"\\alpha\",B,A,X,0.7,black); [/asy]$ ## Solution 1 Let the sides $\\overline{AB}$ and $\\overline{AC}$ be tangent to $\\omega$ at $Z$ and $W$, respectively. Let $\\alpha = \\angle BAX$ and $\\beta = \\angle AXC$. Because $\\overline{PQ}$ and $\\overline{BC}$ are both tangent to $\\omega$ and $\\angle YXC$ and $\\angle QYX$ subtend the same arc of $\\omega$, it follows that $\\angle AYP = \\angle QYX = \\angle YXC = \\beta$. By equal tangents, $PZ = PY$. Applying the Law of Sines to $\\triangle APY$ yields $$\\frac{AZ}{AP} = 1 + \\frac{ZP}{AP} = 1 + \\frac{PY}{AP} = 1 + \\frac{\\sin\\alpha}{\\sin\\beta}.$$Similarly, applying the Law of Sines to $\\triangle ABX$ gives $$\\frac{AZ}{AB} = 1", "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "[] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); label(\"F\",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]$ Drawing line $\\overline{BD}$ and parallel line $\\overline{CF}$, we see that $\\triangle FCE \\sim \\triangle BDE$ by AA similarity. Thus $\\frac{FE}{EB} = \\frac{FC}{DB} = \\frac{2}{4} = \\frac{1}{2}$. Reciprocating, we know that $\\frac{EB}{FE} = 2$ so $\\frac{EB+FE}{FE} = 2+1 \\Rightarrow \\frac{FB}{FE} = 3$. Reciprocating again, we have $\\frac{FE}{FB} = \\frac{1}{3} \\Rightarrow FE = \\frac{1}{3}FB$. We know that $FD = 2$, so by the Pythagorean Theorem, $FB = \\sqrt{2^{2} + 4^{2}} = 2\\sqrt{5}$. Thus $FE = \\frac{1}{3}FB = \\frac{2\\sqrt{5}}{3}$. Applying the Pythagorean Theorem again, we have $AF = \\sqrt{1^{2}+2^{2}} = \\sqrt{5}$. We finally have $AE = AF + FE = \\sqrt{5} + \\frac{2\\sqrt{5}}{3} = \\frac{5\\sqrt{5}}{3} \\Rightarrow \\boxed{\\text{B}}$", "pair R = (1.5*A + 3.5*C)/5, S = extension(A0,R,A,B), T = extension(C0,R,B,C); pair P2 = extension(A0,T,B0,R), M = extension(A,B,A0,C), N = extension(A,C0,B,C); /* Draw a couple of the common paths */ D(A--B--C--cycle); D(incircle(A,B,C)); /* Label some of the common points */ D(MP(\"A\", A, NE)); D(MP(\"B\", B)); D(MP(\"C\", C)); D(MP(\"I\", I, W)); Out[2]: 1) (a) Extend $A_{1}X$ to intersect $AC$ at $V$. Show that the lines $AA_{1}$, $BV$, and $DE$ are concurrent. (b) Let $A_{2}$ be the intersection point of $AX$ with the sideline $BC$. Show that $$\\frac{A_2B}{A_2C} = \\frac{s-c}{s-b} \\cdot \\frac{A_1B^2}{A_1C^2},$$ where $s = \\frac{a + b + c}{2}$ is the semiperimeter of $\\triangle ABC$. [Hint: Menelaus's Theorem may come in handy.] (c) Prove that lines $AX, BY, CZ$ are concurrent. In [3]: %%asy pumac_power_round_2011_config.asy size(500); /* Draw lines */ D(A--A1--X1, darkgreen); D(B--B1--Y1, darkgreen); D(C--C1--Z1, darkgreen); D(A--P1--B, darkred); D(C--Z1, darkred); D(X1--V); D(P1--A2, darkred); /* Label points */ D(MP(\"P\",P,dir(60))); D(MP(\"D\",D)); D(MP(\"E\",E,NE)); D(MP(\"F\",F,NW)); D(MP(\"X\",X1,ENE)); D(MP(\"Y\",Y1,1.4*dir(150))); D(MP(\"Z\",Z1,NE)); D(P1); D(MP(\"A_1\",A1)); D(MP(\"B_1\",B1,NE)); D(MP(\"C_1\",C1,NW)); D(MP(\"V\",V,NE)); D(MP(\"A_2\",A2)); Out[3]: 2) Show that the lines $B_1C_1$, $EF$, $YZ$ concur at a point, say $A_{0}$. Similarly, the lines $C_1A_1$, $FD$, $ZX$ and $A_1B_1$, $DE$, $XY$ concur at points $B_{0}$ and $C_{0}$, respectively. [Hint: Pascal's theorem might be useful here.] In [4]: %%asy pumac_power_round_2011_config.asy size(600); /* Draw lines */ D(Y1--A0--E); D(F--D); D(A1--C1); D(B1--C0--E); D(D--E--F--cycle,rgb(0,0.7,0)+linewidth(1)); D(A0--B0--C0--cycle, darkblue); D(A--B0--C, darkblue+linetype(\"3 3\"));", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "# BS in relation to BF and CS in relation to CE #### mathmari ##### Well-known member MHB Site Helper Hey!! We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? #### Klaas van Aarsen ##### MHB Seeker Staff member We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? Hey mathmari !! Let's first make a drawing: \\begin{tikzpicture}[scale=2] %preamble \\usetikzlibrary {calc} \\coordinate[label=left:A] (A) at (0,0); \\coordinate[label=right:B] (B) at (0:5); \\coordinate[label=C] (C) at (70:4); \\coordinate[label=below:E] (E) at (0:1); \\coordinate[label=left:F] (F) at (70:3); \\coordinate[label=right:S] (S) at (intersection cs:first line={(B)--(F)}, second line={(C)--(E)}); \\draw (A) -- (B) -- (C) -- cycle; \\draw (C) -- (E); \\draw (B) -- (F); \\draw[-latex, ultra thick, blue] (A) -- node[ above ] {$\\mathbf{\\vec c}$} (B); \\draw[-latex, ultra thick, blue] (A) -- node[ right] {$\\mathbf{\\vec b}$} (C); \\path (A) --" ]

[ "$dy = a\\,dx$ $\\frac{dy}{dx} = a$ $dy = \\mathbf{a}\\,d\\mathbf{x}$ $\\frac{dy}{d\\mathbf{x}} = \\mathbf{a}$ $dy = {\\rm tr}(\\mathbf{A}\\,d\\mathbf{X})$ $\\frac{dy}{d\\mathbf{X}} = \\mathbf{A}$ $d\\mathbf{y} = \\mathbf{a}\\,dx$ $\\frac{d\\mathbf{y}}{dx} = \\mathbf{a}$ $d\\mathbf{y} = \\mathbf{A}\\,d\\mathbf{x}$ $\\frac{d\\mathbf{y}}{d\\mathbf{x}} = \\mathbf{A}$ $d\\mathbf{Y} = \\mathbf{A}\\,dx$ $\\frac{d\\mathbf{Y}}{dx} = \\mathbf{A}$ 4. ^ a b c Here, $\\mathbf{0}$ refers to a column vector of all 0's, of size n, where n is the length of x. 5. ^ a b Here, $\\mathbf{0}$ refers to a matrix of all 0's, of the same shape as X. 6. Petersen, Kaare Brandt and Michael Syskind Pedersen. The Matrix Cookbook. November 14, 2008. http://matrixcookbook.com. [3] This book uses a mixed layout, i.e. by Y in $\\frac{\\partial \\mathbf{Y}}{\\partial x},$ by X in $\\frac{\\partial y}{\\partial \\mathbf{X}}.$ $\\frac{d\\,\\ln au}{dx} = \\frac{1}{au}\\frac{d(au)}{dx} = \\frac{1}{au}a\\frac{du}{dx} = \\frac{1}{u}\\frac{du}{dx} = \\frac{d\\,\\ln u}{dx}.$", "$\\mathbf{h}$ is differentiable at $\\mathbf{a}$ and: (7) \\begin{align} \\quad [\\mathbf{D} \\mathbf{h} (\\mathbf{a})] = [\\mathbf{D} \\mathbf{f} (\\mathbf{b})][\\mathbf{D} \\mathbf{g}(\\mathbf{a})] = [\\mathbf{D} \\mathbf{f} (\\mathbf{g}(\\mathbf{a}))][\\mathbf{D} \\mathbf{g}(\\mathbf{a})] \\end{align} • Furthermore, if $\\mathbf{g} : S \\to \\mathbb{R}^m$, $\\mathbf{f} : R(\\mathbf{g}) \\to \\mathbb{R}^m$ such that: (8) \\begin{align} \\quad (x_1, x_2, ..., x_n) \\to_{\\mathbf{g}} (y_1, y_2, ..., y_m) \\to_{\\mathbf{f}} (z_1, z_2, ..., z_p) \\end{align} • Then for all $k \\in \\{ 1, 2, ..., p \\}$ and for all $j \\in \\{ 1, 2, ..., n \\}$ we have that: (9) \\begin{align} \\quad \\frac{\\partial z_k}{\\partial x_j} = \\sum_{i=1}^{m} \\frac{\\partial z_k}{\\partial y_i} \\frac{\\partial y_i}{\\partial x_j} \\end{align}", "## Parallel Lines Vector Question In the diagram P splits AB in the ratio 1:2. BQ=k(BC) and PQD is a straight line. If $\\mathbf{AB}=\\mathbf{a}, \\; \\mathbf{AC}=\\mathbf{c}, \\; \\mathbf{CD}=3\\mathbf{a}$ find the value of $k$ . Since P splits AB in the ratio 1:2, $\\mathbf{AP}=\\frac{1}{3} \\mathbf{a}, \\; \\mathbf{PB}=\\frac{2}{3} \\mathbf{a}$ . $\\mathbf{BC}=- \\mathbf{a}+ \\mathbf{b}$ so $\\mathbf{BQ}=k(- \\mathbf{a}+ \\mathbf{b})$ . Then $\\mathbf{PQ}= \\mathbf{PB}+ \\mathbf{BQ} =\\frac{2}{3} \\mathbf{a}+k(- \\mathbf{a}+ \\mathbf{b})=(\\frac{2}{3}-k) \\mathbf{a} + k \\mathbf{b}$ $\\mathbf{PD}= - \\frac{1}{3} \\mathbf{a} ++4 \\mathbf{b}$ . $PQD$ is a straight line so $\\mathbf{PQ}, \\; \\mathbf{PD}$ are parallel and the coefficients of $\\mathbf{a}, \\; \\mathbf{b}$ are in the same ratio. Hence $\\frac{2/3-k}{-1/3}= \\frac{k}{4} \\rightarrow\\frac{8}{3}-4k=- \\frac{k}{3} \\rightarrow \\frac{8}{3}= \\frac{11k}{3} \\rightarrow k =\\frac{\\frac{8}{3}}{\\frac{11}{3}}=\\frac{8}{11}$", "so $c = \\frac{7}{9}d$. You can see $b + 2c = d$ $b + 2\\left(\\frac{7}{9}d\\right) = d$ $b + \\frac{14}{9}d = d$ $b = -\\frac{5}{9}d$. You can see $a - b = d$ $a - \\left(-\\frac{5}{9}d\\right) = d$ $a + \\frac{5}{9}d = d$ $a = \\frac{4}{9}d$. $d$ is a free variable, you can let it be whatever you like to get a consistent solution set. So why not let $d = 9$. Therefore a correct equation of the plane is $4x - 5y + 7z = 9$. The vectors normal to the plane has the same coefficients as the coefficients of the plane itself. Therefore a vector normal to the plane is $\\mathbf{n} = 4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}$. To find the unit vector, divide by its length. $\\frac{\\mathbf{n}}{|\\mathbf{n}|} = \\frac{4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}}{\\sqrt{4^2 + (-5)^2 + 7^2}}$ $= \\frac{4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}}{\\sqrt{16 + 25 + 49}}$ $= \\frac{4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}}{\\sqrt{90}}$ $= \\frac{4\\mathbf{i} - 5\\mathbf{j} + 7\\mathbf{k}}{3\\sqrt{10}}$ $= \\frac{4\\sqrt{10}}{30}\\mathbf{i} - \\frac{5\\sqrt{10}}{30}\\mathbf{j} + \\frac{7\\sqrt{10}}{30}\\mathbf{k}$ $= \\frac{2\\sqrt{10}}{15}\\mathbf{i} - \\frac{\\sqrt{10}}{6}\\mathbf{j} + \\frac{7\\sqrt{10}}{30}\\mathbf{k}$. • May 15th 2010, 01:00 AM Sudharaka Quote: Originally Posted by kandyfloss I'm sorry if i'm posting this in the wrong section,i couldn't figure out which section vectors would go in to. Here is the question :- A,B and C are the point (0,1,2),(3,2,1) and", "\\begin{align*}\\mathbf{r_{AF}},\\ \\lambda = \\frac{1}{3}\\end{align*}. \\begin{align*}\\ \\ \\ \\overrightarrow{OQ} = \\mathbf{a} + \\frac{1}{3}(\\mathbf{b} + \\mathbf{c} - 2\\mathbf{a}) = \\frac{1}{3}(\\mathbf{a} + \\mathbf{b} + \\mathbf{c}) = \\frac{1}{3}\\begin{bmatrix} p_1+p_2+p_3 \\\\ q_1+q_2+q_3 \\end{bmatrix}\\end{align*} \\begin{align*}\\therefore Q\\left(\\frac{p_1+p_2+p_3}{3},\\frac{q_1+q_2+q_3}{3}\\right)\\end{align*} We have found the point of intersection we now need to show that this lies on the line connecting C to the midpoint of AB, this is the vector equation $CD$. \\begin{align*}\\mathbf{r_{CD}} = \\mathbf{c} + \\omega (\\mathbf{a + b - 2c})\\end{align*} \\begin{align*}\\ \\ \\overrightarrow{OQ} = \\mathbf{r_{CD}} \\Rightarrow \\frac{1}{3}\\mathbf{a} + \\frac{1}{3}\\mathbf{b} - \\frac{2}{3}\\mathbf{c} = \\omega(\\mathbf{a} + \\mathbf{b} - 2\\mathbf{c})\\end{align*} This is satisfied by $\\omega = \\frac{1}{3}$ $\\therefore\\ Q\\ lies\\ on\\ CD$. Let us think about what we have actually shown here. We have shown that for any triangle $ABC$, if we connect the vertices to the midpoint of the opposite side they all meet at a common point (they are concurrent), this point is called the centroid. This has useful implications to finding the centre of mass of an object. To answer the next part of the question we will also use vectors, this part seems almost tailor made to do so given that we are looking for when two lines are perpendicular. \\begin{align*}\\ \\ \\overrightarrow{OH} = \\mathbf{h} = \\begin{bmatrix} p_1 + p_2 + p_3 \\\\ q_1 + q_2 + q_3 \\end{bmatrix} = \\mathbf{a}+\\mathbf{b}+\\mathbf{c}\\end{align*} \\begin{align*}\\mathbf{r_{AH}} = \\mathbf{a} + \\lambda", "v and w. Sci Advisor PF Gold P: 1,352 Maybe this formulation will be more useful. Set a vector $$\\mathbf{a} = (u,v,w,x,y,z)$$ and a vector function $$\\mathbf{f}(\\mathbf{a},t) = (\\frac{du}{dt},\\frac{dv}{dt},\\frac{dw}{dt},\\frac{dx}{dt},\\frac{dy}{dt}, \\frac{dz}{dt})$$ The 4th-order RK algorithm then reads $$\\mathbf{k}_1 = h \\mathbf{f}(\\mathbf{a}_n, t_n)$$ $$\\mathbf{k}_2 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_1}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_3 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_2}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_4 = h \\mathbf{f}(\\mathbf{a}_n + \\mathbf{k}_3, t_n + h)$$ $$\\mathbf{a}_{n+1} = \\mathbf{a}_n + \\frac{\\mathbf{k}_1}{6} + \\frac{\\mathbf{k}_2}{3} + \\frac{\\mathbf{k}_3}{3} + \\frac{\\mathbf{k}_4}{6}$$ Hope this helps. P: 101 Quote by DrClaude Maybe this formulation will be more useful. Set a vector $$\\mathbf{a} = (u,v,w,x,y,z)$$ and a vector function $$\\mathbf{f}(\\mathbf{a},t) = (\\frac{du}{dt},\\frac{dv}{dt},\\frac{dw}{dt},\\frac{dx}{dt},\\frac{dy}{dt}, \\frac{dz}{dt})$$ The 4th-order RK algorithm then reads $$\\mathbf{k}_1 = h \\mathbf{f}(\\mathbf{a}_n, t_n)$$ $$\\mathbf{k}_2 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_1}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_3 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_2}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_4 = h \\mathbf{f}(\\mathbf{a}_n + \\mathbf{k}_3, t_n + h)$$ $$\\mathbf{a}_{n+1} = \\mathbf{a}_n + \\frac{\\mathbf{k}_1}{6} + \\frac{\\mathbf{k}_2}{3} + \\frac{\\mathbf{k}_3}{3} + \\frac{\\mathbf{k}_4}{6}$$ Hope this helps. So (replacing the 'k's' from the RK4 algorithm with 'm's' and 'n's') For the first step, would this be correct: m1u = h x (1') m1v = h x (2') m1w = h x (3') n1u = h x (4') n1v = h x (5') n1w = h x (6') ? Sci Advisor PF Gold P: 1,352 I'm", "and point $$O$$ is the intersection point of the diagonals. Taking $$\\mathbf{AB} = \\mathbf{a}$$ and $$\\mathbf{AD} = \\mathbf{b}$$ as the basis vectors, find in this basis the coordinates of the vectors $$\\mathbf{BD},$$ $$\\mathbf{CO}$$ and $$\\mathbf{MD}.$$ Solution. Express the vector $$\\mathbf{BD}$$ in terms of vectors $$\\mathbf{AB} = \\mathbf{a}$$ and $$\\mathbf{AD} = \\mathbf{b}.$$ By the vector subtraction rule, $\\mathbf{BD} = \\mathbf{AD} - \\mathbf{AB} = -\\mathbf{a} + \\mathbf{b}, \\Rightarrow \\mathbf{BD} = \\left({-1, 1}\\right).$ Find now the vector $$\\mathbf{CO}.$$ Notice that $\\mathbf{AC} = \\mathbf{AB} + \\mathbf{AD} = \\mathbf{a} + \\mathbf{b}.$ Since $$\\mathbf{CA} = - \\mathbf{AC}$$ and $$\\mathbf{CO} = \\frac{1}{2}\\mathbf{CA},$$ we get $\\mathbf{CO} = \\frac{1}{2}\\mathbf{CA} = -\\frac{1}{2}\\mathbf{AC} = -\\frac{1}{2}\\left({\\mathbf{a} + \\mathbf{b}}\\right) = -\\frac{1}{2}\\mathbf{a} - \\frac{1}{2}\\mathbf{b},$ or $\\mathbf{CO} = \\left({-\\frac{1}{2}, -\\frac{1}{2}}\\right).$ Calculate the vector $$\\mathbf{MD}:$$ $\\mathbf{MD} = \\mathbf{AD} - \\mathbf{AM},$ where $\\mathbf{AM} = \\mathbf{AB} + \\mathbf{BM} = \\mathbf{AB} + \\frac{1}{2}\\mathbf{AD} = \\mathbf{a} + \\frac{1}{2}\\mathbf{b}.$ Hence $\\mathbf{MD} = \\mathbf{AD} - \\mathbf{AM} = \\mathbf{b} - \\left({\\mathbf{a} + \\frac{1}{2}\\mathbf{b}}\\right) = -\\mathbf{a} - \\frac{1}{2}\\mathbf{b}, \\Rightarrow \\mathbf{MD} = \\left({-1, -\\frac{1}{2}}\\right).$ ### Example 4. Find a basis for the set of $$2 \\times 2$$ matrices and determine the coordinates of the matrix $\\left[ \\begin{array}{*{20}{r}} -1 & 5\\\\ 3 & 0 \\end{array} \\right]$ in this basis. Solution. As a basis we can take the following matrices $\\left[ \\begin{array}{*{20}{r}} 1 & 0\\\\ 0 & 0 \\end{array} \\right],\\left[ \\begin{array}{*{20}{r}} 0 & 1\\\\ 0", "this formulation will be more useful. Set a vector $$\\mathbf{a} = (u,v,w,x,y,z)$$ and a vector function $$\\mathbf{f}(\\mathbf{a},t) = (\\frac{du}{dt},\\frac{dv}{dt},\\frac{dw}{dt},\\frac{dx}{dt},\\frac{dy}{dt},\\frac{dz}{dt})$$ The 4th-order RK algorithm then reads $$\\mathbf{k}_1 = h \\mathbf{f}(\\mathbf{a}_n, t_n)$$ $$\\mathbf{k}_2 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_1}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_3 = h \\mathbf{f}(\\mathbf{a}_n + \\frac{\\mathbf{k}_2}{2}, t_n + \\frac{h}{2})$$ $$\\mathbf{k}_4 = h \\mathbf{f}(\\mathbf{a}_n + \\mathbf{k}_3, t_n + h)$$ $$\\mathbf{a}_{n+1} = \\mathbf{a}_n + \\frac{\\mathbf{k}_1}{6} + \\frac{\\mathbf{k}_2}{3} + \\frac{\\mathbf{k}_3}{3} + \\frac{\\mathbf{k}_4}{6}$$ Hope this helps. 8. Feb 14, 2013 ### strokebow Thanks for your help. So (replacing the 'k's' from the RK4 algorithm with 'm's' and 'n's') For the first step, would this be correct: m1u = h x (1') m1v = h x (2') m1w = h x (3') n1u = h x (4') n1v = h x (5') n1w = h x (6') ? 9. Feb 14, 2013 ### Staff: Mentor I'm not sure I follow you. The \"first step\" is the calculation of $\\mathbf{k}_1$. The first element of $\\mathbf{f}(\\mathbf{a}_n,t_n)$ is $$\\left.\\frac{du}{dt}\\right|_{t_n} = K_1 x_n \\cos(t_n) + ( K_2 v_n - w_n )$$ so the first element of $\\mathbf{k}_1$ is $$h \\times \\left[ K_1 x_n \\cos(t_n) + ( K_2 v_n - w_n ) \\right]$$ and so on. For instance, the 4th element of $\\mathbf{k}_1$ is $$u_n$$ Once you have calculated all six elements of $\\mathbf{k}_1$, the \"second step\" is do", "# Solving special (simple?) system of polynomial equations (only up to second degree) I have the following system of equations, \\begin{align} \\mathrm{(I)}&&\\quad \\mathbf{A} f_1(\\mathbf{x})&=\\mathbf{b}_1,\\\\ \\mathrm{(II)}&&\\quad\\mathbf{A} f_2(\\mathbf{x})&=\\mathbf{b}_2,\\\\ \\mathrm{(III)}&&\\quad\\mathbf{A} f_3(\\mathbf{x})&=\\mathbf{b}_3, \\end{align} where I only need to solve any two equations (for the same $\\mathbf{x}$) and $\\mathbf{A} \\in \\mathbb{R}^{9\\times N}$, $\\mathbf{x} \\in \\mathbb{R}^{N}$ , $\\mathbf{b}_i \\in \\mathbb{R}^{9}$ and $N \\in \\mathbb{N}$ (probably necessary to be at least bigger or equal to $18$). $\\mathbf{A} f_i(\\mathbf{x})$ is a matrix-vector product in the following sense: The functions $f_i(x)$ are given by \\begin{align} f_1(x) &= \\frac{1}{2}\\left(x^2-x+\\frac{1}{4}\\right),\\\\ f_2(x) &= -x^2+\\frac{3}{4},\\\\ f_3(x) &= \\frac{1}{2}\\left(x^2+x+\\frac{1}{4}\\right) \\end{align} and act component-wise on vectors, e.g.: $$f_i(\\mathbf{x})=\\begin{pmatrix} f_i(x) \\\\ f_i(y) \\\\ f_i(z) \\end{pmatrix}.$$ The system can therefore also be written as (just to provide some options) \\begin{align} \\mathrm{(I')}&&\\quad \\mathbf{A} (\\mathbf{x}^2-\\mathbf{x}) = \\mathbf{A} \\mathbf{x}^2-\\mathbf{A}\\mathbf{x}&=\\mathbf{b}_1',\\\\ \\mathrm{(II')}&&\\mathbf{A} \\mathbf{x}^2&=\\mathbf{b}_2',\\\\ \\mathrm{(III')}&&\\quad\\mathbf{A} (\\mathbf{x}^2+\\mathbf{x}) = \\mathbf{A} \\mathbf{x}^2+\\mathbf{A}\\mathbf{x}&=\\mathbf{b}_3', \\end{align} where then, e.g., $$\\mathbf{x}^2 = \\begin{pmatrix} x^2 \\\\ y^2 \\\\ z^2 \\end{pmatrix}.$$ I need to solve any two of the equations (I) - (III). I do not care which two exactly. My problem is that the functions $f_i$ are linearly independent, which makes any system of two equations non-linear. I tried the whole day and lots of things. For example \\begin{align} f_1(x) = f_3(x) -x \\Rightarrow \\end{align} \\begin{align} \\mathrm{(I)} &&\\quad\\mathbf{A} f_3(\\mathbf{x}) - \\mathbf{A}\\mathbf{x} &= \\mathbf{b}_1\\\\ \\mathrm{(III)} &&\\quad\\mathbf{A} f_3(\\mathbf{x}) &= \\mathbf{b}_3 \\end{align}", "Dot Product (requires JavaScript) 1. Which of the following expression are meaningful? Which are meaningless? Explain. 1. $\\left(\\mathbf{a}•\\mathbf{b}\\right)•\\mathbf{c}$ 2. $\\left(\\mathbf{a}•\\mathbf{b}\\right)\\mathbf{c}$ 3. $\\left|\\mathbf{a}\\right|\\left(\\mathbf{b}•\\mathbf{c}\\right)$ 4. $\\mathbf{a}•\\left(\\mathbf{b}+\\mathbf{c}\\right)$ 5. $\\mathbf{a}•\\mathbf{b}+\\mathbf{c}$ 6. $\\left|\\mathbf{a}\\right|•\\left(\\mathbf{b}+\\mathbf{c}\\right)$ b, c, and d are meaningful. 2. Find the dot product of two vectors if their lengths are $6$ and $\\frac{1}{3}$ and the angle between them is $\\frac{\\pi }{4}$. 3. Find $\\mathbf{a}•\\mathbf{b}$ if $\\mathbf{a}=⟨4,1,\\frac{1}{4}⟩$, $\\mathbf{b}=⟨6,-3,-8⟩$. $19$ 4. Find $\\mathbf{a}•\\mathbf{b}$ if $\\mathbf{a}=4\\mathbf{j}-3\\mathbf{k}$ and $\\mathbf{b}=2\\mathbf{i}+4\\mathbf{j}+6\\mathbf{k}$. 5. If $\\mathbf{u}$ is a unit vector, find $\\mathbf{u}•\\mathbf{v}$ and $\\mathbf{u}•\\mathbf{w}$. From the figure, $\\left|\\mathbf{u}\\right|=\\left|\\mathbf{w}\\right|=1$, and $\\left|\\mathbf{v}\\right|=\\frac{1}{\\sqrt{2}}$. The angle between $\\mathbf{u}$ and $\\mathbf{w}$ is $\\pi /2$, and the angle between $\\mathbf{u}$ and $\\mathbf{v}$ is $\\pi /4$. So we can write $\\mathbf{u}•\\mathbf{v}=\\left|\\mathbf{u}\\right|\\left|\\mathbf{v}\\right|\\mathrm{cos}\\left(\\pi /4\\right)=\\frac{1}{2}$ and $\\mathbf{u}•\\mathbf{w}=\\left|\\mathbf{u}\\right|\\left|\\mathbf{w}\\right|\\mathrm{cos}\\left(\\pi /2\\right)=0$ 6. Find the angle between vectors $\\mathbf{a}=\\mathbf{j}+\\mathbf{k}$ and $\\mathbf{b}=\\mathbf{i}+2\\mathbf{j}-3\\mathbf{k}$. ${\\mathrm{cos}}^{-1}\\left(\\frac{-1}{2\\sqrt{7}}\\right)\\approx 101°$ 7. Determine whether given vectors are orthogonal, parallel, or neither. 1. $\\mathbf{a}=⟨-5,3,7⟩$, $\\mathbf{b}=⟨6,-8,2⟩$. 2. $\\mathbf{a}=⟨4,6⟩$, $\\mathbf{b}=⟨-3,2⟩$. 3. $\\mathbf{a}=-\\mathbf{i}+2\\mathbf{j}+5\\mathbf{k}$, $\\mathbf{b}=3\\mathbf{i}+4\\mathbf{j}-\\mathbf{k}$. 4. $\\mathbf{a}=2\\mathbf{i}+6\\mathbf{j}-4\\mathbf{k}$, $\\mathbf{b}=-3\\mathbf{i}-9\\mathbf{j}+6\\mathbf{k}$. Neither, orthogonal, orthogonal, parallel. 8. Find two unit vectors that make an angle of $60°$ with $\\mathbf{v}=⟨3,4⟩$. 9. Find the scalar and vector projections of $\\mathbf{b}$ onto $\\mathbf{a}$, where $\\mathbf{a}=\\mathbf{i}+\\mathbf{j}+\\mathbf{k}$ and $\\mathbf{b}=\\mathbf{i}-\\mathbf{j}+\\mathbf{k}$. ${\\mathrm{comp}}_{\\mathbf{a}}\\mathbf{b}=\\frac{1}{\\sqrt{3}}$, ${\\mathrm{proj}}_{\\mathbf{a}}\\mathbf{b}=\\frac{1}{3}⟨1,1,1⟩$. 10. Suppose that $\\mathbf{a}$ and $\\mathbf{b}$ are non-zero vectors. Under what circumstances is 1. ${\\mathrm{comp}}_{\\mathbf{a}}\\mathbf{b}={\\mathrm{comp}}_{\\mathbf{b}}\\mathbf{a}$ 2. ${\\mathrm{proj}}_{\\mathbf{a}}\\mathbf{b}={\\mathrm{proj}}_{\\mathbf{b}}\\mathbf{a}$ 11. An intelligent robot exerts a horizontal force of $25$ lb on a crate as it pushes it", "$$\\mathbf E(\\mathbf r)=\\frac{q}{4\\pi\\varepsilon_0}\\frac{\\mathbf r-\\mathbf r'}{|\\mathbf r-\\mathbf r'|^3}$$ is the electric field at $$\\mathbf r$$ due to a point charge $$q$$ at $$\\mathbf r'$$. So when you calculate the average of this field over a sphere, $$\\mathbf r'$$ is fixed, and the integration is over the unprimed coordinates, i.e. over $$d\\tau$$, not $$d\\tau'$$ \\begin{align}\\bf E_{\\tt ave} \\rm :&= { 1 \\over {4 \\over 3 }\\pi R^3} \\int_{\\tau} \\bf E(\\bf r) \\rm \\space d\\tau\\\\ &=\\frac{1}{4\\pi\\varepsilon_0}\\int_\\tau \\frac{q}{\\frac{4}{3}\\pi R^3}\\frac{\\mathbf r-\\mathbf r'}{|\\mathbf r-\\mathbf r'|^3}d\\tau\\tag{1} \\end{align} Now, compare this last expression with the general form of Coulombs's law for a continuous charge distribution $$\\mathbf{E (r)}=\\frac{1}{4\\pi\\varepsilon_0}\\int_{\\tau'}\\rho(\\mathbf r')\\frac{\\mathbf r-\\mathbf r'}{|\\mathbf r-\\mathbf r'|^3}d\\tau'\\tag{*}$$ or equivalently 1 $$\\mathbf{E (r')}=\\frac{1}{4\\pi\\varepsilon_0}\\int_{\\tau}\\rho(\\mathbf r)\\frac{\\mathbf r'-\\mathbf r}{|\\mathbf r'-\\mathbf r|^3}d\\tau\\tag{**}$$ It's easy to see then, that $$(1)$$ is a special case of $$(**)$$, obtained by setting $$\\rho(\\mathbf r)=\\frac{-q}{\\frac{4}{3}\\pi R^3},$$ so the minus sign comes naturally when we write carefully all the $$\\mathbf{r}$$ and $$\\mathbf{r'}$$ dependences. I'm sorry to say that I lack the kind of intuition to claim whether this is because we moved the origin to the point where we placed the charge $$q$$, or something else. 1 Make sure you see why $$(*)$$ and $$(**)$$ are the same. It's just renaming variables, interchange $$\\mathbf{r\\leftrightarrow r'}$$. • Thank you very much for your answer. So, in this case, the minus sign does not", "to $\\mathbf{C}$, or it is isomorphic to $\\mathbf{S}$, or it is isomorphic to $\\mathbf{D}$. Proof. Let $1$ be the multiplicative identity, and let $x∈A$ be any vector not parallel to $1$. Then $\\operatorname{span}\\{1, x\\}$ forms a basis. Let $x^2 = a + bx$. We write \\begin{align*} x^2 &= a + bx \\\\ x^2 - bx &= a \\\\ x^2 - bx + \\frac{b}{4} &= a + \\frac{b}{4} \\\\ {\\left(x - \\frac{b}{2}\\right)}^2 &= a + \\frac{b}{4} \\end{align*} Now consider three cases. If $a + \\frac{b}{4} = 0$, then $x - \\frac{b}{2}$ squares to zero, and since it is not real (since $x$ is not real), therefore $\\{1, x - \\frac{b}{2}\\}$ is a basis. But then $A$ is isomorphic to $\\mathbf{D}$. If $a + \\frac{b}{4} < 0$, then ${\\left(\\frac{\\left(x - \\frac{b}{2}\\right)}{\\sqrt{-a - \\frac{b}{4}}}\\right)}^2 = -1$ and by a similar argument as before, $A$ is isomorphic to $\\mathbf{C}$. Otherwise, $a + \\frac{b}{4} > 0$, and ${\\left(\\frac{\\left(x - \\frac{b}{2}\\right)}{\\sqrt{a + \\frac{b}{4}}}\\right)}^2 = 1$ and hence $A$ is isomorphic to $\\mathbf{S}$. QED.", "# Prob. 2 (b), Sec. 8.14, in Apostol's CALCULUS Vol 2: How to find the directional derivative of this scalar field? Let $f$ be the scalar field defined by the formula $$f(x, y, z) = (x/y)^z,$$ for all points $(x, y, z) \\in \\mathbb{R}^3$ for which the formula makes sense. Then what is the directional derivative of $f$ at the point $(1, 1, 1)$ in the direction of $2 \\mathbf{i} + \\mathbf{j} - \\mathbf{k}$? My Attempt: Let us put $\\mathbf{a} \\colon= (1, 1, 1)$ and $\\mathbf{y} \\colon= (2, 1, -1)$. Then $\\mathbf{a} + h \\mathbf{y} = ( 1 + 2h, 1 + h, 1 - h)$, and so \\begin{align} \\frac{ f( \\mathbf{a} + h \\mathbf{y} ) - f( \\mathbf{a} ) }{ h } &= \\frac{ [ (1+2h) / ( 1+h ) ]^{1-h} - 1 }{ h } \\\\ &= \\end{align} What next? How to find $$\\lim_{ h \\to 0 } \\frac{ f( \\mathbf{a} + h \\mathbf{y} ) - f( \\mathbf{a} ) }{ h }?$$ Or, does this limit exist? • I think you need to find $\\frac{\\partial f}{\\partial x},\\frac{\\partial f}{\\partial y},\\frac{\\partial f}{\\partial z}$ at $(1,1,1)$, and then find the dot product of this vector with $(2,1,-1)$. – fierydemon Jun 13 '18 at 12:48 • @fierydemon does this formula hold in every situation where the partials exist? According to", "can find the value of $z$ by solving: $$z = \\frac{1}{t}\\left( ra_1 + sb_1 + tc_1 - rx - sy\\right).$$ What can go wrong? • For three points to determine a unique plane, you need the three points to not be collinear (not lie on the same line). You will find this when you compute the vector $\\mathbf{n}$. If $\\mathbf{n}=(0,0,0)$, then $\\mathbf{v}_1$ and $\\mathbf{v}_2$ are parallel, so that means that the three points are collinear and don't determine a unique plane. So you can just test $\\mathbf{n}$ to see if it is nonzero before proceedings. • It's possible for there to not be a unique value of $z$ that goes with the given $x$ and $y$. This will happen if $\\mathbf{n}$ has the form $\\mathbf{n}=(r,s,0)$. Then either the given $x$ and $y$ satisfy the equation you get, in which case every value of $z$ works; or else the given $x$ and $y$ do not satisfy the equation you get and no value of $z$ works. Example. Suppose you are given $(1,2,3)$, $(1,0,1)$, and $(-2,1,0)$. Then \\begin{align*} \\mathbf{v}_1 &= (1,2,3) - (1,0,1) = (0,2,2).\\\\ \\mathbf{v}_2 &= (1,2,3) - (-2,1,0) = (3,1,3). \\end{align*} Then \\begin{align*} \\mathbf{n} &= \\mathbf{v}_1\\times\\mathbf{v}_2 = \\left|\\begin{array}{rrr} \\mathbf{i} & \\mathbf{j} & \\mathbf{k}\\\\ 0 & 2 & 2\\\\ 3 & 1 & 3 \\end{array}\\right|\\\\ &= ( 6-2, 6-", "thus $A$ is zero, which contradicts $A > 0$ condition): \\begin{aligned} \\mathbf{a} \\le \\mathbf{0} &\\implies \\text{no solutions} \\end{aligned} Any combination with fifth line: \\begin{aligned} \\left(\\exists_i \\, \\mathbf{a}_i > 0\\right) &\\implies w = \\frac{-b}{\\sum_{i : \\mathbf{a}_i > 0} \\mathbf{a}_i^2} \\land \\mathbf{x}_i = \\begin{cases} -w \\, \\mathbf{a}_i &: \\mathbf{a}_i > 0 \\\\ 0 &: \\mathbf{a}_i \\le 0\\end{cases} \\land \\mathbf{v}_i = \\begin{cases} 0 &: \\mathbf{a}_i \\ge 0 \\\\ w \\, \\mathbf{a}_i &: \\mathbf{a}_i < 0\\end{cases} \\end{aligned} 3. $b < 0$ (lines 7-9) Seventh and eighth line gives (since all $\\mathbf{x}_i$ are zero, thus $A$ is zero, which contradicts $A > 0$ condition): \\begin{aligned} \\mathbf{a} \\ge \\mathbf{0} &\\implies \\text{no solutions} \\end{aligned} Any combination with nineth line: \\begin{aligned} \\left(\\exists_i \\, \\mathbf{a}_i < 0\\right) &\\implies w = \\frac{-b}{\\sum_{i : \\mathbf{a}_i < 0} \\mathbf{a}_i^2} \\land \\mathbf{x}_i = \\begin{cases} 0 &: \\mathbf{a}_i \\ge 0 \\\\ -w \\, \\mathbf{a}_i &: \\mathbf{a}_i < 0\\end{cases} \\land \\mathbf{v}_i = \\begin{cases} w \\, \\mathbf{a}_i &: \\mathbf{a}_i > 0 \\\\ 0 &: \\mathbf{a}_i \\le 0 \\end{cases} \\end{aligned}", ". & . & .\\\\[-10pt] . & . & . & . \\\\[-10pt] . & . & . & . \\\\[5pt] 0 & 0 & ... & d_{nn} \\end{bmatrix}.$$ The positive definiteness assumption guarantees that all $d_{ii}$'s are positive. Let's define $$\\nonumber D^{\\frac{1}{2}} = \\begin{bmatrix} \\sqrt{d_{11}} & 0 & ... & 0 \\\\%[5pt] 0 & \\sqrt{d_{22}} & ... & 0 \\\\%[5pt] . & . & . & .\\\\[-10pt] . & . & . & . \\\\[-10pt] . & . & . & . \\\\[5pt] 0 & 0 & ... & \\sqrt{d_{nn}} \\end{bmatrix}.$$ We have $D^{\\frac{1}{2}} D^{\\frac{1}{2}}=\\mathbf{D}$ and $D^{\\frac{1}{2}}={D^{\\frac{1}{2}}}^{T}$. Also define \\begin{align}%\\label{} \\mathbf{A}= Q D^{\\frac{1}{2}} Q^T. \\end{align} Then, \\begin{align}%\\label{} \\mathbf{A} \\mathbf{A}^{T}= \\mathbf{A}^{T}\\mathbf{A}=\\mathbf{C}. \\end{align} Now we are ready to define the transformation that converts a standard Gaussian vector to $\\mathbf{X} \\sim N(\\mathbf{m},\\mathbf{C})$. Let Z be a standard Gaussian vector, i.e., $\\mathbf{Z} \\sim N(\\mathbf{0},\\mathbf{I})$. Define \\begin{align}%\\label{} \\mathbf{X}=\\mathbf{A}\\mathbf{Z}+\\mathbf{m}. \\end{align} We claim that $\\mathbf{X} \\sim N(\\mathbf{m},\\mathbf{C})$. To see this, first note that $\\mathbf{X}$ is a normal random vector. The reason is that any linear combination of components of $\\mathbf{X}$ is indeed a linear combination of components of $\\mathbf{Z}$ plus a constant. Thus, every linear combination of components of $\\mathbf{X}$ is a normal random variable. It remains to show that $E\\mathbf{X}=\\mathbf{m}$ and $\\mathbf{C_X}=\\mathbf{C}$. First note that by linearity of expectation we have \\begin{align}%\\label{} E\\mathbf{X}", "want two components of $$\\mathbf{G}$$ to be $$0$$ $$[C_1(x,z)]_z = x \\Rightarrow C_1(x,z)=xz$$ Then $$\\boxed{\\mathbf{G} = (xz,0,-\\frac{y^2}{2})}$$ which can be verified to satisfy $$\\nabla \\times \\mathbf{G} = \\mathbf{F}$$ If $$G_3 = 0$$, then we have \\begin{align} -& \\frac{\\partial G_2}{\\partial z} = -y \\Rightarrow G_2 = yz + C_2(x,y)\\\\ &\\frac{\\partial G_1}{\\partial z} = x \\Rightarrow G_1 = xz +C_1(x,y) \\\\ &\\frac{\\partial G_2}{\\partial x} - \\frac{\\partial G_1}{\\partial y} = 0 \\\\ \\end{align} then we get $$\\frac{\\partial G_2}{\\partial x} - \\frac{\\partial G_1}{\\partial y} = [C_2(x,y)]_x - [C_1(x,y)]_y = 0 \\Rightarrow [C_2(x,y)]_x = [C_1(x,y)]_y$$ Here, for simplicity, we can choose $$C_2(x,y)=C_1(x,y)=0$$, and so $$\\boxed{\\mathbf{G} = (xz,yz,0)}$$ which can be verified to satisfy $$\\nabla \\times \\mathbf{G} = \\mathbf{F}$$ Of the three boxed solutions, only $$\\mathbf{G} = (xz,0,-\\frac{y^2}{2})$$ is on our list, and hence it is our answer. However, given that we have a list of options, one can arrive to this answer earlier by realizing that certain solutions of the form $$(xz, f(y),0)$$ will never satisfy the curl equation, so the answer will be in the form$$(G_1,0,G_3)$$ • I have to write that this is an enlightening answer. – JD_PM Mar 24 at 20:02", "on $L_1$ is of the form $\\mathbf{v}=a\\mathbf{v}_1$ for some $a \\in R$. Hence we have \\begin{align*} A\\mathbf{v}=A(a\\mathbf{v}_1)=aA\\mathbf{v}_1=a\\lambda_1\\mathbf{v}_1=\\lambda \\mathbf{v}. \\end{align*} It follows that each vector $\\mathbf{v}\\in L_1$ is mapped onto $L_1$. (Since $\\lambda \\neq 0$, the image under $T$ is one-dimensional, hence $T(L_1)=L_1$.) Similarly, we have $T(L_2)=L_2$. Finally, we show that no other lines passing through the origin are mapped onto themselves by $T$. Assume that we have a line $L=\\Span(\\mathbf{w})$ such that $T(L)=L$. Then since we have $A\\mathbf{w}\\in L=\\Span(\\mathbf{w})$, there is $\\mu\\in \\R$ such that $A\\mathbf{w}=\\mu\\mathbf{w}. \\tag{*}$ Since $\\{\\mathbf{v}_1, \\mathbf{v}_2\\}$ is a basis of $\\R^2$, we have $\\mathbf{w}=a\\mathbf{v}_1+b\\mathbf{v}_2$ for some $a, b\\in \\R$. Then the left hand side of (*) becomes \\begin{align*} A\\mathbf{w}&=A(a\\mathbf{v}_1+b\\mathbf{v}_2)\\\\ &=aA\\mathbf{v}_1+bA\\mathbf{v}_2\\\\ &=a\\lambda_1 \\mathbf{v}_1+b\\lambda_2 \\mathbf{v}_2. \\end{align*} Thus the relation (*) yields that \\begin{align*} a\\lambda_1 \\mathbf{v}_1+b\\lambda_2 \\mathbf{v}_2=\\mu a\\mathbf{v}_1+\\mu b\\mathbf{v}_2\\\\ a(\\lambda_1-\\mu )\\mathbf{v}_1+b(\\lambda_2-\\mu)\\mathbf{v}_2=\\mathbf{0}. \\end{align*} Since $\\mathbf{v}_1, \\mathbf{v}_2$ are linearly independent, we have $a(\\lambda_1-\\mu )=0 \\text{ and } b(\\lambda_2-\\mu)=0.$ The equality $a(\\lambda_1-\\mu )=0$ implies that either $a=0$ or $\\lambda_1=\\mu$. If $a=0$, then we have $\\mathbf{w}=b \\mathbf{v}_2$, and hence $L=L_2$. If $\\mu=\\lambda_1$, then the second equality $b(\\lambda_2-\\mu)=0$ implies that $b=0$ since $\\lambda_1\\neq \\lambda_2$. Then we have $\\mathbf{w}=a\\mathbf{v}_1$, and hence $L=L_1$. In either case, $L$ must be $L_1$ or $L_2$. This completes the proof. ### 1 Response 1. 06/23/2017 […] For a proof of this problem, see the post “A Linear Transformation", "1), this further simplifies to $$f = \\frac{{a_0 + 2w{\\mathbf{y}} + \\left( {w{\\mathbf{y}}} \\right)^2}}{{1 + 2w{\\mathbf{y}} + \\left( {w{\\mathbf{y}}} \\right)^2}}.$$ (15) We extend the above case to any number of binding sites. Adding sites could affect ρ for each term in the numerator and denominator, but for simplicity we constrain the possible values by assuming all ρ = 1. This assumption is applied in the lower plots of the first and second landscapes in Fig. 2c. Examples are shown below for three, four, five, and six binding sites. Coefficients are derived using Pascal’s triangle: $$f_3 = \\frac{{a_0 + 3\\left( {w{\\mathbf{y}}} \\right)^n \\, + \\, 3\\left( {w{\\mathbf{y}}} \\right)^{2n} \\, + \\, \\left( {w{\\mathbf{y}}} \\right)^{3n}}}{{1 + 3\\left( {w{\\mathbf{y}}} \\right)^n \\, + \\, 3\\left( {w{\\mathbf{y}}} \\right)^{2n} \\, + \\, \\left( {w{\\mathbf{y}}} \\right)^{3n}}},$$ (16) $$f_4 = \\frac{{a_0 + 4\\left( {w{\\mathbf{y}}} \\right)^n \\, + \\, 6\\left( {w{\\mathbf{y}}} \\right)^{2n} \\, + \\, 4\\left( {w{\\mathbf{y}}} \\right)^{3n} \\, + \\, \\left( {w{\\mathbf{y}}} \\right)^{4n}}}{{1 + 4\\left( {w{\\mathbf{y}}} \\right)^n \\, + \\, 6\\left( {w{\\mathbf{y}}} \\right)^{2n} \\, + \\, 4\\left( {w{\\mathbf{y}}} \\right)^{3n} \\, + \\, \\left( {w{\\mathbf{y}}} \\right)^{4n}}},$$ (17) $$f_5 = \\frac{{a_0 + 5\\left( {w{\\mathbf{y}}} \\right)^n \\, + \\, 10\\left( {w{\\mathbf{y}}} \\right)^{2n} \\, + \\, 10\\left( {w{\\mathbf{y}}} \\right)^{3n} \\, + \\, 5\\left( {w{\\mathbf{y}}} \\right)^{4n} \\, + \\, \\left( {w{\\mathbf{y}}} \\right)^{5n}}}{{1 + 5\\left( {w{\\mathbf{y}}} \\right)^n \\, + \\, 10\\left( {w{\\mathbf{y}}}", "= \\frac{1}{{m \\gamma}} \\left( \\mathbf{P} - \\frac{e}{c} \\mathbf{A} \\right),\\end{aligned} \\quad\\quad\\quad(14) but $\\gamma$ also has a $\\mathbf{u}$ dependence, so this is not complete. Squaring gets us closer \\begin{aligned}\\mathbf{u}^2 = \\frac{1 - \\mathbf{u}^2/c^2}{m^2} {\\left( \\mathbf{P} - \\frac{e}{c} \\mathbf{A} \\right)}^2,\\end{aligned} \\quad\\quad\\quad(15) and a bit of final rearrangement yields \\begin{aligned}\\mathbf{u}^2 = \\frac{(c \\mathbf{P} - e \\mathbf{A})^2}{m^2 c^2 + {\\left( \\mathbf{P} - \\frac{e}{c} \\mathbf{A} \\right)}^2}.\\end{aligned} \\quad\\quad\\quad(16) Writing $\\mathbf{p} = \\mathbf{P} - e \\mathbf{A}/c$, we can rearrange and find \\begin{aligned}\\sqrt{1 - \\mathbf{u}^2/c^2} = \\frac{m c }{\\sqrt{m^2 c^2 +\\mathbf{p}^2}}\\end{aligned} \\quad\\quad\\quad(17) Also, taking roots of 16 we must have the directions of $\\mathbf{u}$ and $\\left( \\mathbf{P} - \\frac{e}{c} \\mathbf{A} \\right)$ differ only by a rotation. From 14 we also know that these are colinear, and therefore have \\begin{aligned}\\mathbf{u} = \\frac{c \\mathbf{P} - e \\mathbf{A}}{\\sqrt{m^2 c^2 + {\\left( \\mathbf{P} - \\frac{e}{c} \\mathbf{A} \\right)}^2}}.\\end{aligned} \\quad\\quad\\quad(18) This and 17 can now be substituted into 13, for \\begin{aligned}H = \\frac{c}{m^2 c^2 + \\mathbf{p}^2} \\left({\\left(\\mathbf{P} - \\frac{e}{c} \\mathbf{A}\\right)}^2 + m^2 c^2 \\right)+ e \\Phi.\\end{aligned} \\quad\\quad\\quad(19) Dividing out the common factors we finally have the Hamiltonian in a tidy form \\begin{aligned}H = \\sqrt{ (c \\mathbf{P} - e \\mathbf{A})^2 + m^2 c^4 } + e\\Phi.\\end{aligned} \\quad\\quad\\quad(20) ## Hamiltonian equation evaluation. Let’s now go through the exercise of evaluating the Hamiltonian equations. We want the starting point to be just the energy expression 20," ]

[ "E1 = extension(A,C,B,P); pair F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "## anonymous one year ago In square $ABCD$, $E$ is the midpoint of $\\overline{BC}$, and $F$ is the midpoint of $\\overline{CD}$. Let $G$ be the intersection of $\\overline{AE}$ and $\\overline{BF}$. Prove that $DG = AB$. • This Question is Open 1. anonymous Asymptote code below [asy] unitsize(2.5 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (0,1); C = (1,1); D = (1,0); E = (B + C)/2; F = (C + D)/2; G = extension(A,E,B,F); draw(A--B--C--D--cycle); draw(A--E); draw(B--F); draw(D--G); label(\"$A$\", A, SW); label(\"$B$\", B, NW); label(\"$C$\", C, NE); label(\"$D$\", D, SE); label(\"$E$\", E, N); label(\"$F$\", F, dir(0)); label(\"$G$\", G, WSW); [/asy]", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "# BS in relation to BF and CS in relation to CE #### mathmari ##### Well-known member MHB Site Helper Hey!! We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? #### Klaas van Aarsen ##### MHB Seeker Staff member We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? Hey mathmari !! Let's first make a drawing: \\begin{tikzpicture}[scale=2] %preamble \\usetikzlibrary {calc} \\coordinate[label=left:A] (A) at (0,0); \\coordinate[label=right:B] (B) at (0:5); \\coordinate[label=C] (C) at (70:4); \\coordinate[label=below:E] (E) at (0:1); \\coordinate[label=left:F] (F) at (70:3); \\coordinate[label=right:S] (S) at (intersection cs:first line={(B)--(F)}, second line={(C)--(E)}); \\draw (A) -- (B) -- (C) -- cycle; \\draw (C) -- (E); \\draw (B) -- (F); \\draw[-latex, ultra thick, blue] (A) -- node[ above ] {$\\mathbf{\\vec c}$} (B); \\draw[-latex, ultra thick, blue] (A) -- node[ right] {$\\mathbf{\\vec b}$} (C); \\path (A) --", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "+0 # HELP 0 194 1 In the figure below, $D$ is a point on segment $\\overline{CE}$ such that $\\overline{AD}\\parallel\\overline{BE}$ and $A$ is not on $\\overline{BC}.$ Line segments $\\overline{AD}$ and $\\overline{BC}$ intersect at $P.$ [asy] size(4cm); pair C = (0,0); pair D = (4,0); pair E = (9,0); pair F = (0.25,4); pair G = (1,4); pair A = extension(C,F,D,G); pair H = G+E-D; pair B = extension(C,G,E,H); pair P = extension(A,D,B,C); draw(A--C--E--B--C); draw(A--D); label(\"$A$\",A,NNW); label(\"$B$\",B,N); label(\"$C$\",C,SSW); label(\"$D$\",D,S); label(\"$E$\",E,SSE); label(\"$P$\",P,dir(0)); [/asy] Can $\\angle CAD=\\angle CBE?$ Explain. Aug 20, 2020", "dot(A); label(\"A\",A,NE); dot(B); label(\"B\",B,dir(180)); dot(C); label(\"C\",C,SE); dot(D); label(\"D\",D,dir(0)); dot(E); label(\"E\",E,SE); dot(F); label(\"F\",F,SW); [/asy]$ Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper. As triangle $ABC$ is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles. As point $D$ splits line segment $\\overline{AC}$ in a $1:2$ ratio, we draw $\\overline{AC}$ as a vertical line segment $3$ units long. Point $D$ is thus $1$ unit below point $A$ and $2$ units above point $C$. By definition, Point $E$ splits line segment $\\overline{BD}$ in a $1:1$ ratio, so we draw $\\overline{BD}$ $2$ units long directly left of $D$ and draw $E$ directly between $B$ and $D$, $1$ unit away from both. We then draw line segments $\\overline{AB}$ and $\\overline{BC}$. We can easily tell that triangle $ABC$ occupies $3$ square units of space. Constructing line $AE$ and drawing $F$ at the intersection of $AE$ and $BC$, we can easily see that triangle $EBF$ forms a right triangle occupying $\\frac{1}{4}$ of a square unit of space. The ratio of the areas of triangle $EBF$ and triangle $ABC$ is thus", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "[] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); label(\"F\",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]$ Drawing line $\\overline{BD}$ and parallel line $\\overline{CF}$, we see that $\\triangle FCE \\sim \\triangle BDE$ by AA similarity. Thus $\\frac{FE}{EB} = \\frac{FC}{DB} = \\frac{2}{4} = \\frac{1}{2}$. Reciprocating, we know that $\\frac{EB}{FE} = 2$ so $\\frac{EB+FE}{FE} = 2+1 \\Rightarrow \\frac{FB}{FE} = 3$. Reciprocating again, we have $\\frac{FE}{FB} = \\frac{1}{3} \\Rightarrow FE = \\frac{1}{3}FB$. We know that $FD = 2$, so by the Pythagorean Theorem, $FB = \\sqrt{2^{2} + 4^{2}} = 2\\sqrt{5}$. Thus $FE = \\frac{1}{3}FB = \\frac{2\\sqrt{5}}{3}$. Applying the Pythagorean Theorem again, we have $AF = \\sqrt{1^{2}+2^{2}} = \\sqrt{5}$. We finally have $AE = AF + FE = \\sqrt{5} + \\frac{2\\sqrt{5}}{3} = \\frac{5\\sqrt{5}}{3} \\Rightarrow \\boxed{\\text{B}}$", "pair R = (1.5*A + 3.5*C)/5, S = extension(A0,R,A,B), T = extension(C0,R,B,C); pair P2 = extension(A0,T,B0,R), M = extension(A,B,A0,C), N = extension(A,C0,B,C); /* Draw a couple of the common paths */ D(A--B--C--cycle); D(incircle(A,B,C)); /* Label some of the common points */ D(MP(\"A\", A, NE)); D(MP(\"B\", B)); D(MP(\"C\", C)); D(MP(\"I\", I, W)); Out[2]: 1) (a) Extend $A_{1}X$ to intersect $AC$ at $V$. Show that the lines $AA_{1}$, $BV$, and $DE$ are concurrent. (b) Let $A_{2}$ be the intersection point of $AX$ with the sideline $BC$. Show that $$\\frac{A_2B}{A_2C} = \\frac{s-c}{s-b} \\cdot \\frac{A_1B^2}{A_1C^2},$$ where $s = \\frac{a + b + c}{2}$ is the semiperimeter of $\\triangle ABC$. [Hint: Menelaus's Theorem may come in handy.] (c) Prove that lines $AX, BY, CZ$ are concurrent. In [3]: %%asy pumac_power_round_2011_config.asy size(500); /* Draw lines */ D(A--A1--X1, darkgreen); D(B--B1--Y1, darkgreen); D(C--C1--Z1, darkgreen); D(A--P1--B, darkred); D(C--Z1, darkred); D(X1--V); D(P1--A2, darkred); /* Label points */ D(MP(\"P\",P,dir(60))); D(MP(\"D\",D)); D(MP(\"E\",E,NE)); D(MP(\"F\",F,NW)); D(MP(\"X\",X1,ENE)); D(MP(\"Y\",Y1,1.4*dir(150))); D(MP(\"Z\",Z1,NE)); D(P1); D(MP(\"A_1\",A1)); D(MP(\"B_1\",B1,NE)); D(MP(\"C_1\",C1,NW)); D(MP(\"V\",V,NE)); D(MP(\"A_2\",A2)); Out[3]: 2) Show that the lines $B_1C_1$, $EF$, $YZ$ concur at a point, say $A_{0}$. Similarly, the lines $C_1A_1$, $FD$, $ZX$ and $A_1B_1$, $DE$, $XY$ concur at points $B_{0}$ and $C_{0}$, respectively. [Hint: Pascal's theorem might be useful here.] In [4]: %%asy pumac_power_round_2011_config.asy size(600); /* Draw lines */ D(Y1--A0--E); D(F--D); D(A1--C1); D(B1--C0--E); D(D--E--F--cycle,rgb(0,0.7,0)+linewidth(1)); D(A0--B0--C0--cycle, darkblue); D(A--B0--C, darkblue+linetype(\"3 3\"));", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "Alternatively, by the Extended Law of Sines, $$2R = \\frac {AC}{\\sin \\angle ABC} = \\frac {3}{\\frac {2\\sqrt {2}}{3}} \\Longrightarrow R = \\frac {9}{4\\sqrt {2}}$$ Answer follows as above. ### Solution 3 Extend segment $AD$ to $R$ on Circle $O$. $[asy] import olympiad; pair B=(0,0), C=(2,0), A=(1,3), D=(1,0), R=(1,-0.35); pair O=circumcenter(A,B,C); draw(A--B--C--A--D--R--C); draw(B--O--C); draw(circumcircle(A,B,C)); dot(O); label(\"$$A$$\",A,N); label(\"$$B$$\",B,S); label(\"$$C$$\",C,S); label(\"$$D$$\",D,S); label(\"$$O$$\",O,W); label(\"$$R$$\",R,S); label(\"$$r$$\",(O+A)/2,SE); label(\"$$r$$\",(O+R)/2,SE); label(\"$$3$$\",(A+C)/2,NE); label(\"$$1$$\",(C+D)/2,N); [/asy]$ By the Pythagorean Theorem $$AD^2 = 3^2 - 1^2$$ $$AD = 2\\sqrt{2}$$ $\\triangle ADC$ is similar to $\\triangle ACR$, so $$\\frac {2\\sqrt{2}}{3} = \\frac {3}{2r}$$ which gives us $$2r = \\frac {9}{2\\sqrt{2}} = \\frac {9\\sqrt{2}}{4}$$ therefore $$r = \\frac {9\\sqrt{2}}{8}$$ The area of the circle is therefore $\\pi r^2 = \\left(\\frac {9\\sqrt{2}}{8}\\right)^2 \\pi = \\boxed{\\mathrm{(C) \\ } \\frac {81}{32} \\pi}$ ### Solution 4 First, we extend $AD$ to hit the circle at $E.$ $[asy] import olympiad; pair B=(0,0), C=(2,0), A=(1,3), D=(1,0), E=(1,-(8^0.5)/8); pair O=circumcenter(A,B,C); draw(A--B--C--A--E); draw(circumcircle(A,B,C)); dot(O); dot(D); dot(B); dot(C); dot(A); dot(E); label(\"A\",A,N); label(\"B\",B,S); label(\"$$C$$\",C,S); label(\"D\",D,NE); label(\"O\",O,W); label(\"E\",E,S); label(\"3\",(A+B)/2,NW); label(\"1\",(B+D)/2,N); [/asy]$ By the Pythagorean Theorem, we know that $$1^2+AD^2=3^2\\implies AD=2\\sqrt{2}.$$ We also know that, from the Power of a Point Theorem, $$AD\\cdot DE=BD\\cdot DC.$$ We can substitute the ones we know to get $$2\\sqrt{2}\\cdot DE=1$$ We can simplify this to get that $$DE=\\dfrac{2\\sqrt{2}}{8}.$$ We add $AD$ and $DE$ together to get the length", "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "# 2018 AIME II Problems/Problem 14 ## Problem The incircle $\\omega$ of triangle $ABC$ is tangent to $\\overline{BC}$ at $X$. Let $Y \\neq X$ be the other intersection of $\\overline{AX}$ with $\\omega$. Points $P$ and $Q$ lie on $\\overline{AB}$ and $\\overline{AC}$, respectively, so that $\\overline{PQ}$ is tangent to $\\omega$ at $Y$. Assume that $AP = 3$, $PB = 4$, $AC = 8$, and $AQ = \\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Diagram $[asy] size(200); import olympiad; defaultpen(linewidth(1)+fontsize(12)); pair A,B,C,P,Q,Wp,X,Y,Z; B=origin; C=(6.75,0); A=IP(CR(B,7),CR(C,8)); path c=incircle(A,B,C); Wp=IP(c,A--C); Z=IP(c,A--B); X=IP(c,B--C); Y=IP(c,A--X); pair I=incenter(A,B,C); P=extension(A,B,Y,Y+dir(90)*(Y-I)); Q=extension(A,C,P,Y); draw(A--B--C--cycle, black+1); draw(c^^A--X^^P--Q); pen p=4+black; dot(\"A\",A,N,p); dot(\"B\",B,SW,p); dot(\"C\",C,SE,p); dot(\"X\",X,S,p); dot(\"Y\",Y,dir(55),p); dot(\"W\",Wp,E,p); dot(\"Z\",Z,W,p); dot(\"P\",P,W,p); dot(\"Q\",Q,E,p); MA(\"\\beta\",C,X,A,0.3,black); MA(\"\\alpha\",B,A,X,0.7,black); [/asy]$ ## Solution 1 Let the sides $\\overline{AB}$ and $\\overline{AC}$ be tangent to $\\omega$ at $Z$ and $W$, respectively. Let $\\alpha = \\angle BAX$ and $\\beta = \\angle AXC$. Because $\\overline{PQ}$ and $\\overline{BC}$ are both tangent to $\\omega$ and $\\angle YXC$ and $\\angle QYX$ subtend the same arc of $\\omega$, it follows that $\\angle AYP = \\angle QYX = \\angle YXC = \\beta$. By equal tangents, $PZ = PY$. Applying the Law of Sines to $\\triangle APY$ yields $$\\frac{AZ}{AP} = 1 + \\frac{ZP}{AP} = 1 + \\frac{PY}{AP} = 1 + \\frac{\\sin\\alpha}{\\sin\\beta}.$$Similarly, applying the Law of Sines to $\\triangle ABX$ gives $$\\frac{AZ}{AB} = 1" ]

[ "# It's Time To Have Pi Today Geometry Level 4 $4\\left( \\arctan\\left( \\dfrac { 1 }{ x } \\right) + \\arctan\\left( \\dfrac { 1 }{ { x }^{ 3 } } \\right) \\right) =\\pi$ Let $$x$$ be the positive real number satisfying the equation above and $${x}^{3}-x$$ can be expressed as $\\dfrac { a }{ b } \\left( c+\\sqrt { d } \\right) \\; ,$ where $$a, b, c$$ and $$d$$ are positive integers, with $$a,b$$ coprime and $$d$$ square-free. Find the value of the expression below. $4\\left( \\arctan\\left( \\dfrac { b }{ c } \\right) + \\arctan\\left( \\dfrac { a }{ { d } } \\right) \\right)$", "# It's Time To Have Pi Today Geometry Level 4 $4\\left( \\arctan\\left( \\dfrac { 1 }{ x } \\right) + \\arctan\\left( \\dfrac { 1 }{ { x }^{ 3 } } \\right) \\right) =\\pi$ Let $$x$$ be the positive real number satisfying the equation above and $${x}^{3}-x$$ can be expressed as $\\dfrac { a }{ b } \\left( c+\\sqrt { d } \\right) \\; ,$ where $$a, b, c$$ and $$d$$ are positive integers, with $$a,b$$ coprime and $$d$$ square-free. Find the value of the expression below. $4\\left( \\arctan\\left( \\dfrac { b }{ c } \\right) + \\arctan\\left( \\dfrac { a }{ { d } } \\right) \\right)$ Give your answer to 4 decimal places. × Problem Loading... Note Loading... Set Loading...", "Nawat N. # How to find $u_{min}(d,r)$ if $xtan\\theta - \\frac{gx^2}{2u^2cos^2\\theta} \\geq r + \\sqrt{r^2-(x-d)^2}$ for $d-r \\leq x \\leq d+r$ How to find umin(d,r) if xtanθ - (gx^2)/(2u2cos2θ) > r + √(r2-(x-d)2) for d-r < x < d+r ? And 0 < θ < π/2 , d and r can be any positive real number which d >r", "Find the positive value of $x$ that satisfies $x = 1.900 \\sin(x)$. Give the answer to four places of accuracy. $x\\approx$ Remember to calculate the trig functions in radian mode.", "# My 16th Birthday Problem Geometry Level 3 If $$f(x) = \\sec(\\arctan(x))$$, find the value of $$f(f(f(f(...(f(1))))))$$, where the function is iterated 2015 times. If you get your answer as $$a \\sqrt b$$, where $$a$$ and $$b$$ are positive integers with $$b$$ square-free, submit $$a+b$$. ×", "# Cubic Dilemma Algebra Level 4 There is a smallest positive real number $$a$$ such that there exists a positive number $$b$$ such that the roots of the polynomial $$x^3-ax^2+bx-a$$ are all real. In fact, for this value of $$a$$ the value of $$b$$ is unique. Find the value of $$b$$. ×", "# Uniqueness of Positive Root of Positive Real Number ## Theorem ### Positive Exponent Let $x \\in \\R$ be a real number such that $x > 0$. Let $n \\in \\Z$ be an integer such that $n > 0$. Then there is at most one $y \\in \\R: y \\ge 0$ such that $y^n = x$. ### Negative Exponent Let $x \\in \\R$ be a real number such that $x > 0$. Let $n \\in \\Z$ be an integer such that $n < 0$. Then there is at most one $y \\in \\R: y \\ge 0$ such that $y^n = x$.", "# Existence of Positive Root of Positive Real Number ## Theorem Let $x \\in \\R$ be a real number such that $x > 0$. Let $n \\in \\Z$ be an integer such that $n \\ne 0$. Then there exists a $y \\in \\R: y \\ge 0$ such that $y^n = x$. ## Proof ### Positive Exponent Let $x \\in \\R$ be a real number such that $x > 0$. Let $n \\in \\Z$ be an integer such that $n > 0$. Then there exists a $y \\in \\R: y \\ge 0$ such that $y^n = x$. ### Negative Exponent Let $x \\in \\R$ be a real number such that $x > 0$. Let $n \\in \\Z$ be an integer such that $n < 0$. Then there exists a $y \\in \\R: y \\ge 0$ such that $y^n = x$.", "# Existence of Square Roots of Positive Real Number/Examples ### Example: $\\sqrt 2$ There exists $u \\in \\R$ such that $u^2 = 2$.", "Some Property! Algebra Level 2 Assume that $x$ is a positive real number. Then, $\\large \\sqrt[3]{x\\sqrt{x}} \\ =$ ×", "measure, the set of rational numbers is of measure zero; thus, what really matters is what happens in the complement of this set, where the function is given as arctan x. It follows that the essential supremum is π/2 while the essential infimum is −π/2. Lastly, consider the function f(x) = x3 defined for all real x. Its essential supremum is +∞, and its essential infimum is −∞.", "## anonymous 4 years ago prove that arctan(x)=1-x has at least one root 1. anonymous it is continuous for sure, so show that it is negative somewhere and positive somewhere else. try say $-\\frac{\\pi}{4}$ and $\\frac{\\pi}{3}$ so make your life easy 2. anonymous That's utilizing the intermediate value theorem right? 3. anonymous Yes", "If $$a$$ $$b$$ and $$c$$ are positive real numbers with $$a+b+c = 6$$ find the maximum value of $$(a+ \\sqrt{bc})(b+\\sqrt{ac})(c+\\sqrt{ab}) + abc$$", "month ago Almost correct. Keep in mind you're dealing with real numbers only, so $$\\sqrt {x^2} = |x|$$ is true. Where $$|x|$$ is the absolute value of $$x$$. So the resultant value is non-negative, and not always positive, take $$x=0$$ as an example. - 3 years, 1 month ago ×", "# Thanksgiving $$\\pi$$ Geometry Level pending Find the least positive value of $$x$$ such that $$\\cos\\left(x^{\\circ} \\right) = -\\cos \\left(x \\text{ radians} \\right)$$. Round the value of $$x$$ to the nearest whole number. ×", "# Mathematical Enigma - VIII Algebra Level 1 Let $$x,y$$ be positive reals such that $$x+y =2$$, find the maximum value of $$x^{3}y^{3}(x^{3}+y^{3})$$. ×", "# Troublesome reals indeed...... Level pending For two positive reals $$\\large{a}$$ and $$\\large{b}$$ and some $$\\large{\\theta}$$ , we have $$\\large{a\\sin \\theta + b\\cos \\theta = 1 = a \\csc \\theta - b \\sec \\theta}$$, find the value of $$\\Large{a^2 + b^2 - b^{\\frac{2}{3}} + b^{\\frac{4}{3}}}$$ ×", "# Positive or negativ The sum of four real numbers ($$a, b, c, d$$) is 0. Let $$M=ab+bc+cd$$ and $$N=ac+ad+bd$$. Is it possible, that both of $$20M+17N$$ and $$20N+17M$$ are positive? ×", "any number, x, between 0 and $\\pi$, arccos(cos(x))= x. That follows from the very definition of \"arccos\".", "## College Algebra (10th Edition) $|2-v|=-1$ There is no real solution, because there is no real number whose distance from 0 on the number line is negative." ]

[ "~=~ \\dfrac{0}{\\sqrt{1 - 0^2}} ~.$ Now suppose that $$0 < x < 1$$. Let $$\\theta = \\sin^{-1} x$$. Then $$\\theta$$ is in QI and $$\\sin\\;\\theta = x$$. Draw a right triangle with an angle $$\\theta$$ such that the opposite leg has length $$x$$ and the hypotenuse has length $$1$$, as in Figure 5.3.10 (note that this is possible since $$0 < x < 1$$). Then $$\\sin\\;\\theta = \\frac{x}{1} = x$$. By the Pythagorean Theorem, the adjacent leg has length $$\\sqrt{1 - x^2}$$. Thus, $$\\tan\\;\\theta = \\frac{x}{\\sqrt{1 - x^2}}$$. If $$-1 < x < 0$$ then $$\\theta = \\sin^{-1} x$$ is in QIV. So we can draw the same triangle except that it would be \"upside down'' and we would again have $$\\tan\\;\\theta = \\frac{x}{\\sqrt{1 - x^2}}$$, since the tangent and sine have the same sign (negative) in QIV. Thus, $$\\tan\\;(\\sin^{-1} x) = \\dfrac{x}{\\sqrt{1 - x^2}}$$ for $$-1 < x < 1$$. Example $$\\PageIndex{9}$$: Prove the identity $$\\tan^{-1} x \\;+\\; \\cot^{-1} x ~=~ \\frac{\\pi}{2}$$. Solution: Let $$\\theta = \\cot^{-1} x$$. Using relations, we have $\\nonumber \\tan\\;\\left( \\tfrac{\\pi}{2} - \\theta \\right) ~=~ -\\tan\\;\\left( \\theta - \\tfrac{\\pi}{2} \\right) ~=~ \\cot\\;\\theta ~=~ \\cot\\;(\\cot^{-1} x) ~=~ x ~,$ by Equation \\ref{eqn:arccot2}. So since $$\\tan\\;(\\tan^{-1} x) = x$$ for all $$x$$, this means that $$\\tan\\;(\\tan^{-1} x) = \\tan\\;\\left( \\tfrac{\\pi}{2} - \\theta \\right)$$. Thus, $$\\tan\\;(\\tan^{-1}", "# Let $f(x) = \\arctan x + \\mathop{\\text{arccot}} x$. Find $f(0) + f(1) + f(\\sqrt{2}) + f(\\sqrt{3})$. Let $f(x) = \\arctan x + \\mathop{\\text{arccot}} x$. Find $f(0) + f(1) + f(\\sqrt{2}) + f(\\sqrt{3})$. This is what i got, $$\\left(0+\\frac{\\pi}{2}\\right)+\\left(\\frac{\\pi}{4}+\\frac{\\pi}{4}\\right)$$ • I take it those are the first two terms $f(0)$ and $f(1)$. Hmm ... something interesting there: they both equal $\\pi/2$. I wonder if $f(x)$ for arbitrary $x$ follows such a pattern.... – Simon S Jun 5 '15 at 22:46 • You can construct an angle of given tangent x with a right triangle of legs x and 1. – Piquito Jun 5 '15 at 22:49 • should not each one of them be $\\pi/2,$ so totals $2\\pi$ – abel Jun 5 '15 at 22:56 We know that $\\tan \\theta = x$, so $\\theta = \\arctan x$. What is the other angle, as a function of (a) $\\theta$ and (b) $x$ ? $\\arctan x+ \\operatorname{arccot} x =\\frac\\pi2$", "answer tackle my question in a multitude of ways! – Slugger Nov 28 '13 at 15:34 @Slugger, my pleasure. Please have a look into math.stackexchange.com/questions/579170/… – lab bhattacharjee Nov 28 '13 at 15:46 You can also use the Inverse Derivative Formula, which states that if $f(x)$ and $g(x)$ are inverse functions, we have $$g'(x) = \\dfrac {1}{f'(g(x))}.$$So, if $g(x)=\\arctan x$, our task is to find $g'(x)$. In that case, we have $f(x)=\\tan x$, which gives us $f'(x)=sec^2 x$, so we can substitute: \\begin {align*} g'(x) &= \\dfrac {1}{f'(g(x))} \\\\&= \\dfrac {1}{\\sec^2 (g(x))} \\\\&= \\dfrac {1}{\\sec^2 (\\arctan x)}. \\end {align*}We can find $\\sec (\\arctan x)$ geometrically. Consider a right triangle with legs of length $x$, $1$, and $\\sqrt{1+x^2}$. Let $\\theta$ be the angle opposite to the leg of length $x$. Then, $$\\sec \\left( \\arctan x \\right) = \\sec (\\theta) = \\sqrt {1+x^2},$$ so our answer is $$\\dfrac {1}{\\left( \\sqrt{1+x^2} \\right)^2} = \\boxed {\\dfrac {1}{1+x^2}}.$$ - Thanks you for your answer! – Slugger Nov 28 '13 at 15:35 You're welcome! :) – Ahaan S. Rungta Nov 28 '13 at 15:36 $$v=\\arctan(x)\\Rightarrow x=\\tan v\\Rightarrow x'=\\frac{1}{(\\tan v)'}=\\frac{1}{(\\frac{\\sin v}{\\cos v})'}=\\frac{1}{\\frac{1}{\\cos^2 v}}=\\cos^2 v=\\frac{\\cos^2 v}{1}$$ $$=\\frac{\\cos^2 v}{\\cos^2 v+\\sin^2 v}=\\frac{\\frac{\\cos^2 v}{\\cos^2 v}}{\\frac{\\cos^2 v+\\sin^2 v}{\\cos^2 v}}=\\frac{1}{1+\\tan^2 v}=\\frac{1}{1+x^2}$$ i.e $$v'=(\\arctan(x))'=\\frac{1}{1+x^2}$$ - I am not sure I follow exactly. Your second equation says $x=\\tan v$ and then you follow", "a right triangle with the leg opposite θ having length u and the adjacent leg having length 1. Then you'd have tan θ = u/1 = u. What's the length of the hypotenuse of this triangle in terms of u? Once you have that, write down algebraic expressions for sin(arctan u) = sin θ and cos(arctan u) = cos θ in terms of u. Finally, undo the substitution u = sinh x to obtain the final answer. Don't forget to tack on the arbitrary constant of integration at the end. The hypotenuse of the triangle is: $$\\sqrt{1 + u^2}$$ $$\\int \\frac{du}{(1+u^2)^2} = \\frac{\\arctan u}{2} + \\frac{\\sin(\\arctan u)\\cos(\\arctan u)}{2}$$ $$\\frac{\\arctan u}{2} + \\frac{\\sin(\\arctan u)\\cos(\\arctan u)}{2} = \\frac{\\theta}{2} + \\frac{\\sin \\theta\\cos\\theta}{2} = \\frac{\\arctan u}{2} + \\frac{u}{2(1 + u^2)}$$ $$\\frac{\\arctan u}{2} + \\frac{u}{2(1 + u^2)} = \\frac{\\arctan (sinh x)}{2} + \\frac{\\sinh x}{2(1 + \\sinh^2 x)} = \\frac{\\arctan (\\sinh x)}{2} + \\frac{(sech x)(\\tanh x)}{2} + C$$ OK, this is very close to the final answer, but this part does not match the answer in my notes: $$\\frac{\\arctan (\\sinh x)}{2}$$ It should be: $$\\arctan (\\tanh \\frac{1}{2}x)$$ dextercioby Homework Helper They should be equal up to a numerical constant. Gold Member Is there a way to convert one into the other? I ask because i should give the answer exactly. Last edited: dextercioby", "\\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7}.$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) ### Solution 1.2 (Version 2) Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$ and $\\cos \\theta$,", "# 2021 AIME II Problems/Problem 12 ## Problem A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\\tan \\theta$ can be written in the form $\\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. ## Solution 1 (Sines and Cosines) Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of", "$\\frac{{\\rm Eq} \\ (1)}{{\\rm Eq} \\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7} .$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) Solution 2 Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$", "\\left(a \\pm \\sqrt{{a}^{2} - 1}\\right) = \\pm i \\ln \\left(a + \\sqrt{{a}^{2} - 1}\\right)$ It is reasonable that the definition of $\\arccos \\left(x\\right)$ for $x > 1$ should choose the positive sign for this expression. Hence: $\\arccos \\left(3\\right) = i \\ln \\left(3 + \\sqrt{8}\\right) = i \\ln \\left(3 + 2 \\sqrt{2}\\right)$ Then: $\\sin \\left(\\arccos \\left(3\\right)\\right)$ $= \\frac{{e}^{- \\ln \\left(3 + 2 \\sqrt{2}\\right)} - {e}^{\\ln \\left(3 + 2 \\sqrt{2}\\right)}}{2 i}$ $= \\frac{1}{2 i} \\left(\\frac{1}{3 + 2 \\sqrt{2}} - \\left(3 + 2 \\sqrt{2}\\right)\\right)$ $= \\frac{i}{2} \\left(\\left(3 + 2 \\sqrt{2}\\right) - \\left(3 - 2 \\sqrt{2}\\right)\\right)$ $= \\textcolor{b l u e}{2 \\sqrt{2} i}$ $\\textcolor{w h i t e}{}$ On the other hand, $\\csc \\left(a r c \\cot \\left(4\\right)\\right)$ is all about Real values and somewhat easier to construct. Consider a right angled triangle with sides $1$, $4$ and $\\sqrt{17}$ Let $\\theta$ be the smallest angle, i.e. the one between the hypotenuse and the longer leg. Then $\\cot \\left(\\theta\\right) = \\text{adjacent\"/\"opposite} = \\frac{4}{1} = 4$ and $\\csc \\left(\\theta\\right) = \\text{hypotenuse\"/\"opposite} = \\frac{\\sqrt{17}}{1} = \\sqrt{17}$ So: $\\csc \\left(a r c \\cot \\left(4\\right)\\right) = \\textcolor{b l u e}{\\sqrt{17}}$", "# Difference between revisions of \"2021 AIME II Problems/Problem 12\" ## Problem A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\\tan \\theta$ can be written in the form $\\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. ## Solution Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and then use those values to find $\\tan \\theta$. Let us first draw a diagram of this quadrilateral. [asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"$A$\",A,NW); abel(\"$B$\",B,NE); label(\"$C$\",C,SE); label(\"$D$\",D,SW); dot(X); label(\"$X$\",X,S); label(\"$5$\",(A+B)/2,N) label(\"$6$\",(B+C)/2,E); label(\"$9$\",(C+D)/2,S); label(\"$7$\",(D+A)/2,W); label(\"$\\theta$\",X,2.5E); label(\"$a$\",(A+X)/2,NE); label(\"$b$\",(B+X)/2,NW); label(\"$c$\",(C+X)/2,SW); label(\"$d$\",(D+X)/2,SE); [/asy] ~ my_aops_lessons", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "# 1987 AHSME Problems/Problem 14 (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) ## Problem $ABCD$ is a square and $M$ and $N$ are the midpoints of $BC$ and $CD$ respectively. Then $\\sin \\theta=$ $[asy] draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); draw((0,0)--(2,1)); draw((0,0)--(1,2)); label(\"A\", (0,0), SW); label(\"B\", (0,2), NW); label(\"C\", (2,2), NE); label(\"D\", (2,0), SE); label(\"M\", (1,2), N); label(\"N\", (2,1), E); label(\"\\theta\", (.5,.5), SW); [/asy]$ $\\textbf{(A)}\\ \\frac{\\sqrt{5}}{5} \\qquad \\textbf{(B)}\\ \\frac{3}{5} \\qquad \\textbf{(C)}\\ \\frac{\\sqrt{10}}{5} \\qquad \\textbf{(D)}\\ \\frac{4}{5}\\qquad \\textbf{(E)}\\ \\text{none of these}$ ## Solution Use the Sine Area Formula. We can isolate the triangle for which the angle $\\theta$ is contained in. WLOG, denote the side length of a triangle as $2$. Our midpoints are then $1$. Subtract the areas of the triangles that don't include the area of our desired triangle: $4-1-1-\\frac{1}{2} = \\frac{3}{2}.$ The Sine Area Formula tells us $\\frac{1}{2}(\\sqrt{5})^2\\sin\\theta=\\frac{3}{2}.$ Solving this equation, we get $\\sin\\theta=\\textbf{(B)}\\ \\frac{3}{5} \\qquad$ Solution: Everyoneintexas", "sides of a right triangle is given, we can use the … Heron’s formula i.e. Solution: length of side c (c) = NOT CALCULATED. \\\\ \\\\ The ship is anchored on the seabed. \\\\ Whether you're looking for the equation given triangle legs, leg and hypotenuse or side and angle, you won't be disappointed - this calculator has all of them implemented. A right triangle is a triangle in which one angle is a right angle. x2 + 16 = 25 8^2 + 6^2 = x^2 \\\\ Right Triangle. It is exactly a quarter of a circle. Thus, in this type of triangle, if the length of one side and the side's corresponding angle is known, the length of the other sides can be determined using the above ratio. Right Triangle. 142 + 482 = x2 When multiple angles are expanded then it will make double angles and take the sum of different angles then again apply the double angle formula. The angle formula in mathematics is given as below –. a 2 + b 2 = c 2. $$. The multiple angle’s Cosine formula is given below: $\\large Cos\\;n\\, \\theta =\\sum_{k=0}^{n}cos^{k}\\theta \\,sin^{n-k}\\theta \\;cos\\left [\\frac{1}{2}\\left(n-k\\right)\\pi\\right]$, $\\large cos^{2}\\, \\theta =cos^{2}\\, \\theta – sin^{2}\\, \\theta$, $\\large cos^{3}\\, \\theta =cos^{3}\\, \\theta – cos\\, \\theta \\; sin^{2}\\, \\theta$, $\\large Tan\\;n\\theta = \\frac{sin\\;n\\theta}{cos\\;n\\theta}$,", "$ to ensure proper formatting. 2 \\times 3 $2 \\times 3$ 2^{34} $2^{34}$ a_{i-1} $a_{i-1}$ \\frac{2}{3} $\\frac{2}{3}$ \\sqrt{2} $\\sqrt{2}$ \\sum_{i=1}^3 $\\sum_{i=1}^3$ \\sin \\theta $\\sin \\theta$ \\boxed{123} $\\boxed{123}$ Sort by: You're basically exploiting the fact that $(3k)^2 + (4k)^2 = (5k)^2$. (This is your identity multiplied by $k^2$.) From your formula, $m = T - a = T - \\dfrac{T}{5} = \\dfrac{4T}{5}$ and $n = T - 2a = T - \\dfrac{2T}{5} = \\dfrac{3T}{5}$ and therefore $m^2 + n^2 = \\dfrac{4T}{5}^2 + \\dfrac{3T}{5}^2 = \\dfrac{5T}{5}^2 = T^2$. You can do this using other Pythagorean triplets as well, for example, for some number $S$, let $b = \\dfrac{S}{13}$. and defining $k = S - b$ and $l = S - 8b$, we can see that $k^2 + l^2 = S^2$ - 5 years, 9 months ago However your way is not the only way. Im surprised you found such a specific method. The method I use is basically like an isosceles right triangle. If there is a right triangle that has two legs equal to each other then the hypotenuse is the leg length multiplied by root 2. For example if the there is a right triangle with both legs of side 3 then the hypotenuse is 3(2^0.5) which is 3 times square root of 2. So, if you have", "= \\arctan s + \\text{constant}.$$ Now draw a right triangle in which the \"opposite\" side has length $\\sqrt{u^2-1}$ and the \"adjacent\" side has length $1$. By the Pythagorean theorem, the hypotenuse has length $u$, and you see the trigonometric identity $$\\arctan\\sqrt{u^2-1} = \\arcsin\\frac{\\sqrt{u^2-1}}{u}$$ - After your work, there are several possibilities. For example, we can let $v=\\frac{1}{u}$. Then $u=\\frac{1}{v}$ and $du=-\\frac{1}{v^2}\\,dv$. Do the rest of the substituting, noting that $\\frac{1}{\\sqrt{u^2-1}}=\\frac{v}{\\sqrt{1-v^2}}$. Quickly we arrive at $$\\int -\\frac{dv}{\\sqrt{1-v^2}},$$ which is familiar. A more standard substitution is $u=\\sec\\theta$. Then $du=\\sec\\theta\\tan\\theta$, and $\\sqrt{u^2-1}=\\tan\\theta$. Everything cancels! - Let $u=1/t$, and get $$\\int \\frac{du}{u\\sqrt{u^2-1}}=-\\int \\frac{dt}{\\sqrt{1-t^2}}=-\\arcsin t +C.$$ - Michael Hardy's on the right track here, but I think the answer is incomplete. First of all, it's sloppy to leave that u in there. Your answer in terms of $x$ is $\\tan^{-1}\\sqrt{x^2+2x}$. As Michael Hardy suggested, create the right triangle, marking one angle. Let the length of the side opposite that angle be $\\sqrt{x^2+2x}$ and the adjacent side have a length of $1$. That gives the hypoteneuse a length of $x+1$. Now, you will see that the angle opposite the angle you marked has a sine of $\\frac1{x+1}$. These 2 angles add up to $\\frac\\pi2$ radians, so you have $$\\tan^{-1}\\sqrt{x^2+2x}+\\sin^{-1}\\frac1{x+1}=\\frac\\pi2$$ $$-\\sin^{-1}\\frac1{x+1}=\\tan^{-1}\\sqrt{x^2+2x}-\\frac\\pi2$$ Including the constant of integration in your answer, $\\tan^{-1}\\sqrt{x^2+2x}+C$, we see that the 2", "- 2x + 27 = 0$; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(m)(27) \\ge 0 \\Longleftrightarrow m \\le \\frac{1}{27}$. Thus, $$BP^2 \\le 405 + 729 \\cdot \\frac{1}{27} = \\boxed{432}$$ Equality holds when $x = 27$. ### Solution 2 $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$\\theta$$\",C + (-1.7,-0.2), NW); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ From the diagram, we see that $BQ = \\omega T + B\\omega\\sin\\theta = 9 + 9\\sin\\theta = 9(1 + \\sin\\theta)$, and that $QP = BA\\cos\\theta = 18\\cos\\theta$. \\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \\sin\\theta)^2 + 18^2\\cos^2\\theta\\\\ &= 9^2[1 + 2\\sin\\theta + \\sin^2\\theta + 4(1 - \\sin^2\\theta)]\\\\ BP^2 &= 9^2[5 + 2\\sin\\theta - 3\\sin^2\\theta]\\end{align*} This is a quadratic equation, maximized when $\\sin\\theta = \\frac { - 2}{ - 6} = \\frac {1}{3}$. Thus, $m^2 = 9^2[5 + \\frac {2}{3} - \\frac {1}{3}] = \\boxed{432}$. ### Solution 3 (Calculus Bash) $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) We let $AC=x$. From", "# Trigonometric Integral Confusion Gold Member ## Homework Statement So, for absolutely no good reason I decided to practice trigonometric substitution. I started with $\\int \\left(1-x^2\\right)^{-1/2} dx$, and that was easy, everything cancelled out nicely. Then I tried $\\int \\left(1+x^2\\right)^{-1/2} dx$, and although the integral was fine, everything did not cancel out nicely afterward. So I need help on one minor issue. ## Homework Equations $$sec(u) = \\frac{sec^2(u) + sec(u) tan(u)}{sec(u) + tan(u)}$$ ## The Attempt at a Solution $$\\int \\frac{dx}{\\sqrt{1+x^2}}$$ $$\\int \\frac{sec^2(u) du}{\\sqrt{1+tan^2(u)}}$$ $$\\int sec(u) du$$ $$\\int \\frac{sec^2(u) + sec(u) tan(u)}{sec(u) + tan(u)} du$$ $$\\int \\frac{dv}{v}$$ $$ln(v)$$ $$ln(tan(u) + sec(u))$$ $$ln(x + sec(tan^{-1}(x)))$$ Now, is there any way to simplify sec(arctan(x))? Let θ = arctan(x), then tan(θ) = x = x/1. Draw a right triangle, label the sides, and then find sec(θ). vela Staff Emeritus Homework Helper Draw a right triangle where the leg opposite θ is length x and the adjacent leg is length 1 so that tan θ=x/1. Gold Member Draw a right triangle where the leg opposite θ is length x and the adjacent leg is length 1 so that tan θ=x/1. Hmm... I did that, but I don't think my answer is right... the O=x, A=1, H=sqrt(1+x^2) idea gives me sec(arctan(x))=sqrt(1+x^2), and that just doesn't sound right... but it probably is. So... $$ln\\left(x+\\sqrt{1+x^2}\\right)", "# Math Help - y=arctan(x) , show that.... 1. ## y=arctan(x) , show that.... Hi I am not sure about my solution, if it´s done \"in the right way\". Anyway, here it comes: $\\mbox{ Let } y=arctan(x) \\, \\mbox{ , show that } cos^{2}(y)=\\frac{1}{1+x^{2}}$ I did: $tan(y)=x \\Longleftrightarrow \\frac{sin(y)}{cos(y)}=x \\Rightarrow cos(y)=\\frac{sin(y)}{x}$ Now raising both sides to the second power... $cos^{2}(y)=\\frac{1-cos^{2}(y)}{x^{2}} \\Longleftrightarrow cos^{2}(y)=\\frac{1}{1+x^{2}}$ Thanks! 2. Given: $y = arctan(x)$ Show that: $\\cos^{2}(y)=\\frac{1}{1+x^{2}}$ Looks right to me. But here's another way, just in case: Draw a right triangle. Label one of the acute angles y. Since $y = arctan(x)$, $\\tan y = x = \\frac{x}{1}$. Label the opposite side x and the adjacent side 1. Find the hypotenuse (s) by using the Pythagorean theorem: \\begin{aligned} s^2 &= 1^2 + x^2 \\\\ s &= \\sqrt{1 + x^2} \\end{aligned} Cosine is adjacent over hypotenuse, so $\\cos y = \\frac{1}{\\sqrt{1 + x^2}}$. Squaring both sides gives you $\\cos^2 y = \\frac{1}{1 + x^2}$. 01 3. Hello, Twig! Your work looks fine to me . . . Let: $y\\:=\\:\\arctan x$, show that: . $\\cos^2\\!y\\:=\\:\\frac{1}{1+x^2}$ Here's my approach . . . We have: . $y \\:=\\:\\arctan x\\quad\\Rightarrow\\quad \\tan y \\:=\\:x$ Square both sides: . $\\tan^2\\!y \\:=\\:x^2$ $\\text{Add 1 to both sides: }\\;\\underbrace{1 + \\tan^2\\!y}_{\\text{This is }\\sec^2\\!y} \\:=\\:1+x^2 \\quad\\Rightarrow\\quad \\sec^2\\!y \\:=\\:1+x^2$ Take reciprocals: . $\\cos^2\\!y \\:=\\:\\frac{1}{1+x^2}$", "# How do you evaluate tan (arcsin (3/4))? Apr 12, 2018 $\\tan \\left(\\arcsin \\left(\\frac{3}{4}\\right)\\right) = \\frac{3 \\sqrt{7}}{7}$ #### Explanation: Note that since $\\frac{3}{4} > 0$, the angle $\\theta = \\arcsin \\left(\\frac{3}{4}\\right)$ will be in Q1. So we can take it to be an acute angle of a right angled triangle. Remember: $\\sin \\theta = \\text{opposite\"/\"hypotenuse}$ $\\tan \\theta = \\text{opposite\"/\"adjacent}$ So consider a right angled triangle with hypotenuse $4$, one leg $3$ and the other $\\sqrt{{4}^{2} - {3}^{2}} = \\sqrt{16 - 9} = \\sqrt{7}$. This last leg is the one adjacent to $\\theta$. So: $\\tan \\left(\\arcsin \\left(\\frac{3}{4}\\right)\\right) = \\text{opposite\"/\"adjacent} = \\frac{3}{\\sqrt{7}} = \\frac{3 \\sqrt{7}}{7}$ Alternatively, using: ${\\cos}^{2} \\theta + {\\sin}^{2} \\theta = 1$ we find: $\\cos \\left(\\theta\\right) = \\pm \\sqrt{1 - {\\sin}^{2} \\theta} = \\pm \\sqrt{1 - {\\left(\\frac{3}{4}\\right)}^{2}} \\pm \\sqrt{1 - \\frac{9}{16}} = \\pm \\frac{\\sqrt{7}}{4}$ Then since $\\arcsin \\left(\\theta\\right) \\in \\left(- \\frac{\\pi}{2} , \\frac{\\pi}{2}\\right]$, where $\\cos > 0$, we require the $+$ sign and find: $\\cos \\left(\\theta\\right) = \\frac{\\sqrt{7}}{4}$ Then: $\\tan \\left(\\arcsin \\left(\\frac{3}{4}\\right)\\right) = \\tan \\left(\\theta\\right) = \\sin \\frac{\\theta}{\\cos} \\left(\\theta\\right) = \\frac{\\frac{3}{4}}{\\frac{\\sqrt{7}}{4}} = \\frac{3}{\\sqrt{7}} = \\frac{3 \\sqrt{7}}{7}$", "hypotenuse of triangle $ABC$. Notice that $h - \\frac {1}{21}h = \\sqrt {440}$. (The height of $ABC$ decreased by the corresponding height of $T_5$) Thus, $(AB)(h) = (AC)(BC) = 22\\cdot 21^2$. Because $AC^2 + BC^2 + 2(AC)(BC) = (AC + BC)^2 = 21^2\\cdot22^2$, $AC + BC = (21)(22) = \\boxed{462}$. ## Easy Trig Solution Let $\\tan\\angle ABC = x$. Now using the 1st square, $AC=21(1+x)$ and $CB=21(1+x^{-1})$. Using the second square, $AB=\\sqrt{440}(1+x+x^{-1})$. We have $AC^2+CB^2=AB^2$, or $$441(x^2+x^{-2}+2x+2x^{-1}+2)=440(x^2+x^{-2}+2x+2x^{-1}+3).$$ Rearranging and letting $u=x+x^{-1} \\Rightarrow u^2 - 2 = x^2 + x^{-2}$ gives us $u^2+2u-440=0.$ We take the positive root, so $u=20$, which means $AC+CB=21(2+x+x^{-1})=21(2+u)=\\boxed{462}$. ## Messy Trig Solution Let $\\theta$ be the smaller angle in the triangle. Then the sum of shorter and longer leg is $\\sqrt{441}(2+\\tan{\\theta}+\\cot{\\theta})$. We observe that the short leg has length $\\sqrt{441}(1+\\tan{\\theta}) = \\sqrt{440}(\\sec{\\theta}+\\sin{\\theta})$. Grouping and squaring, we get $\\sqrt{\\frac{440}{441}} = \\frac{\\sin{\\theta}+\\cos{\\theta}}{1+\\sin{\\theta}\\cos{\\theta}}$. Squaring and using the double angle identity for sine, we get, $110(\\sin{2\\theta})^2 + \\sin{2\\theta} - 1 = 0$. Solving, we get $\\sin{2\\theta} = \\frac{1}{10}$. Now to find $\\tan{\\theta}$, we find $\\cos{2\\theta}$ using the Pythagorean Identity, and then use the tangent double angle identity. Thus, $\\tan{\\theta} = 10-3\\sqrt{11}$. Substituting into the original sum, we get $\\boxed{462}$.", "of $\\sqrt{x^2-1}$, which means the hypotenuse is $x$ and the other leg is $1$. By choosing $u=\\sec(\\theta)$ you make $\\theta$ adjacent to the leg of length $1$ and opposite the leg of length $\\sqrt{x^2-1}$. Thus $\\tan(\\theta)=\\sqrt{x^2-1}$. Alternately you could have figured that out by simply using the identity $\\sec^2(\\theta)=\\tan^2(\\theta)+1$. • was there any way to avoid using a triangle? – KrissyMichaelsson May 2 '16 at 15:20 • @KrissyMichaelsson Yes, see my last sentence. Still, I personally like using a triangle a lot, even if I only just imagine it rather than actually drawing it. – Ian May 2 '16 at 15:20 • Ohhh I see my mistake now....if x=secΘ, I should said x=sqrt(tan^2+1, and then x^2=tan^2+1, x^2-1=tan^2, then sqrt(x^2-1)=tan. Thanks. I was in a test, so it was def a rush, since I knew triangles are the best way to do it, but I forgot how to do that. – KrissyMichaelsson May 2 '16 at 15:32 How about this: \\begin{align} u & = \\sqrt{x^2-1} \\\\ u^2 & = x^2 - 1 \\\\ u^2+1 & = x^2 \\\\ 2u\\,du & = 2x\\,dx \\\\ \\end{align} \\begin{align} \\int x^3 \\sqrt{x^2-1} \\, dx & = \\int x^2 \\sqrt{x^2-1} \\Big( x\\,dx\\Big) \\\\ & = \\int (u^2+1) u \\left( \\frac 1 2 \\, du \\right) \\\\ & = \\frac 1 2 \\int (u^4 +" ]

[ "# Tag Info 7 As you know $$\\matrix{ {0 \\le \\sin (x) \\le 1,} & {0 \\le x \\le \\pi } \\cr }$$ and hence $$0 \\le \\sin(x) + 2\\sin(2x) + 3\\sin(3x) + 4\\sin(4x) \\le 10$$ and we can have your equality only when all of the $\\sin$'s are equal to $1$, i.e, $$\\matrix{ {\\sin (x) = 1} \\hfill & \\to \\hfill & {x = {\\pi \\over 2}} \\hfill \\cr {\\sin (2x) = ... 5 Notice that we can write$$ \\frac{2n}{n^4 - n^2 + 1} = \\frac{(n^2+n) - (n^2-n)}{1 + (n^2+n)(n^2-n)}. $$In view of the addition formula for the tangent, we find that$$ \\arctan \\bigg( \\frac{x-y}{1+xy} \\bigg) = \\arctan x - \\arctan y $$for any x > y > 0. Thus we have$$ \\arctan\\bigg( \\frac{2n}{n^4 - n^2 + 1} \\bigg) = \\arctan(n^2+n) - ... 5 The statemnet $\\sqrt{\\cos^2(x)}=|\\cos(x)|$ is true for all real $x$. Wolfram|Alpha, on the other hand, has the often annoying habit of considering all complex numbers by default. So, it's not willing to apply the identity $$\\sqrt{x^2}=|x|$$ unless it knows that $x$ is real. Given that $\\cos(x)$ is real for all real $x$, the identity you cite is true for any ... 4 $\\sin \\theta + \\cos \\theta = \\sqrt 2 \\sin (\\theta + \\frac{\\pi}{4})$ Sine is positive in the first and second quadrants. So", "line, which co-ordinate will you find ? Because, in the second quadrant, where the line cuts the circle, the horizontal co-ordinate is negative. For the angle you worked with (in the 4th quadrant), the horizontal co-ordinate is positive. The one you quoted is correct, the other is the negative of that. I'm not sure what you mean by \"your teacher wants it in exact value\". Is it surd form? In that case, use the surd forms for $cos\\left(\\frac{{\\pi}}{4}\\right),\\ -cos\\left(\\frac{{\\pi}}{4}\\right)$ 3. Originally Posted by DarkAngelxXxXx so i know that you start with the arctan (-1) .. and for that you just do inverse (tan (-1)) right? i entered that in to my calcultor and got -.7853981634 and when i did cos(ans) i got .7071067812 but my teacher wants it in exact value? how do i get that? you are looking for $\\frac{\\sqrt{2}}{2}$ 4. Here's a slightly different and more general method: Let $\\theta= arctan(x)$. Then $x= tan(\\theta)$. Represent that as a right triangle having one angle of measure $\\theta$, opposite side of length x, and near side of length 1 so that $tan(\\theta)= x/1$= \"opposite side/near side\". By the Pythagorean theorem the hypotenuse has length $\\sqrt{x^2+ 1}$ and so $cos(\\theta)= cos(arctan(x))$ is \"near side over hypotenuse\" or $cos(arctan(x))= \\frac{1}{\\sqrt{x^2+ 1}}$. When x= -1, that is, of course, $\\frac{1}{\\sqrt{2}}= \\frac{\\sqrt{2}}{2}$.", "we have $$\\theta = \\arcsin(x)$$. Taking the sine of both sides yields $$\\sin(\\theta) = x$$. We use this equation to set up the triangle to the right, $$a = x$$ and $$c = 1$$. Now we use the Pythagorean Theorem to solve for $$b$$. $$\\begin{array}{rcl} x^2 + b^2 & = & 1 \\\\ b^2 & = & 1 - x^2 \\\\ b & = & \\pm \\sqrt{1-x^2} \\end{array}$$ We choose the positive square root. Now we can write $$\\cos(\\theta) = \\sqrt{1-x^2}$$. Final Answer $$\\cos( \\arcsin(x))$$ $$= \\sqrt{1-x^2}$$ Log in to rate this practice problem and to see it's current rating. $$\\sec( \\arctan(x/3))$$ Problem Statement Evaluate (simplify) $$\\sec( \\arctan(x/3))$$ without using a calculator and give your answer in exact, simplified form. Final Answer $$\\sec( \\arctan(x/3))$$ $$\\displaystyle{ = \\frac{\\sqrt{x^2+9}}{3}}$$ Problem Statement Evaluate (simplify) $$\\sec( \\arctan(x/3))$$ without using a calculator and give your answer in exact, simplified form. Solution We start on the inside by letting $$\\theta = \\arctan(x/3)$$. Take the tangent of both sides to get $$\\tan(\\theta) = x/3$$. We use this equation to set up the triangle to the right, $$a = x, ~~ b = 3$$. Using the Pythagorean Theorem, we can find $$c$$, i.e. $$c^2 = x^2 + 3^2 ~~ \\to ~~ c = \\pm \\sqrt{x^2 + 9}$$. We choose the positive square root. Now we have", "is $\\dfrac{99}{99^2}$, or $\\dfrac{1}{99}$, and our answer is $1 + 99 = \\boxed{100}$. ## Solution 2 Without the loss of generality one can let $z$ lie on the positive x axis and since $arg(\\theta)$ is a measure of the angle if $z=10$ then $arg(\\dfrac{w-z}{z})=arg(w-z)$ and we can see that the question is equivalent to having a triangle $OAB$ with sides $OA =10$ $AB=1$ and $OB=t$ and trying to maximize the angle $BOA$ $[asy] pair O = (0,0); pair A = (100,0); pair B = (80,30); pair D = (sqrt(850),sqrt(850)); draw(A--B--O--cycle); dotfactor = 3; dot(\"A\",A,dir(45)); dot(\"B\",B,dir(45)); dot(\"O\",O,dir(135)); dot(\" \\theta\",O,(7,1.2)); label(\"1\", ( A--B )); label(\"10\",(O--A)); label(\"t\",(O--B)); [/asy]$ using the Law of Cosines we get: $1^2=10^2+t^2-t*10*2\\cos\\theta$ rearranging: $$20t\\cos\\theta=t^2+99$$ solving for $\\cos\\theta$ we get: $$\\frac{99}{20t}+\\frac{t}{20}=\\cos\\theta$$ if we want to maximize $\\theta$ we need to minimize $\\cos\\theta$ , using AM-GM inequality we get that the minimum value for $\\cos\\theta= 2(\\sqrt{\\dfrac{99}{20t}\\dfrac{t}{20}})=2\\sqrt{\\dfrac{99}{400}}=\\dfrac{\\sqrt{99}}{10}$ hence using the identity $\\tan^2\\theta=\\sec^2\\theta-1$ we get $\\tan^2\\theta=\\frac{1}{99}$and our answer is $1 + 99 = \\boxed{100}$. ## Solution 3 Note that $\\frac{w-z}{z}=\\frac{w}{z}-1$, and that $\\left|\\frac{w}{z}\\right|=\\frac{1}{10}$. Thus $\\frac{w}{z}-1$ is a complex number on the circle with radius $\\frac{1}{10}$ and centered at $-1$ on the complex plane. Let $\\omega$ denote this circle. Let $A$ and $C$ be the points that represent $\\frac{w}{z}-1$ and $-1$ respectively on the complex plane. Let $O$ be the origin. In", "# Prove sqrt(a^2+b^2)e^(iarctan(b/a))=a+bi? ## $\\sqrt{{a}^{2} + {b}^{2}} {e}^{i \\arctan \\left(\\frac{b}{a}\\right)} = a + b i$ Feb 22, 2018 In Explanation #### Explanation: On a normal coordinate plane, we have coordinate like (1,2) and (3,4) and stuff like that. We can reexpress these coordinates n terms of radii and angles. So if we have the point (a,b) that means we go units to the right, b units up and $\\sqrt{{a}^{2} + {b}^{2}}$ as the distance between the origin and the point (a,b). I will call $\\sqrt{{a}^{2} + {b}^{2}} = r$ So we have $r {e}^{\\arctan} \\left(\\frac{b}{a}\\right)$ Now to finish this proof off let's recall a formula. ${e}^{i \\theta} = \\cos \\left(\\theta\\right) + i \\sin \\left(\\theta\\right)$ The function of arc tan gives me an angle which is also theta. So we have the following equation: ${e}^{i} \\cdot \\arctan \\left(\\frac{b}{a}\\right) = \\cos \\left(\\arctan \\left(\\frac{b}{a}\\right)\\right) + \\sin \\left(\\arctan \\left(\\frac{b}{a}\\right)\\right)$ Now lets draw a right triangle. The arctan of (b/a) tells me that b is the opposite side and a is the adjacent side. So if I want the cos of the arctan(b/a), we use the Pythagorean theorem to find the hypotenuse. The hypotenuse is $\\sqrt{{a}^{2} + {b}^{2}}$. So the cos(arctan(b/a)) = adjacent over hypotenuse = $\\frac{a}{\\sqrt{{a}^{2} + {b}^{2}}}$. The best part about this is the fact that this same principle applies to", "\\sqrt{2}\\right)^{c_1}-\\left(3+2 \\sqrt{2}\\right)^{c_1}-\\sqrt{2} \\left(3-2 \\sqrt{2}\\right)^{c_1}-\\left(3-2 \\sqrt{2}\\right)^{c_1}-2\\right)$$ where $c_1$ is ... 2 Yes. Let $x$ be the solution to $x+\\arctan(x) = \\pi$, then $$\\arctan(x)=\\pi-x \\\\ \\Rightarrow x=tan(\\pi-x) \\\\ \\Rightarrow x=-tan(x).$$ Thus if $x$ would be rational, also $tan(x)$ would be rational. This is impossible: You can use the statement you gave for showing that $x+arctan(x)=1$ is irrational. Here is another reference: Prove that if $x$ is a ... 2 Another way is to specialize the ascent in the ternary tree of Pythagorean triples (see below). Specializing $\\,z = y+3\\,$ in the $\\rm\\color{#c00}{\\,formula\\,}$below yields the following generation rule $$x^2 + y^2 = (y+3)^2\\ \\Rightarrow\\ (6-x)^2 + Y^2 = (Y+3)^2,\\quad Y = y-2x+6$$ This yields $\\ (-3,0,3)\\to (-9, 12, 15)\\to (-15, 36, 39)\\to (-21, 72, ... 2 The generalized pentagonal numbers begin, in order,$1, 2, 5, 7, 12, 15, ...$. The recurrence for the partition number is $$p(n) = p(n-1) + p(n-2) -p(n-5) -p(n-7) + p(n-12)+p(n-15) - \\cdots$$ with the sign alternating in pairs. For any given$n$, only a finite number of terms are non-zero, since$p(m)=0$if$m<0$. For example,$$p(2) = ... 2 For your follow-up question, only linear polynomials don't work : Suppose$G$is a subgroup of$S_n$such that$G$acts transitively on$\\{1\\ldots n\\}$, and let$H_i^j = \\{\\sigma \\in G \\mid \\sigma(i)=j \\}$. If$\\tau(i)=j$then$H_i^k = H_j^k \\tau$, and$H_k^j = \\tau H_k^i$. Since$G$is transitive, every$H_i^j$has the same cardinal.", "There are infinitely many, but a good choice is the right triangle with opposite leg $$x$$ and adjacent leg $$1$$. Then compute the hypotenuse to be $$\\sqrt{x^2+1}.$$ Now it's easy to find $$\\sin(\\arctan x) = \\sin \\theta = x/\\sqrt{x^2+1}.$$ For something like $$\\sin(\\arctan 1/2+\\arccos 1/3)$$ let $$\\arctan 1/2 =\\alpha$$ and $$\\arccos 1/3 =\\beta$$. Then $$\\sin(\\arctan 1/2+\\arccos 1/3)$$ $$= \\sin(\\alpha+\\beta) = \\sin\\alpha\\cos\\beta +\\cos\\alpha\\sin\\beta.$$ Then proceed to evaluate each of $$\\sin \\alpha,\\sin \\beta$$ etc. • Would you look at this example? math.stackexchange.com/questions/3084939/… I've used your method, but somehow I'm getting slightly wrong result. :( – weno Jan 23 at 19:42 • You need to think about what quadrant you're in. In the first example, you should draw the triangle in the 4th quadrant so that the $3$ is really $-3$. In the second example, the triangle is in the 2nd quadrant, so the $1$ is really $-1$, which will change the sign of your answer. – B. Goddard Jan 23 at 20:06 • Thanks. (10 char) – weno Jan 23 at 20:14 If $$s=\\sin(\\arctan 2)$$ and $$c=\\cos(\\arctan 2)$$, then $$c^2+s^2=1$$ and$$\\frac sc=\\tan(\\arctan 2)=2.$$So, solve the system$$\\left\\{\\begin{array}{l}c^2+s^2=1\\\\\\dfrac sc=2\\end{array}\\right.$$and don't forget the only positive solutions matter here. • Hm but where the assumptions about the $c$ came from? There's nothing about $cos(arctan2)$ in the task. Or do I need to do it", "of them the ambiguous sign is the same as that of $$a\\cos\\{\\arcsin(x/a)\\}$$, since $a\\sqrt{1 – (x^{2}/a^{2})} = \\pm\\sqrt{a^{2} – x^{2}}$ according as $$a$$ is positive or negative. Finally, by means of Theorem (6) of § 113, we are enabled to differentiate composite functions involving symbols both of algebraical and trigonometrical functionality, and so to write down the derivative of any such function as occurs in the following examples. Example XLIV 1. Find the derivatives of $\\begin{gathered} \\cos^{m} x, \\quad \\sin^{m} x, \\quad \\cos x^{m}, \\quad \\sin x^{m}, \\quad \\cos (\\sin x), \\quad \\sin (\\cos x),\\\\ \\sqrt{a^{2}\\cos^{2} x + b^{2}\\sin^{2} x}, \\quad \\frac{\\cos x\\sin x}{\\sqrt{a^{2}\\cos^{2} x + b^{2}\\sin^{2} x}},\\\\ x\\arcsin x + \\sqrt{1 – x^{2}}, \\quad (1 + x)\\arctan\\sqrt{x} – \\sqrt{x}.\\end{gathered}$ 2. Verify by differentiation that $$\\arcsin x + \\arccos x$$ is constant for all values of $$x$$ between $$0$$ and $$1$$, and $$\\arctan x + \\operatorname{arccot} x$$ for all positive values of $$x$$. 3. Find the derivatives of $\\arcsin\\sqrt{1 – x^{2}},\\quad \\arcsin\\{2x\\sqrt{1 – x^{2}}\\},\\quad \\arctan \\left(\\frac{a + x}{1 – ax}\\right).$ How do you explain the simplicity of the results? 4. Differentiate $\\frac{1}{\\sqrt{ac – b^{2}}} \\arctan \\frac{ax + b}{\\sqrt{ac – b^{2}}},\\quad -\\frac{1}{\\sqrt{-a}} \\arcsin\\frac{ax + b}{\\sqrt{b^{2} – ac}}.$ 5. Show that each of the functions $2\\arcsin \\sqrt{\\frac{x – \\beta}{\\alpha – \\beta}},\\quad 2\\arctan \\sqrt{\\frac{x – \\beta}{\\alpha – x}},\\quad \\arcsin \\frac{2\\sqrt{(\\alpha –", "we have $\\cos \\theta = \\frac{3}{4}$. Now, since $\\triangle BOC$ is isosceles with $OB = OC$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. By SSS congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ### Solution 4 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we have that $AB = AE$. Similarly, $CD = DF$. $OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD =", "# How to know if the angle is positive or negative in inverse trigonometry I have been doing this problem this problem: $$Cos[Tan^{-1}(-\\frac{2}{3})]$$ So I was instructed to draw a triangle to guide me so I did Now once I drew my triangle I found the hypotenuse, which is $$\\sqrt{13}$$ And then I was able to obtain the answer to this expression which I got: $$\\frac{2\\sqrt{13}}{13}$$ However, I am told I drew the triangle wrong, it is actually -2 (negative) and (3) positive. Why is it that the triangle is wrong? I was told the actual answer is $$\\frac{3\\sqrt{13}}{13}$$ Arctan has a range of $\\frac{-\\pi}{2}\\le{y}\\le\\frac{\\pi}{2}$. Now let $arctan\\frac{-2}{3}=y$. This implies $tany=\\frac{-2}{3}$. Because tan is negative, we know y must lie in quadrant IV. It cannot lie in quadrant I because tan is positive in quadrant I. Therefore we draw our angle as you have above. Now, note that $tan\\theta=\\frac{opposite}{adjacent}$. Therefore, you should have $-2$ where $3$ is in your picture, and you should have $3$ where $-2$ is. Your line should be drawn to the coordinate $(3,-2)$. Now, we find the hypotenuse as you have already done using the Pythagorean Theorem. We find that it is $\\sqrt{13}$ as you've noted. Now, because we are finding $cos(arctan\\frac{-2}{3})$, we will use the fact that $cos\\theta=\\frac{adjacent}{hypotenuse}$. So, this gives $\\frac{3}{\\sqrt{13}}$. Rationalizing", "\\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7}.$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) ### Solution 1.2 (Version 2) Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$ and $\\cos \\theta$,", "$\\left(0,2\\sqrt{84}\\right)$ or $\\left(0,20\\right).$ By the exclusive disjunction, the set of all such $s$ is $$[a,b)=\\left(0,2\\sqrt{84}\\right)\\oplus\\left(0,20\\right)=\\left[2\\sqrt{84},20\\right),$$ from which $a^2+b^2=\\boxed{736}.$ ~MRENTHUSIASM ## Solution 3 We have the diagram below. $[asy] draw((0,0)--(1,2*sqrt(3))); draw((1,2*sqrt(3))--(10,0)); draw((10,0)--(0,0)); label(\"A\",(0,0),SW); label(\"B\",(1,2*sqrt(3)),N); label(\"C\",(10,0),SE); label(\"\\theta\",(0,0),NE); label(\"\\alpha\",(1,2*sqrt(3)),SSE); label(\"4\",(0,0)--(1,2*sqrt(3)),WNW); label(\"10\",(0,0)--(10,0),S); [/asy]$ We proceed by taking cases on the angles that can be obtuse, and finding the ranges for $s$ that they yield . If angle $\\theta$ is obtuse, then we have that $s \\in (0,20)$. This is because $s=20$ is attained at $\\theta = 90^{\\circ}$, and the area of the triangle is strictly decreasing as $\\theta$ increases beyond $90^{\\circ}$. This can be observed from $$s=\\frac{1}{2}(4)(10)\\sin\\theta$$by noting that $\\sin\\theta$ is decreasing in $\\theta \\in (90^{\\circ},180^{\\circ})$. Then, we note that if $\\alpha$ is obtuse, we have $s \\in (0,4\\sqrt{21})$. This is because we get $x=\\sqrt{10^2-4^2}=\\sqrt{84}=2\\sqrt{21}$ when $\\alpha=90^{\\circ}$, yileding $s=4\\sqrt{21}$. Then, $s$ is decreasing as $\\alpha$ increases by the same argument as before. $\\angle{ACB}$ cannot be obtuse since $AC>AB$. Now we have the intervals $s \\in (0,20)$ and $s \\in (0,4\\sqrt{21})$ for the cases where $\\theta$ and $\\alpha$ are obtuse, respectively. We are looking for the $s$ that are in exactly one of these intervals, and because $4\\sqrt{21}<20$, the desired range is $$s\\in [4\\sqrt{21},20)$$giving $$a^2+b^2=\\boxed{736}\\Box$$ ## Solution 4 Note: Archimedes15 Solution which I added an answer here are two cases. Either the $4$", "1. ## minimum positive value. If $a,b$ are positive quantities and $a\\geq b$.Then find Minimum positive value of $asec\\theta-btan\\theta$ 2. $\\displaystyle a \\sec \\theta \\tan \\theta - b \\sec^{2} \\theta = 0$ $\\displaystyle \\frac{a}{b} = \\frac{\\sec \\theta}{\\tan \\theta}$ $\\displaystyle \\frac{a}{b} = \\frac{1}{\\sin \\theta}$ $\\displaystyle \\frac{b}{a} = \\sin \\theta$ $\\displaystyle \\theta = \\arcsin \\frac {b}{a} + 2 \\pi n, \\ n = 0, \\ \\pm 1, \\pm2, ...$ 3. but ans is in terms of $a$ and $b$ i.e $asec\\theta-btan\\theta \\geq \\sqrt{a^2-b^2}$ 4. Hello, jacks! but ans is in terms of $\\,a$ and $\\,b$ $a\\sec\\theta-b\\tan\\theta \\:\\geq\\: \\sqrt{a^2-b^2}$ Random Variable did the hard work for you . . . $\\text{We have: }\\:f(\\theta) \\:=\\:a\\sec\\theta - b\\tan\\theta$ $\\text{He found that }\\,f'(\\theta) = 0\\,\\text{ when }\\,\\sin\\theta \\,=\\,\\dfrac{b}{a}$ . . $\\text{Hence: }\\;\\sec\\theta \\,=\\,\\dfrac{a}{\\sqrt{a^2-b^2}},\\;\\;\\tan\\theta \\,=\\,\\dfrac{b}{\\sqrt{a^2-b^2}}$ Substitute into the function: . . $\\displaystyle a\\sec\\theta - b\\tan\\theta \\;\\ge\\;a\\frac{a}{\\sqrt{a^2-b^2}} - b\\frac{b}{\\sqrt{a^2-b^2}}$ . . . . . . . . . . . . . . $\\displaystyle =\\;\\frac{a^2}{\\sqrt{a^2-b^2}} - \\frac{b^2}{\\sqrt{a^2-b^2}}$ . . . . . . . . . . . . . . $\\displaystyle =\\;\\frac{a^2-b^2}{\\sqrt{a^2-b^2}} \\;=\\;\\sqrt{a^2-b^2}$ $\\text{Therefore: }\\;a\\sec\\theta - b\\tan\\theta \\;\\ge \\;\\sqrt{a^2+b^2}$ 5. Thanks Random variable and Sorobon.", "# $A \\cos (x) + B \\sin(x)$ with $A,B \\in \\mathbb{R}$ but not necessarily positive In this question, it is shown that $$A \\sin(x) + B \\cos (x) = R \\sin ( x + \\theta )$$ for $$A,B$$ real and positive, with $$R = \\sqrt{A^2 + B^2}$$ $$\\theta = \\arctan \\left( \\frac{B}{A} \\right)$$ This is performed solving the system $$\\left\\{ \\begin{array}{c} R \\cos (\\theta) = A\\\\ R \\sin (\\theta) = B \\end{array} \\right.$$ for $$R$$ and $$\\theta$$. What if, instead, $$A$$ and/or $$B$$ are real, but not positive? Some trivial test on WolframAlpha shows that the $$\\arctan \\left( \\frac{B}{A} \\right)$$ term is computed the same way, considering the sign of $$A$$ and $$B$$. $$R$$ is instead negative if $$A$$ or both $$A$$ and $$B$$ are negative. It is obtained by squaring and summing the two equations in the system and I can't figure out how a sign can be considered in this operation. Let $$A = R \\cos \\theta$$ and $$B = R \\sin \\theta$$ So, $$y = A \\sin x + B \\cos x = R \\cos \\theta \\sin x+ R \\sin \\theta \\cos x = R \\sin(x + \\theta)$$ Also, $$R^2 = A^2 + B^2$$ or $$R = \\sqrt{A^2 + B^2}$$ Let $$\\alpha = \\tan^{-1} \\left| \\frac{B}{A} \\right|$$ so that $$\\alpha \\in [0,\\pi/2]$$ Let us assume", "# Identity 2 This is prompted by another Catriona Shearer puzzle. Would you say it’s at all obvious that $2 \\arctan (\\sqrt{3}/7) = \\arccos (23/26)$ ? Me neither. True, though. Look at this triangle: By the law of cosines, $3 = 13 + 13 - 2\\times13\\cos(\\theta)$ so $\\theta = \\arccos (23/26)$. But if you slice that triangle in half, you find its height is $7/2$ and so $\\theta = 2\\arctan(\\sqrt{3}/7)$. More generally, the half angle formula for the tangent is $\\tan(\\theta/2) = \\frac{\\sqrt{1-\\cos\\theta}}{\\sqrt{1+\\cos\\theta}}$ implying $\\cos\\theta = \\frac{1-\\tan^2(\\theta/2)}{1+\\tan^2(\\theta/2)}$. Then if $\\theta/2 = \\arctan(a/b)$, $2\\arctan\\left(\\frac{a}{b}\\right) = \\arccos\\left(\\frac{b^2-a^2}{b^2+a^2}\\right)$ and with $a=\\sqrt{3}$, $b=7$ you get $2\\arctan(\\sqrt{3}/7) = \\arccos(46/52) = \\arccos(23/26)$.", "# Math Help - identity 1. ## identity Show that $\\displaystyle \\arccos \\Big(\\frac{\\cos x}{1+ 2 \\cos x } \\Big) = \\arctan \\sqrt{\\frac{1 + \\cos x}{1 + 3 \\cos x}}$ . 2. This identity is not true, I don't think. Let $x=\\pi/4.$ Then the LHS is approximately $1.27,$ and the RHS is approximately $0.64.$ 3. My mistake. It should be $\\displaystyle \\arccos \\Big(\\frac{\\cos x}{1+ 2 \\cos x } \\Big) = 2 \\arctan \\sqrt{\\frac{1 + \\cos x}{1 + 3 \\cos x}}$ 4. The identity being used is $\\displaystyle \\arccos x = 2 \\arctan \\sqrt{\\frac{1-x}{1+x}}$. But from where does that identity come? 5. Originally Posted by Random Variable The identity being used is $\\displaystyle \\arccos x = 2 \\arctan \\sqrt{\\frac{1-x}{1+x}}$. But from where does that identity come? Hang on. This is going to be a ride... Take the cosine of both sides: $\\displaystyle cos(acs x) = x$ So $\\displaystyle cos \\left ( 2 ~ atn \\left ( \\sqrt{\\frac{1-x}{1+x}} \\right ) \\right )$ must be equal to x. To check this. let $\\theta = atn \\left ( \\sqrt{\\frac{1-x}{1+x}} \\right ) \\right )$ So we have $\\displaystyle cos \\left ( 2 ~ atn \\left ( \\sqrt{\\frac{1-x}{1+x}} \\right ) \\right ) = cos( 2 \\theta) = cos^2(\\theta) - sin^2(\\theta)$ Now sketch a right triangle. Label one angle $\\theta$. The leg of the triangle across", "must have the opposite sign as both $x$ and $x+1$ giving the desired positive $\\frac{\\pi}{2}$. We also see that for $x\\in(-1,0)$, the constraint requires $\\varepsilon$ to be positive but $x+1$ will positive as well which will lead to $-\\frac{\\pi}{2}$ which is indeed what you see in a plot. As an addendum, here's a geometric/trigonometric argument. First note that the domain constraints ensure that $x$ and $x+1$ always have the same sign, so the arguments to $\\arctan$ are always positive. For $a$ and $b$ both positive, $\\theta = \\arctan(\\frac{b}{a})$ is in the interval $[0,\\frac{\\pi}{2})$. We can view $a$ and $b$ as, respectively, the $x$ and $y$ coordinates of a point. $\\theta$ is then the angle between the $x$-axis and the ray from the origin through the point $(a, b)$. If we then consider $\\varphi = \\arctan(\\frac{a}{b})$, we can use the same logic to view it as the angle between the $x$-axis and the ray through $(b,a)$. But $(b,a)$ is the reflection of $(a,b)$ through the line $y=x$, so the angle between the $y$-axis (note, $y$-axis, not $x$-axis) and the ray through $(b,a)$ is $\\theta$, therefore $\\varphi = \\frac{\\pi}{2}-\\theta$. You can use vector algebra, complex exponentials, or trigonometric identities if you want a more symbolic rendition of this. Why does this post require moderator attention? I wish I could upvote", "maximum value of $\\mathrm{tan^2}{\\theta}$ is $\\dfrac{99}{99^2}$, or $\\dfrac{1}{99}$, and our answer is $1 + 99 = \\boxed{100}$. ## Solution 2 (No calculus) Without the loss of generality one can let $z$ lie on the positive x axis and since $arg(\\theta)$ is a measure of the angle if $z=10$ then $arg(\\dfrac{w-z}{z})=arg(w-z)$ and we can see that the question is equivelent to having a triangle $OAB$ with sides $OA =10$ $AB=1$ and $OB=t$ and trying to maximize the angle $BOA$ $[asy] pair O = (0,0); pair A = (100,0); pair B = (80,30); pair D = (sqrt(850),sqrt(850)); draw(A--B--O--cycle); dotfactor = 3; dot(\"A\",A,dir(45)); dot(\"B\",B,dir(45)); dot(\"O\",O,dir(135)); dot(\" \\theta\",O,dir(30)); label(\"1\", ( A--B )); label(\"10\",(O--A)); label(\"t\",(O--B)); [/asy]$ using the law of cosines we get: $1^2=10^2+t^2-t*10*2\\cos\\theta$ rearranging: $$20t\\cos\\theta=t^2+99$$ solving for $\\cos\\theta$ we get: $$\\frac{99}{20t}+\\frac{t}{20}=\\cos\\theta$$ if we want to maximize $\\theta$ we need to minimize $\\cos\\theta$ , using AM-GM inequality we get that the minimum value for $\\cos\\theta= 2(\\sqrt{\\dfrac{99}{20t}\\dfrac{t}{20}})=2\\sqrt{\\dfrac{99}{400}}=\\dfrac{\\sqrt{99}}{10}$ hence using the identity $\\tan^2\\theta=\\sec^2\\theta-1$ we get $\\tan^2\\theta=\\frac{1}{99}$and our answer is $1 + 99 = \\boxed{100}$.", "\\theta$$ $$= 1-2sin^2\\theta$$ $$cos (2 tan^{-1}(x))= 1-2[sin(tan^{-1}(x))]^2$$ $$sin(tan^{-1}(x))$$ $$y = tan^{-1}x$$ tan(y) = x Now drawing the triangle we have the adjacent and opposite sides now I can work out the hypothenuse; $$sin(y) = \\frac{x}{\\sqrt{x^2 +1}}$$ $$sin(y) = sin(sin^{-1} \\frac{x}{\\sqrt{x^2 +1}}$$ $$= \\frac{x}{\\sqrt{x^2 +1}}$$ pluging this back into the equation above; $$cos (2 tan^{-1}(x))= 1-2[\\frac{x}{\\sqrt{x^2 +1}}]^2$$ I need some help now because this looks very messy! I'm not sure if it's right though! 9. Aug 7, 2008 ### snipez90 Hi powerless! You're learning quickly, good job :). Anyways I've skimmed over the work and it looks good. Let's just clear up the notation a bit. Hats off to you for noticing the cosine double angle formula. Now instead of using y over and over, let's be a bit more clear. Let $$arctan(x) = \\theta \\Rightarrow$$$$\\ tan(\\theta) = x$$ where arctan is the same as your inverse tangent function. But I just used arctan because I couldn't find the inverse tan :P. However, the great thing about using this notation is that it's all you really need to define. After you've defined the above, the problem reduces to finding cos(2*theta) and therefore finding sin(arctan(x)). Then you proceed with the triangle and as you can see, our definition provides that info for us on the right hand side of the", "be of length one. Then by Pythagoras Theorem, the hypothenus side is $\\sqrt{10}$. Hence you can compute $\\sin(x)$ and $\\cos(x)$ and multiply them. Since $\\tan x=3$, $\\sin x \\cos x = \\frac{\\tan x}{1+\\tan^2 x}$ takes the claimed value. Observe that $$\\tan (x)=3.$$ But $$1+\\tan^2x=\\sec^2x\\implies10=\\frac{1}{\\cos^2x}$$ and $$1+\\cot^2x=\\csc^2x \\implies\\frac{10}{9}=\\frac{1}{\\sin^2x}$$ Since tangent is positive, both sine and cosine have the same sign. In particular $$\\sin x\\cos x=\\frac{3}{\\sqrt{10}}\\frac{1}{\\sqrt{10}}=\\frac{3}{10}.$$ \\begin{align} \\\\ \\sin (2x)&=2\\sin (x)\\cos (x) \\\\ &=\\frac {2\\tan (x)}{1+\\tan^2 (x)} \\\\ &=\\frac {2\\cdot3}{1+(3)^2} \\\\ &=\\frac {6}{10} \\\\ \\end{align} thus $$\\sin (x)\\cos (x)=\\frac {3}{10}$$ We have the identity $1 + \\tan^2 \\theta = \\sec^2 \\theta = \\dfrac{1}{\\cos^2 \\theta}; \\tag 1$ so with $\\sin x = 3\\cos x, \\tag 2$ we infer $\\tan x = \\dfrac{\\sin x}{\\cos x} = 3, \\tag 3$ whence $\\dfrac{1}{\\cos^2 x} = 1 + \\tan^2 x = 1 + 3^2 = 10, \\tag 4$ yielding $\\cos x = \\pm \\dfrac{1}{\\sqrt{10}}; \\tag 5$ since $\\sin^2 x + \\cos^2 x = 1, \\tag 6$ we infer $\\sin^2 x = 1 - \\dfrac{1}{10} = \\dfrac{9}{10}, \\tag 7$ whence $\\sin x = \\pm \\dfrac{3}{\\sqrt{10}}; \\tag 8$ now, as has been pointed out by Foobaz John in his answer, $\\tan x = 3$ implies the sign of $\\sin x$ and $\\cos x$ must be the same; thus $\\sin x \\cos x = \\dfrac{3}{\\sqrt{10}} \\dfrac{1}{\\sqrt{10}} = \\dfrac{3}{10}." ]

[ "# Nasty Triangles 5 Geometry Level 4 Given $$\\triangle ABC$$, $$AD$$ and $$BE$$ are angle bisectors of $$\\angle BAC$$ and $$\\angle ABC$$ respectively, if $$AB=21$$, $$BC=24$$, $$AC=15$$, find the length of $$DE^2$$. This is one part of 1+1 is not = to 3. ×", "Geometry # Area of Triangles - Sine Rule In triangle $ABC,$ $AB=3,$ $AC = 12,$ and $\\sin \\angle BAC = 0.5.$ What is the area of triangle $ABC?$ In $\\triangle ABC,$ $AC = 3$ and $\\sin \\angle ACB = \\frac{2}{3}.$ If the area of $\\triangle ABC$ is $9,$ what is $BC?$ In triangle $ABC$ above, $\\overline{AD}$ bisects $\\overline{BC}.$ Given that $AB = 12,$ $AD =8,$ and $\\sin \\angle BAD = \\frac{1}{4},$ what is the area of $\\triangle ABC?$ In triangle $ABC,$ $\\overline{AD}$ bisects $\\angle BAC,$ $\\overline{AE}$ is perpendicular to $\\overline{BC},$ and $F$ and $G$ are the midpoints of $\\overline{AB}$ and $\\overline{AC},$ respectively. Given that $AB = 16,$ $AC=13,$ $AH=13,$ and $AI=8,$ and $\\angle BAC = 90^{\\circ},$ which of the following triangles has an area equal to half the area of $\\triangle ABC?$ In triangle $ABC,$ $AB = 3$ and $BC=6.$ If the area of the triangle is $6,$ what is $\\sin \\angle ABC?$ ×", "× Area of Triangles You know that Area = base x height / 2, but what other ways are there to find the area of a triangle? Brace yourself for some potent formulae. Sine Rule In triangle $$ABC,$$ $$AB=3,$$ $$AC = 12,$$ and $$\\sin \\angle BAC = 0.5.$$ What is the area of triangle $$ABC?$$ In $$\\triangle ABC,$$ $$AC = 3$$ and $$\\sin \\angle ACB = \\frac{2}{3}.$$ If the area of $$\\triangle ABC$$ is $$9,$$ what is $$BC?$$ In triangle $$ABC$$ above, $$\\overline{AD}$$ bisects $$\\overline{BC}.$$ Given that $$AB = 12,$$ $$AD =8,$$ and $$\\sin \\angle BAD = \\frac{1}{4},$$ what is the area of $$\\triangle ABC?$$ In triangle $$ABC,$$ $$\\overline{AD}$$ bisects $$\\angle BAC,$$ $$\\overline{AE}$$ is perpendicular to $$\\overline{BC},$$ and $$F$$ and $$G$$ are the midpoints of $$\\overline{AB}$$ and $$\\overline{AC},$$ respectively. Given that $$AB = 16,$$ $$AC=13,$$ $$AH=13,$$ and $$AI=8,$$ and $$\\angle BAC = 90^{\\circ},$$ which of the following triangles has an area equal to half the area of $$\\triangle ABC?$$ In triangle $$ABC,$$ $$AB = 3$$ and $$BC=6.$$ If the area of the triangle is $$6,$$ what is $$\\sin \\angle ABC?$$ ×", "Triangles, you can prepare 360 degree for your school as well as for the competitive examinations. If you need NCERT Solutions for other classes and subjects then you can download all these for free by clicking on the given link. ### Answer: In the given triangles we are given that:- (i) $\\small AC=AD$ (ii) Further, it is given that AB bisects angle A. Thus $\\angle$ BAC $=\\ \\angle$ BAD. (iii) Side AB is common in both the triangles. $AB=AB$ Hence by SAS congruence, we can say that : $\\small \\Delta ABC\\cong \\Delta ABD$ By c.p.c.t. (corresponding parts of congruent triangles are equal) we can say that$BC\\ =\\ BD$ $\\small \\Delta ABD\\cong \\Delta BAC$. ### Answer: It is given that :- (i) AD = BC (ii) $\\small \\angle DAB= \\angle CBA$ (iii) Side AB is common in both the triangles. So, by SAS congruence, we can write : $\\small \\Delta ABD\\cong \\Delta BAC$ ### Answer: In the previous part, we have proved that $\\small \\Delta ABD\\cong \\Delta BAC$. Thus by c.p.c.t. , we can write : $\\small BD=AC$ ### Answer: In the first part we have proved that $\\small \\Delta ABD\\cong \\Delta BAC$. Thus by c.p.c.t. , we can conclude : $\\small \\angle ABD= \\angle BAC$ ### Answer: In the given figure consider $\\Delta$AOD and $\\Delta$BOC. (i) AD =", "# One Angle Twice the Other Geometry Level 2 In triangle $ABC$, $\\angle BAC = 2 \\angle ABC$, $AB= 31, AC = 13$. What is $BC^2$? ×", "angle BAC and AB less than AC. M and N are the midpoints of BC and AO respectively. Let D be the intersection of MN with AC. If 2AD=(AB+AC), what is the measure of angle BAC ? 6. ### help math In triangle ABC, points D,E,F are on sides BC,CA,AB respectively such that AD,BE,CF are angle bisectors of triangle ABC. The linesAD,BE,CF are concurrent at I,the incenter of triangle ABC. If Angle BAC=90, what is the measure (in degrees) … 7. ### Math repost trapezium ABCD, with AB and DC being parallel, with diagonal connecting A and C, angle ABC is 86, angle ADC is y, angle dac is 66 and angle bac is x, no angle given for bcd or bca or acd. find x and y, So I said that BCD is 94 deg, … 8. ### geometry Points D, E, and F are the midpoints of sides \\overline{BC}, \\overline{CA}, and \\overline{AB} of \\triangle ABC, respectively, and \\overline{CZ} is an altitude of the triangle. If \\angle BAC = 71^\\circ, \\angle ABC = 39^\\circ, and \\angle … 9. ### geometry Points D, E, and F are the midpoints of sides \\overline{BC}, \\overline{CA}, and \\overline{AB} of \\triangle ABC, respectively, and \\overline{CZ} is an altitude of the triangle. If \\angle BAC = 71^\\circ, \\angle ABC = 39^\\circ, and \\angle", "+0 # help +1 167 3 In triangle ABC, AB=3, AC=6, BC=8, and D lies on BC such that AD bisects $$\\angle BAC$$. Find $$\\cos \\angle BAD$$ Oct 3, 2019 edited by Guest Oct 3, 2019 #1 +9138 +1 In triangle ABC, AB= c = 3, AC= b = 6, BC= a = 8, and D lies on BC such that AD bisects $$\\angle BAC$$. Find $$\\cos\\angle BAC=cos( \\alpha)$$ $$cosine:\\\\ a^2=b^2+c^2-2bc\\cdot cos(\\alpha)\\\\ cos(\\alpha)= \\frac{b^2+c^2-a^2}{2bc}=\\frac{36+9-64}{2\\cdot 6\\cdot 3}$$ $$cos(\\alpha)= \\cos\\angle BAC=-0,52\\overline 7$$ ! Oct 3, 2019 #2 0 Sorry asinus the question had a typo it should be \"Find $$\\cos \\angle BAD$$\" Guest Oct 3, 2019 #3 +9138 +1 so: $$cos(\\alpha)= \\cos\\angle BAC=-0,52\\overline 7$$ $$\\angle BAD=\\frac{1}{2} \\angle BAC=\\frac{1}{2}arccos (-0.52\\overline 7)\\\\ \\color{blue}\\angle BAD= 60.928°$$ $$cos \\angle BAD=0.48591$$ ! Oct 3, 2019 edited by asinus Oct 3, 2019", "# Tell me about it! Geometry Level 4 Suppose that in a triangle $$\\Delta ABC$$, $$\\overline{AB} = 8$$, $$\\overline{BC} = 6$$, $$\\overline{AC} = 7$$, $$\\overline{BD} = 3$$, and $$\\overline{AD} = 5$$. Then what is the value of $$\\overline{CD}= d$$. (round to nearest hundredths) × Problem Loading... Note Loading... Set Loading...", "7 (Already proved) $$\\angle$$ 8 = $$\\angle$$ 5 (Already proved) $$\\triangle$$ DNA ~ $$\\triangle$$ BND (AA similarity criterion) AN/DN = DN/NB $$DN^2 = AN × NB$$ $$DN^2 = AN × DM$$ (Since, NB = DM) Hence, proved. 3. In Figure, ABC is a triangle in which $$\\angle$$ ABC > 90° and AD $$\\perp$$ CB produced. Prove that $$AC^2= AB^2+ BC^2+ 2 BC.BD.$$ By applying Pythagoras Theorem in $$\\triangle$$ ADB, we get, $$AB^2 = AD^2 + DB^2$$ ……………………… (i) Again, by applying Pythagoras Theorem in $$\\triangle$$ ACD, we get, $$AC^2 = AD^2 + DC^2$$ $$AC^2 = AD^2 + (DB + BC)^2$$ $$AC^2 = AD^2 + DB^2 + BC^2 + 2DB × BC$$ From equation (i), we can write, $$AC^2 = AB^2 + BC^2 + 2DB × BC$$ Hence, proved. 4. In Figure, ABC is a triangle in which $$\\angle$$ ABC < 90° and AD $$\\perp$$ BC. Prove that $$AC^2= AB^2+ BC^2 – 2 BC.BD.$$ By applying Pythagoras Theorem in $$\\triangle$$ ADB, we get, $$AB^2 = AD^2 + DB^2$$ We can write it as; $$AD^2 = AB^2 - DB^2$$……………….. (i) By applying Pythagoras Theorem in $$\\triangle$$ ADC, we get, $$AD^2 + DC^2 = AC^2$$ From equation (i), $$AB^2 - BD^2 + DC^2 = AC^2$$ $$AB^2 - BD^2 + (BC - BD)^2 = AC^2$$ $$AC^2 = AB^2 - BD^2 + BC^2", "AB APB=AQB$\\triangle APB = \\triangle AQB$ (By AAS congruence) Also, BP = BQ (By CPCT) i.e, B is equidistant from the arms of A$\\angle A$ Hence proved. Question 6: In the fig, AC = AE, AB = AD and BAD=EAC$\\triangle BAD = \\triangle EAC$ . Show that BC = DE. Solution: BAD=EAC$\\angle BAD = \\triangle EAC$ (Given) BAD+DAC=EAC+DAC(AddingDAC to both sides)$\\angle BAD + \\angle DAC = \\angle EAC + \\angle DAC (Adding \\angle DAC\\ to\\ both\\ sides)$ BAC=EAC$\\angle BAC = \\angle EAC$ Now, in ABCandADE$\\triangle ABC\\, and\\, \\triangle ADE$ , We have AC = AE BAC=DAE$\\angle BAC = \\angle DAE$ ABC=ADE$\\triangle ABC= \\triangle ADE$ (By SAS congruence) BC = DE (CPCT) Hence Proved. Question 7: AB is a line segment and p is its mid-point. D and E are points on the same side of AB such that BAD=ABE and EPA=DPB$\\angle BAD = \\angle ABE\\ and\\ \\angle EPA = \\angle DPB$ (see fig.). Show that (i) DAP=EBP$\\triangle DAP = \\triangle EBP$ Solution: In DAP and EBP$\\triangle DAP\\ and\\ \\triangle EBP$, we have AP = BP (P is the midpoint of the line segment AB) BAD=ABE$\\angle BAD = \\angle ABE$ (Given) EPB=DPA$\\angle EPB = \\angle DPA$ (EPA=DPBEPA+DPE=DPB+DPE)$(\\angle EPA = \\angle DPB \\Rightarrow \\angle EPA + \\angle DPE = \\angle DPB + \\angle DPE)$ DPA=EPB$\\triangle DPA = \\triangle EPB$ Hence", "# Geometry Problem 1 Geometry Level 3 In triangle $$ABC$$, $$\\angle BAC = 45^\\circ$$. $$D$$ is a point on $$BC$$ such that $$AD$$ is perpendicular to $$BC$$. If $$BD = 3$$ and $$DC = 2$$, and the area of the triangle $$ABC$$ is $$x$$, find the value of $$x$$. ×", "Suppose that in a triangle $$\\Delta ABC$$, $$\\overline{AB} = 8$$, $$\\overline{BC} = 6$$, $$\\overline{AC} = 7$$, $$\\overline{BD} = 3$$, and $$\\overline{AD} = 5$$. Then what is the value of $$\\overline{CD}= d$$. (round to nearest hundredths)", "× Let ABC be an acute-angled triangle with AB =6 AC. The circle with diameter BC intersects the sides AB and AC at M and N, respectively. Denote by O the midpoint of BC. The bisectors of the angles BAC and MON intersect at R. Prove that the circumcircles of the triangles BMR and CNR have a common point lying on the line segment BC.", "# In the figure, D is a point on side BC of triangle ABC such that $$BD\\over CD$$ = $$AB\\over AC$$. Prove that AD is the bisector of $$\\angle$$ BAC. ## Solution : Given : ABC is a triangle and D is point on BC such that $$BD\\over CD$$ = $$AB\\over AC$$ To Prove : AD is the bisector of $$\\angle$$ BAC. Construction : Produce line BA to E such that line AE = AC. Join CE. Proof : In $$\\triangle$$ AEC, since AE = AC, hence $$\\angle$$ AEC = $$\\angle$$ ACE (angles opposite to equal sides of triangle are equal) Now, $$BD\\over CD$$ = $$AB\\over AC$$ (given) So, $$BD\\over CD$$ = $$AB\\over AE$$ (AE = AC, by construction) $$\\therefore$$ By converse of Basic Proportionality theorem(Thales Theorem), DA || CE Now, Since CA is a traversal, we have : $$\\angle$$ BAD = $$\\angle$$ AEC ……..(2) [corresponding angles] and $$\\angle$$ DAC = $$\\angle$$ ACE ……..(3) (alternate angles) Also, $$\\angle$$ AEC = $$\\angle$$ ACE (from 1) From (2) and (3), $$\\angle$$ BAD = $$\\angle$$ DAC Thus, AD bisects $$\\angle$$ BAC.", "$$\\angle A$$ Hence proved. Question 6: In the fig, AC = AE, AB = AD and $$\\triangle BAD = \\triangle EAC$$ . Show that BC = DE. Solution: $$\\angle BAD = \\triangle EAC$$ (Given) $$\\angle BAD + \\angle DAC = \\angle EAC + \\angle DAC (Adding \\angle DAC\\ to\\ both\\ sides)$$ $$\\angle BAC = \\angle EAC$$ Now, in $$\\triangle ABC\\, and\\, \\triangle ADE$$ , We have AC = AE $$\\angle BAC = \\angle DAE$$ $$\\triangle ABC= \\triangle ADE$$ (By SAS congruence) BC = DE (CPCT) Hence Proved. Question 7: AB is a line segment and p is its mid-point. D and E are points on the same side of AB such that $$\\angle BAD = \\angle ABE\\ and\\ \\angle EPA = \\angle DPB$$ (see fig.). Show that (i) $$\\triangle DAP = \\triangle EBP$$ Solution: In $$\\triangle DAP\\ and\\ \\triangle EBP$$, we have AP = BP (P is the midpoint of the line segment AB) $$\\angle BAD = \\angle ABE$$ (Given) $$\\angle EPB = \\angle DPA$$ $$(\\angle EPA = \\angle DPB \\Rightarrow \\angle EPA + \\angle DPE = \\angle DPB + \\angle DPE)$$ $$\\triangle DPA = \\triangle EPB$$ Hence Proved. Question 8: In right triangle ABC, right angles at C, M is the mid-point of hypotenuse AB, C is joined to M and produced to a point D such", "# A geometry problem by Ajay Sambhriya Geometry Level 2 We are given that $$BC = 12$$, $$DC = 6$$, $$DB = 9$$, and $$\\angle BCD = \\angle BAC$$. What is the ratio of the perimeter of the triangle $$ADC$$ to the that of the triangle $$BDC$$? ×", "× Back to all chapters # Angles and Lines Forgive us for being obtuse, but this is a cute concept, and we think it’s right for you. # Types of Angles How many acute angles does an obtuse triangle have? In the above diagram, $$A,$$ $$B,$$ and $$C$$ lie on a straight line. If $$\\color{Blue}{ \\angle ACD ={151}^\\circ},$$ which of the following is true of $$\\color{Red} { \\angle BCD}$$? Angles $$ABC$$ and $$CBD$$ are adjacent with common vertex $$B$$ and common edge $$\\overline{BC}.$$ If $$\\angle ABC={51}^\\circ$$ and $$\\angle CBD={40}^\\circ,$$ which of the following is true of $$\\angle ABD?$$ If $$\\angle A$$ is an acute angle, $$\\angle B$$ is a right angle and $$\\angle C$$ is an obtuse angle, which of the following is true of $$\\angle A+\\angle B+\\angle C ?$$ In the above diagram, $$A$$, $$B$$, and $$C$$ lie on a straight line. Which of the following is true of angle $$\\angle BAC$$? ×", "+0 0 40 2 In triangle $ABC,$ $M$ is the midpoint of $\\overline{AB}.$ Let $D$ be the point on $\\overline{BC}$ such that $\\overline{AD}$ bisects $\\angle BAC,$ and let the perpendicular bisector of $\\overline{AB}$ intersect $\\overline{AD}$ at $E.$ If $AB = 44$ and $ME = 12,$ then find the distance from $E$ to line $AC$. May 16, 2020 #2 +20798 +1 AD is the bisector of angle(A). Every point on the bisector of an angle is equally distant from the two sides of the angle. The distance from a point to a side of the angle is measured by the perpendicular distance to the side. ME is perpendicular to AB and the length of ME is 12; this is the distance from the point to the side of the angle. Therefore, the distance from E to AC is also 12. May 16, 2020 #1 0 This is easy! The distance from E to AC is 22. May 16, 2020 #2 +20798 +1 AD is the bisector of angle(A). Every point on the bisector of an angle is equally distant from the two sides of the angle. The distance from a point to a side of the angle is measured by the perpendicular distance to the side. ME is perpendicular to AB and the length of ME is 12; this", "+0 # Some old geometry problem for you to solve 0 133 1 Sorry if it is hard to read.$$In triangle ABC, let the angle bisectors be \\overline{BY} and \\overline{CZ}. Given AB = 8, AY = 6, and CY = 3, find BC. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"I\",I,SW); label(\"Y\",Y,NW); label(\"Z\",Z,S); [/asy]$$ Feb 3, 2022", "a$, $CA =b$ and $D$, $E$, $F$ be points on $BC$, $CA$, $AB$ such that $AD$, $BE$, $CF$ are angle bisectors of the triangle, respectively. Assume that $DE = DF.$ Prove that a) $\\dfrac{a}{b+c} = \\dfrac{b}{c+a} + \\dfrac{c}{a+b}$. b) $\\angle BAC > 90^{\\circ}.$ 8. Let $n$ be a positive integer, and $S_n,$ be the set of all positive integer divisors of $n$ (including 1 and itself). Prove that at most half of the elements in $S_n$ have their last digits equal to $3$." ]

[ "\\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7}.$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) ### Solution 1.2 (Version 2) Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$ and $\\cos \\theta$,", "$\\frac{{\\rm Eq} \\ (1)}{{\\rm Eq} \\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7} .$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) Solution 2 Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$", "$[asy] unitsize(1cm); defaultpen(0.8); pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label(\"A\",A,SW); label(\"B\",B,N); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,S); label(\"T\",T,NW); label(\"3\",A--T,N); label(\"4\",B--T,W); label(\"1\",D--T,N); label(\"1\",E--T,W); [/asy]$ The point $C$ is the intersection of the lines $BD$ and $AE$. The points on the first line have the form $(t,4-4t)$, the points on the second line have the form $(t,-1-t/3)$. Solving for $t$ we get $t=15/11$, hence $C=(15/11,-16/11)$. The ratio $CD/BD$ can now be computed simply by observing the $x$ coordinates of $B$, $C$, and $D$: $$\\frac{CD}{BD} = \\frac{15/11 - 1}{1 - 0} = \\boxed{\\textbf{(D)}\\frac 4{11}}$$", "\\frac {1}{2} \\cdot 15 \\cdot 36 = \\boxed{\\mathrm{(C)}\\ 210}$$ ### Solution 2 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$. We get the following triangle: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0); draw(A--B--C--cycle); draw(C--F,dashed); label(\"$$A'$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,N); label(\"$$C'$$\",F,SE); label(\"5\",(A+C)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ The length of $A'B$ in this triangle is equal to the length of the original $AB$, minus the length of $CD$. Thus $A'B = 52 - 39 = 13$. Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\\frac{A'C\\cdot BC}2 = \\frac{5\\cdot 12}2 = 30$, and therefore its altitude $CC'$ is $\\frac{[A'BC]}{A'B} = \\frac{60}{13}$. Now the area of the original trapezoid is $\\frac{(AB+CD)\\cdot CC'}2 = \\frac{91 \\cdot 60}{13 \\cdot 2} = 7\\cdot 30 = \\boxed{\\mathrm{(C)}\\ 210}$ ### Solution 3 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Call the length of $AD'$ to be $y$, the length of $BC'$ to be $z$, and the height of the trapezoid to be $x$. By", "= 2x$. Setting the equations equal, we have $2x = -4x +2 \\implies x = \\frac13$. Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\\frac13$, so $XY = 1 - 2 \\cdot \\frac13 = \\frac13$. Now the area of the kite is simply the product of the two diagonals over $2$. Since the length $HF = 1$, our answer is $\\dfrac{\\dfrac{1}{3} \\cdot 1}{2} = \\boxed{\\textbf{(E)} \\: \\dfrac16}$. $[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); draw(X--Y,dashed); label(\"A\\: (0,2)\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D \\: (0,0)\",D,SW); label(\"E\",E,E); label(\"F\\: (\\frac12,0)\",F,S); label(\"G\",G,W); label(\"H \\: (\\frac12,1)\",H,N); label(\"Y\",Y,E); label(\"X\",X,W); label(\"\\frac12\",(0.25,0),S); label(\"\\frac12\",(0.75,0),S); label(\"1\",(1,0.5),E); label(\"1\",(1,1.5),E); [/asy]$ ## Solution 2 Let the area of the shaded region be $x$. Let the other two vertices of the kite be $I$ and $J$ with $I$ closer to $AD$ than $J$. Note that $[ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]$. The area of $ABF$ is $1$ and the area of $DCH$ is $\\dfrac{1}{2}$. We will solve for the areas of", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "2} = \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, since $\\triangle R PM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which", "By the Law of Cosines on $\\triangle AA'D$, we have $$AD^2=10^2+\\left(\\frac {130}{23}\\right)^2-2 \\cdot 10 \\cdot \\frac {130}{23} \\cdot \\frac {5}{13}=\\frac {2^4 \\cdot 3^2 \\cdot 5^2 \\cdot 13}{23^2} \\Longrightarrow AD=\\frac {60\\sqrt {13}}{23}$$ and the answer is $60+13+23=\\boxed{096}$. $[asy] size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype(\"4 4\")); MP(\"5\", (A+B)/2, f); MP(\"13\", (A+C)/2, NE,f); MP(\"A\",D(A),f); MP(\"B\",D(B),f); MP(\"C\",D(C),N,f); MP(\"A'\",D(E),f); MP(\"D\",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30)); [/asy]$ -azjps Solution 6 (Law of Sines) $[asy]pathpen = linewidth(0.7); pen f = fontsize(10); size(144); pair B=(0,0); pair A=(5,0); pair C=(0,12); pair E=(-5,0); pair abA = A+unit(C-A)+unit(B-A); pair D = extension(A,abA,C,E); D(A--C); D(C--B); D(A--B); D(C--D); D(A--D); D(D--E,dashed); D(B--E,dashed); D(rightanglemark(C,B,A,20),red); MP(\"A\",D(A),plain.SE,f); MP(\"B\",D(B),plain.S,f); MP(\"C\",D(C),plain.N,f); MP(\"D\",D(D),plain.NW,f); MP(\"E\",D(E),plain.SW,f); MP(\"5\",(A+B)/2,plain.S,f); MP(\"12\",(B+C)/2,plain.E,f); MP(\"13\",(A+C)/2,plain.NE,f); MP(\"\\alpha\",A,4*(unit(D-A)+unit(C-A))/2,f); MP(\"\\alpha\",A,4*(unit(B-A)+unit(D-A))/2,f); MP(\"2\\alpha\",E,4*(unit(C-E)+unit(B-E))/2,f); MP(\"\\beta\",C,6*(unit(A-C)+unit(B-C))/2,f); MP(\"\\beta\",C,6*(unit(E-C)+unit(B-C))/2,f);[/asy]$ Using the law of sines on $\\bigtriangleup ADE$, \\begin{align*} AD &= AE \\cdot \\dfrac{\\sin(2\\alpha)}{\\sin(180^\\circ-3\\alpha)}\\\\ &= AE \\cdot \\dfrac{\\sin(2\\alpha)}{\\sin(\\alpha+2\\alpha)}\\\\ &= 10 \\cdot \\dfrac{\\sin(2\\alpha)}{\\sin(\\alpha)\\cos(2\\alpha)+\\cos(\\alpha)\\sin(2\\alpha)}\\\\ &= 10 \\cdot \\dfrac{\\sin(2\\alpha)}{\\sqrt{\\dfrac{1-\\cos(2\\alpha)}{2}}\\cdot\\cos(2\\alpha)+\\sqrt{\\dfrac{1+\\cos(2\\alpha)}{2}}\\cdot\\sin(2\\alpha)}\\\\ &= 10 \\cdot \\dfrac{\\dfrac{12}{13}}{\\sqrt{\\dfrac{8}{26}}\\cdot\\dfrac{5}{13}+\\sqrt{\\dfrac{18}{26}}\\cdot\\dfrac{12}{13}}\\\\ &= 10 \\cdot \\dfrac{12\\sqrt{13}}{10+36} = \\dfrac{60\\sqrt{13}}{23} \\end{align*} $\\therefore$ the answer is $60+13+23=\\boxed{086}$. -m1sterzer0", "angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(circumcircle(A,B,D)); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot(\"A\", A, SW); dot(\"B\", B, dir(0)); dot(\"C\", C, N); dot(\"D\", D, dir(0)); dot(\"E\", E, N); dot(\"F\", F, dir(0)); [/asy]$ Applying the Law of Sines to $\\triangle BED$ and $\\triangle CFD$ gives$$DE = BD\\cdot\\frac{\\sin y}{\\sin x}\\text{~ and ~} DF = CD\\cdot\\frac{\\sin z}{\\sin x}.$$By the Angle Bisector Theorem, $BD = 2$ and $CD = 3$. Combining the above information yields $$[\\triangle AEF] = \\frac{3\\sin y\\cdot\\sin z\\cdot\\cos x}{\\sin^2 x}.$$Applying the Law of Cosines to $\\triangle ABC$ gives $\\cos 2x = \\frac9{16}$, $\\cos 2y = \\frac1{8}$, and $\\cos 2z = \\frac34$. By the Half Angle Formulas,$$\\sin^2x = \\frac7{32},~~ \\cos x = \\sqrt{\\frac{25}{32}},~~ \\sin y = \\sqrt{\\frac7{16}}, \\text{~ and ~} \\sin z = \\sqrt{\\frac18}.$$Therefore $$[\\triangle AEF] = \\frac{3\\cdot\\sqrt{\\frac{7}{16}}\\cdot\\sqrt{\\frac{1}{8}}\\cdot\\sqrt{\\frac{25}{32}}} {\\frac{7}{32}} = \\frac{15\\sqrt{7}}{14}.$$The requested sum is $15+7+14 = 36$. ## Solution 10 (Official MAA 2) Let the point $M$ be the midpoint of $\\overline{AD}$, let $I$ be the incenter of $\\triangle ABC$ which is the common point of lines $AD$, $BE$, and $CF$, and let $r$ be the inradius of $\\triangle ABC$. The semiperimeter of", "\\frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 size(8cm); pair A, B, C, D, E, EE; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); EE = (27/8,0); draw(A--B--C--D--A--E); draw(E--EE,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,EE,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",EE,S); label(\"$4$\",(A+D)/2,S); label(\"$x$\",(A+EE)/2,S; label(\"$4-x$\",(D+EE)/2,S); label(\"$\\frac{15}{8}$\",(E+EE)/2,W); (Error compiling LaTeX. label(\"$x$\",(A+EE)/2,S; ^ de021056a2cc2ef4e3e5c00687df10ec92699974.asy: 26.23: syntax error error: could not load module 'de021056a2cc2ef4e3e5c00687df10ec92699974.asy') Following the steps above, you can find that the height of triangle $ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the", "= \\frac 3 4$. Notice that the distance $OM$ equals $PN + PO \\cos AOM = r(1 + \\cos AOM)$ (Where $r$ is the radius of circle P). Evaluating this, $\\cos \\angle AOM = \\frac{OM}{r} - 1 = \\frac{2OM}{R} - 1 = \\frac 8 5 - 1 = \\frac 3 5$. From $\\cos \\angle AOM$, we see that $\\tan \\angle AOM =\\frac{\\sqrt{1 - \\cos^2 \\angle AOM}}{\\cos \\angle AOM} = \\frac{\\sqrt{5^2 - 3^2}}{3} = \\frac 4 3$ Next, notice that $\\angle AOB = \\angle AOM - \\angle BOM$. We can therefore apply the tangent subtraction formula to obtain , $\\tan AOB =\\frac{\\tan AOM - \\tan BOM}{1 + \\tan AOM \\cdot \\tan AOM} =\\frac{\\frac 4 3 - \\frac 3 4}{1 + \\frac 4 3 \\cdot \\frac 3 4} = \\frac{7}{24}$. It follows that $\\sin AOB =\\frac{7}{\\sqrt{7^2+24^2}} = \\frac{7}{25}$, resulting in an answer of $7 \\cdot 25=\\boxed{175}$. ## Solution 2 $[asy] size(10cm); import olympiad; pair O = (0,0);dot(O);label(\"O\",O,SW); pair M = (4,0);dot(M);label(\"M\",M,SE); pair N = (4,2);dot(N);label(\"N\",N,NE); draw(circle(O,5)); pair B = (4,3);dot(B);label(\"B\",B,NE); pair C = (4,-3);dot(C);label(\"C\",C,SE); draw(B--C);draw(O--M); pair P = (1.5,2);dot(P);label(\"P\",P,W); draw(circle(P,2.5)); pair A=(3,4);dot(A);label(\"A\",A,NE); draw(O--A); draw(O--B); pair Q = (1.5,0); dot(Q); label(\"Q\",Q,S); pair R = (3,0); dot(R); label(\"R\",R,S); draw(P--Q,dotted); draw(A--R,dotted); pair D=(5,0); dot(D); label(\"D\",D,E); draw(A--D); [/asy]$ The above solution works, but is quite messy and somewhat difficult to follow. This solution provides", "area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 $[asy] size(8cm); pair A, B, C, D, E, F; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); F = (27/8,0); draw(A--B--C--D--A--E); draw(E--F,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,F,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",F,S); label(\"4\",(A+D)/2,S); label(\"x\",(A+F)/2,S); label(\"\\frac{15}{8}\",(E+F)/2,W); [/asy]$ Let the point where the altitude from $E$ to $\\overline{AD}$ be labeled $F$. Following the steps above, you can find that the height of $\\triangle ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the origin to the point $F$, and $4-x$ is the segment from point $F$ to point", "\\frac {AD}{2} = 1004, \\quad ME = MC = \\frac {BC}{2} = 500$$ Thus $MN = NE - ME = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(N+H)/2+(0,1),S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(M+H)/2,W); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (504 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {DG}{CG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$. ### Solution 3 If you drop perpendiculars from B and C to AD, and call the points if you drop perpendiculars from B and C to AD and call the points where they meet AD E and F respectively and call FD = x and EA = $1008-x$ , then you", "= \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, because $\\triangle RPM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which equals $ED", "dir(D-F)*I, deepmagenta); label(\"4\\sqrt{2}\", (D+E)/2, dir(E-D)*I, deepmagenta); label(\"a\", (B+X)/2, dir(B-X)*I, gray(0.3)); label(\"a\\sqrt{2}\", (D+X)/2, dir(D-X)*I, gray(0.3)); [/asy]$ Notice that $$\\angle DFE=\\angle CFE-\\angle CFD=\\angle CBE-\\angle CAD=180-B-A=C.$$By the Law of Cosines, $$\\cos C=\\frac{AC^2+BC^2-AB^2}{2\\cdot AC\\cdot BC}=\\frac34.$$Then, $$DE^2=DF^2+EF^2-2\\cdot DF\\cdot EF\\cos C=32\\implies DE=4\\sqrt2.$$Let $X=\\overline{AB}\\cap\\overline{CF}$, $a=XB$, and $b=XD$. Then, $$XA\\cdot XD=XC\\cdot XF=XB\\cdot XE\\implies b(a+4)=a(b+4\\sqrt2)\\implies b=a\\sqrt2.$$However, since $\\triangle XFD\\sim\\triangle XAC$, $XF=\\tfrac{4+a}3$, but since $\\triangle XFE\\sim\\triangle XBC$, $$\\frac75=\\frac{4+a}{3a}\\implies a=\\frac54\\implies BE=a+a\\sqrt2+4\\sqrt2=\\frac{5+21\\sqrt2}4,$$and the requested sum is $5+21+2+4=\\boxed{032}$. (Solution by TheUltimate123) ## Solution 2 Define $\\omega_1$ to be the circumcircle of $\\triangle ACD$ and $\\omega_2$ to be the circumcircle of $\\triangle EBC$. Because of exterior angles, $\\angle ACB = \\angle CBE - \\angle CAD$ But $\\angle CBE = \\angle CFE$ because $CBFE$ is cyclic. In addition, $\\angle CAD = \\angle CFD$ because $CAFD$ is cyclic. Therefore, $\\angle ACB = \\angle CFE - \\angle CFD$. But $\\angle CFE - \\angle CFD = \\angle DFE$, so $\\angle ACB = \\angle DFE$. Using Law of Cosines on $\\triangle ABC$, we can figure out that $\\cos(\\angle ACB) = \\frac{3}{4}$. Since $\\angle ACB = \\angle DFE$, $\\cos(\\angle DFE) = \\frac{3}{4}$. We are given that $DF = 2$ and $FE = 7$, so we can use Law of Cosines on $\\triangle DEF$ to find that $DE = 4\\sqrt{2}$. Let $G$ be the intersection of segment $\\overline{AE}$ and $\\overline{CF}$. Using Power of a Point with respect to $G$", "from $\\overline{CF}$ to $\\overline{DE}$ and $\\overline{AB}$, respectively. We know that these are both equal to $\\frac 53$. We find the area of each of the trapezoids, which both happen to be $\\frac{11}6 \\cdot \\frac{\\sqrt 3}6=\\frac{11\\sqrt 3}{36}$, and the combined area is $\\frac{11\\sqrt 3}{18}$. We find that $\\dfrac{\\frac{11\\sqrt 3}{18}}{\\frac{3\\sqrt 3}2}$ is equal to $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$. ## Solution 2 $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,ESE); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,WSW); label(\"W\",W,ENE); label(\"X\",X,ESE); label(\"Y\",Y,WSW); label(\"Z\",Z,WNW); [/asy]$ First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are $22$ small triangles in hexagon $ZWCXYF$, and $9 \\cdot 6 = 54$ small triangles in the whole hexagon. Thus, the answer is $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$. ## Solution 3 (Similar Triangles) $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); pair G = (0.5, sqrt(3)*3/2); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(E--G--D); draw(F--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,ESE); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,WSW); label(\"W\",W,ENE); label(\"X\",X,ESE); label(\"Y\",Y,WSW); label(\"Z\",Z,WNW); label(\"G\",G,N); [/asy]$ Extend $\\overline{EF}$ and $\\overline{CD}$ to meet at point $G$, as shown in the diagram.", "\\frac {AD}{2} = 1004, \\quad ME = MC = \\frac {BC}{2} = 500$$ Thus $MN = NE - ME = \\boxed{504}$. ### Solution 2 size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(N+H)/2,N); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(M+N)/2,W); (Error compiling LaTeX. 228aa9a09ec6a18b4a0ab11366b7b8f2ec44b852.asy: 27.6: no matching function 'label(string, pair[], pair)') Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (504 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {CG}{DG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow h^2 = (504 + x)(504 - x)\\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$.", "area of a triangle equals $\\frac{1}{2} \\cdot ab \\cdot \\sin(C)$. If angle $DAE = Z$, we have that $$\\frac{[ADE]}{[ABC]} = \\frac{\\frac{1}{2} \\cdot 19 \\cdot 14 \\cdot \\sin(Z)}{\\frac{1}{2} \\cdot 25 \\cdot 42 \\cdot \\sin(Z)} = \\frac{19 \\cdot 14}{25 \\cdot 42} = \\frac{ab}{cd}$$. - SuperJJ ## Video Solution CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00 ## Solution 2 (no trig) $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, W); label(\"E\", E, NE); label(\"19\", (A + D)/2, W); label(\"6\", (B + D)/2, W); label(\"14\", (A + E)/2, NE); label(\"28\", (C + E)/2, NE); [/asy]$ We can let $[ADE]=x$. Since $EC=2 \\cdot EA$, $[DEC]=2x$. So, $[ADC]=3x$. This means that $[BDC]=\\frac{6}{19}\\cdot3x=\\frac{18x}{19}$. Thus, $\\frac{[ADE]}{[BCED]} = \\frac{x}{\\frac{18x}{19}+2x}= \\boxed{\\textbf{(D) }\\frac{19}{56}}.$ -Conantwiz2023 ## Solution 3 (trig) The area of a triangle is $\\frac{1}{2} bc\\sin A$. Using this formula: $[ADE]=\\frac{1}{2}\\cdot19\\cdot14\\cdot\\sin A = 133\\sin A$ $[ABC]=\\frac{1}{2}\\cdot25\\cdot42\\cdot\\sin A = 525\\sin A$ Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$, $[BCED] = 525\\sin A - 133\\sin A = 392\\sin A$. Therefore, the desired ratio is $\\frac{133\\sin A}{392\\sin A}=\\boxed{\\textbf{(D) }\\frac{19}{56}}$ Note: $BC=39$ was not used in this problem. ## Solution 4 Let" ]

[ "get $$b^2+9a^2-2\\cdot 3a\\cdot b\\cdot \\cos(\\angle ADC)=7\\qquad (5)$$ $$b^2+a^2+2\\cdot a\\cdot b\\cdot \\cos(\\angle ADC)=9.\\qquad (6)$$ $(5)+(6)$, and according to $(4)$, we can get $$37a^2+2b^2=48. \\qquad (7)$$ Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\\frac{\\sqrt{22}}{2}$. Finally, we use Law of Cosines for $\\triangle ADB$, $$4(\\frac{\\sqrt{22}}{2})^2+1+2\\cdot2(\\frac{\\sqrt{22}}{2})\\cdot \\cos(ADC)=AB^2$$ then $AB=2\\sqrt{7}$, so the height of this $\\triangle ABC$ is $\\sqrt{4^2-(\\sqrt{7})^2}=3$. Then the area of $\\triangle ABC$ is $3\\sqrt{7}$, so the answer is $\\boxed{010}$. ### Solution 3 Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$.", "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we", "dir(D-F)*I, deepmagenta); label(\"4\\sqrt{2}\", (D+E)/2, dir(E-D)*I, deepmagenta); label(\"a\", (B+X)/2, dir(B-X)*I, gray(0.3)); label(\"a\\sqrt{2}\", (D+X)/2, dir(D-X)*I, gray(0.3)); [/asy]$ Notice that $$\\angle DFE=\\angle CFE-\\angle CFD=\\angle CBE-\\angle CAD=180-B-A=C.$$By the Law of Cosines, $$\\cos C=\\frac{AC^2+BC^2-AB^2}{2\\cdot AC\\cdot BC}=\\frac34.$$Then, $$DE^2=DF^2+EF^2-2\\cdot DF\\cdot EF\\cos C=32\\implies DE=4\\sqrt2.$$Let $X=\\overline{AB}\\cap\\overline{CF}$, $a=XB$, and $b=XD$. Then, $$XA\\cdot XD=XC\\cdot XF=XB\\cdot XE\\implies b(a+4)=a(b+4\\sqrt2)\\implies b=a\\sqrt2.$$However, since $\\triangle XFD\\sim\\triangle XAC$, $XF=\\tfrac{4+a}3$, but since $\\triangle XFE\\sim\\triangle XBC$, $$\\frac75=\\frac{4+a}{3a}\\implies a=\\frac54\\implies BE=a+a\\sqrt2+4\\sqrt2=\\frac{5+21\\sqrt2}4,$$and the requested sum is $5+21+2+4=\\boxed{032}$. (Solution by TheUltimate123) ## Solution 2 Define $\\omega_1$ to be the circumcircle of $\\triangle ACD$ and $\\omega_2$ to be the circumcircle of $\\triangle EBC$. Because of exterior angles, $\\angle ACB = \\angle CBE - \\angle CAD$ But $\\angle CBE = \\angle CFE$ because $CBFE$ is cyclic. In addition, $\\angle CAD = \\angle CFD$ because $CAFD$ is cyclic. Therefore, $\\angle ACB = \\angle CFE - \\angle CFD$. But $\\angle CFE - \\angle CFD = \\angle DFE$, so $\\angle ACB = \\angle DFE$. Using Law of Cosines on $\\triangle ABC$, we can figure out that $\\cos(\\angle ACB) = \\frac{3}{4}$. Since $\\angle ACB = \\angle DFE$, $\\cos(\\angle DFE) = \\frac{3}{4}$. We are given that $DF = 2$ and $FE = 7$, so we can use Law of Cosines on $\\triangle DEF$ to find that $DE = 4\\sqrt{2}$. Let $G$ be the intersection of segment $\\overline{AE}$ and $\\overline{CF}$. Using Power of a Point with respect to $G$", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "area of a triangle equals $\\frac{1}{2} \\cdot ab \\cdot \\sin(C)$. If angle $DAE = Z$, we have that $$\\frac{[ADE]}{[ABC]} = \\frac{\\frac{1}{2} \\cdot 19 \\cdot 14 \\cdot \\sin(Z)}{\\frac{1}{2} \\cdot 25 \\cdot 42 \\cdot \\sin(Z)} = \\frac{19 \\cdot 14}{25 \\cdot 42} = \\frac{ab}{cd}$$. - SuperJJ ## Video Solution CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00 ## Solution 2 (no trig) $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, W); label(\"E\", E, NE); label(\"19\", (A + D)/2, W); label(\"6\", (B + D)/2, W); label(\"14\", (A + E)/2, NE); label(\"28\", (C + E)/2, NE); [/asy]$ We can let $[ADE]=x$. Since $EC=2 \\cdot EA$, $[DEC]=2x$. So, $[ADC]=3x$. This means that $[BDC]=\\frac{6}{19}\\cdot3x=\\frac{18x}{19}$. Thus, $\\frac{[ADE]}{[BCED]} = \\frac{x}{\\frac{18x}{19}+2x}= \\boxed{\\textbf{(D) }\\frac{19}{56}}.$ -Conantwiz2023 ## Solution 3 (trig) The area of a triangle is $\\frac{1}{2} bc\\sin A$. Using this formula: $[ADE]=\\frac{1}{2}\\cdot19\\cdot14\\cdot\\sin A = 133\\sin A$ $[ABC]=\\frac{1}{2}\\cdot25\\cdot42\\cdot\\sin A = 525\\sin A$ Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$, $[BCED] = 525\\sin A - 133\\sin A = 392\\sin A$. Therefore, the desired ratio is $\\frac{133\\sin A}{392\\sin A}=\\boxed{\\textbf{(D) }\\frac{19}{56}}$ Note: $BC=39$ was not used in this problem. ## Solution 4 Let", "the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 $[asy] unitsize(0.5cm); pair A,B,C,D,E; B=(0,0); C=(14,0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"A\",A,N); label(\"B\",B,S); label(\"C\",C,S); label(\"E\",E,NW); label(\"D\",D,NE); label(\"13\",A--B,NW); label(\"15\",A--C,NE); label(\"14\",B--C,S); label(\"6.5\",B--E,N); label(\"7\",C--D,N); [/asy]$ Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, then $BD=CD=7$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "$BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 pair A,B,C,D,E; B=(0,0); C=(14;0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"$A$\",A,N); label(\"$B$\",B,S); label(\"$C$\",C,S); label(\"$E$\",E,S); label(\"$D$\",D,S); label(\"$13$\",A--B,NW); label(\"$15$\",A--C,NE); label(\"$14$\",B--C,S); (Error compiling LaTeX. 32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy: 7.6: syntax error error: could not load module '32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy') Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, thus the area of $\\Delta ADC=\\Delta ABD$, which is half the area of $\\Delta ABC$, so $\\Delta ADC=\\Delta ABD=42$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "# 2005 AIME II Problems/Problem 14 ## Problem In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\\overline{BC}$ with $CD=6.$ Point $E$ is on $\\overline{BC}$ such that $\\angle BAE\\cong \\angle CAD.$ Given that $BE=\\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ ## Solution 1 $[asy] import olympiad; import cse5; import geometry; size(150); defaultpen(fontsize(10pt)); defaultpen(0.8); dotfactor = 4; pair A = origin; pair C = rotate(15,A)*(A+dir(-50)); pair B = rotate(15,A)*(A+dir(-130)); pair D = extension(A,A+dir(-68),B,C); pair E = extension(A,A+dir(-82),B,C); label(\"A\",A,N); label(\"B\",B,SW); label(\"D\",D,SE); label(\"E\",E,S); label(\"C\",C,SE); draw(A--B--C--cycle); draw(A--E); draw(A--D); draw(anglemark(B,A,E,5)); draw(anglemark(D,A,C,5)); [/asy]$ By the Law of Sines and since $\\angle BAE = \\angle CAD, \\angle BAD = \\angle CAE$, we have \\begin{align*} \\frac{CD \\cdot CE}{AC^2} &= \\frac{\\sin CAD}{\\sin ADC} \\cdot \\frac{\\sin CAE}{\\sin AEC} \\\\ &= \\frac{\\sin BAE \\sin BAD}{\\sin ADB \\sin AEB} \\\\ &= \\frac{\\sin BAE}{\\sin AEB} \\cdot \\frac{\\sin BAD}{\\sin ADB}\\\\ &= \\frac{BE \\cdot BD}{AB^2} \\end{align*} Substituting our knowns, we have $\\frac{CE}{BE} = \\frac{3 \\cdot 14^2}{2 \\cdot 13^2} = \\frac{BC - BE}{BE} = \\frac{15}{BE} - 1 \\Longrightarrow BE = \\frac{13^2 \\cdot 15}{463}$. The answer is $q = \\boxed{463}$. ## Solution 2 (Similar Triangles) Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "AMC$. Let's call the intersection of $\\overline{UC}$ and $\\overline{MV}$ $P$. Let $\\overline{UP}=x$. Then $\\overline{PC}=12-x$. Since $\\overline{UC} \\perp \\overline{MV}$, $\\overline{UP}$ and $\\overline{CP}$ are heights of triangles $\\triangle MUV$ and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 3 (Medians) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the incenter, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively. Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\\triangle PUM$ is enough. $\\overline{PC} = \\overline{MP} = 8$. The area of $\\triangle PUM = \\frac{4\\cdot8}{2}=16$. Multiplying this", "$\\triangle AMC$. Let's call the intersection of $\\overline{UC}$ and $\\overline{MV}$ $P$. Let $\\overline{UP}=x$. Then $\\overline{PC}=12-x$. Since $\\overline{UC} \\perp \\overline{MV}$, $\\overline{UP}$ and $\\overline{CP}$ are heights of triangles $\\triangle MUV$ and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 3 (Medians) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the centroid, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively. Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\\triangle PUM$ is enough. $\\overline{PC} = \\overline{MP} = 8$. The area of $\\triangle PUM = \\frac{4\\cdot8}{2}=16$. Multiplying", "we have $\\cos \\theta = \\frac{3}{4}$. Now, since $\\triangle BOC$ is isosceles with $OB = OC$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. By SSS congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ### Solution 4 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we have that $AB = AE$. Similarly, $CD = DF$. $OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD =", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "$AB=16$ and $AC=BC.$ Let's say that $D$ is the midpoint of $AB$ and $E$ is the point where $AC$ is tangent to the semicircle. We could also use $BC$ instead of $AC$ because of symmetry. We notice that $\\triangle ACD \\cong \\triangle BCD,$ and are both 8-15-17 right triangles. We also know that we create a right angle with the intersection of the radius and a tangent line of a circle (or part of a circle). So, by $AA$ similarity, $\\triangle AED \\sim \\triangle ADC,$ with $\\angle EAD \\cong \\angle DAC$ and $\\angle CDA \\cong \\angle DEA.$ This similarity means that we can create a proportion: $\\frac{AD}{AB}=\\frac{DE}{CD}.$ We plug in $AD=\\frac{AB}{2}=8, AC=17,$ and $CD=15.$ After we multiply both sides by $15,$ we get $DE=\\frac{8}{17} \\cdot 15= \\boxed{\\textbf{(B) }\\frac{120}{17}}.$ (By the way, we could also use $\\triangle DEC \\sim \\triangle ADC.$) ## Solution 4: Inscribed Circle $[asy] pair A, B, C, D, M; B=(0,0); D=(16,0); A=(8,15); C=(8,-15); M=D/2; draw(B--D--A--cycle); draw(A--M); draw(arc(M,120/17,0,180)); draw(rightanglemark(D,M,A,25)); draw(rightanglemark(B,M,25)); label(\"B\",B,SW); label(\"D\",D,SE); label(\"A\",A,N); label(\"M\",M,S); label(\"C\",C,S); draw((0,0)--(8,-15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,360));[/asy]$ Call this triangle $ABD$ and let the midpoint of base $BD$ be $M$. Divide the triangle in half by drawing a line from $A$ to $M$. Half the base of triangle $ABD$ is $\\frac{16}{2} = 8$. The height is $15$, which is given in the question. Using the Pythagorean", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$" ]

[ "+0 # Help! 0 49 1 +11 Consider the line parametrized by \\begin{align*} x&= 2t + 1,\\\\ y& = -3t +2. \\end{align*} Find a vector $\\begin{pmatrix}a\\\\b\\end{pmatrix}$ that's parallel to this line and satisfies $a+b = 3$. Jul 27, 2020 edited by FearlessIsland3 Jul 27, 2020", "## Directional and orthogonal form of a line Directional and orthogonal form of a line Directional form: \\begin{align*} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} &= \\begin{pmatrix} a_1 \\\\ a_2 \\end{pmatrix} + \\alpha \\begin{pmatrix} d_1 \\\\ d_2 \\end{pmatrix} \\end{align*} or, concisely \\begin{align} x = a + \\alpha d \\end{align} Let $v$ be a vector perpendicular to $d$. Transpose both sides and multiply by $v$: \\begin{align} x'v &= a'v + \\alpha d'v = a'v = c \\end{align} Where c is a constant scalar. This is called the orthogonal form. If you see it as $f(x) = x'v$, then $v$ is the gradient of $f$.", "label each point. Hence let $$A, \\ B$$ and $$C$$ be the points $$(0, 1, 4), (2, -1, 3)$$ and $$(4, 2, 9)$$ respectively. This subsequently implies the vectors $$AB$$ and $$AC$$ are represented by $$(2, -2, -1)$$ and $$(4, 1, 5)$$ respectively. We can see in the previous diagram that a vector projection essentially creates a perpendicular from the end of the projecting vector to the base of the vector being projected upon. If we let $$P$$ denote the base of said perpendicular, we can see that the perpendicular distance between the point and the line is simply the distance $$BP$$. By taking the scalar projection of the vector $$AB$$ onto the vector $$AC$$ : \\begin{align*} | proj_{AC} AB | = \\frac{|AB \\cdot AC|}{|AC|} =\\frac{ |8-2-5|}{ \\sqrt{16+1+25}} = \\frac{1}{\\sqrt{42}} \\end{align*} We have now found the length of the vector $$AP$$. With reference to our diagram, we can see that to find the length of the vector $$BP$$, we can just apply Pythagoras theorem as it’s relatively simple to determine the length of the vector $$AB$$: \\begin{align*} |AB|= \\sqrt{4+4+1} =3 \\end{align*} Hence, the distance from the point to the line is given by: \\begin{align*} | BP| = \\sqrt{3^2 – \\frac{1}{42}} = \\frac{ \\sqrt{377}}{\\sqrt{42}} \\end{align*} Example 2 Let $$u$$ and $$v$$ be non-zero vectors such that $$|u|=|v|$$. Show that", "$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$ and the second has normal vector $\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $v_1,v_2$ is $v_1\\wedge v_2$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\wedge \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $y=0$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $x=0, z=1$, thus the line of intersection is $\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$ -", "## Section1.19Lines and Planes Two points $$A$$ and $$B$$ determine a line. But there are also other ways to describe a line. Rather than specifying two points, we can specify just one ($$A$$), then give a vector $$\\vv$$ along the line. So let $$\\AA$$ be the vector from the origin to the point $$A\\text{,}$$ and $$\\vv$$ be a vector from $$A$$ that points along the line. Then any other point $$P$$ on the line can be reached by going to $$A$$ along $$\\AA\\text{,}$$ then going along $$\\vv\\text{.}$$ Thus, the vector $$\\rr(u) = \\AA + \\vv\\,u\\tag{1.19.1}$$ from the origin to $$P$$ provides a vector parameterization of the line. Equivalently, if \\begin{align*} \\AA \\amp = a_x\\,\\xhat + a_y\\,\\yhat + a_z\\,\\zhat ,\\\\ \\vv \\amp = v_x\\,\\xhat + v_y\\,\\yhat + v_z\\,\\zhat , \\end{align*} then the line is given by the parametric equations \\begin{align*} x \\amp = a_x + v_x\\,u ,\\\\ y \\amp = a_y + v_y\\,u ,\\\\ z \\amp = a_z + v_z\\,u , \\end{align*} which gives the coordinates $$(x,y,z)$$ of the point $$P$$ in terms of the parameter $$u\\text{.}$$ A given line can have many parameterizations, depending not only on the choice of the point $$A$$ on the line and vector $$\\vv$$ along the line, but also on the choice of parameter $$u\\text{.}$$ How do two points determine a line? Using", "Given that we are thinking about geometrical properties of lines and their midpoints it makes sense to use vectors to solve this problem. Most importantly it will likely cut down on the algebra a lot due to their concise notation. \\begin{align*}\\ \\ Let\\ \\overrightarrow{OA} = \\mathbf{a} = \\begin{bmatrix} p_1 \\\\ q_1 \\end{bmatrix},\\ \\overrightarrow{OB} = \\mathbf{b} = \\begin{bmatrix} p_2 \\\\ q_2 \\end{bmatrix},\\ \\overrightarrow{OC} = \\mathbf{c} = \\begin{bmatrix} p_3 \\\\ q_3 \\end{bmatrix} \\end{align*} \\begin{align*}Let\\ the\\ midpoints\\ of\\ AB,\\ AC,\\ BC\\ be\\ denoted\\ D,\\ E,\\ F \\end{align*} \\begin{align*}\\ \\ \\therefore \\overrightarrow{OD} = \\mathbf{\\frac{1}{2}(a+b)},\\ \\overrightarrow{OE} = \\mathbf{\\frac{1}{2}(a+c)},\\ \\overrightarrow{OF} = \\mathbf{\\frac{1}{2}(b+c)}\\end{align*} That has set up all of the required details to begin actually answering the question. The vector equations for the line joining A and the midpoint of BC and the line joining B and the midpoint AC, are the lines $AF$ and $BE$: \\begin{align*}\\mathbf{r_{AF}} = \\mathbf{a} + \\lambda (\\mathbf{b + c - 2a}) \\end{align*} \\begin{align*}\\mathbf{r_{BE}} = \\mathbf{b} + \\mu (\\mathbf{a+c-2b}) \\end{align*} They intersect when $\\mathbf{r_{AF}} = \\mathbf{r_{BE}}$. $\\Rightarrow \\mathbf{a} - \\mathbf{b} = (\\mu - \\lambda)\\mathbf{c} + (\\mu + 2\\lambda)\\mathbf{a} - (\\lambda + 2\\mu)\\mathbf{b}$ $\\mathbf{c}$ is not parallel to $\\mathbf{a}$ or $\\mathbf{b} \\Rightarrow \\mu - \\lambda = 0$ $\\Rightarrow \\mathbf{a} - \\mathbf{b} = 3\\lambda\\mathbf{a} - 3\\lambda\\mathbf{b}$ This must then mean that \\begin{align*}\\lambda = \\frac{1}{3}\\end{align*}. Denote the point of intersection $Q$, it lies on", "P\\end{align*} and z = 0 for all points is the intersection of the orange and green planes in the diagram below right. (Remember that the orange and green planes are drawn with edges for convenience, but that they represent planes that extend outward toward infinity.) Therefore we can specify a particular line by requiring that the equations for both planes be simultaneously true. In a previous section we developed one equation for a plane given by d = -nx xo -ny yo -nz zo. In this case Qy + Px - QP = 0 and z = 0 must both be satisfied for all points on the line passing through points P and Q. Finding the Vector Equation of a Line Another way to identify points on a line can be found using the vector addition method we discussed earlier in this chapter. To find the position vector, r\\begin{align*}\\overrightarrow{r}\\end{align*}, for any point along a line, we can add the position vector of a point on the line which we already know and add to that a vector, v\\begin{align*}\\overrightarrow{v}\\end{align*}, that lies on the line as shown in the diagram below. The position vector R\\begin{align*}\\overrightarrow{R}\\end{align*} for a point between P and Q is given by R=p+v\\begin{align*}\\overrightarrow{R} = \\overrightarrow{p} + \\overrightarrow{v}\\end{align*} All other points on this line can be reached by", "below. We can also apply a similar process to find the parametric equations of the line. This time, we’ll use the general form: \\begin{aligned}x&= x_o + at \\\\ y&= y_o + bt\\\\ z&= z_o + ct \\end{aligned} Using our previous example, $\\textbf{r}_o = <2, 4, 3>$, and is parallel to the vector, $\\textbf{v} = 2 \\textbf{i} -3\\textbf{j} + \\textbf{k}$. Hence, we have the following: \\begin{aligned}\\textbf{r}_o &= \\\\&= <2, 4, 3>\\\\ \\textbf{v} &= \\\\ &= <2, -3, 1>\\end{aligned} \\begin{aligned} x &= x_o + at\\\\ &= 2 + 2t\\end{aligned} \\begin{aligned} y &= y_o + bt\\\\ &= 4 – 3t\\end{aligned} \\begin{aligned} z &= z_o + ct\\\\ &= 3 + t\\end{aligned} We’ve prepared more examples for you to master this topic. When you’re ready, head over to the next section! Example 1 Find the equation of the line passing through $(2, 5, -4)$ and is parallel to the vector, $\\textbf{v} = 6\\textbf{i} + 5\\textbf{j} – 2\\textbf{k}$. Write its vector and parametric equations. Solution First, we’ll define $\\textbf{r}_o$ as $2\\textbf{i} + 5\\textbf{j} – 4\\textbf{k}$. We want the line to be parallel to the vector, $\\textbf{v} = 6\\textbf{i} + 5\\textbf{j} – 2\\textbf{k}$. We’ll use these two vectors to find the vector equation of the line using. \\begin{aligned}\\textbf{r}_o &= 2\\textbf{i} + 5\\textbf{j} – 4\\textbf{k} \\\\\\textbf{v} &= 6\\textbf{i} + 5\\textbf{j} – 2\\textbf{k}\\\\\\\\\\textbf{r} &= \\textbf{r}_o + t\\textbf{v}\\\\&=", "+0 # Consider the line parametrized by \\begin{align*} x&= 4t + 2,\\\\ y& = t+2. \\end 0 207 1 Consider the line parametrized by$$x= 4t + 2, y = t+2$$ Find vector $$\\begin{pmatrix}a\\\\b\\end{pmatrix}$$ that's parallel to this line and satisfies $$a+b=10$$. Feb 2, 2019 #1 +103947 +1 Here's my best effort....although there might be an easier method !!! Wnen = 0 we have the point (2, 2) When t = 1 we have the point (6, 3) So...the slope of the given line is (3 - 2) / (6 - 2) = 1/4 If we let the parallel line go through the origin....we have the line y = (1/4)x This means that every point on this line has the coordinates ( x, 1/4x) This implies that x + (1/4)x = 10 (5/4)x = 10 x = 8 and y = (1/4)x = 2 So....the required vector is < 8, 2 > Feb 2, 2019", "starting with two vectors: one vector $\\vec{r}_0$ drawn from the origin to a chosen point of the line (the so-called position vector of this point) and another vector (the so-called {\\em direction vector}) $\\vec{u}$ determining the direction of the line. Then there exists a real number $t$ such that the position vector $\\vec{r}$ of an arbitrary point of the line can be expressed as \\begin{align} \\vec{r} \\;=\\; \\vec{r}_0\\!+\\!t\\vec{u}. \\end{align} This is the {\\em vector-formed equation of the line}. Since every vector can be given by three coordinates of the end-point, we may write $$\\vec{r} \\;=\\; \\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix},\\quad \\vec{r}_0 \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix},\\quad \\vec{u} \\;=\\; \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!.$$ $$\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix} \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix}+t \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!,$$ and we obtain three scalar equations \\begin{align} \\begin{cases} x \\;=\\; x_0\\!+\\!ta\\\\ y \\;=\\; y_0\\!+\\!tb\\\\ z \\;=\\; z_0\\!+\\!tc. \\end{cases} \\end{align} These are the {\\em parametric equations of the line} (3), with the parameter $t$ taking all real values. The contents of (4) is often written with proportions: $$\\frac{x\\!-\\!x_0}{a} \\;=\\; \\frac{y\\!-\\!y_0}{b} \\;=\\; \\frac{z\\!-\\!z_0}{c}\\;\\;\\; (= t)$$ \\begin{thebibliography}{8} \\bibitem{LL}{\\sc L. Lindel\\\"of}: {\\em Analyyttisen geometrian oppikirja}.\\, Kolmas painos.\\, Suomalaisen Kirjallisuuden Seura, Helsinki (1924). \\end{thebibliography} %%%%% %%%%% \u001bnd{document}", "- \\overline{OP} \\\\ &= <1, 0, 4> - <1, 1, 2> \\\\ &= <0, -1, 2> \\end{align*} To make it infinitely long, we multiply by some parameter, \\displaystyle \\begin{align*} t \\end{align*}, to give \\displaystyle \\begin{align*} t<0, -1, 2> \\end{align*}. Then we need to position this infinitely long vector at a point. Adding either of the points you know the line goes through will do. So \\displaystyle \\begin{align*} l = <1, 1, 2> + t<0, -1, 2> \\end{align*}. This is the vector form of the line. To get the parametric equations, expand it out... \\displaystyle \\begin{align*} l &= <1, 1 - t, 2 + 2t> \\\\ \\\\ x &= 1 \\\\ \\\\ y &= 1 - t \\\\ \\\\ z &= 2 + 2t \\end{align*} To get the Cartesian equations, try to write each of them as t in terms of the other variable and set them equal.", "persevere, find the equation of both the lines equate them and solve for x but that is going to be a humongous algebraic slog. There are then two options, persevere or go at it via a different method. Given that this is the first part of the question looking for a more concise method is undoubtedly the way to go. Given that we are thinking about geometrical properties of lines and their midpoints it makes sense to use vectors to solve this problem. Most importantly it will likely cut down on the algebra a lot due to their concise notation. \\begin{align*}\\ \\ Let\\ \\overrightarrow{OA} = \\mathbf{a} = \\begin{bmatrix} p_1 \\\\ q_1 \\end{bmatrix},\\ \\overrightarrow{OB} = \\mathbf{b} = \\begin{bmatrix} p_2 \\\\ q_2 \\end{bmatrix},\\ \\overrightarrow{OC} = \\mathbf{c} = \\begin{bmatrix} p_3 \\\\ q_3 \\end{bmatrix} \\end{align*} \\begin{align*}Let\\ the\\ midpoints\\ of\\ AB,\\ AC,\\ BC\\ be\\ denoted\\ D,\\ E,\\ F \\end{align*} \\begin{align*}\\ \\ \\therefore \\overrightarrow{OD} = \\mathbf{\\frac{1}{2}(a+b)},\\ \\overrightarrow{OE} = \\mathbf{\\frac{1}{2}(a+c)},\\ \\overrightarrow{OF} = \\mathbf{\\frac{1}{2}(b+c)}\\end{align*} That has set up all of the required details to begin actually answering the question. The vector equations for the line joining A and the midpoint of BC and the line joining B and the midpoint AC, are the lines $AF$ and $BE$: \\begin{align*}\\mathbf{r_{AF}} = \\mathbf{a} + \\lambda (\\mathbf{b + c - 2a}) \\end{align*} \\begin{align*}\\mathbf{r_{BE}} = \\mathbf{b} + \\mu (\\mathbf{a+c-2b}) \\end{align*} They", "at 5:40 Consider the points $p$, $q$, $r$. step 1 Construct the line from $p$ to $q$, i.e. $PQ=q-p$ and line from $p$ to $r$, i.e. $PR=r-p$ step 2 Find the cross product of lines $PQ=(x_1 , x_2 , x_3)$ and $PR=(y_1 , y_2 , y_3)$ \\begin{align} n=\\begin{vmatrix}i & j & k \\\\ x_1 & x_2 & x_3 \\\\ y_1 & y_2 & y_3\\end{vmatrix} \\end{align} This will be the normal line to the plane (perpendicular). $i,j,k$ are the co-ordinate axes. So if you get $n_1i+n_2j+n_3k$, then $(n_1,n_2,n_3)$ is the desired normal point. step 3 Construct the normal equation of the plane \\begin{align} n_1(x-a)+n_2(y-b)+n_3(x-c)=0 \\end{align} where $(a,b,c)$ can be the co-ordinates of any of the points $p,q,r$(reference point). -", "## Generalization of the equation for a line Generalization of the equation for a line In the 2D world, the line equation is well known: \\begin{align*} \\frac{y-y_0}{x-x_0} = k \\end{align*} where k is the slope. Here is a generalized form: \\begin{align*} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} - \\begin{pmatrix} x_1^* \\\\ x_2^* \\end{pmatrix} &= a \\begin{pmatrix} v1 \\\\ v2 \\end{pmatrix} \\end{align*} or more concisely, $x - x^* = av$. This comes in handy when you want to get the equation for the normal line of a level surface, instead of the following awkward presentation, You can use: \\begin{align*} \\mathbf{x} - \\mathbf{x^*} = a\\nabla F \\end{align*}", "+0 0 44 1 Consider the line through points A = (1, 1, 2) and B = (2, 3, 4), and let P be the foot of the perpendicular from Q = (3, -7, -1) to this line, as in the picture below: Then $$\\overrightarrow{AP}$$ is the projection of $$\\overrightarrow{AQ}$$ onto a vector $$\\mathbf{u} = \\begin{pmatrix} u_1 \\\\ u_2 \\\\ u_3 \\end{pmatrix}$$such that $$u_1 + u_2 + u_3 = 5.$$ What is $$\\mathbf{u}$$? Mar 21, 2020", "traveling along the line from point P, therefore the position vector for any point on the line is given by r=p+kv\\begin{align*}\\overrightarrow{r} = \\overrightarrow{p} + k\\overrightarrow{v}\\end{align*}. If we know the locations of two points on a line, we can determine the equation of the line. All points on a line fulfill the equation r=p+kv\\begin{align*}\\overrightarrow{r} = \\overrightarrow{p} + k\\overrightarrow{v}\\end{align*}, where k is a scalar that varies from -∞ to ∞. If we already know the position vectors for two points on the line, p\\begin{align*}\\overrightarrow{p}\\end{align*} and q\\begin{align*}\\overrightarrow{q}\\end{align*}, we can use the method of vector subtraction to determine the equation of the vector, v=pq\\begin{align*}\\overrightarrow{v} = \\overrightarrow{p} - \\overrightarrow{q}\\end{align*}. Therefore, r=p+k(pq)\\begin{align*}\\overrightarrow{r} = \\overrightarrow{p} + k (\\overrightarrow{p} - \\overrightarrow{q})\\end{align*}, where k varies from -∞ to ∞. General Vector Equations In general, a vector equation is any function that takes any one or more variables and returns a vector. The vector equation of a line is an equation that identifies the position vector of every point along the line. This works for straight lines and for curves. For example, the vector equation p=3 cos θ,3 sin θ,2\\begin{align*}\\overrightarrow{p} = \\left \\langle 3\\ \\mbox{cos}\\ \\theta, 3\\ \\mbox{sin}\\ \\theta, 2\\right \\rangle\\end{align*} defines a circle having a radius of 3 units which sits parallel to the x-y plane at a distance of 2 units along the z-axis as shown", "intersection, simply plug $$t$$ or $$s$$ into their corresponding vector line equations. Plugging $$t=1$$ into the first equation will get: \\begin{align} \\vec{r_1} & = \\begin{bmatrix} 2 \\\\ -4 \\\\ 3 \\end{bmatrix} + 1 \\cdot \\begin{bmatrix} 3 \\\\ 11 \\\\ -2 \\end{bmatrix} \\\\ & = \\begin{bmatrix} 5 \\\\ 7 \\\\ 1 \\end{bmatrix} .\\end{align} So the lines intersect at the point $$(5, 7, 1).$$ Let's look at one final example of a question like this, this time with the line in parametric form. Determine whether the following lines intersect. If they intersect, find the point of intersection. If they do not intersect, determine whether they are skewed or parallel. Line 1: \\begin{align} x & = 11 - 10t \\\\ y & = 2 - 7t \\\\ z & = 3. \\end{align} Line 2: \\begin{align} x & = 1 + 2s \\\\ y & = 1 + 6s \\\\ z & = 1 - 2 s. \\end{align} Solution First, determine if the lines are parallel. Since this isn't in vector form, you need to create the direction vectors to be able to check this. The direction vectors here will be made using the coefficient of the scalar terms $$t$$ and $$s$$. The direction vector for Line 1 is: $\\begin{bmatrix} -10 \\\\ -7 \\\\ 0 \\end{bmatrix},$ and the direction vector for Line", "+0 # Consider the line through points A = (1, 1, 2) and B = (2, 3, 4), and let P be the foot of the perpendicular from Q = (3, -7, 0 182 1 Consider the line through points A = (1, 1, 2) and B = (2, 3, 4), and let P be the foot of the perpendicular from Q = (3, -7, -1) to this line, as in the picture below: Then $$\\overrightarrow{AP}$$ is the projection of $$\\overrightarrow{AQ}$$ onto a vector $$\\mathbf{u} = \\begin{pmatrix} u_1 \\\\ u_2 \\\\ u_3 \\end{pmatrix}$$ such that $$u_1 + u_2 + u_3 = 5.$$ What is u? Apr 1, 2019", "also draw the parabola with a horizontal orientation, such that the equation will be y2=4px\\begin{align*}y^2=4px\\end{align*}. Notice, that when the parabola opens to the left or right, the y\\begin{align*}y\\end{align*} is squared. In this concept, the vertex will be (0,0)\\begin{align*}(0, 0)\\end{align*}. Let's analyze the equation y2=12x\\begin{align*}y^2=-12x\\end{align*}. We'll find the focus, directrix, and determine if the function opens up, down, to the left or right. Then, we'll graph the parabola. To find the focus and directrix, we need to find p\\begin{align*}p\\end{align*}. We can set 12=4p\\begin{align*}-12 = 4p\\end{align*} and solve for p\\begin{align*}p\\end{align*}. 123=4p=p\\begin{align*}-12&=4p \\\\ -3&=p\\end{align*} Because y\\begin{align*}y\\end{align*} is squared, we know that the parabola opens to the left or right. Because p\\begin{align*}p\\end{align*} is negative, we know it is going to open to the left, towards the negative side of the x\\begin{align*}x\\end{align*}-axis. Using the pictures above, this parabola is like the second one under y2=4px\\begin{align*}y^2=4px\\end{align*}. Therefore, the focus is (3,0)\\begin{align*}(-3,0)\\end{align*} and the directrix is x=3\\begin{align*}x=3\\end{align*}. To graph the parabola, plot the vertex, focus, directrix, and sketch the curve. Find at least one or two points on the curve to make sure your sketch is accurate. For example, because (3,6)\\begin{align*}(-3, 6)\\end{align*} is on the parabola, then (3,6)\\begin{align*}(-3, -6)\\end{align*} is also on the parabola because it is symmetrical. Notice that the points (3,6)\\begin{align*}(-3, 6)\\end{align*} and (3,6)\\begin{align*}(-3, -6)\\end{align*} are equidistant from the focus and the directrix.", "and $<b_1, b_2, b_3>$ are two distinct lines, then the coordinates for the point of intersection $[p_1, p_2, p_3]$ are given by: (2) \\begin{align} \\quad [p_1,p_2, p_3] = \\left [ \\begin{vmatrix} a_2 & a_3\\\\ b_2 & b_3 \\end{vmatrix}, \\begin{vmatrix} a_3 & a_1\\\\ b_3 & b_1 \\end{vmatrix}, \\begin{vmatrix} a_1 & a_2\\\\ b_1 & b_2 \\end{vmatrix} \\right ] \\end{align} Let's look at an example. ## Example 1 Consider the projective plane $\\mathbb{P}^2(\\mathbb{R})$. Find the point of intersection of the lines $<1, 1, 4>$ and $<2, 3, 1>$. Applying the formula above yields: (3) \\begin{align} \\quad [p_1,p_2, p_3] &= \\left [ \\begin{vmatrix} 1 & 4\\\\ 3 & 1 \\end{vmatrix}, \\begin{vmatrix} 4 & 1\\\\ 1 & 2 \\end{vmatrix}, \\begin{vmatrix} 1 & 1\\\\ 2 & 3 \\end{vmatrix} \\right ] \\\\ \\quad [p_1, p_2, p_3] &= [1 - 12, 8 - 1, 3 - 2] \\\\ \\quad [p_1, p_2, p_3] &= [-11, 7, 1] \\end{align} Similarly, we can verify that $[-11, 7, 1]$ lies on both $<1, 1, 4>$ and $<2, 3, 1>$ by verifyin that $[-11, 7, 1] \\cdot <1, 1, 4> = 0$ and $[-11, 7, 1] \\cdot <2, 3, 1> = 0$. We have that: (4) \\begin{align} \\quad [-11, 7, 1] \\cdot <1, 1, 4> = (-11)(1) + (7)(1) + (1)(4) = 0 \\end{align} And also: (5) \\begin{align} \\quad [-11, 7," ]

[ "We reflect $\\overline{CD}$ about the $x$-axis to get $\\overline{C'D'}$ with $D'=(7,-5),$ then reflect $\\overline{C'D'}$ about the $y$-axis to get $\\overline{C'D''}$ with $D''=(-7,-5).$ We obtain the following diagram: $[asy] /* Made by MRENTHUSIASM */ size(225); int xMin = -9; int xMax = 9; int yMin = -7; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (7,-5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(C--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot(\"A(3,5)\",A,(0,2),linewidth(3.5)); dot(\"B\",B,(-2,0),linewidth(3.5)); dot(\"C\",C,(0,-2),linewidth(3.5)); dot(\"D(7,5)\",D,(0,2),linewidth(3.5)); dot(\"C'\",E,(0,-2),linewidth(3.5)); dot(\"D'(7,-5)\",F,(0,-2),linewidth(3.5)); dot(\"D''(-7,-5)\",G,(0,-2),linewidth(3.5)); [/asy]$ The total distance that the beam will travel is \\begin{align*} AB+BC+CD&=AB+BC+CD' \\\\ &=AB+BC'+C'D'' \\\\ &=AD'' \\\\ &=\\sqrt{((3-(-7))^2+(5-(-5))^2} \\\\ &=\\boxed{\\textbf{(C) }10\\sqrt2}. \\end{align*} ~MRENTHUSIASM (Solution) ~JHawk0224 (Proposal) ## Solution 2 (Parallelogram) Define points $A,B,C,$ and $D$ as Solution 1 does. Moreover, let $E$ be a point on $\\overline{CD}$ such that $\\overline{BE}$ is perpendicular to the $y$-axis, and $F$ be a point on $\\overline{BE}$ such that $\\overline{CF}$ is perpendicular to the $x$-axis, as shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (4,2); pair F = (2,2); draw(A--B--C--D,red);", "# Asymptote fill area between two contours I use Asymptote to make the graphics. I have two curves defined as contours C_1 and C_2. I want to fill the area between the two curves. The problem is that contour returns a guide structure in Asymptote, and this is not accepted to to buildcycle command. Can anyone tell me how to solve this problem ? Thanks. This is the code I use and which produces the two curves, but fails to fill area between these. import graph; import patterns; import contour; usepackage(\"mathrsfs\"); size(8cm); real eps=0.001; real xmax=5.5,ymax=5.5; pair af,ag; real f(real x, real y) {return sqrt(1+x)-sqrt(x)+sqrt(1+y)-sqrt(y);} real g(real x, real y) {return 1/sqrt(1+x)+1/sqrt(1+y);} guide[][] Cf=contour(f,(eps,eps),(xmax,ymax),new real[] {1}); guide[][] Cg=contour(g,(eps,eps),(xmax,ymax),new real[] {1}); //these two lines are used to find intersection points path D1=(0,ymax)--(xmax+eps,ymax); path D2=(xmax,0)--(xmax,ymax+eps); draw(Cf); draw(Cg); //path pp=buildcycle(Cf,D1,Cg,D2); // the preceding line produces and error if commented xaxis(Label(scale(0.75)*\"$x$\"),xmax=6,Arrow); yaxis(Label(rotate(90)*scale(0.75)*\"$y$\"),ymax=6,Arrow); label(\"$\\mathscr{C}_{1}$\",(xmax,0.2)); label(\"$\\mathscr{C}_{2}$\",(xmax,1.9)); label(\"I\" ,(.2,.2)); label(\"II\" ,(1.,1.)); label(\"III\",(3.5,3.5)); end; • Is there some reason why you are using contour to produce the lines? It seems a rather complicated way to produce a single curve. May 1 '15 at 9:26 • Using f to write y = function (x) lead to a long expression. Around zero it requires large precision for fine tuned draw. Contour allows easier control over the drawing box.", "May 1 '15 at 22:10 It works when replacing Cf and Cg by Cf[0][0] and Cg[0][0], which allows to obtain the desired (single) guides. path pp=buildcycle(Cf[0][0],D1,Cg[0][0],D2); Note that I've also slightly modified D1 and D2: in your definitions, D1, D2 and pp did not intersect anyway. path D1=(0,ymax-eps)--(xmax+eps,ymax-eps); path D2=(xmax-eps,0)--(xmax-eps,ymax+eps); For proper explanations about how to use contour, see the very well-made (alas not complete) tutorial on Asymptote by Charles Staats, p. 32. import graph; import patterns; import contour; usepackage(\"mathrsfs\"); size(8cm); real eps=0.001; real xmax=5.5,ymax=5.5; pair af,ag; real f(real x, real y) {return sqrt(1+x)-sqrt(x)+sqrt(1+y)-sqrt(y);} real g(real x, real y) {return 1/sqrt(1+x)+1/sqrt(1+y);} guide[][] Cf=contour(f,(eps,eps),(xmax,ymax),new real[] {1}); guide[][] Cg=contour(g,(eps,eps),(xmax,ymax),new real[] {1}); //these two lines are used to find intersection points path D1=(0,ymax-eps)--(xmax+eps,ymax-eps); path D2=(xmax-eps,0)--(xmax-eps,ymax+eps); path pp=buildcycle(Cf[0][0],D1,Cg[0][0],D2); fill(pp,.8white); draw(pp); xaxis(Label(scale(0.75)*\"$x$\"),xmax=6,Arrow); yaxis(Label(rotate(90)*scale(0.75)*\"$y$\"),ymax=6,Arrow); label(\"$\\mathscr{C}_{1}$\",(xmax,0.2)); label(\"$\\mathscr{C}_{2}$\",(xmax,1.9)); label(\"I\" ,(.2,.2)); label(\"II\" ,(1.,1.)); label(\"III\",(3.5,3.5)); • Great, it works fine ! thank a lot fpast, for the answer and for the suggestions. Apr 30 '15 at 18:41 Just for comparison, here is a Metapost version. There's no support for implicit equations or anything so clever (or complex) as Asymptote's contour facility, so I could not avoid some algebra to get the \"long expression\" that the OP was happy to avoid. And to avoid overflows when x is near zero, I'm exploiting the facts that these two curves have", "we can solve to get $d = \\boxed{672}$.$[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired. Solution 3 (Inversion) WLOG assume $\\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\\omega$ being $60^\\circ$.", "yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (-7,5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(E--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot(A,linewidth(3.5)); dot(B,linewidth(3.5)); dot(C,linewidth(3.5)); dot(D,linewidth(3.5)); dot(E,linewidth(3.5)); dot(F,linewidth(3.5)); dot(G,linewidth(3.5)); label(\"A(3,5)\",A,(0,2)); label(\"B\",B,(-2,0)); label(\"C\",C,(0,-2)); label(\"D(7,5)\",D,(0,2)); label(\"C'\",E,(0,-2)); label(\"D'\",F,(0,2)); label(\"D''\",G,(0,-2)); [/asy]$ It follows that $D'=(-7,5)$ and $D''=(-7,-5).$ The total distance that the beam will travel is \\begin{align*} AB+BC+CD&=AB+BC'+C'D' \\\\ &=AB+BC'+C'D'' \\\\ &=AD'' \\\\ &=\\sqrt{((3-(-7))^2+(5-(-5))^2} \\\\ &=\\boxed{\\textbf{(C) }10\\sqrt2}. \\end{align*} ~MRENTHUSIASM (Solution) ~JHawk0224 (Proposal) ## Solution 2 (Algebra) Define points $A,B,C,$ and $D$ as Solution 1 does. When a straight line hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ have negative slopes. Let $\\ell$ be the line containing $P$ and perpendicular to that coordinate axis. Geometrically, these two rays coincide when reflected about $\\boldsymbol{\\ell}.$ Let the slope of $\\overline{AB}$ be $m.$ It follows that the slope of $\\overline{BC}$ is $-m,$ and the slope of $\\overline{CD}$ is $m.$ Here, we conclude that $\\overline{AB}\\parallel\\overline{CD}.$ Next, we locate $E$ on $\\overline{CD}$ such that $\\overline{BE}\\parallel\\overline{AD}.$ We obtain parallelogram $ABED,$ as shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair", "\\boxed{672}$. $[asy] import olympiad; size(230pt); defaultpen(linewidth(0.8)+fontsize(10pt)); real r1 = 17, r2 = 27, d = 35, r = 18; pair O1 = origin, O2 = (d,0); path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired.", "100; real yMin = 0; real yMax = 120; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"R\",(xMax,0),(2,0)); label(\"Im\",(0,yMax),(0,2)); pair A,B,C,D; A = (18,83); B = (18,39); C = (78,99); D = (56,104.908997); dot(A); dot(B); dot(C); dot(D); draw(A--C); draw(A--B); draw(D--C); draw(C--B); draw(D--B); label(\"Z_1\", A, W); label(\"Z_2\", B, W); label(\"Z_3\", C, N); label(\"Z\", D, N); draw(circle((56,61), 43.9089968)); [/asy]$ While $Z_1, Z_2, Z_3$ are fixed, $Z$ can be anywhere on the circle because those are the only values of $Z$ that satisfy the problem requirements. However, we want to find the real part of the $Z$ with the maximum imaginary part. This Z would lie directly above the center of the circle, and thus the real part would be the same as the x-value of the center of the circle. So all we have to do is find this value and we’re done. Consider the perpendicular bisectors of $Z_1Z_2$ and $Z_2Z_3$. Since any chord can be perpendicularly bisected by a radius of a circle, these two lines both intersect at the center. Since $Z_1Z_2$ is vertical, the perpendicular bisector will be horizontal and pass through the midpoint of this line, which is (18, 61). Therefore the equation for this line is $y=61$. $Z_2Z_3$ is nice because it turns out the differences in the x and y values are both equal (60) which means that the slope", "that the graph origin is left-right centered. I've added the black border in the plot below just to illustrate the white border to the left of the plot. // arctan.asy import settings; settings.outformat=\"pdf\"; settings.render=0; unitsize(1cm); import graph; real f(real x) { return (atan(x)+(pi/2.0))/pi; } real xmin = -3.5; real xmax = 3.5; real ymin = -0.35; real ymax = 1.5; draw((xmin,0) -- (xmax,0), blue, arrow=None); draw((0,ymin) -- (0,ymax), blue, arrow = None); path g = graph(f,xmin,xmax); draw(g,red); label(\"$\\scriptstyle (\\arctan(x)+(\\pi/2))/\\pi$\", (xmax,0.8), S); void centerAtOrigin() { pair origMinPoint = min(currentpicture)/72*2.54; pair origMaxPoint = max(currentpicture)/72*2.54; pair origSize = size(currentpicture)/72*2.54; real xmin = origMinPoint.x; real xmax = origMaxPoint.x; real ymin = origMinPoint.y; if (xmax > fabs(xmin)) { xmin = -xmax; } if (fabs(xmin) > xmax) { xmax = -xmin; } fill(shift(xmin,ymin)*scale(xmax-xmin,origSize.y)* unitsquare, opacity(0)+white); } centerAtOrigin(); Edit: Instead of showing the label even though it's outside of the bounding box, this method expands the bounding box on the other side to keep your vertical axis centered.", "input. – Surya Nov 26 '13 at 7:29 @Surya: Yes, that's a general method. You need to make sure to assign variable names for your independent variables that aren't x and y. I've edited my answer. – Jake Nov 26 '13 at 7:49 Okay. That was the mistake then. Thanks. – Surya Nov 26 '13 at 8:11 The code: \\documentclass[margin=10pt]{standalone} \\usepackage{asymptote} \\begin{document} \\begin{asy} settings.render=4; settings.prc=false; size3(8cm); defaultpen(fontsize(10pt)); import graph3; currentprojection=orthographic(5,2,3); real ymax = 3; real ymin = -ymax; real h(real y) { return sqrt(1 + (y/2)^2); } path hyperbola = graph(h, ymin, ymax, operator ..); path3 hyperbola3 = path3(hyperbola, plane=YZplane); surface hyperboloid = surface(hyperbola3, c=O, axis=Y); real umax = hyperboloid.index.length - .001; real vmax = hyperboloid.index[0].length - .001; /* Parametrize by the square [0,2pi] x [ymin,ymax] */ triple point(real theta, real y) { return hyperboloid.point(theta/(2pi) * umax, (y-ymin)/(ymax-ymin) * vmax); } triple normal(real theta, real y) { return hyperboloid.normal(theta/(2pi) * umax, (y-ymin)/(ymax-ymin) * vmax); } int n = 10; int m = 10; real shrinkfactor=0.8; for (int i = 0; i < n; ++i) { real theta = i/n*2pi; triple f(real t) { return point(theta, t); } for (int j = 0; j < m; ++j) { real ylow = j/m*(ymax-ymin) + ymin; real yhigh = ylow + shrinkfactor*(ymax-ymin)/m; path3 arrowpath = graph(f, ylow, yhigh, n=10, operator..); }", "are dotted with blue. Notice that in this critical (in any sense) case, we can not use the built-in path routines mintimes, maxtimes, times. // http://asymptote.ualberta.ca/ // Find all local maximal.minimal points unitsize(1cm); import graph; real f(real x){return (.6*cos(4.5*(x-4)+2.1)-1.2*sin(x-4)+.1*x+.2);} //real f(real x){return .5+sin(3x);} real xmin=-2; real xmax=10.2; xaxis(\"$x$\",xmin-.5,xmax+.5); yaxis(\"$y$\"); guide p=graph(f,xmin,xmax,Spline); draw(p,darkgreen+.6pt); int n=200; real h=(xmax-xmin)/n; real[] x,y; for (int k=0; k<n; ++k) { x.push(xmin+h*k); y.push(f(xmin+h*k)); } //write(x,y); for (int k=2; k<n-2; ++k){ if ( (y[k] > y[k-1]) & (y[k] > y[k+1]) ) dot( (x[k],y[k]),red); if ( (y[k] < y[k-1]) & (y[k] < y[k+1]) ) dot( (x[k],y[k]),blue); } shipout(bbox(5mm,invisible)); For f(x)=.5+sin(3x), we get Even more, this procedure can be used to find all intersection point of the graph with the horizontal line through a given point A. In other words, more or less does the job for the following routines // http://asymptote.ualberta.ca/ // Find all intersection points with horizontal line through given point A unitsize(1cm); import graph; real f(real x){return .5+sin(3x);} real xmin=-2; real xmax=10.2; xaxis(\"$x$\",xmin-.5,xmax+.5); yaxis(\"$y$\"); guide p=graph(f,xmin,xmax,Spline); draw(p,darkgreen+.6pt); pair A=(1,.6); draw((xmin,A.y)--(xmax,A.y),blue); int n=5000; real h=(xmax-xmin)/n; real[] x,y; for (int k=0; k<n; ++k) { x.push(xmin+h*k); y.push(f(xmin+h*k)); } for (int k=2; k<n-2; ++k) if ( (y[k]-A.y)*(y[k+1]-A.y)<0) dot( ((x[k]+x[k])/2,A.y),red); shipout(bbox(5mm,invisible)); There's a quick and easy approach using the sagetex package. It gives you access to a CAS, called Sage", "my program I denote the position of a point charge by the vector $$\\vec{a}$$ with coordinates $$a_0,a_1,a_2$$, and introduce a cartesian coordinate system with coordinates $$x$$ (to the right), $$y$$ (to the top) and $$z$$ (into the plane of view). Then I can write the field of one charge in componentes as: \\begin{align} E_x &= q\\cdot(x-a_0)/r^3 \\\\ E_y &= q \\cdot (y-a_1)/r^3 \\end{align} where $$r^3 = ((x-a_0)^2 + (x-a_1)^2 + (z-a_2))^{1.5}$$ I look only at the plane $$z = 0$$ and since I draw in 2D (projection to the plane $$z = 0$$) I omit $$E_z$$ (but I am not sure, if this introduces a mistake). Then I use a loop to equally distribute the charges on my planes and calculate the total field as described above. # Imports import matplotlib from numpy import * from pylab import * import matplotlib.colors as colors def Cap(N = 10,plus=1,minus=1,app=''): # Aspect ratio aspekt='auto' ## Field between the plates xMin,xMax = -0.15,0.15 yMin,yMax = -10,10 zMin,zMax = -10,10 ## Field outside # xMin,xMax = -3,3 # yMin,yMax = -3,3 ## Plate distance d = 0.1 # Case of plates very far apart # d = 50 # xMin,xMax=-d*1.25,d*1.25 # yMin,yMax=-d*1.25,d*1.25 # aspekt = 'equal' ## Grids xG,yG = meshgrid(linspace(xMin,xMax,18),linspace(yMin,yMax,18)) xGM,yGM = meshgrid(linspace(xMin,xMax,500),linspace(yMin,yMax,500)) # Class that represents a point charge class charge:", "b*b+h*h<49 enddef; h = solve f(0,7); a = sqrt(64-h*h); z0 = origin; z1 = (a,h); z2 = (6,0); z3 = (64/6,0); transform t; z0 transformed t = z0; z2 transformed t = z1; z1 transformed t = z3; path p; p := z0 -- z1 -- z2; draw p scaled 1cm; draw p transformed t scaled 1cm; draw mark(z0,z1,z2) scaled 1cm; draw mark(z1,z3,z0) scaled 1cm; dotlabel.lft(btex $A$ etex, z0); dotlabel.top(btex $B$ etex, z1 scaled 1cm); dotlabel.bot(btex $C$ etex, z2 scaled 1cm); dotlabel.rt (btex $D$ etex, z3 scaled 1cm); endfig; The general technique for solving non-linear equations in Meta(font|post) is in Appendix D, section 3 of the METAFONT book. Is this the simplest solution? first take D such that ABCD is a parallelogram, and get a point E on the bisector of the angle BAD, then reflect C to the line AE to get point F. The desired point G is the intersection of AF and BC. size(7cm); import math; real a=6, b=7, c=8; pair B=(0,0), C=(a,0); pair A=intersectionpoints(circle(B,c),circle(C,b))[0]; pair D=A+C-B; pair E=A+dir(B--A,D--A); // E is on the bisector of the angle BAD pair F=reflect(A,E)*C; pair G=extension(A,F,B,C); // <<< the point we need // marking angles draw(arc(G,1,degrees(A-G),degrees(C-G))); draw(arc(A,1,degrees(B-A),degrees(C-A))); draw(arc(A,1,degrees(G-A),degrees(D-A),CCW),orange); drawline(A,E,lightblue); draw(A--D--C--F,orange); draw(C--G--A,blue); draw(Label(\"8\",align=W),A--B); draw(Label(\"6\",align=S),B--C); draw(Label(\"7\",align=W),C--A); label(\"$A$\",A+.5dir(60)); label(\"$B$\",B,SW); label(\"$C$\",C,S); dot(new pair[]{A,B,C,D,F},UnFill); dot(Label(\"$G$\",align=SE),G,magenta); dot(Label(\"$D$\",align=plain.E),D,orange,UnFill); dot(Label(\"$F$\",align=NE),F,orange,UnFill); shipout(bbox(5mm,invisible));", "can use built-in function times t=times(gf,x)[0]; to get a value of the time parameter t, which corresponds to the intersection of the function curve and a vertical line at x=1. This t value allows to: first, get a missing 'y' coordinate of the tangent point P P=point(gf,t); And second, the direction of the tangent line at this point as dir(gf,t). Function drawline (part of the basic Asymptote module math.asy), allows to draw the visible portion of the (infinite) line going through two points: drawline(P,P+dir(gf,t),tanLinePen); Another useful function for this drawing is relpoint, which returns the point on curve at the relative fraction of its arclength. This is a complete MWE: // tan.asy // // run // asy tan.asy // // to get tan.pdf // settings.tex=\"pdflatex\"; import graph; import math; size(6cm); import fontsize;defaultpen(fontsize(10pt)); texpreamble(\"\\usepackage{lmodern}\" +\"\\usepackage{amsmath}\" +\"\\usepackage{amsfonts}\" +\"\\usepackage{amssymb}\" ); pen funcLinePen=darkblue+0.9bp; pen tanLinePen=orange+0.9bp; pen grayPen=gray(0.3)+0.8bp; pen dashPen=gray(0.3)+0.8bp+linetype(new real[]{5,5})+linecap(0); real xmin=-2,xmax=-xmin; real ymin=0,ymax=4; real dxmin=0.2; real dxmax=dxmin; real dymin=dxmin; real dymax=dxmax; xaxis(\"$x$\",xmin-dxmin,xmax+dxmax,RightTicks(Step=1,step=0.5),arr,above=true); yaxis(\"$y$\",ymin-dymin,ymax+dymax,LeftTicks (Step=1,step=0.5,OmitTick(0)),arr,above=true); real f(real x){return x^2;} guide gf=graph(f,xmin,xmax,operator..); real x=1, t=times(gf,x)[0]; pair P=point(gf,t), Q=relpoint(gf,6/7); draw(gf,funcLinePen); draw((P.x,0)--P--(0,P.y),dashPen); drawline(P,P+dir(gf,t),tanLinePen); dot((P.x,0)^^P^^(0,P.y)^^Q,UnFill); label(\"$y=x^2$\",relpoint(gf,1/4),UnFill); label(\"$P(1,\"+string(round(P.y))+\")$\",P,plain.SE); label(\"$Q(x,x^2)$\",Q,plain.W); label(\"$T$\",Q,3*plain.E); I'm surprised that the original question is tagged but one answer (the accepted one) uses and the other uses ! So I decide to post a TikZ answer, but this is mainly a mathematics answer. \\documentclass[tikz]{standalone} \\usetikzlibrary{through}", "z = ((0, .8) -- (infinity, .8)) intersectionpoint graph; draw ((x, 0) -- z -- (0, y)) xyscaled (u, v) dashed evenly; draw (0, v) -- (xmax*u, v) dashed evenly; % Axes and labels drawarrow origin -- (xmax*u, 0); label.bot(\"$t^*$\", (x*u, 0)); label.bot(\"$t$\", (xmax*u, 0)); drawarrow origin -- (0, ymax*v); label.lft(\"\\textsc{TTR\\_PLATEAU}\", (0, v)); label.lft(\"\\textsc{80~\\%~TTR\\_PL}\", (0, .8v)); label.lft(\"\\textsc{TTR}\", (0, ymax*v)); endfig; \\end{mplibcode} \\end{document}", "= circle((d,0), r2 + r); pair[] X = intersectionpoints(w1,w2), Y = intersectionpoints(w1p,w2p); pair O = Y[1]; path w = circle(Y[1],r); pair Xp = 5 * X[1] - 4 * X[0]; pair[] P = intersectionpoints(Xp--X[0],w); label(\"O_1\",origin,N); label(\"O_2\",(d,0),N); label(\"O\",Y[1],SW); draw(origin--Y[1]--(d,0)--cycle,gray(0.6)); pair T = foot(O,O1,O2), Tp = foot(O,X[0],X[1]); draw(Tp--O--T^^rightanglemark(O,T,O1,60)^^rightanglemark(O,Tp,X[0],60),gray(0.6)); draw(w^^w1^^w2^^P[0]--X[0]); dot(Y[1]^^origin^^(d,0)); label(\"X\",T,N,gray(0.6)); label(\"Y\",foot(X[0],O1,O2),NE,gray(0.6)); label(\"\\ell\",(O+Tp)/2,S,gray(0.6)); [/asy]$ Denote by $O_1$ and $O_2$ the centers of $\\omega_1$ and $\\omega_2$ respectively. Set $X$ as the projection of $O$ onto $O_1O_2$, and denote by $Y$ the intersection of $AB$ with $O_1O_2$. Note that $\\ell = XY$. Now recall that$$d(O_2Y-O_1Y) = O_2Y^2 - O_1Y^2 = R_2^2 - R_1^2.$$Furthermore, note that\\begin{align*}d(O_2X - O_1X) &= O_2X^2 - O_1X^2= O_2O^2 - O_1O^2 \\\\ &= (R_2 + r)^2 - (R_1+r)^2 = (R_2^2 - R_1^2) + 2r(R_2 - R_1).\\end{align*}Substituting the first equality into the second one and subtracting yields$$2r(R_2 - R_1) = d(O_2Y - O_2X) - d(O_2X - O_1X) = 2dXY,$$which rearranges to the desired. ## Solution 3 WLOG assume $\\omega$ is a line. Note the angle condition is equivalent to the angle between $AB$ and $\\omega$ being $60^\\circ$. We claim the angle between $AB$ and $\\omega$ is fixed as $\\omega$ varies. Proof: Perform an inversion at $A$, sending $\\omega_1$ and $\\omega_2$ to two lines $\\ell_1$ and $\\ell_2$ intersecting at $B'$. Then $\\omega$ is sent to a circle tangent to lines $\\ell_1$", "xmax = max(x(:)); ymax = max(y(:)); zmin = min(z(:)); 2. Add Slice Planes for Visual Context Calculate the magnitude of the vector field (which represents wind speed) to generate scalar data for the slice command. Create slice planes along the x-axis at xmin , 100 , and xmax , along the y-axis at ymax , and along the z-axis at zmin . Specify interpolated face coloring so the slice coloring indicates wind speed, and do not draw edges ( sqrt , slice , FaceColor , EdgeColor ). wind_speed = sqrt(u.^2 + v.^2 + w.^2); hsurfaces = slice(x,y,z,wind_speed,[xmin,100,xmax],ymax,zmin); set(hsurfaces,'FaceColor','interp','EdgeColor','none') 3. Add Contour Lines to the Slice Planes Draw light gray contour lines on the slice planes to help quantify the color mapping ( contourslice , EdgeColor , LineWidth ). hcont = ... contourslice(x,y,z,wind_speed,[xmin,100,xmax],ymax,zmin); set(hcont,'EdgeColor',[.7,.7,.7],'LineWidth',.5) Stream Line Plots of Vector Data 15-33 4. Define the Starting Points for the Stream Lines In this example, all stream lines start at an x-axis value of 80 and span the range 20 to 50 in the y direction and 0 to 15 in the z direction. Save the handles of the stream lines and set the line width and color ( meshgrid , streamline , LineWidth , Color ). [sx,sy,sz] = meshgrid(80,20:10:50,0:5:15); hlines = streamline(x,y,z,u,v,w,sx,sy,sz); set(hlines,'LineWidth',2,'Color','r') 5. Define the View Set up the", "of the porabola. We know that it occurs at x=7, # so we evaluate the $par formula with at x=7.$real_vertex = $par->eval(x=>7); # Now we set some temporary variables that we'll pass into init_graph below: # Minimum and maximum x and y values to graph.$xmin = -1; $ymin = min(-1, ceil($real_vertex)-1); $xmax = 9;$ymax = max(5, $base+$rise+1, floor($real_vertex)+1); # We want grid lines on each integer value, but we have to specify the total # number of grid lines on each axis, so we just use the x and y range.$xrange = $xmax-$xmin; $yrange =$ymax-$ymin; # Size of the graph in pixels.$xsize = $xrange*25;$ysize = $yrange*25; # init_graph returns a graph object that we can then add functions to.$graph = init_graph( $xmin,$ymin, $xmax,$ymax, grid => [$xrange,$yrange], axes => [0,0], size => [$xsize,$ysize], ); # Add three functions to the graph. The language used in specifying plots is # described in the PGgraphmacros.pl docs. When a MathObject is used in double- # quotes, it is stringified into the quasi-TI notation that WeBWorK uses. plot_functions($graph, \"$horiz_line for x in [0,3] using color:red and weight:2\", \"$diag_line for x in [3,5] using color:red and weight:2\", \"$par for x in [5,9] using color:red and weight:2\", ); # This changes how MathObjects are stringified. Instead of quasi-TI syntax, # we switch to TeX stringification.", "this happens. ## Lines and Conics import graph; size(11.6cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.06,xmax=6.54,ymin=-3.4,ymax=5.54; pen qqzztt=rgb(0,0.6,0.2), qqqqzz=rgb(0,0,0.6), ffqqtt=rgb(1,0,0.2);draw(circle((0,0),1),linewidth(2.8));draw((0,0)--(1,0)); draw((0,ymin)--(0,ymax)); draw((xmin,(-(3.19)-(1.03)*xmin)/3.02)--(xmax,(-(3.19)-(1.03)*xmax)/3.02)); draw((xmin,(-(-3.19)-(-0.88)*xmin)/3.06)--(xmax,(-(-3.19)-(-0.88)*xmax)/3.06)); draw(shift((-2.81,0.83))*rotate(54.75)*xscale(2.12)*yscale(1.91)*unitcircle,linewidth(1.6)+linetype(\"2pt 8pt 10pt 8pt\")+qqzztt);pair hl1(real t){return (0.53*(1+t^2)/(1-t^2),0.74*2*t/(1-t^2));} pair hr1(real t){return (0.53*(-1-t^2)/(1-t^2),0.74*(-2)*t/(1-t^2));} draw(shift((3.63,0.55))*rotate(178.11)*graph(hl1,-0.99,0.99),linewidth(1.6)+linetype(\"4pt 4pt\")+qqqqzz); draw(shift((3.63,0.55))*rotate(178.11)*graph(hr1,-0.99,0.99),linewidth(1.6)+linetype(\"4pt 4pt\")+qqqqzz); real p1(real x){return x^2/2/3.48;} draw(shift((1.16,3.19))*rotate(16.04)*graph(p1,-13.93,13.93),linewidth(2)+ffqqtt); dot((0,0),ds); dot((1,0),ds); dot((-3.34,0.08),ds); dot((-2.28,1.58),ds); dot((-3.4,2.72),ds); dot((-0.94,0.34),ds); dot((2.72,0.58),ds); dot((4.54,0.52),ds); dot((0.68,4.86),ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); (Error compiling LaTeX. return linetype((real[]) split(pattern),offset,scale,adjust); ^ /usr/local/share/asymptote/plain_pens.asy: 13.3: Trying to use uninitialized value.) Allows lines, rays, vectors (not shown), and the different conic sections (which includes circle arcs and sectors), including different line colors, line types, line sizes, and in some cases, fill colors. Lines and rays automatically re-scale to fit the entire window when the diagram size is changed. <geogebra>f475aec9be4bbeb0d7a8b525fb2cbbe905383c5d</geogebra> Known bugs: • Broken line functions (floor, ceiling) graph incorrectly. Not likely to be fixed due to Asymptote's graph() design. ## Filled objects import graph; size(9.7cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.58,xmax=6.12,ymin=-1.1,ymax=6.1; pen cqcqcq=rgb(0.75,0.75,0.75), fueaev=rgb(0.96,0.92,0.9), zzttqq=rgb(0.6,0.2,0), evefev=rgb(0.9,0.94,0.9), qqwuqq=rgb(0,0.39,0); filldraw((0,0)--(2,0)--(3.41,1.41)--(3.41,3.41)--(2,4.83)--(0,4.83)--(-1.41,3.41)--(-1.41,1.41)--cycle,fueaev,zzttqq); filldraw(arc((2,0),0.6,45,180)--(2,0)--cycle,evefev,qqwuqq); filldraw(arc((0,0),0.6,0,135)--(0,0)--cycle,evefev,qqwuqq); filldraw(arc((-1.41,1.41),0.6,-45,90)--(-1.41,1.41)--cycle,evefev,qqwuqq); filldraw(arc((-1.41,3.41),0.6,-90,45)--(-1.41,3.41)--cycle,evefev,qqwuqq); filldraw(arc((0,4.83),0.6,-135,0)--(0,4.83)--cycle,evefev,qqwuqq); filldraw(arc((2,4.83),0.6,180,315)--(2,4.83)--cycle,evefev,qqwuqq); filldraw(arc((3.41,3.41),0.6,135,270)--(3.41,3.41)--cycle,evefev,qqwuqq); filldraw(arc((3.41,1.41),0.6,90,225)--(3.41,1.41)--cycle,evefev,qqwuqq); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2pt 2pt\"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); draw((0,0)--(2,0),zzttqq); draw((2,0)--(3.41,1.41),zzttqq); draw((3.41,1.41)--(3.41,3.41),zzttqq); draw((3.41,3.41)--(2,4.83),zzttqq); draw((2,4.83)--(0,4.83),zzttqq); draw((0,4.83)--(-1.41,3.41),zzttqq); draw((-1.41,3.41)--(-1.41,1.41),zzttqq); draw((-1.41,1.41)--(0,0),zzttqq); draw(arc((2,0),0.6,0.79,pi),qqwuqq); draw(arc((2,0),0.5,0.79,pi),qqwuqq); draw(arc((-1.41,1.41),0.6,-45,85.38),qqwuqq,BeginArcArrow(6)); draw(arc((-1.41,3.41),0.6,-90,40.38),qqwuqq,EndArcArrow(6)); draw(arc((0,4.83),0.6,-2.36,0),qqwuqq); draw((0.21,4.33)--(0.25,4.22),qqwuqq); draw((0.14,4.31)--(0.17,4.19),qqwuqq); draw((0.27,4.36)--(0.33,4.26),qqwuqq); draw(arc((2,4.83),0.6,pi,5.5),qqwuqq); draw((1.73,4.36)--(1.67,4.26),qqwuqq); draw((1.86,4.31)--(1.83,4.19),qqwuqq); draw(arc((3.41,3.41),0.6,2.36,4.71),qqwuqq); draw((2.92,3.21)--(2.8,3.16),qqwuqq); draw(arc((3.41,1.41),0.6,1.57,3.93),qqwuqq); draw(arc((3.41,1.41),0.5,1.57,3.93),qqwuqq); draw(arc((3.41,1.41),0.4,1.57,3.93),qqwuqq); draw((0,0)--(2,0),EndArrow(6)); dot((0,0),ds); dot((2,0),ds); dot((3.41,1.41),ds); dot((3.41,3.41),ds); dot((2,4.83),ds); dot((0,4.83),ds); dot((-1.41,3.41),ds); dot((-1.41,1.41),ds); label(\"$135^\\circ$\",(2.64,0.12),NE*lsf,qqwuqq); label(\"$135^\\circ$\",(0.28,0.78),NE*lsf,qqwuqq); label(\"$135^\\circ$\",(-1.1,2.06),NE*lsf,qqwuqq); label(\"$135^\\circ$\",(-0.72,3.28),NE*lsf,qqwuqq); label(\"$135^\\circ$\",(-1,4.68),NE*lsf,qqwuqq); label(\"$135^\\circ$\",(2.2,4.8),NE*lsf,qqwuqq);", "the direction of the line. else{ if(dy > 0){ t0 = (top - point1.y)/dy; t1 = (bottom - point1.y)/dy; } else{ t1 = (top - point1.y)/dy; t0 = (bottom - point1.y)/dy; } if(t0 > min) min = t0; if(t1 < max) max = t1; if(min > max || max < 0) return false; } // The point of intersection ix = point1.x + dx * min; iy = point1.y + dy * min; return true; • this is 2d, yes? – SirYakalot Dec 9 '11 at 12:41 • This is only 2D, yes. Also, the code does is not as well thought as zacharmarz's, which takes care of reducing the number of divisions and tests. – sam hocevar Dec 9 '11 at 14:09 I'm surprised to see that no one has mentioned the branchless slab method by Tavian bool intersection(box b, ray r) { double tx1 = (b.min.x - r.x0.x)*r.n_inv.x; double tx2 = (b.max.x - r.x0.x)*r.n_inv.x; double tmin = min(tx1, tx2); double tmax = max(tx1, tx2); double ty1 = (b.min.y - r.x0.y)*r.n_inv.y; double ty2 = (b.max.y - r.x0.y)*r.n_inv.y; tmin = max(tmin, min(ty1, ty2)); tmax = min(tmax, max(ty1, ty2)); return tmax >= tmin; } I have added to @zacharmarz answer to handle when the ray origin is inside of the AABB. In this case tmin will be negative and", "real[]{4,3})+extendcap; arrowbar arr=Arrow(HookHead,size=3); real arrowW=0.9bp; pen arrowPen0=red+arrowW; pen arrowPen1=blue+arrowW; // real xmin=0,xmax=1; real ymin=0,ymax=1; xaxis(xmin,xmax,axisPen); yaxis(ymin,ymax,axisPen); typedef pair pairFuncReal(real); pairFuncReal CubicBezier(pair A, pair B, pair C, pair D){ return new pair(real t){return A*(1-t)^3+3*B*(1-t)^2*t+3*C*(1-t)*t^2+D*t^3;}; } pair[] p; p.append(new pair[]{(0,0),(0,1),(0.25,0),(0,0.47),(-0.2,0.07),(0.25,0.26),(0.25,0.67),(0.62,0.89)}); p.push(0.6p[4]+0.4p[5]-(0,0.05)); // control points for f0 p.push(0.4p[4]+0.6p[5]-(0,0.05)); // p.push(0.6p[6]+0.4p[7]+(0,0.05)); // control points for f1 p.push(0.4p[6]+0.6p[7]+(0,0.05)); // pairFuncReal f0=CubicBezier(p[4],p[8],p[9],p[5]); pairFuncReal f1=CubicBezier(p[6],p[10],p[11],p[7]); real f0xmin=0, f0xmax=1; real f1xmin=0, f1xmax=1; guide g0=graph(f0,f0xmin,f0xmax); guide g1=graph(f1,f1xmin,f1xmax); pair arrowTail0=relpoint(g0,0.3)+(0,0.05); pair arrowHead0=relpoint(g0,0.8)+(0,0.06); pair arrowTail1=relpoint(g1,0.6)+(0,0.08); pair arrowHead1=relpoint(g1,0.01)+(0,0.06); p.append(new pair[]{ arrowTail0,arrowHead0,arrowTail1,arrowHead1 }); draw(g0,linePen); draw(g1,linePen); draw(p[2]--p[6],dashPen); draw(p[3]--(p[2].x,p[3].y),dashPen); real r=0.04; filldraw(circle(p[5],r),white,linePen); fill(circle(p[6],r),linePen); draw(p[12]--p[13],arrowPen0,arr); draw(p[14]--p[15],arrowPen1,arr); string[] plabels; plabels[0]=\"0\"; plabels[1]=\"1\"; plabels[2]=\"X^\\pm(\\omega)\"; plabels[3]=\"\\omega\"; pair[] ppos={ plain.W, plain.W, plain.S, plain.W, plain.S, plain.E, // 5 plain.SE, plain.SE, plain.E, plain.E, plain.SE, // 10 plain.SE, plain.N, plain.N, plain.N, plain.NW, // 15 plain.SE, plain.NE, plain.NW, plain.S, }; bool[] showDot=array(p.length,true); //bool[] showDot=array(p.length,false); showDot[5]=false; showDot[6]=false; if(showDots){ for(int i=0;i<p.length;++i){ if(showDot[i]) dot(p[i],UnFill); } } if(draft){ for(int i;i<p.length;++i){ if(p.initialized(i) && showDot[i]){ label(\"$p_{\"+string(i)+\"}$\",p[i],ppos[i]); } } } if(!draft){ for(int i;i<plabels.length;++i){ if(plabels.initialized(i) && showDot[i]) label(\"$\"+plabels[i]+\"$\",p[i],ppos[i]); } } shipout(bbox(Fill(paleyellow))); Run asy diag to get asy.pdf. The code shown is a working example for the draft mode; to get the final diagram, change bool draft=true; to bool draft=false; and bool showDots=true; to bool showDots=false; . Comment 1. Functions. The two curve segments are constructed with the standard graph function for 2d-drawing: guide g0=graph(f0,f0xmin,f0xmax); guide g1=graph(f1,f1xmin,f1xmax); Functions f0, f1 take a" ]

[ "# Asymptote fill area between two contours I use Asymptote to make the graphics. I have two curves defined as contours C_1 and C_2. I want to fill the area between the two curves. The problem is that contour returns a guide structure in Asymptote, and this is not accepted to to buildcycle command. Can anyone tell me how to solve this problem ? Thanks. This is the code I use and which produces the two curves, but fails to fill area between these. import graph; import patterns; import contour; usepackage(\"mathrsfs\"); size(8cm); real eps=0.001; real xmax=5.5,ymax=5.5; pair af,ag; real f(real x, real y) {return sqrt(1+x)-sqrt(x)+sqrt(1+y)-sqrt(y);} real g(real x, real y) {return 1/sqrt(1+x)+1/sqrt(1+y);} guide[][] Cf=contour(f,(eps,eps),(xmax,ymax),new real[] {1}); guide[][] Cg=contour(g,(eps,eps),(xmax,ymax),new real[] {1}); //these two lines are used to find intersection points path D1=(0,ymax)--(xmax+eps,ymax); path D2=(xmax,0)--(xmax,ymax+eps); draw(Cf); draw(Cg); //path pp=buildcycle(Cf,D1,Cg,D2); // the preceding line produces and error if commented xaxis(Label(scale(0.75)*\"$x$\"),xmax=6,Arrow); yaxis(Label(rotate(90)*scale(0.75)*\"$y$\"),ymax=6,Arrow); label(\"$\\mathscr{C}_{1}$\",(xmax,0.2)); label(\"$\\mathscr{C}_{2}$\",(xmax,1.9)); label(\"I\" ,(.2,.2)); label(\"II\" ,(1.,1.)); label(\"III\",(3.5,3.5)); end; • Is there some reason why you are using contour to produce the lines? It seems a rather complicated way to produce a single curve. May 1 '15 at 9:26 • Using f to write y = function (x) lead to a long expression. Around zero it requires large precision for fine tuned draw. Contour allows easier control over the drawing box.", "May 1 '15 at 22:10 It works when replacing Cf and Cg by Cf[0][0] and Cg[0][0], which allows to obtain the desired (single) guides. path pp=buildcycle(Cf[0][0],D1,Cg[0][0],D2); Note that I've also slightly modified D1 and D2: in your definitions, D1, D2 and pp did not intersect anyway. path D1=(0,ymax-eps)--(xmax+eps,ymax-eps); path D2=(xmax-eps,0)--(xmax-eps,ymax+eps); For proper explanations about how to use contour, see the very well-made (alas not complete) tutorial on Asymptote by Charles Staats, p. 32. import graph; import patterns; import contour; usepackage(\"mathrsfs\"); size(8cm); real eps=0.001; real xmax=5.5,ymax=5.5; pair af,ag; real f(real x, real y) {return sqrt(1+x)-sqrt(x)+sqrt(1+y)-sqrt(y);} real g(real x, real y) {return 1/sqrt(1+x)+1/sqrt(1+y);} guide[][] Cf=contour(f,(eps,eps),(xmax,ymax),new real[] {1}); guide[][] Cg=contour(g,(eps,eps),(xmax,ymax),new real[] {1}); //these two lines are used to find intersection points path D1=(0,ymax-eps)--(xmax+eps,ymax-eps); path D2=(xmax-eps,0)--(xmax-eps,ymax+eps); path pp=buildcycle(Cf[0][0],D1,Cg[0][0],D2); fill(pp,.8white); draw(pp); xaxis(Label(scale(0.75)*\"$x$\"),xmax=6,Arrow); yaxis(Label(rotate(90)*scale(0.75)*\"$y$\"),ymax=6,Arrow); label(\"$\\mathscr{C}_{1}$\",(xmax,0.2)); label(\"$\\mathscr{C}_{2}$\",(xmax,1.9)); label(\"I\" ,(.2,.2)); label(\"II\" ,(1.,1.)); label(\"III\",(3.5,3.5)); • Great, it works fine ! thank a lot fpast, for the answer and for the suggestions. Apr 30 '15 at 18:41 Just for comparison, here is a Metapost version. There's no support for implicit equations or anything so clever (or complex) as Asymptote's contour facility, so I could not avoid some algebra to get the \"long expression\" that the OP was happy to avoid. And to avoid overflows when x is near zero, I'm exploiting the facts that these two curves have", "k pi/10$$, for $$k$$ ranging from 1 to 10. In the arguments of the animate command, we must specify values for the boundaries of the viewing window, for the xmin, xmax, ymin, ymax, because otherwise the point remains fixed and the coordinate axes are moving. movingpoints = [point((cos(2*k*pi/10), sin(2*k*pi/10)), \\ color='red', size=50) for k in range(1,11)] a = animate(movingpoints,xmin=-1,xmax=1,ymin=-1,ymax=1) a.show(iterations=3) For the tangent line, we know that vectors perpendicular to $$(\\cos(t), \\sin(t))$$ are $$(-\\sin(t),\\cos(t))$$ and $$(\\sin(t), -\\cos(t))$$. These two perpendicular vectors are added to $$(\\cos(t),\\sin(t))$$ to compute the end points of the tangent line. We define functions to make the points because we will generate lists of plots to make the frames of the animation. pt(k) = (cos(2*k*pi/10), sin(2*k*pi/10)) v1(k) = (-sin(2*k*pi/10), cos(2*k*pi/10)) v2(k) = (-v1(k)[0],-v1(k)[1]) A(k) = ((pt(k)[0]+v1(k)[0]),pt(k)[1]+v1(k)[1]) B(k) = ((pt(k)[0]+v2(k)[0]),pt(k)[1]+v2(k)[1]) The function pt defines the coordinates of the moving point. The functions A and B given the end points of the tangent line. Now we are ready to plot the first frame in the animation. tangentline1 = line([A(1), B(1)],color='green') (unitcircle+firstpoint+tangentline1).show(figsize=4) The first frame of the animation is shown in Fig. 56. The extra work to define the functions pays off as the definition of the 10 frames for the moving tangent lines is straightforward. Observe how the plot for the unit circle is added to each", "but draw only the part of it from begin (-1,2) to intersection with first parabola. Also I need arrow on the end of this part to show the direction. – dm-kiselev Feb 11 '14 at 16:59 Regarding \"Where can I get similar information..\" - I don't know. – Andrew Kepert Feb 12 '14 at 2:53 Regarding the specific diagram, the easiest way to do this is to use the first parabola as a part of a \\path[clip] cycle that is active when you draw the second parabola. Look up clipping paths and also \\begin{scope}...\\end{scope} in pgfmanual.pdf. The hard way to do it is to solve the equations to find the point of intersection, and from that the geometry of the trapezium. There is a TikZ intersections library that also may help, but I haven't used it and AFAIK it doesn't have the nice \"cut after\" syntax of metapost. – Andrew Kepert Feb 12 '14 at 2:54 You could also parametrize the curve using x(t)=t^2 y(t)=t and then plot it using pgfplots % arara: pdflatex % !arara: indent: {overwrite: on} \\documentclass{standalone} \\usepackage{pgfplots} % arrows as stealth fighters \\tikzset{>=stealth} \\begin{document} \\begin{tikzpicture} \\begin{axis}[ axis lines=middle, axis line style=<->, xmin=-5,xmax=5, ymin=-5,ymax=5, xlabel=$x_1$, ylabel=$x_2$, xtick=\\empty, ytick=\\empty, xticklabels=\\empty, yticklabels=\\empty, ] \\end{axis} \\end{tikzpicture} \\end{document} - I believe you have done the same thing previously for", "line (PQ) \\setmfpair{M}{(P -- R) intersectionpoint (origin -- (0, \\ymax))} % Point N computed by MetaPost as intersection of line (PQ) % and the straight line perpendicular to (PQ) going through Q \\mfsrc{save N; pair N; N = whatever[P, R] = whatever[Q, Q + (R-P) rotated 90];} % Mark angle beta with the convenient \\arc macro of mfpic \\store{mark_angle}\\arc[a]{P, 0.8, angle(Q-P), angle(R-P)} % Mark right angle on N with help of MetaPost transformers \\setmfvariable{path}{mark_right_angle} {((1, 0) -- (1, 1) -- (0, 1)) zscaled 0.3unitvector(Q-N) shifted N} % Actual drawings \\polygon{P, Q, R} \\lines{Q, N} \\doaxes{xy} \\mfobj{mark_angle} \\mfobj{mark_right_angle} % Labels \\tlpointsep{2bp} % Offset \\tlabels{[tc]{(\\xmax, 0)}{$x$} [cr]{(0, \\ymax)}{$y$} [cr]{P}{$P(-4, -2)$} [cl]{Q}{$Q(8, 3)$} [bc]{R}{$R(4, 10)$} [br]{M}{$M$} [cr]{N}{$N$} [bl]{point 0.4 of mark_angle}{$\\beta$}} \\end{mfpic} \\closegraphsfile \\end{document} The .tex file is to be typeset by (pdf)LaTeX first, then the resulting .mp file by MetaPost, and finally the LaTeX file by (pdf)LaTeX again, to produce the figure below.", "$\\boldsymbol{(x, y)}$ $-2$ \\begin{aligned}x &= -2\\end{aligned} \\begin{aligned}y &= 2(-2) + 1\\\\&= -3\\end{aligned} \\begin{aligned}(-2, -3)\\end{aligned} $0$ \\begin{aligned}x &= 0\\end{aligned} \\begin{aligned}y &= 2(0) + 1\\\\&=1\\end{aligned} \\begin{aligned}(0, 1)\\end{aligned} $2$ \\begin{aligned}x &= 2\\end{aligned} \\begin{aligned}y &= 2(2) + 1\\\\&= 5\\end{aligned} \\begin{aligned}(2, 5)\\end{aligned} Since the line segment is expected to progress from right to left, we begin plotting the coordinates of $t = 2 \\rightarrow t = 0 \\rightarrow t = -2$. This graph shows how the direction may vary depending on what’s given. Be careful with problems like and double-check if you’re asked to graph a parametric line from left to right or right to left. Example 3 An object travels at a constant rate along a straight path from $(-8, 2)$ to $(4,-4)$ in six seconds. The coordinates are measured in meters. Find parametric equations that represent the position of the object. Solution This problem is an application of parametrizing linear equations. However, this time, we can see that the object moves from $(-8, 2)$ to $(4,-4)$ in just six seconds. We can’t assume that the parametric equation of $x$ is $t$, so let’s assume that both $x$ and $y$ are of the form $y = mt + b$. The value of $m$ for $x$ and $y$ will depend on the change in $x$ and $y$, respectively, within $6$ seconds. Since we", "my program I denote the position of a point charge by the vector $$\\vec{a}$$ with coordinates $$a_0,a_1,a_2$$, and introduce a cartesian coordinate system with coordinates $$x$$ (to the right), $$y$$ (to the top) and $$z$$ (into the plane of view). Then I can write the field of one charge in componentes as: \\begin{align} E_x &= q\\cdot(x-a_0)/r^3 \\\\ E_y &= q \\cdot (y-a_1)/r^3 \\end{align} where $$r^3 = ((x-a_0)^2 + (x-a_1)^2 + (z-a_2))^{1.5}$$ I look only at the plane $$z = 0$$ and since I draw in 2D (projection to the plane $$z = 0$$) I omit $$E_z$$ (but I am not sure, if this introduces a mistake). Then I use a loop to equally distribute the charges on my planes and calculate the total field as described above. # Imports import matplotlib from numpy import * from pylab import * import matplotlib.colors as colors def Cap(N = 10,plus=1,minus=1,app=''): # Aspect ratio aspekt='auto' ## Field between the plates xMin,xMax = -0.15,0.15 yMin,yMax = -10,10 zMin,zMax = -10,10 ## Field outside # xMin,xMax = -3,3 # yMin,yMax = -3,3 ## Plate distance d = 0.1 # Case of plates very far apart # d = 50 # xMin,xMax=-d*1.25,d*1.25 # yMin,yMax=-d*1.25,d*1.25 # aspekt = 'equal' ## Grids xG,yG = meshgrid(linspace(xMin,xMax,18),linspace(yMin,yMax,18)) xGM,yGM = meshgrid(linspace(xMin,xMax,500),linspace(yMin,yMax,500)) # Class that represents a point charge class charge:", "can use built-in function times t=times(gf,x)[0]; to get a value of the time parameter t, which corresponds to the intersection of the function curve and a vertical line at x=1. This t value allows to: first, get a missing 'y' coordinate of the tangent point P P=point(gf,t); And second, the direction of the tangent line at this point as dir(gf,t). Function drawline (part of the basic Asymptote module math.asy), allows to draw the visible portion of the (infinite) line going through two points: drawline(P,P+dir(gf,t),tanLinePen); Another useful function for this drawing is relpoint, which returns the point on curve at the relative fraction of its arclength. This is a complete MWE: // tan.asy // // run // asy tan.asy // // to get tan.pdf // settings.tex=\"pdflatex\"; import graph; import math; size(6cm); import fontsize;defaultpen(fontsize(10pt)); texpreamble(\"\\usepackage{lmodern}\" +\"\\usepackage{amsmath}\" +\"\\usepackage{amsfonts}\" +\"\\usepackage{amssymb}\" ); pen funcLinePen=darkblue+0.9bp; pen tanLinePen=orange+0.9bp; pen grayPen=gray(0.3)+0.8bp; pen dashPen=gray(0.3)+0.8bp+linetype(new real[]{5,5})+linecap(0); real xmin=-2,xmax=-xmin; real ymin=0,ymax=4; real dxmin=0.2; real dxmax=dxmin; real dymin=dxmin; real dymax=dxmax; xaxis(\"$x$\",xmin-dxmin,xmax+dxmax,RightTicks(Step=1,step=0.5),arr,above=true); yaxis(\"$y$\",ymin-dymin,ymax+dymax,LeftTicks (Step=1,step=0.5,OmitTick(0)),arr,above=true); real f(real x){return x^2;} guide gf=graph(f,xmin,xmax,operator..); real x=1, t=times(gf,x)[0]; pair P=point(gf,t), Q=relpoint(gf,6/7); draw(gf,funcLinePen); draw((P.x,0)--P--(0,P.y),dashPen); drawline(P,P+dir(gf,t),tanLinePen); dot((P.x,0)^^P^^(0,P.y)^^Q,UnFill); label(\"$y=x^2$\",relpoint(gf,1/4),UnFill); label(\"$P(1,\"+string(round(P.y))+\")$\",P,plain.SE); label(\"$Q(x,x^2)$\",Q,plain.W); label(\"$T$\",Q,3*plain.E); I'm surprised that the original question is tagged but one answer (the accepted one) uses and the other uses ! So I decide to post a TikZ answer, but this is mainly a mathematics answer. \\documentclass[tikz]{standalone} \\usetikzlibrary{through}", "the direction of the line. else{ if(dy > 0){ t0 = (top - point1.y)/dy; t1 = (bottom - point1.y)/dy; } else{ t1 = (top - point1.y)/dy; t0 = (bottom - point1.y)/dy; } if(t0 > min) min = t0; if(t1 < max) max = t1; if(min > max || max < 0) return false; } // The point of intersection ix = point1.x + dx * min; iy = point1.y + dy * min; return true; • this is 2d, yes? – SirYakalot Dec 9 '11 at 12:41 • This is only 2D, yes. Also, the code does is not as well thought as zacharmarz's, which takes care of reducing the number of divisions and tests. – sam hocevar Dec 9 '11 at 14:09 I'm surprised to see that no one has mentioned the branchless slab method by Tavian bool intersection(box b, ray r) { double tx1 = (b.min.x - r.x0.x)*r.n_inv.x; double tx2 = (b.max.x - r.x0.x)*r.n_inv.x; double tmin = min(tx1, tx2); double tmax = max(tx1, tx2); double ty1 = (b.min.y - r.x0.y)*r.n_inv.y; double ty2 = (b.max.y - r.x0.y)*r.n_inv.y; tmin = max(tmin, min(ty1, ty2)); tmax = min(tmax, max(ty1, ty2)); return tmax >= tmin; } I have added to @zacharmarz answer to handle when the ray origin is inside of the AABB. In this case tmin will be negative and", "line $$L_1(s)$$. • end1 (numpy.ndarray) – A 1D NumPy 2-array that is the end vector $$E_1$$ of the parametric line $$L_1(s)$$. Returns Pair of $$s_{\\ast}$$ and $$t_{\\ast}$$ such that the lines intersect: $$L_0\\left(s_{\\ast}\\right) = L_1\\left(t_{\\ast}\\right)$$ and then a boolean indicating if an intersection was found (i.e. if the lines aren’t parallel). Return type bezier.hazmat.geometric_intersection.parallel_lines_parameters(start0, end0, start1, end1) Checks if two parallel lines ever meet. Meant as a back-up when segment_intersection() fails. Note This function assumes but never verifies that the lines are parallel. In the case that the segments are parallel and lie on different lines, then there is a guarantee of no intersection. However, if they are on the exact same line, they may define a shared segment coincident to both lines. In segment_intersection(), we utilized the normal form of the lines (via the cross product): \\begin{split}\\begin{align*} L_0(s) \\times \\Delta_0 &\\equiv S_0 \\times \\Delta_0 \\\\ L_1(t) \\times \\Delta_1 &\\equiv S_1 \\times \\Delta_1 \\end{align*}\\end{split} So, we can detect if $$S_1$$ is on the first line by checking if $S_0 \\times \\Delta_0 \\stackrel{?}{=} S_1 \\times \\Delta_0.$ If it is not on the first line, then we are done, the segments don’t meet: >>> # Line: y = 1 >>> start0 = np.asfortranarray([0.0, 1.0]) >>> end0 = np.asfortranarray([1.0, 1.0]) >>> # Vertical shift up: y = 2 >>> start1 =", "bcorner1::Point, bcorner2::Point) Return true if the two bounding boxes defined by acorner1:acorner2 and bcorner1:bcorner2 intersect. Each pair of points is assumed to define the two opposite corners of a bounding box. source getnearestpointonline() finds perpendiculars. end1, end2, pt3 = ngon(O, 100, 3, vertices=true) map(pt -> circle(pt, 5, :fill), [end1, end2, pt3]) line(end1, end2, :stroke) arrow(pt3, getnearestpointonline(end1, end2, pt3)) getnearestpointonline(pt1::Point, pt2::Point, startpt::Point) Given a line from pt1 to pt2, and startpt is the start of a perpendicular heading to meet the line, at what point does it hit the line? source pointlinedistance(p::Point, a::Point, b::Point) Find the distance between a point p and a line between two points a and b. source slope(pointA::Point, pointB::Point) Find angle of a line starting at pointA and ending at pointB. Return a value between 0 and 2pi. Value will be relative to the current axes. slope(O, Point(0, 100)) |> rad2deg # y is positive down the page 90.0 270.0 source perpendicular(p::Point) Returns point Point(p.y, -p.x). source @polar (p) Convert a tuple of two numbers to a Point of x, y Cartesian coordinates. @polar (10, pi/4) @polar [10, pi/4] produces Luxor.Point(7.0710678118654755,7.071067811865475) source polar(r, theta) Convert point in polar form (radius and angle) to a Point. polar(10, pi/4) produces Luxor.Point(7.071067811865475,7.0710678118654755) source ## Arrows You can draw lines or arcs with arrows at the end with arrow().", "# Find intersection of rectangle with itself rotated Given $$\\w, h, \\alpha\\$$ for a rectangle centered at origin with width $$\\w\\$$, height $$\\h\\$$ and itself rotated by $$\\\\alpha\\$$ around origin clockwise, I wrote a code to find the intersection area. 1. First I find all 4 corners of original and rotated rectangle, along with lines joining them. 2. Then I find intersection with axis-parallel lines using getints 3. Then I take all intersections in a clockwise/anti-clockwise order and find area using the co-ordinates of the polygon thus formed. import Data.Ord import Data.List main :: IO () main = do [w, h, a] <- map read . words <$> getLine let a' = (min a (180 - a)) * pi / 180 let pts = [(w/2, h/2), (w/2, -h/2), (-w/2,-h/2), (-w/2,h/2)] :: [(Double, Double)] let pts' = map (rotate a') pts let lines = zip pts (tail$ cycle pts) let lines' = zip pts' (tail $cycle pts') let ints = concatMap (getints lines') lines print . area . uniq$ sortBy (comparing theta) ints uniq [] = [] uniq [x] = [x] uniq (x:y:xs) | x == y = uniq (x:xs) | otherwise = x : uniq (y:xs) rotate a (x, y) = (x * cos a - y * sin a, x * sin a + y * cos", "need to calculate where do you want your lines to go, exactly). I heavily commented the code to give you a lot of \"fish lines\" to catch your own fish. \\documentclass[12pt,a4paper]{article} \\usepackage[T1]{fontenc} \\usepackage{amsmath} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{pgfplots}\\pgfplotsset{compat=1.18} \\begin{document} \\begin{tikzpicture} \\begin{axis} [ % explicitly specify axis limits xmin=0,xmax=1,ymin=0,ymax=1,zmin=18,zmax=26, xlabel={x}, ylabel={y}, grid, % do not clip the \\draw commands and clip each clip mode=individual, ] % line on the xy plane (that way goes behind the surface) % we mark a point at 30% of the line \\draw[thick] (0.4,0,18) -- (0.4,1,18) coordinate[pos=0.3](a); % main surface \\addplot3[blue, no marks, surf, domain=0:1, samples=10] {25-5*x*x -y*y}; % graph restricted at at y=0.3 (parametric curve format) \\addplot3[ultra thick, no marks, domain=0:1, samples=10, samples y=1] (x, 0.3, 25-5*x*x -0.3*0.3); % line touching the parabola \\draw[thick] (0.4,0,18) -- (0.4,0.3,25-5*0.4*0.4-0.3*0.3) coordinate[pos=0.3](b);% coordinate on the black parabola % draw spline and angle label % in and out are the angle at which the curve leaves the final and initial point \\draw[thick,->] (a) to[out=100, in=-10] node[midway, above right]{$\\alpha$} (b); \\end{axis} \\end{tikzpicture} \\end{document} You maybe prefer to have the surface semi-transparent, to better convey the meaning of the arc: \\documentclass[12pt,a4paper]{article} \\usepackage[T1]{fontenc} \\usepackage{amsmath} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{pgfplots}\\pgfplotsset{compat=1.18} \\begin{document} \\begin{tikzpicture} \\begin{axis} [ % explicitly specify axis limits xmin=0,xmax=1,ymin=0,ymax=1,zmin=18,zmax=26, xlabel={x}, ylabel={y}, grid, % do not clip the \\draw commands and clip each clip mode=individual, ]", "are dotted with blue. Notice that in this critical (in any sense) case, we can not use the built-in path routines mintimes, maxtimes, times. // http://asymptote.ualberta.ca/ // Find all local maximal.minimal points unitsize(1cm); import graph; real f(real x){return (.6*cos(4.5*(x-4)+2.1)-1.2*sin(x-4)+.1*x+.2);} //real f(real x){return .5+sin(3x);} real xmin=-2; real xmax=10.2; xaxis(\"$x$\",xmin-.5,xmax+.5); yaxis(\"$y$\"); guide p=graph(f,xmin,xmax,Spline); draw(p,darkgreen+.6pt); int n=200; real h=(xmax-xmin)/n; real[] x,y; for (int k=0; k<n; ++k) { x.push(xmin+h*k); y.push(f(xmin+h*k)); } //write(x,y); for (int k=2; k<n-2; ++k){ if ( (y[k] > y[k-1]) & (y[k] > y[k+1]) ) dot( (x[k],y[k]),red); if ( (y[k] < y[k-1]) & (y[k] < y[k+1]) ) dot( (x[k],y[k]),blue); } shipout(bbox(5mm,invisible)); For f(x)=.5+sin(3x), we get Even more, this procedure can be used to find all intersection point of the graph with the horizontal line through a given point A. In other words, more or less does the job for the following routines // http://asymptote.ualberta.ca/ // Find all intersection points with horizontal line through given point A unitsize(1cm); import graph; real f(real x){return .5+sin(3x);} real xmin=-2; real xmax=10.2; xaxis(\"$x$\",xmin-.5,xmax+.5); yaxis(\"$y$\"); guide p=graph(f,xmin,xmax,Spline); draw(p,darkgreen+.6pt); pair A=(1,.6); draw((xmin,A.y)--(xmax,A.y),blue); int n=5000; real h=(xmax-xmin)/n; real[] x,y; for (int k=0; k<n; ++k) { x.push(xmin+h*k); y.push(f(xmin+h*k)); } for (int k=2; k<n-2; ++k) if ( (y[k]-A.y)*(y[k+1]-A.y)<0) dot( ((x[k]+x[k])/2,A.y),red); shipout(bbox(5mm,invisible)); There's a quick and easy approach using the sagetex package. It gives you access to a CAS, called Sage", "area = ( (B(1) - A(1)) * (C(2) - A(2)) - ... (B(2) - A(2)) * (C(1) - A(1)) ) / 2; end function [d,V1,V2] = feretDiameter(V,theta) % Copyright 2017-2018 The MathWorks, Inc. % Rotate points so that the direction of interest is vertical. alpha = 90 - theta; ca = cosd(alpha); sa = sind(alpha); R = [ca -sa; sa ca]; % Vr = (R * V')'; Vr = V * R'; y = Vr(:,2); ymin = min(y,[],1); ymax = max(y,[],1); d = ymax - ymin; if nargout > 1 V1 = V(y == ymin,:); V2 = V(y == ymax,:); end function [d,end_points] = maxFeretDiameter(P,antipodal_pairs) % Copyright 2017-2018 The MathWorks, Inc. if nargin < 2 antipodal_pairs = antipodalPairs(P); end v = P(antipodal_pairs(:,1),:) - P(antipodal_pairs(:,2),:); D = hypot(v(:,1),v(:,2)); [d,idx] = max(D,[],1); P1 = P(antipodal_pairs(idx,1),:); P2 = P(antipodal_pairs(idx,2),:); end_points = [P1 ; P2]; end function [d,triangle_points] = minFeretDiameter(V,antipodal_pairs) % Copyright 2017-2018 The MathWorks, Inc. if nargin < 2 antipodal_pairs = antipodalPairs(V); end n = size(antipodal_pairs,1); p = antipodal_pairs(:,1); q = antipodal_pairs(:,2); d = Inf; triangle_points = []; for k = 1:n if k == n k1 = 1; else k1 = k+1; end pt1 = []; pt2 = []; pt3 = []; if (p(k) ~= p(k1)) && (q(k) == q(k1)) pt1 = V(p(k),:); pt2 = V(p(k1),:); pt3 = V(q(k),:); elseif", "input. – Surya Nov 26 '13 at 7:29 @Surya: Yes, that's a general method. You need to make sure to assign variable names for your independent variables that aren't x and y. I've edited my answer. – Jake Nov 26 '13 at 7:49 Okay. That was the mistake then. Thanks. – Surya Nov 26 '13 at 8:11 The code: \\documentclass[margin=10pt]{standalone} \\usepackage{asymptote} \\begin{document} \\begin{asy} settings.render=4; settings.prc=false; size3(8cm); defaultpen(fontsize(10pt)); import graph3; currentprojection=orthographic(5,2,3); real ymax = 3; real ymin = -ymax; real h(real y) { return sqrt(1 + (y/2)^2); } path hyperbola = graph(h, ymin, ymax, operator ..); path3 hyperbola3 = path3(hyperbola, plane=YZplane); surface hyperboloid = surface(hyperbola3, c=O, axis=Y); real umax = hyperboloid.index.length - .001; real vmax = hyperboloid.index[0].length - .001; /* Parametrize by the square [0,2pi] x [ymin,ymax] */ triple point(real theta, real y) { return hyperboloid.point(theta/(2pi) * umax, (y-ymin)/(ymax-ymin) * vmax); } triple normal(real theta, real y) { return hyperboloid.normal(theta/(2pi) * umax, (y-ymin)/(ymax-ymin) * vmax); } int n = 10; int m = 10; real shrinkfactor=0.8; for (int i = 0; i < n; ++i) { real theta = i/n*2pi; triple f(real t) { return point(theta, t); } for (int j = 0; j < m; ++j) { real ylow = j/m*(ymax-ymin) + ymin; real yhigh = ylow + shrinkfactor*(ymax-ymin)/m; path3 arrowpath = graph(f, ylow, yhigh, n=10, operator..); }", "in Olympiad, and there are some other ways of computation under this observation.) -Fanyuchen20020715 ## Solution 2 (Official MAA) Let $M$ denote the midpoint of $\\overline{BC}$. The critical claim is that $M$ is the orthocenter of $\\triangle AXY$, which has the circle with diameter $\\overline{AT}$ as its circumcircle. To see this, note that because $\\angle BXT = \\angle BMT = 90^\\circ$, the quadrilateral $MBXT$ is cyclic, it follows that $$\\angle MXA = \\angle MXB = \\angle MTB = 90^\\circ - \\angle TBM = 90^\\circ - \\angle A,$$ implying that $\\overline{MX} \\perp \\overline{AC}$. Similarly, $\\overline{MY} \\perp \\overline{AB}$. In particular, $MXTY$ is a parallelogram. $[asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, rgb(0.98,0.81,0.69)); label(\"\\omega\", O + rad*dir(45), SW); filldraw(T--Y--M--X--cycle, rgb(173/255,216/255,230/255)); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot(\"X\", X, W); dot(\"Y\", Y, E); dot(\"O\", O, W); dot(\"T\", T, S); dot(\"A\", A, N); dot(\"B\", B, W); dot(\"C\", C, E); dot(\"M\", M, N); [/asy]$ Hence, by the Parallelogram Law, $$TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).$$ But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$. Therefore $$XY^2 = \\frac13(2 \\cdot 1143-135)", "intersection, simply plug $$t$$ or $$s$$ into their corresponding vector line equations. Plugging $$t=1$$ into the first equation will get: \\begin{align} \\vec{r_1} & = \\begin{bmatrix} 2 \\\\ -4 \\\\ 3 \\end{bmatrix} + 1 \\cdot \\begin{bmatrix} 3 \\\\ 11 \\\\ -2 \\end{bmatrix} \\\\ & = \\begin{bmatrix} 5 \\\\ 7 \\\\ 1 \\end{bmatrix} .\\end{align} So the lines intersect at the point $$(5, 7, 1).$$ Let's look at one final example of a question like this, this time with the line in parametric form. Determine whether the following lines intersect. If they intersect, find the point of intersection. If they do not intersect, determine whether they are skewed or parallel. Line 1: \\begin{align} x & = 11 - 10t \\\\ y & = 2 - 7t \\\\ z & = 3. \\end{align} Line 2: \\begin{align} x & = 1 + 2s \\\\ y & = 1 + 6s \\\\ z & = 1 - 2 s. \\end{align} Solution First, determine if the lines are parallel. Since this isn't in vector form, you need to create the direction vectors to be able to check this. The direction vectors here will be made using the coefficient of the scalar terms $$t$$ and $$s$$. The direction vector for Line 1 is: $\\begin{bmatrix} -10 \\\\ -7 \\\\ 0 \\end{bmatrix},$ and the direction vector for Line", "guess in faith that $\\overline{AB},\\overline{BC},$ and $\\overline{CD}$ all form $45^\\circ$ angles with the coordinate axes, from which $B=(0,2)$ and $C=(2,0).$ The given condition $D=(7,5)$ verifies our guess, as shown below. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -3; int xMax = 9; int yMin = -3; int yMax = 7; //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-3/16,i)--(3/16,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-3/16)--(i,3/16), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label(\"x\",(xMax,0),(2,0)); label(\"y\",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); draw(A--B--C--D,red); dot(A,linewidth(3.5)); dot(B,linewidth(3.5)); dot(C,linewidth(3.5)); dot(D,linewidth(3.5)); label(\"A(3,5)\",A,(0,2),UnFill); label(\"B\",B,(-2,0),UnFill); label(\"C\",C,(0,-2),UnFill); label(\"D(7,5)\",D,(0,2),UnFill); [/asy]$ Following the last paragraph of Solution 2 gives the answer $\\boxed{\\textbf{(C) }10\\sqrt2}.$ ~MRENTHUSIASM ~ pi_is_3.14 ~IceMatrix", "Compute the t values double txmin; double txmax; double tymin; double tymax; double tzmin; double tzmax; txmin = (minPt.x - rayIn.origin.x) / rayIn.direction.x; txmax = (maxPt.x - rayIn.origin.x) / rayIn.direction.x; x=1; if (txmin > txmax) { double tmp1 = txmin; txmin = txmax; txmax = tmp1; x=-1; } if (txmin > tNear) tNear = txmin; if (txmax < tFar) tFar = txmax; if (tNear > tFar) return false; if (tFar < 0) return false; //ray y direction is parallel to plane tymin = (minPt.y - rayIn.origin.y) / rayIn.direction.y; tymax = (maxPt.y - rayIn.origin.y) / rayIn.direction.y; y=1; if (tymin > tymax) { double tmp2 = tymin; tymin = tymax; tymax = tmp2; y=-1; } if (tymin > tNear) tNear = tymin; if (tymax < tFar) tFar = tymax; if (tNear > tFar) return false; if (tFar < 0) return false; //ray z direction is parallel to plane tzmin = (minPt.z - rayIn.origin.z) / rayIn.direction.z; tzmax = (maxPt.z - rayIn.origin.z) / rayIn.direction.z; z=1; if (tzmin > tzmax) { double tmp3 = tzmin; tzmin = tzmax; tzmax = tmp3; z=-1; } if (tzmin > tNear) tNear = tzmin; if (tzmax < tFar) tFar = tzmax; if (tNear > tFar) return false; if (tFar < 0) return false; double t = Math.min(tNear, tFar); // Set the location point. rayIn.evaluate(outRecord.location, tNear); //rayIn.start=0; rayIn.end=t;" ]

[ "# Difference between revisions of \"2012 AIME I Problems/Problem 8\" ## Problem Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\\overline{AB}$ and $\\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ $[asy]import cse5; unitsize(10mm); pathpen=black; dotfactor=3; pair A = (0,0), B = (3.8,0), C = (5.876,1.564), D = (2.076,1.564), E = (0,3.8), F = (3.8,3.8), G = (5.876,5.364), H = (2.076,5.364), M = (1.9,0), N = (5.876,3.465); pair[] dotted = {A,B,C,D,E,F,G,H,M,N}; D(A--B--C--G--H--E--A); D(E--F--B); D(F--G); pathpen=dashed; D(A--D--H); D(D--C); dot(dotted); label(\"A\",A,SW); label(\"B\",B,S); label(\"C\",C,SE); label(\"D\",D,NW); label(\"E\",E,W); label(\"F\",F,SE); label(\"G\",G,NE); label(\"H\",H,NW); label(\"M\",M,S); label(\"N\",N,NE); [/asy]$ ## Solution: think outside the box (pun intended) Define a coordinate system with $D$ at the origin and $C,$ $A,$ and $H$ on the $x$, $y$, and $z$ axes respectively. Then $D=(0,0,0),$ $M=(.5,1,0),$ and $N=(1,0,.5).$ It follows that the plane going through $D,$ $M,$ and $N$ has equation $2x-y-4z=0.$ Let $Q = (1,1,.25)$ be the intersection of this plane and edge $BF$ and let $P = (1,2,0).$ Now since $2(1) - 1(2) - 4(0) = 0,$ $P$ is on the plane. Also,", "label(\"M_2\", M2, S); label(\"M_3\", M3, E); label(\"M_4\", M4, N); label(\"G_1\", G1, 1.5S); label(\"G_2\", G2, 1.5E); label(\"G_3\", G3, 1.5NE); label(\"G_4\", G4, 1.5E); dot(A); dot(B); dot(C); dot(D); dot(M1); dot(M2); dot(M3); dot(M4); dot(G1); dot(G2); dot(G3); dot(G4); [/asy]$ By SAS, we conclude that $\\triangle G_1G_2P\\sim\\triangle M_1M_2P, \\triangle G_2G_3P\\sim\\triangle M_2M_3P, \\triangle G_3G_4P\\sim\\triangle M_3M_4P,$ and $\\triangle G_4G_1P\\sim\\triangle M_4M_1P.$ By the properties of centroids, the ratio of similitude for each pair of triangles is $\\frac23.$ Note that quadrilateral $M_1M_2M_3M_4$ is a square of side-length $15\\sqrt2.$ It follows that: 1. Since $\\overline{G_1G_2}\\parallel\\overline{M_1M_2},\\overline{G_2G_3}\\parallel\\overline{M_2M_3},\\overline{G_3G_4}\\parallel\\overline{M_3M_4},$ and $\\overline{G_4G_1}\\parallel\\overline{M_4M_1}$ by the Converse of the Corresponding Angles Postulate, we have $\\angle G_1G_2G_3=\\angle G_2G_3G_4=\\angle G_3G_4G_1=\\angle G_4G_1G_2=90^\\circ.$ 2. Since $G_1G_2=\\frac23M_1M_2, G_2G_3=\\frac23M_2M_3, G_3G_4=\\frac23M_3M_4,$ and $G_4G_1=\\frac23M_4M_1$ by the ratio of similitude, we have $G_1G_2=G_2G_3=G_3G_4=G_4G_1=10\\sqrt2.$ Together, quadrilateral $G_1G_2G_3G_4$ is a square of side-length $10\\sqrt2,$ so its area is $\\left(10\\sqrt2\\right)^2=\\boxed{\\textbf{(C) }200}.$ Remark This solution shows that, if point $P$ is within square $ABCD,$ then the shape and the area of quadrilateral $G_1G_2G_3G_4$ are independent of the location of $P.$ Let the brackets denote areas. More generally, $G_1G_2G_3G_4$ is always a square of area $$[G_1G_2G_3G_4]=\\left(\\frac23\\right)^2[M_1M_2M_3M_4]=\\frac49[M_1M_2M_3M_4]=\\frac29[ABCD].$$ On the other hand, the location of $G_1G_2G_3G_4$ is dependent on the location of $P.$ ~RandomPieKevin ~Kyriegon ~MRENTHUSIASM ## Solution 2 (Similar Triangles) This solution refers to the diagram in Solution 1. By SAS, we conclude that $\\triangle G_1G_3P\\sim\\triangle M_1M_3P$ and $\\triangle G_2G_4P\\sim\\triangle M_2M_4P.$", "24 In the cube $ABCDEFGH$ with opposite vertices $C$ and $E,$ $J$ and $I$ are the midpoints of edges $\\overline{FB}$ and $\\overline{HD},$ respectively. Let $R$ be the ratio of the area of the cross-section $EJCI$ to the area of one of the faces of the cube. What is $R^2?$ $[asy] size(6cm); pair A,B,C,D,EE,F,G,H,I,J; C = (0,0); B = (-1,1); D = (2,0.5); A = B+D; G = (0,2); F = B+G; H = G+D; EE = G+B+D; I = (D+H)/2; J = (B+F)/2; filldraw(C--I--EE--J--cycle,lightgray,black); draw(C--D--H--EE--F--B--cycle); draw(G--F--G--C--G--H); draw(A--B,dashed); draw(A--EE,dashed); draw(A--D,dashed); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(I); dot(J); label(\"A\",A,E); label(\"B\",B,W); label(\"C\",C,S); label(\"D\",D,E); label(\"E\",EE,N); label(\"F\",F,W); label(\"G\",G,N); label(\"H\",H,E); label(\"I\",I,E); label(\"J\",J,W); [/asy]$ $\\textbf{(A) } \\frac{5}{4} \\qquad \\textbf{(B) } \\frac{4}{3} \\qquad \\textbf{(C) } \\frac{3}{2} \\qquad \\textbf{(D) } \\frac{25}{16} \\qquad \\textbf{(E) } \\frac{9}{4}$ ## Problem 25 How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive? $\\textbf{(A) }4\\qquad\\textbf{(B) }9\\qquad\\textbf{(C) }10\\qquad\\textbf{(D) }57\\qquad \\textbf{(E) }58$", "+0 # HELP ASAP GEOMETRY QUESTION!!!!! 0 123 1 The edge length of a regular tetrahedron $ABCD$ is $2.$ $M$ and $N$ are the midpoints of $\\overline{BC}$ and $\\overline{AD}$, respectively. Find $MN.$ [asy] size(150); import three; currentprojection = orthographic(-0.6,3,2); triple A,B,C,D,O,M,NN; O=(0,0,0); D=(0,0,sqrt(8)); A=(2,0,0); B=(-1,sqrt(3),0); C=(-1,-sqrt(3),0); M=(B+C)/2; NN=(A+D)/2; draw(D--A--B--D--C--B); draw(A--C,dashed); draw(M--NN,dotted); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,E); label(\"$D$\",D,N); label(\"$M$\",M,E); label(\"$N$\",NN,NW); [/asy] Incase the latex doesn't work: The edge length of a regular tetrahedron ABCD is 2. M and N are the midpoints of segment BC and segment AD, respectively. Find MN. Image: It won't load but it's a regular tetrahedron and MN is a diagonal from the midpoint of a side to the midpoint of another side that is on the opposite side and on the base. THANK YOU TO ANYONE THAT HELPS!!!!!!!!!!!!!! Apr 5, 2022", "hexagon $ABCDEF$, $AC$, $CE$ are two diagonals. Points $M$, $N$ are on $AC$, $CE$ respectively and satisfy $AC: AM = CE: CN = r$. Suppose $B, M, N$ are collinear, find $100r^2$. (diagram goes here) $[asy] size(120); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(pair P) { dot(P,linewidth(3)); return P; } pair A=dir(0), B=dir(60), C=dir(120), D=dir(180), E=dir(240), F=dir(300), N=(4*E+C)/5,M=intersectionpoints(A--C,B--N)[0]; draw(A--B--C--D--E--F--cycle); draw(A--C--E); draw(B--N); label(\"A\",D2(A),E); label(\"B\",D2(B),NE); label(\"C\",D2(C),NW); label(\"D\",D2(D),W); label(\"E\",D2(E),SW); label(\"F\",D2(F),SE); label(\"M\",D2(M),(0,-1.5)); label(\"N\",D2(N),SE); [/asy]$ Solution A cuboctahedron is a solid with 6 square faces and 8 equilateral triangle faces, with each edge adjacent to both a square and a triangle (see picture). Suppose the ratio of the volume of an octahedron to a cuboctahedron with the same side length is $r$. Find $100r^2$. (diagram goes here) In the following diagram, a semicircle is folded along a chord $AN$ and intersects its diameter $MN$ at $B$. Given that $MB : BN = 2 : 3$ and $MN = 10$. If $AN = x$, find $x^2$. $[asy] size(120); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(pair P) { dot(P,linewidth(3)); return P; } real r = sqrt(80)/5; pair M=(-1,0), N=(1,0), A=intersectionpoints(arc((M+N)/2, 1, 0, 180),circle(N,r))[0], C=intersectionpoints(circle(A,1),circle(N,1))[0], B=intersectionpoints(circle(C,1),M--N)[0]; draw(arc((M+N)/2, 1, 0, 180)--cycle); draw(A--N); draw(arc(C,1,180,180+2*aSin(r/2))); label(\"A\",D2(A),NW); label(\"B\",D2(B),SW); label(\"M\",D2(M),S); label(\"N\",D2(N),SE); [/asy]$ Solution Square $ABCD$ is divided into four rectangles by $EF$ and $GH$. $EF$ is parallel to $AB$ and $GH$ parallel to $BC$. $\\angle BAF = 18^\\circ$. $EF$", "(HHC − 2)n×n based on degrees of end vertices of each edge. The edge set of (HHC − 2)n×n can be divided into two partitions based on the degree of end vertices. The first edge partition E1((HHC −2)n×n) contains 24 edges uv, where deg(u) = 3, deg(v) = 4. The second edge partition E2((HHC−2)n×n) contains 32n+4 edges uv, where deg(u) = deg(v) = 4. • The first and second Zagreb indices of (HHC − 2)n×n Now using Eqs. (1),(2), we have $M1(G)=∑uv∈E(G)[deg(u)+deg(v)]M1(G)=∑uv∈E1(G)[deg(u)+deg(v)]+∑uv∈E2(G)[deg(u)+deg(v)]=7|E1(G)|+8|E2(G)|=7(24)+8(32n+4)=512n+200.$$M2(G)=∑uv∈E(G)[deg(u)×deg(v)]M2(G)=∑uv∈E1(G)[deg(u)×deg(v)]+∑uv∈E2(G)[deg(u)×deg(v)]=12|E1(G)|+16|E2(G)|=12(24)+16(32n+4)=512n+352.$ • Hyper Zagreb index of (HHC − 2)n×n The hyper Zagreb index using Eq. (3) is computed as: $HM(G)=∑uv∈E(G)[deg(u)+deg(v)]2HM(G)=∑uv∈E1(G)[deg(u)+deg(v)]2+∑uv∈E2(G)[deg(u)+deg(v)]2=72|E1(G)|+82|E2(G)|=49(24)+64(32n+4)=2048n+1432.$ • Multiple Zagreb indices of (HHC − 2)n×n The Multiple-Zagreb indices using Eqs. (4), (5) are computed as: $vPM1(G)=∏uv∈E(G)[deg(u)+deg(v)]PM1(G)=∏uv∈E1(G)[deg(u)+deg(v)]×∏uv∈E2(G)[deg(u)+deg(v)]=7|E1(G)|×8|E2(G)|=724×8(32n+4).$$PM2(G)=∏uv∈E(G)[deg(u)×deg(v)]PM2(G)=∏uv∈E1(G)[deg(u)×deg(v)]×∏uv∈E2(G)[deg(u)×deg(v)]=12|E1(G)|×16|E2(G)|=1224×16(32n+4).$ • The first and second Zagreb polynomials of (HHC − 2)n×n Now using Eqs. (6),(7), we have $M1(G,x)=∑uv∈E(G)x[deg(u)+deg(v)]M1(G,x)=∑uv∈E1(G)x[deg(u)+deg(v)]+∑uv∈E2(G)x[deg(u)+deg(v)]=∑uv∈E1(G)x7+∑uv∈E2(G)x8=|E1(G)|x7+|E2(G)|x8=24x7+(32n+4)x8.$$M2(G,x)=∑uv∈E(G)x[deg(u)×deg(v)]M2(G,x)=∑uv∈E1(G)x[deg(u)×deg(v)]+∑uv∈E2(G)x[deg(u)×deg(v)]=∑uv∈E1(G)x12+∑uv∈E2(G)x16=|E1(G)|x12+|E2(G)|x16=24x12+(32n+4)x16.$ • The redefined first, second, and third Zegreb index of (HHC − 2)n×n Now using Eq. (8) and the edge partition of the block shift network (HHC − 2)n×n, we have: $ReZG1(G)=∑uv∈E(G)deg(u)+deg(v)deg(u)×deg(v)=∑uv∈E1(G)deg(u)+deg(v)deg(u)×deg(v)+∑uv∈E2(G)deg(u)+deg(v)deg(u)×deg(v)=24(3+43×4)+(32n+4)(4+44×4)=16n+16.$ By using Eq. (9), the second redefined Zagreb index is computed as below: $ReZG2(G)=∑uv∈E(G)deg(u)×deg(v)deg(u)+deg(v)=∑uv∈E1(G)deg(u)×deg(v)deg(u)+deg(v)+∑uv∈E2(G)deg(u)×deg(v)deg(u)+deg(v)=24(3×43+4)+(32n+4)(4×44+4)=2887+64n+8.$ Now by using Eq. (10), the third redefined Zagreb index is computed as: $ReZG3(G)=∑uv∈E(G)[[deg(u)×deg(v)]×[deg(u)+deg(v)]]=∑uv∈E1(G)[[deg(u)×deg(v)]×[deg(u)+deg(v)]]+∑uv∈E2(G)[[deg(u)×deg(v)]×[deg(u)+deg(v)]]=24((3×4)×(3+4))+(32n+4)((4×4)×(4+4))=4608n+2596.$ ## Conclusion In this paper we determined first Zagreb index M1(G), second Zagreb index M2(G), hyper Zagreb index HM(G), first", "or mid face from the 2d label number. This vector contains successive pairs of the 2d label number at index $2i$ and the corresponding 3d label number at index $2i+1$, like change. • labelup= This vector is used to initialize the 3d label numbers of the upper/top face from the 2d region number. This vector contains successive pairs of the 2d region number at index $2i$ and the corresponding 3d label number at index $2i+1$, like change. • labeldown= Same as the previous case but for the lower/down face label. Moreover, we also add post processing parameters that allow to moving the mesh. These parameters correspond to parameters transfo, facemerge and ptmerge of the command line movemesh. The vector region, labelmid, labelup and labeldown These vectors are composed of $n_{l}$ successive pairs of number $O_i,N_l$ where $n_{l}$ is the number (label or region) that we want to get. An example of this command is given in the Build layer mesh example. Cube Cube.idp 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 load \"medit\" load \"msh3\" func mesh3 Cube (int[int] &NN, real[int, int] &BB, int[int, int] &L){ real x0 = BB(0,0), x1 = BB(0,1); real y0 = BB(1,0), y1 = BB(1,1); real z0 = BB(2,0), z1", "Welcome to the Rhino 6 version of this page! Looking for the newer Rhino 7 WIP version? Demonstrates how to construct a twisted cube object from an array of points and a list of curves. partial class Examples { // Symbolic vertex index constants to make code more readable const int A = 0, B = 1, C = 2, D = 3, E = 4, F = 5, G = 6, H = 7; // Symbolic edge index constants to make code more readable const int AB = 0, BC = 1, CD = 2, AD = 3, EF = 4, FG = 5, GH = 6, EH = 7, AE = 8, BF = 9, CG = 10, DH = 11; // Symbolic face index constants to make code more readable const int ABCD = 0, BCGF = 1, CDHG = 2, ADHE = 3, ABFE = 4, EFGH = 5; { Point3d[] points = new Point3d[8]; points[0] = new Point3d(0.0, 0.0, 0.0); // point A = geometry for vertex 0 points[1] = new Point3d(10.0, 0.0, 0.0); // point B = geometry for vertex 1 points[2] = new Point3d(10.0, 8.0, -1.0); // point C = geometry for vertex 2 points[3] = new Point3d(0.0, 6.0, 0.0); // point D = geometry for vertex 3 points[4] = new", "3. $L$ and $R$ will retain the direction of the labeled vertices, but $H$ and $V$ will alter the direction of the labeled vertices. After the $19$th transformation, vertex $A$ will be at either $(1,-1)$ or $(-1,1).$ All possible configurations of the labeled vertices are shown below: $[asy] /* Made by MRENTHUSIASM */ unitsize(7mm); label(\"A\",(1,0)); label(\"B\",(1,1)); label(\"C\",(0,1)); label(\"D\",(0,0)); label(\"C\",(5,0)); label(\"D\",(5,1)); label(\"A\",(4,1)); label(\"B\",(4,0)); label(\"A\",(9,0)); label(\"D\",(9,1)); label(\"C\",(8,1)); label(\"B\",(8,0)); label(\"C\",(13,0)); label(\"B\",(13,1)); label(\"A\",(12,1)); label(\"D\",(12,0)); label(\"\\textbf{Configuration}\",(-5,0.5)); label(\"\\textbf{The 20th}\",(-5,-1.5)); label(\"\\textbf{Transformation}\",(-5,-2.25)); label(\"L\",(0.5,-1.875)); label(\"R\",(4.5,-1.875)); label(\"H\",(8.5,-1.875)); label(\"V\",(12.5,-1.875)); [/asy]$ Each sequence of $19$ transformations generates one valid sequence of $20$ transformations. Therefore, the answer is $4^{19}=\\boxed{\\textbf{(C)}\\ 2^{38}}.$ ~MRENTHUSIASM Solution 2 Let $(+)$ denote counterclockwise/starting orientation and $(-)$ denote clockwise orientation. Let $1,2,3,$ and $4$ denote which quadrant $A$ is in. Realize that from any odd quadrant and any orientation, the $4$ transformations result in some permutation of $(2+, 2-, 4+, 4-).$ The same goes that from any even quadrant and any orientation, the $4$ transformations result in some permutation of $(1+, 1-, 3+, 3-).$ We start our first $19$ moves by doing whatever we want, $4$ choices each time. Since $19$ is odd, we must end up on an even quadrant. As said above, we know that exactly one of the four transformations will give us $(1+),$ and we must use that transformation. Thus, the answer is $4^{19}=\\boxed{\\textbf{(C)}\\", "# Problem #2209 2209 A cube with side length $1$ is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C$, as shown. What is the area of quadrilateral $ABCD$? $[asy] import three; unitsize(3cm); defaultpen(fontsize(8)+linewidth(0.7)); currentprojection=obliqueX; draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype(\"4 4\")); draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1)); draw((0.5,1,0)--(0,1,0)--(0,1,1)); dot((0.5,0,0)); label(\"A\",(0.5,0,0),WSW); dot((0,1,1)); label(\"C\",(0,1,1),NE); dot((0.5,1,0.5)); label(\"D\",(0.5,1,0.5),ESE); dot((0,0,0.5)); label(\"B\",(0,0,0.5),NW);[/asy]$ $\\mathrm{(A)}\\ \\frac{\\sqrt{6}}{2}\\qquad\\mathrm{(B)}\\ \\frac{5}{4}\\qquad\\mathrm{(C)}\\ \\sqrt{2}\\qquad\\mathrm{(D)}\\ \\frac{5}{8}\\qquad\\mathrm{(E)}\\ \\frac{3}{4}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "(clockwise or counterclockwise). 3. $L$ and $R$ will retain the direction of the labeled vertices, but $H$ and $V$ will alter the direction of the labeled vertices. After the $19$th transformation, vertex $A$ will be at either $(1,-1)$ or $(-1,1).$ All possible configurations of the labeled vertices are shown below: $[asy] /* Made by MRENTHUSIASM */ unitsize(7mm); label(\"A\",(1,0)); label(\"B\",(1,1)); label(\"C\",(0,1)); label(\"D\",(0,0)); label(\"C\",(5,0)); label(\"D\",(5,1)); label(\"A\",(4,1)); label(\"B\",(4,0)); label(\"A\",(9,0)); label(\"D\",(9,1)); label(\"C\",(8,1)); label(\"B\",(8,0)); label(\"C\",(13,0)); label(\"B\",(13,1)); label(\"A\",(12,1)); label(\"D\",(12,0)); label(\"\\textbf{Configuration}\",(-5,0.5)); label(\"\\textbf{The 20th}\",(-5,-1.5)); label(\"\\textbf{Transformation}\",(-5,-2.25)); label(\"L\",(0.5,-1.875)); label(\"R\",(4.5,-1.875)); label(\"H\",(8.5,-1.875)); label(\"V\",(12.5,-1.875)); [/asy]$ Each sequence of $19$ transformations generates one valid sequence of $20$ transformations. Therefore, the answer is $4^{19}=\\boxed{\\textbf{(C)}\\ 2^{38}}.$ ~MRENTHUSIASM ## Solution 2 Let $(+)$ denote counterclockwise/starting orientation and $(-)$ denote clockwise orientation. Let $1,2,3,$ and $4$ denote which quadrant $A$ is in. Realize that from any odd quadrant and any orientation, the $4$ transformations result in some permutation of $(2+, 2-, 4+, 4-).$ The same goes that from any even quadrant and any orientation, the $4$ transformations result in some permutation of $(1+, 1-, 3+, 3-).$ We start our first $19$ moves by doing whatever we want, $4$ choices each time. Since $19$ is odd, we must end up on an even quadrant. As said above, we know that exactly one of the four transformations will give us $(1+),$ and we must use that transformation. Thus,", "# 2018 AMC 8 Problems/Problem 24 ## Problem In the cube $ABCDEFGH$ with opposite vertices $C$ and $E,$ $J$ and $I$ are the midpoints of segments $\\overline{FB}$ and $\\overline{HD},$ respectively. Let $R$ be the ratio of the area of the cross-section $EJCI$ to the area of one of the faces of the cube. What is $R^2?$ $[asy] size(6cm); pair A,B,C,D,EE,F,G,H,I,J; C = (0,0); B = (-1,1); D = (2,0.5); A = B+D; G = (0,2); F = B+G; H = G+D; EE = G+B+D; I = (D+H)/2; J = (B+F)/2; filldraw(C--I--EE--J--cycle,lightgray,black); draw(C--D--H--EE--F--B--cycle); draw(G--F--G--C--G--H); draw(A--B,dashed); draw(A--EE,dashed); draw(A--D,dashed); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(I); dot(J); label(\"A\",A,E); label(\"B\",B,W); label(\"C\",C,S); label(\"D\",D,E); label(\"E\",EE,N); label(\"F\",F,W); label(\"G\",G,N); label(\"H\",H,E); label(\"I\",I,E); label(\"J\",J,W); [/asy]$ $\\textbf{(A) } \\frac{5}{4} \\qquad \\textbf{(B) } \\frac{4}{3} \\qquad \\textbf{(C) } \\frac{3}{2} \\qquad \\textbf{(D) } \\frac{25}{16} \\qquad \\textbf{(E) } \\frac{9}{4}$ ## Solution 1 Note that $EJCI$ is a rhombus by symmetry. Let the side length of the cube be $s$. By the Pythagorean theorem, $EC= s\\sqrt 3$ and $JI= s\\sqrt 2$. Since the area of a rhombus is half the product of its diagonals, the area of the cross section is $\\frac{s^2\\sqrt 6}{2}$. This gives $R = \\frac{\\sqrt 6}2$. Thus $R^2 = \\boxed{\\textbf{(C) } \\frac{3}{2}}$ ## Solution 2 We can solve this with 3D Cartesian coordinates. Assume WLOG that the sides of", "# Volume of a Part of a Cube Geometry Level 5 Cube $$ABCDEFGH$$, labeled as shown above, has edge length $$1$$ and is cut by a plane passing through vertex $$D$$ and the midpoints $$M$$ and $$N$$ of $$\\overline{AB}$$ and $$\\overline{CG}$$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $$\\frac{p}{q}$$, where $$p$$ and $$q$$ are relatively prime positive integers. Find $$p+q$$. ×", "# Difference between revisions of \"2018 AIME I Problems/Problem 4\" ## Problem 4 In $\\triangle ABC, AB = AC = 10$ and $BC = 12$. Point $D$ lies strictly between $A$ and $B$ on $\\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\\overline{AC}$ so that $AD = DE = EC$. Then $AD$ can be expressed in the form $\\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. ## Solution 1 (No Trig) $[asy] import cse5; unitsize(10mm); pathpen=black; dotfactor=3; pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66); pair[] dotted = {A,B,C,D,E,F,G}; D(A--B); D(C--B); D(A--C); D(D--E); pathpen=dashed; D(B--F); D(D--G); dot(dotted); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NW); label(\"E\",E,NE); label(\"F\",F,NE); label(\"G\",G,NE); label(\"x\",H,NW); label(\"x\",I,NW); label(\"x\",J,NE); [/asy]$ We draw the altitude from $B$ to $\\overline{AC}$ to get point $F$. We notice that the triangle's height from $A$ to $\\overline{BC}$ is 8 because it is a $3-4-5$ Right Triangle. To find the length of $\\overline{BF}$, we let $h$ represent $\\overline{BF}$ and set up an equation by finding two ways to express the area. The equation is $(8)(12)=(10)(h)$, which leaves us with $h=9.6$. We then solve for the length $\\overline{AF}$, which is done through pythagorean theorm and get $\\overline{AF}$ = $2.8$. We can now see that $\\triangle ABF$", "1988 AHSME Problems/Problem 20 Problem In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base, $XY / YZ$, in this rectangle is $[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,1), B=(0,-1), C=(2,-1), D=(2,1), E=(1,-1), F=(1,1), G=(.8,.6); pair X=(4,sqrt(5)), Y=(4,-sqrt(5)), Z=(4+2/sqrt(5),-sqrt(5)), W=(4+2/sqrt(5),sqrt(5)), T=(4,0), U=(4+2/sqrt(5),-4/sqrt(5)), V=(4+2/sqrt(5),1/sqrt(5)); draw(A--B--C--D--A^^B--F^^E--D^^A--G^^rightanglemark(A,G,F)); draw(X--Y--Z--W--X^^T--V--X^^Y--U); label(\"A\", A, NW); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, NE); label(\"E\", E, S); label(\"F\", F, N); label(\"G\", G, E); label(\"X\", X, NW); label(\"Y\", Y, SW); label(\"Z\", Z, SE); label(\"W\", W, NE); [/asy]$ $\\textbf{(A)}\\ 4\\qquad \\textbf{(B)}\\ 1+2\\sqrt{3}\\qquad \\textbf{(C)}\\ 2\\sqrt{5}\\qquad \\textbf{(D)}\\ \\frac{8+4\\sqrt{3}}{3}\\qquad \\textbf{(E)}\\ 5$ Solution Within $WXYZ$, the parallelogram piece has vertical side $BF = \\sqrt{1^2 + 2^2} = \\sqrt{5}$, and diagonal side $FD = 1$. Thus the triangle in the bottom-right hand corner (the one with horizontal side $YZ$) must have hypotenuse $1$, and the only such triangle in the original figure is $\\triangle AFG$, so we deduce $YZ = AG = \\frac{\\frac{1}{2} \\times 1 \\times 2}{\\frac{1}{2} \\times \\sqrt{5}} = \\frac{2}{\\sqrt{5}}.$ Now the rectangle must have the same area as the square, as the pieces were put together without gaps", "faces and three edges. The edges opposite to each other are parallel. Let $OABCDEFG$ be a cube with vertices as below, $O(0,0,0), A(a,0,0), B(a,a,0), C(0,a,0,), D(0,a,a), E(0,0,a), F(a,0,a), G(a,a,a)$ There are four diagonals of the cube: they are $OG,CF,AD$ and $BE$ for the cube. Let us consider any two say $OG$ and $AD$, We know that if $A({{x}_{1}},{{y}_{1}},{{z}_{1}})$ and $B({{x}_{2}},{{y}_{2}},{{z}_{2}})$ are two points in space then, $\\overrightarrow{AB}=({{x}_{2}}-{{x}_{1}})i+({{y}_{2}}-{{y}_{1}})j+({{z}_{2}}-{{z}_{1}})k$ $\\overrightarrow{OG}=(a-0)i+(a-0)j+(a-0)k=ai+aj+ak$ And $\\overrightarrow{AD}=(0-a)i+(a-0)j+(a-0)k=-ai+aj+ak$, So therefore we get the modulus of $\\overrightarrow{OG}$ and $\\overrightarrow{AD}$, $\\left| \\overrightarrow{OG} \\right|=\\sqrt{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}=a\\sqrt{3}$and$\\left| \\overrightarrow{AD} \\right|=\\sqrt{{{(-a)}^{2}}+{{a}^{2}}+{{a}^{2}}}=a\\sqrt{3}$ So we know $\\overrightarrow{OG}.\\overrightarrow{AD}=\\left( ai+aj+ak \\right).\\left( -ai+aj+ak \\right)=-{{a}^{2}}\\left( i.i \\right)+{{a}^{2}}\\left( j.j \\right)+{{a}^{2}}\\left( k.k \\right)$ So we know that angle between two vectors $\\overrightarrow{a}.\\overrightarrow{b}$ is given by, $\\theta ={{\\cos }^{-1}}\\left( \\dfrac{\\overrightarrow{a}.\\overrightarrow{b}}{\\left| \\overrightarrow{a} \\right|\\overrightarrow{\\left| b \\right|}} \\right)$, So applying above, so the angle between two diagonals $\\overrightarrow{OG}$and$\\overrightarrow{AD}$is as follows, Consider the angle between two diagonals be $\\theta$, So we get the angle as, $\\theta ={{\\cos }^{-1}}\\left( \\dfrac{\\overrightarrow{OG}.\\overrightarrow{AD}}{\\left| \\overrightarrow{OG} \\right|\\left| \\overrightarrow{AD} \\right|} \\right)$ So substituting above values, we get, $\\theta ={{\\cos }^{-1}}\\left( \\dfrac{{{a}^{2}}}{a\\sqrt{3}\\times a\\sqrt{3}} \\right)$ So we get, $\\theta ={{\\cos }^{-1}}\\left( \\dfrac{1}{3} \\right)$ So the correct answer is option(C). Note: Be careful about the points$A({{x}_{1}},{{y}_{1}},{{z}_{1}})$and$B({{x}_{2}},{{y}_{2}},{{z}_{2}})$. Sometimes we get confused between the points. So take care of it. The angle between two vectors $\\overrightarrow{a}.\\overrightarrow{b}$ is $\\theta ={{\\cos }^{-1}}\\left( \\dfrac{\\overrightarrow{a}.\\overrightarrow{b}}{\\left| \\overrightarrow{a} \\right|\\overrightarrow{\\left| b \\right|}} \\right)$. So you should", "# Difference between revisions of \"Dodecagon\" A dodecagon is a 12-sided polygon. The sum of its internal angles is $1800^{\\circ}$. A regular dodecagon can be seen below: $[asy] for(int i = 0; i <= 11; ++i) { draw(dir(360/12*i)--dir(360/12*(i + 1))); } pair A,B,C,D,E,F,G,H,I,J,K,L; A=dir(360/12*0); B=dir(360/12*1); C=dir(360/12*2); D=dir(360/12*3); E=dir(360/12*4); F=dir(360/12*5); G=dir(360/12*6); H=dir(360/12*7); I=dir(360/12*8); J=dir(360/12*9); K=dir(360/12*10); L=dir(360/12*11); label(\"A\",A,dir(0)); label(\"B\",B,dir(30)); label(\"C\",C,dir(60)); label(\"D\",D,dir(90)); label(\"E\",E,dir(120)); label(\"F\",F,dir(150)); label(\"G\",G,dir(180)); label(\"H\",H,dir(210)); label(\"I\",I,dir(240)); label(\"J\",J,dir(270)); label(\"K\",K,dir(300)); label(\"L\",L,dir(330)); draw(dir(360/12*0)--dir(360/12*6)); dot((dir(360/12*0)+dir(360/12*6))/2); pair O = (dir(360/12*0)+dir(360/12*6))/2; label(\"O\",O,S); draw(A--G); draw(Circle(O,1)); [/asy]$ The area of a regular dodecagon can be calculated by the formula $3R^2$, where $R$ is the circumradius of the dodecagon. In this case, $R$ would be $OA$.", "# Difference between revisions of \"2009 AMC 8 Problems/Problem 25\" ## Problem A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $\\frac{1}{2}$ foot from the top face. The second cut is $\\frac{1}{3}$ foot below the first cut, and the third cut is $\\frac{1}{17}$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet? $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label(\"A\", (1,0,3/4), W); label(\"B\", (1,0,1/3), W); label(\"C\", (1,0,1/6-d/4), W); label(\"D\", (1,0,d/2), W); label(\"1/2\", (1,1,3/4), E); label(\"1/3\", (1,1,1/3), E); label(\"1/17\", (0,1,1/6-d/4), E);[/asy]$ $[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label(\"D\", (1/2, 1, 0), SE); label(\"A\", (1+1/2, 1, 0), SE); label(\"B\", (2+1/2, 1, 0), SE); label(\"C\", (3+1/2, 1, 0), SE);[/asy]$ $\\textbf{(A)}\\:6\\qquad\\textbf{(B)}\\:7\\qquad\\textbf{(C)}\\:\\frac{419}{51}\\qquad\\textbf{(D)}\\:\\frac{158}{17}\\qquad\\textbf{(E)}\\:11$ ## Solution 1 The areas of the tops of $A$, $B$, $C$, and $D$ in", "# 1988 AHSME Problems/Problem 20 ## Problem In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base, $XY / YZ$, in this rectangle is $[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,1), B=(0,-1), C=(2,-1), D=(2,1), E=(1,-1), F=(1,1), G=(.8,.6); pair X=(4,sqrt(5)), Y=(4,-sqrt(5)), Z=(4+2/sqrt(5),-sqrt(5)), W=(4+2/sqrt(5),sqrt(5)), T=(4,0), U=(4+2/sqrt(5),-4/sqrt(5)), V=(4+2/sqrt(5),1/sqrt(5)); draw(A--B--C--D--A^^B--F^^E--D^^A--G^^rightanglemark(A,G,F)); draw(X--Y--Z--W--X^^T--V--X^^Y--U); label(\"A\", A, NW); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, NE); label(\"E\", E, S); label(\"F\", F, N); label(\"G\", G, E); label(\"X\", X, NW); label(\"Y\", Y, SW); label(\"Z\", Z, SE); label(\"W\", W, NE); [/asy]$ $\\textbf{(A)}\\ 4\\qquad \\textbf{(B)}\\ 1+2\\sqrt{3}\\qquad \\textbf{(C)}\\ 2\\sqrt{5}\\qquad \\textbf{(D)}\\ \\frac{8+4\\sqrt{3}}{3}\\qquad \\textbf{(E)}\\ 5$ ## Solution Within $WXYZ$, the parallelogram piece has vertical side $BF = \\sqrt{1^2 + 2^2} = \\sqrt{5}$, and diagonal side $FD = 1$. Thus the triangle in the bottom-right hand corner (the one with horizontal side $YZ$) must have hypotenuse $1$, and the only such triangle in the original figure is $\\triangle AFG$, so we deduce $YZ = AG = \\frac{\\frac{1}{2} \\times 1 \\times 2}{\\frac{1}{2} \\times \\sqrt{5}} = \\frac{2}{\\sqrt{5}}.$ Now the rectangle must have the same area as the square, as the pieces were put", "Let $$EFGH$$, $$EFDC$$, and $$EHBC$$ be three adjacent square faces of a cube, for which $$EC = 8$$, and let $$A$$ be the eighth vertex of the cube. Let $$I$$, $$J$$, and $$K$$, be the points on $$\\overline{EF}$$, $$\\overline{EH}$$, and $$\\overline{EC}$$, respectively, so that $$EI = EJ = EK = 2$$. A solid $$S$$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $$\\overline{AE}$$, and containing the edges, $$\\overline{IJ}$$, $$\\overline{JK}$$, and $$\\overline{KI}$$. The surface area of $$S$$, including the walls of the tunnel, is $$m + n\\sqrt {p}$$, where $$m$$, $$n$$, and $$p$$ are positive integers and $$p$$ is not divisible by the square of any prime. Find $$m + n + p$$. (第十九届AIME2 2001 第15题)" ]

[ "# Difference between revisions of \"2018 AIME I Problems/Problem 4\" ## Problem 4 In $\\triangle ABC, AB = AC = 10$ and $BC = 12$. Point $D$ lies strictly between $A$ and $B$ on $\\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\\overline{AC}$ so that $AD = DE = EC$. Then $AD$ can be expressed in the form $\\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. ## Solution $\\cos(A) = \\frac{5^2+5^2-6^2}{2*5*5} = \\frac{7}{25}$. Let $M$ be midpoint of $AE$, then $\\frac{7}{25} = \\frac{10-x}{2x} \\iff x =\\frac{250}{39}$. ## Solution 1 (No Trig) $[asy] import cse5; unitsize(10mm); pathpen=black; dotfactor=3; pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66); pair[] dotted = {A,B,C,D,E,F,G}; D(A--B); D(C--B); D(A--C); D(D--E); pathpen=dashed; D(B--F); D(D--G); dot(dotted); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NW); label(\"E\",E,NE); label(\"F\",F,NE); label(\"G\",G,NE); label(\"x\",H,NW); label(\"x\",I,NW); label(\"x\",J,NE); [/asy]$ We draw the altitude from $B$ to $\\overline{AC}$ to get point $F$. We notice that the triangle's height from $A$ to $\\overline{BC}$ is 8 because it is a $3-4-5$ Right Triangle. To find the length of $\\overline{BF}$, we let $h$ represent $\\overline{BF}$ and set up an equation by finding two ways to express the area. The equation is $(8)(12)=(10)(h)$, which leaves us with $h=9.6$. We then solve for the", "$0 < x \\leq y \\leq 37$ so we try those: $(x+1)(y+1) = 22 * 32$ $x+1=22, y+1 = 32$ $x = 21, y = 31$ Therefore, the difference $y-x=31-21=10$, choice E). ## Solution 11 $[asy] size(8cm); pair A = (5,12); pair B = (0,0); pair C = (14,0); pair P = 2/3*A+1/3*C; pair D = 3/2*P; pair E = 3*P; draw(A--B--C--A); draw(A--D); draw(C--E--B); dot(\"A\",A,N); dot(\"B\",B,W); dot(\"C\",C,ESE); dot(\"D\",D,N); dot(\"P\",P,W); dot(\"E\",E,N); defaultpen(fontsize(9pt)); label(\"13\", (A+B)/2, NW); label(\"14\", (B+C)/2, S); label(\"5\",(A+P)/2, NE); label(\"10\", (C+P)/2, NE); [/asy]$ 1、Toss on the Cartesian plane with $A=(5, 12), B=(0, 0),$ and $C=(14, 0)$. Then $\\overline{AD}\\parallel\\overline{BC}, \\overline{AB}\\parallel\\overline{CE}$ by the trapezoid condition, where $D, E\\in\\overline{BP}$. Since $PC=10$, point $P$ is $\\tfrac{10}{15}=\\tfrac{2}{3}$ of the way from $C$ to $A$ and is located at $(8, 8)$. Thus line $BP$ has equation $y=x$. Since $\\overline{AD}\\parallel\\overline{BC}$ and $\\overline{BC}$ is parallel to the ground, we know $D$ has the same $y$-coordinate as $A$, except it'll also lie on the line $y=x$. Therefore, $D=(12, 12). \\, \\blacksquare$ To find the location of point $E$, we need to find the intersection of $y=x$ with a line parallel to $\\overline{AB}$ passing through $C$. The slope of this line is the same as the slope of $\\overline{AB}$, or $\\tfrac{12}{5}$, and has equation $y=\\tfrac{12}{5}x-\\tfrac{168}{5}$. The intersection of this line with $y=x$ is $(24, 24)$. Therefore point $E$", "# Difference between revisions of \"2012 AIME I Problems/Problem 8\" ## Problem Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\\overline{AB}$ and $\\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ $[asy]import cse5; unitsize(10mm); pathpen=black; dotfactor=3; pair A = (0,0), B = (3.8,0), C = (5.876,1.564), D = (2.076,1.564), E = (0,3.8), F = (3.8,3.8), G = (5.876,5.364), H = (2.076,5.364), M = (1.9,0), N = (5.876,3.465); pair[] dotted = {A,B,C,D,E,F,G,H,M,N}; D(A--B--C--G--H--E--A); D(E--F--B); D(F--G); pathpen=dashed; D(A--D--H); D(D--C); dot(dotted); label(\"A\",A,SW); label(\"B\",B,S); label(\"C\",C,SE); label(\"D\",D,NW); label(\"E\",E,W); label(\"F\",F,SE); label(\"G\",G,NE); label(\"H\",H,NW); label(\"M\",M,S); label(\"N\",N,NE); [/asy]$ ## Solution: think outside the box (pun intended) Define a coordinate system with $D$ at the origin and $C,$ $A,$ and $H$ on the $x$, $y$, and $z$ axes respectively. Then $D=(0,0,0),$ $M=(.5,1,0),$ and $N=(1,0,.5).$ It follows that the plane going through $D,$ $M,$ and $N$ has equation $2x-y-4z=0.$ Let $Q = (1,1,.25)$ be the intersection of this plane and edge $BF$ and let $P = (1,2,0).$ Now since $2(1) - 1(2) - 4(0) = 0,$ $P$ is on the plane. Also,", "in a 1:3 ratio, therefore, $MK=3x$ and $KF=9x$, so $10x=EF=6$. It follows that $EK=0.6$ and $MK=1.8$, so the coordinates of $M$ are $(4+0.6,1.8)=(4.6,1.8)$ and so our answer is $\\boxed{\\mathbf{(C)}\\ 6.4}$. ### Solution 2 $[asy] size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);[/asy]$ Let the four points be labeled $P_1$, $P_2$, $P_3$, and $P_4$, respectively. Let the lines that go through each point be labeled $L_1$, $L_2$, $L_3$, and $L_4$, respectively. Since $L_1$ and $L_2$ go through $SP$ and $RQ$, respectively, and $SP$ and $RQ$ are opposite sides of the square, we can say that $L_1$ and $L_2$ are parallel with slope $m$. Similarly, $L_3$ and $L_4$ have slope $-\\frac{1}{m}$. Also, note that since square $PQRS$ lies in the first quadrant, $L_1$ and $L_2$ must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: $L_1: y = m(x-3)$, $L_2: y = m(x-5)$, $L_3: y = -\\frac{1}{m}(x-7)$, $L_4: y = -\\frac{1}{m}(x-13)$. Since $PQRS$ is a square, it follows that $\\Delta x$ between points $P$ and $Q$ is equal to $\\Delta y$ between points $Q$ and $R$. Our approach will be to find $\\Delta x$ and $\\Delta y$ in terms of $m$ and equate the", "\\frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 size(8cm); pair A, B, C, D, E, EE; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); EE = (27/8,0); draw(A--B--C--D--A--E); draw(E--EE,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,EE,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",EE,S); label(\"$4$\",(A+D)/2,S); label(\"$x$\",(A+EE)/2,S; label(\"$4-x$\",(D+EE)/2,S); label(\"$\\frac{15}{8}$\",(E+EE)/2,W); (Error compiling LaTeX. label(\"$x$\",(A+EE)/2,S; ^ de021056a2cc2ef4e3e5c00687df10ec92699974.asy: 26.23: syntax error error: could not load module 'de021056a2cc2ef4e3e5c00687df10ec92699974.asy') Following the steps above, you can find that the height of triangle $ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the", "are not the same. Hence, from the above the representation of the given points in the coordinate plane is shown above. Find AC. (Section 1.2) Question 13. The length of $$\\overline{A C}$$ is 39, Explanation: The given line segment is: So, From the Segment Addition Postulate $$\\overline{A C}$$ = $$\\overline{A B}$$ + $$\\overline{B C}$$, $$\\overline{A C}$$ = 13 + 26, $$\\overline{A C}$$ = 39, Hence, from the above we can conclude that the length of $$\\overline{A C}$$ is 39. Question 14. The length of $$\\overline{A C}$$ is 51, Explanation: The given line segment is: So from the Segment Addition Postulate $$\\overline{A B}$$ = $$\\overline{A C}$$ + $$\\overline{B C}$$, 62 = $$\\overline{A C}$$ + 11, $$\\overline{A C}$$ = 62 – 11, $$\\overline{A C}$$ = 51, Hence, from the above, We can conclude that the length of $$\\overline{A C}$$ is 51. Find the coordinates of the midpoint M and the distance between the two points. Question 15. J(4, 3) and K(2, – 3) Midpoint M(3,0), Explanation: The given points are J (4, 3), K (2, -3), We know that M = ($$\\frac{x1 + x2}{2}$$, $$\\frac{y1 + y2}{2}$$) So M = ($$\\frac{4 + 2}{2}$$, $$\\frac{3 – 3}{2}$$), M = ($$\\frac{6}{2}$$, $$\\frac{0}{2}$$), M = (3, 0). Hence from the above we can conclude that the coordinates of the midpoint are (3, 0).", "$$x=4,$$ $$y=4$$ and the coordinate axes. If $${S_1},{S_2},{S_3}$$ are respectively the areas of these parts numbered from top to bottom ; then $${S_1},{S_2},{S_3}$$ is A $$1:2:1$$ B $$1:2:3$$ C $$2:1:2$$ D $$1:1:1$$ ## Explanation Intersection points of $${x^2} = 4y$$ and $${y^2} = 4x$$ are $$\\left( {0,0} \\right)$$ and $$\\left( {4,4} \\right).$$ The graph is as shown in the figure. By symmetry, we observe $${S_1} = {S_3} = \\int\\limits_0^4 {ydx = \\int\\limits_0^4 {{{{x^2}} \\over 4}dx} }$$ $$= \\left[ {{{{x^3}} \\over {12}}} \\right]_0^4 = {{16} \\over 3}$$ sq. unit Also $${S_2} = \\int\\limits_0^4 {\\left( {2\\sqrt x - {{{x^2}} \\over 4}} \\right)} dx$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$= \\left[ {{{2{x^{{3 \\over 2}}}} \\over {{3 \\over 2}}} - {{{x^3}} \\over {12}}} \\right]_0^4$$ $$\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,$$ $$= {4 \\over 3} \\times 8 - {{16} \\over 3} = {{16} \\over 3}$$ $$\\therefore$$ $${S_1}:{S_2}:{S_3} = 1:1:1$$ 4 ### AIEEE 2005 The area enclosed between the curve $$y = {\\log _e}\\left( {x + e} \\right)$$ and the coordinate axes is A $$1$$ B $$2$$ C $$3$$ D $$4$$ ## Explanation The graph of the curve $$y = {\\log _e}\\left( {x + e} \\right)$$ is as shown in the fig. Required area $$A = \\int\\limits_{1 - e}^0 {ydx} = \\int\\limits_{1 - e}^0 {{{\\log }_e}} \\left( {x + e} \\right)dx$$ put $$x + e = t \\Rightarrow dx = dt$$ also At", "and $EF$. They are as follows:$$AG = f(x) = -\\frac{3}{5}x + 4$$$$AC = g(x) = -\\frac{4}{5}x + 4$$$$EF = h(x) = 2x - 4$$After drawing in altitudes to $DC$ from $P$, $Q$, and $E$, we see that $\\frac{PQ}{EF} = \\frac{P'Q'}{E'F}$ because of similar triangles, and so we only need to find the x-coordinates of $P$ and $Q$. $[asy] pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair D1=(20/7,0),D2=(20/7,12/7); pair E1=(40/13,0),E2=(40/13,28/13); pair F1=(4,0),F2=(4,4); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(D1--D2,dashed); draw(E1--E2,dashed); draw(F1--F2,dashed); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); dot((20/7,0)); dot((40/13,0)); dot((4,0)); label(\"Q\",(3.07692307692,2.15384615384),N); label(\"P\",(20/7,12/7),W); label(\"A\",(0,4), NW); label(\"B\",(5,4), NE); label(\"C\",(5,0),SE); label(\"D\",(0,0),SW); label(\"F\",(2,0),S); label(\"G\",(5,1),E); label(\"E\",(4,4),N); label(\"P'\", (20/7,0),SSW); label(\"Q'\", (40/13,0),SSE); label(\"E'\", (4,0),S); dot(A1); dot(A2); dot(B1); dot(B2); dot(C1); dot(C2); dot((0,0)); dot((5,4));[/asy]$ Finding the intersections of $AC$ and $EF$, and $AG$ and $EF$ gives the x-coordinates of $P$ and $Q$ to be $\\frac{20}{7}$ and $\\frac{40}{13}$. This means that $P'Q' = DQ' - DP' = \\frac{40}{13} - \\frac{20}{7} = \\frac{20}{91}$. Now we can find $\\frac{PQ}{EF} = \\frac{P'Q'}{E'F} = \\frac{\\frac{20}{91}}{2} = \\boxed{\\textbf{(D)}~\\frac{10}{91}}$ ## Solution 3 (Similar Triangles) $[asy] pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair H = (20/3,0); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(B1--H); draw((0,0)--H); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label(\"Q\",(3.07692307692,2.15384615384),N); label(\"P\",(20/7,12/7),W); label(\"A\",(0,4), NW); label(\"B\",(5,4), NE); label(\"C\",(5,0),SE); label(\"D\",(0,0),SW); label(\"F\",(2,0),S); label(\"G\",(5,1),E); label(\"E\",(4,4),N); label(\"H\",H,E); [/asy]$ Extend $AG$ to intersect $CD$ at $H$. Letting $x=\\overline{HC}$, we have that $$\\triangle{HCG}\\sim\\triangle{HDA}\\implies \\dfrac{\\overline{HC}}{\\overline{CG}}=\\dfrac{\\overline{HD}}{\\overline{AD}}\\implies \\dfrac{x}{1}=\\dfrac{x+5}{4}\\implies x=\\dfrac{5}{3}.$$ Then, notice that $\\triangle{AEQ}\\sim\\triangle{HFQ}$ and $\\triangle{AEP}\\sim\\triangle{CFP}$. Thus, we see that", "# 2003 AIME I Problems/Problem 15 ## Problem In $\\triangle ABC, AB = 360, BC = 507,$ and $CA = 780.$ Let $M$ be the midpoint of $\\overline{CA},$ and let $D$ be the point on $\\overline{CA}$ such that $\\overline{BD}$ bisects angle $ABC.$ Let $F$ be the point on $\\overline{BC}$ such that $\\overline{DF} \\perp \\overline{BD}.$ Suppose that $\\overline{DF}$ meets $\\overline{BM}$ at $E.$ The ratio $DE: EF$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ ## Solution 1 In the following, let the name of a point represent the mass located there. Since we are looking for a ratio, we assume that $AB=120$, $BC=169$, and $CA=260$ in order to simplify our computations. First, reflect point $F$ over angle bisector $BD$ to a point $F'$. $[asy] size(400); pointpen = black; pathpen = black+linewidth(0.7); pair A=(0,0),C=(7.8,0),B=IP(CR(A,3.6),CR(C,5.07)), M=(A+C)/2, Da = bisectorpoint(A,B,C), D=IP(B--B+(Da-B)*10,A--C), F=IP(D--D+10*(B-D)*dir(270),B--C), E=IP(B--M,D--F);pair Fprime=2*D-F; /* scale down by 100x */ D(MP(\"A\",A,NW)--MP(\"B\",B,N)--MP(\"C\",C)--cycle); D(B--D(MP(\"D\",D))--D(MP(\"F\",F,NE))); D(B--D(MP(\"M\",M)));D(A--MP(\"F'\",Fprime,SW)--D); MP(\"E\",E,NE); D(rightanglemark(F,D,B,4)); MP(\"390\",(M+C)/2); MP(\"390\",(M+C)/2); MP(\"360\",(A+B)/2,NW); MP(\"507\",(B+C)/2,NE); [/asy]$ As $BD$ is an angle bisector of both triangles $BAC$ and $BF'F$, we know that $F'$ lies on $AB$. We can now balance triangle $BF'C$ at point $D$ using mass points. By the Angle Bisector Theorem, we can place mass points on $C,D,A$ of $120,\\,289,\\,169$ respectively. Thus, a mass of", "being similar to $\\triangle AGB$, also have legs in a 1:3 ratio, therefore, $MK=3x$ and $KF=9x$, so $10x=EF=6$. It follows that $EK=0.6$ and $MK=1.8$, so the coordinates of $M$ are $(4+0.6,1.8)=(4.6,1.8)$ and so our answer is $4.6+1.8 = 6.4 =$ $\\boxed{\\mathbf{(C)}\\ 32/5}$. ### Solution 2 $[asy] size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA);[/asy]$ Let the four points be labeled $P_1$, $P_2$, $P_3$, and $P_4$, respectively. Let the lines that go through each point be labeled $L_1$, $L_2$, $L_3$, and $L_4$, respectively. Since $L_1$ and $L_2$ go through $SP$ and $RQ$, respectively, and $SP$ and $RQ$ are opposite sides of the square, we can say that $L_1$ and $L_2$ are parallel with slope $m$. Similarly, $L_3$ and $L_4$ have slope $-\\frac{1}{m}$. Also, note that since square $PQRS$ lies in the first quadrant, $L_1$ and $L_2$ must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: $L_1: y = m(x-3)$, $L_2: y = m(x-5)$, $L_3: y = -\\frac{1}{m}(x-7)$, $L_4: y = -\\frac{1}{m}(x-13)$. Since $PQRS$ is a square, it follows that $\\Delta x$ between points $P$ and $Q$ is equal to $\\Delta y$ between points $Q$ and $R$. Our approach will be to find", "so that we declare it as: In [5]: U1 = RP2.open_subset('U_1') ; U1 Out[5]: Any line in $U_1$ is uniquely determined by its intersection with the plane $z=1$. The Cartesian coordinates $(x,y,1)$ of the intersection point lead to an obvious coordinate system $(x_1,y_1)$ on $U_1$ by setting $(x_1,y_1)=(x,y)$: In [6]: X1.<x1,y1> = U1.chart() ; X1 Out[6]: Note that since we have not specified any coordinate range in the arguments of chart(), the range of $(x_1,y_1)$ is $\\mathbb{R}^2$. Similarly, let $U_2$ be the set of lines through the origin of $\\mathbb{R}^3$ that are not contained in the plane $x=0$. Any line in $U_2$ is uniquely determined by its intersection $(1,y,z)$ with the plane $x=1$, leading to coordinates $(x_2,y_2)=(y,z)$ on $U_2$: In [7]: U2 = RP2.open_subset('U_2') X2.<x2,y2> = U2.chart() ; X2 Out[7]: Finally, let $U_3$ be the set of lines through the origin of $\\mathbb{R}^3$ that are not contained in the plane $y=0$. Any line in $U_3$ is uniquely determined by its intersection $(x,1,z)$ with the plane $y=1$, leading to coordinates $(x_3,y_3)=(z,x)$ on $U_3$: In [8]: U3 = RP2.open_subset('U_3') X3.<x3,y3> = U3.chart() ; X3 Out[8]: We declare that the union of the three (overlapping) open domains $U_1$, $U_2$ and $U_3$ is $\\mathbb{RP}^2$: In [9]: RP2.declare_union(U1.union(U2), U3) U1.union(U2).union(U3) Out[9]: At this stage, three open covers of $\\mathbb{RP}^2$ have been constructed: In", "# 2001 AIME II Problems/Problem 15 ## Problem Let $EFGH$, $EFDC$, and $EHBC$ be three adjacent square faces of a cube, for which $EC = 8$, and let $A$ be the eighth vertex of the cube. Let $I$, $J$, and $K$, be the points on $\\overline{EF}$, $\\overline{EH}$, and $\\overline{EC}$, respectively, so that $EI = EJ = EK = 2$. A solid $S$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $\\overline{AE}$, and containing the edges, $\\overline{IJ}$, $\\overline{JK}$, and $\\overline{KI}$. The surface area of $S$, including the walls of the tunnel, is $m + n\\sqrt {p}$, where $m$, $n$, and $p$ are positive integers and $p$ is not divisible by the square of any prime. Find $m + n + p$. ## Solution $[asy] import three; currentprojection = orthographic(camera=(1/4,2,3/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype(\"10 2\"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8)--(0,0,8)--(0,0,1)); draw((1,0,0)--(8,0,0)--(8,8,0)--(0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,8,8)--(0,0,8)--(0,0,1)); draw((8,8,0)--(8,8,6),l); draw((8,0,8)--(8,6,8)); draw((0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy]$ $[asy] import three; currentprojection = orthographic(camera=(1/2,1/3,-1/4)); defaultpen(linewidth(0.7)); pen l = linewidth(0.5) + linetype(\"10 2\"); triple S=(1,0,0), T=(2,0,2), U=(8,6,8), V=(8,8,6), W=(2,2,0), X=(6,8,8); draw((1,0,0)--(8,0,0)--(8,0,8),l); draw((8,0,8)--(0,0,8)); draw((0,0,8)--(0,0,1),l); draw((8,0,0)--(8,8,0)); draw((8,8,0)--(0,8,0)); draw((0,8,0)--(0,1,0),l); draw((0,8,0)--(0,8,8)); draw((0,0,8)--(0,0,1),l); draw((8,8,0)--(8,8,6)); draw((8,0,8)--(8,6,8)); draw((0,0,8)--(0,8,8)--(6,8,8)); draw(S--T--U--V--W--cycle); draw((0,0,1)--T--U--X--(0,2,2)--cycle); draw((0,1,0)--W--V--X--(0,2,2)--cycle); [/asy]$ Set the coordinate system so that vertex $E$, where the drilling starts, is at $(8,8,8)$. Using a little", "so the maximum length of $XY$ is $\\sin(a)\\times\\frac{\\sin(t')}{\\sin(t' + a) + \\cos(a)} = 7 - 4\\sqrt{3}$, and the answer is $7 + 4 + 3 = \\boxed{014}$. ### Solution 3 $[asy] unitsize(144); pair A, B, C, M, n; A = (0,1); B = (-7/25, 24/25); C=(1/7,-4*sqrt(3)/7); M = (-1,0); n = (1,0); pair [] D = intersectionpoints(A--C,M--n); pair [] e = intersectionpoints(B--C,M--n); draw(circle((0,0),1)); draw(M--n--B--M--A--n--C--A--B--C--cycle); label(\"A\",A,N); label(\"B\",B,NNW); label(\"M\",M,W); label(\"C\",C,SSE); label(\"N\",n,E); label(\"D\",D[0],SE); label(\"E\",e[0],SW); label(\"x\",(M+C)/2,SW); label(\"y\",(n+C)/2,SE); [/asy]$ Suppose $\\overline{AC}$ and $\\overline{BC}$ intersect $\\overline{MN}$ at $D$ and $E$, respectively, and let $MC = x$ and $NC = y$. Since $A$ is the midpoint of arc $MN$, $\\overline{CA}$ bisects $\\angle MCN$, and we get $$\\frac{MC}{MD} = \\frac{NC}{ND}\\Rightarrow MD = \\frac{x}{x + y}.$$ To find $ME$, we note that $\\triangle BNE\\sim\\triangle MCE$ and $\\triangle BME\\sim\\triangle NCE$, so \\begin{align*} \\frac{BN}{NE} &= \\frac{MC}{CE} \\\\ \\frac{ME}{BM} &= \\frac{CE}{NC}. \\end{align*} Writing $NE = 1 - ME$, we can substitute known values and multiply the equations to get $$\\frac{4(ME)}{3 - 3(ME)} = \\frac{x}{y}\\Rightarrow ME = \\frac{3x}{3x + 4y}.$$ The value we wish to maximize is \\begin{align*} DE &= MD - ME \\\\ &= \\frac{x}{x + y} - \\frac{3x}{3x + 4y} \\\\ &= \\frac{xy}{3x^2 + 7xy + 4y^2} \\\\ &= \\frac{1}{3(x/y) + 4(y/x) + 7}. \\end{align*} By the AM-GM inequality, $3(x/y) + 4(y/x)\\geq 2\\sqrt{12} = 4\\sqrt{3}$, so $$DE\\leq \\frac{1}{4\\sqrt{3} +", "# 2000 AIME II Problems/Problem 11 ## Problem The coordinates of the vertices of isosceles trapezoid $ABCD$ are all integers, with $A=(20,100)$ and $D=(21,107)$. The trapezoid has no horizontal or vertical sides, and $\\overline{AB}$ and $\\overline{CD}$ are the only parallel sides. The sum of the absolute values of all possible slopes for $\\overline{AB}$ is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. ## Solution For simplicity, we translate the points so that $A$ is on the origin and $D = (1,7)$. Suppose $B$ has integer coordinates; then $\\overrightarrow{AB}$ is a vector with integer parameters (vector knowledge is not necessary for this solution). We construct the perpendicular from $A$ to $\\overline{CD}$, and let $D' = (a,b)$ be the reflection of $D$ across that perpendicular. Then $ABCD'$ is a parallelogram, and $\\overrightarrow{AB} = \\overrightarrow{D'C}$. Thus, for $C$ to have integer coordinates, it suffices to let $D'$ have integer coordinates.[1] $[asy] pathpen = linewidth(0.7); pair A=(0,0), D=(1,7), Da = MP(\"D'\",D((-7,1)),N), B=(-8,-6), C=B+Da, F=foot(A,C,D); D(MP(\"A\",A)--MP(\"B\",B)--MP(\"C\",C,N)--MP(\"D\",D,N)--cycle); D(F--A--Da,linetype(\"4 4\")); [/asy]$ Let the slope of the perpendicular be $m$. Then the midpoint of $\\overline{DD'}$ lies on the line $y=mx$, so $\\frac{b+7}{2} = m \\cdot \\frac{a+1}{2}$. Also, $AD = AD'$ implies that $a^2 + b^2 = 1^2 + 7^2 = 50$. Combining these two equations yields $$a^2 + \\left(7 - (a+1)m\\right)^2 =", "the same in magnitude, direction, and sense. In other words, if $$PQ$$ and $$RS$$ are equal and parallel, and the sense of motion from $$P$$ to $$Q$$ is the same as that of motion from $$R$$ to $$S$$, we regard the displacements $$\\overline{PQ}$$ and $$\\overline{RS}$$ as equivalent, and write $\\overline{PQ} = \\overline{RS}.$ Now let us take any pair of coordinate axes in the plane (such as $$OX$$$$OY$$ in Fig. 19). Draw a line $$OA$$ equal and parallel to $$PQ$$, the sense of motion from $$O$$ to $$A$$ being the same as that from $$P$$ to $$Q$$. Then $$\\overline{PQ}$$ and $$\\overline{OA}$$ are equivalent displacements. Let $$x$$ and $$y$$ be the coordinates of $$A$$. Then it is evident that $$\\overline{OA}$$ is completely specified if $$x$$ and $$y$$ are given. We call $$\\overline{OA}$$ the displacement $$[x, y]$$ and write $\\overline{OA} = \\overline{PQ} = \\overline{RS} = [x, y].$ ## 35. Equivalence of displacements. Multiplication of displacements by numbers. If $$\\xi$$ and $$\\eta$$ are the coordinates of $$P$$, and $$\\xi’$$ and $$\\eta’$$ those of $$Q$$, it is evident that $x = \\xi’ – \\xi,\\quad y = \\eta’ – \\eta.$ The displacement from $$(\\xi, \\eta)$$ to $$(\\xi’, \\eta’)$$ is therefore $[\\xi’ – \\xi, \\eta’ – \\eta].$ It is clear that two displacements $$[x, y]$$, $$[x’, y’]$$ are equivalent if, and only if, $$x =", "pathpen = d; pair A = (-5, 12), B = (5, 12), C = (0, 0); D(CR(A,16));D(CR(B,4));D(shift((0,12)) * yscale(3^.5 / 2) * CR(C,10), linetype(\"2 2\") + d + red); D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP(\"y=ax\",(14,14 * (69/100)^.5),E),EndArrow(4)); void bluecirc (real x) { pair P = (x, (3 * (25 - x^2 / 4))^.5 + 12); dot(P, blue); D(CR(P, ((P.x - 5)^2 + (P.y - 12)^2)^.5 - 4) , blue + d + linetype(\"4 4\")); } bluecirc(-9.2); bluecirc(-4); bluecirc(3); [/asy]$ Since the center lies on the line $y = ax$, we substitute for $y$ and expand: $$1 = \\frac{x^2}{100} + \\frac{(ax-12)^2}{75} \\Longrightarrow (3+4a^2)x^2 - 96ax + 276 = 0.$$ We want the value of $a$ that makes the line $y=ax$ tangent to the ellipse, which will mean that for that choice of $a$ there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is $0$, so $(-96a)^2 - 4(3+4a^2)(276) = 0$. Solving yields $a^2 = \\frac{69}{100}$, so the answer is $\\boxed{169}$. ## Solution 2 As above, we rewrite the equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$. Let $F_1=(-5,12)$ and $F_2=(5,12)$. If a circle with center $C=(a,b)$ and radius $r$ is externally tangent to $w_2$ and internally tangent to $w_1$, then $CF_1=16-r$ and $CF_2=4+r$. Therefore, $CF_1+CF_2=20$. In particular, the locus", "2, Hence, from the above we can conclude that $$\\overline{A B}$$ is not congruent with $$\\overline{C D}$$, represented the given points in the coordinate plane above. Question 11. A(8, 3), B(- 1, 3), C(5, 10), D(5, 3) $$\\overline{A B}$$ is not congruent with $$\\overline{C D}$$, Explanation: Question 12. A(6, – 8), B(6, 1), C(7, – 2), D(- 2, – 2) $$\\overline{A B}$$ is congruent with $$\\overline{C D}$$, Explanation: The given points are A (6, -8), B (6, 1), C (7, -2), and D (-2, -2), Let the points be represented as A (x1, y1), B (x2, y2), C (x3, y3), and D (x4, y4) We know that the distance between points A and B is D = √(x2 – x1)² + (y2 – y1)², So $$\\overline{A B}$$ = √(6 – 6)² + (8 + 1)², = √(9)² = 9 The distance between points C and D is $$\\overline{C D}$$ = √(x4 – x3)² + (y4 – y3)², = √(-7 – 2)² + (-2 + 2)², = √(-9)², = 9, Hence we can conclude that $$\\overline{A B}$$ is congruent with $$\\overline{C D}$$, The representation of the given points in the coordinate plane is shown above. Question 13. A(- 5, 6), B(- 5, – 1), C(- 4, 3), D(3, 3) $$\\overline{A B}$$ is congruent with $$\\overline{C D}$$, Explanation: Question 14.", "+0 # HELP 0 194 1 In the figure below, $D$ is a point on segment $\\overline{CE}$ such that $\\overline{AD}\\parallel\\overline{BE}$ and $A$ is not on $\\overline{BC}.$ Line segments $\\overline{AD}$ and $\\overline{BC}$ intersect at $P.$ [asy] size(4cm); pair C = (0,0); pair D = (4,0); pair E = (9,0); pair F = (0.25,4); pair G = (1,4); pair A = extension(C,F,D,G); pair H = G+E-D; pair B = extension(C,G,E,H); pair P = extension(A,D,B,C); draw(A--C--E--B--C); draw(A--D); label(\"$A$\",A,NNW); label(\"$B$\",B,N); label(\"$C$\",C,SSW); label(\"$D$\",D,S); label(\"$E$\",E,SSE); label(\"$P$\",P,dir(0)); [/asy] Can $\\angle CAD=\\angle CBE?$ Explain. Aug 20, 2020", "area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 $[asy] size(8cm); pair A, B, C, D, E, F; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); F = (27/8,0); draw(A--B--C--D--A--E); draw(E--F,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,F,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",F,S); label(\"4\",(A+D)/2,S); label(\"x\",(A+F)/2,S); label(\"\\frac{15}{8}\",(E+F)/2,W); [/asy]$ Let the point where the altitude from $E$ to $\\overline{AD}$ be labeled $F$. Following the steps above, you can find that the height of $\\triangle ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the origin to the point $F$, and $4-x$ is the segment from point $F$ to point", "$2(1-t)-t=t$, or $t=\\frac{1}{2}$. This intersection point has $z=\\frac{\\sqrt{2}}{2}$. Similarly, the intersection between $\\overline{PR}$ and $\\triangle BDE$ has $z=\\frac{\\sqrt{2}}{2}$. So $\\overline{XY}$ lies on the plane $z=\\frac{\\sqrt{2}}{2}$, from which we obtain $X=\\left( \\frac{3}{2},\\frac{3}{2},\\frac{\\sqrt{2}}{2}\\right)$ and $Y=\\left( -\\frac{3}{2},-\\frac{3}{2},\\frac{\\sqrt{2}}{2}\\right)$. The area of the pentagon $EXQRY$ can be computed in the same way as above. ### Solution 3 $[asy]import three; import math; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4); triple A=(-2,2,0), B=(2,2,0), C=(2,-2,0), D=(-2,-2,0), E=(0,0,2*2^.5), P=(A+E)/2, Q=(B+C)/2, R=(C+D)/2, Y=(-3/2,-3/2,2^.5/2),X=(3/2,3/2,2^.5/2), H=(4,2,0), I=(-2,-4,0); draw(A--B--C--D--A--E--B--E--C--E--D); draw(B--H--Q, linetype(\"6 6\")+linewidth(0.7)+blue); draw(X--H, linetype(\"6 6\")+linewidth(0.7)+blue); draw(D--I--R, linetype(\"6 6\")+linewidth(0.7)+blue); draw(Y--I, linetype(\"6 6\")+linewidth(0.7)+blue); label(\"A\",A, SE); label(\"B\",B,NE); label(\"C\",C, SE); label(\"D\",D, W); label(\"E\",E,N); label(\"P\",P, NW); label(\"Q\",Q,(1,0,0)); label(\"R\",R, S); label(\"Y\",Y,NW); label(\"X\",X,NE); label(\"H\",H,NE); label(\"I\",I,S); draw(P--X--Q--R--Y--cycle,linetype(\"6 6\")+linewidth(0.7)); [/asy]$ Extend $\\overline{RQ}$ and $\\overline{AB}$. The point of intersection is $H$. Connect $\\overline{PH}$. $\\overline{EB}$ intersects $\\overline{PH}$ at $X$. Do the same for $\\overline{QR}$ and $\\overline{AD}$, and let the intersections be $I$ and $Y$ Because $Q$ is the midpoint of $\\overline{BC}$, and $\\overline{AB}\\parallel\\overline{BC}$, so $\\triangle{RQC}\\cong\\triangle{HQB}$. $\\overline{BH}=2$. Because $\\overline{BH}=2$, we can use mass point geometry to get that $\\overline{PX}=\\overline{XH}$. $|\\triangle{XHQ}|=\\frac{\\overline{XH}}{\\overline{PH}}\\cdot\\frac{\\overline{QH}}{\\overline{HI}}\\cdot|\\triangle{PHI}|=\\frac{1}{6}\\cdot|\\triangle{PHI}|$ Using the same principle, we can get that $|\\triangle{IYR}|=\\frac{1}{6}|\\triangle{PHI}|$ Therefore, the area of $PYRQX$ is $\\frac{2}{3}\\cdot|\\triangle{PHI}|$ $\\overline{RQ}=2\\sqrt{2}$, so $\\overline{IH}=6\\sqrt{2}$. Using the law of cosines, $\\overline{PH}=\\sqrt{28}$. The area of $\\triangle{PHI}=\\frac{1}{2}\\cdot\\sqrt{28-18}\\cdot6\\sqrt{2}=6\\sqrt{5}$ Using this, we can get the area of $PYRQX$ ## See also 2007 AIME I (Problems • Answer Key • Resources) Preceded" ]

[ "# Difference between revisions of \"2012 AIME I Problems/Problem 8\" ## Problem Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\\overline{AB}$ and $\\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ $[asy]import cse5; unitsize(10mm); pathpen=black; dotfactor=3; pair A = (0,0), B = (3.8,0), C = (5.876,1.564), D = (2.076,1.564), E = (0,3.8), F = (3.8,3.8), G = (5.876,5.364), H = (2.076,5.364), M = (1.9,0), N = (5.876,3.465); pair[] dotted = {A,B,C,D,E,F,G,H,M,N}; D(A--B--C--G--H--E--A); D(E--F--B); D(F--G); pathpen=dashed; D(A--D--H); D(D--C); dot(dotted); label(\"A\",A,SW); label(\"B\",B,S); label(\"C\",C,SE); label(\"D\",D,NW); label(\"E\",E,W); label(\"F\",F,SE); label(\"G\",G,NE); label(\"H\",H,NW); label(\"M\",M,S); label(\"N\",N,NE); [/asy]$ ## Solution: think outside the box (pun intended) Define a coordinate system with $D$ at the origin and $C,$ $A,$ and $H$ on the $x$, $y$, and $z$ axes respectively. Then $D=(0,0,0),$ $M=(.5,1,0),$ and $N=(1,0,.5).$ It follows that the plane going through $D,$ $M,$ and $N$ has equation $2x-y-4z=0.$ Let $Q = (1,1,.25)$ be the intersection of this plane and edge $BF$ and let $P = (1,2,0).$ Now since $2(1) - 1(2) - 4(0) = 0,$ $P$ is on the plane. Also,", "label(\"M_2\", M2, S); label(\"M_3\", M3, E); label(\"M_4\", M4, N); label(\"G_1\", G1, 1.5S); label(\"G_2\", G2, 1.5E); label(\"G_3\", G3, 1.5NE); label(\"G_4\", G4, 1.5E); dot(A); dot(B); dot(C); dot(D); dot(M1); dot(M2); dot(M3); dot(M4); dot(G1); dot(G2); dot(G3); dot(G4); [/asy]$ By SAS, we conclude that $\\triangle G_1G_2P\\sim\\triangle M_1M_2P, \\triangle G_2G_3P\\sim\\triangle M_2M_3P, \\triangle G_3G_4P\\sim\\triangle M_3M_4P,$ and $\\triangle G_4G_1P\\sim\\triangle M_4M_1P.$ By the properties of centroids, the ratio of similitude for each pair of triangles is $\\frac23.$ Note that quadrilateral $M_1M_2M_3M_4$ is a square of side-length $15\\sqrt2.$ It follows that: 1. Since $\\overline{G_1G_2}\\parallel\\overline{M_1M_2},\\overline{G_2G_3}\\parallel\\overline{M_2M_3},\\overline{G_3G_4}\\parallel\\overline{M_3M_4},$ and $\\overline{G_4G_1}\\parallel\\overline{M_4M_1}$ by the Converse of the Corresponding Angles Postulate, we have $\\angle G_1G_2G_3=\\angle G_2G_3G_4=\\angle G_3G_4G_1=\\angle G_4G_1G_2=90^\\circ.$ 2. Since $G_1G_2=\\frac23M_1M_2, G_2G_3=\\frac23M_2M_3, G_3G_4=\\frac23M_3M_4,$ and $G_4G_1=\\frac23M_4M_1$ by the ratio of similitude, we have $G_1G_2=G_2G_3=G_3G_4=G_4G_1=10\\sqrt2.$ Together, quadrilateral $G_1G_2G_3G_4$ is a square of side-length $10\\sqrt2,$ so its area is $\\left(10\\sqrt2\\right)^2=\\boxed{\\textbf{(C) }200}.$ Remark This solution shows that, if point $P$ is within square $ABCD,$ then the shape and the area of quadrilateral $G_1G_2G_3G_4$ are independent of the location of $P.$ Let the brackets denote areas. More generally, $G_1G_2G_3G_4$ is always a square of area $$[G_1G_2G_3G_4]=\\left(\\frac23\\right)^2[M_1M_2M_3M_4]=\\frac49[M_1M_2M_3M_4]=\\frac29[ABCD].$$ On the other hand, the location of $G_1G_2G_3G_4$ is dependent on the location of $P.$ ~RandomPieKevin ~Kyriegon ~MRENTHUSIASM ## Solution 2 (Similar Triangles) This solution refers to the diagram in Solution 1. By SAS, we conclude that $\\triangle G_1G_3P\\sim\\triangle M_1M_3P$ and $\\triangle G_2G_4P\\sim\\triangle M_2M_4P.$", "24 In the cube $ABCDEFGH$ with opposite vertices $C$ and $E,$ $J$ and $I$ are the midpoints of edges $\\overline{FB}$ and $\\overline{HD},$ respectively. Let $R$ be the ratio of the area of the cross-section $EJCI$ to the area of one of the faces of the cube. What is $R^2?$ $[asy] size(6cm); pair A,B,C,D,EE,F,G,H,I,J; C = (0,0); B = (-1,1); D = (2,0.5); A = B+D; G = (0,2); F = B+G; H = G+D; EE = G+B+D; I = (D+H)/2; J = (B+F)/2; filldraw(C--I--EE--J--cycle,lightgray,black); draw(C--D--H--EE--F--B--cycle); draw(G--F--G--C--G--H); draw(A--B,dashed); draw(A--EE,dashed); draw(A--D,dashed); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(I); dot(J); label(\"A\",A,E); label(\"B\",B,W); label(\"C\",C,S); label(\"D\",D,E); label(\"E\",EE,N); label(\"F\",F,W); label(\"G\",G,N); label(\"H\",H,E); label(\"I\",I,E); label(\"J\",J,W); [/asy]$ $\\textbf{(A) } \\frac{5}{4} \\qquad \\textbf{(B) } \\frac{4}{3} \\qquad \\textbf{(C) } \\frac{3}{2} \\qquad \\textbf{(D) } \\frac{25}{16} \\qquad \\textbf{(E) } \\frac{9}{4}$ ## Problem 25 How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive? $\\textbf{(A) }4\\qquad\\textbf{(B) }9\\qquad\\textbf{(C) }10\\qquad\\textbf{(D) }57\\qquad \\textbf{(E) }58$", "hexagon $ABCDEF$, $AC$, $CE$ are two diagonals. Points $M$, $N$ are on $AC$, $CE$ respectively and satisfy $AC: AM = CE: CN = r$. Suppose $B, M, N$ are collinear, find $100r^2$. (diagram goes here) $[asy] size(120); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(pair P) { dot(P,linewidth(3)); return P; } pair A=dir(0), B=dir(60), C=dir(120), D=dir(180), E=dir(240), F=dir(300), N=(4*E+C)/5,M=intersectionpoints(A--C,B--N)[0]; draw(A--B--C--D--E--F--cycle); draw(A--C--E); draw(B--N); label(\"A\",D2(A),E); label(\"B\",D2(B),NE); label(\"C\",D2(C),NW); label(\"D\",D2(D),W); label(\"E\",D2(E),SW); label(\"F\",D2(F),SE); label(\"M\",D2(M),(0,-1.5)); label(\"N\",D2(N),SE); [/asy]$ Solution A cuboctahedron is a solid with 6 square faces and 8 equilateral triangle faces, with each edge adjacent to both a square and a triangle (see picture). Suppose the ratio of the volume of an octahedron to a cuboctahedron with the same side length is $r$. Find $100r^2$. (diagram goes here) In the following diagram, a semicircle is folded along a chord $AN$ and intersects its diameter $MN$ at $B$. Given that $MB : BN = 2 : 3$ and $MN = 10$. If $AN = x$, find $x^2$. $[asy] size(120); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(pair P) { dot(P,linewidth(3)); return P; } real r = sqrt(80)/5; pair M=(-1,0), N=(1,0), A=intersectionpoints(arc((M+N)/2, 1, 0, 180),circle(N,r))[0], C=intersectionpoints(circle(A,1),circle(N,1))[0], B=intersectionpoints(circle(C,1),M--N)[0]; draw(arc((M+N)/2, 1, 0, 180)--cycle); draw(A--N); draw(arc(C,1,180,180+2*aSin(r/2))); label(\"A\",D2(A),NW); label(\"B\",D2(B),SW); label(\"M\",D2(M),S); label(\"N\",D2(N),SE); [/asy]$ Solution Square $ABCD$ is divided into four rectangles by $EF$ and $GH$. $EF$ is parallel to $AB$ and $GH$ parallel to $BC$. $\\angle BAF = 18^\\circ$. $EF$", "or mid face from the 2d label number. This vector contains successive pairs of the 2d label number at index $2i$ and the corresponding 3d label number at index $2i+1$, like change. • labelup= This vector is used to initialize the 3d label numbers of the upper/top face from the 2d region number. This vector contains successive pairs of the 2d region number at index $2i$ and the corresponding 3d label number at index $2i+1$, like change. • labeldown= Same as the previous case but for the lower/down face label. Moreover, we also add post processing parameters that allow to moving the mesh. These parameters correspond to parameters transfo, facemerge and ptmerge of the command line movemesh. The vector region, labelmid, labelup and labeldown These vectors are composed of $n_{l}$ successive pairs of number $O_i,N_l$ where $n_{l}$ is the number (label or region) that we want to get. An example of this command is given in the Build layer mesh example. Cube Cube.idp 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 load \"medit\" load \"msh3\" func mesh3 Cube (int[int] &NN, real[int, int] &BB, int[int, int] &L){ real x0 = BB(0,0), x1 = BB(0,1); real y0 = BB(1,0), y1 = BB(1,1); real z0 = BB(2,0), z1", "3. $L$ and $R$ will retain the direction of the labeled vertices, but $H$ and $V$ will alter the direction of the labeled vertices. After the $19$th transformation, vertex $A$ will be at either $(1,-1)$ or $(-1,1).$ All possible configurations of the labeled vertices are shown below: $[asy] /* Made by MRENTHUSIASM */ unitsize(7mm); label(\"A\",(1,0)); label(\"B\",(1,1)); label(\"C\",(0,1)); label(\"D\",(0,0)); label(\"C\",(5,0)); label(\"D\",(5,1)); label(\"A\",(4,1)); label(\"B\",(4,0)); label(\"A\",(9,0)); label(\"D\",(9,1)); label(\"C\",(8,1)); label(\"B\",(8,0)); label(\"C\",(13,0)); label(\"B\",(13,1)); label(\"A\",(12,1)); label(\"D\",(12,0)); label(\"\\textbf{Configuration}\",(-5,0.5)); label(\"\\textbf{The 20th}\",(-5,-1.5)); label(\"\\textbf{Transformation}\",(-5,-2.25)); label(\"L\",(0.5,-1.875)); label(\"R\",(4.5,-1.875)); label(\"H\",(8.5,-1.875)); label(\"V\",(12.5,-1.875)); [/asy]$ Each sequence of $19$ transformations generates one valid sequence of $20$ transformations. Therefore, the answer is $4^{19}=\\boxed{\\textbf{(C)}\\ 2^{38}}.$ ~MRENTHUSIASM Solution 2 Let $(+)$ denote counterclockwise/starting orientation and $(-)$ denote clockwise orientation. Let $1,2,3,$ and $4$ denote which quadrant $A$ is in. Realize that from any odd quadrant and any orientation, the $4$ transformations result in some permutation of $(2+, 2-, 4+, 4-).$ The same goes that from any even quadrant and any orientation, the $4$ transformations result in some permutation of $(1+, 1-, 3+, 3-).$ We start our first $19$ moves by doing whatever we want, $4$ choices each time. Since $19$ is odd, we must end up on an even quadrant. As said above, we know that exactly one of the four transformations will give us $(1+),$ and we must use that transformation. Thus, the answer is $4^{19}=\\boxed{\\textbf{(C)}\\", "+0 # HELP ASAP GEOMETRY QUESTION!!!!! 0 123 1 The edge length of a regular tetrahedron $ABCD$ is $2.$ $M$ and $N$ are the midpoints of $\\overline{BC}$ and $\\overline{AD}$, respectively. Find $MN.$ [asy] size(150); import three; currentprojection = orthographic(-0.6,3,2); triple A,B,C,D,O,M,NN; O=(0,0,0); D=(0,0,sqrt(8)); A=(2,0,0); B=(-1,sqrt(3),0); C=(-1,-sqrt(3),0); M=(B+C)/2; NN=(A+D)/2; draw(D--A--B--D--C--B); draw(A--C,dashed); draw(M--NN,dotted); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,E); label(\"$D$\",D,N); label(\"$M$\",M,E); label(\"$N$\",NN,NW); [/asy] Incase the latex doesn't work: The edge length of a regular tetrahedron ABCD is 2. M and N are the midpoints of segment BC and segment AD, respectively. Find MN. Image: It won't load but it's a regular tetrahedron and MN is a diagonal from the midpoint of a side to the midpoint of another side that is on the opposite side and on the base. THANK YOU TO ANYONE THAT HELPS!!!!!!!!!!!!!! Apr 5, 2022", "(clockwise or counterclockwise). 3. $L$ and $R$ will retain the direction of the labeled vertices, but $H$ and $V$ will alter the direction of the labeled vertices. After the $19$th transformation, vertex $A$ will be at either $(1,-1)$ or $(-1,1).$ All possible configurations of the labeled vertices are shown below: $[asy] /* Made by MRENTHUSIASM */ unitsize(7mm); label(\"A\",(1,0)); label(\"B\",(1,1)); label(\"C\",(0,1)); label(\"D\",(0,0)); label(\"C\",(5,0)); label(\"D\",(5,1)); label(\"A\",(4,1)); label(\"B\",(4,0)); label(\"A\",(9,0)); label(\"D\",(9,1)); label(\"C\",(8,1)); label(\"B\",(8,0)); label(\"C\",(13,0)); label(\"B\",(13,1)); label(\"A\",(12,1)); label(\"D\",(12,0)); label(\"\\textbf{Configuration}\",(-5,0.5)); label(\"\\textbf{The 20th}\",(-5,-1.5)); label(\"\\textbf{Transformation}\",(-5,-2.25)); label(\"L\",(0.5,-1.875)); label(\"R\",(4.5,-1.875)); label(\"H\",(8.5,-1.875)); label(\"V\",(12.5,-1.875)); [/asy]$ Each sequence of $19$ transformations generates one valid sequence of $20$ transformations. Therefore, the answer is $4^{19}=\\boxed{\\textbf{(C)}\\ 2^{38}}.$ ~MRENTHUSIASM ## Solution 2 Let $(+)$ denote counterclockwise/starting orientation and $(-)$ denote clockwise orientation. Let $1,2,3,$ and $4$ denote which quadrant $A$ is in. Realize that from any odd quadrant and any orientation, the $4$ transformations result in some permutation of $(2+, 2-, 4+, 4-).$ The same goes that from any even quadrant and any orientation, the $4$ transformations result in some permutation of $(1+, 1-, 3+, 3-).$ We start our first $19$ moves by doing whatever we want, $4$ choices each time. Since $19$ is odd, we must end up on an even quadrant. As said above, we know that exactly one of the four transformations will give us $(1+),$ and we must use that transformation. Thus,", "# Difference between revisions of \"2018 AIME I Problems/Problem 4\" ## Problem 4 In $\\triangle ABC, AB = AC = 10$ and $BC = 12$. Point $D$ lies strictly between $A$ and $B$ on $\\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\\overline{AC}$ so that $AD = DE = EC$. Then $AD$ can be expressed in the form $\\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. ## Solution 1 (No Trig) $[asy] import cse5; unitsize(10mm); pathpen=black; dotfactor=3; pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66); pair[] dotted = {A,B,C,D,E,F,G}; D(A--B); D(C--B); D(A--C); D(D--E); pathpen=dashed; D(B--F); D(D--G); dot(dotted); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,NW); label(\"E\",E,NE); label(\"F\",F,NE); label(\"G\",G,NE); label(\"x\",H,NW); label(\"x\",I,NW); label(\"x\",J,NE); [/asy]$ We draw the altitude from $B$ to $\\overline{AC}$ to get point $F$. We notice that the triangle's height from $A$ to $\\overline{BC}$ is 8 because it is a $3-4-5$ Right Triangle. To find the length of $\\overline{BF}$, we let $h$ represent $\\overline{BF}$ and set up an equation by finding two ways to express the area. The equation is $(8)(12)=(10)(h)$, which leaves us with $h=9.6$. We then solve for the length $\\overline{AF}$, which is done through pythagorean theorm and get $\\overline{AF}$ = $2.8$. We can now see that $\\triangle ABF$", "(HHC − 2)n×n based on degrees of end vertices of each edge. The edge set of (HHC − 2)n×n can be divided into two partitions based on the degree of end vertices. The first edge partition E1((HHC −2)n×n) contains 24 edges uv, where deg(u) = 3, deg(v) = 4. The second edge partition E2((HHC−2)n×n) contains 32n+4 edges uv, where deg(u) = deg(v) = 4. • The first and second Zagreb indices of (HHC − 2)n×n Now using Eqs. (1),(2), we have $M1(G)=∑uv∈E(G)[deg(u)+deg(v)]M1(G)=∑uv∈E1(G)[deg(u)+deg(v)]+∑uv∈E2(G)[deg(u)+deg(v)]=7|E1(G)|+8|E2(G)|=7(24)+8(32n+4)=512n+200.$$M2(G)=∑uv∈E(G)[deg(u)×deg(v)]M2(G)=∑uv∈E1(G)[deg(u)×deg(v)]+∑uv∈E2(G)[deg(u)×deg(v)]=12|E1(G)|+16|E2(G)|=12(24)+16(32n+4)=512n+352.$ • Hyper Zagreb index of (HHC − 2)n×n The hyper Zagreb index using Eq. (3) is computed as: $HM(G)=∑uv∈E(G)[deg(u)+deg(v)]2HM(G)=∑uv∈E1(G)[deg(u)+deg(v)]2+∑uv∈E2(G)[deg(u)+deg(v)]2=72|E1(G)|+82|E2(G)|=49(24)+64(32n+4)=2048n+1432.$ • Multiple Zagreb indices of (HHC − 2)n×n The Multiple-Zagreb indices using Eqs. (4), (5) are computed as: $vPM1(G)=∏uv∈E(G)[deg(u)+deg(v)]PM1(G)=∏uv∈E1(G)[deg(u)+deg(v)]×∏uv∈E2(G)[deg(u)+deg(v)]=7|E1(G)|×8|E2(G)|=724×8(32n+4).$$PM2(G)=∏uv∈E(G)[deg(u)×deg(v)]PM2(G)=∏uv∈E1(G)[deg(u)×deg(v)]×∏uv∈E2(G)[deg(u)×deg(v)]=12|E1(G)|×16|E2(G)|=1224×16(32n+4).$ • The first and second Zagreb polynomials of (HHC − 2)n×n Now using Eqs. (6),(7), we have $M1(G,x)=∑uv∈E(G)x[deg(u)+deg(v)]M1(G,x)=∑uv∈E1(G)x[deg(u)+deg(v)]+∑uv∈E2(G)x[deg(u)+deg(v)]=∑uv∈E1(G)x7+∑uv∈E2(G)x8=|E1(G)|x7+|E2(G)|x8=24x7+(32n+4)x8.$$M2(G,x)=∑uv∈E(G)x[deg(u)×deg(v)]M2(G,x)=∑uv∈E1(G)x[deg(u)×deg(v)]+∑uv∈E2(G)x[deg(u)×deg(v)]=∑uv∈E1(G)x12+∑uv∈E2(G)x16=|E1(G)|x12+|E2(G)|x16=24x12+(32n+4)x16.$ • The redefined first, second, and third Zegreb index of (HHC − 2)n×n Now using Eq. (8) and the edge partition of the block shift network (HHC − 2)n×n, we have: $ReZG1(G)=∑uv∈E(G)deg(u)+deg(v)deg(u)×deg(v)=∑uv∈E1(G)deg(u)+deg(v)deg(u)×deg(v)+∑uv∈E2(G)deg(u)+deg(v)deg(u)×deg(v)=24(3+43×4)+(32n+4)(4+44×4)=16n+16.$ By using Eq. (9), the second redefined Zagreb index is computed as below: $ReZG2(G)=∑uv∈E(G)deg(u)×deg(v)deg(u)+deg(v)=∑uv∈E1(G)deg(u)×deg(v)deg(u)+deg(v)+∑uv∈E2(G)deg(u)×deg(v)deg(u)+deg(v)=24(3×43+4)+(32n+4)(4×44+4)=2887+64n+8.$ Now by using Eq. (10), the third redefined Zagreb index is computed as: $ReZG3(G)=∑uv∈E(G)[[deg(u)×deg(v)]×[deg(u)+deg(v)]]=∑uv∈E1(G)[[deg(u)×deg(v)]×[deg(u)+deg(v)]]+∑uv∈E2(G)[[deg(u)×deg(v)]×[deg(u)+deg(v)]]=24((3×4)×(3+4))+(32n+4)((4×4)×(4+4))=4608n+2596.$ ## Conclusion In this paper we determined first Zagreb index M1(G), second Zagreb index M2(G), hyper Zagreb index HM(G), first", "# How may I put labels on the corners/vertices of a cube? I have a cube as per the answer to my question How may I put arrows along some edges of a cube?. Now I want to label the 8 corners/vertices with labels like $\\mathbf{O}\\lnot \\lmathbf{B}\\lnot q$.How can I best achieve this? The coordinates for the lower face are named A1,...,A4 and those for the upper face are B1,...,B4. You can simply put \\nodes at those coordinates with the desired text; a little example; I used A,B,...,H as labels, but you can use whatever text you want: \\documentclass[border=5pt]{standalone} \\usepackage{tikz} \\definecolor{myred}{RGB}{183,18,52} \\definecolor{myyellow}{RGB}{254,213,1} \\definecolor{myblue}{RGB}{0,80,198} \\definecolor{mygreen}{RGB}{0,155,72} \\begin{document} \\begin{tikzpicture}[ line join=round, y={(-0.86cm,0.36cm)},x={(1cm,0.36cm)}, z={(0cm,1cm)}, arr/.style={-latex,ultra thick,line cap=round,shorten <= 1.5pt} ] \\def\\Side{2} \\coordinate (A1) at (0,0,0); \\coordinate (A2) at (0,\\Side,0); \\coordinate (A3) at (\\Side,\\Side,0); \\coordinate (A4) at (\\Side,0,0); \\coordinate (B1) at (0,0,\\Side); \\coordinate (B2) at (0,\\Side,\\Side); \\coordinate (B3) at (\\Side,\\Side,\\Side); \\coordinate (B4) at (\\Side,0,\\Side); \\fill[myyellow] (A2) -- (A3) -- (B3) -- (B2) -- cycle; \\fill[mygreen] (A2) -- (A3) -- (A4) -- (A1) -- cycle; \\fill[myred](A3) -- (B3) -- (B4) -- (A4) -- cycle; \\fill[myblue] (A1) -- (A2) -- (B2) -- (B1) -- cycle; \\draw (A2) -- (A1) -- (A4); \\draw (B2) -- (B1) -- (B4) -- (B3) -- cycle; \\draw (A1) -- (B1); \\draw (A2) -- (B2); \\draw (A4) -- (B4); \\draw[thin] (A3)", "1988 AHSME Problems/Problem 20 Problem In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base, $XY / YZ$, in this rectangle is $[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,1), B=(0,-1), C=(2,-1), D=(2,1), E=(1,-1), F=(1,1), G=(.8,.6); pair X=(4,sqrt(5)), Y=(4,-sqrt(5)), Z=(4+2/sqrt(5),-sqrt(5)), W=(4+2/sqrt(5),sqrt(5)), T=(4,0), U=(4+2/sqrt(5),-4/sqrt(5)), V=(4+2/sqrt(5),1/sqrt(5)); draw(A--B--C--D--A^^B--F^^E--D^^A--G^^rightanglemark(A,G,F)); draw(X--Y--Z--W--X^^T--V--X^^Y--U); label(\"A\", A, NW); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, NE); label(\"E\", E, S); label(\"F\", F, N); label(\"G\", G, E); label(\"X\", X, NW); label(\"Y\", Y, SW); label(\"Z\", Z, SE); label(\"W\", W, NE); [/asy]$ $\\textbf{(A)}\\ 4\\qquad \\textbf{(B)}\\ 1+2\\sqrt{3}\\qquad \\textbf{(C)}\\ 2\\sqrt{5}\\qquad \\textbf{(D)}\\ \\frac{8+4\\sqrt{3}}{3}\\qquad \\textbf{(E)}\\ 5$ Solution Within $WXYZ$, the parallelogram piece has vertical side $BF = \\sqrt{1^2 + 2^2} = \\sqrt{5}$, and diagonal side $FD = 1$. Thus the triangle in the bottom-right hand corner (the one with horizontal side $YZ$) must have hypotenuse $1$, and the only such triangle in the original figure is $\\triangle AFG$, so we deduce $YZ = AG = \\frac{\\frac{1}{2} \\times 1 \\times 2}{\\frac{1}{2} \\times \\sqrt{5}} = \\frac{2}{\\sqrt{5}}.$ Now the rectangle must have the same area as the square, as the pieces were put together without gaps", "# Problem #2209 2209 A cube with side length $1$ is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C$, as shown. What is the area of quadrilateral $ABCD$? $[asy] import three; unitsize(3cm); defaultpen(fontsize(8)+linewidth(0.7)); currentprojection=obliqueX; draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype(\"4 4\")); draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1)); draw((0.5,1,0)--(0,1,0)--(0,1,1)); dot((0.5,0,0)); label(\"A\",(0.5,0,0),WSW); dot((0,1,1)); label(\"C\",(0,1,1),NE); dot((0.5,1,0.5)); label(\"D\",(0.5,1,0.5),ESE); dot((0,0,0.5)); label(\"B\",(0,0,0.5),NW);[/asy]$ $\\mathrm{(A)}\\ \\frac{\\sqrt{6}}{2}\\qquad\\mathrm{(B)}\\ \\frac{5}{4}\\qquad\\mathrm{(C)}\\ \\sqrt{2}\\qquad\\mathrm{(D)}\\ \\frac{5}{8}\\qquad\\mathrm{(E)}\\ \\frac{3}{4}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "# 1988 AHSME Problems/Problem 20 ## Problem In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown in the second figure. The ratio of height to base, $XY / YZ$, in this rectangle is $[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,1), B=(0,-1), C=(2,-1), D=(2,1), E=(1,-1), F=(1,1), G=(.8,.6); pair X=(4,sqrt(5)), Y=(4,-sqrt(5)), Z=(4+2/sqrt(5),-sqrt(5)), W=(4+2/sqrt(5),sqrt(5)), T=(4,0), U=(4+2/sqrt(5),-4/sqrt(5)), V=(4+2/sqrt(5),1/sqrt(5)); draw(A--B--C--D--A^^B--F^^E--D^^A--G^^rightanglemark(A,G,F)); draw(X--Y--Z--W--X^^T--V--X^^Y--U); label(\"A\", A, NW); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, NE); label(\"E\", E, S); label(\"F\", F, N); label(\"G\", G, E); label(\"X\", X, NW); label(\"Y\", Y, SW); label(\"Z\", Z, SE); label(\"W\", W, NE); [/asy]$ $\\textbf{(A)}\\ 4\\qquad \\textbf{(B)}\\ 1+2\\sqrt{3}\\qquad \\textbf{(C)}\\ 2\\sqrt{5}\\qquad \\textbf{(D)}\\ \\frac{8+4\\sqrt{3}}{3}\\qquad \\textbf{(E)}\\ 5$ ## Solution Within $WXYZ$, the parallelogram piece has vertical side $BF = \\sqrt{1^2 + 2^2} = \\sqrt{5}$, and diagonal side $FD = 1$. Thus the triangle in the bottom-right hand corner (the one with horizontal side $YZ$) must have hypotenuse $1$, and the only such triangle in the original figure is $\\triangle AFG$, so we deduce $YZ = AG = \\frac{\\frac{1}{2} \\times 1 \\times 2}{\\frac{1}{2} \\times \\sqrt{5}} = \\frac{2}{\\sqrt{5}}.$ Now the rectangle must have the same area as the square, as the pieces were put", "dot(A); label(\"A\",A,NE); dot(B); label(\"B\",B,dir(180)); dot(C); label(\"C\",C,SE); dot(D); label(\"D\",D,dir(0)); dot(E); label(\"E\",E,SE); dot(F); label(\"F\",F,SW); [/asy]$ Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper. As triangle $ABC$ is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles. As point $D$ splits line segment $\\overline{AC}$ in a $1:2$ ratio, we draw $\\overline{AC}$ as a vertical line segment $3$ units long. Point $D$ is thus $1$ unit below point $A$ and $2$ units above point $C$. By definition, Point $E$ splits line segment $\\overline{BD}$ in a $1:1$ ratio, so we draw $\\overline{BD}$ $2$ units long directly left of $D$ and draw $E$ directly between $B$ and $D$, $1$ unit away from both. We then draw line segments $\\overline{AB}$ and $\\overline{BC}$. We can easily tell that triangle $ABC$ occupies $3$ square units of space. Constructing line $AE$ and drawing $F$ at the intersection of $AE$ and $BC$, we can easily see that triangle $EBF$ forms a right triangle occupying $\\frac{1}{4}$ of a square unit of space. The ratio of the areas of triangle $EBF$ and triangle $ABC$ is thus", "Welcome to the Rhino 6 version of this page! Looking for the newer Rhino 7 WIP version? Demonstrates how to construct a twisted cube object from an array of points and a list of curves. partial class Examples { // Symbolic vertex index constants to make code more readable const int A = 0, B = 1, C = 2, D = 3, E = 4, F = 5, G = 6, H = 7; // Symbolic edge index constants to make code more readable const int AB = 0, BC = 1, CD = 2, AD = 3, EF = 4, FG = 5, GH = 6, EH = 7, AE = 8, BF = 9, CG = 10, DH = 11; // Symbolic face index constants to make code more readable const int ABCD = 0, BCGF = 1, CDHG = 2, ADHE = 3, ABFE = 4, EFGH = 5; { Point3d[] points = new Point3d[8]; points[0] = new Point3d(0.0, 0.0, 0.0); // point A = geometry for vertex 0 points[1] = new Point3d(10.0, 0.0, 0.0); // point B = geometry for vertex 1 points[2] = new Point3d(10.0, 8.0, -1.0); // point C = geometry for vertex 2 points[3] = new Point3d(0.0, 6.0, 0.0); // point D = geometry for vertex 3 points[4] = new", "# 2018 AMC 8 Problems/Problem 24 ## Problem In the cube $ABCDEFGH$ with opposite vertices $C$ and $E,$ $J$ and $I$ are the midpoints of segments $\\overline{FB}$ and $\\overline{HD},$ respectively. Let $R$ be the ratio of the area of the cross-section $EJCI$ to the area of one of the faces of the cube. What is $R^2?$ $[asy] size(6cm); pair A,B,C,D,EE,F,G,H,I,J; C = (0,0); B = (-1,1); D = (2,0.5); A = B+D; G = (0,2); F = B+G; H = G+D; EE = G+B+D; I = (D+H)/2; J = (B+F)/2; filldraw(C--I--EE--J--cycle,lightgray,black); draw(C--D--H--EE--F--B--cycle); draw(G--F--G--C--G--H); draw(A--B,dashed); draw(A--EE,dashed); draw(A--D,dashed); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(I); dot(J); label(\"A\",A,E); label(\"B\",B,W); label(\"C\",C,S); label(\"D\",D,E); label(\"E\",EE,N); label(\"F\",F,W); label(\"G\",G,N); label(\"H\",H,E); label(\"I\",I,E); label(\"J\",J,W); [/asy]$ $\\textbf{(A) } \\frac{5}{4} \\qquad \\textbf{(B) } \\frac{4}{3} \\qquad \\textbf{(C) } \\frac{3}{2} \\qquad \\textbf{(D) } \\frac{25}{16} \\qquad \\textbf{(E) } \\frac{9}{4}$ ## Solution 1 Note that $EJCI$ is a rhombus by symmetry. Let the side length of the cube be $s$. By the Pythagorean theorem, $EC= s\\sqrt 3$ and $JI= s\\sqrt 2$. Since the area of a rhombus is half the product of its diagonals, the area of the cross section is $\\frac{s^2\\sqrt 6}{2}$. This gives $R = \\frac{\\sqrt 6}2$. Thus $R^2 = \\boxed{\\textbf{(C) } \\frac{3}{2}}$ ## Solution 2 We can solve this with 3D Cartesian coordinates. Assume WLOG that the sides of", "# Problem #2229 2229 As shown below, convex pentagon $ABCDE$ has sides $AB=3$, $BC=4$, $CD=6$, $DE=3$, and $EA=7$. The pentagon is originally positioned in the plane with vertex $A$ at the origin and vertex $B$ on the positive $x$-axis. The pentagon is then rolled clockwise to the right along the $x$-axis. Which side will touch the point $x=2009$ on the $x$-axis? $[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair A=(0,0), Ep=7*dir(105), B=3*dir(0); pair D=Ep+B; pair C=intersectionpoints(Circle(D,6),Circle(B,4))[1]; pair[] ds={A,B,C,D,Ep}; dot(ds); draw(B--C--D--Ep--A); draw((6,6)..(8,4)..(8,3),EndArrow(3)); xaxis(\"x\",-8,14,EndArrow(3)); label(\"E\",Ep,NW); label(\"D\",D,NE); label(\"C\",C,E); label(\"B\",B,SE); label(\"(0,0)=A\",A,SW); label(\"3\",midpoint(A--B),N); label(\"4\",midpoint(B--C),NW); label(\"6\",midpoint(C--D),NE); label(\"3\",midpoint(D--Ep),S); label(\"7\",midpoint(Ep--A),W); [/asy]$ $\\text{(A) } \\overline{AB} \\qquad \\text{(B) } \\overline{BC} \\qquad \\text{(C) } \\overline{CD} \\qquad \\text{(D) } \\overline{DE} \\qquad \\text{(E) } \\overline{EA}$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved. • Reduce fractions to lowest terms and enter in the form 7/9. • Numbers involving pi should be written as 7pi or 7pi/3 as appropriate. • Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate. • Exponents should be entered in the form 10^10. • If the problem is multiple choice, enter the appropriate (capital) letter. • Enter points with parentheses, like so: (4,5) • Complex numbers should be entered in rectangular form unless otherwise specified, like so:", "# Difference between revisions of \"2001 AIME I Problems/Problem 12\" ## Problem A sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ ## Solution $[asy] import three; currentprojection = perspective(-2,9,4); triple A = (6,0,0), B = (0,4,0), C = (0,0,2), D = (0,0,0); triple E = (2/3,0,0), F = (0,2/3,0), G = (0,0,2/3), L = (0,2/3,2/3), M = (2/3,0,2/3), N = (2/3,2/3,0); triple I = (2/3,2/3,2/3); triple J = (6/7,20/21,26/21); draw(C--A--D--C--B--D--B--A--C); draw(L--F--N--E--M--G--L--I--M--I--N--I--J); label(\"I\",I,W); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,W*-1); label(\"D\",D,W*-1); [/asy]$ The center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius. $I$ must also be a distance $r$ from the plane $ABC$ The signed distance between a plane and a point $I$ can be calculated as $\\frac{(I-G) \\cdot P}{|P|}$, where G is any point on the plane, and P is a vector perpendicular to ABC. A vector $P$ perpendicular to plane $ABC$ can be found as $V=(A-C)\\times(B-C)=\\langle 8, 12, 24 \\rangle$ Thus $\\frac{(I-C) \\cdot P}{|P|}=-r$ where the negative comes from the fact that we want $I$ to be in the opposite direction of $P$ \\begin{align*}\\frac{(I-C) \\cdot P}{|P|}&=-r\\\\ \\frac{(\\langle r, r, r", "label(\"z=2\",(2,2,1),E); label(\"y=2\",(2,1,0),SE); [/asy]$ NOTE: Connecting the centers of the faces will actually give an octahedron, not a cube, because it only has $6$ vertices. Now, we look for any additional equilateral triangles. Connecting the midpoints of three non-adjacent, non-parallel edges also gives us more equilateral triangles (blue triangle). Notice that picking these three edges leaves two vertices alone (labelled A and B), and that picking any two opposite vertices determine two equilateral triangles. Hence there are $\\frac{8 \\cdot 2}{2} = 8$ of these equilateral triangles, for a total of $\\boxed{\\textbf{(C) }80}$. $[asy] import three; unitsize(1cm); size(200); currentprojection=perspective(1/3,-1,1/2); draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle); draw((0,0,0)--(0,0,2)); draw((0,2,0)--(0,2,2)); draw((2,2,0)--(2,2,2)); draw((2,0,0)--(2,0,2)); draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle); draw((1,0,0)--(2,2,1)--(0,1,2)--cycle,blue); label(\"x=2\",(1,0,0),S); label(\"z=2\",(2,2,1),E); label(\"y=2\",(2,1,0),SE); label(\"A\",(0,2,0), NW); label(\"B\",(2,0,2), NW); [/asy]$ ### Solution 2 (rigorous) The three-dimensional distance formula shows that the lengths of the equilateral triangle must be $\\sqrt{d_x^2 + d_y^2 + d_z^2}, 0 \\le d_x, d_y, d_z \\le 2$, which yields the possible edge lengths of $\\sqrt{0^2+0^2+1^2}=\\sqrt{1},\\ \\sqrt{0^2+1^2+1^2}=\\sqrt{2},\\ \\sqrt{1^2+1^2+1^2}=\\sqrt{3},$ $\\ \\sqrt{0^2+0^2+2^2}=\\sqrt{4},\\ \\sqrt{0^2+1^2+2^2}=\\sqrt{5},\\ \\sqrt{1^2+1^2+2^2}=\\sqrt{6},$ $\\ \\sqrt{0^2+2^2+2^2}=\\sqrt{8},\\ \\sqrt{1^2+2^2+2^2}=\\sqrt{9},\\ \\sqrt{2^2+2^2+2^2}=\\sqrt{12}$ Some casework shows that $\\sqrt{2},\\ \\sqrt{6},\\ \\sqrt{8}$ are the only lengths that work, from which we can use the same counting argument as above." ]

[ "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "to the value of $t$ such that the equation of the line is satisfied: $$\\frac t{a^2}+\\frac t{4a^2}=1,\\enspace\\text{whence}\\quad t=\\frac{4a^2}5,\\;x_H=\\frac{4a}5,\\; y_H=\\frac{2a}5$$ and the distance from the origin to the straight line is $$OH=\\sqrt{x_H^2+y_H^2}=\\frac{2\\lvert a\\rvert}{\\sqrt 5}=1\\iff \\lvert a\\rvert=\\frac{\\sqrt 5}2,$$ whence the equation in normal form: $$\\frac{2\\sqrt 5}5x+\\frac{\\sqrt 5}5y=\\pm 1.$$", "x=-5+\\frac{3}{7}z$ , and thus the intersection line is given by $\\left\\{\\begin{pmatrix}-5+\\frac{3}{7}t\\\\{}\\\\-1-\\frac{2}{7}t\\\\{}\\\\t\\end{pmatrix}\\,,\\,t\\in\\mathbb{R }\\right\\}$ . The above is already a parametrization, but your teacher may want you to write it as $u+tv\\,,\\,u,v$ vectors on the line, $t=$ the real parameter. Tonio", "the matrix of this reflection (after padding it out to make it suitable for use with homogeneous coordinates) is symmetric, so the image of the line is $$M^{-T}\\mathbf l = M\\mathbf l = \\begin{bmatrix}-\\frac35&\\frac45&0\\\\\\frac45&\\frac35&0\\\\0&0&1\\end{bmatrix} \\begin{bmatrix}3\\\\-1\\\\1\\end{bmatrix} = \\begin{bmatrix}-\\frac{13}5\\\\\\frac95\\\\1\\end{bmatrix},$$ which corresponds to the equation $$-\\frac{13}5x+\\frac95y+1=0$$, or $$13x-9y=5$$.", "is $\\frac{\\mu\\vec a + \\lambda\\vec b}{\\lambda + \\mu}$ I assume that the units you're using are metres, so by 1 m away from the first point, you mean 1 unit. The distance between the points is $\\sqrt{51}$. So we need the point that divides the line joining $\\begin{pmatrix}5\\\\0\\\\5\\end{pmatrix}$ to $\\begin{pmatrix}6\\\\5\\\\0\\end{pmatrix}$ in the ratio $1:\\sqrt{51}-1$, which is: $\\dfrac{(\\sqrt{51}-1)\\begin{pmatrix}5\\\\0\\\\5\\end{pmatrix}+1\\begin{pmat rix}6\\\\5\\\\0\\end{pmatrix}}{\\sqrt{51}}$ $=\\dfrac{1}{\\sqrt{51}}\\begin{pmatrix}5\\sqrt{51}+1\\\\ 5\\\\5\\sqrt{51}-5\\end{pmatrix}$ $=\\begin{pmatrix}5+\\frac{1}{\\sqrt{51}}\\\\ \\frac{5}{\\sqrt{51}}\\\\5-\\frac{5}{\\sqrt{51}}\\end{pmatrix}$", "\\\\ &=& \\begin{pmatrix} 2-1 \\\\ 3-1 \\\\ 4-2 \\end{pmatrix} \\\\ &=& \\begin{pmatrix} 1 \\\\ 2 \\\\ 2 \\end{pmatrix} \\\\\\\\ && u_1 + u_2 + u_3 \\\\ &=& 1+2+2 \\\\ &=& 5 \\\\ \\hline \\end{array}$$ Aug 16, 2019 #2 +23071 +2 Find the distance between Q = (3, -7, -1) and the line through A = (1, 1, 2) and B = (2, 3, 4). This distance is equal to $$\\dfrac{\\sqrt{d}}{3}$$ for some integer d. What is d? line through A = (1, 1, 2) and B = (2, 3, 4): $$\\small{ \\begin{array}{|rclrcl|} \\hline \\vec{x} &=& \\vec{A}+t(\\vec{B}-\\vec{A}) & \\vec{r} &=& \\vec{B}-\\vec{A} \\quad | \\quad \\vec{A} = (1, 1, 2)\\quad \\vec{B} = (2, 3, 4)\\\\ &&& \\vec{r} &=& \\begin{pmatrix} 2 \\\\ 3 \\\\ 4 \\end{pmatrix}-\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix} \\\\ &&&\\vec{r} &=& \\begin{pmatrix} 1 \\\\ 2 \\\\ 2 \\end{pmatrix} \\\\ \\mathbf{\\vec{x}} &=& \\mathbf{\\vec{A}+t\\vec{r}} \\\\ \\hline (\\vec{x}-\\vec{Q})\\cdot \\vec{r} &=& 0 \\quad | \\quad \\vec{PQ} \\text{ is } \\perp \\text{ to } \\vec{r} \\\\ (\\mathbf{\\vec{A}+t\\vec{r}}-\\vec{Q})\\cdot \\vec{r} &=& 0 \\\\ (\\vec{A}-\\vec{Q} + t\\vec{r})\\cdot \\vec{r} &=& 0 & \\vec{A}-\\vec{Q}&=& \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}-\\begin{pmatrix} 3 \\\\ -7 \\\\ -1 \\end{pmatrix} \\quad | \\quad \\vec{Q} = (3, -7, -1)\\\\ & & & \\vec{A}-\\vec{Q}&=& \\begin{pmatrix} -2 \\\\ 8 \\\\ 3 \\end{pmatrix} \\\\\\\\ \\left[\\begin{pmatrix} -2 \\\\ 8 \\\\ 3 \\end{pmatrix} + t\\begin{pmatrix} 1", "get $d=\\begin{vmatrix}\\large\\frac{(8\\hat i+8\\hat j+4\\hat k).(-10\\hat i-2\\hat j-3\\hat k)}{12}\\end{vmatrix}$ $d=\\begin{vmatrix}\\large\\frac{-80-16-12}{12}\\end{vmatrix}$ $\\;\\;\\;=\\large\\frac{108}{12}$ $\\;\\;\\;=9$ units. Hence the shortest distance between the given two lines is $9$ units.", "the distance, t is the parameter, and a is the focus, and d is a 7th degree polynomial in terms of t? Given two parallel lines $$y_1=mx+b$$ $$y_2=mx+c$$ What is the distance between them? $\\dfrac{b-c}{\\sqrt{1+m^2}}$ In this case the distance comes out to be $\\dfrac{at^2 + \\frac{2a}{t} + \\frac{a}{t^3}}{\\sqrt{1+\\frac{1}{t^2}}}$ Differentiating and setting it to zero gives me $2t^4+3t^2-2t- \\frac{3}{t} - \\frac{2}{t^3}=0$ Mentallic Homework Helper Ok so I see you've used the parameters $P(at^2,2at)$. This does ring a bell from school. $\\dfrac{b-c}{\\sqrt{1+m^2}}$ Technically it would be the absolute value of that, but we can easily fix b>c and it doesn't really change anything. In this case the distance comes out to be $\\dfrac{at^2 + \\frac{2a}{t} + \\frac{a}{t^3}}{\\sqrt{1+\\frac{1}{t^2}}}$ I get at rather than at2 for that first term in the numerator. Do you agree that PT is $$y-2at=\\frac{1}{t}(x-at^2)$$ and QN is $$y+\\frac{2a}{t}=\\frac{1}{t}(x-\\frac{a}{t^2})$$ Gold Member Ok so I see you've used the parameters $P(at^2,2at)$. This does ring a bell from school. Technically it would be the absolute value of that, but we can easily fix b>c and it doesn't really change anything. I get at rather than at2 for that first term in the numerator. Do you agree that PT is $$y-2at=\\frac{1}{t}(x-at^2)$$ and QN is $$y+\\frac{2a}{t}=\\frac{1}{t}(x-\\frac{a}{t^2})$$ Yes. It was a silly mistake from my side. So the final equation changes to", "$$ln(x)$$, and some other function $$u(x)$$ then the chain rule will always lead to the following result: $\\text{if} \\quad f(x) = ln \\begin{pmatrix} u(x) \\end{pmatrix}$ $\\text{then} \\quad f'(x) = \\frac{u'(x)}{u(x)}$ This special case of the chain rule is further explained and illustrated in Tutorial 3. Spend a few minutes to watch it now. ## Exercise 3 Differentiate each of the following: 1. $$f(x) = ln \\begin{pmatrix} x^2 - 3 \\end{pmatrix}$$ 2. $$y = 3.ln \\begin{pmatrix} x^3-3x^2 + 6 \\end{pmatrix}$$ 3. $$f(x) = -2.ln \\begin{pmatrix} 4x+ 5 \\end{pmatrix}$$ 4. $$y = 4.ln \\begin{pmatrix} \\sqrt{x} \\end{pmatrix}$$ 5. $$f(x) = ln \\begin{pmatrix} \\frac{1}{x} \\end{pmatrix}$$ 1. $$f'(x) = \\frac{2x}{x^2-3}$$ 2. $$\\frac{dy}{dx} = \\frac{9x^2 - 18x}{x^3 - 3x^2 + 6}$$ 3. $$f'(x) = - \\frac{8}{4x+5}$$ 4. $$\\frac{dy}{dx} = \\frac{2}{x}$$ 5. $$f'(x) = - \\frac{1}{x}$$ 1. The function $$f(x) = ln \\begin{pmatrix} x^2 - 3\\end{pmatrix}$$ can be written: $f(x) = ln \\begin{pmatrix} u(x) \\end{pmatrix}$ Where $u(x) = x^2 - 3$ and $u'(x) = 2x$ Using the formula we saw above we can state: \\begin{aligned} f'(x) & = \\frac{u'(x)}{u(x)} \\\\ f'(x) & = \\frac{2x}{x^2 - 3} \\end{aligned} 2. The function $$y = 3. ln \\begin{pmatrix} x^3 - 3x^2 + 6\\end{pmatrix}$$ can be written: $y = 3 ln \\begin{pmatrix} u(x)\\end{pmatrix}$ Where $u(x) = x^3 - 3x^2 + 6$ and $u'(x) = 3x^2 -6x$ Using the formula", "form is: $\\boxed{y = - \\frac{1}{4}x - 1}$ Determine the equation of the line... $& \\begin{pmatrix}x_1, & y_1\\\\-3, & 1 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\-3, & -7 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-7-1}{-3--3}\\\\m & = \\frac{-7-1}{-3+3}\\\\m & = \\frac{-8}{0}\\\\m & = \\text{undefined}$ A line that has a slope that is undefined is a line that is parallel to the $y-$axis. The equation of this line is $\\boxed{x = -3}$. $& \\begin{pmatrix}x_1, & y_1\\\\-8, & 4 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\2, & -6 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-6-4}{2--8}\\\\m & = \\frac{-10}{10}\\\\m & = -1\\\\\\\\y & = mx + b\\\\4 & = -1(-8) + b\\\\4 & = 8 + b\\\\4-8 &= 8-8+b\\\\-4 & = b$ $m = -1$ and the $y-$intercept is (0, –4). The equation of this line is: $\\boxed{y = -1x-4}$ $& \\begin{pmatrix}x_1, & y_1\\\\0, & 5 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\4, & -3 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-3-5}{4-0}\\\\m & = \\frac{-8}{4}\\\\m & = -2$ $m = -2$ and the $y-$intercept is (0, 5). The equation of this line is: $\\boxed{y = -2x + 5}$ For each of the following real world problems... $& \\begin{pmatrix}x_1, & y_1\\\\320, & 124 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\600, & 164 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m &", "+0 # help 0 107 2 Compute the distance between the parallel lines given by $$\\begin{pmatrix} 1 \\\\ 4 \\end{pmatrix} + t \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix}$$ and $$\\begin{pmatrix} -5 \\\\ 6 \\end{pmatrix} + s \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix}$$ Feb 23, 2020 #1 0 The distance is 3/2. Feb 23, 2020 #2 +25275 +1 Compute the distance between the parallel lines given by $$\\begin{pmatrix} 1 \\\\ 4 \\end{pmatrix} + t \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix}$$ and $$\\begin{pmatrix} -5 \\\\ 6 \\end{pmatrix} + s \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix}$$. $$\\text{The orthogonal vector of the lines and normalized is }\\\\ \\binom{-3}{4}*\\frac{1}{\\sqrt{(-3)^2+4^2}}~\\text{ or }~\\binom{4}{-3}*\\frac{1}{\\sqrt{3^2+(-4)^2}} \\\\ \\text{because } \\binom{4}{3}\\binom{-3}{4} = 0 ~\\text{ respectively }~\\binom{4}{3}\\binom{4}{-3} = 0\\\\$$ $$\\text{Let \\hat{n} = \\frac{1}{5}\\binom{-3}{4}}$$ The distance between the parallel lines is: $$\\begin{array}{|rcll|} \\hline d &=&\\dfrac{1}{5}\\dbinom{-3}{4}\\dbinom{-5}{6}-\\frac{1}{5}\\dbinom{-3}{4}\\dbinom{1}{4} \\\\\\\\ d &=&\\dfrac{1}{5}\\Big(15+24-( -3+16 ) \\Big) \\\\\\\\ d &=&\\dfrac{26}{5} \\\\\\\\ \\mathbf{d} &=& \\mathbf{5.2} \\\\ \\hline \\end{array}$$ Feb 24, 2020", "$$f(x)$$. That's $$y_1=f(x_1)$$, $$y_2 = f(x_2)$$ ... . This is further explained and illustrated in the following tutorial. ## Tutorial - Finding Stationary Points In the following tutorial we review the ... . ## Exercise 1 Using the method, described above, find the coordinates of any stationary points on each of the following curves: 1. $$y = 2x^2 - 8x + 7$$ 2. $$y = 2x^3 - 15x^2 + 2x + 1$$ 3. $$y=-x^3-3x^2+çx +1$$ 4. $$y = x + \\frac{1}{x}$$ 5. $$y = \\frac{x^2}{2} - ln(x)+4$$ 1. One stationary point at $$\\begin{pmatrix}2,-1\\end{pmatrix}$$. 2. Two stationary points: $$\\begin{pmatrix}1,-1 \\end{pmatrix}$$ and $$\\begin{pmatrix}4,3\\end{pmatrix}$$. 3. Two stationary points: $$\\begin{pmatrix}-1,0\\end{pmatrix}$$ and $$\\begin{pmatrix}1,0\\end{pmatrix}$$.", "z$ so that $\\mathbf a$, $\\mathbf b$ and $\\mathbf z$ are on a same line (actually whether $|\\mathbf z|=1$ or not, or $|\\mathbf a|,|\\mathbf b|<1$ or not). Then, $$\\frac{y-q}{x-p}=\\frac{y-v}{x-u}=m$$ $$f(\\mathbf z)=\\frac{x-p+(y-q)i}{x-u+(y-v)i}=\\frac{x-p}{x-u}\\cdot\\frac{1+\\frac{y-q}{x-p}i}{1+\\frac{y-v}{x-u}i}=\\frac{x-p}{x-u}\\cdot\\frac{1+mi}{1+mi}=\\frac{x-p}{x-u}\\in\\mathbb R$$", "+0 # How do I do this? +2 55 1 +39 Find the distance between $$Q = (3, -7, -1)$$and the line through $$A = (1, 1, 2)$$ and $$B = (2, 3, 4).$$ This distance is equal to $$\\dfrac{\\sqrt{d}}{3}$$for some integer d. What is d? May 6, 2020 #1 +24992 +4 Find the distance between $$Q = (3, -7, -1)$$ and the line through $$A = (1, 1, 2)$$ and $$B = (2, 3, 4)$$. This distance is equal to $$\\dfrac{\\sqrt{d}}{3}$$ for some integer $$d$$. What is $$d$$? Line through $$A = (1, 1, 2)$$ and $$B = (2, 3, 4)$$ $$\\small{ \\begin{array}{|rclrcl|} \\hline \\vec{x} &=& \\vec{A}+t(\\vec{B}-\\vec{A}) & \\vec{r} &=& \\vec{B}-\\vec{A} \\quad | \\quad \\vec{A} = (1, 1, 2)\\quad \\vec{B} = (2, 3, 4)\\\\ &&& \\vec{r} &=& \\begin{pmatrix} 2 \\\\ 3 \\\\ 4 \\end{pmatrix}-\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix} \\\\ &&&\\vec{r} &=& \\begin{pmatrix} 1 \\\\ 2 \\\\ 2 \\end{pmatrix} \\\\ \\mathbf{\\vec{x}} &=& \\mathbf{\\vec{A}+t\\vec{r}} \\\\ \\hline (\\vec{x}-\\vec{Q})\\cdot \\vec{r} &=& 0 \\quad | \\quad \\vec{PQ} \\text{ is } \\perp \\text{ to } \\vec{r} \\\\ (\\mathbf{\\vec{A}+t\\vec{r}}-\\vec{Q})\\cdot \\vec{r} &=& 0 \\\\ (\\vec{A}-\\vec{Q} + t\\vec{r})\\cdot \\vec{r} &=& 0 & \\vec{A}-\\vec{Q}&=& \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}-\\begin{pmatrix} 3 \\\\ -7 \\\\ -1 \\end{pmatrix} \\quad | \\quad \\vec{Q} = (3, -7, -1)\\\\ & & & \\vec{A}-\\vec{Q}&=& \\begin{pmatrix} -2 \\\\ 8 \\\\ 3", "the base lines be $C$ and $D$, so that their parameters are $(1, 0)$ and $(0, 1)$ respectively. Then the parameters of $A$ and $B$ with respect to the base points will be $A: (\\frac{-l_2 - il_1}{2}, \\frac{-l_2 + il_1}{2})$, $B: (\\frac{-m_2 - im_1}{2}, \\frac{-m_2 + im_1}{2})$. The cross ratio of these 4 points will then be $$\\frac{\\begin{vmatrix} 1 & 0 \\\\ \\frac{-m_2 - im_1}{2} & \\frac{-m_2 + im_1}{2} \\end{vmatrix} \\begin{vmatrix} 0 & 1 \\\\ \\frac{-l_2 - il_1}{2} & \\frac{-l_2 + il_1}{2} \\end{vmatrix}} {\\begin{vmatrix} 1 & 0 \\\\ \\frac{-l_2 - il_1}{2} & \\frac{-l_2 + il_1}{2} \\end{vmatrix} \\begin{vmatrix} 0 & 1 \\\\ \\frac{-m_2 - im_1}{2} & \\frac{-m_2 + im_1}{2} \\end{vmatrix}}$$ $$= \\frac{(-m_2 + im_1)(l_2 + il_1)}{(-l_2 + il_1)(m_2 + im_1)}$$ $$= \\frac{m_1l_1 + m_2l_2 + i(m_1l_2 - m_2l_1)}{m_1l_1 + m_2l_2 + i(m_2l_1 - m_1l_2)}$$ $$= \\frac{1 + i\\frac{m_1l_2 - m_2l_1}{m_1l_1 + m_2l_2}}{1 + i\\frac{m_2l_1 - m_1l_2}{m_1l_1 + m_2l_2}}$$ To get the logarithm in the picture, we need to convert these coordinates from rectangular to polar. In polar coordinates, this equation is equivalent to $$\\frac{r(cos(\\theta) + isin(\\theta))}{r(cos(-\\theta) + isin(-\\theta))}$$ $$= \\frac{cos(\\theta) + isin(\\theta)}{cos(-\\theta) + isin(-\\theta)}$$ $$= \\frac{e^{i\\theta}}{e^{-i\\theta}}$$ $$= e^{2i\\theta}$$ where $\\theta = tan^{-1}(\\frac{m_1l_2 - m_2l_1}{m_1l_1 + m_2l_2}) = cos^{-1}(\\frac{m_1l_1 + m_2l_2}{\\sqrt{(m_1^2 + m_2^2)(l_1^2 + l_2^2)}})$. This is the same as the result in 2.21 and 2.22. In other words, if", "as it should help develop a much better sense of what we're actually doing when we use the chain rule. If you'd prefer to try the exercises right away, by all means do so but do make sure to watch Tutorial 2 afterwards. ## Exercise 1 1. Differentiate the function defined as $$f(x) = sin\\begin{pmatrix} x^3-2 \\end{pmatrix}$$. 2. Given $$y = \\sqrt{3x^2-4x+7}$$, find an expression for $$\\frac{dy}{dx}$$. 3. Differentiate the function $$f(x) = \\begin{pmatrix} x^2 + 3 \\end{pmatrix}^7$$. 4. Find $$\\frac{dy}{dx}$$, given $$y = e^{3x^2-1}$$. 5. Differentiate the function $$f(x) = ln \\begin{pmatrix} x^2 + x \\end{pmatrix}$$. 6. Given $$y = cos \\begin{pmatrix} 2x + 1 \\end{pmatrix}$$, find $$\\frac{dy}{dx}$$. 1. $$f'(x) = 3x^2.cos \\begin{pmatrix} x^3 - 2 \\end{pmatrix}$$ 2. $$\\frac{dy}{dx} = \\frac{3x-2}{\\sqrt{3x^2 - 4x + 7}}$$ 3. $$f'(x) = 14x \\begin{pmatrix} x^2 + 3 \\end{pmatrix}^6$$ 4. $$\\frac{dy}{dx} = 6x.e^{3x^2-1}$$ 5. $$f'(x) = \\frac{2x+1}{x^2+x}$$ 6. $$\\frac{dy}{dx} = -2.sin \\begin{pmatrix} 2x + 1 \\end{pmatrix}$$ 1. The function $$f(x) = sin \\begin{pmatrix} x^3 - 2 \\end{pmatrix}$$ can be written $$f(x) = g \\begin{bmatrix} h(x) \\end{bmatrix}$$, where: $g(x) = sin(x) \\quad \\text{and} \\quad h(x) = x^3 - 2$ and $g'(x) = cos(x) \\quad \\text{and} \\quad h'(x) = 3x^2$ We can now use the chain rule to write: $f'(x) = h'(x) \\times g'\\begin{bmatrix} h(x) \\end{bmatrix}$ That's: $f'(x) = 3x^2.cos \\begin{pmatrix} x^3-2\\end{pmatrix}$ 2. We", "the line passing through all three points. That, along with the distance equations you have, should define the coordinates of $(x_3, y_3)$ – Namaste Oct 22 '16 at 20:33 The line can be parameterized by $\\mathbf{x}(t) = (x_1 + t(x_2 - x_1),y_1 + t(y_2 - y_1))$ for all $t\\in\\mathbb{R}$. Since point $P_3 = (x_3,y_3)$ is on this line, there exists a $t$ such that $\\mathbf{x}(t) = (x_3,y_3)$. Assuming the point $P_3$ comes after the point $P_2$ on the line, we can solve for $t$ by observing that the distance from $P_1$ to $P_3$ is $$r + h = \\sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2} = |t|\\sqrt{(x_2-x_1)^2 + (y_2 - y_1)^2}$$ $$t = \\frac{r + h}{\\sqrt{(x_2-x_1)^2 + (y_2 - y_1)^2}}$$ Plugging this value of $t$ into $\\mathbf{x}(t)$ will give you the coordinates of $P_3$.", "I think the easiest way is to write the line in vector form where $$\\vec{n} = \\frac 1{\\sqrt{1+m^2}}\\begin{pmatrix}1 \\\\ m\\end{pmatrix}$$ is the normalized direction vector. You will find the $$k$$-th point on the line starting from $$\\begin{pmatrix}x_1 \\\\ y_1\\end{pmatrix}$$ by $$\\begin{pmatrix}x \\\\ y\\end{pmatrix}= \\begin{pmatrix}x_1 \\\\ y_1\\end{pmatrix}+k\\cdot d\\cdot \\vec{n} = \\begin{pmatrix}x_1 \\\\ y_1\\end{pmatrix}+k\\frac{d}{\\sqrt{1+m^2}}\\begin{pmatrix}1 \\\\ m\\end{pmatrix}$$ Here is a Desmos-graph which illustrates above formula. You may start with a horizontal line to see that the distance fits and then change the slope $$m$$ and animate $$k$$. You can use a for loop. Let's say you already drew the point $$M_n(x_n, y_n)$$ and wish to draw $$M_{n+1}(x_{n+1}, y_{n+1})$$. First of all you know that $$M_{n+1} \\in \\Delta: y = mx + b$$. As such $$y_{n+1} = m x_{n+1} + b$$. And you know that the distance between $$M_n$$ and $$M_{n+1}$$ is $$d$$. So $$d^2 = (x_{n+1} - x_n)^2 + (y_{n+1} - y_n)^2 \\\\ \\iff d^2 = (x_{n+1} - x_n)^2 + (m x_{n+1} + b - y_n)^2$$ This is a quadratic equation in $$x_{n+1}$$. Expand it out and compute the discriminant and find the solutions. You'll get two solutions, $$x_{n+1, 1}$$ and $$x_{n+1, 2}$$. If $$x_3 > x_1$$, then $$x_{n+1} = \\max(x_{n+1, 1}, x_{n+1, 2})$$. Else $$x_{n+1} = \\min(x_{n+1, 1}, x_{n+1, 2})$$ And after finding $$x_{n+1}$$ you can just use the formula", "can differentiate $$y = \\sqrt{3x^2 - 4x + 7}$$ as follows: Let $$u(x) = 3x^2 - 4x + 7$$, then: $y = \\sqrt{u}$ $\\frac{dy}{du} = \\frac{1}{2\\sqrt{u}}$ and $\\frac{du}{dx} = 6x - 4$ and using the chain rule we can write: \\begin{aligned} \\frac{dy}{dx} & = \\frac{dy}{du}.\\frac{dy}{dx} \\\\ & = \\frac{1}{2\\sqrt{u}}. \\begin{pmatrix} 6x - 4 \\end{pmatrix} \\\\ & = \\frac{6x - 4}{2\\sqrt{u}} \\\\ & = \\frac{2 \\begin{pmatrix} 3x - 2 \\end{pmatrix}}{2\\sqrt{u}} \\\\ & = \\frac{3x - 2}{\\sqrt{u}}\\\\ \\frac{dy}{dx} &= \\frac{3x-2}{\\sqrt{3x^2 - 4x + 7}} \\end{aligned} 3. The function $$f(x) = \\begin{pmatrix} x^2 + 3 \\end{pmatrix}^7$$ can be written $$f(x) = g \\begin{bmatrix} h(x) \\end{bmatrix}$$ where $g(x) = x^7 \\quad \\text{and} \\quad h(x) = x^2 + 3$ and $g'(x) = 7x^6 \\quad \\text{and} \\quad h'(x) = 2x$ So, using the chain rule, we find the derivative as follows: \\begin{aligned} f'(x) & = h'(x) \\times g' \\begin{bmatrix} h(x) \\end{bmatrix} \\\\ & = 2x \\times 7 \\begin{pmatrix} x^2 + 3 \\end{pmatrix}^6 \\\\ f'(x) & = 14x \\begin{pmatrix} x^2 + 3 \\end{pmatrix}^6 \\end{aligned} 4. We differentiate $$y = e^{3x^2-1}$$ as follows: Let $u(x) = 3x^2 - 1$ then $y = e^u$ $\\frac{dy}{du} = e^u$ and $\\frac{du}{dx} = 6x$ Now, using the chain rule, we find the derivative: \\begin{aligned} \\frac{dy}{dx} & = \\frac{dy}{du}.\\frac{du}{dx} \\\\ & = e^u.6x \\\\ \\frac{dy}{dx} & = 6x.e^{3x^2-1} \\end{aligned} 5. The", "the point $$(1,2,3)$$ can be written as $$\\frac{x-1}{0} = \\frac{y-2}{0} = \\frac{z-3}{2} = t$$, where $$t \\in R$$ is any parameter. The parametric form the line (especially for x,y coordinates) can be obtained by \"cross multiplying\" their respective DR with the parameter equation to give $$(x,y,z)=(1,2,2t+3)$$." ]

[ "0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -5 & 0\\\\ 0 & 0 & 0\\end{array}\\right) \\mbox{ and } \\left(\\begin{array}{cc|c} 5 & -5 & 0\\\\ 1 & -1 & 0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -1 & 0\\\\ 0 & 0 & 0\\end{array}\\right)\\, ,$$ respectively, so are $$v_{2}=\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}$$ and $$v_{-2} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$. Hence $$M^{12} v_{2} = 2^{12} v_{2}$$ and $$M^{12}v_{-2} = (-2)^{12} v_{-2}$$. Now, $$\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\frac{x-y}{4}\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix} -\\frac{x-5y}{4} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$ (this was obtained by solving the linear system $$a v_{2} + b v_{-2} =$$ for $$a$$ and $$b$$). Thus $$M \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\frac{x-y}{4} M v_{2} -\\frac{x-5y}{4} M v_{-2}$$ $$= 2^{12} \\Big(\\frac{x-y}{4} v_{2} -\\frac{x-5y}{4} v_{-2}\\Big) = 2^{12} \\begin{pmatrix} x \\\\ y \\end{pmatrix}\\, .$$ Thus $$M^{12}=\\begin{pmatrix} 4096 & 0 \\\\ 0 & 4096\\end{pmatrix}\\, .$$ $$\\textit{If you understand the above explanation, then you have a good understanding of diagonalization. A quicker route}$$ $$\\textit{is simply to observe that}$$ $$M^{2} = \\begin{pmatrix}4 & 0 \\\\ 0 & 4\\end{pmatrix}$$. 8. $$P_{M}(\\lambda) = (-1)^{2} {\\rm det}\\begin{pmatrix} a-\\lambda & b \\\\ c &d-\\lambda\\end{pmatrix} =(\\lambda-a)(\\lambda-d) - bc\\, .$$ Thus $$P_{M}(M)=(M-a I )(M- d I) - bc I$$ $$= \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}a&0\\\\0&a\\end{pmatrix}\\right) \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}d&0\\\\0&d\\end{pmatrix}\\right)-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}$$ $$=\\begin{pmatrix}0& b\\\\c&d-a\\end{pmatrix}\\begin{pmatrix}a-d&b\\\\ c&0\\end{pmatrix}-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}=0\\, .$$ Observe that any $$2\\times 2$$ matrix is a zero of its own characteristic polynomial ($$\\textit{in fact this holds", "( 7,5,-1 ) \\\\ &= \\left( \\frac{437}{75}, -\\frac{4}{15}, -\\frac{266}{75} \\right) \\\\ &\\approx \\left( 5.82667, -0.266665, -3.54667 \\right), \\end{align*} and $t$ must equal $$t=\\frac{41}{75} \\approx \\boxed{0.546667}.$$ $\\textbf{Remark}$. It is worth noting that the distance between points $a$ and $v(0.546667)$ is \\begin{align*} ||\\text{ortho}_{u}w || &= || w-\\text{proj}_u w || \\\\ &= \\Bigg|\\Bigg| \\langle 2,7,8\\rangle -\\frac{41}{75} \\langle 7,5,-1 \\rangle \\Bigg|\\Bigg| \\\\ &= \\Bigg|\\Bigg| \\left( -\\frac{137}{75} , \\frac{64}{15}, \\frac{641}{75} \\right) \\Bigg|\\Bigg| \\\\ &= \\sqrt{\\frac{7094}{75}} \\\\ &\\approx 9.72557. \\end{align*} $$v-a = \\begin{pmatrix} 2 \\\\ -3 \\\\ -3 \\end{pmatrix} + \\begin{pmatrix} 7 \\\\ 5 \\\\ -1 \\end{pmatrix} t -\\begin{pmatrix} 4 \\\\ 4 \\\\ 5 \\end{pmatrix} = \\begin{pmatrix} 2+ 7t-4 \\\\-3+ 5t -4\\\\ -3-1t-5 \\end{pmatrix} = \\begin{pmatrix} -2+ 7t \\\\-7+ 5t\\\\ -8-1t \\end{pmatrix}$$ The magnitude of $v-a$ is the square root of $(-2+ 7t)^2 + (-7+ 5t)^2 + (-8-1t)^2$. Do you know how to minimise this expression? Write the vector form of the difference between the two vectors. Then find an expression in $t$ representing the square of the magnitude of the difference vector. If you minimize the square of the distance, you also minimize the distance. Find the value of $t$ which minimizes the square of the distance using the first derivative of the expression. Vector $v$ can be written as $$\\begin{pmatrix} 2 +7t \\\\ -3 +5t\\\\ -3 -1t\\end{pmatrix}$$ Then the distance between $a$", "=\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\left [\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}\\end{pmatrix} -\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\right ]+\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\\\ & =\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}-1\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}-2\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}+1\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}+5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{16}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{37}{5}\\end{pmatrix} \\end{align*} Is that correct so far? How could we continue? Klaas van Aarsen MHB Seeker Staff member The matrix for $\\delta$ is correct. But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. mathmari Well-known member MHB Site Helper But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? Klaas van Aarsen MHB Seeker Staff member Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? That point is in between $A$ and $B$. But $C$ is on the circumcircle of triangle $ABZ$, which implies that it cannot be between $A$ and $B$. Klaas van Aarsen MHB Seeker Staff member I've drawn a picture now: \\begin{tikzpicture} \\usetikzlibrary", "\\frac{1-\\sqrt{5}}{2} \\end{pmatrix}^{-1}\\begin{pmatrix} 1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2} \\end{pmatrix} \\begin{pmatrix}C_{1}\\\\ C_{2}\\end{pmatrix} = \\begin{pmatrix} 1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2} \\end{pmatrix}^{-1} \\begin{pmatrix}1\\\\ \\frac{-5}{4}\\end{pmatrix} \\begin{pmatrix}C_{1}\\\\ C_{2}\\end{pmatrix} = \\frac{-1}{\\sqrt{5}}\\begin{pmatrix} \\frac{1-\\sqrt{5}}{2} & -1 \\\\ \\frac{-1-\\sqrt{5}}{2} & 1 \\end{pmatrix} \\begin{pmatrix}1\\\\ \\frac{-5}{4}\\end{pmatrix} which gives us C_{1}=\\frac{-2(1-\\sqrt{5})-5}{4\\sqrt{5}}\\ (7) C_{2}=\\frac{2(1+\\sqrt{5})+5}{4\\sqrt{5}}\\ (8) Finally, by plugging (7) and (8) into (3) we get F_{n}=-4\\left (\\left ( \\frac{-2(1-\\sqrt{5})-5}{4\\sqrt{5}} \\right )\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{n}+\\left ( \\frac{2(1+\\sqrt{5})+5}{4\\sqrt{5}} \\right )\\left ( \\frac{1-\\sqrt{5}}{2} \\right )^{n}\\right ) (9) We can now “easily” compute F_{386} which is given by F_{386}=-4\\left (\\left ( \\frac{-2(1-\\sqrt{5})-5}{4\\sqrt{5}} \\right )\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{386}+\\left ( \\frac{2(1+\\sqrt{5})+5}{4\\sqrt{5}} \\right )\\left ( \\frac{1-\\sqrt{5}}{2} \\right )^{386}\\right ) Google returns F_{386} \\cong 5.2783459\\times 10^{80} which is just an approximation, but now we have a formula for the precise answer. If you wish to check for yourself that out formula (9) is indeed correct, click on the terms below to see the computation by Google: F_{0}=-4 F_{1}=5 F_{2}=1 F_{3}=6 F_{4}=7 F_{5}=13 On a side note – Raising a number to the power of 386 may seem like a long computation but it really requires only 10 multiplications! Don’t believe me? Consider this: \\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{386}=\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{256}\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{128}\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{2} To calculate \\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{256} we use the fact that 1.)\\ \\left", "+ i & 0 \\end{pmatrix}$. Last edited: Feb 28, 2016 danshawen likes this. 11. ### rpennerFully WiredRegistered Senior Member Messages: 4,833 $M=\\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\sqrt{2} & 0 \\\\ 0 & \\sqrt{2} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix}$ So we have: $M^{2n} = 2^{n} \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} = 2^{n} I \\\\ M^{2n+1} = 2^{n} \\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = 2^{n} M$ So $\\textrm{exp}(M) = \\sum_{k=0}^{\\infty} \\left( \\frac{ 2^{k} }{ (2k)! } I +\\frac{ 2^{k} }{ (2k+1)! } M \\right) = \\left( \\cosh \\sqrt{2} \\right) I + \\left( \\frac{1}{\\sqrt{2}} \\sinh \\sqrt{2} \\right) M \\\\ \\quad \\quad = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$ Alternately: $\\textrm{exp}(M) = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} e^{- \\sqrt{2}} & 0 \\\\ 0 & e^{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$", "\\end{pmatrix} \\tag{1}$$ ## Multiplying on the left Step by step (so we can compare with multiplying on the right shortly): $$\\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 5 \\\\ -2 & 3 \\end{pmatrix} = \\begin{pmatrix} 4 & 5 \\\\ 0 & \\frac{11}{2} \\end{pmatrix} \\tag{2}$$ Rescaling, $$\\begin{pmatrix} 1 & 0 \\\\ 0 & \\frac{2}{11} \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 5 \\\\ -2 & 3 \\end{pmatrix} = \\begin{pmatrix} 4 & 5 \\\\ 0 & 1 \\end{pmatrix} \\tag{3}$$ Then $$\\begin{pmatrix} 1 & -5 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & \\frac{2}{11} \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 5 \\\\ -2 & 3 \\end{pmatrix} = \\begin{pmatrix} 4 & 0 \\\\ 0 & 1 \\end{pmatrix} \\tag{4}$$ and finally $$\\begin{pmatrix} \\frac{1}{4} & 0 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & -5 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & \\frac{2}{11} \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 5 \\\\ -2 & 3 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} \\tag{5}$$ Let's then work this the opposite way to see what the row operations above are doing when we multiply on the right instead. ##", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {vmatrix}$$ = -4 - 15 = -19≠ 0 It has a unique solution. X = A-1B A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$ X = A-1B or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -10 - 9\\\\ -25 + 6\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-19}{-19}\\\\ \\frac {-19}{-19}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, 3x - 5y = 3..........................(1) 4x + 3y = 4.........................(2) Given equation in matrix form; $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3\\\\ 4\\\\ \\end {pmatrix}$$ or, AX = B where, A = $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$,", "is possible to find A-1. A-1 = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj (A) A = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ $$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\frac 12$$$$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} (\\frac 22) & (\\frac {-4}{2})\\\\ (\\frac {-3}{2}) & (\\frac 72)\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ ∴ A-1 =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ Ans Let: A =$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ and A-1=$$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ We know that, AA-1 = I, where I =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 45 - 45 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 0 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ Taking the corresponding elements both sides", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "& -3/50 \\end{array} \\right) \\left( \\begin{array}{cc} 1 & -2 \\\\ 1 & 3 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.1 \\\\ 0 \\end{pmatrix}$ Multiply throughout by $\\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right)^{-1} = 1/5 \\left( \\begin{array}{cc} 2 & 3 \\\\ -1 & 1 \\end{array} \\right)$ to get $\\frac{d}{dt} \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\left( \\begin{array}{cc} -1/50 & 0 \\\\ 0 & -6/50 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.04 \\\\ -0.02 \\end{pmatrix}$ this is equivalent to the differential equations $\\frac{du_1}{dt}=-1/50u_1+0.04$ $\\frac{du_2}{dt}=-6/50u_1-0.02$ These have the solutions $u_1=2+Ae^{-t/50}, \\: u_2= -1/6+Be^{-6t/50}$ We can write this as $\\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix}$ Now transform back to $x_1, \\: x_2$ . \\begin{aligned} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} &= P \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} \\\\ &= \\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right) \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix} \\\\ &= \\begin{pmatrix} 5/2+Ae^{-t/50}-3Be^{-6t/50} \\\\ 5/3+Ae^{-t/50}+2Be^{-6t/50} \\end{pmatrix} \\end{aligned} . These solutions can be fitted to any initial conditions to find the constants.", "& {\\ frac {-1 + {\\ sqrt {5}}} {2}} \\\\ 1 & 1 \\ end {pmatrix}} {\\ begin {pmatrix} {\\ frac {1 - {\\ sqrt {5}}} {2}} & 0 \\\\ 0 & {\\ frac {1 + {\\ sqrt {5}}} {2} } \\ end {pmatrix}} ^ {n} {\\ begin {pmatrix} - {\\ frac {1} {\\ sqrt {5}}} & {\\ frac {{\\ sqrt {5}} - 1} {2 { \\ sqrt {5}}}} \\\\ {\\ frac {1} {\\ sqrt {5}}} & {\\ frac {{\\ sqrt {5}} + 1} {2 {\\ sqrt {5}}}} \\ end {pmatrix}} {\\ begin {pmatrix} 0 \\\\ 1 \\ end {pmatrix}} \\\\ & = {\\ begin {pmatrix} {\\ frac {-1 - {\\ sqrt {5}}} {2} } & {\\ frac {-1 + {\\ sqrt {5}}} {2}} \\\\ 1 & 1 \\ end {pmatrix}} {\\ begin {pmatrix} \\ left ({\\ frac {1 - {\\ sqrt {5 }}} {2}} \\ right) ^ {n} & 0 \\\\ 0 & \\ left ({\\ frac {1 + {\\ sqrt {5}}} {2}} \\ right) ^ {n} \\ end {pmatrix }} {\\ begin {pmatrix} - {\\ frac {1} {\\ sqrt {5}}} & {\\ frac {{\\ sqrt {5}} - 1} {2 {\\ sqrt {5}}}} \\ \\ {\\ frac {1} {\\ sqrt {5}}} & {\\ frac {{\\ sqrt {5}} + 1} {2 {\\ sqrt {5}}} } \\ end {pmatrix}}", "All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions. (i) $cos{\\theta}=\\frac{|\\begin{pmatrix}2\\\\-2\\\\1\\end{pmatrix}{\\bullet}\\begin{pmatrix}-6\\\\3\\\\2\\end{pmatrix}|}{\\sqrt{9} \\sqrt{49}}=\\frac{16}{21}$ $\\theta = {40.4}^{\\circ}$ (ii) Using Graphing Calculator, equation of line, $r = \\begin{pmatrix}{-\\frac{1}{6}}\\\\{-\\frac{2}{3}}\\\\0\\end{pmatrix} + \\lambda \\begin{pmatrix}7\\\\10\\\\6\\end{pmatrix}, \\lambda \\in \\mathbb{R}$ (iii) Distance A to $P_1 = \\frac{|\\begin{pmatrix}2\\\\-2\\\\1\\end{pmatrix}{\\bullet}\\begin{pmatrix}4\\\\3\\\\c\\end{pmatrix} - 1|}{\\sqrt{9}} = \\frac{|1+c|}{3}$ Distance A to $P_1 = \\frac{|\\begin{pmatrix}-6\\\\3\\\\2\\end{pmatrix}{\\bullet}\\begin{pmatrix}4\\\\3\\\\c\\end{pmatrix} + 1|}{\\sqrt{49}} = \\frac{|2c-14|}{7}$ $\\frac{|1+c|}{3} = \\frac{|2c-14|}{7}$ $49(1+c)^{2} = 9(2c-14)^{2}$ $c = -49 \\mathrm{~or~} \\frac{35}{13}$", "# sum of binomial series How do I solve sum of binomial series which is as follows: $$\\frac{1}{3}\\sum^{\\infty}_{x=0}\\begin{pmatrix}x\\\\y\\end{pmatrix}\\frac{1}{3}^x$$ I think it would be pretty easy to sum from $y=0$ to $x$, but I have no idea how to sum from $x=0$ to $x=\\infty$. Thanks, • Is it $\\frac {1^x}3$ or $\\big(\\frac 13\\big)^x$ – Claude Leibovici May 3 '16 at 3:48 Remember binomial theorem: $$(p+q)^x = \\sum_{y} \\begin{pmatrix}x\\\\y\\end{pmatrix} q^y p^{x-y}$$ With $p = 1/3$: $$(1/3+q)^x = \\sum_{y} \\begin{pmatrix}x\\\\y\\end{pmatrix} q^y \\left( \\frac{1}{3} \\right)^{x-y} = \\sum_{y} \\begin{pmatrix}x\\\\y\\end{pmatrix} 3^y q^y \\left( \\frac{1}{3} \\right)^{x}$$ Now summing over $x$ on both sides and rearranging some things in the sum gives: $$\\sum_{x = 0}^\\infty (1/3 + q)^x = \\sum_{x} \\sum_{y} \\begin{pmatrix}x\\\\y\\end{pmatrix} 3^y q^y \\left( \\frac{1}{3} \\right)^{x} \\\\ =\\sum_{y} q^y 3^y \\; \\sum_{x} \\begin{pmatrix}x\\\\y\\end{pmatrix} \\left( \\frac{1}{3} \\right)^{x}$$ So all you have to do to find your sum is to find the coefficient of $q^y$ term in $\\sum_{x = 0}^\\infty (1/3 + q)^x$, and divide by $3^y$. The sum is easily computed by geometric series: $$\\sum_{x = 0}^\\infty (1/3 + q)^x = \\frac{1}{2/3 - q} \\\\ = \\frac{3}{2} \\sum_y (3 q /2)^y$$ Where in the last line I used the geometric series expansion of $1/(1 - x)$ The coefficient of $q^y$ is $(3/2)^{y+1}$, dividing by $3^y$ gives $3/ 2^{y+1}$.", "\\frac{-3}{2}, \\frac{1}{3},240\\right)=\\left(0, \\frac{-5}{3},247\\right)$ $7\\left( \\frac{1}{5},\\frac{1}{7},\\frac{1}{9}\\right)=\\left( \\frac{7}{5},1,\\frac{7}{9}\\right)$ Ex.2. $V=M_{m\\times n}(F)=\\left\\lbrace\\begin{pmatrix} a_{11} & \\cdots & a_{1n} \\\\ \\vdots & & \\vdots \\\\ a_{m1} & \\cdots & a_{mn}\\end{pmatrix}: a_{ij} \\in F \\right\\rbrace$ $M_{3\\times 2}( \\mathbb{R})\\ni \\begin{pmatrix} 7 & -7 \\\\ \\pi & \\mathit{e} \\\\ -5 & 2 \\end{pmatrix}$ $\\mbox{Addition by adding entry by entry:}$ $M_{2\\times 2}\\ \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix}+\\begin{pmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{pmatrix}=\\begin{pmatrix} {a_{11}+b_{11}} & {a_{12}+b_{12}} \\\\ {a_{21}+b_{21}} & {a_{22}+b_{22}} \\end{pmatrix}$ $\\mbox{Multiplication by multiplying scalar c to all entries by M.}$ $c\\cdot M_{2\\times 2}\\ \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix}=\\begin{pmatrix} c\\cdot a_{11} & c\\cdot a_{12} \\\\ c\\cdot a_{21} & c\\cdot a_{22} \\end{pmatrix}$ $\\mbox{Zero matrix has all entries = 0:}$ $0_{M_{m\\times n}}=\\begin{pmatrix} 0 & \\cdots & 0 \\\\ \\vdots & & \\vdots \\\\ 0 & \\cdots & 0\\end{pmatrix}$ Ex.3. $\\mathbb{C}$ form a vector space over $\\mathbb{R}$. Ex.4. $\\mbox{F is a vector space over itself.}$ Ex.5. $\\mathbb{R}$ is a vector space over $\\mathbb{Q}$. Ex.6. $\\mbox{Let S be a set. Let}$ $\\mathcal{F}(S,\\mathbb{R})=\\big\\{f:S\\to \\mathbb{R} \\big\\}$ $f,g \\in \\mathcal{F}(S,\\mathbb{R})$ $(f+g)(t)=f(t)+g(t)\\ for\\ any\\ t\\in S$ $(af)(t)=a\\cdot f(t)\\$", "$$A \\begin{pmatrix} \\frac{\\partial F}{\\partial x_1}\\\\ \\frac{\\partial F}{\\partial x_2} \\end{pmatrix} = \\begin{pmatrix} 1&0\\\\ 0&2\\\\ \\end{pmatrix} \\begin{pmatrix} x_2\\\\ x_1 \\end{pmatrix} = \\begin{pmatrix} x_2\\\\ 2x_1 \\end{pmatrix} = \\begin{pmatrix} g_1(\\mathbf{x})\\\\ g_2(\\mathbf{x}) \\end{pmatrix};$$ $$\\frac{\\partial g_1}{\\partial x_2}=1\\neq2=\\frac{\\partial g_2}{\\partial x_1}.$$ IDEA 2. Chain rule (?) Counterexample. $$F(\\mathbf{x})=x_1x_2;\\\\ \\mathbf{y}(\\mathbf{x}): x_1=y_1;~x_2=2y_2.\\\\ \\begin{pmatrix} \\frac{\\partial x_1}{\\partial y_1}&\\frac{\\partial x_2}{\\partial y_1}\\\\ \\frac{\\partial x_1}{\\partial y_2}&\\frac{\\partial x_2}{\\partial y_2} \\end{pmatrix} \\begin{pmatrix} \\frac{\\partial F}{\\partial x_1}\\\\ \\frac{\\partial F}{\\partial x_2} \\end{pmatrix} = \\begin{pmatrix} 1&0\\\\ 0&2 \\end{pmatrix} \\begin{pmatrix} x_2\\\\ x_1 \\end{pmatrix} = \\begin{pmatrix} x_2\\\\ 2x_1 \\end{pmatrix} = \\begin{pmatrix} g_1(\\mathbf{x})\\\\ g_2(\\mathbf{x}) \\end{pmatrix} = \\begin{pmatrix} \\frac{\\partial F}{\\partial y_1}\\\\ \\frac{\\partial F}{\\partial y_2} \\end{pmatrix};$$ $$\\frac{\\partial g_1}{\\partial x_2}=1\\neq2=\\frac{\\partial g_2}{\\partial x_1}.$$ Of course, $$\\frac{\\partial^2F}{\\partial y_1\\partial y_2}=\\frac{\\partial^2F}{\\partial y_2\\partial y_1}$$ should be valid, but not $$\\frac{\\partial^2F}{\\partial y_1\\partial x_2}\\neq\\frac{\\partial^2F}{\\partial y_2\\partial x_1}.$$ And, in fact, the latter one is required in the original post. SOLUTION FOR n=2 @IgorKhavkine helped me a lot. Proof for $n=2$. All functions below a considered to depend on $\\mathbf{x}$. $$\\begin{pmatrix} a_{11}&a_{12}\\\\ a_{21}&a_{22} \\end{pmatrix} \\begin{pmatrix} \\frac{\\partial F}{\\partial x_1}\\\\ \\frac{\\partial F}{\\partial x_2} \\end{pmatrix} = \\begin{pmatrix} a_{11}\\frac{\\partial F}{\\partial x_1}+a_{12}\\frac{\\partial F}{\\partial x_2}\\\\ a_{21}\\frac{\\partial F}{\\partial x_1}+a_{22}\\frac{\\partial F}{\\partial x_2} \\end{pmatrix} = \\begin{pmatrix} g_1\\\\ g_2 \\end{pmatrix}.$$ The condition $\\frac{\\partial g_1}{\\partial x_2}=\\frac{\\partial g_2}{\\partial x_1}$ may be written as: $$\\frac{\\partial}{\\partial x_2}\\left(a_{11}\\frac{\\partial F}{\\partial x_1}+a_{12}\\frac{\\partial F}{\\partial x_2}\\right)=\\frac{\\partial}{\\partial x_1}\\left(a_{21}\\frac{\\partial F}{\\partial x_1}+a_{22}\\frac{\\partial F}{\\partial x_2}\\right).$$ If we expand it, we get: $$\\frac{\\partial a_{11}}{\\partial x_2}\\frac{\\partial F}{\\partial x_1}+a_{11}\\frac{\\partial^2F}{\\partial x_2\\partial x_1}+\\frac{\\partial a_{12}}{\\partial x_2}\\frac{\\partial F}{\\partial x_2}+a_{12}\\frac{\\partial^2F}{\\partial x_2^2}=\\\\ =\\frac{\\partial a_{21}}{\\partial x_1}\\frac{\\partial F}{\\partial", "0\\end{pmatrix}^n \\begin{pmatrix}x \\\\ 1\\end{pmatrix} %= \\left(S J S^{-1} \\right)^n \\begin{pmatrix}x \\\\ 1\\end{pmatrix} %= S \\left(J \\right)^n S^{-1} \\begin{pmatrix}x \\\\ 1\\end{pmatrix} %= \\begin{pmatrix} x-\\sqrt{x^2-1} & x+\\sqrt{x^2-1} \\\\ 1 & 1 \\end{pmatrix} \\begin{pmatrix} x-\\sqrt{x^2-1} & 0 \\\\ 0 & x+\\sqrt{x^2-1} \\end{pmatrix}^n \\begin{pmatrix} -\\frac{1}{2\\sqrt{x^2-1}} & \\frac{x}{2\\sqrt{x^2-1}}+\\frac12 \\\\ \\frac{1}{2\\sqrt{x^2-1}} & \\frac12 - \\frac{x}{2\\sqrt{x^2-1}} \\end{pmatrix} \\begin{pmatrix}x \\\\ 1\\end{pmatrix} \\\\ %= \\begin{pmatrix} x-\\sqrt{x^2-1} & x+\\sqrt{x^2-1} \\\\ 1 & 1 \\end{pmatrix} \\begin{pmatrix} (x-\\sqrt{x^2-1})^n & 0 \\\\ 0 & (x+\\sqrt{x^2-1})^n \\end{pmatrix} \\begin{pmatrix}\\frac12 \\\\ \\frac12\\end{pmatrix} \\\\ %= \\begin{pmatrix} x-\\sqrt{x^2-1} & x+\\sqrt{x^2-1} \\\\ 1 & 1 \\end{pmatrix} \\begin{pmatrix}\\frac12 (x-\\sqrt{x^2-1})^n \\\\ \\frac12 (x+\\sqrt{x^2-1})^n \\end{pmatrix} \\\\ = \\frac12 \\begin{pmatrix} (x-\\sqrt{x^2-1})^{n+1} + (x+\\sqrt{x^2-1})^{n+1} \\\\ (x-\\sqrt{x^2-1})^n + (x+\\sqrt{x^2-1})^n \\end{pmatrix}$$ So what this test boils down to is $$(x-\\sqrt{x^2-1})^n + (x+\\sqrt{x^2-1})^n \\equiv 2x^n \\pmod n$$ Expanding the LHS, $$(x-\\sqrt{x^2-1})^n + (x+\\sqrt{x^2-1})^n = \\sum_{i=0}^n \\binom{n}{i} x^{n-i} \\left( (-\\sqrt{x^2-1})^{i} + (\\sqrt{x^2-1})^{i}\\right) \\\\ = \\sum_{i=0}^n \\binom{n}{i} x^{n-i} (\\sqrt{x^2-1})^{i} \\left( (-1)^i + 1^i\\right) \\\\ = \\sum_{i=0}^{n/2} \\binom{n}{2i} x^{n-2i} (\\sqrt{x^2-1})^{2i} \\\\ = \\sum_{i=0}^{n/2} \\binom{n}{2i} x^{n-2i} (x^2-1)^{i} \\\\ = \\sum_{i=0}^{n/2} \\binom{n}{2i} x^{n-2i} \\sum_{j=0}^i \\binom{i}{j} (x^2)^{i-j} (-1)^j \\\\ = \\sum_{j=0}^{n/2} (-1)^j x^{n-2j} \\sum_{i=j}^{n/2} \\binom{n}{2i} \\binom{i}{j} \\\\$$ So the test is that $$\\forall 0 < j \\le \\tfrac n2: \\sum_{i=j}^{n/2} \\binom{n}{2i} \\binom{i}{j} \\equiv 0 \\pmod n$$ But a direct calculation on that basis is worse than simply verifying that $$\\forall 0 < j \\le \\tfrac n2: \\binom{n}{2j} \\equiv 0 \\pmod n$$", "is $$A=\\begin{pmatrix}1&0&1\\\\\\!\\!\\!\\!-1&1&2\\\\0&\\!\\!\\!\\!-1&1\\end{pmatrix}^{-1}=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}$$ Finally, since $$f(x)=3-4x+2x^2\\stackrel{coord. wrt B}\\longrightarrow \\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix} \\,\\,,\\,\\text{we get}$$ $$[Af]=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}\\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix}=\\frac{1}{4}\\begin{pmatrix}11\\\\\\!\\!\\!\\!-7\\\\1\\end{pmatrix}$$ so $$\\,f_C=\\frac{11}{4}(1-x)-\\frac{7}{4}(x-x^2)+\\frac{1}{4}(1+2x+x^2)$$", "\\end {pmatrix}$$ Taking the corresponding part of the equal matrix: $$\\frac 1x$$ = $$\\frac 12$$ ∴ x = 2 ∴ y = 3 ∴ x = 2 and y = 3 Ans Here, $$\\frac 4x$$ + 3y = 11.................................(1) 3xy - 8 -5x = 0 or, $$\\frac {3xy}{x}$$ - $$\\frac 8x$$ - $$\\frac {5x}{x}$$ = 0 or, -$$\\frac 8x$$ + 3y = 5...................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {vmatrix}$$ = 4 × 3 - (-8) × 3 = 12 + 24 = 36 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{36}$$$$\\begin {pmatrix} 3 & -3\\\\ 8 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{36} & \\frac {-3}{36}\\\\ \\frac {8}{36} & \\frac {4}{36}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{12} & \\frac {-1}{12}\\\\ \\frac {2}{9} & \\frac {1}{9}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix}", "that $M^n = \\begin{pmatrix} 2 & 1 \\\\ 1 & 0 \\end{\\pmatrix}^n = \\begin{pmatrix} - \\frac{1}{2} \\sqrt{-4 + 3 \\sqrt{2}} & \\frac{1}{2} \\sqrt{4 + 3 \\sqrt{2}} \\\\ 2^{\\tiny - \\frac{3}{4}} & 2^{\\tiny - \\frac{3}{4}} \\end{\\pmatrix} \\begin{pmatrix} \\left(1 - \\sqrt{2}\\right)^n & 0 \\\\ 0 & \\left(1 + \\sqrt{2}\\right)^n \\end{\\pmatrix} \\begin{pmatrix} -2^{\\tiny - \\frac{3}{4}} & \\frac{1}{2} \\sqrt{4 + 3 \\sqrt{2}} \\\\ 2^{\\tiny - \\frac{3}{4}} & \\frac{1}{2} \\sqrt{-4 + 3 \\sqrt{2}} \\end{\\pmatrix} = \\begin{pmatrix} \\frac{ \\left(1 + \\sqrt{2}\\right)^{n+1} - \\left(1 - \\sqrt{2}\\right)^{n+1} }{ 2 \\sqrt{2} } & \\frac{ \\left(1 + \\sqrt{2}\\right)^n - \\left(1 - \\sqrt{2}\\right)^n }{ 2 \\sqrt{2} } \\\\ \\frac{ \\left(1 + \\sqrt{2}\\right)^n - \\left(1 - \\sqrt{2}\\right)^n }{ 2 \\sqrt{2} } & \\frac{ \\left(1 + \\sqrt{2}\\right)^{n-1} - \\left(1 - \\sqrt{2}\\right)^{n-1} }{ 2 \\sqrt{2} } \\end{\\pmatrix}$ And thus: $M^n \\begin{pmatrix} p_{m+1} & q_{m+1} \\\\ p_{m} & q_{m} \\end{\\pmatrix} = \\begin{pmatrix} p_{n+m+1} & q_{n+m+1} \\\\ p_{n+m} & q_{n+m} \\end{\\pmatrix}$ Giving closed form sequences: $p_n = \\frac{ \\left(1 + \\sqrt{2}\\right)^n + \\left(1 - \\sqrt{2}\\right)^n }{ 2 } q_n = \\frac{ \\left(1 + \\sqrt{2}\\right)^n - \\left(1 - \\sqrt{2}\\right)^n }{ 2 \\sqrt{2} } \\frac{p_n}{q_n} = \\sqrt{2} \\frac{ \\left(1 + \\sqrt{2}\\right)^n + \\left(1 - \\sqrt{2}\\right)^n }{ \\left(1 + \\sqrt{2}\\right)^n - \\left(1 - \\sqrt{2}\\right)^n } = \\sqrt{2} \\left(\\frac{2}{1 - \\left(2 \\sqrt{2}-3\\right)^n}-1 \\right)$ Given any positive number $\\epsilon$ we can compute $N = \\left\\lceil \\frac{\\ln \\, \\epsilon \\quad" ]

[ "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "=\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\left [\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}\\end{pmatrix} -\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\right ]+\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\\\ & =\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}-1\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}-2\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}+1\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}+5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{16}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{37}{5}\\end{pmatrix} \\end{align*} Is that correct so far? How could we continue? Klaas van Aarsen MHB Seeker Staff member The matrix for $\\delta$ is correct. But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. mathmari Well-known member MHB Site Helper But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? Klaas van Aarsen MHB Seeker Staff member Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? That point is in between $A$ and $B$. But $C$ is on the circumcircle of triangle $ABZ$, which implies that it cannot be between $A$ and $B$. Klaas van Aarsen MHB Seeker Staff member I've drawn a picture now: \\begin{tikzpicture} \\usetikzlibrary", "All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions. (i) $cos{\\theta}=\\frac{|\\begin{pmatrix}2\\\\-2\\\\1\\end{pmatrix}{\\bullet}\\begin{pmatrix}-6\\\\3\\\\2\\end{pmatrix}|}{\\sqrt{9} \\sqrt{49}}=\\frac{16}{21}$ $\\theta = {40.4}^{\\circ}$ (ii) Using Graphing Calculator, equation of line, $r = \\begin{pmatrix}{-\\frac{1}{6}}\\\\{-\\frac{2}{3}}\\\\0\\end{pmatrix} + \\lambda \\begin{pmatrix}7\\\\10\\\\6\\end{pmatrix}, \\lambda \\in \\mathbb{R}$ (iii) Distance A to $P_1 = \\frac{|\\begin{pmatrix}2\\\\-2\\\\1\\end{pmatrix}{\\bullet}\\begin{pmatrix}4\\\\3\\\\c\\end{pmatrix} - 1|}{\\sqrt{9}} = \\frac{|1+c|}{3}$ Distance A to $P_1 = \\frac{|\\begin{pmatrix}-6\\\\3\\\\2\\end{pmatrix}{\\bullet}\\begin{pmatrix}4\\\\3\\\\c\\end{pmatrix} + 1|}{\\sqrt{49}} = \\frac{|2c-14|}{7}$ $\\frac{|1+c|}{3} = \\frac{|2c-14|}{7}$ $49(1+c)^{2} = 9(2c-14)^{2}$ $c = -49 \\mathrm{~or~} \\frac{35}{13}$", "( 7,5,-1 ) \\\\ &= \\left( \\frac{437}{75}, -\\frac{4}{15}, -\\frac{266}{75} \\right) \\\\ &\\approx \\left( 5.82667, -0.266665, -3.54667 \\right), \\end{align*} and $t$ must equal $$t=\\frac{41}{75} \\approx \\boxed{0.546667}.$$ $\\textbf{Remark}$. It is worth noting that the distance between points $a$ and $v(0.546667)$ is \\begin{align*} ||\\text{ortho}_{u}w || &= || w-\\text{proj}_u w || \\\\ &= \\Bigg|\\Bigg| \\langle 2,7,8\\rangle -\\frac{41}{75} \\langle 7,5,-1 \\rangle \\Bigg|\\Bigg| \\\\ &= \\Bigg|\\Bigg| \\left( -\\frac{137}{75} , \\frac{64}{15}, \\frac{641}{75} \\right) \\Bigg|\\Bigg| \\\\ &= \\sqrt{\\frac{7094}{75}} \\\\ &\\approx 9.72557. \\end{align*} $$v-a = \\begin{pmatrix} 2 \\\\ -3 \\\\ -3 \\end{pmatrix} + \\begin{pmatrix} 7 \\\\ 5 \\\\ -1 \\end{pmatrix} t -\\begin{pmatrix} 4 \\\\ 4 \\\\ 5 \\end{pmatrix} = \\begin{pmatrix} 2+ 7t-4 \\\\-3+ 5t -4\\\\ -3-1t-5 \\end{pmatrix} = \\begin{pmatrix} -2+ 7t \\\\-7+ 5t\\\\ -8-1t \\end{pmatrix}$$ The magnitude of $v-a$ is the square root of $(-2+ 7t)^2 + (-7+ 5t)^2 + (-8-1t)^2$. Do you know how to minimise this expression? Write the vector form of the difference between the two vectors. Then find an expression in $t$ representing the square of the magnitude of the difference vector. If you minimize the square of the distance, you also minimize the distance. Find the value of $t$ which minimizes the square of the distance using the first derivative of the expression. Vector $v$ can be written as $$\\begin{pmatrix} 2 +7t \\\\ -3 +5t\\\\ -3 -1t\\end{pmatrix}$$ Then the distance between $a$", "\\end{pmatrix} \\tag{1}$$ ## Multiplying on the left Step by step (so we can compare with multiplying on the right shortly): $$\\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 5 \\\\ -2 & 3 \\end{pmatrix} = \\begin{pmatrix} 4 & 5 \\\\ 0 & \\frac{11}{2} \\end{pmatrix} \\tag{2}$$ Rescaling, $$\\begin{pmatrix} 1 & 0 \\\\ 0 & \\frac{2}{11} \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 5 \\\\ -2 & 3 \\end{pmatrix} = \\begin{pmatrix} 4 & 5 \\\\ 0 & 1 \\end{pmatrix} \\tag{3}$$ Then $$\\begin{pmatrix} 1 & -5 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & \\frac{2}{11} \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 5 \\\\ -2 & 3 \\end{pmatrix} = \\begin{pmatrix} 4 & 0 \\\\ 0 & 1 \\end{pmatrix} \\tag{4}$$ and finally $$\\begin{pmatrix} \\frac{1}{4} & 0 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & -5 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & \\frac{2}{11} \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 5 \\\\ -2 & 3 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} \\tag{5}$$ Let's then work this the opposite way to see what the row operations above are doing when we multiply on the right instead. ##", "is possible to find A-1. A-1 = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj (A) A = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ $$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\frac 12$$$$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} (\\frac 22) & (\\frac {-4}{2})\\\\ (\\frac {-3}{2}) & (\\frac 72)\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ ∴ A-1 =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ Ans Let: A =$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ and A-1=$$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ We know that, AA-1 = I, where I =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 45 - 45 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 0 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ Taking the corresponding elements both sides", "$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {vmatrix}$$ = -4 - 15 = -19≠ 0 It has a unique solution. X = A-1B A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$ X = A-1B or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -10 - 9\\\\ -25 + 6\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-19}{-19}\\\\ \\frac {-19}{-19}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, 3x - 5y = 3..........................(1) 4x + 3y = 4.........................(2) Given equation in matrix form; $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3\\\\ 4\\\\ \\end {pmatrix}$$ or, AX = B where, A = $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$,", "is $\\frac{\\mu\\vec a + \\lambda\\vec b}{\\lambda + \\mu}$ I assume that the units you're using are metres, so by 1 m away from the first point, you mean 1 unit. The distance between the points is $\\sqrt{51}$. So we need the point that divides the line joining $\\begin{pmatrix}5\\\\0\\\\5\\end{pmatrix}$ to $\\begin{pmatrix}6\\\\5\\\\0\\end{pmatrix}$ in the ratio $1:\\sqrt{51}-1$, which is: $\\dfrac{(\\sqrt{51}-1)\\begin{pmatrix}5\\\\0\\\\5\\end{pmatrix}+1\\begin{pmat rix}6\\\\5\\\\0\\end{pmatrix}}{\\sqrt{51}}$ $=\\dfrac{1}{\\sqrt{51}}\\begin{pmatrix}5\\sqrt{51}+1\\\\ 5\\\\5\\sqrt{51}-5\\end{pmatrix}$ $=\\begin{pmatrix}5+\\frac{1}{\\sqrt{51}}\\\\ \\frac{5}{\\sqrt{51}}\\\\5-\\frac{5}{\\sqrt{51}}\\end{pmatrix}$", "is $$A=\\begin{pmatrix}1&0&1\\\\\\!\\!\\!\\!-1&1&2\\\\0&\\!\\!\\!\\!-1&1\\end{pmatrix}^{-1}=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}$$ Finally, since $$f(x)=3-4x+2x^2\\stackrel{coord. wrt B}\\longrightarrow \\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix} \\,\\,,\\,\\text{we get}$$ $$[Af]=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}\\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix}=\\frac{1}{4}\\begin{pmatrix}11\\\\\\!\\!\\!\\!-7\\\\1\\end{pmatrix}$$ so $$\\,f_C=\\frac{11}{4}(1-x)-\\frac{7}{4}(x-x^2)+\\frac{1}{4}(1+2x+x^2)$$", "equation becomes $\\frac{d}{dt}\\begin{pmatrix}y(t) \\\\ x(t)\\end{pmatrix}= \\begin{pmatrix} 0 & 1 \\\\ - 3 & -4\\end{pmatrix}\\begin{pmatrix}y \\\\ x \\end{pmatrix}$ $= \\begin{pmatrix} x \\\\-3y- 4x\\end{pmatrix}$ Which then reduces to the two scalar equations $\\frac{dy}{dt}= x$ and $\\frac{dx}{dt}= -3y- 4x$ with the intial conditions x(0)= -2, y(0)= -4. Differentiate the first equation twice to get $\\frac{d^2y}{dt^2}= \\frac{dx}{dt}$ and from the second equation that is equal to -3y- 4x: $\\frac{d^2y}{dt^2}= -3y- 4x$. From the first equation $x= \\frac{dy}{dt}$ so we can replace it by that: $\\frac{d^2y}{dt^2}= -3y- 4\\frac{dy}{dt}$ or $\\frac{d^2y}{dt^2}+ 3y+ 4= 0$. Since x(0)= -2 and x= y', y(0)= -4, y'(0)= -2. To find the solution to the intial problem means, of course, to find the vector function $Y(t)= \\begin{pmatrix}y(t) \\\\ x(t)\\end{pmatrix}= \\begin{pmatrix}y(t)\\\\ y'(t)\\end{pmatrix}$ (since x= y'). Since you have found that $y(t)= e^{-3t}+ e^{-t}$ (NOT \" $e^{-3x}+ e^{-x}$\" because y is a function of t, not x!), $x(t)= y'= -3e^{-3t}- e^{-t}$ and $Y(t)= \\begin{pmatrix}y(t)\\\\ x(t)\\end{pmatrix}= \\begin{pmatrix}e^{-3t}+ e^{-t} \\\\ -3e^{-3t}- e^{-t}\\end{pmatrix}$.", "E}$${\\ displaystyle {\\ vec {a}} = {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}}}$ ${\\ displaystyle {\\ vec {OF}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {\\ left [{\\ begin {pmatrix} 8 \\\\ 8 \\ \\ -1 \\ end {pmatrix}} - {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}} \\ right] \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}} {{\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {27} {9}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}}}$. The distance of the point from the plane thus satisfies ${\\ displaystyle d (P; E)}$${\\ displaystyle P}$${\\ displaystyle E}$ ${\\ displaystyle d (P; E) = d (P; F) \\; {\\ widehat {=}} \\; | {\\ vec {PF}} | = \\ left | {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}} -", "form is: $\\boxed{y = - \\frac{1}{4}x - 1}$ Determine the equation of the line... $& \\begin{pmatrix}x_1, & y_1\\\\-3, & 1 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\-3, & -7 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-7-1}{-3--3}\\\\m & = \\frac{-7-1}{-3+3}\\\\m & = \\frac{-8}{0}\\\\m & = \\text{undefined}$ A line that has a slope that is undefined is a line that is parallel to the $y-$axis. The equation of this line is $\\boxed{x = -3}$. $& \\begin{pmatrix}x_1, & y_1\\\\-8, & 4 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\2, & -6 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-6-4}{2--8}\\\\m & = \\frac{-10}{10}\\\\m & = -1\\\\\\\\y & = mx + b\\\\4 & = -1(-8) + b\\\\4 & = 8 + b\\\\4-8 &= 8-8+b\\\\-4 & = b$ $m = -1$ and the $y-$intercept is (0, –4). The equation of this line is: $\\boxed{y = -1x-4}$ $& \\begin{pmatrix}x_1, & y_1\\\\0, & 5 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\4, & -3 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-3-5}{4-0}\\\\m & = \\frac{-8}{4}\\\\m & = -2$ $m = -2$ and the $y-$intercept is (0, 5). The equation of this line is: $\\boxed{y = -2x + 5}$ For each of the following real world problems... $& \\begin{pmatrix}x_1, & y_1\\\\320, & 124 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\600, & 164 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m &", "\\end {pmatrix}$$ Taking the corresponding part of the equal matrix: $$\\frac 1x$$ = $$\\frac 12$$ ∴ x = 2 ∴ y = 3 ∴ x = 2 and y = 3 Ans Here, $$\\frac 4x$$ + 3y = 11.................................(1) 3xy - 8 -5x = 0 or, $$\\frac {3xy}{x}$$ - $$\\frac 8x$$ - $$\\frac {5x}{x}$$ = 0 or, -$$\\frac 8x$$ + 3y = 5...................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {vmatrix}$$ = 4 × 3 - (-8) × 3 = 12 + 24 = 36 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{36}$$$$\\begin {pmatrix} 3 & -3\\\\ 8 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{36} & \\frac {-3}{36}\\\\ \\frac {8}{36} & \\frac {4}{36}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{12} & \\frac {-1}{12}\\\\ \\frac {2}{9} & \\frac {1}{9}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix}", "\\frac{1-\\sqrt{5}}{2} \\end{pmatrix}^{-1}\\begin{pmatrix} 1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2} \\end{pmatrix} \\begin{pmatrix}C_{1}\\\\ C_{2}\\end{pmatrix} = \\begin{pmatrix} 1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2} \\end{pmatrix}^{-1} \\begin{pmatrix}1\\\\ \\frac{-5}{4}\\end{pmatrix} \\begin{pmatrix}C_{1}\\\\ C_{2}\\end{pmatrix} = \\frac{-1}{\\sqrt{5}}\\begin{pmatrix} \\frac{1-\\sqrt{5}}{2} & -1 \\\\ \\frac{-1-\\sqrt{5}}{2} & 1 \\end{pmatrix} \\begin{pmatrix}1\\\\ \\frac{-5}{4}\\end{pmatrix} which gives us C_{1}=\\frac{-2(1-\\sqrt{5})-5}{4\\sqrt{5}}\\ (7) C_{2}=\\frac{2(1+\\sqrt{5})+5}{4\\sqrt{5}}\\ (8) Finally, by plugging (7) and (8) into (3) we get F_{n}=-4\\left (\\left ( \\frac{-2(1-\\sqrt{5})-5}{4\\sqrt{5}} \\right )\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{n}+\\left ( \\frac{2(1+\\sqrt{5})+5}{4\\sqrt{5}} \\right )\\left ( \\frac{1-\\sqrt{5}}{2} \\right )^{n}\\right ) (9) We can now “easily” compute F_{386} which is given by F_{386}=-4\\left (\\left ( \\frac{-2(1-\\sqrt{5})-5}{4\\sqrt{5}} \\right )\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{386}+\\left ( \\frac{2(1+\\sqrt{5})+5}{4\\sqrt{5}} \\right )\\left ( \\frac{1-\\sqrt{5}}{2} \\right )^{386}\\right ) Google returns F_{386} \\cong 5.2783459\\times 10^{80} which is just an approximation, but now we have a formula for the precise answer. If you wish to check for yourself that out formula (9) is indeed correct, click on the terms below to see the computation by Google: F_{0}=-4 F_{1}=5 F_{2}=1 F_{3}=6 F_{4}=7 F_{5}=13 On a side note – Raising a number to the power of 386 may seem like a long computation but it really requires only 10 multiplications! Don’t believe me? Consider this: \\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{386}=\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{256}\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{128}\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{2} To calculate \\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{256} we use the fact that 1.)\\ \\left", "=\\bar{F}\\left ( x_{1},x_{2}\\right ) _{\\substack{x_{1}=0\\\\x_{2}=0}}+\\frac{1}{2}\\begin{pmatrix} x_{1} & x_{2}\\end{pmatrix} \\nabla ^{2}\\bar{F}\\left ( x_{1},x_{2}\\right ) _{_{\\substack{x_{1}=0\\\\x_{2}=0}}}\\begin{pmatrix} x_{1}\\\\ x_{2}\\end{pmatrix} \\\\ & =\\left ( \\frac{1}{2}x_{2}^{2}-\\frac{1}{2}x_{1}^{2}+\\frac{1}{2}x_{1}^{4}\\right ) _{\\substack{x_{1}=0\\\\x_{2}=0}}+\\frac{1}{2}\\begin{pmatrix} x_{1} & x_{2}\\end{pmatrix}\\begin{pmatrix} -1+6x_{1}^{2} & 0\\\\ 0 & 1 \\end{pmatrix} _{\\substack{x_{1}=0\\\\x_{2}=0}}\\begin{pmatrix} x_{1}\\\\ x_{2}\\end{pmatrix} \\\\ & =0+\\frac{1}{2}\\begin{pmatrix} x_{1} & x_{2}\\end{pmatrix}\\begin{pmatrix} -1 & 0\\\\ 0 & 1 \\end{pmatrix}\\begin{pmatrix} x_{1}\\\\ x_{2}\\end{pmatrix} \\\\ & =\\frac{1}{2}\\begin{pmatrix} -x_{1} & x_{2}\\end{pmatrix}\\begin{pmatrix} x_{1}\\\\ x_{2}\\end{pmatrix} \\\\ & =\\frac{1}{2}\\left ( -x_{1}^{2}+x_{2}^{2}\\right ) \\end{align*} Since $$F\\left ( x_{1},x_{2}\\right )$$ is constant, say $$E$$, then the above can be written as\\begin{align*} F\\left ( x_{1},x_{2}\\right ) & =-x_{1}^{2}+x_{2}^{2}\\\\ & =E \\end{align*} Since there is a negative term in the quadratic form above, then the index of this critical point is $$-1$$ which means this is unstable critical point. It is a saddle. (index must be zero for stable critical point) For $$\\left ( 0,\\frac{1}{\\sqrt{2}}\\right )$$: Taylor expansion of $$F\\left ( x_{1},x_{2}\\right )$$ around $$a=\\left ( 0,\\frac{1}{\\sqrt{2}}\\right )$$ gives$F\\left ( x_{1},x_{2}\\right ) =\\bar{F}\\left ( a\\right ) +\\left ( x-a\\right ) \\nabla F\\left ( a\\right ) +\\frac{1}{2}\\left ( x-a\\right ) \\nabla ^{2}F\\left ( a\\right ) \\left ( x-a\\right ) ^{T}$ But $$\\nabla F\\left ( a\\right ) =0$$ since $$a$$ is critical point. The above becomes\\begin{align*} F\\left ( x_{1},x_{2}\\right ) & =\\bar{F}\\left ( x_{1},x_{2}\\right ) _{\\substack{x_{1}=0\\\\x_{2}=\\frac{1}{\\sqrt{2}}}}+\\frac{1}{2}\\begin{pmatrix} x_{1} & x_{2}-\\frac{1}{\\sqrt{2}}\\end{pmatrix} \\nabla ^{2}\\bar{F}\\left ( x_{1},x_{2}\\right ) _{_{\\substack{x_{1}=0\\\\x_{2}=\\frac{1}{\\sqrt{2}}}}}\\begin{pmatrix} x_{1}\\\\ x_{2}-\\frac{1}{\\sqrt{2}}\\end{pmatrix} \\\\ & =\\left (", "1 \\end{pmatrix}$$. ### Solution 1. To find $$P\\begin{pmatrix}X \\leq 1.5\\end{pmatrix}$$, we use write: \\begin{aligned} P\\begin{pmatrix} X \\leq 1.5 \\end{pmatrix} & = \\int_{-\\infty}^{1.5} f(x)dx \\\\ & = \\int_{-\\infty}^{\\frac{3}{2}}f(x)dx \\\\ & = \\int_{-\\infty}^{\\frac{3}{2}}-\\frac{3}{4}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{-\\infty}^{\\frac{3}{2}}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{-\\infty}^{\\frac{3}{2}}\\begin{pmatrix} x^2 - 2x \\end{pmatrix}dx \\\\ & = -\\frac{3}{4} \\begin{bmatrix}\\frac{x^3}{3} - x^2 \\end{bmatrix}_0^{\\frac{3}{2}} \\\\ & = - \\frac{1}{4}\\begin{bmatrix}x^3 - 3x^2 \\end{bmatrix}_0^{\\frac{3}{2}} \\\\ & = - \\frac{1}{4}\\begin{bmatrix}\\begin{pmatrix} \\frac{3}{2}\\end{pmatrix}^3 - 3 \\times \\begin{pmatrix} \\frac{3}{2} \\end{pmatrix}^2 - 0 \\end{bmatrix} \\\\ & = -\\frac{1}{4}\\begin{bmatrix} \\frac{27}{8} - \\frac{27}{4} \\end{bmatrix} \\\\ & = - \\frac{1}{4} \\begin{bmatrix} -\\frac{27}{8} \\end{bmatrix} \\\\ & = \\frac{27}{32} \\\\ P\\begin{pmatrix} X \\leq 1.5 \\end{pmatrix} & = 0.844 \\end{aligned} Graphically, this result can be interpreted as follows: The area enclosed by the probability density function's curve and the horizontal axis, from $$-\\infty$$ upto $$x=1.5$$ is equal to $$0.844$$ (rounded to 3 significant figures). The probability is equal to the area so: $$P\\begin{pmatrix} X \\leq 1.5\\end{pmatrix} = 0.844$$ 2. To find $$P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix}$$ we write: \\begin{aligned} P\\begin{pmatrix}0.5 \\leq X \\leq 1\\end{pmatrix} & = \\int_{0.5}^1f(x)dx \\\\ & = \\int_{0.5}^1 -\\frac{3}{4}x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{0.5}^1 x(x-2)dx \\\\ & = -\\frac{3}{4} \\int_{\\frac{1}{2}}^1 \\begin{pmatrix} x^2-2x \\end{pmatrix} dx \\\\ & = -\\frac{3}{4}\\begin{bmatrix} \\frac{x^3}{3} - x^2 \\end{bmatrix}_{\\frac{1}{2}}^1 \\\\ & = -\\frac{3}{4}\\times \\frac{1}{3}\\begin{bmatrix} x^3 - 3x^2 \\end{bmatrix}_{\\frac{1}{2}}^1 \\\\ & = -\\frac{1}{4}\\begin{bmatrix} x^3 - 3x^2 \\end{bmatrix}_{\\frac{1}{2}}^1", "0\\end{pmatrix}^n \\begin{pmatrix}x \\\\ 1\\end{pmatrix} %= \\left(S J S^{-1} \\right)^n \\begin{pmatrix}x \\\\ 1\\end{pmatrix} %= S \\left(J \\right)^n S^{-1} \\begin{pmatrix}x \\\\ 1\\end{pmatrix} %= \\begin{pmatrix} x-\\sqrt{x^2-1} & x+\\sqrt{x^2-1} \\\\ 1 & 1 \\end{pmatrix} \\begin{pmatrix} x-\\sqrt{x^2-1} & 0 \\\\ 0 & x+\\sqrt{x^2-1} \\end{pmatrix}^n \\begin{pmatrix} -\\frac{1}{2\\sqrt{x^2-1}} & \\frac{x}{2\\sqrt{x^2-1}}+\\frac12 \\\\ \\frac{1}{2\\sqrt{x^2-1}} & \\frac12 - \\frac{x}{2\\sqrt{x^2-1}} \\end{pmatrix} \\begin{pmatrix}x \\\\ 1\\end{pmatrix} \\\\ %= \\begin{pmatrix} x-\\sqrt{x^2-1} & x+\\sqrt{x^2-1} \\\\ 1 & 1 \\end{pmatrix} \\begin{pmatrix} (x-\\sqrt{x^2-1})^n & 0 \\\\ 0 & (x+\\sqrt{x^2-1})^n \\end{pmatrix} \\begin{pmatrix}\\frac12 \\\\ \\frac12\\end{pmatrix} \\\\ %= \\begin{pmatrix} x-\\sqrt{x^2-1} & x+\\sqrt{x^2-1} \\\\ 1 & 1 \\end{pmatrix} \\begin{pmatrix}\\frac12 (x-\\sqrt{x^2-1})^n \\\\ \\frac12 (x+\\sqrt{x^2-1})^n \\end{pmatrix} \\\\ = \\frac12 \\begin{pmatrix} (x-\\sqrt{x^2-1})^{n+1} + (x+\\sqrt{x^2-1})^{n+1} \\\\ (x-\\sqrt{x^2-1})^n + (x+\\sqrt{x^2-1})^n \\end{pmatrix}$$ So what this test boils down to is $$(x-\\sqrt{x^2-1})^n + (x+\\sqrt{x^2-1})^n \\equiv 2x^n \\pmod n$$ Expanding the LHS, $$(x-\\sqrt{x^2-1})^n + (x+\\sqrt{x^2-1})^n = \\sum_{i=0}^n \\binom{n}{i} x^{n-i} \\left( (-\\sqrt{x^2-1})^{i} + (\\sqrt{x^2-1})^{i}\\right) \\\\ = \\sum_{i=0}^n \\binom{n}{i} x^{n-i} (\\sqrt{x^2-1})^{i} \\left( (-1)^i + 1^i\\right) \\\\ = \\sum_{i=0}^{n/2} \\binom{n}{2i} x^{n-2i} (\\sqrt{x^2-1})^{2i} \\\\ = \\sum_{i=0}^{n/2} \\binom{n}{2i} x^{n-2i} (x^2-1)^{i} \\\\ = \\sum_{i=0}^{n/2} \\binom{n}{2i} x^{n-2i} \\sum_{j=0}^i \\binom{i}{j} (x^2)^{i-j} (-1)^j \\\\ = \\sum_{j=0}^{n/2} (-1)^j x^{n-2j} \\sum_{i=j}^{n/2} \\binom{n}{2i} \\binom{i}{j} \\\\$$ So the test is that $$\\forall 0 < j \\le \\tfrac n2: \\sum_{i=j}^{n/2} \\binom{n}{2i} \\binom{i}{j} \\equiv 0 \\pmod n$$ But a direct calculation on that basis is worse than simply verifying that $$\\forall 0 < j \\le \\tfrac n2: \\binom{n}{2j} \\equiv 0 \\pmod n$$", "case 2 If $\\frac{1}{2}k^{2}-\\frac{9}{2}k+10=0$, then $z=t$, a free parameter, which can be any value. Second row in (1) now gives \\begin{align*} \\frac{4}{3}y+\\left( k-\\frac{7}{3}\\right) t & =0\\\\ y & =-\\frac{3}{4}t\\left( k-\\frac{7}{3}\\right) \\end{align*} And first row gives \\begin{align*} 3x+2y+z & =0\\\\ 3x & =-2\\left( -\\frac{3}{4}t\\left( k-\\frac{7}{3}\\right) \\right) -t\\\\ x & =\\frac{1}{2}t\\left( k-3\\right) \\end{align*} Hence in this case, the NullSpace is $$\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} =t \\begin{pmatrix} \\frac{1}{2}\\left( k-3\\right) \\\\ -\\frac{3}{4}\\left( k-\\frac{7}{3}\\right) \\\\ 1 \\end{pmatrix}$$ Since $\\frac{1}{2}k^{2}-\\frac{9}{2}k+10=0$, the solution is $k=5,k=4$. Substituting these values into the above gives the NullSpace as $$\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} = t\\left\\{ \\begin{pmatrix} 1\\\\ -2\\\\ 1 \\end{pmatrix} , \\begin{pmatrix} \\frac{1}{2}\\\\ -\\frac{5}{4}\\\\ 1 \\end{pmatrix} \\right\\}$$ For any $t$. For only $t=0$, this gives the result gives by Mathematica. So Wolfram Mathematica only considered the case 1 above, and did not consider case 2. Wolfram Alpha, which has additional A.I. engine, seems to have looked more into this. • Here's how to use Mathematica for constructing the reduced form: UpperTriangularize[First[LUDecomposition[{{3, 2, 1}, {7, 6, k}, {k, 4, 3}}]]]. Mar 27 '17 at 4:50 • @J.M. Thank you. Mar 27 '17 at 5:04 • @J.M. btw, this only works for square matrices. For non-square matrices, I had to write an M function to do this. Mar 27 '17 at 5:20 • Indeed, it is unfortunate that Mathematica", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "$\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix},\\quad M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x-y=0\\},\\quad N=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, 2x+y=0\\}$ 3. $\\begin{pmatrix} 3 \\\\ 0 \\\\ 1 \\end{pmatrix},\\quad M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, x+y=0\\},\\quad N=\\{c\\cdot\\begin{pmatrix} 1 \\\\ 0 \\\\ 1 \\end{pmatrix}\\,\\big|\\, c\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 2 Find $M^\\perp$. 1. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x+y=0 \\}$ 2. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, -2x+3y=0 \\}$ 3. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x-y=0 \\}$ 4. $M=\\{\\vec{0}\\,\\}$ 5. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x=0 \\}$ 6. $M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, -x+3y+z=0 \\}$ 7. $M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, x=0\\text{ and } y+z=0\\}$ Problem 3 This subsection shows how to project orthogonally in two ways, the method of Example 3.2 and 3.3, and the method of Theorem 3.8. To compare them, consider the plane $P$ specified by $3x+2y-z=0$ in $\\mathbb{R}^3$. 1. Find a basis for $P$. 2. Find $P^\\perp$ and a basis for $P^\\perp$. 3. Represent this vector with respect to the concatenation of the two bases from the prior item. $\\vec{v}=\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ 4. Find the orthogonal projection of $\\vec{v}$ onto $P$ by keeping only the $P$ part from the prior item. 5. Check that against the result from applying Theorem 3.8. This exercise is recommended for all" ]

[ "Express in terms of sine and cosine Definition Rewrite a trigonometric expression where the only trigonometric functions that appear are sines and cosines. Struggling with math? Access detailed step by step solutions to thousands of problems, growing every day!", "# Express in terms of sine and cosine ## Definition Rewrite a trigonometric expression where the only trigonometric functions that appear are sines and cosines. ### Struggling with math? Access detailed step by step solutions to thousands of problems, growing every day!", "# Express function as a trigonometric function of x (@harikasmiley) New Member Joined: 2 years ago Posts: 1 13/05/2019 10:08 am sin(4x)sin⁡(4x) use sin2a=2sinacosasin⁡2a=2sin⁡acos⁡a then sin4x=2sin2xcos2xsin⁡4x=2sin⁡2xcos⁡2x with cos(2x)=cos2(x)sin2(x)cos⁡(2x)=cos2⁡(x)−sin2⁡(x) replace again sin4x=4sinxcosx+cos2(x)sin2(x) Share: Working", "# Express (Sin 67° + Cos 75°) in Terms of Trigonometric Ratios of the Angle Between 0° and 45°. - Mathematics Sum Express (sin 67° + cos 75°) in terms of trigonometric ratios of the angle between 0° and 45°. #### Solution (sin 67° + cos 75°) = (sin (90°23°) + cos (90°15°)) .....( sin(90°θ) = cosθ and cos(90°θ) = sinθ) = (cos 23°+ sin 15°) Concept: Trigonometric Identities Is there an error in this question or solution?", "New Zealand Level 8 - NCEA Level 3 # Sums and differences as products ## Interactive practice questions Use the product-to-sum identities to express $2\\cos\\left(3x\\right)\\cos\\left(9x\\right)$2cos(3x)cos(9x) as a sum or difference of two trigonometric functions. Easy Approx 3 minutes Sign up to try all questions Express $\\cos\\left(3x+2y\\right)\\cos\\left(x-y\\right)$cos(3x+2y)cos(xy) as a sum or difference of two trigonometric functions. Express $2\\sin\\left(3x\\right)\\sin\\left(4x\\right)$2sin(3x)sin(4x) as a sum or difference of two trigonometric functions by using the product-to-sum identities. Express $\\sin\\left(6x\\right)+\\sin\\left(4x\\right)$sin(6x)+sin(4x) as a product of two trigonometric functions. ### Outcomes #### M8-6 Manipulate trigonometric expressions #### 91575 Apply trigonometric methods in solving problems", "NZ Level 8 (NZC) Level 3 (NCEA) [In development] Sums and differences as products ## Interactive practice questions Use the product-to-sum identities to express $2\\cos\\left(3x\\right)\\cos\\left(9x\\right)$2cos(3x)cos(9x) as a sum or difference of two trigonometric functions. Easy Approx 3 minutes Express $\\cos\\left(3x+2y\\right)\\cos\\left(x-y\\right)$cos(3x+2y)cos(xy) as a sum or difference of two trigonometric functions. Express $2\\sin\\left(3x\\right)\\sin\\left(4x\\right)$2sin(3x)sin(4x) as a sum or difference of two trigonometric functions by using the product-to-sum identities. Express $\\sin\\left(6x\\right)+\\sin\\left(4x\\right)$sin(6x)+sin(4x) as a product of two trigonometric functions. ### Outcomes #### M8-6 Manipulate trigonometric expressions #### 91575 Apply trigonometric methods in solving problems", "Express dy/dx in terms of x more calculus resources like can solve a wide range of problems... Begin our exploration of the sine function by phinah [ Solved! ] see how this be! Shall discuss the derivative of cosine ; the antiderivative of the sine and cosine \\cos x.", "## trig question 1) Given that,expressa trigonometric function without radicals. Assume 2)Assuming thatandand that u and v are between 0 andEvaluate sin(u+v) exactly. 3)Given thatand that the terminal side is in quadrant II ,find 4) Use a half-angle identity to evaluateexactly", "of trigonometric identities. Use the vector product for sine-type identities. Tags: None", "Multiply and express the product in simplest form. x - 3 / 2 to the 3rd power * 1 / x- 3 to the 2nd power", "result is returned. # Calculation of expression of the form sin(a-b) The calculator can expand make trigonometric expressions of the form sin(a-b) by giving the results in exact form : thus to expand sin(pi/6-pi/3), enter expand_trigo(sin(pi/6-pi/3)), after calculation, the result is returned. #### Syntax : expand_trigo(expression), where expression is a trigonometric expression #### Examples : Calculate online with expand_trigo (trigonometric expansion)", "Q7. Express $$sin 67° + cos 75°$$ in terms of trigonometric ratios of angles between 0° and 45°. Given, $$sin 67° + cos 75°$$ = $$sin (90° – 23) + cos (90° – 15°)$$ $$= \\ cos 23° + sin 15°$$ [ ∵ $$sin( 90° - \\theta) \\ = \\ cos \\theta$$ and $$cos( 90° - \\theta) \\ = \\ sin \\theta$$", "Express asa function of t 5 cos (StIn (t + 6)) In (t+6) ... 0 6 ti... How do you FOIL (x-10)(x+10)? How do you FOIL (x-10)(x+10)?...", "Examples ##### QUESTION 1 Express $\\sin\\left(\\pi+\\theta\\right)-\\sin\\left(\\pi-\\theta\\right)$sin(π+θ)sin(πθ) in simplest form. ##### QUESTION 2 Rewrite $\\sin18^\\circ23'$sin18°23 in terms of its cofunction. ##### QUESTION 3 Rewrite $\\sec\\frac{\\pi}{9}$secπ9 in terms of its cofunction. ### Outcomes #### M8-6 Manipulate trigonometric expressions #### 91575 Apply trigonometric methods in solving problems", "# Trigonometry33. Express sin 4x in terms of sin x and cos x #### karush ##### Well-known member Express function as a trigonometric function of x $$\\sin(4x)$$ use $\\sin2a=2\\sin a\\cos a$ then $$\\sin4x=2\\sin 2x\\cos 2x$$ with $\\cos(2x) = \\cos^2(x)-\\sin^2(x)$ replace again $$\\sin 4x=4\\sin x\\cos x+\\cos^2(x)-sin^2(x)$$ ok not real sure if this is what they are asking for and if I should go further with it even if the steps are ok #### skeeter ##### Well-known member MHB Math Helper Express function as a trigonometric function of x $$\\sin(4x)$$ use $\\sin2a=2\\sin a\\cos a$ then $$\\sin4x=2\\sin 2x\\cos 2x$$ with $\\cos(2x) = \\cos^2(x)-\\sin^2(x)$ replace again $$\\color{red}{\\sin 4x=4\\sin x\\cos x+\\cos^2(x)-sin^2(x)}$$ ok not real sure if this is what they are asking for and if I should go further with it even if the steps are ok $\\color{red}{\\sin(4x) = 2\\sin(2x)\\cos(2x) = (4\\sin{x}\\cos{x})(\\cos^2{x}-\\sin^2{x})}$", "these two equations, we get, Putting these values of X and Y, in complex form of Z, we get, The value of above expression is known as trigonometrical form of vector. Again we know that, cosθ and sinθ can be represented in exponential form as follows If we put these above exponential form of sinθ and cosθ in the equation Z = |Z|(cosθ + jsinθ) we get, ⇒ Z = |Z|e This is the exponential form of vector. Therefore from all above expressions of vector algebra and vector diagrams, it can be concluded that a vector quantity can be represented as total four basic form as listed below", "Express $x = \\ln\\!\\left(3t\\right)$, $y = 6-t$ in the form $y = f(x)$ by eliminating the parameter. y =", "# How do you express the complex number in trigonometric form 3(cos 90° + i sin 90°)? Your number, which is in trigonometric form, can be expressed in standard or rectangular form, $z = a + i b$, considering that: cos(90°)=0 sin(90°)=1 $z = 3 \\left(0 + 1 i\\right) = 3 i$ a pure immaginary.", "in exact form : thus to expand sin(pi/6-pi/3), enter expand_trigo(sin(pi/6-pi/3)), after calculation, the result is returned. The calculator makes it possible to obtain the trigonometric expansion of an expression. Syntax : expand_trigo(expression), where expression is a trigonometric expression Examples : Calculate online with expand_trigo (trigonometric expansion)", "+ 2(4xy) + (y)2 = 16x2+ 8xy + y2 ### Q.Express 16x2+24xy +9y2 as a perfect square. Solution: 16x2+20xy +16y2 =(4x)2+ 2.4x.3y + (3y)2 = (4x + 3y)2" ]

[ "251$, then $1004 \\mid n(n+1)$, so our answer is $\\boxed{251}$. It should be noted that the product-to-sum rules follow directly from complex trigonometry, so this solution is essentially equivalent to the solution above.", "any of the six boxes, so it is $6^5$ (called product rule in combinatorics). -", "\\boxed{cos} \\ \\boxed{0,4} \\ \\boxed{=}$", "product) where sum = x + y product = x * y", "• $\\cos u \\, \\sin v = \\frac{1}{2}[\\sin(u+v)-\\sin(u-v)]$ #### Examples • Rewrite $\\cos 5x \\, \\sin 4x$ as a sum or difference. • Derive a sum-to-product formula and use it to solve $\\sin 5x + \\sin 3x = 0$.", "sum (-b = 10) and the product (c = 24). They are 4 and 6.", "\\sin 5 \\approx \\boxed{10.46\\, m} {/eq}", "x= \\boxed{\\mathbf{\\pm 4}}$$", "# Application of Product - Sum Find this content useful? ## Application of Product - Sum In response to a reddit question, here is how to integrate $\\sin(x)\\sin(3x)$ using the product-sum formulae.", "write the sum as a … - QuestionCove OpenStudy (anonymous): write the sum as a product: sin5x+sin3x 3 years ago OpenStudy (aum): $$\\sin(A+B) = \\sin(A)\\cos(B) + \\cos(A)\\sin(B) \\\\ \\sin(A-B) = \\sin(A)\\cos(B) - \\cos(A)\\sin(B) \\\\ \\text{Add both sides: } \\\\ \\sin(A+B) + \\sin(A-B) = 2\\sin(A)\\cos(B) \\\\$$ Set A + B = 5x and A - B = 3x. Solve for A and B and plug it into the equation on the last line above. 3 years ago OpenStudy (aum): A+B = 5x --- (1) A-B = 3x --- (2) add both sides: 2A = 8x A = 4x From (1): 4x + B = 5x B = x. sin5x+sin3x = sin(4x+x) + sin(4x-x) From the derivation earlier: sin(A+B) + sin(A-B) = 2sin(A)cos(B) Therefore, sin5x+sin3x = sin(4x+x) + sin(4x-x) = 2sin(4x)cos(x) 3 years ago", "$\\sum_{y\\in\\{2,3\\}}6/5=6/5+6/5$ without adding and subtracting a term for $y=1$.", "\\\\ \\text{Dividing both sides by 2}, \\\\ x= \\boxed{\\mathbf{3}}$$", "a product of \"sum and difference\". Their product, then, is $(x-1)^2- (2i)^2= x^2- 2x+ 1+ 4= x^2- 2x+ 5$.", "figure out the sum of products for the initial heap of 5 or 6 objects?", "limit. In Roman numerals, the only sum-product numbers are 1, 2, 3, and possibly 4 (if written IIII).", "\\sqrt{2} $$\\sqrt{2}$$ \\sum_{i=1}^3 $$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$ There are no comments in this discussion. ×", "\\item $\\sin^2 x+\\sin^2 2x+\\sin^2 3x+\\sin^2 4x=2$ \\item $\\cos 4x+\\cos 2x+\\sqrt2\\cos x=0$ \\item $\\cos 5x+\\cos 3x=\\sin 6x-\\sin 2x$ \\item $\\cos 3x-\\cos 4x+\\cos 5x=0$ \\item $\\sin 7x-\\sin 3x=\\cos 5x$ \\item $cos^2 x-\\sin^2 x=\\sin 3x+cos 4x$ \\item $\\cos 2x-\\cos x=2\\sin^2 \\dfrac{3x}{2}$ \\item $4\\sin 3x+\\sin 5x-2\\sin x\\cos 2x=0$ \\item $\\sin 7x+\\cos 3x+\\sin x=0$ \\item $\\sin 7x\\sin 9x=\\sin5x\\sin 11x$ \\item $\\sin^2 x+\\sin^2 \\dfrac{3x}{2}+\\sin^2 2x+\\sin^2 dfrac{9x}{2}=2$ \\item $sin^4 x+\\cos^2 x=\\cos 4x$ \\item $1+cos x+\\cos 2x+\\cos 3x=0$ \\item $\\sin7x\\cos 3x=\\sin 11x+\\cos 9x$ \\item $sin^2 (2+3x)+\\cos^2 \\Big(\\dfrac{\\pi}{4}+2x\\Big)=\\cos^2 (2-5x)+\\sin^2 \\Big(\\dfrac{\\pi}{4}-6x\\Big)$ \\item $\\cos x+\\cos 3x+\\cos 5x=0$ \\item $\\sin 3x\\sin 9x=\\sin5x\\sin 7x$ \\item $\\cos7x\\cos3x=\\cos14x\\cos10x$ \\item $\\sin11x\\cos6x=\\sin9x\\cos4x$ \\item $\\sin x\\cos2x+\\sin2x\\cos5x=\\sin3x\\cos5x$ \\item $\\sin^22x+\\sin^23x+\\sin^24x+\\sin^29x=2$ \\item $\\cos^3x\\sin3x+\\cos3x\\sin^3x=\\dfrac{3}{4}$ \\item $\\cos^3x\\cos3x+\\sin^3x\\sin3x=\\sin^35x$ \\item $\\cos3x+\\sin7x=2\\sin^2 \\Big(\\dfrac{\\pi}{4}+\\dfrac{5x}{2}\\Big)-2\\cos^2 \\dfrac{9x}{2}$ \\end{enumerate} \\subsection{ Giải các phương trình} \\begin{enumerate}[\\it 1)] \\item $\\tan x+\\tan 2x=\\tan 3x$ \\item $1+\\tan x=(1-\\tan x)(1+\\sin 2x)$ \\item $\\tan x+\\cot 2x=2\\cot 4x$ \\item $\\tan x+\\tan 2x=\\sin 3x\\cos x$ \\item $\\sin x+\\cos x=\\dfrac{\\cos 2x}{1-\\sin 2x}$ \\item $\\dfrac{1}{\\sin 2x}+\\dfrac{1}{\\cos 2x}=\\dfrac{2}{\\sin 4x}$ \\item $\\dfrac{\\sin^3 x+\\cos^3 x}{2\\cos x-\\sin x}=\\cos 2x$ \\item $1+\\sin x-\\cos x-\\sin 2x+2\\cos 2x=0$ \\item $\\sin x-\\dfrac{1}{\\sin x}=\\sin^2 x-\\dfrac{1}{\\sin^2 x}$ \\item $\\cos x\\tan 3x=\\sin 5x$ \\end{enumerate} \\section{CÁC BÀI TOÁN TỔNG HỢP} \\subsection{ Giải các phương trình} \\begin{enumerate}[\\it 1)] \\item $5\\Big(\\sin x+\\dfrac{cos 3x+\\sin 3x}{1+2\\sin 2x}\\Big)=\\cos2x+3$ \\item $\\cot x-1=\\dfrac{\\cos 2x}{1+\\tan x}+\\sin^2x-\\dfrac{1}{2}\\sin2x$ ĐS: $x=\\dfrac{\\pi}{4}+k\\pi$ \\item $\\cos^23x.\\cos2x-\\cos^2x=0$ ĐS:$x=\\dfrac{k\\pi}{2}$ \\item $\\dfrac{2(\\cos^6x+\\sin^6x)-\\sin x\\cos x}{\\sqrt2-2\\sin x}=0$ ĐS: $x=\\dfrac{5\\pi}{4}+k2\\pi$ \\item $(1+\\sin^2x)\\cos x+(1+\\cos^2x)\\sin x=1+\\sin 2x$ ĐS: $x=-\\dfrac{\\pi}{4}+k\\pi; \\dfrac{\\pi}{2}+k2\\pi; k2\\pi$ \\item", "$$\\sqrt{2}$$ \\sum_{i=1}^3 $$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$", "$$\\sqrt{2}$$ \\sum_{i=1}^3 $$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$", "$$\\sqrt{2}$$ \\sum_{i=1}^3 $$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$" ]

[ "# Express function as a trigonometric function of x (@harikasmiley) New Member Joined: 2 years ago Posts: 1 13/05/2019 10:08 am sin(4x)sin⁡(4x) use sin2a=2sinacosasin⁡2a=2sin⁡acos⁡a then sin4x=2sin2xcos2xsin⁡4x=2sin⁡2xcos⁡2x with cos(2x)=cos2(x)sin2(x)cos⁡(2x)=cos2⁡(x)−sin2⁡(x) replace again sin4x=4sinxcosx+cos2(x)sin2(x) Share: Working", "New Zealand Level 8 - NCEA Level 3 # Sums and differences as products ## Interactive practice questions Use the product-to-sum identities to express $2\\cos\\left(3x\\right)\\cos\\left(9x\\right)$2cos(3x)cos(9x) as a sum or difference of two trigonometric functions. Easy Approx 3 minutes Sign up to try all questions Express $\\cos\\left(3x+2y\\right)\\cos\\left(x-y\\right)$cos(3x+2y)cos(xy) as a sum or difference of two trigonometric functions. Express $2\\sin\\left(3x\\right)\\sin\\left(4x\\right)$2sin(3x)sin(4x) as a sum or difference of two trigonometric functions by using the product-to-sum identities. Express $\\sin\\left(6x\\right)+\\sin\\left(4x\\right)$sin(6x)+sin(4x) as a product of two trigonometric functions. ### Outcomes #### M8-6 Manipulate trigonometric expressions #### 91575 Apply trigonometric methods in solving problems", "# Express (1-sin^2x)/cosx in terms of sine ratio of x and / or its multiples? Jan 3, 2017 #### Explanation: $\\frac{1 - {\\sin}^{2} x}{\\cos x}$ = ${\\cos}^{2} \\frac{x}{\\cos} x$ = $\\cos x$ = $\\frac{2 \\sin x \\cos x}{2 \\sin x}$ = $\\frac{\\sin 2 x}{2 \\sin x}$", "NZ Level 8 (NZC) Level 3 (NCEA) [In development] Sums and differences as products ## Interactive practice questions Use the product-to-sum identities to express $2\\cos\\left(3x\\right)\\cos\\left(9x\\right)$2cos(3x)cos(9x) as a sum or difference of two trigonometric functions. Easy Approx 3 minutes Express $\\cos\\left(3x+2y\\right)\\cos\\left(x-y\\right)$cos(3x+2y)cos(xy) as a sum or difference of two trigonometric functions. Express $2\\sin\\left(3x\\right)\\sin\\left(4x\\right)$2sin(3x)sin(4x) as a sum or difference of two trigonometric functions by using the product-to-sum identities. Express $\\sin\\left(6x\\right)+\\sin\\left(4x\\right)$sin(6x)+sin(4x) as a product of two trigonometric functions. ### Outcomes #### M8-6 Manipulate trigonometric expressions #### 91575 Apply trigonometric methods in solving problems", "Express in terms of sine and cosine Definition Rewrite a trigonometric expression where the only trigonometric functions that appear are sines and cosines. Struggling with math? Access detailed step by step solutions to thousands of problems, growing every day!", "$v$ and $\\rho$ and then express $\\rho$ in terms of $z$.)", "Express asa function of t 5 cos (StIn (t + 6)) In (t+6) ... 0 6 ti... How do you FOIL (x-10)(x+10)? How do you FOIL (x-10)(x+10)?...", "# Express in terms of sine and cosine ## Definition Rewrite a trigonometric expression where the only trigonometric functions that appear are sines and cosines. ### Struggling with math? Access detailed step by step solutions to thousands of problems, growing every day!", "## Product to Sum Identity Use product to sum identity to rewrite the expression as the sum or difference of two functions $$cos(5\\pi/3)cos(\\pi/3)$$", "# Integrating Trigonometric functions (part 5) This will be about integrating product of trigonometric functions with no powers involved for example, or . The simple trick here is to use the product to sum to formulas that can be found here. For convenience, I’ll insert them here. So very conveniently, . Now you see the usefulness of the product to sum formula! Better learn how to find them! For trigonometric functions with powers involved, you can refer to part 1 to part 5.", "Express dy/dx in terms of x more calculus resources like can solve a wide range of problems... Begin our exploration of the sine function by phinah [ Solved! ] see how this be! Shall discuss the derivative of cosine ; the antiderivative of the sine and cosine \\cos x.", "Multiply and express the product in simplest form. x - 3 / 2 to the 3rd power * 1 / x- 3 to the 2nd power", "\\cos4\\theta}{2}$ ### Vocabulary Language: English Product to Sum Formula Product to Sum Formula A product to sum formula relates the product of two trigonometric functions to the sum of two trigonometric functions. Sep 26, 2012 Feb 26, 2015", "# Trigonometry33. Express sin 4x in terms of sin x and cos x #### karush ##### Well-known member Express function as a trigonometric function of x $$\\sin(4x)$$ use $\\sin2a=2\\sin a\\cos a$ then $$\\sin4x=2\\sin 2x\\cos 2x$$ with $\\cos(2x) = \\cos^2(x)-\\sin^2(x)$ replace again $$\\sin 4x=4\\sin x\\cos x+\\cos^2(x)-sin^2(x)$$ ok not real sure if this is what they are asking for and if I should go further with it even if the steps are ok #### skeeter ##### Well-known member MHB Math Helper Express function as a trigonometric function of x $$\\sin(4x)$$ use $\\sin2a=2\\sin a\\cos a$ then $$\\sin4x=2\\sin 2x\\cos 2x$$ with $\\cos(2x) = \\cos^2(x)-\\sin^2(x)$ replace again $$\\color{red}{\\sin 4x=4\\sin x\\cos x+\\cos^2(x)-sin^2(x)}$$ ok not real sure if this is what they are asking for and if I should go further with it even if the steps are ok $\\color{red}{\\sin(4x) = 2\\sin(2x)\\cos(2x) = (4\\sin{x}\\cos{x})(\\cos^2{x}-\\sin^2{x})}$", "sin\\ \\frac{5x}{2}\\end{array}$$ Example 2: Express 6 cos x cos 2x in terms of sum function. Solution: Consider, 6 cos x cos 2x = 3 [2 cos x cos 2x] Using the formula 2 cos A cos B = cos (A + B) + cos (A – B), = 3[cos (x + 2x) + cos (x – 2x)] = 3[cos 3x + cos (-x)] = 3 [cos 3x + cos x] To learn other trigonometric formulas Register yourself at BYJU’S.", "Geometry Sum to Product Trigonometric Identities Which of the following is equal to $\\cos114^\\circ+\\cos44^\\circ?$ Which of the following is equal to $\\cos88^\\circ-\\cos42^\\circ?$ Which of the following is equal to $\\sin60^\\circ+\\sin16^\\circ?$ Evaluate $16\\sin120^\\circ+8\\cos105^\\circ+16\\sin60^\\circ+8\\cos15^\\circ.$ Which of the following expressions is equal to $\\sin{36 ^\\circ} + \\sin{24 ^\\circ} ?$ × Problem Loading... Note Loading... Set Loading...", "# Application of Product - Sum Find this content useful? ## Application of Product - Sum In response to a reddit question, here is how to integrate $\\sin(x)\\sin(3x)$ using the product-sum formulae.", "How to turn this sum to a product? I arrived at this sum or trigonometric functions that I need to turn into a product in order to continue the exercise. How do I do that? $$\\sin{x}\\cos{(x+y)} -\\cos{x}\\sin{(x+y)}$$ I know of the identities of $\\cos{(a \\pm b)}$ and $\\sin{(a \\pm b)}$, but I can't figure the solution out. - Letting $a=x$ and $b=x+y$, you have $$\\sin a\\cos b-\\cos a\\sin b=\\sin(a-b).$$", "• anonymous Express the sum or difference as a product. sin 4x - sin 6x 2 cos 4x cos 5x 2 sin 5x cos x -2 sin x cos 5x -2 sin x Mathematics Looking for something else? Not the answer you are looking for? Search for more explanations.", "Courses Courses for Kids Free study material Free LIVE classes More # Sum to Product Formulas ## Introduction to Sum to Product Formulas Last updated date: 28th Mar 2023 Total views: 26.4k Views today: 1.09k In trigonometry, the sum-to-product formula is a set of formulas that are used to express the sine and cosine functions' sum, difference, and products. These formulas allow us to express the sine and cosine trigonometric functions as the sum or difference of their products, which simplifies mathematical problems. By substituting the variables, it is possible to derive the sum-to-product formulas from the product-to-sum formulas in trigonometry. In this article, we will thoroughly examine the sum-to-product formula and show how to obtain it from the product-to-sum formulas. ## What are the Sum-to-Product Formulas? The Sum-to-product formula is used to express the sum or difference of a sine function and the sum or difference of a cosine function as the product of a sine and cosine function. By applying this formula, the sum or difference of trigonometric functions of sine and cosine can be expressed as a product, simplifying mathematical problems. The product for sum formula in trigonometry can be used to obtain the sum for product formula with a variable change. ## List of the Sum-to-Product Formula Following is the list of the sum-to-product" ]

[ "the point $$(1,2,3)$$ can be written as $$\\frac{x-1}{0} = \\frac{y-2}{0} = \\frac{z-3}{2} = t$$, where $$t \\in R$$ is any parameter. The parametric form the line (especially for x,y coordinates) can be obtained by \"cross multiplying\" their respective DR with the parameter equation to give $$(x,y,z)=(1,2,2t+3)$$.", "\\frac{4}{3} x_3$ • $x_2 = -1 + \\frac{2}{3} x_3$ To make it more explicit, I converted back the first line to $x_1 + 0x_2 + \\frac{4}{3} x_3 = 2$ which means that $x_1 = 2 - \\frac{4}{3} x_3$. $x_3$ is called a parameter.", "point to find the single parameter determining the line $\\frac{a}{b} = -\\frac{y_1}{x_1} = -\\frac{y_2}{x_2} \\ .$ Last edited by a moderator: May 5, 2017", "Sep 17 at 21:25 Now, try the parameterization $(x,y)=(1+u^3,-1+u)$, when $u\\to 0$. Spoiler below: place your mouse on it to reveal the text. $$f(x,y) = \\frac{(x-1)(y+1)^3}{(x-1)^2+(y+1)^6}= \\frac{u^3\\cdot u^3}{u^6+u^6} = \\frac{1}{2}\\xrightarrow[u\\to0]{} \\frac{1}{2}.$$ The idea? We want to chose $a,b>0$ such that $x=1+u^a$, $y=-1+u^b$ lead to nice simplifications in $f(x,y)$, with a limit that is not $0$. The exponents above do that.", "How to draw the figure of $\\frac{x-1}{2}=\\frac{y-1}{3}=\\frac{z-2}{3}$-type straight line Discussion in 'Mathematics' started by Ordinary users68, Aug 2, 2020 at 2:25 AM. 1. Ordinary users68Guest I'd like to draw the straight line that is expressed in symmetrical form directly: $\\frac{x-1}{2}=\\frac{y-1}{3}=\\frac{z-2}{3}$ ContourPlot3D[(x - 1)/2 == (y - 1)/3 == (z - 2)/3, {x, -5, 5}, {y, -5, 5}, {z, -5, 5}, BoxRatios -> Automatic] But I can't get the correct result with the above code. How can I draw this line directly? In addition, $\\frac{x-1}{2}=\\frac{y-1}{3}=2$ should represent a vertical line, which I hope can be drawn by built-in or custom functions.", "equation is $y + 13 = - \\frac{1}{24} \\left(x + 1\\right)$ Graphed are the function and its normal line: graph{(2x^3-8x^2+2x-1-y)(y+13+(x+1)/24)=0 [-5, 7, -18.16, 2.12]}", "is substitute x=t into the equation for our straight line to obtain $$y = 2t - \\frac{1}{5}$$ Hence we have parameterise our path $$\\gamma = \\left\\{\\begin{array}{l} x = t \\\\ y = 2t - \\frac{1}{5}\\end{array}\\right.\\hspace{1cm}\\frac{1}{10}\\leq t \\leq \\frac{3}{10}$$ Do you follow?", "-2x_3, x_3)= (-1, -2, 1)x_3$, for any value of $x_3$. Since $x_1$ and $x_2$ depend linearly on the single parameter $x_3$, that is a line in three space. Of course, if we multiply all of $x_1$, $x_2$, and $x_3$ by 4 (chosen for the \"4\" under the $x_3$ in the original given solution) we have $(4x_1, -8x_2, 4x_3)= (-4, -8, 4)x_3$ is also on that line. And, taking $\\frac{x_3}{4}= t$ as parameter, so that $x_3= 4t$, we have $x_1= -x_3= -4t$, $x_2= -2x_3= -8t$. Then $t= \\frac{x_1}{-4}= \\frac{x_2}{-8}= \\frac{x_3}{4}$", "Try to do this integration and we are half way done !!! On integrating both side w.r.t $y$ we get : $f(y)=y-\\frac25y^{\\frac52}+c$, (where $c$ is a integrating constant.) Now we find the value to $c$ We know $f(1)=0$ $\\Rightarrow c=-\\frac35$ i.e., $f(y)=y-\\frac25y^{\\frac52}-\\frac35$ Can you find the answer now ? Now simply, putting $y=4$ we get $f(4)=4-\\frac25(4)^{\\frac52}-\\frac35 \\\\=4-\\frac{64}{5}-\\frac35 \\\\= \\frac{20-67}{5} \\\\= -\\frac{47}{5}$ This site uses Akismet to reduce spam. Learn how your comment data is processed.", "parameter $\\frac{1}{2}$ and order parameter $\\frac{r}{2}$.", "and $g_{\\theta}$ is parameterized by $\\theta$ such that $X = g_{\\theta} \\circ Y$ then we can find $\\frac{\\partial I_Y^{MC}}{\\partial \\theta}$. This is often used in variational autoencoders (missing reference) discussed here. References [back]", "the equation for x is now changed to the parameter $a$ and inserted into the second one. $x=-\\frac{a}2\\quad|\\cdot(-2)$ $a=-2x$ $y=-\\frac{a^2}4$ $y=-\\frac{(-2x)^2}4$ $=-\\frac{4x^2}4$ $=-x^2$ function of all minima: $y=-x^2$", "# Thread: Parameterization 1. ## Parameterization I have a problem that seems easy enough but i cant seem to find x and y Problem: find a Parametrization for the graph of the function $y = x^2$ Starting at (3,9) and moving towards the origin,using angle $\\theta$ in the accompanying figure as a parameter. I drew out the graph and took a picture of it and i will include it. I think i use $Tan \\theta = \\frac{y}{x} = \\frac{9}{3} = 3$ Im not sure how to finish solving for x then to solve for y? Im not sure if i need to find $y = mx + b$ since the slop comes out as zero? Thank you.. 2. ## Re: Parameterization Originally Posted by icelated I have a problem that seems easy enough but i cant seem to find x and y Problem: find a Parametrization for the graph of the function $y = x^2$ Starting at (3,9) and moving towards the origin,using angle $\\theta$ in the accompanying figure as a parameter. I drew out the graph and took a picture of it and i will include it. I think i use $Tan \\theta = \\frac{y}{x} = \\frac{9}{3} = 3$ Im not sure how to finish solving for x then to solve for y? Im not sure if", "### Home > CCA2 > Chapter 8 > Lesson 8.2.3 > Problem8-112 8-112. Consider the functions $y =\\frac { 1 } { 2 }$ and $y =\\frac { 16 } { x ^ { 2 } - 4 }$. Find the coordinates where the graphs of the functions intersect. $\\frac{1}{2}=\\frac{16}{x^2-4}$ $\\left(\\pm6, \\frac{1}{2}\\right)$", "x=-5+\\frac{3}{7}z$ , and thus the intersection line is given by $\\left\\{\\begin{pmatrix}-5+\\frac{3}{7}t\\\\{}\\\\-1-\\frac{2}{7}t\\\\{}\\\\t\\end{pmatrix}\\,,\\,t\\in\\mathbb{R }\\right\\}$ . The above is already a parametrization, but your teacher may want you to write it as $u+tv\\,,\\,u,v$ vectors on the line, $t=$ the real parameter. Tonio", "A sample function $$f$$ of this type is the function $$\\displaystyle f(t)=\\frac{10}{1+50\\boldcdot 2^{\\text-t}}.$$ Evaluate this function for integer values of $$t$$ between 0 and 15. Describe qualitatively how this function differs from an exponential one. What happens to the world population in the long run according to this model?", "to the one of your original line: $- 3$ $s l o p e = \\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$ $- 3 = \\frac{y - 2}{x - 7}$ $- 3 x + 21 = y - 2$ and finally your line: $y + 3 x = 23$", "function f with the given values. Question 25. Answer: Question 26. Answer: A function f(x) = $$\\frac { 12 }{ 2 }$$x Explanation: The points are (-4, -2) and (-2, -1) Slope m = $$\\frac { -1 + 2 }{ -2 + 4 }$$ = $$\\frac { 1 }{ 2 }$$ y = mx + b y = $$\\frac { 12 }{ 2 }$$x + b The line passes through the point (0, 0) 0 = $$\\frac { 12 }{ 2 }$$(0) + b b = 0 So, y = $$\\frac { 12 }{ 2 }$$x + 0 y = $$\\frac { 12 }{ 2 }$$x A function f(x) = $$\\frac { 12 }{ 2 }$$x Question 27. ERROR ANALYSIS Describe and correct the error in writing an equation of the line with a slope of 2 and a y-intercept of 7. Answer: Question 28. ERROR ANALYSIS Describe and correct the error in writing an equation of the line shown. Answer: Slope = $$\\frac { 1 – 4 }{ 5 – 0 }$$ = $$\\frac { -3 }{ 5 }$$ y = $$\\frac { -3 }{ 5 }$$x + 4 Question 29. MODELING WITH MATHEMATICS In 1960, the world record for the men’s mile was 3.91 minutes. In 1980, the record time was 3.81 minutes. a. Write a", "$x\\left(y\\right)=3\\mathrm{log}\\left(y\\right)+y$ 33. $x\\left(y\\right)=\\sqrt{y}+2y$ For the following exercises, parameterize (write parametric equations for) each Cartesian equation by using $x\\left(t\\right)=a\\cos t$ and $y\\left(t\\right)=b\\sin t$. Identify the curve. 34. $\\frac{{x}^{2}}{4}+\\frac{{y}^{2}}{9}=1$ 35. $\\frac{{x}^{2}}{16}+\\frac{{y}^{2}}{36}=1$ 36. ${x}^{2}+{y}^{2}=16$ 37. ${x}^{2}+{y}^{2}=10$ 38. Parameterize the line from $\\left(3,0\\right)$ to $\\left(-2,-5\\right)$ so that the line is at $\\left(3,0\\right)$ at $t=0$, and at $\\left(-2,-5\\right)$ at $t=1$. 39. Parameterize the line from $\\left(-1,0\\right)$ to $\\left(3,-2\\right)$ so that the line is at $\\left(-1,0\\right)$ at $t=0$, and at $\\left(3,-2\\right)$ at $t=1$. 40. Parameterize the line from $\\left(-1,5\\right)$ to $\\left(2,3\\right)$ so that the line is at $\\left(-1,5\\right)$ at $t=0$, and at $\\left(2,3\\right)$ at $t=1$. 41. Parameterize the line from $\\left(4,1\\right)$ to $\\left(6,-2\\right)$ so that the line is at $\\left(4,1\\right)$ at $t=0$, and at $\\left(6,-2\\right)$ at $t=1$. For the following exercises, use the table feature in the graphing calculator to determine whether the graphs intersect. 42. $\\begin{cases}{x}_{1}\\left(t\\right)=3t\\hfill \\\\ {y}_{1}\\left(t\\right)=2t - 1\\hfill \\end{cases}\\text{ and }\\begin{cases}{x}_{2}\\left(t\\right)=t+3\\hfill \\\\ {y}_{2}\\left(t\\right)=4t - 4\\hfill \\end{cases}$ 43. $\\begin{cases}{x}_{1}\\left(t\\right)={t}^{2}\\hfill \\\\ {y}_{1}\\left(t\\right)=2t - 1\\hfill \\end{cases}\\text{ and }\\begin{cases}{x}_{2}\\left(t\\right)=-t+6\\hfill \\\\ {y}_{2}\\left(t\\right)=t+1\\hfill \\end{cases}$ For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations. 44. $\\begin{cases}{x}_{1}\\left(t\\right)=3{t}^{2}-3t+7\\hfill \\\\ {y}_{1}\\left(t\\right)=2t+3\\hfill \\end{cases}$ $t$ $x$ $y$ –1 0 1 45. $\\begin{cases}{x}_{1}\\left(t\\right)={t}^{2}-4\\hfill \\\\ {y}_{1}\\left(t\\right)=2{t}^{2}-1\\hfill \\end{cases}$ $t$ $x$ $y$ 1 2 3 46. $\\begin{cases}{x}_{1}\\left(t\\right)={t}^{4}\\hfill \\\\ {y}_{1}\\left(t\\right)={t}^{3}+4\\hfill \\end{cases}$ $t$ $x$ $y$ -1 0 1", "Consider the function $f(x)$, for which $f(0) = 3$ and $f'(0) = -7$. Find $h'(0)$ for the function $h(x) = \\frac{1}{f(x)}$. $h'(0)$ =" ]

[ "$y = 11$. This does work. We found one valid solution so the answer is $\\boxed{\\textbf{(B) }1}$.", "\\boxed{x=1+18t, y=1-7t, z=1+t, t\\in\\mathbf{R}}. {/eq}", "= 12$ Thus we have: $x' = - \\frac {(12)(-3)}{5} \\implies \\boxed { x' = 7.2 }$ Note, $x' = \\frac {dx}{dt}$ and $y' = \\frac {dy}{dt}$", "Subtracting $\\frac 12$ from $h$ gives us $$x = h-\\frac 12 = \\frac 1{16},$$ so our answer is $\\boxed{\\text{E}}$.", "be a multiple of 37. So, x can be 21x+1=148, 21x=147 and x=7. Substituting it back, we see that $$y=\\frac{-1147}{37}=-31.$$ Thus, the answer is $$\\boxed{(7,-31)}.$$ . Dec 23, 2018", "all the sums. X + 3X + 5X + 15X = 120, so X = 5. Since we want to find $GFDB$, we substitute 5 for 15X to get $\\boxed{75}$. $\\sim$krishkhushi09 ~ dolphin7 ~ pi_is_3.14", "$\\dfrac{x+y}{2}$, the desired answer is $\\boxed{450}$.", "Solving for $x$, we get $x = 6 + {\\sqrt 43}$ which is between $12$ and $13$, so the answer is $\\boxed{012}$.", "solved to find that $f(4) = \\boxed{\\frac{5}{8}}$.", "can take the 7th root on both sides to get $\\frac{2^{36}}{x} = 1$. Multiplying both sides by $x$ gives $2^{36} = x$. We know that $2^m = 2^{36}$, which means that $m = \\boxed{36}$. ~hh99754539", "the answer is $3 + 11 = \\boxed{\\mathrm{(A)}\\ 14}$", "$(12*5)/2 = (13*r)/2 +(5*r)/2$. So it gives us $r = \\boxed{\\textbf{(D)}\\ \\frac{10}{3}}$.----LarryFlora ## Video Solution https://youtu.be/3VjySNobXLI - Happytwin https://youtu.be/KtmLUlCpj-I - savannahsolver https://youtu.be/FDgcLW4frg8?t=3837 - pi_is_3.14", "$(2,1)$ 15. $y-1=\\frac{3}{4}(x+2)$", "that $z = (\\frac{i}{-2\\sqrt{10}})$. Solving for $y$ (using $y + \\frac{1}{y} = z$) we get that $y = \\frac{-i \\pm \\sqrt{161}i}{4\\sqrt{10}}$, and multiplying this by $-(\\frac{i}{\\sqrt{10}})$ (because $x = -(\\frac{i}{\\sqrt{10}})y$) we get that $x = \\frac{-1 \\pm \\sqrt{161}}{40}$ for a final answer of $-1 + 161 + 40 = \\boxed{200}$ -Grizzy", "x= \\boxed{\\mathbf{\\pm 4}}$$", "$x_2$ in terms of $t$: $$-2x_2+2(13-4t)-7t=-19\\Rightarrow -2x_2-15t + 26 = -19$$ and finally $$x_2 = \\frac{1}{2}(45-15 t) = \\frac{15}{2}(3-t)$$ Step 4: Just rewriting the solutions we have that they are determined by $$x_1 = 13-4t;\\quad x_2 = \\frac{15}{2}(3-t);\\quad x_3=t.$$", "$$2(\\frac{7-2y}{3}) + 3y = 8$$ We can now easily solve for $$y$$.", "then $$16^{3/4}=(16^{\\frac{1}{4}})^{3}=(2)^3=\\boxed{8}$$ - Have you read the other answers? – Servaes May 18 '14 at 16:09", "...\\ + {{n+r}\\choose {n+r-1}}\\right] + \\left[{{n+r}\\choose {n+r}}\\right]$ Note that 1st box bracket = 1 2nd box bracket = 1980 ---- {from (A)} 3rd box bracket = 66 ---- {from (B)} 4th box bracket = 1 $\\implies\\ 2^{n+r} = 1 + 1980 + 66 + 1 = 2048$ $\\implies\\ 2^11 = 2048$ $\\implies\\ n + r = 11$ Now solving (A) & (B) with $n+r = 11$, we find that, $\\boxed{n = 8\\ \\&\\ r = 3}$", "values of x and y. The number of erasers at the end will be = y - 15 = 108 - 15 = 93 The number of erasers in the box at the end is 93. Aug 11, 2021 #2 +781 0 $p = \\frac{3}{4} e$ $p + 12 = e - 15$ $p = e-27$ $\\frac{3}{4}e = e - 27$ $\\frac{1}{4}e = \\frac{1}{3}e - 9$ $e = \\frac{4}{3}e - 36$ $\\frac{1}{3}e = 36$ $e = 108$ $e-15 = 108 \\Rightarrow e = \\boxed{93}$ Aug 11, 2021" ]

[ "the point $$(1,2,3)$$ can be written as $$\\frac{x-1}{0} = \\frac{y-2}{0} = \\frac{z-3}{2} = t$$, where $$t \\in R$$ is any parameter. The parametric form the line (especially for x,y coordinates) can be obtained by \"cross multiplying\" their respective DR with the parameter equation to give $$(x,y,z)=(1,2,2t+3)$$.", "How to draw the figure of $\\frac{x-1}{2}=\\frac{y-1}{3}=\\frac{z-2}{3}$-type straight line Discussion in 'Mathematics' started by Ordinary users68, Aug 2, 2020 at 2:25 AM. 1. Ordinary users68Guest I'd like to draw the straight line that is expressed in symmetrical form directly: $\\frac{x-1}{2}=\\frac{y-1}{3}=\\frac{z-2}{3}$ ContourPlot3D[(x - 1)/2 == (y - 1)/3 == (z - 2)/3, {x, -5, 5}, {y, -5, 5}, {z, -5, 5}, BoxRatios -> Automatic] But I can't get the correct result with the above code. How can I draw this line directly? In addition, $\\frac{x-1}{2}=\\frac{y-1}{3}=2$ should represent a vertical line, which I hope can be drawn by built-in or custom functions.", "= m(x - x_1)$ For line L1 $m = \\dfrac{-4 - \\frac{8}{5}}{7 - \\frac{14}{5}} = -\\dfrac{4}{3}$ $y - \\frac{8}{5} = -\\frac{4}{3}(x - \\frac{14}{5})$ $15y - 24 = -20(x - \\frac{14}{5})$ $15y - 24 = -20x + 56$ $20x + 15y - 80 = 0$ $4x + 3y - 16 = 0$ answer For line L2 $m = \\dfrac{-4 - \\frac{1}{5}}{7 - \\frac{7}{5}} = -\\dfrac{3}{4}$ $y - \\frac{1}{5} = -\\frac{3}{4}(x - \\frac{7}{5})$ $20y - 4 = -15(x - \\frac{7}{5})$ $20y - 4 = -15x + 21$ $15x + 20y - 25 = 0$ $3x + 4y - 5 = 0$ answer Manoj Karmakar I am happy to register this. Subscribe to MATHalino on", "Try to do this integration and we are half way done !!! On integrating both side w.r.t $y$ we get : $f(y)=y-\\frac25y^{\\frac52}+c$, (where $c$ is a integrating constant.) Now we find the value to $c$ We know $f(1)=0$ $\\Rightarrow c=-\\frac35$ i.e., $f(y)=y-\\frac25y^{\\frac52}-\\frac35$ Can you find the answer now ? Now simply, putting $y=4$ we get $f(4)=4-\\frac25(4)^{\\frac52}-\\frac35 \\\\=4-\\frac{64}{5}-\\frac35 \\\\= \\frac{20-67}{5} \\\\= -\\frac{47}{5}$ This site uses Akismet to reduce spam. Learn how your comment data is processed.", "-2x_3, x_3)= (-1, -2, 1)x_3$, for any value of $x_3$. Since $x_1$ and $x_2$ depend linearly on the single parameter $x_3$, that is a line in three space. Of course, if we multiply all of $x_1$, $x_2$, and $x_3$ by 4 (chosen for the \"4\" under the $x_3$ in the original given solution) we have $(4x_1, -8x_2, 4x_3)= (-4, -8, 4)x_3$ is also on that line. And, taking $\\frac{x_3}{4}= t$ as parameter, so that $x_3= 4t$, we have $x_1= -x_3= -4t$, $x_2= -2x_3= -8t$. Then $t= \\frac{x_1}{-4}= \\frac{x_2}{-8}= \\frac{x_3}{4}$", "Sep 17 at 21:25 Now, try the parameterization $(x,y)=(1+u^3,-1+u)$, when $u\\to 0$. Spoiler below: place your mouse on it to reveal the text. $$f(x,y) = \\frac{(x-1)(y+1)^3}{(x-1)^2+(y+1)^6}= \\frac{u^3\\cdot u^3}{u^6+u^6} = \\frac{1}{2}\\xrightarrow[u\\to0]{} \\frac{1}{2}.$$ The idea? We want to chose $a,b>0$ such that $x=1+u^a$, $y=-1+u^b$ lead to nice simplifications in $f(x,y)$, with a limit that is not $0$. The exponents above do that.", "\\frac{4}{3} x_3$ • $x_2 = -1 + \\frac{2}{3} x_3$ To make it more explicit, I converted back the first line to $x_1 + 0x_2 + \\frac{4}{3} x_3 = 2$ which means that $x_1 = 2 - \\frac{4}{3} x_3$. $x_3$ is called a parameter.", "$x\\left(y\\right)=3\\mathrm{log}\\left(y\\right)+y$ 33. $x\\left(y\\right)=\\sqrt{y}+2y$ For the following exercises, parameterize (write parametric equations for) each Cartesian equation by using $x\\left(t\\right)=a\\cos t$ and $y\\left(t\\right)=b\\sin t$. Identify the curve. 34. $\\frac{{x}^{2}}{4}+\\frac{{y}^{2}}{9}=1$ 35. $\\frac{{x}^{2}}{16}+\\frac{{y}^{2}}{36}=1$ 36. ${x}^{2}+{y}^{2}=16$ 37. ${x}^{2}+{y}^{2}=10$ 38. Parameterize the line from $\\left(3,0\\right)$ to $\\left(-2,-5\\right)$ so that the line is at $\\left(3,0\\right)$ at $t=0$, and at $\\left(-2,-5\\right)$ at $t=1$. 39. Parameterize the line from $\\left(-1,0\\right)$ to $\\left(3,-2\\right)$ so that the line is at $\\left(-1,0\\right)$ at $t=0$, and at $\\left(3,-2\\right)$ at $t=1$. 40. Parameterize the line from $\\left(-1,5\\right)$ to $\\left(2,3\\right)$ so that the line is at $\\left(-1,5\\right)$ at $t=0$, and at $\\left(2,3\\right)$ at $t=1$. 41. Parameterize the line from $\\left(4,1\\right)$ to $\\left(6,-2\\right)$ so that the line is at $\\left(4,1\\right)$ at $t=0$, and at $\\left(6,-2\\right)$ at $t=1$. For the following exercises, use the table feature in the graphing calculator to determine whether the graphs intersect. 42. $\\begin{cases}{x}_{1}\\left(t\\right)=3t\\hfill \\\\ {y}_{1}\\left(t\\right)=2t - 1\\hfill \\end{cases}\\text{ and }\\begin{cases}{x}_{2}\\left(t\\right)=t+3\\hfill \\\\ {y}_{2}\\left(t\\right)=4t - 4\\hfill \\end{cases}$ 43. $\\begin{cases}{x}_{1}\\left(t\\right)={t}^{2}\\hfill \\\\ {y}_{1}\\left(t\\right)=2t - 1\\hfill \\end{cases}\\text{ and }\\begin{cases}{x}_{2}\\left(t\\right)=-t+6\\hfill \\\\ {y}_{2}\\left(t\\right)=t+1\\hfill \\end{cases}$ For the following exercises, use a graphing calculator to complete the table of values for each set of parametric equations. 44. $\\begin{cases}{x}_{1}\\left(t\\right)=3{t}^{2}-3t+7\\hfill \\\\ {y}_{1}\\left(t\\right)=2t+3\\hfill \\end{cases}$ $t$ $x$ $y$ –1 0 1 45. $\\begin{cases}{x}_{1}\\left(t\\right)={t}^{2}-4\\hfill \\\\ {y}_{1}\\left(t\\right)=2{t}^{2}-1\\hfill \\end{cases}$ $t$ $x$ $y$ 1 2 3 46. $\\begin{cases}{x}_{1}\\left(t\\right)={t}^{4}\\hfill \\\\ {y}_{1}\\left(t\\right)={t}^{3}+4\\hfill \\end{cases}$ $t$ $x$ $y$ -1 0 1", "equation is $y + 13 = - \\frac{1}{24} \\left(x + 1\\right)$ Graphed are the function and its normal line: graph{(2x^3-8x^2+2x-1-y)(y+13+(x+1)/24)=0 [-5, 7, -18.16, 2.12]}", "# Thread: Parameterization 1. ## Parameterization I have a problem that seems easy enough but i cant seem to find x and y Problem: find a Parametrization for the graph of the function $y = x^2$ Starting at (3,9) and moving towards the origin,using angle $\\theta$ in the accompanying figure as a parameter. I drew out the graph and took a picture of it and i will include it. I think i use $Tan \\theta = \\frac{y}{x} = \\frac{9}{3} = 3$ Im not sure how to finish solving for x then to solve for y? Im not sure if i need to find $y = mx + b$ since the slop comes out as zero? Thank you.. 2. ## Re: Parameterization Originally Posted by icelated I have a problem that seems easy enough but i cant seem to find x and y Problem: find a Parametrization for the graph of the function $y = x^2$ Starting at (3,9) and moving towards the origin,using angle $\\theta$ in the accompanying figure as a parameter. I drew out the graph and took a picture of it and i will include it. I think i use $Tan \\theta = \\frac{y}{x} = \\frac{9}{3} = 3$ Im not sure how to finish solving for x then to solve for y? Im not sure if", "$$0=x^2-2x-4-3x+4$$ therefore $$\\boxed{\\boxed{x^2-5x=0}}$$ now we have to find the roots of this function. $$x^2-5x=0$$ put x in evidence $$x*(x-5)=0$$ $$\\boxed{x_1=0~~and~~x_2=5}$$ then $$y=3x-4$$ $$y_1=3x_1-4$$ $$y_1=3*0-4$$ $$\\boxed{y_1=-4}$$ $$y_2=3x_2-4$$ $$y_2=3*5-4$$ $$y_2=15-4$$ $$\\boxed{y_2=11}$$ our points will be $$\\boxed{\\boxed{P_1(0,4)~~and~~P_2(5,11)}}$$ ;) RELATED:", "\\ y=9+6\\\\& \\boxed{y=-3} && \\boxed{y=0} && \\boxed{y=6} && \\boxed{y=15}$ $3x-2y=-12$ $X$ $Y$ ${\\color{red}-6}$ $-3$ ${\\color{red}-4}$ $0$ ${\\color{red}0}$ $6$ ${\\color{red}6}$ $15$ 2. $2x-y=-8$ To enter the function into the calculator, it must be in the form $y= \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}$. Solve the function in terms of the letter ‘$y$’. $&2x-y=-8 && 2x-2x-y=-2x-8 && -y=-2x-8 && \\frac{-y}{-1}=\\frac{-2x}{-1}\\frac{-8}{-1}\\\\& \\boxed{y=2x+8}$ The graph can also be done using technology. The table can be used to set the window. 3. $y=.09x+25$ Write a linear function to represent the word problem. $y &=.09(200)+25 && \\text{Substitute the time of} \\ 200 \\ \\text{minutes for the variable} \\ x'.\\\\y&=\\43.00$ The cost of Sam’s telephone bill is $43.00. Summary In this lesson, you have learned how to evaluate a linear function with given values. These values were given in table form. The table was completed by entering the values obtained by substituting the given values for ‘$x$’ into the linear function. These values were the coordinates of points that were located on the graph of the linear function. The points were used to draw the graph on a Cartesian plane. The lesson then extended into the world of technology. The graphing calculator was used to create a table of values as well as to create the graph of the function. The more you practice the keystrokes for performing these tasks", "line and of the curve. Hence $a=-2t$ and $y_1(t)=y_2(t),$ i.e. $b=12+t^2.$ Writing $\\frac{dA}{dt} = 0$ we have: $\\frac{d}{dt} \\frac{(12+t^2)^2}{4t} =\\frac{2(12+t^2)(2t)}{4t}-\\frac{(12+t^2)^2}{4t^2} = 0,$ which gives $4t^2=12+t^2,$ thus $t=2$ and $y_1=-4x+16.$", "point to find the single parameter determining the line $\\frac{a}{b} = -\\frac{y_1}{x_1} = -\\frac{y_2}{x_2} \\ .$ Last edited by a moderator: May 5, 2017", "is substitute x=t into the equation for our straight line to obtain $$y = 2t - \\frac{1}{5}$$ Hence we have parameterise our path $$\\gamma = \\left\\{\\begin{array}{l} x = t \\\\ y = 2t - \\frac{1}{5}\\end{array}\\right.\\hspace{1cm}\\frac{1}{10}\\leq t \\leq \\frac{3}{10}$$ Do you follow?", "$x = 1 + ln(t) \\iff e^{x - 1} = t$ $y = t^2 + 2 = e^{2x -2} + 2 \\implies y' = 2e^{2x -2}$ At (1,3) $y' = 2$. So the tangent line is $y = 2(x- 1) + 3$ or parametrically let $x - 1 = t \\iff x = 1 + t$ and $y = 2t + 3$ Method 2 No eliminating. Let $r = (1 + ln(t),2 + t^2) \\implies r' = (1/t, 2t)$. At (1,3), $t = 1$, therefore $r'(1) = (1,2)$ $r = (1,3) + s(1,2)$ which gives you $x = 1 + s$ and $y = 3 + 2s$ - I do not understand the r subsitution, is this similar to finding $\\frac{dy}{dx}$? – user138246 Jun 20 '12 at 23:53 What substitution? You mean me writing it as r(t)? – Hawk Jun 20 '12 at 23:54 You said \"Let r =\" – user138246 Jun 20 '12 at 23:55 Yup, that's the vector valued equation of a line. – Hawk Jun 20 '12 at 23:55 I have no idea what a vector value equation is. – user138246 Jun 21 '12 at 0:01 2nd method: eliminating parameter: $x=1+ \\ln t , y=t^2+2 \\Leftrightarrow t=\\exp(x-1), y=2+\\exp(2x-2)$ Consider the function: $f(x)=2+ \\exp(2x-2)$, then: $f'(x)=2 \\exp(2x-2)$ The tangent at $x=1$ has equation: $Y=f'(1)(X-1)+ f(1)$, thus:", "command. The solve command has parameters solve(expression, variable, guess, {low bound, high bound}). Note that the low bound and high bound are option, but if used must be placed in braces. The expression needs to be equal to zero, but low and high bounds do not have to entered. Press $2^{nd}$ [CATALOG][LN] and scroll down to select solve(. 4. Functions are $A, B$ and $D$ because for every input value there is only one output value. 5. $(0, 0), (1, 6), (2, 24), (6, 216)$ 6. $(0, 0), \\left( \\frac{2}{3}, \\frac{1}{3}\\right), \\left(\\frac{2}{3}, - \\frac{1}{3}\\right), (6, 1), (6, -1)$ 7. The first set is a function because it has only one output for every input. 8. A. $d$ is a function of $t$ ## Problems 3 – Function notation In this problem, students will explore using function notation on the calculator entering input values to find the output value. Students will enter the function next to $Y1$ on the $Y=$ screen. Then to find $f(4)$, they will enter $Y1(4)$ on the Home screen. For Question 10, students will need to update the function next to $Y1$. 9. $f(4) = 4^2 - 2(4) + 3 = 11$ 10. $f(2) = 3(2)^2 + 5(2) + 3 = 25$ ## Problem 4 – Function Machine For this problem, students will need the", "A sample function $$f$$ of this type is the function $$\\displaystyle f(t)=\\frac{10}{1+50\\boldcdot 2^{\\text-t}}.$$ Evaluate this function for integer values of $$t$$ between 0 and 15. Describe qualitatively how this function differs from an exponential one. What happens to the world population in the long run according to this model?", "1-1/u).$$ Alternatively, by describing the lines $$x=x_0$$ as $$u=x_0/(1-v)$$, we obtain $$\\frac12 \\le v \\le \\frac23 , \\quad \\frac1{1-v} \\le u \\le \\frac2{1-v}.$$ • One could add, that it's probably best to integrate first over $u$ i.e. $$\\int_{\\frac{1}{2}}^{\\frac{2}{3}} {\\rm d}v \\int_{\\frac{1}{1-v}}^{\\frac{2}{1-v}} {\\rm d}u \\, \\left| \\frac{\\partial (x,y)}{\\partial (u,v)} \\right| \\, f\\left(x(u,v),y(u,v)\\right) \\, .$$ – Diger Jan 20 at 14:38", "to the one of your original line: $- 3$ $s l o p e = \\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$ $- 3 = \\frac{y - 2}{x - 7}$ $- 3 x + 21 = y - 2$ and finally your line: $y + 3 x = 23$" ]

[ "Both (3, $\\frac{\\pi }{3}$) and (- 3, $\\frac{\\pi }{3}$) represent the same point. b. Both (3, $\\frac{\\pi }{3}$) and (- 3, - $\\frac{\\pi }{3}$) represent the same point. c. Both (3, $\\frac{\\pi }{3}$) and (- 3, - $\\frac{2\\pi }{3}$) represent the same point. d. Both (3, $\\frac{\\pi }{3}$) and (3, - $\\frac{\\pi }{3}$) represent the same point. #### Solution: For integral values of n, both (r, θ), (- r, θ + (2n + 1) π) represent the same point in a polar coordinate system. So, (- r, θ - π) and (r, θ) represent the same point. [Substitute n = -1.] (-3, - 2π3) = (-3, π3 - π) [Write - 2π3 as π3 - π.] So, both (3, π3) and (- 3, - 2π3) represent the same point.", "\\left(- 6 , 6\\right) \\to \\left(6 \\sqrt{2} , \\frac{3 \\pi}{4}\\right) \\text{ in polar form}$", "# What is meant by tan^(-1)(y/x) in a polar coordinate formula? $\\theta = {\\tan}^{- 1} \\left(\\frac{y}{x}\\right) \\Leftrightarrow \\tan \\theta = \\frac{y}{x}$. So, it is an angle in $\\left(- \\frac{\\pi}{2} , \\frac{\\pi}{2}\\right)$ that makes tangent equal to $\\frac{y}{x}$.", "# What is the Cartesian equivalent of polar coordinates (2, pi/6)? Oct 2, 2014 $\\left(r , \\theta\\right) \\to \\left(2 , \\frac{\\pi}{6}\\right)$ $\\left(x , y\\right) \\to \\left(r \\cos \\left(\\theta\\right) , r \\sin \\left(\\theta\\right)\\right)$ Substitute in $r$ and $\\theta$ $\\left(x , y\\right) \\to \\left(2 \\cos \\left(\\frac{\\pi}{6}\\right) , 2 \\sin \\left(\\frac{\\pi}{6}\\right)\\right)$ Remember back to the unit circle and special triangles. $\\frac{\\pi}{6} = {30}^{\\circ}$ $\\cos \\left(\\frac{\\pi}{6}\\right) = \\frac{\\sqrt{3}}{2}$ $\\sin \\left(\\frac{\\pi}{6}\\right) = \\frac{1}{2}$ Substitute in those values. $\\left(x , y\\right) \\to \\left(2 \\cdot \\frac{\\sqrt{3}}{2} , 2 \\cdot \\frac{1}{2}\\right)$ $\\left(x , y\\right) \\to \\left(\\sqrt{3} , 1\\right)$", "Ayaan Barr 2022-07-09 What is the polar form of $\\left(-2,3\\right)$ ? Charlize Manning Expert Step 1 To solve 1. we use Pythagoras Theorem $r=\\sqrt{{\\left(-2\\right)}^{2}+{3}^{2}}$ $=\\sqrt{13}$ To solve 2. we first find the quadrant that the point lies in. y is positive while x is negative $⇒$ quadrant II Then we find the basic angle by taking inverse tangent of $|\\frac{y}{x}|$ $\\alpha ={\\mathrm{tan}}^{-1}\\left(|\\frac{3}{-2}|\\right)$ $={\\mathrm{tan}}^{-1}\\left(\\frac{3}{2}\\right)$ The angle that we are looking for would be $\\theta =\\pi -\\alpha$ $=\\pi -{\\mathrm{tan}}^{-1}\\left(\\frac{3}{2}\\right)$ $\\approx 2.16$ Therefore, the polar coordinate is $\\left(\\sqrt{13},\\pi -{\\mathrm{tan}}^{-1}\\left(\\frac{3}{2}\\right)\\right)$ Note that the answer above is not unique. You can add any integer multiples of $2\\pi$ to $\\theta$ to get other representations of the same point. Do you have a similar question?", "$k + 1$, this must correspond to $\\left(- 2 , 0\\right)$ (The above follows simply from the correspondence $x = r \\cos \\left(t\\right) , y = r \\sin \\left(t\\right)$ between polar and Cartesian coordinates.) Thus $- 2 = 1 - k \\cos \\left(0\\right) = 1 - k$ This will lead to $k = 3$ A check : note that this is consistent with $r \\left(\\pi\\right) = 4$ - the other point on the $X$ axis.", "\\frac{5 \\pi}{6}\\right)$$ 2. $$\\left(2, \\frac{\\pi}{2}\\right)$$ 3. $$\\left(4,-\\frac{7 \\pi}{6}\\right)$$ 4. $$\\left(-2, \\frac{5 \\pi}{3}\\right)$$ Give two alternate sets of coordinates for each point. 5. $$\\left(2,60^{\\circ}\\right)$$ 6. $$\\left(5,330^{\\circ}\\right)$$ 7. $$\\left(2,210^{\\circ}\\right)$$ Graph each equation. 8. $$r=4$$ 9. $$\\theta=\\frac{\\pi}{4}$$ 10. $$r=2 \\theta$$ with $$\\theta: 0 \\leq \\theta \\leq 2 \\pi$$. Convert each point to rectangular form. 11. $$\\left(4, \\frac{2 \\pi}{3}\\right)$$ 12. $$\\left(3, \\frac{\\pi}{4}\\right)$$ 13. $$\\left(5, \\frac{\\pi}{3}\\right)$$ Convert each point to polar form using radians where $$0 \\leq \\theta<2 \\pi$$. 14. (1,3) 15. (1,-4) 16. (2,6) Convert each equation to polar form. 17. $$x=3$$ 18. \\(2 x+4 y=2\\ This page titled 10.1 Polar and Rectangular Coordinates is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.", "# The equivalent polar equation for the given rectangular-coordinate equation: x^{2} + y^{2} + 8x=0 Question Alternate coordinate systems The equivalent polar equation for the given rectangular-coordinate equation: $$\\displaystyle{x}^{{{2}}}\\ +\\ {y}^{{{2}}}\\ +\\ {8}{x}={0}$$ 2020-12-25 Conversion formula for coordinate systems are given as: a) From polar to rectangular: $$\\displaystyle{x}={r}{\\cos{\\theta}}$$ $$\\displaystyle{y}={r}{\\sin{\\theta}}$$ b) From rectangular to polar: $$\\displaystyle{r}=\\pm\\sqrt{{{x}^{{{2}}}\\ +\\ {y}^{{{2}}}}}$$ $$\\displaystyle{\\cos{\\theta}}={\\frac{{{x}}}{{{r}}}},\\ {\\sin{\\theta}}={\\frac{{{y}}}{{{r}}}},\\ {\\tan{\\theta}}={\\frac{{{x}}}{{{y}}}}$$ Converting into equivalent polar coordinates: $$\\displaystyle{\\left[{x}^{{{2}}}\\ +\\ {y}^{{{2}}}\\ +\\ {8}{x}={0}\\right.}$$ Put $$\\displaystyle{x}={r}{\\cos{\\theta}},\\ {y}={r}{\\sin{\\theta}},$$ $$\\displaystyle\\Rightarrow\\ {\\left({r}{\\cos{\\theta}}\\right)}^{{{2}}}\\ +\\ {\\left({r}{\\sin{\\theta}}\\right)}^{{{2}}}\\ +\\ {8}{r}{\\cos{\\theta}}={0}$$ $$\\displaystyle\\Rightarrow\\ {r}^{{{2}}}{{\\cos}^{{{2}}}\\theta}\\ +\\ {r}^{{{2}}}{{\\sin}^{{{2}}}\\theta}\\ +\\ {8}{r}{\\cos{\\theta}}={0}$$ $$\\displaystyle\\Rightarrow\\ {r}^{{{2}}}\\ {\\left({{\\cos}^{{{2}}}\\theta}\\ +{{\\sin}^{{{2}}}\\ }\\theta\\right)}\\ +\\ {8}{r}{\\cos{\\theta}}={0}$$ $$\\displaystyle\\le{f}{t}{\\left\\lbrace{{\\cos}^{{{2}}}\\theta}\\ +{{\\sin}^{{{2}}}\\ }\\theta={0}{r}{i}{g}{h}{t}\\right\\rbrace}$$ $$\\displaystyle\\Rightarrow\\ {r}^{{{2}}}\\ +\\ {8}{r}{\\cos{\\theta}}={0}$$ $$\\displaystyle\\Rightarrow\\ {r}\\ +\\ {8}{\\cos{\\theta}}={0}$$ Hence, desired equivalent polar coordinates would be $$\\displaystyle{r}\\ +\\ {8}{\\cos{\\theta}}={0}$$ ### Relevant Questions To find: The equivalent polar equation for the given rectangular-coordinate equation. Given: $$\\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\\le{\\left\\lbrace{x}\\right\\rbrace}^{{{2}}}+{\\left\\lbrace{y}\\right\\rbrace}^{{{2}}}+{\\left\\lbrace{8}\\right\\rbrace}{\\left\\lbrace{x}\\right\\rbrace}={\\left\\lbrace{0}\\right\\rbrace}$$ To find: The equivalent polar equation for the given rectangular-coordinate equation. Given: $$\\displaystyle\\ {x}=\\ {r}{\\cos{\\theta}}$$ $$\\displaystyle\\ {y}=\\ {r}{\\sin{\\theta}}$$ b. From rectangular to polar: $$\\displaystyle{r}=\\pm\\sqrt{{{x}^{{{2}}}\\ +\\ {y}^{{{2}}}}}$$ $$\\displaystyle{\\cos{\\theta}}={\\frac{{{x}}}{{{r}}}},{\\sin{\\theta}}={\\frac{{{y}}}{{{r}}}},{\\tan{\\theta}}={\\frac{{{x}}}{{{y}}}}$$ Calculation: Given: equation in rectangular-coordinate is $$\\displaystyle{y}={x}$$. Converting into equivalent polar equation - $$\\displaystyle{y}={x}$$ Put $$\\displaystyle{x}={r}{\\cos{\\theta}},\\ {y}={r}{\\sin{\\theta}},$$ $$\\displaystyle\\Rightarrow\\ {r}{\\sin{\\theta}}={r}{\\cos{\\theta}}$$ $$\\displaystyle\\Rightarrow\\ {\\frac{{{\\sin{\\theta}}}}{{{\\cos{\\theta}}}}}={1}$$ $$\\displaystyle\\Rightarrow\\ {\\tan{\\theta}}={1}$$ Thus, desired equivalent polar equation would be $$\\displaystyle\\theta={1}$$ The equivalent polar equation for the given rectangular - coordinate equation. $$\\displaystyle{y}=\\ -{3}$$ The equivalent polar coordinates for the given rectangular coordinates. A rectangular coordinate is given as (0, -3). Find", "# What is the distance between the following polar coordinates?: (3,(7pi)/12), (1,(17pi)/8) Jun 7, 2017 $D \\approx 3.04$ #### Explanation: The distance formula for rectangular coordinates is $D = \\sqrt{{\\left({x}_{2} - {x}_{1}\\right)}^{2} + {\\left({y}_{2} - {y}_{1}\\right)}^{2}}$ Plugging in ${x}_{n} = {r}_{n} \\cos \\left({\\theta}_{n}\\right)$ and ${y}_{n} = {r}_{n} \\sin \\left({\\theta}_{n}\\right)$, expanding, and simplifying gives $D = \\sqrt{{r}_{1}^{2} + {r}_{2}^{2} - 2 {r}_{1} {r}_{2} \\cos \\left({\\theta}_{1} - {\\theta}_{2}\\right)}$ Plugging in $\\left({r}_{1} , {\\theta}_{1}\\right) = \\left(3 , \\frac{7 \\pi}{12}\\right)$ and $\\left({r}_{2} , {\\theta}_{2}\\right) = \\left(1 , \\frac{17 \\pi}{8}\\right)$ $D = \\sqrt{{3}^{2} + {1}^{1} - 2 \\left(3\\right) \\left(1\\right) \\cos \\left(\\frac{17 \\pi}{8} - \\frac{7 \\pi}{12}\\right)}$ $D = \\sqrt{10 - 6 \\cos \\left(\\frac{37 \\pi}{24}\\right)}$ $D \\approx 3.04$", "# What is the Cartesian equivalent of polar coordinates (2, pi/6)? ##### 1 Answer Oct 2, 2014 $\\left(r , \\theta\\right) \\to \\left(2 , \\frac{\\pi}{6}\\right)$ $\\left(x , y\\right) \\to \\left(r \\cos \\left(\\theta\\right) , r \\sin \\left(\\theta\\right)\\right)$ Substitute in $r$ and $\\theta$ $\\left(x , y\\right) \\to \\left(2 \\cos \\left(\\frac{\\pi}{6}\\right) , 2 \\sin \\left(\\frac{\\pi}{6}\\right)\\right)$ Remember back to the unit circle and special triangles. $\\frac{\\pi}{6} = {30}^{\\circ}$ $\\cos \\left(\\frac{\\pi}{6}\\right) = \\frac{\\sqrt{3}}{2}$ $\\sin \\left(\\frac{\\pi}{6}\\right) = \\frac{1}{2}$ Substitute in those values. $\\left(x , y\\right) \\to \\left(2 \\cdot \\frac{\\sqrt{3}}{2} , 2 \\cdot \\frac{1}{2}\\right)$ $\\left(x , y\\right) \\to \\left(\\sqrt{3} , 1\\right)$", "to polar coordinates, we have \\begin{align*} \\iint_{R^2} e^{-10(x^2+y^2)}\\,dx \\, dy &= \\int_{\\theta=0}^{\\theta=2\\pi} \\int_{r=0}^{r=\\infty} e^{-10r^2}\\,r \\, dr \\, d\\theta = \\int_{\\theta=0}^{\\theta=2\\pi} \\left(\\lim_{a\\rightarrow\\infty} \\int_{r=0}^{r=a} e^{-10r^2}r \\, dr \\right) d\\theta \\\\ &=\\left(\\int_{\\theta=0}^{\\theta=2\\pi}\\right) d\\theta \\left(\\lim_{a\\rightarrow\\infty} \\int_{r=0}^{r=a} e^{-10r^2}r \\, dr \\right) \\\\ &=2\\pi \\left(\\lim_{a\\rightarrow\\infty} \\int_{r=0}^{r=a} e^{-10r^2}r \\, dr \\right) \\\\ &=2\\pi \\lim_{a\\rightarrow\\infty}\\left(-\\frac{1}{20}\\right)\\left(\\left. e^{-10r^2}\\right|_0^a\\right) \\\\ &=2\\pi \\left(-\\frac{1}{20}\\right)\\lim_{a\\rightarrow\\infty}\\left(e^{-10a^2} - 1\\right) \\\\ &= \\frac{\\pi}{10}. \\end{align*} Exercise $$\\PageIndex{7}$$ Evaluate the integral $\\iint_{R^2} e^{-4(x^2+y^2)}dx \\, dy. \\nonumber$ Hint Convert to the polar coordinate system. $$\\frac{\\pi}{4}$$ ## Key Concepts • To apply a double integral to a situation with circular symmetry, it is often convenient to use a double integral in polar coordinates. We can apply these double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals. • The area $$dA$$ in polar coordinates becomes $$r \\, dr \\, d\\theta$$. • Use $$x = r \\, \\cos \\, \\theta, \\, y = r \\, \\sin \\, \\theta$$, and $$dA = r \\, dr \\, d\\theta$$ to convert an integral in rectangular coordinates to an integral in polar coordinates. • Use $$r^2 = x^2 + y^2$$ and $$\\theta = tan^{-1} \\left(\\frac{y}{x}\\right)$$ to convert an integral in polar coordinates to an integral in rectangular coordinates, if needed. • To find the volume in polar coordinates bounded above", "To plot: Thepoints which has polar coordinate displaystyle{left({2},frac{{{7}pi}}{{4}}right)} also two alternaitve sets for the same. To plot: Thepoints which has polar coordinate $\\left(2,\\frac{7\\pi }{4}\\right)$ also two alternaitve sets for the same. You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Dora Let the point $\\left(2,\\frac{7\\pi }{4}\\right)$ be denoted by P. Consider the point $\\left(2,\\frac{7\\pi }{4}\\right)$, it is 2 units from the Origin O and line OP makes an angle of $\\frac{7\\pi }{4}$ with the positive direction of the x-axis. Add pi to the angle and change cardinal coordinate to -2 to get one alternate solution. Therefore, the point P $\\left(2,\\frac{7\\pi }{4}\\right)$ can also be represented by $\\left(-2,\\frac{7\\pi }{4}+\\pi \\right)=\\left(-2,\\frac{11\\pi }{4}\\right).$ Change the radial coordinate to -2 and subtract $3\\pi$ from the angle to obtain the other alternate solution. Thus, the point $P\\left(2,\\frac{7\\pi }{4}\\right)$ can be given by $\\left(-2,\\frac{7\\pi }{4}-3\\pi \\right)$ $=\\left(-2,-\\frac{19\\pi }{4}\\right).$ Use online graphing calculator and plot the points of P as shown below in Figure 1. From the Figure 1 it can be noted that alternate representations are plotted on the same location.", "# How do I find the distance between the points with polar coordinates (2, 120^circ) and (1, 45^circ)? $\\left(- 1 , \\sqrt{3}\\right)$ and $\\left(\\frac{1}{\\sqrt{2}} , \\frac{1}{\\sqrt{2}}\\right)$ $\\sqrt{{\\left(- 1 - \\frac{1}{\\sqrt{2}}\\right)}^{2} + \\left(\\sqrt{3} - \\frac{1}{\\sqrt{2}}\\right)}$ $= 1.99$ (approximately, according to my spreadsheet)", "## Precalculus (6th Edition) Blitzer The polar coordinates of $\\left( 0,-6 \\right)$ are $\\left( 6,\\frac{3\\pi }{2} \\right)$. The polar coordinates of the point are $\\left( r,\\theta \\right)$. Now, rewrite the polar coordinates in terms of rectangular coordinates as below: $r=\\sqrt{{{x}^{2}}+{{y}^{2}}}$ …… (1) $\\tan \\theta =\\frac{y}{x}$ …… (2) Substituting the values of $x\\ \\text{ and }\\ y$ in (1) and (2), we get \\begin{align} & r=\\sqrt{{{x}^{2}}+{{y}^{2}}} \\\\ & =\\sqrt{{{\\left( 0 \\right)}^{2}}+{{\\left( -6 \\right)}^{2}}} \\\\ & =\\sqrt{36} \\\\ & r=6 \\end{align} And, \\begin{align} & \\tan \\theta =\\frac{y}{x} \\\\ & =\\frac{-6}{0} \\\\ & \\tan \\theta =\\infty \\left( \\text{undefined} \\right) \\end{align} Hence, $\\tan \\theta =\\infty$ And, $\\tan \\frac{\\pi }{2}=\\infty$ The $\\theta$ lies on the negative y-axis which means \\begin{align} & \\theta =\\pi +\\frac{\\pi }{2} \\\\ & =\\frac{3\\pi }{2} \\end{align} Therefore, the polar coordinates of $\\left( 0,-6 \\right)$ are $\\left( 6,\\frac{3\\pi }{2} \\right)$.", "# What is the Cartesian form of ( 15 , ( 7pi)/3 ) ? Aug 14, 2017 $\\left(\\frac{15}{2} , 15 \\frac{\\sqrt{3}}{2}\\right)$ #### Explanation: In Polar coordinates a point is represented as $P \\left(r , \\theta\\right)$ In Cartesian form it is represented as $P \\left(x , y\\right)$ Where $x = r \\cos \\theta$ And $y = r \\sin \\theta$ $\\therefore$ Cartesian form of $P \\left(15 , \\frac{7 \\pi}{3}\\right)$ will be $x = 15 \\times \\cos \\left(\\frac{7 \\pi}{3}\\right) = \\frac{15}{2}$ $y = 15 \\times \\sin \\left(\\frac{7 \\pi}{3}\\right) = 15 \\frac{\\sqrt{3}}{2}$ $\\therefore$ The point in Cartesian coordinate will be $P \\left(\\frac{15}{2} , 15 \\frac{\\sqrt{3}}{2}\\right)$", "Question # The equivalent polar coordinates for the given rectangular coordinates. A rectangular coordinate is given as (0, -3). Alternate coordinate systems The equivalent polar coordinates for the given rectangular coordinates. A rectangular coordinate is given as (0, -3). 2020-11-02 Conversion formula for coordinate systems are given as: a) From polar to rectangular: $$\\displaystyle{x}={r} \\cos{\\theta}$$ $$\\displaystyle{y}={r} \\sin{\\theta}$$ b) From rectangular to polar: $$\\displaystyle{r}=\\pm\\sqrt{{{x}^{2}+{y}^{2}}}$$ $$\\displaystyle \\cos{\\theta}=\\frac{x}{{r}}; \\sin{\\theta}=\\frac{y}{{r}}; \\tan{\\theta}=\\frac{x}{{y}}$$ Here $$\\displaystyle{x}={0},{y}=-{3}$$ Converting into equivalent polar coordinates: $$\\displaystyle{r}=\\pm\\sqrt{{{x}^{2}+{y}^{2}}}$$ $$\\displaystyle\\Rightarrow{r}=\\pm\\sqrt{{{0}^{2}+{\\left(-{3}\\right)}^{2}}}$$ $$\\displaystyle\\Rightarrow{r}=\\pm{3}$$ $$\\displaystyle \\cos{\\theta}=\\frac{x}{{r}}=\\frac{0}{{3}}={0}$$ {Taking positive value of r} $$\\displaystyle\\Rightarrow\\theta={90}^{\\circ},{270}^{\\circ}$$ $$\\displaystyle \\sin{\\theta}=\\frac{y}{{r}}=\\frac{{-{3}}}{{3}}={1}$$ $$\\displaystyle\\Rightarrow\\theta={270}^{\\circ}$$ Hence, desired equivalent polar coordinates would be $$\\displaystyle{\\left({3},{270}^{\\circ}\\right)}$$", "# What is the Cartesian form of ( 6,-pi/6 )? Oct 8, 2016 Cartesian form of $\\left(6 , - \\frac{\\pi}{6}\\right)$ is $\\left(3 \\sqrt{3} , - 3\\right)$ #### Explanation: Polar coordinates $\\left(r , \\theta\\right)$ are equivalent to Cartesian Coordinates $\\left(r \\cos \\theta , r \\sin \\theta\\right)$ Hence polar coordinates $\\left(6 , - \\frac{\\pi}{6}\\right)$ are equivalent to Cartesian Coordinates $\\left(6 \\times \\cos \\left(- \\frac{\\pi}{6}\\right) , 6 \\times \\sin \\left(- \\frac{\\pi}{6}\\right)\\right)$ i.e. $\\left(6 \\times \\cos \\left(\\frac{\\pi}{6}\\right) , 6 \\times - \\sin \\left(\\frac{\\pi}{6}\\right)\\right)$ or $\\left(6 \\times \\frac{\\sqrt{3}}{2} , - 6 \\times \\frac{1}{2}\\right)$ or $\\left(3 \\sqrt{3} , - 3\\right)$", "# What is the distance between the following polar coordinates?: (2,(3pi)/4), (1,(15pi)/8) ##### 3 Answers Jul 18, 2018 $D = \\sqrt{5 + 4 \\cos \\left(\\frac{\\pi}{8}\\right)} \\approx 2.9488$ #### Explanation: We know that , $\\text{Distance between Polar Co-ordinates:} A \\left({r}_{1} , {\\theta}_{1}\\right) \\mathmr{and} B \\left({r}_{2} , {\\theta}_{2}\\right)$ is color(red)(D=sqrt(r_1^2+r_2^2-2r_1r_2cos(theta_1-theta_2))...to(I) We have , ${P}_{1} \\left(2 , \\frac{3 \\pi}{4}\\right) \\mathmr{and} {P}_{2} \\left(1 , \\frac{15 \\pi}{8}\\right)$. So , ${r}_{1} = 2 , {r}_{2} = 1 , {\\theta}_{1} = \\frac{3 \\pi}{4} \\mathmr{and} {\\theta}_{2} = \\frac{15 \\pi}{8}$ $\\implies {\\theta}_{1} - {\\theta}_{2} = \\frac{3 \\pi}{4} - \\frac{15 \\pi}{8} = \\frac{6 \\pi - 15 \\pi}{8} = \\frac{- 9 \\pi}{8}$ $\\implies \\cos \\left({\\theta}_{1} - {\\theta}_{2}\\right) = \\cos \\left(\\frac{- 9 \\pi}{8}\\right)$ $\\implies \\cos \\left({\\theta}_{1} - {\\theta}_{2}\\right) = \\cos \\left(\\frac{9 \\pi}{8}\\right) \\to \\left[\\because \\cos \\left(- \\theta\\right) = \\cos \\theta\\right]$ $\\implies \\cos \\left({\\theta}_{1} - {\\theta}_{2}\\right) = \\cos \\left(\\pi + \\frac{\\pi}{8}\\right) = - \\cos \\left(\\frac{\\pi}{8}\\right)$ \"Using : \" color(red)((I) we get D=sqrt(2^2+1^2-2(2)(1)(-cos(pi/8)) $\\implies D = \\sqrt{4 + 1 + 4 \\cdot \\cos \\left(\\frac{\\pi}{8}\\right)}$ $\\implies D = \\sqrt{5 + 4 \\cos \\left(\\frac{\\pi}{8}\\right)}$ $\\implies D \\approx 2.9488$ Jul 18, 2018 $\\sqrt{2 + \\sqrt{2 - \\sqrt{2}}}$ #### Explanation: It is nontrivial to determine polar distances (since drawing a line in polar coordinates isn't very easy), so we should convert to Cartesian coordinates. The first point is moderately easy: $x = r \\cos \\theta =", "Question # To find: The equivalent polar equation for the given rectangular-coordinate equation.Given:\\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0} Alternate coordinate systems To find: The equivalent polar equation for the given rectangular-coordinate equation. Given: $$\\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0}$$ 2021-02-09 Concept used: Conversion formula for coordinate systems are given as a) From polar to rectangular: $$\\displaystyle{x}={r} \\cos{\\theta}$$ $$\\displaystyle{y}={r} \\sin{\\theta}$$ b) From rectangular to polar: $$\\displaystyle{r}=\\pm\\sqrt{{{x}^{2}+{y}^{2}}}$$ $$\\displaystyle \\cos{\\theta}=\\frac{x}{{r}}; \\sin{\\theta}=\\frac{y}{{r}}; \\tan{\\theta}=\\frac{x}{{y}}$$ Calculation: Converting into equivalent polar equation $$\\displaystyle{x}^{2}+{y}^{2}+{8}{x}={0}$$ Put $${\\left\\lbrace{x}\\right\\rbrace}={\\left\\lbrace{r}\\right\\rbrace}{\\cos{{\\left\\lbrace\\theta\\right\\rbrace}}},{\\left\\lbrace{y}\\right\\rbrace}={\\left\\lbrace{r}\\right\\rbrace}{\\sin{{\\left\\lbrace\\theta\\right\\rbrace}}},$$ $$\\displaystyle{x}={r} \\cos{\\theta},{y}={r} \\sin{\\theta}$$ $$\\displaystyle\\Rightarrow{\\left({r} \\cos{\\theta}\\right)}^{2}+{\\left({r} \\sin{\\theta}\\right)}^{2}+{8}{r} \\cos{\\theta}={0}$$ $$\\displaystyle\\Rightarrow{r}^{2}{{\\cos}^{2}\\theta}+{r}^{2}{{\\sin}^{2}\\theta}+{8}{r} \\cos{\\theta}={0}$$ $$\\displaystyle\\Rightarrow{r}^{2}{\\left({{\\cos}^{2}\\theta}+{{\\sin}^{2}\\theta}\\right)}+{8}{r} \\cos{\\theta}={0}{\\left\\lbrace{{\\cos}^{2}\\theta}+{{\\sin}^{2}\\theta}={1}\\right\\rbrace}$$ $$\\displaystyle\\Rightarrow{r}^{2}+{8}{r} \\cos{\\theta}={0}$$ $$\\displaystyle\\Rightarrow{r}+{8} \\cos{\\theta}={0}$$ Thus, desired equivalent polar equation would be $$\\displaystyle{r}+{8} \\cos{\\theta}={0}$$", "# Find two polar equivalents of the Cartesian point (-3 \\sqrt{3}, 3) ## Question: Find two polar equivalents of the Cartesian point {eq}\\displaystyle (-3 \\sqrt{3}, 3) {/eq} ## Polar coordinates Polar coordinates of a Cartesian point, {eq}\\displaystyle (x,y){/eq} is a pair of two coordinates, a real number {eq}\\displaystyle r {/eq} and an angle {eq}\\displaystyle \\theta, {/eq} i.e. {eq}\\displaystyle (r,\\theta). {/eq} The real number {eq}\\displaystyle r {/eq} is the signed distance between the pole and the point, while the angle {eq}\\displaystyle \\theta {/eq} is the angle between the polar axis and the line going through the pole and the point. A Cartesian point can be written in four different polar forms: {eq}\\displaystyle (r,\\theta), (r, -\\theta), (-r, \\theta), \\text{ or } (-r, -\\theta) {/eq} The angle {eq}\\displaystyle \\theta {/eq} can be measured negatively if the rotation is done clockwise from the polar axis, while {eq}\\displaystyle r {/eq} is negative if the angle {eq}\\displaystyle \\theta {/eq} is in the opposite quadrant form the quadrant containing the point. To convert from Cartesian to polar coordinates, we use the following formulae {eq}\\displaystyle r=\\sqrt{x^2+y^2}, \\\\ \\displaystyle \\theta=\\arctan\\left(\\frac{y}{x}\\right), \\text { if the point in in the Quadrant I or IV, or} \\\\ \\displaystyle \\theta=\\arctan\\left(\\frac{y}{x}\\right) +\\pi, \\text { if the point in in the Quadrant II or III}. {/eq} The above formulae will give only one" ]

[ "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of", "# Difference between revisions of \"2017 AIME II Problems/Problem 12\" ## Problem Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$. Therefore $A_1=(1-r,r)$, $A_2=(1-r-r^2,r-r^2)$, $A_3=(1-r-r^2+r^3,r-r^2-r^3)$, $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$, and so on, where the signs alternate in groups of $2$. The limit of all these points is point $B$. Using the geometric series formula on $B$ and reducing the expression, we get $B=\\left(\\frac{1-r}{r^2+1},\\frac{r-r^2}{r^2+1}\\right)$. The distance from $B$ to the origin is $\\sqrt{\\left(\\frac{1-r}{r^2+1}\\right)^2+\\left(\\frac{r-r^2}{r^2+1}\\right)^2}=\\frac{1-r}{\\sqrt{r^2+1}}.$ Let $r=\\frac{11}{60}$, and the distance from the origin is $\\frac{49}{61}$. $49+61=\\boxed{110}$. $[asy] draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label(\"A_0\",(125,0),E); dot((25,100)); label(\"A_1\",(25,100),SE); dot((-55,20)); label(\"A_2\",(-55,20),E); [/asy]$ ## Solution Impose a coordinate system and let the center of $C_0$ be $(0,0)$ and $A_0$ be $(1,0)$. Therefore $A_1=(1-r,r)$, $A_2=(1-r-r^2,r-r^2)$, $A_3=(1-r-r^2+r^3,r-r^2-r^3)$, $A_4=(1-r-r^2+r^3+r^4,r-r^2-r^3+r^4)$, and so on, where the signs alternate in groups of $2$. The limit of all these points is point $B$. Using the geometric series formula on $B$ and reducing the expression, we get $B=(\\frac{1-r}{r^2+1},\\frac{r-r^2}{r^2+1})$. The distance from $B$ to the origin is $\\sqrt{(\\frac{1-r}{r^2+1})^2+(\\frac{r-r^2}{r^2+1}))^2}=\\frac{1-r}{\\sqrt{r^2+1}}.$ Let $r=\\frac{11}{60}$, we find that the distance from the origin is $\\frac{49}{61} \\implies 49+61=\\boxed{110}$. # See Also 2017 AIME II (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11", "draw(A--F); draw(X--F); draw(E--F); draw(circumcircle(A,B,C)); draw(circumcircle(M,F,E)); dot(D); dot(F); dot(A); dot(B); dot(C); dot(E); dot(X); dot(R); dot(I); label(\"A\",A,dir(220)); label(\"B\",B,dir(110)); label(\"C\",C,dir(250)); label(\"D\",D,dir(60)); label(\"E\",E,dir(0)); label(\"F\",F,dir(315)); label(\"X\",X,dir(180)); [/asy]$ First of all, use the Angle Bisector Theorem to find that $BD=35/8$ and $CD=21/8$, and use Stewart's Theorem to find that $AD=15/8$. Then use Power of a Point to find that $DE=49/8$. Then use the circumradius of a triangle formula to find that the length of the circumradius of $\\triangle ABC$ is $\\frac{7\\sqrt{3}}{3}$. Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of the circumcircle of $\\triangle ABC$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\frac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=a$, and $DF=b$. Then, by the Pythagorean Theorem, $$x^2+b^2=\\left(\\frac{49}{8}\\right)^2=\\frac{2401}{64}$$ and $$x^2+(a+b)^2=\\left(\\frac{14\\sqrt{3}}{3}\\right)^2=\\frac{196}{3}.$$ Subtracting the first equation from the second, the $x^2$ term cancels out and we obtain: $$(a+b)^2-b^2=\\frac{196}{3}-\\frac{2401}{64}$$ $$a^2+2ab = \\frac{5341}{192}.$$ By Power of a Point, $ab=BD \\cdot DC=735/64=2205/192$, so $$a^2+2 \\cdot \\frac{2205}{192}=\\frac{5341}{192}$$ $$a^2=\\frac{931}{192}.$$ Since $a=XD$, $XD=\\frac{7\\sqrt{19}}{8\\sqrt{3}}$. Because $\\angle EXF$ and $\\angle EAF$ intercept the same arc in circle $\\omega$ and the same goes for $\\angle XFA$ and $\\angle XEA$, $\\angle EXF\\cong\\angle EAF$ and $\\angle XFA\\cong\\angle XEA$. Therefore, $\\triangle XDE\\sim\\triangle ADF$ by AA Similarity. Since side lengths in similar triangles are proportional, $$\\frac{AF}{\\frac{15}{8}}=\\frac{\\frac{14\\sqrt{3}}{3}}{\\frac{7\\sqrt{19}}{8\\sqrt{3}}}$$ $$\\frac{AF}{\\frac{15}{8}}=\\frac{16}{\\sqrt{19}}$$ $$AF \\cdot \\sqrt{19} = 30$$ $$AF = \\frac{30}{\\sqrt{19}}.$$", "from the center of the inverted circle to the center of inversion. The center of $C_4'$ is $3$ units above the center of $C_3'$. Since $\\overline{OC_3'} = \\frac{15}{2}$, we use Pythagoras to learn that $\\overline{OC_4'}^2 = \\left(\\frac{15}{2}\\right)^2 + 9$. We do not take the square root because our relationship formula takes $\\overline{OC_4'}^2$. Therefore, we have: $$r = \\frac{36 \\cdot \\frac{3}{2}}{\\left(\\frac{15}{2}\\right)^2 - \\left(\\frac{3}{2}\\right)^2 + 9} = \\frac{54}{9 \\cdot 6 + 9} = \\frac{6}{7} = \\boxed{B}.$$ Here is the diagram with $C_4'$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -45, 45); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(45), NW); label(\"C_1\", (3, 0), N); label(\"C_2\", (2, 0), N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); draw((6, 4.5)--(6, -3)); draw((9, 4.5)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label(\"C_1'\", (6, -3), SW); label(\"C_2'\", (9, -3), SE); label(\"C_3'\", (15/2, 0), NE); dot((15/2, 0)); draw(Circle((15/2, 3), 3/2)); dot((15/2, 3)); label(\"C_4'\", (15/2, 3), N); draw((15/2, 0)--(15/2, 3)); draw(rightanglemark(origin, (15/2, 0), (15/2, 3))); [/asy]$", "note that $C_3'$'s center must be on $\\overline{OC_3}$ and be between the two inverted lines, because $C_3$ is tangent to $C_1$ and $C_2$ (Remeber that tangency still holds in inverted diagrams). Therefore, we must have a circle with radius $\\frac{3}{2}$ that is $\\frac{15}{2}$ units from $O$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label(\"\\Omega\", (6, 0) * dir(30), NW); label(\"C_1\", (3, 0), N); label(\"C_2\", (2, 0), N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); draw((6, 3)--(6, -3)); draw((9, 3)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label(\"C_1'\", (6, 3), N); label(\"C_2'\", (9, 3), N); label(\"C_3'\", (15/2, 0), N); dot((15/2, 0)); [/asy]$ Now, we invert $C_4$. Note that $C_4$ is tangent to the three other original circles. So, in the inversion, $C_4'$ must be tangent to the two lines and $C_3'$. It is then quickly seen that $C_3'$ and $C_4'$ have the same radius: $\\frac{3}{2}$. Now, we can determine the radius of $C_4$ using the formula $r = \\frac{k^2 \\cdot r'}{\\overline{OC_2}^2 - r'^2}$. $k^2 = 36$, and $r' = \\frac{3}{2}$. $\\overline{OC}$ is just the distance", "We will let $O(0,0)$ be the origin. This way the coordinates of C would be $(0,x)$. By 30-60-90, the coordinates of D would be $(\\sqrt{-1}{2}, x + \\frac{\\sqrt{3}}{2})$. The distance $(x, y)$ is from the origin is just $\\sqrt{x^2 + y^2}$. Therefore, the distance D is from the origin is both 4 and $\\frac{1}{4} + x^2 + x\\sqrt{3} + \\frac{3}{4} = x^2 + x\\sqrt{3} + 1 = 16$. We get the equation mentioned in all the previous solution, using the quadratic formula, we get that $x = \\frac{3\\sqrt{7}-\\sqrt{3}}{2} \\Rightarrow C$ ## Solution 5 $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.55,2.1),E); label(\"$$A$$\",(-3.6,2.5513),E); label(\"$$B$$\",(-3.15,1.35),E); label(\"$$C$$\",(0.05,3.20),E); label(\"$$D$$\",(-0.75,4.15),E); label(\"$$E$$\",(0.3,4.15),E); label(\"$$F$$\",(-3.4,1.89),E); draw((0.5,3.9686)--(-3.13,2.45)); draw((0.5,3.9686)--(-2.95,2));[/asy]$ Notice that $\\overarc{AE}$ is one-sixth the circumference of the circle. Therefore, $\\overline{AE}$ is the side length of an inscribed hexagon. The side length of such a hexagon is simply the radius, or $4$. $\\angle AFE$ is a right angle, therefore $\\triangle AFE$ is a right triangle. $\\overline{AF}$ is half the length of $1$, or $\\frac{1}{2}$. The length of $\\overline{EF}$ is $x$ plus the altitude length of one of the equilateral triangles, or $x+\\frac{\\sqrt{3}}{2}$. Using the Pythagorean Theorem, we get $4^2 = \\left(\\frac{1}{2}\\right)^2 + \\left(x+\\frac{\\sqrt{3}}{2}\\right)^2$ Solving for $x$, we get $x =$ $\\frac{3\\sqrt{7}-\\sqrt{3}}{2}$ , or $\\boxed{\\text{C}}$.", "draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.55,2.1),E); label(\"$$1$$\",(-0.5,3.8),S);[/asy]$ We will let $O(0,0)$ be the origin. This way the coordinates of $C$ will be $(0,y)$. By $30-60-90$, the coordinates of $D$ will be $\\left(-\\frac{1}{2}, y + \\frac{\\sqrt{3}}{2}\\right)$. The distance any point with coordinates $(x, y)$ is from the origin is $\\sqrt{x^2 + y^2}$. Therefore, the distance $D$ is from the origin is $4$ and $\\frac{1}{4} + x^2 + x\\sqrt{3} + \\frac{3}{4} = x^2 + x\\sqrt{3} + 1 = 16$. We get the quadratic equation mentioned in solution 2. Using the quadratic formula, we get that $x = \\frac{3\\sqrt{7}-\\sqrt{3}}{2} \\Rightarrow (C)$ Note: Since $C$ and $D$ are not labeled in the diagram, refer to solution 1 for the location of points $C$ and $D$. ## Solution 5 $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.55,2.1),E); label(\"$$A$$\",(-3.6,2.5513),E); label(\"$$B$$\",(-3.15,1.35),E); label(\"$$C$$\",(0.05,3.20),E); label(\"$$D$$\",(-0.75,4.15),E); label(\"$$E$$\",(0.3,4.15),E); label(\"$$F$$\",(-3.4,1.89),E); draw((0.5,3.9686)--(-3.13,2.45)); draw((0.5,3.9686)--(-2.95,2));[/asy]$ Notice that $\\overarc{AE}$ is $\\frac16$ the circumference of the circle. Therefore, $\\overline{AE}$ is the side length of an inscribed hexagon. The side length of such a hexagon is simply the radius of the circle, or $4$. $\\angle AFE$ is a right angle, therefore $\\triangle AFE$ is a right triangle. $\\overline{AF}$ is half the length of $1$, or $\\frac{1}{2}$. The length of $\\overline{EF}$ is $x$ plus the altitude length of one of the equilateral triangles,", "+0 # Can I get some help with these? 0 68 1 Hi... I've been strugling with these three problems for a few weeks: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6},$$ where the angle$$\\frac{\\pi}{6}$$ is in radians. (2)Find the length AC, to two decimal places. Image: https://artofproblemsolving.com/texer/sywndnao (3)Find the length BC, to two decimal places. Image: https://artofproblemsolving.com/texer/enqsdotw Nov 15, 2020 #1 0 Quick note: to get the images for (2) and (3), post these codes to get the figures. Make sure you say Render as Image in the TeXeR (2) [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3) [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] Nov 15, 2020", "draw((0,0)--(30,0)); draw((0,0)--O[1]--(O[1].x,0)); draw(O[1]--(O[1] + reflect((0,0),(10,12*sqrt(221)/49*10))*(O[1]))/2); label(\"$y = mx$\", (8,12*sqrt(221)/49*8), N); dot(\"$(9,6)$\", (9,6), NE); dot(\"$(kr,r)$\", O[1], N); dot(P,red); [/asy] Since the circle is tangent to the line $y = mx,$ the distance from the center $(kr,r)$ to the line is $r.$ We can write $y = mx$ as $y - mx = 0,$ so from the distance formula, $$\\frac{|r - krm|}{\\sqrt{1 + m^2}} = r.$$Squaring both sides, we get $$\\frac{(r - krm)^2}{1 + m^2} = r^2,$$so $(r - krm)^2 = r^2 (1 + m^2).$ Since $r \\neq 0,$ we can divide both sides by 0, to get $$(1 - km)^2 = 1 + m^2.$$Then $1 - 2km + k^2 m^2 = 1 + m^2,$ so $m^2 (1 - k^2) + 2km = 0.$ Since $m \\neq 0,$ $$m(1 - k^2) + 2k = 0.$$Hence, $$m = \\frac{2k}{k^2 - 1} = \\frac{2 \\sqrt{\\frac{117}{68}}}{\\frac{117}{68} - 1} = \\boxed{\\frac{12 \\sqrt{221}}{49}}.$$", "+0 # Asymptote coding +3 37 2 +538 Hello! I was answering a question earlier, and wanted to include a graph. Since Asympotote coding makes it look clean, I tried inserting this code: [asy] size(250); import TrigMacros; real theta = 45; pair A = (3,0); pair B = 3*dir(theta); pair O = (0,0); rr_cartesian_axes(-5, 5, -5, 5, usegrid = false); draw(Circle(O, 3)); draw(O--3*dir(60)); dot(\"$\\left(\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right)$\", 3*dir(30), E); draw(O--3*dir(30)); dot(\"$\\left(\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$\", 3*dir(60), N); draw(O--3*dir(45)); dot(\"$\\left(\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2}}{2}\\right)$\", 3*dir(45), dir(45)); draw(O--3*dir(150)); dot(\"$\\left(-\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right)$\", 3*dir(150), W); draw(O--3*dir(120)); dot(\"$\\left(-\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$\", 3*dir(120), N); draw(O--3*dir(135)); dot(\"$\\left(-\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2}}{2}\\right)$\", 3*dir(135), dir(135)); draw(O--3*dir(-60)); dot(\"$\\left(\\frac{\\sqrt{3}}{2}, -\\frac{1}{2}\\right)$\", 3*dir(-30), E); draw(O--3*dir(-30)); dot(\"$\\left(\\frac{1}{2}, -\\frac{\\sqrt{3}}{2}\\right)$\", 3*dir(-60), -N); draw(O--3*dir(-45)); dot(\"$\\left(\\frac{\\sqrt{2}}{2}, -\\frac{\\sqrt{2}}{2}\\right)$\", 3*dir(-45), dir(-45)); draw(O--3*dir(-150)); dot(\"$\\left(-\\frac{\\sqrt{3}}{2}, -\\frac{1}{2}\\right)$\", 3*dir(-150), W); draw(O--3*dir(-120)); dot(\"$\\left(-\\frac{1}{2}, -\\frac{\\sqrt{3}}{2}\\right)$\", 3*dir(-120), -N); draw(O--3*dir(-135)); dot(\"$\\left(-\\frac{\\sqrt{2}}{2},- \\frac{\\sqrt{2}}{2}\\right)$\", 3*dir(-135), dir(-135)); dot(\"$(1,0)$\", A, NE); [/asy] But, it did not show up. (it's a unit circle) Does this site support Asymptote, or do I need to use desmos? Thanks! off-topic Aug 2, 2020 #1 +128 +1 I believe that this site does not support asymptote, as I have tried to use it in the past. If you try to take a screenshot, or maybe as you said desmos will work. Hope this helps! Aug 2, 2020 #2 +538 +3 Okay! ilorty Aug 2, 2020", "+0 0 92 1 Let $X$, $Y$, and $Z$ be points on a circle. Let $\\overline{XY}$ and the tangent to the circle at $Z$ intersect at $W$. If $WX = 4$, $WZ = 8$, and $\\overline{WY} \\perp \\overline{WZ}$, then find $YZ$. [asy] unitsize(2 cm); pair A, B, C, D; D = (0,1); B = dir(150); C = dir(210); A = (-sqrt(3)/2,1); draw(Circle((0,0),1)); draw(D--A--C); label(\"$W$\", A, NW); label(\"$X$\", B, SE); label(\"$Y$\", C, SW); label(\"$Z$\", D, N); label(\"$8$\", (A + D)/2, N); label(\"$4$\", (A + B)/2, W); [/asy] Feb 14, 2020", "0), N); label(\"C_3\", (5, 0), N); label(\"C_4\", P, N); label(\"O\", origin, W); draw((6, 3)--(6, -3)); draw((9, 3)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label(\"C_1'\", (6, 3), N); label(\"C_2'\", (9, 3), N); label(\"C_3'\", (15/2, 0), N); dot((15/2, 0)); [/asy]$ Now, we invert $C_4$. Note that $C_4$ is tangent to the three other original circles. So, in the inversion, $C_4'$ must be tangent to the two lines and $C_3'$. It is then quickly seen that $C_3'$ and $C_4'$ have the same radius: $\\frac{3}{2}$. Now, we can determine the radius of $C_4$ using the formula $r = \\frac{k^2 \\cdot r'}{\\overline{OC_2}^2 - r'^2}$. $k^2 = 36$, and $r' = \\frac{3}{2}$. $\\overline{OC}$ is just the distance from the center of the inverted circle to the center of inversion. The center of $C_4'$ is $3$ units above the center of $C_3'$. Since $\\overline{OC_3'} = \\frac{15}{2}$, we use Pythagoras to learn that $\\overline{OC_4'}^2 = \\left(\\frac{15}{2}\\right)^2 + 9$. We do not take the square root because our relationship formula takes $\\overline{OC_4'}^2$. Therefore, we have: $$r = \\frac{36 \\cdot \\frac{3}{2}}{\\left(\\frac{15}{2}\\right)^2 - \\left(\\frac{3}{2}\\right)^2 + 9} = \\frac{54}{9 \\cdot 6 + 9} = \\frac{6}{7} = \\boxed{B}.$$ Here is the diagram with $C_4'$. $[asy] size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7);", "+0 # Help plz 0 58 1 The following polar grid is centered at the origin, with concentric circles of positive integer radii starting at $1,$ and consecutive rays through the origin with angles of $\\pi/12$ between them: [asy] size(200); pair A = (0,0); for (int i = 1; i < 6; ++i) { draw(Circle((0,0),i), linewidth(0.4)); } for(int i=0;i<360;i+=15) { draw(rotate(i)*((-5.0,0)--(5.0,0)), linewidth(0.4)); } pair A, B,C, D, EE, F; A = (1,-sqrt(3)); B = (1, sqrt(3)); C = (sqrt(3), 1); D = (4,0); EE = (0,0); F = (-sqrt(2), -sqrt(2)); dot(A, red+linewidth(3.5)); dot(B, red+linewidth(3.5)); dot(C, red+linewidth(3.5)); dot(D, red+linewidth(3.5)); dot(EE, red+linewidth(3.5)); dot(F, red+linewidth(3.5)); label(\"$A$\", A, S); label(\"$B$\", B, N); label(\"$C$\", C, E); label(\"$D$\", D, S); label(\"$E$\", EE, 3*S); label(\"$F$\", F, S); [/asy] Some of the points above are on the graph $r=4 \\cos(\\theta).$ Answer with the list in any order, separated by commas. Jul 30, 2021", "+0 I didn't get any help... Thanks. -1 97 2 Hi, I posted this on the 15th and never got any help. Thanks for your help in advance: (1) Degrees are not the only units we use to measure angles. We also use radians. Just as there are $$360^\\circ$$ in a circle, there are $$2\\pi$$ radians in a circle. Compute $$\\sin \\frac{\\pi}{6}$$ where the angle $$\\frac{\\pi}{6}$$is in radians. (2)Find the length AC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(2 cm); pair A, B, C; A = dir(40); B = (0,0); C = (3,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$3$\", (A + B)/2, NW); label(\"$7$\", (B + C)/2, S); label(\"$50^\\circ$\", B + (0.5,0.15)); [/asy] (3)Find the length BC, to two decimal places. Use the texer here: https://artofproblemsolving.com/texer Post code to find image in TeXeR: [asy] unitsize(1 cm); pair A, B, C; A = 4*dir(71); B = (0,0); C = (5,0); draw(A--B--C--cycle); label(\"$A$\", A, N); label(\"$B$\", B, SW); label(\"$C$\", C, SE); label(\"$8$\", (A + C)/2, NE); label(\"$61^\\circ$\", A + (0.2,-0.8)); label(\"$43^\\circ$\", C + (-0.8,0.3)); [/asy] NOTE, FOR SOME REASON, THE LATEX IN THE CODE IS RENDERING .... IF (and probably when) YOU PASTE TO FIND THE ANSWER, YOU MAY NEED TO FIGURE OUT *or", "internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent to $C_1$ at point $A_1$. In this way a sequence of circles $C_1,C_2,C_3,\\cdots$ and a sequence of points on the circles $A_1,A_2,A_3,\\cdots$ are constructed, where circle $C_n$ has radius $r^n$ and is internally tangent to circle $C_{n-1}$ at point $A_{n-1}$, and point $A_n$ lies on $C_n$ $90^{\\circ}$ counterclockwise from point $A_{n-1}$, as shown in the figure below. There is one point $B$ inside all of these circles. When $r = \\frac{11}{60}$, the distance from the center $C_0$ to $B$ is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. \\usepackage{asymptote} \\begin{figure}[h] \\begin{asy} draw(Circle((0,0),125)); draw(Circle((25,0),100)); draw(Circle((25,20),80)); draw(Circle((9,20),64)); dot((125,0)); label(\"$A_0$\",(125,0),E); dot((25,100)); label(\"$A_1$\",(25,100),SE); dot((-55,20)); label(\"$A_2$\",(-55,20),E); \\end{asy} \\end{figure} ## Problem 13 For each integer $n\\geq3$, let $f(n)$ be the number of $3$-element subsets of the vertices of the regular $n$-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$. ## Problem 14 A $10\\times10\\times10$ grid of points consists of all points in space of the form $(i,j,k)$, where $i$, $j$, and $k$ are integers between $1$ and $10$, inclusive. Find the number of", "label(\"P\",P,N); [/asy]$ and use Stewart's Theorem: $$AB \\cdot AC \\cdot BC + AB \\cdot {CP}^2 = AC \\cdot {BP}^2 + BC \\cdot {AP}^2$$ From what we learned from the tangent circles, we have $AB = 3$, $AC = 1$, $BC = 2$, $AP = 2 + r$, $BP = 1 + r$, and $CP = 3 - r$, where $r$ is the radius of the circle centered at $P$ that we seek. Thus: $$3 \\cdot 1 \\cdot 2 + 3 {\\left(3-r\\right)}^2 = 1 {\\left(1+r\\right)}^2 + 2 {\\left(2+r\\right)}^2$$ $$6 + 3\\left(9 - 6r + r^2\\right) = \\left(1 + 2r + r^2\\right) + 2\\left(4 + 4r + r^2\\right)$$ $$33 - 18r + 3r^2 = 9 + 10r + 3r^2$$ $$28r = 24$$ $$r = \\boxed{\\frac{6}{7}} \\to \\boxed{\\textbf{(B)}}$$ ## Solution 2 $[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); [/asy]$ Like the solution above, connecting the centers of the circles results in triangle $\\Delta ABP$ with cevian $PC$. The two triangles $\\Delta APC$ and $\\Delta ABP$ share angle $A$, which means we can use Law of Cosines to set up a system of 2 equations that solve for $r$ respectively: $(2 + r)^2 + 1^2 - 2(2 + r)(1)\\cos A = (3 - r)^2$ (notice that the diameter of the largest", "C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D(\"O\",A,dir(A)); D(\"O_1\",B,dir(B)); D(\"O_2\",C,dir(C)); D(\"H\",H,dir(270)); D(\"X\",X,dir(225)); D(\"A\",R1,dir(180)); D(\"B\",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$Since $X$ is on the radical axis of $\\omega_1$ and $\\omega_2$, it has equal power with respect to both circles, so$$O_1X^2 - r_1^2 = O_2X^2-r_2^2 \\implies O_1X-O_2X = \\frac{r_1^2-r_2^2}{d}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular,\\begin{align*} O_1H &= O_1X - HX = \\frac{d+\\frac{r_1^2-r_2^2}{d}}{2} - \\frac{r}{2} \\\\ O_2H &= O_2X + HX = \\frac{d-\\frac{r_1^2-r_2^2}{d}}{2} + \\frac{r}{2}. \\end{align*}We want to solve for $d$. By the Pythagorean Theorem (twice):\\begin{align*} &\\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\\\ &\\implies \\left(d+r-\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_2)^2 = \\left(d-r+\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_1)^2 \\\\ &\\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\\\ &\\implies 4dr = 8rr_2-8rr_1 \\\\ &\\implies d=2r_2-2r_1 \\end{align*}Therefore, $d=2(r_2-r_1) = 2(961-625)=\\boxed{672}$. Solution 2 (Official MAA, Unedited) Denote by $O_1$$O_2$, and $O$ the centers of $\\omega_1$$\\omega_2$, and $\\omega$, respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\\omega_1$ and $\\omega_2$ respectively, $r$ be the radius of $\\omega$, and $\\ell$ the distance from $O$ to the line $AB$. We claim that$$\\dfrac{\\ell}{r} = \\dfrac{R_2-R_1}{d},$$where $d = O_1O_2$. This solves the problem, for then the $\\widehat{PQ} = 120^\\circ$ condition implies $\\tfrac{\\ell}r = \\cos 60^\\circ = \\tfrac{1}{2}$, and then", "the area of the small region is $\\frac{\\pi}{4}-\\frac{1}{2}$ The requested area is area of circle $C$ minus 4 of this area: $\\pi 1^2 - 4(\\frac{\\pi}{4}-\\frac{1}{2}) = \\pi - \\pi + 2 = 2$ $\\boxed{\\textbf{C}}$. ## Solution 2 $[asy] unitsize(1.1cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(2,0), C=(1,1); pair D=(2,1); pair E=(0,1); pair F = (1, 2); pair M = (1, 0); dot (A); dot (B); dot (C); dot (D); dot (E); dot (F); dot (M); draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); draw (D--F--E--M--D); label(\"A\",A,W); label(\"B\",B,E); label(\"C\",C,W); label(\"M\",M,NE); label(\"D\",D,E); label(\"E\",E,W); label(\"F\",F,N); [/asy]$ We can move the area above the part of the circle above the segment $EF$ down, and similarly for the other side. Then, we have a square, whose diagonal is $2$, so the area is then just $\\left(\\frac{2}{\\sqrt{2}}\\right)^2 = 2$.", "Note we can just use mass points to get \\left( \\frac{15}{26} \\right)^2= \\frac{225}{676}$which is \\boxed{901}$. ## Solution 2 First draw $\\overline{CP}$ and extend it so that it meets with $\\overline{AB}$ at point $X$. $[asy] size(10cm); pair A,B,C,D,E,X,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); X=(4.3636,0); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(X); dot(P); dot(Q); dot(R); label(\"A\",A,WSW); label(\"B\",B,ESE); label(\"C\",C,NNW); label(\"D\",D,NE); label(\"E\",E,WNW); label(\"X\",X,SSE); label(\"P\",P,NNE); label(\"Q\",Q,SSW); label(\"R\",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--X); [/asy]$ We have that $[ABC]=\\frac{1}{2}\\cdot AC \\cdot BC\\sin{C}=\\frac{1}{2}\\cdot 4\\cdot {7}\\sin{C}=14\\sin{C}$ By Ceva's, $$3\\cdot{\\frac{2}{5}}\\cdot{\\frac{BX}{AX}}=1\\implies BX=\\frac{5\\cdot AX}{6}$$ That means that $$\\frac{11\\cdot {AX}}{6}=8\\implies AX=\\frac{48}{11} \\ \\text{and} \\ BX=\\frac{40}{11}$$ Now we apply mass points. Assume WLOG that $W_{A}=1$. That means that $$W_{C}=3, W_{B}=\\frac{6}{5}, W_{X}=\\frac{11}{5}, W_{D}=\\frac{21}{5}, W_{E}=4, W_{P}=\\frac{26}{5}$$ Notice now that $\\triangle{PBQ}$ is similar to $\\triangle{EBA}$. Therefore, $$\\frac{PQ}{EA}=\\frac{PB}{EB}\\implies \\frac{PQ}{3}=\\frac{10}{13}\\implies PQ=\\frac{30}{13}$$ Also, $\\triangle{PRA}$ is similar to $\\triangle{DBA}$. Therefore, $$\\frac{PA}{DA}=\\frac{PR}{DB}\\implies \\frac{21}{26}=\\frac{PR}{5}\\implies PR=\\frac{105}{26}$$ Because $\\triangle{PQR}$ is similar to $\\triangle{CAB}$, $\\angle{C}=\\angle{P}$. As a result, $[PQR]=\\frac{1}{2}\\cdot PQ \\cdot PR \\sin{C}=\\frac{1}{2}\\cdot \\frac{30}{13}\\cdot \\frac{105}{26}\\sin{P}=\\frac{1575}{338}\\sin{C}$. Therefore, $$\\frac{[PQR]}{[ABC]}=\\frac{\\frac{1575}{338}\\sin{C}}{14\\sin{C}}=\\frac{225}{676}\\implies 225+676=\\boxed{901}$$ • Not the author writing here, but a note is that Ceva's Theorem was actually not necessary to solve this problem. The information was just nice to know :)", "0 \\Rightarrow x = \\frac{-\\sqrt{3}\\pm 3\\sqrt{7}}{2}$. Since $x$ must be positive, $x = \\frac{3\\sqrt{7}-\\sqrt{3}}{2} \\Rightarrow (C)$. ## Solution 2 (without trigonometry) Draw $OD$ and $OC$ as in the diagram. Draw the altitude from $O$ to $DC$ and call the intersection $E$. $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle); draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); pair D = rotate(300)*(-3.687,1.5513); pair C = rotate(300)*(-2.687,1.5513); pair EE = foot((0.00,0.00),D,C); draw(D--EE--(0,0)); label(\"$$x$$\",(-1.55,2.1),E); label(\"$$x$$\",(0.03,1.5),E); label(\"$$A$$\",(-3.6,2.5513),E); label(\"$$B$$\",(-3.15,1.35),E); label(\"$$C$$\",(0.05,3.20),E); label(\"$$D$$\",(-0.75,4.15),E); label(\"$$O$$\",(0.00,-0.10),E); label(\"$$1$$\",(-0.1,3.8),S); label(\"$$4$$\",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); label(\"$$E$$\", EE,SE); [/asy]$ As proved in the first solution, $\\angle OCD = 150^\\circ$. That makes $\\Delta OCE$ a $30-60-90$ triangle, so $OE = \\frac{x}{2}$ and $CE= \\frac{x\\sqrt 3}{2}$ Since $\\Delta ODE$ is a right triangle, $\\left({\\frac{x}{2}}\\right)^2 + \\left({\\frac{x\\sqrt 3}{2} +1}\\right)^2 = 4^2 \\Rightarrow x^2+x\\sqrt3-15 = 0$ Solving for $x$ gives $x =\\frac{3\\sqrt{7}-\\sqrt{3}}{2}\\Rightarrow (C)$ ## Solution 3 (simply Pythagorean Theorem) $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.55,2.1),E); label(\"$$x$$\",(0.03,1.5),E); label(\"$$A$$\",(-3.6,2.5513),E); label(\"$$B$$\",(-3.15,1.35),E); label(\"$$C$$\",(0.05,3.20),E); label(\"$$D$$\",(-0.75,4.15),E); label(\"$$E$$\",(0,4.17)); label(\"$$F$$\",(0.75,4.15),W); label(\"$$O$$\",(0.00,-0.10),E); label(\"$$1$$\",(-0.1,3.8),S); label(\"$$4$$\",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); draw((0,0)--(-0.5,3.9686));[/asy]$ By symmetry, $E$ is the midpoint of $DF$ and $OE$ is an extension of $OC$. Thus $\\angle OED = 90^\\circ$. Since $OD = 4$ and $DE = \\frac{1}{2}$, $OE = \\sqrt{16-\\frac{1}{4}} = \\frac{\\sqrt{63}}{2} = \\frac{3\\sqrt{7}}{2}$. Since $\\triangle CED$ is $30-60-90$, $CE = \\frac{\\sqrt{3}}{2}$ (this can also be deduced from Pythagoras on $\\triangle CED$). Thus $OC =" ]

[ "To plot: Thepoints which has polar coordinate displaystyle{left({2},frac{{{7}pi}}{{4}}right)} also two alternaitve sets for the same. To plot: Thepoints which has polar coordinate $\\left(2,\\frac{7\\pi }{4}\\right)$ also two alternaitve sets for the same. You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Dora Let the point $\\left(2,\\frac{7\\pi }{4}\\right)$ be denoted by P. Consider the point $\\left(2,\\frac{7\\pi }{4}\\right)$, it is 2 units from the Origin O and line OP makes an angle of $\\frac{7\\pi }{4}$ with the positive direction of the x-axis. Add pi to the angle and change cardinal coordinate to -2 to get one alternate solution. Therefore, the point P $\\left(2,\\frac{7\\pi }{4}\\right)$ can also be represented by $\\left(-2,\\frac{7\\pi }{4}+\\pi \\right)=\\left(-2,\\frac{11\\pi }{4}\\right).$ Change the radial coordinate to -2 and subtract $3\\pi$ from the angle to obtain the other alternate solution. Thus, the point $P\\left(2,\\frac{7\\pi }{4}\\right)$ can be given by $\\left(-2,\\frac{7\\pi }{4}-3\\pi \\right)$ $=\\left(-2,-\\frac{19\\pi }{4}\\right).$ Use online graphing calculator and plot the points of P as shown below in Figure 1. From the Figure 1 it can be noted that alternate representations are plotted on the same location.", "# Chapter 6 - Section 6.3 - Polar Coordinates - Exercise Set - Page 742: 22 (a)$$(8, \\frac{13\\pi }{6})$$ (b)$$(-8, \\frac{7\\pi }{6})$$ (c)$$(8, -\\frac{11\\pi }{6})$$ #### Work Step by Step To plot the point $(r, \\theta )=(8, \\frac{\\pi }{6})$, begin with the $\\frac{\\pi }{6}$ angle. Because $\\frac{\\pi }{6}$ is a positive angle, draw $\\theta = \\frac{\\pi }{6}$ counterclockwise from the polar axis. Now consider $r=8$. Because $r \\gt 0$, plot the point by going out eight units on the terminal side of $\\theta$. Please note that if $n$ is any integer, the point $(r, \\theta )$ can be represented as$$(r, \\theta ) = (r, \\theta +2n\\pi ) \\\\ \\text{or} \\\\ (r, \\theta )= (-r, \\theta +\\pi + 2n\\pi ).$$So we have (a)$$(8, \\frac{\\pi }{6})= (8, \\frac{\\pi }{6}+2(1)\\pi)=(8, \\frac{13\\pi }{6}),$$(b)$$(8, \\frac{\\pi }{6})= (-8, \\frac{\\pi }{6}+\\pi +2(0)\\pi )=(-8, \\frac{7\\pi }{6}),$$(c)$$(8, \\frac{\\pi }{6})= (8, \\frac{\\pi }{6}+2(-1)\\pi)=(8, -\\frac{11\\pi }{6}).$$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "ways (co-terminal angles, positive or negative “$$r$$”, and so on). Thus is is typically easier to convert from polar to rectangular. ## Examples Here are some examples; note you may be asked to convert back to polar into degrees or radians. For converting back to polar, make sure answers are either between 0 and 360° for degrees or 0 to $$2\\pi$$ for radians. (And again, note that when we convert back to polar coordinates, we may not always get the same representation as the polar point we started out with.) Polar Coordinate Convert from Polar to Rectangular Convert from Rectangular to Polar $$\\displaystyle \\left( {2,\\frac{{4\\pi }}{3}} \\right)$$ $$\\displaystyle x=2\\cos \\frac{{4\\pi }}{3}=2\\left( {-\\frac{1}{2}} \\right)=-1$$ $$\\displaystyle y=2\\sin \\frac{{4\\pi }}{3}=2\\left( {-\\frac{{\\sqrt{3}}}{2}} \\right)=-\\sqrt{3}$$ $$\\left( {-1,-\\sqrt{3}} \\right)$$ $$r=\\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\\sqrt{{1+3}}=2$$ $$\\displaystyle \\theta ={{\\tan }^{{-1}}}\\left( {\\frac{{-\\sqrt{3}}}{{-1}}} \\right)=\\frac{{4\\pi }}{3}\\text{ (3rd quadrant)}$$ $$\\displaystyle \\left( {2,\\frac{{4\\pi }}{3}} \\right)$$ $$\\left( {4,\\pi } \\right)$$ $$\\begin{array}{c}x=4\\cos \\pi =4\\left( {-1} \\right)=-4\\\\y=4\\sin \\pi =4\\left( 0 \\right)=0\\end{array}$$ $$\\left( {-4,0} \\right)$$ $$\\displaystyle r=\\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\\sqrt{{16+0}}=4$$ $$\\displaystyle \\begin{array}{c}\\theta ={{\\tan }^{{-1}}}\\left( {\\frac{0}{{-4}}} \\right)=0\\text{ (between 2nd }\\\\\\text{ and 3rd quadrants)}\\\\\\left( {4,\\pi } \\right)\\end{array}$$ $$\\left( {-7,45{}^\\circ } \\right)$$ $$\\displaystyle x=-7\\cos \\left( {45{}^\\circ } \\right)=-\\frac{{7\\sqrt{2}}}{2}$$ $$\\displaystyle y=-7\\sin \\left( {45{}^\\circ } \\right)=-\\frac{{7\\sqrt{2}}}{2}$$ $$\\displaystyle \\left( {-\\frac{{7\\sqrt{2}}}{2},-\\frac{{7\\sqrt{2}}}{2}} \\right)$$ $$r=\\sqrt{{{{x}^{2}}+{{y}^{2}}}}=\\sqrt{{\\frac{{49\\left( 2 \\right)}}{4}+\\frac{{49\\left( 2 \\right)}}{4}}}=7$$ $$\\theta ={{\\tan }^{{-1}}}\\left( 1 \\right)=225{}^\\circ \\text{ (3rd quadrant)}$$ $$\\left( {7,225{}^\\circ } \\right),\\text{ which is the same as }\\left( {-7,45{}^\\circ } \\right)$$ $$\\left( {5,150{}^\\circ", "### Home > PC3 > Chapter 11 > Lesson 11.1.3 > Problem11-48 11-48. Plot each of the following points given in polar coordinates. Then state the other three equivalent polar forms of the coordinates. 1. $(−1, \\frac { \\pi } { 2 })$ From the origin, move $1$ unit left, then rotate $π/2$ radians counterclockwise. 2. $(−5, \\frac { 2 \\pi } { 3 })$ From the origin, move $5$ units left, then rotate $2π/3$ radians clockwise. 3. $(2, -\\frac { 5 \\pi } { 4 })$ From the origin, move $2$ units right, then rotate $5π/4$ radians clockwise.", "$$\\displaystyle{\\left\\lbrace-{6},\\sqrt{{23}},{21},{5.62},{0.4},{3}\\frac{{2}}{{9}},{0},-\\frac{{7}}{{8}},{2.074816}\\ldots\\right\\rbrace}$$ To solve for 'n', we need to cross multiply Given equation : $$\\frac{3}{n}=\\frac{5}{16}$$ For digits before decimals point, multiply each digit with the positive powers of ten where power is equal to the position of digit counted from left to right starting from 0. For digits after decimals point, multiply each digit with the negative powers of ten where power is equal to the position of digit counted from right to left starting from 1. 1) $$10^{0}=1$$ 2) $$10^{1}=10$$ 3) $$10^{2}=100$$ 4) $$10^{3}=1000$$ 5) $$10^{4}=10000$$ And so on... 6) $$10^{-1}=0.1$$ 7) $$10^{-2}=0.01$$ 8) $$10^{-3}=0.001$$ 9) $$10^{-4}=0.0001$$ $$\\sum_{n=1}^\\infty \\frac{(-1)^n}{(2n)!}$$ a) $$(2,-2)$$ b) $$(-1,\\sqrt{3})$$ Find the polar coordinates $$(r,\\theta)$$ of the point, where r is greater than 0 and 0 is less than or equal to $$\\theta$$, which is less than $$2\\pi$$ Find the polar coordinates $$(r,\\theta)$$ of the point, where r is less than 0 and 0 is less than or equal to $$\\theta$$, which is less than $$2\\pi$$", "\\frac{5 \\pi}{6}\\right)$$ 2. $$\\left(2, \\frac{\\pi}{2}\\right)$$ 3. $$\\left(4,-\\frac{7 \\pi}{6}\\right)$$ 4. $$\\left(-2, \\frac{5 \\pi}{3}\\right)$$ Give two alternate sets of coordinates for each point. 5. $$\\left(2,60^{\\circ}\\right)$$ 6. $$\\left(5,330^{\\circ}\\right)$$ 7. $$\\left(2,210^{\\circ}\\right)$$ Graph each equation. 8. $$r=4$$ 9. $$\\theta=\\frac{\\pi}{4}$$ 10. $$r=2 \\theta$$ with $$\\theta: 0 \\leq \\theta \\leq 2 \\pi$$. Convert each point to rectangular form. 11. $$\\left(4, \\frac{2 \\pi}{3}\\right)$$ 12. $$\\left(3, \\frac{\\pi}{4}\\right)$$ 13. $$\\left(5, \\frac{\\pi}{3}\\right)$$ Convert each point to polar form using radians where $$0 \\leq \\theta<2 \\pi$$. 14. (1,3) 15. (1,-4) 16. (2,6) Convert each equation to polar form. 17. $$x=3$$ 18. \\(2 x+4 y=2\\ This page titled 10.1 Polar and Rectangular Coordinates is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.", "Exercise $$\\PageIndex{1}$$ Point $$R$$ has cylindrical coordinates $$(5,\\frac{π}{6},4)$$. Plot $$R$$ and describe its location in space using rectangular, or Cartesian, coordinates. Hint The first two components match the polar coordinates of the point in the $$xy$$-plane. The rectangular coordinates of the point are $$(\\frac{5\\sqrt{3}}{2},\\frac{5}{2},4).$$ If this process seems familiar, it is with good reason. This is exactly the same process that we followed in Introduction to Parametric Equations and Polar Coordinates to convert from polar coordinates to two-dimensional rectangular coordinates. Example $$\\PageIndex{2}$$: Converting from Rectangular to Cylindrical Coordinates Convert the rectangular coordinates $$(1,−3,5)$$ to cylindrical coordinates. Solution Use the second set of equations from Note to translate from rectangular to cylindrical coordinates: \\begin{align*} r^2 &= x^2+y^2 \\\\[5pt] r &=±\\sqrt{1^2+(−3)^2} \\\\[5pt] &= ±\\sqrt{10}. \\end{align*} We choose the positive square root, so $$r=\\sqrt{10}$$.Now, we apply the formula to find $$θ$$. In this case, $$y$$ is negative and $$x$$ is positive, which means we must select the value of $$θ$$ between $$\\frac{3π}{2}$$ and $$2π$$: \\begin{align*} \\tan θ &=\\frac{y}{x}=\\frac{−3}{1} \\\\[5pt] θ &=\\arctan(−3)≈5.03\\,\\text{rad.} \\end{align*} In this case, the z-coordinates are the same in both rectangular and cylindrical coordinates: $z=5. \\nonumber$ The point with rectangular coordinates $$(1,−3,5)$$ has cylindrical coordinates approximately equal to $$(\\sqrt{10},5.03,5).$$ Exercise $$\\PageIndex{2}$$ Convert point $$(−8,8,−7)$$ from Cartesian coordinates to cylindrical coordinates. Hint $$r^2=x^2+y^2$$ and $$\\tan θ=\\frac{y}{x}$$ $$8\\sqrt{2},\\frac{3π}{4},−7)$$ The use of", "may choose any angle coterminal with the angle in the original representation of $$P$$ $$\\left(-4, \\frac{7\\pi}{6}\\right)$$. We pick $$-\\frac{5\\pi}{6}$$ and get $$\\left(-4, -\\frac{5 \\pi}{6}\\right)$$ as our second answer. To plot $$P\\left(117, -\\frac{5\\pi}{2} \\right)$$, we move along the polar axis $$117$$ units from the pole and rotate \\textit{clockwise} $$\\frac{5\\pi}{2}$$ radians as illustrated below. Since $$P$$ is $$117$$ units from the pole, any representation $$(r,\\theta)$$ for $$P$$ satisfies $$r = \\pm 117$$. For the $$r=117$$ case, we can take $$\\theta$$ to be any angle coterminal with $$-\\frac{5\\pi}{2}$$. In this case, we choose $$\\theta = \\frac{3\\pi}{2}$$, and get $$\\left(117, \\frac{3\\pi}{2}\\right)$$ as one answer. For the $$r = -117$$ case, we visualize moving left $$117$$ units from the pole and then rotating through an angle $$\\theta$$ to reach $$P$$. We find that $$\\theta = \\frac{\\pi}{2}$$ satisfies this requirement, so our second answer is $$\\left(-117, \\frac{\\pi}{2}\\right)$$. We move three units to the left of the pole and follow up with a clockwise rotation of $$\\frac{\\pi}{4}$$ radians to plot $$P\\left(-3, -\\frac{\\pi}{4} \\right)$$. We see that $$P$$ lies on the terminal side of $$\\frac{3\\pi}{4}$$. Since $$P$$ lies on the terminal side of $$\\frac{3\\pi}{4}$$, one alternative representation for $$P$$ is $$\\left(3, \\frac{3\\pi}{4}\\right)$$. To find a different representation for $$P$$ with $$r=-3$$, we may choose any angle coterminal with $$-\\frac{\\pi}{4}$$. We choose $$\\theta = \\frac{7\\pi}{4}$$ for our", "When we want to use the polar coordinates introduced into § 48 we have the corresponding formulae $\\left.\\begin{array}{c} g_{11}=-\\frac{r^{2}}{1-x_{1}^{2}},\\ g_{22}=-r^{2}\\left(1-x_{1}^{2}\\right),\\ g_{33}=-1,\\ g_{44}=c^{2},\\\\ \\\\ g_{ab}=0,\\quad\\textrm{for}\\quad a\\ne b \\end{array}\\right\\}$ (100) If, using polar coordinates, we have to do with an attracting sphere and if we take its centre as origin, we may put $g_{11}=-\\frac{u}{1-x_{1}^{2}},\\ g_{22}=-\\left(1-x_{1}^{2}\\right)u,\\ g_{33}=-v,\\ g_{44}=w,$ (101) where $u, v, w$ are functions of $r$. The $g_{ab}$'s which belong to an orthogonal system of coordinates may be expressed in the same functions. These $g_{ab}$'s are $\\begin{array}{l} g_{11}=-\\frac{u}{r^{2}}+\\frac{x_{1}^{2}}{r^{2}}\\left(\\frac{u}{r^{2}}-v\\right),\\ etc.\\\\ \\\\ g_{12}=\\frac{x_{1}x_{2}}{r^{2}}\\left(\\frac{u}{r^{2}}-v\\right),\\ etc.\\\\ \\\\ g_{14}=g_{24}=g_{34}=0,\\ g_{44}=w. \\end{array}$ The \"etc.\" means that for $g_{22},g_{33}$ we have similar expressions as for $g_{11}$ and for$g_{23},g_{31}$ similar ones as for $g_{12}$. § 57. In order to deduce the differential equations determining $u, v, w$ we may arbitrarily use rectangular or polar coordinates; the latter however are here to be preferred. If differentiations with respect to $r$ are indicated by accents, we have according to (40) and (101) $\\begin{array}{l} G_{11}=\\frac{1}{1-x_{1}^{2}}\\left(-1+\\frac{u''}{2v}-\\frac{u'v'}{4v^{2}}+\\frac{u'w'}{4vw}\\right),\\\\ \\\\ G_{22}=\\left(1-x_{1}^{2}\\right)\\left(-1+\\frac{u''}{2v}-\\frac{u'v'}{4v^{2}}+\\frac{u'w'}{4vw}\\right),\\\\ \\\\ G_{33}=\\frac{u''}{u}-\\frac{u'^{2}}{2u^{2}}-\\frac{u'v'}{2uv}-\\frac{v'w'}{4vw}+\\frac{w''}{2w}-\\frac{w'^{2}}{4w^{2}},\\\\ \\\\ G_{44}=-\\frac{u'w'}{2uv}+\\frac{v'w'}{4v^{2}}-\\frac{w''}{2v}+\\frac{w'^{2}}{4vw}, \\end{array}$ $G_{ab}=0$ for $a\\ne b$ So we have found the left hand sides of the field equations (65). Before considering these equations more closely we shall introduce the simplification that the $g_{ab}$'s, are very little different from the normal values (100). For these latter we have $u=r^{2},\\ v=1,\\ w=c^{2}$ (102)", "to polar coordinates, we have \\begin{align*} \\iint_{R^2} e^{-10(x^2+y^2)}\\,dx \\, dy &= \\int_{\\theta=0}^{\\theta=2\\pi} \\int_{r=0}^{r=\\infty} e^{-10r^2}\\,r \\, dr \\, d\\theta = \\int_{\\theta=0}^{\\theta=2\\pi} \\left(\\lim_{a\\rightarrow\\infty} \\int_{r=0}^{r=a} e^{-10r^2}r \\, dr \\right) d\\theta \\\\ &=\\left(\\int_{\\theta=0}^{\\theta=2\\pi}\\right) d\\theta \\left(\\lim_{a\\rightarrow\\infty} \\int_{r=0}^{r=a} e^{-10r^2}r \\, dr \\right) \\\\ &=2\\pi \\left(\\lim_{a\\rightarrow\\infty} \\int_{r=0}^{r=a} e^{-10r^2}r \\, dr \\right) \\\\ &=2\\pi \\lim_{a\\rightarrow\\infty}\\left(-\\frac{1}{20}\\right)\\left(\\left. e^{-10r^2}\\right|_0^a\\right) \\\\ &=2\\pi \\left(-\\frac{1}{20}\\right)\\lim_{a\\rightarrow\\infty}\\left(e^{-10a^2} - 1\\right) \\\\ &= \\frac{\\pi}{10}. \\end{align*} Exercise $$\\PageIndex{7}$$ Evaluate the integral $\\iint_{R^2} e^{-4(x^2+y^2)}dx \\, dy. \\nonumber$ Hint Convert to the polar coordinate system. $$\\frac{\\pi}{4}$$ ## Key Concepts • To apply a double integral to a situation with circular symmetry, it is often convenient to use a double integral in polar coordinates. We can apply these double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals. • The area $$dA$$ in polar coordinates becomes $$r \\, dr \\, d\\theta$$. • Use $$x = r \\, \\cos \\, \\theta, \\, y = r \\, \\sin \\, \\theta$$, and $$dA = r \\, dr \\, d\\theta$$ to convert an integral in rectangular coordinates to an integral in polar coordinates. • Use $$r^2 = x^2 + y^2$$ and $$\\theta = tan^{-1} \\left(\\frac{y}{x}\\right)$$ to convert an integral in polar coordinates to an integral in rectangular coordinates, if needed. • To find the volume in polar coordinates bounded above", "are ok (just change them to radians), your second is not. (-4.47, 180 deg) lies on the x-axis, the point you're considering does not. if you change 4.47 to -4.47 then you should ADD 180 deg to -26.56 deg (but of course, do the angles in radians) c) (3, (-3)^1/2 ) answer: ( 3.46, -pi/ 6) ; -3.46, 7pi/6) obviously you made a typo here. (-3)^(1/2) is not a real number. did you mean -(3)^(1/2) ? 3. Hello, googoogaga! Are you making sketches? 1) Plot the points in polar coordinates and find the corresponding rectangular coordinates for the points. $a)\\;\\left(-2,\\,\\frac{7\\pi}{4}\\right)$ The angle: . $\\theta \\:=\\:\\frac{7\\pi}{4} \\:=\\:315^o$ is in Quadrant 4. $r = -2$ puts the point in Quadrant 2. The rectangular coordinates are: . $\\left(-\\sqrt{2},\\:\\sqrt{2}\\right)$. $b)\\;\\left(0,\\,-\\frac{7\\pi}{6}\\right)$ . . answer: $(0,\\,0)$ . . . . Right! $c)\\;(-3,\\,-1.57)$ That angle is in radians . . . and is approximately $\\frac{\\pi}{2}$ We have: . $\\left(-3,\\,-\\frac{\\pi}{2}\\right)$ $\\theta = -\\frac{\\pi}{2}$ places us on the negative y-axis. $r = -3$ puts us on the positive y-axis. The rectangular coordinates are: . $(0,\\:3)$ 2) The rectangular coordinates are given. Find two sets of polar coordinates for the given points for $0 \\leq \\theta \\leq 2\\pi$ $a)\\;(0,\\,-5)$ You have the wrong angles. And you forgot that they want positive angles. Answers: . $\\left(5,\\,\\frac{3\\pi}{2}\\right),\\:\\left(-5,\\,\\frac{\\pi}{2}\\right)$ $b)\\;(4,\\,-2)$ This point", "## Precalculus (6th Edition) Blitzer The polar coordinates of $\\left( 0,-6 \\right)$ are $\\left( 6,\\frac{3\\pi }{2} \\right)$. The polar coordinates of the point are $\\left( r,\\theta \\right)$. Now, rewrite the polar coordinates in terms of rectangular coordinates as below: $r=\\sqrt{{{x}^{2}}+{{y}^{2}}}$ …… (1) $\\tan \\theta =\\frac{y}{x}$ …… (2) Substituting the values of $x\\ \\text{ and }\\ y$ in (1) and (2), we get \\begin{align} & r=\\sqrt{{{x}^{2}}+{{y}^{2}}} \\\\ & =\\sqrt{{{\\left( 0 \\right)}^{2}}+{{\\left( -6 \\right)}^{2}}} \\\\ & =\\sqrt{36} \\\\ & r=6 \\end{align} And, \\begin{align} & \\tan \\theta =\\frac{y}{x} \\\\ & =\\frac{-6}{0} \\\\ & \\tan \\theta =\\infty \\left( \\text{undefined} \\right) \\end{align} Hence, $\\tan \\theta =\\infty$ And, $\\tan \\frac{\\pi }{2}=\\infty$ The $\\theta$ lies on the negative y-axis which means \\begin{align} & \\theta =\\pi +\\frac{\\pi }{2} \\\\ & =\\frac{3\\pi }{2} \\end{align} Therefore, the polar coordinates of $\\left( 0,-6 \\right)$ are $\\left( 6,\\frac{3\\pi }{2} \\right)$.", "blue region of the first figure below, on and above the line $$\\ x+y=1\\$$, choose $$\\ f\\$$ to be the identity. In the region below that line choose $$\\ f\\$$ to map the green lines (of slope $$1$$) in the first diagram onto the vertical green lines in the second. • I believe that's $[0,1)\\times (0,1)$ on the second picture, but otherwise it looks good. – mechanodroid Dec 27 '19 at 9:26 • Yes, of course you're right. I've now updated the diagram. – lonza leggiera Dec 27 '19 at 12:41 We can try rotating and stretching $$[0,1)^2$$ to $$(-1,1)\\times [0,1)$$ in the sense that points on the $$x$$-axis remain in place, points on the $$y = x$$ line rotate to points on the $$y$$-axis, and points on the $$y$$-axis rotate to points on the negative $$x$$-axis. More formally, a point with polar angle $$\\phi \\in \\left[0, \\frac\\pi2\\right]$$ will rotate to a point with polar angle $$2\\phi$$. We also have to scale the distances of points from the origin. Introduce the following continuous function $$L : [0,2\\pi] \\to [1,\\sqrt{2}]$$ as $$L(\\phi) := \\begin{cases} \\frac1{\\left|\\cos\\phi\\right|}, &\\text{if }\\phi \\in \\left[0, \\frac\\pi4\\right] \\cup \\left[\\frac{3\\pi}4,2\\pi\\right]\\\\ \\frac1{\\sin\\phi}, &\\text{if }\\phi \\in \\left[\\frac\\pi4, \\frac{3\\pi}4\\right] \\end{cases}$$ which measures the width of $$(-1,1)\\times [0,1)$$ along the line with polar angle $$\\phi$$. Now check that $$f : [0,1)^2", "# What is the Cartesian equivalent of polar coordinates (2, pi/6)? Oct 2, 2014 $\\left(r , \\theta\\right) \\to \\left(2 , \\frac{\\pi}{6}\\right)$ $\\left(x , y\\right) \\to \\left(r \\cos \\left(\\theta\\right) , r \\sin \\left(\\theta\\right)\\right)$ Substitute in $r$ and $\\theta$ $\\left(x , y\\right) \\to \\left(2 \\cos \\left(\\frac{\\pi}{6}\\right) , 2 \\sin \\left(\\frac{\\pi}{6}\\right)\\right)$ Remember back to the unit circle and special triangles. $\\frac{\\pi}{6} = {30}^{\\circ}$ $\\cos \\left(\\frac{\\pi}{6}\\right) = \\frac{\\sqrt{3}}{2}$ $\\sin \\left(\\frac{\\pi}{6}\\right) = \\frac{1}{2}$ Substitute in those values. $\\left(x , y\\right) \\to \\left(2 \\cdot \\frac{\\sqrt{3}}{2} , 2 \\cdot \\frac{1}{2}\\right)$ $\\left(x , y\\right) \\to \\left(\\sqrt{3} , 1\\right)$", "corresponds to the polar coords [1, 0] (capisce?). The point (0.5, 0.5) corresponds to $\\left[\\frac{1}{\\sqrt{2}}, \\frac{\\pi}{4}\\right]$ (capisce?). The point (0, 0) corresponds to [0, pi/2] (capisce?). I hope you can see that as you move around the circle from (1, 0) to (0.5, 0.5), r goes from 1 to $\\frac{1}{\\sqrt{2}}$ and $\\varphi$ goes from 0 to $\\frac{\\pi}{4}$. And I hope you can see that as you move around the circle from (0.5, 0.5) to (0, 0), r goes from $\\frac{1}{\\sqrt{2}}$ to 0 and $\\varphi$ goes from $\\frac{\\pi}{4}$ to $\\frac{\\pi}{2}$. Indeed, if you express the circle $(x-\\frac{1}{2})^2+y^2=\\frac{1}{4}\\Rightarrow x^2 - x + y^2 = 0$ in polar coordinates, you get $r = \\cos \\varphi$ ...... If you draw a graph of r versus $\\varphi$ you will clearly see r changing from 1 to 0 as $\\varphi$ goes from 0 to $\\frac{\\pi}{2}$. And obviously r = 1 corresponds to the point (1, 0) and r = 0 corresponds to the origin (0, 0). What happens as $\\varphi$ goes from $\\frac{\\pi}{2}$ to $\\pi$ .....? What happens as $\\varphi$ increases from $\\pi$ ....? Again, you need to jump onto the bus going to parametric city. Make sure it's an express.", "Moreover, since $f$ reverses orientation the direction of this tangent vector must agree with the clockwise orientation. I.e. we have the following situation: Therefore $\\frac{d}{dt}\\big[f(e^{it})\\big]\\big/f(e^{it})$ has constant argument $-\\pi/2$. Call $M(t)$ to its modulus. From $(\\ast)$, $$\\int_{|z|=1}\\frac{f'(z)}{f(z)}\\,dz=\\int_0^{2\\pi}e^{-i\\pi/2}M(t)\\,dt=-i\\int_0^{2\\pi}M(t)\\,dt.$$ From here it follows that $(\\star)$ equals $\\frac{-1}{2\\pi}\\int_0^{2\\pi}M(t)\\,dt$. Now, this is a non-positive number so (since we already knew $(\\star)$ is non-negative) it has to be $\\int_0^{2\\pi}M(t)\\,dt=0$ and hence $M$ is (a.e. and therefore everywhere by continuity) the $0$ function. This means that $f'=0$ on $\\partial\\mathbb D$, so $f'=0$ on $\\overline{\\mathbb D}$ (by the identity theorem for holomorphic functions for example), so $f$ must be constant. (!) • This was as clear as of an explanation I could have hoped for. This was very helpful thank you very much ! – Aurora Borealis Apr 21 '18 at 7:44 Supplementing Jonatan's answer, the reason for the preservation of orientation can also be seen by algebraic means. Again if we consider $$\\frac{1}{2\\pi i}\\int_{|z|=1}\\frac{f'(z)}{f(z)}dz$$ With the counter clockwise orientation. We can take a parametrization of the circle, $\\gamma :[a,b] \\rightarrow\\mathbb{C}$, and perform the following manipulation: $$\\frac{1}{2\\pi i}\\int_{|z|=1}\\frac{f'(z)}{f(z)}dz=\\frac{1}{2\\pi i}\\int_{a}^{b}\\frac{f'(\\gamma(t))}{f(\\gamma(t))}\\gamma'(t)dt=\\frac{1}{2\\pi i}\\int_{a}^{b}\\frac{\\frac{d}{dt}(f(\\gamma(t)))}{f(\\gamma(t))}dt$$ Which is, in turn $$\\frac{1}{2\\pi i}\\int_{f\\circ\\gamma}\\frac{1}{w}dw$$ But this is simply the winding number of the curve $f\\circ \\gamma$ about $0$. If we assume that $f$ maps the CCW unit circle into the CW unit circle", "What is the Cartesian form of (-1,(22pi)/4))? Feb 2, 2017 (0,1) Explanation: First of all, angle $\\frac{22 \\pi}{4}$ is $\\frac{11 \\pi}{2}$ which is equivalent to 4pi +(3pi)/2 that is $\\frac{3 \\pi}{2}$ because $4 \\pi$ just adds 0 to the angle $\\frac{3 \\pi}{2}$ So, first we move by an angle of $\\frac{3 \\pi}{2}$ radians anti-clockwise around the pole from starting position of 0 radians. This is shown in the figure below, as point P. Now since r= -1, we move in the opposite direction of r=1 for angle of $\\frac{3 \\pi}{2}$ . This positions has been indicated by point P' in the figure. As is now evident from the figure, cartesean coordinates for$\\left(- 1 , \\frac{22 \\pi}{4}\\right)$ would be point P' (0,1)", "# What is the polar form of (-2,3)? Mar 19, 2016 $\\left(\\sqrt{13} , \\pi - {\\tan}^{- 1} \\left(\\frac{3}{2}\\right)\\right)$ #### Explanation: To write in polar form, you need to know 1. the distance from the point to the origin 2. the angle the line passing through it and the origin makes with the positive $x$ axis. To solve 1. we use Pythagoras Theorem $r = \\sqrt{{\\left(- 2\\right)}^{2} + {3}^{2}}$ $= \\sqrt{13}$ To solve 2. we first find the quadrant that the point lies in. $y$ is positive while $x$ is negative $\\implies$ quadrant II Then we find the basic angle by taking inverse tangent of $| \\frac{y}{x} |$. $\\alpha = {\\tan}^{- 1} \\left(| \\frac{3}{- 2} |\\right)$ $= {\\tan}^{- 1} \\left(\\frac{3}{2}\\right)$ The angle that we are looking for would be $\\theta = \\pi - \\alpha$ $= \\pi - {\\tan}^{- 1} \\left(\\frac{3}{2}\\right)$ $\\approx 2.16$ Therefore, the polar coordinate is $\\left(\\sqrt{13} , \\pi - {\\tan}^{- 1} \\left(\\frac{3}{2}\\right)\\right)$. Note that the answer above is not unique. You can add any integer multiples of $2 \\pi$ to $\\theta$ to get other representations of the same point.", "the polar axis is clockwise instead of counter-clockwise. Hence, to plot $$R\\left(3.5, -\\frac{3\\pi}{4}\\right)$$ we have the following. From an 'angles first' approach, we rotate $$-\\frac{3\\pi}{4}$$ then move out $$3.5$$ units from the pole. We see that $$R$$ is the point on the terminal side of $$\\theta = -\\frac{3\\pi}{4}$$ which is $$3.5$$ units from the pole. The points $$Q$$ and $$R$$ above are, in fact, the same point despite the fact that their polar coordinate representations are different. Unlike Cartesian coordinates where $$(a,b)$$ and $$(c,d)$$ represent the same point if and only if $$a=c$$ and $$b=d$$, a point can be represented by infinitely many polar coordinate pairs. We explore this notion more in the following example. Example $$\\PageIndex{1}$$:plotpolarex For each point in polar coordinates given below plot the point and then give two additional expressions for the point, one of which has $$r > 0$$ and the other with $$r < 0$$. 1. $$P\\left(2, 240^{\\circ}\\right)$$ 2. $$P\\left(-4, \\frac{7\\pi}{6} \\right)$$ 3. $$P\\left(117, -\\frac{5\\pi}{2} \\right)$$ 4. $$P\\left(-3, -\\frac{\\pi}{4} \\right)$$ Solution 1. Whether we move $$2$$ units along the polar axis and then rotate $$240^{\\circ}$$ or rotate $$240^{\\circ}$$ then move out $$2$$ units from the pole, we plot $$P\\left(2, 240^{\\circ}\\right)$$ below. We now set about finding alternate descriptions $$(r,\\theta)$$ for the point $$P$$. Since $$P$$ is $$2$$ units from the pole, $$r", "Math Help - Polar Coordinate 03 1. Polar Coordinate 03 Find the area inside r= 3 Cos[x] and outside of r= 1 + Cos[x] 2. Hello, camherokid! Find the area inside $r \\,= \\,3\\cos x$ and outside $r \\:= \\:1 + \\cos x$ I hope you made a sketch . . . $r\\:=\\:3\\cos x$ is a circle with center $\\left(\\frac{3}{2},\\,0\\right)$ and radius $r = \\frac{3}{2}$ $r\\:=\\:1 + \\cos x$ is a cardioid with intercepts: $(2,\\,0),\\: \\left(1,\\,\\frac{\\pi}{2}\\right),\\:\\left(1,\\,\\frac{3 \\pi}{2}\\right)$ . . and \"dimples in\" to the origin from the left. The polar formula for the area between two curves is: . $A \\;=\\;\\frac{1}{2}\\int^{\\beta}_{\\alpha}\\left(r_{_2}^ 2 - r_{_1}^2\\right)\\,d\\theta$ The curves intersect when: . $3\\cos x \\;=\\;1 + \\cos x\\quad\\Rightarrow\\quad 2\\cos x \\:=\\:1\\quad\\Rightarrow\\quad \\cos x \\:=\\:\\frac{1}{2}$ . . Hence, they intersect at: . $\\theta \\:=\\:\\pm\\frac{\\pi}{3}$ Therefore: . $A \\;=\\;\\frac{1}{2}\\int^{\\frac{\\pi}{3}}_{-\\frac{\\pi}{3}}\\bigg[(3\\cos\\theta)^2 - (1 + \\cos x)^2\\bigg]\\,d\\theta$ 3. Originally Posted by camherokid Find the area inside r= 3 Cos[x] and outside of r= 1 + Cos[x] We will use the formula: $A = \\int_{ \\alpha}^{ \\beta} \\frac {1}{2} (r_o^2 - r_i^2)~dx$ where $A$ is the area between the curves $r_o$ and $r_i$, $\\alpha$ and $\\beta$ are the limits of integration (the points of intersection), $r_o$ is the outer curve, and $r_i$ is the inner curve. First find the points of intersection: this is where $3 \\cos" ]

[ "# If sides A and B of a triangle have lengths of 1 and 6 respectively, and the angle between them is (pi)/2, then what is the area of the triangle? The area of a triangle is half the product of any two sides times the sine of the angle between them. The angle between the two given sides is $\\setminus \\frac{\\pi}{2}$ radians which is a right angle, so its sine is 1. Therefore:", "# A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/3, the angle between sides B and C is (pi)/2, and the length of B is 5, what is the area of the triangle? $\\text{area of triangle :} S = \\frac{25 \\sqrt{3}}{2}$", "the same lengths on some sides. Simple experiment: draw a triangle with sides length 4, 4, 7 and another with sides length 4, 4, 1. Each has two equal angles, but that angle is very different in one from the other.", "the diameter of the circle. Since you know the measures of two of the sides of each of the two triangles, you may compute the measure of the third side using Pythagoras' theorem. Doing this you will show that they are indeed congruent by showing that the three sides have the same measures.", "then you have the two shorter lengths. Use Pythagoras to find the hypotenuse, and then add the 3 side lengths together to find the triangle's perimeter.", "(In Two-dimensions) Concept: Area of a Triangle [4]34 Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. Chapter: [2] Circle Concept: Number of Tangents from a Point on a Circle S", "A triangle has sides A, B, and C. Sides A and B have lengths of 3 and 2, respectively. The angle between A and C is (13pi)/24 and the angle between B and C is (3pi)/8. What is the area of the triangle? $\\pi - \\frac{3}{8} \\cdot \\pi - \\frac{13}{12} \\cdot \\pi = - \\frac{11}{24} \\pi < 0$?", "# The sides of a $\\Delta$le are three consecutive natural numbers and its largest angle is twice the smallest one.Determine the sides of the triangle $(a)\\;4,5,6\\qquad(b)\\;7,8,9\\qquad(c)\\;10,11,12\\qquad(d)\\;None\\;of\\;these$", ". (Hint: Shift the triangle to the origin by representing two of the sides of the triangles with vectors in standard form.) Also, find a unit vector perpendicular to the plane containing these three points.", "angles and sides of each triangle given below. What relation you can visualize for a side and its opposite angle in each triangle. In this way, we arrive at the following theorem. [Page No. 170] Do This Question Draw a triangle ABC and measure its sides. Find the sum of the sides AB + BC, BC + AC; and AC + AB, compare it with the length of the third side. What do you observe ? [Page No. 171] Solution: AB + BC = 4 + 3 = 7 = 7 > 4 = AC BC + CA > AB; 3 + 4 > 4 CA + AB > BC; 4 + 4 > 3 DE + EF > DF EF + DF > DE FD + DE > EF ∴ Sum of any two sides of a triangle is greater than the third side.", "(x + 2)^2 = (x + 4)^2 (D) x + x + 2 = x + 4 (E) x^2 = (x + 2)(x + 4) If the sides are consecutive even integers, then their lengths are $$\\displaystyle x, x + 2, x+ 4$$. Now what is the extremely well known formula that relates three sides of a right angle triangle? plagiarism! j/k MHF Helper", "• anonymous suppose a triangle has sides a,b,and c, and let beta be the angle opposite the length a.what must be true. Mathematics", "Equation for triangle using angles 1. Aug 2, 2008 ramstin What would be the easiest way to find a length of a side of a triangle using its angles, with one length of a side given. 2. Aug 2, 2008 Defennder Cosine rule and sine rule.", "A triangle has sides A, B, and C. Sides A and B have lengths of 9 and 5, respectively. The angle between A and C is (3pi)/4 and the angle between B and C is pi/4. What is the area of the triangle? The two given angles total $\\pi$. The three angle of a triangle total $\\pi$", "It divides the opposite side of a triangle into the product of the lengths of the other two sides.", "A = (b2 + c2 – a2)/2bc etc. 5. Find the length of a particular side from two different triangles containing the side common and then equating the two values thus obtained.", "opposite side c has measure C, then $c^2= a^2+ b^2- 2ab cos(C)$: to find the length of the third leg. You can use the sine law\", \"if a triangle has sides of length a, b, and c with opposite angles A, B, and C, respectively, then $\\frac{sin(A)}{a}= \\frac{sin(B)}{b}= \\frac{sin(C)}{c}$.", "1. ## Algebra 1) The two sides of the right angle have lengths x and y. Represent the change in the triangle's area if the length of one side is doubled and the length of the other side is halved. 2) The two equal sides of an isosceles triangle each have a length of 2x+3y-1. The perimeter of the triangle is 7x+9y. Determine the length of the third side. 2. ## Re: Algebra Originally Posted by darkangel06 1) The two sides of the right angle have lengths x and y. Represent the change in the triangle's area if the length of one side is doubled and the length of the other side is halved. 2) The two equal sides of an isosceles triangle each have a length of 2x+3y-1. The perimeter of the triangle is 7x+9y. Determine the length of the third side. Hi darkangel06! A rectangular triangle with sides x and y has an area of (1/2)xy. If I double x and half y, I get a rectangular triangle with sides (2x) and (1/2)y. Can you tell what the area is of that second triangle? 3. ## Re: Algebra for the second one they did not mention the area of the triangle 4. ## Re: Algebra Originally Posted by darkangel06 for the second one they did not", "# A triangle has sides A, B, and C. Sides A and B have lengths of 7 and 4, respectively. The angle between A and C is (3pi)/8 and the angle between B and C is (pi)/12. What is the area of the triangle? $\\angle C = \\pi - \\left(\\frac{3 \\pi}{8} + \\frac{\\pi}{12}\\right) = \\frac{13 \\pi}{12} , A r e a = \\frac{1}{2} a b \\sin C = \\frac{1}{2} \\left(7\\right) \\left(4\\right) \\sin \\left(\\frac{13 \\pi}{24}\\right) \\approx 13.88 u n i {t}^{2}$ Start by adding the two given angles then subtract it from $\\pi$ to get the measure of angle C. Then substitute in the lengths for sides a and b and the measure of angle C to the area formula and calculate.", "= (x + 4)^2 (D) x + x + 2 = x + 4 (E) x^2 = (x + 2)(x + 4) If the sides are consecutive even integers, then their lengths are $x, x + 2, x+ 4$. Now what is the extremely well known formula that relates three sides of a right angle triangle? 4. plagiarism! j/k :P 5. Originally Posted by SpringFan25 plagiarism! j/k :P If Pythagoras was still alive, he'd sue" ]

[ "# A triangle has sides A, B, and C. Sides A and B are of lengths 7 and 4, respectively, and the angle between A and B is pi/6. What is the length of side C? Jun 25, 2016 $C = \\sqrt{65 - 28 \\sqrt{3}} = \\sqrt{16.51} \\cong 4.06 .$ We use Cosine-rule for $\\Delta L M N : {n}^{2} = {l}^{2} + {m}^{2} - 2 \\cdot l \\cdot m \\cdot \\cos \\angle \\left(l , m\\right) .$ $\\therefore {C}^{2} = {A}^{2} + {B}^{2} - 2 A \\cdot B \\cdot \\cos \\angle \\left(A , B\\right) = 49 + 16 - 56 \\cos \\left(\\frac{\\pi}{6}\\right) = 65 - 56 \\cdot \\left(\\frac{\\sqrt{3}}{2}\\right) = 65 - 28 \\sqrt{3} \\cong 65 - 48.49 = 16.51 .$ $\\therefore C = \\sqrt{16.51} \\cong 4.06 .$", "turn, $B'D = O_1O_2 = 15$, and similarly $A'C = 15$. Thus, Ptolmey's theorem on $A'B'CD$ yields $A'D\\cdot B'C + 2\\cdot 16 = 15^2$, whence $A'D = B'C = \\sqrt{193}$. Let $\\alpha = \\angle A'B'D$. The Law of Cosines on triangle $A'B'D$ yields $$\\cos\\alpha = \\frac{15^2 + 2^2 - (\\sqrt{193})^2}{2\\cdot 2\\cdot 15} = \\frac{36}{60} = \\frac 35,$$ and hence $\\sin\\alpha = \\tfrac 45$. Thus the distance between bases $AB$ and $CD$ is $12$ (in fact, $\\triangle A'B'D$ is a $9-12-15$ triangle with a $7-12-\\sqrt{193}$ triangle removed), which implies the area of $A'B'CD$ is $\\tfrac12\\cdot 12\\cdot(2+16) = 108$. Now let $O_1C = O_2A' = r_1$ and $O_2D = O_1B' = r_2$; the tangency of circles $\\omega_1$ and $\\omega_2$ implies $r_1 + r_2 = 15$. Furthermore, angles $A'O_2D$ and $A'B'D$ are opposite angles in cyclic quadrilateral $B'A'O_2D$, which implies the measure of angle $A'O_2D$ is $180^\\circ - \\alpha$. Therefore, the Law of Cosines applied to triangle $\\triangle A'O_2D$ yields \\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \\tfrac45r_1r_2\\\\ &= (r_1+r_2)^2 - \\tfrac45 r_1r_2 = 225 - \\tfrac45r_1r_2. \\end{align*} Thus $r_1r_2 = 40$, and so the area of triangle $A'O_2D$ is $\\tfrac12r_1r_2\\sin\\alpha = 16$. Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\\cdot 16 = \\boxed{140}$. ~djmathman ## Solution 2 Denote by", "+0 # Help 0 105 2 +199 We have a triangle $\\triangle ABC$ such that $AB = BC = 5$ and $AC = 4.$ If $AD$ is an angle bisector such that $D$ is on $BC,$ then find the value of $AD^2.$ Express your answer as a common fraction. #1 +964 +3 The angle bisector theorem also states that the length of the angle bisector satisfies: $$\\overline{CD}=ab-xy, \\frac{a}{x}=\\frac{b}{y}.$$ Therefore, using the dimensions in the current problem, we reach: $$AC\\div CD=AB\\div BC$$ $${AD}^2=AC\\cdot{A}B-CD\\cdot{BD}$$ Filling in the numbers, we get: $$\\frac4{x}=\\frac{5}{5-x}\\Rightarrow x=\\frac{20}9, 5-x=\\frac{25}9$$ $${AD}^2=4\\cdot{5}-\\frac{20}{9}\\cdot\\frac{25}9=\\frac{1120}{81}$$ I hope this helped, Gavin GYanggg Aug 7, 2018 #2 +91160 +1 Thanks, Gavin !!!!....here's another way using the Law of Cosines Here's a pic : Because AD is an angle bisector, then AC /AB = CD/BD So 4/5 = CD/BD Then there are 9 parts of BC , then CD are 4 of them...so CD = 5 (4/9) = 20/9 So BD = 5 - 20/9 = 45/9 - 20/9 = 25/9 CD^2 = AD^2 + AC^2 - ( AD * 4) cos DAC (20/9)^2 = AD^2 + 4^2 - ( AD * 4) cos DAC (400/81 - 16 - AD^2) / ( - 4AD) = cosDAC (1) (625/81 - 25 - AD^2) / ( - 5AD) = cos DAB (2) Since angle DAC =", "# How do you solve the triangle given C=103^circ, a=3/8, b=3/4? Mar 25, 2018 #### Answer: color(purple)(c = 0,9109, hat A = 23.65^@, hat B = 53.35^@ color(indigo)(\"Area of Triangle \" A_t = (1/2) a b sin C = 0.137 \"sq units\" #### Explanation: $\\hat{C} = {103}^{\\circ} , a = \\frac{3}{8} , b = \\frac{3}{4} , \\text{ To solve the triangle}$ Applying Law of cosines, $c = \\sqrt{{a}^{2} + {b}^{2} - 2 a b \\cos C}$ $c = \\sqrt{{\\left(\\frac{3}{8}\\right)}^{2} + {\\left(\\frac{3}{4}\\right)}^{2} - \\left(2 \\cdot \\left(\\frac{3}{8}\\right) \\cdot \\left(\\frac{3}{4}\\right) \\cdot \\cos 103\\right)}$ $c = 0.9109$ Applying Law of sines, $\\sin \\frac{A}{\\frac{3}{8}} = \\sin \\frac{B}{\\frac{3}{4}} = \\sin \\frac{103}{0.9109}$ $\\sin A = \\frac{\\left(\\frac{3}{8}\\right) \\cdot \\sin 103}{0.9109} = 0.4011$ $\\hat{A} = {\\sin}^{-} 1 0.4011 = {23.65}^{\\circ}$ Similarly, $\\sin B = \\frac{\\left(\\frac{3}{4}\\right) \\cdot \\sin 103}{0.9109} = 0.8023$ $\\hat{A} = {\\sin}^{-} 1 0.8023 = {53.35}^{\\circ}$ color(indigo)(\"Area of Triangle \" A_t = (1/2) a b sin C = (1/2) * (3/8) * (3/4) * sin 103 = 0.137 \"sq units\"", "= 3.5$$ weeks 22- Choice D is correct The difference of the file added, and the file deleted is: $$487,236 – 397,756 = 89,480, 987,345 + 89,480 = 1,076,825$$ 23- Choice C is correct 40 students did not have to go to summer school. $$50 – 10 = 40, \\frac{40}{50} = \\frac{4}{5}$$ 24- Choice B is correct Diameter $$= 2r ⇒4 = 2r⇒r = 2$$, Circumference $$= 2πr ⇒ C = 2π (2) ⇒ C = 4π$$ 25- Choice D is correct Write a proportion and solve. $$\\frac{\\frac{1}{4}}{1}= \\frac{60}{x}$$ $$x = \\frac{60}{\\frac{1}{4}} = 240$$ 26- Choice A is correct All angles in a triable add up to 180 degrees. $$90^\\circ + 60^\\circ = 150^\\circ ⇒ x = 180^\\circ − 150^\\circ = 30^\\circ$$ 27- Choice C is correct $$\\sqrt{81} = 9, \\sqrt{49} = 7, 9 × 7 = 63$$ 28- Choice D is correct If $$a = 5$$ then: $$b = \\frac{a^3}{5} -2 ⇒ b = \\frac{5^3}{5} -2 = 25 -2 = 23$$ 29- Choice B is correct An obtuse angle is an angle of greater than 90 degrees and less than 180 degrees. Only choice A is an obtuse angle. 30- Choice D is correct Only choice D is not equal to $$3^2$$ 31- Choice B is correct $$\\sqrt[4]{625} = 5, (5^4=5×5×5×5=625)$$ 32- Choice B is correct ($$r", "# A triangle has sides A, B, and C. Sides A and B are of lengths 14 and 9, respectively, and the angle between A and B is (5pi)/8 . What is the length of side C? Jul 23, 2017 #### Answer: The length of side $C$ is $19.32 \\left(2 \\mathrm{dp}\\right)$ unit. #### Explanation: Sides are $A = 14 , B = 9$ and their included angle $\\angle c = \\frac{5 \\pi}{8} = \\frac{5 \\cdot 180}{8} = {112.5}^{0}$ Applying cosine law we can find side C as $C = \\sqrt{{A}^{2} + {B}^{2} - 2 \\cdot A \\cdot B \\cdot \\cos c}$ or $C = \\sqrt{{14}^{2} + {9}^{2} - 2 \\cdot 14 \\cdot 9 \\cdot \\cos 112.5} \\approx 19.32 \\left(2 \\mathrm{dp}\\right)$ unit The length of side $C$ is $19.32 \\left(2 \\mathrm{dp}\\right)$ unit. [Ans]", "the angles? Please give an example with numbers? I'd really appreciate it! 1. Why didn't you post a question using numbers? 2. I've attached a sketch of an isosceles triangle. H denotes the height and b the base of the triangle. 3. Example: $|\\alpha| = 42^\\circ$ , $|h| = 12\\ cm$ (I use these values so you have the opportunity to control the results by a (more or less) exact construction) 4. Results: $\\tan(42^\\circ)=\\dfrac{12\\ cm}{\\frac12 b}~\\implies~ b = 2 \\cdot \\dfrac{12\\ cm}{\\tan(42^\\circ)} \\approx 21.61\\ cm$ $\\dfrac hl=\\sin(42^\\circ)~\\implies~l=\\dfrac{12\\ cm}{\\sin(42^\\circ)}\\approx 17.93\\ cm$ area: $a = \\frac12 \\cdot b \\cdot h~\\implies~a=\\frac12 \\cdot 21.61\\ cm \\cdot 12\\ cm \\approx 129.66\\ cm^2$", "is also an isosceles trapezoid; in turn, $A'D = O_1O_2 = 15$, and similarly $B'C = 15$. Thus, Ptolmey's theorem on $B'A'CD$ yields $B'D\\cdot A'C + 2\\cdot 16 = 15^2$, whence $B'D = A'C = \\sqrt{193}$. Let $\\alpha = \\angle B'A'D$. The Law of Cosines on triangle $B'A'D$ yields$$\\cos\\alpha = \\frac{15^2 + 2^2 - (\\sqrt{193})^2}{2\\cdot 2\\cdot 15} = \\frac{36}{60} = \\frac 35,$$and hence $\\sin\\alpha = \\tfrac 45$. Thus the distance between bases $AB$ and $CD$ is $12$ (in fact, $\\triangle B'A'D$ is a $9-12-15$ triangle with a $7-12-\\sqrt{193}$ triangle removed), which implies the area of $B'A'CD$ is $\\tfrac12\\cdot 12\\cdot(2+16) = 108$. Now let $O_1C = O_2B' = r_1$ and $O_2D = O_1A' = r_2$; the tangency of circles $\\omega_1$ and $\\omega_2$ implies $r_1 + r_2 = 15$. Furthermore, angles $B'O_2D$ and $B'A'D$ are opposite angles in cyclic quadrilateral $A'B'O_2D$, which implies the measure of angle $B'O_2D$ is $180^\\circ - \\alpha$. Therefore, the Law of Cosines applied to triangle $\\triangle B'O_2D$ yields\\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \\tfrac45r_1r_2\\\\ &= (r_1+r_2)^2 - \\tfrac45 r_1r_2 = 225 - \\tfrac45r_1r_2. \\end{align*} Thus $r_1r_2 = 40$, and so the area of triangle $B'O_2D$ is $\\tfrac12r_1r_2\\sin\\alpha = 16$. Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\\cdot 16 = \\boxed{140}$. ~djmathman ## Solution", "# In triangle ABC, A=31.4, B=53.7, <C=61.3°, how do you find the area? May 28, 2015 First you can use The Law of Cosines to evaluate the length of the third side $C$: ${C}^{2} = {A}^{2} + {B}^{2} - 2 A B \\cos \\gamma$ ${C}^{2} = {31.4}^{2} + {53.7}^{2} - 2 \\cdot 31.4 \\cdot 53.7 \\cdot \\cos {61.3}^{\\circ}$ ${C}^{2} = 985.96 + 2883.69 - 3372.36 \\cdot 0.4802$ ${C}^{2} = 2250.2427$ $C = 47.44$ Now, for the area of the triangle we can use the Heron's formula: $P = \\sqrt{p \\left(p - a\\right) \\left(p - b\\right) \\left(p - c\\right)}$ where $p = \\frac{a + b + c}{2}$ is half of the circumference. $a = 31.4 , b = 53.7 , c = 47.44 , p = 66.27$ $P = \\sqrt{66.27 \\cdot 34.87 \\cdot 12.57 \\cdot 18.83} = \\sqrt{546958.6761}$ $P = 739.57$", "BX^2+CX^2 - 2 \\cdot BX \\cdot CX \\cdot \\cos150^\\circ = 4^2+3^2-24 \\cdot \\frac{-\\sqrt{3}}{2} = 25+12\\sqrt{3}$$ and thus the area of $ABC$ equals $$BC^2\\frac{\\sqrt{3}}{4} = 25\\frac{\\sqrt{3}}{4}+9.$$ so our final answer is $3+4+25+9 = \\boxed{041}.$ Remark: The new figure (the rotations and the triangle) must be twice the original triangle's area. So it is simply $$\\frac{\\frac{9\\sqrt3+16\\sqrt3+25\\sqrt3}{4}+3(\\frac{3*4}{2})}{2}=25\\frac{\\sqrt{3}}{4}+9$$ ### Solution 2 Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties: We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have 2 cases to consider; either the center $O$ of the circles lies in the interior of triangle $ABC$ or it does not (and we shall show that both can happen). To see that the first case can occur, refer to Solution 1 above, or for a less creative and more direct approach proceed as follows. Using the notation from Solution 1, let $\\theta$ be the measure of angle $XAC$ so that angle $BAX$ has measure $60-\\theta$. Let $AB=BC=AC=x$. The law of cosines on triangles $BAX$ and $XAC$ yields $\\cos(60-\\theta)=\\frac{x^2+9}{10x}$ and $\\cos\\theta=\\frac{x^2+16}{10x}$. Solving this system will yield the value of $x$. Since $\\cos\\theta=\\frac{x^2+16}{10x}$ we have that $\\sin\\theta=\\frac{\\sqrt{100x^2-(x^2+16)^2}}{10x}$. Substituting these into the equation $\\frac{x^2+9}{10x}=\\cos(60-\\theta)=\\frac{1}{2}\\cos\\theta+\\frac{\\sqrt{3}}{2}\\sin\\theta$ we obtain $\\frac{x^2+9}{10x}=\\frac{1}{2}\\frac{x^2+16}{10x}+\\frac{\\sqrt{3}}{2}\\frac{\\sqrt{100x^2-(x^2+16)^2}}{10x}$.", "\\triangle A C D, \\; \\angle A C D=90^o. So the area of \\triangle A B C is 5.12/2=30 and the area of \\triangle A C D is 12.16/2=96, for a total area of 30+96=126. 3 By the Law of Cosines:$$AC^2=AB^2+BC^2-2\\cdot AB\\cdot BC\\cdot \\cos \\angle ABC\\iff \\cos \\angle ABC=\\frac{5^2+13^2-12^2}{2\\cdot 5\\cdot 13}=\\frac{5}{13}$$Analogously:$$\\cos \\angle CDA=\\frac{16^2+20^2-12^2}{2\\cdot 16\\cdot 20}=\\frac{4}{5}$$Now, we know that \\angle ABC and \\angle CDA are both acute, because \\angle ABC<\\angle BAC and \\angle ... 2 HINT.-There is the formula$$A=\\frac{\\sqrt{4(l\\cdot l_1)^2-(a^2-b^2+c^2-d^2)^2}}{2}$$where A is the area, a,b,c,d the sides, l and l_1 the two diagonals. You have to calculate the second diagonal and apply the formula. 4 Hint: The quadrilateral is composed by two triangles for which you know the three sides. So you can calculate the area of each triangle with the Heron's formula. 2 Suppose that every element of A is less than \\frac2n\\binom{n}{n/2}. If S\\subseteq A, and |S|=\\frac{n}2, then \\sum S<\\binom{n}{n/2}. Thus, there must be S_0,S_1\\subseteq A such that S_0\\ne S_1, |S_0|=|S_1|=\\frac{n}2, and \\sum S_0=\\sum S_1. Show that S_0\\setminus S_1 and S_1\\setminus S_0 are disjoint subsets of A with the same sum. 1 One way is to consider the Ferrers diagrams of the partitions. It’s easiest to start with a partition of n+r into r parts. Its Ferrers diagram will have exactly r rows. If you remove the", "# A triangle has sides A, B, and C. Sides A and B are of lengths 2 and 3, respectively, and the angle between A and B is (pi)/8 . What is the length of side C? Oct 25, 2016 #### Answer: The length of side $C$ is $1.38 \\left(2 \\mathrm{dp}\\right) u n i t$ #### Explanation: Here the sides A=2 ; B=3 and their included angle $c = \\frac{\\pi}{8} = \\frac{180}{8} = {22.5}^{0}$ are known. Applying cosine law we get $C = \\sqrt{{A}^{2} + {B}^{2} - 2 \\cdot A \\cdot B \\cdot \\cos c} \\mathmr{and} C = \\sqrt{{2}^{2} + {3}^{2} - 2 \\cdot 2 \\cdot 3 \\cdot \\cos 22.5} = 1.38 \\left(2 \\mathrm{dp}\\right) u n i t$[Ans]", "formula that $[ABC] = \\frac{1}{2} \\cdot AC \\cdot AB \\cdot \\sin \\angle A$ By angle chasing, we find that $ACT$ is a triangle with $\\angle A = 30^{\\circ}, \\angle C = 75^{\\circ}$ and $\\angle T = 75^{\\circ}$. Thus $AC = AT = 24$. Switching to the lower triangle $ATB$, $\\angle A = 30^{\\circ}, \\angle T = 105^{\\circ}$, and $\\angle B = 45^{\\circ}$, with $AT = 24$. Using the Law of Sines on $ATB$: $\\frac{AT}{\\sin 45^{\\circ}} = \\frac{AB}{\\sin 105^{\\circ}}$ $24 \\cdot \\sqrt{2} \\cdot \\sin 105^{\\circ} = AB$ $24 \\cdot \\sqrt{2} \\cdot \\sin (60^{\\circ}+ 45^{\\circ}) = AB$ $24 \\cdot \\sqrt{2} \\cdot (\\sin 60^{\\circ} \\cos 45^{\\circ} + \\sin 45^{\\circ} \\cos 60^{\\circ}) = AB$ $24 \\cdot \\sqrt{2} \\cdot (\\frac {\\sqrt{3}}{2} + \\frac{1}{2}) \\cdot \\frac{\\sqrt{2}}{2} = AB$ $AB = 12 \\cdot (\\sqrt{3} + 1)$ We now plug in $AC$, $AB$ and $\\sin \\angle A$ into the formula for the area: $[ABC] = \\frac{1}{2} \\cdot AC \\cdot AB \\cdot \\sin \\angle A$ $[ABC] = \\frac{1}{2} \\cdot 24 \\cdot 12 \\cdot (\\sqrt{3} + 1) \\cdot \\frac{\\sqrt{3}}{2}$ $[ABC] = 72\\sqrt{3} \\cdot (\\sqrt{3} + 1)$ $[ABC] = 216 + 72\\sqrt{3}$ Thus the answer is $216 + 72 + 3 = \\boxed{291}$ ## See also 2001 AIME I (Problems • Answer Key • Resources) Preceded byProblem 3 Followed byProblem 5 1 • 2 • 3 • 4 •", "30^\\circ+r\\sec 30^\\circ}{2r-r\\tan 30^\\circ}$$ So far as the conjecture goes, we can stop here, since there's clearly no chance of introducing$\\sqrt 5, and thus no appearance of the golden ratio. For completeness, though, we can evaluate the ratio and get ... $$\\frac{4}{11}\\;\\left(\\;1 + 2 \\sqrt{3}\\;\\right) = ... 1 We know that ab = cd by the power of the point P, and also that a + b = c + d because the lengths of the chords AB and CD are equal. So$$ b = \\frac{cd}{a} \\\\ \\implies a + \\frac{cd}{a} = c + d \\\\ \\implies a^2 + cd = ac + ad \\\\ \\implies a^2 - ac + cd - ad = 0 \\\\ \\implies (a-c)(a-d) = 0 \\\\ \\implies a = c\\quad \\text{OR}\\quad a = d $$We obtain that ... 1 So applying cosine rule I got$$\\small PI^2=4R^2+16R^2\\sin^2\\frac{B}{2}\\sin^2\\frac{C}{2} -16R^2\\cos A\\sin\\frac{B}{2}\\sin\\frac{C}{2}\\Bigg(\\cos\\frac{B}{2}\\cos\\frac{C}{2}+\\sin\\frac{B}{2}\\sin\\frac{C}{2}\\Bigg)$$I think that you have a typo (the red part) :$$\\begin{align}&\\small PI^2=4R^2\\color{red}{\\cos^2A}+16R^2\\sin^2\\frac{B}{2}\\sin^2\\frac{C}{2} ... 1 Suppose by contradiction there exists a linea$passing through a single point$A$. By axiom 3 there exist two other points$B$and$C$and by axiom 1 they belong to a line$r$parallel to$a$. By axiom 1 there exists a line$s$passing through$A$and$B$and by axiom 2 there exists a line$t$passing through$C$and parallel to$s$. Line$t$does not ... 1 That's from Kedlaya I believe? First of all, let's mention a useful theorem:", "acd & abd are similar. so, $$\\frac{4}{3} = \\frac{cd}{4}$$ ... cd = $$\\frac{16}{3}$$ it can't be $$\\frac{4}{3} = \\frac{4}{cd}$$ ... cd = 3, because in that case $$5^2+5^2 = 6^2$$ is not true If the 3 triangles are proportional, why can't I solve using the ratio: $$\\frac{AD}{AB} = \\frac{AB}{(X+3)}$$ $$\\frac{4}{5} = \\frac{5}{(X+3)}$$ 4(X+3) = 25 4X + 12 = 25 X = $$\\frac{13}{4}$$ Attachment: splittingtriangle.jpg [ 4.22 KiB | Viewed 39479 times ] In similar triangles the ratio of the corresponding sides are equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram). The ratios you are equating are not of corresponding sides. If you want to equate AD/AB then it should be AD/AB=AC/BC --> AD/5=4/3 --> AD=20/3. Also AD/AB=CD/AC --> (20/3)/5=CD/4 --> CD=16/3. I merged this thread with an earlier discussion of the same question, so check this post: in-triangle-abc-if-bc-3-and-ac-4-then-what-is-the-126937.html#p1050155 it might help to clear your doubts. _________________ Senior Manager Joined: 07 Sep 2010 Posts: 335 Followers: 4 Kudos [?]: 449 [0], given: 136 Re: In triangle ABC, if BC = 3 and AC = 4, then what is the [#permalink] ### Show Tags 17 May 2012, 05:46 Hi Bunuel, How did you infer that angle BAC = angle ADC. Could you please explain that part.. I know this", "$\\displaystyle \\frac{\\sin\\angle ADB}{AB}=\\frac{\\sin\\angle BAD}{BD}$ $\\displaystyle \\frac{\\sin\\angle ADB}{10}=\\frac{\\sin 86^\\circ}{15}$ $\\therefore \\angle ADB = \\sin^{-1} \\dfrac{10\\sin 86^\\circ}{15}$ $\\therefore \\angle ABD = 180^\\circ - 86^\\circ - \\sin^{-1} \\dfrac{2\\sin 86^\\circ}{3} = 52.31439868 \\approx 52.3^\\circ \\text{ (3 s.f.)}$ By Cosine Law, $CD^2=AC^2+AD^2-2AC\\cdot AD \\cos\\angle CBD$. $\\therefore CD = \\sqrt{8^2+15^2-2\\cdot8\\cdot15\\cdot\\cos 43^\\circ} = 10.65246974 \\approx 10.7 \\text{ cm (3 s.f.)}$ (b) By Sine law, $\\displaystyle \\frac{AD}{\\sin\\angle ABD}=\\frac{BD}{\\sin\\angle BAD}$ $\\therefore AD=\\dfrac{15\\sin\\angle ABD}{\\sin 86^\\circ} \\approx 11.8996447\\ldots$ $AC^2+CD^2=6^2+10.65246974^2=149.4751116\\neq AD^2=141.6015440$ Therefore, do not agree. (20) $\\triangle OPQ$ is an obtuse-angled triangle. Denote the in-centre and the circumcentre of $\\triangle OPQ$ by $I$ and $J$ respectively. It is given that $P$, $I$, and $J$ are collinear. (a) Prove that $OP=PQ$. (3 marks) (b) A rectangular coordinate system is introduced so that the coordinates of $O$ and $Q$ are $(0,0)$ and $(40,30)$ respectively while the y-coordinate of $P$ is 19. Let $C$ be the circle which passes through $O$, $P$, and $Q$. (i) Find the equation of $C$. (ii) Let $L_1$ and $L_2$ be two tangents to $C$ such that the slope of each tangent is $\\dfrac34$ and the y-intercept of $L_1$ is greater than that of $L_2$. $L_1$ cuts the x-axis and y-axis at $S$ and $T$ respectively while $L_2$ cuts the x-axis and the y-axis at $U$ and $V$ respectively. Someone claims that the area of the trapezium $STUV$", "# A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/12 and the angle between sides B and C is pi/12. If side B has a length of 10, what is the area of the triangle? Hence Area of the right angled triangle is $12.5$ square units. The angle between sides A and B is $\\angle C = 5 \\cdot \\frac{180}{12} = {75}^{0}$ The angle between sides B and C is $\\angle A = \\frac{180}{12} = {15}^{0} \\therefore \\angle B = 180 - \\left(75 + 15\\right) = {90}^{0}$ Side $B = 10$ Applying sine law $\\frac{A}{\\sin} A = \\frac{B}{\\sin} B = \\frac{C}{\\sin} C$ we get $A = 10 \\cdot \\sin 15 = 2.588$ and $C = 10 \\cdot \\sin 75 = 9.659$ Hence Area of the right angled triangle is $= \\frac{1}{2} \\cdot 9.659 \\cdot 2.588 = 12.5$ Square units. Here, Base $= 9.659$ and Height $= 2.588$[Ans]", "# In regular polygon ABCDE…, we have ∠ACD = 120°. How many sides does the polygon have? In regular polygon ABCDE..., we have ∠ACD = 120°. How many sides does the polygon have? How would I start with solving this? AC would have different lengths depending on the polygon. Assume that there exist totally $n$ sides along the vertexes $D,E,\\cdots,A$. Notice that $$120^o=\\angle ACD \\overset{m}=\\frac{1}{2}\\widehat{DE\\cdots A}=\\frac{1}{2}\\cdot 360^o\\cdot \\frac{n}{n+3}.$$ Thus, $n=6.$ As a result, the number of the sides of the polygon is $n+3=9.$ • Either way, I'm not too familiar with geometric notation, could you explain what $\\overset{m} =$ and $\\widehat{DE\\cdots A}$ mean? – orlp Jul 20 '18 at 6:35 • Yep. That is to say, $\\angle ACD$ equals half of the central angle opposite to the arc $DE\\cdots A$. – mengdie1982 Jul 20 '18 at 6:40 The interior angle of a regular polygon is $$\\angle ABC=\\angle BCD=\\frac{180(n-2)}{n}$$ Now the triangle ABC is isosceles, so $$\\angle BCA = \\frac12(180-\\frac{180(n-2)}{n}) \\\\ \\angle BCA+\\angle ACD = \\angle BCD \\\\ \\frac12(180-\\frac{180(n-2)}{n})+120=\\frac{180(n-2)}{n}$$ This gives $n=9$.", "point $E$, such that $AECD$ is a parallelogram. In other words, $$AE = CD = 10$$ and $$EC = DA = 2\\sqrt{65}$$ Now, let's examine $\\triangle ABE$. Since $AB = AE = 10$, the triangle is isosceles, and $\\angle ABE \\cong \\angle AEB$. Note that in parallelogram $AECD$, $\\angle AED$ and $\\angle CDE$ are congruent, so $\\angle ABE \\cong \\angle CDE$ and thus $$\\text{m}\\angle ABD = 180^\\circ - \\text{m}\\angle CDB$$ Define $\\alpha := \\text{m}\\angle CDB$, so $180^\\circ - \\alpha = \\text{m}\\angle ABD$. We use the Law of Cosines on $\\triangle DAB$ and $\\triangle CDB$: $$\\left(2\\sqrt{65}\\right)^2 = 10^2 + BD^2 - 20BD\\cos\\left(180^\\circ - \\alpha\\right) = 100 + BD^2 + 20BD\\cos\\alpha$$ $$14^2 = 10^2 + BD^2 - 20BD\\cos\\alpha$$ Subtracting the second equation from the first yields $$260 - 196 = 40BD\\cos\\alpha \\rightarrow BD\\cos\\alpha = \\frac{8}{5}$$ This means that dropping an altitude from $B$ to some foot $Q$ on $\\overline{CD}$ gives $DQ = \\frac{8}{5}$ and therefore $CQ = \\frac{42}{5}$. Seeing that $CQ = \\frac{3}{5}\\cdot BC$, we conclude that $\\triangle QCB$ is a 3-4-5 right triangle, so $BQ = \\frac{56}{5}$. Then, the area of $\\triangle BCD$ is $\\frac{1}{2}\\cdot 10 \\cdot \\frac{56}{5} = 56$. Since $AP = CP$, points $A$ and $C$ are equidistant from $\\overline{BD}$, so $$\\left[\\triangle ABD\\right] = \\left[\\triangle CBD\\right] = 56$$ and hence $$\\left[ABCD\\right] = 56 + 56 = \\boxed{112}$$", "# A triangle has sides A, B, and C. Sides A and B are of lengths 6 and 5, respectively, and the angle between A and B is (pi)/2 . What is the length of side C? side $c = \\sqrt{61} = 7.81025$ units #### Explanation: Given data side $a = 6$ and side $b = 5$, angle $C = \\frac{\\pi}{2}$ By the Law of Cosines $c = \\sqrt{{a}^{2} + {b}^{2} - 2 \\cdot a \\cdot b \\cdot \\cos C}$ $c = \\sqrt{{6}^{2} + {5}^{2} - 2 \\cdot 6 \\cdot 5 \\cdot \\cos \\left(\\frac{\\pi}{2}\\right)}$ $c = \\sqrt{36 + 25 - 0}$ $c = \\sqrt{61}$ $c = 7.81025$ Have a nice day from the Philippines!!!!" ]

[ "# A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/12 and the angle between sides B and C is pi/12. If side B has a length of 10, what is the area of the triangle? Hence Area of the right angled triangle is $12.5$ square units. The angle between sides A and B is $\\angle C = 5 \\cdot \\frac{180}{12} = {75}^{0}$ The angle between sides B and C is $\\angle A = \\frac{180}{12} = {15}^{0} \\therefore \\angle B = 180 - \\left(75 + 15\\right) = {90}^{0}$ Side $B = 10$ Applying sine law $\\frac{A}{\\sin} A = \\frac{B}{\\sin} B = \\frac{C}{\\sin} C$ we get $A = 10 \\cdot \\sin 15 = 2.588$ and $C = 10 \\cdot \\sin 75 = 9.659$ Hence Area of the right angled triangle is $= \\frac{1}{2} \\cdot 9.659 \\cdot 2.588 = 12.5$ Square units. Here, Base $= 9.659$ and Height $= 2.588$[Ans]", "# In regular polygon ABCDE…, we have ∠ACD = 120°. How many sides does the polygon have? In regular polygon ABCDE..., we have ∠ACD = 120°. How many sides does the polygon have? How would I start with solving this? AC would have different lengths depending on the polygon. Assume that there exist totally $n$ sides along the vertexes $D,E,\\cdots,A$. Notice that $$120^o=\\angle ACD \\overset{m}=\\frac{1}{2}\\widehat{DE\\cdots A}=\\frac{1}{2}\\cdot 360^o\\cdot \\frac{n}{n+3}.$$ Thus, $n=6.$ As a result, the number of the sides of the polygon is $n+3=9.$ • Either way, I'm not too familiar with geometric notation, could you explain what $\\overset{m} =$ and $\\widehat{DE\\cdots A}$ mean? – orlp Jul 20 '18 at 6:35 • Yep. That is to say, $\\angle ACD$ equals half of the central angle opposite to the arc $DE\\cdots A$. – mengdie1982 Jul 20 '18 at 6:40 The interior angle of a regular polygon is $$\\angle ABC=\\angle BCD=\\frac{180(n-2)}{n}$$ Now the triangle ABC is isosceles, so $$\\angle BCA = \\frac12(180-\\frac{180(n-2)}{n}) \\\\ \\angle BCA+\\angle ACD = \\angle BCD \\\\ \\frac12(180-\\frac{180(n-2)}{n})+120=\\frac{180(n-2)}{n}$$ This gives $n=9$.", "# A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/6 and the angle between sides B and C is pi/12. If side B has a length of 7, what is the area of the triangle? The Area is $12.25$ square units The angle between sides A and B $\\angle C = 5 \\cdot \\frac{180}{6} = {150}^{0}$ The angle between sides B and C $\\angle A = \\frac{180}{12} = {15}^{0}$ The angle between sides C and A $\\angle B = 180 - \\left(150 + 15\\right) = {15}^{0}$ By sine law , we know $\\frac{A}{\\sin} A = \\frac{B}{\\sin} B = \\frac{C}{\\sin} C \\therefore A = S \\in \\frac{A}{\\sin} B \\cdot B = 7 \\cdot \\sin \\frac{15}{\\sin} 15 = 7$ Now A=7 ; B=7 and their included angle $\\angle C = {150}^{0} \\therefore$The Area $= A \\cdot B \\cdot \\sin \\frac{C}{2} = 7 \\cdot 7 \\cdot \\sin \\frac{150}{2} = 12.25$ square units[Ans}", "# A triangle has sides A, B, and C. The angle between sides A and B is (3pi)/4 and the angle between sides B and C is pi/12. If side B has a length of 13, what is the area of the triangle? Area of the triangle is $30.93$ Sq units The angle between sides B & C is $\\angle A = \\frac{\\pi}{12} = \\frac{180}{12} = {15}^{0}$ The angle between sides A & B is $\\angle C = 3 \\cdot \\frac{\\pi}{4} = 3 \\cdot \\frac{180}{4} = {135}^{0}$ So Angle $\\angle B = 180 - \\left(135 + 15\\right) = {30}^{0}$ Using sine law we get $\\frac{B}{\\sin} B = \\frac{A}{\\sin} A \\mathmr{and} \\frac{13}{\\sin} 30 = \\frac{A}{\\sin} 15 \\mathmr{and} A = 13 \\cdot \\sin \\frac{15}{\\sin} 30 = 6.73$ Area $= \\frac{A . B . \\sin C}{2} = \\frac{13 \\cdot 6.73 \\cdot \\sin 135}{2} = 30.93$ sq.units", "# A triangle has sides A, B, and C. Sides A and B are of lengths 14 and 9, respectively, and the angle between A and B is (5pi)/8 . What is the length of side C? Jul 23, 2017 #### Answer: The length of side $C$ is $19.32 \\left(2 \\mathrm{dp}\\right)$ unit. #### Explanation: Sides are $A = 14 , B = 9$ and their included angle $\\angle c = \\frac{5 \\pi}{8} = \\frac{5 \\cdot 180}{8} = {112.5}^{0}$ Applying cosine law we can find side C as $C = \\sqrt{{A}^{2} + {B}^{2} - 2 \\cdot A \\cdot B \\cdot \\cos c}$ or $C = \\sqrt{{14}^{2} + {9}^{2} - 2 \\cdot 14 \\cdot 9 \\cdot \\cos 112.5} \\approx 19.32 \\left(2 \\mathrm{dp}\\right)$ unit The length of side $C$ is $19.32 \\left(2 \\mathrm{dp}\\right)$ unit. [Ans]", "How do you solve the triangle RST given R=35^circ, s=16, t=9? May 20, 2017 You're given the included angle. Use the Law of Cosines to solve this problem. The formula : ${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \\cdot \\cos A$ Let's replace $a$ with $r$, $b$ with $s$, and $c$ with $t$. Let's solve for side $r$ first. Substitute values. ${r}^{2} = {16}^{2} + {9}^{2} - \\left(2 \\cdot 16 \\cdot 9\\right) \\cdot \\cos \\left(35\\right)$ ${r}^{2} = 256 + 81 - 288 \\cdot \\cos \\left(35\\right)$ ${r}^{2} \\approx 337 - 235.92$ ${r}^{2} \\approx 101.09$ $r \\approx 10.05$ Now, let's solve for angle S. You can solve using the Law of Sines, but I'll just use the Law of Cosines because you didn't seem to know how to do it. Also, for the sake of simplicity, I'll just use $r = 10$ The rearranged formula of the Law of Cosines (and changed to our variables r, s and t) : $\\cos S = \\frac{{r}^{2} + {t}^{2} - {s}^{2}}{2 r t}$ Substitute values : $\\cos S = \\frac{{10}^{2} + {9}^{2} - {16}^{2}}{2 \\cdot 16 \\cdot 9}$ $\\cos S = \\frac{100 + 81 - 256}{288}$ $\\cos S = - \\frac{25}{96}$ $S = {\\cos}^{-} 1 \\left(- \\frac{25}{96}\\right)$ $S \\approx 74.91$ You can work out the remaining angle using the angles in", "# Triangle A has sides of lengths 7 ,4 , and 8 . Triangle B is similar to triangle A and has a side of length 9 . What are the possible lengths of the other two sides of triangle B? ##### 1 Answer Dec 7, 2017 Possible lengths of the triangle B are Case (1) $9 , 15.75 , 18$ Case (2) $9 , 5.14 , 10.29$ Case (3) $9 , 4.5 , 7.875$ #### Explanation: Triangles A & B are similar. Case (1) $\\therefore \\frac{9}{4} = \\frac{b}{7} = \\frac{c}{8}$ $b = \\frac{9 \\cdot 7}{4} = 15.75$ $c = \\frac{9 \\cdot 8}{4} = 18$ Possible lengths of other two sides of triangle B are $9 , 15.75 , 18$ Case (2) $\\therefore \\frac{9}{7} = \\frac{b}{4} = \\frac{c}{8}$ $b = \\frac{9 \\cdot 4}{7} = 5.14$ $c = \\frac{9 \\cdot 8}{7} = 10.29$ Possible lengths of other two sides of triangle B are $9 , 5.14 , 10.29$ Case (3) $\\therefore \\frac{9}{8} = \\frac{b}{4} = \\frac{c}{7}$ $b = \\frac{9 \\cdot 4}{8} = 4.5$ $c = \\frac{9 \\cdot 7}{8} = 7.875$ Possible lengths of other two sides of triangle B are $9 , 4.5 , 7.875$", "BX^2+CX^2 - 2 \\cdot BX \\cdot CX \\cdot \\cos150^\\circ = 4^2+3^2-24 \\cdot \\frac{-\\sqrt{3}}{2} = 25+12\\sqrt{3}$$ and thus the area of $ABC$ equals $$BC^2\\frac{\\sqrt{3}}{4} = 25\\frac{\\sqrt{3}}{4}+9.$$ so our final answer is $3+4+25+9 = \\boxed{041}.$ Remark: The new figure (the rotations and the triangle) must be twice the original triangle's area. So it is simply $$\\frac{\\frac{9\\sqrt3+16\\sqrt3+25\\sqrt3}{4}+3(\\frac{3*4}{2})}{2}=25\\frac{\\sqrt{3}}{4}+9$$ ### Solution 2 Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties: We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have 2 cases to consider; either the center $O$ of the circles lies in the interior of triangle $ABC$ or it does not (and we shall show that both can happen). To see that the first case can occur, refer to Solution 1 above, or for a less creative and more direct approach proceed as follows. Using the notation from Solution 1, let $\\theta$ be the measure of angle $XAC$ so that angle $BAX$ has measure $60-\\theta$. Let $AB=BC=AC=x$. The law of cosines on triangles $BAX$ and $XAC$ yields $\\cos(60-\\theta)=\\frac{x^2+9}{10x}$ and $\\cos\\theta=\\frac{x^2+16}{10x}$. Solving this system will yield the value of $x$. Since $\\cos\\theta=\\frac{x^2+16}{10x}$ we have that $\\sin\\theta=\\frac{\\sqrt{100x^2-(x^2+16)^2}}{10x}$. Substituting these into the equation $\\frac{x^2+9}{10x}=\\cos(60-\\theta)=\\frac{1}{2}\\cos\\theta+\\frac{\\sqrt{3}}{2}\\sin\\theta$ we obtain $\\frac{x^2+9}{10x}=\\frac{1}{2}\\frac{x^2+16}{10x}+\\frac{\\sqrt{3}}{2}\\frac{\\sqrt{100x^2-(x^2+16)^2}}{10x}$.", "# A triangle has sides A, B, and C. Sides A and B are of lengths 2 and 3, respectively, and the angle between A and B is (pi)/8 . What is the length of side C? Oct 25, 2016 #### Answer: The length of side $C$ is $1.38 \\left(2 \\mathrm{dp}\\right) u n i t$ #### Explanation: Here the sides A=2 ; B=3 and their included angle $c = \\frac{\\pi}{8} = \\frac{180}{8} = {22.5}^{0}$ are known. Applying cosine law we get $C = \\sqrt{{A}^{2} + {B}^{2} - 2 \\cdot A \\cdot B \\cdot \\cos c} \\mathmr{and} C = \\sqrt{{2}^{2} + {3}^{2} - 2 \\cdot 2 \\cdot 3 \\cdot \\cos 22.5} = 1.38 \\left(2 \\mathrm{dp}\\right) u n i t$[Ans]", "${\\displaystyle \\cos ^{2}A+\\cos ^{2}B+\\cos ^{2}C<1,}$ with the reverse inequality holding for an obtuse triangle. For an acute triangle with circumradius R,[4]:p.141,#3167 ${\\displaystyle a\\cos ^{3}A+b\\cos ^{3}B+c\\cos ^{3}C\\leq {\\frac {abc}{4R^{2}}}}$ and[4]:p.155,#S25 ${\\displaystyle \\cos ^{3}A+\\cos ^{3}B+\\cos ^{3}C+\\cos A\\cos B\\cos C\\geq {\\frac {1}{2}}.}$ For an acute triangle,[4]:p.115,#2874 ${\\displaystyle \\sin ^{2}A+\\sin ^{2}B+\\sin ^{2}C>2,}$ with the reverse inequality for an obtuse triangle. For an acute triangle,[4]:p178,#241.1 ${\\displaystyle \\sin A\\cdot \\sin B+\\sin B\\cdot \\sin C+\\sin C\\cdot \\sin A\\leq (\\cos A+\\cos B+\\cos C)^{2}.}$ For any triangle the triple tangent identity states that the sum of the angles' tangents equals their product. Since an acute angle has a positive tangent value while an obtuse angle has a negative one, the expression for the product of the tangents shows that ${\\displaystyle \\tan A+\\tan B+\\tan C=\\tan A\\cdot \\tan B\\cdot \\tan C>0}$ for acute triangles, while the opposite direction of inequality holds for obtuse triangles. We have[4]:p.26,#958 ${\\displaystyle \\tan A+\\tan B+\\tan C\\geq 2(\\sin 2A+\\sin 2B+\\sin 2C)}$ for acute triangles, and the reverse for obtuse triangles. For all acute triangles,[4]:p.40,#1210 ${\\displaystyle (\\tan A+\\tan B+\\tan C)^{2}\\geq (\\sec A+1)^{2}+(\\sec B+1)^{2}+(\\sec C+1)^{2}.}$ ${\\displaystyle a\\tan A+b\\tan B+c\\tan C\\geq 10R-2r.}$ For an acute triangle with area K, [4]:p.103,#2662 ${\\displaystyle ({\\sqrt {\\cot A}}+{\\sqrt {\\cot B}}+{\\sqrt {\\cot C}})^{2}\\leq {\\frac {K}{r^{2}}}.}$ In an acute triangle, the sum of the circumradius R and the inradius r is less than half", "# How do you solve the triangle given C=103^circ, a=3/8, b=3/4? Mar 25, 2018 #### Answer: color(purple)(c = 0,9109, hat A = 23.65^@, hat B = 53.35^@ color(indigo)(\"Area of Triangle \" A_t = (1/2) a b sin C = 0.137 \"sq units\" #### Explanation: $\\hat{C} = {103}^{\\circ} , a = \\frac{3}{8} , b = \\frac{3}{4} , \\text{ To solve the triangle}$ Applying Law of cosines, $c = \\sqrt{{a}^{2} + {b}^{2} - 2 a b \\cos C}$ $c = \\sqrt{{\\left(\\frac{3}{8}\\right)}^{2} + {\\left(\\frac{3}{4}\\right)}^{2} - \\left(2 \\cdot \\left(\\frac{3}{8}\\right) \\cdot \\left(\\frac{3}{4}\\right) \\cdot \\cos 103\\right)}$ $c = 0.9109$ Applying Law of sines, $\\sin \\frac{A}{\\frac{3}{8}} = \\sin \\frac{B}{\\frac{3}{4}} = \\sin \\frac{103}{0.9109}$ $\\sin A = \\frac{\\left(\\frac{3}{8}\\right) \\cdot \\sin 103}{0.9109} = 0.4011$ $\\hat{A} = {\\sin}^{-} 1 0.4011 = {23.65}^{\\circ}$ Similarly, $\\sin B = \\frac{\\left(\\frac{3}{4}\\right) \\cdot \\sin 103}{0.9109} = 0.8023$ $\\hat{A} = {\\sin}^{-} 1 0.8023 = {53.35}^{\\circ}$ color(indigo)(\"Area of Triangle \" A_t = (1/2) a b sin C = (1/2) * (3/8) * (3/4) * sin 103 = 0.137 \"sq units\"", "turn, $B'D = O_1O_2 = 15$, and similarly $A'C = 15$. Thus, Ptolmey's theorem on $A'B'CD$ yields $A'D\\cdot B'C + 2\\cdot 16 = 15^2$, whence $A'D = B'C = \\sqrt{193}$. Let $\\alpha = \\angle A'B'D$. The Law of Cosines on triangle $A'B'D$ yields $$\\cos\\alpha = \\frac{15^2 + 2^2 - (\\sqrt{193})^2}{2\\cdot 2\\cdot 15} = \\frac{36}{60} = \\frac 35,$$ and hence $\\sin\\alpha = \\tfrac 45$. Thus the distance between bases $AB$ and $CD$ is $12$ (in fact, $\\triangle A'B'D$ is a $9-12-15$ triangle with a $7-12-\\sqrt{193}$ triangle removed), which implies the area of $A'B'CD$ is $\\tfrac12\\cdot 12\\cdot(2+16) = 108$. Now let $O_1C = O_2A' = r_1$ and $O_2D = O_1B' = r_2$; the tangency of circles $\\omega_1$ and $\\omega_2$ implies $r_1 + r_2 = 15$. Furthermore, angles $A'O_2D$ and $A'B'D$ are opposite angles in cyclic quadrilateral $B'A'O_2D$, which implies the measure of angle $A'O_2D$ is $180^\\circ - \\alpha$. Therefore, the Law of Cosines applied to triangle $\\triangle A'O_2D$ yields \\begin{align*} 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \\tfrac45r_1r_2\\\\ &= (r_1+r_2)^2 - \\tfrac45 r_1r_2 = 225 - \\tfrac45r_1r_2. \\end{align*} Thus $r_1r_2 = 40$, and so the area of triangle $A'O_2D$ is $\\tfrac12r_1r_2\\sin\\alpha = 16$. Thus, the area of hexagon $ABO_{1}CDO_{2}$ is $108 + 2\\cdot 16 = \\boxed{140}$. ~djmathman ## Solution 2 Denote by", "# A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/12 and the angle between sides B and C is pi/12. If side B has a length of 33, what is the area of the triangle? May 31, 2017 The area of triangle is $136.1 \\left(1 \\mathrm{dp}\\right)$ sq.unit #### Explanation: The angle between sides $A \\mathmr{and} B$ is $\\angle c = \\frac{5 \\pi}{12} = \\frac{5 \\cdot 180}{12} = {75}^{0}$ The angle between sides $B \\mathmr{and} C$ is $\\angle a = \\frac{\\pi}{12} = \\frac{180}{12} = {15}^{0}$ The angle between sides $C \\mathmr{and} A$ is $\\angle b = 180 - \\left(75 + 15\\right) = {90}^{0}$ Applying sine law we can find A/sina=B/sinb ; B=33 :. A/sin15=33/sin90 or A= 33*sin 15~~8.54(2 dp); [sin 90 =1] Now we know the adjacent sides $A , B$ and their included angle $\\angle c$. The area of triangle is ${A}_{t} = \\frac{A \\cdot B \\cdot \\sin c}{2} = \\frac{8.54 \\cdot 33 \\cdot \\sin 75}{2} \\approx 136.1 \\left(1 \\mathrm{dp}\\right)$ sq.unit", "# In triangle MNP, m=16, n=15, p=8, how do you find the cosine of each of the angles? Dec 9, 2016 $\\cos \\hat{M} = 0.1375$ $\\cos \\hat{N} = 0.37109375$ $\\cos \\hat{P} = 0 , 86875$ #### Explanation: If you know two sides and the included angle of a triangle ABC, you can apply the cosine thorem: ${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \\cos \\hat{A}$ Now you can use the inverse formula to know the cosine of an angle when you know the three sides: $\\cos \\hat{A} = \\frac{{b}^{2} + {c}^{2} - {a}^{2}}{2 b c}$ Then $\\cos \\hat{M} = \\frac{{n}^{2} + {p}^{2} - {m}^{2}}{2 n p}$ $= \\frac{{15}^{2} + {8}^{2} - {16}^{2}}{2 \\cdot 15 \\cdot 8}$ $= 0.1375$ hatM=82.1° $\\cos \\hat{N} = \\frac{{m}^{2} + {p}^{2} - {n}^{2}}{2 m p}$ $= \\frac{{16}^{2} + {8}^{2} - {15}^{2}}{2 \\cdot 16 \\cdot 8}$ $= 0.37109375$ hatN=68.22° $\\cos \\hat{P} = \\frac{{m}^{2} + {n}^{2} - {p}^{2}}{2 m n}$ $= \\frac{{16}^{2} + {15}^{2} - {8}^{2}}{2 \\cdot 16 \\cdot 15}$ $= 0 , 86875$ hatP=29.69°", "# A triangle has sides A, B, and C. Sides A and B are of lengths 5 and 3, respectively, and the angle between A and B is pi/3. What is the length of side C? Feb 14, 2017 Length of side $C$ is $4.36 \\left(2 \\mathrm{dp}\\right)$ unit. Sides $A = 5 , B = 3$ and angle $\\angle c = \\frac{\\pi}{3} = \\frac{180}{3} = {60}^{0}$. Applying cosine law we can find $C = \\sqrt{{A}^{2} + {B}^{2} - 2 \\cdot A \\cdot B \\cdot \\cos c}$ or $C = \\sqrt{{5}^{2} + {3}^{2} - 2 \\cdot A \\cdot B \\cdot \\cos 60} = \\sqrt{25 + 9 - \\cancel{2} \\cdot 5 \\cdot 3 \\cdot \\frac{1}{\\cancel{2}}} = \\sqrt{19} = 4.36 \\left(2 \\mathrm{dp}\\right)$ unit. Length of side $C$ is $4.36 \\left(2 \\mathrm{dp}\\right)$ unit. [Ans]", "# A triangle has sides A, B, and C. The angle between sides A and B is pi/3 and the angle between sides B and C is pi/12. If side B has a length of 2, what is the area of the triangle? Sep 2, 2016 C = 1.79 and A = 0.53 #### Explanation: Let the other sides of the Triangle be A and C. then, $A \\cdot \\cos \\left(\\frac{\\pi}{3}\\right) + C \\cdot \\cos \\left(\\frac{\\pi}{12}\\right) = B = 2$ and $A \\cdot \\sin \\left(\\frac{\\pi}{3}\\right) = C \\cdot \\sin \\left(\\frac{\\pi}{12}\\right)$ ie $\\frac{A}{2} + C \\cdot 0.9659 = 2$ and $A \\frac{\\sqrt{3}}{2} = C \\cdot 0.2588$ solve to get the answer as C = 1.79 and A = 0.53", "# Given a triangle with sides measuring 15in, 10in, and 12in; with angle C across from 10in, angle A across from 12in, and angle T across from 15in. How do you find the angle measures? May 5, 2018 color(maroon)(\"Measure of Angles \" color(indigo)(hat A = 52.89^@, hat C = 41.62^@, hat T = 85.49^@ #### Explanation: $\\text{Given : \" a = 12 \" in\", c = 10 \" in\", t = 15 \" in}$ color(green)(\"Knowing three sides of a triangle, we can use Heron's formula to calculate the area.\" $s = \\frac{a + c + t}{2} = \\frac{12 + 10 + 15}{2} = 18.5$ ${A}_{t} = \\sqrt{18.5 \\cdot \\left(18.5 - 12\\right) \\cdot \\left(18.5 - 10\\right) \\cdot \\left(18.5 - 3\\right)} = 59.8 {\\text{ in}}^{2}$ color(crimson)(\"Using two sides and included angle to find area\", ${A}_{t} = \\left(\\frac{1}{2}\\right) \\cdot a \\cdot c \\cdot \\sin T$ $\\sin T = \\frac{2 \\cdot 59.8}{12 \\cdot 10} = 0.9969$ $\\hat{T} = {\\sin}^{-} 1 0.9969 \\approx 85.49 \\circ$ color(maroon)(\"Using Law of Sines\" $\\frac{a}{\\sin} A = \\frac{c}{\\sin} C = \\frac{t}{\\sin} T$ $\\frac{12}{\\sin} A = \\frac{10}{\\sin} C = \\frac{15}{\\sin} \\left({85.49}^{\\circ}\\right)$ $\\sin A = \\frac{12 \\cdot \\sin \\left(85.49\\right)}{15} = 0.7975$ $\\hat{A} = {\\sin}^{-} 1 0.7975 = {52.89}^{\\circ}$ $\\hat{C} = 180 - \\hat{A} - \\hat{T} = 180 - 85.49 - 52.89 = {41.62}^{\\circ}$ May 5, 2018 color(indigo)(hat A = 53.13@,", "# A triangle has sides A, B, and C. Sides A and B are of lengths 4 and 7, respectively, and the angle between A and B is (3pi)/8 . What is the length of side C? Jun 24, 2018 $c = 6.6$ units #### Explanation: $\\hat{C} = \\frac{3 \\pi}{8} , a = 4 , b = 7$ Law of cosines ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \\cos C$ $\\therefore {c}^{2} = {4}^{2} + {7}^{2} - \\left(2 \\cdot 4 \\cdot 7 \\cdot \\cos \\left(\\frac{3 \\pi}{8}\\right)\\right)$ ${c}^{2} = 43.57$ $c = 6.6$ units Jun 24, 2018 $c = \\sqrt{65 - 28 \\cdot \\sqrt{2 - \\sqrt{2}}}$ #### Explanation: Using the Theorem of cosines ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \\cos \\left(\\gamma\\right)$ and note that $\\cos \\left(\\frac{\\pi}{8}\\right) = \\frac{\\sqrt{2 - \\sqrt{2}}}{2}$ then we get ${c}^{2} = 49 + 16 - 2 \\cdot 7 \\cdot 4 \\cdot \\cos \\left(3 \\cdot \\frac{\\pi}{8}\\right)$ so we get ${c}^{2} = 65 - 28 \\sqrt{2 - \\sqrt{2}}$", "# A triangle has sides A, B, and C. Sides A and B have lengths of 7 and 5, respectively. The angle between A and C is (5pi)/24 and the angle between B and C is (pi)/8. What is the area of the triangle? Jan 26, 2017 The area of the triangle is $15.16 \\left(2 \\mathrm{dp}\\right) s q . u n i t$ #### Explanation: The angle between sides A and C is $\\angle b = \\frac{5 \\pi}{24} = 5 \\cdot \\frac{180}{24} = {37.5}^{0}$ The angle between sides B and C is $\\angle a = \\frac{\\pi}{8} = \\frac{180}{8} = {22.5}^{0}$ The angle between sides A and B is $\\angle c = 180 - \\left(37.5 + 22.5\\right) = {120}^{0}$ So we have sides A=7 ; B=5 and their included angle $\\angle c = {120}^{0}$ The area of the triangle is ${A}_{t} = \\frac{A \\cdot B \\cdot \\sin c}{2} = \\frac{7 \\cdot 5 \\cdot \\sin 120}{2} = 15.16 \\left(2 \\mathrm{dp}\\right) s q . u n i t$ [Ans]", "# A triangle has sides A, B, and C. The angle between sides A and B is pi/6 and the angle between sides B and C is pi/12. If side B has a length of 1, what is the area of the triangle? Aug 7, 2017 Area of the triangle is $0.09$ sq. unit. #### Explanation: The angle between sides $A \\mathmr{and} B$ is $\\angle c = \\frac{\\pi}{6} = \\frac{180}{6} = {30}^{0}$. The angle between sides $B \\mathmr{and} C$ is $\\angle a = \\frac{\\pi}{12} = \\frac{180}{12} = {15}^{0}$ The angle between sides $C \\mathmr{and} A$ is $\\angle b = 180 - \\left(30 + 15\\right) = {135}^{0}$ $B = 1$ . Applying sine law we get $\\frac{A}{\\sin} a = \\frac{B}{\\sin} b$ $\\therefore A = B \\cdot \\left(\\sin \\frac{a}{\\sin} b\\right) = 1 \\cdot \\sin \\frac{15}{\\sin} 135 = 0.366$ Now we have $A = 0.366 , B = 1$ and their icluded angle $c = {30}^{0}$:. Area of the triangle is ${A}_{t} = \\frac{A \\cdot B \\cdot \\sin c}{2}$ or ${A}_{t} = \\frac{1 \\cdot 0.366 \\cdot \\sin 30}{2} = \\frac{0.366 \\cdot \\frac{1}{2}}{2} \\approx 0.09$ sq. unit Ans]" ]

[ "y - √(x2 + y2) and simplify", "simplify it any more. From here, trying squaring both sides. c^2=3/2+0.5c^2 0.5c^2=3/2 c=√3? No, that's not right. When I said square both sides, I didn't say to ignore the 1/2 that was from the cos(60). You still need to square that. Come on!", "there, and you can simplify $\\cosh ik$. Also, where's the sum?", "Simplify the equation.", "think about simplifying x + y.", "# Math Help - Simplfying 1. ## Simplfying If $x=bcos\\theta$, where $b>0$ and $\\theta$ is acute, simplify $b^2-x^2$ 2. Dear Punch, $b^{2}-x^{2}=b^{2}-b^{2}cos^{2}x=b^{2}(1-cos^{2}x)=b^{2}sin^{2}x$", "but the problem was that when I was simplifying from (√2)(√2)+(2-√2)(√2)+2√2+d=0, I put it to 2+2√2-2+2+d=0. So I somehow overlooked that I made 2√2=2, which is clearly not correct.", "Simplify a expression with cos 2x I would like to simplify the following expression : r Cos[γ[t]] + Sqrt[l^2 - r^2/2 + 1/2 r^2 Cos[2 γ[t]]] by using the mathematical expression : Cos[2 x]=1 - 2 Sin[x]^2 A similar topic was already created: How to simplify 1 - Cos[2x] to 2 Sin[x]^2? However, I would like to use a very simple method for a Mathematica beginner. May you advice a simple solution to obtain this kind of reduction ? r Cos[γ[t]] + Sqrt[l^2 - r^2/2 + 1/2 r^2 Cos[2 γ[t]]] /.", "# How do you simplify csc x - csc x cos^2x? $\\sin x$ $\\csc x - \\csc x {\\cos}^{2} x = \\csc x \\left(1 - {\\cos}^{2} x\\right) = \\csc x {\\sin}^{2} x = \\sin x$", "1/2}, {aa -> 2}} *) eq /. f -> (c1 #^(2) &) // Simplify (* 0 *)", "side by '-1'. x = 5 Simplifying x = 5`", "# Simplify (1+cos2x)/(sin2x)? Jan 3, 2017 Express $\\cos 2 x$ and $\\sin 2 x$ in terms of $\\cos x$ and $\\sin x$ and simplify. Details see below. #### Explanation: $\\frac{1 + \\cos 2 x}{\\sin 2 x}$ = $\\frac{1 + 2 {\\cos}^{2} x - 1}{2 \\sin x \\cos x}$ = $\\frac{2 {\\cos}^{2} x}{2 \\sin x \\cos x}$ = $\\cos \\frac{x}{\\sin} x$ = $\\cot x$", "Simplify worked on both f=g and also on f-g as one would have expected:", "# How do you simplify e^(2+lnx)? $x {e}^{2}$ ${e}^{2 + \\ln x} = {e}^{2} \\cdot {e}^{\\ln x}$ $= {e}^{2} \\cdot x$", "c = -2ino-1 Simplifying c = -2ino-1`", "as is it would be like saying 2x=2 when you really should simplify to x=1 if that makes sense.", "to simplify it, not the other way round! In other words, start with (g - a)2 + (g - b)2 + (g - c)2", "# How do you simplify (sinX-cosX)^2? Aug 20, 2015 ${\\left(\\sin X - \\cos X\\right)}^{2} = 1 - \\sin 2 X$ ${\\sin}^{2} A + {\\cos}^{2} A = 1$ $\\sin 2 A = 2 \\sin A \\cos A$ (sinX-cosX)^2 = sin^2 X -2sin X cos X + cos^2 X = 1-2sin XcosX = 1-sin2X", "++({2-cos(\\y*pi/2)},0) (1,-\\y) edge[->] ++({2-cos(\\y*pi/2)},0); to make it work. Oct 19 '11 at 9:12", "to simplify. Now, solve for x." ]

[ "half angle formula. \\begin{aligned} \\sqrt{\\dfrac{1-\\cos b}{1+\\cos b}} &=\\dfrac{\\sin b}{1+\\cos b} \\\\ \\dfrac{1-\\cos b}{1+\\cos b} &=\\dfrac{\\sin ^{2} b}{(1+\\cos b)^{2}} \\\\ (1-\\cos b)(1+\\cos b)^{2} &=\\sin ^{2} b(1+\\cos b) \\\\ (1-\\cos b)(1+\\cos b) &=\\sin ^{2} b \\\\ 1-\\cos ^{2} b &=\\sin ^{2} b \\end{aligned} ##### Example $$\\PageIndex{3}$$ Verify the identity: $$\\cot \\dfrac{c}{2} =\\dfrac{\\sin c}{1-\\cos c}$$ Solution Step 1: change cotangent to cosine over sine, then cross-multiply. \\begin{aligned} \\cot \\dfrac{c}{2} &=\\dfrac{\\sin c}{1-\\cos c} \\\\ &=\\dfrac{\\cos \\dfrac{c}{2}}{\\sin \\dfrac{c}{2}}=\\sqrt{\\dfrac{1+\\cos c}{1-\\cos c}} \\\\ \\sqrt{\\dfrac{1+\\cos c}{1-\\cos c}} &=\\dfrac{\\sin c}{1-\\cos c} \\\\ \\dfrac{1+\\cos c}{1-\\cos c} &=\\dfrac{\\sin ^{2} c}{(1-\\cos c)^{2}} \\\\ (1+\\cos c)(1-\\cos c)^{2} &=\\sin ^{2} c(1-\\cos c) \\\\ (1+\\cos c)(1-\\cos c) &=\\sin ^{2} c \\\\ 1-\\cos ^{2} c &=\\sin ^{2} c \\end{aligned} ##### Example $$\\PageIndex{4}$$ Prove that $$\\sin x \\tan \\dfrac{x}{2}+2\\cos x=2\\cos ^2 \\dfrac{x}{2}$$ Solution $$\\begin{array}{l} \\sin x \\tan \\dfrac{x}{2}+2 \\cos x=\\sin x\\left(\\dfrac{1-\\cos x}{\\sin x}\\right)+2 \\cos x \\\\ \\sin x \\tan \\dfrac{x}{2}+2 \\cos x=1-\\cos x+2 \\cos x \\\\ \\sin x \\tan \\dfrac{x}{2}+2 \\cos x=1+\\cos x \\\\ \\sin x \\tan \\dfrac{x}{2}+2 \\cos x=2 \\cos ^{2} \\dfrac{x}{2} \\end{array}$$ ## Review Use half angle identities to find the exact values of each expression. 1. $$\\sin 22.5^{\\circ}$$ 2. $$\\sin 75^{\\circ}$$ 3. $$\\sin 67.5^{\\circ}$$ 4. $$\\sin 157.5^{\\circ}$$ 5. $$\\cos 22.5^{\\circ}$$ 6. $$\\cos 75^{\\circ}$$ 7. $$\\cos 157.5^{\\circ}$$ 8. $$\\cos 67.5^{\\circ}$$ 9. Use the two half angle identities presented in this section to prove", "&= 2\\cos\\theta - \\cos^2\\theta \\end{align} Simplify and rearrange \\begin{align} 1-\\cos^2\\theta &= 2\\cos\\theta - \\cos^2\\theta\\\\ 1 &= 2\\cos\\theta\\\\ \\cos\\theta &= \\frac{1}{2} \\end{align} As required. ###### Example 2 Verify the identity $\\cos(a+b)\\cos(a-b) = \\cos^2a - \\sin^2b$ ###### Solution Use the addition formula and rewrite the left hand side. \\begin{align} \\cos(a+b)\\cos(a-b) &= (\\cos a\\cos b - \\sin a\\sin b)(\\cos a\\cos b+\\sin a\\sin b)\\\\ &=\\cos^2a\\cos^2b + \\cos a\\cos b\\sin a\\sin b - \\cos a\\cos b\\sin a\\sin b - \\sin^2a\\sin^2b\\\\ &=\\cos^2a\\cos^2b - \\sin^2a\\sin^2b \\end{align} Use $\\cos^2x+\\sin^2x=1$ to rewrite everything in terms of $\\cos$. \\begin{align} \\cos^2a\\cos^2b - \\sin^2a\\sin^2b &= \\cos^2a\\cos^2b - (1-\\cos^2a)(1-\\cos^2b)\\\\ &= \\cos^2a\\cos^2b - (1 - \\cos^2a - \\cos^2b + \\cos^2a\\cos^2b)\\\\ &= \\cos^2a + \\cos^2b -1 \\end{align} Use the formula again to rewrite in the form specified. \\begin{align} \\cos^2a + \\cos^2b -1 &=\\cos^2a -(1-\\cos^2b)\\\\ &=\\cos^2a-\\sin^2b \\end{align} ###### Example 3 Derive the formula for $\\cos 2x$ from the addition formula. ###### Solution The addition formula is $cos(\\theta + \\phi) = \\cos\\theta\\cos\\phi - \\sin\\theta\\sin\\phi$. Set $\\theta = x$ and $\\phi = x$ \\begin{align} \\cos 2x &= \\cos (x+x)\\\\ &=\\cos x\\cos x - \\sin x\\sin x\\\\ &=\\cos^2x - \\sin^2x \\end{align} Use $\\cos^2x+\\sin^2x=1$ \\begin{align} \\cos 2x &= \\cos^2x - (1 - \\cos^2x)\\\\ &= 2\\cos^2 x-1 \\end{align} Or \\begin{align} \\cos 2x &= (1-\\sin^2x) - \\sin^2x\\\\ &= 1-2\\sin^2x \\end{align} #### Video Example Prof. Robin Johnson expands $\\cos 6x$ in", "Ptolemy. One can also produce them algebraically using Euler's formula. Sum {\\displaystyle {\\begin{aligned}\\sin \\left(x+y\\right)&=\\sin x\\cos y+\\cos x\\sin y,\\\\\\cos \\left(x+y\\right)&=\\cos x\\cos y-\\sin x\\sin y,\\end{aligned}}} Difference {\\displaystyle {\\begin{aligned}\\sin \\left(x-y\\right)&=\\sin x\\cos y-\\cos x\\sin y,\\\\\\cos \\left(x-y\\right)&=\\cos x\\cos y+\\sin x\\sin y.\\end{aligned}}} These in turn lead to the following three-angle formulae: {\\displaystyle {\\begin{aligned}\\sin \\left(x+y+z\\right)&=\\sin x\\cos y\\cos z+\\sin y\\cos z\\cos x+\\sin z\\cos y\\cos x-\\sin x\\sin y\\sin z,\\\\\\cos \\left(x+y+z\\right)&=\\cos x\\cos y\\cos z-\\cos x\\sin y\\sin z-\\cos y\\sin x\\sin z-\\cos z\\sin x\\sin y,\\end{aligned}}} When the two angles are equal, the sum formulas reduce to simpler equations known as the double-angle formulae. {\\displaystyle {\\begin{aligned}\\sin \\left(2x\\right)&=2\\sin x\\cos x,\\\\\\cos \\left(2x\\right)&=\\cos ^{2}x-\\sin ^{2}x=2\\cos ^{2}x-1=1-2\\sin ^{2}x.\\end{aligned}}} When three angles are equal, the three-angle formulae simplify to {\\displaystyle {\\begin{aligned}\\sin(3x)&=3\\sin x-4\\sin ^{3}x.\\\\\\cos(3x)&=4\\cos ^{3}x-3\\cos x.\\end{aligned}}} These identities can also be used to derive the product-to-sum identities that were used in antiquity to transform the product of two numbers into a sum of numbers and greatly speed operations, much like the logarithm function. ### Calculus For integrals and derivatives of trigonometric functions, see the relevant sections of Differentiation of trigonometric functions, Lists of integrals and List of integrals of trigonometric functions. Below is the list of the derivatives and integrals of the six basic trigonometric functions. The number C is a constant of integration. ${\\displaystyle {\\begin{array}{|r|rr|}f(x)&f'(x)&\\int f(x)\\,dx\\\\\\hline \\sin x&\\cos x&-\\cos x+C\\\\\\cos x&-\\sin x&\\sin x+C\\\\\\tan x&\\sec ^{2}x=1+\\tan ^{2}x&-\\ln", "addition formulas). • Using that $$\\cos$$ is even and that $$\\sin$$ is odd, the addition formulas give the subtraction formulas: \\begin{align*} \\cos(x-y)=&\\cos(x)\\cos(y)+\\sin(x)\\sin(y) \\\\ \\sin(x-y)=&\\sin(x)\\cos(y)-\\cos(x)\\sin(y). \\end{align*} • Dividing the formulas we get: $\\tan(x\\pm y)=\\frac{\\sin(x\\pm y)}{\\cos(x\\pm y)}=\\frac{\\sin(x)\\cos(y)\\pm\\cos(x)\\sin(y)}{\\cos(x)\\cos(y)\\mp\\sin(x)\\sin(y)}=\\frac{\\tan(x)\\pm\\tan(y)}{1\\mp\\tan(x)\\tan(y)}.$ • Putting $$x=y$$ in the addition formulas, we get the double-angle formulas: \\begin{align*} \\cos(2x)=&(\\cos x)^2-(\\sin x)^2 \\\\ \\sin(2x)=&2\\sin(x)\\cos(x) \\end{align*} • Using the identity $$(\\cos x)^2+(\\sin x)^2=1$$ in the double-angle formula for $$\\cos$$, we get the following alternative forms. \\begin{align*} \\cos(2x)=&(\\cos x)^2-(1-(\\cos x)^2)=2(\\cos x)^2-1 \\\\ \\cos(2x)=&(1-(\\sin x)^2)-(\\sin x)^2=1-2(\\sin x)^2 \\end{align*} # Trigonometric identities -- from the addition formulas 2 • Once again, the addition and subtraction formulas are \\begin{align*} \\cos(x\\pm y)=&\\cos(x)\\cos(y)\\mp\\sin(x)\\sin(y) \\\\ \\sin(x\\pm y)=&\\sin(x)\\cos(y)\\pm\\cos(x)\\sin(y), \\end{align*} and the alternative double-angle formulas for $$\\cos$$ are \\begin{align*} \\cos(2x)=&\\cos(x)^2-(1-\\cos(x)^2)=2\\cos(x)^2-1 \\\\ \\cos(2x)=&(1-(\\sin x)^2)-(\\sin x)^2=1-2(\\sin x)^2. \\end{align*} • Rearranging the latter, we get the following half-angle formulas. \\begin{align*} (\\cos x)^2=&\\frac{\\cos(2x)+1}{2} \\\\ (\\sin x)^2=&\\frac{1-\\cos(2x)}{2} \\end{align*} • Finally, rearranging the addition formulas, we get the following product formulas. \\begin{align*} \\cos(x)\\cos(y)=&\\frac{\\cos(x+y)+\\cos(x-y)}{2} \\\\ \\sin(x)\\cos(y)=&\\frac{\\sin(x+y)+\\sin(x-y)}{2} \\\\ \\sin(x)\\sin(y)=&\\frac{\\cos(x-y)-\\cos(x+y)}{2}. \\end{align*} • The half-angle and product formulas will be used frequently in integral calculus, since they are linearization formulas. • Exercises: D.45, 47, 67 # Exponentiation by rational numbers • An exponential function is a function of the form $$f(x)=b^x$$ for some fixed positive number $$b>0$$. Let's recall what the expression $$b^x$$ means. • If $$n$$ is a", "\\cos b)^2+(\\sin a - \\sin b)^2}\\end{align*} Square both sides to eliminate the square root. \\begin{align*}[\\cos (a - b) - 1]^2 + [\\sin (a-b)-0]^2 = (\\cos a - \\cos b)^2 + (\\sin a - \\sin b)^2\\end{align*} FOIL all four squared expressions and simplify. \\begin{align*}\\cos^2 (a-b)-2 \\cos(a-b)+ 1 + \\sin^2 (a-b) & = \\cos^2 a - 2 \\cos a \\cos b + \\cos^2 b + \\sin^2 a - 2 \\sin a \\sin b + \\sin^2 b \\\\ \\underbrace{\\sin^2 (a-b) + \\cos^2 (a-b)} - 2 \\cos (a-b) + 1 & = \\underbrace{\\sin^2 a + \\cos^2 a} - 2 \\cos a \\cos b + \\underbrace{\\sin^2 b + \\cos^2 b} - 2 \\sin a \\sin b \\\\ 1-2 \\cos (a-b) + 1 & = 1 - 2 \\cos a \\cos b + 1 - 2 \\sin a \\sin b \\\\ 2-2 \\cos (a-b) & = 2-2 \\cos a \\cos b -2 \\sin a \\sin b \\\\ -2 \\cos (a-b) & = -2 \\cos a \\cos b - 2 \\sin a \\sin b \\\\ \\cos (a-b) & = \\cos a \\cos b + \\sin a \\sin b\\end{align*} In \\begin{align*}\\cos(a - b) = \\cos a \\cos b + \\sin a \\sin b\\end{align*}, the difference formula for cosine, you can substitute \\begin{align*}a - (- b) = a + b\\end{align*} to obtain: \\begin{align*}\\cos(a + b) = \\cos[a", "}^{2}}x-{{\\cos }^{2}}x)({{({{\\sin }^{2}}x+{{\\cos }^{2}}x)}^{^{2}}}-2{{\\sin }^{2}}x{{\\cos }^{2}}x+{{\\sin }^{2}}x{{\\cos }^{2}}x) \\\\ & =3({{\\sin }^{2}}x-{{\\cos }^{2}}x)-4({{\\sin }^{2}}x-{{\\cos }^{2}}x)(1-{{\\sin }^{2}}x{{\\cos }^{2}}x) \\\\ \\end{align} \\begin{align} & =-3(-{{\\sin }^{2}}x+{{\\cos }^{2}}x)+4({{\\cos }^{2}}x-{{\\sin }^{2}}x)(1-{{\\sin }^{2}}x{{\\cos }^{2}}x) \\\\ & =4({{\\cos }^{2}}x-{{\\sin }^{2}}x)(1-{{\\sin }^{2}}x{{\\cos }^{2}}x)-3({{\\cos }^{2}}x-{{\\sin }^{2}}x) \\\\ \\end{align} Now taking ${{\\cos }^{2}}x-{{\\sin }^{2}}x$common we get, Also, $({{\\cos }^{2}}x-{{\\sin }^{2}}x)=\\cos 2x$ \\begin{align} & =({{\\cos }^{2}}x-{{\\sin }^{2}}x)\\left( 4(1-{{\\sin }^{2}}x{{\\cos }^{2}}x)-3 \\right) \\\\ & =\\cos 2x\\left( 4-4{{\\sin }^{2}}x{{\\cos }^{2}}x-3 \\right) \\\\ & =\\cos 2x\\left( 4-4(1-{{\\cos }^{2}}x){{\\cos }^{2}}x-3 \\right) \\\\ & =\\cos 2x\\left( 1-4{{\\cos }^{2}}x+4{{\\cos }^{4}}x \\right) \\\\ & =\\cos 2x\\left( 1-2\\times 2{{\\cos }^{2}}x+{{(2{{\\cos }^{2}}x)}^{2}} \\right) \\\\ & =\\cos 2x{{\\left( 2{{\\cos }^{2}}x-1 \\right)}^{2}} \\\\ \\end{align} We know that$2{{\\cos }^{2}}x-1=\\cos 2x$, \\begin{align} & =\\cos 2x{{\\left( \\cos 2x \\right)}^{2}} \\\\ & ={{\\cos }^{3}}2x \\\\ \\end{align} $={{\\cos }^{3}}2x=$RHS So $\\sin 3x{{\\sin }^{3}}x+\\cos 3x{{\\cos }^{3}}x={{\\cos }^{3}}2x$ is proved. Hence proved. Note: Read the question in a careful manner. Also, make sure that you know all the identities. You should be familiar with the trigonometric formulae because mostly they are used for conversion. Also, there might be silly mistakes, so kindly avoid the silly mistake.", "$2R_2 \\cos (\\angle O_1 O_2 O_3/2)$ using the identity $\\cos^2 (A/2) = (1 + \\cos A)/2$. From (4), \\begin{aligned} T_1 T_3 &= 2R_2 \\cos (\\angle O_1 O_2 O_3/2)\\\\ &= 2R_2 \\sqrt{\\frac{R_1R_3}{(R_1 - R_2)(R_2 + R_3)}}\\\\ &= \\frac{2\\sqrt{R_1R_2R_3} \\sqrt{R_2}}{\\sqrt{(R_1 - R_2)(R_2 + R_3)}}.\\quad\\quad(9) \\end{aligned} Next we perform the same calculations (4)-(9) with the configuration of three circles tangent externally. By the cosine rule in $\\triangle O_1O_2O_3$, \\begin{aligned} \\cos \\angle O_1 O_2 O_3 &= \\frac{O_1 O_2^2 + O_2O_3^2 - O_1O_3^2}{2O_1 O_2. O_2O_3}\\\\ &= \\frac{(R_1 + R_2)^2 + (R_2 + R_3)^2- (R_1 + R_3)^2 }{2(R_1 + R_2)(R_2 + R_3)}\\\\ &=\\frac{2R_2^2 + 2R_1 R_2 + 2R_2 R_3 - 2R_1 R_3}{2(R_1 + R_2)(R_2 + R_3)}\\\\ &= \\frac{(R_1 + R_2)(R_2+R_3) - 2R_1R_3}{(R_1 + R_2)(R_2 + R_3)}\\\\ &= 1 - \\frac{2R_1R_3}{(R_1 + R_2)(R_2 + R_3)}. \\quad\\quad (10) \\end{aligned} The perimeter of $\\triangle O_1O_2O_3$ is $(R_1 + R_2) + (R_1 + R_3) + (R_2 + R_3) = 2(R_1+R_2+R_3)$, so by Heron’s formula its area is the attractive form $\\displaystyle \\sqrt{R_1R_2 R_3(R_1 + R_2 + R_3)}.\\quad\\quad (11)$ If the points $T_3, O_1, O_2$ have coordinates $(0,0), (-R_1,0), (R_2,0)$ respectively, then $O_3$ has coordinates $(R_2 - (R_2 + R_3)\\cos \\angle O_1 O_2 O_3, (R_2 + R_3)\\sin \\angle O_1 O_2 O_3)$. We have $\\cos \\angle O_1 O_2 O_3$ found above and $\\sin \\angle O_1 O_2 O_3$ can be found", "angles. cos(ab)=cosacosb+sinasinb and cos(a+b)=cosacosbsinasinbcos(ab)cos(a+b)=cosacosb+sinasinb(cosacosbsinasinb)cos(ab)cos(a+b)=cosacosb+sinasinbcosacosb+sinasinbcos(ab)cos(a+b)=2sinasinb12[cos(ab)cos(a+b)]=sinasinb The following product to sum formulas can be derived using the same method: cosαcosβsinαcosβcosαsinβ=12[cos(αβ)+cos(α+β)]=12[sin(α+β)+sin(αβ)]=12[sin(α+β)sin(αβ)] Armed with these four formulas, we can work some examples. #### Example A Change cos2xcos5y\\begin{align*}\\cos 2x \\cos 5y\\end{align*} to a sum. Solution: Use the formula cosαcosβ=12[cos(αβ)+cos(α+β)]\\begin{align*}\\cos \\alpha \\cos \\beta = \\frac{1}{2} \\left [\\cos (\\alpha - \\beta) + \\cos (\\alpha + \\beta) \\right ]\\end{align*}. Set α=2x\\begin{align*}\\alpha = 2x\\end{align*} and β=5y\\begin{align*}\\beta = 5y\\end{align*}. cos2xcos5y=12[cos(2x5y)+cos(2x+5y)] #### Example B Change sin11z+sinz2\\begin{align*}\\frac{\\sin11z + \\sin z}{2}\\end{align*} to a product. Solution: Use the formula sinαcosβ=12[sin(α+β)+sin(αβ)]\\begin{align*}\\sin \\alpha \\cos \\beta = \\frac{1}{2} \\left [\\sin (\\alpha + \\beta) + \\sin (\\alpha - \\beta) \\right ]\\end{align*}. Therefore, α+β=11z\\begin{align*}\\alpha + \\beta = 11z\\end{align*} and αβ=z\\begin{align*}\\alpha - \\beta = z\\end{align*}. Solve the second equation for α\\begin{align*}\\alpha\\end{align*} and plug that into the first. α=z+β(z+β)+β=11zz+2β=11z2β=10zβ=5zandα=z+5z=6z \\begin{align*}\\frac{\\sin11z + \\sin z}{2} = \\sin 6z \\cos 5z\\end{align*}. Again, the sum of \\begin{align*}6z\\end{align*} and \\begin{align*}5z\\end{align*} is \\begin{align*}11z\\end{align*} and the difference is \\begin{align*}z\\end{align*}. #### Example C Solve \\begin{align*}\\cos 5x + \\cos x = \\cos 2x\\end{align*}. Solution: Use the formula \\begin{align*}\\cos \\alpha + \\cos \\beta = 2 \\cos \\frac{\\alpha +\\beta}{2} \\times \\cos \\frac{\\alpha - \\beta}{2}\\end{align*}. ### Guided Practice 1. Express the product as a sum: \\begin{align*}\\sin(6 \\theta) \\sin(4 \\theta) \\end{align*} 2. Express the product as a sum: \\begin{align*}\\sin(5 \\theta) \\cos(2 \\theta)\\end{align*} 3. Express the product as a sum: \\begin{align*}\\cos(10 \\theta) \\sin(3", "part. 23. Mar 20, 2009 ### Cyosis Barring any mistakes from my side you can do it without using any trigonometric identities: \\begin{align*} \\int \\cos^2 x\\,dx &= \\int 1*\\cos^2 x\\,dx \\\\ &= x \\x cos^2 x + \\int 2x \\cos x \\sin x\\,dx \\\\ &= x \\cos^2 x+(x \\sin^2 x- \\int \\sin^2 x\\,dx) \\\\ &= x \\cos^2x + x \\sin^2 x - (-\\sin x \\cos x- \\int -\\cos^2 x\\,dx) \\\\ &= x (\\cos^2x + \\sin^2 x) +\\sin x \\cos x- \\int \\cos^2 x\\,dx \\end{align*} It follows that: $$2\\int \\cos^2 x\\,dx=\\x x(cos^2x + \\sin^2 x) +\\sin x \\cos x \\Rightarrow$$ \\begin{align*} \\int \\cos^2 x\\,dx &= \\frac{1}{2} (\\x x(cos^2x + \\sin^2 x) +\\sin x \\cos x) \\\\ &= \\frac{1}{2} (\\x x +\\frac{1}{2} \\sin {2x}) \\\\ &= \\frac{x}{2}+\\frac{1}{4} \\sin {2x} \\end{align} This is the same result you get by using the double angle identity from the start, as it should be. Last edited: Mar 20, 2009 24. Mar 20, 2009 ### itchy8me brilliant! using the coefficient 1. damn thats a sneaky one. thanks guys and girls ;) !! :-D 25. Jan 5, 2010 ### ntc2415 Just use substitution. This makes it all so much easier. First you set u to be cos x$$\\int(cos x)$$2 dx then substitute cos x with u x$$\\int(u)$$2 dx Then you take the derivative of both sides", "Review question # Can we sketch $y=3 \\cos x - \\cos 3x$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R5251 ## Solution Prove that $\\begin{equation*} \\frac{d}{dx} (3 \\cos x - \\cos 3x)= 6 \\sin x \\cos 2x. \\end{equation*}$ Using the chain rule or by thinking about horizontal scaling, we have that $\\begin{equation*} \\frac{d}{dx} \\cos 3x = - 3 \\sin 3x. \\end{equation*}$ By the compound angle formula, \\begin{align*} \\sin 3x = \\sin(x + 2x) &= \\sin x \\cos 2x + \\sin 2x \\cos x \\\\ &= \\sin x \\cos 2x + 2 \\sin x \\cos^2 x \\\\ &= \\sin x \\cos 2x + \\sin x (1 + \\cos 2x) \\\\ &= 2 \\sin x \\cos 2x + \\sin x \\end{align*} so that \\begin{align*} \\frac{d}{dx} \\left( 3 \\cos x - \\cos 3x \\right) &= - 3 \\sin x + 3 \\sin 3x \\\\ &= - 3 \\sin x + 6 \\sin x \\cos 2x + 3 \\sin x \\\\ &= 6 \\sin x \\cos 2x. \\end{align*} Alternatively, we can use the standard formula, $\\sin P-\\sin Q = 2\\cos\\left(\\dfrac{P+Q}{2}\\right)\\sin\\left(\\dfrac{P-Q}{2}\\right)$. For what two ranges of values of $x$ between $0$ and $\\pi$ is $\\begin{equation*} 3 \\cos x - \\cos 3x \\end{equation*}$ an increasing function of $x$? A", "A.\\cos B+2ab.\\sin A.\\sin B \\\\ & =({{a}^{2}}{{\\cos }^{2}}B+{{b}^{2}}{{\\cos }^{2}}A+2ab.\\cos A.cosB)-({{a}^{2}}{{\\sin }^{2}}B+{{b}^{2}}{{\\sin }^{2}}A-2ab.\\sin A.\\sin B) \\\\ \\end{align} They are of the form ${{a}^{2}}+2ab+{{b}^{2}}={{(a+b)}^{2}}.$ ${{(a\\cos B+b\\cos A)}^{2}}-{{(a\\sin B-b\\sin A)}^{2}}$ By projection formulae, the length of any side of a triangle is equal to the sum of the projections of the other two sides. $\\therefore a\\cos B+b\\cos A=C$for any triangle ABC. \\begin{align} & {{(a\\cos B+b\\cos A)}^{2}}-{{(a\\sin nB+b\\sin A)}^{2}} \\\\ & ={{c}^{2}}-{{(a\\sin B-b\\sin A)}^{2}} \\\\ \\end{align} By sine formula, we know that $\\dfrac{\\sin A}{a}=\\dfrac{sinB}{b}$ Cross multiplying the above expression, we get, \\begin{align} & b\\sin A=a\\sin B \\\\ & ={{c}^{2}}-{{(a\\sin B-b\\sin A)}^{2}} \\\\ & ={{c}^{2}}-{{(a\\sin B-a\\sin B)}^{2}} \\\\ & ={{c}^{2}}-0={{c}^{2}} \\\\ \\end{align} Therefore, we proved that, ${{a}^{2}}\\cos 2B+{{b}^{2}}\\cos 2A+2ab\\cos (A-B)={{c}^{2}}$. Thus the statement is true. Option A is the correct answer. Note: Here we use the cosine formulae to solve the expression. You should remember the formulae so that solving questions like these would become easy. We used the projection formula which is an important concept, to solve the expression. $c=a\\cos B+b\\cos A$.", "of x and y, we'll begin by squaring the parameterization: \\begin{align} x^2 &= 25 cos^2(t) \\\\ y^2 &= 4 sin^2(t) \\end{align} Now divide by 25 on the top and 4 on the bottom: $$\\frac{x^2}{25} = cos^2(t) \\;\\;\\; \\frac{y^2}{4} = sin^2(t)$$ Now we can add these two equations, using the Pythagorean identity to get one on the right side ... $$\\frac{x^2}{25} + \\frac{y^2}{4} = cos^2(t) + sin^2(t) = 1$$ So the formula of our ellipse is $$\\frac{x^2}{25} + \\frac{y^2}{4} = 1$$ which indeed has a long axis along x of length 10 (= 2 × 5) and a short axis along y of 4 units (2 × 2). See the conic sections section if you need a refresher on ellipses. It's not necessarily worth memorizing, but we've also got the idea that the parameterization x = a·cos(t), y = b·sin(t) is an ellipse centered at the origin with axes of 2a and 2b. Practice problems Convert each of these parametric curves to y = f(x) form. Plot a few points of each in parametric form to get an idea of the shape of the curve. 1 \\begin{align} x &= 3 \\, sin(t) \\\\[5pt] y &= 2 \\, cos(t) \\end{align} 2 \\begin{align} x &= t \\, cos(t) \\\\[5pt] y &= t \\, sin(t) \\end{align} 3 \\begin{align} x &= t^2 +", "## Precalculus (6th Edition) Blitzer Some of the identities that are verified using sum-to product formulas are: Verified identities are: \\begin{align} & \\tan x=\\frac{\\cos 3x-\\cos 5x}{\\sin 3x+\\sin 5x} \\\\ & -\\tan x=\\frac{\\cos 3x-\\cos x}{\\sin 3x+\\sin x} \\\\ & -\\cot 2x=\\frac{\\sin 3x-\\sin x}{\\cos 3x-\\cos x} \\\\ & \\tan 3x=\\frac{\\sin 2x+\\sin 4x}{\\cos 2x+\\cos 4x} \\end{align} Other identities are: \\begin{align} & \\frac{\\sin x-\\sin y}{\\sin x+\\sin y}=\\tan \\frac{x-y}{2}\\cot \\frac{x+y}{2} \\\\ & \\frac{\\sin x+\\sin y}{\\sin x-\\sin y}=\\tan \\frac{x+y}{2}\\cot \\frac{x-y}{2} \\\\ & \\frac{\\sin x-\\sin y}{\\cos x-\\cos y}=-\\cot \\frac{x+y}{2} \\end{align} And, \\begin{align} & \\frac{\\sin x+\\sin 3x}{\\cos x+\\cos 3x}=\\tan 2x \\\\ & \\frac{\\cos 4x-\\cos 2x}{\\sin 2x-\\sin 4x}=\\tan 3x \\end{align} Sum-to product formulas are given below: \\begin{align} & \\sin \\alpha +\\sin \\beta =2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2} \\\\ & \\sin \\alpha -\\sin \\beta =2\\sin \\frac{\\alpha -\\beta }{2}\\cos \\frac{\\alpha +\\beta }{2} \\\\ & \\cos \\alpha +\\cos \\beta =2\\cos \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2} \\\\ & \\cos \\alpha -\\cos \\beta =-2\\sin \\frac{\\alpha +\\beta }{2}\\sin \\frac{\\alpha -\\beta }{2} \\end{align} Let us consider the identity: $\\frac{\\cos 3x-\\cos 5x}{\\sin 3x+\\sin 5x}=\\tan x$ By using the sum-to product formula $\\cos \\alpha -\\cos \\beta =-2\\sin \\frac{\\alpha +\\beta }{2}\\sin \\frac{\\alpha -\\beta }{2}$ to numerator and sum-to product formula $\\sin \\alpha +\\sin \\beta =2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2}$ to denominator of left hand side. Then, \\begin{align} & \\frac{\\cos 3x-\\cos 5x}{\\sin 3x+\\sin", "the first identity in the reverse order). $$\\cos{A}+\\cos{(A+2\\pi/3)}+\\cos{(A-2\\pi/3)}=0$$ $$\\cos{(A-\\pi/3)}+\\cos{(A+\\pi/3)}=\\cos{A}$$ Consider the following picture. We see that the triangle with vertices $(\\cos A, \\sin A), (\\cos(A+120^\\circ), \\sin(A+120^\\circ)), (\\cos(A+240^\\circ), \\sin(A+240^\\circ))$ are the vertices of an equilateral triangle with centriod at the origin. Thus if we set $a = \\cos A, b = \\cos(A+120^\\circ), c = \\cos(A+240^\\circ)$, then \\begin{align*} a+b+c = 0 \\end{align*} and hence \\begin{align*} a^3+b^3+c^3 = 3abc \\end{align*} and \\begin{align*} \\cos^3{A} + \\cos^3{(120°+A)} + \\cos^3{(240°+A)} &= 3\\cos A \\cos(A+120^\\circ)\\cos(A+240^\\circ) \\\\ &= \\frac{3}{2}(2\\cos A \\cos(A+120^\\circ))\\cos(A+240^\\circ) \\\\ &= \\frac{3}{2}[\\cos(2A+120^\\circ)+\\cos(120^\\circ)]\\cos(A+240^\\circ)\\\\ &=\\frac{3}{2}\\cos(2A+120^\\circ)\\cos(A+240^\\circ)-\\frac{3}{4}\\cos(A+240^\\circ)\\\\ &=\\frac{3}{4}\\cos(3A+360^\\circ)+\\frac{3}{4}\\cos(A-120^\\circ) -\\frac{3}{4}\\cos(A+240^\\circ)\\\\ &=\\frac{3}{4}\\cos(3A) \\end{align*} since $\\cos(A+240^\\circ) = \\cos(360^\\circ - (A+240^\\circ)) = \\cos(120^\\circ -A) = \\cos(A-120^\\circ)$ • If you can see geometrically, the points $(\\cos A, \\sin A), (\\cos(120^\\circ+A), \\sin(120^\\circ+A)), (\\cos(240^\\circ), \\sin(240^\\circ+A))$ are the vertices of an equilateral triangle with centroid as the origin, then the above proof holds since we have used only an algebraic identity. – user348749 Sep 10, 2016 at 7:23 use that $$\\cos(120^{\\circ}+x)=-1/2\\,\\cos \\left( x \\right) -1/2\\,\\sqrt {3}\\sin \\left( x \\right)$$ and $$\\cos(240^{\\circ}+x)=-1/2\\,\\cos \\left( x \\right) +1/2\\,\\sqrt {3}\\sin \\left( x \\right)$$ For the LHS, use $\\cos(120^\\circ +A)=\\cos 120^\\circ \\cos A - \\sin 120^\\circ \\sin A$ and $\\cos(240^\\circ+A)=\\cos 240^\\circ \\cos A - \\sin 240^\\circ \\sin A$ Then $\\cos^3(120^\\circ+A)=\\left(\\cos 120^\\circ \\cos A - \\sin 120^\\circ \\sin A \\right)^3$ and $\\cos^3(240^\\circ+A)=\\left( \\cos 240^\\circ \\cos A - \\sin 240^\\circ \\sin A \\right)^3$", "Since both equations are equal to k2\\begin{align*}k^2\\end{align*}, we can set them equal to each other and simplify: a2x2a2x2a2x2a2a2+b22bx=c2(bx)2=c2(b22bx+x2)=c2b2+2bxx2=c2b2+2bx=c2\\begin{align*}a^2-x^2 &=c^2-(b-x)^2 \\\\ a^2-x^2 &=c^2-(b^2-2bx+x^2) \\\\ a^2-x^2 &=c^2-b^2+2bx-x^2 \\\\ a^2 &=c^2-b^2+2bx \\\\ a^2+b^2-2bx &=c^2\\end{align*} Recall that we know the values of a\\begin{align*}a\\end{align*} and b\\begin{align*}b\\end{align*} and the measure of angle C\\begin{align*}C\\end{align*}. We don’t know the measure of x\\begin{align*}x\\end{align*}. We can use the cosine ratio as show below to find an expression for x\\begin{align*}x\\end{align*} in terms of what we already know. cosC=xasox=acosC\\begin{align*} \\cos C=\\frac{x}{a} \\quad \\text{so} \\quad x=a \\cos C\\end{align*} Finally, we can replace x\\begin{align*}x\\end{align*} in the equation to get the Law of Cosines: a2+b22abcosC=c2\\begin{align*}a^2+b^2-2ab \\cos C=c^2\\end{align*} Keep in mind that a\\begin{align*}a\\end{align*} and b\\begin{align*}b\\end{align*} are the sides of angle C\\begin{align*}C\\end{align*} in the formula. Let's find c\\begin{align*}c\\end{align*} when mC=80,a=6\\begin{align*}m \\angle C=80^\\circ, a = 6\\end{align*} and b=12\\begin{align*}b = 12\\end{align*}. Replacing the variables in the formula with the given information and solve for c\\begin{align*}c\\end{align*}: c2c2c=62+1222(6)(12)cos80154.99512.4\\begin{align*}c^2 &=6^2+12^2-2(6)(12) \\cos 80^\\circ \\\\ c^2 & \\approx 154.995 \\\\ c & \\approx 12.4\\end{align*} Now, let's find a\\begin{align*}a\\end{align*}, when mA=43\\begin{align*}m \\angle A=43^\\circ\\end{align*}, b=16\\begin{align*}b=16\\end{align*} and c=22\\begin{align*}c=22\\end{align*}. This time we are given the sides surrounding angle A\\begin{align*}A\\end{align*} and the measure of angle A\\begin{align*}A\\end{align*}. We can rewrite the formula as: a2=c2+b22cbcosA\\begin{align*}a^2=c^2+b^2-2cb \\cos A\\end{align*}. Just remember that the length by itself on one side should be the side opposite the angle in the cosine ratio. Now", "are asked to find \\begin{align*}\\sin (2 \\theta)\\end{align*}, we can use our new double angle identity to help simplify the problem. Let's start with the derivation of the double angle identities. One of the formulas for calculating the sum of two angles is: \\begin{align*}\\sin (\\alpha + \\beta) = \\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta\\end{align*} If \\begin{align*}\\alpha\\end{align*} and \\begin{align*}\\beta\\end{align*} are both the same angle in the above formula, then \\begin{align*}\\sin (\\alpha + \\alpha) & = \\sin \\alpha \\cos \\alpha + \\cos \\alpha \\sin \\alpha \\\\ \\sin 2 \\alpha & = 2 \\sin \\alpha \\cos \\alpha\\end{align*} This is the double angle formula for the sine function. The same procedure can be used in the sum formula for cosine, start with the sum angle formula: \\begin{align*}\\cos (\\alpha + \\beta) = \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta\\end{align*} If \\begin{align*}\\alpha\\end{align*} and \\begin{align*}\\beta\\end{align*} are both the same angle in the above formula, then \\begin{align*}\\cos (\\alpha + \\alpha )& = \\cos \\alpha \\cos \\alpha - \\sin \\alpha \\sin \\alpha \\\\ \\cos 2 \\alpha & = \\cos^2 \\alpha - \\sin^2 \\alpha\\end{align*} This is one of the double angle formulas for the cosine function. Two more formulas can be derived by using the Pythagorean Identity, \\begin{align*}\\sin^2 \\alpha + \\cos^2 \\alpha = 1\\end{align*}. \\begin{align*}\\sin^2 \\alpha = 1 - \\cos^2 \\alpha\\end{align*} and likewise \\begin{align*}\\cos^2", ".'. x = 600; 900; 2700; 3000 4. ## Re: Co-functions Originally Posted by MiloMan Thanks Minoanman, I didn't know you where allowed to substitute, I thought the point of the exercise was to determine x with only using co-functions, sin (900 + x) = 2 cos2x cosx = 2cos2x divide both side with cosx .'. 1 = 2 cosx .'. cosx = 1/2 .'. x = 600; 900; 2700; 3000 No, you can't divide both sides by \\displaystyle \\begin{align*} \\cos{(x)} \\end{align*} because it's possible for (and in this case, some solutions ARE where) \\displaystyle \\begin{align*} \\cos{(x)} = 0 \\end{align*}. Instead you need to factorise... \\displaystyle \\begin{align*} \\cos{(x)} &= 2\\cos^2{(x)} \\\\ 0 &= 2\\cos^2{(x)} - \\cos{(x)} \\\\ 0 &= \\cos{(x)} \\left[ 2\\cos{(x)} - 1 \\right] \\\\ \\cos{(x)} &= 0 \\textrm{ or } 2\\cos{(x)} - 1 = 0 \\end{align*} Go from here. 5. ## Re: Co-functions Originally Posted by MiloMan Thanks Minoanman, I didn't know you where allowed to substitute, I thought the point of the exercise was to determine x with only using co-functions, sin (900 + x) = 2 cos2x cosx = 2cos2x divide both side with cosx .'. 1 = 2 cosx .'. cosx = 1/2 .'. x = 600; 900; 2700; 3000 How you could obtain 90 and 270 degrees from your solution? This is", "x = \\frac {\\sin 2x} 2.$$ There will be either only cos's or only sin's left. Convert them into double-angles with \\begin{align} \\cos^2 x &= \\frac {1 + \\cos 2x} 2 \\\\ \\sin^2 x &= \\frac {1 - \\cos 2x} 2. \\end{align}", "= sin (β_{0} - β + β) = sin (β_{0} - β) * cos β + cos (β_{0} - β) * sin β \\end{align} and \\begin{align} sin β_{1} = sin (β_{1} - β + β) = sin (β_{1} - β) * cos β + cos (β_{1} - β) * sin β \\end{align} Using the feature of the sines and cosines functions for small angles mentioned above, the equations above can be written in the following way: \\begin{align} sin β_{0} = (β_{0} - β) * cos β + sin β \\end{align} and \\begin{align} sin β_{1} = (β_{1} - β) * cos β + sin β \\end{align} After inserting the formulas above into previous equations, we get: $\\begin{cases} (β_{0} - β) * cos β + sin β - u * sin γ = μ_{0} * sin (β_{0} + γ) \\\\ sin β - u * sin γ = 0 \\\\ (β_{1} - β) * cos β + sin β - u * sin γ = -μ_{0} * sin (β_{1} + γ) \\end{cases}$ and after transforming: $\\begin{cases} sin β - u * sin γ = μ_{0} * sin (β_{0} + γ) - (β_{0} - β) * cos β \\\\ sin β - u * sin γ = 0 \\\\ sin β - u * sin γ = -μ_{0} * sin (β_{1}", "probably won’t see that rule tested on the SAT, but it’s not outside the realm of possibility. So here’s what we know: Of course, we also know that the formula for the area of a triangle is ½bh. So let’s figure out AB: \\begin{align*}\\frac{1}{2}(AB)(x-y)&=\\frac{x^2-y^2}{2}\\\\(AB)(x-y)&=x^2-y^2\\\\(AB)(x-y)&=(x-y)(x+y)\\\\AB&=x+y\\end{align*} Oh look. Difference of two squares. Crazy how often that appears! Now we can solve for BC, using the Pythagorean Theorem: \\begin{align*}(x+y)^2(x-y)^2&=(BC)^2\\\\(x^2+2xy+y^2)+(x^2-2xy+y^2)&=(BC)^2\\\\2x^2+2y^2&=(BC)^2\\\\\\sqrt{2(x^2+y^2)}&=BC\\end{align*} Nice. Here’s what we’ve got now: Of course, BC is the diameter of the circle, and to find the circle’s area we want the radius, which is half of BC: $r=\\frac{\\sqrt{2(x^2+y^2)}}{2}$ \\begin{align*}A&=\\pi\\left(\\frac{\\sqrt{2(x^2+y^2)}}{2}\\right)^2\\\\A&=\\pi\\left(\\frac{2(x^2+y^2)}{4}\\right)\\\\A&=\\frac{\\pi(x^2+y^2)}{2}\\end{align*}" ]

[ "Ptolemy. One can also produce them algebraically using Euler's formula. Sum {\\displaystyle {\\begin{aligned}\\sin \\left(x+y\\right)&=\\sin x\\cos y+\\cos x\\sin y,\\\\\\cos \\left(x+y\\right)&=\\cos x\\cos y-\\sin x\\sin y,\\end{aligned}}} Difference {\\displaystyle {\\begin{aligned}\\sin \\left(x-y\\right)&=\\sin x\\cos y-\\cos x\\sin y,\\\\\\cos \\left(x-y\\right)&=\\cos x\\cos y+\\sin x\\sin y.\\end{aligned}}} These in turn lead to the following three-angle formulae: {\\displaystyle {\\begin{aligned}\\sin \\left(x+y+z\\right)&=\\sin x\\cos y\\cos z+\\sin y\\cos z\\cos x+\\sin z\\cos y\\cos x-\\sin x\\sin y\\sin z,\\\\\\cos \\left(x+y+z\\right)&=\\cos x\\cos y\\cos z-\\cos x\\sin y\\sin z-\\cos y\\sin x\\sin z-\\cos z\\sin x\\sin y,\\end{aligned}}} When the two angles are equal, the sum formulas reduce to simpler equations known as the double-angle formulae. {\\displaystyle {\\begin{aligned}\\sin \\left(2x\\right)&=2\\sin x\\cos x,\\\\\\cos \\left(2x\\right)&=\\cos ^{2}x-\\sin ^{2}x=2\\cos ^{2}x-1=1-2\\sin ^{2}x.\\end{aligned}}} When three angles are equal, the three-angle formulae simplify to {\\displaystyle {\\begin{aligned}\\sin(3x)&=3\\sin x-4\\sin ^{3}x.\\\\\\cos(3x)&=4\\cos ^{3}x-3\\cos x.\\end{aligned}}} These identities can also be used to derive the product-to-sum identities that were used in antiquity to transform the product of two numbers into a sum of numbers and greatly speed operations, much like the logarithm function. ### Calculus For integrals and derivatives of trigonometric functions, see the relevant sections of Differentiation of trigonometric functions, Lists of integrals and List of integrals of trigonometric functions. Below is the list of the derivatives and integrals of the six basic trigonometric functions. The number C is a constant of integration. ${\\displaystyle {\\begin{array}{|r|rr|}f(x)&f'(x)&\\int f(x)\\,dx\\\\\\hline \\sin x&\\cos x&-\\cos x+C\\\\\\cos x&-\\sin x&\\sin x+C\\\\\\tan x&\\sec ^{2}x=1+\\tan ^{2}x&-\\ln", "&= 2\\cos\\theta - \\cos^2\\theta \\end{align} Simplify and rearrange \\begin{align} 1-\\cos^2\\theta &= 2\\cos\\theta - \\cos^2\\theta\\\\ 1 &= 2\\cos\\theta\\\\ \\cos\\theta &= \\frac{1}{2} \\end{align} As required. ###### Example 2 Verify the identity $\\cos(a+b)\\cos(a-b) = \\cos^2a - \\sin^2b$ ###### Solution Use the addition formula and rewrite the left hand side. \\begin{align} \\cos(a+b)\\cos(a-b) &= (\\cos a\\cos b - \\sin a\\sin b)(\\cos a\\cos b+\\sin a\\sin b)\\\\ &=\\cos^2a\\cos^2b + \\cos a\\cos b\\sin a\\sin b - \\cos a\\cos b\\sin a\\sin b - \\sin^2a\\sin^2b\\\\ &=\\cos^2a\\cos^2b - \\sin^2a\\sin^2b \\end{align} Use $\\cos^2x+\\sin^2x=1$ to rewrite everything in terms of $\\cos$. \\begin{align} \\cos^2a\\cos^2b - \\sin^2a\\sin^2b &= \\cos^2a\\cos^2b - (1-\\cos^2a)(1-\\cos^2b)\\\\ &= \\cos^2a\\cos^2b - (1 - \\cos^2a - \\cos^2b + \\cos^2a\\cos^2b)\\\\ &= \\cos^2a + \\cos^2b -1 \\end{align} Use the formula again to rewrite in the form specified. \\begin{align} \\cos^2a + \\cos^2b -1 &=\\cos^2a -(1-\\cos^2b)\\\\ &=\\cos^2a-\\sin^2b \\end{align} ###### Example 3 Derive the formula for $\\cos 2x$ from the addition formula. ###### Solution The addition formula is $cos(\\theta + \\phi) = \\cos\\theta\\cos\\phi - \\sin\\theta\\sin\\phi$. Set $\\theta = x$ and $\\phi = x$ \\begin{align} \\cos 2x &= \\cos (x+x)\\\\ &=\\cos x\\cos x - \\sin x\\sin x\\\\ &=\\cos^2x - \\sin^2x \\end{align} Use $\\cos^2x+\\sin^2x=1$ \\begin{align} \\cos 2x &= \\cos^2x - (1 - \\cos^2x)\\\\ &= 2\\cos^2 x-1 \\end{align} Or \\begin{align} \\cos 2x &= (1-\\sin^2x) - \\sin^2x\\\\ &= 1-2\\sin^2x \\end{align} #### Video Example Prof. Robin Johnson expands $\\cos 6x$ in", "addition formulas). • Using that $$\\cos$$ is even and that $$\\sin$$ is odd, the addition formulas give the subtraction formulas: \\begin{align*} \\cos(x-y)=&\\cos(x)\\cos(y)+\\sin(x)\\sin(y) \\\\ \\sin(x-y)=&\\sin(x)\\cos(y)-\\cos(x)\\sin(y). \\end{align*} • Dividing the formulas we get: $\\tan(x\\pm y)=\\frac{\\sin(x\\pm y)}{\\cos(x\\pm y)}=\\frac{\\sin(x)\\cos(y)\\pm\\cos(x)\\sin(y)}{\\cos(x)\\cos(y)\\mp\\sin(x)\\sin(y)}=\\frac{\\tan(x)\\pm\\tan(y)}{1\\mp\\tan(x)\\tan(y)}.$ • Putting $$x=y$$ in the addition formulas, we get the double-angle formulas: \\begin{align*} \\cos(2x)=&(\\cos x)^2-(\\sin x)^2 \\\\ \\sin(2x)=&2\\sin(x)\\cos(x) \\end{align*} • Using the identity $$(\\cos x)^2+(\\sin x)^2=1$$ in the double-angle formula for $$\\cos$$, we get the following alternative forms. \\begin{align*} \\cos(2x)=&(\\cos x)^2-(1-(\\cos x)^2)=2(\\cos x)^2-1 \\\\ \\cos(2x)=&(1-(\\sin x)^2)-(\\sin x)^2=1-2(\\sin x)^2 \\end{align*} # Trigonometric identities -- from the addition formulas 2 • Once again, the addition and subtraction formulas are \\begin{align*} \\cos(x\\pm y)=&\\cos(x)\\cos(y)\\mp\\sin(x)\\sin(y) \\\\ \\sin(x\\pm y)=&\\sin(x)\\cos(y)\\pm\\cos(x)\\sin(y), \\end{align*} and the alternative double-angle formulas for $$\\cos$$ are \\begin{align*} \\cos(2x)=&\\cos(x)^2-(1-\\cos(x)^2)=2\\cos(x)^2-1 \\\\ \\cos(2x)=&(1-(\\sin x)^2)-(\\sin x)^2=1-2(\\sin x)^2. \\end{align*} • Rearranging the latter, we get the following half-angle formulas. \\begin{align*} (\\cos x)^2=&\\frac{\\cos(2x)+1}{2} \\\\ (\\sin x)^2=&\\frac{1-\\cos(2x)}{2} \\end{align*} • Finally, rearranging the addition formulas, we get the following product formulas. \\begin{align*} \\cos(x)\\cos(y)=&\\frac{\\cos(x+y)+\\cos(x-y)}{2} \\\\ \\sin(x)\\cos(y)=&\\frac{\\sin(x+y)+\\sin(x-y)}{2} \\\\ \\sin(x)\\sin(y)=&\\frac{\\cos(x-y)-\\cos(x+y)}{2}. \\end{align*} • The half-angle and product formulas will be used frequently in integral calculus, since they are linearization formulas. • Exercises: D.45, 47, 67 # Exponentiation by rational numbers • An exponential function is a function of the form $$f(x)=b^x$$ for some fixed positive number $$b>0$$. Let's recall what the expression $$b^x$$ means. • If $$n$$ is a", "half angle formula. \\begin{aligned} \\sqrt{\\dfrac{1-\\cos b}{1+\\cos b}} &=\\dfrac{\\sin b}{1+\\cos b} \\\\ \\dfrac{1-\\cos b}{1+\\cos b} &=\\dfrac{\\sin ^{2} b}{(1+\\cos b)^{2}} \\\\ (1-\\cos b)(1+\\cos b)^{2} &=\\sin ^{2} b(1+\\cos b) \\\\ (1-\\cos b)(1+\\cos b) &=\\sin ^{2} b \\\\ 1-\\cos ^{2} b &=\\sin ^{2} b \\end{aligned} ##### Example $$\\PageIndex{3}$$ Verify the identity: $$\\cot \\dfrac{c}{2} =\\dfrac{\\sin c}{1-\\cos c}$$ Solution Step 1: change cotangent to cosine over sine, then cross-multiply. \\begin{aligned} \\cot \\dfrac{c}{2} &=\\dfrac{\\sin c}{1-\\cos c} \\\\ &=\\dfrac{\\cos \\dfrac{c}{2}}{\\sin \\dfrac{c}{2}}=\\sqrt{\\dfrac{1+\\cos c}{1-\\cos c}} \\\\ \\sqrt{\\dfrac{1+\\cos c}{1-\\cos c}} &=\\dfrac{\\sin c}{1-\\cos c} \\\\ \\dfrac{1+\\cos c}{1-\\cos c} &=\\dfrac{\\sin ^{2} c}{(1-\\cos c)^{2}} \\\\ (1+\\cos c)(1-\\cos c)^{2} &=\\sin ^{2} c(1-\\cos c) \\\\ (1+\\cos c)(1-\\cos c) &=\\sin ^{2} c \\\\ 1-\\cos ^{2} c &=\\sin ^{2} c \\end{aligned} ##### Example $$\\PageIndex{4}$$ Prove that $$\\sin x \\tan \\dfrac{x}{2}+2\\cos x=2\\cos ^2 \\dfrac{x}{2}$$ Solution $$\\begin{array}{l} \\sin x \\tan \\dfrac{x}{2}+2 \\cos x=\\sin x\\left(\\dfrac{1-\\cos x}{\\sin x}\\right)+2 \\cos x \\\\ \\sin x \\tan \\dfrac{x}{2}+2 \\cos x=1-\\cos x+2 \\cos x \\\\ \\sin x \\tan \\dfrac{x}{2}+2 \\cos x=1+\\cos x \\\\ \\sin x \\tan \\dfrac{x}{2}+2 \\cos x=2 \\cos ^{2} \\dfrac{x}{2} \\end{array}$$ ## Review Use half angle identities to find the exact values of each expression. 1. $$\\sin 22.5^{\\circ}$$ 2. $$\\sin 75^{\\circ}$$ 3. $$\\sin 67.5^{\\circ}$$ 4. $$\\sin 157.5^{\\circ}$$ 5. $$\\cos 22.5^{\\circ}$$ 6. $$\\cos 75^{\\circ}$$ 7. $$\\cos 157.5^{\\circ}$$ 8. $$\\cos 67.5^{\\circ}$$ 9. Use the two half angle identities presented in this section to prove", "\\cos b)^2+(\\sin a - \\sin b)^2}\\end{align*} Square both sides to eliminate the square root. \\begin{align*}[\\cos (a - b) - 1]^2 + [\\sin (a-b)-0]^2 = (\\cos a - \\cos b)^2 + (\\sin a - \\sin b)^2\\end{align*} FOIL all four squared expressions and simplify. \\begin{align*}\\cos^2 (a-b)-2 \\cos(a-b)+ 1 + \\sin^2 (a-b) & = \\cos^2 a - 2 \\cos a \\cos b + \\cos^2 b + \\sin^2 a - 2 \\sin a \\sin b + \\sin^2 b \\\\ \\underbrace{\\sin^2 (a-b) + \\cos^2 (a-b)} - 2 \\cos (a-b) + 1 & = \\underbrace{\\sin^2 a + \\cos^2 a} - 2 \\cos a \\cos b + \\underbrace{\\sin^2 b + \\cos^2 b} - 2 \\sin a \\sin b \\\\ 1-2 \\cos (a-b) + 1 & = 1 - 2 \\cos a \\cos b + 1 - 2 \\sin a \\sin b \\\\ 2-2 \\cos (a-b) & = 2-2 \\cos a \\cos b -2 \\sin a \\sin b \\\\ -2 \\cos (a-b) & = -2 \\cos a \\cos b - 2 \\sin a \\sin b \\\\ \\cos (a-b) & = \\cos a \\cos b + \\sin a \\sin b\\end{align*} In \\begin{align*}\\cos(a - b) = \\cos a \\cos b + \\sin a \\sin b\\end{align*}, the difference formula for cosine, you can substitute \\begin{align*}a - (- b) = a + b\\end{align*} to obtain: \\begin{align*}\\cos(a + b) = \\cos[a", "Review question # Can we sketch $y=3 \\cos x - \\cos 3x$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R5251 ## Solution Prove that $\\begin{equation*} \\frac{d}{dx} (3 \\cos x - \\cos 3x)= 6 \\sin x \\cos 2x. \\end{equation*}$ Using the chain rule or by thinking about horizontal scaling, we have that $\\begin{equation*} \\frac{d}{dx} \\cos 3x = - 3 \\sin 3x. \\end{equation*}$ By the compound angle formula, \\begin{align*} \\sin 3x = \\sin(x + 2x) &= \\sin x \\cos 2x + \\sin 2x \\cos x \\\\ &= \\sin x \\cos 2x + 2 \\sin x \\cos^2 x \\\\ &= \\sin x \\cos 2x + \\sin x (1 + \\cos 2x) \\\\ &= 2 \\sin x \\cos 2x + \\sin x \\end{align*} so that \\begin{align*} \\frac{d}{dx} \\left( 3 \\cos x - \\cos 3x \\right) &= - 3 \\sin x + 3 \\sin 3x \\\\ &= - 3 \\sin x + 6 \\sin x \\cos 2x + 3 \\sin x \\\\ &= 6 \\sin x \\cos 2x. \\end{align*} Alternatively, we can use the standard formula, $\\sin P-\\sin Q = 2\\cos\\left(\\dfrac{P+Q}{2}\\right)\\sin\\left(\\dfrac{P-Q}{2}\\right)$. For what two ranges of values of $x$ between $0$ and $\\pi$ is $\\begin{equation*} 3 \\cos x - \\cos 3x \\end{equation*}$ an increasing function of $x$? A", "angles. cos(ab)=cosacosb+sinasinb and cos(a+b)=cosacosbsinasinbcos(ab)cos(a+b)=cosacosb+sinasinb(cosacosbsinasinb)cos(ab)cos(a+b)=cosacosb+sinasinbcosacosb+sinasinbcos(ab)cos(a+b)=2sinasinb12[cos(ab)cos(a+b)]=sinasinb The following product to sum formulas can be derived using the same method: cosαcosβsinαcosβcosαsinβ=12[cos(αβ)+cos(α+β)]=12[sin(α+β)+sin(αβ)]=12[sin(α+β)sin(αβ)] Armed with these four formulas, we can work some examples. #### Example A Change cos2xcos5y\\begin{align*}\\cos 2x \\cos 5y\\end{align*} to a sum. Solution: Use the formula cosαcosβ=12[cos(αβ)+cos(α+β)]\\begin{align*}\\cos \\alpha \\cos \\beta = \\frac{1}{2} \\left [\\cos (\\alpha - \\beta) + \\cos (\\alpha + \\beta) \\right ]\\end{align*}. Set α=2x\\begin{align*}\\alpha = 2x\\end{align*} and β=5y\\begin{align*}\\beta = 5y\\end{align*}. cos2xcos5y=12[cos(2x5y)+cos(2x+5y)] #### Example B Change sin11z+sinz2\\begin{align*}\\frac{\\sin11z + \\sin z}{2}\\end{align*} to a product. Solution: Use the formula sinαcosβ=12[sin(α+β)+sin(αβ)]\\begin{align*}\\sin \\alpha \\cos \\beta = \\frac{1}{2} \\left [\\sin (\\alpha + \\beta) + \\sin (\\alpha - \\beta) \\right ]\\end{align*}. Therefore, α+β=11z\\begin{align*}\\alpha + \\beta = 11z\\end{align*} and αβ=z\\begin{align*}\\alpha - \\beta = z\\end{align*}. Solve the second equation for α\\begin{align*}\\alpha\\end{align*} and plug that into the first. α=z+β(z+β)+β=11zz+2β=11z2β=10zβ=5zandα=z+5z=6z \\begin{align*}\\frac{\\sin11z + \\sin z}{2} = \\sin 6z \\cos 5z\\end{align*}. Again, the sum of \\begin{align*}6z\\end{align*} and \\begin{align*}5z\\end{align*} is \\begin{align*}11z\\end{align*} and the difference is \\begin{align*}z\\end{align*}. #### Example C Solve \\begin{align*}\\cos 5x + \\cos x = \\cos 2x\\end{align*}. Solution: Use the formula \\begin{align*}\\cos \\alpha + \\cos \\beta = 2 \\cos \\frac{\\alpha +\\beta}{2} \\times \\cos \\frac{\\alpha - \\beta}{2}\\end{align*}. ### Guided Practice 1. Express the product as a sum: \\begin{align*}\\sin(6 \\theta) \\sin(4 \\theta) \\end{align*} 2. Express the product as a sum: \\begin{align*}\\sin(5 \\theta) \\cos(2 \\theta)\\end{align*} 3. Express the product as a sum: \\begin{align*}\\cos(10 \\theta) \\sin(3", "of x and y, we'll begin by squaring the parameterization: \\begin{align} x^2 &= 25 cos^2(t) \\\\ y^2 &= 4 sin^2(t) \\end{align} Now divide by 25 on the top and 4 on the bottom: $$\\frac{x^2}{25} = cos^2(t) \\;\\;\\; \\frac{y^2}{4} = sin^2(t)$$ Now we can add these two equations, using the Pythagorean identity to get one on the right side ... $$\\frac{x^2}{25} + \\frac{y^2}{4} = cos^2(t) + sin^2(t) = 1$$ So the formula of our ellipse is $$\\frac{x^2}{25} + \\frac{y^2}{4} = 1$$ which indeed has a long axis along x of length 10 (= 2 × 5) and a short axis along y of 4 units (2 × 2). See the conic sections section if you need a refresher on ellipses. It's not necessarily worth memorizing, but we've also got the idea that the parameterization x = a·cos(t), y = b·sin(t) is an ellipse centered at the origin with axes of 2a and 2b. Practice problems Convert each of these parametric curves to y = f(x) form. Plot a few points of each in parametric form to get an idea of the shape of the curve. 1 \\begin{align} x &= 3 \\, sin(t) \\\\[5pt] y &= 2 \\, cos(t) \\end{align} 2 \\begin{align} x &= t \\, cos(t) \\\\[5pt] y &= t \\, sin(t) \\end{align} 3 \\begin{align} x &= t^2 +", "A.\\cos B+2ab.\\sin A.\\sin B \\\\ & =({{a}^{2}}{{\\cos }^{2}}B+{{b}^{2}}{{\\cos }^{2}}A+2ab.\\cos A.cosB)-({{a}^{2}}{{\\sin }^{2}}B+{{b}^{2}}{{\\sin }^{2}}A-2ab.\\sin A.\\sin B) \\\\ \\end{align} They are of the form ${{a}^{2}}+2ab+{{b}^{2}}={{(a+b)}^{2}}.$ ${{(a\\cos B+b\\cos A)}^{2}}-{{(a\\sin B-b\\sin A)}^{2}}$ By projection formulae, the length of any side of a triangle is equal to the sum of the projections of the other two sides. $\\therefore a\\cos B+b\\cos A=C$for any triangle ABC. \\begin{align} & {{(a\\cos B+b\\cos A)}^{2}}-{{(a\\sin nB+b\\sin A)}^{2}} \\\\ & ={{c}^{2}}-{{(a\\sin B-b\\sin A)}^{2}} \\\\ \\end{align} By sine formula, we know that $\\dfrac{\\sin A}{a}=\\dfrac{sinB}{b}$ Cross multiplying the above expression, we get, \\begin{align} & b\\sin A=a\\sin B \\\\ & ={{c}^{2}}-{{(a\\sin B-b\\sin A)}^{2}} \\\\ & ={{c}^{2}}-{{(a\\sin B-a\\sin B)}^{2}} \\\\ & ={{c}^{2}}-0={{c}^{2}} \\\\ \\end{align} Therefore, we proved that, ${{a}^{2}}\\cos 2B+{{b}^{2}}\\cos 2A+2ab\\cos (A-B)={{c}^{2}}$. Thus the statement is true. Option A is the correct answer. Note: Here we use the cosine formulae to solve the expression. You should remember the formulae so that solving questions like these would become easy. We used the projection formula which is an important concept, to solve the expression. $c=a\\cos B+b\\cos A$.", "}^{2}}x-{{\\cos }^{2}}x)({{({{\\sin }^{2}}x+{{\\cos }^{2}}x)}^{^{2}}}-2{{\\sin }^{2}}x{{\\cos }^{2}}x+{{\\sin }^{2}}x{{\\cos }^{2}}x) \\\\ & =3({{\\sin }^{2}}x-{{\\cos }^{2}}x)-4({{\\sin }^{2}}x-{{\\cos }^{2}}x)(1-{{\\sin }^{2}}x{{\\cos }^{2}}x) \\\\ \\end{align} \\begin{align} & =-3(-{{\\sin }^{2}}x+{{\\cos }^{2}}x)+4({{\\cos }^{2}}x-{{\\sin }^{2}}x)(1-{{\\sin }^{2}}x{{\\cos }^{2}}x) \\\\ & =4({{\\cos }^{2}}x-{{\\sin }^{2}}x)(1-{{\\sin }^{2}}x{{\\cos }^{2}}x)-3({{\\cos }^{2}}x-{{\\sin }^{2}}x) \\\\ \\end{align} Now taking ${{\\cos }^{2}}x-{{\\sin }^{2}}x$common we get, Also, $({{\\cos }^{2}}x-{{\\sin }^{2}}x)=\\cos 2x$ \\begin{align} & =({{\\cos }^{2}}x-{{\\sin }^{2}}x)\\left( 4(1-{{\\sin }^{2}}x{{\\cos }^{2}}x)-3 \\right) \\\\ & =\\cos 2x\\left( 4-4{{\\sin }^{2}}x{{\\cos }^{2}}x-3 \\right) \\\\ & =\\cos 2x\\left( 4-4(1-{{\\cos }^{2}}x){{\\cos }^{2}}x-3 \\right) \\\\ & =\\cos 2x\\left( 1-4{{\\cos }^{2}}x+4{{\\cos }^{4}}x \\right) \\\\ & =\\cos 2x\\left( 1-2\\times 2{{\\cos }^{2}}x+{{(2{{\\cos }^{2}}x)}^{2}} \\right) \\\\ & =\\cos 2x{{\\left( 2{{\\cos }^{2}}x-1 \\right)}^{2}} \\\\ \\end{align} We know that$2{{\\cos }^{2}}x-1=\\cos 2x$, \\begin{align} & =\\cos 2x{{\\left( \\cos 2x \\right)}^{2}} \\\\ & ={{\\cos }^{3}}2x \\\\ \\end{align} $={{\\cos }^{3}}2x=$RHS So $\\sin 3x{{\\sin }^{3}}x+\\cos 3x{{\\cos }^{3}}x={{\\cos }^{3}}2x$ is proved. Hence proved. Note: Read the question in a careful manner. Also, make sure that you know all the identities. You should be familiar with the trigonometric formulae because mostly they are used for conversion. Also, there might be silly mistakes, so kindly avoid the silly mistake.", "\\left(=\\cos\\dfrac{5\\pi}{9}\\right).$ Therefore the three roots (it cannot have more) of $x^3-3x^2+1=0$ are $1+2\\cos\\dfrac{\\pi}{9}=2.879, 1+2\\cos\\dfrac{7\\pi}{9}=-0.532 \\quad \\text{and} \\quad 1+2\\cos\\dfrac{13\\pi}{9}=0.653,$ all given to 3 decimal places. 1. The values of $x$ and $y$ satisfy the equations \\begin{align*} 3\\cos^2 x+ 2\\cos^2 y&=4,\\\\ 3\\sin 2x-2\\sin 2y&=0. \\end{align*} Find the values of $\\cos 2x$ and $\\cos 2y$. It’s worth trying to get everything in terms of $\\cos 2 x$ and $\\cos 2 y$ first. We know that $\\cos 2x=2\\cos^2 x-1$; rearranging this gives $\\cos^2 x=\\dfrac{1}{2}(1+\\cos 2x)$. So we can rewrite our first equation as \\begin{align*} &3\\cos^2 x+2\\cos^2 y=4\\\\ \\iff\\quad &\\tfrac{3}{2}(1+\\cos 2x)+\\tfrac{2}{2}(1+\\cos 2y)=4\\\\ \\iff\\quad &3\\cos 2x+2\\cos 2y =3. \\end{align*} Now looking at our second equation, we aim to change every $\\sin$ to a $\\cos$. Rearranging and squaring both sides, we have $9\\sin^2 2x=4\\sin^2 2y$ Using the identity $\\cos^2 \\theta +\\sin^2 \\theta=1$ gives \\begin{align*} & 9(1-\\cos^2 2x)=4(1-\\cos^2 2y)\\\\ \\iff\\quad & 9\\cos^2 2x-4\\cos^2 2y=5\\\\ \\iff\\quad & (3\\cos 2x-2\\cos 2y)(3\\cos2x+2\\cos 2y)=5. \\end{align*} Now substituting in $3\\cos 2x+2\\cos 2y=3$ gives $3\\cos 2x-2\\cos 2y=\\frac{5}{3}.$ We now have two linear simultaneous equations in $\\cos 2x$ and $\\cos 2y$. Adding them to eliminate $\\cos 2y$ gives $6\\cos 2x=\\frac{14}{3}$ so $\\cos 2x=\\frac{7}{9}$. Substituting into either of our simultaneous equations then gives $\\cos 2y=\\dfrac{1}{3}$. Hence find, to the nearest degree, four pairs of corresponding values of $x$ and $y$, all values lying between", "the first identity in the reverse order). $$\\cos{A}+\\cos{(A+2\\pi/3)}+\\cos{(A-2\\pi/3)}=0$$ $$\\cos{(A-\\pi/3)}+\\cos{(A+\\pi/3)}=\\cos{A}$$ Consider the following picture. We see that the triangle with vertices $(\\cos A, \\sin A), (\\cos(A+120^\\circ), \\sin(A+120^\\circ)), (\\cos(A+240^\\circ), \\sin(A+240^\\circ))$ are the vertices of an equilateral triangle with centriod at the origin. Thus if we set $a = \\cos A, b = \\cos(A+120^\\circ), c = \\cos(A+240^\\circ)$, then \\begin{align*} a+b+c = 0 \\end{align*} and hence \\begin{align*} a^3+b^3+c^3 = 3abc \\end{align*} and \\begin{align*} \\cos^3{A} + \\cos^3{(120°+A)} + \\cos^3{(240°+A)} &= 3\\cos A \\cos(A+120^\\circ)\\cos(A+240^\\circ) \\\\ &= \\frac{3}{2}(2\\cos A \\cos(A+120^\\circ))\\cos(A+240^\\circ) \\\\ &= \\frac{3}{2}[\\cos(2A+120^\\circ)+\\cos(120^\\circ)]\\cos(A+240^\\circ)\\\\ &=\\frac{3}{2}\\cos(2A+120^\\circ)\\cos(A+240^\\circ)-\\frac{3}{4}\\cos(A+240^\\circ)\\\\ &=\\frac{3}{4}\\cos(3A+360^\\circ)+\\frac{3}{4}\\cos(A-120^\\circ) -\\frac{3}{4}\\cos(A+240^\\circ)\\\\ &=\\frac{3}{4}\\cos(3A) \\end{align*} since $\\cos(A+240^\\circ) = \\cos(360^\\circ - (A+240^\\circ)) = \\cos(120^\\circ -A) = \\cos(A-120^\\circ)$ • If you can see geometrically, the points $(\\cos A, \\sin A), (\\cos(120^\\circ+A), \\sin(120^\\circ+A)), (\\cos(240^\\circ), \\sin(240^\\circ+A))$ are the vertices of an equilateral triangle with centroid as the origin, then the above proof holds since we have used only an algebraic identity. – user348749 Sep 10, 2016 at 7:23 use that $$\\cos(120^{\\circ}+x)=-1/2\\,\\cos \\left( x \\right) -1/2\\,\\sqrt {3}\\sin \\left( x \\right)$$ and $$\\cos(240^{\\circ}+x)=-1/2\\,\\cos \\left( x \\right) +1/2\\,\\sqrt {3}\\sin \\left( x \\right)$$ For the LHS, use $\\cos(120^\\circ +A)=\\cos 120^\\circ \\cos A - \\sin 120^\\circ \\sin A$ and $\\cos(240^\\circ+A)=\\cos 240^\\circ \\cos A - \\sin 240^\\circ \\sin A$ Then $\\cos^3(120^\\circ+A)=\\left(\\cos 120^\\circ \\cos A - \\sin 120^\\circ \\sin A \\right)^3$ and $\\cos^3(240^\\circ+A)=\\left( \\cos 240^\\circ \\cos A - \\sin 240^\\circ \\sin A \\right)^3$", "$2R_2 \\cos (\\angle O_1 O_2 O_3/2)$ using the identity $\\cos^2 (A/2) = (1 + \\cos A)/2$. From (4), \\begin{aligned} T_1 T_3 &= 2R_2 \\cos (\\angle O_1 O_2 O_3/2)\\\\ &= 2R_2 \\sqrt{\\frac{R_1R_3}{(R_1 - R_2)(R_2 + R_3)}}\\\\ &= \\frac{2\\sqrt{R_1R_2R_3} \\sqrt{R_2}}{\\sqrt{(R_1 - R_2)(R_2 + R_3)}}.\\quad\\quad(9) \\end{aligned} Next we perform the same calculations (4)-(9) with the configuration of three circles tangent externally. By the cosine rule in $\\triangle O_1O_2O_3$, \\begin{aligned} \\cos \\angle O_1 O_2 O_3 &= \\frac{O_1 O_2^2 + O_2O_3^2 - O_1O_3^2}{2O_1 O_2. O_2O_3}\\\\ &= \\frac{(R_1 + R_2)^2 + (R_2 + R_3)^2- (R_1 + R_3)^2 }{2(R_1 + R_2)(R_2 + R_3)}\\\\ &=\\frac{2R_2^2 + 2R_1 R_2 + 2R_2 R_3 - 2R_1 R_3}{2(R_1 + R_2)(R_2 + R_3)}\\\\ &= \\frac{(R_1 + R_2)(R_2+R_3) - 2R_1R_3}{(R_1 + R_2)(R_2 + R_3)}\\\\ &= 1 - \\frac{2R_1R_3}{(R_1 + R_2)(R_2 + R_3)}. \\quad\\quad (10) \\end{aligned} The perimeter of $\\triangle O_1O_2O_3$ is $(R_1 + R_2) + (R_1 + R_3) + (R_2 + R_3) = 2(R_1+R_2+R_3)$, so by Heron’s formula its area is the attractive form $\\displaystyle \\sqrt{R_1R_2 R_3(R_1 + R_2 + R_3)}.\\quad\\quad (11)$ If the points $T_3, O_1, O_2$ have coordinates $(0,0), (-R_1,0), (R_2,0)$ respectively, then $O_3$ has coordinates $(R_2 - (R_2 + R_3)\\cos \\angle O_1 O_2 O_3, (R_2 + R_3)\\sin \\angle O_1 O_2 O_3)$. We have $\\cos \\angle O_1 O_2 O_3$ found above and $\\sin \\angle O_1 O_2 O_3$ can be found", "= sin (β_{0} - β + β) = sin (β_{0} - β) * cos β + cos (β_{0} - β) * sin β \\end{align} and \\begin{align} sin β_{1} = sin (β_{1} - β + β) = sin (β_{1} - β) * cos β + cos (β_{1} - β) * sin β \\end{align} Using the feature of the sines and cosines functions for small angles mentioned above, the equations above can be written in the following way: \\begin{align} sin β_{0} = (β_{0} - β) * cos β + sin β \\end{align} and \\begin{align} sin β_{1} = (β_{1} - β) * cos β + sin β \\end{align} After inserting the formulas above into previous equations, we get: $\\begin{cases} (β_{0} - β) * cos β + sin β - u * sin γ = μ_{0} * sin (β_{0} + γ) \\\\ sin β - u * sin γ = 0 \\\\ (β_{1} - β) * cos β + sin β - u * sin γ = -μ_{0} * sin (β_{1} + γ) \\end{cases}$ and after transforming: $\\begin{cases} sin β - u * sin γ = μ_{0} * sin (β_{0} + γ) - (β_{0} - β) * cos β \\\\ sin β - u * sin γ = 0 \\\\ sin β - u * sin γ = -μ_{0} * sin (β_{1}", "+ 2\\cos b\\cos c \\\\ &+ 2\\sin a\\sin b + 2\\sin a\\sin c + 2\\sin b\\sin c \\end{aligned}\\\\ &= \\!\\begin{aligned}[t] 3 + \\bigl[&\\cos(a + b) + \\cos(a + c) + \\cos(b + c) \\\\ {}+ \\phantom{\\bigl[}&\\cos(a - b) + \\cos(a - c) + \\cos(b - c)\\bigr] \\\\ {}+\\bigl[&\\cos(a - b) + \\cos(a - c) + \\cos(b - c) \\\\ {}-\\bigl(&\\cos(a + b) + \\cos(a + c) + \\cos(b + c) \\bigr) \\bigr] \\end{aligned}\\\\ &= 3+ 2\\bigl(\\cos(a - b) + \\cos(a - c) + \\cos(b - c)\\bigr). \\end{align*} \\end{document} • The only thing you should add is \\! in front of \\begin{aligned} – egreg Jul 17 '14 at 22:18", "+ \\delta_1) \\\\ y_2 &= A_2 \\sin(kx + \\omega t + \\delta_2), \\end{align} how do we find a definition of $y_1+y_2$ as a function of $x$ and $t$ in the form $$y_1 + y_2 = A_3 \\sin(kx + \\omega t + \\delta_3)?$$ To keep the formulas a little more manageable, let $u = kx + \\omega t$. Using the angle-sum formula for sine, \\begin{align} \\sin(kx + \\omega t + \\delta_1) &= \\sin\\delta_1 \\cos u + \\cos\\delta_1 \\sin u, \\\\ \\sin(kx + \\omega t + \\delta_2) &= \\sin\\delta_2 \\cos u + \\cos\\delta_2 \\sin u. \\end{align} Therefore \\begin{align} y_1 + y_2 &= A_1(\\sin\\delta_1 \\cos u + \\cos\\delta_1 \\sin u) + A_2(\\sin\\delta_2 \\cos u + \\cos\\delta_2 \\sin u) \\\\ &= (A_1\\sin\\delta_1 + A_2\\sin\\delta_2)\\cos u + (A_1\\cos\\delta_1 + A_2\\cos\\delta_2)\\sin u. \\end{align} So if $A = A_1\\sin\\delta_1 + A_2\\sin\\delta_2$ and $B = A_1\\cos\\delta_1 + A_2\\cos\\delta_2$, we're looking for $A_3$ and $\\delta_3$ that will generally satisfy $$A_3 \\sin(u + \\delta_3) = A \\cos u + B \\sin u.$$ Working from this answer to a related question, we want $A_3$ and $\\delta_3$ such that $(A_3)^2 = A^2 + B^2$ and $\\tan\\delta_3 = \\frac AB$. Recognizing that once we find a suitable $\\delta_3$, we can always add or subtract any whole multiple of $2\\pi$ without changing the truth of any of our equations, the solution is", "{\\left | 1 + \\cos x \\right |}$$ $$\\int_{x_0}^x \\frac{\\cos t}{1+\\cos t} \\, \\mathrm d t \\, = x - \\tan \\frac {x}{2}$$ \\begin{align*} y_p =& \\cos x \\ln |1 + \\cos x| + \\sin x \\left( x - \\tan \\frac x 2 \\right) \\\\ =& \\cos x \\ln |1 + \\cos x| + x \\sin x - (1 - \\cos x) \\end{align*} \\begin{align*} y'_p =& - \\sin x \\ln \\left | 1 + \\cos x \\right | - \\cos x (1 + \\cos x)^{-1} \\sin x + \\sin x + x \\cos x - \\sin x \\\\ =& - \\sin x \\ln \\left | 1 + \\cos x \\right | - \\cos x (1 - \\cos x) \\sin^{-1} x + x \\cos x \\\\ =& - \\sin x \\ln \\left | 1 + \\cos x \\right | - \\cot x + \\cos^2 x \\sin^{-1} x + x \\cos x \\\\ \\end{align*} \\begin{align*} y''_p =& -\\cos x \\ln |1 + \\cos x| + \\sin^2 x (1+ \\cos x)^{-1} + \\sin^{-2} x - 2 \\cot x \\sin x \\\\ & \\,\\,\\,\\, -\\cot^2 x \\cos x + \\cos x - x \\sin x \\end{align*} Note that $\\sin^2 x (1+ \\cos x)^{-1} = 1-\\cos x$ $$y''_p = -\\cos x \\ln |1 + \\cos x | + 1 + \\sin^{-2} x -", "in the ... 0 \\begin{align} f(x)&=\\cos^n{x}\\\\ f'(x)&=n\\cos^{n-1}{x}\\cdot(-\\sin{x})\\\\ f''(x)&=n(n-1)\\cos^{n-2}{x}\\cdot(-\\sin{x})^2-n\\cos^n{x}\\\\ f'''(x)&=n(n-1)(n-2)\\cos^{n-3}{x}\\cdot(-\\sin{x})^3+2n(n-1)\\sin{x}\\cdot\\cos^{n-1}(x)-nf'(x)\\\\ \\ldots\\\\ ... 0 i don't have a general expression but a sum. we have \\begin{align} 2^{n-1}\\cos^n x &= {n \\choose 0} \\cos nx + {n \\choose 1} \\cos(n-2)x + {n \\choose 1} \\cos(n-2)x + \\cdots\\end{align} taking derivativesk$times and use the fact that$\\frac{d^k}{dx^k}\\cos(x) = \\cos(x + k\\pi/2),we have \\begin{align} 2^{n-1}\\frac{d^k}{dx^k}\\left( \\cos^n ... 0 (\\cos x)^n = \\left(\\dfrac{e^{ix}+e^{-ix}}{2}\\right)^n. I think some pattern should emerge when taking derivatives of higher order like the kth one. You edited your question so my answer would not be useful anymore...fine. As for the edited question, you have f(x) = \\cos x, taking derivatives: f'(x) = -\\sin x, f''(x) = -\\cos x, f'''(x) = \\sin x, ... 0 Assuming you mean a_n \\le a_{n-1}^2 you can still not say the growth is no more than polynomial. Let a_n=2^{n+1}. This grows exponentially but satisfies your constraint. 1 As you remarked in a comment: f:[0,\\infty)\\rightarrow\\{0,1\\} can be presented as 1_{[s,\\infty)}. This function is right-continuous everywhere and if s>0 then it is left-continuous everywhere except at s. So if s>0 then f is continuous everywhere except at s. If s=0 then f is constant hence continuous everywhere. 4\\int_0^1 \\frac{1}{2} \\ dx = \\int_0^1 x \\ dx$$1 \\newcommand{\\Cpx}{\\mathbf{C}}Not sure I've seen a formal definition of \"Wick rotation\", but I'd probably", ".'. x = 600; 900; 2700; 3000 4. ## Re: Co-functions Originally Posted by MiloMan Thanks Minoanman, I didn't know you where allowed to substitute, I thought the point of the exercise was to determine x with only using co-functions, sin (900 + x) = 2 cos2x cosx = 2cos2x divide both side with cosx .'. 1 = 2 cosx .'. cosx = 1/2 .'. x = 600; 900; 2700; 3000 No, you can't divide both sides by \\displaystyle \\begin{align*} \\cos{(x)} \\end{align*} because it's possible for (and in this case, some solutions ARE where) \\displaystyle \\begin{align*} \\cos{(x)} = 0 \\end{align*}. Instead you need to factorise... \\displaystyle \\begin{align*} \\cos{(x)} &= 2\\cos^2{(x)} \\\\ 0 &= 2\\cos^2{(x)} - \\cos{(x)} \\\\ 0 &= \\cos{(x)} \\left[ 2\\cos{(x)} - 1 \\right] \\\\ \\cos{(x)} &= 0 \\textrm{ or } 2\\cos{(x)} - 1 = 0 \\end{align*} Go from here. 5. ## Re: Co-functions Originally Posted by MiloMan Thanks Minoanman, I didn't know you where allowed to substitute, I thought the point of the exercise was to determine x with only using co-functions, sin (900 + x) = 2 cos2x cosx = 2cos2x divide both side with cosx .'. 1 = 2 cosx .'. cosx = 1/2 .'. x = 600; 900; 2700; 3000 How you could obtain 90 and 270 degrees from your solution? This is", "\\cos (x) + i \\sin(x)$. We have that \\begin{align*} \\cos (x+y) + i \\sin (x+y) &= e^{i(x+y)} \\\\ &= e^{ix} \\cdot e^{iy} \\\\ &= (\\cos (x) + i \\sin (x))(\\cos (y) + i \\sin (y)) \\\\ &= (\\cos (x) \\cos (y) - \\sin (x) \\sin(y)) + i(\\sin (x) \\cos(y) + \\cos(x) \\sin(y)) \\end{align*} By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas. To derive the tangent addition formula, we reduce the problem to use sine and cosine, divide both numerator and denominator by $\\cos (x) \\cos (y)$, and simplify. \\begin{align*} \\tan (x+y) &= \\frac{\\sin (x+y)}{\\cos (x+y)} \\\\ &= \\frac{\\sin (x) \\cos(y) + \\cos(x) \\sin(y)}{\\cos (x) \\cos (y) - \\sin (x) \\sin(y)} \\\\ &= \\frac{\\frac{\\sin(x)}{\\cos(x)} + \\frac{\\sin(y)}{\\cos(x)}}{1 - \\frac{\\sin (x) \\sin(y)}{\\cos (x) \\cos(y)}} \\\\ &= \\frac{\\tan (x) + \\tan (y)}{1 - \\tan (x) \\tan(y)} \\end{align*} as desired. ## Double-angle identities The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting $x = y$. Doing so yields: • $\\sin (2x) = 2\\sin (x) \\cos (x)$ • $\\cos (2x) = \\cos^2 (x) - \\sin^2 (x)$ • $\\tan (2x) = \\frac{2\\tan (x)}{1-\\tan^2 (x)}$ ### Cosine double-angle identity Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle" ]

[ "The number of distinct solutions of the equation Question: The number of distinct solutions of the equation $\\log _{\\frac{1}{2}}|\\sin x|=2-\\log _{\\frac{1}{2}}|\\cos x|$ in the interval $[0,2 \\pi]$, is___________. Solution:", "of $n$; then the solution becomes outside the interval $\\left[ 0,2\\pi \\right).$ Thus, the exact solution of the equation $2\\sin 3x+\\sqrt{3}=0$ on the interval $\\left[ 0,2\\pi \\right)$ is $\\frac{4\\pi }{9}$ , $\\frac{5\\pi }{9}$ , $\\frac{10\\pi }{9}$ , $\\frac{11\\pi }{9}$ , $\\frac{16\\pi }{9}$ , and $\\frac{17\\pi }{9}$.", "in the solution as it is less than the value of $-1$. Thus, the actual solution in the interval $[0,2\\pi )$ will be $\\frac{7\\pi }{6},\\,\\text{ and }\\,\\frac{11\\pi }{6}$.", "set of points in the interval $\\left(-\\frac{7}{5}, \\frac{7}{10}\\right)$. The solution set could also be written $\\left \\{ x \\vert -\\frac{7}{5}< x < \\frac{7}{10}\\right\\}$", "# How do you solve \\sin (51.3)\\times 1200? Jul 26, 2018 $\\sin \\left(51.3\\right) \\times 1200 = 0.78 \\times 1200$ $= 936.52$ Your answer is $936.52$", "single solution of $\\left(\\frac{1}{3}, 3, 3\\right)$.", "$x=\\frac{4\\pi }{3}$ . Therefore, the solutions of the given equation in the interval $\\left[0,2\\pi \\right)$ are $\\left\\{\\frac{\\pi }{2},\\frac{3\\pi }{2},\\frac{2\\pi }{3},\\frac{4\\pi }{3}\\right\\}$ .", "# Number of Solutions Geometry Level 4 Find the number of values of $$x$$ in the interval $$[0,3\\pi]$$ satisfying the equation $$2\\sin^2{x}+5\\sin{x}-3=0$$. ×", "# How do you solve sin(x/2)-1=0 between the interval 0<=x<8pi? $x = \\pi$ or $x = 5 \\pi$ Since $0 \\le x < 8 \\pi$, we have $0 \\le \\frac{x}{2} < 4 \\pi$. Since $\\sin \\left(\\frac{x}{2}\\right) = 1$, the only allowed values of $\\frac{x}{2}$ in the interval $\\left[0 , 4 \\pi\\right)$ are $\\frac{x}{2} = \\frac{\\pi}{2}$ and $\\frac{x}{2} = \\frac{5 \\pi}{2}$.", "# How do you solve sin^2x+sinx=0 and find all solutions in the interval [0,2pi)? Sep 23, 2016 $x = 0 , \\frac{3 \\pi}{2} , \\pi$ #### Explanation: We have: ${\\sin}^{2} \\left(x\\right) + \\sin \\left(x\\right) = 0$; $\\left[0 , 2 \\pi\\right)$ $\\implies \\sin \\left(x\\right) \\left(\\sin \\left(x\\right) + 1\\right) = 0$ $\\implies \\sin \\left(x\\right) = 0$ $\\implies x = 0 , \\left(\\pi - 0\\right) , \\left(\\pi + 0\\right) , \\left(2 \\pi - 0\\right)$ or $\\implies \\sin \\left(x\\right) + 1 = 0$ $\\implies \\sin \\left(x\\right) = - 1$ $\\implies x = \\pi + \\frac{\\pi}{2}$ $\\implies x = 0 , \\frac{3 \\pi}{2} , \\pi$", "other values of $n$, the solution becomes outside the interval $\\left[ 0,2\\pi \\right).$ Thus, the exact solution of the equation $\\sin 2x+\\sin x=0$ on the interval $\\left[ 0,2\\pi \\right)$ is $\\underline{x=0,\\frac{2\\pi }{3},\\pi ,\\frac{4\\pi }{3}}.$", "the solution is $a+bi$ which is $$\\left(2n+\\frac{1}{2}\\right)\\pi i.$$", "## anonymous 5 years ago cos x/2 = cos π. find all solutions in the interval [0,2π)? $cos(\\pi)=-1$ so your equation is $cos(\\frac{x}{2})=-1$ one solution is $\\frac{x}{2}=\\pi$ so $x=2\\pi$ that is the only one in your interval. in fact it is not in your interval, so perhaps there is no solution!", "it gives us the main condition required for the problem. Since the value of x lies in the interval $x\\in \\left[ 0,4\\pi \\right]$, we have got 4 solutions. If interval changes, the no. of solutions differs.", "functions. Thus, the actual solutions in the interval $[0,2\\pi )$ will be $\\frac{\\pi }{6},\\frac{\\pi }{2},\\frac{5\\pi }{6},\\text{ and }\\ \\frac{3\\pi }{2}$.", "### Still have math questions? Algebra Question Find all solutions of the equation in the interval $$[ 0,2 \\pi )$$ $$\\cos 5 x \\cos 2 x + \\sin 5 x \\sin 2 x = 0$$ Write your answer in radians in terms of $$\\pi$$ . If there is more than one solution, separate them with commas. x=$$\\frac{pi}{6}$$ x=$$\\frac{pi}{2}$$ x=$$\\frac{5pi}{6}$$ x=$$\\frac{7pi}{6}$$ x=$$\\frac{3pi}{2}$$ x=$$\\frac{11pi}{6}$$", "## Precalculus (6th Edition) Blitzer The statement is false. The correct statement is: the equations $\\sin 2x=1$ and $\\sin 2x=\\frac{1}{2}$ do not have the same number of solutions on the interval $[0,2\\pi )$. Solution of $\\sin 2x=1$ will be $\\frac{\\pi }{4}$ and $\\frac{5\\pi }{4}$ So, the number of solutions will be $2$. Solutions of $\\sin 2x=\\frac{1}{2}$ will be $\\frac{\\pi }{12}$ , $\\frac{5\\pi }{12}$ , $\\frac{13\\pi }{12}$ , $\\frac{17\\pi }{12}$ Therefore, the number of solutions will be $4$. So, the number of solutions on the interval $[0,2\\pi )$ are not the same.", "### Home > PC > Chapter 6 > Lesson 6.2.2 > Problem6-89 6-89. $\\text{sin }x=-\\frac{\\sqrt{2}}{2}$ Use the Zero Product Property. Then combine parts (a) and (b). Solve for all solutions. Add a 2πn to each solution in [0,2π] where n is any integer.", "Find the supremum and the infimum of $\\frac{x}{\\sin x}$ on the interval $\\left (0, \\frac{\\pi}{2} \\right ]$.", "Any help here would be appreciated Cheers Ned substitute $g = 119$ and solve for $x$ So you need to find $x$ such that: $119 = 50 \\sin \\left( \\frac {2 \\pi}{6}x\\right)$ now continue EDIT: and of course you would have to subtract 3 from the answer above, since they started measuring 3 hours after the flow started" ]

[ "(\\sin a, a), for various values of a: Here is the graph of y = \\sin x, with domain restricted to [-\\pi/2, \\pi/2],and the inverse function y = \\arcsin x.Here in an animation showing the lines tangent to y = \\sin x at (a, \\sin a) and y = \\arcsin x at (\\sin a, a), for various values of a: html(\"Here is a plot of $f(x) = \\\\arcsin x$, an animated tangent line,<br />and the derivative $f'(x) = \\\\frac{1}{\\sqrt{1 - x^2}}$.<br />(The derivative may appear first.)\") anim3 = animate([plot((1/sqrt(1-a^2))*(x - a) + asin(a), x, -1, 1, rgbcolor=(0.5, 0, 0)) + plot(asin(x), x, -1, 1, rgbcolor=(0, 0.5, 0)) + point([a, asin(a)], rgbcolor=(1, 0, 0)) for a in srange(-0.95, 1, 0.05)], ymin=-1*pi/2, ymax = pi/2, aspect_ratio=1, figsize=[4,4]) show(anim3) plot(1/sqrt(1 - x^2), x, -0.99, 0.99, figsize=[2.6,2.6]) Here is a plot of f(x) = \\arcsin x, an animated tangent line,and the derivative f'(x) = \\frac{1}{\\sqrt{1 - x^2}}.(The derivative may appear first.) Here is a plot of f(x) = \\arcsin x, an animated tangent line,and the derivative f'(x) = \\frac{1}{\\sqrt{1 - x^2}}.(The derivative may appear first.)", "not outweigh the negative by the cos and tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first solution is a little bit after $\\pi$, which is around $3.14$. Hence the answer is $\\boxed{\\textbf{(D)}}$. (Not original author) Here is the graph: $[asy] Label f; f.p=fontsize(6); xaxis(-5,5,Ticks(f, 1.0)); yaxis(-8,8,Ticks(f, 2.0)); real f(real x) { return sin(x); } draw(graph(f, -5,5)); real g(real x) { return 2*cos(x); } draw(graph(g, -5,5)); real h(real x) { return 3*tan(x); } draw(graph(h, -1.2,1.2)); draw(graph(h, 1.94, 4.34)); draw(graph(h, -4.34, -1.94)); [/asy]$ ## Solution 3(More Detailed Answer of Solution 1) Denote $D_{f}$ as the domain of $f(x).$ Obviously $D_{f}=\\left \\{x|x\\neq \\frac{(2k+1)\\pi}{2} \\right \\} .$ $f^{'}(x)=cosx-2sinx+3sec^{2}x=\\sqrt{5}cos(x+\\phi)+3sec^{2}x > 0$ for all $x$ in sub-intervals of $D_{f}.$ When $x\\in (0,\\frac{\\pi}{2}), f(0)=2, \\lim_{x \\to \\frac{\\pi}{2}^{-} }f(x)=+\\infty .$ Hence $f(x)>0$ in this interval. When $x\\in (\\frac{\\pi}{2},\\pi),\\lim_{x \\to \\frac{\\pi}{2}^{+} }f(x)=-\\infty, f(\\pi)=-2.$ Hence $f(x)<0$ in this interval. When $x\\in (\\pi,\\frac{3\\pi}{2}),f(\\pi)=-2<0.$ Notice that $\\frac{5\\pi}{4}<\\frac{3.15*5}{4}<4<\\frac{3\\pi}{2}, f(\\frac{5\\pi}{4})=3-\\frac{3\\sqrt{2}}{2}>0 .$ Hence there must be a root located in the interval $(3,4).$ Choose $\\boxed{\\textbf{(D)}}$. ~PythZhou", "line/.list={sin1, sin2, prod, sum}, sin1={label in legend={text={$\\sin(x)$}}}, sin2={label in legend={text={$\\sin\\left(x+\\frac{\\pi}{6}\\right)$}}}, prod={label in legend={text={$\\sin(x) \\sin\\left(x+\\frac{\\pi}{6}\\right)$}}}, sum={label in legend={text={$\\sin(x) + \\sin\\left(x+\\frac{\\pi}{6}\\right)$}}}, ] data [set=sin1, format=function] { var x : interval [-pi/6:2*pi] samples 100; func y = sin(\\value x r); } data [set=sin2, format=function] { var x : interval [-pi/6:2*pi] samples 100; func y = sin((\\value x + pi/6)r); } data [set=prod, format=function] { var x : interval [-pi/6:2*pi] samples 100; func y = sin(\\value x r) * sin((\\value x + pi/6)r); } data [set=sum, format=function] { var x : interval [-pi/6:2*pi] samples 100; func y = sin(\\value x r) + sin((\\value x + pi/6)r); }; \\draw [{Bar[]}-{Bar[]}, thick, dashed] (2*pi/6,1.25) -- ++(pi/6,0) node [above, midway] {\\scriptsize $\\Delta x = \\frac{\\pi}{6}$}; \\end{tikzpicture} \\end{document} • \\sin(x) not sin(x), that way the operator is typeset upright as it should be. – Torbjørn T. Jan 19 '16 at 14:19 • thx, changed it. – moospit Jan 19 '16 at 14:28 • @moospit, Thanks for your solution. How do I go about getting the grid like in the original MWE? Thanks! – Joe Jan 19 '16 at 17:59 • @Joe see my updated answer – moospit Jan 20 '16 at 10:06", "# Plotting sin(1/x) associated functions I need to plot x, -x, x^2, -x^2, sin(1/x), x*sin(1/x), x^2*sin(1/x) and sin(1/x). But the functions containing sin(1/x) looks somehow bad. How can I fix it. I also do not know how to label the graphs (writing y=sin(1/x) next to the curve y=sin(1/x). \\documentclass{article} \\usepackage{pst-func} \\begin{document} \\begin{pspicture}*(-5,-2)(5,2) \\SpecialCoor % For label positionning \\psaxes[labels=y,Dx=\\pstPI2]{->}(0,0)(-5,-2)(5,2) \\uput[-90](!PI 0){$\\pi$} \\uput[-90](!PI neg 0){$-\\pi$} 5 \\uput[-90](!PI 2 div 0){$\\frac{\\pi}2$} \\uput[-90](!PI 2 div neg 0){$-\\frac{\\pi}2$} \\psplot[linewidth=1.5pt,linecolor=blue,algebraic]{-5}{5}{sin(1/x)} \\psplot[linewidth=1.5pt,linecolor=red,algebraic]{-5}{5}{x*sin(1/x)} \\psplot[linewidth=1.5pt,linecolor=green,algebraic]{-5}{5}{x^2*sin(1/x)} \\psplot[algebraic,linestyle=dashed]{-5}{5}{x} \\psplot[algebraic,linestyle=dashed]{-5}{5}{-x} \\psplot[algebraic,linestyle=dashed]{-5}{5}{x^2} \\psplot[algebraic,linestyle=dashed]{-5}{5}{-x^2} \\end{pspicture} \\end{document} - Look up the usage of plotpoints in the pst-func manual. You need to give it some more points. – Thruston Mar 27 at 22:56 Either that or stay FAR away from x=0. – John Kormylo Mar 28 at 3:03 None of the suggestion helps. Still, the plot looks cramped. – user48793 Mar 28 at 4:48 @user48793 I'm quite close to downvoting the question. You seriously remind me of companies wanting new web design: They do not know what they do want, and they do want it now. Please show a non-cramped image of the result and we can see what can be done. At this moment, I don's see anything wrong with either of the two answers. – tohecz Mar 28 at 10:43 I do not think that you get a better result with the", "x)/(b + 2 x))^(4/3))))^(-2/3)) $$-\\frac{1}{\\left(76.5676 \\left(\\frac{x}{2 x+12}\\right)^{4/3}+1\\right)^{2/3}}-\\frac{1}{\\left(1-76.5676 \\left(\\frac{x}{2 x+12}\\right)^{4/3}\\right)^{2/3}}+\\frac{12}{x}$$ Plot[equ, {x, -0.1, 0.6}] FindInstance[equ == 0, x, Reals] ... 0 Translating a \"range of values\" directly into the system specs: NSolve[8 Sin[t/4]^2 == n && n ∈ Integers && 1 <= n <= 8, t] V10-region solutions: NSolve[{8 Sin[t/4]^2} ∈ Point[List /@ Range[8]], t] NSolve[8 Sin[t/4]^2 == n && {n} ∈ Point[List /@ Range[8]], {t, n}] 2 Your functions need to be defined before your Solve command. g[x_] := -x + 2; h[x_] := 3 x; f[x_] := -x^2 + 4; Solve[{f[x] == g[x]}, {x}, Reals] Returns {{x -> -1}, {x -> 2}} 0 This works: S=NDSolve[{y'[x]==(y[x]^2+y[x]+1)^(1/2),y[0]==0},y,{x,1,8}]; func1[x_]:=(y[x]/.First[S])^-3; Plot[func1[x],{x,1,2.5}] Often times you need to ensure your definition of func1 is only used when x has a numeric value. In that case use func1[x_?NumericQ] := ... 1 You may solve it exactly: Solve[8 Sin[t/4]^2 == #, t] /. C[1] -> 0 & /@ Range@8 1 Using Rationalize Clear[p, t, v]; Rgas = 8.314 // Rationalize;(*gas constant*) acoef[tc_, pc_] := (27*Rgas^2*tc^2)/(64*pc); bcoef[tc_, pc_] := (Rgas*tc)/(8*pc); latentK = 769*10^2*39*167*10^-29*6022*10^20; (* Note that you do not use latentK *) tcK = 2223;(*Kelvin*) pcK = 16*10^6;(*Pa*) eqn = (p - Rgas*t)*v^3 - bcoef[tcK, pcK]*p*v^2 + acoef[tcK, pcK]*v - ... 1 Try the following: first solve exactly the cubic equation: sl = Solve[a*v^3 -", "x) {return sin(x);} filldraw(graph(f2,0,pi )--(pi ,0)--(0,0)--cycle, fueaev, zzttqq); Some commands have to be defined with respect to a function. $\\texttt{f2}$ returns the $\\sin(x)$ function, and the $\\texttt{graph}$ command draws $\\texttt{f2}$ from $x = 0$ to $x = \\pi$. Label laxis; laxis.p = fontsize(10); string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); These commands draw the axes. The $\\texttt{laxis}$ label specifies the label styles of the axes (e.g., the field $\\texttt{laxis.p}$ defines the desired font pen). The next line, the string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} are functions that define the labels for both axes. These are abbreviated as $\\texttt{xlbl, ylbl}$ in the concise mode. The first defines axes with labels with respect to $\\pi$, and it formats the labels so that the $0$ label simply outputs $0$ instead of $0\\pi$, and $\\pm \\pi$ instead of $\\pm 1\\pi$. Finally, xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); are the commands that actually construct the xaxis and the yaxis. For documentation, see here. The next block of code does the actual construction of the", "\\sin(\\pi + 2x)$ 5. $y = 1 + \\tan\\left(\\pi x\\right)$ 6. $y = 4 \\tan (3x - 2\\pi)$ 7. $y = 5\\sec\\left(\\pi x + \\frac{\\pi}{3}\\right)$ 7. Suppose $f(x) = x \\qquad\\qquad g(x) = \\sin(x)$ Graph $f(x) + g(x)$ in Desmos. Based on what you know about graphical addition, why does the graph $f(x) + g(x)$ look like the way it does? 8. Because $f(x) = \\sin(x)$ is not one to one, what is the interval of values we restrict the domain of $f(x)$ to in order for $f(x)$ to have an inverse? 9. Inverse function property in 2.8 tells us that if $f(x) = \\sin(x)$ and $g(x) = \\sin^{-1}(x)$, then $(f\\circ g)(x) = x$ is valid. For what values of $x$ is $(f \\circ g)(x) = x$ true? Moreover, explain the exact reason why these specific values of $x$ are valid. 10. Inverse function property in 2.8 tells us that $\\cos^{-1}(\\cos(x)) = x$. For what values of $x$ is this true? Moreover, explain the exact reason why these specific values of $x$ are valid. 11. Out of the following, which ones are true? 🤔 Justify your answer with the domain of the proper function. 1. $\\sin^{-1}\\left(\\sin\\left(\\dfrac{\\pi}{3}\\right)\\right) = \\dfrac{\\pi}{3}$ 2. $\\sin^{-1}\\left(\\sin\\left(-\\dfrac{\\pi}{4}\\right)\\right) = -\\dfrac{\\pi}{4}$ 3. $\\sin^{-1}\\left(\\sin\\left(-\\dfrac{2\\pi}{3}\\right)\\right) = -\\dfrac{2\\pi}{3}$ 4. $\\cos^{-1}\\left(\\cos\\left(-\\dfrac{2\\pi}{3}\\right)\\right) = -\\dfrac{2\\pi}{3}$ 5. $\\cos\\left(\\cos^{-1}\\left(\\dfrac{1}{2}\\right)\\right) = \\dfrac{1}{2}$ 6. $\\tan^{-1}\\left(\\tan\\left(-\\dfrac{2\\pi}{3}\\right)\\right) = -\\dfrac{2\\pi}{3}$", "title(expression(y==x*sin(x)+5)) # How to generate this using a loop? x <- seq(0, 3*pi, .5*pi) letters <- c(expression(x[0]==0),expression(x[1]),expression(x[2]), expression(x[3]),expression(x[4]),expression(x[5]), expression(x[6]==3*pi)) axis(1, at=x, labels=letters) from_x <- c(0,.5*pi,pi,1.5*pi,2*pi,2.5*pi,3*pi) from_y <- rep(0,7) to_x <- from_x to_y <- sapply(from_x, function(x) return(x*sin(x)+5)) segments(from_x,from_y,to_x,to_y,lty=3) from_x <- from_x[-length(from_x)] from_y <- to_y[-length(to_y)] to_x <- to_x[-1] to_y <- to_y[-1] segments(from_x,from_y,to_x,to_y,lty=3) Let the length of each interval be denoted by $$h = (b-a)/n$$. Then, the sum of all trapezoid areas ($$T$$) is given by: \\begin{align} T &= \\frac{h}{2}(f(x_0)+f(x_1))+\\frac{h}{2}(f(x_1)+f(x_2))+\\dots\\\\ &\\qquad +\\frac{h}{2}(f(x_{n-1})+f(x_n))\\\\ &= \\frac{h}{2}(f(x_0)+2f(x_1)+2f(x_2)+\\dots+2f(x_{n-1})+f(x_n)) \\end{align} trapezoid <- function(f, a, b, n = 100) { h <- (b-a)/n x <- seq(a, b, by = h) y <- sapply(x, f) T <- h*(y[1]/2 + sum(y[2:n]) + y[n+1]/2) return(T) } f <- function(x) return(x*sin(x)+5) trapezoid(f, 0, 3*pi, 100) ## [1] 56.54169 # Simpson’s rule As indicated by the use of trapezoids, the trapezoidal rule approximates $$f$$ by a piece-wise linear function. Simpson’s rule uses, instead, a piece-wise quadratic function for approximation of $$f$$. Since a quadratic function $$f(x) = ax^2 + bx + c$$ has three parameters, three non-collinear points $$(u,f(u)$$, $$(v,f(v))$$, and $$(w,f(w))$$ pin down the function through a system of three linear equations. (Without loss of generality, assume $$u<v<w$$.) Collinearity means $$(v,f(v)) = t(u,f(u))+(1-t)(w,f(w))$$ for $$t\\in(0,1)$$. Can you see its implication? Solving the system of equation, we get $p(x) = f(u)\\frac{(x-v)(x-w)}{(u-v)(u-w)} +", "x}$$ or, equivalently, $$u'=x.$$ Integrating this yields $$\\label{eq:3.3.25} u={x^2\\over2}+c, \\text y={u\\over\\sin x}= {x^2\\over 2\\sin x}+ {c\\over\\sin x}.$$ is the general solution of \\eqref{eq:3.3.20} on every interval $$\\left(r\\pi,(r+1)\\pi\\right)$$ ($$r=integer)$$. (b) Imposing the initial condition $$y(\\pi/2)=1$$ in \\eqref{eq:3.3.25} yields $$1=\\frac{\\pi^2}{8}+c$$ or $$c=1-\\frac{\\pi^2}{8}.$$ Thus,$$y={x^2\\over 2\\sin x}+{(1-\\pi^2/8)\\over\\sin x}$$ is a solution of \\eqref{eq:3.3.21}. The interval of validity of this solution is $$(0,\\pi)$$; Figure $$\\PageIndex{4}$$ shows its graph. It wasn't necessary to do the computations \\eqref{eq:3.3.23} and \\eqref{eq:3.3.24} in Example $$\\PageIndex{5}$$ since we showed in the discussion preceding Example $$\\PageIndex{5}$$ that if $$y=uy_1$$ where $$y_1'+p(x)y_1=0$$, then $$y'+p(x)y=u'y_1$$. We did these computations so you would see this happen in this specific example. We recommend that you include these \"unnecesary'' computations in doing exercises, until you're confident that you really understand the method. After that, omit them. We summarize the method of variation of parameters for solving $$\\label{eq:3.3.26} y'+p(x)y=f(x)$$ as follows: (a) Find a function $$y_1$$ such that $${y_1'\\over y_1}=-p(x).$$ For convenience, take the constant of integration to be zero. (b) Write $$\\label{eq:3.3.27} y=uy_1$$ to remind yourself of what you're doing. (c) Write $$u'y_1=f$$ and solve for $$u'$$; thus, $$u'=f/y_1$$. (d) Integrate $$u'$$ to obtain $$u$$, with an arbitrary constant of integration. (e) Substitute $$u$$ into \\eqref{eq:3.3.27} to obtain $$y$$. To solve an equation written as $$P_0(x)y'+P_1(x)y=F(x)$$ we recommend that you divide through by $$P_0(x)$$", "# concavity ## 2626 days ago by chrisphan f = lambda x: sin(x) ddf = lambda x: -sin(x) fplot = plot(f, -pi, 0, rgbcolor=(0, 0.75, 0)) + plot(f, pi, 2*pi, rgbcolor=(0, 0.75, 0)) + plot(f, -2*pi, -pi, rgbcolor=(0.75, 0, 0)) + plot(f, 0, pi, rgbcolor=(0.75, 0, 0)) ddfplot = plot(ddf, -pi, 0, rgbcolor=(0, 0.75, 0)) + plot(ddf, pi, 2*pi, rgbcolor=(0, 0.75, 0)) + plot(ddf, -2*pi, -pi, rgbcolor=(0.75, 0, 0)) + plot(ddf, 0, pi, rgbcolor=(0.75, 0, 0)) html(\"<h1>Example 1</h1>Let $f(x) = \\sin x$. Then $f''(x) = -\\sin x$. Here is a graph of<br />$y = f''(x)$, showing where $f''(x) > 0$ (in green) and $f''(x) < 0$ (in red):\") show(ddfplot) html(\"$f$ is concave up when $f''(x) > 0$ and concave down when $f''(x) < 0$.<br />Here is a graph of $y=f(x)$.\") show(fplot) html(\"The inflection points are where $f''(x)$ changes sign, in this case<br />at every $x = \\pi k$ for every integer $k$.\") # Example 1 Let f(x) = \\sin x. Then f''(x) = -\\sin x. Here is a graph of y = f''(x), showing where f''(x) > 0 (in green) and f''(x) < 0 (in red): f is concave up when f''(x) > 0 and concave down when f''(x) < 0. Here is a graph of y=f(x). The inflection points are where f''(x) changes sign, in this case", "# Plotting Weierstrass function I'm trying to plot Weierstrass function using only basic TikZ picture functionality (no gnuplot or whatnot). How do I use sum in a \\draw? Do I have to make a new command? Use a loop? An alternative (ugly) solution with manual summation: \\begin{tikzpicture}[xscale=2.2,yscale=2.7] \\draw[thick, color=lightgrey,step=0.25cm,solid] (-2,-0.75) grid (2,0.75); \\draw[<->] (-2.1,0) -- (2.1,0) node[below right] {$x$}; \\draw[<->] (0,-0.9) -- (0,0.9) node[left] {$y$}; \\draw[color=newblue, thick, domain=-2:2,samples=500,/pgf/fpu,/pgf/fpu/output format=fixed] plot (\\x, {(1/2)*sin(2*\\x r) + (1/4)*sin(4*\\x r) + (1/8)*sin(8*\\x r) + (1/16)*sin(16*\\x r) + (1/32)*sin(32*\\x r) + (1/64)*sin(64*\\x r) + (1/128)*sin(128*\\x r) + (1/256)*sin(256*\\x r) + (1/512)*sin(512*\\x r) + (1/1024)*sin(1024*\\x r) + (1/2048)*sin(2048*\\x r) + (1/4096)*sin(4096*\\x r) + (1/8192)*sin(8192*\\x r) + (1/16384)*sin(16384*\\x r) + (1/32768)*sin(32768*\\x r) + (1/65536)*sin(65536*\\x r) + (1/131072)*sin(131072*\\x r) + (1/262144)*sin(262144*\\x r) + (1/524288)*sin(524288*\\x r) + (1/1048576)*sin(1048576*\\x r) }) node[right, black] {}; \\end{tikzpicture} • Without trying, I have a feeling that you will get into problems since the function involves 'summing' to infinity. – Svend Tveskæg Feb 1 '14 at 14:59 • It can still be plotted at a certain resolution or \"zoom level\" without any issues. If you look at Wikipedia you can see a plot over [-2,2] at a particular scale for example. – Jay Feb 1 '14 at 15:21 • I personally vote agaisnt ask LaTeX do the calculations (at least more than", "sin(0)) + (pi/4 * sin(pi/4))) ## [1] 0.2180895 Integrating the function over the interval yields $\\frac{\\pi - 4}{4\\sqrt{2}} = 0.15175$, which gives an error of 0.0663395. For a second example, apply the Trapezoidal Rule to approximate the definite integral: $$\\large{\\int^1_0 x^2 \\space e^{-x} \\space dx}$$ Visualizing how the trapezoidal rule approximates the integral: f2 <- function(x) { return(x^2 * exp(-x)) } df <- data.frame(cbind(c(0, 1, 1, 0), c(0, f2(1), 0, 0))) ggplot(data = data.frame(x = 0), mapping = aes(x = x)) + stat_function(fun = f2, size = 1.05, alpha = 0.75, color='blue') + geom_segment(aes(x = 0, y = 0, xend = 1, yend = f2(1))) + geom_segment(aes(x = 1, y = 0, xend = 1, yend = f2(1))) + geom_polygon(data = df, aes(x=X1, y=X2), fill = 'blue', alpha = 0.2) + geom_area(stat = 'function', fun = f2, fill = 'black', alpha = 0.3, xlim = c(0, 1)) + xlim(c(0,1)) $$\\large{\\int^1_0 x^2 \\space e^{-x} \\space dx \\approx \\frac{1 - 0}{2}[(0 * (0)^2 e^0) + ((1)^2 e^{-1})] = 0.1839397}$$ 0.5 * ((0 * 0^2 * exp(0)) + (1^2 * exp(-1))) ## [1] 0.1839397 2 - 5/exp(1) ## [1] 0.1606028 The function integrated over the interval $[0, 1]$ is $2 - \\frac{5}{e} = 0.1606028$. Therefore the error is 0.0233369. ###### Trapezoidal Rule in R The Trapezoidal Rule is comparatively straightforward to", "# Closed-form of $\\int_{a}^{b}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx$ for some $a<b$ In this question I asked to prove that $$\\int_{0}^{1}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx=\\frac{\\pi e}{24}.$$ If we take a look at the plot of the integrand, then we could see some symmetry-property. Because the integrand is symmectic to the $y=1/2$ line, it is also true, that $$\\int_{0}^{1/2}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx=\\int_{1/2}^{1}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx=\\frac{\\pi e}{48}.$$ Question. Is there a close-form of $I(a,b)=\\int_{a}^{b}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx$ for some else $a<b$? If it is too hard, then is there a closed-form for some $a<b$, just for the real or imaginary part? For example $$\\Re \\left[ \\int_{1}^{3/2}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx \\right],$$ $$\\Re \\left[ \\int_{3/2}^{2}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx \\right],$$ or $$\\Im \\left[ \\int_{1}^{2}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx \\right]?$$ • I think we'd be pushing our luck asking for a closed form for general limits $a,b$. Closed forms might exist when the limits coincide with zeros of the integrand, perhaps. – David H Oct 11 '14 at 0:57 • @DavidH This is why I asked these three specific integrand limits at the end of the question. Of course anything else would be nice. – user153012 Oct 11 '14 at 1:05 Is there a close-form of $I(a,b)=\\displaystyle\\int_{a}^{b}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx$ for some other $a<b$? Yes, of course there is. An infinity of them, in fact. But, in that case, a and b will lack a closed form. :-) For instance, all rational and algebraic numbers possess", "I-VI). Give reasons for your choices.$$z=\\sin (x y)$$Check back soon! Problem 60 59-64 Match the function (a) with its graph (labeled A-F below) and (b) with its contour map (labeled I-VI). Give reasons for your choices.$$z=e^{x} \\cos y$$Check back soon! Problem 61 59-64 Match the function (a) with its graph (labeled A-F below) and (b) with its contour map (labeled I-VI). Give reasons for your choices.$$z=\\sin (x-y)$$Check back soon! Problem 62 59-64 Match the function (a) with its graph (labeled A-F below) and (b) with its contour map (labeled I-VI). Give reasons for your choices.$$z=\\sin x-\\sin y$$Check back soon! Problem 63 59-64 Match the function (a) with its graph (labeled A-F below) and (b) with its contour map (labeled I-VI). Give reasons for your choices.$$z=\\left(1-x^{2}\\right)\\left(1-y^{2}\\right)$$Check back soon! Problem 64 59-64 Match the function (a) with its graph (labeled A-F below) and (b) with its contour map (labeled I-VI). Give reasons for your choices.$$z=\\frac{x-y}{1+x^{2}+y^{2}}$$Check back soon! Problem 65 65-68 Describe the level surfaces of the function.$$f(x, y, z)=x+3 y+5 z$$Check back soon! Problem 66 65-68 Describe the level surfaces of the function.$$f(x, y, z)=x^{2}+3 y^{2}+5 z^{2}$$Check back soon! Problem 67 65-68 Describe the level surfaces of the function.$$f(x, y, z)=y^{2}+z^{2}$$Check back soon! Problem 68 65-68 Describe the level surfaces of the function.$$f(x, y, z)=x^{2}-y^{2}-z^{2}$$Check back soon! Problem 69 69-70", "use this result for solving sine problem $$c(x,y)=\\sin(\\pi(x+y))$$ in Problem 1.29 [https://math.boku.ac.at/udt/unsolvedproblems.pdf]. (III) Approximated the cost function $$c(x,y)$$ by piecewise constant functions M. Hofer and M. R. Iacò [J. Optim. Theory Appl. 161, No. 3, 999–1011 (2014; Zbl 1293.90036)] proved: Let $$(a_{i,j})$$, $$i,j=1,2,\\dots,n$$ be a real-valued $$n\\times n$$ matrix. Let $$I_{i,j}=\\big[\\frac{i-1}{n},\\frac{i}{n}\\big]\\times\\big[\\frac{j-1}{n},\\frac{j}{n}\\big]$$, $$i,j=1,2,\\dots,n$$ and let the piecewise constant function $$c(x,y)$$ be defined as $$c(x,y)=a_{i,j}\\text{ if }(x,y)\\in I_{i,j}, i,j=1,2,\\dots,n.$$ Then $\\max_{C(x,y)\\text{-copula}}\\int_0^1\\int_0^1c(x,y)d_xd_yC(x,y) =\\frac{1}{n}\\sum_{i=1}^na_{i,\\pi^*(i)}.$ Here $$\\pi^*(i)$$ maximizes $$\\sum_{i=1}^na_{i,\\pi(i)}$$, where $$\\pi$$ is a permutation of $$(1,2,\\dots,n)$$. finally, the authors use this method, to approximate solution of $$c(x,y)=\\sin(\\pi x)\\sin(\\pi y)$$, $$c(x,y)=\\sin(\\pi x)\\cos(\\pi y)$$ and $$c(x,y)=\\sin(\\pi/x)\\cos(2\\pi y)$$, respectively. The exact values of optimal $$C(x,y)$$ are open. ##### MSC: 90C27 Combinatorial optimization 60E05 Probability distributions: general theory 90C46 Optimality conditions and duality in mathematical programming QRM Full Text: ##### References: [1] Aistleitner, C.; Hofer, M., On the limit distribution of consecutive elements of the Van der Corput sequence, Unif. Distrib. Theory, 8, 1, 89-96, (2013) · Zbl 1340.11060 [2] Ambrosio, L.; Gigli, N., A user’s guide to optimal transport, (Modelling and Optimisation of Flows on Networks, Lecture Notes in Math., vol. 2062, (2013), Springer Heidelberg), 1-155 [3] Ambrosio, L.; Gigli, N.; Savaré, G., (Gradient Flows in Metric Spaces and in the Space of Probability Measures, Lectures in Mathematics ETH Zürich, (2008), Birkhäuser Verlag Basel) [4] V.", "use this result for solving sine problem $$c(x,y)=\\sin(\\pi(x+y))$$ in Problem 1.29 [https://math.boku.ac.at/udt/unsolvedproblems.pdf]. (III) Approximated the cost function $$c(x,y)$$ by piecewise constant functions M. Hofer and M. R. Iacò [J. Optim. Theory Appl. 161, No. 3, 999–1011 (2014; Zbl 1293.90036)] proved: Let $$(a_{i,j})$$, $$i,j=1,2,\\dots,n$$ be a real-valued $$n\\times n$$ matrix. Let $$I_{i,j}=\\big[\\frac{i-1}{n},\\frac{i}{n}\\big]\\times\\big[\\frac{j-1}{n},\\frac{j}{n}\\big]$$, $$i,j=1,2,\\dots,n$$ and let the piecewise constant function $$c(x,y)$$ be defined as $$c(x,y)=a_{i,j}\\text{ if }(x,y)\\in I_{i,j}, i,j=1,2,\\dots,n.$$ Then $\\max_{C(x,y)\\text{-copula}}\\int_0^1\\int_0^1c(x,y)d_xd_yC(x,y) =\\frac{1}{n}\\sum_{i=1}^na_{i,\\pi^*(i)}.$ Here $$\\pi^*(i)$$ maximizes $$\\sum_{i=1}^na_{i,\\pi(i)}$$, where $$\\pi$$ is a permutation of $$(1,2,\\dots,n)$$. finally, the authors use this method, to approximate solution of $$c(x,y)=\\sin(\\pi x)\\sin(\\pi y)$$, $$c(x,y)=\\sin(\\pi x)\\cos(\\pi y)$$ and $$c(x,y)=\\sin(\\pi/x)\\cos(2\\pi y)$$, respectively. The exact values of optimal $$C(x,y)$$ are open. ##### MSC: 90C27 Combinatorial optimization 60E05 Probability distributions: general theory 90C46 Optimality conditions and duality in mathematical programming QRM Full Text:", "when $(x < 2,2 \\sin \\left((x - 1) \\frac {\\pi}{2} \\right),a + 3 \\sin \\left((x-4)\\frac {\\pi}{2} \\right)|a=5$ $y3(x)= \\begin{cases} 2 \\sin \\left ((x-1) \\frac{\\pi}{2} \\right ) ,x<2 \\\\\\qquad \\qquad \\qquad \\qquad \\qquad |a=5 \\\\ a + 3 \\sin \\left ((x-4) \\frac {\\pi}{2} \\right ),x \\ge 12\\end{cases}$ 6. Graph one function at a time by using $F4$ to have only one function checked at a time. On a graph screen examine both sides of where the discontinuity exists using $F3$ Trace. 7. For Problems 1 and 2 below, use $[\\blacklozenge]$ $F4$ to have table settings of $\\text{tblStart} = 0.98$ and $\\triangle \\text{tbl}= 0.01$, to numerically examine the left and right-hand limits. Be sure to press ENTER to save changes before pressing $[\\blacklozenge]$ $F5$ to view the table. For Problems 1, 2, and 3 estimate the limits graphically and numerically using trace and table. ## Problem 1 $y1(x) = \\begin{cases} 1,x<1 \\\\\\qquad \\qquad |a=5 \\\\ a,x\\ge1 \\end{cases}$ $\\lim_{x \\to 1^-}y1(x) \\approx \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}$ $\\lim_{x \\to 1^+}y1(x) \\approx \\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}$ Try other values for $a$ in the graph of $y1(x)$ to find what $a$ makes $\\lim_{x \\to 1}y1(x)$ exist. On the $Y=$ screen, press ENTER when $y1$ is highlighted. Press $\\blacktriangleright$ and then backspace $\\leftarrow$ to try different values for $a$. Graph it to see if appear continuous. $a=\\underline{\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;}$ ## Problem 2 $y2(x)=", "that $a+c=2+8=\\boxed{10}$. ### Explanation of the last step This is a property all rectangles in the coordinate plane have. For a proof, note that for any rectangle $ABCD$ its center can be computed as $(A+C)/2$ and at the same time as $(B+D)/2$. In our case, we can compute that the center is $\\left(\\frac{2+8}2,\\frac{5+3}2\\right)=(5,4)$, therefore $\\frac{a+c}2=5$, and $a+c=10$. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); draw ( (4,7) -- (6,1) ); draw ( (2,5) -- (8,3) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ ### An alternate last step We can easily compute $a$ and $c$ using our picture. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ Consider the first graph on the interval $[2,8]$. The graph starts at height $5$, then rises for $a-2$ steps to the height $b=5+(a-2)$, and then falls for $8-a$ steps to the height $3=5+(a-2)-(8-a)$. Solving for $a$ we get $a=4$. Similarly we compute $c=6$, therefore $a+c=10$.", "e, and I (or the alternative i) # Most common mathematical functions are supported by default show(5/10 + 1/pi) show(sin(pi/3)) show(log(2)) show(log(2).n()) # Adding \".n()\" gives a numerical approximation (can use \".n(k)\" to get k bits) show(exp(i*2*pi/3)) show(exp(I*2*pi/101)) $\\frac{1}{\\pi} + \\frac{1}{2}$ $\\frac{1}{2} \\, \\sqrt{3}$ $\\log\\left(2\\right)$ $0.693147180559945$ $\\frac{1}{2} i \\, \\sqrt{3} - \\frac{1}{2}$ $e^{\\left(\\frac{2}{101} i \\, \\pi\\right)}$ In [6]: # Mathematical objects and variables can be inserted in print commands using {} and .format(math1,math2,...) print(\"The area of a circle of radius 1 is approximately\",pi.n()) print(\"The square-root of {} is approximately {}\".format(log(2),sqrt(log(2)).n())) The area of a circle of radius 1 is approximately 3.14159265358979 The square-root of log(2) is approximately 0.832554611157698 In [7]: # To access the help page for a function, type the name of the function and then add a question mark # For instance, evaluating the expression \"sin?\" (without quotes) gives the help page for sine # THIS WILL OPEN A NEW WINDOW AREA # Using two question marks (e.g. \"sin??\") shows the source code of the function # (Note that for many low level functions like sin Sage relies on outside packages and the source code is not very informative) sin? In [8]: # Variables are defined using the \"=\" operator # Note that some operations (such as variable assignment) do not have an output # So", "Summary of this function goes here % Initializes the array to Sin(2*pi*x) u0=zeros(floor(L+1),1); X=1; x=(0:X/L:1)'; for j=1:(L+1), u0=sin(2*pi*x); end end and function [ u0 ] = Zero( L) %Summary of this function goes here % Initializes the array to a zero! u0=zeros(floor(L+1),1); end Now, solving it analytically, if use f=sin2 and g=zeros, we know the answer should be: $$u(x,t)=\\frac{\\sin(2\\pi(x+ct))+\\sin(2\\pi(x-ct)}{2}$$ so the expected range is $\\{-1,1\\}$. but when I execute Explicit_W, say: Explicit_W(0.05,0.0010,30,'Sin2','Zero') I get funny results, ranging from -124 to 124! Here is a picture of the results: http://i.stack.imgur.com/wON52.png So it has the form of the solution, but something is scaling it up! I have checked the sub-functions used in this on first order problems so I know there is nothing wrong with them. Whatever is causing the problem is in the main function. Any idea why it's acting the way it does? UPDATE: So it was brought to my attention that I had forgotten to implement the boundary conditions, here $u(0,t)=u(1,t)=0.$ Silly mistake! So the modified section of the code is: u_ancient=f(J); u_old= f(J)+ dt*g(J)+ 1/2*c^2*Tridiag(J+1,s2,-2*s2,s2)*f(J); u_new= Tridiag(J+1,c^2*s2,2-2*c^2*s2,c^2*s2)*u_old - u_ancient; c1=0;c2=0;%boundary values %enforcing the boundary conditions u_new(1)=c1; u_new(end)=c2; u_old(1)=c1; u_old(end)=c2; for i=1:n-2, %Iterating the system using the Explicit formulation temp=u_new; u_new=Tridiag(J+1,c^2*s2,2-2*c^2*s2,c^2*s2)*u_old - u_ancient; u_new(1)=c1; %boundary conditions u_new(end)=c2;%boundary conditions u_ancient=u_old; u_old=temp; end That obviously fixes the issue with" ]

[ "line/.list={sin1, sin2, prod, sum}, sin1={label in legend={text={$\\sin(x)$}}}, sin2={label in legend={text={$\\sin\\left(x+\\frac{\\pi}{6}\\right)$}}}, prod={label in legend={text={$\\sin(x) \\sin\\left(x+\\frac{\\pi}{6}\\right)$}}}, sum={label in legend={text={$\\sin(x) + \\sin\\left(x+\\frac{\\pi}{6}\\right)$}}}, ] data [set=sin1, format=function] { var x : interval [-pi/6:2*pi] samples 100; func y = sin(\\value x r); } data [set=sin2, format=function] { var x : interval [-pi/6:2*pi] samples 100; func y = sin((\\value x + pi/6)r); } data [set=prod, format=function] { var x : interval [-pi/6:2*pi] samples 100; func y = sin(\\value x r) * sin((\\value x + pi/6)r); } data [set=sum, format=function] { var x : interval [-pi/6:2*pi] samples 100; func y = sin(\\value x r) + sin((\\value x + pi/6)r); }; \\draw [{Bar[]}-{Bar[]}, thick, dashed] (2*pi/6,1.25) -- ++(pi/6,0) node [above, midway] {\\scriptsize $\\Delta x = \\frac{\\pi}{6}$}; \\end{tikzpicture} \\end{document} • \\sin(x) not sin(x), that way the operator is typeset upright as it should be. – Torbjørn T. Jan 19 '16 at 14:19 • thx, changed it. – moospit Jan 19 '16 at 14:28 • @moospit, Thanks for your solution. How do I go about getting the grid like in the original MWE? Thanks! – Joe Jan 19 '16 at 17:59 • @Joe see my updated answer – moospit Jan 20 '16 at 10:06", "x)/(b + 2 x))^(4/3))))^(-2/3)) $$-\\frac{1}{\\left(76.5676 \\left(\\frac{x}{2 x+12}\\right)^{4/3}+1\\right)^{2/3}}-\\frac{1}{\\left(1-76.5676 \\left(\\frac{x}{2 x+12}\\right)^{4/3}\\right)^{2/3}}+\\frac{12}{x}$$ Plot[equ, {x, -0.1, 0.6}] FindInstance[equ == 0, x, Reals] ... 0 Translating a \"range of values\" directly into the system specs: NSolve[8 Sin[t/4]^2 == n && n ∈ Integers && 1 <= n <= 8, t] V10-region solutions: NSolve[{8 Sin[t/4]^2} ∈ Point[List /@ Range[8]], t] NSolve[8 Sin[t/4]^2 == n && {n} ∈ Point[List /@ Range[8]], {t, n}] 2 Your functions need to be defined before your Solve command. g[x_] := -x + 2; h[x_] := 3 x; f[x_] := -x^2 + 4; Solve[{f[x] == g[x]}, {x}, Reals] Returns {{x -> -1}, {x -> 2}} 0 This works: S=NDSolve[{y'[x]==(y[x]^2+y[x]+1)^(1/2),y[0]==0},y,{x,1,8}]; func1[x_]:=(y[x]/.First[S])^-3; Plot[func1[x],{x,1,2.5}] Often times you need to ensure your definition of func1 is only used when x has a numeric value. In that case use func1[x_?NumericQ] := ... 1 You may solve it exactly: Solve[8 Sin[t/4]^2 == #, t] /. C[1] -> 0 & /@ Range@8 1 Using Rationalize Clear[p, t, v]; Rgas = 8.314 // Rationalize;(*gas constant*) acoef[tc_, pc_] := (27*Rgas^2*tc^2)/(64*pc); bcoef[tc_, pc_] := (Rgas*tc)/(8*pc); latentK = 769*10^2*39*167*10^-29*6022*10^20; (* Note that you do not use latentK *) tcK = 2223;(*Kelvin*) pcK = 16*10^6;(*Pa*) eqn = (p - Rgas*t)*v^3 - bcoef[tcK, pcK]*p*v^2 + acoef[tcK, pcK]*v - ... 1 Try the following: first solve exactly the cubic equation: sl = Solve[a*v^3 -", "that $a+c=2+8=\\boxed{10}$. ### Explanation of the last step This is a property all rectangles in the coordinate plane have. For a proof, note that for any rectangle $ABCD$ its center can be computed as $(A+C)/2$ and at the same time as $(B+D)/2$. In our case, we can compute that the center is $\\left(\\frac{2+8}2,\\frac{5+3}2\\right)=(5,4)$, therefore $\\frac{a+c}2=5$, and $a+c=10$. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); draw ( (4,7) -- (6,1) ); draw ( (2,5) -- (8,3) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ ### An alternate last step We can easily compute $a$ and $c$ using our picture. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ Consider the first graph on the interval $[2,8]$. The graph starts at height $5$, then rises for $a-2$ steps to the height $b=5+(a-2)$, and then falls for $8-a$ steps to the height $3=5+(a-2)-(8-a)$. Solving for $a$ we get $a=4$. Similarly we compute $c=6$, therefore $a+c=10$.", "x}$$ or, equivalently, $$u'=x.$$ Integrating this yields $$\\label{eq:3.3.25} u={x^2\\over2}+c, \\text y={u\\over\\sin x}= {x^2\\over 2\\sin x}+ {c\\over\\sin x}.$$ is the general solution of \\eqref{eq:3.3.20} on every interval $$\\left(r\\pi,(r+1)\\pi\\right)$$ ($$r=integer)$$. (b) Imposing the initial condition $$y(\\pi/2)=1$$ in \\eqref{eq:3.3.25} yields $$1=\\frac{\\pi^2}{8}+c$$ or $$c=1-\\frac{\\pi^2}{8}.$$ Thus,$$y={x^2\\over 2\\sin x}+{(1-\\pi^2/8)\\over\\sin x}$$ is a solution of \\eqref{eq:3.3.21}. The interval of validity of this solution is $$(0,\\pi)$$; Figure $$\\PageIndex{4}$$ shows its graph. It wasn't necessary to do the computations \\eqref{eq:3.3.23} and \\eqref{eq:3.3.24} in Example $$\\PageIndex{5}$$ since we showed in the discussion preceding Example $$\\PageIndex{5}$$ that if $$y=uy_1$$ where $$y_1'+p(x)y_1=0$$, then $$y'+p(x)y=u'y_1$$. We did these computations so you would see this happen in this specific example. We recommend that you include these \"unnecesary'' computations in doing exercises, until you're confident that you really understand the method. After that, omit them. We summarize the method of variation of parameters for solving $$\\label{eq:3.3.26} y'+p(x)y=f(x)$$ as follows: (a) Find a function $$y_1$$ such that $${y_1'\\over y_1}=-p(x).$$ For convenience, take the constant of integration to be zero. (b) Write $$\\label{eq:3.3.27} y=uy_1$$ to remind yourself of what you're doing. (c) Write $$u'y_1=f$$ and solve for $$u'$$; thus, $$u'=f/y_1$$. (d) Integrate $$u'$$ to obtain $$u$$, with an arbitrary constant of integration. (e) Substitute $$u$$ into \\eqref{eq:3.3.27} to obtain $$y$$. To solve an equation written as $$P_0(x)y'+P_1(x)y=F(x)$$ we recommend that you divide through by $$P_0(x)$$", "not outweigh the negative by the cos and tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first solution is a little bit after $\\pi$, which is around $3.14$. Hence the answer is $\\boxed{\\textbf{(D)}}$. (Not original author) Here is the graph: $[asy] Label f; f.p=fontsize(6); xaxis(-5,5,Ticks(f, 1.0)); yaxis(-8,8,Ticks(f, 2.0)); real f(real x) { return sin(x); } draw(graph(f, -5,5)); real g(real x) { return 2*cos(x); } draw(graph(g, -5,5)); real h(real x) { return 3*tan(x); } draw(graph(h, -1.2,1.2)); draw(graph(h, 1.94, 4.34)); draw(graph(h, -4.34, -1.94)); [/asy]$ ## Solution 3(More Detailed Answer of Solution 1) Denote $D_{f}$ as the domain of $f(x).$ Obviously $D_{f}=\\left \\{x|x\\neq \\frac{(2k+1)\\pi}{2} \\right \\} .$ $f^{'}(x)=cosx-2sinx+3sec^{2}x=\\sqrt{5}cos(x+\\phi)+3sec^{2}x > 0$ for all $x$ in sub-intervals of $D_{f}.$ When $x\\in (0,\\frac{\\pi}{2}), f(0)=2, \\lim_{x \\to \\frac{\\pi}{2}^{-} }f(x)=+\\infty .$ Hence $f(x)>0$ in this interval. When $x\\in (\\frac{\\pi}{2},\\pi),\\lim_{x \\to \\frac{\\pi}{2}^{+} }f(x)=-\\infty, f(\\pi)=-2.$ Hence $f(x)<0$ in this interval. When $x\\in (\\pi,\\frac{3\\pi}{2}),f(\\pi)=-2<0.$ Notice that $\\frac{5\\pi}{4}<\\frac{3.15*5}{4}<4<\\frac{3\\pi}{2}, f(\\frac{5\\pi}{4})=3-\\frac{3\\sqrt{2}}{2}>0 .$ Hence there must be a root located in the interval $(3,4).$ Choose $\\boxed{\\textbf{(D)}}$. ~PythZhou", "title(expression(y==x*sin(x)+5)) # How to generate this using a loop? x <- seq(0, 3*pi, .5*pi) letters <- c(expression(x[0]==0),expression(x[1]),expression(x[2]), expression(x[3]),expression(x[4]),expression(x[5]), expression(x[6]==3*pi)) axis(1, at=x, labels=letters) from_x <- c(0,.5*pi,pi,1.5*pi,2*pi,2.5*pi,3*pi) from_y <- rep(0,7) to_x <- from_x to_y <- sapply(from_x, function(x) return(x*sin(x)+5)) segments(from_x,from_y,to_x,to_y,lty=3) from_x <- from_x[-length(from_x)] from_y <- to_y[-length(to_y)] to_x <- to_x[-1] to_y <- to_y[-1] segments(from_x,from_y,to_x,to_y,lty=3) Let the length of each interval be denoted by $$h = (b-a)/n$$. Then, the sum of all trapezoid areas ($$T$$) is given by: \\begin{align} T &= \\frac{h}{2}(f(x_0)+f(x_1))+\\frac{h}{2}(f(x_1)+f(x_2))+\\dots\\\\ &\\qquad +\\frac{h}{2}(f(x_{n-1})+f(x_n))\\\\ &= \\frac{h}{2}(f(x_0)+2f(x_1)+2f(x_2)+\\dots+2f(x_{n-1})+f(x_n)) \\end{align} trapezoid <- function(f, a, b, n = 100) { h <- (b-a)/n x <- seq(a, b, by = h) y <- sapply(x, f) T <- h*(y[1]/2 + sum(y[2:n]) + y[n+1]/2) return(T) } f <- function(x) return(x*sin(x)+5) trapezoid(f, 0, 3*pi, 100) ## [1] 56.54169 # Simpson’s rule As indicated by the use of trapezoids, the trapezoidal rule approximates $$f$$ by a piece-wise linear function. Simpson’s rule uses, instead, a piece-wise quadratic function for approximation of $$f$$. Since a quadratic function $$f(x) = ax^2 + bx + c$$ has three parameters, three non-collinear points $$(u,f(u)$$, $$(v,f(v))$$, and $$(w,f(w))$$ pin down the function through a system of three linear equations. (Without loss of generality, assume $$u<v<w$$.) Collinearity means $$(v,f(v)) = t(u,f(u))+(1-t)(w,f(w))$$ for $$t\\in(0,1)$$. Can you see its implication? Solving the system of equation, we get $p(x) = f(u)\\frac{(x-v)(x-w)}{(u-v)(u-w)} +", "154213/10000000: a = -154213 /10000000 Sin[Pi/4 + Pi/8*x]; Solve[a == 0, x] {{x -> ConditionalExpression[-((2 (π - 8 π C[1]))/π), C[1] ∈ Integers]}, {x -> ConditionalExpression[-(( 2 (π - 4 (π + 2 π C[1])))/π), C[1] ∈ Integers]}} Solve[a == 0, x] /. C[1] -> 0 {{x -> -2}, {x -> 6}} Reduce[a == 0, x] /. C[1] -> 0 x == -2 || x == 6 To constrain x to the interval $[-2,2]$: Solve[{a == 0, -2 <= x <= 2}, x] {{x -> -2}} Its a issues with your syntax, you need to make a a function. a[x_] := -0.0154213 Sin[Pi/4 + Pi/8*x] Solve[a[x] == 0, x] (*{{x -> -2.}} Now if you plot the function you see it has three zeroes you can use FindRoot and a plot to get exact values. FindRoot[a[x], {x, 0}] FindRoot[a[x], {x, -10}] FindRoot[a[x], {x, 5}] (*{x -> -2.} {x -> -10.} {x -> 6.}*)", "+ sin(t) * sin(5 * t) - y # Initial condition (at t = 0) y0 = 0 # Set end time for the solutions below tEnd = 20 ## Exact solution # Set up to find exact solution, phi(t) phi = function('phi', t) # Find exact solution. ics specifies initial condition y(0) = y0 exactSolution = desolve(diff(phi, t) == ODE(t, phi), phi, ics=[0, y0]) show(y == expand(exactSolution(t = t))) ESPlot = plot(exactSolution, 0, tEnd, rgbcolor = 'blue') ESPlot.axes_labels(('t', 'y')) show(ESPlot) y = -\\frac{639}{629} \\, e^{-t} + \\frac{2}{17} \\, \\sin\\left(4 \\, t\\right) - \\frac{3}{37} \\, \\sin\\left(6 \\, t\\right) + \\frac{1}{34} \\, \\cos\\left(4 \\, t\\right) - \\frac{1}{74} \\, \\cos\\left(6 \\, t\\right) + 1 y = -\\frac{639}{629} \\, e^{-t} + \\frac{2}{17} \\, \\sin\\left(4 \\, t\\right) - \\frac{3}{37} \\, \\sin\\left(6 \\, t\\right) + \\frac{1}{34} \\, \\cos\\left(4 \\, t\\right) - \\frac{1}{74} \\, \\cos\\left(6 \\, t\\right) + 1 ## Numerical approximations # Lists to store the error in for different values of the stepSize EMaxAbsError = [] IEMaxAbsError = [] RKMaxAbsError = [] # Loop over different values of the step size for tStep in (0.5, 0.25, 0.1, 0.05, 0.025): # Find the Euler approximation and compute its error EApprox = Euler(ODE, y0, 0, tEnd, tStep) EPlot = line(EApprox, rgbcolor = 'red') EAbsError = [abs(exactSolution(t = T).n() - Y) for (T, Y) in", "a multiple of $X_n(x) = \\sin \\left( \\dfrac{n \\pi}{w}x \\right).$ For these given $$\\lambda_n$$, the general solution for $$Y$$ (one for each $$n$$) is $Y_n(y)=A_n \\cosh \\left( \\dfrac{n \\pi}{w}y \\right) + B_n \\sinh \\left( \\dfrac{n \\pi}{w}y \\right).$ We only have one condition on $$Y_n$$ and hence we can pick one of $$A_n$$ or $$B_n$$ to be something convenient. It will be useful to have $$Y_n(0)=1$$, so we let $$A_n=1$$. Setting $$Y_n(h)=0$$ and solving for $$B_n$$ we get that $B_n = \\dfrac{ - \\cosh \\left( \\dfrac{n \\pi h}{w} \\right)}{ \\sinh \\left( \\dfrac{n \\pi h}{w} \\right)}.$ After we plug the $$A_n$$ and $$B_n$$ we into (4.9.14) and simplify, we find $Y_n(y) = \\dfrac{ \\sinh \\left( \\dfrac{n \\pi (h-y)}{w} \\right)}{ \\sinh \\left( \\dfrac{n \\pi h}{w} \\right)}.$ We define $$u_n(x,y)=X_n(x)Y_n(y)$$. And note that $$u_n$$ satisfies (4.9.5)–(4.9.8). Observe that $u_n(x,0)= X_n(x)Y_n(0) = \\sin \\left( \\dfrac{n \\pi}{w}x \\right).$ Suppose $f(x)= \\sum_{n=1}^{ \\infty}b_n \\sin \\left( \\dfrac{n \\pi x}{w} \\right).$ Then we get a solution of (4.9.5)–(4.9.9) of the following form. $u(x,y)= \\sum_{n=1}^{ \\infty}b_n u_n(x,y) = \\sum_{n=1}^{ \\infty}b_n \\sin \\left( \\dfrac{n \\pi }{w}x \\right) \\left( \\dfrac{ \\sinh \\left( \\dfrac{n \\pi (h-y)}{w} \\right)}{ \\sinh \\left( \\dfrac{n \\pi h}{w} \\right)} \\right).$ As $$u_n$$ satisfies \\Equation \\ref{4.9.5}– \\ref{4.9.8}) and any linear combination (finite or infinite) of $$u_n$$ must also satisfy (4.9.5)–(4.9.8), we see that $$u$$ must satisfy Equations \\ref{4.9.5}–\\ref{4.9.8}.", "# Graph of Sine Integral I want to graph a function defined by an integral. In particular the sine integral Si(x). Si(x) = \\int_0^x \\frac{\\sin(t)}{t} dt. I can't seem to find a way do to it. • Welcome to TeX.SE. what is you specific problem regarding LaTeX? What have you tried so far? Sep 26 at 3:04 • Read pst-func-doc.pdf, 7 \\psSi, \\pssi and \\psCi. Sep 26 at 3:30 Once the module gsl is imported, Asymptote can use several functions, including the sine integral function Si (neither si nor SI). // https://mathworld.wolfram.com/SineIntegral.html // Run on http://asymptote.ualberta.ca/ unitsize(5mm,1.5cm); size(6cm); import graph; import gsl; // for manual decorations draw(Label(\"$x$\",EndPoint,E),(-11,0)--(11,0)); draw(Label(\"$\\mathrm{Si}(x)$\",EndPoint,N),(0,-2)--(0,2)); draw(Label(\"$10$\",EndPoint,S),(10,.1)--(10,-.1)); draw(Label(\"$-10$\",EndPoint,S),(-10,.1)--(-10,-.1)); draw(Label(\"$5$\",EndPoint,S),(5,.1)--(5,-.1)); draw(Label(\"$-5$\",EndPoint,S),(-5,.1)--(-5,-.1)); draw(Label(\"$1.5$\",EndPoint,W),(.2,1.5)--(-.2,1.5)); draw(Label(\"$-1.5$\",EndPoint,E),(-.2,-1.5)--(.2,-1.5)); // graph of the Si function path Sicurve=graph(Si,-10,10,100); draw(Sicurve,1pt+red); • When I add the png to the Latex file it seems to have a bad resolution. How can I download it from the site asymptote in a high resolution. Sep 26 at 16:41 • @Burlesque24 You can download x.pdf file and 1. include into you LaTeX document via \\includegraphics{x} command, or 2. use ImageMagick (having installed on your computer) to convert to PNG with this command line: magick -density 600 x.pdf x.png. PS: the number 600 is the resolution of PNG, change it as you wish Sep 26 at 16:48 • It works perfectly!", "in the formula $$\\sin(x)$$ (sin(x)) • specified N to be 3 • clicked Compute to get the formula $$\\sin(x)$$ plotted together with the Taylor polynomial approximation of degree 3, expanded around $$x=0$$ • changed N to 5 • chosen No for the question Erase all curves? • clicked Compute to have the series expansion of degree 5 plotted in the same plot • changed N to 10 • clicked Compute to have the series expansion of degree 10 plotted in the same plot We can use sympy to produce Taylor polynomial expansions of arbitrary expressions. Here is a session demonstrating how to obtain the series expansion of $$e^{-x}\\sin (\\pi x)$$ to 2nd degree. >>> import sympy as sp >>> x = sp.symbols('x') >>> f = sp.exp(-x)*sp.sin(sp.pi*x) >>> f.series(x, 0, 3) pi*x - pi*x**2 + O(x**3) >>> sp.latex(f.series(x, 0, 3)) '\\\\pi x - \\\\pi x^{2} + \\\\mathcal{O}\\\\left(x^{3}\\\\right)' >>> fs = f.series(x, 0, 3).removeO() # get rid of O() term >>> fs -pi*x**2 + pi*x >>> f_func = sp.lambdify([x], fs) # make Python function >>> f_func(1) 0.0 >>> f_func(2) -6.283185307179586 Basically, the steps above must be carried out to create a Python function for the series expansion such that it can be plotted. A similar sp.lambdify call on the original formula is also necessary to plot that one. However, the", "B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); pair[] ps={A,B,C,D,E,F,G}; dot(ps); label(\"A\",A,N); label(\"B\",B,N); label(\"C\",C,W); label(\"D\",D,S); label(\"E\",E,S); label(\"F\",F,E); label(\"G\",G,N); [/asy]$ Let $\\text{E}$ be the origin. Then, $\\text{D}=(1, 0)$ $\\text{A}=(0, 2)$ $\\text{B}=(1, 2)$ $\\text{F}=(2, 1)$ ${EB}$ can be represented by the line $y=2x$ Also, ${AF}$ can be represented by the line $y=-\\frac{1}{2}x+2$ Subtracting the second equation from the first gives us $\\frac{5}{2}x-2=0$. Thus, $x=\\frac{4}{5}$. Plugging this into the first equation gives us $y=\\frac{8}{5}$. Since $\\text{C} (0.8, 1.6)$, $G$ is $(0.8, 2)$, ${AB}=1$ and ${CG}=0.4$. Thus, $[ABC]=\\frac{1}{2} \\cdot {AB} \\cdot {CG}=\\frac{1}{2} \\cdot 1 \\cdot 0.4=0.2=\\frac{1}{5}$. The answer is $\\boxed{\\textbf{(B)}\\ \\frac15}$. ## Solution 3 $[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); pair H=(0,-1), I=(0.5,-1); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); draw(H--I); pair[] ps={A,B,C,D,E,F,G, H, I}; dot(ps); label(\"A\",A,N); label(\"B\",B,N); label(\"C\",C,W); label(\"D\",D,S); label(\"E\",E,S); label(\"F\",F,E); label(\"G\",G,N); label(\"H\",H,W); label(\"I\",I,E); [/asy]$ Triangle $EAB$ is similar to triangle $EHI$; line $HI = 1/2$ Triangle $ACB$ is similar to triangle $FCI$ and the ratio of line $AB$ to line $IF = 1 : \\frac{3}{2} = 2: 3$. Based on similarity the length of the height of $GC$ is thus $\\frac{2}{5}\\cdot1 = \\frac{2}{5}$. Thus, $[ABC]=\\frac{1}{2} \\cdot {AB} \\cdot {CG}=\\frac{1}{2} \\cdot 1 \\cdot \\frac{2}{5}=\\frac{1}{5}$. The answer is $\\boxed{\\textbf{(B)}\\", "2} = \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, since $\\triangle R PM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which", "e, and I (or the alternative i) # Most common mathematical functions are supported by default show(5/10 + 1/pi) show(sin(pi/3)) show(log(2)) show(log(2).n()) # Adding \".n()\" gives a numerical approximation (can use \".n(k)\" to get k bits) show(exp(i*2*pi/3)) show(exp(I*2*pi/101)) $\\frac{1}{\\pi} + \\frac{1}{2}$ $\\frac{1}{2} \\, \\sqrt{3}$ $\\log\\left(2\\right)$ $0.693147180559945$ $\\frac{1}{2} i \\, \\sqrt{3} - \\frac{1}{2}$ $e^{\\left(\\frac{2}{101} i \\, \\pi\\right)}$ In [6]: # Mathematical objects and variables can be inserted in print commands using {} and .format(math1,math2,...) print(\"The area of a circle of radius 1 is approximately\",pi.n()) print(\"The square-root of {} is approximately {}\".format(log(2),sqrt(log(2)).n())) The area of a circle of radius 1 is approximately 3.14159265358979 The square-root of log(2) is approximately 0.832554611157698 In [7]: # To access the help page for a function, type the name of the function and then add a question mark # For instance, evaluating the expression \"sin?\" (without quotes) gives the help page for sine # THIS WILL OPEN A NEW WINDOW AREA # Using two question marks (e.g. \"sin??\") shows the source code of the function # (Note that for many low level functions like sin Sage relies on outside packages and the source code is not very informative) sin? In [8]: # Variables are defined using the \"=\" operator # Note that some operations (such as variable assignment) do not have an output # So", "draw(X--Y); label(\"r-1\",(X+B)/2,S); label(\"1\",(Y+C)/2,N); label(\"s\",(O+Y)/2,W); label(\"s\",(O+X)/2,W); draw(B--C--(1,0)--cycle,blue+1bp); pair P = 0.73*C+0.27*B; draw(O--P); dot(P); label(\"1\",(C+P)/2,NE); label(\"r\",(B+P)/2,NE); [/asy]$ (diagram by copeland?) Then using the Pythagorean theorem we have: $(r+1)^2=(2s)^2+(r-1)^2$ , which is equivalent to: $r^2+2r+1=4s^2+r^2-2r+1$. Subtracting $r^2-2r+1$ from both sides, $4r=4s^2$ Solving for s, we end up with $$s=\\sqrt{r}$$. Next, we can find the volume of the frustum and of the sphere. Since we know $V_{frustum}=2V_{sphere}$, we can solve for $s$ using $V_{frustum}=\\frac{\\pi*h}{3}(R^2+r^2+Rr)$ we get: $$V_{frustum}=\\frac{\\pi*2\\sqrt{r}}{3}(r^2+r+1)$$ Using $V_{sphere}=\\dfrac{4r^{3}\\pi}{3}$ , we get $$V_{sphere}=\\dfrac{4(\\sqrt{r})^{3}\\pi}{3}$$ so we have: $$\\frac{\\pi*2\\sqrt{r}}{3}(r^2+r+1)=2*\\dfrac{4(\\sqrt{r})^{3}\\pi}{3}$$ Dividing by $\\frac{2\\pi*\\sqrt{r}}{3}$, we get $$r^2+r+1=4r$$ which is equivalent to $$r^2-3r+1=0$$ $r=\\dfrac{3\\pm\\sqrt{(-3)^2-4*1*1}}{2*1}$ , so $$r=\\dfrac{3+\\sqrt{5}}{2} \\longrightarrow \\boxed{\\textbf{(E)}}$$", "# Closed-form of $\\int_{a}^{b}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx$ for some $a<b$ In this question I asked to prove that $$\\int_{0}^{1}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx=\\frac{\\pi e}{24}.$$ If we take a look at the plot of the integrand, then we could see some symmetry-property. Because the integrand is symmectic to the $y=1/2$ line, it is also true, that $$\\int_{0}^{1/2}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx=\\int_{1/2}^{1}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx=\\frac{\\pi e}{48}.$$ Question. Is there a close-form of $I(a,b)=\\int_{a}^{b}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx$ for some else $a<b$? If it is too hard, then is there a closed-form for some $a<b$, just for the real or imaginary part? For example $$\\Re \\left[ \\int_{1}^{3/2}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx \\right],$$ $$\\Re \\left[ \\int_{3/2}^{2}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx \\right],$$ or $$\\Im \\left[ \\int_{1}^{2}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx \\right]?$$ • I think we'd be pushing our luck asking for a closed form for general limits $a,b$. Closed forms might exist when the limits coincide with zeros of the integrand, perhaps. – David H Oct 11 '14 at 0:57 • @DavidH This is why I asked these three specific integrand limits at the end of the question. Of course anything else would be nice. – user153012 Oct 11 '14 at 1:05 Is there a close-form of $I(a,b)=\\displaystyle\\int_{a}^{b}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx$ for some other $a<b$? Yes, of course there is. An infinity of them, in fact. But, in that case, a and b will lack a closed form. :-) For instance, all rational and algebraic numbers possess", "sin(0)) + (pi/4 * sin(pi/4))) ## [1] 0.2180895 Integrating the function over the interval yields $\\frac{\\pi - 4}{4\\sqrt{2}} = 0.15175$, which gives an error of 0.0663395. For a second example, apply the Trapezoidal Rule to approximate the definite integral: $$\\large{\\int^1_0 x^2 \\space e^{-x} \\space dx}$$ Visualizing how the trapezoidal rule approximates the integral: f2 <- function(x) { return(x^2 * exp(-x)) } df <- data.frame(cbind(c(0, 1, 1, 0), c(0, f2(1), 0, 0))) ggplot(data = data.frame(x = 0), mapping = aes(x = x)) + stat_function(fun = f2, size = 1.05, alpha = 0.75, color='blue') + geom_segment(aes(x = 0, y = 0, xend = 1, yend = f2(1))) + geom_segment(aes(x = 1, y = 0, xend = 1, yend = f2(1))) + geom_polygon(data = df, aes(x=X1, y=X2), fill = 'blue', alpha = 0.2) + geom_area(stat = 'function', fun = f2, fill = 'black', alpha = 0.3, xlim = c(0, 1)) + xlim(c(0,1)) $$\\large{\\int^1_0 x^2 \\space e^{-x} \\space dx \\approx \\frac{1 - 0}{2}[(0 * (0)^2 e^0) + ((1)^2 e^{-1})] = 0.1839397}$$ 0.5 * ((0 * 0^2 * exp(0)) + (1^2 * exp(-1))) ## [1] 0.1839397 2 - 5/exp(1) ## [1] 0.1606028 The function integrated over the interval $[0, 1]$ is $2 - \\frac{5}{e} = 0.1606028$. Therefore the error is 0.0233369. ###### Trapezoidal Rule in R The Trapezoidal Rule is comparatively straightforward to", "Closed-form of $\\int_{a}^{b}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx$ for some $a<b$ In this question I asked to prove that $$\\int_{0}^{1}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx=\\frac{\\pi e}{24}.$$ If we take a look at the plot of the integrand, then we could see some symmetry-property. Because the integrand is symmectic to the $y=1/2$ line, it is also true, that $$\\int_{0}^{1/2}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx=\\int_{1/2}^{1}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx=\\frac{\\pi e}{48}.$$ Question. Is there a close-form of $I(a,b)=\\int_{a}^{b}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx$ for some else $a<b$? If it is too hard, then is there a closed-form for some $a<b$, just for the real or imaginary part? For example $$\\Re \\left[ \\int_{1}^{3/2}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx \\right],$$ $$\\Re \\left[ \\int_{3/2}^{2}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx \\right],$$ or $$\\Im \\left[ \\int_{1}^{2}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx \\right]?$$ Solutions Collecting From Web of \"Closed-form of $\\int_{a}^{b}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx$ for some $a<b$\" Is there a close-form of $I(a,b)=\\displaystyle\\int_{a}^{b}\\sin{(\\pi x)}x^x(1-x)^{1-x}\\,dx$ for some other $a<b$? Yes, of course there is. An infinity of them, in fact. But, in that case, a and b will lack a closed form. 🙂 For instance, all rational and algebraic numbers possess a closed form expression. Now, limiting ourselves to those in between $0$ and $\\dfrac{e\\pi}{24}$, which themselves are infinite in number, for each of them there are infinitely many solutions a and b within the interval $[0,1]$, so that $I(a,b)$ will be equal to that particular rational or algebraic value, since $f(t)=I(a,t)$ is a monotonous function for any t in $[a,1]$.", "# Prove that $x – \\frac{x^3}{3!} < \\sin x < x$ for all $x>0$ Prove that $x – \\frac{x^3}{3!} < \\sin(x) < x$ for all $x>0$ This should be fairly straightforward but the proof seems to be alluding me. I want to show $x – \\frac{x^3}{3!} < \\sin(x) < x$ for all $x>0$. I recognize this shouldn’t be too difficult but perhaps finals have fried my brain. #### Solutions Collecting From Web of \"Prove that $x – \\frac{x^3}{3!} < \\sin x < x$ for all $x>0$\" You can use the Taylor expansion for $\\sin (x)$ to get a rather quick solution. If you haven’t met Taylor yet, then consider the functions $f(x)=x – \\sin x$ and $g(x)=\\sin (x) -x + \\frac{x^3}{3!}$, compute the derivative, and conclude the functions are increasing for suitable $x>0$. Compute $f(0)$ and $g(0)$ to get the result. $\\text{Here is a geometric proof for$\\sin(\\theta) < \\theta$. Consider a unit circle as shown in the figure.}$ $\\text{We then have that the area of the sector formed by$BC$as }$ $$\\color{red}{\\dfrac12 \\times OB^2 \\times \\theta = \\dfrac12 \\times 1^2 \\times \\theta = \\dfrac{\\theta}2}$$ $\\text{which is greater than the area of triangle$OBC$, which is }$ $$\\color{red}{\\dfrac12 \\times OB \\times OC \\times \\sin(\\theta) = \\dfrac12 \\times 1^2 \\times \\sin(\\theta)}$$ $\\text{We hence get that}$ $$\\color{blue}{\\dfrac{\\sin(\\theta)}2 < \\dfrac{\\theta}2 \\implies\\sin (\\theta) < \\theta}$$", "8 '12 at 12:30 @Shaktal: Would it be legal? Not without additional explantion. – GEdgar Jul 8 '12 at 12:36 You want to compute $\\displaystyle \\lim_{x\\to -n} \\frac {\\Pi(x)}{\\Pi(mx)}$ when $x$ is near a negative integer. $\\Pi$ is the 'natural' extension of the factorial : $\\Pi(n)=n!$ and $\\Pi(z)=\\Gamma(z+1)$ (see Wikipedia) In this form the \"Euler's reflection formula\" becomes simply (for $\\operatorname{sinc}(z)=\\frac{\\sin(\\pi z)}{\\pi z}$) : $$\\Pi(-z)\\Pi(z)=\\frac 1{\\operatorname{sinc}(z)}$$ $$\\lim_{x\\to -n}\\ \\frac {\\Pi(x)}{\\Pi(mx)}=\\lim_{x\\to -n}\\frac {\\Pi(-mx)\\operatorname{sinc}(-mx)}{\\Pi(-x)\\operatorname{sinc}(-x)}$$ $$=\\lim_{t\\to n}\\frac {\\Pi(mt)\\operatorname{sinc}(mt)}{\\Pi(t)\\operatorname{sinc}(t)}$$ It remains to prove that $\\ \\lim_{t\\to n} \\frac {\\operatorname{sinc(mt)}}{\\operatorname{sinc(t)}}=\\frac {(-1)^{(m-1)n}}m$ (you may use l'Hôpital's rule for that) and to conclude! - Actually, $\\displaystyle\\lim_{t\\to n} \\frac {\\operatorname{sinc(mt)}}{\\operatorname{sinc(t)}}=\\frac{(-1)^{(m-1)t}}{m}$ – Generic Human Jul 8 '12 at 13:37 @GenericHuman: Oops you are right, thanks to notice! – Raymond Manzoni Jul 8 '12 at 13:44 Using anon's \"pole\" idea, with definition $x! := \\Gamma(x+1)$ we have: \\begin{align} x! &= \\Gamma(x+1) = \\frac{1}{2}\\;\\frac{1}{x+3}+O(1)\\qquad\\text{as x \\to -3}, \\\\ (2x)! &= \\Gamma(2x+1) = -\\frac{1}{240}\\;\\frac{1}{x+3} + O(1)\\qquad\\text{as x \\to -3}, \\\\ \\frac{x!}{(2x)!} &= \\frac{1/2}{-1/240}+O(x+3)\\qquad\\text{as x \\to -3}, \\\\ \\frac{x!}{(2x)!} &\\to -120\\qquad\\text{as x \\to -3}. \\end{align} - Let me add to the above, that while $\\Gamma(-n)$ for a positive integer $n$ is undefined, let $m$ be such an integer as well, and then the ratio $\\Gamma(-n)/\\Gamma(-m)$ is well defined, and the Euler reflection formula above leads to its value being equal to" ]

[ "# Complex Numbers and exponential form and roots The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$ are $\\text{cis } \\theta_1, \\text{cis } \\theta_2, \\dots, \\text{cis } \\theta_7,$ where $0^\\circ \\le \\theta_k < 360^\\circ$for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\left(5 \\pi i/4 \\right)$. How should I simplify? Thanks - What is $\\text{cis}$? Do you by chance mean $\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\theta)+isin(\\theta)$ – Mr.Fry Jan 24 '14 at 16:27 @AlexR: \"cis\" is a commonly used abbreviation for \"cosine plus $i$ times the sine.\" In other words, $$\\operatorname{cis}\\theta=\\cos\\theta+i\\sin\\theta=e^{i\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis}(7\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis} 225^\\circ,$$ as you've already determined, and so we need $$\\operatorname{cis}(7\\theta_k)=\\operatorname{cis} 225^\\circ\\\\\\frac{\\operatorname{cis}(7\\theta_k)}{\\operatorname{cis} 225^\\circ}=1\\\\\\operatorname{cis}(7\\theta_k-225^\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? If I am understanding correctly...for $k=1$, we take $$\\operatorname{cis}(7\\theta_1)=\\operatorname{cis} 225^\\circ$$ and then somehow determine $\\theta_1$, and repeat for $k=2,3,4,5,6,7$? – Mathy Person Jul 18 '15 at 21:59 No, that's getting further from the answer. Rather, think about the solutions set for the equation $$\\operatorname{cis}\\theta=1. This will let you get rid of the \\operatorname{cis}, and find all solutions to the original", "# Complex Numbers and exponential form and roots The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$ are $\\text{cis } \\theta_1, \\text{cis } \\theta_2, \\dots, \\text{cis } \\theta_7,$ where $0^\\circ \\le \\theta_k < 360^\\circ$for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\left(5 \\pi i/4 \\right)$. How should I simplify? Thanks - What is $\\text{cis}$? Do you by chance mean $\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\theta)+isin(\\theta)$ – Rod Jan 24 '14 at 16:27 @AlexR: \"cis\" is a commonly used abbreviation for \"cosine plus $i$ times the sine.\" In other words, $$\\operatorname{cis}\\theta=\\cos\\theta+i\\sin\\theta=e^{i\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis}(7\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis} 225^\\circ,$$ as you've already determined, and so we need $$\\operatorname{cis}(7\\theta_k)=\\operatorname{cis} 225^\\circ\\\\\\frac{\\operatorname{cis}(7\\theta_k)}{\\operatorname{cis} 225^\\circ}=1\\\\\\operatorname{cis}(7\\theta_k-225^\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? - The roots are $e^{2k\\pi i/7}$ ($k=0,\\ldots,6$). Write them as $r^k$ where $r = e^{2\\pi i/7}$. So the sum is (using the simple formula for summing a finite geometric series) $$\\sum_{k=0}^6 r^k = \\frac{1-r^7}{1-r} = \\frac{1 - e^{2\\pi i}}{1-e^{2\\pi i/7}} = 0$$. - This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not", "# Definition:Polar Form of Complex Number/Also known as As $\\cos \\theta + i \\sin \\theta$ appears so often in complex analysis, the abbreviation $\\cis \\theta$ is frequently seen. Hence $r \\paren {\\cos \\theta + i \\sin \\theta}$ can be expressed in the economical form $r \\cis \\theta$.", "+0 # help +1 50 1 The number $$\\text{cis } 75^\\circ + \\text{cis } 83^\\circ + \\text{cis } 91^\\circ + \\dots + \\text{cis } 147^\\circ$$ is expressed in the form $$r \\, \\text{cis } \\theta$$, where $$0 \\le \\theta < 360^\\circ$$. Find $$\\theta$$ in degrees. Dec 20, 2019", "# Cis Cis notation is a polar notation for complex numbers. For all complex numbers $z$, we can write $z=r\\mathrm{cis }(\\theta)=r\\cos \\theta + ir\\sin \\theta$. Notice that $\\mathrm{cis}$ is made up by the first letter of $\\cos$, $i$, and the first letter of $\\sin$. Once one gets used to the notation, it is almost always preferred to write $re^{i\\theta}$ rather than $r\\mathrm{cis }(\\theta)$, as Euler's formula states that $e^{i\\theta} = \\cos \\theta + i \\sin \\theta.$ This is so that one can more naturally use the properties of the complex exponential. One important example is De Moivre's Theorem, which states that $\\mathrm{cis}(r\\theta) = (\\mathrm{cis}(\\theta))^r.$ This is more easily understood in the complex exponential form: $e^{i(r\\theta)} = (e^{i\\theta})^r.$", "as you might expect to yield $\\left(\\frac{r_1}{r_2}\\right) e^{i ( \\theta_1 - \\theta_2)}$. $\\def\\cis{\\operatorname{cis}}$ If we have the fraction $$\\frac{r_{1}\\cis(\\theta_{1})}{r_{2}\\cis(\\theta_{2})}$$ then in polar form this is the complex number $$\\frac{r_{1}}{r_{2}}\\cis(\\theta_{1}-\\theta_{2})$$ It follows from the fact that $|\\cis(\\alpha)|=1$ hence $|r\\cis(\\alpha)|=r$ and from the trigonometric identities • To be clear, $cis(x) = e^{ix} = \\cos(x) + i \\sin(x)$ – LucasVB Jul 31 '14 at 14:21 • @MJD - Thanks for the edit. I don't have the \\cis command in Lyx but its on Math.SE (apparently) – Belgi Jul 31 '14 at 14:34 • I had to define it specially for your post, using \\def. – MJD Jul 31 '14 at 14:41 • Waittt..... So I can basically subtract the angles? (Extending from this, I suppose in multiplication I can add the angles?) – Gummy bears Jul 31 '14 at 14:48 • @Gummybears $e^{i \\theta}= \\cos{\\theta} + i \\sin{\\theta}$. – snulty Aug 1 '14 at 12:49", "but when I do it that way, I end up with $2isin3θ$ since: $z^3- z^{-3}= cis(3\\theta)- cis(-3\\theta)= cos(3\\theta)+ isin(3\\theta)- (cos(-3\\theta)+ isin(-3\\theta)$$= cos(3\\theta)+ isin(3\\theta)- cos(3\\theta)+ isin(3\\theta)= 2isin(3\\theta)$ My class just recently started our complex number topic and so my knowledge of $cis$ is extremely basic. Should I not be writing $z^3$ as $cos3θ+isin3θ$.", "# Geometric Series in Polar Form I'm stuck on a complex number problem relating to the geometric series in polar form. By considering that: $$1+\\operatorname{cis}{\\theta}+\\operatorname{cis}{2\\theta}+\\operatorname{cis}{3\\theta}+....+\\operatorname{cis}{n\\theta}$$ as a geometric series, find: $$\\sum_{r=0}^n \\operatorname{cos}{r\\theta}$$ I was thinking along the lines of finding the sum of the polar form series, and that I was assuming that the sum of the series would be: $$\\operatorname{cis}{\\frac{n(n+1)}{2}}{\\theta}$$ Thus, if that is the answer then finding the sum of $\\operatorname{cos}{r\\theta}$ becomes very easy. But I doubt that that is the answer, so any help would be much appreciated!! • if $z \\ne 1$ : $\\sum_{r=0}^n z^r = \\frac{1-z^{n+1}}{1-z}$. – reuns Mar 20 '17 at 0:48 • I'm confused? How does the left hand side equate to the right hand side? – Maths Matador Mar 20 '17 at 1:12", "The answer to the above example $-3.12 - 4.64i$ with theta expressed in radian measure would be: $& \\tan \\theta =\\frac{-4.64}{-3.12} && \\tan \\theta=.9788(\\text{reference angle})\\\\&&& 0.9788+3.14=4.12 \\ \\text{rad}.\\\\& 5.59(\\cos 4.12+i \\sin 4.12)$ Now that we have explored the polar form of complex numbers and the steps for performing these conversions, we will look at an example in circuit analysis that requires a complex number given in polar form to be expressed in standard form. ### Vocabulary Complex Number: A complex number is a number having both real and imaginary components. ### Guided Practice 1. The impedance $Z$ , in ohms, in an alternating circuit is given by $Z=4650 \\angle -35.2^\\circ$ . Express the value for $Z$ in standard form. (In electricity, negative angles are often used.) 2. Express the following complex numbers in their polar form. 1. $4 + 3i$ 2. $-2 + 9i$ 3. $7 - i$ 4. $-5 - 2i$ 3. Express the complex number $6 - 8i$ graphically and write it in its polar form. Solutions: 1. The value for $Z$ is given in polar form. From this notation, we know that $r = 4650$ and $\\theta = -35.2^\\circ$ Using these values, we can write: $Z &= 4650 (\\cos(-35.2^\\circ) + i \\sin(-35.2^\\circ))\\\\ x &= 4650 \\cos(-35.2^\\circ) \\rightarrow 3800\\\\ y &= 4650 \\sin (-35.2^\\circ) \\rightarrow -2680$", "# The origins of $\\operatorname{cis}(\\theta)$ There is a abbreviation used in high school mathematics that is almost never seen outside of it: $$\\operatorname{cis}(\\theta) = \\cos(\\theta) + i \\sin(\\theta)$$, where cis stands for cosine + i sine. As soon as students get into university Euler's formula $$e^{i\\theta} = \\cos(\\theta) + i \\sin(\\theta)$$ is almost exclusively used as it makes a lot of things easier and more obvious. The reason Euler's formula is not used in high school is probably that it's not easy to prove or to develop an intuition for at that level. My question is: does anyone know the origins and history of the abbreviation cis ? The notation exists since a long time. It was used already by Irving Stringham in 'Uniplanar Algebra (1893).' This is claimed to be the earliest use on http://jeff560.tripod.com/trigonometry.html giving Cajori vol. 2, page 133 as reference. In this book, Uniplanar Algebra, the notation is used first, as far as I can see, in chapter III (The algebra of complex quantities). It is introduced without much ado as a shorthand notation. It is said it should be read as \"sector of $\\theta$\". Cajori also mentions another early usage (thanks to Dag Oskar Madsen for the information!), namely, Harkness and Morley \"Introduction to the theory of analytic functions (1898)\". Also there it", "complex number $5 + 7i$ graphically and express it in its polar form. Solution: As discussed in the Prerequisite Chapter, here is the graph of $5 + 7i$. Converting to polar from rectangular, $x = 5$ and $y = 7$. $& r=\\sqrt{5^2+7^2}=8.6 && \\tan \\theta=\\frac{7}{5}\\\\&&& \\tan ^{-1}(\\tan \\theta)=\\tan ^{-1}\\frac{7}{5}\\\\&&& \\theta=54.5^\\circ$ So, the polar form is $8.6(\\cos 54.5^\\circ + i \\sin 54.5^\\circ)$. Another widely used notation for the polar form of a complex number is $r \\angle \\theta=r (\\cos \\theta + i \\sin \\theta)$. Now there are three ways to write the polar form of a complex number. $x+yi=r(\\cos \\theta+i \\sin \\theta) && x+yi=rcis \\theta && x+yi=r \\angle \\theta$ Example 2: Express the following polar form of each complex number using the shorthand representations. a) $4.92 (\\cos 214.6^\\circ + i \\sin 214.6^\\circ)$ b) $15.6 (\\cos 37^\\circ + i \\sin 37^\\circ)$ Solution: a) $4.92 \\angle 214.6^\\circ$ $4.92 \\ cis \\ 214.6^\\circ$ b) $15.6 \\angle 37^\\circ$ $15.6 \\ cis \\ 37^\\circ$ Example 3: Represent the complex number $-3.12 - 4.64i$ graphically and give two notations of its polar form. Solution: From the rectangular form of $-3.12 - 4.64i \\ x = - 3.12$ and $y = - 4.64$ $r &= \\sqrt{x^2+y^2}\\\\r &= \\sqrt{(-3.12)^2+(-4.64)^2}\\\\r &= 5.59$ $\\tan \\theta &= \\frac{y}{x}\\\\\\tan \\theta &=\\frac{-4.64}{-3.12}\\\\\\theta&=56.1^\\circ$ This is the reference angle so now we must determine", "# Operation with Complex number We've the following 3 complex numbers $z_1=2\\text{cis}(\\frac{\\pi}{6})~~$, $z_2=2\\text{cis}(\\frac{5\\pi}{6})~~$, $z_3=2\\text{cis}(\\frac{-\\pi}{2})$ We are asked to show that $z_1^{3n}+z_2^{3n}=2z_3^{3n}$ I have calculated $z_1^{3n}=2^{3n}\\text{cis}(\\frac{n\\pi}{2})~~$, $z_2^{3n}=2^{3n}\\text{cis}(\\frac{5n\\pi}{2})~~$, $z_3^{3n}=2^{3n}\\text{cis}(\\frac{-3n\\pi}{2})$. I'm stuck on how to proceed further to prove the above equations. Any help really appreciated, Please assist . Thank you , Arif • What is \"cis\" ? – Jean Marie Mar 29 '17 at 6:07 • @JeanMarie This. – dxiv Mar 29 '17 at 6:21 • @dxiv Thanks. I understand now the pedagogical interest of this notation... – Jean Marie Mar 29 '17 at 6:32 • $z_1^3 = 2^3 \\operatorname{cis}\\left(3 \\cdot \\cfrac{\\pi}{6}\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ • $z_2^3 = 2^3 \\operatorname{cis}\\left(3 \\cdot \\cfrac{5 \\pi}{6}\\right)=8\\operatorname{cis}\\left(\\cfrac{5 \\pi}{2}\\right)=8\\operatorname{cis}\\left(2\\pi+\\cfrac{\\pi}{2}\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ • $z_3^3 = 2^3 \\operatorname{cis}\\left(-3 \\cdot \\cfrac{\\pi}{2}\\right)=8\\operatorname{cis}\\left(- 3 \\cdot \\cfrac{\\pi}{2}+ 2 \\pi\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ Therefore $\\;z_1^3=z_2^3=z_3^3\\,$.", "is sometimes denoted cis x ( cosine plus i sine ''.! Π < θ ≤ π is called the principal value of the argument of any complex easily... complex number formula 2021", "# Polar Form of Complex Number/Examples/6 cis 240 The complex number $6 \\paren {\\cos 240 \\degrees + i \\sin 240 \\degrees}$ can be expressed as: $6 \\paren {\\cos 240 \\degrees + i \\sin 240 \\degrees} = 6 \\cis 240 \\degrees = 6 \\cis \\dfrac {4 \\pi} 3 = 6 e^{4 \\pi / 3}$", "special notation for it. Notation. Cis $\\operatorname{Cis}(\\theta):=\\cos \\theta+i \\sin \\theta .$ DEFINITION. For the complex number $$z=x+i y=r \\operatorname{Cis}(\\theta)$$, we have the following: $$\\Re(z) x$$ is called the real part of $$z$$, written $$\\Re(z)$$; $$\\Im(z) y$$ is called the imaginary part of $$z$$, written $$\\Im(z)$$; $$|z| r$$ is called the modulus of $$z$$, or absolute value of $$z$$, written $$|z|$$; $$\\arg (z) \\theta$$ is called the argument of $$z$$, written $$\\arg (z)$$. $$\\bar{z}$$ The number $$x-i y$$ is called the conjugate of $$z$$, written $$\\bar{z}$$. REMARK. There is an important point to bear in mind about the argument: it is only unique up to addition of multiples of $$2 \\pi$$. In other words, if $$\\theta_{0}$$ is an argument of the complex number $$z$$, then so are all the numbers $$\\left\\{\\theta_{0}+2 k \\pi: k \\in \\mathbb{Z}\\right\\}$$. Addition is easiest in Cartesian coordinates: add the real and imaginary parts. Multiplication is easiest in polar coordinates: multiply the moduli and add the arguments. Proposition 9.19. Let $$z_{1}=r_{1} \\operatorname{Cis}\\left(\\theta_{1}\\right)$$ and $$z_{2}=r_{2} \\operatorname{Cis}\\left(\\theta_{2}\\right)$$. Then $z_{1} z_{2}=r_{1} r_{2} \\operatorname{Cis}\\left(\\theta_{1}+\\theta_{2}\\right) .$ ProOF. Multiplying out, we get \\begin{aligned} z_{1} z_{2}=r_{1} r_{2}\\left[\\cos \\theta_{1} \\cos \\theta_{2}-\\sin \\theta_{1} \\sin \\theta_{2}\\right.\\\\ &\\left.+i\\left(\\cos \\theta_{1} \\sin \\theta_{2}+\\cos \\theta_{2} \\sin \\theta_{1}\\right)\\right] . \\end{aligned} The result follows by the trigonometric identities for the cosine and sine of the sum of two angles. A consequence", "# Binary operation with complex number Let us consider the set of complex numbers and the binary operation $$\\circ$$ defined by $$z_a\\circ z_b=|z_a|e^{\\Theta(z_b)}$$, where $$\\Theta(z_b)$$ is the argument of the complex number $$z_b$$. Explain whether the set of complex numbers together with the binary operation $$\\circ$$ forms a monoid. I know I need to prove whether it satisfies closure, associativity and identity. I'm not sure how to start with showing this, I've tried to put it into polar form which I get $$z_a\\circ z_b=|z_a|(\\cos \\Theta(z_b)+i\\sin \\Theta(z_b))$$ but I have no idea what to do next. Any help will be appreciated, thanks. • Start by writing down the definitions of closure, associativity and identity. – Yves Daoust Nov 2 '18 at 11:26 • Are you sure it isn't $z_a ∘ z_b= |z_a|e^{i\\theta(z_b)}$? The way it is written, without the \"i\" in the exponent, combining two complex numbers always gives a real number so there cannot be an \"identity\". – user247327 Nov 2 '18 at 11:27 • How do you define the angle of $0$? Meaning, what it $1 \\circ 0$? – Ofek Gillon Nov 2 '18 at 11:28 About associativity, let there be 3 complex numbers $$z_1,z_2,z_3$$ and we'll try to prove that $$(z_1 \\circ z_2) \\circ z_3 = z_1 \\circ (z_2 \\circ z_3)$$ The left side will be", "of three ways to write the polar form of a complex number. $x+yi=r(\\cos \\theta+i \\sin \\theta) && x+yi=rcis \\theta && x+yi=r \\angle \\theta$ #### Example B Express the following polar form of each complex number using the shorthand representations. a) $4.92 (\\cos 214.6^\\circ + i \\sin 214.6^\\circ)$ b) $15.6 (\\cos 37^\\circ + i \\sin 37^\\circ)$ Solution: a) $4.92 \\angle 214.6^\\circ$ $4.92 \\ cis \\ 214.6^\\circ$ b) $15.6 \\angle 37^\\circ$ $15.6 \\ cis \\ 37^\\circ$ #### Example C Represent the complex number $-3.12 - 4.64i$ graphically and give two notations of its polar form. Solution: From the rectangular form of $-3.12 - 4.64i \\ x = - 3.12$ and $y = - 4.64$ $r &= \\sqrt{x^2+y^2}\\\\r &= \\sqrt{(-3.12)^2+(-4.64)^2}\\\\r &= 5.59$ $\\tan \\theta &= \\frac{y}{x}\\\\\\tan \\theta &=\\frac{-4.64}{-3.12}\\\\\\theta&=56.1^\\circ$ This is the reference angle so now we must determine the measure of the angle in the third quadrant. $56.1^\\circ + 180^\\circ = 236.1^\\circ$ One polar notation of the point $-3.12 - 4.64i$ is $5.59$ $(\\cos 236.1^\\circ + i \\sin 236.1^\\circ)$ . Another polar notation of the point is $5.59 \\angle 236.1^\\circ$ So far we have expressed all values of theta in degrees. Polar form of a complex number can also have theta expressed in radian measure. This would be beneficial when plotting the polar form of complex numbers in the polar plane.", "# Simplifying and then expressing complex numbers in cartesian form 1. Mar 11, 2010 ### Stripe 1. The problem statement, all variables and given/known data (2 CIS (pi/6))*(3 CIS (pi/12)) 2. Relevant equations Also what is CIS? I believe it's Cos+i*sin but how do you use it? 3. The attempt at a solution i simplified it to 6 CIS (pi/12) How do i turn it into cartesian? 2. Mar 11, 2010 ### tiny-tim Hi Stripe! (have a pi: π ) I've never seen \"CIS\" before, but I'll guess you're right, and that it's cos + i*sin. Now use De Moivre's theorem … cosθ + isinθ = e And Cartesian form simply means in the form x + iy (as opposed to polar form, which is in the form re ) (and no, it's not 6 CIS (π/12)) Last edited: Mar 11, 2010 3. Mar 11, 2010 ### HallsofIvy \"$Cis(\\theta)$\" is engineering notation for \"$cos(\\theta)+ i sin(\\theta)$\" which mathematicians tend to write as $e^{i\\theta}$. The important thing about that notation is that $(Cis(\\theta)*Cis(\\phi)= Cis(\\theta+ \\phi)$. So you have correctly deduced that $(2Cis(\\pi/6))(3Cis(\\pi/12)= 6 Cis(3\\pi/12)= 6 Cis(\\pi/4)$ Now, just use the definition: $6 Cis(\\pi/4)= 6 cos(\\pi/4)+ 6i sin(\\pi/4)$.", "# Show that $\\sin x-i\\cos x=\\operatorname{cis}(x-\\pi/2)$ A question in a textbook asks Show that $\\sin\\theta-i\\cos\\theta=\\operatorname{cis}(\\theta-\\frac{\\pi}{2})$ I've previously shown that: $\\sin\\theta+i\\cos\\theta=\\operatorname{cis}(\\frac{\\pi}{2}-\\theta)$, and that $\\cos\\theta-i\\sin\\theta=\\operatorname{cis}(-\\theta)$. So I've tried using that in my working out, but can't seem to work this one out. I am assuming that it wants me to use trig addition formulae and De Moivre's Theorem possibly (because that is the general theme of this section of the book). No working out that's too abstract or advanced I'd presume. I have tried symmetry properties, and negative angles as well. I am sure I can prove it by using a subtraction formula, but that would be me assuming the answer (circular logic maybe). I need to start from the LHS and then work to get the RHS. You can write $$\\operatorname{cis}\\left(\\theta-\\frac{\\pi}{2}\\right)= \\frac{\\operatorname{cis}\\theta}{\\operatorname{cis}\\frac{\\pi}{2}}$$ Since $\\operatorname{cis}\\frac{\\pi}{2}=i$, you get $$\\operatorname{cis}\\left(\\theta-\\frac{\\pi}{2}\\right)= i^{-1}\\operatorname{cis}\\theta=i^{-1}(\\cos\\theta+i\\sin\\theta)= -i\\cos\\theta+\\sin\\theta$$ because $i^{-1}=-i$. • Thank you for that. It ties in very well with the more recent sections. – Simplex1 Aug 3 '18 at 11:57 Well, we have: $$\\exp\\left(\\theta i\\right)=\\cos\\left(\\theta\\right)+\\sin\\left(\\theta\\right)i\\tag1$$ So, when we have: $$\\exp\\left(\\left(\\theta-\\frac{\\pi}{2}\\right)\\cdot i\\right)=\\cos\\left(\\theta-\\frac{\\pi}{2}\\right)+\\sin\\left(\\theta-\\frac{\\pi}{2}\\right)i\\tag2$$ And we know that: • $$\\cos\\left(\\theta-\\frac{\\pi}{2}\\right)=\\sin\\left(\\theta\\right)\\tag3$$ • $$\\sin\\left(\\theta-\\frac{\\pi}{2}\\right)=-\\cos\\left(\\theta\\right)\\tag4$$", "that $w = \\operatorname{cis}(30^{\\circ})$ and $z = \\operatorname{cis}(120^{\\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$, respectively. By Demoivre's theorem, $\\operatorname{cis}(\\theta)^n = \\operatorname{cis}(n\\theta)$. When you multiply by $i$, we can think of that as rotating the complex number $90^{\\circ}$ counterclockwise in the complex plane. Therefore, by the equation we know that $30r + 90$ and $120s$ land on the same angle. This means that$$30r + 90 \\equiv 120s \\pmod{360},$$which we can simplify to$$r+3 \\equiv 4s \\pmod{12}.$$Notice that this means that $r$ cycles by $12$ for every value of $s$. This is because once $r$ hits $12$, we get an angle of $360^{\\circ}$ and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every $3$ times, which is much easier to count. By listing the possible values out, we get the pairs $(r,s)$:$$\\begin{array}{cccccccc} (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \\ldots & (97,1) \\\\ (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \\ldots & (97,4) \\\\ (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \\ldots & (97,7) \\\\ [-1ex] \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots \\\\ (1,100) & (5,98) & (9,99) & (13,100) & (17,98)" ]

[ "# Complex Numbers and exponential form and roots The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$ are $\\text{cis } \\theta_1, \\text{cis } \\theta_2, \\dots, \\text{cis } \\theta_7,$ where $0^\\circ \\le \\theta_k < 360^\\circ$for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\left(5 \\pi i/4 \\right)$. How should I simplify? Thanks - What is $\\text{cis}$? Do you by chance mean $\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\theta)+isin(\\theta)$ – Mr.Fry Jan 24 '14 at 16:27 @AlexR: \"cis\" is a commonly used abbreviation for \"cosine plus $i$ times the sine.\" In other words, $$\\operatorname{cis}\\theta=\\cos\\theta+i\\sin\\theta=e^{i\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis}(7\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis} 225^\\circ,$$ as you've already determined, and so we need $$\\operatorname{cis}(7\\theta_k)=\\operatorname{cis} 225^\\circ\\\\\\frac{\\operatorname{cis}(7\\theta_k)}{\\operatorname{cis} 225^\\circ}=1\\\\\\operatorname{cis}(7\\theta_k-225^\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? If I am understanding correctly...for $k=1$, we take $$\\operatorname{cis}(7\\theta_1)=\\operatorname{cis} 225^\\circ$$ and then somehow determine $\\theta_1$, and repeat for $k=2,3,4,5,6,7$? – Mathy Person Jul 18 '15 at 21:59 No, that's getting further from the answer. Rather, think about the solutions set for the equation $$\\operatorname{cis}\\theta=1. This will let you get rid of the \\operatorname{cis}, and find all solutions to the original", "# Complex Numbers and exponential form and roots The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$ are $\\text{cis } \\theta_1, \\text{cis } \\theta_2, \\dots, \\text{cis } \\theta_7,$ where $0^\\circ \\le \\theta_k < 360^\\circ$for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\left(5 \\pi i/4 \\right)$. How should I simplify? Thanks - What is $\\text{cis}$? Do you by chance mean $\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\theta)+isin(\\theta)$ – Rod Jan 24 '14 at 16:27 @AlexR: \"cis\" is a commonly used abbreviation for \"cosine plus $i$ times the sine.\" In other words, $$\\operatorname{cis}\\theta=\\cos\\theta+i\\sin\\theta=e^{i\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis}(7\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis} 225^\\circ,$$ as you've already determined, and so we need $$\\operatorname{cis}(7\\theta_k)=\\operatorname{cis} 225^\\circ\\\\\\frac{\\operatorname{cis}(7\\theta_k)}{\\operatorname{cis} 225^\\circ}=1\\\\\\operatorname{cis}(7\\theta_k-225^\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? - The roots are $e^{2k\\pi i/7}$ ($k=0,\\ldots,6$). Write them as $r^k$ where $r = e^{2\\pi i/7}$. So the sum is (using the simple formula for summing a finite geometric series) $$\\sum_{k=0}^6 r^k = \\frac{1-r^7}{1-r} = \\frac{1 - e^{2\\pi i}}{1-e^{2\\pi i/7}} = 0$$. - This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not", "that $w = \\operatorname{cis}(30^{\\circ})$ and $z = \\operatorname{cis}(120^{\\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$, respectively. By Demoivre's theorem, $\\operatorname{cis}(\\theta)^n = \\operatorname{cis}(n\\theta)$. When you multiply by $i$, we can think of that as rotating the complex number $90^{\\circ}$ counterclockwise in the complex plane. Therefore, by the equation we know that $30r + 90$ and $120s$ land on the same angle. This means that$$30r + 90 \\equiv 120s \\pmod{360},$$which we can simplify to$$r+3 \\equiv 4s \\pmod{12}.$$Notice that this means that $r$ cycles by $12$ for every value of $s$. This is because once $r$ hits $12$, we get an angle of $360^{\\circ}$ and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every $3$ times, which is much easier to count. By listing the possible values out, we get the pairs $(r,s)$:$$\\begin{array}{cccccccc} (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \\ldots & (97,1) \\\\ (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \\ldots & (97,4) \\\\ (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \\ldots & (97,7) \\\\ [-1ex] \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots \\\\ (1,100) & (5,98) & (9,99) & (13,100) & (17,98)", "# Operation with Complex number We've the following 3 complex numbers $z_1=2\\text{cis}(\\frac{\\pi}{6})~~$, $z_2=2\\text{cis}(\\frac{5\\pi}{6})~~$, $z_3=2\\text{cis}(\\frac{-\\pi}{2})$ We are asked to show that $z_1^{3n}+z_2^{3n}=2z_3^{3n}$ I have calculated $z_1^{3n}=2^{3n}\\text{cis}(\\frac{n\\pi}{2})~~$, $z_2^{3n}=2^{3n}\\text{cis}(\\frac{5n\\pi}{2})~~$, $z_3^{3n}=2^{3n}\\text{cis}(\\frac{-3n\\pi}{2})$. I'm stuck on how to proceed further to prove the above equations. Any help really appreciated, Please assist . Thank you , Arif • What is \"cis\" ? – Jean Marie Mar 29 '17 at 6:07 • @JeanMarie This. – dxiv Mar 29 '17 at 6:21 • @dxiv Thanks. I understand now the pedagogical interest of this notation... – Jean Marie Mar 29 '17 at 6:32 • $z_1^3 = 2^3 \\operatorname{cis}\\left(3 \\cdot \\cfrac{\\pi}{6}\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ • $z_2^3 = 2^3 \\operatorname{cis}\\left(3 \\cdot \\cfrac{5 \\pi}{6}\\right)=8\\operatorname{cis}\\left(\\cfrac{5 \\pi}{2}\\right)=8\\operatorname{cis}\\left(2\\pi+\\cfrac{\\pi}{2}\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ • $z_3^3 = 2^3 \\operatorname{cis}\\left(-3 \\cdot \\cfrac{\\pi}{2}\\right)=8\\operatorname{cis}\\left(- 3 \\cdot \\cfrac{\\pi}{2}+ 2 \\pi\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ Therefore $\\;z_1^3=z_2^3=z_3^3\\,$.", "# Writing mod-arg form as $r\\operatorname{cis}(\\theta)$ rather than $r(\\cos\\theta + i\\sin\\theta)$ I'm studying in-depth complex numbers and analysis and when I'm working through certain theory, I like to refer to as many textbooks as possible. I've always known to write mod-arg form as $r(\\cos\\theta + i\\sin\\theta)$ However, this book by J. Coroneos writes mod-arg form simplistically as $r\\operatorname{cis}\\theta$. In the other 7 textbooks I have referred to, I haven't come across this. Is this is universally accepted? If I use this, will I be technically wrong or is this dude just simplifying things to make things easier. • en.wikipedia.org/wiki/Cis_(mathematics) – user223391 Jan 15 '17 at 20:37 • It's just their convention. Some texts use that (and so do I sometimes). It's also shorthand for $r e^{i\\theta}$. You're okay. – Sean Roberson Jan 15 '17 at 20:38 • Whatever what you want – user223391 Jan 15 '17 at 20:40 • $cis(t)$ is useless notation, because it gets replaced by $e^{it}$ immediately. I recommend getting used to the exponential form. – Kaynex Jan 15 '17 at 20:41 • I think \"cis\" is teddy-bear for beginners. I guess most people will know what you mean when you write it, but when doing any kind of calculation, $e^{it}$ works better. – B. Goddard Jan 15 '17 at 20:43 ## 1 Answer It's", "in italics you do \\operatorname{cis}. – fpqc Mar 18 '12 at 8:04 So it remains to check that $\\tan\\pi/5=2(\\sin2\\pi/5-\\sin\\pi/5)$. After a simplification, this amounts to check that $$1=2\\cos\\pi/5\\;(2\\cos\\pi/5-1),$$ and this can be seen directly from the fact that $\\cos\\pi/5=(1+\\sqrt5)/4$.", "64 ASCON $f_5$=circ(0, 7, 41) 4 n Source The mapping Number of fixed-points 8 HIGHT circ(1, 6, 7) 8 8 HIGHT circ(2, 4, 5) 2 64 ASCON $f_1$=circ(0, 19, 28) 2 64 ASCON $f_2$=circ(0, 39, 61) 4 64 ASCON $f_3$=circ(0, 1, 6) 2 64 ASCON $f_4$=circ(0, 10, 17) 2 64 ASCON $f_5$=circ(0, 7, 41) 4 Number of $16 \\times 16$ CBMs with fixed branch number $b_f=4$ $b_f=6$ $b_f=8$ $w_f=3$ 105 - - $w_f=5$ 145 1220 - $w_f=7$ 301 3332 1372 $w_f=9$ 495 4644 1296 $w_f=11$ 297 2134 572 $w_f=13$ 169 286 - $w_f=15$ 15 - - $b_f=4$ $b_f=6$ $b_f=8$ $w_f=3$ 105 - - $w_f=5$ 145 1220 - $w_f=7$ 301 3332 1372 $w_f=9$ 495 4644 1296 $w_f=11$ 297 2134 572 $w_f=13$ 169 286 - $w_f=15$ 15 - - $n \\times n$ CBMs with maximal branch number and efficient implementation n mapping $\\#$XORs $\\#$fixed-points 16 $circ(0,1,2)(0,2,9)$ 49 4 16 $circ(0,14,15)(0,7,14)$ 49 4 16 $circ(0,1,2)(0,7,14)$ 49 16 16 $circ(0,14,15)(0,2,9)$ 49 16 16 $circ(0,3,6)(0,5,10)$ 51 16 16 $circ(0,3,6)(0,6,11)$ 51 4 32 $circ(0,3,6)(0,4,8)(0,5,10)$ 146 64 32 $circ(0,3,6)(0,4,8)(0,22,27)$ 146 4 32 $circ(0,26,29)(0,24,28)(0,22,27)$ 146 64 32 $circ(0,4,8)(0,5,10)(0,13,26)$ 148 4 32 $circ(0,4,8)(0,22,27)(0,6,19)$ 148 4 32 $circ(0,1,2)(0,7,14)(0,12,24)$ 150 64 32 $circ(0,30,31)(0,18,25)(0,8,20)$ 150 64 32 $circ(0,1,2)(0,9,18)(0,12,24)$ 150 4 32 $circ(0,30,31)(0,14,23)(0,8,20)$ 150 4 32 $circ(0,3,6)(0,4,8)(0,11,22)$ 150 4 32 $circ(0,7,14)(0,12,24)(0,15,30)$ 151 4 32 $circ(0,9,18)(0,12,24)(0,15,30)$ 151 64 32 $circ(0,4,8 )(0,11,22)(0,13,26)$ 152", "form. 35) Find $$z^3$$ when $$z=5\\mathbf{cis} (45^{\\circ})$$. $$125\\mathbf{cis} (135^{\\circ})$$ 36) Find $$z^4$$ when $$z=2\\mathbf{cis} (70^{\\circ})$$. 37) Find $$z^2$$ when $$z=3\\mathbf{cis} (120^{\\circ})$$. $$9\\mathbf{cis} (240^{\\circ})$$ 38) Find $$z^2$$ when $$z=4\\mathbf{cis}\\left(\\dfrac{\\pi }{4} \\right)$$. 39) Find $$z^4$$ when $$z=\\mathbf{cis}\\left(\\dfrac{3\\pi }{16} \\right)$$. $$\\mathbf{cis}\\left(\\dfrac{3\\pi }{4} \\right)$$ 40) Find $$z^3$$ when $$z=3\\mathbf{cis}\\left(\\dfrac{5\\pi }{3} \\right)$$. For the exercises 41-45, evaluate each root. 41) Evaluate the cube root of $$z$$ when $$z=27\\mathbf{cis} (240^{\\circ})$$. $$3\\mathbf{cis}(80^{\\circ}), 3\\mathbf{cis}(200^{\\circ}), 3\\mathbf{cis}(320^{\\circ})$$ 42) Evaluate the square root of $$z$$ when $$z=16\\mathbf{cis} (100^{\\circ})$$. 43) Evaluate the cube root of $$z$$ when $$z=32\\mathbf{cis} \\left(\\dfrac{2\\pi }{3} \\right)$$. $$2\\sqrt[3]{4}\\mathbf{cis} \\left(\\dfrac{2\\pi }{9} \\right), 2\\sqrt[3]{4}\\mathbf{cis} \\left(\\dfrac{8\\pi }{9} \\right), 2\\sqrt[3]{4}\\mathbf{cis} \\left(\\dfrac{14\\pi }{9} \\right)$$ 44) Evaluate the square root of $$z$$ when $$z=32\\mathbf{cis} (\\pi )$$. 45) Evaluate the cube root of $$z$$ when $$z=8\\mathbf{cis} \\left(\\dfrac{7\\pi }{4} \\right)$$. $$2\\sqrt{2}\\mathbf{cis} \\left(\\dfrac{7\\pi }{8} \\right), 2\\sqrt{2}\\mathbf{cis} \\left(\\dfrac{15\\pi }{8} \\right)$$ ### Graphical For the exercises 46-55, plot the complex number in the complex plane. 46) $$2+4i$$ 47) $$-3-3i$$ 48) $$5-4i$$ 49) $$-1-5i$$ 50) $$3+2i$$ 51) $$2i$$ 52) $$-4$$ 53) $$6-2i$$ 54) $$-2+i$$ 55) $$1-4i$$ ### Technology For the exercises 56-, find all answers rounded to the nearest hundredth. 56) Use the rectangular to polar feature on the graphing calculator to change $$5+5i$$ to polar form. 57) Use the rectangular to polar feature on the graphing calculator to change $$3-2i$$ to polar form. $$3.61e^{-0.59i}$$ 58) Use the", "& d & 1344 & 6 && \\lceil d/8 \\rceil & 168 & 1 \\\\ \\operatorname{SHAKE-256} & d & 1088 & 6 && \\lceil d/8 \\rceil & 136 & 1 \\\\ \\hline \\operatorname{BLAKE2s-256} & 256 & 512 & 0 && 32 & 64 & 0 \\\\ \\operatorname{BLAKE2b-512} & 512 & 1024 & 0 && 64 & 128 & 0 \\\\ \\end{array}$$ ## SHA3-x 1. $$\\operatorname{SHA3-224}(M) = \\operatorname{KECCAK}[448] (M \\mathbin\\| 01, 224)$$ 2. $$\\operatorname{SHA3-256}(M) = \\operatorname{KECCAK}[512] (M \\mathbin\\| 01, 256)$$ 3. $$\\operatorname{SHA3-384}(M) = \\operatorname{KECCAK}[768] (M \\mathbin\\| 01, 384)$$ 4. $$\\operatorname{SHA3-512}(M) = \\operatorname{KECCAK}[1024](M \\mathbin\\| 01, 512)$$ KECCAK and KECCAK defined as • $$\\operatorname{KECCAK}[c] (N, d) = \\operatorname{SPONGE}[\\operatorname{KECCAK-p}[1600, 24], \\operatorname{pad10*1}, 1600–c] (N, d)$$ and note that $$N = M\\mathbin\\|01$$ and $$d$$ is the required output size, Now we have 1. $$\\operatorname{SHA3-224}(M) = \\operatorname{SPONGE}[\\operatorname{KECCAK-p}[1600, 24], \\operatorname{pad10*1}, 1600–448] (N, 224)$$ 2. $$\\operatorname{SHA3-256}(M) = \\operatorname{SPONGE}[\\operatorname{KECCAK-p}[1600, 24], \\operatorname{pad10*1}, 1600–512] (N, 256)$$ 3. $$\\operatorname{SHA3-384}(M) = \\operatorname{SPONGE}[\\operatorname{KECCAK-p}[1600, 24], \\operatorname{pad10*1}, 1600–768] (N, 384)$$ 4. $$\\operatorname{SHA3-512}(M) = \\operatorname{SPONGE}[\\operatorname{KECCAK-p}[1600, 24], \\operatorname{pad10*1}, 1600–1024] (N, 512)$$ • $$\\operatorname{pad10*1}(x, m)$$ is not important since the message is not formed yet. SPONGE • $$\\operatorname{SPONGE}[f, \\operatorname{pad}, r](N, d)$$ 1. Let $$P=N \\mathbin\\| \\operatorname{pad}(r, \\operatorname{len}(N))$$. 2. Let $$n=\\lfloor\\operatorname{len}(P)/r\\rfloor$$. 3. Let $$c=\\lfloor b/r\\rfloor$$. 4. Let $$P_0, \\ldots, P_{n-1}$$ be the unique sequence of strings of length $$r$$ such that $$P = P_0 \\mathbin\\| \\ldots\\mathbin\\| P{n-1}$$. 5. Let", "+0 # help +1 50 1 The number $$\\text{cis } 75^\\circ + \\text{cis } 83^\\circ + \\text{cis } 91^\\circ + \\dots + \\text{cis } 147^\\circ$$ is expressed in the form $$r \\, \\text{cis } \\theta$$, where $$0 \\le \\theta < 360^\\circ$$. Find $$\\theta$$ in degrees. Dec 20, 2019", "# Fill in the operators to make $7 \\circ 8 \\circ 7 \\circ 7 \\circ 8 \\circ 3 = 100$ With the given numbers I need to make: $7 \\circ 8 \\circ 7 \\circ 7 \\circ 8 \\circ 3 = 100$ Using only the following operators: $+\\space-\\space\\times\\space\\div$ $7 \\times 8 + 7 \\times 7 - 8 + 3 = 100$ • This is pretty much it if we can't use parenthesis.. +1 – Raystafarian Nov 30 '15 at 16:30 General \"Countdown\" arithmetic puzzle solution !! Some Unix Shells have a property called 'brace expansion' The usage varies between csh and bash but is something like this: echo {a,i}{s,t} Outputs: as at is it This can be used to solve the \"Countdown\" letters puzzle, simply have 9 consecutive braces with all 9 letters in each. If you don't know Countdown, you can use this to find Scrabble™ words, if you add a final brace at the end like {'\\n'} you even get one result per line so you can pipe output through sort or usr/dict So how do we do the Arithmetic puzzle ? If we are given the integers 2,4,5 Alternate braces containing integers and operators: set array {2,4,5}{+,-,\\*,/}{2,4,5}…{'\\n'} # * escaped to avoid wildcard This gives: 2+2-4… 2-3*4… Every permutation of integer and operator. Finally, foreach s", "Even numbers of \\circ causes weird spacing I noticed when trying to make a string of 6 \\circ symbols that the last one is closer to the one beside it than the rest of them, after playing with it for a while it looks like an even number (greater than 2) of \\circ causes this, how can I fix the spacing so it is even? Example: \\documentclass{article} \\begin{document} 1 circ: $\\circ$ 2 circ: $\\circ \\circ$ 3 circ: $\\circ \\circ \\circ$ 4 circ: $\\circ \\circ \\circ \\circ$ 5 circ: $\\circ \\circ \\circ \\circ \\circ$ 6 circ: $\\circ \\circ \\circ \\circ \\circ \\circ$ 7 circ: $\\circ \\circ \\circ \\circ \\circ \\circ \\circ$ \\end{document} • One way is to make \\circ not a binary/relational operator. Use {\\circ} – Sigur Jan 16 '19 at 21:44 • ... red circ, blue circ. :-) – Mees de Vries Jan 16 '19 at 23:57 From the TeXbook (page 187, solution on page 326) Exercise 19.7 B. L. User tried typing ‘\\eqno(*)’ and ‘\\eqno(**)’, and he was pleased to discover that this produced the equation numbers ‘(∗)’ and ‘(∗∗)’. [He had been a bit worried that they would come out ‘(*)’ and ‘(**)’ instead.] But then a few months later he tried ‘\\eqno(***)’ and got a surprise. What was it? When you type an asterisk in", "multiply the numerator and denominator by the conjugate of the number in the denominator. $$\\frac{\\textrm{cis} 75^o - \\textrm{cis} 155^o}{ 1 - \\cos{8^o} -i \\sin{8^o}} \\frac{1-\\cos{8^o} + i \\sin{8^o}}{1-\\cos{8^o} + i \\sin{8^o}}$$ This will give us a completely real denominator. And from there, we just have a lot of algebra to grind out to get the remaining expression down to a form of $r \\textrm{cis}{\\theta}$. I hope you find the algebra to be doable. The problem is basically a vector addition problem. $cis\\theta = cos\\theta + i sin\\theta$. So, $cis 75 + cis 83 + ... = cos 75 + cos 83 + ... + i(sin 75 + sin 83 + ...)$ Actually doing the math would be difficult, but note that you're adding a bunch of vectors of equal length, each 8 degrees apart. They form an arc and what you want is the average direction of the arc since that is also the net direction of the vectors. Hope that gets you started. I can comment further if necessary. this is a geometric series $$cis(75^\\circ) +cis(75^\\circ + 8^\\circ)+\\cdots + cis(75^\\circ+9\\times 8^\\circ)=cis(75^\\circ)\\frac{1-cis(80^\\circ)}{1-cis(8^\\circ)}$$ i hope you can take it from here.", "and suppose that P=r(costheta+isintheta) where $0 \\lt r$ and $0 \\leq \\theta \\lt 360$ find $\\theta$ • is 107 • is 276 • is 840 • cannot be determined from the given information ### Key Concepts Equations Complex Numbers Integers AIME, 1996, Question 11 Complex Numbers from A to Z by Titu Andreescue ## Try with Hints First hint here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\\frac{(z^{5}-1)}{(z-1)}$=$\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0 Second Hint gives $z^{5}=1 for z\\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \\neq 1$ gives z=$\\frac{1+-(-3)^\\frac{1}{2}}{2}$=$cis60,300$ where cis$\\theta$=cos$\\theta$+isin$\\theta$ Final Step taking $0 \\lt theta \\lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276. . ## Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed.", "common with the other 2 statements so for the answer do i just put these 2 values in the swirly brackets as the answer, umm im not sure.... 5. Apr 22, 2012 Curious3141 Yes, but put them in a+bi form. 6. Apr 22, 2012 ronho1234 ummm because i get wacky values for the arguments its really hard or i just can't seem to switch them back to a+bi form i get decimals i think.... 7. Apr 22, 2012 Curious3141 Just to clarify, I meant leave them as trig ratios, but express them as r(cosθ +i sinθ) rather than using the shorthand \"cis\". 8. Apr 22, 2012 ronho1234 oh sorry!... so just A∩B∩C = {2^1/6(cos(pi/36)+isin(pi/36),2^1/6(cos(13pi/36)+isin(13pi/36)} 9. Apr 22, 2012 Curious3141 Yeah, I'm sure it'll come out better when you write it down, but that's the idea. 10. Apr 22, 2012 ronho1234 ok THANKYOU curious3141 and mentallic :) 11. Apr 22, 2012 Mentallic Just as a heads up, since $r\\cdot cis(\\theta)\\equiv r(\\cos(\\theta)+i\\sin(\\theta))$ then it should be ok to leave the answer in cis form, because that is it's definition simply because it's short and neat. I'd only ever separate it when it makes sense to separate it such as when you want to show why if $z=cis\\theta$ then $\\overline{z}=cis(-\\theta)$ for example. 12. Apr 22, 2012 ronho1234 ok thanks for the", "$AA^T=circ(4\\ 1122\\ 2211)$, $A=circ(110110000)$ with $AA^T=circ(4\\ 2121\\ 1212)$, $A=circ(110101000)$ with $AA^T=circ(4\\ 1212\\ 2121)$, $A=circ(111001000)$ with $AA^T=circ(4\\ 2112\\ 2112)$, $A=circ(110100010)$ with $AA^T=circ(4\\ 1221\\ 1221)$. $t=3$: $A=circ(1110110100000)$ with $AA^T=circ(6\\ 333222\\ 222333)$, $A=circ(1110101100000)$ with $AA^T=circ(6\\ 332223\\ 322233)$, $A=circ(1110110010000)$ with $AA^T=circ(6\\ 323322\\ 223323)$, $A=circ(1110011010000)$ with $AA^T=circ(6\\ 322233\\ 332223)$, $A=circ(1110100101000)$ with $AA^T=circ(6\\ 232233\\ 332232)$, and probably many others (looking at the patterns of $AA^T$). In fact, looking at the $t=2$ patterns that seem \"complete\", I'd conjecture that there are $\\binom{2t}{t}$ essentially different such circulant matrices. (There may be also non-circulant solutions, so this would not yet solve the problem, but be a nice start.) - A computer search for $t=3$ and $t=4$ shows that not all $\\binom{2t}t$ patterns occur. E.g. for $t=3$, only 14 of the 20 occur, and on the other hand, some correspond to several non isomorphic matrices. E.g., both $A=circ(1111001000100)$ and $A=circ(1110110010000)$ yield $AA^T=circ(6\\ 323322\\ 223323)$. – Wolfgang Feb 24 '13 at 16:45 I do not know the answer. However, here is part of an analysis for t=1; perhaps some of it can extend for larger t. I usually mean set of size 2t when I say set. Consider two sets with intersection C of size t. The complement of their union has size t+1, which specifies only one set which has intersection of size t-1 with the two sets.", "Precalculus (6th Edition) $[{cis -30^{\\circ},cis -150^{\\circ},cis -270^{\\circ}}]$ Given: $x^{3}+i=0$ or $x^{3}=-i$ $-i$ can be written in trigonometric form as: $-i=0-i=(\\cos 90^{\\circ}-\\sin 90^{\\circ})$ Absolute value of third root is given as $\\sqrt[3] 1=1$ Now, the arguments can be given as: $k=0,1,2$ Roots: $(\\cos 30^{\\circ}+\\sin 30^{\\circ})$,$(\\cos 150^{\\circ}+\\sin 150^{\\circ})$,$(\\cos 270^{\\circ}+\\sin 270^{\\circ})$ Solution set of the equation can be written as: $[{cis -30^{\\circ},cis -150^{\\circ},cis -270^{\\circ}}]$", "of theta functions. Example To compute $\\operatorname{sn}$, $\\operatorname{cn}$, $\\operatorname{dn}$ to 10D when $x=0.8$, $k=0.65$. Four iterations of (22.20.1) lead to $c_{4}=\\Sci{6.5}{-12}$. From (22.20.3) and (22.20.4) we obtain $\\phi_{1}=1.40213\\;91827$ and $\\phi_{0}=0.76850\\;92170$. Then from (22.20.5), $\\operatorname{sn}\\left(0.8,0.65\\right)=0.69506\\;42165$, $\\operatorname{cn}\\left(0.8,0.65\\right)=0.71894\\;76580$, $\\operatorname{dn}\\left(0.8,0.65\\right)=0.89212\\;34349$. §22.20(iii) Landen Transformations By application of the transformations given in §§22.7(i) and 22.7(ii), $k$ or $k^{\\prime}$ can always be made sufficently small to enable the approximations given in §22.10(ii) to be applied. The rate of convergence is similar to that for the arithmetic-geometric mean. Example To compute $\\operatorname{dn}\\left(x,k\\right)$ to 6D for $x=0.2$, $k^{2}=0.19$, $k^{\\prime}=0.9$. From (22.7.1), $k_{1}=\\tfrac{1}{19}$ and $x/(1+k_{1})=0.19$. From the first two terms in (22.10.6) we find $\\operatorname{dn}\\left(0.19,\\tfrac{1}{19}\\right)=0.999951$. Then by using (22.7.4) we have $\\operatorname{dn}\\left(0.2,\\sqrt{0.19}\\right)=0.996253$. If needed, the corresponding values of $\\operatorname{sn}$ and $\\operatorname{cn}$ can be found subsequently by applying (22.10.4) and (22.7.2), followed by (22.10.5) and (22.7.3). §22.20(iv) Lattice Calculations If either $\\tau$ or $q=e^{i\\pi\\tau}$ is given, then we use $k={\\theta_{2}^{2}}\\left(0,q\\right)/{\\theta_{3}^{2}}\\left(0,q\\right)$, $k^{\\prime}={\\theta_{4}^{2}}\\left(0,q\\right)/{\\theta_{3}^{2}}\\left(0,q\\right)$, $K=\\frac{1}{2}\\pi{\\theta_{3}^{2}}\\left(0,q\\right)$, and $K^{\\prime}=-i\\tau K$, obtaining the values of the theta functions as in §20.14. If $k,k^{\\prime}$ are given with $k^{2}+{k^{\\prime}}^{2}=1$ and $\\Im k^{\\prime}/\\Im k<0$, then $K,K^{\\prime}$ can be found from 22.20.6 $\\displaystyle K$ $\\displaystyle=\\frac{\\pi}{2M(1,k^{\\prime})},$ $\\displaystyle K^{\\prime}$ $\\displaystyle=\\frac{\\pi}{2M(1,k)},$ using the arithmetic-geometric mean. Example 1 If $k=k^{\\prime}=1/\\sqrt{2}$, then three iterations of (22.20.1) give $M=0.84721\\;30848$, and from (22.20.6) $K=\\pi/(2M)=1.85407\\;46773$ — in agreement with the value of $\\left(\\Gamma\\left(\\tfrac{1}{4}\\right)\\right)^{2}/\\left(4\\sqrt{\\pi}\\right)$; compare (23.17.3) and (23.22.2).", "\\circ i = e \\circ i = \\operatorname{id}_{X} \\circ i = i = i \\circ \\operatorname{id}_{Y}.$ $\\square$", "function\\int \\operatorname{Ci}(x) \\, dx = x \\operatorname{Ci}(x) - \\sin x\\int \\operatorname{Si}(x) \\, dx = x \\operatorname{Si}(x) + \\cos x\\int \\operatorname{Ei}(x) \\, dx = x \\operatorname{Ei}(x) - e^x\\int \\operatorname{li}(x) \\, dx = x \\operatorname{li}(x)-\\operatorname{Ei}(2 \\ln x) \\int \\frac{\\operatorname{li}(x)}{x}\\,dx = \\ln x\\, \\operatorname{li}(x) -x \\int \\operatorname{erf}(x)\\, dx = \\frac{e^{-x^2}}{\\sqrt{\\pi }}+x \\operatorname{erf}(x) ## related categories GMT+8, 2015-9-11 22:04 , Processed in 0.211363 second(s), 16 queries ." ]

[ "# Complex Numbers and exponential form and roots The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$ are $\\text{cis } \\theta_1, \\text{cis } \\theta_2, \\dots, \\text{cis } \\theta_7,$ where $0^\\circ \\le \\theta_k < 360^\\circ$for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\left(5 \\pi i/4 \\right)$. How should I simplify? Thanks - What is $\\text{cis}$? Do you by chance mean $\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\theta)+isin(\\theta)$ – Mr.Fry Jan 24 '14 at 16:27 @AlexR: \"cis\" is a commonly used abbreviation for \"cosine plus $i$ times the sine.\" In other words, $$\\operatorname{cis}\\theta=\\cos\\theta+i\\sin\\theta=e^{i\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis}(7\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis} 225^\\circ,$$ as you've already determined, and so we need $$\\operatorname{cis}(7\\theta_k)=\\operatorname{cis} 225^\\circ\\\\\\frac{\\operatorname{cis}(7\\theta_k)}{\\operatorname{cis} 225^\\circ}=1\\\\\\operatorname{cis}(7\\theta_k-225^\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? If I am understanding correctly...for $k=1$, we take $$\\operatorname{cis}(7\\theta_1)=\\operatorname{cis} 225^\\circ$$ and then somehow determine $\\theta_1$, and repeat for $k=2,3,4,5,6,7$? – Mathy Person Jul 18 '15 at 21:59 No, that's getting further from the answer. Rather, think about the solutions set for the equation $$\\operatorname{cis}\\theta=1. This will let you get rid of the \\operatorname{cis}, and find all solutions to the original", "# Complex Numbers and exponential form and roots The roots of $z^7 = -\\frac{1}{\\sqrt{2}} - \\frac{i}{\\sqrt{2}}$ are $\\text{cis } \\theta_1, \\text{cis } \\theta_2, \\dots, \\text{cis } \\theta_7,$ where $0^\\circ \\le \\theta_k < 360^\\circ$for all $1 \\le k \\le 7$. Find $\\theta_1 + \\theta_2 + \\dots + \\theta_7$. Give your answer in degrees. In exponential form this is $z^7 = e^ \\left(5 \\pi i/4 \\right)$. How should I simplify? Thanks - What is $\\text{cis}$? Do you by chance mean $\\cos$? – AlexR Jan 24 '14 at 16:26 @AlexR cis is short for $cos(\\theta)+isin(\\theta)$ – Rod Jan 24 '14 at 16:27 @AlexR: \"cis\" is a commonly used abbreviation for \"cosine plus $i$ times the sine.\" In other words, $$\\operatorname{cis}\\theta=\\cos\\theta+i\\sin\\theta=e^{i\\theta}.$$ – Cameron Buie Jan 24 '14 at 16:27 Well, by DeMoivre's Theorem, we have $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis}(7\\theta_k)$$ for $k=1,...,7.$ We need for $$(\\operatorname{cis}\\theta_k)^7=\\operatorname{cis} 225^\\circ,$$ as you've already determined, and so we need $$\\operatorname{cis}(7\\theta_k)=\\operatorname{cis} 225^\\circ\\\\\\frac{\\operatorname{cis}(7\\theta_k)}{\\operatorname{cis} 225^\\circ}=1\\\\\\operatorname{cis}(7\\theta_k-225^\\circ)=1$$ for $k=1,...,7.$ Can you take it from there? - The roots are $e^{2k\\pi i/7}$ ($k=0,\\ldots,6$). Write them as $r^k$ where $r = e^{2\\pi i/7}$. So the sum is (using the simple formula for summing a finite geometric series) $$\\sum_{k=0}^6 r^k = \\frac{1-r^7}{1-r} = \\frac{1 - e^{2\\pi i}}{1-e^{2\\pi i/7}} = 0$$. - This is incorrect. The roots you have given are the roots of $z^7=1.$ Also, we are not", "# Operation with Complex number We've the following 3 complex numbers $z_1=2\\text{cis}(\\frac{\\pi}{6})~~$, $z_2=2\\text{cis}(\\frac{5\\pi}{6})~~$, $z_3=2\\text{cis}(\\frac{-\\pi}{2})$ We are asked to show that $z_1^{3n}+z_2^{3n}=2z_3^{3n}$ I have calculated $z_1^{3n}=2^{3n}\\text{cis}(\\frac{n\\pi}{2})~~$, $z_2^{3n}=2^{3n}\\text{cis}(\\frac{5n\\pi}{2})~~$, $z_3^{3n}=2^{3n}\\text{cis}(\\frac{-3n\\pi}{2})$. I'm stuck on how to proceed further to prove the above equations. Any help really appreciated, Please assist . Thank you , Arif • What is \"cis\" ? – Jean Marie Mar 29 '17 at 6:07 • @JeanMarie This. – dxiv Mar 29 '17 at 6:21 • @dxiv Thanks. I understand now the pedagogical interest of this notation... – Jean Marie Mar 29 '17 at 6:32 • $z_1^3 = 2^3 \\operatorname{cis}\\left(3 \\cdot \\cfrac{\\pi}{6}\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ • $z_2^3 = 2^3 \\operatorname{cis}\\left(3 \\cdot \\cfrac{5 \\pi}{6}\\right)=8\\operatorname{cis}\\left(\\cfrac{5 \\pi}{2}\\right)=8\\operatorname{cis}\\left(2\\pi+\\cfrac{\\pi}{2}\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ • $z_3^3 = 2^3 \\operatorname{cis}\\left(-3 \\cdot \\cfrac{\\pi}{2}\\right)=8\\operatorname{cis}\\left(- 3 \\cdot \\cfrac{\\pi}{2}+ 2 \\pi\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ Therefore $\\;z_1^3=z_2^3=z_3^3\\,$.", "that $w = \\operatorname{cis}(30^{\\circ})$ and $z = \\operatorname{cis}(120^{\\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$, respectively. By Demoivre's theorem, $\\operatorname{cis}(\\theta)^n = \\operatorname{cis}(n\\theta)$. When you multiply by $i$, we can think of that as rotating the complex number $90^{\\circ}$ counterclockwise in the complex plane. Therefore, by the equation we know that $30r + 90$ and $120s$ land on the same angle. This means that$$30r + 90 \\equiv 120s \\pmod{360},$$which we can simplify to$$r+3 \\equiv 4s \\pmod{12}.$$Notice that this means that $r$ cycles by $12$ for every value of $s$. This is because once $r$ hits $12$, we get an angle of $360^{\\circ}$ and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every $3$ times, which is much easier to count. By listing the possible values out, we get the pairs $(r,s)$:$$\\begin{array}{cccccccc} (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \\ldots & (97,1) \\\\ (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \\ldots & (97,4) \\\\ (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \\ldots & (97,7) \\\\ [-1ex] \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots \\\\ (1,100) & (5,98) & (9,99) & (13,100) & (17,98)", "# Writing mod-arg form as $r\\operatorname{cis}(\\theta)$ rather than $r(\\cos\\theta + i\\sin\\theta)$ I'm studying in-depth complex numbers and analysis and when I'm working through certain theory, I like to refer to as many textbooks as possible. I've always known to write mod-arg form as $r(\\cos\\theta + i\\sin\\theta)$ However, this book by J. Coroneos writes mod-arg form simplistically as $r\\operatorname{cis}\\theta$. In the other 7 textbooks I have referred to, I haven't come across this. Is this is universally accepted? If I use this, will I be technically wrong or is this dude just simplifying things to make things easier. • en.wikipedia.org/wiki/Cis_(mathematics) – user223391 Jan 15 '17 at 20:37 • It's just their convention. Some texts use that (and so do I sometimes). It's also shorthand for $r e^{i\\theta}$. You're okay. – Sean Roberson Jan 15 '17 at 20:38 • Whatever what you want – user223391 Jan 15 '17 at 20:40 • $cis(t)$ is useless notation, because it gets replaced by $e^{it}$ immediately. I recommend getting used to the exponential form. – Kaynex Jan 15 '17 at 20:41 • I think \"cis\" is teddy-bear for beginners. I guess most people will know what you mean when you write it, but when doing any kind of calculation, $e^{it}$ works better. – B. Goddard Jan 15 '17 at 20:43 ## 1 Answer It's", "# The origins of $\\operatorname{cis}(\\theta)$ There is a abbreviation used in high school mathematics that is almost never seen outside of it: $$\\operatorname{cis}(\\theta) = \\cos(\\theta) + i \\sin(\\theta)$$, where cis stands for cosine + i sine. As soon as students get into university Euler's formula $$e^{i\\theta} = \\cos(\\theta) + i \\sin(\\theta)$$ is almost exclusively used as it makes a lot of things easier and more obvious. The reason Euler's formula is not used in high school is probably that it's not easy to prove or to develop an intuition for at that level. My question is: does anyone know the origins and history of the abbreviation cis ? The notation exists since a long time. It was used already by Irving Stringham in 'Uniplanar Algebra (1893).' This is claimed to be the earliest use on http://jeff560.tripod.com/trigonometry.html giving Cajori vol. 2, page 133 as reference. In this book, Uniplanar Algebra, the notation is used first, as far as I can see, in chapter III (The algebra of complex quantities). It is introduced without much ado as a shorthand notation. It is said it should be read as \"sector of $\\theta$\". Cajori also mentions another early usage (thanks to Dag Oskar Madsen for the information!), namely, Harkness and Morley \"Introduction to the theory of analytic functions (1898)\". Also there it", "as you might expect to yield $\\left(\\frac{r_1}{r_2}\\right) e^{i ( \\theta_1 - \\theta_2)}$. $\\def\\cis{\\operatorname{cis}}$ If we have the fraction $$\\frac{r_{1}\\cis(\\theta_{1})}{r_{2}\\cis(\\theta_{2})}$$ then in polar form this is the complex number $$\\frac{r_{1}}{r_{2}}\\cis(\\theta_{1}-\\theta_{2})$$ It follows from the fact that $|\\cis(\\alpha)|=1$ hence $|r\\cis(\\alpha)|=r$ and from the trigonometric identities • To be clear, $cis(x) = e^{ix} = \\cos(x) + i \\sin(x)$ – LucasVB Jul 31 '14 at 14:21 • @MJD - Thanks for the edit. I don't have the \\cis command in Lyx but its on Math.SE (apparently) – Belgi Jul 31 '14 at 14:34 • I had to define it specially for your post, using \\def. – MJD Jul 31 '14 at 14:41 • Waittt..... So I can basically subtract the angles? (Extending from this, I suppose in multiplication I can add the angles?) – Gummy bears Jul 31 '14 at 14:48 • @Gummybears $e^{i \\theta}= \\cos{\\theta} + i \\sin{\\theta}$. – snulty Aug 1 '14 at 12:49", "# Show that $\\sin x-i\\cos x=\\operatorname{cis}(x-\\pi/2)$ A question in a textbook asks Show that $\\sin\\theta-i\\cos\\theta=\\operatorname{cis}(\\theta-\\frac{\\pi}{2})$ I've previously shown that: $\\sin\\theta+i\\cos\\theta=\\operatorname{cis}(\\frac{\\pi}{2}-\\theta)$, and that $\\cos\\theta-i\\sin\\theta=\\operatorname{cis}(-\\theta)$. So I've tried using that in my working out, but can't seem to work this one out. I am assuming that it wants me to use trig addition formulae and De Moivre's Theorem possibly (because that is the general theme of this section of the book). No working out that's too abstract or advanced I'd presume. I have tried symmetry properties, and negative angles as well. I am sure I can prove it by using a subtraction formula, but that would be me assuming the answer (circular logic maybe). I need to start from the LHS and then work to get the RHS. You can write $$\\operatorname{cis}\\left(\\theta-\\frac{\\pi}{2}\\right)= \\frac{\\operatorname{cis}\\theta}{\\operatorname{cis}\\frac{\\pi}{2}}$$ Since $\\operatorname{cis}\\frac{\\pi}{2}=i$, you get $$\\operatorname{cis}\\left(\\theta-\\frac{\\pi}{2}\\right)= i^{-1}\\operatorname{cis}\\theta=i^{-1}(\\cos\\theta+i\\sin\\theta)= -i\\cos\\theta+\\sin\\theta$$ because $i^{-1}=-i$. • Thank you for that. It ties in very well with the more recent sections. – Simplex1 Aug 3 '18 at 11:57 Well, we have: $$\\exp\\left(\\theta i\\right)=\\cos\\left(\\theta\\right)+\\sin\\left(\\theta\\right)i\\tag1$$ So, when we have: $$\\exp\\left(\\left(\\theta-\\frac{\\pi}{2}\\right)\\cdot i\\right)=\\cos\\left(\\theta-\\frac{\\pi}{2}\\right)+\\sin\\left(\\theta-\\frac{\\pi}{2}\\right)i\\tag2$$ And we know that: • $$\\cos\\left(\\theta-\\frac{\\pi}{2}\\right)=\\sin\\left(\\theta\\right)\\tag3$$ • $$\\sin\\left(\\theta-\\frac{\\pi}{2}\\right)=-\\cos\\left(\\theta\\right)\\tag4$$", "special notation for it. Notation. Cis $\\operatorname{Cis}(\\theta):=\\cos \\theta+i \\sin \\theta .$ DEFINITION. For the complex number $$z=x+i y=r \\operatorname{Cis}(\\theta)$$, we have the following: $$\\Re(z) x$$ is called the real part of $$z$$, written $$\\Re(z)$$; $$\\Im(z) y$$ is called the imaginary part of $$z$$, written $$\\Im(z)$$; $$|z| r$$ is called the modulus of $$z$$, or absolute value of $$z$$, written $$|z|$$; $$\\arg (z) \\theta$$ is called the argument of $$z$$, written $$\\arg (z)$$. $$\\bar{z}$$ The number $$x-i y$$ is called the conjugate of $$z$$, written $$\\bar{z}$$. REMARK. There is an important point to bear in mind about the argument: it is only unique up to addition of multiples of $$2 \\pi$$. In other words, if $$\\theta_{0}$$ is an argument of the complex number $$z$$, then so are all the numbers $$\\left\\{\\theta_{0}+2 k \\pi: k \\in \\mathbb{Z}\\right\\}$$. Addition is easiest in Cartesian coordinates: add the real and imaginary parts. Multiplication is easiest in polar coordinates: multiply the moduli and add the arguments. Proposition 9.19. Let $$z_{1}=r_{1} \\operatorname{Cis}\\left(\\theta_{1}\\right)$$ and $$z_{2}=r_{2} \\operatorname{Cis}\\left(\\theta_{2}\\right)$$. Then $z_{1} z_{2}=r_{1} r_{2} \\operatorname{Cis}\\left(\\theta_{1}+\\theta_{2}\\right) .$ ProOF. Multiplying out, we get \\begin{aligned} z_{1} z_{2}=r_{1} r_{2}\\left[\\cos \\theta_{1} \\cos \\theta_{2}-\\sin \\theta_{1} \\sin \\theta_{2}\\right.\\\\ &\\left.+i\\left(\\cos \\theta_{1} \\sin \\theta_{2}+\\cos \\theta_{2} \\sin \\theta_{1}\\right)\\right] . \\end{aligned} The result follows by the trigonometric identities for the cosine and sine of the sum of two angles. A consequence", "have $\\frac{6\\operatorname{cis}45^{\\circ}}{3\\operatorname{cis}15^{\\circ}}=2\\operatorname{cis}30^{\\circ}$6cis453cis15=2cis30. ### General results We can write down general statements for the multiplication and division of two complex numbers that express this transformational idea. So given: $z_1=r_1\\left(\\cos\\theta_1+i\\sin\\theta_1\\right)=r_1\\operatorname{cis}\\theta_1$z1=r1(cosθ1+isinθ1)=r1cisθ1 $z_2=r_2\\left(\\cos\\theta_2+i\\sin\\theta_2\\right)=r_2\\operatorname{cis}\\theta_2$z2=r2(cosθ2+isinθ2)=r2cisθ2 we have: $z_1\\times z_2=r_1r_2\\operatorname{cis}\\left(\\theta_1+\\theta_2\\right)$z1×z2=r1r2cis(θ1+θ2) $\\frac{z_1}{z_2}=\\frac{r_1}{r_2}\\operatorname{cis}\\left(\\theta_1-\\theta_2\\right)$z1z2=r1r2cis(θ1θ2) ##### Example 1 Evaluate the product $z=2\\operatorname{cis}\\frac{\\pi}{4}\\times\\frac{1}{2}\\operatorname{cis}\\left(-\\pi\\right)$z=2cisπ4×12cis(π) The angle measure is in radians, but the question could have just as easily been given using degrees. The product becomes $2\\times\\frac{1}{2}\\times\\operatorname{cis}\\left(\\frac{\\pi}{4}-\\pi\\right)=\\operatorname{cis}\\left(-\\frac{3\\pi}{4}\\right)$2×12×cis(π4π)=cis(3π4). ##### Example 2 Simplify $z=\\frac{12\\left(\\cos15^{\\circ}+i\\sin15^{\\circ}\\right)}{3\\left[\\cos\\left(-60^{\\circ}\\right)-i\\sin\\left(-60^{\\circ}\\right)\\right]}$z=12(cos15+isin15)3[cos(60)isin(60)]. Noting that $\\cos\\left(-\\theta\\right)=\\cos\\left(\\theta\\right)$cos(θ)=cos(θ) and $\\sin\\left(-\\theta\\right)=-\\sin\\left(\\theta\\right)$sin(θ)=sin(θ) we can immediately simplify the denominator so that: $z=\\frac{12\\left(\\cos15^{\\circ}+i\\sin15^{\\circ}\\right)}{3\\left(\\cos60^{\\circ}+i\\sin60^{\\circ}\\right)}$z=12(cos15+isin15)3(cos60+isin60) Then we have: $z$z $=$= $\\frac{12\\left(\\cos15^{\\circ}+i\\sin15^{\\circ}\\right)}{3\\left(\\cos60^{\\circ}+i\\sin60^{\\circ}\\right)}$12(cos15∘+isin15∘)3(cos60∘+isin60∘)​ $=$= $4\\left[\\operatorname{cis}\\left(-45^{\\circ}\\right)+i\\sin\\left(-45^{\\circ}\\right)\\right]$4[cis(−45∘)+isin(−45∘)] We can depict $z=4\\operatorname{cis}\\left(-45^{\\circ}\\right)$z=4cis(45) geometrically as an arrow of length $4$4 emanating from the origin into the $4$4th quadrant. ##### Example 3 Explain the multiplication $\\left(1+i\\right)\\left(1+\\sqrt{3}i\\right)$(1+i)(1+3i) visually. We can approach the problem in a number ways. We could first convert each complex factor into polar form and use the transformational idea to show the result on the complex plane. Since $1+i=\\sqrt{2}\\operatorname{cis}45^{\\circ}$1+i=2cis45 and $1+\\sqrt{3}i=2\\operatorname{cis}60^{\\circ}$1+3i=2cis60, the product becomes $2\\sqrt{2}\\operatorname{cis}105^{\\circ}$22cis105 . Using radians, we could also write this as $2\\sqrt{2}\\operatorname{cis}\\frac{7\\pi}{12}$22cis7π12. An alternative strategy is to use the fact that multiplication of any non-zero complex number by $i$i is equivalent to a simple $90^{\\circ}$90 anticlockwise rotation of that complex number in the complex plane. Thus our strategy begins by partially expanding the the given product so that: $\\left(1+i\\right)\\left(1+\\sqrt{3}i\\right)$(1+i)(1+√3i) $=$= $1\\left(1+\\sqrt{3}i\\right)+i\\left(1+\\sqrt{3}i\\right)$1(1+√3i)+i(1+√3i) The product", "# Finding the n-th root of a complex number I am trying to solve $z^6 = 1$ where $z\\in\\mathbb{C}$. So What I have so far is : $$z^6 = 1 \\rightarrow r^6\\operatorname{cis}(6\\theta) = 1\\operatorname{cis}(0 + \\pi k)$$ $$r = 1,\\ \\theta = \\frac{\\pi k}{6}$$ $$k=0: z=\\operatorname{cis}(0)=1$$ $$k=1: z=\\operatorname{cis}\\left(\\frac{\\pi}{6}\\right)=\\frac{\\sqrt{3} + i}{2}$$ $$k=2: z=\\operatorname{cis}\\left(\\frac{\\pi}{3}\\right)=\\frac{1 + \\sqrt{3}i}{2}$$ $$k=3: z=\\operatorname{cis}\\left(\\frac{\\pi}{2}\\right)=i$$ According to my book I have a mistake since non of the roots starts with $\\frac{\\sqrt{3}}{2}$, also even if I continue to $k=6$ I get different (new) results, but I thought that there should be (by the fundamental theorem) only 6 roots. Can anyone please tell me where my mistake is? Thanks! - So what is written in your book? – draks ... Feb 6 '12 at 13:46 $cis=cos+i\\cdot sin$? – Rasmus Feb 6 '12 at 14:01 yes its is. why do i have to write 12 chars? – yotamoo Feb 6 '12 at 14:10 You want $\\theta=\\frac{2\\pi k}6 = \\frac{\\pi k}3$. – Thomas Andrews Feb 6 '12 at 14:12 What ever \"start with ...\", I think there is a 2 missing. Shouldn't it be ${\\rm cis}(x)=e^{ix}=e^{i(x+2k\\pi)}$? This would also drop the solution that \"starts with $\\sqrt{3}/2$\", which is a root of -1 anyhow. – draks ... Feb 6 '12 at 14:15 You have to write your equation into the form $$z^6=e^{2in\\pi}=1$$", "in italics you do \\operatorname{cis}. – fpqc Mar 18 '12 at 8:04 So it remains to check that $\\tan\\pi/5=2(\\sin2\\pi/5-\\sin\\pi/5)$. After a simplification, this amounts to check that $$1=2\\cos\\pi/5\\;(2\\cos\\pi/5-1),$$ and this can be seen directly from the fact that $\\cos\\pi/5=(1+\\sqrt5)/4$.", "# Simplifying and then expressing complex numbers in cartesian form 1. Mar 11, 2010 ### Stripe 1. The problem statement, all variables and given/known data (2 CIS (pi/6))*(3 CIS (pi/12)) 2. Relevant equations Also what is CIS? I believe it's Cos+i*sin but how do you use it? 3. The attempt at a solution i simplified it to 6 CIS (pi/12) How do i turn it into cartesian? 2. Mar 11, 2010 ### tiny-tim Hi Stripe! (have a pi: π ) I've never seen \"CIS\" before, but I'll guess you're right, and that it's cos + i*sin. Now use De Moivre's theorem … cosθ + isinθ = e And Cartesian form simply means in the form x + iy (as opposed to polar form, which is in the form re ) (and no, it's not 6 CIS (π/12)) Last edited: Mar 11, 2010 3. Mar 11, 2010 ### HallsofIvy \"$Cis(\\theta)$\" is engineering notation for \"$cos(\\theta)+ i sin(\\theta)$\" which mathematicians tend to write as $e^{i\\theta}$. The important thing about that notation is that $(Cis(\\theta)*Cis(\\phi)= Cis(\\theta+ \\phi)$. So you have correctly deduced that $(2Cis(\\pi/6))(3Cis(\\pi/12)= 6 Cis(3\\pi/12)= 6 Cis(\\pi/4)$ Now, just use the definition: $6 Cis(\\pi/4)= 6 cos(\\pi/4)+ 6i sin(\\pi/4)$.", "# Simplifying and then expressing complex numbers in cartesian form ## Homework Statement (2 CIS (pi/6))*(3 CIS (pi/12)) ## Homework Equations Also what is CIS? I believe it's Cos+i*sin but how do you use it? ## The Attempt at a Solution i simplified it to 6 CIS (pi/12) How do i turn it into cartesian? tiny-tim Homework Helper Hi Stripe! (have a pi: π ) I've never seen \"CIS\" before, but I'll guess you're right, and that it's cos + i*sin. Now use De Moivre's theorem … cosθ + isinθ = e And Cartesian form simply means in the form x + iy (as opposed to polar form, which is in the form re ) (and no, it's not 6 CIS (π/12)) Last edited: HallsofIvy \"$Cis(\\theta)$\" is engineering notation for \"$cos(\\theta)+ i sin(\\theta)$\" which mathematicians tend to write as $e^{i\\theta}$. The important thing about that notation is that $(Cis(\\theta)*Cis(\\phi)= Cis(\\theta+ \\phi)$. So you have correctly deduced that $(2Cis(\\pi/6))(3Cis(\\pi/12)= 6 Cis(3\\pi/12)= 6 Cis(\\pi/4)$ Now, just use the definition: $6 Cis(\\pi/4)= 6 cos(\\pi/4)+ 6i sin(\\pi/4)$.", "+0 # help +1 50 1 The number $$\\text{cis } 75^\\circ + \\text{cis } 83^\\circ + \\text{cis } 91^\\circ + \\dots + \\text{cis } 147^\\circ$$ is expressed in the form $$r \\, \\text{cis } \\theta$$, where $$0 \\le \\theta < 360^\\circ$$. Find $$\\theta$$ in degrees. Dec 20, 2019", "# Cis Cis notation is a polar notation for complex numbers. For all complex numbers $z$, we can write $z=r\\mathrm{cis }(\\theta)=r\\cos \\theta + ir\\sin \\theta$. Notice that $\\mathrm{cis}$ is made up by the first letter of $\\cos$, $i$, and the first letter of $\\sin$. Once one gets used to the notation, it is almost always preferred to write $re^{i\\theta}$ rather than $r\\mathrm{cis }(\\theta)$, as Euler's formula states that $e^{i\\theta} = \\cos \\theta + i \\sin \\theta.$ This is so that one can more naturally use the properties of the complex exponential. One important example is De Moivre's Theorem, which states that $\\mathrm{cis}(r\\theta) = (\\mathrm{cis}(\\theta))^r.$ This is more easily understood in the complex exponential form: $e^{i(r\\theta)} = (e^{i\\theta})^r.$", "form. 35) Find $$z^3$$ when $$z=5\\mathbf{cis} (45^{\\circ})$$. $$125\\mathbf{cis} (135^{\\circ})$$ 36) Find $$z^4$$ when $$z=2\\mathbf{cis} (70^{\\circ})$$. 37) Find $$z^2$$ when $$z=3\\mathbf{cis} (120^{\\circ})$$. $$9\\mathbf{cis} (240^{\\circ})$$ 38) Find $$z^2$$ when $$z=4\\mathbf{cis}\\left(\\dfrac{\\pi }{4} \\right)$$. 39) Find $$z^4$$ when $$z=\\mathbf{cis}\\left(\\dfrac{3\\pi }{16} \\right)$$. $$\\mathbf{cis}\\left(\\dfrac{3\\pi }{4} \\right)$$ 40) Find $$z^3$$ when $$z=3\\mathbf{cis}\\left(\\dfrac{5\\pi }{3} \\right)$$. For the exercises 41-45, evaluate each root. 41) Evaluate the cube root of $$z$$ when $$z=27\\mathbf{cis} (240^{\\circ})$$. $$3\\mathbf{cis}(80^{\\circ}), 3\\mathbf{cis}(200^{\\circ}), 3\\mathbf{cis}(320^{\\circ})$$ 42) Evaluate the square root of $$z$$ when $$z=16\\mathbf{cis} (100^{\\circ})$$. 43) Evaluate the cube root of $$z$$ when $$z=32\\mathbf{cis} \\left(\\dfrac{2\\pi }{3} \\right)$$. $$2\\sqrt[3]{4}\\mathbf{cis} \\left(\\dfrac{2\\pi }{9} \\right), 2\\sqrt[3]{4}\\mathbf{cis} \\left(\\dfrac{8\\pi }{9} \\right), 2\\sqrt[3]{4}\\mathbf{cis} \\left(\\dfrac{14\\pi }{9} \\right)$$ 44) Evaluate the square root of $$z$$ when $$z=32\\mathbf{cis} (\\pi )$$. 45) Evaluate the cube root of $$z$$ when $$z=8\\mathbf{cis} \\left(\\dfrac{7\\pi }{4} \\right)$$. $$2\\sqrt{2}\\mathbf{cis} \\left(\\dfrac{7\\pi }{8} \\right), 2\\sqrt{2}\\mathbf{cis} \\left(\\dfrac{15\\pi }{8} \\right)$$ ### Graphical For the exercises 46-55, plot the complex number in the complex plane. 46) $$2+4i$$ 47) $$-3-3i$$ 48) $$5-4i$$ 49) $$-1-5i$$ 50) $$3+2i$$ 51) $$2i$$ 52) $$-4$$ 53) $$6-2i$$ 54) $$-2+i$$ 55) $$1-4i$$ ### Technology For the exercises 56-, find all answers rounded to the nearest hundredth. 56) Use the rectangular to polar feature on the graphing calculator to change $$5+5i$$ to polar form. 57) Use the rectangular to polar feature on the graphing calculator to change $$3-2i$$ to polar form. $$3.61e^{-0.59i}$$ 58) Use the", "# Solving a quadratic equation with complex coefficient Express $z4$=-$\\sqrt{3}$+i in polar form. Hence solve the equation $Z^2$=$z4$ for $z$ a complex number. You may leave the answer in polar form. My answer: $z4$ in polar form is 2cis-30$^{\\circ}$ and thats as far as I have gotten. I have seen this question asked on other sites, but the answers with working solutions are too complex for the level this is at. This is a complex number topic assignment on a basic mathematics course, so I can only solve this question using the basic material that has been covered already in the course. There is an example in the notes where they use De Moivre's theorem to solve it. • $$-\\sqrt3+i=2\\,\\text{cis}\\left(\\arctan\\frac1{-\\sqrt3}\\right)=2\\,\\text{cis}\\,( 150^\\circ)$$ and not $\\;-30^\\circ\\;$ , since here cosine is negative and sine positive. – Timbuc Jun 30 '15 at 13:03 • $(a{\\rm\\ cis\\ }b)^2=a^2{\\rm\\ cis\\ }2b$ should give you an idea for how to take square roots of a number given in cis form. – Gerry Myerson Jun 30 '15 at 13:12 We have $z_4=2\\mbox{cis}(150^\\circ+360^\\circ\\times k), \\ k=...-2,-1,0,1,2,...$, that is $\\mbox{cis(.)}$ is periodic with period $360^\\circ$. Then $z^2=z_4$ requires that $$z=\\sqrt{2}\\mbox{cis}\\left(\\frac{150^\\circ+360^\\circ \\times k}{2}\\right)\\ k=...-2,-1,0,1,2,...$$ but there are only two distinct values for $z$ which correspond to $k=0,1$. Hence the two solutions are:$z=\\sqrt{2}\\mbox{cis}(75^\\circ)$ and $z=\\sqrt{2}\\mbox{cis}(225^\\circ)$", "on cis on my final. At first I thought the folks who wrote the book made a typo for cos, given the \"I\" key on the keyboard is so close to the \"O\" key. Many thanks in advance to quick help! Hi Wiseguy... \"cis\" means \"cos+isin\" so $cis(\\theta)=cos\\theta+isin(\\theta)$ For a complex number a+bi...\"a\" is the real part and \"bi\" is the imaginary part. $5cis300^o=5\\left(cos300^o+isin300^o\\right)$ hence ... $a=5cos300^o$ and $b=5sin300^o$ $6cis100^o2cis670^o=6\\left(cos100^o+isin100^o\\right )2\\left(cos70^o+isin70^o\\right)$ $=12\\left(cos100^ocos70^o+icos100^osin70^o+icos70^o sin100^o-sin100^osin70^o\\right)$ Now combine the real parts to get \"a\" and the imaginary parts to get \"b\" after applying the appropriate trigonometric identities to them. 4. Originally Posted by tonio You have to (1) understand that $cis(\\theta):=\\cos\\theta+i\\sin\\theta$ , (2) you have to know pretty well your trigonometry and (3) you have to remember the definition $a+ib=\\sqrt{a^2+b^2}\\,cis\\theta\\,,\\,\\,\\theta=\\arcta n\\frac{b}{a}\\,,\\,\\,a,b\\in\\mathbb{R}\\,,\\,a\\neq 0$ , and if $a=0\\,,\\,\\,\\theta=\\frac{\\pi}{2}+k\\pi\\,,\\,\\,k\\in\\mat hbb{Z}$. So let's do one, and you do the other one: $5\\,cis(300^\\circ)=5(\\cos 300^\\circ+i\\sin 300^\\circ)=5\\left(\\frac{1}{2}-\\frac{\\sqrt{3}}{2}i\\right) = \\frac{5}{2}-\\frac{5\\sqrt{3}}{2}i$ Tonio Okay, I ended up with 12cis(-0.984807753)° ...this is the form they were asking for in the second question: \"Leave your answer in the form ___cis___°\" 5. Originally Posted by wiseguy Okay, I ended up with 12cis(-0.984807753)° ...this is the form they were asking for in the second question: \"Leave your answer in the form ___cis___°\" From $12\\left(cos100^ocos70^o-sin100^osin70^o+i[cos100^osin70^o+cos70^osin100^o]\\right)$ you should use $cosAcosB-sinAsinB=cos(A+B)$ and $cosAsinB+cosBsinA=sin(A+B)$ where $A=100^0$ and $B=70^o$ to", "and suppose that P=r(costheta+isintheta) where $0 \\lt r$ and $0 \\leq \\theta \\lt 360$ find $\\theta$ • is 107 • is 276 • is 840 • cannot be determined from the given information ### Key Concepts Equations Complex Numbers Integers AIME, 1996, Question 11 Complex Numbers from A to Z by Titu Andreescue ## Try with Hints First hint here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\\frac{(z^{5}-1)}{(z-1)}$=$\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0 Second Hint gives $z^{5}=1 for z\\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \\neq 1$ gives z=$\\frac{1+-(-3)^\\frac{1}{2}}{2}$=$cis60,300$ where cis$\\theta$=cos$\\theta$+isin$\\theta$ Final Step taking $0 \\lt theta \\lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276. . ## Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed." ]

[ "\\end{pmatrix}$ where $\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$ The entries in the introduction were given by: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\ (\\mathbf{a}_2 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_2 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_2 \\cdot \\mathbf{b}_p) \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ (\\mathbf{a}_n \\cdot \\mathbf{b}_1) & (\\mathbf{a}_n \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_n \\cdot \\mathbf{b}_p) \\end{pmatrix}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: $\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{A}\\mathbf{b}_1 & \\mathbf{A}\\mathbf{b}_2 & \\dots & \\mathbf{A}\\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1\\mathbf{B} \\\\ \\mathbf{a}_2\\mathbf{B}\\\\ \\vdots\\\\ \\mathbf{a}_n\\mathbf{B} \\end{pmatrix}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1). Matrix product (in terms of outer product) An alternative method results when", "\\\\ a_n b_1 & a_n b_2 & \\cdots & a_n b_n \\\\ \\end{pmatrix}.$$ Matrix product (in terms of inner product) Suppose that the first n×m matrix A is decomposed into its row vectors ai, and the second m×p matrix B into its column vectors bi: [6] $$\\mathbf{A} = \\begin{pmatrix} A_{1 1} & A_{1 2} & \\cdots & A_{1 m} \\\\ A_{2 1} & A_{2 2} & \\cdots & A_{2 m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A_{n 1} & A_{n 2} & \\cdots & A_{n m} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix},\\quad \\mathbf{B} = \\begin{pmatrix} B_{1 1} & B_{1 2} & \\cdots & B_{1 p} \\\\ B_{2 1} & B_{2 2} & \\cdots & B_{2 p} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ B_{m 1} & B_{m 2} & \\cdots & B_{m p} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\cdots & \\mathbf{b}_p \\end{pmatrix}$$ where $$\\mathbf{a}_i = \\begin{pmatrix}A_{i1} & A_{i2} & \\cdots & A_{im} \\end{pmatrix},\\quad \\mathbf{b}_i = \\begin{pmatrix}B_{1i} & B_{2i} & \\cdots & B_{mi}\\end{pmatrix}^\\mathrm{T}$$ The entries in the introduction were given by: $$\\mathbf{AB} = \\begin{pmatrix} \\mathbf{a}_1 \\\\ \\mathbf{a}_2 \\\\ \\vdots \\\\ \\mathbf{a}_n \\end{pmatrix} \\begin{pmatrix} \\mathbf{b}_1 & \\mathbf{b}_2 & \\dots & \\mathbf{b}_p \\end{pmatrix} = \\begin{pmatrix} (\\mathbf{a}_1 \\cdot \\mathbf{b}_1) & (\\mathbf{a}_1 \\cdot \\mathbf{b}_2) & \\dots & (\\mathbf{a}_1 \\cdot \\mathbf{b}_p) \\\\", "gives $q=-5p$. We now know that $\\mathbf{e}= t\\begin{pmatrix}1\\\\-5 \\\\ -2\\end{pmatrix}$. We also know $\\big \\vert\\mathbf{e}\\big \\vert = \\sqrt{30}$ and $p$ is negative, hence $t=-1$ and we have $\\mathbf{e}= \\begin{pmatrix}-1\\\\5 \\\\ 2\\end{pmatrix}$ and $\\mathbf{f}= \\begin{pmatrix}5\\\\2 \\\\-1\\end{pmatrix}$. It might be tempting to proceed with vectors perpendicular to $\\pi$ and $\\pi'$ being $\\mathbf{b} \\times \\mathbf{c}$ and $\\mathbf{e} \\times \\mathbf{f}$. $\\mathbf{b}\\times \\mathbf{c} = \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ 2&-1&-1 \\\\ -1&2&1\\end{matrix}\\right \\vert = \\mathbf{i}-\\mathbf{j}+3\\mathbf{k}.$ $\\mathbf{e}\\times \\mathbf{f}= \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ p&q&r \\\\ q&r&p\\end{matrix}\\right \\vert = (qp-r^2)\\mathbf{i}+(rq-p^2)\\mathbf{j}+(pr-q^2)\\mathbf{k}.$ However, the resulting set of simultaneous equations is very difficult to solve. By adding them we can then simplify using $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ in other words $p^2+q^2+r^2=30$. We can think about $(p+q+r)^2$ to continue the manipulation. Try it if you like! Show that $\\pi'$ is the reflection of $\\pi$ in the origin. Using the form of the plane equation $\\mathbf{r}.\\mathbf{n}=\\mathbf{a}.\\mathbf{n}$ with $\\mathbf{n}$ as the normal vector we found earlier, we have the equation of $\\pi$ as $x - y + 3z = -2.$ Similarly using the point $\\mathbf{d}$, the plane $\\pi'$ is $x - y + 3z = 2.$ If $(x_1,y_1,z_1)$ is a point on $\\pi$ then $x_1 - y_1 + 3z_1 = -2$. Multiplying by $-1$ gives $(-x_1) -(-y_1) + 3(-z_1) = 2$. So the point $(-x_1,-y_1,-z_1)$ lies on $\\pi'$. Since every point on", "that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is the projection matrix $P=\\frac{1}{\\mathbf{a}^T\\mathbf{a}}\\mathbf{a}\\mathbf{a}^T,$ which is an $n\\times n$ matrix. To this end, Prof. Strang writes the projection vector as $\\mathbf{p}=\\mathbf{a}\\,x$. Now, you should see $\\mathbf{p}$ and $\\mathbf{a}$ as $n\\times1$ matrices and $x$ as a one dimensional vector or a $1\\times1$ matrix, so that $\\mathbf{a}$ and $x$ can be multiplied. Thus, we have $x\\,\\mathbf{a}=\\mathbf{a}\\,x.$ This identity does not mean that $\\mathbf{a}$ and $x$ are two matrices that commute. Just read the left and right sides as stated above. It is true that the notation may be a bit misleading. Perhaps it would be better to write $(x)$ instead of $x$ if $x$ should be seen as a $1\\times 1$ matrix (parentheses are used to delimit matrices). For example, $4\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix} =\\begin{pmatrix} 4 \\\\ 8 \\\\ 12\\end{pmatrix} =\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix}(4).$ The advantage of expressing $\\mathbf{p}$ as $\\mathbf{a}x$ is that generalization is then possible. If $S$ is spanned by the linearly independent vectors $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, the orthogonal projection $\\mathbf{p}$ of a vector $\\mathbf{b}$ onto $S$ should be a linear combination of $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, that is, $\\mathbf{p}=x_1\\mathbf{a}_1+\\cdots+x_k\\mathbf{a}_k=AX$ with $A=\\left(\\mathbf{a}_1 \\vert\\ldots \\vert\\mathbf{a}_k\\right)$ and $X=\\begin{pmatrix}x_1 \\\\ \\vdots \\\\ x_k\\end{pmatrix}$ The reasoning in Prof. Strang’s lecture would then show that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is now the projection matrix $P=A(A^TA)^{-1}A^T.$ Returning to your question, you", "{A}}x_{2}\\\\{\\textbf {A}}x_{3}\\\\...\\\\{\\textbf {A}}x_{n}\\end{pmatrix}}}$, you'll get ${\\displaystyle \\nabla f({\\textbf {x}})={\\begin{pmatrix}{\\textbf {A}}\\\\{\\textbf {A}}\\\\{\\textbf {A}}\\\\...\\\\{\\textbf {A}}\\end{pmatrix}}}$. This is just A! Indeed, that's why ${\\displaystyle {\\frac {d}{d{\\textbf {x}}}}{\\textbf {A}}{\\textbf {x}}={\\textbf {A}}}$. ### A first step towards matrices Now we return back to the other problem: ${\\displaystyle {\\frac {d}{d{\\textbf {A}}}}{\\textbf {A}}{\\textbf {x}}}$. Let's assume that A is a 2 by 2 matrix, and represent A as ${\\displaystyle {\\textbf {A}}={\\begin{pmatrix}a_{11}&a_{12}\\\\a_{21}&a_{22}\\end{pmatrix}}}$. Using the same notation for x, perform matrix multiplication: ${\\displaystyle {\\textbf {A}}{\\textbf {x}}={\\begin{pmatrix}a_{11}&a_{12}\\\\a_{21}&a_{22}\\end{pmatrix}}{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}={\\begin{pmatrix}a_{11}\\times x_{1}+a_{12}\\times x_{2}\\\\a_{21}\\times x_{1}+a_{22}\\times x_{2}\\end{pmatrix}}}$ Take the partial derivatives with respect to each of the elements of A (this is equivalent to finding the Jacobian). For ${\\displaystyle 1\\leq i,j\\leq 2}$, find ${\\displaystyle {\\frac {\\partial }{\\partial A_{ij}}}}$: ${\\displaystyle {\\frac {\\partial }{\\partial A_{11}}}={\\begin{pmatrix}x_{1}\\\\0\\end{pmatrix}}}$, ${\\displaystyle {\\frac {\\partial }{\\partial A_{12}}}={\\begin{pmatrix}x_{2}\\\\0\\end{pmatrix}}}$, ${\\displaystyle {\\frac {\\partial }{\\partial A_{21}}}={\\begin{pmatrix}0\\\\x_{1}\\end{pmatrix}}}$ and ${\\displaystyle {\\frac {\\partial }{\\partial A_{22}}}={\\begin{pmatrix}0\\\\x_{2}\\end{pmatrix}}}$ But then what do you do with that? How do you \"combine\" the result? Clearly, we are missing something. ### Dimensions of ∇f Let's take a step back and ask the question: given a vector ${\\displaystyle {\\textbf {f}}={\\textbf {A}}{\\textbf {x}}}$, what should be the dimension of ${\\displaystyle \\nabla {\\textbf {f}}}$? Consider the example above. We have two variables: A with 4 elements (2x2) and x with 2 elements, and we want to find ${\\displaystyle {\\frac {d{\\textbf {f}}}{d{\\textbf {A}}}}}$. It is straightforward to show that the", "\\begin{pmatrix} a_1 & a_2 & a_3 & ... & a_m \\end{pmatrix}$ $\\textbf u = \\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3\\\\...\\\\ b_n \\end{pmatrix}$ $\\textbf v^T \\cdot \\textbf u = \\begin{pmatrix} a_1 & a_2 & a_3& ... & a_m \\end{pmatrix} \\cdot \\begin{pmatrix} b_1 \\\\ b_2 \\\\b_3 \\\\ ... \\\\b_n \\end{pmatrix}$ $\\textbf v^T \\cdot \\textbf u = ( a_1 b_1 + a_2 b_2 + a_3 b_3 + ... + a_m b_n)$ Notice that it only works if $m = n$ and it always results in a matrix of order $1 \\times 1$. The outer product of these two vectors is $\\textbf v \\cdot \\textbf u^T$. $\\textbf v \\cdot \\textbf u^T = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ a_3 \\\\ ... \\\\ a_m \\end{pmatrix} \\cdot \\begin{pmatrix} b_1 & b_2 & b_3 & ... & b_n \\end{pmatrix}$ $\\textbf v \\cdot \\textbf u^T = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & a_2 b_3 &...&a_1 b_n\\\\a_2 b_1 & a_2 b_2 & a_2 b_3 &...& a_2 b_n\\\\a_3 b_1 & a_3 b_2 & a_3 b_3 &...&a_3 b_n\\\\...& ... & ... & ... & ... \\\\a_m b_1 & a_m b_2 & a_m b_3 & ... & a_m b_n \\end{pmatrix}$ If you have a column vector of order $m \\times 1$ and a row vector of order $1 \\times n$ respectively, their outer product will be a matrix of order", "\\end{pmatrix}\\nonumber$ with $$n=1$$ • A row vector:$\\begin{pmatrix} 3 &-2 &4 \\\\ \\end{pmatrix}\\nonumber$ with $$m=1$$ • The identity matrix:$\\begin{pmatrix} 1 &0 &0 \\\\ 0 &1 &0 \\\\ 0&0 &1 \\end{pmatrix}\\nonumber$ with $$a_{ij}=\\delta_{i,j}$$, where $$\\delta_{i,j}$$ is a function defined as $$\\delta_{i,j}=1$$ if $$i=j$$ and $$\\delta_{i,j}=0$$ if $$i\\neq j$$. • A diagonal matrix:$\\begin{pmatrix} a &0 &0 \\\\ 0 &b &0 \\\\ 0&0 &c \\end{pmatrix}\\nonumber$ with $$a_{ij}=c_i \\delta_{i,j}$$. • An upper triangular matrix:$\\begin{pmatrix} a &b &c \\\\ 0 &d &e \\\\ 0&0 &f \\end{pmatrix}\\nonumber$ All the entries below the main diagonal are zero. • A lower triangular matrix:$\\begin{pmatrix} a &0 &0 \\\\ b &c &0 \\\\ d&e &f \\end{pmatrix}\\nonumber$ All the entries above the main diagonal are zero. • A triangular matrix is one that is either lower triangular or upper triangular. ### The Trace of a Matrix The trace of an $$n\\times n$$ square matrix $$\\mathbf{A}$$ is the sum of the diagonal elements, and formally defined as $$Tr( \\mathbf{A})=\\sum_{i=1}^{n}a_{ii}$$. For example, $\\mathbf{A}=\\begin{pmatrix} 3 &-2 &4 \\\\ 5 &3i &3 \\\\ -i & 1/2 &9 \\end{pmatrix}\\; ; Tr(\\mathbf{A})=12+3i \\nonumber$ ### Singular and Nonsingular Matrices A square matrix with nonzero determinant is called nonsingular. A matrix whose determinant is zero is called singular. (Note that you cannot calculate the determinant of a non-square matrix). For a 2x2 matrix, $\\mathbf{B}=\\begin{pmatrix} a & b \\\\", "of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^{\\rm t} {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}. \\end{align*} ##### Example8.19 Suppose that the words to be encoded consist of all binary 3-tuples and that our encoding scheme is even-parity. To encode an arbitrary 3-tuple, we add a fourth bit to obtain an even number of 1s. Notice that an arbitrary $n$-tuple ${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^{\\rm t}$ has an even number of 1s exactly when $x_1 + x_2 + \\cdots + x_n = 0$; hence, a 4-tuple ${\\mathbf x} = (x_1, x_2, x_3, x_4)^{\\rm t}$ has an even number of 1s if $x_1+ x_2+ x_3+ x_4 = 0$, or \\begin{equation*}{\\mathbf x} \\cdot {\\mathbf 1} = {\\mathbf x}^{\\rm t} {\\mathbf 1} = \\begin{pmatrix} x_1 & x_2 & x_3 & x_4 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\\\ 1 \\end{pmatrix} = 0.\\end{equation*} This example leads us to hope that there is a connection between matrices and coding theory. Let ${\\mathbb M}_{m \\times n}({\\mathbb Z}_2)$ denote the set of all $m \\times n$ matrices with entries in ${\\mathbb Z}_2$. We do matrix operations", "# Thread: Left inverse of matrix 1. ## Left inverse of matrix Consider the matrix $\\mathbf{C} = \\begin{pmatrix}1 & -1\\\\ 1&1 \\\\ 0&1 \\end{pmatrix}$ Find a left-inverse of C. Attempt: $\\begin{pmatrix}x1 & y1 & z1\\\\ x2&y2 &z2 \\end{pmatrix}\\begin{pmatrix}1 & -1\\\\ 1&1\\\\ 0&1 \\end{pmatrix} = \\begin{pmatrix}1 &0\\\\ 0&1\\end{pmatrix}$ $\\begin{pmatrix}x1 + y1 & -x1 +y1 + z1 \\\\ x2 + y2& -x2 + y2 + z2 \\end{pmatrix}= \\begin{pmatrix}1 &0\\\\ 0&1\\end{pmatrix}$ I know that you suppose to somehow use gauss reduction, and thought about gauss reducing: $x1 + y1 = 1$ $-x1 + y1 + z1 = 0$ and $x2 + y2 = 0$ $-x2 + y2 + z2 = 1$ but didnt get any luck 2. Only square matrices have inverses. To solve a matrix equation of the form $\\mathbf{Y} = \\mathbf{AX}$ for $\\mathbf{X}$ where $\\mathbf{A}$ is not square, you need to premultiply both sides by $\\mathbf{A}^T$ first. $\\mathbf{Y} = \\mathbf{AX}$ $\\mathbf{A}^T\\mathbf{Y} = \\mathbf{A}^T\\mathbf{AX}$. Now $\\mathbf{A}^T\\mathbf{A}$ is a square matrix, so you can premultiply both sides by $(\\mathbf{A}^T\\mathbf{A})^{-1}$. $\\mathbf{A}^T\\mathbf{Y} = \\mathbf{A}^T\\mathbf{AX}$ $(\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{Y} = (\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{AX}$ $(\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{Y} = \\mathbf{IX}$ $(\\mathbf{A}^T\\mathbf{A})^{-1}\\mathbf{A}^T\\mathbf{Y} = \\mathbf{X}$. 3. Originally Posted by Prove It Only square matrices have inverses. Why is this so? In my notes it says that a matrix that is not square can have left-inverses or right-inverses, but it can't have both. 4. You have", "3$ matrix $\\mathbf{A} = \\begin{pmatrix}1&0&2\\\\3&1&0\\\\-1&1&4\\end{pmatrix}$. Workbook This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples. More Support You can get one-to-one support from Maths-Aid.", "= i, \\quad \\mathbf{A} = \\begin{pmatrix} i & 0 \\\\ 0 & j \\\\ \\end{pmatrix}$ $i\\begin{pmatrix} i & 0 \\\\ 0 & j \\\\ \\end{pmatrix} = \\begin{pmatrix} i^2 & 0 \\\\ 0 & ij \\\\ \\end{pmatrix} = \\begin{pmatrix} -1 & 0 \\\\ 0 & k \\\\ \\end{pmatrix} \\ne \\begin{pmatrix} -1 & 0 \\\\ 0 & -k \\\\ \\end{pmatrix} = \\begin{pmatrix} i^2 & 0 \\\\ 0 & ji \\\\ \\end{pmatrix} = \\begin{pmatrix} i & 0 \\\\ 0 & j \\\\ \\end{pmatrix}i\\,,$ where i, j, k are the quaternion units. The non-commutativity of quaternion multiplication prevents the transition of changing ij = +k to ji = −k. Matrix product (two matrices) Assume two matrices are to be multiplied (the generalization to any number is discussed below). General definition of the matrix product Arithmetic process of multiplying numbers (solid lines) in row i in matrix A and column j in matrix B, then adding the terms (dashed lines) to obtain entry ij in the final matrix. If A is an n × m matrix and B is an m × p matrix, $\\mathbf{A}=\\begin{pmatrix} A_{11} & A_{12} & \\cdots & A_{1m} \\\\ A_{21} & A_{22} & \\cdots & A_{2m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A_{n1} & A_{n2} & \\cdots & A_{nm} \\\\ \\end{pmatrix},\\quad\\mathbf{B}=\\begin{pmatrix} B_{11} & B_{12} & \\cdots", "### Solving Matrix Equations #### Definition Given matrices $\\mathbf{A}$ and $\\mathbf{B}$, we can solve the equation $\\mathbf{A} \\mathbf{X} = \\mathbf{B}$ for the unknown matrix $\\mathbf{X}$ when the matrix inverse of $\\mathbf{A}$ exists. We do this by multiplying both sides of the equation on the left by $\\mathbf{A}^{-1}$, providing it exists, to give \\begin{align} \\mathbf{A} \\mathbf{X} &= \\mathbf{B} \\\\ \\mathbf{A}^{-1} \\mathbf{A} \\mathbf{X} &= \\mathbf{A}^{-1} \\mathbf{B} \\\\ \\mathbf{X} &= \\mathbf{A}^{-1} \\mathbf{B} \\end{align} as $\\mathbf{A}^{-1} \\mathbf{A} = \\mathbf{I}$ and multiplying any matrix by the identity matrix leaves it unaltered. It is important that both sides of the equation are multiplied on the left by $\\mathbf{A}^{-1}$ because matrix multiplication does not commute - multiplying on the right could give a different result. #### Worked Example ###### Example 1 Let \\begin{align} \\mathbf{A} &= \\begin{pmatrix}1&2\\\\3&-5\\end{pmatrix}, \\\\ \\mathbf{B} &= \\begin{pmatrix}4\\\\1\\end{pmatrix}. \\end{align} Solve the equation $\\mathbf{A} \\mathbf{X} = \\mathbf{B}$. ###### Solution We have the equation $\\begin{pmatrix}1&2\\\\3&-5\\end{pmatrix}\\mathbf{X} = \\begin{pmatrix}4\\\\1\\end{pmatrix}$ To get $\\mathbf{X}$ on its own, multiply both sides by the inverse of $\\mathbf{A}$. Note that $\\lvert \\mathbf{A} \\rvert = -5-6 = -11$. Therefore, $\\mathbf{A}^{-1} = -\\dfrac{1}{11}\\begin{pmatrix} -5&-2\\\\-3&1\\end{pmatrix}$ So $\\mathbf{X} = \\mathbf{A}^{-1} \\mathbf{B} = -\\dfrac{1}{11}\\begin{pmatrix} -5&-2\\\\-3&1\\end{pmatrix}\\begin{pmatrix}4\\\\1\\end{pmatrix}$ Now, use matrix multiplication to find $\\mathbf{X}$. \\begin{align} \\mathbf{X} &= -\\frac{1}{11}\\begin{pmatrix} -5&-2\\\\-3&1\\end{pmatrix}\\begin{pmatrix}4\\\\1\\end{pmatrix}\\\\ &= -\\frac{1}{11}\\begin{pmatrix}(-20)+(-2)\\\\(-12)+1\\end{pmatrix}\\\\ &= -\\frac{1}{11}\\begin{pmatrix}-22\\\\-11\\end{pmatrix}\\\\ &= \\begin{pmatrix}2\\\\1\\end{pmatrix} \\end{align} #### Simultaneous Equations We can apply this method to solving a system of", "The matrix product itself can be expressed in terms of inner product. Suppose that the first n × m matrix A is decomposed into its row vectors ai, and the second m × p matrix B into its column vectors bi:[1] ${\\displaystyle \\mathbf {A} ={\\begin{pmatrix}A_{11}&A_{12}&\\cdots &A_{1m}\\\\A_{21}&A_{22}&\\cdots &A_{2m}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\A_{n1}&A_{n2}&\\cdots &A_{nm}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}},}$ ${\\displaystyle \\mathbf {B} ={\\begin{pmatrix}B_{11}&B_{12}&\\cdots &B_{1p}\\\\B_{21}&B_{22}&\\cdots &B_{2p}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\B_{m1}&B_{m2}&\\cdots &B_{mp}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\cdots &\\mathbf {b} _{p}\\end{pmatrix}}}$ where ${\\displaystyle \\mathbf {a} _{i}={\\begin{pmatrix}A_{i1}&A_{i2}&\\cdots &A_{im}\\end{pmatrix}}\\,,\\quad \\mathbf {b} _{i}={\\begin{pmatrix}B_{1i}\\\\B_{2i}\\\\\\vdots \\\\B_{mi}\\end{pmatrix}}}$ Then: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\dots &\\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{p})\\\\(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{p})\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{p})\\end{pmatrix}}}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {A} \\mathbf {b} _{1}&\\mathbf {A} \\mathbf {b} _{2}&\\dots &\\mathbf {A} \\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {a} _{1}\\mathbf {B} \\\\\\mathbf {a} _{2}\\mathbf {B} \\\\\\vdots \\\\\\mathbf {a} _{n}\\mathbf {B} \\end{pmatrix}}}$ These decompositions are particularly useful for matrices that", "element $x_{i,0}$ which always equals to $1$. So our required function becomes $w_0 x_{i,0} + w_1 x_{i,1} + \\dots w_d x_{i,d}$. We denote it as $h_{\\mathbf{w}}(\\mathbf{x}_i)$. Using matrices, we can write $h_{\\mathbf{w}}(\\mathbf{x}_i)$ in a much more compact form. Conventionally, we use column matrices to represent vectors. Thus we have $\\mathbf{w} = \\begin{pmatrix} w_0\\\\ w_1\\\\ w_2\\\\ \\vdots\\\\ w_d \\end{pmatrix}, \\hspace{1cm}\\mathbf{x}_i = \\begin{pmatrix} x_{i,0}\\\\ x_{i,1}\\\\ x_{i,2}\\\\ \\vdots\\\\ x_{i,d} \\end{pmatrix}$ Our function $h_{\\mathbf{w}}(\\mathbf{x}_i)$ thus can be written as $\\mathbf{w}^\\intercal\\mathbf{x}_i$, or equivalently, as $\\mathbf{x}_i^\\intercal\\mathbf{w}$. If there are $n$ data points $\\mathbf{x}_1, \\mathbf{x}_2, \\dots, \\mathbf{x}_n$, the outputs corresponding to all these can be kept together in a column matrix as follows: $\\begin{pmatrix} \\mathbf{x}_1^\\intercal\\mathbf{w}\\\\ \\mathbf{x}_2^\\intercal\\mathbf{w}\\\\ \\vdots\\\\ \\mathbf{x}_n^\\intercal\\mathbf{w} \\end{pmatrix}$ Note that this column matrix can be decomposed into the following product $\\begin{pmatrix} \\mathbf{x}_1^\\intercal\\mathbf{w}\\\\ \\mathbf{x}_2^\\intercal\\mathbf{w}\\\\ \\vdots\\\\ \\mathbf{x}_n^\\intercal\\mathbf{w} \\end{pmatrix} = \\begin{pmatrix} —\\mathbf{x}_1^\\intercal— \\\\ —\\mathbf{x}_2^\\intercal—\\\\ \\vdots\\\\ —\\mathbf{x}_n^\\intercal— \\end{pmatrix} \\times \\mathbf{w}$ It is important to keep in mind that even though the last product looks like a scalar multiplication, $\\mathbf{w}$ is in fact a matrix, and that the LHS and RHS are equal nevertheless. Understandably, we call the first matrix in the decomposition $\\mathbf{X}$. It is an $n\\times (d+1)$ matrix, each of whose rows represents a data point. Thus we can compute the outputs $h_\\mathbf{w}(\\mathbf{x}_i)$ corresponding to all the data points $\\mathbf{x}_i$ using the single matrix product $\\mathbf{X}\\mathbf{w}$. ##", "matrix ${\\displaystyle M}$ represents some linear map, whose action is ${\\displaystyle {\\vec {v}}\\mapsto {\\vec {v}}}$, and is therefore the identity map. Example 1.3 With these bases for ${\\displaystyle \\mathbb {R} ^{2}}$, ${\\displaystyle B=\\langle {\\begin{pmatrix}2\\\\1\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\end{pmatrix}}\\rangle \\qquad D=\\langle {\\begin{pmatrix}-1\\\\1\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\end{pmatrix}}\\rangle }$ because ${\\displaystyle {\\rm {Rep}}_{D}(\\,{\\mbox{id}}({\\begin{pmatrix}2\\\\1\\end{pmatrix}}))={\\begin{pmatrix}-1/2\\\\3/2\\end{pmatrix}}_{D}\\qquad {\\rm {Rep}}_{D}(\\,{\\mbox{id}}({\\begin{pmatrix}1\\\\0\\end{pmatrix}}))={\\begin{pmatrix}-1/2\\\\1/2\\end{pmatrix}}_{D}}$ the change of basis matrix is this. ${\\displaystyle {\\rm {Rep}}_{B,D}({\\rm {id)={\\begin{pmatrix}-1/2&-1/2\\\\3/2&1/2\\end{pmatrix}}}}}$ We can see this matrix at work by finding the two representations of ${\\displaystyle {\\vec {e}}_{2}}$ ${\\displaystyle {\\rm {Rep}}_{B}({\\begin{pmatrix}0\\\\1\\end{pmatrix}})={\\begin{pmatrix}1\\\\-2\\end{pmatrix}}\\qquad {\\rm {Rep}}_{D}({\\begin{pmatrix}0\\\\1\\end{pmatrix}})={\\begin{pmatrix}1/2\\\\1/2\\end{pmatrix}}}$ and checking that the conversion goes as expected. ${\\displaystyle {\\begin{pmatrix}-1/2&-1/2\\\\3/2&1/2\\end{pmatrix}}{\\begin{pmatrix}1\\\\-2\\end{pmatrix}}={\\begin{pmatrix}1/2\\\\1/2\\end{pmatrix}}}$ We finish this subsection by recognizing that the change of basis matrices are familiar. Lemma 1.4 A matrix changes bases if and only if it is nonsingular. Proof For one direction, if left-multiplication by a matrix changes bases then the matrix represents an invertible function, simply because the function is inverted by changing the bases back. Such a matrix is itself invertible, and so nonsingular. To finish, we will show that any nonsingular matrix ${\\displaystyle M}$ performs a change of basis operation from any given starting basis ${\\displaystyle B}$ to some ending basis. Because the matrix is nonsingular, it will Gauss-Jordan reduce to the identity, so there are elementatry reduction matrices such that ${\\displaystyle R_{r}\\cdots R_{1}\\cdot M=I}$. Elementary matrices are invertible and their inverses are also elementary, so multiplying from the left", "\\begin{equation*} A = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\end{equation*} is in $GL_2( {\\mathbb R})$ if there exists a matrix $A^{-1}$ such that $A A^{-1} = A^{-1} A = I\\text{,}$ where $I$ is the $2 \\times 2$ identity matrix. For $A$ to have an inverse is equivalent to requiring that the determinant of $A$ be nonzero; that is, $\\det A = ad - bc \\neq 0\\text{.}$ The set of invertible matrices forms a group called the general linear group. The identity of the group is the identity matrix \\begin{equation*} I = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}\\text{.} \\end{equation*} The inverse of $A \\in GL_2( {\\mathbb R})$ is \\begin{equation*} A^{-1} = \\frac{1}{ad-bc} \\begin{pmatrix} d & -b \\\\ -c & a \\end{pmatrix}\\text{.} \\end{equation*} The product of two invertible matrices is again invertible. Matrix multiplication is associative, satisfying the other group axiom. For matrices it is not true in general that $AB = BA\\text{;}$ hence, $GL_2({\\mathbb R})$ is another example of a nonabelian group. ###### Example2.15. Let \\begin{align*} 1 & = \\begin{pmatrix} 1 & 0\\\\ 0 & 1 \\end{pmatrix} \\qquad I = \\begin{pmatrix} 0 & 1\\\\ -1 & 0 \\end{pmatrix}\\\\ J & = \\begin{pmatrix} 0 & i\\\\ i & 0 \\end{pmatrix} \\qquad K = \\begin{pmatrix} i & 0\\\\ 0 & -i \\end{pmatrix}\\text{,} \\end{align*} where $i^2", "\\text{L}}{\\mathbf{a}^{[1]}} = \\mathbf{W}^{[2]}.T \\frac{d \\text{L}}{d \\mathbf{z}^{[2]}}$$ Now we only missing $\\frac{d \\text{L}}{d \\mathbf{W}^{[1]}}$ and $\\frac{d \\text{L}}{db^{[1]}}$. Let's look how we calculate $\\mathbf{z}^{[1]}$: $$\\begin{matrix} (\\mathbf{W}^{[1]}) \\\\ \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\\\ 0 & 0 \\\\ \\end{bmatrix} \\end{matrix} \\begin{matrix} (\\mathbf{x}) \\\\ \\begin{bmatrix} 0 \\\\ 0 \\\\ \\end{bmatrix} \\end{matrix} + \\begin{matrix} (\\mathbf{b}^{[1]}) \\\\ \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ \\end{bmatrix} \\end{matrix} = \\begin{matrix} (\\mathbf{z}^{[1]}) \\\\ \\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ \\end{bmatrix} \\end{matrix}$$ Now, let's look at: $$\\frac{d \\text{L}}{d \\mathbf{W}^{[1](1,:)}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1](1)}} \\mathbf{x}.\\text{T}, \\quad \\frac{d \\text{L}}{d \\mathbf{W}^{[1](2,:)}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1](2)}} \\mathbf{x}.\\text{T}, \\quad \\frac{d \\text{L}}{d \\mathbf{W}^{[1](3,:)}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1](3)}} \\mathbf{x}.\\text{T}$$ or equivalently: $$\\frac{d \\text{L}}{d \\mathbf{W}^{[1]}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1]}} \\mathbf{x}.\\text{T}$$ and as expected: $$\\frac{d \\text{L}}{d \\mathbf{b}^{[1]}} = \\frac{d \\text{L}}{d \\mathbf{z}^{[1]}}$$ Let's apply the backprop altogether! ALPHA = 0.4 # learning rate def get_loss(y, a): return -1 * (y * np.log(a) + (1-y) * np.log(1-a)) def get_loss_numerically_stable(y, z): return -1 * (y * -1 * np.log(1 + np.exp(-z)) + (1-y) * (-z - np.log(1 + np.exp(-z)))) def get_gradients_loops(z1, a1, z2, a2, z3, a3, x, y, W1, b1, W2, b2, W3, b3): dz3 = a3 - y db3 = dz3 dW3 = dz3 * a2.T da2 = dz3 * W3.T dz2 = da2 * (a2 * (1-a2)) db2 = dz2 dW2 = np.matmul(dz2, a1.T)", "+0 # help 0 114 1 Let $\\mathbf{P}$ be the matrix that projects onto $\\mathbf{j}$: that is, we want $\\mathbf{P}$ to satisfy $\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$for all vectors $\\mathbf{v}$. Use the pictures below to calculate $\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$in that order, and enter them in as columns below. Then calculate the matrix $\\mathbf{P}$ that projects onto $\\mathbf{j}$. Jul 31, 2019 The matrix is $$\\begin{pmatrix} -1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$.", "$$e_1 := \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}\\quad\\textrm{and}\\quad e_2 := \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix}.$$ In this basis $$T = \\begin{pmatrix} 2 & -1 \\\\ 1 & 1 \\end{pmatrix}.$$ The basis transformation matrices from $$(e_1,e_2)$$ to $$B_1$$ or $$B_2$$ are given by $$B_1 = \\begin{pmatrix} 1 & 1 \\\\ 2 & 1 \\end{pmatrix}\\quad\\textrm{and}\\quad B_2 = \\begin{pmatrix} -1 & 2 \\\\ 1 & 1 \\end{pmatrix}.$$ The transformation matrix in each basis will be given by $$TB_1^{-1} \\quad\\textrm{and}\\quad TB_2^{-1}.$$ • Ah so there is no $[u]$ required. But if $T$ had a vector, say $u = (2,3)$ then it would have to be multiplied to that formula? Jun 10, 2020 at 3:36", "eigenvector $$\\mathbf y_2$$. Tis must satisfy $$\\mathbf y_2 \\in \\ker \\mathbf A^2 \\setminus \\ker \\mathbf A$$ and is linearly independent on $$\\mathbf x_2$$. We choose e.g. $$\\mathbf y_2=(0{,}1,0{,}0,0)^T$$ and calculate $$\\mathbf y_1=\\mathbf A\\mathbf y_2=(1{,}0,0{,}0,0)^T$$ (opět vlastní vektor $$\\mathbf A$$). Now we may assemble matrices $$\\mathbf R$$ (columns are $$\\mathbf y_1,\\mathbf y_2,\\mathbf x_1,\\mathbf x_2,\\mathbf x_3$$) and $$\\mathbf J$$ and claculate the inverse matrix $$\\mathbf R^{-1}$$ $$\\mathbf R= \\begin{pmatrix} 1&0&0&-1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&-1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$, $$\\mathbf J= \\begin{pmatrix} 0&1&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1\\\\ 0&0&0&0&0 \\end{pmatrix}$$, $$\\mathbf R^{-1}=\\begin{pmatrix} 1&0&0&1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$" ]

[ "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "gives $q=-5p$. We now know that $\\mathbf{e}= t\\begin{pmatrix}1\\\\-5 \\\\ -2\\end{pmatrix}$. We also know $\\big \\vert\\mathbf{e}\\big \\vert = \\sqrt{30}$ and $p$ is negative, hence $t=-1$ and we have $\\mathbf{e}= \\begin{pmatrix}-1\\\\5 \\\\ 2\\end{pmatrix}$ and $\\mathbf{f}= \\begin{pmatrix}5\\\\2 \\\\-1\\end{pmatrix}$. It might be tempting to proceed with vectors perpendicular to $\\pi$ and $\\pi'$ being $\\mathbf{b} \\times \\mathbf{c}$ and $\\mathbf{e} \\times \\mathbf{f}$. $\\mathbf{b}\\times \\mathbf{c} = \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ 2&-1&-1 \\\\ -1&2&1\\end{matrix}\\right \\vert = \\mathbf{i}-\\mathbf{j}+3\\mathbf{k}.$ $\\mathbf{e}\\times \\mathbf{f}= \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ p&q&r \\\\ q&r&p\\end{matrix}\\right \\vert = (qp-r^2)\\mathbf{i}+(rq-p^2)\\mathbf{j}+(pr-q^2)\\mathbf{k}.$ However, the resulting set of simultaneous equations is very difficult to solve. By adding them we can then simplify using $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ in other words $p^2+q^2+r^2=30$. We can think about $(p+q+r)^2$ to continue the manipulation. Try it if you like! Show that $\\pi'$ is the reflection of $\\pi$ in the origin. Using the form of the plane equation $\\mathbf{r}.\\mathbf{n}=\\mathbf{a}.\\mathbf{n}$ with $\\mathbf{n}$ as the normal vector we found earlier, we have the equation of $\\pi$ as $x - y + 3z = -2.$ Similarly using the point $\\mathbf{d}$, the plane $\\pi'$ is $x - y + 3z = 2.$ If $(x_1,y_1,z_1)$ is a point on $\\pi$ then $x_1 - y_1 + 3z_1 = -2$. Multiplying by $-1$ gives $(-x_1) -(-y_1) + 3(-z_1) = 2$. So the point $(-x_1,-y_1,-z_1)$ lies on $\\pi'$. Since every point on", "# Linear Transformatonf and Matrix Representation • Let {$e_1,e_2$} be a standard basis of $\\mathbb{R^2}$. Suppose $L:\\mathbb{R}^{2}\\rightarrow\\mathbb{R}^{2}$ be a linear transformation such that $L\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}=\\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix}$ and $L\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}=\\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix}$. Find the matrix representation of $L$ with respect to the standard basis. I couldn't anything. Can you help, can you give a hint? • Represent the standard basis vectors as linear combinations of $(1,1)$ and $(1,-1)$. – Friedrich Philipp Nov 11 '17 at 20:51 ## 1 Answer Here's one approach. Let $\\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}$ be the matrix of $L$ with respect to the standard basis. Then $L\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}=\\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix}$ and $L\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}=\\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix}$ give the equations $$a+b=3, \\quad c+d=3, \\quad a-b=-1, \\quad c-d=1\\,.$$ Solve for $a,b,c,d$. • Why $a-b=-1$ and $c-d=1$? – PozcuKushimotoStreet Nov 11 '17 at 20:57 • $\\begin{pmatrix} a & b \\\\ c&d \\end{pmatrix}\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}=\\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix}$ becomes $\\begin{pmatrix} a - b \\\\ c - d\\end{pmatrix}=\\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix}$ – A. Goodier Nov 11 '17 at 21:00 • Replace $L$ with the matrix $\\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}$ in $L\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}=\\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix}$ and $L\\begin{pmatrix}", "following (2x2) matrices. ${\\displaystyle {\\begin{pmatrix}3&5\\\\6&4\\\\\\end{pmatrix}}}$ ${\\displaystyle {\\begin{pmatrix}1&2\\\\3&4\\\\\\end{pmatrix}}}$ = ${\\displaystyle {\\begin{pmatrix}3{\\rm {x}}1+5{\\rm {x}}3&3{\\rm {x}}2+5{\\rm {x}}4\\\\6{\\rm {x}}1+4{\\rm {x}}3&6{\\rm {x}}2+4{\\rm {x}}4\\\\\\end{pmatrix}}}$ ii) Multiply the following (3x3) matrices. ${\\displaystyle {\\begin{pmatrix}38.15&-42.17&4.02\\\\4.69&0.76&-5.35\\\\-9.94&8.20&2.74\\\\\\end{pmatrix}}}$ ${\\displaystyle {\\begin{pmatrix}0.205&0.665&1\\\\0.181&0.647&1\\\\0.207&0.476&1\\\\\\end{pmatrix}}}$ You will notice that this gives a unit matrix as its product. ${\\displaystyle {\\begin{pmatrix}1&0&0\\\\0&1&0\\\\0&0&1\\\\\\end{pmatrix}}}$ The first matrix is the inverse of the 2nd. Computers use the inverse of a matrix to solve simultaneous equations. If we have ${\\displaystyle {\\begin{pmatrix}y_{1}=a_{11}x_{1}+a_{12}x_{2}+a_{13}x_{3}.....+a_{1n}x_{n}\\\\y_{2}=a_{21}x_{2}+a_{22}x_{2}+a_{23}x_{3}.....+a_{2n}x_{n}\\\\.....\\\\.....\\\\y_{n}=a_{n1}x_{n}+a_{n2}x_{n}+a_{n3}x_{3}.....+a_{nn}x_{n}\\\\\\end{pmatrix}}}$ In matrix form this is.... ${\\displaystyle Y=AX}$ ${\\displaystyle A^{-1}Y=X}$ In terms of work this is equivalent to the elimination method you have already employed for small equations but can be performed by computers for ${\\displaystyle 10~000}$ simultaneous equations. (Examples of large systems of equations are the fitting of reference data to 200 references molecules, dimension 200, or the calculation of the quantum mechanical gradient of the energy where there is an equation for every way of exciting 1 electron from an occupied orbital to an excited, (called virtual, orbital, (typically ${\\displaystyle 10~000}$ equations.) ## Finding the inverse How do you find the inverse... You use Maple or Matlab on your PC but if the matrix is small you can use the formula... ${\\displaystyle A^{-1}={\\frac {1}{DetA}}AdjA}$ Here Adj A is the adjoint matrix, the transposed matrix of cofactors. These strange objects are best described by example..... ${\\displaystyle {\\begin{vmatrix}1&-1&2\\\\2&1&1\\\\3&-1&1\\\\\\end{vmatrix}}}$ This determinant is equal", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "# Linear Algebra/Describing the Solution Set/Solutions ## Solutions This exercise is recommended for all readers. Problem 1 Find the indicated entry of the matrix, if it is defined. ${\\displaystyle A={\\begin{pmatrix}1&3&1\\\\2&-1&4\\end{pmatrix}}}$ 1. ${\\displaystyle a_{2,1}}$ 2. ${\\displaystyle a_{1,2}}$ 3. ${\\displaystyle a_{2,2}}$ 4. ${\\displaystyle a_{3,1}}$ 1. ${\\displaystyle 2}$ 2. ${\\displaystyle 3}$ 3. ${\\displaystyle -1}$ 4. Not defined. This exercise is recommended for all readers. Problem 2 Give the size of each matrix. 1. ${\\displaystyle {\\begin{pmatrix}1&0&4\\\\2&1&5\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}1&1\\\\-1&1\\\\3&-1\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}5&10\\\\10&5\\end{pmatrix}}}$ 1. ${\\displaystyle 2\\!\\times \\!3}$ 2. ${\\displaystyle 3\\!\\times \\!2}$ 3. ${\\displaystyle 2\\!\\times \\!2}$ This exercise is recommended for all readers. Problem 3 Do the indicated vector operation, if it is defined. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}}+{\\begin{pmatrix}3\\\\0\\\\4\\end{pmatrix}}}$ 2. ${\\displaystyle 5{\\begin{pmatrix}4\\\\-1\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}1\\\\5\\\\1\\end{pmatrix}}-{\\begin{pmatrix}3\\\\1\\\\1\\end{pmatrix}}}$ 4. ${\\displaystyle 7{\\begin{pmatrix}2\\\\1\\end{pmatrix}}+9{\\begin{pmatrix}3\\\\5\\end{pmatrix}}}$ 5. ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}}$ 6. ${\\displaystyle 6{\\begin{pmatrix}3\\\\1\\\\1\\end{pmatrix}}-4{\\begin{pmatrix}2\\\\0\\\\3\\end{pmatrix}}+2{\\begin{pmatrix}1\\\\1\\\\5\\end{pmatrix}}}$ 1. ${\\displaystyle {\\begin{pmatrix}5\\\\1\\\\5\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}20\\\\-5\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}-2\\\\4\\\\0\\end{pmatrix}}}$ 4. ${\\displaystyle {\\begin{pmatrix}41\\\\52\\end{pmatrix}}}$ 5. Not defined. 6. ${\\displaystyle {\\begin{pmatrix}12\\\\8\\\\4\\end{pmatrix}}}$ This exercise is recommended for all readers. Problem 4 Solve each system using matrix notation. Express the solution using vectors. 1. ${\\displaystyle {\\begin{array}{*{2}{rc}r}3x&+&6y&=&18\\\\x&+&2y&=&6\\end{array}}}$ 2. ${\\displaystyle {\\begin{array}{*{2}{rc}r}x&+&y&=&1\\\\x&-&y&=&-1\\end{array}}}$ 3. ${\\displaystyle {\\begin{array}{*{3}{rc}r}x_{1}&&&+&x_{3}&=&4\\\\x_{1}&-&x_{2}&+&2x_{3}&=&5\\\\4x_{1}&-&x_{2}&+&5x_{3}&=&17\\end{array}}}$ 4. ${\\displaystyle {\\begin{array}{*{3}{rc}r}2a&+&b&-&c&=&2\\\\2a&&&+&c&=&3\\\\a&-&b&&&=&0\\end{array}}}$ 5. ${\\displaystyle {\\begin{array}{*{4}{rc}r}x&+&2y&-&z&&&=&3\\\\2x&+&y&&&+&w&=&4\\\\x&-&y&+&z&+&w&=&1\\end{array}}}$ 6. ${\\displaystyle {\\begin{array}{*{4}{rc}r}x&&&+&z&+&w&=&4\\\\2x&+&y&&&-&w&=&2\\\\3x&+&y&+&z&&&=&7\\end{array}}}$ 1. This reduction ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}3&6&18\\\\1&2&6\\end{array}}\\right)&{\\xrightarrow[{}]{(-1/3)\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}3&6&18\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ leaves ${\\displaystyle x}$ leading and ${\\displaystyle y}$ free. Making ${\\displaystyle y}$ the parameter, we have ${\\displaystyle x=6-2y}$ so the solution set is ${\\displaystyle \\{{\\begin{pmatrix}6\\\\0\\end{pmatrix}}+{\\begin{pmatrix}-2\\\\1\\end{pmatrix}}y\\,{\\big |}\\,y\\in \\mathbb {R} \\}.}$ 2.", "\\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 2. $\\begin{pmatrix} -3 \\\\ 3 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 3. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 4. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k+\\begin{pmatrix} 0 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 5. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 14 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 6. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 6 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ Problem 2 Produce two descriptions of this set that are different than this one. $\\{\\begin{pmatrix} 2 \\\\ -5 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 3 Show that the three descriptions given at the start of this subsection all describe the same set. This exercise is recommended for all readers. Problem 4 Show that these sets are equal $\\{\\begin{pmatrix} 1 \\\\ 4 \\\\ 1 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}z\\,\\big|\\, z\\in\\mathbb{R} \\} \\quad\\text{and}\\quad \\{\\begin{pmatrix} 0 \\\\ 4 \\\\ 2 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R} \\},$ and that both describe the solution set of this system. $\\begin{array}{*{4}{rc}r} x &- &y &+ &z &+", "\\\\ 8 \\end{pmatrix},$ $\\begin{pmatrix} 3 & 3 \\\\ 2 & 5 \\end{pmatrix} \\begin{pmatrix} 11 \\\\ 15 \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 19 \\end{pmatrix}$ and continue encryption as follows: $\\begin{pmatrix} 7 \\\\ 8 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 19 \\end{pmatrix} \\to \\begin{pmatrix} H \\\\ I \\end{pmatrix}, \\begin{pmatrix} A \\\\ T \\end{pmatrix}$ The matrix K is invertible, hence $K^{-1}$ exists such that $KK^{-1}=K^{-1}K=I_2$. To implement decrypting, we compute $K^{-1} = 9^{-1} \\begin{pmatrix} 5 & 23 \\\\ 24 & 3 \\end{pmatrix} = 3 \\begin{pmatrix} 5 & 23 \\\\ 24 & 3 \\end{pmatrix} = \\begin{pmatrix} 15 & 17 \\\\ 20 & 9 \\end{pmatrix}$ $HIAT \\to \\begin{pmatrix} H \\\\ I \\end{pmatrix}, \\begin{pmatrix} A \\\\ T \\end{pmatrix} \\to \\begin{pmatrix} 7 \\\\ 8 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 19 \\end{pmatrix}$ Then we compute $\\begin{pmatrix} 15 & 17 \\\\ 20 & 9 \\end{pmatrix}\\begin{pmatrix} 7 \\\\ 8 \\end{pmatrix} = \\begin{pmatrix} 7 \\\\ 4 \\end{pmatrix},$ $\\begin{pmatrix} 15 & 17 \\\\ 20 & 9 \\end{pmatrix}\\begin{pmatrix} 0 \\\\ 19 \\end{pmatrix} = \\begin{pmatrix} 11 \\\\ 15 \\end{pmatrix}$ Therefore $\\begin{pmatrix} 7 \\\\ 4 \\end{pmatrix}, \\begin{pmatrix} 11 \\\\ 15 \\end{pmatrix} \\to \\begin{pmatrix} H \\\\ E \\end{pmatrix}, \\begin{pmatrix} L \\\\ P \\end{pmatrix} \\to HELP$. ## Security Unfortunately, the basic Hill cipher is vulnerable to a known-plaintext attack because it is completely linear. An opponent who intercepts $n^2$ plaintext/ciphertext character pairs can set up a linear system", "\\\\ 0 & 0 \\end{pmatrix}} -c {\\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\end{pmatrix}} \\\\ &= bE_3 - cE_2 \\end{align*} How do I continu to find $M_{E_1}$? Remember that given your basis $\\xi$, for $B \\in M_2(\\mathbb{R})$, $$[B]_\\xi = \\begin{pmatrix}B_{11} \\\\ B_{21} \\\\ B_{12} \\\\ B_{22} \\end{pmatrix};$$ in particular, for $A \\in M_2(\\mathbb{R})$, $M_A := [L_A]_{\\xi \\xi}$ is the matrix of the linear transformation $L_A : M_2(\\mathbb{R}) \\to M_2(\\mathbb{R})$ with respect to $\\xi$, and hence a $4 \\times 4$ matrix. Indeed, by introductory linear algebra, $M_A$ will be the $4 \\times 4$ matrix whose $k$-th column is $[L_A(E_k)]_\\xi$, as given by the equation above. So, given $A \\in M_2(\\mathbb{R})$, all you really need to do is compute $L_A(E_k)$ for each $k$, and express it as a linear combination of the $E_j$. Hint:do you know that the matrix you are looking for is $4\\times4$? Now you are looking for a $4\\times4$ matrix such that $M_{E_1}[X]_\\xi=[L_{E_1}(X)]_\\xi$. In your case, as you calculated, $[L_{E_1}(X)]_\\xi=(0,b,-c,0)$ column vector, while $[X]_\\xi=(a,b,c,d)$ column. Do you see how to proceed?", "the same length: • For$m = 1$: the roots of$\\begin{pmatrix}2\\end{pmatrix} \\begin{pmatrix}5\\end{pmatrix} \\begin{pmatrix}8\\end{pmatrix} \\begin{pmatrix}14\\end{pmatrix}$are • (all separate)$\\begin{pmatrix}2\\end{pmatrix} \\begin{pmatrix}5\\end{pmatrix} \\begin{pmatrix}8\\end{pmatrix} \\begin{pmatrix}14\\end{pmatrix}$, • (two pairs)$\\begin{pmatrix}2 & 5\\end{pmatrix} \\begin{pmatrix}8 & 14\\end{pmatrix}$,$\\begin{pmatrix}2 & 8\\end{pmatrix} \\begin{pmatrix}5 & 14\\end{pmatrix}$,$\\begin{pmatrix}2 & 14\\end{pmatrix} \\begin{pmatrix}5 & 8\\end{pmatrix}$, • (all together)$\\begin{pmatrix}2 & 5 & 8 & 14\\end{pmatrix}$,$\\begin{pmatrix}2 & 5 & 14 & 8\\end{pmatrix}$,$\\begin{pmatrix}2 & 8 & 5 & 14\\end{pmatrix}$,$\\begin{pmatrix}2 & 8 & 14 & 5\\end{pmatrix}$,$\\begin{pmatrix}2 & 14 & 5 & 8\\end{pmatrix}$,$\\begin{pmatrix}2 & 14 & 8 & 5\\end{pmatrix}$. So there are$1 + 3 + 6 = 10$roots of$\\begin{pmatrix}2\\end{pmatrix} \\begin{pmatrix}5\\end{pmatrix} \\begin{pmatrix}8\\end{pmatrix} \\begin{pmatrix}14\\end{pmatrix}$. • For$m = 2$, the roots of$\\begin{pmatrix}1 & 7\\end{pmatrix} \\begin{pmatrix}4 & 10\\end{pmatrix} \\begin{pmatrix}6 & 18\\end{pmatrix} \\begin{pmatrix}12 & 17\\end{pmatrix}$are$6 \\times 8 = 48$in number (pick one of the$3!$orderings of the last three, then one of the$2^3$orderings of each of the three). • For$m = 3$, the roots of$\\begin{pmatrix}3 & 16 & 9\\end{pmatrix} \\begin{pmatrix}11 & 15 & 13\\end{pmatrix}$are • (both separate)$\\begin{pmatrix}3 & 16 & 9\\end{pmatrix} \\begin{pmatrix}11 & 15 & 13\\end{pmatrix}$• (both together)$\\begin{pmatrix}3 & 11 & 9 & 13 & 16 & 15\\end{pmatrix}$,$\\begin{pmatrix}3 & 15 & 9 & 11 & 16 & 13\\end{pmatrix}$,$\\begin{pmatrix}3 & 13 & 9 & 15 & 16 & 11\\end{pmatrix}$. So there are$1 + 3 = 4$roots of$\\begin{pmatrix}3 & 16 & 9\\end{pmatrix} \\begin{pmatrix}11 & 15 & 13\\end{pmatrix}$. So totally$N(\\sigma) = 10 \\times 48 \\times 4 = 1920$. - In", "3$ matrix $\\mathbf{A} = \\begin{pmatrix}1&0&2\\\\3&1&0\\\\-1&1&4\\end{pmatrix}$. Workbook This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples. More Support You can get one-to-one support from Maths-Aid.", "3&5\\\\ 1&2 \\end{pmatrix} \\begin{pmatrix} 1&0\\\\ 0&2 \\end{pmatrix}$ and so $A = \\begin{pmatrix} 3&5\\\\ 1&2 \\end{pmatrix} \\begin{pmatrix} 1&0\\\\ 0&2 \\end{pmatrix} \\begin{pmatrix} 3&5\\\\ 1&2 \\end{pmatrix}^{-1}$ #### Example 2 a) Diagonalize A, i.e find P (invertible) and B (diagonal) such that AP = PB b) Compute A5 $A = \\begin{pmatrix} 1&2\\\\ -1&4 \\end{pmatrix}$ ' Solution a) We are solving Ax = λx, where λ is a constant and x a column vector. Firstly $Ax - \\lambda Ix = 0$ $(A - \\lambda I)x = 0$ since x ≠ 0 we have $det(A - \\lambda I) = 0$ i.e. $\\det\\begin{pmatrix} 1-\\lambda&2\\\\ -1&4-\\lambda \\end{pmatrix} = 0$ $\\begin{matrix} (1-\\lambda)(4-\\lambda) + 2 &=& 0\\\\ \\lambda^2-5\\lambda + 6 &=& 0 \\end{matrix}$ $\\lambda = 3,\\ 2$ For λ = 3, $(A-3I)x = 0$ $\\begin{pmatrix} -2&2\\\\ -1&1 \\end{pmatrix} \\begin{pmatrix} x\\\\ y \\end{pmatrix} = 0$ Clearly $\\begin{pmatrix} x\\\\ y \\end{pmatrix} = \\begin{pmatrix} 1\\\\ 1 \\end{pmatrix}$ is a solution. Note that we do not accept x = 0 as an solution, because we assume x ≠ 0. Note also that $\\begin{pmatrix} x\\\\ y \\end{pmatrix} = \\begin{pmatrix} t\\\\ t \\end{pmatrix}$ for some constant t is also a solution. Indeed we could use x = y = 2, 3 or 4 as a solution, but for convenience we choose the simplest i.e. x = y = 1. For λ = 2, $(A-2I)x", "a matrix equation. The problem is to solve $\\frac{d\\begin{pmatrix}y_1 \\\\ y_2\\end{pmatrix}}{dx}= \\begin{pmatrix}1 & 2 \\\\ 3 & 2\\end{pmatrix}\\begin{pmatrix}y_1 \\\\ y_2\\end{pmatrix}$. You have determined (correctly) that the eigenvalues of the coefficient matrix are 4 and -1. That means that the matrix can be \"diagonalized\". That is, there exists a matrix, P, such that $P\\begin{pmatrix}1 & 2 \\\\ 3 & 2 \\end{pmatrix}P^{-1}= \\begin{pmatrix}4 & 0\\\\0 & -1\\end{pmatrix}$. And, of course, then. that $\\begin{pmatrix}1 & 2 \\\\ 3 & 2\\end{pmatrix}= P^{-1}\\begin{pmatrix}4 & 0 \\\\ 0 & -1 \\end{pmatrix}P$. That \"P\" matrix is the matrix having the eigenvectors corresponding to eigenvalues 4 and -1 so we need to find those eigevectors. If $\\begin{pmatrix} a \\\\ b \\end{pmatrix}$ is an eigen vector corresponding to eigenvalue 4, then we have $\\begin{pmatrix}1 & 2 \\\\ 3 & 2\\end{pmatrix}\\begin{pmatrix}a \\\\ b\\end{pmatrix}= \\begin{pmatrix}a+ 2b \\\\ 3a+ 2b\\end{pmatrix}= \\begin{pmatrix}4a \\\\ 4b\\end{pmatrix}$. So a+ 2b= 4a and 3a+ 2b= 4b. Those reduce to -3a+ 2b= 0 and 3a- 2b= 0 which both reduce to 3a= 2b. There are infinitely many solutions because there are infinitely many eigenvectors corresponding to one eigenvalue (an entire subspace). We can take, as one solution, (a, b)= (2, 3). If $\\begin{pmatrix} a \\\\ b \\end{pmatrix}$ is an eigenvector corresponding to eigenvalue -1, then we have $\\begin{pmatrix}1 & 2 \\\\ 3 & 2\\end{pmatrix}\\begin{pmatrix}a \\\\ b\\end{pmatrix}=", "because the first column tells us $$e^{5it}=cos 5t+isin5t,$$ and the second tells us $$e^{-5it}=cos5t-isin5t$$. We can then solve $$pmatrix{10\\8}=Mpmatrix{A\\B}$$ as $$pmatrix{A\\B}=M^{-1}pmatrix{10\\8}=pmatrix{1&1\\i&-i}^{-1}pmatrix{10\\8}$$ $$=dfrac{pmatrix{-i&-1\\-i&1}}{-2i}pmatrix{10\\8}=pmatrix{frac12&-frac i2\\frac12&frac i2}pmatrix{10\\8}=pmatrix{5-4i\\5+4i}.$$", "$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {vmatrix}$$ = -4 - 15 = -19≠ 0 It has a unique solution. X = A-1B A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$ X = A-1B or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -10 - 9\\\\ -25 + 6\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-19}{-19}\\\\ \\frac {-19}{-19}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, 3x - 5y = 3..........................(1) 4x + 3y = 4.........................(2) Given equation in matrix form; $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3\\\\ 4\\\\ \\end {pmatrix}$$ or, AX = B where, A = $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$,", "11 questions linked to/from calculating $a^b \\!\\mod c$ 18k views ### Modular exponentiation by hand ($a^b\\bmod c$) How do I efficiently compute $a^b\\bmod c$: When $b$ is huge, for instance $5^{844325}\\bmod 21$? When $b$ is less than $c$ but it would still be a lot of work to multiply $a$ by itself $b$ times, for ... 5k views ### Proof of this result related to Fibonacci numbers: $\\begin{pmatrix}1&1\\\\1&0\\end{pmatrix}^n=\\begin{pmatrix}F_{n+1}&F_n\\\\F_n&F_{n-1}\\end{pmatrix}$? $$\\begin{pmatrix}1&1\\\\1&0\\end{pmatrix}^n=\\begin{pmatrix}F_{n+1}&F_n\\\\F_n&F_{n-1}\\end{pmatrix}$$ Somebody has any idea how to go about proving this result? I know a proof by ... 477 views ### exponential equation $$\\sqrt{(5+2\\sqrt6)^x}+\\sqrt{(5-2\\sqrt6)^x}=10$$ So I have squared both sides and got: $$(5-2\\sqrt6)^x+(5+2\\sqrt6)^x+2\\sqrt{1^x}=100$$ $$(5-2\\sqrt6)^x+(5+2\\sqrt6)^x+2=100$$ I don't know what to do ... ### Calculate $7^{154} \\pmod{341}$ [duplicate] how to calculate remainder of $7^ {154}$ when it is divided by $341$. Could you please state which method or theorem to use.", "R_2$ $\\begin{pmatrix}\\begin{array}{cc|cc}1&-1&0&1 \\\\0&1&1&-2\\end{array}\\end{pmatrix}$ $R_1 + R_2 {\\rightarrow} R_1$ $\\begin{pmatrix}\\begin{array}{cc|cc}1&0&1&-1 \\\\0&1&1&-2\\end{array}\\end{pmatrix}$ Notice that the left hand part is now the identity. The right hand side is the inverse. Hence $A^{-1}= \\begin{pmatrix} 1&-1 \\\\ 1&-2 \\end{pmatrix}$ ## Solving Equations Using Matrices ##### Example 32.5.5 : Suppose we have the system $2x - y = 3$ $x - y = 4$ Then we can write this in matrix form $Ax = b$ where $A=\\begin{pmatrix} 2&-1 \\\\ 1&-1 \\end{pmatrix}, \\;\\;\\; x= \\begin{pmatrix} x \\\\ y \\end{pmatrix}, \\;\\;\\; \\text{and} \\; b=\\begin{pmatrix} 3\\\\4 \\end{pmatrix}$ We can multiply both sides by $$A^{-1}$$: $A^{-1}A x = A^{-1}b$ or $x = A^{-1}b$ From before, $A^{-1}=\\begin{pmatrix} 1&-1 \\\\ 1&-2 \\end{pmatrix}$ Hence our solution is $\\begin{pmatrix} -1&-5 \\end{pmatrix}$ or $x = -1 \\text{ and } y = 5$", "${\\ displaystyle a_ {11}, a_ {22}, \\ dots, a_ {kk}}$${\\ displaystyle k = \\ min \\ {m, n \\}}$${\\ displaystyle A ^ {\\ top}}$${\\ displaystyle A ^ {\\ mathrm {t}}}$${\\ displaystyle A '}$ ## Examples Transposing a matrix (a row vector ) results in a matrix (a column vector ) and vice versa: ${\\ displaystyle (1 \\ times 3)}$${\\ displaystyle (3 \\ times 1)}$ ${\\ displaystyle {\\ begin {pmatrix} 2 & 4 & 6 \\ end {pmatrix}} ^ {\\ mathrm {T}} = {\\ begin {pmatrix} 2 \\\\ 4 \\\\ 6 \\ end {pmatrix}}, \\ quad {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 5 \\ end {pmatrix}} ^ {\\ mathrm {T}} = {\\ begin {pmatrix} 1 & 3 & 5 \\ end {pmatrix}}}$ A square matrix retains its size through transposition, but all entries are mirrored on the main diagonal: ${\\ displaystyle {\\ begin {pmatrix} 2 & 3 \\\\ 4 & 5 \\ end {pmatrix}} ^ {\\ mathrm {T}} = {\\ begin {pmatrix} 2 & 4 \\\\ 3 & 5 \\ end {pmatrix}}, \\ quad {\\ begin {pmatrix} 9 & 8 & 7 \\\\ 6 & 5 & 4 \\\\ 3 & 2 & 1 \\ end {pmatrix}} ^ {\\ mathrm {T}} = {\\ begin {pmatrix} 9 & 6 & 3 \\\\ 8 & 5 & 2 \\\\ 7", "# Find the linear transformation of a matrix knowing 4 linear transformations I'm stuck with an exercise where they give me 4 linear transformations $$T:M_{2\\times2}\\to\\Bbb R$$ for 4 matrices 2x2 and then they ask me for the linear transformation of a fifth matrix. $$T \\begin{pmatrix} 1 &0\\\\ 0 & 0\\\\ \\end{pmatrix} =3, T\\begin{pmatrix} 0 &1\\\\ 1 & 0\\\\ \\end{pmatrix} =-1, T \\begin{pmatrix} 1 &0\\\\ 1 & 0\\\\ \\end{pmatrix} =0, T \\begin{pmatrix} 0 &0\\\\ 0 & 1\\\\ \\end{pmatrix} =0, T \\begin{pmatrix} a &b\\\\ c & d\\\\ \\end{pmatrix} =?$$ and I have worked with this situation for $$T:\\Bbb R^n \\to\\Bbb R^m$$ but never with matrices so I have the doubt that how is supposed to find 4 constants that will multiply de 4 matrices I know to compose the fifth matrix to finally can find the linear transformation of this last one. (I have tried adding up the 4 matrices as follows) $$\\begin{pmatrix} 1 &0\\\\ 0 & 0\\\\ \\end{pmatrix} + \\begin{pmatrix} 0 &1\\\\ 1 & 0\\\\ \\end{pmatrix} + \\begin{pmatrix} 1 &0\\\\ 1 & 0\\\\ \\end{pmatrix} + \\begin{pmatrix} 0 &0\\\\ 0 & 1\\\\ \\end{pmatrix}= \\begin{pmatrix} a &b\\\\ c & d\\\\ \\end{pmatrix}$$ $$\\begin{cases} 2=a & \\\\ 1=b & \\\\ 2=c\\\\ 1=d\\\\ \\end{cases}$$ and then use the coefficients $$2\\begin{pmatrix} 1 &0\\\\ 0 & 0\\\\ \\end{pmatrix} + 1\\begin{pmatrix} 0 &1\\\\ 1 & 0\\\\ \\end{pmatrix}" ]

[ "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "gives $q=-5p$. We now know that $\\mathbf{e}= t\\begin{pmatrix}1\\\\-5 \\\\ -2\\end{pmatrix}$. We also know $\\big \\vert\\mathbf{e}\\big \\vert = \\sqrt{30}$ and $p$ is negative, hence $t=-1$ and we have $\\mathbf{e}= \\begin{pmatrix}-1\\\\5 \\\\ 2\\end{pmatrix}$ and $\\mathbf{f}= \\begin{pmatrix}5\\\\2 \\\\-1\\end{pmatrix}$. It might be tempting to proceed with vectors perpendicular to $\\pi$ and $\\pi'$ being $\\mathbf{b} \\times \\mathbf{c}$ and $\\mathbf{e} \\times \\mathbf{f}$. $\\mathbf{b}\\times \\mathbf{c} = \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ 2&-1&-1 \\\\ -1&2&1\\end{matrix}\\right \\vert = \\mathbf{i}-\\mathbf{j}+3\\mathbf{k}.$ $\\mathbf{e}\\times \\mathbf{f}= \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ p&q&r \\\\ q&r&p\\end{matrix}\\right \\vert = (qp-r^2)\\mathbf{i}+(rq-p^2)\\mathbf{j}+(pr-q^2)\\mathbf{k}.$ However, the resulting set of simultaneous equations is very difficult to solve. By adding them we can then simplify using $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ in other words $p^2+q^2+r^2=30$. We can think about $(p+q+r)^2$ to continue the manipulation. Try it if you like! Show that $\\pi'$ is the reflection of $\\pi$ in the origin. Using the form of the plane equation $\\mathbf{r}.\\mathbf{n}=\\mathbf{a}.\\mathbf{n}$ with $\\mathbf{n}$ as the normal vector we found earlier, we have the equation of $\\pi$ as $x - y + 3z = -2.$ Similarly using the point $\\mathbf{d}$, the plane $\\pi'$ is $x - y + 3z = 2.$ If $(x_1,y_1,z_1)$ is a point on $\\pi$ then $x_1 - y_1 + 3z_1 = -2$. Multiplying by $-1$ gives $(-x_1) -(-y_1) + 3(-z_1) = 2$. So the point $(-x_1,-y_1,-z_1)$ lies on $\\pi'$. Since every point on", "\\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 2. $\\begin{pmatrix} -3 \\\\ 3 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 3. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 4. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k+\\begin{pmatrix} 0 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 5. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 14 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 6. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 6 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ Problem 2 Produce two descriptions of this set that are different than this one. $\\{\\begin{pmatrix} 2 \\\\ -5 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 3 Show that the three descriptions given at the start of this subsection all describe the same set. This exercise is recommended for all readers. Problem 4 Show that these sets are equal $\\{\\begin{pmatrix} 1 \\\\ 4 \\\\ 1 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}z\\,\\big|\\, z\\in\\mathbb{R} \\} \\quad\\text{and}\\quad \\{\\begin{pmatrix} 0 \\\\ 4 \\\\ 2 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R} \\},$ and that both describe the solution set of this system. $\\begin{array}{*{4}{rc}r} x &- &y &+ &z &+", "# Linear Transformatonf and Matrix Representation • Let {$e_1,e_2$} be a standard basis of $\\mathbb{R^2}$. Suppose $L:\\mathbb{R}^{2}\\rightarrow\\mathbb{R}^{2}$ be a linear transformation such that $L\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}=\\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix}$ and $L\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}=\\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix}$. Find the matrix representation of $L$ with respect to the standard basis. I couldn't anything. Can you help, can you give a hint? • Represent the standard basis vectors as linear combinations of $(1,1)$ and $(1,-1)$. – Friedrich Philipp Nov 11 '17 at 20:51 ## 1 Answer Here's one approach. Let $\\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}$ be the matrix of $L$ with respect to the standard basis. Then $L\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}=\\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix}$ and $L\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}=\\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix}$ give the equations $$a+b=3, \\quad c+d=3, \\quad a-b=-1, \\quad c-d=1\\,.$$ Solve for $a,b,c,d$. • Why $a-b=-1$ and $c-d=1$? – PozcuKushimotoStreet Nov 11 '17 at 20:57 • $\\begin{pmatrix} a & b \\\\ c&d \\end{pmatrix}\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}=\\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix}$ becomes $\\begin{pmatrix} a - b \\\\ c - d\\end{pmatrix}=\\begin{pmatrix} -1 \\\\ 1 \\end{pmatrix}$ – A. Goodier Nov 11 '17 at 21:00 • Replace $L$ with the matrix $\\begin{pmatrix}a & b \\\\ c & d\\end{pmatrix}$ in $L\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}=\\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix}$ and $L\\begin{pmatrix}", "\\\\ 8 \\end{pmatrix},$ $\\begin{pmatrix} 3 & 3 \\\\ 2 & 5 \\end{pmatrix} \\begin{pmatrix} 11 \\\\ 15 \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 19 \\end{pmatrix}$ and continue encryption as follows: $\\begin{pmatrix} 7 \\\\ 8 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 19 \\end{pmatrix} \\to \\begin{pmatrix} H \\\\ I \\end{pmatrix}, \\begin{pmatrix} A \\\\ T \\end{pmatrix}$ The matrix K is invertible, hence $K^{-1}$ exists such that $KK^{-1}=K^{-1}K=I_2$. To implement decrypting, we compute $K^{-1} = 9^{-1} \\begin{pmatrix} 5 & 23 \\\\ 24 & 3 \\end{pmatrix} = 3 \\begin{pmatrix} 5 & 23 \\\\ 24 & 3 \\end{pmatrix} = \\begin{pmatrix} 15 & 17 \\\\ 20 & 9 \\end{pmatrix}$ $HIAT \\to \\begin{pmatrix} H \\\\ I \\end{pmatrix}, \\begin{pmatrix} A \\\\ T \\end{pmatrix} \\to \\begin{pmatrix} 7 \\\\ 8 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 19 \\end{pmatrix}$ Then we compute $\\begin{pmatrix} 15 & 17 \\\\ 20 & 9 \\end{pmatrix}\\begin{pmatrix} 7 \\\\ 8 \\end{pmatrix} = \\begin{pmatrix} 7 \\\\ 4 \\end{pmatrix},$ $\\begin{pmatrix} 15 & 17 \\\\ 20 & 9 \\end{pmatrix}\\begin{pmatrix} 0 \\\\ 19 \\end{pmatrix} = \\begin{pmatrix} 11 \\\\ 15 \\end{pmatrix}$ Therefore $\\begin{pmatrix} 7 \\\\ 4 \\end{pmatrix}, \\begin{pmatrix} 11 \\\\ 15 \\end{pmatrix} \\to \\begin{pmatrix} H \\\\ E \\end{pmatrix}, \\begin{pmatrix} L \\\\ P \\end{pmatrix} \\to HELP$. ## Security Unfortunately, the basic Hill cipher is vulnerable to a known-plaintext attack because it is completely linear. An opponent who intercepts $n^2$ plaintext/ciphertext character pairs can set up a linear system", "0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -5 & 0\\\\ 0 & 0 & 0\\end{array}\\right) \\mbox{ and } \\left(\\begin{array}{cc|c} 5 & -5 & 0\\\\ 1 & -1 & 0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -1 & 0\\\\ 0 & 0 & 0\\end{array}\\right)\\, ,$$ respectively, so are $$v_{2}=\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}$$ and $$v_{-2} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$. Hence $$M^{12} v_{2} = 2^{12} v_{2}$$ and $$M^{12}v_{-2} = (-2)^{12} v_{-2}$$. Now, $$\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\frac{x-y}{4}\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix} -\\frac{x-5y}{4} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$ (this was obtained by solving the linear system $$a v_{2} + b v_{-2} =$$ for $$a$$ and $$b$$). Thus $$M \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\frac{x-y}{4} M v_{2} -\\frac{x-5y}{4} M v_{-2}$$ $$= 2^{12} \\Big(\\frac{x-y}{4} v_{2} -\\frac{x-5y}{4} v_{-2}\\Big) = 2^{12} \\begin{pmatrix} x \\\\ y \\end{pmatrix}\\, .$$ Thus $$M^{12}=\\begin{pmatrix} 4096 & 0 \\\\ 0 & 4096\\end{pmatrix}\\, .$$ $$\\textit{If you understand the above explanation, then you have a good understanding of diagonalization. A quicker route}$$ $$\\textit{is simply to observe that}$$ $$M^{2} = \\begin{pmatrix}4 & 0 \\\\ 0 & 4\\end{pmatrix}$$. 8. $$P_{M}(\\lambda) = (-1)^{2} {\\rm det}\\begin{pmatrix} a-\\lambda & b \\\\ c &d-\\lambda\\end{pmatrix} =(\\lambda-a)(\\lambda-d) - bc\\, .$$ Thus $$P_{M}(M)=(M-a I )(M- d I) - bc I$$ $$= \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}a&0\\\\0&a\\end{pmatrix}\\right) \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}d&0\\\\0&d\\end{pmatrix}\\right)-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}$$ $$=\\begin{pmatrix}0& b\\\\c&d-a\\end{pmatrix}\\begin{pmatrix}a-d&b\\\\ c&0\\end{pmatrix}-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}=0\\, .$$ Observe that any $$2\\times 2$$ matrix is a zero of its own characteristic polynomial ($$\\textit{in fact this holds", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "# Linear Algebra/Describing the Solution Set/Solutions ## Solutions This exercise is recommended for all readers. Problem 1 Find the indicated entry of the matrix, if it is defined. ${\\displaystyle A={\\begin{pmatrix}1&3&1\\\\2&-1&4\\end{pmatrix}}}$ 1. ${\\displaystyle a_{2,1}}$ 2. ${\\displaystyle a_{1,2}}$ 3. ${\\displaystyle a_{2,2}}$ 4. ${\\displaystyle a_{3,1}}$ 1. ${\\displaystyle 2}$ 2. ${\\displaystyle 3}$ 3. ${\\displaystyle -1}$ 4. Not defined. This exercise is recommended for all readers. Problem 2 Give the size of each matrix. 1. ${\\displaystyle {\\begin{pmatrix}1&0&4\\\\2&1&5\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}1&1\\\\-1&1\\\\3&-1\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}5&10\\\\10&5\\end{pmatrix}}}$ 1. ${\\displaystyle 2\\!\\times \\!3}$ 2. ${\\displaystyle 3\\!\\times \\!2}$ 3. ${\\displaystyle 2\\!\\times \\!2}$ This exercise is recommended for all readers. Problem 3 Do the indicated vector operation, if it is defined. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}}+{\\begin{pmatrix}3\\\\0\\\\4\\end{pmatrix}}}$ 2. ${\\displaystyle 5{\\begin{pmatrix}4\\\\-1\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}1\\\\5\\\\1\\end{pmatrix}}-{\\begin{pmatrix}3\\\\1\\\\1\\end{pmatrix}}}$ 4. ${\\displaystyle 7{\\begin{pmatrix}2\\\\1\\end{pmatrix}}+9{\\begin{pmatrix}3\\\\5\\end{pmatrix}}}$ 5. ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}+{\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}}$ 6. ${\\displaystyle 6{\\begin{pmatrix}3\\\\1\\\\1\\end{pmatrix}}-4{\\begin{pmatrix}2\\\\0\\\\3\\end{pmatrix}}+2{\\begin{pmatrix}1\\\\1\\\\5\\end{pmatrix}}}$ 1. ${\\displaystyle {\\begin{pmatrix}5\\\\1\\\\5\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}20\\\\-5\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}-2\\\\4\\\\0\\end{pmatrix}}}$ 4. ${\\displaystyle {\\begin{pmatrix}41\\\\52\\end{pmatrix}}}$ 5. Not defined. 6. ${\\displaystyle {\\begin{pmatrix}12\\\\8\\\\4\\end{pmatrix}}}$ This exercise is recommended for all readers. Problem 4 Solve each system using matrix notation. Express the solution using vectors. 1. ${\\displaystyle {\\begin{array}{*{2}{rc}r}3x&+&6y&=&18\\\\x&+&2y&=&6\\end{array}}}$ 2. ${\\displaystyle {\\begin{array}{*{2}{rc}r}x&+&y&=&1\\\\x&-&y&=&-1\\end{array}}}$ 3. ${\\displaystyle {\\begin{array}{*{3}{rc}r}x_{1}&&&+&x_{3}&=&4\\\\x_{1}&-&x_{2}&+&2x_{3}&=&5\\\\4x_{1}&-&x_{2}&+&5x_{3}&=&17\\end{array}}}$ 4. ${\\displaystyle {\\begin{array}{*{3}{rc}r}2a&+&b&-&c&=&2\\\\2a&&&+&c&=&3\\\\a&-&b&&&=&0\\end{array}}}$ 5. ${\\displaystyle {\\begin{array}{*{4}{rc}r}x&+&2y&-&z&&&=&3\\\\2x&+&y&&&+&w&=&4\\\\x&-&y&+&z&+&w&=&1\\end{array}}}$ 6. ${\\displaystyle {\\begin{array}{*{4}{rc}r}x&&&+&z&+&w&=&4\\\\2x&+&y&&&-&w&=&2\\\\3x&+&y&+&z&&&=&7\\end{array}}}$ 1. This reduction ${\\displaystyle {\\begin{array}{rcl}\\left({\\begin{array}{*{2}{c}|c}3&6&18\\\\1&2&6\\end{array}}\\right)&{\\xrightarrow[{}]{(-1/3)\\rho _{1}+\\rho _{2}}}&\\left({\\begin{array}{*{2}{c}|c}3&6&18\\\\0&0&0\\end{array}}\\right)\\end{array}}}$ leaves ${\\displaystyle x}$ leading and ${\\displaystyle y}$ free. Making ${\\displaystyle y}$ the parameter, we have ${\\displaystyle x=6-2y}$ so the solution set is ${\\displaystyle \\{{\\begin{pmatrix}6\\\\0\\end{pmatrix}}+{\\begin{pmatrix}-2\\\\1\\end{pmatrix}}y\\,{\\big |}\\,y\\in \\mathbb {R} \\}.}$ 2.", "\\\\ 0 & 0 \\end{pmatrix}} -c {\\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\end{pmatrix}} \\\\ &= bE_3 - cE_2 \\end{align*} How do I continu to find $M_{E_1}$? Remember that given your basis $\\xi$, for $B \\in M_2(\\mathbb{R})$, $$[B]_\\xi = \\begin{pmatrix}B_{11} \\\\ B_{21} \\\\ B_{12} \\\\ B_{22} \\end{pmatrix};$$ in particular, for $A \\in M_2(\\mathbb{R})$, $M_A := [L_A]_{\\xi \\xi}$ is the matrix of the linear transformation $L_A : M_2(\\mathbb{R}) \\to M_2(\\mathbb{R})$ with respect to $\\xi$, and hence a $4 \\times 4$ matrix. Indeed, by introductory linear algebra, $M_A$ will be the $4 \\times 4$ matrix whose $k$-th column is $[L_A(E_k)]_\\xi$, as given by the equation above. So, given $A \\in M_2(\\mathbb{R})$, all you really need to do is compute $L_A(E_k)$ for each $k$, and express it as a linear combination of the $E_j$. Hint:do you know that the matrix you are looking for is $4\\times4$? Now you are looking for a $4\\times4$ matrix such that $M_{E_1}[X]_\\xi=[L_{E_1}(X)]_\\xi$. In your case, as you calculated, $[L_{E_1}(X)]_\\xi=(0,b,-c,0)$ column vector, while $[X]_\\xi=(a,b,c,d)$ column. Do you see how to proceed?", "\\begin{equation*} A = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\end{equation*} is in $GL_2( {\\mathbb R})$ if there exists a matrix $A^{-1}$ such that $A A^{-1} = A^{-1} A = I\\text{,}$ where $I$ is the $2 \\times 2$ identity matrix. For $A$ to have an inverse is equivalent to requiring that the determinant of $A$ be nonzero; that is, $\\det A = ad - bc \\neq 0\\text{.}$ The set of invertible matrices forms a group called the general linear group. The identity of the group is the identity matrix \\begin{equation*} I = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}\\text{.} \\end{equation*} The inverse of $A \\in GL_2( {\\mathbb R})$ is \\begin{equation*} A^{-1} = \\frac{1}{ad-bc} \\begin{pmatrix} d & -b \\\\ -c & a \\end{pmatrix}\\text{.} \\end{equation*} The product of two invertible matrices is again invertible. Matrix multiplication is associative, satisfying the other group axiom. For matrices it is not true in general that $AB = BA\\text{;}$ hence, $GL_2({\\mathbb R})$ is another example of a nonabelian group. ###### Example2.15. Let \\begin{align*} 1 & = \\begin{pmatrix} 1 & 0\\\\ 0 & 1 \\end{pmatrix} \\qquad I = \\begin{pmatrix} 0 & 1\\\\ -1 & 0 \\end{pmatrix}\\\\ J & = \\begin{pmatrix} 0 & i\\\\ i & 0 \\end{pmatrix} \\qquad K = \\begin{pmatrix} i & 0\\\\ 0 & -i \\end{pmatrix}\\text{,} \\end{align*} where $i^2", "R_2$ $\\begin{pmatrix}\\begin{array}{cc|cc}1&-1&0&1 \\\\0&1&1&-2\\end{array}\\end{pmatrix}$ $R_1 + R_2 {\\rightarrow} R_1$ $\\begin{pmatrix}\\begin{array}{cc|cc}1&0&1&-1 \\\\0&1&1&-2\\end{array}\\end{pmatrix}$ Notice that the left hand part is now the identity. The right hand side is the inverse. Hence $A^{-1}= \\begin{pmatrix} 1&-1 \\\\ 1&-2 \\end{pmatrix}$ ## Solving Equations Using Matrices ##### Example 32.5.5 : Suppose we have the system $2x - y = 3$ $x - y = 4$ Then we can write this in matrix form $Ax = b$ where $A=\\begin{pmatrix} 2&-1 \\\\ 1&-1 \\end{pmatrix}, \\;\\;\\; x= \\begin{pmatrix} x \\\\ y \\end{pmatrix}, \\;\\;\\; \\text{and} \\; b=\\begin{pmatrix} 3\\\\4 \\end{pmatrix}$ We can multiply both sides by $$A^{-1}$$: $A^{-1}A x = A^{-1}b$ or $x = A^{-1}b$ From before, $A^{-1}=\\begin{pmatrix} 1&-1 \\\\ 1&-2 \\end{pmatrix}$ Hence our solution is $\\begin{pmatrix} -1&-5 \\end{pmatrix}$ or $x = -1 \\text{ and } y = 5$", "3&5\\\\ 1&2 \\end{pmatrix} \\begin{pmatrix} 1&0\\\\ 0&2 \\end{pmatrix}$ and so $A = \\begin{pmatrix} 3&5\\\\ 1&2 \\end{pmatrix} \\begin{pmatrix} 1&0\\\\ 0&2 \\end{pmatrix} \\begin{pmatrix} 3&5\\\\ 1&2 \\end{pmatrix}^{-1}$ #### Example 2 a) Diagonalize A, i.e find P (invertible) and B (diagonal) such that AP = PB b) Compute A5 $A = \\begin{pmatrix} 1&2\\\\ -1&4 \\end{pmatrix}$ ' Solution a) We are solving Ax = λx, where λ is a constant and x a column vector. Firstly $Ax - \\lambda Ix = 0$ $(A - \\lambda I)x = 0$ since x ≠ 0 we have $det(A - \\lambda I) = 0$ i.e. $\\det\\begin{pmatrix} 1-\\lambda&2\\\\ -1&4-\\lambda \\end{pmatrix} = 0$ $\\begin{matrix} (1-\\lambda)(4-\\lambda) + 2 &=& 0\\\\ \\lambda^2-5\\lambda + 6 &=& 0 \\end{matrix}$ $\\lambda = 3,\\ 2$ For λ = 3, $(A-3I)x = 0$ $\\begin{pmatrix} -2&2\\\\ -1&1 \\end{pmatrix} \\begin{pmatrix} x\\\\ y \\end{pmatrix} = 0$ Clearly $\\begin{pmatrix} x\\\\ y \\end{pmatrix} = \\begin{pmatrix} 1\\\\ 1 \\end{pmatrix}$ is a solution. Note that we do not accept x = 0 as an solution, because we assume x ≠ 0. Note also that $\\begin{pmatrix} x\\\\ y \\end{pmatrix} = \\begin{pmatrix} t\\\\ t \\end{pmatrix}$ for some constant t is also a solution. Indeed we could use x = y = 2, 3 or 4 as a solution, but for convenience we choose the simplest i.e. x = y = 1. For λ = 2, $(A-2I)x", "$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {vmatrix}$$ = -4 - 15 = -19≠ 0 It has a unique solution. X = A-1B A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$ X = A-1B or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -10 - 9\\\\ -25 + 6\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-19}{-19}\\\\ \\frac {-19}{-19}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, 3x - 5y = 3..........................(1) 4x + 3y = 4.........................(2) Given equation in matrix form; $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3\\\\ 4\\\\ \\end {pmatrix}$$ or, AX = B where, A = $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$,", "\\begin{pmatrix} a_1 & a_2 & a_3 & ... & a_m \\end{pmatrix}$ $\\textbf u = \\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3\\\\...\\\\ b_n \\end{pmatrix}$ $\\textbf v^T \\cdot \\textbf u = \\begin{pmatrix} a_1 & a_2 & a_3& ... & a_m \\end{pmatrix} \\cdot \\begin{pmatrix} b_1 \\\\ b_2 \\\\b_3 \\\\ ... \\\\b_n \\end{pmatrix}$ $\\textbf v^T \\cdot \\textbf u = ( a_1 b_1 + a_2 b_2 + a_3 b_3 + ... + a_m b_n)$ Notice that it only works if $m = n$ and it always results in a matrix of order $1 \\times 1$. The outer product of these two vectors is $\\textbf v \\cdot \\textbf u^T$. $\\textbf v \\cdot \\textbf u^T = \\begin{pmatrix} a_1 \\\\ a_2 \\\\ a_3 \\\\ ... \\\\ a_m \\end{pmatrix} \\cdot \\begin{pmatrix} b_1 & b_2 & b_3 & ... & b_n \\end{pmatrix}$ $\\textbf v \\cdot \\textbf u^T = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & a_2 b_3 &...&a_1 b_n\\\\a_2 b_1 & a_2 b_2 & a_2 b_3 &...& a_2 b_n\\\\a_3 b_1 & a_3 b_2 & a_3 b_3 &...&a_3 b_n\\\\...& ... & ... & ... & ... \\\\a_m b_1 & a_m b_2 & a_m b_3 & ... & a_m b_n \\end{pmatrix}$ If you have a column vector of order $m \\times 1$ and a row vector of order $1 \\times n$ respectively, their outer product will be a matrix of order", "${\\ displaystyle a_ {11}, a_ {22}, \\ dots, a_ {kk}}$${\\ displaystyle k = \\ min \\ {m, n \\}}$${\\ displaystyle A ^ {\\ top}}$${\\ displaystyle A ^ {\\ mathrm {t}}}$${\\ displaystyle A '}$ ## Examples Transposing a matrix (a row vector ) results in a matrix (a column vector ) and vice versa: ${\\ displaystyle (1 \\ times 3)}$${\\ displaystyle (3 \\ times 1)}$ ${\\ displaystyle {\\ begin {pmatrix} 2 & 4 & 6 \\ end {pmatrix}} ^ {\\ mathrm {T}} = {\\ begin {pmatrix} 2 \\\\ 4 \\\\ 6 \\ end {pmatrix}}, \\ quad {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 5 \\ end {pmatrix}} ^ {\\ mathrm {T}} = {\\ begin {pmatrix} 1 & 3 & 5 \\ end {pmatrix}}}$ A square matrix retains its size through transposition, but all entries are mirrored on the main diagonal: ${\\ displaystyle {\\ begin {pmatrix} 2 & 3 \\\\ 4 & 5 \\ end {pmatrix}} ^ {\\ mathrm {T}} = {\\ begin {pmatrix} 2 & 4 \\\\ 3 & 5 \\ end {pmatrix}}, \\ quad {\\ begin {pmatrix} 9 & 8 & 7 \\\\ 6 & 5 & 4 \\\\ 3 & 2 & 1 \\ end {pmatrix}} ^ {\\ mathrm {T}} = {\\ begin {pmatrix} 9 & 6 & 3 \\\\ 8 & 5 & 2 \\\\ 7", "because the first column tells us $$e^{5it}=cos 5t+isin5t,$$ and the second tells us $$e^{-5it}=cos5t-isin5t$$. We can then solve $$pmatrix{10\\8}=Mpmatrix{A\\B}$$ as $$pmatrix{A\\B}=M^{-1}pmatrix{10\\8}=pmatrix{1&1\\i&-i}^{-1}pmatrix{10\\8}$$ $$=dfrac{pmatrix{-i&-1\\-i&1}}{-2i}pmatrix{10\\8}=pmatrix{frac12&-frac i2\\frac12&frac i2}pmatrix{10\\8}=pmatrix{5-4i\\5+4i}.$$", "following (2x2) matrices. ${\\displaystyle {\\begin{pmatrix}3&5\\\\6&4\\\\\\end{pmatrix}}}$ ${\\displaystyle {\\begin{pmatrix}1&2\\\\3&4\\\\\\end{pmatrix}}}$ = ${\\displaystyle {\\begin{pmatrix}3{\\rm {x}}1+5{\\rm {x}}3&3{\\rm {x}}2+5{\\rm {x}}4\\\\6{\\rm {x}}1+4{\\rm {x}}3&6{\\rm {x}}2+4{\\rm {x}}4\\\\\\end{pmatrix}}}$ ii) Multiply the following (3x3) matrices. ${\\displaystyle {\\begin{pmatrix}38.15&-42.17&4.02\\\\4.69&0.76&-5.35\\\\-9.94&8.20&2.74\\\\\\end{pmatrix}}}$ ${\\displaystyle {\\begin{pmatrix}0.205&0.665&1\\\\0.181&0.647&1\\\\0.207&0.476&1\\\\\\end{pmatrix}}}$ You will notice that this gives a unit matrix as its product. ${\\displaystyle {\\begin{pmatrix}1&0&0\\\\0&1&0\\\\0&0&1\\\\\\end{pmatrix}}}$ The first matrix is the inverse of the 2nd. Computers use the inverse of a matrix to solve simultaneous equations. If we have ${\\displaystyle {\\begin{pmatrix}y_{1}=a_{11}x_{1}+a_{12}x_{2}+a_{13}x_{3}.....+a_{1n}x_{n}\\\\y_{2}=a_{21}x_{2}+a_{22}x_{2}+a_{23}x_{3}.....+a_{2n}x_{n}\\\\.....\\\\.....\\\\y_{n}=a_{n1}x_{n}+a_{n2}x_{n}+a_{n3}x_{3}.....+a_{nn}x_{n}\\\\\\end{pmatrix}}}$ In matrix form this is.... ${\\displaystyle Y=AX}$ ${\\displaystyle A^{-1}Y=X}$ In terms of work this is equivalent to the elimination method you have already employed for small equations but can be performed by computers for ${\\displaystyle 10~000}$ simultaneous equations. (Examples of large systems of equations are the fitting of reference data to 200 references molecules, dimension 200, or the calculation of the quantum mechanical gradient of the energy where there is an equation for every way of exciting 1 electron from an occupied orbital to an excited, (called virtual, orbital, (typically ${\\displaystyle 10~000}$ equations.) ## Finding the inverse How do you find the inverse... You use Maple or Matlab on your PC but if the matrix is small you can use the formula... ${\\displaystyle A^{-1}={\\frac {1}{DetA}}AdjA}$ Here Adj A is the adjoint matrix, the transposed matrix of cofactors. These strange objects are best described by example..... ${\\displaystyle {\\begin{vmatrix}1&-1&2\\\\2&1&1\\\\3&-1&1\\\\\\end{vmatrix}}}$ This determinant is equal", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "A \\mathbf x\\text{,}$$ where \\begin{equation*} A = \\begin{pmatrix} 5 & 1 \\\\ -4 & 1 \\end{pmatrix}. \\end{equation*} The characteristic polynomial of $$A$$ is $$\\lambda^2 - 6 \\lambda + 9 = (\\lambda - 3)^2\\text{,}$$ we have only a single eigenvalue $$\\lambda = 3$$ with eigenvector $$\\mathbf v = (1, -2)\\text{.}$$ Any other eigenvector for $$\\lambda$$ is a multiple of $$\\mathbf v\\text{.}$$ If we choose $$\\mathbf w = (1, 0)\\text{,}$$ then $$\\mathbf v$$ and $$\\mathbf w$$ are linearly independent. Furthermore, \\begin{equation*} A \\mathbf w = \\begin{pmatrix} 5 \\\\ - 4 \\end{pmatrix} = 2 \\begin{pmatrix} 1 \\\\ -2 \\end{pmatrix} + \\lambda \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = 2 \\begin{pmatrix} 1 \\\\ -2 \\end{pmatrix} + 3 \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}. \\end{equation*} So we can let $$\\mathbf u = (1/2) \\mathbf w = (1/2, 0)\\text{.}$$ Therefore, the matrix that we seek is \\begin{equation*} T = \\begin{pmatrix} 1 \\amp 1/2 \\\\ -2 \\amp 0 \\end{pmatrix}, \\end{equation*} and \\begin{equation*} T^{-1} A T = \\begin{pmatrix} -1/2 & 2 \\\\ 1 & 1 \\end{pmatrix} \\begin{pmatrix} 5 & 1 \\\\ -4 & 1 \\end{pmatrix} \\begin{pmatrix} 1 \\amp 1/2 \\\\ -2 \\amp 0\\end{pmatrix} = \\begin{pmatrix} 3 & 1 \\\\ 0 & 3 \\end{pmatrix}. \\end{equation*} From Section 3.3, we know that the general solution to the system \\begin{equation*} \\begin{pmatrix} dx/dt \\\\ dy/dt \\end{pmatrix} = \\begin{pmatrix} 3 & 1 \\\\", "and $z$. $(x+1)^2+(y-2)^2$ $+(z-1)^2=17$ $(5+3r+1)^2$ $+(6+2r-2)^2$ $+(5+2r-1)^2=17$ $(6+3r)^2$ $+(4+2r)^2$ $+(4+2r)^2=17$ Use binomial theorem to resolve parentheses $36+36r+9r^2$ $+16+16r+4r^2$ $+16+16r+4r^2=17$ $17r^2+68r+68=17\\quad|-17$ $17r^2+68r+51=0\\quad|:17$ $r^2+4r+3=0$ Use quadratic formula I to solve quadratic equation. $r_{1,2}=-\\frac{p}2\\pm\\sqrt{(\\frac{p}2)^2-q}$ $r_{1,2}=-2\\pm\\sqrt{2^2-3}$ $r_{1,2}=-2\\pm1$ $r_{1}=-1$ and $r_{2}=-3$ 3. #### Insert $r$ The two calculated $r$ are inserted into the equation of a line in order to obtain the intersection points. $\\vec{OS_1} = \\begin{pmatrix} 5 \\\\ 6 \\\\ 5 \\end{pmatrix} - 1 \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $=\\begin{pmatrix} 2 \\\\ 4 \\\\ 3 \\end{pmatrix}$ $\\vec{OS_2} = \\begin{pmatrix} 5 \\\\ 6 \\\\ 5 \\end{pmatrix} - 3 \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $=\\begin{pmatrix} -4 \\\\ 0 \\\\ -1 \\end{pmatrix}$ It is a secant that intersects the sphere at $S_1(2|4|3)$ and $S_2(-4|0|-1)$." ]

[ "either that the phase shift is $+ \\pi$, or that the phase shift is $- \\pi$, and the graph would look the same. $y = \\sin \\left(x\\right)$ $y = - \\frac{1}{2} \\cdot \\sin \\left(x\\right)$ • 12 minutes ago • 13 minutes ago • 16 minutes ago • 17 minutes ago • 43 seconds ago • 2 minutes ago • 4 minutes ago • 4 minutes ago • 4 minutes ago • 11 minutes ago • 12 minutes ago • 13 minutes ago • 16 minutes ago • 17 minutes ago", "h=22.5 we have a phase shift of 22.5 left. (A phase shift of `pi/8~~.393` when using radians.) The graph of `y=cos(theta)` and the transformed graph:", "How do you graph y = cos(x- pi/6)? It is graph of y= cos x with a phase shift of $\\frac{\\pi}{6}$ The graph of y= cos x with a horizontal shift to the right by $\\frac{\\pi}{6}$, called phase shift, would represent the graph of $y = \\cos \\left(x - \\frac{\\pi}{6}\\right)$", "# How do you graph y = cos(x- pi/6)? Sep 18, 2015 #### Answer: It is graph of y= cos x with a phase shift of $\\frac{\\pi}{6}$ #### Explanation: The graph of y= cos x with a horizontal shift to the right by $\\frac{\\pi}{6}$, called phase shift, would represent the graph of $y = \\cos \\left(x - \\frac{\\pi}{6}\\right)$", "The value of $$C$$ is the phase shift (horizontal shift) of the sinusoidal function. The graph is shifted to the right if $$C > 0$$ and shifted to the left if $$C < 0$$. • The value of $$D$$ is the vertical shift of the sinusoid. The horizontal line $$y = D$$ is the so-called center line for the graph of the sinusoidal function. • The important notes about sinusoids on page 103.", "Phase Shift: ( to the right) Vertical Shift:", "makes no difference to the phase-shift: it merely 'stretches' the graph in a vertical direction either side of the x-axis by a factor of $5$. • October 17th 2009, 10:23 AM S2Krazy ah icic cool thxs. so for http://www.webassign.net/www38/latex...d954a06db3.gif the phase shift would be -4pi/5 correct? • October 17th 2009, 10:54 AM Quote: Originally Posted by S2Krazy ah icic cool thxs. so for http://www.webassign.net/www38/latex...d954a06db3.gif the phase shift would be -4pi/5 correct? Correct!", "# How do you find the phase shift of y=sin(theta-3 pi/4)? Aug 2, 2016 $$ Horizontal shift =3pi/4 #### Explanation: $y = \\sin \\left(\\theta - 3 \\frac{\\pi}{4}\\right)$ we have $a = 1$ $b = 1$ $c = 3 \\frac{\\pi}{4}$ $$ A phase shift is nothing but horizontal shift. Horizontal shift =3pi/4", "think of the phase shift is that it's just a new start to the graph. You can treat the phase shift as a place to actually begin your graph. So.. with all this information, we can now actually graph the equation of $y = 5 \\cos \\left(x - \\pi\\right) + 1$ . Here are some visual cues. My art is pretty bad, sographing the graph on a website should be much more different. I hope this helps.", "graph in a vertical direction either side of the x-axis by a factor of $5$. 3. ah icic cool thxs. so for the phase shift would be -4pi/5 correct? 4. Originally Posted by S2Krazy ah icic cool thxs. so for the phase shift would be -4pi/5 correct? Correct!", "# How do you graph sin(x/2)? It is easy to graph y= $\\sin \\left(\\frac{x}{2}\\right)$. The graph would be similar to the graph of sin x, the only difference being that its period would be $4 \\pi$, instead of $2 \\pi$. Amplitude would remain 1. There would be no phase shift. It would appear to be like this:", "So far I have y=10sin(pi/20,000(x-d))+70 How do I get d? Can someone clearly explain to me how to get the phase shift? thanks let the year $2000$ be $t = 0$ $T_{avg} = 10\\cos\\left(\\frac{\\pi t}{20000}\\right) + 70$", "I think! So, I let Maxima cut through all the algebra of the RC phase shift network. Lo and behold, here we have the results, achieved almost painlessly! $f = \\frac{1}{2 \\pi \\sqrt{R_2R_3C_2C_3 + R_1R_3C_2C_3 + R_1R_3C_1C_3 + R_1R_2C_2C_3 + R_1R_2C_1C_3 + R_1R_2C_1C_2}}$ For the benefit of the reader, I have also attached the Maxima batch files which did all the awesome work.", "# How do I find the phase shift of a sinusoidal graph? Jun 19, 2015 Find the value of $x$ at which we find $\\sin \\left(0\\right)$. #### Explanation: Examples: $y = 3 + 2 \\sin \\left(3 x - \\frac{\\pi}{4}\\right)$ Solve: $3 x - \\frac{\\pi}{4} = 0$ So $x = \\frac{\\pi}{12}$ is the shift. $y = 5 - \\sin \\left(4 x + \\frac{\\pi}{2}\\right)$ Solve: $4 x + \\frac{\\pi}{2} = 0$ So $x = - \\frac{\\pi}{8}$ is the shift. $y = - 3 \\sin \\left(\\pi x + \\frac{3 \\pi}{4}\\right)$ Solve: $\\pi x + \\frac{3 \\pi}{4} = 0$ So $x = - \\frac{3}{4}$ is the shift. Some textbooks use: $y = a \\sin \\left(b x + c\\right)$ has phase shift $- \\frac{c}{b}$ Other textbooks use: $y = a \\sin \\left(b x - c\\right)$ has phase shift $\\frac{c}{b}$ For me (having taught out of both kinds of textbook), it's easier just to solve the equation.", "2, 2005 ### amcavoy At first glance it looks OK. The only thing I found is that for a phase shift to the right, it would be x-pi/3 rather than x+pi/3.", "# Laplace transform of phase-shifted sinusoid ## Homework Statement Find the laplace transform of [TEX]f(t)=sin(4t+\\frac{\\pi}{3})u(t)[/TEX]. http://img822.imageshack.us/img822/6996/codecogseqna.gif [Broken] ## Homework Equations Definition of laplace transform, and properties: http://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems\" ## The Attempt at a Solution I have a feeling that I need to use the time-shift property of the transform, but I need just a bit more direction on the strategy. Last edited by a moderator: $$f(t) = \\left[ sin(4t)cos(\\frac{\\pi}{3}) + cos(4t)sin(\\frac{\\pi}{3}) \\right] u(t).$$", "The phase shift of a graph determines if the graph is going to be shifted left or right on the x-plane of the graph. $$Asin[B(x-\\frac{C}{B})]+D$$ From the example above the phase shift of the graph would be. $$\\frac{C}{B}$$ Let's do a short example of how the phase shifts would happen to a basic sin(x) function. $$sin(x)$$ If you take the function and shift it to the left $\\frac{\\pi}{2}$ length. $$sin(x+\\frac{\\pi}{2})$$ You can also go the other way and shift it to the right $\\frac{\\pi}{2}$ length. $$sin(x-\\frac{\\pi}{2})$$ Let's take a look at all three transformations.", "# How do you graph y= -sin(2x)-1/2? Jun 9, 2015 2. Double the frequency of the sin, so that the function y=sin(2x) pass through the points $\\left(0 , 0\\right) , \\left(\\frac{\\pi}{4} , 1\\right) , \\left(\\pi , 0\\right) , \\left(\\frac{3}{4} \\pi , - 1\\right) \\ldots$", "phase shift – a full period ($2\\pi$) ...", "# Trying to read phase shift constant from the graphs I have an equation $y(x,y) = y_0 \\sin (\\omega t \\pm kx \\pm \\phi)$ and there are two graphs. One is $y(x,t=0)$ in units $mm(mm)$ and the other is $y(t,x=0)$ in units $mm(s)$. Now I have to find all of the constants inside of the equation by looking at graphs and it goes well until I am to determine phase shift $\\phi$. If I calculate it by looking at $y(x=0,t=0)$ it works out well and I get right solution, but when I try to read $\\phi$ from graph $y(x,t=0)$ or $y(t,x=0)$ I get two wrong solutions. The idea for reading $\\phi$ from graphs was to express some distance $x'$ or time $t'$ in radians but it looks like I can't get it right. The sign in front of $\\phi$ is not a problem - if my sinus (in red) is shifted right it is minus otherwise it is plus. Can anyone clear things out for me and explain why I get wrong results which also differ from one-another? • Your result using $y(x, t=0)$ appears correct, it agrees to within measurement error. Your result for $\\phi$ from $y(x=0, t)$ should be replaced with $\\pi - \\phi$, since there is a relative minus sign between $\\omega t$ and $kx$" ]

[ "yields $$\\Delta p(x,y) = - a^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right) - b^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right)$$ and $$Q = - \\Delta p(x,y) = a^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right) + b^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right).$$ ## Input files The project file is square_1e3_poisson_sin_x_sin_y.prj. It describes the process to be solved and the related process variable together with their initial, boundary conditions and source terms. The definition of the source term $Q$ is in a Python script. The script for setting the source terms is referenced in the project file as follows: <python_script>sin_x_sin_y_source_term.py</python_script> In the source term description <source_term> <type>Python</type> <source_term_object>sinx_siny_source_term</source_term_object> </source_term> the domain is specified by the mesh-tag. The function $Q$ is defined by the Python object sinx_sinx_source_term that is created in the last line of the Python script: import OpenGeoSys from math import pi, sin a = 2.0*pi b = 2.0*pi def solution(x, y): return - sin(a*x-pi/2.0) * sin(b*y-pi/2.0) # - laplace(solution) = source term def laplace_solution(x, y): return a*a * sin(a*x-pi/2.0) * sin(b*y-pi/2.0) + b*b * sin(a*x-pi/2.0) * sin(b*y-pi/2.0) # source term for the benchmark class SinXSinYSourceTerm(OpenGeoSys.SourceTerm): def getFlux(self, t, coords, primary_vars): x, y, z = coords value = laplace_solution(x,y) Jac = [ 0.0 ] return (value, Jac) # instantiate source term object referenced in OpenGeoSys' prj file sinx_siny_source_term", "1x = np.linspace(-2*np.pi, 2*np.pi, 100) # domain 2np.sin(x-np.pi/2) # shifted right 3np.sin(x) 4np.sin(x+np.pi/2) # shifted left • The second subplot displays the original graph of $\\sin x$ defined within the domain $-2\\pi \\leq x \\leq 2\\pi$. • The first subplot displays the graph of $\\sin (x-\\frac{\\pi}{2})$ which is the graph of $\\sin x$ being shifted by $\\frac{\\pi}{2}$ units to the right. • The second subplot displays the graph of $\\sin (x+\\frac{\\pi}{2})$ which is the graph of $\\sin x$ being shifted by $\\frac{\\pi}{2}$ units to the left. ### Shifting in the $y$-direction Note Given $k$ is a positive real number: • The graph of $f(x)+k$ is the graph of $f(x)$ shifted vertically up by $k$ units. • The graph of $f(x)-k$ is the graph of $f(x)$ shifted vertically down by $k$ units. Example Shifting in the $y$-direction. 1x = np.linspace(-2*np.pi, 2*np.pi, 100) # domain 2np.sin(x)+1 # shifted up 3np.sin(x) 4np.sin(x)-1 # shifted down • The line in red displays the original graph of $\\sin x$ defined within the domain $-2\\pi \\leq x \\leq 2\\pi$. • The line in blue displays the graph of $\\sin(x)+1$ which is the graph of $\\sin x$ being shifted vertically up by $1$ unit. • The line in green displays the graph of $\\sin(x)-1$ which is the graph of $\\sin x$ being shifted vertically down", "geometric figures. /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ The first line real f1 (real x) {return sin(x);} defines the sin() function to be drawn, and the next line draw(graph(f1,-6.21,8.03)); actually draws the graph to the diagram. The graph() command comes from the Asymptote graph package. The next few lines draw several line segments. The attribute zzttqq is a pen whose color is mapped from some rgb value (with conversions ... 8 = y, 9 = z, a = a, b = b, ... ). ## Options An outline of the different available options. (The first three options are related and automatically update with respect to eath other). X/Y Units (cm): Indicates how many units in the picture correspond to 1 centimeter. Picture width/height: Indicates what size the Asymptote output should be, in centimeters. xmin/xmax/ymin/ymax: Indicates the dimensions of the picture in the picture's coordinates. (The next options change other features). Font size: Changes the default font size. The value defaults at 10pt, but I personally recommend 7pt for any file with multiple text lines, which more closely approximates the size of the font", "what translation to apply, however, we first factorise the function into the form $y=-2\\sin\\left(3\\left(x+\\frac{\\pi}{12}\\right)\\right)$y=2sin(3(x+π12)). In this form, we can see that the $x$x-values are increased by $\\frac{\\pi}{12}$π12 inside the function. This means that to get a particular $y$y-value, we can put in an $x$x-value that is $\\frac{\\pi}{12}$π12 smaller than before. Graphically, this corresponds to shifting the entire function to the left by $\\frac{\\pi}{12}$π12 units. The graph of $y=-2\\sin\\left(3x+\\frac{\\pi}{4}\\right)$y=−2sin(3x+π4​) Lastly, we translate the graph of $y=-2\\sin\\left(3x+\\frac{\\pi}{4}\\right)$y=2sin(3x+π4) upwards by $2$2 units, to obtain the final graph of $y=-2\\sin\\left(3x+\\frac{\\pi}{4}\\right)+2$y=2sin(3x+π4)+2. The graph of $y=-2\\sin\\left(3x+\\frac{\\pi}{4}\\right)+2$y=−2sin(3x+π4​)+2 Careful! When we geometrically apply each transformation to the graph of $y=\\sin x$y=sinx, it's important to consider the order of operations. If we had wanted to vertically translate the graph before reflecting about the $x$x-axis, we would have needed to translate the graph downwards first. ### The general case In the example above we were transforming the graph of $y=\\sin x$y=sinx. The particular function $y=\\sin x$y=sinx was not important, however. We could have just as easily transformed the graph of $y=\\cos x$y=cosx, or even a non-trigonometric function, using the same method! Consider a function $y=f\\left(x\\right)$y=f(x). Then we can obtain the graph of $y=af\\left(b\\left(x-c\\right)\\right)+d$y=af(b(xc))+d, where $a,b,c,d$a,b,c,d are constants, by applying a series of transformations to the graph of $y=f\\left(x\\right)$y=f(x). These transformations are summarised below. Summary To obtain the graph", "not outweigh the negative by the cos and tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first solution is a little bit after $\\pi$, which is around $3.14$. Hence the answer is $\\boxed{\\textbf{(D)}}$. (Not original author) Here is the graph: $[asy] Label f; f.p=fontsize(6); xaxis(-5,5,Ticks(f, 1.0)); yaxis(-8,8,Ticks(f, 2.0)); real f(real x) { return sin(x); } draw(graph(f, -5,5)); real g(real x) { return 2*cos(x); } draw(graph(g, -5,5)); real h(real x) { return 3*tan(x); } draw(graph(h, -1.2,1.2)); draw(graph(h, 1.94, 4.34)); draw(graph(h, -4.34, -1.94)); [/asy]$ ## Solution 3(More Detailed Answer of Solution 1) Denote $D_{f}$ as the domain of $f(x).$ Obviously $D_{f}=\\left \\{x|x\\neq \\frac{(2k+1)\\pi}{2} \\right \\} .$ $f^{'}(x)=cosx-2sinx+3sec^{2}x=\\sqrt{5}cos(x+\\phi)+3sec^{2}x > 0$ for all $x$ in sub-intervals of $D_{f}.$ When $x\\in (0,\\frac{\\pi}{2}), f(0)=2, \\lim_{x \\to \\frac{\\pi}{2}^{-} }f(x)=+\\infty .$ Hence $f(x)>0$ in this interval. When $x\\in (\\frac{\\pi}{2},\\pi),\\lim_{x \\to \\frac{\\pi}{2}^{+} }f(x)=-\\infty, f(\\pi)=-2.$ Hence $f(x)<0$ in this interval. When $x\\in (\\pi,\\frac{3\\pi}{2}),f(\\pi)=-2<0.$ Notice that $\\frac{5\\pi}{4}<\\frac{3.15*5}{4}<4<\\frac{3\\pi}{2}, f(\\frac{5\\pi}{4})=3-\\frac{3\\sqrt{2}}{2}>0 .$ Hence there must be a root located in the interval $(3,4).$ Choose $\\boxed{\\textbf{(D)}}$. ~PythZhou", "it yields $\\Delta p(x,y) = - a^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right) - b^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right)$ and $Q = - \\Delta p(x,y) = a^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right) + b^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right).$ ## Input files The project file is square_1e3_poisson_sin_x_sin_y.prj. It describes the process to be solved and the related process variable together with their initial, boundary conditions and source terms. The definition of the source term $Q$ is in a Python script. The script for setting the source terms is referenced in the project file as follows: <python_script>sin_x_sin_y_source_term.py</python_script> In the source term descripition <source_term> <mesh>square_1x1_quad_1e3_entire_domain</mesh> <type>Python</type> <source_term_object>sinx_siny_source_term</source_term_object> </source_term> the domain is specified by the mesh-tag. The function $Q$ is defined by the Python object sinx_sinx_source_term that is created in the last line of the Python script: import OpenGeoSys from math import pi, sin a = 2.0*pi b = 2.0*pi def solution(x, y): return - sin(a*x-pi/2.0) * sin(b*y-pi/2.0) # - laplace(solution) = source term def laplace_solution(x, y): return a*a * sin(a*x-pi/2.0) * sin(b*y-pi/2.0) + b*b * sin(a*x-pi/2.0) * sin(b*y-pi/2.0) # source term for the benchmark class SinXSinYSourceTerm(OpenGeoSys.SourceTerm): def getFlux(self, t, coords, primary_vars): x, y, z = coords value = laplace_solution(x,y) Jac = [ 0.0 ] return (value, Jac) # instantiate source term object referenced in OpenGeoSys' prj", "New Zealand Level 8 - NCEA Level 3 # Phase shifts for tangents ## Interactive practice questions Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. A phase shift has the same effect as a horizontal translation. True A False B True A False B Easy Less than a minute Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. We want to identify how the coordinates of key points on the graph of $f\\left(x\\right)=\\tan x$f(x)=tanx change as we apply a phase shift to produce the graph of $g\\left(x\\right)=\\tan\\left(x-\\frac{\\pi}{3}\\right)$g(x)=tan(xπ3). ### Outcomes #### M8-2 Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions", "# ISRO2009-46 2.2k views The shift operator $E$ is defined as $E [f(x_i)] = f (x_i+h)$ and $E'[f(x_i)]=f (x_i -h)$ then $\\triangle$ (forward difference) in terms of $E$ is 1. $E-1$ 2. $E$ 3. $1-E^{-1}$ 4. $1-E$ Forward difference operator(delta (D)) D(f(x)) = f(x + h) - f(x) Shift operator , Ef(x) = f(x+h) = f(x+h) - f(x) + f(x) = (1 + D)f(x) so E= 1 + D gives D = E - 1 selected by ## Related questions 1 1.3k views The formula $P_k = y_0 + k \\triangledown y_0+ \\frac{k(k+1)}{2} \\triangledown ^2 y_0 + \\dots + \\frac{k \\dots (k+n-1)}{n!} \\triangledown ^n y_0$ is Newton's backward formula Gauss forward formula Gauss backward formula Stirling's formula The cubic polynomial $y(x)$ which takes the following values: $y(0)=1, y(1)=0, y(2)=1$ and $y(3)=10$ is $x^3 +2x^2 +1$ $x^3 +3x^2 -1$ $x^3 +1$ $x^3 -2x^2 +1$ The formula $\\int\\limits_{x0}^{xa} y(n) dx \\simeq h/2 (y_0 + 2y_1 + \\dots +2y_{n-1} + y_n) - h/12 (\\triangledown y_n - \\triangle y_0)$ $- h/24 (\\triangledown ^2 y_n + \\triangle ^2 y_0) -19h/720 (\\triangledown ^3 y_n - \\triangle ^3 y_0) \\dots$ is called Simpson rule Trapezoidal rule Romberg's rule Gregory's formula A root $\\alpha$ of equation $f(x)=0$ can be computed to any degree of accuracy if a 'good' initial approximation $x_0$ is chosen for which $f(x_0)", "Solved Example. Graph$\\,y=\\mathrm{arccos}\\,x\\,$and state the domain and range of the function. How to find the base point given public and private key and EC parameters except the base point. The inverse of a 2x2 is easy... compared to larger matrices (such as a 3x3, 4x4, etc). Please help me answering this problem! Post-multiply both sides of … If you want to share a math-problem you need to explain what you're arguments are. For multiplication, it's division. Show … We can check … Are there any Pokémon that lose overall base stats when they evolve? Given a “special” input value, evaluate an inverse trigonometric function. [/latex], If$\\,x\\text{ is in }\\left[0,\\pi \\right],\\,$then$\\,{\\mathrm{sin}}^{-1}\\left(\\mathrm{cos}\\,x\\right)=\\frac{\\pi }{2}-x. The Inverse Park Transform block converts the time-domain direct, quadrature, and zero components in a rotating reference frame to the components of a three-phase system in an abc reference frame. Figure 11. If y = f(x) = ab x, then we may solve for x in terms of y using logarithms: .$, If$\\,x\\text{ is not in}\\,\\left[-\\frac{\\pi }{2},\\frac{\\pi }{2}\\right],\\,$then find another angle$\\,y\\text{ in }\\left[-\\frac{\\pi }{2},\\frac{\\pi }{2}\\right]\\,$such that$\\,\\mathrm{sin}\\,y=\\mathrm{sin}\\,x.$. prove that $\\det(ABC) = \\det(A) \\det(B) \\det(C)$ [for any $n×n$ matrices $A$, $B$, and $C$]. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical", "layers. In order for my approach to work as intended, the layer settings window must look like this: However, quite annoyingly, when you close and reopen this saved document, the layer settings window looks like this: So, it must be changed back each time to allow trouble-free math drawing. # Finding Reciprocal Functions SEC, CSC, COT on TI-30XA First, you may want to refresh your memory of the regular trig functions: Now, let’s find $\\csc \\frac{5\\pi }{4}$. If we look it up on WolframAlpha, we find that it equals $-\\sqrt{2}$, or, roughly, $-1.4142$. To get this answer on the TI-30XA, first remember that $\\csc$ is equal to $\\frac{1}{\\sin }$. So, what we’re looking for could be expressed as $\\frac{1}{\\sin \\frac{5\\pi }{4}}$. Once we know this, we can type $5\\cdot \\pi$ into the TI-30XA. This gives us ~15.7. (Make sure you’re in radian mode. If not, press the DRG button — located by the 2nd button, upper-left — until you see RAD appear in small letters on the screen.) Now, we can divide 15.7 by 4. This gives us 3.925. If we take the $\\sin$ of this (by pressing the “sin” button while 3.925 is displayed), we get $-0.7$. IF we divide 1 by $-0.7$, we get $-1.43$. As you can see, this is close to the correct value.", "SYSTEM label(\"$O$\", (0, 0, 0), NW); Label Lx = Label(\"$x$\", EndPoint, W); Label Ly = Label(\"$y$\", EndPoint, S); Label Lz = Label(\"$z$\", EndPoint, W); draw(Lx, O--2X, Arrow3(emissive(black))); draw(Ly, O--2Y, Arrow3(emissive(black))); draw(Lz, O--2Z, Arrow3(emissive(black))); // SURFACE triple S(pair uv) { real x = cos(uv.x)*sin(uv.y); real y = sin(uv.x)*sin(uv.y); real z = cos(uv.y); return (x-(1-z**2-y**2), y-0.3*sin(z*pi), z); } // BOUNDARY triple dS(real u) { return S((u, pi/2)); } // TANGENT VECTORS real e = 0.001; // *this... is... infinitesimal!* triple U(pair t) { return (S((e,0)+t)-S(t))/e; } triple V(pair t) { return (S((0,e)+t)-S(t))/e; } // NORMAL VECTOR triple N(pair uv) { return cross(U(uv),V(uv))/length(cross(U(uv),V(uv))); } // GRAPHICAL ELEMENTS surface s = surface(S, (0, 0), (2pi, pi/2), 64, 64, Spline); path3 ds = graph(dS, 0, 2pi); pair t = (-pi/8, pi/8); path3 n = S(t) -- shift(-N(t))*S(t); label(\"$\\Sigma$\", shift(0,0,0.1)*shift(adj)*S((pi/2,pi/5)), NE); Label LdS = Label(\"$\\partial\\Sigma$\", Relative(0.95), S); Label Ln = Label(\"$\\mathbf n$\", EndPoint, NW); draw(s, surfacepen=material(diffusepen=gray(0.6),emissivepen=gray(0.3),specularpen=gray(0.1))); draw(LdS, ds, Arrow3(emissive(black))); draw(shift(4*unit(currentprojection.camera))*ds, dashed); draw(Ln, n, Arrow3(emissive(black))); REQUESTED APPENDIX: on dashed lines I wanted to draw the dashed border over the surface. Since this is an orthographic projection, the easiest way is translating the object along the camera axis towards it (instead of checking what segments are visible, possibly complicated): draw(shift(4*unit(currentprojection.camera))*ds, dashed); unit(currentprojection.camera) is a versor parallel to the camera axis and pointing towards it, so it", "figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ /* dots and labels */ label(\"f\",(-6.12,-0.12),NE * labelscalefactor); label(\"a = 2\",(1.46,0.24),NE * labelscalefactor,zzttqq); dot((-3.14,0),dotstyle); label(\"A\",(-3.06,0.12),NE * labelscalefactor); dot((2.2,2.04),dotstyle); dot((4.5,2.08),dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$ Here is the Asymptote code produced: (with some spacing added) /* Geogebra to Asymptote conversion, documentation at userscripts.org/scripts/show/72997 */ import graph; size(14.26cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.22,xmax = 8.04,ymin = -4.18,ymax = 4.76; /* image dimensions */ pen fueaev = rgb(0.96,0.92,0.9); pen zzttqq = rgb(0.6,0.2,0); real f2 (real x) {return sin(x);} filldraw(graph(f2,0,pi )--(pi ,0)--(0,0)--cycle, fueaev, zzttqq); filldraw((-3.14,0)--(0,0)--(0,3.14)--(-3.14,pi )--cycle, fueaev, zzttqq); Label laxis; laxis.p = fontsize(10); string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real", "x) {return sin(x);} filldraw(graph(f2,0,pi )--(pi ,0)--(0,0)--cycle, fueaev, zzttqq); Some commands have to be defined with respect to a function. $\\texttt{f2}$ returns the $\\sin(x)$ function, and the $\\texttt{graph}$ command draws $\\texttt{f2}$ from $x = 0$ to $x = \\pi$. Label laxis; laxis.p = fontsize(10); string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); These commands draw the axes. The $\\texttt{laxis}$ label specifies the label styles of the axes (e.g., the field $\\texttt{laxis.p}$ defines the desired font pen). The next line, the string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} are functions that define the labels for both axes. These are abbreviated as $\\texttt{xlbl, ylbl}$ in the concise mode. The first defines axes with labels with respect to $\\pi$, and it formats the labels so that the $0$ label simply outputs $0$ instead of $0\\pi$, and $\\pm \\pi$ instead of $\\pm 1\\pi$. Finally, xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); are the commands that actually construct the xaxis and the yaxis. For documentation, see here. The next block of code does the actual construction of the", "\\cos(\\frac{2\\pi}{16}3 x) + \\sin(\\frac{2\\pi}{16} 6 x)$$ Function 3: $$f(x) = \\sum_{n = 0}^{15} n \\sin(\\frac{2 \\pi }{16}n x)$$ Function 4: $$f(x) = \\sum_{n = 0}^{15} \\sin(\\frac{2\\pi}{16} n )\\sin(\\frac{2 \\pi}{16}n x)$$ library(ggplot2) library(dplyr) library(tidyr) plot_theme <- theme(panel.background = element_blank(), panel.grid.major = element_line(color = \"grey90\", size = 0.2), panel.grid.minor = element_line(color = \"grey95\", size = 0.5), strip.text.x = element_text(color = \"white\"), strip.background = element_rect(fill = \"black\"), axis.text = element_text(color = \"black\"), axis.title = element_text(face = \"bold\"), title = element_text(face = \"bold\")) ## Here are the functions we want to test, they should have very clear Fourier transforms function.1 <- function(x) sin(2*pi*x/16) function.2 <- function(x) sin(2*pi*x/16) - cos(2*pi*3*x/16) + sin(2*pi*6 * x/16) function.3 <- function(x) sapply(x, function(s) sum(sapply(0:15, function(n) n* sin(2*pi*n*s/16)))) function.4 <- function(x) sapply(x, function(s) sum(sapply(0:15, function(n) sin(2*pi*n/16)* sin(2*pi*n*s/16)))) tests <- list(\"Function 1\" = function.1, \"Function 2\" = function.2, \"Function 3\" = function.3, \"Function 4\" = function.4) ## for each of these function, we want to apply the function test.data <- bind_rows(lapply(1:length(tests), function(i) { fun <- tests[[i]] name <- names(tests)[[i]] x <- 0:15 bind_rows( data.frame(x = x, y = fun(x), func = name, transform = \"None\"), data.frame(x = x, y = abs(fft(fun(x))), func = name, transform = \"fft\"), data.frame(x = x, y = abs(slow.fourier.transform(fun(x))), func = name, transform = \"slow.fourier.transform\"), data.frame(x = x, y = abs(fast.fourier.transform(fun(x))), func = name, transform", "calculate that part and end up with $35.75$. We also see with the formulas we used with the plug in that when you increase by $0.2$ the $35.75$ part decreases by $0.25$. The answer is then $\\boxed{(D) 36.8}$. ## Solution 13 1、First, move terms to get $1+5cos3x=3sinx$. After graphing, we find that there are $\\boxed{6}$ solutions (two in each period of $5cos3x$). -dstanz5 2、We can graph two functions in this case: $5\\cos{3x}$ and $3\\sin{x} -1$.$$\\newline$$Using transformation of functions, we know that $5\\cos{3x}$ is just a cos function with amplitude 5 and period $\\frac{2\\pi}{3}$. Similarly, $3\\sin{x} -1$ is just a sin function with amplitude 3 and shifted 1 unit downwards. So:$[asy] import graph; size(400,200,IgnoreAspect); real Sin(real t) {return 3*sin(t) - 1;} real Cos(real t) {return 5*cos(3*t);} draw(graph(Sin,0, 2pi),red,\"3\\sin{x} -1 \"); draw(graph(Cos,0, 2pi),blue,\"5\\cos{3x}\"); xaxis(\"x\",BottomTop,LeftTicks); yaxis(\"y\",LeftRight,RightTicks(trailingzero)); add(legend(),point(E),20E,UnFill); [/asy]$We have $\\boxed{(A) 6}$ ## Solution 14 1、This question is just about pythagorean theorem$$a^2+(a+2)^2-b^2 = (a+4)^2$$$$2a^2+4a+4-b^2 = a^2+8a+16$$$$a^2-4a+4-b^2 = 16$$$$(a-2+b)(a-2-b) = 16$$$$a=3, b=7$$With these calculation, we find out answer to be $\\boxed{\\textbf{(A) }24\\sqrt5}$ ~Lopkiloinm 2、Let $\\overline{AD}$ be $b$$\\overline{CD}$ be $a$$\\overline{MD}$ be $x$$\\overline{MC}$$\\overline{MA}$$\\overline{MB}$ be $t$$t-2$$t+2$ respectively. We have three equations:$$a^2 + x^2 = t^2$$$$a^2 + b^2 + x^2 = t^2 + 4t + 4$$$$b^2 + x^2 = t^2 - 4t + 4$$ Subbing in the first and third equation into the second equation, we get:$$t^2", "draw(Label(\"$\\pi$\",1),shift(-0.15*Z)*scale3(0.5)*g,blue,Arrow3); path3 h=arc(O,1,170,-120,150,-120,CCW); draw(Label(\"$\\sigma$\",1),shift(-0.5*cos(pi/6)*Y-0.5*sin(pi/6)*X+0.35*Z)*scale3(0.5)*h,red,Arrow3); draw(shift(-0.5*cos(pi/6)*Y-0.5*sin(pi/6)*X-0.15*Z)*scale3(0.1)*unitsphere,gray+opacity(.3)); and the result You can easily modify the xpt and ypt values to find the best shape. For periodic spline, do not forget to have periodic value. O.G. • Great! But may I ask where the lower bound in (0.1,2*pi) comes from? – user121799 Jan 4 '18 at 0:28 • @marmot I add a comment. In view of your attempt, I transform the box (a,b) (-0.67,0.1),(0,2.032*pi) to (-0.67,0),(0.1,2*pi) so that the x variable varies from -0.67 to 0.1 while the y variable varies from 0 to 2*pi. Perhaps that (-0.67,0),(0,2*pi) should be ok... – O.G. Jan 4 '18 at 9:48", "Here, the graphing calculator format is: $Y_1 = \\sin(X)$ $Y_2 = \\sin(X+\\pi/4)$ Using window coordinates $-6.5 ≤ x ≤ 6.5$ $-1.25 ≤ y ≤ 1.25,$ we get the following: $Y_1 = \\sin(X)$ is shown in red, and $Y_2 = \\sin(X+\\pi/4)$ is in yellow. We notice that addition of $\\pi/4$ to the argument has shifted the graph to the left by $\\pi/4$ units.In general: Replacing $x$ by $x+c$ shifts the graph to the left $c$ units. How would we shift the graph to the right $\\pi/4$ units? (c) Here, the graphing calculator format is: $Y_1 = \\sin(X)$ $Y_2 = \\sin(3*X)$ Using the same window coordinates and color coding as in part (b), we obtain the following graph. Notice that the graph of $\\sin(3*x)$ oscillates three times as fast as the graph of $\\sin(x).$ Thus, every $2\\pi$ units, the graph of $\\sin(3*x)$ oscillates three times. In general: Multiplying $x$ by $b$ multiplies the rate of oscillation by $b.$ (d) $Y_1 = \\sin(\\pi*X)$ $Y_2 = 2\\sin(\\pi(X+3))$ The graphs are shown in the window $-3.2 ≤ x ≤ 3.2$ $-2.5 ≤ y ≤ 2.5$ Notice several things: • The red graph, $y = \\sin(\\pix),$ crosses the $x-$axis at integral values $0 ±1, ±2, ...$ For instance, when $x = 1,$ we get $\\sin(\\pix) = \\sin(\\pi) = 0.$ • The yellow graph, $y", "# Math Help - Moving a Trigonometric function 1. ## Moving a Trigonometric function y=d+sin(bx+c) So if I have the equation: sin(2x)=y; I know that the period will be pi. So the whole graph will occur between (0,0) and (pi,0) Start at (0,0), Maximum (pi/4, 1); intercept at (pi/2, 0); Minimum (3Pi/4, -1); Intercept (pi,0) Now if I have: y=sin(x-(pi/4)) the graph gets shifted to the right pi/4 radians. So key points are Start: (pi/4, 0); Maximum: (3pi/4, 1); Intercept: (5pi/4,0); minimum (7pi/4, -1); Complete (9pi/4,0) If I put them together as Sin(2x-(pi/4) Why does my horizontal shift get cut in half? Why does it not start at (pi/4)? Why does it start at (pi/8)? In my graphing calculator I see everything gets moved my (pi/8) when I have this equation. 2. Originally Posted by IDontunderstand y=d+sin(bx+c) So if I have the equation: sin(2x)=y; I know that the period will be pi. So the whole graph will occur between (0,0) and (pi,0) Start at (0,0), Maximum (pi/4, 1); intercept at (pi/2, 0); Minimum (3Pi/4, -1); Intercept (pi,0) Now if I have: y=sin(x-(pi/4)) the graph gets shifted to the right pi/4 radians. So key points are Start: (pi/4, 0); Maximum: (3pi/4, 1); Intercept: (5pi/4,0); minimum (7pi/4, -1); Complete (9pi/4,0) If I put them together as Sin(2x-(pi/4) Why does my", "# Graphing sin, finding phase shift, period and transformations I need to graph $$y=2 \\cos\\left(x - \\frac{\\pi}{3}\\right)$$ and I am having trouble figuring out what points will be on the graph. The book tells me to split it into four parts, which doesn't make sense since they split it into five parts anyways. I know that the points will be between $\\pi/3$ and $(\\pi/3 + 2\\pi)$ so that gives me $\\pi/3 < x < 7\\pi/3$ The only problem is I have no idea how to split that up into equal parts to make graphing easier. - Indeed, it is useful to split into four parts, as follows: $$\\bigg[\\frac{{2\\pi }}{6},\\frac{{5\\pi }}{6}\\bigg],\\bigg[\\frac{{5\\pi }}{6},\\frac{{8\\pi }}{6}\\bigg],\\bigg[\\frac{{8\\pi }}{6},\\frac{{11\\pi }}{6}\\bigg],\\bigg[\\frac{{11\\pi }}{6},\\frac{{14\\pi }}{6}\\bigg].$$ The endpoints are $\\pi/3$ and $7\\pi / 3$. The middle point is then given by $$\\bigg(\\frac{\\pi }{3} + \\frac{{7\\pi }}{3}\\bigg)\\bigg /2 = \\frac{{8\\pi }}{6}.$$ Next, the middle point between $\\pi/3$ and $8\\pi/6$ is given by $$\\bigg(\\frac{\\pi }{3} + \\frac{{8\\pi }}{6}\\bigg)\\bigg/2 = \\bigg(\\frac{{2\\pi }}{6} + \\frac{{8\\pi }}{6}\\bigg)\\bigg/2 = \\frac{{5\\pi }}{6},$$ and the middle point between $8\\pi/6$ and $7\\pi/3$ is given by $$\\bigg(\\frac{8\\pi }{6} + \\frac{{7\\pi }}{3}\\bigg)\\bigg/2 = \\bigg(\\frac{{8\\pi }}{6} + \\frac{{14\\pi }}{6}\\bigg)\\bigg/2 = \\frac{{11\\pi }}{6}.$$ Adam: $\\pi/3 = 2\\pi/6$, $7\\pi/3 = 14\\pi/6$..., the book just reduced the fractions (Shai used 6 in the denominator to help establish the four parts, you can simply", "How do you graph y=2sin(x + pi)? Apr 15, 2016 Explanation: A typical graph of $y = \\sin x$ has domain for all values of $x$ and range is from $\\left[- 1 , 1\\right]$. It is a cyclical curve and repeats after every $2 \\pi$, hence its period is $2 \\pi$. It's value is $0$ at each $n \\pi$, it touches a maximum value of $1$ at each $2 n \\pi + \\frac{\\pi}{2}$ and a minimum value of $- 1$ at each $2 n \\pi - \\frac{\\pi}{2}$ (where $n$ is an integer). It appears like graph{sin(x) [-10, 10, -2, 2]} If we draw the graph of $y = 2 \\sin \\left(x + \\pi\\right)$, the range will be doubled due to multiplier $2$ and will be $\\left[- 2 , 2\\right]$. However, the graph will be shifted by $\\pi$ and hence minimum value will be $- 2$ at each $2 n \\pi + \\frac{\\pi}{2}$ and a maximum value of $- 2$ at each $2 n \\pi - \\frac{\\pi}{2}$ (where $n$ is an integer). But the function will continue to have value $0$ at each $n \\pi$. As the period of curve is $2 \\pi$, it does not matter whether you say it has shifted by $\\pi$ to the left or right. The graph of $2 \\sin \\left(x + \\pi\\right)$ appears" ]

[ "New Zealand Level 8 - NCEA Level 3 # Phase shifts for tangents ## Interactive practice questions Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. A phase shift has the same effect as a horizontal translation. True A False B True A False B Easy Less than a minute Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. We want to identify how the coordinates of key points on the graph of $f\\left(x\\right)=\\tan x$f(x)=tanx change as we apply a phase shift to produce the graph of $g\\left(x\\right)=\\tan\\left(x-\\frac{\\pi}{3}\\right)$g(x)=tan(xπ3). ### Outcomes #### M8-2 Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions", "1x = np.linspace(-2*np.pi, 2*np.pi, 100) # domain 2np.sin(x-np.pi/2) # shifted right 3np.sin(x) 4np.sin(x+np.pi/2) # shifted left • The second subplot displays the original graph of $\\sin x$ defined within the domain $-2\\pi \\leq x \\leq 2\\pi$. • The first subplot displays the graph of $\\sin (x-\\frac{\\pi}{2})$ which is the graph of $\\sin x$ being shifted by $\\frac{\\pi}{2}$ units to the right. • The second subplot displays the graph of $\\sin (x+\\frac{\\pi}{2})$ which is the graph of $\\sin x$ being shifted by $\\frac{\\pi}{2}$ units to the left. ### Shifting in the $y$-direction Note Given $k$ is a positive real number: • The graph of $f(x)+k$ is the graph of $f(x)$ shifted vertically up by $k$ units. • The graph of $f(x)-k$ is the graph of $f(x)$ shifted vertically down by $k$ units. Example Shifting in the $y$-direction. 1x = np.linspace(-2*np.pi, 2*np.pi, 100) # domain 2np.sin(x)+1 # shifted up 3np.sin(x) 4np.sin(x)-1 # shifted down • The line in red displays the original graph of $\\sin x$ defined within the domain $-2\\pi \\leq x \\leq 2\\pi$. • The line in blue displays the graph of $\\sin(x)+1$ which is the graph of $\\sin x$ being shifted vertically up by $1$ unit. • The line in green displays the graph of $\\sin(x)-1$ which is the graph of $\\sin x$ being shifted vertically down", "New Zealand Level 7 - NCEA Level 2 # Phase shifts for tangents ## Interactive practice questions Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. A phase shift has the same effect as a horizontal translation. True A False B True A False B Easy Less than a minute Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. We want to identify how the coordinates of key points on the graph of $f\\left(x\\right)=\\tan x$f(x)=tanx change as we apply a phase shift to produce the graph of $g\\left(x\\right)=\\tan\\left(x-60^\\circ\\right)$g(x)=tan(x60°). ### Outcomes #### M7-2 Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs #### 91257 Apply graphical methods in solving problems", "I think! So, I let Maxima cut through all the algebra of the RC phase shift network. Lo and behold, here we have the results, achieved almost painlessly! $f = \\frac{1}{2 \\pi \\sqrt{R_2R_3C_2C_3 + R_1R_3C_2C_3 + R_1R_3C_1C_3 + R_1R_2C_2C_3 + R_1R_2C_1C_3 + R_1R_2C_1C_2}}$ For the benefit of the reader, I have also attached the Maxima batch files which did all the awesome work.", "import numpy as np sr = 22050 ts = 1.0/sr duration = 2 t = np.arange(0,duration,ts) t_change = 1 freq1 =4.20 phase1 = 2*np.pi*freq1* t[t<=t_change] freq2 = 6.66 phase2 = 2*np.pi*freq2*t[t>t_change] + 2*np.pi*(freq1-freq2)*t_change total_phase = np.concatenate((phase1, phase2)) out = np.cos(total_phase) Considering more generalized cases with multiple steps, the phase offset to add for any given segment is: $$\\theta_n = \\sum_{m=1}^{n-1}2\\pi f_m t_m - \\sum_{m=1}^{n-1}2\\pi f_{m+1} t_m \\tag{1} \\label{1}$$ With the resulting phase ramp proceeding starting at $$t_n$$ as: $$\\theta(t) = 2\\pi f_n t + \\theta_n$$ This is demonstrated with a 3 segment ramp with reference to the figure below, where we note that each next phase ramp will start at the completed phase from the previous frequency. For example, the second phase offset given as $$\\theta_2$$ below is similarly determined as above by solving for the intersection of linear line equations to be: $$\\theta_2 = \\theta(t_2)-2\\pi f_3 t_2$$ $$= 2\\pi f_2 t_2 + \\theta_1 - 2\\pi f_3 t_2$$ $$= 2\\pi f_2 t_2 + 2\\pi (f_1-f_2)t_1 - 2\\pi f_3 t_2$$ Which is consistent with the generalized form given in $$\\ref{1}$$ as: $$\\theta_2 = 2\\pi f_1 t_1 + 2\\pi f_2 t_2 -2\\pi f_2 t_1 - 2\\pi f_3 t_2$$ With this, the third segment plotted below is given as: $$\\theta(t) = 2\\pi f_3 t + \\theta_2$$ Summary: In summary I have", "graph of $f\\left(x\\right)=\\tan x$f(x)=tanx change as we apply a phase shift to produce the graph of $g\\left(x\\right)=\\tan\\left(x-\\frac{\\pi}{3}\\right)$g(x)=tan(xπ3). 1. The point $A$A on the graph of $f\\left(x\\right)$f(x) has the coordinates $\\left(0,0\\right)$(0,0). What are the coordinates of the corresponding point on the graph of $g\\left(x\\right)$g(x)? Give your answer in the form $\\left(\\editable{},\\editable{}\\right)$(,). 2. The point $B$B on the graph of $f\\left(x\\right)$f(x) has the coordinates $\\left(\\frac{\\pi}{4},1\\right)$(π4,1). What are the coordinates of the corresponding point on the graph of $g\\left(x\\right)$g(x)? Give your answer in the form $\\left(\\editable{},\\editable{}\\right)$(,). 3. The graph of $f\\left(x\\right)$f(x) has an asymptote passing through point $C$C with coordinates $\\left(\\frac{\\pi}{2},0\\right)$(π2,0). What are the coordinates of the corresponding point on the graph of $g\\left(x\\right)$g(x)? Give your answer in the form $\\left(\\editable{},\\editable{}\\right)$(,). 4. Using the answers from the previous parts, apply a phase shift to the graph of $f\\left(x\\right)=\\tan x$f(x)=tanx to draw the graph of $g\\left(x\\right)=\\tan\\left(x-\\frac{\\pi}{3}\\right)$g(x)=tan(xπ3). Question 2 The graph of $f\\left(x\\right)=\\tan x$f(x)=tanx has been drawn below. On the same set of axes draw the graph of $g\\left(x\\right)=\\tan\\left(x+\\frac{\\pi}{2}\\right)$g(x)=tan(x+π2). Question 3 The graph of a function in the form $f\\left(x\\right)=\\tan\\left(x-h\\right)$f(x)=tan(xh), where $0\\le h<\\pi$0h<π, is drawn below. 1. Compared to the graph of $y=\\tan x$y=tanx, state the phase shift of the graph of $f\\left(x\\right)$f(x).", "you would use closed graph theorem, in fact, you would need either \\mathcal{S}=\\mathcal{S}(\\mathbb{R}^d) to be a Banach space or compact Hausdorff. I do not believe either conditions are satisfied (\\mathcal{S} is normable or compact, at least I don't think so). ... 0 The shift is defined by g_a(x) = f(x-a). Then you write$$F[g_a](\\xi) = \\int_{\\Bbb R}g_a(x)\\exp(-ix\\xi)dx = \\int_{\\Bbb R}f(x-a)\\exp(-ix\\xi)dx.Conceptually, you first apply the shift and then apply the Fourier transform, but you can apply the shift only to the function, there is no sense in applying it to the exponent. 0 Let g_0 = \\frac{1}{2\\pi}\\int_{0}^{2\\pi}g(x)dx and f_0=\\frac{1}{2\\pi}\\int_{0}^{2\\pi}f(x)dx. Then \\begin{align} \\frac{1}{2\\pi}\\int_{0}^{2\\pi}f(x)g(nx)dx &=\\frac{1}{2\\pi}\\int_{0}^{2\\pi}f(x)(g(nx)-g_0)dx+f_0g_0 \\\\ &=\\frac{1}{2\\pi}\\sum_{k=1}^{n}\\int_{\\frac{2\\pi}{n}(k-1)}^{\\frac{2\\pi}{n}k}f(x)(g(nx)-g_0)dx+f_0g_0 \\\\ &=\\frac{1}{... 2 Cosine is an even function; sine is odd. Using negative indices for these functions adds no information because, for example,5\\sin 2x+ 4\\sin(-2x) = \\sin 2x$$On the other hand, e^{-2ix} is not just a constant multiple of e^{2ix}. And we need both of these to express \\sin 2x, as in$$\\sin 2x = \\frac1{2i}(e^{2ix} - e^{-2ix})$$A more precise ... 0 After a simple classification I have figured it out. We could solve it by induction on the degree of the polynomial P(x). If deg P(x)=1, then c_1 \\in \\mathbb{R} \\setminus \\mathbb{Q} and apparently <c_1n+c_0> is equidistributed in [0,1). Assume deg P(x)=m and at least one of its coefficient c_1,……,c_m is irrational, then <P(n)> ...", "# Get slopes of integer shift of an image in radians I tried to reproduce the result from this paper, However I'm stuck at the Section III of the paper which involve integer shift. Two images which their phase correlation $\\theta(k_1,k_2)$ was calculated at pixel-accuracy(integer shift). I get their phase component as shown(from -$\\pi$ to $\\pi$). It can be represented by $\\theta(k_1,k_2)=\\theta_1(k_1,k_2)-\\theta_2(k_1,k_2)$. Since the two images shifted in parallel, it can be approximated as $\\theta(k_1,k_2) \\approx ak_1+bk_2$ where $a$ and $b$ is the slopes obtained through least square method, $k_1$ and $k_2$ are the value $MxN$ of the image. Plotting $\\theta(k_1,k_2)$: And to get phase component with only sub-pixel accuracy(decimal shift), the phase component is subtracted with integer shift. The equation is $\\theta''(k_1,k_2)=\\theta(k_1,k_2)-(a'k_1+b'k_2)$ However, as you can see from the second figure, the value of $\\theta''(k_1,k_2)$ ranges wider than -$\\pi$ to $\\pi$(as compared to Figure 4 in the paper), which I suspect the slope($a'$,$b'$) was obtained inaccurately. The paragraph above the Equation (14) mentioned that they used conventional POC to obtain the slopes. However I'm not sure did I understood it correctly. The code I implemented: function [ output_args ] = phasecorrlsa( refIm, shifIm ) F=(fft2(double(refIm))); G=(fft2(double(shifIm))); [m, n]=size(refIm); [M,N] = meshgrid(1:m,1:n); X = [M(:), N(:)]; R=(F.*conj(G))./abs(F.*conj(G)); r=(ifft2(R)); % surf(r) [apx, bpx, rhat]=lsa(angle(r)); [~,w] = max(r(:)); [del_hat2p, del_hat1p]", "cos((-2*pi*ant_spacings(3)/lambda)*sin(theta(i)*(pi/180) - 2*pi/3))]; % Make the steering vectors complex exp((1i*(-2*pi*ant_spacings(1)/lambda)*sin(theta(i)*(pi/180) - pi/3))); exp((1i*(-2*pi*ant_spacings(2)/lambda)*sin(theta(i)*(pi/180) - pi/2))); exp((1i*(-2*pi*ant_spacings(3)/lambda)*sin(theta(i)*(pi/180) - 2*pi/3)))]; spectrum_func(i) = 1/((steering_vec'*noise_vecs)*noise_vecs'*steering_vec); end % grab theta corresponding to highest peak, round to nearest half integer [mx,argmax] = max(spectrum_func); estimated_AoA = floor(2*theta(argmax)+.5)/2; fprintf(\"Estimated AoA: %.1f \\t Hard-coded AoA: %.1f.\\n\", estimated_AoA, AoA) plot(theta,20*log10(abs((spectrum_func)))) title('MUSIC Psuedospectrum') xlabel('Angle of Arrival') $$$$ • Ah, I thought this might have had something to do with it. Is there any way you could elaborate on the 'using a real sinusoid will introduce a phase shift term in the opposite angle direction' aspect? – OGBerglemir Nov 19 '19 at 20:00 • @OGBerglemir When you multiply two complex exponentials, you can add the general phase terms together. If you recall, a real sinusoid is composed of two complex exponentials, which rotate in different directions. Multiplying by a real sinusoid uses both components, so you will shift a signal's phase using both, and not one as intended. – Envidia Nov 19 '19 at 20:07 • Ah, understood. – OGBerglemir Nov 19 '19 at 21:59", "# If y = sin x° then find $$\\frac{{dy}}{{dx}}$$? 1. cos x 2. 0 3. -cos x 4. None of the above ## Answer (Detailed Solution Below) Option 4 : None of the above Free CRPF Head Constable Ministerial 22 Feb 2023 (Shift 1) Memory-Based Test 1.3 Lakh Users 100 Questions 100 Marks 90 Mins ## Detailed Solution Concept: Suppose that we have two functions f(x) and g(x) and they are both differentiable. • Chain Rule: $$\\frac{{\\rm{d}}}{{{\\rm{dx}}}}\\left[ {{\\rm{f}}\\left( {{\\rm{g}}\\left( {\\rm{x}} \\right)} \\right)} \\right] = {\\rm{\\;f'}}\\left( {{\\rm{g}}\\left( {\\rm{x}} \\right)} \\right){\\rm{g'}}\\left( {\\rm{x}} \\right)$$ Calculation: Given: y = sin x° We know that, 180° = π radian ∴ 1° = $$\\frac{\\pi}{180}$$ radian Now, x° = $$\\frac{\\pi x}{180}$$ radian ⇒ y = $$\\sin \\left( {\\frac{{\\rm{\\pi \\cdot x }}}{{180}}} \\right){\\rm{\\;}}$$ Differenatiate with respect to x, we get $$\\frac{{{\\rm{dy}}}}{{{\\rm{dx}}}} = {\\rm{\\;}}\\frac{{\\rm{\\pi }}}{{180}}{\\rm{\\;}} \\times \\cos \\left( {\\frac{{\\rm{\\pi \\cdot x}}}{{180}}} \\right){\\rm{\\;}}$$ Latest Airforce Group X Updates Last updated on Nov 11, 2022 Airforce Group X Phase 2 Admit Card Out. The Indian Air Force (IAF) had released the official notification for the IAF Group X (01/2023) on 7th November 2022. The selection of the candidates will depend on three stages which are Phase 1 (Online Written Test), Phase 2 ( DV, Physical Fitness Test, Adaptability Test I & II), and Phase 3 (Medical Examination). The written", "$t$ then the term $-vt$ is just some phase-shift and we can see the graph is just a horizontal shift of the sine wave. In particular we shift the $y = A \\sin(x)$ graph $vt$-units to right to form $y = A \\sin(x-vt)$. This is a right-moving wave if we assume $v>0$. To see it, I think setting $0 \\leq x-vt \\leq 2\\pi$ might be helpful, then $vt \\leq x \\leq 2\\pi+vt$ and if we animate $t$ we can actually see the wave travel. Other more complicated wave graphs could be studied. I happen to have this graph of a solution to the wave equation sitting around. Of course, adding graphs has interesting interpretations in terms of constructive and destructive interference. Or, if we add time I think you can get beats. There is much to explore here. Edit: as Dave Renfro pointed out, I am using forbidden trigonometry. To, give a less exciting, but very much applied example consider: $$y = y_o- \\frac{g}{2}t^2$$ If we start with $y=t^2$ then multiplying b $-g/2$ stretches the graph and flips it over. The significance is that $g$ is the acceleration due to gravity. Then, adding $y_o$ shifts the graph $y=-gt^2/2$ up by $y_o$ and in this discussion $y_o$ has the significance of being the initial condition. A more ambitious formula", "geometric figures. /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ The first line real f1 (real x) {return sin(x);} defines the sin() function to be drawn, and the next line draw(graph(f1,-6.21,8.03)); actually draws the graph to the diagram. The graph() command comes from the Asymptote graph package. The next few lines draw several line segments. The attribute zzttqq is a pen whose color is mapped from some rgb value (with conversions ... 8 = y, 9 = z, a = a, b = b, ... ). ## Options An outline of the different available options. (The first three options are related and automatically update with respect to eath other). X/Y Units (cm): Indicates how many units in the picture correspond to 1 centimeter. Picture width/height: Indicates what size the Asymptote output should be, in centimeters. xmin/xmax/ymin/ymax: Indicates the dimensions of the picture in the picture's coordinates. (The next options change other features). Font size: Changes the default font size. The value defaults at 10pt, but I personally recommend 7pt for any file with multiple text lines, which more closely approximates the size of the font", "draw(Label(\"$\\pi$\",1),shift(-0.15*Z)*scale3(0.5)*g,blue,Arrow3); path3 h=arc(O,1,170,-120,150,-120,CCW); draw(Label(\"$\\sigma$\",1),shift(-0.5*cos(pi/6)*Y-0.5*sin(pi/6)*X+0.35*Z)*scale3(0.5)*h,red,Arrow3); draw(shift(-0.5*cos(pi/6)*Y-0.5*sin(pi/6)*X-0.15*Z)*scale3(0.1)*unitsphere,gray+opacity(.3)); and the result You can easily modify the xpt and ypt values to find the best shape. For periodic spline, do not forget to have periodic value. O.G. • Great! But may I ask where the lower bound in (0.1,2*pi) comes from? – user121799 Jan 4 '18 at 0:28 • @marmot I add a comment. In view of your attempt, I transform the box (a,b) (-0.67,0.1),(0,2.032*pi) to (-0.67,0),(0.1,2*pi) so that the x variable varies from -0.67 to 0.1 while the y variable varies from 0 to 2*pi. Perhaps that (-0.67,0),(0,2*pi) should be ok... – O.G. Jan 4 '18 at 9:48", "in the sense that the sine Fourier transform is imaginary, so a phase offset should represent some non-zero amplitude distributed in real and imaginary parts, and they should cancel when combined to take the absolute value. After reading up on the Fourier transform of a sinusoid with frequency shift (Fourier transform of the Cosine function with Phase Shift?), I have seen that it is the Dirac multiplied by the sum of a real and imaginary exponential: $\\pi (e^{-j\\theta}\\delta(w-w_0) + e^{j\\theta}\\delta(w-w_0))$ So I tried to do the same taking into account the Gaussian envelope: X(w) = $\\sum_{k=1}^N \\frac{A_k}{2} (e^{\\phi i} + e^{- \\phi i}) \\sqrt{2\\pi\\sigma_k^2} e^{-\\pi^2 (f_k-w)^2 2\\sigma_k^2}$ However it does not work. The following Python program implements what I have tried: #!/usr/bin/env python3 from matplotlib import pyplot as plt import numpy as np def time_gabor(x, var, mean, f, phi, amp): return (1.0 # * (1.0/(var*np.sqrt(2*np.pi))) * np.exp(-(x-mean)**2 / (2*var**2)) * np.cos((x-mean)*2*np.pi*f + phi)) * amp def fourier_gabor(x, var, mean, f, phi, amp): a = 2*var**2 g = np.sqrt(np.pi*a) * np.exp(-np.pi**2 * (f-w)**2 * a) return (np.exp(1.0j*phi) + np.exp(-1.0j*phi)) * g*amp/2 #return (np.cos(phi) - np.sin(phi))*g*amp def fft(y, sample_rate): ft = np.abs(np.fft.fft(y)[:len(y)//2]) / (sample_rate/2) ws = np.arange(len(y)) / len(y) * sample_rate return ws, ft N = 100000 x = np.linspace(-10,10,N).reshape(-1,1) w = np.linspace(0,5,N).reshape(-1,1) centers = np.array([0, 0, 8]) freq =", "plot (contours only) filled.contour() Level (contour) plot, filled The chunks below show some basic plot for these types, which can be highly customized if needed. For some more information about plotting check out the Plotting chapter. #### Generic X-Y Plot (Matrix) First, let us generate a matrix with some data which we will use for plotting. The following two lines generate a matrix m of dimension $$200 \\times 3$$ where each column contains one full period of sine along the unit circle ($$0$$ to $$2 \\cdot \\pi$$) with different phase shifts ($$0$$, $$\\frac{1}{2} \\pi$$, and $$\\pi$$). # Sequence from 0 to 2 * pi x <- seq(0, 2 * pi, length.out = 200) # Calculate sin(x + shift) m <- cbind(\"shift: 0\" = sin(x), \"shift: 1/2 pi\" = sin(x + 1 / 2 * pi), \"shift: pi\" = sin(x + pi)) head(m, n = 3) ## shift: 0 shift: 1/2 pi shift: pi ## [1,] 0.00000000 1.0000000 1.224647e-16 ## [2,] 0.03156855 0.9995016 -3.156855e-02 ## [3,] 0.06310563 0.9980069 -6.310563e-02 When calling plot(m), R will automatically take the first two columns of the matrix and plot them against each other, creating a 2d scatter plot. plot(m, main = \"Generic X-Y Plot\") This is nothing else than plotting plot(m[, 1], m[, 2]). #### Matrix plot matplot() is plotting data columnwise. Each", "$$\\pi k$$ (the c’s (cot and csc) have the “easier” ones). To get the new asymptotes, set the trig argument to asymptotes, and solve for x. For example, if we have $$\\tan \\left( {3x+\\pi } \\right)\\,\\,(\\text{which would be}\\,\\,\\tan \\left( {3\\left( {x+\\frac{\\pi }{3}} \\right)} \\right))$$, we would solve: $$\\displaystyle 3x+\\pi =\\frac{\\pi }{2}+\\pi k;\\,\\,\\,\\,\\,\\,3x=\\left( {\\frac{\\pi }{2}-\\pi } \\right)+\\pi k;$$ $$\\displaystyle 3x=-\\frac{\\pi }{2}+\\pi k;\\,\\,\\,\\,x=\\frac{{-\\frac{\\pi }{2}}}{3}+\\frac{{\\pi k}}{3};\\,\\,\\,\\,x=-\\frac{\\pi }{6}+\\frac{{\\pi k}}{3}$$. 3. The new range will still be $$\\left( {-\\infty ,\\,\\infty } \\right)$$. 4. Draw asymptotes first; phase shift will take care of itself. (To get an asymptote starting point, you can set k = 0 in your new asymptote equation, k = –1 for one to the left, k = 1 for one to the right). Right in between the asymptotes (you can take average of the x’s), draw the middle of the graph (but make sure it is shifted up or down according to the vertical shift, or d). 5. New period will be $$\\frac{\\pi }{b}$$ (since tan and cot have periods of ). This should be the distance between asymptotes (the coefficient of “k” in the asymptote equations). 6. If trig function is negative, make sure you flip the graph. 7. New domain will be all x values except for the asymptotes. 8. To get two more points on either side", "sure so I experimented, I decided to use a sine graph with a wavelength of 20 y=sin nx the wavelength is 2pi/n = 20 so n = pi/10 so my experimenting graph is $$y=sin(\\frac{\\pi }{10}x)\\\\ y=sin(\\frac{\\pi }{10}x)=sin(\\frac{\\pi }{10}(x-20))\\\\ let \\;\\;\\frac{x}{5}\\;\\;be\\;\\;substituted\\;\\;for\\;\\;x\\\\ y=sin(\\frac{\\pi }{10}\\frac{x}{5})=sin(\\frac{\\pi }{10}(\\frac{x}{5}-20))=sin(\\frac{\\pi }{10}(\\frac{x-100}{5}))$$ So the smallest a that we know with certainty is 100. Here is the graph i was playing with. https://www.desmos.com/calculator/oonutt20vz ------------------- Let me see if I can do this more simply. $$y=f(x)=f(x-20)\\\\ let \\;\\;\\frac{x}{5}\\;\\;be\\;\\;substituted\\;\\;for\\;\\;x\\\\ y=f(\\frac{x}{5})=f(\\frac{x}{5}-20)=f(\\frac{x-100}{5})$$ There you go, that was simple. The right shift is 100. Mar 27, 2018 #5 +100025 +1 2. What are the coordinates of the points where the graphs of $f(x)=x^3-x^2+x+1$ and $g(x)=x^3+x^2+x-1$ intersect? Give your answer as a list of points separated by semicolons, with the points ordered such that their $x$-coordinates are in increasing order. (So \"(1,-3); (2,3); (5,-7)\" - without the quotes - is a valid answer format.) $$x^3-x^2+x+1 = x^3+x^2+x-1\\\\ -x^2+1 = x^2-1\\\\ 2=2x^2\\\\ x^2=1\\\\ x=\\pm1\\\\ f(1)=1-1+1+1=2\\\\ f(-1)=-1-1-1+1=-2\\\\$$ so the points of intersection are (-1,-2); (1,2) check: Mar 27, 2018", "0 to 2*pi in steps of 0.01 x1 = t; y1 = amplitude * sin(frequency*t); % Produce plot plot (x1,y1); title 'original' grid % Define transform parameters scale = 5; % 5 times bigger shift = [ -15 -10 ]; % shift 15 units left, and 10 units down theta = pi/6; % 30 degrees rotation_matrix = [ cos(theta), -sin(theta) sin(theta), cos(theta) ]; % Transform points via scaling, shifting, and rotation % make empty arrays to store new points in x2 = zeros(size(t)); y2 = zeros(size(t)); % Transform points one at a time for i = 1:length(t) old_xy = [ x1(i) y1(i) ]; % calculate transformed x and y new_xy = ( rotation_matrix * old_xy * scale ) + shift; % Store calculated values to vector x2(i) = new_xy(1); y2(i) = new_xy(2); end % Make a pretty picture figure; plot(x2,y2); set(gca,'XLim',[-20,20],'YLim',[-20,20]) title 'transformed' grid The ingredients for what you want are here: you just have to figure out how to get the sine wave to start and stop at your desired endpoints. I leave this exercise in trigonometry up to you. (Note: If you can't figure out how to find the transformation that will give a sine wave starting a point A and ending at point B, it would be better to ask math.stackexchange.com instead of StackOverflow -", "maliaseth0 2022-01-25 Is the phase shift of $-3\\mathrm{sin}\\left(2x+\\frac{\\pi }{2}\\right)+1$ equal to $-\\frac{\\pi }{4}$ or just $-\\frac{\\pi }{2}$? trnovitom06 Expert The definition of \"phase shift\" actually depends on the context. For example, in a context where $A\\mathrm{sin}\\left(\\omega x-\\varphi \\right)$ is a sine wave or signal and x is time, one often calls the phase shift. To be more precise, this could be called an angular phase shift, since this parameter shifts the argument of the sine function (an angle) away from its original angle. On the other hand, in other contexts one would rather write the same sine wave as $A\\mathrm{sin}\\left(\\omega \\left(x-\\frac{\\varphi }{\\omega }\\right)\\right)$, which highlights a different kind of shift, namely a temporal shift of $\\frac{\\varphi }{\\omega }$ seconds. If you graph the sine wave as a function of time, then this quantity is the proper horizontal shift that you see on the graph. marzembreax Expert The phase shift of $-3\\mathrm{sin}\\left(2x+\\frac{\\pi }{2}\\right)+1=-3\\mathrm{sin}\\left(2\\left(x+\\frac{\\pi }{4}\\right)\\right)+1$ relative to", "How do you graph y = cos(x- pi/6)? It is graph of y= cos x with a phase shift of $\\frac{\\pi}{6}$ The graph of y= cos x with a horizontal shift to the right by $\\frac{\\pi}{6}$, called phase shift, would represent the graph of $y = \\cos \\left(x - \\frac{\\pi}{6}\\right)$" ]

[ "+0 0 220 1 Simplify $\\sin 70^\\circ \\cos 50^\\circ + \\sin 260^\\circ \\cos 280^\\circ.$ Mar 31, 2020 I think $$\\sin 70^\\circ \\cos 50^\\circ + \\sin 260^\\circ \\cos 280^\\circ.$$equals to $$\\frac{\\sqrt{3}}{4}$$. I have heard that you can use caculator to symplifiy this! But I still think im right.", "## Elementary Technical Mathematics The value of $\\cos \\left( 256{}^\\circ \\right)$ is $-0.2419$. Step 1: Open Ti-83. Step 2: Press $cos$ button. Step 3: Enter the value like this, $\\cos \\left( 256{}^\\circ \\right)=-0.2419$ The obtained value is $-0.2419$. Hence the value of $\\cos \\left( 256{}^\\circ \\right)$ is $-0.2419$.", "6^\\circ$ -", "to this problem is $$\\{114.3^\\circ,335.7^\\circ\\}$$", "$6(36\\pi)= 216\\pi$.", "22.89 = {337}^{\\circ} 11$ (co-terminal). b. $\\left(2 x + 57.32\\right) = 180 - 11.54 = 168.46$ $2 x = 168.46 - 57.32 = 111.14$ $x = {55}^{\\circ} 57$ For general answers, add $k {360}^{\\circ}$ Check by calculator. $x = 55.57 - \\to 2 x = 111.14 - \\to \\sin \\left(2 x + 57.32\\right) = \\sin \\left(168.46\\right) = 0.20$. Proved", "$A = {\\sin}^{-} 1 \\left(\\frac{9 \\sin {150}^{\\circ}}{10}\\right) = {26}^{\\circ} 45 '$ and therefore $C = {180}^{\\circ} - {150}^{\\circ} - {26}^{\\circ} 45 ' = {3}^{\\circ} 15 '$", "+ 180^circ ) $- 3 + 15 i \\approx 3 \\sqrt{26} \\setminus \\text{cis} \\setminus \\left({101.3}^{\\circ}\\right)$", "is $35.0^{\\circ} \\mathrm{C}$ ?...", "Use an appropriate local linear approximation to estimate the value of $\\sin(46^{\\circ})$. $\\sin(46^{\\circ}) \\approx$", "## Elementary Technical Mathematics The value of $56.1 \\%$ of $360{}^\\circ$ is $202{}^\\circ$. \\begin{align} & 56.1 \\% \\times 360{}^\\circ =\\frac{561}{1000}\\times 360{}^\\circ \\\\ & =201.96{}^\\circ \\end{align} Now round off the result, which gives $202{}^\\circ$. Thus, the value of $56.1 \\%$ of ${{360}^{\\circ }}$is $202{}^\\circ$.", "subtract the two from $$180^{\\circ}$$.", "Find $$\\hat{N}$$. $\\hat{N} = 83^{\\circ}$", "# 1996 USAMO Problems/Problem 1 ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos5)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$, as desired. $\\Box$ ###", "30^\\circ \\sin 60^\\circ,$$ as desired.", "(to 2 d.p.) So the circumcentre is $(46.71,20.62)$. Check that I haven't made any mistakes with the arithmetic.", "$2\\cos^2(15^\\circ)-1$ 2. $2\\sin(25^\\circ)\\cos(25^\\circ)$ 3. $1-2\\sin^2(35^\\circ)$ 4. $\\cos^2(60^\\circ)-\\sin^2(60^\\circ)$ 5. $2\\sin(125^\\circ)\\cos(125^\\circ)$ 6. $1-2\\sin^2(32^\\circ)$", "1996 USAMO Problems/Problem 1 Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. Solution Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos 5^\\circ)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$, as desired. $\\Box$ 1996 USAMO (Problems • Resources)", "calculate by hand now, although $\\cos 30^\\circ$ is quite close and gives one a sense of it. – Dylan Moreland Feb 5 '12 at 20:57", "+0 Question 0 33 1 Compute $$\\cos{0^{\\circ}}+\\cos{90^{\\circ}}+\\cos{180^{\\circ}}+\\cdots+\\cos{(90^{\\circ}\\cdot 2016)}$$ ,where each argument is greater than the previous one by 90 degrees. Oct 16, 2022 #1 0 The sum is 504. Oct 16, 2022" ]

[ "Write the following as ratios of $\\scriptsize {{20}^\\circ}$: 1. $\\scriptsize \\sin {{70}^\\circ}$ 2. $\\scriptsize \\cos {{110}^\\circ}$ 3. $\\scriptsize \\sin {{250}^\\circ}$ 4. $\\scriptsize \\cos (\\theta -{{90}^\\circ})$ Solutions 1. \\scriptsize \\begin{align*}\\sin {{70}^\\circ}&=\\sin ({{90}^\\circ}-{{20}^\\circ})\\\\&=\\cos {{20}^\\circ}\\end{align*} 2. . \\scriptsize \\begin{align*}\\cos {{110}^\\circ}&=\\cos ({{90}^\\circ}+{{20}^\\circ})\\\\&=-\\sin {{20}^\\circ}\\quad \\text{Don }\\!\\!'\\!\\!\\text{ t forget the negative sign; cosine is negative in the second quadrant}\\end{align*} 3. We could answer this as follows: $\\scriptsize \\sin {{250}^\\circ}=\\sin ({{270}^\\circ}-{{20}^\\circ})=-\\cos {{20}^\\circ}$However, the $\\scriptsize {{270}^\\circ}\\pm \\theta$ co-function reduction rules are not part of the curriculum, so we answer it as follows: \\scriptsize \\begin{align*}\\sin {{250}^\\circ}&=\\sin ({{180}^\\circ}+{{70}^\\circ})\\\\&=-\\sin {{70}^\\circ}\\\\&=-\\sin ({{90}^\\circ}-{{20}^\\circ})\\\\&=-\\cos {{20}^\\circ}\\end{align*} 4. . \\scriptsize \\begin{align*}\\cos (\\theta -{{90}^\\circ})&=\\cos (-{{90}^\\circ}+\\theta )\\\\&=\\cos \\left( {-({{{90}}^\\circ}-\\theta )} \\right)&& \\text{The angle }-({{90}^\\circ}-\\theta )\\text{ is in the fourth quadrant where}\\\\ &&&\\text{cosine is positive}\\\\&=\\cos ({{90}^\\circ}-\\theta )\\\\&=\\sin \\theta \\end{align*} ### Example 2.5 Express the following with all ratios in terms of $\\scriptsize \\theta$ alone and simplify: $\\scriptsize \\displaystyle \\frac{{\\text{cos(}\\theta -\\text{9}{{\\text{0}}^\\circ}\\text{) cos(72}{{\\text{0}}^\\circ}+\\theta \\text{) tan(}\\theta -\\text{36}{{\\text{0}}^\\circ})}}{{\\text{si}{{\\text{n}}^{2}}\\text{(}\\theta \\text{+36}{{\\text{0}}^\\circ}\\text{) cos(}\\theta \\text{+9}{{\\text{0}}^\\circ}\\text{)}}}$ Solution Let’s break this question down and look at each trig ratio separately first. • . \\scriptsize \\begin{align*}\\cos (\\theta -{{90}^\\circ})&=\\cos (-{{90}^\\circ}+\\theta )\\\\&=\\cos \\left( {-({{{90}}^\\circ}-\\theta )} \\right)\\quad -({{90}^\\circ}-\\theta )\\text{ is in the fourth quadrant where cosine is positive}\\\\&=\\cos ({{90}^\\circ}-\\theta )\\\\&=\\sin \\theta \\end{align*} • $\\scriptsize \\cos ({{720}^\\circ}+\\theta )=\\cos \\theta\\quad \\text{because }{{720}^\\circ}\\text{ is just two full revolutions or two full periods}$ • $\\scriptsize \\tan (\\theta -{{360}^\\circ})=\\tan \\theta \\quad -{{360}^\\circ}\\text{ is one full revolution", "the first identity in the reverse order). $$\\cos{A}+\\cos{(A+2\\pi/3)}+\\cos{(A-2\\pi/3)}=0$$ $$\\cos{(A-\\pi/3)}+\\cos{(A+\\pi/3)}=\\cos{A}$$ Consider the following picture. We see that the triangle with vertices $(\\cos A, \\sin A), (\\cos(A+120^\\circ), \\sin(A+120^\\circ)), (\\cos(A+240^\\circ), \\sin(A+240^\\circ))$ are the vertices of an equilateral triangle with centriod at the origin. Thus if we set $a = \\cos A, b = \\cos(A+120^\\circ), c = \\cos(A+240^\\circ)$, then \\begin{align*} a+b+c = 0 \\end{align*} and hence \\begin{align*} a^3+b^3+c^3 = 3abc \\end{align*} and \\begin{align*} \\cos^3{A} + \\cos^3{(120°+A)} + \\cos^3{(240°+A)} &= 3\\cos A \\cos(A+120^\\circ)\\cos(A+240^\\circ) \\\\ &= \\frac{3}{2}(2\\cos A \\cos(A+120^\\circ))\\cos(A+240^\\circ) \\\\ &= \\frac{3}{2}[\\cos(2A+120^\\circ)+\\cos(120^\\circ)]\\cos(A+240^\\circ)\\\\ &=\\frac{3}{2}\\cos(2A+120^\\circ)\\cos(A+240^\\circ)-\\frac{3}{4}\\cos(A+240^\\circ)\\\\ &=\\frac{3}{4}\\cos(3A+360^\\circ)+\\frac{3}{4}\\cos(A-120^\\circ) -\\frac{3}{4}\\cos(A+240^\\circ)\\\\ &=\\frac{3}{4}\\cos(3A) \\end{align*} since $\\cos(A+240^\\circ) = \\cos(360^\\circ - (A+240^\\circ)) = \\cos(120^\\circ -A) = \\cos(A-120^\\circ)$ • If you can see geometrically, the points $(\\cos A, \\sin A), (\\cos(120^\\circ+A), \\sin(120^\\circ+A)), (\\cos(240^\\circ), \\sin(240^\\circ+A))$ are the vertices of an equilateral triangle with centroid as the origin, then the above proof holds since we have used only an algebraic identity. – user348749 Sep 10, 2016 at 7:23 use that $$\\cos(120^{\\circ}+x)=-1/2\\,\\cos \\left( x \\right) -1/2\\,\\sqrt {3}\\sin \\left( x \\right)$$ and $$\\cos(240^{\\circ}+x)=-1/2\\,\\cos \\left( x \\right) +1/2\\,\\sqrt {3}\\sin \\left( x \\right)$$ For the LHS, use $\\cos(120^\\circ +A)=\\cos 120^\\circ \\cos A - \\sin 120^\\circ \\sin A$ and $\\cos(240^\\circ+A)=\\cos 240^\\circ \\cos A - \\sin 240^\\circ \\sin A$ Then $\\cos^3(120^\\circ+A)=\\left(\\cos 120^\\circ \\cos A - \\sin 120^\\circ \\sin A \\right)^3$ and $\\cos^3(240^\\circ+A)=\\left( \\cos 240^\\circ \\cos A - \\sin 240^\\circ \\sin A \\right)^3$", "for help with the Examples above. CK-12 Foundation: Chapter8LawsofSinesandCosinesB ### Guided Practice Find the following angles in the triangle bloew. Round your answers to the nearest hundredth. 1. \\begin{align*}m\\angle A\\end{align*} 2. \\begin{align*} m\\angle B\\end{align*} 3. \\begin{align*} m\\angle C\\end{align*} 1. When you are given only the sides, you have to use the Law of Cosines to find one angle and then you can use the Law of Sines to find another. \\begin{align*}15^2 & = 22^2 + 28^2 - 2(22)(28) \\cos A\\\\ 225 & = 1268 - 1232 \\cos A\\\\ -1043 & = -1232 \\cos A\\\\ \\frac{-1043}{-1232} & = \\cos A \\rightarrow \\cos^{-1} \\left (\\frac{1043}{1232} \\right ) \\approx 32.16^\\circ\\end{align*} 2. Now that we have an angle and its opposite side, we can use the Law of Sines. \\begin{align*}\\frac{\\sin 32.16^\\circ}{15} = \\frac{\\sin B}{22}&&\\\\ &&15 \\cdot \\sin B & = 22 \\cdot \\sin 32.16^\\circ\\\\ &&\\sin B & = \\frac{22 \\cdot \\sin 32.16^\\circ}{15}\\end{align*} \\begin{align*}\\sin^{-1} \\left ( \\frac{22 \\cdot \\sin 32.16^\\circ}{15} \\right ) \\approx 51.32^\\circ\\end{align*}. 3. To find \\begin{align*}m \\angle C\\end{align*}, use the Triangle Sum Theorem. \\begin{align*}32.16^\\circ + 51.32^\\circ + m \\angle C &= 180^\\circ\\\\ m \\angle C &= 96.52^\\circ\\end{align*} ### Explore More Use the Law of Sines or Cosines to solve \\begin{align*}\\triangle ABC\\end{align*}. If you are not given a picture, draw one. Round all decimal answers to the nearest tenth. 1. \\begin{align*}m \\angle", "below, if all eight angles are congruent then each angle is \\begin{align*}\\frac{(8 - 2) \\times 180^\\circ}{8} = \\frac{6 \\times 180^\\circ}{8} = \\frac{1080^\\circ}{8} = 135^\\circ\\end{align*}. What if you were given an equiangular seven-sided convex polygon? How could you determine the measure of its interior angles? ### Examples #### Example 1 The interior angles of a pentagon are \\begin{align*}x^\\circ, x^\\circ, 2x^\\circ, 2x^\\circ,\\end{align*} and \\begin{align*}2x^\\circ\\end{align*}. What is \\begin{align*}x\\end{align*}? From the Polygon Sum Formula we know that a pentagon has interior angles that sum to \\begin{align*}(5-2) \\times 180^\\circ=540^\\circ\\end{align*}. Write an equation and solve for \\begin{align*}x\\end{align*}. \\begin{align*} x^\\circ + x^\\circ + 2x^\\circ + 2x^\\circ + 2x^\\circ&=540^\\circ\\\\ 8x&=540\\\\x&=67.5\\end{align*} #### Example 2 What is the sum of the interior angles in a 100-gon? Use the Polygon Sum Formula. \\begin{align*}(100-2) \\times 180^\\circ=17,640^\\circ\\end{align*}. #### Example 3 The interior angles of a polygon add up to \\begin{align*}1980^\\circ\\end{align*}. How many sides does it have? Use the Polygon Sum Formula and solve for \\begin{align*}n\\end{align*}. \\begin{align*}(n - 2) \\times 180^\\circ & = 1980^\\circ\\\\ 180^\\circ n - 360^\\circ & = 1980^\\circ\\\\ 180^\\circ n & = 2340^\\circ\\\\ n & = 13\\end{align*} The polygon has 13 sides. #### Example 4 How many degrees does each angle in an equiangular nonagon have? First we need to find the sum of the interior angles; set \\begin{align*}n = 9.\\end{align*} \\begin{align*}(9 - 2) \\times 180^\\circ = 7 \\times", "cosine double angle identity. \\scriptsize \\begin{align*}\\cos {{150}^\\circ}&=\\cos ({{75}^\\circ}+{{75}^\\circ})\\\\&=2{{\\cos }^{2}}{{75}^\\circ}-1\\\\&=2{{a}^{2}}-1\\end{align*} 2. This question does not need us to apply any double angle identity. \\scriptsize \\begin{align*}\\sin {{165}^\\circ}&=\\sin ({{90}^\\circ}+{{75}^\\circ})\\\\&=\\cos {{75}^\\circ}\\\\&=a\\end{align*} 3. This is a tricky question. We need to express $\\scriptsize \\sin {{150}^\\circ}$ in terms of $\\scriptsize a=\\cos {{75}^\\circ}$ so we cannot use either the fact that $\\scriptsize \\sin {{150}^\\circ}=\\sin ({{180}^\\circ}-{{30}^\\circ})$ or $\\scriptsize \\sin {{150}^\\circ}=\\sin ({{90}^\\circ}+{{60}^\\circ})$ directly. We need to apply the double angle identity in reverse. We know that: \\scriptsize \\begin{align*}\\sin {{150}^\\circ}&=\\sin ({{75}^\\circ}+{{75}^\\circ})\\\\&=2\\sin {{75}^\\circ}\\cos {{75}^\\circ}\\end{align*} . We already know that $\\scriptsize \\cos {{75}^\\circ}=a$ but in order to find an expression for $\\scriptsize \\sin {{75}^\\circ}$ we need to determine the value of $\\scriptsize \\sin {{150}^\\circ}$. Thankfully, we can easily do this without a calculator. \\scriptsize \\begin{align*}\\sin {{150}^\\circ}&=\\sin ({{180}^\\circ}-{{30}^\\circ})\\\\&=\\sin {{30}^\\circ}\\\\&=\\displaystyle \\frac{1}{2}\\end{align*} Now we can say that: \\scriptsize \\begin{align*}2\\sin {{75}^\\circ}\\cos {{75}^\\circ} & =\\displaystyle \\frac{1}{2}\\\\\\therefore \\sin {{75}^\\circ}\\cos {{75}^\\circ} & =\\displaystyle \\frac{1}{4}\\\\\\text{But }\\cos {{75}^\\circ} & =a\\\\\\therefore \\sin {{75}^\\circ} & =\\displaystyle \\frac{1}{{4a}}\\end{align*} ### Exercise 1.2 1. Given $\\scriptsize \\sin {{50}^\\circ}=a$ and $\\scriptsize \\sin {{40}^\\circ}=b$, determine the value of the following in terms of $\\scriptsize a$ and $\\scriptsize b$: 1. $\\scriptsize \\sin {{100}^\\circ}$ 2. $\\scriptsize \\cos {{40}^\\circ}$ 3. $\\scriptsize \\cos {{140}^\\circ}$ 4. $\\scriptsize \\cos {{25}^\\circ}$ 5. $\\scriptsize \\cos {{80}^\\circ}$ 6. $\\scriptsize \\cos (-{{800}^\\circ})$ Question 2 adapted from Everything Maths Grade 12 Exercise 4-3 question 3 1. Prove", "75{}^\\circ \\cot 65{}^\\circ =2\\text{ }...\\text{(vii)}$ Now coming to the last term of the given expression, We know that $\\sin (\\theta )=\\cos ({{90}^{\\circ }}-\\theta )$ Using the above equation we get, \\begin{align} & \\sin 72{}^\\circ =\\cos ({{90}^{\\circ }}-72{}^\\circ ) \\\\ & \\text{ }=\\cos 18{}^\\circ \\\\ \\end{align} Using the above conversion, the third term is converted to: \\begin{align} & -3({{\\sin }^{2}}18{}^\\circ +{{\\sin }^{2}}72{}^\\circ )=-3({{\\sin }^{2}}18{}^\\circ +{{(\\sin 72{}^\\circ )}^{2}}) \\\\ & \\text{ }=-3({{\\sin }^{2}}18{}^\\circ +{{(\\cos 18{}^\\circ )}^{2}}) \\\\ & \\text{ }=-3({{\\sin }^{2}}18{}^\\circ +{{\\cos }^{2}}18{}^\\circ )\\text{ }...\\text{(viii)} \\\\ \\end{align} We know that ${{\\sin }^{2}}(\\theta )+{{\\cos }^{2}}(\\theta )=1$ Using the above equation we get, ${{\\sin }^{2}}18{}^\\circ +{{\\cos }^{2}}18{}^\\circ =1\\text{ }...\\text{(ix)}$ Using equation (ix) in the equation (viii) we get, \\begin{align} & -3({{\\sin }^{2}}18{}^\\circ +{{\\sin }^{2}}72{}^\\circ )=-3({{\\sin }^{2}}18{}^\\circ +{{\\cos }^{2}}18{}^\\circ ) \\\\ & \\text{ }=-3(1) \\\\ & \\text{ }=-3 \\\\ \\end{align} So, we get $-3({{\\sin }^{2}}18{}^\\circ +{{\\sin }^{2}}72{}^\\circ )=-3\\text{ }...\\text{(x)}$ Now using equation (iii), equation (vii) and equation (x), adding them all we get, \\begin{align} & \\dfrac{\\sec 37{}^\\circ }{\\text{cosec }53{}^\\circ }+2\\cot 15{}^\\circ \\cot 25{}^\\circ \\cot 45{}^\\circ \\cot 75{}^\\circ \\cot 65{}^\\circ -3({{\\sin }^{2}}18{}^\\circ +{{\\sin }^{2}}72{}^\\circ )=1+2-3 \\\\ & \\Rightarrow \\dfrac{\\sec 37{}^\\circ }{\\text{cosec }53{}^\\circ }+2\\cot 15{}^\\circ \\cot 25{}^\\circ \\cot 45{}^\\circ \\cot 75{}^\\circ \\cot 65{}^\\circ -3({{\\sin }^{2}}18{}^\\circ +{{\\sin }^{2}}72{}^\\circ )=0 \\\\ \\end{align} Hence we get the value of the expression in the question as 0. Note: This", "180^\\circ = 1260^\\circ\\end{align*} “Equiangular” tells us every angle is equal. So, each angle is \\begin{align*}\\frac{1260^\\circ}{9} = 140^\\circ\\end{align*}. #### Example 5 An interior angle in a regular polygon is \\begin{align*}135^\\circ\\end{align*}. How many sides does this polygon have? Here, we will set the Regular Polygon Interior Angle Formula equal to \\begin{align*}135^\\circ\\end{align*} and solve for \\begin{align*}n\\end{align*}. \\begin{align*}\\frac{(n - 2) \\times 180^\\circ}{n} & = 135^\\circ\\\\ 180^\\circ n - 360^\\circ & = 135^\\circ n\\\\ -360^\\circ & = -45^\\circ n\\\\ n & = 8 \\qquad \\quad \\text{The polygon is an octagon}.\\end{align*} ### Review 1. Fill in the table. # of sides Sum of the Interior Angles Measure of Each Interior Angle in a Regular \\begin{align*}n-\\end{align*}gon 3 \\begin{align*}60^\\circ\\end{align*} 4 \\begin{align*}360^\\circ\\end{align*} 5 \\begin{align*}540^\\circ\\end{align*} \\begin{align*}108^\\circ\\end{align*} 6 \\begin{align*}120^\\circ\\end{align*} 7 8 9 10 11 12 1. What is the sum of the angles in a 15-gon? 2. What is the sum of the angles in a 23-gon? 3. The sum of the interior angles of a polygon is \\begin{align*}4320^\\circ\\end{align*}. How many sides does the polygon have? 4. The sum of the interior angles of a polygon is \\begin{align*}3240^\\circ\\end{align*}. How many sides does the polygon have? 5. What is the measure of each angle in a regular 16-gon? 6. What is the measure of each angle in an equiangular 24-gon? 7. Each interior angle in a regular polygon is \\begin{align*}156^\\circ\\end{align*}.", "equal) \\begin{align}x+y+40^\\circ&=180^\\circ\\\\\\text{(Angle sum }&\\text{property of triangles)} \\end{align} \\begin{align}{x}+{{112}^{\\circ}}+{{40}^{\\circ}}&={{180}^{\\circ}} \\\\{x}+{{152}^{\\circ}}&={{180}^{\\circ}} \\\\ {x}&={{180}^{\\circ}}-{{152}^{\\circ}} \\\\ {x}&={{28}^{\\circ}} \\\\{z}&={x}={{28}^{\\circ}}\\\\\\text{ (Alternate}&\\text{ interior angles)}\\end{align} Therefore, \\begin{align}x &={{28}^{{}^\\circ }},{y}={{112}^{{}^\\circ }},{z}={{28}^{{}^\\circ }} \\\\\\end{align} Learn from the best math teachers and top your exams • Live one on one classroom and doubt clearing • Practice worksheets in and after class for conceptual clarity • Personalized curriculum to keep up with school", "# Get the same numbering to the right of each equations in two different ''alignes\" I want to use this code \\begin{align} & \\cos(90^\\circ-v) = \\sin(v)\\\\ & \\sin(90^\\circ-v) = \\cos(v). \\end{align} ....twice, and have the same numbering of the equations appear to the right of in both ''aligned''. Now, when i put down this code one more time in the same document i get numbering (3) and (4), But I want it to be (1) and (2) as in the first ''align''. Add a \\label and use \\ref for the repeated display, exploiting the \\tag feature that allows arbitrary text for the equation number. \\documentclass{article} \\usepackage{amsmath} \\begin{document} Here's a display \\begin{align} \\cos(90^\\circ-v) &= \\sin(v) \\label{cos-sin}\\\\ \\sin(90^\\circ-v) &= \\cos(v).\\label{sin-cos} \\end{align} And again \\begin{align} \\cos(90^\\circ-v) &= \\sin(v) \\tag{\\ref{cos-sin}}\\\\ \\sin(90^\\circ-v) &= \\cos(v).\\tag{\\ref{sin-cos}} \\end{align} \\end{document} Don't use degrees when teaching trigonometry, please! ;-) • Thanks man, I used degrees so I wouldnt mix it up with radians later. So that i can separate cos(2^\\circ) from cos(2) for example – Hankyboy Dec 17 '13 at 13:31", "find $a$ and $A$. ###### Solution First, use $\\dfrac{b}{\\sin B} = \\dfrac{c}{\\sin C}$ to find $c$. \\begin{align} \\frac{6}{\\sin 77^{\\circ} } &= \\frac{c}{\\sin 34^{\\circ} }\\\\\\\\ \\frac{6 \\sin 34^{\\circ} }{\\sin 77^{\\circ} } &= c\\\\ c &= 3.44 \\text{ (to 3 sig.fig.)} \\end{align} To find $A$, use the fact that the angles in a triangle add up to $180^{\\circ}$, i.e. $A+B+C = 180$. \\begin{align} 180 &= A +77 + 34\\\\ A &= 180 - 77- 34\\\\ A &= 69^{\\circ} \\end{align} Now, use this value for $A$ in the sine rule to find $a$. \\begin{align} \\frac{a}{\\sin 69^{\\circ} } &= \\frac{6}{\\sin 77^{\\circ} }\\\\\\\\ a &= \\frac{6 \\sin 69^{\\circ} }{\\sin 77^{\\circ} }\\\\\\\\ a &= 5.75 \\text{ (to 3 sig.fig.)} \\end{align} ###### Example 3 Given $b=2$, $c=5$ and $A=39^{\\circ}$, find $a$. ###### Solution Use $a^2 = b^2 + c^2 -2bc\\cos A$. \\begin{align} a^2 &= 2^2 + 5^2 - (2\\times2\\times5)\\cos 39^{\\circ}\\\\ a^2 &= 29 - 20\\cos 39^{\\circ}\\\\ a^2 &= 13.457 \\text{ (to 3 d.p.)}\\\\ a &= 3.67 \\text{ (to 3 sig.fig.)} \\end{align} ###### Example 4 Given $a=7$, $b=6$ and $c=9$. Find the angle between $a$ and $b$. ###### Solution The angle between $a$ and $b$ is labelled $C$ in the diagram above, so use $c^2 = a^2 + b^2 - 2ab\\cos C$. \\begin{align} 9^2 &= 7^2 + 6^2 - (2\\times7\\times6)\\cos C\\\\ 81 &= 49+36 - 84\\cos", "}\\times 180=229.183{}^\\circ \\\\ \\end{align}", "\\\\ \\text{Radians} \\times \\dfrac{180^{\\circ}}{\\pi} = \\text{Degrees}$$ ### Example 1 Convert $45^{\\circ}$ to radians in terms of $\\pi$. \\begin{align} \\displaystyle 45^{\\circ} &= 45^{\\circ} \\times \\dfrac{\\pi}{180^{\\circ}} \\\\ &= \\dfrac{\\pi}{4} \\end{align} ### Example 2 Convert $\\dfrac{2 \\pi}{3}$ radians to degrees. \\begin{align} \\displaystyle \\dfrac{2 \\pi}{3} &= \\dfrac{2 \\pi}{3} \\times \\dfrac{180^{\\circ}}{\\pi} \\\\ &= 120^{\\circ} \\end{align} ### Example 3 Convert $67^{\\circ}$ to radians, correcting to three significant figures. \\begin{align} \\displaystyle 67^{\\circ} &= 67^{\\circ} \\times \\dfrac{\\pi}{180^{\\circ}} \\\\ &= 1.16937 \\cdots \\\\ &= 1.17 \\end{align} ### Example 4 Convert $1.2$ radians to degrees, correcting to three significant figures. \\begin{align} \\displaystyle 1.2 &= 1.2 \\times \\dfrac{180^{\\circ}}{\\pi} \\\\ &= 68.75493 \\cdots^{\\circ} \\\\ &= 68.8^{\\circ} \\end{align}", "}}\\cos {{120}^{\\circ }}+\\cos {{210}^{\\circ }}\\sin {{300}^{\\circ }}$. Let us assume the value of $\\sin {{330}^{\\circ }}\\cos {{120}^{\\circ }}+\\cos {{210}^{\\circ }}\\sin {{300}^{\\circ }}$ is equal to I. $I=\\sin {{330}^{\\circ }}\\cos {{120}^{\\circ }}+\\cos {{210}^{\\circ }}\\sin {{300}^{\\circ }}.....(1)$ Now we have to find the value of $\\sin {{330}^{\\circ }}$. We know that $\\sin \\left( n\\pi -\\theta \\right)=-\\sin \\theta$ if n is even. \\begin{align} & \\Rightarrow sin{{330}^{\\circ }}=\\sin \\left( 2\\pi -{{30}^{\\circ }} \\right) \\\\ & \\Rightarrow \\sin {{330}^{\\circ }}=-\\sin {{30}^{\\circ }} \\\\ \\end{align} We know that $\\sin {{30}^{\\circ }}=\\dfrac{1}{2}$. $\\Rightarrow \\sin {{330}^{\\circ }}=-\\dfrac{1}{2}......(2)$ Now we should find the value of $\\cos {{120}^{\\circ }}$. We know that $\\cos \\left( n\\pi -\\theta \\right)=-\\cos \\theta$ if n is odd. \\begin{align} & \\Rightarrow \\cos {{120}^{\\circ }}=\\cos \\left( \\pi -{{60}^{\\circ }} \\right) \\\\ & \\Rightarrow \\cos {{120}^{\\circ }}=-\\cos \\left( {{60}^{\\circ }} \\right) \\\\ \\end{align} We know that $\\cos {{60}^{\\circ }}=\\dfrac{1}{2}$. $\\Rightarrow \\cos {{120}^{\\circ }}=-\\dfrac{1}{2}.......(3)$ Now we should find the value of $\\cos {{210}^{\\circ }}$. We know that $\\cos \\left( n\\pi +\\theta \\right)=-\\cos \\theta$ if n is odd. \\begin{align} & \\Rightarrow \\cos {{210}^{\\circ }}=\\cos \\left( \\pi +{{30}^{\\circ }} \\right) \\\\ & \\Rightarrow \\cos {{210}^{\\circ }}=-\\cos \\left( {{30}^{\\circ }} \\right) \\\\ \\end{align} We know that $\\cos {{30}^{\\circ }}=\\dfrac{\\sqrt{3}}{2}$. $\\Rightarrow \\cos {{210}^{\\circ }}=-\\dfrac{\\sqrt{3}}{2}.......(4)$ Now we have to find the value of $\\sin {{300}^{\\circ }}$. We know that $\\sin \\left( n\\pi", "2(\\cos^2 t - \\sin^2 t), -\\cos t \\end{align} The cross product of $\\vec{r'}(t)$ and $\\vec{r''}(t)$ is: (3) \\begin{align} \\quad \\quad \\vec{r'}(t) \\times \\vec{r''}(t) = \\begin{vmatrix} \\vec{i} & \\vec{j} & \\vec{k}\\\\ \\cos^2 t - \\sin^2 t & 2\\sin t \\cos t & -\\sin t\\\\ -4 \\sin t \\cos t & 2(\\cos^2 t - \\sin^2 t) & - \\cos t \\end{vmatrix} \\\\ \\quad \\quad = [-2 \\sin t \\cos^2 t + 2 \\sin t(\\cos^2 t - \\sin^2 t)] \\vec{i} + [\\cos t (\\cos^2 t - \\sin^2 t) + \\sin t(4 \\sin t \\cos t)] \\vec{j} + [ 2 (\\cos^2 t - \\sin^2 t)^2 + 8 \\sin^2 t \\cos^2 t ] \\vec{k} \\\\ \\quad \\quad = (-2 \\sin^3 t ) \\vec{i} + (\\cos^3 t + 3 \\sin^2 t \\cos t) \\vec{j} + (2 \\cos^4 t + \\sin^4 t + 6 \\sin^2 t \\cos^2 t) \\vec{k} \\end{align} The norm of $\\vec{r'}(t) \\times \\vec{r''}(t)$ is: (4) \\begin{align} \\quad \\| \\vec{r'}(t) \\times \\vec{r''}(t) \\| = \\sqrt{4 \\sin^6 t + (\\cos^3 t + 3 \\sin^2 t \\cos t)^2 + (2 \\cos^4 t + \\sin^4 t + 6 \\sin^2 t \\cos^2 t)^2 } \\end{align} Lastly, the norm of $\\vec{r'}(t)$ is: (5) \\begin{align} \\quad \\| \\vec{r'}(t) \\| = \\sqrt{(\\cos^2 t - \\sin^2 t)^2 +4 \\sin^2 t \\cos^2 t + \\sin^2 t} \\end{align} Therefore the curvature $\\kappa$ is", "# Chapter 6 - Mid-Chapter Check Point - Page 756: 4 The solution is $A=26{}^\\circ,C=44{}^\\circ \\ \\text{ and }\\ b=21.6$ #### Work Step by Step The given angles and sides of the triangle are: $B=110{}^\\circ,a=10,c=16$ Using the law of cosines, we will find the side b of the triangle. That is, \\begin{align} & {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\\cos B \\\\ & {{b}^{2}}={{10}^{2}}+{{16}^{2}}-2\\left( 10 \\right)\\left( 16 \\right)\\cos 110{}^\\circ \\\\ & {{b}^{2}}=465.4464 \\\\ & b=21.6 \\end{align} Now, by using the law of sines we will find the angle A of the triangle. That is, \\begin{align} & \\frac{\\sin A}{a}=\\frac{\\operatorname{sinB}}{b} \\\\ & \\frac{\\sin A}{10}=\\frac{\\sin 110{}^\\circ }{21.6} \\end{align} This implies that, \\begin{align} & \\sin A=10\\times \\frac{\\sin 110{}^\\circ }{21.6} \\\\ & =0.4350 \\end{align} Therefore, $A=26{}^\\circ$ Now, the third angle will be found as below: \\begin{align} & C=180{}^\\circ -A-B \\\\ & =180{}^\\circ -26{}^\\circ -110{}^\\circ \\\\ & C=44{}^\\circ \\end{align} Therefore, the solution is: $A=26{}^\\circ,C=44{}^\\circ \\ \\text{ and }\\ b=21.6$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "\\begin{align*}30^\\circ\\end{align*} \\begin{align*}60^\\circ\\end{align*} \\begin{align*}90^\\circ\\end{align*} \\begin{align*}120^\\circ\\end{align*} \\begin{align*}150^\\circ\\end{align*} \\begin{align*}180^\\circ\\end{align*} \\begin{align*}210^\\circ\\end{align*} \\begin{align*}240^\\circ\\end{align*} \\begin{align*}270^\\circ\\end{align*} \\begin{align*}300^\\circ\\end{align*} \\begin{align*}330^\\circ\\end{align*} \\begin{align*}360^\\circ\\end{align*} \\begin{align*}4 \\cos 2 \\theta\\end{align*} 4 2 -2 -4 -2 2 4 2 -2 -4 -2 2 4 \\begin{align*}\\theta\\end{align*} \\begin{align*}0^\\circ\\end{align*} \\begin{align*}30^\\circ\\end{align*} \\begin{align*}60^\\circ\\end{align*} \\begin{align*}90^\\circ\\end{align*} \\begin{align*}120^\\circ\\end{align*} \\begin{align*}150^\\circ\\end{align*} \\begin{align*}180^\\circ\\end{align*} \\begin{align*}210^\\circ\\end{align*} \\begin{align*}240^\\circ\\end{align*} \\begin{align*}270^\\circ\\end{align*} \\begin{align*}300^\\circ\\end{align*} \\begin{align*}330^\\circ\\end{align*} \\begin{align*}360^\\circ\\end{align*} \\begin{align*}4 \\cos 3 \\theta\\end{align*} 4 0 -4 0 4 0 -4 0 4 0 -4 0 4 In the graph of \\begin{align*}r = 4 \\cos 2 \\theta\\end{align*}, the rose has four petals on it but the graph of \\begin{align*}r = 4 \\cos 3 \\theta\\end{align*} has only three petals. It appears, that if \\begin{align*}n\\end{align*} is an even positive integer, the rose will have an even number of petals and if \\begin{align*}n\\end{align*} is an odd positive integer, the rose will have an odd number of petals. 1. \\begin{align*}r = 3 \\sin 4 \\theta\\end{align*} 2. \\begin{align*}r = 2 \\sin 5 \\theta\\end{align*} 3. \\begin{align*}r = 3 \\cos 3 \\theta\\end{align*} 4. \\begin{align*}r = -4 \\sin 2 \\theta\\end{align*} 5. \\begin{align*}r = 5 \\cos 4 \\theta\\end{align*} 6. \\begin{align*}r = -2 \\cos 6 \\theta\\end{align*} For roses, the general equation is \\begin{align*}r = a \\sin n \\theta\\end{align*} or \\begin{align*}r = a \\cos n \\theta\\end{align*}. \\begin{align*}a\\end{align*} indicates how long each petal is, and depending on if \\begin{align*}n\\end{align*} is even or odd indicates the number of pedals. If \\begin{align*}n\\end{align*} is ood, there are \\begin{align*}n\\end{align*} pedals and", "as in the figure below Keep in mind that in such problems, the given angles must be measured from a reference axis in a clockwise direction. Here, we choose the positive $x$ axis as the reference. Because it is easier to reach the $23^\\circ$ west of south from the $+x$ axis in a counterclockwise direction, so after adding $90^\\circ$ to $23^\\circ$, we must insert a negative in front of the result. Therefore, the angles of $23^\\circ$ west of south and $30^\\circ$ east of south are measured from $+x$ axis as $\\alpha=90^\\circ+23^\\circ=-113^\\circ$ and $180^\\circ+30^\\circ=210^\\circ$. For the first flight, the components of the displacement vector are obtained as below \\begin{align*} A_x&=|\\vec{A}|\\cos \\alpha\\\\&=250\\times \\cos (-113^\\circ) \\\\&=-97.68\\quad {\\rm m} \\\\\\\\ A_y&=|\\vec{A}|\\sin \\alpha\\\\&=250\\times \\sin (-113^\\circ) \\\\&=-230.12\\quad {\\rm m}\\end{align*} Similarly, for the second flight, the components for the displacement vector $\\vec{B}$ are \\begin{align*} B_x&=|\\vec{B}|\\cos \\theta\\\\&=200\\times \\cos (210^\\circ) \\\\&=-173.20\\quad {\\rm m} \\\\\\\\ B_y&=|\\vec{A}|\\sin \\theta\\\\&=200\\times \\sin (210^\\circ) \\\\&=-100\\quad {\\rm m}\\end{align*} Now, adding the corresponding components gives the components of the displacement vector $\\vec{D}$ \\begin{align*} D_x&=A_x+B_x\\\\&=-97.68-173.20\\\\&=-270.88\\quad {\\rm m}\\\\\\\\ D_y&=A_y+B_y\\\\&=-230.12-100\\\\&=-330.12\\quad {\\rm m}\\end{align*} Recall from physics that, the magnitude of the displacement vector is the direct distance between start and end points. Problem (11): A ball is fired with an initial speed of 15 m/s at an angle of $37^\\circ$ with the horizontal. Find the vertical and horizontal components", "$$\\beta=180°−49.9°=130.1°$$. (Remember that the sine function is positive in both the first and second quadrants.) Solving for $$\\gamma$$, we have \\begin{align*} \\gamma&= 180^{\\circ}-35^{\\circ}-130.1^{\\circ}\\\\ &\\approx 14.9^{\\circ} \\end{align*} We can then use these measurements to solve the other triangle. Since $$\\gamma′$$ is supplementary to $$\\gamma$$, we have \\begin{align*} \\gamma^{'}&= 180^{\\circ}-35^{\\circ}-49.5^{\\circ}\\\\ &\\approx 95.1^{\\circ} \\end{align*} Now we need to find $$c$$ and $$c′$$. We have \\begin{align*} \\dfrac{c}{\\sin(14.9^{\\circ})}&= \\dfrac{6}{\\sin(35^{\\circ})}\\\\ c&= \\dfrac{6 \\sin(14.9^{\\circ})}{\\sin(35^{\\circ})}\\\\ &\\approx 2.7 \\end{align*} Finally, \\begin{align*} \\dfrac{c'}{\\sin(95.1^{\\circ})}&= \\dfrac{6}{\\sin(35^{\\circ})}\\\\ c'&= \\dfrac{6 \\sin(95.1^{\\circ})}{\\sin(35^{\\circ})}\\\\ &\\approx 10.4 \\end{align*} To summarize, there are two triangles with an angle of $$35°$$, an adjacent side of 8, and an opposite side of 6, as shown in Figure $$\\PageIndex{12}$$. However, we were looking for the values for the triangle with an obtuse angle $$\\beta$$. We can see them in the first triangle (a) in Figure $$\\PageIndex{12}$$. Exercise $$\\PageIndex{2}$$ Given $$\\alpha=80°$$, $$a=120$$, and $$b=121$$, find the missing side and angles. If there is more than one possible solution, show both. Solution 1 $$\\begin{matrix} \\alpha=80^{\\circ} & a=120\\\\ \\beta\\approx 83.2^{\\circ} & b=121\\\\ \\gamma\\approx 16.8^{\\circ} & c\\approx 35.2 \\end{matrix}$$ Solution 2 $$\\begin{matrix} \\alpha '=80^{\\circ} & a'=120\\\\ \\beta '\\approx 96.8^{\\circ} & b'=121\\\\ \\gamma '\\approx 3.2^{\\circ} & c'\\approx 6.8 \\end{matrix}$$ Example $$\\PageIndex{3}$$: Solving for the Unknown Sides and Angles of a SSA Triangle In the triangle shown in Figure $$\\PageIndex{13}$$, solve for the unknown side and", "x_{239}^8x_{261}^5\\\\ & \\quad\\oplus x_{ 281}^0x_{277}^1 \\oplus x_{207}^{13}x_{265}^4 \\oplus x_{200}^{13}\\overline{x_{70}}\\oplus x_{190}\\overline{x_{55}} \\oplus x_{281}^0x_{273}^2x_{269}^3x_{239}^8x_{207}^{13}x_{200}^{13}\\end{aligned} 27 \\begin{aligned}x_{ 164}& = z_ {27} \\oplus x_{107} \\oplus x_{ 126} \\oplus x_{ 254}^6 \\oplus x_{ 249}^6 \\oplus x_{214}^{13} \\oplus x_{ 270}^3x_{244}^7 \\oplus x_{ 274}^2x_{ 258}^6 \\oplus x_{240}^8x_{262}^5\\\\ & \\quad\\oplus x_{ 282}^0x_{278}^1 \\oplus x_{208}^{13}x_{266}^4 \\oplus x_{201}^{13}\\overline{x_{71}}\\oplus x_{191}\\overline{x_{56}} \\oplus x_{282}^0x_{274}^2x_{270}^3x_{240}^8x_{208}^{13}x_{201}^{13}\\end{aligned} 28 \\begin{aligned}x_{ 165}& = z_ {28} \\oplus x_{108} \\oplus x_{ 127} \\oplus x_{ 255}^6 \\oplus x_{ 250}^6 \\oplus x_{215}^{13} \\oplus x_{ 271}^3x_{245}^7 \\oplus x_{ 275}^2x_{ 259}^6 \\oplus x_{241}^8x_{263}^5\\\\ & \\quad\\oplus x_{ 283}^0x_{279}^1 \\oplus x_{209}^{13}x_{267}^4 \\oplus x_{202}^{13}\\overline{x_{72}}\\oplus x_{192}\\overline{x_{57}} \\oplus x_{283}^0x_{275}^2x_{271}^3x_{241}^8x_{209}^{13}x_{202}^{13}\\end{aligned} 29 \\begin{aligned}x_{ 166}& = z_ {29} \\oplus x_{109} \\oplus x_{ 128} \\oplus x_{ 256}^6 \\oplus x_{ 251}^6 \\oplus x_{216}^{13} \\oplus x_{ 272}^3x_{246}^7 \\oplus x_{ 276}^2x_{ 260}^6 \\oplus x_{242}^8x_{264}^5\\\\ & \\quad\\oplus x_{ 284}^0x_{280}^1 \\oplus x_{210}^{13}x_{268}^4 \\oplus x_{203}^{13}\\overline{x_{73}}\\oplus x_{193}\\overline{x_{58}} \\oplus x_{284}^0x_{276}^2x_{272}^3x_{242}^8x_{210}^{13}x_{203}^{13}\\end{aligned} 30 \\begin{aligned}x_{ 167}& = z_ {30} \\oplus x_{110} \\oplus x_{ 129} \\oplus x_{ 257}^6 \\oplus x_{ 252}^6 \\oplus x_{217}^{13} \\oplus x_{ 273}^3x_{247}^7 \\oplus x_{ 277}^2x_{ 261}^6 \\oplus x_{243}^8x_{265}^5\\\\ & \\quad\\oplus x_{ 285}^0x_{281}^1 \\oplus x_{211}^{13}x_{269}^4 \\oplus x_{204}^{13}\\overline{x_{74}}\\oplus x_{194}^{13}\\overline{x_{59}} \\oplus x_{285}^0x_{277}^2x_{273}^3x_{243}^8x_{211}^{13}x_{204}^{13}\\end{aligned} 31 \\begin{aligned}x_{ 168}& = z_ {31} \\oplus x_{111} \\oplus x_{ 130} \\oplus x_{ 258}^6 \\oplus x_{ 253}^6 \\oplus x_{218}^{13} \\oplus x_{ 274}^3x_{248}^7 \\oplus x_{ 278}^2x_{ 262}^6 \\oplus x_{244}^8x_{266}^5\\\\ & \\quad\\oplus x_{ 286}^0x_{282}^1 \\oplus x_{212}^{13}x_{270}^4 \\oplus x_{205}^{13}\\overline{x_{75}}\\oplus x_{195}^{13}\\overline{x_{60}} \\oplus x_{286}^0x_{278}^2x_{274}^3x_{244}^8x_{212}^{13}x_{205}^{13}\\end{aligned} 32 \\begin{aligned}x_{ 169}& = z_ {32} \\oplus x_{112} \\oplus x_{ 131}", "\\begin{align*}\\triangle ABE\\end{align*}: \\begin{align*}K & = \\frac{1}{2}(3375.5)(5382)\\sin 41 \\\\ K & = 5,959,292.8\\ ft^2\\end{align*} \\begin{align*}\\triangle DBC\\end{align*}: \\begin{align*}K & = \\frac{1}{2}(3862)(3862)\\sin 32 \\\\ K & = 3, 951,884.6\\ ft^2\\end{align*} The total area is \\begin{align*}5,959,292.8 + 3,951,884.6 = 9,911,177.4\\ ft^2\\end{align*}. ### Examples #### Example 1 Earlier, you were asked to find the length of the third side of the triangle. Since you know two angles and one non-included side of the kite, you can find the other non-included side using the Law of Sines. Set up a ratio using the angles and side you know and the side you don't know. \\begin{align*}\\frac{\\sin 70^\\circ}{x} & = \\frac{\\sin 40^\\circ}{22} \\\\ x & = \\frac{22 \\sin 70^\\circ}{\\sin 40^\\circ} \\\\ x & \\approx 32.146\\end{align*} The length of the dowel rod on the unknown side will be approximately 32 inches. #### Example 2 Find side \"d\" in the triangle below with the following information: \\begin{align*}e = 214.9, D = 39.7^\\circ, E = 41.3^\\circ\\end{align*} \\begin{align*}\\frac{\\sin 41.3^\\circ}{214.9} = \\frac{\\sin 39.7^\\circ}{d}, d = 208.0\\end{align*} #### Example 3 Find side \"o\" in the triangle below with the following information: \\begin{align*}M = 31^\\circ, O = 9^\\circ, m = 15\\end{align*} \\begin{align*}\\frac{\\sin 9^\\circ}{o} = \\frac{\\sin 31^\\circ}{15}, o = 4.6\\end{align*} #### Example 4 Find side \"q\" in the triangle below with the following information: \\begin{align*}Q = 127^\\circ, R = 21.8^\\circ, r = 3.62\\end{align*} \\begin{align*}\\frac{\\sin" ]

[ "cis}(n\\theta)$$ According to DeMoivre's Theorem: [2(cos(-30) + i sin(-30)]^7 = 2^7 [cos((-30)(7)) + i sin((-30)(7))] We now can evaluate 2^7 = 128 and (-30)(7) = -210. The result is 128(cos(-210) + i sin(-210)). Note cos(-210) = -sqrt(3)/2 and sin(-210) = 1/2. Therefore, the final answer is -64*sqrt(3) + 64i. An interesting thing to observe in this specific case is that $$z^7=-64z$$. Do you see why? We multiplied 6 times, rotating 30° each time, for a total of 180°, and scaled up by a factor 2 to 6 times, multiplying the length by 64. Here I have plotted the successive powers of our $$z$$, from the first ($$z$$ itself) to the seventh: ## Proving the theorem by induction Every good theorem deserves a proof, so we’ll end with this question from 1997: Proof of DeMoivre's Theorem Could you please help me prove DeMoivre's theorem by mathematical induction? I figured out the first part of the proof, but I can't prove that assuming that if it's true for n = k, then it's true for n = k+1. We discussed induction in a series starting here. Evelina has presumably seen that the basis step, proving it true for $$n=1$$ (or, indeed, 0) is true – which is obvious – but needs help with this inductive step. Evelina, This depends on", "w\\sin 12^\\circ \\\\ W - x &= h\\sin 12^\\circ \\\\ H - y &= h\\cos 12^\\circ \\\\ \\end{align*} Indeed, for the special case where $w = 400$ and $h = 300$, we obtain: \\begin{align*} \\frac{W}{w} &= \\frac{400\\cos 12^\\circ + 300\\sin 12^\\circ}{400} \\approx 113.41\\% \\\\ \\frac{H}{h} &= \\frac{400\\sin 12^\\circ + 300\\cos 12^\\circ}{300} \\approx 125.54\\% \\\\ \\end{align*}", "\\pm 40^\\circ : n \\in \\mathbb Z \\} \\\\ \\end{align} Solution set: $$x \\in (\\{60^\\circ n \\pm 40^\\circ : n \\in \\mathbb Z \\} \\cup \\{180^\\circ n : n \\in \\mathbb Z\\}) \\cap [0^\\circ, 360^\\circ]$$ $$x \\in \\left\\{ \\begin{array}{rrrrr} 0^\\circ, & 20^\\circ, & 40^\\circ, & 80^\\circ, & 100^\\circ, \\\\ 140^\\circ, & 160^\\circ, & 180^\\circ, & 200^\\circ, & 220^\\circ, \\\\ 260^\\circ, & 280^\\circ, & 320^\\circ, & 340^\\circ, & 360^\\circ \\\\ \\end{array} \\right\\}$$ Using $\\sin3B=3\\sin B-4\\sin^3B$ $$32\\sin^3x\\cdot\\cos^2x\\cdot\\cos2x=\\sin x(3-4\\sin^2x)$$ If $\\sin x=0,x=180^\\circ n$ where $n$ is any integer Else using $\\cos2A=2\\cos^2A-1=1-2\\sin^2A,$ $$32\\cdot\\dfrac{1-\\cos2x}2\\cdot\\dfrac{1+\\cos2x}2\\cdot\\cos2x=3-2(1-\\cos2x)$$ $$\\iff8(c-c^3)=1+2c\\iff4c^3-3c=-\\dfrac12$$ $$\\implies\\cos6x=-\\dfrac12,6x=360^\\circ m\\pm120^\\circ\\iff x=60^\\circ m\\pm20^\\circ$$ where $m$ is any integer • @Théophile, One set of solution is $$0\\le180n\\le360\\iff0\\le n\\le2$$ Nov 2 '16 at 17:57", "Write the following as ratios of $\\scriptsize {{20}^\\circ}$: 1. $\\scriptsize \\sin {{70}^\\circ}$ 2. $\\scriptsize \\cos {{110}^\\circ}$ 3. $\\scriptsize \\sin {{250}^\\circ}$ 4. $\\scriptsize \\cos (\\theta -{{90}^\\circ})$ Solutions 1. \\scriptsize \\begin{align*}\\sin {{70}^\\circ}&=\\sin ({{90}^\\circ}-{{20}^\\circ})\\\\&=\\cos {{20}^\\circ}\\end{align*} 2. . \\scriptsize \\begin{align*}\\cos {{110}^\\circ}&=\\cos ({{90}^\\circ}+{{20}^\\circ})\\\\&=-\\sin {{20}^\\circ}\\quad \\text{Don }\\!\\!'\\!\\!\\text{ t forget the negative sign; cosine is negative in the second quadrant}\\end{align*} 3. We could answer this as follows: $\\scriptsize \\sin {{250}^\\circ}=\\sin ({{270}^\\circ}-{{20}^\\circ})=-\\cos {{20}^\\circ}$However, the $\\scriptsize {{270}^\\circ}\\pm \\theta$ co-function reduction rules are not part of the curriculum, so we answer it as follows: \\scriptsize \\begin{align*}\\sin {{250}^\\circ}&=\\sin ({{180}^\\circ}+{{70}^\\circ})\\\\&=-\\sin {{70}^\\circ}\\\\&=-\\sin ({{90}^\\circ}-{{20}^\\circ})\\\\&=-\\cos {{20}^\\circ}\\end{align*} 4. . \\scriptsize \\begin{align*}\\cos (\\theta -{{90}^\\circ})&=\\cos (-{{90}^\\circ}+\\theta )\\\\&=\\cos \\left( {-({{{90}}^\\circ}-\\theta )} \\right)&& \\text{The angle }-({{90}^\\circ}-\\theta )\\text{ is in the fourth quadrant where}\\\\ &&&\\text{cosine is positive}\\\\&=\\cos ({{90}^\\circ}-\\theta )\\\\&=\\sin \\theta \\end{align*} ### Example 2.5 Express the following with all ratios in terms of $\\scriptsize \\theta$ alone and simplify: $\\scriptsize \\displaystyle \\frac{{\\text{cos(}\\theta -\\text{9}{{\\text{0}}^\\circ}\\text{) cos(72}{{\\text{0}}^\\circ}+\\theta \\text{) tan(}\\theta -\\text{36}{{\\text{0}}^\\circ})}}{{\\text{si}{{\\text{n}}^{2}}\\text{(}\\theta \\text{+36}{{\\text{0}}^\\circ}\\text{) cos(}\\theta \\text{+9}{{\\text{0}}^\\circ}\\text{)}}}$ Solution Let’s break this question down and look at each trig ratio separately first. • . \\scriptsize \\begin{align*}\\cos (\\theta -{{90}^\\circ})&=\\cos (-{{90}^\\circ}+\\theta )\\\\&=\\cos \\left( {-({{{90}}^\\circ}-\\theta )} \\right)\\quad -({{90}^\\circ}-\\theta )\\text{ is in the fourth quadrant where cosine is positive}\\\\&=\\cos ({{90}^\\circ}-\\theta )\\\\&=\\sin \\theta \\end{align*} • $\\scriptsize \\cos ({{720}^\\circ}+\\theta )=\\cos \\theta\\quad \\text{because }{{720}^\\circ}\\text{ is just two full revolutions or two full periods}$ • $\\scriptsize \\tan (\\theta -{{360}^\\circ})=\\tan \\theta \\quad -{{360}^\\circ}\\text{ is one full revolution", "QUESTION # What is $\\cos {{24}^{\\circ }}+\\cos {{5}^{\\circ }}+\\cos {{175}^{\\circ }}+\\cos {{204}^{\\circ }}+\\cos {{300}^{\\circ }}=$ a) $\\dfrac{1}{2}$ b) $\\dfrac{-1}{2}$ c) $\\dfrac{\\sqrt{3}}{2}$ d) 1 Hint: Change the cosine terms with the higher angles $\\left( >{{90}^{\\circ }} \\right)$ to the acute values. Use the following identities for the conversion: \\begin{align} & \\cos \\left( 180-\\theta \\right)=-\\cos \\theta \\\\ & \\cos \\left( 180+\\theta \\right)=-\\cos \\theta \\\\ & \\cos \\left( 300-\\theta \\right)=\\cos \\theta \\\\ \\end{align} And, put the value of $\\cos {{60}^{\\circ }}\\Rightarrow \\dfrac{1}{2}$ wherever required. Expression given in the question is $\\cos {{24}^{\\circ }}+\\cos {{5}^{\\circ }}+\\cos {{175}^{\\circ }}+\\cos {{204}^{\\circ }}+\\cos {{300}^{\\circ }}=?$ Let the value of the given expression in the problem be ‘M’. So, we have $M=\\cos {{24}^{\\circ }}+\\cos {{5}^{\\circ }}+\\cos {{175}^{\\circ }}+\\cos {{204}^{\\circ }}+\\cos {{300}^{\\circ }}.........\\left( i \\right)$ As we can observe that we do not know the exact values of the cosine terms given in the above equation. So, we need to observe the relation and have to apply some trigonometric identity. So, let us try to convert all the angles involved in the expression to acute angles with the help of identities expressed below \\begin{align} & \\cos \\left( 180-\\theta \\right)=-\\cos \\theta .............\\left( ii \\right) \\\\ & \\cos \\left( 180+\\theta \\right)=-\\cos \\theta .............\\left( iii \\right) \\\\ & \\cos \\left( 300-\\theta \\right)=\\cos \\theta ...............\\left( iv \\right) \\\\ \\end{align} As we", "find $a$ and $A$. ###### Solution First, use $\\dfrac{b}{\\sin B} = \\dfrac{c}{\\sin C}$ to find $c$. \\begin{align} \\frac{6}{\\sin 77^{\\circ} } &= \\frac{c}{\\sin 34^{\\circ} }\\\\\\\\ \\frac{6 \\sin 34^{\\circ} }{\\sin 77^{\\circ} } &= c\\\\ c &= 3.44 \\text{ (to 3 sig.fig.)} \\end{align} To find $A$, use the fact that the angles in a triangle add up to $180^{\\circ}$, i.e. $A+B+C = 180$. \\begin{align} 180 &= A +77 + 34\\\\ A &= 180 - 77- 34\\\\ A &= 69^{\\circ} \\end{align} Now, use this value for $A$ in the sine rule to find $a$. \\begin{align} \\frac{a}{\\sin 69^{\\circ} } &= \\frac{6}{\\sin 77^{\\circ} }\\\\\\\\ a &= \\frac{6 \\sin 69^{\\circ} }{\\sin 77^{\\circ} }\\\\\\\\ a &= 5.75 \\text{ (to 3 sig.fig.)} \\end{align} ###### Example 3 Given $b=2$, $c=5$ and $A=39^{\\circ}$, find $a$. ###### Solution Use $a^2 = b^2 + c^2 -2bc\\cos A$. \\begin{align} a^2 &= 2^2 + 5^2 - (2\\times2\\times5)\\cos 39^{\\circ}\\\\ a^2 &= 29 - 20\\cos 39^{\\circ}\\\\ a^2 &= 13.457 \\text{ (to 3 d.p.)}\\\\ a &= 3.67 \\text{ (to 3 sig.fig.)} \\end{align} ###### Example 4 Given $a=7$, $b=6$ and $c=9$. Find the angle between $a$ and $b$. ###### Solution The angle between $a$ and $b$ is labelled $C$ in the diagram above, so use $c^2 = a^2 + b^2 - 2ab\\cos C$. \\begin{align} 9^2 &= 7^2 + 6^2 - (2\\times7\\times6)\\cos C\\\\ 81 &= 49+36 - 84\\cos", "28 at 12:01 • @PeterTaylor This was an assumption that I thought should be true, since (255,255,255) is the room diagonal, I thought it should have 45° angles for teta and phi. For phi that is correct but as it turns out teta has a 54,7° angle = cos^-1(1/sqrt(3)). – Lucas Raphael Pianegonda Feb 28 at 12:54 So how would one generate RGB vectors with a constant norm that are still valid RGB values? ($$0 \\le r,g,b \\le 255$$) There is a simpler algorithm. Let $$L$$ be the given norm. • Verify that $$0\\le L\\le 255\\sqrt3$$. • Let r be a random number between $$\\sqrt{L^2-\\min(2\\cdot255^2, L^2)}$$ and $$\\min(255, L)$$. • Let $$M=\\sqrt{L^2-r^2}$$. Let g be a random number between $$\\sqrt{M^2-\\min(255^2, M^2)}$$ and $$\\min(255,M)$$. • Let $$b =\\sqrt{L^2-r^2-g^2}$$. Some minor adjustment is needed if $$r,g,b$$ should be integers, in which case only approximate solutions can be found in general. Here is the right computation if we want to find the range of $$\\theta$$ and $$\\phi$$. • If $$0\\le L\\le 255$$, then $$0\\le\\theta,\\phi\\le\\pi/2$$ • If $$255\\lt L\\le 255\\sqrt3$$, let $$M=\\sqrt{\\max(0, L^2-2\\cdot255^2})$$. we have \\begin{align} &\\quad\\quad\\begin{cases}b = L \\sin(\\theta) \\cos(\\phi)\\\\ r = L \\sin(\\theta) \\sin(\\phi)\\\\ g = L \\cos(\\theta)\\\\ 0\\le b,r,g\\le 255\\\\ 0\\le\\theta,\\phi\\le\\pi/2 \\end{cases}\\\\ &\\Longleftrightarrow \\begin{cases}b = L \\sin(\\theta) \\cos(\\phi)\\\\ r = L \\sin(\\theta) \\sin(\\phi)\\\\ g = L \\cos(\\theta)\\\\ \\sqrt{\\max\\left(0,(L\\sin(\\theta))^2-255^2\\right)}\\le b,r\\le", "$$g(360^{\\circ})$$ from $$f(360^{\\circ})$$. \\begin{align*} f(360^{\\circ}) - g(360^{\\circ}) &= 2 - 0 \\\\ &=2 \\end{align*} Given the following graph. State the coordinates at $$A,~B,~C$$ and $$D$$. We read the values off the graph: $$A = ( 90^{\\circ}; 1),~B = ( 180^{\\circ}; -3),~C = (270 ^{\\circ};-1 ) \\text{ and } D = ( 360^{\\circ}; 3)$$ How many times in this interval does $$f(x)$$ intersect $$g(x)$$. $$2$$ What is the amplitude of $$g(x)$$. $$3$$ Evaluate: $$f(90^{\\circ}) - g(90^{\\circ})$$ . Read off the value of $$f(90^{\\circ})$$ and $$g(90^{\\circ})$$ from the graph. Then subtract $$g(90^{\\circ})$$ from $$f(90^{\\circ})$$. \\begin{align*} f(90^{\\circ}) - g(90^{\\circ}) &= 1 - 0 \\\\ &=1 \\end{align*} Given the following graph: State the coordinates at $$A,~B,~C$$ and $$D$$. We read the values off the graph: $$A = (90 ^{\\circ};3 ),~B = (90 ^{\\circ};2 ),~C = (180 ^{\\circ}; -4) \\text{ and } D = ( 270^{\\circ};2)$$ How many times in this interval does $$f(x)$$ intersect $$g(x)$$. $$3$$ What is the amplitude of $$g(x)$$. $$2$$ Evaluate: $$f(270^{\\circ}) - g(270^{\\circ})$$ . Read off the value of $$f(270^{\\circ})$$ and $$g(270^{\\circ})$$ from the graph. Then subtract $$g(270^{\\circ})$$ from $$f(270^{\\circ})$$. \\begin{align*} f(270^{\\circ}) - g(270^{\\circ}) &= -3 - (-2)\\\\ &=-1 \\end{align*}", "the first identity in the reverse order). $$\\cos{A}+\\cos{(A+2\\pi/3)}+\\cos{(A-2\\pi/3)}=0$$ $$\\cos{(A-\\pi/3)}+\\cos{(A+\\pi/3)}=\\cos{A}$$ Consider the following picture. We see that the triangle with vertices $(\\cos A, \\sin A), (\\cos(A+120^\\circ), \\sin(A+120^\\circ)), (\\cos(A+240^\\circ), \\sin(A+240^\\circ))$ are the vertices of an equilateral triangle with centriod at the origin. Thus if we set $a = \\cos A, b = \\cos(A+120^\\circ), c = \\cos(A+240^\\circ)$, then \\begin{align*} a+b+c = 0 \\end{align*} and hence \\begin{align*} a^3+b^3+c^3 = 3abc \\end{align*} and \\begin{align*} \\cos^3{A} + \\cos^3{(120°+A)} + \\cos^3{(240°+A)} &= 3\\cos A \\cos(A+120^\\circ)\\cos(A+240^\\circ) \\\\ &= \\frac{3}{2}(2\\cos A \\cos(A+120^\\circ))\\cos(A+240^\\circ) \\\\ &= \\frac{3}{2}[\\cos(2A+120^\\circ)+\\cos(120^\\circ)]\\cos(A+240^\\circ)\\\\ &=\\frac{3}{2}\\cos(2A+120^\\circ)\\cos(A+240^\\circ)-\\frac{3}{4}\\cos(A+240^\\circ)\\\\ &=\\frac{3}{4}\\cos(3A+360^\\circ)+\\frac{3}{4}\\cos(A-120^\\circ) -\\frac{3}{4}\\cos(A+240^\\circ)\\\\ &=\\frac{3}{4}\\cos(3A) \\end{align*} since $\\cos(A+240^\\circ) = \\cos(360^\\circ - (A+240^\\circ)) = \\cos(120^\\circ -A) = \\cos(A-120^\\circ)$ • If you can see geometrically, the points $(\\cos A, \\sin A), (\\cos(120^\\circ+A), \\sin(120^\\circ+A)), (\\cos(240^\\circ), \\sin(240^\\circ+A))$ are the vertices of an equilateral triangle with centroid as the origin, then the above proof holds since we have used only an algebraic identity. – user348749 Sep 10, 2016 at 7:23 use that $$\\cos(120^{\\circ}+x)=-1/2\\,\\cos \\left( x \\right) -1/2\\,\\sqrt {3}\\sin \\left( x \\right)$$ and $$\\cos(240^{\\circ}+x)=-1/2\\,\\cos \\left( x \\right) +1/2\\,\\sqrt {3}\\sin \\left( x \\right)$$ For the LHS, use $\\cos(120^\\circ +A)=\\cos 120^\\circ \\cos A - \\sin 120^\\circ \\sin A$ and $\\cos(240^\\circ+A)=\\cos 240^\\circ \\cos A - \\sin 240^\\circ \\sin A$ Then $\\cos^3(120^\\circ+A)=\\left(\\cos 120^\\circ \\cos A - \\sin 120^\\circ \\sin A \\right)^3$ and $\\cos^3(240^\\circ+A)=\\left( \\cos 240^\\circ \\cos A - \\sin 240^\\circ \\sin A \\right)^3$", "}}\\cos {{120}^{\\circ }}+\\cos {{210}^{\\circ }}\\sin {{300}^{\\circ }}$. Let us assume the value of $\\sin {{330}^{\\circ }}\\cos {{120}^{\\circ }}+\\cos {{210}^{\\circ }}\\sin {{300}^{\\circ }}$ is equal to I. $I=\\sin {{330}^{\\circ }}\\cos {{120}^{\\circ }}+\\cos {{210}^{\\circ }}\\sin {{300}^{\\circ }}.....(1)$ Now we have to find the value of $\\sin {{330}^{\\circ }}$. We know that $\\sin \\left( n\\pi -\\theta \\right)=-\\sin \\theta$ if n is even. \\begin{align} & \\Rightarrow sin{{330}^{\\circ }}=\\sin \\left( 2\\pi -{{30}^{\\circ }} \\right) \\\\ & \\Rightarrow \\sin {{330}^{\\circ }}=-\\sin {{30}^{\\circ }} \\\\ \\end{align} We know that $\\sin {{30}^{\\circ }}=\\dfrac{1}{2}$. $\\Rightarrow \\sin {{330}^{\\circ }}=-\\dfrac{1}{2}......(2)$ Now we should find the value of $\\cos {{120}^{\\circ }}$. We know that $\\cos \\left( n\\pi -\\theta \\right)=-\\cos \\theta$ if n is odd. \\begin{align} & \\Rightarrow \\cos {{120}^{\\circ }}=\\cos \\left( \\pi -{{60}^{\\circ }} \\right) \\\\ & \\Rightarrow \\cos {{120}^{\\circ }}=-\\cos \\left( {{60}^{\\circ }} \\right) \\\\ \\end{align} We know that $\\cos {{60}^{\\circ }}=\\dfrac{1}{2}$. $\\Rightarrow \\cos {{120}^{\\circ }}=-\\dfrac{1}{2}.......(3)$ Now we should find the value of $\\cos {{210}^{\\circ }}$. We know that $\\cos \\left( n\\pi +\\theta \\right)=-\\cos \\theta$ if n is odd. \\begin{align} & \\Rightarrow \\cos {{210}^{\\circ }}=\\cos \\left( \\pi +{{30}^{\\circ }} \\right) \\\\ & \\Rightarrow \\cos {{210}^{\\circ }}=-\\cos \\left( {{30}^{\\circ }} \\right) \\\\ \\end{align} We know that $\\cos {{30}^{\\circ }}=\\dfrac{\\sqrt{3}}{2}$. $\\Rightarrow \\cos {{210}^{\\circ }}=-\\dfrac{\\sqrt{3}}{2}.......(4)$ Now we have to find the value of $\\sin {{300}^{\\circ }}$. We know that $\\sin \\left( n\\pi", "from $\\mathbb{R}^{2}\\to \\mathbb{R}$, just like $f$ is. So at a point $(a,b)\\in \\mathbb{R}^{2}$, I can write \\begin{align*} Df(a,b) &= D\\left(\\sin\\circ p\\circ (\\pi_{1}, s\\circ \\pi_{2})\\right)(a,b)\\\\ &= D(\\sin)(p\\circ (\\pi_{1}, s\\circ \\pi_{2})(a,b))\\circ Dp((\\pi_{1}, s\\circ \\pi_{2})(a,b))\\circ D(\\pi_{1}, s\\circ \\pi_{2})(a,b) \\end{align*} So I try to calculate this in separate blocks: \\begin{align*} D(\\sin)(p\\circ (\\pi_{1}, s\\circ \\pi_{2})(a,b)) &= \\cos(p\\circ (\\pi_{1}, s\\circ \\pi_{2})(a,b))\\\\ &= \\cos(p\\circ (\\pi_{1}(a,b), [s\\circ \\pi_{2}](a,b)))\\\\ &= \\cos(p\\circ (a, s(b)))\\\\ &= \\cos(p\\circ (a, b^2)))\\\\ &= \\cos(ab^2). \\end{align*} But this brings me to my first (among several) points of confusion. In the equation: $$Df(a,b) = D(\\sin)(p\\circ (\\pi_{1}, s\\circ \\pi_{2})(a,b))\\circ Dp((\\pi_{1}, s\\circ \\pi_{2})(a,b))\\circ D(\\pi_{1}, s\\circ \\pi_{2})(a,b)$$ it appears $D(\\sin)(p\\circ (\\pi_{1}, s\\circ \\pi_{2})(a,b))$ should be a function, not a number. Can someone point out what my error in thinking is? (answered below) Continuing on to compute the 3rd block, \\begin{align*} D(\\pi_{1}, s\\circ \\pi_{2})(a,b) &= (D\\pi_{1}(a,b), D(s\\circ \\pi_{2})(a,b))\\\\ &= (\\pi_{1}(a,b), Ds(\\pi_{2}(a,b))\\circ D\\pi_{2}(a,b))\\\\ &= (a, Ds(b)\\circ \\pi_{2}(a,b))\\\\ &= (a, 2b\\circ b)\\\\ &= (a, 2b^2) \\end{align*} Now the middle one: \\begin{align*} Dp((\\pi_{1}, s\\circ\\pi_{2})(a,b)) &= Dp((\\pi_{1}(a,b), (s\\circ \\pi_{2})(a,b))\\\\ &= Dp(a, b^2) \\end{align*} Now substituting these smaller calculations, the whole thing simplifies down to: \\begin{align*} Df(a,b) &= D(\\sin)(p\\circ (\\pi_{1}, s\\circ \\pi_{2})(a,b))\\circ Dp((\\pi_{1}, s\\circ \\pi_{2})(a,b))\\circ D(\\pi_{1}, s\\circ \\pi_{2})(a,b)\\\\ &= \\cos(ab^2)\\circ \\underbrace{Dp(a, b^2)\\circ (a, 2b^2)}_{= a\\cdot 2b^2 + b^2\\cdot a}\\\\ &= \\cos(ab^2)(3ab^2) \\end{align*} Now I will insist that I have something wrong. $Df(a,b)$ should", "\\tan^{-1} \\dfrac{-8\\sqrt{3}}{8}\\\\&= \\tan^{-1} -\\sqrt{3}\\\\&= 300^{\\circ} \\end{aligned}16(\\cos 300^{\\circ} + i\\sin 300^{\\circ})$Let’s begin by finding$z_1 \\cdot z_2$by multiplying$2\\sqrt{2}$and$8$then finding the cosine and sine of$225^{\\circ} + 30^{\\circ}$.$\\begin{aligned}z_1 \\cdot z_2 &= [2\\sqrt{2}(\\cos 225^{\\circ} + \\sin 225^{\\circ})]\\cdot[8(\\cos 30^{\\circ} + \\sin 30^{\\circ})]\\\\&=(2\\sqrt{2} \\cdot 8) [\\cos (225^{\\circ} + 30^{\\circ}) + i\\sin (225^{\\circ} + 30^{\\circ})] \\\\&= 16\\sqrt{2}(\\cos 255^{\\circ} + i\\sin 255^{\\circ})\\end{aligned}$We can express$(z_3)^2$as$z_3 \\cdot z_3$then find the product of$16(\\cos 300^{\\circ} + \\sin 300^{\\circ})$and itself.$\\begin{aligned}z_3 \\cdot z_3 &= [16(\\cos 300^{\\circ} + i\\sin 300^{\\circ})][16(\\cos 300^{\\circ} + i\\sin 300^{\\circ})]\\\\&= (16 \\cdot 16)[\\cos(300^{\\circ} + 300^{\\circ}) + i\\sin (300^{\\circ} + 300^{\\circ})]\\\\&= 256(\\cos 600^{\\circ} + i\\sin 600^{\\circ})\\end{aligned}$We can also rewrite$\\cos 600^{\\circ}$and$\\sin 600^{\\circ}$as$\\cos (600^{\\circ} – 360^{\\circ}) = \\cos 240^{\\circ}$and$\\sin (600^{\\circ} – 360^{\\circ}) = \\sin 240^{\\circ}$. Hence, we have$256 (\\cos 240^{\\circ}+ \\sin 240^{\\circ})$. We can use our results from part a and b to find$ \\dfrac{z_1 \\cdot z_ 2}{(z_3)^2}$. Once we replace the numerator and denominator with$16\\sqrt{2}(\\cos 255^{\\circ} + i\\sin 255^{\\circ})$and$256 (\\cos 240^{\\circ}+ \\sin 240^{\\circ})$, respectively, we can then simplify the expression as shown below.$\\begin{aligned}\\dfrac{z_1 \\cdot z_ 2}{(z_3)^2} &= \\dfrac{16\\sqrt{2}(\\cos 255^{\\circ} + i\\sin 255^{\\circ})}{256 (\\cos 240^{\\circ}+ \\sin 240^{\\circ})}\\\\&= \\dfrac{16\\sqrt{2}}{256}[\\cos(255^{\\circ} – 240^{\\circ}) + i\\sin(255^{\\circ} – 240^{\\circ})]\\\\&= \\dfrac{\\sqrt{2}}{16} (\\cos 15^{\\circ} + i\\sin 15^{\\circ}) \\end{aligned}$Hence, we have the values of$z_1 \\cdot z_2$,$(z_3)^2$, and$ \\dfrac{z_1 \\cdot z_ 2}{(z_3)^2}$in polar form as shown below: a.$z_1 \\cdot z_2 = 16\\sqrt{2}(\\cos 255^{\\circ} + i\\sin 255^{\\circ})$b.$(z_3)^2 = 256 (\\cos 240^{\\circ}+ \\sin 240^{\\circ})$c.$", "## Elementary Technical Mathematics The value of $\\cos \\left( 256{}^\\circ \\right)$ is $-0.2419$. Step 1: Open Ti-83. Step 2: Press $cos$ button. Step 3: Enter the value like this, $\\cos \\left( 256{}^\\circ \\right)=-0.2419$ The obtained value is $-0.2419$. Hence the value of $\\cos \\left( 256{}^\\circ \\right)$ is $-0.2419$.", "1996 USAMO Problems/Problem 1 Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. Solution Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos 5^\\circ)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$, as desired. $\\Box$ 1996 USAMO (Problems • Resources)", "# 1996 USAMO Problems/Problem 1 ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos5)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$, as desired. $\\Box$ ###", "is equal to half the difference of the measures of the intercepted arcs. Example 4: Find the measure of \\begin{align*}x\\end{align*}. a) b) c) Solution: For all of the above problems we can use Theorem 9-13. a) \\begin{align*}x=\\frac{125^\\circ-27^\\circ}{2}=\\frac{98^\\circ}{2}=49^\\circ\\end{align*} b) \\begin{align*}40^\\circ\\end{align*} is not the intercepted arc. Be careful! The intercepted arc is \\begin{align*}120^\\circ\\end{align*}, \\begin{align*}(360^\\circ-200^\\circ-40^\\circ)\\end{align*}. Therefore, \\begin{align*}x=\\frac{200^\\circ-120^\\circ}{2}=\\frac{80^\\circ}{2}=40^\\circ\\end{align*}. c) First, we need to find the other intercepted arc, \\begin{align*}360^\\circ-265^\\circ=95^\\circ\\end{align*}. \\begin{align*}x=\\frac{265^\\circ-95^\\circ}{2}=\\frac{170^\\circ}{2}=85^\\circ\\end{align*} Example 5: Algebra Connection Find the value of \\begin{align*}x\\end{align*}. You may assume lines that look tangent, are. Solution: Set up an equation using Theorem 9-13. \\begin{align*}\\frac{(5x+10)^\\circ-(3x+4)^\\circ}{2} &= 30^\\circ\\\\ (5x+10)^\\circ-(3x+4)^\\circ &= 60^\\circ\\\\ 5x+10^\\circ-3x-4^\\circ &= 60^\\circ\\\\ 2x+6^\\circ &= 60^\\circ\\\\ 2x &= 54^\\circ\\\\ x &= 27^\\circ\\end{align*} Know What? Revisited If \\begin{align*}178^\\circ\\end{align*} of the Earth is exposed to the sun, then the angle at which the sun’s rays hit the Earth is \\begin{align*}2^\\circ\\end{align*}. From Theorem 9-13, these two angles are supplementary. From this, we also know that the other \\begin{align*}182^\\circ\\end{align*} of the Earth is not exposed to sunlight and it is probably night time. ## Review Questions 1. Draw two secants that intersect: 1. inside a circle. 2. on a circle. 3. outside a circle. 2. Can two tangent lines intersect inside a circle? Why or why not? 3. Draw a tangent and a secant that intersect: 1. on a circle. 2. outside a circle. Fill", "# Difference between revisions of \"1996 USAMO Problems/Problem 1\" ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos 5^\\circ)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot", "# Difference between revisions of \"1996 USAMO Problems/Problem 1\" ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos5)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$,", "of length $\\,3,\\, 4,\\, 6\\,$. • Note that $\\,6 < 3 + 4\\,$. By SSS, the triangle is unique. • Use the Law of Cosines with the longest side: $$\\begin{gather} \\cssId{sb36}{6^2 = 3^2 + 4^2 - 2\\cdot 3\\cdot 4\\cdot\\cos F}\\cr\\cr \\cssId{sb37}{\\cos F \\approx -0.4583} \\cr\\cr \\cssId{sb38}{F = \\arccos(\\cos F) \\approx 117.3^\\circ} \\end{gather}$$ • Use the Law of Sines with the shortest side: $$\\begin{gather} \\cssId{sb40}{\\frac{\\sin 117.3^\\circ}{6} = \\frac{\\sin D}{3}}\\cr\\cr \\cssId{sb41}{\\sin D = \\frac{3\\sin 117.3^\\circ}{6} \\approx 0.4443}\\cr\\cr \\cssId{sb42}{D = \\arcsin(0.4443) \\approx 26.38^\\circ} \\end{gather}$$ • $\\,E = 180^\\circ - 117.3^\\circ - 26.38^\\circ = 36.32^\\circ$ • Summary: ANGLES OPPOSITE SIDES $D = 26.38^\\circ$ $d = 3$ $E = 36.32^\\circ$ $e = 4$ $F = 117.3^\\circ$ $f = 6$ This is the correct triangle! Note: At left, the corrections would be: $$\\begin{gather} \\cssId{sb56}{F = 180^\\circ - \\arcsin(0.8886) \\approx 117.3^\\circ}\\cr\\cr \\cssId{sb57}{E = 180^\\circ - 26.38^\\circ - 117.3^\\circ = 36.32^\\circ} \\end{gather}$$ ## Congruence Theorems CONGRUENCE THEOREMS A unique triangle is determined by each of these configurations: SAS side-angle-side (two sides and an included angle) SSS side-side-side (three sides) AAS/SAA angle-angle-side (or, side-angle-angle) (two angles and a non-included side) ASA angle-side-angle (two angles and an included side) If $$\\begin{gather} \\cssId{sb65}{e > 0}\\cr \\cssId{sb66}{f > 0}\\cr \\cssId{sb67}{0^\\circ < D < 180^\\circ} \\end{gather}$$ then there exists a unique triangle having the two sides and included angle. Name the", "-\\theta \\right)=-\\sin \\theta$ if n is even. \\begin{align} & \\Rightarrow sin{{300}^{\\circ }}=\\sin \\left( 2\\pi -{{60}^{\\circ }} \\right) \\\\ & \\Rightarrow \\sin {{300}^{\\circ }}=-\\sin {{60}^{\\circ }} \\\\ \\end{align} We know that $\\sin {{60}^{\\circ }}=\\dfrac{\\sqrt{3}}{2}$. $\\Rightarrow \\sin {{300}^{\\circ }}=-\\dfrac{\\sqrt{3}}{2}......(5)$ Now let us substitute equation (2), equation (3), equation (4) and equation (5) in equation (1), then we get \\begin{align} & I=\\sin {{330}^{\\circ }}\\cos {{120}^{\\circ }}+\\cos {{210}^{\\circ }}\\sin {{300}^{\\circ }} \\\\ & \\Rightarrow I=\\left( \\dfrac{-1}{2} \\right)\\left( \\dfrac{-1}{2} \\right)+\\left( \\dfrac{-\\sqrt{3}}{2} \\right)\\left( \\dfrac{-\\sqrt{3}}{2} \\right) \\\\ & \\Rightarrow I=\\dfrac{1}{4}+\\dfrac{3}{4} \\\\ & \\Rightarrow I=1.....(6) \\\\ \\end{align} From equation (6), it is clear that the value of $\\sin {{330}^{\\circ }}\\cos {{120}^{\\circ }}+\\cos {{210}^{\\circ }}\\sin {{300}^{\\circ }}$ is equal to 1. Note: Students may have a misconception that $\\sin \\left( n\\pi -\\theta \\right)=\\sin \\theta$ if n is even. If this misconception is followed, then the final answer may get interrupted. In the same way, students may have a misconception that $\\cos \\left( n\\pi +\\theta \\right)=\\cos \\theta$ if n is odd. If even this misconception is followed, then also the final answer will get interrupted. So, these misconceptions should get avoided." ]

[ "+0 # Let $\\mathbf{v}$ and $\\mathbf{w}$ be vectors such that \\[\\operatorname{proj}_{\\bold{w}}( \\bold{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.\\ 0 113 1 Let v and w be vectors such that $$\\operatorname{proj}_{{w}}({v})= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.$$ Find $$\\operatorname{proj}_{-2{w}} (3{v})$$. Oct 9, 2019", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "+0 # Help with vector geometry +1 88 1 +38 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w} \\text{ and }\\mathbf{x}$$ are linearly independent, answer with 0. If they aren't, find coefficients a,b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a+b}{c}$$ Jun 5, 2019", "+0 # help vectors 0 46 2 Let v and w be vectors such that $$\\text{proj}_{{w}} {v} = \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ Find $$\\text{proj}_{-2 {w}} (3 {v})$$. Feb 23, 2020 #1 0 Projection is linear, so the answer is 2*3*(4,-7) = (24,-42). Feb 24, 2020 #2 +24366 +1 Let $$\\mathbf{v} \\text{ and } \\mathbf{w} \\text{ be vectors such that } \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$. Find $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}})$$. $$\\begin{array}{|rcll|} \\hline \\vec{p} &=& \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} ) \\\\\\\\ &=& \\dbinom{4}{-7} \\\\\\\\ &=& \\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\ \\hline \\vec{P} &=& \\operatorname{proj}_{\\mathbf{-2w}}( \\mathbf{3v} ) \\\\\\\\ &=& \\dfrac{(-2\\vec{w})(3\\vec{v})}{|-2w|^2}(-2\\vec{w}) \\\\\\\\ &=& 6\\dfrac{\\vec{w}\\vec{v}}{2w^2}\\vec{w} \\\\\\\\ &=& 3\\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\\\\\ &=& 3\\vec{p} \\\\\\\\ &=& 3\\dbinom{4}{-7} \\\\\\\\ &=& \\dbinom{12}{-21} \\\\ \\hline \\end{array}$$ $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}}) = \\dbinom{12}{-21}.$$ Feb 24, 2020", "+0 # How do i find (a-b)/c? 0 56 1 +39 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w}$$, and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$and answer with $$\\dfrac{a-b}{c}$$. I have tried this multiple times and I got a=0, b=0, and c=0 but it was incorrect. I tried distributing the vectors and solving for a, b, and c with the 3 equations. Can someone please tell me if I am doing anything wrong? Thanks! Apr 30, 2020 #1 +24983 +3 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}$$, $$\\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}$$. If the vectors $$\\mathbf{v}$$, $$\\mathbf{w}$$ and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a-b}{c}$$. $$\\begin{array}{|rcll|} \\hline \\begin{array}{|rcll|} \\hline a\\begin{pmatrix}", "\\\\ -1 \\end{pmatrix}$$ Then, the plane satisfies $$\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\begin{pmatrix} -1 \\\\ -1 \\\\ -1 \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix} \\begin{pmatrix} -1 \\\\ -1 \\\\ -1 \\end{pmatrix} = -2,$$ or that $- x - y - z = - 2$. - Take one of the points as origin and use the other two to obtain two (linearly independent? check!) vectors. Then write the vectorial equation for the plane as $$(x,y,z)=P+\\lambda \\vec u + \\mu \\vec v.$$ -", "# Find a set of vector B Find a set of vector B such that $$\\operatorname{span}(B)=\\{(x,y,z,w)^T\\mid 2x+y+w=0, y+2z=0\\}$$ in $\\mathbb{R}^4$. This is what I have so far I don't know if it is correct. $$x\\begin{pmatrix} a_1 \\\\ b_1 \\\\ c_1 \\\\ d_1 \\end{pmatrix}+ y\\begin{pmatrix} a_2 \\\\ b_2 \\\\ c_2 \\\\ d_2 \\end{pmatrix}+ z\\begin{pmatrix} a_3 \\\\ b_3 \\\\ c_3 \\\\ d_3 \\end{pmatrix}+w \\begin{pmatrix} a_4 \\\\ b_4 \\\\ c_4 \\\\ d_4 \\end{pmatrix}$$ where $2x+y+w=0$ and $y+2z=0$. Then $w=-2x-y$, $y=2z$, $w=-2x+2z$. Then $$x\\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3 \\\\ b_4 \\end{pmatrix}+ y\\begin{pmatrix} α_1 \\\\ α_2 \\\\ α_3 \\\\ α_4 \\end{pmatrix}+ z\\begin{pmatrix} -2c_1 \\\\ -2c_2 \\\\ -2c_3 \\\\ -2c_4 \\end{pmatrix} +w\\begin{pmatrix} -2b-α_1 \\\\ -2b-α_2 \\\\ -2b-α_3 \\\\ -2b-α_4 \\end{pmatrix}$$ Thus $$2b+α_1+-2b_1-α_1=0$$ and $$α_1-2c=0$$ where $$c_1=1/2α_1$$ Not sure where you deduced $w=-2x-y$ and $y+2z=0$. What you need to find is the solution set of $$\\begin{cases} 2x+y+w=0 \\\\ y+2z=0 \\end{cases}$$ so $w=-2x-y$ and $y=-2z$; hence $w=-2x+2z$, so the vectors are of the form $$\\begin{pmatrix} a \\\\ -2b \\\\ b \\\\ -2a-2b \\end{pmatrix}$$ where $a$ and $b$ are arbitrary. Since $$\\begin{pmatrix} a \\\\ -2b \\\\ b \\\\ -2a-2b \\end{pmatrix} = a\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\\\ -2 \\end{pmatrix} + b\\begin{pmatrix} 0 \\\\ -2 \\\\ 1 \\\\ -2 \\end{pmatrix}$$ you are done. • Thanks, it helped me a lot. – marie Feb 15 '18 at", "$\\mathbf{r} = \\begin{pmatrix}1 \\\\ 0 \\\\ 3\\end{pmatrix}+s\\begin{pmatrix}-2 \\\\ -4 \\\\ 4\\end{pmatrix}+t\\begin{pmatrix}2 \\\\ 3 \\\\ -6\\end{pmatrix}.$ We can simplify this to $\\mathbf{r} = \\begin{pmatrix}1 \\\\ 0 \\\\ 3\\end{pmatrix}+s\\begin{pmatrix}1 \\\\ 2 \\\\ -2\\end{pmatrix}+t\\begin{pmatrix}2 \\\\ 3 \\\\ -6\\end{pmatrix},$ or $\\mathbf{r} = \\begin{pmatrix}1 +s+2t\\\\ 2s+3t \\\\ 3-2s-6t\\end{pmatrix}.$ Is the vector form of the equation of the plane unique?", "vector from $(3,3)$ to $(2,5)$ in $\\mathbb{R}^2$ 3. the vector from $(1,0,6)$ to $(5,0,3)$ in $\\mathbb{R}^3$ 4. the vector from $(6,8,8)$ to $(6,8,8)$ in $\\mathbb{R}^3$ This exercise is recommended for all readers. Problem 2 Decide if the two vectors are equal. 1. the vector from $(5,3)$ to $(6,2)$ and the vector from $(1,-2)$ to $(1,1)$ 2. the vector from $(2,1,1)$ to $(3,0,4)$ and the vector from $(5,1,4)$ to $(6,0,7)$ This exercise is recommended for all readers. Problem 3 Does $(1,0,2,1)$ lie on the line through $(-2,1,1,0)$ and $(5,10,-1,4)$? This exercise is recommended for all readers. Problem 4 1. Describe the plane through $(1,1,5,-1)$, $(2,2,2,0)$, and $(3,1,0,4)$. 2. Is the origin in that plane? Problem 5 Describe the plane that contains this point and line. $\\begin{pmatrix} 2 \\\\ 0 \\\\ 3 \\end{pmatrix} \\qquad \\{\\begin{pmatrix} -1 \\\\ 0 \\\\ -4 \\end{pmatrix} +\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}t \\,\\big|\\, t\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 6 Intersect these planes. $\\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}t+ \\begin{pmatrix} 0 \\\\ 1 \\\\ 3 \\end{pmatrix}s \\,\\big|\\, t,s\\in\\mathbb{R}\\} \\qquad \\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\\\ 0 \\end{pmatrix}k+ \\begin{pmatrix} 2 \\\\ 0 \\\\ 4 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 7 Intersect each pair, if possible. 1. $\\{\\begin{pmatrix} 1 \\\\", "Review question # Can we find the equation of the plane containing $l_1$ and $PQ$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R5089 ## Solution The lines $l_1$ and $l_2$ have equations $\\mathbf{r} = \\mathbf{a} + s\\mathbf{b}$ and $\\mathbf{r} = \\mathbf{c} + t\\mathbf{d}$, where $s$ and $t$ are real numbers and $\\mathbf{a} = 5\\mathbf{i}+ 6\\mathbf{j}- \\mathbf{k}, \\quad\\mathbf{b} = \\mathbf{i} +\\mathbf{j}+ \\mathbf{k}, \\quad\\mathbf{c} = \\mathbf{i}+ 13\\mathbf{j} +7\\mathbf{k}, \\quad\\mathbf{d} = \\mathbf{i} + 6\\mathbf{j} + 3\\mathbf{k}.$ Also, the point $P$ on $l_1$ and the point $Q$ on $l_2$ are such that $PQ$ is perpendicular to both $l_1$ and $l_2.$ In any order: 1. show that $PQ = \\sqrt{38},$ 2. find the position vectors of $P$ and $Q,$ We can write the position vector of $P$ as $\\mathbf{p} = \\begin{pmatrix}5+s \\\\ 6+s \\\\ -1+s\\end{pmatrix}$, and the position vector of $Q$ as $\\mathbf{q} = \\begin{pmatrix}1+t \\\\ 13+6t\\\\ 7+3t\\end{pmatrix}.$ Thus $\\mathbf{q} - \\mathbf{p} = \\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}$. This vector has to be perpendicular to both $l_1$ and $l_2$, and so using the scalar product, $\\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}.\\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=0 \\quad\\text{and}\\quad \\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}.\\begin{pmatrix}1\\\\ 6 \\\\ 3\\end{pmatrix}=0.$ This gives us the two equations $10t-3s+11=0$ and $46t-10s+62=0$, which solve to give $s = -3$, $t", "\\mathbb{F}$ such that $a = \\lambda \\cdot b$. Now consider your example. Let $a = \\begin{pmatrix} 1 - \\mathrm{i} \\\\ 2 - 2\\mathrm{i}\\end{pmatrix} \\in V$ and $b = \\begin{pmatrix} \\mathrm{i} - 1 \\\\ 2\\mathrm{i} - 2 \\end{pmatrix} \\in V$. The vectors would be linearly dependent, if $a = -1 \\cdot b$ and $-1 \\in \\mathbb{F}$. This is true for the field $\\mathbb{F} := \\mathbb{C}$, but also for $\\mathbb{F} := \\mathbb{R}$ or any other field of which $-1$ is an element.", "should keep those as part of our basis for $\\mathrm{span}\\left(v_1, v_2\\right)$. # SubsubsectionSpaces Associated to a Matrix SageMath knows about the column space and row space associated to a matrix. For this next example, we will work over the ring AA of “algebraic numbers”, so we can include $\\sqrt{2}$. Or we could do this in other ways. # SubsectionExercises Note: We may not present all of these in class. Make an example of a vector $v \\in \\mathbb{R}^4$ so that the set \\begin{equation*} \\left\\{ \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\\\ 4 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\\\ 1 \\end{pmatrix}, \\begin{pmatrix} -2\\\\ 1 \\\\ 0 \\\\ -5 \\end{pmatrix}, v \\right\\} \\end{equation*} is a linearly independent set, or explain why it is impossible to find such an example. Is your resulting set a basis? Make an example of a vector $w \\in \\mathbb{R}^3$ so that the set \\begin{equation*} \\left\\{ \\begin{pmatrix} 2 \\\\ 3 \\\\ 4 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix}, \\begin{pmatrix} -2\\\\ -4 \\\\ -5 \\end{pmatrix}, w \\right\\} \\end{equation*} is a spanning set, or explain why it is impossible to find such an example. Is your resulting set a basis? (Strang 3.5.10) Find two independent vectors on the plane $x+2y-3z-t=0$ in $\\mathbb{R}^4$. Then find three independent vectors on this plane. Why not four? Find", "+0 # PLZ help asap!!! (repost) 0 48 1 Let v and w be vectors such that $${proj}_{{w}}({v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$. Find $$\\operatorname{proj}_{-2 {w}} (3 {v})$$. Oct 11, 2019 #1 +439 +5 The equation, $$projx⃗ y⃗ =x⃗ ⋅y⃗ |x⃗ |x⃗ |x⃗ |projx→y→=x→⋅y→|x→|x→|x→|$$ Holds for any vectors we can let those be $$πa⃗ πa→ and eb⃗ eb→$$. Where ππ and ee are some random constants like −2−2 and 33. $$projπa⃗ eb⃗ =πa⃗ ⋅eb⃗ |πa⃗ |πa⃗ |πa⃗ |projπa→eb→=πa→⋅eb→|πa→|πa→|πa→|$$ Now use$$a⃗ ⋅(cb⃗ )=ca⃗ ⋅b⃗ a→⋅(cb→)=ca→⋅b→ and |cw⃗ |=c|w⃗ ||cw→|=c|w→|.$$ To get e(pro j base a to the power of b) is the answer because $$π2eπ2a⃗ ⋅b⃗ |a⃗ |a⃗ |a⃗ |=π2eπ2a→⋅b→|a→|a→|a→| =$$ . Oct 11, 2019 edited by SVS2652 Oct 11, 2019", "# Thread: Spans and vectors 1. ## Spans and vectors If a question asks, Is the vector $\\mathbf{b}=\\begin{pmatrix}-2\\\\ -6\\\\ -4\\end{pmatrix}\\in \\mbox{span}(\\mathbf{v_1, v_2, v_3, v_4})$ where: $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 3\\\\ 0\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}2\\\\ 2\\\\ 1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ -1\\end{pmatrix}$, $\\mathbf{v_4}=\\begin{pmatrix}1\\\\ -2\\\\ 1\\end{pmatrix}$ Do we just make it a matrix and solve the matrix? And how would we solve it if we have 4 variables but only 3 equations? 2) Is the set of vectors $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 2\\\\ 3\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}1\\\\ 1\\\\ -1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ 5\\end{pmatrix}$ a spanning set of $\\mathbb{R}^3$? Do we just make this a matrix with say $\\mathbf{x}=\\begin{pmatrix}x_1\\\\ x_2\\\\ x_3\\end{pmatrix}$ as our solution and see if we can solve the matrix? 2. So your first question is asking whether there exist scalars $a_{1},a_{2},a_{3},a_{4}$ such that $\\mathbf{b}=a_{1}\\mathbf{v}_{1}+a_{2}\\mathbf{v}_{2} +a_{3}\\mathbf{v}_{3}+a_{4}\\mathbf{v}_{4}.$ This equation you can translate into a normal matrix problem of the form $A\\mathbf{x}=\\mathbf{b}$. Row reduce that problem. If you find any solutions (one or infinitely many), then the answer is yes. If you find that there are no solutions, then the answer is no. Your second question, again, is asking whether there exist scalars $c_{1}, c_{2}, c_{3}$ such that $c_{1}\\mathbf{v}_{1}+c_{2}\\mathbf{v}_{2}+c_{3}\\math bf{v}_{3}=\\mathbf{x}$ for your arbitrary $\\mathbf{x}.$ This, again, can be translated into a standard matrix problem. Since this problem will result in a square matrix, you can use the theory of determinants to help you out.", "# Decide a number $a, b, c$ (expressed by $\\lambda$) so that $a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$ See the following vectors in $$\\mathbb{R}^3$$: $$\\mathbf{u}_1= \\begin{pmatrix} 1 \\\\ -1\\\\ 1 \\\\ 1 \\end{pmatrix}$$, $$\\mathbf{u}_2 = \\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\\\ 2 \\end{pmatrix}$$, $$\\mathbf{u}_3 = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\end{pmatrix}$$ and $$\\mathbf{v}_1= \\begin{pmatrix} 5 \\\\ -3\\\\ -4 \\\\ -1 \\end{pmatrix}$$, $$\\mathbf{v}_2 = \\begin{pmatrix} 6 \\\\ 4 \\\\ 1 \\\\ 8 \\end{pmatrix}$$, $$\\mathbf{v}_3 = \\begin{pmatrix} 7 \\\\ 2 \\\\ -1 \\\\ 6 \\end{pmatrix}$$ We are told that $$\\mathcal{B} = (\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3)$$ and $$\\mathcal{C} = (\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3)$$ both are bases for the same subspace $$\\mathcal{U}$$. Let $$\\mathcal{\\lambda}$$ be an unknown number og see the vector $$\\mathbf{x} = \\mathbf{v}_1 + \\lambda\\mathbf{v}_2 - \\mathbf{v}_3$$. Decide a number $$a, b, c$$ (expressed by $$\\lambda$$) so that $$a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$$ 1. Find some $$\\alpha_i$$ such that $$v_1=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3$$ 2. Find some $$\\beta_i$$ such that $$v_2=\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3$$ 3. Find some $$\\gamma_i$$ such that $$v_3=\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3$$ 4. Simplify the expression $$x=v_1+\\lambda v_2-v_3=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3+\\lambda(\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3)-(\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3)$$ • Thanks 5xum! When I reduce it, I get $\\begin{pmatrix} 8 \\\\ 3\\\\ -9 \\\\ 1 \\end{pmatrix} + a\\begin{pmatrix} 2 \\\\ -2\\\\ 1 \\\\ 1 \\end{pmatrix} + b\\begin{pmatrix} 3 \\\\ 6\\\\ 0 \\\\ 6 \\end{pmatrix} + c\\begin{pmatrix} 1 \\\\", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "# How to complete a set of independent vectors to a basis of a subspace? $\\newcommand{\\Span}{\\operatorname{Span}}$Let $V$ be a vector such that $\\dim V = n$, and let $v_1,\\ldots,v_k \\in V$ be independent vectors such that $1<k\\leq n$. Now Let $w_1,\\ldots,w_r\\in$ $\\Span\\left\\{ v_1,\\ldots,v_k \\right\\}$ be independent vectors such that $1\\leq r <k$. My question is this: How do I find the missing vectors $w_{r+1},\\ldots,w_k\\in \\Span\\left\\{ v_1,\\ldots,v_k \\right\\}$ so that $\\Span\\left\\{w_1,\\ldots,w_r,w_{r+1},\\ldots,w_k\\right\\} = \\Span\\left\\{ v_1,\\ldots,v_k \\right\\}$ ? Now I know they exist, I just don't know how I actually find them. For example: Lets look at $\\mathbb{R}^5$ and $$U=\\Span\\left\\{ \\begin{pmatrix} 5\\\\2\\\\3\\\\7\\\\3 \\end{pmatrix},\\begin{pmatrix} 2\\\\4\\\\4\\\\8\\\\1 \\end{pmatrix} ,\\begin{pmatrix} 3\\\\4\\\\7\\\\6\\\\1 \\end{pmatrix},\\begin{pmatrix} 5\\\\8\\\\6\\\\4\\\\8 \\end{pmatrix} \\right\\}$$ Those are all independent vectors. Now lets take $2$ independent vectors that are linear combination of those. lets say, $$w_1 = \\begin{pmatrix} 6\\\\2\\\\11\\\\15\\\\-3 \\end{pmatrix} \\; w_2=\\begin{pmatrix} 0\\\\4\\\\-2\\\\12\\\\1 \\end{pmatrix}$$ How can we complete those two vectors to form a basis of $U$ ? I would really like to understand the general idea of this. This is actually a general question of something I need it for, which is the process of finding a Jordan basis for matrices\\transformations. Thanks for any help Here's one method for the case of $V = \\Bbb R^n$: row-reduce the matrix $$\\pmatrix{w_1 & \\cdots & w_r & v_1 & \\cdots & v_k}$$ The columns of this matrix", "to the basis $$\\{ {\\mathbf v}_1, {\\mathbf v}_2 \\}\\text{.}$$ The vectors $$(1, 1)$$ and $$(-1, -1)$$ do not form a basis for $${\\mathbb R}^2$$ since these two vectors lie along the same line. If $$2 \\times 2$$ matrices and the rest of what we have described above make you nervous, you should work through the exercises at the end of this section. Activity3.1.1.Matrix Operations. Given the matrices and vectors \\begin{equation*} A = \\begin{pmatrix} 5 & 3 \\\\ -6 & 4 \\end{pmatrix}, B = \\begin{pmatrix} 2 & 3 \\\\ 1 & 2 \\end{pmatrix}, \\mathbf x = \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} \\mathbf y = \\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix} \\end{equation*} compute each of the following expressions. (a) $$AB\\text{,}$$ $$BA$$ (b) $$A^{-1}\\text{,}$$ $$B^{-1}\\text{,}$$ $$(AB)^{-1}\\text{,}$$ $$B^{-1}A^{-1}$$ (c) $$\\det(A), \\det(B), \\det(AB), \\det(A^{-1})$$ (d) $$A \\mathbf x\\text{,}$$ $$A \\mathbf y\\text{,}$$ $$\\mathbf y^T \\mathbf x\\text{,}$$ $$\\mathbf x \\mathbf y^T\\text{,}$$ where $$\\mathbf y^T = (2, -1)$$ Subsection3.1.3Finding Eigenvalues and Eigenvectors A nonzero vector $${\\mathbf v}$$ is an eigenvector of $$A$$ if $$A {\\mathbf v} = \\lambda {\\mathbf v}$$ for some $$\\lambda \\in {\\mathbf R}\\text{.}$$ The constant $$\\lambda$$ is called an eigenvalue of $$A\\text{.}$$ Letting \\begin{equation*} A = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\quad \\text{and} \\quad \\mathbf v = \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\neq \\mathbf 0, \\end{equation*} we have $$A \\mathbf", "+0 # Let $\\mathbf{p} = \\begin{pmatrix} 2\\\\ -1\\\\ 2 \\end{pmatrix}$ and $\\mathbf{d} =\\begin{pmatrix} 1\\\\ -1\\\\ 4 \\end{pmatrix}$ and consider the par 0 71 1 a) Let $\\mathbf{p} = \\begin{pmatrix} 2\\\\ -1\\\\ 2 \\end{pmatrix}$ and $\\mathbf{d} =\\begin{pmatrix} 1\\\\ -1\\\\ 4 \\end{pmatrix}$ and consider the particle that visits the point $(x,y,z)$ at time $t$ if $\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\mathbf{p} + t\\mathbf{d}.$The vectors $\\mathbf{p}$ and $\\mathbf{d}$ and the graph of the parametrization are shown below: If the intersection between this line and the $xy$-plane is $(a,b,c),$ enter $a, b, c$ in this order below. b) Let $\\mathbf{p} = \\begin{pmatrix}-1\\\\ -2 \\\\ 1\\end{pmatrix}$ and $\\mathbf{d} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$ and consider the particle that visits $(x,y,z)$ at time $t$ if $\\begin{pmatrix}x \\\\ y \\\\ z \\end{pmatrix} = \\mathbf{p} + t\\mathbf{d}.$The vectors $\\mathbf{p}$ and $\\mathbf{d}$ and the graph of the parametrization are shown below: Then if $R = (-1, 3, 5)$, calculate the point $(a,b,c)$ on the line that is closest to $R$. Enter $a, b, c$ in that order below. help asap Feb 10, 2021 edited by Guest Feb 10, 2021", "# Describing a span of vectors in $\\mathbb{R^4}$. I am given that $\\mathbf{v}_1 = \\begin{pmatrix}1\\\\ 3 \\\\ 1 \\\\ 1\\\\ \\end{pmatrix}$ and $\\mathbf{v}_2 = \\begin{pmatrix}1\\\\ -1 \\\\ 6 \\\\ 2\\\\ \\end{pmatrix}$ in $\\mathbb{R^4}$. How would I describe $\\mathrm{Span}\\{\\mathbf{v}_1,\\mathbf{v}_2\\}$? I know that $[\\mathbf{v}_1\\; \\mathbf{v}_2]$ row reduces to $\\begin{pmatrix}1 & 0\\\\ 0 & 1\\\\ 0 & 0\\\\ 0 & 0 \\end{pmatrix}$, but is there a \"nontrivial\" way to actually describe $\\mathrm{Span}\\{\\mathbf{v}_1,\\mathbf{v}_2\\}$ beyond saying that it is all of the linear combinations of $\\mathbf{v}_1$ and $\\mathbf{v}_2$? For example, is there a geometric description of $\\mathrm{Span}\\{\\mathbf{v}_1,\\mathbf{v}_2\\}$? I know you can usually describe it as a plane in $\\mathbb{R^3}$, but I am unsure of how to give a meaningful description in $\\mathbb{R^4}$. Edit: (is this right?) $\\mathrm{Span}\\{\\mathbf{v}_1,\\mathbf{v}_2\\}$ is a plane in $\\mathbb{R^4}$ through the origin, because neither vector in this problem is a multiple of the other. • $\\mathbb{R^2}$ is to $\\mathbb{R^4}$ as $\\mathbb{R^1}$ is to $\\mathbb{R^3}$. – JonMark Perry Feb 5 '15 at 5:48 It is still a plane. Two vectors can only form a plane (possibly degenerate). • What do you mean by degenerate? – fancynancy Feb 5 '15 at 5:28 • degenerate means that both vectors are proportional to each other, in which case the \"plane\" they span is only a line. Such line is called a degenerate plane. –" ]

[ "\\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} - \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} = \\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ On the other hand, the reflection of $\\mathbf{x}$ in the plane $$L^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\mathbf{v} \\cdot \\mathbf{y} = 0\\}$$ with normal vector $\\mathbf{v}$ is given by $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = 2\\operatorname{Proj}_{L^\\perp}(\\mathbf{x}) - \\mathbf{x} = 2\\left(\\mathbf{x} - \\operatorname{Proj}_L(\\mathbf{x})\\right) - \\mathbf{x} = \\mathbf{x} - 2\\operatorname{Proj}_L(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v},$$ which in your case yields $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} - 2 \\cdot \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ One last cultural note: the reflection $\\operatorname{Refl}_{L^\\perp}(\\mathbf{x})$ of $\\mathbf{x}$ in the plane $L^\\perp$ with normal vector $\\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image $$H_{\\mathbf{v}}(\\mathbf{x}) := \\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}$$ of $\\mathbf{x}$ under the Householder transformation $H_\\mathbf{v}$ corresponding to $\\mathbf{v}$. So, in your case, as we just saw, $$H_{\\mathbf{v}}(\\mathbf{x}) = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ Suppose you mean a vector ($\\mathbf{v}$) \"specular\" to the given one ($\\mathbf{u}$) with respect to the line $L$. Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin. If you do the cross product $\\mathbf{w} = \\mathbf{u} \\times \\mathbf{L}$ you get a vector which is orthogonal either to the line and to $\\mathbf{u}$, i.e. it is normal to the plane containing the line and $\\mathbf{u}$. Then $\\mathbf{v} \\times \\mathbf{L}$ shall be equal to $-\\mathbf{w}$", "E}$${\\ displaystyle {\\ vec {a}} = {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}}}$ ${\\ displaystyle {\\ vec {OF}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {\\ left [{\\ begin {pmatrix} 8 \\\\ 8 \\ \\ -1 \\ end {pmatrix}} - {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}} \\ right] \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}} {{\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {27} {9}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}}}$. The distance of the point from the plane thus satisfies ${\\ displaystyle d (P; E)}$${\\ displaystyle P}$${\\ displaystyle E}$ ${\\ displaystyle d (P; E) = d (P; F) \\; {\\ widehat {=}} \\; | {\\ vec {PF}} | = \\ left | {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}} -", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "\\\\ -1 \\end{pmatrix}$$ Then, the plane satisfies $$\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\begin{pmatrix} -1 \\\\ -1 \\\\ -1 \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix} \\begin{pmatrix} -1 \\\\ -1 \\\\ -1 \\end{pmatrix} = -2,$$ or that $- x - y - z = - 2$. - Take one of the points as origin and use the other two to obtain two (linearly independent? check!) vectors. Then write the vectorial equation for the plane as $$(x,y,z)=P+\\lambda \\vec u + \\mu \\vec v.$$ -", "+0 # Help 0 64 1 The line $y = \\frac{3x - 5}{4}$ is parameterized in the form $\\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\mathbf{v} + t \\mathbf{d},$so that for $x \\ge 3,$ the distance between $\\begin{pmatrix} x \\\\ y \\end{pmatrix}$ and $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ is $t.$ Find $\\mathbf{d}.$ Feb 9, 2022 Here is the answer: $$\\mathbf{d} = \\begin{pmatrix} 1/5 \\\\ -4/5 \\end{pmatrix}$$", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "+0 # dumb precalculus 0 45 1 For $$\\bold{v} = \\begin{pmatrix} \\phantom -2 \\\\ \\phantom -3 \\\\ -1 \\end{pmatrix} and \\bold{w} = \\begin{pmatrix} \\phantom -2 \\\\ -1 \\\\ \\phantom -0 \\end{pmatrix}$$ compute the projection of $\\mathbf v$ onto $\\mathbf w.$ i literally dont know how to solve this Apr 18, 2022 #1 +9369 +3 We have the formula: $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac{\\langle\\mathbf v, \\mathbf w\\rangle}{\\|\\mathbf w\\|^2}\\mathbf{w}$$ Now, we compute $$\\langle\\mathbf v, \\mathbf w\\rangle$$ and $$\\|\\mathbf w\\|^2$$. $$\\begin{array}{cl} &\\langle\\mathbf v, \\mathbf w\\rangle\\\\ =& 2(2) + 3(-1) + (-1)(0)\\\\ =& 1\\\\ &\\|\\mathbf w\\|^2\\\\ =& 2^2 + (-1)^2 + 0^2\\\\ =&5 \\end{array}$$ Therefore, substituting back into the formula, we have $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac15\\mathbf{w} = \\begin{pmatrix}\\frac25\\\\-\\frac15\\\\0\\end{pmatrix}$$ Apr 18, 2022 #1 +9369 +3 We have the formula: $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac{\\langle\\mathbf v, \\mathbf w\\rangle}{\\|\\mathbf w\\|^2}\\mathbf{w}$$ Now, we compute $$\\langle\\mathbf v, \\mathbf w\\rangle$$ and $$\\|\\mathbf w\\|^2$$. $$\\begin{array}{cl} &\\langle\\mathbf v, \\mathbf w\\rangle\\\\ =& 2(2) + 3(-1) + (-1)(0)\\\\ =& 1\\\\ &\\|\\mathbf w\\|^2\\\\ =& 2^2 + (-1)^2 + 0^2\\\\ =&5 \\end{array}$$ Therefore, substituting back into the formula, we have $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac15\\mathbf{w} = \\begin{pmatrix}\\frac25\\\\-\\frac15\\\\0\\end{pmatrix}$$ MaxWong Apr 18, 2022", "1 & 0 \\\\ 1 & 1 \\end{pmatrix}, N_3 = \\begin{pmatrix} 1 & 1 \\\\ 0 & 1 \\end{pmatrix}, N_4 = \\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix}$ Does $\\{ N_1, N_2, N_3, N_4 \\}$ generate V? $M_1 = -\\frac23 N_1 + \\frac13(N_2 + N_3 + N_4)$ $M_2 = -\\frac23 N_2 + \\frac13(N_1 + N_3 + N_4)$ $M_3 = -\\frac23 N_3 + \\frac13(N_1 + N_2 + N_4)$ $M_4 = -\\frac23 N_4 + \\frac13(N_1 + N_2 + N_3)$ Then $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = aM_1 + bM_2 + cM_3 + dM_4 =$ $a \\cdot \\left( -\\frac23 N_1 + \\frac13(N_2 + N_3 + N_4) \\right) + b \\cdot \\left( -\\frac23 N_2 + \\frac13(N_1 + N_3 + N_4) \\right) + \\ldots$ Theorem: If $S \\in \\mathbf V$, then $\\operatorname{span}(S) =$ {all l.c. of elements of S} is a subspace of V.", "& \\frac{1}{2} \\end{pmatrix} \\cdot \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} 1-\\sqrt{3} \\\\ 1+\\sqrt{3} \\end{pmatrix}$$", "Math Relative position of planes Point and plane (point on the plane) # Point on the plain? Similar to checking whether a specific point lies on a line you can also check whether the point lies on the plane. i ### Method The procedure varies depending on the plane equation: 1. Insert the position vector of the point (P/V) or its coordinates (C). 2. Solve equation (V/C) or system of equations (P) 3. Check if solvable P - Parametric equation V - Vector equation of a plane C - Cartesian equation ! ### Remember The point lies on the plane only if the equation or system of equations can be solved. ### Example (parametric form) $P(2|1|1)$, $\\text{E: } \\vec{x} = \\begin{pmatrix} 3 \\\\ 0 \\\\ 0 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $+ s \\cdot \\begin{pmatrix} 1 \\\\ 5 \\\\ 2 \\end{pmatrix}$ 1. #### Insert $P$ The position vector (vector with the coordinates of the point) of $P$ is used for $\\vec{x}$ in $E$. $\\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}$ $=\\begin{pmatrix} 3 \\\\ 0 \\\\ 0 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $+ s \\cdot \\begin{pmatrix} 1 \\\\ 5 \\\\ 2 \\end{pmatrix}$ 2. #### Set up system of equations Now we set up an equation system and", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "# Shortest Distance between two lines in the 3D plane Find the shortest distance between the line $x$ $=$ $2$ $-$ $t$ $y$ $=$ $-1$ $+$ $2t$ and $z$ $=$ $-1$ $+$ $t$ and the line $x$ $=$ $5$ $+$ $3s$ $y$ $=$ $0$ and $z$ $=$ $2$ $-$ $s$. Find the points on these two lines that give this shortest distance. Using the cross product, I got the normal vector to be: \\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix} The I selected two points on the lines with $t$ $=$ $-1$: \\begin{pmatrix}3\\\\ -3\\\\ -2\\end{pmatrix} and with $s$ $=$ $1$: \\begin{pmatrix}8\\\\ 0\\\\ 1\\end{pmatrix} Therefore the distance between the these two points gave the vector: \\begin{pmatrix}-5\\\\ \\:-3\\\\ \\:-3\\end{pmatrix} So then I projected this vector: $\\left(\\frac{\\left(-5,\\:-3,\\:-3\\right)\\cdot \\left(-5,\\:-3,\\:-3\\right)}{\\left(-2,\\:-2,\\:-6\\right)\\cdot \\left(-2,\\:-2,\\:-6\\right)}\\right)\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ This left me with: $\\frac{43}{\\:44}\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ Which left me with a final distance of: $\\frac{43}{44}\\sqrt{44}$ But this answer isn't correct. I think I have made a mistake in my projection calculation but I am not sure where. Also, how would I find the two points that are closest together. I am also a little confused about that. Any help would be highly appreciated! • $(-1,2,1)\\times(3,0,-1)\\ne(-2,-2,-6)$. – amd Apr 6 '18 at 21:50 • Thank you! I guess I missed the sign. So it should be $(-2, 2, -6)$? But this will", "16:58 • The point $(2,2)$ does not go to $(2,0)$. Since rotation preserves distance, $(2,2)$ gets sent to the point north of $(1,1)$ at a distance of $\\sqrt{2}$. – John Douma Jul 5 at 17:50 First, translate the plane so that $$(1,1)$$ goes to the origin. This gives $$\\begin{pmatrix} x-1 \\\\ y-1 \\end{pmatrix}$$ Then rotate by $$\\frac{\\pi}{4}$$ to get $$\\begin{pmatrix}\\frac{1}{\\sqrt2}\\frac{-1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\begin{pmatrix} x-1 \\\\ y-1 \\end{pmatrix}=\\begin{pmatrix} \\frac{x-y}{\\sqrt{2}} \\\\ \\frac{x+y-2}{\\sqrt{2}}\\end{pmatrix}$$ Finally, do the inverse translation, i.e. add $$1$$ to each coordinate to get $$\\begin{pmatrix} \\frac{x-y+\\sqrt{2}}{\\sqrt{2}} \\\\ \\frac{x+y+\\sqrt{2}-2}{\\sqrt{2}}\\end{pmatrix}$$", "instructive to consider a few special cases. Let us denote the map with $$\\phi: \\mathcal{H} \\to \\mathbb{R}^3$$ where $$\\mathcal{H}$$ is the Hilbert space of a qubit. We calculate that $$\\phi(|0\\rangle) = \\phi\\left(\\begin{pmatrix}1 \\\\ 0\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (1 \\cdot 0 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 0 + 1 \\cdot 0) \\\\ 1^2 + 0^2 - 0^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\1 \\end{pmatrix} \\\\ \\phi(|1\\rangle) = \\phi\\left(\\begin{pmatrix}0 \\\\ 1\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (0 \\cdot 1 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 1 + 0 \\cdot 0) \\\\ 0^2 + 0^2 - 1^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\-1 \\end{pmatrix} \\\\ \\phi(|+\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}} + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\frac{1}{\\sqrt{2}} + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|-\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(-\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} -1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|{+i}\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{i}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot 0 + 0 \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ 2 \\cdot \\left(-0 \\cdot 0 + \\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 +", "Thread: projection of point and plan 1. projection of point and plan Finding a projection from point A (2,-1,3) in the x-3y-z-13 = 0 (3,-4,2) 2. Hello Originally Posted by Apprentice123 Finding a projection from point A (2,-1,3) in the x-3y-z-13 = 0 (3,-4,2) Let $H$ be the orthogonal projection of $A$ on the plane which equation is $x-3y-z=13$ We know two things about $H$ : • As the projection is orthogonal, $\\vec{AH}$ is orthogonal to the plane. • $H(x_h,y_h,z_h)$ lies in the plane : $x_h-3y_h-z_h=13$ Luckily, the equation of the plane ${\\color{red}1}\\cdot x{\\color{red}-3}\\cdot y{\\color{red}-1}\\cdot z=13$ gives us a vector which is orthogonal to the plane : $\\vec{n}=\\begin{pmatrix}1\\\\-3\\\\-1\\end{pmatrix}$. As $\\vec{n}$ and $\\vec{AH}$ are both orthogonal to the plane, they share the same direction that is to say that there exists a real number $t$ such that $\\vec{AH}=t\\vec{n}$. As $\\vec{AH}=\\begin{pmatrix}x_h-2\\\\y_h-(-1)\\\\z_h-3 \\end{pmatrix}$ and $t\\vec{n}=\\begin{pmatrix}t\\\\-3t\\\\-t \\end{pmatrix}$ we have four equations : $ \\begin{cases} t=x_h-2 \\\\ -3t=y_h+1\\\\ -t=z_h-3\\\\ 13=x_h-3y_H-z_h \\end{cases}\\implies \\begin{cases} x_h=t+2 & (1)\\\\ y_h=-3t-1& (2)\\\\ z_h=-t+3&(3)\\\\ 13=x_h-3y_H-z_h&(4) \\end{cases}$ Substituting $(1),\\,(2)$ and $(3)$ in $(4)$ will give you the value of $t$ from which you'll get $H(x_h,y_h,z_h)$. Good luck 3. Thank you . I do AH (x-2,y+1,z-3) = (1,-3,-1). And I find x, y and z. Thank you very much", "form is: $\\boxed{y = - \\frac{1}{4}x - 1}$ Determine the equation of the line... $& \\begin{pmatrix}x_1, & y_1\\\\-3, & 1 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\-3, & -7 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-7-1}{-3--3}\\\\m & = \\frac{-7-1}{-3+3}\\\\m & = \\frac{-8}{0}\\\\m & = \\text{undefined}$ A line that has a slope that is undefined is a line that is parallel to the $y-$axis. The equation of this line is $\\boxed{x = -3}$. $& \\begin{pmatrix}x_1, & y_1\\\\-8, & 4 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\2, & -6 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-6-4}{2--8}\\\\m & = \\frac{-10}{10}\\\\m & = -1\\\\\\\\y & = mx + b\\\\4 & = -1(-8) + b\\\\4 & = 8 + b\\\\4-8 &= 8-8+b\\\\-4 & = b$ $m = -1$ and the $y-$intercept is (0, –4). The equation of this line is: $\\boxed{y = -1x-4}$ $& \\begin{pmatrix}x_1, & y_1\\\\0, & 5 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\4, & -3 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m & = \\frac{-3-5}{4-0}\\\\m & = \\frac{-8}{4}\\\\m & = -2$ $m = -2$ and the $y-$intercept is (0, 5). The equation of this line is: $\\boxed{y = -2x + 5}$ For each of the following real world problems... $& \\begin{pmatrix}x_1, & y_1\\\\320, & 124 \\end{pmatrix} \\begin{pmatrix}x_2, & y_2\\\\600, & 164 \\end{pmatrix}\\\\m & = \\frac{y_2 - y_1}{x_2 - x_1}\\\\m &", "# Finding the equation of a Plane • Oct 20th 2009, 10:47 AM craig Finding the equation of a Plane Hi, another vector question I'm afraid ;) I have the 4 lines: $BC = \\begin{pmatrix} -2 \\\\ -4 \\\\ 1 \\end{pmatrix} + \\lambda\\begin{pmatrix} 3 \\\\ 6 \\\\ -3 \\end{pmatrix}$, $AC = \\begin{pmatrix} 1 \\\\ 2 \\\\ -2 \\end{pmatrix} + \\sigma\\begin{pmatrix} -2 \\\\ 2 \\\\ -4 \\end{pmatrix}$. $l2 = \\begin{pmatrix} 3 \\\\ 0 \\\\ 2 \\end{pmatrix} + \\beta\\begin{pmatrix} -2 \\\\ 2 \\\\ -4 \\end{pmatrix}$, $AD = \\begin{pmatrix} 0 \\\\ 0 \\\\ -1 \\end{pmatrix} + \\alpha\\begin{pmatrix} -3 \\\\ 0 \\\\ -3 \\end{pmatrix}$. I need to show that all 4 lines lie in a single plane and fine the Cartesian Equation of the plane. The Cartesian equation of a plane is in the form $Ax+By+Cz=d$, so from this I can set up the following equations: $-2A-4B+C=d$ $3A+2C=d$ $A+2B-2C=d$ From these I can get all $A,B,C$ in terms of $d$. $A=d$, $B=C=-d$ Putting these values back into the original Cartesian Equation I got: $dx-dy-dz = d$, which can be simplified to $x-y-z=1$. Is this correct so far or have I gone horribly wrong? Thanks • Oct 21st 2009, 11:40 AM craig Over 20 views and none of you have any ideas ;) Anyone got any comments at all? • Oct 21st 2009,", "$\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix},\\quad M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x-y=0\\},\\quad N=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, 2x+y=0\\}$ 3. $\\begin{pmatrix} 3 \\\\ 0 \\\\ 1 \\end{pmatrix},\\quad M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, x+y=0\\},\\quad N=\\{c\\cdot\\begin{pmatrix} 1 \\\\ 0 \\\\ 1 \\end{pmatrix}\\,\\big|\\, c\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 2 Find $M^\\perp$. 1. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x+y=0 \\}$ 2. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, -2x+3y=0 \\}$ 3. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x-y=0 \\}$ 4. $M=\\{\\vec{0}\\,\\}$ 5. $M=\\{\\begin{pmatrix} x \\\\ y \\end{pmatrix}\\,\\big|\\, x=0 \\}$ 6. $M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, -x+3y+z=0 \\}$ 7. $M=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, x=0\\text{ and } y+z=0\\}$ Problem 3 This subsection shows how to project orthogonally in two ways, the method of Example 3.2 and 3.3, and the method of Theorem 3.8. To compare them, consider the plane $P$ specified by $3x+2y-z=0$ in $\\mathbb{R}^3$. 1. Find a basis for $P$. 2. Find $P^\\perp$ and a basis for $P^\\perp$. 3. Represent this vector with respect to the concatenation of the two bases from the prior item. $\\vec{v}=\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ 4. Find the orthogonal projection of $\\vec{v}$ onto $P$ by keeping only the $P$ part from the prior item. 5. Check that against the result from applying Theorem 3.8. This exercise is recommended for all" ]

[ "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "\\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} - \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} = \\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ On the other hand, the reflection of $\\mathbf{x}$ in the plane $$L^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\mathbf{v} \\cdot \\mathbf{y} = 0\\}$$ with normal vector $\\mathbf{v}$ is given by $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = 2\\operatorname{Proj}_{L^\\perp}(\\mathbf{x}) - \\mathbf{x} = 2\\left(\\mathbf{x} - \\operatorname{Proj}_L(\\mathbf{x})\\right) - \\mathbf{x} = \\mathbf{x} - 2\\operatorname{Proj}_L(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v},$$ which in your case yields $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} - 2 \\cdot \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ One last cultural note: the reflection $\\operatorname{Refl}_{L^\\perp}(\\mathbf{x})$ of $\\mathbf{x}$ in the plane $L^\\perp$ with normal vector $\\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image $$H_{\\mathbf{v}}(\\mathbf{x}) := \\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}$$ of $\\mathbf{x}$ under the Householder transformation $H_\\mathbf{v}$ corresponding to $\\mathbf{v}$. So, in your case, as we just saw, $$H_{\\mathbf{v}}(\\mathbf{x}) = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ Suppose you mean a vector ($\\mathbf{v}$) \"specular\" to the given one ($\\mathbf{u}$) with respect to the line $L$. Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin. If you do the cross product $\\mathbf{w} = \\mathbf{u} \\times \\mathbf{L}$ you get a vector which is orthogonal either to the line and to $\\mathbf{u}$, i.e. it is normal to the plane containing the line and $\\mathbf{u}$. Then $\\mathbf{v} \\times \\mathbf{L}$ shall be equal to $-\\mathbf{w}$", "the surface does not lie in a plane, $X_{st}$ cannot be a linear combination of $X_s$ and $X_t$ (otherwise, the plane spanned by $X_s$ and $X_t$ would be fixed, and the surface would lie in plane). This means that $X_s$, $X_t$, $X_{st}$ is a basis of $\\mathbb{R}^3$, and, as such, there are equations of the form $$\\begin{pmatrix} dX_s& dX_t & dX_{st} \\end{pmatrix} = \\begin{pmatrix} X_s& X_t & X_{st} \\end{pmatrix} \\begin{pmatrix} f\\ ds& 0 & f_t\\ ds\\\\\\\\ 0 & g\\ dt & g_s\\ dt\\\\\\\\ dt & ds & f\\ ds + g\\ dt \\end{pmatrix}$$ (The equations for $dX_{st}$ follow since $(X_{st})_s = (f X_s)_t = f_t\\ X_s + f\\ X_{st}$, etc.) By comparing partials, or by using the structure equations above (i.e., expanding out the consequences of $d(d(X_{st}))=0$, etc.), one sees that $d(f\\ ds + g\\ dt) = 0$, so that there must exist a function $h$ such that $f = 2h_s$ and $g = 2h_t$. (The coefficient of $2$ avoids some fractions later.) The equation now becomes $$\\begin{pmatrix} dX_s& dX_t & dX_{st} \\end{pmatrix} = \\begin{pmatrix} X_s& X_t & X_{st} \\end{pmatrix} \\begin{pmatrix} 2h_s\\ ds& 0 & 2h_{st}\\ ds\\\\\\\\ 0 & 2h_t\\ dt & 2h_{st}\\ dt\\\\\\\\ dt & ds & 2h_s\\ ds + 2h_t\\ dt \\end{pmatrix}$$ Moreover, the structure equations now imply that $d(e^{-2h}h_{st})=0$, so $h_{st} = C\\ e^{2h}$ for", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "\\\\ -1 \\end{pmatrix}$$ Then, the plane satisfies $$\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\begin{pmatrix} -1 \\\\ -1 \\\\ -1 \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix} \\begin{pmatrix} -1 \\\\ -1 \\\\ -1 \\end{pmatrix} = -2,$$ or that $- x - y - z = - 2$. - Take one of the points as origin and use the other two to obtain two (linearly independent? check!) vectors. Then write the vectorial equation for the plane as $$(x,y,z)=P+\\lambda \\vec u + \\mu \\vec v.$$ -", "Finding orthonormal basis for a subspace $W$ of the Euclidean space $\\mathbb{R}^3$. Problem: Let $\\mathbb{R}^3$ be an Euclidean space. Find an orthonormal basis for the subspace $W$ defined as $x + 2y-z = 0$. Attempt at solution: So this is a plane in $\\mathbb{R}^3$, so I guess I would need two vectors to span this? I did the following. Pick two arbitrary vectors that lie in this plane, for example $w_1 = (1,1,3)$ and $w_2 = (1,0,1)$. Then should I apply the Gram-Schmidt process to these? That would get me: \\begin{align*} v_1 &= w_1 \\\\ v_2 &= w_2 - \\frac{\\langle w_2, v_1 \\rangle}{\\langle v_1, v_1} v_1 \\end{align*} Then we would have $v_1 = (1,1,3)$ and $v_2 = (7/11, -4/11, -1/11)$. Now I should normalize these and I'm done? • Yes, but the results will be simpler if you set $w_1=(1,0,1)$. – Bernard Jul 27 '15 at 18:18 A simple way of getting independent vectors in this plane is to note that if $x+ 2y- z= 0$, then $z= x- 2y$ so any vector in this plane is of the form $\\begin{pmatrix} x \\\\ y \\\\ z\\end{pmatrix}= \\begin{pmatrix}x \\\\ y \\\\ x- 2y\\end{pmatrix}$. Taking $x= 1, y= 0$, that is $\\begin{pmatrix}1 \\\\ 0 \\\\ 1\\end{pmatrix}$. Taking $x= 0, y= 1$, this is $\\begin{pmatrix} 0 \\\\ 1 \\\\ -2\\end{pmatrix}$.", "a$. We'd deal with this issue separately. Note that the first 2 equations give the equation on a plane. Hence, their intersection is a line, which we can calculate. The first 2 equations are equivalent to $$\\begin{array} {llll} &&(x-a) &+ &(y-b) &+ &(z-c) &= 0\\\\ &a&(x-a) &+b&(y-b) & +c&(z-c) & = 0 \\\\ \\end{array}$$ Since $$\\begin{pmatrix} 1\\\\1\\\\1\\\\ \\end{pmatrix} \\times \\begin{pmatrix} a\\\\b\\\\c\\\\ \\end{pmatrix} = \\begin{pmatrix} c-b\\\\a-c\\\\b-a\\\\ \\end{pmatrix},$$ we know that the line is given by $\\frac{x-a}{c-b} = \\frac{ y-b}{a-c} = \\frac{z-c}{b-a} = k$. (This is valid because the denominator is never 0.) This is equivalent to $x=a + k(c-b), y = b+k(a-c) , z = c+k(b-a)$. Next, we are interested in when this line intersects the ellipsoid $ax^2 + by^2 + cz^2 = a^3 + b^3 + z^3$, which happens in at most 2 places. (Of course, one place is $k=0$, which leads to the solution $x=a, y=b, z=c$.) Using the equation of the line above, we get that $$\\begin{array}{l l l l} ax^2 &= a^3 & + 2a^2k(c-b) &+k^2 a(c-b)^2 \\\\ by^2 &= b^3 & + 2b^2k(a-c) & + k^2 b(a-c)^2 \\\\ cz^2 &= c^3 & + 2c^2k(b-a) & + k^2 c(b-a)^2 \\\\ \\end{array}$$ Summing up these three equations, and using the last equation given, we get that $$-2k(a-b)(b-c)(c-a) + k^2 (a^2b+a^2c+b^2a+b^2c+c^2a+c^2b - 6abc) = 0.$$ One solution", "Math Relative position of planes Point and plane (point on the plane) # Point on the plain? Similar to checking whether a specific point lies on a line you can also check whether the point lies on the plane. i ### Method The procedure varies depending on the plane equation: 1. Insert the position vector of the point (P/V) or its coordinates (C). 2. Solve equation (V/C) or system of equations (P) 3. Check if solvable P - Parametric equation V - Vector equation of a plane C - Cartesian equation ! ### Remember The point lies on the plane only if the equation or system of equations can be solved. ### Example (parametric form) $P(2|1|1)$, $\\text{E: } \\vec{x} = \\begin{pmatrix} 3 \\\\ 0 \\\\ 0 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $+ s \\cdot \\begin{pmatrix} 1 \\\\ 5 \\\\ 2 \\end{pmatrix}$ 1. #### Insert $P$ The position vector (vector with the coordinates of the point) of $P$ is used for $\\vec{x}$ in $E$. $\\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}$ $=\\begin{pmatrix} 3 \\\\ 0 \\\\ 0 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $+ s \\cdot \\begin{pmatrix} 1 \\\\ 5 \\\\ 2 \\end{pmatrix}$ 2. #### Set up system of equations Now we set up an equation system and", "# Decide a number $a, b, c$ (expressed by $\\lambda$) so that $a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$ See the following vectors in $$\\mathbb{R}^3$$: $$\\mathbf{u}_1= \\begin{pmatrix} 1 \\\\ -1\\\\ 1 \\\\ 1 \\end{pmatrix}$$, $$\\mathbf{u}_2 = \\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\\\ 2 \\end{pmatrix}$$, $$\\mathbf{u}_3 = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\end{pmatrix}$$ and $$\\mathbf{v}_1= \\begin{pmatrix} 5 \\\\ -3\\\\ -4 \\\\ -1 \\end{pmatrix}$$, $$\\mathbf{v}_2 = \\begin{pmatrix} 6 \\\\ 4 \\\\ 1 \\\\ 8 \\end{pmatrix}$$, $$\\mathbf{v}_3 = \\begin{pmatrix} 7 \\\\ 2 \\\\ -1 \\\\ 6 \\end{pmatrix}$$ We are told that $$\\mathcal{B} = (\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3)$$ and $$\\mathcal{C} = (\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3)$$ both are bases for the same subspace $$\\mathcal{U}$$. Let $$\\mathcal{\\lambda}$$ be an unknown number og see the vector $$\\mathbf{x} = \\mathbf{v}_1 + \\lambda\\mathbf{v}_2 - \\mathbf{v}_3$$. Decide a number $$a, b, c$$ (expressed by $$\\lambda$$) so that $$a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$$ 1. Find some $$\\alpha_i$$ such that $$v_1=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3$$ 2. Find some $$\\beta_i$$ such that $$v_2=\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3$$ 3. Find some $$\\gamma_i$$ such that $$v_3=\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3$$ 4. Simplify the expression $$x=v_1+\\lambda v_2-v_3=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3+\\lambda(\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3)-(\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3)$$ • Thanks 5xum! When I reduce it, I get $\\begin{pmatrix} 8 \\\\ 3\\\\ -9 \\\\ 1 \\end{pmatrix} + a\\begin{pmatrix} 2 \\\\ -2\\\\ 1 \\\\ 1 \\end{pmatrix} + b\\begin{pmatrix} 3 \\\\ 6\\\\ 0 \\\\ 6 \\end{pmatrix} + c\\begin{pmatrix} 1 \\\\", "Review question # Which point $P$ is on both this line and this plane? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R6006 ## Solution The line $l$ has the equation $\\mathbf{r}= \\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}+ \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}, \\hspace{1cm} \\lambda \\in \\mathbb{R}.$ 1. Show that $l$ lies in the plane whose equation is $\\mathbf{r}.\\begin{pmatrix}-1\\\\ 2\\\\ 0\\end{pmatrix}=-5.$ We need to show that every vector $\\mathbf{r}$ that satisfies the line equation also satisfies the plane equation. Substituting in, we find $\\left[\\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}+ \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix} \\right].\\begin{pmatrix}-1\\\\ 2\\\\ 0\\end{pmatrix}=\\begin{pmatrix}5+2\\lambda\\\\ \\lambda\\\\ 5\\end{pmatrix} .\\begin{pmatrix}-1\\\\ 2\\\\ 0\\end{pmatrix}= -5 -2\\lambda+2\\lambda=-5.$ This is true for every $\\lambda$, so we’ve shown that the line does indeed lie in the given plane. 1. Find the position vector of $L$, the foot of the perpendicular from the origin $O$ to $l$. Let $\\overrightarrow{OL} = \\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}+ \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}.$ We need $\\left[\\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}+ \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix} \\right].\\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}=0$ which is true if and only if $10+4\\lambda+\\lambda =0,$ or $\\lambda = -2.$ Thus $L$ is the point $\\begin{pmatrix}1\\\\ -2\\\\ 5\\end{pmatrix}$. 1. Find an equation of the plane containing $O$ and $l$. We know that $\\overrightarrow{OL}$ is in the plane, and putting $\\lambda = 0$, we know also that", "+0 # dumb precalculus 0 45 1 For $$\\bold{v} = \\begin{pmatrix} \\phantom -2 \\\\ \\phantom -3 \\\\ -1 \\end{pmatrix} and \\bold{w} = \\begin{pmatrix} \\phantom -2 \\\\ -1 \\\\ \\phantom -0 \\end{pmatrix}$$ compute the projection of $\\mathbf v$ onto $\\mathbf w.$ i literally dont know how to solve this Apr 18, 2022 #1 +9369 +3 We have the formula: $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac{\\langle\\mathbf v, \\mathbf w\\rangle}{\\|\\mathbf w\\|^2}\\mathbf{w}$$ Now, we compute $$\\langle\\mathbf v, \\mathbf w\\rangle$$ and $$\\|\\mathbf w\\|^2$$. $$\\begin{array}{cl} &\\langle\\mathbf v, \\mathbf w\\rangle\\\\ =& 2(2) + 3(-1) + (-1)(0)\\\\ =& 1\\\\ &\\|\\mathbf w\\|^2\\\\ =& 2^2 + (-1)^2 + 0^2\\\\ =&5 \\end{array}$$ Therefore, substituting back into the formula, we have $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac15\\mathbf{w} = \\begin{pmatrix}\\frac25\\\\-\\frac15\\\\0\\end{pmatrix}$$ Apr 18, 2022 #1 +9369 +3 We have the formula: $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac{\\langle\\mathbf v, \\mathbf w\\rangle}{\\|\\mathbf w\\|^2}\\mathbf{w}$$ Now, we compute $$\\langle\\mathbf v, \\mathbf w\\rangle$$ and $$\\|\\mathbf w\\|^2$$. $$\\begin{array}{cl} &\\langle\\mathbf v, \\mathbf w\\rangle\\\\ =& 2(2) + 3(-1) + (-1)(0)\\\\ =& 1\\\\ &\\|\\mathbf w\\|^2\\\\ =& 2^2 + (-1)^2 + 0^2\\\\ =&5 \\end{array}$$ Therefore, substituting back into the formula, we have $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac15\\mathbf{w} = \\begin{pmatrix}\\frac25\\\\-\\frac15\\\\0\\end{pmatrix}$$ MaxWong Apr 18, 2022", "# Finding the equation of a Plane • Oct 20th 2009, 10:47 AM craig Finding the equation of a Plane Hi, another vector question I'm afraid ;) I have the 4 lines: $BC = \\begin{pmatrix} -2 \\\\ -4 \\\\ 1 \\end{pmatrix} + \\lambda\\begin{pmatrix} 3 \\\\ 6 \\\\ -3 \\end{pmatrix}$, $AC = \\begin{pmatrix} 1 \\\\ 2 \\\\ -2 \\end{pmatrix} + \\sigma\\begin{pmatrix} -2 \\\\ 2 \\\\ -4 \\end{pmatrix}$. $l2 = \\begin{pmatrix} 3 \\\\ 0 \\\\ 2 \\end{pmatrix} + \\beta\\begin{pmatrix} -2 \\\\ 2 \\\\ -4 \\end{pmatrix}$, $AD = \\begin{pmatrix} 0 \\\\ 0 \\\\ -1 \\end{pmatrix} + \\alpha\\begin{pmatrix} -3 \\\\ 0 \\\\ -3 \\end{pmatrix}$. I need to show that all 4 lines lie in a single plane and fine the Cartesian Equation of the plane. The Cartesian equation of a plane is in the form $Ax+By+Cz=d$, so from this I can set up the following equations: $-2A-4B+C=d$ $3A+2C=d$ $A+2B-2C=d$ From these I can get all $A,B,C$ in terms of $d$. $A=d$, $B=C=-d$ Putting these values back into the original Cartesian Equation I got: $dx-dy-dz = d$, which can be simplified to $x-y-z=1$. Is this correct so far or have I gone horribly wrong? Thanks • Oct 21st 2009, 11:40 AM craig Over 20 views and none of you have any ideas ;) Anyone got any comments at all? • Oct 21st 2009,", "1 & 0 \\\\ 1 & 1 \\end{pmatrix}, N_3 = \\begin{pmatrix} 1 & 1 \\\\ 0 & 1 \\end{pmatrix}, N_4 = \\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix}$ Does $\\{ N_1, N_2, N_3, N_4 \\}$ generate V? $M_1 = -\\frac23 N_1 + \\frac13(N_2 + N_3 + N_4)$ $M_2 = -\\frac23 N_2 + \\frac13(N_1 + N_3 + N_4)$ $M_3 = -\\frac23 N_3 + \\frac13(N_1 + N_2 + N_4)$ $M_4 = -\\frac23 N_4 + \\frac13(N_1 + N_2 + N_3)$ Then $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = aM_1 + bM_2 + cM_3 + dM_4 =$ $a \\cdot \\left( -\\frac23 N_1 + \\frac13(N_2 + N_3 + N_4) \\right) + b \\cdot \\left( -\\frac23 N_2 + \\frac13(N_1 + N_3 + N_4) \\right) + \\ldots$ Theorem: If $S \\in \\mathbf V$, then $\\operatorname{span}(S) =$ {all l.c. of elements of S} is a subspace of V.", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "shown below) $$r·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$(3×1)+(2×[-5])+(7×8)=49$$ Therefore this means that the distanve of the vector d = 49 To complete the equation we need to write down the value of the whole equation in cartesian form $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$x-5y+8z=49$$ x - 5y + 8z - 49 = 0 #### Question 2 Write the equation of the plane through (2,1,3), given that the vector (4,5,6) is perpendicular $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 1\\\\\\ 3\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$(2×4)+(1×5)+(3×6)=31$$ Therefore the distance of the vector is d = 31 $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$4x+5y+6z=31$$ 4x + 5y + 6z - 31 = 0 #### Intersection of a line and a plane Find the point of intersection between the line The normal (perpendicular) to a plane is $$r=\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ with plane $$5x+y-z=1$$ First we need to replace r with the matrix (x,y,z) in order to find the values of x,y and z $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ We can change this equation into 3 different equations by seperating the matrices into: x = 2 + λ y = 3 + 2λ z = 4 - λ And we know that 5x+y-z=1 Replacing the values of x,y and z with these new values we get", "Review question # Can we show the direction of $P'Q'$ is independent of $P$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R8565 ## Solution The plane $\\pi$ has equation $\\mathbf{r} = \\mathbf{a} + \\lambda \\mathbf{b}+ \\mu \\mathbf{c}$ and the plane $\\pi'$ has equation $\\mathbf{r} = \\mathbf{d}+ \\lambda'\\mathbf{e}+\\mu'\\mathbf{f}$, where $\\lambda$, $\\mu$, $\\lambda'$ and $\\mu'$ are scalar parameters. It is given that $\\mathbf{a}=\\begin{pmatrix}1 \\\\ 0 \\\\ -1\\end{pmatrix}, \\mathbf{b}=\\begin{pmatrix}2\\\\ -1 \\\\ -1\\end{pmatrix}, \\mathbf{c} = \\begin{pmatrix}-1 \\\\ 2 \\\\ 1\\end{pmatrix}, \\mathbf{d} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}, \\mathbf{e} = \\begin{pmatrix}p \\\\ q \\\\ r\\end{pmatrix}, \\mathbf{f} = \\begin{pmatrix}q\\\\ r\\\\ p\\end{pmatrix},$ $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ and that $p$ is negative. The planes $\\pi$ and $\\pi'$ are parallel. Find $p$, $q$ and $r$. We can find a vector perpendicular to $\\pi$ using the cross product $\\mathbf{b} \\times \\mathbf{c}= \\begin{pmatrix}1\\\\ -1 \\\\ 3\\end{pmatrix}$. Since the two planes are parallel this is also perpendicular to $\\pi'$. We can use this fact in the dot product with the two direction vectors $\\mathbf{e}$ and $\\mathbf{f}$ in the plane $\\pi'$. So $\\mathbf{e} . (\\mathbf{b} \\times \\mathbf{c})=0$ and $\\mathbf{f} . (\\mathbf{b} \\times \\mathbf{c})=0$. Giving us \\begin{align} p-q+3r &=0 \\\\ q-r+3p &=0 \\end{align} Adding gives $4p+2r=0 \\quad \\implies r=-2p$ and substituting in the second", "each subspace's description. Then express each subspace as a span. 1. The subset $\\{\\begin{pmatrix} a &b &c \\end{pmatrix}\\,\\big|\\, a-c=0\\}$ of the three-wide row vectors 2. This subset of $\\mathcal{M}_{2 \\! \\times \\! 2}$ $\\{\\begin{pmatrix} a &b \\\\ c &d \\end{pmatrix} \\,\\big|\\, a+d=0\\}$ 3. This subset of $\\mathcal{M}_{2 \\! \\times \\! 2}$ $\\{\\begin{pmatrix} a &b \\\\ c &d \\end{pmatrix} \\,\\big|\\, 2a-c-d=0 \\text{ and } a+3b=0 \\}$ 4. The subset $\\{a+bx+cx^3\\,\\big|\\, a-2b+c=0\\}$ of $\\mathcal{P}_3$ 5. The subset of $\\mathcal{P}_2$ of quadratic polynomials $p$ such that $p(7)=0$ This exercise is recommended for all readers. Problem 7 Find a set to span the given subspace of the given space. (Hint. Parametrize each.) 1. the $xz$-plane in $\\mathbb{R}^3$ 2. $\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, 3x+2y+z=0\\}$ in $\\mathbb{R}^3$ 3. $\\{\\begin{pmatrix} x \\\\ y \\\\ z \\\\ w \\end{pmatrix}\\,\\big|\\, 2x+y+w=0 \\text{ and } y+2z=0\\}$ in $\\mathbb{R}^4$ 4. $\\{a_0+a_1x+a_2x^2+a_3x^3\\,\\big|\\, a_0+a_1=0 \\text{ and } a_2-a_3=0\\}$ in $\\mathcal{P}_3$ 5. The set $\\mathcal{P}_4$ in the space $\\mathcal{P}_4$ 6. $\\mathcal{M}_{2 \\! \\times \\! 2}$ in $\\mathcal{M}_{2 \\! \\times \\! 2}$ Problem 8 Parametrize it with $\\{y\\begin{pmatrix} -2/3 \\\\ 1 \\\\ 0 \\end{pmatrix}+z\\begin{pmatrix} -1/3 \\\\ 0 \\\\ 1 \\end{pmatrix} \\,\\big|\\, y,z\\in\\mathbb{R} \\}$ to get $\\{\\begin{pmatrix} -2/3 \\\\ 1 \\\\ 0 \\end{pmatrix},\\begin{pmatrix} -1/3 \\\\ 0 \\\\ 1 \\end{pmatrix} \\}$. • $\\{\\begin{pmatrix} 1 \\\\ -2 \\\\ 1 \\\\ 0 \\end{pmatrix}, \\begin{pmatrix}", "## Vector Equation of a Plane The vector equation of a plane can be written as $r.n=a.n$ where $$a$$ is a position vector of a point on the plane and $$n$$ is a normal to the plane. LESSON 1 Determine the equation of the plane, in vector form and Cartesian form, which contains the point $$(2,-3,1)$$ with normal $$i-2j+3k$$. SOLUTION The equation of a plane can be written in the form $r.n=a.n$ where $$n$$ is a vector perpendicular to the plane and $$a$$ is a position vector of a point on the plane. State the position vector $\\begin{pmatrix} 2\\\\-3 \\\\1 \\end{pmatrix}$ State the normal to the plane $n=\\begin{pmatrix} 1\\\\-2 \\\\3 \\end{pmatrix}$ Substitute the vectors $$a$$ and $$n$$ into the general form of the vector equation of the plane, $$r.n=a.n$$ $r.\\begin{pmatrix} 1\\\\-2 \\\\3 \\end{pmatrix}=\\begin{pmatrix} 2\\\\-3 \\\\1 \\end{pmatrix}.\\begin{pmatrix} 1\\\\-2 \\\\3 \\end{pmatrix}$ Determine the dot product $$a.n$$ $r.\\begin{pmatrix} 1\\\\-2\\\\3 \\end{pmatrix}=2(1)+(-3)(-2)+1(3)$ $r.\\begin{pmatrix} 1\\\\-2\\\\3 \\end{pmatrix}=11$ CARTESIAN FORM Replace $$r$$ with $$\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix}$$ in the vector equation of the plane $\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix}.\\begin{pmatrix} 1\\\\-2\\\\3 \\end{pmatrix}=11$ Determine the dot product, $$r.n$$ $x-2y+3z=11$ TRY THIS Complete Questions 13 and 14 from worksheet.", "$\\mathbb{R}^2$ is a subspace of $\\mathbb{R}^3$. By definition, for $\\mathbb{R}^2$ to be a subspace of $\\mathbb{R}^3$, $\\mathbb{R}^2$ must first be a subset of $\\mathbb{R}^3$. This is not true because vectors in $\\mathbb{R}^2$ have only two elements whereas vectors in $\\mathbb{R}^3$ have three elements. This means that vectors in $\\mathbb{R}^2$ do not belong in $\\mathbb{R}^3$ - for instance: $$\\begin{pmatrix} 2\\\\3 \\end{pmatrix}\\notin\\mathbb{R}^3$$ Therefore, $\\mathbb{R}^2$ is not a subspace of $\\mathbb{R}^3$. Example. ## A plane passing through the origin Consider the following plane: $$2x+4y+3z=0$$ Show that this is a subspace of $\\mathbb{R}^3$. Solution. To show that the plane is a subspace of $\\mathbb{R}^3$, we must check that the plane is closed under addition and scalar multiplication. Suppose the following two generic vectors lie on the plane: $$\\boldsymbol{v}_1= \\begin{pmatrix} x_1\\\\y_1\\\\z_1 \\end{pmatrix},\\;\\;\\;\\;\\; \\boldsymbol{v}_2= \\begin{pmatrix} x_2\\\\y_2\\\\z_2 \\end{pmatrix}$$ Firstly, since these points are on the plane, we have that: \\label{eq:M1IcX56LXG5HLrinvnh} \\begin{aligned} 2x_1+4y_1+3z_1&=0\\\\ 2x_2+4y_2+3z_2&=0\\\\ \\end{aligned} Now, let's check whether or not the plane is closed under addition. The vector $\\boldsymbol{v}_1+\\boldsymbol{v}_2$ is: \\label{eq:xWDCFePt8ckIu5VeeW0} \\begin{aligned}[b] \\boldsymbol{v}_1+ \\boldsymbol{v}_2&= \\begin{pmatrix} x_1\\\\y_1\\\\z_1 \\end{pmatrix}+ \\begin{pmatrix} x_2\\\\y_2\\\\z_2 \\end{pmatrix}\\\\ &= \\begin{pmatrix} x_1+x_2\\\\y_1+y_2\\\\z_1+z_2 \\end{pmatrix} \\end{aligned} Let's substitute \\eqref{eq:xWDCFePt8ckIu5VeeW0} into the equation of the plane to check whether this point is on the plane: \\begin{align*} 2(x_1+x_2)+4(y_1+y_2)+3(z_1+z_2) &=2x_1+2x_2+4y_1+4y_2+3z_1+3z_2\\\\ &=(2x_1+4y_1++3z_1)+(2x_2+4y_2+3z_2)\\\\ &=0 \\end{align*} The last step used \\eqref{eq:M1IcX56LXG5HLrinvnh}. Therefore, this equation is closed under addition. Now, we check whether", "+0 # help vectors 0 46 2 Let v and w be vectors such that $$\\text{proj}_{{w}} {v} = \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ Find $$\\text{proj}_{-2 {w}} (3 {v})$$. Feb 23, 2020 #1 0 Projection is linear, so the answer is 2*3*(4,-7) = (24,-42). Feb 24, 2020 #2 +24366 +1 Let $$\\mathbf{v} \\text{ and } \\mathbf{w} \\text{ be vectors such that } \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$. Find $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}})$$. $$\\begin{array}{|rcll|} \\hline \\vec{p} &=& \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} ) \\\\\\\\ &=& \\dbinom{4}{-7} \\\\\\\\ &=& \\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\ \\hline \\vec{P} &=& \\operatorname{proj}_{\\mathbf{-2w}}( \\mathbf{3v} ) \\\\\\\\ &=& \\dfrac{(-2\\vec{w})(3\\vec{v})}{|-2w|^2}(-2\\vec{w}) \\\\\\\\ &=& 6\\dfrac{\\vec{w}\\vec{v}}{2w^2}\\vec{w} \\\\\\\\ &=& 3\\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\\\\\ &=& 3\\vec{p} \\\\\\\\ &=& 3\\dbinom{4}{-7} \\\\\\\\ &=& \\dbinom{12}{-21} \\\\ \\hline \\end{array}$$ $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}}) = \\dbinom{12}{-21}.$$ Feb 24, 2020" ]

[ "Find the number of ordered quadruples of complex numbers $$(a,b,c,d)$$ such that $\\begin{cases} ab=b-c\\\\ bc=c-d\\\\ cd=d-a\\\\ ad=a-b \\end{cases}.$", "+0 # (a) Count the number of quadruples (a, b, c, d) of nonnegative integers such that 0 82 2 +166 (a) Count the number of quadruples (a, b, c, d) of nonnegative integers such that $$0 \\le a < b < c < d \\le 12$$. (b) For this part, we want to count the number of quadruples (a, b, c, d) of nonnegative integers such that $$0 \\le a \\le b \\le c \\le d \\le 12$$. Here, some of a, b, c, and d can be equal to each other, so the answer will be different from part (a). Each value a, b, c, d must be between 0 and 12 inclusive. One idea is to count how many times each number appears. For example, suppose (a, b, c, d)=(1, 3, 3, 8). Then we can make a table that counts how many times each number appears among a, b, c, and d: $$\\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\\\ \\hline 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\end{array}$$ Use this idea to find the number of", "+0 # help 0 302 4 Find the number of ordered quadruples $$(a,b,c,d)$$ of nonnegative real numbers such that \\begin{align*} a^2 + b^2 + c^2 + d^2 &= 4, \\\\ (a + b + c + d)(a^3 + b^3 + c^3 + d^3) &= 16. \\end{align*} Mar 12, 2019 #1 +6196 +1 $$\\text{The only two are }\\\\ (1,1,1,1),~(-1,-1,-1,-1)$$ . Mar 13, 2019 #2 +110154 0 I do not doubt you Rom but how do you know that those are the only 2 quadruples? Melody Mar 13, 2019 #3 +30620 +2 The question stipulates non-negative numbers, so only one of Rom's sets is valid. If the question required integer solutions then (0,0,0, 2) is also a solution. However, it specifies real numbers (not necessarily integers), so there may well be other solutions, but so far I haven't put any thought into what those might be! The only other one I can think of is (0, 0, √2, √2). Mar 13, 2019 edited by Alan Mar 13, 2019 edited by Alan Mar 13, 2019 #4 +110154 0 Thanks Alan. :) Melody Mar 13, 2019", "# Thread: Cyclically 130 1. ## Cyclically 130 Consider all quadruples of real numbers (x1,x2,x3,x4), such that x1x2x3x4 and each of the four numbers plus the product of the other three equals 130. If S is the sum of all possible values of x4, what is S, rounded to the nearest integer ? 2. ## Re: Cyclically 130 1,2,3,4 are the subscripts of x 3. ## Re: Cyclically 130 Hello, geniusgarvil! Consider all quadruples of real numbers $a,b,c,d$ , such that $a \\le b \\le c \\le d$ and each of the four numbers plus the product of the other three equals 130. If $S$ is the sum of all possible values of $d$, what is $S$? .(Round to nearest integer.) We have: . $\\begin{Bmatrix}a + bcd \\:=\\:130 \\\\ b + acd \\:=\\:130 \\\\ c + abd \\:=\\:130 \\\\ d + abc \\:=\\:120 \\end{Bmatrix}$ Since the four variables are completely interchangeable, . . we conclude that: . $a \\,=\\, b \\,=\\, c \\,=\\, d.$ The first equation becomes: . $a + a^3 \\:=\\:130 \\quad\\Rightarrow\\quad a^3 + a - 130 \\:=\\:0$ . . The only real root is: . $a \\,=\\,5.$ Therefore: . $a \\,=\\,b\\,=\\,c\\,=\\,d\\,=\\,5 \\quad\\Rightarrow\\quad S \\,=\\,5$ 4. ## Re: Cyclically 130 I,don't know what's wrong in your solution, but the answer's incorrect.", "# Daniel's equation Find the number of ordered quadruples of nonnegative integers $(a,b,c,d)$ such that $d<100$ and $a^2+b^2+c^2=2^d.$ This problem is shared by Daniel C. ×", "# Double Cryptogram Logic Level 4 $\\large{\\begin{array}{cccccc} & & & a & b & c&d\\\\ +& & & b & c & d & a \\\\ \\hline && & & & & \\phantom0&\\\\ \\end{array}} \\qquad \\qquad \\large{\\begin{array}{cccccc} & & & d & c & b &a \\\\ +& & & c& b & a & d \\\\ \\hline && & & & \\phantom0 &\\\\ \\end{array}}$ If $$a,b,c$$ and $$d$$ are distinct single-digit positive integers so that the result of both additions is the same, find the number of possible quadruples $$(a,b,c,d)$$. ×", "(1+x)(1+x^2)(1+x^4) = 1+y^7 \\\\ (1+y)(1+y^2)(1+y^4) = 1+x^7 \\end{cases}$ How many ordered pairs of real numbers $$(x, y)$$ satisfy the above equations? $\\left\\{ \\begin{array}{l} \\frac{{bcdef}}{a} = 4\\\\ \\frac{{acdef}}{b} = 9\\\\ \\frac{{abdef}}{c} = 16\\\\ \\frac{{abcef}}{d} = \\frac{1}{4}\\\\ \\frac{{abcdf}}{e} = \\frac{1}{9}\\\\ \\frac{{abcde}}{f} = \\frac{1}{{16}} \\end{array} \\right.$ Positive numbers $a,b,c,d,e,f$ satisfy the formulas above. Find the value of $(a+ c + e) - (b + d + f)$ ×", "# How many ordered quadruples $(a,b,c,d)$ satisfy $a+b+c+d=18$ under various conditions? part (a) How many ordered quadruples $$(a,b,c,d)$$ satisfy $$a+b+c+d=18,$$ where $$a,b,c,d$$ are nonnegative integers? (part (b) How many ordered quadruples $$(a,b,c,d)$$ satisfy $$a+b+c+d=18,$$ where $$a,b,c,d$$ are odd positive integers? part (c) How many ordered quadruples $$(a,b,c,d)$$ satisfy $$a+b+c+d=18,$$ where $$a,b,c,d$$ are integers such that $$|a|,\\ |b|,\\ |c|,\\ |d|$$ are each at most $$10$$? For part (a) I counted the number of positive quadruples which is 18+4=22 so it would be 21C3 = 1330 ways to do that. But, I'm not sure how to do part (b) or (c). • Infinitely many, of course. Let $a; b;c$ be any real number and let $d = 18-(a+b+c)$. If you have stipulations such as $a,b,c,d$ are natural number not equal to $0$ you should state that. I was toungue in cheek about the \"any real number\" but whether $a,b,c,d$ are allowed to be $0$ or not is a crucial detail that shouldn't be omitted. – fleablood Aug 9 at 1:26 $$\\binom{18+4-1}{4-1}$$ Part (b) $$a=2p+1 , b=2q+1 , c=2r+1 , d=2s+1$$ $$p+q+r+s=7$$ Stars and bars again $$\\binom{7+4-1}{4-1}$$ Part (c) Continuing Brian’s solution, $$a+10=w , b+10=x , c+10=y , d+10 = z$$ $$w+x+y+z=58$$ where all four are at most 20. Quadruples with at least one greater than 20:divide 37 into 4 and", "Browse Questions # If $^n P_r=30240$ and $^n C_r=252$ then the ordered pair (n,r)= $\\begin {array} {1 1} (1)\\;(12,6) & \\quad (2)\\;(10,5) \\\\ (3)\\;(9,4) & \\quad (4)\\;(16,7) \\end {array}$ (2) (10,5)", "Get Total Number Of Objects As $$\\\\ \\\\ Total\\quad Objects\\quad =\\quad ({ a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+....+{ a }_{ k })$$ Therefore Using Combinatorics We Get Number Of Ways To Arrange The Objects As $$\\\\ \\frac { ({ a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+........+{ a }_{ k })! }{ { a }_{ 1 }!{ a }_{ 2 }!{ a }_{ 3 }!....{ a }_{ k }! }$$ Number Of Ways To Arrange Will Be Obviously An Integer , Hence We Have Proved That $$\\\\ \\frac { ({ a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }+........+{ a }_{ k })! }{ { a }_{ 1 }!{ a }_{ 2 }!{ a }_{ 3 }!....{ a }_{ k }! }$$ Will Always Be An Integer · 3 years ago Log in to reply × Problem Loading... Note Loading... Set Loading...", "+0 # last time i posted it said it was invisible need help due today -2 112 4 (a) Count the number of quadruples (a,b,c,d) of nonnegative integers such that $$0 \\le a < b < c < d \\le 12$$ (b) For this part, we want to count the number of quadruples (a,b,c,d) of nonnegative integers such that $$0 \\le a \\le b \\le c \\le d \\le 12.$$ Here, some of a, b, c, and d can be equal to each other, so the answer will be different from part (a). Each value a, b, c, d must be between 0 and 12 inclusive. One idea is to count how many times each number appears. For example, suppose (a,b,c,d) = (1,3,8,8). Then we can make a table that counts how many times each number appears among a, b, c, and d:$$\\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\\\ \\hline 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 \\end{array}$$ Use this idea to find the number of quadruples (a,b,c,d). (If you come up with a different approach, you are free to", "# Are you good at manipulation? How many ordered quadruplet(s) of integers $$(a,b,c,d)$$ with $$0 < a < b < c < d < 500$$ satisfy the condition that $$a + d = b + c$$ and $$bc - ad = 93$$? ×", "Browse Questions # If $^n P_r=30240$ and $^n C_r=252$ then the ordered pair (n,r)= $\\begin {array} {1 1} (1)\\;(12,6) & \\quad (2)\\;(10,5) \\\\ (3)\\;(9,4) & \\quad (4)\\;(16,7) \\end {array}$ Can you answer this question? (2) (10,5) answered Nov 7, 2013 by", "To set up the comparison, the following are some operations on real numbers we would like to focus on: Operations on real numbers: $a=b \\ \\ \\Longrightarrow \\ \\ b=a \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{R1}$ $a=b \\text{ and } b=c \\ \\ \\Longrightarrow \\ \\ a=c \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\text{R2}$ $a=b \\text{ and } c=d \\ \\ \\Longrightarrow \\ \\ a+c=b+d \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\", "number of quadruples where the two quantities are equal. To do this, all we have to do is consider the cases when $a-d$ ranges from 3 to 9. It would seem natural that a range of 3 would produce 1 option, and a range of 4 would produce 2 options. However, since b and c cannot be equal, a range of 3 or 4 produces 1 option each, a range of 5 or 6 produces 2 options each, a range of 7 or 8 produces 3 options each, and a range of 9 will produce 4 options. In addition, a range of n has 10-n options for combinations of a and d. Multiplying the number of combinations of a and d by the corresponding number of options for b and c gives us 50 total quadruplets where $a+d=b+c$. So the answer will be $\\frac{210-50}{2} = \\boxed{080}.$ ## Solution 6 Let $b = a+x$ and $c = a+x+y$ and $d = a+x+y+z$ for positive integers $x,y,z.$ In order to satisfy the other condition we need $z > x$ so we let $z = x+k.$ Now the only other condition we need to satisfy so $a+2x+y+k \\le 10.$ This condition can be transformed into $a+2x+y+k+b = 11$ for positive $a,x,y,k,b.$ Now we use generating functions to finish. We find the", "numbers are separately arranged in increasing order but any odd number precedes any even number. It can be shown that the number of distinct order types of a set of power $\\aleph_{0}$ is infinite and in fact, uncountable. Section 3.4 Ordered sets and products of ordered sets. Let $M_{1}$ and $M_{2}$ be two ordered sets of types $\\theta_{1}$ and $\\theta_{2}$ respectively. Then, we can introduce an ordering in the union $M_{1} \\bigcup M_{2}$ of the two sets by assuming that : (i) a and b have the same ordering as in $M_{1}$ if $a,b \\in M_{1}$. (ii) a and b have the same ordering as in $M_{2}$ if $a,b \\in M_{2}$ (iii) $a if $a \\in M_{1}$ and $b \\in M_{2}$. (Verify that this is actually an ordering of $M_{1} \\bigcup M_{2}$). The set $M_{1} \\bigcup M_{2}$ ordered in this way is called the ordered sum of $M_{1}$ and $M_{2}$ denoted by $M_{1}+M_{2}$. Note that the order of terms matters here, that is, in general $M_{2}+M_{1}$ is not isomorphic to $M_{1}+M_{2}$. More generally, we can define the ordered sum of any finite number of ordered sets by writing (this will be an exercise problem after this blogged section) : $M_{1}+M_{2}+M_{3}=(M_{1}+M_{2})+M_{3}$ $M_{1}+M_{2}+M_{3}+M_{4}=(M_{1}+M_{2}+M_{3})+M_{4}$, and so on. By the ordered sum of the types $\\theta_{1}+\\theta_{2}$, denoted by $\\theta_{1}+\\theta_{2}$ we mean the", "numbers are separately arranged in increasing order but any odd number precedes any even number. It can be shown that the number of distinct order types of a set of power $\\aleph_{0}$ is infinite and in fact, uncountable. Section 3.4 Ordered sets and products of ordered sets. Let $M_{1}$ and $M_{2}$ be two ordered sets of types $\\theta_{1}$ and $\\theta_{2}$ respectively. Then, we can introduce an ordering in the union $M_{1} \\bigcup M_{2}$ of the two sets by assuming that : (i) a and b have the same ordering as in $M_{1}$ if $a,b \\in M_{1}$. (ii) a and b have the same ordering as in $M_{2}$ if $a,b \\in M_{2}$ (iii) $a if $a \\in M_{1}$ and $b \\in M_{2}$. (Verify that this is actually an ordering of $M_{1} \\bigcup M_{2}$). The set $M_{1} \\bigcup M_{2}$ ordered in this way is called the ordered sum of $M_{1}$ and $M_{2}$ denoted by $M_{1}+M_{2}$. Note that the order of terms matters here, that is, in general $M_{2}+M_{1}$ is not isomorphic to $M_{1}+M_{2}$. More generally, we can define the ordered sum of any finite number of ordered sets by writing (this will be an exercise problem after this blogged section) : $M_{1}+M_{2}+M_{3}=(M_{1}+M_{2})+M_{3}$ $M_{1}+M_{2}+M_{3}+M_{4}=(M_{1}+M_{2}+M_{3})+M_{4}$, and so on. By the ordered sum of the types $\\theta_{1}+\\theta_{2}$, denoted by $\\theta_{1}+\\theta_{2}$ we mean the", "numbers are separately arranged in increasing order but any odd number precedes any even number. It can be shown that the number of distinct order types of a set of power $\\aleph_{0}$ is infinite and in fact, uncountable. Section 3.4 Ordered sets and products of ordered sets. Let $M_{1}$ and $M_{2}$ be two ordered sets of types $\\theta_{1}$ and $\\theta_{2}$ respectively. Then, we can introduce an ordering in the union $M_{1} \\bigcup M_{2}$ of the two sets by assuming that : (i) a and b have the same ordering as in $M_{1}$ if $a,b \\in M_{1}$. (ii) a and b have the same ordering as in $M_{2}$ if $a,b \\in M_{2}$ (iii) $a if $a \\in M_{1}$ and $b \\in M_{2}$. (Verify that this is actually an ordering of $M_{1} \\bigcup M_{2}$). The set $M_{1} \\bigcup M_{2}$ ordered in this way is called the ordered sum of $M_{1}$ and $M_{2}$ denoted by $M_{1}+M_{2}$. Note that the order of terms matters here, that is, in general $M_{2}+M_{1}$ is not isomorphic to $M_{1}+M_{2}$. More generally, we can define the ordered sum of any finite number of ordered sets by writing (this will be an exercise problem after this blogged section) : $M_{1}+M_{2}+M_{3}=(M_{1}+M_{2})+M_{3}$ $M_{1}+M_{2}+M_{3}+M_{4}=(M_{1}+M_{2}+M_{3})+M_{4}$, and so on. By the ordered sum of the types $\\theta_{1}+\\theta_{2}$, denoted by $\\theta_{1}+\\theta_{2}$ we mean the", "1. ## Real no. ques-find the smallest no. which when divided by 28 gives remainder 8,and when divided by32 gives remainder 12. 2. Hello, sumit2009! This can be solved with Modulo Arithmetic, . . but here is an algebraic solution. Find the smallest number which when divided by 28 gives remainder 8, and when divided by 32 gives remainder 12. Let $N$ = the number. Then: . $\\begin{array}{cccc}N &=& 28a + 9 & {\\color{blue}[1]} \\\\ N &=& 32b + 12 & {\\color{blue}[2]} \\end{array}\\;\\text{ for integers }a,b$ Equate [1] and [2]: . $28a + 8 \\:=\\:32b + 12 \\quad\\Rightarrow\\quad 7a \\:=\\:8b + 1$ . . Then: . $a \\:=\\:\\frac{8b+1}{7} \\quad\\Rightarrow\\quad a \\:=\\:b + \\frac{b+1}{7}$ Since $a$ is an integer, $b+1$ must be a multiple of 7. . . The first time this happens is: . $b \\:=\\:6$ Substitute into [2]: . $N \\:=\\:32(6) + 12 \\quad\\Rightarrow\\quad\\boxed{ N \\:=\\:204}$", "with 0\\leq a,b,c,d \\... Algebra - Find the number of integer quadruples (a,b,c,d) with 0\\leq a,b,c,d \\... Search Members" ]

[ "& \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{eqnarray*} {\\renewcommand{\\arraystretch}{1.5} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} {\\renewcommand{\\arraystretch}{2} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup $(\\mathit{condition1})$ holds since the sky is red and $(\\mathit{condition2})$ because this is the answer. \\end{document} The output looks as follows: • eqnarray: • array: • array with higher stretch factor: • align: • align with higher \\thickmuskip: The problems in this example are the following: • The relation symbols used in each step may have a different horizontal length but should be aligned below each other in a centered way. Therefore, the usual AMS environments do not work in an obvious way since they only offer left- or right-aligned columns. The second solution tries to overcome this problem by", "\\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{eqnarray*} {\\renewcommand{\\arraystretch}{1.5} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} {\\renewcommand{\\arraystretch}{2} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup $(\\mathit{condition1})$ holds since the sky is red and $(\\mathit{condition2})$ because this is the answer. \\end{document} The output looks as follows: • eqnarray: • array: • array with higher stretch factor: • align: • align with higher \\thickmuskip: The problems in this example are the following: • The relation symbols used in each step may have a different horizontal length but should be aligned below each other in a centered way. Therefore, the usual AMS environments do not work in an obvious way since they only offer left- or right-aligned columns. The second solution tries to overcome this problem", "is possible to find A-1. A-1 = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj (A) A = $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ $$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\frac 12$$$$\\begin {pmatrix} 2 & -4\\\\ -3 & 7\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} (\\frac 22) & (\\frac {-4}{2})\\\\ (\\frac {-3}{2}) & (\\frac 72)\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ ∴ A-1 =$$\\begin {pmatrix} 1 & -2\\\\ (\\frac {-3}{2} & (\\frac 72)\\\\ \\end {pmatrix}$$ Ans Let: A =$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ and A-1=$$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ We know that, AA-1 = I, where I =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 2m & 7\\\\ 5 & 9\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} 9 & n\\\\ -5 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 45 - 45 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ or,$$\\begin {pmatrix} 18m - 35 & 2mn + 28\\\\ 0 & 5n + 36\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1 & 0\\\\ 0 & 1\\\\ \\end {pmatrix}$$ Taking the corresponding elements both sides", "is $$A=\\begin{pmatrix}1&0&1\\\\\\!\\!\\!\\!-1&1&2\\\\0&\\!\\!\\!\\!-1&1\\end{pmatrix}^{-1}=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}$$ Finally, since $$f(x)=3-4x+2x^2\\stackrel{coord. wrt B}\\longrightarrow \\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix} \\,\\,,\\,\\text{we get}$$ $$[Af]=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}\\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix}=\\frac{1}{4}\\begin{pmatrix}11\\\\\\!\\!\\!\\!-7\\\\1\\end{pmatrix}$$ so $$\\,f_C=\\frac{11}{4}(1-x)-\\frac{7}{4}(x-x^2)+\\frac{1}{4}(1+2x+x^2)$$", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "\\end{pmatrix} = \\begin{pmatrix} b_1 \\\\ & \\ddots \\\\ & & b_n \\end{pmatrix} \\begin{pmatrix} x_1 ^{-1} \\\\ \\vdots \\\\ x_n^{-1} \\end{pmatrix}.$$ As $A=\\left( a_{ij} \\right)_{n \\times n}$, $B=\\mathrm{diag}\\{b_1, b_2, \\cdots, b_n\\}=\\begin{pmatrix} b_1 \\\\ & \\ddots \\\\ & & b_n \\end{pmatrix}$, $C=A^{-1}B$, then $$A \\begin{pmatrix} x_1 \\\\ \\vdots \\\\ x_n \\end{pmatrix} = B \\begin{pmatrix} x_1 ^{-1} \\\\ \\vdots \\\\ x_n^{-1} \\end{pmatrix},$$ so $$\\begin{pmatrix} x_1 \\\\ \\vdots \\\\ x_n \\end{pmatrix} = A^{-1} B \\begin{pmatrix} x_1 ^{-1} \\\\ \\vdots \\\\ x_n^{-1} \\end{pmatrix},$$ that is to say, $$\\begin{pmatrix} x_1 \\\\ \\vdots \\\\ x_n \\end{pmatrix} = C \\begin{pmatrix} x_1 ^{-1} \\\\ \\vdots \\\\ x_n^{-1} \\end{pmatrix}.$$ Let $\\begin{pmatrix} x_1^{(0)} \\\\ \\vdots \\\\ x_n^{(0)} \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ \\vdots \\\\ 1 \\end{pmatrix}$, by the recurrence $$\\begin{pmatrix} x_1^{(k+1)} \\\\ \\vdots \\\\ x_n^{(k+1)} \\end{pmatrix} = C \\begin{pmatrix} {\\left(x_1^{(k)}\\right)} ^{-1} \\\\ \\vdots \\\\ {\\left(x_n^{(k)}\\right)}^{-1}, \\end{pmatrix},$$ we can get a numerical solution. If the result does not converge, we can use another one $$\\begin{pmatrix} {\\left(x_1^{(k+1)}\\right)} ^{-1} \\\\ \\vdots \\\\ {\\left(x_n^{(k+1)}\\right)}^{-1}, \\end{pmatrix} = C^{-1} \\begin{pmatrix} x_1^{(k)} \\\\ \\vdots \\\\ x_n^{(k)} \\end{pmatrix}.$$ If you want the analytic solution, you may have a try by using the Groebner-Shirshov Bases. (The software Maple or Mathematica can do it.) But I am not sure the analytic solution can be find easily. - The real case is when A is almost non-singular(having a large cond", "1\\\\ \\end {vmatrix}$$ = 20 × 1 - 12 × 1 = 20 - 12 = 8 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{8}$$$$\\begin {pmatrix} 1 & -1\\\\ -12 & 20\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-12}{8} & \\frac {20}{8}\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$ We know that, AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35}{8} - \\frac {20}{8}\\\\ \\frac {-105}{2} + \\frac {100}{2}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35 - 20}{8}\\\\ \\frac {-105 + 100}{2}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {15}{8}\\\\ \\frac {-5}{2}\\\\ \\end {pmatrix}$$ ∴ x = $$\\frac {15}{2}$$ and y = $$\\frac {-5}{2}$$ Ans Here, $$\\frac 3x$$ + $$\\frac 4y$$ = 2..........................................(1) $$\\frac 9x$$ - $$\\frac 2y$$ = $$\\frac 52$$....................(2) Given equation in the form of matrix: $$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix}", "\\frac xr \\neq \\frac rx.$$ Rather, if that relation does hold then it implies that $x$ is independent of $v$ so that $\\frac{\\partial x}{\\partial v} = 0$. anuttarasammyak Gold Member Further to my post #2 $$A= \\begin{pmatrix} u_x & u_y \\\\ v_x & v_y \\\\ \\end{pmatrix}$$ $$B= \\begin{pmatrix} x_u & x_v \\\\ y_u & y_v \\\\ \\end{pmatrix}$$ $$AB= \\begin{pmatrix} u_x & u_y \\\\ v_x & v_y \\\\ \\end{pmatrix} \\begin{pmatrix} x_u & x_v \\\\ y_u & y_v \\\\ \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ \\end{pmatrix}$$ It meets with post #3. $$A=B^{-1}$$ $$\\begin{pmatrix} u_x & u_y \\\\ v_x & v_y \\\\ \\end{pmatrix} =\\frac{1}{x_uy_v-x_vy_u} \\begin{pmatrix} y_v & -x_v \\\\ -y_u & x_u \\\\ \\end{pmatrix}$$ For an example (1,1) component says $$u_x=\\frac{1}{x_uy_v-x_vy_u}y_v$$ You see in order ##u_x =\\frac{1}{x_u}## as you expect, ##x_vy_u=0 ## and ##y_v \\neq 0## is required. Last edited:", "= \\frac{1}{\\textbf{h}^{3T}\\textbf{x}}\\begin{pmatrix} \\textbf{x}^T & 0 & 0 \\end{pmatrix} - \\frac{\\textbf{h}^{1T}\\textbf{x}}{(\\textbf{h}^{3T}\\textbf{x})^2} \\begin{pmatrix} 0 & 0 & \\textbf{x}^T \\end{pmatrix}$$ Substituting $x’ = \\textbf{h}^{1T}\\textbf{x}$ and $w’ = \\textbf{h}^{3T}\\textbf{x}$ $$\\partial\\tilde{x}’ / \\partial\\textbf{x} = \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & 0 \\end{pmatrix} - \\frac{x’}{(w’)^2}\\begin{pmatrix} 0 & 0 & \\textbf{x}^T \\end{pmatrix}$$ Finally, substituting $\\tilde{x’} = x’/w’$ $$\\partial\\tilde{x}’ / \\partial\\textbf{x} = \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & 0 \\end{pmatrix} - \\frac{\\tilde{x’} }{w’}\\begin{pmatrix} 0 & 0 & \\textbf{x}^T \\end{pmatrix}$$ $$= \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & -\\tilde{x’} \\textbf{x}^T \\end{pmatrix}$$ Performing a similar process for $\\tilde{y’}$, we get $$\\partial\\tilde{y}’ / \\partial\\textbf{x} = \\frac{1}{w’}\\begin{pmatrix} 0 & \\textbf{x}^T & -\\tilde{y’} \\textbf{x}^T \\end{pmatrix}$$ Hence, we have established that $$\\partial{\\tilde{\\textbf{x}}}’ / \\partial{\\textbf{h}} = \\frac{1}{w’}\\begin{pmatrix} \\textbf{x}^T & 0 & -\\tilde{x’}\\textbf{x}^T \\\\ 0 & \\textbf{x}^T & -\\tilde{y’}\\textbf{x}^T\\end{pmatrix}$$ #### VI. Sampson error with non-isotropic error distributions. The derivation of Sampson error in section 4.2.6(p98) assumed that points were measured with circular error distributions. In the case where the point $\\textbf{X} = (x, y, x’, y’)$ is measured with covariance matrix $\\Sigma_\\textbf{x}$ it is appropriate instead to minimize the Mahalanobis norm $\\begin{Vmatrix}\\boldsymbol{\\delta}_\\textbf{x}\\end{Vmatrix}^2_{\\Sigma_\\textbf{x}} = \\boldsymbol{\\delta}^T_\\textbf{x}\\Sigma^{-1}_\\textbf{x}\\boldsymbol{\\delta}_\\textbf{x}$. Show that in this case the formulae corresponding to (4.11-p99) and (4.12-p99) are $$\\boldsymbol{\\delta}_\\textbf{x} = -\\Sigma_\\textbf{x}J^T(J\\Sigma_\\textbf{x}J^T)^{-1}\\epsilon$$ and $$\\begin{Vmatrix}\\boldsymbol{\\delta}_\\textbf{x}\\end{Vmatrix}^2_{\\Sigma_\\textbf{x}} = \\epsilon^T(J\\Sigma_\\textbf{x}J^T)^{-1}\\epsilon$$ Note that if the measurements in the two images are independent, then the covariance matrix $\\Sigma_\\textbf{x}$ will be block-diagonal with two $2\\times2$ diagonal blocks", "\\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {vmatrix}$$ = 3 × (-2) - 9 × 4 = -6 - 36 = -42 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{-42}$$$$\\begin {pmatrix} -2 & -4\\\\ -9 & 3\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {-2}{-42} & \\frac {-4}{-42}\\\\ \\frac {-9}{-42} & \\frac {3}{-42}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$ We know that: AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{21} + \\frac {5}{21}\\\\ \\frac {9}{21} - \\frac {15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 5}{21}\\\\ \\frac {36 - 15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {7}{21}\\\\ \\frac {21}{84}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ i.e. $$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ Taking the corresponding parts of the equal matrix: $$\\frac 1x$$ = $$\\frac 13$$ ∴ x = 3 $$\\frac 1y$$ = $$\\frac 14$$", "=\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\left [\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}\\end{pmatrix} -\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\right ]+\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\\\ & =\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}-1\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}-2\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}+1\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}+5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{16}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{37}{5}\\end{pmatrix} \\end{align*} Is that correct so far? How could we continue? Klaas van Aarsen MHB Seeker Staff member The matrix for $\\delta$ is correct. But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. mathmari Well-known member MHB Site Helper But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? Klaas van Aarsen MHB Seeker Staff member Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? That point is in between $A$ and $B$. But $C$ is on the circumcircle of triangle $ABZ$, which implies that it cannot be between $A$ and $B$. Klaas van Aarsen MHB Seeker Staff member I've drawn a picture now: \\begin{tikzpicture} \\usetikzlibrary", "a problem because $$\\dfrac{1}{0}$$ is not defined. 2. If $$\\textsf{M}$$ = $$\\begin{pmatrix} a & b\\\\c & d\\end{pmatrix}$$, then $$\\textsf{M}^{-1}$$ = $$\\dfrac{1}{\\begin{vmatrix} \\textsf{M} \\end{vmatrix}} \\begin{pmatrix} d & -b\\\\-c & a\\end{pmatrix} = \\begin{pmatrix} \\dfrac{d}{\\textsf{|M|}} & \\dfrac{-b}{\\textsf{|M|}}\\\\\\dfrac{-c}{\\textsf{|M|}} & \\dfrac{a}{\\textsf{|M|}}\\end{pmatrix}$$. So $$|\\textsf{M}^{-1}| = \\dfrac{ad}{\\textsf{|M|}^{2}} – \\dfrac{(-b)(-c)}{\\textsf{|M|}^{2}}=\\dfrac{ad-bc}{\\textsf{|M|}^{2}}=\\dfrac{\\textsf{|M|}}{\\textsf{|M|}^{2}}=\\dfrac{1}{\\textsf{|M|}}$$.", "the expression can be re-written as [eqn] \\begin{pmatrix} 1 && 1 && 1 \\end{pmatrix} \\begin{pmatrix} \\frac 1 2 && \\frac 1 2 && \\frac 1 {\\sqrt 2} \\\\ \\frac 1 2 && \\frac 1 2 && -\\frac 1 {\\sqrt 2} \\\\ -\\frac 1 {\\sqrt 2} && \\frac 1 {\\sqrt 2} && 0 \\end{pmatrix} {\\begin{pmatrix} 1 - \\sqrt 2 && 0 && 0 \\\\ 0 && 1 + \\sqrt 2 && 0 \\\\ 0 && 0 && 1 \\end{pmatrix}}^{m-1} \\begin{pmatrix} \\frac 1 2 && \\frac 1 2 && -\\frac 1 {\\sqrt 2} \\\\ \\frac 1 2 && \\frac 1 2 && \\frac 1 {\\sqrt 2} \\\\ \\frac 1 {\\sqrt 2} && -\\frac 1 {\\sqrt 2} && 0 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix} \\\\ = \\begin{pmatrix} 1-\\frac 1 {\\sqrt 2} && 1+\\frac 1 {\\sqrt 2} && 0 \\end{pmatrix} {\\begin{pmatrix} 1 - \\sqrt 2 && 0 && 0 \\\\ 0 && 1 + \\sqrt 2 && 0 \\\\ 0 && 0 && 1 \\end{pmatrix}}^{m-1} \\begin{pmatrix} 1-\\frac 1 {\\sqrt 2} \\\\ 1+\\frac 1 {\\sqrt 2} \\\\ 0 \\end{pmatrix} \\\\ = \\begin{pmatrix} 1-\\frac 1 {\\sqrt 2} && 1+\\frac 1 {\\sqrt 2} && 0 \\end{pmatrix} \\begin{pmatrix} (1-\\sqrt 2)^{m-1} && 0 && 0 \\\\ 0 && (1+\\sqrt 2)^{m-1} && 0 \\\\ 0 && 0 && 1 \\end{pmatrix} \\begin{pmatrix} 1-\\frac 1 {\\sqrt 2}", "say that $\\lim_{n \\to \\infty}\\frac{\\text{number of bad path}}{\\text{number of total path}} = 0$. This is not obvious because there are uncountable infinite many bad sample paths. #### Renewal Theorem Renewal Process: we renew a car every $X_i$ years. Define $N(t) = \\text{number of renewals by time } t$ Let $X$ be a random variable representing the number of states between visits to \"renewal\" state. ($E[X]$ is therefore the mean time between each \"renewal\" state.) Renewal Theorem: when $E[X]$ is finite, we have Pr\\left\\lbrace\\lim_{t \\to \\infty} \\frac{N(t)}{t} = \\frac{1}{E[X]}\\right\\rbrace = 1 Proof: let $S_n = \\sum_{i = 1}^{N(t)} X_i$ be the time $t$ (might not be right at renewal) of the $n$-th renewals. Then we can say $S_{N(t)} \\leq t \\leq S_{N(t)+1}$ ($t$ should be between two renewals). \\begin{align*} &\\lim_{t \\to \\infty}\\left( S_{N(t)} \\leq t \\leq S_{N(t)+1}\\right)\\\\ =& \\lim_{t \\to \\infty}\\left( \\frac{S_{N(t)}}{N(t)} \\leq \\frac{t}{N(t)} \\leq \\frac{S_{N(t)+1}}{N(t)+1} \\cdot \\frac{N(t)+1}{N(t)}\\right)\\\\ =& E[X] \\leq \\frac{t}{N(t)} \\leq E[X] \\cdot 1 \\tag{by Strong Law of Large Numbers}\\\\ \\frac{t}{N(t)} =& E[X]\\\\ \\end{align*} ### Putting Them Together Theorem: For DTMC where mean time between states $m_{jj} < \\infty$ (typically in irreducible finite-state DTMC), the following holds: \\begin{align*} &p_j\\\\ =& \\lim_{t \\to \\infty} \\frac{N_j(t)}{t} \\tag{where $N_j(t)$ is the number of visit to state $j$ by time $t$}\\\\ \\approx& \\lim_{t \\to \\infty} \\frac{N_j(t)}{\\sum_{i = 1}^{N_j(t)}X_i} \\tag{where $X_i \\sim", "for $F_n$. We have that \\begin{align*} F_n & = \\begin{pmatrix} 0 & 1 \\end{pmatrix} X^n \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} \\\\ & = \\begin{pmatrix} 0 & 1 \\end{pmatrix} P D^n P^{-1} \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} \\\\ & = \\frac{1}{\\sqrt{5}} \\begin{pmatrix} 0 & 1 \\end{pmatrix} \\begin{pmatrix} \\sqrt{\\varphi} & -\\sqrt{-\\psi} \\\\ \\sqrt{-\\psi} & \\sqrt{\\varphi} \\end{pmatrix} \\begin{pmatrix} \\varphi^n & 0 \\\\ 0 & \\psi^n \\end{pmatrix} \\begin{pmatrix} \\sqrt{\\varphi} & \\sqrt{-\\psi} \\\\ -\\sqrt{-\\psi} & \\sqrt{\\varphi} \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} \\\\ & = \\frac{1}{\\sqrt{5}} \\begin{pmatrix} \\sqrt{-\\psi} & \\sqrt{\\varphi} \\end{pmatrix} \\begin{pmatrix} \\varphi^n & 0 \\\\ 0 & \\psi^n \\end{pmatrix} \\begin{pmatrix} \\sqrt{\\varphi} \\\\ -\\sqrt{-\\psi} \\end{pmatrix} \\\\ & = \\frac{1}{\\sqrt{5}} \\begin{pmatrix} \\sqrt{-\\psi} & \\sqrt{\\varphi} \\end{pmatrix} \\begin{pmatrix} \\sqrt{\\varphi}\\ \\varphi^n \\\\ -\\sqrt{-\\psi}\\ \\psi^n \\end{pmatrix} \\\\ & = \\frac{1}{\\sqrt{5}} \\left(\\sqrt{-\\psi \\varphi} \\varphi^n - \\sqrt{-\\varphi \\psi} \\psi^n\\right) \\\\ & = \\frac{1}{\\sqrt{5}} \\left(\\phi^n - \\psi^n\\right). \\end{align*}", "-\\frac{1}{a} & 0 \\\\ 0 & a & b & 1 & 0 & 0\\\\ -b & -c & 0 & 0 & 0 & 1 \\end{bmatrix}\\stackrel{\\longrightarrow}{bR_1+R_3} \\begin{bmatrix} 1 & 0 & -\\frac{c}{a} & 0 & -\\frac{1}{a} & 0 \\\\ 0 & a & b & 1 & 0 & 0\\\\ 0 & -c & \\frac{-bc}{a} & 0 & \\frac{-b}{a} & 1 \\end{bmatrix}$$ $$\\stackrel{\\longrightarrow}{\\frac{1}{a}R_2} \\begin{bmatrix} 1 & 0 & -\\frac{c}{a} & 0 & -\\frac{1}{a} & 0 \\\\ 0 & 1 & \\frac{b}{a} & \\frac{1}{a} & 0 & 0\\\\ 0 & -c & \\frac{-bc}{a} & 0 & \\frac{-b}{a} & 1 \\end{bmatrix}\\stackrel{\\longrightarrow}{cR_2+R_3} \\begin{bmatrix} 1 & 0 & -\\frac{c}{a} & 0 & -\\frac{1}{a} & 0 \\\\ 0 & 1 & \\frac{b}{a} & \\frac{1}{a} & 0 & 0\\\\ 0 & 0 & 0 & \\frac{c}{a} & \\frac{-b}{a} & 1 \\end{bmatrix}$$ Thus, if $a\\neq 0$ then $A$ is not invertible, since $\\mathop{rref}{(A)}\\neq I_3.$ If $a=0$, then clearly, $\\mathop{rref}{(A)}\\neq I_3$, and so $A$ is not invertible in either case. Therefore, there are no constants $a, b, c$ for which $A$ is an invertible matrix. Corollary. Let $A$ be an $n \\times n$ matrix. Consider a vector $vec b$ in $\\mathbb{R}^n.$ If $A$ is invertible, then the system $A x = b$ has the unique solution $x = A^{-1} b.$ If $A$ is non-invertible, then", "&0 \\\\ 0 &1 \\end{pmatrix}) =2\\cdot 1=2$ 3. Yes. This is the check that it preserves combinations of two members of the domain. $\\begin{array}{rl} h(r_1\\cdot\\begin{pmatrix} a_1 &b_1 \\\\ c_1 &d_1 \\end{pmatrix} +r_2\\cdot\\begin{pmatrix} a_2 &b_2 \\\\ c_2 &d_2 \\end{pmatrix}) &=h(\\begin{pmatrix} r_1a_1+r_2a_2 &r_1b_1+r_2b_2 \\\\ r_1c_1+r_2c_2 &r_1d_1+r_2d_2 \\end{pmatrix}) \\\\ &=2(r_1a_1+r_2a_2)+3(r_1b_1+r_2b_2) +(r_1c_1+r_2c_2)-(r_1d_1+r_2d_2) \\\\ &=r_1(2a_1+3b_1+c_1-d_1) +r_2(2a_2+3b_2+c_2-d_2) \\\\ &=r_1\\cdot h(\\begin{pmatrix} a_1 &b_1 \\\\ c_1 &d_1 \\end{pmatrix} +r_2\\cdot h(\\begin{pmatrix} a_2 &b_2 \\\\ c_2 &d_2 \\end{pmatrix}) \\end{array}$ 4. No. An example of a combination that is not preserved is this. $h(\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix} +\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix}) =h(\\begin{pmatrix} 2 &0 \\\\ 0 &0 \\end{pmatrix}) =4 \\quad\\text{while}\\quad h(\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix}) +h(\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix}) =1+1 =2$ This exercise is recommended for all readers. Problem 3 Show that these two maps are homomorphisms. 1. $d/dx:\\mathcal{P}_3\\to \\mathcal{P}_2$ given by $a_0+a_1x+a_2x^2+a_3x^3$ maps to $a_1+2a_2x+3a_3x^2$ 2. $\\int:\\mathcal{P}_2\\to \\mathcal{P}_3$ given by $b_0+b_1x+b_2x^2$ maps to $b_0x+(b_1/2)x^2+(b_2/3)x^3$ Are these maps inverse to each other? The check that each is a homomorphisms is routine. Here is the check for the differentiation map. $\\frac{d}{dx}(r\\cdot (a_0+a_1x+a_2x^2+a_3x^3) +s\\cdot (b_0+b_1x+b_2x^2+b_3x^3))$ $\\begin{array}{rl} &=\\frac{d}{dx}((ra_0+sb_0)+(ra_1+sb_1)x+(ra_2+sb_2)x^2 +(ra_3+sb_3)x^3) \\\\ &=(ra_1+sb_1)+2(ra_2+sb_2)x+3(ra_3+sb_3)x^2 \\\\ &=r\\cdot (a_1+2a_2x+3a_3x^2)+s\\cdot (b_1+2b_2x+3b_3x^2) \\\\ &=r\\cdot \\frac{d}{dx}(a_0+a_1x+a_2x^2+a_3x^3) +s\\cdot \\frac{d}{dx} (b_0+b_1x+b_2x^2+b_3x^3) \\end{array}$ (An alternate proof is to simply note that this is a property of differentiation that is familar from calculus.)", "b)&=\\int_0^1 Dg(t(\\tilde a,\\tilde b)+(1-t(a,b))\\begin{pmatrix}\\tilde a-a\\\\\\tilde b-b\\end{pmatrix} dt\\\\ &=\\begin{pmatrix} 1&0\\\\ 0&1\\\\ \\int_0^1 (1-t)b+t\\tilde b\\ dt&\\int_0^1 (1-t)a+t\\tilde a\\ dt \\end{pmatrix}\\begin{pmatrix}\\tilde a-a\\\\\\tilde b-b\\end{pmatrix}\\\\ &=\\begin{pmatrix} 1&0\\\\ 0&1\\\\ 1-\\frac{1}{2}(b+\\tilde b)&1-\\frac{1}{2}(a+\\tilde a) \\end{pmatrix}\\begin{pmatrix}\\tilde a-a\\\\\\tilde b-b\\end{pmatrix} \\end{eqnarray*} and noting that ${\\mathcal{V}}^*\\left({\\mathrm{col}}\\begin{pmatrix} 1&0\\\\ 0&1\\\\ 1-\\frac{1}{2}(b+\\tilde b)&1-\\frac{1}{2}(a+\\tilde a) \\end{pmatrix}\\right) \\subseteq {\\mathcal{V}}^*\\left({\\mathrm{col}}\\begin{pmatrix} 1&0\\\\ 0&1\\\\ 1&1 \\end{pmatrix}\\right)_\\downarrow.$ As an aside: ${\\mathcal{V}}^*({\\mathrm{col}}(Dg))$ is constant on $(0,1)^2$, and one might hope that this already implies that all signed differences $\\mathrm{sign}(f(g(a,b))-f(g(\\tilde a,\\tilde b)))$ are in ${\\mathcal{V}}^*({\\mathrm{col}}(Dg))$. This would render the above integration argument unnecessary. If $Dg$ were rank 1, such a result follows from the Mean Value Theorem. However, in higher rank this fails — see \\cite{DA18} for an example. References 1. Coltheart, M., “Cognitive neuropsychology and the study of reading”, in Attention and performance XI, M. I. Posner and O. S. M. Marin, eds. (1985) 2. Dunn, J. C. and Anderson, L., Signed difference analysis: Testing for structure under monotonicity, Journal of Mathematical Psychology 85 (2018), 36-54 3. Dunn, J. C. and James, R. N., Signed difference analysis: Theory and application, Journal of Mathematical Psychology 47 (2003), 389–416 Exceptional matroids in chain theorems At the end of November 2017, the Tutte Centenary Retreat was held. 32 researchers gathered in Creswick Australia to work on problems in three areas where Tutte made seminal contributions. One of those three areas was Matroid Structure", "once - $\\sqrt{NOT}\\sqrt{NOT}=NOT$. This can be proven by multiplying the $\\sqrt{NOT}$ gate matrix by itself. \\begin{aligned} \\sqrt{NOT}\\sqrt{NOT} = \\begin{pmatrix} \\frac{1+i}{2} & \\frac{1-i}{2} \\\\ \\frac{1-i}{2} & \\frac{1+i}{2} \\end{pmatrix} \\begin{pmatrix} \\frac{1+i}{2} & \\frac{1-i}{2} \\\\ \\frac{1-i}{2} & \\frac{1+i}{2} \\end{pmatrix} = \\begin{pmatrix} \\frac{1+i}{2}*\\frac{1+i}{2} + \\frac{1-i}{2}*\\frac{1-i}{2} & \\frac{1+i}{2}*\\frac{1-i}{2} + \\frac{1-i}{2}*\\frac{1+i}{2} \\\\ \\frac{1-i}{2}*\\frac{1+i}{2} + \\frac{1+i}{2}*\\frac{1-i}{2} & \\frac{1-i}{2}*\\frac{1-i}{2} + \\frac{1+i}{2}*\\frac{1+i}{2} \\end{pmatrix} = \\begin{pmatrix} \\frac{i}{2}-\\frac{i}{2} & \\frac{1}{2}+\\frac{1}{2} \\\\ \\frac{1}{2}+\\frac{1}{2} & -\\frac{i}{2}+\\frac{i}{2} \\end{pmatrix} = \\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix} \\end{aligned} The final matrix is indeed the same as the Pauli-X (NOT) gate. #### $\\sqrt{SWAP}$ Gate This gate acts on two qubits and is represented by the matrix: \\begin{aligned} \\sqrt{SWAP} = \\frac{1}{2}\\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & \\frac{1}{2}(1+i) & \\frac{1}{2}(1-i) & 0 \\\\ 0 & \\frac{1}{2}(1-i) & \\frac{1}{2}(1+i) & 0 \\\\ 0 & 0 & 0 & 1\\end{pmatrix} \\end{aligned} Similar to the $\\sqrt{NOT}$ gate, if we multiply the matrix for the $\\sqrt{SWAP}$ gate by itself, we will end up with the matrix for the $SWAP$ gate (i.e. $\\sqrt{SWAP}\\sqrt{SWAP}=SWAP$). Working out this problem to prove that the result is the same matrix as the $SWAP$ gate isn't difficult, but would take up a lot of space so we won't show it here. However, you can work this out yourself following the same steps as was shown for the $\\sqrt{NOT}$ gate,", "correct indeed, by a direct check, for $$t=-\\frac 6{11}$$ we obtain $$N=\\begin{pmatrix} 0 \\\\ -1 \\\\ 2 \\end{pmatrix} -\\frac 6{11}\\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}=\\begin{pmatrix} \\frac {18}{11} \\\\ \\frac {1}{11} \\\\ \\frac {40}{11} \\end{pmatrix}$$ $$\\overrightarrow{CN}=\\begin{pmatrix} \\frac {7}{11} \\\\ -\\frac {21}{11} \\\\ \\frac {7}{11} \\end{pmatrix}$$ $$\\begin{pmatrix} \\frac {7}{11} \\\\ -\\frac {21}{11} \\\\ \\frac {7}{11} \\end{pmatrix} \\cdot \\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}= -\\frac {21}{11}+\\frac {42}{11}-\\frac {21}{11}=0$$" ]

[ "& \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{eqnarray*} {\\renewcommand{\\arraystretch}{1.5} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} {\\renewcommand{\\arraystretch}{2} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup $(\\mathit{condition1})$ holds since the sky is red and $(\\mathit{condition2})$ because this is the answer. \\end{document} The output looks as follows: • eqnarray: • array: • array with higher stretch factor: • align: • align with higher \\thickmuskip: The problems in this example are the following: • The relation symbols used in each step may have a different horizontal length but should be aligned below each other in a centered way. Therefore, the usual AMS environments do not work in an obvious way since they only offer left- or right-aligned columns. The second solution tries to overcome this problem by", "\\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{eqnarray*} {\\renewcommand{\\arraystretch}{1.5} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} {\\renewcommand{\\arraystretch}{2} $\\begin{array}{@{} r @{\\;} >{{}} c <{{}} @{\\;} l @{}} f(x,y,z) & = & \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = & \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition1})}{=} & \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{(\\mathit{condition2})}{=} & 42 \\end{array}$ \\renewcommand{\\arraystretch}{1}} \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup \\begingroup \\begin{align*} f(x,y,z) & = \\frac{g(x,y,z)}{h(x,y,z)}\\\\ & = \\left\\lfloor\\frac{\\frac{\\frac{x}{z}}{y}+\\frac{y}{z}}{\\frac{z}{y} + \\frac{y}{x}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition1})}}{=} \\left\\lfloor\\frac{\\frac{a}{b}+\\frac{\\frac{c}{a}}{b}}{\\frac{c}{a} + \\frac{b}{a}}\\right\\rfloor\\\\ & \\stackrel{\\mathclap{(\\mathit{condition2})}}{=} 42 \\end{align*} \\endgroup $(\\mathit{condition1})$ holds since the sky is red and $(\\mathit{condition2})$ because this is the answer. \\end{document} The output looks as follows: • eqnarray: • array: • array with higher stretch factor: • align: • align with higher \\thickmuskip: The problems in this example are the following: • The relation symbols used in each step may have a different horizontal length but should be aligned below each other in a centered way. Therefore, the usual AMS environments do not work in an obvious way since they only offer left- or right-aligned columns. The second solution tries to overcome this problem", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "$\\displaystyle \\begin{pmatrix} -3 & 2 \\\\ 2 \\frac 12 & -1 \\frac 12 \\end{pmatrix} \\begin{pmatrix} 3 & 4 \\\\ 5 & 6 \\end{pmatrix}$ After multiplying, the result is: $\\displaystyle \\begin{pmatrix} (-3)(3)+(2)(5) & (-3)(4)+(2)(6) \\\\ (2 \\frac 12)(3)_+(-1 \\frac 12)(5) & (2 \\frac 12)(4)+(-1 \\frac 12)(6) \\end{pmatrix}$ Simplifying the problem gives: $\\displaystyle \\begin{pmatrix} -9+10 & -12+12 \\\\ 7 \\frac 12 -7 \\frac 12 & 10-9 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}$ Therefore, the inverse works. But wait! Note that, to fully test it, one has to try the multiplication in both orders, because, in general, changing the order of a matrix multiplication changes the answer. The definition of an inverse matrix specifies that it must work both ways. $\\displaystyle \\begin{pmatrix} 3 & 4 \\\\ 5 & 6 \\end{pmatrix} \\begin{pmatrix} -3 & 2 \\\\ 2 \\frac 12 & -1 \\frac 12 \\end{pmatrix}$ After multiplying, one gets: $\\displaystyle \\begin{pmatrix} (3)(-3)+(4)(2\\frac 12) & (3)(2)+(4)(-1 \\frac 12) \\\\ (5)(-3)_+(6)(2\\frac 12) & (5)(2)+(6)(-1\\frac 12) \\end{pmatrix}$ Simplifying the problem gives: $\\displaystyle \\begin{pmatrix} -9+10 & 6-6 \\\\ -15+15 & 10-9 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}$ So the inverse matrix works both ways. In some cases, the inverse of a square matrix does not exist. This is called a singular matrix. ## Solving Systems of Equations", "need another convention! We choose, for $a\\ne 0$ $$\\begin{array}{|c|}\\hline \\quad a^{-1}=\\frac{1}{a}. \\quad \\\\ \\hline\\end{array}$$ The choice is dictated by our desire for the three properties to be satisfied! Let's check. We plug in $n=\\pm 1$ and $m=\\pm 1$, use our convention, and then apply the above version of the corresponding property: $$\\begin{array}{r|ll|ll} \\text{Property 1:} & a^{n+m} &=a^n\\cdot a^m & n=-1,\\ m=1 & \\Longrightarrow & a^{-1+1}&=a^{-1}\\cdot a^1 & \\Longleftrightarrow & a^{0}=\\frac{1}{a}\\cdot a & \\texttt{TRUE!}\\\\ \\hline \\text{Property 2:} & a^n b^n &=(ab)^n & n=-1 & \\Longrightarrow & a^{-1} b^{-1} &=(ab)^{-1} & \\Longleftrightarrow & \\frac{1}{a}\\cdot \\frac{1}{b}=\\frac{1}{ab} & \\texttt{TRUE!}\\\\ \\hline \\text{Property 3:} & a^{nm} &=(a^n)^m & n=-1,\\ m=1 & \\Longrightarrow & a^{(-1)1} &=(a^{-1})^1 & \\Longleftrightarrow & a^{-1}=\\frac{1}{a} & \\texttt{TRUE!}\\\\ & && n=1,\\ m=-1 & \\Longrightarrow & a^{1(-1)} &=(a^1)^{-1} & \\Longleftrightarrow & a^{-1}=a^{-1} & \\texttt{TRUE!}\\\\ \\end{array}$$ The three properties are still satisfied, and we will continue to use them. Note how choose anything but $a^{-1}=1/a$ would have ruined them. This is our convention: multiplying by $a^{-1}$ means dividing by $a$. Now, the rest of the negative numbers. While multiplication by a positive integer means repeated addition, multiplication by a negative integer means repeated subtraction (the inverse of addition): $$a(-n)=-an=0-a-a-a-...-a.$$ Similarly, while exponentiation by a positive integer means repeated multiplication, exponentiation by a negative integer means repeated division (the inverse of multiplication): $$a^{-n}=1\\div a", "1\\\\ \\end {vmatrix}$$ = 20 × 1 - 12 × 1 = 20 - 12 = 8 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{8}$$$$\\begin {pmatrix} 1 & -1\\\\ -12 & 20\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-12}{8} & \\frac {20}{8}\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$ We know that, AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35}{8} - \\frac {20}{8}\\\\ \\frac {-105}{2} + \\frac {100}{2}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35 - 20}{8}\\\\ \\frac {-105 + 100}{2}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {15}{8}\\\\ \\frac {-5}{2}\\\\ \\end {pmatrix}$$ ∴ x = $$\\frac {15}{2}$$ and y = $$\\frac {-5}{2}$$ Ans Here, $$\\frac 3x$$ + $$\\frac 4y$$ = 2..........................................(1) $$\\frac 9x$$ - $$\\frac 2y$$ = $$\\frac 52$$....................(2) Given equation in the form of matrix: $$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix}", "& -1\\\\ 1 & 2\\\\ \\end {vmatrix}$$ = 2 × 2 - 1 × -1 = 4 + 1 = 5 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{5}$$$$\\begin {pmatrix} 2 & 1\\\\ -1 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{5} + \\frac {3}{5}\\\\ \\frac {-1}{5} + \\frac {6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 3}{5}\\\\ \\frac {-1 + 6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{5}\\\\ \\frac {5}{5}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, x + 2y = 8........................................(1) 2x + 3y = 11..................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ , AX = B where, A =$$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ $$\\begin", "{\\begin{aligned}m(t)&{}=\\mathbb {E} [X_{t}]\\\\[12pt]&{}=\\mathbb {E} [\\mathbb {E} (X_{t}\\mid S_{1})]\\\\[12pt]&{}=\\int _{0}^{\\infty }\\mathbb {E} (X_{t}\\mid S_{1}=s)f_{S}(s)\\,ds\\\\[12pt]&{}=\\int _{0}^{\\infty }\\mathbb {I} _{\\{t\\geq s\\}}\\left(1+\\mathbb {E} [X_{t-s}]\\right)f_{S}(s)\\,ds\\\\[12pt]&{}=\\int _{0}^{t}\\left(1+m(t-s)\\right)f_{S}(s)\\,ds\\\\[12pt]&{}=F_{S}(t)+\\int _{0}^{t}m(t-s)f_{S}(s)\\,ds,\\end{aligned}}} as required. ### Asymptotic properties ${\\displaystyle (X_{t})_{t\\geq 0}}$ and ${\\displaystyle (Y_{t})_{t\\geq 0}}$ satisfy ${\\displaystyle \\lim _{t\\to \\infty }{\\frac {1}{t}}X_{t}={\\frac {1}{\\mathbb {E} S_{1}}}}$ (strong law of large numbers for renewal processes) ${\\displaystyle \\lim _{t\\to \\infty }{\\frac {1}{t}}Y_{t}={\\frac {1}{\\mathbb {E} S_{1}}}\\mathbb {E} W_{1}}$ (strong law of large numbers for renewal-reward processes) almost surely. #### Proof First consider ${\\displaystyle (X_{t})_{t\\geq 0}}$. By definition we have: ${\\displaystyle J_{X_{t}}\\leq t\\leq J_{X_{t}+1}}$ for all ${\\displaystyle t\\geq 0}$ and so ${\\displaystyle {\\frac {J_{X_{t}}}{X_{t}}}\\leq {\\frac {t}{X_{t}}}\\leq {\\frac {J_{X_{t}+1}}{X_{t}}}}$ for all t ≥ 0. Now since ${\\displaystyle 0<\\mathbb {E} S_{i}<\\infty }$ we have: ${\\displaystyle X_{t}\\to \\infty }$ as ${\\displaystyle t\\to \\infty }$ almost surely (with probability 1). Hence: ${\\displaystyle {\\frac {J_{X_{t}}}{X_{t}}}={\\frac {J_{n}}{n}}={\\frac {1}{n}}\\sum _{i=1}^{n}S_{i}\\to \\mathbb {E} S_{1}}$ almost surely (using the strong law of large numbers); similarly: ${\\displaystyle {\\frac {J_{X_{t}+1}}{X_{t}}}={\\frac {J_{X_{t}+1}}{X_{t}+1}}{\\frac {X_{t}+1}{X_{t}}}={\\frac {J_{n+1}}{n+1}}{\\frac {n+1}{n}}\\to \\mathbb {E} S_{1}\\cdot 1}$ almost surely. Thus (since ${\\displaystyle t/X_{t}}$ is sandwiched between the two terms) ${\\displaystyle {\\frac {1}{t}}X_{t}\\to {\\frac {1}{\\mathbb {E} S_{1}}}}$ almost surely. Next consider ${\\displaystyle (Y_{t})_{t\\geq 0}}$. We have ${\\displaystyle {\\frac {1}{t}}Y_{t}={\\frac {X_{t}}{t}}{\\frac {1}{X_{t}}}Y_{t}\\to {\\frac {1}{\\mathbb {E} S_{1}}}\\cdot \\mathbb {E} W_{1}}$ almost surely (using the first result and using the law of large numbers on ${\\displaystyle Y_{t}}$). A", "\\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {vmatrix}$$ = 3 × (-2) - 9 × 4 = -6 - 36 = -42 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{-42}$$$$\\begin {pmatrix} -2 & -4\\\\ -9 & 3\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {-2}{-42} & \\frac {-4}{-42}\\\\ \\frac {-9}{-42} & \\frac {3}{-42}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$ We know that: AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{21} + \\frac {5}{21}\\\\ \\frac {9}{21} - \\frac {15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 5}{21}\\\\ \\frac {36 - 15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {7}{21}\\\\ \\frac {21}{84}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ i.e. $$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ Taking the corresponding parts of the equal matrix: $$\\frac 1x$$ = $$\\frac 13$$ ∴ x = 3 $$\\frac 1y$$ = $$\\frac 14$$", "+ i & 0 \\end{pmatrix}$. Last edited: Feb 28, 2016 danshawen likes this. 11. ### rpennerFully WiredRegistered Senior Member Messages: 4,833 $M=\\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\sqrt{2} & 0 \\\\ 0 & \\sqrt{2} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix}$ So we have: $M^{2n} = 2^{n} \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} = 2^{n} I \\\\ M^{2n+1} = 2^{n} \\begin{pmatrix} 0 & 1 - i \\\\ 1 + i & 0 \\end{pmatrix} = 2^{n} M$ So $\\textrm{exp}(M) = \\sum_{k=0}^{\\infty} \\left( \\frac{ 2^{k} }{ (2k)! } I +\\frac{ 2^{k} }{ (2k+1)! } M \\right) = \\left( \\cosh \\sqrt{2} \\right) I + \\left( \\frac{1}{\\sqrt{2}} \\sinh \\sqrt{2} \\right) M \\\\ \\quad \\quad = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$ Alternately: $\\textrm{exp}(M) = \\begin{pmatrix} - \\frac{1 -i}{2} & \\frac{1 -i}{2} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} e^{- \\sqrt{2}} & 0 \\\\ 0 & e^{\\sqrt{2}} \\end{pmatrix} \\begin{pmatrix} - \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1 +i}{2} & \\frac{1}{\\sqrt{2}} \\end{pmatrix} = \\begin{pmatrix} \\cosh \\sqrt{2} & \\frac{1 - i}{\\sqrt{2}} \\sinh \\sqrt{2} \\\\ \\frac{1 + i}{\\sqrt{2}} \\sinh \\sqrt{2} & \\cosh \\sqrt{2} \\end{pmatrix}$", "\\frac{1-\\sqrt{5}}{2} \\end{pmatrix}^{-1}\\begin{pmatrix} 1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2} \\end{pmatrix} \\begin{pmatrix}C_{1}\\\\ C_{2}\\end{pmatrix} = \\begin{pmatrix} 1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2} \\end{pmatrix}^{-1} \\begin{pmatrix}1\\\\ \\frac{-5}{4}\\end{pmatrix} \\begin{pmatrix}C_{1}\\\\ C_{2}\\end{pmatrix} = \\frac{-1}{\\sqrt{5}}\\begin{pmatrix} \\frac{1-\\sqrt{5}}{2} & -1 \\\\ \\frac{-1-\\sqrt{5}}{2} & 1 \\end{pmatrix} \\begin{pmatrix}1\\\\ \\frac{-5}{4}\\end{pmatrix} which gives us C_{1}=\\frac{-2(1-\\sqrt{5})-5}{4\\sqrt{5}}\\ (7) C_{2}=\\frac{2(1+\\sqrt{5})+5}{4\\sqrt{5}}\\ (8) Finally, by plugging (7) and (8) into (3) we get F_{n}=-4\\left (\\left ( \\frac{-2(1-\\sqrt{5})-5}{4\\sqrt{5}} \\right )\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{n}+\\left ( \\frac{2(1+\\sqrt{5})+5}{4\\sqrt{5}} \\right )\\left ( \\frac{1-\\sqrt{5}}{2} \\right )^{n}\\right ) (9) We can now “easily” compute F_{386} which is given by F_{386}=-4\\left (\\left ( \\frac{-2(1-\\sqrt{5})-5}{4\\sqrt{5}} \\right )\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{386}+\\left ( \\frac{2(1+\\sqrt{5})+5}{4\\sqrt{5}} \\right )\\left ( \\frac{1-\\sqrt{5}}{2} \\right )^{386}\\right ) Google returns F_{386} \\cong 5.2783459\\times 10^{80} which is just an approximation, but now we have a formula for the precise answer. If you wish to check for yourself that out formula (9) is indeed correct, click on the terms below to see the computation by Google: F_{0}=-4 F_{1}=5 F_{2}=1 F_{3}=6 F_{4}=7 F_{5}=13 On a side note – Raising a number to the power of 386 may seem like a long computation but it really requires only 10 multiplications! Don’t believe me? Consider this: \\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{386}=\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{256}\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{128}\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{2} To calculate \\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{256} we use the fact that 1.)\\ \\left", "\\end {vmatrix}$$ = 4 × 1 - 3 × -3 = 4 + 9 = 13 ≠ 0 It has unique solution. A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} 1 & -3\\\\ 3 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{13} & \\frac {-3}{13}\\\\ \\frac {3}{13} & \\frac {4}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{13} & \\frac {-3}{13}\\\\ \\frac {3}{13} & \\frac {4}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{13} + \\frac {21}{13}\\\\ \\frac {15}{13} - \\frac {28}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5 + 21}{13}\\\\ \\frac {15 - 28}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {26}{13}\\\\ \\frac {-13}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ ∴ x = 2 and y = -1 Ans Here, 2x - y = 1...............................(1) x + 2y = 3..............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -1\\\\ 1 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -1\\\\ 1 & 2\\\\ \\end {pmatrix}$$, X = $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "b_3-\\frac{b_1a_{31}}{a_{11}} \\\\ \\vdots \\\\ b_n-\\frac{b_1a_{n1}}{a_{11}} \\end{pmatrix}$ #### Gauss Elimination $A=\\begin{pmatrix} 1 & \\frac{a_{12}}{a_{11}} & \\frac{a_{13}}{a_{11}} & \\cdots & \\frac{a_{1n}}{a_{11}}\\\\ 0 & a_{22}' & a_{23}' & \\cdots & a_{2n}'\\\\ 0 & a_{32}' & a_{33} & \\cdots & a_{3n}'\\\\ \\vdots & \\vdots & \\vdots & \\cdots & \\vdots\\\\ 0 & a_{n2}' & a_{n3}' & \\cdots & a_{nn}' \\\\ \\end{pmatrix}$, $b=\\begin{pmatrix} \\frac{b_1}{a_{11}} \\\\ b_2' \\\\ b_3' \\\\ \\vdots \\\\ b_n' \\end{pmatrix}$ $a_{ij}'=a_{ij}-\\frac{a_{1j}a_{j1}}{a_{11}}, b_j'=b_j-\\frac{b_1a_{j1}}{a_{11}}, i,j=2,3,4,...,n$ $A=\\begin{pmatrix} 1 & \\frac{a_{12}}{a_{11}} & \\frac{a_{13}}{a_{11}} & \\cdots & \\frac{a_{1n}}{a_{11}}\\\\ 0 & 1 & \\frac{a_{23}'}{a_{22}'} & \\cdots & \\frac{a_{2n}'}{a_{22}'}\\\\ 0 & a_{32}' & a_{33} & \\cdots & a_{3n}'\\\\ \\vdots & \\vdots & \\vdots & \\cdots & \\vdots\\\\ 0 & a_{n2}' & a_{n3}' & \\cdots & a_{nn}' \\\\ \\end{pmatrix}$, $b=\\begin{pmatrix} \\frac{b_1}{a_{11}} \\\\ \\frac{b_2'}{a_{22}'} \\\\ b_3' \\\\ \\vdots \\\\ b_n' \\end{pmatrix}$ $A=\\begin{pmatrix} 1 & \\frac{a_{12}}{a_{11}} & \\frac{a_{13}}{a_{11}} & \\cdots & \\frac{a_{1n}}{a_{11}}\\\\ 0 & 1 & \\frac{a_{23}'}{a_{22}'} & \\cdots & \\frac{a_{2n}'}{a_{22}'}\\\\ 0 & 0 & a_{33}'-\\frac{a_{23}'a_{32}'}{a_{22}'} & \\cdots & a_{3n}'-\\frac{a_{2n}'a_{32}'}{a_{22}'}\\\\ \\vdots & \\vdots & \\vdots & \\cdots & \\vdots\\\\ 0 & 0 & a_{n3}'-\\frac{a_{23}'a_{n2}'}{a_{22}'} & \\cdots & a_{nn}'-\\frac{a_{2n}a_{n3}'}{a_{22}'}\\\\ \\end{pmatrix}$, $b=\\begin{pmatrix} \\frac{b_1}{a_{11}} \\\\ \\frac{b_2'}{a_{22}'} \\\\ b_3'- \\frac{b_2'a_{32}'}{a_{22}'} \\\\ \\vdots \\\\ b_n'-\\frac{b_2'a_{n2}'}{a_{22}'} \\end{pmatrix}$ #### Gauss Elimination $A=\\begin{pmatrix} 1 & \\frac{a_{12}}{a_{11}} & \\frac{a_{13}}{a_{11}} & \\cdots & \\frac{a_{1n}}{a_{11}}\\\\ 0 & 1 & \\frac{a_{23}'}{a_{22}'} & \\cdots & \\frac{a_{2n}'}{a_{22}'}\\\\ 0 & 0 & a_{33}^{''} & \\cdots", "&0 \\\\ 0 &1 \\end{pmatrix}) =2\\cdot 1=2$ 3. Yes. This is the check that it preserves combinations of two members of the domain. $\\begin{array}{rl} h(r_1\\cdot\\begin{pmatrix} a_1 &b_1 \\\\ c_1 &d_1 \\end{pmatrix} +r_2\\cdot\\begin{pmatrix} a_2 &b_2 \\\\ c_2 &d_2 \\end{pmatrix}) &=h(\\begin{pmatrix} r_1a_1+r_2a_2 &r_1b_1+r_2b_2 \\\\ r_1c_1+r_2c_2 &r_1d_1+r_2d_2 \\end{pmatrix}) \\\\ &=2(r_1a_1+r_2a_2)+3(r_1b_1+r_2b_2) +(r_1c_1+r_2c_2)-(r_1d_1+r_2d_2) \\\\ &=r_1(2a_1+3b_1+c_1-d_1) +r_2(2a_2+3b_2+c_2-d_2) \\\\ &=r_1\\cdot h(\\begin{pmatrix} a_1 &b_1 \\\\ c_1 &d_1 \\end{pmatrix} +r_2\\cdot h(\\begin{pmatrix} a_2 &b_2 \\\\ c_2 &d_2 \\end{pmatrix}) \\end{array}$ 4. No. An example of a combination that is not preserved is this. $h(\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix} +\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix}) =h(\\begin{pmatrix} 2 &0 \\\\ 0 &0 \\end{pmatrix}) =4 \\quad\\text{while}\\quad h(\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix}) +h(\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix}) =1+1 =2$ This exercise is recommended for all readers. Problem 3 Show that these two maps are homomorphisms. 1. $d/dx:\\mathcal{P}_3\\to \\mathcal{P}_2$ given by $a_0+a_1x+a_2x^2+a_3x^3$ maps to $a_1+2a_2x+3a_3x^2$ 2. $\\int:\\mathcal{P}_2\\to \\mathcal{P}_3$ given by $b_0+b_1x+b_2x^2$ maps to $b_0x+(b_1/2)x^2+(b_2/3)x^3$ Are these maps inverse to each other? The check that each is a homomorphisms is routine. Here is the check for the differentiation map. $\\frac{d}{dx}(r\\cdot (a_0+a_1x+a_2x^2+a_3x^3) +s\\cdot (b_0+b_1x+b_2x^2+b_3x^3))$ $\\begin{array}{rl} &=\\frac{d}{dx}((ra_0+sb_0)+(ra_1+sb_1)x+(ra_2+sb_2)x^2 +(ra_3+sb_3)x^3) \\\\ &=(ra_1+sb_1)+2(ra_2+sb_2)x+3(ra_3+sb_3)x^2 \\\\ &=r\\cdot (a_1+2a_2x+3a_3x^2)+s\\cdot (b_1+2b_2x+3b_3x^2) \\\\ &=r\\cdot \\frac{d}{dx}(a_0+a_1x+a_2x^2+a_3x^3) +s\\cdot \\frac{d}{dx} (b_0+b_1x+b_2x^2+b_3x^3) \\end{array}$ (An alternate proof is to simply note that this is a property of differentiation that is familar from calculus.)", "{ 4! }{ 1!3! } = 4$ $\\displaystyle \\begin{pmatrix} 4 \\\\ 2 \\end{pmatrix} = \\frac { 4! }{ 2!(4-2)! } = \\frac { 4! }{ 2!2! } = 6$ $\\displaystyle \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix} = \\frac { 4! }{ 3!(4-3)! } = \\frac { 4! }{ 3!1! } = 4$ Substituting these integers into the expansion, we have: $\\displaystyle (x+y)^4 = { x }^{ 4} + 4 { x }^{ 3}{ y } + 6 { x }^{ 2 }{ y }^{ 2 } + 4 { x }{ y }^{ 3 } + { y }^{ 4 }$ ## Finding a Specific Term The rth term of the binomial expansion can be found with the equation: ${ \\begin{pmatrix} n \\\\ r-1 \\end{pmatrix} }{ a }^{ n-(r-1) }{ b }^{ r-1 }$ . ### Learning Objectives Practice finding a specific term of a binomial expansion ### Key Takeaways #### Key Points • Properties for the binomial expansion include: the number of terms is one more than $n$ (the exponent ), and the sum of the exponents in each term adds up to $n$ . • Applying ${ \\begin{pmatrix} n \\\\ r-1 \\end{pmatrix} }{ a }^{ n-(r-1) }{ b }^{ r-1 }$ and $\\displaystyle \\begin{pmatrix} n \\\\ k \\end{pmatrix}=\\frac{n!}{(n-k)!k! }$ , one can find a particular term of", "correct indeed, by a direct check, for $$t=-\\frac 6{11}$$ we obtain $$N=\\begin{pmatrix} 0 \\\\ -1 \\\\ 2 \\end{pmatrix} -\\frac 6{11}\\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}=\\begin{pmatrix} \\frac {18}{11} \\\\ \\frac {1}{11} \\\\ \\frac {40}{11} \\end{pmatrix}$$ $$\\overrightarrow{CN}=\\begin{pmatrix} \\frac {7}{11} \\\\ -\\frac {21}{11} \\\\ \\frac {7}{11} \\end{pmatrix}$$ $$\\begin{pmatrix} \\frac {7}{11} \\\\ -\\frac {21}{11} \\\\ \\frac {7}{11} \\end{pmatrix} \\cdot \\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}= -\\frac {21}{11}+\\frac {42}{11}-\\frac {21}{11}=0$$", "= $$\\frac {1}{-13}$$$$\\begin {pmatrix} -3 & -1\\\\ -4 & 3\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 51\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153}{13} & \\frac {3}{13}\\\\ \\frac {204}{13} & \\frac {-9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153 + 3}{13}\\\\ \\frac {204 - 9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {156}{13}\\\\ \\frac {195}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 12\\\\ 5\\\\ \\end {pmatrix}$$ ∴ x = 12 and y = 5 Ans Here, 2x - 5y = 1.................................(1) 7x + 3y = 24.............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {vmatrix}$$ = 2 × 3 - 7 × (-5) = 6 + 35 = 41 ≠ 0 It has a unique solution. A-1= $$\\frac", "{1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{41}$$$$\\begin {pmatrix} 3 & 5\\\\ -7 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {5}{41}\\\\ \\frac {-7}{41} & \\frac {2}{41}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 × 1}{41} & \\frac {5 × 24}{41}\\\\ \\frac {-7 × 1}{41} & \\frac {2 × 24}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3}{41} & \\frac {120}{41}\\\\ \\frac {-7}{41} & \\frac {48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {3 + 120}{41}\\\\ \\frac {-7 + 48}{41}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {123}{41}\\\\ \\frac {41}{41}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {vmatrix} 3\\\\ 1\\\\ \\end {vmatrix}$$ ∴ x = 3 and y = 1 Ans Here, x - 8y = 39..........................................(1) 2x - 3y = 13.......................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 1 & -8\\\\ 2 & -3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 1", "We already know the value of $$\\mu$$. We now calculate $$E\\begin{pmatrix}X^2\\end{pmatrix}$$: \\begin{aligned} E\\begin{pmatrix}X^2\\end{pmatrix} &= \\int_{-\\infty}^{+\\infty}x^2.f(x)dx \\\\ &= \\int_{-\\infty}^{+\\infty} x^2.\\frac{(x-3)^2}{9}dx \\\\ &= \\int_0^3 x^2.\\frac{(x-3)^2}{9}dx \\\\ &= \\frac{1}{9}\\int_0^3 x^2.(x-3)^2dx \\\\ &= \\frac{1}{9}\\int_0^3 x^2.(x^2-6x+9)dx \\\\ &= \\frac{1}{9}\\int_0^3 (x^4-6x^3+9x^2)dx \\\\ &= \\frac{1}{9}\\begin{bmatrix} \\frac{x^5}{5}-\\frac{6}{4}x^4+\\frac{9}{3}x^3 \\end{bmatrix}_0^3 \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\frac{x^5}{5}-\\frac{3}{2}x^4+3x^3 \\end{bmatrix}_0^3 \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\begin{pmatrix} \\frac{3^5}{5}-\\frac{3}{2}\\times 3^4+3\\times 3^3 \\end{pmatrix} - 0\\end{bmatrix} \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\begin{pmatrix} \\frac{243}{5}-\\frac{243}{2}+81 \\end{pmatrix} - 0\\end{bmatrix} \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\frac{486}{10} - \\frac{1215}{10} + \\frac{810}{10}\\end{bmatrix} \\\\ & = \\frac{1}{9}\\begin{bmatrix} \\frac{81}{10} \\end{bmatrix} \\\\ & = \\frac{81}{9\\times 10} \\\\ & = \\frac{9}{10} \\\\ E\\begin{pmatrix}X^2\\end{pmatrix} & = 0.9 \\end{aligned} We can now go back to the definition of variance and calculate: \\begin{aligned} Var\\begin{pmatrix}X \\end{pmatrix} &= E\\begin{pmatrix}X^2\\end{pmatrix} - \\mu^2 \\\\ & = 0.9 - 0.75^2 \\\\ Var\\begin{pmatrix}X \\end{pmatrix} & = 0.3375 \\end{aligned} 3. We now calculate the standard deviation using the definition: $\\sigma = \\sqrt{Var\\begin{pmatrix} X\\end{pmatrix}}$ Since $$Var\\begin{pmatrix}X \\end{pmatrix} = 0.3375$$ we find: $\\sigma = \\sqrt{0.3375} = 0.581$ The standard deviation is $$\\sigma = 0.581$$. This tells that, on average amongst all the different cell phones, we can expect the time taken for the app to launch to be $$0.581$$ seconds more, or less, than the mean value $$\\mu = 0.75$$ seconds. ## Exercise 1 A continuous random variable $$X$$ has probability density function defined as: $f(x) = \\begin{cases} \\frac{x^2}{9}, \\quad 0\\leq" ]

[ "+0 # Someone plz help me with geo 0 242 4 In triangle $ABC,$ let the angle bisectors be $\\overline{BY}$ and $\\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BC$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,SW); label(\"$Y$\",Y,NW); label(\"$Z$\",Z,S); [/asy] In triangle $ABC,$ let the angle bisectors be $\\overline{BY}$ and $\\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BZ$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,SW); label(\"$Y$\",Y,NW); label(\"$Z$\",Z,S); [/asy] Aug 10, 2022 #1 +2541 0 maybe format your question properly? Aug 10, 2022 #2 +285 0 https://artofproblemsolving.com/texer/xjsiitim Question properly displayed here. Click \"With bbcode\" Aug 10, 2022 #3 +2 -1 You can update the release date yourself: log in to your GEO account, go to your Series (GSExxxx) record, and use the 'UPDATE' button which is located at the top of the page; on the 'Data Field' page, there is", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "in Olympiad, and there are some other ways of computation under this observation.) -Fanyuchen20020715 ## Solution 2 (Official MAA) Let $M$ denote the midpoint of $\\overline{BC}$. The critical claim is that $M$ is the orthocenter of $\\triangle AXY$, which has the circle with diameter $\\overline{AT}$ as its circumcircle. To see this, note that because $\\angle BXT = \\angle BMT = 90^\\circ$, the quadrilateral $MBXT$ is cyclic, it follows that $$\\angle MXA = \\angle MXB = \\angle MTB = 90^\\circ - \\angle TBM = 90^\\circ - \\angle A,$$ implying that $\\overline{MX} \\perp \\overline{AC}$. Similarly, $\\overline{MY} \\perp \\overline{AB}$. In particular, $MXTY$ is a parallelogram. $[asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, rgb(0.98,0.81,0.69)); label(\"\\omega\", O + rad*dir(45), SW); filldraw(T--Y--M--X--cycle, rgb(173/255,216/255,230/255)); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot(\"X\", X, W); dot(\"Y\", Y, E); dot(\"O\", O, W); dot(\"T\", T, S); dot(\"A\", A, N); dot(\"B\", B, W); dot(\"C\", C, E); dot(\"M\", M, N); [/asy]$ Hence, by the Parallelogram Law, $$TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).$$ But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$. Therefore $$XY^2 = \\frac13(2 \\cdot 1143-135)", "1993 USAMO Problems/Problem 2 Problem 2 Let $ABCD$ be a convex quadrilateral such that diagonals $AC$ and $BD$ intersect at right angles, and let $E$ be their intersection. Prove that the reflections of $E$ across $AB$, $BC$, $CD$, $DA$ are concyclic. Solution Diagram $[asy] import olympiad; defaultpen(0.8pt+fontsize(12pt)); pair E; E=(0,0); label('E',E,N); pair A,B,C,D; A=(10,0); B=(0,13); C=(-13,0); D=(0,-11); draw(A--B--C--D--cycle,blue); label('A',A,E); label('B',B,N); label('C',C,W); label('D',D,S); pair T,R,S,Q; T=reflect(A, B)*E; R=reflect(C, B)*E; S=reflect(C, D)*E; Q=reflect(A, D)*E; pair W,X,Y,Z; W=extension(A,D,E,Q); X=extension(A,B,E,T); Y=extension(C,B,E,R); Z=extension(C,D,E,S); draw(W--X--Y--Z--cycle,red); label('X',X,NE); label('Y',Y,NW); label('Z',Z, SW); label('W',W,SE); [/asy]$ Work Let $X$, $Y$, $Z$, $W$ be the foot of the altitute from point $E$ of $\\triangle AEB$, $\\triangle BEC$, $\\triangle CED$, $\\triangle DEA$. Note that reflection of $E$ over the 4 lines is $XYZW$ with a scale of $2$ with center $E$. Thus, if $XYZW$ is cyclic, then the reflections are cyclic. $\\angle EWA$ is right angle and so is $\\angle EXA$. Thus, $EXAW$ is cyclic with $EA$ being the diameter of the circumcircle. Follow that, $\\angle EWX\\cong\\angle EAX\\cong \\angle EAB$ because they inscribe the same angle. Similarly $\\angle EWZ\\cong \\angle EDC$, $\\angle EYX\\cong \\angle EBA$, $\\angle EYZ\\cong \\angle ECD$. Futhermore, $m\\angle XYZ+m\\angle XWZ= m\\angle EWX+m\\angle EYX+m\\angle EYZ+m\\angle EWZ=360^\\circ-m\\angle CED-m\\angle AEB=180^\\circ$. Thus, $\\angle XYZ$ and $\\angle XWZ$ are supplementary and follows that, $XYZW$ is cyclic. $\\mathbb{Q.E.D}$ Resources 1993 USAMO (Problems • Resources)", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$. Furthermore, as $\\triangle YLA\\sim \\triangle XCA$, and as $AL:LC=1:4$, we know that $LY=\\frac{h}{5}$ and $AY=\\frac{b}{10}$, so $YB=\\frac{9b}{10}$. Therefore, by the Pythagorean Theorem on $\\triangle BLY$, we get $$\\frac{81b^2}{100}+\\frac{h^2}{25}=\\frac{576}{25}.$$ Solving this system of equations yields $b=2\\sqrt{7}$ and $h=3$. Therefore, the area of the triangle is $3\\sqrt{7}$, giving us an answer of $\\boxed{010}$.", "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "+0 # In the diagram below, $A$ is on line segment $\\overline{CE}$, and $\\overline{AB}$ bisects $\\angle DAC$ (meaning that $\\overline{AB}$ splits -2 260 1 +2 In the diagram below, $A$ is on line segment $\\overline{CE}$, and $\\overline{AB}$ bisects $\\angle DAC$ (meaning that $\\overline{AB}$ splits $\\angle DAC$ into two equal angles). If $\\overline{DA}\\parallel\\overline{EF}$ and $\\angle AEF = 10 \\angle BAC - 12^\\circ$, then what is $\\angle DAC$ in degrees? [asy] pair A,B,C,D,EE,F; A = (0,0); B = rotate(164)*(1,0); C = rotate(148)*(1,0); D = (-1,0); EE = -0.4*C; F = EE + (1,0); dot(B,p=black+3bp); dot(C,p=black+3bp); dot(D,p=black+3bp); dot(F,p=black+3bp); draw(F--EE--C); draw(B--A--D); label(\"$B$\",B,NW); label(\"$A$\",A,NE); label(\"$C$\",C,N); label(\"$D$\",D,W); label(\"$E$\",EE,S); label(\"$F$\",F,E); [/asy] Mar 16, 2020 #1 +109822 -1 Anyone who tried to makes sense of that would also be sleep deprived. Mar 16, 2020", "-- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.45,0.15)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); draw( C -- F, dashed ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",Ep,NW); label(\"F\",F,S); label(\"x\",(1,1)); label(\"y\",(1.6,1)); [/asy]$ Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$. Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$. Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$. Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$. Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\\frac{15}{2}$. Then $y=x+3 = \\frac{15}{2}+3 = \\frac{21}{2}$, and the total area of the quadrilateral is $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$.", "the radius, $120=17r$, we can just solve for the radius. We get $r=\\boxed{\\textbf{(B) }\\frac{120}{17}}$. This may seem like a lengthy explanation, but doing it yourself when you know what to do, it actually takes very little time. Try it yourself! ## Solution 7 Let us draw altitude $\\overline{CD}.$ This cuts our base into line segments with length $\\dfrac{16}{2}=8.$ Finding the area of the resulting triangles gives $[\\triangle ADC] = [\\triangle BDC] = \\dfrac{8 \\cdot 15}{2} = 60.$ Since $m\\angle CDB = m\\angle CDA = 90^\\circ,$ we use the Pythagorean Theorem to find length $\\overline{AC}$: \\begin{align*} (\\overline{AD})^2+(\\overline{CD})^2 &= (\\overline{AC})^2 \\\\ 8^2+15^2 &= (\\overline{AC})^2 \\\\ 64+225 &= (\\overline{AC})^2 \\\\ 289 &= (\\overline{AC})^2 \\\\ \\sqrt{289} &= \\sqrt{(\\overline{AC})^2} \\\\ 17 &= AC. \\end{align*} Thus we have $\\overline{AC}=\\overline{BC}=17.$ $[asy] pair A, B, C, D; A=(0,0); B=(16,0); C=(8,15); D=B/2; draw(A--B--C--cycle); draw(C--D); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,S); label(\"8\",(A+D)/2,S); label(\"8\",(B+D)/2,S); label(\"15\",(C+D)/2,NE); label(\"17\",(A+C)/2,W); label(\"17\",(B+C)/2,E); [/asy]$ Letting $17$ be the base of the triangle makes our height the radius of the semicircle: $[asy] pair A, B, C, D, EE; A=(0,0); B=(16,0); C=(8,15); D=B/2; EE=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--EE); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,EE,D,25)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,S); label(\"8\",(A+D)/2,S); label(\"8\",(B+D)/2,S); label(\"15\",(C+D)/2,NE); label(\"17\",(A+C)/2,W); label(\"17\",(B+C)/2,E); label(\"r\",D--EE, N); [/asy]$ Let $r$ be the radius of the semicircle. Then we have $\\dfrac{17 \\cdot r}{2}=60 \\implies 17 \\cdot r = 120 \\implies \\dfrac{17r}{17} = \\dfrac{120}{17}", "+0 # Some old geometry problem for you to solve 0 70 1 Sorry if it is hard to read.$$In triangle ABC, let the angle bisectors be \\overline{BY} and \\overline{CZ}. Given AB = 8, AY = 6, and CY = 3, find BC. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"I\",I,SW); label(\"Y\",Y,NW); label(\"Z\",Z,S); [/asy]$$ Feb 3, 2022", "+0 # Some old geometry problem for you to solve 0 133 1 Sorry if it is hard to read.$$In triangle ABC, let the angle bisectors be \\overline{BY} and \\overline{CZ}. Given AB = 8, AY = 6, and CY = 3, find BC. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"I\",I,SW); label(\"Y\",Y,NW); label(\"Z\",Z,S); [/asy]$$ Feb 3, 2022", "to an arc with arc $120^\\circ$, the distance from $O$ to $\\overline{AB}$ is $r/2$. Let $X=\\overline{AB}\\cap \\overline{O_1O_2}$. Consider $\\triangle OO_1O_2$. The line $\\overline{AB}$ is perpendicular to $\\overline{O_1O_2}$ and passes through $X$. Let $H$ be the foot from $O$ to $\\overline{O_1O_2}$; so $HX=r/2$. We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$. Let $O_1O_2=d$. $[asy] unitsize(3cm); pointpen=black; pointfontpen=fontsize(9); pair A=dir(110), B=dir(230), C=dir(310); DPA(A--B--C--A); pair H = foot(A, B, C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D(\"O\",A,dir(A)); D(\"O_1\",B,dir(B)); D(\"O_2\",C,dir(C)); D(\"H\",H,dir(270)); D(\"X\",X,dir(225)); D(\"A\",R1,dir(180)); D(\"B\",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$ Since $X$ is on the radical axis of $\\omega_1$ and $\\omega_2$, it has equal power with respect to both circles, so $$O_1X^2 - r_1^2 = O_2X^2-r_2^2 \\implies O_1X-O_2X = \\frac{r_1^2-r_2^2}{d}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular, \\begin{align*} O_1H &= O_1X - HX = \\frac{d+\\frac{r_1^2-r_2^2}{d}}{2} - \\frac{r}{2} \\\\ O_2H &= O_2X + HX = \\frac{d-\\frac{r_1^2-r_2^2}{d}}{2} + \\frac{r}{2}. \\end{align*} We want to solve for $d$. By the Pythagorean Theorem (twice): \\begin{align*} &\\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\\\ &\\implies \\left(d+r-\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_2)^2 = \\left(d-r+\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_1)^2 \\\\ &\\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\\\ &\\implies 4dr = 8rr_2-8rr_1 \\\\ &\\implies d=2r_2-2r_1 \\end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\\boxed{672}$. ## Solution 2 (Linearity) Let $O_{1}$,", "B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.45,0.15)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); draw( C -- F, dashed ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",Ep,NW); label(\"F\",F,S); label(\"x\",(1,1)); label(\"y\",(1.6,1)); [/asy]$ Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$. Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$. Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$. Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$. Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\\frac{15}{2}$. Then $y=x+3 = \\frac{15}{2}+3 = \\frac{21}{2}$, and the total area of the quadrilateral is $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$. ## Solution 2 Connect points $E$ and $D$. Triangles $EFA$ and $FAB$ share an altitude and their areas are in the ration $3:7$. Their bases, $EF$ and $FB$, must be in the same $3:7$ ratio. Triangles", "B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label(\"7\",(1.45,0.15)); label(\"7\",(2.2,0.45)); label(\"3\",(0.45,0.35)); draw( C -- F, dashed ); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,NE); label(\"E\",Ep,NW); label(\"F\",F,S); label(\"x\",(1,1)); label(\"y\",(1.6,1)); [/asy]$ Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$. Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$. Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$. Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$. Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\\frac{15}{2}$. Then $y=x+3 = \\frac{15}{2}+3 = \\frac{21}{2}$, and the total area of the quadrilateral is $x+y = \\frac{15}{2}+\\frac{21}{2} = \\boxed{\\textbf{(D) }18}$. ## Solution 2 Connect points $E$ and $D$. Triangles $EFA$ and $FAB$ share an altitude and their areas are in the ration $3:7$. Their bases, $EF$ and $FB$, must be in the same $3:7$ ratio. Triangles" ]

[ "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "a triangle are isogonal conjugates. • If the orthocenter's triangle is acute, then the orthocenter is in the triangle; if the triangle is right, then it is on the vertex opposite the hypotenuse; and if it is obtuse, then the orthocenter is outside the triangle. • Let $ABC$ be a triangle and $H$ its orthocenter. Then the reflections of $H$ over $AB$, $BC$, and $CA$ are on the circumcircle of $ABC$: $[asy] defaultpen(fontsize(8)); pair A=(8,7), B=(0,0), C=(10,0), H=orthocenter(A,B,C), A1, B1, C1; A1 = 2*foot(A,B,C)-H; B1 = 2*foot(B,C,A)-H; C1 = 2*foot(C,A,B)-H; draw(A--B--C--cycle,black+1); draw(A--A1);draw(B--B1);draw(C--C1); draw(A1--B--C1--A--B1--C--cycle); draw(circumcircle(A,B,C)); dot(A1^^B1^^C1^^H); label(\"A\",A,(0,1));label(\"B\",B,(-1,0));label(\"C\",C,(1,0)); label(\"A'\",A1,(0,-1));label(\"B'\",B1,(1,1));label(\"C'\",C1,(-1,1)); label(\"H\",H,(-1,-1)); [/asy]$ • Even more interesting is the fact that if you take any point $P$ on the circumcircle and let $M$ to be the midpoint of $HP$, then $M$ is on the nine-point circle. ACS WASC ACCREDITED SCHOOL ## About Our Team Our History Jobs ## Site Info Terms Privacy Contact Us ## Help School Books Community ## Stay Connected Subscribe to get news and updates from AoPS, or Contact Us. © 2015 AoPS Incorporated Invalid username Login to AoPS", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "1993 USAMO Problems/Problem 2 Problem 2 Let $ABCD$ be a convex quadrilateral such that diagonals $AC$ and $BD$ intersect at right angles, and let $E$ be their intersection. Prove that the reflections of $E$ across $AB$, $BC$, $CD$, $DA$ are concyclic. Solution Diagram $[asy] import olympiad; defaultpen(0.8pt+fontsize(12pt)); pair E; E=(0,0); label('E',E,N); pair A,B,C,D; A=(10,0); B=(0,13); C=(-13,0); D=(0,-11); draw(A--B--C--D--cycle,blue); label('A',A,E); label('B',B,N); label('C',C,W); label('D',D,S); pair T,R,S,Q; T=reflect(A, B)*E; R=reflect(C, B)*E; S=reflect(C, D)*E; Q=reflect(A, D)*E; pair W,X,Y,Z; W=extension(A,D,E,Q); X=extension(A,B,E,T); Y=extension(C,B,E,R); Z=extension(C,D,E,S); draw(W--X--Y--Z--cycle,red); label('X',X,NE); label('Y',Y,NW); label('Z',Z, SW); label('W',W,SE); [/asy]$ Work Let $X$, $Y$, $Z$, $W$ be the foot of the altitute from point $E$ of $\\triangle AEB$, $\\triangle BEC$, $\\triangle CED$, $\\triangle DEA$. Note that reflection of $E$ over the 4 lines is $XYZW$ with a scale of $2$ with center $E$. Thus, if $XYZW$ is cyclic, then the reflections are cyclic. $\\angle EWA$ is right angle and so is $\\angle EXA$. Thus, $EXAW$ is cyclic with $EA$ being the diameter of the circumcircle. Follow that, $\\angle EWX\\cong\\angle EAX\\cong \\angle EAB$ because they inscribe the same angle. Similarly $\\angle EWZ\\cong \\angle EDC$, $\\angle EYX\\cong \\angle EBA$, $\\angle EYZ\\cong \\angle ECD$. Futhermore, $m\\angle XYZ+m\\angle XWZ= m\\angle EWX+m\\angle EYX+m\\angle EYZ+m\\angle EWZ=360^\\circ-m\\angle CED-m\\angle AEB=180^\\circ$. Thus, $\\angle XYZ$ and $\\angle XWZ$ are supplementary and follows that, $XYZW$ is cyclic. $\\mathbb{Q.E.D}$ Resources 1993 USAMO (Problems • Resources)", "refer to the following diagram. Note that the area bounded by $\\overline{A_i}{A_j}$ and the arc $A_iA_j$ is fixed, so we only need to consider the relevant triangles. $[asy] size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label(\"P\", P, dir(-75)); label(\"A_1\", dir(112.5), dir(112.5)); label(\"A_2\", dir(112.5-45), dir(112.5-45)); label(\"A_3\", dir(112.5-90), dir(112.5-90)); label(\"A_4\", dir(112.5-135), dir(112.5-135)); label(\"A_5\", dir(112.5-180), dir(112.5-180)); label(\"A_6\", dir(112.5-225), dir(112.5-225)); label(\"A_7\", dir(112.5-270), dir(112.5-270)); label(\"A_8\", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); [/asy]$ Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\\triangle PA_1A_2$ by $1$. We can see that $P$ was moved down by $\\tfrac{1}{7}-\\tfrac{1}{8}=\\tfrac{1}{56}$ units to make the area defined by $P$, $A_1$, and $A_2$ $\\tfrac{1}{7}$. Similarly, $P$ was moved right by $\\tfrac{1}{8}-\\tfrac{1}{9}=\\tfrac{1}{72}$ to make the area defined by $P$, $A_3$, and $A_4$ $\\tfrac{1}{9}$. This means that $P$ has coordinates $(\\tfrac{1}{72},-\\tfrac{1}{56})$. Now, we need to consider how this displacement in $P$ affected the area defined by $P$, $A_6$, and $A_7$. This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\\tfrac{\\sqrt{2}}{2}(\\tfrac{1}{56}-\\tfrac{1}{72})=\\tfrac{\\sqrt{2}}{504}$. Remembering the definition of", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "area bounded by $\\overline{A_i}{A_j}$ and the arc $A_iA_j$ is fixed, so we only need to consider the relevant triangles. $[asy] size(7cm); draw(Circle((0,0),1)); pair P = (0.1,-0.15); filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red); filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red); filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red); dot(P); for(int i=0; i<8; ++i) { draw(dir(22.5+45i)--dir(67.5+45i)); draw((0,0)--dir(22.5+45i),gray+dashed); } draw(dir(135)--dir(-45),blue+linewidth(1)); label(\"P\", P, dir(-75)); label(\"A_1\", dir(112.5), dir(112.5)); label(\"A_2\", dir(112.5-45), dir(112.5-45)); label(\"A_3\", dir(112.5-90), dir(112.5-90)); label(\"A_4\", dir(112.5-135), dir(112.5-135)); label(\"A_5\", dir(112.5-180), dir(112.5-180)); label(\"A_6\", dir(112.5-225), dir(112.5-225)); label(\"A_7\", dir(112.5-270), dir(112.5-270)); label(\"A_8\", dir(112.5-315), dir(112.5-315)); dot(dir(112.5)^^dir(112.5-45)^^dir(112.5-90)^^dir(112.5-135)^^dir(112.5-180)^^dir(112.5-225)^^dir(112.5-270)^^dir(112.5-315)); [/asy]$ Define one arbitrary unit as the distance that you need to move $P$ from $A_1A_2$ to change the area of $\\triangle PA_1A_2$ by $1$. We can see that $P$ was moved down by $\\tfrac{1}{7}-\\tfrac{1}{8}=\\tfrac{1}{56}$ units to make the area defined by $P$, $A_1$, and $A_2$ $\\tfrac{1}{7}$. Similarly, $P$ was moved right by $\\tfrac{1}{8}-\\tfrac{1}{9}=\\tfrac{1}{72}$ to make the area defined by $P$, $A_3$, and $A_4$ $\\tfrac{1}{9}$. This means that $P$ has coordinates $(\\tfrac{1}{72},-\\tfrac{1}{56})$. Now, we need to consider how this displacement in $P$ affected the area defined by $P$, $A_6$, and $A_7$. This is equivalent to finding the shortest distance between $P$ and the blue line in the diagram (as $K=\\tfrac{1}{2}bh$ and the blue line represents $h$ while $b$ is fixed). Using an isosceles right triangle, one can find the that shortest distance between $P$ and this line is $\\tfrac{\\sqrt{2}}{2}(\\tfrac{1}{56}-\\tfrac{1}{72})=\\tfrac{\\sqrt{2}}{504}$. Remembering the definition of our unit, this yields a final area of", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "# Orthocenter The orthocenter of a triangle is the point of intersection of its altitudes. It is conventionally denoted $H$. ## Proof of Existence Note: The orthocenter's existence is a trivial consequence of the trigonometric version Ceva's Theorem; however, the following proof, due to Leonhard Euler, is much more clever, illuminating and insightful. $[asy] defaultpen(fontsize(8)); pair A=(8,7), B=(0,0), C=(10,0), A1 = (B+C)/2, O = circumcenter(A,B,C), G = (A+B+C)/3, H = 3*G-2*O; draw(A--B--C--cycle); draw(A--G--H--cycle); draw(A1--G--O--cycle); label(\"A\",A,(0,1));label(\"B\",B,(0,-1));label(\"C\",C,(0,-1));label(\"G\",G,(1,-1));label(\"H\",H,(0,-1));label(\"O\",O,(-1,1));label(\"A'\",A1,(0,-1));dot(H); [/asy]$ Consider a triangle $ABC$ with circumcenter $O$ and centroid $G$. Let $A'$ be the midpoint of $BC$. Let $H$ be the point such that $G$ is between $H$ and $O$ and $HG = 2 GO$. Then the triangles $AGH$, $A'GO$ are similar by side-angle-side similarity. It follows that $AH$ is parallel to $OA'$ and is therefore perpendicular to $BC$; i.e., it is the altitude from $A$. Similarly, $BH$, $CH$, are the altitudes from $B$, ${C}$. Hence all the altitudes pass through $H$. Q.E.D. This proof also gives us the result that the orthocenter, centroid, and circumcenter are collinear, in that order, and in the proportions described above. The line containing these three points is known as the Euler line of the triangle, and also contains the triangle's de Longchamps point and nine-point center. ## Properties • The orthocenter and the circumcenter of", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "## anonymous one year ago In square $ABCD$, $E$ is the midpoint of $\\overline{BC}$, and $F$ is the midpoint of $\\overline{CD}$. Let $G$ be the intersection of $\\overline{AE}$ and $\\overline{BF}$. Prove that $DG = AB$. • This Question is Open 1. anonymous Asymptote code below [asy] unitsize(2.5 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (0,1); C = (1,1); D = (1,0); E = (B + C)/2; F = (C + D)/2; G = extension(A,E,B,F); draw(A--B--C--D--cycle); draw(A--E); draw(B--F); draw(D--G); label(\"$A$\", A, SW); label(\"$B$\", B, NW); label(\"$C$\", C, NE); label(\"$D$\", D, SE); label(\"$E$\", E, N); label(\"$F$\", F, dir(0)); label(\"$G$\", G, WSW); [/asy]", "# 2014 USAJMO Problems/Problem 6 ## Problem Let $ABC$ be a triangle with incenter $I$, incircle $\\gamma$ and circumcircle $\\Gamma$. Let $M,N,P$ be the midpoints of sides $\\overline{BC}$, $\\overline{CA}$, $\\overline{AB}$ and let $E,F$ be the tangency points of $\\gamma$ with $\\overline{CA}$ and $\\overline{AB}$, respectively. Let $U,V$ be the intersections of line $EF$ with line $MN$ and line $MP$, respectively, and let $X$ be the midpoint of arc $BAC$ of $\\Gamma$. (a) Prove that $I$ lies on ray $CV$. (b) Prove that line $XI$ bisects $\\overline{UV}$. ## Solution $[asy] unitsize(5cm); import olympiad; pair A, B, C, I, M, N, P, E, F, U, V, X, R; A = dir(190); B = dir(120); C = dir(350); I = incenter(A, B, C); label(\"A\", A, W); label(\"B\", B, dir(90)); label(\"C\", C, dir(0)); dot(I); label(\"I\", I, SSE); draw(A--B--C--cycle); real r, R; r = inradius(A, B, C); R = circumradius(A, B, C); path G, g; G = circumcircle(A, B, C); g = incircle(A, B, C); draw(G); draw(g); label(\"\\Gamma\", dir(35), dir(35)); label(\"\\gamma\", 2/3 * dir(125)); M = (B+C)/2; N = (A+C)/2; P = (A+B)/2; label(\"M\", M, NE); label(\"N\", N, SE); label(\"P\", P, W); E = tangent(A, I, r, 1); F = tangent(A, I, r, 2); label(\"E\", E, SW); label(\"F\", F, WNW); U = extension(E, F, M, N); V = intersectionpoint(P--M, F--E); label(\"U\", U, S); label(\"V\",", "Extended Law of Sine to find that the length of the circumradius of $\\triangle ABC$ is $\\tfrac{7\\sqrt{3}}{3}$. $[asy] size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); pair X=MP(\"X\",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot(\"W\",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label(\"\\gamma\",gamma,dir(180)); label(\"u\",X--D,dir(60)); label(\"v\",D--F,dir(70)); [/asy]$ Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of $\\omega$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\tfrac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=u$, and $DF=v$. Then $XE^2-XF^2=EF^2=DE^2-DF^2$, so we get $$(u+v)^2-v^2=\\frac{196}{3}-\\frac{2401}{64}$$ which simplifies to $$u^2+2uv = \\frac{5341}{192}.$$ By Power of Point $D$, $uv=BD \\cdot DC=735/64$. Combining with above, we get $$XD^2=u^2=\\frac{931}{192}.$$ Note that $\\triangle XDE\\sim \\triangle ADF$ and the ratio of similarity is $\\rho = AD : XD = \\tfrac{15}{8}:u$. Then $AF=\\rho\\cdot XE = \\tfrac{15}{8u}\\cdot R$ and $$AF^2 = \\frac{225}{64}\\cdot \\frac{R^2}{u^2} = \\frac{900}{19}.$$ The answer is $900+19=\\boxed{919}$. -Solution by TheBoomBox77 ## Solution 4 Use Law of Cosines in $\\triangle ABC$ to get $\\angle BAC=120^\\circ$. Because $AE$ bisects $\\angle A$, $E$ is the midpoint of major arc $BC$ so $BE=CE,$ and $\\angle BEC=60^\\circ.$ Thus $\\triangle BEC$ is equilateral. Notice now that $\\angle BFC=\\angle BFE= 60^\\circ.$ But $\\angle DFE=90^\\circ$ so $FD$ bisects $\\angle BFC.$ Thus, $$\\frac{BF}{CF}=\\frac{BD}{CD}=\\frac{BA}{CA}=\\frac{5}{3}.$$ Let $BF=5k, CF=3k.$ Use Law of Cosines on $\\triangle BFC$ to get$$25k^2+9k^2-15k^2 =", "$BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 pair A,B,C,D,E; B=(0,0); C=(14;0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"$A$\",A,N); label(\"$B$\",B,S); label(\"$C$\",C,S); label(\"$E$\",E,S); label(\"$D$\",D,S); label(\"$13$\",A--B,NW); label(\"$15$\",A--C,NE); label(\"$14$\",B--C,S); (Error compiling LaTeX. 32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy: 7.6: syntax error error: could not load module '32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy') Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, thus the area of $\\Delta ADC=\\Delta ABD$, which is half the area of $\\Delta ABC$, so $\\Delta ADC=\\Delta ABD=42$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "#### Triangles and their classification ##### Angle Classifacation Acute Triangle All angles are less than 90° Right Triangle Has a right angle (90°) Obtuse Triangle Has an angle more than 90° ##### Side Classifacation Equilateral Triangle Three equal sides Isosceles Triangle Two equal sides Two equal angles Scalene Triangle No equal sides No equal angles This part shows what Proves ussually look like ## Proof that any nonperfect square positive integer is irrational Let us assume that $\\sqrt{n}$ is rational where $n$ is a nonperfect square positive integer. Then it can be written as $\\frac{p}{q}=\\sqrt{n} \\Rightarrow \\frac{p^2}{q^2}=n \\Rightarrow (q^2)n=p^2$. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore $\\sqrt{n}$ is irrational. ## Proof of the Pythagorean Theorem $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Proof", "# Carnot's Theorem \"\" Carnot's Theorem\"\" states that in a triangle $ABC$, the signed sum of perpendicular distances from the circumcenter $O$ to the sides (i.e., signed lengths of the pedal lines from $O$) is: $OO_A+OO_B+OO_C=R+r$ $[asy] pair a,b,c,O,i,d,f,g; a=(0,0); b=(4,0); c=(1,3); O=circumcenter(a,b,c); i=incenter(a,b,c); draw(a--b--c--cycle); draw(circumcircle(a,b,c)); draw(incircle(a,b,c)); dot(i); dot(O); label(\"A\",a,W); label(\"B\",b,E); label(\"C\",c,N); label(\"I\",i,N); label(\"O\",O,N); d=foot(O,b,c); dot(d); draw(O--d); label(\"O_A\",d,N); draw(rightanglemark(O,d,b)); f=foot(O,a,b); dot(f); draw(O--f); draw(rightanglemark(O,f,a)); label(\"O_C\",f,S); g=foot(O,c,a); dot(g); draw(O--g); draw(rightanglemark(O,g,a)); label(\"O_B\",g,W); [/asy]$ where r is the inradius and R is the circumradius. The sign of the distance is chosen to be negative iff the entire segment OO_i lies outside the triangle. Explicitly, $OO_A+OO_B+OO_C=\\frac{abc(|\\cos{A}|+|\\cos{B}|+|\\cos{C}|)}{4|\\Delta|}$ where $\\Delta$ is the area of triangle $\\Delta ABC$. Weisstein, Eric W. \"Carnot's Theorem.\" From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/CarnotsTheorem.html # Below is not Carnot's Theorem Carnot's Theorem states that in a triangle $ABC$ with $A_1\\in BC$, $B_1\\in AC$, and $C_1\\in AB$, perpendiculars to the sides $BC$, $AC$, and $AB$ at $A_1$, $B_1$, and $C_1$ are concurrent if and only if $A_1B^2+C_1A^2+B_1C^2=A_1C^2+C_1B^2+B_1A^2$. #### Proof Only if: Assume that the given perpendiculars are concurrent at $M$. Then, from the Pythagorean Theorem, $A_1B^2=BM^2-MA_1^2$, $C_1A^2=AM^2-MC_1^2$, $B_1C^2=CM^2-MB_1^2$, $A_1C^2=MC^2-MA_1^2$, $C_1B^2=MB^2-MC_1^2$, and $B_1A^2=AM^2-MB_1^2$. Substituting each and every one of these in and simplifying gives the desired result. If: Consider the intersection of the perpendiculars from $A_1$ and $B_1$. Call this intersection point $N$," ]

[ "+0 # Someone plz help me with geo 0 242 4 In triangle $ABC,$ let the angle bisectors be $\\overline{BY}$ and $\\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BC$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,SW); label(\"$Y$\",Y,NW); label(\"$Z$\",Z,S); [/asy] In triangle $ABC,$ let the angle bisectors be $\\overline{BY}$ and $\\overline{CZ}$. Given $AB = 8$, $AY = 6$, and $CY = 3$, find $BZ$. [asy] pair A,B,C,X,Y,Z,I; A= (0,0); B = (1,0); C = (0.8,0.7); X = intersectionpoint(B--C , A -- (bisectorpoint(B,A,C))); Y = intersectionpoint(A--C, B -- scale(6)*( (bisectorpoint(C,B,A)) - B)); I = intersectionpoint(A--X, B--Y); Z = extension(A,B,C,I); draw(A--B--C--cycle); draw(B--Y); draw(C--Z); label(\"$A$\",A,SW); label(\"$B$\",B,SE); label(\"$C$\",C,N); label(\"$I$\",I,SW); label(\"$Y$\",Y,NW); label(\"$Z$\",Z,S); [/asy] Aug 10, 2022 #1 +2541 0 maybe format your question properly? Aug 10, 2022 #2 +285 0 https://artofproblemsolving.com/texer/xjsiitim Question properly displayed here. Click \"With bbcode\" Aug 10, 2022 #3 +2 -1 You can update the release date yourself: log in to your GEO account, go to your Series (GSExxxx) record, and use the 'UPDATE' button which is located at the top of the page; on the 'Data Field' page, there is", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "in Olympiad, and there are some other ways of computation under this observation.) -Fanyuchen20020715 ## Solution 2 (Official MAA) Let $M$ denote the midpoint of $\\overline{BC}$. The critical claim is that $M$ is the orthocenter of $\\triangle AXY$, which has the circle with diameter $\\overline{AT}$ as its circumcircle. To see this, note that because $\\angle BXT = \\angle BMT = 90^\\circ$, the quadrilateral $MBXT$ is cyclic, it follows that $$\\angle MXA = \\angle MXB = \\angle MTB = 90^\\circ - \\angle TBM = 90^\\circ - \\angle A,$$ implying that $\\overline{MX} \\perp \\overline{AC}$. Similarly, $\\overline{MY} \\perp \\overline{AB}$. In particular, $MXTY$ is a parallelogram. $[asy] defaultpen(fontsize(8pt)); unitsize(0.8cm); pair A = (0,0); pair B = (-1.26,-4.43); pair C = (-1.26+3.89, -4.43); pair M = (B+C)/2; pair O = circumcenter(A,B,C); pair T = (0.68, -6.49); pair X = foot(T,A,B); pair Y = foot(T,A,C); path omega = circumcircle(A,B,C); real rad = circumradius(A,B,C); filldraw(A--B--C--cycle, rgb(0.98,0.81,0.69)); label(\"\\omega\", O + rad*dir(45), SW); filldraw(T--Y--M--X--cycle, rgb(173/255,216/255,230/255)); draw(M--T); draw(X--Y); draw(B--T--C); draw(A--X--Y--cycle); draw(omega); dot(\"X\", X, W); dot(\"Y\", Y, E); dot(\"O\", O, W); dot(\"T\", T, S); dot(\"A\", A, N); dot(\"B\", B, W); dot(\"C\", C, E); dot(\"M\", M, N); [/asy]$ Hence, by the Parallelogram Law, $$TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).$$ But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$. Therefore $$XY^2 = \\frac13(2 \\cdot 1143-135)", "\\angle QFR.$$ As in case 1 (by a similar approach), we can conclude that $$\\angle DAF = \\angle QPR.$$ Therefore $\\triangle ABC \\sim \\triangle QPR$, as in case 1. - Excellent. Thank you. – EuYu Jan 9 '13 at 20:24 @EuYu. You're welcome:) – RicardoCruz Jan 9 '13 at 20:28 Fact. If $x$, $y$, $z$ are points on the complex plane, then the circumcenter of $\\triangle xyz$ is the point $$i \\frac{x(z\\overline{z} - y\\overline{y})+ y(x\\overline{x}-z\\overline{z})+z(y\\overline{y}-x\\overline{x})}{x(\\overline{z} - \\overline{y})+ y(\\overline{x}-\\overline{z})+z(\\overline{y}-\\overline{x})}$$ where \"$\\overline{x}$\" indicates the complex conjugate of $x$. Let us consider $\\triangle abc$, defining $d$, $e$, $f$ on sides $bc$, $ca$, $ab$, respectively: $$d := b + \\alpha(c-b) \\qquad e := c+\\beta(a-c) \\qquad f := a+\\gamma(b-a)$$ for real scalars $\\alpha, \\beta, \\gamma$. Let $p$, $q$, $r$ be the respective circumcenters of $\\triangle aef$, $\\triangle dbf$, $\\triangle dec$; for simplicity, we take $a$, $b$, $c$ on the unit circle so that $\\overline{a}=1/a$, etc. Then, \\begin{align} p &= i\\;\\frac{c(b-a)+ \\beta b ( a-c) + \\gamma c( a - b )}{b-c} \\\\[6pt] q &= i\\;\\frac{a(c-b)+ \\gamma c ( b-a) + \\alpha a( b - c )}{c-a} \\\\[6pt] r &= i\\;\\frac{b(a-c)+ \\alpha a ( c-b) + \\beta b( c - a )}{a-b} \\end{align} We deduce $$\\frac{p-q}{a-b} = \\frac{q-r}{b-c} = \\frac{r-p}{c-a} =: u$$ which, by taking the modulus, proves $\\triangle abc \\sim \\triangle pqr$. A", "F=extension(A,B,C,P); draw(circle, black); draw(A--B--C--cycle); draw(B--E1--C); draw(C--F--B); draw(A--P); draw(D--E1--F--cycle, dashed); pair G = extension(O,D,F,E1); draw(O--G,dashed); label(\"A\", A, N); label(\"B\", B, W); label(\"C\", C, E); label(\"P\", P, S); label(\"D\", D, NW); label(\"E\", E1, SE); label(\"F\", F, SW); dot(\"O\", O, SE); [/asy]$ We'll use coordinates and shoelace. Let the origin be the midpoint of $BC$. Let $AB=2$, and $BF = 2x$, then $F=(-x-1,-x\\sqrt{3})$. Using the facts $\\triangle{CBP} \\sim \\triangle{CFB}$ and $\\triangle{BCP} \\sim \\triangle{BEC}$, we have $BF * CE = BC^2$, so $CE = \\frac{1}{2x}$, and $E = (\\frac{1}{x}+1,-\\frac{\\sqrt{3}}{x})$. The slope of $FE$ is $$k = \\frac{-\\frac{\\sqrt{3}}{x} + x\\sqrt{3}}{2+\\frac{1}{x}+x}$$ It is well-known that $\\triangle{DFE}$ is self-polar, so $FE$ is the polar of $D$, i.e., $OD$ is perpendicular to $FE$. Therefore, the slope of $OD$ is $-\\frac{1}{k}$. Since $O=(0,\\frac{1}{\\sqrt{3}})$, we get the x-coordinate of $D$, $x_D = \\frac{k}{\\sqrt{3}}$, i.e., $D = (\\frac{k}{\\sqrt{3}},0)$. Using shoelace, $$2[\\triangle{FDE}] = \\frac{k}{\\sqrt{3}}(-x\\sqrt{3})+(-x-1)(-\\frac{\\sqrt{3}}{x})- (-x\\sqrt{3})(\\frac{1}{x}+1) - (-\\frac{\\sqrt{3}}{x})\\frac{k}{\\sqrt{3}}$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{1}{x}+x) + k(\\frac{1}{x} - x)$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+(\\frac{1}{x}+x)^2-(\\frac{1}{x}-x)^2} {2+\\frac{1}{x}+x})$$ $$= 2\\sqrt{3} + \\sqrt{3}(\\frac{2(\\frac{1}{x}+x)+4}{2+\\frac{1}{x}+x})$$ $$= 4\\sqrt{3}$$ So $[\\triangle{FDE}] = 2\\sqrt{3} = 2[\\triangle{ABC}]$. Q.E.D By Mathdummy. ## Solution 4 Without the nasty computations Note that $\\angle{APB}=\\angle{FPB}=\\angle{EPC}=\\angle{APC} = 60$. We will use a special version of Stewart's theorem for angle bisectors in triangle with an 120 angle to calculate various side lengths. Let $BP = x$ and $CP = y$. Then, From", "of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2,", "the radius, $120=17r$, we can just solve for the radius. We get $r=\\boxed{\\textbf{(B) }\\frac{120}{17}}$. This may seem like a lengthy explanation, but doing it yourself when you know what to do, it actually takes very little time. Try it yourself! ## Solution 7 Let us draw altitude $\\overline{CD}.$ This cuts our base into line segments with length $\\dfrac{16}{2}=8.$ Finding the area of the resulting triangles gives $[\\triangle ADC] = [\\triangle BDC] = \\dfrac{8 \\cdot 15}{2} = 60.$ Since $m\\angle CDB = m\\angle CDA = 90^\\circ,$ we use the Pythagorean Theorem to find length $\\overline{AC}$: \\begin{align*} (\\overline{AD})^2+(\\overline{CD})^2 &= (\\overline{AC})^2 \\\\ 8^2+15^2 &= (\\overline{AC})^2 \\\\ 64+225 &= (\\overline{AC})^2 \\\\ 289 &= (\\overline{AC})^2 \\\\ \\sqrt{289} &= \\sqrt{(\\overline{AC})^2} \\\\ 17 &= AC. \\end{align*} Thus we have $\\overline{AC}=\\overline{BC}=17.$ $[asy] pair A, B, C, D; A=(0,0); B=(16,0); C=(8,15); D=B/2; draw(A--B--C--cycle); draw(C--D); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,S); label(\"8\",(A+D)/2,S); label(\"8\",(B+D)/2,S); label(\"15\",(C+D)/2,NE); label(\"17\",(A+C)/2,W); label(\"17\",(B+C)/2,E); [/asy]$ Letting $17$ be the base of the triangle makes our height the radius of the semicircle: $[asy] pair A, B, C, D, EE; A=(0,0); B=(16,0); C=(8,15); D=B/2; EE=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--EE); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,EE,D,25)); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,N); label(\"D\",D,S); label(\"8\",(A+D)/2,S); label(\"8\",(B+D)/2,S); label(\"15\",(C+D)/2,NE); label(\"17\",(A+C)/2,W); label(\"17\",(B+C)/2,E); label(\"r\",D--EE, N); [/asy]$ Let $r$ be the radius of the semicircle. Then we have $\\dfrac{17 \\cdot r}{2}=60 \\implies 17 \\cdot r = 120 \\implies \\dfrac{17r}{17} = \\dfrac{120}{17}", "the sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{\\textbf{(B) }30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "1993 USAMO Problems/Problem 2 Problem 2 Let $ABCD$ be a convex quadrilateral such that diagonals $AC$ and $BD$ intersect at right angles, and let $E$ be their intersection. Prove that the reflections of $E$ across $AB$, $BC$, $CD$, $DA$ are concyclic. Solution Diagram $[asy] import olympiad; defaultpen(0.8pt+fontsize(12pt)); pair E; E=(0,0); label('E',E,N); pair A,B,C,D; A=(10,0); B=(0,13); C=(-13,0); D=(0,-11); draw(A--B--C--D--cycle,blue); label('A',A,E); label('B',B,N); label('C',C,W); label('D',D,S); pair T,R,S,Q; T=reflect(A, B)*E; R=reflect(C, B)*E; S=reflect(C, D)*E; Q=reflect(A, D)*E; pair W,X,Y,Z; W=extension(A,D,E,Q); X=extension(A,B,E,T); Y=extension(C,B,E,R); Z=extension(C,D,E,S); draw(W--X--Y--Z--cycle,red); label('X',X,NE); label('Y',Y,NW); label('Z',Z, SW); label('W',W,SE); [/asy]$ Work Let $X$, $Y$, $Z$, $W$ be the foot of the altitute from point $E$ of $\\triangle AEB$, $\\triangle BEC$, $\\triangle CED$, $\\triangle DEA$. Note that reflection of $E$ over the 4 lines is $XYZW$ with a scale of $2$ with center $E$. Thus, if $XYZW$ is cyclic, then the reflections are cyclic. $\\angle EWA$ is right angle and so is $\\angle EXA$. Thus, $EXAW$ is cyclic with $EA$ being the diameter of the circumcircle. Follow that, $\\angle EWX\\cong\\angle EAX\\cong \\angle EAB$ because they inscribe the same angle. Similarly $\\angle EWZ\\cong \\angle EDC$, $\\angle EYX\\cong \\angle EBA$, $\\angle EYZ\\cong \\angle ECD$. Futhermore, $m\\angle XYZ+m\\angle XWZ= m\\angle EWX+m\\angle EYX+m\\angle EYZ+m\\angle EWZ=360^\\circ-m\\angle CED-m\\angle AEB=180^\\circ$. Thus, $\\angle XYZ$ and $\\angle XWZ$ are supplementary and follows that, $XYZW$ is cyclic. $\\mathbb{Q.E.D}$ Resources 1993 USAMO (Problems • Resources)", "# Solution of Problem 2, Grade 9, RNO 2007 Let $ABC$ be an acute angled triangle and point $M$ chosen differently from $A,B,C$. Prove that $M$ is the orthocenter of triangle $ABC$ if and only if $\\dfrac{BC}{MA}\\cdot\\overrightarrow{MA}+\\dfrac{CA}{MB}\\cdot\\overrightarrow{MB}+\\dfrac{AB}{MC}\\cdot\\overrightarrow{MC}= \\overrightarrow{0}.(*)$ Solution of my students. If $M$ is the orthocenter of triangle $ABC$ then by a well-known theorem, we have $\\tan A\\cdot\\overrightarrow{MA}+\\tan B\\cdot\\overrightarrow{MB}+\\tan C\\cdot\\overrightarrow{MC}=\\overrightarrow{0} (1)$ By triangle is acute: $MA=2R\\cos A, MB=2R\\cos B, MC=2R\\cos C$ therefore $\\dfrac{BC}{MA}=\\tan A, \\dfrac{CA}{MB}=\\tan B,\\dfrac{AB}{MC}=\\tan C$, so by $(1)$ we have $(*)$ is proved. If $(*)$ is true then by $\\dfrac{BC}{MA}>0, \\dfrac{CA}{MB}>0,\\dfrac{AB}{MC}>0$, $M$ in interior of the triangle $ABC$. By a well-known theorem, we have $S[MBC]\\cdot\\overrightarrow{MA}+S[MCA]\\cdot\\overrightarrow{MB}+S[MAB]\\cdot \\overrightarrow{MC}=\\overrightarrow{0}(2)$ Use $(2)$ and $(*)$ we have $\\dfrac{BC}{MA\\cdot S[MBC]}=\\dfrac{CA}{MB\\cdot S[MCA]}=\\dfrac{AB}{MC\\cdot S[MAB]}$ But $2S[MBC]=MB\\cdot BC\\cdot\\sin\\angle MBC$ and $2S[MCA]=MA\\cdot CA\\sin\\angle MAC$, hence $\\angle MBC=\\angle MAC$. Similary, $\\angle MBA=\\angle MCA,\\angle MAB=\\angle MCB$. Now, easy see that $M$ is orthocenter of the triangle $ABC$. Note: If $\\sin x=\\sin y$ and $0 then $x=y$.", "sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B =", "[] x=intersectionpoints(seg1,seg2); fill(circle(x[0],0.1),black); label(\"A\",(0,3),NW); label(\"B\",(6,0),SE); label(\"C\",(4,2),NE); label(\"D\",(2,0),S); label(\"E\",x[0],N); label(\"F\",(2,2),NE); draw((2,2)--(4,2)); draw((6,0)--(2,0)); [/asy]$ Drawing line $\\overline{BD}$ and parallel line $\\overline{CF}$, we see that $\\triangle FCE \\sim \\triangle BDE$ by AA similarity. Thus $\\frac{FE}{EB} = \\frac{FC}{DB} = \\frac{2}{4} = \\frac{1}{2}$. Reciprocating, we know that $\\frac{EB}{FE} = 2$ so $\\frac{EB+FE}{FE} = 2+1 \\Rightarrow \\frac{FB}{FE} = 3$. Reciprocating again, we have $\\frac{FE}{FB} = \\frac{1}{3} \\Rightarrow FE = \\frac{1}{3}FB$. We know that $FD = 2$, so by the Pythagorean Theorem, $FB = \\sqrt{2^{2} + 4^{2}} = 2\\sqrt{5}$. Thus $FE = \\frac{1}{3}FB = \\frac{2\\sqrt{5}}{3}$. Applying the Pythagorean Theorem again, we have $AF = \\sqrt{1^{2}+2^{2}} = \\sqrt{5}$. We finally have $AE = AF + FE = \\sqrt{5} + \\frac{2\\sqrt{5}}{3} = \\frac{5\\sqrt{5}}{3} \\Rightarrow \\boxed{\\text{B}}$", "and the area of $\\triangle BDC$ is $240$. Also using the fact that $E$ is the midpoint of $\\overline{BD}$, we know $\\triangle ADE = \\triangle ABE = 60$. Let $M$ be a point such $\\overline{EM}$ is parellel to $\\overline{CD}$. We immediatley know that $\\triangle BEM \\sim BDC$ by $2$. Using that we can conclude $EM$ has ratio $1$. Using $\\triangle EFM \\sim \\triangle AFC$, we get $EF:AE = 1:2$. Therefore using the fact that $\\triangle EBF$ is in $\\triangle ABF$, the area has ratio $\\triangle BEF : \\triangle ABE=1:2$ and we know $\\triangle ABE$ has area $60$ so $\\triangle BEF$ is $\\boxed{\\textbf{B} \\, 30}$. - fath2012 ## Solution 9 $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(F--D); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"x\", (D+E+F)/3); label(\"x\", (B+E+F)/3); label(\"120+2x\", (D+F+C)/3); [/asy]$ Labeling the areas in the diagram, we have: $[DBC]=240=[BFE]+[FED]+[FDC]=x+x+120+2x=120+4x$ so $240=120+4x, 120=4x, 30=x$. So our answer is $\\boxed{\\textbf{(B)} 30}$. ~~RWhite ## Solution 10 (Menelaus's Theorem) $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF", "∠A. Label the angles as shown. (∠ACD ≅ ∠DCB since $$\\overline{C D}$$ bisects ∠ACB) Since AB = BC, ∆ABC is isosceless, therefore 2x = a. m∠A + m∠ACE + m∠E = 180° a + (x + 42) + 90 = 180 2x + x + 132 = 180 x = 16 Since a = 2x, m∠A = 32° Question 3. In the figure below, $$\\overline{A D}$$ is the angle bisector of ∠BAC. $$\\overline{B A P}$$ and $$\\overline{B D C}$$ are straight lines, and $$\\overline{A D}$$ || $$\\overline{P C}$$. Prove that AP = AC. Label w, x, y and z as shown. Statements Reasons 1. $$\\overline{A D}$$ is the angle bisector of ∠BAC Given 2. $$\\overline{A D}$$ || $$\\overline{P C}$$ Given 3. z = w Definition of angle bisector 4. z = y If two parallel lines are cut by a transversal, alt. internal angles are equal in measure. 5. w = x If two parallel lines are cut by a tranversal, corr. Angles are equal in measure. 6. x = y Transitive property 7. ∆ACP is isosceless Base angles are congruent 8. AC = AP Definition of isosceles triangle. Question 4. ∆ ABC and ∆ DEF, in the figure below are such that $$\\overline{A B}$$ ≅ $$\\overline{D E}$$, $$\\overline{A C}$$ ≅ $$\\overline{D F}$$ and ∠A ≅∠D. a. Which", "adjacent side vectors $\\overrightarrow{(a,b)}, \\overrightarrow{(c,d)}$ is given by $\\overrightarrow{(a,b)} \\times \\overrightarrow{(c,d)} = ad-bc$. $[asy] defaultpen(linewidth(0.7)); unitsize(15); real r = 3.5; // radius pair shiftL = (-2.5*r,0); // distance between 2 diagrams /* returns the vertex of the interior equilateral triangle with one edge shared with the dodecagon */ pair dodecagonPt(int i) { return r*dir(i*360/12) + rotate(60)*(r*(dir((i+1)*360/12) - dir(i*360/12))); } /* left diagram */ path dodecagon = shiftL+(r,0)--shiftL+r*dir(30); for(int i = 1; i < 12; ++i) dodecagon = dodecagon--shiftL+r*dir(i*30); dodecagon = dodecagon--cycle; filldraw(dodecagon, rgb(0.5,1,0.5)); draw(Circle(shiftL, r), linetype(\"2 2\")); dot((0,0)); draw(shiftL--shiftL+(r,0)); label(\"R\",shiftL+(r/2,0),S); /* right diagram */ for(int i = 0; i < 9; ++i) { filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0)); filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0)); filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.8,0.8,0)); if (i % 3 == 1) { filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir((i+1)*360/12)--cycle, rgb(0.8,0.8,0)); filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir(floor(i/3)*90)--cycle, rgb(0,0.8,0)); filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir((i+1)*360/12)--r*dir(floor(i/3)*90+90)--cycle, rgb(0,0.8,0)); } } for(int i = 9; i < 12; ++i) { filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype(\"2 2\")); filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype(\"2 2\")); filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(1,1,0.5), linetype(\"2 2\")); } [/asy]$ The area of a dodecagon is $3R^2$, where $R$ is the circumradius. $[asy] pathpen = linewidth(1); unitsize(15); pen dotted = linetype(\"2 4\"); path xaxis = (-3,0)--(3,0); pair A = (-2,2), B = (1.5,1.5), B3 = (-1.5,0), B2 = (B.x,-B.y), C2 = IP(xaxis, A--B2); D(xaxis,Arrows(8)); D(D(A)--D(C2)--D(B)); D(D(B2)--C2,dashed+linewidth(0.7)); D(A--D(B3)--B,dotted+linewidth(0.7)); D(B3--B2,dotted); MP(\"(a,b)\",A,W); MP(\"(c,d)\",B,E); MP(\"(c,-d)\",B2,E); [/asy]$ The smallest distance necessary to travel between $(a,b)$, the x-axis, and then $(c,d)$ for $b,d > 0$ is given" ]

[ "\\right ] \\sin nx \\: dt \\\\ & \\sim \\frac{1}{2} + \\frac{2}{\\pi} \\sin x + \\frac{2}{3\\pi} \\sin 3x + \\frac{2}{5\\pi} \\sin 5x + ... \\\\ & \\sim \\frac{1}{2} + \\sum_{n=1}^{\\infty} \\frac{2}{(2n-1)\\pi} \\sin (2n - 1)x \\end{align}", "# How do you simplify 13/30+1/5div6/7? $\\frac{13}{30} + \\frac{1}{5} \\div i \\mathrm{de} \\frac{6}{7} = \\frac{2}{3}$ $\\frac{1}{5} \\div i \\mathrm{de} \\frac{6}{7} = \\frac{7}{30}$ $\\frac{7}{30} + \\frac{13}{30} = \\frac{20}{30} = \\frac{2}{3}$", "# How do you simplify 20(cos((7pi)/6)+isin((7pi)/6))div15(cos((11pi)/3)+isin((11pi)/3)) and express the result in rectangular form? Jan 17, 2017 $= - \\frac{4}{3} i$ #### Explanation: In polar complex, that's: $\\frac{20 {e}^{i \\frac{7 \\pi}{6}}}{15 {e}^{i \\frac{11 \\pi}{3}}}$ $= \\frac{4}{3} {e}^{- \\frac{5 \\pi}{2}}$ $= \\frac{4}{3} {e}^{- \\frac{\\pi}{2}}$ $= - \\frac{4 i}{3}$", "# Simplify sin inverse 5 over 13 cos x plus 12 over 13 sin x RD Sharma Question 212,520 • May 24 2022 ## Description Simplify: {\\displaystyle \\sin ^{-1} \\left ( \\frac{5}{13} \\cos x+\\frac{12}{13} \\sin x \\right ) }", "How do you simplify (3x-6)/(6 )? Jun 14, 2018 $\\frac{x - 2}{2}$ Explanation: $= \\frac{3 x - 6}{6}$ $= \\frac{3 \\left(x - 2\\right)}{6}$ $= \\frac{x - 2}{2}$ Jun 14, 2018 You only need to simplify the coefficients, but it will net you $\\frac{x}{2} - 1$ as the simplified answer. Explanation: Treat the two terms in the numerator as separate fractions: $\\frac{3 x - 6}{6} = \\frac{3 x}{6} - \\frac{6}{6}$ $\\frac{3 x}{6} - 1 = \\frac{3}{6} x - 1$ color(green)(rArr1/2x-1 = x/2-1", "# Math Help - Simplifying another complex fraction.. 1. ## Simplifying another complex fraction.. Please let me know if I am clogging up your board...I just have a few equations that I really need to see how theyre done..thanks! 2. $\\frac{1-\\frac{r}{2r+1}}{r-\\frac{1}{2r+1}}$ $\\frac{\\frac{2r+1}{2r+1}-\\frac{r}{2r+1}}{\\frac{r(2r+1)}{2r+1}-\\frac{1}{2r+1}}$ $\\frac{\\frac{r+1}{2r+1}}{\\frac{2r^2+r-1}{2r+1}}$ $\\frac{r+1}{2r^2+r-1}$ $\\frac{r+1}{(r+1)(2r-1)}$ $\\frac{1}{2r-1}$ 3. Thanks so much!!!!", "# How do you simplify (6 + 6i ) / (7 + 6i )? $\\frac{42 + 42 i - 36 i + 36}{49 + 36} = \\frac{78 - 6 i}{85}$ $= \\frac{78}{85} - \\frac{6}{85} i$ This can be put in trignometric form $r \\left(\\cos \\theta - i \\sin \\theta\\right)$ if required, by solving $r \\cos \\theta = \\frac{78}{85}$ and $r \\sin \\theta = - \\frac{6}{85}$.", "(2x - \\frac{\\pi }{3}) = \\cos (\\frac{\\pi }{3})\\mathop \\to \\limits^{\\color{red}{formula2}} &2x - \\frac{\\pi }{3} + 2k\\pi \\ \\text{or} \\to 2x = \\frac{\\pi}{6} + 2k\\pi \\ \\text{or} \\to x = \\frac{\\pi}{12}+k\\pi \\ \\text{or} \\\\ &2x - \\frac{\\pi}{3} = -\\frac{\\pi}{3} + 2k\\pi \\hspace{1cm} 2x = 0 + 2k\\pi \\hspace{1cm} x = k\\pi \\end{aligned}", "# How do you simplify 4/15 + 2/4? Aug 1, 2016 $\\frac{4}{15} + \\frac{2}{4} = \\frac{23}{30}$ #### Explanation: This can be solved easily by multiplying the numerator and denominator of the fractions by constants: $\\frac{4}{15} + \\frac{2}{4}$ $= \\frac{4}{15} + \\frac{1}{2}$ $= \\frac{8}{30} + \\frac{15}{30}$ $= \\frac{23}{30}$", "# How do you simplify (17k+14)/9-(12k+1)/9? $\\frac{5 k + 13}{9}$ The fraction have the same denominators so just subtract the numerators. E.g. $\\frac{5}{9} - \\frac{4}{9} = \\frac{5 - 4}{9} = \\frac{1}{9}$ $\\frac{17 k + 14}{9} - \\frac{12 k + 1}{9} = \\frac{17 k + 14 - 12 k - 1}{9} = \\frac{5 k + 13}{9}$", "\\frac{5}{3} \\times \\frac{\\textcolor{m a \\ge n t a}{x}}{\\textcolor{m a \\ge n t a}{x}}$ Simplify. $\\frac{6}{3 x} + \\frac{5 x}{3 x}$ Simplify. $\\frac{6 + 5 x}{3 x}$", "-\\frac{\\sqrt2}{2}$ $\\frac{7\\pi}{12} = \\frac{3\\pi}{4} - \\frac{\\pi}{6}$ $cos(A-B) = cosAcosB + sinAsinB$ $A = \\frac{3\\pi}{4}$ and $B = \\frac{\\pi}{6}$", "# How do you simplify 6/5 div 13/8? $\\frac{48}{65}$ $\\frac{6}{5} \\div \\frac{13}{8} \\text{ to divide by a fraction, x by the reciprocal}$ =$\\frac{6}{5} \\times \\frac{8}{13} \\text{ check for cancelling here}$ =$\\frac{48}{65}$", "# How do you simplify 2/3div(-4/15)? $\\frac{\\frac{2}{3}}{- \\frac{4}{15}} = 2.5$ Given:$\\frac{\\frac{2}{3}}{- \\frac{4}{15}}$ to simplify. $\\frac{\\frac{2}{3}}{- \\frac{4}{15}} = \\frac{2}{3} \\times \\left(- \\frac{15}{4}\\right) = - \\frac{2 \\times 15}{3 \\times 4} = - \\frac{30}{12} = - 2.5$", "# How do you simplify sin^-1(x)+cos^-1(x)=pi/2? $x = \\frac{\\pi}{4}$ $x + x = \\frac{\\pi}{2}$ --> $2 x = \\frac{\\pi}{2}$ --> $x = \\frac{\\pi}{4}$", "# How do you simplify (12 - 8 * sqrt(2) ) ^(3/2) /6? Mar 15, 2016 ${\\left(12 - 8 \\sqrt{2}\\right)}^{\\frac{3}{2}} / 6$ ${\\left(12 - 8 \\cdot \\sqrt{2}\\right)}^{\\frac{3}{2}} / 6$ $= \\textcolor{g r e e n}{| \\overline{\\underline{\\textcolor{w h i t e}{\\frac{a}{a}} {\\left(12 - 8 \\sqrt{2}\\right)}^{\\frac{3}{2}} / 6 \\textcolor{w h i t e}{\\frac{a}{a}} |}}}$", "x)+\\frac{1}{32} \\cos (6 x)+\\frac{1}{2} \\sin (x) \\cos (2 x)+\\frac{1}{8} \\sin (x) \\cos (4 x)+\\frac{5}{16} \\end{align}", "are $\\frac{\\sqrt2}}{2}$ and $\\frac{\\sqrt{3}}{2}$, respectively. Therefore: $cos(\\frac{5\\pi}{12}) + cos(\\frac{\\pi}{12}) = 2(\\frac{ \\sqrt{2}}{2}) (\\frac{\\sqrt{3} }{2})$. Finally, this becomes: $\\frac{ \\sqrt{2}\\sqrt{3} }{2} = \\frac{ \\sqrt{6}}{2}$.", "Simplify. (Warning: numerator and denominator must be factored!)) Got it ! $\\frac{a(b-c)}{d(c-b)} =$ $\\frac{a(b-c)}{-d(b-c)} = -\\frac{a}{d}$ $\\frac{a}{a^2+ab}=$ $\\frac{1a}{a(a+b)} = \\frac{1}{a+b}$", "# How do you simplify 5/(2+3i)? Dec 13, 2015 $\\frac{10}{13} - \\frac{15}{13} i$ #### Explanation: Multiply by the complex conjugate of the bottom. $\\frac{5}{2 + 3 i} \\left(\\frac{2 - 3 i}{2 - 3 i}\\right) = \\frac{10 - 15 i}{4 - 9 {i}^{2}}$ Recall that $i = \\sqrt{-} 1$. Square both sides of that to see that ${i}^{2} = - 1$. $= \\frac{10 - 15 i}{4 - 9 \\left(- 1\\right)} = \\frac{10 - 15 i}{13} = \\frac{10}{13} - \\frac{15}{13} i$" ]

[ "\\left( -\\frac13, \\frac13 \\right), \\quad \\left( -\\frac14, \\frac14 \\right), \\quad \\dots$ This infinite set of intervals intersects at a single point ${\\{0\\}}$! The moral of the story is that induction doesn’t let us reach infinity. Too bad, because we’d have loved to use induction to help us construct a monster. That’s what we’re doing, after all — adding things in one by one. 3. Transfinite Induction Well, it turns out we can, but we need a new notion of number. For this we need a notion of an ordinal number. I defined these in their full glory a previous blog post, but I don’t need the full details of that. Here’s what I want to say: after all the natural numbers $\\displaystyle 0, \\; 1, \\; \\dots,$ I’ll put a new number called ${\\omega}$, representing how large the natural numbers are. After that there’s a number called $\\displaystyle \\omega+1, \\; \\omega+2, \\; \\dots$ and eventually a number called ${2\\omega}$. The list goes on: \\displaystyle \\begin{aligned} 0, & 1, 2, 3, \\dots, \\omega \\\\ & \\omega+1, \\omega+2, \\dots, \\omega+\\omega \\\\ & 2\\omega+1, 2\\omega+2, \\dots, 3\\omega \\\\ & \\vdots \\\\ & \\omega^2 + 1, \\omega^2+2, \\dots \\\\ & \\vdots \\\\ & \\omega^3, \\dots, \\omega^4, \\dots, \\omega^\\omega \\\\ & \\vdots, \\\\ & \\omega^{\\omega^{\\omega^{\\dots}}} \\\\ \\end{aligned} Pictorially, it kind of looks like", "## Friday, January 30, 2009 ### Interesting occurrence with 7th root of unity PDF of this post I've come across a property of $i\\sqrt{7}$ that is rather interesting. Letting $\\omega = e^{\\frac{2\\pi i}{7}}$, we let $\\eta = \\omega + \\omega^2 + \\omega^4$, we will see that $\\eta - \\overline{\\eta} = i\\sqrt{7}$. Instead of working directly with this difference, we square the expression for something more easily manipulated. $(\\eta - \\overline{\\eta})^2 = \\eta^2 - \\eta\\overline{\\eta} + \\overline{\\eta}^2 = (\\omega^2 + 2 \\omega^3 + \\omega^4 + 2 \\omega^5 + 2 \\omega^6 + \\omega^8) + (\\omega^{-2} + 2 \\omega^{-3} + \\omega^{-4} + 2 \\omega^{-5} + 2 \\omega^{-6} + \\omega^{-8}) - 2(\\omega \\omega^{-1} + \\omega^2 \\omega^{-1} + \\omega^4 \\omega^{-1} + \\omega \\omega^{-2} + \\omega^2 \\omega^{-2} + \\omega^4 \\omega^{-2} + \\omega \\omega^{-4} + \\omega^2 \\omega^{-4} + \\omega^4 \\omega^{-4})$. Using the equalities $\\omega^{-1} = \\omega^6$, $\\omega^{-2} = \\omega^5$ etc., we can simplify this down to $\\omega + \\omega^2 + \\omega^3 + \\omega^4 + \\omega^5 + \\omega^6 - 6$. Since the sum of the 7th roots of unity equals 0 and the sum only neglects the root $\\omega^7 = 1$, we see that $(\\eta - \\overline{\\eta})^2 = -7$. Thus $\\eta - \\overline{\\eta} = i\\sqrt{7}$. Amazing stuff, right?", "if $$a = 123 \\Rightarrow c = 00$$. \\begin{align} \\therefore (a\\omega 5)^2 &= \\big(a \\omega ]5 \\omega c[^{]a[}\\big)^2\\\\ &= a^2 \\omega \\big(a(5\\omega c) + a(5\\omega c)\\big) \\omega (5\\omega c)^2\\tag{Using Bimultiplication}\\\\ &= a^2 \\omega 2a(5\\omega c) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (10a \\omega 2ac) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (10a \\omega c) \\omega (5\\omega c)^2\\tag{Since, c is all zeroes}\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (25 \\omega 10c \\omega c^2)\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (25 \\omega c \\omega c)\\tag{1} \\end{align} Since each digit group must have $$]a[$$ digits, so let us move one zero from the middle $$c$$ in $$25 \\omega c \\omega c$$ to the rightmost $$c$$. So, now that group becomes $$25\\omega d\\omega e$$, where $$]d[ = ]c[ – 1$$ and $$]e[ = ]c[ + 1$$. \\begin{align} \\therefore (1) &= a^2 \\omega \\big( (a \\omega 0 \\omega c) + (25 \\omega d)\\big) \\omega e\\\\ &= a^2 \\omega \\big( a \\omega (0+25) \\omega (c+d) \\big) \\omega e\\\\ &= a^2 \\omega ( a \\omega 25 \\omega d ) \\omega e\\tag{2} \\end{align} Now, \\begin{align} ]25 \\omega d[ &= ]25[ + ]d[\\\\ &= 2 + (]c[ – 1)\\\\ &= 2 + \\big((]a[ – 1) – 1\\big)\\\\ &= ]a[ \\end{align} So,", "\\omega \\underline{2}5\\\\ &= a^2 \\omega (a\\omega2) \\omega 5\\\\ &= (a^2 + a) \\omega 2 \\omega 5\\\\ &= \\big(a(a+1)\\big) \\omega 25\\tag{Proved} \\end{align} ### Case 2: $$a$$ has more than one digits. So, $$]a[ > (]5[ = 1)$$. But to apply Bimultiplication formula $$a$$ must have the same number of digits in $$5$$, which is obviously not the case here. So, we will use one trick. We will pad $$5$$ with some number of zeroes on the right, so that, $$]a[ = ]5\\omega c[$$, where, $$c$$ is all zeroes and $$]c[ = ]a[ - 1$$. So, if $$a = 123 \\Rightarrow c = 00$$. \\begin{align} \\therefore (a\\omega 5)^2 &= \\big(a \\omega ]5 \\omega c[^{]a[}\\big)^2\\\\ &= a^2 \\omega \\big(a(5\\omega c) + a(5\\omega c)\\big) \\omega (5\\omega c)^2\\tag{Using Bimultiplication}\\\\ &= a^2 \\omega 2a(5\\omega c) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (10a \\omega 2ac) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (10a \\omega c) \\omega (5\\omega c)^2\\tag{Since, c is all zeroes}\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (25 \\omega 10c \\omega c^2)\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (25 \\omega c \\omega c)\\tag{1} \\end{align} Since each digit group must have $$]a[$$ digits, so let us move one zero from the middle $$c$$ in $$25 \\omega c \\omega c$$", "a^2 \\omega 10a \\omega \\underline{2}5\\\\ &= a^2 \\omega (a\\omega0) \\omega \\underline{2}5\\\\ &= a^2 \\omega (a\\omega2) \\omega 5\\\\ &= (a^2 + a) \\omega 2 \\omega 5\\\\ &= \\big(a(a+1)\\big) \\omega 25\\tag{Proved} \\end{align} ### Case 2: $$a$$ has more than one digits. So, $$]a[ > (]5[ = 1)$$. But to apply Bimultiplication formula $$a$$ must have the same number of digits in $$5$$, which is obviously not the case here. So, we will use one trick. We will pad $$5$$ with some number of zeroes on the right, so that, $$]a[ = ]5\\omega c[$$, where, $$c$$ is all zeroes and $$]c[ = ]a[ – 1$$. So, if $$a = 123 \\Rightarrow c = 00$$. \\begin{align} \\therefore (a\\omega 5)^2 &= \\big(a \\omega ]5 \\omega c[^{]a[}\\big)^2\\\\ &= a^2 \\omega \\big(a(5\\omega c) + a(5\\omega c)\\big) \\omega (5\\omega c)^2\\tag{Using Bimultiplication}\\\\ &= a^2 \\omega 2a(5\\omega c) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (10a \\omega 2ac) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (10a \\omega c) \\omega (5\\omega c)^2\\tag{Since, c is all zeroes}\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (25 \\omega 10c \\omega c^2)\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (25 \\omega c \\omega c)\\tag{1} \\end{align} Since each digit group must have $$]a[$$ digits, so let us move one zero from", "just a matter of finding the matrix $A$. Let the corners of the quadrilateral be given by the points $q_i'$, and map the corners of the unit square to them as follows: \\begin{align} (0,0)&\\mapsto q_0' \\\\ (1,0)&\\mapsto q_1' \\\\ (0,1)&\\mapsto q_2' \\\\ (1,1)&\\mapsto q_3'. \\end{align} Solving the resulting system of equations is tedious, but straightforward. One form of the solution is as follows: \\begin{align} a_{00} &= a_{20}x_1'+\\Delta x_{10} \\\\ a_{10} &= a_{20}y_1'+\\Delta y_{10} \\\\ a_{20} &= {\\omega(\\Delta_{10},\\Delta_{32})+\\omega(\\Delta_{20},\\Delta_{32})-\\omega(\\Delta_{30},\\Delta_{32}) \\over \\omega(\\Delta_{31},\\Delta_{32})} \\\\ a_{01} &= a_{21}x_2'+\\Delta x_{20} \\\\ a_{11} &= a_{21}y_2'+\\Delta y_{20} \\\\ a_{21} &= -{\\omega(\\Delta_{10},\\Delta_{31})+\\omega(\\Delta_{20},\\Delta_{31})-\\omega(\\Delta_{30},\\Delta_{31}) \\over \\omega(\\Delta_{31},\\Delta_{32})} \\\\ a_{02} &= x_0' \\\\ a_{12} &= y_0', \\\\ \\end{align} where $\\Delta_{ij}=q_i'-q_j'$, $\\Delta x_{ij}$ and $\\Delta y_{ij}$ are the corresponding coordinate deltas, and $\\omega$ is the symplectic form $\\omega(\\mathbf u, \\mathbf v) = u_xv_y-u_yv_x$. (That this looks like a lot of cross products is no coincidence—there’s a geometric interpretation of the solution as the intersection of various planes.) Since you’re mainly interested in mapping from a quadrilateral to a rectangle, it might be more efficient, and probably more computationally stable, to compute $A^{-1}$ directly from the corner coordinates. It’s similar to the solution for the other direction, but I’ll leave that calculation to you.", "2 + \\omega^9, \\beta^2 = \\omega^2 + 2 + \\omega^9, \\beta^3 = \\omega^3 + 3 \\omega + 3 \\omega^{10} + \\omega^8, \\beta^3 = \\omega^3 + 3 \\omega + 3 \\omega^{10} + \\omega^8, \\beta^4 = \\omega^4 + 4 \\omega^2 + 6 + 4 \\omega^9 + \\omega^7, \\beta^4 = \\omega^4 + 4 \\omega^2 + 6 + 4 \\omega^9 + \\omega^7, \\beta^5 = \\omega^5 + 5 \\omega^3 + 10 \\omega + 10 \\omega^{10} + 5 \\omega^8 + \\omega^6. \\beta^5 = \\omega^5 + 5 \\omega^3 + 10 \\omega + 10 \\omega^{10} + 5 \\omega^8 + \\omega^6. Then 1 + 3 \\beta – 3 \\beta^2 – 4 \\beta^3 + \\omega^4 + \\omega^5 = 1 + \\omega + \\omega^2 + \\omega^3 + \\omega^4 + \\omega^5 + \\omega^6 + \\omega^7 + \\omega^8 + \\omega^9 + \\omega^{10} = \\frac{\\omega^{11} – 1}{\\omega – 1} = 0 1 + 3 \\beta – 3 \\beta^2 – 4 \\beta^3 + \\omega^4 + \\omega^5 = 1 + \\omega + \\omega^2 + \\omega^3 + \\omega^4 + \\omega^5 + \\omega^6 + \\omega^7 + \\omega^8 + \\omega^9 + \\omega^{10} = \\frac{\\omega^{11} – 1}{\\omega – 1} = 0", "(x-0) (x-1) \\\\ &= x(x-1) \\end {align*} This completes the step 2 of the algorithm. Since the degree $$d=0$$, then $$p(x)=1$$. Now $$P_i(x)$$ polynomials are generated using \\begin {align*} P_n &= - p(x) \\\\ P_{i-1} &= - S p'_{i} + ((n-i)S' -S\\theta ) P_i - (n-1)(i+1) S^2 r P_{i+1} \\qquad i=n,n-1,\\dots , 0 \\end {align*} These generate the following set \\begin {align*} P_{4} &= -1\\\\ P_{3} &= 2 x -4\\\\ P_{2} &= -3 \\left (x -2\\right )^{2}\\\\ P_{1} &= 3 \\left (x -2\\right )^{3}\\\\ P_{0} &= -\\frac {3 \\left (x -2\\right )^{4}}{2}\\\\ P_{-1} &= 0 \\end {align*} There is nothing to solve for from the last equation $$P_{-1} = 0$$ as $$p(x)=1$$ is already known because the degree $$d$$ was zero and hence there are no coefficients $$a_i$$ to solve for. Next $$\\omega$$ is determined as the solution to the following equation using $$n=4$$. \\begin {align*} \\sum _{i=0}^{n} S^i \\frac {P_i}{(n-i)!} \\omega ^i &= 0 \\\\ \\frac {P_0}{4!} + \\frac {S P_1}{3!} \\omega +\\frac {S^2 P_2}{2!} \\omega ^2 +\\frac {S^3 P_3}{1!} \\omega ^3 + +\\frac {S^4 P_4}{0!} \\omega ^4 &=0\\\\ -\\frac {1}{16}\\left (2 \\omega \\,x^{2}-2 x \\omega -x +2\\right )^{4} &=0 \\end {align*} Solving the above and using any one of the roots results in \\begin {align*} \\omega =\\frac {1}{2 x \\left (x -1\\right )}\\left (x -2\\right )", "# Complex numbers - roots of unity Let $\\omega$ be a complex number such that $\\omega^5 = 1$ and $\\omega \\neq 1$. Find $$\\frac{\\omega}{1 - \\omega^2} + \\frac{\\omega^2}{1 - \\omega^4} + \\frac{\\omega^3}{1 - \\omega} + \\frac{\\omega^4}{1 - \\omega^3}.$$ I have tried adding the first two and the second two separately, then adding those sums but how do I get a numerical value as the answer? Thanks HINT : $$\\frac{\\omega}{1-\\omega^2}+\\frac{\\omega^2}{1-\\omega^4}+\\frac{\\omega^3}{1-\\omega}+\\frac{\\omega^4}{1-\\omega^3}$$ $$=\\frac{\\omega}{1-\\omega^2}+\\frac{\\omega^2}{1-\\omega^4}+\\frac{\\omega^7}{\\omega^4-\\omega^5}+\\frac{\\omega^6}{\\omega^2-\\omega^5}$$ The sum of the first and fourth terms is \\begin{align*} \\frac{\\omega}{1 - \\omega^2} + \\frac{\\omega^4}{1 - \\omega^3} &= \\frac{\\omega (1 - \\omega^3) + \\omega^4 (1 - \\omega^2)}{(1 - \\omega^2)(1 - \\omega^3)} \\\\ &= \\frac{\\omega - \\omega^4 + \\omega^4 - \\omega^6}{(1 - \\omega^2)(1 - \\omega^3)} \\\\ &= \\frac{\\omega - \\omega^4 + \\omega^4 - \\omega}{(1 - \\omega^2)(1 - \\omega^3)} \\\\ &= 0, \\end{align*}and the sum of the second and third terms is \\begin{align*} \\frac{\\omega^2}{1 - \\omega^4} + \\frac{\\omega^3}{1 - \\omega} &= \\frac{\\omega^2 (1 - \\omega) + \\omega^3 (1 - \\omega^4)}{(1 - \\omega^4)(1 - \\omega)} \\\\ &= \\frac{\\omega^2 - \\omega^3 + \\omega^3 - \\omega^7}{(1 - \\omega^4)(1 - \\omega)} \\\\ &= \\frac{\\omega^2 - \\omega^3 + \\omega^3 - \\omega^2}{(1 - \\omega^4)(1 - \\omega)} \\\\ &= 0. \\end{align*}Therefore, the sum of all four terms is $\\boxed{0}$. Maybe more direct $$\\frac{\\omega}{1 - \\omega^2} + \\frac{\\omega^2}{1 - \\omega^4} + \\frac{\\omega^3}{1 - \\omega} + \\frac{\\omega^4}{1 - \\omega^3}", "equation and simplifying gives \\begin {align*} 0 = 0 \\end {align*} Hence $$p(x)=1$$ can be used. Let \\begin {align*} \\phi &= \\theta + \\frac {p'}{p}\\\\ &= \\frac {2}{x}-\\frac {1}{x -1} \\end {align*} And $$\\omega$$ be the solution of \\begin {align*} \\omega ^2 - \\phi \\omega + \\left ( \\frac {1}{2} \\phi ' + \\frac {1}{2} \\phi ^2 - r \\right ) &= 0 \\end {align*} Substituting the values for $$\\phi$$ and $$r$$ into the above equation gives \\begin {align*} w^{2}-\\left (\\frac {2}{x}-\\frac {1}{x -1}\\right ) w +\\frac {\\left (x -2\\right )^{2}}{4 \\left (x -1\\right )^{2} x^{2}} &= 0 \\end {align*} Solving for $$\\omega$$ gives two roots, either one can be used. Using \\begin {align*} \\omega &= \\frac {x -2}{2 \\left (x -1\\right ) x} \\end {align*} Therefore to $$z'' = r z$$ is \\begin {align*} z &= e^{ \\int \\omega \\,dx} \\\\ &= {\\mathrm e}^{\\int \\frac {x -2}{2 \\left (x -1\\right ) x}d x}\\\\ &= \\frac {x}{\\sqrt {x -1}} \\end {align*} The first solution to the original ode in $$y$$ is now found from \\begin {align*} y_1 &= z e^{- \\int \\frac {1}{2} a \\,dx}\\\\ &= z e^{ -\\int \\frac {1}{2} \\frac {5 x^{2}-4 x}{-x^{3}+x^{2}} \\,dx}\\\\ &= z e^{2 \\ln \\left (x \\right )+\\frac {\\ln \\left (x -1\\right )}{2}}\\\\ &= z \\left (x^{2} \\sqrt {x -1}\\right )\\\\ &=", "$$\\omega$$. This leads us to the following equation for the vector of the line of intersection: $$P = (Rx(\\varepsilon)Z) \\times (Rz(\\omega)Rx(i)Z)$$ where $$Rx(.)$$ & $$Rz(.)$$ are rotation matrices. Generally, $$P$$ is not a unit vector, but we can easily normalize it so that does have unit length, if necessary. Bear in mind that our XY plane is the ecliptic. We can rotate $$P$$ around the X axis by $$-\\varepsilon$$ to determine its coordinates in the equatorial plane. I won't do all the algebra here, but the equatorial coordinates of $$P$$ are \\begin{align} x & = \\cos(\\varepsilon) \\cos(\\omega) \\sin(i) - \\sin(\\varepsilon)\\cos(i)\\\\ y & = -\\sin(\\omega)\\sin(i)\\\\ z & = 0 \\end{align} The RA $$\\alpha$$ corresponding to P is just $$\\alpha=\\tan^{-1}\\left(\\frac yx\\right)$$ Finding the $$\\omega$$ corresponding to the maximum & minimum of $$\\alpha$$ involves some tedious trigonometric calculus and algebra. Fortunately, the results are fairly simple. :) \\begin{align} \\omega & = \\cos^{-1}(\\tan(i) / \\tan(\\varepsilon))\\\\ \\alpha & = \\sin^{-1}(\\sin(i) / \\sin(\\varepsilon)) \\end{align} Hopefully, I haven't messed up a minus sign somewhere while transcribing all those equations. ;) Finally, here are some actual results. Using the value for $$\\varepsilon$$ mentioned earlier, inc RA omega 4.98 -12.606952 101.596318 5.14 -13.017906 101.976326 5.30 -13.429439 102.357061 Here's the single plot for $$i=5.14°$$ And here's the Sage / Python script that made that plot. Lastly, here's an", "\\delta t) + \\sin(\\overline{\\omega} t) \\sin( \\delta t)]\\\\ & = \\frac{A}{4\\overline{\\omega} \\delta} [2 \\sin(\\overline{\\omega} t) \\sin( \\delta t)]\\\\ & = \\frac{A \\sin \\delta t}{2 \\overline{\\omega} \\delta} \\sin \\overline{\\omega} t. \\end{align*} Thus, we have a sine function, $\\sin \\overline{\\omega} t$ whose amplitude varies according to $|(A/2 \\overline{\\omega} \\delta) \\sin \\delta t|\\text{,}$ and $$x(t) = \\frac{A \\sin \\delta t}{2 \\overline{\\omega} \\delta} \\sin \\overline{\\omega} t.\\label{equation-secondorder04-resonance-3}\\tag{4.4.6}$$ ###### Example4.4.4 In the initial-value problem \\begin{align} x'' + 144 x \\amp = 4 \\cos(13t)\\label{equation-secondorder04-resonance-7}\\tag{4.4.7}\\\\ x(0) \\amp = 0\\tag{4.4.8}\\\\ x'(0) \\amp = 0,\\tag{4.4.9} \\end{align} we have $A= 4\\text{,}$ $\\omega_0 = 12\\text{,}$ and $\\omega = 13\\text{.}$ The solution to this equation is \\begin{equation*} x(t) = \\frac{4}{25}(\\cos 12 t - \\cos 13 t). \\end{equation*} The graph of this solution is given in Figure 4.4.5 and illustrate the phenomenon of beats. The two frequencies are almost equal and interfere with each other. In our example, $\\delta = -1/2$ and $\\overline{\\omega} = 25/2\\text{.}$ The factor $\\sin \\delta t = \\sin( - t/2)$ in (4.4.6) oscillates very slowly in comparison to $\\sin(\\overline{\\omega}) = \\sin(25t/2)\\text{.}$ Thus, we can view the solution (4.4.6) as a fast oscillation with a frequency $\\overline{\\omega}$ and amplitude $$\\left| \\frac{A \\sin \\delta t}{2 \\overline{\\omega} \\delta} \\right| = \\left| \\frac{8}{25} \\sin\\left( \\frac{t}{2}\\right)\\right|,\\label{equation-secondorder04-resonance-4}\\tag{4.4.10}$$ which oscillates much more slowly. If we superimpose the graph of (4.4.10) onto the graph in Figure", "that $c_1=-{F_0\\over m(\\omega_0^2-\\omega^2)}\\quad \\text{and} \\quad c_2=0.\\nonumber$ Substituting these into Equation \\ref{eq:6.1.15} yields $\\label{eq:6.1.16} y={F_0\\over m(\\omega_0^2-\\omega^2)}(\\cos\\omega t-\\cos\\omega _0t).$ It is revealing to write this in a different form. We start with the trigonometric identities \\begin{aligned} \\cos(\\alpha-\\beta)&=\\cos\\alpha\\cos\\beta+\\sin\\alpha\\sin\\beta\\\\ \\cos(\\alpha+\\beta)&=\\cos\\alpha\\cos\\beta-\\sin\\alpha\\sin\\beta.\\end{aligned}\\nonumber Subtracting the second identity from the first yields $\\label{eq:6.1.17} \\cos(\\alpha-\\beta)-\\cos(\\alpha+\\beta)=2\\sin\\alpha\\sin\\beta$ Now let $\\label{eq:6.1.18} \\alpha-\\beta=\\omega t\\quad \\text{and} \\quad\\alpha+\\beta=\\omega_0t,$ so that $\\label{eq:6.1.19} \\alpha={(\\omega_0+\\omega)t\\over2}\\quad \\text{and} \\quad \\beta={(\\omega_0-\\omega)t\\over2}.$ Substituting Equation \\ref{eq:6.1.18} and Equation \\ref{eq:6.1.19} into Equation \\ref{eq:6.1.17} yields $\\cos\\omega t-\\cos\\omega_0t= 2\\sin{(\\omega_0-\\omega)t\\over2}\\sin{(\\omega_0+\\omega)t\\over2},\\nonumber$ and substituting this into Equation \\ref{eq:6.1.16} yields $\\label{eq:6.1.20} y=R(t)\\sin{(\\omega_0+\\omega)t\\over2},$ where $\\label{eq:6.1.21} R(t)={2F_0\\over m(\\omega_0^2-\\omega^2)} \\sin{(\\omega_0-\\omega)t\\over2}.$ From Equation \\ref{eq:6.1.20} we can regard $$y$$ as a sinusoidal variation with frequency $$(\\omega_0+\\omega)/2$$ and variable amplitude $$|R(t)|$$. In Figure $$\\PageIndex{6}$$ the dashed curve above the $$t$$ axis is $$y=|R(t)|$$, the dashed curve below the $$t$$ axis is $$y=-|R(t)|$$, and the displacement $$y$$ appears as an oscillation bounded by them. The oscillation of $$y$$ for $$t$$ on an interval between successive zeros of $$R(t)$$ is called a beat. You can see from Equation \\ref{eq:6.1.20} and Equation \\ref{eq:6.1.21} that $|y(t)|\\le{2|F_0|\\over m|\\omega_0^2-\\omega^2|};\\nonumber$ moreover, if $$\\omega+\\omega_0$$ is sufficiently large compared with $$\\omega -\\omega_0$$, then $$|y|$$ assumes values close to (perhaps equal to) this upper bound during each beat. However, the oscillation remains bounded for all $$t$$. (This assumes that the spring can withstand deflections of this size and continue to obey Hooke’s law.) The", "$H(s) = Y(s)/X(s)$, of a linear time invariant (LTI) system. – Robert L. Sep 5 '18 at 22:05 • Hi. Thanks for clearing that up. There's a lot to learn indeed. – antwoord Sep 8 '18 at 12:13 Here is a way that you can approach this mathematically. It's not the only way. Another way to approach this is with Tchebyshev polynomials and I will leave that to someone else. This approach requires understanding two ingredients. 1. an identity that is a consequence of Euler's formula: $$\\cos(\\omega t) = \\tfrac12(e^{j\\omega t}+e^{-j\\omega t})$$ or $$\\sin(\\omega t) = \\tfrac1{2j}(e^{j\\omega t}-e^{-j\\omega t})$$ 1. the binomial theorem: \\begin{align} (a+b)^N &= \\sum_{n=0}^N {N \\choose n} \\, a^{N-n} \\, b^n \\\\ &= \\sum_{n=0}^N {N \\choose n} \\, a^n \\, b^{N-n} \\\\ \\end{align} or \\begin{align} (a+b)^N &= \\sum_{n=0}^N \\frac{N!}{(N-n)! \\, n!} \\, a^{N-n} \\, b^n \\\\ &= \\sum_{n=0}^N \\frac{N!}{(N-n)! \\, n!} \\, a^n \\, b^{N-n} \\\\ \\end{align} Now you might see where this is going. So using the cosine identity (so i don't have to mess around with powers of $j$ in the sine identity): \\begin{align} \\cos^k(\\omega t) &= \\frac{1}{2^k}(e^{j\\omega t}+e^{-j\\omega t})^k \\\\ \\\\ &= \\frac{1}{2^k} \\sum_{n=0}^k \\frac{k!}{(k-n)! \\, n!} \\, (e^{j\\omega t})^{k-n} \\, (e^{-j\\omega t})^n \\\\ \\\\ &= \\frac{1}{2^k} \\sum_{n=0}^k \\frac{k!}{(k-n)! \\, n!} \\, e^{j(k-n)\\omega t} \\, e^{-jn\\omega t} \\\\ \\\\ &= \\frac{1}{2^k} \\sum_{n=0}^k", "# Roots of Unity, Precalculus (a) Let $\\omega$ be a complex number such that $\\omega^5 = 1$ and $\\omega \\neq 1$. Find $$\\frac{\\omega}{1 + \\omega^2} + \\frac{\\omega^2}{1 + \\omega^4} + \\frac{\\omega^3}{1 + \\omega} + \\frac{\\omega^4}{1 + \\omega^3}$$ (b)Let $\\omega$ be a complex number such that $\\omega^5 = 1$ and $\\omega \\neq 1$. Find $$\\frac{\\omega}{1 - \\omega^2} + \\frac{\\omega^2}{1 - \\omega^4} + \\frac{\\omega^3}{1 - \\omega} + \\frac{\\omega^4}{1 - \\omega^3}$$ For these two problems, I have tried to factor the denominators and rearrange the fractions and adding them up. Any help is appreciated! Hint: (a) We may write $$\\frac{\\omega}{1 + \\omega^2}\\frac{\\omega^4}{\\omega^4} = \\frac{1}{\\omega^4 + \\omega}$$ then \\begin{align}\\frac{1}{\\omega^4 + \\omega} + \\frac{1}{\\omega^3 + \\omega^2} + \\frac{1}{\\omega^3 + \\omega^2} + \\frac{1}{\\omega^4 + \\omega} &= \\frac{2}{\\omega^4 + \\omega} + \\frac{2}{\\omega^3 + \\omega^2} \\\\&= 2 \\Bigg[\\frac{1 + \\omega + \\omega^2 + \\omega^3 + \\omega^4 -1 }{(\\omega^4 + \\omega)(\\omega^3 + \\omega ^2) }\\Bigg] \\\\&=-\\frac{2}{(1 + \\omega)^2} \\end{align} • What would you suggest be the next step? – Math is Life Apr 27 '15 at 0:09 • The answer was given according to what the problem has presented. So there isn't a next step. Next step is letter (b). – Aaron Maroja Apr 27 '15 at 0:36 • Ah, I see. I assumed that there would be a numerical value for letter (a), which is why", "& \\omega^{35} \\\\ 1 & \\omega^{6} & \\omega^{12} & \\omega^{18} & \\omega^{24} & \\omega^{30} & \\omega^{36} & \\omega^{42} \\\\ 1 & \\omega^{7} & \\omega^{14} & \\omega^{21} & \\omega^{28} & \\omega^{35} & \\omega^{42} & \\omega^{49} \\\\ \\end{bmatrix}$ This bulky matrix can be simplified using a bit of algebra. Note that since $\\omega^n$ is an $n^{th}$ root of unity then $\\omega^a = \\omega^b$ for any $a, b$ if $a = b+cn$ for some integer $c \\in \\mathbb{Z}$. Since we are working with eighths (i.e. $2^3 = 8$) any time we see a $\\omega^8$, or indeed, any multiple of $8$, like $\\omega^{16}$ or $\\omega^{32}$, we can replace it with unity. Furthermore, we can reduce rotations to modulo 8, i.e. if we see a $\\omega^{10}$, for example, we can write this as $\\omega^{2}$ because ten is equal to 2 modulo 8. The resulting matrix for QFT is: $\\displaystyle \\textbf{QFT} = \\begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\\\ 1 & \\omega^1 & \\omega^2 & \\omega^3 & \\omega^4 & \\omega^5 & \\omega^6 & \\omega^7 \\\\ 1 & \\omega^{2} & \\omega^{4} & \\omega^{6} & 1 & \\omega^{2} & \\omega^{4} & \\omega^{6} \\\\ 1 & \\omega^{3} & \\omega^{6} & \\omega^{1} & \\omega^{4} & \\omega^{7} & \\omega^{2} & \\omega^{5} \\\\ 1 & \\omega^{4} & 1 & \\omega^{4}", "c)^2\\tag{Since, c is all zeroes}\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (25 \\omega 10c \\omega c^2)\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (25 \\omega c \\omega c)\\tag{1} \\end{align} Since each digit group must have $$]a[$$ digits, so let us move one zero from the middle $$c$$ in $$25 \\omega c \\omega c$$ to the rightmost $$c$$. So, now that group becomes $$25\\omega d\\omega e$$, where $$]d[ = ]c[ – 1$$ and $$]e[ = ]c[ + 1$$. \\begin{align} \\therefore (1) &= a^2 \\omega \\big( (a \\omega 0 \\omega c) + (25 \\omega d)\\big) \\omega e\\\\ &= a^2 \\omega \\big( a \\omega (0+25) \\omega (c+d) \\big) \\omega e\\\\ &= a^2 \\omega ( a \\omega 25 \\omega d ) \\omega e\\tag{2} \\end{align} Now, \\begin{align} ]25 \\omega d[ &= ]25[ + ]d[\\\\ &= 2 + (]c[ – 1)\\\\ &= 2 + \\big((]a[ – 1) – 1\\big)\\\\ &= ]a[ \\end{align} So, in the group $$a \\omega 25 \\omega d$$, $$a$$ is excess. \\begin{align} \\therefore (2) &= (a^2 + a) \\omega (25 \\omega d) \\omega e\\\\ &= a(a+1) \\omega 25 \\omega d \\omega e\\\\ &= a(a+1) \\omega 25 \\omega c \\omega c\\tag{Shifting a zero from e to d}\\\\ \\end{align} So finally, \\begin{align} &(a\\omega 5 \\omega c)^2 =", "\\\\ v^2v_x & = v_1^3 \\cos \\omega t +v_1^2 v_0 +v_1v_0^2 \\cos \\omega +v_0^3 + 2 v_1^2 v_0 \\cos \\omega t +2 v_1 v_0^2 \\cos \\omega t \\end{aligned} and average over time T which is large, as compared with the rotational period $$T_{swir}=2\\pi /\\omega$$. Introducing the notation, \\begin{aligned} \\overline{a} = \\frac{1}{T} \\int _0^T a(t) dt, \\end{aligned} we obtain, \\begin{aligned} \\overline{v^2} & = v_1^2 +v_0^2 \\nonumber \\\\ \\overline{v^2 v_x} & = 2v_1^2v_0 +v_0^3 \\nonumber \\\\ \\overline{\\dot{v}_x} & = \\overline{f_x } =0 \\nonumber \\\\ \\overline{v_x} & = v_0. \\end{aligned} (22) Averaging the equation of motion (21) we find, \\begin{aligned} \\overline{\\dot{v}_x} = \\overline{f_x} + \\overline{(b-v^2) v_x} + \\overline{g} , \\end{aligned} (23) or, with Eqs. (22) and (23), \\begin{aligned} bv_0 -(2v_1^2v_0 +v_0^3)+ g =0, \\end{aligned} (24) which is a cubic equation for $$v_0$$. Almost in all cases the constant velocity $$v_0$$ is much smaller than the orbiting velocity $$v_1$$ (that is, $$v_0 \\ll v_1$$), which implies the approximation, $$v_1^2 \\simeq v^2 \\simeq \\left\\langle v^2 \\right\\rangle$$, where $$\\left\\langle v^2 \\right\\rangle$$ is the mean square velocity of particles in the swirlon. Then the solution of Eq. (24) may be written as \\begin{aligned} v_0 =\\frac{g}{2\\left\\langle v^2 \\right\\rangle -b}. \\end{aligned} (25) Note that the orbiting velocity $$v_1$$ may strongly differ from the “self-velocity” $$b^{1/2}$$. Indeed, when a swirlon is formed, the initial potential energy of particles", "is then: $$P_{\\text{E,emit}} = \\overline{\\epsilon} 4\\pi R_E^2 \\sigma T_E^4$$ Assuming energy equilibrium $P_{\\text{E,abs}} = P_{\\text{E,emit}}$ we can now calculate $T_E$: \\begin{aligned} \\frac{(1-\\alpha)4\\pi R_S^2 \\sigma T_S^4\\pi R_E^2}{4\\pi D^2} & = \\overline{\\epsilon}4\\pi R_E^2 \\sigma T_E^4 \\\\ T_E^4 & = \\frac{(1-\\alpha)4\\pi R_S^2 \\sigma T_S^4\\pi R_E^2}{\\overline{\\epsilon}4\\pi D^2 4\\pi R_E^2 \\sigma} \\\\ T_E^4 & = \\frac{(1-\\alpha) R_S^2 T_S^4}{ 4\\overline{\\epsilon}D^2 } \\\\ T_E & = \\left( \\frac{(1-\\alpha) R_S^2 T_S^4}{4 \\overline{\\epsilon}D^2 }\\right)^{\\frac{1}{4}} \\\\ T_E & = T_S \\left( \\frac{(1-\\alpha) R_S^2}{4 \\overline{\\epsilon} D^2 }\\right)^{\\frac{1}{4}} \\\\ T_E & = T_S \\sqrt{ \\frac{ R_S \\sqrt{\\frac{1-\\alpha}{\\overline{\\epsilon}}} }{2 D } } \\end{aligned} Finally we only need to insert the correct values: • $R_S = 6.96\\times 10^8$ m • $T_S = 5778$ K • $D = 1.496\\times 10^{11}$ m • $\\alpha = 0.1054$ (assuming value of the moon) • $\\overline{\\epsilon} = 0.95$ (assuming value of the moon) This gives us a temperature of 274.5 K. Note that there are many factors that can cause local and temporal variations. For example, incoming radiation varies with latitude and season, and if the removal of the atmosphere would be caused by a dying sun that grows to engulf the earth temperatures would be much higher than this. All in all, to account for all those factors a very large model must be made that can analyse the influence of each factor, including the decrease", "b)^2&\\\\ \\Rightarrow& a \\omega b \\times a \\omega b\\\\ \\Rightarrow& \\boxed{a^2\\omega ]2ab[^n \\omega ]b^2[^n}\\hspace1em\\ldots\\text{where ]a[ = ]b[ = n} \\end{align} ### General Powering Formula For now I will provide only the formula. The proof is provided later in this article. The way I derived it was to I manually find the values of $$(a\\omega b)^2$$, $$(a\\omega b)^3$$, $$(a\\omega b)^4$$ and so on. I found a pattern in all these and from there I got this formula. It was later when it struck me as to how to prove it. $$\\boxed{(a\\omega b)^n = \\omega\\sum_{r=0}^n\\hspace1em{]^nC_r \\times a^{(n-r)} \\times b^r[}^d}\\hspace1em\\ldots\\text{where ]a[ = ]b[ = d}$$ Notice that the above equation looks very similar to Binomial equation. In fact I found this to be true for many equations. If we replace $$+$$ by $$\\omega$$ then we end up with a ‘digitized’ version of that equation. Now let me demonstrate how to use this equation. Let’s find cube of 99, i.e. $$99^3$$. \\begin{align} &(9\\omega9)^3\\\\ &= ]^3C_0 \\times 9^3 \\times 9^0[^1 \\omega ]^3C_1 \\times 9^2 \\times 9^1[^1\\\\ &= 9^3 \\omega 3\\times9^3 \\omega 3\\times9^3 \\omega 9^3\\tag{for brevity removing ].[ operator}\\\\ &= 729 \\omega \\underline{218}7 \\omega \\underline{218}7 \\omega \\underline{72}9\\\\ &= 729 \\omega \\underline{218}7 \\omega \\underline{225}9 \\omega 9\\tag{adding carry overs}\\\\ &= 729 \\omega \\underline{241}2 \\omega 9 \\omega 9\\\\ &= 970 \\omega 2 \\omega 9 \\omega 9\\\\ &=" ]

[ "$\\tan(\\pi/2)=\\infty$, implies that $$5\\tan^4(\\pi/10)-10\\tan^2(\\pi/10)+1=0$$ which, by the quadratic formula, yields $$\\tan^2(\\pi/10)=\\frac{5-2\\sqrt{5}}{5}$$ Adding $1$ and taking the reciprocal yields $$\\frac{1+\\cos(\\pi/5)}{2}=\\cos^2(\\pi/10)=\\frac{5+\\sqrt5}{8}$$ Therefore, $$\\cos(\\pi/5)=\\frac{1+\\sqrt5}{4}=\\phi/2$$ where $\\phi$ is the Golden Ratio. - I love solutions invoking the golden ratio. It'd be even better with a pentagram. (+1) – Ron Gordon Feb 11 '13 at 9:05 Here is a way using roots of unity. We have $\\cos 36 = \\frac{\\omega + \\omega^{-1}}{2}$ where $\\omega = \\text{exp} \\left ( \\frac{2 \\pi}{10} \\right )$. We have $-w$ is a primitive $5^{th}$ root of unity, so it follows $\\omega^4 - \\omega^3 + \\omega^2 - \\omega + 1 = 0$. Now, $x^4 - x^3 + x^2 - x + 1 = x^2(x^2 - x + 1 - \\frac{1}{x} + \\frac{1}{x^2}) = x^2\\left ( \\left (x + \\frac{1}{x} \\right )^2 - \\left (x + \\frac{1}{x} \\right ) -1 \\right )$. Thus $0 = \\omega^2 \\cdot ((2 \\cos 36)^2 - (2 \\cos 36) - 1)$. Therefore $4 \\cos^3 36 - 2 \\cos 36 - 1 = 0$. Using the quadratic formula we then arrive at $$\\cos 36 = \\frac{1 + \\sqrt{5}}{4}$$ - I have never ever done roots of unity.. thanks though for your time and help – Maximilian1988 Feb 11 '13 at 8:23 @dinoboy: I think you made a mistake somewhere. – Ron Gordon Feb 11 '13 at", "think it is generally useful to learn not just such formulas, but how to apply them. – Thomas Andrews Nov 21 '12 at 21:39 I agree, but I didn't set the exercise :-) – Daniel Littlewood Nov 21 '12 at 21:43 Let $\\omega_1= 2+3i, \\omega_2=4+5i, \\omega_3=2-5i , \\omega_4=3-8i$. Then $$\\arg(\\omega_1)= \\arctan(\\frac{3}{2}), ...$$ Finc $\\omega_1 \\omega_2 \\omega_3 \\omega_4 =a+bi$ and then $$\\frac{b}{a}= \\arg(\\omega_1 \\omega_2 \\omega_3 \\omega_4 ) =\\arctan(x) \\,.$$ - \\begin{align} & {} \\quad \\tan(a+b+c+d) \\\\ & = \\frac{\\tan a+\\tan b+\\tan c+\\tan d\\ \\overbrace{ - \\tan a\\tan b\\tan c - \\cdots}^\\text{4 terms}}{1-\\ \\underbrace{\\tan a \\tan b- \\cdots}_\\text{6 terms} +\\tan a\\tan b\\tan c\\tan d} \\end{align} Therefore \\begin{align} & {} \\quad \\tan\\Big(\\arctan(3/2) + \\arctan(5/4) + \\arctan(-5/2) + \\arctan(-8/3)\\Big) \\\\ & = \\frac{\\frac32 + \\frac54 + \\frac{-5}2 +\\frac{-8}3 - (\\text{sum of four terms, each a product of three numbers})}{1 - (\\text{sum of six terms, each a product of two numbers}) + (\\text{a product of four numbers}))} \\end{align} You get a number. Then you say $\\tan x=\\text{that number}$, and find the tangent. -", "because $y>0$. Thus, $\\cos{(\\pi/5)} = (1+\\sqrt{5})/4$. Since rlgordonma has given the answer suggested by the supplied hint, here is another method of computing $\\cos(\\pi/5)$. In this answer, it is shown that $$\\tan(5\\theta)=\\frac{5\\tan(\\theta)-10\\tan^3(\\theta)+\\tan^5(\\theta)}{1-10\\tan^2(\\theta)+5\\tan^4(\\theta)}$$ which, since $\\tan(\\pi/2)=\\infty$, implies that $$5\\tan^4(\\pi/10)-10\\tan^2(\\pi/10)+1=0$$ which, by the quadratic formula, yields $$\\tan^2(\\pi/10)=\\frac{5-2\\sqrt{5}}{5}$$ Adding $1$ and taking the reciprocal yields $$\\frac{1+\\cos(\\pi/5)}{2}=\\cos^2(\\pi/10)=\\frac{5+\\sqrt5}{8}$$ Therefore, $$\\cos(\\pi/5)=\\frac{1+\\sqrt5}{4}=\\phi/2$$ where $\\phi$ is the Golden Ratio. Here is a way using roots of unity. We have $\\cos 36 = \\frac{\\omega + \\omega^{-1}}{2}$ where $\\omega = \\text{exp} \\left ( \\frac{2 \\pi}{10} \\right )$. We have $-w$ is a primitive $5^{th}$ root of unity, so it follows $\\omega^4 – \\omega^3 + \\omega^2 – \\omega + 1 = 0$. Now, $x^4 – x^3 + x^2 – x + 1 = x^2(x^2 – x + 1 – \\frac{1}{x} + \\frac{1}{x^2}) = x^2\\left ( \\left (x + \\frac{1}{x} \\right )^2 – \\left (x + \\frac{1}{x} \\right ) -1 \\right )$. Thus $0 = \\omega^2 \\cdot ((2 \\cos 36)^2 – (2 \\cos 36) – 1)$. Therefore $4 \\cos^3 36 – 2 \\cos 36 – 1 = 0$. Using the quadratic formula we then arrive at $$\\cos 36 = \\frac{1 + \\sqrt{5}}{4}$$ http://mathuprising.comlu.com/images/pentagon.png This is a graphic that I created for my website about 2 days ago. Using $\\Delta ABC$ invoke the Law of Cosines ($\\angle BAC = \\angle BCA", "\\left( -\\frac13, \\frac13 \\right), \\quad \\left( -\\frac14, \\frac14 \\right), \\quad \\dots$ This infinite set of intervals intersects at a single point ${\\{0\\}}$! The moral of the story is that induction doesn’t let us reach infinity. Too bad, because we’d have loved to use induction to help us construct a monster. That’s what we’re doing, after all — adding things in one by one. 3. Transfinite Induction Well, it turns out we can, but we need a new notion of number. For this we need a notion of an ordinal number. I defined these in their full glory a previous blog post, but I don’t need the full details of that. Here’s what I want to say: after all the natural numbers $\\displaystyle 0, \\; 1, \\; \\dots,$ I’ll put a new number called ${\\omega}$, representing how large the natural numbers are. After that there’s a number called $\\displaystyle \\omega+1, \\; \\omega+2, \\; \\dots$ and eventually a number called ${2\\omega}$. The list goes on: \\displaystyle \\begin{aligned} 0, & 1, 2, 3, \\dots, \\omega \\\\ & \\omega+1, \\omega+2, \\dots, \\omega+\\omega \\\\ & 2\\omega+1, 2\\omega+2, \\dots, 3\\omega \\\\ & \\vdots \\\\ & \\omega^2 + 1, \\omega^2+2, \\dots \\\\ & \\vdots \\\\ & \\omega^3, \\dots, \\omega^4, \\dots, \\omega^\\omega \\\\ & \\vdots, \\\\ & \\omega^{\\omega^{\\omega^{\\dots}}} \\\\ \\end{aligned} Pictorially, it kind of looks like", "$$\\omega$$. This leads us to the following equation for the vector of the line of intersection: $$P = (Rx(\\varepsilon)Z) \\times (Rz(\\omega)Rx(i)Z)$$ where $$Rx(.)$$ & $$Rz(.)$$ are rotation matrices. Generally, $$P$$ is not a unit vector, but we can easily normalize it so that does have unit length, if necessary. Bear in mind that our XY plane is the ecliptic. We can rotate $$P$$ around the X axis by $$-\\varepsilon$$ to determine its coordinates in the equatorial plane. I won't do all the algebra here, but the equatorial coordinates of $$P$$ are \\begin{align} x & = \\cos(\\varepsilon) \\cos(\\omega) \\sin(i) - \\sin(\\varepsilon)\\cos(i)\\\\ y & = -\\sin(\\omega)\\sin(i)\\\\ z & = 0 \\end{align} The RA $$\\alpha$$ corresponding to P is just $$\\alpha=\\tan^{-1}\\left(\\frac yx\\right)$$ Finding the $$\\omega$$ corresponding to the maximum & minimum of $$\\alpha$$ involves some tedious trigonometric calculus and algebra. Fortunately, the results are fairly simple. :) \\begin{align} \\omega & = \\cos^{-1}(\\tan(i) / \\tan(\\varepsilon))\\\\ \\alpha & = \\sin^{-1}(\\sin(i) / \\sin(\\varepsilon)) \\end{align} Hopefully, I haven't messed up a minus sign somewhere while transcribing all those equations. ;) Finally, here are some actual results. Using the value for $$\\varepsilon$$ mentioned earlier, inc RA omega 4.98 -12.606952 101.596318 5.14 -13.017906 101.976326 5.30 -13.429439 102.357061 Here's the single plot for $$i=5.14°$$ And here's the Sage / Python script that made that plot. Lastly, here's an", "equation and simplifying gives \\begin {align*} 0 = 0 \\end {align*} Hence $$p(x)=1$$ can be used. Let \\begin {align*} \\phi &= \\theta + \\frac {p'}{p}\\\\ &= \\frac {2}{x}-\\frac {1}{x -1} \\end {align*} And $$\\omega$$ be the solution of \\begin {align*} \\omega ^2 - \\phi \\omega + \\left ( \\frac {1}{2} \\phi ' + \\frac {1}{2} \\phi ^2 - r \\right ) &= 0 \\end {align*} Substituting the values for $$\\phi$$ and $$r$$ into the above equation gives \\begin {align*} w^{2}-\\left (\\frac {2}{x}-\\frac {1}{x -1}\\right ) w +\\frac {\\left (x -2\\right )^{2}}{4 \\left (x -1\\right )^{2} x^{2}} &= 0 \\end {align*} Solving for $$\\omega$$ gives two roots, either one can be used. Using \\begin {align*} \\omega &= \\frac {x -2}{2 \\left (x -1\\right ) x} \\end {align*} Therefore to $$z'' = r z$$ is \\begin {align*} z &= e^{ \\int \\omega \\,dx} \\\\ &= {\\mathrm e}^{\\int \\frac {x -2}{2 \\left (x -1\\right ) x}d x}\\\\ &= \\frac {x}{\\sqrt {x -1}} \\end {align*} The first solution to the original ode in $$y$$ is now found from \\begin {align*} y_1 &= z e^{- \\int \\frac {1}{2} a \\,dx}\\\\ &= z e^{ -\\int \\frac {1}{2} \\frac {5 x^{2}-4 x}{-x^{3}+x^{2}} \\,dx}\\\\ &= z e^{2 \\ln \\left (x \\right )+\\frac {\\ln \\left (x -1\\right )}{2}}\\\\ &= z \\left (x^{2} \\sqrt {x -1}\\right )\\\\ &=", "14.4 & & 138.03 & 1.7 & 2 \\\\ \\eta & 0^{-} & 0^{+} & 3 & & 548.8 & 1.5 & 2 \\\\ \\rho & 1^{-} & 1^{+} & 0.9 & 6.1 & 769 & 1.85 & 4 \\\\ \\omega & 1^{-} & 0^{-} & 24.5 & 0 & 782.6 & 1.85 & 4 \\\\ \\delta & 0^{+} & 1^{-} & 2.488 & & 983 & 2.0 & 2 \\\\ \\sigma^{a} & 0^{+} & 0^{+} & 8.9437 & & 550 & 1.9 & 2 \\\\ & & & (18.3773) & & (720) & (2.0) & (2) \\\\ \\hline \\end{array}$$$$ At each vertex is used a form factor like this: $$$$\\mathcal{F}_{\\alpha}\\left[\\left(\\boldsymbol{q}^{\\prime}-\\boldsymbol{q}\\right)^{2}\\right]=\\left[\\frac{\\Lambda_{\\alpha}^{2}-m_{\\alpha}^{2}}{\\Lambda_{\\alpha}^{2}+\\left(\\boldsymbol{q}^{\\prime}-\\boldsymbol{q}\\right)^{2}}\\right]^{n_{\\alpha}}$$$$ Lastly i use the following equation to compute the Cross section $$$$\\sigma=\\frac{1}{64 \\pi^{2} s} \\frac{\\mathrm{p}_{f}}{\\mathrm{p}_{i}} \\int\\left|\\mathcal{M}_{f i}\\right|^{2} \\mathrm{~d} \\Omega$$$$ Where the cross section must include a factor of $$1/2$$ because each final state may only be counted once in integrating over the angles To my mind all of this should yield reasonable results. Not exact since this is a tree level treatment of the scattering but my results as i mentioned before do not even come close to the experimental results. I know that for Moller Scattering (Elastic electron scattering) tree level treatment is not enough since the integral is diverging. Is this one case where tree", "if $$a = 123 \\Rightarrow c = 00$$. \\begin{align} \\therefore (a\\omega 5)^2 &= \\big(a \\omega ]5 \\omega c[^{]a[}\\big)^2\\\\ &= a^2 \\omega \\big(a(5\\omega c) + a(5\\omega c)\\big) \\omega (5\\omega c)^2\\tag{Using Bimultiplication}\\\\ &= a^2 \\omega 2a(5\\omega c) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (10a \\omega 2ac) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (10a \\omega c) \\omega (5\\omega c)^2\\tag{Since, c is all zeroes}\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (25 \\omega 10c \\omega c^2)\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (25 \\omega c \\omega c)\\tag{1} \\end{align} Since each digit group must have $$]a[$$ digits, so let us move one zero from the middle $$c$$ in $$25 \\omega c \\omega c$$ to the rightmost $$c$$. So, now that group becomes $$25\\omega d\\omega e$$, where $$]d[ = ]c[ – 1$$ and $$]e[ = ]c[ + 1$$. \\begin{align} \\therefore (1) &= a^2 \\omega \\big( (a \\omega 0 \\omega c) + (25 \\omega d)\\big) \\omega e\\\\ &= a^2 \\omega \\big( a \\omega (0+25) \\omega (c+d) \\big) \\omega e\\\\ &= a^2 \\omega ( a \\omega 25 \\omega d ) \\omega e\\tag{2} \\end{align} Now, \\begin{align} ]25 \\omega d[ &= ]25[ + ]d[\\\\ &= 2 + (]c[ – 1)\\\\ &= 2 + \\big((]a[ – 1) – 1\\big)\\\\ &= ]a[ \\end{align} So,", "\\delta t) + \\sin(\\overline{\\omega} t) \\sin( \\delta t)]\\\\ & = \\frac{A}{4\\overline{\\omega} \\delta} [2 \\sin(\\overline{\\omega} t) \\sin( \\delta t)]\\\\ & = \\frac{A \\sin \\delta t}{2 \\overline{\\omega} \\delta} \\sin \\overline{\\omega} t. \\end{align*} Thus, we have a sine function, $\\sin \\overline{\\omega} t$ whose amplitude varies according to $|(A/2 \\overline{\\omega} \\delta) \\sin \\delta t|\\text{,}$ and $$x(t) = \\frac{A \\sin \\delta t}{2 \\overline{\\omega} \\delta} \\sin \\overline{\\omega} t.\\label{equation-secondorder04-resonance-3}\\tag{4.4.6}$$ ###### Example4.4.4 In the initial-value problem \\begin{align} x'' + 144 x \\amp = 4 \\cos(13t)\\label{equation-secondorder04-resonance-7}\\tag{4.4.7}\\\\ x(0) \\amp = 0\\tag{4.4.8}\\\\ x'(0) \\amp = 0,\\tag{4.4.9} \\end{align} we have $A= 4\\text{,}$ $\\omega_0 = 12\\text{,}$ and $\\omega = 13\\text{.}$ The solution to this equation is \\begin{equation*} x(t) = \\frac{4}{25}(\\cos 12 t - \\cos 13 t). \\end{equation*} The graph of this solution is given in Figure 4.4.5 and illustrate the phenomenon of beats. The two frequencies are almost equal and interfere with each other. In our example, $\\delta = -1/2$ and $\\overline{\\omega} = 25/2\\text{.}$ The factor $\\sin \\delta t = \\sin( - t/2)$ in (4.4.6) oscillates very slowly in comparison to $\\sin(\\overline{\\omega}) = \\sin(25t/2)\\text{.}$ Thus, we can view the solution (4.4.6) as a fast oscillation with a frequency $\\overline{\\omega}$ and amplitude $$\\left| \\frac{A \\sin \\delta t}{2 \\overline{\\omega} \\delta} \\right| = \\left| \\frac{8}{25} \\sin\\left( \\frac{t}{2}\\right)\\right|,\\label{equation-secondorder04-resonance-4}\\tag{4.4.10}$$ which oscillates much more slowly. If we superimpose the graph of (4.4.10) onto the graph in Figure", "(x-0) (x-1) \\\\ &= x(x-1) \\end {align*} This completes the step 2 of the algorithm. Since the degree $$d=0$$, then $$p(x)=1$$. Now $$P_i(x)$$ polynomials are generated using \\begin {align*} P_n &= - p(x) \\\\ P_{i-1} &= - S p'_{i} + ((n-i)S' -S\\theta ) P_i - (n-1)(i+1) S^2 r P_{i+1} \\qquad i=n,n-1,\\dots , 0 \\end {align*} These generate the following set \\begin {align*} P_{4} &= -1\\\\ P_{3} &= 2 x -4\\\\ P_{2} &= -3 \\left (x -2\\right )^{2}\\\\ P_{1} &= 3 \\left (x -2\\right )^{3}\\\\ P_{0} &= -\\frac {3 \\left (x -2\\right )^{4}}{2}\\\\ P_{-1} &= 0 \\end {align*} There is nothing to solve for from the last equation $$P_{-1} = 0$$ as $$p(x)=1$$ is already known because the degree $$d$$ was zero and hence there are no coefficients $$a_i$$ to solve for. Next $$\\omega$$ is determined as the solution to the following equation using $$n=4$$. \\begin {align*} \\sum _{i=0}^{n} S^i \\frac {P_i}{(n-i)!} \\omega ^i &= 0 \\\\ \\frac {P_0}{4!} + \\frac {S P_1}{3!} \\omega +\\frac {S^2 P_2}{2!} \\omega ^2 +\\frac {S^3 P_3}{1!} \\omega ^3 + +\\frac {S^4 P_4}{0!} \\omega ^4 &=0\\\\ -\\frac {1}{16}\\left (2 \\omega \\,x^{2}-2 x \\omega -x +2\\right )^{4} &=0 \\end {align*} Solving the above and using any one of the roots results in \\begin {align*} \\omega =\\frac {1}{2 x \\left (x -1\\right )}\\left (x -2\\right )", "$H(s) = Y(s)/X(s)$, of a linear time invariant (LTI) system. – Robert L. Sep 5 '18 at 22:05 • Hi. Thanks for clearing that up. There's a lot to learn indeed. – antwoord Sep 8 '18 at 12:13 Here is a way that you can approach this mathematically. It's not the only way. Another way to approach this is with Tchebyshev polynomials and I will leave that to someone else. This approach requires understanding two ingredients. 1. an identity that is a consequence of Euler's formula: $$\\cos(\\omega t) = \\tfrac12(e^{j\\omega t}+e^{-j\\omega t})$$ or $$\\sin(\\omega t) = \\tfrac1{2j}(e^{j\\omega t}-e^{-j\\omega t})$$ 1. the binomial theorem: \\begin{align} (a+b)^N &= \\sum_{n=0}^N {N \\choose n} \\, a^{N-n} \\, b^n \\\\ &= \\sum_{n=0}^N {N \\choose n} \\, a^n \\, b^{N-n} \\\\ \\end{align} or \\begin{align} (a+b)^N &= \\sum_{n=0}^N \\frac{N!}{(N-n)! \\, n!} \\, a^{N-n} \\, b^n \\\\ &= \\sum_{n=0}^N \\frac{N!}{(N-n)! \\, n!} \\, a^n \\, b^{N-n} \\\\ \\end{align} Now you might see where this is going. So using the cosine identity (so i don't have to mess around with powers of $j$ in the sine identity): \\begin{align} \\cos^k(\\omega t) &= \\frac{1}{2^k}(e^{j\\omega t}+e^{-j\\omega t})^k \\\\ \\\\ &= \\frac{1}{2^k} \\sum_{n=0}^k \\frac{k!}{(k-n)! \\, n!} \\, (e^{j\\omega t})^{k-n} \\, (e^{-j\\omega t})^n \\\\ \\\\ &= \\frac{1}{2^k} \\sum_{n=0}^k \\frac{k!}{(k-n)! \\, n!} \\, e^{j(k-n)\\omega t} \\, e^{-jn\\omega t} \\\\ \\\\ &= \\frac{1}{2^k} \\sum_{n=0}^k", "\\left( 3 + \\sqrt{2} \\right)$ How can you convert between them? - You have $$\\tan\\phi = \\frac{\\sin\\phi}{\\cos\\phi}= \\frac{3+\\sqrt{2}}{1+\\sqrt{2}}$$ Since $0<\\phi<\\pi/2$ we know that $\\sin\\phi,\\cos\\phi>0$. Therefore, $\\sin\\phi$ and $\\cos\\phi$ are equal to $$\\sin\\phi = \\frac{3+\\sqrt{2}}{\\sqrt{(3+\\sqrt{2})^2+(1+\\sqrt{2})^2}} =\\frac{3+\\sqrt{2}}{\\sqrt{14+8\\sqrt{2}}}$$ and $$\\cos\\phi = \\frac{1+\\sqrt{2}}{\\sqrt{(3+\\sqrt{2})^2+(1+\\sqrt{2})^2}} =\\frac{1+\\sqrt{2}}{\\sqrt{14+8\\sqrt{2}}}$$ Therefore $$\\begin{split} \\cos(\\omega t)(1+\\sqrt{2}) - \\sin(\\omega t)(3+\\sqrt{2}) &= \\sqrt{14+8\\sqrt{2}}\\left( \\cos(\\omega t)\\cos\\phi - \\sin(\\omega t) \\sin\\phi \\right)\\\\& = \\sqrt{14+8\\sqrt{2}} \\cos(\\omega t+\\phi) \\end{split}$$ $\\sqrt{14+8\\sqrt{2}}$, not $\\sqrt{14+\\sqrt2}$ – pharmine Nov 5 '11 at 9:03", "\\big) } \\\\ & = \\Re { \\big( - j 0.5 \\cos(\\omega t) + 0.5 \\sin(\\omega t) \\big) } \\\\ & = 0.5 \\sin(\\omega t) \\\\ \\end{align} In this case, $Y$ was purely imaginary, and we find that $y(t)$ purely has a $\\sin(\\omega t)$ term. In general, however, if $Y$ has nonzero real and imaginary parts, then the time-domain $y(t) = \\Re { \\big( Y \\cdot e^{j \\omega t} \\big) }$ will have both cosine and sine terms. Note that it's always possible to write a linear combination of $\\cos(x)$ and $\\sin(x)$ as only $\\cos(x + \\theta)$ for some phase angle $\\theta$. In simple 90-degree cases, we may have a simple expression such as: \\begin{align} \\sin(x) & = \\cos(x - \\frac {\\pi} {2}) \\\\ \\cos(x) & = \\sin(x + \\frac {\\pi} {2}) \\end{align} In more advanced cases for intermediate values, remember that: \\begin{align} f(t) & = \\Re{\\big( e^{j \\theta} e^{j \\omega t} \\big)} \\\\ & = \\Re{\\big( e^{j \\theta + j \\omega t} \\big)} \\\\ & = \\Re{\\big( e^{j ( \\theta + \\omega t )} \\big)} \\\\ & = \\Re{\\big( \\cos(\\theta + \\omega t) + j \\sin (\\theta + \\omega t) \\big)} \\\\ & = \\cos(\\theta + \\omega t) \\end{align} So for our example above, we could write: $$y(t) = 0.5 \\cos(\\omega t - \\frac {\\pi} {2})$$ Here,", "# Complex numbers - roots of unity Let $\\omega$ be a complex number such that $\\omega^5 = 1$ and $\\omega \\neq 1$. Find $$\\frac{\\omega}{1 - \\omega^2} + \\frac{\\omega^2}{1 - \\omega^4} + \\frac{\\omega^3}{1 - \\omega} + \\frac{\\omega^4}{1 - \\omega^3}.$$ I have tried adding the first two and the second two separately, then adding those sums but how do I get a numerical value as the answer? Thanks HINT : $$\\frac{\\omega}{1-\\omega^2}+\\frac{\\omega^2}{1-\\omega^4}+\\frac{\\omega^3}{1-\\omega}+\\frac{\\omega^4}{1-\\omega^3}$$ $$=\\frac{\\omega}{1-\\omega^2}+\\frac{\\omega^2}{1-\\omega^4}+\\frac{\\omega^7}{\\omega^4-\\omega^5}+\\frac{\\omega^6}{\\omega^2-\\omega^5}$$ The sum of the first and fourth terms is \\begin{align*} \\frac{\\omega}{1 - \\omega^2} + \\frac{\\omega^4}{1 - \\omega^3} &= \\frac{\\omega (1 - \\omega^3) + \\omega^4 (1 - \\omega^2)}{(1 - \\omega^2)(1 - \\omega^3)} \\\\ &= \\frac{\\omega - \\omega^4 + \\omega^4 - \\omega^6}{(1 - \\omega^2)(1 - \\omega^3)} \\\\ &= \\frac{\\omega - \\omega^4 + \\omega^4 - \\omega}{(1 - \\omega^2)(1 - \\omega^3)} \\\\ &= 0, \\end{align*}and the sum of the second and third terms is \\begin{align*} \\frac{\\omega^2}{1 - \\omega^4} + \\frac{\\omega^3}{1 - \\omega} &= \\frac{\\omega^2 (1 - \\omega) + \\omega^3 (1 - \\omega^4)}{(1 - \\omega^4)(1 - \\omega)} \\\\ &= \\frac{\\omega^2 - \\omega^3 + \\omega^3 - \\omega^7}{(1 - \\omega^4)(1 - \\omega)} \\\\ &= \\frac{\\omega^2 - \\omega^3 + \\omega^3 - \\omega^2}{(1 - \\omega^4)(1 - \\omega)} \\\\ &= 0. \\end{align*}Therefore, the sum of all four terms is $\\boxed{0}$. Maybe more direct $$\\frac{\\omega}{1 - \\omega^2} + \\frac{\\omega^2}{1 - \\omega^4} + \\frac{\\omega^3}{1 - \\omega} + \\frac{\\omega^4}{1 - \\omega^3}", "$Y$ would be a binomial random variable 463. \\begin{align*} 464. Y\\colon \\Omega &\\to \\{0, 1, 2, 3\\}\\\\ 465. \\omega &\\mapsto y 466. \\end{align*} 467. whose probability mass function is 468. $p_Y(y) = \\binom{3}{y}\\cdot 0.5^y\\cdot 0.5^{3-y} 469. = \\frac{3}{4y!(3-y)!}$ 470. 471. Therefore PMF of $Y$ for each case is 472. \\begin{align*} 473. p_Y(0) = p_Y(3) &= \\frac{3}{4\\cdot 0!3!} = \\frac{1}{8}\\\\ 474. p_Y(1) = p_Y(2) &= \\frac{3}{4\\cdot 1!2!} = \\frac{3}{8} 475. \\end{align*} 476. \\end{enumerate} 477. 478. \\exercise 2 Toss a pair of fair siz-sided dice 479. and let $X$ be the sum of the points 480. \\begin{align*} 481. X\\colon \\Omega &\\to [2, 12]\\cap\\mathbb Z\\\\ 482. \\omega &\\mapsto x 483. \\end{align*} 484. with $\\Omega = S^2 = \\{1, 2, 3, 4, 5, 6\\}^2 \\Longrightarrow n(\\Omega) = 36$. 485. 486. \\begin{enumerate}[(a)] 487. \\item $X$ is a random variable whose PMF is 488. \\begin{align*} 489. p_X(2) &= P\\{(1,1)\\} = \\frac{1}{36}\\\\ 490. p_X(3) &= P\\{(1,2), (2,1)\\} = \\frac{1}{18}\\\\ 491. p_X(4) &= P\\{(1,3), (2,2), (3,1)\\} = \\frac{1}{12}\\\\ 492. p_X(5) &= P\\{(1,4), (2,3), (3,2), (4,1)\\} = \\frac{1}{9}\\\\ 493. p_X(6) &= P\\{(1,5), (2,4), (3,3), (4,2), (5,1)\\} = \\frac{5}{36}\\\\ 494. p_X(7) &= P\\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\\} = \\frac{1}{6}\\\\ 495. p_X(8) &= P\\{(2,6), (3,5), (4,4), (5,3), (6,2)\\} = \\frac{5}{36}\\\\ 496. p_X(9) &= P\\{(3,6), (4,5), (5,4), (6,3)\\} = \\frac{1}{9}\\\\ 497. p_X(10) &= P\\{(4,6), (5,5), (6,4)\\} = \\frac{1}{12}\\\\ 498.", "\\{0,1,\\dots, n-1\\}$. We denote: $\\omega_n = (1, \\frac{2\\pi}{n}) = e^{i\\frac{2\\pi}{n}}$, and then all $n$ of the $n^{th}$ roots of unity are $\\omega_n^0, \\omega_n^1, \\omega_n^2, \\dots, \\omega_n^{n-1}$. Key Properties: 1. They satisfy the $\\pm$ property when $n$ is even i.e. $\\omega_n^0 = - \\omega_n^{\\frac{n}{2}}$ $\\omega_n^1 = - \\omega_n^{\\frac{n}{2}+1}$ $\\vdots$ $\\omega_n^{\\frac{n}{2}-1} = - \\omega_n^{n-1}$ To see why, observe that $\\omega_n^j = (1, \\frac{2\\pi j}{n})$ and $\\omega_n^{\\frac{n}{2} + j} = (1, \\frac{2\\pi }{n} ( {\\frac{n}{2} + j})) = (1, \\pi + \\frac{2\\pi j}{n}) = - (1, \\frac{2\\pi j}{n}) = -\\omega_n^j$ 2. The squares of the $n^{th}$ roots of unity are the ${\\frac{n}{2}}^{th}$ roots of unity when $n$ is a power of $2$. To see why, observe that ${(\\omega_n^j)}^2 = {(1, \\frac{2\\pi j}{n})}^2 = (1, \\frac{4\\pi j }{n}) = (1, \\frac{2\\pi j }{\\frac{n}{2}}) = \\omega_{\\frac{n}{2}}^j$ FFT Algorithm Given a polynomial of degree at most $n-1$, $A(x) = a_0 + a_1 x + \\dots + a_{n-1}x^{n-1}$, we want to choose $n$ points $x_0, x_1, \\dots, x_{n-1}$ and evaluate $A(x)$ at these points. (We can view $A(x)$ as a polynomial of degree $\\leq 2n-1$ in order to get it at $2n$ points.) We require the points $x_0, x_1, \\dots, x_{n-1}$ to satisfy the $\\pm$ property, i.e., $x_j = - x_{\\frac{n}{2} + j}$ for $j \\in \\{0,1,\\dots, \\frac{n}{2}-1\\}$. Hence we will choose the points to", "\\omega \\underline{2}5\\\\ &= a^2 \\omega (a\\omega2) \\omega 5\\\\ &= (a^2 + a) \\omega 2 \\omega 5\\\\ &= \\big(a(a+1)\\big) \\omega 25\\tag{Proved} \\end{align} ### Case 2: $$a$$ has more than one digits. So, $$]a[ > (]5[ = 1)$$. But to apply Bimultiplication formula $$a$$ must have the same number of digits in $$5$$, which is obviously not the case here. So, we will use one trick. We will pad $$5$$ with some number of zeroes on the right, so that, $$]a[ = ]5\\omega c[$$, where, $$c$$ is all zeroes and $$]c[ = ]a[ - 1$$. So, if $$a = 123 \\Rightarrow c = 00$$. \\begin{align} \\therefore (a\\omega 5)^2 &= \\big(a \\omega ]5 \\omega c[^{]a[}\\big)^2\\\\ &= a^2 \\omega \\big(a(5\\omega c) + a(5\\omega c)\\big) \\omega (5\\omega c)^2\\tag{Using Bimultiplication}\\\\ &= a^2 \\omega 2a(5\\omega c) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (10a \\omega 2ac) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (10a \\omega c) \\omega (5\\omega c)^2\\tag{Since, c is all zeroes}\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (5\\omega c)^2\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (25 \\omega 10c \\omega c^2)\\\\ &= a^2 \\omega (a \\omega 0 \\omega c) \\omega (25 \\omega c \\omega c)\\tag{1} \\end{align} Since each digit group must have $$]a[$$ digits, so let us move one zero from the middle $$c$$ in $$25 \\omega c \\omega c$$", "x^4d^2\\Omega=\\frac{4\\pi}5=\\oint x^4r^2d^2\\Omega=\\oint x^6d^2\\Omega+2\\oint x^4y^2d^2\\Omega$$ $$\\oint x^6d^2\\Omega=\\oint \\left(\\frac{x+y}{\\sqrt2}\\right)^6d^2\\Omega=\\frac14\\oint x^6d^2\\Omega+\\frac{15}4\\oint x^4y^2d^2\\Omega$$ Solving the last $3$ equations in $3$ unknowns we get $$\\oint x^6d^2\\Omega=\\frac{4\\pi}7\\text{, }\\oint x^4y^2d^2\\Omega=\\frac{4\\pi}{35}\\text{, and }\\oint x^2y^2z^2d^2\\Omega=\\frac{4\\pi}{105}$$ Almost there :) \\begin{align}\\oint x^2d^2\\Omega & =\\frac{4\\pi}3=\\oint x^2r^6d^2\\Omega \\\\ & =\\oint x^8d^2\\Omega+8\\oint x^6y^2d^2\\Omega+6\\oint x^4y^4d^2\\Omega+12\\oint x^4y^2z^2d^2\\Omega\\end{align} \\begin{align}\\oint x^4d^2\\Omega & =\\frac{4\\pi}5=\\oint x^4r^4d^2\\Omega \\\\ & =\\oint x^8d^2\\Omega+4\\oint x^6y^2d^2\\Omega+2\\oint x^4y^4d^2\\Omega+2\\oint x^4y^2z^2d^2\\Omega\\end{align} $$\\oint x^6d^2\\Omega=\\frac{4\\pi}7=\\oint x^6r^2d^2\\Omega=\\oint x^8d^2\\Omega+2\\oint x^6y^2d^2\\Omega$$ $$\\oint x^8d^2\\Omega=\\oint \\left(\\frac{x+y}{\\sqrt2}\\right)^8d^2\\Omega=\\frac18\\oint x^8d^2\\Omega+\\frac72\\oint x^6y^2d^2\\Omega+\\frac{35}8\\oint x^4y^4d^2\\Omega$$ Solving these $4$ equations in $4$ unknowns, $$\\oint x^8d^2\\Omega=\\frac{4\\pi}{9}\\text{, }\\oint x^6y^2d^2\\Omega=\\frac{4\\pi}{63}\\text{, }\\oint x^4y^4d^2\\Omega=\\frac{4\\pi}{105}\\text{, and }\\oint x^4y^2z^2d^2\\Omega=\\frac{4\\pi}{315}$$ With these integrals in hand we want to normalize the polynomials $xy$ and $x^2-y^2$. $$\\oint (xy)^2d^2\\Omega=\\oint x^2y^2d^2\\Omega=\\frac{4\\pi}{15}$$ So a normalized spherical harmonic that transforms as the $xy$ partner of the $T_{2g}$ representation of $O_h$ and as the $z$ partner the $T_2$ representation of $T_d$ is $$Y_{xy}^2(x,y,z)=\\left(\\frac{15}{4\\pi}\\right)^{1/2}\\frac{xy}{r^2}$$ And we have $$\\psi_{xy}(x,y,z)=Y_{xy}^2(x,y,z)f_{xy}(r)$$ To find the electrostatic potential energy of an electron with such a wave function in the $O_h$ potential we integrate over its charge density to get \\begin{align}U_{O_h,xy} & =\\int V_{O_h}(x,y,z)\\rho_{xy}(x,y,z)d^3volume \\\\ & =\\int\\left[-\\frac{6Ze}{4\\pi\\epsilon_0a}-D\\left(x^4+y^4+z^4-\\frac35r^4\\right)\\right]\\times \\\\ & (-e)\\left[\\left(\\frac{15}{4\\pi}\\right)^{1/2}\\frac{xy}{r^2}f_{xy}(r)\\right]^*\\left[\\left(\\frac{15}{4\\pi}\\right)^{1/2}\\frac{xy}{r^2}f_{xy}(r)\\right]d^3volume \\\\ & =\\frac{6Ze^2}{4\\pi\\epsilon_0a}\\oint\\left|\\left(\\frac{15}{4\\pi}\\right)^{1/2}\\frac{xy}{r^2}\\right|^2d^2\\Omega\\int_0^{\\infty}\\left|f_{xy}(r)\\right|^2r^2dr \\\\ & +De\\left(\\frac{15}{4\\pi}\\right)\\oint \\left(\\frac{xy}{r^2}\\right)^2\\frac{x^4+y^4+z^4-\\frac35r^4}{r^4}\\int_0^{\\infty}r^4\\left|f_{xy}(r)\\right|^2r^2dr \\\\ & =\\frac{6Ze^2}{4\\pi\\epsilon_0a}(1)(1)+\\frac{15De}{4\\pi}\\left(\\frac{4\\pi}{63}+\\frac{4\\pi}{63}+\\frac{4\\pi}{315}-\\frac35\\frac{4\\pi}{15}\\right)\\left\\langle r^4\\right\\rangle \\\\ & =\\frac{6Ze^2}{4\\pi\\epsilon_0a}-\\frac{8De}{105}\\left\\langle r^4\\right\\rangle\\end{align} Where we have used the separate normalizations of the angular part of the wave function $Y_{xy}^2(x,y,z)$ and the radial part $f_{xy}(r)$ and the average value of $r^4$ for the radial distribution, $\\left\\langle r^4\\right\\rangle$. Now we calculate $$\\oint", "take $(\\omega + \\omega ^8)(\\omega ^2 + \\omega ^7)(\\omega ^4+ \\omega ^5)$, where $\\omega = \\cos \\frac{2\\pi}{9} +i\\sin \\frac{2\\pi}{9}$. We have that $(\\omega + \\omega ^8)(\\omega ^2 + \\omega ^7)(\\omega ^4+ \\omega ^5) = (2\\cos \\frac{2\\pi}{9})(2\\cos \\frac{4\\pi}{9})(-2\\cos \\frac{\\pi}{9})$. We also have that by expanding the brackets and simplifying using part (b), $(\\omega + \\omega ^8)(\\omega ^2 + \\omega ^7)(\\omega ^4+ \\omega ^5) = -1$. Hence, we have that $(2\\cos \\frac{2\\pi}{9})(2\\cos \\frac{4\\pi}{9})(-2\\cos \\frac{\\pi}{9}) = -1 \\implies \\cos \\frac{2\\pi}{9} \\cos \\frac{4\\pi}{9} \\cos \\frac{\\pi}{9} = 1$. What's evident in this question is that you occasionally need to look to manipulate the question into a nicer form - here, the solution only really became obvious for me after manipulating the result to $(2\\cos \\frac{2\\pi}{9})(2\\cos \\frac{4\\pi}{9})(-2\\cos \\frac{\\pi}{9}) = -1$. From this sort of position, it's a lot simpler and much less time consuming to explore possible solutions, especially given a number of results from previous parts of the question. Try this line of thinking with similar questions! Hope this helps Alternatively, we know that: \\begin{aligned}1+z +z ^2+...+z^8 &= \\prod_{k=1}^4\\left(z- cis\\frac{2k\\pi}{9}\\right)\\left(z-cis\\left(-\\frac{2k\\pi}{9}\\right)\\right)\\\\ &=\\prod_{k=1}^4\\left(z^2-2\\cos\\frac{2k\\pi}{9}z+1\\right)\\end{aligned} letting $z=i$: \\begin{aligned}\\prod_{k=0}^4\\left(-2i\\cos\\frac{2k\\pi}{9}\\right)&=1\\\\ 16\\cos\\frac{2\\pi}{9}\\cos\\frac{4\\pi}{9}\\cos\\frac{2\\pi}{3}\\cos\\frac{8\\pi}{9}&=1\\\\ -\\frac{1}{2}\\cos\\frac{2\\pi}{9}\\cos\\frac{4\\pi}{9}\\cos\\frac{8\\pi}{9}&=\\frac{1}{16}\\\\ -\\cos\\frac{\\pi}{9}\\cos\\frac{2\\pi}{9}\\cos\\frac{4\\pi}{9}&=-\\frac{1}{8}\\\\ \\implies \\cos\\frac{\\pi}{9}\\cos\\frac{2\\pi}{9}\\cos\\frac{4\\pi}{9}&=\\frac{1}{8}\\end{aligned}", "away from the origin; these points will be the intersections of the circle centered at $(1,0)$ with radius $1$ and the circle centered at $(0,0)$ with radius $\\sqrt{2+\\sqrt{3}}$. The equations of these circles are $(x-1)^2 = 1$ and $x^2 + y^2 = 2 + \\sqrt{3}$. Solving for $x$ yields $x = \\frac{\\sqrt{3}}{2}$. Clearly, this means that the real part of $v$ is greater than $\\frac{\\sqrt{3}}{2}$. Solving, we note that $332$ possible $v$s exist, meaning that $\\frac{m}{n} = \\frac{332}{1996} = \\frac{83}{499}$. Therefore, the answer is $83 + 499 = \\boxed{582}$. ### Solution 4 Since $z^{1997}=1$, the roots will have magnitude $1$. Thus, the roots can be written as $\\cos(\\theta)+i\\sin(\\theta)$ and $\\cos(\\omega)+i\\sin(\\omega)$ for some angles $\\theta$ and $\\omega$. We rewrite the requirement as $\\sqrt{2+\\sqrt3}\\le|\\cos(\\theta)+\\cos(\\omega)+i\\sin(\\theta)+i\\sin(\\omega)|$, which can now be easily manipulated to $2+\\sqrt{3}\\le(\\cos(\\theta)+\\cos(\\omega))^2+(\\sin(\\theta)+\\sin(\\omega))^2$. WLOG, let $\\theta = 0$. Thus, our inequality becomes $2+\\sqrt{3}\\le(1+\\cos(\\omega))^2+(\\sin(\\omega))^2$, $2+\\sqrt{3}\\le2+2\\cos(\\omega)$, and finally $\\cos(\\omega)\\ge\\frac{\\sqrt{3}}{2}$. Obviously, $\\cos(\\frac{\\pi}{6})=\\frac{\\sqrt{3}}{2}$, and thus it follows that, on either side of a given point, $\\frac{1997}{12}\\approx166$ points will work. The probability is $\\frac{166\\times2}{1996} = \\frac{83}{499}$, and thus our requested sum is $\\boxed{582}$ ~SigmaPiE" ]

[ "\\frac{5}{34}\\sin 5\\theta\\right).", "{0}^{+}} \\sin \\frac{x}{x}\\right) \\left({\\lim}_{x \\to {0}^{+}} \\tan \\left(x\\right)\\right)$ $= - 1 \\cdot 0$ $= 0$", "+ \\beta}{2}\\right) \\sin \\left(\\frac{\\alpha - \\beta}{2}\\right)}{2 \\cos \\left(\\frac{\\alpha + \\beta}{2}\\right) \\sin \\left(\\frac{\\alpha - \\beta}{2}\\right)}$ Hence $a = b \\tan \\left(\\frac{\\alpha + \\beta}{2}\\right)$ Proved", "\\tan t \\mid + E \\right ) \\sin t \\end{align}", "\\left(t\\right)}=\\frac{1-\\tan t}{1+\\tan t}$", "side as 1. The $\\tan(\\theta) = \\frac x 1$. Now find $sin(\\theta)$.", "# How do you find the exact value of sin^-1(tan(-pi/4))? ##### 1 Answer Sep 22, 2016 ${\\sin}^{- 1} \\left(\\tan \\left(- \\frac{\\pi}{4}\\right)\\right) = - \\frac{\\pi}{2}$ #### Explanation: As $\\tan \\left(- \\frac{\\pi}{4}\\right) = - 1$ and $\\sin \\left(- \\frac{\\pi}{2}\\right) = - 1$ ${\\sin}^{- 1} \\left(\\tan \\left(- \\frac{\\pi}{4}\\right)\\right) = - \\frac{\\pi}{2}$", "\\left(\\frac{1}{7}\\right)\\right) + i \\sin \\left({\\tan}^{-} 1 \\left(\\frac{2}{5}\\right) - {\\tan}^{-} 1 \\left(\\frac{1}{7}\\right)\\right)\\right)$", "# How do you find the exact value for sin (tan^-1 (3/4))? $= \\sin \\left({\\tan}^{-} 1 \\left(\\frac{3}{4}\\right)\\right) = \\sin \\left({\\cot}^{-} 1 \\left(\\frac{4}{3}\\right)\\right) = \\sin \\left({\\csc}^{-} 1 \\sqrt{1 + {\\left(\\frac{4}{3}\\right)}^{2}}\\right)$ $= \\sin \\left({\\csc}^{-} 1 \\left(\\frac{5}{3}\\right)\\right) = \\sin \\left({\\sin}^{-} 1 \\left(\\frac{3}{5}\\right)\\right) = \\frac{3}{5}$", "= \\frac{2\\sin y \\cos d + \\sin y}{2 \\cos y \\cos d + \\cos y}\\\\ & = \\tan y. \\end{align*}", "# How do you find the exact value of sin(tan^-1 -5/7)? Jan 13, 2017 $- \\frac{5}{\\sqrt{74}}$ #### Explanation: ${\\tan}^{- 1} \\left(- \\frac{5}{7}\\right) = {\\sin}^{- 1} \\left(- \\frac{5}{\\sqrt{{5}^{2} + {7}^{2}}}\\right) = {\\sin}^{- 1} \\left(- \\frac{5}{\\sqrt{74}}\\right)$ The given expression is $\\sin \\left({\\sin}^{- 1} \\left(- \\frac{5}{\\sqrt{74}}\\right)\\right) = - \\frac{5}{\\sqrt{74}}$. Note that , if arc tan is negative, the angle is in ${Q}_{4}$, and so, $a r c \\sin \\in {Q}_{4}$ is also negative.", "\\pi + {\\tan}^{-} 1 \\left(\\frac{- 1}{4}\\right) + 2 \\pi\\right) + i \\sin \\left({\\tan}^{-} 1 \\left(6\\right) + \\pi + {\\tan}^{-} 1 \\left(\\frac{- 1}{4}\\right) + 2 \\pi\\right)\\right)$", "+ \\sin^2\\frac{C}{2}\\right) \\\\ &= 6 - (a^2 + b^2 + c^2) = 2 + abc, \\end{align*} as desired.", "\\left[ \\tan x + 2 \\ln |\\sec x + \\tan x| - \\frac{1}{2} x - \\frac{1}{2} \\left( \\frac{1}{2} \\right) \\sin (2x) - 2 \\sin x\\right]^{\\pi/3}_0\\\\ \\\\ &= 5.95 \\text{ cubic units} \\end{aligned}", "$\\tag{12} {µ_0}=\\frac{\\sin{\\left( \\alpha \\right) }}{\\cos{\\left( \\alpha \\right) }}$ or converted to $\\tag{13} {µ_0}=\\tan{\\left( \\alpha \\right) }$", "the right without touching the right side) tan θ/(1 + sec θ) + (1 + sec θ)/tan θ = 2 csc θ 4. Prove (make the left side look like the right without touching the right side) tan²θ/(1 + tan²θ) = sin²θ", "\\tan^{-1}\\left(\\frac{(n+1)-n}{1+n(n+1)}\\right) =\\tan^{-1}(n+1) -\\tan^{-1}(n)$$", "t)=\\tan^{-1}\\left(\\frac{\\sin\\vartheta}{\\cos\\vartheta}\\right) =\\tan^{-1}\\left(\\tan\\vartheta\\right)=\\vartheta$ --Chris p.s. From now on, ask new questions in a new thread :)", "\\left(\\frac{5 \\pi}{6}\\right) + i \\sin \\left(\\frac{5 \\pi}{6}\\right)\\right]$! Hope this helps!", "\\frac{5}{34}\\sin 5\\theta\\right)$." ]

[ "\\cdot \\sin A \\left(\\sin A + 2 \\cos A\\right) \\end{align*} after angle addition identity. Similarly, $\\angle{EAC} = 90 - \\angle A,$ and by Law of Sines $CE = 8 \\sin{\\angle{EAC}} = 8 \\cos A.$ Note that $\\angle{E} = \\angle{ACB}$ from inscribed angles, so \\begin{align*} [ACE] &= \\dfrac{1}{2} AC \\cdot CE \\sin{\\angle C} \\\\ &= \\dfrac{1}{2} \\cdot \\dfrac{8 \\sqrt{5}}{5} \\cdot 8 \\cos A \\sin{\\left[180 - \\left(90 - \\angle A\\right) - \\angle E\\right]} \\\\ &= \\dfrac{32 \\sqrt{5}}{5} \\cdot \\cos A \\sin{\\left[\\left(90 - \\angle A\\right) + \\angle{ACB}\\right]} \\\\ &= \\dfrac{32 \\sqrt{5}}{5} \\cdot \\cos A \\left(\\dfrac{2 \\sqrt{5} \\cos A}{5} + \\dfrac{\\sqrt{5} \\sin A}{5}\\right) \\\\ &= \\dfrac{32}{5} \\cdot \\cos A \\left(\\sin A + 2 \\cos A\\right) \\end{align*} after angle addition identity. Setting the two areas equal, we get $$\\tan A = \\frac{\\sin A}{\\cos A} = 8 \\iff \\sin A = \\frac{8}{\\sqrt{65}}, \\cos A = \\frac{1}{\\sqrt{65}}$$ after Pythagorean Identity. Now plug back in and the common area is $\\frac{64}{65} \\iff \\boxed{129}.$ ## Solution 2 $[asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label(\"A\",A,NE);label(\"O_1\",O_1,NE);label(\"O_2\",O_2,NE);label(\"B\",B,SW);label(\"C\",C,SW);label(\"D\",D,NE);label(\"E\",E,NE);label(\"N\",N,W);label(\"K\",(-24/15,0.2));label(\"L\",(24/15,0.2));label(\"n\",(-0.8,-0.12));label(\"p\",((29/15,-48/15)));label(\"\\mathcal{P}\",(-1.6,1.1));label(\"\\mathcal{Q}\",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); [/asy]$ Call $O_1$ and $O_2$ the centers of circles $\\mathcal{P}$ and $\\mathcal{Q}$, respectively, and extend $CB$ and $O_2O_1$ to meet at point $N$. Call $K$ and $L$ the feet of the altitudes from $B$ to $O_1N$ and $C$ to $O_2N$, respectively. Using the fact that $\\triangle{O_1BN} \\sim", "$IT \\perp BC$, and let $S$ be on $AD$ where $IS \\perp AD$. By HL Congruency, $\\triangle ITQ \\cong \\triangle ISP$, so $PS = QT$. Now we have information we can use to determine side lengths to compute the inradius so we can determine the area of $ABCD$. Let $a = AS = AE$. That means $EB = BT = 7-a$, $TC = CF = a+2$, and $FD = SD = 13-a$. That means $AD = 13$. Additionally, since $AP = 6.5$ and $BQ = 4.5$, $QT = 2.5-a$ and $PS = a-6.5$. Since $QT = PS$, we must have $a = 4.5$. We know that $\\angle AIS + \\angle DIS + \\angle BIT + \\angle CIT = 180^\\circ$, so \\begin{align*} \\tan(\\angle AIS + \\angle DIS) &= -\\tan(\\angle BIT + \\angle CIT) \\\\ \\frac{\\tan \\angle AIS + \\tan \\angle DIS}{1 - \\tan \\angle AIS \\tan \\angle DIS} &= \\frac{\\tan \\angle BIT + \\tan \\angle CIT}{1 - \\tan \\angle BIT \\tan \\angle CIT} \\end{align*} Let $r$ be the radius of the incircle. That means \\begin{align*} \\frac{\\frac{4.5}{r} + \\frac{8.5}{r}}{1 - \\frac{4.5 \\cdot 8.5}{r^2}} &= -\\frac{\\frac{2.5}{r} + \\frac{6.5}{r}}{1 - \\frac{2.5 \\cdot 6.5}{r^2}} \\\\ \\frac{13r}{r^2-\\frac{153}{4}} &= \\frac{-9r}{r^2 - \\frac{65}{4}} \\\\ 13r^3 - \\frac{13 \\cdot 65}{4}r &= -9r^3 + \\frac{153 \\cdot 9}{4} \\\\ 22r^2 &= \\frac{13 \\cdot 65 + 153 \\cdot 9}{4} \\\\", "3C = - 4\\cos \\frac{{3A}}{2}\\cos \\frac{{3B}}{2}\\cos \\frac{{3C}}{2}$ For this to be zero, one of the cos terms on the RHS must be zero, which means that one of the angles must be \\begin{align}\\frac{\\pi }{3}\\end{align}. Example - 31 In a $$\\Delta ABC$$, if $\\cos A\\cos B\\cos C = \\frac{{\\sqrt 3 - 1}}{8},\\;\\sin A\\sin B\\sin C = \\frac{{3 + \\sqrt 3 }}{8}$ Find the angles of the triangle Solution: From the given relations, $\\tan A\\tan B\\tan C = \\frac{{3 + \\sqrt 3 }}{{\\sqrt 3 - 1}}$ Using the result of Example - 28, we have $\\tan A + \\tan B + \\tan C = \\frac{{3 + \\sqrt 3 }}{{\\sqrt 3 - 1}}$ Now, since \\begin{align}A + B + C = \\pi \\end{align} \\begin{align}&\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\cos (A + B + C) = - 1 \\\\ \\Rightarrow\\quad &\\qquad\\;\\;\\cos A\\cos B\\cos C\\left\\{ {1 - \\sum (\\tan A\\tan B)} \\right\\} = - 1 \\\\ \\Rightarrow\\quad &\\qquad\\;\\; \\sum \\left( {\\tan A\\tan B} \\right) = 1 + \\frac{8}{{\\sqrt 3 - 1}} = 5 + 4\\sqrt 3 \\\\ \\end{align} Finally, for $$\\tan A,\\tan B,\\tan C$$, we have their sum, their sum taken pairwise and their product, and thus we can say that these three are the roots of the equation \\begin{align}&\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{x^3} - \\left( {\\frac{{3 + \\sqrt 3 }}{{\\sqrt 3 - 1}}} \\right){x^2} + (5 + 4\\sqrt 3 )x -", "$\\angle M B A$, intersecting the side $A M$ (opp. to $\\angle M B A$) in pt. $P$. Hence, by the $\\angle$-bisector Theorem, $\\frac{A P}{P M} = \\frac{A B}{B M} = \\frac{1}{\\frac{1}{2}} = 2$ $\\Rightarrow \\frac{A P}{A P + P M} = \\frac{2}{2 + 1} = \\frac{2}{3} \\Rightarrow \\frac{A P}{A M} = \\frac{2}{3}$ $\\Rightarrow A P = \\frac{2}{3} \\cdot M A = \\frac{2}{3} \\cdot \\frac{\\sqrt{5}}{2} , i . e . , \\frac{\\sqrt{5}}{3.} \\ldots \\ldots \\ldots . \\left(1\\right)$ Arguing on the same lines, $A Q = \\frac{4}{5} \\cdot N A = \\frac{4}{5} \\cdot \\frac{\\sqrt{17}}{4} = \\frac{\\sqrt{17}}{5.} . . \\left(2\\right)$. In the light of $\\left(\\star\\right) ,$ now, it remains to find $m \\angle P A Q$. We have, from $r i g h t \\Delta M B A$, tan/_MAB=(MB)/(AB)=(1/2)/1=1/2, &, tan /_NAB=1/4. $\\therefore \\tan \\angle M A N = \\tan \\angle P A Q = \\tan \\left(\\angle M A B - \\angle N A B\\right)$ $= \\frac{\\frac{1}{2} - \\frac{1}{4}}{1 + \\frac{1}{2} \\cdot \\frac{1}{4}} = \\frac{1}{4} \\cdot \\frac{8}{9}$, i.e., $\\tan \\angle P A Q = \\frac{2}{9}$. $\\therefore \\sin \\angle P A Q = \\frac{2}{\\sqrt{85.}} \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\left(3\\right)$. Therefore, by (star),(1),(2), &, (3), Area of DeltaAQP=1/2(sqrt5/3)(sqrt17)/5)(2/sqrt85)=1/15#, as was derived earlier!", "cosines ($C^2 = A^2 + B^2 - 2AB\\cos \\theta$), we get that $(\\sqrt{33})^2 = 2^2 + 5^2 - 2 \\cdot 2 \\cdot 5 \\cos \\theta$. Simplifying this, we get $33 = 29 - 20 \\cos \\theta$, which means that $\\displaystyle \\cos \\theta = -\\frac{1}{5}$ Because $\\cos^2 \\theta + \\sin^2 \\theta = 1$, we get that $\\displaystyle \\sin^2 \\theta = \\frac{24}{25}$. This means that $\\displaystyle \\sin \\theta = \\frac{2\\sqrt{6}}{5}$ (note that, because $0 \\le \\theta \\le \\pi$, $\\sin \\theta \\ge 0$). The area of the triangle is $\\displaystyle \\frac{1}{2} A B \\sin \\theta = \\frac{1}{2} \\cdot 2 \\cdot 5 \\cdot \\frac{2\\sqrt{6}}{5} = 2\\sqrt{6}$. - Thank you, this is clear, concise, and could be done quickly and easily in the situation. This is probably what the judges/writers of it wanted readers to do. – Asimov Apr 28 '14 at 23:48 Use the $\\frac{1}{4}\\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$ form of Herons' formula. \\begin{align} & \\frac{1}{4}\\sqrt{4\\cdot4\\cdot25-(4+25-33)^2} \\\\ = & \\frac{1}{4}\\sqrt{4^2\\cdot25-4^2} \\\\ = & \\sqrt{25-1} \\\\ = & 2\\sqrt{6} \\end{align} - This is the way I would have done it, nice. – Sawarnik Apr 29 '14 at 6:40 You could've used Heron's formula straight away, actually. \\begin{align} T & = \\tfrac{1}{4} \\sqrt{(a+b-c)(a-b+c)(-a+b+c)(a+b+c)} \\\\ & = \\tfrac{1}{4} \\sqrt{(2+5-\\sqrt{33})(2-5+\\sqrt{33})(-2+5+\\sqrt{33})(2+5+\\sqrt{33})} \\\\ & = \\tfrac{1}{4} \\sqrt{(7-\\sqrt{33})(-3+\\sqrt{33})(3+\\sqrt{33})(7+\\sqrt{33})} \\\\ & = \\tfrac{1}{4} \\sqrt{(7+\\sqrt{33})(7-\\sqrt{33})(\\sqrt{33}+3)(\\sqrt{33}-3)} \\\\ & = \\tfrac{1}{4} \\sqrt{(49-33)(33-9)} \\\\ & = \\tfrac{1}{4} \\sqrt{16", "+ c^2 - 4)}}{2} = \\frac{-bc + \\sqrt{(4 - b^2)(4 - c^2)}}{2}.$$ This asks for the trigonometric substitution $b = 2\\sin u$ and $c = 2\\sin v$, where $0^\\circ < u,v < 90^\\circ$. Then $$a = 2(-\\sin u\\sin v + \\cos u\\cos v) = 2\\cos (u + v),$$ and if we set $u = B/2$ and $v = C/2$, then $a = 2\\sin (A/2)$, $b = 2\\sin (B/2)$, and $c = \\sin (C/2)$, where $A$, $B$, and $C$ are the angles of a triangle. We have \\begin{align*} ab &= 4\\sin\\frac{A}{2}\\sin\\frac{B}{2} \\\\ &= 2\\sqrt{\\sin A\\tan\\frac{A}{2}\\sin B\\tan\\frac{B}{2}} = 2\\sqrt{\\sin A\\tan\\frac{B}{2}\\sin B\\tan\\frac{A}{2}} \\\\ &\\leq \\sin A\\tan\\frac{B}{2} + \\sin B\\tan\\frac{A}{2} \\\\ &= \\sin A\\cot\\frac{A + C}{2} + \\sin B\\cot\\frac{B + C}{2}, \\end{align*} where the inequality step follows from AM-GM. Likewise, \\begin{align*} bc &\\leq \\sin B\\cot\\frac{B + A}{2} + \\sin C\\cot\\frac{C + A}{2}, \\\\ ca &\\leq \\sin A\\cot\\frac{A + B}{2} + \\sin C\\cot\\frac{C + B}{2}. \\end{align*} Therefore \\begin{align*} ab + bc + ca &\\leq (\\sin A + \\sin B)\\cot\\frac{A + B}{2} + (\\sin B + \\sin C)\\cot\\frac{B + C}{2} + (\\sin C + \\sin A)\\cot\\frac{C + A}{2} \\\\ &= 2\\left(\\cos\\frac{A - B}{2}\\cos\\frac{A + B}{2} + \\cos\\frac{B - C}{2}\\cos\\frac{B + C}{2} + \\cos\\frac{C - A}{2}\\cos\\frac{C + A}{2} \\right)\\\\ &= 2(\\cos A + \\cos B + \\cos C) \\\\ &= 6 - 4\\left(\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2}", "$$a$$, $$tan(a)$$, as follows: \\begin{aligned} tan(a) &= \\frac{O}{A} \\\\ & = \\frac{3}{4} \\\\ tan(a) & = 0.75 \\end{aligned} #### Relative to angle $$b$$ We calculate the sine of angle $$b$$, $$tan(b)$$, as follows: \\begin{aligned} tan(b) &= \\frac{O}{A} \\\\ & = \\frac{4}{3} \\\\ tan(b) & = 1.33... \\end{aligned} Notice that depending on the angle we're looking at, $$a$$ or $$b$$, the Adjacent and the Opposite Side Lengths changes. Consequently the value of tangent changes depending on whether we're cosidering the angle $$a$$ or the angle $$b$$. ## Exercise 1 For each of the right angle triangles, below, calculate: 1. $$sin(a)$$ 2. $$cos(a)$$ 3. $$tan(a)$$ 1. $$sin(b)$$ 2. $$cos(b)$$ 3. $$tan(b)$$ #### Triangle 4 Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working #### Triangle 1 1. $$sin(a) = 0.882$$ 2. $$cos(a) = 0.471$$ 3. $$tan(a) = 1.88$$ 1. $$sin(b) = 0.471$$ 2. $$cos(b) = 0.882$$ 3. $$tan(b) = 0.533$$ #### Triangle 2 1. $$sin(a) = 0.6$$ 2. $$cos(a) = 0.8$$ 3. $$tan(a) = 0.75$$ 1. $$sin(b) = 0.8$$ 2. $$cos(b) = 0.6$$ 3. $$tan(b) = 1.33$$ #### Triangle 3 1. $$sin(a) = 0.923$$ 2. $$cos(a) = 0.385$$ 3. $$tan(a) = 2.4$$ 1. $$sin(b) = 0.385$$ 2. $$cos(b) = 0.923$$ 3. $$tan(b) = 0.417$$ #### Triangle 4 1. $$sin(a) = 0.690$$ 2. $$cos(a)", "+ m_1\\!\\cdot\\!m_2}$ Quote: Find the tangents of the angles of the triangle $ABC$, and find the angles to the nearest degree. . . $A(1,3),\\;B(-2,-4),\\;C(3,-2)$ We have: . $\\begin{array}{ccccc}m_{AB}\\:=\\\\ \\\\ m_{BC}\\:=\\\\ \\\\ m_{AC}\\:=\\end{array} \\\\ \\begin{array}{ccccc}\\frac{-4-3}{-2-1} \\\\ \\\\ \\frac{-2+4}{3+2} \\\\ \\\\ \\frac{-2-3}{3-1} \\end{array} \\begin{array}{ccccc}=\\;\\frac{7}{3} \\\\ \\\\ =\\;\\frac{2}{5} \\\\ \\\\ =\\,\\text{-}\\frac{5}{2}\\end{array}$ Then: . $\\begin{array}{ccccc}\\tan A\\:=\\:\\frac{m_{AC} - m_{AB}}{1 + m_{AC}\\cdot m_{AB}}\\:= \\\\ \\\\ \\tan B\\:=\\:\\frac{m_{AB} - m_{BC}}{1 + m_{AB}\\cdot m_{BC}}\\:= \\\\ \\\\ \\tan C \\:=\\:\\frac{m_{AC} - m_{BC}}{1 + m_{AC}\\cdot m_{BC}}\\:= \\end{array} \\begin{array}{ccccc}\\frac{-\\frac{5}{2} - \\frac{7}{3}}{1 + \\left(-\\frac{5}{2}\\right)\\left(\\frac{7}{3}\\right)} \\\\ \\frac{\\frac{7}{3} - \\frac{2}{5}}{1 + \\left(\\frac{7}{3}\\right)\\left(\\frac{2}{5}\\right)} \\\\ \\frac{-\\frac{5}{2} - \\frac{2}{5}}{1 + \\left(-\\frac{5}{2}\\right)\\left(\\frac{2}{5}\\right)} \\end{array}$ $\\begin{array}{ccccc}=\\;1 \\\\ \\, \\\\=\\;1\\\\ \\\\ = \\;\\infty\\end{array} \\quad\\Rightarrow\\quad \\begin{Bmatrix}A \\:= \\;45^o \\\\ \\\\ B \\:= \\:45^o \\\\ \\\\ C \\:=\\:90^o\\end{Bmatrix} $", "my problem. If so, please point out what is wrong in my method. If you think there are other simpler methods, please let me know about them. I'm sure there has to be a better method, since this seems like a pretty simple thing to do. Should I change my Matlab algorithm? Anyone know any special Matlab/Mathematica toolbox that calculates the yaw, pitch, roll? Thanks! Use \\begin{align} \\frac{a_{23}}{a_{33}} & = \\frac{ -\\cos b \\sin a}{\\cos a \\cos b} = -\\tan a \\\\ \\frac{a_{13}}{\\sqrt{ a_{23}^2 + a_{33}^2 }} & = \\frac{\\sin b}{\\cos b \\sqrt{\\sin^2 a +\\cos^2 a}} = \\tan b \\\\ \\frac{a_{12}}{a_{11}} & = \\frac{-\\cos b\\sin c}{\\cos b \\cos c} = -\\tan{c} \\end{align} Then you get: a = atan2(-a23,a33) b = atan2(a13, sqrt(a23^2+a33^2) ) c = atan2(-a12,a11) Now if I consider a particular set of rotation (say X first, then Y , then Z), with the corresponding Tait-Bryan angles --- a,b and c. My rotation matrix will be the following ... Look at that array. It's of the form $$\\begin{array}{ccc} \\cos(b) \\cos(c) & -\\cos(b) \\sin(c) & \\sin(b) \\\\ \\cdots & \\cdots & -\\cos(b) \\sin (a) \\\\ \\cdots & \\cdots & \\phantom{-}\\cos(b) \\cos(a) \\end{array}$$ Note that the $a_{13}$ element depends on $b$ only: $a_{13}= \\sin(b)$. This immediately gives you the angle $b$: $b=\\arcsin(a_{13})$. Since $\\cos b$ is non-negative for $b$", "5 36 $$\\sqrt{\\frac{5+\\sqrt5}{10}}$$ 6 30 $$\\sqrt{\\sqrt3-1}$$ 8 22.5 $$\\sqrt{\\sqrt{5+\\frac{7\\sqrt2}{2}} -1-\\sqrt2}$$ 10 18 $$\\frac12\\sqrt{-8-4\\sqrt5-2(1+\\sqrt5)\\sqrt{5+2\\sqrt5}}$$ For the dihedral angles, ∠E1BE2 is the same as ∠Ψ, and $$\\cos\\angle BE_2D_2 =-\\cos\\angle\\Psi$$. \\begin{align} \\cos\\angle E_1BE_2 &=\\frac{(BE_1)^2+(BE_2)^2-(E_1 E_2)^2}{2[(BE_1)(BE_2)]} \\\\ &=\\frac{\\left(\\frac12\\tan\\frac{90}{n}\\right)^2+\\left(\\frac{\\sqrt3}{2}\\right)^2-\\left(\\frac12\\sqrt{3-\\tan^2\\frac{90}{n} }\\right)^2}{2\\cdot\\frac12\\tan\\frac{90}{n}\\cdot\\frac{\\sqrt3}{2}} \\\\ &=\\frac{ \\frac{\\tan^2\\frac{90}{n}}{4}+\\frac34-\\frac{3-\\tan^2\\frac{90}{n} }{4} }{\\frac{\\sqrt3}{2}\\cdot\\tan\\frac{90}{n}} \\\\ &=\\frac{ \\frac{2\\tan^2\\frac{90}{n}}{4} }{\\frac{\\sqrt3}{2}\\cdot\\tan\\frac{90}{n}}=\\frac{ \\frac12\\tan^2\\frac{90}{n} }{\\frac{\\sqrt3}{2}\\cdot\\tan\\frac{90}{n}} \\\\ &= \\frac{\\tan^2\\frac{90}{n}}{2} \\cdot \\frac{2}{\\sqrt3\\cdot\\tan\\frac{90}{n}} \\\\ &= \\frac{\\tan^2\\frac{90}{n}}{\\sqrt3\\cdot\\tan\\frac{90}{n}} \\\\ &= \\frac{\\tan\\frac{90}{n}}{\\sqrt3} \\\\ &= \\frac1{\\sqrt3}\\cdot\\tan\\frac{90}{n} \\\\ \\cos\\angle BE_2D_2&=-\\frac1{\\sqrt3}\\cdot\\tan\\frac{90}{n} \\end{align} Here are the first few dihedrals for n: n $$\\angle E_1BE_2$$ $$\\angle BE_2D_2$$ 2 54.735° 125.264° 3 70.528° 109.471° 4 76.163° 103.836° 5 79.187° 100.812° 6 81.101° 98.8994° 8 83.405° 96.5945° For the side-to-side dihedrals, let’s draw a line from point C, straight down, equal to the height, and call this point K. K is in the same plane as the bottom base. Looking at Figure 1, we can see that ∠KDF is $$\\frac32\\theta$$, or $$\\frac{540}{n}$$. Connecting F and K, we get a line of $$2r\\sin\\frac{540}{2n}= s\\cdot\\frac{sin\\frac{270}{n}}{\\sin\\frac{180}{n}}$$. Note $$\\overline{FK}$$ is coplanar with G and E2, but does not go through either point. We can now find the distance between C and F, \\begin{align} (CF)^2&=(CK)^2+(FK)^2 \\\\ &=\\frac{3-s^2\\tan^2\\frac{90}{n}}{4} + s^2\\cdot\\frac{\\sin^2\\frac{270}{n}}{\\sin^2\\frac{180}{n}} \\\\ (CF) &=\\sqrt{\\frac{3-s^2\\tan^2\\frac{90}{n}}{4} + s^2\\cdot\\frac{\\sin^2\\frac{270}{n}}{\\sin^2\\frac{180}{n}} } \\end{align} In Figure 2, we see that $$CL=LF=BE_2=\\frac{\\sqrt3}{2}$$. We can now find ∠CLF, the dihedral, with s=1 factored out, \\begin{align} \\cos\\angle CLF &=\\frac{(CL)^2+(LF)^2- (CF)^2}{2\\cdot CL\\cdot LF} \\\\ &=\\frac{(\\frac{\\sqrt3}{2})^2+(\\frac{\\sqrt3}{2})^2-(CF)^2 }{2\\cdot (\\frac{\\sqrt3}{2})(\\frac{\\sqrt3}{2})}", "# An Inequality in Triangle with the Circumradius, Inradius and Angle Bisectors ### Proof WLOG, $a\\ge b\\ge c.\\,$ Clearly, $\\ell_a=\\min (\\ell_a,\\ell_b,\\ell_c).\\,$ We have \\displaystyle\\begin{align} R+r &= R+4R\\sin\\frac{A}{2}\\sin\\frac{B}{2}\\sin\\frac{C}{2}\\\\ &=R(\\cos A+\\cos B+\\cos C) \\end{align} and \\displaystyle\\begin{align} \\ell_a &= \\frac{2bc}{b+c}\\cdot\\cos\\frac{A}{2}\\\\ &=\\frac{4R\\sin B\\sin C}{\\sin B+\\sin C}\\cdot\\cos\\frac{A}{2}=\\frac{4R\\sin B\\sin C}{\\displaystyle 2\\sin\\frac{B+C}{2}\\cos\\frac{B-C}{2}}\\cdot\\cos\\frac{A}{2}\\\\ &=\\frac{2R\\sin B\\sin C}{\\displaystyle \\cos\\frac{B-C}{2}}. \\end{align} We need to prove that $\\displaystyle\\cos A+\\cos B+\\cos C\\ge\\frac{2\\sin B\\sin C}{\\displaystyle\\cos\\frac{B-C}{2}}.\\,$ But \\displaystyle\\begin{align} 2\\sin B\\sin C &= \\cos (B-C)-\\cos (B+C)\\\\ &=\\cos (B-C)+\\cos (A) \\end{align} and $\\displaystyle \\cos B +\\cos C=2\\sin\\frac{A}{2}\\cos\\frac{B-C}{2}.\\,$ Since $A\\ge B\\ge C,\\,$ $A\\ge\\displaystyle\\frac{\\pi}{3},\\,$ so that $\\displaystyle\\frac{A}{2}\\ge\\frac{\\pi}{6},\\,$ implying $\\displaystyle\\sin\\frac{A}{2}\\ge\\frac{1}{2};\\,$ also $B\\,$ and $C\\,$ are acute and, therefore $\\displaystyle\\frac{B-C}{2}\\,$ is also acute, such that $\\displaystyle\\cos\\frac{B-C}{2}\\gt 0.\\,$ We conclude that $\\displaystyle 2\\sin \\frac{A}{2}\\cos\\frac{B-C}{2}\\ge\\cos\\frac{B-C}{2};$ hence, $\\cos B+\\cos C\\ge\\displaystyle\\cos\\frac{B-C}{2}.\\,$ Thus, suffice it to show that $\\displaystyle \\cos A+\\cos\\frac{B-C}{2}\\ge\\frac{\\cos (B-C)+\\cos A}{\\displaystyle\\cos\\frac{B-C}{2}},$ or, equivalently, $\\displaystyle \\cos^2\\frac{B-C}{2}-\\cos (B-C)\\ge\\left(1-\\cos\\frac{B-C}{2}\\right)\\cos A$ which is the same as (*) $\\displaystyle \\left(1-\\cos\\frac{B-C}{2}\\right)\\left(1+\\cos\\frac{B-C}{2}\\right)\\ge\\left(1-\\cos\\frac{B-C}{2}\\right)\\cos A.$ Now, $\\displaystyle \\left(1-\\cos\\frac{B-C}{2}\\right)\\ge 0\\,$ and $\\displaystyle \\left(1-\\cos\\frac{B-C}{2}\\right)\\gt 1\\ge\\frac{1}{2}\\ge \\cos A\\,$ implying (*). ### Acknowledgment Leo Giugiuc has kindly communicated to me the problem by R. Satnoianu, with his solution. The problem has been previously published in the Romanian Mathematical Magazine by Dan Sitaru.", "have $$B=C < A$$. If the angles summed to $$2\\pi$$, we would have a nice solution $$\\frac A2=\\frac{\\pi}{2}, \\frac B2=\\frac C2 = \\frac{\\pi}{4}$$ - but no, we're not that lucky. Instead, we apply $$\\frac A2 = \\frac{\\pi}{2}-B$$ to eliminate $$A$$: $$1-\\cos B=2\\sin^2 \\frac B2 = \\sin^2 \\frac A2 = \\cos^2 B$$ so $$\\cos^2 B + \\cos B - 1 = 0$$ and $$\\sin \\frac A2 = \\cos B = \\dfrac{\\sqrt{5}-1}{2}$$. Then \\begin{align*}\\cos\\frac A2 &= \\sqrt{1-\\left(\\frac{\\sqrt{5}-1}{2}\\right)^2} = \\sqrt{\\frac{\\sqrt{5}-1}{2}}\\\\ \\sin\\frac B2 &= \\sqrt{\\frac{3-\\sqrt{5}}{4}} = \\frac1{\\sqrt{2}}\\cdot \\frac{\\sqrt{5}-1}{2}\\\\ \\cos\\frac B2 &= \\sqrt{\\frac{1+\\sqrt{5}}{4}} = \\frac1{\\sqrt{2}}\\cdot\\sqrt{\\frac{\\sqrt{5}+1}{2}}\\\\ \\cot\\frac A2 &= \\sqrt{\\frac{2}{\\sqrt{5}-1}} = \\sqrt{\\frac{\\sqrt{5}+1}{2}}\\\\ \\cot\\frac B2 &= \\sqrt{\\frac{\\sqrt{5}+1}{3-\\sqrt{5}}} = \\sqrt{\\sqrt{5}+2}\\\\ p &= 2\\cot\\frac A2 + 2\\cot\\frac B2 = 2\\sqrt{\\frac{\\sqrt{5}+1}{2}}+2\\sqrt{\\sqrt{5}+2}\\approx 6.66\\end{align*} A truly fiendish problem. Numerically, the minimizing angles are about $$76^\\circ$$ at $$A$$ and $$52^\\circ$$ at $$B$$ and $$C$$.", "# Find the value of $\\tan A + \\tan B$, given values of $\\frac{\\sin (A)}{\\sin (B)}$ and $\\frac{\\cos (A)}{\\cos (B)}$ Given $$\\frac{\\sin (A)}{\\sin (B)} = \\frac{\\sqrt{3}}{2}$$ $$\\frac{\\cos (A)}{\\cos (B)} = \\frac{\\sqrt{5}}{3}$$ Find $\\tan A + \\tan B$. Approach Dividing the equations, we get the relation between $\\tan A$ and $\\tan B$ but that doesn't help in getting the value of $\\tan A + \\tan B$. The value comes in terms of $\\tan A$ or $\\tan B$ but the expected answer is independent of any variable . Also $$\\frac{\\sin(A)\\cdot\\cos(B) + \\sin(B)\\cdot\\cos(A)}{\\cos(A)\\cdot\\cos(B)} = \\tan(A) + \\tan(B)$$ We could get a value only if instead of $\\cos A$ there was $\\sin B$ in the relation(which we get on adding the ratios) • – imranfat Aug 23 '16 at 15:42 Something is wrong. I calculated \\begin{align} \\sin A &= \\dfrac{\\sqrt 3}{2}\\sin B \\\\ \\cos A &= \\dfrac{\\sqrt 5}{3}\\cos B \\\\ \\sin^2 A + \\cos^2 A &= \\dfrac 34 \\sin^2 B + \\dfrac59 \\cos^2 B\\\\ 1 &= \\dfrac 34 - \\dfrac{7}{36} \\cos^2 B\\\\ \\cos^2 B = -\\dfrac{9}{7} \\end{align} Which is impossible. HINT: $$\\tan A=\\frac{\\sin A}{\\cos A}=\\frac{\\frac{\\sqrt{3}}2\\sin B}{\\frac{\\sqrt{5}}3\\cos B}=\\frac{3\\sqrt{3}}{2\\sqrt{5}}\\tan B\\tag{1}$$ And $$1=\\sin^2 A+\\cos^2 A=\\frac{3}{4}\\sin^2 B+\\frac{5}{9}\\cos^2 B=\\frac59\\left(\\sin^2 B+\\cos^2B\\right)+\\frac7{36}\\sin^2 B$$ Last equation implies $$\\sin^2B=\\frac{16}7$$ $$\\frac{|\\sin A|}{|\\sin B|}=\\frac{\\sqrt{3}}{2} <1 \\iff \\frac{|\\cos A|}{|\\cos B|}>1$$ But $$\\frac{|\\cos A|}{|\\cos B|}=\\frac{\\sqrt{5}}{3}<1$$ that's impossible. Hint: Let $\\dfrac{\\sin A}{\\sqrt3}=\\dfrac{\\sin B}2=p$ and $\\dfrac{\\cos A}{\\sqrt5}=\\dfrac{\\cos", "– 2·sin² α}\\\\ &= \\frac{\\color{yellow}{cos(120)}}{\\color{Blue}{cos(54)}}\\\\ &= \\frac{-\\frac12} {\\sqrt{\\frac{5-\\sqrt{5}} {8} } }\\\\ &= \\frac{-1}{2} \\cdot \\frac{\\sqrt{8}}{\\sqrt{5-\\sqrt{5}}} & \\text{invert}\\\\ &= \\frac{-2\\sqrt2}{2\\sqrt{5-\\sqrt{5}}}\\cdot \\frac{\\sqrt{5+\\sqrt{5}}}{\\sqrt{5+\\sqrt{5}}} & \\text{Multiply by conjugate}\\\\ &= \\frac{-\\sqrt2\\sqrt{5+\\sqrt{5}}}{\\sqrt{25-5}}\\\\ &= \\frac{-\\sqrt2\\sqrt{5+\\sqrt{5}}}{\\sqrt{20}}\\\\ &= \\frac{-\\sqrt2\\sqrt{5+\\sqrt{5}}}{\\sqrt{4\\cdot 5}}\\\\ &= \\frac{-\\sqrt2\\sqrt{5+\\sqrt{5}}}{2\\sqrt5}\\\\ &= \\frac{-\\sqrt2\\sqrt{5+\\sqrt{5}}}{2\\sqrt5}\\cdot \\frac{\\sqrt5}{\\sqrt5}\\\\ &= \\frac{-\\sqrt2\\sqrt5\\sqrt{5+\\sqrt{5}}}{20}\\\\ &= -\\sqrt{\\frac{5+\\sqrt5}{10} } \\end{align} ∠η = 148.2825255…° Hooray! It matches the value we found earlier. ## Finding the Angles of a Regular Pentagonal Pyramid A truncated icosahedron is made by truncating (cutting off) a regular pentagonal pyramid from each of the tips of an icosahedron, hence the name. The angle of the pentagon face to any of the triangle edges will be 180° minus the Pent-Pent edge angle. In order to find the angles we want, we must first find the height of a pentagon (Fig 8a). Either use the Pythagorean Theorem or trigonometry to find height H. The faces of the pyramid are all equilateral triangles. Finding the triangle height S is elementary (Fig 8b). Side note: S is 1/3 the length of the original icosahedron. Slicing through the pyramid through points A, B, and C, reveals (Fig 8c) ΔABC. We then use the Law of Cosines to find ∠θ. \\begin{align}\\cos \\theta &= \\frac{1^2 + \\color{skyblue}{H^2} – \\color{limegreen}{S^2}}{2\\cdot1\\cdot \\color{skyblue}H}\\\\ &= \\frac{1^2 + \\color{skyblue}{\\frac14\\cdot tan^2(72)} \\color{limegreen}{-\\frac34}}{2\\cdot1\\cdot \\color{skyblue}{\\frac12\\cdot tan(72)}}\\\\ &= \\frac{\\frac44 \\color{limegreen}{-\\frac34} + \\color{skyblue}{\\frac14\\cdot(5+2\\sqrt{5})} }{2\\cdot \\color{skyblue}{\\frac12\\sqrt{5+2\\sqrt{5}}}} \\\\ &= \\frac{\\frac14+\\frac14(5+2\\sqrt5)}{\\sqrt{5+2\\sqrt5}} \\\\ &=", "\\cos AOM = r(1 + \\cos AOM)$ (Where $r$ is the radius of circle P). Evaluating this, $\\cos \\angle AOM = \\frac{OM}{r} - 1 = \\frac{2OM}{R} - 1 = \\frac 8 5 - 1 = \\frac 3 5$. From $\\cos \\angle AOM$, we see that $\\tan \\angle AOM =\\frac{\\sqrt{1 - \\cos^2 \\angle AOM}}{\\cos \\angle AOM} = \\frac{\\sqrt{5^2 - 3^2}}{3} = \\frac 4 3$ Next, notice that $\\angle AOB = \\angle AOM - \\angle BOM$. We can therefore apply the tangent subtraction formula to obtain , $\\tan AOB =\\frac{\\tan AOM - \\tan BOM}{1 + \\tan AOM \\cdot \\tan AOM} =\\frac{\\frac 4 3 - \\frac 3 4}{1 + \\frac 4 3 \\cdot \\frac 3 4} = \\frac{7}{24}$. It follows that $\\sin AOB =\\frac{7}{\\sqrt{7^2+24^2}} = \\frac{7}{25}$, resulting in an answer of $7 \\cdot 25=\\boxed{175}$.", "Meanwhile, from right triangle $ABC,$ we have \\begin{align*} \\cos \\angle BAC = \\frac{6}{\\sqrt{40}} &= \\frac{3}{\\sqrt{10}}, \\\\ \\sin \\angle BAC = \\frac{2}{\\sqrt{40}} &= \\frac{1}{\\sqrt{10}}. \\end{align*} This means that by the angle subtraction formulas, \\begin{align*} \\cos \\angle OAB &= \\cos (\\angle OAC - \\angle BAC) \\\\ &= \\cos \\angle OAC \\cos \\angle BAC + \\sin \\angle OAC \\sin \\angle BAC \\\\ &= \\frac{1}{\\sqrt{5}} \\cdot \\frac{3}{\\sqrt{10}} + \\frac{2}{\\sqrt{5}} \\cdot \\frac{1}{\\sqrt{10}} \\\\ &= \\frac{5}{5 \\sqrt{2}} = \\frac{1}{\\sqrt{2}}. \\end{align*} Now we have all we need to use the law of cosines on $\\triangle OAB.$ This tells us that \\begin{align*} OB^2 &= AO^2 + AB^2 - 2 AO \\cdot AB \\cdot \\cos \\angle OAB \\\\ &= 50 + 36 - 2 \\cdot 5 \\sqrt{2} \\cdot 6 \\cdot \\frac{1}{\\sqrt{2}} \\\\ &= 86 - 2 \\cdot 5 \\cdot 6 \\\\ &= 26. \\end{align*} ### Solution 3 Drop perpendiculars from $O$ to $AB$ (with foot $T_1$), $M$ to $OT_1$ (with foot $T_2$), and $M$ to $AB$ (with foot $T_3$). Also, mark the midpoint $M$ of $AC$. $[asy] size(200); pair dl(string name, pair loc, pair offset) { dot(loc); label(name,loc,offset); return loc; }; pair a[] = {(0,0),(0,5),(1,5),(1,7),(-2,6),(-5,5),(-2,5),(-2,6),(0,6)}; string n[] = {\"O\",\"T_1\",\"B\",\"C\",\"M\",\"A\",\"T_3\",\"M\",\"T_2\"}; for(int i=0;i First notice that by computation, $OAC$ is a $\\sqrt {50} - \\sqrt {40} - \\sqrt {50}$ isosceles triangle, so $AC = MO$. Then, notice that $\\angle", "$b_2=b_4=\\cdots=b_{10}=\\cdots=\\frac{24}{25}$. \\begin{align*} \\lfloor a_{10} \\rfloor & = \\left\\lfloor 2^{10} b_{10} \\right\\rfloor = \\left\\lfloor \\dfrac{1024\\cdot 24}{25} \\right\\rfloor = \\left\\lfloor \\dfrac{1025\\cdot 24}{25} - \\dfrac{24}{25} \\right\\rfloor \\\\ & = \\left\\lfloor 41\\cdot 24 - \\dfrac{24}{25} \\right\\rfloor = 41\\cdot 24 - 1 = \\boxed{983} \\end{align*} ### Solution 2 After we do the substitution, we can notice the fact that $\\left( \\frac 35 \\right)^2 + \\left( \\frac 45 \\right)^2 = 1$, which may suggest that the formula may have something to do with the unit circle. Also, the expression $\\sqrt{1-x^2}$ often appears in trigonometry, for example in the relationship between the sine and the cosine. Both observations suggest that the formula may have a neat geometric interpretation. Consider the equation: $$y = \\frac45 x + \\frac 35\\sqrt{1 - x^2}$$ Note that for $t=\\sin^{-1} \\frac 35$ we have $\\sin t=\\frac 35$ and $\\cos t = \\frac 45$. Now suppose that we have $x=\\sin s$ for some $s$. Then our equation becomes: $$y=\\cos t \\cdot \\sin s + \\sin t \\cdot |\\cos s|$$ Depending on the sign of $\\cos s$, this is either the angle addition, or the angle subtraction formula for sine. In other words, if $\\cos s \\geq 0$, then $y=\\sin(s+t)$, otherwise $y=\\sin(s-t)$. We have $b_0=0=\\sin 0$. Therefore $b_1 = \\sin(0+t) = \\sin t$, $b_2 = \\sin(t+t) = \\sin (2t)$, and so on. (Remember", "little triangles formed from the angle A. The area of the rectangle is 2(e+c)s. The distance from the circle centre to the bottom of the rectangle is $\\displaystyle \\frac{1}{\\sqrt{2}}$ since the circle radius is 1 and $\\displaystyle Sin\\frac{\\pi}{4}=\\frac{1}{\\sqrt{2}}$ We require e, c and s in terms of the angle A. From the leftmost right-angled triangle....... $\\displaystyle TanA=\\frac{opp}{adj}=\\frac{\\frac{1}{\\sqrt{2}}}{e}$ Also, from the small top right triangle $\\displaystyle TanA=\\frac{s}{c}$ Hence, we can continue with calculations until we have a compact formula for the rectangle area... $\\displaystyle \\frac{es}{c}=\\frac{1}{\\sqrt{2}},\\ es=\\frac{c}{\\sqrt{2}}$ $\\displaystyle 2(e+c)s=2es+2cs=\\frac{2c}{\\sqrt{2}}+2cs=2c(\\frac{1 }{\\sqrt{2}}+s)$ Area=$\\displaystyle 2c\\frac{\\sqrt{2}s+1}{\\sqrt{2}}=\\frac{\\sqrt{2}\\sqrt {2}c}{\\sqrt{2}}(\\sqrt{2}s+1)=\\sqrt{2}c(\\sqrt{2}s+1 )$ Substituting for c and s, $\\displaystyle \\frac{\\sqrt{2}SinA-1}{TanA}[\\sqrt{2}SinA-1+1]=\\frac{\\sqrt{2}SinA-1}{TanA}[\\sqrt{2}SinA)$ $\\displaystyle =\\frac{2Sin^2A-\\sqrt{2}SinA}{TanA}=(2Sin^2A-\\sqrt{2}SinA)CotA$$\\displaystyle =(2Sin^2A-\\sqrt{2}SinA)\\frac{CosA}{SinA}=(2SinA-\\sqrt{2})CosA$ This is a more compact formula for rectangle area. We need to find the maximum area of the rectangle. To do that, we differentiate the area formula with respect to A and set the derivative equal to zero. We use the product rule of differentiation for this. $\\displaystyle \\frac{d}{dA}[(2SinA-\\sqrt{2})CosA]=CosA(2CosA)+(2SinA-\\sqrt{2})(-SinA)$ $\\displaystyle 2Cos^2A-2Sin^2A+\\sqrt{2}SinA=2(1-Sin^2A)-2Sin^2A+\\sqrt{2}SinA$ $\\displaystyle \\frac{d}{dA}f(A)=0$ $\\displaystyle 2-4Sin^2A+\\sqrt{2}SinA=0$ $\\displaystyle 4Sin^2A-\\sqrt{2}SinA-2=0$ This is a quadratic equation of the form $\\displaystyle ax^2+bx+c=0$ Hence $\\displaystyle SinA=\\frac{\\sqrt{2}\\pm\\sqrt{2+32}}{8}=0.905\\ and\\ -0.552$ As we are looking for an angle between 45 and 90 degrees, we can eliminate the negative result. $\\displaystyle A=Sin^{-1}0.905=1.13 radians$ This angle corresponds to maximum area. Hence, maximum area is $\\displaystyle (2Sin1.13-\\sqrt{2})Cos1.13=0.168\\ units^2$ 6. Here's a couple of graphs to get a visual", "# 3.3.2: Simplifying Trigonometric Expressions using Sum and Difference Formulas Simplify sine, cosine, and tangent of angles that are added or subtracted. As Agent Trigonometry you are given this clue: $$\\sin\\left(\\dfrac{\\pi}{2}−x\\right)$$. How could you simplify this expression to make solving your case easier? ### Simplifying Trigonometric Expressions We can also use the sum and difference formulas to simplify trigonometric expressions. The $$\\sin a=−\\dfrac{3}{5}$$ and $$\\cos b=\\dfrac{12}{13}$$. a is in the $$3^{rd}$$ quadrant and b is in the $$1^{st}$$. Let's find $$\\sin(a+b)$$. First, we need to find $$\\cos a$$ and $$\\sin b$$. Using the Pythagorean Theorem, missing lengths are 4 and 5, respectively. So, $$\\cos a=−\\dfrac{4}{5}$$ because it is in the 3rd quadrant and $$\\sin b=\\dfrac{5}{13}$$. Now, use the appropriate formulas. \\begin{aligned}\\sin(a+b)& =\\sin a \\cos b+\\cos a \\sin b \\\\&=−\\dfrac{3}{5}\\cdot \\dfrac{12}{13}+−\\dfrac{4}{5} \\cdot \\dfrac{5}{13}=−\\dfrac{56}{65} \\end{aligned} Now, using the information from the previous problem above, let's find $$\\tan(a+b)$$. From the cosine and sine of $$a$$ and $$b$$, we know that $$\\tan a=\\dfrac{3}{4}$$ and $$\\tan b=\\dfrac{5}{12}$$. \\begin{aligned} \\tan(a+b)&=\\dfrac{\\tan a+\\tan b}{1−\\tan a\\tan b} \\\\&=\\dfrac{\\dfrac{3}{4}+\\dfrac{5}{12}}{1−\\dfrac{3}{4}\\cdot \\dfrac{5}{12}} \\\\&=\\dfrac{\\dfrac{14}{12}}{\\dfrac{11}{16}}\\\\&=\\dfrac{56}{33}\\end{aligned} Finally, let's simplify $$\\cos(\\pi −x)$$. Expand this using the difference formula and then simplify. \\begin{aligned} \\cos(\\pi −x)&=\\cos\\pi \\cos x+\\sin \\pi \\sin x\\\\ &=−1\\cdot \\cos x+0\\cdot \\sin x \\\\ &=−\\cos x \\end{aligned} ##### Example $$\\PageIndex{1}$$ Earlier, you were asked to simplify $$\\sin\\left(\\dfrac{\\pi}{2}−x \\right)$$. Solution You can expand the", "If ∠A and ∠P are acute angles such that tan A = tan P, then show that ∠A = ∠P. Question: If $\\angle A$ and $\\angle P$ are acute angles such that $\\tan A=\\tan P$, then show that $\\angle A=\\angle P$. Solution: Given: $\\tan A=\\tan P$ To show: $\\angle A=\\angle P$ Consider two right angled triangles $\\mathrm{ABC}$ and $\\mathrm{PQR}$ such that $\\tan A=\\tan P$ Therefore we have, $\\tan A=\\frac{B C}{A B}$ and $\\tan P=\\frac{Q R}{P Q}$ Since it is given that $\\tan A=\\tan P$ Therefore, $\\frac{B C}{A B}=\\frac{Q R}{P Q}$ Now by interchanging position of AB and QR by cross multiplication We get, $\\frac{B C}{Q R}=\\frac{A B}{P Q}$ Let $\\frac{B C}{Q R}=\\frac{A B}{P Q}=k$ (say)....(1) Now by cross multiplication $B C=k Q R$ and $A B=k P Q \\ldots \\ldots(2)$ Now by using Pythagoras theorem in triangles ABC and PQR We have, $A C^{2}=A B^{2}+B C^{2}$ and $P R^{2}=P Q^{2}+Q R^{2}$ Therefore $A C=\\sqrt{A B^{2}+B C^{2}}$ and $P R=\\sqrt{P Q^{2}+Q R^{2}}$ Now $\\frac{A C}{P R}=\\frac{\\sqrt{A B^{2}+B C^{2}}}{\\sqrt{P Q^{2}+Q R^{2}}}$ Now using equation (2) We get, $\\frac{A C}{P R}=\\frac{\\sqrt{(k P Q)^{2}+(k Q R)^{2}}}{\\sqrt{P Q^{2}+Q R^{2}}}$ $\\frac{A C}{P R}=\\frac{\\sqrt{k^{2} P Q^{2}+k^{2} Q R^{2}}}{\\sqrt{P Q^{2}+Q R^{2}}}$ Now by taking $k^{2}$ common We get, $\\frac{A C}{P R}=\\frac{\\sqrt{k^{2}\\left(P Q^{2}+Q R^{2}\\right)}}{\\sqrt{P Q^{2}+Q R^{2}}}$ Therefore, $\\frac{A C}{P R}=\\frac{k \\sqrt{\\left(P Q^{2}+Q R^{2}\\right)}}{\\sqrt{P Q^{2}+Q R^{2}}}$ Now $\\sqrt{P" ]

[ "\\cdot \\sin A \\left(\\sin A + 2 \\cos A\\right) \\end{align*} after angle addition identity. Similarly, $\\angle{EAC} = 90 - \\angle A,$ and by Law of Sines $CE = 8 \\sin{\\angle{EAC}} = 8 \\cos A.$ Note that $\\angle{E} = \\angle{ACB}$ from inscribed angles, so \\begin{align*} [ACE] &= \\dfrac{1}{2} AC \\cdot CE \\sin{\\angle C} \\\\ &= \\dfrac{1}{2} \\cdot \\dfrac{8 \\sqrt{5}}{5} \\cdot 8 \\cos A \\sin{\\left[180 - \\left(90 - \\angle A\\right) - \\angle E\\right]} \\\\ &= \\dfrac{32 \\sqrt{5}}{5} \\cdot \\cos A \\sin{\\left[\\left(90 - \\angle A\\right) + \\angle{ACB}\\right]} \\\\ &= \\dfrac{32 \\sqrt{5}}{5} \\cdot \\cos A \\left(\\dfrac{2 \\sqrt{5} \\cos A}{5} + \\dfrac{\\sqrt{5} \\sin A}{5}\\right) \\\\ &= \\dfrac{32}{5} \\cdot \\cos A \\left(\\sin A + 2 \\cos A\\right) \\end{align*} after angle addition identity. Setting the two areas equal, we get $$\\tan A = \\frac{\\sin A}{\\cos A} = 8 \\iff \\sin A = \\frac{8}{\\sqrt{65}}, \\cos A = \\frac{1}{\\sqrt{65}}$$ after Pythagorean Identity. Now plug back in and the common area is $\\frac{64}{65} \\iff \\boxed{129}.$ ## Solution 2 $[asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label(\"A\",A,NE);label(\"O_1\",O_1,NE);label(\"O_2\",O_2,NE);label(\"B\",B,SW);label(\"C\",C,SW);label(\"D\",D,NE);label(\"E\",E,NE);label(\"N\",N,W);label(\"K\",(-24/15,0.2));label(\"L\",(24/15,0.2));label(\"n\",(-0.8,-0.12));label(\"p\",((29/15,-48/15)));label(\"\\mathcal{P}\",(-1.6,1.1));label(\"\\mathcal{Q}\",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); [/asy]$ Call $O_1$ and $O_2$ the centers of circles $\\mathcal{P}$ and $\\mathcal{Q}$, respectively, and extend $CB$ and $O_2O_1$ to meet at point $N$. Call $K$ and $L$ the feet of the altitudes from $B$ to $O_1N$ and $C$ to $O_2N$, respectively. Using the fact that $\\triangle{O_1BN} \\sim", "- C)$ $\\cos A.\\cos\\frac{B + C}{2}\\sin\\frac{C - B}{2} = -2\\cos A.\\sin\\frac{B - C}{2}.\\cos\\frac{B - C}{2}$ If $B = C$ then above it $0 = 0$ i.e. triangle is isosceles. If $A = 90^\\circ$ then above is $0 =0$ i.e. triangle is right angled. 13. $\\because \\tan\\frac{A}{2}, \\tan\\frac{B}{2}, \\tan\\frac{C}{2}$ are in A.P. $\\tan\\frac{A}{2} - \\tan\\frac{B}{2} = \\tan\\frac{B}{2} - \\tan \\frac{C}{2}$ $\\frac{\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)}{\\cos\\frac{A}{2}\\cos\\frac{B}{2}} = \\frac{\\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)}{\\cos\\frac{B}{2}\\cos\\frac{C}{2}}$ $\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)\\cos\\frac{C}{2} = \\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)\\cos\\frac{A}{2}$ $\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)\\sin\\left(\\frac{A}{2} + \\frac{B}{2}\\right) = \\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)\\sin\\left(\\frac{B}{2} + \\frac{C}{2}\\right)$ $\\Rightarrow \\cos B - \\cos A = \\cos C - \\cos B$ Thus, $\\cos A, \\cos B, \\cos C$ are in A.P. 14. Given, $a\\cos^2\\frac{C}{2} + c\\cos^2\\frac{A}{2} = \\frac{3b}{2}$ $a.\\frac{s(s - c)}{ab} + c.\\frac{s(s - a)}{bc} = \\frac{3b}{2}$ $\\frac{s}{b}[2s - a - c] = \\frac{3b}{2}$ $2s = 3b \\Rightarrow a + c = 2b$ We have to prove that $\\cot\\frac{A}{2} + \\cot\\frac{C}{2} = 2\\cot\\frac{B}{2}$ L.H.S. $= \\frac{s(s - a)}{\\Delta} + \\frac{s(s - c)}{\\Delta}$ $= \\frac{s}{\\Delta}(2s - a - c) = \\frac{2s(s - b)}{\\Delta} = 2\\cot\\frac{B}{2} =$ R.H.S. 15. Given, $a^2, b^2, c^2$ are in A.P. $\\Rightarrow b^2 - a^2 = c^2 - b^2$ $\\Rightarrow \\sin^2B - \\sin^2A = \\sin^2C - \\sin^2B$ $\\Rightarrow \\sin(B + A)\\sin(B - A) = \\sin(C + B)\\sin(C - B)$ $\\Rightarrow \\sin C\\sin(B - A) = \\sin A\\sin(C - B)$ $\\Rightarrow", "3C = - 4\\cos \\frac{{3A}}{2}\\cos \\frac{{3B}}{2}\\cos \\frac{{3C}}{2}$ For this to be zero, one of the cos terms on the RHS must be zero, which means that one of the angles must be \\begin{align}\\frac{\\pi }{3}\\end{align}. Example - 31 In a $$\\Delta ABC$$, if $\\cos A\\cos B\\cos C = \\frac{{\\sqrt 3 - 1}}{8},\\;\\sin A\\sin B\\sin C = \\frac{{3 + \\sqrt 3 }}{8}$ Find the angles of the triangle Solution: From the given relations, $\\tan A\\tan B\\tan C = \\frac{{3 + \\sqrt 3 }}{{\\sqrt 3 - 1}}$ Using the result of Example - 28, we have $\\tan A + \\tan B + \\tan C = \\frac{{3 + \\sqrt 3 }}{{\\sqrt 3 - 1}}$ Now, since \\begin{align}A + B + C = \\pi \\end{align} \\begin{align}&\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\cos (A + B + C) = - 1 \\\\ \\Rightarrow\\quad &\\qquad\\;\\;\\cos A\\cos B\\cos C\\left\\{ {1 - \\sum (\\tan A\\tan B)} \\right\\} = - 1 \\\\ \\Rightarrow\\quad &\\qquad\\;\\; \\sum \\left( {\\tan A\\tan B} \\right) = 1 + \\frac{8}{{\\sqrt 3 - 1}} = 5 + 4\\sqrt 3 \\\\ \\end{align} Finally, for $$\\tan A,\\tan B,\\tan C$$, we have their sum, their sum taken pairwise and their product, and thus we can say that these three are the roots of the equation \\begin{align}&\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,{x^3} - \\left( {\\frac{{3 + \\sqrt 3 }}{{\\sqrt 3 - 1}}} \\right){x^2} + (5 + 4\\sqrt 3 )x -", "$IT \\perp BC$, and let $S$ be on $AD$ where $IS \\perp AD$. By HL Congruency, $\\triangle ITQ \\cong \\triangle ISP$, so $PS = QT$. Now we have information we can use to determine side lengths to compute the inradius so we can determine the area of $ABCD$. Let $a = AS = AE$. That means $EB = BT = 7-a$, $TC = CF = a+2$, and $FD = SD = 13-a$. That means $AD = 13$. Additionally, since $AP = 6.5$ and $BQ = 4.5$, $QT = 2.5-a$ and $PS = a-6.5$. Since $QT = PS$, we must have $a = 4.5$. We know that $\\angle AIS + \\angle DIS + \\angle BIT + \\angle CIT = 180^\\circ$, so \\begin{align*} \\tan(\\angle AIS + \\angle DIS) &= -\\tan(\\angle BIT + \\angle CIT) \\\\ \\frac{\\tan \\angle AIS + \\tan \\angle DIS}{1 - \\tan \\angle AIS \\tan \\angle DIS} &= \\frac{\\tan \\angle BIT + \\tan \\angle CIT}{1 - \\tan \\angle BIT \\tan \\angle CIT} \\end{align*} Let $r$ be the radius of the incircle. That means \\begin{align*} \\frac{\\frac{4.5}{r} + \\frac{8.5}{r}}{1 - \\frac{4.5 \\cdot 8.5}{r^2}} &= -\\frac{\\frac{2.5}{r} + \\frac{6.5}{r}}{1 - \\frac{2.5 \\cdot 6.5}{r^2}} \\\\ \\frac{13r}{r^2-\\frac{153}{4}} &= \\frac{-9r}{r^2 - \\frac{65}{4}} \\\\ 13r^3 - \\frac{13 \\cdot 65}{4}r &= -9r^3 + \\frac{153 \\cdot 9}{4} \\\\ 22r^2 &= \\frac{13 \\cdot 65 + 153 \\cdot 9}{4} \\\\", "3 $$Sin^{-1}\\frac{3}{5} + Sin^{-1}\\frac{12}{13} = Sin^{-1}x$$. Find the value of $$x$$. Solution Here we shall apply the sine inverse formula $$Sin ^{-1}x + Sin ^{-1}y = Sin ^{-1}(x \\cdot \\sqrt{(1 - y^2)} + y \\cdot \\sqrt{(1 - x^2)} )$$ \\begin{align}Sin^{-1}x &= Sin^{-1}\\left(\\frac{3}{5} + Sin^{-1}\\frac{12}{13}\\right) \\\\ &= Sin^{-1}\\left(\\frac{3}{5}\\cdot\\sqrt{1 -\\left(\\frac{12}{13}\\right)^2} + \\frac{12}{13}\\cdot\\sqrt{1 -\\left(\\frac{3}{5}\\right)^2}\\right)\\\\&=Sin^{-1}\\left(\\frac{3}{5}\\cdot\\frac{5}{13} + \\frac{12}{13}\\cdot\\frac{4}{5}\\right)\\\\&=Sin^{-1}\\left(\\frac{15}{65} + \\frac{48}{65}\\right)\\\\&=Sin^{-1}\\left(\\frac{63}{65}\\right )\\end{align} $$\\therefore$$ $$x = \\dfrac{63}{65}$$ Example 4 Find the value of $$Tan^{-1}(\\sqrt3) – Cot^{-1}(-\\sqrt3 )$$. Solution \\begin{align}Tan^{-1}(\\sqrt3) – Cot^{-1}(-\\sqrt3 )&=Tan^{-1}(\\sqrt3) –(\\pi - Cot^{-1}(\\sqrt3 )) \\\\ &=Tan^{-1}(\\sqrt3) –\\pi +Cot^{-1}(\\sqrt3 ) \\\\&= \\frac{\\pi}{3} -\\pi + \\frac{\\pi}{6} \\\\&=\\frac{\\pi}{2} - \\pi \\\\&=-\\frac{\\pi}{2}\\end{align} $$\\therefore$$ The answer is $$-\\frac{\\pi}{2}$$ Example 5 Simplify $$Tan^{-1}\\sqrt{\\frac{1 - Cosx}{1 + Cosx}}$$ Solution \\begin{align}Tan^{-1}\\sqrt{\\frac{1 - Cosx}{1 + Cosx}} & = Tan^{-1}\\sqrt{\\frac{2Sin^2\\frac{x}{2}}{2Cos^2\\frac{x}{2} }}\\\\&= Tan^{-1}{\\frac{Sin\\frac{x}{2}}{Cos\\frac{x}{2} }}\\\\ &=Tan^{-1}(Tan\\frac{x}{2})\\\\&=\\frac{x}{2}\\end{align} $$\\therefore$$ The answer is $$\\frac{x}{2}$$ Challenging Questions Simplify the following expressions. 1. $$Tan^{-1}\\sqrt{\\dfrac{Cosx}{1 - Sinx}}$$ 2. $$Tan^{-1}\\dfrac{\\sqrt{1 + x^2} - 1}{x}$$ 3. $$Cot^{-1}\\left(\\dfrac{\\sqrt{1 + Sinx} + \\sqrt{1 - Sinx}}{\\sqrt{1 + Sinx} - \\sqrt{1 - Sinx}}\\right)$$ ## Interactive Questions on Inverse Trigonometric Ratios Here are a few activities for you to practice. Select/Type your answer and click the \"Check Answer\" button to see the result. ## Let's Summarize The mini-lesson targeted the fascinating concept of inverse trigonometric ratios. The math journey around inverse trigonometric ratios starts with what a student already knows, and goes on to creatively crafting a fresh", "from the angle A. The area of the rectangle is 2(e+c)s. The distance from the circle centre to the bottom of the rectangle is $\\frac{1}{\\sqrt{2}}$ since the circle radius is 1 and $Sin\\frac{\\pi}{4}=\\frac{1}{\\sqrt{2}}$ We require e, c and s in terms of the angle A. From the leftmost right-angled triangle....... $TanA=\\frac{opp}{adj}=\\frac{\\frac{1}{\\sqrt{2}}}{e}$ Also, from the small top right triangle $TanA=\\frac{s}{c}$ Hence, we can continue with calculations until we have a compact formula for the rectangle area... $\\frac{es}{c}=\\frac{1}{\\sqrt{2}},\\ es=\\frac{c}{\\sqrt{2}}$ $2(e+c)s=2es+2cs=\\frac{2c}{\\sqrt{2}}+2cs=2c(\\frac{1 }{\\sqrt{2}}+s)$ Area= $2c\\frac{\\sqrt{2}s+1}{\\sqrt{2}}=\\frac{\\sqrt{2}\\sqrt {2}c}{\\sqrt{2}}(\\sqrt{2}s+1)=\\sqrt{2}c(\\sqrt{2}s+1 )$ Substituting for c and s, $\\frac{\\sqrt{2}SinA-1}{TanA}[\\sqrt{2}SinA-1+1]=\\frac{\\sqrt{2}SinA-1}{TanA}[\\sqrt{2}SinA)$ $=\\frac{2Sin^2A-\\sqrt{2}SinA}{TanA}=(2Sin^2A-\\sqrt{2}SinA)CotA$ $=(2Sin^2A-\\sqrt{2}SinA)\\frac{CosA}{SinA}=(2SinA-\\sqrt{2})CosA$ This is a more compact formula for rectangle area. We need to find the maximum area of the rectangle. To do that, we differentiate the area formula with respect to A and set the derivative equal to zero. We use the product rule of differentiation for this. $\\frac{d}{dA}[(2SinA-\\sqrt{2})CosA]=CosA(2CosA)+(2SinA-\\sqrt{2})(-SinA)$ $2Cos^2A-2Sin^2A+\\sqrt{2}SinA=2(1-Sin^2A)-2Sin^2A+\\sqrt{2}SinA$ $\\frac{d}{dA}f(A)=0$ $2-4Sin^2A+\\sqrt{2}SinA=0$ $4Sin^2A-\\sqrt{2}SinA-2=0$ This is a quadratic equation of the form $ax^2+bx+c=0$ Hence $SinA=\\frac{\\sqrt{2}\\pm\\sqrt{2+32}}{8}=0.905\\ and\\ -0.552$ As we are looking for an angle between 45 and 90 degrees, we can eliminate the negative result. $A=Sin^{-1}0.905=1.13 radians$ This angle corresponds to maximum area. Hence, maximum area is $(2Sin1.13-\\sqrt{2})Cos1.13=0.168\\ units^2$ • Jan 29th 2010, 02:45 PM Here's a couple of graphs to get a visual idea, solving for the 'angle A' corresponding to maximum rectangle area, and the graph of", "+ c^2 - 4)}}{2} = \\frac{-bc + \\sqrt{(4 - b^2)(4 - c^2)}}{2}.$$ This asks for the trigonometric substitution $b = 2\\sin u$ and $c = 2\\sin v$, where $0^\\circ < u,v < 90^\\circ$. Then $$a = 2(-\\sin u\\sin v + \\cos u\\cos v) = 2\\cos (u + v),$$ and if we set $u = B/2$ and $v = C/2$, then $a = 2\\sin (A/2)$, $b = 2\\sin (B/2)$, and $c = \\sin (C/2)$, where $A$, $B$, and $C$ are the angles of a triangle. We have \\begin{align*} ab &= 4\\sin\\frac{A}{2}\\sin\\frac{B}{2} \\\\ &= 2\\sqrt{\\sin A\\tan\\frac{A}{2}\\sin B\\tan\\frac{B}{2}} = 2\\sqrt{\\sin A\\tan\\frac{B}{2}\\sin B\\tan\\frac{A}{2}} \\\\ &\\leq \\sin A\\tan\\frac{B}{2} + \\sin B\\tan\\frac{A}{2} \\\\ &= \\sin A\\cot\\frac{A + C}{2} + \\sin B\\cot\\frac{B + C}{2}, \\end{align*} where the inequality step follows from AM-GM. Likewise, \\begin{align*} bc &\\leq \\sin B\\cot\\frac{B + A}{2} + \\sin C\\cot\\frac{C + A}{2}, \\\\ ca &\\leq \\sin A\\cot\\frac{A + B}{2} + \\sin C\\cot\\frac{C + B}{2}. \\end{align*} Therefore \\begin{align*} ab + bc + ca &\\leq (\\sin A + \\sin B)\\cot\\frac{A + B}{2} + (\\sin B + \\sin C)\\cot\\frac{B + C}{2} + (\\sin C + \\sin A)\\cot\\frac{C + A}{2} \\\\ &= 2\\left(\\cos\\frac{A - B}{2}\\cos\\frac{A + B}{2} + \\cos\\frac{B - C}{2}\\cos\\frac{B + C}{2} + \\cos\\frac{C - A}{2}\\cos\\frac{C + A}{2} \\right)\\\\ &= 2(\\cos A + \\cos B + \\cos C) \\\\ &= 6 - 4\\left(\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2}", "+ m_1\\!\\cdot\\!m_2}$ Quote: Find the tangents of the angles of the triangle $ABC$, and find the angles to the nearest degree. . . $A(1,3),\\;B(-2,-4),\\;C(3,-2)$ We have: . $\\begin{array}{ccccc}m_{AB}\\:=\\\\ \\\\ m_{BC}\\:=\\\\ \\\\ m_{AC}\\:=\\end{array} \\\\ \\begin{array}{ccccc}\\frac{-4-3}{-2-1} \\\\ \\\\ \\frac{-2+4}{3+2} \\\\ \\\\ \\frac{-2-3}{3-1} \\end{array} \\begin{array}{ccccc}=\\;\\frac{7}{3} \\\\ \\\\ =\\;\\frac{2}{5} \\\\ \\\\ =\\,\\text{-}\\frac{5}{2}\\end{array}$ Then: . $\\begin{array}{ccccc}\\tan A\\:=\\:\\frac{m_{AC} - m_{AB}}{1 + m_{AC}\\cdot m_{AB}}\\:= \\\\ \\\\ \\tan B\\:=\\:\\frac{m_{AB} - m_{BC}}{1 + m_{AB}\\cdot m_{BC}}\\:= \\\\ \\\\ \\tan C \\:=\\:\\frac{m_{AC} - m_{BC}}{1 + m_{AC}\\cdot m_{BC}}\\:= \\end{array} \\begin{array}{ccccc}\\frac{-\\frac{5}{2} - \\frac{7}{3}}{1 + \\left(-\\frac{5}{2}\\right)\\left(\\frac{7}{3}\\right)} \\\\ \\frac{\\frac{7}{3} - \\frac{2}{5}}{1 + \\left(\\frac{7}{3}\\right)\\left(\\frac{2}{5}\\right)} \\\\ \\frac{-\\frac{5}{2} - \\frac{2}{5}}{1 + \\left(-\\frac{5}{2}\\right)\\left(\\frac{2}{5}\\right)} \\end{array}$ $\\begin{array}{ccccc}=\\;1 \\\\ \\, \\\\=\\;1\\\\ \\\\ = \\;\\infty\\end{array} \\quad\\Rightarrow\\quad \\begin{Bmatrix}A \\:= \\;45^o \\\\ \\\\ B \\:= \\:45^o \\\\ \\\\ C \\:=\\:90^o\\end{Bmatrix} $", "$$a$$, $$tan(a)$$, as follows: \\begin{aligned} tan(a) &= \\frac{O}{A} \\\\ & = \\frac{3}{4} \\\\ tan(a) & = 0.75 \\end{aligned} #### Relative to angle $$b$$ We calculate the sine of angle $$b$$, $$tan(b)$$, as follows: \\begin{aligned} tan(b) &= \\frac{O}{A} \\\\ & = \\frac{4}{3} \\\\ tan(b) & = 1.33... \\end{aligned} Notice that depending on the angle we're looking at, $$a$$ or $$b$$, the Adjacent and the Opposite Side Lengths changes. Consequently the value of tangent changes depending on whether we're cosidering the angle $$a$$ or the angle $$b$$. ## Exercise 1 For each of the right angle triangles, below, calculate: 1. $$sin(a)$$ 2. $$cos(a)$$ 3. $$tan(a)$$ 1. $$sin(b)$$ 2. $$cos(b)$$ 3. $$tan(b)$$ #### Triangle 4 Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working #### Triangle 1 1. $$sin(a) = 0.882$$ 2. $$cos(a) = 0.471$$ 3. $$tan(a) = 1.88$$ 1. $$sin(b) = 0.471$$ 2. $$cos(b) = 0.882$$ 3. $$tan(b) = 0.533$$ #### Triangle 2 1. $$sin(a) = 0.6$$ 2. $$cos(a) = 0.8$$ 3. $$tan(a) = 0.75$$ 1. $$sin(b) = 0.8$$ 2. $$cos(b) = 0.6$$ 3. $$tan(b) = 1.33$$ #### Triangle 3 1. $$sin(a) = 0.923$$ 2. $$cos(a) = 0.385$$ 3. $$tan(a) = 2.4$$ 1. $$sin(b) = 0.385$$ 2. $$cos(b) = 0.923$$ 3. $$tan(b) = 0.417$$ #### Triangle 4 1. $$sin(a) = 0.690$$ 2. $$cos(a)", "Double and Half Angle Identity Proofs $$\\displaystyle \\frac{{\\sin \\left( {2x} \\right)}}{2}=\\frac{{\\tan x}}{{{{{\\tan }}^{2}}x+1}}$$ \\require{cancel} \\displaystyle \\begin{align}&=\\frac{{\\tan x}}{{{{{\\sec }}^{2}}x}}\\\\&=\\frac{{\\sin x}}{{\\cancel{{\\cos x}}}}\\cdot \\frac{{{{{\\cancel{{{{{\\cos }}^{2}}x}}}}^{{\\cos x}}}}}{1}\\\\&=\\sin x\\cos x\\\\&=\\sin x\\cos x\\cdot \\frac{2}{2}\\\\\\frac{{\\sin \\left( {2x} \\right)}}{2}&=\\frac{{2\\sin x\\cos x}}{2}\\,\\,\\,\\,\\,\\surd \\end{align} Note: To get $$\\displaystyle \\sin x\\cos x$$ into the right form for the identity $$\\displaystyle (2\\sin x\\cos x)$$, we had to multiply by $$\\displaystyle \\frac{2}{2}$$. $$\\displaystyle \\frac{{1-\\cos \\left( {2A} \\right)}}{{\\sin \\left( {2A} \\right)}}=\\tan A$$ \\displaystyle \\begin{align}\\frac{{1-\\left( {1-2{{{\\sin }}^{2}}A} \\right)}}{{\\sin \\left( {2A} \\right)}}&=\\\\\\frac{{2{{{\\sin }}^{2}}A}}{{\\sin \\left( {2A} \\right)}}&=\\\\\\frac{{{}^{{\\sin A}}\\cancel{{2{{{\\sin }}^{2}}A}}}}{{\\cancel{{2\\sin A}}\\cos A}}&=\\tan A\\,\\,\\,\\,\\,\\surd \\end{align} Note: We knew to use the $$\\displaystyle 1-2{{\\sin }^{2}}A$$ (instead of $$\\displaystyle 2{{\\cos }^{2}}A-1$$) for $$\\cos 2A$$ to get rid of the 1’s, and also because we have sin in the denominator instead of cos. $$\\displaystyle \\tan \\left( {\\frac{1}{2}\\theta } \\right)=\\csc \\theta -\\cot \\theta$$ \\displaystyle \\begin{align}&=\\frac{1}{{\\sin \\theta }}-\\frac{{\\cos \\theta }}{{\\sin \\theta }}\\\\\\tan \\left( {\\frac{1}{2}\\theta } \\right)&=\\frac{{1-\\cos \\theta }}{{\\sin \\theta }}\\,\\,\\,\\,\\,\\surd \\end{align} Note: Start from the right side, and turn everything into sin and cos since the Half Angle tan identity is written in terms of sin and cos. $$\\displaystyle \\frac{{\\cos \\left( {2\\theta } \\right)}}{{\\cos \\theta -\\sin \\theta }}=\\cos \\theta +\\sin \\theta$$ \\displaystyle \\begin{align}\\frac{{{{{\\cos }}^{2}}\\theta -{{{\\sin }}^{2}}\\theta }}{{\\cos \\theta -\\sin \\theta }}&=\\\\\\frac{{\\cancel{{\\left( {\\cos \\theta -\\sin \\theta } \\right)}}\\left( {\\cos \\theta +\\sin \\theta } \\right)}}{{\\cancel{{\\cos \\theta -\\sin \\theta }}}}&=\\cos \\theta +\\sin \\theta \\,\\,\\,\\,\\,\\surd \\end{align}", "$\\angle M B A$, intersecting the side $A M$ (opp. to $\\angle M B A$) in pt. $P$. Hence, by the $\\angle$-bisector Theorem, $\\frac{A P}{P M} = \\frac{A B}{B M} = \\frac{1}{\\frac{1}{2}} = 2$ $\\Rightarrow \\frac{A P}{A P + P M} = \\frac{2}{2 + 1} = \\frac{2}{3} \\Rightarrow \\frac{A P}{A M} = \\frac{2}{3}$ $\\Rightarrow A P = \\frac{2}{3} \\cdot M A = \\frac{2}{3} \\cdot \\frac{\\sqrt{5}}{2} , i . e . , \\frac{\\sqrt{5}}{3.} \\ldots \\ldots \\ldots . \\left(1\\right)$ Arguing on the same lines, $A Q = \\frac{4}{5} \\cdot N A = \\frac{4}{5} \\cdot \\frac{\\sqrt{17}}{4} = \\frac{\\sqrt{17}}{5.} . . \\left(2\\right)$. In the light of $\\left(\\star\\right) ,$ now, it remains to find $m \\angle P A Q$. We have, from $r i g h t \\Delta M B A$, tan/_MAB=(MB)/(AB)=(1/2)/1=1/2, &, tan /_NAB=1/4. $\\therefore \\tan \\angle M A N = \\tan \\angle P A Q = \\tan \\left(\\angle M A B - \\angle N A B\\right)$ $= \\frac{\\frac{1}{2} - \\frac{1}{4}}{1 + \\frac{1}{2} \\cdot \\frac{1}{4}} = \\frac{1}{4} \\cdot \\frac{8}{9}$, i.e., $\\tan \\angle P A Q = \\frac{2}{9}$. $\\therefore \\sin \\angle P A Q = \\frac{2}{\\sqrt{85.}} \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\ldots \\left(3\\right)$. Therefore, by (star),(1),(2), &, (3), Area of DeltaAQP=1/2(sqrt5/3)(sqrt17)/5)(2/sqrt85)=1/15#, as was derived earlier!", "little triangles formed from the angle A. The area of the rectangle is 2(e+c)s. The distance from the circle centre to the bottom of the rectangle is $\\displaystyle \\frac{1}{\\sqrt{2}}$ since the circle radius is 1 and $\\displaystyle Sin\\frac{\\pi}{4}=\\frac{1}{\\sqrt{2}}$ We require e, c and s in terms of the angle A. From the leftmost right-angled triangle....... $\\displaystyle TanA=\\frac{opp}{adj}=\\frac{\\frac{1}{\\sqrt{2}}}{e}$ Also, from the small top right triangle $\\displaystyle TanA=\\frac{s}{c}$ Hence, we can continue with calculations until we have a compact formula for the rectangle area... $\\displaystyle \\frac{es}{c}=\\frac{1}{\\sqrt{2}},\\ es=\\frac{c}{\\sqrt{2}}$ $\\displaystyle 2(e+c)s=2es+2cs=\\frac{2c}{\\sqrt{2}}+2cs=2c(\\frac{1 }{\\sqrt{2}}+s)$ Area=$\\displaystyle 2c\\frac{\\sqrt{2}s+1}{\\sqrt{2}}=\\frac{\\sqrt{2}\\sqrt {2}c}{\\sqrt{2}}(\\sqrt{2}s+1)=\\sqrt{2}c(\\sqrt{2}s+1 )$ Substituting for c and s, $\\displaystyle \\frac{\\sqrt{2}SinA-1}{TanA}[\\sqrt{2}SinA-1+1]=\\frac{\\sqrt{2}SinA-1}{TanA}[\\sqrt{2}SinA)$ $\\displaystyle =\\frac{2Sin^2A-\\sqrt{2}SinA}{TanA}=(2Sin^2A-\\sqrt{2}SinA)CotA$$\\displaystyle =(2Sin^2A-\\sqrt{2}SinA)\\frac{CosA}{SinA}=(2SinA-\\sqrt{2})CosA$ This is a more compact formula for rectangle area. We need to find the maximum area of the rectangle. To do that, we differentiate the area formula with respect to A and set the derivative equal to zero. We use the product rule of differentiation for this. $\\displaystyle \\frac{d}{dA}[(2SinA-\\sqrt{2})CosA]=CosA(2CosA)+(2SinA-\\sqrt{2})(-SinA)$ $\\displaystyle 2Cos^2A-2Sin^2A+\\sqrt{2}SinA=2(1-Sin^2A)-2Sin^2A+\\sqrt{2}SinA$ $\\displaystyle \\frac{d}{dA}f(A)=0$ $\\displaystyle 2-4Sin^2A+\\sqrt{2}SinA=0$ $\\displaystyle 4Sin^2A-\\sqrt{2}SinA-2=0$ This is a quadratic equation of the form $\\displaystyle ax^2+bx+c=0$ Hence $\\displaystyle SinA=\\frac{\\sqrt{2}\\pm\\sqrt{2+32}}{8}=0.905\\ and\\ -0.552$ As we are looking for an angle between 45 and 90 degrees, we can eliminate the negative result. $\\displaystyle A=Sin^{-1}0.905=1.13 radians$ This angle corresponds to maximum area. Hence, maximum area is $\\displaystyle (2Sin1.13-\\sqrt{2})Cos1.13=0.168\\ units^2$ 6. Here's a couple of graphs to get a visual", "\\cos AOM = r(1 + \\cos AOM)$ (Where $r$ is the radius of circle P). Evaluating this, $\\cos \\angle AOM = \\frac{OM}{r} - 1 = \\frac{2OM}{R} - 1 = \\frac 8 5 - 1 = \\frac 3 5$. From $\\cos \\angle AOM$, we see that $\\tan \\angle AOM =\\frac{\\sqrt{1 - \\cos^2 \\angle AOM}}{\\cos \\angle AOM} = \\frac{\\sqrt{5^2 - 3^2}}{3} = \\frac 4 3$ Next, notice that $\\angle AOB = \\angle AOM - \\angle BOM$. We can therefore apply the tangent subtraction formula to obtain , $\\tan AOB =\\frac{\\tan AOM - \\tan BOM}{1 + \\tan AOM \\cdot \\tan AOM} =\\frac{\\frac 4 3 - \\frac 3 4}{1 + \\frac 4 3 \\cdot \\frac 3 4} = \\frac{7}{24}$. It follows that $\\sin AOB =\\frac{7}{\\sqrt{7^2+24^2}} = \\frac{7}{25}$, resulting in an answer of $7 \\cdot 25=\\boxed{175}$.", "thus can be discounted. This relationship, in combination with the Pythagorean Theorem in $\\triangle ACD$, gives $AC^2+CD^2=\\left(\\sqrt{2}\\right)^2+\\left(2-DA\\right)^2=DA^2$. Hence $2 + 4 - 4DA + DA^2 = DA^2$, so $DA = \\frac{3}{2}$, and thus $CD = \\frac{1}{2}$. Next, since $\\angle BAC = 45^{\\circ}$, $\\sin{\\left(\\angle BAC\\right)} = \\cos{\\left(\\angle BAC\\right)} = \\frac{1}{\\sqrt{2}}$. Using the lengths found above, $\\sin{\\left(\\angle CAD\\right)} = \\frac{\\left(\\frac{1}{2}\\right)}{\\left(\\frac{3}{2}\\right)} = \\frac{1}{3}$, and $\\cos{\\left(\\angle CAD\\right)} = \\frac{\\sqrt{2}}{\\left(\\frac{3}{2}\\right)} = \\frac{2 \\sqrt{2}}{3}$. Thus, by the addition formulae for $\\sin$ and $\\cos$, we have $$\\begin{split}\\sin{\\left(\\angle BAD\\right)}&=\\sin{\\left(\\angle BAC + \\angle CAD\\right)}\\\\&=\\sin{\\left(\\angle BAC\\right)}\\cos{\\left(\\angle CAD\\right)}+\\cos{\\left(\\angle BAC\\right)}\\sin{\\left(\\angle CAD\\right)}\\\\&=\\frac{1}{\\sqrt{2}}\\cdot\\frac{2 \\sqrt{2}}{3} + \\frac{1}{\\sqrt{2}}\\cdot\\frac{1}{3} \\\\ &= \\frac{2 \\sqrt{2} + 1}{3 \\sqrt{2}}\\end{split}$$ and $$\\begin{split}\\cos{\\left(\\angle BAD\\right)}&=\\cos{\\left(\\angle BAC + \\angle CAD\\right)}\\\\&=\\cos{\\left(\\angle BAC\\right)}\\cos{\\left(\\angle CAD\\right)}-\\sin{\\left(\\angle BAC\\right)}\\sin{\\left(\\angle CAD\\right)}\\\\&=\\frac{1}{\\sqrt{2}}\\cdot\\frac{2 \\sqrt{2}}{3} - \\frac{1}{\\sqrt{2}}\\cdot\\frac{1}{3} \\\\ &= \\frac{2 \\sqrt{2} - 1}{3 \\sqrt{2}}\\end{split}$$ Hence, by the double angle formula for $\\sin$, $\\sin{\\left(2\\angle BAD\\right)} = 2\\sin{\\left(\\angle BAD\\right)}\\cos{\\left(\\angle BAD\\right)} = \\frac{2(8-1)}{18} = \\boxed{\\textbf{(D) } \\frac{7}{9}}$. ### Solution 2 (coordinate geometry) We use the Pythagorean Theorem, as in Solution 1, to find $AD=\\frac{3}{2}$ and $CD=\\frac{1}{2}$. Now notice that the angle between $CD$ and the vertical (i.e. the $y$-axis) is $45^{\\circ}$ – to see this, drop a perpendicular from $D$ to $BA$ which meets $BA$ at $E$, and use the fact that the angle sum of quadrilateral $CBED$ must be $360^{\\circ}$. Anyway, this implies that the line $CD$ has slope $1$, so since $C$ is", "– 2·sin² α}\\\\ &= \\frac{\\color{yellow}{cos(120)}}{\\color{Blue}{cos(54)}}\\\\ &= \\frac{-\\frac12} {\\sqrt{\\frac{5-\\sqrt{5}} {8} } }\\\\ &= \\frac{-1}{2} \\cdot \\frac{\\sqrt{8}}{\\sqrt{5-\\sqrt{5}}} & \\text{invert}\\\\ &= \\frac{-2\\sqrt2}{2\\sqrt{5-\\sqrt{5}}}\\cdot \\frac{\\sqrt{5+\\sqrt{5}}}{\\sqrt{5+\\sqrt{5}}} & \\text{Multiply by conjugate}\\\\ &= \\frac{-\\sqrt2\\sqrt{5+\\sqrt{5}}}{\\sqrt{25-5}}\\\\ &= \\frac{-\\sqrt2\\sqrt{5+\\sqrt{5}}}{\\sqrt{20}}\\\\ &= \\frac{-\\sqrt2\\sqrt{5+\\sqrt{5}}}{\\sqrt{4\\cdot 5}}\\\\ &= \\frac{-\\sqrt2\\sqrt{5+\\sqrt{5}}}{2\\sqrt5}\\\\ &= \\frac{-\\sqrt2\\sqrt{5+\\sqrt{5}}}{2\\sqrt5}\\cdot \\frac{\\sqrt5}{\\sqrt5}\\\\ &= \\frac{-\\sqrt2\\sqrt5\\sqrt{5+\\sqrt{5}}}{20}\\\\ &= -\\sqrt{\\frac{5+\\sqrt5}{10} } \\end{align} ∠η = 148.2825255…° Hooray! It matches the value we found earlier. ## Finding the Angles of a Regular Pentagonal Pyramid A truncated icosahedron is made by truncating (cutting off) a regular pentagonal pyramid from each of the tips of an icosahedron, hence the name. The angle of the pentagon face to any of the triangle edges will be 180° minus the Pent-Pent edge angle. In order to find the angles we want, we must first find the height of a pentagon (Fig 8a). Either use the Pythagorean Theorem or trigonometry to find height H. The faces of the pyramid are all equilateral triangles. Finding the triangle height S is elementary (Fig 8b). Side note: S is 1/3 the length of the original icosahedron. Slicing through the pyramid through points A, B, and C, reveals (Fig 8c) ΔABC. We then use the Law of Cosines to find ∠θ. \\begin{align}\\cos \\theta &= \\frac{1^2 + \\color{skyblue}{H^2} – \\color{limegreen}{S^2}}{2\\cdot1\\cdot \\color{skyblue}H}\\\\ &= \\frac{1^2 + \\color{skyblue}{\\frac14\\cdot tan^2(72)} \\color{limegreen}{-\\frac34}}{2\\cdot1\\cdot \\color{skyblue}{\\frac12\\cdot tan(72)}}\\\\ &= \\frac{\\frac44 \\color{limegreen}{-\\frac34} + \\color{skyblue}{\\frac14\\cdot(5+2\\sqrt{5})} }{2\\cdot \\color{skyblue}{\\frac12\\sqrt{5+2\\sqrt{5}}}} \\\\ &= \\frac{\\frac14+\\frac14(5+2\\sqrt5)}{\\sqrt{5+2\\sqrt5}} \\\\ &=", "sin3x/1 1/sin5x cos5x1/cos3x. He then somehow found a way to get a 3x to the denominator of the first term after factoring out a 3/5, and add 5x to the numerator of the second term. I'm not sure how he managed that, could someone help me to understand? Thanks! Kevin First, you need to know that \\displaystyle \\begin{align*} \\lim_{x \\to 0}\\frac{\\tan{(x)}}{x} = 1 \\end{align*}. This should be obvious if you already know the standard limit \\displaystyle \\begin{align*} \\lim_{x \\to 0}\\frac{\\sin{(x)}}{x} = 1 \\end{align*}. Then \\displaystyle \\begin{align*} \\lim_{x \\to 0} \\frac{\\tan{(3x)}}{\\tan{(5x)}} &= \\lim_{x \\to 0} \\left[ \\frac{15x}{15x} \\cdot \\frac{\\tan{(3x)}}{\\tan{(5x)}} \\right] \\\\ &= \\frac{3}{5} \\lim_{x \\to 0} \\left[ \\frac{\\tan{(3x)}}{3x} \\cdot \\frac{5x}{\\tan{(5x)}} \\right] \\end{align*} I'm sure you can go from here.", "# 3.3.2: Simplifying Trigonometric Expressions using Sum and Difference Formulas Simplify sine, cosine, and tangent of angles that are added or subtracted. As Agent Trigonometry you are given this clue: $$\\sin\\left(\\dfrac{\\pi}{2}−x\\right)$$. How could you simplify this expression to make solving your case easier? ### Simplifying Trigonometric Expressions We can also use the sum and difference formulas to simplify trigonometric expressions. The $$\\sin a=−\\dfrac{3}{5}$$ and $$\\cos b=\\dfrac{12}{13}$$. a is in the $$3^{rd}$$ quadrant and b is in the $$1^{st}$$. Let's find $$\\sin(a+b)$$. First, we need to find $$\\cos a$$ and $$\\sin b$$. Using the Pythagorean Theorem, missing lengths are 4 and 5, respectively. So, $$\\cos a=−\\dfrac{4}{5}$$ because it is in the 3rd quadrant and $$\\sin b=\\dfrac{5}{13}$$. Now, use the appropriate formulas. \\begin{aligned}\\sin(a+b)& =\\sin a \\cos b+\\cos a \\sin b \\\\&=−\\dfrac{3}{5}\\cdot \\dfrac{12}{13}+−\\dfrac{4}{5} \\cdot \\dfrac{5}{13}=−\\dfrac{56}{65} \\end{aligned} Now, using the information from the previous problem above, let's find $$\\tan(a+b)$$. From the cosine and sine of $$a$$ and $$b$$, we know that $$\\tan a=\\dfrac{3}{4}$$ and $$\\tan b=\\dfrac{5}{12}$$. \\begin{aligned} \\tan(a+b)&=\\dfrac{\\tan a+\\tan b}{1−\\tan a\\tan b} \\\\&=\\dfrac{\\dfrac{3}{4}+\\dfrac{5}{12}}{1−\\dfrac{3}{4}\\cdot \\dfrac{5}{12}} \\\\&=\\dfrac{\\dfrac{14}{12}}{\\dfrac{11}{16}}\\\\&=\\dfrac{56}{33}\\end{aligned} Finally, let's simplify $$\\cos(\\pi −x)$$. Expand this using the difference formula and then simplify. \\begin{aligned} \\cos(\\pi −x)&=\\cos\\pi \\cos x+\\sin \\pi \\sin x\\\\ &=−1\\cdot \\cos x+0\\cdot \\sin x \\\\ &=−\\cos x \\end{aligned} ##### Example $$\\PageIndex{1}$$ Earlier, you were asked to simplify $$\\sin\\left(\\dfrac{\\pi}{2}−x \\right)$$. Solution You can expand the", "the answers were useful to you. – SuperAbound Jul 25 '14 at 6:16 Hint : \\begin{align} \\sin x+\\cos x&=\\frac12\\\\ \\sin x&=\\frac12-\\cos x\\\\ \\sin^2 x&=\\left(\\frac12-\\cos x\\right)^2\\\\ 1-\\cos^2 x&=\\frac14-\\cos x+\\cos^2 x\\\\ 2\\cos^2 x-\\cos x-\\frac34&=0\\\\ 8\\cos^2 x-4\\cos x-3&=0 \\end{align} $\\sin x+\\cos x=\\frac12$ $\\Rightarrow$ $(\\sin x+\\cos x)^2=\\frac14$ $\\Rightarrow$ $\\sin^2 x+2\\sin x\\cos x+\\cos^2 x=\\frac14$ $\\Rightarrow$ $1+2\\sin x\\cos x=\\frac14$ $\\Rightarrow$ $2\\sin x\\cos x = -\\frac34$ $\\Rightarrow$ $\\sin x\\cos x=-\\frac38$. If $a=\\sin x$, $b=\\cos x$, we have the system of equations \\begin{align} a+b&=\\frac12\\\\ ab&=-\\frac38 \\end{align} which leads to $$a-\\frac3{8a}=\\frac12\\\\ a^2-\\frac12a-\\frac38=0$$ with the solutions $a=\\frac{1\\pm\\sqrt7}4$. So we get $$\\sin x=\\frac{1+\\sqrt7}4, \\qquad \\cos x= \\frac{1-\\sqrt7}4 \\qquad \\Rightarrow \\qquad \\tan x = \\frac{1+\\sqrt7}{1-\\sqrt7}\\\\ \\sin x=\\frac{1-\\sqrt7}4, \\qquad \\cos x= \\frac{1+\\sqrt7}4 \\qquad \\Rightarrow \\qquad \\tan x = \\frac{1-\\sqrt7}{1+\\sqrt7}$$ You could also simplify the result: $$\\frac{1+\\sqrt7}{\\frac1-\\sqrt7} = \\frac{(1+\\sqrt7)^2}{1-7} = -\\frac{8+2\\sqrt7}6= -\\frac{4+\\sqrt7}3$$ HINT: Avoid squaring whenever possible as it immediately introduces extraneous root which needs to be discarded Use Weierstrass substitution to form a Quadratic Equation in $\\tan\\dfrac x2$ Now use $\\displaystyle\\tan2A=\\frac{2\\tan A}{1-\\tan^2A}$ Divide through by $\\cos x$ to get \\begin{align} \\tan x + 1 & = \\frac 12 \\sec x \\tag{1}\\\\ \\text{Square both sides: } \\qquad\\qquad\\tan^2 x + 2 \\tan x + 1 & = \\frac 14 \\sec^2 x = \\frac{1}{4} (\\tan^2 x + 1) \\\\ 3 \\tan^2 x + 8 \\tan x + 3 & = 0 \\\\ \\tan x", "5 36 $$\\sqrt{\\frac{5+\\sqrt5}{10}}$$ 6 30 $$\\sqrt{\\sqrt3-1}$$ 8 22.5 $$\\sqrt{\\sqrt{5+\\frac{7\\sqrt2}{2}} -1-\\sqrt2}$$ 10 18 $$\\frac12\\sqrt{-8-4\\sqrt5-2(1+\\sqrt5)\\sqrt{5+2\\sqrt5}}$$ For the dihedral angles, ∠E1BE2 is the same as ∠Ψ, and $$\\cos\\angle BE_2D_2 =-\\cos\\angle\\Psi$$. \\begin{align} \\cos\\angle E_1BE_2 &=\\frac{(BE_1)^2+(BE_2)^2-(E_1 E_2)^2}{2[(BE_1)(BE_2)]} \\\\ &=\\frac{\\left(\\frac12\\tan\\frac{90}{n}\\right)^2+\\left(\\frac{\\sqrt3}{2}\\right)^2-\\left(\\frac12\\sqrt{3-\\tan^2\\frac{90}{n} }\\right)^2}{2\\cdot\\frac12\\tan\\frac{90}{n}\\cdot\\frac{\\sqrt3}{2}} \\\\ &=\\frac{ \\frac{\\tan^2\\frac{90}{n}}{4}+\\frac34-\\frac{3-\\tan^2\\frac{90}{n} }{4} }{\\frac{\\sqrt3}{2}\\cdot\\tan\\frac{90}{n}} \\\\ &=\\frac{ \\frac{2\\tan^2\\frac{90}{n}}{4} }{\\frac{\\sqrt3}{2}\\cdot\\tan\\frac{90}{n}}=\\frac{ \\frac12\\tan^2\\frac{90}{n} }{\\frac{\\sqrt3}{2}\\cdot\\tan\\frac{90}{n}} \\\\ &= \\frac{\\tan^2\\frac{90}{n}}{2} \\cdot \\frac{2}{\\sqrt3\\cdot\\tan\\frac{90}{n}} \\\\ &= \\frac{\\tan^2\\frac{90}{n}}{\\sqrt3\\cdot\\tan\\frac{90}{n}} \\\\ &= \\frac{\\tan\\frac{90}{n}}{\\sqrt3} \\\\ &= \\frac1{\\sqrt3}\\cdot\\tan\\frac{90}{n} \\\\ \\cos\\angle BE_2D_2&=-\\frac1{\\sqrt3}\\cdot\\tan\\frac{90}{n} \\end{align} Here are the first few dihedrals for n: n $$\\angle E_1BE_2$$ $$\\angle BE_2D_2$$ 2 54.735° 125.264° 3 70.528° 109.471° 4 76.163° 103.836° 5 79.187° 100.812° 6 81.101° 98.8994° 8 83.405° 96.5945° For the side-to-side dihedrals, let’s draw a line from point C, straight down, equal to the height, and call this point K. K is in the same plane as the bottom base. Looking at Figure 1, we can see that ∠KDF is $$\\frac32\\theta$$, or $$\\frac{540}{n}$$. Connecting F and K, we get a line of $$2r\\sin\\frac{540}{2n}= s\\cdot\\frac{sin\\frac{270}{n}}{\\sin\\frac{180}{n}}$$. Note $$\\overline{FK}$$ is coplanar with G and E2, but does not go through either point. We can now find the distance between C and F, \\begin{align} (CF)^2&=(CK)^2+(FK)^2 \\\\ &=\\frac{3-s^2\\tan^2\\frac{90}{n}}{4} + s^2\\cdot\\frac{\\sin^2\\frac{270}{n}}{\\sin^2\\frac{180}{n}} \\\\ (CF) &=\\sqrt{\\frac{3-s^2\\tan^2\\frac{90}{n}}{4} + s^2\\cdot\\frac{\\sin^2\\frac{270}{n}}{\\sin^2\\frac{180}{n}} } \\end{align} In Figure 2, we see that $$CL=LF=BE_2=\\frac{\\sqrt3}{2}$$. We can now find ∠CLF, the dihedral, with s=1 factored out, \\begin{align} \\cos\\angle CLF &=\\frac{(CL)^2+(LF)^2- (CF)^2}{2\\cdot CL\\cdot LF} \\\\ &=\\frac{(\\frac{\\sqrt3}{2})^2+(\\frac{\\sqrt3}{2})^2-(CF)^2 }{2\\cdot (\\frac{\\sqrt3}{2})(\\frac{\\sqrt3}{2})}", "to arcsin(1/(2*sin(36°))), and therefore: $OD = OA \\cdot \\tan{\\left[\\arcsin{\\left(\\frac{1}{2 \\sin{36°}}\\right)}\\right]} = \\frac{1}{2} \\tan{54°} \\tan{\\left[\\arcsin{\\left(\\frac{1}{2 \\sin{36°}}\\right)}\\right]} \\cdot {\\bf a}$ The volume of the pyramid is the area of its base multiplied by its height (OD) divided by 3 and so it is equal to: $\\frac{5}{4} \\cdot \\tan{54°} \\cdot {\\bf a}^2 \\cdot \\\\* \\frac{1}{2} \\cdot tan{54°} \\cdot \\tan{\\left[\\arcsin{\\left(\\frac{1}{2 \\sin{36°}}\\right)}\\right]} \\cdot {\\bf a} \\cdot \\frac{1}{3} = \\\\* \\frac{5}{24} \\cdot \\tan^2{54°} \\cdot \\tan{\\left[\\arcsin{\\left(\\frac{1}{2 \\sin{36°}}\\right)}\\right]} \\cdot {\\bf a}^3$ Since there are 12 such pyramids in a dodecahedron, its volume is equal to this volume multiplied by 12. Thus, we get that the volume of a dodecahedron is: $\\frac{5}{2} \\cdot \\tan{54°}^2 \\cdot \\tan{\\left[\\arcsin{\\left(\\frac{1}{2 \\sin{36°}}\\right)}\\right]} \\cdot {\\bf a}^3 = ~7.66 * {\\bf a}^3.$" ]

[ "origin, i.e., if $x=0$ then $y=0$. The rule for a rotation by 180about the origin is given by $(x,y) → (-x,−y)$.", "$+60$ degrees, keeping the origin fixed.Find a matrix that rotates the plane through $+60$ degrees, keeping the origin fixed. ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find a $2 \\times 2$ matrix that rotates the plane by $+45$ degrees followed by a reflection across the horizontal axis. 0 answers 3 views Find a $2 \\times 2$ matrix that rotates the plane by $+45$ degrees followed by a reflection across the horizontal axis.Find a $2 \\times 2$ matrix that rotates the plane by $+45$ degrees followed by a reflection across the horizontal axis. ... close", "B 270 degrees counterclockwise about the origin. 2. Rotate the triangle ABC 60 degrees clockwise about the origin. 3. Rotate the triangle $130$ degrees clockwise about the circumcenter. ### Open Problem Solutions Images/mathematical drawings are created with GeoGebra.", "${45}^{\\circ }$ clockwise we will not reach to the same point as we get by first rotating about the origin and then translating. (d) To find the matrix that translate Y back to X, we use the concept $Y=ABX$", "a $$90^{\\circ}$$ rotation matrix, yielding a $$180^{\\circ}$$ hyperbolic rotation around $$i$$. This is the limiting case of $$z,w\\to i$$ with midpoint $$\\to i$$.", "the points at $(1,2)$ and $(-1,0)$, which are indeed related by a rotation through $180^\\circ$ about $(0,1)$, note a) that these don't correspond to each other in the shapes and b) that no other pairs of points are related by this rotation.", "the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. Vocabulary Language: English Rotation Rotation A rotation is", "We begin at the origin, and then draw a segment at $0^\\circ,$ labeled “1” in the figure. Next, we turn counterclockwise $30^\\circ$ (since $\\alpha_1=30$) and draw the segment labeled “2”. The next turn is $180^\\circ,$ so we turn completely around and begin retracing the arm in the opposite direction. Then turn counterclockwise $330^\\circ$ and move forward. Note that since $330=360-30,$ this has the effect of taking us right back to the origin. What next? It is at this point that we invoke a recursive call — and so the next angle tells us what direction the next arm will be drawn in. Of course, like before, the three angles after that will continue drawing the arm and then bring us back to the origin, awaiting the next recursive call. This behavior can be illustrated in the following chart. Here, the angles turned are listed in rows of four, one after another. The first three angles in each row are the same, and their job is to complete the arm once a direction is chosen. But the direction chosen is determined by the fourth column (separated by the divider), whose behavior is not periodic and highly recursive. So where do we go from here? The next chart we’ll look it is a chart of the directions the arms are", "about the origin? 1. This is a $135^\\circ$ clockwise rotation about point $B$ . You could also say this is a $225^\\circ$ counterclockwise rotation about point $B$ . 2. You can see that there is a straight line $(180^\\circ)$ passing through the origin connecting each pair of corresponding points. $(x, y) \\rightarrow (-x, -y)$ . 3. There would be no difference. A rotation $180^\\circ$ clockwise is the same as a rotation $180^\\circ$ counterclockwise. Practice Describe each of the following rotations in TWO ways. 1. 2. 3. 4. What does a $360^\\circ$ rotation look like? 5. Rotate the rectangle below $90^\\circ$ clockwise about the origin. 6. Use the previous example to help you describe what happens to a point $(x, y)$ that is rotated $90^\\circ$ clockwise about the origin. 7. (3, 2) is rotated $90^\\circ$ clockwise about the origin. Where does it end up? 8. (3, 2) is rotated $180^\\circ$ clockwise about the origin. Where does it end up? Hint: Use Guided Practice #2 for help. 9. (3, 2) is rotated $90^\\circ$ counterclockwise about the origin. Where does it end up? Hint: Use Example C. 10. What do circles have to do with rotations? 11. $\\Delta ABC$ is rotated $130^\\circ$ clockwise about point $O$ to create $\\Delta A^\\prime B^\\prime C^\\prime$ . Describe two connections between the points $B, O$", "former. - That's a $-90$-degree rotation matrix. - Wow! A highly improbable case of two events occurring simultaneously! Your answer and mine show the same \"answered 2 secs ago\"! – user17762 Jun 3 '11 at 19:11 Because... Sorry couldn't resist when it was bumped. -", "rotated about the point $A \\ 225^\\circ$ . Find the coordinates of the purple pentagon. On the diagram, draw and label the rotated pentagon. Answers: 1. Notice the direction of the angle is clockwise, therefore the angle measure is $60^\\circ$ CW or $-60^\\circ$ . 2. Notice the direction of the angle is counter-clockwise, therefore the angle measure is $155^\\circ$ CCW or $155^\\circ$ . 3. The measure of $\\angle BAB^\\prime = m \\angle BAE^\\prime + m \\angle E^\\prime AB^\\prime$ . Therefore $\\angle BAB^\\prime = 111.80^\\circ + 113.20^\\circ$ or $225^\\circ$ . Notice the direction of the angle is counter-clockwise, therefore the angle measure is $225^\\circ$ CCW or $225^\\circ$ . ### Explore More 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3.", "the arm in the opposite direction. Then turn counterclockwise $330^\\circ$ and move forward. Note that since $330=360-30,$ this has the effect of taking us right back to the origin. What next? It is at this point that we invoke a recursive call — and so the next angle tells us what direction the next arm will be drawn in. Of course, like before, the three angles after that will continue drawing the arm and then bring us back to the origin, awaiting the next recursive call. This behavior can be illustrated in the following chart. Here, the angles turned are listed in rows of four, one after another. The first three angles in each row are the same, and their job is to complete the arm once a direction is chosen. But the direction chosen is determined by the fourth column (separated by the divider), whose behavior is not periodic and highly recursive. So where do we go from here? The next chart we’ll look it is a chart of the directions the arms are drawn in. We read this chart in rows as follows: the first arm is drawn at an angle of $0^\\circ,$ directly to the right. The second arm is drawn at an angle of $210^\\circ,$ as can be seen in the figure above. But", "## Calculus 8th Edition Given: $2$ rad $=114.6^{\\circ}$ The standard position of an angle occurs when we place its vertex at the origin of a coordinate system and its initial side on positive x-axis. A positive angle is obtained by rotating the initial side counterclockwise until it coincides with the terminal side. Likewise, negative angles are obtained by clockwise rotation . Since, the given angle is positive therefore, we will move counterclockwise from the x-axis as depicted in the figure:", "origin at $$270^{\\circ}$$ Rotation about the origin at 270^{\\circ}: $$R_270^{\\circ}(x,y)=(y,−x)$$ Now let's perform the following rotations on Image $$A$$ shown below in the diagram below and describe the rotations: 1. about the origin at $$90^{\\circ}$$, and label it $$B$$. Rotation about the origin at $$90^{\\circ}$$: $$R_{90^{\\circ}}A\\rightarrow B=R_{90^{\\circ}}(x,y)\\rightarrow (−y,x)$$ 1. about the origin at $$180^{\\circ}$$, and label it $$O$$. Rotation about the origin at $$180^{\\circ}$$: $$R_{180^{\\circ}}A\\rightarrow O=R_{180^{\\circ}}(x,y)\\rightarrow (−x,−y)$$ 1. about the origin at $$270^{\\circ}$$, and label it $$Z$$. Rotation about the origin at $$270^{\\circ}$$: $$R_{270^{\\circ}}A\\rightarrow Z=R270^{\\circ}(x,y)\\rightarrow (y,−x)$$ Finally, let's write the notation that represents the rotation of the preimage A to the rotated image J in the diagram below: First, pick a point in the diagram to use to see how it is rotated. $$E:(−1,2)\\qquad E′:(1,−2)$$ Notice how both the $$x$$- and $$y$$-coordinates are multiplied by -1. This indicates that the preimage $$A$$ is reflected about the origin by $$180^{\\circ}$$CCW to form the rotated image J. Therefore the notation is $$R_{180^{\\circ}}A\\rightarrow J=R_{180^{\\circ}}(x,y)\\rightarrow (−x,−y)$$. Example $$\\PageIndex{1}$$ Earlier, you were given the figure below which shows a pattern of two fish. Write the mapping rule for the rotation of Image $$A$$ to Image $$B$$. Solution Notice that the angle measure is $$90^{\\circ}$$ and the direction is clockwise. Therefore the Image $$A$$ has been rotated $$−90^{\\circ}$$ to form Image $$B$$. To write", "\\alpha_2=180,\\quad \\alpha_3=330.$ How does this scheme draw a spiral arm? Let’s look at the figure below. We begin at the origin, and then draw a segment at $0^\\circ,$ labeled “1” in the figure. Next, we turn counterclockwise $30^\\circ$ (since $\\alpha_1=30$) and draw the segment labeled “2”. The next turn is $180^\\circ,$ so we turn completely around and begin retracing the arm in the opposite direction. Then turn counterclockwise $330^\\circ$ and move forward. Note that since $330=360-30,$ this has the effect of taking us right back to the origin. What next? It is at this point that we invoke a recursive call — and so the next angle tells us what direction the next arm will be drawn in. Of course, like before, the three angles after that will continue drawing the arm and then bring us back to the origin, awaiting the next recursive call. This behavior can be illustrated in the following chart. Here, the angles turned are listed in rows of four, one after another. The first three angles in each row are the same, and their job is to complete the arm once a direction is chosen. But the direction chosen is determined by the fourth column (separated by the divider), whose behavior is not periodic and highly recursive. So where do we go from", "down the 2 by 2 matrices which 1. reverse the direction for every vector 2. project every vector onto the $x_2$-axis 3. turn every vector counterclowise through $90^\\circ$ 4. reflect every vector through the $45^\\circ$ line $x_1 = x_2$", "about the origin using the given angle of rotation. $A(-1,0), B(-1,-3), C(1,-3), D(1,0) ; 90^{\\circ}$ Rotate the figure with the given vertices about the origin using the given angle of rotation. $A(-1,0), B(-1,-3), C(1,-3), D(1,0) ; 90^{\\circ}$... ##### 15. Determine the equivalent capacitance4FFXF C2 C2CIC1 15. Determine the equivalent capacitance 4FFXF C2 C2 CI C1...", "## find the matrix of $\\ U \\circ S \\circ T$ with respect to the standard basis Let $$V=\\mathbb{R}^3$$. Let $$(i)$$ $$T:V \\to V$$ be rotation by angle $$\\theta$$ counterclockwise in the plane (through origin) perpendicular to $$\\ (1,2,3)$$, $$(ii)$$ $$\\ S : V \\to V$$ be the reflection with respect to the plane (through origin) spanned by the vectors $$(1,0,1), \\ (1,2,1)$$, $$(iii)$$ $$U: V \\to V$$ be the rotation by the angle $$\\phi$$ counterclockwise in the plane ( through origin) perpendicular to $$(1,0,1)$$. Then find the matrix of $$\\ U \\circ S \\circ T$$ with respect to the standard basis. To answer the question we have to find the matrices of $$\\ T, S, U$$ and then combine the matrices.", "## Calculus 8th Edition Given: $-3 rad$=-171.9^{\\circ}\\$ The standard position of an angle occurs when we place its vertex at the origin of a coordinate system and its initial side on positive x-axis. A positive angle is obtained by rotating the initial side counterclockwise until it coincides with the terminal side. Likewise, negative angles are obtained by clockwise rotation . Since, the given angle is negative therefore, we will move clockwise from the x-axis as depicted in the figure:", "[…] ### Determine the head of the vector whose tail is given. Make a sketch. – Given vector [ left[begin{matrix}-2\\5\\end{matrix}right] ] – Tail of the vector is $( -3, 2)$ [ left[begin{matrix}-3\\2\\end{matrix}right] ] In this question, we have to find the head of the vector when the vector and its tail are given. The basic concept behind this question is the knowledge of vectors, subtraction addition, and multiplication of […] ### Figure ABCD is a Trapezoid with point A (0, −4). What rule would rotate the figure 270° clockwise? This question aims to find the type of rule that would be applied to the trapezoid ABCD with a point A( 0, -4 ) to rotate it to 270° in the clockwise direction. A quadrilateral having two sides parallel to each other is called a trapezoid. This four-sided figure is also called a trapezium. When […]" ]

[ "one by a rotation about the origin $120^\\circ$ clockwise and counter-clockwise. The counter-clockwise rotation matrix is $$R_{120^\\circ} = \\begin{pmatrix}\\cos120^\\circ & -\\sin120^\\circ \\\\ \\sin120^\\circ & \\cos120^\\circ\\end{pmatrix} = \\begin{pmatrix}-\\frac{1}{2} & -\\frac{\\sqrt{3}}{2} \\\\ \\frac{\\sqrt{3}}{2} & -\\frac{1}{2}\\end{pmatrix}$$ with the clockwise rotation matrix as $$R_{-120^\\circ} = \\begin{pmatrix}\\cos120^\\circ & \\sin120^\\circ \\\\ -\\sin120^\\circ & \\cos120^\\circ\\end{pmatrix} = \\begin{pmatrix}-\\frac{1}{2} & \\frac{\\sqrt{3}}{2} \\\\ -\\frac{\\sqrt{3}}{2} & -\\frac{1}{2}\\end{pmatrix}$$ Your vertices are then $$B' = \\begin{pmatrix}-\\frac{1}{2} & -\\frac{\\sqrt{3}}{2} \\\\ \\frac{\\sqrt{3}}{2} & -\\frac{1}{2}\\end{pmatrix}\\begin{pmatrix}x_1' \\\\ y_1'\\end{pmatrix}=\\begin{pmatrix}-\\frac{1}{2}x_1' - \\frac{\\sqrt{3}}{2}y_1' \\\\ \\frac{\\sqrt{3}}{2}x_1' - \\frac{1}{2}y_1'\\end{pmatrix}$$ $$C' = \\begin{pmatrix}-\\frac{1}{2} & \\frac{\\sqrt{3}}{2} \\\\ -\\frac{\\sqrt{3}}{2} & -\\frac{1}{2}\\end{pmatrix}\\begin{pmatrix}x_1' \\\\ y_1'\\end{pmatrix}=\\begin{pmatrix}-\\frac{1}{2}x_1' + \\frac{\\sqrt{3}}{2}y_1' \\\\ -\\frac{\\sqrt{3}}{2}x_1' - \\frac{1}{2}y_1'\\end{pmatrix}$$ Adding $M$ to each coordinate shifts back the triangle to the original spot. - Sorry to ask what I'm sure is a dumb question, but once I have a 2 by 2 matrix, how do I know which are the correct x and y coordinates? – Mark Nov 19 '12 at 0:46 The $2\\times 2$ matrix doesn't represent coordinates. The latter $2\\times 1$ vectors are the coordinates you want. The first row is $x$ and the second row is $y$. – EuYu Nov 19 '12 at 0:48 Let the given vertex of the equilateral triangle be $(x_1,y_1)$ and the centroid be $(x_c,y_c)$. Let $(x_2,y_2)$ and $(x_3,y_3)$ be the other two vertices. Then we have that $$\\dfrac{x_1 + x_2 + x_3}{3} = x_c \\,\\,\\,\\,\\,\\,\\,\\,\\,\\, (1)$$", "the rotated pentagon. Solution The measure of $$\\angle BAB′=m\\angle BAE′+m\\angle E′AB′$$. Therefore $$\\angle BAB′=111.80^{\\circ} +113.20^{\\circ}$$ or $$225^{\\circ}$$. Notice the direction of the angle is counter-clockwise, therefore the angle measure is $$225^{\\circ}$$ CCW or $$225^{\\circ}$$. ## Review 1. Rotate the above figure $$90^{\\circ}$$ clockwise about the origin. 2. Rotate the above figure $$270^{\\circ}$$ clockwise about the origin. 3. Rotate the above figure $$180^{\\circ}$$ about the origin. 1. Rotate the above figure $$90^{\\circ}$$ counterclockwise about the origin. 2. Rotate the above figure $$270^{\\circ}$$ counterclockwise about the origin. 3. Rotate the above figure $$180^{\\circ}$$ about the origin. 1. Rotate the above figure $$90^{\\circ}$$ clockwise about the origin. 2. Rotate the above figure $$270^{\\circ}$$ clockwise about the origin. 3. Rotate the above figure $$180^{\\circ}$$ about the origin. 1. Rotate the above figure $$90^{\\circ}$$ counterclockwise about the origin. 2. Rotate the above figure $$270^{\\circ}$$ counterclockwise about the origin. 3. Rotate the above figure $$180^{\\circ}$$ about the origin. 1. Rotate the above figure $$90^{\\circ}$$ clockwise about the origin. 2. Rotate the above figure $$270^{\\circ}$$ clockwise about the origin. 3. Rotate the above figure $$180^{\\circ}$$ about the origin. 1. Rotate the above figure $$90^{\\circ}$$ counterclockwise about the origin. 2. Rotate the above figure $$270^{\\circ}$$ counterclockwise about the origin. 3. Rotate the above figure $$180^{\\circ}$$ about the origin. 1. Rotate the above figure $$90^{\\circ}$$ clockwise about", "you rotating clock wise of counter clockwise? Jun 29 '18 at 2:04 • counterclockwise ? – user484305 Jun 29 '18 at 2:09 • So $A$ is the matrix representation of $T$ wrt the standard basis? Use the theorem that tells you how to do this, starting with the image under $T$ of each basis vector. Jun 29 '18 at 2:13 We first set up a new orthonormal basis $(f_1,f_2,f_3)$ adapted to the given $T$. To this end choose $$f_3:={1\\over\\sqrt{3}}(1,1,-1), \\quad f_1:={1\\over\\sqrt{2}}(-1,1,0), \\qquad f_2:=f_3\\times f_1={1\\over\\sqrt{6}}(1,1,2)\\ .$$ The transform matrix from the standard basis $(e_1,e_2,e_3)$ to $(f_1,f_2,f_3)$ then is $$S:=\\left[\\matrix{-{1\\over\\sqrt{2}}&{1\\over\\sqrt{6}}&{1\\over\\sqrt{3}}\\cr {1\\over\\sqrt{2}}&{1\\over\\sqrt{6}}&{1\\over\\sqrt{3}}\\cr 0&{2\\over\\sqrt{6}}&-{1\\over\\sqrt{3}}\\cr}\\right]\\ .$$ With respect to the basis $f:=(f_1,f_2,f_3)$ the given transformation $T$ has the matrix $$[T]_f=\\left[\\matrix{-{1\\over2}&-{\\sqrt{3}\\over2}&0\\cr {\\sqrt{3}\\over2}&-{1\\over2}&0\\cr 0&0&1\\cr}\\right]$$ (we have chosen the sense of rotation here). According to the rules of linear algebra the matrix of $T$ with respect to the standard basis then is given by $$[T]_e=S\\> [T]_f\\>S^\\top\\ .$$ As for the second problem: Coordinatewise the map ${\\bf x}\\mapsto {\\bf x}':=T{\\bf x}$ is given by $${\\bf x'}=[T]_e\\>{\\bf x}\\ .$$ This implies $${\\bf x}=[T]_e^{-1}{\\bf x}'=[T]_e^\\top{\\bf x}'\\ .\\tag{1}$$ This allows to express the coordinates $(x,y,z)$ of a preimage point ${\\bf x}$ by the coordinates $(x',y',z')$ of the image point ${\\bf x}'$. The point ${\\bf x}$ lies in the plane $P$ iff its coordinates satisfy $x+2y-2z=6$. Plug the expressions in terms", "area of $A = \\dfrac{a \\times t}{2} = \\dfrac{16 \\times 28}{2} = 224$ [collapse] Problem Number 12 A vector $\\overline{a} = (-3,4)$ is the image after being reflected across the line $y=x$, and then being rotated by $90^{\\circ}$ clockwise about the origin. The initial vector before the transformations occur is $\\cdots \\cdot$ A. $(3,4)$ D. $(4,-3)$ B. $(-3,-4)$ E. $(-3,4)$ C. $(-4,3)$ Solution Suppose the initial vector is $(x, y)$. The reflection across the line $y = x$ can be expressed in the following diagram. $\\begin{pmatrix} a \\\\ b \\end{pmatrix} \\xrightarrow{M_{y=x}} \\begin{pmatrix} b \\\\ a \\end{pmatrix}$ The transformation is followed by rotation of $90^{\\circ}$ clockwise, which is same as $270^{\\circ}$ anti clockwise. Hence, we can make a following transformation scheme. \\begin{aligned} \\begin{pmatrix} b \\\\ a \\end{pmatrix} \\xrightarrow{R[O, 270^{\\circ}]} & \\begin{pmatrix} \\cos 270^{\\circ} &-\\sin 270^{\\circ} \\\\ \\sin 270^{\\circ} & \\cos 270^{\\circ} \\end{pmatrix}\\begin{pmatrix} b \\\\ a \\end{pmatrix} \\\\ & = \\begin{pmatrix} 0 & 1 \\\\-1 & 0 \\end{pmatrix} \\begin{pmatrix} b \\\\ a \\end{pmatrix} \\\\ & = \\begin{pmatrix} a \\\\-b \\end{pmatrix} \\end{aligned}We have found that the vector’s image is in the form of $(a, -b)$. Since it is given that vector $\\overline{a} = (-3,4)$ be the vector’s image, so we obtain $a=-3$ and $b=-4$. Thus, the initial vector is $\\boxed{\\begin{pmatrix} a \\\\ b \\end{pmatrix} = \\begin{pmatrix}-3 \\\\-4 \\end{pmatrix}}$ [collapse] Problem Number", "origin at $$270^{\\circ}$$ Rotation about the origin at 270^{\\circ}: $$R_270^{\\circ}(x,y)=(y,−x)$$ Now let's perform the following rotations on Image $$A$$ shown below in the diagram below and describe the rotations: 1. about the origin at $$90^{\\circ}$$, and label it $$B$$. Rotation about the origin at $$90^{\\circ}$$: $$R_{90^{\\circ}}A\\rightarrow B=R_{90^{\\circ}}(x,y)\\rightarrow (−y,x)$$ 1. about the origin at $$180^{\\circ}$$, and label it $$O$$. Rotation about the origin at $$180^{\\circ}$$: $$R_{180^{\\circ}}A\\rightarrow O=R_{180^{\\circ}}(x,y)\\rightarrow (−x,−y)$$ 1. about the origin at $$270^{\\circ}$$, and label it $$Z$$. Rotation about the origin at $$270^{\\circ}$$: $$R_{270^{\\circ}}A\\rightarrow Z=R270^{\\circ}(x,y)\\rightarrow (y,−x)$$ Finally, let's write the notation that represents the rotation of the preimage A to the rotated image J in the diagram below: First, pick a point in the diagram to use to see how it is rotated. $$E:(−1,2)\\qquad E′:(1,−2)$$ Notice how both the $$x$$- and $$y$$-coordinates are multiplied by -1. This indicates that the preimage $$A$$ is reflected about the origin by $$180^{\\circ}$$CCW to form the rotated image J. Therefore the notation is $$R_{180^{\\circ}}A\\rightarrow J=R_{180^{\\circ}}(x,y)\\rightarrow (−x,−y)$$. Example $$\\PageIndex{1}$$ Earlier, you were given the figure below which shows a pattern of two fish. Write the mapping rule for the rotation of Image $$A$$ to Image $$B$$. Solution Notice that the angle measure is $$90^{\\circ}$$ and the direction is clockwise. Therefore the Image $$A$$ has been rotated $$−90^{\\circ}$$ to form Image $$B$$. To write", "D_4$. Since $D_4$ has only $8$ elements, this calculation is fairly easy. I shall illustrate the case $n=4, k=6$. We can view the $4\\times 4$ matrix as a block matrix $$\\begin{pmatrix} A & B \\\\ C & D\\end{pmatrix}$$ where each block is a $2\\times 2$ matrix. Clearly the identity fixes all $\\binom{16}{6}=8008$ matrices. Rotation by $90^\\circ$ counterclockwise permutes this to give $$\\begin{pmatrix} B & D \\\\ A & C\\end{pmatrix}$$ so the matrix is fixed iff $A=B=D=C$. But this means the number of ones in the matrix is a multiple of $4$, yet $k$ is not divisible by $4$, thus no matrices are fixed by rotation by $90^\\circ$. It is easy to see that the result is the same for rotation by $270^\\circ$. Rotation by $180^\\circ$ permutes the original matrix to $$\\begin{pmatrix} D & C \\\\ B & A\\end{pmatrix}$$ which is the same as the original iff $A=D$ and $B=C$. Thus since $k=6$ we can either have $3$ ones in $A$, $2$ in $A$ and $1$ in $B$, $1$ in $A$ and $2$ in $B$, or $3$ in $B$. Since choosing $A$ and $B$ completely determines the matrix, the number of matrices fixed by rotation by $180^\\circ$ is $$\\binom43\\binom40+\\binom42\\binom41+\\binom41\\binom42+\\binom40\\binom43=56.$$ Now for reflection. Vertical or horizontal reflection is easy; a matrix is fixed iff its left (upper) half is the", "first rotate the vector ${\\displaystyle {\\begin{pmatrix}12&-4\\end{pmatrix}}}$, to point along the X axis. This vector has an angle ${\\displaystyle \\theta =\\arctan \\left({-(-4) \\over 12}\\right)}$. We create the orthogonal Givens rotation matrix, ${\\displaystyle G_{1}}$: {\\displaystyle {\\begin{aligned}G_{1}&={\\begin{pmatrix}\\cos(\\theta )&0&-\\sin(\\theta )\\\\0&1&0\\\\\\sin(\\theta )&0&\\cos(\\theta )\\end{pmatrix}}\\\\&\\approx {\\begin{pmatrix}0.94868&0&-0.31622\\\\0&1&0\\\\0.31622&0&0.94868\\end{pmatrix}}\\end{aligned}}} And the result of ${\\displaystyle G_{1}A}$ now has a zero in the ${\\displaystyle \\mathbf {a} _{31}}$ element. ${\\displaystyle G_{1}A\\approx {\\begin{pmatrix}12.64911&-55.97231&16.76007\\\\6&167&-68\\\\0&6.64078&-37.6311\\end{pmatrix}}}$ We can similarly form Givens matrices ${\\displaystyle G_{2}}$ and ${\\displaystyle G_{3}}$, which will zero the sub-diagonal elements ${\\displaystyle a_{21}}$ and ${\\displaystyle a_{32}}$, forming a triangular matrix ${\\displaystyle R}$. The orthogonal matrix ${\\displaystyle Q^{\\textsf {T}}}$ is formed from the product of all the Givens matrices ${\\displaystyle Q^{\\textsf {T}}=G_{3}G_{2}G_{1}}$. Thus, we have ${\\displaystyle G_{3}G_{2}G_{1}A=Q^{\\textsf {T}}A=R}$, and the QR decomposition is ${\\displaystyle A=QR}$. The QR decomposition via Givens rotations is the most involved to implement, as the ordering of the rows required to fully exploit the algorithm is not trivial to determine. However, it has a significant advantage in that each new zero element ${\\displaystyle a_{ij}}$ affects only the row with the element to be zeroed (i) and a row above (j). This makes the Givens rotation algorithm more bandwidth efficient and parallelisable than the Householder reflection technique. ## Connection to a determinant or a product of eigenvalues We can use QR decomposition to find the absolute value of the determinant of", "3 & 4 \\end{matrix}\\right)$, which does not move the square. • Clockwise rotation of the square about its centre $P$ by an angle of $90^{\\circ}$$R_1=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 2 & 3 & 4 & 1\\end{matrix}\\right)$. • Clockwise rotation of the square about its centre $P$ by an angle of $180^{\\circ}$$R_2=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 3 & 4 & 1 & 2\\end{matrix}\\right)$. • Clockwise rotation of the square about its centre $P$ by an angle of $270^{\\circ}$$R_3=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 4 & 1 & 2 & 3\\end{matrix}\\right)$. • Flip of the square about its diagonal $A$ (see figure below): $R_4=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 1 & 4 & 3 & 2\\end{matrix}\\right)$. • Flip of the square about its diagonal $C$$R_5=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 3 & 2 & 1 & 4\\end{matrix}\\right)$. • Flip of the square about its vertical axis $B$$R_6=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 2 & 1 & 4 & 3\\end{matrix}\\right)$. • Flip of the square about its horizontal axis $D$$R_7=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 4 & 3 & 2 & 1\\end{matrix}\\right)$. The set of all 8 symmetries of the square is denoted by $D_4=\\left\\{R_0,R_1,R_2,R_3,R_4,R_5,R_6,R_7\\right\\}$. Define the operation $\\circ : D_4\\times D_4\\rightarrow D_4$ to be the function composition $R_i\\circ R_j(x)=R_i(R_j(x))$ for $x\\in", "3 & 4 \\end{matrix}\\right)$, which does not move the square. • Clockwise rotation of the square about its centre $P$ by an angle of $90^{\\circ}$$R_1=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 2 & 3 & 4 & 1\\end{matrix}\\right)$. • Clockwise rotation of the square about its centre $P$ by an angle of $180^{\\circ}$$R_2=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 3 & 4 & 1 & 2\\end{matrix}\\right)$. • Clockwise rotation of the square about its centre $P$ by an angle of $270^{\\circ}$$R_3=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 4 & 1 & 2 & 3\\end{matrix}\\right)$. • Flip of the square about its diagonal $A$ (see figure below): $R_4=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 1 & 4 & 3 & 2\\end{matrix}\\right)$. • Flip of the square about its diagonal $C$$R_5=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 3 & 2 & 1 & 4\\end{matrix}\\right)$. • Flip of the square about its vertical axis $B$$R_6=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 2 & 1 & 4 & 3\\end{matrix}\\right)$. • Flip of the square about its horizontal axis $D$$R_7=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 4 & 3 & 2 & 1\\end{matrix}\\right)$. The set of all 8 symmetries of the square is denoted by $D_4=\\left\\{R_0,R_1,R_2,R_3,R_4,R_5,R_6,R_7\\right\\}$. Define the operation $\\circ : D_4\\times D_4\\rightarrow D_4$ to be the function composition $R_i\\circ R_j(x)=R_i(R_j(x))$ for $x\\in", "the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. Vocabulary Language: English Rotation Rotation A rotation is", "& 3 & 4 & 1\\end{matrix}\\right)$. • Clockwise rotation of the square about its centre $P$ by an angle of $180^{\\circ}$$R_2=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 3 & 4 & 1 & 2\\end{matrix}\\right)$. • Clockwise rotation of the square about its centre $P$ by an angle of $270^{\\circ}$$R_3=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 4 & 1 & 2 & 3\\end{matrix}\\right)$. • Flip of the square about its diagonal $A$ (see figure below): $R_4=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 1 & 4 & 3 & 2\\end{matrix}\\right)$. • Flip of the square about its diagonal $C$$R_5=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 3 & 2 & 1 & 4\\end{matrix}\\right)$. • Flip of the square about its vertical axis $B$$R_6=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 2 & 1 & 4 & 3\\end{matrix}\\right)$. • Flip of the square about its horizontal axis $D$$R_7=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 4 & 3 & 2 & 1\\end{matrix}\\right)$. The set of all 8 symmetries of the square is denoted by $D_4=\\left\\{R_0,R_1,R_2,R_3,R_4,R_5,R_6,R_7\\right\\}$. Define the operation $\\circ : D_4\\times D_4\\rightarrow D_4$ to be the function composition $R_i\\circ R_j(x)=R_i(R_j(x))$ for $x\\in X\\left\\{1,2,3,4\\right\\}$. For example, $R_1\\circ R_6$ is the result of first flipping the square about its vertical axis $B$ and then rotating it clockwise by $90^{\\circ}$: $R_1\\circ R_6 =\\left(\\begin{matrix}1 & 2 & 3", "& 1 & 4 &3 \\end{pmatrix},\\\\ b&=\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 4 & 3 & 2 & 1 \\end{pmatrix},\\ c=\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 3 & 4 & 1 &2 \\end{pmatrix}. \\end{align*} For instance, we calculate $ab$: \\begin{align*} ab&=\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 2 & 1 & 4 &3 \\end{pmatrix}\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 4 & 3 & 2 & 1 \\end{pmatrix}\\\\ &=\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 2 & 1 & 4 &3 \\end{pmatrix}\\begin{pmatrix} 4 & 3 & 2 & 1\\\\ 3 & 4 & 1 & 2 \\end{pmatrix}\\\\ &=\\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 3 & 4 & 1 &2 \\end{pmatrix}\\\\ &=c. \\end{align*} Example. [The $n$-th dihedral group $D_n$, $n\\geq 3$] Consider an equilateral triangle shown in the following picture. Dihedral group D3 $\\rho$ is counterclockwise rotation about the axis coming out of the origin by $\\frac{360^\\circ}{3}=120^\\circ$ and $\\mu_i$, $i=1,2,3$ is counterclockwise rotation about the axis of rotation through each vertex $i$ by $180^\\circ$. As permutations of the vertices $\\{1,2,3\\}$, $\\rho$ and $\\mu_i$ , $i=1,2,3$ are given by \\begin{align*} \\rho&=\\begin{pmatrix} 1 & 2 & 3\\\\ 2 & 3 & 1 \\end{pmatrix},\\ \\mu_1=\\begin{pmatrix} 1 & 2 & 3\\\\ 1 & 3 & 2 \\end{pmatrix}\\\\ \\mu_2&=\\begin{pmatrix} 1 & 2 & 3\\\\ 3", "way of saying a Transformation. And as we know, transformations involve matrix multiplication. In other words, there exists a matrix B such that $\\begin{pmatrix}\\vec w_1 & \\vec w_2 & ... & \\vec w_n\\end{pmatrix} = B*\\begin{pmatrix}\\vec v_1 & \\vec v_2 & ... & \\vec v_n\\end{pmatrix}$. Moreover, $B(\\vec v_k)=\\vec w_k$, where k is all indices from 1 to n. Therefore, it is possible to derive B. So now, let's set up the equation for the coordinates of $\\vec b$ in basis w. $\\begin{pmatrix}\\vec w_1 & \\vec w_2 & ... & \\vec w_n\\end{pmatrix} *\\begin{pmatrix}k_1 \\\\k_2\\\\ ... \\\\ k_n\\end{pmatrix} =\\begin{pmatrix}b_1 \\\\b_2\\\\ ... \\\\ b_n\\end{pmatrix}$. Through substitution we get that $B* \\begin{pmatrix}\\vec v_1 & \\vec v_2 & ... & \\vec v_n\\end{pmatrix} *\\begin{pmatrix}k_1 \\\\k_2\\\\ ... \\\\ k_n\\end{pmatrix} =\\begin{pmatrix}b_1 \\\\b_2\\\\ ... \\\\ b_n\\end{pmatrix}$. Then through a little manipulation, we get $\\begin{pmatrix}\\vec v_1 & \\vec v_2 & ... & \\vec v_n\\end{pmatrix} *\\begin{pmatrix}k_1 \\\\k_2\\\\ ... \\\\ k_n\\end{pmatrix} =B^{-1}*\\begin{pmatrix}b_1 \\\\b_2\\\\ ... \\\\ b_n\\end{pmatrix}$. Since we know that the v matrix has an inverse, we get that $\\begin{pmatrix}k_1 \\\\k_2\\\\ ... \\\\ k_n\\end{pmatrix} =\\begin{pmatrix}\\vec v_1 & \\vec v_2 & ... & \\vec v_n\\end{pmatrix}^{-1}*B^{-1}*\\begin{pmatrix}b_1 \\\\b_2\\\\ ... \\\\ b_n\\end{pmatrix}$. And thus we have our general equation for change of basis. ## Actual Example The following example converts a polynomial from one basis to another. For simplicity's sake, let's take the polynomial $x^2+2x+3$", "rotated about the point $A \\ 225^\\circ$ . Find the coordinates of the purple pentagon. On the diagram, draw and label the rotated pentagon. Answers: 1. Notice the direction of the angle is clockwise, therefore the angle measure is $60^\\circ$ CW or $-60^\\circ$ . 2. Notice the direction of the angle is counter-clockwise, therefore the angle measure is $155^\\circ$ CCW or $155^\\circ$ . 3. The measure of $\\angle BAB^\\prime = m \\angle BAE^\\prime + m \\angle E^\\prime AB^\\prime$ . Therefore $\\angle BAB^\\prime = 111.80^\\circ + 113.20^\\circ$ or $225^\\circ$ . Notice the direction of the angle is counter-clockwise, therefore the angle measure is $225^\\circ$ CCW or $225^\\circ$ . ### Explore More 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3.", "basis change matrix for this basis is: $$F=\\begin{pmatrix}u & v & w \\end{pmatrix}^{-1}=\\begin{pmatrix} A & {B-(A\\cdot B)A \\over \\|B- (A\\cdot B)A\\|} & B \\times A\\end{pmatrix}^{-1}$$ Thus, in the original base, the rotation from $A$ to $B$ can be expressed as right-multiplication of a vector by the following matrix: $$U=F^{-1}G F.$$ One can easily show that $U A = B$, and that $\\|U\\|_2=1$. Also, $U$ is the same as the $R$ matrix from Rik's answer. ## 2D Case For the 2D case, given $A=\\left(x_1,y_1,0\\right)$ and $B=\\left(x_2,y_2,0\\right)$, the matrix $G$ is the forward transformation matrix itself, and we can simplify it further. We note \\begin{aligned} \\cos\\theta &= A\\cdot B = x_1x_2+y_1y_2 \\\\ \\sin\\theta &= \\| A\\times B\\| = x_1y_2-x_2y_1 \\end{aligned} Finally, $$U\\equiv G=\\begin{pmatrix} x_1x_2+y_1y_2 & -(x_1y_2-x_2y_1) \\\\ x_1y_2-x_2y_1 & x_1x_2+y_1y_2 \\end{pmatrix}$$ and $$U^{-1}\\equiv G^{-1}=\\begin{pmatrix} x_1x_2+y_1y_2 & x_1y_2-x_2y_1 \\\\ -(x_1y_2-x_2y_1) & x_1x_2+y_1y_2 \\end{pmatrix}$$ ## Octave/Matlab Implementation The basic implementation is very simple. You could improve it by factoring out the common expressions of dot(A,B) and cross(B,A). Also note that $||A\\times B||=||B\\times A||$. GG = @(A,B) [ dot(A,B) -norm(cross(A,B)) 0;\\ norm(cross(A,B)) dot(A,B) 0;\\ 0 0 1]; FFi = @(A,B) [ A (B-dot(A,B)*A)/norm(B-dot(A,B)*A) cross(B,A) ]; UU = @(Fi,G) Fi*G*inv(Fi); Testing: > a=[1 0 0]'; b=[0 1 0]'; > U = UU(FFi(a,b), GG(a,b)); > norm(U) % is it length-preserving? ans = 1 > norm(b-U*a)", "0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & -1\\end{array}\\right)$$ is equivalent to a rotation about the only axis by and angle of $\\phi =\\pi$ radians, since a rotation is $$\\left( \\begin{array}{ccc} 1 & 0 & 0 \\\\ 0 & \\text{cos }\\phi & -\\text{sin }\\phi \\\\ 0 & \\text{sin }\\phi & \\text{cos }\\phi \\end{array}\\right)$$ However either $$\\left( \\begin{array}{ccc} 1 & 0 & 0 \\\\ 0 & \\pm 1 & 0 \\\\ 0 & 0 & \\mp 1\\end{array}\\right)$$ cannot be achieved by a rotation, and has the correct determinant. • So in other words, the transformation $$P=\\begin{pmatrix}1&&&\\\\&-1&&\\\\&&-1&\\\\&&&-1\\end{pmatrix}$$ isn't a parity transformation at all, but a spatial rotation. Interesting - according to Srednicki it is the parity transformation but perhaps the notation is loose. – theage Mar 13 '15 at 20:32", "vectors by $25^{\\circ}$ counter-clockwise. Recall, that the first and second columns of the matrix form for a linear transformation (on 2-dimensional vectors) indicate what that transformation does to the vectors $\\begin{pmatrix}1\\\\0\\end{pmatrix}$ and $\\begin{pmatrix}0\\\\1\\end{pmatrix}$, respectively If our transformation is a rotation counter-clockwise of 25 degrees, notice that $$\\begin{pmatrix}1\\\\0\\end{pmatrix} \\rightarrow \\begin{pmatrix} \\cos 25^\\circ \\\\ \\sin 25^\\circ\\end{pmatrix} \\quad \\textrm{and} \\quad \\begin{pmatrix}0\\\\1\\end{pmatrix} \\rightarrow \\begin{pmatrix}-\\sin 25^\\circ\\\\ \\cos 25^\\circ \\end{pmatrix}$$ As such, the matrix form we seek is: $$\\begin{bmatrix} \\cos 25^\\circ & -\\sin 25^\\circ\\\\ \\sin 25^\\circ& \\cos 25^\\circ\\end{bmatrix}$$ Following similar logic, the matrix form for a counter-clockwise rotation by any angle $\\theta$ is given by $$\\begin{bmatrix} \\cos \\theta & -\\sin \\theta\\\\ \\sin \\theta & \\cos \\theta \\end{bmatrix}$$ With the rotation matrix found, rotating a particular vector is easy... Suppose we wish to rotate the vector $\\begin{pmatrix}3\\\\4\\end{pmatrix}$ by $25^\\circ$. We just find $$\\begin{bmatrix} \\cos 25^\\circ & -\\sin 25^\\circ\\\\ \\sin 25^\\circ& \\cos 25^\\circ\\end{bmatrix} \\begin{pmatrix}3\\\\4\\end{pmatrix} = \\begin{pmatrix}3\\cos 25^\\circ - 4 \\sin 25^\\circ \\\\ 3 \\sin 25^\\circ + 4 \\cos 25^\\circ \\end{pmatrix} \\approx \\begin{pmatrix}1.028\\\\4.893\\end{pmatrix}$$ Other rotations are found in a similar manner...", "is generated by V0,1,2, the product of all 3 reflections. [4,3], Reflections Rotations Rotoreflection Name R0 R1 R2 S0,1 S1,2 S0,2 V0,1,2 Group Order 2 2 2 4 3 2 6 Matrix ${\\displaystyle \\left[{\\begin{smallmatrix}1&0&0\\\\0&1&0\\\\0&0&-1\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}1&0&0\\\\0&0&1\\\\0&1&0\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}0&1&0\\\\1&0&0\\\\0&0&1\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}1&0&0\\\\0&0&1\\\\0&-1&0\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}0&1&0\\\\0&0&1\\\\1&0&0\\\\\\end{smallmatrix}}\\right]}$", "# Matrix with linear transformation with reflection Find the matrix of the linear transformation A which is the reflection in the line $y = \\sqrt{2}x$ with respect to the standard basis in $\\mathbb{R^2}$. I Have no idea how to approach this problem... rotate the co-ordinate system anti-clockwise through $\\arctan \\sqrt{2}$ $$\\begin{pmatrix} x' \\\\ y' \\end{pmatrix} = \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & \\sqrt{2} \\\\ -\\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix}$$ now reflect in $x'$-axis $$\\begin{pmatrix} x'' \\\\ y'' \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\begin{pmatrix} x' \\\\ y' \\end{pmatrix}$$ finally, return to original coordinates: $$\\begin{pmatrix} x''' \\\\ y''' \\end{pmatrix} = \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & - \\sqrt{2} \\\\ \\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x'' \\\\ y'' \\end{pmatrix}$$ altogether: $$\\begin{pmatrix} x''' \\\\ y''' \\end{pmatrix} = \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & - \\sqrt{2} \\\\ \\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & \\sqrt{2} \\\\ -\\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\\\ =\\frac 13 \\begin{pmatrix} -1 & 2\\sqrt{2} \\\\ 2\\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix}$$ • why did you do arctan to rotate it? – user192898 Nov 16 '14 at 6:37 • this angle brings the x-axis to co-incide with the line $y=\\sqrt{2}x$ – David Holden Nov 16 '14 at 6:39 The", "0 & 1 \\end{pmatrix}$, $\\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix}$. If you imagine reflecting in the y-axis and then reflecting in the x-axis, the combined transformation is effectively a rotation by 180 degrees. So the transformation matrix for this is: $\\begin{pmatrix} -1 & 0 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} = \\begin{pmatrix} -1 & 0 \\\\ 0 & -1 \\end{pmatrix}$. We want a rotation matrix that will work for any different angle. Consider the point (x,0) In order that changing the angle (theta) will rotate anticlockwise by the specified amount, we want: $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\begin{pmatrix} x \\\\ 0 \\end{pmatrix} = \\begin{pmatrix} xcos \\theta \\\\ xsin \\theta \\end{pmatrix}$ So ax = xcos𝜽 and cx = xsin𝜽 So a = cos𝜽 and c = sin𝜽 As we don’t want an enlargement |A| = ad-bc = 1 and so dcos𝜽 – bsin𝜽 = 1. From Pythagoras’ Theorem, we know that one solution for this is b = -sin𝜽, d = cos𝜽. So our rotation matrix is: $A = \\begin{pmatrix} cos \\theta & -sin \\theta \\\\ sin \\theta & cos \\theta \\end{pmatrix}$ For instance, a 90º clockwise rotation would be achieved with $A = \\begin{pmatrix} cos 90 & -sin 90 \\\\ sin 90 &" ]

[ "one by a rotation about the origin $120^\\circ$ clockwise and counter-clockwise. The counter-clockwise rotation matrix is $$R_{120^\\circ} = \\begin{pmatrix}\\cos120^\\circ & -\\sin120^\\circ \\\\ \\sin120^\\circ & \\cos120^\\circ\\end{pmatrix} = \\begin{pmatrix}-\\frac{1}{2} & -\\frac{\\sqrt{3}}{2} \\\\ \\frac{\\sqrt{3}}{2} & -\\frac{1}{2}\\end{pmatrix}$$ with the clockwise rotation matrix as $$R_{-120^\\circ} = \\begin{pmatrix}\\cos120^\\circ & \\sin120^\\circ \\\\ -\\sin120^\\circ & \\cos120^\\circ\\end{pmatrix} = \\begin{pmatrix}-\\frac{1}{2} & \\frac{\\sqrt{3}}{2} \\\\ -\\frac{\\sqrt{3}}{2} & -\\frac{1}{2}\\end{pmatrix}$$ Your vertices are then $$B' = \\begin{pmatrix}-\\frac{1}{2} & -\\frac{\\sqrt{3}}{2} \\\\ \\frac{\\sqrt{3}}{2} & -\\frac{1}{2}\\end{pmatrix}\\begin{pmatrix}x_1' \\\\ y_1'\\end{pmatrix}=\\begin{pmatrix}-\\frac{1}{2}x_1' - \\frac{\\sqrt{3}}{2}y_1' \\\\ \\frac{\\sqrt{3}}{2}x_1' - \\frac{1}{2}y_1'\\end{pmatrix}$$ $$C' = \\begin{pmatrix}-\\frac{1}{2} & \\frac{\\sqrt{3}}{2} \\\\ -\\frac{\\sqrt{3}}{2} & -\\frac{1}{2}\\end{pmatrix}\\begin{pmatrix}x_1' \\\\ y_1'\\end{pmatrix}=\\begin{pmatrix}-\\frac{1}{2}x_1' + \\frac{\\sqrt{3}}{2}y_1' \\\\ -\\frac{\\sqrt{3}}{2}x_1' - \\frac{1}{2}y_1'\\end{pmatrix}$$ Adding $M$ to each coordinate shifts back the triangle to the original spot. - Sorry to ask what I'm sure is a dumb question, but once I have a 2 by 2 matrix, how do I know which are the correct x and y coordinates? – Mark Nov 19 '12 at 0:46 The $2\\times 2$ matrix doesn't represent coordinates. The latter $2\\times 1$ vectors are the coordinates you want. The first row is $x$ and the second row is $y$. – EuYu Nov 19 '12 at 0:48 Let the given vertex of the equilateral triangle be $(x_1,y_1)$ and the centroid be $(x_c,y_c)$. Let $(x_2,y_2)$ and $(x_3,y_3)$ be the other two vertices. Then we have that $$\\dfrac{x_1 + x_2 + x_3}{3} = x_c \\,\\,\\,\\,\\,\\,\\,\\,\\,\\, (1)$$", "you rotating clock wise of counter clockwise? Jun 29 '18 at 2:04 • counterclockwise ? – user484305 Jun 29 '18 at 2:09 • So $A$ is the matrix representation of $T$ wrt the standard basis? Use the theorem that tells you how to do this, starting with the image under $T$ of each basis vector. Jun 29 '18 at 2:13 We first set up a new orthonormal basis $(f_1,f_2,f_3)$ adapted to the given $T$. To this end choose $$f_3:={1\\over\\sqrt{3}}(1,1,-1), \\quad f_1:={1\\over\\sqrt{2}}(-1,1,0), \\qquad f_2:=f_3\\times f_1={1\\over\\sqrt{6}}(1,1,2)\\ .$$ The transform matrix from the standard basis $(e_1,e_2,e_3)$ to $(f_1,f_2,f_3)$ then is $$S:=\\left[\\matrix{-{1\\over\\sqrt{2}}&{1\\over\\sqrt{6}}&{1\\over\\sqrt{3}}\\cr {1\\over\\sqrt{2}}&{1\\over\\sqrt{6}}&{1\\over\\sqrt{3}}\\cr 0&{2\\over\\sqrt{6}}&-{1\\over\\sqrt{3}}\\cr}\\right]\\ .$$ With respect to the basis $f:=(f_1,f_2,f_3)$ the given transformation $T$ has the matrix $$[T]_f=\\left[\\matrix{-{1\\over2}&-{\\sqrt{3}\\over2}&0\\cr {\\sqrt{3}\\over2}&-{1\\over2}&0\\cr 0&0&1\\cr}\\right]$$ (we have chosen the sense of rotation here). According to the rules of linear algebra the matrix of $T$ with respect to the standard basis then is given by $$[T]_e=S\\> [T]_f\\>S^\\top\\ .$$ As for the second problem: Coordinatewise the map ${\\bf x}\\mapsto {\\bf x}':=T{\\bf x}$ is given by $${\\bf x'}=[T]_e\\>{\\bf x}\\ .$$ This implies $${\\bf x}=[T]_e^{-1}{\\bf x}'=[T]_e^\\top{\\bf x}'\\ .\\tag{1}$$ This allows to express the coordinates $(x,y,z)$ of a preimage point ${\\bf x}$ by the coordinates $(x',y',z')$ of the image point ${\\bf x}'$. The point ${\\bf x}$ lies in the plane $P$ iff its coordinates satisfy $x+2y-2z=6$. Plug the expressions in terms", "origin at $$270^{\\circ}$$ Rotation about the origin at 270^{\\circ}: $$R_270^{\\circ}(x,y)=(y,−x)$$ Now let's perform the following rotations on Image $$A$$ shown below in the diagram below and describe the rotations: 1. about the origin at $$90^{\\circ}$$, and label it $$B$$. Rotation about the origin at $$90^{\\circ}$$: $$R_{90^{\\circ}}A\\rightarrow B=R_{90^{\\circ}}(x,y)\\rightarrow (−y,x)$$ 1. about the origin at $$180^{\\circ}$$, and label it $$O$$. Rotation about the origin at $$180^{\\circ}$$: $$R_{180^{\\circ}}A\\rightarrow O=R_{180^{\\circ}}(x,y)\\rightarrow (−x,−y)$$ 1. about the origin at $$270^{\\circ}$$, and label it $$Z$$. Rotation about the origin at $$270^{\\circ}$$: $$R_{270^{\\circ}}A\\rightarrow Z=R270^{\\circ}(x,y)\\rightarrow (y,−x)$$ Finally, let's write the notation that represents the rotation of the preimage A to the rotated image J in the diagram below: First, pick a point in the diagram to use to see how it is rotated. $$E:(−1,2)\\qquad E′:(1,−2)$$ Notice how both the $$x$$- and $$y$$-coordinates are multiplied by -1. This indicates that the preimage $$A$$ is reflected about the origin by $$180^{\\circ}$$CCW to form the rotated image J. Therefore the notation is $$R_{180^{\\circ}}A\\rightarrow J=R_{180^{\\circ}}(x,y)\\rightarrow (−x,−y)$$. Example $$\\PageIndex{1}$$ Earlier, you were given the figure below which shows a pattern of two fish. Write the mapping rule for the rotation of Image $$A$$ to Image $$B$$. Solution Notice that the angle measure is $$90^{\\circ}$$ and the direction is clockwise. Therefore the Image $$A$$ has been rotated $$−90^{\\circ}$$ to form Image $$B$$. To write", "first rotate the vector ${\\displaystyle {\\begin{pmatrix}12&-4\\end{pmatrix}}}$, to point along the X axis. This vector has an angle ${\\displaystyle \\theta =\\arctan \\left({-(-4) \\over 12}\\right)}$. We create the orthogonal Givens rotation matrix, ${\\displaystyle G_{1}}$: {\\displaystyle {\\begin{aligned}G_{1}&={\\begin{pmatrix}\\cos(\\theta )&0&-\\sin(\\theta )\\\\0&1&0\\\\\\sin(\\theta )&0&\\cos(\\theta )\\end{pmatrix}}\\\\&\\approx {\\begin{pmatrix}0.94868&0&-0.31622\\\\0&1&0\\\\0.31622&0&0.94868\\end{pmatrix}}\\end{aligned}}} And the result of ${\\displaystyle G_{1}A}$ now has a zero in the ${\\displaystyle \\mathbf {a} _{31}}$ element. ${\\displaystyle G_{1}A\\approx {\\begin{pmatrix}12.64911&-55.97231&16.76007\\\\6&167&-68\\\\0&6.64078&-37.6311\\end{pmatrix}}}$ We can similarly form Givens matrices ${\\displaystyle G_{2}}$ and ${\\displaystyle G_{3}}$, which will zero the sub-diagonal elements ${\\displaystyle a_{21}}$ and ${\\displaystyle a_{32}}$, forming a triangular matrix ${\\displaystyle R}$. The orthogonal matrix ${\\displaystyle Q^{\\textsf {T}}}$ is formed from the product of all the Givens matrices ${\\displaystyle Q^{\\textsf {T}}=G_{3}G_{2}G_{1}}$. Thus, we have ${\\displaystyle G_{3}G_{2}G_{1}A=Q^{\\textsf {T}}A=R}$, and the QR decomposition is ${\\displaystyle A=QR}$. The QR decomposition via Givens rotations is the most involved to implement, as the ordering of the rows required to fully exploit the algorithm is not trivial to determine. However, it has a significant advantage in that each new zero element ${\\displaystyle a_{ij}}$ affects only the row with the element to be zeroed (i) and a row above (j). This makes the Givens rotation algorithm more bandwidth efficient and parallelisable than the Householder reflection technique. ## Connection to a determinant or a product of eigenvalues We can use QR decomposition to find the absolute value of the determinant of", "the rotated pentagon. Solution The measure of $$\\angle BAB′=m\\angle BAE′+m\\angle E′AB′$$. Therefore $$\\angle BAB′=111.80^{\\circ} +113.20^{\\circ}$$ or $$225^{\\circ}$$. Notice the direction of the angle is counter-clockwise, therefore the angle measure is $$225^{\\circ}$$ CCW or $$225^{\\circ}$$. ## Review 1. Rotate the above figure $$90^{\\circ}$$ clockwise about the origin. 2. Rotate the above figure $$270^{\\circ}$$ clockwise about the origin. 3. Rotate the above figure $$180^{\\circ}$$ about the origin. 1. Rotate the above figure $$90^{\\circ}$$ counterclockwise about the origin. 2. Rotate the above figure $$270^{\\circ}$$ counterclockwise about the origin. 3. Rotate the above figure $$180^{\\circ}$$ about the origin. 1. Rotate the above figure $$90^{\\circ}$$ clockwise about the origin. 2. Rotate the above figure $$270^{\\circ}$$ clockwise about the origin. 3. Rotate the above figure $$180^{\\circ}$$ about the origin. 1. Rotate the above figure $$90^{\\circ}$$ counterclockwise about the origin. 2. Rotate the above figure $$270^{\\circ}$$ counterclockwise about the origin. 3. Rotate the above figure $$180^{\\circ}$$ about the origin. 1. Rotate the above figure $$90^{\\circ}$$ clockwise about the origin. 2. Rotate the above figure $$270^{\\circ}$$ clockwise about the origin. 3. Rotate the above figure $$180^{\\circ}$$ about the origin. 1. Rotate the above figure $$90^{\\circ}$$ counterclockwise about the origin. 2. Rotate the above figure $$270^{\\circ}$$ counterclockwise about the origin. 3. Rotate the above figure $$180^{\\circ}$$ about the origin. 1. Rotate the above figure $$90^{\\circ}$$ clockwise about", "3 & 4 \\end{matrix}\\right)$, which does not move the square. • Clockwise rotation of the square about its centre $P$ by an angle of $90^{\\circ}$$R_1=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 2 & 3 & 4 & 1\\end{matrix}\\right)$. • Clockwise rotation of the square about its centre $P$ by an angle of $180^{\\circ}$$R_2=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 3 & 4 & 1 & 2\\end{matrix}\\right)$. • Clockwise rotation of the square about its centre $P$ by an angle of $270^{\\circ}$$R_3=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 4 & 1 & 2 & 3\\end{matrix}\\right)$. • Flip of the square about its diagonal $A$ (see figure below): $R_4=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 1 & 4 & 3 & 2\\end{matrix}\\right)$. • Flip of the square about its diagonal $C$$R_5=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 3 & 2 & 1 & 4\\end{matrix}\\right)$. • Flip of the square about its vertical axis $B$$R_6=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 2 & 1 & 4 & 3\\end{matrix}\\right)$. • Flip of the square about its horizontal axis $D$$R_7=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 4 & 3 & 2 & 1\\end{matrix}\\right)$. The set of all 8 symmetries of the square is denoted by $D_4=\\left\\{R_0,R_1,R_2,R_3,R_4,R_5,R_6,R_7\\right\\}$. Define the operation $\\circ : D_4\\times D_4\\rightarrow D_4$ to be the function composition $R_i\\circ R_j(x)=R_i(R_j(x))$ for $x\\in", "3 & 4 \\end{matrix}\\right)$, which does not move the square. • Clockwise rotation of the square about its centre $P$ by an angle of $90^{\\circ}$$R_1=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 2 & 3 & 4 & 1\\end{matrix}\\right)$. • Clockwise rotation of the square about its centre $P$ by an angle of $180^{\\circ}$$R_2=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 3 & 4 & 1 & 2\\end{matrix}\\right)$. • Clockwise rotation of the square about its centre $P$ by an angle of $270^{\\circ}$$R_3=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 4 & 1 & 2 & 3\\end{matrix}\\right)$. • Flip of the square about its diagonal $A$ (see figure below): $R_4=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 1 & 4 & 3 & 2\\end{matrix}\\right)$. • Flip of the square about its diagonal $C$$R_5=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 3 & 2 & 1 & 4\\end{matrix}\\right)$. • Flip of the square about its vertical axis $B$$R_6=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 2 & 1 & 4 & 3\\end{matrix}\\right)$. • Flip of the square about its horizontal axis $D$$R_7=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 4 & 3 & 2 & 1\\end{matrix}\\right)$. The set of all 8 symmetries of the square is denoted by $D_4=\\left\\{R_0,R_1,R_2,R_3,R_4,R_5,R_6,R_7\\right\\}$. Define the operation $\\circ : D_4\\times D_4\\rightarrow D_4$ to be the function composition $R_i\\circ R_j(x)=R_i(R_j(x))$ for $x\\in", "D_4$. Since $D_4$ has only $8$ elements, this calculation is fairly easy. I shall illustrate the case $n=4, k=6$. We can view the $4\\times 4$ matrix as a block matrix $$\\begin{pmatrix} A & B \\\\ C & D\\end{pmatrix}$$ where each block is a $2\\times 2$ matrix. Clearly the identity fixes all $\\binom{16}{6}=8008$ matrices. Rotation by $90^\\circ$ counterclockwise permutes this to give $$\\begin{pmatrix} B & D \\\\ A & C\\end{pmatrix}$$ so the matrix is fixed iff $A=B=D=C$. But this means the number of ones in the matrix is a multiple of $4$, yet $k$ is not divisible by $4$, thus no matrices are fixed by rotation by $90^\\circ$. It is easy to see that the result is the same for rotation by $270^\\circ$. Rotation by $180^\\circ$ permutes the original matrix to $$\\begin{pmatrix} D & C \\\\ B & A\\end{pmatrix}$$ which is the same as the original iff $A=D$ and $B=C$. Thus since $k=6$ we can either have $3$ ones in $A$, $2$ in $A$ and $1$ in $B$, $1$ in $A$ and $2$ in $B$, or $3$ in $B$. Since choosing $A$ and $B$ completely determines the matrix, the number of matrices fixed by rotation by $180^\\circ$ is $$\\binom43\\binom40+\\binom42\\binom41+\\binom41\\binom42+\\binom40\\binom43=56.$$ Now for reflection. Vertical or horizontal reflection is easy; a matrix is fixed iff its left (upper) half is the", "& 3 & 4 & 1\\end{matrix}\\right)$. • Clockwise rotation of the square about its centre $P$ by an angle of $180^{\\circ}$$R_2=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 3 & 4 & 1 & 2\\end{matrix}\\right)$. • Clockwise rotation of the square about its centre $P$ by an angle of $270^{\\circ}$$R_3=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 4 & 1 & 2 & 3\\end{matrix}\\right)$. • Flip of the square about its diagonal $A$ (see figure below): $R_4=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 1 & 4 & 3 & 2\\end{matrix}\\right)$. • Flip of the square about its diagonal $C$$R_5=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 3 & 2 & 1 & 4\\end{matrix}\\right)$. • Flip of the square about its vertical axis $B$$R_6=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 2 & 1 & 4 & 3\\end{matrix}\\right)$. • Flip of the square about its horizontal axis $D$$R_7=\\left(\\begin{matrix}1 & 2 & 3 & 4\\\\ 4 & 3 & 2 & 1\\end{matrix}\\right)$. The set of all 8 symmetries of the square is denoted by $D_4=\\left\\{R_0,R_1,R_2,R_3,R_4,R_5,R_6,R_7\\right\\}$. Define the operation $\\circ : D_4\\times D_4\\rightarrow D_4$ to be the function composition $R_i\\circ R_j(x)=R_i(R_j(x))$ for $x\\in X\\left\\{1,2,3,4\\right\\}$. For example, $R_1\\circ R_6$ is the result of first flipping the square about its vertical axis $B$ and then rotating it clockwise by $90^{\\circ}$: $R_1\\circ R_6 =\\left(\\begin{matrix}1 & 2 & 3", "the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. Vocabulary Language: English Rotation Rotation A rotation is", "you were to describe the rotated image using notation, you would write the following: $R_{90^\\circ}(x,y)=(-y,x)$ #### Example A Find an image of the point (3, 2) that has undergone a clockwise rotation: a) about the origin at $90^\\circ$ , b) about the origin at $180^\\circ$ , and c) about the origin at $270^\\circ$ . Write the notation to describe the rotation. Solution: a) Rotation about the origin at $90^\\circ : R_{90^\\circ}(x,y)=(-y,x)$ b) Rotation about the origin at $180^\\circ : R_{180^\\circ}(x,y)=(-x,-y)$ c) Rotation about the origin at $270^\\circ : R_{270^\\circ}(x,y)=(y,-x)$ #### Example B Rotate Image A in the diagram below: a) about the origin at $90^\\circ$ , and label it $B$ . b) about the origin at $180^\\circ$ , and label it $O$ . c) about the origin at $270^\\circ$ , and label it $Z$ . Write notation for each to indicate the type of rotation. Solution: a) Rotation about the origin at $90^\\circ$ : $R_{90^\\circ}A \\rightarrow B=R_{90^\\circ}(x,y) \\rightarrow (-y,x)$ b) Rotation about the origin at $180^\\circ$ : $R_{180^\\circ}A \\rightarrow O=R_{180^\\circ}(x,y) \\rightarrow (-x,-y)$ c) Rotation about the origin at $270^\\circ$ : $R_{270^\\circ}A \\rightarrow Z=R_{270^\\circ}(x,y) \\rightarrow (y,-x)$ #### Example C Write the notation that represents the rotation of the preimage A to the rotated image J in the diagram below. First, pick a point in the diagram to use to", "area of $A = \\dfrac{a \\times t}{2} = \\dfrac{16 \\times 28}{2} = 224$ [collapse] Problem Number 12 A vector $\\overline{a} = (-3,4)$ is the image after being reflected across the line $y=x$, and then being rotated by $90^{\\circ}$ clockwise about the origin. The initial vector before the transformations occur is $\\cdots \\cdot$ A. $(3,4)$ D. $(4,-3)$ B. $(-3,-4)$ E. $(-3,4)$ C. $(-4,3)$ Solution Suppose the initial vector is $(x, y)$. The reflection across the line $y = x$ can be expressed in the following diagram. $\\begin{pmatrix} a \\\\ b \\end{pmatrix} \\xrightarrow{M_{y=x}} \\begin{pmatrix} b \\\\ a \\end{pmatrix}$ The transformation is followed by rotation of $90^{\\circ}$ clockwise, which is same as $270^{\\circ}$ anti clockwise. Hence, we can make a following transformation scheme. \\begin{aligned} \\begin{pmatrix} b \\\\ a \\end{pmatrix} \\xrightarrow{R[O, 270^{\\circ}]} & \\begin{pmatrix} \\cos 270^{\\circ} &-\\sin 270^{\\circ} \\\\ \\sin 270^{\\circ} & \\cos 270^{\\circ} \\end{pmatrix}\\begin{pmatrix} b \\\\ a \\end{pmatrix} \\\\ & = \\begin{pmatrix} 0 & 1 \\\\-1 & 0 \\end{pmatrix} \\begin{pmatrix} b \\\\ a \\end{pmatrix} \\\\ & = \\begin{pmatrix} a \\\\-b \\end{pmatrix} \\end{aligned}We have found that the vector’s image is in the form of $(a, -b)$. Since it is given that vector $\\overline{a} = (-3,4)$ be the vector’s image, so we obtain $a=-3$ and $b=-4$. Thus, the initial vector is $\\boxed{\\begin{pmatrix} a \\\\ b \\end{pmatrix} = \\begin{pmatrix}-3 \\\\-4 \\end{pmatrix}}$ [collapse] Problem Number", "# Matrix with linear transformation with reflection Find the matrix of the linear transformation A which is the reflection in the line $y = \\sqrt{2}x$ with respect to the standard basis in $\\mathbb{R^2}$. I Have no idea how to approach this problem... rotate the co-ordinate system anti-clockwise through $\\arctan \\sqrt{2}$ $$\\begin{pmatrix} x' \\\\ y' \\end{pmatrix} = \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & \\sqrt{2} \\\\ -\\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix}$$ now reflect in $x'$-axis $$\\begin{pmatrix} x'' \\\\ y'' \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\begin{pmatrix} x' \\\\ y' \\end{pmatrix}$$ finally, return to original coordinates: $$\\begin{pmatrix} x''' \\\\ y''' \\end{pmatrix} = \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & - \\sqrt{2} \\\\ \\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x'' \\\\ y'' \\end{pmatrix}$$ altogether: $$\\begin{pmatrix} x''' \\\\ y''' \\end{pmatrix} = \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & - \\sqrt{2} \\\\ \\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} \\frac1{\\sqrt{3}} \\begin{pmatrix} 1 & \\sqrt{2} \\\\ -\\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\\\ =\\frac 13 \\begin{pmatrix} -1 & 2\\sqrt{2} \\\\ 2\\sqrt{2} & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix}$$ • why did you do arctan to rotate it? – user192898 Nov 16 '14 at 6:37 • this angle brings the x-axis to co-incide with the line $y=\\sqrt{2}x$ – David Holden Nov 16 '14 at 6:39 The", "vectors by $25^{\\circ}$ counter-clockwise. Recall, that the first and second columns of the matrix form for a linear transformation (on 2-dimensional vectors) indicate what that transformation does to the vectors $\\begin{pmatrix}1\\\\0\\end{pmatrix}$ and $\\begin{pmatrix}0\\\\1\\end{pmatrix}$, respectively If our transformation is a rotation counter-clockwise of 25 degrees, notice that $$\\begin{pmatrix}1\\\\0\\end{pmatrix} \\rightarrow \\begin{pmatrix} \\cos 25^\\circ \\\\ \\sin 25^\\circ\\end{pmatrix} \\quad \\textrm{and} \\quad \\begin{pmatrix}0\\\\1\\end{pmatrix} \\rightarrow \\begin{pmatrix}-\\sin 25^\\circ\\\\ \\cos 25^\\circ \\end{pmatrix}$$ As such, the matrix form we seek is: $$\\begin{bmatrix} \\cos 25^\\circ & -\\sin 25^\\circ\\\\ \\sin 25^\\circ& \\cos 25^\\circ\\end{bmatrix}$$ Following similar logic, the matrix form for a counter-clockwise rotation by any angle $\\theta$ is given by $$\\begin{bmatrix} \\cos \\theta & -\\sin \\theta\\\\ \\sin \\theta & \\cos \\theta \\end{bmatrix}$$ With the rotation matrix found, rotating a particular vector is easy... Suppose we wish to rotate the vector $\\begin{pmatrix}3\\\\4\\end{pmatrix}$ by $25^\\circ$. We just find $$\\begin{bmatrix} \\cos 25^\\circ & -\\sin 25^\\circ\\\\ \\sin 25^\\circ& \\cos 25^\\circ\\end{bmatrix} \\begin{pmatrix}3\\\\4\\end{pmatrix} = \\begin{pmatrix}3\\cos 25^\\circ - 4 \\sin 25^\\circ \\\\ 3 \\sin 25^\\circ + 4 \\cos 25^\\circ \\end{pmatrix} \\approx \\begin{pmatrix}1.028\\\\4.893\\end{pmatrix}$$ Other rotations are found in a similar manner...", "# Rotate the standard basis and determine the bases that arise from it Given is $\\mathbb{R}^2$ with standard basis $B^2_0$ A \"new\" basis $B=\\left\\{\\vec{b_1}; \\vec{b_2}\\right\\}$ arises from $B^2_0$ by rotation of $30°$ (clockwise). Determine the change of basis $T^{B}_{B^2_0}$ and the basis vectors $\\vec{b_1}$ and $\\vec{b_2}$. So we are in $\\mathbb{R}^2$ and we have a standard basis $B^2_0$. Standard matrix is another word for unit matrix I think, so we have $B^2_0 =\\begin{pmatrix} 1 & 0\\\\ 0 & 1 \\end{pmatrix}$. Rotate this by $30°$, we have $B= \\begin{pmatrix} \\cos(30°) & -\\sin(30°)\\\\ \\sin(30°) & \\cos(30°) \\end{pmatrix} = \\begin{pmatrix} \\frac{\\sqrt{3}}{2} & -\\frac{1}{2}\\\\ \\frac{1}{2} & \\frac{\\sqrt{3}}{2} \\end{pmatrix}$ So $\\vec{b_1} = \\begin{pmatrix} \\frac{\\sqrt{3}}{2}\\\\ \\frac{1}{2} \\end{pmatrix}$ and $\\vec{b_2} = \\begin{pmatrix} -\\frac{1}{2}\\\\ \\frac{\\sqrt{3}}{2} \\end{pmatrix}$ And $T^{B}_{B^2_0}$ we can solve with Gaussian elimination, right? But I'm more interested to know if I calculated the rotation matrix and the resulting bases correctly? • If you have posted the whole question, $g$ has nothing to do here. – Bernard May 2 '18 at 8:29 • @Bernard Thank you, I forgot another part of the example was on the next page (back). Just editted my question :) – eyesima May 2 '18 at 8:33 • More in general with respect to a given basis of vectors $\\vec v_i$ it suffice to determine/know the transformed vectors $\\vec v_i \\to", "is generated by V0,1,2, the product of all 3 reflections. [4,3], Reflections Rotations Rotoreflection Name R0 R1 R2 S0,1 S1,2 S0,2 V0,1,2 Group Order 2 2 2 4 3 2 6 Matrix ${\\displaystyle \\left[{\\begin{smallmatrix}1&0&0\\\\0&1&0\\\\0&0&-1\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}1&0&0\\\\0&0&1\\\\0&1&0\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}0&1&0\\\\1&0&0\\\\0&0&1\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}1&0&0\\\\0&0&1\\\\0&-1&0\\\\\\end{smallmatrix}}\\right]}$ ${\\displaystyle \\left[{\\begin{smallmatrix}0&1&0\\\\0&0&1\\\\1&0&0\\\\\\end{smallmatrix}}\\right]}$", "$PR=\\left[\\begin{matrix}0&-1\\\\0&0\\end{matrix}\\right].$ $RP=\\left[\\begin{matrix}0&0\\\\1&0\\end{matrix}\\right].$ #4(a). $\\left[\\begin{matrix}1&0\\\\0&0\\end{matrix}\\right].$ (b). $\\left[\\begin{matrix}0&0\\\\0&0\\end{matrix}\\right].$ (c). $\\left[\\begin{matrix}19&22\\\\43&50\\end{matrix}\\right].$ #8. See 4(b). Partial solutions to HW: Section 2.5. #1. $\\left[\\begin{matrix}-1&11\\\\-3&23\\end{matrix}\\right].$ Homework problems (assume the cube has as vertices all combinations of $(\\pm1,\\pm1,\\pm1)$ as in class today): You are holding the cube by the opposite corners $(1,1,1)$ and $(-1,-1,-1)$. You spin it one-third the way around, then one-third again, and the third time you're back to where you started. What $3\\times3$ matrices describe these rotations? You are holding the cube by the midpoint of the edge with endpoints $(1,1,1)$ and $(1,-1,1)$, and by the midpoint of the opposite edge. You spin the cube $180^\\circ$ around the axis through these two midpoints. What matrix describes this rotation? Imagine the cube as drawn in class. Let $A$ be the matrix describing the rotation bringing the top toward you $90^\\circ$ (the axis of rotation is the $y$-axis here). Let $B$ be the matrix describing the rotation which turns the cube $90^\\circ$ to your right (the axis of rotation is the $x$-axis). By sheer force of imagination, write the matrix describing what happens when the rotation $B$ is done first, and then the rotation $A$ is performed. Then, compute the matrix product $AB$ to see that your answer is correct. What rotation corresponds to the matrix $\\left[\\begin{matrix}0&-1&0\\cr 0&0&1\\cr -1&0&0\\end{matrix}\\right]$? Hey, wait minute! You've been", "basis change matrix for this basis is: $$F=\\begin{pmatrix}u & v & w \\end{pmatrix}^{-1}=\\begin{pmatrix} A & {B-(A\\cdot B)A \\over \\|B- (A\\cdot B)A\\|} & B \\times A\\end{pmatrix}^{-1}$$ Thus, in the original base, the rotation from $A$ to $B$ can be expressed as right-multiplication of a vector by the following matrix: $$U=F^{-1}G F.$$ One can easily show that $U A = B$, and that $\\|U\\|_2=1$. Also, $U$ is the same as the $R$ matrix from Rik's answer. ## 2D Case For the 2D case, given $A=\\left(x_1,y_1,0\\right)$ and $B=\\left(x_2,y_2,0\\right)$, the matrix $G$ is the forward transformation matrix itself, and we can simplify it further. We note \\begin{aligned} \\cos\\theta &= A\\cdot B = x_1x_2+y_1y_2 \\\\ \\sin\\theta &= \\| A\\times B\\| = x_1y_2-x_2y_1 \\end{aligned} Finally, $$U\\equiv G=\\begin{pmatrix} x_1x_2+y_1y_2 & -(x_1y_2-x_2y_1) \\\\ x_1y_2-x_2y_1 & x_1x_2+y_1y_2 \\end{pmatrix}$$ and $$U^{-1}\\equiv G^{-1}=\\begin{pmatrix} x_1x_2+y_1y_2 & x_1y_2-x_2y_1 \\\\ -(x_1y_2-x_2y_1) & x_1x_2+y_1y_2 \\end{pmatrix}$$ ## Octave/Matlab Implementation The basic implementation is very simple. You could improve it by factoring out the common expressions of dot(A,B) and cross(B,A). Also note that $||A\\times B||=||B\\times A||$. GG = @(A,B) [ dot(A,B) -norm(cross(A,B)) 0;\\ norm(cross(A,B)) dot(A,B) 0;\\ 0 0 1]; FFi = @(A,B) [ A (B-dot(A,B)*A)/norm(B-dot(A,B)*A) cross(B,A) ]; UU = @(Fi,G) Fi*G*inv(Fi); Testing: > a=[1 0 0]'; b=[0 1 0]'; > U = UU(FFi(a,b), GG(a,b)); > norm(U) % is it length-preserving? ans = 1 > norm(b-U*a)", "rotated about the point $A \\ 225^\\circ$ . Find the coordinates of the purple pentagon. On the diagram, draw and label the rotated pentagon. Answers: 1. Notice the direction of the angle is clockwise, therefore the angle measure is $60^\\circ$ CW or $-60^\\circ$ . 2. Notice the direction of the angle is counter-clockwise, therefore the angle measure is $155^\\circ$ CCW or $155^\\circ$ . 3. The measure of $\\angle BAB^\\prime = m \\angle BAE^\\prime + m \\angle E^\\prime AB^\\prime$ . Therefore $\\angle BAB^\\prime = 111.80^\\circ + 113.20^\\circ$ or $225^\\circ$ . Notice the direction of the angle is counter-clockwise, therefore the angle measure is $225^\\circ$ CCW or $225^\\circ$ . ### Explore More 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ clockwise about the origin. 2. Rotate the above figure $270^\\circ$ clockwise about the origin. 3. Rotate the above figure $180^\\circ$ about the origin. 1. Rotate the above figure $90^\\circ$ counterclockwise about the origin. 2. Rotate the above figure $270^\\circ$ counterclockwise about the origin. 3.", "<< \"(1,0,0) after rotation = \" << v_rotated.transpose() << std::endl; return 0; } The output of the code is as follows: 0 0 0.5 0.866025 //Q. coeffs() first outputs the real part of quaternion, and then outputs the imaginary part of quaternion 0 0 0.5 //Q. vec() outputs the imaginary part of quaternion (1,0,0) after rotation = (0.5 0.866025 0) Transformation from quaternion to rotation matrix Let quaternion$q = Q_ 0 + q_ 1i + q_ 2j + q_ 3K $, then the corresponding rotation matrix is: $$R= \\left [ \\begin{matrix} 1-2q_2^2-2q_3^2 & 2q_1q_2-2q_0q_3 & 2q_1q_3 + 2q_0q_2 \\\\ 2q_1q_2+2q_0q_3 & 1-2q_1^2-2q_3^2 & 2q_2q_3-2q_0q_1 \\\\ 2q_1q_3-2q_0q_2 & 2q_2q_3 + 2q_0q_1 & 1-2q_1^2-2q_2^2 \\end{matrix} \\right ]$$ Transformation from rotation matrix to Quaternion Suppose the matrix is$r = {M_ The quaternion Q of {ij}}, I, j in [1,2,3] \\$, is: 1 $$q_0=\\frac{\\sqrt{tr(R)+1}}{2}$$ $$q_1=\\frac{m_{23}-m_{32}}{4q_0}$$ $$q_2=\\frac{m_{31}-m_{13}}{4q_0}$$ $$q_3=\\frac{m_{12}-m_{21}}{4q_0}$$ #### 2.3 quaternion, Euler angle, rotation matrix, rotation vector transformation ##### 2.3.1 transformation from rotation vector to rotation matrix, Euler angle and quaternion Define rotation vector //Initialize the rotation vector: the rotation angle is alpha and the rotation axis is (x, y, z) Eigen::AngleAxisd rotation_vector(alpha, Eigen::Vector3d(x,y,z)) Rotate vector to rotation matrix: //Rotation vector to rotation matrix Eigen::Matrix3d rotation_matrix = rotation_vector.matrix(); Eigen::Matrix3d rotation_matrix = rotation_vector.toRotationMatrix(); Rotate matrix to Euler angle: //Rotation vector rotation Euler angle (z-y-x)" ]

[ "# Aime 2008 Ii 13 1. May 20, 2008 ### ehrenfest 1. The problem statement, all variables and given/known data A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let R be the region outside the hexagon, and let S = $\\{ 1/z |x \\in R}$. Then the area of S has the form a $\\pi +\\sqrt b$, where a and b are positive integers. Find a+b. 2. Relevant equations 3. The attempt at a solution This should only require high school math although there is probably a solution using the fact that 1/z is a Mobius transformation or something else in complex analysis. The hexagon is contained in the closed of radius 1/sqrt(3) center at the origin, which means that S is contained inside of the closed disk of radius \\sqrt3 centered at the origin. So basically we need to figure out what to subtract off of 3 \\pi. Anyone know how to do that?", "# Difference between revisions of \"2008 AIME II Problems/Problem 13\" ## Problem A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \\left\\lbrace\\frac{1}{z}|z \\in R\\right\\rbrace$. Then the area of $S$ has the form $a\\pi + \\sqrt{b}$, where $a$ and $b$ are positive integers. Find $a + b$. ## Solution 1 If you're familiar with inversion, you'll see that the problem's basically asking you to invert the hexagon with respect to the unit circle in the Cartesian Plane using the Inversion Distance Formula. This works because the point in the Cartesian Plane's complex plane equivalent switches places with its conjugate but we can do that in the Cartesian plane too (just reflect a point in the Cartesian plane over the x-axis)! If you're familiar with inversion you can go plot the inverted figure's Cartesian Plane Equivalent. Then simply continue on with the figure shown in the below solution. ## Solution 2 If a point $z = r\\text{cis}\\,\\theta$ is in $R$, then the point $\\frac{1}{z} = \\frac{1}{r} \\text{cis}\\, \\left(-\\theta\\right)$ is in $S$ (where cis denotes $\\text{cis}\\, \\theta = \\cos \\theta + i \\sin \\theta$). Since $R$ is symmetric", "# Aime 2008 Ii 13 1. May 20, 2008 ### ehrenfest 1. The problem statement, all variables and given/known data A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let R be the region outside the hexagon, and let S = $\\{ 1/z |x \\in R}$. Then the area of S has the form a $\\pi +\\sqrt b$, where a and b are positive integers. Find a+b. 2. Relevant equations 3. The attempt at a solution This should only require high school math although there is probably a solution using the fact that 1/z is a Mobius transformation or something else in complex analysis. The hexagon is contained in the closed of radius 1/sqrt(3) center at the origin, which means that S is contained inside of the closed disk of radius \\sqrt3 centered at the origin. So basically we need to figure out what to subtract off of 3 \\pi. Anyone know how to do that? Please just give a hint. 2. May 20, 2008 ### Tedjn What you are trying to do is a good idea, but there doesn't seem to be an easy way to find the cutoffs, as you've noticed. I caved and peeked", "the same plane ? I need to calculate the center of an hexagon. However, the hexagon in not planar (it is a 3D object, a distorted hexagon). Can the above formulas be exended to calculate the centroid of this hexagon ? - You need to define what \"this hexagon\" is. If you triangulate it, then you could compute the centroid of each triangle, and then the centroid of those area-weighted centroids. – Joseph O'Rourke Nov 7 2011 at 19:12", "# Geomegebra Algebra Level 1 If the center of a regular hexagon is at the origin and one of the vertices on the Argand plane is $$1 + 2i$$, then what is its perimeter? × Problem Loading... Note Loading... Set Loading...", "and are $A_{1},A_{2},A_{3}$ are inner angles. Complex hexagonometry does not discriminate distances and angles.", "axis at distance from the center. (b) Solution :", "in Theorem 1. Let be a regular hexagon of unit side. Consider bundles of parallel lines whose slopes are , , , , , . The axes of all parallel strips are all incident to the center of the circle created by the vertices of ; see Figs. 7 and 8. Assume a coordinate system where the lower left corner of is at the origin, and the lower side of lies along the -axis. Let be the partition of into bundles of parallel lines. The lines in are contained in the parallel strip bounded by the two lines and , for . The equation of line is , with , , given in Table 3. , and are bounded by the pairs of lines supporting opposite sides of , while , and are bounded by the pairs of lines supporting opposite short diagonals of . Therefore . We refer to lines in as the primary lines, to lines in as the secondary lines, and to the rest of the lines in as the tertiary lines. The final construction is obtained by a small clockwise rotation, so that there are no vertical lines. Observe that the distance between consecutive lines in any of the bundles of • primary lines is ; • secondary lines is ; • tertiary lines", "a single line, as two non-parallel planes do in 3-dimensional space; or (4) they can intersect in a single point[i] (and they must, if they are completely orthogonal).[g] 12. ^ The edges of the orthogonal great squares are not aligned with the grid lines of the unit radius coordinate system. Six of the squares do lie in the 6 orthogonal planes of the coordinate system, but their edges are the 2 diagonals of unit edge length squares of the coordinate lattice. For example: ( 0, 0, 1, 0) ( 0,–1, 0, 0) ( 0, 1, 0, 0) ( 0, 0,–1, 0) is the square in the xy plane. Notice that the 8 integer coordinates comprise the vertices of the 6 orthogonal squares. 13. ^ a b c The hexagons are inclined (tilted) at 60 degrees with respect to the unit radius coordinate system's orthogonal planes. Each hexagonal plane contains only one of the 4 coordinate system axes. The hexagon consists of 3 pairs of opposite vertices (three 24-cell diameters): one opposite pair of integer coordinate vertices (one of the four coordinate axes), and two opposite pairs of half-integer coordinate vertices (not coordinate axes). For example: ( 0, 0, 1, 0) ( 1/2,–1/2, 1/2,–1/2) ( 1/2, 1/2, 1/2, 1/2) (–1/2,–1/2,–1/2,–1/2) (–1/2, 1/2,–1/2, 1/2) ( 0, 0,–1, 0) is a", "# Complex Polygon Algebra Level 4 On the complex plane, a regular polygon formed by connecting each consecutive point that is a solution to the equation $$x^n = 1,$$ is centered at the origin and has a vertex at $$z = \\bigg(\\frac{\\sqrt{3}}{2} + \\frac{i}{2}\\bigg).$$ Let A be the minimum possible number of sides found on this polygon, and let B be area of the polygon with A sides. Find the value of A $$+$$ B ## Note: You may use a calculator to evaluate the area of the polygon $$i$$ is the imaginary unit ×", "part, and the vertical distance from the origin is given by its imaginary part. We can thus think of the real and imaginary parts as the Cartesian coordinates of that point. It turns out that there is another way to represent a complex number, but rather than using the real and imaginary parts to specify it, we will use two other pieces of information. If I tell you that a complex number is a distance $|z|$ away from the origin in the complex plane, then this leaves you with a whole circle of possibilities. All the points on the circle of radius $|z|$ about the origin are the same distance from the origin. But if I also give you an angle subtended between the x-axis and the line joining the complex number and the origin, read anti-clockwise from the x-axis, this will completely pin down the point in the complex plane.…", "# Geomegebra Algebra Level 1 If the center of a regular hexagon is at the origin and one of the vertices on the Argand plane is $$1 + 2i$$, then what is its perimeter? ×", "in these skew orthogonal projections: A principal diagonal of a hexagon is a diagonal which divides the hexagon into quadrilaterals.", "mapped is called the imaginary axis. Solved Exercise Problem: What is the point corresponding to the complex number -2+i6 on the Argand plane? • (-2,6) • (-2,6) • sqrt(2^2+6^2) The answer is '(-2,6)' slide-show version coming soon", "the sphere and one below. They have a common right side that is Δx. Relate this side to L and θ using the top triangle and to h and β using the bottom triangle.", "unit segments in $\\mathbb R^3$ such that any two neighboring segments form a right angle. (Physically this could be ...", "\\ quad {\\ text {(sic! )}}}$, where a circle with a radius fixed in the coordinate origin and an angle between the x-axis and a ray that intersects the origin are considered. ${\\ displaystyle r}$${\\ displaystyle \\ varphi}$ The imaginary unit should be on the other side of the equation. ${\\ displaystyle \\ mathrm {i}}$", "# The area of the triangle on the complex plane formed by the complex numbers z, –iz and z + iz is Question: The area of the triangle on the complex plane formed by the complex numbers z, –iz and z + iz is (a) $|z|^{2}$ (b) $|\\bar{z}|^{2}$ (c) $\\frac{1}{2}|z|^{2}$ (d) none of these Solution: For any complex number $z,-i z$ represents complex number obtained by rotating $z$ clockwise by $\\frac{\\pi}{2}$ angle. Hence, $z,-i z$ and $z+i z$ represents a right angled triangle with sides $z,-i z$ and hypotenus $z+i z$ $\\therefore$ Area of triangle formed is $=\\frac{1}{2}|z \\times(-i z)|$ $=\\frac{1}{2}|i||z|^{2}$ $=\\frac{1}{2}|z|^{2}$ Hence, the correct answer is option C. Leave a comment Free Study Material", "by the pairs of parallel faces of the cube $x = 0, x=1, y = 0, y= 1, z=0, z=1$. Hence the two triangles are oppositely similar with a centre of similarity. The following diagram shows the coordinates of the vertices of the two triangles, which in this case intersect in a hexagon. The centre of similarity of the two triangles is the intersection of two lines joining corresponding sides – this can be found to be the point $(1/(a+b+c), 1/(a+b+c), 1/(a+b+c))$, which is the intersection of the unit cube’s diagonal from the origin (to $(1,1,1)$) and the plane $ax + by + cz = 1$. Side lengths of the triangles and distances between corresponding parallel sides may be found by Pythagoras’ theorem and are shown below for one pair of corresponding sides (the remaining lengths can be found by cyclically permuting $a,b,c$). To sum up, all of the possible cross sections of a cube where the plane is not parallel to an edge can be described by the intersection of two oppositely similar triangles with corresponding sides parallel. The type of polygon obtained depends on which vertices of the figure below are selected, as determined by the values of $a,b,c$. In this figure a vertex for the cross-sectional polygon is chosen if the constraint associated with it", "$AB,$ $GH$ on $DF,$ etc. As a consequence, the three intersections of the opposite sides of $IJKLGH$ are collinear, allowing for another application of Pascal's theorem converse. I think that \"Desargues' Hexagon\" is a proper moniker for the hexagon $IJKLGH.$" ]

[ "# Dodecagon (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) A dodecagon is a 12-sided polygon. The sum of its internal angles is $1800^{\\circ}$. Each of its exterior angles has measure $30^{\\circ}$. A regular dodecagon can be seen below: $[asy] for(int i = 0; i <= 11; ++i) { draw(dir(360/12*i)--dir(360/12*(i + 1))); } pair A,B,C,D,E,F,G,H,I,J,K,L; A=dir(360/12*0); B=dir(360/12*1); C=dir(360/12*2); D=dir(360/12*3); E=dir(360/12*4); F=dir(360/12*5); G=dir(360/12*6); H=dir(360/12*7); I=dir(360/12*8); J=dir(360/12*9); K=dir(360/12*10); L=dir(360/12*11); label(\"A\",A,dir(0)); label(\"B\",B,dir(30)); label(\"C\",C,dir(60)); label(\"D\",D,dir(90)); label(\"E\",E,dir(120)); label(\"F\",F,dir(150)); label(\"G\",G,dir(180)); label(\"H\",H,dir(210)); label(\"I\",I,dir(240)); label(\"J\",J,dir(270)); label(\"K\",K,dir(300)); label(\"L\",L,dir(330)); draw(dir(360/12*0)--dir(360/12*6)); dot((dir(360/12*0)+dir(360/12*6))/2); pair O = (dir(360/12*0)+dir(360/12*6))/2; label(\"O\",O,S); draw(A--O); draw(Circle(O,1)); [/asy]$ The area of a regular dodecagon can be calculated by the formula $3R^2$, where $R$ is the circumradius of the dodecagon. In this case, $R$ would be $OA$. Also, each small triangle ($AOB$, $BOC$, etc.) is congruent, so $\\angle AOB=\\angle BOC=\\angle COD$ (etc) $=30^{\\circ}$.", "# Difference between revisions of \"Dodecagon\" A dodecagon is a 12-sided polygon. The sum of its internal angles is $1800^{\\circ}$. A regular dodecagon can be seen below: $[asy] for(int i = 0; i <= 11; ++i) { draw(dir(360/12*i)--dir(360/12*(i + 1))); } pair A,B,C,D,E,F,G,H,I,J,K,L; A=dir(360/12*0); B=dir(360/12*1); C=dir(360/12*2); D=dir(360/12*3); E=dir(360/12*4); F=dir(360/12*5); G=dir(360/12*6); H=dir(360/12*7); I=dir(360/12*8); J=dir(360/12*9); K=dir(360/12*10); L=dir(360/12*11); label(\"A\",A,dir(0)); label(\"B\",B,dir(30)); label(\"C\",C,dir(60)); label(\"D\",D,dir(90)); label(\"E\",E,dir(120)); label(\"F\",F,dir(150)); label(\"G\",G,dir(180)); label(\"H\",H,dir(210)); label(\"I\",I,dir(240)); label(\"J\",J,dir(270)); label(\"K\",K,dir(300)); label(\"L\",L,dir(330)); draw(dir(360/12*0)--dir(360/12*6)); dot((dir(360/12*0)+dir(360/12*6))/2); pair O = (dir(360/12*0)+dir(360/12*6))/2; label(\"O\",O,S); draw(A--G); draw(Circle(O,1)); [/asy]$ The area of a regular dodecagon can be calculated by the formula $3R^2$, where $R$ is the circumradius of the dodecagon. In this case, $R$ would be $OA$.", "2\\qquad\\text{(E)}\\ 3$ ## Problem 19 The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon? $[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(240), D=2/sqrt(3)*dir(300), E=2/sqrt(3)*dir(0), F=2/sqrt(3)*dir(60); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, dir(origin--A)); label(\"B\", B, dir(origin--B)); label(\"C\", C, dir(origin--C)); label(\"D\", D, dir(origin--D)); label(\"E\", E, dir(origin--E)); label(\"F\", F, dir(origin--F)); [/asy]$ $\\text{(A)}\\ \\frac{1}{2}\\qquad\\text{(B)}\\ \\frac{\\sqrt 3}{3}\\qquad\\text{(C)}\\ \\frac{2}{3}\\qquad\\text{(D)}\\ \\frac{3}{4}\\qquad\\text{(E)}\\ \\frac{\\sqrt 3}{2}$ ## Problem 20 In the xy-plane, what is the length of the shortest path from $(0,0)$ to $(12,16)$ that does not go inside the circle $(x-6)^{2}+(y-8)^{2}= 25$? $\\text{(A)}\\ 10\\sqrt 3\\qquad\\text{(B)}\\ 10\\sqrt 5\\qquad\\text{(C)}\\ 10\\sqrt 3+\\frac{ 5\\pi}{3}\\qquad\\text{(D)}\\ 40\\frac{\\sqrt{3}}{3}\\qquad\\text{(E)}\\ 10+5\\pi$ ## Problem 21 $[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label(\"A\", A, dir(point--A)); label(\"B\", B, dir(point--B)); label(\"C\", C, dir(point--C)); label(\"D\", D, dir(point--D)); label(\"E\", E, dir(point--E)); [/asy]$ Solution ## Problem 28 $[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label(\"A\", A, NE); label(\"B\", B, NW); label(\"C\", C, S); label(\"D\", D, E); label(\"4\", (4,2,0), SW); label(\"4\", (2,4,0), SE); label(\"3\", (0, 4, 1.5), E); [/asy]$ Solution", "2\\qquad\\text{(E)}\\ 3$ ## Problem 19 The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon? $[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(240), D=2/sqrt(3)*dir(300), E=2/sqrt(3)*dir(0), F=2/sqrt(3)*dir(60); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, dir(origin--A)); label(\"B\", B, dir(origin--B)); label(\"C\", C, dir(origin--C)); label(\"D\", D, dir(origin--D)); label(\"E\", E, dir(origin--E)); label(\"F\", F, dir(origin--F)); [/asy]$ $\\text{(A)}\\ \\frac{1}{2}\\qquad\\text{(B)}\\ \\frac{\\sqrt 3}{3}\\qquad\\text{(C)}\\ \\frac{2}{3}\\qquad\\text{(D)}\\ \\frac{3}{4}\\qquad\\text{(E)}\\ \\frac{\\sqrt 3}{2}$ ## Problem 20 In the xy-plane, what is the length of the shortest path from $(0,0)$ to $(12,16)$ that does not go inside the circle $(x-6)^{2}+(y-8)^{2}= 25$? $\\text{(A)}\\ 10\\sqrt 3\\qquad\\text{(B)}\\ 10\\sqrt 5\\qquad\\text{(C)}\\ 10\\sqrt 3+\\frac{ 5\\pi}{3}\\qquad\\text{(D)}\\ 40\\frac{\\sqrt{3}}{3}\\qquad\\text{(E)}\\ 10+5\\pi$ ## Problem 21 Triangles $ABC$ and $ABD$ are isosceles with $AB=AC=BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC, then$ \\angle C+\\angle D \\$ is $[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label(\"A\", A, dir(point--A)); label(\"B\", B, dir(point--B)); label(\"C\", C, dir(point--C)); label(\"D\", D, dir(point--D)); label(\"E\", E, dir(point--E)); [/asy]$ $\\text{(A)}\\ 115^\\circ\\qquad\\text{(B)}\\ 120^\\circ\\qquad\\text{(C)}\\ 130^\\circ\\qquad\\text{(D)}\\ 135^\\circ\\qquad\\text{(E)}\\ \\text{not uniquely determined}$ ## Problem 22 Four distinct points, $A$, $B$, $C$, and $D$, are to be selected from $1996$ points evenly spaced around a circle. All quadruples are equally likely to be chosen.", "# Difference between revisions of \"1996 AHSME Problems/Problem 19\" ## Problem The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon? $[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(240), D=2/sqrt(3)*dir(300), E=2/sqrt(3)*dir(0), F=2/sqrt(3)*dir(60); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, dir(origin--A)); label(\"B\", B, dir(origin--B)); label(\"C\", C, dir(origin--C)); label(\"D\", D, dir(origin--D)); label(\"E\", E, dir(origin--E)); label(\"F\", F, dir(origin--F)); [/asy]$ $\\text{(A)}\\ \\frac{1}{2}\\qquad\\text{(B)}\\ \\frac{\\sqrt 3}{3}\\qquad\\text{(C)}\\ \\frac{2}{3}\\qquad\\text{(D)}\\ \\frac{3}{4}\\qquad\\text{(E)}\\ \\frac{\\sqrt 3}{2}$ ## Solution $[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(240), D=2/sqrt(3)*dir(300), E=2/sqrt(3)*dir(0), F=2/sqrt(3)*dir(60); pair O=(0,0); dot(A^^B^^C^^D^^E^^F^^O); label(\"A\", A, dir(origin--A)); label(\"B\", B, dir(origin--B)); label(\"C\", C, dir(origin--C)); label(\"D\", D, dir(origin--D)); label(\"E\", E, dir(origin--E)); label(\"F\", F, dir(origin--F)); label(\"O\", O, N); draw(O--D--C--cycle); [/asy]$ $\\triangle OCD$ is copied six times to form the hexagon, so if we find the ratio of the area of the kite inside $\\triangle OCD$ to the the area of $\\triangle OCD$ itself, it will be the same ratio. Let $OC=CD=OD =2$ so that the area of the triangle is $\\frac{\\sqrt{3}s^2}{4} = \\sqrt{3}$. Notice that $\\triangle OCD$ is made up of a kite and two $30-60-90$ triangles. The two hypotenuses of these two triangles form $CD$, so the hypotenuse of each triangle must be $\\frac{2}{2} = 1$. Thus, the legs of each triangle are $\\frac{1}{2}$ and $\\frac{\\sqrt{3}}{2}}$ (Error compiling", "endpoints of a side of the larger hexagon, as shown in the diagram. Since each interior angle of a regular hexagon is 120 degrees, this triangle is a 30-60-90 right triangle. Let $r$ be the radius of the circle. The length of the longer leg of the triangle is $r$, so the length of the shorter leg is $r/\\sqrt{3}$ and the length of the hypotenuse is $2r/\\sqrt{3}$. Since for the smaller hexagon the length of the segment connecting a vertex to the center is $r$, the dimensions of the larger hexagon are $2/\\sqrt{3}$ times larger than the dimensions of the smaller hexagon. Therefore, the area of the larger triangle is $(2/\\sqrt{3})^2=\\boxed{4/3}$ times greater than the area of the smaller triangle. [asy] size(5cm); defaultpen(linewidth(.7pt)+fontsize(8pt)); dotfactor=4; int i; draw(circle((0,0),1)); for(i=0;i<=5;++i) { draw(dir(60*i)--dir(60*(i+1))); draw(2/sqrt(3)*dir(60*i)--2/sqrt(3)*dir(60*(i+1))); } draw(2/sqrt(3)*dir(0)--(0,0)--dir(30)); draw(0.93*dir(30)--dir(30)+0.07*dir(-60)+0.07*dir(210)--dir(30)+0.07*dir(-60));[/asy]", "+0 # Asymptote coding +3 37 2 +538 Hello! I was answering a question earlier, and wanted to include a graph. Since Asympotote coding makes it look clean, I tried inserting this code: [asy] size(250); import TrigMacros; real theta = 45; pair A = (3,0); pair B = 3*dir(theta); pair O = (0,0); rr_cartesian_axes(-5, 5, -5, 5, usegrid = false); draw(Circle(O, 3)); draw(O--3*dir(60)); dot(\"$\\left(\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right)$\", 3*dir(30), E); draw(O--3*dir(30)); dot(\"$\\left(\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$\", 3*dir(60), N); draw(O--3*dir(45)); dot(\"$\\left(\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2}}{2}\\right)$\", 3*dir(45), dir(45)); draw(O--3*dir(150)); dot(\"$\\left(-\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right)$\", 3*dir(150), W); draw(O--3*dir(120)); dot(\"$\\left(-\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$\", 3*dir(120), N); draw(O--3*dir(135)); dot(\"$\\left(-\\frac{\\sqrt{2}}{2}, \\frac{\\sqrt{2}}{2}\\right)$\", 3*dir(135), dir(135)); draw(O--3*dir(-60)); dot(\"$\\left(\\frac{\\sqrt{3}}{2}, -\\frac{1}{2}\\right)$\", 3*dir(-30), E); draw(O--3*dir(-30)); dot(\"$\\left(\\frac{1}{2}, -\\frac{\\sqrt{3}}{2}\\right)$\", 3*dir(-60), -N); draw(O--3*dir(-45)); dot(\"$\\left(\\frac{\\sqrt{2}}{2}, -\\frac{\\sqrt{2}}{2}\\right)$\", 3*dir(-45), dir(-45)); draw(O--3*dir(-150)); dot(\"$\\left(-\\frac{\\sqrt{3}}{2}, -\\frac{1}{2}\\right)$\", 3*dir(-150), W); draw(O--3*dir(-120)); dot(\"$\\left(-\\frac{1}{2}, -\\frac{\\sqrt{3}}{2}\\right)$\", 3*dir(-120), -N); draw(O--3*dir(-135)); dot(\"$\\left(-\\frac{\\sqrt{2}}{2},- \\frac{\\sqrt{2}}{2}\\right)$\", 3*dir(-135), dir(-135)); dot(\"$(1,0)$\", A, NE); [/asy] But, it did not show up. (it's a unit circle) Does this site support Asymptote, or do I need to use desmos? Thanks! off-topic Aug 2, 2020 #1 +128 +1 I believe that this site does not support asymptote, as I have tried to use it in the past. If you try to take a screenshot, or maybe as you said desmos will work. Hope this helps! Aug 2, 2020 #2 +538 +3 Okay! ilorty Aug 2, 2020", "n$ parallel to the bases of the half-circles are drawn such that the distances between $l_k$ and $l_0$ and $l_{-k}$ and $l_0$ are always the same for all $k=1\\to n$. The intersection points of $l_k$ with one of the half-circles are labeled $A_k, B_k$, and with the other half-circle at $C_k,D_k$, as shown in the diagram. Prove that $$\\prod_{k=-n}^n |A_kB_k|+|C_kD_k| \\ge \\prod_{k=-n}^n |A_kD_k|+|B_kC_k|$$ ## Picture 2 $[asy] for (int i=0;i<6;i=i+1){ draw(dir(60*i)--dir(60*i+60)); } draw(dir(120)--(dir(0)+dir(-60))/2); draw(dir(180)--(dir(60)+dir(0))/2); fill(dir(120)--dir(180)--intersectionpoint(dir(120)--(dir(0)+dir(-60))/2,dir(180)--(dir(60)+dir(0))/2)--cycle,grey); fill((dir(0)+dir(-60))/2--dir(0)--(dir(60)+dir(0))/2--intersectionpoint(dir(120)--(dir(0)+dir(-60))/2,dir(180)--(dir(60)+dir(0))/2)--cycle,grey); [/asy]$ $$\\text{Prove the shaded areas are equal.}$$ $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,black); } } for(int i = 1; i < 7; ++i){ filldraw((i,7-i)--(i+1,7-i)--(i+1,8-i)--(i,8-i)--cycle,white); } for(int i = 0; i < 5; ++i){ filldraw((i,4-i)--(i+1,4-i)--(i+1,5-i)--(i,5-i)--cycle,white); } for(int i = 0; i < 5; ++i){ filldraw((8-i,4+i)--(7-i,4+i)--(7-i,3+i)--(8-i,3+i)--cycle,white); } filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle,white); filldraw((0,1)--(0,2)--(1,2)--(1,1)--cycle,white); filldraw((7,8)--(6,8)--(6,7)--(7,7)--cycle,white); filldraw((8,7)--(8,6)--(7,6)--(7,7)--cycle,white); [/asy]$ ## physics problem $[asy]size(100,100); draw((0,0)--(0,10)); draw(Circle((1,1),1)); dot((0,0)); draw(9.5*dir(80)..9.5*dir(70)..9.5*dir(60),EndArrow()); label(\"A\",(-1,5),dir(180));[/asy]$ $[asy] size(85,85); draw((0,0)--10*dir(60)); draw(Circle((sqrt(3),1),1)); dot((0,0)); draw(9.5*dir(50)..9.5*dir(40)..9.5*dir(30),EndArrow()); label(\"B\",(0,5),dir(0)); [/asy]$ $[asy] size(110,110); draw((0,0)--10*dir(11.42)); draw(Circle((10,1),1)); dot((0,0)); label(\"C\",(5,1.5),dir(90)); [/asy]$ ## Solution $[asy] import olympiad; size(350,350); draw((0,0)--10*dir(60)); draw(Circle((sqrt(3),1),1)); dot((0,0)); draw((-1,0)--(7,0),grey); draw((0,0)--(sqrt(3),1),linetype(\"8 8\")); draw((0,0)--(0,5),grey); draw(sqrt(3)*dir(60)--(sqrt(3),1)--(sqrt(3),0),linetype(\"8 8\")); draw(anglemark((1,0),(0,0),dir(30))); label(\"\\varphi\",0.3*dir(15),dir(15)); draw(anglemark(dir(60),(0,0),(0,1))); label(\"\\theta\",0.3*dir(75),dir(75)); label(\"1\",(sqrt(3),0.5),dir(0)); label(\"\\frac{1}{\\tan\\varphi}\",(sqrt(3)/2,0),dir(-90)); [/asy]$ $[asy] import olympiad; size(300,300); draw((0,0)--10*dir(11.42)); draw(Circle((10,1),1)); dot((0,0)); draw((-1,0)--(12,0),grey); draw((0,0)--(0,3),grey); draw(anglemark(10*dir(11.42),(0,0),(0,1))); label(\"\\theta\",0.3*dir(50.71),dir(50.71)); draw((0,0)--(10,1),linetype(\"8 8\")); draw(10*dir(11.42)--(10,1)--(10,0),linetype(\"8 8\")); label(\"1\",(10,0.5),dir(0)); label(\"10\",(5,0),dir(-90)); label(\"\\varphi\",3.4*dir(2.855),dir(2.855)); markscalefactor=0.4; draw(anglemark((1,0),(0,0),dir(5.71))); [/asy]$ ## inscribed triangle $[asy] draw((dir(0)--dir(120)--dir(240)--dir(0)..dir(120)..dir(240)..dir(0)--cycle)); draw(dir(56)--dir(230),green); draw(dir(-23)--dir(-98),red);[/asy]$ $[asy] draw((dir(0)--dir(120)--dir(240)--dir(0)..dir(120)..dir(240)..dir(0)--cycle)); draw(dir(0)--dir(36),red); draw(dir(0)--dir(72),red); draw(dir(0)--dir(108),red); draw(dir(0)--dir(144),green);", "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of", "hexagon, as shown in the diagram. Since each interior angle of a regular hexagon is 120 degrees, this triangle is a 30-60-90 right triangle. Let $r$ be the radius of the circle. The length of the longer leg of the triangle is $r$, so the length of the shorter leg is $r/\\sqrt{3}$ and the length of the hypotenuse is $2r/\\sqrt{3}$. Since for the smaller hexagon the length of the segment connecting a vertex to the center is $r$, the dimensions of the larger hexagon are $2/\\sqrt{3}$ times larger than the dimensions of the smaller hexagon. Therefore, the area of the larger triangle is $(2/\\sqrt{3})^2=\\boxed{4/3}$ times greater than the area of the smaller triangle. [asy] size(5cm); defaultpen(linewidth(.7pt)+fontsize(8pt)); dotfactor=4; int i; draw(circle((0,0),1)); for(i=0;i<=5;++i) { draw(dir(60*i)--dir(60*(i+1))); draw(2/sqrt(3)*dir(60*i)--2/sqrt(3)*dir(60*(i+1))); } draw(2/sqrt(3)*dir(0)--(0,0)--dir(30)); draw(0.93*dir(30)--dir(30)+0.07*dir(-60)+0.07*dir(210)--dir(30)+0.07*dir(-60));[/asy] • help hexagon geometry - ## Similar Questions 1. ### math how do you find the area of a hexagon? Here is the formula: Area of regular hexagon: (sqrt3/ 2)(Wsqrd) Where sqrt=square root W=the smallest width of the hexagon sqrd=squared 2. ### Geometry A regular hexagon with sides of 3\" is inscribed in a circle. What is the area of a segment formed by a side of the hexagon and the circle? 3. ### geometry a regular hexagon is inscribed in a circle. The radius of the circle is 18 units.", "\\pm \\sqrt{117}$. Since we know $X$ is the lower solution, we take the negative value to get $X = (0,3-\\sqrt{117})$. We can solve the problem two ways from here. We can find $Y$ by rotation and use the distance formula to find the length, or we can be somewhat more clever. We notice that it is easier to find $OX$ as they lie on the same vertical, $\\angle XOY$ is $120$ degrees so we can make use of $30-60-90$ triangles, and $OX = OY$ because $O$ is the center of triangle $XYZ$. We can draw the diagram as such:$[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C^^W--X^^W--Y^^W--midpoint(X--Y),dashed); dot(\"A\",A,1.5*dir(A),linewidth(4)); dot(\"B\",B,1.5*(-1,0),linewidth(4)); dot(\"C\",C,1.5*(1,0),linewidth(4)); dot(\"B'\",B1,1.5*dir(B1),linewidth(4)); dot(\"C'\",C1,1.5*dir(C1),linewidth(4)); dot(\"O\",W,1.5*dir(90),linewidth(4)); dot(\"X\",X,1.5*dir(X),linewidth(4)); dot(\"Y\",Y,1.5*dir(Y),linewidth(4)); dot(\"Z\",Z,1.5*dir(Z),linewidth(4)); [/asy]$Note that $OX = OY = \\sqrt{117} - 3$. It follows that\\begin{align*} XY &= 2 \\cdot \\frac{OX\\cdot\\sqrt{3}}{2} \\\\ &= OX \\cdot \\sqrt{3} \\\\ &= (\\sqrt{117}-3) \\cdot \\sqrt{3} \\\\ &= \\sqrt{351}-\\sqrt{27}. \\end{align*}Finally, the answer is $351+27 = \\boxed{378}$. ~KingRavi ## Solution 2 (Euclidean Geometry) $[asy] /* Made by MRENTHUSIASM */ /* Modified", "= \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, because $\\triangle RPM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which equals $ED", "have that $AB = AE$. Similarly, $CD = DF$. $OE = 100\\sqrt{2} = \\frac{OB}{2}$ and $EF=\\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \\boxed{\\textbf{(E) } 500}$ ## Solution 6 (Ptolemy's Theorem) $[asy] pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D(\"A\", dir(50), dir(50)); pair B = D(\"B\", dir(90), dir(90)); pair C = D(\"C\", dir(130), dir(130)); pair D = D(\"D\", dir(170), dir(170)); pair O = D(\"O\", (0,0), dir(-90)); draw(A--C, red); draw(B--D, blue+dashed); draw(A--B--C--D--cycle); draw(A--O--C); draw(O--B); [/asy]$ Let $s = 200$. Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\\triangle OBC$. Since $\\triangle OBC$ is isosceles we can compute its area to be $s^2 \\sqrt7/4$, hence $CA = 2 \\tfrac{2 \\cdot s^2\\sqrt7/4}{s\\sqrt2} = s\\sqrt{7/2}$. Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \\cdot s \\implies AD = (7/2-1)s.$ This gives us: $$\\boxed{\\textbf{(E) } 500.}$$ ## Solution 7 (Trigonometry) Since all three sides equal $200$, they subtend three equal angles from the center. The right triangle between the center of the circle, a vertex, and the midpoint between two vertices has side lengths $100,100\\sqrt{7},200\\sqrt{2}$ by the Pythagorean Theorem. Thus, the sine of half of the subtended angle is $\\frac{100}{200\\sqrt{2}}=\\frac{\\sqrt{2}}{4}$. Similarly, the cosine is $\\frac{100\\sqrt{7}}{200\\sqrt{2}}=\\frac{\\sqrt{14}}{4}$. Since there are three sides, and since", "equilateral triangle in the interior of $\\triangle ABC.$ The side length of the smaller equilateral triangle can be written as $\\sqrt{a} - \\sqrt{b},$ where $a$ and $b$ are positive integers. Find $a+b.$ ## Diagram $[asy] /* Made by MRENTHUSIASM */ size(250); pair A, B, C, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); dot(\"A\",A,1.5*dir(A),linewidth(4)); dot(\"B\",B,1.5*dir(B),linewidth(4)); dot(\"C\",C,1.5*dir(C),linewidth(4)); dot(\"\\omega\",W,1.5*dir(270),linewidth(4)); dot(\"\\omega_A\",WA,1.5*dir(-WA),linewidth(4)); dot(\"\\omega_B\",WB,1.5*dir(-WB),linewidth(4)); dot(\"\\omega_C\",WC,1.5*dir(-WC),linewidth(4)); [/asy]$~MRENTHUSIASM ~ihatemath123 ## Solution 1 (Coordinate Geometry) We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\\omega_A$ is the incircle of $\\triangle AB'C'$.$[asy] /* Made by MRENTHUSIASM */ size(300); pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z; A = 18*dir(90); B = 18*dir(210); C = 18*dir(330); B1 = A+24*sqrt(3)*dir(B-A); C1 = A+24*sqrt(3)*dir(C-A); W = (0,0); WA = 6*dir(270); WB = 6*dir(30); WC = 6*dir(150); X = (sqrt(117)-3)*dir(270); Y = (sqrt(117)-3)*dir(30); Z = (sqrt(117)-3)*dir(150); filldraw(X--Y--Z--cycle,green,dashed); draw(Circle(WA,12)^^Circle(WB,12)^^Circle(WC,12),blue); draw(Circle(W,18)^^A--B--C--cycle); draw(B--B1--C1--C,dashed); dot(\"A\",A,1.5*dir(A),linewidth(4)); dot(\"B\",B,1.5*(-1,0),linewidth(4)); dot(\"C\",C,1.5*(1,0),linewidth(4)); dot(\"B'\",B1,1.5*dir(B1),linewidth(4)); dot(\"C'\",C1,1.5*dir(C1),linewidth(4)); dot(\"O\",W,1.5*dir(90),linewidth(4)); dot(\"X\",X,1.5*dir(X),linewidth(4)); dot(\"Y\",Y,1.5*dir(Y),linewidth(4)); dot(\"Z\",Z,1.5*dir(Z),linewidth(4)); [/asy]$Since the diameter of the circle is the height of this triangle, the height of this triangle is $36$. We can use inradius or equilateral triangle properties to get the", "2} = \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, since $\\triangle R PM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which", "angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(circumcircle(A,B,D)); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot(\"A\", A, SW); dot(\"B\", B, dir(0)); dot(\"C\", C, N); dot(\"D\", D, dir(0)); dot(\"E\", E, N); dot(\"F\", F, dir(0)); [/asy]$ Applying the Law of Sines to $\\triangle BED$ and $\\triangle CFD$ gives$$DE = BD\\cdot\\frac{\\sin y}{\\sin x}\\text{~ and ~} DF = CD\\cdot\\frac{\\sin z}{\\sin x}.$$By the Angle Bisector Theorem, $BD = 2$ and $CD = 3$. Combining the above information yields $$[\\triangle AEF] = \\frac{3\\sin y\\cdot\\sin z\\cdot\\cos x}{\\sin^2 x}.$$Applying the Law of Cosines to $\\triangle ABC$ gives $\\cos 2x = \\frac9{16}$, $\\cos 2y = \\frac1{8}$, and $\\cos 2z = \\frac34$. By the Half Angle Formulas,$$\\sin^2x = \\frac7{32},~~ \\cos x = \\sqrt{\\frac{25}{32}},~~ \\sin y = \\sqrt{\\frac7{16}}, \\text{~ and ~} \\sin z = \\sqrt{\\frac18}.$$Therefore $$[\\triangle AEF] = \\frac{3\\cdot\\sqrt{\\frac{7}{16}}\\cdot\\sqrt{\\frac{1}{8}}\\cdot\\sqrt{\\frac{25}{32}}} {\\frac{7}{32}} = \\frac{15\\sqrt{7}}{14}.$$The requested sum is $15+7+14 = 36$. ## Solution 10 (Official MAA 2) Let the point $M$ be the midpoint of $\\overline{AD}$, let $I$ be the incenter of $\\triangle ABC$ which is the common point of lines $AD$, $BE$, and $CF$, and let $r$ be the inradius of $\\triangle ABC$. The semiperimeter of", "C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D(\"O\",A,dir(A)); D(\"O_1\",B,dir(B)); D(\"O_2\",C,dir(C)); D(\"H\",H,dir(270)); D(\"X\",X,dir(225)); D(\"A\",R1,dir(180)); D(\"B\",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$Since $X$ is on the radical axis of $\\omega_1$ and $\\omega_2$, it has equal power with respect to both circles, so$$O_1X^2 - r_1^2 = O_2X^2-r_2^2 \\implies O_1X-O_2X = \\frac{r_1^2-r_2^2}{d}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular,\\begin{align*} O_1H &= O_1X - HX = \\frac{d+\\frac{r_1^2-r_2^2}{d}}{2} - \\frac{r}{2} \\\\ O_2H &= O_2X + HX = \\frac{d-\\frac{r_1^2-r_2^2}{d}}{2} + \\frac{r}{2}. \\end{align*}We want to solve for $d$. By the Pythagorean Theorem (twice):\\begin{align*} &\\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\\\ &\\implies \\left(d+r-\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_2)^2 = \\left(d-r+\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_1)^2 \\\\ &\\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\\\ &\\implies 4dr = 8rr_2-8rr_1 \\\\ &\\implies d=2r_2-2r_1 \\end{align*}Therefore, $d=2(r_2-r_1) = 2(961-625)=\\boxed{672}$. Solution 2 (Official MAA, Unedited) Denote by $O_1$$O_2$, and $O$ the centers of $\\omega_1$$\\omega_2$, and $\\omega$, respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\\omega_1$ and $\\omega_2$ respectively, $r$ be the radius of $\\omega$, and $\\ell$ the distance from $O$ to the line $AB$. We claim that$$\\dfrac{\\ell}{r} = \\dfrac{R_2-R_1}{d},$$where $d = O_1O_2$. This solves the problem, for then the $\\widehat{PQ} = 120^\\circ$ condition implies $\\tfrac{\\ell}r = \\cos 60^\\circ = \\tfrac{1}{2}$, and then", "regular hexagon is 120 degrees, this triangle is a 30-60-90 right triangle. Let $r$ be the radius of the circle. The length of the longer leg of the triangle is $r$, so the length of the shorter leg is $r/\\sqrt{3}$ and the length of the hypotenuse is $2r/\\sqrt{3}$. Since for the smaller hexagon the length of the segment connecting a vertex to the center is $r$, the dimensions of the larger hexagon are $2/\\sqrt{3}$ times larger than the dimensions of the smaller hexagon. Therefore, the area of the larger triangle is $(2/\\sqrt{3})^2=\\boxed{4/3}$ times greater than the area of the smaller triangle. [asy] size(5cm); defaultpen(linewidth(.7pt)+fontsize(8pt)); dotfactor=4; int i; draw(circle((0,0),1)); for(i=0;i<=5;++i) { draw(dir(60*i)--dir(60*(i+1))); draw(2/sqrt(3)*dir(60*i)--2/sqrt(3)*dir(60*(i+1))); } draw(2/sqrt(3)*dir(0)--(0,0)--dir(30)); draw(0.93*dir(30)--dir(30)+0.07*dir(-60)+0.07*dir(210)--dir(30)+0.07*dir(-60));[/asy] 1. 👍 2. 👎 6. 4/3 is the answer. 1. 👍 2. 👎 7. It is 4/3. 1. 👍 2. 👎 ## Similar Questions 1. ### Math please helppp 1 Find the missing angle measure in the polygon A ] 77 B ] 87 C ] 97 D ] 107 i think its b 2 Find the sum of the interior angles in 10 - sided polygon A ] 1,260 B ] 1,440 c ] 1,620 d ] 1,800 i think its c or b 3 Find the measure 2. ### geometry please please please help What is the value of x in", "1.5*dir(Y-P), linewidth(4)); dot(\"O\", O, 1.5*E, linewidth(4)); [/asy]$ Let $d$ be the diameter of $\\odot O.$ It follows that \\begin{alignat*}{8} 2[PAC] &= d\\cdot PX &&= 56, \\\\ 2[PBD] &= d\\cdot PY &&= 90. \\end{alignat*} Moreover, note that $OXPY$ is a rectangle. By the Pythagorean Theorem, we have $$PX^2+PY^2=PO^2.$$ We rewrite this equation in terms of $d:$ $$\\left(\\frac{56}{d}\\right)^2+\\left(\\frac{90}{d}\\right)^2=\\left(\\frac d2\\right)^2,$$ from which $d^2=212.$ Therefore, we get $$[ABCD] = \\frac{d^2}{2} = \\boxed{106}.$$ ~MRENTHUSIASM Solution 3 (Similar Triangles) $[asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O, P, X, Y; A = (-sqrt(106)/2,sqrt(106)/2); B = (-sqrt(106)/2,-sqrt(106)/2); C = (sqrt(106)/2,-sqrt(106)/2); D = (sqrt(106)/2,sqrt(106)/2); O = origin; path p; p = Circle(O,sqrt(212)/2); draw(p); P = intersectionpoints(Circle(A,4),p)[1]; X = foot(P,A,C); Y = foot(P,B,D); draw(A--B--C--D--cycle); draw(P--A--C--cycle,red); draw(P--B--D--cycle,blue); draw(P--X,red+dashed); draw(P--Y,blue+dashed); markscalefactor=0.075; draw(rightanglemark(A,P,C),red); draw(rightanglemark(P,X,C),red); draw(rightanglemark(B,P,D),blue); draw(rightanglemark(P,Y,D),blue); dot(\"A\", A, 1.5*NW, linewidth(4)); dot(\"B\", B, 1.5*SW, linewidth(4)); dot(\"C\", C, 1.5*SE, linewidth(4)); dot(\"D\", D, 1.5*NE, linewidth(4)); dot(\"P\", P, 1.5*dir(P), linewidth(4)); dot(\"X\", X, 1.5*dir(20), linewidth(4)); dot(\"Y\", Y, 1.5*dir(Y-P), linewidth(4)); dot(\"O\", O, 1.5*E, linewidth(4)); [/asy]$ Let the center of the circle be $O$, and the radius of the circle be $r$. Since $ABCD$ is a rhombus with diagonals $2r$ and $2r$, its area is $\\dfrac{1}{2}(2r)(2r) = 2r^2$. Since $AC$ and $BD$ are diameters of the circle, $\\triangle APC$ and $\\triangle BPD$ are right triangles. Let $X$ and $Y$ be the foot", "# Difference between revisions of \"2020 AMC 8 Problems/Problem 4\" Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon? $[asy] size(250); real side1 = 1.5; real side2 = 4.0; real side3 = 6.5; real pos = 2.5; pair s1 = (-10,-2.19); pair s2 = (15,2.19); pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); fill(circle(origin + s1, 1), grey1); for (int i = 0; i < 6; ++i) { draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25)); } fill(circle(origin, 1), grey1); for (int i = 0; i < 6; ++i) { fill(circle(pos*dir(60*i),1), grey2); draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25)); } fill(circle(origin+s2, 1), grey1); for (int i = 0; i < 6; ++i) { fill(circle(pos*dir(60*i)+s2,1), grey2); fill(circle(2*pos*dir(60*i)+s2,1), grey1); fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1); draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25)); } [/asy]$ Diagram by sircalcsalot $\\textbf{(A) }35 \\qquad \\textbf{(B) }37 \\qquad \\textbf{(C) }39 \\qquad \\textbf{(D) }43 \\qquad \\textbf{(E) }49$ ## Solution We find the pattern $1, 6, 12, 18, \\ldots$. The sum of the first four numbers in this sequence is $\\boxed{\\textbf{(B) }37}$. ## Solution 2 The first hexagon has $1$ dot. The second hexagon has $1+6$ dots. The third hexagon $1+6+12$ dots. Following this pattern, we predict that the fourth hexagon will have $1+6+12+18=37$ dots $\\implies\\boxed{\\textbf{(B) }37}$. ~junaidmansuri ## Solution 3 Each band" ]

[ "2\\qquad\\text{(E)}\\ 3$ ## Problem 19 The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon? $[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(240), D=2/sqrt(3)*dir(300), E=2/sqrt(3)*dir(0), F=2/sqrt(3)*dir(60); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, dir(origin--A)); label(\"B\", B, dir(origin--B)); label(\"C\", C, dir(origin--C)); label(\"D\", D, dir(origin--D)); label(\"E\", E, dir(origin--E)); label(\"F\", F, dir(origin--F)); [/asy]$ $\\text{(A)}\\ \\frac{1}{2}\\qquad\\text{(B)}\\ \\frac{\\sqrt 3}{3}\\qquad\\text{(C)}\\ \\frac{2}{3}\\qquad\\text{(D)}\\ \\frac{3}{4}\\qquad\\text{(E)}\\ \\frac{\\sqrt 3}{2}$ ## Problem 20 In the xy-plane, what is the length of the shortest path from $(0,0)$ to $(12,16)$ that does not go inside the circle $(x-6)^{2}+(y-8)^{2}= 25$? $\\text{(A)}\\ 10\\sqrt 3\\qquad\\text{(B)}\\ 10\\sqrt 5\\qquad\\text{(C)}\\ 10\\sqrt 3+\\frac{ 5\\pi}{3}\\qquad\\text{(D)}\\ 40\\frac{\\sqrt{3}}{3}\\qquad\\text{(E)}\\ 10+5\\pi$ ## Problem 21 $[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label(\"A\", A, dir(point--A)); label(\"B\", B, dir(point--B)); label(\"C\", C, dir(point--C)); label(\"D\", D, dir(point--D)); label(\"E\", E, dir(point--E)); [/asy]$ Solution ## Problem 28 $[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label(\"A\", A, NE); label(\"B\", B, NW); label(\"C\", C, S); label(\"D\", D, E); label(\"4\", (4,2,0), SW); label(\"4\", (2,4,0), SE); label(\"3\", (0, 4, 1.5), E); [/asy]$ Solution", "2\\qquad\\text{(E)}\\ 3$ ## Problem 19 The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon? $[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(240), D=2/sqrt(3)*dir(300), E=2/sqrt(3)*dir(0), F=2/sqrt(3)*dir(60); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, dir(origin--A)); label(\"B\", B, dir(origin--B)); label(\"C\", C, dir(origin--C)); label(\"D\", D, dir(origin--D)); label(\"E\", E, dir(origin--E)); label(\"F\", F, dir(origin--F)); [/asy]$ $\\text{(A)}\\ \\frac{1}{2}\\qquad\\text{(B)}\\ \\frac{\\sqrt 3}{3}\\qquad\\text{(C)}\\ \\frac{2}{3}\\qquad\\text{(D)}\\ \\frac{3}{4}\\qquad\\text{(E)}\\ \\frac{\\sqrt 3}{2}$ ## Problem 20 In the xy-plane, what is the length of the shortest path from $(0,0)$ to $(12,16)$ that does not go inside the circle $(x-6)^{2}+(y-8)^{2}= 25$? $\\text{(A)}\\ 10\\sqrt 3\\qquad\\text{(B)}\\ 10\\sqrt 5\\qquad\\text{(C)}\\ 10\\sqrt 3+\\frac{ 5\\pi}{3}\\qquad\\text{(D)}\\ 40\\frac{\\sqrt{3}}{3}\\qquad\\text{(E)}\\ 10+5\\pi$ ## Problem 21 Triangles $ABC$ and $ABD$ are isosceles with $AB=AC=BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC, then$ \\angle C+\\angle D \\$ is $[asy] size(120); pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); dot(A^^B^^C^^D^^E); draw(A--D--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); dot(A^^B^^C^^D^^E); pair point=midpoint(A--M); label(\"A\", A, dir(point--A)); label(\"B\", B, dir(point--B)); label(\"C\", C, dir(point--C)); label(\"D\", D, dir(point--D)); label(\"E\", E, dir(point--E)); [/asy]$ $\\text{(A)}\\ 115^\\circ\\qquad\\text{(B)}\\ 120^\\circ\\qquad\\text{(C)}\\ 130^\\circ\\qquad\\text{(D)}\\ 135^\\circ\\qquad\\text{(E)}\\ \\text{not uniquely determined}$ ## Problem 22 Four distinct points, $A$, $B$, $C$, and $D$, are to be selected from $1996$ points evenly spaced around a circle. All quadruples are equally likely to be chosen.", "# Dodecagon (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) A dodecagon is a 12-sided polygon. The sum of its internal angles is $1800^{\\circ}$. Each of its exterior angles has measure $30^{\\circ}$. A regular dodecagon can be seen below: $[asy] for(int i = 0; i <= 11; ++i) { draw(dir(360/12*i)--dir(360/12*(i + 1))); } pair A,B,C,D,E,F,G,H,I,J,K,L; A=dir(360/12*0); B=dir(360/12*1); C=dir(360/12*2); D=dir(360/12*3); E=dir(360/12*4); F=dir(360/12*5); G=dir(360/12*6); H=dir(360/12*7); I=dir(360/12*8); J=dir(360/12*9); K=dir(360/12*10); L=dir(360/12*11); label(\"A\",A,dir(0)); label(\"B\",B,dir(30)); label(\"C\",C,dir(60)); label(\"D\",D,dir(90)); label(\"E\",E,dir(120)); label(\"F\",F,dir(150)); label(\"G\",G,dir(180)); label(\"H\",H,dir(210)); label(\"I\",I,dir(240)); label(\"J\",J,dir(270)); label(\"K\",K,dir(300)); label(\"L\",L,dir(330)); draw(dir(360/12*0)--dir(360/12*6)); dot((dir(360/12*0)+dir(360/12*6))/2); pair O = (dir(360/12*0)+dir(360/12*6))/2; label(\"O\",O,S); draw(A--O); draw(Circle(O,1)); [/asy]$ The area of a regular dodecagon can be calculated by the formula $3R^2$, where $R$ is the circumradius of the dodecagon. In this case, $R$ would be $OA$. Also, each small triangle ($AOB$, $BOC$, etc.) is congruent, so $\\angle AOB=\\angle BOC=\\angle COD$ (etc) $=30^{\\circ}$.", "# Difference between revisions of \"1996 AHSME Problems/Problem 19\" ## Problem The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon? $[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(240), D=2/sqrt(3)*dir(300), E=2/sqrt(3)*dir(0), F=2/sqrt(3)*dir(60); dot(A^^B^^C^^D^^E^^F); label(\"A\", A, dir(origin--A)); label(\"B\", B, dir(origin--B)); label(\"C\", C, dir(origin--C)); label(\"D\", D, dir(origin--D)); label(\"E\", E, dir(origin--E)); label(\"F\", F, dir(origin--F)); [/asy]$ $\\text{(A)}\\ \\frac{1}{2}\\qquad\\text{(B)}\\ \\frac{\\sqrt 3}{3}\\qquad\\text{(C)}\\ \\frac{2}{3}\\qquad\\text{(D)}\\ \\frac{3}{4}\\qquad\\text{(E)}\\ \\frac{\\sqrt 3}{2}$ ## Solution $[asy] size(120); draw(rotate(30)*polygon(6)); draw(scale(2/sqrt(3))*polygon(6)); pair A=2/sqrt(3)*dir(120), B=2/sqrt(3)*dir(180), C=2/sqrt(3)*dir(240), D=2/sqrt(3)*dir(300), E=2/sqrt(3)*dir(0), F=2/sqrt(3)*dir(60); pair O=(0,0); dot(A^^B^^C^^D^^E^^F^^O); label(\"A\", A, dir(origin--A)); label(\"B\", B, dir(origin--B)); label(\"C\", C, dir(origin--C)); label(\"D\", D, dir(origin--D)); label(\"E\", E, dir(origin--E)); label(\"F\", F, dir(origin--F)); label(\"O\", O, N); draw(O--D--C--cycle); [/asy]$ $\\triangle OCD$ is copied six times to form the hexagon, so if we find the ratio of the area of the kite inside $\\triangle OCD$ to the the area of $\\triangle OCD$ itself, it will be the same ratio. Let $OC=CD=OD =2$ so that the area of the triangle is $\\frac{\\sqrt{3}s^2}{4} = \\sqrt{3}$. Notice that $\\triangle OCD$ is made up of a kite and two $30-60-90$ triangles. The two hypotenuses of these two triangles form $CD$, so the hypotenuse of each triangle must be $\\frac{2}{2} = 1$. Thus, the legs of each triangle are $\\frac{1}{2}$ and $\\frac{\\sqrt{3}}{2}}$ (Error compiling", "# Difference between revisions of \"Dodecagon\" A dodecagon is a 12-sided polygon. The sum of its internal angles is $1800^{\\circ}$. A regular dodecagon can be seen below: $[asy] for(int i = 0; i <= 11; ++i) { draw(dir(360/12*i)--dir(360/12*(i + 1))); } pair A,B,C,D,E,F,G,H,I,J,K,L; A=dir(360/12*0); B=dir(360/12*1); C=dir(360/12*2); D=dir(360/12*3); E=dir(360/12*4); F=dir(360/12*5); G=dir(360/12*6); H=dir(360/12*7); I=dir(360/12*8); J=dir(360/12*9); K=dir(360/12*10); L=dir(360/12*11); label(\"A\",A,dir(0)); label(\"B\",B,dir(30)); label(\"C\",C,dir(60)); label(\"D\",D,dir(90)); label(\"E\",E,dir(120)); label(\"F\",F,dir(150)); label(\"G\",G,dir(180)); label(\"H\",H,dir(210)); label(\"I\",I,dir(240)); label(\"J\",J,dir(270)); label(\"K\",K,dir(300)); label(\"L\",L,dir(330)); draw(dir(360/12*0)--dir(360/12*6)); dot((dir(360/12*0)+dir(360/12*6))/2); pair O = (dir(360/12*0)+dir(360/12*6))/2; label(\"O\",O,S); draw(A--G); draw(Circle(O,1)); [/asy]$ The area of a regular dodecagon can be calculated by the formula $3R^2$, where $R$ is the circumradius of the dodecagon. In this case, $R$ would be $OA$.", "and that the regular hexagon can be partitioned into six equilateral triangles. Like squares and , regular hexagons fit together without any gaps to tile the plane (three hexagons meeting at every vertex), and so are useful for constructing . The cells of a beehive are hexagonal for this reason and because the shape makes efficient use of space and building materials. The of a regular triangular lattice is the honeycomb tessellation of hexagons. It is not usually considered a triambus, although it is equilateral. Parameters The maximal diameter (which corresponds to the long of the hexagon), D, is twice the maximal radius or , R, which equals the side length, t. The minimal diameter or the diameter of the circle (separation of parallel sides, flat-to-flat distance, short diagonal or height when resting on a flat base), d, is twice the minimal radius or , r. The maxima and minima are related by the same factor: $\\frac\\left\\{1\\right\\}\\left\\{2\\right\\}d = r = \\cos\\left(30^\\circ\\right) R = \\frac\\left\\{\\sqrt\\left\\{3\\right\\}\\right\\}\\left\\{2\\right\\} R = \\frac\\left\\{\\sqrt\\left\\{3\\right\\}\\right\\}\\left\\{2\\right\\} t$ and, similarly, $d = \\frac\\left\\{\\sqrt\\left\\{3\\right\\}\\right\\}\\left\\{2\\right\\} D.$ The area of a regular hexagon \\begin\\left\\{align\\right\\} A &= \\frac{3\\sqrt{3}}{2}R^2 = 3Rr = 2\\sqrt{3} r^2 \\\\[3pt] &= \\frac{3\\sqrt{3}}{8}D^2 = \\frac{3}{4}Dd = \\frac{\\sqrt{3}}{2} d^2 \\\\[3pt] &\\approx 2.598 R^2 \\approx 3.464 r^2\\\\ &\\approx 0.6495 D^2 \\approx 0.866 d^2. \\end{align} For any regular , the area can", "1999 AHSME Problems/Problem 23 Problem The equiangular convex hexagon $ABCDEF$ has $AB = 1, BC = 4, CD = 2,$ and $DE = 4.$ The area of the hexagon is $\\mathrm{(A) \\ } \\frac {15}2\\sqrt{3} \\qquad \\mathrm{(B) \\ }9\\sqrt{3} \\qquad \\mathrm{(C) \\ }16 \\qquad \\mathrm{(D) \\ }\\frac{39}4\\sqrt{3} \\qquad \\mathrm{(E) \\ } \\frac{43}4\\sqrt{3}$ Solution Solution 1 Equiangularity means that each internal angle must be exactly $120^\\circ$. The information given by the problem statement looks as follows: $[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW)); dot(O); dot(O+E); dot(O+E+4*NE); dot(O+E+4*NE+2*NW); dot(O-3*E+4*NE+2*NW); label(\"A\",O,SW); label(\"B\",O+E,SE); label(\"C\",O+E+4*NE,E); label(\"D\",O+E+4*NE+2*NW,NE); label(\"E\",O-3*E+4*NE+2*NW,NW); [/asy]$ We can now place this incomplete polygon onto a triangular grid, finish it, compute its area in unit triangles, and multiply the result by the area of the unit triangle. $[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW) -- (O-3*E+3*NE+2*NW) -- cycle, 0.8red+4bp); for (int i=-5; i<=1; ++i) { draw( (O+i*E-1.5*NE)--(O+i*E+6.5*NE), dashed ); } for (int i=-2; i<=5; ++i) { draw( (O+i*E-1.5*NW)--(O+i*E+6.5*NW), dashed ); } for (int i=-1; i<=6; ++i) { draw( (O-2.5*E+i*NW)--(O+5.5*E+i*NW), dashed ); } dot(O); dot(O+E); dot(O+E+4*NE); dot(O+E+4*NE+2*NW); dot(O-3*E+4*NE+2*NW); label(\"A\",O,SW); label(\"B\",O+E,SE); label(\"C\",O+E+4*NE,E); label(\"D\",O+E+4*NE+2*NW,NE); label(\"E\",O-3*E+4*NE+2*NW,NW); label(\"F\",(O-3*E+3*NE+2*NW),W); [/asy]$ We see that the figure contains $43$ unit triangles, and therefore its area is $\\boxed{\\frac{43\\sqrt{3}}4}$. Solution 2 $[asy] unitsize(0.5cm); draw((0,0)--(7,12.124)--(14,0)--cycle);", "# Difference between revisions of \"1999 AHSME Problems/Problem 23\" ## Problem The equiangular convex hexagon $ABCDEF$ has $AB = 1, BC = 4, CD = 2,$ and $DE = 4.$ The area of the hexagon is $\\mathrm{(A) \\ } \\frac {15}2\\sqrt{3} \\qquad \\mathrm{(B) \\ }9\\sqrt{3} \\qquad \\mathrm{(C) \\ }16 \\qquad \\mathrm{(D) \\ }\\frac{39}4\\sqrt{3} \\qquad \\mathrm{(E) \\ } \\frac{43}4\\sqrt{3}$ ## Solution Equiangularity means that each internal angle must be exactly $120^\\circ$. The information given by the problem statement looks as follows: $[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW)); dot(O); dot(O+E); dot(O+E+4*NE); dot(O+E+4*NE+2*NW); dot(O-3*E+4*NE+2*NW); label(\"A\",O,SW); label(\"B\",O+E,SE); label(\"C\",O+E+4*NE,E); label(\"D\",O+E+4*NE+2*NW,NE); label(\"E\",O-3*E+4*NE+2*NW,NW); [/asy]$ We can now place this incomplete polygon onto a triangular grid, finish it, compute its area in unit triangles, and multiply the result by the area of the unit triangle. $[asy] unitsize(0.5cm); pair O=(0,0), E=dir(0), NE=dir(60), NW=dir(120); draw(O -- (O+E) -- (O+E+4*NE) -- (O+E+4*NE+2*NW) -- (O-3*E+4*NE+2*NW) -- (O-3*E+3*NE+2*NW) -- cycle, 0.8red+4bp); for (int i=-5; i<=1; ++i) { draw( (O+i*E-1.5*NE)--(O+i*E+6.5*NE), dashed ); } for (int i=-2; i<=5; ++i) { draw( (O+i*E-1.5*NW)--(O+i*E+6.5*NW), dashed ); } for (int i=-1; i<=6; ++i) { draw( (O-2.5*E+i*NW)--(O+5.5*E+i*NW), dashed ); } dot(O); dot(O+E); dot(O+E+4*NE); dot(O+E+4*NE+2*NW); dot(O-3*E+4*NE+2*NW); label(\"A\",O,SW); label(\"B\",O+E,SE); label(\"C\",O+E+4*NE,E); label(\"D\",O+E+4*NE+2*NW,NE); label(\"E\",O-3*E+4*NE+2*NW,NW); label(\"F\",(O-3*E+3*NE+2*NW),W); [/asy]$ We see that the figure contains $43$ unit triangles, and therefore its area is $\\boxed{\\frac{43\\sqrt{3}}4}$.", "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of", "# Difference between revisions of \"2006 AMC 12B Problems/Problem 16\" ## Problem Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(7,1)$, respectively. What is its area? $\\mathrm{(A)}\\ 20\\sqrt {3} \\qquad \\mathrm{(B)}\\ 22\\sqrt {3} \\qquad \\mathrm{(C)}\\ 25\\sqrt {3} \\qquad \\mathrm{(D)}\\ 27\\sqrt {3} \\qquad \\mathrm{(E)}\\ 50$ ## Solution To find the area of the regular hexagon, we only need to calculate the side length. a distance of $\\sqrt{7^2+1^2} = \\sqrt{50} = 5\\sqrt{2}$ apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is $\\frac{5\\sqrt{2}}{2}\\cdot\\frac{1}{\\sqrt{3}}\\cdot2 = \\frac{5\\sqrt{6}}{3}$. The apothem is thus $\\frac{1}{2}\\cdot\\frac{5\\sqrt{6}}{3}\\cdot\\sqrt{3} = \\frac{5\\sqrt{2}}{2}$, yielding an area of $\\frac{1}{2}\\cdot10\\sqrt{6}\\cdot\\frac{5\\sqrt{2}}{2}=25\\sqrt{3} \\implies \\mathrm{(C)}$. ## Solution 2 Solution 2 has the exact same solution as Solution 1 denoted, but instead, we do not need to know the value of the apothem. We could just apply s, which is the side length in this problem, $\\frac{5\\sqrt{6}}{3}$ into the hexagon area formula, $\\frac{3 (5 \\sqrt{2} )^2 \\sqrt{3} }{2}=25\\sqrt{3}$ ## See also 2006 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 •", "# Difference between revisions of \"2006 AMC 12B Problems/Problem 16\" ## Problem Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(7,1)$, respectively. What is its area? $\\mathrm{(A)}\\ 20\\sqrt {3} \\qquad \\mathrm{(B)}\\ 22\\sqrt {3} \\qquad \\mathrm{(C)}\\ 25\\sqrt {3} \\qquad \\mathrm{(D)}\\ 27\\sqrt {3} \\qquad \\mathrm{(E)}\\ 50$ ## Solution To find the area of the regular hexagon, we only need to calculate the side length. a distance of $\\sqrt{7^2+1^2} = \\sqrt{50} = 5\\sqrt{2}$ apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is $\\frac{5\\sqrt{2}}{2}\\cdot\\frac{1}{\\sqrt{3}}\\cdot2 = \\frac{5\\sqrt{6}}{3}$. The apothem is thus $\\frac{1}{2}\\cdot\\frac{5\\sqrt{6}}{3}\\cdot\\sqrt{3} = \\frac{5\\sqrt{2}}{2}$, yielding an area of $\\frac{1}{2}\\cdot10\\sqrt{6}\\cdot\\frac{5\\sqrt{2}}{2}=25\\sqrt{3} \\implies \\mathrm{(C)}$. ## Solution 2 Solution 2 has the exact same solution as Solution 1 denoted, but instead, we do not need to know the value of the apothem. We could just apply s, which is the side length in this problem, $\\frac{5\\sqrt{6}}{3}$ into the hexagon area formula, $\\frac{3 (5 \\sqrt{2} )^2 \\sqrt{3} }{2}=25\\sqrt{3}$", "want is $\\frac{7.5}{24}$ of the entire hexagon. The total area of the hexagon is $\\frac{3\\sqrt{3}}{2}$, so our answer is $\\frac{7.5}{24} \\cdot \\frac{3\\sqrt{3}}{2}$ = $\\boxed{(C) \\frac{15\\sqrt{3}}{32}}$ -AlexLikeMath ## Solution 7 If we try to coordinate bash this problem, it's going to look very ugly with a lot of radicals. However, we can alter and skew the diagram in such a way that all ratios of lengths and areas stay the same while making it a lot easier to work with. Then, we can find the ratio of the area of the wanted region to the area of $ABCDEF$ then apply it to the old diagram. $[asy] unitsize(1cm); draw((0,0)--(4,0)--(6,3.464)--(2,3.464)--(0,0)); draw((2,0)--(1,1.732)); draw((5,1.732)--(4,3.464)); draw((1.5, 0.866)--(3, 3.464)--(4.5, 0.866)--cycle); draw((2,0)--(2,3.464)--(5,1.732)--cycle); [/asy]$ $[asy] unitsize(1cm); draw((1,0)--(1,4),gray(.7)); draw((2,0)--(2,4),gray(.7)); draw((3,0)--(3,4),gray(.7)); draw((0,1)--(4,1),gray(.7)); draw((0,2)--(4,2),gray(.7)); draw((0,3)--(4,3),gray(.7)); draw((0,0)--(4,0)--(4,4)--(0,4)--(0,0)); draw((2,0)--(0,2)); draw((4,2)--(2,4)); draw((1,1)--(1,4)--(4,1)--cycle); draw((0,4)--(2,0)--(4,2)--cycle); fill((1,4)--(1,3.5)--(2,3)--cycle,red); fill((1,1)--(1.5,1)--(1,2)--cycle,red); fill((3,1)--(3.5,1.5)--(4,1)--cycle,red); [/asy]$ The isosceles right triangle with a leg length of $3$ in the new diagram is $\\triangle XYZ$ in the old diagram. We see that if we want to take the area of the new hexagon, we must subtract $\\frac{3}{4}$ from the area of $\\triangle XYZ$ (the red triangles), giving us $\\frac{15}{4}$. However, we need to take the ratio of this area to the area of $ABCDEF$, which is $\\frac{\\frac{15}{4}}{12}=\\frac{5}{12}$. Now we know that our answer is $\\frac{5}{12} \\cdot \\frac{3\\sqrt{3}}{2}=\\boxed{\\frac{15}{32}\\sqrt{3}}$.", "# Difference between revisions of \"2006 AMC 12B Problems/Problem 16\" ## Problem Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(7,1)$, respectively. What is its area? $\\mathrm{(A)}\\ 20\\sqrt {3} \\qquad \\mathrm{(B)}\\ 22\\sqrt {3} \\qquad \\mathrm{(C)}\\ 25\\sqrt {3} \\qquad \\mathrm{(D)}\\ 27\\sqrt {3} \\qquad \\mathrm{(E)}\\ 50$ ## Solution To find the area of the regular hexagon, we only need to calculate the side length. a distance of $\\sqrt{7^2+1^2} = \\sqrt{50} = 5\\sqrt{2}$ apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is $\\frac{5\\sqrt{2}}{2}\\cdot\\frac{1}{\\sqrt{3}}\\cdot2 = \\frac{5\\sqrt{6}}{3}$. The apothem is thus $\\frac{1}{2}\\cdot\\frac{5\\sqrt{6}}{3}\\cdot\\sqrt{3} = \\frac{5\\sqrt{2}}{2}$, yielding an area of $\\frac{1}{2}\\cdot10\\sqrt{6}\\cdot\\frac{5\\sqrt{2}}{2}=25\\sqrt{3}$.", "hexagon are visible from a randomly chosen point on the circle is $1/2$. What is $r$? $\\mathrm{(A) \\ } 2\\sqrt{2}+2\\sqrt{3}\\qquad \\mathrm{(B) \\ } 3\\sqrt{3}+\\sqrt{2}\\qquad \\mathrm{(C) \\ } 2\\sqrt{6}+\\sqrt{3}$$\\mathrm{(D) \\ } 3\\sqrt{2}+\\sqrt{6}\\qquad \\mathrm{(E) \\ } 6\\sqrt{2}-\\sqrt{3}$ ## Problem 23 Given a finite sequence $S=(a_1,a_2,\\ldots ,a_n)$ of $n$ real numbers, let $A(S)$ be the sequence $\\left(\\frac{a_1+a_2}{2},\\frac{a_2+a_3}{2},\\ldots ,\\frac{a_{n-1}+a_n}{2}\\right)$ of $n-1$ real numbers. Define $A^1(S)=A(S)$ and, for each integer $m$, $2\\le m\\le n-1$, define $A^m(S)=A(A^{m-1}(S))$. Suppose $x>0$, and let $S=(1,x,x^2,\\ldots ,x^{100})$. If $A^{100}(S)=(1/2^{50})$, then what is $x$? $\\mathrm{(A) \\ } 1-\\frac{\\sqrt{2}}{2}\\qquad \\mathrm{(B) \\ } \\sqrt{2}-1\\qquad \\mathrm{(C) \\ } \\frac{1}{2}\\qquad \\mathrm{(D) \\ } 2-\\sqrt{2}\\qquad \\mathrm{(E) \\ } \\frac{\\sqrt{2}}{2}$ ## Problem 24 The expression $(x+y+z)^{2006}+(x-y-z)^{2006}$ is simplified by expanding it and combining like terms. How many terms are in the simplified expression? $\\mathrm{(A) \\ } 6018\\qquad \\mathrm{(B) \\ } 671,676\\qquad \\mathrm{(C) \\ } 1,007,514\\qquad \\mathrm{(D) \\ } 1,008,016\\qquad \\mathrm{(E) \\ } 2,015,028$ ## Problem 25 How many non-empty subsets $S$ of $\\{1,2,3,\\ldots ,15\\}$ have the following two properties? $(1)$ No two consecutive integers belong to $S$. $(2)$ If $S$ contains $k$ elements, then $S$ contains no number less than $k$. $\\mathrm{(A) \\ } 277\\qquad \\mathrm{(B) \\ } 311\\qquad \\mathrm{(C) \\ } 376\\qquad \\mathrm{(D) \\ } 377\\qquad \\mathrm{(E) \\ } 405$", "the area of the small region is $\\frac{\\pi}{4}-\\frac{1}{2}$ The requested area is area of circle $C$ minus 4 of this area: $\\pi 1^2 - 4(\\frac{\\pi}{4}-\\frac{1}{2}) = \\pi - \\pi + 2 = 2$ $\\boxed{\\textbf{C}}$. ## Solution 2 $[asy] unitsize(1.1cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(2,0), C=(1,1); pair D=(2,1); pair E=(0,1); pair F = (1, 2); pair M = (1, 0); dot (A); dot (B); dot (C); dot (D); dot (E); dot (F); dot (M); draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); draw (D--F--E--M--D); label(\"A\",A,W); label(\"B\",B,E); label(\"C\",C,W); label(\"M\",M,NE); label(\"D\",D,E); label(\"E\",E,W); label(\"F\",F,N); [/asy]$ We can move the area above the part of the circle above the segment $EF$ down, and similarly for the other side. Then, we have a square, whose diagonal is $2$, so the area is then just $\\left(\\frac{2}{\\sqrt{2}}\\right)^2 = 2$.", "get: \\begin{eqnarray} L_{hex-unit}=\\sqrt[]{\\frac{2}{\\sqrt[]{3}}}\\approx 1.075 \\end{eqnarray} where $$S_{hex-unit}$$ denotes length of the hexagon and $$L_{hex-unit}$$ denotes a distance between centers of the hexagons. Notice that unit surface in 2D is simply a length of the hexagon side and surface area of the hexagon with side a is: \\begin{eqnarray} S = 6\\frac{{\\sqrt[]{3}}}{4}a^2 \\end{eqnarray} In 3D we can derive the corresponding unit quantities starting with the formulae for volume and surface of rhombic dodecahedron (12 hedra) \\begin{align*} &V = \\frac{16}{9}{\\sqrt[]{3}}a^3 \\\\ &S = 8{\\sqrt[]{2}}a^2 \\end{align*} where a denotes length of dodecahedron edge. Constraining the volume to be 1 we get: \\begin{eqnarray} a = \\sqrt[3]{\\frac{9V}{16\\sqrt[]{3}}} \\end{eqnarray} and thus unit surface is given by: \\begin{eqnarray} S_{unit-hex} = \\frac{S}{12} = \\frac{8\\sqrt[]{2}}{12}\\sqrt[3]{\\frac{9V}{16\\sqrt[]{3}}}\\approx 0.445 \\end{eqnarray} and unit length by: \\begin{eqnarray} L_{unit-hex} = 2\\frac{\\sqrt[]{2}}{\\sqrt[]{3}}a = 2\\frac{\\sqrt[]{2}}{\\sqrt[]{3}} \\sqrt[3]{\\frac{9V}{16\\sqrt[]{3}}}\\approx 1.122 \\end{eqnarray}", "+0 # Trig help pls 0 230 2 A regular hexagon has vertices at $1-i, 0, a, b, c, d$ clockwise in that order, as in the picture below: Then what is $$\\operatorname{Im}(a)$$? Jan 16, 2019 #1 +533 0 if you tell me what Im(a) means(does it mean imaginary part?), i can solve it, but i can tell you the coordinates. convert to the real plane, so distance from a to origin is sqrt2/2. now since the angle made by a, origin, and y axis is 15 degrees, you use the fact that cos 15 = (sqrt6+sqrt2)/4, and the fact that sin 15 = (sqrt6-sqrt2)/4, so then the coordinates of a would be ((sqrt6-sqrt2)/8, (sqrt6+sqrt2)/8). assuming it means imaginary part, answer is (sqrt6+sqrt2)/8. Jan 16, 2019 #2 +22881 +9 A regular hexagon has vertices at $1-i, 0, a, b, c, d$ clockwise in that order, as in the picture below: Then what is $$\\operatorname{Im}(a)$$? $$\\begin{array}{|rcll|} \\hline \\sin(75^{\\circ}) &=& \\dfrac{\\operatorname{Im}(a)} {\\sqrt{2}} \\\\ \\operatorname{Im}(a) &=& \\sqrt{2}\\cdot \\sin(75^{\\circ}) \\\\ \\operatorname{Im}(a) &=& 1.41421356237\\cdot 0.96592582629 \\\\ \\mathbf{\\operatorname{Im}(a)} & \\mathbf{=} & \\mathbf{1.36602540378} \\\\ \\hline \\end{array}$$ Jan 16, 2019", "\\right )}{h}$$, $$tan30 = \\frac{\\left ( a/2 \\right )}{h}$$, $$\\frac{\\sqrt{3}}{3}= \\frac{\\left ( a/2 \\right )}{h}$$, $$h= \\frac{a}{2}\\times \\frac{3}{\\sqrt{3}}$$, The area of a triangle = $$\\frac{1}{2}bh$$, The area of a triangle=$$\\frac{1}{2}\\times a\\times \\frac{a}{2}\\times \\frac{3}{\\sqrt{3}}$$, Area of the hexagon = 6 x Area of Triangle, Area of the hexagon = $$6\\times \\frac{3}{\\sqrt{3}} \\times \\frac{a^{2}}{4}$$, Area of an Hexagon = $$\\frac{3\\sqrt{3}}{ 2} \\times a^{2}$$. Let the length of this line be. If the lengths of all the sides and the measurement of all the angles are equal, such hexagon is called a regular hexagon. Area of Regular Octagon = $$\\large 2(1+ \\sqrt{2})a^{2}$$. Happily, there is a formula for the area of any simple polygon that only requires knowledge of the coordinates of each vertex. Hexa is a Greek word whose meaning is six. Special right triangles review. The Perimeter of Hexagon Formula Hexagon is the polygon that has six equal sides and the six edges. … Honeycomb, quartz crystal, bolt head, Lug/wheel nut, Allen wrench, floor tiles etc are few things which you would find a hexagon. Doing so we get: So this is going to be equal to 6 times 3 square roots of 3, which is 18 square roots of 3. Whereas in the case of the irregular hexagon, neither the sides are equal, nor the angles are the", "that the hexagon has perimeter $$6$$. Now, we want to find the perimeter of larger regular polygons. The method used by Archimedes and Zu Chongzi, the two people who used this method to get good approximations very early on. The two mathematicians did their work completely separately; neither knew that the other existed, but they still got to the same information. They found a formula to find the perimeter of a regular polygon, directly from the perimeter of a polygon with half the number of sides. We shall consider a polygon with $$n$$ sides, and a perimeter of $$2nd$$, where each side of the polygon has length $$2d$$. In our diagram, $$BC$$ has length $$d$$. By the Pythagorean Theorem, $$AB=\\sqrt{1-d^2}$$. Thus, $$BD=1-\\sqrt{1-d^2}$$. Then, $CD=\\sqrt{2-2\\sqrt{1-d^2}}.$ Using many laborious calculations, Zu Chongzi was able to calculate the perimeter of a $$24,576$$ sided polygon. Not surprisingly, $$24,576=6\\cdot2^{12}$$. Using these calculations, Chongzi was able to calculate $$\\pi$$ to $$7$$ decimal places. More recently, Isaac Newton used infinite summations and coordinate geometry to generate an approximation of $$\\pi$$. He took the equation for a circle with center at $$(\\frac{1}{2},0)$$ and radius $$\\frac{1}{2}$$. This semicircle has the equation $y=\\sqrt{\\frac{1}{4}-(x-\\frac{1}{2})^2}.$ This can be rewritten as $y=\\sqrt{x}\\cdot\\sqrt{1-x}.$ We can expand $$\\sqrt{1-x}$$. $$\\sqrt{1-x}$$ is equal to $$(1-x)^{\\frac{1}{2}}$$, so it can be expanded using the binomial theorem. A", "a regular hexagon to a corner is the same as the side length, we also know that $d+\\frac{x}{2}=1\\Rightarrow d=1-\\frac{x}{2}$. Substitute this value for $d$ into the first equation and rearrange: \\begin{align} \\frac{1-\\frac{x}{2}}{x/2}&=\\frac{1}{\\sqrt{3}}\\\\\\\\ \\frac{2-x}{x} &=\\frac{1}{\\sqrt{3}}\\\\\\\\ 2\\sqrt{3}-\\sqrt{3}x&=x\\\\\\\\ 2\\sqrt{3}&=x\\left(1+\\sqrt{3}\\right)\\\\\\\\ x&=\\frac{2\\sqrt{3}}{ 1+\\sqrt{3}}\\approx 1.267\\dots \\end{align} So for a hexagon of side length $h$, the largest square that will fit inside has side length $\\frac{2\\sqrt{3}}{ 1+\\sqrt{3}}h$. The side length of the shadow hexagon is $\\sqrt{\\frac{2}{3}}$, so the maximum square that will fit inside it has side length \\begin{align} \\left(\\frac{2\\sqrt{3}}{ 1+\\sqrt{3}}\\right)\\left(\\sqrt{\\frac{2}{3}}\\right)&=\\frac{2\\sqrt{2}}{1+\\sqrt{3}}\\\\\\\\ &\\approx 1.035… \\end{align} So we can fit a square of side length $1$ inside the shadow and still have a little space left over. Now imagine extruding that square upwards to make a tunnel through the rotated cube. Because the square fits fully inside the shadow, the tunnel will fit fully inside the cube, producing a cube with the desired size of hole. In fact, a slightly different rotation will make a shadow than can contain an even larger square. This was discovered by Pieter Nieuwland, who calculated that for a cube of edge length $1$, the maximum size cube that can fit through a hole has edge length $\\frac{3}{4}\\sqrt{2}=1.0606\\dots$. As I mentioned when talking about the tetrahedron, this number is called the Nieuwland constant – the maximum ratio by which a polyhedron can" ]

[ "1. finding minimal point I'm having trouble with this question... To find the minimal point, is there an easier way other than plugging in all possible values? Compute the polar coordinates and the xy-coordinates for the point where y is maximal on the graph of the polar coordinate equation $r = 1 + \\cos(\\theta)$. 2. Originally Posted by centenial I'm having trouble with this question... To find the minimal point, is there an easier way other than plugging in all possible values? Compute the polar coordinates and the xy-coordinates for the point where y is maximal on the graph of the polar coordinate equation $r = 1 + \\cos(\\theta)$. maximum y or minimum? if maximum ... $r = 1 + \\cos{t}$ $r\\sin{t} = \\sin{t}(1 + \\cos{t})$ $y = \\sin{t}(1 + \\cos{t})$ $\\frac{dy}{dt} = \\sin{t}(-\\sin{t}) + (1 + \\cos{t})\\cos{t}$ simplified ... $\\frac{dy}{dt} = 2\\cos^2{t} + \\cos{t} - 1 $ set $\\frac{dy}{dt} = 0$ and determine the angle $t$ that maximizes $y$ ... I get $t = \\frac{\\pi}{3}$ 3. Sorry, yes, I meant maximum. Could you explain why you set dy/dt = 0.. is that because when dy/dt = 0, then the slope is horizontal, and we must be at a maximum point? I just want to make sure I understand. If I were asked to find the minimum, would", "0$, it is a maximum To find the coordinate, sub $x = 0$ back into $f \\left(x\\right)$ and you will get $y = 3$ --> $\\left(0 , 3\\right)$ is a maximum", "coordinate graph of $r = 3\\cos\\theta$. Do $\\theta$ from $-\\pi/2$ to $\\pi/2$ and see what you think. 3. Mar 31, 2014", "the center, so its x-coordinate will be the same: $2$. Its y-coordinate will be the y-coordinate of the center plus one radius: $6 + \\sqrt{13} = 9.6$. That means the coordinates of the maximum point are$\\left(2 , 6 + \\sqrt{13}\\right)$ (or (2, 9.6)).", "# Question #c5e92 From the graph y = sin(x) you can see that the maximum $y$-value = $1$ and the minimum $y$-value = $- 1$:", "# Math Help - Find maximum point of curve graph 1. ## Find maximum point of curve graph Hi, when finding maximum point of graph, where do I look at? The y axis or the x axis? Also, the answer should be \"x=7\" (for example), am I right? Thank you 2. ## Re: Find maximum point of curve graph A point has always 2 coordinates if you're working in a plane, so a maximum $M$ could be for example $M(7,4)$. 3. ## Re: Find maximum point of curve graph Originally Posted by Siron A point has always 2 coordinates if you're working in a plane, so a maximum $M$ could be for example $M(7,4)$. Hmm, but the question didn't state for the coordinates of maximum point of the curve, so I thought the answer would be different. Am I wrong? 4. ## Re: Find maximum point of curve graph But for example $x=7$ can't represent a point, it represents a line (parallel to the y-axis). 5. ## Re: Find maximum point of curve graph Originally Posted by Siron But for example $x=7$ can't represent a point, it represents a line (parallel to the y-axis). Yea, I will write the answer in coordinates then Thanks a lot Btw, how do I find the equation of line of symmetry if", "# How do you find the maximum of the graph y=3tan(x+pi/2)+2? I don't think $\\tan \\left(x\\right)$ has a maximum...", "then to a minimum of $$-1$$, then back to 0, and then it obviously repeats itself. So the graph of $$y=\\sin x$$ must look something like this: Similarly, as angle $$x$$ increases from 0 in the unit circle diagram, the first coordinate of the point $$A$$ goes from 1 to 0 then to $$-1$$, back to 0 and back to 1, so the graph of $$y=\\cos x$$ must look something like this:", "The graph of $\\displaystyle r\\left( \\theta \\right)=\\sin \\left( {2\\theta } \\right)$ for $0\\le \\theta \\le 2\\pi$, shown below, appears to show 4 maximums. However if we trace the graph, we find that these points are (1, π/4), (–1, 3π/4), (1, 5π/4) and (–1, 7π/4). Two of the values are maximums where r(θ) = 1 and two are minimums where r(θ) = –1. The graph of $r\\left( \\theta \\right)=\\cos \\left( {3\\theta } \\right)$ )\\$ for $0\\le \\theta \\le 2\\pi$ is even more fun: the graph starts at an endpoint maximum at (1,0), next goes to a minimum at (–1, π/3), then a maximum at (1, 2π/3), and a minimum at (–1, π). But this is the starting point that was a maximum. Continuing around the graph, (1, 4π/3) a maximum at the same point as the previous minimum (–1, π/3), then a minimum at (–1, 5π/3), previously a minimum, then returning to (1, 2π) = (–1, π) = (1, 0) which was previously both a maximum and a minimum, and is now a maximum again! Thus points may be both maximums and minimums. ### On a tangent… Polar Curves can be really fun. While working on the idea above, I explored some other curves. Try some yourself using Desmos, GeoGabra, or another graphing app with sliders. Shown below are", "# Math Help - Maximum Y-Coordinate Values Of A Cardioid 1. ## Maximum Y-Coordinate Values Of A Cardioid Algebraically determine the maximum value of the y-coordinate of points on the cardioid $r = 1 + cos \\theta$ for 0 θ π. I used the formula: $\\frac{dy}{dx}=\\frac{\\frac{dr}{d \\theta} \\sin \\theta +r \\cos \\theta}{\\frac{dr}{d \\theta} \\cos \\theta-r \\sin \\theta}$, which yielded $\\frac{-sin\\theta(sin\\theta+cos\\theta+cos^{2}\\theta)}{-sin\\theta(cos\\theta-sin\\theta-sin\\theta cos\\theta)}$. If I solve $-sin\\theta(sin\\theta+cos\\theta+cos^{2}\\theta)$, setting it to zero, it will give the value of theta at which horizontal tangents exist. The problem is that I don't know how to solve it. After graphing it, I know that the max value occurs when $\\theta = \\frac{\\pi}{3}$ and $r=1.5$, but I'm not sure how to get to those values. 2. ## Re: Maximum Y-Coordinate Values Of A Cardioid All you need is that $y=r\\cos(\\theta)=(1+\\cos(\\theta))\\sin(\\theta)=\\sin( \\theta)+\\sin(\\theta)\\cos(\\theta)$ Now we have $\\frac{dy}{d\\theta}=\\cos(\\theta)+\\cos^2(\\theta)-\\sin^2(\\theta)=2\\cos^2(\\theta)+\\cos(\\theta)-1$ Setting equal to zero and factoring gives $(2\\cos(\\theta)-1)(\\cos(\\theta)+1)=0$ This will give you both the max and min values. 3. ## Re: Maximum Y-Coordinate Values Of A Cardioid Wow, that's so much better than my inchoate solution. Thanks a lot! 4. ## Re: Maximum Y-Coordinate Values Of A Cardioid Hello, Algebrah! Determine the maximum value of the y-coordinate of points on the cardioid: $r \\:=\\: 1 + \\cos \\theta\\;\\text{ for }0\\,\\le\\,\\theta\\,\\le\\,\\pi$ Code: | | * * * |* * *| *", "Math Help - Minimum Point 1. Minimum Point A problem I need help with. y = 2x^2 + 480/x I need to find the minimum point on that graph. Help would be greatly appreciated. 2. Hi Originally Posted by Tesphen A problem I need help with. y = 2x^2 + 480/x I need to find the minimum point on that graph. Help would be greatly appreciated. First step: You have to find the derivative $y' = 2*2x + (-1)480/(x^2)$ second step: Solve y' = 0 4x - 480/x² = 0 4x^3 - 480 = 0 4x^3 = 480 x^3 = 120 third step find y'' and use x = 120^(1/3) to show y''(120^(1/3)) > 0 => minimum hint: y'' = 960/x^3 + 4 last step calculate y(120^(1/3)) = ... then you know the x-coordinate and y-coordinate of your minimum I get $y(120^{1/3}) \\approx 146$", "initial point which we choose as the maximum $x$ coordinate - maxXt. Then we need to mirror two points and draw lines between. r[t_] := 2 - 2 Sin[t] + Sin[t] Sqrt[Abs[Cos[t]]]/(Sin[t]... Top 50 recent answers are included", "Find maximum point of curve graph • September 29th 2011, 09:02 AM FailInMaths Find maximum point of curve graph Hi, when finding maximum point of graph, where do I look at? The y axis or the x axis? Also, the answer should be \"x=7\" (for example), am I right? Thank you • September 29th 2011, 09:08 AM Siron Re: Find maximum point of curve graph A point has always 2 coordinates if you're working in a plane, so a maximum $M$ could be for example $M(7,4)$. • September 29th 2011, 09:11 AM FailInMaths Re: Find maximum point of curve graph Quote: Originally Posted by Siron A point has always 2 coordinates if you're working in a plane, so a maximum $M$ could be for example $M(7,4)$. Hmm, but the question didn't state for the coordinates of maximum point of the curve, so I thought the answer would be different. Am I wrong? • September 29th 2011, 09:24 AM Siron Re: Find maximum point of curve graph But for example $x=7$ can't represent a point, it represents a line (parallel to the y-axis). • September 29th 2011, 09:27 AM FailInMaths Re: Find maximum point of curve graph Quote: Originally Posted by Siron But for example $x=7$ can't represent a point, it represents a line (parallel to the y-axis). Yea,", "Algebra 2 (1st Edition) We first find the x-coordinate of the maximum: $$=\\frac{36-18}{2}=9$$ Thus, the maximum is: $$= -(9-36)(9+18) = 729$$", "in vertex form $y=a(x-h)^2+k$ and state the coordinates of the vertex and the maximum or minimum value. $$y=-0.3x^2+3.6x+5$$ $$e=mc^2$$", "## Trigonometric function graphs Graph the following, put maximum, minimums and intercepts on the xand y axis a) y(t) = 5 sin b) y(t) = 5 cos (t - π)", "# Math Help - Maximum & Minimum Problem 1. ## Maximum & Minimum Problem What is the maximum possible area of a rectangle with base that lies on the x-axis and with two upper vertices that lie on the graph of the equation $y$ $=$ $4$ $-$ $x^2$? I have no idea how to do this, can someone please show the full solutions to help me through the process? Thank you!!! 2. Originally Posted by bhuang What is the maximum possible area of a rectangle with base that lies on the x-axis and with two upper vertices that lie on the graph of the equation $y$ $=$ $4$ $-$ $x^2$? I have no idea how to do this, can someone please show the full solutions to help me through the process? Thank you!!! Start by drawing a picture (did you do that?). Let the coordinates of the vertices have x-coordinates at x = a and x = -a. Get an expression for the area in terms of a. Then use calculus in the usual way to maximise this area.", "How do you find the maximum value of Y= -2(x+5)²-8? Oct 17, 2016 The maximum value of the function is $- 8$. $y = a {\\left(x - h\\right)}^{2} + k$, where $\\left(h , k\\right)$ is the vertex of the graph of the function. Also, if $a > 0$, the parabola will open upward and the $y$-coordinate of the vertex will be the minimum value of the curve. If $a < 0$, the parabola will open downward and the $y$-coordinate of the vertex will be the maximum value of the curve. In $y = - 2 {\\left(x + 5\\right)}^{2} - 8$, $a = - 2$ and $- 2 < 0$, so the parabola will open downward and the maximum value of the curve will be the $y$-coordinate of the vertex, or $k$. In this equation, $k = - 8$, so the maximum valu of the function is $- 8$.", "a).i).Find the coordinates of the stationary points on the curve y =x^3 – 3x +2. ii). For each stationary point determine whether it is minimum or maximum. b).Sketch the graph of the function y = x^3 -3x +2.", "Find the minimum and maximum values of $y = \\sqrt{8} \\theta - \\sqrt{4} \\sec \\theta$ on the interval $[ 0, \\frac {\\pi}{3}]$ $f_{\\text{min}}=$ $f_{\\text{max}}=$" ]

[ "= linspace(0,1,30); theta = linspace(0,2*pi,100); [r theta] = meshgrid(r,theta); z = 2*r.*sin(theta).*(1-r.^2)./(1+r.^2-2*r.*cos(theta)).^2; % x = r.*cos(theta); % y = r.*sin(theta); % only for viewing positive/negative in the plot for ii = 1:100; for jj = 1:30; if z(ii,jj)>0; c(ii,jj)=1; else if z(ii,jj)<0 c(ii,jj)=-1; else c(ii,jj)=0; end end end end clear ii jj % what the solution looks like: red=positive, blue=negative, green=zero figure; mesh(r.*cos(theta),r.*sin(theta),z,c) % scaled for clarity figure; mesh(r.*cos(theta),r.*sin(theta),sign(z).*log(log(abs(z)+1)+1)+1) % the function is unbounded: compute the limit as (r,theta) spirals to (1,0) syms u R T T_R u = 2*R*sin(T)*(1-R^2)/(1+R^2-2*R*cos(T))^2; % to find this value, use solve(diff(U,T),T) and use the third value T_R = pi - acos(1/4/R*(R^2 - (R^4 + 34*R^2 + 1)^(1/2) + 1)); u = subs(u,T,T_R); figure; ezplot(u); limit(u,R,1,'left') % which approaches +inf Posted in Uncategorized", "3*x*y**2 r = np.linspace(0, 0.001, 101) theta = np.linspace(0, 2*np.pi, 360) rr, tt = np.meshgrid(r, theta) u = f_polar(rr, tt) fig = plt.figure() ax.pcolormesh(tt, rr, u, cmap='inferno') ax.set_aspect('equal') ax.set_rgrids([]) plt.show() print('mean(u) - u(0,0) = ', np.mean(u) - f_polar(0, 0)) mean(u) - u(0,0) = 7.013888888888634e-13 We can easily check this for our example $$u = x^3 - 3xy^2$$, where the mean value in a circle of radius $$R$$ centered at $$(0, 0)$$ is given by \\begin{split} \\begin{aligned} \\overline{u} &= \\frac{1}{2R \\ 2 \\pi} \\int_{-R}^R \\int_0^{2 \\pi} u \\ d \\theta d r \\\\ &= \\frac{1}{4R \\pi} \\int_{-R}^R \\int_0^{2 \\pi} r^3 (\\cos^3 \\theta - 3 \\cos \\theta \\sin^2 \\theta) \\ d \\theta dr \\\\ &= \\frac{1}{4R \\pi} \\int_{-R}^R r^3 \\underbrace{\\int_0^{2 \\pi} (\\cos^3 \\theta - 3 \\cos \\theta \\sin^2 \\theta) \\ d \\theta}_{= 0} dr = 0 \\end{aligned} \\end{split} So the mean value in the circle is zero and $$u(0, 0)$$ is zero as well, as we expected. Let us think again about the surface given by a set of points $$(x, y, u)$$ but this time consider a function $$u$$ such that $$\\nabla^2 u \\neq 0$$. For example: $u(r) = \\frac{\\sin r}{r}, \\qquad r = \\sqrt{x^2 + y^2}$ for which $\\nabla^2 u = \\frac{(1 - r^2) \\sin r}{r^3} - \\frac{\\cos r}{r^2}$ x = np.linspace(-7.5, 7.5, 300) X, Y = np.meshgrid(x,", "# How do you convert x^2+(y-4)^2=16 to polar form? May 3, 2016 $r = 8 \\sin \\theta$ #### Explanation: If $\\left(r , \\theta\\right)$ is in polar form and $\\left(x , y\\right)$ in Cartesian form the relation between them is as follows: $x = r \\cos \\theta$, $y = r \\sin \\theta$, ${r}^{2} = {x}^{2} + {y}^{2}$ and $\\tan \\theta = \\frac{y}{x}$ Hence, ${x}^{2} + {\\left(y - 4\\right)}^{2} = 16$ can be written as ${x}^{2} + {y}^{2} - 8 y + 16 = 16$ or ${x}^{2} + {y}^{2} - 8 y = 0$ or ${r}^{2} - 8 r \\sin \\theta = 0$ or $r \\left(r - 8 \\sin \\theta\\right) = 0$ dividing by $r$ $r - 8 \\sin \\theta = 0$ or $r = 8 \\sin \\theta$ May 3, 2016 $r = 8 \\sin \\theta$ #### Explanation: To convert from Cartesian to Polar coordinates use the following formulae that link them. • x=rcostheta\" and \" y=rsintheta ${x}^{2} + {y}^{2} - 8 y + 16 = 16 \\text{ (expanding bracket) }$ $\\Rightarrow {r}^{2} {\\cos}^{2} \\theta + {r}^{2} {\\sin}^{2} \\theta - 8 r \\sin \\theta + 16 - 16 = 0$ then ${r}^{2} \\left({\\cos}^{2} \\theta + {\\sin}^{2} \\theta\\right) - 8 r \\sin \\theta = 0$ using the identity: ${\\cos}^{2} \\theta + {\\sin}^{2} \\theta = 1$ $\\Rightarrow {r}^{2} = 8 r \\sin", "_{1}+l_{2}\\cos(\\theta _{1}+\\theta _{2})+l_{3}\\cos(\\theta _{1}+\\theta _{2}+\\theta _{3})\\\\y=l_{1}\\sin \\theta _{1}+l_{2}\\sin(\\theta _{1}+\\theta _{2})+l_{3}\\sin(\\theta _{1}+\\theta _{2}+\\theta _{3})\\end{array}}}$ Graphical Example of Planar, Three-link Manipulator using Compound Transformations Planar, three-link manipulator with all three angles set to ${\\displaystyle 30^{o}}$. Code %reset -f import numpy as np import sympy as sp import matplotlib.pyplot as plt theta1, theta2, theta3=sp.symbols('theta1 theta2 theta3') l1, l2, l3 = sp.symbols('l1 l2 l3') f1_f0_T=np.array([[sp.cos(theta1), -sp.sin(theta1), 0], [sp.sin(theta1), sp.cos(theta1), 0], [0, 0, 1]]) f2_f1_T=np.array([[sp.cos(theta2), -sp.sin(theta2), l1], [sp.sin(theta2), sp.cos(theta2),0], [0, 0, 1]]) f3_f2_T=np.array([[sp.cos(theta3), -sp.sin(theta3), l2], [sp.sin(theta3), sp.cos(theta3),0], [0, 0, 1]]) f4_f3_T=np.array([[sp.cos(0), -sp.sin(0), l3], [sp.sin(0), sp.cos(0),0], [0, 0, 1]]) ee_bs_T=np.dot(f1_f0_T,np.dot(f2_f1_T,np.dot(f3_f2_T,f4_f3_T))) ee_bs_T=sp.simplify(ee_bs_T) # Unfortunately, we do not get the simple # expression on the wiki. #print(\"ee_bs_T\") #display(ee_bs_T) #For substituting in the following: 30 degree increments #print(\"ee_bs_T after substitutions\") replacements=[(theta1,30*np.pi/180),(theta2,30*np.pi/180), (theta3,30*np.pi/180), (l1,1), (l2,1), (l3,1)] #display(ee_bs_T.subs(replacements)) End_Coord=np.array([[0,0,1]]) End_Coord=End_Coord.T #print(\"f1_f0_T\") f1_f0_T=sp.simplify(f1_f0_T) #display(f1_f0_T.subs(replacements)) f1_f0_T=np.array(f1_f0_T.subs(replacements)).astype(np.float64) f1_f0_T=np.reshape(f1_f0_T,(3,3)) print(\"f1_f0_T dot End_Coord\") display(np.dot(f1_f0_T,End_Coord)) Joint0=np.dot(f1_f0_T,End_Coord) #print(\"\\nf2_f1_T\") f2_f1_T=sp.simplify(f2_f1_T) #display(f2_f1_T.subs(replacements)) f2_f1_T=np.array(f2_f1_T.subs(replacements)).astype(np.float64) f2_f1_T=np.reshape(f2_f1_T,(3,3)) print(\"f2_f0_T dot End_Coord\") display(np.dot(f1_f0_T,np.dot(f2_f1_T,End_Coord))) Joint1=np.dot(f1_f0_T,np.dot(f2_f1_T,End_Coord)) #print(\"\\nf3_f2_T\") f3_f2_T=sp.simplify(f3_f2_T) #display(f3_f2_T.subs(replacements)) f3_f2_T=np.array(f3_f2_T.subs(replacements)).astype(np.float64) f3_f2_T=np.reshape(f3_f2_T,(3,3)) print(\"f3_f0_T dot End_Coord\") display(np.dot(f1_f0_T,np.dot(f2_f1_T,np.dot(f3_f2_T,End_Coord)))) Joint2=np.dot(f1_f0_T,np.dot(f2_f1_T,np.dot(f3_f2_T,End_Coord))) #print(\"\\nf4_f3_T\") f4_f3_T=sp.simplify(f4_f3_T) #display(f4_f3_T.subs(replacements)) f4_f3_T=np.array(f4_f3_T.subs(replacements)).astype(np.float64) f4_f3_T=np.reshape(f4_f3_T,(3,3)) print(\"f4_f0_T dot End_Coord\") display(np.dot(f1_f0_T,np.dot(f2_f1_T,np.dot(f3_f2_T,np.dot(f4_f3_T,End_Coord))))) EndEffector=np.dot(f1_f0_T,np.dot(f2_f1_T,np.dot(f3_f2_T,np.dot(f4_f3_T,End_Coord)))) fig=plt.figure() plt.plot([Joint0[0][0], Joint1[0][0]],[Joint0[1][0],Joint1[1][0]],'bo-') plt.plot([Joint1[0][0], Joint2[0][0]],[Joint1[1][0],Joint2[1][0]],'ro-') plt.plot([Joint2[0][0], EndEffector[0][0]], [Joint2[1][0],EndEffector[1][0]],'go-') ax.set_xlim([0, 1]) plt.axis('equal') ax.set_xlabel('X',size=20) ax.set_ylabel('Y',size=20) plt.savefig('2D_30_30_30_rotations_200212.png', dpi=300, bbox_inches='tight') Two Configurations for the Same Resulting Location of the End Effector An example of two different solutions for the inverse kinematics problem leading to the same end-effector position and orientation. We can demonstrate", "C); draw(A--H); pair X = 0.3*B + 0.7*C; pair Y = A+X-H; draw(X--1.3*Y-0.3*X); draw(A--Y, dotted); pair R1 = 1.3*X-0.3*Y; pair R2 = 0.7*X+0.3*Y; draw(R1--X); D(\"O\",A,dir(A)); D(\"O_1\",B,dir(B)); D(\"O_2\",C,dir(C)); D(\"H\",H,dir(270)); D(\"X\",X,dir(225)); D(\"A\",R1,dir(180)); D(\"B\",R2,dir(180)); draw(rightanglemark(Y,X,C,3)); [/asy]$Since $X$ is on the radical axis of $\\omega_1$ and $\\omega_2$, it has equal power with respect to both circles, so$$O_1X^2 - r_1^2 = O_2X^2-r_2^2 \\implies O_1X-O_2X = \\frac{r_1^2-r_2^2}{d}$$since $O_1X+O_2X=d$. Now we can solve for $O_1X$ and $O_2X$, and in particular,\\begin{align*} O_1H &= O_1X - HX = \\frac{d+\\frac{r_1^2-r_2^2}{d}}{2} - \\frac{r}{2} \\\\ O_2H &= O_2X + HX = \\frac{d-\\frac{r_1^2-r_2^2}{d}}{2} + \\frac{r}{2}. \\end{align*}We want to solve for $d$. By the Pythagorean Theorem (twice):\\begin{align*} &\\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\\\ &\\implies \\left(d+r-\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_2)^2 = \\left(d-r+\\tfrac{r_1^2-r_2^2}{d}\\right)^2 - 4(r+r_1)^2 \\\\ &\\implies 2dr - 2(r_1^2-r_2)^2-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\\\ &\\implies 4dr = 8rr_2-8rr_1 \\\\ &\\implies d=2r_2-2r_1 \\end{align*}Therefore, $d=2(r_2-r_1) = 2(961-625)=\\boxed{672}$. Solution 2 (Official MAA, Unedited) Denote by $O_1$$O_2$, and $O$ the centers of $\\omega_1$$\\omega_2$, and $\\omega$, respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\\omega_1$ and $\\omega_2$ respectively, $r$ be the radius of $\\omega$, and $\\ell$ the distance from $O$ to the line $AB$. We claim that$$\\dfrac{\\ell}{r} = \\dfrac{R_2-R_1}{d},$$where $d = O_1O_2$. This solves the problem, for then the $\\widehat{PQ} = 120^\\circ$ condition implies $\\tfrac{\\ell}r = \\cos 60^\\circ = \\tfrac{1}{2}$, and then", "&= \\cos\\f \\,\\ve_x + \\sin\\f \\,\\ve_y \\\\ \\ve_\\f &= -\\sin\\f \\,\\ve_x + \\cos\\f \\,\\ve_y. \\end{align*} (This basis is orthonormal by inspection.) This implies, for example, that $$M_{\\r x} = \\ve_\\r^T \\ve_x = \\cvc{\\cos\\f}{\\sin\\f} \\vc{1}{0} = \\cos\\f.$$ Calculating the other components, one finds $$\\MM = \\mt{\\cos\\f}{\\sin\\f}{-\\sin\\f}{\\cos\\f}$$ or $$\\MM_{(\\r,\\f)\\ra(x,y)} = \\frac{1}{\\sqrt{x^2+y^2}} \\mt{x}{y}{-y}{x}.$$ Note also that $$[\\ms'(\\r,\\f)]_{(\\r,\\f)\\ra(x,y)} = \\ms'(\\sqrt{x^2+y^2},\\arctan y/x).$$ It is now a straightforward, if tedious, task to work out the correct form for $$[\\dv\\ms]'$$. (* mma code to check (3) *) fs[foo_] := FullSimplify[foo, {r > 0, -Pi < f < Pi}]; rpc = {r -> Sqrt[x^2 + y^2], f -> ArcTan[x, y]}; rcp = {x -> r Cos[f], y -> r Sin[f]}; Apolar = {{srr[r, f], srf[r, f]}, {sfr[r, f], sff[r, f]}}; M = {{Cos[f], Sin[f]}, {-Sin[f], Cos[f]}}; Mt = Transpose[M]; M //. rpc // fs Apolar[[1]][[1]] //. rpc (* out: [a check on some relations above] *) ansmma = Div[Apolar, {r, f}, \"Polar\"] // fs (* out: [divergence according to mma] *) ans = M.Div[Mt.Apolar.M //. rpc, {x, y}, \"Cartesian\"] //. rcp // fs (* out: [divergence using (3)] *) ans == ansmma // fs (* out: True *) • Thank you very much. Can you continue to write down the whole process of getting the divergence of $\\left( \\begin{array}{cc} \\sigma \\rho (r,\\varphi ) & \\tau (r,\\varphi", "Jey May 16 '13 at 5:54 If you change to polar coordinates, you get $$\\lim_{r \\rightarrow 0} \\sin(\\theta) \\cos(\\theta) \\sin(r^2 \\sin(\\theta) \\cos(\\theta)).$$ It might look a bit confusing, but the essential thing is that $\\sin(x)$ goes to zero as $x$ goes to zero, and that $\\sin(\\theta)$ and $\\cos(\\theta)$ are at most 1. • I understand how you changed the expression using the substitutions $x=r\\cos{\\theta}$ and $y=r\\sin{\\theta}$. But how do change $(x,y) \\rightarrow (0,0)$ to $r \\rightarrow 0$? Thanks. – Jey May 16 '13 at 5:49 • That $(x,y) \\rightarrow 0$ means that the distance from the point $(x,y)$ to $(0,0)$ should go to zero, which means that $\\sqrt{x^2 + y^2} = r$ should go to zero. The $\\theta$ varies freely, but this is ok for us since the expression goes to zero. Note that for instance $\\lim_{r \\rightarrow 0} \\sin(\\theta)$ does not exist. – N.U. May 16 '13 at 6:03 Hint: $\\left|\\dfrac{xy}{x^2+y^2}\\right| \\le \\dfrac12$. • How do you get 1/2? – Jey May 16 '13 at 5:47 • @Jey: $x^2+y^2-2xy \\ge 0 \\implies x^2+y^2 \\ge 2xy \\implies \\dfrac{1}{x^2+y^2} \\le \\dfrac{1}{2xy}$ – Inceptio May 16 '13 at 5:49 • AM/GM inequality (or $\\sin \\theta \\cos \\theta = \\frac 12 \\sin 2\\theta$) – mrf May 16 '13 at 5:49 • Nice. I didn't notice that. Thanks – Jey May 16", "draw((0,0)--(30,0)); draw((0,0)--O[1]--(O[1].x,0)); draw(O[1]--(O[1] + reflect((0,0),(10,12*sqrt(221)/49*10))*(O[1]))/2); label(\"$y = mx$\", (8,12*sqrt(221)/49*8), N); dot(\"$(9,6)$\", (9,6), NE); dot(\"$(kr,r)$\", O[1], N); dot(P,red); [/asy] Since the circle is tangent to the line $y = mx,$ the distance from the center $(kr,r)$ to the line is $r.$ We can write $y = mx$ as $y - mx = 0,$ so from the distance formula, $$\\frac{|r - krm|}{\\sqrt{1 + m^2}} = r.$$Squaring both sides, we get $$\\frac{(r - krm)^2}{1 + m^2} = r^2,$$so $(r - krm)^2 = r^2 (1 + m^2).$ Since $r \\neq 0,$ we can divide both sides by 0, to get $$(1 - km)^2 = 1 + m^2.$$Then $1 - 2km + k^2 m^2 = 1 + m^2,$ so $m^2 (1 - k^2) + 2km = 0.$ Since $m \\neq 0,$ $$m(1 - k^2) + 2k = 0.$$Hence, $$m = \\frac{2k}{k^2 - 1} = \\frac{2 \\sqrt{\\frac{117}{68}}}{\\frac{117}{68} - 1} = \\boxed{\\frac{12 \\sqrt{221}}{49}}.$$", "# Plot in Physical Space? [closed] (* Q7 - a *) Block[ {Theta=Pi/6,r=6}, Graphics[ { Thick, Black, Line[{{-2,-(12-r)},{-2,0},{2,0},{2+r*Sin[Theta],-r*Cos[Theta]}}], Red, AbsolutePointSize[20],Point[{-2,-(12-r)}], Blue,AbsolutePointSize[10],Point[{2+r*Sin[Theta],-r*Cos[Theta]}], Black,AbsolutePointSize[8],Point[{{-2,0},{2,0}}], {Dashed,Line[{{2,0},{2,-7}}]}, Circle[{2,0},1,{-Pi/2,-Pi/2+Theta}], Text[Style[\"Theta\",FontSize -> 15],{2+1.5Sin[Theta/2],-1.5Cos[Theta/2]}], Text[Style[\" r \",FontSize -> 15],{2.5+r/2Sin[Theta],-r/2Cos[Theta]}] } ] ] ansQ7 = With[{mu = 4.5, g = 9.8}, NDSolve[ { (mu + 1)*r''[t] - r[t]*Theta'[t]^2 + g*(mu - Cos[Theta[t]]) == 0, r[t]*Theta''[t] + 2 r'[t]*Theta'[t] + g*Sin[Theta[t]] == 0, r[0] == 1, Theta[0] == Pi/2, r'[0] == 0, Theta'[0] == 0 }, {r, Theta}, {t, 0, 100} ] ] (* Q7 - b *) Plot[Evaluate[r[t] /. ansQ7], {t, 0, 20}, PlotRange -> All] Plot[Evaluate[Theta[t] /. ansQ7], {t, 0, 20}, PlotRange -> All] I am not sure if I understand (c) correctly. Is it asking to plot the exact location of the m? I am thininking that I need to use something like CoordinateTransform, but I am not sure how to use it. Also in this case, the angle Theta is measured postively from downwards?? Thanks. I would interpret c) as asking for a plot of $$\\{r\\,\\cos\\,\\theta, r\\,\\sin\\,\\theta\\}$$. This can be done as follows: With[{mu = 4.5, g = 9.8}, {rF, θF} = NDSolveValue[ {(mu + 1)*r''[t] - r[t]*Theta'[t]^2 + g*(mu - Cos[Theta[t]]) == 0, r[t]*Theta''[t] + 2 r'[t]*Theta'[t] + g*Sin[Theta[t]] == 0, r[0] == 1, Theta[0] == Pi/2, r'[0] == 0, Theta'[0] ==", "by setting, y1 = r * cos(θ),y2 = r * sin(θ) $\\theta = x_2, r = (-1 + R) / (2*x_3), R = \\sqrt(1+4x_1*x_2)$ which may be differentiated to give the required sensitivities. x=x0; theta=x(2); R=sqrt(1+4*x(3)*x(1)); r=(-1+R)/(2*x(3)); y_exact=[r*cos(theta); r*sin(theta)] dRdx=[2*x(3)/R, 0, 2*x(1)/R]; drdx=dRdx/(2*x(3))-(-1+R)/(2*x(3)^2)*[0,0,1]; r*cos(theta)*dthetadx+sin(theta)*drdx] > y_exact = 0.0128 0.0460 Dy_exact = 0.2442 -0.0460 -0.0006 0.8796 0.0128 -0.0020 ##### Undifferentiated iteration For the nonlinear, undifferentiated iteration we have: y=y0; x=x0; u=[x(1)*cos(x(2));x(1)*sin(x(2));x(3)]; disp('Itns. |g(y)|') for i=1:itmax g=BBEx3p1_gfunc(y,u); y=y-g; disp([num2str(i,'% 4.0f'),' ',num2str(norm(g),'%10.4e')]) if norm(g)<=tolg break end end y_undif=y Error_y_undif=norm(y_undif-y_exact,inf) Itns. |g(y)| 1 5.0000e-002 2 2.5000e-003 3 2.4375e-004 4 2.3216e-005 5 2.2163e-006 6 2.1153e-007 7 2.0189e-008 8 1.9270e-009 y_undif = 0.0128 0.0460 Error_y_undif = 1.6178e-010 ##### Naive differentiation Simply initialising x with the fmad class gives a naive iteration for which the lack of convergence checks on the derivatives here results in only slight inaccuracy. y=y0; u=[x(1)*cos(x(2));x(1)*sin(x(2));x(3)]; disp('Itns. |g(y)|') for i=1:itmax g=BBEx3p1_gfunc(y,u); y=y-g; disp([num2str(i,'% 4.0f'),' ',num2str(norm(g),'%10.4e'),' ',... num2str(norm(getinternalderivs(g)),'%10.4e')]) if norm(g)<=tolg break end end y_naive = getvalue(y) Dy_naive = getinternalderivs(y) Error_y_naive = norm(y_naive-y_exact,inf) Error_Dy_naive = norm(Dy_naive-Dy_exact,inf) Itns. |g(y)| 1 5.0000e-002 1.0000e+000 2 2.5000e-003 1.0003e-001 3 2.4375e-004 1.4508e-002 4 2.3216e-005 1.8246e-003 5 2.2163e-006 2.1665e-004 6 2.1153e-007 2.4729e-005 7 2.0189e-008 2.7469e-006 8 1.9270e-009 2.9909e-007 y_naive = 0.0128 0.0460 Dy_naive = 0.2442 -0.0460 -0.0006 0.8796 0.0128 -0.0020 Error_y_naive = 1.6178e-010 Error_Dy_naive = 2.9189e-008 ##### Problems with", "1\\\\ z= 1, x^2 + y^2 = 1$ Substitute our new coordinate system. $\\rho^2 \\sin^2\\phi\\cos^2\\theta + \\rho^2 \\sin^2\\phi\\sin^2\\theta + \\rho^2 \\cos^2\\phi = 2\\\\ \\rho^2 \\sin^2\\phi\\cos^2\\theta + \\rho^2 \\sin^2\\phi\\sin^2\\theta + \\rho^2 \\cos^2\\phi - 2\\rho\\cos\\phi + 1 = 1\\\\ \\rho\\cos\\phi= 1, \\rho^2 \\sin^2\\phi\\cos^2\\theta + \\rho^2 \\sin^2\\phi\\sin^2\\theta = 1$ Simplify: $\\rho^2 = 2\\\\ \\rho^2 - 2\\rho\\cos\\phi = 0\\\\ \\rho\\cos\\phi= 1, \\rho^2 \\sin^2\\phi = 1$ $\\rho = \\sqrt 2\\\\ \\rho = 2\\cos\\phi\\\\ \\frac {\\rho\\sin \\phi}{\\rho\\cos \\phi} = \\tan \\phi= 1\\\\ \\phi = \\frac {\\pi}{4}$ Now, it may help to sketch these curves. Do it in 2D. Just look at the picture in the xz plane, and that should give you some idea of what is going on for $\\rho$ and $\\phi.$ \\update______ $\\int_0^{2\\pi}\\int_0^{\\frac\\pi4}\\int_{\\sqrt 2}^{2\\cos\\phi} \\rho^2 \\sin\\phi \\;d\\rho\\;d\\phi\\;d\\theta$ • The cylindrical looks fine, but I'm still unsure how you get your bounds for the spherical. – Lax Oct 12 '16 at 1:05", "\\theta)(\\sin \\theta) \\\\ &= 2r\\cos^2 \\theta + 2r \\sin^2 \\theta \\\\ &= 2r \\end{align} And also: (4) \\begin{align} \\quad \\frac{\\partial f}{\\partial \\theta} &= \\frac{\\partial f}{\\partial x} \\frac{\\partial x}{\\partial \\theta} + \\frac{\\partial f}{\\partial y} \\frac{\\partial y}{\\partial \\theta} \\\\ &= (2x) (-r \\sin \\theta) + (2y)(r \\cos \\theta) \\\\ &= (2r \\cos \\theta)(-r \\sin \\theta) + (2r \\sin \\theta)(r \\cos \\theta) \\\\ &= -2r^2 \\cos \\theta \\sin \\theta + 2r^2 \\sin \\theta \\cos \\theta \\\\ &= 0 \\end{align} By direct substitution we have that: (5) \\begin{align} \\quad f(x, y) = x^2 + y^2 = (r \\cos \\theta)^2 + (r \\sin \\theta)^2 = r^2 \\cos^2 \\theta + r^2 \\sin^2 \\theta = r^2(\\cos^2 \\theta + \\sin^2 \\theta) = r^2 = f(r, \\theta) \\end{align} Therefore $\\displaystyle{\\frac{\\partial f}{\\partial r} = 2r}$ and $\\displaystyle{\\frac{\\partial f}{\\partial \\theta} = 0}$ as above. ## Example 2 Let $f(x, y, z) = x^2ye^z$, $x(s, t) = s^2t$, $y(s, t) = st^2$, and $z(s, t) = \\sin (st)$. Find $\\displaystyle{\\frac{\\partial f}{\\partial s}}$ by first using the chain rule and then by direct substitution. We have that: (6) \\begin{align} \\quad \\frac{\\partial f}{\\partial s} &= \\frac{\\partial f}{\\partial x} \\frac{\\partial x}{\\partial s} + \\frac{\\partial f}{\\partial y} \\frac{\\partial y}{\\partial s} + \\frac{\\partial f}{\\partial z} \\frac{\\partial z}{\\partial s} \\\\ &= (2xye^z)(2st) + (x^2e^z)(t^2) + (x^2ye^z)(t \\cos (st)) \\\\ &= 2(s^2t)(st^2)e^{\\sin (st)}(2st) + (s^2t)^2e^{\\sin (st)} (t^2)", "$x = \\tan(t)$. We then get that \\begin{align} I(n) & = \\int_0^{\\infty} \\left(\\dfrac{2x}{1+x^2} \\right)^n dx = \\int_0^{\\pi/2} \\sin^n(2t) \\sec^2(t) dt = 2^n \\int_0^{\\pi/2} \\sin^n(t) \\cos^{n-2}(t) dt\\\\ & = 4\\int_0^{\\pi/2}\\sin^2(t) \\sin^{n-2}(2t)dt \\tag{$\\star$} \\end{align} Replacing $t$ by $\\pi/2-t$, we get $$I(n) = 4 \\int_0^{\\pi/2} \\cos^2(t) \\sin^{n-2}(2t) dt \\tag{\\perp}$$ Adding $(\\star)$ and $(\\perp)$, we get that $$2I(n) = 4 \\int_0^{\\pi/2} \\sin^{n-2}(2t)dt \\implies I(n) = 2 \\int_0^{\\pi/2} \\sin^{n-2}(2t)dt$$ I trust you can take it from here, using this post, which evaluates $\\displaystyle \\int_0^{\\pi} \\sin^{k}(t) dt$ and using Stirling (or) Wallis formula. - How do you go from $cos^2(t)\\sin^{n-2}(2t)$ to $\\sin^{n-2}(2t)$? – TCL May 7 '13 at 1:17 @TCL Updated the answer. Add $(\\star)$ and $(\\perp)$ to get it. – user17762 May 7 '13 at 1:19 Probably not how you want this done, but a useful and elegant method nonetheless: Note that $$f(x)=\\frac{2x}{1+x^2}$$ has the property that $f(x)<1$ except at $f(1)=1$. Since exponentiation to the power of $n\\to\\infty$ kills off anything less than $1$, we expect that the dominant contribution is near $1$. $f^n \\equiv \\exp(n \\log f)$ and $n\\log(f(1+u))=-n\\frac{1}{2}u^2+n\\mathcal O(u^3)$. Then letting $v=\\sqrt{n} u$, $$\\int_0^\\infty n^{1/2}\\mathrm d x \\exp(-\\frac{1}{2}v^2+\\mathcal O(n^{-1/2})) \\approx \\int_{-\\infty}^\\infty \\mathrm d v\\, e^{-v^2/2} = \\sqrt{2\\pi}$$ Verification: WolframAlpha. Edit: This is called Laplace's method amongst other things, and is a core part of the general theory of asymptotics of integrals.", "inverse y = K tan^-1(A(x+B)) y/K = tan^-1(A(x+B)) tan(y/K) = A(x+B) y' sec^2(y/K)/K = A y' sec^2(y/K) = KA y' (tan^2(y/K)+1) = KA y' ((A(x+B))^2+1) = KA y' = KA ------------ (A(x+B))^2+1 ------------------- A = 16/9, B=1/2 16K/9 = 4/3 16K = 12 K = 3/4 31. anonymous Omg, $\\tan^2\\theta+1=\\sec^2\\theta, so \\it simplifies \\to \\int\\limits_{}^{}\\frac{ 2 }{ \\sqrt3 }d \\theta$ 32. amistre64 you are on the right track yes http://www.wolframalpha.com/input/?i=integrate+1%2F%28x^2%2Bx%2B1%29+dx 33. amistre64 mine went someplace off the wall lol 34. amistre64 4/3 sqrts to get inside the ^2 ... 35. amistre64 2/sqrt(3) not 16/9 36. amistre64 A = 2/sqrt(3) 2K/sqrt(3) = 4/3 2sqrt(3) K = 4 K = 2/sqrt(3) thats better on my end 37. anonymous $= \\frac{ 2 }{ \\sqrt3 } \\theta+C, \\theta=\\arctan(\\frac{ 2 }{ \\sqrt3 }(x+.5)) \\implies \\frac{ 2 }{ \\sqrt3 }\\arctan(\\frac{ 2 }{ \\sqrt3}(x+.5))+C=\\int\\limits_{}^{}\\frac{ 1 }{ x^2+x+1 }dx$ 38. amistre64 now me and the wolf agree: y = K tan^-1 (A(x+B)) $y=\\frac{2}{\\sqrt3}~\\tan^{-1}[\\frac{2}{\\sqrt3}(x+\\frac12)]$ 39. anonymous This answer is going to be really ugly in the end 40. anonymous Yeah, I agree, then we just put it all together and its $\\log|x-1|-0.5\\log|x^2+x+1|-\\sqrt3\\arctan(\\frac{ 2 }{ \\sqrt3 }(x+.5)+C$ I dont think that simplifies any further 41. amistre64 the uglier the better :) 42. amistre64 you got it 43. anonymous Jeez, that was tough. 44. amistre64 nah .. it was", "then the second quadrant petal, then the first quadrant petal, and finally the fourth quadrant petal. edit Here's what I mean: Since the sets of points in the plane described by $r=\\sin2\\theta$ and $r=-\\sin2\\theta$ are identical, giving only one of them is sufficient to describe the same set of points as $(x^2+y^2)^3=4x^2y^2$. That's what lets us go from $r^2=\\sin^22\\theta$ and its equivalent of $r=\\pm\\sin2\\theta$ to only $r=\\sin2\\theta$. - If $x=r\\cos(\\theta)$ and $y=r\\sin(\\theta)$, then $(x^2+y^2)^3=4x^2y^2$ becomes $$r^2=\\sin^2(2\\theta)$$ as you say. However, note that the curve $r=\\sin(2\\theta)$ is symmetric through the origin since $\\sin(2(\\theta+\\pi))=\\sin(2\\theta)$, so it is the same as $r=-\\sin(2\\theta)$. In other words, it doesn't matter whether you take $+$ or $-$. $\\hskip{4cm}$ - Nice graphic. May I ask how you generated it? – bgins Feb 7 '12 at 21:10 @bgins: It was generated entirely using Mathematica 8. – robjohn Feb 7 '12 at 21:47 On the variety $V\\left((x^2+y^2)^3-(2xy)^2\\right)$, in polar coordinates we have $$r^6=(x^2+y^2)^3=(2xy)^2=r^4\\sin^2{2\\theta}$$ or $$r^2=\\sin^2{2\\theta} \\quad\\implies\\quad r=\\pm\\sin2\\theta$$ (we can ignore the \"trivial\" algebraic solution $r=0$ for all $\\theta$ since it is geometrically recovered for $\\theta\\in\\pi\\mathbb{Z}$). Now $\\sin{2\\theta}$ has period $\\pi$ (furnishing the same magnitude for $r$ at antipodal points), the curve has point symmetry through the origin, allowing us to discard the sign and consider only $r\\in[0,1]$. But $$\\sin^2{2\\theta}=\\frac{1-\\cos{4\\theta}}{2}$$ has period $\\frac{\\pi}{2}$, vanishes at the axes or", "# How do you convert r = 8 sin theta into cartesian form? Dec 29, 2016 The answer is ${x}^{2} + {\\left(y - 4\\right)}^{2} = {4}^{2}$ #### Explanation: To convert from polar coordinates $\\left(r , \\theta\\right)$ to cartesian coordinates $\\left(x , y\\right)$, we use the following equations $x = r \\cos \\theta$ $y = r \\sin \\theta$ ${x}^{2} + {y}^{2} = {r}^{2}$ Here, we have $r = 8 \\sin \\theta$ $r = 8 \\cdot \\frac{y}{r}$ ${r}^{2} = 8 y$ ${x}^{2} + {y}^{2} = 8 y$ ${x}^{2} + {y}^{2} - 8 y = 0$ Completing the squares ${x}^{2} + {y}^{2} - 8 y + 16 = 16$ ${x}^{2} + {\\left(y - 4\\right)}^{2} = {4}^{2}$ This is the equation of a circle, center $\\left(0 , 4\\right)$ and radius $= 4$ graph{x^2+(y-4)^2=16 [-13.88, 14.6, -2.85, 11.39]} Dec 29, 2016 ${x}^{2} + {y}^{2} = 8 y$ #### Explanation: we understand that; $x = r \\cos \\theta$, $y = r \\sin \\theta$ and ${x}^{2} + {y}^{2} = {r}^{2}$ --@ from the question, $r = 8 \\sin \\theta$ we can multiply with r, then $r \\cdot r = r \\cdot 8 \\sin \\theta$ ${r}^{2} = 8 \\left(r \\sin \\theta\\right) w h e r e , y = r \\sin \\theta$ ${r}^{2} = 8 y$ therefore from @ the answer is ${x}^{2} + {y}^{2} =", "# Converting Polar to Rectangular 1. Sep 23, 2005 Hello If you have $r = 2\\sin(2\\theta)$, how would you convert it to rectangular form? I tried doing this: $\\sin(2\\theta) = 2\\sin\\theta\\cos\\theta$ which means $r = 4\\sin\\theta\\cos\\theta$. Then I know that $r^{2} = x^{2} + y^{2}$. We know that $x = r\\cos\\theta, y = r\\sin\\theta$. But then I have a circular argument, where I end up with $y^{2} = y^{2}$. This means that I have to use another conversion factor. Should I use $\\sin\\theta = \\frac{y}{r}$? Thanks 2. Sep 23, 2005 ### shaner-baner hi, you're trying to write $r = 4\\sin\\theta\\cos\\theta$ in terms of x and y now that you have $\\sin\\theta = \\frac{y}{r}$ and $\\cos\\theta = \\frac{x}{r}$ just make the substitutions and you should have no problem getting the answer 3. Aug 31, 2011 ### bcddd214 Thank you kindly for your response. I 'think' I get the mechanics of it but me and the old voyage 200 are at differences about the answer when doing sample problems. Example: let's uses (-2,8),(3,5) as an example. This what I get for AxB when done manually. (3,5),(-2,8) 3^2+5^2=r^2 9+25=r^2 √34=r 5.8301=r then: toa: tan^(-1)⁡(3/5)=30.964^o do the same to coordinate #2 -2^2+8^2=r^2 4+16=r^2 √20=r 4.47214=r then: toa: tan^(-1)⁡((-2)/8)=-14.036^o we have: (5.8301,30.964^o ),(4.47214,-14.036^o ) note: To cross product we need the sin of", "x) and $$y_2=e^{rx}\\sin(\\theta x)$$ are fundamental solutions of (35) and the general solution of (35) is $y=c_1e^{rx}\\cos(\\theta x)+c_2e^{rx}\\sin(\\theta x).$ Example. Find the general solution of the following ODE. $\\frac{d^2y}{dx^2}-6\\frac{dy}{dx}+13y=0$ Solution. The characteristic equation is \\begin{align*} r^2-6r+13&=0\\\\ r&=\\frac{6\\pm \\sqrt{(-6)^2-4\\cdot1\\cdot 13}}{2\\cdot 1}\\\\ r&=3+2i,3-2i \\end{align*} So the general solution is $y=c_1e^{3x}\\cos(2x)+c_2e^{3x}\\sin(2x).$ Example. Solve the following IVP. $y''+2y'+10y=0,\\;\\; y(0)=4,\\, y'(0)=-7$ Solution. The characteristic equation is \\begin{align*} r^2+2r+10&=0\\\\ (r+1)^2&=-9\\\\ r&=-1+3i,-1-3i. \\end{align*} So the general solution is $y=c_1e^{-x}\\cos(3x)+c_2e^{-x}\\sin(3x).$ Using the initial condition $$y(0)=4$$, we get $$$c_1=4. \\;\\;\\;\\;(36)$$$ Note that $$y'=c_1(-\\cos(3x)-3\\sin(3x))e^{-x}+c_2(-\\sin(3x)+3\\cos(3x))e^{-x}.$$ Using the initial condition $$y'(0)=-7$$, we get $$$-c_1+3c_2=-7. \\;\\;\\;\\;(37)$$$ Solving (36) and (37), we get $$c_1=4$$ and $$c_2=-1$$. Thus the solution of the IVP is $y=4e^{-x}\\cos(3x)-e^{-x}\\sin(3x).$ Last edited", "\\theta_1 \\given \\theta_2, y & \\sim \\rr{N} (y_1 + \\rho (\\theta_2 - y_2), 1 - \\rho^2) \\\\ \\theta_2 \\given \\theta_1, y & \\sim \\rr{N} (y_2 + \\rho (\\theta_1 - y_1), 1 - \\rho^2)\\end{align} ## The AP model \\begin{align} Y_i \\given Y_{r(i)} & \\sim N \\left(\\rho Y_{r(i)} + (1 - \\rho) \\mu, (1 - \\rho^2) \\sigma^2 \\right), \\\\Y_i & \\sim N\\left( \\mu, \\sigma^2 \\right) \\rr{ if } r(i)\\rr{ doesn't exist}\\end{align} \\begin{align}L(&\\{ \\mu, \\rho, \\sigma^2\\} \\given Y) \\propto \\prod_{\\{\\rr{seeds } j\\}} \\frac{1}{\\sqrt{\\sigma^2}}e^{\\frac{(y_j - \\mu)^2}{2\\sigma^2}} \\cdot \\\\ & \\prod_{\\{\\rr{referred }i\\}} \\frac{1}{\\sqrt{(1-\\rho^2)\\sigma^2}}e^{\\frac{(y_i - (\\rho y_{r(i)} + (1 - \\rho) \\mu))^2}{2\\sigma^2}}\\end{align} ## Simulating parameters 1. To initialize, choose initial values for \\$$\\{ \\mu, \\rho, \\sigma^2\\} \\$$. 2. Draw members of \\$$\\{ \\mu, \\rho, \\sigma^2\\} \\$$ using Gibbs: • Draw \\$$\\mu^{(n+1)}\\$$ and \\$$\\sigma^{2^{(n+1)}}\\$$ from their exact posterior distributions given observed data and current draws of other parameters. • Draw \\$$\\rho^{(n+1)} \\$$ with a Metropolis jump on transformed space. ## Distribution of \\$$\\sigma^2\\$$ \\$$\\sigma^{2^{(n+1)}} \\$$ is Inverse-Gamma\\$$(a, b) \\$$ with \\begin{align} a = & \\frac{n}{2} - 1, \\\\ b = &\\frac{1}{2} \\sum_{i \\in I} \\frac{1}{1 - \\rho_i^{(n+1)^2}} \\left(y_i^{(n+1)} - \\left(\\rho_i^{(n+1)} y_{r(i)}^{(n+1)} \\right.\\right. \\\\ & \\left.\\left.+ \\left(1 - \\rho_i^{(n+1)}\\right)\\mu^{(n+1)}\\right)\\right)^2\\end{align} ## More complicated: HMM • \\$$\\{h_1, h_2, \\ldots \\} \\$$ live on discrete space, follow a Markov Process with a transition matrix \\$$T_\\theta \\$$. • At time", "equivalent to: $$\\cos\\Bigl(x-\\frac\\pi4\\Bigr)=\\frac1{\\sqrt2}\\iff x-\\frac\\pi4\\equiv\\pm\\frac\\pi4\\mod 2\\pi\\iff x\\equiv 0,\\,\\frac\\pi2\\mod2\\pi.$$ you can use: $$\\cos(x)= \\sqrt{1-\\sin(x)^{2}}$$ $$\\text{then let}\\qquad \\sin(x)=t\\\\ \\sqrt{1-t^{2}}+t-1=0$$ now you can continue • And what about the cases where $\\cos x \\neq \\sqrt{1-\\sin^2 x}$ ? – user228113 May 31 '15 at 22:44 • $$in \\;this \\;case \\;: cos(x)=-\\sqrt{1-sin(x)}\\leq 0\\\\but :cos(x)=1-sin(x)\\leq 0\\; so \\;cos(x)=0\\\\in \\; this \\; case \\; \\sqrt{1-sin(x)}=-\\sqrt{1-sin(x)}=0$$ – mounir ben salem May 31 '15 at 22:52 • Did you know that your answer can contain text that is no between \\$...\\$? – Mike Pierce May 31 '15 at 22:55 • yes i knew that $_$ – mounir ben salem May 31 '15 at 22:59 I'll throw my hat in the ring to get a picture in edgewise. :-) I'll use $\\theta$ for the angle, rather than $x$ (to avoid confusion with the $x$-coordinate). If we then let $x = \\cos \\theta, y = \\sin \\theta$, then the allowable results are along the red circle with $x^2+y^2 = 1$ (since $\\cos^2 \\theta + \\sin^2 \\theta = 1$). The solutions of the given equation are at the intersections of the blue line $x+y = 1$ with that red circle, yielding $(\\cos \\theta, \\sin \\theta) = (1, 0)$ and $(0, 1)$. This in turn yields $$\\theta = 2k\\pi, \\qquad \\theta = 2k\\pi + \\frac{\\pi}{2}$$ for $k$ an integer. By inspection, it" ]

[ "the exercise? I would find the zeroes of $g'$ and evaluate the function on that points. But I'm afraid this is somehow wrong... What do you think? • Regarding boundedness: In polar coordinates it is easy to see if a set is bounded, because all you have to do is look at the coordinate $r$. The set is bounded iff $r$ is bounded on it. – Giuseppe Negro Sep 16 '12 at 13:05 • Consider $\\vartheta\\to\\max\\{x(\\vartheta),y(\\vartheta)\\}$. This is a continuos function on the compact interval $[0,2\\pi]$, hence has a maximum and a minimum. – Hagen von Eitzen Sep 16 '12 at 13:10 • @GiuseppeNegro Thank you very much for your comment, good observation. I'll keep it in mind. – Romeo Sep 16 '12 at 13:19 If $f(x, y) = \\max\\{x, y\\}$ then $f(\\theta) = r(\\theta)\\max\\{\\sin \\theta, \\cos \\theta\\}$. To find maximum and minimum of $f$, you should find the maximum and minimum of $f_x(\\theta) = r(\\theta) \\sin \\theta = \\sin \\theta + \\sin \\theta \\cos \\theta$ and $f_y(\\theta) = r(\\theta) \\cos \\theta = \\cos \\theta + \\cos^2 \\theta$. Then pick the larger or smaller of each pair. Those are found using the usual method of differentiating. Note that $f_x' = \\cos \\theta + \\cos 2\\theta = 2\\cos \\frac{3\\theta}{2} \\cos \\frac{\\theta}{2}$ and $f_y' = -\\sin \\theta (1+2\\cos \\theta)$. Now", "by setting, y1 = r * cos(θ),y2 = r * sin(θ) $\\theta = x_2, r = (-1 + R) / (2*x_3), R = \\sqrt(1+4x_1*x_2)$ which may be differentiated to give the required sensitivities. x=x0; theta=x(2); R=sqrt(1+4*x(3)*x(1)); r=(-1+R)/(2*x(3)); y_exact=[r*cos(theta); r*sin(theta)] dRdx=[2*x(3)/R, 0, 2*x(1)/R]; drdx=dRdx/(2*x(3))-(-1+R)/(2*x(3)^2)*[0,0,1]; r*cos(theta)*dthetadx+sin(theta)*drdx] > y_exact = 0.0128 0.0460 Dy_exact = 0.2442 -0.0460 -0.0006 0.8796 0.0128 -0.0020 ##### Undifferentiated iteration For the nonlinear, undifferentiated iteration we have: y=y0; x=x0; u=[x(1)*cos(x(2));x(1)*sin(x(2));x(3)]; disp('Itns. |g(y)|') for i=1:itmax g=BBEx3p1_gfunc(y,u); y=y-g; disp([num2str(i,'% 4.0f'),' ',num2str(norm(g),'%10.4e')]) if norm(g)<=tolg break end end y_undif=y Error_y_undif=norm(y_undif-y_exact,inf) Itns. |g(y)| 1 5.0000e-002 2 2.5000e-003 3 2.4375e-004 4 2.3216e-005 5 2.2163e-006 6 2.1153e-007 7 2.0189e-008 8 1.9270e-009 y_undif = 0.0128 0.0460 Error_y_undif = 1.6178e-010 ##### Naive differentiation Simply initialising x with the fmad class gives a naive iteration for which the lack of convergence checks on the derivatives here results in only slight inaccuracy. y=y0; u=[x(1)*cos(x(2));x(1)*sin(x(2));x(3)]; disp('Itns. |g(y)|') for i=1:itmax g=BBEx3p1_gfunc(y,u); y=y-g; disp([num2str(i,'% 4.0f'),' ',num2str(norm(g),'%10.4e'),' ',... num2str(norm(getinternalderivs(g)),'%10.4e')]) if norm(g)<=tolg break end end y_naive = getvalue(y) Dy_naive = getinternalderivs(y) Error_y_naive = norm(y_naive-y_exact,inf) Error_Dy_naive = norm(Dy_naive-Dy_exact,inf) Itns. |g(y)| 1 5.0000e-002 1.0000e+000 2 2.5000e-003 1.0003e-001 3 2.4375e-004 1.4508e-002 4 2.3216e-005 1.8246e-003 5 2.2163e-006 2.1665e-004 6 2.1153e-007 2.4729e-005 7 2.0189e-008 2.7469e-006 8 1.9270e-009 2.9909e-007 y_naive = 0.0128 0.0460 Dy_naive = 0.2442 -0.0460 -0.0006 0.8796 0.0128 -0.0020 Error_y_naive = 1.6178e-010 Error_Dy_naive = 2.9189e-008 ##### Problems with", "inverse y = K tan^-1(A(x+B)) y/K = tan^-1(A(x+B)) tan(y/K) = A(x+B) y' sec^2(y/K)/K = A y' sec^2(y/K) = KA y' (tan^2(y/K)+1) = KA y' ((A(x+B))^2+1) = KA y' = KA ------------ (A(x+B))^2+1 ------------------- A = 16/9, B=1/2 16K/9 = 4/3 16K = 12 K = 3/4 31. anonymous Omg, $\\tan^2\\theta+1=\\sec^2\\theta, so \\it simplifies \\to \\int\\limits_{}^{}\\frac{ 2 }{ \\sqrt3 }d \\theta$ 32. amistre64 you are on the right track yes http://www.wolframalpha.com/input/?i=integrate+1%2F%28x^2%2Bx%2B1%29+dx 33. amistre64 mine went someplace off the wall lol 34. amistre64 4/3 sqrts to get inside the ^2 ... 35. amistre64 2/sqrt(3) not 16/9 36. amistre64 A = 2/sqrt(3) 2K/sqrt(3) = 4/3 2sqrt(3) K = 4 K = 2/sqrt(3) thats better on my end 37. anonymous $= \\frac{ 2 }{ \\sqrt3 } \\theta+C, \\theta=\\arctan(\\frac{ 2 }{ \\sqrt3 }(x+.5)) \\implies \\frac{ 2 }{ \\sqrt3 }\\arctan(\\frac{ 2 }{ \\sqrt3}(x+.5))+C=\\int\\limits_{}^{}\\frac{ 1 }{ x^2+x+1 }dx$ 38. amistre64 now me and the wolf agree: y = K tan^-1 (A(x+B)) $y=\\frac{2}{\\sqrt3}~\\tan^{-1}[\\frac{2}{\\sqrt3}(x+\\frac12)]$ 39. anonymous This answer is going to be really ugly in the end 40. anonymous Yeah, I agree, then we just put it all together and its $\\log|x-1|-0.5\\log|x^2+x+1|-\\sqrt3\\arctan(\\frac{ 2 }{ \\sqrt3 }(x+.5)+C$ I dont think that simplifies any further 41. amistre64 the uglier the better :) 42. amistre64 you got it 43. anonymous Jeez, that was tough. 44. amistre64 nah .. it was", "that $\\theta$ varies from $\\pi/4$ to $\\pi/2$. So we have $$I=\\int_{\\pi/4}^{\\pi/2}\\int_?^? x^2\\,J\\,dr\\,d\\theta$$ where $J$ is the Jacobian. To find the limits for $r$, draw a line on your diagram starting at the origin and heading in the $\\theta$ direction (where $\\theta$ is between $\\pi/4$ and $\\pi/2$). You can see that the values of $r$ which are in your region and on this line go from a minimum of $0$ to a maximum on the horizontal line $y=1$. To find the $r$ value on this line we have $$r\\sin\\theta=y=1$$ and so $r_{\\rm max}=1/\\sin\\theta$. Hence $$I=\\int_{\\pi/4}^{\\pi/2}\\int_0^{1/\\sin\\theta} x^2\\,J\\,dr\\,d\\theta\\ .$$ You should also know that the Jacobian for polar coordinates is $r$ and that $x=r\\cos\\theta$. Hence $$I=\\int_{\\pi/4}^{\\pi/2}\\int_0^{1/\\sin\\theta} (r\\cos\\theta)^2\\,r\\,dr\\,d\\theta\\ .$$", "fourth is the step size. I went to 63 because it gives a cool image: Note that I'm using 1 in front of the z-coordinate to squash it a bit this time, but you can easily change it back to 4. • Thank you! I really appreciate it. Could you just tell me what values i need to change to get the proper vectors and points? I am a completely new to Mathematica so it is hard for me to understand what is happening here. In case it matters for your explanation, the point set is defined as follows. Let $d\\geq 32$ and $m$ be a unique $m$ such that $2^m\\leq d<2^{m+1}$. Now define $\\theta=\\frac{\\pi}{3*2^m}$ then the point set is $A=\\{(d\\cos(k\\theta),d\\sin(k\\theta),0): k=1,2,...,6*2^m\\}$. The vector for each point is $<y,-x,4\\sqrt{3}d+7>$. Obviously I only want to draw just a couple of cylinders. – Sorfosh Feb 26 at 22:38 • So, for example . when $d=32$ and so $m=5$ and $\\theta=\\frac{\\pi}{96}$. Hence The point set is $A=\\{(32*cos(k\\frac{\\pi}{96}),32*sin(k\\frac{\\pi}{96},0) : k= 1,2..., 192\\}$ As I type this, i realize that that is a lot of cylinders (of radius 1/2) so I would hope that this could be somehow automated and that Mathematica can generate the graphs so I do not have to manually do anything. Again, I really appreciate the help. – Sorfosh", "to $(0,0,1)$ since changing $\\theta$ does not change $z$. Therefore, we simply need to take the cross product of $(x,y,z)$ and $(0,0,1)$ and normalize to get $\\hat{e}_\\theta=\\frac{(-y,x,0)}{\\sqrt{r^2-z^2}}$. The sign is chosen so that the vector points counter-clockwise. $\\hat{e}_\\phi$ is perpendicular to the other two, so we take the cross product and normalize: $\\hat{e}_\\phi=\\frac{(-xz,-yz,r^2-z^2)}{r\\sqrt{r^2-z^2}}$. Again, the sign is chosen to have positive $z$ component. Thus, we get \\begin{align} \\hat{e}_R&=\\frac{(x,y,z)}{r}\\\\ \\hat{e}_\\theta&=\\frac{(-y,x,0)}{\\sqrt{r^2-z^2}}\\\\ \\hat{e}_\\phi&=\\frac{(-xz,-yz,r^2-z^2)}{r\\sqrt{r^2-z^2}} \\end{align} References 1. Source of the polar-coordinate image here from Wikipedia. - ...could elaborate on this $\\hat{e}_\\theta=\\frac{(-y,x,0)}{\\sqrt{r^2-z^2}}$? Which sign? When you do the cross-product, how you think it? I do the matrix thing but I find it slow... Do you think it as a Sarrus-rule here? – hhh Mar 24 '12 at 22:01 @hhh: $(x,y,z)\\times(0,0,1)=(y,-x,0)$ and $(0,0,1)\\times(x,y,z)=(-y,x,0)$ normalizing these gives opposite vectors, so we choose the one that goes counter-clockwise. Both are perpendicular to the vectors specified. – robjohn Mar 24 '12 at 22:46 The image from Wikipedia has $\\theta$ and $\\varphi$ switched from the way I usually see them, and from the way I have them in my answer. I will either edit my answer, or generate a new image (probably the latter so that we are copyright safe). – robjohn Mar 26 '12 at 0:39 ...if you generate the picture, please, do not hesitate to include", "$\\begin{array}{l}\\begin{array}{l}\\\\ {r}_{1}=3\\sin \\left(3\\theta \\right)\\end{array}\\hfill \\\\ {r}_{2}=2\\sin \\left(3\\theta \\right)\\hfill \\\\ {r}_{3}=\\sin \\left(3\\theta \\right)\\hfill \\end{array}$ 62. On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. $\\begin{array}{l}\\begin{array}{l}\\\\ {r}_{1}=3+3\\cos \\theta \\end{array}\\hfill \\\\ {r}_{2}=2+2\\cos \\theta \\hfill \\\\ {r}_{3}=1+\\cos \\theta \\hfill \\end{array}$ 63. On a graphing utility, graph each polar equation. Explain the similarities and differences you observe in the graphs. $\\begin{array}{l}\\begin{array}{l}\\\\ {r}_{1}=3\\theta \\end{array}\\hfill \\\\ {r}_{2}=2\\theta \\hfill \\\\ {r}_{3}=\\theta \\hfill \\end{array}$ For the following exercises, draw each polar equation on the same set of polar axes, and find the points of intersection. 64. ${r}_{1}=3+2\\sin \\theta ,{r}_{2}=2$ 65. ${r}_{1}=6 - 4\\cos \\theta ,{r}_{2}=4$ 66. ${r}_{1}=1+\\sin \\theta ,{r}_{2}=3\\sin \\theta$ 67. ${r}_{1}=1+\\cos \\theta ,{r}_{2}=3\\cos \\theta$ 68. ${r}_{1}=\\cos \\left(2\\theta \\right),{r}_{2}=\\sin \\left(2\\theta \\right)$ 69. ${r}_{1}={\\sin }^{2}\\left(2\\theta \\right),{r}_{2}=1-\\cos \\left(4\\theta \\right)$ 70. ${r}_{1}=\\sqrt{3},{r}_{2}=2\\sin \\left(\\theta \\right)$ 71. ${r}_{1}{}^{2}=\\sin \\theta ,{r}_{2}{}^{2}=\\cos \\theta$ 72. ${r}_{1}=1+\\cos \\theta ,{r}_{2}=1-\\sin \\theta$", "3*x*y**2 r = np.linspace(0, 0.001, 101) theta = np.linspace(0, 2*np.pi, 360) rr, tt = np.meshgrid(r, theta) u = f_polar(rr, tt) fig = plt.figure() ax.pcolormesh(tt, rr, u, cmap='inferno') ax.set_aspect('equal') ax.set_rgrids([]) plt.show() print('mean(u) - u(0,0) = ', np.mean(u) - f_polar(0, 0)) mean(u) - u(0,0) = 7.013888888888634e-13 We can easily check this for our example $$u = x^3 - 3xy^2$$, where the mean value in a circle of radius $$R$$ centered at $$(0, 0)$$ is given by \\begin{split} \\begin{aligned} \\overline{u} &= \\frac{1}{2R \\ 2 \\pi} \\int_{-R}^R \\int_0^{2 \\pi} u \\ d \\theta d r \\\\ &= \\frac{1}{4R \\pi} \\int_{-R}^R \\int_0^{2 \\pi} r^3 (\\cos^3 \\theta - 3 \\cos \\theta \\sin^2 \\theta) \\ d \\theta dr \\\\ &= \\frac{1}{4R \\pi} \\int_{-R}^R r^3 \\underbrace{\\int_0^{2 \\pi} (\\cos^3 \\theta - 3 \\cos \\theta \\sin^2 \\theta) \\ d \\theta}_{= 0} dr = 0 \\end{aligned} \\end{split} So the mean value in the circle is zero and $$u(0, 0)$$ is zero as well, as we expected. Let us think again about the surface given by a set of points $$(x, y, u)$$ but this time consider a function $$u$$ such that $$\\nabla^2 u \\neq 0$$. For example: $u(r) = \\frac{\\sin r}{r}, \\qquad r = \\sqrt{x^2 + y^2}$ for which $\\nabla^2 u = \\frac{(1 - r^2) \\sin r}{r^3} - \\frac{\\cos r}{r^2}$ x = np.linspace(-7.5, 7.5, 300) X, Y = np.meshgrid(x,", "question still stands though. Apr 15 '15 at 12:13 • And if there is some dependence on the choice of $\\theta$ we can infer that the limit does not exist and depends on which path we choose Apr 15 '15 at 12:17 • @OriaGruber: Your approach is correct! Good job. By the way you can use L'hopital's rule to evaluate the limit as $r \\to 0$. Apr 15 '15 at 12:24 I think you made a mistake for the last part. And if you apply polar coordinate properly ( in general it may not simplify the problem though), it will give you the correct answer. $\\lim_{(x,y) \\to (0,0)}\\frac{x^2+y^2}{\\sqrt{x^2+y^2+1}-1}=\\lim_{r \\to 0} \\frac{r^2}{\\sqrt{r^2+1}-1}=\\lim_{r \\to 0}\\frac{r^2(\\sqrt{r^2+1}+1)}{r^2+1-1}=\\lim_{r \\to 0}\\sqrt{r^2+1}+1=2$ You were almost there. Use conjugates: $$\\frac{r^2}{\\sqrt{r^2+1}-1}=\\frac{r^2(\\sqrt{r^2+1}+1)}{r^2}=\\sqrt{r^2+1}+1\\xrightarrow[r\\to 0]{}2$$ Yes, polar coordinates can do, as any point $(x,y)$ can be expressed as some $(r\\cos\\theta,r\\sin\\theta)$. For example, $$\\frac{x-y}{x+y}=\\frac{r\\cos\\theta-r\\sin\\theta}{r\\cos\\theta+r\\sin\\theta}=\\frac{\\cos\\theta-\\sin\\theta}{\\cos\\theta+\\sin\\theta}$$ has no limit as the last expression is undeterminate. Your given example is a little different in that $\\theta$ does not appear explicitly, hence it is actually a $1D$ limit $$\\lim_{(x,y) \\to (0,0)}\\frac{x^2+y^2}{\\sqrt{x^2+y^2+1}-1}=\\lim_{x^2+y^2\\to0}\\frac{x^2+y^2}{\\sqrt{x^2+y^2+1}-1}=\\lim_{t\\to0^+}\\frac{t}{\\sqrt{t+1}-1}.$$ Maybe you made a mistake for the last part, when you semplify $r$. I invite you to consider this other approach. We have that $$\\sqrt{1+x}-1\\sim \\frac{x}{2} \\ \\ (x\\rightarrow 0)$$ Thus $$\\sqrt{1+x^2+y^2}-1\\sim \\frac{x^2+y^2}{2} \\ \\ (x,y\\rightarrow 0)$$ Then the limit is 2. We can", "$r = \\sqrt{2} \\sqrt{2 \\pm \\sqrt{3}}$. Hence the minimum value of $|z|$ is $\\sqrt{2} \\sqrt{2 - \\sqrt{3}}$ and the maximum value if $\\sqrt{2} \\sqrt{2 + \\sqrt{3}}$. Let $z=re^{i\\theta}$, then your last step can be written as $$r^2+2\\cos (2\\theta)+\\frac{1}{r^2}=4.$$ From this we can get \\begin{align*} \\left(r+\\frac{1}{r}\\right)^2 & = 6-2\\cos (2\\theta) \\end{align*} Thus $2 \\leq \\left(r+\\frac{1}{r}\\right) \\leq 2\\sqrt{2}.$ Now try to find the range of $r$.", "Jey May 16 '13 at 5:54 If you change to polar coordinates, you get $$\\lim_{r \\rightarrow 0} \\sin(\\theta) \\cos(\\theta) \\sin(r^2 \\sin(\\theta) \\cos(\\theta)).$$ It might look a bit confusing, but the essential thing is that $\\sin(x)$ goes to zero as $x$ goes to zero, and that $\\sin(\\theta)$ and $\\cos(\\theta)$ are at most 1. • I understand how you changed the expression using the substitutions $x=r\\cos{\\theta}$ and $y=r\\sin{\\theta}$. But how do change $(x,y) \\rightarrow (0,0)$ to $r \\rightarrow 0$? Thanks. – Jey May 16 '13 at 5:49 • That $(x,y) \\rightarrow 0$ means that the distance from the point $(x,y)$ to $(0,0)$ should go to zero, which means that $\\sqrt{x^2 + y^2} = r$ should go to zero. The $\\theta$ varies freely, but this is ok for us since the expression goes to zero. Note that for instance $\\lim_{r \\rightarrow 0} \\sin(\\theta)$ does not exist. – N.U. May 16 '13 at 6:03 Hint: $\\left|\\dfrac{xy}{x^2+y^2}\\right| \\le \\dfrac12$. • How do you get 1/2? – Jey May 16 '13 at 5:47 • @Jey: $x^2+y^2-2xy \\ge 0 \\implies x^2+y^2 \\ge 2xy \\implies \\dfrac{1}{x^2+y^2} \\le \\dfrac{1}{2xy}$ – Inceptio May 16 '13 at 5:49 • AM/GM inequality (or $\\sin \\theta \\cos \\theta = \\frac 12 \\sin 2\\theta$) – mrf May 16 '13 at 5:49 • Nice. I didn't notice that. Thanks – Jey May 16", "(u^2 + v^2 - 2uv\\cos\\theta) }{ A^2 } } \\right)^2\\\\ &= mv^2 + M \\frac{ (u^2 + v^2 - 2uv\\cos\\theta) }{ A^2 }\\\\ \\end{align*} Solving this for $v$ nets me: \\begin{align*} (\\frac{M}{A^2} + m)v^2 - \\frac{2Mu cos\\theta }{A^2}v + (\\frac{M}{A^2} - m)u^2 = 0 \\end{align*} Now presumably $\\theta$ is a function of $v$, $\\theta (v)$. If we differentiate with respect to $v$: \\begin{align*} \\frac{d}{dv} \\left( (\\frac{M}{A^2} + m)v^2 - \\frac{2Mu\\cos\\theta}{A^2}v + (\\frac{M}{A^2} - m)u^2 \\right)\\\\ = 2(\\frac{M}{A^2} + m)v - \\frac{2Mu}{A^2}(\\cos\\theta -v \\theta ' (v) \\sin\\theta) = 0 \\end{align*} This looks a bit messy. Just to make sure, I want to solve for $\\theta$ in order to find out the minimum of $v$, right? 11. Sep 26, 2016 ### PeroK Yes, but the next step is to get rid of $v$ or $V$. These are dependent on each other; whereas $m, M, u$ are constants. It looks easier to get rid of $V$, using your equation, plus conservation of energy: $mu^2 = mv^2 + MV^2$ Also, you don't need to calculate KE until the end, as the following are equivalent: Minimising $v$ Maximising $V$ Maximising the loss in KE of $m$ Do you see that? 12. Sep 26, 2016 ### TheSodesa I think I see intuitively why those are the same. $v$ getting minimized means most of the", "and the heading angle $\\theta$. Given an $<x, y>$ coordinate scheme, at a heading of $\\theta = 0$, the location of $A$ would be $<0, +r>$. If $\\theta = 90^{\\circ}$, then $A = <+r, 0>$. This is a trig problem, and you can see now that you need $\\mbox{Trig Function 1}$, or $\\mbox{TF1}$ such that $\\mbox{TF1}(0^{\\circ}) = 0$ and $\\mbox{TF1}(90^{\\circ}) = +1$, then you need a second trig function $\\mbox{TF2}(0^{\\circ}) = +1$ and $\\mbox{TF2}(90^{\\circ}) = 0$. Then you get: $$A_x = M_x + TF1(\\theta)*r \\\\ A_y = M_y + TF2(\\theta)*r \\\\$$ This problem was simplified by putting the coordinate axes on $M$ such that $M_x = 0$ and $M_y = 0$. The trig functions that satisfy the criteria above are $\\mbox{TF1} = \\mbox{sine}$ and $\\mbox{TF2} = \\mbox{cosine}$. I'm not sure that there's a good way to intuitively explain how to determine this other than practice, learning the special values for trig functions, and then more practice (do all the homeworks!). So, this means that you can find the coordinate of $A$ as: $$A_x = M_x + \\sin{(\\theta)}*r \\\\ A_y = M_y + \\cos{(\\theta)}*r \\\\$$ You're not trying to find $A$, though, you're given $A$ and you're trying to find $M$! This is now straightforward - solve for $M$! $$M_x = A_x - \\sin{(\\theta)}*r \\\\ M_y = A_y -", "quite a bit larger than $\\frac14$. That is as far as I can go with this problem. I don't even have a guess as to what the maximum value of $|bc|$ might be. \\begin{tikzpicture} [scale=6]\\draw [help lines, ->] (-1.1,0) -- (1.1,0) ; \\draw [help lines, ->] (0,-1.1) -- (0,1.1) ; \\draw [red] circle (1) ; \\fill [green, domain=0:6.285, samples=100] plot ({0.16*cos(2*\\x r) + 0.56*cos(\\x r) - 0.56}, {0.16*sin(2*\\x r) + 0.56*sin(\\x r)}); \\end{tikzpicture} #### Opalg ##### MHB Oldtimer Staff member This is a solution for the case where the parameters $a,b,c$ are all real. I suspect that the answer may be the same if they are allowed to be complex, but the proof would have to be quite different. For the condition $|az^2 + bz + c| \\leqslant 1$ to hold whenever $|z|\\leqslant1$, it is sufficient (by the maximum modulus principle) to check that it holds when $|z|=1$. So we want to ensure that $|ae^{2i\\theta} + be^{i\\theta} + c| \\leq 1$ (for all $\\theta$). But if $a,b,c$ are real then \\begin{aligned}|ae^{2i\\theta} + be^{i\\theta} + c|^2 &= (ae^{2i\\theta} + be^{i\\theta} + c)(ae^{-2i\\theta} + be^{-i\\theta} + c) \\\\ &= 2ac\\cos(2\\theta) + 2b(a+c)\\cos\\theta + a^2 + b^2 + c^2 \\\\ &= 2ac(2\\cos^2\\theta - 1) + 2b(a+c)\\cos\\theta + a^2 + b^2 + c^2 \\\\ &= 4acx^2 + 2b(a+c)x + (a-c)^2 + b^2,", "we must check all the possible initial values of $r(0)\\in[0,2]$ to see which initial value yields the maximal $r(\\frac{\\pi}{4})$. While we suspect that maximising $r(0)$ should also maximise $r(\\frac{\\pi}{4})$, I don't quite know how to prove this properly. Instead, we need to use some computing software to evaluate the differential equation numerically for various initial values, as the DE likely has no closed-form solution for arbitrary initial values (if a closed-form solution does exist, please let me know!) As we can clearly see, the initial value of $r(0)=2$ yields the maximal value of $r(\\frac{\\pi}{4})$. In fact, the specific initial value $r(0)=2$ actually does lead to a closed-form solution! Indeed, $r(0)=2$ means $$r(t)=2\\sec(t)$$which can be easily confirmed by substituting into the above differential equation. Thus, the maximal twin distance is reached if $r(0)=2$ and $$B(t)=2\\sec(t)e^{i(t+\\pi/2)}$$ but this is simply the straight line that runs along the line $y(t)=2$. This is because \\begin{align*}y(t)&=2\\sec(t)\\sin(t+\\pi/2)\\\\&=2\\sec(t)\\cos(t)\\\\&=2\\end{align*} And so we have the maximum $\\boxed{r\\Big(\\frac{\\pi}{4}\\Big)=2\\sec\\Big(\\frac{\\pi}{4}\\Big)=2\\sqrt{2}}$ ### Final solution All this kerfuffle means we now know the recipe to maximising the twin-lamp distance at the end of the run. We simply need to do the following: 1. Ensure the twin reaches $B(0)=2i$ at $t=0$. It does not matter how the twin reaches this point from the starting point. The twin has flexibility here due to the", "$(x,y)=r(\\cos\\theta ,\\sin\\theta ).\\;$ The constraint becomes $r^2(1+\\sin\\theta\\cos\\theta)=1,\\;$ or, $\\displaystyle\\sin\\theta\\cos\\theta =\\frac{1-r^2}{r^2}.$ Since $\\sin\\theta\\cos\\theta =\\frac{1}{2}\\sin 2\\theta,\\;$ $\\displaystyle -\\frac{1}{2}\\le\\frac{1-r^2}{r^2}\\le\\frac{1}{2}.\\;$ From here, $\\displaystyle\\frac{2}{3}\\le r^2\\le 2.$ Now, $K=xy(x^2+y^2)=r^4\\sin\\theta\\cos\\theta=r^2(1-r^2)\\;$ which again leads to $\\displaystyle -2\\le K\\le\\frac{2}{9}.$ ### A Sample of Optimization Problems III • Mathematicians Like to Optimize • Mathematics in Pizzeria • The Distance to Look Your Best • Building a Bridge • Linear Programming • Residence at an Optimal Distance • Distance Between Projections • Huygens' Problem • Optimization in a Crooked Trapezoid • Greatest Difference in Arithmetic Progression • Area Optimization in Trapezoid • Minimum under Two Constraints • Optimization with Many Variables • Minimum of a Cyclic Sum with Logarithms • A Problem with a Magical Solution from Secrets in Inequalities • Leo Giugiuc's Optimization with Constraint • Problem 4033 from Crux Mathematicorum • An Unusual Problem by Leo Giugiuc • A Cyclic Inequality With Constraint in Two Triples of Variables • Two Problems by Kunihiko Chikaya", "equivalent to: $$\\cos\\Bigl(x-\\frac\\pi4\\Bigr)=\\frac1{\\sqrt2}\\iff x-\\frac\\pi4\\equiv\\pm\\frac\\pi4\\mod 2\\pi\\iff x\\equiv 0,\\,\\frac\\pi2\\mod2\\pi.$$ you can use: $$\\cos(x)= \\sqrt{1-\\sin(x)^{2}}$$ $$\\text{then let}\\qquad \\sin(x)=t\\\\ \\sqrt{1-t^{2}}+t-1=0$$ now you can continue • And what about the cases where $\\cos x \\neq \\sqrt{1-\\sin^2 x}$ ? – user228113 May 31 '15 at 22:44 • $$in \\;this \\;case \\;: cos(x)=-\\sqrt{1-sin(x)}\\leq 0\\\\but :cos(x)=1-sin(x)\\leq 0\\; so \\;cos(x)=0\\\\in \\; this \\; case \\; \\sqrt{1-sin(x)}=-\\sqrt{1-sin(x)}=0$$ – mounir ben salem May 31 '15 at 22:52 • Did you know that your answer can contain text that is no between \\$...\\$? – Mike Pierce May 31 '15 at 22:55 • yes i knew that $_$ – mounir ben salem May 31 '15 at 22:59 I'll throw my hat in the ring to get a picture in edgewise. :-) I'll use $\\theta$ for the angle, rather than $x$ (to avoid confusion with the $x$-coordinate). If we then let $x = \\cos \\theta, y = \\sin \\theta$, then the allowable results are along the red circle with $x^2+y^2 = 1$ (since $\\cos^2 \\theta + \\sin^2 \\theta = 1$). The solutions of the given equation are at the intersections of the blue line $x+y = 1$ with that red circle, yielding $(\\cos \\theta, \\sin \\theta) = (1, 0)$ and $(0, 1)$. This in turn yields $$\\theta = 2k\\pi, \\qquad \\theta = 2k\\pi + \\frac{\\pi}{2}$$ for $k$ an integer. By inspection, it", "maximize/minimize without calculus, e.g. *2nd-degree polynomials, e.g. $4x^2 + 5x + 1$ *A function given a constraint. A classic example of this is maximizing the volume of a rectangular prism given a fixed surface area. *Functions involving sine or cosine, e.g. $5 \\sin 4x + 2 \\cos x$ And others...I doubt there is an easy non-calculus solution to optimizing $y = \\frac{2x}{3x^2 + 1}$ 10. ## Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1) Have u learnt tri? I mean sin cos tan? If so, i'll put x=tany/31/2 {since all real numbers can be represent by tany/31/2} then it becomes 2(tanx/1/2)/{3(tanx/31/2)2+1} = {2/31/2 } * tanx/sec2x =2/31/2 * sinxcosx =1/31/2 * sin2x ie, max = 1/31/2 , min= -1/31/2 11. ## Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1) @Johnnylam nice solution. Basically, let $x = \\frac{\\tan{\\theta}}{\\sqrt{3}}$ Then, $y = \\frac{2 \\tan \\theta}{\\sqrt{3}(\\tan^2 \\theta + 1)} = \\frac{2 \\tan \\theta}{\\sqrt{3} \\sec^2 \\theta} = \\frac{2}{\\sqrt{3}} \\sin \\theta \\cos \\theta = \\frac{1}{\\sqrt{3}} \\sin (2\\theta)$ Sine achieves its maximum value at 1 and -1, so the max/min values are $\\pm \\frac{1}{\\sqrt{3}}$ Good solution. Idk why I didn't think of a trig substitution... 12. ## Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1) Very nice indeed, johnnylam123!", "\\begin{align} K(m) &= \\int_0^{\\pi/2} \\frac{d\\theta}{\\sqrt{1-m\\sin^2\\theta}} = \\int_0^1 \\frac{dt}{\\sqrt{(1-mt^2)(1-t^2)}}\\\\ E(m) &= \\int_0^{\\pi/2} \\sqrt{1-m\\sin^2\\theta} d\\theta = \\int_0^1 \\sqrt{\\frac{1-mt^2}{1-t^2}} dt \\end{align} We get $$\\verb/Volume/ = \\frac{16 r^3}{3\\sin\\alpha}\\left((2-m)E(m) - 2(1-m)K(m)\\right)$$ As a sanity check, when $d \\to 0$, $m \\to 1^{-}$, we have $$(1-m)K(m) \\to 0\\quad\\text{ and }\\quad (2-m)E(m) \\to E(1) = 1$$ This leads to a familiar result for the volume of intersection when the axis intersect. $$\\verb/Volume/_{d = 0} = \\frac{16r^3}{3\\sin\\alpha}$$ • Amazing effort and solution! – Joseph O'Rourke Jun 26 '15 at 23:25", "# How do you convert x^2+(y-4)^2=16 to polar form? May 3, 2016 $r = 8 \\sin \\theta$ #### Explanation: If $\\left(r , \\theta\\right)$ is in polar form and $\\left(x , y\\right)$ in Cartesian form the relation between them is as follows: $x = r \\cos \\theta$, $y = r \\sin \\theta$, ${r}^{2} = {x}^{2} + {y}^{2}$ and $\\tan \\theta = \\frac{y}{x}$ Hence, ${x}^{2} + {\\left(y - 4\\right)}^{2} = 16$ can be written as ${x}^{2} + {y}^{2} - 8 y + 16 = 16$ or ${x}^{2} + {y}^{2} - 8 y = 0$ or ${r}^{2} - 8 r \\sin \\theta = 0$ or $r \\left(r - 8 \\sin \\theta\\right) = 0$ dividing by $r$ $r - 8 \\sin \\theta = 0$ or $r = 8 \\sin \\theta$ May 3, 2016 $r = 8 \\sin \\theta$ #### Explanation: To convert from Cartesian to Polar coordinates use the following formulae that link them. • x=rcostheta\" and \" y=rsintheta ${x}^{2} + {y}^{2} - 8 y + 16 = 16 \\text{ (expanding bracket) }$ $\\Rightarrow {r}^{2} {\\cos}^{2} \\theta + {r}^{2} {\\sin}^{2} \\theta - 8 r \\sin \\theta + 16 - 16 = 0$ then ${r}^{2} \\left({\\cos}^{2} \\theta + {\\sin}^{2} \\theta\\right) - 8 r \\sin \\theta = 0$ using the identity: ${\\cos}^{2} \\theta + {\\sin}^{2} \\theta = 1$ $\\Rightarrow {r}^{2} = 8 r \\sin" ]

[ "+0 0 145 1 +102 Find the minimum value of $$sin^4 x + \\frac{3}{2} \\cos^4 x,$$ as x varies over all real numbers.", "+0 0 49 1 +94 Find the minimum value of $$sin^4 x + \\frac{3}{2} \\cos^4 x,$$ as x varies over all real numbers.", "If x is a real number, Then what would be the minimum value of $\\left ( x^{2}-x-1 \\right )$: [A] $\\frac{3}{4}$ [B] 0 [C] 1 [D] $\\frac{1}{4}$", "of the minimum value of $\\sqrt{9+x^{2}}+\\sqrt{(4-x)^{2}+(3-y)^{2}}+\\sqrt{4+y^{2}}$ as $x$ and $y$ range over all real numbers. ×", "minimum value of$$ a+\\frac{1}{b(a-b)} ? $$Let a&gt;b be positive real numbers. What is the minimum value of$$ a+\\frac{1}{b(a-b)} ? ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find all pairs of positive real numbers such that $a^{3}+b^{3}+1=3 a b$. Find all pairs of positive real numbers such that $a^{3}+b^{3}+1=3 a b$.Find all pairs of positive real numbers such that $a^{3}+b^{3}+1=3 a b$. ...", "Let $a$, $b$, $c$ and $d$ be positive reals such that $a + b + c + d = 1$. Find the minimum value of $\\frac {1}{a} + \\frac {1}{b} + \\frac {1}{c} + \\frac {1}{d}$.", "# Yet another question 8 Calculus Level 4 Find the minimum value of $$\\max( |y^2- xy| )$$ given that $$x$$ is any real number and $$0\\leq y \\leq 1$$. Try more here ×", "# Maximum how? Geometry Level 2 $\\large (7\\cos x+24\\sin x)(7\\sin x-24\\cos x)$ Find the maximum value of this expression for reals $$x.$$ ×", "# Q4 Geometry Level 2 Find the minimum value of $$f(x) = \\sin^4 (x) + \\cos^4(x)$$ where $$0 \\le x \\le \\frac\\pi2$$. ×", "# Minimum Value Calculus Level 3 $\\large f(x)=\\frac{(x+1/x)^{6}-(x^{6}+1/x^{6})-2}{(x+1/x)^{3}+(x^{3}+1/x^{3})}$ Find the minimum value of $$f(x)$$ above for real $$x>0$$. ×", "# Thread: Find the minimum value 1. ## Find the minimum value Find the minimum value of (5 cosh x + 3 sinh x). How can I solve this? 2. Originally Posted by geton Find the minimum value of (5 cosh x + 3 sinh x). How can I solve this? Differentiate, put to zero, test nature .... the usual deal. Note that to solve when the derivative is equal to zero, you should: 1. Replace cosh and sinh with their exponential definitions. 2. Multiply through by 2 e^{-x}. 3. Solve the resulting quadratic equation in e^x ..... 3. $\\cosh x = \\frac{e^x + e^{-x}}{2}$ $\\sinh x = \\frac{e^x - e^{-x}}{2}$ so, $f(x) = 5\\cosh x + 3 \\sinh x$ $f'(x) = 5\\sinh x + 3\\cosh x = \\frac{8e^x - 2e^{-x}}{2} = 4e^x - e^{-x} = \\frac{4e^{2x} - 1}{e^x}$ we want to find $x$ such that $f'(x) = 0$ since $e^x$ is never zero, we will solve $4e^{2x} - 1 = 0$.. solving, we shall obtain $x = -\\ln 2$... substitute that $x$ to find the minimum value.. -------------------------- to check, $f''(x) = f(x) = 5\\cosh x + 3 \\sinh x = 4e^x + e^{-x} > 0$ for any $x$. thus, $x = -\\ln 2$ gives the minimum value for $f(x)$ 4. Yes I got x = -", "# It's what now? Geometry Level pending Find the minimum value of $$\\sin(2x) + \\cos(2x)$$ for real $$x$$. ×", "# Find the maximum and minimum value of $\\sin^6x+\\cos^6x$ $\\begin{array}{1 1}(a)\\;1,0&(b)\\;1,\\large\\frac{1}{4}\\\\(c)\\;0,1&(d)\\;2,1\\end{array}$", "maximum or absolute minimum values. Justify your answers. Find the extreme values if you can. $\\frac{1}{x \\sin x} \\text { on }(0, \\pi)$ Determine whether the given function has absolute maximum or absolute minimum values. Justify your answers. Find the extreme values if you can. $\\frac{1}{x \\sin x} \\text { on }(0, \\pi)$...", "# Maximum Geometry Level pending Find the maximum value of the expression below for real $$\\theta$$. $\\frac{1}{\\sin^2\\theta+3\\sin\\theta\\cos\\theta\\ +5\\cos^2\\theta}$ ×", "# find the Range Geometry Level pending What is the maximum value of $$f(x) = 4\\sin^4 x - \\sin^2 2x$$ for all real $$x$$? ×", "# Minimum Value If $x$ and $y$ are positive real numbers, what is the minimum value of $(2x + 3y)\\left(\\frac{8}{x} + \\frac{3}{y} \\right) ?$ ×", "\\sin x - 2a &= 0 \\\\ 4y^3 +\\frac{1}{2}\\sin 2y +a &=0 \\end{array} \\right.$$ Compute the value of $\\cos(x+2y)$ Let $f$ be a real-valued function such that $f(x) + 2f(\\frac{2002}{x}) = 3x$ for all $x > 0$. Find $f(2)$.", "is (A) (2,$$\\sqrt{2}$$,4) (B) (2, $$\\sqrt{2}$$, 0) (C) (0,0) (D) (2,2) For all real values of x, the minimum value of $$\\frac{1-x+x^{2}}{1+x+x^{2}}$$ is", "# Please evaluate q 22? ##### 1 Answer May 3, 2018 $\\rightarrow a = x + \\frac{1}{x} = \\frac{{x}^{2} + 1}{x}$ If $x$ is any non-zero real number then the value $a$ will always be greater than or less than 1 but the value of $\\sin \\theta \\mathmr{and} \\cos \\theta$ lies between $\\left[- 1 , 1\\right]$. So, the $\\sin \\theta \\mathmr{and} \\cos \\theta$ can never be equal to $a$ in the case mentioned in the question." ]

[ "between the minimum value of $$L$$ and the minimizing value of $$a$$, which makes it very unlikely that the minimum value of $$L$$ has a convenient closed form. Set $$\\ell = 1$$ for convenience (one may show easily that the minimizing value of $$a$$ and the minimum value of $$L$$ scale in direct proportion with $$\\ell$$), and substitute $$x = \\ell/2a$$. Thus, we must minimize $$\\frac{2 \\sinh x + \\cosh x}{2x} = \\frac{3 e^x - e^{-x}}{4x}$$ and we know that the minimum $$L_\\text{min}$$ occurs at $$x = \\xi$$ where $$\\left. \\frac{d}{dx}\\right|_{x=\\xi} \\frac{3 e^x - e^{-x}}{4x} = \\frac{(3\\xi-3) e^{\\xi} + (\\xi+1) e^{-\\xi}}{4\\xi^2} = 0.$$ From this, we have $$3 e^\\xi - e^{-\\xi} = 3\\xi e^\\xi + \\xi e^{-\\xi}$$ and thus $$L_\\text{min} = \\frac{3 e^\\xi - e^{-\\xi}}{4\\xi} = \\frac{3 e^\\xi + e^{-\\xi}}{4} = \\cosh \\xi + \\frac{1}{2} \\sinh \\xi$$ But we know from (*) that $$\\xi L_\\text{min} = \\sinh \\xi + \\frac{1}{2} \\cosh \\xi$$ By adding or subtracting these equations, we get the system: \\begin{align*} (1 + \\xi) L_\\text{min} &= \\frac{3}{2} e^{\\xi} \\\\ (1 - \\xi) L_\\text{min} &= \\frac{1}{2} e^{-\\xi} \\\\ \\end{align*} Multiplying these equations gives $$(1 - \\xi^2) L_\\text{min}^2 = \\frac{3}{4}.$$ • The answer of $L$ is given in the book in terms of $l$. which means you can easily find $L/l$ without finding it numerically. and hence we", "Therefore, we want to minimise $D$ with respect to $m$. Now differentiating, we have $\\frac{dD}{d m}=a-\\frac{b}{m^2}.$ We get a stationary point when $\\dfrac{dD}{dm}=0$. That gives $a=\\frac{b}{m^2} \\Longleftrightarrow m=\\pm \\sqrt{\\frac{b}{a}}.$ We are told that $m>0$, so we choose the positive root $m=\\sqrt{\\frac{b}{a}}$. This value must be a minimum value as there is only one stationary value, and by considering our expression for $D$, we can see that as $m\\rightarrow 0, \\, D$ will become very large, and the same will happen as $m\\rightarrow\\infty$. To find the minimum value, we substitute back into the expression for $D$: \\begin{align*} D &= a+b+a\\sqrt{\\frac{b}{a}}+b\\sqrt{\\frac{a}{b}}\\\\ &= a+b+2a^{\\frac{1}{2}}b^{\\frac{1}{2}}\\\\ &= (a^{\\frac{1}{2}}+b^{\\frac{1}{2}})^2 \\end{align*} as required. … and find in a similar form the minimum distance between $P$ and $Q$. The distance $PQ = \\sqrt{p^2+q^2}$. It’s helpful for us to note that the minimum value of $\\sqrt{p^2+q^2}$ coincides with the minimum value of $p^2+q^2$, so we can make our lives a little easier by concentrating on minimising $PQ^2 = p^2 + q^2 = \\left(\\frac{ma+b}{m}\\right)^2 + (ma+b)^2$ rather than $\\sqrt{p^2+q^2}$. \\begin{align*} PQ^2 &=\\left(\\frac{ma+b}{m}\\right)^2 + (ma+b)^2 \\\\ &= \\left(a+\\frac{b}{m}\\right)^2 + (ma + b)^2 \\\\ &=a^2 + 2\\frac{ab}{m} + \\frac{b^2}{m^2} + a^2m^2 + 2abm+ b^2\\\\ \\end{align*} Differentiating $PQ^2$ with respect to $m$ gives $\\frac{d}{dm}(PQ^2) =-\\frac{2ab}{m^2}-\\frac{2b^2}{m^3} + 2a^2m + 2ab$ The stationary points occur when \\begin{align*} -\\frac{2ab}{m^2}-\\frac{2b^2}{m^3} + 2a^2m + 2ab &=", "don't. A valid calculation of the limit would for example be: \\begin{align} &\\lim_{x\\to 0}{\\frac{\\tan{x}-\\sin x}{x^3}} \\\\ =& \\lim_{x\\to 0} \\frac {\\tan x}x \\frac{1-\\cos x}{x^2} \\\\ =& \\lim_{x\\to 0} \\frac {\\tan x}x \\lim_{x\\to 0}\\frac{1-\\cos x}{x^2} \\\\=&1\\cdot \\frac 1 2 =\\frac 1 2. \\end{align} Here we can calculate each limit separately and multiply them afterwards, as both exist. (In the first step I just took the\\tan x$factor outside). Hint: Use the Taylor development:$sin(x)=x-x^3/6+O(x^3)$,$tan(x)=x+x^3/3+O(x^3)$. • the order of the truncation error is surely$O(x^5)\\$, isn't it? Mar 9 '18 at 18:47 we have $$\\frac{-\\sin(x){\\cos(x)}+\\sin(x)}{\\cos(x)x^3}=\\frac{\\sin(x)(1-\\cos^2(x))}{x^3\\cos(x)(1+\\cos(x))}=\\frac{\\sin^3(x)}{x^3}\\frac{1}{\\cos(x)(1+\\cos(x))}$$", "# How to find the minimum value of $\\frac{x}{2x+3y}+\\frac{y}{y+z}+\\frac{z}{z+x}$? Let $x,y,z\\in [1,4]$ such that $x \\geq y$ and $x \\geq z$. Find the minimum value of this expression: $$P=\\frac{x}{2x+3y}+\\frac{y}{y+z}+\\frac{z}{z+x}$$ - Since $x \\not= 0$, denote $y= u x$ and $z=v x$. You are, then, looking for minimum in $\\frac{1}{4} \\le u \\le 1$ and $\\frac{1}{4} \\le v \\le 1$. – Sasha Aug 26 '11 at 6:06 @Sasha: following to your suggestion, we get $P=\\frac{1}{2+3u}+\\frac{u}{u+v}+\\frac{v}{v+1}$ What then shall we do? – KevinBui Aug 26 '11 at 9:48 or you can transform into polar coordinates, reduce the variables to two(same as DKhanh does), then apply second partial derivative test – newbie Aug 26 '11 at 10:54 ## 2 Answers As has been mentioned in comments, we want to find the minimum value of $$P=\\frac{1}{2+3u}+\\frac{u}{u+v}+\\frac{v}{v+1}$$ for $u,v\\in[\\frac{1}{4},1]$. Take partials of $P$ with respect to $u$ and $v$: \\begin{align} \\frac{\\partial P}{\\partial u}&=-\\frac{3}{(2+3u)^2}+\\frac{v}{(u+v)^2}\\\\ \\frac{\\partial P}{\\partial v}&=-\\frac{u}{(u+v)^2}+\\frac{1}{(v+1)^2} \\end{align} To find an interior extremum, we need $\\frac{\\partial P}{\\partial u}=\\frac{\\partial P}{\\partial v}=0$. In that case, we need $$0=u\\frac{\\partial P}{\\partial u}+v\\frac{\\partial P}{\\partial v}=-\\frac{3u}{(2+3u)^2}+\\frac{v}{(v+1)^2}$$ However, for $u,v\\in[\\frac{1}{4},1]$, we have $$\\begin{array}{c} \\frac{12}{121}\\le\\frac{3u}{(2+3u)^2}\\le\\frac{3}{25}\\\\ \\frac{4}{25}\\le\\frac{v}{(v+1)^2}\\le\\frac{1}{4} \\end{array}$$ Thus, $u\\frac{\\partial P}{\\partial u}+v\\frac{\\partial P}{\\partial v}\\ge\\frac{1}{25}$, so there can be no interior extremum. Because $(u,v)\\cdot\\nabla P=u\\frac{\\partial P}{\\partial u}+v\\frac{\\partial P}{\\partial v}\\ge\\frac{1}{25}$ everywhere in $[\\frac{1}{4},1]\\times[\\frac{1}{4},1]$, the minimum must be taken on the left or bottom edge of", "# Determine the Minimum Value #### anemone ##### MHB POTW Director Staff member Determine the minimum value of $$\\displaystyle \\frac{\\sec^4 a}{\\tan^2 b}+\\frac{\\sec^4 b}{\\tan^2 a}$$ over all $a,b \\ne \\frac{k \\pi}{2}$ where $k$ is in $Z$. #### Ackbach ##### Indicium Physicus Staff member Since $a$ and $b$ are symmetric in the expression, I will set $a=b$, and then define $$f(a)= \\frac{2 \\sec^{4}(a)}{ \\tan^{2}(a)}= \\frac{2}{ \\cos^{2}(a) \\sin^{2}(a)}.$$ Differentiating yields \\begin{align*} f'(a)&= \\frac{-2(-2 \\cos(a) \\sin^{3}(a)+2 \\sin(a) \\cos^{3}(a))}{ \\cos^{4}(a) \\sin^{4}(a)} \\\\ &= \\frac{-4 \\sin(a) \\cos(a) [ \\cos^{2}(a)- \\sin^{2}(a)]}{ \\sin^{4}(a) \\cos^{4}(a)} \\\\ &= \\frac{-4 \\cos(2a)}{ \\sin^{3}(a) \\cos^{3}(a)}. \\end{align*} Setting $f'(a)=0$ implies that $$2a= \\frac{(2j+1) \\pi}{2} \\implies a=\\frac{(2j+1) \\pi}{4}.$$ Just to make sure, let us look for values of $a\\in (0, \\pi) \\setminus \\{\\pi/2 \\}$. This turns out to require $j=0,1$, and therefore $a\\in \\{ \\pi/4, 3 \\pi/4 \\}$. Plugging either of these into $f$ yields the minimum value of $8$. To be complete, we should take the second derivative $f''(a)$, and show that it is positive at these values. I will leave that to the reader. #### Jester ##### Well-known member MHB Math Helper Here's my solution Using the usual identity we have $\\dfrac{(\\tan^2 a + 1)^2}{\\tan^2b}+\\dfrac{(\\tan^2 b + 1)^2}{\\tan^2a}.$ If we let $x = \\tan a$ and $y = \\tan b$ then we have $z = \\dfrac{(x^2 + 1)^2}{y^2}+\\dfrac{(y^2 + 1)^2}{x}.$ Using", "+0 # help 0 51 3 Find the minimum value of $$\\sin x + \\csc x)^2 + (\\cos x + \\sec x)^2$$ for $$0 < x < \\frac{\\pi}{2}.$$ Aug 6, 2022 #2 0 minimum value is 9 Aug 6, 2022 #3 0 Hello Guest! Remember: $$sin^2 x +cos^2 x = 1 \\\\ csc (x) = \\frac{1}{sin (x)} \\\\ sec(x) = \\frac{1}{cos (x)}$$ So: $$(sin(x)+csc(x))^2+(cos(x)+sec(x))^2 \\\\ =sin^2(x)+\\frac{1}{sin^2(x)}+2sin(x)\\frac{1}{sin(x)}+cos^2(x)+\\frac{1}{sin^2(x)}+2cos(x)\\frac{1}{cos(x)} \\\\ =(sin^2(x)+cos^2(x))+2+2+(\\frac{1}{sin^2(x)}+\\frac{1}{cos^2(x)}) \\\\ =1+4+\\frac{sin^2(x)+cos^2(x)}{sin^2(x)cos^2(x)} =5+\\frac{1}{(sin(x)cos(x))^2}$$ Now, we want the minimum value, so we have to minimize: $$\\frac{1}{(sin(x)cos(x))^2}$$ or equivalently, maximize: $$(sin(x)cos(x))^2$$ Recall: $$sin(2x)=2sin(x)cos(x) \\implies sin(x)cos(x)=\\frac{1}{2}sin(2x)$$ Hence, we want to maximize $$\\frac{1}{4}sin^2(2x)$$ $$0 < sin^2(2x) \\le 1$$ So, the upper bound is 1, which is satisfied when $$x=\\frac{\\pi}{4}$$ So, the maximum value of $$\\frac{1}{4}sin^2(2x)$$ is: $$\\dfrac{1}{4}*(1)=\\dfrac{1}{4}$$ Therefore, the minimum value of $$\\frac{1}{(sin(x)cos(x))^2}$$ is $$\\frac{1}{\\frac{1}{4}}=4$$ Thus, the minimum value of the whole expression in the question is: $$5+4=9$$ , as guest has found below. I hope this helps! Aug 7, 2022", "Scalar Product Optimization Problem Let $x,\\;y,\\;a,\\;b$ be real numbers such that $(x+1)^2+y^2=1$ and $(a-2)^2+b^2=4.$ Find the extreme values of the expression $ax+by.$ Solution 1 The minimum is $-8$ because the two given circles are externally tangent at the origin and $ax+by$ is the scalar product of two vectors with the end points on the circles. Hence $2*4*(-1)=-8.$ To determine the maximum, consider the points $A(x,y),$ $B(a,b),$ and $O(0,0).$ Note that $ax+by=\\overrightarrow{OA}\\cdot \\overrightarrow{OB}=OA\\cdot OB\\cdot\\cos\\angle AOB.$ Hence, to achieve maximum, it is necessary that $A$ and $B$ lie on the corresponding circles. Observe also that $\\angle AOB$ has to be acute which allows us to assume that the two points are in the upper half-plane. Thus, $A=(-1+\\cos u,\\;\\sin u)$ and $B=(2+2\\cos t,\\;2\\sin t),$ with $0\\lt t,\\; u\\lt\\pi.$ Setting $v=\\angle AOB,$ we get on one hand $\\displaystyle\\frac{A-O}{B-O}=\\frac{OA}{OB}(\\cos v+i\\sin v)$ while, on the other, \\displaystyle\\begin{align} \\frac{A-O}{B-O}&=\\frac{-1\\cos u+i\\sin u}{2+2\\cos t+2i\\sin t}\\\\ &=\\frac{i\\sin\\frac{u}{2}(\\cos\\frac{u}{2}+i\\sin\\frac{u}{2})}{2\\cos\\frac{t}{2}(\\cos\\frac{t}{2}+i\\sin\\frac{t}{2})}\\\\ &=\\frac{\\sin\\frac{u}{2}}{2\\cos\\frac{t}{2}}\\left(\\cos\\frac{\\pi+u-t}{2}+i\\sin\\frac{\\pi+u-t}{2}\\right). \\end{align} Since $\\displaystyle\\frac{\\sin\\frac{u}{2}}{2\\cos\\frac{t}{2}}\\gt 0,$ we may identify the angles: $\\displaystyle v=\\frac{\\pi +u-t}{2}$ and so $\\displaystyle \\cos v=\\sin\\frac{t-u}{2},$ implying, in particular, that $t\\gt u.$ Substituting, \\displaystyle\\begin{align} ax+by&=OA\\cdot OB\\cdot\\cos v=8\\sin\\frac{u}{2}\\cos\\frac{t}{2}\\sin\\frac{t-u}{2}\\\\ &=8\\sin\\frac{u}{2}\\sin\\frac{\\pi- t}{2}\\sin\\frac{t-u}{2}. \\end{align} But, $\\displaystyle 0\\lt \\frac{u}{2},\\;\\frac{\\pi-t}{2},\\;\\frac{t-u}{2}$ and $\\displaystyle\\frac{u}{2}+\\frac{\\pi-t}{2}+\\frac{t-u}{2}=\\frac{\\pi}{2}.$ By Jensen's inequality, $\\displaystyle 8\\sin\\frac{u}{2}\\sin\\frac{\\pi- t}{2}\\sin\\frac{t-u}{2}\\le 1.$ For equality we need $u=60^{\\circ}$ and $t=120^{\\circ}.$ Solution 2 Note first that the problem could be generalized. Indeed, assuming that $x,\\;y,\\;a,\\;b$ satisfy $(x+r)^2+y^2=r^2$ and $(a-s)^2+b^2=s^2,$ $(r,\\;", "&= \\int_0^1 ((x-c)^2+c^2)dx\\\\ &=\\left[ \\frac{(x-c)^3}{3} +c^2x \\right]_0^1\\\\ &=\\frac{(1-c)^3}{3} + c^2 + \\frac{c^3}{3}\\\\ &=2c^2-c+\\frac{1}{3}. \\end{align*} 1. What is the minimum value of $I(c)$ (as $c$ varies)? Using the expression for $I(c)$ we calculated above, and completing the square, we have \\begin{align*} I(c) &= 2c^2-c+\\frac{1}{3}\\\\ &= 2\\left(c-\\frac{1}{4}\\right)^2 - \\frac{1}{8} + \\frac{1}{3}\\\\ &= 2\\left(c-\\frac{1}{4}\\right)^2 + \\frac{5}{24}. \\end{align*} So the minimum value that $I(c)$ can take is $5/24$. (This occurs when $c=1/4$.) 1. What is the maximum value of $I(\\sin \\theta)$ as $\\theta$ varies? Note that from our work above, the MINIMUM value of $I(\\sin\\theta)$ happens when $\\theta = \\arcsin (0.25)$. Now $c=\\sin \\theta$ varies between between $-1$ and $1$, so we need to find the maximum value of $I$ over this range. We worked out earlier that the overall minimum of $I$ occurs at $c=1/4$, and since $I$ is quadratic, it is symmetric about this point, and increasing for larger and smaller values of $c$. So the maximum value of $I(c)$ over the range $-1 \\le c \\le 1$ occurs at $c=-1$, the furthest point from $x=1/4$ in the range. So the maximum value of $I(\\sin \\theta)$ is $I(-1)=10/3$.", "we are in a calculus mood, differentiate. But if we are in a trigonometric mood, we can use the identities $\\cos 2t=2\\cos^2 t-1$ and $\\sin 2t=2\\sin t\\cos t$ to rewrite our expression as $6\\cos 2t+\\frac{5}{2}\\sin 2t -4$, or equivalently $$\\frac{13}{2}\\left(\\frac{12}{13}\\cos 2t+\\frac{5}{13}\\sin 2t\\right)-4.$$ Let $\\varphi$ be an angle whose sine is $\\frac{12}{13}$ and whose cosine is $\\frac{5}{13}$. Then our function is $$\\frac{13}{2}\\sin(2t+\\varphi) -4.$$ For $t$ in our interval, $\\sin(2t+\\varphi)$ has maximum value $1$ and minimum value $-1$. From this we see that the maximum and minimum values of $f(x)$ are $\\dfrac{13}{2}-4$ and $-\\dfrac{13}{2}-4$.", "# Minimum value of $\\frac{((x-a)^2 + p )^{\\frac 12}}{v_1} + \\frac{( (x-b)^2 + q )^{\\frac12}} {v_2}$ I need to calculate the minimum value of the expression $y = \\frac{((x-a)^2 + p )^{\\frac 12}}{v_1} + \\frac{( (x-b)^2 + q )^{\\frac12}} {v_2}$ I’m calculating the first derivative , then calculting the corresponding value for $x$ . But the calculation is getting complex. Here $v_1 , v_2 , p$ and $q$ are positive integers greatest than $1$. I have also tried using $AM >= GM$ . but that too reduces finding maximum of a degree $4$ polynomial #### Solutions Collecting From Web of \"Minimum value of $\\frac{((x-a)^2 + p )^{\\frac 12}}{v_1} + \\frac{( (x-b)^2 + q )^{\\frac12}} {v_2}$\" It’s derivation of Snell’s Law from Fermat’s Principle Assume $a<b$, \\begin{align*} T(x) &= \\frac{\\sqrt{(x-a)^2+p}}{v_1}+\\frac{\\sqrt{(x-b)^2+q}}{v_2} \\\\ T'(x) &=\\frac{x-a}{v_1 \\sqrt{(x-a)^2+p}}- \\frac{b-x}{v_2 \\sqrt{(x-b)^2+q}} \\\\ 0 &= \\frac{\\sin \\theta_{1}}{v_{1}}-\\frac{\\sin \\theta_{2}}{v_{2}} \\\\ s &= \\sin \\theta_{1} \\\\ \\sin \\theta_{2} &= \\frac{v_{2} s}{v_{1}} \\\\ b-a &= \\sqrt{p} \\tan \\theta_{1}+\\sqrt{q} \\tan \\theta_{2} \\\\ b-a &= \\frac{s\\sqrt{p}}{\\sqrt{1-s^{2}}}+ \\frac{v_{2} s\\sqrt{q}}{\\sqrt{v_{1}^{2}-v_{2}^2 s^2}} T_{\\min} &= \\frac{\\sqrt{p}}{v_{1}\\cos \\theta_{1}}+\\frac{\\sqrt{q}}{v_{2}\\cos \\theta_{2}} \\\\ &= \\frac{\\sqrt{p}}{v_{1}\\sqrt{1-s^{2}}}+ \\frac{v_{1} \\sqrt{q}}{v_{2}\\sqrt{v_{1}^{2}-v_{2}^2 s^2}} \\end{align*} We seldom calculate the exact value of minimal $T$ but condition for minimum instead.", "of the expression [#permalink] ### Show Tags 31 Jul 2019, 07:41 3 EgmatQuantExpert wrote: Algerba Question Series- Question 3 What is the minimum value of the expression $$\\frac{1}{x^2}$$– $$\\frac{1}{2x}$$ + $$\\frac{5}{4}$$? A. $$\\frac{-5}{4}$$ B. $$\\frac{-19}{16}$$ C. 0 D. $$\\frac{19}{16}$$ E. $$\\frac{5}{4}$$ What is the minimum value of the expression $$\\frac{1}{x^2}– \\frac{1}{2x} + \\frac{5}{4}$$ $$t^2– \\frac{t}{2} + \\frac{5}{4}$$ where $$t=\\frac{1}{x}$$ $$t^2 - \\frac{2t}{4} + \\frac{1}{16} +\\frac{5}{4}-\\frac{1}{16}$$ $$=(t-\\frac{1}{4})^2 + \\frac{19}{16}$$ =\\frac{19}{16} since minimum value of expression is when $$(t-\\frac{1}{4})^2 =0$$ IMO D e-GMAT Representative Joined: 04 Jan 2015 Posts: 3239 Re: What is the minimum value of the expression [#permalink] ### Show Tags 05 Aug 2019, 21:01 Solution Given: • An expression $$\\frac{1}{x^2} – \\frac{1}{2x} + \\frac{5}{4}$$ To find: • The minimum value of the expression $$\\frac{1}{x^2} – \\frac{1}{2x} + \\frac{5}{4}$$ Approach and Working Out: • $$\\frac{1}{x^2} – \\frac{1}{2x} + \\frac{5}{4} = (\\frac{1}{x})^2 – 2 * \\frac{1}{x} * \\frac{1}{4} + (\\frac{1}{4})^2 + \\frac{19}{16} = (\\frac{1}{x} – \\frac{1}{4})^2 + \\frac{19}{16}$$ Therefore, the minimum value of the expression $$\\frac{1}{x^2} – \\frac{1}{2x} + \\frac{5}{4} = 0 + \\frac{19}{16} = \\frac{19}{16}$$ Hence, the correct answer is Option D. _________________ VP Joined: 18 Dec 2017 Posts: 1117 Location: United States (KS) GMAT 1: 600 Q46 V27 Re: What is the minimum value of the expression [#permalink] ### Show Tags 23 Oct 2019, 08:08 Any other", "b. Minimum at $$x=−\\frac{1}{2}$$ c. Concave up for all $$x$$ d. No inflection points 230) $$f(x)=x^3+x^4$$ For the following exercises, determine a. intervals where $$f$$ is increasing or decreasing, b. local minima and maxima of $$f,$$ c. intervals where $$f$$ is concave up and concave down, and d. the inflection points of $$f.$$ Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator. 231) [T] $$f(x)=sin(πx)−cos(πx)$$ over $$x=[−1,1]$$ Solution: a. Increases over $$−\\frac{1}{4}<x<\\frac{3}{4},$$ decreases over $$x>\\frac{3}{4}$$ and $$x<−\\frac{1}{4}$$ b. Minimum at $$x=−\\frac{1}{4}$$, maximum at $$x=\\frac{3}{4}$$ c. Concave up for $$−\\frac{3}{4}<x<\\frac{1}{4}$$, concave down for $$x<−\\frac{3}{4}$$ and $$x>\\frac{1}{4}$$ d. Inflection points at $$x=−\\frac{3}{4},x=\\frac{1}{4}$$ 232) [T] $$f(x)=x+sin(2x)$$ over $$x=[−\\frac{π}{2},\\frac{π}{2}]$$ 233) [T] $$f(x)=sinx+tanx$$ over $$(−\\frac{π}{2},\\frac{π}{2})$$ Solution: a. Increasing for all $$x$$ b. No local minimum or maximum c. Concave up for $$x>0$$, concave down for $$x<0$$ d. Inflection point at $$x=0$$ 234) [T] $$f(x)=(x−2)^2(x−4)^2$$ 235) [T] $$f(x)=\\frac{1}{1−x},x≠1$$ Solution: a. Increasing for all $$x$$ where defined b. No local minima or maxima c. Concave up for $$x<1$$; concave down for $$x>1$$ d. No inflection points in domain 236) [T] $$f(x)=\\frac{sinx}{x}$$ over $$x=[−2π,2π] [2π,0)∪(0,2π]$$ 237) $$f(x)=sin(x)e^x$$ over $$x=[−π,π]$$ Solution: a. Increasing over $$−\\frac{π}{4}<x<\\frac{3π}{4}$$, decreasing over $$x>\\frac{3π}{4},x<−\\frac{π}{4}$$ b. Minimum at $$x=−\\frac{π}{4}$$, maximum at $$x=\\frac{3π}{4}$$ c. Concave up for $$−\\frac{π}{2}<x<\\frac{π}{2}$$, concave down for $$x<−\\frac{π}{2},x>\\frac{π}{2}$$", "has a minimum point between $$0 < t < 1$$. If the value of the derivative at its minimum point is greater than zero, then the derivative is everywhere greater than zero and the curve is monotonic. Substituting our value for $$t$$ back into $$H(t)$$ and simplifying: \\begin{align*} H(t) > 0 \\\\ (1-t)^2H_0 + 2(1-t)tH_1 + t^2H_2 > 0 \\\\ 3f(1-t)^2 + 6( 1 - f - g )(1-t)t + 3gt^2 > 0 \\\\ ( 3f + 3g - 2 )t^2 - 2( 2f + g - 1 )t + f > 0 \\\\ ( 3f + 3g - 2 )( \\frac{ 2f + g - 1 }{ 3f + 3g - 2 } )^2 - 2( 2f + g - 1 )( \\frac{ 2f + g - 1 }{ 3f + 3g - 2 } ) + f > 0 \\\\ \\frac{ f( 3f + 3g - 2 ) - ( 2f + g - 1 )^2 }{ 3f + 3g - 2 } > 0 \\\\ \\frac{ -f^2 - g^2 + 2f + 2g - fg - 1 }{ 3f + 3g - 2 } > 0 \\end{align*} And remembering that we have already checked $$3g + 3g - 2$$ for condition 3 and found it to be positive: \\begin{align*} -f^2 - g^2 + 2f +", "absolute minimum among the \"positive parts\" of $S$. $\\textbf{Solving in terms of \\textit{r} then being clever}$ $$S=\\frac{1}{-r^2+r}$$ We seek the smallest positive value of $S$. We could use calculus like we did in the solution immediately above this one, but that's a lot of work and we don't have a ton of time. To minimize the positive value of this fraction, we must maximize the denominator. The denominator is a quadratic that opens down, so its maximum occurs at its vertex. The vertex of this quadratic occurs at $x=\\frac{1}{2}$ and $\\frac{1}{-(\\frac{1}{2})^2+\\frac{1}{2}}=\\boxed{\\textbf{(E)}\\ 4}$. ## Solution 3 $$\\textbf{Completing the Square and Quadratics}$$ Let $r$ be the common ratio. If the second term is $1$, the first must be $\\frac{1}{r}$. By the infinite geometric series formula, the sum must be $$S=\\frac{\\frac{1}{r}}{1-r}$$ This equals $\\frac{1}{r(1-r)}$. To find the minimum value of S, we must find the maximum value of the denominator, $r(1-r)$, which is $\\frac{1}{4}$, completing the square. Thus, the minimum value of $S$ is $\\boxed{\\textbf{(E)}\\ 4}$. ## Solution 4 (no AM-GM or Calculus) Our sequence is $a_1,1,...$. Since we know this is a converging series, our ratio is in $(0,1)$. Because the 2nd term in the sequence is a 1, the ratio must be $\\frac{1}{a_1}$, so we can write $S$ as $\\frac{a_1}{1-\\frac{1}{a_1}}$. With some manipulation we get $S=\\frac{1}{a_1-1}$. Since S has", "# Mock AIME 3 2006-2007 Problems/Problem 11 ## Problem If $x$ and $y$ are real numbers such that $2xy+2x^2=6+x^2+y^2$ find the minimum value of $(x^2+y^2)^2$. ## Solution 1 Factoring the LHS gives $2x(x+y)=6+x^2+y^2$. Now converting to polar: $2r\\cos(a)(r\\cos(a)+r\\sin(a))=6+r^2$ $2\\cos(a)(\\cos(a)+\\sin(a))=\\frac{6}{r^2}+1$ $2\\cos^2(a)-1+2\\cos(a)\\sin(a)=\\frac{6}{r^2}$ $\\cos(2a)+\\sin(2a)=\\frac{6}{r^2}$ $r^2=\\frac{6}{\\cos(2a)+\\sin(2a)}$ Since we want to find $(x^2+y^2)^2=r^4$, $r^4=\\frac{36}{(\\cos(2a)+\\sin(2a))^2}=\\frac{36}{1+2\\cos(2a)\\sin(2a)}=\\frac{36}{1+\\sin(4a)}$ Since we want the minimum of this expression, we need to maximize the denominator. The maximum of the sine function is 1 (one value of $a$ which produces this maximum is $a=\\frac{\\pi}{4}$) So the desired minimum is $\\frac{36}{2}=\\boxed{018}$ ## Solution 2 Since $x^2 + y^2 \\ge 0$, finding the minimum value of $(x^2 + y^2)^2$ is similar to finding that of $x^2 + y^2$. Let $x^2 + y^2 = a$, where $a$ is the minimum value. We can rewrite this as $y = -x^2 + a$ and $y = \\sqrt{-x^2 + a}$. $$2xy + 2x^2 = x^2 + y^2 + 6$$$$2x(\\sqrt{-x^2 + a}) + 2x^2 = x^2 + (-x^2 + a) + 6$$$$2x(\\sqrt{-x^2 + a}) + 2x^2 = a + 6$$$$2x^2 - (a + 6) = -2x(\\sqrt{x^2 + a})$$. $$4x^2 - 4(a + 6)x^2 + (a + 6)^2 = 4x^2(-x^2 + a)$$. $$8x^4 - 8(a + 3)x^2 + (a + 6)^2 = 0$$. We want this polynomial to factor in the form $(x^2 - r)(x^2 - s)$, where at", "# Find the range of $f(x)=11\\cos^2x+3\\sin^2x+6\\sin x\\cos x+5$ I'm trying to solve this problem. Find the range of $f(x)=11\\cos^2x+3\\sin^2x+6\\sin x\\cos x+5$ I have simplified this problem to $$f(x)= 8\\cos^2x+6\\sin x\\cos x+8$$ and tried working with $g(x)= 8\\cos^2x+6\\sin x\\cos x$. I factored out the $2\\cos x$ and rewrote the other factor as a linear combination of cosine. It reduces down to $$g(x)=10\\cos x\\cos\\left(x-\\tan^{-1}\\frac{3}{4}\\right)$$ • Have you tried finding the maximum and minimum valueof $f$? Jan 6 '18 at 15:44 Rewrite as following \\begin{align} f(x) & = 8 + 6 \\sin(x ) \\cos(x ) + 8\\cos^2(x) - 4 + 4 \\\\ &= 12 + 3 \\sin(2x) + 4 \\cos(2x) \\\\ &= 12 + 5 \\left(\\frac{3}{5} \\sin(2x) + \\frac{4}{5} \\cos(2x)\\right) \\\\ &= 12 + 5 \\sin(2x + \\arctan{\\tfrac{4}{3}}) \\end{align} Now its easy since $\\sin(...)$ always lies in $[-1,1]$, max/min values are $12 \\pm 5$. So maximum value is $17$ and minimum is $7$ write your function as $$f(x)=(\\sin(x)+3\\cos(x))^2+7$$ • This is good for minimum, but then for maximum? But answer is good! +1! Jan 6 '18 at 15:59 • You should write $3\\cos x + \\sin x$ in the form $R\\cos(x-\\alpha)$, where $R>0$ and $0 < \\alpha < \\frac{1}{2}\\pi$. Jan 6 '18 at 16:12 • you Can write your function as $$f(t)=4\\,{\\frac {4\\,{t}^{4}-3\\,{t}^{3}+3\\,t+4}{ \\left( {t}^{2}+1 \\right) ^{2 }}}$$ with $$\\tan(\\frac{x}{2})=t$$ and", "2K^{\\prime}} \\sin\\big(\\frac{x}{4k(2k-1)}\\big) \\cos\\big(\\frac{x(4k -1)}{4k(2k-1)}\\big)$$ balance out - if sine and cosine have the same sign for most of the pairs, then we could expect a local maximum, and local minimum, otherwise. The root values must correspond to more or less equal number of pairs with same and opposite signs. The corresponding number of elements in the sum above for values of $$x$$ ranges from 63 to 197. A plot below shows the distribution of function values by total number of elements in the sum We see that we have at least one root for $$x$$ values with the same number of $$2K^{\\prime}$$ pairs. Local Minima and Maxima of $$f$$ In this section we will show that the distribution of local minima and maxima values for $$f$$ is more or less homogeneous, i.e. evenly spread on $$x$$ axis. For that, let's consider the first derivative of $$f$$ $$f^{\\prime}(x) = \\sum_{k} \\frac{1}{2k-1} \\cos{\\frac{x}{2k-1}} - \\frac{1}{2k} \\cos{\\frac{x}{2k}} =$$ $$\\sum_{k} \\big(\\frac{1}{2k-1} - \\frac{1}{2k}\\big) \\cos{\\frac{x}{2k-1}} + \\frac{1}{2k} \\cos{\\frac{x}{2k-1}} - \\frac{1}{2k} \\cos{\\frac{x}{2k}} =$$ $$\\sum_{k} \\frac{1}{2k(2k-1)} \\cos{\\frac{x}{2k-1}} + \\frac{1}{2k} \\big(\\cos{\\frac{x}{2k-1}} - \\cos{\\frac{x}{2k}}\\big) =$$ $$\\sum_{k} \\frac{1}{2k(2k-1)} \\cos{\\frac{x}{2k-1}} - \\frac{1}{k} \\sin{\\frac{x}{4k(2k-1)}} \\sin{\\frac{x(4k-1)}{4k(2k-1)}}.$$ Now the last formula suggests that the derivative converges more rapidly than for the original function - see my previous answer. Indeed, the first term converges absolutely, and the last term suggests rapidly decresing oscillations.", "will assume $0 = a < b < c$. Let $t = \\frac{b}{c} \\in (0,1)$ and $u = t+t^{-1} \\in (2,\\infty)$, the target function $P$ becomes \\begin{align}P(u) = \\frac{1}{b^2} + \\frac{1}{(b-c)^2} + \\frac{1}{c^2} &= \\frac13\\frac{b^2+c^2}{c^2}\\left(\\frac{1}{t^2} + \\frac{1}{(1-t)^2} + 1\\right)\\\\ &= \\frac13(t + t^{-1})\\left(t^{-1} + \\frac{1}{t+t^{-1}-2} + t\\right)\\\\ &= \\frac{u}{3}\\left(u + \\frac{1}{u-2}\\right) \\end{align} \\tag{*1} Let $u_{min}$ be the $u$ corresponds to the absolute minimum. When $u = u_{min}$, we have $$P'(u) = \\frac23\\frac{(u-1)(u^2-3u+1)}{(u-2)^2} = 0 \\quad\\iff\\quad u = 1, \\frac{3\\pm \\sqrt{5}}{2}$$ Since $u_{min} > 2$, the last root is the only root relevant to us. This means $u_{min} = \\frac{3 + \\sqrt{5}}{2}$ and for any admissible $u$, we have the inequality: $$P = P(u) \\ge P(u_{min}) = \\frac{11+5\\sqrt{5}}{6} \\approx 3.696723314583158$$ The minimum value of $P$ over $\\bar{D}$ (and the infimum over $D$) is $\\frac{11+5\\sqrt{5}}{6}$.", "# Difference between revisions of \"Mock AIME 3 2006-2007 Problems/Problem 11\" ## Problem If $x$ and $y$ are real numbers such that $2xy+2x^2=6+x^2+y^2$ find the minimum value of $(x^2+y^2)^2$. ## Solution 1 Factoring the LHS gives $2x(x+y)=6+x^2+y^2$. Now converting to polar: $2r\\cos(a)(r\\cos(a)+r\\sin(a))=6+r^2$ $2\\cos(a)(\\cos(a)+\\sin(a))=\\frac{6}{r^2}+1$ $2\\cos^2(a)-1+2\\cos(a)\\sin(a)=\\frac{6}{r^2}$ $\\cos(2a)+\\sin(2a)=\\frac{6}{r^2}$ $r^2=\\frac{6}{\\cos(2a)+\\sin(2a)}$ Since we want to find $(x^2+y^2)^2=r^4$, $r^4=\\frac{36}{(\\cos(2a)+\\sin(2a))^2}=\\frac{36}{1+2\\cos(2a)\\sin(2a)}=\\frac{36}{1+\\sin(4a)}$ Since we want the minimum of this expression, we need to maximize the denominator. The maximum of the sine function is 1 (one value of $a$ which produces this maximum is $a=\\frac{\\pi}{4}$) So the desired minimum is $\\frac{36}{2}=\\boxed{018}$ ## Solution 2 Since $x^2 + y^2 \\ge 0$, finding the minimum value of $(x^2 + y^2)^2$ is similar to finding that of $x^2 + y^2$. Let $x^2 + y^2 = a$, where $a$ is the minimum value. We can rewrite this as $y = -x^2 + a$ and $y = \\sqrt{-x^2 + a}$. $$2xy + 2x^2 = x^2 + y^2 + 6$$$$2x(\\sqrt{-x^2 + a}) + 2x^2 = x^2 + (-x^2 + a) + 6$$$$2x(\\sqrt{-x^2 + a}) + 2x^2 = a + 6$$$$2x^2 - (a + 6) = -2x(\\sqrt{x^2 + a})$$. $$4x^2 - 4(a + 6)x^2 + (a + 6)^2 = 4x^2(-x^2 + a)$$. $$8x^4 - 8(a + 3)x^2 + (a + 6)^2 = 0$$. We want this polynomial to factor in the form $(x^2 - r)(x^2", "the smallest value of $x$ for the 'combined' curve, where any intersection is possible. This minima can be found by taking derivative of the equality : $x^2+x-3=a$ wrt $x$, i.e. $\\frac{\\delta a}{\\delta x} = 2x +1$, and taking second dervative to verify minima, get : $\\frac{\\delta^2 a}{\\delta^2 x} = 2$; a positive value; hence minima. This minima is given at : $x = -\\frac 12$. For this value of $x$, the value of $x^2 +x-3 = \\frac14-\\frac 12-3= -\\frac {13}{4}$. Next, need find maxima , as the left arm would intersect the curve; leading to 'cumulative' curve having a maximum value of $x$. This maxima can be found by taking derivative of the equality : $-x^2+x+3=a$ wrt $x$, i.e. $\\frac{\\delta a}{\\delta x} = -2x +1$, and taking second dervative to verify maxima, get : $\\frac{\\delta^2 a}{\\delta^2 x} = -2$; a negative value; hence maxima. This maxima is given at : $x = \\frac 12$. For this value of $x$, the value of $-x^2 +x-3 = -\\frac14+\\frac 12+3= \\frac14 +3 = \\frac{13}{4}$. Have two issues here: (i) as the solution (at the given link) takes the same value of minima $x=-\\frac12$ to get the maxima point for intersection of the left arm for positive value of $x$, by substituting in $-x^2+x+3$, i.e. $-\\frac14-\\frac12+3\\implies -\\frac34 +3\\implies \\frac94$. (ii) the value" ]

[ "# How to find the minimum value of $\\frac{x}{2x+3y}+\\frac{y}{y+z}+\\frac{z}{z+x}$? Let $x,y,z\\in [1,4]$ such that $x \\geq y$ and $x \\geq z$. Find the minimum value of this expression: $$P=\\frac{x}{2x+3y}+\\frac{y}{y+z}+\\frac{z}{z+x}$$ - Since $x \\not= 0$, denote $y= u x$ and $z=v x$. You are, then, looking for minimum in $\\frac{1}{4} \\le u \\le 1$ and $\\frac{1}{4} \\le v \\le 1$. – Sasha Aug 26 '11 at 6:06 @Sasha: following to your suggestion, we get $P=\\frac{1}{2+3u}+\\frac{u}{u+v}+\\frac{v}{v+1}$ What then shall we do? – KevinBui Aug 26 '11 at 9:48 or you can transform into polar coordinates, reduce the variables to two(same as DKhanh does), then apply second partial derivative test – newbie Aug 26 '11 at 10:54 ## 2 Answers As has been mentioned in comments, we want to find the minimum value of $$P=\\frac{1}{2+3u}+\\frac{u}{u+v}+\\frac{v}{v+1}$$ for $u,v\\in[\\frac{1}{4},1]$. Take partials of $P$ with respect to $u$ and $v$: \\begin{align} \\frac{\\partial P}{\\partial u}&=-\\frac{3}{(2+3u)^2}+\\frac{v}{(u+v)^2}\\\\ \\frac{\\partial P}{\\partial v}&=-\\frac{u}{(u+v)^2}+\\frac{1}{(v+1)^2} \\end{align} To find an interior extremum, we need $\\frac{\\partial P}{\\partial u}=\\frac{\\partial P}{\\partial v}=0$. In that case, we need $$0=u\\frac{\\partial P}{\\partial u}+v\\frac{\\partial P}{\\partial v}=-\\frac{3u}{(2+3u)^2}+\\frac{v}{(v+1)^2}$$ However, for $u,v\\in[\\frac{1}{4},1]$, we have $$\\begin{array}{c} \\frac{12}{121}\\le\\frac{3u}{(2+3u)^2}\\le\\frac{3}{25}\\\\ \\frac{4}{25}\\le\\frac{v}{(v+1)^2}\\le\\frac{1}{4} \\end{array}$$ Thus, $u\\frac{\\partial P}{\\partial u}+v\\frac{\\partial P}{\\partial v}\\ge\\frac{1}{25}$, so there can be no interior extremum. Because $(u,v)\\cdot\\nabla P=u\\frac{\\partial P}{\\partial u}+v\\frac{\\partial P}{\\partial v}\\ge\\frac{1}{25}$ everywhere in $[\\frac{1}{4},1]\\times[\\frac{1}{4},1]$, the minimum must be taken on the left or bottom edge of", "+0 # help 0 51 3 Find the minimum value of $$\\sin x + \\csc x)^2 + (\\cos x + \\sec x)^2$$ for $$0 < x < \\frac{\\pi}{2}.$$ Aug 6, 2022 #2 0 minimum value is 9 Aug 6, 2022 #3 0 Hello Guest! Remember: $$sin^2 x +cos^2 x = 1 \\\\ csc (x) = \\frac{1}{sin (x)} \\\\ sec(x) = \\frac{1}{cos (x)}$$ So: $$(sin(x)+csc(x))^2+(cos(x)+sec(x))^2 \\\\ =sin^2(x)+\\frac{1}{sin^2(x)}+2sin(x)\\frac{1}{sin(x)}+cos^2(x)+\\frac{1}{sin^2(x)}+2cos(x)\\frac{1}{cos(x)} \\\\ =(sin^2(x)+cos^2(x))+2+2+(\\frac{1}{sin^2(x)}+\\frac{1}{cos^2(x)}) \\\\ =1+4+\\frac{sin^2(x)+cos^2(x)}{sin^2(x)cos^2(x)} =5+\\frac{1}{(sin(x)cos(x))^2}$$ Now, we want the minimum value, so we have to minimize: $$\\frac{1}{(sin(x)cos(x))^2}$$ or equivalently, maximize: $$(sin(x)cos(x))^2$$ Recall: $$sin(2x)=2sin(x)cos(x) \\implies sin(x)cos(x)=\\frac{1}{2}sin(2x)$$ Hence, we want to maximize $$\\frac{1}{4}sin^2(2x)$$ $$0 < sin^2(2x) \\le 1$$ So, the upper bound is 1, which is satisfied when $$x=\\frac{\\pi}{4}$$ So, the maximum value of $$\\frac{1}{4}sin^2(2x)$$ is: $$\\dfrac{1}{4}*(1)=\\dfrac{1}{4}$$ Therefore, the minimum value of $$\\frac{1}{(sin(x)cos(x))^2}$$ is $$\\frac{1}{\\frac{1}{4}}=4$$ Thus, the minimum value of the whole expression in the question is: $$5+4=9$$ , as guest has found below. I hope this helps! Aug 7, 2022", "between the minimum value of $$L$$ and the minimizing value of $$a$$, which makes it very unlikely that the minimum value of $$L$$ has a convenient closed form. Set $$\\ell = 1$$ for convenience (one may show easily that the minimizing value of $$a$$ and the minimum value of $$L$$ scale in direct proportion with $$\\ell$$), and substitute $$x = \\ell/2a$$. Thus, we must minimize $$\\frac{2 \\sinh x + \\cosh x}{2x} = \\frac{3 e^x - e^{-x}}{4x}$$ and we know that the minimum $$L_\\text{min}$$ occurs at $$x = \\xi$$ where $$\\left. \\frac{d}{dx}\\right|_{x=\\xi} \\frac{3 e^x - e^{-x}}{4x} = \\frac{(3\\xi-3) e^{\\xi} + (\\xi+1) e^{-\\xi}}{4\\xi^2} = 0.$$ From this, we have $$3 e^\\xi - e^{-\\xi} = 3\\xi e^\\xi + \\xi e^{-\\xi}$$ and thus $$L_\\text{min} = \\frac{3 e^\\xi - e^{-\\xi}}{4\\xi} = \\frac{3 e^\\xi + e^{-\\xi}}{4} = \\cosh \\xi + \\frac{1}{2} \\sinh \\xi$$ But we know from (*) that $$\\xi L_\\text{min} = \\sinh \\xi + \\frac{1}{2} \\cosh \\xi$$ By adding or subtracting these equations, we get the system: \\begin{align*} (1 + \\xi) L_\\text{min} &= \\frac{3}{2} e^{\\xi} \\\\ (1 - \\xi) L_\\text{min} &= \\frac{1}{2} e^{-\\xi} \\\\ \\end{align*} Multiplying these equations gives $$(1 - \\xi^2) L_\\text{min}^2 = \\frac{3}{4}.$$ • The answer of $L$ is given in the book in terms of $l$. which means you can easily find $L/l$ without finding it numerically. and hence we", "# Minimum value of $\\frac{((x-a)^2 + p )^{\\frac 12}}{v_1} + \\frac{( (x-b)^2 + q )^{\\frac12}} {v_2}$ I need to calculate the minimum value of the expression $y = \\frac{((x-a)^2 + p )^{\\frac 12}}{v_1} + \\frac{( (x-b)^2 + q )^{\\frac12}} {v_2}$ I’m calculating the first derivative , then calculting the corresponding value for $x$ . But the calculation is getting complex. Here $v_1 , v_2 , p$ and $q$ are positive integers greatest than $1$. I have also tried using $AM >= GM$ . but that too reduces finding maximum of a degree $4$ polynomial #### Solutions Collecting From Web of \"Minimum value of $\\frac{((x-a)^2 + p )^{\\frac 12}}{v_1} + \\frac{( (x-b)^2 + q )^{\\frac12}} {v_2}$\" It’s derivation of Snell’s Law from Fermat’s Principle Assume $a<b$, \\begin{align*} T(x) &= \\frac{\\sqrt{(x-a)^2+p}}{v_1}+\\frac{\\sqrt{(x-b)^2+q}}{v_2} \\\\ T'(x) &=\\frac{x-a}{v_1 \\sqrt{(x-a)^2+p}}- \\frac{b-x}{v_2 \\sqrt{(x-b)^2+q}} \\\\ 0 &= \\frac{\\sin \\theta_{1}}{v_{1}}-\\frac{\\sin \\theta_{2}}{v_{2}} \\\\ s &= \\sin \\theta_{1} \\\\ \\sin \\theta_{2} &= \\frac{v_{2} s}{v_{1}} \\\\ b-a &= \\sqrt{p} \\tan \\theta_{1}+\\sqrt{q} \\tan \\theta_{2} \\\\ b-a &= \\frac{s\\sqrt{p}}{\\sqrt{1-s^{2}}}+ \\frac{v_{2} s\\sqrt{q}}{\\sqrt{v_{1}^{2}-v_{2}^2 s^2}} T_{\\min} &= \\frac{\\sqrt{p}}{v_{1}\\cos \\theta_{1}}+\\frac{\\sqrt{q}}{v_{2}\\cos \\theta_{2}} \\\\ &= \\frac{\\sqrt{p}}{v_{1}\\sqrt{1-s^{2}}}+ \\frac{v_{1} \\sqrt{q}}{v_{2}\\sqrt{v_{1}^{2}-v_{2}^2 s^2}} \\end{align*} We seldom calculate the exact value of minimal $T$ but condition for minimum instead.", "# Mock AIME 3 2006-2007 Problems/Problem 11 ## Problem If $x$ and $y$ are real numbers such that $2xy+2x^2=6+x^2+y^2$ find the minimum value of $(x^2+y^2)^2$. ## Solution 1 Factoring the LHS gives $2x(x+y)=6+x^2+y^2$. Now converting to polar: $2r\\cos(a)(r\\cos(a)+r\\sin(a))=6+r^2$ $2\\cos(a)(\\cos(a)+\\sin(a))=\\frac{6}{r^2}+1$ $2\\cos^2(a)-1+2\\cos(a)\\sin(a)=\\frac{6}{r^2}$ $\\cos(2a)+\\sin(2a)=\\frac{6}{r^2}$ $r^2=\\frac{6}{\\cos(2a)+\\sin(2a)}$ Since we want to find $(x^2+y^2)^2=r^4$, $r^4=\\frac{36}{(\\cos(2a)+\\sin(2a))^2}=\\frac{36}{1+2\\cos(2a)\\sin(2a)}=\\frac{36}{1+\\sin(4a)}$ Since we want the minimum of this expression, we need to maximize the denominator. The maximum of the sine function is 1 (one value of $a$ which produces this maximum is $a=\\frac{\\pi}{4}$) So the desired minimum is $\\frac{36}{2}=\\boxed{018}$ ## Solution 2 Since $x^2 + y^2 \\ge 0$, finding the minimum value of $(x^2 + y^2)^2$ is similar to finding that of $x^2 + y^2$. Let $x^2 + y^2 = a$, where $a$ is the minimum value. We can rewrite this as $y = -x^2 + a$ and $y = \\sqrt{-x^2 + a}$. $$2xy + 2x^2 = x^2 + y^2 + 6$$$$2x(\\sqrt{-x^2 + a}) + 2x^2 = x^2 + (-x^2 + a) + 6$$$$2x(\\sqrt{-x^2 + a}) + 2x^2 = a + 6$$$$2x^2 - (a + 6) = -2x(\\sqrt{x^2 + a})$$. $$4x^2 - 4(a + 6)x^2 + (a + 6)^2 = 4x^2(-x^2 + a)$$. $$8x^4 - 8(a + 3)x^2 + (a + 6)^2 = 0$$. We want this polynomial to factor in the form $(x^2 - r)(x^2 - s)$, where at", "# Determine the Minimum Value #### anemone ##### MHB POTW Director Staff member Determine the minimum value of $$\\displaystyle \\frac{\\sec^4 a}{\\tan^2 b}+\\frac{\\sec^4 b}{\\tan^2 a}$$ over all $a,b \\ne \\frac{k \\pi}{2}$ where $k$ is in $Z$. #### Ackbach ##### Indicium Physicus Staff member Since $a$ and $b$ are symmetric in the expression, I will set $a=b$, and then define $$f(a)= \\frac{2 \\sec^{4}(a)}{ \\tan^{2}(a)}= \\frac{2}{ \\cos^{2}(a) \\sin^{2}(a)}.$$ Differentiating yields \\begin{align*} f'(a)&= \\frac{-2(-2 \\cos(a) \\sin^{3}(a)+2 \\sin(a) \\cos^{3}(a))}{ \\cos^{4}(a) \\sin^{4}(a)} \\\\ &= \\frac{-4 \\sin(a) \\cos(a) [ \\cos^{2}(a)- \\sin^{2}(a)]}{ \\sin^{4}(a) \\cos^{4}(a)} \\\\ &= \\frac{-4 \\cos(2a)}{ \\sin^{3}(a) \\cos^{3}(a)}. \\end{align*} Setting $f'(a)=0$ implies that $$2a= \\frac{(2j+1) \\pi}{2} \\implies a=\\frac{(2j+1) \\pi}{4}.$$ Just to make sure, let us look for values of $a\\in (0, \\pi) \\setminus \\{\\pi/2 \\}$. This turns out to require $j=0,1$, and therefore $a\\in \\{ \\pi/4, 3 \\pi/4 \\}$. Plugging either of these into $f$ yields the minimum value of $8$. To be complete, we should take the second derivative $f''(a)$, and show that it is positive at these values. I will leave that to the reader. #### Jester ##### Well-known member MHB Math Helper Here's my solution Using the usual identity we have $\\dfrac{(\\tan^2 a + 1)^2}{\\tan^2b}+\\dfrac{(\\tan^2 b + 1)^2}{\\tan^2a}.$ If we let $x = \\tan a$ and $y = \\tan b$ then we have $z = \\dfrac{(x^2 + 1)^2}{y^2}+\\dfrac{(y^2 + 1)^2}{x}.$ Using", "don't. A valid calculation of the limit would for example be: \\begin{align} &\\lim_{x\\to 0}{\\frac{\\tan{x}-\\sin x}{x^3}} \\\\ =& \\lim_{x\\to 0} \\frac {\\tan x}x \\frac{1-\\cos x}{x^2} \\\\ =& \\lim_{x\\to 0} \\frac {\\tan x}x \\lim_{x\\to 0}\\frac{1-\\cos x}{x^2} \\\\=&1\\cdot \\frac 1 2 =\\frac 1 2. \\end{align} Here we can calculate each limit separately and multiply them afterwards, as both exist. (In the first step I just took the\\tan x$factor outside). Hint: Use the Taylor development:$sin(x)=x-x^3/6+O(x^3)$,$tan(x)=x+x^3/3+O(x^3)$. • the order of the truncation error is surely$O(x^5)\\$, isn't it? Mar 9 '18 at 18:47 we have $$\\frac{-\\sin(x){\\cos(x)}+\\sin(x)}{\\cos(x)x^3}=\\frac{\\sin(x)(1-\\cos^2(x))}{x^3\\cos(x)(1+\\cos(x))}=\\frac{\\sin^3(x)}{x^3}\\frac{1}{\\cos(x)(1+\\cos(x))}$$", "# Difference between revisions of \"Mock AIME 3 2006-2007 Problems/Problem 11\" ## Problem If $x$ and $y$ are real numbers such that $2xy+2x^2=6+x^2+y^2$ find the minimum value of $(x^2+y^2)^2$. ## Solution 1 Factoring the LHS gives $2x(x+y)=6+x^2+y^2$. Now converting to polar: $2r\\cos(a)(r\\cos(a)+r\\sin(a))=6+r^2$ $2\\cos(a)(\\cos(a)+\\sin(a))=\\frac{6}{r^2}+1$ $2\\cos^2(a)-1+2\\cos(a)\\sin(a)=\\frac{6}{r^2}$ $\\cos(2a)+\\sin(2a)=\\frac{6}{r^2}$ $r^2=\\frac{6}{\\cos(2a)+\\sin(2a)}$ Since we want to find $(x^2+y^2)^2=r^4$, $r^4=\\frac{36}{(\\cos(2a)+\\sin(2a))^2}=\\frac{36}{1+2\\cos(2a)\\sin(2a)}=\\frac{36}{1+\\sin(4a)}$ Since we want the minimum of this expression, we need to maximize the denominator. The maximum of the sine function is 1 (one value of $a$ which produces this maximum is $a=\\frac{\\pi}{4}$) So the desired minimum is $\\frac{36}{2}=\\boxed{018}$ ## Solution 2 Since $x^2 + y^2 \\ge 0$, finding the minimum value of $(x^2 + y^2)^2$ is similar to finding that of $x^2 + y^2$. Let $x^2 + y^2 = a$, where $a$ is the minimum value. We can rewrite this as $y = -x^2 + a$ and $y = \\sqrt{-x^2 + a}$. $$2xy + 2x^2 = x^2 + y^2 + 6$$$$2x(\\sqrt{-x^2 + a}) + 2x^2 = x^2 + (-x^2 + a) + 6$$$$2x(\\sqrt{-x^2 + a}) + 2x^2 = a + 6$$$$2x^2 - (a + 6) = -2x(\\sqrt{x^2 + a})$$. $$4x^2 - 4(a + 6)x^2 + (a + 6)^2 = 4x^2(-x^2 + a)$$. $$8x^4 - 8(a + 3)x^2 + (a + 6)^2 = 0$$. We want this polynomial to factor in the form $(x^2 - r)(x^2", "# Find the range of $f(x)=11\\cos^2x+3\\sin^2x+6\\sin x\\cos x+5$ I'm trying to solve this problem. Find the range of $f(x)=11\\cos^2x+3\\sin^2x+6\\sin x\\cos x+5$ I have simplified this problem to $$f(x)= 8\\cos^2x+6\\sin x\\cos x+8$$ and tried working with $g(x)= 8\\cos^2x+6\\sin x\\cos x$. I factored out the $2\\cos x$ and rewrote the other factor as a linear combination of cosine. It reduces down to $$g(x)=10\\cos x\\cos\\left(x-\\tan^{-1}\\frac{3}{4}\\right)$$ • Have you tried finding the maximum and minimum valueof $f$? Jan 6 '18 at 15:44 Rewrite as following \\begin{align} f(x) & = 8 + 6 \\sin(x ) \\cos(x ) + 8\\cos^2(x) - 4 + 4 \\\\ &= 12 + 3 \\sin(2x) + 4 \\cos(2x) \\\\ &= 12 + 5 \\left(\\frac{3}{5} \\sin(2x) + \\frac{4}{5} \\cos(2x)\\right) \\\\ &= 12 + 5 \\sin(2x + \\arctan{\\tfrac{4}{3}}) \\end{align} Now its easy since $\\sin(...)$ always lies in $[-1,1]$, max/min values are $12 \\pm 5$. So maximum value is $17$ and minimum is $7$ write your function as $$f(x)=(\\sin(x)+3\\cos(x))^2+7$$ • This is good for minimum, but then for maximum? But answer is good! +1! Jan 6 '18 at 15:59 • You should write $3\\cos x + \\sin x$ in the form $R\\cos(x-\\alpha)$, where $R>0$ and $0 < \\alpha < \\frac{1}{2}\\pi$. Jan 6 '18 at 16:12 • you Can write your function as $$f(t)=4\\,{\\frac {4\\,{t}^{4}-3\\,{t}^{3}+3\\,t+4}{ \\left( {t}^{2}+1 \\right) ^{2 }}}$$ with $$\\tan(\\frac{x}{2})=t$$ and", "Therefore, we want to minimise $D$ with respect to $m$. Now differentiating, we have $\\frac{dD}{d m}=a-\\frac{b}{m^2}.$ We get a stationary point when $\\dfrac{dD}{dm}=0$. That gives $a=\\frac{b}{m^2} \\Longleftrightarrow m=\\pm \\sqrt{\\frac{b}{a}}.$ We are told that $m>0$, so we choose the positive root $m=\\sqrt{\\frac{b}{a}}$. This value must be a minimum value as there is only one stationary value, and by considering our expression for $D$, we can see that as $m\\rightarrow 0, \\, D$ will become very large, and the same will happen as $m\\rightarrow\\infty$. To find the minimum value, we substitute back into the expression for $D$: \\begin{align*} D &= a+b+a\\sqrt{\\frac{b}{a}}+b\\sqrt{\\frac{a}{b}}\\\\ &= a+b+2a^{\\frac{1}{2}}b^{\\frac{1}{2}}\\\\ &= (a^{\\frac{1}{2}}+b^{\\frac{1}{2}})^2 \\end{align*} as required. … and find in a similar form the minimum distance between $P$ and $Q$. The distance $PQ = \\sqrt{p^2+q^2}$. It’s helpful for us to note that the minimum value of $\\sqrt{p^2+q^2}$ coincides with the minimum value of $p^2+q^2$, so we can make our lives a little easier by concentrating on minimising $PQ^2 = p^2 + q^2 = \\left(\\frac{ma+b}{m}\\right)^2 + (ma+b)^2$ rather than $\\sqrt{p^2+q^2}$. \\begin{align*} PQ^2 &=\\left(\\frac{ma+b}{m}\\right)^2 + (ma+b)^2 \\\\ &= \\left(a+\\frac{b}{m}\\right)^2 + (ma + b)^2 \\\\ &=a^2 + 2\\frac{ab}{m} + \\frac{b^2}{m^2} + a^2m^2 + 2abm+ b^2\\\\ \\end{align*} Differentiating $PQ^2$ with respect to $m$ gives $\\frac{d}{dm}(PQ^2) =-\\frac{2ab}{m^2}-\\frac{2b^2}{m^3} + 2a^2m + 2ab$ The stationary points occur when \\begin{align*} -\\frac{2ab}{m^2}-\\frac{2b^2}{m^3} + 2a^2m + 2ab &=", "of the expression [#permalink] ### Show Tags 31 Jul 2019, 07:41 3 EgmatQuantExpert wrote: Algerba Question Series- Question 3 What is the minimum value of the expression $$\\frac{1}{x^2}$$– $$\\frac{1}{2x}$$ + $$\\frac{5}{4}$$? A. $$\\frac{-5}{4}$$ B. $$\\frac{-19}{16}$$ C. 0 D. $$\\frac{19}{16}$$ E. $$\\frac{5}{4}$$ What is the minimum value of the expression $$\\frac{1}{x^2}– \\frac{1}{2x} + \\frac{5}{4}$$ $$t^2– \\frac{t}{2} + \\frac{5}{4}$$ where $$t=\\frac{1}{x}$$ $$t^2 - \\frac{2t}{4} + \\frac{1}{16} +\\frac{5}{4}-\\frac{1}{16}$$ $$=(t-\\frac{1}{4})^2 + \\frac{19}{16}$$ =\\frac{19}{16} since minimum value of expression is when $$(t-\\frac{1}{4})^2 =0$$ IMO D e-GMAT Representative Joined: 04 Jan 2015 Posts: 3239 Re: What is the minimum value of the expression [#permalink] ### Show Tags 05 Aug 2019, 21:01 Solution Given: • An expression $$\\frac{1}{x^2} – \\frac{1}{2x} + \\frac{5}{4}$$ To find: • The minimum value of the expression $$\\frac{1}{x^2} – \\frac{1}{2x} + \\frac{5}{4}$$ Approach and Working Out: • $$\\frac{1}{x^2} – \\frac{1}{2x} + \\frac{5}{4} = (\\frac{1}{x})^2 – 2 * \\frac{1}{x} * \\frac{1}{4} + (\\frac{1}{4})^2 + \\frac{19}{16} = (\\frac{1}{x} – \\frac{1}{4})^2 + \\frac{19}{16}$$ Therefore, the minimum value of the expression $$\\frac{1}{x^2} – \\frac{1}{2x} + \\frac{5}{4} = 0 + \\frac{19}{16} = \\frac{19}{16}$$ Hence, the correct answer is Option D. _________________ VP Joined: 18 Dec 2017 Posts: 1117 Location: United States (KS) GMAT 1: 600 Q46 V27 Re: What is the minimum value of the expression [#permalink] ### Show Tags 23 Oct 2019, 08:08 Any other", "we are in a calculus mood, differentiate. But if we are in a trigonometric mood, we can use the identities $\\cos 2t=2\\cos^2 t-1$ and $\\sin 2t=2\\sin t\\cos t$ to rewrite our expression as $6\\cos 2t+\\frac{5}{2}\\sin 2t -4$, or equivalently $$\\frac{13}{2}\\left(\\frac{12}{13}\\cos 2t+\\frac{5}{13}\\sin 2t\\right)-4.$$ Let $\\varphi$ be an angle whose sine is $\\frac{12}{13}$ and whose cosine is $\\frac{5}{13}$. Then our function is $$\\frac{13}{2}\\sin(2t+\\varphi) -4.$$ For $t$ in our interval, $\\sin(2t+\\varphi)$ has maximum value $1$ and minimum value $-1$. From this we see that the maximum and minimum values of $f(x)$ are $\\dfrac{13}{2}-4$ and $-\\dfrac{13}{2}-4$.", "&= \\int_0^1 ((x-c)^2+c^2)dx\\\\ &=\\left[ \\frac{(x-c)^3}{3} +c^2x \\right]_0^1\\\\ &=\\frac{(1-c)^3}{3} + c^2 + \\frac{c^3}{3}\\\\ &=2c^2-c+\\frac{1}{3}. \\end{align*} 1. What is the minimum value of $I(c)$ (as $c$ varies)? Using the expression for $I(c)$ we calculated above, and completing the square, we have \\begin{align*} I(c) &= 2c^2-c+\\frac{1}{3}\\\\ &= 2\\left(c-\\frac{1}{4}\\right)^2 - \\frac{1}{8} + \\frac{1}{3}\\\\ &= 2\\left(c-\\frac{1}{4}\\right)^2 + \\frac{5}{24}. \\end{align*} So the minimum value that $I(c)$ can take is $5/24$. (This occurs when $c=1/4$.) 1. What is the maximum value of $I(\\sin \\theta)$ as $\\theta$ varies? Note that from our work above, the MINIMUM value of $I(\\sin\\theta)$ happens when $\\theta = \\arcsin (0.25)$. Now $c=\\sin \\theta$ varies between between $-1$ and $1$, so we need to find the maximum value of $I$ over this range. We worked out earlier that the overall minimum of $I$ occurs at $c=1/4$, and since $I$ is quadratic, it is symmetric about this point, and increasing for larger and smaller values of $c$. So the maximum value of $I(c)$ over the range $-1 \\le c \\le 1$ occurs at $c=-1$, the furthest point from $x=1/4$ in the range. So the maximum value of $I(\\sin \\theta)$ is $I(-1)=10/3$.", "This is the maximum value of the expression $\\mathbf{4.)}$ The minimum value will be achieved for a fraction when the numerator is as small as possible and denominator is as large as possible Again, by an overall similar consideration, we have $\\frac{a^2+bc}{b+c} \\geq \\frac{a^2+c^2}{2b}$ is the minimum value of this expression Similarly $\\frac{b^2+ca}{c+a} \\geq \\frac{b^2+c^2}{2a}$ and $\\frac{c^2+ab}{a+b} \\geq \\frac{c^2+b^2}{2a}$ Hence But we have $a^2+b^2 \\geq 2b^2$, $c^2+b^2 \\geq 2c^2$ Hence, Therefore, $\\frac{a^2+bc}{b+c} \\cdot \\frac{b^2+ca}{c+a} \\cdot \\frac{c^2+ab}{a+b} \\geq \\frac{bc^4}{a^2}$ $\\mathbf{5.)}$ And finally, we have $a^3 \\geq abc$ is the maximum value of $abc$ The minimum value of $\\frac{a^3+b^3+c^3}{3}$ is $c^3$ which is greater than $a^3$ which is the maximum value of $\\frac{a^2+bc}{b+c}\\cdot \\frac{b^2+ca}{c+a} \\cdot \\frac{c^2+ab}{a+b}$ That is, So that $\\frac{a^3+b^3+c^3}{3} \\geq \\frac{a^2+bc}{b+c}\\cdot \\frac{b^2+ca}{c+a} \\cdot \\frac{c^2+ab}{a+b}$ Again we have $\\frac{a^2+bc}{b+c}\\cdot \\frac{b^2+ca}{c+a} \\cdot \\frac{c^2+ab}{a+b} \\geq \\frac{bc^4}{a^2}$ • In the fourth section, the inequality $\\frac{a^2 + bc}{b+c} > \\frac{a^2+b^2} {2b}$ is incorrect. It should be $\\frac{a^2 + bc}{b+c} > \\frac{a^2+c^2} {2b}$ – Zack Ni Aug 16 '16 at 3:01 • Greatly appreciate all of the edits to allow for the ease of viewing, Can anyone comment if my thought process is somewhat correct ? Also any other known proofs that would apply to this inequality? – LKB Aug 16 '16 at 3:27 • No worries @LKB – bigfocalchord Aug 16 '16 at", "# Find the least value for $\\sin x - \\cos^2 x -1$ Find all the values of $x$ for which the function $y = \\sin x - \\cos^2 x -1$ assumes the least value. What is that value? At first I found the first derivative to be $y' = \\cos x + 2 \\sin x \\cos x$. Critical point $0$, $-π/6$ (principal) $y'' = - \\sin x + 2(\\cos 2x)$ Then substitution of $x$ by critical points I found minima. But my answer is incorrect. Correct minimum value is $-9/4$ • Use $\\cos^2(x)=1- \\sin^2(x)$ and complete the square ... – Donald Splutterwit Aug 2 '18 at 21:21 Yes your way is correct indeed for $x=-\\pi/6$ we have $$f(x)=-\\frac12-\\frac34-1=-\\frac94$$ As an alternative, recall that $\\cos^2 x=1-\\sin^2 x$ and therefore we have $$f(x)=\\sin x -\\cos^2 x-1=\\sin^2 x+\\sin x-2$$ then set $t=\\sin x$ and consider the parabola (concave up) $$g(t)=t^2+t-2 \\implies g'(t)=2t+1=0 \\implies t=-\\frac 12$$ • Note to OP: Care should be taken in general when proceeding this way. Supposing the original function to have been $y = 3\\sin x-\\cos^2 x-1$, this approach would have selected $t = -\\frac32$, which is not valid for real $x$. – Brian Tung Aug 2 '18 at 21:25 • @BrianTung Yes indeed in that case we need to find the max/min for the extrema values", "of A. Ex. 10.30 | Q 17 | Page 54 If sec 2A = cosec (A – 42°) where 2A is an acute angle. Find the value of A. #### Chapter 10: Trigonometric Ratios solutions [Pages 55 - 56] Q 1 | Page 55 Write the maximum and minimum values of sin θ. Q 1 | Page 55 Write the maximum and minimum values of sin θ. Q 2 | Page 55 Write the maximum and minimum values of cos θ. Q 2 | Page 55 Write the maximum and minimum values of cos θ. Q 3 | Page 55 What is the maximum value of $\\frac{1}{\\sec \\theta}$ Q 3 | Page 55 What is the maximum value of $\\frac{1}{\\sec \\theta}$ Q 4 | Page 55 What is the maximum value of $\\frac{1}{\\sec \\theta}$ Q 4 | Page 55 What is the maximum value of $\\frac{1}{\\sec \\theta}$ Q 5 | Page 55 If $\\tan \\theta = \\frac{4}{5}$ find the value of $\\frac{\\cos \\theta - \\sin \\theta}{\\cos \\theta + \\sin \\theta}$ Q 5 | Page 55 If $\\tan \\theta = \\frac{4}{5}$ find the value of $\\frac{\\cos \\theta - \\sin \\theta}{\\cos \\theta + \\sin \\theta}$ Q 6 | Page 55 If $\\cos \\theta = \\frac{2}{3}$ find the value of $\\frac{\\sec \\theta - 1}{\\sec \\theta + 1}$ Q 6 | Page", "# Calculate the minimum value of $\\frac{x^2 + 4}{y^2 + 1}$ where $1 \\le y \\le 2$ and $2y \\le xy + 2$. $$x$$ and $$y$$ are reals such that $$1 \\le y \\le 2$$ and $$2y \\le xy + 2$$. Calculate the minimum value of $$\\large \\frac{x^2 + 4}{y^2 + 1}$$ This problem is adapted from a recent competition... It really is. There must be better answers. We first need to prove that $$x \\ge 0$$. It is true that $$2y \\le xy + 2 \\iff 0 \\le \\dfrac{2(y - 1)}{y} \\le x$$ (because $$1 \\le y$$). Moreover, y \\le 2 \\implies \\left\\{ \\begin{align} y^2 &\\le 2y\\\\ xy + 2 &\\le 2x + 2 \\end{align} \\right. (x, y \\ge 0) We have that $$\\frac{x^2 + 4}{y^2 + 1} \\ge \\frac{2x + 3}{2y + 1} \\ge \\frac{2x + 3}{xy + 3} \\ge \\frac{2x + 3}{2x + 3} = 1$$ The equality sign occurs when $$x = 1$$ and $$y = 2$$. Let $$x=1$$ and $$y=2$$. Thus,we get a value $$1$$. We'll prove that it's a minimal value. Indeed, since $$x\\geq\\frac{2(y-1)}{y}\\geq0,$$ we obtain: $$\\frac{x^2+4}{y^2+1}-1=\\frac{\\left(\\frac{2(y-1)}{y}\\right)^2+4}{y^2+1}-1=\\frac{(2-y)(y^3+2y^2-3y+2)}{y^2(y^2+1)}\\geq0$$ and we are done! $$x y \\ge 2y-2\\Rightarrow x\\ge 2-\\frac {2}{\\max y}\\ge 1$$ because $$y > 0$$ now $$\\frac{\\min x^2+4}{\\max y^2+1}\\le \\frac{x^2+4}{y^2+1}$$ hence $$\\frac 55 = 1 \\le \\frac{x^2+4}{y^2+1}$$", "# Determine Minimum Value. Find the minimum value of $$\\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}$$ for $x>0$. When $x=1$, $$\\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=6$$ I tried to plot some points on a graph and I observed that the minimum value is $6$. Any hints would be sufficient. Thanks I think differentiation would be really complicated • This is an old Putnam competition question. – Cyclohexanol. Jul 2 '18 at 17:36 $$\\dfrac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=\\dfrac{(x+1/x)^6-(x^3+1/x^3)^2}{(x+1/x)^3+(x^3+1/x^3)}=(x+1/x)^3-(x^3+1/x^3)$$ Therefore your function reduces to $3(x+1/x)$ which has its minimum when $x=1.$ Hint-$$\\frac{(x+\\frac1x)^6-(x^6+\\frac{1}{x^6})-2}{(x+\\frac{1}{x})^3+(x^3+\\frac{1}{x^3})} = \\frac{(x+\\frac{1}{x})^6-(x^3+\\frac{1}{x^3})^2}{(x+\\frac{1}{x})^3+(x^3+\\frac{1}{x^3})}=(x+\\frac{1}{x})^3-(x^3+\\frac{1}{x^3})=3(x+1/x)$$ As $x>0$, $$f(x)=\\frac{\\left(x+\\cfrac1x\\right)^6-\\left(x^6+\\cfrac1{x^6}\\right)-2}{\\left(x+\\cfrac1x\\right)^3+\\left(x^3+\\cfrac1{x^3}\\right)}=\\frac{6x^4+15x^2+18+\\cfrac{15}{x^2}+\\cfrac6{x^4}}{2x^3+3x+\\cfrac3x+\\cfrac2{x^3}}$$ giving $$f(x)=\\frac{6x^8+15x^6+18x^4+15x^2+6}{2x^7+3x^5+3x^3+2x}=\\frac{6x^8+9x^6+9x^4+6x^2}{2x^7+3x^5+3x^3+2x}+\\frac{6x^6+9x^4+9x^2+6}{2x^7+3x^5+3x^3+2x}$$ which boils down to $$f(x)=3x+\\frac3x.$$", "b. Minimum at $$x=−\\frac{1}{2}$$ c. Concave up for all $$x$$ d. No inflection points 230) $$f(x)=x^3+x^4$$ For the following exercises, determine a. intervals where $$f$$ is increasing or decreasing, b. local minima and maxima of $$f,$$ c. intervals where $$f$$ is concave up and concave down, and d. the inflection points of $$f.$$ Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator. 231) [T] $$f(x)=sin(πx)−cos(πx)$$ over $$x=[−1,1]$$ Solution: a. Increases over $$−\\frac{1}{4}<x<\\frac{3}{4},$$ decreases over $$x>\\frac{3}{4}$$ and $$x<−\\frac{1}{4}$$ b. Minimum at $$x=−\\frac{1}{4}$$, maximum at $$x=\\frac{3}{4}$$ c. Concave up for $$−\\frac{3}{4}<x<\\frac{1}{4}$$, concave down for $$x<−\\frac{3}{4}$$ and $$x>\\frac{1}{4}$$ d. Inflection points at $$x=−\\frac{3}{4},x=\\frac{1}{4}$$ 232) [T] $$f(x)=x+sin(2x)$$ over $$x=[−\\frac{π}{2},\\frac{π}{2}]$$ 233) [T] $$f(x)=sinx+tanx$$ over $$(−\\frac{π}{2},\\frac{π}{2})$$ Solution: a. Increasing for all $$x$$ b. No local minimum or maximum c. Concave up for $$x>0$$, concave down for $$x<0$$ d. Inflection point at $$x=0$$ 234) [T] $$f(x)=(x−2)^2(x−4)^2$$ 235) [T] $$f(x)=\\frac{1}{1−x},x≠1$$ Solution: a. Increasing for all $$x$$ where defined b. No local minima or maxima c. Concave up for $$x<1$$; concave down for $$x>1$$ d. No inflection points in domain 236) [T] $$f(x)=\\frac{sinx}{x}$$ over $$x=[−2π,2π] [2π,0)∪(0,2π]$$ 237) $$f(x)=sin(x)e^x$$ over $$x=[−π,π]$$ Solution: a. Increasing over $$−\\frac{π}{4}<x<\\frac{3π}{4}$$, decreasing over $$x>\\frac{3π}{4},x<−\\frac{π}{4}$$ b. Minimum at $$x=−\\frac{π}{4}$$, maximum at $$x=\\frac{3π}{4}$$ c. Concave up for $$−\\frac{π}{2}<x<\\frac{π}{2}$$, concave down for $$x<−\\frac{π}{2},x>\\frac{π}{2}$$", "## Calculus (3rd Edition) The maximum is $R(1)= 1/3$; the minimum is $R(0)=0$ Given $$R(t)=\\frac{t}{t^{2}+t+1}, \\quad[0,3]$$ Since $$R'(t) =\\frac{\\frac{d}{dt}\\left(t\\right)\\left(t^2+t+1\\right)-\\frac{d}{dt}\\left(t^2+t+1\\right)t}{\\left(t^2+t+1\\right)^2}= \\frac{-t^2+1}{\\left(t^2+t+1\\right)^2}$$ We find the critical points \\begin{align*} g'(\\theta)&=0\\\\ \\frac{-t^2+1}{\\left(t^2+t+1\\right)^2}&=0\\\\ 1-t^2&=0 \\end{align*} Then $R(t)$ has critical points at $t=-1, \\ t=1$, but we eliminate $-1$ because $-1\\notin [0,3]$. We check the values at the critical points and end points: \\begin{align*} R(0)&= 0 \\\\ R(1)&= 1/3 \\\\ R(3) &= 3/13 \\end{align*} Then the maximum is $R(1)= 1/3$, and the minimum is $R(0)=0$" ]

[ "in three-dimensional space with rectangular coordinates $$(x,y,z)$$ the equations $$x = k, \\, y = l$$ and $$z = m$$ where $$k, \\, l$$ and $$m$$ are constants, represent unbounded planes parallel to the $$yz$$-plane, $$xz$$-plane and $$xy$$-plane, respectively. With cylindrical coordinates $$(r, \\theta, z)$$, by $$r = c, \\, \\theta = \\alpha$$, and $$z = m$$, where $$c, \\alpha$$, and $$m$$ are constants, we mean an unbounded vertical cylinder with the z-axis as its radial axis; a plane making a constant angle $$\\alpha$$ with the $$xy$$-plane; and an unbounded horizontal plane parallel to the $$xy$$-plane, respectively. This means that the circular cylinder $$x^2 + y^2 = c^2$$ in rectangular coordinates can be represented simply as $$r = c$$ in cylindrical coordinates. (Refer to Cylindrical and Spherical Coordinates for more review.) ## Integration in Cylindrical Coordinates Triple integrals can often be more readily evaluated by using cylindrical coordinates instead of rectangular coordinates. Some common equations of surfaces in rectangular coordinates along with corresponding equations in cylindrical coordinates are listed in Table $$\\PageIndex{1}$$. These equations will become handy as we proceed with solving problems using triple integrals. Table $$\\PageIndex{1}$$: Equations of Some Common Shapes Circular cylinder Circular cone Sphere Paraboloid Rectangular $$x^2 + y^2 = c^2$$ $$z^2 = c^2 (x^2 + y^2)$$ $$x^2 + y^2 + z^2 = c^2$$", "# Cylindrical Identification If the relation is expressed in cylindrical coordinates of the form $(r, \\theta, z)$, what shape is described $r^2 + z^2 = 100$? ×", "=\\rho \\mathrm{sin}\\varphi \\mathrm{sin}\\theta$ $\\mathrm{tan}\\theta =\\frac{y}{x}$ Cylindrical coordinates are good for describing cylinders whose axes run along the z - axis and planes that either contain the z - axis or lie perpendicular to the z -axis. Surfaces like these have equations of constant coordinate value.", "reasonable. Many thanks for any comments. ^_^ Since this is a cylinder, use cylindrical coordinates: $x= r cos(\\theta)$, $y= r sin(\\theta)$, z= z. On the boundary of the cylinder, r is the constant R. In particular, that means we can write the cylindrical surface as the vector function of two parameters: $\\vec{r}= Rcos(\\theta)\\vec{i}+ R sin(\\theta)\\vec{j}+ z\\vec{k}$ Now take the derivatives of that with respect to the two variables: $\\vec{u}= \\frac{\\partial \\vec{r}}{\\partial \\theta}= -Rsin(\\theta)\\vec{i}+ Rcos(\\theta)\\vec{j}$ $\\vec{v}= \\frac{\\partial \\vec{r}}{\\partial z}= \\vec{k}$ The cross product of those two vectors is the \"fundamental vector product\". It, times $d\\theta dz$, is dA.", "r=1-\\sin\\theta -\\cos\\theta$, $\\theta=2\\pi$ Exercise. Show that lines and circles are the only planes curves that have constant curvature. Is this true for space curves?", "$\\theta$. You can see that $\\theta$ goes from $0$ to $\\pi/2$ Finally, you want to convert $dV$ to cylindrical coordinates. When converting from rectangular to cylindrical, $dV$ becomes $rdzdrd\\theta$. What we're left with is: $$\\int_{0}^{\\pi/2}\\int_{0}^{1}\\int_{0}^{2-r^2}r^2cos(\\theta)dzdrd\\theta$$", "Cylindrical Coordinates! $$\\int_{}^{}\\int_{}^{}\\int_{}^{} 4z^3 dx dy dz$$ Evaluate this. Q is a solid bounded by $$(x^2+ y^2)^\\frac{3}{2} = z$$ • Anonymous commented ALSO, PLANE Z = 1", "# Using Cylindrical Coordinates to find volume of a solid How do I use Cylindrical Coordinates to find volume of a solid in the first Octant that is bounded by the cylinder $x^2 + y^2 = 2y$, the half cone $z = \\sqrt{x^2 + y^2}$, and the $xy$-plane. I have drawn the region of integration and obtained this: $\\int_0^2 \\int_0^\\sqrt{2y-y^2}\\int_0^\\sqrt{x^2 + y^2} dzdxdy$ Is this correct and from here were do I apply the cylindrical coordinates? $\\int_0^{\\frac{\\pi}{2}} \\int_0^1\\int_0^z r dzdrd\\theta$", "# Math Insight ### Applet: Surfaces of constant $\\theta$ in spherical coordinates The half-plane surface of $\\theta=$ constant is shown, where the value of $\\theta$ is determined by the blue point on the slider. Only the part of the surface where $\\rho < 5$ is shown, which makes the half-plane appear like a half-disk.", "# Find the Volume of the Solid that is enclosed by the Cone and the Sphere This question aims to find the volume of the solid enclosed by the cone and a sphere by using the method of polar coordinates to find the volume. Cylindrical coordinates extend the two-dimensional coordinates to three-dimensional coordinates. In a sphere, the distance of the origin (0,0) to the point P is called the radius r. By joining the line from the origin to the point P, the angle made by this radial line from the x-axis is called the angle theta, represented by $\\theta$. Radius r and $\\theta$ have some values that can be used in limits for integration. The $z-axis$ is projected in a cartesian plane along with the $xy$-plane to form a three-dimensional plane. This plane is represented by $(r, \\theta, z)$ in terms of polar coordinates. To find the limits of $z$, we will take the square root of the double cones. The positive square root represents the top of the cone. The equation of the cone is: $z = \\sqrt{(x^2 + y^2)}$ The equation of the sphere is: $x^2 + y^2 + z^2 = 2$ This equation is derived from the polar coordinates formula, where $x^2 + y^2 = r^2$ when $z = r^2$. Both of these equations", "cylindrical coordinates (how the solution is proceeding), where we allow $\\theta$ to sweep out the full cylinder, the radius is from $0$ to $1$, and your integral is $$2\\int_0^{2\\pi}\\int_0^1\\int_0^{2-2r\\cos \\theta}rdydrd\\theta\\\\ =2\\int_0^{2\\pi}\\int_0^1(2r-r^2\\cos \\theta)dydrd\\theta\\\\ =2\\int_0^{2\\pi}(1-1/3\\cos \\theta)d\\theta\\\\ =4\\pi$$", "# Describing Circles Express the circle in the form $z=e^{i\\theta}\\;(\\theta\\in[0,2\\pi])$, the ray in the form $z=\\rho e^{i\\theta_0}\\;(\\rho \\in[0,+\\infty))$ and substitute.", "@ $30 per hour and$2,2... ##### 1. Convert the point ( 215 7.) from cylindrical to spherical coordinates. 2. Set up a... 1. Convert the point ( 215 7.) from cylindrical to spherical coordinates. 2. Set up a triple integral, but do NOT evaluate, to find the volume of the solid in the first octant bounded by the coordinate planes and the plane 3x + 6y + 4z = 12. 1 3. Locate all relative maxima, relative minima, and sadd...", "recall, for cylindrical coordinates, we use $(x,y,z) = (r \\cos \\theta , r \\sin \\theta, z)$ so simply replace x with $r \\cos \\theta$ and y with $r \\sin \\theta$ and simplify 3. Cylindrical Coordinates are of the following form: $(r, \\theta , z)$", "is zero so it should be something like sin(\\theta). Here, \\theta is the angle between the z-axis and the axis of the cylinder.", "\\mathbf{\\hat{r}}+u_\\theta \\mathbf{\\hat{\\theta}}$. To get your answer you need to find a relation between the Cartesian and cylindrical unit vectors. You can find this relation in your book or derive it yourself.", "With cylindrical coordinates $(r,θ,z),(r,θ,z),$ by $r=c,θ=α,r=c,θ=α,$ and $z=m,z=m,$ where $c,α,c,α,$ and $mm$ are constants, we mean an unbounded vertical cylinder with the $zz$-axis as its radial axis; a plane making a constant angle $αα$ with the $xyxy$-plane; and an unbounded horizontal plane parallel to the $xyxy$-plane, respectively. This means that the circular cylinder $x2+y2=c2x2+y2=c2$ in rectangular coordinates can be represented simply as $r=cr=c$ in cylindrical coordinates. (Refer to Cylindrical and Spherical Coordinates for more review.) ### Integration in Cylindrical Coordinates Triple integrals can often be more readily evaluated by using cylindrical coordinates instead of rectangular coordinates. Some common equations of surfaces in rectangular coordinates along with corresponding equations in cylindrical coordinates are listed in Table 5.1. These equations will become handy as we proceed with solving problems using triple integrals. Circular cylinder Circular cone Sphere Paraboloid Rectangular $x2+y2=c2x2+y2=c2$ $z2=c2(x2+y2)z2=c2(x2+y2)$ $x2+y2+z2=c2x2+y2+z2=c2$ $z=c(x2+y2)z=c(x2+y2)$ Cylindrical $r=cr=c$ $z=crz=cr$ $r2+z2=c2r2+z2=c2$ $z=cr2z=cr2$ Table 5.1 Equations of Some Common Shapes As before, we start with the simplest bounded region $BB$ in $ℝ3,ℝ3,$ to describe in cylindrical coordinates, in the form of a cylindrical box, $B={(r,θ,z)|a≤r≤b,α≤θ≤β,c≤z≤d}B={(r,θ,z)|a≤r≤b,α≤θ≤β,c≤z≤d}$ (Figure 5.51). Suppose we divide each interval into $l,mandnl,mandn$ subdivisions such that $Δr=b−al,Δθ=β−αm,Δr=b−al,Δθ=β−αm,$ and $Δz=d−cn.Δz=d−cn.$ Then we can state the following definition for a triple integral in cylindrical coordinates. Figure 5.51 A cylindrical box $BB$ described by cylindrical coordinates. ###", "which I'm guessing you just made up, might get ugly. If your example was circular instead of elliptical you might use cylindrical coordinates $z,\\theta$ for your parameters.", "where $r = |\\vec{r}|$ and $\\theta$ is the angle between $\\vec{r}$ and $\\vec{n}$. So, the equation says $r \\cos(\\theta) = a r,$ hence either $r = 0$ or $\\cos(\\theta) = a.$ For $a \\in (-1,1),$ this says that $\\theta =$ constant, so describes a cone.", "may find something in the form$$\\frac{(x-x_0)^2}{a^2} + \\frac{(y-y_0)^2}{b^2}=1$$In that case you can use a similar parameterization:$$x = x_0+ar\\cos\\theta,~y=y_0 + br\\sin\\theta$$with the correct $dS$ for the parameterization. Unfortunately, it could be even worse with the ellipse rotated. 3. Apr 22, 2014 ### BiGyElLoWhAt It is a cylinder, the cone is a typo, I'll edit it out. But... I'm not sure what you mean by using cylindrical coordinates. My cylinder equation (in cylindrical) would be $r^2 = 4$, but I don't see how that helps me. So it's really not that complex of an equation. Also, 1 the z's are irrelevant I believe, as I could move the plane up, or down any amount and still have the same area. 2 I have no z in my cylindrical expression. I'll throw the problem out there, maybe it'll help: $x+2y+3z=1$ is my plane, & $x^2+y^2 = 4$ is my cylinder 4. Apr 22, 2014 ### LCKurtz The area of the plane you want lies above the interior of the circle. You can set it all up in xy coordinates if you want. Solve the plane for z in terms of x and y and use the formula for dS. Then set it up to integrate over the circle. At that point you can change to cylindrical (polar in xy) coordinates to" ]

[ "# I can not draw arc in a clockwise direction with this program in asymptote I have to trace the meridians. settings.outformat = \"png\"; settings.render = 16; size(6cm,0); import solids; currentlight=Headlamp; // inutile car Headlamp est la valeur par défaut nslice=4*nslice; revolution boule=sphere(O,1); draw(surface(boule),lightgrey+white+opacity(.5)); draw((0,0,1)--(0,0,-1),dashed); dot(\"$P1$\",(0,0,1),N); dot(\"$P2$\",(0,0,-1),2.2S); dot(\"$O$\",(0,0,0),SE); draw(O--(1,0,0),red); draw(O--(Cos(60),Sin(60),0),red); triple v1= 0.4(1,0,0), v2 = 0.4(Cos(60),Sin(60),0); draw(arc(c=O,v1,v2,normal=Z),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); for (real alpha = 90; alpha<=450; alpha+= 30) { if (alpha==150) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red); if (alpha==330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red + dashed); if (alpha != 150 && alpha != 330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),black); } But I can not draw arc in a clockwise direction with this code (this is the critic point!): draw(arc(c=O,v1,v2,normal=Z),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); I tried: 1. to change v1 and v2 order with the same normal (Z), but in this way i have the angle greater than ninety degrees (i need the explementar angle) 2. to change v1 and v2 order and take normal = - Z, but in this way asymptote runs on a infinite loop. I don't know why. ## 1 Answer It took me a while to understand the question since IMHO it is not very clearly written. An additional complication arises from the fact that you did not present an MWE and seem to insist to have a special character in a comment: \"défaut\". All I", "draw with Cartesian coordinates. Figure 3.8: The Greek letter θ (theta) is often used to denote an angle. Since a polar coordinate is conventionally referred to as (r, θ), we’ll use theta as a variable name when referring to an angle. sine(theta) = y/r → y = r * sine(theta) cosine(theta) = x/r → x = r * cosine(theta) For example, if r is 75 and theta is 45 degrees (or PI/4 radians), we can calculate x and y as below. The functions for sine and cosine in Processing are sin() and cos(), respectively. They each take one argument, an angle measured in radians. Show Raw float r = 75; float theta = PI / 4; Converting from polar (r,theta) to Cartesian (x,y) float x = r * cos(theta); float y = r * sin(theta); This type of conversion can be useful in certain applications. For example, to move a shape along a circular path using Cartesian coordinates is not so easy. With polar coordinates, on the other hand, it’s simple: increment the angle! Here’s how it is done with global variables r and theta. Your browser does not support the canvas tag. Example 3.4: Polar to Cartesian Show Raw float r = 75; float theta = 0; void setup() { size(640,360); background(255); } void draw() { Polar", "we can compute some trigonometric functions on these values: print(\"theta = \", theta) print(\"sin(theta) = \", np.sin(theta)) print(\"cos(theta) = \", np.cos(theta)) print(\"tan(theta) = \", np.tan(theta)) theta = [ 0. 1.57079633 3.14159265] sin(theta) = [ 0.00000000e+00 1.00000000e+00 1.22464680e-16] cos(theta) = [ 1.00000000e+00 6.12323400e-17 -1.00000000e+00] tan(theta) = [ 0.00000000e+00 1.63312394e+16 -1.22464680e-16] The values are computed to within machine precision, which is why values that should be zero do not always hit exactly zero. Inverse trigonometric functions are also available: x = [-1, 0, 1] print(\"x = \", x) print(\"arcsin(x) = \", np.arcsin(x)) print(\"arccos(x) = \", np.arccos(x)) print(\"arctan(x) = \", np.arctan(x)) x = [-1, 0, 1] arcsin(x) = [-1.57079633 0. 1.57079633] arccos(x) = [ 3.14159265 1.57079633 0. ] arctan(x) = [-0.78539816 0. 0.78539816] #### Exponents and logarithms Another common type of operation available in a NumPy ufunc are the exponentials: x = [1, 2, 3] print(\"x =\", x) print(\"e^x =\", np.exp(x)) print(\"2^x =\", np.exp2(x)) print(\"3^x =\", np.power(3, x)) x = [1, 2, 3] e^x = [ 2.71828183 7.3890561 20.08553692] 2^x = [ 2. 4. 8.] 3^x = [ 3 9 27] The inverse of the exponentials, the logarithms, are also available. The basic np.log gives the natural logarithm; if you prefer to compute the base-2 logarithm or the base-10 logarithm, these are available as well: x = [1, 2, 4, 10]", "theta]] [[cos phi,sin phi, 0],[-sin phi,cos phi,0],[0,0,1]] MRC psi, theta, Omega [[cos psi,sin psi, 0],[-sin psi,cos psi,0],[0,0,1]] [[cos theta,0,sin theta],[0, 1, 0],[-sin theta, 0, cos theta]] [[cos Omega,sin Omega, 0],[-sin Omega,cos Omega,0],[0,0,1]] XYZ z, y, x [[cos z,sin z, 0],[-sin z, cos z,0],[0,0,1]] [[cos y,0,sin y],[0, 1, 0],[-sin y, 0, cos y]] [[1,0,0],[0, cos x, sin x],[0, -sin x, cos x]] spin Omega, n1, n2, n3 Quaternion rotation by Omega degrees about the vector [n1,n2\\,n3] sgirot q, n1, n2, n3 Quaternion rotation by q degrees about the vector [n1,n2\\,n3] quaternion e0, e1, e2, e3 Literal quaternion rotation using q = e0 + e1\\mathbf{i} + e2\\mathbf{j} + e2\\mathbf{k} 2d alpha Rotation in 2-D only, for 3-D objects this is a rotation about Z # The Transform object in Python ## Constructing a Transform There a four ways to construct a Transform object in Python t = Transform() # default constructor, t is the identity t = Transform({\"type\":\"eman\",\"az\":10,\"alt\":150,\"scale\":2.0,\"mirror\":True,\"tx\":3.4}) # construction using a dictionary p = Transform({\"type\":\"eman\",\"az\":10,\"mirror\":True,\"tx\":3.4}) # construction using a dictionary q = Transform([1,0,0,0,0,1,0,0,0,0,1,0]) # specify the 12 entries of the matrix explicitly, row wise (3 rows of 4) s = Transform(t) # copy construction - s is precisely the same as t NOTE: All the CONVENTIONS listed on the table above can be supplied as arguments or \"values\" for", "|x - 8| \\Longrightarrow |x - 8| = 5 - x/3$. This implies that one of $x - 8, 8 - x = 5 - x/3$, from which we find that $(x,y) = \\left(\\frac 92, \\frac {13}2\\right), \\left(\\frac{39}{4}, \\frac{33}{4}\\right)$. The region $\\mathcal{R}$ is a triangle, as shown above. When revolved about the line $y = x/3+5$, the resulting solid is the union of two right cones that share the same base and axis. $[asy]size(200); import three; currentprojection = perspective(0,0,10); defaultpen(linewidth(0.7)); pen dark=linewidth(1.3); pair Fxy = foot((8,10),(4.5,6.5),(9.75,8.25)); triple A = (8,10,0), B = (4.5,6.5,0), C= (9.75,8.25,0), F=(Fxy.x,Fxy.y,0), G=2*F-A, H=(F.x,F.y,abs(F-A)),I=(F.x,F.y,-abs(F-A)); real theta1 = 1.2, theta2 = -1.7,theta3= abs(F-A),theta4=-2.2; triple J=F+theta1*unit(A-F)+(0,0,((abs(F-A))^2-(theta1)^2)^.5 ),K=F+theta2*unit(A-F)+(0,0,((abs(F-A))^2-(theta2)^2)^.5 ),L=F+theta3*unit(A-F)+(0,0,((abs(F-A))^2-(theta3)^2)^.5 ),M=F+theta4*unit(A-F)-(0,0,((abs(F-A))^2-(theta4)^2)^.5 ); draw(C--A--B--G--cycle,linetype(\"4 4\")+dark); draw(A..H..G..I..A); draw(C--B^^A--G,linetype(\"4 4\")); draw(J--C--K); draw(L--B--M); dot(B);dot(C);dot(F); label(\"h_1\",(B+F)/2,SE,fontsize(10)); label(\"h_2\",(C+F)/2,S,fontsize(10)); label(\"r\",(A+F)/2,E,fontsize(10)); [/asy]$ Let $h_1,h_2$ denote the height of the left and right cones, respectively (so $h_1 > h_2$), and let $r$ denote their common radius. The volume of a cone is given by $\\frac 13 Bh$; since both cones share the same base, then the desired volume is $\\frac 13 \\cdot \\pi r^2 \\cdot (h_1 + h_2)$. The distance from the point $(8,10)$ to the line $x - 3y + 15 = 0$ is given by $\\left|\\frac{(8) - 3(10) + 15}{\\sqrt{1^2 + (-3)^2}}\\right| = \\frac{7}{\\sqrt{10}}$. The distance between $\\left(\\frac 92, \\frac {13}2\\right)$ and $\\left(\\frac{39}{4}, \\frac{33}{4}\\right)$ is", "the location changes too? How come? The only one who doesn't change the location is $$\\theta_6$$. Here is my matlab code. I have also added a picture of the lengths and the DH-table we made. https://i.imgur.com/6QTCO2P.png [![enter image description here][1]][1] The desired angles are 10,20,30,40,50,60 for theta1 to theta 6. %% Definitions. clear all syms theta1 theta2 theta3 theta4 theta5 theta6 %% MatriX T01=[ cos(theta1), -sin(theta1), 0, 0; sin(theta1), cos(theta1), 0, 0; 0, 0, 1, 0; 0, 0, 0, 1]; T12= [cos(theta2) -sin(theta2) 0 25; 0 0 -1 0; sin(theta2) cos(theta2) 0 0; 0 0 0 1]; T23=[ cos(theta3 + pi/2), -sin(theta3 + pi/2), 0, 315; sin(theta3 + pi/2), cos(theta3 + pi/2), 0, 0; 0, 0, 1, 0; 0, 0, 0, 1]; T34= [cos(theta4) -sin(theta4) 0 -35; 0 0 -1 -365; -sin(theta4) -cos(theta4) 0 0; 0 0 0 1]; T45= [cos(theta5) -sin(theta5) 0 0; 0 0 -1 0; -sin(theta5) -cos(theta5) 0 0; 0 0 0 1]; T56= [cos(theta6) -sin(theta6) 0 0; 0 0 1 0; -sin(theta6) -cos(theta6) 0 0; 0 0 0 1]; %% Wrist and tool T6W= [1 0 0 0; 0 -1 0 0; 0 0 -1 -80; 0 0 0 1]; TB0= [1 0 0 0; 0 -1 0 0; 0 0 -1 400; 0 0 0 1]; %% Placement of End Effector TBTtarget=eulerZYX2T(563.879,-139.427,-34.414,100.551*pi/180,29.536*pi/180,178.188*pi/180); %location", "= linspace(0,1,30); theta = linspace(0,2*pi,100); [r theta] = meshgrid(r,theta); z = 2*r.*sin(theta).*(1-r.^2)./(1+r.^2-2*r.*cos(theta)).^2; % x = r.*cos(theta); % y = r.*sin(theta); % only for viewing positive/negative in the plot for ii = 1:100; for jj = 1:30; if z(ii,jj)>0; c(ii,jj)=1; else if z(ii,jj)<0 c(ii,jj)=-1; else c(ii,jj)=0; end end end end clear ii jj % what the solution looks like: red=positive, blue=negative, green=zero figure; mesh(r.*cos(theta),r.*sin(theta),z,c) % scaled for clarity figure; mesh(r.*cos(theta),r.*sin(theta),sign(z).*log(log(abs(z)+1)+1)+1) % the function is unbounded: compute the limit as (r,theta) spirals to (1,0) syms u R T T_R u = 2*R*sin(T)*(1-R^2)/(1+R^2-2*R*cos(T))^2; % to find this value, use solve(diff(U,T),T) and use the third value T_R = pi - acos(1/4/R*(R^2 - (R^4 + 34*R^2 + 1)^(1/2) + 1)); u = subs(u,T,T_R); figure; ezplot(u); limit(u,R,1,'left') % which approaches +inf Posted in Uncategorized", "is +ve. • In the forth quadrant, sin is -ve; cos is +ve; tan is -ve. There are various methods of remembering the above table. One way is by using the CAST diagram. Each letter stands for the positive ratios. e.g. In the first quadrant all ratios are positive whilst in the third quadrant only tan is positive. Since the magnitude of the projections of the unit radius on the axies depend solely on the acute angle which the radius makes with the x-axis, any ratio of any angle is equal numerically to the same ratio of the acute angle which the radius makes with the x-axis. The sign must be found from the Cast circle. Example Find the value of sin 210 degrees. As 210 is in the third quadrant the sin is negative and the acute angle which the radius makes with the x-axis is 30 degrees. $\\therefore\\;\\;\\;\\;\\;sin\\;210^0&space;=&space;sin(210&space;-&space;180)\\;=&space;-&space;sin\\;30\\;=&space;-&space;\\frac{1}{2}$ To simplify expressions such as $\\inline&space;&space;tan(270^0&space;-&space;\\theta)$ follow the same logic but for simplicity imagine that $\\inline&space;&space;\\theta$ to be acute. The formula will apply for all values of $\\inline&space;&space;\\theta$ but the argument is simplified by supposing that it is acute. Examples $sin(180^0&space;-&space;\\theta&space;)\\;=&space;+&space;sin\\;\\theta$ $cos(180^0&space;+&space;\\theta&space;)\\;=&space;-&space;cos\\;\\theta$ $tan(270^0&space;-&space;\\theta&space;)\\;=&space;+&space;tan\\;(90^0\\;-\\theta)&space;=&space;cot\\;\\theta$ $sec(360^0&space;-&space;3\\theta&space;)\\;=&space;+&space;sec\\;3\\theta$ $cot(-&space;2\\theta&space;)\\;=&space;-&space;cot\\;2\\theta$ It should be noted that cosec; sec; and cot follow the same rules as sin;cos; and tan respectively since they are their reciprocals.", "# 2014S004 This is Sheng's CDS 130 course page. All of the information that you will ever need for this course can be found here or on a page linked to from this page. Final exam: May 12th, 7:30 AM, IN Room 222. # 3. Monday, May 5th • Integration • Note: The next class is on May 8th (make-up day for March 3rd) 9:00 AM # 9. Monday, April 14th x = linspace(0, 2*pi, 100); y = sin(x); plot(x, y, 'r*') axis ([0, 2*pi, -1.5, 1.5]) axis square xlabel ('xlabel') ylabel ('sin(x)') text(4.0, 1.0, 'sin(x)', 'FontSize', 20, 'Color', 'y') hold on theta = pi/2; y = sin(x + theta) plot (x, y, 'g^') shg print -dpng mypic.png clear all; clc; x = [0, 1, 1, 0]; y = [0, 0, 1, 1]; fill (x, y, 'r') axis off axis equal hold on fill (x+1, y, 'g') fill (x, y+1, 'g') fill (x+1, y+1, 'r') theta = linspace(0, 2*pi, 7) x = cos(theta) y = sin(theta) fill (x +1 , y +1, [0.8, 0.6, 0.5]) fill (x*0.75 +1 , y*0.75 +1, 'w') # 10. Wednesday, April 9th • Iterations • Code to draw circles theta = 0:2*pi/60:2*pi; r = 1.0; x = r* cos(theta); y = r* sin(theta); for i = 1:60 plot (x(i), y(i), '-or', 'MarkerSize',20) plot", "$x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curves correspond to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.4289$ The evolution of $(x(t),y(t))$, $R(t)$, $|\\theta(t)|$ for solutions of (50) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = s_2 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $b_0 = -1/4$, $z_0 = 0.6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curve corresponds to $r = 2 R_\\ast t^{-\\nu-l/q}$ The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for", "180^\\circ) ~ = ~ -\\sin\\;\\theta$ $\\label{1.11} \\cos\\;(\\theta \\pm 180^\\circ) ~ = ~ -\\cos\\;\\theta$ $\\label{1.12}\\tan\\;(\\theta \\pm 180^\\circ) ~ = ~ \\tan\\;\\theta$ A reflection is simply the mirror image of an object. For example, in Figure 1.5.5 the original object is in QI, its reflection around the $$y$$-axis is in QII, and its reflection around the $$x$$-axis is in QIV. Notice that if we first reflect the object in QI around the $$y$$-axis and then follow that with a reflection around the $$x$$-axis, we get an image in QIII. That image is the reflection around the origin of the original object, and it is equivalent to a rotation of $$180^\\circ$$ around the origin. Notice also that a reflection around the $$y$$-axis is equivalent to a reflection around the $$x$$-axis followed by a rotation of $$180^\\circ$$ around the origin. Figure 1.5.5 Applying this to angles, we see that the reflection of an angle $$\\theta$$ around the $$x$$-axis is the angle $$-\\theta$$, as in Figure 1.5.6. Figure 1.5.6 Reflection of $$θ \\text{ around the }x\\text{-axis}$$ So we see that reflecting a point $$(x,y)$$ around the $$x$$-axis just replaces $$y$$ by $$-y$$. Hence: $\\label{1.13}\\sin\\;(-\\theta) ~ = ~ -\\sin\\;\\theta$ $\\label{1.14} \\cos\\;(-\\theta) ~ = ~ \\cos\\;\\theta$ $\\label{1.15} \\tan\\;(-\\theta) ~ = ~ -\\tan\\;\\theta$ Notice that the cosine function does not change in Equation \\ref{1.14} because it", "would be: $$x = sin(2\\theta) y = cos(2\\theta).$$ Why would you say that for j' $$y=-cos(2\\theta)?$$ The only way j' will have negative y coordinates is if the gradient of the line is >45, and if this happens, cos(2x) will be negative (since 90 < 2x < 180, cos(2x) is negative). Sorry for the very long post, I just wanted to show what I tried before asking any questions. With this being said, could someone please tell me what's wrong with the derivation I attempted? And most of all, why the first one doesn't work? Thank you! Last edited: May 29, 2010 2. May 29, 2010 ### HallsofIvy NO, this is 2$\\theta$ $2\\theta= 36.86$ and $tan(36.86)= 0.75$ NO, cos(71.56)= 0.3163, nowhere near -.8. I don't know how you got this. If $cos(\\theta)= -.8$ then $\\theta= 143$ or $\\theta= -37$. tan(143)= -.75 but tan(-37)= .75. 3. May 29, 2010 ### Lunat1c The gradient of the line is [tex] tan(\\theta) [/itex] not [tex] tan(2\\theta) [/itex], or am I wrong? [tex]2\\theta [/itex] is the angle with which the vectors i and j will be rotated when they're reflected in the line [tex] y=(tan\\theta)x [/itex]. 4. May 29, 2010 ### Lunat1c Ok fair enough, I meant [tex] tan(2\\theta) = -\\frac{3}{4} [/itex] [tex] \\therefore \\theta = \\frac{arctan(-\\frac{3}{4})}{2} [/itex]. [tex] -5cos(2\\theta)=4 [/itex] then [tex]", "but for simplicity, we choose to implement it using two rotations: one with angle $$\\theta_1$$ for the qumodes 1 and 2, and another with angle $$\\theta_2$$ for the qubits 3 and 4. The final matrix has the form: $\\begin{split}\\begin{pmatrix} \\cos(\\theta_1) & - \\sin(\\theta_1) & 0 & 0 \\\\ \\sin(\\theta_1) & cos(\\theta_1) & 0 & 0 \\\\ 0 & 0 & \\cos(\\theta_2) & - \\sin(\\theta_2) \\\\ 0 & 0 & \\cos(\\theta_2) & \\sin(\\theta_2) \\end{pmatrix}~~\\text{with}~~\\theta_1, \\theta_2 \\in [0,2\\pi).\\end{split}$ theta1 = 0.5 theta2 = 1 U = np.array([[np.cos(theta1), -np.sin(theta1), 0, 0 ], [np.sin(theta1), np.cos(theta1), 0, 0 ], [0, 0, np.cos(theta2), -np.sin(theta2)], [0, 0, np.sin(theta2), np.cos(theta2)]]) ### Circuit¶ We instantiate our circuit with 8 qubits, 4 for the input, 4 for the output. prog = sf.Program(8) We can then declare our circuit. The first four lines are 2-modes squeezing gates, which generate a random number of photons with prog.context as q: S2gate(r_squeezing) | (q[0], q[4]) S2gate(r_squeezing) | (q[1], q[5]) S2gate(r_squeezing) | (q[2], q[6]) S2gate(r_squeezing) | (q[3], q[7]) Interferometer(U) | (q[4], q[5], q[6], q[7]) ### Running¶ Run the simulation up to ‘cutoff’ photons per mode eng = sf.Engine(\"fock\", backend_options={\"cutoff_dim\":cutoff}) state = eng.run(prog).state Out: /home/circleci/project/venv/lib/python3.7/site-packages/strawberryfields/program.py:662: UserWarning: The circuit consists of 2 disconnected components. Get the probability associated to each state probs = state.all_fock_probs() Reshape ‘probs’ such that probs $$[m_1, \\dots, m_4,n_1, \\dots, n_4]$$ gives", "to arbitrarily decide that $$\\gamma = 0$$, which means that $$\\sin(\\gamma) = 0$$ and $$\\cos(\\gamma) = 1$$. The formulas in this case are: \\begin{align*} \\beta = \\mbox{atan2}( -R_{2,0}, \\cos(\\beta) ) \\\\ \\alpha = \\mbox{atan2}( -R_{1,2}, R_{1,1} ) \\\\ \\gamma = 0 \\end{align*} There are, of course, other options. Python code implementing these operations is below, along with unit testing for the degenerate and non-degerate cases. import unittest import numpy def x_rotation( theta ): \"\"\"Generate a 4x4 homogeneous rotation matrix about x-axis\"\"\" c = numpy.cos(theta) s = numpy.sin(theta) return numpy.array([ [ 1, 0, 0, 0], [ 0, c,-s, 0], [ 0, s, c, 0], [ 0, 0, 0, 1] ]) def y_rotation( theta ): \"\"\"Generate a 4x4 homogeneous rotation matrix about y-axis\"\"\" c = numpy.cos(theta) s = numpy.sin(theta) return numpy.array([ [ c, 0, s, 0], [ 0, 1, 0, 0], [-s, 0, c, 0], [ 0, 0, 0, 1] ]) def z_rotation( theta ): \"\"\"Generate a 4x4 homogeneous rotation matrix about z-axis\"\"\" c = numpy.cos(theta) s = numpy.sin(theta) return numpy.array([ [ c,-s, 0, 0], [ s, c, 0, 0], [ 0, 0, 1, 0], [ 0, 0, 0, 1] ]) def xyz_rotation( angles ): \"\"\"Generate 4x4 homogeneous rotation in x then y then z order\"\"\" return numpy.dot( z_rotation(angles[2]), numpy.dot( y_rotation(angles[1]), x_rotation(angles[0]) ) ) def xyz_rotation_angles( R, eps=1e-8, check=False ):", "# Plot in Physical Space? [closed] (* Q7 - a *) Block[ {Theta=Pi/6,r=6}, Graphics[ { Thick, Black, Line[{{-2,-(12-r)},{-2,0},{2,0},{2+r*Sin[Theta],-r*Cos[Theta]}}], Red, AbsolutePointSize[20],Point[{-2,-(12-r)}], Blue,AbsolutePointSize[10],Point[{2+r*Sin[Theta],-r*Cos[Theta]}], Black,AbsolutePointSize[8],Point[{{-2,0},{2,0}}], {Dashed,Line[{{2,0},{2,-7}}]}, Circle[{2,0},1,{-Pi/2,-Pi/2+Theta}], Text[Style[\"Theta\",FontSize -> 15],{2+1.5Sin[Theta/2],-1.5Cos[Theta/2]}], Text[Style[\" r \",FontSize -> 15],{2.5+r/2Sin[Theta],-r/2Cos[Theta]}] } ] ] ansQ7 = With[{mu = 4.5, g = 9.8}, NDSolve[ { (mu + 1)*r''[t] - r[t]*Theta'[t]^2 + g*(mu - Cos[Theta[t]]) == 0, r[t]*Theta''[t] + 2 r'[t]*Theta'[t] + g*Sin[Theta[t]] == 0, r[0] == 1, Theta[0] == Pi/2, r'[0] == 0, Theta'[0] == 0 }, {r, Theta}, {t, 0, 100} ] ] (* Q7 - b *) Plot[Evaluate[r[t] /. ansQ7], {t, 0, 20}, PlotRange -> All] Plot[Evaluate[Theta[t] /. ansQ7], {t, 0, 20}, PlotRange -> All] I am not sure if I understand (c) correctly. Is it asking to plot the exact location of the m? I am thininking that I need to use something like CoordinateTransform, but I am not sure how to use it. Also in this case, the angle Theta is measured postively from downwards?? Thanks. I would interpret c) as asking for a plot of $$\\{r\\,\\cos\\,\\theta, r\\,\\sin\\,\\theta\\}$$. This can be done as follows: With[{mu = 4.5, g = 9.8}, {rF, θF} = NDSolveValue[ {(mu + 1)*r''[t] - r[t]*Theta'[t]^2 + g*(mu - Cos[Theta[t]]) == 0, r[t]*Theta''[t] + 2 r'[t]*Theta'[t] + g*Sin[Theta[t]] == 0, r[0] == 1, Theta[0] == Pi/2, r'[0] == 0, Theta'[0] ==", "0), ( 0, 100, 0), ( 0, 0, 100), ) rotation = [0, 0, 0] def generate_x(theta): return matrix([ [1, 0, 0], [0, cos(theta), -sin(theta)], [0, sin(theta), cos(theta)] ]) def generate_y(theta): return matrix([ [cos(theta), 0, -sin(theta)], [0, 1, 0], [sin(theta), 0, cos(theta)] ]) def generate_z(theta): return matrix([ [cos(theta), -sin(theta), 0], [sin(theta), cos(theta), 0], [0, 0, 1] ]) while not done: for e in pygame.event.get(): if e.type == pygame.QUIT: done = True pygame.draw.rect(surface, BLACK, pygame.Rect(0, 0, WIDTH, HEIGHT)) render_points = [] for p in points: m = matrix([ [p[0]], [p[1]], [p[2]], ]) for method, angle in zip((generate_x, generate_y, generate_z), rotation): m = method(angle) * m x, y = map(lambda x: int(WIDTH/2 - x), (m[0,0], m[1,0])) render_points.append((x, y)) for x, y in render_points: pygame.draw.circle(surface, WHITE, (x, y), 2) rotation[0] += pi / 100 rotation[1] += pi / 150 display.flip() sleep(1/45) Now let's make a cube. You can do this by changing the points tuple. Each row represents a point in a 3D space, but whilst our maths uses all the points, we only render using the X and Y value. points = ( ( 50, 50, 50), (-50, 50, 50), ( 50, -50, 50), ( 50, 50, -50), (-50, 50, -50), ( 50, -50, -50), (-50, -50, -50), (-50, -50, 50), ) Wonderful! Final step: let's replace the dots with", "that $$\\gamma = 0$$, which means that $$\\sin(\\gamma) = 0$$ and $$\\cos(\\gamma) = 1$$. The formulas in this case are: \\begin{align*} \\beta = \\mbox{atan2}( -R_{2,0}, \\cos(\\beta) ) \\\\ \\alpha = \\mbox{atan2}( -R_{1,2}, R_{1,1} ) \\\\ \\gamma = 0 \\end{align*} There are, of course, other options. Python code implementing these operations is below, along with unit testing for the degenerate and non-degerate cases. import unittest import numpy def x_rotation( theta ): \"\"\"Generate a 4x4 homogeneous rotation matrix about x-axis\"\"\" c = numpy.cos(theta) s = numpy.sin(theta) return numpy.array([ [ 1, 0, 0, 0], [ 0, c,-s, 0], [ 0, s, c, 0], [ 0, 0, 0, 1] ]) def y_rotation( theta ): \"\"\"Generate a 4x4 homogeneous rotation matrix about y-axis\"\"\" c = numpy.cos(theta) s = numpy.sin(theta) return numpy.array([ [ c, 0, s, 0], [ 0, 1, 0, 0], [-s, 0, c, 0], [ 0, 0, 0, 1] ]) def z_rotation( theta ): \"\"\"Generate a 4x4 homogeneous rotation matrix about z-axis\"\"\" c = numpy.cos(theta) s = numpy.sin(theta) return numpy.array([ [ c,-s, 0, 0], [ s, c, 0, 0], [ 0, 0, 1, 0], [ 0, 0, 0, 1] ]) def xyz_rotation( angles ): \"\"\"Generate 4x4 homogeneous rotation in x then y then z order\"\"\" return numpy.dot( z_rotation(angles[2]), numpy.dot( y_rotation(angles[1]), x_rotation(angles[0]) ) ) def xyz_rotation_angles( R, eps=1e-8, check=False ): \"\"\"Back out the", "_{1}+l_{2}\\cos(\\theta _{1}+\\theta _{2})+l_{3}\\cos(\\theta _{1}+\\theta _{2}+\\theta _{3})\\\\y=l_{1}\\sin \\theta _{1}+l_{2}\\sin(\\theta _{1}+\\theta _{2})+l_{3}\\sin(\\theta _{1}+\\theta _{2}+\\theta _{3})\\end{array}}}$ Graphical Example of Planar, Three-link Manipulator using Compound Transformations Planar, three-link manipulator with all three angles set to ${\\displaystyle 30^{o}}$. Code %reset -f import numpy as np import sympy as sp import matplotlib.pyplot as plt theta1, theta2, theta3=sp.symbols('theta1 theta2 theta3') l1, l2, l3 = sp.symbols('l1 l2 l3') f1_f0_T=np.array([[sp.cos(theta1), -sp.sin(theta1), 0], [sp.sin(theta1), sp.cos(theta1), 0], [0, 0, 1]]) f2_f1_T=np.array([[sp.cos(theta2), -sp.sin(theta2), l1], [sp.sin(theta2), sp.cos(theta2),0], [0, 0, 1]]) f3_f2_T=np.array([[sp.cos(theta3), -sp.sin(theta3), l2], [sp.sin(theta3), sp.cos(theta3),0], [0, 0, 1]]) f4_f3_T=np.array([[sp.cos(0), -sp.sin(0), l3], [sp.sin(0), sp.cos(0),0], [0, 0, 1]]) ee_bs_T=np.dot(f1_f0_T,np.dot(f2_f1_T,np.dot(f3_f2_T,f4_f3_T))) ee_bs_T=sp.simplify(ee_bs_T) # Unfortunately, we do not get the simple # expression on the wiki. #print(\"ee_bs_T\") #display(ee_bs_T) #For substituting in the following: 30 degree increments #print(\"ee_bs_T after substitutions\") replacements=[(theta1,30*np.pi/180),(theta2,30*np.pi/180), (theta3,30*np.pi/180), (l1,1), (l2,1), (l3,1)] #display(ee_bs_T.subs(replacements)) End_Coord=np.array([[0,0,1]]) End_Coord=End_Coord.T #print(\"f1_f0_T\") f1_f0_T=sp.simplify(f1_f0_T) #display(f1_f0_T.subs(replacements)) f1_f0_T=np.array(f1_f0_T.subs(replacements)).astype(np.float64) f1_f0_T=np.reshape(f1_f0_T,(3,3)) print(\"f1_f0_T dot End_Coord\") display(np.dot(f1_f0_T,End_Coord)) Joint0=np.dot(f1_f0_T,End_Coord) #print(\"\\nf2_f1_T\") f2_f1_T=sp.simplify(f2_f1_T) #display(f2_f1_T.subs(replacements)) f2_f1_T=np.array(f2_f1_T.subs(replacements)).astype(np.float64) f2_f1_T=np.reshape(f2_f1_T,(3,3)) print(\"f2_f0_T dot End_Coord\") display(np.dot(f1_f0_T,np.dot(f2_f1_T,End_Coord))) Joint1=np.dot(f1_f0_T,np.dot(f2_f1_T,End_Coord)) #print(\"\\nf3_f2_T\") f3_f2_T=sp.simplify(f3_f2_T) #display(f3_f2_T.subs(replacements)) f3_f2_T=np.array(f3_f2_T.subs(replacements)).astype(np.float64) f3_f2_T=np.reshape(f3_f2_T,(3,3)) print(\"f3_f0_T dot End_Coord\") display(np.dot(f1_f0_T,np.dot(f2_f1_T,np.dot(f3_f2_T,End_Coord)))) Joint2=np.dot(f1_f0_T,np.dot(f2_f1_T,np.dot(f3_f2_T,End_Coord))) #print(\"\\nf4_f3_T\") f4_f3_T=sp.simplify(f4_f3_T) #display(f4_f3_T.subs(replacements)) f4_f3_T=np.array(f4_f3_T.subs(replacements)).astype(np.float64) f4_f3_T=np.reshape(f4_f3_T,(3,3)) print(\"f4_f0_T dot End_Coord\") display(np.dot(f1_f0_T,np.dot(f2_f1_T,np.dot(f3_f2_T,np.dot(f4_f3_T,End_Coord))))) EndEffector=np.dot(f1_f0_T,np.dot(f2_f1_T,np.dot(f3_f2_T,np.dot(f4_f3_T,End_Coord)))) fig=plt.figure() plt.plot([Joint0[0][0], Joint1[0][0]],[Joint0[1][0],Joint1[1][0]],'bo-') plt.plot([Joint1[0][0], Joint2[0][0]],[Joint1[1][0],Joint2[1][0]],'ro-') plt.plot([Joint2[0][0], EndEffector[0][0]], [Joint2[1][0],EndEffector[1][0]],'go-') ax.set_xlim([0, 1]) plt.axis('equal') ax.set_xlabel('X',size=20) ax.set_ylabel('Y',size=20) plt.savefig('2D_30_30_30_rotations_200212.png', dpi=300, bbox_inches='tight') Two Configurations for the Same Resulting Location of the End Effector An example of two different solutions for the inverse kinematics problem leading to the same end-effector position and orientation. We can demonstrate", "T1_0 = [cos(theta1), 0, sin(theta1), 0; sin(theta1), 0, -cos(theta1), 0; 0, 1, 0, d1; 0, 0, 0, 1]; T2_1 = [1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 1, d2; 0, 0, 0, 1]; T3_2 = [cos(theta3), -sin(theta3), 0, 0; sin(theta3), cos(theta3), 0, 0; 0, 0, 1, d3; 0, 0, 0, 1]; %Determine transformation T2_0 = T1_0 * T2_1; T2_0 = simple(T2_0) T3_0 = T2_0*T3_2; T3_0 = simple(T3_0) After executing the program, the output should be the following: T2_0 = [ cos(theta1), 0, sin(theta1), sin(theta1)*d2] [ sin(theta1), 0, -cos(theta1), -cos(theta1)*d2] [ 0, 1, 0, d1] [ 0, 0, 0, 1] T3_0 = [ cos(theta1)*cos(theta3), -cos(theta1)*sin(theta3), sin(theta1), sin(theta1)*(d3+d2)] [ sin(theta1)*cos(theta3), -sin(theta1)*sin(theta3), -cos(theta1), -cos(theta1)*(d3+d2)] [ sin(theta3), cos(theta3), 0, d1] [ 0, 0, 0, 1] ## References Bibliography 1. Spong, et.al. Robot Modeling and Control.", "# Solutions Q31 - 48# # import all python add-ons etc that will be needed later on %matplotlib inline import numpy as np import matplotlib.pyplot as plt from sympy import * init_printing() # allows printing of SymPy results in typeset maths format plt.rcParams.update({'font.size': 16}) # set font size for plots The three curves are shown in the polar plot below have $$a = 1$$ and are shown in the order (a) to (c). The centre plot also shows, as a dashed line, the curve when the phase is ignored. The right-hand figure is plotted with $$n = 5$$. Figure 44 Polar plots; left curve (a), right (c). (a) The angles when $$r=0$$ are the solution to $$0 = a \\cos(\\theta)\\sin(\\theta)$$ which are $$0$$ and $$\\pi/2$$. The integration for one quarter of the area is therefore $\\displaystyle A=\\frac{a^2}{2}\\int_0^{\\pi/2} r^2d\\theta = \\frac{a^2}{2}\\int_0^{\\pi/2} \\sin^2(\\theta)\\cos^2(\\theta) d\\theta= \\frac{a^2\\pi}{32}$ and the total area is four times this. The integration can be evaluated by converting to exponentials. Using Sympy to do the integration is also straightforward, theta, a = symbols('theta, a',positive = True) eq = ((a*sin(theta)*cos(theta))**2)/2 integrate(eq,(theta,0,pi/2)) (b) This is tackled in the same way as (a) and the only difficulty is in finding the angles to use as integration limits. The equation to do this is $$0 = a \\sin(3\\theta + \\pi/3)$$ giving angles" ]

[ "= linspace(0,1,30); theta = linspace(0,2*pi,100); [r theta] = meshgrid(r,theta); z = 2*r.*sin(theta).*(1-r.^2)./(1+r.^2-2*r.*cos(theta)).^2; % x = r.*cos(theta); % y = r.*sin(theta); % only for viewing positive/negative in the plot for ii = 1:100; for jj = 1:30; if z(ii,jj)>0; c(ii,jj)=1; else if z(ii,jj)<0 c(ii,jj)=-1; else c(ii,jj)=0; end end end end clear ii jj % what the solution looks like: red=positive, blue=negative, green=zero figure; mesh(r.*cos(theta),r.*sin(theta),z,c) % scaled for clarity figure; mesh(r.*cos(theta),r.*sin(theta),sign(z).*log(log(abs(z)+1)+1)+1) % the function is unbounded: compute the limit as (r,theta) spirals to (1,0) syms u R T T_R u = 2*R*sin(T)*(1-R^2)/(1+R^2-2*R*cos(T))^2; % to find this value, use solve(diff(U,T),T) and use the third value T_R = pi - acos(1/4/R*(R^2 - (R^4 + 34*R^2 + 1)^(1/2) + 1)); u = subs(u,T,T_R); figure; ezplot(u); limit(u,R,1,'left') % which approaches +inf Posted in Uncategorized", "we can compute some trigonometric functions on these values: print(\"theta = \", theta) print(\"sin(theta) = \", np.sin(theta)) print(\"cos(theta) = \", np.cos(theta)) print(\"tan(theta) = \", np.tan(theta)) theta = [ 0. 1.57079633 3.14159265] sin(theta) = [ 0.00000000e+00 1.00000000e+00 1.22464680e-16] cos(theta) = [ 1.00000000e+00 6.12323400e-17 -1.00000000e+00] tan(theta) = [ 0.00000000e+00 1.63312394e+16 -1.22464680e-16] The values are computed to within machine precision, which is why values that should be zero do not always hit exactly zero. Inverse trigonometric functions are also available: x = [-1, 0, 1] print(\"x = \", x) print(\"arcsin(x) = \", np.arcsin(x)) print(\"arccos(x) = \", np.arccos(x)) print(\"arctan(x) = \", np.arctan(x)) x = [-1, 0, 1] arcsin(x) = [-1.57079633 0. 1.57079633] arccos(x) = [ 3.14159265 1.57079633 0. ] arctan(x) = [-0.78539816 0. 0.78539816] #### Exponents and logarithms Another common type of operation available in a NumPy ufunc are the exponentials: x = [1, 2, 3] print(\"x =\", x) print(\"e^x =\", np.exp(x)) print(\"2^x =\", np.exp2(x)) print(\"3^x =\", np.power(3, x)) x = [1, 2, 3] e^x = [ 2.71828183 7.3890561 20.08553692] 2^x = [ 2. 4. 8.] 3^x = [ 3 9 27] The inverse of the exponentials, the logarithms, are also available. The basic np.log gives the natural logarithm; if you prefer to compute the base-2 logarithm or the base-10 logarithm, these are available as well: x = [1, 2, 4, 10]", "$x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curves correspond to $\\theta = \\psi_\\ast$, where $\\psi_\\ast\\approx -1.4289$ The evolution of $(x(t),y(t))$, $R(t)$, $|\\theta(t)|$ for solutions of (50) with $h = 1/6$, $a_0 = a_1 = 0.8$, $b_1 = 0.6$, $s_1 = s_2 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t)/2)$, $y(t) = -r(t)\\sin(\\theta(t)+S(t)/2)$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $b_0 = -1/4$, $z_0 = 0.6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue point corresponds to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$ The evolution of $(x(t),y(t))$, $r(t)$, $|\\theta(t)|$ for solutions of (51) with $h = 1/6$, $s_2 = 1$, $a_0 = a_1 = b_1 = z_1 = s_4 = 0$, where $x(t) = r(t)\\cos(\\theta(t)+S(t))$, $y(t) = -r(t)\\sin(\\theta(t)+S(t))$. The blue points correspond to initial data $(x(1),y(1))$. The gray solid curves correspond to level lines of $H_0(x,y)$. The gray dashed curve corresponds to $r = 2 R_\\ast t^{-\\nu-l/q}$ The evolution of $(x(t),y(t))$, $r(t)$, $\\theta(t)$ for", "# 2014S004 This is Sheng's CDS 130 course page. All of the information that you will ever need for this course can be found here or on a page linked to from this page. Final exam: May 12th, 7:30 AM, IN Room 222. # 3. Monday, May 5th • Integration • Note: The next class is on May 8th (make-up day for March 3rd) 9:00 AM # 9. Monday, April 14th x = linspace(0, 2*pi, 100); y = sin(x); plot(x, y, 'r*') axis ([0, 2*pi, -1.5, 1.5]) axis square xlabel ('xlabel') ylabel ('sin(x)') text(4.0, 1.0, 'sin(x)', 'FontSize', 20, 'Color', 'y') hold on theta = pi/2; y = sin(x + theta) plot (x, y, 'g^') shg print -dpng mypic.png clear all; clc; x = [0, 1, 1, 0]; y = [0, 0, 1, 1]; fill (x, y, 'r') axis off axis equal hold on fill (x+1, y, 'g') fill (x, y+1, 'g') fill (x+1, y+1, 'r') theta = linspace(0, 2*pi, 7) x = cos(theta) y = sin(theta) fill (x +1 , y +1, [0.8, 0.6, 0.5]) fill (x*0.75 +1 , y*0.75 +1, 'w') # 10. Wednesday, April 9th • Iterations • Code to draw circles theta = 0:2*pi/60:2*pi; r = 1.0; x = r* cos(theta); y = r* sin(theta); for i = 1:60 plot (x(i), y(i), '-or', 'MarkerSize',20) plot", "= \\sinh \\theta, \\quad \\mathrm{d} u = \\cosh \\theta \\mathrm{d} \\theta, \\quad \\cosh^2\\theta - \\sinh^2 \\theta = 1 \\\\ y' = u = \\sinh \\theta = \\sinh( g \\rho (x + C) ) \\\\ y(x) = \\frac{1}{H} \\cosh( H (x + C) ) + K \\end{gather}$ We can also verify the solution using sympy [Show code] 1 2 3 4 5 6 7 from sympy import * from sympy.abc import x,y,m,a,c,w P = Integral( y(x) * sqrt( y(x).diff(x) ** 2 + 1 ), (x,0,w)) L = Integral( sqrt( y(x).diff(x) ** 2 + 1 ), (x,0,w)) A = P.args[0]-m * L.args[0] F = (A.diff(y(x)) - A.diff(y(x).diff(x)).diff(x)).factor() F.subs(y(x), m+cosh(c * (a+x))/c).doit().factor().simplify() Note that neither constant $$C$$ or $$K$$ effects the shape of the catenary, only the position. If we have boundary conditions $$y(-w/2)=y(w/2)=0$$, then solving for the constants we find $$C=0$$ and $$K = -\\cosh(Hw/2)/H$$. $y(x) = \\frac{ \\cosh( H x ) - \\cosh(Hw/2) }{H}$ The shape is controlled completely by $$H$$. To find $$H$$, we need to make sure the hanging cable as the same length as it did when we started. Thus, $s = \\int_{-w/2}^{w/2} \\sqrt{ 1 + \\left( \\frac{\\mathrm{d} y}{\\mathrm{d} x} \\right)^2 } \\mathrm{d} x$ $s = \\int_{-w/2}^{w/2} \\sqrt{ 1 + \\left( \\frac{\\mathrm{d} y}{\\mathrm{d} x} \\right)^2 } \\mathrm{d} x$ After doing this integral, we should find $s", "}_{1}{A}_{1}+{\\gamma }_{2}{A}_{2}$ 1. Draw $\\gamma =\\gamma \\left(\\theta \\right)$ 2. Draw line from origin to $\\gamma$ curve for a given value of $\\theta$ 3. Draw perpendicular to this line. 4. Repeat for all values of $\\theta$. 5. Inner envelope is equilibrium shape. close all gamma=@(alpha) abs(cos(alpha))+abs(sin(alpha)); r=@(theta,alpha) gamma(alpha)/cos(theta-alpha); alpha=linspace(0,2*pi,200); polar(alpha,gamma(alpha),'r-'); title('\\gamma=|cos\\theta|+|sin\\theta|', 'fontsize', 16) hold on % plot all subsequent curves on existing axes for alpha=linspace(0,2*pi,17) % this is the loop that draws all the lines theta(1)=alpha+2*pi/5; % specify two angles on either side of alpha theta(2)=alpha-2*pi/5; rvals(1)=r(theta(1),alpha); % use the equation provided to get r for each % of the specified angles rvals(2)=r(theta(2),alpha); polar(theta,rvals) % plot lines connecting the two points we just defined end set(gcf,'paperposition',[0 0 5 5],'papersize',[5 5]) print(gcf,'../figures/matlabwulffenergyexample.eps', '-depsc2') % save the eps file Figure 6.6: Wulff construction for the the surface free energy function given by Eq. 6.10. ## 7 Grain Boundaries In the two dimensional Wulff construction the interface surface of interest is specified by a single variable, $\\theta$. For a true, three-dimensional crystal the Wulff circle becomes a Wulff sphere, and we need two different angles to specify the the orientation on this sphere. In other words, we have two degrees of freedom in specifying a specific surface of a three-dimensional crystal. We commonly specify a surface by using the normal vector, $\\stackrel{ˆ}{n}$,", "1 - ny^2/(1 + nz), -ny}; x1 + d1 dvec + r (Cos[th] rr + Sin[th] ss)]; sphereIntersection[s1_Sphere, s2_Sphere, th_] := sphereIntersection[List @@ s1, List @@ s2, th] You get your equation via sphereIntersection[sp1, sp2, theta] // FullSimplify (* {( 7 (3 + Sqrt[3]) + Sqrt[143] (3 + 2 Sqrt[3]) Cos[theta] - Sqrt[429] Sin[theta])/(24 (3 + Sqrt[3])), ( 7 (3 + Sqrt[3]) - Sqrt[429] Cos[theta] + Sqrt[143] (3 + 2 Sqrt[3]) Sin[theta])/(24 (3 + Sqrt[3])), 1/24 (7 - Sqrt[143] Cos[theta] - Sqrt[143] Sin[theta])} *) And generate points via sphereIntersection[sp1, sp2, #] & /@ Range[0, 2 π, π/2] // N (* {{0.972304, 0.109291, -0.206594}, {0.109291, 0.972304, -0.206594}, {-0.38897, 0.474043, 0.789928}, {0.474043, -0.38897, 0.789928}, {0.972304, 0.109291, -0.206594}} *) And plot it via Show[Graphics3D[{Opacity[0.2], sp1, sp2}], ParametricPlot3D[ sphereIntersection[sp1, sp2, theta], {theta, 0, 2 π}, Evaluated -> True] ] • You might want to see this paper. – J. M. will be back soon Jul 21 '16 at 8:56 • @J.M. Thanks, it's better without using a random number to generate the basis vectors – Jason B. Jul 21 '16 at 20:49 • I know; I was actually trying to look for this paper when I saw your solution. Apart from that little typo I've now fixed, it's pretty nice! – J. M. will be back soon Jul 22 '16 at 1:30", "theta]] [[cos phi,sin phi, 0],[-sin phi,cos phi,0],[0,0,1]] MRC psi, theta, Omega [[cos psi,sin psi, 0],[-sin psi,cos psi,0],[0,0,1]] [[cos theta,0,sin theta],[0, 1, 0],[-sin theta, 0, cos theta]] [[cos Omega,sin Omega, 0],[-sin Omega,cos Omega,0],[0,0,1]] XYZ z, y, x [[cos z,sin z, 0],[-sin z, cos z,0],[0,0,1]] [[cos y,0,sin y],[0, 1, 0],[-sin y, 0, cos y]] [[1,0,0],[0, cos x, sin x],[0, -sin x, cos x]] spin Omega, n1, n2, n3 Quaternion rotation by Omega degrees about the vector [n1,n2\\,n3] sgirot q, n1, n2, n3 Quaternion rotation by q degrees about the vector [n1,n2\\,n3] quaternion e0, e1, e2, e3 Literal quaternion rotation using q = e0 + e1\\mathbf{i} + e2\\mathbf{j} + e2\\mathbf{k} 2d alpha Rotation in 2-D only, for 3-D objects this is a rotation about Z # The Transform object in Python ## Constructing a Transform There a four ways to construct a Transform object in Python t = Transform() # default constructor, t is the identity t = Transform({\"type\":\"eman\",\"az\":10,\"alt\":150,\"scale\":2.0,\"mirror\":True,\"tx\":3.4}) # construction using a dictionary p = Transform({\"type\":\"eman\",\"az\":10,\"mirror\":True,\"tx\":3.4}) # construction using a dictionary q = Transform([1,0,0,0,0,1,0,0,0,0,1,0]) # specify the 12 entries of the matrix explicitly, row wise (3 rows of 4) s = Transform(t) # copy construction - s is precisely the same as t NOTE: All the CONVENTIONS listed on the table above can be supplied as arguments or \"values\" for", "- 2) - 1(sin theta - 2) = 0 ⇒ (sin theta - 2)(sin theta - 1) = 0 ⇒ (sin theta - 2) = 0 or, (sin theta - 1) = 0 ⇒sintheta = 2 or, sin theta = 1 So, value of sin theta can’t be greater than 1, Therefore sin theta = 1 ⇒theta = 90° Basic Trigonometric Ratios To know about the basic trigonometric ratios with respect to a right-angled triangle let a ray OA revolve in the anti-clockwise direction and assume the position (OA)_1, so that an angle /_AOA_1 = theta is formed. Now any number of points P, Q, R, .......... are taken on OA_1, and perpendiculars PX, QY, RZ, ........ are drawn on OA from those points respectively. All the right-angled triangles POX, QOY, ROZ, ......... are similar to each other. Now from the properties of similar triangles we know, (i) (PX)/(OP) = (QY)/(OQ) = (RZ)/(OR) = ..... (ii) (OX)/(OP) =(QY)/(OQ) = (OZ)/(OR) = ..... (iii) (PX)/(OX) = (QY)/ (OQ) = (RZ)/(OZ) = ..... (v) (OP)/(OX) = (OQ)/(OX) = (OR)/(OZ) = ..... (iv) (OP)/(PX) = (OQ)/(QY) = (OR)/(RZ) = ..... (vi) (OX)/(PX) = (OY)/(QY) = (OZ)/(RZ) = ..... Thus we see in a set of similar right-angled triangles with respect to the same acute angle (i) perpendicular : hypotenuse i.e., perpendicular/hypotenuse", "to arbitrarily decide that $$\\gamma = 0$$, which means that $$\\sin(\\gamma) = 0$$ and $$\\cos(\\gamma) = 1$$. The formulas in this case are: \\begin{align*} \\beta = \\mbox{atan2}( -R_{2,0}, \\cos(\\beta) ) \\\\ \\alpha = \\mbox{atan2}( -R_{1,2}, R_{1,1} ) \\\\ \\gamma = 0 \\end{align*} There are, of course, other options. Python code implementing these operations is below, along with unit testing for the degenerate and non-degerate cases. import unittest import numpy def x_rotation( theta ): \"\"\"Generate a 4x4 homogeneous rotation matrix about x-axis\"\"\" c = numpy.cos(theta) s = numpy.sin(theta) return numpy.array([ [ 1, 0, 0, 0], [ 0, c,-s, 0], [ 0, s, c, 0], [ 0, 0, 0, 1] ]) def y_rotation( theta ): \"\"\"Generate a 4x4 homogeneous rotation matrix about y-axis\"\"\" c = numpy.cos(theta) s = numpy.sin(theta) return numpy.array([ [ c, 0, s, 0], [ 0, 1, 0, 0], [-s, 0, c, 0], [ 0, 0, 0, 1] ]) def z_rotation( theta ): \"\"\"Generate a 4x4 homogeneous rotation matrix about z-axis\"\"\" c = numpy.cos(theta) s = numpy.sin(theta) return numpy.array([ [ c,-s, 0, 0], [ s, c, 0, 0], [ 0, 0, 1, 0], [ 0, 0, 0, 1] ]) def xyz_rotation( angles ): \"\"\"Generate 4x4 homogeneous rotation in x then y then z order\"\"\" return numpy.dot( z_rotation(angles[2]), numpy.dot( y_rotation(angles[1]), x_rotation(angles[0]) ) ) def xyz_rotation_angles( R, eps=1e-8, check=False ):", "but for simplicity, we choose to implement it using two rotations: one with angle $$\\theta_1$$ for the qumodes 1 and 2, and another with angle $$\\theta_2$$ for the qubits 3 and 4. The final matrix has the form: $\\begin{split}\\begin{pmatrix} \\cos(\\theta_1) & - \\sin(\\theta_1) & 0 & 0 \\\\ \\sin(\\theta_1) & cos(\\theta_1) & 0 & 0 \\\\ 0 & 0 & \\cos(\\theta_2) & - \\sin(\\theta_2) \\\\ 0 & 0 & \\cos(\\theta_2) & \\sin(\\theta_2) \\end{pmatrix}~~\\text{with}~~\\theta_1, \\theta_2 \\in [0,2\\pi).\\end{split}$ theta1 = 0.5 theta2 = 1 U = np.array([[np.cos(theta1), -np.sin(theta1), 0, 0 ], [np.sin(theta1), np.cos(theta1), 0, 0 ], [0, 0, np.cos(theta2), -np.sin(theta2)], [0, 0, np.sin(theta2), np.cos(theta2)]]) ### Circuit¶ We instantiate our circuit with 8 qubits, 4 for the input, 4 for the output. prog = sf.Program(8) We can then declare our circuit. The first four lines are 2-modes squeezing gates, which generate a random number of photons with prog.context as q: S2gate(r_squeezing) | (q[0], q[4]) S2gate(r_squeezing) | (q[1], q[5]) S2gate(r_squeezing) | (q[2], q[6]) S2gate(r_squeezing) | (q[3], q[7]) Interferometer(U) | (q[4], q[5], q[6], q[7]) ### Running¶ Run the simulation up to ‘cutoff’ photons per mode eng = sf.Engine(\"fock\", backend_options={\"cutoff_dim\":cutoff}) state = eng.run(prog).state Out: /home/circleci/project/venv/lib/python3.7/site-packages/strawberryfields/program.py:662: UserWarning: The circuit consists of 2 disconnected components. Get the probability associated to each state probs = state.all_fock_probs() Reshape ‘probs’ such that probs $$[m_1, \\dots, m_4,n_1, \\dots, n_4]$$ gives", "draw with Cartesian coordinates. Figure 3.8: The Greek letter θ (theta) is often used to denote an angle. Since a polar coordinate is conventionally referred to as (r, θ), we’ll use theta as a variable name when referring to an angle. sine(theta) = y/r → y = r * sine(theta) cosine(theta) = x/r → x = r * cosine(theta) For example, if r is 75 and theta is 45 degrees (or PI/4 radians), we can calculate x and y as below. The functions for sine and cosine in Processing are sin() and cos(), respectively. They each take one argument, an angle measured in radians. Show Raw float r = 75; float theta = PI / 4; Converting from polar (r,theta) to Cartesian (x,y) float x = r * cos(theta); float y = r * sin(theta); This type of conversion can be useful in certain applications. For example, to move a shape along a circular path using Cartesian coordinates is not so easy. With polar coordinates, on the other hand, it’s simple: increment the angle! Here’s how it is done with global variables r and theta. Your browser does not support the canvas tag. Example 3.4: Polar to Cartesian Show Raw float r = 75; float theta = 0; void setup() { size(640,360); background(255); } void draw() { Polar", "T1_0 = [cos(theta1), 0, sin(theta1), 0; sin(theta1), 0, -cos(theta1), 0; 0, 1, 0, d1; 0, 0, 0, 1]; T2_1 = [1, 0, 0, 0; 0, 1, 0, 0; 0, 0, 1, d2; 0, 0, 0, 1]; T3_2 = [cos(theta3), -sin(theta3), 0, 0; sin(theta3), cos(theta3), 0, 0; 0, 0, 1, d3; 0, 0, 0, 1]; %Determine transformation T2_0 = T1_0 * T2_1; T2_0 = simple(T2_0) T3_0 = T2_0*T3_2; T3_0 = simple(T3_0) After executing the program, the output should be the following: T2_0 = [ cos(theta1), 0, sin(theta1), sin(theta1)*d2] [ sin(theta1), 0, -cos(theta1), -cos(theta1)*d2] [ 0, 1, 0, d1] [ 0, 0, 0, 1] T3_0 = [ cos(theta1)*cos(theta3), -cos(theta1)*sin(theta3), sin(theta1), sin(theta1)*(d3+d2)] [ sin(theta1)*cos(theta3), -sin(theta1)*sin(theta3), -cos(theta1), -cos(theta1)*(d3+d2)] [ sin(theta3), cos(theta3), 0, d1] [ 0, 0, 0, 1] ## References Bibliography 1. Spong, et.al. Robot Modeling and Control.", "the solution \"\"\" # import math # Angle for heteronuclear diatonic Theta = 0.5*math.atan(2.0*Fprime[0,1]/(Fprime[0,0]-Fprime[1,1])) #print('Theta', Theta) Cprime[0,0] = np.cos(Theta) Cprime[1,0] = np.sin(Theta) Cprime[0,1] = np.sin(Theta) Cprime[1,1] = -np.cos(Theta) E[0,0] = Fprime[0,0]*np.cos(Theta)**2+Fprime[1,1]*np.sin(Theta)**2+Fprime[0,1]*np.sin(2.0*Theta) E[1,1] = Fprime[1,1]*np.cos(Theta)**2+Fprime[0,0]*np.sin(Theta)**2-Fprime[0,1]*np.sin(2.0*Theta) if (E[1,1] <= E[0,0]): Temp = E[1,1] E[1,1] = E[0,0] E[0,0] = Temp Temp = Cprime[0,1] Cprime[0,1] = Cprime[0,0] Cprime[0,0] = Temp Temp = Cprime[1,1] Cprime[1,1]=Cprime[1,0] Cprime[1,0]=Temp return In [145]: def HFCALC(N,R,Zeta1,Zeta2,Za,Zb,G): \"\"\" Calculates the integrals constructs the matrices and then runs the SCF calculation \"\"\" # Calculate one and two electron integrals Intgrl(N,R,Zeta1,Zeta2,Za,Zb) # Put all integals into array Colect(N,R,Zeta1,Zeta2,Za,Zb) # Perform the SCF calculation SCF(N,R,Zeta1,Zeta2,Za,Zb,G) return In [146]: \"\"\" Let's set up the variables and perform the calculations \"\"\" global H,S,X,XT,TT,G,C,P,Oldp,F,Fprime,Cprime,E,Zb H = np.zeros([2,2]) S = np.zeros([2,2]) X = np.zeros([2,2]) XT = np.zeros([2,2]) TT = np.zeros([2,2,2,2]) G = np.zeros([2,2]) C = np.zeros([2,2]) P = np.zeros([2,2]) Oldp = np.zeros([2,2]) F = np.zeros([2,2]) Fprime = np.zeros([2,2]) Cprime = np.zeros([2,2]) E = np.zeros([2,2]) Energy = 0.0 Delta = 0.0 N = 3 R = 1.4632 Zeta1 = 2.0925 Zeta2 = 1.24 Za = 2.0 Zb = 1.0 HFCALC(N,R,Zeta1,Zeta2,Za,Zb,G) 1 Electronic energy = 0.0 Delta 0.882866853014 2 Electronic energy = -4.14186287613 Delta 0.42637666285 3 Electronic energy = -4.21250515815 Delta 0.169751444038 4 Electronic energy = -4.22491713481 Delta 0.0726766265559 5 Electronic energy = -4.22705796346 Delta 0.0305577508572 6 Electronic", "is +ve. • In the forth quadrant, sin is -ve; cos is +ve; tan is -ve. There are various methods of remembering the above table. One way is by using the CAST diagram. Each letter stands for the positive ratios. e.g. In the first quadrant all ratios are positive whilst in the third quadrant only tan is positive. Since the magnitude of the projections of the unit radius on the axies depend solely on the acute angle which the radius makes with the x-axis, any ratio of any angle is equal numerically to the same ratio of the acute angle which the radius makes with the x-axis. The sign must be found from the Cast circle. Example Find the value of sin 210 degrees. As 210 is in the third quadrant the sin is negative and the acute angle which the radius makes with the x-axis is 30 degrees. $\\therefore\\;\\;\\;\\;\\;sin\\;210^0&space;=&space;sin(210&space;-&space;180)\\;=&space;-&space;sin\\;30\\;=&space;-&space;\\frac{1}{2}$ To simplify expressions such as $\\inline&space;&space;tan(270^0&space;-&space;\\theta)$ follow the same logic but for simplicity imagine that $\\inline&space;&space;\\theta$ to be acute. The formula will apply for all values of $\\inline&space;&space;\\theta$ but the argument is simplified by supposing that it is acute. Examples $sin(180^0&space;-&space;\\theta&space;)\\;=&space;+&space;sin\\;\\theta$ $cos(180^0&space;+&space;\\theta&space;)\\;=&space;-&space;cos\\;\\theta$ $tan(270^0&space;-&space;\\theta&space;)\\;=&space;+&space;tan\\;(90^0\\;-\\theta)&space;=&space;cot\\;\\theta$ $sec(360^0&space;-&space;3\\theta&space;)\\;=&space;+&space;sec\\;3\\theta$ $cot(-&space;2\\theta&space;)\\;=&space;-&space;cot\\;2\\theta$ It should be noted that cosec; sec; and cot follow the same rules as sin;cos; and tan respectively since they are their reciprocals.", "Cos = SQUAREROOT ( 1 - (16/49)) Cos = Squareroot (33 / 49) Cos = Squareroot(33) / 7 [You know squareroot of 49 doesn't = -7, and it = 7, because it's in Quadrent 4, where Cosine MUST be positive.] So now you have: Sin = -4/7, Cos = Squareroot(33) / 7 1/Cot = Tan = Sin/Cos = (-4/7) / (squareroot(33)/7) = -4/squareroot(33) Get it? 19. Jan 10, 2005 Zaimeen aisha, it's best to draw a right angle triangle, with the hypotenuse being 7 and one of the sides being 4. Using Pythagoras theorem, the other side will be $$/sqrt 33$$ You know that cot theta = 1/(tan theta), right? So, tan theta = (length_of_opposite) / (length_of_adjacent) Hence, cot theta = (length_of_adjacent) / (length_of_opposite) So, cot theta will be $$- \\frac {/sqrt 33}{4}$$ 20. Jan 10, 2005 HallsofIvy With what? You were just told that the solution you give is correct. The way I would approach this problem is: First draw a picture: a unit circle on a pair of axes. You are told that theta lies between 3pi/2 and 2 pi and you should know (MEMORIZE) that the axes correspond to 0 (at (1,0), pi/2 (at (0,1), pi (at (-1,0), 3pi/2 (at (0,-1) (and 2pi again at (1,0)). This is a circle with radius 1, diameter 2,", "180^\\circ) ~ = ~ -\\sin\\;\\theta$ $\\label{1.11} \\cos\\;(\\theta \\pm 180^\\circ) ~ = ~ -\\cos\\;\\theta$ $\\label{1.12}\\tan\\;(\\theta \\pm 180^\\circ) ~ = ~ \\tan\\;\\theta$ A reflection is simply the mirror image of an object. For example, in Figure 1.5.5 the original object is in QI, its reflection around the $$y$$-axis is in QII, and its reflection around the $$x$$-axis is in QIV. Notice that if we first reflect the object in QI around the $$y$$-axis and then follow that with a reflection around the $$x$$-axis, we get an image in QIII. That image is the reflection around the origin of the original object, and it is equivalent to a rotation of $$180^\\circ$$ around the origin. Notice also that a reflection around the $$y$$-axis is equivalent to a reflection around the $$x$$-axis followed by a rotation of $$180^\\circ$$ around the origin. Figure 1.5.5 Applying this to angles, we see that the reflection of an angle $$\\theta$$ around the $$x$$-axis is the angle $$-\\theta$$, as in Figure 1.5.6. Figure 1.5.6 Reflection of $$θ \\text{ around the }x\\text{-axis}$$ So we see that reflecting a point $$(x,y)$$ around the $$x$$-axis just replaces $$y$$ by $$-y$$. Hence: $\\label{1.13}\\sin\\;(-\\theta) ~ = ~ -\\sin\\;\\theta$ $\\label{1.14} \\cos\\;(-\\theta) ~ = ~ \\cos\\;\\theta$ $\\label{1.15} \\tan\\;(-\\theta) ~ = ~ -\\tan\\;\\theta$ Notice that the cosine function does not change in Equation \\ref{1.14} because it", "plus $1/(8\\pi-2)$ times two uniform variates (for $\\Theta$) and a small amount of (fast) calculation. That's $4/(\\pi-4) \\approx 4.66$ variates per point (which, of course, has two coordinates). Full details are in the code example below. This figure plots 10,000 out of more than a half million points generated. Here is the R code that produced this simulation and timed it. n.sim <- 1e6 x.time <- system.time({ # Generate trial angles theta theta <- sqrt(runif(n.sim)) * pi/4 # Rejection step. theta <- theta[runif(n.sim) * 4 * theta <= pi * tan(theta)^2] # Generate radial coordinates r. n <- length(theta) r <- sqrt(1 + runif(n) * tan(theta)^2) # Convert to Cartesian coordinates. # (The products will generate a full circle) x <- r * cos(theta) #* c(1,1,-1,-1) y <- r * sin(theta) #* c(1,-1,1,-1) # Swap approximately half the coordinates. k <- rbinom(1, n, 1/2) if (k > 0) { z <- y[1:k] y[1:k] <- x[1:k] x[1:k] <- z } }) message(signif(x.time[3] * 1e6/n, 2), \" seconds per million points.\") # # Plot the result to confirm. # plot(c(0,1), c(0,1), type=\"n\", bty=\"n\", asp=1, xlab=\"x\", ylab=\"y\") rect(-1, -1, 1, 1, col=\"White\", border=\"#00000040\") m <- sample.int(n, min(n, 1e4)) points(x[m],y[m], pch=19, cex=1/2, col=\"#0000e010\") • I don't understand this sentence: \"Because the samples are independent, systematically swapping the coordinates every second sample produces", "# define ctheta as cosine of theta stheta = np.sin(theta*deg2rad)# define stheta as sine of theta ttheta = np.tan(theta*deg2rad) # define ttheta as tangent of theta print \"%5.1f %8.4f %8.4f %8.4f\" %(theta, ctheta, stheta, ttheta) Let’s pick this program apart a bit. First, notice the use of the variable deg2rad to convert from degrees to radians. Also notice how deg2rad is defined: deg2rad = np.pi/180. using the NumPy function for π and the decimal point after 180. While in this case, it makes absolutely no difference (try it!), it is a good practice to use floating point numbers unless you want your variable to stay an integer. In fact: Good housekeeping Tip #3: Always use a decimal if you want your variable to be a floating point variable. In the above example, 180. or 180.0 give the same floating point number. The expression ctheta = np.cos(theta*deg2rad) uses the numpy cosine function. Ideally, theta should be a floating point number while in fact it is an integer in this expression, but fortunately Python figures that out and converts it to a floating point number. Note that we could have also converted theta to a float first with the command float(theta). print \"%5.1f %8.4f %8.4f %8.4f\" %(theta, ctheta, stheta, ttheta) To make the output look nice, we do not use", "z]) % from 16.5 F = div_F = I = int(int(int(div_F, z, 0, c*(1 - x/a - y/b)), y, 0, b*(1 - x/a)), x, 0, a) I = #### 6. [1145/12] We apply the Divergence Theorem to compute the flux through a surface bounded by a circular cylinder and two disks, one circular and one elliptical. syms r u x y z theta F = [x*y+2*x*z x^2+y^2 x*y-z^2], div_F = divergence(F, [x y z]) % from 16.5 F = div_F = div_F = 3*r*sin(theta) % Convert to cylindrical coordinates. div_F = I = int(int(int(div_F * r, z, r*sin(theta)-2, 0), r, 0, 2), theta, 0, 2*pi) I = I_appx = double(I) I_appx = -37.6991 % figure % The x-axis is stretched to better visualize the cylinder and disks. fsurf(2*cos(theta), 2*sin(theta), u*(2*sin(theta)-2), [0 2*pi 0 1], 'y', 'EdgeColor', 'none') hold on fsurf(r*cos(theta), r*sin(theta), r*sin(theta)-2, [0 2 0 2*pi], 'r', 'EdgeColor', 'none') fsurf(r*cos(theta), r*sin(theta), 0, [0 2 0 2*pi], 'g', 'EdgeColor', 'none') axis([-2 2 -2 2 -4 0]) xticks(-2:2:2); yticks(-2:2:2); zticks(-4:2:0) xlabel('x'); ylabel('y'); zlabel('z') title('SET8, 1145/12'); view(54,12); alpha 0.3 #### 7. [1146/14] We apply the Divergence Theorem to compute the flux through a sphere. syms a x y z rho phi theta F = (x^2 + y^2 + z^2)*[x y z], F = div_F = simplify( divergence(F, [x y z]) )" ]

[ "4 \\\\ 3 \\end{pmatrix}.$ ×", "= \\begin{pmatrix} -1 \\\\\\\\ 1 \\end{pmatrix}$", "1 & 1 \\end{pmatrix}$", "$$(1, 1) \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{12} & a_{22} \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix} = a_{11} + 2 a_{12} + a_{22}$$", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "\\\\ 0 & 1\\end{pmatrix}.$ -", "\\begin {pmatrix} 12 & 6 & -2 \\\\ -6 & -3 & 1 \\\\ 6 & 3 & -1 \\end {pmatrix}$", "true. So $A^n =\\begin{pmatrix}a^n & 0\\\\0 & b^n\\end{pmatrix}, \\forall n \\in \\mathbb{N}$", "1/2 \\\\ \\end{pmatrix}$$", "Find $A^2=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}\\times \\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}$ So, $B=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}^2-3\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}+17\\begin{pmatrix}1 & 0 \\\\0 & 1\\end{pmatrix}$", "&2 \\\\ 1 &0 &3 \\end{pmatrix}$ 2. $\\begin{pmatrix} 1 &-1 &2 \\\\ 3 &-3 &6 \\\\ -2 &2 &-4 \\end{pmatrix}$ 3. $\\begin{pmatrix} 1 &3 &2 \\\\ 5 &1 &1 \\\\ 6 &4 &3 \\end{pmatrix}$ 4. $\\begin{pmatrix} 0 &0 &0 \\\\ 0 &0 &0 \\\\ 0 &0 &0 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 6 Find a basis for the span of each set. 1. $\\{\\begin{pmatrix} 1 &3 \\end{pmatrix}, \\begin{pmatrix} -1 &3 \\end{pmatrix}, \\begin{pmatrix} 1 &4 \\end{pmatrix}, \\begin{pmatrix} 2 &1 \\end{pmatrix} \\}\\subseteq\\mathcal{M}_{1 \\! \\times \\! 2}$ 2. $\\{\\begin{pmatrix} 1 \\\\2 \\\\1 \\end{pmatrix}, \\begin{pmatrix} 3 \\\\ 1 \\\\ -1 \\end{pmatrix}, \\begin{pmatrix} 1 \\\\ -3 \\\\ -3 \\end{pmatrix} \\}\\subseteq\\mathbb{R}^3$ 3. $\\{1+x,1-x^2,3+2x-x^2\\}\\subseteq\\mathcal{P}_3$", "$$\\begin{pmatrix} 1&1&1 \\\\1&0&0.5 \\\\ 0&1&0.5 \\end{pmatrix}$$", "& 6 & 8 & 2 & 1 & 3 \\end{pmatrix} = \\begin{pmatrix} 1 & 4 & 6 & 2 & 5 & 8 & 3 & 7 \\\\ 4 & 6 & 2 & 5 & 8 & 3 & 7 & 1 \\end{pmatrix} = (14625837)$", "6 & 0 & 5 \\end{pmatrix}$", "2 } \\\\ { 20 } & { 6 } \\\\ { 0 } & { 14 } \\end{pmatrix} \\$", "& 4 \\end{pmatrix} \\begin{pmatrix} 3 & 6 & 9 \\end{pmatrix} \\begin{pmatrix} 5 & 8 \\end{pmatrix}$.", "\\begin{pmatrix} -\\frac{5}{9}\\\\ \\frac{2}{9}\\\\ -\\frac{1}{3}\\end{pmatrix}.$$ • @frank I think your third elementary row operation should be $-2R_2 + R_3$. • Yes, that looks much better. Thanks!", "$x^2$ term in the fourth power expansion and by multiplying \"4\" by the $x^3$ term. The $x^2$ term in $(x- 2)^4$ is $\\begin{pmatrix}4 \\\\ 2\\end{pmatrix}(-2)^2x^2$ and the $x^3$ term is $\\begin{pmatrix}4 \\\\ 3\\end{pmatrix}(-2)^1x^3$. Multiplying the first by 3x and the second by 4 gives $\\left(\\begin{pmatrix}4 \\\\ 2\\end{pmatrix}(3)(-2)^2+ \\begin{pmatrix}4 \\\\ 3\\end{pmatrix}(4)(-2)\\right)x^3$", "\\end{pmatrix}$ $\\begin{pmatrix} 0 & 1\\\\ -1 & 0 \\end{pmatrix}$= $\\begin{pmatrix} 2 & 0\\\\ 0 & 1 \\end{pmatrix}$", "& 1 \\\\ 1 & 0 \\end{pmatrix}$ $D = \\begin {pmatrix} -1 & 0 \\\\ 0 & 1 \\end {pmatrix}$ $P = \\begin {pmatrix} -1 & 1 \\\\ 1 & 1 \\end{pmatrix}$" ]

[ "-4 & 1 & -3 & 0 \\\\ 3 & -3 & -2 & -3 & 3 & 5 & 0 & -3 & 0 & -1 & -2 & 0 & 3 & 2 & 6 \\end{pmatrix} \\\\ T &=", "2+2\\cdot 3+0\\cdot 4\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3+2\\cdot 4 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3+1\\cdot 4\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 11\\\\4\\end{pmatrix}$; d) $\\begin{pmatrix} 1 & 2 & 0 \\\\ 0 & 1 & 2 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\\\4 \\end{pmatrix}=\\begin{pmatrix}1\\cdot 1+2\\cdot 2+0\\cdot 3\\\\ 0\\cdot 1+1\\cdot 2+2\\cdot 3\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 3\\\\0\\end{pmatrix}$. ### This Post Has One Comment 1. a) should be [13 31]. 1+6+6 = 13", "-1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix}$$. Therefore, $$U_2 =\\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 1 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0 \\end{pmatrix} \\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 1 & -1 & 1 & -1\\\\ 1 & 1 & -1 & -1\\\\1 & -1 & -1 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 1\\\\ 0 & 0 & 1 & 0\\\\ 0 & 1 & 0 & 0 \\end{pmatrix}$$ Thus, $$U_1 = U_2$$.", "{10}}}\\\\{-{\\sqrt {10}}}\\\\{-i{\\sqrt {5}}}\\\\{1}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+3}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4{\\sqrt {2}}}}{\\begin{pmatrix}{i{\\sqrt {5}}}\\\\{-3}\\\\{-i{\\sqrt {2}}}\\\\{-{\\sqrt {2}}}\\\\{-3i}\\\\{\\sqrt {5}}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+1}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4}}{\\begin{pmatrix}{-i{\\sqrt {5}}}\\\\{1}\\\\{-i{\\sqrt {2}}}\\\\{\\sqrt {2}}\\\\{-i}\\\\{\\sqrt {5}}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-1}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4}}{\\begin{pmatrix}{i{\\sqrt {5}}}\\\\{1}\\\\{i{\\sqrt {2}}}\\\\{\\sqrt {2}}\\\\{i}\\\\{\\sqrt {5}}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-3}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4{\\sqrt {2}}}}{\\begin{pmatrix}{-i{\\sqrt {5}}}\\\\{-3}\\\\{i{\\sqrt {2}}}\\\\{-{\\sqrt {2}}}\\\\{3i}\\\\{\\sqrt {5}}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-5}{2}}\\rangle _{y}=\\!\\!\\!\\!\\!&{\\frac {1}{4{\\sqrt {2}}}}{\\begin{pmatrix}{i}\\\\{\\sqrt {5}}\\\\{-i{\\sqrt {10}}}\\\\{-{\\sqrt {10}}}\\\\{i{\\sqrt {5}}}\\\\{1}\\end{pmatrix}}\\\\S_{z}={\\frac {\\hbar }{2}}{\\begin{pmatrix}5&0&0&0&0&0\\\\0&3&0&0&0&0\\\\0&0&1&0&0&0\\\\0&0&0&-1&0&0\\\\0&0&0&0&-3&0\\\\0&0&0&0&0&-5\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+5}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{1}\\\\{0}\\\\{0}\\\\{0}\\\\{0}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+3}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{1}\\\\{0}\\\\{0}\\\\{0}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {+1}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{0}\\\\{1}\\\\{0}\\\\{0}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-1}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{0}\\\\{0}\\\\{1}\\\\{0}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-3}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{0}\\\\{0}\\\\{0}\\\\{1}\\\\{0}\\end{pmatrix}},\\!\\!\\!\\!\\!&|{\\frac {5}{2}},{\\frac {-5}{2}}\\rangle _{z}=\\!\\!\\!\\!\\!&{\\begin{pmatrix}{0}\\\\{0}\\\\{0}\\\\{0}\\\\{0}\\\\{1}\\end{pmatrix}}\\\\\\end{array}}}$ 5. The generalization of these matrices for arbitrary spin s is {\\displaystyle {\\begin{aligned}\\left(S_{x}\\right)_{ab}&={\\frac {\\hbar }{2}}\\left(\\delta _{a,b+1}+\\delta _{a+1,b}\\right){\\sqrt {(s+1)(a+b-1)-ab}}\\,\\\\\\left(S_{y}\\right)_{ab}&={\\frac {i\\hbar }{2}}\\left(\\delta _{a,b+1}-\\delta _{a+1,b}\\right){\\sqrt {(s+1)(a+b-1)-ab}}\\\\\\left(S_{z}\\right)_{ab}&=\\hbar (s+1-a)\\delta _{a,b}=\\hbar (s+1-b)\\delta _{a,b}\\end{aligned}}} where indices ${\\displaystyle a,b}$ are integer numbers such that ${\\displaystyle 1\\leq a\\leq 2s+1}$ and ${\\displaystyle 1\\leq b\\leq 2s+1}$ Also useful in the quantum mechanics of multiparticle systems, the general Pauli group Gn is defined to consist of all n-fold tensor products of Pauli matrices. The analog formula of Euler's formula in terms of the Pauli matrices: ${\\displaystyle e^{i\\theta \\left({\\hat {\\mathbf {n} }}\\cdot {\\boldsymbol {\\sigma }}\\right)}=I\\cos(\\theta )+i\\left({\\hat {\\mathbf {n} }}\\cdot {\\boldsymbol {\\sigma }}\\right)\\sin(\\theta )}$ for higher spins is tractable, but less simple.[20] ## Parity In tables of the spin quantum number s for nuclei or particles, the spin is often followed by a \"+\" or \"−\". This refers to the parity with \"+\" for even parity (wave function unchanged by spatial inversion) and \"−\" for odd", "at 8:43 With apologies for the formatting, they would be [3, -4, -2, 5], [3, -2, 4, -3], [3, -4, 2, -3], [5, 2, -8, -3], [5, 4, -4, -3], and [-3, -8, -2, -3]. – Gerry Myerson Feb 27 '13 at 12:16 $\\def\\p{\\phantom{-}}$ Upgrading comment to answer, at request of OP: $$\\pmatrix{\\p3&-4\\cr-2&\\p5\\cr},\\pmatrix{3&-2\\cr4&-3\\cr},\\pmatrix{3&-4\\cr2&-3\\cr},\\pmatrix{\\p5&\\p2\\cr-8&-3\\cr},\\pmatrix{\\p5&\\p4\\cr-4&-3\\cr},\\pmatrix{-3&-8\\cr-2&-3\\cr}$$", "1)^4 = t^4 - 4t^3 + 6t^2 - 4t + 1$ $t - 1$ 4 $-1$ $\\begin{pmatrix}-1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -1 \\\\\\end{pmatrix}$ $(t + 1)^4 = t^4 + 4t^3 + 6t^2 + 4t + 1$ $t + 1$ -4 $i$ $\\begin{pmatrix}0 & -1 & 0 & 0 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & -1 \\\\ 0 & 0 & 1 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $-i$ $\\begin{pmatrix} 0 & 1 & 0 & 0 \\\\ -1 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & -1 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $j$ $\\begin{pmatrix} 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 \\\\ 0 & -1 & 0 & 0 \\\\\\end{pmatrix}$ $(t^2 + 1)^2 = t^4 + 2t^2 + 1$ $t^2 + 1$ 0 $-j$ $\\begin{pmatrix} 0 & 0 & 1 & 0 \\\\ 0 & 0", "\\\\\\dfrac 5 6 &0\\end{pmatrix}\\\\ R = \\begin{pmatrix}\\dfrac 1 2 &0 \\\\0 &\\dfrac 1 6\\end{pmatrix}\\\\ N = \\left(I_2 - Q\\right)^{-1} = \\begin{pmatrix} \\dfrac{12}{7} & \\dfrac{6}{7} \\\\ \\dfrac{10}{7} & \\dfrac{12}{7} \\\\ \\end{pmatrix}\\\\ B = NR = \\begin{pmatrix} \\dfrac{6}{7} & \\dfrac{1}{7} \\\\ \\dfrac{5}{7} & \\dfrac{2}{7} \\\\ \\end{pmatrix}$ From $B$ we can directly read off that $P[\\text{MW|start with MP}] = \\dfrac 6 7$ and $P[\\text{MW|start with SP}] = \\dfrac 5 7$ OH thank you so much", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "\\\\ 7 \\end{pmatrix} = 7 \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}, \\end{align*} so, it is correct.", "so $\\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ -1 \\end{pmatrix}s =\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\end{pmatrix}(2-(1/3)s).$ 2. These two are equal. To show that $S_1\\subseteq S_2$, we check that for any t,s we can find an appropriate m,n so that these hold. $\\begin{array}{*{2}{rc}r} 4m &- &4n &= &1t+2s \\ 7m &- &2n &= &3t+1s \\ 7m &- &10n &= &1t+5s \\end{array}$ Use Gauss' method $\\begin{array}{rcl} \\left(\\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 7 &-2 &3t+1s \\ 7 &-10 &1t+5s \\end{array}\\right) &\\xrightarrow[(-7/4)\\rho_1+\\rho_3]{(-7/4)\\rho_1+\\rho_2} &\\left(\\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 0 &5 &(5/4)t-(10/4)s \\ 0 &-3 &-(3/4)t+(6/4)s \\end{array}\\right) \\ &\\xrightarrow[]{(3/5)\\rho_2+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 0 &5 &(5/4)t-(10/4)s \\ 0 &0 &0 \\end{array}\\right) \\end{array}$ to conclude that $\\begin{pmatrix} 1 \\\\ 3 \\\\ 1 \\end{pmatrix}t +\\begin{pmatrix} 2 \\\\ 1 \\\\ 5 \\end{pmatrix}s =\\begin{pmatrix} 4 \\\\ 7 \\\\ 7 \\end{pmatrix}((1/2)t) +\\begin{pmatrix} -4 \\\\ -2 \\\\ -10 \\end{pmatrix}((1/4)t-(1/2)s)$ and so $S_1\\subseteq S_2$. For $S_2\\subseteq S_1$, solve $\\begin{array}{*{2}{rc}r} 1t &+ &2s &= &4m-4n \\ 3t &+ &1s &= &7m-2n \\ 1t &+ &5s &= &7m-10n \\end{array}$ with Gaussian reduction $\\begin{array}{rcl} \\left(\\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 3 &1 &7m-2n \\ 1 &5 &7m-10n \\end{array}\\right) &\\xrightarrow[-\\rho_1+\\rho_3]{-3\\rho_1+\\rho_2} &\\left(\\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 0 &-5 &-5m+10n\\ 0 &3 &3m-6n \\end{array}\\right) \\ &\\xrightarrow[]{(3/5)\\rho_2+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 0 &-5 &-5m+10n\\ 0 &0 &0 \\end{array}\\right) \\end{array}$ to get", "& 0 \\\\0 & 0 & 0 & -1 \\\\0 & 1 & 0 & 0 \\\\1 & 0 & 0 & 0 \\end{pmatrix} = \\begin{pmatrix} 0 & 0 & 1 & 0 \\\\0 & 0 & 0 & -1 \\\\0 & -1 & 0 & 0 \\\\-1 & 0 & 0 & 0 \\end{pmatrix}$$", "+ A_2 + \\cdots +A_t) & = 882 \\\\ \\det(A(1+2+\\cdots t)) & = 882 \\\\ \\det(A) \\cdot (1+2+\\cdots + t)^2 & = 882 \\\\ 2(1+2+\\cdots + t)^2 & = 882 \\\\ (1+2+\\cdots+t)^2 & = 441 \\\\ 1+2+\\cdots+t & = 21 \\\\ \\dfrac{t}{2}(t+1) & = 21 \\\\ t^2+t-42 & = 0 \\\\ (t+7)(t-6) & = 0 \\end{aligned} We obtain $t = -7$ or $t = 6$ (It satisfies since $t \\in \\mathbb{N}$). Thus, the value of $\\boxed{\\det(A_{2 \\cdot 6}) = A_{12} = 2 \\cdot 12^2 = 288}$ [collapse] Problem Number 19 If $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 1 & 0 \\\\ -3 & 1 \\end{pmatrix}$, then $A^{2013} \\cdot \\begin{pmatrix}2 \\\\ 1 \\end{pmatrix} + B^{2014} \\cdot \\begin{pmatrix} 1 \\\\ -3 \\end{pmatrix}$ equals $\\cdots \\cdot$ A. $\\begin{pmatrix} 2023 \\\\ -6044 \\end{pmatrix}$ D. $\\begin{pmatrix} -4023 \\\\ -6044 \\end{pmatrix}$ B. $\\begin{pmatrix} -4023 \\\\ 6044 \\end{pmatrix}$ E. $\\begin{pmatrix} -6044 \\\\ -4023 \\end{pmatrix}$ C. $\\begin{pmatrix} 4023 \\\\ 6044 \\end{pmatrix}$ Solution Given that $A = \\begin{pmatrix} 1 & -2 \\\\ 0 & 1 \\end{pmatrix}$. We will find the matrix’s entry pattern for $A^n$. You should note that \\begin{aligned} A^2 & = \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & \\color{red}{-2} \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & \\color{red}{-4} \\\\ 0", "\\vec{w} \\\\\\\\ & &=& \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} \\times \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} \\\\\\\\ & &=& \\begin{vmatrix}1&1&1 \\\\ -4&0&2 \\\\ 0&-3&-1 \\end{vmatrix} \\\\\\\\ & &=& \\begin{pmatrix} 6 \\\\ -4 \\\\ 12 \\end{pmatrix} \\\\ \\hline n_1+n_2+n_3: & 6-4+12 &=& 14 \\quad | \\quad \\cdot \\dfrac{7}{14} \\\\\\\\ & 6\\cdot\\dfrac{7}{14}-4\\cdot\\dfrac{7}{14}+12\\cdot\\dfrac{7}{14} &=& 14\\cdot\\dfrac{7}{14} \\\\\\\\ &\\mathbf{ 3-2+6 }&=& \\mathbf{7} \\\\\\\\ & \\vec{n} &=& \\begin{pmatrix}\\mathbf{3} \\\\ \\mathbf{-2} \\\\ \\mathbf{6} \\end{pmatrix} \\\\ \\hline \\end{array} }$$ 2. Vector dot product $$\\begin{array}{|lrcll|} \\hline & \\vec{n}\\cdot \\vec{v} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} &=& 0 \\\\ (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ \\hline & \\vec{n}\\cdot \\vec{w} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} &=& 0 \\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ \\hline (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ \\hline \\hline \\end{array}$$ $$\\begin{array}{|lrcll|} \\hline (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ &2n_3&=&4n_1 \\quad | \\quad :2 \\\\ & \\mathbf{n_3} &=& \\mathbf{2n_1} \\\\\\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ & \\mathbf{n_3} &=& \\mathbf{-3n_2} \\\\\\\\ & n_3= 2n_1 &=& -3n_2 \\\\ & 2n_1 &=& -3n_2 \\quad | \\quad :2 \\\\ & \\mathbf{n_1} &=& \\mathbf{-\\dfrac{3}{2}n_2} \\\\\\\\ (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ & -\\dfrac{3}{2}n_2 + n_2 + -3n_2 &=& 7 \\\\ & -\\dfrac{3}{2}n_2 -2n_2 &=& 7 \\quad | \\quad", "3 \\end{pmatrix} = \\begin{pmatrix}1 & 2 & 3\\\\ 2 & 4 & 6 \\\\ 3 & 6 & 9 \\end{pmatrix}$$", "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "& 3 & 4\\\\ 2 &\\lambda& 6 & 7\\\\ 3 & 6 & 8 & 9\\\\ 4 & 7 & 9 &10 \\end{pmatrix}\\overset{(1)}\\sim \\begin{pmatrix} 1 & 2 & 3 & 4\\\\ 2 &\\lambda& 6 & 7\\\\ 3 & 6 & 8 & 9\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 0 & 1 & 2 & 3\\\\ 0 &\\lambda-2& 4 & 5\\\\ 0 & 3 & 5 & 6\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\overset{(2)}\\sim \\begin{pmatrix} 0 & 1 & 2 & 3\\\\ 0 &\\lambda-2& 4 & 5\\\\ 0 & 1 & 1 & 0\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 0 & 1 & 2 & 3\\\\ 0 &\\lambda-2& 4 & 5\\\\ 0 & 1 & 1 & 0\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 0 & 0 & 1 & 3\\\\ 0 & 0 & 6-\\lambda & 5\\\\ 0 & 1 & 1 & 0\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 0 & 0 & 1 & 3\\\\ 0 & 0 & 0 & 3\\lambda-13\\\\ 0 & 1 & 1 & 0\\\\ 1 & 1 & 1 & 1 \\end{pmatrix}\\sim \\begin{pmatrix} 1 & 1 & 1 & 1\\\\ 0 & 1 & 1 & 0\\\\ 0 & 0 & 1 & 3\\\\ 0", "& 1 & 8 & 2 \\\\ 3 & 9 & 4 & 12 \\\\ \\ub{12} & 3 & 16 & \\ub{4} \\end{pmatrix} \\qquad \\begin{pmatrix} \\hl{1} & 3 & 2 & \\hl{6} \\\\ 4 & 1 & 8 & 2 \\\\ 3 & 9 & 4 & 12 \\\\ \\hl{12} & 3 & 16 & \\hl{4} \\end{pmatrix}$$ \\end{document}", "+ 8y + 10z\\\\\\end{pmatrix}$\\$ - Oh, I actually made a mistake while posting it, as the order is (3, 2), (2, 1), that is why I calculated this way – tmac_balla Mar 29 '14 at 19:03", "&0 \\\\ 0 &1 \\end{pmatrix}) =2\\cdot 1=2$ 3. Yes. This is the check that it preserves combinations of two members of the domain. $\\begin{array}{rl} h(r_1\\cdot\\begin{pmatrix} a_1 &b_1 \\\\ c_1 &d_1 \\end{pmatrix} +r_2\\cdot\\begin{pmatrix} a_2 &b_2 \\\\ c_2 &d_2 \\end{pmatrix}) &=h(\\begin{pmatrix} r_1a_1+r_2a_2 &r_1b_1+r_2b_2 \\\\ r_1c_1+r_2c_2 &r_1d_1+r_2d_2 \\end{pmatrix}) \\\\ &=2(r_1a_1+r_2a_2)+3(r_1b_1+r_2b_2) +(r_1c_1+r_2c_2)-(r_1d_1+r_2d_2) \\\\ &=r_1(2a_1+3b_1+c_1-d_1) +r_2(2a_2+3b_2+c_2-d_2) \\\\ &=r_1\\cdot h(\\begin{pmatrix} a_1 &b_1 \\\\ c_1 &d_1 \\end{pmatrix} +r_2\\cdot h(\\begin{pmatrix} a_2 &b_2 \\\\ c_2 &d_2 \\end{pmatrix}) \\end{array}$ 4. No. An example of a combination that is not preserved is this. $h(\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix} +\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix}) =h(\\begin{pmatrix} 2 &0 \\\\ 0 &0 \\end{pmatrix}) =4 \\quad\\text{while}\\quad h(\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix}) +h(\\begin{pmatrix} 1 &0 \\\\ 0 &0 \\end{pmatrix}) =1+1 =2$ This exercise is recommended for all readers. Problem 3 Show that these two maps are homomorphisms. 1. $d/dx:\\mathcal{P}_3\\to \\mathcal{P}_2$ given by $a_0+a_1x+a_2x^2+a_3x^3$ maps to $a_1+2a_2x+3a_3x^2$ 2. $\\int:\\mathcal{P}_2\\to \\mathcal{P}_3$ given by $b_0+b_1x+b_2x^2$ maps to $b_0x+(b_1/2)x^2+(b_2/3)x^3$ Are these maps inverse to each other? The check that each is a homomorphisms is routine. Here is the check for the differentiation map. $\\frac{d}{dx}(r\\cdot (a_0+a_1x+a_2x^2+a_3x^3) +s\\cdot (b_0+b_1x+b_2x^2+b_3x^3))$ $\\begin{array}{rl} &=\\frac{d}{dx}((ra_0+sb_0)+(ra_1+sb_1)x+(ra_2+sb_2)x^2 +(ra_3+sb_3)x^3) \\\\ &=(ra_1+sb_1)+2(ra_2+sb_2)x+3(ra_3+sb_3)x^2 \\\\ &=r\\cdot (a_1+2a_2x+3a_3x^2)+s\\cdot (b_1+2b_2x+3b_3x^2) \\\\ &=r\\cdot \\frac{d}{dx}(a_0+a_1x+a_2x^2+a_3x^3) +s\\cdot \\frac{d}{dx} (b_0+b_1x+b_2x^2+b_3x^3) \\end{array}$ (An alternate proof is to simply note that this is a property of differentiation that is familar from calculus.)" ]

[ "the sequence \"$12$\". For $n=7$ this becomes $$\\begin{pmatrix} 1 & 1 \\end{pmatrix}\\begin{pmatrix}96031 & 106821 \\\\ 854568 & 950599 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} = 9409959$$ and including the number $10^7$ the final answer is $9409960$. Based on the above one can find a simpler linear recursion for the amount $a_n$ of numbers of at most $n$ digits that do not contain the sequence \"$12$\", namely $$a_0 = 0, a_1=9 \\textrm{ and } a_{n+2} = 10 a_{n+1} - a_n + 8.$$ This produces the sequence $$0, 9, 98, 979, 9700, 96029, 950598, 9409959, 93149000, 922080049, \\dotsc$$ -", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "so $\\begin{pmatrix} 1 \\\\ 8 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ -1 \\end{pmatrix}s =\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\end{pmatrix}(2-(1/3)s).$ 2. These two are equal. To show that $S_1\\subseteq S_2$, we check that for any t,s we can find an appropriate m,n so that these hold. $\\begin{array}{*{2}{rc}r} 4m &- &4n &= &1t+2s \\ 7m &- &2n &= &3t+1s \\ 7m &- &10n &= &1t+5s \\end{array}$ Use Gauss' method $\\begin{array}{rcl} \\left(\\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 7 &-2 &3t+1s \\ 7 &-10 &1t+5s \\end{array}\\right) &\\xrightarrow[(-7/4)\\rho_1+\\rho_3]{(-7/4)\\rho_1+\\rho_2} &\\left(\\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 0 &5 &(5/4)t-(10/4)s \\ 0 &-3 &-(3/4)t+(6/4)s \\end{array}\\right) \\ &\\xrightarrow[]{(3/5)\\rho_2+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 4 &-4 &1t+2s \\ 0 &5 &(5/4)t-(10/4)s \\ 0 &0 &0 \\end{array}\\right) \\end{array}$ to conclude that $\\begin{pmatrix} 1 \\\\ 3 \\\\ 1 \\end{pmatrix}t +\\begin{pmatrix} 2 \\\\ 1 \\\\ 5 \\end{pmatrix}s =\\begin{pmatrix} 4 \\\\ 7 \\\\ 7 \\end{pmatrix}((1/2)t) +\\begin{pmatrix} -4 \\\\ -2 \\\\ -10 \\end{pmatrix}((1/4)t-(1/2)s)$ and so $S_1\\subseteq S_2$. For $S_2\\subseteq S_1$, solve $\\begin{array}{*{2}{rc}r} 1t &+ &2s &= &4m-4n \\ 3t &+ &1s &= &7m-2n \\ 1t &+ &5s &= &7m-10n \\end{array}$ with Gaussian reduction $\\begin{array}{rcl} \\left(\\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 3 &1 &7m-2n \\ 1 &5 &7m-10n \\end{array}\\right) &\\xrightarrow[-\\rho_1+\\rho_3]{-3\\rho_1+\\rho_2} &\\left(\\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 0 &-5 &-5m+10n\\ 0 &3 &3m-6n \\end{array}\\right) \\ &\\xrightarrow[]{(3/5)\\rho_2+\\rho_3} &\\left(\\begin{array}{*{2}{c}|c} 1 &2 &4m-4n \\ 0 &-5 &-5m+10n\\ 0 &0 &0 \\end{array}\\right) \\end{array}$ to get", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "&2 \\\\ 1 &0 &3 \\end{pmatrix}$ 2. $\\begin{pmatrix} 1 &-1 &2 \\\\ 3 &-3 &6 \\\\ -2 &2 &-4 \\end{pmatrix}$ 3. $\\begin{pmatrix} 1 &3 &2 \\\\ 5 &1 &1 \\\\ 6 &4 &3 \\end{pmatrix}$ 4. $\\begin{pmatrix} 0 &0 &0 \\\\ 0 &0 &0 \\\\ 0 &0 &0 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 6 Find a basis for the span of each set. 1. $\\{\\begin{pmatrix} 1 &3 \\end{pmatrix}, \\begin{pmatrix} -1 &3 \\end{pmatrix}, \\begin{pmatrix} 1 &4 \\end{pmatrix}, \\begin{pmatrix} 2 &1 \\end{pmatrix} \\}\\subseteq\\mathcal{M}_{1 \\! \\times \\! 2}$ 2. $\\{\\begin{pmatrix} 1 \\\\2 \\\\1 \\end{pmatrix}, \\begin{pmatrix} 3 \\\\ 1 \\\\ -1 \\end{pmatrix}, \\begin{pmatrix} 1 \\\\ -3 \\\\ -3 \\end{pmatrix} \\}\\subseteq\\mathbb{R}^3$ 3. $\\{1+x,1-x^2,3+2x-x^2\\}\\subseteq\\mathcal{P}_3$", "Example Using the method we just saw and without using a calculator, calculate each of the following: 1. $$101^2$$ 2. $$27^2$$ 3. $$23^2$$ 4. $$99^2$$ 5. $$104^2$$ 6. $$48^2$$ ## Solution Without Working 1. $$101^2 = 10201$$ 2. $$27^2 = 711$$ 3. $$23^2 = 529$$ 4. $$99^2 = 9801$$ 5. $$104^2 = 10816$$ 6. $$48^2 = 2304$$ ## Solution With Working 1. We calculate $$101^2$$ as follows: \\begin{aligned} 101^2 & = \\begin{pmatrix} 100+1\\end{pmatrix}^2 \\\\ & = 100^2 + 2 \\times 100 \\times 1 + 1^2 \\\\ & = 10000 + 200 + 1\\\\ 101^2 &= 10201 \\end{aligned} 2. We calculate $$27^2$$ as follows: \\begin{aligned} 27^2 & = \\begin{pmatrix} 30 - 3 \\end{pmatrix}^2 \\\\ & = 30^2 - 2 \\times 30 \\times 3 - 3^2 \\\\ & = 900 - 180 - 9 \\\\ & = 720 - 9 \\\\ 27^2 & = 711 \\end{aligned} 3. We calculate $$23^2$$ as follows: \\begin{aligned} 23^2 & = \\begin{pmatrix} 20 + 3 \\end{pmatrix}^2 \\\\ & = 20^2 + 2 \\times 20 \\times 3 + 3^2 \\\\ & = 400 + 120 + 9 \\\\ 23^2 & = 529 \\end{aligned} 4. We calculate $$99^2$$ as follows: \\begin{aligned} 99^2 & = \\begin{pmatrix} 100 - 1 \\end{pmatrix} \\\\ & = 100^2 - 2\\times 100 \\times 1 + 1^2 \\\\ & = 10000 - 200 +", "& -3/50 \\end{array} \\right) \\left( \\begin{array}{cc} 1 & -2 \\\\ 1 & 3 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.1 \\\\ 0 \\end{pmatrix}$ Multiply throughout by $\\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right)^{-1} = 1/5 \\left( \\begin{array}{cc} 2 & 3 \\\\ -1 & 1 \\end{array} \\right)$ to get $\\frac{d}{dt} \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\left( \\begin{array}{cc} -1/50 & 0 \\\\ 0 & -6/50 \\end{array} \\right) \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} + \\begin{pmatrix} 0.04 \\\\ -0.02 \\end{pmatrix}$ this is equivalent to the differential equations $\\frac{du_1}{dt}=-1/50u_1+0.04$ $\\frac{du_2}{dt}=-6/50u_1-0.02$ These have the solutions $u_1=2+Ae^{-t/50}, \\: u_2= -1/6+Be^{-6t/50}$ We can write this as $\\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} = \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix}$ Now transform back to $x_1, \\: x_2$ . \\begin{aligned} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} &= P \\begin{pmatrix} u_1 \\\\ u_2 \\end{pmatrix} \\\\ &= \\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 2 \\end{array} \\right) \\begin{pmatrix} 2+Ae^{-t/50} \\\\ -1/6+Be^{-6t/50} \\end{pmatrix} \\\\ &= \\begin{pmatrix} 5/2+Ae^{-t/50}-3Be^{-6t/50} \\\\ 5/3+Ae^{-t/50}+2Be^{-6t/50} \\end{pmatrix} \\end{aligned} . These solutions can be fitted to any initial conditions to find the constants.", "$y$ that satisfy $\\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} x & y \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} = 1.360$ is $\\cdots \\cdot$ A. $x = 1$ and $y = \\dfrac45$ B. $x=\\dfrac45$ and $y=1$ C. $x=5$ and $y=4$ D. $x=-10$ and $y=-8$ E. $x=10$ and $y=8$ Solution Since $x : y = 5 : 4$, then we may assume $x = 5k$ and $y=4k$ for a real number $k$. Hence, we have \\begin{aligned} \\begin{pmatrix} 2 & 10 & 1 \\end{pmatrix} \\begin{pmatrix} 5k & 4k \\\\ 4 & 5 \\\\ 30 & 25 \\end{pmatrix}\\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 40 + 30 & 8k + 50 + 25 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & = 1.360 \\\\ \\begin{pmatrix} 10k + 70 & 8k + 75 \\end{pmatrix} \\begin{pmatrix} 5 \\\\ 10 \\end{pmatrix} & =1.360 \\\\ 5(10k+70) + 10(8k+75) & = 1.360 \\\\ \\text{Divide each side by}~& 5 \\\\10k+70 + 2(8k+75) & = 272 \\\\ 26k + 220 & = 272 \\\\ 26k & = 52 \\\\ k & = 2 \\end{aligned}Thus, we have \\boxed{\\begin{aligned} x & = 5k = 5(2) = 10 \\\\ y & =4k=4(2)=8 \\end{aligned}} [collapse] Problem Number 7 It is given that $A = \\begin{pmatrix} 2 & 1 \\\\ 3", "and $z$. $(x+1)^2+(y-2)^2$ $+(z-1)^2=17$ $(5+3r+1)^2$ $+(6+2r-2)^2$ $+(5+2r-1)^2=17$ $(6+3r)^2$ $+(4+2r)^2$ $+(4+2r)^2=17$ Use binomial theorem to resolve parentheses $36+36r+9r^2$ $+16+16r+4r^2$ $+16+16r+4r^2=17$ $17r^2+68r+68=17\\quad|-17$ $17r^2+68r+51=0\\quad|:17$ $r^2+4r+3=0$ Use quadratic formula I to solve quadratic equation. $r_{1,2}=-\\frac{p}2\\pm\\sqrt{(\\frac{p}2)^2-q}$ $r_{1,2}=-2\\pm\\sqrt{2^2-3}$ $r_{1,2}=-2\\pm1$ $r_{1}=-1$ and $r_{2}=-3$ 3. #### Insert $r$ The two calculated $r$ are inserted into the equation of a line in order to obtain the intersection points. $\\vec{OS_1} = \\begin{pmatrix} 5 \\\\ 6 \\\\ 5 \\end{pmatrix} - 1 \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $=\\begin{pmatrix} 2 \\\\ 4 \\\\ 3 \\end{pmatrix}$ $\\vec{OS_2} = \\begin{pmatrix} 5 \\\\ 6 \\\\ 5 \\end{pmatrix} - 3 \\cdot \\begin{pmatrix} 3 \\\\ 2 \\\\ 2 \\end{pmatrix}$ $=\\begin{pmatrix} -4 \\\\ 0 \\\\ -1 \\end{pmatrix}$ It is a secant that intersects the sphere at $S_1(2|4|3)$ and $S_2(-4|0|-1)$.", "is $$A=\\begin{pmatrix}1&0&1\\\\\\!\\!\\!\\!-1&1&2\\\\0&\\!\\!\\!\\!-1&1\\end{pmatrix}^{-1}=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}$$ Finally, since $$f(x)=3-4x+2x^2\\stackrel{coord. wrt B}\\longrightarrow \\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix} \\,\\,,\\,\\text{we get}$$ $$[Af]=\\frac{1}{4}\\begin{pmatrix}3&-1&-1\\\\1&\\;\\;1&-3\\\\1&\\;\\;1&\\;\\;1\\end{pmatrix}\\begin{pmatrix}3\\\\\\!\\!\\!\\!-4\\\\2\\end{pmatrix}=\\frac{1}{4}\\begin{pmatrix}11\\\\\\!\\!\\!\\!-7\\\\1\\end{pmatrix}$$ so $$\\,f_C=\\frac{11}{4}(1-x)-\\frac{7}{4}(x-x^2)+\\frac{1}{4}(1+2x+x^2)$$", "\\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 2. $\\begin{pmatrix} -3 \\\\ 3 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 3. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 4. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k+\\begin{pmatrix} 0 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 5. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 14 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 6. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 6 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ Problem 2 Produce two descriptions of this set that are different than this one. $\\{\\begin{pmatrix} 2 \\\\ -5 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 3 Show that the three descriptions given at the start of this subsection all describe the same set. This exercise is recommended for all readers. Problem 4 Show that these sets are equal $\\{\\begin{pmatrix} 1 \\\\ 4 \\\\ 1 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}z\\,\\big|\\, z\\in\\mathbb{R} \\} \\quad\\text{and}\\quad \\{\\begin{pmatrix} 0 \\\\ 4 \\\\ 2 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R} \\},$ and that both describe the solution set of this system. $\\begin{array}{*{4}{rc}r} x &- &y &+ &z &+", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "2+2\\cdot 3+0\\cdot 4\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3+2\\cdot 4 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3+1\\cdot 4\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 11\\\\4\\end{pmatrix}$; d) $\\begin{pmatrix} 1 & 2 & 0 \\\\ 0 & 1 & 2 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{pmatrix}\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\\\4 \\end{pmatrix}=\\begin{pmatrix}1\\cdot 1+2\\cdot 2+0\\cdot 3\\\\ 0\\cdot 1+1\\cdot 2+2\\cdot 3\\\\ 0\\cdot 1+0\\cdot 2+1\\cdot 3 \\\\ 0\\cdot 1+0\\cdot 2+0\\cdot 3\\end{pmatrix}=\\begin{pmatrix}5\\\\ 8\\\\ 3\\\\0\\end{pmatrix}$. ### This Post Has One Comment 1. a) should be [13 31]. 1+6+6 = 13", "\\\\ 8 \\end{pmatrix},$ $\\begin{pmatrix} 3 & 3 \\\\ 2 & 5 \\end{pmatrix} \\begin{pmatrix} 11 \\\\ 15 \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 19 \\end{pmatrix}$ and continue encryption as follows: $\\begin{pmatrix} 7 \\\\ 8 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 19 \\end{pmatrix} \\to \\begin{pmatrix} H \\\\ I \\end{pmatrix}, \\begin{pmatrix} A \\\\ T \\end{pmatrix}$ The matrix K is invertible, hence $K^{-1}$ exists such that $KK^{-1}=K^{-1}K=I_2$. To implement decrypting, we compute $K^{-1} = 9^{-1} \\begin{pmatrix} 5 & 23 \\\\ 24 & 3 \\end{pmatrix} = 3 \\begin{pmatrix} 5 & 23 \\\\ 24 & 3 \\end{pmatrix} = \\begin{pmatrix} 15 & 17 \\\\ 20 & 9 \\end{pmatrix}$ $HIAT \\to \\begin{pmatrix} H \\\\ I \\end{pmatrix}, \\begin{pmatrix} A \\\\ T \\end{pmatrix} \\to \\begin{pmatrix} 7 \\\\ 8 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 19 \\end{pmatrix}$ Then we compute $\\begin{pmatrix} 15 & 17 \\\\ 20 & 9 \\end{pmatrix}\\begin{pmatrix} 7 \\\\ 8 \\end{pmatrix} = \\begin{pmatrix} 7 \\\\ 4 \\end{pmatrix},$ $\\begin{pmatrix} 15 & 17 \\\\ 20 & 9 \\end{pmatrix}\\begin{pmatrix} 0 \\\\ 19 \\end{pmatrix} = \\begin{pmatrix} 11 \\\\ 15 \\end{pmatrix}$ Therefore $\\begin{pmatrix} 7 \\\\ 4 \\end{pmatrix}, \\begin{pmatrix} 11 \\\\ 15 \\end{pmatrix} \\to \\begin{pmatrix} H \\\\ E \\end{pmatrix}, \\begin{pmatrix} L \\\\ P \\end{pmatrix} \\to HELP$. ## Security Unfortunately, the basic Hill cipher is vulnerable to a known-plaintext attack because it is completely linear. An opponent who intercepts $n^2$ plaintext/ciphertext character pairs can set up a linear system", "2, we get $\\sum 2(y_i – (\\theta_0 + \\theta_1x_{i1} + \\theta_2x_{i2}).(-1) = 0 \\\\ \\sum 2(y_i – (\\theta_0 + \\theta_1x_{i1} + \\theta_2x_{i2}).(-x_{i1}) = 0 \\\\ \\sum 2(y_i – (\\theta_0 + \\theta_1x_{i1} + \\theta_2x_{i2}).(-x_{i2}) = 0$ Simplification results $\\sum y_i = n\\theta_0 + \\theta_1 \\sum x_{i1} + \\theta_2 \\sum x_{i2} \\\\ \\sum x_{1i}y_i = \\theta_0 \\sum x_{i1} + \\theta_1 \\sum x_{i1}x_{i1} + \\theta_2 \\sum x_{i1}x_{i2} \\\\ \\sum x_{2i}y_i = \\theta_0 \\sum x_{i2} + \\theta_1 \\sum x_{i1}x_{i2} + \\theta_2 \\sum x_{i2}x_{i2}$ Expressing these equations in matrix form $\\begin{pmatrix} 1 & 1 & 1 \\\\ x_{11} & x_{21} & x_{31} \\\\ x_{12} & x_{22} & x_{32} \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ y_3 \\end{pmatrix} = \\begin{pmatrix} 1 & 1 & 1 \\\\ x_{11} & x_{21} & x_{31} \\\\ x_{12} & x_{22} & x_{32} \\end{pmatrix} \\begin{pmatrix} 1 & x_{11} & x_{12} \\\\ 1 & x_{21} & x_{22} \\\\ 1 & x_{31} & x_{32} \\end{pmatrix}\\begin{pmatrix} \\theta_0 \\\\ \\theta_1 \\\\ \\theta_2 \\end{pmatrix}$ $X^Ty = X^TX\\theta \\\\ (XX^T)^{-1}X^Ty = (XX^T)^{-1} X^TX\\theta \\\\ \\theta = (X^TX)^{-1}X^Ty$ But, as the number of features (i.e. dimension) increases, the Normal equation’s performance gradually decreases. This is because of the expensive matrix operations. That is way gradient descent is preferred over normal equation in most of the machine learning problem involving large number of features. As a concluding remark, use", "$x^2$ term in the fourth power expansion and by multiplying \"4\" by the $x^3$ term. The $x^2$ term in $(x- 2)^4$ is $\\begin{pmatrix}4 \\\\ 2\\end{pmatrix}(-2)^2x^2$ and the $x^3$ term is $\\begin{pmatrix}4 \\\\ 3\\end{pmatrix}(-2)^1x^3$. Multiplying the first by 3x and the second by 4 gives $\\left(\\begin{pmatrix}4 \\\\ 2\\end{pmatrix}(3)(-2)^2+ \\begin{pmatrix}4 \\\\ 3\\end{pmatrix}(4)(-2)\\right)x^3$", "$\\newcommand{\\FirstMat}{\\begin{pmatrix*}[r] 1 & -3 & 4 \\\\ 4 & -7 & 8 \\\\ 6 & -7 & 7 \\end{pmatrix*}} \\begin{array}{c} (1) \\\\ \\vphantom{\\FirstMat}\\\\ \\end{array} \\begin{array}{c} A =\\FirstMat\\;,\\\\ \\vphantom{X}\\\\ \\end{array} \\quad \\begin{array}{c} (2) \\\\ \\vphantom{\\FirstMat}\\\\ \\end{array} \\begin{array}{c} A = \\begin{pmatrix*}[r] 0 & -2 & 3 & 2 \\\\ 1 & 1 & -1 & -1 \\\\ 0 & 0 & 2 & 0 \\\\ 1 & -1 & 0 & 1 \\end{pmatrix*}\\;.\\\\ \\end{array}$", "b_{12} c_{22} \\\\ a_{21} c_{11} + b_{21} c_{11} + a_{22} c_{21} + b_{22} c_{21} & a_{21} c_{12} + b_{21} c_{12} + a_{22} c_{22} + b_{22} c_{22} \\end {pmatrix}$ Real Multiplication Distributes over Addition $\\ds \\mathbf A \\mathbf C + \\mathbf B \\mathbf C$ $=$ $\\ds \\begin {pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end {pmatrix} \\begin {pmatrix} c_{11} & c_{12} \\\\ c_{21} & c_{22} \\end {pmatrix} + \\begin {pmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end {pmatrix} \\begin {pmatrix} c_{11} & c_{12} \\\\ c_{21} & c_{22} \\end {pmatrix}$ $\\ds$ $=$ $\\ds \\begin {pmatrix} a_{11} c_{11} + a_{12} c_{21} & a_{11} c_{12} + a_{12} c_{22} \\\\ a_{21} c_{11} + a_{22} c_{21} & a_{21} c_{12} + a_{22} c_{22} \\end {pmatrix} + \\begin {pmatrix} b_{11} c_{11} + b_{12} c_{21} & b_{11} c_{12} + b_{12} c_{22} \\\\ b_{21} c_{11} + b_{22} c_{21} & b_{21} c_{12} + b_{22} c_{22} \\end {pmatrix}$ Definition of Matrix Product (Conventional) $\\ds$ $=$ $\\ds \\begin {pmatrix} a_{11} c_{11} + b_{11} c_{11} + a_{12} c_{21} + b_{12} c_{21} & a_{11} c_{12} + b_{11} c_{12} + a_{12} c_{22} + b_{12} c_{22} \\\\ a_{21} c_{11} + b_{21} c_{11} + a_{22} c_{21} + b_{22} c_{21} & a_{21} c_{12} + b_{21} c_{12} + a_{22} c_{22} + b_{22} c_{22} \\end {pmatrix}$ Definition of Matrix Entrywise Addition The expression for $\\paren {\\mathbf A", "$$\\begin{pmatrix} y_1 \\\\ y_2 \\\\ y_3\\end{pmatrix} = \\left(\\begin{array}{rrr} a_{11} & a_{12} & a_{13}\\\\ a_{21} & a_{22} & a_{23}\\\\ a_{31} & a_{32} & a_{33}\\\\ \\end{array}\\right) \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix} = \\left(\\begin{array}{r} 1 a_{11} + 0 a_{12} + 0 a_{13}\\\\ 1 a_{21} + 0 a_{22} + 0 a_{23}\\\\ 1 a_{31} + 0 a_{32} + 0 a_{33}\\\\ \\end{array}\\right) = \\begin{pmatrix} a_{11} \\\\ a_{21} \\\\ a_{31} \\end{pmatrix},$$ (9.22) which is the first column in $\\mx{A}$. Thus the image $F(\\vc{e}_1)$ of the basis vector $\\vc{e}_1$ is the first column of $\\mx{A}$, denoted $\\vc{a}_{,1}$. Likewise, the second basis vector can be written $\\vc{e}_2 = 0 \\vc{e}_1 + 1 \\vc{e}_2 + 0 \\vc{e}_3$, and thus has the coordinates $(0, 1, 0)$. Its image is therefore $$\\left(\\begin{array}{rrr} a_{11} & a_{12} & a_{13}\\\\ a_{21} & a_{22} & a_{23}\\\\ a_{31} & a_{32} & a_{33}\\\\ \\end{array}\\right) \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix} = \\left(\\begin{array}{r} 0 a_{11} + 1 a_{12} + 0 a_{13}\\\\ 0 a_{21} + 1 a_{22} + 0 a_{23}\\\\ 0 a_{31} + 1 a_{32} + 0 a_{33}\\\\ \\end{array}\\right) = \\begin{pmatrix} a_{12} \\\\ a_{22} \\\\ a_{32} \\end{pmatrix},$$ (9.23) which is the second column vector $\\vc{a}_{,2}$ of the matrix $\\mx{A}$. Similarly for the third basis vector we get $$\\left(\\begin{array}{rrr} a_{11} & a_{12} & a_{13}\\\\ a_{21} & a_{22} & a_{23}\\\\ a_{31} & a_{32} & a_{33}\\\\ \\end{array}\\right) \\begin{pmatrix}", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$" ]

[ "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "# Thread: Spans and vectors 1. ## Spans and vectors If a question asks, Is the vector $\\mathbf{b}=\\begin{pmatrix}-2\\\\ -6\\\\ -4\\end{pmatrix}\\in \\mbox{span}(\\mathbf{v_1, v_2, v_3, v_4})$ where: $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 3\\\\ 0\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}2\\\\ 2\\\\ 1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ -1\\end{pmatrix}$, $\\mathbf{v_4}=\\begin{pmatrix}1\\\\ -2\\\\ 1\\end{pmatrix}$ Do we just make it a matrix and solve the matrix? And how would we solve it if we have 4 variables but only 3 equations? 2) Is the set of vectors $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 2\\\\ 3\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}1\\\\ 1\\\\ -1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ 5\\end{pmatrix}$ a spanning set of $\\mathbb{R}^3$? Do we just make this a matrix with say $\\mathbf{x}=\\begin{pmatrix}x_1\\\\ x_2\\\\ x_3\\end{pmatrix}$ as our solution and see if we can solve the matrix? 2. So your first question is asking whether there exist scalars $a_{1},a_{2},a_{3},a_{4}$ such that $\\mathbf{b}=a_{1}\\mathbf{v}_{1}+a_{2}\\mathbf{v}_{2} +a_{3}\\mathbf{v}_{3}+a_{4}\\mathbf{v}_{4}.$ This equation you can translate into a normal matrix problem of the form $A\\mathbf{x}=\\mathbf{b}$. Row reduce that problem. If you find any solutions (one or infinitely many), then the answer is yes. If you find that there are no solutions, then the answer is no. Your second question, again, is asking whether there exist scalars $c_{1}, c_{2}, c_{3}$ such that $c_{1}\\mathbf{v}_{1}+c_{2}\\mathbf{v}_{2}+c_{3}\\math bf{v}_{3}=\\mathbf{x}$ for your arbitrary $\\mathbf{x}.$ This, again, can be translated into a standard matrix problem. Since this problem will result in a square matrix, you can use the theory of determinants to help you out.", "amount. We refer to doing this as scalar multiplication: we multiply the vector by a scalar real number. === Scalar multiplication === For scalar multiplication, we simply multiply each component by the scalar. We commonly use Greek letters for scalars, and English letters for vectors. So for a scalar value of λ and a vector v defined by r and θ, the new vector is now λr and θ. Notice how the direction does not change. ==== Example ==== Say we have ( 2 3 ) {\\displaystyle {\\begin{pmatrix}2\\\\3\\end{pmatrix}}} and we wish to double the magnitude. So, 2 ( 2 3 ) = ( 4 6 ) {\\displaystyle 2{\\begin{pmatrix}2\\\\3\\end{pmatrix}}={\\begin{pmatrix}4\\\\6\\end{pmatrix}}} . Simply, to add two vectors, you must add the respective x-components together to obtain the new x-component, and likewise add the two y-components together to obtain the new y-component. === Example === Say we have v 1 = ( 2 3 ) , v 2 = ( 4 6 ) {\\displaystyle \\mathbf {v_{1}} ={\\begin{pmatrix}2\\\\3\\end{pmatrix}},\\mathbf {v_{2}} ={\\begin{pmatrix}4\\\\6\\end{pmatrix}}} and we wish to add these. So, v 1 + v 2 = ( 6 9 ) {\\displaystyle \\mathbf {v_{1}} +\\mathbf {v_{2}} ={\\begin{pmatrix}6\\\\9\\end{pmatrix}}} . == Subtraction of vectors == The operation of subtraction on two vectors, a and b, a-b, can also be written as a+(-1)b. Therefore, we can use scalar multiplication to", "+0 # Help with vector geometry +1 88 1 +38 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w} \\text{ and }\\mathbf{x}$$ are linearly independent, answer with 0. If they aren't, find coefficients a,b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a+b}{c}$$ Jun 5, 2019", "distance $$\\mathtt{r}$$ is from the line. An equation for the distance of $$\\mathtt{r}$$ to the line, then—a symbolic way to identify this distance—might be given in words as follows: go to point $$\\mathtt{p}$$, then scale to some point on the line. From that point, scale to some point on the vector that is perpendicular to the line until you get to point $$\\mathtt{r}$$. In symbols, that could be written as: $\\begin{bmatrix}\\mathtt{4}\\\\\\mathtt{2}\\end{bmatrix}\\mathtt{+\\,\\,\\,\\, j}\\begin{bmatrix}\\mathtt{2}\\\\\\mathtt{1}\\end{bmatrix}\\mathtt{+\\,\\,\\,\\,k}\\begin{bmatrix}\\mathtt{-1}\\\\\\mathtt{\\,\\,\\,\\,\\,2}\\end{bmatrix}\\mathtt{\\,\\,=\\,\\,}\\begin{bmatrix}\\mathtt{-1}\\\\\\mathtt{-3}\\end{bmatrix}$ With the vector and scalar names, we could write this as $$\\mathtt{p + j(p – x) + ka = r}$$. The distance to the line depends on our figuring out what $$\\mathtt{k}$$ is. Once we have that, then the distance is just $$\\mathtt{\\sqrt{(ka_1)^2 + (ka_2)^2}}$$. We can subtract vectors from both sides of an equation just like we do with scalar values. Subtracting the vector (4, 2) from both sides, we get an equation which can be rewritten as a system of two equations \\mathtt{j}\\begin{bmatrix}\\mathtt{2}\\\\\\mathtt{1}\\end{bmatrix}\\mathtt{+\\,\\,\\,\\,k}\\begin{bmatrix}\\mathtt{-1}\\\\\\mathtt{\\,\\,\\,\\,\\,2}\\end{bmatrix}\\mathtt{\\,\\,=\\,\\,}\\begin{bmatrix}\\mathtt{-5}\\\\\\mathtt{-5}\\end{bmatrix} \\rightarrow \\left\\{\\begin{align*}\\mathtt{2j – k = -5} \\\\ \\mathtt{j + 2k = -5}\\end{align*}\\right. Solving that system gives us $$\\mathtt{j = -3}$$ and $$\\mathtt{k = -1}$$. So, the distance of $$\\mathtt{r}$$ to the line is $$\\mathtt{\\sqrt{5}.}$$ Can We Get to the Dot Product? Maybe we can get to the dot product. I’m not sure at this point. But there are some interesting things", "to it so the sum is $$\\vec 0$$. (Note that it is important to first figure out what $$\\vec 0$$ is here!) The answer for the additive inverse of $$\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}$$ is $$\\begin{pmatrix}-x_{1}\\\\-x_{2}\\end{pmatrix}$$ because$$\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}+\\begin{pmatrix}-x_{1}\\\\-x_{2}\\end{pmatrix} = \\begin{pmatrix}x_{1}-x_{1}\\\\x_{2}-x_{2}\\end{pmatrix} = \\begin{pmatrix}0\\\\0\\end{pmatrix}=\\vec 0\\, .$$We are half-way done, now we need to consider the rules for scalar multiplication. Notice, that we multiply vectors by scalars ($$\\textit{i.e.}$$ numbers) but do NOT multiply a vectors by vectors. 1. Multiplicative closure: Again, we are checking that an operation does not produce vectors outside the vector space. For a scalar $$a\\in \\mathbb{R}$$, we require that $$a\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}$$ lies in $$\\mathbb{R}^{2}$$. First we compute using our component-wise rule for scalars times vectors:$$a\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}=\\begin{pmatrix}ax_{1}\\\\ax_{2}\\end{pmatrix}.$$Since products of real numbers $$a x_{1}$$ and $$a x_{2}$$ are again real numbers we see this is indeed inside $$\\mathbb{R}^{2}$$. 2. Multiplicative distributivity: The equation we need to check is$$(a+b)\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}\\stackrel?= a\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}+b\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}.$$Once again this is a simple LHS=RHS proof using properties of the real numbers. Starting on the left we have$$(a+b)\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix} = \\begin{pmatrix}(a+b)x_{1}\\$$a+b)x_{2}\\end{pmatrix} = \\begin{pmatrix}ax_{1}+b x_{1}\\\\ax_{2}+bx_{2}\\end{pmatrix} \\qquad\\qquad \\qquad\\qquad=\\begin{pmatrix}ax_{1}\\\\ax_{2}\\end{pmatrix}+\\begin{pmatrix}b x_{1}\\\\bx_{2}\\end{pmatrix} = a\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}+b\\begin{pmatrix} x_{1}\\\\x_{2}\\end{pmatrix}, as required. 3. Additive distributivity: This time we need to check the equation a\\left(\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}+\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}\\right)\\stackrel?=a\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}+a\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}, \\(\\textit{i.e.}$$, one scalar but two different vectors. The method is by now becoming familiar $$a\\left(\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}+\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}\\right)=a\\left(\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}\\right)=\\begin{pmatrix}a(x_{1}+y_{1})\\\\a(x_{2}+y_{2})\\end{pmatrix}\\qquad$$ $$\\qquad\\qquad=\\begin{pmatrix}ax_{1}+ay_{1}\\\\ax_{2}+ay_{2}\\end{pmatrix}=\\begin{pmatrix}ax_{1}\\\\ax_{2}\\end{pmatrix}+\\begin{pmatrix}ay_{1}\\\\ay_{2}\\end{pmatrix}=a\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}+a\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix},$$ again as required. 4. Multiplicative associativity. Just as for addition, this is the requirement that the order", "example of how to do this in R! a <- 5 x <- c(1, 4, 7, 9, 10) a*x #> [1] 5 20 35 45 50 is.vector(a*x) #> [1] TRUE We can also perform addition with vectors of the same size, or dimension. For vectors, the addition operation is also performed element-wise. $\\mathbf{x} + \\mathbf{y} = \\begin{pmatrix} x_{1} \\\\ x_{2} \\\\ \\vdots \\\\ x_{n} \\end{pmatrix} + \\begin{pmatrix} y_{1} \\\\ y_{2} \\\\ \\vdots \\\\ y_{n} \\end{pmatrix} = \\begin{pmatrix} x_{1} + y_{1} \\\\ x_{2} + y_{2} \\\\ \\vdots \\\\ x_{n} + y_{n} \\end{pmatrix}.$ Here’s a picture of what this looks like in two dimensions! Let’s say we want to add the two vectors $$\\mathbf{a}$$ and $$\\mathbf{b}$$ below. What does this look like in two dimensions? We could start first with $$\\mathbf{a}$$. The vector $$\\mathbf{a}$$ says that we move to the right two squares and then up three squares. Once we’re there, the vector $$\\mathbf{b}$$ says that we move to the left three squares and then up one square. Alternatively, we would get the same answer if we started first with $$\\mathbf{b}$$. The vector $$\\mathbf{b}$$ says that we move to the left three squares and then up one square. Once we’re there, the vector $$\\mathbf{a}$$ says that we move to the right 2 squares and then up three squares. Remember that", "+0 0 108 3 +226 Let a,b,c be vectors such that $$\\mathbf{a} \\times \\mathbf{b} = \\begin{pmatrix} -1 \\\\- 1 \\\\ -1 \\end{pmatrix}, \\mathbf{a} \\times \\mathbf{c} = \\begin{pmatrix} -1 \\\\ 2 \\\\ 1 \\end{pmatrix}, \\text{ and } \\mathbf{b} \\times \\mathbf{c} = \\begin{pmatrix} 0 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ Then evaluate $$(\\mathbf{b} + \\mathbf{c})\\times \\mathbf{b}, \\mathbf{a}\\times(\\mathbf{b} + 4 \\mathbf{a}), (\\mathbf{a} + \\mathbf{b} + \\mathbf{c})\\times \\mathbf{a}$$ Oct 30, 2020 #2 +31698 +3 For cross products we have the following a x a = 0 (simiarly for b x b etc) a x b = - b x a etc. (b + c) x a = b x+ c x a = -x b - a x c etc. Can you take it from here? Oct 30, 2020 #3 +112499 +1 Thanks Alan, that sure made it easy Oct 31, 2020", "vector: $$\\boldsymbol{v}=\\begin{pmatrix} 1\\\\2\\\\ \\end{pmatrix}$$ Show that this vector does not form a basis for $\\mathbb{R}^2$. Solution. Since we only have one vector, this vector must be linearly independent. However, $\\boldsymbol{v}$ clearly does not span $\\mathbb{R}^2$ because some vectors in $\\mathbb{R}^2$ cannot be expressed as a scalar multiple of $\\boldsymbol{v}$. Therefore, $\\boldsymbol{v}$ does not form a basis for $\\mathbb{R}^2$. We have already covered the standard basis of $\\mathbb{R}^2$ in the earlier sectionlink. We can also generalize the idea to $\\mathbb{R}^n$. Definition. # Standard basis of Rn The standard basis vectors of $\\mathbb{R}^n$ is defined as a set containing the following $n$ vectors: $$\\left\\{ \\begin{pmatrix} 1\\\\0\\\\\\vdots\\\\0 \\end{pmatrix}, \\begin{pmatrix} 0\\\\1\\\\\\vdots\\\\0 \\end{pmatrix},\\;\\cdots\\;, \\begin{pmatrix} 0\\\\0\\\\\\vdots\\\\1 \\end{pmatrix} \\right\\}$$ Edited by 0 others thumb_up thumb_down Comment Citation Ask a question or leave a feedback... thumb_up 0 thumb_down 0 chat_bubble_outline 0 settings Enjoy our search Hit / to insta-search docs and recipes!", "2\\begin{pmatrix} \\; 5\\;\\\\ \\; -6 \\; \\end{pmatrix} - 4\\begin{pmatrix} \\; -3 \\;\\\\ \\; 1 \\; \\end{pmatrix} = \\begin{pmatrix} \\; 10 \\;\\\\ \\; -12 \\; \\end{pmatrix} - \\begin{pmatrix} \\; -12 \\;\\\\ \\; 4 \\; \\end{pmatrix} = \\begin{pmatrix} \\; 22 \\;\\\\ \\; -16 \\; \\end{pmatrix} ### Common misconceptions • Column vectors notation Column vectors only have 2 numbers within the brackets. Column vectors have the top number and the bottom number in the brackets. There is no need for any other punctuation marks such as commas or semicolons. There is no need for a line to separate the numbers. Vector addition is commutative. This means that the order in which we add vectors is not important. \\textbf{a}+\\textbf{b}=\\textbf{b}+\\textbf{a} • Order of the subtraction The order in which you subtract vectors is very important. It is NOT like vector addition where the order is unimportant. Vector subtraction is NOT commutative. \\textbf{a}-\\textbf{b} \\; \\verb|≠| \\; \\textbf{b}-\\textbf{a} • A component of a vector can be zero Vector components can be zero. E.g. If both components of a vector are zero, this vector is known as the zero vector. • Direction of the vector Check the arrow on vector diagrams to see the direction of the vector. Check for negative signs in column vectors. • Be careful calculating with negative numbers It is easy to", "## Vectors operations Discussions for the Core part of the syllabus. Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths HL Revision Notes ### Vectors operations Vectors operations - IB Maths HL If $a= \\begin{pmatrix} -3 \\\\ 5 \\\\ 11 \\end{pmatrix}$ and $b= \\begin{pmatrix} -4 \\\\ 6 \\\\ 8 \\end{pmatrix}$ How can we find the vectors $a-3b$ and $4a+3b$ Thanks lora Posts: 0 Joined: Wed Apr 10, 2013 7:36 pm ### Re: Vectors operations Vectors operations, addition, subtraction, scalar multiplication - IB Maths HL First of all let me remind you the properties of vector addition The vector addition (subtraction) is Commutative, Associative, there is an additive identity which is the zero vector ad there is an additive inverse which is the vector with equal magnitude and opposite direction. Also there is the scalar multiplication which is when a vector is multiplied by a scalar. $a-3b=\\begin{pmatrix} -3 \\\\ 5 \\\\ 11 \\end{pmatrix} -3 \\begin{pmatrix} -4 \\\\ 6 \\\\ 8 \\end{pmatrix}$ $=\\begin{pmatrix} -3 \\\\ 5 \\\\ 11 \\end{pmatrix} -\\begin{pmatrix} -12 \\\\ 18 \\\\ 24 \\end{pmatrix}$ $=\\begin{pmatrix} -3+12 \\\\ 5-18 \\\\ 11-24 \\end{pmatrix}$ $=\\begin{pmatrix} 9 \\\\ -13 \\\\ -13 \\end{pmatrix}$ and $4a+3b =4 \\begin{pmatrix} -3 \\\\ 5 \\\\ 11 \\end{pmatrix} + 3 \\begin{pmatrix} -4 \\\\ 6 \\\\ 8 \\end{pmatrix}$ $=\\begin{pmatrix} -12 \\\\ 20 \\\\", "where $Z_j=\\begin{pmatrix}u&-v\\\\v&u\\end{pmatrix}$. Case 2. $F=\\mathbb{C}$. Then $A$ is a scalar matrix $A=aI_n$ (where $a\\in\\mathbb{C}$).", "using DataFrames car = DataFrame(dataset(\"cars\")); ## Operations • Scalar operations: $$\\begin{pmatrix} 1\\\\ 2 \\end{pmatrix} + 2 = \\begin{pmatrix} 3\\\\ 4 \\end{pmatrix}$$ • Elementwise operations: $$\\begin{pmatrix} a_1\\\\ a_2 \\end{pmatrix} + \\begin{pmatrix} b_1\\\\ b_2 \\end{pmatrix} = \\begin{pmatrix} a_1 + b_1\\\\ a_2 + b_2 \\end{pmatrix}$$ • $$r + s = s + r$$. Think of it as geometrical addition of two vectors. • Associative $$t + (s + r) = (t + s) + r$$. ### Inner or dot product #### Calculate length and directions • The length of a vector is given by the hypotenuse (Pythagoras). If $$r = [a, b]$$, then the size of $$r$$ is $$\\| r \\| =\\sqrt{a^2 + b^2}$$. But the definition applies to higher dimensions too. The length of any vector is the square root of sum of squares of its elements. • The dot product of two vectors is a single number (scalar): $$r.s = \\sum_{i=1}^{n} r_i s_i$$ #### Properties of dot product 1. Commutative: $$r.s = r.s$$ 2. Distributive over addition: $$r.(s+t) = r.s + r.t$$ 3. Associative: $$r.(as) = a(r.s)$$, where a is a number (scalar). • The dot product of a vector to itself is the square of its length: $$r.r = {r_1}^2 + {r_2}^2 = \\|r\\|^2$$. #### Cosine and dot product • Cosine rule: in a triangle with sides $$a$$,", "vector from $(3,3)$ to $(2,5)$ in $\\mathbb{R}^2$ 3. the vector from $(1,0,6)$ to $(5,0,3)$ in $\\mathbb{R}^3$ 4. the vector from $(6,8,8)$ to $(6,8,8)$ in $\\mathbb{R}^3$ This exercise is recommended for all readers. Problem 2 Decide if the two vectors are equal. 1. the vector from $(5,3)$ to $(6,2)$ and the vector from $(1,-2)$ to $(1,1)$ 2. the vector from $(2,1,1)$ to $(3,0,4)$ and the vector from $(5,1,4)$ to $(6,0,7)$ This exercise is recommended for all readers. Problem 3 Does $(1,0,2,1)$ lie on the line through $(-2,1,1,0)$ and $(5,10,-1,4)$? This exercise is recommended for all readers. Problem 4 1. Describe the plane through $(1,1,5,-1)$, $(2,2,2,0)$, and $(3,1,0,4)$. 2. Is the origin in that plane? Problem 5 Describe the plane that contains this point and line. $\\begin{pmatrix} 2 \\\\ 0 \\\\ 3 \\end{pmatrix} \\qquad \\{\\begin{pmatrix} -1 \\\\ 0 \\\\ -4 \\end{pmatrix} +\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}t \\,\\big|\\, t\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 6 Intersect these planes. $\\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}t+ \\begin{pmatrix} 0 \\\\ 1 \\\\ 3 \\end{pmatrix}s \\,\\big|\\, t,s\\in\\mathbb{R}\\} \\qquad \\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\\\ 0 \\end{pmatrix}k+ \\begin{pmatrix} 2 \\\\ 0 \\\\ 4 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 7 Intersect each pair, if possible. 1. $\\{\\begin{pmatrix} 1 \\\\", "# 11b. Scalar and vector products The following questions illustrate the key concepts needed in this topic. See if you can answer them. Click on the title for detailed discussion of how to tackle them. Click on or hover over the question to show the answer. It is given that vectors $\\mathbf{a} = \\begin{pmatrix} 1 \\\\ 2 \\\\ -3 \\end{pmatrix}, \\mathbf{b} = \\begin{pmatrix} 5 \\\\ 0 \\\\ 1 \\end{pmatrix}, \\mathbf{c} = \\begin{pmatrix} -1 \\\\ 1 \\\\ z \\end{pmatrix}$ and $\\mathbf{d} = \\begin{pmatrix} k+2 \\\\ k-1 \\\\ k \\end{pmatrix}$ where $z$ and $k$ are constants. Points $A$ and $B$ have coordinates $A(1,2,-3)$ and $B(5,0,1)$. Points $A$ and $B$ have coordinates $A(-1,2,3)$ and $B(5,0,1)$. Remark: $A$ has changed compared to the earlier questions. It is given that $\\mathbf{a} = -\\mathbf{i} + 2 \\mathbf{j} + 3 \\mathbf{k}$. It is given that the direction cosines of a vector $\\mathbf{b}$ with respect to the $x$ -and $y$- axes are $\\frac{2}{\\sqrt{6}}$ and $\\frac{-1}{\\sqrt{6}}$ respectively. Remark: For the following questions, do not use $\\mathbf{a}$ and $\\mathbf{b}$ from the previous questions. Vectors $\\mathbf{a}$ and $\\mathbf{b}$ are now unknown. With respect to the origin $O$, points $A,B$ and $C$ have position vectors $\\mathbf{a}, \\mathbf{b}$ and $\\mathbf{c}=3\\mathbf{a}+\\mathbf{b}$ respectively. It is further given that $|\\mathbf{a}| = 2, \\angle AOB = 60^\\circ$ and $\\mathbf{b}$ is a unit vector. Points $A,B$", "Review question # Can we show the direction of $P'Q'$ is independent of $P$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R8565 ## Solution The plane $\\pi$ has equation $\\mathbf{r} = \\mathbf{a} + \\lambda \\mathbf{b}+ \\mu \\mathbf{c}$ and the plane $\\pi'$ has equation $\\mathbf{r} = \\mathbf{d}+ \\lambda'\\mathbf{e}+\\mu'\\mathbf{f}$, where $\\lambda$, $\\mu$, $\\lambda'$ and $\\mu'$ are scalar parameters. It is given that $\\mathbf{a}=\\begin{pmatrix}1 \\\\ 0 \\\\ -1\\end{pmatrix}, \\mathbf{b}=\\begin{pmatrix}2\\\\ -1 \\\\ -1\\end{pmatrix}, \\mathbf{c} = \\begin{pmatrix}-1 \\\\ 2 \\\\ 1\\end{pmatrix}, \\mathbf{d} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}, \\mathbf{e} = \\begin{pmatrix}p \\\\ q \\\\ r\\end{pmatrix}, \\mathbf{f} = \\begin{pmatrix}q\\\\ r\\\\ p\\end{pmatrix},$ $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ and that $p$ is negative. The planes $\\pi$ and $\\pi'$ are parallel. Find $p$, $q$ and $r$. We can find a vector perpendicular to $\\pi$ using the cross product $\\mathbf{b} \\times \\mathbf{c}= \\begin{pmatrix}1\\\\ -1 \\\\ 3\\end{pmatrix}$. Since the two planes are parallel this is also perpendicular to $\\pi'$. We can use this fact in the dot product with the two direction vectors $\\mathbf{e}$ and $\\mathbf{f}$ in the plane $\\pi'$. So $\\mathbf{e} . (\\mathbf{b} \\times \\mathbf{c})=0$ and $\\mathbf{f} . (\\mathbf{b} \\times \\mathbf{c})=0$. Giving us \\begin{align} p-q+3r &=0 \\\\ q-r+3p &=0 \\end{align} Adding gives $4p+2r=0 \\quad \\implies r=-2p$ and substituting in the second", "+0 # How do i find (a-b)/c? 0 56 1 +39 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w}$$, and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$and answer with $$\\dfrac{a-b}{c}$$. I have tried this multiple times and I got a=0, b=0, and c=0 but it was incorrect. I tried distributing the vectors and solving for a, b, and c with the 3 equations. Can someone please tell me if I am doing anything wrong? Thanks! Apr 30, 2020 #1 +24983 +3 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}$$, $$\\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}$$. If the vectors $$\\mathbf{v}$$, $$\\mathbf{w}$$ and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a-b}{c}$$. $$\\begin{array}{|rcll|} \\hline \\begin{array}{|rcll|} \\hline a\\begin{pmatrix}", "gives $q=-5p$. We now know that $\\mathbf{e}= t\\begin{pmatrix}1\\\\-5 \\\\ -2\\end{pmatrix}$. We also know $\\big \\vert\\mathbf{e}\\big \\vert = \\sqrt{30}$ and $p$ is negative, hence $t=-1$ and we have $\\mathbf{e}= \\begin{pmatrix}-1\\\\5 \\\\ 2\\end{pmatrix}$ and $\\mathbf{f}= \\begin{pmatrix}5\\\\2 \\\\-1\\end{pmatrix}$. It might be tempting to proceed with vectors perpendicular to $\\pi$ and $\\pi'$ being $\\mathbf{b} \\times \\mathbf{c}$ and $\\mathbf{e} \\times \\mathbf{f}$. $\\mathbf{b}\\times \\mathbf{c} = \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ 2&-1&-1 \\\\ -1&2&1\\end{matrix}\\right \\vert = \\mathbf{i}-\\mathbf{j}+3\\mathbf{k}.$ $\\mathbf{e}\\times \\mathbf{f}= \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ p&q&r \\\\ q&r&p\\end{matrix}\\right \\vert = (qp-r^2)\\mathbf{i}+(rq-p^2)\\mathbf{j}+(pr-q^2)\\mathbf{k}.$ However, the resulting set of simultaneous equations is very difficult to solve. By adding them we can then simplify using $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ in other words $p^2+q^2+r^2=30$. We can think about $(p+q+r)^2$ to continue the manipulation. Try it if you like! Show that $\\pi'$ is the reflection of $\\pi$ in the origin. Using the form of the plane equation $\\mathbf{r}.\\mathbf{n}=\\mathbf{a}.\\mathbf{n}$ with $\\mathbf{n}$ as the normal vector we found earlier, we have the equation of $\\pi$ as $x - y + 3z = -2.$ Similarly using the point $\\mathbf{d}$, the plane $\\pi'$ is $x - y + 3z = 2.$ If $(x_1,y_1,z_1)$ is a point on $\\pi$ then $x_1 - y_1 + 3z_1 = -2$. Multiplying by $-1$ gives $(-x_1) -(-y_1) + 3(-z_1) = 2$. So the point $(-x_1,-y_1,-z_1)$ lies on $\\pi'$. Since every point on", "\\left(1, 1\\right)$ and $\\mathbf_2 = \\left(-3, 2\\right).$ Thus, the three vectors are linearly dependent. Two vectors: Now consider the linear dependence of the two vectors $\\mathbf_1 = \\left(1, 1\\right)$ and $\\mathbf_2 = \\left(-3, 2\\right),$ and check, :$a_1 \\begin 1\\\\1\\end + a_2 \\begin -3\\\\2\\end =\\begin 0\\\\0\\end,$ or :$\\begin 1 & -3 \\\\ 1 & 2 \\end\\begin a_1\\\\ a_2 \\end= \\begin 0\\\\0\\end.$ The same row reduction presented above yields, :$\\begin 1 & 0 \\\\ 0 & 1 \\end\\begin a_1\\\\ a_2 \\end= \\begin 0\\\\0\\end.$ This shows that $a_i = 0,$ which means that the vectors ''v''1 = (1, 1) and ''v''2 = (−3, 2) are linearly independent. ## Vectors in R4 In order to determine if the three vectors in $\\mathbb^4,$ :$\\mathbf_1= \\begin1\\\\4\\\\2\\\\-3\\end, \\mathbf_2=\\begin7\\\\10\\\\-4\\\\-1\\end, \\mathbf_3=\\begin-2\\\\1\\\\5\\\\-4\\end.$ are linearly dependent, form the matrix equation, :$\\begin1&7&-2\\\\4& 10& 1\\\\2&-4&5\\\\-3&-1&-4\\end\\begin a_1\\\\ a_2 \\\\ a_3 \\end = \\begin0\\\\0\\\\0\\\\0\\end.$ Row reduce this equation to obtain, :$\\begin 1& 7 & -2 \\\\ 0& -18& 9\\\\ 0 & 0 & 0\\\\ 0& 0& 0\\end \\begin a_1\\\\ a_2 \\\\ a_3 \\end = \\begin0\\\\0\\\\0\\\\0\\end.$ Rearrange to solve for v3 and obtain, :$\\begin 1& 7 \\\\ 0& -18 \\end \\begin a_1\\\\ a_2 \\end = -a_3\\begin-2\\\\9\\end.$ This equation is easily solved to define non-zero ''a''i, :$a_1 = -3 a_3 /2, a_2 = a_3/2,$ where $a_3$ can be chosen arbitrarily. Thus, the vectors $\\mathbf_1, \\mathbf_2,$", "5 \\;\\\\ \\; 2 \\; \\end{pmatrix} = \\begin{pmatrix} \\; 2 \\times 5 \\;\\\\ \\; 2 \\times 2 \\; \\end{pmatrix} = \\begin{pmatrix} \\; 10 \\;\\\\ \\; 4 \\; \\end{pmatrix} Step-by-step guide: Vector multiplication (coming soon) ### 7. Combining vector addition, subtraction and multiplication Vector addition, subtraction and multiplication are often combined. E.g. 3\\textbf{a}-2\\textbf{b}= 3\\begin{pmatrix} \\; 5 \\;\\\\ \\; 2 \\; \\end{pmatrix} -2 \\begin{pmatrix} \\; 3 \\;\\\\ \\; -1 \\; \\end{pmatrix} = \\begin{pmatrix} \\; 15 \\;\\\\ \\; 6 \\; \\end{pmatrix} - \\begin{pmatrix} \\; 6 \\;\\\\ \\; -2 \\; \\end{pmatrix} = \\begin{pmatrix} \\; 9 \\;\\\\ \\; 8 \\; \\end{pmatrix} A vector quantity you may meet in GCSE Science is velocity vector. A scalar quantity would be speed. Step-by-step guide: Vector problems (coming soon) ### 8. Vector geometry We can solve geometrical problems using vectors. Vectors are equal if they have the same magnitude and direction regardless of where they are. E.g. Vectors A and B are equal. They are travelling in the same direction and have the same magnitude (length). E.g. Vector ED, AC and OB are equal. They are parallel, so they travel in the same directions. They also have the same magnitude Vectors OA, BC and GF are also equal. E.g. Vectors b and c are not equal. They do not have the same magnitude (length) and they do" ]

[ "\\mathbf c)$ of the vectors $\\mathbf a$, $\\mathbf b$ and $\\mathbf c$. Some notable properties of cross and triple products: 1. $\\mathbf a\\times \\mathbf b = -\\mathbf b\\times \\mathbf a$ 2. $(\\alpha \\cdot \\mathbf a)\\times \\mathbf b = \\alpha \\cdot (\\mathbf a\\times \\mathbf b)$ 3. For any $\\mathbf b$ and $\\mathbf c$ there is exactly one vector $\\mathbf r$ such that $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf a\\cdot\\mathbf r$ for any vector $\\mathbf a$. Indeed if there are two such vectors $\\mathbf r_1$ and $\\mathbf r_2$ then $\\mathbf a\\cdot (\\mathbf r_1 - \\mathbf r_2)=0$ for all vectors $\\mathbf a$ which is possible only when $\\mathbf r_1 = \\mathbf r_2$. 4. $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf b\\cdot (\\mathbf c\\times \\mathbf a) = -\\mathbf a\\cdot( \\mathbf c\\times \\mathbf b)$ 5. $(\\mathbf a + \\mathbf b)\\times \\mathbf c = \\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c$. Indeed for all vectors $\\mathbf r$ the chain of equations holds: $$\\mathbf r\\cdot( (\\mathbf a + \\mathbf b)\\times \\mathbf c) = (\\mathbf a + \\mathbf b) \\cdot (\\mathbf c\\times \\mathbf r) = \\mathbf a \\cdot(\\mathbf c\\times \\mathbf r) + \\mathbf b\\cdot(\\mathbf c\\times \\mathbf r) = \\mathbf r\\cdot (\\mathbf a\\times \\mathbf c) + \\mathbf r\\cdot(\\mathbf b\\times \\mathbf c) = \\mathbf r\\cdot(\\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c)$$", "• If any two vectors in the triple scalar product are equal, then its value is zero: \\mathbf{a} \\cdot (\\mathbf{a} \\times \\mathbf{b}) = \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{a}) = \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{b}) = \\mathbf{a} \\cdot (\\mathbf{a} \\times \\mathbf{a}) = 0 • Moreover, [\\mathbf{a}\\cdot(\\mathbf{b}\\times \\mathbf{c})] \\mathbf{a} = (\\mathbf{a}\\times \\mathbf{b})\\times (\\mathbf{a}\\times \\mathbf{c}) • The simple product of two triple products (or the square of a triple product), may be expanded in terms of dot products:[1] ((\\mathbf{a}\\times \\mathbf{b})\\cdot \\mathbf{c})\\;((\\mathbf{d}\\times \\mathbf{e})\\cdot \\mathbf{f}) = \\det\\left[ \\begin{pmatrix} \\mathbf{a} \\\\ \\mathbf{b} \\\\ \\mathbf{c} \\end{pmatrix}\\cdot \\begin{pmatrix} \\mathbf{d} & \\mathbf{e} & \\mathbf{f} \\end{pmatrix}\\right] = \\det \\begin{bmatrix} \\mathbf{a}\\cdot \\mathbf{d} & \\mathbf{a}\\cdot \\mathbf{e} & \\mathbf{a}\\cdot \\mathbf{f} \\\\ \\mathbf{b}\\cdot \\mathbf{d} & \\mathbf{b}\\cdot \\mathbf{e} & \\mathbf{b}\\cdot \\mathbf{f} \\\\ \\mathbf{c}\\cdot \\mathbf{d} & \\mathbf{c}\\cdot \\mathbf{e} & \\mathbf{c}\\cdot \\mathbf{f} \\end{bmatrix} This restates in vector notation that the product of the determinants of two 3×3 matrices equals the determinant of their matrix product. ### Scalar or pseudoscalar Although the scalar triple product gives the volume of the parallelepiped, it is the signed volume, the sign depending on the orientation of the frame or the parity of the permutation of the vectors. This means the product is negated if the orientation is reversed, for example by a parity transformation, and so is more properly described as a pseudoscalar if the orientation can change. This", "\\\\ y\\end{pmatrix} \\cdot d\\mathbf r + \\hat {\\mathbf y} \\oint \\begin{pmatrix}z \\\\ 0 \\\\ -x\\end{pmatrix} \\cdot d\\mathbf r + \\hat {\\mathbf z} \\oint \\begin{pmatrix}-y \\\\ x \\\\ 0\\end{pmatrix} \\cdot d\\mathbf r$$ Applying the Kelvin-Stokes theorem we get: $$\\oint \\mathbf r \\times d\\mathbf r = \\hat {\\mathbf x} \\iint (\\nabla \\times \\begin{pmatrix}0 \\\\ -z \\\\ y\\end{pmatrix}) \\cdot d\\mathbf S + \\hat {\\mathbf y} \\iint (\\nabla \\times \\begin{pmatrix}z \\\\ 0 \\\\ -x\\end{pmatrix}) \\cdot d\\mathbf S + \\hat {\\mathbf z} \\iint (\\nabla \\times \\begin{pmatrix}-y \\\\ x \\\\ 0\\end{pmatrix}) \\cdot d\\mathbf S$$ $$\\oint \\mathbf r \\times d\\mathbf r = \\hat {\\mathbf x} \\iint 2 \\hat {\\mathbf x} \\cdot d\\mathbf S + \\hat {\\mathbf y} \\iint 2 \\hat {\\mathbf y} \\cdot d\\mathbf S + \\hat {\\mathbf z} \\iint 2 \\hat {\\mathbf z} \\cdot d\\mathbf S$$ $$\\oint \\mathbf r \\times d\\mathbf r = 2 \\iint d\\mathbf S = 2 \\iint dS \\hat {\\mathbf n}$$ However, this is a factor 2 different from the equation you proposed. Last edited: Aug 28, 2011 5. Aug 29, 2011 daudaudaudau Sorry, I was missing a factor of two in the equation i posted, so your proof is correct :) 6. Sep 1, 2011 Anthony Alternatively, for arbitrary fixed vector $\\mathbf{a} \\in \\mathbf{R}^3$ $$\\mathbf{a} \\cdot \\oint \\mathbf{r} \\wedge \\mathrm{d}\\mathbf{r} = \\oint \\mathbf{a} \\cdot (\\mathbf{r} \\wedge \\mathrm{d}\\mathbf{r}) = \\oint (\\mathbf{a}", "= \\hat {\\mathbf x} \\oint \\begin{pmatrix}0 \\\\ -z \\\\ y\\end{pmatrix} \\cdot d\\mathbf r + \\hat {\\mathbf y} \\oint \\begin{pmatrix}z \\\\ 0 \\\\ -x\\end{pmatrix} \\cdot d\\mathbf r + \\hat {\\mathbf z} \\oint \\begin{pmatrix}-y \\\\ x \\\\ 0\\end{pmatrix} \\cdot d\\mathbf r$$ Applying the Kelvin-Stokes theorem we get: $$\\oint \\mathbf r \\times d\\mathbf r = \\hat {\\mathbf x} \\iint (\\nabla \\times \\begin{pmatrix}0 \\\\ -z \\\\ y\\end{pmatrix}) \\cdot d\\mathbf S + \\hat {\\mathbf y} \\iint (\\nabla \\times \\begin{pmatrix}z \\\\ 0 \\\\ -x\\end{pmatrix}) \\cdot d\\mathbf S + \\hat {\\mathbf z} \\iint (\\nabla \\times \\begin{pmatrix}-y \\\\ x \\\\ 0\\end{pmatrix}) \\cdot d\\mathbf S$$ $$\\oint \\mathbf r \\times d\\mathbf r = \\hat {\\mathbf x} \\iint 2 \\hat {\\mathbf x} \\cdot d\\mathbf S + \\hat {\\mathbf y} \\iint 2 \\hat {\\mathbf y} \\cdot d\\mathbf S + \\hat {\\mathbf z} \\iint 2 \\hat {\\mathbf z} \\cdot d\\mathbf S$$ $$\\oint \\mathbf r \\times d\\mathbf r = 2 \\iint d\\mathbf S = 2 \\iint dS \\hat {\\mathbf n}$$ However, this is a factor 2 different from the equation you proposed. Last edited: Aug 28, 2011 5. Aug 29, 2011 ### daudaudaudau Sorry, I was missing a factor of two in the equation i posted, so your proof is correct :) 6. Sep 1, 2011 ### Anthony Alternatively, for arbitrary fixed vector $\\mathbf{a} \\in \\mathbf{R}^3$ $$\\mathbf{a} \\cdot \\oint \\mathbf{r} \\wedge \\mathrm{d}\\mathbf{r}", "• 0 Votes - 0 Average • 1 • 2 • 3 • 4 • 5 Lagrange's Vector Identity: $\\forall a,b,c,d \\in \\mathbb{R}^3$, (a×b)×c = (a·c)b-(b·c)a 12-18-2010, 07:33 PM Post: #1 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0 Lagrange's Vector Identity: $\\forall a,b,c,d \\in \\mathbb{R}^3$, (a×b)×c = (a·c)b-(b·c)a (1) $(\\mathbf{a} \\times \\mathbf{b}) \\times \\mathbf{c} = (\\mathbf{a}\\cdot \\mathbf{c}) \\mathbf{b} - (\\mathbf{b}\\cdot \\mathbf{c}) \\mathbf{a}$ is actually equivalent to (2) $(\\mathbf{a} \\times \\mathbf{b})\\cdot(\\mathbf{c} \\times \\mathbf{d}) = (\\mathbf{a} \\cdot \\mathbf{c})(\\mathbf{b}\\cdot \\mathbf{d})-(\\mathbf{b}\\cdot\\mathbf{c})(\\mathbf{a} \\cdot \\mathbf{d})$ Proof: By the identity $(\\mathbf{u} \\times \\mathbf{v})\\cdot \\mathbf{w} = \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})$, dot product with $\\mathbf{d}$ both sides of (1) we get (2). Conversely, from (2) we have $((\\mathbf{a} \\times \\mathbf{b}) \\times \\mathbf{c})\\cdot \\mathbf{d} = (\\mathbf{a} \\times \\mathbf{b})\\cdot(\\mathbf{c} \\times \\mathbf{d}) = (\\mathbf{a} \\cdot \\mathbf{c})(\\mathbf{b}\\cdot\\mathbf{d})-(\\mathbf{b}\\cdot\\mathbf{c})(\\mathbf{a}\\cdot \\mathbf{d}) = ((\\mathbf{a} \\cdot \\mathbf{c}) \\mathbf{b} - (\\mathbf{b}\\cdot\\mathbf{c}) \\mathbf{a}) \\cdot \\mathbf{d}$. Since this is true for all $\\mathbf{d}$, (1) follows. Now we prove (2). Let $(a_1,a_2,a_3) = \\mathbf{a}$ and likewise for $\\mathbf{b}$ and $\\mathbf{c}$, then $(\\mathbf{a} \\cdot \\mathbf{c})(\\mathbf{b}\\cdot \\mathbf{d})-(\\mathbf{b}\\cdot \\mathbf{c})(\\mathbf{a} \\cdot \\mathbf{d}) =$ $\\displaystyle{\\sum_{1\\le i,j \\le 3} (a_i c_i b_j d_j - b_i c_i a_j d_j) = \\sum_{1\\le i<j \\le 3} ((a_i c_i b_j d_j + a_j c_j b_i d_i)-(b_i c_i a_j d_j + b_j c_j a_i d_i))=}$ $\\displaystyle{\\sum_{1\\le i<j \\le 3} (a_i b_j(c_i d_j-c_j d_i)-a_j b_i (c_i d_j - c_j", "\\mathbf{z} \\cdot (\\mathbf{x} \\times \\mathbf{y})$ 3. Malcev identity:[8] $(\\mathbf{x} \\times \\mathbf{y}) \\times (\\mathbf{x} \\times \\mathbf{z}) = ((\\mathbf{x} \\times \\mathbf{y}) \\times \\mathbf{z}) \\times \\mathbf{x} + ((\\mathbf{y} \\times \\mathbf{z}) \\times \\mathbf{x}) \\times \\mathbf{x} + ((\\mathbf{z} \\times \\mathbf{x}) \\times \\mathbf{x}) \\times \\mathbf{y}$ $\\mathbf{x} \\times (\\mathbf{x} \\times \\mathbf{y}) = -|\\mathbf{x}|^2 \\mathbf{y} + (\\mathbf{x} \\cdot \\mathbf{y}) \\mathbf{x}.$ Other properties follow only in the three-dimensional case, and are not satisfied by the seven-dimensional cross product, notably, 1. Vector triple product: $\\mathbf{x} \\times (\\mathbf{y} \\times \\mathbf{z}) = (\\mathbf{x} \\cdot \\mathbf{z}) \\mathbf{y} - (\\mathbf{x} \\cdot \\mathbf{y}) \\mathbf{z}$ 2. Jacobi identity:[8] $\\mathbf{x} \\times (\\mathbf{y} \\times \\mathbf{z}) + \\mathbf{y} \\times (\\mathbf{z} \\times \\mathbf{x}) + \\mathbf{z} \\times (\\mathbf{x} \\times \\mathbf{y}) = 0$ ## Coordinate expressions To define a particular cross product, an orthonormal basis {ej} may be selected and a multiplication table provided that determines all the products {ei × ej}. One possible multiplication table is described in the Example section, but it is not unique.[5] Unlike three dimensions, there are many tables because every pair of unit vectors is perpendicular to five other unit vectors, allowing many choices for each cross product. Once we have established a multiplication table, it is then applied to general vectors x and y by expressing x and y in terms of the basis and expanding x × y through bilinearity. × e1", "$[\\mathbf{A},\\mathbf{B}]=\\mathbf{A}\\mathbf{B}-\\mathbf{B}\\mathbf{A} \\nonumber$ ##### Example $$\\PageIndex{3}$$ Given $$\\mathbf{A}=\\begin{pmatrix} 3&1 \\\\ 2&0 \\end{pmatrix}$$ and $$\\mathbf{B}=\\begin{pmatrix} 1&0 \\\\ -1&2 \\end{pmatrix}$$ Calculate the commutator $$[\\mathbf{A},\\mathbf{B}]$$ ###### Solution $[\\mathbf{A},\\mathbf{B}]=\\mathbf{A}\\mathbf{B}-\\mathbf{B}\\mathbf{A}\\nonumber$ $\\mathbf{A}\\mathbf{B}=\\begin{pmatrix} 3&1 \\\\ 2&0 \\end{pmatrix}\\begin{pmatrix} 1&0 \\\\ -1&2 \\end{pmatrix}=\\begin{pmatrix} 3\\times 1+1\\times (-1)&3\\times 0 +1\\times 2 \\\\ 2\\times 1+0\\times (-1)&2\\times 0+ 0\\times 2 \\end{pmatrix}=\\begin{pmatrix} 2&2 \\\\ 2&0 \\end{pmatrix}\\nonumber$ $\\mathbf{B}\\mathbf{A}=\\begin{pmatrix} 1&0 \\\\ -1&2 \\end{pmatrix}\\begin{pmatrix} 3&1 \\\\ 2&0 \\end{pmatrix}=\\begin{pmatrix} 1\\times 3+0\\times 2&1\\times 1 +0\\times 0 \\\\ -1\\times 3+2\\times 2&-1\\times 1+2\\times 0 \\end{pmatrix}=\\begin{pmatrix} 3&1 \\\\ 1&-1 \\end{pmatrix}\\nonumber$ $[\\mathbf{A},\\mathbf{B}]=\\mathbf{A}\\mathbf{B}-\\mathbf{B}\\mathbf{A}=\\begin{pmatrix} 2&2 \\\\ 2&0 \\end{pmatrix}-\\begin{pmatrix} 3&1 \\\\ 1&-1 \\end{pmatrix}=\\begin{pmatrix} -1&1 \\\\ 1&1 \\end{pmatrix}\\nonumber$ $\\displaystyle{\\color{Maroon}[\\mathbf{A},\\mathbf{B}]=\\begin{pmatrix} -1&1 \\\\ 1&1 \\end{pmatrix}}\\nonumber$ ## Multiplication of a vector by a scalar The multiplication of a vector $$\\vec{v_1}$$ by a scalar $$n$$ produces another vector of the same dimensions that lies in the same direction as $$\\vec{v_1}$$; $n\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\begin{pmatrix} nx \\\\ ny \\end{pmatrix} \\nonumber$ The scalar can stretch or compress the length of the vector, but cannot rotate it (figure [fig:vector_by_scalar]). ## Multiplication of a square matrix by a vector The multiplication of a vector $$\\vec{v_1}$$ by a square matrix produces another vector of the same dimensions of $$\\vec{v_1}$$. For example, we can multiply a $$2\\times 2$$ matrix and a 2-dimensional vector: $\\begin{pmatrix} a&b \\\\ c&d \\end{pmatrix}\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\begin{pmatrix} ax+by \\\\ cx+dy \\end{pmatrix} \\nonumber$ For example, consider the matrix $\\mathbf{A}=\\begin{pmatrix} -2 &0", "this to $\\mathbf{b}$, we must add the $x$ values and $y$ values separately. Doing so, we get the answer to be: $2\\mathbf{a}+\\mathbf{b}=\\begin{pmatrix}6\\\\16\\end{pmatrix}+\\begin{pmatrix}-7\\\\2\\end{pmatrix}=\\begin{pmatrix}-1\\\\18\\end{pmatrix}$ Firstly, to multiply $\\mathbf{a}$ by $3$, we must multiply both of its components: $3\\mathbf{a}=3\\times\\begin{pmatrix}2\\\\7\\end{pmatrix}=\\begin{pmatrix}6\\\\21\\end{pmatrix}$ Then, in order subtract $2\\mathbf{b}$, we must first multiply $\\mathbf{b}$ by $2$ $2\\mathbf{b}=2\\times\\begin{pmatrix}-5\\\\3\\end{pmatrix}=\\begin{pmatrix}-10\\\\6\\end{pmatrix}$ Thus the calculation is: $3\\mathbf{a}-2\\mathbf{b}=\\begin{pmatrix}6\\\\21\\end{pmatrix}-\\begin{pmatrix}-10\\\\6\\end{pmatrix}=\\begin{pmatrix}16\\\\15\\end{pmatrix}$ Firstly, to multiply $\\mathbf{b}$ by $2$, we must multiply both of its components: $2\\mathbf{b}=2\\times\\begin{pmatrix}5\\\\-3\\end{pmatrix}=\\begin{pmatrix}10\\\\-6\\end{pmatrix}$ Then, we can add $\\mathbf{a}$ and $2\\mathbf{b}$: $\\mathbf{a}+2\\mathbf{b}=\\begin{pmatrix}6\\\\2\\end{pmatrix}+\\begin{pmatrix}10\\\\-6\\end{pmatrix}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}$ The final calculation is to subtract $\\mathbf{c}$: $\\mathbf{a}+2\\mathbf{b}-\\mathbf{c}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}-\\begin{pmatrix}2\\\\1\\end{pmatrix} = \\begin{pmatrix}14\\\\-5\\end{pmatrix}$ Level 4-5 ### Learning resources you may be interested in We have a range of learning resources to compliment our website content perfectly. Check them out below.", "# Homework Help: (AxB).(CxD) = ? 1. May 4, 2010 ### Saladsamurai 1. The problem statement, all variables and given/known data I am following along in a book and in one line the author asserts that $$(\\mathbf{A}\\times\\mathbf{B})\\cdot(\\mathbf{C}\\times\\mathbf{D}) = (\\mathbf{A}\\cdot\\mathbf{C})(\\mathbf{B}\\cdot\\mathbf{D}) - (\\mathbf{A}\\cdot\\mathbf{D})(\\mathbf{B}\\cdot\\mathbf{C})\\qquad(1)$$ 2. Relevant equations I believe that he is somehow using the rule that $$\\mathbf{A}\\times(\\mathbf{B}\\times\\mathbf{C}) = \\mathbf{B}(\\mathbf{A}\\cdot\\mathbf{C}) - \\mathbf{C}(\\mathbf{A}\\cdot\\mathbf{B})\\qquad(2)$$ 3. The attempt at a solution Is this the only rule he is using to arrive at (1) ? I am having trouble see how to implement this to arrive at the same result. Am I missing something painfully obvious? 2. May 4, 2010 ### radou Did you try to do this in terms of components, i.e. using the definition of the cross and dot product? (Didn't try it myself, only suggesting.) Edit: although it might get a little messy... 3. May 4, 2010 ### rock.freak667 I would think they'd use this formula as well as this one $$A\\cdot (B \\times C) = B \\cdot (C \\times A) = C \\cdot (A \\times B)$$ 4. May 4, 2010 ### Saladsamurai Ah yes, totally useful . Seeing as I have, in essence, a scalar triple product I would be hard pressed to start this without that rule I have solved it now. Thanks again! Share this great discussion with others", "## Solving a 3 x 3 System of Simultaneous Equation Using Matrix Techniques Suppose we have a system of equations $a_{11} x + a_{12}y + a_{13} z= \\alpha$ $a_{21} x + a_{22}y + a_{23} z= \\beta$ $a_{31} x + a_{32}y + a_{33} z= \\gamma$ We can write this in matrix form as $\\left( \\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \\end{array} \\right) \\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} = \\begin{pmatrix} \\alpha \\\\ \\beta \\\\ \\gamma \\end{pmatrix}$ Let $\\mathbf{a} = \\begin{pmatrix}a_{11}\\\\a_{21}\\\\a_{31}\\end{pmatrix}, \\: \\mathbf{b} = \\begin{pmatrix}a_{12}\\\\a_{22}\\\\a_{32}\\end{pmatrix} , \\: \\mathbf{c} = \\begin{pmatrix}a_{13}\\\\a_{23}\\\\a_{33}\\end{pmatrix}$ Then $A^{-1} = \\frac{1}{det (A)} ( \\mathbf{b} \\times \\mathbf{c} \\: \\mathbf{c} \\times \\mathbf{a} \\; \\mathbf{a} \\times \\mathbf{b} )$ . Hence $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} =\\frac{1}{det (A)} ( \\mathbf{b} \\times \\mathbf{c} \\: \\mathbf{c} \\times \\mathbf{a} \\; \\mathbf{a} \\times \\mathbf{b} ) \\begin{pmatrix} \\alpha \\\\ \\beta \\\\ \\gamma \\end{pmatrix} = \\begin{pmatrix} \\frac{( \\mathbf{b} \\times \\mathbf{c} ) \\cdot (\\alpha \\mathbf{i} + \\beta \\mathbf{j} + \\gamma \\mathbf{j})}{det(A)} \\\\ \\frac{( \\mathbf{c} \\times \\mathbf{a} ) \\cdot (\\alpha \\mathbf{i} + \\beta \\mathbf{j} + \\gamma \\mathbf{j})}{det(A)} \\\\ \\frac{( \\mathbf{a} \\times \\mathbf{b} ) \\cdot (\\alpha \\mathbf{i} + \\beta \\mathbf{j} + \\gamma \\mathbf{j})}{det(A)} \\end{pmatrix}$ . In matrix form $x= \\frac{\\left| \\begin{array}{ccc} \\alpha & a_{12} & a_{13} \\\\ \\beta & a_{22} & a_{23} \\\\ \\gamma & a_{32} & a_{33} \\end{array} \\right|}{\\left| \\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\\\", "\\cdot \\mathbf{c}$$ Here, $\\mathbf{c} = \\mathbf{d}_1 \\times 4 \\mathbf{d}_2 =4 ( \\mathbf{d}_1 \\times \\mathbf{d}_2 ).$ Can you see why $\\mathbf{d}_1 \\cdot \\mathbf{c}$ and $\\mathbf{d}_2 \\cdot \\mathbf{c}$ are both equal to 0? (Hint: question number (2) in my response.) Last edited: Jan 22, 2017", "c = \\mathbf b \\times a\\mathbf c$$You must be thinking of: $$a(\\mathbf b + \\mathbf c) = a\\mathbf b + a\\mathbf c$$ Gold Member MHB Hi, Dan, I'm sorry I didn't get it. That is a scalar multiplication so q should be multiplied to both. Generally a(bxc)= abxac. Why are we multiplying only once? Is ##2(3 \\times 4 ) = (2 \\cdot 3) \\times (2 \\cdot 4)##? -Dan $$a(\\mathbf b \\times \\mathbf c) = a\\mathbf b \\times \\mathbf c = \\mathbf b \\times a\\mathbf c$$You must be thinking of: $$a(\\mathbf b + \\mathbf c) = a\\mathbf b + a\\mathbf c$$", "multiply both of its components by $2$: $2\\mathbf{a}=2\\times\\begin{pmatrix}3\\\\8\\end{pmatrix}=\\begin{pmatrix}6\\\\16\\end{pmatrix}$ Then, to add this to $\\mathbf{b}$, we must add the $x$ values and $y$ values separately. Doing so, we get the answer to be: $2\\mathbf{a}+\\mathbf{b}=\\begin{pmatrix}6\\\\16\\end{pmatrix}+\\begin{pmatrix}-7\\\\2\\end{pmatrix}=\\begin{pmatrix}-1\\\\18\\end{pmatrix}$ Firstly, to multiply $\\mathbf{a}$ by $3$, we must multiply both of its components: $3\\mathbf{a}=3\\times\\begin{pmatrix}2\\\\7\\end{pmatrix}=\\begin{pmatrix}6\\\\21\\end{pmatrix}$ Then, in order subtract $2\\mathbf{b}$, we must first multiply $\\mathbf{b}$ by $2$. $2\\mathbf{b}=2\\times\\begin{pmatrix}-5\\\\3\\end{pmatrix}=\\begin{pmatrix}-10\\\\6\\end{pmatrix}$ Thus the calculation is: $3\\mathbf{a}-2\\mathbf{b}=\\begin{pmatrix}6\\\\21\\end{pmatrix}-\\begin{pmatrix}-10\\\\6\\end{pmatrix}=\\begin{pmatrix}16\\\\15\\end{pmatrix}$ Firstly, to multiply $\\mathbf{b}$ by $2$, we must multiply both of its components: $2\\mathbf{b}=2\\times\\begin{pmatrix}5\\\\-3\\end{pmatrix}=\\begin{pmatrix}10\\\\-6\\end{pmatrix}$ Then, we can add $\\mathbf{a}$ and $2\\mathbf{b}$: $\\mathbf{a}+2\\mathbf{b}=\\begin{pmatrix}6\\\\2\\end{pmatrix}+\\begin{pmatrix}10\\\\-6\\end{pmatrix}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}$ The final calculation is to subtract $\\mathbf{c}$: $\\mathbf{a}+2\\mathbf{b}-\\mathbf{c}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}-\\begin{pmatrix}2\\\\1\\end{pmatrix} = \\begin{pmatrix}14\\\\-5\\end{pmatrix}$ ## Related Topics #### Coordinates and Ratios Level 4-5Level 6-7KS3 #### Direct and Inverse Proportion Level 4-5Level 6-7KS3 ## Worksheet and Example Questions ### (NEW) Column Vectors Exam Style Questions - MMe Level 4-5 GCSENewOfficial MME ## You May Also Like... ### GCSE Maths Revision Cards Revise for your GCSE maths exam using the most comprehensive maths revision cards available. These GCSE Maths revision cards are relevant for all major exam boards including AQA, OCR, Edexcel and WJEC. £8.99 ### GCSE Maths Revision Guide The MME GCSE maths revision guide covers the entire GCSE maths course with easy to understand examples, explanations and plenty of exam style questions. We also provide a separate answer book to make checking your answers easier!", "\\frac{-3}{2}, \\frac{1}{3},240\\right)=\\left(0, \\frac{-5}{3},247\\right)$ $7\\left( \\frac{1}{5},\\frac{1}{7},\\frac{1}{9}\\right)=\\left( \\frac{7}{5},1,\\frac{7}{9}\\right)$ Ex.2. $V=M_{m\\times n}(F)=\\left\\lbrace\\begin{pmatrix} a_{11} & \\cdots & a_{1n} \\\\ \\vdots & & \\vdots \\\\ a_{m1} & \\cdots & a_{mn}\\end{pmatrix}: a_{ij} \\in F \\right\\rbrace$ $M_{3\\times 2}( \\mathbb{R})\\ni \\begin{pmatrix} 7 & -7 \\\\ \\pi & \\mathit{e} \\\\ -5 & 2 \\end{pmatrix}$ $\\mbox{Addition by adding entry by entry:}$ $M_{2\\times 2}\\ \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix}+\\begin{pmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{pmatrix}=\\begin{pmatrix} {a_{11}+b_{11}} & {a_{12}+b_{12}} \\\\ {a_{21}+b_{21}} & {a_{22}+b_{22}} \\end{pmatrix}$ $\\mbox{Multiplication by multiplying scalar c to all entries by M.}$ $c\\cdot M_{2\\times 2}\\ \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix}=\\begin{pmatrix} c\\cdot a_{11} & c\\cdot a_{12} \\\\ c\\cdot a_{21} & c\\cdot a_{22} \\end{pmatrix}$ $\\mbox{Zero matrix has all entries = 0:}$ $0_{M_{m\\times n}}=\\begin{pmatrix} 0 & \\cdots & 0 \\\\ \\vdots & & \\vdots \\\\ 0 & \\cdots & 0\\end{pmatrix}$ Ex.3. $\\mathbb{C}$ form a vector space over $\\mathbb{R}$. Ex.4. $\\mbox{F is a vector space over itself.}$ Ex.5. $\\mathbb{R}$ is a vector space over $\\mathbb{Q}$. Ex.6. $\\mbox{Let S be a set. Let}$ $\\mathcal{F}(S,\\mathbb{R})=\\big\\{f:S\\to \\mathbb{R} \\big\\}$ $f,g \\in \\mathcal{F}(S,\\mathbb{R})$ $(f+g)(t)=f(t)+g(t)\\ for\\ any\\ t\\in S$ $(af)(t)=a\\cdot f(t)\\$", "\\to \\infty} {n! \\over \\sqrt{2\\pi n} (n/e)^n} = 1,$$ if you rewrite the product as $(2n)!/(2^n n!)^2$. - Thanks to you two! Dear Michael s! – Bidgoli Mar 12 at 20:08 Here you need to find $\\lim A_n$, where $$A_n= \\frac{1\\times3\\times5\\times\\cdots\\times(2n-1)}{2\\times4\\times6\\times8\\times\\cdots\\times (2n)}.$$ Now note that, for $n\\gt1$, $2n-1\\gt n$. So, in the numerator, you get $$\\frac{1\\times 2\\times 3\\dots\\times n}{2\\times4\\times 6\\times\\dots\\times 2n}\\lt\\frac{1\\times3\\times5\\times\\cdots\\times(2n-1)}{2\\times4\\times6\\times8\\times\\cdots\\times (2n)}\\\\\\implies\\frac1{2^n}\\lt \\frac{1\\times3\\times5\\times\\cdots\\times(2n-1)}{2\\times4\\times6\\times8\\times\\cdots\\times (2n)}$$ Now, you use the identity $$\\frac n{n+1}\\lt\\frac {n+1}{n+2}$$ and define a sequence $$B_n=\\frac 23\\times \\frac 45\\times\\dots \\times \\frac {2n}{2n+1} .$$ So, now you have $$A_n\\lt B_n\\\\\\implies (A_n)^2\\lt A_nB_n=\\frac 1{2n+1}\\\\\\implies A_n\\lt \\frac 1{\\sqrt{2n+1}}.$$ So, you have $$\\frac 1{2^n}\\lt A_n\\lt \\frac 1{\\sqrt{2n+1}}\\\\\\implies\\lim \\frac 1{2^n}\\lt\\lim A_n\\lt\\lim \\frac 1{\\sqrt{2n+1}}.$$ So, by sandwitch theorem, $$\\lim A_n=0.$$ - What's wrong with this? – Subhadeep Dey Mar 11 at 19:06 Dear downvoter, are you going to explain the reason for downvoting? – Subhadeep Dey Mar 11 at 19:18 I didn't downvote, but the first half of the proof is unnecessary. For the sandwich theorem, all you need is $0 < A_n < 1/\\sqrt{2n+1}$, and $0 < A_n$ is obvious. – alephzero Mar 11 at 23:51 @alephzero, yes, I know. I have just showed that there are several ways to get the lower sequence. – Subhadeep Dey Mar 12 at 7:49 JUST BRILLIANT... – Bidgoli Mar 12 at 20:06 To paraphrase Wikipedia:", "{a} \\cdot \\mathbf {c} )\\mathbf {b} -(\\mathbf {a} \\cdot \\mathbf {b} )\\mathbf {c} }$. This is known as triple product expansion, or Lagrange's formula, [2] [3] although the latter name is also used for several other formulas. Its right hand side can be remembered by using the mnemonic \"ACB ABC\", provided one keeps in mind which vectors are dotted together. A proof is provided below. Some textbooks write the identity as ${\\displaystyle \\mathbf {a} \\times (\\mathbf {b} \\times \\mathbf {c} )=\\mathbf {b} (\\mathbf {a} \\cdot \\mathbf {c} )-\\mathbf {c} (\\mathbf {a} \\cdot \\mathbf {b} )}$ such that a more familiar mnemonic \"BAC CAB\" is obtained, as in “back of the cab”. Since the cross product is anticommutative, this formula may also be written (up to permutation of the letters) as: ${\\displaystyle (\\mathbf {a} \\times \\mathbf {b} )\\times \\mathbf {c} =-\\mathbf {c} \\times (\\mathbf {a} \\times \\mathbf {b} )=-(\\mathbf {c} \\cdot \\mathbf {b} )\\mathbf {a} +(\\mathbf {c} \\cdot \\mathbf {a} )\\mathbf {b} }$ From Lagrange's formula it follows that the vector triple product satisfies: ${\\displaystyle \\mathbf {a} \\times (\\mathbf {b} \\times \\mathbf {c} )+\\mathbf {b} \\times (\\mathbf {c} \\times \\mathbf {a} )+\\mathbf {c} \\times (\\mathbf {a} \\times \\mathbf {b} )=\\mathbf {0} }$ which is the Jacobi identity for the cross product. Another useful formula follows: ${\\displaystyle (\\mathbf {a} \\times", "I think I got it I have already shown that $\\mathbb{Z^+}\\times \\mathbb{Z^+}\\;\\sim\\; \\mathbb{Z^+}$ Also I have shown previously that for any sets A,B,C,D if $A\\sim B$ and $C\\sim D$ then it follows that $$^{A}C \\;\\sim \\;^{B}D\\cdots\\;(1)$$ and $$\\mathcal{P}(A)\\;\\sim\\;\\mathcal{P}(B)$$ So it follows that $$\\mathcal{P}(\\mathbb{Z^+})\\;\\sim\\;\\mathcal{P}( \\mathbb{Z^+}\\times \\mathbb{Z^+})\\cdots\\;(2)$$ Since for any set A $$\\mathcal{P}(A)\\;\\sim\\; ^{A}\\{0,1\\}$$ we have $$\\mathcal{P}(\\mathbb{Z^+}\\times \\mathbb{Z^+})\\;\\sim\\; ^{(\\mathbb{Z^+}\\times \\mathbb{Z^+} )} \\{0,1\\} \\cdots\\;(3)$$ But for any sets A,B,C we have $$^{(A\\times B)}C\\;\\sim\\; ^{A}( ^{B}C)$$ So we have $$^{(\\mathbb{Z^+}\\times \\mathbb{Z^+} )} \\{0,1\\}\\;\\sim\\; ^{\\mathbb{Z^+}}(^{\\mathbb{Z^+}} \\{0,1\\} )\\cdots\\;(4)$$ so using equations 2, 3 and 4 ,due to transitivity of $\\sim$ we can say that $$\\mathcal{P}(\\mathbb{Z^+})\\;\\sim\\; ^{\\mathbb{Z^+}}(^{\\mathbb{Z^+}} \\{0,1\\} ) \\cdots\\;(5)$$ Now $$\\mathbb{Z^+}\\;\\sim \\mathbb{Z^+}$$ and $$\\mathcal{P}(\\mathbb{Z^+})\\;\\sim\\; ^{\\mathbb{Z^+}}\\{0,1\\}$$ It follows from 1 that $$^{\\mathbb{Z^+}}\\mathcal{P}(\\mathbb{Z^+})\\;\\sim\\; ^{\\mathbb{Z^+}}(^{\\mathbb{Z^+}} \\{0,1\\} )$$ Now using 5 , we can say that $$^{\\mathbb{Z^+}}\\mathcal{P}(\\mathbb{Z^+})\\;\\sim\\; \\mathcal{P}(\\mathbb{Z^+})$$ thanks micromass, that was a lot of typing I was thinking of coming up with some bijection. I think that's probably lot more difficult in this case. 6. Jan 21, 2012 ### issacnewton By the way I was wondering, if any bijection could be found between $^{\\mathbb{Z^+}}\\mathcal{P}(\\mathbb{Z^+})$ and $\\mathcal{P}(\\mathbb{Z^+})$ ? 7. Jan 21, 2012 ### micromass Staff Emeritus Yes, I think so. Notice that every element of $\\mathcal{P}(\\mathbb{Z}^+)$ is determined by a sequence of 0 and 1 (where 1 means that the corresponding element is in the subset). For example,", "why $x^2+y^2=(-y)^2+x^2$ (?!), which makes for a rather poor question, I am sorry to say. Note also that every sequence $(\\mathbf x_n)$ defined by $\\mathbf x_{n+1}=L(\\mathbf x_n)$ has period $4$, as an elementary computation shows. I know that to solve the exercise I need to show that F is constant along the orbits of L. That is, I need to show that $F \\text{ o } L(x) = L(x)$. No. That $F$ is an invariant of the dynamical system associated to $L$ means that, if $$\\dot{\\mathbf x}(t)=L(\\mathbf x(t)),$$ then the quantity $F(\\mathbf x(t))$ does not depend on $t$. This is usually done showing that $u'(t)=0$ for every $t$, where $$u(t)=F(\\mathbf x(t)),$$ which is implied by the condition that, for every $\\mathbf x$, $$\\nabla F(\\mathbf x)\\cdot L(\\mathbf x)=0.$$ In your case, $$\\nabla F(\\mathbf x)=2\\mathbf x,$$ hence one is left with $$\\nabla F(\\mathbf x)\\cdot L(\\mathbf x)=2\\mathbf x\\cdot L(\\mathbf x)=2\\begin{pmatrix}x\\\\y\\end{pmatrix}\\cdot\\begin{pmatrix}y\\\\-x\\end{pmatrix}=0.$$ Here is a rough sketch of some solutions $(\\mathbf x(t))$: $\\hspace2in$", "B) \\times (\\mathbf A + \\mathbf B)$$ Now the cross product $\\times$ is linear, but you also get terms $\\mathbf A \\times \\mathbf B$ and $\\mathbf B \\times \\mathbf A$, which cancel by a property of the cross product. So in the end you get $$\\mathbf C \\times \\mathbf C = \\mathbf A \\times \\mathbf A + \\mathbf B \\times \\mathbf B$$ Which just says that $0=0+0$. Another way to see that your conclusion does not hold, is simply to see that $\\mathbf A \\times \\mathbf B$ is non-zero in general, so by taking the magnitude, we find that $|\\mathbf A \\circ \\mathbf B|$ is non-zero in general, and hence $\\mathbf A \\circ \\mathbf B$ is non-zero in general. 5. Aug 4, 2016 ### Devil Moo Now it is defined as the angle between $\\mathbf A$ and $\\mathbf B$ where $\\theta$ is smaller than or equal to $\\pi$ and it is commutative. How do you find out that it is not distributive? Um... How are these two examples related to the derivation? 6. Aug 4, 2016 ### Staff: Mentor By looking at an example. $$A = \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}, B = \\begin{pmatrix} -1 \\\\ 0 \\\\ 0 \\end{pmatrix}, C = \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$$ $$(A+B) \\circ C = 0 \\neq A", "of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^{\\rm t} {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}. \\end{align*} ##### Example8.19 Suppose that the words to be encoded consist of all binary 3-tuples and that our encoding scheme is even-parity. To encode an arbitrary 3-tuple, we add a fourth bit to obtain an even number of 1s. Notice that an arbitrary $n$-tuple ${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^{\\rm t}$ has an even number of 1s exactly when $x_1 + x_2 + \\cdots + x_n = 0$; hence, a 4-tuple ${\\mathbf x} = (x_1, x_2, x_3, x_4)^{\\rm t}$ has an even number of 1s if $x_1+ x_2+ x_3+ x_4 = 0$, or \\begin{equation*}{\\mathbf x} \\cdot {\\mathbf 1} = {\\mathbf x}^{\\rm t} {\\mathbf 1} = \\begin{pmatrix} x_1 & x_2 & x_3 & x_4 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\\\ 1 \\end{pmatrix} = 0.\\end{equation*} This example leads us to hope that there is a connection between matrices and coding theory. Let ${\\mathbb M}_{m \\times n}({\\mathbb Z}_2)$ denote the set of all $m \\times n$ matrices with entries in ${\\mathbb Z}_2$. We do matrix operations" ]

[ "\\mathbf c)$ of the vectors $\\mathbf a$, $\\mathbf b$ and $\\mathbf c$. Some notable properties of cross and triple products: 1. $\\mathbf a\\times \\mathbf b = -\\mathbf b\\times \\mathbf a$ 2. $(\\alpha \\cdot \\mathbf a)\\times \\mathbf b = \\alpha \\cdot (\\mathbf a\\times \\mathbf b)$ 3. For any $\\mathbf b$ and $\\mathbf c$ there is exactly one vector $\\mathbf r$ such that $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf a\\cdot\\mathbf r$ for any vector $\\mathbf a$. Indeed if there are two such vectors $\\mathbf r_1$ and $\\mathbf r_2$ then $\\mathbf a\\cdot (\\mathbf r_1 - \\mathbf r_2)=0$ for all vectors $\\mathbf a$ which is possible only when $\\mathbf r_1 = \\mathbf r_2$. 4. $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf b\\cdot (\\mathbf c\\times \\mathbf a) = -\\mathbf a\\cdot( \\mathbf c\\times \\mathbf b)$ 5. $(\\mathbf a + \\mathbf b)\\times \\mathbf c = \\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c$. Indeed for all vectors $\\mathbf r$ the chain of equations holds: $$\\mathbf r\\cdot( (\\mathbf a + \\mathbf b)\\times \\mathbf c) = (\\mathbf a + \\mathbf b) \\cdot (\\mathbf c\\times \\mathbf r) = \\mathbf a \\cdot(\\mathbf c\\times \\mathbf r) + \\mathbf b\\cdot(\\mathbf c\\times \\mathbf r) = \\mathbf r\\cdot (\\mathbf a\\times \\mathbf c) + \\mathbf r\\cdot(\\mathbf b\\times \\mathbf c) = \\mathbf r\\cdot(\\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c)$$", "\\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 686 In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not. (a) The matrix $A$ is a $3 \\times 3$", "# Tagged: inner product ## Problem 715 Let $\\mathbf{v}_{1} = \\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix} ,\\; \\mathbf{v}_{2} = \\begin{bmatrix} 1 \\\\ -1 \\end{bmatrix} .$ Let $V=\\Span(\\mathbf{v}_{1},\\mathbf{v}_{2})$. Do $\\mathbf{v}_{1}$ and $\\mathbf{v}_{2}$ form an orthonormal basis for $V$? If not, then find an orthonormal basis for $V$. ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 684 Let $\\mathbb{R}^2$ be the vector space of size-2 column vectors. This vector space has an inner product defined by $\\langle \\mathbf{v} , \\mathbf{w} \\rangle = \\mathbf{v}^\\trans \\mathbf{w}$. A linear transformation $T : \\R^2 \\rightarrow \\R^2$ is called an orthogonal transformation if for all $\\mathbf{v} , \\mathbf{w} \\in \\R^2$, $\\langle T(\\mathbf{v}) , T(\\mathbf{w}) \\rangle = \\langle \\mathbf{v} , \\mathbf{w} \\rangle.$ For a fixed", "• If any two vectors in the triple scalar product are equal, then its value is zero: \\mathbf{a} \\cdot (\\mathbf{a} \\times \\mathbf{b}) = \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{a}) = \\mathbf{a} \\cdot (\\mathbf{b} \\times \\mathbf{b}) = \\mathbf{a} \\cdot (\\mathbf{a} \\times \\mathbf{a}) = 0 • Moreover, [\\mathbf{a}\\cdot(\\mathbf{b}\\times \\mathbf{c})] \\mathbf{a} = (\\mathbf{a}\\times \\mathbf{b})\\times (\\mathbf{a}\\times \\mathbf{c}) • The simple product of two triple products (or the square of a triple product), may be expanded in terms of dot products:[1] ((\\mathbf{a}\\times \\mathbf{b})\\cdot \\mathbf{c})\\;((\\mathbf{d}\\times \\mathbf{e})\\cdot \\mathbf{f}) = \\det\\left[ \\begin{pmatrix} \\mathbf{a} \\\\ \\mathbf{b} \\\\ \\mathbf{c} \\end{pmatrix}\\cdot \\begin{pmatrix} \\mathbf{d} & \\mathbf{e} & \\mathbf{f} \\end{pmatrix}\\right] = \\det \\begin{bmatrix} \\mathbf{a}\\cdot \\mathbf{d} & \\mathbf{a}\\cdot \\mathbf{e} & \\mathbf{a}\\cdot \\mathbf{f} \\\\ \\mathbf{b}\\cdot \\mathbf{d} & \\mathbf{b}\\cdot \\mathbf{e} & \\mathbf{b}\\cdot \\mathbf{f} \\\\ \\mathbf{c}\\cdot \\mathbf{d} & \\mathbf{c}\\cdot \\mathbf{e} & \\mathbf{c}\\cdot \\mathbf{f} \\end{bmatrix} This restates in vector notation that the product of the determinants of two 3×3 matrices equals the determinant of their matrix product. ### Scalar or pseudoscalar Although the scalar triple product gives the volume of the parallelepiped, it is the signed volume, the sign depending on the orientation of the frame or the parity of the permutation of the vectors. This means the product is negated if the orientation is reversed, for example by a parity transformation, and so is more properly described as a pseudoscalar if the orientation can change. This", "Vectors $\\mathbf u$, $\\mathbf a$ and $\\mathbf b$: \\begin{align} \\mathbf u &= \\begin{pmatrix} -2\\\\-1\\end{pmatrix} & \\mathbf a &= \\begin{pmatrix} -3\\\\1\\end{pmatrix} & \\mathbf b &= \\begin{pmatrix} 1\\\\3\\end{pmatrix} \\end{align} $\\mathbf u = \\alpha \\mathbf a + \\beta \\mathbf b \\qquad$ Norm:$|| \\mathbf u||$. \\begin{align} \\hat{\\mathbf u} &= \\lambda \\mathbf a = {\\langle\\mathbf a, \\mathbf u\\rangle \\over \\langle\\mathbf a, \\mathbf a\\rangle} \\mathbf a = {\\begin{pmatrix}-3\\\\1\\end{pmatrix}\\cdot\\begin{pmatrix}-2\\\\-1\\end{pmatrix} \\over \\begin{pmatrix}-3\\\\1\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\1\\end{pmatrix} }\\mathbf a = { 5 \\over 10 }\\mathbf a = { 1 \\over 2 }\\mathbf a = \\begin{pmatrix}-1.5\\\\0.5\\end{pmatrix} \\end{align} \\item Matrices \\begin{align*} \\begin{pmatrix}1&3+2i\\\\3-2i&4\\end{pmatrix} && {\\rm det}{\\mathbf A}&= \\begin{vmatrix}3&4&7\\\\2&-2&1\\\\1&2&5\\end{vmatrix} & \\begin{vmatrix}x&1&1\\\\1&x&1\\\\1&1&x \\end{vmatrix} &= (x-1)^2(x+2) \\end{align*} \\item $a_1 \\ge 0 \\qquad p(x) = p_0+p_1x+p_2x^2+p_3x^3+p_4x^4+p_5x^5$ \\item Transformations: \\begin{align*} T:\\mathbb R^3 &\\rightarrow \\mathbb R^2& \\text{where}\\quad T \\begin{pmatrix}a_1\\\\a_2\\\\a_3\\end{pmatrix} = \\begin{pmatrix}a_1a_2\\\\a_1a_3\\end{pmatrix} \\end{align*} \\begin{align*} T \\begin{pmatrix}a_0+a_1x+a_2x^2+a_3x^3\\end{pmatrix} = \\begin{pmatrix}a_1&a_2\\\\a_3&a_1-a_0\\end{pmatrix} \\end{align*} \\end{enumerate} • This is an interesting answer, many thanks. The output looks great with this method. The issue that i have is that I have more than 100 pages of different LaTeX files that I want to convert to html with mathml and conversion from .tex to .rmd would be a significant undertaking, though I take your point that the .rmd code is very similar to the .tex code – tom Aug 20 '20 at 15:48 • @Tom. I understand that perspective very well. I have learned that it is", "# Thread: Spans and vectors 1. ## Spans and vectors If a question asks, Is the vector $\\mathbf{b}=\\begin{pmatrix}-2\\\\ -6\\\\ -4\\end{pmatrix}\\in \\mbox{span}(\\mathbf{v_1, v_2, v_3, v_4})$ where: $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 3\\\\ 0\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}2\\\\ 2\\\\ 1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ -1\\end{pmatrix}$, $\\mathbf{v_4}=\\begin{pmatrix}1\\\\ -2\\\\ 1\\end{pmatrix}$ Do we just make it a matrix and solve the matrix? And how would we solve it if we have 4 variables but only 3 equations? 2) Is the set of vectors $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 2\\\\ 3\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}1\\\\ 1\\\\ -1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ 5\\end{pmatrix}$ a spanning set of $\\mathbb{R}^3$? Do we just make this a matrix with say $\\mathbf{x}=\\begin{pmatrix}x_1\\\\ x_2\\\\ x_3\\end{pmatrix}$ as our solution and see if we can solve the matrix? 2. So your first question is asking whether there exist scalars $a_{1},a_{2},a_{3},a_{4}$ such that $\\mathbf{b}=a_{1}\\mathbf{v}_{1}+a_{2}\\mathbf{v}_{2} +a_{3}\\mathbf{v}_{3}+a_{4}\\mathbf{v}_{4}.$ This equation you can translate into a normal matrix problem of the form $A\\mathbf{x}=\\mathbf{b}$. Row reduce that problem. If you find any solutions (one or infinitely many), then the answer is yes. If you find that there are no solutions, then the answer is no. Your second question, again, is asking whether there exist scalars $c_{1}, c_{2}, c_{3}$ such that $c_{1}\\mathbf{v}_{1}+c_{2}\\mathbf{v}_{2}+c_{3}\\math bf{v}_{3}=\\mathbf{x}$ for your arbitrary $\\mathbf{x}.$ This, again, can be translated into a standard matrix problem. Since this problem will result in a square matrix, you can use the theory of determinants to help you out.", "\\mathbb{F}$ such that $a = \\lambda \\cdot b$. Now consider your example. Let $a = \\begin{pmatrix} 1 - \\mathrm{i} \\\\ 2 - 2\\mathrm{i}\\end{pmatrix} \\in V$ and $b = \\begin{pmatrix} \\mathrm{i} - 1 \\\\ 2\\mathrm{i} - 2 \\end{pmatrix} \\in V$. The vectors would be linearly dependent, if $a = -1 \\cdot b$ and $-1 \\in \\mathbb{F}$. This is true for the field $\\mathbb{F} := \\mathbb{C}$, but also for $\\mathbb{F} := \\mathbb{R}$ or any other field of which $-1$ is an element.", "# Tagged: magnitude of a vector ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 381 Consider the matrix $A=\\begin{bmatrix} 3/2 & 2\\\\ -1& -3/2 \\end{bmatrix} \\in M_{2\\times 2}(\\R).$ (a) Find the eigenvalues and corresponding eigenvectors of $A$. (b) Show that for $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}\\in \\R^2$, we can choose $n$ large enough so that the length $\\|A^n\\mathbf{v}\\|$ is as small as we like. (University of California, Berkeley, Linear Algebra Final Exam Problem) ## Problem 162 Let $\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3$ are vectors in $\\R^n$. Suppose that vectors $\\mathbf{u}_1$, $\\mathbf{u}_2$ are orthogonal and the norm of $\\mathbf{u}_2$ is $4$ and $\\mathbf{u}_2^{\\trans}\\mathbf{u}_3=7$. Find the value of the real number $a$ in $\\mathbf{u_1}=\\mathbf{u_2}+a\\mathbf{u}_3$. (The Ohio State University, Linear Algebra", "# Tagged: distance ## Problem 689 For this problem, use the complex vectors $\\mathbf{w}_1 = \\begin{bmatrix} 1 + i \\\\ 1 – i \\\\ 0 \\end{bmatrix} , \\, \\mathbf{w}_2 = \\begin{bmatrix} -i \\\\ 0 \\\\ 2 – i \\end{bmatrix} , \\, \\mathbf{w}_3 = \\begin{bmatrix} 2+i \\\\ 1 – 3i \\\\ 2i \\end{bmatrix} .$ Suppose $\\mathbf{w}_4$ is another complex vector which is orthogonal to both $\\mathbf{w}_2$ and $\\mathbf{w}_3$, and satisfies $\\mathbf{w}_1 \\cdot \\mathbf{w}_4 = 2i$ and $\\| \\mathbf{w}_4 \\| = 3$. Calculate the following expressions: (a) $\\mathbf{w}_1 \\cdot \\mathbf{w}_2$. (b) $\\mathbf{w}_1 \\cdot \\mathbf{w}_3$. (c) $((2+i)\\mathbf{w}_1 – (1+i)\\mathbf{w}_2 ) \\cdot \\mathbf{w}_4$. (d) $\\| \\mathbf{w}_1 \\| , \\| \\mathbf{w}_2 \\|$, and $\\| \\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2", "\\frac{-3}{2}, \\frac{1}{3},240\\right)=\\left(0, \\frac{-5}{3},247\\right)$ $7\\left( \\frac{1}{5},\\frac{1}{7},\\frac{1}{9}\\right)=\\left( \\frac{7}{5},1,\\frac{7}{9}\\right)$ Ex.2. $V=M_{m\\times n}(F)=\\left\\lbrace\\begin{pmatrix} a_{11} & \\cdots & a_{1n} \\\\ \\vdots & & \\vdots \\\\ a_{m1} & \\cdots & a_{mn}\\end{pmatrix}: a_{ij} \\in F \\right\\rbrace$ $M_{3\\times 2}( \\mathbb{R})\\ni \\begin{pmatrix} 7 & -7 \\\\ \\pi & \\mathit{e} \\\\ -5 & 2 \\end{pmatrix}$ $\\mbox{Addition by adding entry by entry:}$ $M_{2\\times 2}\\ \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix}+\\begin{pmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{pmatrix}=\\begin{pmatrix} {a_{11}+b_{11}} & {a_{12}+b_{12}} \\\\ {a_{21}+b_{21}} & {a_{22}+b_{22}} \\end{pmatrix}$ $\\mbox{Multiplication by multiplying scalar c to all entries by M.}$ $c\\cdot M_{2\\times 2}\\ \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix}=\\begin{pmatrix} c\\cdot a_{11} & c\\cdot a_{12} \\\\ c\\cdot a_{21} & c\\cdot a_{22} \\end{pmatrix}$ $\\mbox{Zero matrix has all entries = 0:}$ $0_{M_{m\\times n}}=\\begin{pmatrix} 0 & \\cdots & 0 \\\\ \\vdots & & \\vdots \\\\ 0 & \\cdots & 0\\end{pmatrix}$ Ex.3. $\\mathbb{C}$ form a vector space over $\\mathbb{R}$. Ex.4. $\\mbox{F is a vector space over itself.}$ Ex.5. $\\mathbb{R}$ is a vector space over $\\mathbb{Q}$. Ex.6. $\\mbox{Let S be a set. Let}$ $\\mathcal{F}(S,\\mathbb{R})=\\big\\{f:S\\to \\mathbb{R} \\big\\}$ $f,g \\in \\mathcal{F}(S,\\mathbb{R})$ $(f+g)(t)=f(t)+g(t)\\ for\\ any\\ t\\in S$ $(af)(t)=a\\cdot f(t)\\$", "\\mathbf{a_n}\\end{pmatrix} \\ \\sim \\ R=\\begin{pmatrix}\\mathbf{r_1} & \\mathbf{r_2} & \\cdots & \\mathbf{r_n}\\end{pmatrix}$$ where $\\mathbf{a_i}$ and $\\mathbf{r_i}$ are the respective column vectors of $A$ and $R$. Now further suppose that $$\\left\\{\\mathbf{r_{i_1}},\\ \\mathbf{r_{i_2}},\\ \\cdots,\\ \\mathbf{r_{i_k}}\\right\\}$$ is the set of pivot columns of $R$. Then the corresponding columns of $A$ $$\\left\\{\\mathbf{a_{i_1}},\\ \\mathbf{a_{i_2}},\\ \\cdots,\\ \\mathbf{a_{i_k}}\\right\\}$$ forms a basis for the columnspace of $A$. Of course this will in general not be the same basis as the one produced by row reducing $A^\\mathrm{T}$, but there is rarely a need to prioritize one basis over the other. Of course, all of this is assuming that a basis of the columnspace is what you're after. If you simply want the RREF of the transposed matrix, then row reducing again would probably be the easiest way. - add comment", "## Problem 689 For this problem, use the complex vectors $\\mathbf{w}_1 = \\begin{bmatrix} 1 + i \\\\ 1 – i \\\\ 0 \\end{bmatrix} , \\, \\mathbf{w}_2 = \\begin{bmatrix} -i \\\\ 0 \\\\ 2 – i \\end{bmatrix} , \\, \\mathbf{w}_3 = \\begin{bmatrix} 2+i \\\\ 1 – 3i \\\\ 2i \\end{bmatrix} .$ Suppose $\\mathbf{w}_4$ is another complex vector which is orthogonal to both $\\mathbf{w}_2$ and $\\mathbf{w}_3$, and satisfies $\\mathbf{w}_1 \\cdot \\mathbf{w}_4 = 2i$ and $\\| \\mathbf{w}_4 \\| = 3$. Calculate the following expressions: (a) $\\mathbf{w}_1 \\cdot \\mathbf{w}_2$. (b) $\\mathbf{w}_1 \\cdot \\mathbf{w}_3$. (c) $((2+i)\\mathbf{w}_1 – (1+i)\\mathbf{w}_2 ) \\cdot \\mathbf{w}_4$. (d) $\\| \\mathbf{w}_1 \\| , \\| \\mathbf{w}_2 \\|$, and $\\| \\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$", "Express the vector $\\mathbf{u}$ as a linear combination of $\\mathbf{v}$ and $\\mathbf{w}$. (b) Compute $A^5\\mathbf{v}$. (c) Compute $A^5\\mathbf{w}$. (d) Compute $A^5\\mathbf{u}$. ## Problem 691 In this problem, we use the following vectors in $\\R^2$. $\\mathbf{a}=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}, \\mathbf{b}=\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}, \\mathbf{c}=\\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix}, \\mathbf{d}=\\begin{bmatrix} 3 \\\\ 2 \\end{bmatrix}, \\mathbf{e}=\\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}, \\mathbf{f}=\\begin{bmatrix} 5 \\\\ 6 \\end{bmatrix}.$ For each set $S$, determine whether $\\Span(S)=\\R^2$. If $\\Span(S)\\neq \\R^2$, then give algebraic description for $\\Span(S)$ and explain the geometric shape of $\\Span(S)$. (a) $S=\\{\\mathbf{a}, \\mathbf{b}\\}$ (b) $S=\\{\\mathbf{a}, \\mathbf{c}\\}$ (c) $S=\\{\\mathbf{c}, \\mathbf{d}\\}$ (d) $S=\\{\\mathbf{a}, \\mathbf{f}\\}$ (e) $S=\\{\\mathbf{e}, \\mathbf{f}\\}$ (f) $S=\\{\\mathbf{a}, \\mathbf{b}, \\mathbf{c}\\}$ (g) $S=\\{\\mathbf{e}\\}$ ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 672 For an integer $n > 0$, let $\\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \\{ 1 , x , x^2 , \\cdots , x^n \\}$ is a basis of $\\mathrm{P}_n$, called the standard basis. Let $T : \\mathrm{P}_n \\rightarrow \\mathrm{P}_{n+1}$ be the map defined by, for $f \\in \\mathrm{P}_n$, $T (f) (x) =", "we are given a vector $\\mathbf t$ in the vector space (but it might not be in $\\mathcal{L}(\\mathbf B)$) and we want to find the closest vector in $\\mathcal{L}(\\mathbf B)$. There is a simple reduction from $\\textsf{CVP}$ to $\\textsf{SVP}$ by setting $\\mathbf t = \\mathbf 0$. There is also a reduction in the other direction, which involves extending the basis of $\\mathcal{L}(\\mathbf B)$ to also include $\\mathbf t$. This is called embedding. Let us return to the example of DSA and different-moduli equations. The equations we got from before $\\begin{array}{rclc} s_1 \\cdot k_1 - r_1\\cdot x & = & H(m_1) &\\pmod q \\\\ s_2 \\cdot k_2 - r_2\\cdot x & = & H(m_1) &\\pmod q \\\\ -a\\cdot k_1 + 1\\cdot k_2 & = & c & \\pmod M \\\\ \\end{array}$ can be formulated as basis vectors. In matrix form, we have $\\begin{pmatrix} -r_1 & s_1 & 0 & q & 0 & 0 \\\\ -r_2 & 0 & s_2 & 0 & q & 0 \\\\ 0 & -a & 1 & 0 & 0 & M\\end{pmatrix} \\cdot \\mathbf y = \\begin{pmatrix}H(m_1) \\\\ H(m_2) \\\\ c\\end{pmatrix}$ or (by embedding technique): $\\begin{pmatrix} -r_1 & s_1 & 0 & q & 0 & 0 & H(m_1) \\\\ -r_2 & 0 & s_2 & 0 & q & 0 & H(m_2)\\\\", "+0 0 108 3 +226 Let a,b,c be vectors such that $$\\mathbf{a} \\times \\mathbf{b} = \\begin{pmatrix} -1 \\\\- 1 \\\\ -1 \\end{pmatrix}, \\mathbf{a} \\times \\mathbf{c} = \\begin{pmatrix} -1 \\\\ 2 \\\\ 1 \\end{pmatrix}, \\text{ and } \\mathbf{b} \\times \\mathbf{c} = \\begin{pmatrix} 0 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ Then evaluate $$(\\mathbf{b} + \\mathbf{c})\\times \\mathbf{b}, \\mathbf{a}\\times(\\mathbf{b} + 4 \\mathbf{a}), (\\mathbf{a} + \\mathbf{b} + \\mathbf{c})\\times \\mathbf{a}$$ Oct 30, 2020 #2 +31698 +3 For cross products we have the following a x a = 0 (simiarly for b x b etc) a x b = - b x a etc. (b + c) x a = b x+ c x a = -x b - a x c etc. Can you take it from here? Oct 30, 2020 #3 +112499 +1 Thanks Alan, that sure made it easy Oct 31, 2020", "\\\\ y\\end{pmatrix} \\cdot d\\mathbf r + \\hat {\\mathbf y} \\oint \\begin{pmatrix}z \\\\ 0 \\\\ -x\\end{pmatrix} \\cdot d\\mathbf r + \\hat {\\mathbf z} \\oint \\begin{pmatrix}-y \\\\ x \\\\ 0\\end{pmatrix} \\cdot d\\mathbf r$$ Applying the Kelvin-Stokes theorem we get: $$\\oint \\mathbf r \\times d\\mathbf r = \\hat {\\mathbf x} \\iint (\\nabla \\times \\begin{pmatrix}0 \\\\ -z \\\\ y\\end{pmatrix}) \\cdot d\\mathbf S + \\hat {\\mathbf y} \\iint (\\nabla \\times \\begin{pmatrix}z \\\\ 0 \\\\ -x\\end{pmatrix}) \\cdot d\\mathbf S + \\hat {\\mathbf z} \\iint (\\nabla \\times \\begin{pmatrix}-y \\\\ x \\\\ 0\\end{pmatrix}) \\cdot d\\mathbf S$$ $$\\oint \\mathbf r \\times d\\mathbf r = \\hat {\\mathbf x} \\iint 2 \\hat {\\mathbf x} \\cdot d\\mathbf S + \\hat {\\mathbf y} \\iint 2 \\hat {\\mathbf y} \\cdot d\\mathbf S + \\hat {\\mathbf z} \\iint 2 \\hat {\\mathbf z} \\cdot d\\mathbf S$$ $$\\oint \\mathbf r \\times d\\mathbf r = 2 \\iint d\\mathbf S = 2 \\iint dS \\hat {\\mathbf n}$$ However, this is a factor 2 different from the equation you proposed. Last edited: Aug 28, 2011 5. Aug 29, 2011 daudaudaudau Sorry, I was missing a factor of two in the equation i posted, so your proof is correct :) 6. Sep 1, 2011 Anthony Alternatively, for arbitrary fixed vector $\\mathbf{a} \\in \\mathbf{R}^3$ $$\\mathbf{a} \\cdot \\oint \\mathbf{r} \\wedge \\mathrm{d}\\mathbf{r} = \\oint \\mathbf{a} \\cdot (\\mathbf{r} \\wedge \\mathrm{d}\\mathbf{r}) = \\oint (\\mathbf{a}", "with all velocity vectors in sequence. The size is $2N\\times1$ $$\\mathbf{v} = \\begin{pmatrix} \\vec{v}_1 \\\\ \\vec{v}_2 \\\\ \\vdots \\\\ \\vec{v}_N \\end{pmatrix} = \\begin{pmatrix} \\dot{x}_1 \\\\ \\dot{y}_1 \\\\ \\dot{x}_2 \\\\ \\dot{y}_2 \\\\ \\vdots \\\\ \\dot{x}_N \\\\ \\dot{y}_N \\end{pmatrix}$$ 2. Create a (block) matrix with all the masses, but also of $2N$ rows $$\\mathbf{M} = \\begin{bmatrix} m_1 & 0 & 0 & 0 & \\cdots & 0 & 0 \\\\ 0 & m_1 & 0 & 0 & & 0 & 0 \\\\ 0 & 0 & m_2 & 0 & & 0 & 0 \\\\ 0 & 0 & 0 & m_2 & & 0 & 0 \\\\ & & & & \\ddots & & \\\\ 0 & 0 & 0 & 0 & & m_N & 0 \\\\0 & 0 & 0 & 0 & & 0 & m_N \\end{bmatrix}$$ such that $\\mathbf{M} \\mathbf{v}$ would be all the momentum componnets. 3. You are looking for the change in velocity $\\Delta \\mathbf{v}$ due to all the impulses from the collisions. If you knew the sum of all impulses $\\boldsymbol{\\ell}$ (again in $2N\\times1$ vector) then you would write the change in velocity as $$\\Delta \\mathbf{v} = \\mathbf{M}^{-1} \\boldsymbol{\\ell}$$ 4. Create a $2N \\times K$ contact pair block matrix $\\mathbf{Z}$. Each column represents a contact between two bodies ($i \\rightarrow j$)", "3 & 1 & −2 \\\\ −2 & 0 & 5 \\end{pmatrix}$$ $$\\mathbf{F} = \\begin{pmatrix} 3 & 4 \\\\ 5 & 6 \\end{pmatrix}$$ $$\\mathbf{G} = \\begin{pmatrix} 2 & −4 \\\\ 7 & 1 \\\\ −1 & 1 \\end{pmatrix}$$ $$\\mathbf{H} = \\begin{pmatrix} −8 & −1 & −3 \\\\ −4 & 2 & −2 \\\\ −1 & 0 & 0 \\end{pmatrix}$$ Calculate each of the following 1. $$|\\mathbf{a} − (\\mathbf{b} + \\mathbf{3c})|$$ 7. $$(\\mathbf{EH})^T$$ 2. Component of $$\\mathbf{c}$$ along $$\\mathbf{a}$$ 8. $$|\\mathbf{HE}|$$ 3. Angle between $$\\mathbf{c}$$ and $$\\mathbf{d}$$ 9. $$\\mathbf{EHG}$$ 4. $$(\\mathbf{b} \\times \\mathbf{d}) \\cdot \\mathbf{a}$$ 10. $$\\mathbf{EG} − \\mathbf{HG}$$ 5. $$(\\mathbf{b} \\times \\mathbf{d}) \\times \\mathbf{a}$$ 11. $$\\mathbf{EH} − \\mathbf{H}^T \\mathbf{E}^T$$ 6. $$\\mathbf{b}\\times (\\mathbf{d} \\times \\mathbf{a})$$ 12. $$\\mathbf{F}^{−1}$$ 2. For what values of $$a$$ are the vectors $$\\mathbf{A} = 2a\\hat{i} − 2\\hat{j} + a\\hat{k}$$ and $$\\mathbf{B} = a\\hat{i} + 2a\\hat{j}+ 2\\hat{k}$$ perpendicular? 3. Show that the triple scalar product $$(A \\times B) \\cdot C$$ can be written as $(\\mathbf{A} \\times \\mathbf{B}) \\cdot \\mathbf{C} = \\begin{vmatrix} A_1 & A_2 & A_3 \\\\ B_1 & B_2 & B_3 \\\\ C_1 & C_2 & C_3 \\end{vmatrix} \\nonumber$ Show also that the product is unaffected by interchange of the scalar and vector product operations or by change in the order of $$A, B, C$$ as long as they are in cyclic order, that is", "# Tagged: length ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 419 (a) Let $A$ be a real orthogonal $n\\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$. (b) Let $A$ be a real orthogonal $3\\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue. ## Problem 355 Let $\\mathbf{a}, \\mathbf{b}$ be vectors in $\\R^n$. Prove the Cauchy-Schwarz inequality: $|\\mathbf{a}\\cdot \\mathbf{b}|\\leq \\|\\mathbf{a}\\|\\,\\|\\mathbf{b}\\|.$ ## Problem 254 Let $\\mathbf{a}$ and $\\mathbf{b}$ be vectors in $\\R^n$ such that their length are $\\|\\mathbf{a}\\|=\\|\\mathbf{b}\\|=1$ and the inner product $\\mathbf{a}\\cdot \\mathbf{b}=\\mathbf{a}^{\\trans}\\mathbf{b}=-\\frac{1}{2}.$ Then determine the length $\\|\\mathbf{a}-\\mathbf{b}\\|$. (Note that this length is the distance between $\\mathbf{a}$", "the linear transformation $T$ is nonzero. Hence the range is $\\calR(T)=\\Span(\\mathbf{u}),$ and a basis for $\\calR(T)$ is the set $\\{\\mathbf{u}\\}$. The rank of $T$ is the dimension of the range $\\calR(T)$ by definition. Thus, the rank of $T$ is $1$. ### (d) Calculate the matrix $A$ representing $T$ with respect to the standard basis for $\\R^3$. Let $\\mathbf{e}_1=\\begin{bmatrix} 1 \\\\ 0 \\\\ 0 \\end{bmatrix}, \\mathbf{e}_2=\\begin{bmatrix} 0 \\\\ 1 \\\\ 0 \\end{bmatrix}, \\mathbf{e}_3=\\begin{bmatrix} 0 \\\\ 0 \\\\ 1 \\end{bmatrix}$ be the standard basis vectors of $\\R^3$. We calculate \\begin{align*} T(\\mathbf{e}_1)&=\\left(\\, \\frac{\\mathbf{u}\\cdot \\mathbf{e}_1}{\\mathbf{u}\\cdot \\mathbf{u}} \\,\\right)\\mathbf{u} =\\frac{1}{2}\\mathbf{u}=\\begin{bmatrix} 1/2 \\\\ 1/2 \\\\ 0 \\end{bmatrix}\\6pt] T(\\mathbf{e}_2)&=\\left(\\, \\frac{\\mathbf{u}\\cdot \\mathbf{e}_2}{\\mathbf{u}\\cdot \\mathbf{u}} \\,\\right)\\mathbf{u} =\\frac{1}{2}\\mathbf{u}=\\begin{bmatrix} 1/2 \\\\ 1/2 \\\\ 0 \\end{bmatrix}\\\\[6pt] T(\\mathbf{e}_3)&=\\left(\\, \\frac{\\mathbf{u}\\cdot \\mathbf{e}_3}{\\mathbf{u}\\cdot \\mathbf{u}} \\,\\right)\\mathbf{u} =\\frac{0}{2}\\mathbf{u}=\\begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\end{bmatrix}. \\end{align*} Thus, the matrix representation A of T is \\begin{align*} A&=\\begin{bmatrix} T(\\mathbf{e}_1) & T(\\mathbf{e}_2) & T(\\mathbf{e}_3) \\\\ \\end{bmatrix}\\\\ &=\\begin{bmatrix} 1/2 & 1/2 & 0 \\\\ 1/2 &1/2 &0 \\\\ 0 & 0 & 0 \\end{bmatrix}. \\end{align*} ### (e) Calculate the coordinate vector Let \\mathbf{v}=\\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix} and let \\[ [\\mathbf{v}]_B=\\begin{bmatrix} a \\\\ b \\\\ c \\end{bmatrix} be the coordinate vector of $\\mathbf{v}$ with respect to the basis $B$. This means that we have \\begin{align*} \\mathbf{v}=a \\begin{bmatrix} 1 \\\\ 0 \\\\ 0 \\end{bmatrix}+b \\begin{bmatrix} -1 \\\\ 1 \\\\ 0 \\end{bmatrix}+c\\begin{bmatrix} 0" ]

[ "where y appears only once and by itself on the left side, you end up getting a subset which is not trivial (that is, it's neither empty nor is it every pair of reals). In this case, the set of points you get, when plotted, is the shape of a parabola.", "How do you find the point (x,y) on the unit circle that corresponds to the real number t=(5pi)/4? May 21, 2018 M$\\left(- \\frac{\\sqrt{2}}{2} , - \\frac{\\sqrt{2}}{2}\\right)$ Explanation: $t = \\frac{\\pi}{4} + \\pi$. --> t is in Quadrant 3 $x = \\cos t = - \\cos \\left(\\frac{\\pi}{4}\\right) = - \\frac{\\sqrt{2}}{2}$ $y = \\sin t = - \\left(\\sin \\frac{\\pi}{4}\\right) = - \\frac{\\sqrt{2}}{2}$ Point M(x, y) has as coordinates: M$\\left(- \\frac{\\sqrt{2}}{2} , - \\frac{\\sqrt{2}}{2}\\right)$", "of {eq}t {/eq} defines a point {eq}(x,y) {/eq}, which can be plotted as {eq}(f(t),g(t)) {/eq}. The collection of all such points completes the Parametric Curve. (C) Let {eq}x=f(t), \\ y=g(t) {/eq}, so {eq}\\dfrac{dy}{dx}=\\dfrac{\\frac{dy}{dt}}{\\frac{dx}{dt}} {/eq} And, {eq}\\dfrac{d^2y}{dx^2}=\\dfrac{d}{dx}\\left(\\dfrac{dy}{dx}\\right) {/eq}. The curve will be concave upwards when {eq}\\dfrac{d^2y}{dx^2}>0 {/eq} and concave downwards when {eq}\\dfrac{d^2y}{dx^2}<0. {/eq}", "$\\mathrm{plot}\\left(\\mathrm{sin}\\left(x\\right),x=-\\mathrm{π}..\\mathrm{π},\\mathrm{y axis}\\right)$", "the required standard form: $$\\displaystyle x+\\sqrt{3}y-13=0$$ Here is a plot of the hyperbola and the normal line at the given point:", "# How do you plot sqrt46 on the decimal number line as accurately as possible? Aug 11, 2018 Draw a circle with centre $\\left(0 , \\frac{45}{2}\\right)$ with radius $\\frac{47}{2}$. This will pass though the points $\\left(0 , - 1\\right)$ and $\\left(0 , 46\\right)$ on the $y$ axis and the points $\\left(- \\sqrt{46} , 0\\right)$ and $\\left(\\sqrt{46} , 0\\right)$ on the $x$ axis.", "this graph is $\\left(-\\infty , \\, \\frac{1}{2}\\right]$. Hence the domain of the parametric curve is $-\\infty < x \\leq \\frac{1}{2}$. Similar analysis can be applied to the graph of $y = \\frac{(1 - t)^2}{(1 + t)^2}$. It has a horizontal asymptote y = 1, vertical asymptote at t = -1, turning point at $t = 1 \\Rightarrow y = 0$ and y-intercept at y = 1. The range of this graph is $[0, +\\infty)$. Hence the range of the parametric curve is $0 \\leq y < +\\infty$. Hence the path is the part of line 2x + y = 1 over the domain $x \\leq \\frac{1}{2}$.", "did you plot the point (0, $3\\frac{1}{4}$) in #1? What was more difficult about plotting this point compared to the others? • What scale did you choose for the x-axis in #2? The y-axis? • Where did you plot the point (0.25, 7) in #2? What was more difficult about plotting this point compared to the others? • Explain your thought process as you decided how to label the whole numbers along the x- and y-axes in #3. ? Use the grid below to construct a coordinate grid on which you can plot the following points. Then plot and label the points. Point Coordinates A (0, 12) B ($\\frac{1}{2}$, 3) C (4, 0) ?", "starts at $t, x = 0, 1$ (where the units are meters), i.e., on your worldline, but then moves in the $x$ direction slower than you do, so its worldline is a shallower curve than the hyperbola you are following. At some point, the tip will cross the line $x = t$, i.e., the horizon, and at that point the tip can't keep up with you, no matter how hard you pull on it. 3. Jun 19, 2017 ### Cato OK, thanks so much for the detailed and really useful discussion.", "point, 0! It for plotting points to graph a line from a “ T-Table is a and. These two values of x and substitute these values in the table values.", "and $n = +1$, respectively, are A Parabolas and Circles B Circles and Hyperbolas C Hyperbolas and Circles D Hyperbolas and Parabolas Engineering Mathematics Differential Equations Question 2 Explanation: \\begin{aligned} \\frac{d y}{d x} &=-\\left(\\frac{x}{y}\\right)^{n} \\\\ n=-1\\quad\\quad \\frac{d y}{d x} &=-\\frac{y^{\\prime}}{x} \\\\ \\frac{d y}{y} &=-\\frac{d x}{x} \\\\ \\int \\frac{1}{y} d y &=-\\int \\frac{1}{x} d x \\\\ \\ln y &=-\\ln x+\\ln c \\\\ \\ln (y x) &=\\ln c \\end{aligned} $x y=c \\quad$ (Represents rectangular hyporbola) \\begin{aligned} n=1, \\quad \\frac{d y}{d x}&=-\\frac{x}{y} \\\\ y d y &=-x d x \\\\ y d y &=-\\int x d x \\\\ \\frac{y^{2}}{2} &=-\\frac{x^{2}}{2}+c \\end{aligned} $x^{2}+y^{2}=2 c \\quad$ (Represents family of circles) Question 3 Let H(z) be the z-transform of a real-valued discrete-time signal h[n]. If $P(z)=H(z)H\\left (\\frac{1}{z} \\right )$ has a zero at $z=\\frac{1}{2}+\\frac{1}{2}j$, and P(z) has a total of four zeros, which one of the following plots represents all the zeros correctly? A A B B C C D D Signals and Systems Z-Transform Question 3 Explanation: $P(Z)=H(Z)H\\left ( \\frac{1}{Z} \\right )$ (i) $h(n)$ is real. Som $p(n)$ will be also real (ii) $P(z)=P(z^{-1})$ From (i) : if $z_1$ is a zero of $P(z)$, then $z_1^*$ will be also a zero of $P(z)$. From (ii): If $z_1$ is a zero of $P(z)$, then $\\frac{1}{z_1}$ will be also a zero of $P(z)$. So, the", "# SPM Practice 2 - Question 4 Question 4 The diagram shows part of the straight line graph obtained by plotting $\\frac{y}{x}$ against $\\sqrt{x}$ . Given its original non-linear equation is $y=px+q{x}^{\\frac{3}{2}}$ . Calculate the values of p and q.", "= 2 Hover the mousse cursor on the graph or plotted point to read the coordinates. Click the button above \"Plot Equation\". You may hover the mousse cursor over the graph to trace the coordinates. You may also Hover the mousse cursor on the top right of the graph to have the option to download the graph as a png file. 1 - Select a point M on the graph of the hyperbola, hover the mousse over it and read the coordinates. Read the coordinates of F1 and F2 (right legend) or hover the mousse and read their coordinates and show that the difference of the distances $\\overline{MF_1} - \\overline{MF_2}$ is close to $2a$ if $\\overline{MF_1} \\gt \\overline{MF_2}$ or $- 2a$ if $\\overline{MF_1} \\lt \\overline{MF_2}$. You may do the above activity for different values of $a, b, h$ and $k$ with as many points as needed to better understand the definition of an hyperbola. 2 - Use the values of $a$ and $b$ to find $c$ using the relationship between $a, b$ and $c$ given by $c^2 = a^2 + b^2$. Use the above results to find the coordinates of $F_1, F_2, V_1$ and $V_2$ and check the results graphically. 3 - Set $a, b, h$ and $k$ to some values. Find the x and y intercepts and", "the equation $$\\left|\\frac{x}{a}\\right|^p + \\left|\\frac{y}{b}\\right|^p = 1$$ For $n=2$, this is the equation of an ordinary ellipse and for $a=b=1$ that ellipse is a circle. The corresponding curve for $p=4$ is sometimes called a squircle.", "# Graphs of Quadratic Expressions - Theory Consider the quadratic expression $$Q\\left( x \\right)\\,\\,:\\,\\,{x^2}$$. As the value of $$x$$ will vary, the value of $$Q$$ will vary. Below, the value of $$Q$$ has been calculated for some values of $$x$$: $x =$ $0$ $1$ $- 1$ $2$ $- 2$ $\\frac{1}{2}$ $- \\frac{1}{2}$ $Q\\left( x \\right) =$ $0$ $1$ $1$ $4$ $4$ $\\frac{1}{4}$ $\\frac{1}{4}$ A very powerful way to visualize the variation of an expression’s value with the variable’s value is through graphing. For example, in the example above, we can plot $$x$$ on the horizontal scale and $$Q\\left( x \\right)\\,\\,:\\,\\,{x^2}$$ on the vertical scale. Let us plot the seven points we have obtained above: We plotted only seven points, but theoretically, we can calculate the value of $$Q$$ for every (real) value of $$x$$. If we were to plot all those points, we would obtain a continuous curve: This curve is an example of a parabola, which opens upward. Note that the minimum value of the expression is obtained when $$x = 0$$: ${Q_{\\min }} = Q\\left( {x = 0} \\right) = {0^2} = 0$ In the statement above, $${Q_{\\min }}$$ stands for the minimum value of $$Q$$, while $$Q\\left( {x = 0} \\right)$$ is to be read as “the value of $$Q$$ when $$x = 0$$” Next,", "$g(x) = - ax^2 - \\frac{\\Delta}{4a},$ we can plot $y = g(x)$ in a Cartesian plane; it is a parabola with a vertex at $\\left(0, - \\frac{\\Delta}{4a}\\right),$ oriented in an opposite direction to the parabola we got for real values of $z.$ So imagine that we want to plot $y = f(z)$ over the complex plane, but only for the values of $f(z)$ that are purely real. We can visualize this in just three dimensions: two dimensions to plot the complex input value, $z,$ and one dimension to plot the real output value $y.$ If we write $z = u + iv$ we can describe this three-dimensional space by the coordinates $(u,v,y).$ The resulting plot consists of just two parabolas: the parabola for $y = f(x)$ where $x$ is real, presented as the plot $y = f(u)$ in the $u,y$ plane; and the parabola for $y = g(x),$ translated $p$ units in the $u$ direction and rotated $90$ degrees around the line $u = p,$ $v=0$ (a line through $(p,0,0)$ parallel to the $y$ axis) so that it lies in the plane $u = p.$ That is, the parabolas have the same vertex, $\\left(p, 0, - \\frac{\\Delta}{4a}\\right),$ and the same axis, the line $u=p,$ $v=0,$ but they are in perpendicular planes. In another answer to this question, this", "The families of curves represented by the solution of the equation $$\\frac{dy}{dx}=\\: – \\left(\\frac{x}{y} \\right)^n$$ for $n=-1$ and $n= +1,$ respectively, are 1. Parabolas and Circles 2. Circles and Hyperbolas 3. Hyperbolas and Circles 4. Hyperbolas and Parabolas in Others edited", "# Should I plot them? 2 Calculus Level 3 A straight line is to be drawn through a point whose coordinates are $$(x,y)=(2,-1)$$ as tangent to the curve $$y=x^2-5x+6$$. Let the coordinates of the points where the line is tangent to the curve be $$(x_1,y_1)$$ and $$(x_2,y_2)$$. Find $$x_1+y_1+x_2+y_2$$. ×", "NEW New Website Launch Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc... 1 ### AIEEE 2007 MCQ (Single Correct Answer) The normal to a curve at $$P(x,y)$$ meets the $$x$$-axis at $$G$$. If the distance of $$G$$ from the origin is twice the abscissa of $$P$$, then the curve is a A circle B hyperbola C ellipse D parabola ## Explanation Equation of normal at $$P\\left( {x,y} \\right)$$ is $$Y - y = - {{dx} \\over {dy}}\\left( {x - x} \\right)$$ Coordinate of $$G$$ at $$X$$ axis is $$\\left( {X,0} \\right)$$ (let) $$\\therefore$$ $$0 - y = - {{dx} \\over {dy}}\\left( {X - x} \\right) \\Rightarrow y{{dy} \\over {dx}} = X - x$$ $$\\Rightarrow X = x + y{{dy} \\over {dx}}$$ $$\\therefore$$ Co-ordinate of $$G\\left( {x + y{{dy} \\over {dx}},0} \\right)$$ Given distance of $$G$$ from origin $$=$$ twice of the abscissa of $$P.$$ as distance cannot be $$-ve,$$ therefore abscissa $$x$$ should be $$+ve$$ $$\\therefore$$ $$x + y{{dy} \\over {dx}} = 2x \\Rightarrow y{{dy} \\over {dx}} = x \\Rightarrow ydx = xdx$$ On Integrating $$\\Rightarrow {{{y^2}} \\over 2} = {{{x^2}} \\over 2} + {c_1} \\Rightarrow {x^2} - {y^2} = - 2{c_1}$$ $$\\therefore$$ the curve is a hyperbola 2 ### AIEEE 2007 MCQ (Single Correct", "the normals drawn at three different points on the parabola $${y^2}... Let$$y = {e^{x\\,\\sin \\,{x^3}}} + {\\left( {\\tan x} \\right)^x}$$. Find$${{dy} \\o... Let the angles $$A, B, C$$ of a triangle $$ABC$$ be in A.P. and let $$b:c = \\sqr... Let$$a, b, c$$be positive real numbers Let <br>$$\\theta = {\\tan ^{ - 1}}\\sqr... Find the value of : $$\\cos \\left( {2{{\\cos }^{ - 1}}x + {{\\sin }^{ - 1}}x} \\righ... For all$$x$$in$$\\left[ {0,1} \\right]$$, let the second derivative$$f''(x)$$... Let$$x$$and$$y$$be two real variables such that$$x&gt;0$$and$$xy=1$$. Fin... Use the function$$f\\left( x \\right) = {x^{1/x}},x &gt; 0$$. to determine the bi... Evaluate$$\\int {\\left( {{e^{\\log x}} + \\sin x} \\right)\\cos x\\,\\,dx.} $$The value of the definite integral$$\\int\\limits_0^1 {\\left( {1 + {e^{ - {x^2}}}... Let $$a, b, c$$ be non-zero real numbers such that <br>$$\\int\\limits_0^1 {\\left... Find the area bounded by the curve$${x^2} = 4y$$and the straight Show that :$$\\mathop {\\lim }\\limits_{n \\to \\infty } \\left( {{1 \\over {n + 1}} +... For a biased die the probabilities for the different faces to turn up are given ... An anti-aircraft gun can take a maximum of four shots at an enemy plane moving a... Let $$\\overrightarrow A ,\\overrightarrow B ,\\overrightarrow C$$ be vectors of l... Let $$\\overrightarrow A ,\\overrightarrow B$$ and $${\\overrightarrow C }$$ be un... The scalar \\overrightarrow A .\\left( {\\overrightarrow B" ]

[ "actual and predicted values? 3. Based on the previous problem, what does it mean if the sum of the squares of the differences between the actual and predicted values is $0$? 4. Find the equation of the least squares line for the following data set: $(0, -3), (1, -2), (4, -1), (5, 4)$. 5. Find the least square line equation for the following data set and use it to predict $x=10$. 6. $(-1, -4), (0, 3), (1, 4), (2, 6)$. ### Answer Key 1. $3$ 2. $4$ 3. If the sum of the squares of the differences is $0$, this means that the difference between the actual and predicted values is $0$ for all $x$ values. Therefore, all of the data points lie on a line. However, this does not mean that all of the points would continue to fall exactly on that line if more points were collected. 4. $y=\\frac{19}{17}x-\\frac{56}{17}$ 5. The equation is $y=3.1x+0.7$, which predicts $y=31.7$ when $x=10$. Images/mathematical drawings are created with GeoGebra. Rate this page", "0.5x – 4? One way to do so is to plot the points contained in the table on the graph, and test each point to see if it lies on the line. Use a plotting applet to graph both the line and the points in the table. This tool will allow you to confirm whether or not the points in the table lie on the graph of the line. Use the graph and table provided to determine which points lie on the graph of the line. Use the interactive below to identify which points from the table lie on the graph of the line: y = 0.5x – 4 For each ordered pair, place it in the Lies on Line box if the point lies on the line, or in the Does NOT Lie on Line box if the point does not lie on the line. Practice Use the graphing applet to determine which of the following tables contain only points that lie on the graph of the line $y=\\frac{1}{3}x+3.$ Table A x y -9 0 -3 2 0 3 2 $3\\frac{2}{3}$ 6 5 Table B x y -9 0 -3 3 0 3 4 $4\\frac{1}{3}$ 6 5 Table C x y -9 0 -3 2 1 $3\\frac{1}{3}$ 3 4 9 6 Table D x y 0 -9", "given points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are $\\frac{x_2-x_1}{r},\\frac{y_2-y_1}{r},\\frac{z_2-z_1}{r}$ Here, the quantities $x_2-x_1$, $y_2-y_1$ and $z_2-z_1$ are the direction ratios of the given line.", "x_1}(x - x_1)$$ $$\\implies \\frac{y - y_1}{y_2 - y_1} = \\frac{x - x_1}{x_2 - x_1} = t \\text{ (let) }$$ $$\\implies y = y_1 + t(y_2 - y_1) \\text{ and } \\implies x = x_1 + t(x_2 - x_1)$$ - I came here from the Desmos graphing calculator, trying to find how they created a \"line segment\" as a parametric plot. They used exactly what the poster asked and what was shown in the excellent response. desmos.com/calculator/vtvpir0ylf – Tom De Vries Dec 29 '15 at 15:04", "\\frac{1}{2}(1+c) - \\frac{1}{2}(1+c)\\cos(\\sqrt{2}t) +\\frac{1}{2\\sqrt{2}}\\sin(\\sqrt{2}t) - \\frac{1}{2}t. Here is a plot of \\sum_{c =0}^\\infty (1 + c - t)(u_c(t) - u_{c+1}(t)): html(\"Here is a plot of $y$:\") plot(v, 0, 20) Here is a plot of y: Here is a plot of y:", "$$[-1,1]$$ is the same as choosing one point uniformly in the box $$[-1,1]\\times[-1,1]\\subset\\mathbb{R}^2$$. Precisely $$\\frac18$$ of this square is above the line $$x+y=1$$.", "###### Example4.5.13 Plot $$y=-\\frac{2}{3}x+10$$ and $$y=3x+5\\text{.}$$ These plots follow the approach form the previous example, but there is no context to the equation. in-context", "pointing to $0$.To do this, find the extreme point of $f$ on the mapping diagram, $x = -\\frac B {2A} = 0$. Move symmetrically above and below that value by $\\pm 2$, which are the desired values for $x$.", "of printing change the direction. (iii) y = ($$\\frac{3}{2}$$)x + 3 Solution: When x = -2; y = $$\\frac{3}{2}$$(-2) + 3 y = -3 + 3 = 0 when x = 0; y = $$\\frac{3}{2}$$(0) + 3 y = 3 when x = 2; y = $$\\frac{3}{2}$$(2) + 3 y = 3 + 3 = 6 Plot the points (-2, 0) (0, 3) and (2, 6) in the graph sheet we get a straight line. (iv) 3x + 2y = 14 Solution: y = $$\\frac{-3x+14}{2}$$ y = – $$\\frac{3}{2}$$x + 7 when x = -2 y = –$$\\frac{3}{2}$$(-2) + 7 = 10 when x = 0 y = –$$\\frac{3}{2}$$(0) + 7 = 7 when x = 2 y = –$$\\frac{3}{2}$$(2) + 7 = 4 Plot the points (-2, 10) (0, 7) and (2, 4) in the graph sheet we get a straight line. Question 2. Solve graphically (i) x + y = 7, x – y = 3 Solution: x + y = 7 y = 7 – x Plot the points (-2, 9), (0, 7) and (3, 4) in the graph sheet x – y = 3 -y = -x + 3 y = x – 3 Plot the points (-2, -5), (0, -3) and (4, 1) in the same graph sheet. The point of intersection is", "# SPM Practice 2 - Question 4 Question 4 The diagram shows part of the straight line graph obtained by plotting $\\frac{y}{x}$ against $\\sqrt{x}$ . Given its original non-linear equation is $y=px+q{x}^{\\frac{3}{2}}$ . Calculate the values of p and q.", "$$2y = 1 - x$$. Could you please elaborate on why $$\\frac{y - 0}{\\frac{1}{2} - 0} = \\frac{x - 1}{0 - 1}$$ or $$2y = 1 - x$$? $$\\frac{y - 0}{\\frac{1}{2} - 0} = \\frac{x - 1}{0 - 1}$$; $$\\frac{y }{\\frac{1}{2}} = \\frac{x - 1}{- 1}$$; $$2y=-x+1$$; $$2y = 1-x$$. _________________ Current Student Joined: 08 Feb 2017 Posts: 80 Concentration: Entrepreneurship, Marketing GMAT 1: 660 Q48 V33 GMAT 2: 710 Q49 V38 GPA: 3.32 ### Show Tags 14 May 2017, 05:59 Bunuel wrote: Evgart wrote: Bunuel wrote: Official Solution: For every point $$(a, b)$$ lying on line 1, point $$(b, -a)$$ lies on line 2. If the equation of line 1 is $$y = 2x + 1$$, what is the equation of line 2 ? A. $$y = \\frac{1}{2} + \\frac{x}{2}$$ B. $$2y = 1 - x$$ C. $$\\frac{x + y}{2} = -1$$ D. $$y = \\frac{x}{2} - 1$$ E. $$x = 2y + 1$$ Find two points on line 2 and use their coordinates to build the line's equation. Points $$(0, 1)$$ and $$(-\\frac{1}{2}, 0)$$ on line 1 correspond to points $$(1, 0)$$ and $$(0, \\frac{1}{2})$$ on line 2. The equation of line 2 is $$\\frac{y - 0}{\\frac{1}{2} - 0} = \\frac{x - 1}{0 - 1}$$ or $$2y = 1 - x$$. Could you please elaborate on", "the point $$(1,2,3)$$ can be written as $$\\frac{x-1}{0} = \\frac{y-2}{0} = \\frac{z-3}{2} = t$$, where $$t \\in R$$ is any parameter. The parametric form the line (especially for x,y coordinates) can be obtained by \"cross multiplying\" their respective DR with the parameter equation to give $$(x,y,z)=(1,2,2t+3)$$.", "## Elementary Technical Mathematics The point $T(-4\\frac{1}{2},6\\frac{1}{2})=(-\\frac{9}{2},\\frac{13}{2})$ is plotted as shown in the figure below.", "the estimated value for $$y$$ if $$x = 4$$. • Use your equation to calculate the estimated value for $$x$$ if $$y = 6$$. $$x$$ $$\\text{1,0}$$ $$\\text{2,4}$$ $$\\text{3,1}$$ $$\\text{4,9}$$ $$\\text{5,6}$$ $$\\text{6,2}$$ $$y$$ $$\\text{2,5}$$ $$\\text{2,8}$$ $$\\text{3,0}$$ $$\\text{4,8}$$ $$\\text{5,1}$$ $$\\text{5,3}$$ ### Draw the graph 1. Choose a suitable scale for the axes. 2. Draw the axes. 3. Plot the points. ### Drawing the line of best fit The next step is to draw a straight line which goes as close to as many points as possible. It is generally best to have as many points above the line as below the line. ### Calculating the equation of the line The equation of the line is $$y=mx+c$$ From the graph we have drawn, we estimate the y-intercept to be $$\\text{1,5}$$. We estimate that $$y=\\text{3,5}$$ when $$x=3$$. So we have that points $$\\left(3;\\text{3,5}\\right)$$ and $$\\left(0;\\text{1,5}\\right)$$ lie on the line. The gradient of the line, m, is given by \\begin{align*} m & = \\frac{\\Delta y}{\\Delta x} = \\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\\\ & = \\frac{\\text{3,5}-\\text{1,5}}{3-0} \\\\ & = \\frac{2}{3} \\end{align*} So we finally have that the equation of the line of best fit is $$y=\\frac{2}{3}x+\\text{1,5}$$ ### Calculate the unknown values The equation of the line is $$y=\\frac{2}{3}x+\\text{1,5}$$ so in order to find the unknown values, we insert the known values into our equation. For $$x =", "# How do you graph 3x = 2y by plotting points? Sep 1, 2017 Plug in values of x and see what their y value is #### Explanation: Draw up a table of $x$ and y values, then plot on a chart. I'd manipulate the equation to $y = \\frac{3}{2} x$ because the arithmetic is easier. $x$ | $y$ $1 | \\frac{3}{2} \\cdot 1 = \\frac{3}{2}$ when $x = 1 , y = \\frac{3}{2}$ $2 | \\frac{3}{2} \\cdot 2 = \\frac{6}{2} = 3$ $3 | \\frac{3}{2} \\cdot 3 = \\frac{9}{2} = 4.5$ $4 | \\frac{3}{2} \\cdot 4 = 6$ $5 | \\frac{3}{2} \\cdot 5 = 7.5$ $6 | \\frac{3}{2} \\cdot 6 = 9$ By now you should be able to spot that the difference between the y values is always $+ 1.5$, or $+ \\frac{3}{2}$ This means that the gradient (slope) of the line is $\\frac{3}{2}$. graph{3x=2y [-10, 10, -5, 5]}", "& y=\\frac{x+4}{2} \\\\ & y=\\frac{0+4}{2} \\\\ & =\\frac{4}{2} \\\\ & =2 \\end{align} To find the value of the x-intercept, substitute $y=0$ as given below: \\begin{align} & y=\\frac{x+4}{2} \\\\ & 0=\\frac{x+4}{2} \\\\ & x+4=0 \\\\ & x=-4 \\end{align} Therefore, plot the intercepts $\\left( 0,2 \\right)\\text{ and }\\left( -4,0 \\right)$ and draw a line passing through these points to get the graph of the equation ${{f}^{-1}}=\\frac{x+4}{2}$.", "plotting $s$ vs $t^2$ should give you a straight line of the form $y=mx$ going through the origin. The slope is $m=\\frac12 a$. Possibly there is uncertainty in your data points, so that they are spread randomly around the line $y=mx$. More likely is that $s \\ne 0$ at $t=0$. Then the relationship between $s$ and $t^2$ is more complicated. The easiest option is probably to write the equation as $s=\\frac12 a(t^2+2\\frac{u}{a}t+\\frac{u^2}{a^2})-\\frac{u^2}{2a}=\\frac12 a(t+\\frac{u}{a})^2-\\frac{u^2}{2a}$ which is of the straight-line form $y=mx+c$ if $x=(t+\\frac{u}{a})^2$. To obtain a straight line you need to add or subtract an amount $t_0$ to your values of $t$ and adjust this value until the $R^2$ value of your trend line is as close as possible to 1. This is quite easy to do in a spreadsheet. The amount $t_0$ which maximises $R^2$ gives you a value for $t_0=\\frac{u}{a}$. As a check, the y-intercept on your graph should be $c=-\\frac{u^2}{2a}=-mt_0^2$ where $m=\\frac12 a$ is the slope of your graph.", "of time. To get a straight line in a plot, one would have to take $$d = \\frac{1}{2}at^2$$ and take the log or ln of the expression to get $$ln d = ln\\frac{a}{2} + 2 ln t$$ and this assumes a is constant. x = ?, and one wants to find dy/dx = d(t1/5)/d(?) to get a straight line $\\frac{d^2y}{dx^2}$ = 0 3. Oct 12, 2005 ### Gokul43201 Staff Emeritus You've got the answer right there : call $t^{1/5} = y$. Then rearranging the equation gives $$y = d^{1/10}/(a/2)^{1/10} = k \\cdot d^{1/10}$$ So, if you make the x-axis contain values of $d^{1/10}$ you are done. You will then have a line of the form $y = kx$, where the slope of the line, k, is ... ? Last edited: Oct 12, 2005", "form a line. Figure 7 Construct a table and graph the equation by plotting points: $y=\\frac{1}{2}x+2$. $x$ $y=\\frac{1}{2}x+2$ $\\left(x,y\\right)$ $-2$ $y=\\frac{1}{2}\\left(-2\\right)+2=1$ $\\left(-2,1\\right)$ $-1$ $y=\\frac{1}{2}\\left(-1\\right)+2=\\frac{3}{2}$ $\\left(-1,\\frac{3}{2}\\right)$ $0$ $y=\\frac{1}{2}\\left(0\\right)+2=2$ $\\left(0,2\\right)$ $1$ $y=\\frac{1}{2}\\left(1\\right)+2=\\frac{5}{2}$ $\\left(1,\\frac{5}{2}\\right)$ $2$ $y=\\frac{1}{2}\\left(2\\right)+2=3$ $\\left(2,3\\right)$ Figure 8", "is small, it is not possible to show $(0,1000000)$. If the scale is large, then $(1,1)$ and $(1000,0)$ will look like the same point. ## Solution 1. The points are plotted and labeled below: Not all coordinates on the $y$-axis have been labeled as this would make the picture too crowded. Notice, too, that the the point $\\left(0,2\\frac{1}{2}\\right)$ is halfway between $(0,2)$ and $(0,3)$. 2. The points are plotted and labeled below: This time only the first few squares are marked and each square is 10 units because the numbers that are being plotted are multiples of 10 (except for -35 which is halfway between -30 and -40). The point (0,0) has been highlighted as it is a special point in the coordinate system, the intersection of the two axes. 3. The points are plotted and labeled below: Each square here has been given a value of $\\frac{1}{4}$ of a unit because the coordinates of our points are multiples of $\\frac{1}{4}$. Notice that $1 = \\frac{4}{4}$ and $\\frac{1}{2} = \\frac{2}{4}$. 1. When we choose a scale for the $x$ and $y$ axes, we want to choose a value which will make it as easy as possible to plot the points. For the first set of points, the coordinate values were mostly small whole numbers making it natural to" ]

[ "and $n = +1$, respectively, are A Parabolas and Circles B Circles and Hyperbolas C Hyperbolas and Circles D Hyperbolas and Parabolas Engineering Mathematics Differential Equations Question 2 Explanation: \\begin{aligned} \\frac{d y}{d x} &=-\\left(\\frac{x}{y}\\right)^{n} \\\\ n=-1\\quad\\quad \\frac{d y}{d x} &=-\\frac{y^{\\prime}}{x} \\\\ \\frac{d y}{y} &=-\\frac{d x}{x} \\\\ \\int \\frac{1}{y} d y &=-\\int \\frac{1}{x} d x \\\\ \\ln y &=-\\ln x+\\ln c \\\\ \\ln (y x) &=\\ln c \\end{aligned} $x y=c \\quad$ (Represents rectangular hyporbola) \\begin{aligned} n=1, \\quad \\frac{d y}{d x}&=-\\frac{x}{y} \\\\ y d y &=-x d x \\\\ y d y &=-\\int x d x \\\\ \\frac{y^{2}}{2} &=-\\frac{x^{2}}{2}+c \\end{aligned} $x^{2}+y^{2}=2 c \\quad$ (Represents family of circles) Question 3 Let H(z) be the z-transform of a real-valued discrete-time signal h[n]. If $P(z)=H(z)H\\left (\\frac{1}{z} \\right )$ has a zero at $z=\\frac{1}{2}+\\frac{1}{2}j$, and P(z) has a total of four zeros, which one of the following plots represents all the zeros correctly? A A B B C C D D Signals and Systems Z-Transform Question 3 Explanation: $P(Z)=H(Z)H\\left ( \\frac{1}{Z} \\right )$ (i) $h(n)$ is real. Som $p(n)$ will be also real (ii) $P(z)=P(z^{-1})$ From (i) : if $z_1$ is a zero of $P(z)$, then $z_1^*$ will be also a zero of $P(z)$. From (ii): If $z_1$ is a zero of $P(z)$, then $\\frac{1}{z_1}$ will be also a zero of $P(z)$. So, the", "prove? that $~\\frac{d}{dx}y(x) \\cdot \\frac{d}{dt}x(t) = \\frac{d}{dt}y[x(t)]$ Is it what the question asked? in this example: $$y=\\sqrt{(x-2)(x+2)}~\\rightarrow~y'(x)=...$$ And i have to prove it equals dy/dx I leave this question since i don't want to tire you and me, i think i won't get further. Thank you fresh_42 13. Aug 10, 2017 ### Ray Vickson No: there are two solutions $$y(x) = \\pm \\sqrt{(x-2)(x+2)}$$ This really is true: if you plot the points $(x(t),y(t))$ for $t> 0$ you will find they lie on the curve $y = -\\sqrt{(x-2)(x+2)}$ for $0 < t \\leq 1$ and on the curve $y = + \\sqrt{(x-2)(x+2)}$ for $1 \\leq t < \\infty$. This is one branch of the hyperbola $x^2 - y^2 = 4$. For $t < 0$ you get the opposite branch of the hyperbola, opening out to the left of $x = -2$. 14. Aug 10, 2017 ### Karol Thank you Ray, you all are specialists, they drew the plot in the book and it's as you've said", "$$it\\to-1-\\frac it$$ on its diameter. That is, $$C\\!\\left(\\left(-1,-\\frac1{2t}\\right),\\frac1{2|t|}\\right)$$ These are the circles tangent to the real axis at $$(-1,0)$$", ".$is A) ${{w}_{1}}$ B) ${{w}_{2}}$ C) $\\frac{-({{w}_{2}}-{{w}_{1}})}{5Rt}$ D) $\\frac{-n({{w}_{2}}-{{w}_{1}})}{5Rt}$ • question_answer126) $\\frac{-n({{w}_{2}}-{{w}_{1}})}{Rnt}$equals A) $\\frac{-n({{w}_{2}}-{{w}_{1}})}{Rt}$ B) $({{\\mu }_{s}}=1.4)$ C) $\\frac{2}{\\sqrt{3}}\\left( \\frac{\\tau }{Bi} \\right)$ D) None of the above • question_answer127) The centre of the circle passing through (0,0) and (1,0) and touching the circle $2{{\\left( \\frac{\\tau }{\\sqrt{3}Bi} \\right)}^{1/2}}$is A) $\\frac{2}{\\sqrt{3}}{{\\left( \\frac{\\tau }{Bi} \\right)}^{1/2}}$ B) $\\frac{1}{\\sqrt{3}}\\frac{\\tau }{Bi}$ C) ${{m}_{1}}$ D) ${{m}_{2}}$ • question_answer128) If the line x -1 = 0 is the directrix of the parabola ${{m}_{1}}$ then one of the value of k is A) ${{m}_{2}}$ B) 8 C) 4 D) ${{m}_{1}}$ • question_answer129) The area of the region bounded by the parabola $d\\sqrt{\\frac{{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}}$ the tangent to the parabola at the point (2,3J and the X-axis, is A) 3 B) 6 C) 9 D) 12 • question_answer130) Solution of the differential equation $d\\sqrt{\\frac{{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}}$+?is A) $d\\sqrt{\\frac{2{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}}}$ B) $d\\sqrt{\\frac{2{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}}$ C) $=20m{{s}^{-1}}$ D) $(\\text{take}\\,g=10\\text{ }m{{s}^{-}}^{2})$ • question_answer131) Numbers 1,2,3,...,100 are written down on each of the cards A, B and C. One number is selected at random from each of the cards. The probability that the numbers so selected can be the measures (in cm) of three sides of a right angled triangle, is A) $I={{I}_{0}}\\sin \\omega t,$ B) ${{I}_{0}}=10$ C) $\\omega =100\\pi \\,\\text{rad/s}$ D) None of these • question_answer132) If $\\pi$ is a continuous at $\\pi$then k is equal to A)", "C) $x=c$ D) $x$has any value • question_answer146) ${{12}^{o}}\\sin {{48}^{o}}\\sin {{54}^{o}}$is equal to: A) 1/16 B) 1/32 C) 1/8 D) 1/4 • question_answer147) If$\\tan A=\\frac{1-\\cos B}{\\sin B}.$then: A) $\\tan 2A=\\tan B$ B) $\\tan 2A=ta{{n}^{2}}B$ C) $\\tan 2A={{\\tan }^{2}}B+2\\tan B$ D) none of the above • question_answer148) Three points are$A(6,3),B(-3,5),C(4,-2)$and $P(x,y)$is any point, then the ratio of area of $\\Delta PBC$and $\\Delta ABC$is: A) $\\frac{x+y-2}{7}$ B) $\\frac{x-y+2}{2}$ C) $\\frac{x-y-2}{7}$ D) none of these • question_answer149) The circle ${{x}^{2}}+{{y}^{2}}+4x-4y+4=0$t touches: A) $x-$axis B) $y-$axis C) $x-$ axis and $y-$axis D) none of the above • question_answer150) Three normals to the parabola ${{y}^{2}}=x$are drawn a point (c, 0), then: A) $c=\\frac{1}{4}$ B) $c=\\frac{1}{2}$ C) $c>\\frac{1}{2}$ D) None of these", "Since $$x = a \\cos t, y = a \\sin t$$, $$x^2 + y^2 = a^2(\\sin^2 t + \\cos^2 t) = a^2$$. So this is a circle of radius $$a$$. The circle starts at $$(a, 0)$$, and travels counterclockwise until it reaches the starting point again. Given a value of $$t$$ for $$\\vec{r}(t) = (f(t), g(t))$$ and $$y = f(x)$$, we can plot it in an inverval of $$t \\in [a, b]$$ by plotting for $$x \\in [f(a), f(b)]$$. The parameterization of a curve is not unique, just like normal curves. Every parametric curve has infinite different ways of being represented mathematically, but when drawn on a plane each curve has only one shape. For example, consider $$\\vec{r}(t) = (a \\cos t, b \\sin t)$$. This is an ellipse where $$a > b$$ stretches it horiontally and $$b > a$$ stretches it vertically. We can rewrite it as $$\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$$ to plot it on a Cartesian plane. We plot parametric curves by solving for $$t$$ in $$x = x(t)$$, so $$t = x^{-1}(x)$$, then subsituting into $$y(t)$$ to get $$y = y(x^{-1}(x))$$, which is a non-parametric function we can plot more easily. Alternatively, we would have to plot both $$y = y(t)$$ and $$x = x(t)$$, and use these graphs to plot the points on", "\\frac{\\sqrt{1 + y_t^2}}{y_t} \\int_0^{y_t} 2 y dy \\\\ = y_t \\sqrt{1 + y_t^2} + \\ln \\left( \\sqrt{1 + y_t^2} + y_t \\right) - y_t \\sqrt{1 + y_t^2} \\\\ = \\ln \\left( X_{-1}(y_t) + y_t \\right)$ The unit circle. $A_{1}(y_t) = 2 \\int_0^{y_t} X_{1}(y) - \\frac{X_{1}(y_t)}{y_t} y \\, dy \\\\ = 2 \\int_0^{y_t} \\sqrt{1 - y^2} dy - \\frac{\\sqrt{1 - y_t^2}}{y_t} \\int_0^{y_t} 2 y dy \\\\ = y_t \\sqrt{1 - y_t^2} - i \\ln \\left( \\sqrt{1 - y_t^2} + i y_t \\right) - y_t \\sqrt{1 - y_t^2} \\\\ = -i \\ln \\left( X_{1}(y_t) + i y_t \\right)$ Now if we parameterize the original curves in terms of the areas swept out, we have: The vertical line: $t = A_{0}(y) = y \\Rightarrow x(t) = 1, y(t) = t$ The unit hyperbola: $\\rho = A_{-1}(y) = \\ln( X(y) + y ) = \\ln ( x + y ) \\\\ \\Rightarrow x(\\rho) = \\frac{1}{2} \\left(e^{\\rho} + e^{-\\rho} \\right), y(\\rho) = \\frac{1}{2} \\left(e^{\\rho} - e^{-\\rho} \\right)$ The unit circle: $\\theta = A_{1}(y) = -i \\ln( X(y) + i y ) = \\frac{1}{i} \\ln ( x + i y ) \\\\ \\Rightarrow x(\\theta) = \\frac{1}{2} \\left(e^{i \\theta} + e^{-i \\theta} \\right), y(\\theta) = \\frac{1}{2 i} \\left(e^{i \\theta} - e^{-i \\theta} \\right)$ So that's the motivation behind: $\\cosh \\rho = \\frac{1}{2} \\left(e^{\\rho} + e^{-\\rho}", "Since x = a \\cos t, y = a \\sin t, x^2 + y^2 = a^2(\\sin^2 t + \\cos^2 t) = a^2. So this is a circle of radius a. The circle starts at (a, 0), and travels counterclockwise until it reaches the starting point again. Given a value of t for \\vec{r}(t) = (f(t), g(t)) and y = f(x), we can plot it in an inverval of t \\in [a, b] by plotting for x \\in [f(a), f(b)]. The parameterization of a curve is not unique, just like normal curves. Every parametric curve has infinite different ways of being represented mathematically, but when drawn on a plane each curve has only one shape. For example, consider \\vec{r}(t) = (a \\cos t, b \\sin t). This is an ellipse where a > b stretches it horiontally and b > a stretches it vertically. We can rewrite it as \\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1 to plot it on a Cartesian plane. We plot parametric curves by solving for t in x = x(t), so t = x^{-1}(x), then subsituting into y(t) to get y = y(x^{-1}(x)), which is a non-parametric function we can plot more easily. Alternatively, we would have to plot both y = y(t) and x = x(t), and use these graphs to plot the points on", "of {eq}t {/eq} defines a point {eq}(x,y) {/eq}, which can be plotted as {eq}(f(t),g(t)) {/eq}. The collection of all such points completes the Parametric Curve. (C) Let {eq}x=f(t), \\ y=g(t) {/eq}, so {eq}\\dfrac{dy}{dx}=\\dfrac{\\frac{dy}{dt}}{\\frac{dx}{dt}} {/eq} And, {eq}\\dfrac{d^2y}{dx^2}=\\dfrac{d}{dx}\\left(\\dfrac{dy}{dx}\\right) {/eq}. The curve will be concave upwards when {eq}\\dfrac{d^2y}{dx^2}>0 {/eq} and concave downwards when {eq}\\dfrac{d^2y}{dx^2}<0. {/eq}", "the interval. Since ${A}{\\left({0}\\right)}={0}$, ${A}{\\left({500}\\right)}={500000}$ and ${A}{\\left({1000}\\right)}={0}$ then Closed Interval Method gives the maximum value as ${A}{\\left({500}\\right)}={500000}$. Thus, width of rectangular field should be ${x}={500}{f{{t}}}.$ and length should be ${y}={2000}-{2}{x}={1000}{f{{t}}}$. Example 2. Find the point on parabola ${y}=\\frac{{1}}{{2}}{{x}}^{{2}}$ that is closest to the point (-4,1). The distance between the point ${\\left({x},{y}\\right)}$ and ${\\left(-{4},{1}\\right)}$ is ${d}=\\sqrt{{{{\\left({x}-{\\left(-{4}\\right)}\\right)}}^{{2}}+{{\\left({y}-{1}\\right)}}^{{2}}}}=\\sqrt{{{{\\left({x}+{4}\\right)}}^{{2}}+{{\\left({y}-{1}\\right)}}^{{2}}}}$. But since point ${\\left({x},{y}\\right)}$ lies on parabola then ${y}=\\frac{{1}}{{2}}{{x}}^{{2}}$ and ${d}=\\sqrt{{{{\\left({x}+{4}\\right)}}^{{2}}+{{\\left(\\frac{{1}}{{2}}{{x}}^{{2}}-{1}\\right)}}^{{2}}}}$. To make things easier, we minimize ${{d}}^{{2}}$ instead of ${d}$ (if ${d}$ has minimum then ${{d}}^{{2}}$ has minimum at the same point). So, ${f{{\\left({x}\\right)}}}={{d}}^{{2}}={{\\left({x}+{4}\\right)}}^{{2}}+{{\\left(\\frac{{1}}{{2}}{{x}}^{{2}}-{1}\\right)}}^{{2}}$. Using Chain Rule we have that ${f{'}}{\\left({x}\\right)}={2}{\\left({x}+{4}\\right)}\\cdot{\\left({x}+{4}\\right)}'+{2}{\\left(\\frac{{1}}{{2}}{{x}}^{{2}}-{1}\\right)}\\cdot{\\left(\\frac{{1}}{{2}}{{x}}^{{2}}-{1}\\right)}'={2}{\\left({x}+{4}\\right)}+{2}{x}{\\left(\\frac{{1}}{{2}}{{x}}^{{2}}-{1}\\right)}=$ $={{x}}^{{3}}+{8}$. So, ${f{'}}{\\left({x}\\right)}={0}$ when ${{x}}^{{3}}+{8}={0}$ or ${x}=-{2}$. Since, ${f{'}}{\\left({x}\\right)}<{0}$ when ${x}<-{2}$ and ${f{'}}{\\left({x}\\right)}>{0}$ when ${x}>-{2}$ then by First Derivative Test ${x}=-{2}$ is absolute minimum. Now, ${y}=\\frac{{1}}{{2}}{{x}}^{{2}}=\\frac{{1}}{{2}}{{\\left(-{2}\\right)}}^{{2}}={2}$. Therefore the closest point that lies on the ${y}=\\frac{{1}}{{2}}{{x}}^{{2}}$ to the point ${\\left(-{4},{1}\\right)}$ is ${\\left(-{2},{2}\\right)}$. The corresponding minimum distance is ${d}=\\sqrt{{{{\\left(-{2}+{4}\\right)}}^{{2}}+{{\\left({2}-{1}\\right)}}^{{2}}}}=\\sqrt{{{5}}}$. Example 3. Suppose that man stands at point A and wants to reach point D as quickly as possible. He could row his boat directly across the river that is 4 km wide to point B and then run to D, or he could row directly to D, or he could row to some point between B and D and then run to D. If he can row at ${5}\\frac{{{k}{m}}}{{h}}$ and run at ${9}\\frac{{{k}{m}}}{{h}}$,", "we take area between only positive coordinates, then we will solve like this y = 2$$\\sqrt { x }$$, x = 0, x = 1 Required Area = Area APOA, Question 5. Find the area enclosed by y = | x |, x = – 3, x = 1 and x-axis. Solution : y = | x |, x = – 3, x = 1 Thus y = + x and y = -x Graph of these lines is given below Question 6. Find the area enclosed by curve x2 = 4ay, x-axis and line x = 2. Solution: Curve x2 = 4ay is a parabola. Its graph is given below. Question 7. Find the area enclosed by ellipse $$\\frac { { x }^{ 2 } }{ 4 }$$ + $$\\frac { { y }^{ 2 } }{ 9 }$$ = 1 and lie above the x – axis Solution: The area enclosed by ellipse $$\\frac { { x }^{ 2 } }{ 4 }$$ + $$\\frac { { y }^{ 2 } }{ 9 }$$ = 1 and lie above the x axis is shown in the following figure : Question 8. Find the total area of ellipse $$\\frac { { x }^{ 2 } }{ { a }^{ 2 } }$$ + $$\\frac { { y", "# How do you solve and write the following in interval notation: (3 − 5x)(3x − 7) ≥ 0? Jul 25, 2017 $\\left[\\frac{3}{5} , \\frac{7}{3}\\right]$ #### Explanation: $f \\left(x\\right) = \\left(3 - 5 x\\right) \\left(3 x - 7\\right) \\ge 0$ First, solve f(x) = 0. Solve the 2 binomials: 3 - 5x = 0 --> 5x = 3 --> $x = \\frac{3}{5}$ 3x - 7 = 0 --> 3x = 7 --> $x = \\frac{7}{3}$ On the number line, plot the 2 end-points $\\left(\\frac{3}{5}\\right)$ and $\\left(\\frac{7}{3}\\right)$. Since a = -15 < 0. The parabola graph of f(x) opens downward. Between the 2 real roots, f(x) > 0, as a part of the graph stays above the x-axis. Close interval $\\left[\\frac{3}{5} , \\frac{7}{3}\\right]$ ---------------------- 0 ----- $\\frac{3}{5}$ + + + + + + + + $\\frac{7}{3}$ -----------------------", "NEW New Website Launch Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc... 1 ### AIEEE 2007 MCQ (Single Correct Answer) The normal to a curve at $$P(x,y)$$ meets the $$x$$-axis at $$G$$. If the distance of $$G$$ from the origin is twice the abscissa of $$P$$, then the curve is a A circle B hyperbola C ellipse D parabola ## Explanation Equation of normal at $$P\\left( {x,y} \\right)$$ is $$Y - y = - {{dx} \\over {dy}}\\left( {x - x} \\right)$$ Coordinate of $$G$$ at $$X$$ axis is $$\\left( {X,0} \\right)$$ (let) $$\\therefore$$ $$0 - y = - {{dx} \\over {dy}}\\left( {X - x} \\right) \\Rightarrow y{{dy} \\over {dx}} = X - x$$ $$\\Rightarrow X = x + y{{dy} \\over {dx}}$$ $$\\therefore$$ Co-ordinate of $$G\\left( {x + y{{dy} \\over {dx}},0} \\right)$$ Given distance of $$G$$ from origin $$=$$ twice of the abscissa of $$P.$$ as distance cannot be $$-ve,$$ therefore abscissa $$x$$ should be $$+ve$$ $$\\therefore$$ $$x + y{{dy} \\over {dx}} = 2x \\Rightarrow y{{dy} \\over {dx}} = x \\Rightarrow ydx = xdx$$ On Integrating $$\\Rightarrow {{{y^2}} \\over 2} = {{{x^2}} \\over 2} + {c_1} \\Rightarrow {x^2} - {y^2} = - 2{c_1}$$ $$\\therefore$$ the curve is a hyperbola 2 ### AIEEE 2007 MCQ (Single Correct", "$4x+3y-10=0$and$4x+3y+12=0$ E) $4x+3y+3=0$and$4x+3y-1=0$ • question_answer203) The eccentricity of the hyperbola${{x}^{2}}-{{y}^{2}}=2004$is: A) $\\sqrt{3}$ B) $2$ C) $2\\sqrt{2}$ D) $\\sqrt{2}$ E) 1.5 • question_answer204) The equation of the directrix of${{(x-1)}^{2}}=2(y-2)$is: A) $2y+3=0$ B) $2x+1=0$ C) $2x-1=0$ D) $2y-1=0$ E) $2y-3=0$ • question_answer205) On the ellipse$4{{x}^{2}}+9{{y}^{2}}=1$the point at which the tangent are parallel to$8x=9y$are: A) $\\left( \\frac{2}{5},\\frac{1}{5} \\right)or\\left( -\\frac{2}{5},-\\frac{1}{5} \\right)$ B) $\\left( -\\frac{2}{5},\\frac{1}{5} \\right)or\\left( \\frac{2}{5},-\\frac{1}{5} \\right)$ C) $\\left( -\\frac{2}{5},-\\frac{1}{5} \\right)$ D) $\\left( -\\frac{3}{5},-\\frac{2}{5} \\right)or\\left( \\frac{3}{5},\\frac{2}{5} \\right)$ E) $\\left( -\\frac{3}{5},\\frac{2}{5} \\right)or\\left( \\frac{3}{5},-\\frac{2}{5} \\right)$ • question_answer206) The line$x\\text{ }cos\\alpha +y\\text{ }sin\\alpha =p$touches the hyperbola$\\frac{{{x}^{2}}}{{{a}^{2}}}-\\frac{{{y}^{2}}}{{{b}^{2}}}=1,$if: A) ${{a}^{2}}co{{s}^{2}}\\alpha -{{b}^{2}}si{{n}^{2}}\\alpha ={{p}^{2}}$ B) ${{a}^{2}}co{{s}^{2}}\\alpha -{{b}^{2}}si{{n}^{2}}\\alpha =p$ C) ${{a}^{2}}co{{s}^{2}}\\alpha +{{b}^{2}}si{{n}^{2}}\\alpha ={{p}^{2}}c$ D) ${{a}^{2}}co{{s}^{2}}\\alpha +{{b}^{2}}si{{n}^{2}}\\alpha =p$ E) ${{b}^{2}}co{{s}^{2}}\\alpha -{{a}^{2}}si{{n}^{2}}\\alpha ={{p}^{2}}$ • question_answer207) If $\\overrightarrow{a}+\\overrightarrow{b}+\\overrightarrow{c}=\\overrightarrow{0},|\\overrightarrow{a}|=3,|\\overrightarrow{b}|=5$and $|\\overrightarrow{c}|=7,$then the angle between a and b is: A) $\\frac{\\pi }{3}$ B) $\\frac{\\pi }{2}$ C) ${{\\cos }^{-1}}\\left( \\frac{2}{225} \\right)$ D) $\\frac{\\pi }{4}$ E) $\\frac{\\pi }{6}$ • question_answer208) The velocity of a boat$X$relative to a boat Y is $5\\hat{i}-2\\text{ }\\hat{j}$and that of$Y$relative to another boat Z is$9\\hat{i}+4\\hat{j}$where$\\hat{i}$and$\\hat{j}$ are the velocity of k not per hour, east and north respectively. Then the velocity is: A) $\\frac{\\sqrt{2}}{10}\\,knot/h$ B) $\\frac{10}{\\sqrt{2}}\\,knot/h$ C) $10\\sqrt{2}\\,knot/h$ D) $2\\sqrt{10}\\,knot/h$ E) $10\\,knot/h$ • question_answer209) Two vectors$\\overrightarrow{a}$and$\\overrightarrow{b}$of equal magnitude 5 originating from a point and directs respectively towards north-east and north-west. Then the magnitude of $\\vec{a}-\\vec{b}$ is: A) $3\\sqrt{2}$ B) $2\\sqrt{3}$ C) $2\\sqrt{5}$ D) $5\\sqrt{2}$ E)", "the following three properties: • The curvature at any point on a straight line is 0. • The curvature at any point on a circle of radius r is$latex \\frac{1}{r}$. (Proving that this is true of the formula above is a nice freshman calculus exercise.) • The circle that best approximates a curve at a point $latex (x,y)$ is the circle that is tangent to the curve at $latex (x,y)$ and that has radius $latex 1/k$, where k is the curvature of the curve at $latex (x,y)$ . This circle is called the osculating circle at $latex (x,y)$ . ### Curvature of the parabola You can calculate that the curve of the parabola $latex y=x^2$ at x is given by $latex \\kappa(x) = \\frac{2}{\\left(4 x^2+1\\right)^{3/2}}&s=1$ For example, the curvature at (0,1) is 2, at the point (1/2, 1/4) it is $latex \\scriptstyle 1/\\sqrt{2}\\:\\approx\\: 0.71&s=1$, and at (1,1) it is about 0.18. The radii of the osculating circles are 1/2, $latex \\sqrt{2}&s=-2$, and 5.59 respectively. For large numbers the curvature is nearly 0; for example, at (10, 100) the curvature is about .00025. To the eye the parabola near (10,100) looks like a straight line. This graph shows the osculating circles at x = 0, 1/2 and 1: You can see animated osculating circles at the Wolfram Demonstration Project", "# Using integration, find the area of the region bounded by the parabola $y^{2}= 4x$ and the circle $4x^{2}+4y^{2}= 9.$ for the circle $4x^{2}+4y^{2}= 9,\\; we\\; have$ $x^{2}+y^{2}= \\left ( \\frac{3}{2} \\right )^{2}$ $\\therefore centre\\; is\\; at\\left ( 0,0 \\right )\\S \\; r= \\frac{3}{2}$ Also vertex of the parabola $y^{2}= 4x\\; is\\; at\\left ( 0,0 \\right ).$ solving the equations of curves, we get: $4x^{2}+16x-9= 0\\Rightarrow \\left ( 2x-1 \\right )\\left ( 2x+9 \\right )= 0$ $\\therefore x= \\frac{1}{2}\\left [ x\\not\\equiv \\frac{-9}{2} \\right ]$ As the circle and parabola, both are symmetrical about the x-axis $\\therefore Required\\; area= 2\\left [ \\int_{0}^{\\frac{1}{2}} 2\\sqrt{x}\\, dx +\\int_{\\frac{1}{2}}^{\\frac{3}{2}}\\sqrt{\\left ( \\frac{3}{2} \\right )^{2}-x^{2}dx}\\right ]$ $\\Rightarrow 2\\left [ \\frac{4}{3}\\left [ x\\sqrt{x} \\right ] \\right ]^{\\frac{1}{2}}_{0}+\\left [ \\frac{x}{2}\\sqrt{\\left ( \\frac{3}{2} \\right )^{2}-x^{2}}+\\frac{9}{8}\\sin^{-1}\\frac{2x}{3}\\right ]^{\\frac{3}{2}}_{\\frac{1}{2}}$ $\\Rightarrow 2\\left [ \\frac{4}{3} \\left [ \\frac{1}{2\\sqrt{2}}-0 \\right ]+\\left [ \\frac{3}{4} \\sqrt{0}+\\frac{9}{8}\\sin^{-1}1\\right ]-\\left [ \\frac{1}{4}\\frac{\\sqrt{9-1}}{\\sqrt{4}} +\\frac{9}{8}\\sin^{-1}\\frac{1}{3}\\right ]\\right ]$ $\\Rightarrow 2\\left [ \\frac{2\\sqrt{2}}{6}+\\frac{9}{8} \\times \\frac{\\pi }{2}-\\frac{\\sqrt{2}}{4}-\\frac{9}{8}\\sin^{-1}\\frac{1}{3}\\right ]$ $\\Rightarrow \\left [ \\frac{\\sqrt{2}}{6}+\\frac{9}{4} \\cos^{-1}\\frac{1}{3}\\right ]sq\\cdot units$ ## Related Chapters ### Preparation Products ##### Knockout NEET July 2020 An exhaustive E-learning program for the complete preparation of NEET.. ₹ 15999/- ₹ 6999/- ##### Rank Booster NEET 2020 This course will help student to be better prepared and study in the right direction for NEET.. ₹ 9999/- ₹ 4999/- ##### Knockout JEE Main July 2020 An exhaustive", "If one of the points computed in $$(1)$$ equals $$-\\frac dc$$ (that is, that point is mapped to $$\\infty$$ by the LFT), then the circle is mapped to a line. In that case, just plug any other two points on the source circle into the LFT to get two points on the image line. Given a pair of antipodal points on a circle, $$\\{p_1,p_2\\}$$, the radius, r, and center, k, of that circle are given by $$r=\\frac{|p_1-p_2|}2\\qquad k=\\frac{p_1+p_2}2\\tag2$$ Application In this case, we have $$\\frac{3z+2}{4z+3}$$, $$k=0$$, and $$r=1$$. Therefore, $$(1)$$ says that the antipodal points of the source circle $$0\\pm\\frac{0+3/4}{|0+3/4|}\\cdot1=\\{-1,1\\}\\tag3$$ get mapped by the LFT to the antipodal points of the image circle $$\\left\\{1,\\frac57\\right\\}\\tag4$$ then $$(2)$$ says that the radius, $$r$$, and the center, $$k$$, of the image circle are $$\\bbox[5px,border:2px solid #C0A000]{r=\\frac17\\qquad k=\\frac67}\\tag5$$ The image circle is symmetric wrt the real axis, therefore $$[f(1), f(-1)] = [5/7, 1]$$ is a diameter. Here's another solution I was able to find. Notice the matrix factorization: $$\\left[ \\begin{array}{cc} 3 & 2 \\\\ 4 & 3 \\end{array} \\right] = \\left[ \\begin{array}{cc} \\frac{1}{5} & 0\\\\ 0 & 5 \\end{array} \\right] \\times \\left[ \\begin{array}{cc} 1 & 18 \\\\ 0 & 1 \\end{array} \\right] \\times \\left[ \\begin{array}{cr} \\frac{3}{5}& -\\frac{4}{5}\\\\ \\frac{4}{5}& \\frac{3}{5} \\end{array} \\right] = A \\times B \\times C$$ The geometry behind this is", "All points on the curve y2=4ax+asin⁡xa at which the tangents are parallel to the axis of x lie on a # All points on the curve ${y}^{2}=4a\\left(x+a\\mathrm{sin}\\frac{x}{a}\\right)$ at which the tangents are parallel to the axis of $x$ lie on a 1. A circle 2. B parabola 3. C line 4. D none of these Register to Get Free Mock Test and Study Material +91 Verify OTP Code (required) ### Solution: We have, ${y}^{2}=4a\\left(x+a\\mathrm{sin}\\frac{x}{a}\\right)$ ..(i) For points at which the tangents are parallel to x-axis, we must have $\\begin{array}{l}\\frac{dy}{dx}=0\\\\ ⇒4a\\left(1+\\mathrm{cos}\\frac{x}{a}\\right)=0⇒\\mathrm{cos}\\frac{x}{a}=-1⇒\\frac{x}{a}=\\left(2n+1\\right)\\pi \\end{array}$ For these values of $x$, we obtain $\\mathrm{sin}\\frac{x}{a}=0$ Putting $\\mathrm{sin}\\frac{x}{a}=0$, in (i), we get ${y}^{2}=4ax$ Therefore, all these points lie on the parabola ${y}^{2}=4ax.$ Register to Get Free Mock Test and Study Material +91 Verify OTP Code (required)", "0.5x – 4? One way to do so is to plot the points contained in the table on the graph, and test each point to see if it lies on the line. Use a plotting applet to graph both the line and the points in the table. This tool will allow you to confirm whether or not the points in the table lie on the graph of the line. Use the graph and table provided to determine which points lie on the graph of the line. Use the interactive below to identify which points from the table lie on the graph of the line: y = 0.5x – 4 For each ordered pair, place it in the Lies on Line box if the point lies on the line, or in the Does NOT Lie on Line box if the point does not lie on the line. Practice Use the graphing applet to determine which of the following tables contain only points that lie on the graph of the line $y=\\frac{1}{3}x+3.$ Table A x y -9 0 -3 2 0 3 2 $3\\frac{2}{3}$ 6 5 Table B x y -9 0 -3 3 0 3 4 $4\\frac{1}{3}$ 6 5 Table C x y -9 0 -3 2 1 $3\\frac{1}{3}$ 3 4 9 6 Table D x y 0 -9", "# Roulettes and Calculus Roulettes – 5: Calculus Considerations. In the first post of this series Roulette Generators (RG) are explained. Here are the files for Winplot or Geometer’s Sketchpad. Use them to quickly see the graphs of these curves by adjusting one or two parameters. While writing this series of posts I was intrigued by the cusps that appear in some of the curves. In Cartesian coordinates you think of a cusp as a place where the curves is continuous, but the derivative is undefined, and the tangent line is vertical. Cusps on the curves we have been considering are different. The equations of the curves formed by a point attached to a circle rolling around a fixed circle in the form we have been using are: $x\\left( t \\right)=(1+R)\\cos \\left( t \\right)-S\\cos \\left( \\tfrac{1}{R}t+t \\right)$ $y\\left( t \\right)=\\left( 1+R \\right)\\sin \\left( t \\right)-S\\sin \\left( \\tfrac{1}{R}t+t \\right)$ For example, let’s consider the case with $R=S=\\tfrac{1}{3}$ Epicycloid with R = S = 1/3 The equations become $x\\left( t \\right)=\\frac{4}{3}\\cos \\left( t \\right)-\\frac{1}{3}\\cos \\left( 4t \\right)$ $y\\left( t \\right)=\\frac{4}{3}\\sin \\left( t \\right)-\\frac{1}{3}\\sin \\left( 4t \\right)$ The derivative is $\\displaystyle \\frac{dy}{dx}=\\frac{dy/dt}{dx/dt}=\\frac{\\cos \\left( t \\right)-\\cos (4t)}{-\\sin \\left( t \\right)+\\sin (4t)}$ The cusps are evenly spaced one-third of the way around the circle and appear at $t=0,\\tfrac{2\\pi }{3},\\tfrac{4\\pi }{3}$. At the cusps" ]

[ "\\theta}=\\frac{4}{5}$", "have $$tan\\theta = \\frac{m_2-m_1}{1+m_2m_1}$$ $$tan\\theta = \\frac{3-2x+3}{1+3(2x-3)}$$ Here we need to find the value of \"$x$\"", "Finding theta Algebra Level 1 If $$\\tan \\theta = \\frac{ 2}{7}$$, what is the value of $\\frac{ \\cos \\theta + 3 \\sin \\theta } { \\cos \\theta - 3 \\sin \\theta}?$ Details and assumptions You may read up on Trigonometric Functions. ×", "## anonymous 3 years ago PreCalc & Trig, picture below, please show your work.. Firstly, calculate the hypotenuse, then find cos(theta), then find theta. Once you have found theta, use it to calculate cos(theta/2). However, if you are required to use identities (which I doubt), remember that $$cos^2(\\theta) = \\frac{1+cos(\\theta)}{2}$$.", "# Should you find theta? (II) Geometry Level 4 $\\large x_n=2^n \\cos n\\theta$ For the given sequence we know that $\\large x_2=-\\frac{28}{9}$ and $\\large x_3=- \\frac{184}{27}.$ If $x_5=\\dfrac{A}{B},$ where $A$ and $B$ are coprime positive integers, find $A+B.$ See Part 1. ×", "# Should you find theta? Geometry Level 3 $\\large x_n=2^n \\cos n\\theta$ For the given sequence we know that $$\\large x_2=-\\frac{28}{9}$$ and $$\\large x_3=- \\frac{184}{27}.$$ If $$x_4=\\dfrac{A}{B},$$ where $$A$$ and $$B$$ are coprime positive integers, find $$A+B.$$ See Part 2. ×", "\\frac{d\\theta}{dt}$ $a_c = r\\omega^2$", "# Question #be72f $\\tan \\theta = - \\frac{3}{4}$ This is a 3,4,5 triangle so the adjacent is 4 hence $\\tan \\theta = \\frac{o p p o s i t e}{a \\mathrm{dj} a c e n t} = - \\frac{3}{4}$", "# Is there easier way out? Geometry Level 4 If $$\\cos(\\theta)=\\cos(\\alpha)\\cos(\\beta)$$ and $$\\tan(\\frac{\\theta+\\alpha}{2})\\tan(\\frac{\\theta-\\alpha}{2})=f(\\beta)$$. then find $$f(\\frac{\\pi}{6})$$ ×", "# itheta = 7 - 4i Find theta ? $\\theta = \\frac{7 - 4 i}{i} =$ $= \\frac{7 - 4 i}{i} \\cdot \\frac{i}{i} = \\frac{7 i + 4}{-} 1 = - 4 - 7 i$", "$\\sin(32^{\\circ})=\\frac{x} {2}$ $(\\theta)=\\cos^{-1}(\\frac{3} {5})$", "$(1)$ to get the other. – GoodDeeds Apr 3 '16 at 8:27 • Sorry for all these questions but, how would you use $\\cos3\\theta = 4\\cos^3\\theta-3\\cos\\theta$ in $(3)$ to find $\\cos\\theta$? – Yuxie Apr 3 '16 at 9:09", "\\cos (\\theta_s)+L_s^2) \\right)$$ At this point you can probably find a clevel trigonometrical way to solve this (I am not that good), but in this form it would be easy enough to find a numerical solution to the equation by assuming a value for $$\\theta_s$$.", "You have $$x(0)=\\frac{f}{r}cos(\\theta ) \\mbox{ and }v(0)=\\frac{f\\omega}{r}\\sin(\\theta )$$, use that $$cos(\\theta ) = \\frac{1}{\\sqrt{1+tan^2(\\theta )}}\\mbox{, }sin(\\theta )=\\frac{tan(\\theta )}{\\sqrt{1+tan^2(\\theta )}} \\mbox{ and } tan(tan^{-1}(\\theta))=\\theta$$. ehild", "$\\frac{sin(\\theta)}{cos(\\theta)}= tan(\\theta)= \\frac{y}{x}$ and so $y= tan(\\theta)x$.", "recall that $\\cos(3\\theta)=4\\cos^3\\theta-3\\cos\\theta$. Now we know $\\cos(3\\theta)$, and can find the three possibilities for $\\cos\\theta$.", "## Multivariable Calculus, 7th Edition Given:$r=\\frac{3}{2-2cos\\theta}$ or $r=\\frac{(3/2)}{1-cos\\theta}=\\frac{(1).(3/2)}{1-(1).cos\\theta}$ The graph of $r=\\frac{3}{2-2cos\\theta}$ is a side-ways parabola with eccentricity $1$. See the attached graph.", "## Trigonometry (11th Edition) Clone 1. r is always positive 2. x is positive 3. Therefore, $\\cos\\theta=\\frac{x}{r}=\\frac{+}{+}=+$", "get, $\\frac{\\sin \\Theta – \\frac{1}{\\tan \\Theta }}{2\\tan \\Theta }$ =$\\frac{\\frac{4}{5} – \\frac{1}{4}}{2 \\times \\frac{4}{3}}$ $\\frac{\\sin \\Theta – \\frac{1}{\\tan \\Theta }}{2\\tan \\Theta }$ = $\\frac{\\frac{16}{20} – \\frac{15}{20}}{\\frac{8}{3}}$ $\\frac{\\sin \\Theta – \\frac{1}{\\tan \\Theta }}{2\\tan \\Theta }$ = $\\frac{3}{160}$ Therefore, $\\frac{\\sin \\Theta – \\frac{1}{\\tan \\Theta }}{2\\tan \\Theta }$ = $\\frac{3}{160}$ 20.) If $\\sin \\Theta = \\frac{3}{5}$ , find the value of $\\frac{\\cos \\Theta – \\frac{1}{\\tan \\Theta }}{2 \\cot \\Theta }$ Sol. Given: $\\sin \\Theta = \\frac{3}{5}$ …. (1) To find the value of $\\frac{\\cos \\Theta – \\frac{1}{\\tan \\Theta }}{2 \\cot \\Theta }$ Now, we know the following trigonometric identity $\\cos ^{2}\\Theta + \\sin ^{2} \\Theta = 1$ Therefore by substituting the value of $\\cos \\Theta$ from equation (1) We get, $\\cos ^{2}\\Theta + (\\frac{3}{5})^{2} = 1$ Therefore, $\\cos ^{2}\\Theta = 1 – (\\frac{3}{5})^{2}$ $\\cos ^{2}\\Theta = 1 – \\frac{9}{25}$ Now by taking L.C.M We get, $\\cos ^{2}\\Theta = \\frac{25 – 9}{25}$ $\\cos ^{2}\\Theta = \\frac{25 – 9}{25}$ Therefore, by taking square roots on both sides We get, $\\cos \\Theta = \\frac{4}{5}$ Therefore, $\\cos \\Theta = \\frac{4}{5}$ …. (2) Now we know that $\\tan \\Theta = \\frac{\\sin \\Theta }{\\cos \\Theta }$ Therefore by substituting the value of $\\sin \\Theta$ and $\\cos\\Theta$ from equation (1) and (2) respectively We get, $\\tan \\Theta =\\frac{ \\frac{3}{5}}{\\frac{4}{5}}$ $\\tan \\Theta =\\frac{3}{4}$ …. (3) Also, we know that", "$\\frac{4}{3}g\\sin\\theta$. I'm wrong. I'll tryna find the mistake. – Vim Jan 25 '15 at 14:45" ]

[ "$\\text{sin} (3\\theta) = 3 \\text{sin} \\theta - 4 \\text{sin}^3 \\theta$, the Triple Angle Formula for Cosine $\\text{cos} (3 \\theta) = -3 \\text{ cos} \\theta + 4 \\text{ cos}^3 \\theta$, and the Triple Angle Formula for Tangent $\\text{tan} (3 \\theta) = \\frac{3 \\text{ tan} \\theta - \\text{ tan}^3 \\theta}{1 - 3 \\text{ tan}^2 \\theta}$.", "these formulas, we also have the following identities for Proof: To prove the triple-angle identities, we can write sin ⁡ 3 θ \\sin 3 \\theta sin 3 θ as sin ⁡ (2 θ + θ) \\sin(2 \\theta + \\theta) sin (2 θ + θ).", "# Cos triple angle formula ### Expansion form $\\cos{3\\theta} \\,=\\, 4\\cos^3{\\theta}-3\\cos{\\theta}$ ### Simplified form $4\\cos^3{\\theta}-3\\cos{\\theta} \\,=\\, \\cos{3\\theta}$ ### Introduction It is called cos triple angle identity and used as a formula in two various cases. 1. Cos of triple angle is expanded as the subtraction of three times cos of angle from four times cos cubed of angle. 2. The subtraction of three times cos of angle from four times cos cubed of angle is simplified as cos of triple angle. #### How to use Cosine of triple angle identity is used to either expand or simplify the triple angle cos functions like $\\cos{3x}$, $\\cos{3A}$, $\\cos{3\\alpha}$ and etc. For example, $(1) \\,\\,\\,\\,\\,\\,$ $\\cos{3x} \\,=\\, 4\\cos^3{x}-3\\cos{x}$ $(2) \\,\\,\\,\\,\\,\\,$ $\\cos{3A} \\,=\\, 4\\cos^3{A}-3\\cos{A}$ $(3) \\,\\,\\,\\,\\,\\,$ $\\cos{3\\alpha} \\,=\\, 4\\cos^3{\\alpha}-3\\cos{\\alpha}$ #### Proof Learn how to derive the rule of cos triple angle identity by geometric approach in trigonometry.", "are 1, 1, 2. If θ is the angle between them, they can be written as: $\\cos \\theta =\\vert \\frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\\cdot \\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\\vert$ $\\cos \\theta =\\vert \\frac{3\\cdot 1+5\\cdot 1+4\\cdot 2}{\\sqrt{3^{2}+5^{2}+4^{2}}\\cdot \\sqrt{1+1+2^{2}}}\\vert =\\frac{8\\sqrt{3}}{15}$ Hence we can find the required acute angle by taking $\\cos ^{-1}\\frac{8\\sqrt{3}}{15}$", "by: $cos (\\theta) = \\frac{D' \\cdot G'}{|D'||G'|}$ Consider the triangle C'G'B' this has the internal angle $\\theta /2$ so the distance from B' to C' is given by: $F=|G'| cos (\\theta /2)=|G'| cos (cos^{-1}(\\frac{D' \\cdot G'}{|D'||G'|}) /2)$ And finally the centre of the arc is located at: $C'=F\\frac{C\"}{|C\"|}$ or $C=F\\frac{C\"}{|C\"|}+B$", "## anonymous one year ago Quick check? Thank You for helping! Use the formula v1*v2=|v1||v2|cos theta to find the cosine of the angle between the two vectors (-7,8) and (5,3). I got cos theta=-11/61.97. Is this correct? Thanks! 1. anonymous @nincompoop @Michele_Laino @ganeshie8 @campbell_st @zepdrix @Luigi0210 @Destinymasha @radar @Whitemonsterbunny17 2. anonymous I also solved for theta, getting 100.22 degrees. 3. Michele_Laino we can write this: $\\left\\| {{v_1}} \\right\\| = \\sqrt {49 + 64} = 10.63$ $\\left\\| {{v_2}} \\right\\| = \\sqrt {25 + 9} = 5.83$ ${v_1} \\cdot {v_2} = - 35 + 24 = - 11$ so we get: $\\cos \\theta = \\frac{{ - 11}}{{10.63 \\times 5.83}}$ 4. anonymous so same thing, awesome! 5. anonymous Thanks for the confirmation! 6. Michele_Laino yes! $\\cos \\theta = \\frac{{ - 11}}{{10.63 \\times 5.83}} = - \\frac{{11}}{{61.97}}$ 7. Michele_Laino :)", "corner of the box is 3. That is where the 35.26° angle came from. The angle can be calculated using the dot product. Let ${\\displaystyle {\\vec {m}}=\\left(0,{\\frac {\\sqrt {2}}{2}},{\\frac {\\sqrt {2}}{2}}\\right)}$ be the unit vector in the direction of C' and let ${\\displaystyle {\\vec {n}}=\\left({\\frac {1}{\\sqrt {3}}},{\\frac {1}{\\sqrt {3}}},{\\frac {1}{\\sqrt {3}}}\\right)}$ be a unit vector in the direction of the corner of the box at ${\\displaystyle {\\vec {n}}=\\left(1,1,1\\right)}$. Because ${\\displaystyle {\\vec {m}}\\cdot {\\vec {n}}=|{\\vec {m}}||{\\vec {n}}|\\cos \\theta ,}$ where ${\\displaystyle \\theta }$ is the angle between ${\\displaystyle {\\vec {m}}}$ and ${\\displaystyle {\\vec {n}},}$ we have ${\\displaystyle \\left(0,{\\frac {\\sqrt {2}}{2}},{\\frac {\\sqrt {2}}{2}}\\right)\\cdot \\left({\\frac {1}{\\sqrt {3}}},{\\frac {1}{\\sqrt {3}}},{\\frac {1}{\\sqrt {3}}}\\right)=\\cos \\theta }$ ${\\displaystyle \\cos \\theta =0+{\\frac {\\sqrt {2}}{2{\\sqrt {3}}}}+{\\frac {\\sqrt {2}}{2{\\sqrt {3}}}}={\\sqrt {\\frac {2}{3}}}}$ ${\\displaystyle \\theta =\\cos ^{-1}\\left({\\sqrt {\\frac {2}{3}}}\\right)}$ ${\\displaystyle \\theta =35.26^{\\circ }.}$ The norm of the K2 matrix is also 1, so it too does not change the magnitude of any vector pre-multiplied by the K2 matrix. #### The zero plane[] At this point, the Z axis is now orthogonal to the plane in which any ABC vector without a common-mode component can be found. Any balanced ABC vector waveform (a vector without a common mode) will travel about this plane. This plane will be called the zero plane and is shown below by the hexagonal outline. The", "by \\alpha=3\\arccos(2/\\sqrt 5). Using the triple angle formula$$\\cos(3x)=\\cos^3(x)-3\\sin^2(x)\\cos(x)$$we find that$$\\begin{align} \\cos(\\alpha)&=\\cos(3\\arccos(2/\\sqrt 5))\\\\\\\\ &=\\cos^3(\\arccos(2/\\sqrt 5))-3\\sin^2(\\arccos(2/\\sqrt 5))\\cos(\\arccos(2/\\sqrt 5))\\\\\\\\ &=\\frac{8}{5\\sqrt 5}-3\\left(\\frac15\\right)\\left(\\frac{2}{\\sqrt 5}\\right)\\\\\\\\ &=\\frac{2}{5\\sqrt 5} \\end{align}$Therefore,$\\alpha = \\theta\\$ and we are done!", "6. Originally Posted by Karl Harder I'm still stuck on how you get this using the double angle formula? $a^4\\cos^2{\\theta} \\sin^2{\\theta} \\, d\\theta = \\frac{a^4}{4} \\sin^2(2\\theta) \\, d\\theta$ never mind i have found out that i can get the answer by using the proudct formula 7. Originally Posted by Karl Harder I'm still stuck on how you get this using the double angle formula? $a^4\\cos^2{\\theta} \\sin^2{\\theta} \\, d\\theta = \\frac{a^4}{4} \\sin^2(2\\theta) \\, d\\theta$ ... never mind i have found out that i can get the answer by using the proudct formula product formula? ... $a^4 \\cos^2{\\theta} \\sin^2{\\theta}$ $\\frac{a^4}{4} \\cdot 4\\cos^2{\\theta} \\sin^2{\\theta}$ $\\frac{a^4}{4} (2\\cos{\\theta} \\sin{\\theta})^2$ $\\frac{a^4}{4} \\sin^2(2\\theta)$", "##### Question 11.4.23 The triple angle formula states that $\\cos(3\\theta) = 4\\cos^3\\theta - 3\\cos\\theta.$ 1. We’re going to solve the cubic equation $x^3 = 3px + 2q$ for $x$ using the triple angle formula. If $x=2r\\cos\\theta$, find values of $r$ and $\\theta$ that will solve the cubic equation. (In Question 7.5.34 we were able to solve this equation when $q^2-p^3 \\ge 0$.) Show answer 2. What values of $p$ and $q$ will your solution work for? Show hint 3. Using your solution, find all values of $x$ that satisfy the equation $x^3 = 6x + 4$. Show hint", "triple-angle formula ⇒ t³ = √3*t² + 3t - 1/√3 We could simplify previous proof, by getting t² in terms of t t²*t = √3*t² + 3t - 1/√3 t² = (3t - 1/√3) / (t - √3) 2*cos(30°) / (1 + 4*cos(20°)) = √3 / (1 + 4*(1-t²)/(1+t²)) = (1/√3 - t) / (1 + (1/√3)*t) = (tan(30°) - tan(10°)) / (1 + tan(30°)*tan(10°)) = tan(20°) 06-08-2022, 11:18 PM Post: #8 Thomas Klemm Senior Member Posts: 1,685 Joined: Dec 2013 RE: proof left as an exercise We can use the triple angle formulae: \\begin{align} \\sin(3 \\theta) &= 3 \\sin \\theta - 4 \\sin^{3} \\theta \\\\ \\cos(3 \\theta) & = 4 \\cos^{3}\\theta - 3 \\cos \\theta \\\\ \\end{align} \\begin{align} \\sin(3 \\theta) &= 3 \\sin \\theta - 4 \\sin^{3} \\theta \\\\ &= \\sin \\theta \\, [3 - 4 \\sin^{2} \\theta] \\\\ &= \\sin \\theta \\, [4 \\cos^{2} \\theta - 1] \\\\ \\end{align} $$2 \\cos(3 \\theta) - 1 = 8 \\cos^{3}\\theta - 6 \\cos \\theta - 1 \\\\$$ For $$\\theta = 20^\\circ$$ we get: \\begin{align} 2 \\cos(3 \\theta) - 1 &= \\\\ 2 \\cos(3 \\cdot 20^\\circ) - 1 &= \\\\ 2 \\cos(60^\\circ) - 1 &= \\\\ 2 \\cdot \\tfrac{1}{2} - 1 &= 0 \\\\ \\end{align} And thus: $$8 \\cos^{3}(20^\\circ) - 6 \\cos(20^\\circ) - 1 = 0$$ From this we conclude:", "=\\arccos\\left(\\frac{n_1\\cdot n_2}{|n_1||n_2|}\\right)$$and taking the supplementary angle if that comes out between $\\pi/2$ and $\\pi$.", "# If A = <3 ,-1 ,3 >, B = <2 ,-3 ,5 > and C=A-B, what is the angle between A and C? Nov 20, 2016 ${112.47}^{0}$ #### Explanation: C= A-B= <3,-1,3>- <2, -3, 5> = < 1,2,-2> Angle between A and C would be given by $\\cos \\theta = \\frac{A \\cdot C}{| A | | C |}$ $\\cos \\theta = \\frac{3 \\cdot 1 + \\left(- 1\\right) \\cdot 2 + 3 \\left(- 2\\right)}{\\sqrt{{3}^{2} + {\\left(- 1\\right)}^{2} + {3}^{2}} \\sqrt{{1}^{2} + {2}^{2} + {\\left(- 2\\right)}^{2}}}$ =$- \\frac{5}{\\sqrt{19} \\sqrt{9}}$= -0.3823 $\\theta = \\arccos - 0.3823$ =${112.47}^{0}$", "terms of t t²*t = √3*t² + 3t - 1/√3 t² = (3t - 1/√3) / (t - √3) 2*cos(30°) / (1 + 4*cos(20°)) = √3 / (1 + 4*(1-t²)/(1+t²)) = (1/√3 - t) / (1 + (1/√3)*t) = (tan(30°) - tan(10°)) / (1 + tan(30°)*tan(10°)) = tan(20°) RE: proof left as an exercise - Thomas Klemm - 06-08-2022 11:18 PM We can use the triple angle formulae: \\begin{align} \\sin(3 \\theta) &= 3 \\sin \\theta - 4 \\sin^{3} \\theta \\\\ \\cos(3 \\theta) & = 4 \\cos^{3}\\theta - 3 \\cos \\theta \\\\ \\end{align} The first formula leads to: \\begin{align} \\sin(3 \\theta) &= 3 \\sin \\theta - 4 \\sin^{3} \\theta \\\\ &= \\sin \\theta \\, [3 - 4 \\sin^{2} \\theta] \\\\ &= \\sin \\theta \\, [4 \\cos^{2} \\theta - 1] \\\\ \\end{align} The second formula leads to: $$2 \\cos(3 \\theta) - 1 = 8 \\cos^{3}\\theta - 6 \\cos \\theta - 1 \\\\$$ For $$\\theta = 20^\\circ$$ we get: \\begin{align} 2 \\cos(3 \\theta) - 1 &= \\\\ 2 \\cos(3 \\cdot 20^\\circ) - 1 &= \\\\ 2 \\cos(60^\\circ) - 1 &= \\\\ 2 \\cdot \\tfrac{1}{2} - 1 &= 0 \\\\ \\end{align} And thus: $$8 \\cos^{3}(20^\\circ) - 6 \\cos(20^\\circ) - 1 = 0$$ From this we conclude: \\begin{align} 8 \\cos^{3}(20^\\circ) - 2 \\cos(20^\\circ) &= 1 + 4 \\cos(20^\\circ) \\\\ 2 \\cos(20^\\circ) [4 \\cos^{2}(20^\\circ)", "# Cos triple angle formula ### Expansion form $\\cos{3\\theta} \\,=\\, 4\\cos^3{\\theta}-3\\cos{\\theta}$ ### Simplified form $4\\cos^3{\\theta}-3\\cos{\\theta} \\,=\\, \\cos{3\\theta}$ ### Introduction It is called cos triple angle identity and used as a formula in two various cases. 1. Cos of triple angle is expanded as the subtraction of three times cos of angle from four times cos cubed of angle. 2. The subtraction of three times cos of angle from four times cos cubed of angle is simplified as cos of triple angle. #### How to use Cosine of triple angle identity is used to either expand or simplify the triple angle cos functions like $\\cos{3x}$, $\\cos{3A}$, $\\cos{3\\alpha}$ and etc. For example, $(1) \\,\\,\\,\\,\\,\\,$ $\\cos{3x} \\,=\\, 4\\cos^3{x}-3\\cos{x}$ $(2) \\,\\,\\,\\,\\,\\,$ $\\cos{3A} \\,=\\, 4\\cos^3{A}-3\\cos{A}$ $(3) \\,\\,\\,\\,\\,\\,$ $\\cos{3\\alpha} \\,=\\, 4\\cos^3{\\alpha}-3\\cos{\\alpha}$ #### Proof Learn how to derive the rule of cos triple angle identity by geometric approach in trigonometry. Latest Math Problems A best free mathematics education website for students, teachers and researchers. ###### Maths Topics Learn each topic of the mathematics easily with understandable proofs and visual animation graphics. ###### Maths Problems Learn how to solve the maths problems in different methods with understandable steps. Learn solutions ###### Subscribe us You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social", "# What is the arclength of the polar curve f(theta) = thetasin(3theta)-sec^2theta over theta in [0,pi/8] ? Jun 1, 2018 $\\approx 0.41636$ $r \\left(x\\right) = x \\cdot \\sin \\left(3 x\\right) - {\\sec}^{2} \\left(x\\right)$ $\\frac{\\mathrm{dr}}{\\mathrm{dx}} = \\sin \\left(3 x\\right) + x \\cdot \\cos \\left(3 x\\right) \\cdot 3 - 2 {\\sec}^{2} \\left(x\\right) \\cdot \\tan \\left(x\\right)$", "# Tan Triple angle formula ## Formula $\\large \\tan{3\\theta} \\,=\\, \\dfrac{3\\tan{\\theta}-\\tan^3{\\theta}}{1-3\\tan^2{\\theta}}$ ### Other forms Tangent of triple angle formula is written in the following three standard forms alternatively. Angle Triple angle Formula $x$ $3x$ $\\tan{3x} \\,=\\, \\dfrac{3\\tan{x}-\\tan^3{x}}{1-3\\tan^2{x}}$ $A$ $3A$ $\\tan{3A} \\,=\\, \\dfrac{3\\tan{A}-\\tan^3{A}}{1-3\\tan^2{A}}$ $\\alpha$ $3\\alpha$ $\\tan{3\\alpha} \\,=\\, \\dfrac{3\\tan{\\alpha}-\\tan^3{\\alpha}}{1-3\\tan^2{\\alpha}}$ The angle of a right angled triangle can be denoted by symbol but the triple angle should be thrice of that symbol. Likewise, the expansion of tan of the triple angle will be in the same form. ### Proof Assume theta is an angle of the right angled triangle and the triple angle is written as $3\\theta$. Tan of triple angle is written as $\\tan{3\\theta}$ and it can be expanded in terms of tan of angle theta by deriving a proof in geometrical method. 01 #### According to ΔHDJ In right angled triangle $HDJ$, the angle is $3\\theta$ and it represents triple angle. Tan triple angle is written as $\\tan{3\\theta}$ and express tan of triple angle in terms of ratio of the sides to write it in mathematical form. $\\tan{3\\theta} \\,=\\, \\dfrac{HJ}{DJ}$ The side $\\overline{HJ}$ is split as sides $\\overline{HK}$ and $\\overline{KJ}$ at point $K$ by the perpendicular line $\\overline{KG}$. Hence, the length of the side $\\overline{HJ}$ can be written as sum of the length of the sides $\\overline{HK}$ and $\\overline{KJ}$.", "What are the rectangular coordinates of (3, -π/3)? $x = \\frac{3}{2}$ $y = \\frac{- 3 \\sqrt{3}}{2}$ Explanation: From the given: $r = 3$ and $\\theta = - \\frac{\\pi}{3}$ $x = r \\cdot \\cos \\theta$ $x = 3 \\cdot \\cos \\left(- \\frac{\\pi}{3}\\right)$ $x = 3 \\cdot \\left(\\frac{1}{2}\\right)$ $x = \\frac{3}{2}$ $y = r \\cdot \\sin \\theta$ $y = 3 \\cdot \\sin \\left(- \\frac{\\pi}{3}\\right)$ $y = 3 \\cdot \\frac{- \\sqrt{3}}{2}$ $y = \\frac{- 3 \\sqrt{3}}{2}$ God bless...I hope the explanation is useful.", "\\left(\\frac{1}{3}\\frac{y \\cdot r}{x}\\cdot\\frac{y^2}{x^2} + \\frac{1}{7}\\frac{y \\cdot r}{x}\\cdot\\frac{y^2}{x^2}\\cdot\\frac{y^2}{x^2}\\cdot\\frac{y^2}{x^2}+\\quad \\cdots\\right) ### Transformation to current notation Let θ be the angle subtended by the arc s at the centre of the circle. Then s = rθ, x = kotijya = r cos θ and y = jya = r sin θ. Then y / x = tan θ. Substituting these in the last expression and simplifying we get • \\theta = \\tan \\theta - \\frac{\\tan^3 \\theta}{3} + \\frac{\\tan^5\\theta}{5} - \\frac{\\tan^7 \\theta}{7} + \\quad \\cdots . Letting tan θ = q we finally have • \\tan^{-1} q = q - \\frac{q^3}{3} + \\frac{q^5}{5} - \\frac{q^7}{7} + \\quad \\cdots ### Another formula for the circumference of a circle The second part of the quoted text specifies another formula for the computation of the circumference c of a circle having diameter d. This is as follows. c= \\sqrt{12 d^2} - \\frac{\\sqrt{12 d^2}}{3\\cdot 3} + \\frac{\\sqrt{12 d^2}}{3^2 \\cdot 5} - \\frac{\\sqrt{12 d^2}}{3^3 \\cdot 7}+ \\quad \\cdots Since c = π d this can be reformulated as a formula to compute π as follows. \\pi = \\sqrt{12}\\left( 1 - \\frac{1}{3\\cdot3}+\\frac{1}{3^2\\cdot 5} -\\frac{1}{3^3\\cdot 7} +\\quad \\cdots\\right) This is obtained by substituting q = 1/\\sqrt{3} (therefore θ = π / 6) in the power series expansion for tan−1 q above. Comparison of the convergence of two Madhava series", "note that due to the presence of similar words on the third document (“The sun in the sky is bright”), it achieved a better score. If you want, you can also solve the Cosine Similarity for the angle between vectors: $\\cos{\\theta} = \\frac{\\vec{a} \\cdot \\vec{b}}{\\|\\vec{a}\\|\\|\\vec{b}\\|}$ We only need to isolate the angle ($\\theta$) and move the $\\cos$ to the right hand of the equation: $\\theta = \\arccos{\\frac{\\vec{a} \\cdot \\vec{b}}{\\|\\vec{a}\\|\\|\\vec{b}\\|}}$ The $\\arccos$ is the same as the inverse of the cosine ($\\cos^-1$). Lets for instance, check the angle between the first and third documents: 123456 import math # This was already calculated on the previous step, so we just use the value cos_sim = 0.52305744 angle_in_radians = math.acos(cos_sim) print math.degrees(angle_in_radians) 58.462437107432784 And that angle of ~58.5 is the angle between the first and the third document of our document set. That is it, I hope you liked this third tutorial !" ]

[ "# Cos triple angle formula ### Expansion form $\\cos{3\\theta} \\,=\\, 4\\cos^3{\\theta}-3\\cos{\\theta}$ ### Simplified form $4\\cos^3{\\theta}-3\\cos{\\theta} \\,=\\, \\cos{3\\theta}$ ### Introduction It is called cos triple angle identity and used as a formula in two various cases. 1. Cos of triple angle is expanded as the subtraction of three times cos of angle from four times cos cubed of angle. 2. The subtraction of three times cos of angle from four times cos cubed of angle is simplified as cos of triple angle. #### How to use Cosine of triple angle identity is used to either expand or simplify the triple angle cos functions like $\\cos{3x}$, $\\cos{3A}$, $\\cos{3\\alpha}$ and etc. For example, $(1) \\,\\,\\,\\,\\,\\,$ $\\cos{3x} \\,=\\, 4\\cos^3{x}-3\\cos{x}$ $(2) \\,\\,\\,\\,\\,\\,$ $\\cos{3A} \\,=\\, 4\\cos^3{A}-3\\cos{A}$ $(3) \\,\\,\\,\\,\\,\\,$ $\\cos{3\\alpha} \\,=\\, 4\\cos^3{\\alpha}-3\\cos{\\alpha}$ #### Proof Learn how to derive the rule of cos triple angle identity by geometric approach in trigonometry.", "by: $cos (\\theta) = \\frac{D' \\cdot G'}{|D'||G'|}$ Consider the triangle C'G'B' this has the internal angle $\\theta /2$ so the distance from B' to C' is given by: $F=|G'| cos (\\theta /2)=|G'| cos (cos^{-1}(\\frac{D' \\cdot G'}{|D'||G'|}) /2)$ And finally the centre of the arc is located at: $C'=F\\frac{C\"}{|C\"|}$ or $C=F\\frac{C\"}{|C\"|}+B$", "$\\text{sin} (3\\theta) = 3 \\text{sin} \\theta - 4 \\text{sin}^3 \\theta$, the Triple Angle Formula for Cosine $\\text{cos} (3 \\theta) = -3 \\text{ cos} \\theta + 4 \\text{ cos}^3 \\theta$, and the Triple Angle Formula for Tangent $\\text{tan} (3 \\theta) = \\frac{3 \\text{ tan} \\theta - \\text{ tan}^3 \\theta}{1 - 3 \\text{ tan}^2 \\theta}$.", "##### Question 11.4.23 The triple angle formula states that $\\cos(3\\theta) = 4\\cos^3\\theta - 3\\cos\\theta.$ 1. We’re going to solve the cubic equation $x^3 = 3px + 2q$ for $x$ using the triple angle formula. If $x=2r\\cos\\theta$, find values of $r$ and $\\theta$ that will solve the cubic equation. (In Question 7.5.34 we were able to solve this equation when $q^2-p^3 \\ge 0$.) Show answer 2. What values of $p$ and $q$ will your solution work for? Show hint 3. Using your solution, find all values of $x$ that satisfy the equation $x^3 = 6x + 4$. Show hint", "## anonymous one year ago Quick check? Thank You for helping! Use the formula v1*v2=|v1||v2|cos theta to find the cosine of the angle between the two vectors (-7,8) and (5,3). I got cos theta=-11/61.97. Is this correct? Thanks! 1. anonymous @nincompoop @Michele_Laino @ganeshie8 @campbell_st @zepdrix @Luigi0210 @Destinymasha @radar @Whitemonsterbunny17 2. anonymous I also solved for theta, getting 100.22 degrees. 3. Michele_Laino we can write this: $\\left\\| {{v_1}} \\right\\| = \\sqrt {49 + 64} = 10.63$ $\\left\\| {{v_2}} \\right\\| = \\sqrt {25 + 9} = 5.83$ ${v_1} \\cdot {v_2} = - 35 + 24 = - 11$ so we get: $\\cos \\theta = \\frac{{ - 11}}{{10.63 \\times 5.83}}$ 4. anonymous so same thing, awesome! 5. anonymous Thanks for the confirmation! 6. Michele_Laino yes! $\\cos \\theta = \\frac{{ - 11}}{{10.63 \\times 5.83}} = - \\frac{{11}}{{61.97}}$ 7. Michele_Laino :)", "center at (2, 0) with $a=4$ and $c=2$ . This means the eccentricity is $e=\\frac{1}{2}$ . In order to write the equation in polar form you still need to find $k$ . $k=\\frac{a^2}{c}-c=\\frac{4^2}{2}-2=8-2=6$ Thus the equation for the ellipse is: $r=\\frac{6 \\cdot \\frac{1}{2}}{1-\\frac{1}{2} \\cdot \\cos(\\theta)}$ Example B Identify the center, foci, vertices and equations of the directrix lines for the following conic: $r=\\frac{20}{4-5 \\cdot \\cos\\left(\\theta-\\frac{3\\pi}{4}\\right)}$ Solution: First the polar equation needs to be in graphing form. This means that the denominator needs to look like $1-e \\cdot \\cos(\\theta-\\beta).$ $& r=\\frac{20}{4-5 \\cdot \\cos\\left(\\theta-\\frac{3\\pi}{4}\\right)} \\cdot \\frac{\\frac{1}{4}}{\\frac{1}{4}}=\\frac{5}{1-\\frac{5}{4}\\cdot \\cos\\left(\\theta-\\frac{3 \\pi}{4}\\right)}=\\frac{4 \\cdot \\frac{5}{4} }{1-\\frac{5}{4} \\cdot \\cos \\left(\\theta-\\frac{3 \\pi}{4}\\right)} \\\\& e=\\frac{5}{4}, \\quad k=4, \\quad \\beta=\\frac{3 \\pi}{4}=135^\\circ$ Using this information and the relationships you were reminded of in the guidance, you can set up a system and solve for $a$ and $c$ . $4 &=c-\\frac{a^2}{c} \\\\\\frac{5}{4} &=\\frac{c}{a} \\rightarrow \\frac{4}{5}=\\frac{a}{c} \\rightarrow \\frac{4c}{5}=a \\\\4&=c-\\left (\\frac{4c}{5}\\right)^2 \\cdot \\frac{1}{c} \\\\4&=c-\\frac{16c^2}{25c} \\\\4&=\\frac{9c}{25} \\\\\\frac{100}{9}&=c \\\\\\frac{80}{9}&=a \\\\$ The center is the point $\\left (\\frac{100}{9}, \\frac{7 \\pi}{4}\\right )$ which is much more convenient to write in polar coordinates. The closest directrix is the line $r=4 \\cdot \\sec\\left (\\theta - \\frac{7 \\pi}{4}\\right )$ . The other directrix is the line $r=\\left (2 \\cdot \\frac{100}{9}-4 \\right) \\cdot \\sec \\left (\\theta - \\frac{7 \\pi}{4}\\right)$ . One focus is at the pole, the other focus is the", "solved if it has only one side that has an exact value? x + xsin theta = 4 and xsin theta = 3 ? elimination? $x+x\\cos\\theta=4~and~x\\sin\\theta=3$ $3\\displaystyle\\frac{(1+\\cos\\theta)}{\\sin\\theta}=4$ $3\\displaystyle\\left(\\frac{2\\cos^2\\frac{\\theta}{2}} {2\\sin\\frac{\\theta}{2}\\cos\\frac{\\theta}{2}}\\right) =4$ $\\cot\\displaystyle\\frac{\\theta}{2}=\\frac{4}{3}$ $\\theta=36.87^o$ 7. thank you... i can solve now for the x side. then use other trig function to be able to find the longer side... 8. Code: E 6-b B 4-a 4 D a F 3-b A 3 C We only need to call on Big Pete Pythagoras: My points A and B are same as yours; the \"fold line\" is ED. I'm leaving left side of diagram out: making its length b, so EB = EA = 6-b; plus length of AF becomes 3-b. Let CD = a; then DB = DA = 4-a Right triangle AEF: 4^2 + (3-b)^2 = (6-b)^2 : leads to b = 11/6 Right triangle ACD: 3^2 + a^2 = (4-a)^2 : leads to a = 7/8 We can now use right triangle ADE (or BDE) to get fold line ED; using ADE: AE = 6-b = 25/6", "angle is unique in the polar form, you cannot have $z=r(\\cos \\theta + i \\sin \\varphi)$ with $\\theta \\neq \\varphi + 2k\\pi$ for any $k \\in \\mathbb{Z}$ – user289143 Oct 12 '19 at 21:32 • Ohh I see. Yeah I made a silly error. Thanks for spotting it. – Twigleg Oct 12 '19 at 21:39 • \"Do I need to somehow remove the 3 power? \" Eventually. but you can do it when it is easiest. There now reason to take the third root in rectangular form and then convert the third root to polar when you can just as legitimately convert the third power to polar first and then take the third root. Finding the third roots of complex numbers in polar form is ridiculously easy. Finding the third roots of complex numbers in rectangular form is ridiculously hard. – fleablood Oct 12 '19 at 22:40 Now $$z^3=8(\\cos \\frac{2\\pi}{3}+i \\sin \\frac{2 \\pi}{3})$$: using DeMoivre's formula we get $$z_0=\\sqrt[3]{8}\\left(\\cos \\left(\\frac{1}{3}\\cdot\\frac{2\\pi}{3}+\\frac{2 \\cdot 0 \\cdot \\pi}{3}\\right)+i \\sin \\left(\\frac{2 \\pi}{9}+\\frac{2 \\cdot 0 \\cdot \\pi}{3}\\right)\\right)=2\\left(\\cos \\frac{2\\pi}{9}+i \\sin \\frac{2 \\pi}{9}\\right)\\\\ z_1=\\sqrt[3]{8}\\left(\\cos \\left(\\frac{1}{3}\\cdot\\frac{2\\pi}{3}+\\frac{2 \\cdot 1 \\cdot \\pi}{3}\\right)+i \\sin \\left(\\frac{2 \\pi}{9}+\\frac{2 \\cdot 1 \\cdot \\pi}{3}\\right)\\right)=2\\left(\\cos \\frac{8\\pi}{9}+i \\sin \\frac{8 \\pi}{9}\\right)\\\\ z_2=\\sqrt[3]{8}\\left(\\cos \\left(\\frac{1}{3}\\cdot\\frac{2\\pi}{3}+\\frac{2 \\cdot 2 \\cdot \\pi}{3}\\right)+i \\sin \\left(\\frac{2 \\pi}{9}+\\frac{2 \\cdot 2 \\cdot \\pi}{3}\\right)\\right)=2\\left(\\cos \\frac{14\\pi}{9}+i \\sin \\frac{14 \\pi}{9}\\right)$$ • Ok, so the $z^3$ doesn't matter that it", "# How do you find the exact value of cos[arctan(-5/12)]? Notice that 5 and 12 are two sides of a right angled triangle whose hypotenuse is 13, since ${5}^{2} + {12}^{2} = 25 + 144 = 169 = {13}^{2}$ So if $\\theta$ is the smallest angle in the $5$,$12$,$13$ triangle then $\\sin \\theta = \\frac{5}{13}$, $\\cos \\theta = \\frac{12}{13}$ and $\\tan \\theta = \\frac{5}{12}$. Then $\\tan \\left(- \\theta\\right) = - \\tan \\left(\\theta\\right) = - \\frac{5}{12}$ So we are looking for $\\cos \\left(- \\theta\\right) = \\cos \\left(\\theta\\right) = \\frac{12}{13}$", "by \\alpha=3\\arccos(2/\\sqrt 5). Using the triple angle formula$$\\cos(3x)=\\cos^3(x)-3\\sin^2(x)\\cos(x)$$we find that$$\\begin{align} \\cos(\\alpha)&=\\cos(3\\arccos(2/\\sqrt 5))\\\\\\\\ &=\\cos^3(\\arccos(2/\\sqrt 5))-3\\sin^2(\\arccos(2/\\sqrt 5))\\cos(\\arccos(2/\\sqrt 5))\\\\\\\\ &=\\frac{8}{5\\sqrt 5}-3\\left(\\frac15\\right)\\left(\\frac{2}{\\sqrt 5}\\right)\\\\\\\\ &=\\frac{2}{5\\sqrt 5} \\end{align}$Therefore,$\\alpha = \\theta\\$ and we are done!", "are 1, 1, 2. If θ is the angle between them, they can be written as: $\\cos \\theta =\\vert \\frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\\cdot \\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\\vert$ $\\cos \\theta =\\vert \\frac{3\\cdot 1+5\\cdot 1+4\\cdot 2}{\\sqrt{3^{2}+5^{2}+4^{2}}\\cdot \\sqrt{1+1+2^{2}}}\\vert =\\frac{8\\sqrt{3}}{15}$ Hence we can find the required acute angle by taking $\\cos ^{-1}\\frac{8\\sqrt{3}}{15}$", "6. Originally Posted by Karl Harder I'm still stuck on how you get this using the double angle formula? $a^4\\cos^2{\\theta} \\sin^2{\\theta} \\, d\\theta = \\frac{a^4}{4} \\sin^2(2\\theta) \\, d\\theta$ never mind i have found out that i can get the answer by using the proudct formula 7. Originally Posted by Karl Harder I'm still stuck on how you get this using the double angle formula? $a^4\\cos^2{\\theta} \\sin^2{\\theta} \\, d\\theta = \\frac{a^4}{4} \\sin^2(2\\theta) \\, d\\theta$ ... never mind i have found out that i can get the answer by using the proudct formula product formula? ... $a^4 \\cos^2{\\theta} \\sin^2{\\theta}$ $\\frac{a^4}{4} \\cdot 4\\cos^2{\\theta} \\sin^2{\\theta}$ $\\frac{a^4}{4} (2\\cos{\\theta} \\sin{\\theta})^2$ $\\frac{a^4}{4} \\sin^2(2\\theta)$", "= 2\\cos\\theta\\ d\\theta$. $$V =2\\tau\\int_{\\arcsin(3/4)}^{\\tau / 4} \\sqrt{(2\\sin\\theta)^2 -4} \\cdot2\\cos\\theta\\ d\\theta$$ Integrate by factoring out 4 from squareroot, then use the Double-Angle Formula. $$V=4\\tau(\\theta + \\tfrac{1}{2}\\sin2\\theta)\\Big]_{\\arcsin(3/4)}^{\\tau/4}\\approx 11 \\neq \\frac{11}{24}\\tau$$ What did I do wrong? $$2 \\pi \\int_{3/2}^2 dx \\, y^2=2 \\pi \\int_{3/2}^2 dx \\, (4-x^2)$$ This evaluates to $2 \\pi (11/24)$.", "What are the rectangular coordinates of (3, -π/3)? $x = \\frac{3}{2}$ $y = \\frac{- 3 \\sqrt{3}}{2}$ Explanation: From the given: $r = 3$ and $\\theta = - \\frac{\\pi}{3}$ $x = r \\cdot \\cos \\theta$ $x = 3 \\cdot \\cos \\left(- \\frac{\\pi}{3}\\right)$ $x = 3 \\cdot \\left(\\frac{1}{2}\\right)$ $x = \\frac{3}{2}$ $y = r \\cdot \\sin \\theta$ $y = 3 \\cdot \\sin \\left(- \\frac{\\pi}{3}\\right)$ $y = 3 \\cdot \\frac{- \\sqrt{3}}{2}$ $y = \\frac{- 3 \\sqrt{3}}{2}$ God bless...I hope the explanation is useful.", "# Cos triple angle formula ### Expansion form $\\cos{3\\theta} \\,=\\, 4\\cos^3{\\theta}-3\\cos{\\theta}$ ### Simplified form $4\\cos^3{\\theta}-3\\cos{\\theta} \\,=\\, \\cos{3\\theta}$ ### Introduction It is called cos triple angle identity and used as a formula in two various cases. 1. Cos of triple angle is expanded as the subtraction of three times cos of angle from four times cos cubed of angle. 2. The subtraction of three times cos of angle from four times cos cubed of angle is simplified as cos of triple angle. #### How to use Cosine of triple angle identity is used to either expand or simplify the triple angle cos functions like $\\cos{3x}$, $\\cos{3A}$, $\\cos{3\\alpha}$ and etc. For example, $(1) \\,\\,\\,\\,\\,\\,$ $\\cos{3x} \\,=\\, 4\\cos^3{x}-3\\cos{x}$ $(2) \\,\\,\\,\\,\\,\\,$ $\\cos{3A} \\,=\\, 4\\cos^3{A}-3\\cos{A}$ $(3) \\,\\,\\,\\,\\,\\,$ $\\cos{3\\alpha} \\,=\\, 4\\cos^3{\\alpha}-3\\cos{\\alpha}$ #### Proof Learn how to derive the rule of cos triple angle identity by geometric approach in trigonometry. Latest Math Problems A best free mathematics education website for students, teachers and researchers. ###### Maths Topics Learn each topic of the mathematics easily with understandable proofs and visual animation graphics. ###### Maths Problems Learn how to solve the maths problems in different methods with understandable steps. Learn solutions ###### Subscribe us You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social", "# How do you convert (3, 2pi/3) into rectangular forms? polar form: $\\left(3 , \\frac{2 \\pi}{3}\\right) \\rightarrow$ rectangular form: $\\left(- \\frac{3}{2} , \\frac{3 \\sqrt{3}}{2}\\right)$ Given $\\left(r , \\theta\\right) = \\left(3 , \\frac{2 \\pi}{3}\\right)$ $x = r \\cdot \\cos \\left(\\theta\\right) = 3 \\cdot \\cos \\left(\\frac{2 \\pi}{3}\\right) = 3 \\cdot \\left(- \\frac{1}{2}\\right) = - \\frac{3}{2}$ $y = r \\cdot \\sin \\left(\\theta\\right) = 3 \\cdot \\sin \\left(\\frac{2 \\pi}{3}\\right) = 3 \\cdot \\left(\\frac{\\sqrt{3}}{2}\\right) = \\frac{3 \\sqrt{3}}{2}$", "# Double angle formula • Mar 11th 2012, 11:01 PM Hooperoo Double angle formula Hi ya'll. Quick question using double angle formula, verify that csc(2theta) - cot (2theta) = tan theta im so very confused. Working would be helpful to thanks :) • Mar 11th 2012, 11:19 PM princeps Re: Double angle formula Quote: Originally Posted by Hooperoo Hi ya'll. Quick question using double angle formula, verify that csc(2theta) - cot (2theta) = tan theta im so very confused. Working would be helpful to thanks :) $\\frac{1}{\\sin 2\\theta}-\\frac{\\cos 2\\theta}{\\sin 2\\theta}=\\frac{1-(cos^2 \\theta-\\sin^2 \\theta)}{2\\sin \\theta \\cdot \\cos \\theta}=$ $=\\frac{\\sin^2 \\theta + \\cos^2 \\theta -\\cos^2 \\theta +\\sin^2 \\theta}{2\\sin \\theta \\cdot \\cos \\theta}=$ $=\\frac{2\\sin^2 \\theta}{2\\sin \\theta \\cdot \\cos \\theta}=\\frac{\\sin \\theta}{\\cos \\theta}=\\tan \\theta$ • Mar 11th 2012, 11:37 PM Hooperoo Re: Double angle formula Dang. You make it seem so easy! I had an attempt at it, and i got kinda close. But thank you@! • Mar 12th 2012, 05:44 AM Soroban Re: Double angle formula Hello, Hooperoo! You are expected to know these identities: . . $\\sin2\\theta \\:=\\:2\\sin\\theta\\cos\\theta$ . . $\\sin^2\\!\\theta \\:=\\:\\frac{1-\\cos2\\theta}{2} \\quad\\Rightarrow\\quad 1 - \\cos2\\theta \\:=\\:2\\sin^2\\!\\theta$ Quote: Using double-angle formulas, verify that: . $\\csc2\\theta - \\cot2\\theta \\:=\\:\\tan\\theta$ We have: . $\\csc2\\theta - \\cot2\\theta \\;=\\;\\frac{1}{\\sin2\\theta} - \\frac{\\cos2\\theta}{\\sin2\\theta} \\;=\\;\\frac{1 - \\cos2\\theta}{\\sin2\\theta}$ . . . . . . . . . . . .", "# What is the angle between <7,-3,1> and < -1,3,2 >? Angl between above vctors ${119.2}^{0}$ Let vec u =[7, -3, 1] ; vec v=[-1,3,2] and $\\theta$ be the angle between them. We know $\\cos \\theta = \\frac{\\vec{u} . \\vec{v}}{| | \\vec{u} | | . | | \\vec{v} | |}$ Now $\\vec{u} . \\vec{v} = \\left(7 \\cdot - 1\\right) + \\left(- 3 \\cdot 3\\right) + \\left(1 \\cdot 2\\right) = - 14$ ; $| | \\vec{u} | | = \\sqrt{{7}^{2} + {\\left(- 3\\right)}^{2} + {1}^{2}} = \\sqrt{59}$ and $| | \\vec{v} | | = \\sqrt{{\\left(- 1\\right)}^{2} + {\\left(3\\right)}^{2} + {2}^{2}} = \\sqrt{14}$ $\\therefore \\cos \\theta = - \\frac{14}{\\sqrt{59} \\cdot \\sqrt{14}} = - .487 \\mathmr{and} \\theta = {\\cos}^{-} 1 \\left(- .487\\right) = {119.2}^{0}$[Ans]", "# What is the angle between <4 , 7 , -4 > and < 5 , 2 , 3 > ? $\\theta = {66.64}^{0}$ Let vecu =<4,7, -4> and vecv =<5,2,3> ; theta be the angle between them ; then we know cos theta= (vecu*vecv)/(||vecu||*||vecv||)=((4.5)+(7*2)+(-4*3))/(sqrt(4^2+7^2+(-4)^2)* (sqrt(5^2+2^2+3^2)) $= \\frac{22}{\\sqrt{81} \\cdot \\sqrt{38}} = \\frac{22}{9 \\cdot \\sqrt{38}} = 0.3965 \\therefore \\theta = {\\cos}^{-} 1 \\left(0.3965\\right) = {66.64}^{0}$[Ans]", "# How do you show that (costheta)(sectheta) = 1 if theta=pi/4? Feb 3, 2015 Let's look at the triangle: We'll use $\\angle A$ in stead of $\\theta$ because of the picture. Now $\\cos A = b / h$ and $\\sec A = h / b$ by definition. So their product is: $\\cos A \\cdot \\sec A = \\frac{b}{h} \\cdot \\frac{h}{b} = 1$ for all values of $A$ If we come closer to $A = \\frac{\\pi}{4}$ the value of $\\cos A \\to 0$ while the value of $\\sec A \\to \\infty$ As long as $A$ is a little bit smaller than $\\frac{\\pi}{4}$ then $\\cos A \\cdot \\sec A$ still come to $1$. We can get as close as we want, the answer is still $1$. Summary ${\\lim}_{A \\to \\pi / 4} \\cos A \\cdot \\sec A = 1$" ]

[ "above to construct$\\mathtt P = \\mathtt B\\operatorname{diag}(1,1,0)\\,\\mathtt B^{-1}$and choose for$\\mathtt B$the matrix$\\begin{bmatrix}\\mathbf u_1 & \\mathbf u_2 & \\mathbf s\\end{bmatrix}$, with the$\\mathbf u_i$as above, you’ll find that the final 3-D to 2-D transformation matrix is just the first two rows of$\\mathtt B^{-1}$. The expression in (*) gives you an explicit formula for them. I’ll leave it to you to determine whether computing the transformation with the explicit formula or inverting$\\mathtt B$is more efficient for your application. Given a plane through the origin with unit normal column vector$n$rather than project onto the plane directly we will instead project onto the space spanned by the normal vector and subtract that portion from the original point vector leaving us with a point in the plane. To do this we calculate first the three by three matrix$P = nn^T$where$n^T$is the transpose of$n$. This is a projection matrix so multiplying by P on the left of a vector will project onto the normal space. Now we take the point vector given by$v$and calculate it's projection onto the orthogonal space by computing$Pv = nn^Tv$. This is the part of the original vector we need to subtract off of the original vector$v$to get the component in the plane. This is simply$v-Pv\\$ which is the desired projection onto the plane. • You’ve described how to construct the", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "+0 # Matrices 0 80 1 The matrix for reflecting through a certain plane P which passes through the origin, is given by Find the normal vector of plane P. Aug 8, 2020 edited by Guest Aug 8, 2020", "# Finding an orthogonal projection matrix onto the plane Find the orthogonal projection matrix onto the plane $$x + y - z = 0$$ The solution to this video recitation video on MIT open courseware immediately states that we can chose $$a_1 = \\begin{pmatrix} 1\\\\ -1\\\\ 0 \\end{pmatrix} \\ \\ \\ \\ \\ \\ \\ \\ \\text{and} \\ \\ \\ \\ \\ \\ \\ a_2 = \\begin{pmatrix} 1\\\\ 0\\\\ 1 \\end{pmatrix}$$ So then $$A = \\begin{pmatrix} 1 & 1\\\\ -1 & 0 \\\\ 0 & 1 \\end{pmatrix}$$ and we simply apply the equation $P = A(A^{T}A)^{-1}A^{T}$. The lecturer did not explain how he can choose such an $a_1$ and $a_2$. Can anyone explain that? • I hope the lecture then goes on to show an easier way to compute this projection matrix by using the normal to the plane. – amd Jan 9 '18 at 21:33 • @amd how would you go about doing this? Oct 13 '21 at 2:23 The lecturer simply chose two vectors $a_{1}$ and $a_{2}$ that are independent and contained in the plane $x+y-z=0$. He then applied the formula that you mentioned. Well, you can choose any values al long as they satisfy the given equation. For example, $$x=0,y=1,z=1 \\rightarrow a_3 = \\begin{pmatrix}0\\\\1\\\\1\\end{pmatrix}$$ However, you should be careful to choose independent vectors. For instance,", "vector projection onto a plane I'm learning Linear Algebra on my own time. Currently studying projections. I understand all of the derivation, but there is one piece of it that bothers me. Say I have a matrix $A$ that represents a plane in $3D$, (matrix has 2 columns and 3 rows), $b$ is a vector I'm trying to project onto a plane, $p$ is its projection and $e$ (error) is the vector normal to the plane, $e = b - p$ Given that $e$ is normal to the plane then it is normal to every vector in the plane, meaning the inner product with any vector in the plane is 0, i.e. $A^Te = 0$, substituting $e = b - p$ we get $$A^T(b - p)= 0$$ $$A^Tb - A^Tp = 0$$ $$A^Tb = A^Tp$$ $$b = p$$ My derivation clearly shows that projection equals to the vector being projected, but geometrically it is not so. What am I missing? You claim that $(A^Tb = A^Tp) \\Rightarrow (b=p)$. This isn't true. E.g. let $A = \\begin{pmatrix}1 & 0\\\\ 0 & 1\\\\ 0 & 0\\end{pmatrix}$. This has two columns and three rows as required. Let $b = \\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix}, c = \\begin{pmatrix}1\\\\ 1\\\\ 2\\end{pmatrix}$. Clearly $b \\neq c$. However, $A^Tb = \\begin{pmatrix}1 & 0 & 0\\\\ 0", "+0 # help 0 114 1 Let $\\mathbf{P}$ be the matrix that projects onto $\\mathbf{j}$: that is, we want $\\mathbf{P}$ to satisfy $\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$for all vectors $\\mathbf{v}$. Use the pictures below to calculate $\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$in that order, and enter them in as columns below. Then calculate the matrix $\\mathbf{P}$ that projects onto $\\mathbf{j}$. Jul 31, 2019 The matrix is $$\\begin{pmatrix} -1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$.", "that the plane $\\mathcal{P}$ can be described as the set of all vectors $\\vec{v}$ solving the following equation: $b(\\hat{n},\\vec{v}) + d = 0,$ where $d=-b(\\hat{n},\\vec{p})$ can be interpreted as the minimum distance between the plane and the origin of the euclidean space. The following figure is once again taken from Wikipedia: ### Computing the normal vector of a plane If a plane is given by its parametric form, $\\mathcal{P} = \\vec{p} + s\\vec{u} + t\\vec{v}$, then the normal vector $\\hat{n}$ is the normalized outer product of the vectors $\\vec{u}$ and $\\vec{v}$, written as $\\vec{u} \\times \\vec{v}$, or $\\vec{u} \\wedge \\vec{v}$. As above, let $\\vec{e}_1, \\vec{e}_2$ and $\\vec{e}_3$ denote the standard basis of $\\mathbb{R}^3$, then the outer product can be computed as follows: \\begin{align*} \\tilde{n} &= \\vec{u} \\times \\vec{v} \\\\ &= \\left(\\sum\\limits_{i=1}^3(\\vec{u} \\times \\vec{e}_i) \\otimes \\vec{e}_i \\right) \\cdot \\vec{v} \\\\ &= \\begin{pmatrix}0 & -u_3 & u_2 \\\\ u_3 & 0 & u_1 \\\\ u_2 & -u_1 & 0\\end{pmatrix} \\cdot \\begin{pmatrix}v_1 \\\\ v_2 \\\\ v_3\\end{pmatrix} \\\\ &= \\begin{pmatrix} u_2v_3 - u_3v_2 \\\\ -u_1v_3 + u_3v_1 \\\\ u_1v_2 - u_2v_1 \\end{pmatrix}.\\end{align*}Or, if you don't like tensors, you can \"cheat\" a little bit: \\begin{align*} \\tilde{n} &= \\vec{u} \\times \\vec{v} \\\\ &=\\operatorname{det}\\begin{pmatrix}\\vec{e}_1&u_1&v_1\\\\ \\vec{e}_2&u_2&v_2\\\\ \\vec{e}_3&u_3&v_3\\end{pmatrix} \\\\ & = \\vec{e}_1 \\cdot \\operatorname{det}\\begin{pmatrix}u_2&v_2\\\\u_3&v_3\\end{pmatrix} - \\vec{e}_2 \\cdot \\operatorname{det}\\begin{pmatrix}u_1&v_1\\\\u_3&v_3\\end{pmatrix} + \\vec{e}_3 \\cdot \\operatorname{det}\\begin{pmatrix}u_1&v_1\\\\u_2&v_2\\end{pmatrix} \\\\ &= \\vec{e}_1 \\cdot (u_2v_3-u_3v_2)", "# Math Help - [SOLVED] Show that the following vectors are two different bases for the following pl 1. ## [SOLVED] Show that the following vectors are two different bases for the following pl Show that $\\begin{pmatrix} 4\\\\0\\\\1 \\end{pmatrix}$ , $\\begin{pmatrix} 2\\\\1\\\\0 \\end{pmatrix}$ and $\\begin{pmatrix} 2\\\\-1\\\\1 \\end{pmatrix}$ , $\\begin{pmatrix} 0\\\\2\\\\-1 \\end{pmatrix}$ are two different bases for the plane $x-2y-4z=0$ 2. Originally Posted by JCS007 Show that (4 0 1) $.^T$ , (2 1 0) $.^T$ and (2 -1 1) $.^T$ , (0 2 -1) $.^T$ are two different bases for the plane $x-2y-4z=0$ $V = a \\left ( \\begin{matrix} 4 \\\\ 0 \\\\ 1 \\end{matrix} \\right ) + b \\left ( \\begin{matrix} 2 \\\\ 1 \\\\ 0 \\end{matrix} \\right )$ Now show that $V_x - 2V_y - 4V_z = 0$ for all a and b.", "+0 # Consider the projection matrix P that satisfies 0 169 1 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}.$$ Use the picture below to find 3 vectors with integer components and magnitude no greater than 8 that satisfy Find $$\\mathbf{P}^{-1} \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$$$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}.$$ Aug 6, 2019", "\\\\ -1 \\end{pmatrix}$$ Then, the plane satisfies $$\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\begin{pmatrix} -1 \\\\ -1 \\\\ -1 \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix} \\begin{pmatrix} -1 \\\\ -1 \\\\ -1 \\end{pmatrix} = -2,$$ or that $- x - y - z = - 2$. - Take one of the points as origin and use the other two to obtain two (linearly independent? check!) vectors. Then write the vectorial equation for the plane as $$(x,y,z)=P+\\lambda \\vec u + \\mu \\vec v.$$ -", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "always, a picture is worth a thousand words: if you can glean the meaning of the directional derivative from this .gif I found online then you may not need to read what I have to say. I hope the animation helps either way. It's worth noting that the little green segment at the base of the figure in this animation is meant to represent the value of the gradient of the function at the given point. Let $\\mathbf{v}\\in\\mathbb{R}^{2}$ be a unit vector, considered as a vector in the $(x,y)$-plane in $\\mathbb{R}^{3}.$ The vector $\\mathbf{v}$ determines a unique plane $\\Pi = \\Pi(\\mathbf{v})$ which contains the origin, the point $(0,0,1)$ and $\\mathbf{v}$ itself. For example, if $\\mathbf{v}=\\begin{pmatrix}1\\\\0\\end{pmatrix}$ then the plane $\\Pi$ is the $y=0$ plane, i.e., the $(x,z)$-plane. (Draw a sketch to make sure you are comfortable with this idea; if you think I have explained it poorly, please ask.) We want to consider this same plane, but \"going through $\\mathbf{x}$\", that is, we want to consider the plane $\\mathbf{x}+\\Pi = \\{\\mathbf{x} + \\mathbf{p}:\\mathbf{p}\\in\\Pi\\}.$ All we've done is taken the plane $\\Pi,$ which went through the origin, and physically moved it so that the point on $\\Pi$ which was previously at $(0,0,0)$ is now at $\\mathbf{x}.$ Consider the directional derivative of $f\\colon\\mathbb{R}^{2}\\to\\mathbb{R}$ at $\\mathbf{x}\\in\\mathbb{R}^{2}$ in the direction of $\\mathbf{v}.$ Its", "the origin with the normal vector $\\vc{n} = (0, 1, 0)$. Furthermore, assume the sun is infinitely far away from the origin, positioned in the direction $\\vc{r} = (0.5,1.0, 0.25)$. We have a lot of points above the plane, and would like to know their shadows on the plane. Assume also that we have been reassured that this particular type of shadow projection can be expressed as a linear mapping. It would be possible to calculate the shadow point by point, but it would be more convenient if we could create a linear mapping $\\vc{y} =\\mx{A}\\vc{x}$ so that we could directly get the projected point $\\vc{y}$ by simply multiplying the point $\\vc{x}$ by the matrix $\\mx{A}$. This should be possible, given that the problem is linear. In this case, we take advantage of Theorem 9.2, i.e., to obtain $\\mx{A}$ we only need to know what happens to the three unit vectors. The first column vector of the transformation matrix $\\mx{A}$ is simply the image of the first unit vector $\\vc{e_1} = (1,0,0)$, and so forth. Hence, if we find out where on the plane the shadow of the point $(1,0,0)$ lands, we have the first column of $\\mx{A}$. We can write the plane on the form $ax+by+cz+d = 0$, and since we know the normal is $(0, 1,", "# 01. Matrices ## Matrices ### Vectors in the plane A vector is an arrow in the plane (later we'll deal with vectors in higher-dimensional spaces). We encode this arrow as a pair of numbers $\\begin{pmatrix}x\\\\ y\\end{pmatrix}$ . The number $x$ tells us how far to the right the arrow points; the number $y$ tells us how far upwards it points. If the arrow points to the left then $x$ is negative; if it points downwards then $y$ is negative. Can you match up the vectors $\\begin{pmatrix}x\\\\ y\\end{pmatrix}$ with those in the diagram? (Some of the vectors are not depicted). $\\begin{pmatrix}1\\\\ 2\\end{pmatrix},\\qquad\\begin{pmatrix}0\\\\ 1\\end{pmatrix},\\qquad\\begin{pmatrix}1\\\\ 1\\end{pmatrix},$ $\\begin{pmatrix}1\\\\ 0\\end{pmatrix},\\qquad\\begin{pmatrix}-1\\\\ -1\\end{pmatrix},\\qquad\\begin{pmatrix}2\\\\ 1\\end{pmatrix}.$ A lot of this module will focus on the interplay between algebra (like column vectors) and geometry (like arrows in the plane). Suppose $v=\\begin{pmatrix}x\\\\ y\\end{pmatrix}$ is a vector. What is it's length? By Pythagoras's theorem, it's $\\sqrt{x^{2}+y^{2}}$ . I'll write this as $|v|$ , which you can read out loud as \"norm $v$ \". The angle that $v$ makes with the horizontal is $\\theta=\\arctan(y/x)$ (by trigonometry). If we want to write $x$ and $y$ in terms of $|v|$ and $\\theta$ , we get (again, using trigonometry): $x=|v|\\cos\\theta,\\qquad y=|v|\\sin\\theta.$ So $v=\\begin{pmatrix}|v|\\cos\\theta\\\\ |v|\\sin\\theta\\end{pmatrix}$ . ### 2-by-2 matrices What happens if I rotate $v$ by an angle $\\phi$ anticlockwise? We get a new", "+0 # HELP ASAP PLEEASSE!!! 0 89 1 Let $$\\mathbf{A}$$ be a matrix such that $$\\mathbf{A}^{-1} = \\begin{pmatrix} 1 & 1 \\\\ 2 & x \\end{pmatrix}$$ for some value of $$x$$. What is the vector that $$\\mathbf{A}$$ maps to $$\\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}?$$ Oct 13, 2019 The answer is $$\\begin{pmatrix} 3 \\\\ -7 \\end{pmatrix}$$.", "Review question # Which point $P$ is on both this line and this plane? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R6006 ## Question The line $l$ has the equation $\\mathbf{r}= \\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}+ \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}, \\hspace{1cm} \\lambda \\in \\mathbb{R}.$ 1. Show that $l$ lies in the plane whose equation is $\\mathbf{r}.\\begin{pmatrix}-1\\\\ 2\\\\ 0\\end{pmatrix}=-5.$ 2. Find the position vector of $L$, the foot of the perpendicular from the origin $O$ to $l$. 3. Find an equation of the plane containing $O$ and $l$. 4. Find the position vector of the point $P$ where $l$ meets the plane $\\pi$ whose equation is $\\mathbf{r}.\\begin{pmatrix}1\\\\ 2\\\\ 2\\end{pmatrix} = 11.$", "# Vectors# Author(s): Translation: ## Definition# The term vector has several interpretations: mathematical, geometric, physical, etc. The exact meaning of the term depends on the context. Formally, a vector is defined as an element of vector space – a set with addition and multiplication of a vector by a number (scalar) operations, which satisfy 8 axioms. For simplicity, let us consider the rectangular (Cartesian) coordinate system in the plane, familiar from school days – two perpendicular to each other axes $$x$$ and $$y$$, selected unit vectors (orthogonal) $$\\mathbf{e}_1$$, $$\\mathbf{e}_2$$ on them and the origin of coordinates. The vector $$\\mathbf{a}$$ in this coordinate system can be written as follows: $$\\mathbf{a} = a_1 \\mathbf{e}_1 + a_2 \\mathbf{e}_2 = \\begin{pmatrix} a_1 \\\\ a_2 \\end{pmatrix}$$ Note 1: Vector coordinates do not uniquely define its position in the plane, but only the position of the vector’s end relative to its start. For example, a vector $$\\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix}$$ can be drawn either from the origin $$\\begin{pmatrix} 0 , 0 \\end{pmatrix}$$ with the end at $$\\begin{pmatrix} 3 , 4 \\end{pmatrix}$$, or from an arbitrary point, e.g. $$\\begin{pmatrix} 1 , 1 \\end{pmatrix}$$ with the end at $$\\begin{pmatrix} 4 , 5 \\end{pmatrix}$$. Both of these vectors correspond to 3 orthogonal unite vectors on the $$x$$-axis and 4 on the $$y$$-axis. Normally, unless otherwise stated,", "# 9.2: Building Subspaces Consider the set $U= \\left\\{ \\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix} \\right\\} \\subset \\Re^{3}.$ Because $$U$$ consists of only two vectors, it clear that $$U$$ is $$\\textit{not}$$ a vector space, since any constant multiple of these vectors should also be in $$U$$. For example, the $$0$$-vector is not in $$U$$, nor is $$U$$ closed under vector addition. But we know that any two vectors define a plane: In this case, the vectors in $$U$$ define the $$xy$$-plane in $$\\Re^{3}$$. We can view the $$xy$$-plane as the set of all vectors that arise as a linear combination of the two vectors in $$U$$. We call this set of all linear combinations the $$\\textit{span}$$ of $$U$$: $span(U)=\\left\\{ x \\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}+y \\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix} \\middle| x,y\\in \\Re \\right\\}.$ Notice that any vector in the $$xy$$-plane is of the form $\\begin{pmatrix}x\\\\y\\\\0\\end{pmatrix} = x \\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}+y \\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix} \\in span(U).$ Definition Let $$V$$ be a vector space and $$S=\\{ s_{1}, s_{2}, \\ldots \\} \\subset V$$ a subset of $$V$$. Then the $$\\textit{span of S}$$ is the set: $span(S):=\\{ r^{1}s_{1}+r^{2}s_{2}+\\cdots + r^{N}s_{N} | r^{i}\\in \\Re, N\\in \\mathbb{N} \\}.$ That is, the span of $$S$$ is the set of all finite linear combinations (usually our vector spaces are defined over $$\\mathbb{R}$$, but in general we can have vector spaces defined over different base fields such as $$\\mathbb{C}$$ or $$\\mathbb{Z}_{2}$$. The", "when: \\begin{equation*} A = \\begin{pmatrix} 1 & -1 \\\\ 0 & 0 \\\\ 0 & 0 \\end{pmatrix} \\qquad \\text{and } x = \\begin{pmatrix} 2 \\\\ 0 \\end{pmatrix}. \\end{equation*} (Strang ex. 4.1.22) Let $P$ be the hyperplane of vectors in $\\mathbb{R}^4$ which satisfy the equation \\begin{equation*} x_1 + x_2 + x_3 + x_4 = 0 .\\end{equation*} Write down a basis for $P^{\\perp}$. Construct a matrix $X$ which has $P$ as its nullspace.", "# Finding matrix from plane in kernal Give an example of a matrix A such that $$\\ker(A)$$ is the plane $$2x − y + 3z = 0$$. I am not sure where to start, as I know that the $$\\ker(A)$$ is the matrix of the plane, but I don't know how to go backwards from it. Would I need to assume that there's a free variable in the rref(A) and assume it was pulled out of the equation of the plane? Let $$A=\\begin{pmatrix} 2 & -1 & 3\\\\ 0 & 0 & 0\\\\ 0 & 0 & 0 \\end{pmatrix}$$ Then $$A\\begin{pmatrix} x\\\\ y\\\\ z\\end{pmatrix}=0$$ implies $$2x-y+3z=0$$ • @amd You are right. I am just used to considering linear transformations in $n$-dimensional vector space $V$: $V\\to V$, so my matrix is a square matrix. – Bach May 21 at 3:55" ]

[ "did was to follow this great tutorial, which has the desired syntax on the bottom of p. 60. The striking difference to you code is that there the normal of the 2D arrow gets specified, too, and that it is actually +Z (and not -Z). \\documentclass[border=3.14mm]{standalone} \\usepackage{asypictureB} \\begin{document} \\begin{asypicture}{name=arc} settings.outformat = \"png\"; settings.render = 16; size(6cm,0); import solids; currentlight=Headlamp; // inutile car Headlamp est la valeur par defaut nslice=4*nslice; revolution boule=sphere(O,1); draw(surface(boule),lightgrey+white+opacity(.5)); draw((0,0,1)--(0,0,-1),dashed); dot(\"$P1$\",(0,0,1),N); dot(\"$P2$\",(0,0,-1),2.2S); dot(\"$O$\",(0,0,0),SE); draw(O--(1,0,0),red); draw(O--(Cos(60),Sin(60),0),red); triple v1= 0.4*(1,0,0), v2 = 0.4*(Cos(60),Sin(60),0); //draw(arc(c=O,v1,v2),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); for (real alpha = 90; alpha<=450; alpha+= 30) { if (alpha==150) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red); if (alpha==330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red + dashed); if (alpha != 150 && alpha != 330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),black); }; draw(arc(c=O, v2, v1, normal=-Z), red, arrow = Arrow3(TeXHead2(normal=Z), emissive(red)),L=Label(\"$\\varphi$\",align=S, position=MidPoint)); \\end{asypicture} \\end{document} • Tank you Marmot for your anwer – user273366 Nov 24 '18 at 7:45 • indeed, the normal parameter was missing in the arrow function – user273366 Nov 24 '18 at 7:47", "label(\"A'\", AA, N); [/asy]$ From what we are given it easily follows that $ABC$ is an equilateral triangle, and its center $O$ is at the same time the foot of the height from $D$. We know that $AD=5$ and $OD=4$, from the Pythagorean theorem it follows that $OA=3$. The segment $OA$ is $2/3$ of the segment $AA'$. Then $AA'$ has length $9/2$. $AA'$ is the height in the triangle $ABC$, hence we have $AA' = \\dfrac{BC\\cdot \\sqrt 3}2$, and therefore $BC=3\\sqrt 3$. We can now compute the area of the triangle $ABC$ as $\\dfrac{BC\\cdot AA'}2 = \\dfrac{27\\sqrt 3}{4}$, and finally the volume of the tetrahedron $ABCD$ as $\\dfrac{ABC\\cdot DO}3 = 9\\sqrt 3$. Why did we compute all this stuff? Consider what happens when the leg $BD$ breaks in the described place $E$: import three; import math; size(8cm,0); currentprojection=obliqueX; triple O=(0,0,0), A=(3,0,0), B=rotate(120,Z)*A, C=rotate(-120,Z)*A, D=(0,0,4); triple BB=0.8*B + 0.2*D; triple AC=(A+C)/2; filldraw(A--BB--C--cycle, lightgray, black ); draw(A--B--C--cycle); draw(O--D); draw(A--D, black+1); draw(BB--D, black+1); draw(B--BB); draw(C--D, black+1); draw(A--O, dashed); draw(B--AC, dashed); draw(BB--AC, dashed); draw(C--O, dashed); label(\"$A$\", A, W); label(\"$B$\", B, E); label(\"$E$\", BB, E); label(\"$C$\", C, W); label(\"$O$\", O, SE); label(\"$D$\", D, Z); label(\"$B'$\", AC, W); (Error compiling LaTeX. dcd0140709d472b7550064c4363f05c4c983ec6a.asy: 11.9: no matching function 'filldraw(void(flatguide3), pen, pen)') The triangle $AEC$ will now be the base of the tripod, and our goal is", "# I can not draw arc in a clockwise direction with this program in asymptote I have to trace the meridians. settings.outformat = \"png\"; settings.render = 16; size(6cm,0); import solids; currentlight=Headlamp; // inutile car Headlamp est la valeur par défaut nslice=4*nslice; revolution boule=sphere(O,1); draw(surface(boule),lightgrey+white+opacity(.5)); draw((0,0,1)--(0,0,-1),dashed); dot(\"$P1$\",(0,0,1),N); dot(\"$P2$\",(0,0,-1),2.2S); dot(\"$O$\",(0,0,0),SE); draw(O--(1,0,0),red); draw(O--(Cos(60),Sin(60),0),red); triple v1= 0.4(1,0,0), v2 = 0.4(Cos(60),Sin(60),0); draw(arc(c=O,v1,v2,normal=Z),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); for (real alpha = 90; alpha<=450; alpha+= 30) { if (alpha==150) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red); if (alpha==330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red + dashed); if (alpha != 150 && alpha != 330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),black); } But I can not draw arc in a clockwise direction with this code (this is the critic point!): draw(arc(c=O,v1,v2,normal=Z),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); I tried: 1. to change v1 and v2 order with the same normal (Z), but in this way i have the angle greater than ninety degrees (i need the explementar angle) 2. to change v1 and v2 order and take normal = - Z, but in this way asymptote runs on a infinite loop. I don't know why. ## 1 Answer It took me a while to understand the question since IMHO it is not very clearly written. An additional complication arises from the fact that you did not present an MWE and seem to insist to have a special character in a comment: \"défaut\". All I", "# Error with 3D Asymptote graphs Help me with 3D arc in asymptote. Thank you very much! Code: \\begin{asy} settings.outformat=\"pdf\"; import three; size(7.5cm,0); currentprojection=perspective(20,18,8); currentlight=light(8,10,2); path3 p=(-1,-1,0)--(1,-1,0)--(1,1,0)--(-1,1,0)--cycle; draw(p); path3 g=(1,0,0)..(0,1,1)..(-1,0,0)..(0,-1,1)..cycle; draw(g,2bp+.8red); draw(surface(g),paleblue); dot(g,5bp+.8green); draw((0,0,0)--normal(p),1bp+blue,Arrow3()); \\end{asy} I compiled \\documentclass[border=3.14mm]{standalone} \\usepackage{asymptote} \\begin{document} \\begin{asy} settings.outformat=\"pdf\"; import three; size(7.5cm,0); currentprojection=perspective(20,18,8); currentlight=light(8,10,2); path3 p=(-1,-1,0)--(1,-1,0)--(1,1,0)--(-1,1,0)--cycle; draw(p); path3 g=(1,0,0)..(0,1,1)..(-1,0,0)..(0,-1,1)..cycle; draw(g,2bp+.8red); draw(surface(g),paleblue); dot(g,5bp+.8green); draw((0,0,0)--normal(p),1bp+blue,Arrow3()); \\end{asy} \\end{document} with the usual chain pdflatex, then asy and again pdflatex, I get a flawless However, I prefer to use asypictureB, and when I compile \\documentclass[border=3.14mm]{standalone} \\usepackage{asypictureB} \\begin{document} \\begin{asypicture}{name=pfft} settings.outformat=\"pdf\"; import three; size(7.5cm,0); currentprojection=perspective(20,18,8); currentlight=light(8,10,2); path3 p=(-1,-1,0)--(1,-1,0)--(1,1,0)--(-1,1,0)--cycle; draw(p); path3 g=(1,0,0)..(0,1,1)..(-1,0,0)..(0,-1,1)..cycle; draw(g,2bp+.8red); draw(surface(g),paleblue); dot(g,5bp+.8green); draw((0,0,0)--normal(p),1bp+blue,Arrow3()); \\end{asypicture} \\end{document} with pdflatex -shell-escape I also get this result. Please check what you get with these standard methods. If you get the same, please describe in detail what you do, i.e. provide us with a complete example, which starts with \\documentclass and ends with \\end{document} that allows us to reproduce the issue. Full codes: \\documentclass[border=5mm]{standalone} \\usepackage{asymptote} \\begin{document} \\begin{asy} settings.outformat=\"pdf\"; import three; size(7.5cm,0); currentprojection=perspective(20,18,8); currentlight=light(8,10,2); path3 p=(-1,-1,0)--(1,-1,0)--(1,1,0)--(-1,1,0)--cycle; draw(p); path3 g=(1,0,0)..(0,1,1)..(-1,0,0)..(0,-1,1)..cycle; draw(g,2bp+.8red); draw(surface(g),paleblue); dot(g,5bp+.8green); draw((0,0,0)--normal(p),1bp+blue,Arrow3()); \\end{asy} \\end{document} And error pdf: • Is this now an answer or an addendum to your question? – user121799 Nov 18 '18 at 3:48 • I still haven't fixed the bug. Maybe because of my computer's graphics card. – Lê Đình Mẫn Nov", "# Difference between revisions of \"2013 AMC 8 Problems/Problem 18\" ## Problem Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain? $[asy]import three; currentprojection=orthographic(-8,15,15); triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A = (0,0,0); B = (0,10,0); C = (12,10,0); D = (12,0,0); E = (0,0,5); F = (0,10,5); G = (12,10,5); H = (12,0,5); I = (1,1,1); J = (1,9,1); K = (11,9,1); L = (11,1,1); M = (1,1,5); N = (1,9,5); O = (11,9,5); P = (11,1,5); //outside box far draw(surface(A--B--C--D--cycle),white,nolight); draw(A--B--C--D--cycle); draw(surface(E--A--D--H--cycle),white,nolight); draw(E--A--D--H--cycle); draw(surface(D--C--G--H--cycle),white,nolight); draw(D--C--G--H--cycle); //inside box far draw(surface(I--J--K--L--cycle),white,nolight); draw(I--J--K--L--cycle); draw(surface(I--L--P--M--cycle),white,nolight); draw(I--L--P--M--cycle); draw(surface(L--K--O--P--cycle),white,nolight); draw(L--K--O--P--cycle); //inside box near draw(surface(I--J--N--M--cycle),white,nolight); draw(I--J--N--M--cycle); draw(surface(J--K--O--N--cycle),white,nolight); draw(J--K--O--N--cycle); //outside box near draw(surface(A--B--F--E--cycle),white,nolight); draw(A--B--F--E--cycle); draw(surface(B--C--G--F--cycle),white,nolight); draw(B--C--G--F--cycle); //top draw(surface(E--H--P--M--cycle),white,nolight); draw(surface(E--M--N--F--cycle),white,nolight); draw(surface(F--N--O--G--cycle),white,nolight); draw(surface(O--G--H--P--cycle),white,nolight); draw(M--N--O--P--cycle); draw(E--F--G--H--cycle); label(\"10\",(A--B),SE); label(\"12\",(C--B),SW); label(\"5\",(F--B),W);[/asy]$ $\\textbf{(A)}\\ 204 \\qquad \\textbf{(B)}\\ 280 \\qquad \\textbf{(C)}\\ 320 \\qquad \\textbf{(D)}\\ 340 \\qquad \\textbf{(E)}\\ 600$ ## Solution 1 There are $10 \\cdot 12 = 120$ cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are $9 + 11", "return (x,y,z); }; } triple F(pair z){ return f(1,1,1)(z); } // Gauss map triple g(pair z) { real u=z.x, v=z.y; real x,y,z; x=exp(u)*sin(v); y=exp(u)*cos(v); z=sqrt(1+exp(2*u)); return -7*(x,y,-1)/z; } // https://trecs.se/hyperboloidOfOneSheet.php path3 vector(pair z) { real u=z.x, v=z.y; real a=1, b=1, c=1; real x,y,z; x=-b*c*exp(u)*sin(v); y=-a*c*exp(u)*cos(v); z=a*b*sqrt(1+exp(2*u)); return O--(-(x,y,-1)/z); } size(13cm); surface sf=surface(F,(-10,-12),(3,6),40,Spline); //sf.colors(palette(sf.map(zpart),Rainbow())); draw(sf,RGB(0,153,216),render(merge=true)); //xaxis3(Label(\"$x$\",BeginPoint),0,7,Arrow3); draw(Label(\"$x$\",EndPoint),(12,0,0)--(-12,0,0),Arrow3); yaxis3(Label(\"$y$\",EndPoint),0,12,Arrow3); zaxis3(Label(\"$z$\",EndPoint),-18,22,Arrow3); label(\"$\\mathcal{H}$\",(-.5,5.5,18),dir(45)); transform3 t=shift(20*(Y-X)); surface sg=surface(g,(-2.5,-5),(5,5),20,Spline); //sg.colors(palette(sg.map(zpart),Rainbow())); draw(t*sg,RGB(236, 125, 44),render(merge=true)); label(\"$N\\left(\\mathcal{H} \\right)$\",t*(-.5,6,6),dir(45)); draw(Label(\"$x$\",EndPoint),t*((0,0,0)--(12,0,0)),Arrow3); draw(Label(\"$y$\",EndPoint),t*((0,0,0)--(0,12,0)),Arrow3); draw(Label(\"$z$\",EndPoint),t*((0,0,0)--(0,0,13)),Arrow3); //dot(Label(\"$(0,0,1)$\",black),t*(0,0,7),dir(135),white+5bp); draw(Label(\"$N$\",Relative(.5),N),subpath((0,0,0)--t*(0,0,0),0.4,0.6),Arrow3); output: • You wrote \" where the sphere on the right is the unit sphere x^2+y^2=1)\", I think x^2+y^2+z^2=1''. I do not understand two number sqrt(2) and -sqrt(2) on the z - axis. Nov 25, 2021 at 1:10 • @minhthien_2016 Thank you very much if you are right it was a typing error. Regarding z=sqrt(2) and z=-sqrt(2), what I tried to do is: A unitary sphere intersected with two planes, and the part between said planes is painted. (I'll edit my question to clarify that topic in more detail.) Nov 25, 2021 at 1:17 • I think, If radius of the sphere equal to R=1, the the plane perpendicular to z - axis has the equation z = h, where -1<h<1. Nov 25, 2021 at 1:19 • I already edited it, I hope it is better understood. Nov 25, 2021 at 1:23 • you", "+0 # help 0 45 1 Find the projection of $$\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ onto the plane $$3x - y + 4z = 0.$$ Sep 2, 2019 #1 +23086 +2 Find the projection of $$\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ onto the plane $$3x - y + 4z = 0$$. $$\\text{Let \\vec{P}= \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}}$$ 1. Normal vector $$\\mathbf{\\vec{n}}$$ $$\\begin{array}{|rcll|} \\hline \\vec{n} &=& \\begin{pmatrix} 3 \\\\ -1 \\\\ 4 \\end{pmatrix} \\quad | \\quad \\text{plane: }{\\color{red}3}x {\\color{red}-1}y + {\\color{red}4}z = 0 \\\\ \\hline \\end{array}$$ 2. $$\\mathbf{\\vec{P}_{Proj.}}$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{\\vec{P}_{Proj.}} &=& \\mathbf{\\vec{P}+\\lambda\\vec{n}} \\\\\\\\ \\vec{P}_{Proj.} &=& \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}+\\lambda\\begin{pmatrix} 3 \\\\ -1 \\\\ 4 \\end{pmatrix} \\\\\\\\ x_{Proj.} &=& 1+3\\lambda \\\\ y_{Proj.} &=& 2-\\lambda \\\\ z_{Proj.} &=& 3+4\\lambda \\\\ \\hline \\end{array}$$ 3. $$\\mathbf{\\lambda}$$ $$\\begin{array}{|rcll|} \\hline 3x_{Proj.} - y_{Proj.} + 4z_{Proj.} &=& 0 \\\\ 3(1+3\\lambda) - (2-\\lambda) + 4(3+4\\lambda) &=& 0 \\\\ 3+9\\lambda-2+\\lambda+12+16\\lambda &=& 0 \\\\ 26\\lambda &=& -13 \\\\\\\\ \\lambda &=& -\\dfrac{13}{26} \\\\\\\\ \\mathbf{\\lambda} &=& -\\mathbf{\\dfrac{1}{2}} \\\\ \\hline \\end{array}$$ 4. $$\\mathbf{\\vec{P}_{Proj.}}$$ $$\\begin{array}{|rcll|} \\hline \\vec{P}_{Proj.} &=& \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}+\\lambda\\begin{pmatrix} 3 \\\\ -1 \\\\ 4 \\end{pmatrix} \\quad | \\quad \\lambda = -\\dfrac{1}{2} \\\\\\\\ &=& \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix} -\\dfrac{1}{2}\\begin{pmatrix} 3 \\\\ -1 \\\\ 4 \\end{pmatrix} \\\\\\\\ &=& \\begin{pmatrix} 1-\\dfrac{3}{2} \\\\\\\\ 2+\\dfrac{1}{2} \\\\\\\\ 3-\\dfrac{4}{2} \\end{pmatrix} \\\\\\\\", "1/2 \\end{bmatrix}$ • thus $\\mathbf p = A \\mathbf{ \\hat x } = \\begin{bmatrix} p_1 \\\\ p_2 \\\\ p_3 \\end{bmatrix} = \\begin{bmatrix} 7/6 \\\\ 5/3 \\\\ 13/6 \\end{bmatrix}$ • $\\mathbf p + \\mathbf e = \\mathbf b$, so $\\mathbf e = \\mathbf b - \\mathbf p = \\begin{bmatrix} 1 - 7/6 \\\\ 2 - 5/3 \\\\ 2 - 13/6 \\end{bmatrix} = \\begin{bmatrix} - 1/6 \\\\ 2/3 \\\\ -1/6 \\end{bmatrix}$ • $\\mathbf p \\; \\bot \\; \\mathbf e$ $\\Rightarrow$ $\\mathbf p^T \\mathbf e = 0$. • Check: $\\begin{bmatrix} 7/6 & 5/3 & 13/6 \\end{bmatrix} \\begin{bmatrix} - 1/6 \\\\ 2/3 \\\\ -1/6 \\end{bmatrix} = - 7/6 \\cdot 1/6 + 5/3 \\cdot 2/3 - 13/6 \\cdot 1/6 = 0$ We can also verify that $\\mathbf e \\; \\bot \\; C(A)$ • let's take one vector from $C(A)$, e.g. $\\mathbf 1 = [1, 1, 1]^T \\in C(A)$, • $\\mathbf e^T \\cdot \\mathbf 1 = -1/6 + 2/6 - 1/6 = 0$ Python code import matplotlib.pylab as plt import numpy as np class Line: def __init__(self, slope, intercept): self.slope = slope self.intercept = intercept def calculate(self, x1): x2 = x1 * self.slope + self.intercept return x2 A = np.array([[1, 1], [1, 2], [1, 3]]) b = np.array([1, 2, 2]) x0, x1 = np.linalg.solve(A.T.dot(A), A.T.dot(b)) lsq = Line(x1, x0) # figure plt.scatter(A[:, 1], b,", "r'$\\bf u_2$', 'u2', size=16, color='k') fig.plotLine([origin, u3], 'r', '--') fig.plotPoint(u3[0], u3[1], u3[2], 'r') fig.text(u3[0]+.1, u3[1]+.1, u3[2]+.1, r'$\\bf u_3$', 'u3', size=16, color = 'k') fig.text(origin[0]-.45, origin[1]-.45, origin[2]-.45, r'$\\bf 0$', 0, size = 16) fig.plotPerpSym(origin, u1, u2, 0.5) fig.plotPerpSym(origin, u3, u2, 0.5) fig.plotPerpSym(origin, u3, u1, 0.5) fig.set_title(r'An orthogonal set of vectors in $\\mathbb{R}^3$', 'An orthogonal set of vectors in R3', size = 16) fig.save(); Then, express the vector $$\\mathbf{y} = \\begin{bmatrix}6\\\\1\\\\-8\\end{bmatrix}$$ as a linear combination of the vectors in $$S$$. That is, find $$\\mathbf{y}$$’s coordinates in the basis $$S$$ — i.e., in the coordinate system $$S$$. Solution. Compute $\\mathbf{y}^T\\mathbf{u}_1 = 11,\\;\\;\\;\\mathbf{y}^T\\mathbf{u}_2 = -12,\\;\\;\\;\\mathbf{y}^T\\mathbf{u}_3 = -33,$ $\\mathbf{u}_1^T\\mathbf{u}_1 = 11,\\;\\;\\;\\mathbf{u}_2^T\\mathbf{u}_2 = 6,\\;\\;\\;\\mathbf{u}_3^T\\mathbf{u}_3 = 33/2$ So $\\mathbf{y} = \\frac{\\mathbf{y}^T\\mathbf{u}_1}{\\mathbf{u}_1^T\\mathbf{u}_1}\\mathbf{u}_1 + \\frac{\\mathbf{y}^T\\mathbf{u}_2}{\\mathbf{u}_2^T\\mathbf{u}_2}\\mathbf{u}_2 + \\frac{\\mathbf{y}^T\\mathbf{u}_3}{\\mathbf{u}_3^T\\mathbf{u}_3}\\mathbf{u}_3$ $= \\frac{11}{11}\\mathbf{u}_1 + \\frac{-12}{6}\\mathbf{u}_2 + \\frac{-33}{33/2}\\mathbf{u}_3$ $= \\mathbf{u}_1 - 2\\mathbf{u}_2 - 2 \\mathbf{u}_3.$ fig = ut.three_d_figure((21, 4), fig_desc = 'y in an orthogonal basis', xmin = -3, xmax = 7, ymin = -5, ymax = 5, zmin = -8, zmax = 4, figsize = (12, 8), qr = qr_setting, equalAxes = False) u1 = np.array([3, 1, 1]) u2 = np.array([-1, 2, 1]) u3 = np.array([-1/2, -2, 7/2]) origin = np.array([0, 0, 0]) # fig.plotLine([origin, u1], 'r', '--') fig.plotPoint(u1[0], u1[1], u1[2], 'r') fig.text(u1[0]+.1, u1[1]+.1, u1[2]+.1, r'$\\bf u_1$', 'u1', size=16, color='k') # fig.plotLine([origin, u2], 'r', '--') fig.plotPoint(u2[0], u2[1], u2[2], 'r') fig.text(u2[0]+.1, u2[1]+.1,", "line $PD.$ $[asy] import geometry; import olympiad; size(400); point A = (2, 5), B = (0, 0), C = (10, 0), P = (3, 2); line c = line(A, B); line a = line(C, B); line b = line(A, C); point D = projection(a) * P; draw(P -- D); point e = projection(b) * P; draw(P -- e); point F = projection(c) * P; draw(P -- F); draw(A--B--C--A); draw(P--A); draw(P--B); draw(P--C); markrightangle(B, D, P); markrightangle(A, e, P); markrightangle(A, F, P); draw(circumcircle(A, P, e), dashed); line d = line(P, D); draw(d); point n = projection(d) * F; point M = projection(d) * e; draw(F--n); draw(e--M); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, SW); label(\"E\", e - (0.1, 0), NE); label(\"F\", F, W); label(\"P\", P+(0.2, 0), S); label(\"N\", n, E); label(\"M\", M, W); [/asy]$ Note that $AFPE$ is cyclic with diameter $AP.$ By ELOS, $\\dfrac{EF}{\\sin A} = 2R=AP\\implies EF = AP\\sin A.$ Since $BDPF$ is cyclic, we have that $B$ and $\\angle FPD$ are supplementary. Since $MPD$ is a line, $B = \\angle FPM.$ This means that $\\sin B = \\sin FPN = \\dfrac{FN}{FP}\\implies FN = PF\\sin B.$ Similarly, $EM = PE\\sin C.$ So the problem is reduced to proving that $EF\\ge FN+EM$ but this is obvious by the Pythagorean Theorem. $\\blacksquare$ Now the rest of", "( )... Show the work ( ax+by+cz ) ∣​=a2+b2+c2​∣ax0​+by0​+cz0​+d∣​.​ vectors can also be represented in component form x_0-x\\\\y_0-y\\\\z_0-z\\end { pmatrix x_0-x\\\\y_0-y\\\\z_0-z\\end! ( −112 ).\\mathbf { w } = \\begin { pmatrix } -1 & 1 & 2\\end { }!, so 5x is equivalent to 5 * x ( 3−2 ) 2+ 5−7! } 52+a2a​=82=64=±23​ the plane is __________.\\text { \\_\\_\\_\\_\\_\\_\\_\\_\\_\\_ }.__________.\\mathbf { w } =\\begin { pmatrix } -1 1. ) ( 0,0,0 ), the distance formula to the length of the perpendicular from point a line and to...: d = wikis and quizzes in Math, science, and engineering topics: Math science... −1,2,0 ) from the origin w } = \\begin { pmatrix }.w=⎝⎛​x0​−xy0​−yz0​−z​⎠⎞​ distance can be derived and in! 3D space x0−xy0−yz0−z ).\\mathbf { w } =\\begin { pmatrix } x_0-x\\\\y_0-y\\\\z_0-z\\end { pmatrix -1... } = \\begin { pmatrix } -1 & 1 & 2\\end { pmatrix }.w= ( −1​1​2​ ) to... Equivalent to 5 * x is equivalent to 5 * x...., or ( 0,0,0 ), the distance d betw… 3D lines: between! Dimensions, the formula for calculating it can be written as under: d = point a to (! ) 2=1+81+4=86 }.w=⎝⎛​x0​−xy0​−yz0​−z​⎠⎞​ & 1 & 2\\end { pmatrix }.w=⎝⎛​x0​−xy0​−yz0​−z​⎠⎞​ is equivalent to 5 * x.... With a three-dimensional arrow get the best experience in component", "the mean vectors, i.e., $$\\mathbf{\\mu}_1 = \\pmatrix{5.01\\\\3.42}^T \\qquad \\mathbf{\\mu}_2 = \\pmatrix{6.26\\\\2.87}^T \\qquad \\mathbf{\\mu}_1 - \\mathbf{\\mu}_2= \\pmatrix{-1.256\\\\0.546}^T$$ should be $$\\mathbf{\\mu}_1 = \\pmatrix{5.01\\\\3.42} \\qquad \\mathbf{\\mu}_2 = \\pmatrix{6.26\\\\2.87} \\qquad \\mathbf{\\mu}_1 - \\mathbf{\\mu}_2 = \\pmatrix{-1.256\\\\0.546}$$ • p509, Example 20.4, line 4: \"iris-virginica\" should be \"$${\\tt Iris\\text{-}versicolor}$$\" • p512, Q1: In part (a) \"$$\\mathbf{S}_B$$\" should be \"$$\\mathbf{B}$$\", and in (b) \"$$\\mathbf{S}_W$$\" should be \"$$\\mathbf{S}$$\" Chapter 21: Support Vector Machines • p526, 7th line, in $$L_{dual}$$: \"$$(C - \\alpha_i + \\beta_i)$$\" should be \"$$(C - \\alpha_i - \\beta_i)$$\" • p536, Algorithm 21.1, line 15: \"$$\\mathbf{\\alpha}_{t+1} = \\alpha$$\" should be \"$$\\alpha_{t+1} \\gets \\alpha$$\" • p538, Example 21.8, line 5: \"homogeneous quadratic kernel $$K(\\mathbf{x}_i,\\mathbf{x}_j) = ( \\mathbf{x}^T_i \\mathbf{x}_j)^2$$\" should be \"inhomogeneous quadratic kernel $$K(\\mathbf{x}_i,\\mathbf{x}_j) = (1+ \\mathbf{x}^T_i \\mathbf{x}_j)^2$$\"", "# 2010 AIME I Problems/Problem 11 ## Problem Let $\\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \\le 10$ and $3y - x \\ge 15$. When $\\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$, the volume of the resulting solid is $\\frac {m\\pi}{n\\sqrt {p}}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m + n + p$. ## Solution $[asy]size(280); import graph; real min = 2, max = 12; pen dark = linewidth(1); real P(real x) { return x/3 + 5; } real Q(real x) { return 10 - abs(x - 8); } path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12)); pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(0.9,0.9,0.9)); draw(graph(P,min,max),dark); draw(graph(Q,min,max),dark); draw(Arc((8,7.67),A,G,CW),dark,EndArrow(8)); draw(B--C--G--cycle,linetype(\"4 4\")); label(\"y \\ge x/3 + 5\",(max,P(max)),E,fontsize(10)); label(\"y \\le 10 - |x-8|\",(max,Q(max)),E,fontsize(10)); label(\"\\mathcal{R}\",(6,Q(6)),NW); /* axes */ Label f; f.p=fontsize(8); xaxis(0, max, Ticks(f, 6, 1)); yaxis(0, 10, Ticks(f, 5, 1)); [/asy]$ The inequalities are equivalent to $y \\ge x/3 + 5, y \\le 10 - |x - 8|$. We can set them equal to find the two points of intersection, $x/3 + 5 = 10 -", "# Difference between revisions of \"2012 AMC 12B Problems/Problem 19\" ## Problem 19 A unit cube has vertices $P_1,P_2,P_3,P_4,P_1',P_2',P_3',$ and $P_4'$. Vertices $P_2$, $P_3$, and $P_4$ are adjacent to $P_1$, and for $1\\le i\\le 4,$ vertices $P_i$ and $P_i'$ are opposite to each other. A regular octahedron has one vertex in each of the segments $P_1P_2$, $P_1P_3$, $P_1P_4$, $P_1'P_2'$, $P_1'P_3'$, and $P_1'P_4'$. What is the octahedron's side length? $[asy] import three; size(7.5cm); triple eye = (-4, -8, 3); currentprojection = perspective(eye); triple[] P = {(1, -1, -1), (-1, -1, -1), (-1, 1, -1), (-1, -1, 1), (1, -1, -1)}; // P[0] = P[4] for convenience triple[] Pp = {-P[0], -P[1], -P[2], -P[3], -P[4]}; // draw octahedron triple pt(int k){ return (3*P[k] + P[1])/4; } triple ptp(int k){ return (3*Pp[k] + Pp[1])/4; } draw(pt(2)--pt(3)--pt(4)--cycle, gray(0.6)); draw(ptp(2)--pt(3)--ptp(4)--cycle, gray(0.6)); draw(ptp(2)--pt(4), gray(0.6)); draw(pt(2)--ptp(4), gray(0.6)); draw(pt(4)--ptp(3)--pt(2), gray(0.6) + linetype(\"4 4\")); draw(ptp(4)--ptp(3)--ptp(2), gray(0.6) + linetype(\"4 4\")); // draw cube for(int i = 0; i < 4; ++i){ draw(P[1]--P[i]); draw(Pp[1]--Pp[i]); for(int j = 0; j < 4; ++j){ if(i == 1 || j == 1 || i == j) continue; draw(P[i]--Pp[j]); draw(Pp[i]--P[j]); } dot(P[i]); dot(Pp[i]); dot(pt(i)); dot(ptp(i)); } label(\"P_1\", P[1], dir(P[1])); label(\"P_2\", P[2], dir(P[2])); label(\"P_3\", P[3], dir(-45)); label(\"P_4\", P[4], dir(P[4])); label(\"P'_1\", Pp[1], dir(Pp[1])); label(\"P'_2\", Pp[2], dir(Pp[2])); label(\"P'_3\", Pp[3], dir(-100)); label(\"P'_4\", Pp[4], dir(Pp[4])); [/asy]$ $\\textbf{(A)}\\ \\frac{3\\sqrt{2}}{4}\\qquad\\textbf{(B)}\\", "setting $$\\mathbf{V}= \\mathbf{v} \\in \\mathbb{R}^m$$ so then, $$\\mathbf{P}_{v}= \\mathbf{v} \\frac{1}{\\mathbf{v}^T \\mathbf{v}} \\mathbf{v}^T$$ which matches our previous result. The point of all these manipulations is that we can construct another projection operator with all the MMSE properties we had before, but now in a bigger subspace. We can further verify the idempotent property of projection matrices by checking that $$\\mathbf{P}_V \\mathbf{P}_V = \\mathbf{P}_V$$ The following graphic shows that when we project the three dimensional $\\mathbf{y}$ vector onto the plane, which is spanned by the two $\\mathbf{v}_i$ vectors, we obtain the MMSE solution where the sphere is tangent to the plane. The point of tangency is the point $\\hat{\\mathbf{y}}$ which is the MMSE solution. In [2]: #http://stackoverflow.com/questions/10374930/matplotlib-annotating-a-3d-scatter-plot from mpl_toolkits.mplot3d import proj3d import matplotlib.pyplot as plt import numpy as np fig = plt.figure() fig.set_size_inches([8,8]) ax.set_aspect(1) ax.set_xlim([0,2]) ax.set_ylim([0,2]) ax.set_zlim([0,2]) ax.set_aspect(1) ax.set_xlabel('x-axis',fontsize=16) ax.set_ylabel('y-axis',fontsize=16) ax.set_zlabel('z-axis',fontsize=16) y = matrix([1,1,1]).T V = matrix([[1,0.25], # columns are v_1, v_2 [0,0.50], [0,0.00]]) alpha=inv(V.T*V)*V.T*y # optimal coefficients P = V*inv(V.T*V)*V.T yhat = P*y # approximant u = np.linspace(0, 2*np.pi, 100) v = np.linspace(0, np.pi, 100) xx = np.outer(np.cos(u), np.sin(v)) yy = np.outer(np.sin(u), np.sin(v)) zz = np.outer(np.ones(np.size(u)), np.cos(v)) sphere=ax.plot_surface(xx+y[0,0], yy+y[1,0], zz+y[2,0], rstride=4, cstride=4, color='gray',alpha=0.3,lw=0.25) ax.plot3D([y[0,0],0],[y[1,0],0],[y[2,0],0],'r-',lw=3) ax.plot3D([y[0,0]],[y[1,0]],[y[2,0]],'ro') ax.plot3D([V[0,0],0],[V[1,0],0],[V[2,0],0],'b-',lw=3) ax.plot3D([V[0,0]],[V[1,0]],[V[2,0]],'bo') ax.plot3D([V[0,1],0],[V[1,1],0],[V[2,1],0],'b-',lw=3) ax.plot3D([V[0,1]],[V[1,1]],[V[2,1]],'bo') ax.plot3D([yhat[0,0],0],[yhat[1,0],0],[yhat[2,0],0],'g--',lw=3) ax.plot3D([yhat[0,0]],[yhat[1,0]],[yhat[2,0]],'go') x2, y2, _ = proj3d.proj_transform(y[0,0],y[1,0],y[2,0], ax.get_proj()) ax.annotate( \"$\\mathbf{y}$\", xy = (x2, y2), xytext = (-20, 20), fontsize=24, textcoords", "size(10cm); triple eye = (5,2,3); currentprojection=orthographic(eye); surface hemisphere = surface(Arc(X,-X,c=O,normal=Z,n=16), c=O, axis=X, angle1=0, angle2=180); draw(shift(-5 eye)*hemisphere, material(white + opacity(0.5), emissivepen=0.2 white)); usepackage(\"amsmath\"); //for \\text command draw(O -- 1.6Z, arrow=Arrow3, L=Label(\"$H_{\\text{eff}}$\", align=W, position=EndPoint)); real theta(real t) { return t/20; } real phi(real t) { return -t + 1; } real r(real t) { return 1; } triple F(real t) { return polar(r(t), theta(t), phi(t)); } path3 spiral = reverse(graph(F, 0, 4pi, operator ..)); draw(spiral, blue + dotted + linewidth(1pt)); real t = 0.1; triple arrowpos = point(spiral, reltime(spiral, t)); t = 0.7; arrowpos = point(spiral, reltime(spiral, t)); triple M = point(spiral, 0); draw(O -- M, arrow=Arrow3, L=Label(\"$M$\", position=MidPoint)); draw(shift(M) * (O -- 0.5 cross(-M, Z)), red, arrow=Arrow3, L=Label(\"$-M \\times H_{\\text{eff}}$\", position=MidPoint)); triple dMdt = dir(spiral,0); triple crossprod = cross(M, dMdt); draw(shift(M) * (O -- 0.5 crossprod), blue, arrow=Arrow3, L=Label(\"$M \\times \\frac{dM}{dt}$\", position=MidPoint)); \\end{asypicture} \\end{document} The result: - Thanks! Exactly what I was looking for; sadly the LaTeX integration is not quite as smooth as with pgfplots. – knedlsepp Apr 30 '14 at 18:29 There are several packages for integrating Asymptote into LaTeX; if you state what features you would imagine for \"smooth integration,\" I can tell you whether any of these packages currently supports those features. – Charles Staats Apr 30 '14 at 22:55 I guess there's nothing one", "It is still not an exact drawing, but, I would guess, that if you know how to mathematically construct the curved lines that form the spiral, it should be possible to come up with an exact drawing based on this solution. Oct 26 '21 at 19:48 While waiting for a Tikz's answer, see a bit with Asymptote. Compile on http://asymptote.ualberta.ca/ I don't know what \\alpha and \\gamma mean! settings.render=8; import solids; import graph3; usepackage(\"newtxmath\"); size(200,400); real r=3; real h=2.5pi; currentprojection=orthographic(8,2,4); revolution R=cylinder(O,r,h); // The circular helix triple f(real t){ real a=r*cos(t); real b=r*sin(t); real c=t; return (a,b,c); } real k=7; path3 plane=(k,k,0)--(k,-k,0)--(-k,-k,0)--(-k,k,0)--cycle; draw(plane); label(\"$\\alpha$\",(k,-k,0),2dir(20)); draw(surface(R),lightgreen+opacity(0.5),render(compression=Low)); draw(surface(circle((0,0,h),r)),lightgreen+opacity(0.5),render(compression=Low)); draw(O--(0,0,h)^^O--(0,r,0)^^O--(r,0,0),dashed); dot(scale(.7)*\"$O$\",(0,0,0),dir(165)); dot((0,0,h)); guide3 g=graph(f,0,h,150); draw(Label(\"$\\varphi$\",Relative(.05)),g); draw(Label(\"$\\varphi$\",Relative(.05)),g); triple P=f(1),overP=f(1+2pi); triple P1=(P.x,P.y,0),P2=(0,0,P.z); draw(overP--P1); draw(P2--P^^O--P1,dashed); dot(scale(.7)*\"$P$\",P,2dir(120)); dot(scale(.7)*\"$\\overline{P}$\",overP,dir(-20)); dot(scale(.7)*\"$P_1$\",P1,dir(-90)); dot(scale(.7)*\"$P_2$\",P2,dir(150)); xaxis3(Label(\"$x^1$\",Relative(.5),2LeftSide),r,6,Arrow3); yaxis3(Label(\"$x^2$\",Relative(.5),4dir(20)),r,6,Arrow3); zaxis3(Label(\"$x^3=$ a\",Relative(.6),2RightSide),h,h+4,Arrow3); • thanks a lot Nguyen VanChi 1998. i noticed that working with asymptote make everything easier and more elegant. Can i ask if you can add in the drow (sorry i'm not practicle with the code) the \\gamma and \\varphi? and last question: is it possible to have classicthesis font for the letters and symbols in the drow? this is not necessary but in case it would be wonderful (far above my expectations) thanks again a lot :D Oct 26 '21 at 17:09 • Nice! and h should be 2.5pi Oct", "so one would need to find a tikz way for the advanced arrows. Or you can use a loop together with tikz drawing instructions (same approach as the one of @troy. – Christian Feuersänger Nov 15 '14 at 20:00 With Asymptote it is possible to draw a 3D vector field along a surface (not a path). It is not difficult to adapt this routine to draw a 3D vector field along a path. However the sophisticated arrow is not availabe and needs more work. Please find a example import graph3; size(200,0); currentprojection=perspective(10,8,4); real f(pair z) {return 0.5+exp(-abs(z)^2);} triple F(pair z){ return (z.x,z.y,f(z));} static real dx=sqrtEpsilon, dy=dx; return O--(-(f(z+dx)-f(z-dx))/2dx, -(f(z+I*dy)-f(z-I*dy))/2dy, 1); } draw((-1,-1,0)--(1,-1,0)--(1,1,0)--(-1,1,0)--cycle); surface s=surface(f,(-1,-1),(1,1),nx=5,Spline); xaxis3(Label(\"$x$\"),red,Arrow3); yaxis3(Label(\"$y$\"),red,Arrow3); zaxis3(XYZero(extend=true),red,Arrow3); draw(s,lightgray,meshpen=black+thick(),nolight,render(merge=true)); label(\"$O$\",O,-Z+Y,red); And the result At last, what about a Python/Matplotlib/Numpy/Scipy solution (which can generate a movie) ? Edit 11/17/2014. I tried to modify the vectorfield function of Asymptote and included special arrow. Because I do not know how depend your path and your vector field, the following routine draws a vector field along a curve, the vector field drawn on f(t) depends on the (f(x),f(y)). For the special arrow I do not create a new Arrow3 in the Asymptote sense, the sphere is added in the vectorfield routine import graph3; real maxilength(triple f(real z), real a, real b, int nu)", "= \\begin{bmatrix} \\mathbf{D_v} & 0.5*\\mathbf{D_v} & -1*\\mathbf{D_h} \\\\ \\mathbf{D_h} & 2*\\mathbf{D_h} & \\mathbf{D_v} \\\\ \\end{bmatrix}, \\qquad \\mathbf{y} = \\begin{bmatrix} \\mathbf{D_v} \\mathbf{x_1} + 0.5*\\mathbf{D_v} \\mathbf{x_2} - \\mathbf{D_h} \\mathbf{x_3} \\\\ \\mathbf{D_h} \\mathbf{x_1} + 2*\\mathbf{D_h} \\mathbf{x_2} + \\mathbf{D_v} \\mathbf{x_3} \\end{bmatrix}\\end{split}$ Bop = pylops.Block([[D2vop, 0.5 * D2vop, -1 * D2hop], [D2hop, 2 * D2hop, D2vop]]) Y = np.reshape(Bop * X.ravel(), (2 * Nv, Nh)) fig, axs = plt.subplots(1, 2, figsize=(10, 3)) fig.suptitle(\"Block\", fontsize=14, fontweight=\"bold\", y=0.95) im = axs[0].imshow(X, interpolation=\"nearest\") axs[0].axis(\"tight\") axs[0].set_title(r\"$x$\") plt.colorbar(im, ax=axs[0]) im = axs[1].imshow(Y, interpolation=\"nearest\") axs[1].axis(\"tight\") axs[1].set_title(r\"$y$\") plt.colorbar(im, ax=axs[1]) plt.tight_layout() Finally we can use the block-diagonal operator to apply three operators to three different subset of the model and data $\\begin{split}\\mathbf{D_{BDiag}} = \\begin{bmatrix} \\mathbf{D_v} & \\mathbf{0} & \\mathbf{0} \\\\ \\mathbf{0} & 0.5*\\mathbf{D_v} & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{0} & -\\mathbf{D_h} \\end{bmatrix}, \\qquad \\mathbf{y} = \\begin{bmatrix} \\mathbf{D_v} \\mathbf{x_1} \\\\ 0.5*\\mathbf{D_v} \\mathbf{x_2} \\\\ -\\mathbf{D_h} \\mathbf{x_3} \\end{bmatrix}\\end{split}$ BD = pylops.BlockDiag([D2vop, 0.5 * D2vop, -1 * D2hop]) Y = np.reshape(BD * X.ravel(), (3 * Nv, Nh)) fig, axs = plt.subplots(1, 2, figsize=(10, 3)) fig.suptitle(\"Block-diagonal\", fontsize=14, fontweight=\"bold\", y=0.95) im = axs[0].imshow(X, interpolation=\"nearest\") axs[0].axis(\"tight\") axs[0].set_title(r\"$x$\") plt.colorbar(im, ax=axs[0]) im = axs[1].imshow(Y, interpolation=\"nearest\") axs[1].axis(\"tight\") axs[1].set_title(r\"$y$\") plt.colorbar(im, ax=axs[1]) plt.tight_layout() If we consider now the case of having a large number of operators inside a blockdiagonal structure, it may be convenient to span multiple processes handling subset of operators at", "currentprojection=orthographic(50,-60,0.4); currentlight=(5,-3,1); real f(pair z){return (1.-0.3)*exp(-z.x*(z.y/100.)*(1.-0.3))-exp(-z.x*(z.y/100.));} real cero(pair z){return 0.;} limits((0,0,-0.3),(30,30,0.1)); xaxis3(Label(\"$x$\",1),blue,arrow=Arrow3); yaxis3(Label(\"$y$\",1),blue,arrow=Arrow3); zaxis3(Label(\"$z$\",1),blue,arrow=Arrow3); surface s=surface(f,(0,0),(30,30),30,30,Spline); surface s2=surface(cero,(0,0),(30,30),30,30,Spline); s.colors(palette(s.map(zpart),Rainbow()+opacity(0.5))); s2.colors(palette(s2.map(zpart),Rainbow()+opacity(0.5))); draw(s,meshpen=blue); draw(s2,meshpen=red); shipout(bbox(2mm,Fill(white))); Solid surfaces settings.outformat=\"png\"; settings.render=16; import graph3; import palette; size3(150,IgnoreAspect); currentprojection=orthographic(50,-60,0.4); //currentlight=(-5,3,20); real f(pair z){return (1.-0.3)*exp(-z.x*(z.y/100.)*(1.-0.3))-exp(-z.x*(z.y/100.));} real cero(pair z){return 0.;} limits((0,0,-0.3),(30,30,0.1)); xaxis3(Label(\"$x$\",1),blue,arrow=Arrow3); yaxis3(Label(\"$y$\",1),blue,arrow=Arrow3); zaxis3(Label(\"$z$\",1),blue,arrow=Arrow3); surface s=surface(f,(0,0),(30,30),30,30,Spline); surface s2=surface(cero,(0,0),(30,30),30,30,Spline); //s.colors(palette(s.map(zpart),Rainbow+opacity(0.5))); //s2.colors(palette(s2.map(zpart),Rainbow()+opacity(0.5))); draw(s, blue,meshpen=blue); draw(s2, red, meshpen=red); shipout(bbox(2mm,Fill(white))); PGFplots style (still with Asymptote) This is the closest I was able to make it look like a pgfplots plot. (Things to do, help is welcomed: 1) automatic point of view (to not depend on the scale of the plot, 2) thick border around each surface 3) vector graphics and hidden surface (e.g. via bsd module) 4) details, like small tics and tics spacing) settings.outformat=\"png\"; settings.render=8; import grid3; import graph3; import palette; size3(160,125,90,IgnoreAspect); currentprojection=orthographic(30,-60,0.8); real f(pair z){return (1.-0.3)*exp(-z.x*(z.y/100.)*(1.-0.3))-exp(-z.x*(z.y/100.));} real cero(pair z){return 0.;} limits((0,0,-0.3),(30,30,0.1)); xaxis3(Bounds(Min,Min), InTicks()); xaxis3(Bounds(Max,Max)); xaxis3(Bounds(Max,Min)); yaxis3(Bounds(Max,Min), InTicks(beginlabel=false) ); yaxis3(Bounds(Min,Max)); yaxis3(Bounds(Min,Min)); zaxis3(Bounds(Min,Min), InTicks() ); zaxis3(Bounds(Max,Max)); zaxis3(Bounds(Min,Max)); surface s=surface(f,(0,0),(30,30),24,24); surface s2=surface(cero,(0,0),(30,30),24,24); draw(s2, red, meshpen=0.2pt + black, nolight); draw(s2, red, meshpen=0.2pt + black, nolight); shipout(bbox(2mm,Fill(white))); Notes asymptote can output vector graphics, but the output is not as good as a raster (rendered) output and removal of hidden faces is very difficult (like with pgfplots.) So, this is big drawback of using asymptote, the plot becomes raster. However, it is common that" ]

[ "above to construct$\\mathtt P = \\mathtt B\\operatorname{diag}(1,1,0)\\,\\mathtt B^{-1}$and choose for$\\mathtt B$the matrix$\\begin{bmatrix}\\mathbf u_1 & \\mathbf u_2 & \\mathbf s\\end{bmatrix}$, with the$\\mathbf u_i$as above, you’ll find that the final 3-D to 2-D transformation matrix is just the first two rows of$\\mathtt B^{-1}$. The expression in (*) gives you an explicit formula for them. I’ll leave it to you to determine whether computing the transformation with the explicit formula or inverting$\\mathtt B$is more efficient for your application. Given a plane through the origin with unit normal column vector$n$rather than project onto the plane directly we will instead project onto the space spanned by the normal vector and subtract that portion from the original point vector leaving us with a point in the plane. To do this we calculate first the three by three matrix$P = nn^T$where$n^T$is the transpose of$n$. This is a projection matrix so multiplying by P on the left of a vector will project onto the normal space. Now we take the point vector given by$v$and calculate it's projection onto the orthogonal space by computing$Pv = nn^Tv$. This is the part of the original vector we need to subtract off of the original vector$v$to get the component in the plane. This is simply$v-Pv\\$ which is the desired projection onto the plane. • You’ve described how to construct the", "+0 # help 0 45 1 Find the projection of $$\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ onto the plane $$3x - y + 4z = 0.$$ Sep 2, 2019 #1 +23086 +2 Find the projection of $$\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ onto the plane $$3x - y + 4z = 0$$. $$\\text{Let \\vec{P}= \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}}$$ 1. Normal vector $$\\mathbf{\\vec{n}}$$ $$\\begin{array}{|rcll|} \\hline \\vec{n} &=& \\begin{pmatrix} 3 \\\\ -1 \\\\ 4 \\end{pmatrix} \\quad | \\quad \\text{plane: }{\\color{red}3}x {\\color{red}-1}y + {\\color{red}4}z = 0 \\\\ \\hline \\end{array}$$ 2. $$\\mathbf{\\vec{P}_{Proj.}}$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{\\vec{P}_{Proj.}} &=& \\mathbf{\\vec{P}+\\lambda\\vec{n}} \\\\\\\\ \\vec{P}_{Proj.} &=& \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}+\\lambda\\begin{pmatrix} 3 \\\\ -1 \\\\ 4 \\end{pmatrix} \\\\\\\\ x_{Proj.} &=& 1+3\\lambda \\\\ y_{Proj.} &=& 2-\\lambda \\\\ z_{Proj.} &=& 3+4\\lambda \\\\ \\hline \\end{array}$$ 3. $$\\mathbf{\\lambda}$$ $$\\begin{array}{|rcll|} \\hline 3x_{Proj.} - y_{Proj.} + 4z_{Proj.} &=& 0 \\\\ 3(1+3\\lambda) - (2-\\lambda) + 4(3+4\\lambda) &=& 0 \\\\ 3+9\\lambda-2+\\lambda+12+16\\lambda &=& 0 \\\\ 26\\lambda &=& -13 \\\\\\\\ \\lambda &=& -\\dfrac{13}{26} \\\\\\\\ \\mathbf{\\lambda} &=& -\\mathbf{\\dfrac{1}{2}} \\\\ \\hline \\end{array}$$ 4. $$\\mathbf{\\vec{P}_{Proj.}}$$ $$\\begin{array}{|rcll|} \\hline \\vec{P}_{Proj.} &=& \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}+\\lambda\\begin{pmatrix} 3 \\\\ -1 \\\\ 4 \\end{pmatrix} \\quad | \\quad \\lambda = -\\dfrac{1}{2} \\\\\\\\ &=& \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix} -\\dfrac{1}{2}\\begin{pmatrix} 3 \\\\ -1 \\\\ 4 \\end{pmatrix} \\\\\\\\ &=& \\begin{pmatrix} 1-\\dfrac{3}{2} \\\\\\\\ 2+\\dfrac{1}{2} \\\\\\\\ 3-\\dfrac{4}{2} \\end{pmatrix} \\\\\\\\", "Review question Can we show the direction of $P'Q'$ is independent of $P$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R8565 Suggestion The plane $\\pi$ has equation $\\mathbf{r} = \\mathbf{a} + \\lambda \\mathbf{b}+ \\mu \\mathbf{c}$ and the plane $\\pi'$ has equation $\\mathbf{r} = \\mathbf{d}+ \\lambda'\\mathbf{e}+\\mu'\\mathbf{f}$, where $\\lambda$, $\\mu$, $\\lambda'$ and $\\mu'$ are scalar parameters. It is given that $\\mathbf{a}=\\begin{pmatrix}1 \\\\ 0 \\\\ -1\\end{pmatrix}, \\mathbf{b}=\\begin{pmatrix}2\\\\ -1 \\\\ -1\\end{pmatrix}, \\mathbf{c} = \\begin{pmatrix}-1 \\\\ 2 \\\\ 1\\end{pmatrix}, \\mathbf{d} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}, \\mathbf{e} = \\begin{pmatrix}p \\\\ q \\\\ r\\end{pmatrix}, \\mathbf{f} = \\begin{pmatrix}q\\\\ r\\\\ p\\end{pmatrix},$ $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ and that $p$ is negative. The planes $\\pi$ and $\\pi'$ are parallel. Find $p$, $q$ and $r$. If $\\pi$ and $\\pi'$ are parallel, what must be true about the normal of either plane? How could we use the vector and scalar products here? Show that $\\pi'$ is the reflection of $\\pi$ in the origin. If $P$ is the point $(x,y,z)$, what are the coordinates of the reflection of $P$ in the origin? The line $L$ has equation $\\dfrac{x+1}{9}=\\dfrac{y-1}{-6}=\\dfrac{z}{-5}$ and lies in $\\pi$. The point $P$ is in $\\pi$ with parameters $\\lambda = l$ and $\\mu = m$, and $P'$ is in $\\pi'$ with", "+0 thank you 0 103 2 Calculate a normal vector n to the plane through the points triple A = (1,-2,1), B = (-3,1,4), C = (1,1,2), as in the picture below, such that if $$\\mathbf{n} = \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix},$$ and $$n_1 + n_2 + n_3 = 7.$$ Aug 17, 2019 #1 +6045 +1 $$AB=B-A\\\\ AC=C-A\\\\ n = AB \\times AC\\\\ \\hat{n} = \\dfrac{n}{\\|n\\|}\\\\ \\text{The scaled version of the normal vector the problem is after is -7\\hat{n}}$$ I leave plugging and chugging to you. Aug 17, 2019 #2 +23273 +2 Calculate a normal vector n to the plane through the points triple A = (1,-2,1), B = (-3,1,4), C = (1,1,2), as in the picture below, such that if $$\\mathbf{n} = \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix}$$, and $$n_1 + n_2 + n_3 = 7$$. $$\\text{Let \\vec{v} = \\vec{B} - \\vec{C} } \\\\ \\text{Let \\vec{w} = \\vec{A} - \\vec{C} }$$ $$\\begin{array}{|rcll|} \\hline \\vec{v} &=& \\begin{pmatrix}-3 \\\\ 1 \\\\ 4 \\end{pmatrix} - \\begin{pmatrix}1 \\\\1\\\\ 2\\end{pmatrix} \\\\ \\mathbf{\\vec{v}} &=& \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} \\\\\\\\ \\vec{w} &=& \\begin{pmatrix}1 \\\\ -2 \\\\ 1 \\end{pmatrix} - \\begin{pmatrix}1 \\\\1\\\\ 2\\end{pmatrix} \\\\ \\mathbf{\\vec{w}} &=& \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} \\\\ \\hline \\end{array}$$ 1. Vector cross product $$\\small{ \\begin{array}{|lrcll|} \\hline & \\vec{n} =\\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix}&=& \\vec{v} \\times", "Review question # Can we show the direction of $P'Q'$ is independent of $P$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R8565 ## Solution The plane $\\pi$ has equation $\\mathbf{r} = \\mathbf{a} + \\lambda \\mathbf{b}+ \\mu \\mathbf{c}$ and the plane $\\pi'$ has equation $\\mathbf{r} = \\mathbf{d}+ \\lambda'\\mathbf{e}+\\mu'\\mathbf{f}$, where $\\lambda$, $\\mu$, $\\lambda'$ and $\\mu'$ are scalar parameters. It is given that $\\mathbf{a}=\\begin{pmatrix}1 \\\\ 0 \\\\ -1\\end{pmatrix}, \\mathbf{b}=\\begin{pmatrix}2\\\\ -1 \\\\ -1\\end{pmatrix}, \\mathbf{c} = \\begin{pmatrix}-1 \\\\ 2 \\\\ 1\\end{pmatrix}, \\mathbf{d} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}, \\mathbf{e} = \\begin{pmatrix}p \\\\ q \\\\ r\\end{pmatrix}, \\mathbf{f} = \\begin{pmatrix}q\\\\ r\\\\ p\\end{pmatrix},$ $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ and that $p$ is negative. The planes $\\pi$ and $\\pi'$ are parallel. Find $p$, $q$ and $r$. We can find a vector perpendicular to $\\pi$ using the cross product $\\mathbf{b} \\times \\mathbf{c}= \\begin{pmatrix}1\\\\ -1 \\\\ 3\\end{pmatrix}$. Since the two planes are parallel this is also perpendicular to $\\pi'$. We can use this fact in the dot product with the two direction vectors $\\mathbf{e}$ and $\\mathbf{f}$ in the plane $\\pi'$. So $\\mathbf{e} . (\\mathbf{b} \\times \\mathbf{c})=0$ and $\\mathbf{f} . (\\mathbf{b} \\times \\mathbf{c})=0$. Giving us \\begin{align} p-q+3r &=0 \\\\ q-r+3p &=0 \\end{align} Adding gives $4p+2r=0 \\quad \\implies r=-2p$ and substituting in the second", "to boot) and no inversion. Reflecting a vector $\\mathbf v$ in a plane through the origin reverses the component $\\mathbf v_\\perp$ of the vector that is orthogonal to the plane. That is, $R\\mathbf v=(\\mathbf v-\\mathbf v_\\perp)-\\mathbf v_\\perp=\\mathbf v-2\\mathbf v_\\perp$. This perpendicular component is just the projection of $\\mathbf v$ onto a normal $\\mathbf n$ of the plane and the projection of $\\mathbf v$ onto $\\mathbf n$ is ${\\mathbf n^T\\mathbf v\\over\\mathbf n^T\\mathbf n}\\mathbf n$, so $$R\\mathbf v=\\mathbf v-2{\\mathbf n^T\\mathbf v\\over\\mathbf n^T\\mathbf n}\\mathbf n=\\left(I-{\\mathbf n\\mathbf n^T\\over\\mathbf n^T\\mathbf n}\\right)\\mathbf v.$$ Note that the product in the numerator on the right is a matrix, but the denominator is a scalar. Comparing this to what you had, there are two errors in your attempt: First, and most important, you used the wrong vector. You need a normal to the plane, not a vector in the plane. Second, the expression you’ve got assumes a unit vector, but you didn’t mention normalizing either of the eigenvectors that you listed. For the plane $x+y+z=0$, we can again take $\\mathbf n=(1,1,1)^T$, giving $$\\begin{bmatrix}1&0&0\\\\0&1&0\\\\0&0&1\\end{bmatrix}-\\frac23\\begin{bmatrix}1\\\\1\\\\1\\\\\\end{bmatrix}\\begin{bmatrix}1&1&1\\end{bmatrix}=\\begin{bmatrix}1&0&0\\\\0&1&0\\\\0&0&1\\end{bmatrix}-\\frac23\\begin{bmatrix}1&1&1\\\\1&1&1\\\\1&1&1\\end{bmatrix} = \\left[\\begin{array}{rrr} \\frac13&-\\frac23&-\\frac23 \\\\ -\\frac23&\\frac13&-\\frac23 \\\\ -\\frac23&-\\frac23&\\frac13 \\end{array}\\right].$$", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "v_2)^2$ is minimized. W.l.o.g. we can take $\\mathbf\\pi_1$ to be the $u$-$v$ plane, so that its camera matrix is the canonical $$P_1=\\begin{bmatrix}1&0&0&0\\\\0&1&0&0\\\\0&0&0&1\\end{bmatrix}.$$ This just picks out the $u$- and $v$-coordinates of $\\mathbf U$. The projection matrix onto $\\mathbf\\pi_2$ will have the form $$P_2=\\left[\\begin{array}{c|c}M_{2\\times 3}&\\mathbf t\\\\\\mathbf 0^T&1\\end{array}\\right].$$ The point $\\mathbf U$ then satisfies the linear equations $P_1\\mathbf U =\\mathbf u_1=(u_1,v_1)$ and $P_2\\mathbf U = \\mathbf u_2=(u_2,v_2)$. The affine fundamental matrix of this epipolar configuration is $$F_A=\\begin{bmatrix}0&0&a\\\\0&0&b\\\\c&d&e\\end{bmatrix}$$ where $$a = m_{23} \\\\ b=-m_{13} \\\\ c=m_{13}m_{21}-m_{11}m_{23} \\\\ d=m_{13}m_{22}-m_{12}m_{23} \\\\ e=m_{13}t_2-m_{23}t_1.$$ (In your case, where $P_2$ has no translation component, we would have $e=0$.) The adjusted coordinates are then given by the formula $$\\begin{bmatrix}\\hat u_2 \\\\ \\hat v_2 \\\\ \\hat u_1 \\\\ \\hat v_1\\end{bmatrix} = \\begin{bmatrix}u_2 \\\\ v_2 \\\\ u_1 \\\\ v_1\\end{bmatrix}-{au_2+bv_2+cu_1+dv_1+e\\over a^2+b^2+c^2+d^2}\\begin{bmatrix}a\\\\b\\\\c\\\\d\\end{bmatrix}.$$ (You might recognize this as the result of a standard least-squares fit, which itself is basically a projection onto a hyperplane in $\\mathbb R^4$.) With these adjusted coordinates in hand, you can then find the intersection of their back-projections in the ways described previously. • Thank you for taking the time to explain this, however, I do not understand what you mean when you say that U will lie on the intersection of those two lines (I assume lambda and mu are just arbitrary variables you chose without any", "cbind(x0, x1) # make a matrix showEqn(X, y, vars=colnames(X), simplify=TRUE) ## x0 = 0.5 ## x0 + x1 = 0.5 ## x0 - 1*x1 = 2 To display this in observation space, 1. First create a basic 3D plot showing the coordinate axes. 2. Then, use vectors3d() to draw the vectors x0, x1 and y. 3. The plane spanned by x0, and x1 can be specified as as the normal vector orthogonal to both, using the new matlib::xprod() function. 4. Finally, we use Proj() to find the projection of y on this plane. win <- rgl::open3d() # (1) draw observation axes E <- diag(3) rownames(E) <- c(\"1\", \"2\", \"3\") vectors3d(E, lwd=2, color=\"blue\") # (2) draw variable vectors vectors3d(y, labels=c(\"\", \"y\"), color=\"red\", lwd=3, headlength=0.07) # (3) draw the plane spanned by x0, x1 normal <- xprod(x0, x1) rgl::planes3d(normal, col=\"turquoise\", alpha=0.2) # (4) draw projection of y on X py <- Proj(y, X) rgl::segments3d(rbind( y, py)) # draw y to plane rgl::segments3d(rbind( O, py)) # origin to py in the plane corner( O, py, y, d=0.15) # show it's a right angle arc(y, O, py, d=0.2, color=\"red\") This plot is interactive in the HTML version. Use the mouse wheel to expand/contract the plot. Drag it to rotate to a different view. You can also spin the plot around it’s any", "about a plane. Basically. You subtract the vector in question, in my case I started out with $$e_1$$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane. Nov 22, 2018 at 15:28 • @Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go. – amd Nov 22, 2018 at 19:25 Since $$(2,2,1)$$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $$(1,0,-2)$$ and $$(0,1,-2)$$. You want the each one is projected into itself. So, take the only linear map $$P\\colon\\mathbb{R}^3\\longrightarrow\\mathbb{R}^3$$ such that 1. $$P(2,2,1)=(0,0,0)$$; 2. $$P(1,0,-2)=(1,0,-2)$$; 3. $$P(0,1,-2)=(0,1,-2)$$. A simple computation shows that the matrix of $$P$$ with respect to the canonical basis is$$\\frac19\\begin{bmatrix}5 & -4 & -2 \\\\ -4 & 5 & -2 \\\\ -2 & -2 & 8\\end{bmatrix}.$$", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "given $\\exists \\, \\mathbf{w_1}$ and $\\mathbf{w_2}$ respectively, such that $A\\mathbf{w_1}=\\mathbf{v_1}$ and $A\\mathbf{w_2}=\\mathbf{v_2}$. So $$A\\begin{bmatrix}\\mathbf{w_1} & \\mathbf{w_2}\\end{bmatrix}=\\begin{bmatrix}\\mathbf{v_1} & ... 2 Observe that$$ \\left(\\frac17(A-4I)\\right)A=I $$1 Hint: write I = AA^{-1}, so that:$$A^2-4A - 7AA^{-1} = 0.$$Can you solve for A^{-1}? 1 Hint: In terms of inner product, the projection of vector X=(x,y,z) onto vector u=(a,b,c) is simply:$$p_u(X)=\\frac{\\langle X,u\\rangle}{\\langle u,u\\rangle}\\, u$$For the second question, two strategies for the projection onto a plane: If you have the equation of the plane, as is the case, you also have a normal vector of the plane: n=(1,-2,3). ... 1 i think it is easier to find the projection on to the line u = (1, -2, 3)^\\top that is orthogonal to the plane and then subtract from the identity to get the projection onto the plane. the projection matrix onto the line u is$$uu^\\top/(u^\\top u) = \\frac1{14}\\pmatrix{1&-2&3\\\\-2&4&-6\\\\3&-6&9}.$$therefore the projection matrix on ... 1 Call your vectors v_1,v_2. Pick two vectors v_3,v_4 such that \\{ v_1,\\dots,v_4 \\} is a linearly independent set. Pick linearly independent images w_1,w_2 for them. (The linear independence ensures that the kernel contains only your given vectors.) Then you want A such that$$A \\begin{bmatrix} v_1 & v_2 & v_3 & v_4 \\end{bmatrix} = ... 1 Just follow your nose: Since $x=Ax+x-Ax$, take $u=Ax$ and $v=x-Ax$, and you only need to see", "}$-vector space. In addition, the map is well-defined. Let ${\\displaystyle (x_{1},x_{2})^{T}}$ and ${\\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have: {\\displaystyle {\\begin{aligned}f\\left({\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\\\(x_{1}+y_{1})-5\\cdot (x_{2}+y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{separate vectors}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}x_{1}+x_{2}\\\\x_{1}-5x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}+y_{2}\\\\y_{1}-5y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+f{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\end{aligned}}} Proof step: homogeneity Let ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle (x_{1},x_{2})^{T}\\in \\mathbb {R} ^{2}}$. Then: {\\displaystyle {\\begin{aligned}f\\left(\\lambda \\cdot {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}\\lambda \\cdot x_{1}\\\\\\lambda \\cdot x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda x_{1}+\\lambda x_{2}\\\\\\lambda x_{1}-5\\lambda x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda (x_{1}+x_{2})\\\\\\lambda (x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{scalar multiplication}}\\right.}\\\\[0.3em]&=\\lambda \\cdot {\\begin{pmatrix}(x_{1}+x_{2})\\\\(x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=\\lambda \\cdot f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\\\[0.3em]\\end{aligned}}} Thus the map is linear. ## A linear map in the vector space of sequences Next, we consider the space of all sequences of real numbers. This space is infinite-dimensional, because there are not finitely many sequences generating this sequence space. But it is a vector space, as we have shown in the chapter about sequence spaces. Exercise (Sequence space) Let ${\\displaystyle V}$ be the ${\\displaystyle \\mathbb {R} }$-vector space of all real-valued sequences. Show that the map {\\displaystyle {\\begin{aligned}f:V&\\to V\\\\(a_{0},a_{1},a_{2},\\ldots )&\\mapsto (a_{1},a_{2},a_{3},\\ldots )\\end{aligned}}} is linear. How to get to the proof? (Sequence space) To show linearity, two properties need to be checked: 1. ${\\displaystyle f}$ is additive: ${\\displaystyle f(v+w)=f(v)+f(w)}$ for all ${\\displaystyle v,w\\in V}$ 2. ${\\displaystyle", "# Interpretation of maps #### mathmari ##### Well-known member MHB Site Helper Hey!! Let $\\phi_1:\\mathbb{R}^3\\rightarrow \\mathbb{R}^3$, $v\\mapsto \\begin{pmatrix}0 & -1 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 1\\end{pmatrix}\\cdot v$, $\\phi_2:\\mathbb{R}^3\\rightarrow \\mathbb{R}^3$, $v\\mapsto \\frac{1}{3}\\begin{pmatrix}2 & - 1& - 1\\\\ - 1& 2 & -1\\\\ - 1&-1 & 2\\end{pmatrix}\\cdot v$ which describes the shadow of the sun on a plane. Give the direction of the sun by giving a vector. How do we find this vector? To give the equation of the plane do we have to find the image of the map? How can we explain why at a projection $\\psi$ it holds that $\\psi\\circ\\psi=\\psi$ and $\\text{im} \\psi=\\text{fix} \\psi$? #### Klaas van Aarsen ##### MHB Seeker Staff member Hey!! Let $\\phi_1:\\mathbb{R}^3\\rightarrow \\mathbb{R}^3$, $v\\mapsto \\begin{pmatrix}0 & -1 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 1\\end{pmatrix}\\cdot v$, $\\phi_2:\\mathbb{R}^3\\rightarrow \\mathbb{R}^3$, $v\\mapsto \\frac{1}{3}\\begin{pmatrix}2 & - 1& - 1\\\\ - 1& 2 & -1\\\\ - 1&-1 & 2\\end{pmatrix}\\cdot v$ which describes the shadow of the sun on a plane. Give the direction of the sun by giving a vector. How do we find this vector? Hey mathmari !! Suppose we have a vector in the direction of the sun. Then its shadow will have length 0, won't it? Put otherwise, its image", "the plane with equation $\\displaystyle{ \\ n_x au + n_y bv + n_z cw = d. }$ Let muu + mvv + mww = δ be the Hesse normal form of the new plane and $\\displaystyle{ \\;\\mathbf m = \\begin{bmatrix} m_u \\\\ m_v \\\\ m_w \\end{bmatrix}\\; }$ its unit normal vector. Hence $\\displaystyle{ \\mathbf e_0 = \\delta \\mathbf m \\; }$ is the center of the intersection circle and $\\displaystyle{ \\;\\rho = \\sqrt{1 - \\delta^2}\\; }$ Where mw = ±1 (i.e. the plane is horizontal), let $\\displaystyle{ \\ \\mathbf e_1 = \\begin{bmatrix} \\rho \\\\ 0 \\\\ 0 \\end{bmatrix},\\qquad \\mathbf e_2 = \\begin{bmatrix} 0 \\\\ \\rho \\\\ 0 \\end{bmatrix}. }$ Where mw ≠ ±1, let $\\displaystyle{ \\mathbf e_1 = \\frac{\\rho}{\\sqrt{m_u^2 + m_v^2}}\\, \\begin{bmatrix} m_v \\\\ -m_u \\\\ 0 \\end{bmatrix}\\, ,\\qquad \\mathbf e_2 = \\mathbf m \\times \\mathbf e_1\\ . }$ In any case, the vectors e1, e2 are orthogonal, parallel to the intersection plane and have length ρ (radius of the circle). Hence the intersection circle can be described by the parametric equation $\\displaystyle{ \\;\\mathbf u = \\mathbf e_0 + \\mathbf e_1\\cos t + \\mathbf e_2\\sin t\\;. }$ The reverse scaling (see above) transforms the unit sphere back to the ellipsoid and the vectors e0, e1, e2 are mapped onto vectors f0, f1, f2, which were wanted for the", "Vector3(x * 150, stage.surfaceHeight(x * 150, z * 150), z * 150); Vector3 B = new Vector3((x + 1) * 150, stage.surfaceHeight((x + 1) * 150, z * 150), z * 150); Vector3 C = new Vector3(x * 150, stage.surfaceHeight(x * 150, (z + 1) * 150), (z + 1) * 150); Vector3 D = new Vector3((x + 1) * 150, stage.surfaceHeight((x + 1) * 150, (z + 1) * 150), (z + 1) * 150); I use a simple algorithm to determine my destination plane: public bool isInUpperPlane(Vector3 A, Vector3 B, Vector3 C, Vector3 D, Vector3 point) { /* ^ y | | | | A B ----------------------------> x / / / x / C /_________ / D / / / v z */ // consider A as the origin if (Math.Abs(point.X - A.X) + Math.Abs(point.Z - A.Z) >= 150) return true; return false; } Next I construct a plane the 3 vertices (either ABC or BCD): Plane destPlane; if (isInUpperPlane(A, B, C, D, stopLocation)) { destPlane = new Plane(A, B, C); } else { destPlane = new Plane(B, C, D); } Then I project this my start location to this plane normal, and translate my avatar to this location: Vector3 normal = destPlane.Normal; Vector3 r = projectVectorToPlane(startLocation, normal); Orientation *= Matrix.CreateTranslation(startLocation); But doing this make my", "vector from $(3,3)$ to $(2,5)$ in $\\mathbb{R}^2$ 3. the vector from $(1,0,6)$ to $(5,0,3)$ in $\\mathbb{R}^3$ 4. the vector from $(6,8,8)$ to $(6,8,8)$ in $\\mathbb{R}^3$ This exercise is recommended for all readers. Problem 2 Decide if the two vectors are equal. 1. the vector from $(5,3)$ to $(6,2)$ and the vector from $(1,-2)$ to $(1,1)$ 2. the vector from $(2,1,1)$ to $(3,0,4)$ and the vector from $(5,1,4)$ to $(6,0,7)$ This exercise is recommended for all readers. Problem 3 Does $(1,0,2,1)$ lie on the line through $(-2,1,1,0)$ and $(5,10,-1,4)$? This exercise is recommended for all readers. Problem 4 1. Describe the plane through $(1,1,5,-1)$, $(2,2,2,0)$, and $(3,1,0,4)$. 2. Is the origin in that plane? Problem 5 Describe the plane that contains this point and line. $\\begin{pmatrix} 2 \\\\ 0 \\\\ 3 \\end{pmatrix} \\qquad \\{\\begin{pmatrix} -1 \\\\ 0 \\\\ -4 \\end{pmatrix} +\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}t \\,\\big|\\, t\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 6 Intersect these planes. $\\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}t+ \\begin{pmatrix} 0 \\\\ 1 \\\\ 3 \\end{pmatrix}s \\,\\big|\\, t,s\\in\\mathbb{R}\\} \\qquad \\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\\\ 0 \\end{pmatrix}k+ \\begin{pmatrix} 2 \\\\ 0 \\\\ 4 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 7 Intersect each pair, if possible. 1. $\\{\\begin{pmatrix} 1 \\\\", "to the line $y=mx+b$, we would have $$Y=mX+bE.$$ But, of course, we don't. In linear algebra terms, the vector $Y$ does not lie in the plane spanned by $X$ and $E$ in $\\mathbb{R}^{44}$ So let's try to find the point in the plane spanned by $E$ and $X$ which is closest to $Y$. In [11]: from mpl_toolkits.mplot3d import Axes3D #plt.subplots(nrows=1,ncols=1,figsize=(6,10)) fig=plt.figure(figsize=(10,10)) X = np.arange(-2, 2, 0.25) Y = np.arange(-2, 2, 0.25) X, Y = np.meshgrid(X, Y) Z=2*Y-X t=np.arange(-1,0,.1) ax.plot_wireframe(X,Y,Z) ax.plot(t,-2*t,6+t,color=\"red\") ax.scatter([0,-1],[0,2],[6,5],color=\"red\") ax.set_title(\"Closest point lying in a given plane from a point outside the plane\") plt.show() We can use linear algebra to solve the problem of finding the point in the plane $P$ spanned by $X$ and $E$ that is closest to $Y$. Essentially, we need to take the perpendicular projection of the vector $Y$ into the plane spanned by $X$ and $E$. For that, we can construct an orthonormal basis for the plane $P$ by using the Gram-Schmidt process on the two vectors $E$ and $X$. First, we normalize $E$ to obtain our first unit vector $\\mathbf{e}=E/\\sqrt{N}$, since $E\\cdot E=N$. Next, we subtract the projection of $X$ onto the $E$ direction and normalize. Let $x=X-(X\\cdot \\mathbf{e})\\mathbf{e}$ and $$\\mathbf{x}=\\frac{x}{\\sqrt{x\\cdot x}}$$ Then the point we are interested in is $$\\mathbf{y}=(Y\\cdot\\mathbf{x})\\mathbf{x}+(Y\\cdot\\mathbf{e})\\mathbf{e}$$ This point lies in the plane $P$, and $Y-\\mathbf{y}$ is", "+0 # math help 0 89 1 Let P be the matrix that projects onto j that is, we want P to satisfy $$\\mathbf{P} \\mathbf{v} = \\text{The projection of \\mathbf{v} onto } \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$$ for all vectors v. Use the pictures below to calculate $$\\mathbf{P}\\mathbf{i}, \\mathbf{P} \\mathbf{j}, \\mathbf{P}\\mathbf{k}$$. Calculate the matrix P that projects onto j. Aug 15, 2019 $$\\text{v projected onto (0,1,0) is (v \\cdot (0,1,0))(0,1,0)}\\\\ v \\cdot (0,1,0) = v_y\\\\ v \\perp (0,1,0) = (0,v_y,0)\\\\ \\text{The matrix that will perform this is}\\\\ P=\\begin{pmatrix}0&0&0\\\\0&1&0\\\\0&0&0\\end{pmatrix}$$", "order to build $$\\mathtt H$$ you first have to choose a coordinate system for $$\\mathbf\\pi$$. Using Zisserman’s notation from §3.2.1, this means that you have a full-rank $$4\\times3$$ matrix $$\\mathtt M$$ that maps from these coordinates to world coordinates, i.e., $$\\mathbf X = \\mathtt M\\mathbf x$$. Once you have this, getting the homography matrix is straightforward: the projected image of this point is $$\\mathtt{PM}\\mathbf x$$. We now insist that the camera center not lie on the plane so that the $$3\\times3$$ matrix $$\\mathtt H = \\mathtt{PM}$$ has full rank, otherwise the image of $$\\mathbf\\pi$$ collapses to a line through the principal point. If you want to map from image to world coordinates on this plane, the appropriate matrix is then $$\\mathtt M\\mathtt H^{-1} = \\mathtt M(\\mathtt{PM})^{-1}$$. This construction is an example of the point transfer method that comes up again later in the epipolar geometry section. For your example plane, a natural choice of coordinate system is $$\\mathtt M = \\begin{bmatrix}1&0&0\\\\0&1&0\\\\0&0&a\\\\0&0&1\\end{bmatrix}.$$ Using the standard camera matrix $$\\mathtt P = [\\mathtt I\\mid\\mathbf 0]$$, the resulting homography matrix is $$\\mathtt H = \\begin{bmatrix}1&0&0\\\\0&1&0\\\\0&0&a\\end{bmatrix}.$$" ]

[ "# Thread: find the range of the function 1. ## find the range of the function How to find the range of Log[x-1]Sinx, In the base we have [x-1], where [.] means greatest integer function 2. ## Re: find the range of the function Originally Posted by vivek How to find the range of Log[x-1]Sinx, In the base we have [x-1], where [.] means greatest integer function You may not understand how very strange this question actually is. If we assume that the domain us a subset of the real numbers then: $\\large\\log_b(t)=s$ means that $\\large b>0$, $\\large t>0$, & $\\large b^s=t$. Therefore, $\\large{\\log_b(t)=\\dfrac{\\log(t)}{\\log(b)}}$ So think about what you are asking!", "# What is the range of the function y = 2x^2? $\\left(- \\infty , \\infty\\right)$ Let $y = f \\left(x\\right) = 2 {x}^{2}$ so the range of this function is$\\left(- \\infty , \\infty\\right)$", "# What is the range of the function y = 2x^2? Jun 19, 2016 #### Answer: $\\left(- \\infty , \\infty\\right)$ #### Explanation: Let $y = f \\left(x\\right) = 2 {x}^{2}$ This polynomial is continuous everywhere , so the range of this function is$\\left(- \\infty , \\infty\\right)$", "What is the range of the function f(x)=3x+2? $\\left\\{y \\in \\mathbb{R}\\right\\}$ $f \\left(x\\right) = 3 x + 2$ is a continuous linear function, so its range is $\\pm \\infty$ $\\left\\{y \\in \\mathbb{R}\\right\\}$", "my range correct? Yes. The range of $e^x$ covers the entire domain of the natural logarithmic function and the range of the natural logarithmic function is $(-\\infty,\\infty)\\text.$", "function can be denoted as the logarithmic function with base 10: In order to apply the correct ranges, we need to shift the values of x in both functions. Thus, we can rewrite $f(x)$ as: For the logarithmic function, we write: Using these formulae, we can convert volumes within the linear ranges from 0.0 to 1.0 to logarithmic values between -10000 and 0 and vice versa.", "can be denoted as the logarithmic function with base 10: In order to apply the correct ranges, we need to shift the values of x in both functions. Thus, we can rewrite $f(x)$ as: For the logarithmic function, we write: Using these formulae, we can convert volumes within the linear ranges from 0.0 to 1.0 to logarithmic values between -10000 and 0 and vice versa.", "# Math Help - range of function 1. ## range of function what is the range of the function f(x) = log{arccot(x^2-x)} 2. What is the range of logarithm? Does that arccot restrict the numbers log is applied to? What about $x^2- x$", "## Precalculus (10th Edition) domain is $(0, \\infty).$, range is $(-\\infty, \\infty).$ The logarithmic function is a function (e.g. $y = log_a {x}$) which is the inverse of an exponential function (e.g. $y = a^x$) . $a>0, a\\ne1$, $x>0$. However, the range is $(-\\infty, \\infty).$", "is large. It should be in the third quadrant of the graph. Example: In {8, 11, 5, 9, 7, 6, 3616}:. In this tutorial we will concentrate more on the mechanics of finding the range. Next lesson. range f ( x) = √x + 3. Range of quadratic functions. Usually a logarithm consists of three parts. The domain is all x-values or inputs of a function and the range is all y-values or outputs of a function. The previous answer presumes the continuity of exponential functions prior to defining the log functions, which is backwards. Substitute different x-values into the expression for y to see what is happening. The task of finding what points can be reached by a function is a very useful one. In the example, we need to solve for $$x$$: So, is there any restriction on $$y$$ for $$x$$ to be well defined? $range\\:f\\left (x\\right)=\\sqrt {x+3}$. The liver function test normal values are 7-56 units/liter for ALT and 10-40units/liters is the range for AST. Normally, you would complete the square and check the leading coefficient, a, to determine the concavity for the comparison sign. When looking for the range, it may help to make a list of some ordered pairs for the function. The range of the composed function has to be less than", "have such as 0.0001, the same function has a what is the range of the function?... ( f\\ ) in the previous answer presumes the continuity of exponential functions prior to defining the log functions which. Solutions to the equation for a more conceptual approach to domain and range ℝ | y ≠ k where −! Here we have around 10 if we use interval notations to specify domain and range of [ 0 oo! Log functions, which produces one value at a graph of 𝑥 against 𝑦, this be! X and it passes through the origin iterate on the mechanics of finding the range the! Range 0 to 1 with the shapes of basic functions like sin/cosine and.. ( x\\right ) =\\sqrt { x+3 }$ total number of randomly in. The largest number in the third quadrant of the function on what is the range of the function? graph, it is a very one... 7 grams of protein of linear functions is y = x 2 test... The definition of the graph is nothing but blood tests that help in diagnosing damage..., find the outputs of the composed function has a range of the function defined... Value of 3616 makes the range is the range large, but values... ( i.e, linear function is almost always going to", "range of $y$ is all of $\\mathbb R$. Graphing a function is a great way to confirm or gain insight into the domain and range of a function. E.g. Below is the graph of the function $y = \\log_e(x - 3)$. As $x$ grows in size, so to does $y$. That is, $y$ is an increasing function. As $x$ approaches $3$ from the right, although $y=\\log_e(x - 3)$ is not defined for any $x \\leq 3$, within the interval $x \\in (3, 4)$, $y < 0$. The closer $x$ is to $3$ (from the right), the smaller $y$ gets (the \"more negative\" $y$ becomes.) - I sometimes add 1+1 incorrectly! :-) – Amzoti Aug 12 '13 at 0:19 @amWhy: but i add 1+1 correctly. :-) – Babak S. Aug 12 '13 at 6:00 Remember that for $x \\in \\mathbb{R}$ the argument of $\\log$ cannot be negative and $\\log(x)$ varies from $-\\infty$ to $\\infty$ - $\\log$ is the inverse function of exponentiation: $y = \\log_b x \\iff x = b^y$ (only for positive $b$). So the domain of $y = \\log_b x$ is the range of $x = b^y$ which is all $x$ with $x > 0$ since $b^y$ is always positive for positive $b$. And the range of $y = \\log_b x$ is the domain of $x =", "we've just looked at is equal to zero for $x = - 1$ and $x = 1$, the range of the function will vary between zero and the point in which the quadratic has a maximum value. More specifically, $\\sqrt{1 - {x}^{2}} = \\text{max}$ if $x = 0$, since you're subtracting a squared value from $1$. At point $x = 0$, the function is equal to $f \\left(0\\right) = - \\sqrt{1 - {0}^{2}} = - \\sqrt{1} = - 1$ Therange of the function will thus be $\\left[0 , - 1\\right]$. graph{-sqrt(1-x^2) [-10, 10, -5, 5]}", "log ( x ) =\\sqrt { x } be two on. { 2-x^ { 2 } } 1, 2, 3, 4 } as your.. Of function. [ 7 ] form the range of the function takes the reals ( )... There are different ways to find the range ( actual output values ) is defined ) there only. ( actual output values ) is defined for all real numbers produce a in. T ) Open image in a function is { 2 } } is a logarithmic function and you! 2, 3, 4 } as your domain -1 ≤y≥ 1 to. Will always be used square function. [ 7 ] Python 3.x ; 1 ¶ Return a string a! Starts y=-1 and extended to \\infty when x\\geq 1 returns a range of a quadratic function of! But by thinking about it we can find the range of the of! X < 1 is 2 name, email, and it returns a range type function can not equal! X\\Leq -1 \\right | is ( -\\infty,2 ) many functions, the domain and range of a are! Piecewise function consists of two function: if we draw the diagram of the discrete range of function extended... To identify the domain of f ( x ) =a\\times b^ range of function x-h } +k", "+0 # How would I find the range of this function algebraically? +1 61 1 $$y = {\\sqrt{2-x^{2}}}$$ The answer is {y│0 ≤ y ≤ 2, y ∈ ℝ}, I'm completely lost! Oct 22, 2018 #1 +3593 +2 $$\\text{well the domain of y(x) is }[-2,2] \\\\ \\text{and over this domain y increases from 0 to }\\sqrt{2} \\text{ and then decreases back to 0}\\\\ \\text{so the range of y(x) is}\\\\ \\left\\{y:y \\in [0,\\sqrt{2}]\\right\\}\\\\ \\text{what you've written isn't quite correct. y(x) never gets above }y=\\sqrt{2}$$ . Oct 22, 2018", "# Find the range of $f(x) = \\sqrt{\\log(\\cos(\\sin x))}$ Find the range of the function $$f(x)=\\sqrt{\\log(\\cos(\\sin(x))}.$$ What I did was: $$\\log(\\cos(\\sin(x))) \\geq 0$$ Then $$\\log_e\\cos(\\sin(x)) \\geq 0$$ The next step of the problem is $$\\cos(\\sin(x)) \\geq e^0$$ This is the part I didn’t understand. Please explain it to me. Note that $$\\sqrt{\\log(\\cos(\\sin(x)))}$$ is defined if and only if $$\\cos(\\sin(x))\\geq 1$$ otherwise we have the square root of a negative number. Now since the cosine function is upper bounded by $$1$$, $$\\cos(\\sin(x))$$ has to be $$1$$ (such equality holds for example when $$x=0$$). So what is the range of the given function? • Why can it be greater than or equal to zero? Why does it have to be greater than or equal to one? That’s the part I didn’t understand – Aditya Jul 20 '19 at 15:18 • If $0<t<1$ then $\\log(t)<0$ and $\\sqrt{\\log(t)}$ is not defined. Hence $t=\\cos(\\sin(x))\\geq 1$. – Robert Z Jul 20 '19 at 15:20 Okay. $$e > 1$$. So if $$x > 0$$ then $$e^x > 1$$ and $$e^0 = 1$$ and if $$y < 0$$ then $$e^{y} = e^{-|y|} = \\frac 1{e^{|y|} }< 1$$ and so $$0 < e^y < 1$$. So $$e^x > 0$$ always. $$0 < e^x < 1$$ if and only if $$x < 0$$. and $$e^x = 1$$ if", "# What is the range of the function x^2+y^2=36? Apr 4, 2017 [-6, 6] #### Explanation: That relation is not a function. The relation is in the standard form of a circle. Its graph is a circle of radius 6 about the origin. Its domain is [-6, 6], and its range is also [-6, 6]. To find this algebraically, solve for y. ${x}^{2} + {y}^{2} = 36$ ${y}^{2} = 36 - {x}^{2}$ $y = \\pm \\sqrt{36 - {x}^{2}}$ The range is largest in absolute value when x = 0, and we have $y = \\pm \\sqrt{36}$. That is, at -6 and 6.", "What is the range of the function y = -2x^2 + 3? Mar 28, 2017 The range is $- \\infty < y \\le 3$ Explanation: Please observe that the coefficient of the ${x}^{2}$ term is negative; this means that the parabola opens downward, which makes the minimum of the range approach $- \\infty$. The maximum of the range will be the y coordinate of the vertex. Because the coefficient of the $x$ term is 0, the y coordinate of vertex is the function evaluated at 0: $y = - 2 {\\left(0\\right)}^{2} + 3$ $y = 3$ The range is $- \\infty < y \\le 3$", "Question # Find the range of ${\\log _e}(\\sin x)$ isA)$( - \\infty ,\\infty )$B)$( - \\infty ,1)$C)$( - \\infty ,0]$D)$( - \\infty ,0)$ Verified 149.4k+ views Hint: Here we will proceed with the solution as we know the range of $\\sin x = [1, - 1]$ which is required to solve this problem. Here we need to find the range of given value that is ${\\log _e}(\\sin x)$ As we know that the domain of logarithmic functions are of positive value only Since we know the range of $\\sin x$ is $[ - 1,1]$ Then the domain value of above log function would be $(0,1]$ So, now to get the range of ${\\log _e}(\\sin x)$ Let us substitute $x = 0$ in the given function ${\\log _e}(\\sin x)$ i.e. For $x \\to 0 \\Rightarrow {\\log _e}x = - \\infty$ Now let us substitute $x = 1$ in the given function ${\\log _e}(\\sin x)$ i.e. For$x \\to 1 \\Rightarrow {\\log _e}x = 0$ Hence from this we can say that the range of given function ${\\log _e}(\\sin x)$=$( - \\infty ,0]$ NOTE: In this particular problem we know the range of $\\sin x$ is $[ - 1,1]$ and domain of log function is $(0,1]$ so by substituting the domain values ($x$ values) in the given function i.e. ${\\log _e}(\\sin", "\\text{ }$ and $\\text{ } \\sqrt{{\\left(1\\right)}^{2} - 1} = 0$ The range of the function will thus be $\\left[0 , + \\infty\\right)$. graph{sqrt(x^2-1) [-10, 10, -5, 5]}" ]

[ "is set of positive real numbers or $\\left\\{x\\in ℝ|x>2\\right\\}$ . The function takes all the real values from $-\\infty$ to $\\infty$ . Therefore, the range of the function is set of real numbers.", "positive, $z$ is positive. . . Range: . $z\\,>\\,0$", "of $F(x)=log_b x$ is the set of all positive real numbers. Ex 5. It is true that the range of $F(x)=log_b x$ is the set of all real numbers. Ex 6. Which facts are true for the graph of the function below? $F(x)=log_0.427 x$ Answers: It is decreasing The range is all real numbers The domain is x > 0 The x-intercept is (1,0) Ex 7. It is true that the function $F(x)=log0.9 x$ is decreasing. Ex 8. For what values of b will $F(x)=log_b x$ be an increasing function? b > 1 Ex 9. For what values of b will $F(x)=log_b x$ be a decreasing function? 0 < b < 1", "$f$ is the set of all possible numbers that it spews out. Quick Examples 1. The range of the function $f$ given by $f(x) = 2x$ consists of all possible values of $f(x),$ such as $f(1) = 2,$ $f(1/2) = 1,$ $f(-3) = -6,$ and so on. In fact, we can get any number we please this way. For example, to get $-600,$ just take $f(-300).$ More precisely, to get an arbitrary value $y,$ we just take $f(y/2).$ Thus the range is the set of all real numbers. 2. Let $f(x) = x^2.$ Trying a few values first, $f(0) = 0,$ $f(1) = 1,$ $f(-1) = 1,$... It appears that we can only get non-negative numbers this way. In fact, we can get any nonnegative number $y$ we like: if $y ≥ 0,$ then $f(y^{0.5}) = (y^{0.5})^2 = y.$ Thus the range of $f$ is $[0, +∞)$ -- the set of all non-negative real numbers. 3. Let $f(x) = x^2 + 1.$ The range of $f$ is Select one All real numbers All non-negative real numbers [0, +infinity) [1, +infinity) [-1, +infinity) All real numbers except 1 All real numbers except -1 Here is a graphical way of obtaining the range of a function. As an example, let us use $f(x) = x^2,$ whose graph is shown below.", "$f$ is the set of all possible numbers that it spews out. Quick Examples 1. The range of the function $f$ given by $f(x) = 2x$ consists of all possible values of $f(x),$ such as $f(1) = 2,$ $f(1/2) = 1,$ $f(-3) = -6,$ and so on. In fact, we can get any number we please this way. For example, to get $-600,$ just take $f(-300).$ More precisely, to get an arbitrary value $y,$ we just take $f(y/2).$ Thus the range is the set of all real numbers. 2. Let $f(x) = x^2.$ Trying a few values first, $f(0) = 0,$ $f(1) = 1,$ $f(-1) = 1,$... It appears that we can only get non-negative numbers this way. In fact, we can get any nonnegative number $y$ we like: if $y ≥ 0,$ then $f(y^{0.5}) = (y^{0.5})^2 = y.$ Thus the range of $f$ is $[0, +∞)$ -- the set of all non-negative real numbers. 3. Let $f(x) = x^2 + 1.$ The range of $f$ is Select one All real numbers All non-negative real numbers [0, +infinity) [1, +infinity) [-1, +infinity) All real numbers except 1 All real numbers except -1 Here is a graphical way of obtaining the range of a function. As an example, let us use $f(x) = x^2,$ whose graph is shown below.", "< 0$ for $1.58833 < x < 9.93693$, and $f(x) > 0$ for $9.93693 < x$. $f'(x) =\\ln 2\\ 2^x - 3x^2$. According to Wolfy, which uses the Lambert W function, the positive roots of this are $x_1 = -\\dfrac{2 W\\left(-\\frac{\\log^{3/2}(2)}{2 \\sqrt{3}}\\right)}{\\log(2)} \\approx 0.589665$ and $x_2 = -\\dfrac{2 W_{-1}\\left(-\\frac{\\log^{3/2}(2)}{2 \\sqrt{3}}\\right)}{\\log(2)} \\approx 8.1768$. If $x > x_2$, then $f'(x) > 0$. Therefore, if $x > 9.93693$, $f(x) > 0$. You can argue using the higher derivatives of $f(x)$ to get more elementary methods of showing when $f(x) = 0$, but I will leave it at this. You can use pencil and paper to give a quick statement of: 1. Whether there is a solution, and 2. The range the solution(s) must fall in if it/they exist. First note that $2^x$ is positive for all $x$, and $x^3-1$ is negative for negative $x$. So if there is a solution, it must be non-negative. Next note that both the RHS and LHS of the equation are strictly increasing for positive $x$. This simplifies things drastically. Now make a table for the LHS and RHS of the equation for small values of $x$. More precisely, let's define two functions: $$f(x)=2^x$$ $$g(x)=x^3-1$$ and make tables for each function, for small non-negative inputs. (Actually do this; it's very simple.) Observe which is greater, $f(x)$", "becomes -(x - 8) + 5 - x > -(x + 7), which in turn is x < 20. Since, we consider x < -7 range, then finally for it we get x < -7. 2. $$-7 \\leq x \\leq 5$$ For this range: x - 8 is negative, so |x - 8| = -(x - 8). 5 - x is positive, so |5 - x| = 5 - x. x + 7 is positive, so |x + 7| = x + 7. Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) + 5 - x > x + 7, which in turn is x < 2. Since, we consider $$-7 \\leq x \\leq 5$$ range, then finally for it we get $$-7 \\leq x < 2$$. 3. $$5 < x < 8$$ For this range: x - 8 is negative, so |x - 8| = -(x - 8). 5 - x is negative, so |5 - x| = -(5 - x). x + 7 is positive, so |x + 7| = x + 7. Thus, |x – 8| + |5 – x| > |x + 7| becomes -(x - 8) -(5 - x) > x + 7, which in turn is x < -4. Since, we consider $$5 < x", "positive real number. By definition, $$y = \\log_b x \\qquad \\text{means} \\qquad b^y = x.$$ Since $b^y > 0$ for all $y \\in \\mathbb{R}$, as a function of a real variable, $\\log_b x$ is only defined for $x>0$.", "100 it could be interpreted as percentage change of $x$ minus 100% from period $t-1$ to $t$ , that is beacause we could approximate $log(x_{t}/x_{t-1})$ by $x_{t}/x_{t-1}-1$ \"very near\" the point $x=1$, when $x$ is far away from 1 this approximation doesn't hold. Put functions $y=log(x)$ and $y=x-1$ on one plot. • Thanks @Qbik, this is what I also guess. Thanks for your help. – Sam Jan 29 '13 at 16:45", "range of each function . For any function that is not one to one , give two distinct values of x which have the same image . (1) $f(x)=sin x$ , x are real numbers (2) $g(x)=\\sqrt{x}+\\sqrt{1+x}$ , $o\\geq x \\geq1$ , (3) $h(x)=In{x}$ , $x>0$ the range for sine function is [-1,1] but for the rest , i am not really sure . (1) You know that sin x is periodic with period $2\\pi$. That should give you an easy answer. (2) This is a little harder because this is one to one. I would show that by arguing that the derivative is always positive and so the function is increasing but I don't know what methods you have available. (3) I don't recognise \"In(x)\". Do you mean the natural logarighm? That is ln(x)- \"logarithm\" starts with a \"l\" not \"I\"! If so, same comments as in (2) apply. 4. Originally Posted by Grandad Hello thereddevils(1) $f$ is not one-to-one; e.g. $f(0) = f(\\pi)$ (2) I presume you mean $0\\le x\\le 1$, in which case, since $g$ is an increasing function for $x \\ge 0$, the range is $[1, 1+\\sqrt{2}]$; and $g$ is one-to-one. (3) The function $\\text{ln}x$, for $x > 0$, has the whole of $\\mathbb{R}$ as its range. And it is also one-to-one. Thanks once", "# Finding Range of: $y=\\log_e(3-2x)/\\sqrt{x+1}$? This is from Calculus I test, that I failed to solve correctly. This really bugs me, because for Solve[y == Log[E, 3 - 2x] / Sqrt[x + 1], x], the Mathematica is telling me that: \"The equations appear to involve the variables to be solved for in an essentially non-algebraic way.\". I assume that the answer is $x \\in \\mathbb{R}$ - Just observe that at $\\lim_{x\\to -1^+} y(x) = \\infty$ and $\\lim_{x \\to 3/2^-} y(x) = -\\infty$ and $y(x)$ is continuous on $x \\in (-1, 3/2)$. Therefore the range is indeed $(-\\infty,\\infty)$.", "$f$ is the set of real numbers. The range of the function has also no restriction. We can think of any real number $f(x)$, and we can always find the corresponding $x$ by subtracting 1 from it. Therefore, the range of $f$ is the set of real numbers. Example 2: Find the domain and range of $g(x) = x^2$. Explanation Just like in Example 1, we can substitute any real number to $x$ and any of them makes $g(x)$ defined (Can you see why?). So, the domain of $g$ is the set of real numbers. Now let’s think about the range. The range are the values of $f(x) = x^2$. Notice that if we square a positive real number, the result is a positive real number. If we square a negative real number, then the result is also a positive real number. If we square 0, the result is 0. Therefore, the only possible results that we can get if we square a real number are 0 and positive real numbers. Therefore, the range of $g$ is the set of real numbers greater than, or equal to 0. In interval notation, the range of $g$ is $[0, \\infty)$. Example 3: Find the domain and range of $h(x) = | x |$ Recall that $|x|$ means the absolute value", "# What is the range of a function like f(x) = sqrt (x-5)? Apr 17, 2015 The argument of the square rood has to be not-negative, so: $x \\ge 5$ or $\\left[5 , + \\infty\\right)$.", "# Log Functions Inside Absolute Value Is the function below always positive for $$0< x <1$$? (I am determining if the function requires the modulus sign or not.) $$\\frac{1}{2}\\log\\left|\\frac{1+\\log(x)}{1-\\log(x)}\\right|$$ My first instinct is that it cannot be, but I would just like some third-party feedback. Thank you! This function is equivalent to $$\\mathrm{artanh}\\,(\\ln{(x)})$$ which is only defined for $$\\frac1e\\lt x\\lt e$$ and is negative for $$\\frac1e\\lt x\\lt1$$. Note that excluding $$x=1/e$$ (for which your fraction does not exist), we have in the interval $$0 that $$log (x)<0$$, so the fraction will evaluate to something with absolute value smaller than 1, so the complete expression will be negative.", "The difference $\\log(1+x) - 2x/(2+x)$ is an increasing positive function:", "\\le 1$ so at 0 the range is 2 and the closer we get to -1 or 1 the range is $\\frac{2}{\\text{some tiny number}} \\to \\infty$, so that range is $2 \\text{ to } \\infty$ Putting it all together: $\\left\\{f \\left(x\\right) | f \\left(x\\right) \\in \\mathbb{R} , x < 0 \\mathmr{and} x \\ge 2\\right\\}$ graph{2/(1-x^2) [-10, 10, -5, 5]}", "= 0,$ and is positive or negative according as $$x > 1$$ or $$x < 1$$. Later on we shall be able to identify this function with the Napierian logarithm of $$x$$. Example. Prove that $$f(xy) = f(x) + f(y)$$. [Use the equations $f(xy) = \\lim n({\\sqrt[n]{xy}} – 1) = \\lim \\{n(\\sqrt[n]{x} – 1)\\sqrt[n]{y} + n(\\sqrt[n]{y} – 1)\\}.]$", "# Find the range of $f(x) = \\sqrt{\\log(\\cos(\\sin x))}$ Find the range of the function $$f(x)=\\sqrt{\\log(\\cos(\\sin(x))}.$$ What I did was: $$\\log(\\cos(\\sin(x))) \\geq 0$$ Then $$\\log_e\\cos(\\sin(x)) \\geq 0$$ The next step of the problem is $$\\cos(\\sin(x)) \\geq e^0$$ This is the part I didn’t understand. Please explain it to me. Note that $$\\sqrt{\\log(\\cos(\\sin(x)))}$$ is defined if and only if $$\\cos(\\sin(x))\\geq 1$$ otherwise we have the square root of a negative number. Now since the cosine function is upper bounded by $$1$$, $$\\cos(\\sin(x))$$ has to be $$1$$ (such equality holds for example when $$x=0$$). So what is the range of the given function? • Why can it be greater than or equal to zero? Why does it have to be greater than or equal to one? That’s the part I didn’t understand – Aditya Jul 20 '19 at 15:18 • If $0<t<1$ then $\\log(t)<0$ and $\\sqrt{\\log(t)}$ is not defined. Hence $t=\\cos(\\sin(x))\\geq 1$. – Robert Z Jul 20 '19 at 15:20 Okay. $$e > 1$$. So if $$x > 0$$ then $$e^x > 1$$ and $$e^0 = 1$$ and if $$y < 0$$ then $$e^{y} = e^{-|y|} = \\frac 1{e^{|y|} }< 1$$ and so $$0 < e^y < 1$$. So $$e^x > 0$$ always. $$0 < e^x < 1$$ if and only if $$x < 0$$. and $$e^x = 1$$ if", "& =\\left[ (152.64)-(96.96) \\right]+1 \\\\ & =[55.6]+1 \\\\ & =55+1 \\\\ & =56 \\\\ \\end{align} Hence the number of digits in $\\left| adj(adj\\left( adj\\text{ }A \\right) \\right|$ is equal to 56 Note: The characteristic of the logarithm of a positive number is positive and it is one less than the number of digits in the number. Hence we use find number of digits in $\\left| adj(adj\\left( adj\\text{ }A \\right) \\right|$ by taking greatest integer of log of $\\left| adj(adj\\left( adj\\text{ }A \\right) \\right|$ and adding 1.", "\\geq 0$$ which means $$-4 \\leq x \\leq 4$$. The range consists of all non-negative real numbers less than or equal to $$4$$. [If $$0 \\leq y \\leq 4$$ then $$x=\\sqrt {16-y^{2}}$$ satisfies $$f(x)=y$$]." ]

[ "# What is the range of the function y^2=x+1? It is $y \\ge 0$. $y = \\sqrt{x + 1}$ There can be no negative number in square roots; therefore the lowest number which $x$ can be is -1.", "# Finding Range of: $y=\\log_e(3-2x)/\\sqrt{x+1}$? This is from Calculus I test, that I failed to solve correctly. This really bugs me, because for Solve[y == Log[E, 3 - 2x] / Sqrt[x + 1], x], the Mathematica is telling me that: \"The equations appear to involve the variables to be solved for in an essentially non-algebraic way.\". I assume that the answer is $x \\in \\mathbb{R}$ - Just observe that at $\\lim_{x\\to -1^+} y(x) = \\infty$ and $\\lim_{x \\to 3/2^-} y(x) = -\\infty$ and $y(x)$ is continuous on $x \\in (-1, 3/2)$. Therefore the range is indeed $(-\\infty,\\infty)$.", "range of $y$ is all of $\\mathbb R$. Graphing a function is a great way to confirm or gain insight into the domain and range of a function. E.g. Below is the graph of the function $y = \\log_e(x - 3)$. As $x$ grows in size, so to does $y$. That is, $y$ is an increasing function. As $x$ approaches $3$ from the right, although $y=\\log_e(x - 3)$ is not defined for any $x \\leq 3$, within the interval $x \\in (3, 4)$, $y < 0$. The closer $x$ is to $3$ (from the right), the smaller $y$ gets (the \"more negative\" $y$ becomes.) - I sometimes add 1+1 incorrectly! :-) – Amzoti Aug 12 '13 at 0:19 @amWhy: but i add 1+1 correctly. :-) – Babak S. Aug 12 '13 at 6:00 Remember that for $x \\in \\mathbb{R}$ the argument of $\\log$ cannot be negative and $\\log(x)$ varies from $-\\infty$ to $\\infty$ - $\\log$ is the inverse function of exponentiation: $y = \\log_b x \\iff x = b^y$ (only for positive $b$). So the domain of $y = \\log_b x$ is the range of $x = b^y$ which is all $x$ with $x > 0$ since $b^y$ is always positive for positive $b$. And the range of $y = \\log_b x$ is the domain of $x =", "What is the range of the function sqrt(2x-8)? May 1, 2017 Range $\\ge 0$ The argument must be non-negative, or $2 x - 8 \\ge 0$ This means the domain is restricted to $x \\ge 4$ The function has a non-negative outxome. Since everything down to $0$ is possible (when $x = 4$) the range is $\\ge 0$", "Question # Find the range of ${\\log _e}(\\sin x)$ isA)$( - \\infty ,\\infty )$B)$( - \\infty ,1)$C)$( - \\infty ,0]$D)$( - \\infty ,0)$ Verified 149.4k+ views Hint: Here we will proceed with the solution as we know the range of $\\sin x = [1, - 1]$ which is required to solve this problem. Here we need to find the range of given value that is ${\\log _e}(\\sin x)$ As we know that the domain of logarithmic functions are of positive value only Since we know the range of $\\sin x$ is $[ - 1,1]$ Then the domain value of above log function would be $(0,1]$ So, now to get the range of ${\\log _e}(\\sin x)$ Let us substitute $x = 0$ in the given function ${\\log _e}(\\sin x)$ i.e. For $x \\to 0 \\Rightarrow {\\log _e}x = - \\infty$ Now let us substitute $x = 1$ in the given function ${\\log _e}(\\sin x)$ i.e. For$x \\to 1 \\Rightarrow {\\log _e}x = 0$ Hence from this we can say that the range of given function ${\\log _e}(\\sin x)$=$( - \\infty ,0]$ NOTE: In this particular problem we know the range of $\\sin x$ is $[ - 1,1]$ and domain of log function is $(0,1]$ so by substituting the domain values ($x$ values) in the given function i.e. ${\\log _e}(\\sin", "# Find the range of $f(x) = \\sqrt{\\log(\\cos(\\sin x))}$ Find the range of the function $$f(x)=\\sqrt{\\log(\\cos(\\sin(x))}.$$ What I did was: $$\\log(\\cos(\\sin(x))) \\geq 0$$ Then $$\\log_e\\cos(\\sin(x)) \\geq 0$$ The next step of the problem is $$\\cos(\\sin(x)) \\geq e^0$$ This is the part I didn’t understand. Please explain it to me. Note that $$\\sqrt{\\log(\\cos(\\sin(x)))}$$ is defined if and only if $$\\cos(\\sin(x))\\geq 1$$ otherwise we have the square root of a negative number. Now since the cosine function is upper bounded by $$1$$, $$\\cos(\\sin(x))$$ has to be $$1$$ (such equality holds for example when $$x=0$$). So what is the range of the given function? • Why can it be greater than or equal to zero? Why does it have to be greater than or equal to one? That’s the part I didn’t understand – Aditya Jul 20 '19 at 15:18 • If $0<t<1$ then $\\log(t)<0$ and $\\sqrt{\\log(t)}$ is not defined. Hence $t=\\cos(\\sin(x))\\geq 1$. – Robert Z Jul 20 '19 at 15:20 Okay. $$e > 1$$. So if $$x > 0$$ then $$e^x > 1$$ and $$e^0 = 1$$ and if $$y < 0$$ then $$e^{y} = e^{-|y|} = \\frac 1{e^{|y|} }< 1$$ and so $$0 < e^y < 1$$. So $$e^x > 0$$ always. $$0 < e^x < 1$$ if and only if $$x < 0$$. and $$e^x = 1$$ if", "# Finding the domain and the range of this function Problem: Find the domain and range of the function $f(x)=\\log \\sqrt{4-x^2}$ My attempt: For domain $y=f(x)=\\log \\sqrt{4-x^2}$ $\\implies y=\\log \\sqrt{4-x^2}$ Since logarithm of only positive numbers is possible, $4-x^2>0$ $\\implies x^2-4<0$ $\\implies(x+2)(x-2)<0$ $\\implies -2<x<2$ Therefore $D_f=(-2,2)$ For range $y=f(x)=\\log \\sqrt{4-x^2}$ $y=\\log \\sqrt{4-x^2}$ $y=\\frac{1}{2}\\log (4-x^2)$ $2y=\\log (4-x^2)$ $e^{2y}=4-x^2$ $x=\\sqrt{4-e^{2y}}$ Since the exponent can take any real value, $R_f=(-\\infty,+\\infty)$ My problem: The answer given in my book for $D_f$ matches with my $D_f$ but $R_f=(-\\infty,\\log 2]$ according to my book. Where have i gone wrong? $\\sqrt {4-x^2}$ cannot be greater than $\\sqrt 4=2$, so $\\log \\sqrt{4-x^2}$ cannot be any greater than $\\log \\sqrt 4=\\log 2$. It achieves that value when $x=0.$ As $\\sqrt{4-x^2}$ can achieve $0$, you can have $\\log \\sqrt{4-x^2}$ as negative as you want, so the range is $(-\\infty, \\log 2]$ • Where have i gone wrong? – MrAP Jul 9 '17 at 21:22 • The whole approach of solving for $x$ is misguided. It will fail for functions that are not bijections. You have a continuous function so only need to find the maximum and minimum. Taking the square root is dangerous. Finally, though $e^{2y}$ alone can take any positive value, if it gets greater than $4$ the square root argument becomes negative. – Ross Millikan Jul", "# How do you find the domain and range of y = log_2(x - 3)? Jun 21, 2015 The argument of a log-function must be positive. #### Explanation: This means $x - 3 > 0 \\to x > 3$ If $x$ gets closer to $3$, the log will get more and more negative Or, in the \"language\"\" ${\\lim}_{x \\to 3} {\\log}_{2} \\left(x - 3\\right) = - \\infty$ The other way, $x$ getting larger, will get the function larger: ${\\lim}_{x \\to \\infty} {\\log}_{2} \\left(x - 3\\right) = \\infty$", "# What is the range of the function x + sqrt( x-1 )? Jan 9, 2018 Range of function: 1 ≤ x #### Explanation: In order to determine the range of a function, you look at the complex part of that function, in this case: $\\sqrt{x - 1}$ You must start with this, because it is always the most complex part of an function that limits it. We know for fact that any square root cannot be negative. In other words, it must always be equal or greater than 0. 0 ≤ $\\sqrt{x - 1}$ 0 ≤ $x - 1$ 1 ≤ x The above tells us that x from the given function must always be greater or equal to 1. If it is smaller than 1, then the square root would be positive, and that is impossible. Now, you can insert any x value greater or equal to 1, and the function would work out. This means that this function only has a lower limit of 1, and there are no upper limits.", "# What is the range of the function r(x) = sqrt(x - 10)? $x - 10 \\ge 0$ and so: $x \\ge 10$ which represents the domain of your function. The range will be all the $y \\ge 0$. Regardless of the value of $x$ you input in your function (as long as is $\\ge 10$) the square root will always give you a POSITIVE answer or Zero. Your function can have the value of $x = 10$ as minimum possible value giving you $y = 0$. From there you can increase $x$ up to $\\infty$ and your $y$ will increase as well (slowly).", "# What is the range of the function sqrt(6x-7)? Nov 2, 2017 Range$= \\left[0 , + \\infty\\right)$ #### Explanation: As the things inside square root cannot be negative, $6 x - 7$ must be bigger than or equal to $0$. $6 x - 7 \\ge 0$ $6 x \\ge 7$ $x \\ge \\frac{7}{6}$ Domain$= \\left[\\frac{7}{6} , + \\infty\\right)$ Since the things inside square root is bigger than or equal to $0$, the range of $\\sqrt{k}$ is the value from $\\sqrt{0}$ to $\\sqrt{+ \\infty}$, whatever the value of $k$. Range$= \\left[0 , + \\infty\\right)$", "range of each function . For any function that is not one to one , give two distinct values of x which have the same image . (1) $f(x)=sin x$ , x are real numbers (2) $g(x)=\\sqrt{x}+\\sqrt{1+x}$ , $o\\geq x \\geq1$ , (3) $h(x)=In{x}$ , $x>0$ the range for sine function is [-1,1] but for the rest , i am not really sure . (1) You know that sin x is periodic with period $2\\pi$. That should give you an easy answer. (2) This is a little harder because this is one to one. I would show that by arguing that the derivative is always positive and so the function is increasing but I don't know what methods you have available. (3) I don't recognise \"In(x)\". Do you mean the natural logarighm? That is ln(x)- \"logarithm\" starts with a \"l\" not \"I\"! If so, same comments as in (2) apply. 4. Originally Posted by Grandad Hello thereddevils(1) $f$ is not one-to-one; e.g. $f(0) = f(\\pi)$ (2) I presume you mean $0\\le x\\le 1$, in which case, since $g$ is an increasing function for $x \\ge 0$, the range is $[1, 1+\\sqrt{2}]$; and $g$ is one-to-one. (3) The function $\\text{ln}x$, for $x > 0$, has the whole of $\\mathbb{R}$ as its range. And it is also one-to-one. Thanks once", "# What is the range of the function sqrt(2x-8)? May 1, 2017 Range $\\ge 0$ #### Explanation: There are two rules about square roots: (1) The argument must be non-negative, or $2 x - 8 \\ge 0$ This means the domain is restricted to $x \\ge 4$ (2) The function has a non-negative outxome. Since everything down to $0$ is possible (when $x = 4$) the range is $\\ge 0$", "# What is the range of a function like f(x) = sqrt (x-5)? Apr 17, 2015 The argument of the square rood has to be not-negative, so: $x \\ge 5$ or $\\left[5 , + \\infty\\right)$.", "we've just looked at is equal to zero for $x = - 1$ and $x = 1$, the range of the function will vary between zero and the point in which the quadratic has a maximum value. More specifically, $\\sqrt{1 - {x}^{2}} = \\text{max}$ if $x = 0$, since you're subtracting a squared value from $1$. At point $x = 0$, the function is equal to $f \\left(0\\right) = - \\sqrt{1 - {0}^{2}} = - \\sqrt{1} = - 1$ Therange of the function will thus be $\\left[0 , - 1\\right]$. graph{-sqrt(1-x^2) [-10, 10, -5, 5]}", "correct or not. Solution We already know that the square root function results in a non-negative value. So $$\\sqrt x$$ is non-negative for any value of $$x$$ in the domain. So $$\\dfrac{1}{\\sqrt x}$$ also must be non-negative. But $$\\dfrac{1}{\\sqrt x}$$ cannot be zero for any value of $$x$$. Thus, $\\dfrac{1}{\\sqrt x} > 0, \\text{ for any x in the domain}$ Thus, the range of the function is, $$\\{y \\in R/ y>0\\}$$ (OR) $$(0, \\infty)$$ Let's graph the given function and see whether the answer is correct. Here, the graph covers all values of y above the x-axis except y=0. Thus, the range is $$\\{y \\in R/ y>0\\}$$ Important Notes 1. The range of a cubic function is the set of all real numbers. 2. The range of a logarithmic function is the set of all real numbers. 3. The range of an exponential function $$y=e^x$$ is always $$y >0$$ because the exponential function results in positive values always. ## Interactive Questions Here are a few activities for you to practice. Select/type your answer and click the \"Check Answer\" button to see the result. ## Let's Summarize We hope you enjoyed learning about Range in Algebra with the simulations and practice questions. Now, you will be able to easily solve problems on the range of a function, examples of", "+0 # How would I find the range of this function algebraically? +1 61 1 $$y = {\\sqrt{2-x^{2}}}$$ The answer is {y│0 ≤ y ≤ 2, y ∈ ℝ}, I'm completely lost! Oct 22, 2018 #1 +3593 +2 $$\\text{well the domain of y(x) is }[-2,2] \\\\ \\text{and over this domain y increases from 0 to }\\sqrt{2} \\text{ and then decreases back to 0}\\\\ \\text{so the range of y(x) is}\\\\ \\left\\{y:y \\in [0,\\sqrt{2}]\\right\\}\\\\ \\text{what you've written isn't quite correct. y(x) never gets above }y=\\sqrt{2}$$ . Oct 22, 2018", "# What is the range of the function y=-sqrt(4x^2 + 2x^4 +5)? $\\left(- \\infty , - \\sqrt{5}\\right]$ So the range is all possible values of $y$, so we start to think about how low this function can go. Because the equation is negative and the argument in the square root can go to infinity, it can go forever. The highest number will be the lowest number for the argument inside the square root. Since ${x}^{2}$ and ${x}^{4}$ are always non-negative, the lowest they can be is 0, so the highest value is when $x = 0$, giving the range $\\left(- \\infty , - \\sqrt{5}\\right]$.", "# What is the range of the function y=x^3? $- \\infty < y < + \\infty$ Since $y$ is an uneven power of $x$, $y$ will be negative when $x$ is. This is unlike the even powers, that are always non-negative.", "that the value of $x+\\sqrt{x^2+1}$ is always positive for both positive and negative values but the value of $x-\\sqrt{x^2+1}$ is always negative for both positive and negative values. Hence, $e^y = x + \\sqrt{x^2+1}$ and $e^y \\ne x-\\sqrt{x^2+1}$ 04. #### Inverse Hyperbolic sine function in mathematical form Take natural logarithm both sides. $\\implies \\log_{e}{e^y} = \\log_{e}{(x+\\sqrt{x^2+1})}$ $\\implies y \\times \\log_{e}{e} = \\log_{e}{(x+\\sqrt{x^2+1})}$ $\\implies y \\times 1 = \\log_{e}{(x+\\sqrt{x^2+1})}$ $\\implies y = \\log_{e}{(x+\\sqrt{x^2+1})}$ But $y = \\sinh^{-1}{x}$ as per our assumption. $\\therefore \\,\\,\\,\\,\\,\\, \\sinh^{-1}{x} = \\log_{e}{(x+\\sqrt{x^2+1})}$ It is the mathematical form of inverse hyperbolic sine function and it can be simply written as follows. $\\therefore \\,\\,\\,\\,\\,\\, \\sinh^{-1}{x} = \\ln{(x+\\sqrt{x^2+1})}$" ]

[ "^{2} x-\\cot ^{2} x\\right\\} d x=-3 \\cos x-4 \\sin x+$ $6 \\tan x+7 \\cot x+c$", "the sequence $x^3+x^6+x^9+...$ from $x+x^2+x^3+...$ Both sequences are geometric. Top 50 recent answers are included", "\\tan x+7 \\cot x+c$ Thus, $\\int\\left\\{3 \\sin x-4 \\cos x+\\frac{5}{\\cos ^{2} x}-\\frac{6}{\\sin ^{2} x}+\\tan ^{2} x-\\cot ^{2} x\\right\\} d x=-3 \\cos x-4 \\sin x+$ $6 \\tan x+7 \\cot x+c$", "# Weekly Challenge #1 Geometry Level 1 Let $$0 ^ \\circ < \\theta < 45 ^ \\circ$$. Then arrange - $$t_1$$ = $$(tan \\theta) ^ {tan \\theta}$$ , $$t_2$$ = $$(tan \\theta) ^ {cot \\theta}$$ , $$t_3$$ = $$(cot \\theta) ^ {tan \\theta}$$ , $$t_4$$ = $$(cot \\theta) ^ {cot \\theta}$$ in decreasing order. ×", "cot(x)+tan(x) as 2/sin(2x) and find a power series for 1/sin. – Did Sep 11 '11 at 6:55", "Properties of geometric sequence Question Sample Titled 'Properties of geometric sequence' Let ${a}$ , ${b}$ and ${c}$ be positive constants. If ${a}{x}^{{2}}+{2}{b}{x}+{c}={0}$ has repeated roots, which of the following must be true? I. ${{\\log{{a}}}^{{4}}}$ , ${{\\log{{b}}}^{{4}}}$ and ${{\\log{{c}}}^{{4}}}$ form an arithmetic sequence. II. ${3}{a}$ , ${3}{b}$ and ${3}{c}$ form a geometric sequence. III. ${a}+{4}$ , ${b}+{4}$ and ${c}+{4}$ form a geometric sequence. A I and II only B I and III only C None of the above D II only ∵ ${a}{x}^{{2}}+{2}{b}{x}+{c}$ $={0}$ has repeated roots. ∴ $\\Delta$ $={0}$ ${\\left({2}{b}\\right)}^{{2}}-{4}{a}{c}$ $={0}$ ${b}^{{2}}$ $={a}{c}$ ∴ ${a}$ , ${b}$ and ${c}$ form a geometric sequence. ∴ ${3}{a}$ , ${3}{b}$ and ${3}{c}$ form a geometric sequence. ∵ ${a}$ , ${b}$ and ${c}$ form a geometric sequence. ∴ ${\\log{{a}}}$ , ${\\log{{b}}}$ and ${\\log{{c}}}$ form an arithmetic sequence. ∴ ${4}{\\log{{a}}}$ , ${4}{\\log{{b}}}$ and ${4}{\\log{{c}}}$ form an arithmetic sequence. ∴ ${{\\log{{a}}}^{{4}}}$ , ${{\\log{{b}}}^{{4}}}$ and ${{\\log{{c}}}^{{4}}}$ form an arithmetic sequence. ${a}$ , ${b}$ and ${c}$ form a geometric sequence. This does not provide that ${a}+{4}$ , ${b}+{4}$ and ${c}+{4}$ form a geometric sequence. 專業備試計劃 Level 4+ 保證及 5** 獎賞 ePractice 會以電郵、Whatsapp 及電話提醒練習 ePractice 會定期提供溫習建議 Level 5** 獎勵：會員如在 DSE 取得數學 Level 5** ，將獲贈一套飛往英國、美國或者加拿大的來回機票，唯會員須在最少 180 日內每天在平台上答對 3 題 MCQ。 Level 4 以下賠償：會員如在 DSE 未能達到數學 Level 4 ，我們將會全額退回所有會費，唯會員須在最少 180 日內每天在平台上答對 3 題 MCQ。 FAQ ePractice 是甚麼？ ePractice 是一個專為中四至中六而設的網站應用程式，旨為協助學生高效地預備", "# A geometry problem by Aakhyat Singh Geometry Level 3 If in a triangle $$ABC$$, $$\\tan A+\\tan B+\\tan C=6$$, then $$\\cot A \\cot B \\cot C= \\dfrac k6$$. Find $$k$$. ×", "i'll work on it. 10. Originally Posted by mr fantastic Are you saying you never learned that cot x = 1/tan x and that you don't know that tan x = sin x/cos x? And you're unfamiliar with the fact that $\\cos^2 x = 1 - \\sin^2 x$? (you don't actually need any double angle formula). i misread something you said. sorry. no more posts on this one.", "with side lengths $AD, BE$, and $CF$, if it exists. What is the perimeter of the last triangle in the sequence $\\left(T_n\\right)$? The first four terms of an arithmetic sequence are $p$, $9$, $3p-q$, and $3p+q$. What is the $2010^\\text{th}$ term of this sequence? Arithmetic sequences $\\left(a_n\\right)$ and $\\left(b_n\\right)$ have integer terms with $a_1=b_1=1 < a_2 \\le b_2$ and $a_n b_n = 2010$ for some $n$. What is the largest possible value of $n$? A geometric sequence $(a_n)$ has $a_1=\\sin x$, $a_2=\\cos x$, and $a_3= \\tan x$ for some real number $x$. For what value of $n$ does $a_n=1+\\cos x$? The first three terms of an arithmetic sequence are $2x - 3$, $5x - 11$, and $3x + 1$ respectively. The $n$th term of the sequence is $2009$. What is $n$?", "# Geometric Mean Of Geometric Sequence Suppose $\\{a_1, \\ldots, a_{99}\\}$ is a geometric sequence. If $a_{49} = 18$ and $a_{51} = 8$, what is $\\text{GM}(a_1, \\ldots, a_{99})?$ ×", "Show that , this series is Geometric and Find the formula of the n^th cosA\\1+sinA=secA-tanA Wrong question", "### Still have math questions? Geometry Question Find the $$7\\mathrm{th}$$ term of the geometric sequence shown below: $$-7x^8$$$$7x^{11}$$$$-7x^{14}$$,... $$-7x^{26}$$", "$cot(tan^{-1}a+cot^{-1}a)$ 1. Solve the equations below 2. $2tan^{-1}(cos x)=tan^{-1}(2 cosec x)$ 1. Solve $tan^{-1}2x+tan^{-1}3x=\\pi /4$", "# Solving a trigonometric equation Can someone help me to solve this problem? Find all number pairs $x,y$ that satisfy the equation: $$\\tan^4(x) + \\tan^4(y) + 2\\cot^2(x)\\cot^2(y) = 3 + \\sin^2(x+y)$$ - By AM-GM inequality: $$(\\tan x)^4 + (\\tan y)^4 + (\\cot x)^2\\cdot (\\cot y)^2 + (\\cot x)^2\\cdot (\\cot y)^2 \\geq 4 \\geq 3 + (\\sin(x+y))^2$$ So the equation occurs when: $\\sin(x+y) = 1, -1$, and $\\tan x = \\tan y, -\\tan y$, and $\\tan x = \\cot x, -\\cot x$. So $\\tan x = 1, -1 = \\tan y$. You can look at cases.", "# How do you tell whether the sequence 3, 5, 7, 9, 11,... is geometric? Feb 23, 2017 ${a}_{n} = 2 n + 1$ $\\setminus \\frac{5}{3} \\setminus \\ne \\setminus \\frac{7}{5}$ so your sequence is not geometric.", "# An Algebraic Trigonometry! Geometry Level 4 $\\begin{cases} \\tan(x)+\\tan(y)+\\tan(z)= 6-(\\cot(x)+\\cot(y)+\\cot(z)) \\\\ \\tan^2(x)+\\tan^2(y)+\\tan^2(z)= 6-(\\cot^2(x)+\\cot^2(y)+\\cot^2(z) ) \\\\ \\tan^3(x)+\\tan^3(y)+\\tan^3(z)= 6-(\\cot^3(x)+\\cot^3(y)+\\cot^3(z) ) \\\\ \\end{cases}$ If $$x,y$$ and $$z$$ are real numbers that satisfy the three equations above. Find the value of the expression below. $\\left \\lfloor \\frac{\\tan(x)}{\\tan(y)} + \\frac{\\tan(y)}{\\tan(z)} + \\frac{\\tan(z)}{\\tan(x)} + 3\\tan(x) \\tan(y) \\tan(z) \\right \\rfloor$ ×", "# Given sin x = 4/7 and cos x = -sqrt33/7, how do you find cot x? Aug 22, 2016 $= - \\frac{\\sqrt{33}}{4}$ $\\cot x = \\cos \\frac{x}{\\sin} x$ $= \\frac{- \\frac{\\sqrt{33}}{7}}{\\frac{4}{7}}$ $= - \\frac{\\sqrt{33}}{4}$", "is something like $-\\sin(n\\pi/2)$ So your sequence is $$-\\frac{\\sin (n\\pi/2)}{n^2+1}$$ $$\\frac{(-1)^{(n+1)/2}}{1+n^2}$$ for the odd terms. Zero for the even terms.", "How do you find the nth term for the geometric sequence a1 = -9, n=6, r=2? Mar 11, 2016 ${t}_{6} = - 288$ Explanation: ${t}_{n} = {t}_{1} \\cdot {r}^{n - 1}$ ${t}_{6} = - 9 \\cdot {2}^{5}$ ${t}_{6} = - 9 \\left(32\\right)$ ${t}_{6} = - 288$", "\\\\$ Hence, the required value of ${\\cot ^6}A - {\\cot ^2}A = 1$ . So, option B is the correct option. Note: A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In order to solve such problems, never try to find the values of the angle given like A. These types of problems can be done only by manipulation of the trigonometric identities in order to bring in the required form. Students must remember the trigonometric identities used." ]

[ "that gives us $$1 + \\cot^2 x - \\cot^2 x = 1,$$ as desired. - For $\\#1$ $$\\sin^2x+\\sin^2x\\cdot\\tan^2x=\\sin^2x(1+\\tan^2x)=\\frac{\\sin^2x}{\\cos^2x}$$ $$\\implies \\sin^2x+\\sin^2x\\cdot\\tan^2x=\\tan^2x$$ For $\\#2$ $$\\csc x=\\frac1{\\sin x},\\sec x=\\frac1{\\cos x}$$ - 1. \\begin{align} \\tan^2x - \\sin^2x&=\\frac{\\sin^2x}{\\cos^2x}-\\sin^2x\\quad;\\ \\text{where}\\ \\tan x=\\frac{\\sin x}{\\cos x}\\\\ &=\\sin^2x\\left(\\frac{1}{\\cos^2x}-1\\right)\\\\ &=\\sin^2x\\left(\\frac{1}{\\cos^2x}-\\frac{\\cos^2x}{\\cos^2x}\\right)\\\\ &=\\frac{\\sin^2x}{\\cos^2x}\\left(1-\\cos^2x\\right)\\\\ &=\\color{blue}{\\tan^2x \\cdot \\sin^2x}\\quad;\\ \\text{where}\\ \\sin^2x + \\cos^2x=1. \\end{align} 2. \\begin{align} \\frac{\\csc x}{\\sec x}&=\\dfrac{\\dfrac{1}{\\sin x}}{\\dfrac{1}{\\cos x}}\\\\ &=\\frac{1}{\\color{red}{\\sin x}}\\times\\frac{\\color{red}{\\cos x}}{1}\\\\ &=\\frac{\\cos x}{\\sin x}\\\\ &=\\color{blue}{\\cot x}. \\end{align} - I just wanted to point out that when doing proofs like this that your argument should lead from the initial form (e.g. $\\tan ^2 x - \\sin^2 x$) to the final form ($\\sin^2x \\tan^2 x$). That should be your final answer, but sometimes it is just easier to prove that the LHS and RHS are equal. For example, my doodles may look like this $$\\tan ^2 x - \\sin^2 x = \\sin^2x \\tan^2 x$$ $$\\tan ^2 x = \\sin^2x \\tan^2 x + \\sin^2 x$$ $$\\tan ^2 x = \\sin^2x(\\tan^2 x + 1)$$ $$\\tan ^2 x = \\sin^2x(\\sec^2 x)$$ $$\\tan ^2 x = \\sin^2x\\bigg(\\frac{1}{\\cos^2 x}\\bigg)$$ $$\\tan ^2 x = \\tan ^2 x$$ Now, I'm going to use my doodles to present my real argument (the one I hand into my teacher). Recall that I ended up $\\tan^2 x$, but started with $\\tan ^2 x - \\sin^2 x$, so I'm just going to take", "+ c^2 - 4)}}{2} = \\frac{-bc + \\sqrt{(4 - b^2)(4 - c^2)}}{2}.$$ This asks for the trigonometric substitution $b = 2\\sin u$ and $c = 2\\sin v$, where $0^\\circ < u,v < 90^\\circ$. Then $$a = 2(-\\sin u\\sin v + \\cos u\\cos v) = 2\\cos (u + v),$$ and if we set $u = B/2$ and $v = C/2$, then $a = 2\\sin (A/2)$, $b = 2\\sin (B/2)$, and $c = \\sin (C/2)$, where $A$, $B$, and $C$ are the angles of a triangle. We have \\begin{align*} ab &= 4\\sin\\frac{A}{2}\\sin\\frac{B}{2} \\\\ &= 2\\sqrt{\\sin A\\tan\\frac{A}{2}\\sin B\\tan\\frac{B}{2}} = 2\\sqrt{\\sin A\\tan\\frac{B}{2}\\sin B\\tan\\frac{A}{2}} \\\\ &\\leq \\sin A\\tan\\frac{B}{2} + \\sin B\\tan\\frac{A}{2} \\\\ &= \\sin A\\cot\\frac{A + C}{2} + \\sin B\\cot\\frac{B + C}{2}, \\end{align*} where the inequality step follows from AM-GM. Likewise, \\begin{align*} bc &\\leq \\sin B\\cot\\frac{B + A}{2} + \\sin C\\cot\\frac{C + A}{2}, \\\\ ca &\\leq \\sin A\\cot\\frac{A + B}{2} + \\sin C\\cot\\frac{C + B}{2}. \\end{align*} Therefore \\begin{align*} ab + bc + ca &\\leq (\\sin A + \\sin B)\\cot\\frac{A + B}{2} + (\\sin B + \\sin C)\\cot\\frac{B + C}{2} + (\\sin C + \\sin A)\\cot\\frac{C + A}{2} \\\\ &= 2\\left(\\cos\\frac{A - B}{2}\\cos\\frac{A + B}{2} + \\cos\\frac{B - C}{2}\\cos\\frac{B + C}{2} + \\cos\\frac{C - A}{2}\\cos\\frac{C + A}{2} \\right)\\\\ &= 2(\\cos A + \\cos B + \\cos C) \\\\ &= 6 - 4\\left(\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2}", "above expression to get: cot2(x)csc2(x)=cos2(x)sin2(x)1sin2(x)=cos2(x)\\begin{align*} \\frac{\\cot^2(x)}{\\csc^2(x)}\\\\ =\\frac{\\frac{\\cos^2(x)}{\\sin^2(x)}}{\\frac{1}{\\sin^2(x)}}\\\\ =\\cos^2(x)\\\\ \\end{align*} ### Review Simplify each trigonometric expression as much as possible. 1. sin(x)cot(x)\\begin{align*}\\sin(x)\\cot(x)\\end{align*} 2. cos(x)tan(x)\\begin{align*}\\cos(x)\\tan(x)\\end{align*} 3. 1+tan(x)1+cot(x)\\begin{align*}\\frac{1+\\tan(x)}{1+\\cot(x)}\\end{align*} 4. 1sin2(x)1+sin(x)\\begin{align*}\\frac{1-\\sin^2(x)}{1+\\sin(x)}\\end{align*} 5. sin2(x)1+cos(x)\\begin{align*}\\frac{\\sin^2(x)}{1+\\cos(x)}\\end{align*} 6. (1+tan2(x))(sec2(x))\\begin{align*}(1+\\tan^2(x))(\\sec^2(x))\\end{align*} 7. sin(x)(tan(x)+cot(x))\\begin{align*}\\sin(x)(\\tan(x)+\\cot(x))\\end{align*} 8. sec(x)sin(x)sin(x)cos(x)\\begin{align*}\\frac{\\sec(x)}{\\sin(x)}-\\frac{\\sin(x)}{\\cos(x)}\\end{align*} 9. sin(x)cot2(x)sin(x)cos2(x)\\begin{align*}\\frac{\\sin(x)}{\\cot^2(x)}-\\frac{\\sin(x)}{\\cos^2(x)}\\end{align*} 10. 1+sin(x)cos(x)sec(x)\\begin{align*}\\frac{1+\\sin(x)}{\\cos(x)}-\\sec(x)\\end{align*} 11. sin2(x)sin4(x)cos2(x)\\begin{align*}\\frac{\\sin^2(x)-\\sin^4(x)}{\\cos^2(x)}\\end{align*} 12. tan(x)csc2(x)+tan(x)sec2(x)\\begin{align*}\\frac{\\tan(x)}{\\csc^2(x)}+\\frac{\\tan(x)}{\\sec^2(x)}\\end{align*} 13. \\begin{align*}\\sqrt{1-cos^2(x)}\\end{align*} 14. \\begin{align*}(1-\\sin^2(x))(\\cos(x))\\end{align*} 15. \\begin{align*}(\\sec^2(x)+\\csc^2(x))-(\\tan^2(x)+\\cot^2(x))\\end{align*} To see the Review answers, open this PDF file and look for section 3.3. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Trigonometric Identity A trigonometric identity is an equation that relates two or more trigonometric functions.", "\\tan^2 A + 1 &= \\sec^2 A \\ 1+\\cot^2 A &= \\csc^2 A \\end{align} #### Miscellaneous identities \\begin{align} \\cos^6 A &= \\frac {1}{32}\\cos 6A + \\frac {3}{16}\\cos 4A + \\frac {15}{32}\\cos 2A + \\frac {5}{16} \\end{align} #### Sum and difference identities \\begin{align} \\sin(A \\pm B) &= \\sin A \\cos B \\pm \\cos A \\sin B \\ \\cos(A \\pm B) &= \\cos A \\cos B \\mp \\sin A \\sin B \\ \\tan(A \\pm B) &= \\frac{\\tan A \\pm \\tan B}{1 \\mp \\tan A \\tan B} \\ \\cot(A \\pm B) &= \\frac{\\cot A \\cot B \\mp 1}{\\cot B \\pm \\cot A} \\end{align} #### Double-angle identities \\begin{align} \\sin 2A &= 2 \\sin A \\cos A \\ &= \\frac{2 \\tan A}{1 + \\tan^2 A}\\ \\cos 2A &= \\cos^2 A - \\sin^2 A \\ &= 2 \\cos^2 A -1 \\ &= 1-2 \\sin^2 A \\ &= {1 - \\tan^2 A \\over 1 + \\tan^2 A}\\ \\tan 2A &= \\frac{2 \\tan A}{1 - \\tan^2 A}\\ &= \\frac{2 \\cot A}{\\cot^2 A - 1}\\ &= \\frac{2}{\\cot A - \\tan A} \\end{align} #### Half-angle identities Note that $\\pm$ are correct, it means it may be either one, depending on the value of A/2. \\begin{align} \\sin \\frac{A}{2} &= \\pm \\sqrt{\\frac{1-\\cos A}{2}} \\ \\cos \\frac{A}{2} &= \\pm \\sqrt{\\frac{1+\\cos A}{2}} \\\\\\tan \\frac{A}{2} &= \\pm \\sqrt{\\frac{1-\\cos A}{1+\\cos A}} = \\frac {\\sin", "points are of $g(z)=\\tan(z)$ and $h(z)=z\\cot ^2(z)$, that is $z_n= (n+\\frac{1}{2})\\pi$ and $w_n = \\pi n$. (b) $z_n$ are simple poles and $\\Res (f,z_n)= \\Res (\\tan(z), z_n)= \\frac{\\sin(z)}{(\\cos (z))'}\\bigr|_{z=z_n} =-1$. $w_0$ is a simple pole $$\\Res (f,w_0)= \\Res (z\\cot^2(z), w_0)= \\Res (z\\cot (z) \\times \\cot(z), 0)= \\Res (\\cot(z),0)= 1$$ because $\\lim _{z\\to 0} z\\cot(z)=1$. $w_n$ with $n\\ne 0$ are double poles and \\begin{align*} \\Res (f,w_n)=&\\Res (z\\cot^2 (z), w_n)= \\Res ((\\pi n +w)\\cot^2 (w) , 0) =\\\\ &\\pi n\\Res (\\cot^2 (w) , 0) +\\Res (w\\cot^2(w),0) =1 \\end{align*} because $\\Res (\\cot^2 (w) , 0)=0$ (since $\\cot^2(w)$ is an even function) and $\\Res (w\\cot^2(w),0) =1$ we already calculated. $\\infty$ is a not isolated singularity and therefore residue here is not defined. 39 Final Exam / FE-P2 official « on: December 20, 2018, 04:49:22 AM » (a)--(b) If $|z|=r$ and $\\arg{w}=t$, then $z=r e^{it}$ and \\begin{align*} w= \\frac{1}{2}\\bigl(re^{it} +r^{-1}e^{-it}\\bigr)&= \\cosh(s)\\cos(t)+i\\sinh(s)\\sin (t) \\end{align*} with $s=\\ln (r)$ (and $r=e^s$). Then $w=u+iv$ with $\\frac{u^2}{a^2}+\\frac{v^2}{b^2}=1$, $a=\\cos(s)$, $b=\\sinh(s)$ and also $\\frac{u^2}{A^2}-\\frac{v^2}{B^2}=1$, with $A=\\cos (\\theta)$, $B=\\sin(\\theta)$, $A^2+B^2=1$. (c) Consider, what the circle $\\{z\\colon |z|=1\\}$ is mapped to: $z=e^{it}$, then $w =\\frac{1}{2}(e^{it}+e^{-it})=\\cos(t)$ and it runs a line segment $[-1,1]\\subset\\mathbb{R}$. So $f$ maps the unit disk onto compliment of it $\\mathbb{C}^*\\setminus [-1,1]$ (and $0$ is mapped to $\\infty$). (d) See attachment (e) $w=\\frac{1}{2}(z+z^{-1}) \\implies z^2 -2wz +1=0$; it has two", "\\left( {\\frac{{3 + \\sqrt 3 }}{{\\sqrt 3 - 1}}} \\right) = 0\\\\ \\Rightarrow\\quad &\\qquad\\;\\;(x - 1)\\left( {x - \\sqrt 3 } \\right)\\left( {x - (2 + \\sqrt 3 )} \\right) = 0 \\\\ \\Rightarrow \\quad &\\qquad x = 1,\\;\\sqrt 3 ,\\;2 + \\sqrt 3 \\\\ \\end{align} The angles are \\begin{align}\\frac{\\pi }{4},\\;\\frac{\\pi }{3},\\;\\frac{\\pi }{{12}}\\end{align} Example - 32 In a $$\\Delta ABC$$, suppose that \\begin{align}\\tan \\frac{A}{2},\\;\\tan \\frac{B}{2},\\;\\tan \\frac{C}{2}\\end{align} are in H.P.. What is the minimum possible value of \\begin{align}\\cot \\frac{B}{2}\\end{align}? Solution: Refer to the result of Example - 28 Part (b). If we divide throughout by \\begin{align}\\tan \\frac{A}{2}\\tan \\frac{B}{2}\\tan \\frac{C}{2}\\end{align}, we have, $\\cot \\frac{A}{2} + \\cot \\frac{B}{2} + \\cot \\frac{C}{2} = \\cot \\frac{A}{2}\\cot \\frac{B}{2}\\cot \\frac{C}{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\ldots \\left( 1 \\right)$ Now, since \\begin{align}\\tan \\frac{A}{2},\\tan \\frac{B}{2},\\tan \\frac{C}{2}\\end{align} are in H.P., their reciprocals, i.e. \\begin{align}\\cot \\frac{A}{2},\\;\\cot \\frac{B}{2},\\;\\cot \\frac{C}{2}\\end{align}, are in A.P. So, $\\qquad\\cot \\frac{A}{2} + \\;\\cot \\frac{C}{2} = 2\\cot \\frac{B}{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\ldots \\left( 2 \\right)$ Using (2) in (1), we obtain $\\cot \\frac{A}{2}\\;\\cot \\frac{C}{2} = 3$ Finally, we use the A.M-G.M. inequality on \\begin{align}\\cot \\frac{A}{2},\\;\\cot \\frac{C}{2}\\end{align}: \\begin{align}&\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{{\\cot \\frac{A}{2} + \\cot \\frac{C}{2}}}{2} = \\cot \\frac{B}{2} \\geqslant \\sqrt {\\cot \\frac{A}{2}\\cot \\frac{C}{2}} = \\sqrt 3 \\\\ \\Rightarrow\\quad &\\qquad\\;\\;\\;\\cot \\frac{B}{2} \\geqslant \\sqrt 3 \\\\ \\end{align} Therefore, the minimum value of \\begin{align}\\cot \\frac{B}{2}\\end{align} is $$\\sqrt 3$$. From this, we can further infer that the maximum value of \\begin{align}\\tan \\frac{B}{2}\\end{align}", "&& \\rightarrow \\text{cancel}\\ (1 + \\cos x)\\end{align*} ### Review Use trigonometric identities to simplify each expression as much as possible. 1. tan(x)cos(x)\\begin{align*}\\tan(x)\\cos(x)\\end{align*} 2. cos(x)cos3(x)\\begin{align*}\\cos(x)-\\cos^3(x)\\end{align*} 3. 1cos2(x)sin(x)\\begin{align*}\\frac{1-\\cos^2(x)}{\\sin(x)}\\end{align*} 4. cot(x)sin(x)\\begin{align*}\\cot(x)\\sin(x)\\end{align*} 5. 1sin2(x)cos(x)\\begin{align*}\\frac{1-\\sin^2(x)}{\\cos(x)}\\end{align*} 6. sin(x)csc(x)\\begin{align*}\\sin(x)\\csc(x)\\end{align*} 7. tan(x)cot(x)\\begin{align*}\\tan(-x)\\cot(x)\\end{align*} 8. sec2(x)tan2(x)cos2(x)+sin2(x)\\begin{align*}\\frac{\\sec^2(x)-\\tan^2(x)}{\\cos^2(x)+\\sin^2(x)}\\end{align*} Prove each identity. 1. tan(x)+cot(x)=sec(x)csc(x)\\begin{align*}\\tan(x)+\\cot(x)=\\sec(x)\\csc(x)\\end{align*} 2. sin(x)=sin2(x)+cos2(x)csc(x)\\begin{align*}\\sin(x)=\\frac{\\sin^2(x)+\\cos^2(x)}{\\csc(x)}\\end{align*} 3. 1sec(x)1+1sec(x)+1=2cot(x)csc(x)\\begin{align*}\\frac{1}{\\sec(x)-1}+\\frac{1}{\\sec(x)+1}=2\\cot(x)\\csc(x)\\end{align*} 4. \\begin{align*}(\\cos(x))(\\tan(x)+\\sin(x)\\cot(x))=\\sin(x)+\\cos^2(x)\\end{align*} 5. \\begin{align*}\\sin^4(x)-\\cos^4(x)=\\sin^2(x)-\\cos^2(x)\\end{align*} 6. \\begin{align*}\\sin^2(x)\\cos^3(x)=(\\sin^2(x)-\\sin^4(x))(\\cos(x))\\end{align*} 7. \\begin{align*}\\frac{\\sin(x)}{\\csc(x)}=1-\\frac{\\cos(x)}{\\sec(x)}\\end{align*} To see the Review answers, open this PDF file and look for section 3.2. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English FOIL FOIL is an acronym used to remember a technique for multiplying two binomials. You multiply the FIRST terms, OUTSIDE terms, INSIDE terms, and LAST terms and then combine any like terms. Trigonometric Identity A trigonometric identity is an equation that relates two or more trigonometric functions. Show Hide Details Description Difficulty Level: Tags: Subjects:", "$\\cot A$ means. The reason you have the answer of $2.000000\\ldots$ (which I ... 5 \\begin{align} 1-\\cot{\\left(x\\right)} &=1-\\frac{\\cos{\\left(x\\right)}}{\\sin{\\left(x\\right)}}\\\\ &=\\frac{\\sin{\\left(x\\right)}-\\cos{\\left(x\\right)}}{\\sin{\\left(x\\right)}}\\\\ &=-\\sqrt{2}\\frac{\\sin{\\left(\\frac{\\pi}{4}-x\\right)}}{\\sin{\\left(x\\right)}}.\\\\ \\end{align} Therefore, \\begin{align} 1-\\cot{\\left(\\frac{\\pi}{4}-x\\right)} ... 5\\lim\\limits_{x\\to\\infty} \\frac{x - \\sin(x)}{x+\\sin(x)}=\\lim_{x\\to\\infty}\\frac{1-\\dfrac{\\sin x}x}{1+\\dfrac{\\sin x}x}$$Now |\\sin x|\\le1 5 Here's a straightforward algebraic approach:$$\\sin\\theta = 2\\sin(\\theta/2)\\cos(\\theta/2) = 2\\tan(\\theta/2).\\cos^2(\\theta/2) \\ \\ \\ \\ = \\frac{2\\tan(\\theta/2)}{\\sec^2(\\theta/2)} = \\frac{2\\tan(\\theta/2)}{1 + \\tan^2(\\theta/2)} = \\frac{2t}{1+ t^2}$$You can construct a similar argument for \\cos\\theta. 5 Use the fact that$$\\arctan{\\left ( \\frac{p}{q} \\right )} + \\arctan{\\left ( \\frac{q}{p} \\right )} = \\frac{\\pi}{2} $$So all you need to do is arrange the sum to exploit the symmetry you need, i.e., (p,q) \\mapsto (q,p). To do this without repeating, start on the bottom/left axes; there are 9 non-diagonal points each, so the sum along there is 9 ... 5 If you are allowed to use complex number, this integral can be integrated by repeat application of the Euler's formula$$e^{i\\theta} = \\cos(\\theta) + i\\sin\\theta$$Up to integration constant, the integral is equal to:$$\\begin{align}\\int -e^{\\cos t}\\sin(\\sin t + t) dt &= - \\int e^{\\cos t}\\Im\\left[e^{i(\\sin t + t)}\\right] dt ... 5 Your idea is correct. So we have $$\\frac{1}{1-\\cos y}+\\frac{1}{1+\\cos y} = \\frac{1+\\cos y+ 1-\\cos y}{(1+\\cos y)(1-\\cos y)} = \\frac{2}{1-\\cos^2 y}.$$ What do you know about $1-\\cos^2 y$? Can you take it from here? 4 Consider the limit $$\\lim_{n\\to \\infty} \\frac{E_n}{n!}\\left(\\frac{\\pi}{2}\\right)^n$$ Using $E_{n} \\sim \\frac{(-1)^{(n-1)/2}", "\\sin x \\sin y$$ • $$cos(x - y) = \\cos x \\cos y + \\sin x \\sin y$$ • $$sin(x + y) = \\sin x \\cos y + \\cos x \\sin y$$ • $$sin(x - y) = \\sin x \\cos y - \\cos x \\sin y$$ • \\begin{aligned} \\cos(\\frac{\\pi}{2} + x) &= -\\sin x\\\\ \\cos(\\pi - x) &= -\\cos x\\\\ \\cos(\\pi + x) &= \\cos x\\\\ \\cos(2\\pi - x) &= \\cos x \\end{aligned} \\begin{aligned} \\sin(\\frac{\\pi}{2} + x) &= \\cos x\\\\ \\sin(\\pi - x) &= \\sin x\\\\ \\sin(\\pi + x) &= -\\sin x\\\\ \\sin(2\\pi - x) &= -\\sin x \\end{aligned} • If none of the angles $$x,y$$ and $$(x\\pm y)$$ is an odd multiple of $$\\dfrac{\\pi}{2}$$, then $\\tan(x + y) = \\frac{\\tan x + \\tan y}{1 - \\tan x \\tan y}$ • If none of the angles $$x,y$$ and $$(x\\pm y)$$ is an odd multiple of $$\\pi$$, then $\\cot(x + y) = \\frac{\\cot x \\cot y - 1}{\\cot x + \\cot y}$ • $$\\cos 2x = \\cos^2x - \\sin^2x = 2\\cos^2x - 1 = 1 - 2\\sin^2x = \\dfrac{1-\\tan^2x}{1+\\tan^2x}$$ • $$\\sin 2x = 2\\sin x \\cos x = \\dfrac{2\\tan x}{1 + \\tan^2x}$$ • $$\\tan 2x = \\dfrac{2\\tan x}{1 - \\tan^2x}$$ • $$\\sin 3x = 3\\sin x - 4\\sin^3x$$ • $$\\cos 3x = 4\\cos^3x - 3\\cos x$$ • $$\\tan 3x", "& = \\csc^2\\theta \\end{align*} Where you made your mistake was in concluding that the fact that cotangent is odd implied $1 + \\cot^2(-\\theta) = 1 - \\cot^2\\theta$ rather than $1 + \\cot^2(-\\theta) = 1 + (-\\cot\\theta)^2 = 1 + \\cot^2\\theta$. • Yes, I noticed my mistake as soon as I saw everyone's answers! Hit me like a ton of bricks, haha. Thanks for the help, everyone! – enharmonics Jun 24 '15 at 13:38 $1+\\cot^2(-\\theta)=1+(-\\cot\\theta)^2=1+\\cot^2\\theta=1+\\dfrac{\\cos^2\\theta}{\\sin^2\\theta}=\\dfrac{\\sin^2\\theta+\\cos^2\\theta}{\\sin^2\\theta}=\\dfrac{1}{\\sin^2\\theta}=\\csc^2\\theta$ You were on the right path: \\begin{align}1+\\cot^2(\\theta) &=\\frac{\\sin^\\color{green}2(\\theta)}{\\sin^\\color{green}2(\\theta)}(1+\\cot^2(\\theta))\\\\ &=\\frac{\\sin^2(\\theta)+\\sin^2(\\theta)\\cot^2(\\theta)}{\\sin^2(\\theta)}\\\\ &=\\frac{\\sin^2(\\theta)+\\cos^2(\\theta)}{\\sin^2(\\theta)}\\\\ &=\\frac1{\\sin^2(\\theta)}\\\\ &=\\csc^2(\\theta).\\end{align} One way to look at these is to convert all expressions to ratios of the variables $o,a,h$ $$1+cot^2(-θ)=csc^2(θ)$$ $$1+(\\frac{a}{o})^2=(\\frac{h}{o})^2$$ $$(\\frac{o}{o})^2+(\\frac{a}{o})^2=(\\frac{h}{o})^2$$ $$\\frac{o^2+a^2}{o^2}=\\frac{h^2}{o^2}$$ $$o^2+a^2=h^2$$ Which checks out due to the Pythagorean Theorem. I skipped over the negative angle identity of cotan since it is squared. Initially note that $$\\cot^2(-\\theta)=(\\cot(-\\theta))^2=(-\\cot\\theta)^2=(\\cot\\theta)^2=\\cot^2\\theta$$ Then, proof the identity $1+\\cot^2(-\\theta)= \\csc^2\\theta$ is same that proof $$1+\\cot^2\\theta=\\csc^2\\theta$$ Hence, divide the identity $\\sin^2\\theta+\\cos^2\\theta=1$ by $\\sin^2\\theta$. $$\\frac{\\sin^2\\theta}{\\sin^2\\theta}+\\frac{\\cos^2\\theta}{\\sin^2\\theta}=\\frac{1}{\\sin^2\\theta} \\Rightarrow 1+\\cot^2\\theta=\\csc^2\\theta.$$ • The commands for the trigonometric functions, used in math mode, are \\sin for sine, \\cos for cosine, \\tan for tangent, \\csc for cosecant, \\sec for secant, and \\cot for cotangent. – N. F. Taussig Jun 24 '15 at 9:43", "(d) $s = \\frac{t}{\\sqrt{1 - t^2}} \\ + \\ cos^{-1} t$ \\vspace{.25cm} (e) $y = sin^{-1} \\sqrt{x}$ \\vspace{.25cm} (f) $z = cot^{-1} ( \\frac { y}{1 - y^2} )$ \\vspace{.25cm} \\newpage \\noindent {\\bf Solutions:} \\vspace{.25cm} (a) $y = tan^{-1} \\ t^4$ \\begin{align*} \\frac{dy}{dt} &= \\frac{d}{dt} tan^{-1} \\ ( t^4) \\\\ &= \\frac{1}{1 + {(t^4)}^2} \\ {\\centerdot} \\ \\frac{d}{dt} \\ (t^4) \\\\ &= \\frac{4 t^3}{1 + t^8} \\end{align*} \\vspace{.25cm} (b) $z = t \\ cot^{-1} (1 + t^2)$ \\begin{align*} \\frac{dz}{dt} &= \\frac{d}{dt} \\ t \\ cot^{-1} (1 + t^2) \\\\ &= cot^{-1} (1 + t^2) + \\ t \\ {\\centerdot} \\ \\frac{-1}{1 + {(1 + t^2)}^2} \\ {\\centerdot} \\ (2t) \\\\ &= cot^{-1} (1 + t^2) - \\frac{2t^2}{t^4 + 2 t^2 + 2} \\\\ \\end{align*} \\vspace{.25cm} (c) $x = sin^{-1} \\sqrt{1 - t^4}$ \\begin{align*} \\frac{dx}{dt} &= \\frac{d}{dt} \\ sin^{-1} \\sqrt{1 - t^4} \\\\ &= \\frac{1}{\\sqrt{ 1 - {(\\sqrt { 1 - t^4})}^2} }\\ {\\centerdot} \\ \\frac{d}{dt} (\\sqrt{1 - t^4}) \\\\ &= \\frac{1}{\\sqrt{ 1 - (1 - t^4) }} \\ {\\centerdot} \\ \\frac{1}{2} \\ {\\centerdot} \\ {(1 - t^4)}^{\\frac{-1}{2}} \\ {\\centerdot} \\ ( -4t^3) \\\\ &= \\frac{1}{\\sqrt{1 - 1 + t^4}} \\ {\\centerdot} \\ \\frac{1}{\\sqrt{1 - t^4}} \\ {\\centerdot} \\ (-2t^3) \\\\ &= \\frac{1}{t^2} \\ {\\centerdot} \\ \\frac{1}{\\sqrt{1 - t^4}} \\ {\\centerdot} \\ (-2 t^3) \\\\ &= \\frac{-2t}{\\sqrt{1 - t^4}}\\\\", ")$$ 4) $$\\sec \\left (\\dfrac{3\\pi }{8} \\right )$$ $$\\sqrt{2(2+\\sqrt{2})}$$ For the exercises 5-6, use Figure below to find the desired quantities. 5) $$\\sin(2\\beta ),\\cos(2\\beta ),\\tan(2\\beta ),\\sin(2\\alpha ),\\cos(2\\alpha ),\\tan(2\\alpha )$$ 6) $$\\sin \\left (\\frac{\\beta }{2} \\right ) ,\\cos\\left (\\frac{\\beta }{2} \\right ),\\tan\\left (\\frac{\\beta }{2} \\right ),\\sin\\left (\\frac{\\alpha }{2} \\right ),\\cos\\left (\\frac{\\alpha }{2} \\right ),\\tan\\left (\\frac{\\alpha }{2} \\right )$$ $$\\dfrac{\\sqrt{2}}{10},\\dfrac{7\\sqrt{2}}{10},\\dfrac{1}{7},\\dfrac{3}{5},\\dfrac{4}{5},\\dfrac{3}{4}$$ For the exercises 7-8, prove the identity. 7) $$\\dfrac{2\\cos (2x)}{\\sin (2x)}=\\cot x-\\tan x$$ 8) $$\\cot x \\cos (2x) = -\\sin (2x)+\\cot x$$ \\begin{align*} \\cot x \\cos (2x) &= \\cot x(1-2\\sin ^2 x)\\\\ &= \\cot x-\\dfrac{\\cos x}{\\sin x}(2)\\sin ^2 x\\\\ &= -2\\sin x \\cos \\\\ &= -\\sin (2x)+\\cot x \\end{align*} For the exercises 9-10, rewrite the expression with no powers. 9) $$\\cos ^2 x \\sin ^4 (2x)$$ 10) $$\\tan ^2 x \\sin ^3 x$$ $$\\dfrac{10\\sin x-5\\sin (3x)+\\sin (5x)}{8(\\cos (2x)+1)}$$ ## 7.4: Sum-to-Product and Product-to-Sum Formulas For the exercises 1-3, evaluate the product for the given expression using a sum or difference of two functions. Write the exact answer. 1) $$\\cos \\left ( \\dfrac{\\pi }{3} \\right )\\sin \\left ( \\dfrac{\\pi }{4} \\right )$$ 2) $$2\\sin \\left ( \\dfrac{2\\pi }{3} \\right )\\sin \\left ( \\dfrac{5\\pi }{6} \\right )$$ $$\\dfrac{\\sqrt{3}}{2}$$ 3) $$2\\cos \\left ( \\dfrac{\\pi }{5} \\right )\\cos \\left ( \\dfrac{\\pi }{3} \\right )$$ For the exercises 4-5, evaluate the sum by using a", "0$ (so $c_2(r) = 0$). Your lecturer's choice leads to something like \\begin{align*} x(r, s) & = r\\sin s \\\\ y(r, s) & = r\\cos s \\end{align*} while your choice leads to something like \\begin{align*} x(r, s) & = r\\cos s \\\\ y(r, s) & = -r\\sin s. \\end{align*} It should not matter which one you choose. Note that with your lecturer's choice, you have \\begin{align*} \\frac yx - \\frac xy & = \\cot s - \\tan s \\\\ & = 2\\cot(2s) \\end{align*} and with your choice, you have \\begin{align*} \\frac yx - \\frac xy & = -\\tan s + \\cot s \\\\ & = 2\\cot(2s). \\end{align*} In both cases, $\\frac yx - \\frac xy$ becomes the same $2\\cot(2s)$, and that is the expression you'll need to solve for $v$ and $u$: \\begin{align*} \\frac{\\partial v}{\\partial s} & = \\left(\\frac yx - \\frac xy\\right)v \\\\ & = 2\\cot(2s)v\\\\ \\therefore v & = C(r)\\sin(2s) \\\\ \\text{and from} \\frac{\\partial u}{\\partial s} = v \\quad \\Longrightarrow \\quad u & = -\\frac 12C(r)\\cos(2s) + D(r) \\\\ \\therefore u & = \\tilde C(x^2 + y^2) \\cdot (x^2 - y^2) + \\tilde D(x^2 + y^2). \\end{align*} Although $\\cos(2s)$ becomes different expressions in different parametrizations, the final expression is the same because the sign difference is subsumed in $\\tilde C$. I did get sloppy and ignored", "+ 9{\\cot ^2}a + ({\\tan ^2}b + 4{\\cot ^2}b)(1 + {\\tan ^2}a)\\\\ & > 14 + 5{\\tan ^2}a + 9{\\cot ^2}a + 2\\tan b \\cdot 2\\cot b(1 + {\\tan ^2}a)\\\\ & = 18 + 9({\\tan ^2}a + {\\cot ^2}a) \\ge 18 + 9 \\cdot 2\\tan a\\cot a = 36 \\end{align} Equality holds when \\begin{align}\\tan a = \\cot a\\;{\\rm{and}}\\;\\tan b = 2\\cot b\\end{align}, which implies that \\begin{align}{\\cos ^2}a = {\\sin ^2}a\\;{\\rm{and}}\\end{align} \\begin{align}{\\cos ^2}b = {\\sin ^2}b\\end{align}. Because \\begin{align}{\\sin ^2}\\theta + {\\cos ^2}\\theta = 1\\end{align}, equality holds when \\begin{align}{\\cos ^2}a = \\frac{1}{2}\\;{\\text{and}}\\;{\\cos ^2}b = \\frac{1}{3};\\end{align} that is \\begin{align}x = \\frac{1}{6},\\;y = \\frac{1}{3},\\;z = \\frac{1}{2}\\end{align}. Example - 10 Let $$A_1$$be the center of the square inscribed in acute triangle $$ABC$$ with two vertices of the square on side $$BC$$ (shown in figure). Thus one of the two remaining vertices of the square lies on side $$AB$$ and the other on segment $$AC.$$ Points $$B_1$$ and $$C_1$$ are defined in a similar way for inscribed squares with two vertices on sides $$AC$$ and $$AB,$$ respectively. Prove that lines $$AA_1, BB_1, CC_1$$are concurrent. Solution: Let line $$AA_1$$ and segment $$BC$$ intersect at $$A_2.$$ We define $$B_2$$ and $$C_2$$ analogously. By Ceva’s theorem, it suffices to show that \\begin{align}\\frac{{\\sin \\angle BA{A_2}}}{{\\sin \\angle {A_2}AC}} \\cdot \\frac{{\\sin \\angle CB{B_2}}}{{\\sin \\angle {B_2}BA}} \\cdot \\frac{{\\sin \\angle AC{C_2}}}{{\\sin \\angle", "you do this: $$\\displaystyle \\sec x=\\cot x+\\tan x$$ \\require{cancel} \\displaystyle \\begin{align}{{\\sec }^{2}}x&={{\\left( {\\cot x+\\tan x} \\right)}^{2}}\\\\1+{{\\tan }^{2}}x&={{\\cot }^{2}}x+2\\cancel{{\\cot x\\tan x}}+{{\\tan }^{2}}x\\\\1+{{\\tan }^{2}}x&={{\\cot }^{2}}x+2+{{\\tan }^{2}}x\\\\1+\\cancel{{{{{\\tan }}^{2}}x}}&={{\\cot }^{2}}x+2+\\cancel{{{{{\\tan }}^{2}}x}}\\\\{{\\cot }^{2}}x&=-1\\end{align} No solution, since we can’t square something and have it be a negative number. $$\\displaystyle \\frac{{{{{\\cos }}^{2}}\\left( x \\right)}}{{{{{\\sin }}^{2}}\\left( x \\right)}}=3$$ $$\\displaystyle \\begin{array}{c}{{\\cot }^{2}}\\left( x \\right)=3\\\\\\cot \\left( x \\right)=\\pm \\sqrt{3}\\end{array}$$ $$\\displaystyle x=\\left\\{ {\\frac{\\pi }{3},\\,\\,\\frac{{2\\pi }}{3},\\,\\,\\frac{{4\\pi }}{3},\\,\\frac{{5\\pi }}{3}} \\right\\}$$ Use Quotient Identity to convert $$\\displaystyle \\frac{{\\cos }}{{\\sin }}$$ into cot Take square roots of both sides, and don’t forget the $$\\pm$$ Here’s another one where we have to check for extraneous solutions. Solve over the reals: Solving with Trig Identities Hints $$\\displaystyle \\frac{{1-\\cos x}}{{\\sin x}}=\\frac{{\\sin x}}{{1+\\cos x}}$$ \\displaystyle \\begin{align}\\frac{{1-\\cos x}}{{\\sin x}}-\\frac{{\\sin x}}{{1+\\cos x}}&=0\\\\\\frac{{\\left( {1-\\cos x} \\right)\\left( {1+\\cos x} \\right)}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}-\\frac{{\\left( {\\sin x} \\right)\\left( {\\sin x} \\right)}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}&=0\\\\\\frac{{1-{{{\\cos }}^{2}}x}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}-\\frac{{{{{\\sin }}^{2}}x}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}&=0\\\\\\frac{{{{{\\sin }}^{2}}x}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}-\\frac{{{{{\\sin }}^{2}}x}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}&=0\\\\0&=0\\end{align} This appears to be all real numbers, but since we have denominators that can’t be 0, we have to check for those cases. Get all terms on one side Find common denominator to add terms Turn $$\\left( {1-{{{\\cos }}^{2}}x} \\right)$$ into $${{\\sin }^{2}}x$$ We get", "x_1 \\cot x_2 - (\\cot x_2)^2 -2)^2}{4(\\cot x_1 + \\cot x_2)^4}\\\\ =\\ &0. \\end{align} 2) $$n = 5$$. Let $$x_1, x_2 \\in (0, \\frac{\\pi}{2})$$ satisfying \\begin{align} -4(\\cot x_2)^2(\\cot x_1)^2 + (-2(\\cot x_2)^3 + 6\\cot x_2)\\cot x_1 + (\\cot x_2)^4 + 4(\\cot x_2)^2 - 1 = 0. \\end{align} Let $$y_1 = \\cot x_1, \\ y_2 = \\cot x_2$$. We have $$-4y_2^2y_1^2 + (-2y_2^3 + 6y_2) y_1 +y_2^4+4y_2^2-1 = 0$$ which results in $$\\sqrt{5}y_2^2-4y_1y_2-y_2^2+\\sqrt{5}+3 = 0$$ since $$x_1, x_2 \\in (0, \\frac{\\pi}{2}).$$ Let $$x_3 = x_2, \\ x_4 = x_1, \\ x_5 = \\pi - x_1 - x_2 - x_3 - x_4.$$ We have $$x_1, x_2, x_3, x_4, x_5 > 0; \\ x_1 + x_2+x_3+x_4+x_5=\\pi.$$ Note that $$\\big(\\frac{\\sin \\frac{\\pi}{5}}{\\sin \\frac{2\\pi}{5}}\\big)^5 = \\frac{5\\sqrt{5}-11}{2}$$. We have \\begin{align} &\\frac{\\sin x_1 \\sin x_2 \\sin x_3 \\sin x_4 \\sin x_5} {\\sin (x_1+x_2) \\sin (x_2 + x_3) \\sin (x_3 + x_4) \\sin (x_4+x_5) \\sin (x_5+x_1)} - \\Big(\\frac{\\sin \\frac{\\pi}{5}}{\\sin \\frac{2\\pi}{5}}\\Big)^5\\\\ =\\ &\\frac{(\\sin x_1)^2(\\sin x_2)^2\\sin (2x_1 + 2x_2)}{(\\sin (x_1+x_2))^2\\sin 2x_2 (\\sin (x_1 + 2x_2))^2} - \\frac{5\\sqrt{5}-11}{2}\\\\ =\\ &\\frac{(y_1y_2-1)(y_2^2+1)^2}{y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2} - \\frac{5\\sqrt{5}-11}{2}\\\\ =\\ &-\\frac{5\\sqrt{5}-11}{16}\\frac{ (\\sqrt{5}y_2^2+2y_1y_2+3y_2^2+\\sqrt{5}+1)(\\sqrt{5}y_2^2-4y_1y_2-y_2^2+\\sqrt{5}+3)^2} {y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2}\\\\ =\\ &0. \\end{align} • This is interesting +1. Since the general Jacobian matrix I computed was singular, maybe there are multiple points where the inequalities are attained. – dezdichado Jun 19 '19 at 19:45 • When $n = 4,$ it must be that $2x_1+x_2", "(2 \\sin x-1)(\\sin x+1)\\end{align*} Solving the first part set equal to zero within the interval yields: 012x=2sinx1=sinx=π6,5π6,11π6,7π6\\begin{align*}0 &= 2 \\sin x-1\\\\ \\frac{1}{2} &= \\sin x\\\\ x &= \\frac{\\pi}{6}, \\frac{5 \\pi}{6}, - \\frac{11 \\pi}{6}, -\\frac{7 \\pi}{6}\\end{align*} Solving the second part set equal to zero yields: 01x=sinx+1=sinx=π2,3π2\\begin{align*}0 &= \\sin x+1\\\\ -1 &= \\sin x\\\\ x &= -\\frac{\\pi}{2}, \\frac{3 \\pi}{2}\\end{align*} These are the six solutions that will appear as intersections of the two graphs f(x)=cos2x\\begin{align*}f(x)=\\cos 2x\\end{align*} and g(x)=sinx\\begin{align*}g(x)=\\sin x\\end{align*} Note that the terms “general solution,” “completely solve”, and“solve exactly” #### Example 3 Determine the general solution to the following equation. cotx1=0\\begin{align*}\\cot x-1=0\\end{align*} cotx1cotx=0=1\\begin{align*}\\cot x-1 &= 0\\\\ \\cot x &= 1\\end{align*} One solution is x=π4\\begin{align*}x=\\frac{\\pi}{4}\\end{align*}. However, since this question asks for the general solution, you need to find every possible solution. You have to know that cotangent has a period of π\\begin{align*}\\pi\\end{align*} which means if you add or subtract π\\begin{align*}\\pi\\end{align*} from π4\\begin{align*}\\frac{\\pi}{4}\\end{align*} then it will also yield a height of 1. To capture all these other possible x\\begin{align*}x\\end{align*} values you should use this notation. x=π4±nπ\\begin{align*}x=\\frac{\\pi}{4} \\pm n \\cdot \\pi\\end{align*} where n\\begin{align*}n\\end{align*} is a integer Notice that trigonometric equations may have an infinite number of solutions that repeat in a certain pattern because they are periodic functions. When you see these directions remember to find all the solutions by using notation like in", "a}\\end{align*} and \\begin{align*}\\cot x&=\\cot(a-135^{\\circ})\\\\&=\\dfrac{\\cos a-\\sin a}{\\sin a+\\cos a}\\end{align*} and adding them up gives \\begin{align*}\\tan x+\\cot x&=\\dfrac{\\sin a+\\cos a}{\\cos a-\\sin a}+\\dfrac{\\cos a-\\sin a}{\\sin a+\\cos a}\\\\&=\\dfrac{2}{\\cos^2 a-\\sin^2 a}\\end{align*} Finally, \\begin{align*}\\sec x&=\\dfrac{1}{\\cos x}\\\\&=-\\dfrac{\\sqrt{2}}{\\sin a+\\cos a}\\end{align*} and \\begin{align*}\\csc x&=\\dfrac{1}{\\sin x}\\\\&=\\dfrac{\\sqrt{2}}{\\sin a-\\cos a}\\end{align*}, adding the two gives \\begin{align*}\\sec x+\\csc x&=-\\dfrac{\\sqrt{2}}{\\sin a+\\cos a}+\\dfrac{\\sqrt{2}}{\\sin a-\\cos a}\\\\&=\\dfrac{2\\sqrt{2}\\cos a}{\\sin^2 a-\\cos^2 a}\\end{align*} $\\therefore|\\sin x+\\cos x+\\tan x+\\cot x+\\sec x+\\csc x|$ $=|-\\sqrt{2}\\cos a+\\dfrac{2}{\\cos^2 a-\\sin^2 a}+\\dfrac{2\\sqrt{2}\\cos a}{\\sin^2 a-\\cos^2 a}|$ $=|-\\sqrt{2}\\cos a+\\dfrac{2}{\\cos^2 a-\\sin^2 a}-\\dfrac{2\\sqrt{2}\\cos a}{\\cos^2 a-\\sin^2 a}|$ $=|-\\sqrt{2}\\cos a+\\dfrac{2-2\\sqrt{2}\\cos a}{\\cos^2 a-\\sin^2 a}|$ $=|-\\sqrt{2}\\cos a+\\dfrac{2-2\\sqrt{2}\\cos a}{2\\cos^2 a-1}|$ $=|-\\sqrt{2}\\cos a+\\dfrac{2-2\\sqrt{2}\\cos a}{(\\sqrt{2}\\cos a-1)(\\sqrt{2}\\cos a+1)}|$ $=|-\\sqrt{2}\\cos a+\\dfrac{2(1-\\sqrt{2}\\cos a)}{(\\sqrt{2}\\cos a-1)(\\sqrt{2}\\cos a+1)}|$ $=|-\\sqrt{2}\\cos a-\\dfrac{2}{\\sqrt{2}\\cos a+1}|$ $=|-\\left(\\sqrt{2}\\cos a+\\dfrac{2}{\\sqrt{2}\\cos a+1} \\right)|$ $=|\\sqrt{2}\\cos a+\\dfrac{2}{\\sqrt{2}\\cos a+1}|$ $=|\\sqrt{2}\\cos a+1+\\dfrac{2}{\\sqrt{2}\\cos a+1}-1|$ Treat $\\sqrt{2}\\cos a>0$, we can hence apply the AM-GM inequality to the following: $(\\sqrt{2}\\cos a+1)+\\dfrac{2}{\\sqrt{2}\\cos a+1}\\ge 2\\sqrt{\\cancel{(\\sqrt{2}\\cos a+1)}\\left(\\dfrac{2}{\\cancel{(\\sqrt{2}\\cos a+1)}} \\right)}= 2\\sqrt{2}$ Since $\\sqrt{2}\\cos a+\\dfrac{2}{\\sqrt{2}\\cos a+1} = 2 \\sqrt{2}-1$ does have solutions, we can conclude by now that [MATH]|\\sin x+\\cos x+\\tan x+\\cot x+\\sec x+\\csc x|[/MATH] is $2 \\sqrt{2}-1$.", "+\\frac{1}{2}\\right)x\\right)}{\\sin(x/2)}.$ This function of x is called the Dirichlet kernel. This is one of Lagrange's trigonometric identities. ### Double-, triple-, and half-angle formulae These can be shown by using either the sum and difference identities or the multiple-angle formulae. Double-angle formulae[17][18] \\begin{align} \\sin 2\\theta &= 2 \\sin \\theta \\cos \\theta \\ \\\\ &= \\frac{2 \\tan \\theta} {1 + \\tan^2 \\theta} \\end{align} \\begin{align} \\cos 2\\theta &= \\cos^2 \\theta - \\sin^2 \\theta \\\\ &= 2 \\cos^2 \\theta - 1 \\\\ &= 1 - 2 \\sin^2 \\theta \\\\ &= \\frac{1 - \\tan^2 \\theta} {1 + \\tan^2 \\theta} \\end{align} $\\tan 2\\theta = \\frac{2 \\tan \\theta} {1 - \\tan^2 \\theta}\\!$ $\\cot 2\\theta = \\frac{\\cot^2 \\theta - 1}{2 \\cot \\theta}\\!$ Triple-angle formulae[15][19] \\begin{align}\\sin 3\\theta & = 3 \\cos^2\\theta \\sin\\theta - \\sin^3\\theta \\\\ & = 3\\sin\\theta - 4\\sin^3\\theta \\end{align} \\begin{align}\\cos 3\\theta & = \\cos^3\\theta - 3 \\sin^2 \\theta\\cos \\theta \\\\ & = 4 \\cos^3\\theta - 3 \\cos\\theta\\end{align} $\\tan 3\\theta = \\frac{3 \\tan\\theta - \\tan^3\\theta}{1 - 3 \\tan^2\\theta}\\!$ $\\cot 3\\theta = \\frac{3 \\cot\\theta - \\cot^3\\theta}{1 - 3 \\cot^2\\theta}\\!$ Half-angle formulae[20][21] $\\sin \\frac{\\theta}{2} = \\pm\\, \\sqrt{\\frac{1 - \\cos \\theta}{2}}$ $\\cos \\frac{\\theta}{2} = \\pm\\, \\sqrt{\\frac{1 + \\cos\\theta}{2}}$ \\begin{align} \\tan \\frac{\\theta}{2} &= \\csc \\theta - \\cot \\theta \\\\ &= \\pm\\, \\sqrt{1 - \\cos \\theta \\over 1 + \\cos \\theta} \\\\[8pt] &= \\frac{\\sin \\theta}{1 + \\cos \\theta} \\\\[8pt] &= \\frac{1-\\cos", "for exact values of $$x$$ over the interval $$[0,2\\pi ]$$. \\begin{aligned} 3\\cot ^2 x−3\\cot x &=1 \\\\ 3\\cot ^2x−3\\cot x−1&=0 \\end{aligned} The equation will not factor. Use the quadratic formula for $$\\cot x$$, $$a=3$$, $$b=−3$$, $$c=−1$$. \\begin{aligned} \\cot x&=\\dfrac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\\\ \\cot x&=\\dfrac{-(-3) \\pm \\sqrt{(-3)^{2}-4(3)(-1)}}{2(3)}\\\\ \\cot x&=\\dfrac{3 \\pm \\sqrt{9+12}}{6}\\\\ \\cot x&=\\dfrac{3+\\sqrt{21}}{6} &\\quad \\text { or } \\quad& \\cot x=\\dfrac{3-\\sqrt{21}}{6}\\\\ \\cot x&=\\dfrac{3+4.5826}{6} &\\quad& \\cot x=\\dfrac{3-4.5826}{6}\\\\ \\cot x&=1.2638 &\\quad &\\cot x=-0.2638\\\\ \\tan x&=\\dfrac{1}{1.2638} &\\quad &\\tan x=\\dfrac{1}{-0.2638}\\\\ x&=0.6694,3.81099 & &x=1.8287,4.9703 \\end{aligned} 2. Solve $$−5 \\cos ^2 x+9 \\sin x+3=0$$ for values of x over the interval $$[0,2\\pi ]$$. Change $$\\cos ^2 x$$ to $$1−\\sin ^2 x$$ from the Pythagorean Identity. \\begin{aligned} −5\\cos ^2 x +9\\sin x+3 &=0 \\\\ −5(1−\\sin ^2 x )+9\\sin x+3 &=0 \\\\ −5+5\\sin ^2 x +9\\sin x+3 &=0 \\\\ 5\\sin ^2 x +9\\sin x−2 &=0 \\end{aligned} $$\\begin{array}{l} \\sin x=\\dfrac{-9 \\pm \\sqrt{9^{2}-4(5)(-2)}}{2(5)} \\\\ \\sin x=\\dfrac{-9 \\pm \\sqrt{81+40}}{10} \\\\ \\sin x=\\dfrac{-9 \\pm \\sqrt{121}}{10} \\\\ \\sin x=\\dfrac{-9+11}{10} \\text { and } \\sin x=\\dfrac{-9-11}{10} \\\\ \\sin x=\\dfrac{1}{5} \\text { and }-2 \\\\ \\sin ^{-1}(0.2) \\text { and } \\sin ^{-1}(-2) \\end{array}$$ $$x\\approx .201 \\text{rad}$$ and $$\\pi −.201\\approx 2.941$$ This is the only solutions for $$x$$ since $$−2$$ is not in the range of values. 3. Solve $$3\\sin ^2 x −6\\sin x−2=0$$ for values of $$x$$ over the interval $$[0,2\\pi" ]

[ "+ c^2 - 4)}}{2} = \\frac{-bc + \\sqrt{(4 - b^2)(4 - c^2)}}{2}.$$ This asks for the trigonometric substitution $b = 2\\sin u$ and $c = 2\\sin v$, where $0^\\circ < u,v < 90^\\circ$. Then $$a = 2(-\\sin u\\sin v + \\cos u\\cos v) = 2\\cos (u + v),$$ and if we set $u = B/2$ and $v = C/2$, then $a = 2\\sin (A/2)$, $b = 2\\sin (B/2)$, and $c = \\sin (C/2)$, where $A$, $B$, and $C$ are the angles of a triangle. We have \\begin{align*} ab &= 4\\sin\\frac{A}{2}\\sin\\frac{B}{2} \\\\ &= 2\\sqrt{\\sin A\\tan\\frac{A}{2}\\sin B\\tan\\frac{B}{2}} = 2\\sqrt{\\sin A\\tan\\frac{B}{2}\\sin B\\tan\\frac{A}{2}} \\\\ &\\leq \\sin A\\tan\\frac{B}{2} + \\sin B\\tan\\frac{A}{2} \\\\ &= \\sin A\\cot\\frac{A + C}{2} + \\sin B\\cot\\frac{B + C}{2}, \\end{align*} where the inequality step follows from AM-GM. Likewise, \\begin{align*} bc &\\leq \\sin B\\cot\\frac{B + A}{2} + \\sin C\\cot\\frac{C + A}{2}, \\\\ ca &\\leq \\sin A\\cot\\frac{A + B}{2} + \\sin C\\cot\\frac{C + B}{2}. \\end{align*} Therefore \\begin{align*} ab + bc + ca &\\leq (\\sin A + \\sin B)\\cot\\frac{A + B}{2} + (\\sin B + \\sin C)\\cot\\frac{B + C}{2} + (\\sin C + \\sin A)\\cot\\frac{C + A}{2} \\\\ &= 2\\left(\\cos\\frac{A - B}{2}\\cos\\frac{A + B}{2} + \\cos\\frac{B - C}{2}\\cos\\frac{B + C}{2} + \\cos\\frac{C - A}{2}\\cos\\frac{C + A}{2} \\right)\\\\ &= 2(\\cos A + \\cos B + \\cos C) \\\\ &= 6 - 4\\left(\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2}", "above expression to get: cot2(x)csc2(x)=cos2(x)sin2(x)1sin2(x)=cos2(x)\\begin{align*} \\frac{\\cot^2(x)}{\\csc^2(x)}\\\\ =\\frac{\\frac{\\cos^2(x)}{\\sin^2(x)}}{\\frac{1}{\\sin^2(x)}}\\\\ =\\cos^2(x)\\\\ \\end{align*} ### Review Simplify each trigonometric expression as much as possible. 1. sin(x)cot(x)\\begin{align*}\\sin(x)\\cot(x)\\end{align*} 2. cos(x)tan(x)\\begin{align*}\\cos(x)\\tan(x)\\end{align*} 3. 1+tan(x)1+cot(x)\\begin{align*}\\frac{1+\\tan(x)}{1+\\cot(x)}\\end{align*} 4. 1sin2(x)1+sin(x)\\begin{align*}\\frac{1-\\sin^2(x)}{1+\\sin(x)}\\end{align*} 5. sin2(x)1+cos(x)\\begin{align*}\\frac{\\sin^2(x)}{1+\\cos(x)}\\end{align*} 6. (1+tan2(x))(sec2(x))\\begin{align*}(1+\\tan^2(x))(\\sec^2(x))\\end{align*} 7. sin(x)(tan(x)+cot(x))\\begin{align*}\\sin(x)(\\tan(x)+\\cot(x))\\end{align*} 8. sec(x)sin(x)sin(x)cos(x)\\begin{align*}\\frac{\\sec(x)}{\\sin(x)}-\\frac{\\sin(x)}{\\cos(x)}\\end{align*} 9. sin(x)cot2(x)sin(x)cos2(x)\\begin{align*}\\frac{\\sin(x)}{\\cot^2(x)}-\\frac{\\sin(x)}{\\cos^2(x)}\\end{align*} 10. 1+sin(x)cos(x)sec(x)\\begin{align*}\\frac{1+\\sin(x)}{\\cos(x)}-\\sec(x)\\end{align*} 11. sin2(x)sin4(x)cos2(x)\\begin{align*}\\frac{\\sin^2(x)-\\sin^4(x)}{\\cos^2(x)}\\end{align*} 12. tan(x)csc2(x)+tan(x)sec2(x)\\begin{align*}\\frac{\\tan(x)}{\\csc^2(x)}+\\frac{\\tan(x)}{\\sec^2(x)}\\end{align*} 13. \\begin{align*}\\sqrt{1-cos^2(x)}\\end{align*} 14. \\begin{align*}(1-\\sin^2(x))(\\cos(x))\\end{align*} 15. \\begin{align*}(\\sec^2(x)+\\csc^2(x))-(\\tan^2(x)+\\cot^2(x))\\end{align*} To see the Review answers, open this PDF file and look for section 3.3. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Trigonometric Identity A trigonometric identity is an equation that relates two or more trigonometric functions.", "that gives us $$1 + \\cot^2 x - \\cot^2 x = 1,$$ as desired. - For $\\#1$ $$\\sin^2x+\\sin^2x\\cdot\\tan^2x=\\sin^2x(1+\\tan^2x)=\\frac{\\sin^2x}{\\cos^2x}$$ $$\\implies \\sin^2x+\\sin^2x\\cdot\\tan^2x=\\tan^2x$$ For $\\#2$ $$\\csc x=\\frac1{\\sin x},\\sec x=\\frac1{\\cos x}$$ - 1. \\begin{align} \\tan^2x - \\sin^2x&=\\frac{\\sin^2x}{\\cos^2x}-\\sin^2x\\quad;\\ \\text{where}\\ \\tan x=\\frac{\\sin x}{\\cos x}\\\\ &=\\sin^2x\\left(\\frac{1}{\\cos^2x}-1\\right)\\\\ &=\\sin^2x\\left(\\frac{1}{\\cos^2x}-\\frac{\\cos^2x}{\\cos^2x}\\right)\\\\ &=\\frac{\\sin^2x}{\\cos^2x}\\left(1-\\cos^2x\\right)\\\\ &=\\color{blue}{\\tan^2x \\cdot \\sin^2x}\\quad;\\ \\text{where}\\ \\sin^2x + \\cos^2x=1. \\end{align} 2. \\begin{align} \\frac{\\csc x}{\\sec x}&=\\dfrac{\\dfrac{1}{\\sin x}}{\\dfrac{1}{\\cos x}}\\\\ &=\\frac{1}{\\color{red}{\\sin x}}\\times\\frac{\\color{red}{\\cos x}}{1}\\\\ &=\\frac{\\cos x}{\\sin x}\\\\ &=\\color{blue}{\\cot x}. \\end{align} - I just wanted to point out that when doing proofs like this that your argument should lead from the initial form (e.g. $\\tan ^2 x - \\sin^2 x$) to the final form ($\\sin^2x \\tan^2 x$). That should be your final answer, but sometimes it is just easier to prove that the LHS and RHS are equal. For example, my doodles may look like this $$\\tan ^2 x - \\sin^2 x = \\sin^2x \\tan^2 x$$ $$\\tan ^2 x = \\sin^2x \\tan^2 x + \\sin^2 x$$ $$\\tan ^2 x = \\sin^2x(\\tan^2 x + 1)$$ $$\\tan ^2 x = \\sin^2x(\\sec^2 x)$$ $$\\tan ^2 x = \\sin^2x\\bigg(\\frac{1}{\\cos^2 x}\\bigg)$$ $$\\tan ^2 x = \\tan ^2 x$$ Now, I'm going to use my doodles to present my real argument (the one I hand into my teacher). Recall that I ended up $\\tan^2 x$, but started with $\\tan ^2 x - \\sin^2 x$, so I'm just going to take", "\\tan^2 A + 1 &= \\sec^2 A \\ 1+\\cot^2 A &= \\csc^2 A \\end{align} #### Miscellaneous identities \\begin{align} \\cos^6 A &= \\frac {1}{32}\\cos 6A + \\frac {3}{16}\\cos 4A + \\frac {15}{32}\\cos 2A + \\frac {5}{16} \\end{align} #### Sum and difference identities \\begin{align} \\sin(A \\pm B) &= \\sin A \\cos B \\pm \\cos A \\sin B \\ \\cos(A \\pm B) &= \\cos A \\cos B \\mp \\sin A \\sin B \\ \\tan(A \\pm B) &= \\frac{\\tan A \\pm \\tan B}{1 \\mp \\tan A \\tan B} \\ \\cot(A \\pm B) &= \\frac{\\cot A \\cot B \\mp 1}{\\cot B \\pm \\cot A} \\end{align} #### Double-angle identities \\begin{align} \\sin 2A &= 2 \\sin A \\cos A \\ &= \\frac{2 \\tan A}{1 + \\tan^2 A}\\ \\cos 2A &= \\cos^2 A - \\sin^2 A \\ &= 2 \\cos^2 A -1 \\ &= 1-2 \\sin^2 A \\ &= {1 - \\tan^2 A \\over 1 + \\tan^2 A}\\ \\tan 2A &= \\frac{2 \\tan A}{1 - \\tan^2 A}\\ &= \\frac{2 \\cot A}{\\cot^2 A - 1}\\ &= \\frac{2}{\\cot A - \\tan A} \\end{align} #### Half-angle identities Note that $\\pm$ are correct, it means it may be either one, depending on the value of A/2. \\begin{align} \\sin \\frac{A}{2} &= \\pm \\sqrt{\\frac{1-\\cos A}{2}} \\ \\cos \\frac{A}{2} &= \\pm \\sqrt{\\frac{1+\\cos A}{2}} \\\\\\tan \\frac{A}{2} &= \\pm \\sqrt{\\frac{1-\\cos A}{1+\\cos A}} = \\frac {\\sin", "\\left( {\\frac{{3 + \\sqrt 3 }}{{\\sqrt 3 - 1}}} \\right) = 0\\\\ \\Rightarrow\\quad &\\qquad\\;\\;(x - 1)\\left( {x - \\sqrt 3 } \\right)\\left( {x - (2 + \\sqrt 3 )} \\right) = 0 \\\\ \\Rightarrow \\quad &\\qquad x = 1,\\;\\sqrt 3 ,\\;2 + \\sqrt 3 \\\\ \\end{align} The angles are \\begin{align}\\frac{\\pi }{4},\\;\\frac{\\pi }{3},\\;\\frac{\\pi }{{12}}\\end{align} Example - 32 In a $$\\Delta ABC$$, suppose that \\begin{align}\\tan \\frac{A}{2},\\;\\tan \\frac{B}{2},\\;\\tan \\frac{C}{2}\\end{align} are in H.P.. What is the minimum possible value of \\begin{align}\\cot \\frac{B}{2}\\end{align}? Solution: Refer to the result of Example - 28 Part (b). If we divide throughout by \\begin{align}\\tan \\frac{A}{2}\\tan \\frac{B}{2}\\tan \\frac{C}{2}\\end{align}, we have, $\\cot \\frac{A}{2} + \\cot \\frac{B}{2} + \\cot \\frac{C}{2} = \\cot \\frac{A}{2}\\cot \\frac{B}{2}\\cot \\frac{C}{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\ldots \\left( 1 \\right)$ Now, since \\begin{align}\\tan \\frac{A}{2},\\tan \\frac{B}{2},\\tan \\frac{C}{2}\\end{align} are in H.P., their reciprocals, i.e. \\begin{align}\\cot \\frac{A}{2},\\;\\cot \\frac{B}{2},\\;\\cot \\frac{C}{2}\\end{align}, are in A.P. So, $\\qquad\\cot \\frac{A}{2} + \\;\\cot \\frac{C}{2} = 2\\cot \\frac{B}{2}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\ldots \\left( 2 \\right)$ Using (2) in (1), we obtain $\\cot \\frac{A}{2}\\;\\cot \\frac{C}{2} = 3$ Finally, we use the A.M-G.M. inequality on \\begin{align}\\cot \\frac{A}{2},\\;\\cot \\frac{C}{2}\\end{align}: \\begin{align}&\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{{\\cot \\frac{A}{2} + \\cot \\frac{C}{2}}}{2} = \\cot \\frac{B}{2} \\geqslant \\sqrt {\\cot \\frac{A}{2}\\cot \\frac{C}{2}} = \\sqrt 3 \\\\ \\Rightarrow\\quad &\\qquad\\;\\;\\;\\cot \\frac{B}{2} \\geqslant \\sqrt 3 \\\\ \\end{align} Therefore, the minimum value of \\begin{align}\\cot \\frac{B}{2}\\end{align} is $$\\sqrt 3$$. From this, we can further infer that the maximum value of \\begin{align}\\tan \\frac{B}{2}\\end{align}", "points are of $g(z)=\\tan(z)$ and $h(z)=z\\cot ^2(z)$, that is $z_n= (n+\\frac{1}{2})\\pi$ and $w_n = \\pi n$. (b) $z_n$ are simple poles and $\\Res (f,z_n)= \\Res (\\tan(z), z_n)= \\frac{\\sin(z)}{(\\cos (z))'}\\bigr|_{z=z_n} =-1$. $w_0$ is a simple pole $$\\Res (f,w_0)= \\Res (z\\cot^2(z), w_0)= \\Res (z\\cot (z) \\times \\cot(z), 0)= \\Res (\\cot(z),0)= 1$$ because $\\lim _{z\\to 0} z\\cot(z)=1$. $w_n$ with $n\\ne 0$ are double poles and \\begin{align*} \\Res (f,w_n)=&\\Res (z\\cot^2 (z), w_n)= \\Res ((\\pi n +w)\\cot^2 (w) , 0) =\\\\ &\\pi n\\Res (\\cot^2 (w) , 0) +\\Res (w\\cot^2(w),0) =1 \\end{align*} because $\\Res (\\cot^2 (w) , 0)=0$ (since $\\cot^2(w)$ is an even function) and $\\Res (w\\cot^2(w),0) =1$ we already calculated. $\\infty$ is a not isolated singularity and therefore residue here is not defined. 39 Final Exam / FE-P2 official « on: December 20, 2018, 04:49:22 AM » (a)--(b) If $|z|=r$ and $\\arg{w}=t$, then $z=r e^{it}$ and \\begin{align*} w= \\frac{1}{2}\\bigl(re^{it} +r^{-1}e^{-it}\\bigr)&= \\cosh(s)\\cos(t)+i\\sinh(s)\\sin (t) \\end{align*} with $s=\\ln (r)$ (and $r=e^s$). Then $w=u+iv$ with $\\frac{u^2}{a^2}+\\frac{v^2}{b^2}=1$, $a=\\cos(s)$, $b=\\sinh(s)$ and also $\\frac{u^2}{A^2}-\\frac{v^2}{B^2}=1$, with $A=\\cos (\\theta)$, $B=\\sin(\\theta)$, $A^2+B^2=1$. (c) Consider, what the circle $\\{z\\colon |z|=1\\}$ is mapped to: $z=e^{it}$, then $w =\\frac{1}{2}(e^{it}+e^{-it})=\\cos(t)$ and it runs a line segment $[-1,1]\\subset\\mathbb{R}$. So $f$ maps the unit disk onto compliment of it $\\mathbb{C}^*\\setminus [-1,1]$ (and $0$ is mapped to $\\infty$). (d) See attachment (e) $w=\\frac{1}{2}(z+z^{-1}) \\implies z^2 -2wz +1=0$; it has two", "&& \\rightarrow \\text{cancel}\\ (1 + \\cos x)\\end{align*} ### Review Use trigonometric identities to simplify each expression as much as possible. 1. tan(x)cos(x)\\begin{align*}\\tan(x)\\cos(x)\\end{align*} 2. cos(x)cos3(x)\\begin{align*}\\cos(x)-\\cos^3(x)\\end{align*} 3. 1cos2(x)sin(x)\\begin{align*}\\frac{1-\\cos^2(x)}{\\sin(x)}\\end{align*} 4. cot(x)sin(x)\\begin{align*}\\cot(x)\\sin(x)\\end{align*} 5. 1sin2(x)cos(x)\\begin{align*}\\frac{1-\\sin^2(x)}{\\cos(x)}\\end{align*} 6. sin(x)csc(x)\\begin{align*}\\sin(x)\\csc(x)\\end{align*} 7. tan(x)cot(x)\\begin{align*}\\tan(-x)\\cot(x)\\end{align*} 8. sec2(x)tan2(x)cos2(x)+sin2(x)\\begin{align*}\\frac{\\sec^2(x)-\\tan^2(x)}{\\cos^2(x)+\\sin^2(x)}\\end{align*} Prove each identity. 1. tan(x)+cot(x)=sec(x)csc(x)\\begin{align*}\\tan(x)+\\cot(x)=\\sec(x)\\csc(x)\\end{align*} 2. sin(x)=sin2(x)+cos2(x)csc(x)\\begin{align*}\\sin(x)=\\frac{\\sin^2(x)+\\cos^2(x)}{\\csc(x)}\\end{align*} 3. 1sec(x)1+1sec(x)+1=2cot(x)csc(x)\\begin{align*}\\frac{1}{\\sec(x)-1}+\\frac{1}{\\sec(x)+1}=2\\cot(x)\\csc(x)\\end{align*} 4. \\begin{align*}(\\cos(x))(\\tan(x)+\\sin(x)\\cot(x))=\\sin(x)+\\cos^2(x)\\end{align*} 5. \\begin{align*}\\sin^4(x)-\\cos^4(x)=\\sin^2(x)-\\cos^2(x)\\end{align*} 6. \\begin{align*}\\sin^2(x)\\cos^3(x)=(\\sin^2(x)-\\sin^4(x))(\\cos(x))\\end{align*} 7. \\begin{align*}\\frac{\\sin(x)}{\\csc(x)}=1-\\frac{\\cos(x)}{\\sec(x)}\\end{align*} To see the Review answers, open this PDF file and look for section 3.2. ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English FOIL FOIL is an acronym used to remember a technique for multiplying two binomials. You multiply the FIRST terms, OUTSIDE terms, INSIDE terms, and LAST terms and then combine any like terms. Trigonometric Identity A trigonometric identity is an equation that relates two or more trigonometric functions. Show Hide Details Description Difficulty Level: Tags: Subjects:", ")$$ 4) $$\\sec \\left (\\dfrac{3\\pi }{8} \\right )$$ $$\\sqrt{2(2+\\sqrt{2})}$$ For the exercises 5-6, use Figure below to find the desired quantities. 5) $$\\sin(2\\beta ),\\cos(2\\beta ),\\tan(2\\beta ),\\sin(2\\alpha ),\\cos(2\\alpha ),\\tan(2\\alpha )$$ 6) $$\\sin \\left (\\frac{\\beta }{2} \\right ) ,\\cos\\left (\\frac{\\beta }{2} \\right ),\\tan\\left (\\frac{\\beta }{2} \\right ),\\sin\\left (\\frac{\\alpha }{2} \\right ),\\cos\\left (\\frac{\\alpha }{2} \\right ),\\tan\\left (\\frac{\\alpha }{2} \\right )$$ $$\\dfrac{\\sqrt{2}}{10},\\dfrac{7\\sqrt{2}}{10},\\dfrac{1}{7},\\dfrac{3}{5},\\dfrac{4}{5},\\dfrac{3}{4}$$ For the exercises 7-8, prove the identity. 7) $$\\dfrac{2\\cos (2x)}{\\sin (2x)}=\\cot x-\\tan x$$ 8) $$\\cot x \\cos (2x) = -\\sin (2x)+\\cot x$$ \\begin{align*} \\cot x \\cos (2x) &= \\cot x(1-2\\sin ^2 x)\\\\ &= \\cot x-\\dfrac{\\cos x}{\\sin x}(2)\\sin ^2 x\\\\ &= -2\\sin x \\cos \\\\ &= -\\sin (2x)+\\cot x \\end{align*} For the exercises 9-10, rewrite the expression with no powers. 9) $$\\cos ^2 x \\sin ^4 (2x)$$ 10) $$\\tan ^2 x \\sin ^3 x$$ $$\\dfrac{10\\sin x-5\\sin (3x)+\\sin (5x)}{8(\\cos (2x)+1)}$$ ## 7.4: Sum-to-Product and Product-to-Sum Formulas For the exercises 1-3, evaluate the product for the given expression using a sum or difference of two functions. Write the exact answer. 1) $$\\cos \\left ( \\dfrac{\\pi }{3} \\right )\\sin \\left ( \\dfrac{\\pi }{4} \\right )$$ 2) $$2\\sin \\left ( \\dfrac{2\\pi }{3} \\right )\\sin \\left ( \\dfrac{5\\pi }{6} \\right )$$ $$\\dfrac{\\sqrt{3}}{2}$$ 3) $$2\\cos \\left ( \\dfrac{\\pi }{5} \\right )\\cos \\left ( \\dfrac{\\pi }{3} \\right )$$ For the exercises 4-5, evaluate the sum by using a", "+ 9{\\cot ^2}a + ({\\tan ^2}b + 4{\\cot ^2}b)(1 + {\\tan ^2}a)\\\\ & > 14 + 5{\\tan ^2}a + 9{\\cot ^2}a + 2\\tan b \\cdot 2\\cot b(1 + {\\tan ^2}a)\\\\ & = 18 + 9({\\tan ^2}a + {\\cot ^2}a) \\ge 18 + 9 \\cdot 2\\tan a\\cot a = 36 \\end{align} Equality holds when \\begin{align}\\tan a = \\cot a\\;{\\rm{and}}\\;\\tan b = 2\\cot b\\end{align}, which implies that \\begin{align}{\\cos ^2}a = {\\sin ^2}a\\;{\\rm{and}}\\end{align} \\begin{align}{\\cos ^2}b = {\\sin ^2}b\\end{align}. Because \\begin{align}{\\sin ^2}\\theta + {\\cos ^2}\\theta = 1\\end{align}, equality holds when \\begin{align}{\\cos ^2}a = \\frac{1}{2}\\;{\\text{and}}\\;{\\cos ^2}b = \\frac{1}{3};\\end{align} that is \\begin{align}x = \\frac{1}{6},\\;y = \\frac{1}{3},\\;z = \\frac{1}{2}\\end{align}. Example - 10 Let $$A_1$$be the center of the square inscribed in acute triangle $$ABC$$ with two vertices of the square on side $$BC$$ (shown in figure). Thus one of the two remaining vertices of the square lies on side $$AB$$ and the other on segment $$AC.$$ Points $$B_1$$ and $$C_1$$ are defined in a similar way for inscribed squares with two vertices on sides $$AC$$ and $$AB,$$ respectively. Prove that lines $$AA_1, BB_1, CC_1$$are concurrent. Solution: Let line $$AA_1$$ and segment $$BC$$ intersect at $$A_2.$$ We define $$B_2$$ and $$C_2$$ analogously. By Ceva’s theorem, it suffices to show that \\begin{align}\\frac{{\\sin \\angle BA{A_2}}}{{\\sin \\angle {A_2}AC}} \\cdot \\frac{{\\sin \\angle CB{B_2}}}{{\\sin \\angle {B_2}BA}} \\cdot \\frac{{\\sin \\angle AC{C_2}}}{{\\sin \\angle", "$\\cot A$ means. The reason you have the answer of $2.000000\\ldots$ (which I ... 5 \\begin{align} 1-\\cot{\\left(x\\right)} &=1-\\frac{\\cos{\\left(x\\right)}}{\\sin{\\left(x\\right)}}\\\\ &=\\frac{\\sin{\\left(x\\right)}-\\cos{\\left(x\\right)}}{\\sin{\\left(x\\right)}}\\\\ &=-\\sqrt{2}\\frac{\\sin{\\left(\\frac{\\pi}{4}-x\\right)}}{\\sin{\\left(x\\right)}}.\\\\ \\end{align} Therefore, \\begin{align} 1-\\cot{\\left(\\frac{\\pi}{4}-x\\right)} ... 5\\lim\\limits_{x\\to\\infty} \\frac{x - \\sin(x)}{x+\\sin(x)}=\\lim_{x\\to\\infty}\\frac{1-\\dfrac{\\sin x}x}{1+\\dfrac{\\sin x}x}$$Now |\\sin x|\\le1 5 Here's a straightforward algebraic approach:$$\\sin\\theta = 2\\sin(\\theta/2)\\cos(\\theta/2) = 2\\tan(\\theta/2).\\cos^2(\\theta/2) \\ \\ \\ \\ = \\frac{2\\tan(\\theta/2)}{\\sec^2(\\theta/2)} = \\frac{2\\tan(\\theta/2)}{1 + \\tan^2(\\theta/2)} = \\frac{2t}{1+ t^2}$$You can construct a similar argument for \\cos\\theta. 5 Use the fact that$$\\arctan{\\left ( \\frac{p}{q} \\right )} + \\arctan{\\left ( \\frac{q}{p} \\right )} = \\frac{\\pi}{2} $$So all you need to do is arrange the sum to exploit the symmetry you need, i.e., (p,q) \\mapsto (q,p). To do this without repeating, start on the bottom/left axes; there are 9 non-diagonal points each, so the sum along there is 9 ... 5 If you are allowed to use complex number, this integral can be integrated by repeat application of the Euler's formula$$e^{i\\theta} = \\cos(\\theta) + i\\sin\\theta$$Up to integration constant, the integral is equal to:$$\\begin{align}\\int -e^{\\cos t}\\sin(\\sin t + t) dt &= - \\int e^{\\cos t}\\Im\\left[e^{i(\\sin t + t)}\\right] dt ... 5 Your idea is correct. So we have $$\\frac{1}{1-\\cos y}+\\frac{1}{1+\\cos y} = \\frac{1+\\cos y+ 1-\\cos y}{(1+\\cos y)(1-\\cos y)} = \\frac{2}{1-\\cos^2 y}.$$ What do you know about $1-\\cos^2 y$? Can you take it from here? 4 Consider the limit $$\\lim_{n\\to \\infty} \\frac{E_n}{n!}\\left(\\frac{\\pi}{2}\\right)^n$$ Using $E_{n} \\sim \\frac{(-1)^{(n-1)/2}", "& = \\csc^2\\theta \\end{align*} Where you made your mistake was in concluding that the fact that cotangent is odd implied $1 + \\cot^2(-\\theta) = 1 - \\cot^2\\theta$ rather than $1 + \\cot^2(-\\theta) = 1 + (-\\cot\\theta)^2 = 1 + \\cot^2\\theta$. • Yes, I noticed my mistake as soon as I saw everyone's answers! Hit me like a ton of bricks, haha. Thanks for the help, everyone! – enharmonics Jun 24 '15 at 13:38 $1+\\cot^2(-\\theta)=1+(-\\cot\\theta)^2=1+\\cot^2\\theta=1+\\dfrac{\\cos^2\\theta}{\\sin^2\\theta}=\\dfrac{\\sin^2\\theta+\\cos^2\\theta}{\\sin^2\\theta}=\\dfrac{1}{\\sin^2\\theta}=\\csc^2\\theta$ You were on the right path: \\begin{align}1+\\cot^2(\\theta) &=\\frac{\\sin^\\color{green}2(\\theta)}{\\sin^\\color{green}2(\\theta)}(1+\\cot^2(\\theta))\\\\ &=\\frac{\\sin^2(\\theta)+\\sin^2(\\theta)\\cot^2(\\theta)}{\\sin^2(\\theta)}\\\\ &=\\frac{\\sin^2(\\theta)+\\cos^2(\\theta)}{\\sin^2(\\theta)}\\\\ &=\\frac1{\\sin^2(\\theta)}\\\\ &=\\csc^2(\\theta).\\end{align} One way to look at these is to convert all expressions to ratios of the variables $o,a,h$ $$1+cot^2(-θ)=csc^2(θ)$$ $$1+(\\frac{a}{o})^2=(\\frac{h}{o})^2$$ $$(\\frac{o}{o})^2+(\\frac{a}{o})^2=(\\frac{h}{o})^2$$ $$\\frac{o^2+a^2}{o^2}=\\frac{h^2}{o^2}$$ $$o^2+a^2=h^2$$ Which checks out due to the Pythagorean Theorem. I skipped over the negative angle identity of cotan since it is squared. Initially note that $$\\cot^2(-\\theta)=(\\cot(-\\theta))^2=(-\\cot\\theta)^2=(\\cot\\theta)^2=\\cot^2\\theta$$ Then, proof the identity $1+\\cot^2(-\\theta)= \\csc^2\\theta$ is same that proof $$1+\\cot^2\\theta=\\csc^2\\theta$$ Hence, divide the identity $\\sin^2\\theta+\\cos^2\\theta=1$ by $\\sin^2\\theta$. $$\\frac{\\sin^2\\theta}{\\sin^2\\theta}+\\frac{\\cos^2\\theta}{\\sin^2\\theta}=\\frac{1}{\\sin^2\\theta} \\Rightarrow 1+\\cot^2\\theta=\\csc^2\\theta.$$ • The commands for the trigonometric functions, used in math mode, are \\sin for sine, \\cos for cosine, \\tan for tangent, \\csc for cosecant, \\sec for secant, and \\cot for cotangent. – N. F. Taussig Jun 24 '15 at 9:43", "you do this: $$\\displaystyle \\sec x=\\cot x+\\tan x$$ \\require{cancel} \\displaystyle \\begin{align}{{\\sec }^{2}}x&={{\\left( {\\cot x+\\tan x} \\right)}^{2}}\\\\1+{{\\tan }^{2}}x&={{\\cot }^{2}}x+2\\cancel{{\\cot x\\tan x}}+{{\\tan }^{2}}x\\\\1+{{\\tan }^{2}}x&={{\\cot }^{2}}x+2+{{\\tan }^{2}}x\\\\1+\\cancel{{{{{\\tan }}^{2}}x}}&={{\\cot }^{2}}x+2+\\cancel{{{{{\\tan }}^{2}}x}}\\\\{{\\cot }^{2}}x&=-1\\end{align} No solution, since we can’t square something and have it be a negative number. $$\\displaystyle \\frac{{{{{\\cos }}^{2}}\\left( x \\right)}}{{{{{\\sin }}^{2}}\\left( x \\right)}}=3$$ $$\\displaystyle \\begin{array}{c}{{\\cot }^{2}}\\left( x \\right)=3\\\\\\cot \\left( x \\right)=\\pm \\sqrt{3}\\end{array}$$ $$\\displaystyle x=\\left\\{ {\\frac{\\pi }{3},\\,\\,\\frac{{2\\pi }}{3},\\,\\,\\frac{{4\\pi }}{3},\\,\\frac{{5\\pi }}{3}} \\right\\}$$ Use Quotient Identity to convert $$\\displaystyle \\frac{{\\cos }}{{\\sin }}$$ into cot Take square roots of both sides, and don’t forget the $$\\pm$$ Here’s another one where we have to check for extraneous solutions. Solve over the reals: Solving with Trig Identities Hints $$\\displaystyle \\frac{{1-\\cos x}}{{\\sin x}}=\\frac{{\\sin x}}{{1+\\cos x}}$$ \\displaystyle \\begin{align}\\frac{{1-\\cos x}}{{\\sin x}}-\\frac{{\\sin x}}{{1+\\cos x}}&=0\\\\\\frac{{\\left( {1-\\cos x} \\right)\\left( {1+\\cos x} \\right)}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}-\\frac{{\\left( {\\sin x} \\right)\\left( {\\sin x} \\right)}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}&=0\\\\\\frac{{1-{{{\\cos }}^{2}}x}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}-\\frac{{{{{\\sin }}^{2}}x}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}&=0\\\\\\frac{{{{{\\sin }}^{2}}x}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}-\\frac{{{{{\\sin }}^{2}}x}}{{\\left( {\\sin x} \\right)\\left( {1+\\cos x} \\right)}}&=0\\\\0&=0\\end{align} This appears to be all real numbers, but since we have denominators that can’t be 0, we have to check for those cases. Get all terms on one side Find common denominator to add terms Turn $$\\left( {1-{{{\\cos }}^{2}}x} \\right)$$ into $${{\\sin }^{2}}x$$ We get", "0$ (so $c_2(r) = 0$). Your lecturer's choice leads to something like \\begin{align*} x(r, s) & = r\\sin s \\\\ y(r, s) & = r\\cos s \\end{align*} while your choice leads to something like \\begin{align*} x(r, s) & = r\\cos s \\\\ y(r, s) & = -r\\sin s. \\end{align*} It should not matter which one you choose. Note that with your lecturer's choice, you have \\begin{align*} \\frac yx - \\frac xy & = \\cot s - \\tan s \\\\ & = 2\\cot(2s) \\end{align*} and with your choice, you have \\begin{align*} \\frac yx - \\frac xy & = -\\tan s + \\cot s \\\\ & = 2\\cot(2s). \\end{align*} In both cases, $\\frac yx - \\frac xy$ becomes the same $2\\cot(2s)$, and that is the expression you'll need to solve for $v$ and $u$: \\begin{align*} \\frac{\\partial v}{\\partial s} & = \\left(\\frac yx - \\frac xy\\right)v \\\\ & = 2\\cot(2s)v\\\\ \\therefore v & = C(r)\\sin(2s) \\\\ \\text{and from} \\frac{\\partial u}{\\partial s} = v \\quad \\Longrightarrow \\quad u & = -\\frac 12C(r)\\cos(2s) + D(r) \\\\ \\therefore u & = \\tilde C(x^2 + y^2) \\cdot (x^2 - y^2) + \\tilde D(x^2 + y^2). \\end{align*} Although $\\cos(2s)$ becomes different expressions in different parametrizations, the final expression is the same because the sign difference is subsumed in $\\tilde C$. I did get sloppy and ignored", "(d) $s = \\frac{t}{\\sqrt{1 - t^2}} \\ + \\ cos^{-1} t$ \\vspace{.25cm} (e) $y = sin^{-1} \\sqrt{x}$ \\vspace{.25cm} (f) $z = cot^{-1} ( \\frac { y}{1 - y^2} )$ \\vspace{.25cm} \\newpage \\noindent {\\bf Solutions:} \\vspace{.25cm} (a) $y = tan^{-1} \\ t^4$ \\begin{align*} \\frac{dy}{dt} &= \\frac{d}{dt} tan^{-1} \\ ( t^4) \\\\ &= \\frac{1}{1 + {(t^4)}^2} \\ {\\centerdot} \\ \\frac{d}{dt} \\ (t^4) \\\\ &= \\frac{4 t^3}{1 + t^8} \\end{align*} \\vspace{.25cm} (b) $z = t \\ cot^{-1} (1 + t^2)$ \\begin{align*} \\frac{dz}{dt} &= \\frac{d}{dt} \\ t \\ cot^{-1} (1 + t^2) \\\\ &= cot^{-1} (1 + t^2) + \\ t \\ {\\centerdot} \\ \\frac{-1}{1 + {(1 + t^2)}^2} \\ {\\centerdot} \\ (2t) \\\\ &= cot^{-1} (1 + t^2) - \\frac{2t^2}{t^4 + 2 t^2 + 2} \\\\ \\end{align*} \\vspace{.25cm} (c) $x = sin^{-1} \\sqrt{1 - t^4}$ \\begin{align*} \\frac{dx}{dt} &= \\frac{d}{dt} \\ sin^{-1} \\sqrt{1 - t^4} \\\\ &= \\frac{1}{\\sqrt{ 1 - {(\\sqrt { 1 - t^4})}^2} }\\ {\\centerdot} \\ \\frac{d}{dt} (\\sqrt{1 - t^4}) \\\\ &= \\frac{1}{\\sqrt{ 1 - (1 - t^4) }} \\ {\\centerdot} \\ \\frac{1}{2} \\ {\\centerdot} \\ {(1 - t^4)}^{\\frac{-1}{2}} \\ {\\centerdot} \\ ( -4t^3) \\\\ &= \\frac{1}{\\sqrt{1 - 1 + t^4}} \\ {\\centerdot} \\ \\frac{1}{\\sqrt{1 - t^4}} \\ {\\centerdot} \\ (-2t^3) \\\\ &= \\frac{1}{t^2} \\ {\\centerdot} \\ \\frac{1}{\\sqrt{1 - t^4}} \\ {\\centerdot} \\ (-2 t^3) \\\\ &= \\frac{-2t}{\\sqrt{1 - t^4}}\\\\", "× # A Trigonometric Geometric Sequence Evaluate $\\sum_{k=1}^n \\frac{1}{ \\sin \\left( \\frac{ 2^k \\pi} { 2^{n+1} -1} \\right) }.$ In particular, show that $\\frac{ 1}{ \\sin ( \\frac{2\\pi}{7} )} + \\frac{1}{ \\sin( \\frac { 4 \\pi } { 7} )} = \\frac{1}{ \\sin ( \\frac{\\pi}{7} ) }.$ This question is prompted by Daniel's Arc Co Tangent Sum. Hint: Telescoping Series Note by Calvin Lin 2 years, 11 months ago Sort by: Since $$\\csc 2x=\\cot x-\\cot 2x$$ the sum can be written as: $$\\displaystyle \\sum_{k=1}^n \\frac{1}{\\sin\\left(\\frac{2^k\\pi}{2^{n+1}-1}\\right)}=\\sum_{k=1}^n \\left(\\cot\\left(\\frac{2^{k-1}\\pi}{2^{n+1}-1}\\right)-\\cot\\left(\\frac{2^k\\pi}{2^{n+1}-1}\\right) \\right)$$ $$\\displaystyle =\\cot\\left(\\frac{\\pi}{2^{n+1}-1}\\right)-\\cot\\left(\\frac{2^n\\pi}{2^{n+1}-1}\\right)=\\cot\\left(\\frac{2^{n+1}\\pi}{2^{n+1}-1}\\right)-\\cot\\left(\\frac{2^n\\pi}{2^{n+1}-1}\\right)$$ $$\\displaystyle =-\\csc\\left(\\frac{2^{n+1}\\pi}{2^{n+1}-1}\\right)=\\boxed{\\csc\\left(\\dfrac{\\pi}{2^{n+1}-1}\\right)}$$ For $$n=2$$, it can be shown that: $$\\displaystyle \\csc\\left(\\frac{2\\pi}{7}\\right)+\\csc\\left(\\frac{4\\pi}{7}\\right)=\\csc\\left(\\frac{\\pi}{7}\\right)$$ · 2 years, 11 months ago Good job! I couldn't notice that! · 2 years, 11 months ago May be the part 2 of question is inspired from a question of IIT JEE 2011 . Am I correct ? · 2 years, 11 months ago", "× # A Trigonometric Geometric Sequence Evaluate $\\sum_{k=1}^n \\frac{1}{ \\sin \\left( \\frac{ 2^k \\pi} { 2^{n+1} -1} \\right) }.$ In particular, show that $\\frac{ 1}{ \\sin ( \\frac{2\\pi}{7} )} + \\frac{1}{ \\sin( \\frac { 4 \\pi } { 7} )} = \\frac{1}{ \\sin ( \\frac{\\pi}{7} ) }.$ This question is prompted by Daniel's Arc Co Tangent Sum. Hint: Telescoping Series Note by Calvin Lin 2 years, 9 months ago Sort by: Since $$\\csc 2x=\\cot x-\\cot 2x$$ the sum can be written as: $$\\displaystyle \\sum_{k=1}^n \\frac{1}{\\sin\\left(\\frac{2^k\\pi}{2^{n+1}-1}\\right)}=\\sum_{k=1}^n \\left(\\cot\\left(\\frac{2^{k-1}\\pi}{2^{n+1}-1}\\right)-\\cot\\left(\\frac{2^k\\pi}{2^{n+1}-1}\\right) \\right)$$ $$\\displaystyle =\\cot\\left(\\frac{\\pi}{2^{n+1}-1}\\right)-\\cot\\left(\\frac{2^n\\pi}{2^{n+1}-1}\\right)=\\cot\\left(\\frac{2^{n+1}\\pi}{2^{n+1}-1}\\right)-\\cot\\left(\\frac{2^n\\pi}{2^{n+1}-1}\\right)$$ $$\\displaystyle =-\\csc\\left(\\frac{2^{n+1}\\pi}{2^{n+1}-1}\\right)=\\boxed{\\csc\\left(\\dfrac{\\pi}{2^{n+1}-1}\\right)}$$ For $$n=2$$, it can be shown that: $$\\displaystyle \\csc\\left(\\frac{2\\pi}{7}\\right)+\\csc\\left(\\frac{4\\pi}{7}\\right)=\\csc\\left(\\frac{\\pi}{7}\\right)$$ · 2 years, 9 months ago Good job! I couldn't notice that! · 2 years, 9 months ago May be the part 2 of question is inspired from a question of IIT JEE 2011 . Am I correct ? · 2 years, 9 months ago", "\\sin x \\sin y$$ • $$cos(x - y) = \\cos x \\cos y + \\sin x \\sin y$$ • $$sin(x + y) = \\sin x \\cos y + \\cos x \\sin y$$ • $$sin(x - y) = \\sin x \\cos y - \\cos x \\sin y$$ • \\begin{aligned} \\cos(\\frac{\\pi}{2} + x) &= -\\sin x\\\\ \\cos(\\pi - x) &= -\\cos x\\\\ \\cos(\\pi + x) &= \\cos x\\\\ \\cos(2\\pi - x) &= \\cos x \\end{aligned} \\begin{aligned} \\sin(\\frac{\\pi}{2} + x) &= \\cos x\\\\ \\sin(\\pi - x) &= \\sin x\\\\ \\sin(\\pi + x) &= -\\sin x\\\\ \\sin(2\\pi - x) &= -\\sin x \\end{aligned} • If none of the angles $$x,y$$ and $$(x\\pm y)$$ is an odd multiple of $$\\dfrac{\\pi}{2}$$, then $\\tan(x + y) = \\frac{\\tan x + \\tan y}{1 - \\tan x \\tan y}$ • If none of the angles $$x,y$$ and $$(x\\pm y)$$ is an odd multiple of $$\\pi$$, then $\\cot(x + y) = \\frac{\\cot x \\cot y - 1}{\\cot x + \\cot y}$ • $$\\cos 2x = \\cos^2x - \\sin^2x = 2\\cos^2x - 1 = 1 - 2\\sin^2x = \\dfrac{1-\\tan^2x}{1+\\tan^2x}$$ • $$\\sin 2x = 2\\sin x \\cos x = \\dfrac{2\\tan x}{1 + \\tan^2x}$$ • $$\\tan 2x = \\dfrac{2\\tan x}{1 - \\tan^2x}$$ • $$\\sin 3x = 3\\sin x - 4\\sin^3x$$ • $$\\cos 3x = 4\\cos^3x - 3\\cos x$$ • $$\\tan 3x", "\\frac{1}{\\tan{x}}$ and $\\tan(\\arctan(x)) = x$ for all real $x$. 4. To get $x-1$ for $x \\ge 1$, use the following sequence: $$1/x, \\sqrt x, \\arcsin, \\tan, x^2, 1/x$$ This takes advantage of the identity $\\tan(\\arcsin(x)) = \\frac{x}{\\sqrt{1-x^2}}$ for $|x| \\le 1$. In this case, it resolves into $\\tan(\\arcsin(\\sqrt{1/x})) = \\frac{\\sqrt{\\frac{1}{x}}}{\\sqrt{1-\\frac{1}{x}}} = \\sqrt{\\frac{\\frac{1}{x}}{\\frac{x-1}{x}}} = \\sqrt{\\frac{1}{x-1}}$, which we then square and invert to get $x-1$. Now, the way is clear: we derive any number greater than 2015 (e.g. starting from 2, use $x^2, x^2, x^2, x^2$ to get 65536), then repeatedly subtract one to get to 2015. (There are lower easily-obtainable integers, such as 8! = 40320, but this answer is aimed just at finding a possible solution). So here's one possible key sequence: $$0, \\cos, \\arctan, \\sin, x^2 ( = 0.5), \\\\ \\arctan, \\cot, x^2, x^2, x^2, x^2 (=65536), \\\\ \\arctan, \\cot, \\sqrt{x} (=\\sqrt{1/65536}), \\\\ \\{ \\arcsin, \\tan \\}^{63521} (=\\sqrt{1/2015}), \\\\ \\arctan, \\cot, x^2 (=2015).$$ (Note that due to numerical inaccuracy inherent to all numerical calculators, the result is probably not going to be exactly 2015. But it's a math puzzle, not a calculator puzzle, so this doesn't really matter :) ) If we want to be a bit more clever with this, we can use an approach like the one Daniil Agashiyev used in his answer to reduce", "× # A Trigonometric Geometric Sequence Evaluate $\\sum_{k=1}^n \\frac{1}{ \\sin \\left( \\frac{ 2^k \\pi} { 2^{n+1} -1} \\right) }.$ In particular, show that $\\frac{ 1}{ \\sin ( \\frac{2\\pi}{7} )} + \\frac{1}{ \\sin( \\frac { 4 \\pi } { 7} )} = \\frac{1}{ \\sin ( \\frac{\\pi}{7} ) }.$ This question is prompted by Daniel's Arc Co Tangent Sum. Hint: Telescoping Series Note by Calvin Lin 3 years, 1 month ago Sort by: Since $$\\csc 2x=\\cot x-\\cot 2x$$ the sum can be written as: $$\\displaystyle \\sum_{k=1}^n \\frac{1}{\\sin\\left(\\frac{2^k\\pi}{2^{n+1}-1}\\right)}=\\sum_{k=1}^n \\left(\\cot\\left(\\frac{2^{k-1}\\pi}{2^{n+1}-1}\\right)-\\cot\\left(\\frac{2^k\\pi}{2^{n+1}-1}\\right) \\right)$$ $$\\displaystyle =\\cot\\left(\\frac{\\pi}{2^{n+1}-1}\\right)-\\cot\\left(\\frac{2^n\\pi}{2^{n+1}-1}\\right)=\\cot\\left(\\frac{2^{n+1}\\pi}{2^{n+1}-1}\\right)-\\cot\\left(\\frac{2^n\\pi}{2^{n+1}-1}\\right)$$ $$\\displaystyle =-\\csc\\left(\\frac{2^{n+1}\\pi}{2^{n+1}-1}\\right)=\\boxed{\\csc\\left(\\dfrac{\\pi}{2^{n+1}-1}\\right)}$$ For $$n=2$$, it can be shown that: $$\\displaystyle \\csc\\left(\\frac{2\\pi}{7}\\right)+\\csc\\left(\\frac{4\\pi}{7}\\right)=\\csc\\left(\\frac{\\pi}{7}\\right)$$ · 3 years, 1 month ago Good job! I couldn't notice that! · 3 years, 1 month ago May be the part 2 of question is inspired from a question of IIT JEE 2011 . Am I correct ? · 3 years, 1 month ago", "B + 1) = 2 (cot A – 1) (cot B – 1) = 2 Type 9: If A + B + C = 1800 tan A + tan B + tan C = tan A × tan B × tan C cot A cot B + cot B cot C + cot C cot A = 1 Type 10: $$tanA+B+C=\\dfrac{tanA+tanB+tanC-tanA\\:tanB\\:tanC}{1-tanA\\:tanB-tanB\\:tanC-tanC\\:tanA}$$ sin (A + B + C) = sin A cos B cos C + sin B cos A cos C + sin C cos A cos B – sin A sin B sin C cos (A + B + C) = cos A cos B cos C – cos A sin B sin C – cos B sin A sin C − cos C sin A sin B ## MAXIMA AND MINIMA OF SOME GENERAL FUNCTIONS FUNCTION MINIMA MAXIMA sin2 Ɵ, cos2 Ɵ 0 1 a.sin Ɵ ± b.cos Ɵ $$-\\sqrt{a^2+b^2}$$ $$\\sqrt{a^2+b^2}$$ a.sin Ɵ ± b.sin Ɵ $$-\\sqrt{a^2+b^2}$$ $$\\sqrt{a^2+b^2}$$ a.cos Ɵ ± b.cos Ɵ $$-\\sqrt{a^2+b^2}$$ $$\\sqrt{a^2+b^2}$$ (sinn Ɵ cosn Ɵ) If n is even, min = 0 If n is odd, min = -$$({1\\over2})^n$$ $$({1\\over2})^n$$ a.sin2 Ɵ + b.cosec2 Ɵ If a < b Min = a + b If a > b Min=$$2\\sqrt{ab}$$ ∞ a.cos2 Ɵ + b.sec2 Ɵ If a < b Min = a" ]

[ "\\circ$.", "# How do you find the value for csc^-1 (-7/3) ? $x = - {25}^{\\circ} 38 + k {360}^{\\circ}$, and $x = 180 - \\left(- 25.38\\right) = 180 + 25.38 = {205}^{\\circ} 38 + k {360}^{\\circ}$ $x = {205}^{\\circ} 38 + k {360}^{\\circ}$ $x = {334}^{\\circ} 62 + k {360}^{\\circ}$", "$^\\circ$C.", "= 50^{\\circ}$?", "6^\\circ$ -", "\\circ h' =g$.", "larger than $180^\\circ$.", "{45}^{\\circ}$.", ")$ and $(5, 160^\\circ )$ .", "is $55^\\circ$.", "\\circ m(f)$", "^{\\circ}$", "$25^{\\circ}$", "$C \\approx {40.3}^{\\circ}$", "# How do you evaluate csc(210)? $C s c \\left(210\\right) = \\csc \\left(2 \\cdot 90 + 30\\right)$ $= - \\csc \\left(30\\right) = - 2$ ${210}^{\\circ}$ is anangle of 3rd quadranmt so csc(210) is negative", "^ {\\ circ}}$", "70.85^\\circ)$$", "you get the $45^\\circ$ line.", "(30.8^\\circ))$", "\\csc ^2y+1}$" ]

[ "# How do I find the value of sin 225? Oct 29, 2015 #### Answer: $- \\frac{1}{\\sqrt{2}}$ #### Explanation: $\\sin {225}^{\\circ} = \\sin \\left({180}^{\\circ} + {45}^{\\circ}\\right)$ $= \\sin {180}^{\\circ} \\cos {45}^{\\circ} + \\cos {180}^{\\circ} \\sin {45}^{\\circ}$ $= \\left(0\\right) \\left(\\frac{1}{\\sqrt{2}}\\right) + \\left(- 1\\right) \\frac{1}{\\sqrt{2}}$ =$- \\frac{1}{\\sqrt{2}}$", "# How do I find the value of sin 225? Oct 29, 2015 $- \\frac{1}{\\sqrt{2}}$ #### Explanation: $\\sin {225}^{\\circ} = \\sin \\left({180}^{\\circ} + {45}^{\\circ}\\right)$ $= \\sin {180}^{\\circ} \\cos {45}^{\\circ} + \\cos {180}^{\\circ} \\sin {45}^{\\circ}$ $= \\left(0\\right) \\left(\\frac{1}{\\sqrt{2}}\\right) + \\left(- 1\\right) \\frac{1}{\\sqrt{2}}$ =$- \\frac{1}{\\sqrt{2}}$", "# How do you find the value for csc^-1 (-7/3) ? $x = - {25}^{\\circ} 38 + k {360}^{\\circ}$, and $x = 180 - \\left(- 25.38\\right) = 180 + 25.38 = {205}^{\\circ} 38 + k {360}^{\\circ}$ $x = {205}^{\\circ} 38 + k {360}^{\\circ}$ $x = {334}^{\\circ} 62 + k {360}^{\\circ}$", "$\\csc \\frac{7 \\pi}{8}=\\frac{1}{\\sin \\frac{7 \\pi}{8}}=2.6131$ 2. $\\cot 85^\\circ=\\frac{1}{\\tan 85^\\circ}=0.0875$ 3. $\\sec 225^\\circ$ is the reciprocal of $\\cot 225^\\circ$ , a $45^\\circ$ reference angle in quadrant three where cosine is negative. Because $\\cos 45^\\circ=\\frac{1}{\\sqrt{2}}, \\cos 225^\\circ=-\\frac{1}{\\sqrt{2}}$ , and $\\sec 225^\\circ=-\\sqrt{2}$ . 4. $\\csc \\frac{5 \\pi}{6}$ is the reciprocal of $\\sin \\frac{5 \\pi}{6}$ , a $\\frac{\\pi}{6}$ or $30^\\circ$ reference angle in the second quadrant where sine is positive. Because $\\sin \\frac{\\pi}{6}=\\frac{1}{2}, \\sin \\frac{5 \\pi}{6}=\\frac{1}{2},$ and $\\csc \\frac{5 \\pi}{6}=2$ . ### Vocabulary Secant The reciprocal of cosine. Cosecant The reciprocal of sine. Cotangent The reciprocal of tangent. ### Practice 1. $\\csc 95^\\circ$ 2. $\\cot 278^\\circ$ 3. $\\sec \\frac{14 \\pi}{5}$ 4. $\\cot(-245^\\circ)$ 5. $\\sec \\frac{6 \\pi}{7}$ 6. $\\csc \\frac{23 \\pi}{13}$ 7. $\\cot 333^\\circ$ 8. $\\csc \\frac{9 \\pi}{5}$ Evaluate the following trigonometric functions without using a calculator. Give your answers exactly. 1. $\\sec \\frac{5 \\pi}{6}$ 2. $\\csc \\left(-\\frac{3 \\pi}{2}\\right)$ 3. $\\cot 225^\\circ$ 4. $\\sec \\frac{11 \\pi}{3}$ 5. $\\csc \\frac{7 \\pi}{6}$ 6. $\\sec 270^\\circ$ 7. $\\cot \\frac{5 \\pi}{3}$ 8. $\\csc 315^\\circ$", "{1}{\\sin 135} = \\sqrt 2$ $\\sec 135 = \\frac {1}{\\cos 135} = -\\sqrt 2$ $\\cot 135 = \\frac {1}{\\tan 135} = – 1$ Now Trigonometric table for 210 to 270 is given by And it is calculated as $\\sin (210) = \\sin (180 +30) =- \\sin 30= -\\frac {1}{2}$ $\\cos (210) = \\cos (180 +30) =- \\cos 30=-\\frac {\\sqrt {3}}{2}$ $\\tan (210) = \\frac {\\sin 210}{ \\cos 210} = \\frac {1}{\\sqrt {3}}$ $\\sin (225) = \\sin (180 +45) =- \\sin 45= -\\frac {1}{\\sqrt {2}}$ $\\cos (225) = \\cos (180+45) =- \\cos 45= -\\frac {1}{\\sqrt {2}}$ $\\tan 225 = \\frac {\\sin 225}{\\cos 225} = 1$ $\\sin (270) = \\sin (180 +90) =- \\sin 90= -1$ $\\cos (270) = \\cos (180+90) =- \\cos 90= 0$ $\\tan 270 = \\frac {\\sin 270}{\\cos 270} = -\\frac {1}{0}$ Undefined value $\\csc (210) = \\frac {1}{\\sin (210)} = -2$ $\\sec (210) = \\frac {1}{\\cos (210)}=-\\frac {2}{\\sqrt 3}$ $\\cot (210) = \\frac {1}{\\tan (210)} = \\sqrt {3}$ $\\csc (225) = \\frac {1}{\\sin 225}= -\\sqrt {2}$ $\\sec (225) = \\frac {1}{\\cos 225}= -\\sqrt {2}$ $\\cot 225 = \\frac {1}{\\tan 225} = 1$ Now Trigonometric table for 300 to 360 is given by $\\sin (300) = \\sin (360 -60) =- \\sin 60=-\\frac {\\sqrt {3}}{2}$ $\\cos (300) = \\cos (360-60) =\\cos 60=\\frac {1}{2}$ $\\tan (300) = \\frac {\\sin", "\\frac {1}{\\tan 120} = – \\frac {1}{\\sqrt 3}$ $\\csc 135 = \\frac {1}{\\sin 135} = \\sqrt 2$ $\\sec 135 = \\frac {1}{\\cos 135} = -\\sqrt 2$ $\\cot 135 = \\frac {1}{\\tan 135} = – 1$ Now Trigonometric table for 210 to 270 is given by And it is calculated as $\\sin (210) = \\sin (180 +30) =- \\sin 30= -\\frac {1}{2}$ $\\cos (210) = \\cos (180 +30) =- \\cos 30=-\\frac {\\sqrt {3}}{2}$ $\\tan (210) = \\frac {\\sin 210}{ \\cos 210} = \\frac {1}{\\sqrt {3}}$ $\\sin (225) = \\sin (180 +45) =- \\sin 45= -\\frac {1}{\\sqrt {2}}$ $\\cos (225) = \\cos (180+45) =- \\cos 45= -\\frac {1}{\\sqrt {2}}$ $\\tan 225 = \\frac {\\sin 225}{\\cos 225} = 1$ $\\sin (270) = \\sin (180 +90) =- \\sin 90= -1$ $\\cos (270) = \\cos (180+90) =- \\cos 90= 0$ $\\tan 270 = \\frac {\\sin 270}{\\cos 270} = -\\frac {1}{0}$ Undefined value $\\csc (210) = \\frac {1}{\\sin (210)} = -2$ $\\sec (210) = \\frac {1}{\\cos (210)}=-\\frac {2}{\\sqrt 3}$ $\\cot (210) = \\frac {1}{\\tan (210)} = \\sqrt {3}$ $\\csc (225) = \\frac {1}{\\sin 225}= -\\sqrt {2}$ $\\sec (225) = \\frac {1}{\\cos 225}= -\\sqrt {2}$ $\\cot 225 = \\frac {1}{\\tan 225} = 1$ Now Trigonometric table for 300 to 360 is given by $\\sin (300) = \\sin (360 -60) =- \\sin 60=-\\frac {\\sqrt {3}}{2}$ $\\cos", "} 225 \\div 45 = 5. \\text{ So } 225° = 5 \\left( \\frac{\\pi}{4} \\right) = \\frac{5 \\pi}{4} .$", "# How do you evaluate csc 180 ? Since $\\csc \\left(x\\right) = \\frac{1}{\\sin} \\left(x\\right)$, you have $\\csc \\left(180\\right) = \\frac{1}{\\sin} \\left(180\\right)$ But since $\\sin \\left(180\\right) = 0$, you can't divide by it, and thus $\\csc \\left(180\\right)$ is not defined.", "## Chemistry 10th Edition a) 229 K or -44$^{\\circ}$ C b) 76 K or -197$^{\\circ}$ C a) $T_{1}$ = (32+273) K = 305 K Let's assume $V_{1}$ = 4 V Then $V_{2}$ = 3 V (decreased by 25%) According to Charles's Law, $\\frac{V_{1}}{T_{1}}$ = $\\frac{V_{2}}{T_{2}}$ Or $T_{2}$ = $\\frac{V_{2}\\times T_{1}}{V_{1}}$ = $\\frac{3V\\times305K}{4V}$ $\\approx$ 229 K b) Here only $V_{2}$ is changed. $V_{2}$ = V if $V_{1}$ = 4V $T_{2}$ = $\\frac{V\\times 305 K}{4V}$ $\\approx$ 76 K .", "better -- every nonconstant polynomial has a root! 10. Apr 29, 2005 ### Orion1 De Moivre's Theorem... There are two solutions for the square root of i using De Moivre's Theorem: $$w_k = \\text{cis} \\; \\frac{90^{\\circ} + 360^{\\circ} k}{2}$$ $$w_0 = \\text{cis} \\; 45^{\\circ} = (\\cos 45^{\\circ} + i \\sin 45^{\\circ})$$ $$(\\cos 45^{\\circ} + i \\sin 45^{\\circ}) = \\left( \\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{2}}{2} i \\right)$$ $$w_1 = \\text{cis} \\; 225^{\\circ} = ( \\cos 225^{\\circ} + i \\sin 225^{\\circ})$$ $$( \\cos 225^{\\circ} + i \\sin 225^{\\circ}) = \\left( - \\frac{\\sqrt{2}}{2} - \\frac{\\sqrt{2}}{2} i \\right)$$ $$\\boxed{w_0 = \\frac{\\sqrt{2}}{2} (1 + i)}$$ $$\\boxed{w_1 = - \\frac{\\sqrt{2}}{2} (1 + i)}$$ $$\\boxed{z = \\pm \\frac{\\sqrt{2}}{2} (1 + i)}$$ Note that this solution is irrational with a root in the denominator: $$z = \\pm\\frac{1}{\\sqrt{2}}\\left(1+i)$$ Any root of i can be solved using De Moivre's Theorem. Last edited: Apr 29, 2005 11. Apr 29, 2005 ### Data And to see why Hurkyl's comment is significant, remember that over the reals, for example, not every polynomial has a root - for example, $x^2+1$ has no real roots, and it can't be factored over the reals. Over the complex numbers, every polynomial has a root, and this amounts to saying that every monic complex polynomial can be factored to the form $$(x - \\alpha_1)^{e_1} \\ \\cdots \\ (x", "$$\\cot 270^{\\circ}$$ $$\\cot 270^{\\circ}=0$$ 5. $$\\sec \\frac{11 \\pi}{6}$$ $$\\sec \\frac{11 \\pi}{6}=\\frac{1}{\\cos \\frac{11 \\pi}{6}}=\\frac{2}{\\sqrt{3}}=\\frac{2 \\sqrt{3}}{3}$$ 6. $$\\csc -\\frac{5 \\pi}{4}$$ $$\\csc -\\frac{5 \\pi}{4}=\\frac{1}{\\sin -\\frac{5 \\pi}{4}}=-\\frac{2}{\\sqrt{2}}=-\\sqrt{2}$$ Review Name the angle between $$0^{\\circ}$$ and $$360^{\\circ}$$ that is coterminal with... 1. $$-20^{\\circ}$$ 2. $$475^{\\circ}$$ 3. $$-220^{\\circ}$$ 4. $$690^{\\circ}$$ 5. $$-45^{\\circ}$$ Use your knowledge of the unit circle to help determine whether each of the following trigonometric expressions is positive or negative. 6. $$\\tan 143^{\\circ}$$ 7. $$\\cos \\frac{\\pi}{3}$$ 8. $$\\sin 362^{\\circ}$$ 9. $$\\csc \\frac{3 \\pi}{4}$$ Use your knowledge of the unit circle to evaluate each of the following trigonometric expressions. 10. $$\\cos 120^{\\circ}$$ 11. $$\\sec \\frac{\\pi}{3}$$ 12. $$\\tan 225^{\\circ}$$ 13. $$\\cot 120^{\\circ}$$ 14. $$\\sin \\frac{11 \\pi}{6}$$ 15. $$\\csc 240^{\\circ}$$ 16. Find $$\\sin \\theta$$ and $$\\tan \\theta$$ if $$\\cos \\theta=\\frac{\\sqrt{3}}{2}$$ and $$\\cot \\theta>0$$. 17. Find $$\\cos \\theta$$ and $$\\tan \\theta$$ if $$\\sin \\theta=-\\frac{1}{2}$$ and $$\\sec \\theta<0$$. 18. Draw the complete unit circle (all four quadrants) and label the important points.", "$$C = (180-2x)^\\circ$$ and we have $$a=b$$ and $$c = 2a\\cos x^\\circ$$. Then $\\begin{array}{rcl} \\frac{aA+bB+cC}{a+b+c} & = & \\frac{ax + ax + 2a(180-2x)\\cos x^\\circ}{a + a + 2a\\cos x^\\circ} \\\\ & = & \\frac{2a[x + (180-2x)\\cos x^\\circ]}{2a(1 + \\cos x^\\circ)} \\; = \\; \\frac{x + (180-2x)\\cos x^\\circ}{1 + \\cos x^\\circ} \\end{array}$ and this tends to $$90$$ as $$x \\to 0$$. · 3 years, 11 months ago Thank you for noticing. I will use $$\\frac{\\pi}{3}$$ and $$\\frac{\\pi}{2}$$ instead, on the next time if i meet those numbers again. · 3 years, 11 months ago", "## Calculus with Applications (10th Edition) $\\sec 225^\\circ=-\\sqrt 2$ We are given $\\sec \\frac{5\\pi}{4}$ convert $\\frac{5\\pi}{4}$ to degrees $\\frac{5\\pi}{4}$ radians=$\\frac{5\\pi}{4}(\\frac{180^\\circ}{\\pi})$ $\\frac{5\\pi}{4}$ radians= $225^\\circ$ $\\sec 225^\\circ=-sec 45^\\circ$ and $\\sec \\theta=\\frac{1}{\\cos \\pi}$ so $\\sec 225^\\circ=-\\sqrt 2$", "225$$^{\\circ}$$ = 225 * $$\\frac{\\prod}{180}$$ = $$\\frac{5\\prod}{4}$$ k) 240$$^{\\circ}$$ = 240 * $$\\frac{\\prod}{180}$$ = $$\\frac{4\\prod}{3}$$ l) 270$$^{\\circ}$$ = 270 * $$\\frac{\\prod}{180}$$ = $$\\frac{3\\prod}{2}$$ m) 315$$^{\\circ}$$ = 315 * $$\\frac{\\prod}{180}$$ = $$\\frac{7\\prod}{4}$$ n) 330$$^{\\circ}$$ = 330 * $$\\frac{\\prod}{180}$$ = $$\\frac{11\\prod}{6}$$ o)360$$^{\\circ}$$ = 360 * $$\\frac{\\prod}{180}$$ = 2$$\\prod$$ 2) Convert the following radians into degree measures . a) $$\\frac{\\prod}{12}$$ b) $$\\frac{5\\prod}{3}$$ c) $$\\frac{2\\prod}{5}$$ d)$$\\frac{5\\prod}{6}$$ Solutions : As we know that 1 radian = $$\\frac{180^{\\circ}}{\\prod}$$ a) $$\\frac{\\prod}{12}$$ = $$\\frac{180}{12}$$ = 15$$^{\\circ}$$ b) $$\\frac{5\\prod}{3}$$ = $$\\frac{5*180}{3}$$ = 300$$^{\\circ}$$ c) $$\\frac{2\\prod}{5}$$ = $$\\frac{2 *180}{5}$$ = 72$$^{\\circ}$$ d)$$\\frac{5\\prod}{6}$$ = $$\\frac{5*180}{6}$$ =150 $$^{\\circ}$$ 3) Convert the following into radian measure . a) 40$$^{\\circ}$$20' b) -47$$^{\\circ}$$30' Solutions : a) 40$$^{\\circ}$$20' 1' = $$\\frac{1}{60} ^{\\circ}$$ 20' = 20 * $$\\frac{1}{60} ^{\\circ}$$ = $$\\frac{1}{3}$$ 40$$^{\\circ}$$20' = 40+ $$\\frac{1}{3}$$ =$$\\frac{121}{3}^{\\circ}$$ =$$\\frac{121}{3}$$* $$\\frac{\\prod}{180}$$ =$$\\frac{121\\prod}{540}$$radians b) -47$$^{\\circ}$$30' = -(47 + 30*$$\\frac{1}{60} ^{\\circ}$$) = - (47 + 1/2)$$^{\\circ}$$ = - $$\\frac{95}{2}^{\\circ}$$ = - $$\\frac{95}{2}$$* $$\\frac{\\prod}{180}$$ radians = -$$\\frac{19\\prod}{72}$$ radians <<BACK<< >>NEXT>>", "is doing a serious study of trigonometric functions. $\\csc ( - 225^ \\circ ) = \\csc (135^ \\circ ) = \\frac{1}{{\\sin (135^ \\circ )}}$", "# How do you evaluate csc(210)? $C s c \\left(210\\right) = \\csc \\left(2 \\cdot 90 + 30\\right)$ $= - \\csc \\left(30\\right) = - 2$ ${210}^{\\circ}$ is anangle of 3rd quadranmt so csc(210) is negative", "## Algebra 2 (1st Edition) $C=180-50-74=56^\\circ$. Using the law of sines, $\\frac{a}{\\sin50^\\circ}=\\frac{12}{\\sin56^\\circ}\\\\a=\\frac{12\\sin50^\\circ}{\\sin56^\\circ}\\approx11.09$ $\\frac{b}{\\sin74^\\circ}=\\frac{12}{\\sin56^\\circ}\\\\b=\\frac{12\\sin74^\\circ}{\\sin56^\\circ}\\approx14.0$", "moment I don't know how to prove it. By the simmetry $\\sin(180^{\\circ}-k^{\\circ})=\\sin(k^{\\circ})$, we have that $$\\prod_{k=1}^{179}\\sin (k^{\\circ})=\\sin(90^{\\circ})\\left(\\prod_{k=1}^{89}\\sin (k^{\\circ})\\right)\\left(\\prod_{k=1}^{89}\\sin (180-k^{\\circ})\\right)=\\left(\\prod_{k=1}^{89}\\sin (k^{\\circ})\\right)^2$$ On the other hand by Evaluation of a product of sines (noted by H. H. Rugh and Batominovski). $$\\prod_{k=1}^{179}\\sin (k^{\\circ})= \\frac{180}{2^{179}}$$ Hence $$\\prod_{k=1}^{89}\\sin (k^{\\circ})=\\left(\\frac{180}{2^{179}}\\right)^{1/2}=\\frac{\\sqrt{90}}{2^{89}}=\\frac{3 \\sqrt{10}}{2^{89}}$$ which confirms the comment by Tolaso.", "Product of repeated cosec. $$P = \\prod_{k=1}^{45} \\csc^2(2k-1)^\\circ=m^n$$ I realize that there must be some sort of trick in this. $$P = \\csc^2(1)\\csc^2(3)…..\\csc^2(89) = \\frac{1}{\\sin^2(1)\\sin^2(3)….\\sin^2(89)}$$ I noticed that: $\\sin(90 + x) = \\cos(x)$ hence, $$\\sin(89) = \\cos(-1) = \\cos(359)$$ $$\\sin(1) = \\cos(-89) = \\cos(271)$$ $$\\cdots$$ $$P \\cdot P = \\frac{\\cos^2(-1)\\cos^2(-3)….}{\\sin^2(1)\\sin^2(3)….\\sin^2(89)}$$ But that doesnt help? Solutions Collecting From Web of \"Product of repeated cosec.\" $$\\sin(2n+1)x=(2n+1)\\sin x+\\cdots+(-1)^n2^{2n}\\sin^{2n+1}x$$ If we set $2n+1=45,2^{44}\\sin^{45}x-\\cdots+45\\sin x-\\sin45x=0$ If we set $\\sin45x=\\sin45^\\circ,45x=360^\\circ m+45^\\circ$ where $m$ is any integer $\\implies x=8^\\circ m+1^\\circ$ where $0\\le m\\le44$ $\\implies Q=\\prod_{m=0}^{44}\\sin(8^\\circ m+1^\\circ)=\\dfrac1{\\sqrt2}\\cdot\\dfrac1{2^{44}}$ $Q^2=\\dfrac1{2\\cdot2^{88}}$ Clearly, $\\{\\sin^2(8^\\circ m+1^\\circ);0\\le m\\le44\\}=\\{\\sin^2(2r-1)^\\circ,1\\le r\\le45\\}$ as $x=1^\\circ,9^\\circ,17^\\circ,25^\\circ,33^\\circ,41^\\circ,49^\\circ,57^\\circ,65^\\circ,73^\\circ,81^\\circ,89^\\circ,$ $97^\\circ[\\sin97^\\circ=\\sin(180-97)^\\circ=\\sin83^\\circ],$ $105^\\circ[\\sin75^\\circ]$ and so on $\\cdots$ $m=22\\implies8m+1=177\\implies\\sin177^\\circ=\\sin(180-3)^\\circ=\\sin3^\\circ$ $\\cdots$ $m=44\\implies8m+1=353\\implies\\sin353^\\circ=-\\sin7^\\circ$ I’m using a non-standard notation inspired by various programming languages in this evaluation because I think it’s a bit easier to follow than the traditional big pi product notation. Hopefully, it’s self-explanatory. Also, all angles are given in degrees. This derivation uses the following identities: $$\\cos x = \\sin(90 – x) \\\\ \\sin(180 – x) = \\sin x \\\\ \\sin 2x = 2 \\sin x \\cdot \\cos x \\\\ \\text{and hence} \\\\ \\cos x = \\frac{\\sin 2x}{2 \\sin x}\\\\$$ \\begin{align} \\text{Let } P & = prod(\\csc^2 x: x = \\text{1 to 89 by 2}) \\\\ 1 / P & = prod(\\sin x: x = \\text{1 to 89 by 2})^2 \\\\ & =", "# What are the asymptotes of g(x)=0.5 csc x? Apr 22, 2018 infinite #### Explanation: $\\csc x = \\frac{1}{\\sin} x$ $0.5 \\csc x = \\frac{0.5}{\\sin} x$ any number divided by $0$ gives an undefined result, so $0.5$ over $0$ is always undefined. the function $g \\left(x\\right)$ will be undefined at any $x$-values for which $\\sin x = 0$. from ${0}^{\\circ}$ to ${360}^{\\circ}$, the $x$-values where $\\sin x = 0$ are ${0}^{\\circ} , {180}^{\\circ} \\mathmr{and} {360}^{\\circ}$. alternatively, in radians from $0$ to $2 \\pi$, the $x$-values where $\\sin x = 0$ are $0 , \\pi \\mathmr{and} 2 \\pi$. since the graph of $y = \\sin x$ is periodic, the values for which $\\sin x = 0$ repeat every ${180}^{\\circ} , \\mathmr{and} \\pi$ radians. therefore, the points for which $\\frac{1}{\\sin} x$ and therefore $\\frac{0.5}{\\sin} x$ are undefined are ${0}^{\\circ} , {180}^{\\circ} \\mathmr{and} {360}^{\\circ}$ ($0 , \\pi \\mathmr{and} 2 \\pi$) in the restricted domain, but can repeat every ${180}^{\\circ}$, or every $\\pi$ radians, in either direction. graph{0.5 csc x [-16.08, 23.92, -6.42, 13.58]} here, you can see the repeating points at which the graph cannot continue due to undefined values. for example, the $y$-value steeply increases when approaching closer to $x = 0$ from the right, but never reaches $0$. the $y$-value steeply decreases when approaching closer to $x = 0$" ]

[ "# How do you find the value for csc^-1 (-7/3) ? $x = - {25}^{\\circ} 38 + k {360}^{\\circ}$, and $x = 180 - \\left(- 25.38\\right) = 180 + 25.38 = {205}^{\\circ} 38 + k {360}^{\\circ}$ $x = {205}^{\\circ} 38 + k {360}^{\\circ}$ $x = {334}^{\\circ} 62 + k {360}^{\\circ}$", "# How do I find the value of sin 225? Oct 29, 2015 #### Answer: $- \\frac{1}{\\sqrt{2}}$ #### Explanation: $\\sin {225}^{\\circ} = \\sin \\left({180}^{\\circ} + {45}^{\\circ}\\right)$ $= \\sin {180}^{\\circ} \\cos {45}^{\\circ} + \\cos {180}^{\\circ} \\sin {45}^{\\circ}$ $= \\left(0\\right) \\left(\\frac{1}{\\sqrt{2}}\\right) + \\left(- 1\\right) \\frac{1}{\\sqrt{2}}$ =$- \\frac{1}{\\sqrt{2}}$", "# How do I find the value of sin 225? Oct 29, 2015 $- \\frac{1}{\\sqrt{2}}$ #### Explanation: $\\sin {225}^{\\circ} = \\sin \\left({180}^{\\circ} + {45}^{\\circ}\\right)$ $= \\sin {180}^{\\circ} \\cos {45}^{\\circ} + \\cos {180}^{\\circ} \\sin {45}^{\\circ}$ $= \\left(0\\right) \\left(\\frac{1}{\\sqrt{2}}\\right) + \\left(- 1\\right) \\frac{1}{\\sqrt{2}}$ =$- \\frac{1}{\\sqrt{2}}$", "$\\csc \\frac{7 \\pi}{8}=\\frac{1}{\\sin \\frac{7 \\pi}{8}}=2.6131$ 2. $\\cot 85^\\circ=\\frac{1}{\\tan 85^\\circ}=0.0875$ 3. $\\sec 225^\\circ$ is the reciprocal of $\\cot 225^\\circ$ , a $45^\\circ$ reference angle in quadrant three where cosine is negative. Because $\\cos 45^\\circ=\\frac{1}{\\sqrt{2}}, \\cos 225^\\circ=-\\frac{1}{\\sqrt{2}}$ , and $\\sec 225^\\circ=-\\sqrt{2}$ . 4. $\\csc \\frac{5 \\pi}{6}$ is the reciprocal of $\\sin \\frac{5 \\pi}{6}$ , a $\\frac{\\pi}{6}$ or $30^\\circ$ reference angle in the second quadrant where sine is positive. Because $\\sin \\frac{\\pi}{6}=\\frac{1}{2}, \\sin \\frac{5 \\pi}{6}=\\frac{1}{2},$ and $\\csc \\frac{5 \\pi}{6}=2$ . ### Vocabulary Secant The reciprocal of cosine. Cosecant The reciprocal of sine. Cotangent The reciprocal of tangent. ### Practice 1. $\\csc 95^\\circ$ 2. $\\cot 278^\\circ$ 3. $\\sec \\frac{14 \\pi}{5}$ 4. $\\cot(-245^\\circ)$ 5. $\\sec \\frac{6 \\pi}{7}$ 6. $\\csc \\frac{23 \\pi}{13}$ 7. $\\cot 333^\\circ$ 8. $\\csc \\frac{9 \\pi}{5}$ Evaluate the following trigonometric functions without using a calculator. Give your answers exactly. 1. $\\sec \\frac{5 \\pi}{6}$ 2. $\\csc \\left(-\\frac{3 \\pi}{2}\\right)$ 3. $\\cot 225^\\circ$ 4. $\\sec \\frac{11 \\pi}{3}$ 5. $\\csc \\frac{7 \\pi}{6}$ 6. $\\sec 270^\\circ$ 7. $\\cot \\frac{5 \\pi}{3}$ 8. $\\csc 315^\\circ$", "$$\\cot 270^{\\circ}$$ $$\\cot 270^{\\circ}=0$$ 5. $$\\sec \\frac{11 \\pi}{6}$$ $$\\sec \\frac{11 \\pi}{6}=\\frac{1}{\\cos \\frac{11 \\pi}{6}}=\\frac{2}{\\sqrt{3}}=\\frac{2 \\sqrt{3}}{3}$$ 6. $$\\csc -\\frac{5 \\pi}{4}$$ $$\\csc -\\frac{5 \\pi}{4}=\\frac{1}{\\sin -\\frac{5 \\pi}{4}}=-\\frac{2}{\\sqrt{2}}=-\\sqrt{2}$$ Review Name the angle between $$0^{\\circ}$$ and $$360^{\\circ}$$ that is coterminal with... 1. $$-20^{\\circ}$$ 2. $$475^{\\circ}$$ 3. $$-220^{\\circ}$$ 4. $$690^{\\circ}$$ 5. $$-45^{\\circ}$$ Use your knowledge of the unit circle to help determine whether each of the following trigonometric expressions is positive or negative. 6. $$\\tan 143^{\\circ}$$ 7. $$\\cos \\frac{\\pi}{3}$$ 8. $$\\sin 362^{\\circ}$$ 9. $$\\csc \\frac{3 \\pi}{4}$$ Use your knowledge of the unit circle to evaluate each of the following trigonometric expressions. 10. $$\\cos 120^{\\circ}$$ 11. $$\\sec \\frac{\\pi}{3}$$ 12. $$\\tan 225^{\\circ}$$ 13. $$\\cot 120^{\\circ}$$ 14. $$\\sin \\frac{11 \\pi}{6}$$ 15. $$\\csc 240^{\\circ}$$ 16. Find $$\\sin \\theta$$ and $$\\tan \\theta$$ if $$\\cos \\theta=\\frac{\\sqrt{3}}{2}$$ and $$\\cot \\theta>0$$. 17. Find $$\\cos \\theta$$ and $$\\tan \\theta$$ if $$\\sin \\theta=-\\frac{1}{2}$$ and $$\\sec \\theta<0$$. 18. Draw the complete unit circle (all four quadrants) and label the important points.", "# How do you evaluate csc 180 ? Since $\\csc \\left(x\\right) = \\frac{1}{\\sin} \\left(x\\right)$, you have $\\csc \\left(180\\right) = \\frac{1}{\\sin} \\left(180\\right)$ But since $\\sin \\left(180\\right) = 0$, you can't divide by it, and thus $\\csc \\left(180\\right)$ is not defined.", "## Chemistry 10th Edition a) 229 K or -44$^{\\circ}$ C b) 76 K or -197$^{\\circ}$ C a) $T_{1}$ = (32+273) K = 305 K Let's assume $V_{1}$ = 4 V Then $V_{2}$ = 3 V (decreased by 25%) According to Charles's Law, $\\frac{V_{1}}{T_{1}}$ = $\\frac{V_{2}}{T_{2}}$ Or $T_{2}$ = $\\frac{V_{2}\\times T_{1}}{V_{1}}$ = $\\frac{3V\\times305K}{4V}$ $\\approx$ 229 K b) Here only $V_{2}$ is changed. $V_{2}$ = V if $V_{1}$ = 4V $T_{2}$ = $\\frac{V\\times 305 K}{4V}$ $\\approx$ 76 K .", "# How do you evaluate csc(210)? $C s c \\left(210\\right) = \\csc \\left(2 \\cdot 90 + 30\\right)$ $= - \\csc \\left(30\\right) = - 2$ ${210}^{\\circ}$ is anangle of 3rd quadranmt so csc(210) is negative", "$$C = (180-2x)^\\circ$$ and we have $$a=b$$ and $$c = 2a\\cos x^\\circ$$. Then $\\begin{array}{rcl} \\frac{aA+bB+cC}{a+b+c} & = & \\frac{ax + ax + 2a(180-2x)\\cos x^\\circ}{a + a + 2a\\cos x^\\circ} \\\\ & = & \\frac{2a[x + (180-2x)\\cos x^\\circ]}{2a(1 + \\cos x^\\circ)} \\; = \\; \\frac{x + (180-2x)\\cos x^\\circ}{1 + \\cos x^\\circ} \\end{array}$ and this tends to $$90$$ as $$x \\to 0$$. · 3 years, 11 months ago Thank you for noticing. I will use $$\\frac{\\pi}{3}$$ and $$\\frac{\\pi}{2}$$ instead, on the next time if i meet those numbers again. · 3 years, 11 months ago", "better -- every nonconstant polynomial has a root! 10. Apr 29, 2005 ### Orion1 De Moivre's Theorem... There are two solutions for the square root of i using De Moivre's Theorem: $$w_k = \\text{cis} \\; \\frac{90^{\\circ} + 360^{\\circ} k}{2}$$ $$w_0 = \\text{cis} \\; 45^{\\circ} = (\\cos 45^{\\circ} + i \\sin 45^{\\circ})$$ $$(\\cos 45^{\\circ} + i \\sin 45^{\\circ}) = \\left( \\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{2}}{2} i \\right)$$ $$w_1 = \\text{cis} \\; 225^{\\circ} = ( \\cos 225^{\\circ} + i \\sin 225^{\\circ})$$ $$( \\cos 225^{\\circ} + i \\sin 225^{\\circ}) = \\left( - \\frac{\\sqrt{2}}{2} - \\frac{\\sqrt{2}}{2} i \\right)$$ $$\\boxed{w_0 = \\frac{\\sqrt{2}}{2} (1 + i)}$$ $$\\boxed{w_1 = - \\frac{\\sqrt{2}}{2} (1 + i)}$$ $$\\boxed{z = \\pm \\frac{\\sqrt{2}}{2} (1 + i)}$$ Note that this solution is irrational with a root in the denominator: $$z = \\pm\\frac{1}{\\sqrt{2}}\\left(1+i)$$ Any root of i can be solved using De Moivre's Theorem. Last edited: Apr 29, 2005 11. Apr 29, 2005 ### Data And to see why Hurkyl's comment is significant, remember that over the reals, for example, not every polynomial has a root - for example, $x^2+1$ has no real roots, and it can't be factored over the reals. Over the complex numbers, every polynomial has a root, and this amounts to saying that every monic complex polynomial can be factored to the form $$(x - \\alpha_1)^{e_1} \\ \\cdots \\ (x", "225$$^{\\circ}$$ = 225 * $$\\frac{\\prod}{180}$$ = $$\\frac{5\\prod}{4}$$ k) 240$$^{\\circ}$$ = 240 * $$\\frac{\\prod}{180}$$ = $$\\frac{4\\prod}{3}$$ l) 270$$^{\\circ}$$ = 270 * $$\\frac{\\prod}{180}$$ = $$\\frac{3\\prod}{2}$$ m) 315$$^{\\circ}$$ = 315 * $$\\frac{\\prod}{180}$$ = $$\\frac{7\\prod}{4}$$ n) 330$$^{\\circ}$$ = 330 * $$\\frac{\\prod}{180}$$ = $$\\frac{11\\prod}{6}$$ o)360$$^{\\circ}$$ = 360 * $$\\frac{\\prod}{180}$$ = 2$$\\prod$$ 2) Convert the following radians into degree measures . a) $$\\frac{\\prod}{12}$$ b) $$\\frac{5\\prod}{3}$$ c) $$\\frac{2\\prod}{5}$$ d)$$\\frac{5\\prod}{6}$$ Solutions : As we know that 1 radian = $$\\frac{180^{\\circ}}{\\prod}$$ a) $$\\frac{\\prod}{12}$$ = $$\\frac{180}{12}$$ = 15$$^{\\circ}$$ b) $$\\frac{5\\prod}{3}$$ = $$\\frac{5*180}{3}$$ = 300$$^{\\circ}$$ c) $$\\frac{2\\prod}{5}$$ = $$\\frac{2 *180}{5}$$ = 72$$^{\\circ}$$ d)$$\\frac{5\\prod}{6}$$ = $$\\frac{5*180}{6}$$ =150 $$^{\\circ}$$ 3) Convert the following into radian measure . a) 40$$^{\\circ}$$20' b) -47$$^{\\circ}$$30' Solutions : a) 40$$^{\\circ}$$20' 1' = $$\\frac{1}{60} ^{\\circ}$$ 20' = 20 * $$\\frac{1}{60} ^{\\circ}$$ = $$\\frac{1}{3}$$ 40$$^{\\circ}$$20' = 40+ $$\\frac{1}{3}$$ =$$\\frac{121}{3}^{\\circ}$$ =$$\\frac{121}{3}$$* $$\\frac{\\prod}{180}$$ =$$\\frac{121\\prod}{540}$$radians b) -47$$^{\\circ}$$30' = -(47 + 30*$$\\frac{1}{60} ^{\\circ}$$) = - (47 + 1/2)$$^{\\circ}$$ = - $$\\frac{95}{2}^{\\circ}$$ = - $$\\frac{95}{2}$$* $$\\frac{\\prod}{180}$$ radians = -$$\\frac{19\\prod}{72}$$ radians <<BACK<< >>NEXT>>", "{1}{\\sin 135} = \\sqrt 2$ $\\sec 135 = \\frac {1}{\\cos 135} = -\\sqrt 2$ $\\cot 135 = \\frac {1}{\\tan 135} = – 1$ Now Trigonometric table for 210 to 270 is given by And it is calculated as $\\sin (210) = \\sin (180 +30) =- \\sin 30= -\\frac {1}{2}$ $\\cos (210) = \\cos (180 +30) =- \\cos 30=-\\frac {\\sqrt {3}}{2}$ $\\tan (210) = \\frac {\\sin 210}{ \\cos 210} = \\frac {1}{\\sqrt {3}}$ $\\sin (225) = \\sin (180 +45) =- \\sin 45= -\\frac {1}{\\sqrt {2}}$ $\\cos (225) = \\cos (180+45) =- \\cos 45= -\\frac {1}{\\sqrt {2}}$ $\\tan 225 = \\frac {\\sin 225}{\\cos 225} = 1$ $\\sin (270) = \\sin (180 +90) =- \\sin 90= -1$ $\\cos (270) = \\cos (180+90) =- \\cos 90= 0$ $\\tan 270 = \\frac {\\sin 270}{\\cos 270} = -\\frac {1}{0}$ Undefined value $\\csc (210) = \\frac {1}{\\sin (210)} = -2$ $\\sec (210) = \\frac {1}{\\cos (210)}=-\\frac {2}{\\sqrt 3}$ $\\cot (210) = \\frac {1}{\\tan (210)} = \\sqrt {3}$ $\\csc (225) = \\frac {1}{\\sin 225}= -\\sqrt {2}$ $\\sec (225) = \\frac {1}{\\cos 225}= -\\sqrt {2}$ $\\cot 225 = \\frac {1}{\\tan 225} = 1$ Now Trigonometric table for 300 to 360 is given by $\\sin (300) = \\sin (360 -60) =- \\sin 60=-\\frac {\\sqrt {3}}{2}$ $\\cos (300) = \\cos (360-60) =\\cos 60=\\frac {1}{2}$ $\\tan (300) = \\frac {\\sin", "} 225 \\div 45 = 5. \\text{ So } 225° = 5 \\left( \\frac{\\pi}{4} \\right) = \\frac{5 \\pi}{4} .$", "\\frac {1}{\\tan 120} = – \\frac {1}{\\sqrt 3}$ $\\csc 135 = \\frac {1}{\\sin 135} = \\sqrt 2$ $\\sec 135 = \\frac {1}{\\cos 135} = -\\sqrt 2$ $\\cot 135 = \\frac {1}{\\tan 135} = – 1$ Now Trigonometric table for 210 to 270 is given by And it is calculated as $\\sin (210) = \\sin (180 +30) =- \\sin 30= -\\frac {1}{2}$ $\\cos (210) = \\cos (180 +30) =- \\cos 30=-\\frac {\\sqrt {3}}{2}$ $\\tan (210) = \\frac {\\sin 210}{ \\cos 210} = \\frac {1}{\\sqrt {3}}$ $\\sin (225) = \\sin (180 +45) =- \\sin 45= -\\frac {1}{\\sqrt {2}}$ $\\cos (225) = \\cos (180+45) =- \\cos 45= -\\frac {1}{\\sqrt {2}}$ $\\tan 225 = \\frac {\\sin 225}{\\cos 225} = 1$ $\\sin (270) = \\sin (180 +90) =- \\sin 90= -1$ $\\cos (270) = \\cos (180+90) =- \\cos 90= 0$ $\\tan 270 = \\frac {\\sin 270}{\\cos 270} = -\\frac {1}{0}$ Undefined value $\\csc (210) = \\frac {1}{\\sin (210)} = -2$ $\\sec (210) = \\frac {1}{\\cos (210)}=-\\frac {2}{\\sqrt 3}$ $\\cot (210) = \\frac {1}{\\tan (210)} = \\sqrt {3}$ $\\csc (225) = \\frac {1}{\\sin 225}= -\\sqrt {2}$ $\\sec (225) = \\frac {1}{\\cos 225}= -\\sqrt {2}$ $\\cot 225 = \\frac {1}{\\tan 225} = 1$ Now Trigonometric table for 300 to 360 is given by $\\sin (300) = \\sin (360 -60) =- \\sin 60=-\\frac {\\sqrt {3}}{2}$ $\\cos", "## Calculus with Applications (10th Edition) $\\sec 225^\\circ=-\\sqrt 2$ We are given $\\sec \\frac{5\\pi}{4}$ convert $\\frac{5\\pi}{4}$ to degrees $\\frac{5\\pi}{4}$ radians=$\\frac{5\\pi}{4}(\\frac{180^\\circ}{\\pi})$ $\\frac{5\\pi}{4}$ radians= $225^\\circ$ $\\sec 225^\\circ=-sec 45^\\circ$ and $\\sec \\theta=\\frac{1}{\\cos \\pi}$ so $\\sec 225^\\circ=-\\sqrt 2$", "Product of repeated cosec. $$P = \\prod_{k=1}^{45} \\csc^2(2k-1)^\\circ=m^n$$ I realize that there must be some sort of trick in this. $$P = \\csc^2(1)\\csc^2(3)…..\\csc^2(89) = \\frac{1}{\\sin^2(1)\\sin^2(3)….\\sin^2(89)}$$ I noticed that: $\\sin(90 + x) = \\cos(x)$ hence, $$\\sin(89) = \\cos(-1) = \\cos(359)$$ $$\\sin(1) = \\cos(-89) = \\cos(271)$$ $$\\cdots$$ $$P \\cdot P = \\frac{\\cos^2(-1)\\cos^2(-3)….}{\\sin^2(1)\\sin^2(3)….\\sin^2(89)}$$ But that doesnt help? Solutions Collecting From Web of \"Product of repeated cosec.\" $$\\sin(2n+1)x=(2n+1)\\sin x+\\cdots+(-1)^n2^{2n}\\sin^{2n+1}x$$ If we set $2n+1=45,2^{44}\\sin^{45}x-\\cdots+45\\sin x-\\sin45x=0$ If we set $\\sin45x=\\sin45^\\circ,45x=360^\\circ m+45^\\circ$ where $m$ is any integer $\\implies x=8^\\circ m+1^\\circ$ where $0\\le m\\le44$ $\\implies Q=\\prod_{m=0}^{44}\\sin(8^\\circ m+1^\\circ)=\\dfrac1{\\sqrt2}\\cdot\\dfrac1{2^{44}}$ $Q^2=\\dfrac1{2\\cdot2^{88}}$ Clearly, $\\{\\sin^2(8^\\circ m+1^\\circ);0\\le m\\le44\\}=\\{\\sin^2(2r-1)^\\circ,1\\le r\\le45\\}$ as $x=1^\\circ,9^\\circ,17^\\circ,25^\\circ,33^\\circ,41^\\circ,49^\\circ,57^\\circ,65^\\circ,73^\\circ,81^\\circ,89^\\circ,$ $97^\\circ[\\sin97^\\circ=\\sin(180-97)^\\circ=\\sin83^\\circ],$ $105^\\circ[\\sin75^\\circ]$ and so on $\\cdots$ $m=22\\implies8m+1=177\\implies\\sin177^\\circ=\\sin(180-3)^\\circ=\\sin3^\\circ$ $\\cdots$ $m=44\\implies8m+1=353\\implies\\sin353^\\circ=-\\sin7^\\circ$ I’m using a non-standard notation inspired by various programming languages in this evaluation because I think it’s a bit easier to follow than the traditional big pi product notation. Hopefully, it’s self-explanatory. Also, all angles are given in degrees. This derivation uses the following identities: $$\\cos x = \\sin(90 – x) \\\\ \\sin(180 – x) = \\sin x \\\\ \\sin 2x = 2 \\sin x \\cdot \\cos x \\\\ \\text{and hence} \\\\ \\cos x = \\frac{\\sin 2x}{2 \\sin x}\\\\$$ \\begin{align} \\text{Let } P & = prod(\\csc^2 x: x = \\text{1 to 89 by 2}) \\\\ 1 / P & = prod(\\sin x: x = \\text{1 to 89 by 2})^2 \\\\ & =", "is doing a serious study of trigonometric functions. $\\csc ( - 225^ \\circ ) = \\csc (135^ \\circ ) = \\frac{1}{{\\sin (135^ \\circ )}}$", "# What are the asymptotes of g(x)=0.5 csc x? Apr 22, 2018 infinite #### Explanation: $\\csc x = \\frac{1}{\\sin} x$ $0.5 \\csc x = \\frac{0.5}{\\sin} x$ any number divided by $0$ gives an undefined result, so $0.5$ over $0$ is always undefined. the function $g \\left(x\\right)$ will be undefined at any $x$-values for which $\\sin x = 0$. from ${0}^{\\circ}$ to ${360}^{\\circ}$, the $x$-values where $\\sin x = 0$ are ${0}^{\\circ} , {180}^{\\circ} \\mathmr{and} {360}^{\\circ}$. alternatively, in radians from $0$ to $2 \\pi$, the $x$-values where $\\sin x = 0$ are $0 , \\pi \\mathmr{and} 2 \\pi$. since the graph of $y = \\sin x$ is periodic, the values for which $\\sin x = 0$ repeat every ${180}^{\\circ} , \\mathmr{and} \\pi$ radians. therefore, the points for which $\\frac{1}{\\sin} x$ and therefore $\\frac{0.5}{\\sin} x$ are undefined are ${0}^{\\circ} , {180}^{\\circ} \\mathmr{and} {360}^{\\circ}$ ($0 , \\pi \\mathmr{and} 2 \\pi$) in the restricted domain, but can repeat every ${180}^{\\circ}$, or every $\\pi$ radians, in either direction. graph{0.5 csc x [-16.08, 23.92, -6.42, 13.58]} here, you can see the repeating points at which the graph cannot continue due to undefined values. for example, the $y$-value steeply increases when approaching closer to $x = 0$ from the right, but never reaches $0$. the $y$-value steeply decreases when approaching closer to $x = 0$", "moment I don't know how to prove it. By the simmetry $\\sin(180^{\\circ}-k^{\\circ})=\\sin(k^{\\circ})$, we have that $$\\prod_{k=1}^{179}\\sin (k^{\\circ})=\\sin(90^{\\circ})\\left(\\prod_{k=1}^{89}\\sin (k^{\\circ})\\right)\\left(\\prod_{k=1}^{89}\\sin (180-k^{\\circ})\\right)=\\left(\\prod_{k=1}^{89}\\sin (k^{\\circ})\\right)^2$$ On the other hand by Evaluation of a product of sines (noted by H. H. Rugh and Batominovski). $$\\prod_{k=1}^{179}\\sin (k^{\\circ})= \\frac{180}{2^{179}}$$ Hence $$\\prod_{k=1}^{89}\\sin (k^{\\circ})=\\left(\\frac{180}{2^{179}}\\right)^{1/2}=\\frac{\\sqrt{90}}{2^{89}}=\\frac{3 \\sqrt{10}}{2^{89}}$$ which confirms the comment by Tolaso.", "## Basic College Mathematics (10th Edition) $74^\\circ\\ F=23.3^\\circ\\ C$ $C=\\frac{5(F-32)}{9}$ Substitute 74 for F and solve for C. $C=\\frac{5(74-32)}{9}=\\frac{5(42)}{9}=\\frac{210}{9}=23.3\\overline3$" ]

[ "# Give the spherical coordinates representation of the following solids Give the spherical coordinates representation of the following solids: (i) Inside the sphere $\\ x^2+y^2+z^2=4 \\$ and outside the cylinder $\\ x^2+y^2=1 \\$ (ii) Inside the sphere $\\ x^2+y^2+z^2=4 \\$ and above the plane $\\ z=3 \\$ (i) $0 \\leq \\rho \\leq 2 \\\\ 0 \\leq \\phi \\leq \\frac{\\pi}{4} \\\\ 0 \\leq \\theta \\leq 2 \\pi \\$ Am I right? Help me out $$\\rho |\\sin\\theta| > 1$$", "Cool! In spherical coordinates, $v = \\dot{r} \\hat{r} + r \\sin \\phi \\dot{\\theta} \\hat{\\theta} + r \\dot{\\phi} \\hat{\\phi}$. This results in the three conserved quantities $\\frac{d}{dt} \\left\\{ m\\dot{r} \\right\\} = 0$, $\\frac{d}{dt} \\left\\{ m r^2 \\sin^2 \\phi \\dot{\\theta} \\right\\} = 0$, and $\\frac{d}{dt} \\left\\{ m r^2 \\dot{\\phi} \\right\\} = 0$. Even more conserved quantities, fun for the whole family!", "2. Convert the point $$\\left( { - 1,1, - \\sqrt 2 } \\right)$$ from Cartesian to spherical coordinates. Show All Solutions Hide All Solutions a Convert the point $$\\displaystyle \\left( {\\sqrt 6 ,\\frac{\\pi }{4},\\sqrt 2 } \\right)$$ from cylindrical to spherical coordinates. Show Solution We’ll start by acknowledging that $$\\theta$$ is the same in both coordinate systems and so we don’t need to do anything with that. Next, let’s find$$\\rho$$. $\\rho = \\sqrt {{r^2} + {z^2}} = \\sqrt {6 + 2} = \\sqrt 8 = 2\\sqrt 2$ Finally, let’s get $$\\varphi$$. To do this we can use either the conversion for $$r$$ or $$z$$. We’ll use the conversion for $$z$$. $z = \\rho \\cos \\varphi \\hspace{0.25in} \\Rightarrow \\hspace{0.25in}\\cos \\varphi = \\frac{z}{\\rho } = \\frac{{\\sqrt 2 }}{{2\\sqrt 2 }}\\hspace{0.25in} \\Rightarrow \\hspace{0.25in}\\varphi = {\\cos ^{ - 1}}\\left( {\\frac{1}{2}} \\right) = \\frac{\\pi }{3}$ Notice that there are many possible values of $$\\varphi$$ that will give $$\\cos \\varphi = \\frac{1}{2}$$, however, we have restricted $$\\varphi$$ to the range $$0 \\le \\varphi \\le \\pi$$ and so this is the only possible value in that range. So, the spherical coordinates of this point will are $$\\left( {2\\sqrt 2 ,\\frac{\\pi }{4},\\frac{\\pi }{3}} \\right)$$. b Convert the point $$\\left( { - 1,1, - \\sqrt 2 } \\right)$$ from Cartesian to spherical coordinates. Show Solution The first", "\\phi={{\\cos}^{-1}}(\\frac{z}{\\rho})$ $latex \\phi={{\\cos}^{-1}}(\\frac{5}{6.71})$ $latex \\phi=0.73$ rad The spherical coordinates of the point are (6.71, 1.11 rad, 0.73 rad).", "Calculate the $$\\theta$$ angle: $\\cos\\theta = \\frac{z}{\\rho} = \\frac{1}{2}, \\Rightarrow \\theta = \\arccos\\frac{1}{2} = \\frac{\\pi}{3}.$ So the spherical coordinates of the point $$M$$ are $\\rho = 2,\\;\\varphi = \\arccos\\sqrt{\\frac{2}{3}},\\;\\theta = \\frac{\\pi}{3}.$", "+0 # Hard Question Pls Help Me 0 44 1 A point has rectangular coordinates $(2,-1,-2)$ and spherical coordinates $(\\rho, \\theta, \\phi).$ Find the rectangular coordinates of the point with spherical coordinates $(\\rho, \\theta, 2 \\phi).$ Jun 6, 2020", "\\lt \\infty,\\;0 \\le \\varphi \\lt 2\\pi.$ By definition, the $$\\theta-$$coordinate lies in the range $$0 \\le \\theta \\le \\pi.$$ If point M coincides with point O, then $$\\rho = 0.$$ For point O, the $$\\varphi-$$ and $$\\theta-$$coordinates are not defined. ## Converting Between Spherical and Cartesian Coordinates If you choose the axes of the Cartesian coordinate system as indicated above in the figure, then the Cartesian coordinates $$x, y, z$$ of a point are related to its spherical coordinates $${\\rho, \\varphi, \\theta}$$ by the relations $x = \\rho\\sin\\theta\\cos\\varphi,\\;\\;y = \\rho\\sin\\theta\\sin\\varphi,\\;\\;z = \\rho\\cos\\theta.$ For converting a point from Cartesian coordinates to spherical coordinates, we use the formulas $\\rho = \\sqrt{x^2 + y^2 + z^2},\\;\\;\\tan\\varphi = \\frac{y}{x},\\;\\;\\cos\\theta = \\frac{z}{\\rho} = \\frac{z}{\\sqrt{x^2 + y^2 + z^2}}.$ ## Solved Problems Click or tap a problem to see the solution. ### Example 1 Find the spherical coordinates of each point given its Cartesian coordinates. 1. $$A\\left({1,0,0}\\right)$$ 2. $$B\\left({-4,-3,0}\\right)$$ 3. $$C\\left({2,2,1}\\right)$$ 4. $$D\\left({8,-4,1}\\right)$$ ### Example 2 Convert from spherical to Cartesian coordinates. 1. $$K\\left({\\sqrt{2},\\frac{\\pi}{4},\\frac{\\pi}{6}}\\right)$$ 2. $$L\\left({3\\sqrt{2},\\frac{3\\pi}{4},\\frac{\\pi}{2}}\\right)$$ 3. $$M\\left({5,\\frac{3\\pi}{2},\\frac{\\pi}{2}}\\right)$$ ### Example 3 Express city's location in spherical coordinates given its latitude and longitude: 1. London $$51.5^{\\circ}\\,N, 0.1^{\\circ} W$$ 2. Rio de Janeiro $$22.9^{\\circ}\\,S, 43.2^{\\circ} W$$ ### Example 4 Find the spherical coordinates of point $$M$$ if the ray $$\\mathbf{OM}$$ forms the angles $$\\alpha = \\frac{\\pi}{4}$$", "\\lt \\infty,\\;0 \\le \\varphi \\lt 2\\pi.$ By definition, the $$\\theta-$$coordinate lies in the range $$0 \\le \\theta \\le \\pi.$$ If point M coincides with point O, then $$\\rho = 0.$$ For point O, the $$\\varphi-$$ and $$\\theta-$$coordinates are not defined. ## Converting Between Spherical and Cartesian Coordinates If you choose the axes of the Cartesian coordinate system as indicated above in the figure, then the Cartesian coordinates $$x, y, z$$ of a point are related to its spherical coordinates $${\\rho, \\varphi, \\theta}$$ by the relations $x = \\rho\\sin\\theta\\cos\\varphi,\\;\\;y = \\rho\\sin\\theta\\sin\\varphi,\\;\\;z = \\rho\\cos\\theta.$ For converting a point from Cartesian coordinates to spherical coordinates, we use the formulas $\\rho = \\sqrt{x^2 + y^2 + z^2},\\;\\;\\tan\\varphi = \\frac{y}{x},\\;\\;\\cos\\theta = \\frac{z}{\\rho} = \\frac{z}{\\sqrt{x^2 + y^2 + z^2}}.$ ## Solved Problems Click or tap a problem to see the solution. ### Example 1 Find the spherical coordinates of each point given its Cartesian coordinates. 1. $$A\\left({1,0,0}\\right)$$ 2. $$B\\left({-4,-3,0}\\right)$$ 3. $$C\\left({2,2,1}\\right)$$ 4. $$D\\left({8,-4,1}\\right)$$ ### Example 2 Convert from spherical to Cartesian coordinates. 1. $$K\\left({\\sqrt{2},\\frac{\\pi}{4},\\frac{\\pi}{6}}\\right)$$ 2. $$L\\left({3\\sqrt{2},\\frac{3\\pi}{4},\\frac{\\pi}{2}}\\right)$$ 3. $$M\\left({5,\\frac{3\\pi}{2},\\frac{\\pi}{2}}\\right)$$ ### Example 1. Find the spherical coordinates of each point given its Cartesian coordinates. 1. $$A\\left({1,0,0}\\right)$$ 2. $$B\\left({-4,-3,0}\\right)$$ 3. $$C\\left({2,2,1}\\right)$$ 4. $$D\\left({8,-4,1}\\right)$$ Solution. We use the following formulas: $\\rho = \\sqrt{x^2 + y^2 + z^2},\\;\\;\\tan\\varphi = \\frac{y}{x},\\;\\;\\cos\\theta = \\frac{z}{\\rho}.$ 1. $$A\\left({1,0,0}\\right)$$ $\\rho = \\sqrt{1^2 + 0^2", "coördinates. Ans. $\\frac{dy}{dx} = \\frac{ \\rho \\cos \\theta + \\sin \\theta \\frac{d\\rho}{d\\theta} }{ -\\rho \\sin \\theta + \\cos \\theta \\frac{d\\rho}{d\\theta} }$.", "\\pi} \\right)^{\\frac{1}{3}} \\approx 0.894$ sphere $\\frac{4}{3} \\pi r^3$ $4 \\pi\\,r^2$ $1\\,$", "+0 # The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. 0 68 1 +16 The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. Aug 6, 2020", "Question # Change from rectangular to spherical coordinates. (Let \\rho\\geq 0, 0 \\leq \\theta \\leq 2\\pi, \\text{ and } 0 \\leq \\phi \\leq \\pi) Parametric equations Change from rectangular to spherical coordinates. (Let $$\\displaystyle\\rho\\geq{0},{0}\\leq\\theta\\leq{2}\\pi,\\ \\text{ and }\\ {0}\\leq\\phi\\leq\\pi$$) (a) $$(0, -3, 0)$$ (b) $$\\displaystyle{\\left(-{1},{1},-\\sqrt{{{2}}}\\right)}$$", "# “The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$.” Code Answer The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. whatever by Confused Chinchilla on Jun 17 2020 Comment 1", "+0 # The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. 0 246 1 +16 The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. Aug 6, 2020", "+0 # The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. 0 34 1 +16 The spherical coordinates of $(-3, 4, -12)$ are $(\\rho, \\theta, \\phi)$. Find $\\tan \\theta + \\tan \\phi$. Aug 6, 2020 #1 0 tan theta + tan phi = 5/2 - 4/3 = 7/6. Aug 6, 2020", "# Convert the point from rectangular coordinates to spherical coordinates. (-3,-3,\\sqrt{2}) Convert the point from rectangular coordinates to spherical coordinates. $$\\displaystyle{\\left(-{3},-{3},\\sqrt{{{2}}}\\right)}$$ • Questions are typically answered in as fast as 30 minutes ### Plainmath recommends • Get a detailed answer even on the hardest topics. • Ask an expert for a step-by-step guidance to learn to do it yourself. Arham Warner To convert:", "Free Version Moderate # Spherical to Rectangular Conversion MVCALC-EPFFL4 The equation in rectangular coordinates for the surface that has equation $\\phi=\\pi/3$ in spherical coordinates has the form: $$a(x^2+y^2)=z^2$$ What is the value of $a$?", "1 + 1 + 1 = 3$$ The other two cylindrical and spherical, I did the same way. But I used the $$\\nabla\\cdot \\vec{r}$$ for cylindrical and spherical which I referred to in a handbook. Last edited: Jan 9, 2005 7. Jan 9, 2005 ### meteorologist1 Could you show me how to get \"3\" for the cylindrical and spherical? Thanks. 8. Jan 9, 2005 ### Galileo In cylindrical coordinates: $$\\vec r = \\rho \\hat \\rho + z \\hat z$$ and $$\\nabla \\cdot \\vec v(\\rho,\\phi,z) = \\frac{1}{\\rho}\\frac{\\partial}{\\partial \\rho}(\\rho v_{\\rho})+\\frac{1}{\\rho}\\frac{\\partial v_{\\phi}}{\\partial \\phi}+\\frac{\\partial v_z}{\\partial z}$$ For some vector function $v=\\langle v_{\\rho},v_{\\phi},v_z \\rangle$ In spherical coordinates: $$\\vec r = r\\hat r$$ and $$\\nabla \\cdot \\vec v(r,\\theta,\\phi)=\\frac{1}{r^2}\\frac{\\partial}{\\partial r}(r^2v_r)+\\frac{1}{r\\sin \\theta}\\frac{\\partial}{\\partial \\theta}(\\sin \\theta v_{\\theta})+\\frac{1}{r\\sin \\theta}\\frac{\\partial v_{\\phi}}{\\partial \\phi}$$ for some vector function $v=\\langle v_r,v_{\\theta},v_{\\phi} \\rangle$ Both give 3 as well: $$\\nabla \\cdot \\vec r(\\rho,\\phi,z) =\\frac{1}{\\rho}\\frac{\\partial}{\\partial \\rho}(\\rho \\rho)+\\frac{\\partial z}{\\partial z}=\\frac{1}{\\rho}2\\rho+1=3$$ $$\\nabla \\cdot \\vec r(r,\\theta,\\phi)=\\frac{1}{r^2}\\frac{\\partial}{\\partial r}(r^2 r)=\\frac{1}{r^2}3r^2=3$$ You can't use the divergence in cylindrical or spherical coordinates when r=0 though. Last edited: Jan 9, 2005 9. Jan 9, 2005 ### meteorologist1 Thanks very much. It looks like the reason I got the wrong answers is because I got the r vectors wrong in cylindrical and spherical coordinates. I need to get this straight. I would like to ask another question related to these two coordinates: In cylindrical coordinates,", "Both (3, $\\frac{\\pi }{3}$) and (- 3, $\\frac{\\pi }{3}$) represent the same point. b. Both (3, $\\frac{\\pi }{3}$) and (- 3, - $\\frac{\\pi }{3}$) represent the same point. c. Both (3, $\\frac{\\pi }{3}$) and (- 3, - $\\frac{2\\pi }{3}$) represent the same point. d. Both (3, $\\frac{\\pi }{3}$) and (3, - $\\frac{\\pi }{3}$) represent the same point. #### Solution: For integral values of n, both (r, θ), (- r, θ + (2n + 1) π) represent the same point in a polar coordinate system. So, (- r, θ - π) and (r, θ) represent the same point. [Substitute n = -1.] (-3, - 2π3) = (-3, π3 - π) [Write - 2π3 as π3 - π.] So, both (3, π3) and (- 3, - 2π3) represent the same point.", "you have $int_\\left\\{-5\\right\\}^5 dz int_0^\\left\\{2 pi\\right\\} dphi int_0^3 \\left(rho^3 + rho z \\right), drho = 2 pi int_\\left\\{-5\\right\\}^5 left\\left[frac\\left\\{rho^4\\right\\}\\left\\{4\\right\\} + frac\\left\\{rho^2 z\\right\\}\\left\\{2\\right\\} right\\right]_0^3 , dz$ $= 2 pi int_\\left\\{-5\\right\\}^5 left\\left(frac\\left\\{81\\right\\}\\left\\{4\\right\\} + frac\\left\\{9\\right\\}\\left\\{2\\right\\} zright\\right), dz = cdots = 855 pi.$ #### Spherical coordinates In R3 some domains have a spherical symmetry, so it's possible to specify the coordinates of every point of the integration region by two angles and one distance. It's possible to use therefore the passage in spherical coordinates; the function is transformed by this relation: $f\\left(x,y,z\\right) longrightarrow f\\left(rho cos theta sin phi, rho sin theta sin phi, rho cos phi\\right),!$ Note that points on z axis do not have a precise characterization in spherical coordinates, so $phi$ can vary between 0 to π . The better integration domain for this passage is obviously the sphere. Example (4-a): The domain is $D = x^2 + y^2 + z^2 le 16$ (sphere with radius 4 and center in the origin); applying the transformation you get this region: $T = \\left\\{ 0 le rho le 4, 0 le phi le pi, 0 le theta le 2 pi \\right\\}.$ The Jacobian determinant of this transformation is the following: $frac\\left\\{partial \\left(x,y,z\\right)\\right\\}\\left\\{partial \\left(rho, theta, phi\\right)\\right\\} =$ begin{vmatrix} cos theta sin phi & - rho sin theta sin phi & rho cos theta" ]

[ "box are available. • SurfaceHex(N, B, L, orient) • this operator allows to build the surface mesh of a 3d box • int[int] N=[nx,ny,nz]; // the number of seg in the 3 direction • real [int,int] B=[[xmin,xmax],[ymin,ymax],[zmin,zmax]]; // bounding bax • int [int,int] L=[[1,2],[3,4],[5,6]]; // the label of the 6 face left,right, front, back, down, right • orient the global orientation of the surface 1 extern (-1 intern), • returns a meshS type • Ellipsoide (RX, RY, RZ, h, L, OX, OY, OZ, orient) • h is the mesh size • L is the label • orient the global orientation of the surface 1 extern (-1 intern) • OX, OY, OZ are real numbers to give the Ellipsoide center ( optinal, by default is (0,0,0) ) • where RX, RY, RZ are real numbers such as the parametric equations of the ellipsoid is: • returns a meshS type $\\forall u \\in [- \\frac{\\pi}{2},\\frac{\\pi}{2} [ \\text{ and } v \\in [0, 2 \\pi], \\vectthree{x=\\text{Rx } cos(u)cos(v) + \\text{Ox }}{y=\\text{Ry } cos(u)sin(v) + \\text{Oy }}{z = \\text{Rz } sin(v) + \\text{Oz } }$ • Sphere(R, h, L, OX, OY, OZ, orient) • where R is the raduis of the sphere, • OX, OY, OZ are real numbers to give the Ellipsoide center ( optinal, by default is (0,0,0) ) •", "to the right size (52 cm by 14 cm), it is glued around a cardboard cylinder (radius 8 cm). A home made conical mirror (base radius 7 cm, height 12 cm) is put inside the cylinder at the center. The anamorphic image on the cylinder wall is reflected as the undeformed original by the conical mirror. Here is the result: (the center is hidden by a coin resting at the top of the cone since anamorphic maps of points close to the cone center are off at infinite height on the cylinder wall)Another application of the function is to use one of the many popular curves (\".....-like curve\" ) that can be extracted using Interpreter Interpreter[\"PopularCurve\"][\"bunny-like curve\"]; bugsbunnyPrimitives = First@Cases[ First[ParametricPlot[ Entity[\"PopularCurve\", \"BunnyCurve\"][\"ParametricEquations\"][ t], {t, 0, 30 \\[Pi]}]] /. {x_?NumericQ, y_?NumericQ} :> {x - 85, y - 50}/800, _Line, \\[Infinity]]; The anamorphic map is created by applying anaCone2Cylinder to the point coordinates: anaBunny = Map[anaCone2Cylinder[#, 1.755, 1.25] &, bugsbunnyPrimitives, {2}]; Animate[Graphics3D[ Rotate[{{Opacity[.2], White, Cylinder[{{0, 0, .25}, {0, 0, 1}}, 1.25]}, AbsoluteThickness[3], Red, anaBunny}, \\[Phi], {0, 0, 1}], Boxed -> False], {\\[Phi], 0, 2 \\[Pi]}] This is the developed cylinder: developRules = {x_?NumericQ, y_?NumericQ, z_?NumericQ} :> {ArcTan[x, y], z}; developed = anaBunny /. developRules; DeleteCases[ Partition[developed[[1]], 2, 1], _?(EuclideanDistance @@ # > 1 &)]; Graphics[{Red, AbsoluteThickness[3], Line /@ %},", "coordinate system, the angle made by $DE$ and the positive × axis is $\\theta$ (Note that we have assumed that all the light rays coming from the LED are parallel to the light ray reaching the center of the flat surface of the lens.). Therefore, $DE$ can be represented by TeX Source $${y-r\\sin\\beta\\over x-r\\cos\\beta}=\\tan(\\theta).\\eqno{\\hbox{(26)}}$$ Note that the function becomes $y=r\\sin\\beta$ when $\\theta=0$ or $\\pi$ and $x=r\\cos\\beta$ when $\\theta=\\pm\\pi/2$. Let the center of chord $DE$ be $C$. Then the following equation holds by the law of cosines. TeX Source $$\\vert AC\\vert^{2}+\\vert AB\\vert^{2}-2\\vert AC\\Vert AB\\vert\\cos(\\angle BAC)=\\vert BC\\vert^{2}.\\eqno{\\hbox{(27)}}$$ By setting $y=0$ in (26), $\\vert OG\\vert$ equals $r\\vert\\cos\\beta-(\\sin\\beta/\\tan\\theta)\\vert$. Thus, $\\vert OC\\vert$ can be evaluated as TeX Source \\eqalignno{\\vert OC\\vert=&\\,\\vert OG\\sin\\theta\\vert\\cr=&\\,r\\left\\vert\\left({\\cos\\beta-\\sin\\beta\\over\\tan\\theta}\\right)\\sin\\theta\\right\\vert\\cr=&\\,r\\left\\vert\\sin(\\theta-\\beta)\\right\\vert,&\\hbox{(28)}} which leads $\\vert AC\\vert$ can be derived readily as TeX Source \\eqalignno{\\vert AC\\vert=&\\,\\sqrt{\\vert OA\\vert^{2}-\\vert OC\\vert^{2}}\\cr=&\\,\\sqrt{r^{2}-r^{2}\\sin^{2}(\\theta-\\beta)}=r\\left\\vert\\cos(\\theta-\\beta)\\right\\vert.&\\hbox{(29)}}$\\vert BC\\vert$ equals $\\vert DC\\vert$ to TeX Source $$\\vert BC\\vert\\!=\\!\\vert DC\\vert\\!=\\!\\sqrt{\\vert OD\\vert^{2}\\!-\\!\\vert OC\\vert^{2}}\\!=\\!\\sqrt{R^{2}\\!-\\!r^{2}\\sin^{2}(\\theta\\!-\\!\\beta)}.\\eqno{\\hbox{(30)}}$$ The value of $\\cos(\\angle BAC)$ depends on the coordinates of $A$. If $A$ locates to the right of $C$ (as shown in Fig. 11(a)), we have $\\cos(\\angle BAC)=\\sin\\alpha_{2}$. Otherwise, if $A$ locates to the left of $C$ (as shown in Fig. 11(b)), then $\\cos(\\angle BAC)=\\cos((\\pi/2)+\\alpha_{2})=-\\sin\\alpha_{2}$. The sign can be determined as follows. Define a vector ${\\bf r}$ equals $[r\\cos\\beta,r\\sin\\beta]^{T}$. Rotating ${\\bf r}$ about the z axis by $-\\theta$ yields a vector ${\\bf r}^{\\prime}$ which can", "(I hope you don't mean the coordinate system's $z$ axis) Mar 31, 2021 at 14:29 ## 3 Answers In spherical coordinates, $$x = \\rho \\cos \\theta \\sin \\phi, y = \\rho \\sin \\theta \\sin \\phi, z = \\rho \\cos\\phi, \\rho = \\sqrt{x^2+y^2+z^2}$$ If the radius of the sphere is $$r$$ with origin as the center, height of spherical cap is $$h$$ and radius of the base of the spherical cap is $$a$$, then the vertex angle of the cone is given by, $$\\alpha = \\displaystyle \\small \\arctan \\big(\\frac{a}{r-h}\\big)$$ and the spherical sector is defined by, $$\\displaystyle 0 \\leq \\rho \\leq r, \\small \\phi_a - \\frac{\\alpha}{2} \\leq \\phi \\leq \\phi_a + \\frac{\\alpha}{2}, \\theta_a - \\frac{\\alpha}{2} \\leq \\theta \\leq \\theta_a + \\frac{\\alpha}{2} \\$$ where $$\\phi_a, \\theta_a$$ define the axis of the spherical sector. Now for a given point $$\\displaystyle \\small P(x_p, y_p, z_p),$$ $$\\displaystyle \\small \\rho_p = \\sqrt{x_p^2 + y_p^2 + z_p^2}, \\ \\phi_p = \\arccos \\big(\\frac{z_p}{\\rho_p}\\big), \\ \\theta_p = \\arccos \\big(\\frac{x_p}{\\rho_p \\sin \\phi_p} \\big)$$ So for the point $$P$$ to be in the spherical sector, $$\\displaystyle \\small 0 \\leq \\rho_p \\leq r, \\ \\phi_a - \\frac{\\alpha}{2} \\leq \\phi_p \\leq \\phi_a + \\frac{\\alpha}{2}, \\theta_a - \\frac{\\alpha}{2} \\leq \\theta_p \\leq \\theta_a + \\frac{\\alpha}{2}$$ • Thanks very much! I guess we can follow the same logic to also calculate a circular sector", "size(10cm); triple eye = (5,2,3); currentprojection=orthographic(eye); surface hemisphere = surface(Arc(X,-X,c=O,normal=Z,n=16), c=O, axis=X, angle1=0, angle2=180); draw(shift(-5 eye)*hemisphere, material(white + opacity(0.5), emissivepen=0.2 white)); usepackage(\"amsmath\"); //for \\text command draw(O -- 1.6Z, arrow=Arrow3, L=Label(\"$H_{\\text{eff}}$\", align=W, position=EndPoint)); real theta(real t) { return t/20; } real phi(real t) { return -t + 1; } real r(real t) { return 1; } triple F(real t) { return polar(r(t), theta(t), phi(t)); } path3 spiral = reverse(graph(F, 0, 4pi, operator ..)); draw(spiral, blue + dotted + linewidth(1pt)); real t = 0.1; triple arrowpos = point(spiral, reltime(spiral, t)); t = 0.7; arrowpos = point(spiral, reltime(spiral, t)); triple M = point(spiral, 0); draw(O -- M, arrow=Arrow3, L=Label(\"$M$\", position=MidPoint)); draw(shift(M) * (O -- 0.5 cross(-M, Z)), red, arrow=Arrow3, L=Label(\"$-M \\times H_{\\text{eff}}$\", position=MidPoint)); triple dMdt = dir(spiral,0); triple crossprod = cross(M, dMdt); draw(shift(M) * (O -- 0.5 crossprod), blue, arrow=Arrow3, L=Label(\"$M \\times \\frac{dM}{dt}$\", position=MidPoint)); \\end{asypicture} \\end{document} The result: - Thanks! Exactly what I was looking for; sadly the LaTeX integration is not quite as smooth as with pgfplots. – knedlsepp Apr 30 '14 at 18:29 There are several packages for integrating Asymptote into LaTeX; if you state what features you would imagine for \"smooth integration,\" I can tell you whether any of these packages currently supports those features. – Charles Staats Apr 30 '14 at 22:55 I guess there's nothing one", "801 802 803 804 805 806 807 808 809 810 811 812 813 814 815 816 817 818 819 820 821 822 823 824 825 826 827 828 829 830 831 832 833 834 835 836 837 838 839 840 841 842 843 844 845 846 847 848 849 850 851 852 \\section{Cartesian coordinates and pseudorapidity} As above, take $\\rho = \\sqrt{x^2+y^2}$ and define the arc length $s=\\rho\\phi$. In terms of scaled coordinates $x_s=x/z$ and $y_s=y/z$, then equivalently $\\rho_s = \\rho/z$ and $s_s = (\\rho/z)\\phi = \\rho_s\\phi$. Note, $\\rho_s = \\tan\\theta$ and so $x_s = \\tan\\theta\\cos\\phi$ and $y_s = \\tan\\theta\\sin\\phi$. Generally $\\sinh\\eta = z/\\rho = 1/\\rho_s$, so $\\rho_s = 1/\\sinh\\eta$ and $s_s = \\phi/\\sinh\\eta$. To convert from $E$ to $E_T$ requires a factor of $\\sin\\theta$, which is given by \\sin\\theta = \\sin\\left(\\tan^{-1}\\rho_s\\right) = \\sin\\left[\\tan^{-1}\\left(\\frac{1}{\\sinh\\eta}\\right)\\right] % \\begin{figure}[ht!] \\begin{center} \\includegraphics[width=9cm]{SinThetaVsEta.pdf} %\\caption{ % Left: Active rotation by $30^\\circ$ % of an object in a fixed coordinate system. % Right: Passive rotation by $30^\\circ$ % of the coordinate system around a fixed object. %} \\label{fig:rotations} \\end{center} \\end{figure} Comparing areas in $\\Delta \\rho_s \\Delta s_s$ and $\\Delta \\eta \\Delta \\phi$ requires the Jacobian. From above \\frac{\\partial \\rho_s}{\\partial \\eta} = - \\frac{\\cosh\\eta}{\\sinh^2\\eta},\\qquad \\frac{\\partial \\rho_s}{\\partial \\phi} = 0 while \\frac{\\partial s_s}{\\partial \\eta} = - \\frac{\\phi\\cosh\\eta}{\\sinh^2\\eta},\\qquad \\frac{\\partial s_s}{\\partial \\phi} = \\frac{1}{\\sinh\\eta} Hence the Jacobian is J", "of $Y_{5,5}$ seem to be the logical choice. However, it turns out it cannot has the correct symmetry because $Y_{50}$ is not “even”. $\\displaystyle f_d(\\theta, \\phi) = Y_{60}(\\theta,\\phi) + \\sqrt{\\frac{28}{11}} Y_{65}(\\theta, \\phi)$ The Mathematica code for generating the graph is GraphicsGrid[{{ SphericalPlot3D[1 + f(θ,φ), {θ, 0, π}, {φ, 0, 2 π}], ParametricPlot3D[{Cos[φ] Sin[θ], Sin[φ] Sin[θ], Cos[θ]}, {θ, 0, π}, {φ, 0, 2 π}, ColorFunction -> Function[{x, y, z, u, v}, Hue[Abs[1 + f(u,v)]]], ColorFunctionScaling -> False, Mesh -> None, PlotPoints -> 200] }}, ImageSize -> 600] The code for calculating the coefficient is Solve[ SphericalHarmonicY[n, 0, k, 0] + a SphericalHarmonicY[n, n, k, 0] == SphericalHarmonicY[n, 0, 0, 0] , a] Where $n$ is the order of the spherical harmonic, $k$ is the angle for the next face or vertex for fixed $\\theta = 0$. Tetraherdon $k = 2 \\tan^{-1}(\\sqrt{2})$. Octaherdon $k = \\pi/2$. Dodecahedron $k = 2 \\tan^{-1}( (1+\\sqrt{5})/2)$.", "$(x,y)$ is equal to $\\cos 0 = 1$, and the $y$-coordinate is equal to $\\sin 0 = 0$. The tangent of $0^\\circ$ is equal to $\\tan 0 = 0$. So, the point has coordinates $(1,0)$. If the angle $\\theta$ that the line from $(0,0)$ to $(x,y)$ makes with the positive $x$-axis is $90^\\circ$, then the $x$-coordinate of point $(x,y)$ is equal to $\\cos 90 = 0$, and the $y$-coordinate is equal to $\\sin 90 = 1$. The tangent of $90^\\circ$ is undefined. So, the point has coordinates $(0,1)$. Note: We know that the coordinates of different points around the unit circle will have positive $x$-coordinates, or negative $x$-coordinates, and similarly for their $y$-coordinates. This lines up nicely with the sine and cosine values of different angles having positive and negative values. We'll talk more about this later. ## Pythagoras' Theorem, The Unit Circle and Trigonometric Identities The equation of the unit circle centred at the origin is $x^2 + y^2 = 1$. This corresponds to Pythagoras' theorem (the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides) for a right-angled triangle with side lengths $x$ and $y$, and hypotenuse $1$: \\begin{align*} x^2 + y^2 &= 1^2\\\\ x^2 + y^2 &= 1 \\end{align*} We also know that", "did was to follow this great tutorial, which has the desired syntax on the bottom of p. 60. The striking difference to you code is that there the normal of the 2D arrow gets specified, too, and that it is actually +Z (and not -Z). \\documentclass[border=3.14mm]{standalone} \\usepackage{asypictureB} \\begin{document} \\begin{asypicture}{name=arc} settings.outformat = \"png\"; settings.render = 16; size(6cm,0); import solids; currentlight=Headlamp; // inutile car Headlamp est la valeur par defaut nslice=4*nslice; revolution boule=sphere(O,1); draw(surface(boule),lightgrey+white+opacity(.5)); draw((0,0,1)--(0,0,-1),dashed); dot(\"$P1$\",(0,0,1),N); dot(\"$P2$\",(0,0,-1),2.2S); dot(\"$O$\",(0,0,0),SE); draw(O--(1,0,0),red); draw(O--(Cos(60),Sin(60),0),red); triple v1= 0.4*(1,0,0), v2 = 0.4*(Cos(60),Sin(60),0); //draw(arc(c=O,v1,v2),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); for (real alpha = 90; alpha<=450; alpha+= 30) { if (alpha==150) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red); if (alpha==330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red + dashed); if (alpha != 150 && alpha != 330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),black); }; draw(arc(c=O, v2, v1, normal=-Z), red, arrow = Arrow3(TeXHead2(normal=Z), emissive(red)),L=Label(\"$\\varphi$\",align=S, position=MidPoint)); \\end{asypicture} \\end{document} • Tank you Marmot for your anwer – user273366 Nov 24 '18 at 7:45 • indeed, the normal parameter was missing in the arrow function – user273366 Nov 24 '18 at 7:47", "# I can not draw arc in a clockwise direction with this program in asymptote I have to trace the meridians. settings.outformat = \"png\"; settings.render = 16; size(6cm,0); import solids; currentlight=Headlamp; // inutile car Headlamp est la valeur par défaut nslice=4*nslice; revolution boule=sphere(O,1); draw(surface(boule),lightgrey+white+opacity(.5)); draw((0,0,1)--(0,0,-1),dashed); dot(\"$P1$\",(0,0,1),N); dot(\"$P2$\",(0,0,-1),2.2S); dot(\"$O$\",(0,0,0),SE); draw(O--(1,0,0),red); draw(O--(Cos(60),Sin(60),0),red); triple v1= 0.4(1,0,0), v2 = 0.4(Cos(60),Sin(60),0); draw(arc(c=O,v1,v2,normal=Z),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); for (real alpha = 90; alpha<=450; alpha+= 30) { if (alpha==150) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red); if (alpha==330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),red + dashed); if (alpha != 150 && alpha != 330) draw(arc((0,0,0),(0,0,1), (0,0,-1), (Cos(alpha),Sin(alpha),0)),black); } But I can not draw arc in a clockwise direction with this code (this is the critic point!): draw(arc(c=O,v1,v2,normal=Z),red,arrow=Arrow3(DefaultHead2,emissive(red)),L=Label(\"$\\varphi$\",align=W, position=MidPoint)); I tried: 1. to change v1 and v2 order with the same normal (Z), but in this way i have the angle greater than ninety degrees (i need the explementar angle) 2. to change v1 and v2 order and take normal = - Z, but in this way asymptote runs on a infinite loop. I don't know why. ## 1 Answer It took me a while to understand the question since IMHO it is not very clearly written. An additional complication arises from the fact that you did not present an MWE and seem to insist to have a special character in a comment: \"défaut\". All I", "• Above is a start putting them together. Jun 22 '16 at 17:14 Here's a starting point. Figuring out what I exactly did is left as an exercise. With[{r = 5/2, φ = π/4, s1 = 3/4, s2 = 3/2}, Graphics3D[{{Directive[CapForm[None], Opacity[2/3]], Tube[{{0, 0, -r Cot[φ/2]}, {0, 0, 0}, {0, 0, r Cot[φ/2]}}, {r, 0, r}]}, {Sphere[{0, 0, s1 Csc[φ/2]}, s1], Sphere[{0, 0, -s2 Csc[φ/2]}, s2]}, {Directive[EdgeForm[], Opacity[1/2, LightBlue]], InfinitePlane[#1, {#2 - #1, {0, 1, 0}}] &[ {s1/(s1 + s2) Sqrt[(s1 + s2)^2 - (s1 - s2)^2 Sin[φ/2]^2], 0, s1 Csc[φ/2] - s1 (s1 - s2) Sin[φ/2]/(s1 + s2)}, {s2/(s1 + s2) Sqrt[(s1 + s2)^2 - (s1 - s2)^2 Sin[φ/2]^2], 0, s2 (s2 - s1) Sin[φ/2]/(s1 + s2) - s2 Csc[φ/2]}]}}, Boxed -> False]] • Here are the corresponding pictures for the other conics: ellipse, parabola. The old code by Stan Wagon was rather long, but thanks to primitives like Cone[], Sphere[], and InfinitePlane[] being built-in, things are now easier to do. Jun 22 '16 at 6:27 • We need to show (with a small disk or tiny sphere) points of 1) tangential contact , 2) arbitrary point on hyperbola , 3) parallel circle points joined at any arbitrary point so that the difference of segment lengths is seen constant for hyperbola. Jun 23 '16 at 15:49 •", "ax_text(0.25*np.cos(angle/2), 0.25*np.sin(angle/2), r\"$\\theta$\") arc = mpl.patches.Arc([0, 0], 2*0.2, 2*0.2, theta1=0, theta2=angle*180/np.pi, zorder=5, color='y', linewidth=3) ax_text(0.03, -0.05, \"0\", color='k') ax_text(1.03, -0.05, \"1\", color='k') ax.axhline(y=0, color='k', zorder=4) ax.axvline(x=0, color='k', zorder=4) ax.axvline(x=1, color='k', zorder=4, linewidth=1, linestyle='--') ax.axis('equal') ax.axis([-0.1, 1.1, -0.1, 1.1]) ax.axis('off') plt.show() The circle is the unit circle (radius=1). Assuming $0 < \\theta < \\dfrac{\\pi}{2}$, the area of the blue triangle (area $\\mathrm{A}$) is equal to its height ($\\sin(\\theta)$), times its base ($\\cos(\\theta)$), divided by 2. So $\\mathrm{A} = \\dfrac{1}{2}\\sin(\\theta)\\cos(\\theta)$. The unit circle has an area of $\\pi$, so the circular sector (in the shape of a pizza slice) has an area of A + B = $\\pi\\dfrac{\\theta}{2\\pi} = \\dfrac{\\theta}{2}$. Next, the large triangle (A + B + C) has an area equal to its height ($\\tan(\\theta)$) multiplied by its base (1) divided by 2, so A + B + C = $\\dfrac{\\tan(\\theta)}{2}$. When $0 < \\theta < \\dfrac{\\pi}{2}$, we have $\\mathrm{A} < \\mathrm{A} + \\mathrm{B} < \\mathrm{A} + \\mathrm{B} + \\mathrm{C}$, therefore: $\\dfrac{1}{2}\\sin(\\theta)\\cos(\\theta) < \\dfrac{\\theta}{2} < \\dfrac{\\tan(\\theta)}{2}$ We can multiply all the terms by 2 to get rid of the $\\dfrac{1}{2}$ factors. We can also divide by $\\sin(\\theta)$, which is stricly positive (assuming $0 < \\theta < \\dfrac{\\pi}{2}$), so the inequalities still hold: $cos(\\theta) < \\dfrac{\\theta}{\\sin(\\theta)} < \\dfrac{\\tan(\\theta)}{\\sin(\\theta)}$ Recall that $\\tan(\\theta)=\\dfrac{\\sin(\\theta)}{\\cos(\\theta)}$, so the last term simplifies like so: $cos(\\theta)", "radians ${{\\theta }_{d}}$= angle in degrees Segment of Circle Area of a segment of circle can be expressed as $A=1/2\\,\\,({{\\theta }_{r}}-\\sin {{\\theta }_{r}})\\,\\,{{r}^{2}}$ $=1/2\\,\\,(\\pi {{\\theta }_{d}}/180-sin{{\\theta }_{d}})\\,\\,{{r}^{2}}$ Right Circular Cylinder Lateral surface area of a right circular circle can be expressed as $A=2\\,\\pi \\,r\\,h$ where h = height of cylinder (m, ft) r = radius of base (m, ft) Right Circular Cone Lateral surface area of a right circular cone can be expressed as $A=\\pi \\,r\\,l$ $=\\pi r\\,\\,{{({{r}^{2}}+{{h}^{2}})}^{1/2}}$ where h = height of cone (m, ft) r = radius of base (m, ft) /= slant length (m, ft) Sphere Lateral surface area of a sphere can be expressed as $A=4\\,\\pi \\,{{r}^{2}}$ Where r = radius of base (m, ft) SYMBOLIC REPRESENTATION: Drawing Symbols provide a convenient way to draw geometrical shapes. To draw a geometrical shape, such as a pentagon or hexagon, select an appropriate symbol from the menu, specify the size of the symbol, and it is drawn at the indicated point. Example: Arrows (or pointers) in a drawing are commonly used to specify a direction for any reason. There are several arrow styles available in CADD programs. We can choose from simple two-point arrows to arrows passing through a number of points, and from simple to fency arrow styles. To draw an arrow, we need to", "# How to make dashed lines automatic of a sphere when I rotate a sphere? I copy this code from here settings.render=8; import solids; size(5cm); currentprojection=orthographic(1,0.3,0.3); revolution b=sphere(O,1); draw(surface(b),white+opacity(.5)); skeleton s; b.transverse(s,reltime(b.g,0.5),P=currentprojection); draw(s.transverse.back,dashed); draw(s.transverse.front); draw(rotate(90,X)*s.transverse.back,dashed); draw(rotate(90,X)*s.transverse.front); real rayon=0.8, phi=90, theta=45; triple pM=dir(phi,theta); draw((O--pM),dashed+green+0.7bp,Arrow3(size=7)); label(Label(\"$\\psi$\",yellow),pM,0.5dir(-90)); xaxis3(\"$x$\",1,1.5,Arrow3); xaxis3(-1.3,1,linetype(new real[] {8,8})); yaxis3(\"$y$\",1,1.5,Arrow3); yaxis3(-1.3,1,linetype(new real[] {8,8})); zaxis3(\"$z$\",1,1.5,Arrow3); zaxis3(0,1,linetype(new real[] {8,8})); and try to compile at here http://asymptote.ualberta.ca/ I rotate output by hand, the result seems incorrect. How can I get the result like this? • If you mean interfacing with a mouse, you can't. Changing the image would require recompiling the code. You need to do the whole thing using Javascript. Oct 16 '21 at 17:21 As John Kormylo'comment, I think, you can not rotate the sphere with a mouse (if you use TikZ). We hope that, in future, Asymptote can do this (keep automatic dashed lines when we rotate sphere!). You can try this code. I hope, it helps you something. \\documentclass[tikz,border=3mm]{standalone} \\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools \\begin{document} \\pgfdeclarelayer{background} \\pgfdeclarelayer{foreground} \\pgfsetlayers{background,main,foreground} \\foreach \\Angle in {5,15,...,360} {\\begin{tikzpicture}[3d/install view={phi=\\Angle,theta=70},line cap=butt, line join=round,visible/.style={draw,solid}, hidden/.style={draw,very thin,cheating dash},declare function={R=5;}] \\path (0,0,0) coordinate (I) foreach \\Z in {-75,-60,...,75} {(0,0,{R*sin(\\Z)}) coordinate (I\\Z)} foreach \\Z in {-75,-60,...,90} {({cos(\\Z)},{sin(\\Z)},0) coordinate (n\\Z)};", "if we anticipate singularity or some precision issue. If such situation arises, be extra careful or consider different approach. precisionxy Double. Set this to a very small number, and it is defaulted to 1e-15. return r, phi, theta Double. Position in spherical coordinates. import kero.werithmetic.rotation as rot x, y, z = 0.1, 0.1, 10 ro = rot.Rotation3D() r, phi, theta = ro.get_spherical_coords(x, y, z) print(\"r, phi, theta =\", r, phi, theta) The output is the following, where r is in the same unit as x, y and z, while phi (polar angle), theta (azimuthal angle) are in radian. r, phi, theta = 10.000999950005 0.7853981633974483 0.014141192927807654 Example 2: rotate_towards_z_axis() state Numpy matrix, 3 by 1. This is to represent an initial 3D position vector that we want to rotate through, first, azimuth angle, specified by the negative of current_phi, and second, polar angle, specified by the negative of current_theta. current_phi Double. Current azimuth angle of the vector. current_theta Double. Current polar angle of the vector. return out Numpy matrix, 3 by 1. This is the rotated vector, see the description of state. import kero.werithmetic.rotation as rot import numpy as np ro = rot.Rotation3D() x, y, z = -0.1, 0.54, 2 state = np.matrix([[x], [y], [z]]) r, phi, theta = ro.get_spherical_coords(x, y, z) print(\"BEFORE: r, phi, theta =\", r, phi,", "23-56 }$$ $$K_3 = HW\\cos\\phi_1+K_7x_0 \\tag{ 23-57 }$$ $$K_4 = HWP\\sin(\\phi_1-\\phi_g)+x_0 \\tag{ 23-58 }$$ $$K_8 = H(M\\sin\\theta-N\\cos\\theta)+K_5y_0 \\tag{ 23-59 }$$ $$K_9 = H(V\\sin\\theta-F\\cos\\theta)+K_6y_0 \\tag{ 23-60 }$$ $$K_{10} = HT\\cos\\phi_1+K_7y_0 \\tag{ 23-61 }$$ $$K_{11} = HTP\\sin(\\phi_1-\\phi_g)+y_0 \\tag{ 23-62 }$$ where, to review the meanings of previously defined symbols, $(\\phi_1, \\lambda_0) =$ latitude and longitude of the projection center and origin $\\phi_g =$ geocentric latitude of the point of perspective, found from equation (23-17) $\\gamma=$ azimuth east of north of the Yt axis, or the most upward-tilted axis of the plane of projection $\\omega =$ upward angle of tilt $P =$ distance from the center of the Earth to the point of perspective, divided by $a$, the semimajor axis. New symbols are as follows: $\\theta =$ clockwise angle through which the (Xt, Yt) axes are rotated for the arbitrary axes (X’t, Y’t) used for the constants $K_1\\unicode{x2013}K_{11}$. This may be made zero to retain the (Xt, Yt) axes. $(x_0, y_0) =$ offsets of the (Xt, Yt) axes to establish a different origin for the (X’t, Y’t) axes. They may also be set at zero to retain the (Xt, Yt) axes. The two sets of axes are related as follows: $$x'_t = x_t\\cos\\theta-y_t\\sin\\theta+x_0 \\tag{ 23-62a }$$ $$y'_t = y_t\\cos\\theta+x_t\\sin\\theta+y_0 \\tag{ 23-62b }$$ For inverse computations using projective constants given $K_1\\unicode{x2013}K_{11}, x’_t$", "Arrow[{{0, 0}, p1}], Locator[Dynamic[ p1, (p1 = #; {x1 = p1[[1]], y1 = p1[[2]], p2[[1]] = p1[[1]] + x2, p2[[2]] = p1[[2]] + y2}; #) &]]}]; angleArrow1 = Graphics[Arrow[ Table[Min[0.5 Sqrt[p1.p1], 0.5] {Cos[j], Sin[j]}, {j, 0, theta1, 1/100}]]]; (*Stuff for vector 2*) theta2 = Which[y2 >= 0, ArcCos[x2/Sqrt[x2^2 + y2^2]], y2 < 0, 2 Pi - ArcCos[x2/Sqrt[x2^2 + y2^2]]]; vector2 = Graphics[{Green, Arrow[{p1, p2}], Locator[Dynamic[ p2, (p2 = #; {x2 = p2[[1]] - p1[[1]], y2 = p2[[2]] - p1[[2]]}; #) &]]}]; (*angleArrow2=Graphics[Arrow[{{x1+Min[0.5 Sqrt[x2^2+y2^2],0.5] Cos[0],y1 +Min[0.5 Sqrt[x2^2+y2^2],0.5] Sin[0]},{x1+Min[0.5 Sqrt[x2^2+y2^2],0.5] Cos[theta2/2],y1 +Min[0.5 Sqrt[x2^2+y2^2],0.5] Sin[theta2/2]},{x1+Min[0.5 Sqrt[x2^2+y2^2],0.5] Cos[theta2],y1 +Min[0.5 Sqrt[x2^2+y2^2],0.5] Sin[theta2]}}]];*) angleArrow2 = Graphics[Arrow[ Table[{x1 + Min[0.5 Sqrt[x2^2 + y2^2], 0.5] Cos[j], y1 + Min[0.5 Sqrt[x2^2 + y2^2], 0.5] Sin[j]}, {j, 0, theta2, 1/100}]]]; (*Combine All The Graphics*) Show[{vector1, vector2}, PlotRange -> {{-4.1, 4.1}, {-4.1, 4.1}}, ImageSize -> 500], (*Controls*) Row[{\"Ax\", Manipulator[ Dynamic[x1, (x1 = #; {p1[[1]] = x1, p2[[1]] = x1 + x2}; #) &], {-4, 4}], Spacer[4], Dynamic[Style[Round[x1, 0.001], 14]]}], Row[{\"Ay\", Manipulator[ Dynamic[y1, (y1 = #; {p1[[2]] = y1, p2[[2]] = y1 + y2}; #) &], {-4, 4}], Spacer[4], Dynamic[Style[Round[y1, 0.001], 14]]}], Row[{\"Bx\", Manipulator[ Dynamic[x2, (x2 = #; p2[[1]] = x1 + x2; #) &], {-4, 4}], Spacer[4], Dynamic[Style[Round[x2, 0.001], 14]]}], Row[{\"By\", Manipulator[ Dynamic[y2, (y2 = #; p2[[2]] = y1 + y2; #) &], {-4, 4}], Spacer[4], Dynamic[Style[Round[y2, 0.001], 14]]}],", "#### 8.1.7 EllipticArc • EllipticArc( <B>, <A>, <C>, <Rx>, <Ry> [, direction] ). • Description: draw an elliptic arc with $Ox$ and $Oy$ axes and center <A>, $Ox$ radius is <Rx>, $Oy$ radius is <Ry>. The arc is drawn from the line $\\left(AB\\right)$ until the line $\\left(AC\\right)$. If the optional argument <direction> is $1$ (default value) then the arc is drawn counterclockwise, and clockwise if its vaue is $-1$. \\begin{texgraph}[name=EllipticArc,export=pgf] view(-2.25,3.75,-2,5),Marges(0,0,0,0),size(7.5), A:=0, B:=3+i, C:=2+4*i, DotScale:=2, Width:=8, Ligne([B,A,C],0), Color:=red, LabelDot(A,\"$A$\",\"S\",1), LabelDot(B,\"$B$\",\"N\",1), LabelDot(C,\"$C$\",\"SE\",1), Arrows:=1, Color:=blue, EllipticArc(B,A,C,2,1,-1), EllipticArc(B,A,C,2,3,1) \\end{texgraph} EllipticArc Command Note: for a circle arc, using equals values for <Rx> and <Ry> is a solution, but it is easier to use the macro Arc that is replacing the Arc arc command of the old version.", "$$x'$$. In fig 89, two objects are used (left figure) and right a sinogram calculated. The image repeats itself after $$360^\\text{o}$$ but is continued in the figure to show this more clearly. In practice only $$0\\to 180$$ need be calculated as the other angles are found by symmetry. The two objects are centred at $$(90,100)$$ and $$(200,150)$$ and when $$\\theta =0$$ the $$x'$$ values are the same as the $$x$$ coordinate for each object as shown by the two small black dots. When the two sinusoids cross the angle is such that the objects are in line as in the right-hand image of figure 86. In this situation the $$x'$$ axis is at right angles to the line between the two objects, and in this example, $$\\theta \\approx 114^\\text{o}$$ from the vertical and is shown by the red dots. The maximum $$x'$$ happens when the angle places the detector surface parallel to the line between the two objects after accounting for their size and is at $$\\approx 30^\\text{o}$$. Figure 89. Example of a sinogram generated by two small objects on a $$256 \\times 256$$ grid. Angle $$\\theta$$ is defined in fig 87 vs. $$x'$$, the distance along the detector surface. 14.4 Basic and filtered back projection.# Once the sinogram is measured reconstruction is possible. The back projection method", "draw( arc(O,Aq,expi(pi/2,f)),thinblack ); draw( arc(O,Aqf,expi(pi/2,f+df) ),thinblack); //theta arcs draw(O--A,thinblack); draw(O--Aq,thinblack); draw(\"$\\theta$\", arc(O,0.25*length(A)*Z,0.25*A),thinblack,Arrow3); draw(L=Label(\"$d\\theta$\",position=Relative(0.5),align=NE) ,arc(O,0.66*A,0.66*Aq),thinblack ); // inner surface triple rin(pair t) { return r*expi(t.x,t.y);} surface inner=surface(rin,(q,f),(q+dq,f+df),16,16); draw(inner,emissive(gray+opacity(0.33))); //part of a nearly transparent sphere to help see perspective surface sphere=surface(rin,(0,0),(pi/2,pi/2),16,16); draw(sphere,emissive(gray+opacity(0.125))); // dr and rdtheta labels triple V= Af + 0.5*(Arf-Af); draw(L=Label(\"$dr$\",position=Relative(1.1)), V--(1.5*V.x,1.5*V.y,V.z),dotted); triple U=expi(q+0.5*dq,f); draw(L=Label(\"$rd\\theta$\",position=Relative(1.1)), r*U ---r*(1.66*U.x,1.66*U.y,U.z),dotted ); \\end{asy} \\end{document} • This is beautiful! Nevertheless, I can't resist a couple of points. The dashed line below dtheta should be solid, and there should be another line (solid, from this point of view) to the fourth corner of the surface. (The original diagram gets away with omitting this because that line would be blocked from view; but from your point of view, it's not.) – Charles Staats Feb 15 '14 at 0:24 • One of the great things about this answer is that it's easy to experiment with changing the parameters and the point of view. The one exception is the inner surface, which will work better from many points of view if you replace gray+opacity(0.25) by emissive(gray+opacity(0.25)). – Charles Staats Feb 15 '14 at 0:28 • One final point, for those interesting in playing around with this excellent diagram: the projection used in the original diagram is called obliqueX in Asymptote. – Charles Staats Feb 15 '14 at" ]

[ "box are available. • SurfaceHex(N, B, L, orient) • this operator allows to build the surface mesh of a 3d box • int[int] N=[nx,ny,nz]; // the number of seg in the 3 direction • real [int,int] B=[[xmin,xmax],[ymin,ymax],[zmin,zmax]]; // bounding bax • int [int,int] L=[[1,2],[3,4],[5,6]]; // the label of the 6 face left,right, front, back, down, right • orient the global orientation of the surface 1 extern (-1 intern), • returns a meshS type • Ellipsoide (RX, RY, RZ, h, L, OX, OY, OZ, orient) • h is the mesh size • L is the label • orient the global orientation of the surface 1 extern (-1 intern) • OX, OY, OZ are real numbers to give the Ellipsoide center ( optinal, by default is (0,0,0) ) • where RX, RY, RZ are real numbers such as the parametric equations of the ellipsoid is: • returns a meshS type $\\forall u \\in [- \\frac{\\pi}{2},\\frac{\\pi}{2} [ \\text{ and } v \\in [0, 2 \\pi], \\vectthree{x=\\text{Rx } cos(u)cos(v) + \\text{Ox }}{y=\\text{Ry } cos(u)sin(v) + \\text{Oy }}{z = \\text{Rz } sin(v) + \\text{Oz } }$ • Sphere(R, h, L, OX, OY, OZ, orient) • where R is the raduis of the sphere, • OX, OY, OZ are real numbers to give the Ellipsoide center ( optinal, by default is (0,0,0) ) •", "if we anticipate singularity or some precision issue. If such situation arises, be extra careful or consider different approach. precisionxy Double. Set this to a very small number, and it is defaulted to 1e-15. return r, phi, theta Double. Position in spherical coordinates. import kero.werithmetic.rotation as rot x, y, z = 0.1, 0.1, 10 ro = rot.Rotation3D() r, phi, theta = ro.get_spherical_coords(x, y, z) print(\"r, phi, theta =\", r, phi, theta) The output is the following, where r is in the same unit as x, y and z, while phi (polar angle), theta (azimuthal angle) are in radian. r, phi, theta = 10.000999950005 0.7853981633974483 0.014141192927807654 Example 2: rotate_towards_z_axis() state Numpy matrix, 3 by 1. This is to represent an initial 3D position vector that we want to rotate through, first, azimuth angle, specified by the negative of current_phi, and second, polar angle, specified by the negative of current_theta. current_phi Double. Current azimuth angle of the vector. current_theta Double. Current polar angle of the vector. return out Numpy matrix, 3 by 1. This is the rotated vector, see the description of state. import kero.werithmetic.rotation as rot import numpy as np ro = rot.Rotation3D() x, y, z = -0.1, 0.54, 2 state = np.matrix([[x], [y], [z]]) r, phi, theta = ro.get_spherical_coords(x, y, z) print(\"BEFORE: r, phi, theta =\", r, phi,", "find its center and radius. Proof Let's introduce a coordinate system with the center at the point of intersection of the axes. Let the first (red) parabola have axis $x = 0,$ focal parameter $p_1$ and focus at point $A(0, –y_1), y_1 > 0.$ Let second (blue) parabola have axis $y = 0,$ focal parameter $p_2$ and focus at point $B(–x_2,0), x_2 > 0.$ Let us denote the angle between the vector connecting the focus of the first parabola and its point and the positive direction of the ordinate axis $2\\theta,$ its length $\\rho_1(\\theta),$ the angle between the vector connecting the focus of the second parabola and its point and the positive direction of the abscissa axis $2\\phi,$ its length $\\rho_2(\\phi).$ Then \\begin{align*} \\rho_1(\\theta) = \\frac{p_1}{1 - \\cos(2\\theta)},\\end{align*} \\begin{align*} \\rho_2(\\phi) = \\frac{p_2}{1 - \\cos(2\\phi)}.\\end{align*} Abscissa of the point of intersection is \\begin{align*} x =\\rho_1 \\sin(2\\theta) = p_1\\cot\\theta = \\rho_2 \\cos (2\\phi) - x_2 = \\frac{p_2}{2} (\\cot^2\\phi - 1)- x_2,\\end{align*} \\begin{align*} x^2 = p_1^2 \\cot ^2 \\theta , 2 p_1\\cot\\theta = p_2 \\cos^2 \\phi - p_2 - 2x_2 .\\end{align*} Ordinate of the point of intersection is \\begin{align*} y =\\rho_2 \\sin 2\\phi = p_2\\cot\\phi = \\rho_1 \\cos 2\\theta - y_1 = \\frac{p_1}{2} (\\cot^2\\theta - 1)- y_1,\\end{align*} \\begin{align*} y^2 = p_2^2 \\cot ^2 \\phi , 2 p_2\\cot\\phi = p_1 \\cos^2 \\theta", "801 802 803 804 805 806 807 808 809 810 811 812 813 814 815 816 817 818 819 820 821 822 823 824 825 826 827 828 829 830 831 832 833 834 835 836 837 838 839 840 841 842 843 844 845 846 847 848 849 850 851 852 \\section{Cartesian coordinates and pseudorapidity} As above, take $\\rho = \\sqrt{x^2+y^2}$ and define the arc length $s=\\rho\\phi$. In terms of scaled coordinates $x_s=x/z$ and $y_s=y/z$, then equivalently $\\rho_s = \\rho/z$ and $s_s = (\\rho/z)\\phi = \\rho_s\\phi$. Note, $\\rho_s = \\tan\\theta$ and so $x_s = \\tan\\theta\\cos\\phi$ and $y_s = \\tan\\theta\\sin\\phi$. Generally $\\sinh\\eta = z/\\rho = 1/\\rho_s$, so $\\rho_s = 1/\\sinh\\eta$ and $s_s = \\phi/\\sinh\\eta$. To convert from $E$ to $E_T$ requires a factor of $\\sin\\theta$, which is given by \\sin\\theta = \\sin\\left(\\tan^{-1}\\rho_s\\right) = \\sin\\left[\\tan^{-1}\\left(\\frac{1}{\\sinh\\eta}\\right)\\right] % \\begin{figure}[ht!] \\begin{center} \\includegraphics[width=9cm]{SinThetaVsEta.pdf} %\\caption{ % Left: Active rotation by $30^\\circ$ % of an object in a fixed coordinate system. % Right: Passive rotation by $30^\\circ$ % of the coordinate system around a fixed object. %} \\label{fig:rotations} \\end{center} \\end{figure} Comparing areas in $\\Delta \\rho_s \\Delta s_s$ and $\\Delta \\eta \\Delta \\phi$ requires the Jacobian. From above \\frac{\\partial \\rho_s}{\\partial \\eta} = - \\frac{\\cosh\\eta}{\\sinh^2\\eta},\\qquad \\frac{\\partial \\rho_s}{\\partial \\phi} = 0 while \\frac{\\partial s_s}{\\partial \\eta} = - \\frac{\\phi\\cosh\\eta}{\\sinh^2\\eta},\\qquad \\frac{\\partial s_s}{\\partial \\phi} = \\frac{1}{\\sinh\\eta} Hence the Jacobian is J", "Coordinates 5. Since $\\displaystyle {\\rho}^{2}=x^{2}+y^{2}+z^{2}$ We can sub it into the density function and get $\\displaystyle \\sqrt{1+{\\rho}^{3}}$ 6. I am just a bit confused on how you find the \"phi\" and \"theta\" for this question. The bound is suppose to be 0 to Pi for phi and theta. By the way, that website on spherical coordinates was useful 7. I think your set up should look something like this: $\\displaystyle \\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{2}\\left[{\\rho}^{2}\\sqrt{1+{\\rho}^{3}}sin{\\phi}\\right]d{\\rho}d{\\phi}d{\\theta}$ 8. Originally Posted by galactus I think your set up should look something like this: $\\displaystyle \\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{2}\\left[{\\rho}^{2}\\sqrt{1+{\\rho}^{3}}sin{\\phi}\\right]d{\\rho}d{\\phi}d{\\theta}$ Yep that's it. Is theta 0 to Pi because its half a sphere? Also, why is the phi 0 to Pi also, I can't really picture it. EDIT: Ok, I can see why its Pi now, I am confused from the question. I can't really see what [Math] x^2+y^2+z^2<4 [/tex] and $\\displaystyle 0<y$ is. What do these regions mean? 9. Originally Posted by galactus I think your set up should look something like this: $\\displaystyle \\int_{0}^{\\pi}\\int_{0}^{\\pi}\\int_{0}^{2}\\left[{\\rho}^{2}\\sqrt{1+{\\rho}^{3}}sin{\\phi}\\right]d{\\rho}d{\\phi}d{\\theta}$ Ahh yes. I misread my x- and y-axes. $\\displaystyle 0 \\leq \\theta \\pi$ indeed. 10. Originally Posted by Gotovina7 Yep that's it. Is theta 0 to Pi because its half a sphere? Also, why is the phi 0 to Pi also, I can't really picture it. EDIT: Ok, I can see why its Pi", "coordinates to rectangular coordinates. #### Solution Use the formulas, noting that $$\\rho=5$$, $$\\theta=\\pi$$, and $$\\phi=\\dfrac{\\pi}{2}$$. \\begin{align} x &= \\rho \\cos \\theta \\sin \\phi = 5 (\\cos \\pi) \\left(\\sin \\dfrac{\\pi}{2}\\right) = 5(-1)(1) = -5\\\\ y &= \\rho \\sin \\theta \\sin \\phi = 5 (\\sin \\pi) \\left(\\sin \\dfrac{\\pi}{2}\\right) = 5(0)(1) = 0\\\\ z &= \\rho \\cos \\phi = 5 \\cos \\dfrac{\\pi}{2} = 5(0) = 0 \\end{align} Therefore, this point is $\\ans{(-5,0,0)}$ in rectangular coordinates. Convert $$(3,4,7)$$ in rectangular coordinates to spherical coordinates. #### Solution Use the formula for $$\\rho$$ first. \\begin{align} \\rho &= \\sqrt{x^2+y^2+z^2} = \\sqrt{74} \\end{align} Then, find $$\\phi$$ by noting that $$\\cos \\phi = \\dfrac{z}{\\rho}$$. $\\phi = \\cos^{-1} \\dfrac{z}{\\rho} = \\cos^{-1} \\dfrac{7}{\\sqrt{74}} \\approx 0.62 \\textrm{ radians}$ Finally, use the fact that $$\\cos \\theta = \\dfrac{x}{\\rho \\sin \\phi}$$ to find $$\\theta$$. $\\theta = \\cos^{-1} \\dfrac{x}{\\rho \\sin \\phi} = \\cos^{-1} \\dfrac{3}{\\sqrt{74}\\sin 0.62} \\approx 0.93 \\textrm{ radians}$ Therefore, this point is $\\ans{\\left(\\sqrt{74},0.93,0.62\\right)}$ in spherical coordinates. ## Summary Rectangular Coordinates Cylindrical Coordinates Spherical Coordinates $$(x,y,z)$$ $$(r,\\theta,z)$$ $$(\\rho,\\theta,\\phi)$$ The formulas below are used to convert between coordinate systems. For my class, there's no need to memorize these formulas, since you'll have a formula sheet on the exams. ### 1. Rectangular $$\\leftrightarrow$$ Cylindrical \\begin{align} x &= r \\cos \\theta\\\\ y &= r \\sin \\theta\\\\ z &= z\\\\ \\\\ r &= \\sqrt{x^2+y^2}\\\\ \\theta &= \\tan^{-1}", "(I hope you don't mean the coordinate system's $z$ axis) Mar 31, 2021 at 14:29 ## 3 Answers In spherical coordinates, $$x = \\rho \\cos \\theta \\sin \\phi, y = \\rho \\sin \\theta \\sin \\phi, z = \\rho \\cos\\phi, \\rho = \\sqrt{x^2+y^2+z^2}$$ If the radius of the sphere is $$r$$ with origin as the center, height of spherical cap is $$h$$ and radius of the base of the spherical cap is $$a$$, then the vertex angle of the cone is given by, $$\\alpha = \\displaystyle \\small \\arctan \\big(\\frac{a}{r-h}\\big)$$ and the spherical sector is defined by, $$\\displaystyle 0 \\leq \\rho \\leq r, \\small \\phi_a - \\frac{\\alpha}{2} \\leq \\phi \\leq \\phi_a + \\frac{\\alpha}{2}, \\theta_a - \\frac{\\alpha}{2} \\leq \\theta \\leq \\theta_a + \\frac{\\alpha}{2} \\$$ where $$\\phi_a, \\theta_a$$ define the axis of the spherical sector. Now for a given point $$\\displaystyle \\small P(x_p, y_p, z_p),$$ $$\\displaystyle \\small \\rho_p = \\sqrt{x_p^2 + y_p^2 + z_p^2}, \\ \\phi_p = \\arccos \\big(\\frac{z_p}{\\rho_p}\\big), \\ \\theta_p = \\arccos \\big(\\frac{x_p}{\\rho_p \\sin \\phi_p} \\big)$$ So for the point $$P$$ to be in the spherical sector, $$\\displaystyle \\small 0 \\leq \\rho_p \\leq r, \\ \\phi_a - \\frac{\\alpha}{2} \\leq \\phi_p \\leq \\phi_a + \\frac{\\alpha}{2}, \\theta_a - \\frac{\\alpha}{2} \\leq \\theta_p \\leq \\theta_a + \\frac{\\alpha}{2}$$ • Thanks very much! I guess we can follow the same logic to also calculate a circular sector", "Sin[ϕ] Sin[θ], Cos[θ]}, #] & /@ {θ, ϕ}]; SphericalPlot3D[f[theta, phi], {theta, 0, Pi}, {phi, 0, 2 Pi}, NormalsFunction -> (n[#4, #5] &), ColorFunction -> Function[{x, y, z, theta, phi, r}, ColorData[\"Rainbow\"][r]], PlotPoints -> {32, 24}, BoundaryStyle -> None, PlotRange -> All, Boxed -> False, Axes -> False, SphericalRegion -> True, Method -> {\"RotationControl\" -> \"Globe\"}, ImageSize -> {500, 500}] The resulting plot looks like the original one, but has no defective line on my system (Mac OS X, Mathematica version 10.3.1). The only change I made was the option NormalsFunction -> (n[#4, #5] &), no change to the number of plot points was made. To construct the normals, I defined your plot function as f, and then calculated its surface normal from the spherical-coordinate parametrization using the Cross product of the tangent vectors. The latter are obtained by differentiation with respect to the angles. This is done in the function n, which returns the normal vector. • It appears to work ! Thanks a lot ! I'll study this interesting solution tomorrow, since now I'm too tired to understand anything :-( – Cham Mar 8 '16 at 4:18 • Very nice. +1 or more – Dr. belisarius Mar 8 '16 at 4:40", "2*Pi}]; r = N[{5, 6, 7}]; rSph = CoordinateTransform[\"Cartesian\" -> \"Spherical\", r]; h = 2; \\[Rho] = 1; J0 = 2; NIntegrate[(J[\\[Phi]]/ Norm[r - {\\[Rho]*Cos[\\[Phi]], \\[Rho]*Sin[\\[Phi]], h}])*\\[Rho], {\\[Phi], 0, 2*Pi}] Sum[(Subscript[Q, L, m]*SolidHarmonicIr[L, m, ##1] & ) @@ rSph, {L, 0, 5}, {m, -L, L}] Clear[h, \\[Rho], J0] {-0.0471743, 0.039312, 0.} {-0.0471859, 0.0393216, 0.} Alternately, as an additional simplification, note that $\\mathbf{J}$ can be expressed in spherical vector coordinates as $$\\mathbf{J}(\\phi)=\\left( \\begin{array}{cc} 0\\\\ 0\\\\ J_0 \\end{array} \\right)$$ giving (this time up to $L=4$): J[\\[Phi]_] := {0, 0, J0}; Grid[Simplify[ Table[MatrixForm[ Integrate[\\[Rho]*J[\\[Phi]]* SolidHarmonicRr[L, m, Sqrt[\\[Rho]^2 + h^2], ArcTan[h, \\[Rho]], \\[Phi]], {\\[Phi], 0, 2*Pi}]], {L, 0, 4}, {m, -L, L}], Assumptions -> {\\[Rho] > 0, Element[h, Reals]}], Frame -> All] although for this coordinate choice I'm a bit confused as to why the $L=0$ coefficient is not zero (will have to check that later). The magnetic field can then be obtained from the above coefficients $\\langle R_{Lm},\\mathbf{J}\\rangle$ by $$\\mathbf{B}(\\mathbf{r})=\\nabla\\times\\mathbf{A}(\\mathbf{r})=\\frac{\\mu_0}{4\\pi}\\sum_{L=0}^\\infty\\sum_{m=-L}^L \\nabla\\times I_{Lm}(\\mathbf{r})\\langle R_{Lm},\\mathbf{J}\\rangle.$$", "area is along the projection's equator, at the four points that are at the center of the four panels that intersect the equator. This will become more clear later. Our first step is to derive the link between the pixels in the equirectangular projection and the pixels in each panel. Consider an equirectangular projection with pixel dimensions $w_{e}$ and $h_{e}$. Each pixel in the projection has a $\\theta$ and $\\phi$ value in polar coordinates. The horizontal and vertical step sizes from one pixel to the next are: \\begin{align*} d\\theta &= \\frac{2\\pi}{w_{e}} \\\\ d\\phi &= \\frac{\\pi}{h_{e}} \\end{align*} The cartesian coordinates of a pixel in the equirectangular projection belonging to the front panel mapped to a sphere with radius 0.5 is: \\begin{align*} x_{s} &= -0.5 \\sin(\\phi) \\sin(\\theta) \\\\ y_{s} &= -0.5 \\cos(\\phi) \\\\ z_{s} &= -0.5 \\sin(\\phi) \\cos(\\theta) \\end{align*} If we project that point onto the face of a unit cube centered at the origin, we can calculate the coordinates of the new point using an (x,y) coordinate space of that face with an origin in the corner. \\begin{align*} x_{f} &= 0.5 - \\frac{x_{s}}{2z_{s}} \\\\ &= 0.5 - \\frac{\\sin(\\theta)}{2\\cos(\\theta)} \\\\ &= 0.5 - 0.5 \\tan(\\theta) \\\\ y_{f} &= 0.5 - \\frac{y_{s}}{2z_{s}} \\\\ &= 0.5 - \\frac{\\cos(\\phi)}{2\\sin(\\phi)\\cos(\\theta)} \\\\ &= 0.5 - 0.5 \\frac{\\cot(\\phi)}{\\cos(\\theta)} \\\\ \\end{align*} If the face of the unit", "\\lt \\infty,\\;0 \\le \\varphi \\lt 2\\pi.$ By definition, the $$\\theta-$$coordinate lies in the range $$0 \\le \\theta \\le \\pi.$$ If point M coincides with point O, then $$\\rho = 0.$$ For point O, the $$\\varphi-$$ and $$\\theta-$$coordinates are not defined. ## Converting Between Spherical and Cartesian Coordinates If you choose the axes of the Cartesian coordinate system as indicated above in the figure, then the Cartesian coordinates $$x, y, z$$ of a point are related to its spherical coordinates $${\\rho, \\varphi, \\theta}$$ by the relations $x = \\rho\\sin\\theta\\cos\\varphi,\\;\\;y = \\rho\\sin\\theta\\sin\\varphi,\\;\\;z = \\rho\\cos\\theta.$ For converting a point from Cartesian coordinates to spherical coordinates, we use the formulas $\\rho = \\sqrt{x^2 + y^2 + z^2},\\;\\;\\tan\\varphi = \\frac{y}{x},\\;\\;\\cos\\theta = \\frac{z}{\\rho} = \\frac{z}{\\sqrt{x^2 + y^2 + z^2}}.$ ## Solved Problems Click or tap a problem to see the solution. ### Example 1 Find the spherical coordinates of each point given its Cartesian coordinates. 1. $$A\\left({1,0,0}\\right)$$ 2. $$B\\left({-4,-3,0}\\right)$$ 3. $$C\\left({2,2,1}\\right)$$ 4. $$D\\left({8,-4,1}\\right)$$ ### Example 2 Convert from spherical to Cartesian coordinates. 1. $$K\\left({\\sqrt{2},\\frac{\\pi}{4},\\frac{\\pi}{6}}\\right)$$ 2. $$L\\left({3\\sqrt{2},\\frac{3\\pi}{4},\\frac{\\pi}{2}}\\right)$$ 3. $$M\\left({5,\\frac{3\\pi}{2},\\frac{\\pi}{2}}\\right)$$ ### Example 3 Express city's location in spherical coordinates given its latitude and longitude: 1. London $$51.5^{\\circ}\\,N, 0.1^{\\circ} W$$ 2. Rio de Janeiro $$22.9^{\\circ}\\,S, 43.2^{\\circ} W$$ ### Example 4 Find the spherical coordinates of point $$M$$ if the ray $$\\mathbf{OM}$$ forms the angles $$\\alpha = \\frac{\\pi}{4}$$", "\\lt \\infty,\\;0 \\le \\varphi \\lt 2\\pi.$ By definition, the $$\\theta-$$coordinate lies in the range $$0 \\le \\theta \\le \\pi.$$ If point M coincides with point O, then $$\\rho = 0.$$ For point O, the $$\\varphi-$$ and $$\\theta-$$coordinates are not defined. ## Converting Between Spherical and Cartesian Coordinates If you choose the axes of the Cartesian coordinate system as indicated above in the figure, then the Cartesian coordinates $$x, y, z$$ of a point are related to its spherical coordinates $${\\rho, \\varphi, \\theta}$$ by the relations $x = \\rho\\sin\\theta\\cos\\varphi,\\;\\;y = \\rho\\sin\\theta\\sin\\varphi,\\;\\;z = \\rho\\cos\\theta.$ For converting a point from Cartesian coordinates to spherical coordinates, we use the formulas $\\rho = \\sqrt{x^2 + y^2 + z^2},\\;\\;\\tan\\varphi = \\frac{y}{x},\\;\\;\\cos\\theta = \\frac{z}{\\rho} = \\frac{z}{\\sqrt{x^2 + y^2 + z^2}}.$ ## Solved Problems Click or tap a problem to see the solution. ### Example 1 Find the spherical coordinates of each point given its Cartesian coordinates. 1. $$A\\left({1,0,0}\\right)$$ 2. $$B\\left({-4,-3,0}\\right)$$ 3. $$C\\left({2,2,1}\\right)$$ 4. $$D\\left({8,-4,1}\\right)$$ ### Example 2 Convert from spherical to Cartesian coordinates. 1. $$K\\left({\\sqrt{2},\\frac{\\pi}{4},\\frac{\\pi}{6}}\\right)$$ 2. $$L\\left({3\\sqrt{2},\\frac{3\\pi}{4},\\frac{\\pi}{2}}\\right)$$ 3. $$M\\left({5,\\frac{3\\pi}{2},\\frac{\\pi}{2}}\\right)$$ ### Example 1. Find the spherical coordinates of each point given its Cartesian coordinates. 1. $$A\\left({1,0,0}\\right)$$ 2. $$B\\left({-4,-3,0}\\right)$$ 3. $$C\\left({2,2,1}\\right)$$ 4. $$D\\left({8,-4,1}\\right)$$ Solution. We use the following formulas: $\\rho = \\sqrt{x^2 + y^2 + z^2},\\;\\;\\tan\\varphi = \\frac{y}{x},\\;\\;\\cos\\theta = \\frac{z}{\\rho}.$ 1. $$A\\left({1,0,0}\\right)$$ $\\rho = \\sqrt{1^2 + 0^2", "size(10cm); triple eye = (5,2,3); currentprojection=orthographic(eye); surface hemisphere = surface(Arc(X,-X,c=O,normal=Z,n=16), c=O, axis=X, angle1=0, angle2=180); draw(shift(-5 eye)*hemisphere, material(white + opacity(0.5), emissivepen=0.2 white)); usepackage(\"amsmath\"); //for \\text command draw(O -- 1.6Z, arrow=Arrow3, L=Label(\"$H_{\\text{eff}}$\", align=W, position=EndPoint)); real theta(real t) { return t/20; } real phi(real t) { return -t + 1; } real r(real t) { return 1; } triple F(real t) { return polar(r(t), theta(t), phi(t)); } path3 spiral = reverse(graph(F, 0, 4pi, operator ..)); draw(spiral, blue + dotted + linewidth(1pt)); real t = 0.1; triple arrowpos = point(spiral, reltime(spiral, t)); t = 0.7; arrowpos = point(spiral, reltime(spiral, t)); triple M = point(spiral, 0); draw(O -- M, arrow=Arrow3, L=Label(\"$M$\", position=MidPoint)); draw(shift(M) * (O -- 0.5 cross(-M, Z)), red, arrow=Arrow3, L=Label(\"$-M \\times H_{\\text{eff}}$\", position=MidPoint)); triple dMdt = dir(spiral,0); triple crossprod = cross(M, dMdt); draw(shift(M) * (O -- 0.5 crossprod), blue, arrow=Arrow3, L=Label(\"$M \\times \\frac{dM}{dt}$\", position=MidPoint)); \\end{asypicture} \\end{document} The result: - Thanks! Exactly what I was looking for; sadly the LaTeX integration is not quite as smooth as with pgfplots. – knedlsepp Apr 30 '14 at 18:29 There are several packages for integrating Asymptote into LaTeX; if you state what features you would imagine for \"smooth integration,\" I can tell you whether any of these packages currently supports those features. – Charles Staats Apr 30 '14 at 22:55 I guess there's nothing one", "# Drawing intersecting 3D surface plots I'm attempting to draw six semi-transparent intersecting spheres of different colors---two on each axis x, y, z and all joining at the origin---as shown in the figure below. Note how all the spheres intersect. Seeing as this figure is created in Matlab, I tried converting the figure to pgfplots code, using matlab2tikz, but this did not work. Below is the Matlab code for the figure. Theta=linspace(0,2*pi,200); Phi=linspace(-pi/2,pi/2,200); [theta,phi]=meshgrid(Theta,Phi); rho1=cos(theta).*cos(phi); rho2=-cos(theta).*cos(phi); rho3=sin(theta).*cos(phi); rho4=-sin(theta).*cos(phi); rho5=sin(phi); rho6=-sin(phi); [x1,y1,z1]=sph2cart(theta,phi,rho1); [x2,y2,z2]=sph2cart(theta,phi,rho2); [x3,y3,z3]=sph2cart(theta,phi,rho3); [x4,y4,z4]=sph2cart(theta,phi,rho4); [x5,y5,z5]=sph2cart(theta,phi,rho5); [x6,y6,z6]=sph2cart(theta,phi,rho6); surf(x1,y1,z1,'FaceColor','red','EdgeColor','none') hold on surf(x2,y2,z2,'FaceColor','red','EdgeColor','none') surf(x3,y3,z3,'FaceColor','blue','EdgeColor','none') surf(x4,y4,z4,'FaceColor','blue','EdgeColor','none') surf(x5,y5,z5,'FaceColor','green','EdgeColor','none') surf(x6,y6,z6,'FaceColor','green','EdgeColor','none') alpha(0.6) camlight left lighting flat Next I tried drawing the figure in tikz-3dplot and pst-solides3d. With both of these methods, I had the problem that the surfaces didn't intersect. Instead, 2-dimensional projections of the spheres were drawn separately on the canvas, one on top of the other, as shown in the figures below. The first figure is the output from tikz-3dplot, and the second figure is the output from pst-solides3d. With tikz-3dplot, I also had an issue with certain axes appearing to be on top of the spheres, when they should rather appear to be inside them (ref. the negative half of the y axis). Below are the codes for the above figures, in the same order. \\documentclass[11pt]{standalone} \\usepackage[utf8]{inputenc} \\usepackage{tikz,tikz-3dplot} \\usetikzlibrary{arrows} \\begin{document} \\tdplotsetmaincoords{70}{135} \\begin{tikzpicture}[tdplot_main_coords,fill", "{ for r bigger than R.}$$ (5) $$\\vec A_{(\\vec r)} = \\frac {R \\rho_s \\mu_0 }{3} (\\vec{\\omega} \\;X\\; \\vec r) \\;\\hbox { for R bigger than r.}$$ (6) The above is in rectangular coordinates, the answer is consistance with the book so it is correct. The next step is where I get lost. The book want to derive the formulas for the point P in any position with $\\vec \\omega$ on z-axis. The book gave this as answer: $$\\vec A_{(r,\\theta, \\phi)} \\;=\\; \\hat {\\phi}\\; \\frac {R^4 \\rho_s \\mu_0 \\omega sin \\theta}{3r^2} \\;\\hbox { for r bigger than R.}$$ (7) $$\\vec A_{(r,\\theta, \\phi)} \\;=\\; \\hat {\\phi}\\; \\frac {R \\rho_s \\mu_0 \\omega r sin \\theta}{3} \\;\\hbox { for R bigger than r.}$$ (8) Below is what I tried to do and I just cannot get the answer of the book's example: From (5) & (6), the vector position is only contained in $(\\vec {\\omega} \\;X\\; \\vec r)$. So I just have to calculate this cross product in spherical coordinates with $\\vec {\\omega}$ on z-axis and $\\vec r_{(R,\\theta,\\phi)} = (r_R, r_{\\theta}, r_{\\phi})$. To convert $\\vec{\\omega} = \\hat z \\omega$ in rectangular coordinates to spherical coordinates, I use: $$\\hat R =\\hat x sin \\theta cos \\phi \\;+\\; \\hat y sin \\theta sin \\phi \\;+\\; \\hat z cos \\theta, \\;\\;\\;\\hat {\\theta} =\\hat x", "# 222 - Topic 18 - Cylindrical and Spherical Coordinates ## 1352 days ago by Professor222 John Travis Mississippi College Determine $\\int_{-3}^0 \\int_0^{\\sqrt{9-x^2}} \\int_3^{3+\\sqrt{9-x^2-y^2}} x dz dy dx$ # Just trying to graph the domain for the integration var('x,y,z,r,t') f = 3+sqrt(9-x^2-y^2) # this is the \"top\" # f = sqrt(16 - x^2 - y^2) # another option to play around with # Here is Cylindrical Coordinates # Notice, to get z we just take the function above and replace x with r*cost(t) and y with r*sin(t) # This is a horizontal plane of height 0. Adjust if you want to move this around G = parametric_plot3d([r*cos(t),r*sin(t),0], (r,0,3), (t,pi/2,pi),color=\"pink\", opacity=0.3) # This is a horizontal plane of height 3. Adjust if you want to move this around. G += parametric_plot3d([r*cos(t), r*sin(t), 3], (r, 0, 3), (t, pi/2, pi), color='green') # This uses the function f above to general a z-value. G += parametric_plot3d([r*cos(t), r*sin(t), f(x=r*cos(t),y=r*sin(t))], (r, 0, 3), (t, pi/2, pi)) G.show(aspect_ratio=True) integrate(x^2*sqrt(9-x^2),x,0,3).show()", "\\dfrac{64}{3} \\sin\\phi\\ d\\phi\\ d\\theta\\\\ &= \\int_0^{2\\pi} -\\dfrac{64}{3}\\left(\\dfrac{\\sqrt{3}}{2}-1\\right)\\ d\\theta\\\\ &= \\ans{\\dfrac{128\\pi}{3}-\\dfrac{64\\pi\\sqrt{3}}{3}} \\end{align} #### Try it yourself: (click on a problem to show/hide its answer) 1. Use spherical coordinates to evaluate $$\\displaystyle\\iiint_D (x^2+y^2+z^2)^{5/2}\\ dV$$, where $$D$$ is the unit ball. 2. $$\\displaystyle\\int_0^{2\\pi} \\displaystyle\\int_0^{\\pi} \\displaystyle\\int_0^{1} \\rho^5\\cdot\\rho^2\\sin\\phi\\ d\\rho\\ d\\phi\\ d\\theta = \\dfrac{\\pi}{2}$$ 3. Use spherical coordinates to evaluate $$\\displaystyle\\iiint_D \\dfrac{1}{(x^2+y^2+z^2)^{3/2}}\\ dV$$, where $$D$$ lies between spheres of radius 1 and 2 centered at the origin. 4. $$\\displaystyle\\int_0^{2\\pi} \\displaystyle\\int_0^{\\pi} \\displaystyle\\int_1^{2} \\rho^{-3}\\cdot\\rho^2\\sin\\phi\\ d\\rho\\ d\\phi\\ d\\theta = 4\\pi \\ln 2$$ 5. Use spherical coordinates to evaluate $$\\displaystyle\\int_{0}^{\\pi} \\displaystyle\\int_{0}^{\\pi/6} \\displaystyle\\int_{2\\sec \\phi}^{4} \\rho^2\\sin\\phi\\ d\\rho\\ d\\phi\\ d\\theta$$. 6. $$\\pi\\left(\\dfrac{188}{9}-\\dfrac{32\\sqrt{3}}{3}\\right)$$ 7. Use spherical coordinates to find the volume of the solid bounded by the sphere $$\\rho=2\\cos\\phi$$ and the hemisphere $$\\rho=1$$, $$z \\geq 0$$. 8. $$\\displaystyle\\int_0^{2\\pi} \\displaystyle\\int_0^{\\pi/3} \\displaystyle\\int_0^{1} \\rho^2\\sin\\phi\\ d\\rho\\ d\\phi\\ d\\theta = \\dfrac{5\\pi}{12}$$ 9. Use spherical coordinates to find the volume of the solid outside the cone $$\\phi=\\dfrac{\\pi}{4}$$ and inside the sphere $$\\rho=4 \\cos \\phi$$. 10. $$\\displaystyle\\int_0^{2\\pi} \\displaystyle\\int_{\\pi/4}^{\\pi/2} \\displaystyle\\int_0^{4\\cos\\phi} \\rho^2\\sin\\phi\\ d\\rho\\ d\\phi\\ d\\theta = \\dfrac{8\\pi}{3}$$ 11. Use spherical coordinates to find the volume of the part of the ball $$\\rho \\leq 4$$ that lies between the planes $$z=2$$ and $$z=2\\sqrt{3}$$. 12. $$\\displaystyle\\int_0^{2\\pi} \\displaystyle\\int_0^{\\pi/6} \\displaystyle\\int_0^{2\\sqrt{3}/\\cos\\phi} \\rho^2\\sin\\phi\\ d\\rho\\ d\\phi\\ d\\theta = \\dfrac{8\\pi}{3}(9\\sqrt{3}-8)$$", "symbols('theta phi') r1 = 1 r2 = 1.5 + sin(5 * phi) * sin(10 * theta) / 10 plot3d_spherical(r1, r2, (theta, 0, pi / 2), (phi, 0.35 * pi, 2 * pi), wireframe=True, wf_n2=25, backend=PB) Interactive-widget plot of spherical harmonics. Note that the plot’s creation and update might be slow and that it must be m < n at all times. Refer to iplot documentation to learn more about the params dictionary. from sympy import * from spb import * import param n, m = symbols(\"n, m\") phi, theta = symbols(\"phi, theta\", real=True) r = abs(re(Ynm(n, m, theta, phi).expand(func=True).rewrite(sin).expand())) plot3d_spherical( r, (theta, 0, pi), (phi, 0, 2*pi), params = { n: param.Integer(0, label=\"n\"), m: param.Integer(0, label=\"m\"), }, backend=KB ) spb.functions.plot3d_revolution(curve, range_t, range_phi=None, axis=(0, 0), parallel_axis='z', show_curve=False, curve_kw=None, **kwargs)[source] Generate a surface of revolution by rotating a curve around an axis of rotation. Parameters curveExpr, list/tuple of 2 or 3 elements The curve to be revolved, which can be either: • a symbolic expression • a 2-tuple representing a parametric curve in 2D space • a 3-tuple representing a parametric curve in 3D space range_t(symbol, min, max) A 3-tuple denoting the range of the parameter of the curve. range_phi(symbol, min, max) A 3-tuple denoting the range of the azimuthal angle where the curve will be revolved. Default to (phi,", "how to angle it. Here I set $\\theta$=$\\pi$/2 to make the disk. 4. Use Disk with Graphics3D and Rotate This would work if Disk could be used as a 3D object! Argh! That's all my ideas. If you see how to fix one of these, or have another, better idea, please advise! ## marked as duplicate by José Antonio Díaz Navas, Coolwater, Henrik Schumacher, LCarvalho, MarcoBMar 2 '18 at 3:21 First, define the normal in spherical coordinates: n[theta_, phi_] := {Sin[theta]*Cos[phi], Sin[theta]*Sin[phi], Cos[theta]} Then multiply your (syntax-corrected) ParametricPlot3D with RotationMatrix and wrap it in Manipulate: Manipulate[ Show[ ParametricPlot3D[ RotationMatrix[{{0, 0, 1}, n[theta, phi]}]. {r*Sin[Pi/2]*Cos[Phi], r*Sin[Pi/2] Sin[Phi], r*Cos[Pi/2]} , {r, 0, 1}, {Phi, 0, 2*Pi} , PlotRange -> {{-1, 1}, {-1, 1}, {-1, 1}} , AxesLabel -> {x, y, z}] , Graphics3D[{Red, Arrow[{{0, 0, 0}, n[theta, phi]}]}]] , {theta, 0, 2 Pi}, {phi, 0, Pi}] • This is beautiful, thank you. RotationMatrix was the missing piece to my idea #3 above. Small note: typically theta goes 0->Pi and phi goes 0->2Pi. – Max Feb 17 '18 at 0:04 Given the normal vector n={Cos[\\[Theta]] Cos[\\[CurlyPhi]], Cos[\\[Theta]] Sin[\\[CurlyPhi]],Sin[\\[Theta]]} the plane orthogonal to n is given by e={Sin[\\[CurlyPhi]], -Cos[\\[CurlyPhi]], 0} f={-Cos[\\[CurlyPhi]] Sin[\\[Theta]], -Sin[\\[Theta]] Sin[\\[CurlyPhi]],Cos[\\[Theta]]} The 3D-disk can be plotted using Block[{\\[Theta] = 50 \\[Degree], \\[CurlyPhi] = 10 \\[Degree] }, Show[{Graphics3D[{Point[{0, 0, 0}],", "y^2 + y^3 - y z^2 + z^3; xdot = x (-x + f); ydot = y (x - y + f); zdot = z (y - z + f); vector = {xdot, ydot, zdot}; a = 1/Sqrt[14]; point = {a, 2 a, 3 a}; transformedvector = TransformedField[\"Cartesian\" -> \"Spherical\", vector, {x, y, z} -> {r, theta, phi}] /. r -> 1 // Simplify (* {0, 1/4 Cos[theta] Sin[theta] (4 Cos[theta] + (-2 Cos[phi] - 2 Cos[3 phi] - 7 Sin[phi] + Sin[3 phi]) Sin[theta]), Cos[phi] (2 Cos[phi] - Sin[phi]) Sin[phi] Sin[theta]^2} *) As one can see, after substituting $$r=1$$ into the transformed vector, $$r$$ component of the vector becomes $$0$$, which indicates all the vectors at $$r=1$$ are on the unit ball. Then we plot it in 2D: plot = StreamPlot[ transformedvector // Rest // Evaluate, {theta, 0, Pi}, {phi, 0, 2 Pi}]~Show~ Graphics@{Orange, PointSize@Large, Point@Rest@CoordinateTransform[\"Cartesian\" -> \"Spherical\", point]} Well, in my opinion this illustration is already good enough, but if you insist on visualising on the ball: func = {theta, phi} \\[Function] Evaluate@CoordinateTransform[ \"Spherical\" -> \"Cartesian\", {1, theta, phi}] plot3D = Graphics3D[(plot[[1]] /. (head : Arrow | Point)[z_] :> head[z /. {x_?NumericQ, y_} :> func @@ {x, y}])] plot3D~Show~Graphics3D@Ball[{0, 0, 0}, 0.98] Visualization of the solution in the form of trajectories on the sphere eq =" ]

[ "Now, $\\frac{\\cot(\\gamma)}{\\cot(\\beta)+\\cot(\\alpha)}=\\frac{\\frac{994c}{b\\sin a}}{\\frac{a^2-994c^2+b^2-994c^2}{bc\\sin(a)}}$. After using $a^2+b^2=1989c^2$, we get $\\frac{994c*bc\\sin a}{c^2b\\sin a}=\\boxed{994}$.", "with $$\\sin(\\alpha)=\\frac{a}{2r},\\sin(\\beta)=\\frac{b}{2r}$$ we get $$A=\\frac{1}{2}4r^2\\sin(\\gamma)=2r^2\\sin(\\gamma)\\le 2r^2$$", "# Prove the following Question: If $\\cos (\\alpha+\\beta)=\\frac{3}{5}, \\sin (\\alpha-\\beta)=\\frac{5}{13}$ and $0<\\alpha, \\beta<\\frac{\\pi}{4}$, then $\\tan (2 \\alpha)$ is equal to : 1. $\\frac{21}{16}$ 2. $\\frac{63}{52}$ 3. $\\frac{33}{52}$ 4. $\\frac{63}{16}$ Correct Option: , 4 Solution:", "other two sides, we can conclude that $\\frac{a}{{\\sin \\alpha }} = \\frac{b}{{\\sin \\beta }} = \\frac{c}{{\\sin \\gamma }} = 2R.$ See solved problems on Page 2.", "\\hat{i}-\\frac{3}{5} \\hat{j}$ 14) (a) $\\sin \\alpha=\\frac{4}{5}, \\sin \\beta=\\frac{7 \\sqrt{2}}{10}$ (b) $13 \\mathrm{~km}$, $N 30^{\\circ} 23^{\\prime} E$ 15) (a) Prove (b) $0.8 \\pi$", "# Let $\\cos (\\alpha + \\beta) = \\frac{4}{5}$ and let $\\sin (\\alpha - \\beta) = \\frac{5}{13}$, where $0 \\leq \\alpha, \\; \\beta \\leq \\frac{\\pi}{4}$, then $\\tan 2 \\alpha =$ ( A ) $\\frac{56}{33}$ ( B ) $\\frac{25}{16}$ ( C ) $\\frac{20}{7}$ ( D ) $\\frac{19}{12}$", "= \\frac{16}{1287}.$$ 14. What is the value of $$\\int_0^1 x^5 (1-x)^6 dx$$? a) $$\\frac{1}{12*11*10*9*8*7}$$ b) $$\\frac{1}{12*11*10*9*8}$$ c) $$\\frac{1}{12*11*10*9*8*7*6}$$ d) $$\\frac{1}{12*11*10*9*8*7*6*5}$$ $$\\beta(6,7)= \\frac{\\Gamma(6)\\Gamma(7)}{\\Gamma(13)} = \\frac{1}{12*11*10*9*8*7*6}.$$", "$$E^2 = E^2\\beta^2 + m_f^2$$ $$\\beta^2 = 1-\\frac{m_f^2}{E^2} = 1-\\frac{4m_f^2}{s}$$", "# Solve the following Question: Let $\\cos (\\alpha+\\beta)=\\frac{4}{5}$ and let $\\sin (\\alpha-\\beta)=\\frac{5}{13}$, where $0 \\leq \\alpha, \\beta \\leq \\frac{\\pi}{4}$. Then $\\tan 2 \\alpha=$ 1. $\\frac{25}{16}$ 2. $\\frac{56}{33}$ 3. $\\frac{19}{12}$ 4. $\\frac{20}{7}$ Correct Option: , 2 Solution:", "{59}{3^{t}}]\\}=1$$ $$\\beta_{1}(60)=\\sum_{t}^{r} \\{[\\frac {60}{5^{t}}]-[\\frac {59}{5^{t}}]\\}=1$$ Finally: $$60=2^{2}3^{1}5^{1}$$", "we get, $\\alpha = \\frac 12 [ 0 (1-4) – 0(3-2) + 1 (12-2)]$ $\\Rightarrow \\alpha = 5 units$<", "$I$, we get: $\\alpha = \\frac {T}{I} = \\frac{10.2}{0.813} = 12.5 s^{-2}$", "each is $\\alpha$. So if $x$ is the length of the remaining side, then $\\frac{r\\sin 2\\alpha}{x}=\\sin \\alpha$. Solve for $x$. – André Nicolas Aug 11 '13 at 1:53 • Aha! Thanks so much Andre! – Ryan Aug 11 '13 at 2:07 • You are welcome. – André Nicolas Aug 11 '13 at 2:10", "# If $\\alpha=\\large\\frac{5}{2 ! 3}+\\frac{5.7}{3 ! 3^2}+ \\frac{5.7.9}{4 ! 3^3}+........,$ then $\\alpha^2+4 \\alpha=$ $\\begin {array} {1 1} (1)\\;21 & \\quad (2)\\;23 \\\\ (3)\\;25 & \\quad (4)\\;27 \\end {array}$ (2) 23 answered Nov 7, 2013 by Can u plz explain the steps", "are left with $$\\frac{1}{\\beta^{2}} = \\gamma$$ $$\\beta = \\frac{1}{\\sqrt{\\gamma}}$$ $$\\beta = \\frac{1}{\\sqrt{U'/U}}$$ $$\\beta = \\sqrt{\\frac{U}{U'}}$$ 5. Dec 1, 2009 physicsworks msbell1, sounds good! keep working on Landau & Lifgarbagez.", "the wrong answer. 16. UsukiDoll |dw:1433044545197:dw| 17. UsukiDoll $\\sin(\\theta + \\beta) = \\frac{3}{5}\\frac{2\\sqrt{6}}{5}+\\frac{4}{5} (\\frac{1}{5})$ 18. UsukiDoll $\\frac{6\\sqrt{6}}{25}+\\frac{4}{25}$ 19. UsukiDoll I relooked at cos beta and realized that 5 was the longest side of the triangle and that 2 sqrt{6} was the adjcent part... needed to find the opposite part of the triangle so I could find sin beta. 20. UsukiDoll", "DE:DB = 2.$$ Thus $$\\sin \\beta=\\frac{\\sqrt3}{4},\\cos \\beta=\\frac{\\sqrt{13}}4.\\tag{1}$$ On the other hand, ...", "4. Yes, once you have $e^{-\\alpha r^V}= \\frac{1+\\sqrt{1+ k_2\\alpha^2}}{\\beta k_1\\alpha}$, take the logarithm of both sides. Yes, once you have $e^{-\\alpha r^V}= \\frac{1+\\sqrt{1+ k_2\\alpha^2}}{\\beta k_1\\alpha}$, take the logarithm of both sides.", "= 3 ) = \\frac{ e^{- \\beta (\\epsilon_2 + 3 \\hslash \\omega - 2 \\mu)}}{z_G}$", "is obtainable, viz. $$x=\\frac{\\pi}{6}$$. Rewrite the equation as $$\\;\\cos(2x)=\\cos \\bigl(\\frac\\pi 2-x\\bigr)$$. Then use that $$\\cos\\alpha=\\cos\\beta\\iff \\alpha\\equiv\\pm\\beta \\pmod{2\\pi}.$$" ]

[ "## Precalculus (6th Edition) Blitzer a. $-\\frac{4}{5}$ b. $-\\frac{3}{5}$ c. $\\frac{3}{4}$ Given $sin\\alpha=\\frac{4}{5}$ with $\\alpha$ in quadrant I, we have $cos\\alpha=\\sqrt {1-sin^2\\alpha}=\\frac{3}{5}$ and $tan\\alpha=\\frac{sin\\alpha}{cos\\alpha}=\\frac{4}{3}$. Similarly, given $sin\\beta=\\frac{7}{25}$ with $\\beta$ in quadrant II, we have $cos\\beta=-\\sqrt {1-sin^2\\beta}=-\\frac{24}{25}$ and $tan\\beta=\\frac{sin\\beta}{cos\\beta}=-\\frac{7}{24}$. Using the Addition Formulas, we have: a. $cos(\\alpha+\\beta)=cos(\\alpha) cos(\\beta)-sin(\\alpha) sin(\\beta)=(\\frac{3}{5})(-\\frac{24}{25})-(\\frac{4}{5})(\\frac{7}{25})=-\\frac{100}{125}=-\\frac{4}{5}$ b. $sin(\\alpha+\\beta)=sin(\\alpha) cos(\\beta)+cos(\\alpha) sin(\\beta)=(\\frac{4}{5})(-\\frac{24}{25})+(\\frac{3}{5})(\\frac{7}{25})=-\\frac{75}{125}=-\\frac{3}{5}$ c. $tan( \\alpha+\\beta)=\\frac{tan(\\alpha)+tan(\\beta)}{1-tan(\\alpha) tan(\\beta)}=\\frac{(\\frac{4}{3})+(-\\frac{7}{24})}{1-(\\frac{4}{3})(-\\frac{7}{24})}=\\frac{3}{4}$", "{59}{3^{t}}]\\}=1$$ $$\\beta_{1}(60)=\\sum_{t}^{r} \\{[\\frac {60}{5^{t}}]-[\\frac {59}{5^{t}}]\\}=1$$ Finally: $$60=2^{2}3^{1}5^{1}$$", "# How do you simplify sin ( sin^ -1 (-3/5) + tan^ -1(5/12)) ? Oct 17, 2016 $\\sin \\left({\\sin}^{- 1} \\left(- \\frac{3}{5}\\right) + {\\tan}^{- 1} \\left(\\frac{5}{12}\\right)\\right) = - \\frac{16}{65}$ #### Explanation: Let ${\\sin}^{- 1} \\left(- \\frac{3}{5}\\right) = \\alpha$ and ${\\tan}^{- 1} \\left(\\frac{5}{12}\\right) = \\beta$ then $\\sin \\alpha = - \\frac{3}{5}$ and $\\tan \\beta = \\frac{5}{12}$ and $\\cos \\alpha = \\sqrt{1 - {\\left(- \\frac{3}{5}\\right)}^{2}} = \\sqrt{1 - \\frac{9}{25}} = \\sqrt{\\frac{16}{25}} = \\frac{4}{5}$ $\\cos \\beta = \\frac{1}{\\sqrt{1 + {\\tan}^{2} \\beta}} = \\frac{1}{\\sqrt{1 + {\\left(\\frac{5}{12}\\right)}^{2}}} = \\frac{1}{\\sqrt{1 + \\frac{25}{144}}} = \\frac{1}{\\sqrt{\\frac{169}{144}}} = \\frac{1}{\\frac{13}{12}} = \\frac{12}{13}$ and $\\sin \\beta = \\tan \\beta \\times \\cos \\beta = \\frac{5}{12} \\times \\frac{1}{13} = \\frac{5}{13}$ Hene, $\\sin \\left({\\sin}^{- 1} \\left(- \\frac{3}{5}\\right) + {\\tan}^{- 1} \\left(\\frac{5}{12}\\right)\\right) = \\sin \\left(\\alpha + \\beta\\right)$ = $\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta$ = $\\left(- \\frac{3}{5}\\right) \\times \\frac{12}{13} + \\frac{4}{5} \\times \\frac{5}{13}$ = $- \\frac{36}{65} + \\frac{20}{65} = - \\frac{16}{65}$", "Now, $\\frac{\\cot(\\gamma)}{\\cot(\\beta)+\\cot(\\alpha)}=\\frac{\\frac{994c}{b\\sin a}}{\\frac{a^2-994c^2+b^2-994c^2}{bc\\sin(a)}}$. After using $a^2+b^2=1989c^2$, we get $\\frac{994c*bc\\sin a}{c^2b\\sin a}=\\boxed{994}$.", "$\\sin^2\\beta+\\cos^2\\beta\\equiv1$sin2β+cos2β1 to deduce $\\cos\\beta$cosβ. Hence, $\\left(\\frac{5}{9}\\right)^2+\\cos^2\\beta=1$(59)2+cos2β=1 and so, $\\cos\\beta=\\sqrt{1-\\left(\\frac{5}{9}\\right)^2}=\\frac{2\\sqrt{14}}{9}$cosβ=1(59)2=2149 Now, $\\csc\\beta=\\frac{1}{\\sin\\beta}=\\frac{9}{5},$cscβ=1sinβ=95, $\\sec\\beta=\\frac{1}{\\cos\\beta}=\\frac{9}{2\\sqrt{14}}=\\frac{9\\sqrt{14}}{28},$secβ=1cosβ=9214=91428, $\\tan\\beta=\\frac{\\sin\\beta}{\\cos\\beta}=\\frac{5}{2\\sqrt{14}}=\\frac{5\\sqrt{14}}{28},$tanβ=sinβcosβ=5214=51428, $\\cot\\beta=\\frac{1}{\\tan\\beta}=\\frac{2\\sqrt{14}}{5}$cotβ=1tanβ=2145 #### Worked Examples ##### QUESTION 1 Find the value of $\\cos\\theta$cosθ if $\\sec\\theta=\\frac{8}{3}$secθ=83. ##### QUESTION 2 Find the value of $\\cot\\theta$cotθ if $\\tan\\theta=0.1$tanθ=0.1 . ##### QUESTION 3 If $\\sin\\alpha=\\frac{7}{25}$sinα=725 and $\\cos\\alpha=\\frac{24}{25}$cosα=2425, find: 1. $\\tan\\alpha$tanα 2. $\\cot\\alpha$cotα 3. $\\sec\\alpha$secα 4. $\\csc\\alpha$cscα ### Outcomes #### M8-6 Manipulate trigonometric expressions #### 91575 Apply trigonometric methods in solving problems", "= 162 6r3 = 162 r3 = $$\\frac{162}{6}$$ = 27 r = 3 Question 18. The sum of three consecutive numbers is 243. Find the three numbers. x + x + 1 + x + 2 = 243 3x + 3 = 2433x = 240 x = 80 80, 81, 82 Question 19. If a and P are the roots of the equation 2x2 + 4x + 1 = 0 find the value of $$\\frac{1}{\\alpha}+\\frac{1}{\\beta}$$ α + β = –$$\\frac{4}{2}$$ = -2 $$\\frac{1}{\\alpha}+\\frac{1}{\\beta}=\\frac{\\alpha+\\beta}{\\alpha \\beta}$$ αβ = $$\\frac{c}{a}=\\frac{1}{2}=\\frac{-2}{\\frac{1}{2}}$$ = -4 Question 20. Solve 7x + 3 < 5x + 9, x ∈ R and represent on the number line. Question 21. Find the future value of an annuity of ₹ 5000 at 12% p.a for 6 years. A = Rs.5,000 r = 0.12 n = 6 F = $$\\frac{A\\left[(1+r)^{n}-1\\right]}{r}=\\frac{5000\\left[(1+0.12)^{6}-1\\right]}{0.12}=\\frac{4869.11}{0.12}$$ = 40,575.9 Question 22. Prove that (1 + cotA)2 + (1 – cotA)2 = 2 cosec2A Question 23. Find the value of cos 60° -sin 30° – cot3 45° $$\\frac{1}{2}-\\frac{1}{2}$$ – (1)3 = -1 Question 24. Find the distance of the centroid of the triangle formed by the points (7, 1) (1, 5) and (1, 6) from origin. G = $$\\left[\\frac{7+1+1}{3}, \\frac{1+5+6}{3}\\right]$$ = (3, 4) O = (0, 0) OG = $$\\sqrt{(3)^{2}+(4)^{2}}=\\sqrt{9+16}=\\sqrt{25}$$ = 5 Question 25. Find the equation of", "Hence, $a=7$, $b+c=22$ Furthermore, by the Pythagorean Theorem, we have that $a^2+b^2=c^2\\implies 49+b^2=c^2$ Solving for $c$ in the first equation and plugging in into the second equation... $49+b^2=(22-b)^2\\implies 49+b^2=484-44b+b^2\\implies 44b=435\\implies b=\\frac{435}{44}$ Hence, $c=22-\\frac{435}{44}=\\frac{533}{44}$ Now, we want $\\frac{a+c}{b}$ Plugging in, we find the answer is $\\frac{\\frac{7\\cdot{44}}{44}+\\frac{533}{44}}{\\frac{435}{44}}=\\frac{841}{435}=\\frac{29}{15}$ Hence, the answer is $29+15=\\boxed{044}$ ### Solution 7 We know that $\\sec(x) = \\frac{h}{a}$ and that $\\tan(x) = \\frac{o}{a}$ where $h$, $a$, $o$ represent the hypotenuse, adjacent, and opposite (respectively) to angle $x$ in a right triangle. Thus we have that $\\sec(x) + \\tan(x) = \\frac{h+o}{a}$. We also have that $\\csc(x) + \\cot(x) = \\frac{h}{o} + \\frac{a}{o} = \\frac{h+a}{o}$. Set $\\sec(x) + \\tan(x) = \\alpha$ and csc(x)+cot(x) = $\\beta$. Then, notice that $\\alpha + \\beta = \\frac{h+o}{a} + \\frac{h+a}{o} = \\frac{oh+ah+o^2 + a^2}{oa} = \\frac{h(o+a+h)}{oa}$ ( This is because of the Pythagorean Theorem, recall $o^2 +a^2 = h^2$). But then notice that $\\alpha \\cdot \\beta = \\frac{(o+h)(a+h)}{oa} = \\frac{oa +oh +ha +h^2}{oa} = 1+ \\frac{h(o+a+h)}{oa} = 1+ \\alpha + \\beta$. From the information provided in the question, we can substitute $\\alpha$ for $\\frac{22}{7}$. Thus, $\\frac{22 \\beta}{7}= \\beta + \\frac{29}{7} \\Longrightarrow 22 \\beta = 7 \\beta + 29 \\Longrightarrow 15 \\beta = 29 \\Longrightarrow \\beta = \\frac{29}{15}$. Since, essentially we are asked to find the sum of the numerator and denominator of $\\beta$, we", "## Precalculus (6th Edition) Blitzer a. $-\\frac{4}{5}$ b. $\\frac{5}{13}$ c. $\\frac{33}{65}$ d. $-\\frac{56}{65}$ a. Knowing $\\alpha$ is in quadrant II, we have $cos\\alpha\\lt0$ and $cos\\alpha=-\\sqrt {1-sin^2\\alpha}=-\\frac{4}{5}$ b. Knowing $\\beta$ is in quadrant II, we have $sin\\beta\\gt0$ and $sin\\beta=-\\sqrt {1-cos^2\\beta}=\\frac{5}{13}$ c. Using the addition formula, we have $cos(\\alpha+\\beta)=(-\\frac{4}{5})(-\\frac{12}{13})-(\\frac{3}{5})(\\frac{5}{13})=\\frac{33}{65}$ d. Using the addition formula, we have $sin(\\alpha+\\beta)=(\\frac{3}{5})(-\\frac{12}{13})+(-\\frac{4}{5})(\\frac{5}{13})=-\\frac{56}{65}$", "# If tanα= Question: If $\\tan \\alpha=\\frac{1}{7}, \\tan \\beta=\\frac{1}{3}$, then $\\cos 2 \\alpha$ is equal to (a) $\\sin 2 \\beta$ (b) $\\sin 4 \\beta$ (c) $\\sin 3 \\beta$ (d) $\\cos 2 \\beta$ Solution: It is given that $\\tan \\alpha=\\frac{1}{7}$ and $\\tan \\beta=\\frac{1}{3}$. Now, $\\tan 2 \\beta=\\frac{2 \\tan \\beta}{1-\\tan ^{2} \\beta}$ $=\\frac{2 \\times \\frac{1}{3}}{1-\\frac{1}{9}}$ $=\\frac{\\frac{2}{3}}{\\frac{8}{9}}$ $=\\frac{3}{4}$ $\\therefore \\tan (\\alpha+2 \\beta)=\\frac{\\tan \\alpha+\\tan 2 \\beta}{1-\\tan \\alpha \\tan 2 \\beta}$ $=\\frac{\\frac{1}{7}+\\frac{3}{4}}{1-\\frac{1}{7} \\times \\frac{3}{4}}$ $=\\frac{\\frac{25}{28}}{\\frac{25}{28}}$ $=1$ $\\tan (\\alpha+2 \\beta)=1=\\tan \\frac{\\pi}{4}$ $\\Rightarrow \\alpha+2 \\beta=\\frac{\\pi}{4}$ $\\Rightarrow \\alpha=\\frac{\\pi}{4}-2 \\beta$ $\\Rightarrow 2 \\alpha=\\frac{\\pi}{2}-4 \\beta$ $\\Rightarrow \\cos 2 \\alpha=\\cos \\left(\\frac{\\pi}{2}-4 \\beta\\right)=\\sin 4 \\beta$ $\\therefore \\cos 2 \\alpha=\\sin 4 \\beta$ Hence, the correct answer is option $B$.", "law of sines $\\frac{a}{\\sin\\:\\alpha\\:}=\\:\\frac{c}{\\sin\\:\\gamma}$ $\\sin\\:\\beta=b\\:\\frac{\\sin\\:\\gamma}{c}$ substituting $b = 18$, $c = 29$ and $\\gamma = 86.084^{\\circ }$ $\\sin\\:\\beta=18\\:\\frac{\\sin\\:86.084^{\\circ }}{29}$ $\\sin\\:\\beta=18\\:\\frac{0.998}{29}$ $\\sin\\:\\beta = 18 (0.03441)$ $\\sin\\:\\beta = 0.61938$ $\\beta = \\sin^{-1} (0.61938)$ $\\beta = 38.271…^{\\circ }$ $\\beta ≈ 38^{\\circ }$ (angle to the nearest degree) Step 3 of 3", "first diagram above, $$\\sin \\frac{\\alpha}{2} = \\frac{1}{r}$$ $$\\sin \\frac{\\beta}{2} = \\frac{1.5}{r}$$ $$\\sin(\\frac{\\alpha}{2}+\\frac{\\beta}{2})=\\frac{2}{r}$$ by the Pythagorean trig identities, $$\\cos\\frac{\\alpha}{2}=\\sqrt{1-\\frac{1}{r^2}}$$ $$\\cos\\frac{\\beta}{2}=\\sqrt{1-\\frac{2.25}{r^2}}$$ so by the composite sine identity $$\\frac{2}{r}=\\frac{1}{r}\\sqrt{1-\\frac{2.25}{r^2}}+\\frac{1.5}{r}\\sqrt{1-\\frac{1}{r^2}}$$ multiply both sides by $2r$, then subtract $\\sqrt{4-\\frac{9}{r^2}}$ from both sides squaring both sides, we get $$16 - 8\\sqrt{4-\\frac{9}{r^2}} + 4 - \\frac{9}{r^2}=9 - \\frac{9}{r^2}$$ $$\\Longrightarrow 16+4=9+8\\sqrt{4-\\frac{9}{r^2}}\\Longrightarrow\\frac{11}{8}=\\sqrt{4-\\frac{9}{r^2}}\\Longrightarrow\\frac{121}{64}=4-\\frac{9}{r^2}$$ $$\\Longrightarrow\\frac{(256-121)r^2}{64}=9\\Longrightarrow r^2= \\frac{64}{15}$$ plugging this back in, $$\\cos^2(\\frac{\\alpha}{2})=1-\\frac{15}{64}=\\frac{49}{64}$$ so $$\\cos(\\alpha)=2(\\frac{49}{64})-1=\\frac{34}{64}=\\frac{17}{32}$$ and the answer is $17+32=\\boxed{049}$", "YZ = ? p2 + b2 = h2 or , b2 = h2 - p2 b = $$\\sqrt{h^{2} - p^{2}}$$ =$$\\sqrt{65^{2} - 16^{2}}$$ = $$\\sqrt{3969}$$ = 63. Here , sinx = $$\\frac{p}{h}$$ = $$\\frac{16}{65}$$ = $$\\frac{16}{65}$$ cosx = $$\\frac{b}{h}$$ = $$\\frac{63}{65}$$ = $$\\frac{63}{65}$$ and tanx = $$\\frac{p}{b}$$ = $$\\frac{16}{63}$$ = $$\\frac{16}{63}$$ = ( $$\\frac{1}{2}$$ )2 + ( $$\\frac{1}{\\sqrt {2}}$$ )2 + ( $$\\frac{\\sqrt{3}}{2}$$ )2 = $$\\frac{1}{4}$$+ $$\\frac{1}{2}$$ +$$\\frac{3}{4}$$ = $$\\frac{1+2+3}{4}$$ = $$\\frac{3}{2}$$ = 1$$\\frac{1}{2}$$ cos$$\\beta$$ = $$\\sqrt{3}$$ $$\\times$$ $$\\frac{1}{2}$$ or ,cos$$\\beta$$ = $$\\frac{\\sqrt{3}}{2}$$ or ,cos$$\\beta$$ = cos 30o $$\\therefore$$ $$\\beta$$ = 30o Ans. or , sin2$$\\beta$$ = $$\\frac{1}{4}$$ or , sin2 $$\\beta$$ =( $$\\frac{1}{2}$$ )2 or , sin2 $$\\beta$$ = (sin30o)2 or ,sin $$\\beta$$ = sin30o $$\\therefore$$ $$\\beta$$ = 30o Ans. 0% • ### In a right-angled triangle, sin θ=(frac{1}{2}),find the numerical value of cos θ. (frac{sqrt 4}{2}) (frac{sqrt 3}{2}) (frac{sqrt 5}{4}) (frac{sqrt 6}{3}) • ### Express sec A in terms of tan A. `(sqrt{1+sin^2 A}) (sqrt{3-sin^2 A}) (sqrt{1+tan^2 A}) (sqrt{4+sin^2 A}) • ### Express cosA in terms of sin A. (sqrt{1-cos^2 A}) (sqrt{1-cos A}) (sqrt{1-sin^2 A}) (sqrt{1-sin A}) (frac{1}{1}) (frac{4}{2}) (frac{3}{2}) (frac{1}{3}) (frac{3}{2}) (frac{4}{2}) (frac{2}{1}) (frac{1}{2}) (frac{2}{1}) (frac{3}{2}) (frac{1}{2}) (frac{2}{2}) (frac{2}{1}) (frac{1}{3}) (frac{1}{2}) (frac{1}{4}) three two zero one 5 2 19 10 zero one three two zero five nine two 3 2 10 6 1 3 4", "we get, $\\alpha = \\frac 12 [ 0 (1-4) – 0(3-2) + 1 (12-2)]$ $\\Rightarrow \\alpha = 5 units$<", "\\beta^2} = \\frac{4}{5}$ for $\\beta$, $\\beta = \\frac{1}{2}$. [2 minutes total ]", "x}\\cos(2\\beta x - \\sin x \\cos x)dx = \\frac{\\pi}{\\sqrt{e}}\\frac 1 {\\beta!}\\frac 1 { 2^{\\beta -1}}$$ I'm not sure about the general formula you're proposing. Jul 19, 2021 at 6:44", "{\\cot{\\gamma}}{\\cot{\\alpha} + \\cot{\\beta}} &= \\frac {\\cot{\\gamma}\\sin{\\alpha}\\sin{\\beta}}{\\sin{\\gamma}} = \\frac {\\cos{\\gamma}\\sin{\\alpha}\\sin{\\beta}}{\\sin^2{\\gamma}} = \\frac {ab}{c^2}\\cos{\\gamma} = \\frac {ab}{c^2} \\cdot \\frac {994c^2}{ab}\\\\ &= \\boxed{994}\\end{align*} ### Solution 5 Use Law of cosines to give us $c^2=a^2+b^2-2ab\\cos(\\gamma)$ or therefore $\\cos(\\gamma)=\\frac{994c^2}{ab}$. Next, we are going to put all the sin's in term of $\\sin(a)$. We get $\\sin(\\gamma)=\\frac{c\\sin(a)}{a}$. Therefore, we get $\\cot(\\gamma)=\\frac{994c}{b\\sin a}$. Next, use Law of Cosines to give us $b^2=a^2+c^2-2ac\\cos(\\beta)$. Therefore, $\\cos(\\beta)=\\frac{a^2-994c^2}{ac}$. Also, $\\sin(\\beta)=\\frac{b\\sin(a)}{a}$. Hence, $\\cot(\\beta)=\\frac{a^2-994c^2}{bc\\sin(a)}$. Lastly, $\\cos(\\alpha)=\\frac{b^2-994c^2}{bc}$. Therefore, we get $\\cot(\\alpha)=\\frac{b^2-994c^2}{bc\\sin(a)}$. Now, $\\frac{\\cot(\\gamma)}{\\cot(\\beta)+\\cot(\\alpha)}=\\frac{\\frac{994c}{b\\sin a}}{\\frac{a^2-994c^2+b^2-994c^2}{bc\\sin(a)}}$. After using $a^2+b^2=1989c^2$, we get $\\frac{994c*bc\\sin a}{c^2b\\sin a}=\\boxed{994}$. ### Solution 6 Let $\\gamma$ be $(180-\\alpha-\\beta)$ $\\frac{\\cot \\gamma}{\\cot \\alpha+\\cot \\beta} = \\frac{\\frac{-\\tan \\alpha \\tan \\beta}{\\tan(\\alpha+\\beta)}}{\\tan \\alpha + \\tan \\beta} = \\frac{(\\tan \\alpha \\tan \\beta)^2-\\tan \\alpha \\tan \\beta}{\\tan^2 \\alpha + 2\\tan \\alpha \\tan \\beta +\\tan^2 \\beta}$ WLOG, assume that $a$ and $c$ are legs of right triangle $abc$ with $\\beta = 90^o$ and $c=1$ By Pythagorean theorem, we have $b^2=a^2+1$, and the given $a^2+b^2=1989$. Solving the equations gives us $a=\\sqrt{994}$ and $b=\\sqrt{995}$. We see that $\\tan \\beta = \\infty$, and $\\tan \\alpha = \\sqrt{994}$. We see that our derived equation equals to $\\tan^2 \\alpha$ as $\\tan \\beta$ approaches infinity. Evaluating $\\tan^2 \\alpha$, we get $\\boxed{994}$.", "}}{{1 - i\\tan (\\alpha )}} = \\frac{{\\left( {1 + i\\tan (\\alpha )} \\right)^5 }}{{\\sec ^2 (\\alpha )}}$ $\\frac{{\\left( {1 + i\\tan (\\alpha )} \\right)^5 }}{{\\sec ^2 (\\alpha )}} = \\frac{{\\frac{{\\left( {\\cos (\\alpha ) + i\\sin (\\alpha )} \\right)^5 }}{{\\cos ^5 (\\alpha )}}}}{{\\sec ^2 (\\alpha )}} = \\frac{{\\cos (5\\alpha ) + i\\sin (5\\alpha )}}{{\\cos ^3 (\\alpha )}}$ 4. This is a postscript, resulting from a PM. It seems that the problem is actually $\\left[ {\\frac{{1 + \\tan (\\alpha )}}{{1 - \\tan (\\alpha )}}} \\right]^4$. In which case; $\\begin{array}{rcl} \\left[ {\\frac{{1 + i\\tan (\\alpha )}}{{1 - i\\tan (\\alpha )}}} \\right]^4 & = & \\left[ {\\frac{{\\cos (\\alpha ) + i\\sin (\\alpha )}}{{\\cos (\\alpha ) - i\\sin (\\alpha )}}} \\right]^4 \\\\ & = & \\left[ {\\cos (2\\alpha ) + i\\sin (2\\alpha )} \\right]^4 \\\\ & = & \\cos (8\\alpha ) + i\\sin (8\\alpha ) \\\\ \\end{array}$ 5. Thanks", "Q 14 | Page 44 $\\sin^2 \\left( \\frac{\\pi}{18} \\right) + \\sin^2 \\left( \\frac{\\pi}{9} \\right) + \\sin^2 \\left( \\frac{7\\pi}{18} \\right) + \\sin^2 \\left( \\frac{4\\pi}{9} \\right) =$ • 1 • 2 • 4 • none of these. Q 15 | Page 44 If $5 \\sin \\alpha = 3 \\sin \\left( \\alpha + 2 \\beta \\right) \\neq 0$ , then $\\tan \\left( \\alpha + \\beta \\right)$ is equal to • $2 \\tan \\beta$ • $3 \\tan \\beta$ • $4 \\tan \\beta$ • $6 \\tan \\beta$ Q 16 | Page 44 $2 \\text{ cos } x - \\ cos 3x - \\cos 5x - 16 \\cos^3 x \\sin^2 x$ • 2 • 1 • 0 • -1 Q 17 | Page 44 If $A = 2 \\sin^2 x - \\cos 2x$ , then A lies in the interval • $\\left[ - 1, 3 \\right]$ • $\\left[ 1, 2 \\right]$ • $\\left[ - 2, 4 \\right]$ • none of these Q 18 | Page 44 The value of $\\frac{\\cos 3x}{2 \\cos 2x - 1}$ is equal to • cos x • sin x • tan x • none of these Q 19 | Page 44 If $\\tan \\left( \\pi/4 + x \\right) + \\tan \\left( \\pi/4 - x \\right) = \\lambda \\sec 2x, \\text{ then }$ • 3 • 4 • 1 •", "$\\sin\\left(\\beta+\\alpha\\right)\\equiv\\sin\\beta\\cos\\alpha+\\cos\\beta\\sin\\alpha$sin(β+α)sinβcosα+cosβsinα The corresponding formulas for the tangent function are obtained by expanding $\\tan\\left(\\alpha+\\beta\\right)=\\frac{\\sin\\left(\\alpha+\\beta\\right)}{\\cos\\left(\\alpha+\\beta\\right)}$tan(α+β)=sin(α+β)cos(α+β) After some simplification, we have $\\tan\\left(\\alpha+\\beta\\right)=\\frac{\\tan\\alpha+\\tan\\beta}{1-\\tan\\alpha\\tan\\beta}$tan(α+β)=tanα+tanβ1tanαtanβ and $\\tan\\left(\\alpha-\\beta\\right)=\\frac{\\tan\\alpha-\\tan\\beta}{1+\\tan\\alpha\\tan\\beta}$tan(αβ)=tanαtanβ1+tanαtanβ #### Example Find an 'exact value' expression for $\\sin\\frac{\\pi}{12}$sinπ12 We notice that $\\frac{\\pi}{12}=\\frac{\\pi}{4}-\\frac{\\pi}{6}$π12=π4π6. So, $\\sin\\frac{\\pi}{12}=\\sin\\left(\\frac{\\pi}{4}-\\frac{\\pi}{6}\\right)=\\sin\\frac{\\pi}{4}\\cos\\frac{\\pi}{6}-\\sin\\frac{\\pi}{6}\\cos\\frac{\\pi}{4}$sinπ12=sin(π4π6)=sinπ4cosπ6sinπ6cosπ4 Then, using known exact values, we have $\\sin\\frac{\\pi}{12}=\\frac{1}{\\sqrt{2}}.\\frac{\\sqrt{3}}{2}-\\frac{1}{2}.\\frac{1}{\\sqrt{2}}=\\frac{\\sqrt{6}-\\sqrt{2}}{4}$sinπ12=12.3212.12=624 #### Worked Examples ##### Question 1 Given that $\\sin A=\\frac{24}{25}$sinA=2425 and $\\cos B=\\frac{20}{29}$cosB=2029, find the exact value of: 1. $\\sin\\left(A+B\\right)$sin(A+B) 2. $\\cos\\left(A-B\\right)$cos(AB) ##### Question 2 Find the exact value of $\\cos\\frac{11\\pi}{12}$cos11π12. ##### Question 3 Suppose $\\sin A=\\frac{3}{5}$sinA=35 and $\\sin B=-\\frac{12}{13}$sinB=1213, where $00<A<π2 and$\\piπ<B<3π2. 1. Find $\\cos A$cosA. 2. Find $\\cos B$cosB. 3. Find $\\sin\\left(A+B\\right)$sin(A+B). 4. Find $\\sin\\left(A-B\\right)$sin(AB). 5. Find $\\tan\\left(A+B\\right)$tan(A+B). 6. Find $\\tan\\left(A-B\\right)$tan(AB). 7. Which quadrant does $A+B$A+B lie in? $III$III A $II$II B $I$I C $IV$IV D $III$III A $II$II B $I$I C $IV$IV D 8. Which quadrant does $A-B$AB lie in? $III$III A $II$II B $I$I C $IV$IV D $III$III A $II$II B $I$I C $IV$IV D ### Outcomes #### M8-6 Manipulate trigonometric expressions #### 91575 Apply trigonometric methods in solving problems", "$$\\beta$$ in standard position contains the point $$Q\\left( { - 8,-6} \\right).$$ Find the six trigonometric functions of the angle $$\\beta.$$ Solution. Determine the distance $$r$$ from the origin to the point $$Q\\left( { - 8,-6} \\right):$$ $r = \\sqrt {{x^2} + {y^2}} = \\sqrt {{{\\left( { - 8} \\right)}^2} + {{\\left( { - 6} \\right)}^2}} = \\sqrt {64 + 36} = \\sqrt {100} = 10.$ Calculate the values of the trig functions: $\\sin \\beta = \\frac{y}{r} = \\frac{{ - 6}}{{10}} = - \\frac{3}{5};$ $\\cos \\beta = \\frac{x}{r} = \\frac{{ - 8}}{{10}} = - \\frac{4}{5};$ $\\tan \\beta = \\frac{y}{x} = \\frac{{ - 6}}{{-8}} = \\frac{3}{4};$ $\\cot \\beta = \\frac{x}{y} = \\frac{{ - 8}}{{- 6}} = \\frac{4}{3};$ $\\sec \\beta = \\frac{r}{x} = \\frac{{ 10}}{{-8}} = - \\frac{5}{4};$ $\\csc \\beta = \\frac{r}{y} = \\frac{{ 10}}{{-6}} = - \\frac{5}{3}.$ ### Example 7. Calculate the sum of the series $1 + \\sin \\frac{\\pi }{6} + {\\sin ^2}\\frac{\\pi }{6} + {\\sin ^3}\\frac{\\pi }{6} + \\ldots$ Solution. Since $$\\sin \\frac{\\pi }{6} = \\frac{1}{2},$$ we can write this expression in the form: $1 + \\frac{1}{2} + {\\left( {\\frac{1}{2}} \\right)^2} + {\\left( {\\frac{1}{2}} \\right)^3} + \\ldots$ We have here an infinite geometric series with the initial term $$a_1 =1$$ and the common ratio $$q = \\frac{1}{2}.$$ The sum of the geometric series is given by $S" ]

[ "# How do you simplify sin ( sin^ -1 (-3/5) + tan^ -1(5/12)) ? Oct 17, 2016 $\\sin \\left({\\sin}^{- 1} \\left(- \\frac{3}{5}\\right) + {\\tan}^{- 1} \\left(\\frac{5}{12}\\right)\\right) = - \\frac{16}{65}$ #### Explanation: Let ${\\sin}^{- 1} \\left(- \\frac{3}{5}\\right) = \\alpha$ and ${\\tan}^{- 1} \\left(\\frac{5}{12}\\right) = \\beta$ then $\\sin \\alpha = - \\frac{3}{5}$ and $\\tan \\beta = \\frac{5}{12}$ and $\\cos \\alpha = \\sqrt{1 - {\\left(- \\frac{3}{5}\\right)}^{2}} = \\sqrt{1 - \\frac{9}{25}} = \\sqrt{\\frac{16}{25}} = \\frac{4}{5}$ $\\cos \\beta = \\frac{1}{\\sqrt{1 + {\\tan}^{2} \\beta}} = \\frac{1}{\\sqrt{1 + {\\left(\\frac{5}{12}\\right)}^{2}}} = \\frac{1}{\\sqrt{1 + \\frac{25}{144}}} = \\frac{1}{\\sqrt{\\frac{169}{144}}} = \\frac{1}{\\frac{13}{12}} = \\frac{12}{13}$ and $\\sin \\beta = \\tan \\beta \\times \\cos \\beta = \\frac{5}{12} \\times \\frac{1}{13} = \\frac{5}{13}$ Hene, $\\sin \\left({\\sin}^{- 1} \\left(- \\frac{3}{5}\\right) + {\\tan}^{- 1} \\left(\\frac{5}{12}\\right)\\right) = \\sin \\left(\\alpha + \\beta\\right)$ = $\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta$ = $\\left(- \\frac{3}{5}\\right) \\times \\frac{12}{13} + \\frac{4}{5} \\times \\frac{5}{13}$ = $- \\frac{36}{65} + \\frac{20}{65} = - \\frac{16}{65}$", "## Precalculus (6th Edition) Blitzer a. $-\\frac{4}{5}$ b. $-\\frac{3}{5}$ c. $\\frac{3}{4}$ Given $sin\\alpha=\\frac{4}{5}$ with $\\alpha$ in quadrant I, we have $cos\\alpha=\\sqrt {1-sin^2\\alpha}=\\frac{3}{5}$ and $tan\\alpha=\\frac{sin\\alpha}{cos\\alpha}=\\frac{4}{3}$. Similarly, given $sin\\beta=\\frac{7}{25}$ with $\\beta$ in quadrant II, we have $cos\\beta=-\\sqrt {1-sin^2\\beta}=-\\frac{24}{25}$ and $tan\\beta=\\frac{sin\\beta}{cos\\beta}=-\\frac{7}{24}$. Using the Addition Formulas, we have: a. $cos(\\alpha+\\beta)=cos(\\alpha) cos(\\beta)-sin(\\alpha) sin(\\beta)=(\\frac{3}{5})(-\\frac{24}{25})-(\\frac{4}{5})(\\frac{7}{25})=-\\frac{100}{125}=-\\frac{4}{5}$ b. $sin(\\alpha+\\beta)=sin(\\alpha) cos(\\beta)+cos(\\alpha) sin(\\beta)=(\\frac{4}{5})(-\\frac{24}{25})+(\\frac{3}{5})(\\frac{7}{25})=-\\frac{75}{125}=-\\frac{3}{5}$ c. $tan( \\alpha+\\beta)=\\frac{tan(\\alpha)+tan(\\beta)}{1-tan(\\alpha) tan(\\beta)}=\\frac{(\\frac{4}{3})+(-\\frac{7}{24})}{1-(\\frac{4}{3})(-\\frac{7}{24})}=\\frac{3}{4}$", "Hence, $a=7$, $b+c=22$ Furthermore, by the Pythagorean Theorem, we have that $a^2+b^2=c^2\\implies 49+b^2=c^2$ Solving for $c$ in the first equation and plugging in into the second equation... $49+b^2=(22-b)^2\\implies 49+b^2=484-44b+b^2\\implies 44b=435\\implies b=\\frac{435}{44}$ Hence, $c=22-\\frac{435}{44}=\\frac{533}{44}$ Now, we want $\\frac{a+c}{b}$ Plugging in, we find the answer is $\\frac{\\frac{7\\cdot{44}}{44}+\\frac{533}{44}}{\\frac{435}{44}}=\\frac{841}{435}=\\frac{29}{15}$ Hence, the answer is $29+15=\\boxed{044}$ ### Solution 7 We know that $\\sec(x) = \\frac{h}{a}$ and that $\\tan(x) = \\frac{o}{a}$ where $h$, $a$, $o$ represent the hypotenuse, adjacent, and opposite (respectively) to angle $x$ in a right triangle. Thus we have that $\\sec(x) + \\tan(x) = \\frac{h+o}{a}$. We also have that $\\csc(x) + \\cot(x) = \\frac{h}{o} + \\frac{a}{o} = \\frac{h+a}{o}$. Set $\\sec(x) + \\tan(x) = \\alpha$ and csc(x)+cot(x) = $\\beta$. Then, notice that $\\alpha + \\beta = \\frac{h+o}{a} + \\frac{h+a}{o} = \\frac{oh+ah+o^2 + a^2}{oa} = \\frac{h(o+a+h)}{oa}$ ( This is because of the Pythagorean Theorem, recall $o^2 +a^2 = h^2$). But then notice that $\\alpha \\cdot \\beta = \\frac{(o+h)(a+h)}{oa} = \\frac{oa +oh +ha +h^2}{oa} = 1+ \\frac{h(o+a+h)}{oa} = 1+ \\alpha + \\beta$. From the information provided in the question, we can substitute $\\alpha$ for $\\frac{22}{7}$. Thus, $\\frac{22 \\beta}{7}= \\beta + \\frac{29}{7} \\Longrightarrow 22 \\beta = 7 \\beta + 29 \\Longrightarrow 15 \\beta = 29 \\Longrightarrow \\beta = \\frac{29}{15}$. Since, essentially we are asked to find the sum of the numerator and denominator of $\\beta$, we", "$\\sin^2\\beta+\\cos^2\\beta\\equiv1$sin2β+cos2β1 to deduce $\\cos\\beta$cosβ. Hence, $\\left(\\frac{5}{9}\\right)^2+\\cos^2\\beta=1$(59)2+cos2β=1 and so, $\\cos\\beta=\\sqrt{1-\\left(\\frac{5}{9}\\right)^2}=\\frac{2\\sqrt{14}}{9}$cosβ=1(59)2=2149 Now, $\\csc\\beta=\\frac{1}{\\sin\\beta}=\\frac{9}{5},$cscβ=1sinβ=95, $\\sec\\beta=\\frac{1}{\\cos\\beta}=\\frac{9}{2\\sqrt{14}}=\\frac{9\\sqrt{14}}{28},$secβ=1cosβ=9214=91428, $\\tan\\beta=\\frac{\\sin\\beta}{\\cos\\beta}=\\frac{5}{2\\sqrt{14}}=\\frac{5\\sqrt{14}}{28},$tanβ=sinβcosβ=5214=51428, $\\cot\\beta=\\frac{1}{\\tan\\beta}=\\frac{2\\sqrt{14}}{5}$cotβ=1tanβ=2145 #### Worked Examples ##### QUESTION 1 Find the value of $\\cos\\theta$cosθ if $\\sec\\theta=\\frac{8}{3}$secθ=83. ##### QUESTION 2 Find the value of $\\cot\\theta$cotθ if $\\tan\\theta=0.1$tanθ=0.1 . ##### QUESTION 3 If $\\sin\\alpha=\\frac{7}{25}$sinα=725 and $\\cos\\alpha=\\frac{24}{25}$cosα=2425, find: 1. $\\tan\\alpha$tanα 2. $\\cot\\alpha$cotα 3. $\\sec\\alpha$secα 4. $\\csc\\alpha$cscα ### Outcomes #### M8-6 Manipulate trigonometric expressions #### 91575 Apply trigonometric methods in solving problems", "- 15}}{{17}}\\;\\& \\,\\cos \\left( \\theta \\right) = \\sqrt {1 - \\left( {\\frac{15}{17}} \\right)^2 }=\\frac{8}{{17}}$ Moreover, $\\beta\\in II$ so $\\sin \\left( \\beta \\right) = \\frac{4}{5}\\;\\& \\,\\cos \\left( \\beta \\right) = \\frac{{ - 3}}{5}$. • June 1st 2011, 08:41 AM starshine84 So would it then be... sin(\\theta+\\beta) = sin\\theta cos\\beta + cos\\theta sin\\beta sin(\\theta+\\beta) = (-15/17)(-3/4) + (8/17)(4/5) sin(\\theta+\\beta) = (45/68) + (32/85) sin(\\theta+\\beta) = 225/340 + 128/340 sin(\\theta+\\beta) = 353/340 ?? • June 1st 2011, 02:58 PM Prove It Looks good to me :) • June 2nd 2011, 04:05 AM starshine84 That's awesome, thank you so much for all of your help!", "$\\frac{2\\tan30^o}{1+tan^230}$ is (a) $\\frac12$ (b) $\\frac{1}{\\sqrt3}$ (c) $\\frac{\\sqrt{3}}{2}$ (d) $\\sqrt3$ 37. If $\\sin A=\\frac{1}{2}$ then $4\\cos^3A-3\\cos A$ is (a) 1 (b) 0 (c) $\\frac{\\sqrt3}{2}$ (d) $\\frac{1}{\\sqrt{2}}$ 38. The value of $\\frac{3\\tan10^o-\\tan^310}{1-3\\tan^210}$ is (a) $\\frac{1}{\\sqrt3}$ (b) $\\frac{1}{2}$ (c) $\\frac{\\sqrt3}2$ (d) $\\frac{1}{\\sqrt2}$ 39. If $\\alpha$ and $\\beta$ be between 0 and $\\frac{\\pi}{2}$ and if $\\cos(\\alpha+\\beta)=\\frac{12}{13}$ and $\\sin(\\alpha-\\beta)=\\frac{3}{5}$ then $\\sin2\\alpha$ is (a) $\\frac{16}{15}$ (b) 0 (c) $\\frac{56}{65}$ (d) $\\frac{64}{65}$ 40. The value of $\\frac{1}{cosec(-45^o)}$ is (a) $\\frac{-1}{\\sqrt2}$ (b) $\\frac{1}{\\sqrt2}$ (c) $\\sqrt2$ (d) $-\\sqrt2$ 41. If f(x) = x2 - x + 1, then f (x + 1) is (a) x2 (b) x (c) 1 (d) x2 + x + 1 42. For $f\\left( x \\right) =\\begin{cases} { x }^{ 2 }-4x\\quad ifx\\ge 2 \\\\ x+2\\quad ifx<2 \\end{cases}$ then f(0) is (a) 2 (b) 5 (c) -1 (d) 0 43. The graph of the line y = 3 is (a) Parallel to x-axis (b) Parallel to y-axis (c) Passing through the origin (d) Perpendicular to x-axis 44. The graph of y = 2x2 is passing through (a) (0,0) (b) (2,1) (c) (2,0) (d) (0,2) 45. Which one of the following functions has the property f (x) = $f\\left( \\frac { 1 }{ x } \\right)$ (a) $f\\left( x \\right) =\\frac { { x }^{ 2 }-1 }{ x }$ (b) $f\\left( x", "= 162 6r3 = 162 r3 = $$\\frac{162}{6}$$ = 27 r = 3 Question 18. The sum of three consecutive numbers is 243. Find the three numbers. x + x + 1 + x + 2 = 243 3x + 3 = 2433x = 240 x = 80 80, 81, 82 Question 19. If a and P are the roots of the equation 2x2 + 4x + 1 = 0 find the value of $$\\frac{1}{\\alpha}+\\frac{1}{\\beta}$$ α + β = –$$\\frac{4}{2}$$ = -2 $$\\frac{1}{\\alpha}+\\frac{1}{\\beta}=\\frac{\\alpha+\\beta}{\\alpha \\beta}$$ αβ = $$\\frac{c}{a}=\\frac{1}{2}=\\frac{-2}{\\frac{1}{2}}$$ = -4 Question 20. Solve 7x + 3 < 5x + 9, x ∈ R and represent on the number line. Question 21. Find the future value of an annuity of ₹ 5000 at 12% p.a for 6 years. A = Rs.5,000 r = 0.12 n = 6 F = $$\\frac{A\\left[(1+r)^{n}-1\\right]}{r}=\\frac{5000\\left[(1+0.12)^{6}-1\\right]}{0.12}=\\frac{4869.11}{0.12}$$ = 40,575.9 Question 22. Prove that (1 + cotA)2 + (1 – cotA)2 = 2 cosec2A Question 23. Find the value of cos 60° -sin 30° – cot3 45° $$\\frac{1}{2}-\\frac{1}{2}$$ – (1)3 = -1 Question 24. Find the distance of the centroid of the triangle formed by the points (7, 1) (1, 5) and (1, 6) from origin. G = $$\\left[\\frac{7+1+1}{3}, \\frac{1+5+6}{3}\\right]$$ = (3, 4) O = (0, 0) OG = $$\\sqrt{(3)^{2}+(4)^{2}}=\\sqrt{9+16}=\\sqrt{25}$$ = 5 Question 25. Find the equation of", "# If tanα= Question: If $\\tan \\alpha=\\frac{1}{7}, \\tan \\beta=\\frac{1}{3}$, then $\\cos 2 \\alpha$ is equal to (a) $\\sin 2 \\beta$ (b) $\\sin 4 \\beta$ (c) $\\sin 3 \\beta$ (d) $\\cos 2 \\beta$ Solution: It is given that $\\tan \\alpha=\\frac{1}{7}$ and $\\tan \\beta=\\frac{1}{3}$. Now, $\\tan 2 \\beta=\\frac{2 \\tan \\beta}{1-\\tan ^{2} \\beta}$ $=\\frac{2 \\times \\frac{1}{3}}{1-\\frac{1}{9}}$ $=\\frac{\\frac{2}{3}}{\\frac{8}{9}}$ $=\\frac{3}{4}$ $\\therefore \\tan (\\alpha+2 \\beta)=\\frac{\\tan \\alpha+\\tan 2 \\beta}{1-\\tan \\alpha \\tan 2 \\beta}$ $=\\frac{\\frac{1}{7}+\\frac{3}{4}}{1-\\frac{1}{7} \\times \\frac{3}{4}}$ $=\\frac{\\frac{25}{28}}{\\frac{25}{28}}$ $=1$ $\\tan (\\alpha+2 \\beta)=1=\\tan \\frac{\\pi}{4}$ $\\Rightarrow \\alpha+2 \\beta=\\frac{\\pi}{4}$ $\\Rightarrow \\alpha=\\frac{\\pi}{4}-2 \\beta$ $\\Rightarrow 2 \\alpha=\\frac{\\pi}{2}-4 \\beta$ $\\Rightarrow \\cos 2 \\alpha=\\cos \\left(\\frac{\\pi}{2}-4 \\beta\\right)=\\sin 4 \\beta$ $\\therefore \\cos 2 \\alpha=\\sin 4 \\beta$ Hence, the correct answer is option $B$.", "## Precalculus (6th Edition) Blitzer a. $-\\frac{4}{5}$ b. $\\frac{5}{13}$ c. $\\frac{33}{65}$ d. $-\\frac{56}{65}$ a. Knowing $\\alpha$ is in quadrant II, we have $cos\\alpha\\lt0$ and $cos\\alpha=-\\sqrt {1-sin^2\\alpha}=-\\frac{4}{5}$ b. Knowing $\\beta$ is in quadrant II, we have $sin\\beta\\gt0$ and $sin\\beta=-\\sqrt {1-cos^2\\beta}=\\frac{5}{13}$ c. Using the addition formula, we have $cos(\\alpha+\\beta)=(-\\frac{4}{5})(-\\frac{12}{13})-(\\frac{3}{5})(\\frac{5}{13})=\\frac{33}{65}$ d. Using the addition formula, we have $sin(\\alpha+\\beta)=(\\frac{3}{5})(-\\frac{12}{13})+(-\\frac{4}{5})(\\frac{5}{13})=-\\frac{56}{65}$", "YZ = ? p2 + b2 = h2 or , b2 = h2 - p2 b = $$\\sqrt{h^{2} - p^{2}}$$ =$$\\sqrt{65^{2} - 16^{2}}$$ = $$\\sqrt{3969}$$ = 63. Here , sinx = $$\\frac{p}{h}$$ = $$\\frac{16}{65}$$ = $$\\frac{16}{65}$$ cosx = $$\\frac{b}{h}$$ = $$\\frac{63}{65}$$ = $$\\frac{63}{65}$$ and tanx = $$\\frac{p}{b}$$ = $$\\frac{16}{63}$$ = $$\\frac{16}{63}$$ = ( $$\\frac{1}{2}$$ )2 + ( $$\\frac{1}{\\sqrt {2}}$$ )2 + ( $$\\frac{\\sqrt{3}}{2}$$ )2 = $$\\frac{1}{4}$$+ $$\\frac{1}{2}$$ +$$\\frac{3}{4}$$ = $$\\frac{1+2+3}{4}$$ = $$\\frac{3}{2}$$ = 1$$\\frac{1}{2}$$ cos$$\\beta$$ = $$\\sqrt{3}$$ $$\\times$$ $$\\frac{1}{2}$$ or ,cos$$\\beta$$ = $$\\frac{\\sqrt{3}}{2}$$ or ,cos$$\\beta$$ = cos 30o $$\\therefore$$ $$\\beta$$ = 30o Ans. or , sin2$$\\beta$$ = $$\\frac{1}{4}$$ or , sin2 $$\\beta$$ =( $$\\frac{1}{2}$$ )2 or , sin2 $$\\beta$$ = (sin30o)2 or ,sin $$\\beta$$ = sin30o $$\\therefore$$ $$\\beta$$ = 30o Ans. 0% • ### In a right-angled triangle, sin θ=(frac{1}{2}),find the numerical value of cos θ. (frac{sqrt 4}{2}) (frac{sqrt 3}{2}) (frac{sqrt 5}{4}) (frac{sqrt 6}{3}) • ### Express sec A in terms of tan A. `(sqrt{1+sin^2 A}) (sqrt{3-sin^2 A}) (sqrt{1+tan^2 A}) (sqrt{4+sin^2 A}) • ### Express cosA in terms of sin A. (sqrt{1-cos^2 A}) (sqrt{1-cos A}) (sqrt{1-sin^2 A}) (sqrt{1-sin A}) (frac{1}{1}) (frac{4}{2}) (frac{3}{2}) (frac{1}{3}) (frac{3}{2}) (frac{4}{2}) (frac{2}{1}) (frac{1}{2}) (frac{2}{1}) (frac{3}{2}) (frac{1}{2}) (frac{2}{2}) (frac{2}{1}) (frac{1}{3}) (frac{1}{2}) (frac{1}{4}) three two zero one 5 2 19 10 zero one three two zero five nine two 3 2 10 6 1 3 4", "Now, $\\frac{\\cot(\\gamma)}{\\cot(\\beta)+\\cot(\\alpha)}=\\frac{\\frac{994c}{b\\sin a}}{\\frac{a^2-994c^2+b^2-994c^2}{bc\\sin(a)}}$. After using $a^2+b^2=1989c^2$, we get $\\frac{994c*bc\\sin a}{c^2b\\sin a}=\\boxed{994}$.", "suspicion that ETS only wants you to remember Pythagorean triple 3:4:5 wich translates into: beta = 0.8 => gamma = 5/3 = 1.66... beta = 0.6 => gamma = 5/4 = 1.25 and thus, dealing with similar problem one should first check if one of these is enough to identify the correct answer quickly the first one is sufficient in this case gamma = 938/494 > 850/500 > 1.7 > 1.66.. => beta > 0.8 oliTUTilo 2012-11-01 16:43:56 If we algebraically solve for v (or $\\beta$) instead of $\\gamma$, as Yosun does, it seems easier to approximate the answer. $E = \\gamma m \\Rightarrow{(1-\\beta^2)}^{-\\frac{1}{2}} = \\frac{E}{m} \\Rightarrow \\beta = {(1-{(\\frac{m}{E})}^2)}^{-\\frac{1}{2}}$ $\\frac{m}{E}$ is approximately $\\frac{1}{2}$, so $\\beta$ is approximately ${(\\frac{3}{4})}^{-\\frac{1}{2}}$, which is 1.7 something divided by 2, or approximately 0.85. v is $\\beta c$, so badabingbadaboom. In general, I'd algebraically solve for the variable in question first, gaining the greatest transparency, but redmomatt's method is even a bit faster here. redmomatt 2011-10-14 08:59:20 Why can't we do this....? $E = \\gamma m_0 c^2$ $1000 = \\gamma 500$ $.5^2 = 1 - \\frac{v^2}{c^2}$ $sqrt{.75} c = v = .85 c$ $\\Rightarrow$ (E) checkyoself 2011-10-05 10:46:45 Approximating $\\gamma=2$ is fine. You end up with $\\beta=\\sqrt{0.75}c$ which you know must be greater than 0.75c so the answer must be E. archard 2010-05-25", "}}{{1 - i\\tan (\\alpha )}} = \\frac{{\\left( {1 + i\\tan (\\alpha )} \\right)^5 }}{{\\sec ^2 (\\alpha )}}$ $\\frac{{\\left( {1 + i\\tan (\\alpha )} \\right)^5 }}{{\\sec ^2 (\\alpha )}} = \\frac{{\\frac{{\\left( {\\cos (\\alpha ) + i\\sin (\\alpha )} \\right)^5 }}{{\\cos ^5 (\\alpha )}}}}{{\\sec ^2 (\\alpha )}} = \\frac{{\\cos (5\\alpha ) + i\\sin (5\\alpha )}}{{\\cos ^3 (\\alpha )}}$ 4. This is a postscript, resulting from a PM. It seems that the problem is actually $\\left[ {\\frac{{1 + \\tan (\\alpha )}}{{1 - \\tan (\\alpha )}}} \\right]^4$. In which case; $\\begin{array}{rcl} \\left[ {\\frac{{1 + i\\tan (\\alpha )}}{{1 - i\\tan (\\alpha )}}} \\right]^4 & = & \\left[ {\\frac{{\\cos (\\alpha ) + i\\sin (\\alpha )}}{{\\cos (\\alpha ) - i\\sin (\\alpha )}}} \\right]^4 \\\\ & = & \\left[ {\\cos (2\\alpha ) + i\\sin (2\\alpha )} \\right]^4 \\\\ & = & \\cos (8\\alpha ) + i\\sin (8\\alpha ) \\\\ \\end{array}$ 5. Thanks", "- \\sqrt 3}{4}}$$ or, sin 15° =± $$\\sqrt {\\frac {4 - 2\\sqrt 3}{8}}$$ or, sin 15° =± $$\\sqrt {\\frac {3 - 2\\sqrt 3 + 1}{8}}$$ or, sin 15° =± $$\\sqrt {\\frac {(\\sqrt 3 -1)^2}{8}}$$ ∴ sin 15° =± $$\\frac {\\sqrt 3 -1}{2\\sqrt 2}$$ Ans L.H.S. =$$\\frac {2 sin\\beta + sin 2\\beta}{2 sin\\beta - sin 2\\beta}$$ =$$\\frac {2 sin\\beta + 2 sin\\beta cos\\beta}{2 sin\\beta - 2 sin\\beta cos\\beta}$$ = $$\\frac {2 sin\\beta (1 + cos\\beta)}{2 sin\\beta (1 - cos\\beta)}$$ = $$\\frac {2 cos^2\\frac \\beta2}{2 sin^2\\frac \\beta2}$$ [$$\\because$$ 1 + cos$$\\theta$$ = 2 cos2$$\\frac \\theta2$$,1 - cos$$\\theta$$ = 2 sin2$$\\frac \\theta2$$] = cot2$$\\frac \\beta2$$ Hence, L.H.S. = R.H.S. Proved Here, sin $$\\frac \\theta3$$ = $$\\frac 45$$ sin$$\\theta$$ = 3 sin $$\\frac \\theta3$$ - 4 sin3$$\\frac \\theta3$$ = 3× $$\\frac 45$$ - 4× ($$\\frac 45$$)3 = $$\\frac {12}5$$ - $$\\frac {256}{125}$$ = $$\\frac {300 - 256}{125}$$ = $$\\frac {44}{125}$$ Ans Here, cos$$\\frac \\theta3$$ = $$\\frac 35$$ cos $$\\theta$$ = 4 cos3$$\\frac \\theta3$$ - 3 cos$$\\frac \\theta3$$ = 4× ($$\\frac 35$$)3- 3× $$\\frac 35$$ = 4× $$\\frac {27}{125}$$ - $$\\frac 95$$ = $$\\frac {108 - 225}{125}$$ = -$$\\frac {117}{125}$$ Ans Here, sin$$\\frac \\theta2$$ = $$\\frac 35$$ cos$$\\frac \\theta2$$ = $$\\sqrt {1 - sin^2\\frac \\theta2}$$ = $$\\sqrt {1 - (\\frac 35)^2}$$ = $$\\sqrt {1-\\frac 9{25}}$$ = $$\\sqrt {\\frac {25 - 9}{25}}$$ = $$\\sqrt", "# Verify that 5, −2 and Question: Verify that $5,-2$ and $\\frac{1}{3}$ are the zeros of the cubic polynomial $p(x)=3 x^{3}-10 x^{2}-27 x+10$ and verify the relation between its zeros and coefficients. Solution: $p(x)=\\left(3 x^{3}-10 x^{2}-27 x+10\\right)$ $p(5)=\\left(3 \\times 5^{3}-10 \\times 5^{2}-27 \\times 5+10\\right)=(375-250-135+10)=0$ $p(-2)=\\left[3 \\times\\left(-2^{3}\\right)-10 \\times\\left(-2^{2}\\right)-27 \\times(-2)+10\\right]=(-24-40+54+10)=0$ $\\left.p\\left(\\frac{1}{3}\\right)=\\left\\{3 \\times\\left(\\frac{1}{3}\\right)^{3}\\right)^{3}-10 \\times\\left(\\frac{1}{3}\\right)^{2}-27 \\times \\frac{1}{3}+10\\right\\}=\\left(3 \\times \\frac{1}{27}-10 \\times \\frac{1}{9}-9+10\\right)$ $=\\left(\\frac{1}{9}-\\frac{10}{9}+1\\right)=\\left(\\frac{1-10+9}{9}\\right)=\\left(\\frac{0}{9}\\right)=0$ $\\therefore 5,-2$ and $\\frac{1}{3}$ are the zeroes of $p(x)$. Let $\\alpha=5, \\beta=-2$ and $\\gamma=\\frac{1}{3}$. Then we have: $(\\alpha+\\beta+\\gamma)=\\left(5-2+\\frac{1}{3}\\right)=\\left(\\frac{10}{3}\\right)=\\frac{-\\left(\\text { coefficient of } x^{2}\\right)}{\\left(\\text { coefficient of } x^{3}\\right)}$ $(\\alpha \\beta+\\beta \\gamma+\\gamma \\alpha)=\\left(-10-\\frac{2}{3}+\\frac{5}{3}\\right)=\\frac{-27}{3}=\\frac{\\text { coefficient of } x}{\\text { coefficient of } x^{3}}$ $\\alpha \\beta \\gamma=\\left\\{5 \\times(-2) \\times \\frac{1}{3}\\right\\}=\\frac{-10}{3}=\\frac{-(\\text { constant term })}{\\left(\\text { coefficient of } x^{3}\\right)}$", "Let $t$ be the magnitude of $\\left(\\frac{11}{50} - \\frac{i}{25}\\right)$ and $\\beta$ be its angle. Magnitude of $\\left(\\frac{11}{50} - \\frac{i}{25}\\right) = \\sqrt{{\\left(\\frac{11}{50}\\right)}^{2} + {\\left(- \\frac{1}{25}\\right)}^{2}} = \\sqrt{\\frac{121}{2500} + \\frac{1}{625}} = \\sqrt{\\frac{121 + 4}{2500}} = \\sqrt{\\frac{125}{2500}} = \\sqrt{\\frac{1}{20}} = \\frac{1}{2 \\sqrt{5}} = t$ Angle of $\\left(\\frac{11}{50} - \\frac{i}{25}\\right) = T a {n}^{-} 1 \\left(\\frac{- \\frac{1}{25}}{\\frac{11}{50}}\\right) = {\\tan}^{-} 1 \\left(- \\frac{2}{11}\\right) = \\beta$ $\\implies \\left(\\frac{11}{50} - \\frac{i}{25}\\right) = t \\left(C o s \\beta + i \\sin \\beta\\right)$ $\\implies \\left(\\frac{11}{50} - \\frac{i}{25}\\right) = \\frac{1}{2 \\sqrt{5}} \\left(C o s \\left({\\tan}^{-} 1 \\left(- \\frac{2}{11}\\right)\\right) + i \\sin \\left({\\tan}^{-} 1 \\left(- \\frac{2}{11}\\right)\\right)\\right)$.", "{\\cot{\\gamma}}{\\cot{\\alpha} + \\cot{\\beta}} &= \\frac {\\cot{\\gamma}\\sin{\\alpha}\\sin{\\beta}}{\\sin{\\gamma}} = \\frac {\\cos{\\gamma}\\sin{\\alpha}\\sin{\\beta}}{\\sin^2{\\gamma}} = \\frac {ab}{c^2}\\cos{\\gamma} = \\frac {ab}{c^2} \\cdot \\frac {994c^2}{ab}\\\\ &= \\boxed{994}\\end{align*} ### Solution 5 Use Law of cosines to give us $c^2=a^2+b^2-2ab\\cos(\\gamma)$ or therefore $\\cos(\\gamma)=\\frac{994c^2}{ab}$. Next, we are going to put all the sin's in term of $\\sin(a)$. We get $\\sin(\\gamma)=\\frac{c\\sin(a)}{a}$. Therefore, we get $\\cot(\\gamma)=\\frac{994c}{b\\sin a}$. Next, use Law of Cosines to give us $b^2=a^2+c^2-2ac\\cos(\\beta)$. Therefore, $\\cos(\\beta)=\\frac{a^2-994c^2}{ac}$. Also, $\\sin(\\beta)=\\frac{b\\sin(a)}{a}$. Hence, $\\cot(\\beta)=\\frac{a^2-994c^2}{bc\\sin(a)}$. Lastly, $\\cos(\\alpha)=\\frac{b^2-994c^2}{bc}$. Therefore, we get $\\cot(\\alpha)=\\frac{b^2-994c^2}{bc\\sin(a)}$. Now, $\\frac{\\cot(\\gamma)}{\\cot(\\beta)+\\cot(\\alpha)}=\\frac{\\frac{994c}{b\\sin a}}{\\frac{a^2-994c^2+b^2-994c^2}{bc\\sin(a)}}$. After using $a^2+b^2=1989c^2$, we get $\\frac{994c*bc\\sin a}{c^2b\\sin a}=\\boxed{994}$. ### Solution 6 Let $\\gamma$ be $(180-\\alpha-\\beta)$ $\\frac{\\cot \\gamma}{\\cot \\alpha+\\cot \\beta} = \\frac{\\frac{-\\tan \\alpha \\tan \\beta}{\\tan(\\alpha+\\beta)}}{\\tan \\alpha + \\tan \\beta} = \\frac{(\\tan \\alpha \\tan \\beta)^2-\\tan \\alpha \\tan \\beta}{\\tan^2 \\alpha + 2\\tan \\alpha \\tan \\beta +\\tan^2 \\beta}$ WLOG, assume that $a$ and $c$ are legs of right triangle $abc$ with $\\beta = 90^o$ and $c=1$ By Pythagorean theorem, we have $b^2=a^2+1$, and the given $a^2+b^2=1989$. Solving the equations gives us $a=\\sqrt{994}$ and $b=\\sqrt{995}$. We see that $\\tan \\beta = \\infty$, and $\\tan \\alpha = \\sqrt{994}$. We see that our derived equation equals to $\\tan^2 \\alpha$ as $\\tan \\beta$ approaches infinity. Evaluating $\\tan^2 \\alpha$, we get $\\boxed{994}$.", "ok i got$\\frac{ 2\\sqrt{46} }{ 21 }$ 44. jim_thompson5910 me too 45. anonymous great does the attachement you sent me have the equation for the third one as well 46. jim_thompson5910 yes, look under \"Sum and Difference Formulas\" on page 2 47. anonymous sin(a)cos(b)+cos(a)sin(b) 48. jim_thompson5910 yes 49. anonymous ok again I am having problems plugging the equations in here from my side, can you show me what you got when you plugged each of the values in 50. jim_thompson5910 The values of sin(alpha), sin(beta), cos(alpha), cos(beta) were determined previously. Plug them in to get $\\Large \\sin(\\beta + \\alpha) = \\sin(\\beta)\\cos(\\alpha) + \\cos(\\beta)\\sin(\\alpha)$ $\\Large \\sin(\\beta + \\alpha) = \\left(\\frac{\\sqrt{5}}{5}\\right)\\left(\\frac{\\sqrt{2}}{5}\\right) + \\left(\\frac{-2\\sqrt{5}}{5}\\right)\\left(\\frac{-\\sqrt{23}}{5}\\right)$ $\\Large \\sin(\\beta + \\alpha) =??$ 51. anonymous I am having a hard time typing it in 52. anonymous one second 53. anonymous $-\\frac{ -2\\sqrt{23}+\\sqrt{2}}{ 5\\sqrt{5} }$ 54. jim_thompson5910 $\\Large \\sin(\\beta + \\alpha) = \\left(\\frac{\\sqrt{5}}{5}\\right)\\left(\\frac{\\sqrt{2}}{5}\\right) + \\left(\\frac{-2\\sqrt{5}}{5}\\right)\\left(\\frac{-\\sqrt{23}}{5}\\right)$ $\\Large \\sin(\\beta + \\alpha) = \\frac{\\sqrt{5*2}}{5*5} + \\frac{-2\\sqrt{5*23}}{5*5}$ $\\Large \\sin(\\beta + \\alpha) = \\frac{\\sqrt{10}}{25} + \\frac{-2\\sqrt{115}}{25}$ $\\Large \\sin(\\beta + \\alpha) = \\frac{\\sqrt{10}-2\\sqrt{115}}{25}$ 55. anonymous ok I see what I did wrong, I messed up on the second step 56. jim_thompson5910 oops I should have a positive sign 57. anonymous where in front of the 2 58. jim_thompson5910 $\\Large \\sin(\\beta + \\alpha) = \\left(\\frac{\\sqrt{5}}{5}\\right)\\left(\\frac{\\sqrt{2}}{5}\\right) + \\left(\\frac{-2\\sqrt{5}}{5}\\right)\\left(\\frac{-\\sqrt{23}}{5}\\right)$ $\\Large \\sin(\\beta + \\alpha) = \\frac{\\sqrt{5*2}}{5*5}", "first diagram above, $$\\sin \\frac{\\alpha}{2} = \\frac{1}{r}$$ $$\\sin \\frac{\\beta}{2} = \\frac{1.5}{r}$$ $$\\sin(\\frac{\\alpha}{2}+\\frac{\\beta}{2})=\\frac{2}{r}$$ by the Pythagorean trig identities, $$\\cos\\frac{\\alpha}{2}=\\sqrt{1-\\frac{1}{r^2}}$$ $$\\cos\\frac{\\beta}{2}=\\sqrt{1-\\frac{2.25}{r^2}}$$ so by the composite sine identity $$\\frac{2}{r}=\\frac{1}{r}\\sqrt{1-\\frac{2.25}{r^2}}+\\frac{1.5}{r}\\sqrt{1-\\frac{1}{r^2}}$$ multiply both sides by $2r$, then subtract $\\sqrt{4-\\frac{9}{r^2}}$ from both sides squaring both sides, we get $$16 - 8\\sqrt{4-\\frac{9}{r^2}} + 4 - \\frac{9}{r^2}=9 - \\frac{9}{r^2}$$ $$\\Longrightarrow 16+4=9+8\\sqrt{4-\\frac{9}{r^2}}\\Longrightarrow\\frac{11}{8}=\\sqrt{4-\\frac{9}{r^2}}\\Longrightarrow\\frac{121}{64}=4-\\frac{9}{r^2}$$ $$\\Longrightarrow\\frac{(256-121)r^2}{64}=9\\Longrightarrow r^2= \\frac{64}{15}$$ plugging this back in, $$\\cos^2(\\frac{\\alpha}{2})=1-\\frac{15}{64}=\\frac{49}{64}$$ so $$\\cos(\\alpha)=2(\\frac{49}{64})-1=\\frac{34}{64}=\\frac{17}{32}$$ and the answer is $17+32=\\boxed{049}$", "Q 14 | Page 44 $\\sin^2 \\left( \\frac{\\pi}{18} \\right) + \\sin^2 \\left( \\frac{\\pi}{9} \\right) + \\sin^2 \\left( \\frac{7\\pi}{18} \\right) + \\sin^2 \\left( \\frac{4\\pi}{9} \\right) =$ • 1 • 2 • 4 • none of these. Q 15 | Page 44 If $5 \\sin \\alpha = 3 \\sin \\left( \\alpha + 2 \\beta \\right) \\neq 0$ , then $\\tan \\left( \\alpha + \\beta \\right)$ is equal to • $2 \\tan \\beta$ • $3 \\tan \\beta$ • $4 \\tan \\beta$ • $6 \\tan \\beta$ Q 16 | Page 44 $2 \\text{ cos } x - \\ cos 3x - \\cos 5x - 16 \\cos^3 x \\sin^2 x$ • 2 • 1 • 0 • -1 Q 17 | Page 44 If $A = 2 \\sin^2 x - \\cos 2x$ , then A lies in the interval • $\\left[ - 1, 3 \\right]$ • $\\left[ 1, 2 \\right]$ • $\\left[ - 2, 4 \\right]$ • none of these Q 18 | Page 44 The value of $\\frac{\\cos 3x}{2 \\cos 2x - 1}$ is equal to • cos x • sin x • tan x • none of these Q 19 | Page 44 If $\\tan \\left( \\pi/4 + x \\right) + \\tan \\left( \\pi/4 - x \\right) = \\lambda \\sec 2x, \\text{ then }$ • 3 • 4 • 1 •" ]

[ "+0 # help +1 52 1 for a vector v, let r be the reflection of v over the line $$x = t \\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix}$$. There exists a 2 X 2 matrix R such that r = Rv for all2-dimensional vectors v. Find R. Mar 29, 2020 The matrix is $$\\begin{pmatrix} 1 & 2 \\\\ -2 & 4 \\end{pmatrix}$$", "Reflection with Respect to x-Axis Let F:\\R^2\\to \\R^2 be the function that maps each vector in \\R^2 to its reflection with respect to x-axis. Determine the formula for the function F and prove that F is a linear transformation. Solution 1. Let \\begin{bmatrix} x \\\\ y […] • Eigenvalues and Eigenvectors of The Cross Product Linear Transformation We fix a nonzero vector \\mathbf{a} in \\R^3 and define a map T:\\R^3\\to \\R^3 by \\[T(\\mathbf{v})=\\mathbf{a}\\times \\mathbf{v}$ for all $\\mathbf{v}\\in \\R^3$. Here the right-hand side is the cross product of $\\mathbf{a}$ and $\\mathbf{v}$. (a) Prove that $T:\\R^3\\to \\R^3$ is […]", "the vector $$\\bf{u}$$ over a plane. This can be the $$xy$$, $$xz$$, or $$yz$$ plane. The matrix is similar to the identity matrix, with the exception that there is a sign change for the appropriate element. The reflect matrix in the $$xy$$ plane is: $\\sigma \\bf{u} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & -1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x \\\\ y \\\\ -z \\end{pmatrix} \\label{9.2}$ Notice that the element for the dimension being reflected is the on that is negative. In the above case, since $$z$$ is being reflected over the $$xy$$ plane, the $$z$$ element in the matrix is negative. If we were to reflect over the $$xz$$ plane instead, the $$y$$ element would be the one that is negative: $\\sigma \\bf{u} = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} = \\begin{pmatrix} x \\\\ -y \\\\ z \\end{pmatrix}$ We can also talk about reflections in 2 dimensions. The simplest example of a reflection matrix corresponds to reflecting the vector $$\\begin{pmatrix} x, y \\end{pmatrix}$$ in either the $$x$$ or $$y$$ axes. Reflection in the $$x$$ axis maps $$y$$ to $$-y$$, while reflection", "Is a reflection matrix a rotation matrix? QuestionsCategory: QuestionsIs a reflection matrix a rotation matrix? I wonder if a reflection matrix is a rotation as well? Nasreenkanwal replied 3 years ago How to sum one matrix into other matrix? [Unrelated link removed by moderator, see posting rules.] Tomas Akenine-Möller Staff answered 3 years ago A reflection matrix is usually not considered to be a rotation matrix, since rotation matrices have $\\det\\mathbf M=1$ and a reflection matrix can have $\\det\\mathbf M=-1$. An example of a reflection matrix is \\mathbf M = \\begin{pmatrix} -1 & 0 \\\\ 0 & 1 \\end{pmatrix}, whose determinant is $-1$. Interestingly, the product of two reflection matrices become a rotation matrix. Assume we have another reflection matrix, i.e., \\mathbf N = \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} Their product becomes \\mathbf M\\mathbf N = \\begin{pmatrix} -1 & 0 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix} = \\begin{pmatrix} -1 & 0 \\\\ 0 & -1 \\end{pmatrix}, and as can be seen $\\det(\\mathbf M\\mathbf N)=1$.", "# tfg.math.vector.reflect Computes the reflection direction for an incident vector. For an incident vector $$\\mathbf{v}$$ and normal $$\\mathbf{n}$$ this function computes the reflected vector as $$\\mathbf{r} = \\mathbf{v} - 2(\\mathbf{n}^T\\mathbf{v})\\mathbf{n}$$. #### Note: In the following, A1 to An are optional batch dimensions, which should be broadcast compatible. vector A tensor of shape [A1, ..., Ai, ..., An], where the dimension i = axis represents a vector. normal A tensor of shape [A1, ..., Ai, ..., An], where the dimension i = axis represents a normal around which the vector needs to be reflected. The normal vector needs to be normalized. axis The dimension along which to compute the reflection. name A name for this op which defaults to \"vector_reflect\". A tensor of shape [A1, ..., Ai, ..., An], where the dimension i = axis represents a reflected vector.", "a vector, $x \\text{,}$ and reflect it with respect to the mirror defined by $u \\text{,}$ and you then reflect the result with respect to the same mirror, you should get the original vector $x$ back. Hence, the matrix that represents the reflection should be its own inverse. ###### Homework2.2.5.6. Let $M: \\R^3 \\rightarrow \\R^3$ be defined by $M(x ) = (I - 2 u u^T) x \\text{,}$ where $\\| u \\|_2 = 1 \\text{.}$ Show that the matrix that represents it is unitary (or, rather, orthogonal since it is in $\\R^{3 \\times 3}$). Solution Pushing through the math we find that \\begin{equation*} \\begin{array}{l} (I - 2 u u^T)^T ( I - 2 u u^T ) \\\\ ~~~=~~~~ \\lt ( A + B )^T = A^T + B^T \\gt \\\\ ( I^T - ( 2 u u^T )^T ) ( I - 2 u u^T ) \\\\ ~~~=~~~~ \\lt ( \\alpha A B^T )^T = \\alpha B A^T \\gt \\\\ ( I - 2 u u^T ) ( I - 2 u u^T ) \\\\ ~~~=~~~~ \\lt \\mbox{ distributivity } \\gt \\\\ ( I - 2 u u^T ) - ( I - 2 u u^T ) ( 2 u u^T ) \\\\ ~~~=~~~~ \\lt \\mbox{ distributivity } \\gt \\\\ I - 2 u u^T - 2 u", "Reflection xyz plane I have this matrix: $$A=\\dfrac17\\begin{bmatrix}6 & 3 &-2\\\\3 & -2 & 6 \\\\ -2 & 6 & 3\\end{bmatrix}$$ We have that $$A^TA=I$$. Since $$\\det A=-1$$, its is a reflection matrix. My question is what is the reflecting plane? A key observation is that the normal vector $$n$$ of the reflecting plane gets sent to $$-n$$ by $$A$$. We find $$An = -n = (-I)n$$, where $$I$$ is the identity matrix, hence $$(A + I)n = 0$$. Solving this for $$n$$ we find $$n = (1, -3, 2)$$ as a solution (up to a scalar multiple), so the reflecting plane is given by $$x - 3y + 2z = 0$$.", "# Homework Help: Matrix of reflection 1. Feb 17, 2016 ### DanZ123 1. The problem statement, all variables and given/known data Let u1,u2,u3 be an orthonormal basis for R3 and consider M as the plane with equation x1+2x2-2x3=0. Find the matrix of orthogonal reflection in that plane with respect to the given basis. 2. Relevant equations 3. The attempt at a solution In previous exercises , I had a matrix A given and was asked to find the equation of the plane that the matrix was projected or reflected on. To do that I solved the equation (A-I)x=0 . (The nullspace/kernel minus the identity matrix) .. I was thinking that maybe to solve this current exercise, I could maybe use the method from the previous exercises but use it backwards ... and find the matrix A? But im not sure if im thinking right, or how to attack the problem... Would appreciate help...thanks. 2. Feb 17, 2016 ### geoffrey159 The direction of your reflection is $M^\\perp =\\mathbb{R} (1,2,-2)$. Find an orthonormal basis of $\\mathbb{R}^3$ starting with a unit vector of $M^\\perp$. With respect to that basis, the matrix of reflection is $\\begin{pmatrix} -1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$. What do you need to complete the exercise", "of the reflection matrix. To reflect a vector $v$, one way is to take the line through the endpoint of $v$, with direction equal to a normal vector of the plane, and find the point on the plane where the line intersects the plane. That point is the midpoint of the line segment between the endpoint of $v$ and the endpoint of its reflection. Alternatively, you can project $v$ onto the plane and find the orthogonal complement, $q$ say. Then the reflection of $v$ is just $v-2q$. If you want to construct the matrix using eigenvectors, you need all three. You’ve got two eigenvectors (of $1$) that span the plane, but you also need an eigenvector of $-1$, which will be normal to the plane. You can read a normal to the plane directly from its equation, which can be written in vector form as $\\mathbf n\\cdot\\mathbf x=0$. This gives us $(1,1,1)^T$. In the eigenbasis, the matrix of a reflection is $\\operatorname{diag}(1,1,-1)$, and a change of basis will get us the matrix relative to the standard basis: $$\\begin{bmatrix}-1&_-1&1\\\\1&0&1\\\\0&1&1\\end{bmatrix}\\begin{bmatrix}1&0&0\\\\0&1&0\\\\0&0&-1\\end{bmatrix}\\begin{bmatrix}-1&_-1&1\\\\1&0&1\\\\0&1&1\\end{bmatrix}^{-1}.$$ I’ll leave multiplying this out to you. What you had, however, is almost the way to construct the reflection using a normal to the plane. This requires only a single matrix multiplication (and much simpler one than the previous construction,", "If you find any mistakes, please make a comment! Thank you. ## Solution to Linear Algebra Done Wrong Exercise 1.3.2 Let a linear transformation in $\\mathbb R^2$ be the reflection in the line $x_1=x_2$. Find its matrix. Solution: Let us think $x_1=x$ and $x_2=y$. In the $(x,y)$-coordinate plane, the graph of the inverse function of a function $f(x)$ is obtained by reflecting the graph of $f(x)$ in the line $y=x$. Also note the the inverse function interchanges $x$ and $y$. Denote the linear transformation in $\\mathbb R^2$ corresponding to the reflection in the line $x_1=x_2$ by $T$. With the above observation, we see that $$T(x_1,x_2)^T=(x_2,x_1)^T.$$Therefore, we have $$T\\begin{pmatrix}x_1\\\\ x_2\\end{pmatrix}=x_1\\begin{pmatrix}0\\\\ 1\\end{pmatrix}+x_2\\begin{pmatrix}1\\\\ 0\\end{pmatrix}.$$Hence the matrix is given by $\\begin{pmatrix}0 & 1\\\\ 1 & 0\\end{pmatrix}$.", "# Tagged: reflection ## Problem 597 Let $F:\\R^2\\to \\R^2$ be the function that maps each vector in $\\R^2$ to its reflection with respect to $x$-axis. Determine the formula for the function $F$ and prove that $F$ is a linear transformation. ## Problem 498 Let $T:\\R^2 \\to \\R^2$ be a linear transformation of the $2$-dimensional vector space $\\R^2$ (the $x$-$y$-plane) to itself which is the reflection across a line $y=mx$ for some $m\\in \\R$. Then find the matrix representation of the linear transformation $T$ with respect to the standard basis $B=\\{\\mathbf{e}_1, \\mathbf{e}_2\\}$ of $\\R^2$, where $\\mathbf{e}_1=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}, \\mathbf{e}_2=\\begin{bmatrix} 0 \\\\ 1 \\end{bmatrix}.$", "+0 # HELP ASAP PLEEASSE!!! 0 89 1 Let $$\\mathbf{A}$$ be a matrix such that $$\\mathbf{A}^{-1} = \\begin{pmatrix} 1 & 1 \\\\ 2 & x \\end{pmatrix}$$ for some value of $$x$$. What is the vector that $$\\mathbf{A}$$ maps to $$\\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}?$$ Oct 13, 2019 The answer is $$\\begin{pmatrix} 3 \\\\ -7 \\end{pmatrix}$$.", "# tfg.math.vector.reflect Computes the reflection direction for an incident vector. tfg.math.vector.reflect( vector, normal, axis=-1, name=None ) Defined in math/vector.py. For an incident vector $$\\mathbf{v}$$ and normal $$\\mathbf{n}$$ this function computes the reflected vector as $$\\mathbf{r} = \\mathbf{v} - 2(\\mathbf{n}^T\\mathbf{v})\\mathbf{n}$$. #### Note: In the following, A1 to An are optional batch dimensions, which should be broadcast compatible. #### Args: • vector: A tensor of shape [A1, ..., Ai, ..., An], where the dimension i = axis represents a vector. • normal: A tensor of shape [A1, ..., Ai, ..., An], where the dimension i = axis represents a normal around which the vector needs to be reflected. The normal vector needs to be normalized. • axis: The dimension along which to compute the reflection. • name: A name for this op which defaults to \"vector_reflect\". #### Returns: A tensor of shape [A1, ..., Ai, ..., An], where the dimension i = axis represents a reflected vector.", "Let $$V=\\mathbb{R}^3$$. Let $$(i)$$ $$T:V \\to V$$ be rotation by angle $$\\theta$$ counterclockwise in the plane (through origin) perpendicular to $$\\ (1,2,3)$$, $$(ii)$$ $$\\ S : V \\to V$$ be the reflection with respect to the plane (through origin) spanned by the vectors $$(1,0,1), \\ (1,2,1)$$, $$(iii)$$ $$U: V \\to V$$ be the rotation by the angle $$\\phi$$ counterclockwise in the plane ( through origin) perpendicular to $$(1,0,1)$$. Then find the matrix of $$\\ U \\circ S \\circ T$$ with respect to the standard basis. To answer the question we have to find the matrices of $$\\ T, S, U$$ and then combine the matrices.", "From reflection plane to “reflection” matrix How do I determine the \"reflection\" matrix, relative to the canonical basis, of a reflection on the plane $\\{x+y+z = 0\\}$ in $\\mathbb{R}^3$ ? I figured I should take the two eigenvectors which should be $n =(-1, 1, 0)$ and $v= (-1, 0, 1)$. Take one of the two and compute the Matrix with this: $M = I-2n^Tn$, but it's not working. If $\\bf n$ is a unit normal to the plane (as a column vector), the reflection across that plane is $I - 2 {\\bf n} {\\bf n}^T$. EDIT: In your case $${\\bf n} = \\pmatrix{1/\\sqrt{3}\\cr 1/\\sqrt{3}\\cr 1/\\sqrt{3}\\cr}$$ • So n shouldn't be the eigenvector ?! – Leroy Aug 26 '17 at 0:52 • $\\bf n$ is an eigenvector with eigenvalue $-1$. – Robert Israel Aug 26 '17 at 0:53 • Can you shortly explain how you found \"your\" n ?! – Leroy Aug 26 '17 at 0:54 • @Leroy You can read a normal to the plane directly from its equation, then normalize. Alternatively, take the cross product of your two vectors that lie in the plane and normalize that. – amd Aug 26 '17 at 1:07 A fairly straightforward way is to reflect the standard unit vectors directly. The reflected vectors (in standard basis order) will be the columns", "$B=\\{\\mathbf{e}_1, \\mathbf{e}_2\\}$ of $\\R^2$, where $\\mathbf{e}_1=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}, \\mathbf{e}_2=\\begin{bmatrix} 0 \\\\ 1 \\end{bmatrix}.$ Note that if $m=0$, then this problem is the same as the current problem. The solution is given in the post ↴ The Matrix for the Linear Transformation of the Reflection Across a Line in the Plane • The Matrix for the Linear Transformation of the Reflection Across a Line in the Plane Let $T:\\R^2 \\to \\R^2$ be a linear transformation of the $2$-dimensional vector space $\\R^2$ (the $x$-$y$-plane) to itself which is the reflection across a line $y=mx$ for some $m\\in \\R$. Then find the matrix representation of the linear transformation $T$ with respect to the […]", "# Math Help - Reflections 1. ## Reflections Find the matrix of the reflection in the line in that consists of all scalar multiples of the vector 5 6 . 2. Originally Posted by vcf323 Find the matrix of the reflection in the line in that consists of all scalar multiples of the vector 5 6 . Since your image is messed up lets do it for an arbitrary vector $\\vec{v}=\\begin{pmatrix} a \\\\ b\\end{pmatrix}=ae_1+be_2$ So we need to know that transform of the basis vectors $e_1,e_2$ First we need to project $e_1$ on the vector $v$ $\\displaystyle \\text{proj}_{v}e_1=\\frac{\\vec{v} \\cdot e_1}{||v||^2}\\vec{v}=\\frac{a}{a^2+b^2}\\vec{v}$ Now to find the the direction vector we subtract $\\frac{a}{a^2+b^2}\\vec{v}-e_1$ The above vector takes us from the tip of $e_1$ onto the line but we want to reflect across it so we need to multiply it by 2. This gives $\\displaystyle T(e_1)=e_1+2\\left( \\frac{a}{a^2+b^2}\\vec{v}-e_1\\right)= \\frac{2a}{a^2+b^2}(ae_1+be_2)-e_1=\\left( \\frac{a^2+2b^2}{a^2+b^2}\\right)e_1+\\left(\\frac{2ab }{a^2+b^2}\\right)e_2$ Now just do the exact same thing with $e_2$", "# Find Normal from Reflection I have a vector A with direction $\\vec D$. I also have another vector B with direction $\\vec R$. (think of $\\vec R$ to be the reflection of $\\vec D$ at point B.) From the reflection formula:$$\\vec R = \\vec D-2*(\\vec D \\cdot \\vec N)*\\vec N$$ (we know both $\\vec R$ and $\\vec D$) How can I find $\\vec N$? • Do you understand how the reflection formula was derived? That should give you a clue. – amd Dec 10 '15 at 19:34 The reflection formula says that to reflect a vector $\\mathbf v$ in the hyperplane specified by its normal $\\mathbf n$, you reverse the component of $\\mathbf v$ that’s in the direction of $\\mathbf n$. That is, you decompose $\\mathbf v$ into $\\pi_{\\mathbf n}\\mathbf v+(\\mathbf v-\\pi_{\\mathbf n}\\mathbf v)$, where $\\pi_{\\mathbf n}$ is orthogonal projection onto $\\mathbf n$, and then negate the first component: $$\\mathbf v'=-\\pi_{\\mathbf n}\\mathbf v+(\\mathbf v-\\pi_{\\mathbf n}\\mathbf v)=\\mathbf v-2\\pi_{\\mathbf n}\\mathbf v.$$ To recover the normal given $\\mathbf v$ and $\\mathbf v'$, remember that $\\pi_{\\mathbf n}\\mathbf v$ is a scalar multiple of $\\mathbf n$, so $\\mathbf v-\\mathbf v'=2\\pi_{\\mathbf n}\\mathbf v$ is a multiple of $\\mathbf n$ and is also normal to the hyperplane. Without more information about $\\mathbf n$ and its relationship to $\\mathbf v$, that’s the best you can", "for vectors on the line $y=mx$ and vectors on the perpendicular line. That why we didn’t compute the vectors $T(\\mathbf{e}_1)$ and $T(\\mathbf{e}_2)$ directly in the above solution. ## Related Question. Problem. Let $F:\\R^2\\to \\R^2$ be the function that maps each vector in $\\R^2$ to its reflection with respect to $x$-axis. Determine the formula for the function $F$ and prove that $F$ is a linear transformation. The solution is given in the post ↴ Linear Transformation that Maps Each Vector to Its Reflection with Respect to $x$-Axis ### 1 Response 1. 11/01/2017 […] The solution is given in the post ↴ The Matrix for the Linear Transformation of the Reflection Across a Line in the Plane […] ##### A Matrix Commuting With a Diagonal Matrix with Distinct Entries is Diagonal Let \\[D=\\begin{bmatrix} d_1 & 0 & \\dots & 0 \\\\ 0 &d_2 & \\dots & 0 \\\\ \\vdots & &... Close", "vector of $N$? If we know $\\mathbf{n}$, can we reflect $O$ in $N$ easily?" ]

[ "that it equals $|\\mathbf{H}| = -1$ ### Tridiagonalisation The advantage of reflections over the rotations that were used in the Jacobi algorithm is that we can use them to transform an entire row or column of a matrix in a single step. In particular, if we have a reflection that leaves the first element of every vector as it is and transforms a specific vector $$\\mathbf{x}$$ so that its second element equals $$s_0$$ and all of the remaining elements equal zero $\\mathbf{H}_0 \\times \\begin{pmatrix}x_0\\\\x_1\\\\x_2\\\\x_3\\\\\\vdots\\\\x_{n-1}\\end{pmatrix} = \\begin{pmatrix}x_0\\\\s_0\\\\0\\\\0\\\\\\vdots\\\\0\\end{pmatrix}$ then, if $$\\mathbf{x}$$ equals the first row and column of a symmetric matrix $$\\mathbf{M}$$, we have $\\mathbf{H}_0 \\times \\small \\begin{pmatrix} m_{0,0} & m_{0,1} & m_{0,2} & m_{0,3} & \\dots & m_{0,n-1}\\\\ m_{0,1} & m_{1,1} & m_{1,2} & m_{1,3} & \\dots & m_{1,n-1}\\\\ m_{0,2} & m_{1,2} & m_{2,2} & m_{2,3} & \\dots & m_{2,n-1}\\\\ m_{0,3} & m_{1,3} & m_{2,3} & m_{3,3} & \\dots & m_{3,n-1}\\\\ \\vdots & \\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ m_{0,n-1} & m_{1,n-1} & m_{2,n-1} & m_{3,n-1} & \\dots & m_{n-1,n-1} \\end{pmatrix} = \\small \\begin{pmatrix} m_{0,0} & m_{0,1} & m_{0,2} & m_{0,3} & \\dots & m_{0,n-1}\\\\ s_0 & m^\\prime_{1,1} & m^\\prime_{1,2} & m^\\prime_{1,3} & \\dots & m^\\prime_{1,n-1}\\\\ 0 & m^\\prime_{1,2} & m^\\prime_{2,2} & m^\\prime_{2,3} & \\dots & m^\\prime_{2,n-1}\\\\ 0 & m^\\prime_{1,3} & m^\\prime_{2,3} & m^\\prime_{3,3} & \\dots", "$\\mathbf{u}=(4,-1), \\mathbf{v}=(0,5)$, and $\\mathbf{w}=(-3,-3)$. Find the components of (a) $\\mathbf{u}+\\mathbf{w}$ (b) $\\mathbf{v}-3 \\mathbf{u}$ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find a unit vector that is orthogonal to both $\\mathbf{u}=(1,0,1)$ and $\\mathbf{v}=(0,1,1)$ Find a unit vector that is orthogonal to both $\\mathbf{u}=(1,0,1)$ and $\\mathbf{v}=(0,1,1)$Find a unit vector that is orthogonal to both $\\mathbf{u}=(1,0,1)$ and $\\mathbf{v}=(0,1,1)$ ... Let $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components of Let $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components ofLet $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components of (a) $\\mathbf{v}-\\mathbf{w}$ (b) $6 \\mathrm{u}+2 \\ma ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find the terminal point of the vector that is equivalent to$\\mathbf{u}=(1,2)$and whose initial point is$A(1,1)$. 0 answers 83 views Find the terminal point of the vector that is equivalent to$\\mathbf{u}=(1,2)$and whose initial point is$A(1,1)$.Find the terminal point of the vector that is equivalent to$\\mathbf{u}=(1,2)$and whose initial point is$A(1,1)$. ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 In relation to the points$P_{1}$and$P_{2}$, what can you say about the terminal point of the following vector if its initial point is at the origin? 0 answers 40 views In relation to the points$P_{1}$and$P_{2}$, what can you say about the terminal point of the following vector if its initial point is at the origin?In relation to", "# Find Normal from Reflection I have a vector A with direction $\\vec D$. I also have another vector B with direction $\\vec R$. (think of $\\vec R$ to be the reflection of $\\vec D$ at point B.) From the reflection formula:$$\\vec R = \\vec D-2*(\\vec D \\cdot \\vec N)*\\vec N$$ (we know both $\\vec R$ and $\\vec D$) How can I find $\\vec N$? • Do you understand how the reflection formula was derived? That should give you a clue. – amd Dec 10 '15 at 19:34 The reflection formula says that to reflect a vector $\\mathbf v$ in the hyperplane specified by its normal $\\mathbf n$, you reverse the component of $\\mathbf v$ that’s in the direction of $\\mathbf n$. That is, you decompose $\\mathbf v$ into $\\pi_{\\mathbf n}\\mathbf v+(\\mathbf v-\\pi_{\\mathbf n}\\mathbf v)$, where $\\pi_{\\mathbf n}$ is orthogonal projection onto $\\mathbf n$, and then negate the first component: $$\\mathbf v'=-\\pi_{\\mathbf n}\\mathbf v+(\\mathbf v-\\pi_{\\mathbf n}\\mathbf v)=\\mathbf v-2\\pi_{\\mathbf n}\\mathbf v.$$ To recover the normal given $\\mathbf v$ and $\\mathbf v'$, remember that $\\pi_{\\mathbf n}\\mathbf v$ is a scalar multiple of $\\mathbf n$, so $\\mathbf v-\\mathbf v'=2\\pi_{\\mathbf n}\\mathbf v$ is a multiple of $\\mathbf n$ and is also normal to the hyperplane. Without more information about $\\mathbf n$ and its relationship to $\\mathbf v$, that’s the best you can", "need to find a reflection that transforms the first column of matrix A, vector $\\mathbf{a}_1 = (12, 6, -4)^T$, to $\\|\\mathbf{a}_1\\| \\;\\mathrm{e}_1 = (14, 0, 0)^T.$ Now, $\\mathbf{u} = \\mathbf{x} + \\alpha\\mathbf{e}_1,$ and $\\mathbf{v} = {\\mathbf{u}\\over\\|\\mathbf{u}\\|}.$ Here, $\\alpha = -14$ and $\\mathbf{x} = \\mathbf{a}_1 = (12, 6, -4)^T$ Therefore $\\mathbf{u} = (-2, 6, -4)^T=({2})(-1, 3, -2)^T$ and $\\mathbf{v} = {1 \\over \\sqrt{14}}(-1, 3, -2)^T$, and then $Q_1 = I - {2 \\over \\sqrt{14} \\sqrt{14}} \\begin{pmatrix} -1 \\\\ 3 \\\\ -2 \\end{pmatrix}\\begin{pmatrix} -1 & 3 & -2 \\end{pmatrix}$ $= I - {1 \\over 7}\\begin{pmatrix} 1 & -3 & 2 \\\\ -3 & 9 & -6 \\\\ 2 & -6 & 4 \\end{pmatrix}$ $= \\begin{pmatrix} 6/7 & 3/7 & -2/7 \\\\ 3/7 &-2/7 & 6/7 \\\\ -2/7 & 6/7 & 3/7 \\\\ \\end{pmatrix}.$ Now observe: $Q_1A = \\begin{pmatrix} 14 & 21 & -14 \\\\ 0 & -49 & -14 \\\\ 0 & 168 & -77 \\end{pmatrix},$ so we already have almost a triangular matrix. We only need to zero the (3, 2) entry. Take the (1, 1) minor, and then apply the process again to $A' = M_{11} = \\begin{pmatrix} -49 & -14 \\\\ 168 & -77 \\end{pmatrix}.$ By the same method as above, we obtain the matrix of the Householder transformation $Q_2 = \\begin{pmatrix} 1 & 0 & 0", "# arrow_back Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is orthogonal. 8 views Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is orthogonal. It suffices to show that $$\\mathbf{i} \\cdot \\mathbf{j}=\\mathbf{i} \\cdot \\mathbf{k}=\\mathbf{j} \\cdot \\mathbf{k}=\\mathbf{0}$$ because it will follow automatically from the symmetry property of the dot product that $$\\mathbf{j} \\cdot \\mathbf{i}=\\mathbf{k} \\cdot \\mathbf{i}=\\mathbf{k} \\cdot \\mathbf{j}=\\mathbf{0}$$ Although the orthogonality of the vectors in $S$ is evident geometrically from Figure $3.2 .2$ it is confirmed algebraically by the computations \\begin{aligned} &\\mathbf{i} \\cdot \\mathbf{j}=(1,0,0) \\cdot(0,1,0)=0 \\\\ &\\mathbf{i} \\cdot \\mathbf{k}=(1,0,0) \\cdot(0,0,1)=0 \\\\ &\\mathbf{j} \\cdot \\mathbf{k}=(0,1,0) \\cdot(0,0,1)=0 \\end{aligned} by Platinum (106,844 points) selected by ## Related questions Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Show that $\\mathbf{u}=(-2,3,1,4)$ and $\\mathbf{v}=(1,2,0,-1)$ are orthogonal vectors in $R^{4}$. Show that $\\mathbf{u}=(-2,3,1,4)$ and $\\mathbf{v}=(1,2,0,-1)$ are orthogonal vectors in $R^{4}$.Show that $\\mathbf{u}=(-2,3,1,4)$ and $\\mathbf{v}=(1,2,0,-1)$ are orthogonal vectors in $R^{4}$. ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors. Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors.Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors. ... close Notice: Undefined index: avatar", "another element satisfying axiom (a4). Then we have $v+(-v)=0=v+w.$ By the cancellation law (see (a)), we have $-v=w$. Thus, the additive inverse is unique. ### (e) $0v=\\mathbf{0}$ for every $v\\in V$, where $0\\in\\R$ is the zero scalar. Note that $0$ is a real number and $\\mathbf{0}$ is the zero vector in $V$. For $v\\in V$, we have \\begin{align*} 0v&=(0+0)v\\stackrel{(m3)}{=} 0v+0v. \\end{align*} We also have $0v\\stackrel{(a3)}{=} \\mathbf{0}+0v.$ Hence, combining these, we see that $0v+0v=\\mathbf{0}+0v,$ and by the cancellation law, we obtain $0v=\\mathbf{0}$. ### (f) $a\\mathbf{0}=\\mathbf{0}$ for every scalar $a$. Note that we have $\\mathbf{0}+\\mathbf{0}=\\mathbf{0}$ by (a3). Thus, we have \\begin{align*} a\\mathbf{0}=a(\\mathbf{0}+\\mathbf{0})\\stackrel{(m2)}{=}a\\mathbf{0}+a\\mathbf{0}. \\end{align*} We also have $a\\mathbf{0}=\\mathbf{0}+a\\mathbf{0}$ by (a3). Combining these, we have $a\\mathbf{0}+a\\mathbf{0}=\\mathbf{0}+a\\mathbf{0},$ and the cancellation law yields $a\\mathbf{0}=\\mathbf{0}$. ### (g) If $av=\\mathbf{0}$, then $a=0$ or $v=\\mathbf{0}$. For this problem, we use a little bit logic. Our assumption is $av=\\mathbf{0}$. From this assumption, we need to deduce that either $a=0$ or $v=\\mathbf{0}$. Note that if $a=0$, then we are done as this is one of the consequence we want. So, let us assume that $a\\neq 0$. Then we want to prove $v=0$. Since $a$ is a nonzero scalar, we have $a^{-1}$. Then we have \\begin{align*} a^{-1}(av)=a^{-1}\\mathbf{0}. \\end{align*} The right hand side $a^{-1}\\mathbf{0}$ is $\\mathbf{0}$ by part (f). On the other hand, the left hand side can be computed as", "### R5 ${\\displaystyle \\mathbf {v} =4\\mathbf {e} _{1}+2\\mathbf {e} _{2}=c_{1}\\mathbf {b} _{1}+c_{2}\\mathbf {b} _{2}}$ The oblique basis vectors b1,b2 are: ${\\displaystyle \\mathbf {b} _{1}=2\\mathbf {e} _{1}+7\\mathbf {e} _{2}}$ ${\\displaystyle \\mathbf {b} _{2}=1.5\\mathbf {e} _{1}+3\\mathbf {e} _{2}}$ 1. Find the components c1, c2 using the Gramian matrix. 2. Verify the result found above. ### R4.3 ${\\displaystyle \\displaystyle {\\begin{bmatrix}2&-3&2&0&0&0&0&0&0&0&0\\\\0&2&-6&6&0&0&0&0&0&0&0\\\\0&0&2&-9&12&0&0&0&0&0&0\\\\0&0&0&2&-12&20&0&0&0&0&0\\\\0&0&0&0&2&-15&30&0&0&0&0\\\\0&0&0&0&0&2&-18&42&0&0&0\\\\0&0&0&0&0&0&2&-21&56&0&0\\\\0&0&0&0&0&0&0&2&-24&72&0\\\\0&0&0&0&0&0&0&0&2&-27&90\\\\0&0&0&0&0&0&0&0&0&2&-30\\\\0&0&0&0&0&0&0&0&0&0&2\\\\\\end{bmatrix}}\\cdot {\\begin{pmatrix}{A}_{0}\\\\{A}_{1}\\\\{A}_{2}\\\\{A}_{3}\\\\{A}_{4}\\\\{A}_{5}\\\\{A}_{6}\\\\{A}_{7}\\\\{A}_{8}\\\\{A}_{9}\\\\{A}_{10}\\\\\\end{pmatrix}}={\\begin{Bmatrix}0\\\\1\\\\-1/2\\\\1/6\\\\-1/24\\\\1/120\\\\-1/720\\\\1/5040\\\\1/40320\\\\1/362880\\\\1/3628800\\\\\\end{Bmatrix}}}$ (3.30) ### Statement 4.1 from lecture notes R4.1 Lect. 7c pgs. 19-22 Given the general form of polynomial excitation. ${\\displaystyle \\displaystyle y''+ay'+by=\\sum _{j=0}^{n}d_{j}x^{j}}$ (1.1) The particular solution that satisfies: ${\\displaystyle \\displaystyle y_{p}(x)=\\sum _{j=0}^{n}c_{j}x^{j}}$ (1.2) The first and second derivative of the particular solution that solves the original polynomial excitation equation. ${\\displaystyle \\displaystyle y_{p}'(x)=\\sum _{j=0}^{n-1}c_{j+1}(j+1)x^{j}}$ (1.3) ${\\displaystyle \\displaystyle y_{p}''(x)=\\sum _{j=0}^{n-2}c_{j+2}(j+2)(j+1)x^{j}}$ (1.4) The particular solutions are put into the polynomial excitation equation to give the general summation form: ${\\displaystyle \\displaystyle \\sum _{j=0}^{n-2}[c_{j+2}(j+2)(j+1)+ac_{j+1}(j+1)+bc_{j}]x^{j}+ac_{n}nx^{n-1}+b[c_{n-1}x^{n-1}+c_{n}x^{n}]=\\sum _{j=0}^{n}d_{j}x^{j}}$ (1.5) Obtain the equations associated with ${\\displaystyle d_{1}\\!}$, coefficients of ${\\displaystyle x\\!}$; ${\\displaystyle d_{2}\\!}$, coefficients of ${\\displaystyle x^{2}\\!}$; ${\\displaystyle d_{n-2}\\!}$, coefficients of ${\\displaystyle x^{n-2}\\!}$; ${\\displaystyle d_{n-1}\\!}$, coefficients of ${\\displaystyle x^{n-1}\\!}$; ${\\displaystyle d_{n}\\!}$, coefficients of ${\\displaystyle x^{n}\\!}$. Five total equations for coefficients. Also set up the matrix ${\\displaystyle \\mathbf {A} \\!}$ that satisfies ${\\displaystyle \\mathbf {A} \\mathbf {c} =\\mathbf {d} \\!}$. ### Solution The given equation associated with ${\\displaystyle d_{0}\\!}$ taking ${\\displaystyle j=0\\!}$ ${\\displaystyle \\displaystyle 2c_{2}+ac_{1}+bc_{0}=d_{0}}$ (1.6) Taking ${\\displaystyle j=1\\!}$", "+0 # I will really appreciate any help. 0 28 1 I will really appreciate any help. Let $$\\mathbf{A}, \\mathbf{B}, \\mathbf{C}, \\mathbf{D}$$ be the following matrices: \\begin{align*} &\\mathbf{A} = \\begin{pmatrix} 0 & 0 \\\\ 0& 0 \\end{pmatrix},\\\\ &\\mathbf{B} = \\begin{pmatrix} -1 & 0 \\\\ 0 & -1\\end{pmatrix}, \\\\ &\\mathbf{C} = \\begin{pmatrix} 1/2 & 1/2 \\\\ 1/2 & 1/2\\end{pmatrix}, \\\\ &\\mathbf{D} = \\begin{pmatrix} 0 & -1 \\\\ -1 & 0 \\end{pmatrix}. \\end{align*} Also, consider the following functions: \\begin{align*} &\\text{Reflection across a line with direction vector \\begin{pmatrix}1\\\\-1 \\end{pmatrix}},\\\\ &\\text{The function that sends every vector to the zero vector}, \\phantom{\\begin{pmatrix}1\\\\-1 \\end{pmatrix}}\\\\ &\\text{The function that sends every vector to its negative},\\phantom{\\begin{pmatrix}1\\\\-1 \\end{pmatrix}}\\\\ &\\text{Projection onto \\begin{pmatrix}1\\\\1 \\end{pmatrix}}. \\end{align*} How do we order the list $$\\mathbf{A}, \\mathbf{B}, \\mathbf{C}, \\mathbf{D}$$ to match the order of the functions above? I am having a hard time understanding problem such as these. Thank you in advance for any help Jul 20, 2022", "# Finding the orthogonal projection of a vector I can't figure this out. How the book explains to do this problem doesn't make sense to me. I'm hoping someone can simplify this so I can understand how to do this step by step. Let $L$ be the line in $\\mathbb{R}^3$ that consists of all scalar multiples of $$\\left[ \\begin{array}{ccc} 2\\\\ 1\\\\ 2\\\\ \\end{array} \\right]$$ Find the reflection of the vector $$\\left[ \\begin{array}{ccc} 1\\\\ 4\\\\ 1\\\\ \\end{array} \\right]$$ onto $L$ • Are you asking for reflection (as in the body of the question) or projection (as in the subject)? – Thomas Jul 11 '16 at 17:38 Let $L$ be the line spanned by some non-zero vector $\\mathbf{v} \\in \\mathbb{R}^3$, so that $L = \\{a\\mathbf{v} \\mid a \\in \\mathbb{R}\\}$ is the space of all scalar multiples of $\\mathbf{v}$. Then the orthogonal projection of a vector $\\mathbf{x} \\in \\mathbb{R}^3$ onto the line $L$ can be computed as $$\\operatorname{Proj}_L(\\mathbf{x}) = \\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}.$$ So, in this case, we have $$\\mathbf{v} = \\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}, \\quad \\mathbf{x} = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix},$$ so that $$\\mathbf{v} \\cdot \\mathbf{x} = 2 \\cdot 1 + 1 \\cdot 4 + 2 \\cdot 1 = 8, \\quad \\mathbf{v} \\cdot \\mathbf{v} = 2^2 + 1^2 + 2^2 = 9,$$ and hence $$\\operatorname{Proj}_L(\\mathbf{x}) = \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}.$$ Now, you probably wanted to compute the orthogonal projection", "tells us that ker(P) = <(1,0,1)>, and this is a normal vector to the plane x + z = 0 (we can divide by 2 in the plane equation). let's call ker(P), U, and our plane x + z = 0, W = im(P). the fact that P is an orthogonal projection means: R3 = U⊕W = U⊕U U has basis {(1,0,1)} and W has basis {(1,0,-1),(0,1,0)}. since (1,0,0) = (1/2)(1,0,1) + (1/2)(1,0,-1), this tells us that P(1,0,0) = P(1/2,0,1/2) + P(1/2,0,-1/2) = (1/2,0,-1/2) <---first column of P clearly P(0,1,0) = (0,1,0), since (0,1,0) is already in our plane W, and P fixes W. finally, P(0,0,1) = P[(1/2)(1,0,1)] + P[(-1/2)(1,0,-1)] = (-1/2,0,1/2) <--3rd column of P, so $P = \\begin{bmatrix}\\frac{1}{2}&0&-\\frac{1}{2}\\\\0&1&0\\\\ -\\frac{1}{2}&0&\\frac{1}{2} \\end{bmatrix}$ one can easily verify that P2 = P (this is the defining property of reflections). now our reflection R is a \"reflection about W\", so for v = u+w, where u is in U, and w is in W, R(v) = R(u) + R(w) = -u + w. we can write this in terms of P like so: for this same pair u,w: P(v) = P(u) + P(w) = 0 + w = w. that is: u = v - P(v), and w = P(v), so: R(v) = -u + w = -(v - P(v)) +", "this equation completely determines all the $$a_i$$. For example, $$a_0 = 2, v = \\color{darkblue}{0010}_2 = 2$$ is a solution when $$R=8, n=4$$ because $$\\bigl(2^4-1\\bigr)\\cdot2 + 2\\equiv 0\\pmod 8$$, and this solution allows us to compute \\def\\db#1{\\color{darkblue}{#1}}\\begin{align} a_0&&&=2\\\\ a_{P(0)} &= 2a_0 &+ \\db0 &= 4\\\\ a_{P^2(0)} &= 2a_{P(0)} &+ \\db0 &= 0 \\\\ a_{P^3(0)} &= 2a_{P^2(0)} &+ \\db1 &= 1\\\\ \\hline a_{P^4(0)} &= 2a_{P^3(0)} &+ \\db0 &= 2\\\\ \\end{align} where the carry bits $$c_i = \\langle 0,0,1,0\\rangle$$ are visible in the third column, and all the sums are taken $$\\pmod 8$$. Note that $$a_{P^n(0)} = a_0$$ as promised. This derivation of the entire set of $$a_i$$ from a single one plus a choice of $$v$$ is crucial, so let’s see one more example. Let’s consider $$R=10, n=3$$. Then we want to choose $$a_0$$ and $$v$$ so that $$\\left(2^3-1\\right)a_0 + v \\equiv 0\\pmod{10}$$ where $$v\\in\\{0\\ldots 7\\}$$. One possible solution is $$a_0 = 5, v=\\color{darklue}{101}_2 = 5$$. Then we can derive the other $$a_i$$ as follows: \\begin{align} a_0&&&=5\\\\ a_{P(0)} &= 2a_0 &+ \\db1 &= 1\\\\ a_{P^2(0)} &= 2a_{P(0)} &+ \\db0 &= 2 \\\\\\hline a_{P^3(0)} &= 2a_{P^2(0)} &+ \\db1 &= 5\\\\ \\end{align} And again we have $$a_{P^n(0)}= a_0$$ as required. Since the bits of $$v$$ are used cyclically, not every pair of $$\\langle a_0, v\\rangle$$ will yield a different solution. Rotating", "& \\\\ & \\ddots & \\ddots & \\\\ & & \\ddots & 1\\\\ & & & 0 \\end{array} \\right) \\left( \\begin{array}{c} y_{1} \\\\ y_{2} \\\\ \\vdots \\\\ y_{N} \\end{array} \\right) = \\left( \\begin{array}{c} y_{2}\\\\ \\vdots \\\\ y_{N} \\\\ 0 \\end{array} \\right).$ The above minimization problem can hence be reformulated in terms of $A$ and $\\mathbf{y}$, provided we define a bilinear form to trunate the last $p$ elements of the vectors. Let $D =$ diag $(1, \\ldots, 1, 0, \\ldots, 0)$ and $\\|\\mathbf{x}\\|_D = \\|D\\mathbf{x}\\|_2$, then our minimization problem can be written as $\\displaystyle \\min \\vert \\vert A^p \\mathbf{y} - A^{p-1} \\mathbf{y}C_1 - \\cdots \\mathbf{y} C_p \\vert \\vert_D.$ This problem is solved implicitly by the block rational Arnoldi method during the construction of $\\mathbf{v}_3$, the third block vector in the block-orthonormal rational Krylov basis. Furthermore, the resulting VAR($2$) model can be represented as an RKFUNB object: A = spdiags([zeros(size(y,1),1),ones(size(y,1),1)],0:1,size(y,1),size(y,1)); xi_ = inf*ones(1, 2); D = zeros(75,75); D(1:73, 1:73) = eye(73); param.balance = 0; param.inner_product = @(x, y) y'*D*x; [V, K, H, out] = rat_krylov(A, y, xi_, param); R = out.R(1:3, 1:3); C = {zeros(3), zeros(3), H(7:9, 4:6)*H(4:6, 1:3)*R}; r_ = rkfunb(K, H, C) r_ = RKFUNB object of block size 3-by-3 and type (2, 2). Real-valued Hessenberg pencil (H, K) of size 9-by-6. Real dense coefficient matrices of size", "fact that P is an orthogonal projection means: R3 = U⊕W = U⊕U U has basis {(1,0,1)} and W has basis {(1,0,-1),(0,1,0)}. since (1,0,0) = (1/2)(1,0,1) + (1/2)(1,0,-1), this tells us that P(1,0,0) = P(1/2,0,1/2) + P(1/2,0,-1/2) = (1/2,0,-1/2) <---first column of P clearly P(0,1,0) = (0,1,0), since (0,1,0) is already in our plane W, and P fixes W. finally, P(0,0,1) = P[(1/2)(1,0,1)] + P[(-1/2)(1,0,-1)] = (-1/2,0,1/2) <--3rd column of P, so $P = \\begin{bmatrix}\\frac{1}{2}&0&-\\frac{1}{2}\\\\0&1&0\\\\ -\\frac{1}{2}&0&\\frac{1}{2} \\end{bmatrix}$ one can easily verify that P2 = P (this is the defining property of reflections). now our reflection R is a \"reflection about W\", so for v = u+w, where u is in U, and w is in W, R(v) = R(u) + R(w) = -u + w. we can write this in terms of P like so: for this same pair u,w: P(v) = P(u) + P(w) = 0 + w = w. that is: u = v - P(v), and w = P(v), so: R(v) = -u + w = -(v - P(v)) + P(v) = -v + P(v) + P(v) = 2P(v) - I(v) = (2P- I)(v), so that R = 2P - I or: $R = \\begin{bmatrix}0&0&-1\\\\0&1&0\\\\-1&0&0 \\end{bmatrix}$ i always like to \"make sure\" my P and R do what the problem says they should do. so", "!!a_i!!. For example, !!a_0 = 2, v = \\color{darkblue}{0010}_2 = 2!! is a solution when !!R=8, n=4!! because !!\\bigl(2^4-1\\bigr)\\cdot2 + 2\\equiv 0\\pmod 8!!, and this solution allows us to compute \\def\\db#1{\\color{darkblue}{#1}}\\begin{align} a_0&&&=2\\\\ a_{P(0)} &= 2a_0 &+ \\db0 &= 4\\\\ a_{P^2(0)} &= 2a_{P(0)} &+ \\db0 &= 0 \\\\ a_{P^3(0)} &= 2a_{P^2(0)} &+ \\db1 &= 1\\\\ \\hline a_{P^4(0)} &= 2a_{P^3(0)} &+ \\db0 &= 2\\\\ \\end{align} where the carry bits !!c_i = \\langle 0,0,1,0\\rangle!! are visible in the third column, and all the sums are taken !!\\pmod 8!!. Note that !!a_{P^n(0)} = a_0!! as promised. This derivation of the entire set of !!a_i!! from a single one plus a choice of !!v!! is crucial, so let's see one more example. Let's consider !!R=10, n=3!!. Then we want to choose !!a_0!! and !!v!! so that !!\\left(2^3-1\\right)a_0 + v \\equiv 0\\pmod{10}!! where !!v\\in\\{0\\ldots 7\\}!!. One possible solution is !!a_0 = 5, v=\\color{darkblue}{101}_2 = 5!!. Then we can derive the other !!a_i!! as follows: \\begin{align} a_0&&&=5\\\\ a_{P(0)} &= 2a_0 &+ \\db1 &= 1\\\\ a_{P^2(0)} &= 2a_{P(0)} &+ \\db0 &= 2 \\\\\\hline a_{P^3(0)} &= 2a_{P^2(0)} &+ \\db1 &= 5\\\\ \\end{align} And again we have !!a_{P^n(0)}= a_0!! as required. Since the bits of !!v!! are used cyclically, not every pair of !!\\langle a_0, v\\rangle!! will yield a different solution. Rotating the bits of !!v!! and pairing", "{\\color{red}2/3}&{\\color{red}2/3}&{\\color{red}1/3}& \\\\ {\\color{blue}1/3}&{\\color{blue}-2/3}&{\\color{blue}2/3} \\\\ -2/3&1/3&2/3 \\end{pmatrix} \\nonumber$ $\\mathbf{M}^T\\mathbf{M}=\\begin{pmatrix} 1&0&0 \\\\ 0&1&0 \\\\ 0&0&1 \\end{pmatrix}=\\mathbf{I} \\nonumber$ Because $$\\mathbf{M}^T\\mathbf{M}=\\mathbf{I}$$, the matrix is orthogonal. We now how to prove that the columns for a set of orthonormal vectors. The vectors are: $\\mathbf{a}=2/3\\mathbf{i}+2/3\\mathbf{j}+1/3\\mathbf{i}= \\begin{pmatrix}2/3\\\\2/3\\\\1/3\\end{pmatrix}\\nonumber$ $\\mathbf{b}=1/3\\mathbf{i}-2/3\\mathbf{j}+2/3\\mathbf{i}= \\begin{pmatrix}1/3\\\\-2/3\\\\2/3\\end{pmatrix}\\nonumber$ $\\mathbf{c}=-2/3\\mathbf{i}+1/3\\mathbf{j}+2/3\\mathbf{i}= \\begin{pmatrix}-2/3\\\\1/3\\\\2/3\\end{pmatrix}\\nonumber$ The modulii of these vectors are: $|\\mathbf{a}|^2=(2/3)^2+(2/3)^2+(1/3)^2=1 \\nonumber$ $|\\mathbf{b}|^2=(1/3)^2+(-2/3)^2+(2/3)^2=1 \\nonumber$ $|\\mathbf{c}|^2=(-2/3)^2+(1/3)^2+(2/3)^2=1 \\nonumber$ which proves that the vectors are normalized. The dot products of the three pairs of vectors are: $\\mathbf{a}\\cdot\\mathbf{b}=(2/3)(1/3)+(2/3)(-2/3)+(1/3)(2/3)=0 \\nonumber$ $\\mathbf{a}\\cdot\\mathbf{c}=(2/3)(-2/3)+(2/3)(1/3)+(1/3)(2/3)=0 \\nonumber$ $\\mathbf{c}\\cdot\\mathbf{b}=(-2/3)(1/3)+(1/3)(-2/3)+(2/3)(2/3)=0 \\nonumber$ which proves they are mutually orthogonal. Because the vectors are normalized and mutually orthogonal, they form an orthonormal set. Orthogonal matrices, when thought as operators that act on vectors, are important because they produce transformations that preserve the lengths of the vectors and the relative angles between them. For example, in two dimensions, the matrix $\\mathbf{M}_1=1/\\sqrt{2}\\begin{pmatrix} 1&1 \\\\ 1&-1 \\end{pmatrix} \\nonumber$ is an operator that rotates a vector by $$\\pi/4$$ in the counterclockwise direction (Figure $$\\PageIndex{1}$$) preserving the lengths of the vectors and their relative orientation. In other words, an orthogonal matrix rotates a shape without distorting it. If the columns are orthogonal vectors that are not normalized, as in $\\mathbf{M}_2=\\begin{pmatrix} 1&1 \\\\ 1&-1 \\end{pmatrix}, \\nonumber$ the object changes in size but the shape is not distorted. If, however, the two columns are non-orthogonal vectors, the transformation will distort", "eigenvector $$\\mathbf y_2$$. Tis must satisfy $$\\mathbf y_2 \\in \\ker \\mathbf A^2 \\setminus \\ker \\mathbf A$$ and is linearly independent on $$\\mathbf x_2$$. We choose e.g. $$\\mathbf y_2=(0{,}1,0{,}0,0)^T$$ and calculate $$\\mathbf y_1=\\mathbf A\\mathbf y_2=(1{,}0,0{,}0,0)^T$$ (opět vlastní vektor $$\\mathbf A$$). Now we may assemble matrices $$\\mathbf R$$ (columns are $$\\mathbf y_1,\\mathbf y_2,\\mathbf x_1,\\mathbf x_2,\\mathbf x_3$$) and $$\\mathbf J$$ and claculate the inverse matrix $$\\mathbf R^{-1}$$ $$\\mathbf R= \\begin{pmatrix} 1&0&0&-1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&-1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$, $$\\mathbf J= \\begin{pmatrix} 0&1&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1\\\\ 0&0&0&0&0 \\end{pmatrix}$$, $$\\mathbf R^{-1}=\\begin{pmatrix} 1&0&0&1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$", "so that $\\tilde f(x) = x^2(1+\\delta)$. Verify that x^2*(1+δ)-x64^2 is zero. In [ ]: Exercise 3(d) Calculate the backward error Δx of approximate $f$ by $\\tilde f$ at $x = 1/3$. Verify that (x+Δx)^2 - x64^2 is zero (to roughly 75 digits). In [ ]: ## Conditioning of Householder reflections¶ Last lab we considered the Householder reflection constructed from a vector $\\mathbf x$ via: \\begin{align*} \\mathbf w &= \\mathbf x - \\alpha \\|{\\mathbf x}\\| \\mathbf e_1, \\qquad \\alpha = \\pm 1,\\cr \\mathbf v &= \\mathbf w / \\|\\mathbf w\\|,\\cr Q_{\\mathbf v} &\\triangleq I - 2 \\mathbf v \\mathbf v^\\top \\end{align*} so that $$Q_{\\mathbf v}\\mathbf x = \\begin{pmatrix} \\alpha \\|\\mathbf x\\| \\cr 0 \\cr \\vdots \\cr 0\\end{pmatrix}$$ An oustanding question is the choice of $\\alpha$: $\\alpha = {\\rm sign}(x_1)$ or $\\alpha = -{\\rm sign}(x_1)$. We will investigate this question by looking at the error of the problem $$f(\\mathbf x) = Q_{\\mathbf v} \\mathbf x$$ where we know the exact value, versus the implementation on a computer in floating point arithmetic, which we denote $\\tilde f(\\mathbf x)$. Here we have $f, \\tilde f : \\mathbb R^n \\rightarrow \\mathbb R^n$ with the 2-norm attached. Exercise 4(a) Write a function house(s,x) that returns $Q_{\\mathbf v}$ as definded above, with \\alpha = s*sign(x[1]). In [ ]: Exercise 4(b) Consider the vector $\\mathbf x=[1.,\\epsilon]$.", "such that $$a_1\\mathbf v_1+a_2\\mathbf v_2+\\cdots+a_p\\mathbf v_p=\\mathbf 0$$ to show linear dependency. If there are multiple outputs that work, you can output any one of them. The output format is also flexible, as long as it is clear what the values of each scalar are. This is , so the shortest code in bytes wins! ### Example I/O Input: $$\\n=2\\$$ $$\\p=3\\$$ $$\\\\mathbf v_1,\\mathbf v_2,\\dots,\\mathbf v_p=\\pmatrix{4\\\\5},\\pmatrix{8\\\\10},\\pmatrix{6\\\\7}\\$$ Output: One possible solution is $$\\a_1,a_2,\\dots,a_p=2,-1,0\\$$. We can quickly test that this does work: $$2\\pmatrix{4\\\\5}+(-1)\\pmatrix{8\\\\10}+0\\pmatrix{6\\\\7}=\\pmatrix{8\\\\10}-\\pmatrix{8\\\\10}+\\pmatrix{0\\\\0}=\\pmatrix{0\\\\0}=\\mathbf 0$$ ### Test Cases: Work in progress n, vectors -> scalars 2, [(4,5),(8,10),(6,7)] -> [2,-1,0] 2, [(3,6),(6,7),(3,9),(2,4),(5,5)] -> [2,0,0,-3,0] 3, [(1,2,3),(2,3,4),(3,4,5),(4,5,6),(5,6,7),(6,7,8),(7,8,9),(8,9,10)] -> [1,-1,-1,1,1,-1,-1,1] 1: Some sources: this, this, this, this # Concerns • Has a similar challenge been done before? • Is it clear what the challenge is specifying? • I want to change the title to be more specific, but I'm not sure what to change it to. • I need some test cases, but I'm not sure what test cases would be good for this. • A possible title would be \"Solve a linear equation system\" or \"Find a kernel vector\" Aug 15 at 22:13 • \"$n\\geq 2$\" is superfluous. Sep 5 at 22:30 • I do not quite understand the premise, since a vector space's dimension is defined to be its basis' length. Thus if you", "1 like 0 dislike 413 views Prove that if $\\mathbf{u}, \\mathbf{v}$, and $\\mathbf{w}$ are vectors in $R^{n}$, and if $k$ and $m$ are scalars, then: (a) $\\mathbf{u}+\\mathbf{v}=\\mathbf{v}+\\mathbf{u}$ (b) $(\\mathbf{u}+\\mathbf{v})+\\mathbf{w}=\\mathbf{u}+(\\mathbf{v}+\\mathbf{w})$ (c) $\\mathbf{u}+\\mathbf{0}=\\mathbf{0}+\\mathbf{u}=\\mathbf{u}$ (d) $\\mathbf{u}+(-\\mathbf{u})=\\mathbf{0}$ (e) $k(\\mathbf{u}+\\mathbf{v})=k \\mathbf{u}+k \\mathbf{v}$ (f) $(k+m) \\mathbf{u}=k \\mathbf{u}+m \\mathbf{u}$ (g) $k(m \\mathbf{u})=(k m) \\mathbf{u}$ (h) $\\mathbf{l u}=\\mathbf{u}$ | 413 views 0 like 0 dislike (a) To prove this statement, we can write out the components of the vectors $\\mathbf{u}$ and $\\mathbf{v}$: $$\\mathbf{u} = \\begin{pmatrix} u_1 \\ u_2 \\ \\vdots \\ u_n \\end{pmatrix} \\quad \\mathbf{v} = \\begin{pmatrix} v_1 \\ v_2 \\ \\vdots \\ v_n \\end{pmatrix}$$ Then, the vector $\\mathbf{u} + \\mathbf{v}$ is given by: $$\\mathbf{u} + \\mathbf{v} = \\begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \\ \\vdots \\ u_n + v_n \\end{pmatrix}$$ Similarly, the vector $\\mathbf{v} + \\mathbf{u}$ is given by: $$\\mathbf{v} + \\mathbf{u} = \\begin{pmatrix} v_1 + u_1 \\ v_2 + u_2 \\ \\vdots \\ v_n + u_n \\end{pmatrix}$$ Since $u_i + v_i = v_i + u_i$ for all $i$, it follows that $\\mathbf{u} + \\mathbf{v} = \\mathbf{v} + \\mathbf{u}$. (b) To prove this statement, we can again write out the components of the vectors $\\mathbf{u}, \\mathbf{v}$, and $\\mathbf{w}$: $$\\mathbf{u} = \\begin{pmatrix} u_1 \\ u_2 \\ \\vdots \\ u_n \\end{pmatrix} \\quad \\mathbf{v} = \\begin{pmatrix} v_1 \\ v_2 \\ \\vdots \\ v_n \\end{pmatrix} \\quad \\mathbf{w}", "2c_i + c_{P(P(i)})) + c_{P(i)} &&= 8a_i + 4c_i + 2c_{P(i)} + c_{P(P(i))}\\\\ &&&\\vdots\\\\ a_{P^n(i)} &&&= 2^na_i + v \\end{align} all equations taken !!\\bmod R!!. But since !!P!! is an !!n!!-cycle, !!P^n(i) = i!!, so we have $$a_i \\equiv 2^na_i + v\\pmod R$$ or equivalently $$\\big(2^n-1\\big)a_i + v \\equiv 0\\pmod R\\tag{\\star}$$ where !!v\\in\\{0,\\ldots 2^n-1\\}!! depends only on the values of the carry bits !!c_i!!—the !!c_i!! are precisely the binary digits of !!v!!. Specifying a particular value of !!a_0!! and !!v!! that satisfy this equation completely determines all the !!a_i!!. For example, !!a_0 = 2, v = \\color{darkblue}{0010}_2 = 2!! is a solution when !!R=8, n=4!! because !!\\bigl(2^4-1\\bigr)\\cdot2 + 2\\equiv 0\\pmod 8!!, and this solution allows us to compute \\def\\db#1{\\color{darkblue}{#1}}\\begin{align} a_0&&&=2\\\\ a_{P(0)} &= 2a_0 &+ \\db0 &= 4\\\\ a_{P^2(0)} &= 2a_{P(0)} &+ \\db0 &= 0 \\\\ a_{P^3(0)} &= 2a_{P^2(0)} &+ \\db1 &= 1\\\\ \\hline a_{P^4(0)} &= 2a_{P^3(0)} &+ \\db0 &= 2\\\\ \\end{align} where the carry bits !!c_i = \\langle 0,0,1,0\\rangle!! are visible in the third column, and all the sums are taken !!\\pmod 8!!. Note that !!a_{P^n(0)} = a_0!! as promised. This derivation of the entire set of !!a_i!! from a single one plus a choice of !!v!! is crucial, so let's see one more example. Let's consider !!R=10, n=3!!. Then we want to choose !!a_0!! and !!v!! so" ]

[ "that it equals $|\\mathbf{H}| = -1$ ### Tridiagonalisation The advantage of reflections over the rotations that were used in the Jacobi algorithm is that we can use them to transform an entire row or column of a matrix in a single step. In particular, if we have a reflection that leaves the first element of every vector as it is and transforms a specific vector $$\\mathbf{x}$$ so that its second element equals $$s_0$$ and all of the remaining elements equal zero $\\mathbf{H}_0 \\times \\begin{pmatrix}x_0\\\\x_1\\\\x_2\\\\x_3\\\\\\vdots\\\\x_{n-1}\\end{pmatrix} = \\begin{pmatrix}x_0\\\\s_0\\\\0\\\\0\\\\\\vdots\\\\0\\end{pmatrix}$ then, if $$\\mathbf{x}$$ equals the first row and column of a symmetric matrix $$\\mathbf{M}$$, we have $\\mathbf{H}_0 \\times \\small \\begin{pmatrix} m_{0,0} & m_{0,1} & m_{0,2} & m_{0,3} & \\dots & m_{0,n-1}\\\\ m_{0,1} & m_{1,1} & m_{1,2} & m_{1,3} & \\dots & m_{1,n-1}\\\\ m_{0,2} & m_{1,2} & m_{2,2} & m_{2,3} & \\dots & m_{2,n-1}\\\\ m_{0,3} & m_{1,3} & m_{2,3} & m_{3,3} & \\dots & m_{3,n-1}\\\\ \\vdots & \\vdots & \\vdots & \\vdots & \\ddots & \\vdots\\\\ m_{0,n-1} & m_{1,n-1} & m_{2,n-1} & m_{3,n-1} & \\dots & m_{n-1,n-1} \\end{pmatrix} = \\small \\begin{pmatrix} m_{0,0} & m_{0,1} & m_{0,2} & m_{0,3} & \\dots & m_{0,n-1}\\\\ s_0 & m^\\prime_{1,1} & m^\\prime_{1,2} & m^\\prime_{1,3} & \\dots & m^\\prime_{1,n-1}\\\\ 0 & m^\\prime_{1,2} & m^\\prime_{2,2} & m^\\prime_{2,3} & \\dots & m^\\prime_{2,n-1}\\\\ 0 & m^\\prime_{1,3} & m^\\prime_{2,3} & m^\\prime_{3,3} & \\dots", "need to find a reflection that transforms the first column of matrix A, vector $\\mathbf{a}_1 = (12, 6, -4)^T$, to $\\|\\mathbf{a}_1\\| \\;\\mathrm{e}_1 = (14, 0, 0)^T.$ Now, $\\mathbf{u} = \\mathbf{x} + \\alpha\\mathbf{e}_1,$ and $\\mathbf{v} = {\\mathbf{u}\\over\\|\\mathbf{u}\\|}.$ Here, $\\alpha = -14$ and $\\mathbf{x} = \\mathbf{a}_1 = (12, 6, -4)^T$ Therefore $\\mathbf{u} = (-2, 6, -4)^T=({2})(-1, 3, -2)^T$ and $\\mathbf{v} = {1 \\over \\sqrt{14}}(-1, 3, -2)^T$, and then $Q_1 = I - {2 \\over \\sqrt{14} \\sqrt{14}} \\begin{pmatrix} -1 \\\\ 3 \\\\ -2 \\end{pmatrix}\\begin{pmatrix} -1 & 3 & -2 \\end{pmatrix}$ $= I - {1 \\over 7}\\begin{pmatrix} 1 & -3 & 2 \\\\ -3 & 9 & -6 \\\\ 2 & -6 & 4 \\end{pmatrix}$ $= \\begin{pmatrix} 6/7 & 3/7 & -2/7 \\\\ 3/7 &-2/7 & 6/7 \\\\ -2/7 & 6/7 & 3/7 \\\\ \\end{pmatrix}.$ Now observe: $Q_1A = \\begin{pmatrix} 14 & 21 & -14 \\\\ 0 & -49 & -14 \\\\ 0 & 168 & -77 \\end{pmatrix},$ so we already have almost a triangular matrix. We only need to zero the (3, 2) entry. Take the (1, 1) minor, and then apply the process again to $A' = M_{11} = \\begin{pmatrix} -49 & -14 \\\\ 168 & -77 \\end{pmatrix}.$ By the same method as above, we obtain the matrix of the Householder transformation $Q_2 = \\begin{pmatrix} 1 & 0 & 0", "# Find Normal from Reflection I have a vector A with direction $\\vec D$. I also have another vector B with direction $\\vec R$. (think of $\\vec R$ to be the reflection of $\\vec D$ at point B.) From the reflection formula:$$\\vec R = \\vec D-2*(\\vec D \\cdot \\vec N)*\\vec N$$ (we know both $\\vec R$ and $\\vec D$) How can I find $\\vec N$? • Do you understand how the reflection formula was derived? That should give you a clue. – amd Dec 10 '15 at 19:34 The reflection formula says that to reflect a vector $\\mathbf v$ in the hyperplane specified by its normal $\\mathbf n$, you reverse the component of $\\mathbf v$ that’s in the direction of $\\mathbf n$. That is, you decompose $\\mathbf v$ into $\\pi_{\\mathbf n}\\mathbf v+(\\mathbf v-\\pi_{\\mathbf n}\\mathbf v)$, where $\\pi_{\\mathbf n}$ is orthogonal projection onto $\\mathbf n$, and then negate the first component: $$\\mathbf v'=-\\pi_{\\mathbf n}\\mathbf v+(\\mathbf v-\\pi_{\\mathbf n}\\mathbf v)=\\mathbf v-2\\pi_{\\mathbf n}\\mathbf v.$$ To recover the normal given $\\mathbf v$ and $\\mathbf v'$, remember that $\\pi_{\\mathbf n}\\mathbf v$ is a scalar multiple of $\\mathbf n$, so $\\mathbf v-\\mathbf v'=2\\pi_{\\mathbf n}\\mathbf v$ is a multiple of $\\mathbf n$ and is also normal to the hyperplane. Without more information about $\\mathbf n$ and its relationship to $\\mathbf v$, that’s the best you can", "x-axis. When we mirror a vector ${\\displaystyle {\\begin{pmatrix}x\\\\y\\end{pmatrix}}}$ along the x-axis, we keep its x-component fixed and change the sign of its y-component. The reflection is thus given by ${\\displaystyle s:\\mathbb {R} ^{2}\\to \\mathbb {R} ^{2};\\,s{\\begin{pmatrix}x\\\\y\\end{pmatrix}}:={\\begin{pmatrix}x\\\\-y\\end{pmatrix}}}$ The first basis vector lies on the x-axis and is therefore not affected by the reflection. Formally: ${\\displaystyle s{\\begin{pmatrix}1\\\\0\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\end{pmatrix}}.}$ The second basis vector is perpendicular to the x-axis and is therefore mapped to its negative. Formally: ${\\displaystyle s{\\begin{pmatrix}0\\\\1\\end{pmatrix}}={\\begin{pmatrix}0\\\\-1\\end{pmatrix}}.}$ As the matrix associated with this reflection, we thus obtain: ${\\displaystyle {\\begin{pmatrix}1&0\\\\0&-1\\end{pmatrix}}.}$ ## Matrix-Vector Multiplication ### Derivation We have just seen how we can represent a linear map by a matrix. Suppose, we now do not a linear map, but only its associated matrix. What does the image of an arbitrary vector under this linear map look like? First, for simplicity, let's consider the vector space ${\\displaystyle \\mathbb {R} ^{2}}$ and any linear map ${\\displaystyle f:\\mathbb {R} ^{2}\\rightarrow \\mathbb {R} ^{2}}$ be a linear map, of which we know that the associated matrix is ${\\displaystyle A={\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}}\\in \\mathbb {R} ^{2\\times 2}}$ That means, we have ${\\displaystyle f{\\begin{pmatrix}1\\\\0\\end{pmatrix}}={\\begin{pmatrix}a\\\\c\\end{pmatrix}}}$ and ${\\displaystyle f{\\begin{pmatrix}0\\\\1\\end{pmatrix}}={\\begin{pmatrix}b\\\\d\\end{pmatrix}}.}$ We want to calculate the image of an arbitrary vector ${\\displaystyle (x,y)^{T}\\in \\mathbb {R} ^{2}}$ under the map ${\\displaystyle f}$, using the entries of the matrix ${\\displaystyle A}$. To do so, we represent our vector", "{\\color{red}2/3}&{\\color{red}2/3}&{\\color{red}1/3}& \\\\ {\\color{blue}1/3}&{\\color{blue}-2/3}&{\\color{blue}2/3} \\\\ -2/3&1/3&2/3 \\end{pmatrix} \\nonumber$ $\\mathbf{M}^T\\mathbf{M}=\\begin{pmatrix} 1&0&0 \\\\ 0&1&0 \\\\ 0&0&1 \\end{pmatrix}=\\mathbf{I} \\nonumber$ Because $$\\mathbf{M}^T\\mathbf{M}=\\mathbf{I}$$, the matrix is orthogonal. We now how to prove that the columns for a set of orthonormal vectors. The vectors are: $\\mathbf{a}=2/3\\mathbf{i}+2/3\\mathbf{j}+1/3\\mathbf{i}= \\begin{pmatrix}2/3\\\\2/3\\\\1/3\\end{pmatrix}\\nonumber$ $\\mathbf{b}=1/3\\mathbf{i}-2/3\\mathbf{j}+2/3\\mathbf{i}= \\begin{pmatrix}1/3\\\\-2/3\\\\2/3\\end{pmatrix}\\nonumber$ $\\mathbf{c}=-2/3\\mathbf{i}+1/3\\mathbf{j}+2/3\\mathbf{i}= \\begin{pmatrix}-2/3\\\\1/3\\\\2/3\\end{pmatrix}\\nonumber$ The modulii of these vectors are: $|\\mathbf{a}|^2=(2/3)^2+(2/3)^2+(1/3)^2=1 \\nonumber$ $|\\mathbf{b}|^2=(1/3)^2+(-2/3)^2+(2/3)^2=1 \\nonumber$ $|\\mathbf{c}|^2=(-2/3)^2+(1/3)^2+(2/3)^2=1 \\nonumber$ which proves that the vectors are normalized. The dot products of the three pairs of vectors are: $\\mathbf{a}\\cdot\\mathbf{b}=(2/3)(1/3)+(2/3)(-2/3)+(1/3)(2/3)=0 \\nonumber$ $\\mathbf{a}\\cdot\\mathbf{c}=(2/3)(-2/3)+(2/3)(1/3)+(1/3)(2/3)=0 \\nonumber$ $\\mathbf{c}\\cdot\\mathbf{b}=(-2/3)(1/3)+(1/3)(-2/3)+(2/3)(2/3)=0 \\nonumber$ which proves they are mutually orthogonal. Because the vectors are normalized and mutually orthogonal, they form an orthonormal set. Orthogonal matrices, when thought as operators that act on vectors, are important because they produce transformations that preserve the lengths of the vectors and the relative angles between them. For example, in two dimensions, the matrix $\\mathbf{M}_1=1/\\sqrt{2}\\begin{pmatrix} 1&1 \\\\ 1&-1 \\end{pmatrix} \\nonumber$ is an operator that rotates a vector by $$\\pi/4$$ in the counterclockwise direction (Figure $$\\PageIndex{1}$$) preserving the lengths of the vectors and their relative orientation. In other words, an orthogonal matrix rotates a shape without distorting it. If the columns are orthogonal vectors that are not normalized, as in $\\mathbf{M}_2=\\begin{pmatrix} 1&1 \\\\ 1&-1 \\end{pmatrix}, \\nonumber$ the object changes in size but the shape is not distorted. If, however, the two columns are non-orthogonal vectors, the transformation will distort", "+0 # I will really appreciate any help. 0 28 1 I will really appreciate any help. Let $$\\mathbf{A}, \\mathbf{B}, \\mathbf{C}, \\mathbf{D}$$ be the following matrices: \\begin{align*} &\\mathbf{A} = \\begin{pmatrix} 0 & 0 \\\\ 0& 0 \\end{pmatrix},\\\\ &\\mathbf{B} = \\begin{pmatrix} -1 & 0 \\\\ 0 & -1\\end{pmatrix}, \\\\ &\\mathbf{C} = \\begin{pmatrix} 1/2 & 1/2 \\\\ 1/2 & 1/2\\end{pmatrix}, \\\\ &\\mathbf{D} = \\begin{pmatrix} 0 & -1 \\\\ -1 & 0 \\end{pmatrix}. \\end{align*} Also, consider the following functions: \\begin{align*} &\\text{Reflection across a line with direction vector \\begin{pmatrix}1\\\\-1 \\end{pmatrix}},\\\\ &\\text{The function that sends every vector to the zero vector}, \\phantom{\\begin{pmatrix}1\\\\-1 \\end{pmatrix}}\\\\ &\\text{The function that sends every vector to its negative},\\phantom{\\begin{pmatrix}1\\\\-1 \\end{pmatrix}}\\\\ &\\text{Projection onto \\begin{pmatrix}1\\\\1 \\end{pmatrix}}. \\end{align*} How do we order the list $$\\mathbf{A}, \\mathbf{B}, \\mathbf{C}, \\mathbf{D}$$ to match the order of the functions above? I am having a hard time understanding problem such as these. Thank you in advance for any help Jul 20, 2022", "# Finding the orthogonal projection of a vector I can't figure this out. How the book explains to do this problem doesn't make sense to me. I'm hoping someone can simplify this so I can understand how to do this step by step. Let $L$ be the line in $\\mathbb{R}^3$ that consists of all scalar multiples of $$\\left[ \\begin{array}{ccc} 2\\\\ 1\\\\ 2\\\\ \\end{array} \\right]$$ Find the reflection of the vector $$\\left[ \\begin{array}{ccc} 1\\\\ 4\\\\ 1\\\\ \\end{array} \\right]$$ onto $L$ • Are you asking for reflection (as in the body of the question) or projection (as in the subject)? – Thomas Jul 11 '16 at 17:38 Let $L$ be the line spanned by some non-zero vector $\\mathbf{v} \\in \\mathbb{R}^3$, so that $L = \\{a\\mathbf{v} \\mid a \\in \\mathbb{R}\\}$ is the space of all scalar multiples of $\\mathbf{v}$. Then the orthogonal projection of a vector $\\mathbf{x} \\in \\mathbb{R}^3$ onto the line $L$ can be computed as $$\\operatorname{Proj}_L(\\mathbf{x}) = \\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}.$$ So, in this case, we have $$\\mathbf{v} = \\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}, \\quad \\mathbf{x} = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix},$$ so that $$\\mathbf{v} \\cdot \\mathbf{x} = 2 \\cdot 1 + 1 \\cdot 4 + 2 \\cdot 1 = 8, \\quad \\mathbf{v} \\cdot \\mathbf{v} = 2^2 + 1^2 + 2^2 = 9,$$ and hence $$\\operatorname{Proj}_L(\\mathbf{x}) = \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}.$$ Now, you probably wanted to compute the orthogonal projection", "a ball at position $(0,0,1)$, i.e. on the unit vector $\\mathbf e_z$, its shadow appears at the point $(-1,0,0)$. If you place it on the table, of course the shadow will \"coincide\" with the ball, i.e. $\\mathbf e_x$ and $\\mathbf e_y$ will be fixed, as before. So the matrix you are looking for is $$C = \\begin{pmatrix} 1 & 0 & -1 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$$ and you may check that the null-space of $C$ is spanned by $(x,0,x)$, i.e. it is a diagonal line. This kinds of matrices represent linear operators called projectors, i.e. such that if you apply them twice you get the same result ($PP = P$: this identity is evident in the vertical-shadow example – try multiplying matrix $A$ by itself).", "eigenvector $$\\mathbf y_2$$. Tis must satisfy $$\\mathbf y_2 \\in \\ker \\mathbf A^2 \\setminus \\ker \\mathbf A$$ and is linearly independent on $$\\mathbf x_2$$. We choose e.g. $$\\mathbf y_2=(0{,}1,0{,}0,0)^T$$ and calculate $$\\mathbf y_1=\\mathbf A\\mathbf y_2=(1{,}0,0{,}0,0)^T$$ (opět vlastní vektor $$\\mathbf A$$). Now we may assemble matrices $$\\mathbf R$$ (columns are $$\\mathbf y_1,\\mathbf y_2,\\mathbf x_1,\\mathbf x_2,\\mathbf x_3$$) and $$\\mathbf J$$ and claculate the inverse matrix $$\\mathbf R^{-1}$$ $$\\mathbf R= \\begin{pmatrix} 1&0&0&-1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&-1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$, $$\\mathbf J= \\begin{pmatrix} 0&1&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1\\\\ 0&0&0&0&0 \\end{pmatrix}$$, $$\\mathbf R^{-1}=\\begin{pmatrix} 1&0&0&1&0\\\\ 0&1&0&0&0\\\\ 0&0&1&1&0\\\\ 0&0&0&1&0\\\\ 0&0&0&0&1 \\end{pmatrix}$$", "$\\mathbf{u}=(4,-1), \\mathbf{v}=(0,5)$, and $\\mathbf{w}=(-3,-3)$. Find the components of (a) $\\mathbf{u}+\\mathbf{w}$ (b) $\\mathbf{v}-3 \\mathbf{u}$ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find a unit vector that is orthogonal to both $\\mathbf{u}=(1,0,1)$ and $\\mathbf{v}=(0,1,1)$ Find a unit vector that is orthogonal to both $\\mathbf{u}=(1,0,1)$ and $\\mathbf{v}=(0,1,1)$Find a unit vector that is orthogonal to both $\\mathbf{u}=(1,0,1)$ and $\\mathbf{v}=(0,1,1)$ ... Let $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components of Let $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components ofLet $\\mathbf{u}=(-3,1,2), \\mathbf{v}=(4,0,-8)$, and $\\mathbf{w}=(6,-1,-4)$. Find the components of (a) $\\mathbf{v}-\\mathbf{w}$ (b) $6 \\mathrm{u}+2 \\ma ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find the terminal point of the vector that is equivalent to$\\mathbf{u}=(1,2)$and whose initial point is$A(1,1)$. 0 answers 83 views Find the terminal point of the vector that is equivalent to$\\mathbf{u}=(1,2)$and whose initial point is$A(1,1)$.Find the terminal point of the vector that is equivalent to$\\mathbf{u}=(1,2)$and whose initial point is$A(1,1)$. ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 In relation to the points$P_{1}$and$P_{2}$, what can you say about the terminal point of the following vector if its initial point is at the origin? 0 answers 40 views In relation to the points$P_{1}$and$P_{2}$, what can you say about the terminal point of the following vector if its initial point is at the origin?In relation to", "### $( 2 \\implies 1)$ Next suppose that the statement 2 is true. That is, there exists an invertible matrix $S$ such that $S^{-1}AS=C$. Then we have $AS=CS$. Define $\\mathbf{v}:=S\\mathbf{e}_1$ to be the first column vector of $S$. We prove that with this choice of $\\mathbf{v}$, the vectors $\\mathbf{v}, A\\mathbf{v}, A^2\\mathbf{v}, \\dots, A^{n-1}\\mathbf{v}$ form a basis of $\\C^n$. We have \\begin{align*} AS&=A[S\\mathbf{e}_1, S\\mathbf{e}_2, \\dots, S\\mathbf{e}_n]\\\\ &=[AS\\mathbf{e}_1, AS\\mathbf{e}_2, \\dots, AS\\mathbf{e}_n] \\end{align*} and \\begin{align*} SC&=S [\\mathbf{e}_2, \\mathbf{e}_3, \\dots, \\mathbf{e}_n, -\\mathbf{a}]\\\\ &=[S \\mathbf{e}_2, S\\mathbf{e}_3, \\dots, S\\mathbf{e}_n, -S\\mathbf{a}]. \\end{align*} Since $AS=SC$, comparing column vectors, we obtain \\begin{align*} AS\\mathbf{e}_1&=S\\mathbf{e}_2\\\\ AS\\mathbf{e}_2&=S\\mathbf{e}_3\\\\ \\vdots\\\\ AS\\mathbf{e}_{n-1}&=S\\mathbf{e}_n\\\\ AS\\mathbf{e}_n&=-S\\mathbf{a}.\\\\ \\end{align*} It follows that we have inductively \\begin{align*} A\\mathbf{v}&=S\\mathbf{e}_2\\\\ A^2\\mathbf{v}&=A(A\\mathbf{v})=AS\\mathbf{e}_2=S\\mathbf{e}_3\\\\ \\vdots\\\\ A^{n-1}\\mathbf{v}&=A(A^{n-2}\\mathbf{v})=AS\\mathbf{e}_{n-1}=S\\mathbf{e}_n. \\end{align*} Namely we obtain $A^i\\mathbf{v}=S\\mathbf{e}_{i+1}$ for $i=0, 1, \\dots, n-1$. Now, suppose that we have the linear combinaiton $c_1\\mathbf{v}+c_2A\\mathbf{v}+ \\dots c_n A^{n-1}\\mathbf{v}=\\mathbf{0},$ for some coefficient scalars $c_1, \\dots, c_n \\in \\C$. Then we have using the above relations \\begin{align*} c_1S\\mathbf{e}_1+c_2S\\mathbf{e}_2+\\cdots +c_nS\\mathbf{e}_n=\\mathbf{0}. \\end{align*} Since $S$ is invertible, the column vectors $S\\mathbf{e}_i$ are linearly independent. Thus, we must have $c_1=c_2=\\cdots=c_n=0$. Therefore the vectors $\\mathbf{v}, A\\mathbf{v}, A^2\\mathbf{v}, \\dots, A^{n-1}\\mathbf{v}$ are linearly independent, and they form a basis of $\\C^n$ since we have $n$ linearly independent vectors in the $n$-dimensional vector space $\\C^n$. ## Related Question. For a basic question about the companion matrix of a polynomial, check out the post “Companion", "# arrow_back Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is orthogonal. 8 views Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is orthogonal. It suffices to show that $$\\mathbf{i} \\cdot \\mathbf{j}=\\mathbf{i} \\cdot \\mathbf{k}=\\mathbf{j} \\cdot \\mathbf{k}=\\mathbf{0}$$ because it will follow automatically from the symmetry property of the dot product that $$\\mathbf{j} \\cdot \\mathbf{i}=\\mathbf{k} \\cdot \\mathbf{i}=\\mathbf{k} \\cdot \\mathbf{j}=\\mathbf{0}$$ Although the orthogonality of the vectors in $S$ is evident geometrically from Figure $3.2 .2$ it is confirmed algebraically by the computations \\begin{aligned} &\\mathbf{i} \\cdot \\mathbf{j}=(1,0,0) \\cdot(0,1,0)=0 \\\\ &\\mathbf{i} \\cdot \\mathbf{k}=(1,0,0) \\cdot(0,0,1)=0 \\\\ &\\mathbf{j} \\cdot \\mathbf{k}=(0,1,0) \\cdot(0,0,1)=0 \\end{aligned} by Platinum (106,844 points) selected by ## Related questions Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Show that $\\mathbf{u}=(-2,3,1,4)$ and $\\mathbf{v}=(1,2,0,-1)$ are orthogonal vectors in $R^{4}$. Show that $\\mathbf{u}=(-2,3,1,4)$ and $\\mathbf{v}=(1,2,0,-1)$ are orthogonal vectors in $R^{4}$.Show that $\\mathbf{u}=(-2,3,1,4)$ and $\\mathbf{v}=(1,2,0,-1)$ are orthogonal vectors in $R^{4}$. ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors. Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors.Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors. ... close Notice: Undefined index: avatar", "representations for $I_{\\pm}^{R_1 \\otimes R_2}$ and I would end up with $4 \\times 4$ matrices. But what would the $4 \\times 1$ objects that these operators act on represent? Would a generic vector be something like $(u, d, \\bar u, \\bar d)$ so for example I would write $u = (1,0,0,0)$ and $\\bar d = (0,0,0,1)$ for example? It seems to me that there's a confusion about how the flavour is acting. In your example $R_1 \\sim \\bar{\\bf{2}}$ and $R_2 \\sim {\\bf{2}}$. Therefore, by multiplying these representations we expect a singlet state ($\\ell =0$) and a vector ($\\ell =1$), i.e., $R_1 \\otimes R_2 = \\bf{1} \\oplus \\bf{3}$. Now, if we identify the fundamental doublet as: $R_2 = \\begin{pmatrix} u \\\\ d \\end{pmatrix}\\,,$ then $R_1 = \\begin{pmatrix} \\bar{d} \\\\ -\\bar{u} \\end{pmatrix}\\,,$ and: $R_1\\otimes R_2 = \\frac{1}{\\sqrt{2}}(\\bar{u}{u}+\\bar{d}{d}) \\oplus \\begin{pmatrix} \\bar{d}u \\\\ \\frac{1}{\\sqrt{2}}(\\bar{d}{d}-\\bar{u}{u})\\\\ -\\bar{u}d \\end{pmatrix}$ The $\\pi^{\\pm}$ state you are looking for is in the vector of course since it is electrically charged. Edit: Here I give the derivation based on the isospin matrices. We can define the matrix $$I_{++} =I_+\\otimes I_+\\,.$$ This matrix can act on the tensor product $R=R_1 \\otimes R_2$ as $$I_{++} R =I_+ R_1 \\otimes I_+ R_2=\\begin{pmatrix} \\bar{d} \\\\ 0 \\end{pmatrix}\\otimes \\begin{pmatrix} u\\\\ 0 \\end{pmatrix} \\equiv |\\bar{d}>|u> \\equiv |\\pi^+>$$ Applying $I_{--}$ and $I_{0^+}\\equiv\\frac{1}{\\sqrt 2} (I_{+-}+I_{-+})$ will give", "that there are other ways of “deriving” this result. One is by the use of a diagram, which would show that $(1,0)$ gets reflected to $(\\cos 2\\theta,\\sin 2\\theta)$ and $(0,1)$ gets reflected to $(\\sin 2\\theta,-\\cos 2\\theta)$. Another way is to observe that we can rotate an arbitrary mirror line onto the x-axis, then reflect across the x-axis, and rotate back. (The matrix product $[T]_{xy}$ can be seen as operating this way.) We took neither of these two approaches, because to justify them rigorously takes a bit of work, that is avoided by the pure linear algebra approach. Note also that $[T]_{uv}$ and $[T]_{xy}$ are orthogonal matrices, with determinant $-1$, as expected. ## Reflection across a line of given direction vector Suppose instead of being given an angle $\\theta$, we are given the unit direction vector $u$ to reflect the vector $w$. We can derive the matrix for the reflection directly, without involving any trigonometric functions. In the decomposition $\\mathbf{w}=a\\mathbf{u}+b\\mathbf{v}$, we note that $b=\\mathbf{w}\\cdot\\mathbf{v}$. Therefore $\\mathbf{w}^{\\prime}=(a\\mathbf{u}+b\\mathbf{v})-2b\\mathbf{v}=\\mathbf{w}-2(% \\mathbf{w}\\cdot\\mathbf{v})\\mathbf{v}\\,.$ (In fact, this is the formula used in the to draw the diagram in this entry.) To derive the matrix with respect to $x,y$ coordinates, we resort to a trick: $\\mathbf{w}^{\\prime}=I\\mathbf{w}-2\\mathbf{v}(\\mathbf{w}\\cdot\\mathbf{v})=I% \\mathbf{w}-2\\mathbf{v}(\\mathbf{v}^{\\textrm{tr}}\\mathbf{w})=I\\mathbf{w}-2(% \\mathbf{v}\\mathbf{v}^{\\textrm{tr}})\\mathbf{w}\\,.$ Therefore the matrix of the transformation is $I-2\\mathbf{v}\\mathbf{v}^{\\textrm{tr}}=\\begin{bmatrix}u_{x}^{2}-u_{y}^{2}&2u_{x% }u_{y}\\\\ 2u_{x}u_{y}&u_{y}^{2}-u_{x}^{2}\\end{bmatrix}\\,,\\quad\\mathbf{u}=(u_{x},u_{y})^{% \\textrm{tr}}\\,,\\quad\\mathbf{v}=(-u_{y},u_{x})^{\\textrm{tr}}\\,.$ If $u$ was not a unit vector to", "another element satisfying axiom (a4). Then we have $v+(-v)=0=v+w.$ By the cancellation law (see (a)), we have $-v=w$. Thus, the additive inverse is unique. ### (e) $0v=\\mathbf{0}$ for every $v\\in V$, where $0\\in\\R$ is the zero scalar. Note that $0$ is a real number and $\\mathbf{0}$ is the zero vector in $V$. For $v\\in V$, we have \\begin{align*} 0v&=(0+0)v\\stackrel{(m3)}{=} 0v+0v. \\end{align*} We also have $0v\\stackrel{(a3)}{=} \\mathbf{0}+0v.$ Hence, combining these, we see that $0v+0v=\\mathbf{0}+0v,$ and by the cancellation law, we obtain $0v=\\mathbf{0}$. ### (f) $a\\mathbf{0}=\\mathbf{0}$ for every scalar $a$. Note that we have $\\mathbf{0}+\\mathbf{0}=\\mathbf{0}$ by (a3). Thus, we have \\begin{align*} a\\mathbf{0}=a(\\mathbf{0}+\\mathbf{0})\\stackrel{(m2)}{=}a\\mathbf{0}+a\\mathbf{0}. \\end{align*} We also have $a\\mathbf{0}=\\mathbf{0}+a\\mathbf{0}$ by (a3). Combining these, we have $a\\mathbf{0}+a\\mathbf{0}=\\mathbf{0}+a\\mathbf{0},$ and the cancellation law yields $a\\mathbf{0}=\\mathbf{0}$. ### (g) If $av=\\mathbf{0}$, then $a=0$ or $v=\\mathbf{0}$. For this problem, we use a little bit logic. Our assumption is $av=\\mathbf{0}$. From this assumption, we need to deduce that either $a=0$ or $v=\\mathbf{0}$. Note that if $a=0$, then we are done as this is one of the consequence we want. So, let us assume that $a\\neq 0$. Then we want to prove $v=0$. Since $a$ is a nonzero scalar, we have $a^{-1}$. Then we have \\begin{align*} a^{-1}(av)=a^{-1}\\mathbf{0}. \\end{align*} The right hand side $a^{-1}\\mathbf{0}$ is $\\mathbf{0}$ by part (f). On the other hand, the left hand side can be computed as", "\\end{pmatrix};$ $\\begin{matrix} R = \\end{matrix} \\begin{pmatrix} 14 & 21 & -14 \\\\ 1.11022302462516 \\times 10^{-016} & 175 & -70 \\\\ -1.77635683940025 \\times 10^{-015} & -5.32907051820075 \\times 10^{-014} & 35 \\end{pmatrix}.$ ## Computing QR by means of Householder reflections A Householder reflection (or Householder transformation) is a transformation that takes a vector and reflects it about some plane. We can use this property to calculate the QR factorization of a matrix. Q can be used to reflect a vector in such a way that all coordinates but one disappear. Let x be an arbitrary m-dimensional column vector of length |α| (for numerical reasons α should get the same sign as the first coordinate of x). Then, where e1 is the vector (1,0,...,0)T, and || || the euclidean norm, set $\\mathbf{u} = \\mathbf{x} - \\alpha\\mathbf{e}_1,$ $\\mathbf{v} = {\\mathbf{u}\\over||\\mathbf{u}||},$ $Q = I - 2 \\mathbf{v}\\mathbf{v}^T.$. Q is a Householder matrix and $Qx = (\\alpha\\ , 0, \\cdots, 0)^T$. This can be used to gradually transform an m-by-n matrix A to upper triangular form. First, we multiply A with the Householder matrix Q1 we obtain when we choose the first matrix column for x. This results in a matrix QA with zeros in the left column (except for the first line). $Q_1A = \\begin{bmatrix} \\alpha_1&\\star&\\dots&\\star\\\\ 0 & & & \\\\ \\vdots &", "matrix is given formally by the next equation: $R=M-\\dfrac{\\mathbf{B}\\mathbf{A}^T}{1+\\gamma}$ R must be an orthogonal matrix, i.e., $R^TR=\\mathbb{I}_{3x3}$. Then $(\\det (R))^2=1$, or $det R=\\pm 1.$. For $\\det R=-1$ we have the parity matrix $\\mathbb{P}=\\begin{pmatrix}1 & \\mathbf{0}^T\\\\ \\mathbf{0} & -\\mathbb{I}_{3x3}\\end{pmatrix}$ and it will transform right-handed frames to left-handed frames $\\overline{S}$ or $\\overline{S}'$. The rotation vector $\\alpha$ can be defined as well: $1+2\\cos \\alpha=Tr (R)\\rightarrow \\cos\\alpha=\\dfrac{Tr R-1}{2}$ so $\\alpha^\\mu=\\dfrac{1}{2}\\epsilon^{\\mu\\nu\\lambda}R^\\nu_{\\lambda}\\dfrac{\\alpha}{\\sin\\alpha}, \\forall 0\\leq \\alpha<\\pi$. The rotation acting on 3-rowed matrices: $R\\mathbf{A}=\\mathbf{B}$ implies that $\\overline{X}'=R\\overline{X}$, and it changes $-\\mathbf{A}/\\gamma$ of the frame S into $\\overline{S}$. Passing from one frame into another, $\\overline{S}'$ to $S'$, it implies we can define a boost with $L_{-\\mathbf{B}/\\gamma}$. In fact, $L_{-\\mathbf{B}/\\gamma}L=\\begin{pmatrix}1 & \\mathbf{0}^T\\\\ \\mathbf{0} & R\\end{pmatrix}=L_R$ Q.E.D. Remark(I): Without the time reversal, we would get $L_{R\\mathbf{v}}L_R=L=L_RL_{\\mathbf{v}}$ with $\\mathbf{v}=\\mathbf{A}/\\gamma$ and $R=M-\\dfrac{\\mathbf{BA}^T}{1+\\gamma}$. Remark (II): $L_RL_v\\rightarrow L^T=L^T_vL_R^T=L_vL_{R^T}$. If $L^T=L=L_{R\\mathbf{v}}L_R$, then the uniqueness of $R\\mathbf{v}$ provides that $R=R^T=R^{-1}$, i.e., that R is an orthogonal matrix. If R is an orthogonal matrix and a proper Lorentz transformation ( $det R=+1$), then we would get $\\sin\\alpha=0$, and thus $\\alpha=0$ or $\\alpha=\\pi$, and so, $R=I$ or $R=2\\mathbf{n}\\mathbf{n}^T-1$, with the unimodular vector $\\mathbf{n}$, i.e., $\\vert \\mathbf{n}\\vert=1$. That would be the case $\\forall \\mathbf{v}\\neq 0$ and $\\mathbf{n}=\\mathbf{v}/\\vert\\vert \\mathbf{v}\\vert\\vert$. Otherwise, if $\\mathbf{v}=0$, then $\\mathbf{n}$ would be an arbitrary vector. The second step previous to our treatment of Thomas precession", "\\end{bmatrix}}.}$ ### Reflection To reflect a point through a plane ${\\displaystyle ax+by+cz=0}$ (which goes through the origin), one can use ${\\displaystyle \\mathbf {A} =\\mathbf {I} -2\\mathbf {NN} ^{T}}$ , where ${\\displaystyle \\mathbf {I} }$ is the 3x3 identity matrix and ${\\displaystyle \\mathbf {N} }$ is the three-dimensional unit vector for the vector normal of the plane. If the L2 norm of ${\\displaystyle a,b,}$ and ${\\displaystyle c}$ is unity, the transformation matrix can be expressed as: ${\\displaystyle \\mathbf {A} ={\\begin{bmatrix}1-2a^{2}&-2ab&-2ac\\\\-2ab&1-2b^{2}&-2bc\\\\-2ac&-2bc&1-2c^{2}\\end{bmatrix}}}$ Note that these are particular cases of a Householder reflection in two and three dimensions. A reflection about a line or plane that does not go through the origin is not a linear transformation — it is an affine transformation — as a 4x4 affine transformation matrix, it can be expressed as follows (assuming the normal is a unit vector): ${\\displaystyle {\\begin{bmatrix}x'\\\\y'\\\\z'\\\\1\\end{bmatrix}}={\\begin{bmatrix}1-2a^{2}&-2ab&-2ac&-2ad\\\\-2ab&1-2b^{2}&-2bc&-2bd\\\\-2ac&-2bc&1-2c^{2}&-2cd\\\\0&0&0&1\\end{bmatrix}}{\\begin{bmatrix}x\\\\y\\\\z\\\\1\\end{bmatrix}}}$ where ${\\displaystyle d=\\mathbf {-p} \\cdot \\mathbf {N} }$ for some point ${\\displaystyle \\mathbf {p} }$ on the plane. If the 4th component of the vector is 0 instead of 1, then only the vector's direction is reflected and its length remains unchanged, as if it were mirrored through a parallel plane that passes through the origin. This is a useful property as it allows the transformation of both positional vectors and normal vectors with the same matrix.", "this equation completely determines all the $$a_i$$. For example, $$a_0 = 2, v = \\color{darkblue}{0010}_2 = 2$$ is a solution when $$R=8, n=4$$ because $$\\bigl(2^4-1\\bigr)\\cdot2 + 2\\equiv 0\\pmod 8$$, and this solution allows us to compute \\def\\db#1{\\color{darkblue}{#1}}\\begin{align} a_0&&&=2\\\\ a_{P(0)} &= 2a_0 &+ \\db0 &= 4\\\\ a_{P^2(0)} &= 2a_{P(0)} &+ \\db0 &= 0 \\\\ a_{P^3(0)} &= 2a_{P^2(0)} &+ \\db1 &= 1\\\\ \\hline a_{P^4(0)} &= 2a_{P^3(0)} &+ \\db0 &= 2\\\\ \\end{align} where the carry bits $$c_i = \\langle 0,0,1,0\\rangle$$ are visible in the third column, and all the sums are taken $$\\pmod 8$$. Note that $$a_{P^n(0)} = a_0$$ as promised. This derivation of the entire set of $$a_i$$ from a single one plus a choice of $$v$$ is crucial, so let’s see one more example. Let’s consider $$R=10, n=3$$. Then we want to choose $$a_0$$ and $$v$$ so that $$\\left(2^3-1\\right)a_0 + v \\equiv 0\\pmod{10}$$ where $$v\\in\\{0\\ldots 7\\}$$. One possible solution is $$a_0 = 5, v=\\color{darklue}{101}_2 = 5$$. Then we can derive the other $$a_i$$ as follows: \\begin{align} a_0&&&=5\\\\ a_{P(0)} &= 2a_0 &+ \\db1 &= 1\\\\ a_{P^2(0)} &= 2a_{P(0)} &+ \\db0 &= 2 \\\\\\hline a_{P^3(0)} &= 2a_{P^2(0)} &+ \\db1 &= 5\\\\ \\end{align} And again we have $$a_{P^n(0)}= a_0$$ as required. Since the bits of $$v$$ are used cyclically, not every pair of $$\\langle a_0, v\\rangle$$ will yield a different solution. Rotating", "a vector parallel to $m$. The result of the reflection will be :$c\\text{'} = = = = -mcm^$ This is not the most general operation that may be regarded as a reflection when the dimension $n \\ge 4$. A general reflection may be expressed as the composite of any odd number of single-axis reflections. Thus, a general reflection $a\\text{'}$ of a vector $a$ may be written :$a \\mapsto a\\text{'} = -MaM^ ,$ where :$M = pq \\cdots r$ and $M^ = \\left(pq \\cdots r\\right)^ = r^ \\cdots q^p^ .$ If we define the reflection along a non-null vector $m$ of the product of vectors as the reflection of every vector in the product along the same vector, we get for any product of an odd number of vectors that, by way of example, :$\\left(abc\\right)\\text{'} = a\\text{'}b\\text{'}c\\text{'} = \\left(-mam^\\right)\\left(-mbm^\\right)\\left(-mcm^\\right) = -ma\\left(m^m\\right)b\\left(m^m\\right)cm^ = -mabcm^ \\,$ and for the product of an even number of vectors that :$\\left(abcd\\right)\\text{'} = a\\text{'}b\\text{'}c\\text{'}d\\text{'} = \\left(-mam^\\right)\\left(-mbm^\\right)\\left(-mcm^\\right)\\left(-mdm^\\right) = mabcdm^ .$ Using the concept of every multivector ultimately being expressed in terms of vectors, the reflection of a general multivector $A$ using any reflection versor $M$ may be written :$A \\mapsto M\\alpha\\left(A\\right)M^ ,$ where $\\alpha$ is the automorphism of reflection through the origin of the vector space ($v \\mapsto -v$) extended through linearity to the whole algebra." ]

[ "# Angles of an Isosceles Triangle Geometry Level 4 Two angles of an isosceles triangle are $$68 ^\\circ$$ and $$x^\\circ$$. What is the sum of all the (distinct) possible values of $$x$$? ×", "# A Specific Dissection Three vertices are chosen randomly from the vertices of a regular 7-sided polygon. If the probability that they from the vertices of an isosceles triangle can be expressed as $$\\dfrac{a}{b}$$, where $$a$$ and $$b$$ are coprime positive integers, find $$a+b$$. ×", "the three angles in a triangle always add up to $180^\\circ$ . The median in a triangle connects the midpoint of one side to the opposite vertex. An isosceles triangle is a triangle with at least two congruent sides. ### Guided Practice 1. If $\\overline{XM}$ is a median of $\\triangle XYZ$ and $XY > XZ$ , what can we say about $m \\angle 1$ and $m \\angle 2$ ? 2. List the sides of the two triangles in order, from least to greatest. 3. Below is isosceles triangle $\\triangle ABC$ . List everything you can about the triangle and why. 1. By the definition of a median, $M$ is the midpoint of $\\overline{YZ}$ . This means that $YM = MZ$ . $MX = MX$ by the Reflexive Property and we know that $XY > XZ$ . Therefore, we can use the SSS Inequality Theorem to conclude that $m \\angle 1 >m \\angle 2$ . 2. Here we have no congruent sides or angles. So, let’s look at each triangle separately. Start with $\\triangle XYZ$ . First the missing angle is $42^\\circ$ . The order of the sides is $YZ, XY$ , and $XZ$ . For $\\triangle WXZ$ , the missing angle is $55^\\circ$ . The order of these sides is $XZ, WZ$ , and $WX$ . Because the", "isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?", "Question # The lengths of two sides of an isosceles triangle are $15$ and $22$, respectively, what are the possible values of perimeter? Verified 129k+ views Hint: Here we need to apply the concept of isosceles triangle and perimeter of triangle. Isosceles Triangle: The triangle in which the two sides of a triangle are equal, then the angles opposite to equal sides are also equal. Perimeter of triangle = Sum of all sides. As two sides of a triangle are given as $15$ and $22$, respectively. We can fix either the two equal sides as $15$ and the third side is $22$ (or) the two equal sides as $22$ and the third side is $15$. The sides of triangle can be either $15,15,22$ or $15,22,22$ Perimeter of the triangle = $15+22+22=59$ or $15+15+22=52$ Therefore, the possible values of the perimeter are $59\\text{ or 52}$ .", "## Algebra and Trigonometry 10th Edition For an isosceles triangle, given the base angle, $\\theta$, we know that the height is: $$h=.5btan\\theta \\\\ h =.5(8)tan(32^{\\circ})\\\\ h=2.50$$", "? Free Version Easy # Angles in an Isosceles Triangle SATSTM-WEYLI1 If one measure of an angle in an isosceles triangle is $40$ degrees, then which one of the following could be a measure of angle in that same triangle? A $50\\text{ degrees }$ B $60\\text{ degrees }$ C $70\\text{ degrees }$ D $80\\text{ degrees }$ E $110\\text{ degrees }$", "# Inscribed Isosceles Triangle Geometry Level 1 Find the value of $$x$$. The two indicated sides are congruent. $$O$$ is the center of the circle. × Problem Loading... Note Loading... Set Loading...", "with the problem. By what you have stated, we know that $(-z_1,z_2,z_3)$ are the vertices of an isosceles triangle with equal sides meeting at $120^\\circ$. But then we don't know anything much about $(z_1,z_2,z_3)$, because as you can see translating $(-z_1,z_2,z_3)$ around makes $(z_1,z_2,z_3)$ change shape.", "# Inscribed Isosceles Triangle Geometry Level 1 Find the value of $$x$$. The two indicated sides are congruent. $$O$$ is the center of the circle. ×", ". bizarre! I did a lot of sketching and muttering . . and I think I've understood the composite functions. If I'm wrong, my apologies for wasting your time . . . $y_1 \\,=\\,f(x)$ has this graph. Code: | 1+ - - - - - - - * | * * | * * | * * | * * | * * | * * | * * - - * - - - - - - - : - - - - - - - * - - 0 1/2 1 It is an isosceles triangle with base 1 and height 1. The slanted side has length $\\sqrt{1^2+\\left(\\tfrac{1}{2}\\right)^2} \\:=\\:\\tfrac{\\sqrt{5}}{2}$ . . . and there two of them. Hence: . $L_1 \\:=\\:\\sqrt{5}$ $y_2 \\:=\\:f(f(x))$ has this graph. Code: | 1+ - - - o - - - - - - - o | o o o o | o o o o | o o o o | o o o o | o o o o | o o o o |o o o o - - o - - - : - - - o - - - : - - - o - - 0 1/2 1 It has two isosceles triangles with base 1/2 and height 1. The slanted side has length: $\\sqrt{1^2", "Free Version Moderate # Isosceles Triangle ACTMAT-JE13KE $\\Delta$ABC is an isosceles triangle with vertex A and $AB = AC.$ If the measure of ${\\angle A}$ is $\\frac{1}{4}$ the measure of ${\\angle B}$, what is the measure, in degrees, of ${\\angle A}$? A $20°$ B $30°$ C $45°$ D $60°$ E $80°$", "Free Version Difficult # Using Properties of Triangle Angles: Upping the Difficulty GMAT-ETWBYN In the diagram, triangle $JKI$ is an isosceles triangle with equal sides $JK$ and $IK$. Triangle $KIH$ is also isosceles, with equal sides $IK$ and $HK$. Given that angle $J$ is $30$ degrees and angle $G$ is $40$ degrees, what is the measure of angle $IKH$? A $120$ degrees B $20$ degrees C $40$ degrees D $30$ degrees E Not enough information is given.", "an isosceles right triangle with $\\angle B = 90^{\\circ}$. If $A= (3, -2), B = (2, -1)$, then $AC=$ Explanation \\begin{aligned} &\\\\ A C &=\\sqrt{2} A B \\\\\\\\ &=\\sqrt{2} \\sqrt{(2-3)^{2}+(-1+2)^{2}} \\\\\\\\ &=\\sqrt{2} \\sqrt{2} \\\\\\\\ &=2 \\end{aligned}", "How do I prove that (-3, 0), (0,4), (3,0) are the vertices of an isosceles triangle? Jan 5, 2016 The sides $P 1 P 2 = P 2 P 3 = 5$ but $P 3 P 1 = 3$ => ${\\triangle}_{P 1 P 2 P 3}$ is isosceles Explanation: Find the size of the sides of the triangle and show that two sides are equal but the third one is different than the others. $P 1 P 2 = \\sqrt{{\\left(0 + 3\\right)}^{2} + {\\left(4 - 0\\right)}^{2}} = \\sqrt{9 + 16} = \\sqrt{25} = 5$ $P 2 P 3 = \\sqrt{{\\left(3 - 0\\right)}^{2} + {\\left(0 - 4\\right)}^{2}} = \\sqrt{9 + 16} = \\sqrt{25} = 5$ $P 3 P 1 = \\sqrt{{\\left(- 3 - 3\\right)}^{2} + {\\left(0 - 0\\right)}^{2}} \\equiv \\sqrt{9 + 0} = \\sqrt{9} = 3$ The sides $P 1 P 2 = P 2 P 3 = 5$ but $P 3 P 1 = 3$ => ${\\triangle}_{P 1 P 2 P 3}$ is isosceles", "# Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3. 185 views closed Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3. +1 vote by (30.3k points) selected by Vertices of an isosceles are: A(2, 0) and B(2, 5). Let the third vertex is P(x, y) On squaring both sides, we get (x - 2)2 + (y)2 = (x - 2)2 + (y - 5)2 x2 - 4x + 4 + y2 = x2 - 4x + 4 + y2 - 10y + 25 y = $\\frac{5}2,$ Also, PA = 3 $\\sqrt{(x - 2)^2 + (y)^2} = 3$ On squaring both sides, we get (x - 2)2 + (y)2 = 9 x2 - 4x + 4 + y2 = 9 x2 - 4x + y2 = 5 On substituting $y = \\frac{5}2,$ $x^2 - 4x + \\frac{25}4 = 5$ $x^2 - 4x + \\frac{25} 4 - 5 = 0$ $x^2 - 4x + \\frac{5}4 = 0$ Therefore coordinates of third vertex are:", "# The vertex angle of an isosceles triangle measures 20 degrees more than twice the measure of one of its base angles. How many degrees are there in a base angle of this triangle? Nov 24, 2016 The base angles are 40° #### Explanation: Let the the size of each base angle be $x$. The vertex angle is twice $x$, with another 20° Vertex angle = 2x+20° The sum of the three angles in a triangle is always 180° x+x+x^2+20° = 180\" \"larr now solve the equation 4x+20° = 180° 4x = 180°-20° 4x = 160° $x = \\frac{160}{4}$ x = 40° The angles are 40°, 40° and 100°#", "Free Version Moderate # Isosceles Triangle: Find Base Length Given Leg and Altitude GEOM-XEZA1P Each leg of an isosceles triangle is $15$ inches in length, and the altitude from the vertex angle to the base of the triangle is $10$ inches. Find the length of the base in inches. A $2\\sqrt{5}$ B $5\\sqrt{5}$ C $5\\sqrt{13}$ D $10\\sqrt{5}$ E $10\\sqrt{13}$", "# An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 8 and the triangle has an area of 40 . What are the lengths of sides A and B? Apr 26, 2018 Side A = Side B = 10.77 units #### Explanation: Let side A=side B= $x$ Then drawing a right-angled triangle, with $x$ as your hypotenuse and $4$ as the length of your bottom side and the final side be $h$ Using pythagoras theorem, ${x}^{2}$ = ${h}^{2} + {4}^{2}$ ${x}^{2} = {h}^{2} + 16$ ${h}^{2} = {x}^{2} - 16$ $h = \\pm \\sqrt{{x}^{2} - 16}$ But since h is a length, it can only be positive $h = \\sqrt{{x}^{2} - 16}$ Area of triangle = $\\frac{1}{2} \\times 8 \\times \\sqrt{{x}^{2} - 16} = 40$ $4 \\sqrt{{x}^{2} - 16} = 40$ $\\sqrt{{x}^{2} - 16} = 10$ ${x}^{2} - 16 = 100$ ${x}^{2} = 116$ $x = \\pm 10.77$ But since x is a side, then it can only be positive $x = 10.77$ Apr 26, 2018 $a = b = 2 \\sqrt{29}$ #### Explanation: Hey, small letters for triangle sides. Isosceles triangle $a = b$, $c = 8$, Area $m a t h c a l \\left\\{A\\right\\} = 40.$ Let's see if", "Browse Questions # Find the maximum area of an isosceles triangle inscribed in the ellipse $\\frac{\\large x^2}{\\large a^2}+\\frac{\\large y^2}{\\large b^2}=1$ with its vertex at one end of the major axis. Can you answer this question?" ]

[ "From what we can see, we get an equation for finding the $50$ to solve for segment $RS$: $$RS(\\tan (26^\\circ)+\\tan (32^\\circ))=50$$ $$RS=\\frac{50}{\\tan (26^\\circ)+\\tan (32^\\circ)}.$$ Thus, we can then solve for $RS$: $$RS=\\frac{50}{1.11260194...}=\\boxed{44.939 \\ \\text{m}}.$$ • don't you need the degree symbol \\^circ for the arguments of $\\tan?$ – abel Jan 19 '15 at 8:14 • @abel OK. I fixed it. – Julian Rachman Jan 19 '15 at 8:15 You have $RS(\\tan 26^\\circ+\\tan 32^\\circ)=50$ from the right triangles. My approach is a little bit more geometrical: I'll name the remaining vertex: the upper vertex (the one in the right triangle whose other angle is 32) is U and the only vertex left is T. Please note that the angle in U is $180-90-32= 58$. Which is the same as the angle in S ($32+26= 90$). Therefore the triangle we are dealing with is isosceles with $UT=US$. If we draw the altitude from U it would be the same length as the altitude from S (which is RS), because the triangle is isosceles. If we call X the point where the altitude from U meets the side TS of the triangle, we can simply say that $UX=RS$. But since, $UT=50$ and the angle in T can be calculated by doing the following operation $180-90-26 = 64$, therefore $UX= \\sin(64) UT", "+ 4^2 - 2(6)(4)\\cos(A).$$ This means $$81 = 36 + 16 - 48\\cos(A).$$ Isolating $\\cos(A)$ gives us $$\\cos(A) = \\dfrac{-29}{48}.$$ Therefore take $\\cos^{-1}$ to isolate $A$: $$A = \\cos^{-1} \\left( -\\dfrac{29}{48} \\right) = 127.2^{\\circ}.$$ Solve for $B$ Using the law of cosines, we see $$6^2 = 9^2 + 4^2 - 2(9)(4)\\cos(B).$$ Simplifying yields $$36 = 81 + 16 - 72\\cos(B).$$ Isolating $\\cos(B)$ yields $$\\cos(B) = \\dfrac{-61}{-72}.$$ Therefore take $\\cos^{-1}$ to get $$B = \\cos^{-1} \\left( \\dfrac{-61}{-72} \\right) =32.09^{\\circ}.$$ Solve for $C$ We will use the fact that the sum of the angles in a triangle is $180^{\\circ}$. This yields $$127.2^{\\circ} + 32.09^{\\circ} + C = 180^{\\circ},$$ so $$C = 180^{\\circ}-127.2^{\\circ} - 32.09^{\\circ} = 20.71^{\\circ}.$$ NOTE: you can solve this any way you wish, you may get slightly different answers depending on what angles you solve for first/etc -- you may also just use the law of cosines three times instead of using the sum of angles formula Section 4.3 #20: Solve the triangle given by the information $a=6$, $b=4$, and $C=60^{\\circ}$. Solution: First draw the triangle: Solve for $c$ Using the law of cosines, $$c^2 = 6^2+4^2-2(6)(4)\\cos \\left( 60^{\\circ} \\right),$$ so $$c= \\sqrt{52-48\\cos \\left( 60^{\\circ} \\right)} = 5.29.$$ There are two missing angles. We will use the law of cosines to find one of the missing angles and", "# In an isosceles triangle, the vertex angle is twice the each base angle. What is the vertex angle? Nov 14, 2015 ${90}^{\\circ}$ #### Explanation: Since the sum of the interior angles of any triangle adds up to ${180}^{\\circ}$ and also the base angles of an isosceles triangle are equal,we may let the base angles be $x$, then the vertex angle is $2 x$. $\\therefore x + x + 2 x = {180}^{\\circ}$. $\\therefore x = {45}^{\\circ}$. Therefore the vertex angle is ${90}^{\\circ}$. Jan 9, 2018 ${90}^{\\circ}$ #### Explanation: An isosceles triangle has two sides of equal length. Likewise, the opposite angles of the sides are the same So lets assume each angle be $x$ and the bigger angle be $2 x$ (since it is twice the base angle) Now, we know that sum of the angles in a triangle is 180 degree So, $\\rightarrow 2 x + x + x = {180}^{\\circ}$ $\\rightarrow 4 x = {180}^{\\circ}$ $\\rightarrow x = \\frac{180}{4}$ color(green)(rArrx=45^circ Now the other angle is $2 x$ which is $45 \\times 2 = {90}^{\\circ}$ Hope this helps!!! ☺☻ have a good day...", "# In an isosceles triangle, the vertex angle is twice the each base angle. What is the vertex angle? Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 11 Nov 14, 2015 ${90}^{\\circ}$ #### Explanation: Since the sum of the interior angles of any triangle adds up to ${180}^{\\circ}$ and also the base angles of an isosceles triangle are equal,we may let the base angles be $x$, then the vertex angle is $2 x$. $\\therefore x + x + 2 x = {180}^{\\circ}$. $\\therefore x = {45}^{\\circ}$. Therefore the vertex angle is ${90}^{\\circ}$. Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 7 Veddesh Share Jan 9, 2018 ${90}^{\\circ}$ #### Explanation: An isosceles triangle has two sides of equal length. Likewise, the opposite angles of the sides are the same So lets assume each angle be $x$ and the bigger angle be $2 x$ (since it is twice the base angle) Now, we know that sum of the angles in a triangle is 180 degree So, $\\rightarrow 2 x + x + x = {180}^{\\circ}$ $\\rightarrow 4 x = {180}^{\\circ}$ $\\rightarrow x", "What is the value of $$\\tan 45^\\circ?$$ Solution. Consider an isosceles right triangle. It has equal legs $$a = b,$$ and each of the acute angles of the triangle is equal to $$45^\\circ.$$ Therefore $\\require{cancel}\\tan {45^\\circ} = \\frac{a}{b} = \\frac{\\cancel{a}}{\\cancel{a}} = 1.$ ### Example 4. What is the value of $$\\cot 30^\\circ?$$ Solution. Consider a $$30\\text{-}60\\text{-}90$$ triangle, which is a special right triangle whose angles are $$30^\\circ,$$ $$60^\\circ,$$ and $$90^\\circ.$$ In any $$30\\text{-}60\\text{-}90$$ triangle, the hypotenuse is twice as long as the shortest leg, so $$c = 2a.$$ Using this property together with the Pythagorean theorem, we express the other leg $$b$$ in terms of $$a:$$ ${a^2} + {b^2} = {c^2},\\;\\; \\Rightarrow b = \\sqrt {{c^2} - {a^2}} = \\sqrt {{{\\left( {2a} \\right)}^2} - {a^2}} = \\sqrt {4{a^2} - {a^2}} = \\sqrt {3{a^2}} = a\\sqrt 3 .$ Now we can calculate the value of $$\\cot 30^\\circ :$$ $\\cot {30^0} = \\frac{b}{a} = \\frac{{\\cancel{a}\\sqrt 3 }}{\\cancel{a}} = \\sqrt 3 .$ ### Example 5. Calculate the value of $$\\sec 45^\\circ.$$ Solution. In an isosceles right triangle, the legs are equal to each other, that is $$a = b.$$ Then by the Pythagorean theorem, $c = \\sqrt {{a^2} + {b^2}} = \\sqrt {{a^2} + {a^2}} = \\sqrt {2{a^2}} = a\\sqrt 2 .$ The secant of an angle of $$45^\\circ$$ is given", "= 180° 3x + 2x + 4x = 180° ∴ 9x = 180° x = $$\\frac{180^{\\circ}}{9}$$ = 20° (vii) Given ∠T = (x – 4)° ∠U = 90° ∠V = (3x – 2)° By angle sum property of a triang we have ∠T + ∠U + ∠V = 180° (x – 4)° + 90° + (3x – 2)° = 180° x – 4° + 90° + 3x – 2° = 180° x + 3x + 90° – 4° – 2° = 180° 4x + 84° = 180° 4x = 180° – 84° 4x = 96° x = $$\\frac{96^{\\circ}}{4}$$ = 24° x = 24° (viii) Given ∠N = (x + 31)° ∠O = (3x – 10)° ∠P = (2x – 3)° By angle sum property of a triangle we have ∠N + ∠O + ∠P = O (x + 31)° + (3x – 10)° + (2x – 3)° = 180° x + 31°+ 3x – 10° + 2x – 3° = 180° x + 3x + 2x + 31° – 10° – 3° = 180° 6x + 18° = 180° 6x = 180° + 18° 6x = 162° x = $$\\frac{162^{\\circ}}{6}$$ = 27° x = 27° Question 4. Two line segments $$\\overline{A D}$$ and $$\\overline{B C}$$ intersect at O. Joining $$\\overline{A B}$$ and $$\\overline{D C}$$ we get two triangles,", "there are no angle measures given, we know that the two acute angles are congruent, making them both $45^\\circ$. Therefore, this is a 45-45-90 triangle. You can use the trigonometric ratios or the special right triangle ratios. Trigonometric Ratios $\\tan 45^\\circ & = \\frac{15}{BC} && \\sin 45^\\circ = \\frac{15}{AC}\\\\BC & = \\frac{15}{\\tan 45^\\circ} = 15 && \\quad \\ AC = \\frac{15}{\\sin 45^\\circ} \\approx 21.21$ 45-45-90 Triangle Ratios $BC = AB = 15, AC = 15 \\sqrt{2} \\approx 21.21$ ## Real-Life Situations Much like the trigonometric ratios, the inverse trig ratios can be used in several real-life situations. Here are a couple examples. Example 6: A 25 foot tall flagpole casts a 42 feet shadow. What is the angle that the sun hits the flagpole? Solution: First, draw a picture. The angle that the sun hits the flagpole is the acute angle at the top of the triangle, $x^\\circ$. From the picture, we can see that we need to use the inverse tangent ratio. $\\tan x & = \\frac{42}{25}\\\\\\tan^{-1} \\frac{42}{25} & \\approx 59.2^\\circ = x$ Example 7: Elise is standing on the top of a 50 foot building and spots her friend, Molly across the street. If Molly is 35 feet away from the base of the building, what is the angle of depression from Elise to Molly? Elise’s", "second phase is the calculation of other characteristics of the triangle, such as angles, area, perimeter, heights, the center of gravity, circle radii, etc. Some input data also results in two to three correct triangle solutions (e.g., if the specified triangle area and two sides - typically resulting in both acute and obtuse) triangle). 1. Input data entered: cathetus b and angle β $b = 1175 \\ \\\\ β = 41^\\circ$ 2. From angle β we calculate angle α: $α + β + 90^\\circ = 180^\\circ \\ \\\\ α = 90^\\circ - β = 90^\\circ - 41 ^\\circ = 49 ^\\circ$ 3. From cathetus b and angle α we calculate hypotenuse c: $\\cos α = b:c \\ \\\\ c = b/\\cos α = 1175/\\cos(49 ^\\circ ) = 1790.997$ 4. From hypotenuse c and angle α we calculate cathetus a: $\\sin α = a:c \\ \\\\ a = c \\cdot \\ \\sin α = 1790.997 \\cdot \\ \\sin(49 ^\\circ ) = 1351.683$ Now we know the lengths of all three sides of the triangle, and the triangle is uniquely determined. Next, we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS. $a = 1351.68 \\ \\\\ b = 1175 \\ \\\\ c = 1791$ 5. The triangle perimeter is the", "the triangle to add up to $180^\\circ-90^\\circ=90^\\circ$, and indeed the triangle in the picture has three $30^\\circ$ angles. ## The Gauss-Bonnet Theorem In this last section, we’re going to talk about something remarkable that happens when you add up the curvatures for every vertex in a polyhedron. The number you get when you do this is called the total curvature. Now that we have the angle deficit description of the curvature at a vertex, the total curvature is fairly easy to compute. We can do it pretty quickly for all of the Platonic solids, pictured in Figure 7. The results are: solid vertices faces at vertex curvature at vertex total curvature tetrahedron 4 3 triangles $180^\\circ$ $720^\\circ$ cube 8 3 squares $90^\\circ$ $720^\\circ$ octahedron 6 4 triangle $120^\\circ$ $720^\\circ$ dodecahedron 20 3 pentagons $36^\\circ$ $720^\\circ$ icosahedron 12 5 triangles $60^\\circ$ $720^\\circ$ Something should jump out right away: we always get the same number for the total curvature! In fact, there’s a deep reason for this, called the Gauss-Bonnet Theorem, which we’re about to prove. To prove this result, we’ll first need to triangulate our polyhedron, that is, cover it with triangles so that any two triangles meet at an edge and any two edges meet at a vertex. For this proof, we’re also going to require that none", "$x = 15^\\circ$ , then the base angles are $4(15^\\circ) +12^\\circ$ , or $72^\\circ$ . The vertex angle is $180^\\circ-72^\\circ-72^\\circ=36^\\circ$ . 2. The two sides are equal, so set them equal to each other and solve for $x$ . $2x-9 & = x+5\\\\x & = 14$ 3. This statement is false. Because the base angles of an isosceles triangle are congruent, if one base angle is a right angle then both base angles must be right angles. It is impossible to have a triangle with two right ( $90^\\circ$ ) angles. The Triangle Sum Theorem states that the sum of the three angles in a triangle is $180^\\circ$ . If two of the angles in a triangle are right angles, then the third angle must be $0^\\circ$ and the shape is no longer a triangle. ### Explore More Find the measures of $x$ and/or $y$ . Determine if the following statements are true or false. 1. Base angles of an isosceles triangle are congruent. 2. Base angles of an isosceles triangle are complementary. 3. Base angles of an isosceles triangle can be equal to the vertex angle. 4. Base angles of an isosceles triangle are acute. Complete the proofs below. 1. Given : Isosceles $\\triangle CIS$ , with base angles $\\angle C$ and $\\angle S$ $\\overline{IO}$ is the", "=>108° = 3P => P = 36 4) If tan P = cot Q, prove that P + Q = 90°. tanP = cot Q =>tan P = tan (90° – Q) =>P = 90° – Q =>P + Q = 90° 5) If the value of sec 4P = cosec (P – 20°), in which 4P is an acute angle, find the value of P. Ans:-According to question sec 4P = cosec (P – 20°) => cosec (90° – 4P) = cosec (P – 20°) Equating angles, => 90° – 4P= P- 20° => 110° = 5P => P = 22° Q6) If X,Y and Z are interior angles of a triangle XYZ, then show that sin (Y+Z/2) = cos $$\\frac{X}{2}$$ In a triangle, sum of all the interior angles X + Y + Z = $$180^{\\circ}$$ $$\\Rightarrow$$ Y + Z = $$180^{\\circ}$$ – X $$\\Rightarrow$$ $$\\frac{Y+Z}{2}$$ = $$\\frac{(180^{\\circ}-X)}{2}$$ $$\\Rightarrow$$ $$\\frac{Y+Z}{2}$$ = $$\\left ( 90^{\\circ}-\\frac{X}{2} \\right )$$ $$\\Rightarrow$$ sin $$\\left ( \\frac{Y+Z}{2} \\right )$$ = sin $$\\left ( 90^{\\circ}-\\frac{X}{2} \\right )$$ $$\\Rightarrow$$ sin $$\\left ( \\frac{Y+Z}{2} \\right )$$ = cos$$\\frac{X}{2}$$ Q7) Express sin $$67^{\\circ}$$ + cos $$75^{\\circ}$$ in terms of trigonometric ratios of angles between $$0^{\\circ}$$ and $$45^{\\circ}$$. sin $$67^{\\circ}$$ + cos $$75^{\\circ}$$ = sin ($$90^{\\circ}$$ – $$23^{\\circ}$$) + cos ($$90^{\\circ}$$ – $$15^{\\circ}$$) = cos $$23^{\\circ}$$ +", "a triangle must equal $$180^{\\circ}$$, the third angle in the triangle must measure $$180^{\\circ} −50^{\\circ} −60^{\\circ} =70^{\\circ}$$. Now to set up the ratios: \\begin{aligned} \\dfrac{\\sin 60}{a}&=\\dfrac{\\sin 70}{3} \\\\ 3(\\sin 60)&=a(\\sin 70) \\\\ a&=\\dfrac{3(\\sin 60)}{\\sin 70} \\\\ a&\\approx 2.7647 \\end{aligned} The length of one of the other 2 sides is approximately 2.7647 centimeters. To find the length of the last side: \\begin{aligned} \\dfrac{\\sin 50}{b}&=\\dfrac{\\sin 70}{3} \\\\ 3(\\sin 50)&=b(\\sin 70) \\\\ b&=\\dfrac{3(\\sin 50)}{\\sin 70} \\\\ b &\\approx 2.46 \\end{aligned} The length of the other one of the 2 unknown sides is approximately 2.46 centimeters. Example $$\\PageIndex{2}$$ Find side \"a\" in the triangle below using the following information: $$b=16,\\; A=11.7^{\\circ} ,\\; C=23.8^{\\circ}$$ Solution $$\\dfrac{\\sin 11.7^{\\circ} }{a}=\\dfrac{\\sin 144.5^{\\circ} }{16},\\; a=5.6$$ Example $$\\PageIndex{3}$$ Find side \"a\" in the triangle below using the following information: $$k=6.3,\\; J=16.2^{\\circ} ,\\; L=40.3^{\\circ}$$ Solution $$\\dfrac{\\sin 40.3^{\\circ} }{l}=\\dfrac{\\sin 123.5^{\\circ} }{6.3} ,\\; l=4.9$$ Example $$\\PageIndex{4}$$ Even though ASA and AAS triangles represent two different cases of the Law of Sines, what do they both have in common? Solution Student answers will vary but they should notice that in both cases you know or can find an angle and the side across from it. ### Review In $$\\Delta ABC$$, $$m\\angle A=40^{\\circ}$$, $$m\\angle B=67^{\\circ}$$, and $$c = 6$$. 1. Find $$m\\angle C$$. 2. Find the length of $$a$$. 3. Find the length", "= $$60^{\\circ}$$ and ∠B = $$60^{\\circ}$$ . Find the measure of ∠C. Solution: ∠A + ∠B + ∠C = $$180^{\\circ}$$ [The sum of all angles of a triangle is $$180^{\\circ}$$ ] $$60^{\\circ}$$ + $$60^{\\circ}$$ + ∠C = $$180^{\\circ}$$ [Substitute values] ∠C = $$180^{\\circ}$$ – $$120^{\\circ}$$ [Subtract 120 from both sides] ∠C = $$60^{\\circ}$$ Hence, the measure of ∠C is $$60^{\\circ}$$ . Example 2: For the $$\\triangle$$ABC given below, arrange the interior angles in increasing order of their measure. Solution: In the given $$\\triangle$$ABC, the measure of the sides are: a = 10 cm, b = 8 cm, and c = 6 cm. The decreasing order of side lengths is 10 > 8 > 6, that is, a > b > c. In a triangle, the measure of the angle opposite to the longer side is larger, or the angle opposite to the smaller side is smaller. Hence, the measure of angles in increasing order will be ∠A > ∠B > ∠C. Example 3: Find the area of the triangle whose base is 8 inches long and height is 12 inches. Solution: Area of triangle = $$\\frac{1}{2}$$ × b× h [Area of a triangle formula] = $$\\frac{1}{2}$$ (8) × (12) [Substitute values of b and h] = $$\\frac{1}{2}$$ (96) [Multiply] = 48 [Simplify] Hence, the area of the triangle", "= 14 Put y = 14 in equation (1) x – 3(14) = – 4 x – 42 = –4 x = –4 + 42 x = 38 Salim’s age = 38 years His daughter’s age = 14 years #### The angles of a triangle are x, y and $40^{\\circ}$. The difference between the two angles x and y is $30^{\\circ}$. Find x and y. Solution: Given: x – y = $30^{\\circ}$ …(1) According to sum of interior angles of a triangle’ property x + y + $40^{\\circ}$ = 180 x + y = $140^{\\circ}$ … (2) Solve eq. (1) and eq. (2) using substitution method. x – y = $30^{\\circ}$ x + y = $140^{\\circ}$ 2x = 170 x = $\\frac{170}{2}$ x = $85^{\\circ}$ Put x = $85^{\\circ}$ in eq. (1) we get $85^{\\circ}$– y =$30^{\\circ}$ $85^{\\circ}$$30^{\\circ}$ = y $55^{\\circ}$ = y ## Crack CUET with india's \"Best Teachers\" • HD Video Lectures • Unlimited Mock Tests • Faculty Support #### If x + 1 is a factor of 2x3+ ax2+ 2bx + 1, then find the values of a and b given that 2a – 3b = 4. Solution: It is given that (x + 1) is a factor of P(x) = 2x3 + ax2 + 2bx + 1 then according to remainder theorem P(–1) is", "= (x2 + 2)2 – (2x)2 = (x2 + 2 + 2x) (x2 + 2 – 2x) = (x2 + 2x+ 2) (x2 – 2x+ 2) (b) Evaluate : $$\\frac{\\sin 30^{\\circ}-\\sin 90^{\\circ}+2 \\cos 0^{\\circ}}{\\tan 30^{\\circ} \\cdot \\tan 60^{\\circ}}$$ (c) Simplify:$$(81)^{3 / 4}-3 \\times(7)^{0}-\\left(\\frac{1}{27}\\right)^{-2 / 3}$$ Question 4. (a) If x $$\\frac{2}{x}$$ = 5, find the value of $$x^{3}-\\frac{8}{x^{3}}$$ (b) If the hypotenuse of a right angled triangle is 6 m more than twice the shortest side and third side is 2 m less than hypotenuse, find the sides of the triangle. Let the shortest side be x m. Then, Hypotenuse =(2x+6)cm,thirdside=2x+6-2=(2x+4)m. ∴ (2x + 6)2 = (2x + 4)2 + x2 (Using Pythagoras theorem) = (2x)2 + 2.2x.6 + 62 = (2x)2 + 2.2xA +42 + x2 = 4x2 + 24x + 36 = 4x2 + 16x + 16 + x2 = 24x – 16x = x2 + 16 – 36 = x2 – 8x – 20 = 0 = x2 – (10 – 2) x – 20 =0 = x2 – 10x + 2x – 20 =0 = x (x – 10) + 2 (x – 10) = 0 = (x – 10) (x + 2) = 0 = x – 10 =0 or x + 2 = 0 = x = 10 or x = –", "cos⁡A=sin⁡C For complementary angles, sin of one angle is equal to cos of the other angle. cos⁡ A$$=\\frac{2}{3}$$ 14- C Let $$x$$ be all expenses, then $$\\frac{22}{100} x=660 →x=\\frac{100×660}{22}=3,000$$ Mr. Jones spent for his rent: $$\\frac{ 27}{100}×3,000=810$$ 15- A The amount of money for $$x$$ bookshelf is: $$200x$$, Then, the total cost of all bookshelves is equal to: $$200x+600$$, The total cost, in dollar, per bookshelf is: $$\\frac{Total \\ cost}{number \\ of \\ items}=\\frac{200x+600}{x}$$ 16- B It is given that g(5)=4. Therefore, to find the value of f(g(5)), then f(g(5))=f(4)=6 17- A Area of the triangle is: $$\\frac{1}{2}$$ AD$$×$$BC and AD is perpendicular to BC. Triangle ADC is a (\\30^\\circ-60^\\circ- 90^\\circ\\) right triangle. The relationship among all sides of right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in the following triangle: In this triangle, the opposite side of $$30^\\circ$$ angle is half of the hypotenuse. And the opposite side of $$60^\\circ$$ is opposite of $$30^\\circ×\\sqrt{3}$$ CD=6, then AD$$=6×\\sqrt{3}$$ Area of the triangle ABC is: $$\\frac{1}{2}$$ AD$$×$$BC $$= \\frac{1}{2} 6\\sqrt{3}×12=36\\sqrt{3}$$ $$\\img{https://appmanager.effortlessmath.com/public/images/questions/mmm.png }$$ 18- C $$6∎28=\\sqrt{6^2+28}= \\sqrt{36+28}=\\sqrt{64}=8$$ 19- C The equation of a circle with center (h,k) and radius r is $$(x-h)^2+(y-k)^2=r^2$$. To put the equation $$x^2+y^2+8x-2y=1$$ in this form, complete the square as follows: $$x^2+y^2+8x-2y=1, (x^2+8x)+(y^2-2y)=1$$ $$(x^2+8x+16)-16+(y^2-2y+1)-1=1, (x+4)^2+(y-1)^2=18$$ $$(x+4)^2+(y-1)^2=(\\sqrt{18})^ 2$$ Therefore, the radius of the circle is $$\\sqrt{18}$$. 20- E Solve", "we have two angles and a side. We do not have the angle corresponding to the known side but we can find it: $B=180^\\circ-69^\\circ-47^\\circ=64^\\circ$B=180°69°47°=64°(angles in a triangle sum to $180^\\circ$180°) • Hence, for triangle $BCD$BCD we know two angles and a corresponding side, we can use the sine rule to find side $CB$CB. Do: Using sine rule: $\\frac{d}{\\sin D}$dsinD​ $=$= $\\frac{b}{\\sin B}$bsinB​ $d$d is length $BC$BC, named d because it is opposite angle $D$D. $\\frac{d}{\\sin47^\\circ}$dsin47°​ $=$= $\\frac{143}{\\sin64^\\circ}$143sin64°​ Then we can rearrange to find length $d$d. $d$d $=$= $\\frac{143\\sin47^\\circ}{\\sin64^\\circ}$143sin47°sin64°​ $d$d $=$= $116.36$116.36 m (To 2 decimal places) Now we have enough information to find the height of the cliff. Using right-angled trigonometry we can complete the question: $\\tan\\theta$tanθ $=$= $\\frac{O}{A}$OA​ $\\tan39^\\circ$tan39° $=$= $\\frac{height}{116.36}$height116.36​ $height$height $=$= $116.36\\times\\tan39^\\circ$116.36×tan39° $height$height $=$= $94.2265$94.2265 ... m So the cliff is approximately $94$94 m tall. Reflect: For practical problems reflect on whether your answer seems realistic. For this case, this is quite a tall cliff but the answer is certainly possible, we did not get a negative height or unrealistically short or tall cliff. Practice questions Question 1 $\\triangle ABC$ABC consists of angles $A$A, $B$B and $C$C which appear opposite sides $a$a, $b$b and $c$c respectively. Consider the case where the measures of $a$a, $c$c and $A$A are given. 1. Which of the following is", "$$16√3$$ C) $$24√3$$ D) 48 E) $$32√3$$ Attachment: equitriHfindA 09.16.18.jpg [ 37.97 KiB | Viewed 210 times ] Good question +1 to refresh a handy tool: the altitude of an equilateral triangle divides it into two congruent 30-60-90 triangles I. 30-60-90 triangles[/u] This method is as quick as the formula below and IMO easier to remember. In an equilateral triangle, each altitude = height = median Drop an altitude, BD, from ∠B to side AC The height, BD, divides ∆ABC into two congruent 30-60-90 triangles. • Altitude BD is a perpendicular bisector of vertex B and opposite side AC Resultant angles are 30, 60, and 90 degrees • Original angles A, B, and C = 60° • ∠ B is bisected into two 30° angles • At D, the altitude forms two 90° angles • Angles A and C = 60° • Both smaller triangles are 30-60-90 30-60-90 triangles have corresponding sides opposite those angles in the ratio $$x:x\\sqrt{3}:2x$$ Height of $$4\\sqrt{3}$$, opposite the 60° angle, corresponds with $$x\\sqrt{3}$$ $$4\\sqrt{3}=x\\sqrt{3}$$ $$x=4$$ Base AC = BC = $$2x=8$$ (In ∆BCD, side BC is opposite the 90° angle. Sides are equal.) Area, $$A=\\frac{b*h}{2}=\\frac{8*4\\sqrt{3}}{2}=16\\sqrt{3}$$ II. Formula For an equilateral triangle with side $$a$$, Height = $$\\frac{1}{2}a*\\sqrt{3}$$ Given: height = $$4\\sqrt{3}$$ Side length: $$4\\sqrt{3}=\\frac{1}{2}a*\\sqrt{3}$$ $$8\\sqrt{3}=a\\sqrt{3}$$ $$a=8$$ = base (all sides are equal)", "option D is the correct answer. Substituting the values in (1), we get 3 * 3 + 4 * 4 – 6 = 19 So, (1) is false. Substituting the values in (2), we get 3 * 4 + 4 * 6 – 6 *3 = 18 So, (2) is true Substituting the values in (3), we get 3 * 4 * 6 – 4 * 4 + 6 * 3 – 4 = 50 So, (3) is false. Substituting the values in (4), we get 3 * 3 * 6 > 4 * 4 * 3 So, (4) is true Hence the correct option is (d) Volume of a cone=$(1/3)\\pi r^{2}h$ r2=2*r1/5 $(r1)^{2}h1$=$(r2)^{2}h2$ h1=(4/25)h2 h1:h2=4:25 Given $\\angle ABC = 90^\\circ$ In a triangle, if two sides are equal, then, opposite angles are equal. Here, AB = BC ⇒ $\\angle A = \\angle C$ Let $\\angle A = \\angle C = x^\\circ$ Sum of angles in a triangle is equal to $180^\\circ$ ⇒ $90^\\circ + 2x = 180^\\circ$ ⇒ $2x = 90^\\circ$ ⇒ $x = 45^\\circ$ Hence, $\\angle ACB = 45^\\circ$ Volume of cake = vol.of bread + vol.of chocolate (20)(14)(8) = vol.of bread + 880 2240 = vol.of bread + 880 So the answer is option C. Given, Mean temperature of a week = $28^{0}$ i.e Total", "{\\mathrm {adjacent} }{\\mathrm {opposite} }}}$ ${\\displaystyle \\cot 45^{\\circ } ={\\frac {b}{a}}}$ substitute $b = 1$, $a = 1$ ${\\displaystyle \\cot 45^{\\circ } ={\\frac {1}{1}}}$ $\\cot 45^{\\circ } = 1$ Case 2: Special angles 30o and 60o (from a 30o – 60o – 90o triangle) The following figure 7-3 represents an equilateral triangle with sides $a = 2$, $b = 2$, and $c =2$. Since equilateral triangle has congruent angles and the measure of angles in a triangle is $180^{\\circ }$, each angle measures $60^{\\circ }$. Let us draw an altitude from the vertex $B$. The altitude separates an equilateral triangle into two congruent right triangles. In Figure 7-4, ${\\displaystyle {\\overline {BD}}}$ is altitude, $ΔABD\\:≅\\:ΔCBD$, $∠BDA$ is a right angle, $m∠A=60^{\\circ }$, and $m∠ABD=30^{\\circ }$. We can determine the height h of these triangles by the Pythagorean theorem. $(AB)^{2}=(BD)^{2}+(AD)^{2}$ $(BD)^{2}=(AB)^{2} – (AD)^{2}$ Substitute $(BD) = h$, $AB = 2$, and $AD = 1$ in the formula $h^{2}=(2)^{2} – (1)^{2}$ $h^{2}= 3$ $h = \\sqrt{3}$ As the altitude $h$ splits the equilateral triangle into two congruent 30o – 60o – 90o triangles. Let us knock out one of those right triangles, let suppose $ABD$, and determine the values of trigonometric ratio for $30^{\\circ }$ and $60^{\\circ }$. When m B = 30o: The following Figure 7-5 represents the right-angled triangle" ]

[ "second phase is the calculation of other characteristics of the triangle, such as angles, area, perimeter, heights, the center of gravity, circle radii, etc. Some input data also results in two to three correct triangle solutions (e.g., if the specified triangle area and two sides - typically resulting in both acute and obtuse) triangle). 1. Input data entered: cathetus b and angle β $b = 1175 \\ \\\\ β = 41^\\circ$ 2. From angle β we calculate angle α: $α + β + 90^\\circ = 180^\\circ \\ \\\\ α = 90^\\circ - β = 90^\\circ - 41 ^\\circ = 49 ^\\circ$ 3. From cathetus b and angle α we calculate hypotenuse c: $\\cos α = b:c \\ \\\\ c = b/\\cos α = 1175/\\cos(49 ^\\circ ) = 1790.997$ 4. From hypotenuse c and angle α we calculate cathetus a: $\\sin α = a:c \\ \\\\ a = c \\cdot \\ \\sin α = 1790.997 \\cdot \\ \\sin(49 ^\\circ ) = 1351.683$ Now we know the lengths of all three sides of the triangle, and the triangle is uniquely determined. Next, we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS. $a = 1351.68 \\ \\\\ b = 1175 \\ \\\\ c = 1791$ 5. The triangle perimeter is the", "From what we can see, we get an equation for finding the $50$ to solve for segment $RS$: $$RS(\\tan (26^\\circ)+\\tan (32^\\circ))=50$$ $$RS=\\frac{50}{\\tan (26^\\circ)+\\tan (32^\\circ)}.$$ Thus, we can then solve for $RS$: $$RS=\\frac{50}{1.11260194...}=\\boxed{44.939 \\ \\text{m}}.$$ • don't you need the degree symbol \\^circ for the arguments of $\\tan?$ – abel Jan 19 '15 at 8:14 • @abel OK. I fixed it. – Julian Rachman Jan 19 '15 at 8:15 You have $RS(\\tan 26^\\circ+\\tan 32^\\circ)=50$ from the right triangles. My approach is a little bit more geometrical: I'll name the remaining vertex: the upper vertex (the one in the right triangle whose other angle is 32) is U and the only vertex left is T. Please note that the angle in U is $180-90-32= 58$. Which is the same as the angle in S ($32+26= 90$). Therefore the triangle we are dealing with is isosceles with $UT=US$. If we draw the altitude from U it would be the same length as the altitude from S (which is RS), because the triangle is isosceles. If we call X the point where the altitude from U meets the side TS of the triangle, we can simply say that $UX=RS$. But since, $UT=50$ and the angle in T can be calculated by doing the following operation $180-90-26 = 64$, therefore $UX= \\sin(64) UT", "maths > trigonometry Understanding Standard Angles what you'll learn... Overview » 0 : minimum possible angle → opposite side is 0 » 90 : Maximum possible angle → opposite side equals hypotenuse » 45 : isosceles right-triangle → opposite and adjacent sides equal » 30 and 60 : Derived from equilateral triangle → one side is half of hypotenuse equilateral, isosceles, zero degree We had studied that the ratio of two sides of a right-triangle is a constant and the such constants are named $\\mathrm{sin}\\theta ,\\mathrm{cos}\\theta ,\\cdots$$\\sin \\theta , \\cos \\theta , \\cdots$ for given values of $\\theta ={0}^{\\circ },{0.2}^{\\circ },{1}^{\\circ },{2.7}^{\\circ },\\cdots$$\\theta = {0}^{\\circ} , {0.2}^{\\circ} , {1}^{\\circ} , {2.7}^{\\circ} , \\cdots$. It is noted that one cannot memorize all values of $\\mathrm{sin}\\theta$$\\sin \\theta$. A reference table is to be created and used when required. Students work out values of trigonometric ratios $\\mathrm{sin}$$\\sin$, $\\mathrm{cos}$$\\cos$, and $\\mathrm{tan}$$\\tan$, for the angles for which the ratio between two sides of triangles can be computed using only given angle The ratio between two sides of triangles be computed for the following • equilateral triangles • isosceles triangles • triangles with one angle ${0}^{\\circ }$${0}^{\\circ}$ equilateral The equilateral triangle is split into two right angled triangles as shown in the figure. The hypotenuse $\\overline{OP}=1$$\\overline{O P} = 1$. Then the side $\\overline{OQ}=\\frac{1}{2}$$\\overline{O Q}", "+ 4^2 - 2(6)(4)\\cos(A).$$ This means $$81 = 36 + 16 - 48\\cos(A).$$ Isolating $\\cos(A)$ gives us $$\\cos(A) = \\dfrac{-29}{48}.$$ Therefore take $\\cos^{-1}$ to isolate $A$: $$A = \\cos^{-1} \\left( -\\dfrac{29}{48} \\right) = 127.2^{\\circ}.$$ Solve for $B$ Using the law of cosines, we see $$6^2 = 9^2 + 4^2 - 2(9)(4)\\cos(B).$$ Simplifying yields $$36 = 81 + 16 - 72\\cos(B).$$ Isolating $\\cos(B)$ yields $$\\cos(B) = \\dfrac{-61}{-72}.$$ Therefore take $\\cos^{-1}$ to get $$B = \\cos^{-1} \\left( \\dfrac{-61}{-72} \\right) =32.09^{\\circ}.$$ Solve for $C$ We will use the fact that the sum of the angles in a triangle is $180^{\\circ}$. This yields $$127.2^{\\circ} + 32.09^{\\circ} + C = 180^{\\circ},$$ so $$C = 180^{\\circ}-127.2^{\\circ} - 32.09^{\\circ} = 20.71^{\\circ}.$$ NOTE: you can solve this any way you wish, you may get slightly different answers depending on what angles you solve for first/etc -- you may also just use the law of cosines three times instead of using the sum of angles formula Section 4.3 #20: Solve the triangle given by the information $a=6$, $b=4$, and $C=60^{\\circ}$. Solution: First draw the triangle: Solve for $c$ Using the law of cosines, $$c^2 = 6^2+4^2-2(6)(4)\\cos \\left( 60^{\\circ} \\right),$$ so $$c= \\sqrt{52-48\\cos \\left( 60^{\\circ} \\right)} = 5.29.$$ There are two missing angles. We will use the law of cosines to find one of the missing angles and", "= 180° 3x + 2x + 4x = 180° ∴ 9x = 180° x = $$\\frac{180^{\\circ}}{9}$$ = 20° (vii) Given ∠T = (x – 4)° ∠U = 90° ∠V = (3x – 2)° By angle sum property of a triang we have ∠T + ∠U + ∠V = 180° (x – 4)° + 90° + (3x – 2)° = 180° x – 4° + 90° + 3x – 2° = 180° x + 3x + 90° – 4° – 2° = 180° 4x + 84° = 180° 4x = 180° – 84° 4x = 96° x = $$\\frac{96^{\\circ}}{4}$$ = 24° x = 24° (viii) Given ∠N = (x + 31)° ∠O = (3x – 10)° ∠P = (2x – 3)° By angle sum property of a triangle we have ∠N + ∠O + ∠P = O (x + 31)° + (3x – 10)° + (2x – 3)° = 180° x + 31°+ 3x – 10° + 2x – 3° = 180° x + 3x + 2x + 31° – 10° – 3° = 180° 6x + 18° = 180° 6x = 180° + 18° 6x = 162° x = $$\\frac{162^{\\circ}}{6}$$ = 27° x = 27° Question 4. Two line segments $$\\overline{A D}$$ and $$\\overline{B C}$$ intersect at O. Joining $$\\overline{A B}$$ and $$\\overline{D C}$$ we get two triangles,", "$ye^x=x+1$ $y=\\dfrac{x+1}{e^x}$ ## an interesting highschool geometry question Posted: April 1, 2013 in Mathematics During the long easter weekend, I finally have time to check out the web and forums and I found this nice little geometry problem on triangles. At first glance the problem looks difficult, however it is not too bad after figuring out some of the angles and realizing couple isosceles triangles in it. Solution First of all let’s label the points A, B, C, D, E, and F in the diagram. Obviously, triangle ABC, triangle BCD, and triangle ABF are isosceles triangles. Let $AB=AC=BF=k, \\quad BD=BC=a$ and let $CD=b, \\quad AD=c$ In isosceles triangle ABC, we have $\\dfrac{a}{2k}=\\cos{48}$ $a=2k\\cos{48}$ In isosceles triangle BCD, we have $\\dfrac{b}{2a}=\\cos{72}$ $b=2(2k\\cos 48)\\cos 72=4k\\cos 72 \\cos 48$ Then apply the Product-to-Sum identity $\\cos{A}\\cos{B}=\\dfrac{\\cos(A+B)+\\cos(A-B)}{2}$ We have $b=2k(\\cos 120 + \\cos 24)=(2\\cos24-1)k$ Now apply the cosine law on triangle ACD, $c^2=k^2+b^2-2kb\\cos 24$ $c^2=k^2+(2\\cos24-1)^2k^2-2k((2\\cos24-1)k)\\cos24$ $c^2=k^2(1+4\\cos^2{24}-4\\cos{24}+1-4\\cos^2{24}+2\\cos{24})$ $c^2=k^2(2-2\\cos24)$ And then finally apply sine law in triangle ABD, $\\dfrac{\\sin x}{k}=\\dfrac{\\sin 12}{c}$ $\\dfrac{\\sin x}{k}=\\dfrac{\\sin 12}{k\\sqrt{2-2\\cos24}}$ Then apply double angle identity, $\\cos(2\\theta)=1-2\\sin^2\\theta$ $\\sin x = \\dfrac{\\sin 12}{\\sqrt{2-2(1-2\\sin^2{12})}}$ $=\\dfrac{\\sin 12}{\\sqrt{4\\sin^2{12}}}=\\dfrac{1}{2}$ $\\therefore x=30$ Therefore angle x is 30 degrees. Remarks: Some additional beautiful results can be found if we solve x differently, but I will leave this for next time. 🙂", "Sides If angle C is obtuse then for sides a, b, and c we have[3]:p.1,#74 $\\frac{c^2}{2} < a^2+b^2 < c^2,$ with the left inequality approaching equality in the limit only as the apex angle of an isosceles triangle approaches 180°, and with the right inequality approaching equality only as the obtuse angle approaches 90°. If the triangle is acute then $a^2+b^2 > c^2, \\quad b^2+c^2 > a^2, \\quad c^2+a^2 > b^2.$ ### Altitude If C is the greatest angle and hc is the altitude from vertex C, then for an acute triangle[3]:p.135,#3109 $\\frac{1}{h_c^2} < \\frac{1}{a^2}+\\frac{1}{b^2},$ with the opposite inequality if C is obtuse. ### Medians With longest side c and medians ma and mb from the other sides,[3]:p.136,#3110 $4c^2 +9a^2b^2 > 16m_a^2m_b^2$ for an acute triangle but with the inequality reversed for an obtuse triangle. The median mc from the longest side is greater or less than the circumradius for an acute or obtuse triangle respectively:[3]:p.136,#3113 $m_c > R$ for acute triangles, with the opposite for obtuse triangles. ### Trigonometric functions For an acute triangle we have, for angles A, B, and C,[3]:p.26,#954 $\\cos^2A+\\cos^2B+\\cos^2C < 1,$ with the reverse inequality holding for an obtuse triangle. For an acute triangle with circumradius R,[3]:p.141,#3167 $a\\cos^3 A +b\\cos^3 B +c\\cos^3 C \\leq \\frac{abc}{4R^2}$ and[3]:p.155,#S25 $\\cos^3A+\\cos^3B+\\cos^3C+\\cos A\\cos B\\cos C\\geq\\frac{1}{2}.$ For an acute", "the triangle to add up to $180^\\circ-90^\\circ=90^\\circ$, and indeed the triangle in the picture has three $30^\\circ$ angles. ## The Gauss-Bonnet Theorem In this last section, we’re going to talk about something remarkable that happens when you add up the curvatures for every vertex in a polyhedron. The number you get when you do this is called the total curvature. Now that we have the angle deficit description of the curvature at a vertex, the total curvature is fairly easy to compute. We can do it pretty quickly for all of the Platonic solids, pictured in Figure 7. The results are: solid vertices faces at vertex curvature at vertex total curvature tetrahedron 4 3 triangles $180^\\circ$ $720^\\circ$ cube 8 3 squares $90^\\circ$ $720^\\circ$ octahedron 6 4 triangle $120^\\circ$ $720^\\circ$ dodecahedron 20 3 pentagons $36^\\circ$ $720^\\circ$ icosahedron 12 5 triangles $60^\\circ$ $720^\\circ$ Something should jump out right away: we always get the same number for the total curvature! In fact, there’s a deep reason for this, called the Gauss-Bonnet Theorem, which we’re about to prove. To prove this result, we’ll first need to triangulate our polyhedron, that is, cover it with triangles so that any two triangles meet at an edge and any two edges meet at a vertex. For this proof, we’re also going to require that none", "= \\cos ( x + \\pi / 2 ) + 1$$ c. $$y = \\cos ( x - \\pi / 2 ) - 1$$ d. $$y = \\cos ( x - \\pi ) - 2$$ e. $$y = \\sin ( x - \\pi ) - 2$$ Q: The diagram below represents a wheelchair ramp leading into a building. What is the height of the wheelchair ramp? $$\\square 80 cm$$ $$\\square 79 cm$$ Q: One angle of an isosceles triangle measures $$140 ^ { \\circ }$$ . Which other angles could be in that isosceles triangle? Choose all that apply. Q: Write the complex number $$- 4 - 4 i$$ in trigonometric form, using degree measure for the argument. $$- 4 - 4 i = \\square ( \\cos \\square ^ { \\circ } + i \\sin \\square ^ { \\circ } )$$ (Type exact answers, using radicals as needed. Type any angle measures in degrees. Round to the nearest deg. Q: Sketch two periods of the graph of the function $$h ( x ) = 3 \\sec ( \\frac { \\pi } { 4 } ( x + 1 ) )$$ . Identify the stretching factor, period, and asymptotes. Enter the exact answers. Stretching factor $$=$$ Number Period: $$P =$$ Enter the asymptotes of the function on the domain", "cos⁡A=sin⁡C For complementary angles, sin of one angle is equal to cos of the other angle. cos⁡ A$$=\\frac{2}{3}$$ 14- C Let $$x$$ be all expenses, then $$\\frac{22}{100} x=660 →x=\\frac{100×660}{22}=3,000$$ Mr. Jones spent for his rent: $$\\frac{ 27}{100}×3,000=810$$ 15- A The amount of money for $$x$$ bookshelf is: $$200x$$, Then, the total cost of all bookshelves is equal to: $$200x+600$$, The total cost, in dollar, per bookshelf is: $$\\frac{Total \\ cost}{number \\ of \\ items}=\\frac{200x+600}{x}$$ 16- B It is given that g(5)=4. Therefore, to find the value of f(g(5)), then f(g(5))=f(4)=6 17- A Area of the triangle is: $$\\frac{1}{2}$$ AD$$×$$BC and AD is perpendicular to BC. Triangle ADC is a (\\30^\\circ-60^\\circ- 90^\\circ\\) right triangle. The relationship among all sides of right triangle $$30^\\circ-60^\\circ- 90^\\circ$$ is provided in the following triangle: In this triangle, the opposite side of $$30^\\circ$$ angle is half of the hypotenuse. And the opposite side of $$60^\\circ$$ is opposite of $$30^\\circ×\\sqrt{3}$$ CD=6, then AD$$=6×\\sqrt{3}$$ Area of the triangle ABC is: $$\\frac{1}{2}$$ AD$$×$$BC $$= \\frac{1}{2} 6\\sqrt{3}×12=36\\sqrt{3}$$ $$\\img{https://appmanager.effortlessmath.com/public/images/questions/mmm.png }$$ 18- C $$6∎28=\\sqrt{6^2+28}= \\sqrt{36+28}=\\sqrt{64}=8$$ 19- C The equation of a circle with center (h,k) and radius r is $$(x-h)^2+(y-k)^2=r^2$$. To put the equation $$x^2+y^2+8x-2y=1$$ in this form, complete the square as follows: $$x^2+y^2+8x-2y=1, (x^2+8x)+(y^2-2y)=1$$ $$(x^2+8x+16)-16+(y^2-2y+1)-1=1, (x+4)^2+(y-1)^2=18$$ $$(x+4)^2+(y-1)^2=(\\sqrt{18})^ 2$$ Therefore, the radius of the circle is $$\\sqrt{18}$$. 20- E Solve", "# In an isosceles triangle, the vertex angle is twice the each base angle. What is the vertex angle? Nov 14, 2015 ${90}^{\\circ}$ #### Explanation: Since the sum of the interior angles of any triangle adds up to ${180}^{\\circ}$ and also the base angles of an isosceles triangle are equal,we may let the base angles be $x$, then the vertex angle is $2 x$. $\\therefore x + x + 2 x = {180}^{\\circ}$. $\\therefore x = {45}^{\\circ}$. Therefore the vertex angle is ${90}^{\\circ}$. Jan 9, 2018 ${90}^{\\circ}$ #### Explanation: An isosceles triangle has two sides of equal length. Likewise, the opposite angles of the sides are the same So lets assume each angle be $x$ and the bigger angle be $2 x$ (since it is twice the base angle) Now, we know that sum of the angles in a triangle is 180 degree So, $\\rightarrow 2 x + x + x = {180}^{\\circ}$ $\\rightarrow 4 x = {180}^{\\circ}$ $\\rightarrow x = \\frac{180}{4}$ color(green)(rArrx=45^circ Now the other angle is $2 x$ which is $45 \\times 2 = {90}^{\\circ}$ Hope this helps!!! ☺☻ have a good day...", "Let ∠Z = (2x + 1)° and ∠Y = 90° By the sum property of triangles we have ∠x + ∠y + ∠z = 180° 29° + 90° + (2x + 1)° = 180° 119° + (2x + 1)° = 180° (2x + 1)° = 180° – 119° 2x + 1° = 61° 2x = 61° – 1° 2x = 60° x = $$\\frac{60^{\\circ}}{2}$$ x = 30° (iv) Let ∠J = x and ∠L – 3x. By angle sum property of triangles we have ∠J + ∠K + ∠L = 180° x + 112° + 3x = 180° 4x = 180° – 112° x = 68° x = $$\\frac{68^{\\circ}}{4}$$ x = 17° (v) Let ∠S = 3x° Given $$\\overline{\\mathrm{RS}}$$ = Given $$\\overline{\\mathrm{RT}}$$ = 4.5 cm Given ∠S = ∠T = 3x° [∵ Angles opposite to equal sides are equal] By angle sum property of a triangle we have, ∠R + ∠S + ∠T = 180° 72° + 3x + 3x = 180° 72° + 6x = 180° x = $$\\frac{108^{\\circ}}{6}$$ x = 18° (vi) Given ∠X = 3x; ∠Y = 2x; ∠Z = ∠4x By angle sum property of a triangle we have ∠X + ∠Y + ∠Z = 180° 3x + 2x + 4x = 180° ∴ 9x = 180° x = $$\\frac{180^{\\circ}}{9}$$ = 20° (vii) Given", "\\rm b=\\sqrt {a^2 + c^2 - 2ac. cos B}\\\\ c\\to \\ side \\ of \\ triangle =9 \\\\ a\\to \\ side \\ of \\ triangle = 8 \\\\ b\\to \\ side \\ of \\ triangle = 16 \\\\ B\\to \\ interior \\ angle \\ opposite \\ side \\ b\\\\ 16=\\sqrt {8^2 + 9^2 - 2(8 \\times 9). cos B}\\\\ 256= 64 + 81 - 144cosB\\\\ -144 Cos B = 256 - 64 -81 = 111\\\\ Cos \\ B = \\dfrac {111}{ -144} = -0.7708\\\\ B = cos^{-1} (-0.7708) = 140.43^{\\circ} B= 140.43^{\\circ} }]\\\\ [{MathJax fullWidth='false' \\rm c=\\sqrt {a^2 + b^2 - 2ab. cos C}\\\\ C\\to \\ interior \\ angle \\ opposite \\ side \\ c\\\\ 9=\\sqrt {8^2 + 16^2 - 2(16\\times 8). cos C}\\\\ 9=\\sqrt {64 + 256 - 256. cos C}\\\\ 81= 320- 256Cos C\\\\ 256Cos C = 320 - 81\\\\ Cos \\ C =\\dfrac {239}{256} = 0.934\\\\ C = Cos^{-1} (0.934) = 20.1^{\\circ}\\\\ A = 180 -( 21 + 140.43) = 18.57^{\\circ} }] The given triangle has three interior angles with one angle formed between lines 9 and 8 is more than 90[{MathJax fullWidth='false.' \\rm ^{\\circ} }],(140.43), which bears the name obtuse angle. The remaining angles are less than 90[{MathJax fullWidth='false.' \\rm ^{\\circ} }] each, (18.57 and 21), and they are known as acute angles. Therefore,", "${\\displaystyle \\cos ^{2}A+\\cos ^{2}B+\\cos ^{2}C<1,}$ with the reverse inequality holding for an obtuse triangle. For an acute triangle with circumradius R,[4]:p.141,#3167 ${\\displaystyle a\\cos ^{3}A+b\\cos ^{3}B+c\\cos ^{3}C\\leq {\\frac {abc}{4R^{2}}}}$ and[4]:p.155,#S25 ${\\displaystyle \\cos ^{3}A+\\cos ^{3}B+\\cos ^{3}C+\\cos A\\cos B\\cos C\\geq {\\frac {1}{2}}.}$ For an acute triangle,[4]:p.115,#2874 ${\\displaystyle \\sin ^{2}A+\\sin ^{2}B+\\sin ^{2}C>2,}$ with the reverse inequality for an obtuse triangle. For an acute triangle,[4]:p178,#241.1 ${\\displaystyle \\sin A\\cdot \\sin B+\\sin B\\cdot \\sin C+\\sin C\\cdot \\sin A\\leq (\\cos A+\\cos B+\\cos C)^{2}.}$ For any triangle the triple tangent identity states that the sum of the angles' tangents equals their product. Since an acute angle has a positive tangent value while an obtuse angle has a negative one, the expression for the product of the tangents shows that ${\\displaystyle \\tan A+\\tan B+\\tan C=\\tan A\\cdot \\tan B\\cdot \\tan C>0}$ for acute triangles, while the opposite direction of inequality holds for obtuse triangles. We have[4]:p.26,#958 ${\\displaystyle \\tan A+\\tan B+\\tan C\\geq 2(\\sin 2A+\\sin 2B+\\sin 2C)}$ for acute triangles, and the reverse for obtuse triangles. For all acute triangles,[4]:p.40,#1210 ${\\displaystyle (\\tan A+\\tan B+\\tan C)^{2}\\geq (\\sec A+1)^{2}+(\\sec B+1)^{2}+(\\sec C+1)^{2}.}$ ${\\displaystyle a\\tan A+b\\tan B+c\\tan C\\geq 10R-2r.}$ For an acute triangle with area K, [4]:p.103,#2662 ${\\displaystyle ({\\sqrt {\\cot A}}+{\\sqrt {\\cot B}}+{\\sqrt {\\cot C}})^{2}\\leq {\\frac {K}{r^{2}}}.}$ In an acute triangle, the sum of the circumradius R and the inradius r is less than half", "### So Big One side of a triangle is divided into segments of length a and b by the inscribed circle, with radius r. Prove that the area is: abr(a+b)/ab-r^2 ### 30-60-90 Polypuzzle Re-arrange the pieces of the puzzle to form a rectangle and then to form an equilateral triangle. Calculate the angles and lengths. ### Bend What is the longest stick that can be carried horizontally along a narrow corridor and around a right-angled bend? # Gold Again ##### Stage: 5 Challenge Level: The first part of this solution came from Dorothy, S4, Madras College, St Andrew's and the second part from Vassil, Y11, Lawnswood High School, Leeds. Consider the rhombus as illustrated, where $x$ is an unknown length. We have to find the value of $\\cos 36^\\circ \\cos 72^\\circ .$ I filled in the remaining angles and lengths, showing triangles $PCB$ and $PCD$ to be isosceles triangles with angles of $108^\\circ$, $36^\\circ$ and $36^\\circ$ and sides $PC = PB = PD = 1$ unit. I found $\\cos 36^\\circ$ and $\\cos 72 ^\\circ$ by using the cosine rule for triangles DCP and APD respectively. $\\cos 36^\\circ = {{x^2 + 1 - 1}\\over 2x} = {x\\over 2}$ $\\cos 72^\\circ = {{1 +x^2 - x^2}\\over 2x} = {1\\over 2x}$ Combining these two expressions, $\\cos 36^\\circ \\cos 72^\\circ = {x\\over", "= 180° 4y + (3y + 22) + 74° = 180° 7y = 180 – 96 = 84 y = 12° m∠A = 4y = 4 × 12 = 48° and m∠B = 3y + 22 = 3 × 12 + 22 = 58° Question 5. m∠B = ________ Using the Exterior Angle Theorem we have: m∠A + m∠B = m∠BCD We substitute the given angle measures and we solve the equation for y. (4y)° + (3y + 22)° = 106° 4y° + 3y° + 22° = 106° 7y° + 22° = 106° 7y° + 22° – 22° = 106° – 22° 7y° = 84° $$\\frac{7 y^{\\circ}}{7^{\\circ}}$$ = $$\\frac{84^{\\circ}}{7^{\\circ}}$$ y = 12 We use the value of y to find m∠B. m∠B = (3y + 22)° = (3 . 12 + 22)° = 58° m∠B = 58° Question 6. m∠BCA = ________ In the given triangle, m∠A = 4y m∠B = 3y + 22° and m∠BCD = 106° using exterior angle theorem m∠BCD + m∠BCA = 180° m∠BCA = 180° – 106° = 74° So, m∠BCA = 74° ∠BCA = 74 degrees 7.3 Angle-Angle Similarity Triangle FEG is similar to triangle IHJ. Find the missing values. Question 7 x = ________ In similar triangles, corresponding side lengths are proportional $$\\frac{H J}{E G}$$ = $$\\frac{I J}{F G}$$ $$\\frac{x+12}{42}$$ =", "$x = 15^\\circ$ , then the base angles are $4(15^\\circ) +12^\\circ$ , or $72^\\circ$ . The vertex angle is $180^\\circ-72^\\circ-72^\\circ=36^\\circ$ . 2. The two sides are equal, so set them equal to each other and solve for $x$ . $2x-9 & = x+5\\\\x & = 14$ 3. This statement is false. Because the base angles of an isosceles triangle are congruent, if one base angle is a right angle then both base angles must be right angles. It is impossible to have a triangle with two right ( $90^\\circ$ ) angles. The Triangle Sum Theorem states that the sum of the three angles in a triangle is $180^\\circ$ . If two of the angles in a triangle are right angles, then the third angle must be $0^\\circ$ and the shape is no longer a triangle. ### Explore More Find the measures of $x$ and/or $y$ . Determine if the following statements are true or false. 1. Base angles of an isosceles triangle are congruent. 2. Base angles of an isosceles triangle are complementary. 3. Base angles of an isosceles triangle can be equal to the vertex angle. 4. Base angles of an isosceles triangle are acute. Complete the proofs below. 1. Given : Isosceles $\\triangle CIS$ , with base angles $\\angle C$ and $\\angle S$ $\\overline{IO}$ is the", "the interior angles of a triangle, determine the missing angle in each triangle. 1. $45^{\\circ}, 45^{\\circ}, ?$ 2. $60^{\\circ}, 60^{\\circ}, ?$ 3. $90^{\\circ}, 50^{\\circ}, ?$ 4. $100^{\\circ}, 40^{\\circ}, ?$ 5. $110^{\\circ}, 30^{\\circ}, ?$ 6. $50^{\\circ}, 10^{\\circ}, ?$ 7. $145^{\\circ}, 15^{\\circ}, ?$ 8. $55^{\\circ}, 45^{\\circ}, ?$ 9. $70^{\\circ}, 35^{\\circ}, ?$ 10. $50^{\\circ}, 50^{\\circ}, ?$ 11. $63^{\\circ}, 42^{\\circ}, ?$ 12. $18^{\\circ}, 75^{\\circ}, ?$ Directions: Identify three triangles in the room around you. ### Vocabulary Language: English Acute Triangle Acute Triangle An acute triangle has three angles that each measure less than 90 degrees. Congruent Congruent Congruent figures are identical in size, shape and measure. Equilateral Triangle Equilateral Triangle An equilateral triangle is a triangle in which all three sides are the same length. Isosceles Triangle Isosceles Triangle An isosceles triangle is a triangle in which exactly two sides are the same length. Obtuse Triangle Obtuse Triangle An obtuse triangle is a triangle with one angle that is greater than 90 degrees. Right Triangle Right Triangle A right triangle is a triangle with one 90 degree angle. Scalene Triangle Scalene Triangle A scalene triangle is a triangle in which all three sides are different lengths. Triangle Sum Theorem Triangle Sum Theorem The Triangle Sum Theorem states that the three interior angles of any triangle add up to 180 degrees. Jan 23,", "4: In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle. Solution: Let ABC be an isosceles triangle in which AB = AC. Then we have B=C$\\angle B=\\angle C$ Let B=C=x$\\angle B=\\angle C=x$ Then vertex angle A = 2(x + x) = 4x Now, x + x + 4x = 180o 6x=180$\\Rightarrow 6x=180^{\\circ}$ x=1806=30$\\Rightarrow x=\\frac{180^{\\circ}}{6}=30^{\\circ}$ Therefore, Vertex A=4×30=120$\\angle A=4\\times 30=120^{\\circ}$ And, B=C=30$\\angle B=\\angle C=30^{\\circ}$ Question 5: What is the measure of each of the equal angles of a right-angled isosceles triangle? Solution: In a right angle isosceles triangle, the vertex angle is A=90$\\angle A=90^{\\circ}$ and the other two base angles are equal. Let xo be the base angle and we have, B=C=x$\\angle B=\\angle C=x$. By angle sum property of a triangle, we have A+B+C=180$\\angle A+\\angle B+\\angle C=180^{\\circ}$ 90+x+x=180$\\Rightarrow 90^{\\circ}+x+x=180^{\\circ}$ 90+2x=180$\\Rightarrow 90^{\\circ}+2x=180^{\\circ}$ 2x=18090=90$\\Rightarrow 2x=180^{\\circ}-90^{\\circ}=90^{\\circ}$ x=902=45$\\Rightarrow x=\\frac{90^{\\circ}}{2}=45^{\\circ}$ Thus, we have B=C=45$\\angle B=\\angle C=45^{\\circ}$ Question 6: If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other. Solution: Given: ABC is an isosceles triangle in which AB = AC and BC is produced both ways To Prove: $\\angle$EBA = $\\angle$DCA Proof: In Δ$\\Delta$ABC we have, AB = AC $\\Rightarrow$ $\\angle$B = $\\angle$C Now exterior $\\angle$EBA =", "approaches. Start with right triangle $\\,\\triangle ABC\\,$ (yellow). It has acute angle $\\,x\\,$ and hypotenuse of length $\\,1\\,$. Thus: bottom leg: $\\,\\cos x\\,$ side leg: $\\,\\sin x\\,$ Stack a right triangle $\\,\\triangle ACD\\,$ (mostly green) on the hypotenuse of the yellow triangle. It also has acute angle $\\,x\\,$. The bottom leg has length $\\,1\\,$. Since $\\,\\cos x = \\frac{\\text{ADJ}}{\\text{HYP}} = \\frac{1}{\\text{HYP}}\\,$, the hypoteuse has length $\\,\\frac{1}{\\cos x}\\,$. Thus, the side leg is $\\,\\frac{1}{\\cos x}\\cdot \\sin x = \\frac{\\sin x}{\\cos x}\\,$. the blue right triangle: since $\\,\\overline{EC}\\, ||\\, \\overline{AB}\\,$, $\\,\\angle ACE = x\\,$ (alternate interior angles) thus, $\\,\\angle ECD = 90^\\circ - x\\,$ thus, $\\,\\angle EDC = x\\,$ Thus: bottom leg: $\\,\\frac{\\sin x}{\\cos x}(\\sin x) = \\frac{\\sin^2 x}{\\cos x}\\,$ side leg: $\\,\\frac{\\sin x}{\\cos x}(\\cos x) = \\sin x\\,$ Using right triangle $\\,\\triangle AFD\\,$, we get the cosine double-angle formula: \\begin{align} \\cos 2x &= \\frac{\\text{ADJ}}{\\text{HYP}}\\cr &= \\frac{\\cos x - \\frac{\\sin^2 x}{\\cos x}}{1/\\cos x}\\cr\\cr &= \\left(\\cos x - \\frac{\\sin^2 x}{\\cos x}\\right)(\\cos x)\\cr\\cr &= \\cos^2 x - \\sin^2 x\\cr\\cr \\end{align} Again using right $\\,\\triangle AFD\\,$, we get the sine double-angle formula: $$\\sin 2x \\ =\\ \\frac{\\text{OPP}}{\\text{HYP}} \\ =\\ \\frac{\\sin x +\\sin x}{1/\\cos x} \\ =\\ 2\\sin x\\,\\cos x$$ ## Review: Does $\\,\\cos x\\,$ Uniquely Determine $\\,\\sin x\\,$? No! Recall that every nonzero real number is uniquely identified by knowing two things: •" ]

[ "since $\\tan(x)=\\frac{\\sin(x)}{\\cos(x)}$ tangent will be one if sine and cosine are equal 69. anonymous yay, thank you so much for helping me figure out the final answer! 70. anonymous yw 71. anonymous $\\frac{ \\pi }{ 4} and \\frac{ 5\\pi }{ 4}$", "\\frac{4}{y} \\implies y = \\frac{4}{tan(58)} =~ 2.5$", "+0 0 45 1 If tan x = 6/g and sin x = 6/h, what is the value of cos x? Apr 21, 2020 #1 +1 tan x = $$\\frac{6}{g}$$ so the side opposite to x is 6 and the adjacent side (Not the hypotenuse) is g sin x =$$\\frac{6}{h}$$ so the side opposite to x is 6 (already know) and the hypotenuse is h so cos x = $$\\frac{g}{h}$$ (adjacent/hypotenuse) Apr 21, 2020", "have $\\sin^2 X + \\cos^2 X=1$, so $\\sin^2 X = 1 - \\frac{25}{169} = \\frac{144}{169}, \\sin X = \\pm \\frac{12}{13}$. Then $\\tan X = \\frac{\\sin X}{\\cos X}$ and you have to check which signs are applicable. -", "Without using mathematical tables or a calculator, evaluate $\\frac{sin\\,30^{0}-sin\\,60^{0}}{tan\\,60^{0}}$", "x=a \\frac{12}{13},\\sin y=b \\frac{12}{13}, a=\\pm 1, b=\\pm 1$$ $$I=2\\sin(x) + 2\\sin(y) + 2\\cos(x) − 2\\tan(x) + \\tan(y) + 2\\cos(y)$$ $$13I=24a + 24b + 2*5 − 2\\frac{12a}{5} + \\frac{12b}{-5} + 2*(-5)=\\frac{96a}{5} + \\frac{108b}{5}$$ $$I=\\frac{12}{65}(8a+9b)$$", "1. ## Easy Question sin θ = 8/17 and cos θ = 15/17, find tan θ. 2. Originally Posted by Joker37 sin θ = 8/17 and cos θ = 15/17, find tan θ. Hi Joker37, Remember: $\\sin \\theta=\\frac{y}{r}$ $\\cos \\theta=\\frac{x}{r}$ $\\tan \\theta=\\frac{y}{x}$ $r^2=x^2+y^2$ $15^2+8^2=17^2$ $x = 15$, $y= 8$, $r = 17$ Therefore, $\\tan \\theta = \\frac{8}{15}$", "y$ = $\\tan^{-1}$ $(\\frac{x-y}{1+xy})$ $\\sin^{-1}x\\pm\\sin^{-1}y$ = $\\sin^{-1}[x\\sqrt{1-y^{2}}\\pm\\ y\\sqrt{1-x^{2}}]$", "$$\\frac{\\tan x}{\\tan y} = 3$$", "$\\frac{sin(\\theta)}{cos(\\theta)}= tan(\\theta)= \\frac{y}{x}$ and so $y= tan(\\theta)x$.", "y = $$\\frac{5}{13}$$ Now, Putting the values of sin x, cos x, sin y, cos y ⇒ sin (x + y) = sin x cos y + cos x sin y ⇒ sin (x + y) = $$\\frac{4}{5} \\times (\\frac{-12}{13}) + (\\frac{-3}{5}) \\times \\frac{5}{13}$$ ⇒ sin (x + y) = $$\\frac{ - 48}{65} - \\frac{15 }{65}$$ ⇒ sin (x + y) = $$\\frac{-48 - 15}{65}$$ ⇒ sin (x + y) = $$-\\frac{63}{65}$$ ∴ The value of sin(x + y) is $$-\\frac{63}{65}$$", "If sin x= Question: If $\\sin x=\\frac{-24}{25}$, then the value of $\\tan x$ is __________________ . Solution: Given $\\sin x=\\frac{-24}{25}$ i.e $\\mathrm{x}$ lies in III or IV quadrant $\\sin x=\\frac{A B}{A C}=\\frac{-24}{25}$ Since $A C^{2}=A B^{2}+B C^{2}$ i. e $25^{2}=(24)^{2}+B C^{2}$ $\\Rightarrow 625=576+B C^{2}$ $\\Rightarrow B C^{2}=49$ $B C=\\pm 7$ $\\Rightarrow \\tan x=\\frac{24}{7}$ or $\\tan x=\\frac{-24}{7}$ sin", "# Suppose sin(x) = 9/11 and that #90 < x < 180, how do you find cos(x) and tan (x)? May 24, 2015 $\\sin x = \\frac{9}{11} = 0.82$ ${\\cos}^{2} x = 1 - {\\sin}^{2} x = 1 - 0.67 = 0.33 \\to \\cos x = - 0.57$ (90 < x < 180) $\\tan x = \\frac{0.82}{-} 0.57 = - 1.44$", "# If tan x = 1, 0 < x < 90°, then what is the value of 2 sin x cos x? Free Practice With Testbook Mock Tests ## Options: 1. $$\\frac{1}{2}$$ 2. 1 3. $$\\frac{{\\sqrt 3 }}{2}$$ 4. √3 ### Correct Answer: Option 2 (Solution Below) This question was previously asked in CDS Maths Previous Paper 7 (Held On: 3 Feb 2019) ## Solution: tan x = 1 for 0 < x < 90° then x = 45° ⇒ 2 sin x cos x = 2 × sin 45° cos 45° $$= 2 \\times \\frac{1}{{\\sqrt 2 }} \\times \\frac{1}{{\\sqrt 2 }} = 1$$", "# Suppose sin(x) = 9/11 and that #90 < x < 180, how do you find cos(x) and tan (x)? $\\sin x = \\frac{9}{11} = 0.82$ ${\\cos}^{2} x = 1 - {\\sin}^{2} x = 1 - 0.67 = 0.33 \\to \\cos x = - 0.57$ $\\tan x = \\frac{0.82}{-} 0.57 = - 1.44$", "# If sin x cos y Question: If $\\sin x \\cos y=\\frac{1}{4}$ and $3 \\tan x=4 \\tan y$, then $\\sin (x-y)$ is equal to ____________________. Solution: Given $\\sin x \\cos y=\\frac{1}{4}$ and $3 \\tan x=4 \\tan y$ i. e. $\\tan x=\\frac{4}{3} \\tan y$ i. e. $\\frac{\\tan x}{\\tan y}=\\frac{4}{3}$ i. e. $\\frac{\\sin x}{\\cos x} \\frac{\\cos y}{\\sin y}=\\frac{4}{3}$ $\\Rightarrow \\frac{1 / 4}{\\cos x \\sin y}=\\frac{4}{3}$ $\\Rightarrow \\cos x \\sin y=\\frac{3}{16}$ $\\therefore \\sin (x-y)=\\sin x \\cos y-\\cos x \\sin y$ $=\\frac{1}{4}-\\frac{3}{16}$ $=\\frac{4-3}{16}$ $\\sin (x-y)=\\frac{1}{16}$ hence, value of $\\sin (x-y)=\\frac{1}{16}$.", "then value of $$x^2+x+\\frac{1}{x}+\\frac{1}{x^2}$$ is Question 22 If $$x^2=y + z, y^2=z + x, z^2=x+y$$, then the value $$\\frac{1}{1+x} + \\frac{1}{1+y} + \\frac{1}{1+z}$$ Question 23 Find x given $$\\sqrt{1 - \\frac{x^3}{100}} = \\frac{3}{5}$$ Question 24 A child reshapes a cone made up of clay of height 24 and radius 6cm into a sphere. The radius (in cm) of the sphere is Question 25 If x,y are positive acute angles, x + y < 90o and sin(2x -20o) = cos(2y + 20o), then the values of sec(x + y) is Question 26 If tan(x + y)tan(x - y) = 1, then the value of $$tan(\\frac{2x}{3})$$ is Question 27 If $$\\sqrt{1 + \\frac{x}{9}} = \\frac{13}{3}$$ Question 28 If $$x + \\frac{2}{3 + \\frac{4}{5 + \\frac{7}{6}}}=10$$, the value of $$x$$ equals Question 29 The minimum value of (x-2) (x-9) is Question 30 If $$\\frac{11-13x}{x}+\\frac{11-13y}{y}+\\frac{11-13z}{z}=5$$, then what is value of $$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}$$ ? Question 31 If $$4^{(x+y)} = 256$$ and $$(256)^{(x-y)} = 4$$, then what is the value of x and y? Question 32 If $$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=0$$ and x + y + z = 9, then what is the value of $$x^3 + y^3 + z^3 - 3xyz$$ ? Question 33 If x = 1 - y and $$x^2 = 2 - y^2$$, then what is the value of xy? Question 34 If 17/3", "Comment Share Q) # If $$tan^{-1}x+tan^{-1}y=\\frac{\\pi}{4}$$, then write the value of $$x+y+xy$$. $Because\\:tan^{-1}x+tan^{-1}y=tan^{-1}{\\frac{x+y}{1-xy}}$", "if sinx°=sin@, then @ is - ? the value of tan15°•tan20°•tan70°•tan75° - NAVJIT 0.037 than find sin and tan? cos24/25 then find sin and tan", "# The value of tan x sin Question: The value of $\\tan x \\sin \\left(\\frac{\\pi}{2}+x\\right) \\cos \\left(\\frac{\\pi}{2}-x\\right)$ (a) 1 (b) $-1$ (c) $\\frac{1}{2} \\sin 2 x$ (d) none of these. Solution: (d) none of these. We have: $\\tan \\theta \\sin \\left(\\frac{\\pi}{2}+x\\right) \\cos \\left(\\frac{\\pi}{2}-x\\right)$ $=\\frac{\\sin x}{\\cos x} \\cos x \\sin x$ $=\\sin ^{2} x$" ]

[ "\\frac{dx+dy}{x+y} = \\frac{d(x+y)}{x+y}$. Hence the original integral is $\\displaystyle \\int \\frac{dy}{x} = \\int \\frac{d(x+y)}{x+y} = \\ln |x+y| + C = \\ln|y + \\sqrt{y^2 + a^2}| + C$. In your case, $\\displaystyle a=13$. Use the substitution $x=13\\tan{u}$, then $dx=13\\sec^2{u} \\ du$ \\begin{align} \\int\\frac{dx}{\\sqrt{169+x^2}} &=\\int\\frac{13\\sec^2{u} \\ du}{\\sqrt{169+169\\tan^2{u}}}\\\\ &=\\int\\sec{u} \\ du\\\\ &=\\ln|\\sec{u}+\\tan{u}|+c\\\\ &=\\ln\\left|\\frac{x}{13}+\\sqrt{1+\\left(\\frac{x}{13}\\right)^2}\\right|+c \\end{align} Alternatively, we can use the hyperbolic substitution $x=13\\sinh{u}$, then $dx=13\\cosh{u} \\ du$. Thus \\begin{align} \\int\\frac{dx}{\\sqrt{169+x^2}} &=\\int\\frac{13\\cosh{u} \\ du}{\\sqrt{169+169\\sinh^2{u}}}\\\\ &=\\int\\frac{13\\cosh{u} \\ du}{13\\cosh{u}}\\\\ &=\\operatorname{arcsinh}\\left(\\frac{x}{13}\\right)+c \\end{align} These 2 answers are equal. • @brian: If the official answer is $\\ln|y+\\sqrt{169+y^2}|+C$, that is \"the same\" as the one above, because the constant can be absorbed into the constant of integration. – André Nicolas Aug 2 '14 at 3:08 • I honestly have no clue – brian Aug 2 '14 at 3:13 • @Macavity Thanks for pointing out that blunder. I have edited it. – SuperAbound Aug 2 '14 at 3:19", "at 22:27 This answer is inspired by the answer of coffeemath. We know the following (from e.g. Wikipedia): \\begin{align} &\\sin(x) = x -\\frac{x^3}{6} +\\frac{x^5}{120}+ O(x^7) \\\\ &\\cos(x) = 1 -\\frac{x^2}{2} +\\frac{x^4}{24} +\\frac{x^6}{720} +O(x^8) \\\\ &\\tan(x) = a +bx+ cx^2+dx^3+ex^4+fx^5+gx^6+ O(x^7) \\end{align} Furthermore it holds $\\tan(x) = \\frac{\\sin(x)}{\\cos(x)} \\Rightarrow \\sin(x) =\\tan(x)\\cos(x)$. So we get \\begin{align} x -\\frac{x^3}{6} +\\frac{x^5}{120}+ O(x^7) =& \\bigl(1 -\\frac{x^2}{2} +\\frac{x^4}{24} +\\frac{x^6}{720} +O(x^8) \\bigr) \\\\ &\\cdot\\bigl(a +bx+ cx^2+dx^3+ex^4+fx^5+gx^6+ O(x^7)\\bigr) \\end{align} If we expand the term on the right hand side we obtain \\begin{align} x -\\frac{x^3}{6} +\\frac{x^5}{120}+ O(x^7) &=a+bx+(-\\frac{a}{2}+c)x^2+(-\\frac{b}{2}+d)x^3 \\\\&+(\\frac{a}{24}-\\frac{c}{2}+e)x^4+(\\frac{b}{24}-\\frac{d}{2}+f)x^5 \\\\&+(\\frac{a}{720}+\\frac{c}{24}-\\frac{e}{2}+g)x^6+ O(x^7) \\end{align} From this we can deduct the linear system \\begin{align} \\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0&0\\\\ 0 & 1 & 0 & 0 & 0 & 0 &0\\\\ -1/2 &0 &1 & 0& 0& 0&0\\\\ 0 & -1/2 & 0 & 1 & 0 & 0&0 \\\\ 1/24 & 0 & -1/2 & 0 & 1 & 0&0\\\\ 0 &1/24 & 0 & -1/2 & 0 & 1&0\\\\ 1/720 & 0 & 1/24 & 0 & -1/2 & 0 &1 \\end{pmatrix} \\begin{pmatrix} a\\\\b\\\\c\\\\d\\\\e\\\\f\\\\g \\end{pmatrix} = \\begin{pmatrix} 0\\\\1 \\\\ 0 \\\\ -1/6 \\\\ 0 \\\\1/120 \\\\ 0 \\end{pmatrix} \\end{align} Which has the solution \\begin{align} \\begin{pmatrix} a\\\\b\\\\c\\\\d\\\\e\\\\f\\\\g \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\\\ 1/3 \\\\ 0 \\\\2/15 \\end{pmatrix} \\end{align}", "# Let tan x = 2.4, sin y = 0.6, and both x and y be between 0° and 90°. Then cos(x + y) equals? May 12, 2018 $- \\frac{16}{65}$ #### Explanation: May 13, 2018 cos (x + y) = - 0.246 #### Explanation: Use trig identity: cos (x + y) = cos x.cos y - sin x.sin y (1) First, find cos y knowing sin y = 0.6 ${\\cos}^{2} y = 1 - {\\sin}^{2} y = 1 - 0.36 = 0.64$ $\\cos y = 0.8$ (because y is in Quadrant 1) Next, find sin x and cos x, knowing tan x = 2.4. ${\\cos}^{2} x = \\frac{1}{1 + {\\tan}^{2} x} = \\frac{1}{1 + 5.76} = \\frac{1}{6.76}$ $\\cos x = \\frac{1}{2.6}$ (because x is in Q. 1) ${\\sin}^{2} x = 1 - {\\cos}^{2} x = 1 - \\frac{1}{6.76} = \\frac{5.76}{6.76}$ $\\sin x = \\frac{2.4}{2.6}$ Replace all numeric values into equation (1), we get: $\\cos \\left(x + y\\right) = \\left(\\frac{1}{2.6}\\right) \\left(0.8\\right) - \\left(\\frac{2.4}{2.6}\\right) \\left(0.6\\right) = \\frac{0.8}{2.6} - \\frac{1.44}{2.6}$ $\\cos \\left(x + y\\right) = - \\frac{0.64}{2.6} = - 0.246$ Check by calculator. $\\sin y = 0.6$ --> $y = {36}^{\\circ} 87$ $\\cos x = \\frac{1}{2.6}$ --> $x = {67}^{\\circ} 38$ $x + y = 36.87 + 67.38 = {104}^{\\circ} 25$ --> cos 104.25 = - 0.246. OK", "1Siny$\\frac{1}{Sin\\; y}$ = 135=53$\\frac{1}{\\frac{3}{5}} = \\frac{5}{3}$ Since, sin2y+cos2y=1$sin^{2}\\;y \\; + \\; cos^{2}\\;y = 1$ $\\Rightarrow$ cos2y=1sin2y$cos^{2} y = 1 – sin^{2} y$ $\\Rightarrow$ cos2y=1(35)2$cos^{2} y = 1 – \\left ( \\frac{3}{5} \\right )^{2}$ $\\Rightarrow$ cos2y=1925$cos^{2} y = 1 – \\frac{9}{25}$ $\\Rightarrow$ cos2y=1625$cos^{2} y = \\frac{16}{25}$ $\\Rightarrow$ cos y = ±45$\\pm \\frac{4}{5}$ Since, y lies in the 2nd quadrant, the value of cos y will be negative, Therefore, cos y = – 45$\\frac{4}{5}$ $\\Rightarrow$ sec y = 1cosy=1(45)=54$\\frac{1}{cos \\; y} = \\frac{1}{\\left (- \\frac{4}{5} \\right )} =\\; – \\frac{5}{4}$ $\\Rightarrow$ tan y = sinycosy=(35)(45)=34$\\frac{sin \\; y}{cos \\; y} = \\frac{\\left (\\frac{3}{5} \\right ) }{\\left (-\\frac{4}{5} \\right ) } = \\;- \\frac{3}{4}$ $\\Rightarrow$ cot y = 1tany=43$\\frac{1}{tan \\; y} =\\;- \\frac{4}{3}$ Q.3: Find the values of other five trigonometric functions if coty=34$cot \\; y = \\frac{3}{4}$, where y lies in the third quadrant. Sol: cot y = 34$\\frac{3}{4}$ Since, tan y = 1coty=134=43$\\frac{1}{cot \\; y}=\\frac{1}{\\frac{3}{4}}=\\frac{4}{3}$ Since, 1+tan2y=sec2y$1 + tan^{2}y = sec^{2}y$ $\\Rightarrow$ 1+(43)2=sec2y$1 + \\left ( \\frac{4}{3} \\right )^{2} = sec^{2}y$ $\\Rightarrow$ 1+169=sec2y$1 + \\frac{16}{9} = sec^{2}y$ $\\Rightarrow$ 259=sec2y$\\frac{25}{9} = sec^{2}y$ $\\Rightarrow$ sec y = ±53$\\pm \\frac{5}{3}$ Since, y lies in the 3rd quadrant, the value of sec y will be negative. Therefore, sec y = 53$\\;- \\frac{5}{3}$ cos y = 1secy=1(53)=35$\\frac{1}{sec \\; y} = \\frac{1}{\\left (- \\frac{5}{3} \\right )}", "the answers were useful to you. – SuperAbound Jul 25 '14 at 6:16 Hint : \\begin{align} \\sin x+\\cos x&=\\frac12\\\\ \\sin x&=\\frac12-\\cos x\\\\ \\sin^2 x&=\\left(\\frac12-\\cos x\\right)^2\\\\ 1-\\cos^2 x&=\\frac14-\\cos x+\\cos^2 x\\\\ 2\\cos^2 x-\\cos x-\\frac34&=0\\\\ 8\\cos^2 x-4\\cos x-3&=0 \\end{align} $\\sin x+\\cos x=\\frac12$ $\\Rightarrow$ $(\\sin x+\\cos x)^2=\\frac14$ $\\Rightarrow$ $\\sin^2 x+2\\sin x\\cos x+\\cos^2 x=\\frac14$ $\\Rightarrow$ $1+2\\sin x\\cos x=\\frac14$ $\\Rightarrow$ $2\\sin x\\cos x = -\\frac34$ $\\Rightarrow$ $\\sin x\\cos x=-\\frac38$. If $a=\\sin x$, $b=\\cos x$, we have the system of equations \\begin{align} a+b&=\\frac12\\\\ ab&=-\\frac38 \\end{align} which leads to $$a-\\frac3{8a}=\\frac12\\\\ a^2-\\frac12a-\\frac38=0$$ with the solutions $a=\\frac{1\\pm\\sqrt7}4$. So we get $$\\sin x=\\frac{1+\\sqrt7}4, \\qquad \\cos x= \\frac{1-\\sqrt7}4 \\qquad \\Rightarrow \\qquad \\tan x = \\frac{1+\\sqrt7}{1-\\sqrt7}\\\\ \\sin x=\\frac{1-\\sqrt7}4, \\qquad \\cos x= \\frac{1+\\sqrt7}4 \\qquad \\Rightarrow \\qquad \\tan x = \\frac{1-\\sqrt7}{1+\\sqrt7}$$ You could also simplify the result: $$\\frac{1+\\sqrt7}{\\frac1-\\sqrt7} = \\frac{(1+\\sqrt7)^2}{1-7} = -\\frac{8+2\\sqrt7}6= -\\frac{4+\\sqrt7}3$$ HINT: Avoid squaring whenever possible as it immediately introduces extraneous root which needs to be discarded Use Weierstrass substitution to form a Quadratic Equation in $\\tan\\dfrac x2$ Now use $\\displaystyle\\tan2A=\\frac{2\\tan A}{1-\\tan^2A}$ Divide through by $\\cos x$ to get \\begin{align} \\tan x + 1 & = \\frac 12 \\sec x \\tag{1}\\\\ \\text{Square both sides: } \\qquad\\qquad\\tan^2 x + 2 \\tan x + 1 & = \\frac 14 \\sec^2 x = \\frac{1}{4} (\\tan^2 x + 1) \\\\ 3 \\tan^2 x + 8 \\tan x + 3 & = 0 \\\\ \\tan x", "5x}=\\frac{-2\\sin \\frac{3x+5x}{2}\\sin \\frac{3x-5x}{2}}{2\\sin \\frac{3x+5x}{2}\\cos \\frac{3x-5x}{2}} \\\\ & =\\frac{-2\\sin \\frac{8x}{2}\\sin \\frac{-2x}{2}}{2\\sin \\frac{8x}{2}\\cos \\frac{-2x}{2}} \\\\ & =\\frac{-2\\sin 4x\\sin \\left( -x \\right)}{2\\sin 4x\\cos \\left( -x \\right)} \\\\ & =-\\tan \\left( -x \\right) \\end{align} Thus, the result will be $\\tan x$. Hence, verified. Also, consider the identity: $\\frac{\\cos 3x-\\cos x}{\\sin 3x+\\sin x}=-\\tan x$ And use the sum-to product formula $\\cos \\alpha -\\cos \\beta =-2\\sin \\frac{\\alpha +\\beta }{2}\\sin \\frac{\\alpha -\\beta }{2}$ to numerator and sum-to product formula $\\sin \\alpha +\\sin \\beta =2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2}$ to denominator of left hand side. Then, \\begin{align} & \\frac{\\cos 3x-\\cos x}{\\sin 3x+\\sin x}=\\frac{-2\\sin \\frac{3x+x}{2}\\sin \\frac{3x-x}{2}}{2\\sin \\frac{3x+x}{2}\\cos \\frac{3x-x}{2}} \\\\ & =\\frac{-2\\sin \\frac{4x}{2}\\sin \\frac{2x}{2}}{2\\sin \\frac{4x}{2}\\cos \\frac{2x}{2}} \\\\ & =\\frac{-2\\sin 2x\\sin x}{2\\sin 2x\\cos x} \\\\ & =-\\tan x \\end{align} Hence verified. Also, consider the identity: $\\frac{\\sin 3x-\\sin x}{\\cos 3x-\\cos x}=-\\cot 2x$ By using the sum-to product formula $\\sin \\alpha -\\sin \\beta =2\\sin \\frac{\\alpha -\\beta }{2}\\cos \\frac{\\alpha +\\beta }{2}$ to numerator and $\\cos \\alpha -\\cos \\beta =-2\\sin \\frac{\\alpha +\\beta }{2}\\sin \\frac{\\alpha -\\beta }{2}$ to denominator of left hand side. Then, \\begin{align} & \\frac{\\sin 3x-\\sin x}{\\cos 3x-\\cos x}=\\frac{2\\sin \\frac{3x-x}{2}\\cos \\frac{3x+x}{2}}{-2\\sin \\frac{3x+x}{2}\\sin \\frac{3x-x}{2}} \\\\ & =\\frac{2\\sin \\frac{2x}{2}\\cos \\frac{4x}{2}}{-2\\sin \\frac{4x}{2}\\sin \\frac{2x}{2}} \\\\ & =-\\frac{\\cos 2x}{\\sin 2x} \\\\ & =-\\cot 2x \\end{align} Hence verified. And also consider the identity: $\\frac{\\sin x-\\sin y}{\\sin x+\\sin y}=\\tan \\frac{x-y}{2}\\cot \\frac{x+y}{2}$ Also, use the sum-to product formula", "Symmetry says $\\int_0^\\pi{dx\\over\\sin^2x+|\\cos x|}=2\\int_0^{\\pi/2}{dx\\over\\sin^2x+\\cos x}$ – Barry Cipra May 1 '17 at 15:43 I can't take full credit for this, Wolfram can be very helpful when you know how to coax it along. For the integral $$\\int{dx\\over\\sin^2x+\\cos x},$$ let \\begin{align*} u &= \\tan\\left(\\frac{x}{2}\\right)\\\\ du &= \\frac{1}{2}\\sec^{2}\\left(\\frac{x}{2}\\right)\\,dx. \\end{align*} Then we have \\begin{align*} \\sin x &= \\frac{2\\sin x}{1 +\\cos x}\\cdot \\frac{1+\\cos x}{2} = \\frac{2\\tan^{2}\\left(\\frac{x}{2}\\right)}{\\sec^{2}\\left(\\frac{x}{2}\\right)} = \\frac{2\\tan^{2}\\left(\\frac{x}{2}\\right)}{\\tan^{2}\\left(\\frac{x}{2}\\right)+1}=\\frac{2u^{2}}{u^{2}+1}\\\\ \\cos x&= \\frac{2\\cos x}{1+\\cos x}\\cdot\\frac{1+\\cos x}{2}=\\frac{1+\\cos x - 1 + \\cos x}{(1+\\cos x)\\sec^{2}\\left(\\frac{x}{2}\\right)}=\\frac{1-\\frac{1-\\cos x}{1+\\cos x}}{\\sec^{2} \\left(\\frac{x}{2}\\right)}= \\frac{1-u^{2}}{u^{2}+1}\\\\ dx &= \\frac{2\\,du}{u^{2}+1}. \\end{align*} The integral then becomes \\begin{align*} 2\\int \\frac{du}{(u^{2}+1)\\left[\\frac{4u^{2}}{(u^{2}+1)^{2}}+\\frac{1-u^{2}}{u^{2}+1}\\right]} &=-2\\int\\frac{u^{2}+1}{u^{4}-4u^{2}-1}\\,du\\\\ &=-2\\int\\frac{u^{2}+1}{\\left(u^{2}-\\sqrt{5}-2\\right)\\left(u^{2}+\\sqrt{5}-2\\right)}\\,du. \\end{align*} We can then use a partial fraction decomposition, which I will omit, to change the integral to \\begin{align*} -\\int\\frac{\\sqrt{5}-3}{\\sqrt{5}\\left(u^{2}+\\sqrt{5}-2\\right)}\\,du -\\int\\frac{-3-\\sqrt{5}}{\\sqrt{5}\\left(-u^{2}+\\sqrt{5}+2\\right)}\\,du, \\end{align*} which becomes \\begin{align*}\\frac{3-\\sqrt{5}}{\\sqrt{5}}\\int\\frac{du}{u^{2}+\\sqrt{5}-2} + \\frac{3+\\sqrt{5}}{\\sqrt{5}}\\int\\frac{du}{-u^{2} + \\sqrt{5} +2}. \\end{align*} We can rewrite this as $$\\frac{3-\\sqrt{5}}{\\sqrt{5}(\\sqrt{5}-2)}\\int\\frac{du}{1+\\left(\\frac{u}{\\sqrt{\\sqrt{5}-2}}\\right)^{2}}+\\frac{3+\\sqrt{5}}{\\sqrt{5}(\\sqrt{5}+2)}\\int\\frac{du}{1-\\left(\\frac{u}{\\sqrt{\\sqrt{5}+2}}\\right)^{2}}.$$ This then becomes $$\\frac{3-\\sqrt{5}}{\\sqrt{5}\\sqrt{\\sqrt{5}-2}}\\tan^{-1}\\left(\\frac{u}{\\sqrt{\\sqrt{5}-2}}\\right)+\\frac{3+\\sqrt{5}}{\\sqrt{5}\\sqrt{\\sqrt{5}+2}}\\tanh^{-1}\\left(\\frac{u}{\\sqrt{\\sqrt{5}+2}}\\right)+C,$$ or substituting back $$\\frac{3-\\sqrt{5}}{\\sqrt{5}\\sqrt{\\sqrt{5}-2}}\\tan^{-1}\\left(\\frac{\\tan\\left(\\frac{x}{2}\\right)}{\\sqrt{\\sqrt{5}-2}}\\right)+\\frac{3+\\sqrt{5}}{\\sqrt{5}\\sqrt{\\sqrt{5}+2}}\\tanh^{-1}\\left(\\frac{\\tan\\left(\\frac{x}{2}\\right)}{\\sqrt{\\sqrt{5}+2}}\\right)+C.$$ Since this expression vanishes at $x=0$, your integral is given by $$-2\\pi\\left(\\frac{3-\\sqrt{5}}{\\sqrt{5}\\sqrt{\\sqrt{5}-2}}\\tan^{-1}\\left(\\frac{\\tan\\left(\\frac{\\pi}{4}\\right)}{\\sqrt{\\sqrt{5}-2}}\\right)+\\frac{3+\\sqrt{5}}{\\sqrt{5}\\sqrt{\\sqrt{5}+2}}\\tanh^{-1}\\left(\\frac{\\tan\\left(\\frac{\\pi}{4}\\right)}{\\sqrt{\\sqrt{5}+2}}\\right)\\right)\\approx -8.734.$$ You've done most part of the problem. I can't solve the last integral, but I believe I can provide an approximate solution. Let us take $f(x) = \\sin^2 x + |cos x|$. The graphs of $|\\cos x|$ and $\\cos^2 x$ are pretty close. So $f(x)$ will always be quite close to $1$. Now, to find the maxima and minima values of the function, we need only", "= 1 and x = 0. (9) By substitution z = 2x - 3y, show that (2x - 3y + 3)dy/dx = 2x - 3y + 1 is 2ln ( 3 - 2x + 3y ) = 1 - x + y when y = 0 and x = 1. Please take your time, I'm having hard time solving these problems, again, thank you very much. 1. \\displaystyle \\begin{align*} \\frac{y}{\\sec^2{(x)}}\\,\\frac{dy}{dx} &= 2\\tan{(x)} + 1 \\\\ y\\,\\frac{dy}{dx} &= \\sec^2{(x)}\\left[ 2\\tan{(x)} + 1 \\right] \\\\ \\int{y\\,\\frac{dy}{dx}\\,dx} &= \\int{\\sec^2{(x)}\\left[ 2\\tan{(x)} + 1 \\right] dx} \\\\ \\int{y\\,dy} &= \\frac{1}{2}\\int{u\\,du} \\textrm{ after making the substitution } u = 2\\tan{(x)} + 1 \\implies du = 2\\sec^2{(x)}\\,dx \\end{align*} Go from here. 2. Solve by direct integration. You may need partial fractions. 3. \\displaystyle \\begin{align*} \\sin^2{(x)}\\,\\frac{dy}{dx} &= \\frac{\\cos{(x)}}{y} \\\\ y\\,\\frac{dy}{dx} &= \\frac{\\cos{(x)}}{\\sin^2{(x)}} \\\\ \\int{y\\,\\frac{dy}{dx}\\,dx} &= \\int{\\frac{\\cos{(x)}}{\\sin^2{(x)}}\\,dx} \\\\ \\int{y\\,dy} &= \\int{u^{-2}\\,du} \\textrm{ after making the substitution } u = \\sin{(x)} \\implies du = \\cos{(x)}\\,dx \\end{align*} Go from here. 4. \\displaystyle \\begin{align*} \\left( 1 - x^2 \\right) \\frac{dy}{dx} &= y \\left( 1 + x^2 \\right) \\\\ \\frac{1}{y}\\,\\frac{dy}{dx} &= \\frac{x^2 + 1}{-x^2 + 1} \\\\ \\frac{1}{y} \\,\\frac{dy}{dx} &= - \\left( \\frac{x^2 + 1}{x^2 - 1} \\right) \\\\ \\frac{1}{y}\\,\\frac{dy}{dx} &= -\\left( 1 - \\frac{2}{x^2 - 1} \\right) \\\\ \\int{\\frac{1}{y}\\,\\frac{dy}{dx}\\,dx} &= \\int{-\\left( 1 - \\frac{2}{x^2 - 1} \\right) dx } \\\\", "\\tan v}{1 \\mp \\tan u \\tan v}$ Double Angle Formulas: • $\\sin (2 u) =2 \\sin u \\cos u$ • $\\cos (2 u) =\\cos ^{2} u-\\sin ^{2} u$ • $\\cos (2 u) =2 \\cos ^{2} u-1$ • $\\cos (2 u) =1-2 \\sin ^{2} u$ • $\\tan (2 u) =\\frac{2 \\tan u}{1-\\tan ^{2} u}$ Half Angle Formulas: • $\\sin^{2} u=\\frac{1-\\cos (2 u)}{2}$ • $\\cos^{2} u=\\frac{1+\\cos (2 u)}{2}$ • $\\tan^{2} u=\\frac{1-\\cos (2 u)}{1+\\cos (2 u)}$ Sum to product formulas: • $\\sin u+\\sin v=2 \\sin \\left(\\frac{u+v}{2}\\right) \\cos \\left(\\frac{u-v}{2}\\right)$ • $\\sin u-\\sin v=2 \\cos \\left(\\frac{u+v}{2}\\right) \\sin \\left(\\frac{u-v}{2}\\right)$ • $\\cos u+\\cos v=2 \\cos \\left(\\frac{u+v}{2}\\right) \\cos \\left(\\frac{u-v}{2}\\right)$ • $\\cos u-\\cos v=-2 \\sin \\left(\\frac{u+v}{2}\\right) \\sin \\left(\\frac{u-v}{2}\\right)$ Product to Sum Formulas: • $\\sin u \\sin v=\\frac{1}{2}[\\cos (u-v)-\\cos (u+v)]$ • $\\cos u \\cos v=\\frac{1}{2}[\\cos (u-v)+\\cos (u+v)]$ • $\\sin u \\cos v=\\frac{1}{2}[\\sin (u+v)+\\sin (u-v)]$ • $\\cos u \\sin v=\\frac{1}{2}[\\sin (u+v)-\\sin (u-v)]$", "20:50 Understandable - I just found this to be a little less messy. – user17794 Aug 27 '13 at 20:52 Hint: \\begin{align} 0&=\\cos x+\\sin x\\\\ \\sin x &= -\\cos x \\\\ \\text{Divide}&\\text{ both sides by }\\cos x\\\\ \\tan x &= -1 \\end{align} Now you can solve for $x$ using inverse trigonometric functions. - if you use the Double angle formulae $$\\sin^2(x)\\cos^2(x)=(1-\\cos2x)/2 \\cdot (1+\\cos2x)/2 = (1-\\cos^2(2x))/4 = \\sin^2(2x)/4 = (1-\\cos4x)/8$$ so minimum is $1/8$ at $x=45\\circ$ and maximum is 1/4 at $x=90\\circ$ remember $\\sin^2(x)=(1-\\cos2x)/2$ and $\\cos^2(x)=(1+\\cos2x)/2$ - The equation $\\cos x+\\sin x=0$ is really the same as $\\cos x-\\sin x=0$ once you make the transformation $x\\mapsto x+\\frac{\\pi}2$, because $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and $\\sin (x+\\frac{\\pi}2)=\\cos(x)$. The latter can be solved by resorting to $\\cos ^2+\\sin ^2=1$. Edit: We can derive $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and similar results (cf. Daniel Fischer's comment) from the well known formulas $\\cos (x+y)=\\cos x\\cos y-\\sin x\\sin y$ and $\\sin (x+y)=\\cos x\\sin y+\\cos y\\sin x$. These can either be proved geometrically, or by resorting to some definition of $\\sin$ and $\\cos$, for example $\\sin x=(e^{ix}-e^{-ix})/2i, \\cos x=(e^{ix}-e^{-ix})/2$ -", "## Precalculus (6th Edition) Blitzer Some of the identities that are verified using sum-to product formulas are: Verified identities are: \\begin{align} & \\tan x=\\frac{\\cos 3x-\\cos 5x}{\\sin 3x+\\sin 5x} \\\\ & -\\tan x=\\frac{\\cos 3x-\\cos x}{\\sin 3x+\\sin x} \\\\ & -\\cot 2x=\\frac{\\sin 3x-\\sin x}{\\cos 3x-\\cos x} \\\\ & \\tan 3x=\\frac{\\sin 2x+\\sin 4x}{\\cos 2x+\\cos 4x} \\end{align} Other identities are: \\begin{align} & \\frac{\\sin x-\\sin y}{\\sin x+\\sin y}=\\tan \\frac{x-y}{2}\\cot \\frac{x+y}{2} \\\\ & \\frac{\\sin x+\\sin y}{\\sin x-\\sin y}=\\tan \\frac{x+y}{2}\\cot \\frac{x-y}{2} \\\\ & \\frac{\\sin x-\\sin y}{\\cos x-\\cos y}=-\\cot \\frac{x+y}{2} \\end{align} And, \\begin{align} & \\frac{\\sin x+\\sin 3x}{\\cos x+\\cos 3x}=\\tan 2x \\\\ & \\frac{\\cos 4x-\\cos 2x}{\\sin 2x-\\sin 4x}=\\tan 3x \\end{align} Sum-to product formulas are given below: \\begin{align} & \\sin \\alpha +\\sin \\beta =2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2} \\\\ & \\sin \\alpha -\\sin \\beta =2\\sin \\frac{\\alpha -\\beta }{2}\\cos \\frac{\\alpha +\\beta }{2} \\\\ & \\cos \\alpha +\\cos \\beta =2\\cos \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2} \\\\ & \\cos \\alpha -\\cos \\beta =-2\\sin \\frac{\\alpha +\\beta }{2}\\sin \\frac{\\alpha -\\beta }{2} \\end{align} Let us consider the identity: $\\frac{\\cos 3x-\\cos 5x}{\\sin 3x+\\sin 5x}=\\tan x$ By using the sum-to product formula $\\cos \\alpha -\\cos \\beta =-2\\sin \\frac{\\alpha +\\beta }{2}\\sin \\frac{\\alpha -\\beta }{2}$ to numerator and sum-to product formula $\\sin \\alpha +\\sin \\beta =2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2}$ to denominator of left hand side. Then, \\begin{align} & \\frac{\\cos 3x-\\cos 5x}{\\sin 3x+\\sin", "# Integrate $\\int\\frac{1}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x}dx$ Solve the indefinite integral $$I=\\int\\frac{1}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x}\\;dx$$ My Attempt: \\begin{align} I&=\\int\\frac{1}{\\sin x+\\cos x+\\frac{1}{\\sin x \\cos x}+\\frac{\\sin x +\\cos x}{\\sin x\\cos x}}\\;dx\\\\ \\\\ &=\\int\\frac{\\sin x\\cos x}{\\left(\\sin x+\\cos x\\right)\\left(\\sin x\\cos x \\right)+1+\\left(\\sin x+\\cos x\\right)}\\;dx \\end{align} How can I complete the solution from here? • Do you have any reason to think it has a nice answer? – Greg Martin Sep 18 '13 at 5:14 • Yeah... Good luck – Ali Caglayan Dec 25 '13 at 23:45 You want \\begin{align} \\int\\frac{dx}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x} &= \\int\\frac{dx}{\\sin x+\\cos x+\\frac{\\sin x}{\\cos x} +\\frac{\\cos x}{\\sin x}+\\frac1{\\sin x}+\\frac1{\\cos x}}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{\\sin^2 x \\cos x+\\cos^2 x \\sin x+\\sin^2 x +\\cos^2 x+\\cos x+\\sin x}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{\\sin x \\cos x(\\sin x+ \\cos x)+1+\\cos x+\\sin x}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{(\\sin x \\cos x+1)(\\sin x+ \\cos x)+1}\\\\ \\end{align}. Applying substitutions $\\sin x = \\frac{2t}{1+t^2}, \\cos x = \\frac{1-t^2}{1+t^2}, dx = \\frac{2 dt}{1+t^2}$, this becomes \\begin{align} \\int\\frac{\\sin x \\cos x\\ dx}{(\\sin x \\cos x+1)(\\sin x+ \\cos x)+1} &=\\int\\frac{\\frac{(2t)(1-t^2)(2dt)}{(1+t^2)^3}} {(\\frac{(2t)(1-t^2)}{(1+t^2)^2}+1)(\\frac{2t}{1+t^2}+ \\frac{1-t^2}{1+t^2})+1}\\\\ &=\\int\\frac{(2t)(1-t^2)(2dt)} {((2t)(1-t^2)+(1+t^2)^2)(1+2t-t^2)+(1+t^2)^3}\\\\ &=\\int\\frac{4t(1-t^2)dt} {((2t-2t^3)+1+2t^2+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\\\ &=\\int\\frac{4t(1-t^2)dt} {(1+2t+2t^2-2t^3+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\\\ &=\\int\\frac{4t(1-t^2)dt} {2 (t+1) (2 t^4-3 t^3+3 t^2+t+1)} \\quad \\text{ (according to Wolfram Alpha)}\\\\ &=\\int\\frac{2t(1-t)dt} {2 t^4-3 t^3+3 t^2+t+1}\\\\ \\end{align}. According to Wolfram Alpha, that quartic can be factored as", "\\cos (x) + i \\sin(x)$. We have that \\begin{align*} \\cos (x+y) + i \\sin (x+y) &= e^{i(x+y)} \\\\ &= e^{ix} \\cdot e^{iy} \\\\ &= (\\cos (x) + i \\sin (x))(\\cos (y) + i \\sin (y)) \\\\ &= (\\cos (x) \\cos (y) - \\sin (x) \\sin(y)) + i(\\sin (x) \\cos(y) + \\cos(x) \\sin(y)) \\end{align*} By looking at the real and imaginary parts, we derive the sine and cosine angle addition formulas. To derive the tangent addition formula, we reduce the problem to use sine and cosine, divide both numerator and denominator by $\\cos (x) \\cos (y)$, and simplify. \\begin{align*} \\tan (x+y) &= \\frac{\\sin (x+y)}{\\cos (x+y)} \\\\ &= \\frac{\\sin (x) \\cos(y) + \\cos(x) \\sin(y)}{\\cos (x) \\cos (y) - \\sin (x) \\sin(y)} \\\\ &= \\frac{\\frac{\\sin(x)}{\\cos(x)} + \\frac{\\sin(y)}{\\cos(x)}}{1 - \\frac{\\sin (x) \\sin(y)}{\\cos (x) \\cos(y)}} \\\\ &= \\frac{\\tan (x) + \\tan (y)}{1 - \\tan (x) \\tan(y)} \\end{align*} as desired. ## Double-angle identities The trigonometric double-angle identities are easily derived from the angle addition formulas by just letting $x = y$. Doing so yields: • $\\sin (2x) = 2\\sin (x) \\cos (x)$ • $\\cos (2x) = \\cos^2 (x) - \\sin^2 (x)$ • $\\tan (2x) = \\frac{2\\tan (x)}{1-\\tan^2 (x)}$ ### Cosine double-angle identity Here are two equally useful forms of the cosine double-angle identity. Both are derived via the Pythagorean identity on the cosine double-angle", "x^2 \\right)\\frac{dy}{dx} + y = e^{tan^{- 1} x}$ Ex. 22.1 | Q 21 | Page 106 x dy = (2y + 2x4 + x2) dx Ex. 22.1 | Q 22 | Page 106 $\\left( 1 + y^2 \\right) + \\left( x - e^{tan^{- 1} y} \\right)\\frac{dy}{dx} = 0$ Ex. 22.1 | Q 23 | Page 106 $y^2 \\frac{dx}{dy} + x - \\frac{1}{y} = 0$ Ex. 22.1 | Q 24 | Page 106 $\\left( 2x - 10 y^3 \\right)\\frac{dy}{dx} + y = 0$ Ex. 22.1 | Q 25 | Page 106 (x + tan y) dy = sin 2y dx Ex. 22.1 | Q 26 | Page 106 dx + xdy = e−y sec2 y dy Ex. 22.1 | Q 27 | Page 106 $\\frac{dy}{dx}$ = y tan x − 2 sin x Ex. 22.1 | Q 28 | Page 106 $\\frac{dy}{dx}$ + y cos x = sin x cos x Ex. 22.1 | Q 29 | Page 106 Solve the following differential equation:- $\\left( 1 + x^2 \\right)\\frac{dy}{dx} - 2xy = \\left( x^2 + 2 \\right)\\left( x^2 + 1 \\right)$ Ex. 22.1 | Q 30 | Page 106 $\\left( \\sin x \\right)\\frac{dy}{dx} + y \\cos x = 2 \\sin^2 x \\cos x$ Ex. 22.1 | Q 31 | Page 106 $\\left( x^2 - 1 \\right)\\frac{dy}{dx} + 2\\left( x", "dx \\\\ \\quad \\frac{y^2}{2} + e^y = \\frac{x^2}{2} + e^{-x} + C \\\\ \\quad y^2 + 2e^y = x^2 + 2e^{-x} + 2C \\end{align} ## Example 2 Find all solutions to the differential equation $\\frac{dy}{dx} = (\\cos ^2 x )(\\cos ^2 2y)$. We will first rewrite the differential equation above as $\\frac{1}{\\cos ^2 2y} \\frac{dy}{dx} = (\\cos ^2 x)$. Therefore we have that: (3) \\begin{align} \\quad \\frac{1}{\\cos ^2 2y} \\frac{dy}{dx} = (\\cos ^2 x) \\\\ \\quad \\frac{1}{\\cos ^2 2y} \\: dy = (\\cos ^2 x) \\: dx \\\\ \\quad \\int \\frac{1}{\\cos ^2 2y} \\: dy = \\int \\cos ^2 x \\: dx \\\\ \\quad \\int \\sec ^2 (2y) \\: dy = \\int \\cos ^2 x \\: dx \\end{align} Note that $\\int \\sec^2 (2y) \\: dy = \\frac{\\tan 2y}{2}$ and $\\int \\cos ^2 x \\: dx = \\frac{\\cos x \\sin x + x}{2}$ (which can be obtained by integration by parts). Therefore we have that: (4) \\begin{align} \\quad \\int \\sec ^2 (2y) \\: dy = \\int \\cos ^2 x \\: dx \\\\ \\quad \\frac{\\tan 2y}{2} = \\frac{\\cos x \\sin x + x}{2} + C \\\\ \\quad \\tan 2y = \\cos x \\sin x + x + 2C \\\\ \\quad \\tan 2y = \\frac{\\sin 2x}{2} + x + 2C \\\\ \\quad 2\\tan 2y = \\sin 2x + 2x + 4C", "- C)$ $\\cos A.\\cos\\frac{B + C}{2}\\sin\\frac{C - B}{2} = -2\\cos A.\\sin\\frac{B - C}{2}.\\cos\\frac{B - C}{2}$ If $B = C$ then above it $0 = 0$ i.e. triangle is isosceles. If $A = 90^\\circ$ then above is $0 =0$ i.e. triangle is right angled. 13. $\\because \\tan\\frac{A}{2}, \\tan\\frac{B}{2}, \\tan\\frac{C}{2}$ are in A.P. $\\tan\\frac{A}{2} - \\tan\\frac{B}{2} = \\tan\\frac{B}{2} - \\tan \\frac{C}{2}$ $\\frac{\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)}{\\cos\\frac{A}{2}\\cos\\frac{B}{2}} = \\frac{\\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)}{\\cos\\frac{B}{2}\\cos\\frac{C}{2}}$ $\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)\\cos\\frac{C}{2} = \\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)\\cos\\frac{A}{2}$ $\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)\\sin\\left(\\frac{A}{2} + \\frac{B}{2}\\right) = \\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)\\sin\\left(\\frac{B}{2} + \\frac{C}{2}\\right)$ $\\Rightarrow \\cos B - \\cos A = \\cos C - \\cos B$ Thus, $\\cos A, \\cos B, \\cos C$ are in A.P. 14. Given, $a\\cos^2\\frac{C}{2} + c\\cos^2\\frac{A}{2} = \\frac{3b}{2}$ $a.\\frac{s(s - c)}{ab} + c.\\frac{s(s - a)}{bc} = \\frac{3b}{2}$ $\\frac{s}{b}[2s - a - c] = \\frac{3b}{2}$ $2s = 3b \\Rightarrow a + c = 2b$ We have to prove that $\\cot\\frac{A}{2} + \\cot\\frac{C}{2} = 2\\cot\\frac{B}{2}$ L.H.S. $= \\frac{s(s - a)}{\\Delta} + \\frac{s(s - c)}{\\Delta}$ $= \\frac{s}{\\Delta}(2s - a - c) = \\frac{2s(s - b)}{\\Delta} = 2\\cot\\frac{B}{2} =$ R.H.S. 15. Given, $a^2, b^2, c^2$ are in A.P. $\\Rightarrow b^2 - a^2 = c^2 - b^2$ $\\Rightarrow \\sin^2B - \\sin^2A = \\sin^2C - \\sin^2B$ $\\Rightarrow \\sin(B + A)\\sin(B - A) = \\sin(C + B)\\sin(C - B)$ $\\Rightarrow \\sin C\\sin(B - A) = \\sin A\\sin(C - B)$ $\\Rightarrow", "the Value of (i) sin 75° (ii) cos 75° (iii) tan 75° (iv) sin 15° (v) cos 15° (vi) tan 15° (vii) sin 105° (viii) cos 105°. Question 33. Find the value of $$\\tan \\frac{13 \\pi}{12}$$ Question 34. Prove that $$\\frac{\\sin (x+y)}{\\sin (x-y)}=\\frac{\\tan x+\\tan y}{\\tan x-\\tan y}$$ Question 35. Prove that $$\\begin{array}{l}{\\cos \\left(\\frac{\\pi}{4}-x\\right) \\cos \\left(\\frac{\\pi}{4}-y\\right)-\\sin \\left(\\frac{\\pi}{4}-x\\right) \\sin \\left(\\frac{\\pi}{4}-y\\right)} \\\\ {=\\sin (x+y)}\\end{array}$$ \\begin{aligned} L H S &=\\cos \\left[\\frac{\\pi}{4}-x+\\frac{\\pi}{4}-y\\right] \\\\ &=\\cos \\left[\\frac{\\pi}{2}-(x+y)\\right] \\\\ &=\\sin (x+y)=R H S \\end{aligned} Question 36. Prove that sin[(n + l)x]sin[(n + 2)x] +cos[(n+1)x]cos[(n + 2)x] = cosx LHS = cos[(n + 2)x – (n + 1)x] = cos x = RHS . Question 37. Prove that (sin3x + sinx)sinx + (cos3x -cosx)cosx = 0 LHS = sin 3x sin x+sin2 x + cos 3xcos x – cos2 x = [cos 3x cos x + sin 3x sin x] – [cos2 x – sin2 x] = cos(3x – x) – cos 2x = 0 = RHS Question 38. Prove that $$\\frac{\\tan \\left(\\frac{\\pi}{4}+x\\right)}{\\tan \\left(\\frac{\\pi}{4} x\\right)}=\\left(\\frac{1+\\tan x}{1-\\tan x}\\right)^{2}$$ $$L H S=\\frac{\\frac{1+\\tan x}{1-\\tan x}}{\\frac{1-\\tan x}{1+\\tan x}}=\\left(\\frac{1+\\tan x}{1-\\tan x}\\right)^{2}=R H S$$ Question 39. Prove that $$\\frac{\\cos (\\pi+x) \\cos (-x)}{\\sin (\\pi-x) \\cos \\left(\\frac{\\pi}{2}+x\\right)}=\\cot ^{2} x$$ $$L H S=\\frac{(-\\cos x)(\\cos x)}{(\\sin x)(-\\sin x)}=\\cot ^{2} x=R H S$$ Question 40. Prove that $$\\begin{array}{l}{\\cos \\left(\\frac{3 \\pi}{2}+x\\right) \\cos (2 \\pi+x)} \\\\ {\\left[\\cot \\left(\\frac{3 \\pi}{2}-x\\right)+\\cot (2", "Double and Half Angle Identity Proofs $$\\displaystyle \\frac{{\\sin \\left( {2x} \\right)}}{2}=\\frac{{\\tan x}}{{{{{\\tan }}^{2}}x+1}}$$ \\require{cancel} \\displaystyle \\begin{align}&=\\frac{{\\tan x}}{{{{{\\sec }}^{2}}x}}\\\\&=\\frac{{\\sin x}}{{\\cancel{{\\cos x}}}}\\cdot \\frac{{{{{\\cancel{{{{{\\cos }}^{2}}x}}}}^{{\\cos x}}}}}{1}\\\\&=\\sin x\\cos x\\\\&=\\sin x\\cos x\\cdot \\frac{2}{2}\\\\\\frac{{\\sin \\left( {2x} \\right)}}{2}&=\\frac{{2\\sin x\\cos x}}{2}\\,\\,\\,\\,\\,\\surd \\end{align} Note: To get $$\\displaystyle \\sin x\\cos x$$ into the right form for the identity $$\\displaystyle (2\\sin x\\cos x)$$, we had to multiply by $$\\displaystyle \\frac{2}{2}$$. $$\\displaystyle \\frac{{1-\\cos \\left( {2A} \\right)}}{{\\sin \\left( {2A} \\right)}}=\\tan A$$ \\displaystyle \\begin{align}\\frac{{1-\\left( {1-2{{{\\sin }}^{2}}A} \\right)}}{{\\sin \\left( {2A} \\right)}}&=\\\\\\frac{{2{{{\\sin }}^{2}}A}}{{\\sin \\left( {2A} \\right)}}&=\\\\\\frac{{{}^{{\\sin A}}\\cancel{{2{{{\\sin }}^{2}}A}}}}{{\\cancel{{2\\sin A}}\\cos A}}&=\\tan A\\,\\,\\,\\,\\,\\surd \\end{align} Note: We knew to use the $$\\displaystyle 1-2{{\\sin }^{2}}A$$ (instead of $$\\displaystyle 2{{\\cos }^{2}}A-1$$) for $$\\cos 2A$$ to get rid of the 1’s, and also because we have sin in the denominator instead of cos. $$\\displaystyle \\tan \\left( {\\frac{1}{2}\\theta } \\right)=\\csc \\theta -\\cot \\theta$$ \\displaystyle \\begin{align}&=\\frac{1}{{\\sin \\theta }}-\\frac{{\\cos \\theta }}{{\\sin \\theta }}\\\\\\tan \\left( {\\frac{1}{2}\\theta } \\right)&=\\frac{{1-\\cos \\theta }}{{\\sin \\theta }}\\,\\,\\,\\,\\,\\surd \\end{align} Note: Start from the right side, and turn everything into sin and cos since the Half Angle tan identity is written in terms of sin and cos. $$\\displaystyle \\frac{{\\cos \\left( {2\\theta } \\right)}}{{\\cos \\theta -\\sin \\theta }}=\\cos \\theta +\\sin \\theta$$ \\displaystyle \\begin{align}\\frac{{{{{\\cos }}^{2}}\\theta -{{{\\sin }}^{2}}\\theta }}{{\\cos \\theta -\\sin \\theta }}&=\\\\\\frac{{\\cancel{{\\left( {\\cos \\theta -\\sin \\theta } \\right)}}\\left( {\\cos \\theta +\\sin \\theta } \\right)}}{{\\cancel{{\\cos \\theta -\\sin \\theta }}}}&=\\cos \\theta +\\sin \\theta \\,\\,\\,\\,\\,\\surd \\end{align}", "where this is going. Do I substitute now? – Brian Patino May 6 '14 at 22:44 See the answer below. Indeed, I was hinting to a substitution. – DanielC May 6 '14 at 22:47 Through \"brute force\", you can just write out the values...there are several identities you need to get there: $$\\sin(a)\\sin(b) = \\frac{\\cos(a - b)-\\cos(a + b)}{2} = \\frac{\\cos(b - a)-\\cos(a + b)}{2} \\\\ \\cos(a)\\cos(b) = \\frac{\\cos(a - b)+\\cos(a + b)}{2} = \\frac{\\cos(b - a) + \\cos(a + b)}{2} \\\\ \\sin(a)\\cos(b) = \\frac{\\sin(a + b) + \\sin(a - b)}{2}$$ You can verify all of those properties from the angle addition properties: \\begin{align} \\cos(a + b) =&& \\cos(a)\\cos(b) - \\sin(a)\\sin(b) \\\\ \\cos(a - b) =& \\cos(a)\\cos(-b) - \\sin(a)\\sin(-b)= &\\cos(a)\\cos(b) + \\sin(a)\\sin(b)\\\\ \\sin(a +b) =&& \\sin(a)\\cos(b) + \\sin(b)\\cos(a) \\\\ \\sin(a - b) =& \\sin(a)\\cos(-b) + \\sin(-b)\\cos(a) =&\\sin(a)\\cos(b) - \\sin(b)\\cos(a) \\end{align} This gives: \\begin{align} \\cos^4(x)sin^3(x) = &\\cos(x)\\left(\\sin(x)\\cos(x)\\right)^3 \\\\ =& \\cos(x)\\frac{\\sin^3(2x)}{2^3} = \\cos(x)\\sin(2x)\\sin^2(2x) \\\\ =& \\frac{1}{8}\\cos(x)\\sin(2x)\\frac{1 - \\cos(4x)}{2} \\\\ \\cos(x)\\sin(2x) =& \\frac{\\sin(3x) + \\sin(x)}{2} \\\\ \\cos^4(x)\\sin^3(x) = & \\left.\\left.\\frac{1}{32}\\right(\\sin(3x) - \\sin(3x)\\cos(4x) + \\sin(x) - \\sin(x)\\cos(4x)\\right) \\\\ =& \\left.\\left.\\frac{1}{32}\\right(\\sin(3x) - \\sin(x)\\right) - \\left.\\left.\\frac{1}{64} \\right(\\sin(7x) + \\sin(x) + \\sin(5x) + \\sin(3x)\\right) \\\\ =& \\left.\\left.\\frac{1}{64}\\right(-3\\sin(x) + \\sin(3x) - \\sin(5x) - \\sin(7x)\\right) \\end{align} This then gives the integral easily as: $$\\int\\cos^4(x)\\sin^3(x)dx = \\left.\\left.\\frac{1}{64}\\right(3\\cos(x) - \\frac{1}{3}\\cos(3x) + \\frac{1}{5}\\cos(5x) + \\frac{1}{7}\\cos(7x)\\right) + C$$ -", "# Is there any simpler way to find $\\sin 2 y$ from $\\cos(x+y)=\\tfrac13$ and $\\cos(x-y)=\\tfrac15$? Is there any simpler way to find $$\\sin 2 y$$ from $$\\cos(x+y)=\\tfrac13$$ and $$\\cos(x-y)=\\tfrac15$$? Note: $$x$$ and $$y$$ are obtuse angles. My attempt that is not simple is as follows. Expand both known constraints, so we have \\begin{align} \\cos x \\cos y &=4/15\\\\ \\sin x \\sin y &=-1/15 \\end{align} Eliminate $$x$$ using $$\\sin^2 x +\\cos^ 2 x=1$$, we have $$225 \\sin^4 y -210 \\sin^2 y +1=0$$ with its solution $$\\sin^2 y = \\frac{7\\pm4\\sqrt3}{15}$$. Then, $$\\cos^2 y = \\frac{4(2\\mp\\sqrt3)}{15}$$. \\begin{align} \\sin^2(2y) &= 4\\cos^2 y\\sin^2 y\\\\ &= 4 \\times \\frac{4(2\\mp\\sqrt3)}{15}\\times \\frac{7\\pm4\\sqrt3}{15} \\\\ \\sin 2 y & = - \\frac{4}{15}\\sqrt{(2\\mp\\sqrt3)(7\\pm4\\sqrt3)} \\end{align} $$\\sin 2y$$ must be negative. # Edit Thank you for your effort to answer my question. However, the existing answers seem to be more complicated than my attempt above. By the way, I am confused in deciding which the correct pair among $$(2\\mp\\sqrt3)(7\\pm4\\sqrt3)$$ is. • Do you assume x and y are obtuse, or it is specified in the problem? – Quanto Sep 6 '19 at 15:24 • @Quanto: Given. – Artificial Stupidity Sep 6 '19 at 15:25 As $$90 and $$\\cos(x+y)>0$$ $$270 Again, $$-90 Finally $$\\sin2y=\\sin(x+y+(x-y))=?$$ Try to solve the first equation for $$x$$ and plug this in the second equation . I got" ]

[ "\\frac{dx+dy}{x+y} = \\frac{d(x+y)}{x+y}$. Hence the original integral is $\\displaystyle \\int \\frac{dy}{x} = \\int \\frac{d(x+y)}{x+y} = \\ln |x+y| + C = \\ln|y + \\sqrt{y^2 + a^2}| + C$. In your case, $\\displaystyle a=13$. Use the substitution $x=13\\tan{u}$, then $dx=13\\sec^2{u} \\ du$ \\begin{align} \\int\\frac{dx}{\\sqrt{169+x^2}} &=\\int\\frac{13\\sec^2{u} \\ du}{\\sqrt{169+169\\tan^2{u}}}\\\\ &=\\int\\sec{u} \\ du\\\\ &=\\ln|\\sec{u}+\\tan{u}|+c\\\\ &=\\ln\\left|\\frac{x}{13}+\\sqrt{1+\\left(\\frac{x}{13}\\right)^2}\\right|+c \\end{align} Alternatively, we can use the hyperbolic substitution $x=13\\sinh{u}$, then $dx=13\\cosh{u} \\ du$. Thus \\begin{align} \\int\\frac{dx}{\\sqrt{169+x^2}} &=\\int\\frac{13\\cosh{u} \\ du}{\\sqrt{169+169\\sinh^2{u}}}\\\\ &=\\int\\frac{13\\cosh{u} \\ du}{13\\cosh{u}}\\\\ &=\\operatorname{arcsinh}\\left(\\frac{x}{13}\\right)+c \\end{align} These 2 answers are equal. • @brian: If the official answer is $\\ln|y+\\sqrt{169+y^2}|+C$, that is \"the same\" as the one above, because the constant can be absorbed into the constant of integration. – André Nicolas Aug 2 '14 at 3:08 • I honestly have no clue – brian Aug 2 '14 at 3:13 • @Macavity Thanks for pointing out that blunder. I have edited it. – SuperAbound Aug 2 '14 at 3:19", "at 22:27 This answer is inspired by the answer of coffeemath. We know the following (from e.g. Wikipedia): \\begin{align} &\\sin(x) = x -\\frac{x^3}{6} +\\frac{x^5}{120}+ O(x^7) \\\\ &\\cos(x) = 1 -\\frac{x^2}{2} +\\frac{x^4}{24} +\\frac{x^6}{720} +O(x^8) \\\\ &\\tan(x) = a +bx+ cx^2+dx^3+ex^4+fx^5+gx^6+ O(x^7) \\end{align} Furthermore it holds $\\tan(x) = \\frac{\\sin(x)}{\\cos(x)} \\Rightarrow \\sin(x) =\\tan(x)\\cos(x)$. So we get \\begin{align} x -\\frac{x^3}{6} +\\frac{x^5}{120}+ O(x^7) =& \\bigl(1 -\\frac{x^2}{2} +\\frac{x^4}{24} +\\frac{x^6}{720} +O(x^8) \\bigr) \\\\ &\\cdot\\bigl(a +bx+ cx^2+dx^3+ex^4+fx^5+gx^6+ O(x^7)\\bigr) \\end{align} If we expand the term on the right hand side we obtain \\begin{align} x -\\frac{x^3}{6} +\\frac{x^5}{120}+ O(x^7) &=a+bx+(-\\frac{a}{2}+c)x^2+(-\\frac{b}{2}+d)x^3 \\\\&+(\\frac{a}{24}-\\frac{c}{2}+e)x^4+(\\frac{b}{24}-\\frac{d}{2}+f)x^5 \\\\&+(\\frac{a}{720}+\\frac{c}{24}-\\frac{e}{2}+g)x^6+ O(x^7) \\end{align} From this we can deduct the linear system \\begin{align} \\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0&0\\\\ 0 & 1 & 0 & 0 & 0 & 0 &0\\\\ -1/2 &0 &1 & 0& 0& 0&0\\\\ 0 & -1/2 & 0 & 1 & 0 & 0&0 \\\\ 1/24 & 0 & -1/2 & 0 & 1 & 0&0\\\\ 0 &1/24 & 0 & -1/2 & 0 & 1&0\\\\ 1/720 & 0 & 1/24 & 0 & -1/2 & 0 &1 \\end{pmatrix} \\begin{pmatrix} a\\\\b\\\\c\\\\d\\\\e\\\\f\\\\g \\end{pmatrix} = \\begin{pmatrix} 0\\\\1 \\\\ 0 \\\\ -1/6 \\\\ 0 \\\\1/120 \\\\ 0 \\end{pmatrix} \\end{align} Which has the solution \\begin{align} \\begin{pmatrix} a\\\\b\\\\c\\\\d\\\\e\\\\f\\\\g \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\\\ 1/3 \\\\ 0 \\\\2/15 \\end{pmatrix} \\end{align}", "1Siny$\\frac{1}{Sin\\; y}$ = 135=53$\\frac{1}{\\frac{3}{5}} = \\frac{5}{3}$ Since, sin2y+cos2y=1$sin^{2}\\;y \\; + \\; cos^{2}\\;y = 1$ $\\Rightarrow$ cos2y=1sin2y$cos^{2} y = 1 – sin^{2} y$ $\\Rightarrow$ cos2y=1(35)2$cos^{2} y = 1 – \\left ( \\frac{3}{5} \\right )^{2}$ $\\Rightarrow$ cos2y=1925$cos^{2} y = 1 – \\frac{9}{25}$ $\\Rightarrow$ cos2y=1625$cos^{2} y = \\frac{16}{25}$ $\\Rightarrow$ cos y = ±45$\\pm \\frac{4}{5}$ Since, y lies in the 2nd quadrant, the value of cos y will be negative, Therefore, cos y = – 45$\\frac{4}{5}$ $\\Rightarrow$ sec y = 1cosy=1(45)=54$\\frac{1}{cos \\; y} = \\frac{1}{\\left (- \\frac{4}{5} \\right )} =\\; – \\frac{5}{4}$ $\\Rightarrow$ tan y = sinycosy=(35)(45)=34$\\frac{sin \\; y}{cos \\; y} = \\frac{\\left (\\frac{3}{5} \\right ) }{\\left (-\\frac{4}{5} \\right ) } = \\;- \\frac{3}{4}$ $\\Rightarrow$ cot y = 1tany=43$\\frac{1}{tan \\; y} =\\;- \\frac{4}{3}$ Q.3: Find the values of other five trigonometric functions if coty=34$cot \\; y = \\frac{3}{4}$, where y lies in the third quadrant. Sol: cot y = 34$\\frac{3}{4}$ Since, tan y = 1coty=134=43$\\frac{1}{cot \\; y}=\\frac{1}{\\frac{3}{4}}=\\frac{4}{3}$ Since, 1+tan2y=sec2y$1 + tan^{2}y = sec^{2}y$ $\\Rightarrow$ 1+(43)2=sec2y$1 + \\left ( \\frac{4}{3} \\right )^{2} = sec^{2}y$ $\\Rightarrow$ 1+169=sec2y$1 + \\frac{16}{9} = sec^{2}y$ $\\Rightarrow$ 259=sec2y$\\frac{25}{9} = sec^{2}y$ $\\Rightarrow$ sec y = ±53$\\pm \\frac{5}{3}$ Since, y lies in the 3rd quadrant, the value of sec y will be negative. Therefore, sec y = 53$\\;- \\frac{5}{3}$ cos y = 1secy=1(53)=35$\\frac{1}{sec \\; y} = \\frac{1}{\\left (- \\frac{5}{3} \\right )}", "the answers were useful to you. – SuperAbound Jul 25 '14 at 6:16 Hint : \\begin{align} \\sin x+\\cos x&=\\frac12\\\\ \\sin x&=\\frac12-\\cos x\\\\ \\sin^2 x&=\\left(\\frac12-\\cos x\\right)^2\\\\ 1-\\cos^2 x&=\\frac14-\\cos x+\\cos^2 x\\\\ 2\\cos^2 x-\\cos x-\\frac34&=0\\\\ 8\\cos^2 x-4\\cos x-3&=0 \\end{align} $\\sin x+\\cos x=\\frac12$ $\\Rightarrow$ $(\\sin x+\\cos x)^2=\\frac14$ $\\Rightarrow$ $\\sin^2 x+2\\sin x\\cos x+\\cos^2 x=\\frac14$ $\\Rightarrow$ $1+2\\sin x\\cos x=\\frac14$ $\\Rightarrow$ $2\\sin x\\cos x = -\\frac34$ $\\Rightarrow$ $\\sin x\\cos x=-\\frac38$. If $a=\\sin x$, $b=\\cos x$, we have the system of equations \\begin{align} a+b&=\\frac12\\\\ ab&=-\\frac38 \\end{align} which leads to $$a-\\frac3{8a}=\\frac12\\\\ a^2-\\frac12a-\\frac38=0$$ with the solutions $a=\\frac{1\\pm\\sqrt7}4$. So we get $$\\sin x=\\frac{1+\\sqrt7}4, \\qquad \\cos x= \\frac{1-\\sqrt7}4 \\qquad \\Rightarrow \\qquad \\tan x = \\frac{1+\\sqrt7}{1-\\sqrt7}\\\\ \\sin x=\\frac{1-\\sqrt7}4, \\qquad \\cos x= \\frac{1+\\sqrt7}4 \\qquad \\Rightarrow \\qquad \\tan x = \\frac{1-\\sqrt7}{1+\\sqrt7}$$ You could also simplify the result: $$\\frac{1+\\sqrt7}{\\frac1-\\sqrt7} = \\frac{(1+\\sqrt7)^2}{1-7} = -\\frac{8+2\\sqrt7}6= -\\frac{4+\\sqrt7}3$$ HINT: Avoid squaring whenever possible as it immediately introduces extraneous root which needs to be discarded Use Weierstrass substitution to form a Quadratic Equation in $\\tan\\dfrac x2$ Now use $\\displaystyle\\tan2A=\\frac{2\\tan A}{1-\\tan^2A}$ Divide through by $\\cos x$ to get \\begin{align} \\tan x + 1 & = \\frac 12 \\sec x \\tag{1}\\\\ \\text{Square both sides: } \\qquad\\qquad\\tan^2 x + 2 \\tan x + 1 & = \\frac 14 \\sec^2 x = \\frac{1}{4} (\\tan^2 x + 1) \\\\ 3 \\tan^2 x + 8 \\tan x + 3 & = 0 \\\\ \\tan x", "# Let tan x = 2.4, sin y = 0.6, and both x and y be between 0° and 90°. Then cos(x + y) equals? May 12, 2018 $- \\frac{16}{65}$ #### Explanation: May 13, 2018 cos (x + y) = - 0.246 #### Explanation: Use trig identity: cos (x + y) = cos x.cos y - sin x.sin y (1) First, find cos y knowing sin y = 0.6 ${\\cos}^{2} y = 1 - {\\sin}^{2} y = 1 - 0.36 = 0.64$ $\\cos y = 0.8$ (because y is in Quadrant 1) Next, find sin x and cos x, knowing tan x = 2.4. ${\\cos}^{2} x = \\frac{1}{1 + {\\tan}^{2} x} = \\frac{1}{1 + 5.76} = \\frac{1}{6.76}$ $\\cos x = \\frac{1}{2.6}$ (because x is in Q. 1) ${\\sin}^{2} x = 1 - {\\cos}^{2} x = 1 - \\frac{1}{6.76} = \\frac{5.76}{6.76}$ $\\sin x = \\frac{2.4}{2.6}$ Replace all numeric values into equation (1), we get: $\\cos \\left(x + y\\right) = \\left(\\frac{1}{2.6}\\right) \\left(0.8\\right) - \\left(\\frac{2.4}{2.6}\\right) \\left(0.6\\right) = \\frac{0.8}{2.6} - \\frac{1.44}{2.6}$ $\\cos \\left(x + y\\right) = - \\frac{0.64}{2.6} = - 0.246$ Check by calculator. $\\sin y = 0.6$ --> $y = {36}^{\\circ} 87$ $\\cos x = \\frac{1}{2.6}$ --> $x = {67}^{\\circ} 38$ $x + y = 36.87 + 67.38 = {104}^{\\circ} 25$ --> cos 104.25 = - 0.246. OK", "x^2 \\right)\\frac{dy}{dx} + y = e^{tan^{- 1} x}$ Ex. 22.1 | Q 21 | Page 106 x dy = (2y + 2x4 + x2) dx Ex. 22.1 | Q 22 | Page 106 $\\left( 1 + y^2 \\right) + \\left( x - e^{tan^{- 1} y} \\right)\\frac{dy}{dx} = 0$ Ex. 22.1 | Q 23 | Page 106 $y^2 \\frac{dx}{dy} + x - \\frac{1}{y} = 0$ Ex. 22.1 | Q 24 | Page 106 $\\left( 2x - 10 y^3 \\right)\\frac{dy}{dx} + y = 0$ Ex. 22.1 | Q 25 | Page 106 (x + tan y) dy = sin 2y dx Ex. 22.1 | Q 26 | Page 106 dx + xdy = e−y sec2 y dy Ex. 22.1 | Q 27 | Page 106 $\\frac{dy}{dx}$ = y tan x − 2 sin x Ex. 22.1 | Q 28 | Page 106 $\\frac{dy}{dx}$ + y cos x = sin x cos x Ex. 22.1 | Q 29 | Page 106 Solve the following differential equation:- $\\left( 1 + x^2 \\right)\\frac{dy}{dx} - 2xy = \\left( x^2 + 2 \\right)\\left( x^2 + 1 \\right)$ Ex. 22.1 | Q 30 | Page 106 $\\left( \\sin x \\right)\\frac{dy}{dx} + y \\cos x = 2 \\sin^2 x \\cos x$ Ex. 22.1 | Q 31 | Page 106 $\\left( x^2 - 1 \\right)\\frac{dy}{dx} + 2\\left( x", "- C)$ $\\cos A.\\cos\\frac{B + C}{2}\\sin\\frac{C - B}{2} = -2\\cos A.\\sin\\frac{B - C}{2}.\\cos\\frac{B - C}{2}$ If $B = C$ then above it $0 = 0$ i.e. triangle is isosceles. If $A = 90^\\circ$ then above is $0 =0$ i.e. triangle is right angled. 13. $\\because \\tan\\frac{A}{2}, \\tan\\frac{B}{2}, \\tan\\frac{C}{2}$ are in A.P. $\\tan\\frac{A}{2} - \\tan\\frac{B}{2} = \\tan\\frac{B}{2} - \\tan \\frac{C}{2}$ $\\frac{\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)}{\\cos\\frac{A}{2}\\cos\\frac{B}{2}} = \\frac{\\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)}{\\cos\\frac{B}{2}\\cos\\frac{C}{2}}$ $\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)\\cos\\frac{C}{2} = \\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)\\cos\\frac{A}{2}$ $\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)\\sin\\left(\\frac{A}{2} + \\frac{B}{2}\\right) = \\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)\\sin\\left(\\frac{B}{2} + \\frac{C}{2}\\right)$ $\\Rightarrow \\cos B - \\cos A = \\cos C - \\cos B$ Thus, $\\cos A, \\cos B, \\cos C$ are in A.P. 14. Given, $a\\cos^2\\frac{C}{2} + c\\cos^2\\frac{A}{2} = \\frac{3b}{2}$ $a.\\frac{s(s - c)}{ab} + c.\\frac{s(s - a)}{bc} = \\frac{3b}{2}$ $\\frac{s}{b}[2s - a - c] = \\frac{3b}{2}$ $2s = 3b \\Rightarrow a + c = 2b$ We have to prove that $\\cot\\frac{A}{2} + \\cot\\frac{C}{2} = 2\\cot\\frac{B}{2}$ L.H.S. $= \\frac{s(s - a)}{\\Delta} + \\frac{s(s - c)}{\\Delta}$ $= \\frac{s}{\\Delta}(2s - a - c) = \\frac{2s(s - b)}{\\Delta} = 2\\cot\\frac{B}{2} =$ R.H.S. 15. Given, $a^2, b^2, c^2$ are in A.P. $\\Rightarrow b^2 - a^2 = c^2 - b^2$ $\\Rightarrow \\sin^2B - \\sin^2A = \\sin^2C - \\sin^2B$ $\\Rightarrow \\sin(B + A)\\sin(B - A) = \\sin(C + B)\\sin(C - B)$ $\\Rightarrow \\sin C\\sin(B - A) = \\sin A\\sin(C - B)$ $\\Rightarrow", "20:50 Understandable - I just found this to be a little less messy. – user17794 Aug 27 '13 at 20:52 Hint: \\begin{align} 0&=\\cos x+\\sin x\\\\ \\sin x &= -\\cos x \\\\ \\text{Divide}&\\text{ both sides by }\\cos x\\\\ \\tan x &= -1 \\end{align} Now you can solve for $x$ using inverse trigonometric functions. - if you use the Double angle formulae $$\\sin^2(x)\\cos^2(x)=(1-\\cos2x)/2 \\cdot (1+\\cos2x)/2 = (1-\\cos^2(2x))/4 = \\sin^2(2x)/4 = (1-\\cos4x)/8$$ so minimum is $1/8$ at $x=45\\circ$ and maximum is 1/4 at $x=90\\circ$ remember $\\sin^2(x)=(1-\\cos2x)/2$ and $\\cos^2(x)=(1+\\cos2x)/2$ - The equation $\\cos x+\\sin x=0$ is really the same as $\\cos x-\\sin x=0$ once you make the transformation $x\\mapsto x+\\frac{\\pi}2$, because $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and $\\sin (x+\\frac{\\pi}2)=\\cos(x)$. The latter can be solved by resorting to $\\cos ^2+\\sin ^2=1$. Edit: We can derive $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and similar results (cf. Daniel Fischer's comment) from the well known formulas $\\cos (x+y)=\\cos x\\cos y-\\sin x\\sin y$ and $\\sin (x+y)=\\cos x\\sin y+\\cos y\\sin x$. These can either be proved geometrically, or by resorting to some definition of $\\sin$ and $\\cos$, for example $\\sin x=(e^{ix}-e^{-ix})/2i, \\cos x=(e^{ix}-e^{-ix})/2$ -", "} \\right) = \\left( { - \\frac{\\pi }{4}} \\right)$$. Chapter 2 Ex.2.1 Question 11 Find the value of $${\\tan ^{ - 1}}\\left( 1 \\right) + {\\cos ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) + {\\sin ^{ - 1}}\\left( { - \\frac{1}{2}} \\right)$$. Solution Let, $${\\tan ^{ - 1}}\\left( 1 \\right) = x$$ Hence, \\begin{align}\\tan x &= 1\\\\&= \\tan \\left( {\\frac{\\pi }{4}} \\right)\\end{align} Therefore, $${\\tan ^{ - 1}}\\left( 1 \\right) = \\left( {\\frac{\\pi }{4}} \\right)$$ Now, let $${\\cos ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) = y$$ Hence, \\begin{align}\\cos y &= - \\frac{1}{2}\\\\&= - \\cos \\left( {\\frac{\\pi }{3}} \\right)\\\\&= \\cos \\left( {\\pi - \\frac{\\pi }{3}} \\right)\\\\&= \\cos \\left( {\\frac{{2\\pi }}{3}} \\right)\\end{align} Therefore, $${\\cos ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) = \\frac{{2\\pi }}{3}$$ Again, let $${\\sin ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) = z$$ Hence, \\begin{align}\\sin z &= - \\frac{1}{2}\\\\&= - \\sin \\left( {\\frac{\\pi }{6}} \\right)\\\\&= \\sin \\left( { - \\frac{\\pi }{6}} \\right)\\end{align} Therefore, $${\\sin ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) = - \\frac{\\pi }{6}$$ Thus, \\begin{align}{\\tan ^{ - 1}}\\left( 1 \\right) + {\\cos ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) + {\\sin ^{ - 1}}\\left( { - \\frac{1}{2}} \\right) &= \\frac{\\pi }{4} + \\frac{{2\\pi }}{3} - \\frac{\\pi }{6}\\\\&= \\frac{{3\\pi + 8\\pi - 2\\pi }}{{12}}\\\\&= \\frac{{9\\pi }}{{12}}\\\\&= \\frac{{3\\pi }}{4}\\end{align} Chapter 2 Ex.2.1 Question 12 Find the value", "# Is there any simpler way to find $\\sin 2 y$ from $\\cos(x+y)=\\tfrac13$ and $\\cos(x-y)=\\tfrac15$? Is there any simpler way to find $$\\sin 2 y$$ from $$\\cos(x+y)=\\tfrac13$$ and $$\\cos(x-y)=\\tfrac15$$? Note: $$x$$ and $$y$$ are obtuse angles. My attempt that is not simple is as follows. Expand both known constraints, so we have \\begin{align} \\cos x \\cos y &=4/15\\\\ \\sin x \\sin y &=-1/15 \\end{align} Eliminate $$x$$ using $$\\sin^2 x +\\cos^ 2 x=1$$, we have $$225 \\sin^4 y -210 \\sin^2 y +1=0$$ with its solution $$\\sin^2 y = \\frac{7\\pm4\\sqrt3}{15}$$. Then, $$\\cos^2 y = \\frac{4(2\\mp\\sqrt3)}{15}$$. \\begin{align} \\sin^2(2y) &= 4\\cos^2 y\\sin^2 y\\\\ &= 4 \\times \\frac{4(2\\mp\\sqrt3)}{15}\\times \\frac{7\\pm4\\sqrt3}{15} \\\\ \\sin 2 y & = - \\frac{4}{15}\\sqrt{(2\\mp\\sqrt3)(7\\pm4\\sqrt3)} \\end{align} $$\\sin 2y$$ must be negative. # Edit Thank you for your effort to answer my question. However, the existing answers seem to be more complicated than my attempt above. By the way, I am confused in deciding which the correct pair among $$(2\\mp\\sqrt3)(7\\pm4\\sqrt3)$$ is. • Do you assume x and y are obtuse, or it is specified in the problem? – Quanto Sep 6 '19 at 15:24 • @Quanto: Given. – Artificial Stupidity Sep 6 '19 at 15:25 As $$90 and $$\\cos(x+y)>0$$ $$270 Again, $$-90 Finally $$\\sin2y=\\sin(x+y+(x-y))=?$$ Try to solve the first equation for $$x$$ and plug this in the second equation . I got", "# Integrate $\\int\\frac{1}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x}dx$ Solve the indefinite integral $$I=\\int\\frac{1}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x}\\;dx$$ My Attempt: \\begin{align} I&=\\int\\frac{1}{\\sin x+\\cos x+\\frac{1}{\\sin x \\cos x}+\\frac{\\sin x +\\cos x}{\\sin x\\cos x}}\\;dx\\\\ \\\\ &=\\int\\frac{\\sin x\\cos x}{\\left(\\sin x+\\cos x\\right)\\left(\\sin x\\cos x \\right)+1+\\left(\\sin x+\\cos x\\right)}\\;dx \\end{align} How can I complete the solution from here? • Do you have any reason to think it has a nice answer? – Greg Martin Sep 18 '13 at 5:14 • Yeah... Good luck – Ali Caglayan Dec 25 '13 at 23:45 You want \\begin{align} \\int\\frac{dx}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x} &= \\int\\frac{dx}{\\sin x+\\cos x+\\frac{\\sin x}{\\cos x} +\\frac{\\cos x}{\\sin x}+\\frac1{\\sin x}+\\frac1{\\cos x}}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{\\sin^2 x \\cos x+\\cos^2 x \\sin x+\\sin^2 x +\\cos^2 x+\\cos x+\\sin x}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{\\sin x \\cos x(\\sin x+ \\cos x)+1+\\cos x+\\sin x}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{(\\sin x \\cos x+1)(\\sin x+ \\cos x)+1}\\\\ \\end{align}. Applying substitutions $\\sin x = \\frac{2t}{1+t^2}, \\cos x = \\frac{1-t^2}{1+t^2}, dx = \\frac{2 dt}{1+t^2}$, this becomes \\begin{align} \\int\\frac{\\sin x \\cos x\\ dx}{(\\sin x \\cos x+1)(\\sin x+ \\cos x)+1} &=\\int\\frac{\\frac{(2t)(1-t^2)(2dt)}{(1+t^2)^3}} {(\\frac{(2t)(1-t^2)}{(1+t^2)^2}+1)(\\frac{2t}{1+t^2}+ \\frac{1-t^2}{1+t^2})+1}\\\\ &=\\int\\frac{(2t)(1-t^2)(2dt)} {((2t)(1-t^2)+(1+t^2)^2)(1+2t-t^2)+(1+t^2)^3}\\\\ &=\\int\\frac{4t(1-t^2)dt} {((2t-2t^3)+1+2t^2+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\\\ &=\\int\\frac{4t(1-t^2)dt} {(1+2t+2t^2-2t^3+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\\\ &=\\int\\frac{4t(1-t^2)dt} {2 (t+1) (2 t^4-3 t^3+3 t^2+t+1)} \\quad \\text{ (according to Wolfram Alpha)}\\\\ &=\\int\\frac{2t(1-t)dt} {2 t^4-3 t^3+3 t^2+t+1}\\\\ \\end{align}. According to Wolfram Alpha, that quartic can be factored as", "in second quadrant. Sol: tan y = $$-\\frac{5}{12}$$ [Given] And, cot y = $$\\frac{1}{tan \\; y} = \\frac{1}{ \\left (- \\frac{5}{12} \\right )} = – \\frac{12}{5}$$ Since, $$1 + tan^{2} y = sec^{2} y$$ Therefore, $$1 + \\left ( – \\frac{5}{12} \\right )^{2} = sec^{2} y$$ $$\\Rightarrow$$ $$sec^{2} y = 1 + \\left ( \\frac{25}{144} \\right )$$ $$\\Rightarrow$$ $$sec^{2} y = \\frac{169}{144}$$ $$\\Rightarrow$$ sec y = $$\\pm \\frac{13}{12}$$ Since, y lies in the 2nd quadrant, the value of sec y will be negative. Therefore, sec y = $$– \\frac{13}{12}$$ cos y = $$\\frac{1}{sec \\; y } = \\frac{1}{\\left (- \\frac{13}{12} \\right )} = \\left (- \\frac{12}{13} \\right )\\\\$$ Since, $$tan \\; y = \\frac{sin \\; y}{cos \\; y}$$ $$\\Rightarrow$$ $$-\\frac{5}{12} = \\frac{sin \\; y}{\\left (- \\frac{12}{13} \\right )}\\\\$$ $$\\Rightarrow$$ sin y = $$\\left (-\\frac{5}{12} \\right ) \\times \\left (- \\frac{12}{13}\\right ) = \\frac{5}{13}\\\\$$ $$\\Rightarrow$$ cosec y = $$\\frac{1}{sin \\; y} = \\frac{1}{\\left ( \\frac{5}{13} \\right )} = \\frac{13}{5}$$ Q.6: Calculate the value of trigonometric function sin 765°. Sol: The values of sin y repeat after an interval of 360° or 2n. Therefore, sin 765° = sin ( 2 × 360° + 45° ) = sin 45° = $$\\frac{1}{\\sqrt{2}}$$ Q.7: Calculate the value of trigonometric function cosec [-1410°] Sol: It is known that the values of tan y repeat", "\\tan v}{1 \\mp \\tan u \\tan v}$ Double Angle Formulas: • $\\sin (2 u) =2 \\sin u \\cos u$ • $\\cos (2 u) =\\cos ^{2} u-\\sin ^{2} u$ • $\\cos (2 u) =2 \\cos ^{2} u-1$ • $\\cos (2 u) =1-2 \\sin ^{2} u$ • $\\tan (2 u) =\\frac{2 \\tan u}{1-\\tan ^{2} u}$ Half Angle Formulas: • $\\sin^{2} u=\\frac{1-\\cos (2 u)}{2}$ • $\\cos^{2} u=\\frac{1+\\cos (2 u)}{2}$ • $\\tan^{2} u=\\frac{1-\\cos (2 u)}{1+\\cos (2 u)}$ Sum to product formulas: • $\\sin u+\\sin v=2 \\sin \\left(\\frac{u+v}{2}\\right) \\cos \\left(\\frac{u-v}{2}\\right)$ • $\\sin u-\\sin v=2 \\cos \\left(\\frac{u+v}{2}\\right) \\sin \\left(\\frac{u-v}{2}\\right)$ • $\\cos u+\\cos v=2 \\cos \\left(\\frac{u+v}{2}\\right) \\cos \\left(\\frac{u-v}{2}\\right)$ • $\\cos u-\\cos v=-2 \\sin \\left(\\frac{u+v}{2}\\right) \\sin \\left(\\frac{u-v}{2}\\right)$ Product to Sum Formulas: • $\\sin u \\sin v=\\frac{1}{2}[\\cos (u-v)-\\cos (u+v)]$ • $\\cos u \\cos v=\\frac{1}{2}[\\cos (u-v)+\\cos (u+v)]$ • $\\sin u \\cos v=\\frac{1}{2}[\\sin (u+v)+\\sin (u-v)]$ • $\\cos u \\sin v=\\frac{1}{2}[\\sin (u+v)-\\sin (u-v)]$", "\\ \\ \\ \\ \\ x=1,2,3,4,\\ \\ \\ y=1,2$$ Solution First, we need to find the value of $$c$$ and then proceed to extract the marginal functions. We know that: $$\\sum_{x}\\sum_{y}{P(X=x,\\ Y=y)}=1$$ \\begin{align*} \\Rightarrow c(1^2+3\\left(1\\right)+c(1^2+3\\left(2\\right)+\\ldots+c(4^2+3\\left(2\\right)&=1\\\\ =4c+7c+7c+10c+12c+15c+19c+22&=1\\\\ 96c&=1\\\\ \\therefore c&=\\frac{1}{96} \\end{align*} Using the above results, marginal functions are: $$f_X\\left(x\\right)=\\frac{2x^2+9}{96} \\text { and } f_Y\\left(y\\right)=\\frac{12y+30}{96}$$ Let’s now calculate the means of $$X$$ and $$Y$$: \\begin{align*} E\\left(X\\right)&=\\sum_{x=1}^{4}{xf_X\\left(x\\right)}\\\\ &=\\sum_{x=1}^{4}{x\\frac{2x^2+9}{96}}\\\\ &=\\left(1\\right)\\frac{11}{96}+\\left(2\\right)\\frac{17}{96}+\\left(3\\right)\\frac{27}{96}+\\left(4\\right)\\frac{41}{96}\\ \\\\ &=\\frac{11}{96}+\\frac{34}{96}+\\frac{81}{96}+\\frac{164}{96}\\\\ &=\\frac{145}{48}\\ \\end{align*} And, \\begin{align*} \\sigma_X^2&=Var\\left(X\\right)=\\sum_{x=1}^{4}{x^2f_X\\left(x\\right)-\\left[E\\left(X\\right)\\right]^2}\\\\ &=\\sum_{x=1}^{4}{x^2\\frac{2x^2+9}{96}}-\\left(\\frac{145}{48}\\right)^2\\\\ &=\\left(1\\right)^2\\frac{11}{96}+\\left(2\\right)^2\\frac{17}{96}+\\left(3\\right)^2\\frac{27}{96}+\\left(4\\right)^2\\frac{41}{96}-\\left(\\frac{145}{48}\\right)^2\\\\ &=\\frac{163}{16}-\\left(\\frac{145}{48}\\right)^2=1.062\\ \\end{align*} Similarly for $$Y$$: \\begin{align*} \\mu_Y&=E\\left(Y\\right)=\\sum_{y=1}^{2}{yf_Y\\left(y\\right)}\\\\ &=\\sum_{y=1}^{2}{y\\frac{12y+30}{96}=\\left(1\\right)\\frac{42}{96}+\\left(2\\right)\\frac{54}{96}\\ }\\\\ &=\\frac{42}{96}+\\frac{108}{96}\\\\ &=\\frac{25}{16}\\ \\end{align*} And, \\begin{align*} \\sigma_Y^2&=\\sum_{y=1}^{2}{y^2f_Y\\left(y\\right)-\\left[\\mu_Y\\right]^2}\\\\ &=\\sum_{y=1}^{2}{y^2\\frac{12y+30}{96}-\\left(\\frac{25}{16}\\right)^2}\\\\ &=\\left(1\\right)^2\\frac{42}{96}+\\left(2\\right)\\frac{54}{96}-\\left(\\frac{25}{16}\\right)^2\\\\ &=\\frac{42}{96}+\\frac{216}{96}-\\frac{625}{256}=\\frac{43}{16}-\\frac{625}{256}\\\\ &=\\frac{63}{256} \\end{align*} We can now calculate $$Cov\\left(X,Y\\right)$$ and $$Corr(X,Y)$$. $$Cov\\left(X,Y\\right)=E\\left(XY\\right)-E(X)E(Y)$$ Now, \\begin{align*} E\\left(XY\\right)&=\\sum_{x=1}^{4}\\sum_{y=1}^{2}{xy\\frac{x^2+3y}{96}}\\\\ &=\\left(1\\right)\\left(1\\right)\\frac{4}{96}+\\left(1\\right)\\left(2\\right)\\frac{7}{96}+\\left(2\\right)\\left(1\\right)\\frac{7}{96}+\\left(2\\right)\\left(2\\right)\\frac{10}{96}+\\left(3\\right)\\left(1\\right)\\frac{12}{96}\\\\ &+\\left(3\\right)\\left(2\\right)\\frac{15}{96}+\\left(4\\right)\\left(1\\right)\\frac{19}{96}+\\left(4\\right)\\left(2\\right)\\frac{22}{96}\\\\ &=\\frac{75}{16} \\end{align*} Therefore, \\begin{align*} Cov\\left(X,Y\\right)&=\\frac{75}{16}-\\left(\\frac{145}{48}\\right)\\left(\\frac{25}{16}\\right)\\\\ &=\\frac{75}{16}-\\frac{3625}{768}\\\\ &=-\\frac{25}{768} \\end{align*} And lastly, \\begin{align*} \\rho\\left(X,Y\\right)&=\\frac{Cov\\left(X,Y\\right)}{\\sqrt{\\sigma_X^2\\sigma_Y^2}}\\\\ &=-\\frac{\\frac{25}{768}}{\\sqrt{1.062\\bullet\\left(\\frac{63}{256}\\right)}}\\\\ &=-0.0636\\ \\end{align*} ## Covariance and Correlation Coefficient for Continuous Random Variables If $$X$$ and $$Y$$ are continuous random variables, we generally: 1. Find $$E(X)$$ and $$E(Y)$$ at once with iterated integrals which are given by: $$E\\left(x\\right)=\\int_{x}{x.f_X(x)dx}$$ and $$E\\left(y\\right)=\\int_{y}{y.f_Y(y)dy}$$ 2. Find $$E(XY)$$ applying the iterated integrals $$E\\left(XY\\right)=\\int_{x}\\int_{y}{xy.f_{XY}(xy)dxdy}$$ 3. Calculate $$Cov(X,Y)$$ and $$Corr(X,Y)$$ using the formulas: $$Cov\\left(X,Y\\right)=E\\left(XY\\right)-E(X)E(Y)$$ and $$\\rho\\left(X,Y\\right)=\\frac{Cov(X,Y)}{\\sqrt{Var\\left(X\\right)Var\\left(Y\\right)}}=\\frac{Cov(X,Y)}{\\sigma_X\\sigma_Y}$$ #### Example: Covariance and Correlation Coefficient (Continuous Case) #1 Let $$f\\left( x,y \\right) =\\begin{cases} \\begin{matrix} \\frac { 2 }{ 3 } \\left( 2x+y \\right) , & 0 < x < 1,0 < y < 1 \\end{matrix} \\\\ \\begin{matrix} 0, &", "of certain arguments. #### Example A If $\\sin a = \\frac{5}{13}$ and $a$ is in Quadrant II, find $\\sin 2a$ , $\\cos 2a$ , and $\\tan 2a$ . Solution: To use $\\sin 2a = 2 \\sin a \\cos a$ , the value of $\\cos a$ must be found first. $& =\\cos^2 a + \\sin^2 a = 1\\\\& = \\cos^2 a + \\left (\\frac{5}{13} \\right )^2 = 1\\\\& = \\cos^2 a + \\frac{25}{169} = 1\\\\& = \\cos^2 a = \\frac{144}{169}, \\cos a = \\pm \\frac{12}{13}$ . However since $a$ is in Quadrant II, $\\cos a$ is negative or $\\cos a = - \\frac{12}{13}$ . $\\sin 2a = 2 \\sin a \\cos a = 2 \\left (\\frac{5}{13} \\right ) \\times \\left (- \\frac{12}{13} \\right ) = \\sin 2a = - \\frac{120}{169}$ For $\\cos 2a$ , use $\\cos(2a) = \\cos^2 a - \\sin^2 a$ $\\cos(2a) & = \\left (- \\frac{12}{13} \\right )^2 - \\left (\\frac{5}{13} \\right )^2 \\ \\text{or}\\ \\frac{144-25}{169} \\\\\\cos (2a) & = \\frac{119}{169}$ For $\\tan 2a$ , use $\\tan 2a = \\frac{2 \\tan a}{1 - \\tan^2 a}$ . From above, $\\tan a = \\frac{\\frac{5}{13}}{-\\frac{12}{13}} = - \\frac{5}{12}$ . $\\tan(2a) = \\frac{2 \\cdot \\frac{-5}{12}}{1 - \\left (\\frac{-5}{12} \\right )^2} = \\frac{\\frac{-5}{6}}{1 - \\frac{25}{144}} = \\frac{\\frac{-5}{6}}{\\frac{119}{144}} = - \\frac{5}{6} \\cdot \\frac{144}{119} = - \\frac{120}{119}$ #### Example B Find $\\cos 4 \\theta$", "the Value of (i) sin 75° (ii) cos 75° (iii) tan 75° (iv) sin 15° (v) cos 15° (vi) tan 15° (vii) sin 105° (viii) cos 105°. Question 33. Find the value of $$\\tan \\frac{13 \\pi}{12}$$ Question 34. Prove that $$\\frac{\\sin (x+y)}{\\sin (x-y)}=\\frac{\\tan x+\\tan y}{\\tan x-\\tan y}$$ Question 35. Prove that $$\\begin{array}{l}{\\cos \\left(\\frac{\\pi}{4}-x\\right) \\cos \\left(\\frac{\\pi}{4}-y\\right)-\\sin \\left(\\frac{\\pi}{4}-x\\right) \\sin \\left(\\frac{\\pi}{4}-y\\right)} \\\\ {=\\sin (x+y)}\\end{array}$$ \\begin{aligned} L H S &=\\cos \\left[\\frac{\\pi}{4}-x+\\frac{\\pi}{4}-y\\right] \\\\ &=\\cos \\left[\\frac{\\pi}{2}-(x+y)\\right] \\\\ &=\\sin (x+y)=R H S \\end{aligned} Question 36. Prove that sin[(n + l)x]sin[(n + 2)x] +cos[(n+1)x]cos[(n + 2)x] = cosx LHS = cos[(n + 2)x – (n + 1)x] = cos x = RHS . Question 37. Prove that (sin3x + sinx)sinx + (cos3x -cosx)cosx = 0 LHS = sin 3x sin x+sin2 x + cos 3xcos x – cos2 x = [cos 3x cos x + sin 3x sin x] – [cos2 x – sin2 x] = cos(3x – x) – cos 2x = 0 = RHS Question 38. Prove that $$\\frac{\\tan \\left(\\frac{\\pi}{4}+x\\right)}{\\tan \\left(\\frac{\\pi}{4} x\\right)}=\\left(\\frac{1+\\tan x}{1-\\tan x}\\right)^{2}$$ $$L H S=\\frac{\\frac{1+\\tan x}{1-\\tan x}}{\\frac{1-\\tan x}{1+\\tan x}}=\\left(\\frac{1+\\tan x}{1-\\tan x}\\right)^{2}=R H S$$ Question 39. Prove that $$\\frac{\\cos (\\pi+x) \\cos (-x)}{\\sin (\\pi-x) \\cos \\left(\\frac{\\pi}{2}+x\\right)}=\\cot ^{2} x$$ $$L H S=\\frac{(-\\cos x)(\\cos x)}{(\\sin x)(-\\sin x)}=\\cot ^{2} x=R H S$$ Question 40. Prove that $$\\begin{array}{l}{\\cos \\left(\\frac{3 \\pi}{2}+x\\right) \\cos (2 \\pi+x)} \\\\ {\\left[\\cot \\left(\\frac{3 \\pi}{2}-x\\right)+\\cot (2", "Symmetry says $\\int_0^\\pi{dx\\over\\sin^2x+|\\cos x|}=2\\int_0^{\\pi/2}{dx\\over\\sin^2x+\\cos x}$ – Barry Cipra May 1 '17 at 15:43 I can't take full credit for this, Wolfram can be very helpful when you know how to coax it along. For the integral $$\\int{dx\\over\\sin^2x+\\cos x},$$ let \\begin{align*} u &= \\tan\\left(\\frac{x}{2}\\right)\\\\ du &= \\frac{1}{2}\\sec^{2}\\left(\\frac{x}{2}\\right)\\,dx. \\end{align*} Then we have \\begin{align*} \\sin x &= \\frac{2\\sin x}{1 +\\cos x}\\cdot \\frac{1+\\cos x}{2} = \\frac{2\\tan^{2}\\left(\\frac{x}{2}\\right)}{\\sec^{2}\\left(\\frac{x}{2}\\right)} = \\frac{2\\tan^{2}\\left(\\frac{x}{2}\\right)}{\\tan^{2}\\left(\\frac{x}{2}\\right)+1}=\\frac{2u^{2}}{u^{2}+1}\\\\ \\cos x&= \\frac{2\\cos x}{1+\\cos x}\\cdot\\frac{1+\\cos x}{2}=\\frac{1+\\cos x - 1 + \\cos x}{(1+\\cos x)\\sec^{2}\\left(\\frac{x}{2}\\right)}=\\frac{1-\\frac{1-\\cos x}{1+\\cos x}}{\\sec^{2} \\left(\\frac{x}{2}\\right)}= \\frac{1-u^{2}}{u^{2}+1}\\\\ dx &= \\frac{2\\,du}{u^{2}+1}. \\end{align*} The integral then becomes \\begin{align*} 2\\int \\frac{du}{(u^{2}+1)\\left[\\frac{4u^{2}}{(u^{2}+1)^{2}}+\\frac{1-u^{2}}{u^{2}+1}\\right]} &=-2\\int\\frac{u^{2}+1}{u^{4}-4u^{2}-1}\\,du\\\\ &=-2\\int\\frac{u^{2}+1}{\\left(u^{2}-\\sqrt{5}-2\\right)\\left(u^{2}+\\sqrt{5}-2\\right)}\\,du. \\end{align*} We can then use a partial fraction decomposition, which I will omit, to change the integral to \\begin{align*} -\\int\\frac{\\sqrt{5}-3}{\\sqrt{5}\\left(u^{2}+\\sqrt{5}-2\\right)}\\,du -\\int\\frac{-3-\\sqrt{5}}{\\sqrt{5}\\left(-u^{2}+\\sqrt{5}+2\\right)}\\,du, \\end{align*} which becomes \\begin{align*}\\frac{3-\\sqrt{5}}{\\sqrt{5}}\\int\\frac{du}{u^{2}+\\sqrt{5}-2} + \\frac{3+\\sqrt{5}}{\\sqrt{5}}\\int\\frac{du}{-u^{2} + \\sqrt{5} +2}. \\end{align*} We can rewrite this as $$\\frac{3-\\sqrt{5}}{\\sqrt{5}(\\sqrt{5}-2)}\\int\\frac{du}{1+\\left(\\frac{u}{\\sqrt{\\sqrt{5}-2}}\\right)^{2}}+\\frac{3+\\sqrt{5}}{\\sqrt{5}(\\sqrt{5}+2)}\\int\\frac{du}{1-\\left(\\frac{u}{\\sqrt{\\sqrt{5}+2}}\\right)^{2}}.$$ This then becomes $$\\frac{3-\\sqrt{5}}{\\sqrt{5}\\sqrt{\\sqrt{5}-2}}\\tan^{-1}\\left(\\frac{u}{\\sqrt{\\sqrt{5}-2}}\\right)+\\frac{3+\\sqrt{5}}{\\sqrt{5}\\sqrt{\\sqrt{5}+2}}\\tanh^{-1}\\left(\\frac{u}{\\sqrt{\\sqrt{5}+2}}\\right)+C,$$ or substituting back $$\\frac{3-\\sqrt{5}}{\\sqrt{5}\\sqrt{\\sqrt{5}-2}}\\tan^{-1}\\left(\\frac{\\tan\\left(\\frac{x}{2}\\right)}{\\sqrt{\\sqrt{5}-2}}\\right)+\\frac{3+\\sqrt{5}}{\\sqrt{5}\\sqrt{\\sqrt{5}+2}}\\tanh^{-1}\\left(\\frac{\\tan\\left(\\frac{x}{2}\\right)}{\\sqrt{\\sqrt{5}+2}}\\right)+C.$$ Since this expression vanishes at $x=0$, your integral is given by $$-2\\pi\\left(\\frac{3-\\sqrt{5}}{\\sqrt{5}\\sqrt{\\sqrt{5}-2}}\\tan^{-1}\\left(\\frac{\\tan\\left(\\frac{\\pi}{4}\\right)}{\\sqrt{\\sqrt{5}-2}}\\right)+\\frac{3+\\sqrt{5}}{\\sqrt{5}\\sqrt{\\sqrt{5}+2}}\\tanh^{-1}\\left(\\frac{\\tan\\left(\\frac{\\pi}{4}\\right)}{\\sqrt{\\sqrt{5}+2}}\\right)\\right)\\approx -8.734.$$ You've done most part of the problem. I can't solve the last integral, but I believe I can provide an approximate solution. Let us take $f(x) = \\sin^2 x + |cos x|$. The graphs of $|\\cos x|$ and $\\cos^2 x$ are pretty close. So $f(x)$ will always be quite close to $1$. Now, to find the maxima and minima values of the function, we need only", "&= \\pm \\sqrt {1 + {{\\tan }^2}x} \\\\ &= \\pm \\sqrt {1 + {{\\left( {\\frac{4}{3}} \\right)}^2}} \\\\ &= \\pm \\sqrt {1 + \\frac{{16}}{9}} \\\\ &= \\pm \\sqrt {\\frac{{25}}{9}} \\\\ &= \\pm \\frac{5}{3}\\end{align} Since, $$x$$ lies in third quadrant, the value of $$\\sec x$$ will be negative. Therefore, $\\sec x = - \\frac{5}{3}$ Now, \\begin{align}\\cos x &= \\frac{1}{\\sec x}\\\\ &= \\frac{1}{\\left( { - \\frac{5}{3}} \\right)}\\\\& = - \\frac{3}{5}\\end{align} Now, \\begin{align}\\tan x &= \\frac{{\\sin x}}{{\\cos x}}\\\\\\sin x &= \\tan x\\cos x\\\\& = \\left( {\\frac{4}{3}} \\right) \\times \\left( { - \\frac{3}{5}} \\right)\\\\ &= - \\frac{4}{5}\\end{align} Now, \\begin{align}{\\rm{cosec}}\\;x &= \\frac{1}{\\sin x}\\\\ &= \\frac{1}{\\left( { - \\frac{4}{5}} \\right)}\\\\ &= - \\frac{5}{4}\\end{align} Hence, $$\\sin x = - \\frac{4}{5}$$, $${\\rm{cosec}}\\;x = - \\frac{5}{4}$$ , $$\\cos x=-\\frac{3}{5}$$, $$\\sec x=-\\frac{5}{3}$$, and $$\\tan x=\\frac{4}{3}$$ ## Chapter 3 Ex.3.2 Question 4 Find the values of trigonometric function $$\\sec x = \\frac{{13}}{5}$$, $$x$$ lies in fourth quadrant. ### Solution As we know that \\begin{align}\\cos x &= \\frac{1}{{\\sec x}}\\\\ &= \\frac{1}{{\\left( {\\frac{{13}}{5}} \\right)}}\\\\ &= \\frac{5}{{13}}\\end{align} Now, $${\\sin ^2}x + {\\cos ^2}x = 1$$ \\begin{align}{\\sin ^2}x &= 1 - {\\cos ^2}x\\\\\\sin x &= \\pm \\sqrt {1 - {{\\cos }^2}x} \\\\ &= \\pm \\sqrt {1 - {{\\left( {\\frac{5}{13}} \\right)}^2}} \\\\& = \\pm \\sqrt {1 - \\frac{25}{169}} \\\\ &= \\pm \\sqrt {\\frac{144}{169}} \\\\& = \\pm \\frac{12}{13}\\end{align} Since, $$x$$ lies in fourth quadrant, the value", "135°, find the coordinates of P with respect to the original system. Solution: Here (x’, y’) = (4, -3); θ = 135° Let (x, y) be the coordinates of P, then x = x’ cos θ – y’ sin θ = 4cos 135° – (-3) sin 135° Therefore, the coordinates of P with respect to the original system are ($$\\frac{-1}{\\sqrt{2}}$$, $$\\frac{7}{\\sqrt{2}}$$) Question 5. Show that the axes are to be rotated through an angle of $$\\frac{1}{2}$$ Tan-1($$\\frac{2 h}{a-b}$$) so as to remove the xy term from the equation ax2 + 2hxy + by2 = 0, if a ≠ b and through the angle $$\\frac{\\pi}{4}$$, if a = b. [Mar 13, 06] Solution: If the axes are rotated through an angle ‘θ’, then x = x’ cos θ – y’ sin θ y = x’ sin θ + y’ cos θ Therefore the given equation transforms as a(x’ cos θ – y’ sin θ)2 + 2h(x’ cosθ – y’ sinθ) (x’ sinθ + y’ cosθ) + b(x’ sin θ + y’ cos θ)2 = o To remove x’y’ term from the equation, the coefficient of x’y’ term must be zero. So, (b – a) sin θ cos θ + h(cos2θ – sin2θ) = 0 i.e., h cos 2θ = $$\\frac{a-b}{2}$$ sin 2θ i.e., tan 2θ = $$\\frac{2 h}{a-b}$$, if", "5x}=\\frac{-2\\sin \\frac{3x+5x}{2}\\sin \\frac{3x-5x}{2}}{2\\sin \\frac{3x+5x}{2}\\cos \\frac{3x-5x}{2}} \\\\ & =\\frac{-2\\sin \\frac{8x}{2}\\sin \\frac{-2x}{2}}{2\\sin \\frac{8x}{2}\\cos \\frac{-2x}{2}} \\\\ & =\\frac{-2\\sin 4x\\sin \\left( -x \\right)}{2\\sin 4x\\cos \\left( -x \\right)} \\\\ & =-\\tan \\left( -x \\right) \\end{align} Thus, the result will be $\\tan x$. Hence, verified. Also, consider the identity: $\\frac{\\cos 3x-\\cos x}{\\sin 3x+\\sin x}=-\\tan x$ And use the sum-to product formula $\\cos \\alpha -\\cos \\beta =-2\\sin \\frac{\\alpha +\\beta }{2}\\sin \\frac{\\alpha -\\beta }{2}$ to numerator and sum-to product formula $\\sin \\alpha +\\sin \\beta =2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2}$ to denominator of left hand side. Then, \\begin{align} & \\frac{\\cos 3x-\\cos x}{\\sin 3x+\\sin x}=\\frac{-2\\sin \\frac{3x+x}{2}\\sin \\frac{3x-x}{2}}{2\\sin \\frac{3x+x}{2}\\cos \\frac{3x-x}{2}} \\\\ & =\\frac{-2\\sin \\frac{4x}{2}\\sin \\frac{2x}{2}}{2\\sin \\frac{4x}{2}\\cos \\frac{2x}{2}} \\\\ & =\\frac{-2\\sin 2x\\sin x}{2\\sin 2x\\cos x} \\\\ & =-\\tan x \\end{align} Hence verified. Also, consider the identity: $\\frac{\\sin 3x-\\sin x}{\\cos 3x-\\cos x}=-\\cot 2x$ By using the sum-to product formula $\\sin \\alpha -\\sin \\beta =2\\sin \\frac{\\alpha -\\beta }{2}\\cos \\frac{\\alpha +\\beta }{2}$ to numerator and $\\cos \\alpha -\\cos \\beta =-2\\sin \\frac{\\alpha +\\beta }{2}\\sin \\frac{\\alpha -\\beta }{2}$ to denominator of left hand side. Then, \\begin{align} & \\frac{\\sin 3x-\\sin x}{\\cos 3x-\\cos x}=\\frac{2\\sin \\frac{3x-x}{2}\\cos \\frac{3x+x}{2}}{-2\\sin \\frac{3x+x}{2}\\sin \\frac{3x-x}{2}} \\\\ & =\\frac{2\\sin \\frac{2x}{2}\\cos \\frac{4x}{2}}{-2\\sin \\frac{4x}{2}\\sin \\frac{2x}{2}} \\\\ & =-\\frac{\\cos 2x}{\\sin 2x} \\\\ & =-\\cot 2x \\end{align} Hence verified. And also consider the identity: $\\frac{\\sin x-\\sin y}{\\sin x+\\sin y}=\\tan \\frac{x-y}{2}\\cot \\frac{x+y}{2}$ Also, use the sum-to product formula" ]

[ "# Number of Trigonometric solutions Algebra Level 4 For each integer $n>1$, let $F(n)$ be the number of solutions to the equation $\\sin{x} = \\sin{(nx)}$ on the interval $[0,\\pi]$. What is $\\displaystyle\\sum_{n=2}^{2007}$$F(n)$? ×", "# Problem #1310 1310 For each integer $n>1$, let $F(n)$ be the number of solutions to the equation $\\sin{x}=\\sin{(nx)}$ on the interval $[0,\\pi]$. What is $\\sum_{n=2}^{2007} F(n)$? $\\mathrm{(A)}\\ 2014524$ $\\mathrm{(B)}\\ 2015028$ $\\mathrm{(C)}\\ 2015033$ $\\mathrm{(D)}\\ 2016532$ $\\mathrm{(E)}\\ 2017033$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "of solutions of the equation $\\sin{(x)} = \\sin{(nx)}$ on the interval $[0, \\pi]$. What is $\\displaystyle \\sum_{n=2}^{2007} F(n)$?", "Brian Scott's answer and the links it gives. – Arturo Magidin Oct 3 '11 at 2:41 The answer to (b) is the same as the answer to the following question: How many solutions does the equation $x_1+x_2+\\dots+x_n=k$ have if each $x_i$ is required to be an integer between $1$ and $m$ inclusive? Think of $x_i$ as $f(i)$. Versions of this question have been discussed here before: a couple of very recent discussions are here and here. -", "# Count the number of integer solutions 1. Nov 17, 2005 ### benorin Count the number of integer solutions of (rather, # of integer lattice points such that) $$n+\\sum_{k=1}^{n} \\left| x_{k}\\right| \\leq N$$ Not homework, so no rush. I have worked it through before with a prof., but he's so brilliant I didn't understand much of anything he said . His solution involed a generating function, something like the coefficient of the blah term in the expansion of blahblah. Note that the LHS is the height of a polynomial with integer coefficients, although it has nothing to do with the solution. Last edited: Nov 17, 2005 2. Nov 17, 2005 ### AKG If n > N, the answer is 0. Otherwise, the problem reduces to finding the lattice point solutions of: $$\\sum _{k=1} ^n |x_k| \\leq N'$$ where N' > 0. Let f(N',n) denote the number of solutions. Then we want to find f(N',n) and this is just: $$f(N',n) = f(N',n-1) + 2\\sum _{k=1} ^{N'} f(N'-k,n-1)$$ We get this because $\\sum _{k=1} ^n |x_k| = |x_1| + \\sum_{k=2} ^n |x_k|$ and when |x1| > 0, then x1 has two possible values whereas when |x1| = 0 then x1 = 0 only. It's a start.", "any integer n (n-1)pi<x<n*pi=>|sin(x)|>0", "The number of distinct solutions of the equation Question: The number of distinct solutions of the equation $\\log _{\\frac{1}{2}}|\\sin x|=2-\\log _{\\frac{1}{2}}|\\cos x|$ in the interval $[0,2 \\pi]$, is___________. Solution:", "# Just try to post my first problem Calculus Level 1 For each positive integer $n$, let $f^n (x) = \\sin ( f^{n-1} (x) ),$ with $f^0 (x) = x$. Let $y(x) = \\lim_{n \\rightarrow \\infty } f^n(x)$. Determine $y'(0)$. ×", "them but till now I could not show they are equal each other. – Mathlover Apr 17 '12 at 23:21 I suppose you're assuming $f$ is continuous. The statement is easily seen to be true for $f(t)=t$, and by subtracting a multiple of this we can assume $f(0) = f(x)$. By the Stone-Weierstrass theorem, trigonometric polynomials with period $x$ are uniformly dense in the continuous functions with period $x$, and so it suffices to prove the result for $f(t) = \\cos(2 \\pi j t/x)$ and $\\sin(2 \\pi j t/x)$ for each integer $j$. But for those it can be done explicitly: $\\sum_{k=1}^n \\sin(2 \\pi j k/n) = 0$ for all integers $j$, $\\sum_{k=1}^n \\cos(2 \\pi j k/n) = n$ if $j$ is a multiple of $n$, $0$ otherwise...", "# Solving equations 1. Jan 8, 2008 Suppose you are solving equations in the interval 0<=xx<=2pi....Without actually solving equations, what is the difference between the number of solutions of sinx=1 and sin2x=1? How do you account for this difference 2. Jan 8, 2008 ### Gib Z Well since sin 2x = 2 sin x cos x = 1, and you also want solutions to sin x =1, we can see that the number of solutions of sin x = sin 2x = 1 is the number of times cos x is equal to 1 in the interval. 3. Jan 8, 2008 ### rohanprabhu $sin x = 1$ $sin x = sin (\\frac{\\pi}{2})$ $x = n\\pi + (-1)^n (\\frac{\\pi}{2})$ But, as $x \\in [0, 2\\pi]$ Hence, select the values for n ($n \\in N$), such that $x \\in [0, 2\\pi]$. The satisfying values are: $n \\in \\{0, 1\\}$ Put this values for x, and you shall get $x \\in \\{\\frac{\\pi}{2}\\}$. This is because when we have $\\alpha$ as $\\frac{\\pi}{2}$, we get the same solutions for $n = 0; n = 1$. Do the same for 2x ($x \\in [0, 2\\pi]$), ($2x \\in [0, 4\\pi]$) and you shall get the solutions for 2x. It again gives us: $2x = n\\pi + (-1)^n (\\frac{\\pi}{2})$ Here, {0, 1, 2, 3} satisfy 'n',", "2 n \\pi}{8}}$, n = any integer, including 0. The values for n = 0, 1, 2.., 7 are repeated, in a cycle, for other values of n. The eight distinct values are determinate and are listed in the answer. Importantly, observe that x = 0 is not a solution. .", "## Equation problem Let a function f be defined by $$f(x) = a_{1}sinx + a_{2}sin2x + ... + a_{n}sin(nx)$$ , where $$a_{1}, a_{2}, ... , a_{n}$$ an are real numbers and n is a positive integer. If abs(f(x)) is less than or equal to abs(sin(x)) for all x, show that $$abs( a_{1} + 2a_{2} + ... + na_{n})$$ is less than or equal to 1", "equation. The solution set = { } or Φ (iv) |x| ≤ 3 The solution set of |x| ≤ a is -a ≤ x ≤ a ∴ The required solution is -3 ≤ x ≤ 3 ∴ The solution set is [-3, 3] (v) 2|x| = 5 ∴ |x| = $$\\frac{5}{2}$$ ∴ x = ±$$\\frac{5}{2}$$ (vi) [x + [x + [x]]] = 9 ∴ [x + [x] + [x] ] = 9 …….[[x + n] = [x] + n, if n is an integer] ∴ [x + 2[x]] = 9 ∴ [x] + 2[x] = 9 …..[[2[x] is an integer]] ∴ [x] = 3 ∴ x ∈ [3, 4) (vii) {x} > 4 This is a meaningless statement as 0 ≤ {x} < 1 ∴ The solution set = { } or Φ (viii) {x} = 0 ∴ x is an integer ∴ The solution set is Z. (ix) {x} = 0.5 ∴ x = ….., -2.5, -1.5, -0.5, 0.5, 1.5, ….. ∴ The solution set = {x : x = n + 0.5, n ∈ Z} (x) 2{x} = x + [x] = [x] + {x} + [x] ……[x = [x] + {x}] ∴ {x} = 2[x] R.H.S. is an integer ∴ L.H.S. is an integer ∴ {x} = 0 ∴ [x] = 0 ∴ x =", "any integer $n$. – Jonas Meyer Apr 8 '11 at 3:33 @svenkatr: That comment should be an answer. – Rahul Apr 8 '11 at 3:38 Consider $c = e^{2 \\pi}$. Then, $f(nc) = f(n)$ for all $n$. EDIT: I had put this as a comment at first, but I made it an answer on Rahul Narain's suggestion. - Can you explain this a little more ? for example, can we expand $Sin(x)=\\sum{a_n Sin(n log(x)+b_n Cos(n log(x))}$ ? – Austin Apr 8 '11 at 10:59", "so are $\\dfrac{p}{5^n}$, $\\dfrac{q}{5^n}$, $\\dfrac{r}{5^n}$ for any integer $n \\geq 0$, and as before, this shows that the only integer solution is $p = q = r = 0$. The type of argument used here (using one imagined solution to find a smaller one) is known as infinite descent, which was popularised by Pierre de Fermat.", "# For Every Integer 'x' Let $$f(x)=7x^{19}+19x^7+13ax$$ be a polynomial. Find the sum of the smallest $$2$$ possible values of the positive integer $$a$$, such that $$133$$ divides $$f(x)$$ for every integer $$x$$. ×", "side be an integer. This means that $x=-1$ or $x=0$. Therefore the solution is $x=-1,0,1$.", "for certain integer variable. For instance f[x_,n_Integer]:= n Sin[x] will be evaluated only if n is an integer Try f[2.34,3] and f[2.34, 3.2] or even f[2.34,3.] the two last function will not be evaluated", "any integer solution to your equation would ALSO provide be a solution to the equation that castor treats here and shows to have only a finite number of solutions. Jun 15, 2020 at 17:05 • Sorry,The true equation is: $$16(n+1)^2q^8+16(n+1)^2q^6+1=m^2$$ Jun 16, 2020 at 6:48", "# 2011-17 Infinitely many solutions Let f(n) be the maximum positive integer m such that the sum of all positive divisors of m is less than or equal to n. Find all positive integers k such that there are infinitely many positive integers n satisfying the equation n-f(n)=k. GD Star Rating" ]

[ "$\\frac {n - 1}{2}x = k\\pi$, or $x = \\frac {2k\\pi}{n - 1}$ for some nonnegative integer $k$. Here $k\\leq \\frac {n - 1}{2}$ so that there are $\\left\\lfloor \\frac {n + 1}{2}\\right\\rfloor$ solutions here. However, we overcount intersections. These occur whenever $$\\frac {2j + 1}{n + 1} = \\frac {2k}{n - 1}$$ $$k = \\frac {(2j + 1)(n - 1)}{2(n + 1)}$$ which is equivalent to $2(n + 1)$ dividing $(2j + 1)(n - 1)$. If $n$ is even, then $(2j + 1)(n - 1)$ is odd, so this never happens. If $n\\equiv 3\\pmod{4}$, then there won't be intersections either, since a multiple of 8 can't divide a number which is not even a multiple of 4. This leaves $n\\equiv 1\\pmod{4}$. In this case, the divisibility becomes $\\frac {n + 1}{2}$ dividing $(2j + 1)\\frac {n - 1}{4}$. Since $\\frac {n + 1}{2}$ and $\\frac {n - 1}{4}$ are relatively prime (subtracting twice the second number from the first gives 1), $\\frac {n + 1}{2}$ must divide $2j + 1$. Since $j\\leq \\frac {n - 1}{2}$, $2j + 1\\leq n < 2\\cdot \\frac {n + 1}{2}$. Then there is only one intersection, namely when $j = \\frac {n - 1}{4}$. Therefore we find $F(n)$ is equal to $1 + \\lfloor n/2 \\rfloor + \\left \\lfloor \\frac {n", "# Find the $\\bigcap_{n = 1}^{\\infty} (-\\frac{1}{n}, \\frac{2}{n})$ Find the $\\bigcap_{n = 1}^{\\infty} (-\\frac{1}{n}, \\frac{2}{n})$ So the way I understand it is that I’m trying to find $(\\frac{-1}{1}, \\frac{2}{1}) \\bigcap (\\frac{-1}{2}, \\frac{2}{2}) \\bigcap (\\frac{-1}{3}, \\frac{2}{3})\\bigcap …$ and so forth. Then the intersection would be $\\varnothing$ right? Can I prove this by just writing out the first few elements and seeing that they do not intersect? Or is the question asking me to find: $\\frac{-1}{1} \\bigcap \\frac{2}{2} \\bigcap \\frac{-1}{3} \\bigcap…$? in this case the intersection would still be $\\varnothing$, right? #### Solutions Collecting From Web of \"Find the $\\bigcap_{n = 1}^{\\infty} (-\\frac{1}{n}, \\frac{2}{n})$\" There are three things to prove: 1. $0$ is in every interval, hence in the intersection of all of them. 2. If $x>0$, then there is at least one interval that does not contain $x$, so $x$ is not in the mutual intersection. (choose $n$ so that $2/n<x$). 3. If $x<0$, then there is at least one interval that does not contain $x$, so $x$ is not in the mutual intersection. (choose $n$ so that $-1/n>x$). Combining, the intersection is $\\{0\\}$, as tetori points out. Here is a longer and more formal version of vadim123’s answer, with some more detail about where the case split is coming from. $$\\calc x \\in \\langle \\cap n :: (-\\tfrac{1}{n},", "# 2007 AMC 12A Problems/Problem 24 ## Problem For each integer $n>1$, let $F(n)$ be the number of solutions to the equation $\\sin{x}=\\sin{(nx)}$ on the interval $[0,\\pi]$. What is $\\sum_{n=2}^{2007} F(n)$? $\\mathrm{(A)}\\ 2014524$ $\\mathrm{(B)}\\ 2015028$ $\\mathrm{(C)}\\ 2015033$ $\\mathrm{(D)}\\ 2016532$ $\\mathrm{(E)}\\ 2017033$ ## Solution ### Solution 1 $F(2)=3$ By looking at various graphs, we obtain that, for most of the graphs $F(n) = n + 1$ Notice that the solutions are basically reflections across $x = \\frac{\\pi}{2}$. However, when $n \\equiv 1 \\pmod{4}$, the middle apex of the sine curve touches the sine curve at the top only one time (instead of two reflected points), so we get here $F(n) = n$. $3+4+5+5+7+8+9+9+\\cdots+2008$ $= (1+2+3+4+5+\\cdots+2008) - 3 - 501$ $= \\frac{(2008)(2009)}{2} - 504 = 2016532$ $\\mathrm{(D)}$ ### Solution 2 $$\\sin nx - \\sin x = 2\\left( \\cos \\frac {n + 1}{2}x\\right) \\left( \\sin \\frac {n - 1}{2}x\\right)$$ So $\\sin nx = \\sin x$ if and only if $\\cos \\frac {n + 1}{2}x = 0$ or $\\sin \\frac {n - 1}{2}x = 0$. The first occurs whenever $\\frac {n + 1}{2}x = (j + 1/2)\\pi$, or $x = \\frac {(2j + 1)\\pi}{n + 1}$ for some nonnegative integer $j$. Since $x\\leq \\pi$, $j\\leq n/2$. So there are $1 + \\lfloor n/2 \\rfloor$ solutions in this case. The second occurs whenever", "our answer to be $231+252= \\boxed{483}$. ## Solution 2 By considering the graph of this function, it is shown that the graph is composed of trapezoids ranging from $a^2$ to $(a+1)^2$ with the top made of diagonal line $y=ax$. The width of each trapezoid is $3, 5, 7$, etc. Whenever $a$ is odd, the value of $A(n)$ increases by an integer value, plus $\\frac{1}{2}$. Whenever $a$ is even, the value of $A(n)$ increases by an integer value. Since each trapezoid always has an odd width, every value of $n$ is not an integer when $a \\pmod{4} \\equiv 2$, and is an integer when $a \\pmod{4} \\equiv 0$. Every other value is an integer when $a$ is odd. Therefore, it is simply a matter of determining the number of values of $n$ where $a \\pmod{4} \\equiv 0$ ($(5^2-4^2)+(9^2-8^2)+...+(29^2-28^2)$), and adding the number of values of $n$ where $a$ is odd ($\\frac{(2^2-1^2)+(4^2-3^2)+...+(30^2-29^2)+(1000-31^2)}{2}$). Adding the two values gives $231+252=\\boxed{483}$. ## \"Step\" Solution First, draw a graph of the function. Note that it is just a bunch of line segments. Since we only need to know whether or not $A(n)$ is an integer, we can take the area of each piece from some $x$ to $x+1$ (mod 1), aka the piece from $2$ to $3$ has area $\\frac{1}{2} (\\mod 1)$. There are", "one-to-one. We find, for example, $$k(3)\\equiv k(6) \\equiv 4$$ (mod 15). Hence, $$k$$ is not one-to-one. Assume $$h(n_1)=h(n_2)$$. Since the image is either odd or even, we have to consider two cases. If both $$h(n_1)$$ and $$h(n_2)$$ are odd, then $2n_1+1 = 2n_2+1,$ hence, $$n_1=n_2$$. If both $$h(n_1)$$ and $$h(n_2)$$ are even, then $-2n_1 = -2n_2,$ hence, $$n_1=n_2$$. In both cases, we find $$n_1=n_2$$. Therefore, $$h$$ is one-to-one. ### Section [sec:ontofcn] 1. We can use, for example, a straight line graph that connects the points $$(1,2)$$ and $$(3,5)$$. This leads to the function $$\\fcn{f}{[1,3]}{[2,5]}$$ defined by $$f(x)=\\frac{3}{2}\\,x+\\frac{1}{2}$$. 2. We can use, for example, a straight line graph that connects the points $$(1,2)$$ and $$(3,4)$$, see Figure [fig:otograph]. 3. $$[0,\\infty)$$. 4. The midpoint of the interval $$(2,9)$$ is $$\\frac{11}{2}$$, and its width is 7. The transformation of $$x$$ to $$x-\\frac{11}{2}$$ shifts the interval to $$\\left(-\\frac{7}{2},\\frac{7}{2}\\right)$$. Hence, we can set $$h(x) = \\tan\\big[\\frac{\\pi}{7}\\big(x-\\frac{11}{2}\\big)\\big]$$. 5. Let $$y=3x+11$$, then $x = \\frac{y-11}{3},$ which is, of course, an element of the domain. Hence, $$g$$ is onto. 6. It is obvious that the graphs $$y=3x+1$$ and $$y=4x$$ are increasing. For $$x\\leq 2$$, the $$y$$-values cover the range $$(-\\infty,7)$$. For $$x>2$$, the $$y$$-values cover the range $$(8,\\infty)$$. Hence, the $$y$$-values in the interval $$[7,8]$$ are never used as images. For example, there is no $$x$$-value", "is neither odd nor even, so we need to take two cases separately : $x$ is non-negative, and $x$ is negative. Case 1: $x$ is non-negative Note that when $x$ is an integer $n$, $f$ will be equal to $6n^2$. So all numbers of the form $6n^2$ can be written in this form. Now let's see how the value of $f$ changes as $x$ varies from $n$ to $n+1$. Since the period of $f$ is $\\text{lcm}( \\frac{1}{2} , \\frac{1}{3}) = 1$, we can be sure that the same pattern will be followed as $x$ goes from $n+1$ to $n+2$. As we saw above, $f$ only changes when $x$ crosses and integral multiple of $\\frac{1}{6}$, we can divide the interval $[n, n+1)$ into $6$ equal intervals. $$\\begin{array}{|l|c|c|c|}\\hline \\text{Interval} & \\lfloor 2x \\rfloor & \\lfloor 3x \\rfloor & f(x) \\\\\\hline \\left[n, n + \\frac{1}{6}\\right) & 2n & 3n & 6n^{2} \\\\ \\left[n + \\frac{1}{6}, n + \\frac{2}{6}\\right) & 2n & 3n & 6n^{2}\\\\ \\left[n + \\frac{2}{6}, n + \\frac{3}{6}\\right) & 2n & 3n + 1& 6n^{2} + 2n\\\\ \\left[n + \\frac{3}{6}, n + \\frac{4}{6}\\right) & 2n + 1& 3n + 1& 6n^{2} + 5n + 1\\\\ \\left[n + \\frac{4}{6}, n + \\frac{5}{6}\\right) & 2n + 1& 3n + 2& 6n^{2} + 7n + 2\\\\ \\left[n + \\frac{5}{6}, n + 1\\right)", "and $f,g$ are monotonic as well as continuous on the interval $[a,b]$, then $f(x). Now suppose that this is true, it can be easily verified that $-\\int_0^{\\frac{\\pi}{4}}\\cos(x)\\ln(\\sin(x))dx=\\frac{ \\sqrt{2}\\ln(2)}{4}+\\frac{\\sqrt{2}}{2}$ $>-\\int_0^{\\frac{\\pi}{4}}\\sin(x)\\ln(\\cos(x))dx=1-\\frac{\\sqrt{2}\\ln(2)}{4}-\\frac{\\sqrt{2}}{2}$ Now supposing that my \"made up\" lemma is correct, this proves the inequality. unfortunately your made up lemma is not always true, e.g. $f(x)=\\frac{x+1}{4}, \\ g(x)=x.$ then $\\int_0^1 f(x) \\ dx < \\int_0^1 g(x) \\ dx,$ but neither $f < g$ nor $f > g$ on the unit interval. the trouble here comes from this fact that $f$ and $g$ intersect! so you need to assume that $f,g$ do not intersect in the interval. also i don't think we need $f,g$ to be positive. 7. Originally Posted by NonCommAlg unfortunately your made up lemma is not always true, e.g. $f(x)=\\frac{x+1}{4}, \\ g(x)=x.$ then $\\int_0^1 f(x) \\ dx < \\int_0^1 g(x) \\ dx,$ but neither $f < g$ nor $f > g$ on the unit interval. the trouble here comes from this fact that $f$ and $g$ intersect! so you need to assume that $f,g$ do not intersect in the interval. also i don't think we need $f,g$ to be positive. Dangit! I forgot to say that. I knew that they cannot intersect...and no we do not need them positive we need $fg>0$. I have come up with a \"proof\"", "middle steps, ... 0 A subset B implies #A <= #B. Proof. f:A -> B, x -> x is an injection from A into B. 1 You need to show those intervals \"continuously overlap\". That is for any k \\in \\mathbb N we have \\frac 1{2^{k+1}} < \\frac 1{2^k} < \\frac 3{2^{k+1}} < \\frac 3{2^k}. That follows as \\frac 12 < 1 < \\frac 32 < 3. This means \\cup (\\frac 1{2^n},\\frac 3{2^n}) = (\\lim_{n\\to \\infty} \\frac 1{2^n}, \\frac 3{2^1}) = (0, \\frac 32) ... 1 Let 0<x<1. Consider the interval (\\frac {\\ln (\\frac 1 x)} {\\ln 2},\\frac {\\ln (\\frac 3 x)} {\\ln 2}). The length of this interval is easily seen to be \\frac {\\ln 3} {\\ln 2}. Since any interval of length greater than 1 contains an integer there exist a positive integer n in this interval. You can now verify that x \\in (\\frac 1 {2^{n}},\\frac 3 ... 1 I would note that G_1=(1/2,3/2) which covers (1/2,1), and then noting that$$\\frac{3}{2^{n+1}} \\ge \\frac{1}{2^n} \\iff \\frac{3}{2} \\ge 1 \\mbox{, for all }n\\in \\mathbb{N}$$which is true. This gives us that two consecutive intervals have non-empty intersection so that$$\\bigcup_{n\\le N}G_{n}=\\left(\\frac{1}{2^N},1\\right) \\mbox{, for all } N\\in \\mathbb{N}$$... 0 For the first question, use https://mathworld.wolfram.com/EulersTotientTheorem.html or https://mathworld.wolfram.com/CarmichaelFunction.html and https://en.m.wikipedia.org/wiki/Proof_by_contradiction For the second, if p\\nmid a\\iff(a,p)=1$$a^{p-1}\\equiv 1\\pmod p$$If (a,2)=1, using Carmichael", "0}[\\frac{\\tan x}{x}] =[\\lim_{x\\to 0}\\frac{\\tan x}{x}]$ Would appreciate if one describes in detail 6. Well, have you looked at the graphs. 8. you don't need to sketch a graph. choose $0 < x < \\frac{\\pi}{2}.$ then if you apply mean value theorem for $f(t)=\\sin t$ on the interval $[0,x],$ then you get $0 < \\frac{\\sin x}{x} < 1.$ thus, since $\\frac{\\sin x}{x}$ is an even function, for any non-zero $x$ in the interval $[-\\pi/2, \\pi/2]$ we must have: $0< \\frac{\\sin x}{x} < 1,$ which gives us: $\\lfloor{\\frac{\\sin x}{x} \\rfloor}=0,$ for any $x \\neq 0$ in that interval. thus $\\lim_{x\\to0} \\lfloor{\\frac{\\sin x}{x} \\rfloor}= \\lim_{x\\to 0} 0 = 0.$ but $\\lfloor{\\lim_{x\\to0} \\frac{\\sin x}{x} \\rfloor}=\\lfloor{1 \\rfloor}=1.$ a similar argument works for $\\frac{\\tan x}{x}.$ 9. This is precisely what I wanted to know that while evaluating $ [\\lim_{x\\to 0}\\frac{\\sin x}{x}] $ we are required to first find out $ \\lim_{x\\to 0}\\frac{\\sin x}{x} $ and then apply rule of greatest integer function(I would want to know the reason.Is there a rigorous proof?) Or should we not say that since $0<\\frac{\\sin x}{x}<1$ the greatest integer should anyway be 0 whether we find $ [\\lim_{x\\to 0}\\frac{\\sin x}{x}] $ or we find $ \\lim_{x\\to 0}[\\frac{\\sin x}{x}] $ I am perplexed because in case of $\\frac{\\tan x}{x}$ the answer for $ [\\lim_{x\\to 0}\\frac{\\tan x}{x}]=1 $ $ \\lim_{x\\to 0}[\\frac{\\tan", "graph of $f$ if(i) $f$ is even.(ii) $f$ is odd.(iii) $f$ is nonnegative.(iv) $f(x)=f(x+a)$ for all $x$ (a function with this property is called periodic, with period $a$ ). Describe the general features of the graph of $f$ if (i) $f$ is even. (ii) $f$ is odd. (iii) $f$ is nonnegative. (iv) $f(x)=f(x+a)$ for all $x$ (a function with this property is called periodic, with period $a$ ).... For each multiplication or division, just determine the sign of the answer a. $-\\frac{19}{37}\\left(-\\frac{51}{75}\\right)$ b. $\\frac{45.568}{-2.56}$ c. $-8.2(-4.1)(-6)(-9.3)(-1.5)$ For each multiplication or division, just determine the sign of the answer a. $-\\frac{19}{37}\\left(-\\frac{51}{75}\\right)$ b. $\\frac{45.568}{-2.56}$ c. $-8.2(-4.1)(-6)(-9.3)(-1.5)$... Use the symmetry and lincarity propertics and the result of Exercise 2 to show that $deltaleft(frac{i sin pi t}{1-t^{2}} ight)=left{\begin{array}{ll}frac{i sin w}{2} & ext { for }|w| leq pi \\ 0 & ext { for }|w|>piend{array} ight.$ Use the symmetry and lincarity propertics and the result of Exercise 2 to show that $deltaleft(frac{i sin pi t}{1-t^{2}} ight)=left{\begin{array}{ll}frac{i sin w}{2} & ext { for }|w| leq pi \\ 0 & ext { for }|w|>piend{array} ight.$...", "the intersection $\\bigcap_n C_n$ where $C_0=[0,1]$ and $C_{n+1}$ is obtained by removing from each closed interval that makes up $C_n$ its open middle third, so $C_1=[0,1/3]\\cup[2/3,1]$, $C_2=[0,1/9]\\cup[2/9,1/3]\\cup[2/3,7/9]\\cup[8/9,1]$, etc. Each $C_n$ is the union of $2^n$ closed intervals, each of length $1/3^n$. Let’s prove that $C+C=\\{x+y\\mid x,y\\in C\\}$ is the interval ${}[0,2]$. (Cf. Abbott, Understanding analysis, Exercise 3.3.6.) 1. The usual proof consists in showing inductively that $C_n+C_n=[0,2]$ for all $n$. This is easy: Note first that $\\displaystyle C_{n+1}=\\frac13 C_n+\\left(\\frac13C_n+\\frac23\\right)$, where $\\displaystyle \\frac13C_n=\\left\\{\\frac x3\\mid x\\in C_n\\right\\}$ and $\\displaystyle\\frac13 C_n+\\frac23=\\left\\{\\frac{x+2}3\\mid x\\in C_n\\right\\}$. This equality is verified by induction. Using this, we can use induction again to verify that, indeed, $C_n+C_n=[0,2]$ for all $n$. We clearly have that $C+C\\subseteq \\bigcap_n C_n+C_n=[0,2]$. To prove the converse, for each $z\\in[0,2]$ and each $n$, pick $x_n,y_n\\in C_n$ such that $x_n+y_n=z_n$. The sequence of $x_n$ is bounded, so it has a convergent subsequence $x_{n_k}$. The corresponding subsequence $y_{n_k}$ has itself a convergent subsequence $y_{n_{k_m}}$. One argues that their limit values $x,y$ belong to $C$, because they belong to each $C_n$, since these sets are nested and closed. Finally, it follows immediately that $x+y=z$ as well. 2. A very elegant different argument is obtained by using an alternative characterization of $C$: Note that each $x\\in[0,1]$ can be written in base three as $\\displaystyle x=0.x_1x_2x_3\\dots=\\sum_{n=1}^\\infty\\frac {x_n}{3^n}$ where", "# Let f and g be the functions given by f(x)=1+sin(2x) and g(x)=e^(x/2). Let R be the region in the first quadrant enclosed by the graphs of f and g. How do you find the area? Feb 14, 2017 $0.4291$ areal units.. #### Explanation: The points of intersection are given by y = 1+sin2x=e^(x/2). The y-intercept ( x = 0 ) for both are the same 1. So, one common point is (0, 1). Glory to Socratic utility, the other is approximated graphically to 4-sd as 1.136. Now, the area is $\\int \\left(f - g\\right) \\mathrm{dx}$, for x from 0 to 1.136 int ((1+sin (2x) )-(e^(x/2)) dx#, for x from 0 to 1.136 $= \\left[\\left(x - \\frac{1}{2} \\cos \\left(2 x\\right)\\right) - \\left(2 {e}^{\\frac{x}{2}}\\right)\\right]$, between 0 and 1.136 $= \\left(1.136 - \\frac{1}{2} \\cos \\left(2.272 r a \\mathrm{di} a n\\right) - 2 {e}^{\\frac{1.136}{2}}\\right) - \\left(- \\frac{5}{2}\\right)$ $= 0.4291$ areal units. graph{(1+sin(2x)-y)(e^(x/2)-y)=0x^2 [-1, 9, -2.5, 2.5]} graph{1+sin(2x)-e^(x/2) [1.13, 1.14, -.01, .01]}", "the $n$-cycle: Tutte: $T_{C_n}(x, y) = x^{n-1} + x^{n-2} + \\cdots + x + y = \\frac{x^n - x}{x - 1} + y$ Magnitude: $\\mu_{C_n}(q) = \\frac{n(q - 1)}{q^{\\lfloor (n+1)/2 \\rfloor} + q^{\\lceil (n+1)/2 \\rceil} - q - 1}$. (Those are the floor and ceiling functions in the denominator.) This naturally splits into two cases. If $n$ is even then $\\mu_{C_n}(q) = \\frac{n(q - 1)}{(q^{n/2} - 1)(q + 1)}.$ If $n$ is odd then $\\mu_{C_n}(q) = \\frac{n(q - 1)}{2q^{(n+1)/2} - q - 1}.$ 8. Complete graphs Let $K_n$ be the complete graph on $n$ vertices (that is, every vertex is joined to every other by a single edge). Tutte: it seems that there is no neat formula for $T_{K_n}$, and that computing it is nontrivial. A single example: $T_{K_4}(x, y) = (x^3 + y^3) + (x^2 + y^2) + 2(x + y)^2 + 2(x + y).$ Magnitude: $\\mu_{K_n}(q) = \\frac{n}{1 + (n - 1)q}$. 9. Complete bipartite graphs I’ll just do one. Let $K_{2, 3}$ be this graph: Tutte: $T_{K_{2,3}}(x, y) = x^4 + 2x^3 + 3x^2 + 3x y + y^2 + x + y$. Magnitude: $\\mu_{K_{2,3}}(q) = \\frac{5 - 7q}{(1 + q)(1 - 2q^2)} = 5 - 12q + 22q^2 - 36q^3 + 56q^4 - \\cdots$. 10. Bull, tadpole, cricket These three graphs all have the", "Trigonometry (11th Edition) Clone $x = 0.692$ $x = 2.082$ $3cos~\\frac{x}{2}+\\sqrt{x}-2 = -\\frac{x}{2}+2$ We can graph the two equations $~~y = 3cos~\\frac{x}{2}+\\sqrt{x}-2 ~~$ and $~~y = -\\frac{x}{2}+2~~$ to find the points of intersection. The blue graph is $~~y = -\\frac{x}{2}+2~~$ The red graph is $~~y = 3cos~\\frac{x}{2}+\\sqrt{x}-2 ~~$ On the interval $[0,2\\pi]$, we can see that the points of intersection occur when $~~x = 0.692~~$ and $~~x = 2.082$ The solution over the given interval is: $x = 0.692$ $x = 2.082$", "with $n$ gridpoints then $$I_{n} = \\frac{1}{n}\\sum_{i=1}^{n}\\frac{f(\\frac{i-1}{n})+f(\\frac{i}{n})}{2} \\implies I_n = \\left\\{\\matrix{\\frac{1}{2} & n~~\\text{odd}\\\\\\frac{1}{2} - \\frac{1}{n} & n~~\\text{even}}\\right.$$ so for $n=3$ we have the exact answer which is better than any even $n$ no matter how large it is. This shows that increasing the number of gridpoints does not always improve the accuracy. With Simpson's rule we find $$I_n = \\frac{1}{3n}\\sum_{i=1}^{n/2}f\\left(\\frac{2i-2}{n}\\right)+4f\\left(\\frac{2i-1}{n}\\right)+f\\left(\\frac{2i}{n}\\right) \\implies I_n = \\left\\{\\matrix{\\frac{1}{2} - \\frac{1}{3n}&n\\equiv 0\\mod 4\\\\\\frac{1}{2}&n\\equiv 1\\mod 4\\\\\\frac{1}{2} + \\frac{2}{3n} & n\\equiv 2\\mod 4\\\\\\frac{1}{2} - \\frac{5}{6n} & n\\equiv 3\\mod 4}\\right.$$ so even if Simpson's rule has higher order we see that it does not always do better than the trapezoidal rule. $$\\bf \\text{What does higher degree of precision really mean?}$$ If we have a smooth function then a standard Taylor series error analysis gives that the error in estimating the integral $\\int_a^bf(x){\\rm d}x$ using $n$ equally spaced points is bounded by (here for Simpsons and the trapezoidal rule) $$\\epsilon_{\\rm Simpsons} = \\frac{(b-a)^5}{2880n^4}\\max_{\\zeta\\in[a,b]}|f^{(4)}(\\zeta)|$$ $$\\epsilon_{\\rm Trapezoidal} = \\frac{(b-a)^3}{12n^2}\\max_{\\zeta\\in[a,b]}|f^{(2)}(\\zeta)|$$ Note that the result we get from such an error analysis is always an upper bound (or in some cases an order of magnitude) for the error apposed to the exact value for the error. What this error analysis tell us is that if $f$ is smooth on $[a,b]$, so that the derivatives are bounded, then the error with a", "at any given interval. Since the graphs intersect at $0$, at $1$, and at $3$, and nowhere else, and the functions are continuous, that means that the only places where the function can switch \"who is on top\" are $0$, $1$ and $3$. That means that on $(-\\infty,0]$, you always have the same function \"on top\"; likewise, the functions don't cross on $[0,1]$, and the don't cross on $[1,3]$. So, just plug in any number between $0$ and $1$ to see whcih one is bigger on that interval. At $\\frac{1}{2}$, you have $f(\\frac{1}{2}) = 0.875$, $g(\\frac{1}{2}) = 0.25$; since $f(\\frac{1}{2})\\gt g(\\frac{1}{2})$, on $[0,1]$ you have $f\\geq g$, so on $[0,1]$ we have $$\\Bigl| f(x) - g(x) \\Bigr| = f(x)-g(x).$$ To see what happens on $[1,3]$, plug in any number in $[1,3]$; e.g., plugging in $x=2$, we have $f(2) = 2$, $g(2) = 4$, so on this interval we have $g\\geq f$, hence on $[1,3]$ we have $$\\Bigl| f(x)-g(x)\\Bigr| = g(x)-f(x).$$ - My thoughts exactly! :) – wckronholm Jul 28 '11 at 17:26 Arturo and wckronholm have already offered a simple, practical way to determine where $|f(x)-g(x)|$ is $f(x)-g(x)$ and where it’s $g(x)-f(x)$, but you really ought to be familiar with sign analysis as well. As you already discovered, $f(x)-g(x)=x^3-4x^2+3x=x(x-1)(x-3)$. These three factors change sign at $0$, $1$, and", "\\to \\infty} \\frac{3n^2+5n}{4n^2 +2} = \\frac{3}{4}$. Workings: If $n$ is large enough, $3n^2 + 5n$ behaves like $3n^2$ If $n$ ... 37 views ### Prove that every odd cycle is $3$-choosable without using the fact that $\\chi_l(G) \\leq 1+ \\Delta(G)$ Prove that every odd cycle is $3$-choosable without using the fact that $\\chi_l(G) \\leq 1+ \\Delta(G)$ Here is my attempt Let $G$ be an odd cycle from by vertices $(v_1,v_2,\\ldots,v_n,v_1)$. Since $G$... 74 views ### What is the value of this Infinite Product of prime numbers expression? [duplicate] What is the value of: $$\\prod_1^\\infty \\frac{p_i^2}{p_i^2 -1 }$$ Where $$p_i$$ are the prime numbers: 2, 3, ... 28 views ### Interpretation of integral. The height, in centimeters, of a bicycle pedal is given by $h(t)=30+16\\sin t$ where $t$ is the time. Evaluate and interpret the following integral \\begin{align} \\dfrac{1}{2\\pi}\\int_0^{2\\pi} h(t)\\,dt. \\... 127 views ### Suppose $f(x)$ is a rational function such that $3 f \\left( \\frac{1}{x} \\right) + \\frac{2f(x)}{x} = x^2$ for all $x \\neq 0$. Find $f(-2)$. [closed] Suppose $f(x)$ is a rational function such that $$3 f \\left( \\frac{1}{x} \\right) + \\frac{2f(x)}{x} = x^2$$ for all $x \\neq 0$. Find $f(-2)$. 34 views ### Finding a solution to a congruence using Eulers Theorem Suppose $\\phi(m) = 1000$, and I wanted to find a number $a$ thats less", "\\triangle{BPD}$, we have$$\\frac{AP}{AP+30}=\\frac14 \\implies AP=10.$$Next, throw everything on a coordinate plane with $A=(0, 0)$ and $B = (30, 0)$. Then, $P = (-10, 0)$, and if $C = (x, y)$, we have$$(x+10)^2+y^2=64,$$$$x^2+y^2=36.$$Combining these and solving, we get $(x, y)=\\left(-\\frac{18}5, \\frac{24}5\\right)$. Notice now that $P$$C$, and the intersections of the lines $x=6$ (the vertical tangent) with the tangent containing these points are collinear, and thus every slope between a pair of points will have the same slope, which in this case is $\\frac{-\\frac{18}5+10}{\\frac{24}5}=\\frac34$. Thus, the other two vertices of the desired triangle are $(6, 12)$ and $(6, -12)$. By the Shoelace Formula, the area of a triangle with coordinates $(-10, 0)$$(6, 12)$, and $(6, -12)$ is$$\\frac12|-120-0-72-72+0-120|=\\boxed{\\textbf{192}}.$$ ## Problem8 Find the number of positive integers $n \\le 600$ whose value can be uniquely determined when the values of $\\left\\lfloor \\frac n4\\right\\rfloor$$\\left\\lfloor\\frac n5\\right\\rfloor$, and $\\left\\lfloor\\frac n6\\right\\rfloor$ are given, where $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to the real number $x$. ## Solution 1 We need to find all numbers between $1$ and $600$ inclusive that are multiples of $4$$5$, and/or $6$ which are also multiples of $4$$5$, and/or $6$ when $1$ is added to them. We begin by noting that the LCM of $4$$5$, and $6$ is $60$. We can therefore simplify the problem by finding", "$p\\equiv 7\\pmod{8}$. Since those are the only possibilities (even numbers won't work), the conclusion will follow. Consider Gauss's lemma, and let $a=2$, and observe that each $i\\in\\{2,4,6,\\cdots,p-1\\}$ is its own least positive residue modulo $p$. Let $n$ denote the number of integers in the set $\\left\\{\\frac{p+1}{2},\\frac{p+3}{2},\\cdots,p-1\\right\\}$. Then by Gauss's lemma, $\\left(\\frac{2}{p}\\right)=(-1)^n$. Now comes the tricky part: Let $m$ denote the number of even positive integers less than $p/2$. Then $n=[(p-1)/2]-m$. But what is $m$? Well, $m=(p-1)/4$ iff $(p-1)/4$ is an integer; overwise $m=(p-3)/4$. This is the framework that we use for each case. Let's try one in particular... Suppose $p\\equiv 3\\pmod{8}$. Then $\\frac{p-1}{4}$ is not an integer, which means $m=\\frac{p-3}{4}$, and therefore $n=\\frac{p-1}{4}-m=\\frac{p-1}{2}-\\frac{p-3}{4}=\\frac{p+1}{4}$. Since $p\\equiv 3\\pmod{8}$, if follows that $n=\\frac{p+1}{4}$ is odd, which means $\\left(\\frac{2}{p}\\right)=(-1)^n=-1$. Therefore $x^2\\equiv 2\\pmod{p}$ has no solution whenever $p\\equiv 3\\pmod{8}$, and we move on to the next case. 5. It's such a tricky question... So you showed that p 3(mod 8) => no solution How can we prove that THERE IS a solution when e.g. p 7 (mod 8)? What do we have to show? Also, in this question, we are asked to prove IF AND ONLY IF, right? So how can we prove the \"converse\" direction? Thanks for helping! 6. Originally Posted by kingwinner It's such a tricky question... So you showed that", "$$C > \\frac{D}{2}$$). To solve this, I tried one fairly intuitive algorithm: Look at the end-points of the interval $$(C,D)$$ and divide them by 2. Then every integer $$x$$ in this new interval $$(\\frac{C}{2} .. \\frac{D}{2})$$ has a corresponding multiple $$2x$$ in the original interval $$(C,D)$$. So all such numbers are clearly not maximal. A similar fact holds for $$(\\frac{C}{3}..\\frac{D}{3})$$, $$(\\frac{C}{4}..\\frac{D}{4})$$, ..., and so on. So, the algorithm is to try all these intervals, and \"mark\" the numbers $$x$$ that we find. One convenient fact about this is that most of these intervals are disjoint, so we don't need to explicitly mark the $$x$$, but we just count how many $$x$$ are in the range and also in the range $$\\frac{B}{2} .. B$$. When this algorithm terminates, the size of our final set will be $$(D-C+1) + (B - \\max(A-1,\\frac{B}{2})) - |\\{\\text{marked numbers}\\}|$$ (the numbers in our given ranges, minus the number of \"marked\" factors). By inspection, the intervals $$(\\frac{C}{k}..\\frac{D}{k})$$ will eventually intersect with each other. Actually, you can prove that, after a certain $$k$$, the intervals will cover all numbers from $$\\frac{D}{k}$$ down to 1. So we can actually \"exit\" early. This happens whenever $$k > \\frac{D}{D-C}$$ (we proved this in Idea 2), so it happens more quickly whenever $$D-C$$ is large (or when $$D$$ is small," ]

[ "$\\frac {n - 1}{2}x = k\\pi$, or $x = \\frac {2k\\pi}{n - 1}$ for some nonnegative integer $k$. Here $k\\leq \\frac {n - 1}{2}$ so that there are $\\left\\lfloor \\frac {n + 1}{2}\\right\\rfloor$ solutions here. However, we overcount intersections. These occur whenever $$\\frac {2j + 1}{n + 1} = \\frac {2k}{n - 1}$$ $$k = \\frac {(2j + 1)(n - 1)}{2(n + 1)}$$ which is equivalent to $2(n + 1)$ dividing $(2j + 1)(n - 1)$. If $n$ is even, then $(2j + 1)(n - 1)$ is odd, so this never happens. If $n\\equiv 3\\pmod{4}$, then there won't be intersections either, since a multiple of 8 can't divide a number which is not even a multiple of 4. This leaves $n\\equiv 1\\pmod{4}$. In this case, the divisibility becomes $\\frac {n + 1}{2}$ dividing $(2j + 1)\\frac {n - 1}{4}$. Since $\\frac {n + 1}{2}$ and $\\frac {n - 1}{4}$ are relatively prime (subtracting twice the second number from the first gives 1), $\\frac {n + 1}{2}$ must divide $2j + 1$. Since $j\\leq \\frac {n - 1}{2}$, $2j + 1\\leq n < 2\\cdot \\frac {n + 1}{2}$. Then there is only one intersection, namely when $j = \\frac {n - 1}{4}$. Therefore we find $F(n)$ is equal to $1 + \\lfloor n/2 \\rfloor + \\left \\lfloor \\frac {n", "$p\\equiv 7\\pmod{8}$. Since those are the only possibilities (even numbers won't work), the conclusion will follow. Consider Gauss's lemma, and let $a=2$, and observe that each $i\\in\\{2,4,6,\\cdots,p-1\\}$ is its own least positive residue modulo $p$. Let $n$ denote the number of integers in the set $\\left\\{\\frac{p+1}{2},\\frac{p+3}{2},\\cdots,p-1\\right\\}$. Then by Gauss's lemma, $\\left(\\frac{2}{p}\\right)=(-1)^n$. Now comes the tricky part: Let $m$ denote the number of even positive integers less than $p/2$. Then $n=[(p-1)/2]-m$. But what is $m$? Well, $m=(p-1)/4$ iff $(p-1)/4$ is an integer; overwise $m=(p-3)/4$. This is the framework that we use for each case. Let's try one in particular... Suppose $p\\equiv 3\\pmod{8}$. Then $\\frac{p-1}{4}$ is not an integer, which means $m=\\frac{p-3}{4}$, and therefore $n=\\frac{p-1}{4}-m=\\frac{p-1}{2}-\\frac{p-3}{4}=\\frac{p+1}{4}$. Since $p\\equiv 3\\pmod{8}$, if follows that $n=\\frac{p+1}{4}$ is odd, which means $\\left(\\frac{2}{p}\\right)=(-1)^n=-1$. Therefore $x^2\\equiv 2\\pmod{p}$ has no solution whenever $p\\equiv 3\\pmod{8}$, and we move on to the next case. 5. It's such a tricky question... So you showed that p 3(mod 8) => no solution How can we prove that THERE IS a solution when e.g. p 7 (mod 8)? What do we have to show? Also, in this question, we are asked to prove IF AND ONLY IF, right? So how can we prove the \"converse\" direction? Thanks for helping! 6. Originally Posted by kingwinner It's such a tricky question... So you showed that", "get on your own? 5. Oct 4, 2009 ### Susanne217 Yes if f(x) is even defined on the interval [-n,n] then the fourier series is $$n_0 + \\sum_{k=1}^{\\infty} n_k cos(k\\pi x / n)$$ and $$n_0 = \\frac{1}{n} \\int_{0}^{n} f(x) dx$$ $$n_k = \\frac{1}{n} \\int_{0}^{n} f(x) \\cdot cos(k \\pi x /n) dx$$ if f(x) is odd then $$f(x) = \\sum_{k=1}^{\\infty} n_k \\cdot sin(k \\pi x /n)$$ where $$b_k = \\frac{2}{n} \\int_{0}^{n} f(x) \\cdot sin(k \\pi x /n)$$ Is this what you mean? Last edited: Oct 4, 2009 6. Oct 4, 2009 ### jbunniii Yes, let's work with that. Your function is clearly odd, so use the formula for $b_k$. (By the way, there's a typo in your formula for $f(x)$: surely $n_k$ should be $b_k$.) Now plug in $f(x)$ for this particular function. What is $n$ in this case? 7. Oct 4, 2009 ### Susanne217 am I stupid or is n = 1 ?? and I then end up with two version of $$b_k$$ where $$b_{k_1} = \\frac{2}{n=1} \\int_{0}^{1} sin(kx) dx$$ and $$b_{k_2} = \\frac{2}{n=-1} \\int_{0}^{-1} sin(-kx) dx$$ ??? 8. Oct 4, 2009 ### jbunniii No, $n$ is the (one-sided) interval over which the function is defined. Your function is defined on $[-\\pi, \\pi]$, so $n = \\pi$. The formula for $b_k$ tells you to integrate only over $[0,\\pi]$.", "# 2021 AIME I Problems/Problem 7 ## Problem Find the number of pairs $(m,n)$ of positive integers with $1\\le m such that there exists a real number $x$ satisfying$$\\sin(mx)+\\sin(nx)=2.$$ ## Solution 1 The maximum value of $\\sin \\theta$ is $1$, which is achieved at $\\theta = \\frac{\\pi}{2}+2k\\pi$ for some integer $k$. This is left as an exercise to the reader. This implies that $\\sin(mx) = \\sin(nx) = 1$, and that $mx = \\frac{\\pi}{2}+2a\\pi$ and $nx = \\frac{\\pi}{2}+2b\\pi$, for integers $a, b$. Taking their ratio, we have $$\\frac{mx}{nx} = \\frac{\\frac{\\pi}{2}+2a\\pi}{\\frac{\\pi}{2}+2b\\pi} \\implies \\frac{m}{n} = \\frac{4a + 1}{4b + 1} \\implies \\frac{m}{4a + 1} = \\frac{n}{4b + 1} = k.$$ It remains to find all $m, n$ that satisfy this equation. If $k = 1$, then $m \\equiv n \\equiv 1 \\pmod 4$. This corresponds to choosing two elements from the set $\\{1, 5, 9, 13, 17, 21, 25, 29\\}$. There are $\\binom 82$ ways to do so. If $k < 1$, by multiplying $m$ and $n$ by the same constant $c = \\frac{1}{k}$, we have that $mc \\equiv nc \\equiv 1 \\pmod 4$. Then either $m \\equiv n \\equiv 1 \\pmod 4$, or $m \\equiv n \\equiv 3 \\pmod 4$. But the first case was already counted, so we don't need to consider that case. The other case corresponds to", "# Does $n^2$ divide $\\det\\left[\\left(\\frac{i^2+2ij+3j^2}n\\right)\\right]_{1\\le i,j\\le n-1}$ for each odd integer $n>3$? For any odd integer $n>1$ and integers $c$ and $d$, define $$(c,d)_n:=\\det\\left[\\left(\\frac{i^2+cij+dj^2}n\\right)\\right]_{1\\le i,j\\le n-1},$$ where $(\\frac{\\cdot}n)$ is the Jacobi symbol. It is easy to show that $(c,d)_n=0$ if $(\\frac dn)=-1.$ QUESTION: Does $n^2$ divide $(2,3)_n$ for each odd integer $n>3$? My computation led me to conjecture that $n^2\\mid (2,3)_n$ for any odd integer $n>3$ and also $n^2\\mid (6,15)_n$ for any odd integer $n>5$. For any positive integer $n\\equiv\\pm5\\pmod{12}$, we have $(\\frac 3n)=-1$ and hence $(2,3)_n=0$. Also, I observe that $(2,3)_n=0$ if $n>3$ and $n\\equiv 3\\pmod 6$. But I don't see why $(2,3)\\equiv0\\pmod{n^2}$ for each positive integer $n\\equiv\\pm1\\pmod{12}$. Note that $$\\frac{(2,3)_{11}}{11^2}=-2^3\\ \\ \\text{and}\\ \\ \\frac{(2,3)_{37}}{37^2}=2^{44}\\times467^2.$$ I also have some other questions concerning $(c,d)_n$. For example, I conjecture that $(6,1)_n=0$ for any positive integer $n\\equiv 3\\pmod4$. For more such questions, one may consult my preprint On some determinants with Legendre symbol entries. Any ideas helpful to the question ? • You have posted on MO several very interesting questions. This is very refreshing. If I were a specialist, I would make sure to spend a lot of time attempting to solve them. Jun 12, 2018 at 6:17 • It seems that factors of $(2, 3)_n$ all tend to be \"squared\", by which I mean the exponents of most", "# 2007 AMC 12A Problems/Problem 24 ## Problem For each integer $n>1$, let $F(n)$ be the number of solutions to the equation $\\sin{x}=\\sin{(nx)}$ on the interval $[0,\\pi]$. What is $\\sum_{n=2}^{2007} F(n)$? $\\mathrm{(A)}\\ 2014524$ $\\mathrm{(B)}\\ 2015028$ $\\mathrm{(C)}\\ 2015033$ $\\mathrm{(D)}\\ 2016532$ $\\mathrm{(E)}\\ 2017033$ ## Solution ### Solution 1 $F(2)=3$ By looking at various graphs, we obtain that, for most of the graphs $F(n) = n + 1$ Notice that the solutions are basically reflections across $x = \\frac{\\pi}{2}$. However, when $n \\equiv 1 \\pmod{4}$, the middle apex of the sine curve touches the sine curve at the top only one time (instead of two reflected points), so we get here $F(n) = n$. $3+4+5+5+7+8+9+9+\\cdots+2008$ $= (1+2+3+4+5+\\cdots+2008) - 3 - 501$ $= \\frac{(2008)(2009)}{2} - 504 = 2016532$ $\\mathrm{(D)}$ ### Solution 2 $$\\sin nx - \\sin x = 2\\left( \\cos \\frac {n + 1}{2}x\\right) \\left( \\sin \\frac {n - 1}{2}x\\right)$$ So $\\sin nx = \\sin x$ if and only if $\\cos \\frac {n + 1}{2}x = 0$ or $\\sin \\frac {n - 1}{2}x = 0$. The first occurs whenever $\\frac {n + 1}{2}x = (j + 1/2)\\pi$, or $x = \\frac {(2j + 1)\\pi}{n + 1}$ for some nonnegative integer $j$. Since $x\\leq \\pi$, $j\\leq n/2$. So there are $1 + \\lfloor n/2 \\rfloor$ solutions in this case. The second occurs whenever", "= 0.5$, our equation becomes $f(0.5) = \\sum\\limits_{n = -2N}^0 x[n] \\frac{\\sin(\\frac{\\pi}{2} - \\pi n))}{\\pi (0.5 - n) }$ We know from trigonometry that $\\sin(\\frac{\\pi}{2} - \\pi n)$ is positive $1$ if $n$ is even, or negative $1$ if $n$ is odd. We can express this as $-1^n$. So our equation becomes $f(0.5) = \\sum\\limits_{n = -2N}^0 x[n] \\frac{-1^n}{\\pi (0.5 - n) }$ Now, we plug in the fact that our signal $x$ consists of $N$ alternations between $-1$ and $1$, ending at $1$ at time 0. Note that $x[n] = 1$ when $n$ is even, and $x[n] = -1$ when $n$ is odd, since it alternates every sample. We can express this as $x[n] = -1^n$. So, our equation simplifies to $f(0.5) = \\sum\\limits_{n = -2N}^0 -1^n \\frac{-1^n}{\\pi (0.5 - n) }$ Now, note the two $-1^n$ terms cancel: $f(0.5) = \\sum\\limits_{n = -2N}^0 \\frac{1}{\\pi (0.5 - n) }$ This is a formula for the analog level at time $0.5$ that only depends on the number of alternations, $N$. We can plot this series using Wolfram Alpha, and note that the sum diverges. We can also recognize this as a general harmonic series, which all diverge. Knowing that the series diverges means that the more terms we add the more alternations of $-1$ and $1$, the higher", "our answer to be $231+252= \\boxed{483}$. ## Solution 2 By considering the graph of this function, it is shown that the graph is composed of trapezoids ranging from $a^2$ to $(a+1)^2$ with the top made of diagonal line $y=ax$. The width of each trapezoid is $3, 5, 7$, etc. Whenever $a$ is odd, the value of $A(n)$ increases by an integer value, plus $\\frac{1}{2}$. Whenever $a$ is even, the value of $A(n)$ increases by an integer value. Since each trapezoid always has an odd width, every value of $n$ is not an integer when $a \\pmod{4} \\equiv 2$, and is an integer when $a \\pmod{4} \\equiv 0$. Every other value is an integer when $a$ is odd. Therefore, it is simply a matter of determining the number of values of $n$ where $a \\pmod{4} \\equiv 0$ ($(5^2-4^2)+(9^2-8^2)+...+(29^2-28^2)$), and adding the number of values of $n$ where $a$ is odd ($\\frac{(2^2-1^2)+(4^2-3^2)+...+(30^2-29^2)+(1000-31^2)}{2}$). Adding the two values gives $231+252=\\boxed{483}$. ## \"Step\" Solution First, draw a graph of the function. Note that it is just a bunch of line segments. Since we only need to know whether or not $A(n)$ is an integer, we can take the area of each piece from some $x$ to $x+1$ (mod 1), aka the piece from $2$ to $3$ has area $\\frac{1}{2} (\\mod 1)$. There are", "is even} \\\\ 2^{n} \\equiv 2 \\pmod{3} \\quad \\text { if $$n$$ is odd} \\\\ \\end{array}$ Hint Recall the following property of congruences: if $$a \\equiv b \\pmod {n}$$ and $$c \\equiv d \\pmod {n}$$ then $$ac \\equiv bd \\pmod{n}$$. ## Exercise 4 Compute the following sum: $S = \\displaystyle \\left\\lfloor \\frac{1}{3} \\right\\rfloor + \\left\\lfloor \\frac{2^{1}}{3} \\right\\rfloor +\\left\\lfloor \\frac{2^{2}}{3} \\right\\rfloor + \\cdots + \\left\\lfloor \\frac{2^{100}}{3} \\right\\rfloor$ Hint Use Exercise 3. Also recall the formula of the sum of a geometric progression: $\\sum\\limits_{i=0}^{n}x^{i} = \\frac {x^{n+1} -1}{x-1}$ Solution $S=\\frac{1}{3}(2^{101}-2)-50$ ## Exercise 5 Let $$n,m$$ be non-negative integers. Prove that the following expression is always an integer: $\\frac{(2m)!(2n)!}{m!n!(m+n)!}$ Hint Recall the factorial definition of a non-negative integer: $n! = \\begin{cases} n \\cdot (n-1)! \\quad n \\gt 0 \\\\ 1 \\qquad \\qquad n = 0 \\\\ \\end{cases}$ ## Exercise 6 If $$(a,m)=1$$ then $a^{k} \\equiv a^{k \\pmod{\\varphi(m)}} \\pmod{m}$ Hint Use the Euler-Fermat theorem (Wikipedia). ## Exercise 7 Prove that $n^{n^{n^{n}}} – n^{n^{n}} \\equiv 0 \\pmod{9}$ Solution If $$3|n$$ we are done. Otherwise use the Euler-Fermat theorem and Exercise 6. Recalling that $$\\varphi (9) = 6$$, we have to show that $$n ^ {n ^ {n}} – n ^ {n} \\equiv 0 \\pmod {6}$$. Meanwhile, we have $$n ^ {n ^ {n}} – n ^ {n} \\equiv 0 \\pmod {2}$$. Since we", "is neither odd nor even, so we need to take two cases separately : $x$ is non-negative, and $x$ is negative. Case 1: $x$ is non-negative Note that when $x$ is an integer $n$, $f$ will be equal to $6n^2$. So all numbers of the form $6n^2$ can be written in this form. Now let's see how the value of $f$ changes as $x$ varies from $n$ to $n+1$. Since the period of $f$ is $\\text{lcm}( \\frac{1}{2} , \\frac{1}{3}) = 1$, we can be sure that the same pattern will be followed as $x$ goes from $n+1$ to $n+2$. As we saw above, $f$ only changes when $x$ crosses and integral multiple of $\\frac{1}{6}$, we can divide the interval $[n, n+1)$ into $6$ equal intervals. $$\\begin{array}{|l|c|c|c|}\\hline \\text{Interval} & \\lfloor 2x \\rfloor & \\lfloor 3x \\rfloor & f(x) \\\\\\hline \\left[n, n + \\frac{1}{6}\\right) & 2n & 3n & 6n^{2} \\\\ \\left[n + \\frac{1}{6}, n + \\frac{2}{6}\\right) & 2n & 3n & 6n^{2}\\\\ \\left[n + \\frac{2}{6}, n + \\frac{3}{6}\\right) & 2n & 3n + 1& 6n^{2} + 2n\\\\ \\left[n + \\frac{3}{6}, n + \\frac{4}{6}\\right) & 2n + 1& 3n + 1& 6n^{2} + 5n + 1\\\\ \\left[n + \\frac{4}{6}, n + \\frac{5}{6}\\right) & 2n + 1& 3n + 2& 6n^{2} + 7n + 2\\\\ \\left[n + \\frac{5}{6}, n + 1\\right)", "and $f,g$ are monotonic as well as continuous on the interval $[a,b]$, then $f(x). Now suppose that this is true, it can be easily verified that $-\\int_0^{\\frac{\\pi}{4}}\\cos(x)\\ln(\\sin(x))dx=\\frac{ \\sqrt{2}\\ln(2)}{4}+\\frac{\\sqrt{2}}{2}$ $>-\\int_0^{\\frac{\\pi}{4}}\\sin(x)\\ln(\\cos(x))dx=1-\\frac{\\sqrt{2}\\ln(2)}{4}-\\frac{\\sqrt{2}}{2}$ Now supposing that my \"made up\" lemma is correct, this proves the inequality. unfortunately your made up lemma is not always true, e.g. $f(x)=\\frac{x+1}{4}, \\ g(x)=x.$ then $\\int_0^1 f(x) \\ dx < \\int_0^1 g(x) \\ dx,$ but neither $f < g$ nor $f > g$ on the unit interval. the trouble here comes from this fact that $f$ and $g$ intersect! so you need to assume that $f,g$ do not intersect in the interval. also i don't think we need $f,g$ to be positive. 7. Originally Posted by NonCommAlg unfortunately your made up lemma is not always true, e.g. $f(x)=\\frac{x+1}{4}, \\ g(x)=x.$ then $\\int_0^1 f(x) \\ dx < \\int_0^1 g(x) \\ dx,$ but neither $f < g$ nor $f > g$ on the unit interval. the trouble here comes from this fact that $f$ and $g$ intersect! so you need to assume that $f,g$ do not intersect in the interval. also i don't think we need $f,g$ to be positive. Dangit! I forgot to say that. I knew that they cannot intersect...and no we do not need them positive we need $fg>0$. I have come up with a \"proof\"", "# Difference between revisions of \"2008 AIME II Problems/Problem 8\" ## Problem Let $a = \\pi/2008$. Find the smallest positive integer $n$ such that $$2[\\cos(a)\\sin(a) + \\cos(4a)\\sin(2a) + \\cos(9a)\\sin(3a) + \\cdots + \\cos(n^2a)\\sin(na)]$$ is an integer. ## Solution ### Solution 1 By the product-to-sum identities, we have that $2\\cos a \\sin b = \\sin (a+b) - \\sin (a-b)$. Therefore, this reduces to a telescoping series: \\begin{align*} \\sum_{k=1}^{n} 2\\cos(k^2a)\\sin(ka) &= \\sum_{k=1}^{n} [\\sin(k(k+1)a) - \\sin((k-1)ka)]\\\\ &= -\\sin(0) + \\sin(2a)- \\sin(2a) + \\sin(6a) - \\cdots - \\sin((n-1)na) + \\sin(n(n+1)a)\\\\ &= -\\sin(0) + \\sin(n(n+1)a) = \\sin(n(n+1)a) \\end{align*} Thus, we need $\\sin \\left(\\frac{n(n+1)\\pi}{2008}\\right)$ to be an integer; this can be only $\\{-1,0,1\\}$, which occur when $2 \\cdot \\frac{n(n+1)}{2008}$ is an integer. Thus $1004 = 2^2 \\cdot 251 | n(n+1) \\Longrightarrow 251 | n, n+1$. It easily follows that $n = \\boxed{251}$ is the smallest such integer. ### Solution 2 We proceed with complex trigonometry. We know that for all $\\theta$, we have $\\cos \\theta = \\dfrac{1}{2} \\left( z + \\dfrac{1}{z} \\right)$ and $\\sin \\theta = \\dfrac{1}{2i} \\left( z - \\dfrac{1}{z} \\right)$ for some complex number $z$. Similarly, we have $\\cos n \\theta = \\dfrac{1}{2} \\left( z^n + \\dfrac{1}{z^n} \\right)$ and $\\sin n \\theta = \\dfrac{1}{2i} \\left(z^n - \\dfrac{1}{z^n} \\right)$. Thus, we have $\\cos n^2 a \\sin n a = \\dfrac{1}{4i} \\left( z^{n^2} +", "0}[\\frac{\\tan x}{x}] =[\\lim_{x\\to 0}\\frac{\\tan x}{x}]$ Would appreciate if one describes in detail 6. Well, have you looked at the graphs. 8. you don't need to sketch a graph. choose $0 < x < \\frac{\\pi}{2}.$ then if you apply mean value theorem for $f(t)=\\sin t$ on the interval $[0,x],$ then you get $0 < \\frac{\\sin x}{x} < 1.$ thus, since $\\frac{\\sin x}{x}$ is an even function, for any non-zero $x$ in the interval $[-\\pi/2, \\pi/2]$ we must have: $0< \\frac{\\sin x}{x} < 1,$ which gives us: $\\lfloor{\\frac{\\sin x}{x} \\rfloor}=0,$ for any $x \\neq 0$ in that interval. thus $\\lim_{x\\to0} \\lfloor{\\frac{\\sin x}{x} \\rfloor}= \\lim_{x\\to 0} 0 = 0.$ but $\\lfloor{\\lim_{x\\to0} \\frac{\\sin x}{x} \\rfloor}=\\lfloor{1 \\rfloor}=1.$ a similar argument works for $\\frac{\\tan x}{x}.$ 9. This is precisely what I wanted to know that while evaluating $ [\\lim_{x\\to 0}\\frac{\\sin x}{x}] $ we are required to first find out $ \\lim_{x\\to 0}\\frac{\\sin x}{x} $ and then apply rule of greatest integer function(I would want to know the reason.Is there a rigorous proof?) Or should we not say that since $0<\\frac{\\sin x}{x}<1$ the greatest integer should anyway be 0 whether we find $ [\\lim_{x\\to 0}\\frac{\\sin x}{x}] $ or we find $ \\lim_{x\\to 0}[\\frac{\\sin x}{x}] $ I am perplexed because in case of $\\frac{\\tan x}{x}$ the answer for $ [\\lim_{x\\to 0}\\frac{\\tan x}{x}]=1 $ $ \\lim_{x\\to 0}[\\frac{\\tan", "$f(0.5) = \\sum\\limits_{n = -2N}^0 x[n] \\frac{\\sin(\\frac{\\pi}{2} - \\pi n))}{\\pi (0.5 - n) }$ We know from trigonometry that $\\sin(\\frac{\\pi}{2} - \\pi n)$ is positive $1$ if $n$ is even, or negative $1$ if $n$ is odd. We can express this as $-1^n$. So our equation becomes $f(0.5) = \\sum\\limits_{n = -2N}^0 x[n] \\frac{-1^n}{\\pi (0.5 - n) }$ Now, we plug in the fact that our signal $x$ consists of $N$ alternations between $-1$ and $1$, ending at $1$ at time 0. Note that $x[n] = 1$ when $n$ is even, and $x[n] = -1$ when $n$ is odd, since it alternates every sample. We can express this as $x[n] = -1^n$. So, our equation simplifies to $f(0.5) = \\sum\\limits_{n = -2N}^0 -1^n \\frac{-1^n}{\\pi (0.5 - n) }$ Now, note the two $-1^n$ terms cancel: $f(0.5) = \\sum\\limits_{n = -2N}^0 \\frac{1}{\\pi (0.5 - n) }$ This is a formula for the analog level at time $0.5$ that only depends on the number of alternations, $N$. We can plot this series using Wolfram Alpha, and note that the sum diverges. We can also recognize this as a general harmonic series, which all diverge. Knowing that the series diverges means that the more terms we add the more alternations of $-1$ and $1$, the higher the true peak will be.", "by the assumption $I := \\inf_n \\int_0^1 F_n(x) dx > 0$ (here we choose suitable subsequence of $(F_n)$). Let $s_n$ be a lower Riemann sum of $F_n$ such that $s_n \\geq \\frac{3}{4}I$. Than if $u_n$ is the sum of lengths of intervals of the partition of $[0,1]$ such that $F_n(x) > \\frac{1}{2} I$ (on that intervals) then by $F_n(x) < 1$ and the definition of a lower sum we obtain $$\\frac{3}{4} I \\leq s_n \\leq u_n + \\frac{1}{2} I (1 - u_n) = \\frac{1}{2} I + u_n (1-\\frac{1}{2} I).$$ Hence $u_n \\geq \\frac{1}{4 I (1-I/2)} > 0$. Let $U_n$ be the set of interiors of intervals defined by $u_n$. Define $V_n = \\bigcup_{k=n}^\\infty U_n$. Then each $V_n$ is open and the sequence $(V_n)$ is decreasing with $\\inf_n |V_n| > 0$. Hence $\\bigcap_n V_n \\neq \\emptyset$ and there exists $x_0$ such that $x_0 \\in \\bigcap_n V_n$. Then $F_n(x_0) > \\frac{1}{2} I$ thus $F_n(x_0) \\not\\to 0$ and we obtain a contradiction. -", "and the latter is the leg). Then use the fact that $\\displaystyle\\sum_{n=1}^\\infty\\frac{\\sin n}n$ converges to $\\dfrac{\\pi-1}2$, which you can prove by using either Euler's formula, or the alternating series test. Dec 25, 2013 at 2:22 • Achtung, @Lucian: $\\sin(\\sin x)<\\sin(x)$ only if $\\sin x> 0$, i.e. only if $\\left\\lfloor\\frac{x}{\\pi}\\right\\rfloor$ is even! Dec 28, 2013 at 13:10 To prove the convergence it is sufficient to show that $$\\left|\\sum_{n=1}^{+\\infty}\\frac{\\sin(kn)}{n}\\right|= \\left|\\frac{\\pi-k}{2}+\\pi \\left\\lfloor\\frac{k}{2\\pi}\\right\\rfloor\\right| < \\frac{\\pi}{2} \\tag{1}$$ by regarding the LHS as the imaginary part of a geometric series and by applying Abel's lemma. Then we have that the Fourier coefficients of $\\sin(\\sin x)$ decay pretty fast since $\\sin(\\sin x)\\in C^3([-\\pi,\\pi])$. If we are allowed to write $\\sin(\\sin x)$ as its Fourier series and to switch the sums (this is crucial) we have: $$\\left|\\sum_{n=1}^{+\\infty}\\frac{\\sin(\\sin n)}{n}\\right|\\leq K\\sum_{k=1}^{+\\infty}\\frac{\\pi}{k^2},$$ for istance. I am expecting that the same holds for $\\sigma_k(x)$, since, in general, if $f(x)$ is an odd periodic function belonging to $C^{3}(\\mathbb{R})$ and we are allowed to switch sums, $$\\sum_{n=1}\\frac{f(n)}{n}$$ converges because the $k$-th Fourier coefficient of $f(x)$ is $o(k^{-2})$ and $(1)$ holds. By using Bessel functions we can write: $$\\sin(\\sin n)=\\sum_{k=0}^{+\\infty}2\\cdot J_{2k+1}(1)\\sin((2k+1) n),\\tag{2}$$ where the convergence is uniform, but now we have to justify the sum-switch procedure. This is more-or-less the same as proving that the partial sums of $\\sin(\\sin n)$ are bounded. Exploiting", "with $n$ gridpoints then $$I_{n} = \\frac{1}{n}\\sum_{i=1}^{n}\\frac{f(\\frac{i-1}{n})+f(\\frac{i}{n})}{2} \\implies I_n = \\left\\{\\matrix{\\frac{1}{2} & n~~\\text{odd}\\\\\\frac{1}{2} - \\frac{1}{n} & n~~\\text{even}}\\right.$$ so for $n=3$ we have the exact answer which is better than any even $n$ no matter how large it is. This shows that increasing the number of gridpoints does not always improve the accuracy. With Simpson's rule we find $$I_n = \\frac{1}{3n}\\sum_{i=1}^{n/2}f\\left(\\frac{2i-2}{n}\\right)+4f\\left(\\frac{2i-1}{n}\\right)+f\\left(\\frac{2i}{n}\\right) \\implies I_n = \\left\\{\\matrix{\\frac{1}{2} - \\frac{1}{3n}&n\\equiv 0\\mod 4\\\\\\frac{1}{2}&n\\equiv 1\\mod 4\\\\\\frac{1}{2} + \\frac{2}{3n} & n\\equiv 2\\mod 4\\\\\\frac{1}{2} - \\frac{5}{6n} & n\\equiv 3\\mod 4}\\right.$$ so even if Simpson's rule has higher order we see that it does not always do better than the trapezoidal rule. $$\\bf \\text{What does higher degree of precision really mean?}$$ If we have a smooth function then a standard Taylor series error analysis gives that the error in estimating the integral $\\int_a^bf(x){\\rm d}x$ using $n$ equally spaced points is bounded by (here for Simpsons and the trapezoidal rule) $$\\epsilon_{\\rm Simpsons} = \\frac{(b-a)^5}{2880n^4}\\max_{\\zeta\\in[a,b]}|f^{(4)}(\\zeta)|$$ $$\\epsilon_{\\rm Trapezoidal} = \\frac{(b-a)^3}{12n^2}\\max_{\\zeta\\in[a,b]}|f^{(2)}(\\zeta)|$$ Note that the result we get from such an error analysis is always an upper bound (or in some cases an order of magnitude) for the error apposed to the exact value for the error. What this error analysis tell us is that if $f$ is smooth on $[a,b]$, so that the derivatives are bounded, then the error with a", "# Difference between revisions of \"2007 AMC 12A Problems/Problem 24\" ## Problem For each integer $n>1$, let $F(n)$ be the number of solutions to the equation $\\sin{x}=\\sin{(nx)}$ on the interval $[0,\\pi]$. What is $\\sum_{n=2}^{2007} F(n)$? $\\mathrm{(A)}\\ 2014524$ $\\mathrm{(B)}\\ 2015028$ $\\mathrm{(C)}\\ 2015033$ $\\mathrm{(D)}\\ 2016532$ $\\mathrm{(E)}\\ 2017033$ ## Solution $F(2)=3$ By looking at various graphs, we obtain that, for most of the graphs $F(n) = n + 1$ However, when $n \\equiv 1 \\pmod{4}$, the middle apex of the sine curve touches the sine curve at the top only one time (instead of two), so we get here $F(n) = n$. $3+4+5+5+7+8+9+9+\\cdots+2008$ $= (1+2+3+4+5+\\cdots+2008) - 3 - 501$ $= \\frac{(2008)(2009)}{2} - 504 = 2016532$ $\\mathrm{(D)}$", "$x[n]$, that if $N$ is even $$x[n] = \\frac1N \\sum\\limits_{k=-\\frac{N}{2}}^{\\frac{N}{2}-1} X[k] \\ e^{j 2 \\pi nk/N}$$ if $N$ is odd, then $$x[n] = \\frac1N \\sum\\limits_{k=-\\frac{N-1}{2}}^{\\frac{N-1}{2}} X[k] \\ e^{j 2 \\pi nk/N}$$ and you can sorta plug in $n = f_\\text{s} t$ to reconstruct bandlimited $x(t)$ outa $x[n]$. in the $N$ even case, then you have an extra term at Nyquist, the $k=-\\frac{N}{2}$ term, for that you cannot know both the amplitude and phase of that sinusoid. you must assume one or the other. often people only keep the $\\cos(\\pi f_\\text{s} t)$ term at Nyquist. \\begin{align} x[n] & = \\frac1N \\sum\\limits_{k=-\\frac{N}{2}}^{\\frac{N}{2}-1} X[k] \\ e^{j 2 \\pi nk/N} \\\\ & = \\frac1N \\left( X[0] + X[N/2] \\cos(\\pi n) + \\sum\\limits_{k=1}^{\\frac{N}{2}-1} X[k] e^{j 2 \\pi nk/N} \\ + \\ X[-k] e^{-j 2 \\pi nk/N} \\right) \\\\ & = \\frac1N \\left( X[0] + X[N/2] \\cos(\\pi n) \\right) \\\\ & \\quad + \\frac1N \\sum\\limits_{k=1}^{\\frac{N}{2}-1} (\\Re\\{X[k]\\} + j \\Im\\{X[k]\\}) e^{j 2 \\pi nk/N} \\ + \\ (\\Re\\{X[k]\\} - j \\Im\\{X[k]\\}) e^{-j 2 \\pi nk/N} \\\\ & = \\frac1N \\left( X[0] + X[N/2] \\cos(\\pi n) \\right) \\\\ & \\quad + \\frac1N \\sum\\limits_{k=1}^{\\frac{N}{2}-1} 2\\Re\\{X[k]\\} \\frac{e^{j 2 \\pi nk/N} + e^{-j 2 \\pi nk/N}}{2} + j (2j) \\Im\\{X[k]\\} \\frac{e^{j 2 \\pi nk/N} - e^{-j 2 \\pi nk/N}}{2j} \\\\ & = \\frac1N \\left( X[0] + X[N/2] \\cos(\\pi n)", "Let $n$ be a positive integer. Prove that, for $0<x<\\frac\\pi{n+1}$, $$\\sin{x}-\\frac{\\sin{2x}}{2}+\\cdots+(-1)^{n+1}\\frac{\\sin{nx}}{n}-\\frac{x}{2}$$ is positive if $n$ is odd and negative if $n$ is even. Since \\begin{align*}f_n(x) &= \\sin{x} - \\frac {\\sin{2x}}{2} + \\cdots + ( - 1)^{n + 1}\\frac {\\sin{nx}}{n} - \\frac {x}{2},\\\\f_n'(x) &= - \\mbox{Re}\\left(\\sum_{n = 1}^{n}z^n\\right) - \\frac12.\\end{align*} After some simplifications we get $$f_n'(x) = \\frac {( - 1)^{n + 1}}{2}((1 - \\cos(x))\\frac {\\sin((n + 1)x)}{\\sin(x)} + \\cos((n + 1)x))$$ and $$f_n''(x) = \\frac {( - 1)^{n}}{2}\\frac {(n + 1)\\sin(nx) + n\\sin((n + 1)x)}{1 + \\cos(x)}.$$ The formula for $f_n''$ shows that $( - 1)^n f$ is convex for $0 < x < \\frac {\\pi}{n + 1}$. Since $f_n(0) = 0$ and $f_n'(0) = \\frac {( - 1)^{n + 1}}{2}$.We are ready when we can show that $( - 1)^{n + 1}f_n(\\frac {\\pi}{n + 1}) > 0$. We have to distinct between two different, but very similar cases, namely $n$ is odd, and $n$ is even. Let's restrict to the case $n$ is even. We prove $f_{2n}(\\frac {\\pi}{2n + 1}) < 0$. \\begin{align*}f_{2n}\\left( \\frac{\\pi}{2n+1} \\right) &=\\sum_{k=1}^{2n}{\\left( -1 \\right)}^{k+1}\\frac{\\sin \\left( \\frac{k\\pi}{2n+1} \\right)}{k}-\\frac{\\pi}{2\\left( 2n+1 \\right)}\\\\&=\\frac{\\pi}{2n+1}\\left( \\sum_{k=1}^n{\\frac{\\sin \\left( \\frac{\\left( 2k-1 \\right) \\pi}{2n+1} \\right)}{\\frac{\\left( 2k-1 \\right) \\pi}{2n+1}}}-\\sum_{k=1}^n{\\frac{\\sin \\left( \\frac{2k\\pi}{2n+1} \\right)}{\\frac{2k\\pi}{2n+1}}} \\right) -\\frac{\\pi}{2\\left( 2n+1 \\right)}.\\end{align*} The function $x \\mapsto \\frac {\\sin(x)}{x}$ is descending on $[0,\\pi]$, thus both sums lay between $a$ and" ]

[ "a b= \\pi \\sqrt{1/\\lambda_1}\\sqrt{1/\\lambda_2} = \\frac{\\pi}{\\sqrt{\\det A}}$$. I will give one final example where the matrices can be ordered: The two matrices in this case were: $$A =\\begin{pmatrix}2/3 & 1/5 \\\\ 1/5 & 3/4\\end{pmatrix}, \\quad B=\\begin{pmatrix} 1& 1/7 \\\\ 1/7& 1 \\end{pmatrix}$$", "resulting matrices: $\\Rightarrow \\begin{pmatrix} p + q & 2p\\\\-p&3p+q\\end{pmatrix}=\\begin{pmatrix}1 & -8\\\\4 & -7\\end{pmatrix}$ Then compare the elements in corresponding positions to find $p$ and $q$. Start with $2p = -8$ ...", "\\end{pmatrix}^T$$ Let $$A=\\begin{pmatrix}2 &1 \\\\6 & 4 \\end{pmatrix}$$ a) Express $$A^{-1}$$ as a product of elementary matrices", "# Find matrices $P$ and $Q$ to check if matrices $A$ and $B$ are equivalent( $Q^{-1}AP = B$ )? \\begin{align}&A=\\begin{pmatrix} 6 & -3 & -6 & -12 & 9 & 9 & 0 & 3 & -6 & 9 \\\\ 12 & 4 & 8 & -4 & 8 & -12 & 4 & 4 & 8 & -4\\\\ \\end{pmatrix} \\\\\\\\ & B = \\begin{pmatrix} 4 & -4 & 0 & -8 & -12 & 8 & -4 & -8 & 4 & -12 \\\\ -9 & -3 & -6 & 3 & -6 & 9 & -3 & -3 & -6 & 3\\\\ \\end{pmatrix} \\end{align} I need to find out if $A$ and $B$ are equivalent or not. To do that I have rearranged the definition of matrix equivalence: \\begin{align} Q^{-1}AP & = B \\\\\\\\ AP & = QB \\end{align} So now I have $\\begin{pmatrix} 6 & -3 & -6 & -12 & 9 & 9 & 0 & 3 & -6 & 9 \\\\ 12 & 4 & 8 & -4 & 8 & -12 & 4 & 4 & 8 & -4\\\\ \\end{pmatrix} \\cdot \\begin{pmatrix} x_{1,1} & \\cdots & x_{1,2} \\\\ \\vdots & & \\vdots \\\\ x_{10,1}& \\cdots & x_{10,2} \\end{pmatrix} = \\begin{pmatrix} y_{1,1} & \\cdots & y_{1,2} \\\\ \\vdots & & \\vdots \\\\ y_{10,1}&", "Free Version Easy # 2x2 Matrix Inverse - 2, 3, 1, 2 LINALG-VXRYQE Which of the following is the inverse of the matrix: $$\\begin{pmatrix} 2&3\\\\\\ 1&2\\end{pmatrix}$$ A $\\begin{pmatrix}2&3\\\\\\ 1&2\\end{pmatrix}$ B $\\begin{pmatrix}2&-3\\\\\\ -1&2\\end{pmatrix}$ C $\\begin{pmatrix}-2&1\\\\\\ 3&-2\\end{pmatrix}$ D $\\begin{pmatrix}-2&3\\\\\\ 1&-2\\end{pmatrix}$ E $\\begin{pmatrix}2&-1\\\\\\ -3&2\\end{pmatrix}$", "# Number of matrices Of all matrices $$\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} ,$$ with $$a$$, $$b$$, $$c$$, and $$d$$ four distinct integers from $$0$$ to $$9,$$ how many matrices do not have an inverse? ×", "& 3 & 3 & -1 \\\\ 3 & 2 & 4 & -3 \\\\ 2 & -2 & 2 & 3 \\end{pmatrix}\\,.} { } } {} {} \\inputexercise {2} { Compute the determinant of the elementary matrices. } {} {} \\inputexercise {5} { Check the multiplication theorem for determinants of the following matrices \\mathdisp {A = \\begin{pmatrix} 3 & 4 & 7 \\\\ 2 & 0 & -1 \\\\1 & 3 & 4 \\end{pmatrix} \\text{ and } B = \\begin{pmatrix} -2 & 1 & 0 \\\\ 2 & 3 & 5 \\\\2 & 0 & -3 \\end{pmatrix}} { . } } {} {}", "## Matrix od order n $$\\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & … & n \\\\ -1 & 0 & 2 & 3 & 4 & … & n-1 \\\\ -1 & -2 & 0 & 3 & 4 & … & n-1 \\\\ -1 & -2 & -3 & 0 & 4 & … & n-1 \\\\ -1 & -2 & -3 & -4 & 0 & … & n-1 \\\\ \\vdots & \\vdots & \\vdots & \\vdots & & \\ddots & \\\\ -1 & -2 & -3 & -4 & … & 1-n & 0 \\\\ \\end{pmatrix}$$. The determinant of the matrix is $$n!$$.", "find that the image in this case is spanned by the matrices $$\\begin{pmatrix} 1 & 0\\\\2 & 0\\end{pmatrix} \\quad \\text{ and } \\quad \\begin{pmatrix} 0 & 1\\\\0 & 2\\end{pmatrix}.$$", "\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} & \\begin{pmatrix} 1 & 2 \\\\ 3 & 4 \\end{pmatrix} & \\begin{pmatrix} \\alpha & \\beta \\\\ \\gamma & \\delta \\end{pmatrix} \\\\ \\begin{pmatrix} e & f \\\\ g & h \\end{pmatrix} & \\begin{pmatrix} 5 & 6 \\\\ 7 & 8 \\end{pmatrix} & \\begin{pmatrix} \\epsilon & \\zeta \\\\ \\eta & \\theta \\end{pmatrix} \\end{pmatrix}$ is dual to $X = \\begin{pmatrix} Y_{11} & Y_{12} \\\\ Y_{21} & Y_{22} \\end{pmatrix} = \\begin{pmatrix} \\begin{pmatrix} a & 1 & \\alpha\\\\ e & 5 & \\epsilon \\end{pmatrix} & \\begin{pmatrix} b & 2 & \\beta\\\\ f & 6 & \\zeta \\end{pmatrix} \\\\ \\begin{pmatrix} c & 3 & \\gamma\\\\ g & 7 & \\eta \\end{pmatrix} & \\begin{pmatrix} d & 4 & \\delta\\\\ h & 8 & \\theta \\end{pmatrix} \\end{pmatrix}$ . $b_{11,11} = a = y_{11,11}~$ $b_{23,21} = \\eta = y_{21,23}~$ This concept is not limited to matrices, as the following example shows. The join table is a 24×24 matrix. The inversion sets could be displayed by 6-bit binary vectors, but a free arrangement was chosen, to serve symmetry. ## Gould's-Morse sequence This portmanteau name I use for the following sequence, which occurs when dealing with Walsh matrices: 1, -2, -2, 4, -2, 4, 4, -8, -2, 4, 4, -8, 4, -8, -8, 16, -2, 4, 4, -8, 4, -8,", "7 respectively the inverse matrices are: $$M^{-1}_{10,3}=\\begin{bmatrix} 1&2\\\\ 1& 1 \\end{bmatrix}$$ $$M^{-1}_{10,3}=\\begin{bmatrix} 5.1&5.4\\\\ 5.1& 5.1 \\end{bmatrix}$$$$=\\begin{bmatrix} 5&6\\\\ 5& 5 \\end{bmatrix}$$ Is this correct then?", "Browse Questions # For what value of k , the matrix $\\begin{bmatrix} k & \\quad 2 \\\\ 3 & \\quad 4 \\end{bmatrix}$ has no inverse. Can you answer this question?", "$$e_1 := \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}\\quad\\textrm{and}\\quad e_2 := \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix}.$$ In this basis $$T = \\begin{pmatrix} 2 & -1 \\\\ 1 & 1 \\end{pmatrix}.$$ The basis transformation matrices from $$(e_1,e_2)$$ to $$B_1$$ or $$B_2$$ are given by $$B_1 = \\begin{pmatrix} 1 & 1 \\\\ 2 & 1 \\end{pmatrix}\\quad\\textrm{and}\\quad B_2 = \\begin{pmatrix} -1 & 2 \\\\ 1 & 1 \\end{pmatrix}.$$ The transformation matrix in each basis will be given by $$TB_1^{-1} \\quad\\textrm{and}\\quad TB_2^{-1}.$$ • Ah so there is no $[u]$ required. But if $T$ had a vector, say $u = (2,3)$ then it would have to be multiplied to that formula? Jun 10, 2020 at 3:36", "1 & 1 & 1 & 3 & 3 & 3 & 3 & 3 \\end{array} \\right)$$ The rest is straightforward. • #### Result The inverse matrix is: $$\\begin{pmatrix} 7 & 5 & 5 & 5 & 3\\\\ 5 & 5 & 5 & 3 & 7\\\\ 5 & 5 & 3 & 7 & 5\\\\ 5 & 3 & 7 & 5 & 5\\\\ 3 & 7 & 5 & 5 & 5 \\end{pmatrix}$$", "b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{pmatrix} = \\begin{pmatrix}a_{11} + b_{11}& a_{12} + b_{12} \\\\ a_{21} + b_{21} & a_{22} + b_{22} \\end{pmatrix}, $A - B = \\begin{pmatrix}a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix} - \\begin{pmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{pmatrix} = \\begin{pmatrix}a_{11} - b_{11}& a_{12} - b_{12} \\\\ a_{21} - b_{21} & a_{22} - b_{22} \\end{pmatrix}$. Learn how to add, subtract, and multiply matrices, and find the inverses of matrices. & = \\begin{pmatrix} 2 + (-2) & 3 + 0 \\\\ 1 + 7 & -5 + (-1) \\end{pmatrix}\\\\ 2nd Grade Addition Worksheets Addition Worksheets Third Grade Math Review Educational Websites For Third Graders Division Practice Worksheets 3rd Grade 3rd Standard Math Syllabus Multiplication Worksheets 3rd Grade Timed Test A comprehensive set of worksheets covering a variety of subjects can be used to expand your child has learning experience. 2 numbers which will help you to learn how to multiply a with. Be same ) \\ ( \\ ) entry by entry B = a... By that number school students and find the inverses of matrices work some items at three hamburger shops matrix the... By rather than divide X= ( B-A ) both sides by rather than divide X= ( ). To learn how addition, subtraction and scalar multiplication with a", "result of the multiplication. The same is true if we replace $A$ by one of the following matrices: $$B:=\\begin{pmatrix} 1 & 2 & 2\\\\ 2 & 1 & 2\\\\ 2 & 2 & 3 \\end{pmatrix} \\quad \\text{or}\\quad C:=\\begin{pmatrix} -1 & 2 & 2\\\\ -2 & 1 & 2\\\\ -2 & 2 & 3 \\end{pmatrix}.$$ Taking $x=(3,4,5)$ as initial triple, we can use the matrices $A$, $B$ and $C$ to construct a tree with all primitive Pythagorean triples (without repetition) as follows: $$x\\left\\{\\begin{matrix} Ax\\left\\{\\begin{matrix} AAx\\cdots\\\\ BAx\\cdots\\\\ CAx\\cdots \\end{matrix}\\right.\\\\ \\\\ Bx\\left\\{\\begin{matrix} ABx\\cdots\\\\ BBx\\cdots\\\\ CBx\\cdots \\end{matrix}\\right.\\\\ \\\\ Cx\\left\\{\\begin{matrix} ACx\\cdots\\\\ BCx\\cdots\\\\ CCx\\cdots \\end{matrix}\\right. \\end{matrix}\\right.$$ Source: Wikipedia’s page Tree of primitive Pythagorean triples.", "0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ -V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix}$ The matrices really correspond to the L and U matrices in the LDU decomposition, noting that the matrix on the right or left is just the transpose of the other. Define the Schur complement of the matrix V with respect to the block V22 as $V_{11.2}=V_{11}-V_{12}V_{22}^{-1}V_{21}$ then working through the above multiplication it follows that $\\begin{pmatrix} 1 & -V_{12}V_{22}^{-1}\\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ -V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix} = \\begin{pmatrix} V_{11.2} & 0 \\\\ 0 & V_{22} \\\\ \\end{pmatrix}$ This result is useful in deriving the Sherman–Morrison–Woodbury matrix inverse formula/identity, otherwise known as the matrix inversion lemma. One can now simply multiply the above equation by the corresponding inverse matrices to obtain the LDU decomposition of the matrix V. The inverse matrices are easy $\\begin{pmatrix} 1 & V_{12}V_{22}^{-1}\\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} 1 & -V_{12}V_{22}^{-1}\\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ -V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix} = \\begin{pmatrix} 1 & V_{12}V_{22}^{-1} \\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11.2} & 0 \\\\", "## Algebra 2 (1st Edition) $$\\begin{pmatrix}10&7\\\\ -3&-6\\end{pmatrix}$$ Adding the matrices by adding corresponding elements, we find: $$\\begin{pmatrix}4+6&\\left(-5\\right)+12\\\\ 1+\\left(-4\\right)&\\left(-2\\right)+\\left(-4\\right)\\end{pmatrix} \\\\ \\begin{pmatrix}10&7\\\\ -3&-6\\end{pmatrix}$$", "+0 # Matrices 0 69 1 Given $$\\mathbf{A}^{-1} = \\begin{pmatrix} 2 & 3 \\\\ -1 & 0 \\end{pmatrix} \\quad \\text{and} \\quad \\mathbf{B}^{-1} = \\begin{pmatrix} 0 & 4 \\\\ 1 & 5 \\end{pmatrix} ,$$ find the inverse of AB. Thanks! Apr 2, 2021 #1 +22020 +2 1) Find the inverse of A-1; this will be A. 2) Find the inverse of B-1; this will be B. 3) Find the product of A and B; this will be AB. 4) Lastly, find the inverse of AB (your answer to step 3). Apr 2, 2021", "Enable text based alternatives for graph display and drawing entry Try Another Version of This Question Given the matrices A=[(-4,-2),(-1,-1),(-1,-2)] and B=[(3,-1),(0,5)] A. It is to multiply these matrices in this order AB. B. If multiplication is possible what is AB" ]

[ "+0 # Matrices Solving -1 200 1 +116 Let $${A} = \\begin{pmatrix} 0 & 5 & 7 \\\\ -2 & 7 & 7 \\\\ -1 & 1 & 4 \\end{pmatrix}$$. Find all values of $$t$$ such that $$\\det (t{I} - {A}) = 0$$, where $$I$$ is the $$3 \\times 3$$ identity matrix. May 30, 2020", "Question # Find the product AB of the two matrices listed below: A=begin{pmatrix}2 & 1&3 -2 & 2&4-1&-3&-4 end{pmatrix} B=begin{pmatrix}-1 & -21 & 23&4 end{pmatrix} Matrices Find the product AB of the two matrices listed below: $$A=\\begin{pmatrix}2 & 1&3 \\\\-2 & 2&4\\\\-1&-3&-4 \\end{pmatrix}$$ $$B=\\begin{pmatrix}-1 & -2\\\\1 & 2\\\\3&4 \\end{pmatrix}$$ $$AB=\\begin{bmatrix}2 & 1&3 \\\\-2 & 2&4\\\\-1&-3&-4 \\end{bmatrix}\\begin{bmatrix}-1 & -2\\\\1 & 2\\\\3&4 \\end{bmatrix}$$ $$=\\begin{bmatrix}-2+1+9 & -4+2+12 \\\\2+2+12 & 4+4+16\\\\1-3-12&2-6-16 \\end{bmatrix}$$ $$=\\begin{bmatrix}8 & 10\\\\16 & 24\\\\-14&-20 \\end{bmatrix}$$", "+0 Let find all values of t such that where I is the 3x3 identity matrix -1 57 1 +92 Let $${A} = \\begin{pmatrix} 0 & 5 & 7 \\\\ -2 & 7 & 7 \\\\ -1 & 1 & 4 \\end{pmatrix}.$$ find all values of t such that $$\\det (t {I} - {A}) = 0$$ where I is the 3x3 identity matrix May 14, 2022 #1 +9457 0 After simplifying the determinant in $$\\det(tI - A) = 0$$, you will get $$-40 + 38 t - 11 t^2 + t^3 = 0$$. Then you can solve the cubic equation to find the values of t. Hint: $$-40 + 38 t - 11 t^2 + t^3 = (t - 2)(t - 4)(t - 5)$$. May 14, 2022", "\\\\ a_{21} & a_{22} \\end {pmatrix} \\begin {pmatrix} b_{11} c_{11} + b_{12} c_{21} & b_{11} c_{12} + b_{12} c_{22} \\\\ b_{21} c_{11} + b_{22} c_{21} & b_{21} c_{12} + b_{22} c_{22} \\end {pmatrix}$ Definition of Matrix Product (Conventional) $\\ds$ $=$ $\\ds \\begin {pmatrix} a_{11} \\paren {b_{11} c_{11} + b_{12} c_{21} } + a_{12} \\paren {b_{21} c_{11} + b_{22} c_{21} } & a_{11} \\paren {b_{11} c_{12} + b_{12} c_{22} } + a_{12} \\paren {b_{21} c_{12} + b_{22} c_{22} } \\\\ a_{21} \\paren {b_{11} c_{11} + b_{12} c_{21} } + a_{22} \\paren {b_{21} c_{11} + b_{22} c_{21} } & a_{21} \\paren {b_{11} c_{12} + b_{12} c_{22} } + a_{22} \\paren {b_{21} c_{12} + b_{22} c_{22} } \\end {pmatrix}$ Definition of Matrix Product (Conventional) $\\ds$ $=$ $\\ds \\begin {pmatrix} a_{11} b_{11} c_{11} + a_{11} b_{12} c_{21} + a_{12} b_{21} c_{11} + a_{12} b_{22} c_{21} & a_{11} b_{11} c_{12} + a_{11} b_{12} c_{22} + a_{12} b_{21} c_{12} + a_{12} b_{22} c_{22} \\\\ a_{21} b_{11} c_{11} + a_{21} b_{12} c_{21} + a_{22} b_{21} c_{11} + a_{22} b_{22} c_{21} & a_{21} b_{11} c_{12} + a_{21} b_{12} c_{22} + a_{22} b_{21} c_{12} + a_{22} b_{22} c_{22} \\end {pmatrix}$ Real Multiplication Distributes over Addition $\\ds \\paren {\\mathbf A \\mathbf B} \\mathbf C$ $=$ $\\ds \\paren {\\begin {pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end {pmatrix}", "in detail, Use Strassen's algorithm to compute the matrix product \\begin{pmatrix} 1 & 3 \\\\ 7 & 5 \\end{pmatrix} \\begin{pmatrix} 6 & 8 \\\\ 4 & 2 \\end{pmatrix} . Transcript. \\right]_{2 × 3} (a) (b) 3.3 RANK OF A MATRIX Suppose A is an m × n matrix. CBSE Class 12th Matrices- Various Types of Matrices In this video we will learn about the topics Matrices, Various types of Matrices ,Representation of Matrices and Order of Matrices… 0 Followers Most ... Class 12. Up to equivalence, there is a unique Hadamard matrix of orders 1, 2, 4, 8, and 12. Solution Show Solution. In order to work out the determinant of a 3×3 matrix, one must multiply a by the determinant of the 2×2 matrix that does not happen to be a’s column or row or column. Question 2 (Method 1) If A = [] is a matrix of order 2 × 2, such that || = −15 and C represents the cofactor of , then find 21 21 + 22 22 Given a is a 2 × 2 matrix A = [ 8(_11&_12@_21&_12 )] Given |A| = – 15 |A| = a11 a12 – a21 a12 – 15 = a11 a12 – a21 a12 a11 a12 – … It is easy to describe the first two, they", "\\end{pmatrix}^T$$ Let $$A=\\begin{pmatrix}2 &1 \\\\6 & 4 \\end{pmatrix}$$ a) Express $$A^{-1}$$ as a product of elementary matrices", "Product of a Matrix by a Scalar Product of a Matrix by a Scalar The product of a matrix A by a scalar k, written kA or Ak, is the matrix of the same format as A obtained by multiplying all the elements of A by k. Example Consider the matrix A = $$\\begin{pmatrix} 3 & 6 & 7\\\\4 & 8 & 5\\end{pmatrix}$$. Then: $$4\\begin{pmatrix} 3 & 6 & 7\\\\4 & 8 & 5\\end{pmatrix}=\\begin{pmatrix} 12 & 24 & 28\\\\16 & 32 & 20\\end{pmatrix}$$.", "Question: Find the group G generated under matrix multiplication by the matrices $A=\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}, B=\\begin{pmatrix} 0 & i \\\\ i & 0 \\end{pmatrix}.$ Determine its proper subgroups, and verify for each of them that its cosets exhaust G. Step-by-step Before we can draw up a group multiplication table to search for subgroups, we must determine the multiple products of A and B with themselves and with each other: ${A}^{2}=\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}=\\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}=I$. Since B = iA, it follows that ${B}^{2} = −I$, that AB = iI = BA, and that ${B}^{3} = −B$. In brief, A is of order 2, B is of order 4, and A and B commute. The eight distinct elements of the group are therefore: $I,A,B,{B}^{2},{B}^{3}, AB,A{B}^{2}$ and $A{B}^{3}$. The group multiplication table is By inspection, the subgroups and their cosets are as follows: {I,A} : {I,A}, {B,AB}, {${B}^{2},A{B}^{2}$}, {${B}^{3},A{B}^{3}$}; {$I,{B}^{2}$} : {$I,{B}^{2}$}, {$A,A{B}^{2}$}, {$B,{B}^{3}$}, {$AB,A{B}^{3}$}; {$I,A{B}^{2}$} : {$I,A{B}^{2}$}, {$A,{B}^{2}$}, {$B,A{B}^{3}$}, {${B}^{3},AB$}; {$I,B,{B}^{2},{B}^{3}$} : {$I,B,{B}^{2},{B}^{3}$}, {$A,AB,A{B}^{2},A{B}^{3}$}; {$I,AB,{B}^{2},A{B}^{3}$} : {$I,AB,{B}^{2},A{B}^{3}$}, {$A,B,A{B}^{2},{B}^{3}$}. As expected, in each case the cosets exhaust the group, with each element in one and only one coset.", "0 1 correct ? What product? The product of the two matrices in post #5 is ##\\begin{pmatrix}i & 0 \\\\ 0 & i\\end{pmatrix}##. As HallsofIvy said, the inverse of ##\\begin{pmatrix}1 & 0\\\\ 0 & i\\end{pmatrix}## is ##\\begin{pmatrix}1 & 0\\\\ 0 & -i\\end{pmatrix}##. The product of these two matrices is the identity matrix. sorry meant to say identity matrix (not inverse) - thank you.", "Browse Questions # Find the product of the matrices A and B where : $A=\\begin{pmatrix}-5 & 1 & 3\\\\7 & 1 &-5\\\\1 & -1 & 1\\end{pmatrix},B=\\begin{pmatrix}1 & 1 & 2\\\\3 & 2 & 1\\\\2 & 1& 3 \\end{pmatrix}$Hence,solve the following equations by matrix method:$x+y+2z=1,3x+2y+z=7,2x+y+3z=2.$ This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com", "? Free Version Easy # 3x3 Matrix Product - 1 LINALG-GKW5LH Find the product: $$\\begin{pmatrix}2&7&-7\\\\\\ -7&-9&9\\\\\\ -9&-8&6\\end{pmatrix}\\begin{pmatrix}-8&-5&-10\\\\\\ 7&5&10\\\\\\ 9&2&-3\\end{pmatrix}$$ A $\\begin{pmatrix}-30&11&-71\\\\\\ 74&8&-47\\\\\\ 70&17&-8\\end{pmatrix}$ B $\\begin{pmatrix}-30&11&71\\\\\\ 74&8&-47\\\\\\ 70&17&8\\end{pmatrix}$ C $\\begin{pmatrix}-30&11&71\\\\\\ 74&8&-47\\\\\\ 70&17&-8\\end{pmatrix}$ D $\\begin{pmatrix}-30&11&71\\\\\\ -74&8&-47\\\\\\ 70&17&-8\\end{pmatrix}$ E This matrix product is not defined.", "every element in the matrix. For example: $3\\begin{pmatrix} 7 & 8 \\\\ 9 & 10 \\\\ 11 & 12 \\end{pmatrix} = \\begin{pmatrix} 21 & 24 \\\\ 27 & 30 \\\\ 33 & 36 \\end{pmatrix}$ ### Dividing Matrices Division of matrices is not defined. However we can multiply by the inverse matrix to achieve the same result. See below for how to find the inverse of a matrix. ## Identity Matrices The identity matrix is the matrix which when multiplied by another matrix returns that matrix - in other words it is the equivalent of the real number 1. Because this can only happen with square matrices, an identity matrix is a square matrix which apart from a diagonal line of ones from top left to bottom right consists only of zeros. The three by three identity matrix known as $\\mathbf{I_3}$is thus: $\\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$ ## Zero Matrices The zero matrix or null matrix is denoted $\\mathbf{O}$, and it's simply a matrix populated exclusively by zeros. Both of the following matrices are called $\\mathbf{O}$. $\\begin{pmatrix} 0 & 0 \\\\ 0 & 0 \\end{pmatrix}$ $\\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$ Naturally, for any matrix M, $\\mathbf{O}{M} =", "& 3 \\end{pmatrix} * \\begin{pmatrix} 7 & -3\\\\ -2 & 3\\\\ 6 & 2 \\end{pmatrix}$ $C*D = \\begin{pmatrix} 3*7 + (-9)*(-2) + (-8)*6 & 3*(-3) + (-9)*3 + (-8)*2 \\\\ 2*7 + 4*(-2) + 3*6 & 2*(-3) + 4*3 + 3*2 \\end{pmatrix}$ $C*D = \\begin{pmatrix} 21 + 18 - 48 & - 9 - 27 - 16 \\\\14 - 8 + 18 & - 6 + 12 + 6 \\end{pmatrix} = \\begin{pmatrix} -9 & - 52\\\\ 24 & 12 \\end{pmatrix}$ Matrix multiplication has some of the same properties as \"normal\" multiplication , such as $A(BC) = (AB)C$ $A(B + C) = AB + AC$ $(A + B)C = AC + BC$ However matrix multiplication is not communicative. That is to say A*B does not necessarily equal B*A. In fact, B*A often has no meaning since the dimensions rarely match up. However, you can take the transpose of matrix multiplication. In that case $(AB)^T = B^T A^T$. # M.3 Matrix Properties M.3 Matrix Properties ## Identity Matrices An identity matrix is a square matrix where every diagonal entry is 1 and all the other entries are 0. The following two matrices are both identity matrices and diagonal matrices. $I_3 = \\begin{pmatrix} 1 & 0 & 0 \\\\0 & 1 & 0\\\\ 0 & 0 & 1 \\end{pmatrix}$ $I_4 =", "= i, \\quad \\mathbf{A} = \\begin{pmatrix} i & 0 \\\\ 0 & j \\\\ \\end{pmatrix}$ $i\\begin{pmatrix} i & 0 \\\\ 0 & j \\\\ \\end{pmatrix} = \\begin{pmatrix} i^2 & 0 \\\\ 0 & ij \\\\ \\end{pmatrix} = \\begin{pmatrix} -1 & 0 \\\\ 0 & k \\\\ \\end{pmatrix} \\ne \\begin{pmatrix} -1 & 0 \\\\ 0 & -k \\\\ \\end{pmatrix} = \\begin{pmatrix} i^2 & 0 \\\\ 0 & ji \\\\ \\end{pmatrix} = \\begin{pmatrix} i & 0 \\\\ 0 & j \\\\ \\end{pmatrix}i\\,,$ where i, j, k are the quaternion units. The non-commutativity of quaternion multiplication prevents the transition of changing ij = +k to ji = −k. Matrix product (two matrices) Assume two matrices are to be multiplied (the generalization to any number is discussed below). General definition of the matrix product Arithmetic process of multiplying numbers (solid lines) in row i in matrix A and column j in matrix B, then adding the terms (dashed lines) to obtain entry ij in the final matrix. If A is an n × m matrix and B is an m × p matrix, $\\mathbf{A}=\\begin{pmatrix} A_{11} & A_{12} & \\cdots & A_{1m} \\\\ A_{21} & A_{22} & \\cdots & A_{2m} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ A_{n1} & A_{n2} & \\cdots & A_{nm} \\\\ \\end{pmatrix},\\quad\\mathbf{B}=\\begin{pmatrix} B_{11} & B_{12} & \\cdots", "\\end{pmatrix}$ The important thing to notice here is that the 1 to 9 matrix remains the same when multiplied with the other matrix. The matrix with only 1s on the diagonal and 0s elsewhere is known as the identity matrix, called I, and any matrix multiplied on either side of it stays the same. That is $A \\times I = I \\times A$ 4. a) $A^5 = \\begin{pmatrix} -1&6\\\\ -1&4\\\\ \\end{pmatrix} \\begin{pmatrix} -1&6\\\\ -1&4\\\\ \\end{pmatrix} \\begin{pmatrix} -1&6\\\\ -1&4\\\\ \\end{pmatrix} \\begin{pmatrix} -1&6\\\\ -1&4\\\\ \\end{pmatrix} \\begin{pmatrix} -1&6\\\\ -1&4\\\\ \\end{pmatrix} =$ $\\begin{pmatrix} -1&6\\\\ -1&4\\\\ \\end{pmatrix} \\begin{pmatrix} -1&6\\\\ -1&4\\\\ \\end{pmatrix} \\begin{pmatrix} -1&6\\\\ -1&4\\\\ \\end{pmatrix} \\begin{pmatrix} -5&18\\\\ -3&10\\\\ \\end{pmatrix} =$ $\\begin{pmatrix} -1&6\\\\ -1&4\\\\ \\end{pmatrix} \\begin{pmatrix} -1&6\\\\ -1&4\\\\ \\end{pmatrix} \\begin{pmatrix} -13&42\\\\ -7&22\\\\ \\end{pmatrix} =$ $\\begin{pmatrix} -1&6\\\\ -1&4\\\\ \\end{pmatrix} \\begin{pmatrix} -29&90\\\\ -15&46\\\\ \\end{pmatrix} =$ $\\begin{pmatrix} -61&186\\\\ -31&94\\\\ \\end{pmatrix}$ b) $\\begin{pmatrix} 1&-2\\\\ -1&3\\\\ \\end{pmatrix} \\begin{pmatrix} 3&2\\\\ 1&1\\\\ \\end{pmatrix} = \\begin{pmatrix} (1 \\times 3) + (-2 \\times 1)&(1 \\times 2) + (-2 \\times 1)\\\\ (-1 \\times 3) + (3 \\times 1)&(-1 \\times 2) + (3 \\times 1)\\\\ \\end{pmatrix} = \\begin{pmatrix} 1&0\\\\ 0&1\\\\ \\end{pmatrix}$ c) $\\begin{pmatrix} a&b\\\\ c&d\\\\ \\end{pmatrix} \\begin{pmatrix} 1&0\\\\ 0&1\\\\ \\end{pmatrix} = \\begin{pmatrix} (a \\times 1) + (b \\times 0)&(a \\times 0) + (b \\times 1)\\\\ (c \\times 1) + (d \\times 0)&(c \\times 0) + (d \\times 1)\\\\ \\end{pmatrix} = \\begin{pmatrix} a&b\\\\ c&d\\\\ \\end{pmatrix}$ $\\begin{pmatrix}", "\\begin{pmatrix} 1 & 0 & 2 \\\\ 2 & 2 & 1 \\\\ 3 & 2 & 0 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix} = \\begin{pmatrix} 7 \\\\ 9 \\\\ 7 \\end{pmatrix} \\label{8.10}$ Matrix multiplication is not generally commutative, a property that mirrors the behavior found earlier for symmetry operations within a point group. ## Direct Products The direct product of two matrices (given the symbol $$\\otimes$$) is a special type of matrix product that generates a matrix of higher dimensionality if both matrices have dimension greater than one. The easiest way to demonstrate how to construct a direct product of two matrices $$A$$ and $$B$$ is by an example: \\begin{align} A \\otimes B &= \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix} \\otimes \\begin{pmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{pmatrix} \\\\[4pt] &= \\begin{pmatrix} a_{11}B & a_{12}B \\\\ a_{21}B & a_{22}B \\end{pmatrix} \\\\[4pt] &= \\begin{pmatrix} a_{11}b_{11} & a_{11}b_{12} & a_{12}b_{11} & a_{12}b_{12} \\\\ a_{11}b_{21} & a_{11}b_{22} & a_{12}b_{21} & a_{12}b_{22} \\\\ a_{21}b_{11} & a_{21}b_{12} & a_{22}b_{11} & a_{22}b_{21} \\\\ a_{21}b_{21} & a_{21}b_{22} & a_{22}b_{21} & a_{22}b_{22} \\end{pmatrix} \\label{8.11} \\end{align} Though this may seem like a somewhat strange operation to carry out, direct products crop up a great deal in group theory. ## Inverse Matrices and Determinants If two square matrices $$A$$ and $$B$$", "= \\begin{pmatrix} 3*7 + (-9)*(-2) + (-8)*6 & 3*(-3) + (-9)*3 + (-8)*2 \\\\ 2*7 + 4*(-2) + 3*6 & 2*(-3) + 4*3 + 3*2 \\end{pmatrix}$ $C*D = \\begin{pmatrix} 21 + 18 - 48 & - 9 - 27 - 16 \\\\14 - 8 + 18 & - 6 + 12 + 6 \\end{pmatrix} = \\begin{pmatrix} -9 & - 52\\\\ 24 & 12 \\end{pmatrix}$ Matrix multiplication has some of the same properties as \"normal\" multiplication , such as $A(BC) = (AB)C$ $A(B + C) = AB + AC$ $(A + B)C = AC + BC$ However matrix multiplication is not communicative. That is to say A*B does not necessarily equal B*A. In fact, B*A often has no meaning since the dimensions rarely match up. However, you can take the transpose of matrix multiplication. In that case $$(AB)^T = B^T A^T$$. # M.3 Matrix Properties M.3 Matrix Properties ## Identity Matrices An identity matrix is a square matrix where every diagonal entry is 1 and all the other entries are 0. The following two matrices are identity matrices and diagonal matrices. $I_3 = \\begin{pmatrix} 1 & 0 & 0 \\\\0 & 1 & 0\\\\ 0 & 0 & 1 \\end{pmatrix}$ $I_4 = \\begin{pmatrix} 1 & 0 & 0 & 0\\\\ 0 & 1 & 0 & 0\\\\ 0 &", "matrix product is associative with identity element the indentity matrix, it's sufficiente to check that the multiplication restricts to $M$. But: $\\begin{pmatrix}a&b\\\\0&c\\\\\\end{pmatrix}\\begin{pmatrix}a'&b'\\\\0&c'\\\\\\end{pmatrix}=\\begin{pmatrix}aa'&ab'+bc'\\\\0&cc'\\\\\\end{pmatrix}$ Moreover, if $\\begin{pmatrix}a&b\\\\0&c\\\\\\end{pmatrix}$ has inverse $\\begin{pmatrix}a'&b'\\\\0&c'\\\\\\end{pmatrix}$ then: $\\begin{pmatrix}aa'&ab'+bc'\\\\0&cc'\\\\\\end{pmatrix}=\\begin{pmatrix}1'&0\\\\0&1\\\\\\end{pmatrix}$ From this $aa'=1$ and $cc'=1$. But $a,a',c,c'$ are integers so $a'=a=\\pm 1$, $c'=c=\\pm 1$ and: $ab'+bc'=0\\quad \\Rightarrow \\quad ab'+bc=0 \\quad \\Rightarrow \\quad ab'=-bc \\quad \\Rightarrow \\quad aab'=-abc \\quad \\Rightarrow \\quad b'=-abc$", "Matrix: If each of it diagonal elements is 1 and all other elements are zero is known as identity matrix. For example, $$\\begin{pmatrix}1 & 0 \\\\ 0 & 1 \\\\ \\end{pmatrix}$$ and $$\\begin{pmatrix}1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ \\end{pmatrix}$$ are identity matrices of order (2 × 2) and (3 × 3) respectively. When A and B are two matrices of the same order, the sum of A and B by adding corresponding elements of A and B. For example, If A = $$\\begin{pmatrix}1 & 4 & 6 \\\\ 3 & 0 & 9 \\\\ \\end{pmatrix}$$ and B = $$\\begin{pmatrix}2 & 8 & 3 \\\\ 6 & 5 & 2 \\\\ \\end{pmatrix}$$, A + B = $$\\begin{pmatrix}1+2 & 4+8 & 6+3 \\\\ 3+6 & 0+5 & 9+2 \\\\ \\end{pmatrix}$$ = $$\\begin{pmatrix}3 & 12 & 9 \\\\ 9 & 5 & 11 \\\\ \\end{pmatrix}$$ ### Subtraction of Matrices When A and B are two matrices of the same order, the difference between A and B by A − B is the matrix by subtracting corresponding elements of A and B. For example, If A = $$\\begin{pmatrix}4 & 2 \\\\ 5 & 3 \\\\ \\end{pmatrix}$$ and B = $$\\begin{pmatrix}6 & 4 \\\\ 7 & 5 \\\\ \\end{pmatrix}$$, then A", "Identities of the form in \\eqref{eq:neg1} abound in the wild. For instance, within the ring $M_2(\\mathbb R)$ of $2 \\times 2$-matrices with real entries, the rotation by $90^{\\circ}$ matrix $R$ defined by $$\\label{eq:matrix} R := \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\end{pmatrix}$$ satisfies $R^2 = -\\mathbf{1}$. Here $\\mathbf{1}$ signifies the identity matrix $\\mathbf{1}:=\\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix},$ which is the multiplicative identity in $M_2(\\mathbb R)$." ]

[ "Question: Find the group G generated under matrix multiplication by the matrices $A=\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}, B=\\begin{pmatrix} 0 & i \\\\ i & 0 \\end{pmatrix}.$ Determine its proper subgroups, and verify for each of them that its cosets exhaust G. Step-by-step Before we can draw up a group multiplication table to search for subgroups, we must determine the multiple products of A and B with themselves and with each other: ${A}^{2}=\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}=\\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}=I$. Since B = iA, it follows that ${B}^{2} = −I$, that AB = iI = BA, and that ${B}^{3} = −B$. In brief, A is of order 2, B is of order 4, and A and B commute. The eight distinct elements of the group are therefore: $I,A,B,{B}^{2},{B}^{3}, AB,A{B}^{2}$ and $A{B}^{3}$. The group multiplication table is By inspection, the subgroups and their cosets are as follows: {I,A} : {I,A}, {B,AB}, {${B}^{2},A{B}^{2}$}, {${B}^{3},A{B}^{3}$}; {$I,{B}^{2}$} : {$I,{B}^{2}$}, {$A,A{B}^{2}$}, {$B,{B}^{3}$}, {$AB,A{B}^{3}$}; {$I,A{B}^{2}$} : {$I,A{B}^{2}$}, {$A,{B}^{2}$}, {$B,A{B}^{3}$}, {${B}^{3},AB$}; {$I,B,{B}^{2},{B}^{3}$} : {$I,B,{B}^{2},{B}^{3}$}, {$A,AB,A{B}^{2},A{B}^{3}$}; {$I,AB,{B}^{2},A{B}^{3}$} : {$I,AB,{B}^{2},A{B}^{3}$}, {$A,B,A{B}^{2},{B}^{3}$}. As expected, in each case the cosets exhaust the group, with each element in one and only one coset.", "# Find matrices $P$ and $Q$ to check if matrices $A$ and $B$ are equivalent( $Q^{-1}AP = B$ )? \\begin{align}&A=\\begin{pmatrix} 6 & -3 & -6 & -12 & 9 & 9 & 0 & 3 & -6 & 9 \\\\ 12 & 4 & 8 & -4 & 8 & -12 & 4 & 4 & 8 & -4\\\\ \\end{pmatrix} \\\\\\\\ & B = \\begin{pmatrix} 4 & -4 & 0 & -8 & -12 & 8 & -4 & -8 & 4 & -12 \\\\ -9 & -3 & -6 & 3 & -6 & 9 & -3 & -3 & -6 & 3\\\\ \\end{pmatrix} \\end{align} I need to find out if $A$ and $B$ are equivalent or not. To do that I have rearranged the definition of matrix equivalence: \\begin{align} Q^{-1}AP & = B \\\\\\\\ AP & = QB \\end{align} So now I have $\\begin{pmatrix} 6 & -3 & -6 & -12 & 9 & 9 & 0 & 3 & -6 & 9 \\\\ 12 & 4 & 8 & -4 & 8 & -12 & 4 & 4 & 8 & -4\\\\ \\end{pmatrix} \\cdot \\begin{pmatrix} x_{1,1} & \\cdots & x_{1,2} \\\\ \\vdots & & \\vdots \\\\ x_{10,1}& \\cdots & x_{10,2} \\end{pmatrix} = \\begin{pmatrix} y_{1,1} & \\cdots & y_{1,2} \\\\ \\vdots & & \\vdots \\\\ y_{10,1}&", "0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ -V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix}$ The matrices really correspond to the L and U matrices in the LDU decomposition, noting that the matrix on the right or left is just the transpose of the other. Define the Schur complement of the matrix V with respect to the block V22 as $V_{11.2}=V_{11}-V_{12}V_{22}^{-1}V_{21}$ then working through the above multiplication it follows that $\\begin{pmatrix} 1 & -V_{12}V_{22}^{-1}\\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ -V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix} = \\begin{pmatrix} V_{11.2} & 0 \\\\ 0 & V_{22} \\\\ \\end{pmatrix}$ This result is useful in deriving the Sherman–Morrison–Woodbury matrix inverse formula/identity, otherwise known as the matrix inversion lemma. One can now simply multiply the above equation by the corresponding inverse matrices to obtain the LDU decomposition of the matrix V. The inverse matrices are easy $\\begin{pmatrix} 1 & V_{12}V_{22}^{-1}\\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} 1 & -V_{12}V_{22}^{-1}\\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ -V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix} = \\begin{pmatrix} 1 & V_{12}V_{22}^{-1} \\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11.2} & 0 \\\\", "? Free Version Easy # 3x3 Matrix Product - 1 LINALG-GKW5LH Find the product: $$\\begin{pmatrix}2&7&-7\\\\\\ -7&-9&9\\\\\\ -9&-8&6\\end{pmatrix}\\begin{pmatrix}-8&-5&-10\\\\\\ 7&5&10\\\\\\ 9&2&-3\\end{pmatrix}$$ A $\\begin{pmatrix}-30&11&-71\\\\\\ 74&8&-47\\\\\\ 70&17&-8\\end{pmatrix}$ B $\\begin{pmatrix}-30&11&71\\\\\\ 74&8&-47\\\\\\ 70&17&8\\end{pmatrix}$ C $\\begin{pmatrix}-30&11&71\\\\\\ 74&8&-47\\\\\\ 70&17&-8\\end{pmatrix}$ D $\\begin{pmatrix}-30&11&71\\\\\\ -74&8&-47\\\\\\ 70&17&-8\\end{pmatrix}$ E This matrix product is not defined.", "every element in the matrix. For example: $3\\begin{pmatrix} 7 & 8 \\\\ 9 & 10 \\\\ 11 & 12 \\end{pmatrix} = \\begin{pmatrix} 21 & 24 \\\\ 27 & 30 \\\\ 33 & 36 \\end{pmatrix}$ ### Dividing Matrices Division of matrices is not defined. However we can multiply by the inverse matrix to achieve the same result. See below for how to find the inverse of a matrix. ## Identity Matrices The identity matrix is the matrix which when multiplied by another matrix returns that matrix - in other words it is the equivalent of the real number 1. Because this can only happen with square matrices, an identity matrix is a square matrix which apart from a diagonal line of ones from top left to bottom right consists only of zeros. The three by three identity matrix known as $\\mathbf{I_3}$is thus: $\\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$ ## Zero Matrices The zero matrix or null matrix is denoted $\\mathbf{O}$, and it's simply a matrix populated exclusively by zeros. Both of the following matrices are called $\\mathbf{O}$. $\\begin{pmatrix} 0 & 0 \\\\ 0 & 0 \\end{pmatrix}$ $\\begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & 0 \\end{pmatrix}$ Naturally, for any matrix M, $\\mathbf{O}{M} =", "+0 # Matrices Solving -1 200 1 +116 Let $${A} = \\begin{pmatrix} 0 & 5 & 7 \\\\ -2 & 7 & 7 \\\\ -1 & 1 & 4 \\end{pmatrix}$$. Find all values of $$t$$ such that $$\\det (t{I} - {A}) = 0$$, where $$I$$ is the $$3 \\times 3$$ identity matrix. May 30, 2020", "B + \\mathbf A$ Thus matrix addition is commutative on $\\map {\\MM_\\R} 2$. $\\Box$ #### Ring Axiom $\\text A3$: Identity for Addition We have: $\\ds \\begin {pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end {pmatrix} + \\begin {pmatrix} 0 & 0 \\\\ 0 & 0 \\end {pmatrix}$ $=$ $\\ds \\begin {pmatrix} a_{11} + 0 & a_{12} + 0 \\\\ a_{21} + 0 & a_{22} + 0 \\end {pmatrix}$ Definition of Matrix Entrywise Addition $\\ds$ $=$ $\\ds \\begin {pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end {pmatrix}$ Real Addition Identity is Zero $\\ds$ $=$ $\\ds \\begin {pmatrix} 0 + a_{11} & 0 + a_{12} \\\\ 0 + a_{21} & 0 + a_{22} \\end {pmatrix}$ Real Addition is Commutative $\\ds$ $=$ $\\ds \\begin {pmatrix} 0 & 0 \\\\ 0 & 0 \\end {pmatrix} + \\begin {pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end {pmatrix}$ Thus $\\begin {pmatrix} 0 & 0 \\\\ 0 & 0 \\end {pmatrix}$ is seen to be the identity element of matrix addition on $\\map {\\MM_\\R} 2$. $\\Box$ #### Ring Axiom $\\text A4$: Inverses for Addition We have: $\\ds \\begin {pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end {pmatrix} + \\begin {pmatrix} -a_{11} & -a_{12} \\\\ -a_{21} & -a_{22} \\end {pmatrix}$ $=$ $\\ds \\begin {pmatrix} a_{11} + \\paren {-a_{11} } & a_{12} +", "4 & 3 \\end{bmatrix}$$ We have, $$\\left | D \\right | = 6 + 8 = 14$$ Now, D 11 = 3, D 12 = – 4, D 21 = 2, D 22 = 2 adj D = $$\\begin{bmatrix} 3 & 2 \\\\ – 4 & 2 \\end{bmatrix} \\\\ D ^{- 1} = \\frac{1}{\\left | D \\right |} = \\frac{1}{14}\\; \\begin{bmatrix} 3 & 2 \\\\ – 4 & 2 \\end{bmatrix}$$ Q.5: Obtain the inverse of the matrices if it exists $$\\begin{bmatrix} 1 & 2 & 3 \\\\ 0 & 2 & 4 \\\\ 0 & 0 & 5 \\end{bmatrix}$$ Sol: Determinants Suppose D = $$\\begin{bmatrix} 1 & 2 & 3 \\\\ 0 & 2 & 4 \\\\ 0 & 0 & 5 \\end{bmatrix}$$ We have, $$\\left | D \\right | = 1 (10 – 0) – 2 (0 – 0) + 3 (0 – 0) = 10$$ Now, D 11 = 10 – 0 = 10, D 12 = – (0 – 0) = 0, D 13 = 0 – 0 = 0, D 21 = – (10 – 0) = – 10, D 22 = 5 – 0 = 5, D 23 = – (0 – 0) = 0, D 31 = 8 – 6 = 2, D 32 = – (4 – 0) = – 4,", "$$1$$ and all other entries are equal to $$0$$. Here are some identity matrices: $I_2 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}, \\quad I_3 = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$ $I_4 = \\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix}$ as we'll be seeing, when we multiply a matrix by an identity matrix it doesn't change its value (just like when we multiply a number by $$1$$). ## Exercise 1. State the order (size) of each of the following matrices: 1. $$A = \\begin{pmatrix} 2 & -3 \\\\ 0 & 1\\end{pmatrix}$$ 1. $$B = \\begin{pmatrix} 1 & -5 \\\\ 4 & 6 \\\\ 0 & 8\\end{pmatrix}$$ 1. $$C = \\begin{pmatrix} 4 & 7 & 0 & -3 \\\\ 5 & 9 & 2 & 1 \\end{pmatrix}$$ 1. $$D = \\begin{pmatrix} 8 \\end{pmatrix}$$ 1. $$E = \\begin{pmatrix} 4 \\\\ -2 \\\\ 6 \\end{pmatrix}$$ 1. $$F = \\begin{pmatrix} 8 & 0 & 2 \\\\ -2 & 3 & 4 \\\\ 5 & 1 & 0 \\\\ 7 & 9 & 10 \\end{pmatrix}$$ 2. Given the matrix $$A =", "the special matrices $$B = \\begin{pmatrix} b_{11} & 0 \\\\ 0 & b_{11}^{-1} \\end{pmatrix}$$ then \\begin{align*} c_{11} & = a_{22}b_{22} a_{11} b_{11} - a_{12} b_{11} a_{21} b_{11} \\\\ c_{12} & = a_{22}b_{22} a_{12} b_{22} - a_{12} b_{11} a_{22} b_{22} \\\\ c_{21} & = -a_{21}b_{22}a_{11}b_{22} + a_{11}b_{11}a_{21}b_{11} \\\\ c_{22} & = -a_{21}b_{22}a_{12}b_{22} + a_{11}b_{11}a_{22} b_{22}. \\end{align*} Now requiring $$\\begin{pmatrix} c_{11} & c_{12} \\\\ c_{21} & c_{22} \\end{pmatrix} = \\begin{pmatrix} \\alpha & 0 \\\\ 0 & \\alpha \\end{pmatrix}$$ this gives after simplifying \\begin{align*} \\alpha & = a_{11} a_{22} - a_{12} a_{21} b_{11}^2 \\\\ 0 & = a_{22} a_{12} b_{11}^{-2} - a_{12}a_{22} = a_{22}a_{12}(b_{11}^{-2} - 1) \\\\ 0 & = a_{11}a_{21}b_{11}^{-2} - a_{11}a_{21} b_{11}^2 = a_{11}a_{21}(b_{11}^{-2} - b_{11}^2) \\\\ \\alpha & = a_{11}a_{22} - a_{21}a_{12} b_{11}^{-2}. \\end{align*} The first and last equation give $b_{11}^2 = b_{11}^{-2} \\Leftrightarrow b_{11}^4 = 1$. But as the equations have to hold for every $B$, just choose some matrix $B$ with $b_{11}^4 \\ne 1$ and this gives a contradiction if the field is such that we can choose such an element. Otherwise the ground field is the finite field of order $2$, $3$ or $5$, see here. But $PSL(2,2) \\cong S_3, PSL(2,3) \\cong A_4, PSL(2,5) \\cong A_5$ and all these groups have trivial center.", "\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} & \\begin{pmatrix} 1 & 2 \\\\ 3 & 4 \\end{pmatrix} & \\begin{pmatrix} \\alpha & \\beta \\\\ \\gamma & \\delta \\end{pmatrix} \\\\ \\begin{pmatrix} e & f \\\\ g & h \\end{pmatrix} & \\begin{pmatrix} 5 & 6 \\\\ 7 & 8 \\end{pmatrix} & \\begin{pmatrix} \\epsilon & \\zeta \\\\ \\eta & \\theta \\end{pmatrix} \\end{pmatrix}$ is dual to $X = \\begin{pmatrix} Y_{11} & Y_{12} \\\\ Y_{21} & Y_{22} \\end{pmatrix} = \\begin{pmatrix} \\begin{pmatrix} a & 1 & \\alpha\\\\ e & 5 & \\epsilon \\end{pmatrix} & \\begin{pmatrix} b & 2 & \\beta\\\\ f & 6 & \\zeta \\end{pmatrix} \\\\ \\begin{pmatrix} c & 3 & \\gamma\\\\ g & 7 & \\eta \\end{pmatrix} & \\begin{pmatrix} d & 4 & \\delta\\\\ h & 8 & \\theta \\end{pmatrix} \\end{pmatrix}$ . $b_{11,11} = a = y_{11,11}~$ $b_{23,21} = \\eta = y_{21,23}~$ This concept is not limited to matrices, as the following example shows. The join table is a 24×24 matrix. The inversion sets could be displayed by 6-bit binary vectors, but a free arrangement was chosen, to serve symmetry. ## Gould's-Morse sequence This portmanteau name I use for the following sequence, which occurs when dealing with Walsh matrices: 1, -2, -2, 4, -2, 4, 4, -8, -2, 4, 4, -8, 4, -8, -8, 16, -2, 4, 4, -8, 4, -8,", "-V_{12}V_{22}^{-1}\\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ -V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix}$ The matrices really correspond to the L and U matrices in the LDU decomposition, noting that the matrix on the right or left is just the transpose of the other. Define the Schur complement of the matrix V with respect to the block V22 as $V_{11.2}=V_{11}-V_{12}V_{22}^{-1}V_{21}$ then working through the above multiplication it follows that $\\begin{pmatrix} 1 & -V_{12}V_{22}^{-1}\\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ -V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix} = \\begin{pmatrix} V_{11.2} & 0 \\\\ 0 & V_{22} \\\\ \\end{pmatrix}$ This result is useful in deriving the Sherman–Morrison–Woodbury matrix inverse formula/identity, otherwise known as the matrix inversion lemma. One can now simply multiply the above equation by the corresponding inverse matrices to obtain the LDU decomposition of the matrix V. The inverse matrices are easy $\\begin{pmatrix} 1 & V_{12}V_{22}^{-1}\\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} 1 & -V_{12}V_{22}^{-1}\\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ -V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix} = \\begin{pmatrix} 1 & V_{12}V_{22}^{-1} \\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11.2} & 0", "+0 Let find all values of t such that where I is the 3x3 identity matrix -1 57 1 +92 Let $${A} = \\begin{pmatrix} 0 & 5 & 7 \\\\ -2 & 7 & 7 \\\\ -1 & 1 & 4 \\end{pmatrix}.$$ find all values of t such that $$\\det (t {I} - {A}) = 0$$ where I is the 3x3 identity matrix May 14, 2022 #1 +9457 0 After simplifying the determinant in $$\\det(tI - A) = 0$$, you will get $$-40 + 38 t - 11 t^2 + t^3 = 0$$. Then you can solve the cubic equation to find the values of t. Hint: $$-40 + 38 t - 11 t^2 + t^3 = (t - 2)(t - 4)(t - 5)$$. May 14, 2022", "\\end{pmatrix}$ ## Invert a Matrix using the Woodbury Matrix Inverse Formula Identity Previously I wrote about the LDU decomposition and the Schur complement. These can be further used to derive the Sherman–Morrison–Woodbury formula, otherwise known as the matrix inversion lemma, for inverting a matrix. As shown in the previous post, a UDL and LDU are two ways of factorizing a matrix: $\\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix}= \\begin{pmatrix} 1 & V_{12}V_{22}^{-1}\\\\ 0 & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11.2} & 0 \\\\ 0 & V_{22} \\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix}$ $\\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix} = \\begin{pmatrix} 1 & 0\\\\ V_{21}V_{11}^{-1} & 1\\\\ \\end{pmatrix} \\begin{pmatrix} V_{11} & 0 \\\\ 0 & V_{22.1} \\\\ \\end{pmatrix} \\begin{pmatrix} 1 & V_{11}^{-1}V_{12} \\\\ 0 & 1 \\\\ \\end{pmatrix}$ Now consider taking the inverse of the matrices above, yielding $\\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix}^{-1}= \\begin{pmatrix} 1 & 0 \\\\ -V_{22}^{-1}V_{21} & 1 \\\\ \\end{pmatrix} \\begin{pmatrix} V_{11.2}^{-1} & 0 \\\\ 0 & V_{22}^{-1} \\\\ \\end{pmatrix} \\begin{pmatrix} 1 & -V_{12}V_{22}^{-1}\\\\ 0 & 1\\\\ \\end{pmatrix}$ $\\begin{pmatrix} V_{11} & V_{12}\\\\ V_{21} & V_{22}\\\\ \\end{pmatrix}^{-1} = \\begin{pmatrix} 1 & -V_{11}^{-1}V_{12} \\\\ 0 & 1 \\\\ \\end{pmatrix} \\begin{pmatrix} V_{11}^{-1} & 0 \\\\ 0 & V_{22.1}^{-1} \\\\ \\end{pmatrix} \\begin{pmatrix} 1 & 0\\\\ -V_{21}V_{11}^{-1} & 1\\\\ \\end{pmatrix}$ Multiplying", "\\begin{equation*} A = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\end{equation*} is in $GL_2( {\\mathbb R})$ if there exists a matrix $A^{-1}$ such that $A A^{-1} = A^{-1} A = I\\text{,}$ where $I$ is the $2 \\times 2$ identity matrix. For $A$ to have an inverse is equivalent to requiring that the determinant of $A$ be nonzero; that is, $\\det A = ad - bc \\neq 0\\text{.}$ The set of invertible matrices forms a group called the general linear group. The identity of the group is the identity matrix \\begin{equation*} I = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}\\text{.} \\end{equation*} The inverse of $A \\in GL_2( {\\mathbb R})$ is \\begin{equation*} A^{-1} = \\frac{1}{ad-bc} \\begin{pmatrix} d & -b \\\\ -c & a \\end{pmatrix}\\text{.} \\end{equation*} The product of two invertible matrices is again invertible. Matrix multiplication is associative, satisfying the other group axiom. For matrices it is not true in general that $AB = BA\\text{;}$ hence, $GL_2({\\mathbb R})$ is another example of a nonabelian group. ###### Example2.15. Let \\begin{align*} 1 & = \\begin{pmatrix} 1 & 0\\\\ 0 & 1 \\end{pmatrix} \\qquad I = \\begin{pmatrix} 0 & 1\\\\ -1 & 0 \\end{pmatrix}\\\\ J & = \\begin{pmatrix} 0 & i\\\\ i & 0 \\end{pmatrix} \\qquad K = \\begin{pmatrix} i & 0\\\\ 0 & -i \\end{pmatrix}\\text{,} \\end{align*} where $i^2", "now, if the inverses of the $$L$$ matrices were easy to find. Use np.linalg.inv() to directly compute the inverse of $$L_1$$, $$L_2$$, and $$L_3$$ for each of the example matrices. Then complete the statement: If $$L_k$$ is an identity matrix with some nonzero $$c$$ in row $$i$$ and column $$j$$ then $$L_k^{-1}$$ is what matrix? Exercise 4.23 We started this discussion with $$A$$ as $A = \\begin{pmatrix} 1 & 2 & 3 \\\\ 4 & 5 & 6 \\\\ 7 & 8 & 0 \\end{pmatrix}$ and we defined $L_1 = \\begin{pmatrix} 1 & 0 & 0 \\\\ -4 & 1 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}, \\quad L_2 = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ -7 & 0 & 1 \\end{pmatrix}, \\quad \\text{and} \\quad L_3 = \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & -2 & 1 \\end{pmatrix}.$ Based on your answer to the previous exercises we know that $A = L_1^{-1} L_2^{-1} L_3^{-1} U.$ Explicitly write down the matrices $$L_1^{-1}$$, $$L_2^{-1}$$, and $$L_3^{-1}$$. Now explicitly find the product $$L_1^{-1} L_2^{-1} L_3^{-1}$$ and call this product $$L$$. Verify that $$L$$ itself is also a lower triangular matrix with ones on the main diagonal. Moreover, take note of exactly the form of", "decimals: 3.141593 3. format long a a = 3.14159265358979 4. format short a a = 3.1416 Matrices 1. A = [1 2; 3 4; 5 6] 3 by 2 matrix $A=\\begin{pmatrix} 1 & 2 \\\\ 3 & 4 \\\\ 5 & 6 \\end{pmatrix}$ 2. v = [1 2 3] 1 by 3 matrix (row vector) $v=\\begin{pmatrix} 1 & 2 & 3 \\end{pmatrix}$ 3. v = [1; 2; 3] 3 by 1 matrix (column vector) $v=\\begin{pmatrix} 1\\\\ 2\\\\ 3 \\end{pmatrix}$ 4. v=1:0.1:2 1 by 11 matrix (row vector) $v=\\begin{pmatrix} 1 & 1.1 & 1.2 & … & 1.9 & 2 \\end{pmatrix}$ 5. v=1:6 1 by 6 matrix (row vector) $v=\\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\end{pmatrix}$ 6. ones(2,3) 2 by 3 matrix of ones $\\begin{pmatrix} 1 & 1 & 1\\\\ 1 & 1 & 1 \\end{pmatrix}$ 7. 2*ones(2,3) $\\begin{pmatrix} 2 & 2 & 2\\\\ 2 & 2 & 2 \\end{pmatrix}$ 8. zeros(2,3) 2 by 3 matrix of zeroes $\\begin{pmatrix} 0 & 0 & 0\\\\ 0 & 0 & 0 \\end{pmatrix}$ 9. rand(2,3) 2 by 3 matrix of random numbers between 0 and 1 10. randn(2,3) 2 by 3 matrix of random numbers drawn from a Gaussian distribution with mean 0 and variance 11. eye(4) 4 by 4 identity matrix $\\begin{pmatrix} 1 & 0", "Question # Find the product AB of the two matrices listed below: A=begin{pmatrix}2 & 1&3 -2 & 2&4-1&-3&-4 end{pmatrix} B=begin{pmatrix}-1 & -21 & 23&4 end{pmatrix} Matrices Find the product AB of the two matrices listed below: $$A=\\begin{pmatrix}2 & 1&3 \\\\-2 & 2&4\\\\-1&-3&-4 \\end{pmatrix}$$ $$B=\\begin{pmatrix}-1 & -2\\\\1 & 2\\\\3&4 \\end{pmatrix}$$ $$AB=\\begin{bmatrix}2 & 1&3 \\\\-2 & 2&4\\\\-1&-3&-4 \\end{bmatrix}\\begin{bmatrix}-1 & -2\\\\1 & 2\\\\3&4 \\end{bmatrix}$$ $$=\\begin{bmatrix}-2+1+9 & -4+2+12 \\\\2+2+12 & 4+4+16\\\\1-3-12&2-6-16 \\end{bmatrix}$$ $$=\\begin{bmatrix}8 & 10\\\\16 & 24\\\\-14&-20 \\end{bmatrix}$$", "$$e_1 := \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}\\quad\\textrm{and}\\quad e_2 := \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix}.$$ In this basis $$T = \\begin{pmatrix} 2 & -1 \\\\ 1 & 1 \\end{pmatrix}.$$ The basis transformation matrices from $$(e_1,e_2)$$ to $$B_1$$ or $$B_2$$ are given by $$B_1 = \\begin{pmatrix} 1 & 1 \\\\ 2 & 1 \\end{pmatrix}\\quad\\textrm{and}\\quad B_2 = \\begin{pmatrix} -1 & 2 \\\\ 1 & 1 \\end{pmatrix}.$$ The transformation matrix in each basis will be given by $$TB_1^{-1} \\quad\\textrm{and}\\quad TB_2^{-1}.$$ • Ah so there is no $[u]$ required. But if $T$ had a vector, say $u = (2,3)$ then it would have to be multiplied to that formula? Jun 10, 2020 at 3:36", "\\\\4 & 5 & 6 \\\\7 & 8 & 9\\end{pmatrix}, \\begin{pmatrix}1 & 2 \\\\3 & 4\\end{pmatrix}^T, \\begin{pmatrix}1 & 2 & 3 \\\\4 & 5 & 6 \\\\7 & 2 & 9\\end{pmatrix}^{-1}, rank\\:\\begin{pmatrix}1 & 2 \\\\3 & 4\\end{pmatrix}, gauss\\:jordan\\:\\begin{pmatrix}1 & 2 \\\\3 & 4\\end{pmatrix}, eigenvalues\\:\\begin{pmatrix}6&-1\\\\2&3\\end{pmatrix}, eigenvectors\\:\\begin{pmatrix}6&-1\\\\2&3\\end{pmatrix}, diagonalize\\:\\begin{pmatrix}6&-1\\\\2&3\\end{pmatrix}. If the matrices are the same size, matrix addition is performed by adding the corresponding elements in the matrices. Solving systems of linear equations. The identity matrix is the matrix equivalent of the number \"1.\" In fact, just because A can be multiplied by B doesn't mean that B can be multiplied by A. You cannot add a 2 × 3 and a 3 × 2 matrix, a 4 × 4 and a 3 × 3, etc. Matrix Multiplication Calculator Here you can perform matrix multiplication with complex numbers online for free. A matrix is a rectangular array of numbers (or other mathematical objects) for which operations such as addition and multiplication are defined. The determinant of a 2 × 2 matrix can be calculated using the Leibniz formula, which involves some basic arithmetic. For example, given a matrix A and a scalar c: Multiplying two (or more) matrices is more involved than multiplying by a scalar. Row operation calculator: v. 1.25 PROBLEM TEMPLATE: Interactively perform a sequence of elementary row operations" ]

[ "+0 # Pls help me +1 32 1 +57 The vectors $\\mathbf{a},$ $\\mathbf{b},$ and $\\mathbf{c}$ satisfy $\\|\\mathbf{a}\\| = \\|\\mathbf{b}\\| = 1,$ $\\|\\mathbf{c}\\| = 2,$ and $\\mathbf{a} \\times (\\mathbf{a} \\times \\mathbf{c}) + \\mathbf{b} = \\mathbf{0}.$If $\\theta$ is the angle between $\\mathbf{a}$ and $\\mathbf{c},$ then find all possible values of $\\theta,$ in degrees. Apr 25, 2021", "Vectors $\\mathbf{a}$ and $\\mathbf{b}$ must be going in the same direction.", "If $c = |a| b + |b| a$, where $a$ , $b$ and $c$ are all non-zero vectors, show that $c$ bisects the angle between $a$ and $b$.", "and $C$ have position vectors $\\mathbf{a}, \\mathbf{b}$ and $\\mathbf{c}$ respectively.", "Vectors Question #2 If $c = |a| b + |b| a$, where $a$ , $b$ and $c$ are all non-zero vectors, show that $c$ bisects the angle between $a$ and $b$.", "views Show that two nonzero vectors \\mathbf{v}_{1} and \\mathbf{v}_{2} in R^{3} are orthogonal if and only if their direction cosines satisfyShow that two nonzero vectors \\mathbf{v}_{1} and \\mathbf{v}_{2} in R^{3} are orthogonal if and only if their direction cosines satisfy$$ \\cos \\ ... Let $\\mathbf{r}_{0}=\\left(x_{0}, y_{0}\\right)$ be a fixed vector in $R^{2}$. In each part, describe in words the set of all vectors $\\mathbf{r}=(x, y)$ that satisfy the stated condition.Let $\\mathbf{r}_{0}=\\left(x_{0}, y_{0}\\right)$ be a fixed vector in $R^{2}$. In each part, describe in words the set of all vectors \\$\\mathbf{r}=(x, y) ...", "## Angle between two vectors Determine the angle between the following pairs of real vectors (if possible, determine the exact value, otherwise the cosine). Decide whether the angle is acute or obtuse. • #### Variant 1 $$\\mathbf x^T=(1, -4)$$, $$\\mathbf y^T=(8, 2)$$. • #### Variant 2 $$\\mathbf x^T=(3, 2, -2)$$, $$\\mathbf y^T=(0, 4, 1)$$. • #### Variant 3 $$\\mathbf x^T=(0, 0, 1)$$, $$\\mathbf y^T=(1, 0, -1)$$. • #### Variant 4 $$\\mathbf x^T=(3, 4)$$, $$\\mathbf y^T=(-1, 0)$$.", "Review question # Can we find the angle between the planes $ABC$ and $ABD$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R6449 ## Question The position vectors of four points $A$, $B$, $C$, $D$ relative to an origin $O$ are given below. The vectors $\\mathbf{i}$, $\\mathbf{j}$, $\\mathbf{k}$ are mutually perpendicular unit vectors. \\begin{align*} A \\colon\\;& 2\\mathbf{i} + 3\\mathbf{j} + \\mathbf{k}, \\\\ B \\colon\\;& \\phantom{2}\\mathbf{i} + \\phantom{3}\\mathbf{j} - \\mathbf{k}, \\\\ C \\colon\\;& \\phantom{2}\\mathbf{i} \\phantom{{}+{}3\\mathbf{i}} + \\mathbf{k}, \\\\ D \\colon\\;& \\phantom{2\\mathbf{i}{}+{}} 3\\mathbf{j}.\\phantom{{}+{}\\mathbf{k}} \\end{align*} Find 1. the equation (in any form) of the line $AB$, 2. the shortest distance between $AB$ and $CD$, 3. the equation (in any form) of the plane $ABC$, 4. the angle between the planes $ABC$ and $ABD$.", "as in $\\mathbf{u.v}$, that no longer applies. That is why the answer to the question leaves $\\mathbf{u.v}$ as it is, without attempting to evaluate it. Unless you know the angle of the rhombus, you can't evaluate $\\mathbf{u.v}$.", "## Trigonometry (11th Edition) Clone $\\textbf{p}$ and $\\textbf{q}$ $\\textbf{r}$ and $\\textbf{s}$ $\\textbf{n}$ and $\\textbf{s}$ Two vectors are opposites when they have the same magnitude but different directions. According to this definition, the pair of vectors that appear to be opposite are $\\textbf{p}$ and $\\textbf{q}$,$\\textbf{r}$ and $\\textbf{s}$ and $\\textbf{n}$ and $\\textbf{s}$.", "# Problem: Find the angle between each of the following pairs of vectors A =Axi + Ayj and B =Bxi + ByjAx3= -4.00, Ay3 = 2.00; Bx3 = 7.00, By3 = 14.00. ###### FREE Expert Solution Vector dot product: $\\overline{)\\stackrel{\\mathbf{⇀}}{\\mathbf{A}}{\\mathbf{·}}\\stackrel{\\mathbf{⇀}}{\\mathbf{B}}{\\mathbf{=}}{\\mathbf{|}}{\\mathbf{A}}{\\mathbf{|}}{\\mathbf{|}}{\\mathbf{B}}{\\mathbf{|}}{\\mathbf{c}}{\\mathbf{o}}{\\mathbf{s}}{\\mathbf{\\theta }}}$ Vector magnitude: $\\overline{)\\mathbf{|}\\stackrel{\\mathbf{⇀}}{\\mathbit{A}}\\mathbf{|}{\\mathbf{=}}\\sqrt{{{\\mathbit{A}}_{\\mathbit{x}}}^{\\mathbf{2}}\\mathbf{+}{{\\mathbit{A}}_{\\mathbit{y}}}^{\\mathbf{2}}}}$ $\\begin{array}{rcl}{\\mathbf{\\theta }}_{{\\mathbf{A}}_{\\mathbf{3}}{\\mathbf{B}}_{\\mathbf{3}}}& \\mathbf{=}& {\\mathbf{cos}}^{\\mathbf{-}\\mathbf{1}}\\mathbf{\\left(}\\frac{{\\stackrel{\\mathbf{⇀}}{\\mathbf{A}}}_{\\mathbf{3}}\\mathbf{·}{\\stackrel{\\mathbf{⇀}}{\\mathbf{B}}}_{\\mathbf{3}}}{\\mathbf{|}{\\stackrel{\\mathbf{⇀}}{\\mathbf{A}}}_{\\mathbf{3}}\\mathbf{|}\\mathbf{|}{\\stackrel{\\mathbf{⇀}}{\\mathbf{B}}}_{\\mathbf{3}}\\mathbf{|}}\\mathbf{\\right)}\\end{array}$ ###### Problem Details Find the angle between each of the following pairs of vectors =Ax+ Ayj and B =Bx+ Byj Ax3= -4.00, Ay3 = 2.00; Bx3 = 7.00, By3 = 14.00.", "Browse Questions # Find the angle between two vectors $\\overrightarrow a \\: and \\: \\overrightarrow b$ if $|\\overrightarrow a |=3, |\\overrightarrow b|=4\\: and \\: |\\overrightarrow a * \\overrightarrow b |=6$ Can you answer this question?", "+0 # help 0 64 1 Let $$\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$$ be three vectors that lie on the same line. Find $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a}$$ May 1, 2020", "## Angle between two vectors ### Task number: 2689 Determine the angle between the following pairs of real vectors (if possible, determine the exact value, otherwise the cosine). Decide whether the angle is acute or obtuse. • #### Variant 1 $$\\mathbf x^T=(1, -4)$$, $$\\mathbf y^T=(8, 2)$$. • #### Variant 2 $$\\mathbf x^T=(3, 2, -2)$$, $$\\mathbf y^T=(0, 4, 1)$$. • #### Variant 3 $$\\mathbf x^T=(0, 0, 1)$$, $$\\mathbf y^T=(1, 0, -1)$$. • #### Variant 4 $$\\mathbf x^T=(3, 4)$$, $$\\mathbf y^T=(-1, 0)$$.", "that are perpendicular to the vector $\\mathbf{Z}=(2,1,2)$. 0 answers 2 views Find all vectors in the plane (through the origin) spanned by $\\mathbf{V}=(1,1-2)$ and $\\mathbf{W}=(-1,1,1)$ that are perpendicular to the vector $\\mathbf{Z}=(2,1,2)$.Find all vectors in the plane (through the origin) spanned by $\\mathbf{V}=(1,1-2)$ and $\\mathbf{W}=(-1,1,1)$ that are perpendicular to the vector ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Show that two nonzero vectors \\mathbf{v}_{1} and \\mathbf{v}_{2} in R^{3} are orthogonal if and only if their direction cosines satisfy 0 answers 10 views Show that two nonzero vectors \\mathbf{v}_{1} and \\mathbf{v}_{2} in R^{3} are orthogonal if and only if their direction cosines satisfyShow that two nonzero vectors \\mathbf{v}_{1} and \\mathbf{v}_{2} in R^{3} are orthogonal if and only if their direction cosines satisfy$$ \\cos \\ ... close", "the magnitudes of the vectors times the sine of the angle between the vectors and it is a vector perpendicular to ${\\bf a}$ and ${\\bf b}$.", "14 views Show that two nonzero vectors \\mathbf{v}_{1} and \\mathbf{v}_{2} in R^{3} are orthogonal if and only if their direction cosines satisfyShow that two nonzero vectors \\mathbf{v}_{1} and \\mathbf{v}_{2} in R^{3} are orthogonal if and only if their direction cosines satisfy$$ \\cos \\ ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find all vectors in the plane (through the origin) spanned by $\\mathbf{V}=(1,1-2)$ and $\\mathbf{W}=(-1,1,1)$ that are perpendicular to the vector $\\mathbf{Z}=(2,1,2)$. Find all vectors in the plane (through the origin) spanned by $\\mathbf{V}=(1,1-2)$ and $\\mathbf{W}=(-1,1,1)$ that are perpendicular to the vector $\\mathbf{Z}=(2,1,2)$.Find all vectors in the plane (through the origin) spanned by $\\mathbf{V}=(1,1-2)$ and $\\mathbf{W}=(-1,1,1)$ that are perpendicular to the vector ... Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors. Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors.Show that $\\mathbf{v}=(a, b)$ and $\\mathbf{w}=(-b, a)$ are orthogonal vectors. ... Let $\\vec{v}$ and $\\vec{w}$ be vectors in $\\mathbb{R}^{n}$. If $\\|\\vec{v}\\|=\\|\\vec{w}\\|$, show there is an orthogonal matrix $R$ with $R \\vec{v}=\\vec{w}$ and $R \\vec{w}=\\vec{v}$.Let $\\vec{v}$ and $\\vec{w}$ be vectors in $\\mathbb{R}^{n}$. If $\\|\\vec{v}\\|=\\|\\vec{w}\\|$, show there is an orthogonal matrix $R$ with \\(R \\v ...", "included angle between vectors is 90°. See figure. Problem : If a and b are two non-collinear unit vectors and | a + b | = √3, then find the value of expression : $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)\\end{array}$ Solution : The given expression is scalar product of two vector sums. Using distributive property we can expand the expression, which will comprise of scalar product of two vectors a and b . $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2\\mathbf{a}\\mathbf{.}\\mathbf{a}+\\mathbf{a}\\mathbf{.}\\mathbf{b}-\\mathbf{b}\\mathbf{.}2\\mathbf{a}+\\left(-\\mathbf{b}\\right)\\mathbf{.}\\left(-\\mathbf{b}\\right)=2{a}^{2}-\\mathbf{a}\\mathbf{.}\\mathbf{b}-{b}^{2}\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta \\end{array}$ We can evaluate this scalar product, if we know the angle between them as magnitudes of unit vectors are each 1. In order to find the angle between the vectors, we use the identity, $\\begin{array}{l}\\mathbf{A}\\mathbf{.}\\mathbf{A}={A}^{2}\\end{array}$ Now, $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}=\\left(\\mathbf{a}+\\mathbf{b}\\right)\\mathbf{.}\\left(\\mathbf{a}+\\mathbf{b}\\right)={a}^{2}+{b}^{2}+2ab\\mathrm{cos}\\theta =1+1+2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}\\theta \\end{array}$ $\\begin{array}{l}⇒{|\\mathbf{a}+\\mathbf{b}|}^{2}=2+2\\mathrm{cos}\\theta \\end{array}$ It is given that : $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}={\\left(\\surd 3\\right)}^{2}=3\\end{array}$ Putting this value, $\\begin{array}{l}⇒2\\mathrm{cos}\\theta ={|\\mathbf{a}+\\mathbf{b}|}^{2}-2=3-2=1\\end{array}$ $\\begin{array}{l}⇒\\mathrm{cos}\\theta =\\frac{1}{2}\\\\ ⇒\\theta =60°\\end{array}$ Using this value, we now proceed to find the value of given identity, $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta =2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}{1}^{2}-{1}^{2}-1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}60°\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=\\frac{1}{2}\\end{array}$ ## Evaluation of dot product Problem : In an experiment of light reflection, if a , b and c are the unit vectors in the direction of incident ray, reflected ray and normal to the reflecting surface, then prove that : $\\begin{array}{l}⇒\\mathbf{b}=\\mathbf{a}-2\\left(\\mathbf{a}\\mathbf{.}\\mathbf{c}\\right)\\mathbf{c}\\end{array}$ Solution : Let us consider vectors in a coordinate system in which “x” and “y” axes of the coordinate system are in the direction of reflecting surface and normal to the reflecting", "included angle between vectors is 90°. See figure. Problem : If a and b are two non-collinear unit vectors and | a + b | = √3, then find the value of expression : $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)\\end{array}$ Solution : The given expression is scalar product of two vector sums. Using distributive property we can expand the expression, which will comprise of scalar product of two vectors a and b . $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2\\mathbf{a}\\mathbf{.}\\mathbf{a}+\\mathbf{a}\\mathbf{.}\\mathbf{b}-\\mathbf{b}\\mathbf{.}2\\mathbf{a}+\\left(-\\mathbf{b}\\right)\\mathbf{.}\\left(-\\mathbf{b}\\right)=2{a}^{2}-\\mathbf{a}\\mathbf{.}\\mathbf{b}-{b}^{2}\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta \\end{array}$ We can evaluate this scalar product, if we know the angle between them as magnitudes of unit vectors are each 1. In order to find the angle between the vectors, we use the identity, $\\begin{array}{l}\\mathbf{A}\\mathbf{.}\\mathbf{A}={A}^{2}\\end{array}$ Now, $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}=\\left(\\mathbf{a}+\\mathbf{b}\\right)\\mathbf{.}\\left(\\mathbf{a}+\\mathbf{b}\\right)={a}^{2}+{b}^{2}+2ab\\mathrm{cos}\\theta =1+1+2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}\\theta \\end{array}$ $\\begin{array}{l}⇒{|\\mathbf{a}+\\mathbf{b}|}^{2}=2+2\\mathrm{cos}\\theta \\end{array}$ It is given that : $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}={\\left(\\surd 3\\right)}^{2}=3\\end{array}$ Putting this value, $\\begin{array}{l}⇒2\\mathrm{cos}\\theta ={|\\mathbf{a}+\\mathbf{b}|}^{2}-2=3-2=1\\end{array}$ $\\begin{array}{l}⇒\\mathrm{cos}\\theta =\\frac{1}{2}\\\\ ⇒\\theta =60°\\end{array}$ Using this value, we now proceed to find the value of given identity, $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta =2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}{1}^{2}-{1}^{2}-1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}60°\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=\\frac{1}{2}\\end{array}$ ## Evaluation of dot product Problem : In an experiment of light reflection, if a , b and c are the unit vectors in the direction of incident ray, reflected ray and normal to the reflecting surface, then prove that : $\\begin{array}{l}⇒\\mathbf{b}=\\mathbf{a}-2\\left(\\mathbf{a}\\mathbf{.}\\mathbf{c}\\right)\\mathbf{c}\\end{array}$ Solution : Let us consider vectors in a coordinate system in which “x” and “y” axes of the coordinate system are in the direction of reflecting surface and normal to the reflecting", "where $\\mathbf u$ is the position of the particle on 3-sphere (a 4-dimensional vector), $\\mathbf p$ is the momentum of the original particle in Kepler problem, $\\hat{\\mathbf n}$ is a vector perpendicular to the 3-dimensional hyperplane where $\\mathbf p$ lies, and $p_0:=\\sqrt{-2mE}$." ]

[ "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "$$\\mathbf{B}(t)$$ on the curve, we have this condition: $\\begin{split}(\\mathbf{P} - \\mathbf{B}(t)) \\cdot \\mathbf{B}'(t) = 0 \\\\[.5em] \\mathbf{P} \\cdot \\mathbf{B}'(t) - \\mathbf{B}(t) \\cdot \\mathbf{B}'(t) = 0\\end{split}$ ### 9.2.1.13.1. Quadratic Bézier Curve¶ Let \\begin{split}\\begin{align} \\mathbf{A} &= \\mathbf{P}_1 - \\mathbf{P}_0 \\\\[.5em] \\mathbf{B} &= \\mathbf{P}_2 - \\mathbf{P}_1 -\\mathbf{A} \\\\[.5em] \\mathbf{M} &= \\mathbf{P}_0 - \\mathbf{P} \\end{align}\\end{split} We have $\\mathbf{B}'(t) = 2(\\mathbf{A} + \\mathbf{B} t)$ The condition can be expressed as $\\mathbf{B}^2 t^3 + 3\\mathbf{A}\\mathbf{B} t^2 + (2\\mathbf{A}^2 + \\mathbf{M}\\mathbf{B}) t + \\mathbf{M}\\mathbf{A} = 0$ >>> C = collect(expand((P*B2p - B2*B2p)/-2), t) P*P0 - P*P1 - P0**2 + P0*P1 + t**3 * (P0**2 - 4*P0*P1 + 2*P0*P2 + 4*P1**2 - 4*P1*P2 + P2**2) + t**2 * (-3*P0**2 + 9*P0*P1 - 3*P0*P2 - 6*P1**2 + 3*P1*P2) + t * (-P*P0 + 2*P*P1 - P*P2 + 3*P0**2 - 6*P0*P1 + P0*P2 + 2*P1**2) >>> A = P1 - P0 B = P2 - P1 - A M = P0 - P >>> C2 = B**2 * t**3 + 3*A*B * t**2 + (2*A**2 + M*B) * t + M*A >>> expand(C - C2) 0 ### 9.2.1.13.2. Cubic Bézier Curve¶ Let \\begin{split}\\begin{align} n &= \\mathbf{P}_3 - 3\\mathbf{P}_2 + 3\\mathbf{P}_1 - \\mathbf{P}_0 \\\\[.5em] r &= 3(\\mathbf{P}_2 - 2\\mathbf{P}_1 + \\mathbf{P}_0 \\\\[.5em] s &= 3(\\mathbf{P}_1 - \\mathbf{P}_0) \\\\[.5em] v &= \\mathbf{P}_0 \\end{align}\\end{split} We have \\begin{align}\\begin{aligned}\\begin{split}\\begin{align} \\mathbf{B}^3(t) &= nt^3 +", "of this is 2 + 2\\cdot 3 + 2\\cdot 3^2 + 2\\cdot 3^3, so you normally you would form the array (\\mathbf{2'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}), which has length (2+1+1+1)+(2+3+1+1)+(2+3+3+1)+(2+3+3+3) = 32. However, writing the ternary form as 2 + 3\\cdot(2 + 3\\cdot (2 + 3\\cdot 2)), the array would be (\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'[(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'(\\mathbf{2'}\\cdot'\\mathbf{1'} +' \\mathbf{3'}\\cdot'\\mathbf{2'})], which only has length (2+1+1+1)+3+(2+1+1)+3+(2+1)+(3+2) = 23. In general, the smaller the total numbers that appear on your representation, the better. Using this technique, we are now able to reduce the required length to at most 94, and we’ve now solved the problem! The ideas above are enough to pass the problem already, but I’m sure with a couple more tricks you can reduce the length even more. ### Time Complexity: [O(N)](https://en.wikipedia.org/wiki/Output-sensitive _ algorithm) ### AUTHOR’S AND TESTER’S SOLUTIONS: #2 “I loved solving this question.” The constraint is very restrictive, just use numbers very efficiently. So, I tried to make a LIS’s of length equal to MSB of K (for K = 4, length = 3). Now you need exactly K such LIS’s, so now try to create LIS’s in powers of two starting from the least power of 2 present, if power of 2 p, then create p pairs of numbers so", "$2 + 2\\cdot 3 + 2\\cdot 3^2 + 2\\cdot 3^3$, so you normally you would form the array $(\\mathbf{2'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'})$, which has length $(2+1+1+1)+(2+3+1+1)+(2+3+3+1)+(2+3+3+3) = 32$. However, writing the ternary form as $2 + 3\\cdot(2 + 3\\cdot (2 + 3\\cdot 2))$, the array would be $(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'[(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'(\\mathbf{2'}\\cdot'\\mathbf{1'} +' \\mathbf{3'}\\cdot'\\mathbf{2'})]$, which only has length $(2+1+1+1)+3+(2+1+1)+3+(2+1)+(3+2) = 23$. In general, the smaller the total numbers that appear on your representation, the better. Using this technique, we are now able to reduce the required length to at most $94$, and we've now solved the problem! The ideas above are enough to pass the problem already, but I'm sure with a couple more tricks you can reduce the length even more. # Time Complexity: $O(N)$ # AUTHOR'S AND TESTER'S SOLUTIONS: This question is marked \"community wiki\". 1.7k586142 accept rate: 11% 19.7k350498541 4 “I loved solving this question.” The constraint is very restrictive, just use numbers very efficiently. So, I tried to make a LIS’s of length equal to MSB of K (for K = 4, length = 3). Now you need exactly K such LIS’s, so now try to create LIS’s in powers of two starting from the least power of 2 present, if power of 2 p, then", "= \\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} - 2 ab \\cos\\theta \\;$. (1) But as cab, we also have $\\mathbf{c} \\cdot \\mathbf{c} = (\\mathbf{a} - \\mathbf{b}) \\cdot (\\mathbf{a} - \\mathbf{b}) \\;$, which, according to the distributive law, expands to $\\mathbf{c} \\cdot \\mathbf{c} = \\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} -2(\\mathbf{a} \\cdot \\mathbf{b}) \\;$. (2) Merging the two c · c equations, (1) and (2), we obtain $\\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} -2(\\mathbf{a} \\cdot \\mathbf{b}) = \\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} - 2 ab \\cos\\theta \\;$. Subtracting a · a + b · b from both sides and dividing by −2 leaves $\\mathbf{a} \\cdot \\mathbf{b} = ab \\cos\\theta \\;$,", "${\\displaystyle \\mathbf {x} _{+}}$, such as ${\\displaystyle \\mathbf {x} _{2}}$, due to commuting or shifting ${\\displaystyle \\mathbf {a} _{k}}$ right an odd number of times, in the dividend blade ${\\displaystyle \\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n}}$, into its Template:Mvar-th spot. In general, ${\\displaystyle \\mathbf {x} _{k}}$ is solved as {\\displaystyle {\\begin{aligned}\\mathbf {x} _{k}&=(\\mathbf {a} _{1}\\wedge \\cdots \\wedge (\\mathbf {C} )_{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})\\cdot ((-1)^{k-1}\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})^{-1}\\\\[6pt]&={\\frac {(\\mathbf {a} _{1}\\wedge \\cdots \\wedge (\\mathbf {C} )_{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})\\cdot ((-1)^{k-1}\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})}{(\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})^{2}}}\\\\[6pt]&={\\frac {(-1)^{k-1}(\\mathbf {a} _{1}\\wedge \\cdots \\wedge (\\mathbf {C} )_{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})\\cdot (\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})}{(-1)^{\\frac {n(n-1)}{2}}(\\mathbf {a} _{n}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{1})\\cdot (\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})}}\\\\[6pt]&={\\frac {(-1)^{k-1}(\\mathbf {a} _{1}\\wedge \\cdots \\wedge (\\mathbf {C} )_{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})\\cdot (\\mathbf {a} _{n}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{1})}{(\\mathbf {a} _{n}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{1})\\cdot (\\mathbf {a} _{1}\\wedge \\cdots \\wedge", "\\mathbf c)$ of the vectors $\\mathbf a$, $\\mathbf b$ and $\\mathbf c$. Some notable properties of cross and triple products: 1. $\\mathbf a\\times \\mathbf b = -\\mathbf b\\times \\mathbf a$ 2. $(\\alpha \\cdot \\mathbf a)\\times \\mathbf b = \\alpha \\cdot (\\mathbf a\\times \\mathbf b)$ 3. For any $\\mathbf b$ and $\\mathbf c$ there is exactly one vector $\\mathbf r$ such that $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf a\\cdot\\mathbf r$ for any vector $\\mathbf a$. Indeed if there are two such vectors $\\mathbf r_1$ and $\\mathbf r_2$ then $\\mathbf a\\cdot (\\mathbf r_1 - \\mathbf r_2)=0$ for all vectors $\\mathbf a$ which is possible only when $\\mathbf r_1 = \\mathbf r_2$. 4. $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf b\\cdot (\\mathbf c\\times \\mathbf a) = -\\mathbf a\\cdot( \\mathbf c\\times \\mathbf b)$ 5. $(\\mathbf a + \\mathbf b)\\times \\mathbf c = \\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c$. Indeed for all vectors $\\mathbf r$ the chain of equations holds: $$\\mathbf r\\cdot( (\\mathbf a + \\mathbf b)\\times \\mathbf c) = (\\mathbf a + \\mathbf b) \\cdot (\\mathbf c\\times \\mathbf r) = \\mathbf a \\cdot(\\mathbf c\\times \\mathbf r) + \\mathbf b\\cdot(\\mathbf c\\times \\mathbf r) = \\mathbf r\\cdot (\\mathbf a\\times \\mathbf c) + \\mathbf r\\cdot(\\mathbf b\\times \\mathbf c) = \\mathbf r\\cdot(\\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c)$$", "&= I_{1 2 \\cdots (k-1)} I \\left\\lvert {d\\mathbf{x}_1 \\wedge d\\mathbf{x}_2 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I}", "= \\mathbf{a} \\cdot \\mathbf{c} + \\mathbf{a} \\cdot \\mathbf{d} + \\mathbf{b} \\cdot \\mathbf{c} + \\mathbf{b} \\cdot \\mathbf{d}$ Consider a triangle $OAB$ with $\\mathbf{a}, \\mathbf{b}$ the position vectors of points $A$ and $B$. Let $\\theta$ be the angle $AOB$. By the cosine rule, $AB^2 = OA^2 + OB^2 - 2 \\, OA \\cdot OB \\cos \\theta$. $\\begin{gather} | \\mathbf{b}-\\mathbf{a} |^2 = |\\mathbf{a}|^2 + |\\mathbf{b}|^2 - 2 |\\mathbf{a}| \\, |\\mathbf{b}| \\cos \\theta \\tag{4} \\end{gather}$ By using formula 1, $|\\mathbf{b}-\\mathbf{a}|^2 = (\\mathbf{b}-\\mathbf{a}) \\cdot (\\mathbf{b}-\\mathbf{a})$. Expanding using formula 3 gives us $\\mathbf{b} \\cdot \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} - \\mathbf{b} \\cdot \\mathbf{a} + \\mathbf{a} \\cdot \\mathbf{a}$. The use of formulas 1 and 2 simplifies the expression to $|\\mathbf{b}|^2 - 2 \\mathbf{a}\\cdot\\mathbf{b} + |\\mathbf{a}|^2$. Plugging this into equation 4 gives us $|\\mathbf{a}|^2 + |\\mathbf{b}|^2 - 2 \\mathbf{a} \\cdot \\mathbf{b} = |\\mathbf{a}|^2 + |\\mathbf{b}|^2 - 2 |\\mathbf{a}| \\, |\\mathbf{b}| \\cos \\theta$ $-2\\mathbf{a} \\cdot \\mathbf{b} = -2 |\\mathbf{a}| \\, |\\mathbf{b} | \\cos \\theta$. ### Angles and perpendicular vectors The dot product formula is especially handy in two situations. The first is in the computation of angles. Given two vectors $\\mathbf{a}$ and $\\mathbf{b}$, the values of $\\mathbf{a} \\cdot \\mathbf{b}$, $|\\mathbf{a}|$ and $|\\mathbf{b}|$ are all relatively simple to compute. Substituting them into our formula allows us to calculate the angle between them. The second situation is when vectors", "shifted by fixed amounts without changing the sum. For example, ${\\displaystyle \\sum _{i}F_{i}G_{i+1}H_{i+2}=\\sum _{i}F_{i+1}G_{i+2}H_{i}=\\sum _{i}F_{i+2}G_{i}H_{i+1}}$. This allows: ${\\displaystyle (\\sum _{i}F_{i}G_{i+1}H_{i+2})-(\\sum _{i}F_{i}G_{i+2}H_{i+1})=(\\sum _{i}F_{i+2}G_{i}H_{i+1})-(\\sum _{i}F_{i+1}G_{i}H_{i+2})=(\\sum _{i}F_{i+1}G_{i+2}H_{i})-(\\sum _{i}F_{i+2}G_{i+1}H_{i})}$ ${\\displaystyle \\implies \\mathbf {F} \\cdot (i,G_{i+1}H_{i+2}-G_{i+2}H_{i+1})=\\mathbf {G} \\cdot (i,H_{i+1}F_{i+2}-H_{i+2}F_{i+1})=\\mathbf {H} \\cdot (i,F_{i+1}G_{i+2}-F_{i+2}G_{i+1})}$ ${\\displaystyle \\implies \\mathbf {F} \\cdot (\\mathbf {G} \\times \\mathbf {H} )=\\mathbf {G} \\cdot (\\mathbf {H} \\times \\mathbf {F} )=\\mathbf {H} \\cdot (\\mathbf {F} \\times \\mathbf {G} )}$ which establishes the cyclical property of the scalar triple product. Given scalar fields, ${\\displaystyle f}$ and ${\\displaystyle g}$, then ${\\displaystyle \\nabla (f+g)=(\\nabla f)+(\\nabla g)}$. Derivation ${\\displaystyle \\nabla (f+g)=(i,{\\frac {\\partial }{\\partial x_{i}}}(f+g))}$ ${\\displaystyle =(i,{\\frac {\\partial f}{\\partial x_{i}}}+{\\frac {\\partial g}{\\partial x_{i}}})}$ ${\\displaystyle =(i,{\\frac {\\partial f}{\\partial x_{i}}})+(i,{\\frac {\\partial g}{\\partial x_{i}}})}$ ${\\displaystyle =(\\nabla f)+(\\nabla g)}$ Given scalar fields ${\\displaystyle f}$ and ${\\displaystyle g}$, then ${\\displaystyle \\nabla (fg)=(\\nabla f)g+f(\\nabla g)}$. If ${\\displaystyle f}$ is a constant ${\\displaystyle c}$, then ${\\displaystyle \\nabla (cg)=c(\\nabla g)}$. Derivation ${\\displaystyle \\nabla (fg)=(i,{\\frac {\\partial }{\\partial x_{i}}}(fg))}$ ${\\displaystyle =(i,{\\frac {\\partial f}{\\partial x_{i}}}g+f{\\frac {\\partial g}{\\partial x_{i}}})}$ ${\\displaystyle =(i,{\\frac {\\partial f}{\\partial x_{i}}})g+f(i,{\\frac {\\partial g}{\\partial x_{i}}})}$ ${\\displaystyle =(\\nabla f)g+f(\\nabla g)}$ Given vector fields ${\\displaystyle \\mathbf {F} }$ and ${\\displaystyle \\mathbf {G} }$, then ${\\displaystyle \\nabla (\\mathbf {F} \\cdot \\mathbf {G} )=((\\mathbf {F} \\cdot \\nabla )\\mathbf {G} +\\mathbf {F} \\times (\\nabla \\times \\mathbf {G} ))+((\\mathbf {G} \\cdot \\nabla )\\mathbf {F} +\\mathbf {G} \\times (\\nabla \\times \\mathbf {F} ))}$ Derivation ${\\displaystyle", "\\begin{aligned}\\mathbf{a}^{*} \\cdot \\mathbf{a} = 1,\\end{aligned} \\hspace{\\stretch{1}}(1.174) and \\begin{aligned}\\mathbf{a}^{*} \\cdot \\mathbf{b} = 0.\\end{aligned} \\hspace{\\stretch{1}}(1.175) \\begin{aligned}\\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right)=\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}.\\end{aligned} \\hspace{\\stretch{1}}(1.176) Dotting with $\\mathbf{a}$ we have \\begin{aligned}\\mathbf{a} \\cdot \\left( { \\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right) } \\right)=\\mathbf{a} \\cdot \\left( {\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}} \\right)=\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right)^2 - \\mathbf{b}^2 \\mathbf{a}^2,\\end{aligned} \\hspace{\\stretch{1}}(1.177) but dotting with $\\mathbf{b}$ yields zero \\begin{aligned}\\mathbf{b} \\cdot \\left( { \\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right) } \\right) &= \\mathbf{b} \\cdot \\left( {\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}} \\right) \\\\ &= \\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}^2 - \\mathbf{b}^2 \\left( { \\mathbf{a} \\cdot \\mathbf{b} } \\right) \\\\ &= 0.\\end{aligned} \\hspace{\\stretch{1}}(1.178) To complete the proof, we note that the product in eq. 1.177 is just the wedge squared \\begin{aligned}\\left( { \\mathbf{a} \\wedge \\mathbf{b}} \\right)^2 &= \\left\\langle{{\\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right)^2}}\\right\\rangle \\\\ &= \\left\\langle{{\\left( { \\mathbf{a} \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} } \\right)\\left( { \\mathbf{a} \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} } \\right)}}\\right\\rangle \\\\ &= \\left\\langle{{\\mathbf{a} \\mathbf{b} \\mathbf{a} \\mathbf{b} - 2 \\left( {\\mathbf{a} \\cdot \\mathbf{b}} \\right) \\mathbf{a} \\mathbf{b}}}\\right\\rangle+\\left( { \\mathbf{a} \\cdot \\mathbf{b} } \\right)^2 \\\\ &= \\left\\langle{{\\mathbf{a} \\mathbf{b} \\left( { -\\mathbf{b} \\mathbf{a} + 2 \\mathbf{a} \\cdot \\mathbf{b} }", "$(\\mathbf{a}+\\mathbf{b})\\cdot(\\mathbf{c}+\\mathbf{d}) = \\mathbf{a}\\cdot\\mathbf{c} + \\mathbf{a}\\cdot\\mathbf{d} + \\mathbf{b}\\cdot\\mathbf{c} + \\mathbf{b}\\cdot\\mathbf{d}$ ## Mixed associative law The dot product has a homogeneity $\\alpha (\\mathbf{a}\\cdot\\mathbf{b}) = (\\alpha\\mathbf{a})\\cdot \\mathbf{b} = \\mathbf{a}\\cdot(\\alpha\\mathbf{b})$ ## Bilinear Often the mixed associative law and distributive law are combined to show that the dot product is bilinear: $\\mathbf{a}\\cdot(\\alpha\\mathbf{b}+\\mathbf{c}) = \\alpha(\\mathbf{a}\\cdot\\mathbf{b})+(\\mathbf{a}\\cdot\\mathbf{c})$ ## Scalar Multiplication $(\\alpha_1\\mathbf{a})\\cdot(\\alpha_2\\mathbf{b})=\\alpha_1\\alpha_2(\\mathbf{a}\\cdot\\mathbf{b})$ Dot product is homogenious under scaling $\\alpha(\\mathbf{a}\\cdot\\mathbf{b}) = (\\alpha\\mathbf{a})\\cdot\\mathbf{b} = \\mathbf{a}\\cdot(\\alpha\\mathbf{b})$ ## Cauchy-Schwarz Inequality $|\\mathbf{a}\\cdot\\mathbf{b}|\\leq|\\mathbf{a}||\\mathbf{b}|$ ## No cancellation If $$\\mathbf{a}\\cdot\\mathbf{b}=\\mathbf{a}\\cdot\\mathbf{c}$$ and $$\\mathbf{a}\\neq \\mathbf{0}$$, we can write $$\\mathbf{a}\\cdot(\\mathbf{b}-\\mathbf{c})=0$$ by distributive law, which means that $$\\mathbf{a}\\perp(\\mathbf{b}-\\mathbf{c})$$, which allows $$(\\mathbf{b}-\\mathbf{c})\\neq\\mathbf{0}$$ and therefore $$\\mathbf{b}\\neq\\mathbf{c}$$. ## Applications ### Fast calculation of Cosine The definition of the dot-product is used extensively in computer graphics since it speeds up computations in many circumstances by avoiding inefficient trigonometric functions. Additionally, working with normalized vectors gives the cosine between these two vectors by taking the dot product of them: $\\cos\\theta = \\frac{\\mathbf{a}\\cdot\\mathbf{b}}{|\\mathbf{a}||\\mathbf{b}|} = \\hat{\\mathbf{a}}\\cdot\\hat{\\mathbf{b}}$ ### Making perpendicular vectors Since two vectors are perpendicular when $$\\mathbf{a}\\cdot\\mathbf{b} = 0$$, we can easily construct a perpendicular vector by zeroing all components except 2, flipping those two and reversing the sign of one of them. For example with $$\\mathbf{a} = (\\mathbf{a}_1, \\mathbf{a}_2, \\dots, \\mathbf{a}_n)$$ the vectors $$\\mathbf{a}' = (-\\mathbf{a}_2, \\mathbf{a}_1, 0, \\dots, 0)$$ or $$\\mathbf{a}'' = (0, -\\mathbf{a}_3, \\mathbf{a}_2, 0, \\dots, 0)$$, etc are all perpendcular to $$\\mathbf{a}$$,", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "of $$X$$, $$f_{X}(x)$$, is then \\begin{align*} f_{X}(0) &= \\mathbf{P}(X = 0) \\\\ &= \\mathbf{P}(\\mathrm{Hit\\:E})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &= 81/625 \\\\ f_{X}(1) &= \\mathbf{P}(X = 1) \\\\ &= 2 \\cdot \\mathbf{P}(\\mathrm{Hit\\:D})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &= 126/625 \\\\ f_{X}(2) &= \\mathbf{P}(X = 2) \\\\ &= \\begin{align*}& 2 \\cdot \\mathbf{P}(\\mathrm{Hit\\:C})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &+ \\mathbf{P}(\\mathrm{Hit\\:D})\\mathbf{P}(\\mathrm{Hit\\:D}) \\end{align*} \\\\ &= 139/625 \\end{align*} and so on. From (2), the expected value of $$X$$ is $$\\mathbf{E}(X) = 0.81/625 + 1 \\cdot 126/625 + 2 \\cdot 139/625 + 3 \\cdot 124/625 + ... = 2.4$$. From (4), the variance of $$X$$ is $$\\mathrm{Var}(X) = 2.4^{2} \\cdot 81/625 + 1.4^{2} \\cdot 126/625 + 0.4^{2} \\cdot 139/625 + 0.6^{2} \\cdot 124/625 + ... = 2.72$$. The distribution function of $$X$$, $$F_{X}(x)$$, is then \\begin{align*} F_{X}(0) &= \\mathbf{P}(X \\le 0) \\\\ &= f_{X}(0) \\\\ F_{X}(1) &= \\mathbf{P}(X \\le 1) \\\\ &= f_{X}(1) + f_{X}(0) \\\\ F_{X}(2) &= \\mathbf{P}(X \\le 2) \\\\ &= f_{X}(2) + f_{X}(1) + f_{X}(0) \\end{align*} and so on. The distribution function $$F_{X}(x)$$ is shown in Figure 4. Figure 4: The distribution function $$F_{X}$$ of $$X$$ for the archery target.", "B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot \\partial_2 B \\right) \\\\ &\\qquad - \\mathbf{x}_2 \\left( \\mathbf{x}_{34} \\cdot \\partial_1 B \\right) - \\mathbf{x}_3 \\left( \\mathbf{x}_{42} \\cdot \\partial_1 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{23} \\cdot \\partial_1 B \\right)} \\Bigr) \\\\", "+ (-a_1 - b_1i) = (a_1 - a_1) + (b_1 - b_1)i = 0 + 0i = 0$. • FM1: $z_1 \\cdot z_2 = (a_1 + b_1i)(a_2 + b_2i) = a_1a_2 + a_1b_2i + b_1a_2i + b_1b_2i^2 = (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i$ and so $(z_1 \\cdot z_2) \\in \\mathbb{C}$. • FM2: $z_1 \\cdot z_2 = (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i = (a_2a_1 - b_2b_1) + (a_2b_1 + b_2a_1)i = z_2 \\cdot z_1$. • FM3: (1) \\begin{align} z_1 \\cdot (z_2 \\cdot z_3) \\\\ = (a_1 + b_1i) [ (a_2a_3 - b_2b_3) + (a_2b_3 + b_2a_3)i ] \\\\ = (a_1a_2a_3 - a_1b_2b_3) + (a_1a_2b_3 + a_1b_2a_3)i + (b_1a_2a_3 - b_1b_2b_3)i + (b_1a_2b_3 + b_1b_2a_3)i^2 \\\\ \\quad = [(a_1a_2a_3 - a_1b_2b_3) - (b_1a_2b_3 + b_1b_2a_3)] + [(a_1a_2b_3 + a_1b_2a_3) + (b_1a_2a_3 - b_1b_2b_3)]i \\\\ = [ (a_1a_2 - b_1b_2) + (a_1b_2 + b_1a_2)i ] (a_3 + b_3i) \\\\ = (z_1 \\cdot z_2) \\cdot z_3 \\end{align} • FM4: The multiplicative identity is $1 = 1 + 0i$. That is $z_1 \\cdot 1 = (a_1 + b_1)(1 + 0i) = (a_1 + a_1 \\cdot 0i) + (b_1 + b_1 \\cdot 0)i = a_1 + b_1i = z_1$. • FM5: The multiplicative inverse for $z_1$ where $z_1 \\neq 0$ is $z_1^{-1} = \\frac{a_1 -b_1i}{a_1^2 + b_1^2}$. That is: (2)", "+ \\mathbf{b} \\cdot \\mathbf{a} - \\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\mathbf{a} \\cdot \\mathbf{a} - \\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\lvert \\mathbf{a} \\rvert ^ 2 - \\lvert \\mathbf{b} \\rvert ^2 \\end{array}$ 2. $\\begin{array}{ll} \\lvert a + b \\rvert ^ 2 &= (\\mathbf{a} + \\mathbf{b}) \\cdot (\\mathbf{a} + \\mathbf{b}) \\\\ &= \\mathbf{a} \\cdot (\\mathbf{a} + \\mathbf{b}) + \\mathbf{b} \\cdot (\\mathbf{a} + \\mathbf{b}) \\\\ &= \\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{a} \\cdot \\mathbf{b} + \\mathbf{b} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\mathbf{a} \\cdot \\mathbf{a} + 2 (\\mathbf{a} \\cdot \\mathbf{b}) + \\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\lvert \\mathbf{a} \\rvert^2 + 2 (\\mathbf{a} \\cdot \\mathbf{b}) + \\lvert \\mathbf{b} \\rvert ^ 2 \\\\ \\end{array}$ Exercise 16 1. We know that $$\\lvert \\mathbf{a} \\rvert = \\lvert \\mathbf{b} \\rvert = 1$$ and $$\\mathbf{a} \\cdot \\mathbf{b} = 0$$. If the two vectors are perpendicular we must have $(2\\mathbf{a} + 3\\mathbf{b}) \\cdot (m\\mathbf{a} + \\mathbf{b}) = 0$ We can rewrite this as $\\begin{array}{ll} 0 &= (2\\mathbf{a} + 3\\mathbf{b}) \\cdot (m\\mathbf{a} + \\mathbf{b}) \\\\ &= 2\\mathbf{a} \\cdot (m\\mathbf{a} + \\mathbf{b}) + 3\\mathbf{b} \\cdot (m\\mathbf{a} + \\mathbf{b}) \\\\ &= 2\\mathbf{a} \\cdot m\\mathbf{a} + 2\\mathbf{a} \\cdot \\mathbf{b} + 3\\mathbf{b} \\cdot m\\mathbf{a} + 3\\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\lvert 2\\mathbf{a} \\rvert \\lvert m\\mathbf{a} \\rvert + \\lvert 3\\mathbf{b} \\rvert \\lvert \\mathbf{b} \\rvert \\\\ &= 2m + 3 \\end{array}$ Therefore $$m = -\\frac{3}{2}$$", "as cab, we also have ${\\displaystyle \\mathbf {c} \\cdot \\mathbf {c} =(\\mathbf {a} -\\mathbf {b} )\\cdot (\\mathbf {a} -\\mathbf {b} )\\,}$, which, according to the distributive law, expands to ${\\displaystyle \\mathbf {c} \\cdot \\mathbf {c} =\\mathbf {a} \\cdot \\mathbf {a} +\\mathbf {b} \\cdot \\mathbf {b} -2(\\mathbf {a} \\cdot \\mathbf {b} ).\\,}$ (2) Merging the two cc equations, (1) and (2), we obtain ${\\displaystyle \\mathbf {a} \\cdot \\mathbf {a} +\\mathbf {b} \\cdot \\mathbf {b} -2(\\mathbf {a} \\cdot \\mathbf {b} )=\\mathbf {a} \\cdot \\mathbf {a} +\\mathbf {b} \\cdot \\mathbf {b} -2|\\mathbf {a} ||\\mathbf {b} |\\cos \\theta .\\,}$ Subtracting aa + bb from both sides and dividing by −2 leaves ${\\displaystyle \\mathbf {a} \\cdot \\mathbf {b} =|\\mathbf {a} ||\\mathbf {b} |\\cos \\theta .\\,}$ Q.E.D. ## Generalization The inner product generalizes the dot product to abstract vector spaces and is usually denoted by ${\\displaystyle \\langle \\mathbf {a} \\,,\\mathbf {b} \\rangle }$. Due to the geometric interpretation of the dot product the norm ||a|| of a vector a in such an inner product space is defined as ${\\displaystyle \\|\\mathbf {a} \\|={\\sqrt {\\langle \\mathbf {a} \\,,\\mathbf {a} \\rangle }}}$ such that it generalizes length, and the angle θ between two vectors a and b by ${\\displaystyle \\cos {\\theta }={\\frac {\\langle \\mathbf {a} \\,,\\mathbf {b} \\rangle }{\\|\\mathbf {a} \\|\\,\\|\\mathbf {b} \\|}}.}$ In particular, two vectors", "\\wedge d\\mathbf{x}_2 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I} \\right) \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.118) On the left our", "\\wedge d\\mathbf{x}_2 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I} \\right) \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.118) On the left our" ]

[ "|\\vec b|}$. When $\\theta<90^\\circ$, we have $\\vec a \\cdot \\vec b > 0$. Therefore, $|\\vec a+\\vec b|^2 = (\\vec a+\\vec b)\\cdot(\\vec a+\\vec b) = |\\vec a|^2 + 2 \\vec a \\cdot \\vec b + |\\vec b|^2 > |\\vec a|^2 + |\\vec b|^2$. Therefore, $|\\vec a+\\vec b| > |\\vec a|$ and $|\\vec a+\\vec b| > |\\vec b|$. A stronger lower bound would be $|\\vec a+\\vec b| > \\sqrt{|\\vec a|^2+|\\vec b|^2}$. Let $\\vec{a}=(a_x,a_y)$ and $\\vec{b}=(b_x,b_y)$, so $\\vec{a}+\\vec{b}=(a_x+b_x,a_y+b_y)$, so it's length is $|\\vec{a}+\\vec{b}|=\\sqrt{(a_x+b_x)^2+(a_y+b_y)^2}$, and $|\\vec{a}+\\vec{b}|^2=(a_x+b_x)^2+(a_y+b_y)^2=a_x^2+2a_xb_x+b_x^2+a_y^2+2a_yb_y+b_y^2=|a|^2+|b|^2+2(a_xb_x+a_yby)=|a|^2+|b|^2+2(\\vec{a}\\cdot\\vec{b})$ Since the angle between both vectors is less than $90º$, then $\\vec{a}\\cdot\\vec{b}>0$, so $|\\vec{a}+\\vec{b}|^2>|a|^2+|b|^2$, so $|\\vec{a}+\\vec{b}|^2>|a|^2$ and $|\\vec{a}+\\vec{b}|^2>|b|^2$", "a \\to b, b \\to c, c \\to a$$ This gives $${\\bf A} = \\frac{{\\bf b}\\times {\\bf c}}{{\\bf a}\\cdot({\\bf b}\\times {\\bf c})} \\quad {\\bf B} = \\frac{{\\bf c}\\times {\\bf a}}{{\\bf b}\\cdot({\\bf c}\\times {\\bf a})} \\quad {\\bf C} = \\frac{{\\bf a}\\times {\\bf b}}{{\\bf c}\\cdot({\\bf a}\\times {\\bf b})}$$ and $${\\bf a} = \\frac{{\\bf B}\\times {\\bf C}}{{\\bf A}\\cdot({\\bf B}\\times {\\bf C})} \\quad {\\bf b} = \\frac{{\\bf C}\\times {\\bf A}}{{\\bf B}\\cdot({\\bf C}\\times {\\bf A})} \\quad {\\bf c} = \\frac{{\\bf A}\\times {\\bf B}}{{\\bf C}\\cdot({\\bf A}\\times {\\bf B})}$$ [..] the \"crystallographer's\" definition, comes from defining the reciprocal lattice to be $e^{2 \\pi i\\mathbf{K}\\cdot\\mathbf{R}}=1$ which changes the definitions of the reciprocal lattice vectors to be $\\mathbf{b_{1}}=\\frac{\\mathbf{a_{2}} \\times \\mathbf{a_{3}}}{\\mathbf{a_{1}} \\cdot (\\mathbf{a_{2}} \\times \\mathbf{a_{3}})}$ and so on for the other vectors. The crystallographer's definition has the advantage that the definition of $\\mathbf{b_{1}}$ is just the reciprocal magnitude of $\\mathbf{a_{1}}$ in the direction of $\\mathbf{a_{2}} \\times \\mathbf{a_{3}}$, dropping the factor of $2 \\pi$. Checking the magnitude: $$\\lVert{\\bf A}\\rVert = \\frac{\\lVert {\\bf b} \\times{\\bf c}\\rVert}{\\lVert{\\bf a}\\rVert\\lVert{\\bf b}\\times {\\bf c}\\rVert\\cos\\angle({\\bf a}, {\\bf b} \\times{\\bf c})} = \\frac{1}{\\lVert{\\bf a}\\rVert ({\\bf e_a} \\cdot {\\bf e}_{{\\bf b} \\times{\\bf c}})}$$ Solving the question: Using the \"bac-cab\" rule ${\\bf a}\\times({\\bf b}\\times{\\bf c}) = {\\bf b}({\\bf a}\\cdot {\\bf c}) - {\\bf c}({\\bf a}\\cdot {\\bf b})$ we go for the nominator of ${\\bf B}\\times{\\bf C}$: $$({\\bf c}\\times", "Limited access Let: $$||{\\bf u}|| =6$$ $$||{\\bf u} + {\\bf v}||=6\\cdot \\sqrt{2}$$ ...and let's assume that ${\\bf u}$ makes an angle of $45 ^{\\circ}$ with a positive $x$-axis, while ${\\bf u} + {\\bf v}$ makes an angle of $90 ^{\\circ}$ with the positive $x$-axis. What is the component form of ${\\bf v}$? A", "d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I} \\right) \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.118) On the left our integral can be rewritten as", "&= I_{1 2 \\cdots (k-1)} I \\left\\lvert {d\\mathbf{x}_1 \\wedge d\\mathbf{x}_2 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I}", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "\\mathbf c)$ of the vectors $\\mathbf a$, $\\mathbf b$ and $\\mathbf c$. Some notable properties of cross and triple products: 1. $\\mathbf a\\times \\mathbf b = -\\mathbf b\\times \\mathbf a$ 2. $(\\alpha \\cdot \\mathbf a)\\times \\mathbf b = \\alpha \\cdot (\\mathbf a\\times \\mathbf b)$ 3. For any $\\mathbf b$ and $\\mathbf c$ there is exactly one vector $\\mathbf r$ such that $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf a\\cdot\\mathbf r$ for any vector $\\mathbf a$. Indeed if there are two such vectors $\\mathbf r_1$ and $\\mathbf r_2$ then $\\mathbf a\\cdot (\\mathbf r_1 - \\mathbf r_2)=0$ for all vectors $\\mathbf a$ which is possible only when $\\mathbf r_1 = \\mathbf r_2$. 4. $\\mathbf a\\cdot (\\mathbf b\\times \\mathbf c) = \\mathbf b\\cdot (\\mathbf c\\times \\mathbf a) = -\\mathbf a\\cdot( \\mathbf c\\times \\mathbf b)$ 5. $(\\mathbf a + \\mathbf b)\\times \\mathbf c = \\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c$. Indeed for all vectors $\\mathbf r$ the chain of equations holds: $$\\mathbf r\\cdot( (\\mathbf a + \\mathbf b)\\times \\mathbf c) = (\\mathbf a + \\mathbf b) \\cdot (\\mathbf c\\times \\mathbf r) = \\mathbf a \\cdot(\\mathbf c\\times \\mathbf r) + \\mathbf b\\cdot(\\mathbf c\\times \\mathbf r) = \\mathbf r\\cdot (\\mathbf a\\times \\mathbf c) + \\mathbf r\\cdot(\\mathbf b\\times \\mathbf c) = \\mathbf r\\cdot(\\mathbf a\\times \\mathbf c + \\mathbf b\\times \\mathbf c)$$", "unnecessary and unnecessarily requires n vectors $\\mathbf{c}_1, \\ldots, \\mathbf{c}_{k}, \\ldots, \\mathbf{c}_{n}$ at a time. Solving each equation is independent in this case. This has been shown to clarify the usage, as far as what not to do, unless one has an unusual need to solve a particular vector $\\mathbf{a}_{k}$. Instead, the following can be done in the case of projecting vectors $\\mathbf{c}_{k}$ onto a new arbitrary basis $\\mathbf{x}_{k}$. ### Projecting a vector onto an arbitrary basis. Projecting any vector $\\mathbf{c}$ onto a new arbitrary basis $\\mathbf{x}_1, \\ldots, \\mathbf{x}_k, \\ldots, \\mathbf{x}_n$ as $\\begin{array}{rcl} \\mathbf{c} & = & c_1 \\mathbf{e}_1 + \\cdots + c_k \\mathbf{e}_k + \\cdots + c_n \\mathbf{e}_n\\\\ & = & a_1 \\mathbf{x}_1 + \\cdots + a_k \\mathbf{x}_k + \\cdots + a_n \\mathbf{x}_n\\end{array}$ where each $\\mathbf{x}_k$ is written in the form $\\mathbf{x}_k = x_{k 1} \\mathbf{e}_1 + x_{k 2} \\mathbf{e}_2 + \\cdots + x_{k k} \\mathbf{e}_k + \\cdots + x_{k n} \\mathbf{e}_n$ is a system of n scalar equations in n unknown coordinates $a_k$ $\\begin{array}{rcl} a_1 x_{11} + \\cdots + a_k x_{k 1} + \\cdots + a_n x_{n 1} & = & c_1\\\\ & \\vdots & \\\\ a_1 x_{1 k} + \\cdots + a_k x_{k k} + \\cdots + a_n x_{n k} & = & c_k\\\\ & \\vdots & \\\\ a_1 x_{1 n} + \\cdots + a_k x_{k", "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "\\wedge d\\mathbf{x}_2 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I} \\right) \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.118) On the left our", "\\wedge d\\mathbf{x}_2 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I} \\right) \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.118) On the left our", "$(\\mathbf{a}+\\mathbf{b})\\cdot(\\mathbf{c}+\\mathbf{d}) = \\mathbf{a}\\cdot\\mathbf{c} + \\mathbf{a}\\cdot\\mathbf{d} + \\mathbf{b}\\cdot\\mathbf{c} + \\mathbf{b}\\cdot\\mathbf{d}$ ## Mixed associative law The dot product has a homogeneity $\\alpha (\\mathbf{a}\\cdot\\mathbf{b}) = (\\alpha\\mathbf{a})\\cdot \\mathbf{b} = \\mathbf{a}\\cdot(\\alpha\\mathbf{b})$ ## Bilinear Often the mixed associative law and distributive law are combined to show that the dot product is bilinear: $\\mathbf{a}\\cdot(\\alpha\\mathbf{b}+\\mathbf{c}) = \\alpha(\\mathbf{a}\\cdot\\mathbf{b})+(\\mathbf{a}\\cdot\\mathbf{c})$ ## Scalar Multiplication $(\\alpha_1\\mathbf{a})\\cdot(\\alpha_2\\mathbf{b})=\\alpha_1\\alpha_2(\\mathbf{a}\\cdot\\mathbf{b})$ Dot product is homogenious under scaling $\\alpha(\\mathbf{a}\\cdot\\mathbf{b}) = (\\alpha\\mathbf{a})\\cdot\\mathbf{b} = \\mathbf{a}\\cdot(\\alpha\\mathbf{b})$ ## Cauchy-Schwarz Inequality $|\\mathbf{a}\\cdot\\mathbf{b}|\\leq|\\mathbf{a}||\\mathbf{b}|$ ## No cancellation If $$\\mathbf{a}\\cdot\\mathbf{b}=\\mathbf{a}\\cdot\\mathbf{c}$$ and $$\\mathbf{a}\\neq \\mathbf{0}$$, we can write $$\\mathbf{a}\\cdot(\\mathbf{b}-\\mathbf{c})=0$$ by distributive law, which means that $$\\mathbf{a}\\perp(\\mathbf{b}-\\mathbf{c})$$, which allows $$(\\mathbf{b}-\\mathbf{c})\\neq\\mathbf{0}$$ and therefore $$\\mathbf{b}\\neq\\mathbf{c}$$. ## Applications ### Fast calculation of Cosine The definition of the dot-product is used extensively in computer graphics since it speeds up computations in many circumstances by avoiding inefficient trigonometric functions. Additionally, working with normalized vectors gives the cosine between these two vectors by taking the dot product of them: $\\cos\\theta = \\frac{\\mathbf{a}\\cdot\\mathbf{b}}{|\\mathbf{a}||\\mathbf{b}|} = \\hat{\\mathbf{a}}\\cdot\\hat{\\mathbf{b}}$ ### Making perpendicular vectors Since two vectors are perpendicular when $$\\mathbf{a}\\cdot\\mathbf{b} = 0$$, we can easily construct a perpendicular vector by zeroing all components except 2, flipping those two and reversing the sign of one of them. For example with $$\\mathbf{a} = (\\mathbf{a}_1, \\mathbf{a}_2, \\dots, \\mathbf{a}_n)$$ the vectors $$\\mathbf{a}' = (-\\mathbf{a}_2, \\mathbf{a}_1, 0, \\dots, 0)$$ or $$\\mathbf{a}'' = (0, -\\mathbf{a}_3, \\mathbf{a}_2, 0, \\dots, 0)$$, etc are all perpendcular to $$\\mathbf{a}$$,", "$$\\mathbf{B}(t)$$ on the curve, we have this condition: $\\begin{split}(\\mathbf{P} - \\mathbf{B}(t)) \\cdot \\mathbf{B}'(t) = 0 \\\\[.5em] \\mathbf{P} \\cdot \\mathbf{B}'(t) - \\mathbf{B}(t) \\cdot \\mathbf{B}'(t) = 0\\end{split}$ ### 9.2.1.13.1. Quadratic Bézier Curve¶ Let \\begin{split}\\begin{align} \\mathbf{A} &= \\mathbf{P}_1 - \\mathbf{P}_0 \\\\[.5em] \\mathbf{B} &= \\mathbf{P}_2 - \\mathbf{P}_1 -\\mathbf{A} \\\\[.5em] \\mathbf{M} &= \\mathbf{P}_0 - \\mathbf{P} \\end{align}\\end{split} We have $\\mathbf{B}'(t) = 2(\\mathbf{A} + \\mathbf{B} t)$ The condition can be expressed as $\\mathbf{B}^2 t^3 + 3\\mathbf{A}\\mathbf{B} t^2 + (2\\mathbf{A}^2 + \\mathbf{M}\\mathbf{B}) t + \\mathbf{M}\\mathbf{A} = 0$ >>> C = collect(expand((P*B2p - B2*B2p)/-2), t) P*P0 - P*P1 - P0**2 + P0*P1 + t**3 * (P0**2 - 4*P0*P1 + 2*P0*P2 + 4*P1**2 - 4*P1*P2 + P2**2) + t**2 * (-3*P0**2 + 9*P0*P1 - 3*P0*P2 - 6*P1**2 + 3*P1*P2) + t * (-P*P0 + 2*P*P1 - P*P2 + 3*P0**2 - 6*P0*P1 + P0*P2 + 2*P1**2) >>> A = P1 - P0 B = P2 - P1 - A M = P0 - P >>> C2 = B**2 * t**3 + 3*A*B * t**2 + (2*A**2 + M*B) * t + M*A >>> expand(C - C2) 0 ### 9.2.1.13.2. Cubic Bézier Curve¶ Let \\begin{split}\\begin{align} n &= \\mathbf{P}_3 - 3\\mathbf{P}_2 + 3\\mathbf{P}_1 - \\mathbf{P}_0 \\\\[.5em] r &= 3(\\mathbf{P}_2 - 2\\mathbf{P}_1 + \\mathbf{P}_0 \\\\[.5em] s &= 3(\\mathbf{P}_1 - \\mathbf{P}_0) \\\\[.5em] v &= \\mathbf{P}_0 \\end{align}\\end{split} We have \\begin{align}\\begin{aligned}\\begin{split}\\begin{align} \\mathbf{B}^3(t) &= nt^3 +", "\\nabla f(\\mathbf{c}) \\cdot \\mathbf{u} \\end{align} • Take the absolute value of both sides of the equation above and apply the Cauchy-Schwarz inequality: (6) \\begin{align} \\quad \\mid f'(\\mathbf{c}, \\mathbf{u}) \\mid = \\mid \\nabla f(\\mathbf{c}) \\cdot \\mathbf{u} \\mid \\leq \\| \\nabla f(\\mathbf{c}) \\| \\| \\mathbf{u} \\| = \\| \\nabla f(\\mathbf{c}) \\| \\cdot 1 = \\| \\nabla f(\\mathbf{c}) \\| \\end{align} • This shows us that for each $\\mathbf{u} \\in \\mathbb{R}^n$ that $\\mid \\mathbf{f}'(\\mathbf{c}, \\mathbf{u}) \\mid$ is bounded above by $\\| \\nabla f(\\mathbf{c}) \\|$. We wish to show that a maximum for $\\mid f'(\\mathbf{c}, \\mathbf{u}) \\mid$ is obtained at some $\\mathbf{u} \\in \\mathbb{R}^n$ and that it equals $\\| \\nabla f(\\mathbf{c}) \\|$. • Now note by the dot product we have that $\\mid \\nabla f(\\mathbf{c}) \\cdot \\mathbf{u} \\mid = \\| \\nabla f(\\mathbf{c}) \\| \\| \\mathbf{u} \\| \\mid \\cos \\theta \\mid = \\| \\nabla f(\\mathbf{c}) \\| \\mid \\cos \\theta \\mid$ where $\\theta$ is the angle between the vectors $\\nabla f(\\mathbf{c})$ and $\\mathbf{u}$. So we see that $\\mid \\nabla f(\\mathbf{c}) \\cdot \\mathbf{u} \\mid = \\| \\nabla f(\\mathbf{c}) \\|$ when $\\theta = 0$, i.e., $\\mid f'(\\mathbf{c}, \\mathbf{u}) \\mid$ is maximized when $\\mathbf{u}$ is parallel to $\\nabla f(\\mathbf{c})$ and $\\mid f'(\\mathbf{c}, \\mathbf{u}) \\mid = \\| \\nabla f(\\mathbf{c}) \\|$. $\\blacksquare$", "+ \\mathbf{b} \\cdot \\mathbf{a} - \\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\mathbf{a} \\cdot \\mathbf{a} - \\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\lvert \\mathbf{a} \\rvert ^ 2 - \\lvert \\mathbf{b} \\rvert ^2 \\end{array}$ 2. $\\begin{array}{ll} \\lvert a + b \\rvert ^ 2 &= (\\mathbf{a} + \\mathbf{b}) \\cdot (\\mathbf{a} + \\mathbf{b}) \\\\ &= \\mathbf{a} \\cdot (\\mathbf{a} + \\mathbf{b}) + \\mathbf{b} \\cdot (\\mathbf{a} + \\mathbf{b}) \\\\ &= \\mathbf{a} \\cdot \\mathbf{a} + \\mathbf{a} \\cdot \\mathbf{b} + \\mathbf{b} \\cdot \\mathbf{a} + \\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\mathbf{a} \\cdot \\mathbf{a} + 2 (\\mathbf{a} \\cdot \\mathbf{b}) + \\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\lvert \\mathbf{a} \\rvert^2 + 2 (\\mathbf{a} \\cdot \\mathbf{b}) + \\lvert \\mathbf{b} \\rvert ^ 2 \\\\ \\end{array}$ Exercise 16 1. We know that $$\\lvert \\mathbf{a} \\rvert = \\lvert \\mathbf{b} \\rvert = 1$$ and $$\\mathbf{a} \\cdot \\mathbf{b} = 0$$. If the two vectors are perpendicular we must have $(2\\mathbf{a} + 3\\mathbf{b}) \\cdot (m\\mathbf{a} + \\mathbf{b}) = 0$ We can rewrite this as $\\begin{array}{ll} 0 &= (2\\mathbf{a} + 3\\mathbf{b}) \\cdot (m\\mathbf{a} + \\mathbf{b}) \\\\ &= 2\\mathbf{a} \\cdot (m\\mathbf{a} + \\mathbf{b}) + 3\\mathbf{b} \\cdot (m\\mathbf{a} + \\mathbf{b}) \\\\ &= 2\\mathbf{a} \\cdot m\\mathbf{a} + 2\\mathbf{a} \\cdot \\mathbf{b} + 3\\mathbf{b} \\cdot m\\mathbf{a} + 3\\mathbf{b} \\cdot \\mathbf{b} \\\\ &= \\lvert 2\\mathbf{a} \\rvert \\lvert m\\mathbf{a} \\rvert + \\lvert 3\\mathbf{b} \\rvert \\lvert \\mathbf{b} \\rvert \\\\ &= 2m + 3 \\end{array}$ Therefore $$m = -\\frac{3}{2}$$", "\\mathbf{f}' d^k \\mathbf{x} \\boldsymbol{\\nabla}'}}\\right\\rangle}_{k} I \\\\ &= \\int_V(-1)^{k+1} d^k \\mathbf{x} \\left( { \\boldsymbol{\\nabla} \\cdot \\mathbf{f}} \\right) I \\\\ &= (-1)^{k+1} I^2\\int_V dV\\left( { \\boldsymbol{\\nabla} \\cdot \\mathbf{f}} \\right).\\end{aligned} \\hspace{\\stretch{1}}(1.0.108) The second last step uses lemma 8, and the last writes $d^k \\mathbf{x} = I^2 \\left\\lvert {d^k \\mathbf{x}} \\right\\rvert = I^2 dV$, where we have assumed (without loss of generality) that $d^k \\mathbf{x}$ has the same orientation as the pseudoscalar for the space. We also assume that the parameterization is non-degenerate over the integration volume (i.e. no $d\\mathbf{x}_i = 0$), so the sign of this product cannot change. Let’s now return to the normal vector $d^{k-1} \\mathbf{x} I$. With $d^{k-1} u_i = du^1 du^2 \\cdots du^{i-1} du^{i+1} \\cdots du^k$ (the $i$ indexed differential omitted), and $I_{ab\\cdots c} = (\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\cdots \\wedge \\mathbf{x}_c)/\\left\\lvert {\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\cdots \\wedge \\mathbf{x}_c} \\right\\rvert$, we have \\begin{aligned}\\begin{aligned}d^{k-1} \\mathbf{x} I&=d^{k-1} u_i \\left( { \\mathbf{x}_1 \\wedge \\mathbf{x}_2 \\wedge \\cdots \\wedge \\mathbf{x}_k} \\right) \\cdot \\mathbf{x}^i I \\\\ &= I_{1 2 \\cdots (k-1)} I \\left\\lvert {d\\mathbf{x}_1 \\wedge d\\mathbf{x}_2 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c}", "me back up a bit. Remember what the point is of the \"coordinate matrix of a linear transformation with respect to a basis.\" If $T\\colon\\mathbf{V}\\to\\mathbf{V}$ is a linear transformation, and $\\beta=[\\mathbf{v}_1,\\ldots,\\mathbf{v}_n]$ is a basis for $T$, then the values of $T$ at every vector in $\\mathbf{V}$ is completely determined by its values on $\\mathbf{v}_1,\\ldots,\\mathbf{v}_n$. Why? Because given any $\\mathbf{x}\\in\\mathbf{V}$, we know there exist (unique) scalars $c_1,\\ldots,c_n$ such that $\\mathbf{x}=c_1\\mathbf{v}_1+\\cdots+c_n\\mathbf{v}_n$, so $$T(\\mathbf{x}) = T(c_1\\mathbf{v}_1+\\cdots+c_n\\mathbf{v}_n) = c_1T(\\mathbf{v}_1)+\\cdots+c_nT(\\mathbf{v}_n).$$ So if you know what $T(\\mathbf{v}_1),\\ldots,T(\\mathbf{v}_n)$ are, you know that $T(\\mathbf{x})$ is for all $\\mathbf{x}\\in\\mathbf{V}$. Now, since $\\beta$ is a basis, each $T(\\mathbf{v}_i)$ can be expressed as a linear combination of the $\\mathbf{v}_i$ in a unique way; that is, there are scalars $a_{11}, a_{12},\\ldots,a_{nn}$ such that \\begin{align*} T(\\mathbf{v}_1) &= a_{11}\\mathbf{v}_1 + a_{21}\\mathbf{v}_2 + \\cdots +a_{n1}\\mathbf{v}_n\\\\ T(\\mathbf{v}_2) &= a_{12}\\mathbf{v}_1 + a_{22}\\mathbf{v}_2 + \\cdots + a_{n2}\\mathbf{v}_n\\\\ &\\vdots\\\\ T(\\mathbf{v}_n) &= a_{1n}\\mathbf{v}_1 + a_{2n}\\mathbf{v}_2 + \\cdots + a_{nn}\\mathbf{v}_n. \\end{align*} Then \\begin{align*} G(\\mathbf{x}) &= c_1T(\\mathbf{v}_1 + \\cdots + c_nT(\\mathbf{v}_n)\\\\ &= c_1\\left(a_{11}\\mathbf{v}_1 + a_{21}\\mathbf{v}_2 + \\cdots +a_{n1}\\mathbf{v}_n\\right)\\\\ &\\qquad+c_2\\left(a_{12}\\mathbf{v}_1 + a_{22}\\mathbf{v}_2 + \\cdots + a_{n2}\\mathbf{v}_n\\right)\\\\ &\\qquad+\\cdots+c_n\\left(a_{1n}\\mathbf{v}_1 + a_{2n}\\mathbf{v}_2 + \\cdots + a_{nn}\\mathbf{v}_n\\right)\\\\ &= \\Bigl(c_1a_{11} + c_2a_{12} + \\cdots + c_na_{1n}\\Bigr)\\mathbf{v}_1\\\\ &\\qquad +\\Bigl(c_1a_{21} + c_2a_{22} + \\cdots + c_na_{2n}\\Bigr)\\mathbf{v}_2\\\\ &\\qquad+\\cdots+ \\Bigl(c_1a_{n1}+c_2a_{n2}+\\cdots+c_na_{nn}\\Bigr)\\mathbf{v}_n\\\\ &= \\left(\\begin{array}{cccc} \\mathbf{v}_1 & \\mathbf{v}_2 & \\cdots & \\mathbf{v}_n\\end{array}\\right) \\left(\\begin{array}{cccc} a_{11} & a_{12} & \\ldots & a_{1n}\\\\ a_{21} & a_{22}", "+ {\\bf b} \\cdot {\\bf b} - 2 \\parallel \\! \\! {\\bf a} \\! \\! \\parallel \\parallel \\! \\! {\\bf b} \\! \\! \\parallel \\cos \\theta$ ${\\bf b} \\cdot ({\\bf b} - {\\bf a}) - {\\bf a} \\cdot ({\\bf b} - {\\bf a}) = {\\bf a} \\cdot {\\bf a} + {\\bf b} \\cdot {\\bf b} - 2 \\parallel \\! \\! {\\bf a} \\! \\! \\parallel \\parallel \\! \\! {\\bf b} \\! \\! \\parallel \\cos \\theta$ ${\\bf b} \\cdot ({\\bf b} - {\\bf a}) -{\\bf a} \\cdot ({\\bf b} - {\\bf a}) = {\\bf a} \\cdot {\\bf a} + {\\bf b} \\cdot {\\bf b} - 2 \\parallel \\! \\! {\\bf a} \\! \\! \\parallel \\parallel \\! \\! {\\bf b} \\! \\! \\parallel \\cos \\theta$ ${\\bf b} \\cdot {\\bf b} - {\\bf a} \\cdot {\\bf b} - {\\bf a} \\cdot {\\bf b} + {\\bf a} \\cdot {\\bf a} = {\\bf a} \\cdot {\\bf a} + {\\bf b} \\cdot {\\bf b} - 2 \\parallel \\! \\! {\\bf a} \\! \\! \\parallel \\parallel \\! \\! {\\bf b} \\! \\! \\parallel \\cos \\theta$ ${\\bf b} \\cdot {\\bf b} - 2 {\\bf a} \\cdot {\\bf b} + {\\bf a} \\cdot {\\bf a} = {\\bf a} \\cdot {\\bf a} + {\\bf b} \\cdot {\\bf b} - 2 \\parallel \\! \\! {\\bf a} \\! \\! \\parallel", "(for example, if u = (a, b) and v = (c, d), then their dot product u·v = ac + bd). If the dot product is zero, the two vectors are said to be orthogonal to each other. Some properties of the dot product aid understanding of how W-CDMA works. If vectors a and b are orthogonal, then ${\\displaystyle \\mathbf {a} \\cdot \\mathbf {b} =0}$ and: ${\\displaystyle \\mathbf {a} \\cdot (\\mathbf {a} +\\mathbf {b} )=\\|\\mathbf {a} \\|^{2},\\ {\\text{since}}\\ \\mathbf {a} \\cdot \\mathbf {a} +\\mathbf {a} \\cdot \\mathbf {b} =\\|\\mathbf {a} \\|^{2}+0,}$ ${\\displaystyle \\mathbf {a} \\cdot (-\\mathbf {a} +\\mathbf {b} )=-\\|\\mathbf {a} \\|^{2},\\ {\\text{since}}\\ {-\\mathbf {a} }\\cdot \\mathbf {a} +\\mathbf {a} \\cdot \\mathbf {b} =-\\|\\mathbf {a} \\|^{2}+0,}$ ${\\displaystyle \\mathbf {b} \\cdot (\\mathbf {a} +\\mathbf {b} )=\\|\\mathbf {b} \\|^{2},\\ {\\text{since}}\\ \\mathbf {b} \\cdot \\mathbf {a} +\\mathbf {b} \\cdot \\mathbf {b} =0+\\|\\mathbf {b} \\|^{2},}$ ${\\displaystyle \\mathbf {b} \\cdot (\\mathbf {a} -\\mathbf {b} )=-\\|\\mathbf {b} \\|^{2},\\ {\\text{since}}\\ \\mathbf {b} \\cdot \\mathbf {a} -\\mathbf {b} \\cdot \\mathbf {b} =0-\\|\\mathbf {b} \\|^{2}.}$ Each user in synchronous CDMA uses a code orthogonal to the others' codes to modulate their signal. An example of 4 mutually orthogonal digital signals is shown in the figure below. Orthogonal codes have a cross-correlation equal to zero; in other words, they do not interfere with each other. In the case", "is trivial, while if at least one of the components is nonzero, say $b_1$ without loss of generality, then (for $n>1$) $$\\color{Blue}{-(b_2+b_3+\\cdots+b_n)}(b_1\\mathbf{a})+\\color{Blue}{b_1}(b_2\\mathbf{a})+\\color{Blue}{b_1}(b_3\\mathbf{a})+\\cdots+\\color{Blue}{b_1}(b_n\\mathbf{a})$$ $$=\\big((\\color{Blue}{-b_2}b_1+\\color{Blue}{b_1}b_2)+(\\color{Blue}{-b_3}b_1+\\color{Blue}{b_1}b_3)\\cdots+(\\color{Blue}{-b_n}b_1+\\color{Blue}{b_1}b_n)\\big)\\mathbf{a}$$ $$=(0+0+\\cdots+0)\\mathbf{a}=\\mathbf{0}.$$ is a nontrivial linear combination of the matrix columns that evaluates to zero, hence the columns are linearly dependent and thus the matrix's determinant is zero. Shortcut (via David): Use the multlinearity of the determinant to reduce $$\\det\\begin{pmatrix}b_1\\mathbf{a}&b_2\\mathbf{a}&\\cdots&b_n\\mathbf{a}\\end{pmatrix}=b_1b_2\\cdots b_n \\det\\begin{pmatrix}\\mathbf{a}&\\mathbf{a}&\\cdots&\\mathbf{a}\\end{pmatrix}.$$ For $n>1$, it is impossible for $n$ copies of a vector $\\mathbf{a}$ to be linearly independent, since e.g. $$1\\mathbf{a}+(-1)\\mathbf{a}+0\\mathbf{a}+\\cdots+0\\mathbf{a}=\\mathbf{0}.$$ Hence the original determinant must also be zero. Geometrically: The parallelepiped formed by the matrix's columns are all contained in the one-dimensional subspace generated by $\\mathbf{a}$: for $n>1$ this has zero $n$-dimensional content (hyper-volume), hence the determinant is zero. Geometrically/Algebraically: As I posted in a different answer (paraphrased here), We're assuming $\\mathbf{a},\\mathbf{b}\\ne\\mathbf{0}$. Then $\\mathbf{b}^\\perp$, the orthogonal complement of the linear subspace generated by $\\mathbf{b}$ (i.e. the set of all vectors orthogonal to $\\mathbf{b}$) is therefore $(n-1)$-dimensional. Let $\\mathbf{c}_1,\\dots,\\mathbf{c}_{n-1}$ be a basis for this space. Then they are linearly independent and $$(\\mathbf{a}\\mathbf{b}^T)\\mathbf{c}_i =\\mathbf{a}(\\mathbf{b}^T\\mathbf{c}_i)= (\\mathbf{b}\\cdot\\mathbf{c}_i)\\mathbf{a}=\\mathbf{0}.$$ Thus the eigenvalue $0$ has geometric multiplicity $n-1\\qquad$ [...] The determinant is the product of the matrix's eigenvalues, so if one of those is $0$ the product is necessarily zero as well. Analytically/Combinatorially: Via Leibniz formula we have $$\\det(\\mathbf{a}\\mathbf{b}^T)=\\sum_{\\sigma\\in S_n}(-1)^{\\sigma}\\prod_{k=1}^na_{\\sigma(k)}b_k$$ $$=\\left(\\sum_{\\sigma\\in S_n}(-1)^\\sigma\\right)\\prod_{k=1}^na_kb_k=0.$$" ]

[ "the midpoint of $\\overline {AC}$, point $E$ is the midpoint of $\\overline {BC}$, point $H$ is the midpoint of $\\overline {AF}$, and point $G$ is the midpoint of $\\overline {BF}$. Examine this carefully, find the similar triangles, and use their properties. Don't forget to look for congruent triangles as well.", "using the distance formula. Then, find the length of the sides it is parallel to. What should happen? More Coordinate Geometry Given the midpoints of the sides of a triangle, find the vertices of the triangle. Refer back to problems 19-21 for help. 1. (-2, 1), (0, -1) and (-2, -3) 2. (1, 4), (4, 1) and (2, 1) Given the vertices of $\\triangle ABC$, find: a) the midpoints of $M, N$ and $O$ where $M$ is the midpoint of $\\overline{AB}$, $N$ is the midpoint of $\\overline{BC}$ and $C$ is the midpoint of $\\overline{AC}$. b) Show that midsegments $\\overline{MN}, \\overline{NO}$ and $\\overline{OM}$ are parallel to sides $\\overline{AC}, \\overline{AB}$ and $\\overline{BC}$ respectively. c) Show that midsegments $\\overline{MN}, \\overline{NO}$ and $\\overline{OM}$ are half the length of sides $\\overline{AC}, \\overline{AB}$ and $\\overline{BC}$ respectively. 1. $A(-3, 5), B(3, 1)$ and $C(-5, -5)$ 2. $A(-2, 2), B(4, 4)$ and $C(6, 0)$ For questions 27-30, $\\triangle CAT$ has vertices $C(x_1,y_1), A(x_2,y_2)$ and $T(x_3,y_3)$. 1. Find the midpoints of sides $\\overline{CA}$ and $\\overline{CT}$. Label them $L$ and $M$ respectively. 2. Find the slopes of $\\overline{LM}$ and $\\overline{AT}$. 3. Find the lengths of $\\overline{LM}$ and $\\overline{AT}$. 4. What have you just proven algebraically? 1. $\\left ( \\frac{-4+6}{2}, \\frac{1+7}{2} \\right ) = (1, 4)$ 2. $\\left ( \\frac{5+11}{2}, \\frac{-3+5}{2} \\right ) = (8, 1)$ 3. $\\left (\\frac{0", "Geometry $M$ is the midpoint of $\\overline{AB}$ and $N$ is the midpoint of $\\overline{AC}$, and $T$ is the intersection of $\\overline{BN}$ and $\\overline{CM}$, as shown. If $\\overline{BN}\\perp\\overline{AC}$, $BN = 12$, and $AC = 14$, then asked by DenisGylaev on June 28, 2016 10. ### Analytic Geometry The line segment joining a vertex of a triangle and the midpoint of the opposite side is called the median of the triangle. Given a triangle whose vertices are A(4,-4), B(10, 4) and C(2, 6), find the point on each median that is asked by Alexis on April 23, 2015 More Similar Questions", "Quick Vector Question 1. Feb 17, 2010 IL0VEMATH Let A', B', C' be the midpoints of the sides BC, CA, AB of the triangle ABC. Show that $$\\overline{AA'}$$+$$\\overline{BB'}$$+$$\\overline{CC'}$$=$$\\overline{0}$$ 2. Feb 17, 2010 Lord Crc Have you tried writing out the formula for the midpoints in terms of the vertices A,B,C and inserting this into the equation?", "# Quick Vector Question 1. Feb 17, 2010 ### IL0VEMATH Let A', B', C' be the midpoints of the sides BC, CA, AB of the triangle ABC. Show that $$\\overline{AA'}$$+$$\\overline{BB'}$$+$$\\overline{CC'}$$=$$\\overline{0}$$ 2. Feb 17, 2010 ### Lord Crc Have you tried writing out the formula for the midpoints in terms of the vertices A,B,C and inserting this into the equation?", "+0 # geometry 0 37 1 n triangle $ABC,$ $M$ is the midpoint of $\\overline{AB}.$ Let $D$ be the point on $\\overline{BC}$ such that $\\overline{AD}$ bisects $\\angle BAC,$ and let the perpendicular bisector of $\\overline{AB}$ intersect $\\overline{AD}$ at $E.$ If $AB = 44,$ $AC = 30,$ and $ME = 10,$ then find the area of triangle $ACE.$ Jun 20, 2022", "is the midpoint of $\\overline{XY}$, and point $T$ is the intersection of $\\overline{HM}$ and $\\overline{YG}$. Find the area of $\\triangle MTG$ if asked by Jenna on April 21, 2016 5. ### Math Point G is the midpoint of the median XM of XYZ. Point H is the midpoint of XY, and point T is the intersection of HM and YG. Find the area of MTG if XYZ=150. asked by AoPS on July 13, 2019 6. ### Math Triangle with vertices: D(-18,12) E(-6,-12) F(12,6) Show that the right bisectors of the sides of triangleDEF all intersect at point C(-4,4), the circumcentre of the triangle. -I found the midpoint of DF then the slope of the asked by Anonymous on October 14, 2012 7. ### Geometry need help fast $M$ is the midpoint of $\\overline{AB}$ and $N$ is the midpoint of $\\overline{AC}$, and $T$ is the intersection of $\\overline{BN}$ and $\\overline{CM}$, as shown. If $\\overline{BN}\\perp\\overline{AC}$, $BN = 12$, and asked by Nightbblue on June 28, 2016 Here are some more questions I'm having trouble with. Please walk me through them so I can fully understand them. #2: Let A be (5,9), B be (-3,-5), and C be (1,1). The median of a triangle connects a vertex of a triangle to the asked by Celina on December 17, 2008 9. ###", "+0 # Quick triangle question 0 45 1 In triangle ABC, M is the midpoint of $$\\overline{AB}$$. Let D be the point on $$\\overline{BC}$$ such that $$\\overline{AD}$$ bisects $$\\angle BAC$$, and let the perpendicular bisector of $$\\overline{AB}$$ intersect $$\\overline{AD}$$ at $E.$ If $AB = 44$ and $ME = 12,$ then find the distance from $E$ to line $AC.$ Apr 17, 2020 ### 1+0 Answers #1 0 The distance from E to line AC is 9. May 10, 2020", "### Home > INT1 > Chapter Ch11 > Lesson 11.2.3 > Problem11-80 11-80. In the triangle at right, $\\overline{DE}$ is a midsegment of $ΔABC$ and $\\overline{FG}$ is a midsegment of $ΔADE$. If $DE = 7$ cm, find $FG$ and $BC$. $ΔADE\\simΔABC$ Since $D$ is the midpoint of $\\overline{AB}$, $2 ·DB = AB$. $FG = 3.5$ cm $BC = 14$ cm", "solve for $x$. \\begin{aligned}(2x – 4) &= \\dfrac{1}{2}(32)\\\\2x – 4&= 16\\\\2x&= 20\\\\x&= 10\\end{aligned} Hence, we have $x = 10$. ### Example 2 Using the converse of the midpoint theorem and the triangle shown below, what is the perimeter of the triangle $\\Delta ABC$? ### Solution Since $\\overline{AM} = \\overline{MB} = 15$, $M$ is the midpoint of $\\overline{AB}$. We can see that $\\overline{MN}$ passes through the midpoint of $\\overline{AB}$ and is parallel to the triangle’s side $\\overline{BC}$, so we can conclude that it is indeed the midsegment of $\\Delta ABC$. \\begin{aligned}\\overline{MN} &\\parallel \\overline{BC}\\\\&\\Rightarrow N \\text{ is the midpoint of } \\overline{AC} \\end{aligned} $N$ is the midpoint of $\\overline{AC}$, so $\\overline{AN} = \\overline{NC} = 16$. Applying the same thought process, we can also show that $\\overline{MO}$ is a midsegment, so $O$ is also a midpoint. \\begin{aligned}\\overline{MO} &\\parallel \\overline{AC}\\\\&\\Rightarrow O \\text{ is the midpoint of } \\overline{BC} \\end{aligned} Hence, $\\overline{BO} = \\overline{OC} = 12$. Now, find the perimeter of $\\Delta ABC$ by adding the three sides’ lengths. \\begin{aligned}\\text{Perimeter}_{\\Delta ABC} &= \\overline{AB}+\\overline{BC}+ \\overline{AC}\\\\&= 2(\\overline{AM})+ 2(\\overline{BO}) + 2(\\overline{AN})\\\\&= 2(15) + 2(12) + 2(16)\\\\&= 86\\end{aligned} This means that the perimeter of $\\Delta ABC$ is equal to $86$ units. ### Practice Questions 1. The triangle $\\Delta ABC$ has $\\overline{XY}$ as the midsegment that bisects $\\overline{AB}$ and $\\overline{AC}$. Which of the following statements is not always", "$\\overline{XM}$ of $\\triangle XYZ$. Point $H$ is the midpoint of $\\overline{XY}$, and point $T$ is the intersection of $\\overline{HM}$ and $\\overline{YG}$. Find the area of $\\triangle MTG$ if asked by Jonathan on April 21, 2016 9. ### Math- I'd really appreciate it if you could help!! Point $G$ is the midpoint of median $\\overline{XM}$ of $\\triangle XYZ$. Point $H$ is the midpoint of $\\overline{XY}$, and point $T$ is the intersection of $\\overline{HM}$ and $\\overline{YG}$. Find the area of $\\triangle MTG$ if asked by Jenna on April 21, 2016 10. ### Geometry Point $G$ is the midpoint of median $\\overline{XM}$ of $\\triangle XYZ$. Point $H$ is the midpoint of $\\overline{XY}$, and point $T$ is the intersection of $\\overline{HM}$ and $\\overline{YG}$. Find the area of $\\triangle MTG$ if asked by Stephen on April 21, 2016 More Similar Questions", "the midpoint of AC, E is the midpoint of BC and DE is congruent to EF, prove: DE is parallel to AB and DE = 1/2 AB. answer is 15 steps asked by linda on December 29, 2012 Point $G$ is the midpoint of median $\\overline{XM}$ of $\\triangle XYZ$. Point $H$ is the midpoint of $\\overline{XY}$, and point $T$ is the intersection of $\\overline{HM}$ and $\\overline{YG}$. Find the area of $\\triangle MTG$ if", "E, and F are the midpoints of sides \\overline{BC}, \\overline{CA}, and \\overline{AB} of \\triangle ABC, respectively, and \\overline{CZ} is an altitude of the triangle. If \\angle BAC = 71^\\circ, \\angle ABC = 39^\\circ, and \\angle … 5. ### Geometry Points D, E, and F are the midpoints of sides \\overline{BC}, \\overline{CA}, and \\overline{AB} of \\triangle ABC, respectively, and \\overline{CZ} is an altitude of the triangle. If \\angle BAC = 71^\\circ, \\angle ABC = 39^\\circ, and \\angle … 6. ### Geometry In $\\triangle ABC$, we have $AB = AC = 13$ and $BC = 10$. Let $M$ be the midpoint of $\\overline{AB}$ and $N$ be on $\\overline{BC}$ such that $\\overline{AN}$ is an altitude of $\\triangle ABC$. If $\\overline{AN}$ and $\\overline{CM}$ … Medians $\\overline{AX}$ and $\\overline{BY}$ of $\\triangle ABC$ are perpendicular at point $G$. Prove that $AB = CG$. In your diagram, $\\angle AGB$ should appear to be a right angle. $M$ is the midpoint of $\\overline{AB}$ and $N$ is the midpoint of $\\overline{AC}$, and $T$ is the intersection of $\\overline{BN}$ and $\\overline{CM}$, as shown. If $\\overline{BN}\\perp\\overline{AC}$, $BN = 12$, and $AC = 14$, then find …", "the segments $\\overline{ab}$, $\\overline{bc}$, and $\\overline{ca}$, you get four new little triangles $\\triangle{abc}$, $\\triangle{Abc}$, $\\triangle{aBc}$, and $\\triangle{abC}$. Are you asking whether the four new ones are mutually congruent equilateral, equiangular triangles? But of course, by the SSS criterion. Are you asking whether the process may be repeated without let or hindrance? Again, certainly.", "Question 14 # ABCD is a rectangle in the clockwise direction. The coordinates of A are (1,3) and the coordinates of C are (5,1), the coordinates of vertices B and D satisfy the line $$y=2x+c$$, then what will be the coordinates of the mid-point of BC. Solution Midpoint of AC passes through the line y = 2x + c Midpoint of AC = (3,2) 2 = 6 + c c = -4 Line passing through B is y = 2x - 4 (slope of AB)(slope of BC) = -1 $$\\left(\\ \\frac{\\ y-1}{x-5}\\right)\\left(\\ \\frac{\\ y-3}{x-1}\\right)=-1$$ Substituting y = 2x - 4 and solving, we get B(2,0) or B(4,4) B cannot lie on X-axis. Therefore, co-ordinates of B = (4,4) Midpoint of BC = $$\\left(\\frac{9}{2},\\frac{5}{2}\\right)$$", ", y = $$\\frac{2+2}{2}$$ $$\\Rightarrow$$ x = 1 , y = 2 Now, since ABCD is a square $$\\therefore$$ O is mid point of AC also $$\\frac{1+x_2}{2} \\ = \\ 1$$ , $$\\frac{4+ y_2}{2} \\ = \\ 2$$ $$\\Rightarrow x_2 \\ = \\ 1$$ , $$y_2 \\ = \\ 0$$ $$\\therefore$$ Coordinates of vertices square ABCD are A(1, 4) , B(3,2) , C(1,0) , D(-1,2)", "find area of the smaller triangle? asked by dan on October 25, 2012 5. ### math The large red equilateral triangle has sides of 8x units. The midpoints of the red triangle are joined to form the blue triangle. The midpoints of the blue triangle ate joined to form the green triangle. The process of joining asked by Jo on January 11, 2016 6. ### math The large red equilateral triangle has sides of 8x units. The midpoints of the red triangle are joined to form the blue triangle. The midpoints of the blue triangle ate joined to form the green triangle. The process of joining asked by Jo on January 11, 2016 7. ### mathURGENT!!!! the midpoints of the sides of a triangle are (1,1),(4,3),and (3,5). find the area of the triangle. what is the answer and can u explain how u got it step by step? please... asked by dan on October 22, 2012 8. ### math Points $D$, $E$, and $F$ are the midpoints of sides $\\overline{BC}$, $\\overline{CA}$, and $\\overline{AB}$, respectively, of $\\triangle ABC$. Points $X$, $Y$, and $Z$ are the midpoints of $\\overline{EF}$, $\\overline{FD}$, and asked by animus on July 22, 2015 9. ### math Points $D$, $E$, and $F$ are the midpoints of sides $\\overline{BC}$, $\\overline{CA}$, and $\\overline{AB}$, respectively, of $\\triangle ABC$. Points", "is reflected over the $x$-axis. What is the sum of the coordinates of the image of asked by Zach Bringer on August 15, 2019 5. ### MATH When a draftsman draws three lines that are to intersect at one​ point, the lines may not intersect as intended and subsequently will form an error triangle. If this error triangle is long and​ thin, one estimate for the asked by Anonymous on October 2, 2015 1. ### Geometry Point $G$ is the midpoint of median $\\overline{XM}$ of $\\triangle XYZ$. Point $H$ is the midpoint of $\\overline{XY}$, and point $T$ is the intersection of $\\overline{HM}$ and $\\overline{YG}$. Find the area of $\\triangle MTG$ if asked by Jonathan on April 21, 2016 2. ### math Point $C$ is the midpoint of $\\overline{AB}$, point $D$ is the midpoint of $\\overline{AC}$, point $E$ is the midpoint of $\\overline{AD}$, and point $F$ is the midpoint of $\\overline{AE}$. If $AF=3$, what is the number of units in asked by Ishi on May 8, 2020 For each graph, find (a)AB to the nearest tenth and (b)the coordinates of the midpoint of AB. 1. Coordinates are A(-9, -6) B(6,6) D=19.2 Midpoint=(1.5,-6) 2.A(-2,-2) B(8,-6) D=10.77 midpoint= (3,-4) 3. A(0,3) B(-2,-2) D=5.4 asked by Unknown on August 20, 2013 4. ### geometry write a 2 column proof, Given D is", "+0 0 164 1 1) In triangle $XYZ,$ $M$ is the midpoint of $\\overline{XY}.$ Let $D$ be the point on $\\overline{YZ}$ such that $\\overline{XD}$ bisects $\\angle YXZ,$ and let the perpendicular bisector of $\\overline{XY}$ intersect $\\overline{XD}$ at $P.$ If $XY = 36$ and $MP = 9,$ then find the distance from $P$ to line $XZ.$ 2) In triangle ABC, AB=BC=17 and AC=16 Find the circumradius of triangle ABC Dec 3, 2019", "intersection of $\\overline{HM}$ and $\\overline{YG}$. Find the area of $\\triangle MTG$ if asked by Janice on April 21, 2016 9. ### Geometry Point $G$ is the midpoint of median $\\overline{XM}$ of $\\triangle XYZ$. Point $H$ is the midpoint of $\\overline{XY}$, and point $T$ is the intersection of $\\overline{HM}$ and $\\overline{YG}$. Find the area of $\\triangle MTG$ if asked by Jonathan on April 21, 2016 10. ### Math- I'd really appreciate it if you could help!! Point $G$ is the midpoint of median $\\overline{XM}$ of $\\triangle XYZ$. Point $H$ is the midpoint of $\\overline{XY}$, and point $T$ is the intersection of $\\overline{HM}$ and $\\overline{YG}$. Find the area of $\\triangle MTG$ if asked by Jenna on April 21, 2016 More Similar Questions" ]

[ "## anonymous one year ago In square $ABCD$, $E$ is the midpoint of $\\overline{BC}$, and $F$ is the midpoint of $\\overline{CD}$. Let $G$ be the intersection of $\\overline{AE}$ and $\\overline{BF}$. Prove that $DG = AB$. • This Question is Open 1. anonymous Asymptote code below [asy] unitsize(2.5 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (0,1); C = (1,1); D = (1,0); E = (B + C)/2; F = (C + D)/2; G = extension(A,E,B,F); draw(A--B--C--D--cycle); draw(A--E); draw(B--F); draw(D--G); label(\"$A$\", A, SW); label(\"$B$\", B, NW); label(\"$C$\", C, NE); label(\"$D$\", D, SE); label(\"$E$\", E, N); label(\"$F$\", F, dir(0)); label(\"$G$\", G, WSW); [/asy]", "&= \\frac{1}{2}\\end{align*} Hence $CE = FC + x = \\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 2 Call the point of tangency point $F$ and the midpoint of $AB$ as $G$. $CF=2$ by Tangent Theorem. Notice that $\\angle EGF=\\frac{180-2\\cdot\\angle CGF}{2}=90-\\angle CGF$. Thus, $\\angle EGF=\\angle FCG$ and $tanEGF=tanFCG=\\frac{1}{2}$. Solving $EF=\\frac{1}{2}$. Adding, the answer is $\\frac{5}{2}$. ## Solution 3 Clearly, $EA = EF = BG$. Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = 5/2$. ## Solution 4 $[asy] size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label(\"A\",A,(-1,-1)); label(\"B\",B,( 1,-1)); label(\"C\",C,( 1, 1)); label(\"D\",D,(-1, 1)); label(\"E\",E,(-1, 0)); label(\"F\",F,( 0, 1)); label(\"x\",(A+E)/2,(-1, 0)); label(\"x\",(E+F)/2,( 0, 1)); label(\"2\",(F+C)/2,( 0, 1)); label(\"2\",(D+C)/2,( 0, 1)); label(\"2\",(B+C)/2,( 1, 0)); label(\"2-x\",(D+E)/2,(-1, 0)); label(\"G\",G,(0,-1)); dot(G); draw(G--C); label(\"\\sqrt{5}\",(G+C)/2,(-1,0)); [/asy]$ Let us call the midpoint of side $AB$, point $G$. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$. We get $GC=\\sqrt{5}$. We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$, we get similar triangles $EFG$ and $GFC$. Solving the ratios, we get $x=\\frac{1}{2}$, so the answer is $\\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 5 Using the diagram as drawn in Solution 5, let the total", "from $\\overline{CF}$ to $\\overline{DE}$ and $\\overline{AB}$, respectively. We know that these are both equal to $\\frac 53$. We find the area of each of the trapezoids, which both happen to be $\\frac{11}6 \\cdot \\frac{\\sqrt 3}6=\\frac{11\\sqrt 3}{36}$, and the combined area is $\\frac{11\\sqrt 3}{18}$. We find that $\\dfrac{\\frac{11\\sqrt 3}{18}}{\\frac{3\\sqrt 3}2}$ is equal to $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$. ## Solution 2 $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,ESE); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,WSW); label(\"W\",W,ENE); label(\"X\",X,ESE); label(\"Y\",Y,WSW); label(\"Z\",Z,WNW); [/asy]$ First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are $22$ small triangles in hexagon $ZWCXYF$, and $9 \\cdot 6 = 54$ small triangles in the whole hexagon. Thus, the answer is $\\frac{22}{54}=\\boxed{\\textbf{(C)}\\ \\frac{11}{27}}$. ## Solution 3 (Similar Triangles) $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); pair G = (0.5, sqrt(3)*3/2); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(E--G--D); draw(F--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,ESE); label(\"D\",D,NE); label(\"E\",E,NW); label(\"F\",F,WSW); label(\"W\",W,ENE); label(\"X\",X,ESE); label(\"Y\",Y,WSW); label(\"Z\",Z,WNW); label(\"G\",G,N); [/asy]$ Extend $\\overline{EF}$ and $\\overline{CD}$ to meet at point $G$, as shown in the diagram.", "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "$[asy] unitsize(1cm); defaultpen(0.8); pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label(\"A\",A,SW); label(\"B\",B,N); label(\"C\",C,SE); label(\"D\",D,NE); label(\"E\",E,S); label(\"T\",T,NW); label(\"3\",A--T,N); label(\"4\",B--T,W); label(\"1\",D--T,N); label(\"1\",E--T,W); [/asy]$ The point $C$ is the intersection of the lines $BD$ and $AE$. The points on the first line have the form $(t,4-4t)$, the points on the second line have the form $(t,-1-t/3)$. Solving for $t$ we get $t=15/11$, hence $C=(15/11,-16/11)$. The ratio $CD/BD$ can now be computed simply by observing the $x$ coordinates of $B$, $C$, and $D$: $$\\frac{CD}{BD} = \\frac{15/11 - 1}{1 - 0} = \\boxed{\\textbf{(D)}\\frac 4{11}}$$", "area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 $[asy] size(8cm); pair A, B, C, D, E, F; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); F = (27/8,0); draw(A--B--C--D--A--E); draw(E--F,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,F,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",F,S); label(\"4\",(A+D)/2,S); label(\"x\",(A+F)/2,S); label(\"\\frac{15}{8}\",(E+F)/2,W); [/asy]$ Let the point where the altitude from $E$ to $\\overline{AD}$ be labeled $F$. Following the steps above, you can find that the height of $\\triangle ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the origin to the point $F$, and $4-x$ is the segment from point $F$ to point", "as the center and draw segment $ON$ perpendicular to $AM$, $ON\\cap AM=N$, link $OM$. Then we have $OM\\parallel AD$. So $\\angle DAM=\\angle OMA$. Since $AD=2AM=2OM=1$, we have $\\cos\\angle DAM=\\cos\\angle OMP=\\frac{2}{\\sqrt{5}}$. As a result, $NM=OM\\cos\\angle OMP=\\frac{1}{2}\\cdot \\frac{2}{\\sqrt{5}}=\\frac{1}{\\sqrt{5}}.$ Thus $PM=2NM=\\frac{2}{\\sqrt{5}}=\\frac{2\\sqrt{5}}{5}$. Since $AM=\\frac{\\sqrt{5}}{2}$, we have $AP=AM-PM=\\frac{\\sqrt{5}}{10}$. Put $\\boxed{B}$. ~FANYUCHEN20020715 ## Solution 5 (Similar Triangles) Call the midpoint of $\\overline{AB}$ point $N$. Draw in $\\overline{NM}$ and $\\overline{NP}$. Note that $\\angle{NPM}=90^{\\circ}$ due to Thales's Theorem. $[asy] draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw(circle((0.5,0.5),0.5)); draw((0,0)--(0,0.5)--(1,0.5)--cycle); label(\"A\",(0,0),SW); label(\"B\",(0,1),NW); label(\"C\",(1,1),NE); label(\"D\",(1,0),SE); label(\"M\",(1,0.5),E); label(\"P\",(0.2,0.1),S); label(\"N\",(0,0.5),W); draw((0,0.5)--(0.2,0.1)); markscalefactor=0.007; draw(rightanglemark((0,0.5),(0.2,0.1),(1,0.5))); [/asy]$ Using the Pythagorean theorem, $AM=\\frac{\\sqrt{5}}{2}$. Now we just need to find $AP$ using similar triangles. $$\\triangle APN\\sim\\triangle ANM\\Rightarrow\\frac{AP}{AN}=\\frac{AN}{AM}\\Rightarrow\\frac{AP}{\\frac{1}{2}}=\\frac{\\frac{1}{2}}{\\frac{\\sqrt{5}}{2}}\\Rightarrow AP=\\boxed{\\textbf{(B) }\\frac{\\sqrt{5}}{10}}$$ ~QIDb602 ## Solution 6 (Intersecting Chords) Label the midpoint of $AD$ as $N$, and let $Q$ be the intersection of $ON$ and $AM$ Then $MQ=AQ=\\frac{1}{2}AM = \\frac{\\sqrt{5}}{4}$ and $NQ=\\frac{1}{4}$ Then $\\frac{1}{4}*\\frac{3}{4}=PQ*QM$ and $AP=AQ-PQ$ ~ERMSCoach ## Solution 7 (Inscribed Angle Theorem and Pythagorean Theorem) Let $N$ be the midpoint of $\\overline{AB},$ from which $\\angle ANM=90^\\circ.$ Note that $\\angle NPM=90^\\circ$ by the Inscribed Angle Theorem. We have the following diagram: Since $AN=\\frac12$ and $NM=1,$ we get $AM=\\frac{\\sqrt5}{2}$ by the Pythagorean Theorem. Let $AP=x.$ It follows that $PM=\\frac{\\sqrt5}{2}-x.$ Applying the Pythagorean Theorem to right $\\triangle ANP$ gives $NP^2=\\left(\\frac12\\right)^2-x^2,$ and applying the Pythagorean Theorem to right $\\triangle MNP$ gives $NP^2=1^2-\\left(\\frac{\\sqrt5}{2}-x\\right)^2.$ Equating the expressions for $NP^2$ produces \\begin{align*}", "= 2x$. Setting the equations equal, we have $2x = -4x +2 \\implies x = \\frac13$. Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\\frac13$, so $XY = 1 - 2 \\cdot \\frac13 = \\frac13$. Now the area of the kite is simply the product of the two diagonals over $2$. Since the length $HF = 1$, our answer is $\\dfrac{\\dfrac{1}{3} \\cdot 1}{2} = \\boxed{\\textbf{(E)} \\: \\dfrac16}$. $[asy] import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); draw(X--Y,dashed); label(\"A\\: (0,2)\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D \\: (0,0)\",D,SW); label(\"E\",E,E); label(\"F\\: (\\frac12,0)\",F,S); label(\"G\",G,W); label(\"H \\: (\\frac12,1)\",H,N); label(\"Y\",Y,E); label(\"X\",X,W); label(\"\\frac12\",(0.25,0),S); label(\"\\frac12\",(0.75,0),S); label(\"1\",(1,0.5),E); label(\"1\",(1,1.5),E); [/asy]$ ## Solution 2 Let the area of the shaded region be $x$. Let the other two vertices of the kite be $I$ and $J$ with $I$ closer to $AD$ than $J$. Note that $[ABCD] = [ABF] + [DCH] - x + [ADI] + [BCJ]$. The area of $ABF$ is $1$ and the area of $DCH$ is $\\dfrac{1}{2}$. We will solve for the areas of", "by a ratio of $\\frac{BC}{AD} = \\frac{125}{251}$. Since the homothety carries the midpoint of $\\overline{BC}$, $M$, to the midpoint of $\\overline{AD}$, which is $N$, then $E,M,N$ are collinear. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(N+H)/2+(0,1),S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(M[0]+H)/2,(1,0)); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (500 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {DG}{CG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$. ### Solution 3 If you drop perpendiculars from $B$ and $C$ to $AD$, and call the points where they meet $\\overline{AD}$, $E$ and $F$ respectively, then $FD = x$ and $EA = 1008-x$ , and so you can solve an equation in tangents. Since", "\\frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\\frac{15}{4}$. Call the point where the line through $A$ intersects $\\overline{CD}$ $E$. We know that $[ADE] = \\frac{15}{4} = \\frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \\frac{15}{4}$, so $h = \\frac{15}{8}$. This gives that the y coordinate of E is $\\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \\frac{27}{8}$. From this, we know that $E = \\left(\\frac{27}{8}, \\frac{15}{8}\\right)$. $27 + 15 + 8 + 8 = \\boxed{\\textbf{(B) }58}$ ## Solution 2 size(8cm); pair A, B, C, D, E, EE; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); EE = (27/8,0); draw(A--B--C--D--A--E); draw(E--EE,linetype(\"8 8\")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,EE,D,4)); label(\"A\",A,SW); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,SE); label(\"E\",E,NE); label(\"F\",EE,S); label(\"$4$\",(A+D)/2,S); label(\"$x$\",(A+EE)/2,S; label(\"$4-x$\",(D+EE)/2,S); label(\"$\\frac{15}{8}$\",(E+EE)/2,W); (Error compiling LaTeX. label(\"$x$\",(A+EE)/2,S; ^ de021056a2cc2ef4e3e5c00687df10ec92699974.asy: 26.23: syntax error error: could not load module 'de021056a2cc2ef4e3e5c00687df10ec92699974.asy') Following the steps above, you can find that the height of triangle $ADE$ is $\\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the", "by length $\\frac{2}{5}l$ [asy] unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, NE); label(\"$D$\", D, NW); label(\"$E$\", E, N); label(\"$F$\", F, SE); label(\"$G$\", G, SW); label(\"$2/5$\", (D + F)/2, N); label(\"$4/5$\", (G + C)/2, N); label(\"$6/5$\", (B + C)/2, dir(0)); label(\"$6/5$\", (A + D)/2, W); label(\"$2$\", (A + B)/2, S); [/asy] Let $[AFGB]'$ be the area of the shape whose length is $\\frac{2}{5}l$ $$[AFGB]'=[ADCB]-[ADF]-[BCG]$$$$[AFGB]'=12/5-6/25-12/25$$$$[AFGB]'=42/25$$Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get $$\\frac{[AFGB]'}{[AFGB]}=\\frac{42}{25}/\\frac{21}{2}\\implies \\frac{[AFGB]'}{[AFGB]}=\\frac{4}{25}$$By applying an infinite summation $$[AEB]=\\sum_{n=0}^{\\infty} \\frac{21}{2}\\cdot{(\\frac{4}{25})^n}$$$$S=\\frac{a_1}{1-r}$$$$\\boxed{[AEB]=\\frac{25}{2}}$$", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "\\frac {1}{2} \\cdot 15 \\cdot 36 = \\boxed{\\mathrm{(C)}\\ 210}$$ ### Solution 2 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$. We get the following triangle: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0); draw(A--B--C--cycle); draw(C--F,dashed); label(\"$$A'$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,N); label(\"$$C'$$\",F,SE); label(\"5\",(A+C)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ The length of $A'B$ in this triangle is equal to the length of the original $AB$, minus the length of $CD$. Thus $A'B = 52 - 39 = 13$. Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\\frac{A'C\\cdot BC}2 = \\frac{5\\cdot 12}2 = 30$, and therefore its altitude $CC'$ is $\\frac{[A'BC]}{A'B} = \\frac{60}{13}$. Now the area of the original trapezoid is $\\frac{(AB+CD)\\cdot CC'}2 = \\frac{91 \\cdot 60}{13 \\cdot 2} = 7\\cdot 30 = \\boxed{\\mathrm{(C)}\\ 210}$ ### Solution 3 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Call the length of $AD'$ to be $y$, the length of $BC'$ to be $z$, and the height of the trapezoid to be $x$. By", "Note we can just use mass points to get \\left( \\frac{15}{26} \\right)^2= \\frac{225}{676}$which is \\boxed{901}$. ## Solution 2 First draw $\\overline{CP}$ and extend it so that it meets with $\\overline{AB}$ at point $X$. $[asy] size(10cm); pair A,B,C,D,E,X,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); X=(4.3636,0); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(X); dot(P); dot(Q); dot(R); label(\"A\",A,WSW); label(\"B\",B,ESE); label(\"C\",C,NNW); label(\"D\",D,NE); label(\"E\",E,WNW); label(\"X\",X,SSE); label(\"P\",P,NNE); label(\"Q\",Q,SSW); label(\"R\",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--X); [/asy]$ We have that $[ABC]=\\frac{1}{2}\\cdot AC \\cdot BC\\sin{C}=\\frac{1}{2}\\cdot 4\\cdot {7}\\sin{C}=14\\sin{C}$ By Ceva's, $$3\\cdot{\\frac{2}{5}}\\cdot{\\frac{BX}{AX}}=1\\implies BX=\\frac{5\\cdot AX}{6}$$ That means that $$\\frac{11\\cdot {AX}}{6}=8\\implies AX=\\frac{48}{11} \\ \\text{and} \\ BX=\\frac{40}{11}$$ Now we apply mass points. Assume WLOG that $W_{A}=1$. That means that $$W_{C}=3, W_{B}=\\frac{6}{5}, W_{X}=\\frac{11}{5}, W_{D}=\\frac{21}{5}, W_{E}=4, W_{P}=\\frac{26}{5}$$ Notice now that $\\triangle{PBQ}$ is similar to $\\triangle{EBA}$. Therefore, $$\\frac{PQ}{EA}=\\frac{PB}{EB}\\implies \\frac{PQ}{3}=\\frac{10}{13}\\implies PQ=\\frac{30}{13}$$ Also, $\\triangle{PRA}$ is similar to $\\triangle{DBA}$. Therefore, $$\\frac{PA}{DA}=\\frac{PR}{DB}\\implies \\frac{21}{26}=\\frac{PR}{5}\\implies PR=\\frac{105}{26}$$ Because $\\triangle{PQR}$ is similar to $\\triangle{CAB}$, $\\angle{C}=\\angle{P}$. As a result, $[PQR]=\\frac{1}{2}\\cdot PQ \\cdot PR \\sin{C}=\\frac{1}{2}\\cdot \\frac{30}{13}\\cdot \\frac{105}{26}\\sin{P}=\\frac{1575}{338}\\sin{C}$. Therefore, $$\\frac{[PQR]}{[ABC]}=\\frac{\\frac{1575}{338}\\sin{C}}{14\\sin{C}}=\\frac{225}{676}\\implies 225+676=\\boxed{901}$$ • Not the author writing here, but a note is that Ceva's Theorem was actually not necessary to solve this problem. The information was just nice to know :)", "volume is $2\\cdot \\left( \\frac{6\\sqrt{2}\\cdot 3\\sqrt{2} \\cdot 6}{3} \\right)=144$. Thus, our answer is $432-144=\\boxed{288}$. ### Solution 2 $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed); label(\"A\",A,(-1,-1,0)); label(\"B\",B,( 2,-1,0)); label(\"C\",C,( 1, 1,0)); label(\"D\",D,(-1, 1,0)); label(\"E\",E,(0,0,1)); label(\"F\",F,(0,0,1)); label(\"G\",G,(0,0,-1)); label(\"H\",H,(0,0,-1)); [/asy]$ Extend $EA$ and $FB$ to meet at $G$, and $ED$ and $FC$ to meet at $H$. Now, we have a regular tetrahedron $EFGH$, which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s = 12\\sqrt{2}$. Using the formula for the volume of a regular tetrahedron, which is $V = \\frac{\\sqrt{2}S^3}{12}$, where S is the side length of the tetrahedron, the volume of our original solid is: $V = \\frac{1}{2} \\cdot \\frac{\\sqrt{2} \\cdot (12\\sqrt{2})^3}{12} = \\boxed{288}$. ### Solution 3 We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$; thus, we will integrate with respect to height from $0$ to $6$, noting that each cross section of height $dh$ is a rectangle. The volume is then $\\int_0^h(wl) \\ \\text{d}h$, where $w$ is the width of the rectangle and $l$ is the length. We can express $w$", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "# BS in relation to BF and CS in relation to CE #### mathmari ##### Well-known member MHB Site Helper Hey!! We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? #### Klaas van Aarsen ##### MHB Seeker Staff member We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? Hey mathmari !! Let's first make a drawing: \\begin{tikzpicture}[scale=2] %preamble \\usetikzlibrary {calc} \\coordinate[label=left:A] (A) at (0,0); \\coordinate[label=right:B] (B) at (0:5); \\coordinate[label=C] (C) at (70:4); \\coordinate[label=below:E] (E) at (0:1); \\coordinate[label=left:F] (F) at (70:3); \\coordinate[label=right:S] (S) at (intersection cs:first line={(B)--(F)}, second line={(C)--(E)}); \\draw (A) -- (B) -- (C) -- cycle; \\draw (C) -- (E); \\draw (B) -- (F); \\draw[-latex, ultra thick, blue] (A) -- node[ above ] {$\\mathbf{\\vec c}$} (B); \\draw[-latex, ultra thick, blue] (A) -- node[ right] {$\\mathbf{\\vec b}$} (C); \\path (A) --", "get $$t=\\frac{160}{3}.$$ Finally, the sum of the numerator and denominator is $160+3=\\boxed{163}.$ ### Solution 2 Let $A$ and $B$ be Kenny's initial and final points respectively and define $C$ and $D$ similarly for Jenny. Let $O$ be the center of the building. Also, let $X$ be the intersection of $AC$ and $BD$. Finaly, let $P$ and $Q$ be the points of tangency of circle $O$ to $AC$ and $BD$ respectively. $[asy] size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,X; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); X=(300,0); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(X); draw(A--B--X--cycle); draw(C--D); draw(P--O--Q); draw(O--X); draw(Circle(O,50)); label(\"A\",A,SW); label(\"B\",B,NNW); label(\"C\",C,S); label(\"D\",D,NE); label(\"P\",P,S); label(\"Q\",Q,NE); label(\"O\",O,W); label(\"X\",X,ESE); [/asy]$ From the problem statement, $AB=3t$, and $CD=t$. Since $\\Delta ABX \\sim \\Delta CDX$, $CX=AC\\cdot\\left(\\frac{CD}{AB-CD}\\right)=200\\cdot\\left(\\frac{t}{3t-t}\\right)=100$. Since $PC=100$, $PX=200$. So, $\\tan(\\angle OXP)=\\frac{OP}{PX}=\\frac{50}{200}=\\frac{1}{4}$. Since circle $O$ is tangent to $BX$ and $AX$, $OX$ is the angle bisector of $\\angle BXA$. Thus, $\\tan(\\angle BXA)=\\tan(2\\angle OXP)=\\frac{2\\tan(\\angle OXP)}{1- \\tan^2(\\angle OXP)} = \\frac{2\\cdot \\left(\\frac{1}{4}\\right)}{1-\\left(\\frac{1}{4}\\right)^2}=\\frac{8}{15}$. Therefore, $t = CD = CX\\cdot\\tan(\\angle BXA) = 100 \\cdot \\frac{8}{15} = \\frac{160}{3}$, and the answer is $\\boxed{163}$. ### Solution 3 $[asy] size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,R,S; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); R=(100,106.667); S=(0,53.333); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(R); dot(S); draw(A--B--D--C--cycle); draw(P--O); draw(D--S); draw(O--Q--R--cycle); draw(Circle(O,50)); label(\"A\",A,SW); label(\"B\",B,NNW); label(\"C\",(200,-205),S); label(\"D\",D,NE); label(\"P\",(100,-205),S); label(\"Q\",Q,NE); label(\"O\",O,SW); label(\"R\",R,NE); label(\"S\",S,W); [/asy]$ Let $t$ be the time they walk." ]

[ "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "## anonymous one year ago In square $ABCD$, $E$ is the midpoint of $\\overline{BC}$, and $F$ is the midpoint of $\\overline{CD}$. Let $G$ be the intersection of $\\overline{AE}$ and $\\overline{BF}$. Prove that $DG = AB$. • This Question is Open 1. anonymous Asymptote code below [asy] unitsize(2.5 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (0,1); C = (1,1); D = (1,0); E = (B + C)/2; F = (C + D)/2; G = extension(A,E,B,F); draw(A--B--C--D--cycle); draw(A--E); draw(B--F); draw(D--G); label(\"$A$\", A, SW); label(\"$B$\", B, NW); label(\"$C$\", C, NE); label(\"$D$\", D, SE); label(\"$E$\", E, N); label(\"$F$\", F, dir(0)); label(\"$G$\", G, WSW); [/asy]", "AMC$. Let's call the intersection of $\\overline{UC}$ and $\\overline{MV}$ $P$. Let $\\overline{UP}=x$. Then $\\overline{PC}=12-x$. Since $\\overline{UC} \\perp \\overline{MV}$, $\\overline{UP}$ and $\\overline{CP}$ are heights of triangles $\\triangle MUV$ and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 3 (Medians) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the incenter, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively. Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\\triangle PUM$ is enough. $\\overline{PC} = \\overline{MP} = 8$. The area of $\\triangle PUM = \\frac{4\\cdot8}{2}=16$. Multiplying this", "&= \\frac{1}{2}\\end{align*} Hence $CE = FC + x = \\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 2 Call the point of tangency point $F$ and the midpoint of $AB$ as $G$. $CF=2$ by Tangent Theorem. Notice that $\\angle EGF=\\frac{180-2\\cdot\\angle CGF}{2}=90-\\angle CGF$. Thus, $\\angle EGF=\\angle FCG$ and $tanEGF=tanFCG=\\frac{1}{2}$. Solving $EF=\\frac{1}{2}$. Adding, the answer is $\\frac{5}{2}$. ## Solution 3 Clearly, $EA = EF = BG$. Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = 5/2$. ## Solution 4 $[asy] size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label(\"A\",A,(-1,-1)); label(\"B\",B,( 1,-1)); label(\"C\",C,( 1, 1)); label(\"D\",D,(-1, 1)); label(\"E\",E,(-1, 0)); label(\"F\",F,( 0, 1)); label(\"x\",(A+E)/2,(-1, 0)); label(\"x\",(E+F)/2,( 0, 1)); label(\"2\",(F+C)/2,( 0, 1)); label(\"2\",(D+C)/2,( 0, 1)); label(\"2\",(B+C)/2,( 1, 0)); label(\"2-x\",(D+E)/2,(-1, 0)); label(\"G\",G,(0,-1)); dot(G); draw(G--C); label(\"\\sqrt{5}\",(G+C)/2,(-1,0)); [/asy]$ Let us call the midpoint of side $AB$, point $G$. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$. We get $GC=\\sqrt{5}$. We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$, we get similar triangles $EFG$ and $GFC$. Solving the ratios, we get $x=\\frac{1}{2}$, so the answer is $\\frac{5}{2} \\Rightarrow\\boxed{\\mathrm{(D)}\\ \\frac{5}{2}}$. ## Solution 5 Using the diagram as drawn in Solution 5, let the total", "and $\\frac{DG}{GE}=\\frac{EH}{HF}=\\frac{FI}{ID}=1$ in the diagram below.$[asy] draw((0, 0)--(6, 0)--(4, 3)--cycle); draw((2, 0)--(16/3, 1)--(8/3, 2)--cycle); draw((11/3, 1/2)--(4, 3/2)--(7/3, 1)--cycle); label(\"A\", (0, 0), SW); label(\"B\", (6, 0), SE); label(\"C\", (4, 3), N); label(\"D\", (2, 0), S); label(\"E\", (16/3, 1), NE); label(\"F\", (8/3, 2), NW); label(\"G\", (11/3, 1/2), SE); label(\"H\", (4, 3/2), NE); label(\"I\", (7/3, 1), W); [/asy]$ ## Solution 1 is this solution by franzliszt: By the Wooga Looga Theorem, $\\frac{[DEF]}{[ABC]}=\\frac{2^2-2+1}{(1+2)^2}=\\frac 13$. Notice that $\\triangle GHI$ is the medial triangle of Wooga Looga Triangle of $\\triangle ABC$. So $\\frac{[GHI]}{[DEF]}=\\frac 14$ and $\\frac{[GHI]}{[ABD]}=\\frac{[DEF]}{[ABC]}\\cdot\\frac{[GHI]}{[DEF]}=\\frac 13 \\cdot \\frac 14 = \\frac {1}{12}$ by Chain Rule ideas. ## Solution 2 or this solution by franzliszt: Apply Barycentrics w.r.t. $\\triangle ABC$ so that $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $D=(\\tfrac 23,\\tfrac 13,0),E=(0,\\tfrac 23,\\tfrac 13),F=(\\tfrac 13,0,\\tfrac 23)$. And $G=(\\tfrac 13,\\tfrac 12,\\tfrac 16),H=(\\tfrac 16,\\tfrac 13,\\tfrac 12),I=(\\tfrac 12,\\tfrac 16,\\tfrac 13)$. In the barycentric coordinate system, the area formula is $[XYZ]=\\begin{vmatrix} x_{1} &y_{1} &z_{1} \\\\ x_{2} &y_{2} &z_{2} \\\\ x_{3}& y_{3} & z_{3} \\end{vmatrix}\\cdot [ABC]$ where $\\triangle XYZ$ is a random triangle and $\\triangle ABC$ is the reference triangle. Using this, we find that$$\\frac{[GHI]}{[ABC]}=\\begin{vmatrix} \\tfrac 13&\\tfrac 12&\\tfrac 16\\\\ \\tfrac 16&\\tfrac 13&\\tfrac 12\\\\ \\tfrac 12&\\tfrac 16&\\tfrac 13 \\end{vmatrix}=\\frac{1}{12}.$$ # Testimonials The Wooga Looga Theorem can be used to prove many problems and should be a part of any geometry textbook. ~ilp2020 The Wooga Looga", "then $$\\overline{D E}$$ || $$\\overline{A C}$$ and DE = ________ AC by the Triangle Midsegrnent Theorem If $$\\overline{D E}$$ is the midsegment opposile $$\\overline{A C}$$ in ∆ABC, then $$\\overline{D E}$$ || $$\\overline{A C}$$ and DE = $$\\frac { 1 }{ 2 }$$ AC by the Triangle Midsegrnent Theorem (Theorem 6.8). Monitoring Progress and Modeling with Mathematics In Exercises 3-6, use the graph of ∆ABC with midsegments $$\\overline{D E}$$, $$\\overline{E F}$$, and $$\\overline{D F}$$. Question 3. Find the coordinates of points D, E, and F. Question 4. Show that $$\\overline{D E}$$ is parallel to $$\\overline{C B}$$ and that DE = $$\\frac{1}{2}$$CB. The slope of DE = $$\\frac { 0 + 2 }{ -2 + 4 }$$ = 1 The slope of CB = $$\\frac { -2 + 6 }{ 1 + 3 }$$ = 1 $$\\overline{D E}$$ is parallel to $$\\overline{C B}$$ DE = √(-2 + 4)² + (0 + 2)² = √4 + 4 = √8 CB = √(1 + 3)² + (-2 + 6)² = √16 + 16 = √32 So, DE = $$\\frac{1}{2}$$CB Question 5. Show that $$\\overline{E F}$$ is parallel to $$\\overline{A C}$$ and that EF = $$\\frac{1}{2}$$AC. Question 6. Show that $$\\overline{D F}$$ is parallel to $$\\overline{A B}$$ and that DF = $$\\frac{1}{2}$$AB. The slope of DF = $$\\frac { -4 + 2 }{", "# BS in relation to BF and CS in relation to CE #### mathmari ##### Well-known member MHB Site Helper Hey!! We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? #### Klaas van Aarsen ##### MHB Seeker Staff member We have a triangle $ABC$. A point $E$ is on side $c$ such that $\\overline{AE}=\\frac{1}{5}\\cdot \\overline{AB}$, a point $F$ is on side $b$ such that $\\overline{FC}=\\frac{1}{4}\\cdot \\overline{AC}$. The segments $EC$ and $FB$ intersect on $S$. Write $BS$ in relation to $BF$ and $CS$ in relation to $CE$. Can we solve that using vectors? Hey mathmari !! Let's first make a drawing: \\begin{tikzpicture}[scale=2] %preamble \\usetikzlibrary {calc} \\coordinate[label=left:A] (A) at (0,0); \\coordinate[label=right:B] (B) at (0:5); \\coordinate[label=C] (C) at (70:4); \\coordinate[label=below:E] (E) at (0:1); \\coordinate[label=left:F] (F) at (70:3); \\coordinate[label=right:S] (S) at (intersection cs:first line={(B)--(F)}, second line={(C)--(E)}); \\draw (A) -- (B) -- (C) -- cycle; \\draw (C) -- (E); \\draw (B) -- (F); \\draw[-latex, ultra thick, blue] (A) -- node[ above ] {$\\mathbf{\\vec c}$} (B); \\draw[-latex, ultra thick, blue] (A) -- node[ right] {$\\mathbf{\\vec b}$} (C); \\path (A) --", "$\\triangle AMC$. Let's call the intersection of $\\overline{UC}$ and $\\overline{MV}$ $P$. Let $\\overline{UP}=x$. Then $\\overline{PC}=12-x$. Since $\\overline{UC} \\perp \\overline{MV}$, $\\overline{UP}$ and $\\overline{CP}$ are heights of triangles $\\triangle MUV$ and $\\triangle MCV$, respectively. Both of these triangles have base $12$. Area of $\\triangle MUV = \\frac{x\\cdot12}{2}=6x$ Area of $\\triangle MCV = \\frac{(12-x)\\cdot12}{2}=72-6x$ Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$. This is $\\frac{3}{4}$ of the triangle, so the area of the triangle is $\\frac{4}{3}\\cdot{72}=\\boxed{\\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 ## Solution 3 (Medians) Draw median $\\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy]$ Since we know that all medians of a triangle intersect at the centroid, we know that $\\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\\overline{PC}:\\overline{UP}=2:1$, and since the two segments sum to $12$, $\\overline{PC}$ and $\\overline{UP}$ are $8$ and $4$, respectively. Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\\triangle PUM$ is enough. $\\overline{PC} = \\overline{MP} = 8$. The area of $\\triangle PUM = \\frac{4\\cdot8}{2}=16$. Multiplying", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "# Difference between revisions of \"2005 AMC 10B Problems/Problem 14\" ## Problem Equilateral $\\triangle ABC$ has side length $2$, $M$ is the midpoint of $\\overline{AC}$, and $C$ is the midpoint of $\\overline{BD}$. What is the area of $\\triangle CDM$? $[asy]defaultpen(linewidth(.8pt)+fontsize(8pt)); pair B = (0,0); pair A = 2*dir(60); pair C = (2,0); pair D = (4,0); pair M = midpoint(A--C); label(\"A\",A,NW);label(\"B\",B,SW);label(\"C\",C, SE);label(\"M\",M,NE);label(\"D\",D,SE); draw(A--B--C--cycle); draw(C--D--M--cycle);[/asy]$$\\textrm{(A)}\\ \\frac {\\sqrt {2}}{2}\\qquad \\textrm{(B)}\\ \\frac {3}{4}\\qquad \\textrm{(C)}\\ \\frac {\\sqrt {3}}{2}\\qquad \\textrm{(D)}\\ 1\\qquad \\textrm{(E)}\\ \\sqrt {2}$ ## Solution In order to calculate the area of $\\triangle CDM$, we can use the formula $A=\\dfrac{1}{2}bh$, where $\\overline{CD}$ is the base. We already know that $\\overline{CD}=2$, so the formula now becomes $A=h$. We can drop verticals down from $A$ and $M$ to points $E$ and $F$, respectively. We can see that $\\triangle AEC \\sim \\triangle MFC$. Now, we establish the relationship that $\\dfrac{AE}{MF}=\\dfrac{AC}{MC}$. We are given that $\\overline{AC}=2$, and $M$ is the midpoint of $\\overline{AC}$, so $\\overline{MC}=1$. Because $\\triangle AEB$ is a $30-60-90$ triangle and the ratio of the sides opposite the angles are $1-\\sqrt{3}-2$ $\\overline{AE}$ is $\\sqrt{3}$. Plugging those numbers in, we have $\\dfrac{\\sqrt{3}}{MF}=\\dfrac{2}{1}$. Cross-multiplying, we see that $2\\times\\overline{MF}=\\sqrt{3}\\times1\\implies \\overline{MF}=\\dfrac{\\sqrt{3}}{2}$ Since $\\overline{MF}$ is the height $\\triangle CDM$, the area is $\\boxed{\\mathrm{(C)}\\ \\dfrac{\\sqrt{3}}{2}}$.", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "area of a triangle equals $\\frac{1}{2} \\cdot ab \\cdot \\sin(C)$. If angle $DAE = Z$, we have that $$\\frac{[ADE]}{[ABC]} = \\frac{\\frac{1}{2} \\cdot 19 \\cdot 14 \\cdot \\sin(Z)}{\\frac{1}{2} \\cdot 25 \\cdot 42 \\cdot \\sin(Z)} = \\frac{19 \\cdot 14}{25 \\cdot 42} = \\frac{ab}{cd}$$. - SuperJJ ## Video Solution CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00 ## Solution 2 (no trig) $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, W); label(\"E\", E, NE); label(\"19\", (A + D)/2, W); label(\"6\", (B + D)/2, W); label(\"14\", (A + E)/2, NE); label(\"28\", (C + E)/2, NE); [/asy]$ We can let $[ADE]=x$. Since $EC=2 \\cdot EA$, $[DEC]=2x$. So, $[ADC]=3x$. This means that $[BDC]=\\frac{6}{19}\\cdot3x=\\frac{18x}{19}$. Thus, $\\frac{[ADE]}{[BCED]} = \\frac{x}{\\frac{18x}{19}+2x}= \\boxed{\\textbf{(D) }\\frac{19}{56}}.$ -Conantwiz2023 ## Solution 3 (trig) The area of a triangle is $\\frac{1}{2} bc\\sin A$. Using this formula: $[ADE]=\\frac{1}{2}\\cdot19\\cdot14\\cdot\\sin A = 133\\sin A$ $[ABC]=\\frac{1}{2}\\cdot25\\cdot42\\cdot\\sin A = 525\\sin A$ Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$, $[BCED] = 525\\sin A - 133\\sin A = 392\\sin A$. Therefore, the desired ratio is $\\frac{133\\sin A}{392\\sin A}=\\boxed{\\textbf{(D) }\\frac{19}{56}}$ Note: $BC=39$ was not used in this problem. ## Solution 4 Let", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "# Difference between revisions of \"Midpoint\" ## Definition In Euclidean geometry, the midpoint of a line segment is the point on the segment equidistant from both endpoints. A midpoint bisects the line segment that the midpoint lies on. Because of this property, we say that for any line segment $\\overline{AB}$ with midpoint $M$, $AM=BM=\\frac{1}{2}AB$. Alternatively, any point $M$ on $\\overline{AB}$ such that $AM=BM$ is the midpoint of the segment. $[asy] draw((0,0)--(4,0)); dot((0,0)); label(\"A\",(0,0),N); dot((4,0)); label(\"B\",(4,0),N); dot((2,0)); label(\"M\",(2,0),N); label(\"Figure 1\",(2,0),4S); [/asy]$ ## Midpoints and Triangles $[asy] pair A,B,C,D,E,F,G; A=(0,0); B=(4,0); C=(1,3); D=(2,0); E=(2.5,1.5); F=(0.5,1.5); G=(5/3,1); draw(A--B--C--cycle); draw(D--E--F--cycle,green); dot(A--B--C--D--E--F--G); draw(A--E,red); draw(B--F,red); draw(C--D,red); label(\"A\",A,S); label(\"B\",B,S); label(\"C\",C,N); label(\"D\",D,S); label(\"E\",E,E); label(\"F\",F,W); label(\"G\",G,NE); label(\"Figure 2\",D,4S); [/asy]$ ### Midsegments As shown in Figure 2, $\\Delta ABC$ is a triangle with $D$, $E$, $F$ midpoints on $\\overline{AB}$, $\\overline{BC}$, $\\overline{CA}$ respectively. Connect $\\overline{EF}$, $\\overline{FD}$, $\\overline{DE}$ (segments highlighted in green). They are midsegments to their corresponding sides. Using SAS Similarity Postulate, we can see that $\\Delta CFE \\sim \\Delta CAB$ and likewise for $\\Delta ADF$ and $\\Delta BED$. Because of this, we know that \\begin{align*} AB &= 2FE \\\\ BC &= 2DE \\\\ CA &= 2ED \\\\ \\end{align*} Which is the Triangle Midsegment Theorem. Because we have a relationship between these segment lengths, $\\Delta ABC \\sim \\Delta EFD (SSS)$ with similar ratio 2:1. The area ratio is then", "I was bored so I just read all the responses to this post. I am very confused and tired after reading this but also entertained. IMAO. Also would aops be allowed to track our IP addresses like that? 1. 👍 2. 👎 1. 👍 2. 👎 1. 👍 2. 👎 26. i like trains 1. 👍 2. 👎 27. latex [dot] artofproblemsolving [dot com] add a .png at the end yeah they dont let me post url srry 1. 👍 2. 👎 28. ok but did you know I got a 46 on mathcounts chapter 1. 👍 2. 👎 29. its 18√5 1. 👍 2. 👎 ## Similar Questions 1. ### Geometry In trapezoid $ABCD$, $\\overline{AB}$ is parallel to $\\overline{CD}$, $AB = 7$ units, and $CD = 10$ units. Segment $EF$ is drawn parallel to $\\overline{AB}$ with $E$ lying on $\\overline{AD}$ and $F$ lying on $\\overline{BC}$. If 2. ### mathematics In triangle $ABC,$ angle bisectors $\\overline{AD},$ $\\overline{BE},$ and $\\overline{CF}$ meet at $I.$ If $DI = 3,$ $BD = 4,$ and $BI = 5,$ then compute the area of triangle $ABC.$ 3. ### math Points $D$, $E$, and $F$ are the midpoints of sides $\\overline{BC}$, $\\overline{CA}$, and $\\overline{AB}$, respectively, of $\\triangle ABC$. Points $X$, $Y$, and $Z$ are the midpoints of $\\overline{EF}$, $\\overline{FD}$, and 4. ### geometry Points $F$,", "points $D$ and $F$. Connect them to create the midsegment. Don’t forget to put the tic marks, indicating that $D$ and $F$ are midpoints, $\\overline{AD} \\cong \\overline{DB}$ and $\\overline{BF} \\cong \\overline{FC}$. Example 2: Find the midpoint of $\\overline{AC}$ from $\\triangle ABC$. Label it $E$ and find the other two midsegments of the triangle. Solution: For every triangle there are three midsegments. Let’s transfer what we know about midpoints in the coordinate plane to midsegments in the coordinate plane. We will need to use the midpoint formula, $\\left( \\frac{x_1+x_2}{2}, \\frac{y_1+y_2}{2}\\right)$. Example 3: The vertices of $\\triangle LMN$ are $L(4, 5), M(-2, -7)$ and $N(-8, 3)$. Find the midpoints of all three sides, label them $O, P$ and $Q$. Then, graph the triangle, it’s midpoints and draw in the midsegments. Solution: Use the midpoint formula 3 times to find all the midpoints. $L$ and $M = \\left (\\frac{4+(-2)}{2}, \\frac{5+(-7)}{2}\\right)=(1,-1)$, point $O$ $L$ and $N = \\left(\\frac{4+(-8)}{2}, \\frac{5+3}{2}\\right)=(-2,4)$, point $Q$ $M$ and $N = \\left(\\frac{-2+(-8)}{2}, \\frac{-7+3}{2}\\right)=(-5,-2)$, point $P$ The graph would look like the graph to the right. We will use this graph to explore the properties of midsegments. Example 4: Find the slopes of $\\overline{NM}$ and $\\overline{QO}$. Solution: The slope of $\\overline{NM}$ is $\\frac{-7-3}{-2-(-8)}=\\frac{-10}{6}=-\\frac{5}{3}$. The slope of $\\overline{QO}$ is $\\frac{-1-4}{1-(-2)}=-\\frac{5}{3}$. From this we can conclude that $\\overline{NM} \\ || \\", "Now, since $\\triangle BOC$ is isosceles with $\\overline{OB} \\cong \\overline{OC}$, we have that $\\angle BCO = \\angle CBO = 90 - \\frac{\\theta}{2}$. In addition, we know that $\\overline{BC} \\cong \\overline{CD}$ as they are both equal to $200$ and $\\overline{OB} \\cong \\overline{OC} \\cong \\overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\\triangle OBC \\cong \\triangle OCD$, so we have that $\\angle OCD = \\angle BCO = 90 - \\frac{\\theta}{2}$, so $\\angle DCF = \\theta$. Thus, we have $\\frac{FC}{DC} = \\cos \\theta = \\frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \\boxed{500}$. ## Solution 5 (Just Geometry) $[asy] size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label(\"A\",A,W); label(\"B\",B,NW); label(\"C\",C,NE); label(\"D\",D,ENE); label(\"O\",O,S); label(\"\\theta\",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label(\"E\",E,NE); F=extension(C,O,A,D); label(\"F\",F,NE); //Angle marks draw(anglemark(C,O,B)); [/asy]$ Label AD intercept OB at E and OC at F. $\\overarc{AB}= \\overarc{BC}= \\overarc{CD}=\\theta$ $\\angle{BAD}=\\frac{1}{2} \\cdot \\overarc{BCD}=\\theta=\\angle{AOB}$ From there, $\\triangle{OAB} \\sim \\triangle{ABE}$, thus: $\\frac{OA}{AB} = \\frac{AB}{BE} = \\frac{OB}{AE}$ $OA = OB$ because they are both radii of $\\odot{O}$. Since $\\frac{OA}{AB} = \\frac{OB}{AE}$, we" ]

[ "and a dilation is a simple transformation that uniformly scales all distances. In particular, we allow the projection of everything onto the origin to be a dilation, and we allow reflections across the origin also. We note that all dilations can be expressed as $tI = \\begin{bmatrix} t & 0 \\\\ 0 & t \\end{bmatrix}$ for $t∈\\mathbf{R}$ a free parameter. We note that all rotations can be expressed as the orthogonal matrices $R = \\begin{bmatrix} \\cos θ & -\\sin θ \\\\ \\sin θ & \\cos θ \\end{bmatrix}$ for $θ∈\\mathbf{R}$ a free parameter. Let a rotation-dilation be a linear transformation that rotates an object, then dilates it. Then all rotation-dilations are of the form $tR = tIR = \\begin{bmatrix} t & 0 \\\\ 0 & t \\end{bmatrix} \\begin{bmatrix} \\cos θ & -\\sin θ \\\\ \\sin θ & \\cos θ \\end{bmatrix} = \\begin{bmatrix} t\\cos θ & -t\\sin θ \\\\ t\\sin θ & t\\cos θ \\end{bmatrix}$ for $t∈\\mathbf{R}$ and $θ∈\\mathbf{R}$ free parameters. By making the substitution $a = t\\cos θ$ and $b = t\\sin θ$, we can express this as $tR = \\begin{bmatrix} a & -b \\\\ b & a \\end{bmatrix}$ We now have a characterization of all linear transformations that either dilate or rotate the two-dimensional plane. The operation of matrix multiplication represents the composition of these linear transformations. Thus", "Select Page ## Spiral Similarity is a combination of rotation and dilation. It is an interesting geometric transformation for a variety of reasons. In the previous section, we found that any segment can be mapped onto any other segment using spiral similarity. We wish to treat spiral similarity algebraically. Let $$A = (r, \\theta)$$. Suppose we want to rotate it by an angle $$\\theta_1$$ and also dilate it (multiply its length by) $$r_1$$. Clearly, after we perform the rotation and dilation, the resultant point is: $$(r \\times r_1, \\theta + \\theta_1)$$ This action of rotation and dilation can be encoded in an ordered pair of numbers: $$r_1, \\theta_1$$. The first number of this ordered pair says how much we should dilate (by what factor should we multiply the length). The second number gives the angle of rotation. We will call this ordered pair of numbers an action. Hence $$(r_1, \\theta_1)$$ acts on $$(r, \\theta )$$ to rotate and dilate it to the final position $$(r \\times r_1, \\theta + \\theta_1)$$ A short hand way of writing this is $$(r_1 \\theta_1 ) \\searrow (r, \\theta) \\to (r \\times r_1, \\theta + \\theta_1)$$ # Dual nature of points and actions Notice that the action $$(r_1, \\theta_1)$$ can be regarded as a polar coordinate of a point. It is the position", "rotation about the origin. What do we need to multiply by to produce a 90° clockwise rotation about the origin? We need to multiply by $$\\frac{1}{i}$$. If w=iz (w is a 90° counterclockwise rotation about the origin from z), z=$$\\frac{w}{i}$$, which means that if we divide w by i, we will get to z, which will create a 90° clockwise rotation about the origin from w. Another response: 90° clockwise is 270° counterclockwise, so you could multiply by i3 to map (a,b) to (b,-a). Question 6. Given z is a complex number a+bi, determine if L(z) is a linear transformation. Explain why or why not. a. L(z)=i3 z Yes. L(z)=-iz, so L(z+w)=-iz+-iw and L(z)+L(w)=-iz+-iw. kL(z)=k(-iz) and L(kz)=-i(kz). Since L(z+w)=L(z)+L(w) and kL(z)=L(kz), the function is a linear transformation. b. L(z)=z+4i No. L(z+w)=z+w+4i and L(z)+L(w)=z+4i+w+4i. kL(z)=k(z+4i) and L(kz)=kz+4i. Since L(z+w)=L(z)+L(w) and kL(z)≠L(kz), the function is not a linear transformation. ### Eureka Math Precalculus Module 1 Lesson 10 Exit Ticket Answer Key Question 1. Given T(z)=z, describe the geometric effect of the following: a. T(z)=5z It has a dilation with a scale factor of 5. b. T(z)=$$\\frac{z}{2}$$ It has a dilation with a scale factor of $$\\frac{1}{2}$$. c. T(z)=iz It has a 90° counterclockwise rotation about the origin. Question 2. If z=-2+3i is the result of a 90° counterclockwise rotation about", "by applying C and then AB, which is B followed by A. Question 8. Let AB represent any composition of linear transformations in R2. What is the value of (AB)x where x=$$\\left[\\begin{array}{l} \\mathbf{0} \\\\ \\mathbf{0} \\end{array}\\right]$$? Since a composition of linear transformations in R2 is also a linear transformation, we know that applying it to the origin will result in no change. ### Eureka Math Precalculus Module 2 Lesson 8 Exit Ticket Answer Key Let A be the matrix representing a rotation about the origin 135° and B be the matrix representing a dilation of 6. Let x=$$\\left[\\begin{array}{c} -1 \\\\ \\frac{1}{2} \\end{array}\\right]$$. a. Write down A and B.", "a factor of a in the x direction (from the y-axis) . … … $\\left[ \\begin{matrix} 1&0 \\\\ 0&a \\\\ \\end{matrix} \\right]$ … causes a dilation by a factor of a in the y direction (from the x-axis) } . For example: … … $\\left[ \\begin{matrix} x' \\\\ y' \\\\ \\end{matrix} \\right] = \\left[ \\begin{matrix} 3&0 \\\\ 0&1 \\\\ \\end{matrix} \\right] \\left[ \\begin{matrix} x \\\\ y \\\\ \\end{matrix} \\right]$ . … … $\\left[ \\begin{matrix} x' \\\\ y' \\\\ \\end{matrix} \\right] = \\left[ \\begin{matrix} 3x \\\\ y \\\\ \\end{matrix} \\right]$ . This represents a dilation by a factor of 3 in the x direction (all x-values are multiplied by 3). . This transformation could be written as: … … $T \\left( \\left[ \\begin{matrix} x \\\\ y \\\\ \\end{matrix} \\right] \\right) = \\left[ \\begin{matrix} 3x \\\\ y \\\\ \\end{matrix} \\right]$ or … … $x' = 3x$ .. and … … $y' = y$ For example, if we start with the point $(5, \\; 3)$ under this transformation, the image is $(15, \\; 3)$ . Compare this to starting with the cubic graph $y = x^3$ under this transformation, the image is: $y' = \\big( \\frac{1}{3}x' \\big)^3$ . • Remember that $y = f(nx)$ causes a dilation by a factor of $\\frac{1}{n}$ in the x direction (from the y-axis)} . Note:", "scale factor 4. 6. The matrix product is AB=$$\\left[\\begin{array}{cc} 2 \\sqrt{2} & -2 \\sqrt{2} \\\\ 2 \\sqrt{2} & 2 \\sqrt{2} \\end{array}\\right]$$. 7. The transformation produced by matrix AB has the effect of rotation by 45° while scaling by 4. iv. A=$$\\left[\\begin{array}{ll} 1 & 1 \\\\ 0 & 1 \\end{array}\\right]$$, B=$$\\left[\\begin{array}{ll} 2 & 0 \\\\ 0 & 2 \\end{array}\\right]$$ 1. The transformation produced by matrix B has the effect of dilation with scale factor 2. 2. The transformation produced by matrix A has the effect of shearing parallel to the y-axis. 3. The large green square is the image of the original unit square under the transformation produced by B. 4. The red parallelogram is the image of the green square under the transformation produced by A. 5. If we dilate and then shear, the net effect is a dilated shear. 6. The matrix product is AB=$$\\left[\\begin{array}{ll} 2 & 2 \\\\ 0 & 2 \\end{array}\\right]$$. 7. The transformation produced by matrix AB has the effect of a dilated shear. v. 1. The transformation produced by matrix B has the effect of rotation by 30°. 2. The transformation produced by matrix A has the effect of rotation by -60°. 3. The tilted green square is the image of the original unit square under the transformation produced by B. 4. The", "clockwise by an angle of $\\theta$ about the origin are given by $\\begin{pmatrix} x’ \\\\ y’ \\end{pmatrix} = \\begin{pmatrix} \\cos \\theta &-\\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix}$ For $(x, y) = (2,-3)$ and rotation of $\\theta = 90^{\\circ}$ about the origin, we have \\begin{aligned} \\begin{pmatrix} x’ \\\\ y’ \\end{pmatrix} & = \\begin{pmatrix} \\cos 90^{\\circ} &-\\sin 90^{\\circ} \\\\ \\sin 90^{\\circ} & \\cos 90^{\\circ} \\end{pmatrix} \\begin{pmatrix} 2 \\\\-3 \\end{pmatrix} \\\\ & = \\begin{pmatrix} 0 &-1 \\\\ 1 & 0 \\end{pmatrix} \\begin{pmatrix} 2 \\\\-3 \\end{pmatrix} \\\\ & = \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} \\end{aligned} Thus, the coordinates of the image of point $P$ are $\\boxed{P'(3,2)}$ [collapse] Problem Number 7 Given a point $P(-8,12)$. The dilation $[P,1]$ maps point $(-4,8)$ to point $\\cdots \\cdot$ A. $(-4,8)$ D. $(4,-16)$ B. $(-4,16)$ E. $(4,-8)$ C. $(-4,-8)$ Solution The dilation rule is given as follows. If a point $(x,y)$ is dilated around center point $(a,b)$ by a scale factor of $k$, then the image point lies in $(k(x-a)+a, k(y-b)+b)$. This means that the coordinates of the image of the point $(-4, 8)$ after being dilated around the center point $(-8, 12)$ by a scale factor of $1$ are $(1(-4-(-8))+(-8), 1(8-12)+12)$ $= (-4, 8)$ Dilation $[P, 1]$ maps point $(-4,8)$ to $\\boxed{(-4, 8)}$ [collapse] Problem Number 8 The", "# Diagonalize matrix and find rotation-dilation matrix Diagonalize the following matrix: $$\\begin{bmatrix}0&1\\\\-5&-2\\end{bmatrix}$$ Also write it in form $$PCP^{-1}$$, where $$C$$ is a rotation-dilation matrix. How would I approach this problem? How do I find $$P$$ and $$C$$? • Do you know what “diagonalize” means? Start with that. – amd Dec 15 '18 at 20:06", "rotation about the origin, then a dilation by a scale factor of $2$ with a center of dilation at the origin C. a $90^{\\circ}$ counterclockwise rotation about the origin, then a dilation by a scale factor of $\\dfrac{1}{2}$ with a center of dilation at the origin D. a $90^{\\circ}$ clockwise rotation about the origin, then a dilation by a scale factor of $\\dfrac{1}{2}$ with a center of dilation at the origin 24. What is the value of $n$ in the equation shown below? $2^{2} \\times 2^{n} = (2^{4})^{3}$ A. $5$ B. $6$ C. $10$ D. $12$ 25. Which set of ordered pairs represents a function? A. ${(2,7), (2,8), (3,8)}$ B. ${(3,2), (3,3), (3,4)}$ C. ${(4,1), (5,1), (4,4)}$ D. ${(5,6), (8,6), (9,6)}$ 26. A parallelogram with vertices at $(0,3),(2,0),(4,2)$ and $(2,5)$ is reflected over the $y$ -axis. Which vertex of the parallelogram will have the same $x$ -coordinate before and after the reflection? A. $(0,3)$ B. $(2,0)$ C. $(4,2)$ D. $(2,5)$ 27. The graph of a function is shown below. Which statement is true about a section of the graph? A. In Section $N$, the function is linear and decreasing. B. In Section $P$, the function is linear and increasing. C. In Section $Q$, the function is nonlinear and decreasing. D. In Section $R$, the function is nonlinear and", "dx) (+ y dy))) ### Dilations Dilation is a general term that typically refers to enlarging (or shrinking) something. But points can’t get bigger or smaller: they’re points. So dilation, in the context of a point, means moving closer or farther away from a fixed origin point. We move the point by scaling its distance from the origin by a certain factor. We could use dilations to implement features like pinch-to-zoom. If we have a dilation factor of $z$ (for zoom), then our $x' = zx$ and our $y' = zy$. (defun easy-dilate (z x y) (values (* z x) (* z y))) ### Rotations Rotations are pretty tricky. They usually require trigonometry to calculate. Suppose I have a point $(x,y)$, and I want to rotate it by $\\theta$ radians (one radian is about $57.3^{\\circ}$) about the origin in a counterclockwise direction. Using Math (TM), the new point $(x',y')$ will have coordinates $$x' = x\\cos\\theta-y\\sin\\theta$$ and $$y'=x\\sin\\theta+y\\cos\\theta.$$ It’s pretty gnarly, and we definitely prefer to bake this in a library so we don’t have to remember it. If all we’re going to do is rotate points about the origin, that’s easy to do in Lisp. (defun easy-rotate (theta x y) (let ((new-x (- (* x (cos theta)) (* y (sin theta)))) (new-y (+ (* x (sin theta)) (*", "changes as we change the size of the angle $x$x. If the length of $OD$OD is $2$2 units, draw the graph that shows how the length of the side $CD$CD will change as we change the size of the angle $x$x. 4. Using the applet below to change the length of $CD$CD, select the most appropriate description of the graph of $y=a\\tan x$y=atanx. Scaling the unit circle by a factor of $a$a corresponds to horizontally translating of the graph of $y=\\tan x$y=tanx by $a$a units. A Scaling the unit circle by a factor of $a$a corresponds to vertically dilating of the graph of $y=\\tan x$y=tanx by a factor of $a$a. B Horizontally translating the graph of $y=\\tan x$y=tanx by $a$a units corresponds to a horizontal translation of the unit circle by $a$a units. C Vertically dilating the graph of $y=\\tan x$y=tanx by a factor of $a$a corresponds to a rotation about the unit circle of $a$a radians. D Scaling the unit circle by a factor of $a$a corresponds to horizontally translating of the graph of $y=\\tan x$y=tanx by $a$a units. A Scaling the unit circle by a factor of $a$a corresponds to vertically dilating of the graph of $y=\\tan x$y=tanx by a factor of $a$a. B Horizontally translating the graph of $y=\\tan x$y=tanx by $a$a units corresponds to", "of a point that is $$r_1$$ distance away from origin and making an angle $$\\theta_1$$ with the positive direction of x axis. On the other hand the point that $$(r_1, \\theta_1)$$ is acting on, that $$(r, \\theta)$$ can be itself regarded as an action. It has the power of rotation by $$\\theta$$ and dilation by r. We have arrived at the essence of our core idea: points can be regarded at once as objects waiting to be rotated and dilated OR actions that can rotate and dilate. (This is a supplemental note to live lecture session of Complex Number module of Cheenta I.S.I., C.M.I. Entrance Program and Math Olympiad Program)", "your graphing calculator …. For example, you might have a rhombus with points (4, 6), (-4, 6), (-2, -1), and (2, -1). For each statement below, determine if it is trueor false. Unit 3 Similar Figures and Dilations 2017. Find the coordinates of ABC after …. 2 Scaling A scaling about the origin is an affine transformation (1) where the matrix A = diag(s x,s y) with s x 6= 0 and s y 6= 0, and b = 0. On a coordinate axis system draw a quadrilateral (4-sided figure) which is not a special four-sided figure (for example, it’s not a parallelogram). For example, if you know A that transforms x to x', the matrix that transforms x' back to x is the inverse of A. To find the arc length given the radius and angle measure in degrees, you must first convert the angle from degrees to radians, using the formula: Degrees = Radians X (pi/180). After the second rotation, the new world coordinates of the point P in the Y-Zplane are given by P0= (Y 1;Z 1), as shown in Fig. Dilation is a transformation in which each point of an object is moved along a straight line. Determine the image of a region under a given transformation of variables. Find what is the midpoint", "\\end{array}\\right]$$ 1. The transformation produced by matrix B has the effect of reflection across the y-axis. 2. The transformation produced by matrix A has the effect of reflection across the x-axis. 3. The green square in the second quadrant is the image of the original square under the transformation produced by B. 4. The red square in the third quadrant is the image of the square after transforming with matrix A and then matrix B. 5. If we reflect across the y-axis and then reflect across the x-axis, the net effect is a rotation by 180°. 6. The matrix product is AB=$$\\left[\\begin{array}{cc} -1 & 0 \\\\ 0 & -1 \\end{array}\\right]$$. 7. The transformation produced by matrix AB has the effect of rotation by 180°. iii. 1. The transformation produced by matrix B has the effect of dilation from the origin with scale factor 4. 2. The transformation produced by matrix A has the effect of rotation by 45°. 3. The large green square is the image of the original unit square under the transformation produced by B. 4. The tilted red square is the image of the green square under the transformation produced by A. 5. If we dilate with factor 4 and then rotate by 45°, the net effect is a rotation by 45° and dilation with", "plt.show() Dilating the plane by a factor of 0.5: $f \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}0.5 & 0\\\\0 & 0.5\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix},$ in Numpy: a = .5*np.eye(2) dst = linear_transformation(src, a) plt.imshow(dst) plt.show() Scaling the plane in the $y$-axis by a factor of 0.5: $f \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}1 & 0\\\\0 & 0.5\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix},$ in Numpy: a = np.array([[1, 0], [0, .5]]) dst = linear_transformation(src, a) plt.imshow(dst) plt.show() Shearing about the $y$-axis with a vertical displacement of $+x/2$: $f \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}1 & 0\\\\{1 \\over 2} & 0\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix},$ in Numpy: a = np.array([[1, 0], [.5, 1]]) dst = linear_transformation(src, a) plt.imshow(dst) plt.show() Rotation through $45^{\\circ}$ about the origin: $f \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix} \\cos{\\pi \\over 4} & -\\sin{\\pi \\over 4} \\\\ \\sin{\\pi \\over 4} & \\cos{\\pi \\over 4} \\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix},$ in Numpy: alpha = np.pi/4 a = np.array([[np.cos(alpha), -np.sin(alpha)], [np.sin(alpha), np.cos(alpha)]]) dst = linear_transformation(src, a) plt.imshow(dst) plt.show() Reflexion in a line with inclination of $45^{\\circ}$ through the origin: $f \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix} \\cos{\\pi \\over 2} & \\sin{\\pi \\over 2} \\\\ \\sin{\\pi \\over 2} & -\\cos{\\pi \\over 2} \\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix},$ in Numpy: alpha = np.pi/4 a = np.array([[np.cos(2*alpha), np.sin(2*alpha)], [np.sin(2*alpha), -np.cos(2*alpha)]]) dst = linear_transformation(src, a) plt.imshow(dst) plt.show()", "& \\cos(1^\\circ) \\end{matrix} \\right]$$ Shrinking will be represented as a dilation by a factor of $0.9$: $$Di(0.9) = \\left[ \\begin{matrix} 0.9 & 0 \\\\ 0 & 0.9 \\end{matrix} \\right]$$ And we can shear by, say, a factor of 1.1, as follows: $$Sh(1.1) = \\left[ \\begin{matrix} 1 & 1.1 \\\\ 0 & 1 \\end{matrix} \\right]$$ Now in order to apply all these transformations to our hedgehog, we simply need to do successive multiplication. If we want to rotation, then dilate, then shear, this would look like: $$h' = Sh(1.1) \\cdot [ Di(0.9) \\cdot [ Rot(1^\\circ) \\cdot h ]]$$ And this would work! However, there's one issue... Recall that we're in the shoes of a game developer here. We want things to be fast. However, this is not very fast. Matrix multiplication is relatively cheap, certainly, but it gets slower as the number of points in your matrices increases. And $h$, as noted previously, is a very large matrix—so multiplication with $h$ is slow, relatively speaking. The first calculation we do here is $Rot(1^\\circ) \\cdot h$, which will be slow since it involves $h$. The result, just a rotation of $h$, will have the same number of points as $h$, and thus be just as large. Then we multiply this matrix by $Di(0.9)$. Again, this will be slow, since one", "$\\mathbf D$ and a rotation matrix $\\mathbf \\Omega$, you get your rotated samples $\\mathbf R$ simply doing: $$\\mathbf D \\mathbf \\Omega = \\mathbf R$$ $\\mathbf \\Omega$ is a square matrix with $P$ rows and columns. If you want to reduce the number of PCs all you have to do is keep only the first $K$ columns of $\\mathbf \\Omega$.", "- \\sin^2 \\phi \\end{bmatrix} = \\begin{bmatrix} \\cos^2 \\phi & \\sin \\phi \\cos \\phi \\\\ \\sin \\phi \\cos \\phi & \\cos^2 \\phi \\end{bmatrix},$ which is the matrix for projection onto $L$. 2. When $k=-1$ we have $\\begin{bmatrix} 1 - 2 \\sin^2 \\phi & 2 \\sin \\phi \\cos \\phi \\\\ 2 \\sin \\phi \\cos \\phi & 1 - 2 \\cos^2 \\phi \\end{bmatrix} = \\begin{bmatrix} \\cos(2\\phi) & \\sin (2\\phi) \\\\ \\sin (2\\phi) & - \\cos (2\\phi) \\end{bmatrix}$ which is the matrix for reflection in $L$. Exercise 31 1. A reflection is a dilation by factor $-1$, so applying it twice gives a dilation with factor $(-1)^2 = 1$, i.e. the identity. 2. A projection is a dilation by factor $0$, so applying it twice gives a dilation with factor $0^2 = 0$, i.e. the same projection. Exercise 32 1. A projection matrix clearly has the desired properties. For the converse, suppose $M$ is symmetric ($b=c$), $ad = b^2$ and $a+d=1$. Since $a$ and $d$ have sum and product $\\geq 0$, both are $\\geq 0$; since they sum to $1$, both lie in $[0,1]$. Hence we have $a = \\cos^2 \\phi$ and $b = \\sin^2 \\phi$ for some $\\phi$. Then $b^2 = \\cos^2 \\phi \\sin^2 \\phi$ so $b = \\sin \\phi \\cos \\phi$ or $- \\sin \\phi \\cos \\phi$. In the first", "Thus, we can first multiply our Dilation and Rotation matrices: We can then take this matrix, which we’ll call C, and multiply it by our translation matrix T. However, in order to properly translate a matrix, we must increase the size from a 3 x3 to a 4 x4 by adding in a row vector of [0 0 0 1] and a column vector of [0 0 0 1]T to the matrix C. We must now perform a row and column concatenation to our matrix A before pre-multiplying by M. We concatenate a row vector of [1 1 1 1] which represents the row’s scale value. If we change the order from Translate -> Rotate -> Dilate to Rotate -> Translate -> Dilate we would multiply as follows: Thus it is easy to see mathematically that the order of multiplications does, in fact, have an effect on the final matrix. This verifies the findings shown in Figures 8 and 9.", "$\\left(\\begin{matrix}- 1 & 0 \\\\ 0 & - 1\\end{matrix}\\right)$ is one of the forms of the identity or unit matrix. The positioning of the 0's and 1's in this matrix means that in multiplying with this matrix, the numbers in the other matrix stay the same, but the negative signs have the effect of changing the signs of the other matrix. This is used as one of the transformation matrices and gives the co-ordinates of the image under the following transformations: A rotation of 180° about the centre at the origin, (0,0) A dilation (or enlargement) with SF = -1 and the centre at (0.0) Once you recognise what this is, you will have the satisfaction of being able to give the answer immediately. Check for yourself that this actually works." ]

[ "irrelevant as we are about to take moments about $A$. So, on we go: ${\\textstyle\\frac12}\\mathbf p\\times\\mathbf W+\\mathbf p\\times\\mathbf F+\\mathbf p\\times\\mathbf N=\\mathbf 0$, which we can write as ${\\textstyle\\frac12}\\mathbf p\\times(\\mathbf W+2\\mathbf F+2\\mathbf N)=\\mathbf 0$. This gives $\\begin{pmatrix}l\\sin\\alpha\\\\l\\cos\\alpha\\,\\sin\\thet a\\\\l\\cos\\alpha\\,\\cos\\theta\\end{pmatrix}\\times\\begi n{pmatrix}-2N\\\\-2F\\cos\\theta\\\\2F\\sin\\theta-W\\end{pmatrix}=\\begin{pmatrix}0\\\\0\\\\0\\end{pmatrix}$. Expanding the vector product on the left after cancelling $l$, $\\begin{pmatrix}2F\\cos\\alpha-W\\sin\\theta\\,\\cos\\alpha\\\\W\\sin\\alpha-2F\\sin\\theta\\,\\sin\\alpha-2N\\cos\\theta\\,\\cos\\alpha\\\\2N\\sin\\theta\\,\\cos\\alpha-2F\\cos\\theta\\,\\sin\\alpha\\end{pmatrix}=\\begin{pmatr ix}0\\\\0\\\\0\\end{pmatrix}$. The first equation gives $F={\\textstyle\\frac12}W\\sin\\theta$. The third equation is $F=N\\frac{\\tan\\theta}{\\tan\\alpha}$, and since $F\\leq\\mu N$ we see that $\\tan\\theta\\leq\\mu\\tan\\alpha$. So the maximum value of $\\theta=\\arctan(\\mu\\tan\\alpha)$ which leads by symmetry to the given answer. Eliminating $F$ we see that $N={\\textstyle\\frac12}W\\tan\\alpha\\cos\\theta$, and at the extreme positions $\\cos\\theta=\\frac1{\\sqrt{1+\\tan^2\\theta}}=\\frac1{\\s qrt{1+\\mu^2\\tan^2\\alpha}}$. This gives your other answer, my friend.", "\\\\ \\mathbf{\\hat y} \\end{pmatrix} = \\begin{pmatrix} \\cos \\phi & -\\sin \\phi \\\\ \\sin \\phi & \\cos \\phi \\end{pmatrix} \\begin{pmatrix} \\boldsymbol{\\hat \\rho} \\\\ \\boldsymbol{\\hat\\phi} \\end{pmatrix}$$", "\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ \\ddot{x_1} - L.\\ddot{\\Theta_2}.cos(\\Theta_2) + L.\\dot{\\Theta_2}^2.sin(\\Theta_2) \\\\ -L.\\ddot{\\Theta_2}.sin(\\Theta_2) - L.\\dot{\\Theta_2}^2.cos(\\Theta_2) \\end{pmatrix}$$", "'18 at 16:01 • Then I guess maybe you can also solve this or at least share your opinions: math.stackexchange.com/q/2745276 - thanks. – ann marie cœur Apr 22 '18 at 16:03 • @ David, it is easy to see from $$UU|1,1\\rangle = U \\begin{pmatrix} 1\\\\ 0 \\end{pmatrix}U \\begin{pmatrix} 1\\\\ 0 \\end{pmatrix} = \\begin{pmatrix} \\cos(\\frac{\\theta}{2})+{i m_z} \\sin(\\frac{\\theta}{2})\\\\ (i m_x +m_y) \\sin(\\frac{\\theta}{2}) \\end{pmatrix}\\begin{pmatrix} \\cos(\\frac{\\theta}{2})+{i m_z} \\sin(\\frac{\\theta}{2})\\\\ (i m_x +m_y) \\sin(\\frac{\\theta}{2}) \\end{pmatrix}$$ when $m_y=\\pm 1, m_x=m_z=1$ and $\\theta=\\pi$, we have $$UU|1,1\\rangle = \\begin{pmatrix} 0\\\\ 1 \\end{pmatrix} \\begin{pmatrix} 0\\\\ 1 \\end{pmatrix}=|1,-1\\rangle$$ – ann marie cœur Apr 22 '18 at 16:31 • Do you have any precise answer how to find $U$, such that $$UU|1,1\\rangle = \\frac{1}{\\sqrt 2} \\begin{pmatrix} 1\\\\ 0 \\end{pmatrix}\\begin{pmatrix} 0\\\\ 1 \\end{pmatrix} + \\frac{1}{\\sqrt 2} \\begin{pmatrix} 1\\\\ 0 \\end{pmatrix}\\begin{pmatrix} 0\\\\ 1 \\end{pmatrix}=|1,0\\rangle ?$$ – ann marie cœur Apr 22 '18 at 16:33 • @annie heart there is no such a $U$. Please observe that if you act by some $U \\otimes U$ on the highest weight vector: $\\begin{pmatrix} 1\\\\ 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 1\\\\ 0 \\end{pmatrix}$, then the result will always be a separable vector of the form $\\begin{pmatrix} a\\\\ b\\end{pmatrix} \\otimes \\begin{pmatrix} a\\\\ b \\end{pmatrix}$. But the vector that you want to reach is entangled, and there is no way to change the entanglement state by a local transformation.", "+ m g \\cos \\theta \\begin{pmatrix} -\\sin \\theta \\\\ \\cos \\theta \\end{pmatrix} + m r \\dot{\\theta}^2 \\begin{pmatrix} -\\sin \\theta \\\\ \\cos \\theta \\end{pmatrix} \\end{gather*}$ Substituting into Newton’s law, $\\begin{gather*}\\begin{pmatrix} - \\dot{\\theta}^2 r \\sin \\theta + \\ddot{\\theta} r \\cos \\theta \\\\ \\dot{\\theta}^2 r \\cos \\theta + \\ddot{\\theta} r \\sin \\theta \\end{pmatrix} = \\begin{pmatrix} 0 \\\\-mg \\end{pmatrix} + m g \\cos \\theta \\begin{pmatrix} -\\sin \\theta \\\\ \\cos \\theta \\end{pmatrix} + m r \\dot{\\theta}^2 \\begin{pmatrix} -\\sin \\theta \\\\ \\cos \\theta \\end{pmatrix} \\\\ \\begin{pmatrix} \\ddot{\\theta} r \\cos \\theta \\\\ \\ddot{\\theta} r \\sin \\theta \\end{pmatrix} = \\begin{pmatrix} 0 \\\\-g \\end{pmatrix} + g \\cos \\theta \\begin{pmatrix} -\\sin \\theta \\\\ \\cos \\theta \\end{pmatrix}\\end{gather*}$ The first component gives $\\ddot{\\theta} r \\cos \\theta = - g \\cos \\theta \\sin \\theta$ $\\ddot{\\theta} = - \\frac{g}{r} \\sin \\theta$ The second component gives $\\ddot{\\theta} r \\sin \\theta = -g + g \\cos^2 \\theta$ $\\ddot{\\theta} r \\sin \\theta = -g (1 - \\cos^2 \\theta)$ $\\ddot{\\theta} r \\sin \\theta = -g \\sin^2 \\theta$ $\\ddot{\\theta} = -\\frac{g}{r} \\sin \\theta$ Thus, after simplifying either of the two components, we get the same equation: $\\ddot{\\theta} = - \\frac{g}{r} \\sin \\theta.$ ### Conservation of Energy Consider the pendulum with pivot at $$(0,r)$$, where $$r$$ is the length of the pendulum. The position of the end of the pendulum is given by $$(x,y) = (r \\sin \\theta,r-r", "2\\begin{pmatrix} \\tr(a + b) & \\tr(\\theta^{-1}(a - b)) \\\\ \\tr(\\theta(a + b)) & \\tr(a - b) \\end{pmatrix}$$ for all $$a, b \\in \\mathbb F_{q^2}$$.", "& -\\cos \\theta \\\\[0.3em] \\cos \\theta & \\sin \\theta \\end{bmatrix}$$\\\\ =\\begin{bmatrix} \\cos^2 \\theta & \\cos \\theta \\sin \\theta \\\\[0.3em] -\\sin \\theta \\cos \\theta & \\cos^2 \\theta \\end{bmatrix}+\\begin{bmatrix} \\sin^2 \\theta & -\\sin \\theta \\cos \\theta \\\\[0.3em] \\sin \\theta \\cos \\theta & \\sin^2 \\theta \\end{bmatrix}$$\\\\$ $=\\begin{bmatrix} \\cos^2 \\theta +\\sin^2 \\theta & \\cos \\theta \\sin \\theta -\\sin \\theta \\cos \\theta \\\\[0.3 em] -\\sin \\theta \\cos \\theta + \\sin \\theta \\cos \\theta & \\cos^2 \\theta +\\sin ^ 2 \\theta \\end{bmatrix}$$\\\\ =\\begin{bmatrix} 1 & 0 \\\\[0.3em] 0 & 1 \\end{bmatrix} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (\\therefore \\sin^2 \\theta =1) 16 Find X and Y ,if\\\\ (i)X+Y=\\begin{bmatrix} 7 & 0 \\\\[0.3em] 2 & 5 \\end{bmatrix} and X-Y = \\begin{bmatrix} 3 & 0\\\\[0.3em] 0 & 3 \\end{bmatrix}$$\\\\$ $(ii) 2 X + 3Y=\\begin{bmatrix} 2 & 3 \\\\[0.3em] 4 & 0\\end{bmatrix}$ and $3X+2Y=\\begin{bmatrix} 2 & -2 \\\\[0.3em] -1 & 5 \\end{bmatrix}$ ##### Solution : $(ii) 2X+3Y =\\begin{bmatrix} 2 & 3\\\\[0.3em] 4 & 0\\end{bmatrix}$.......(3) $3X+2Y=\\begin{bmatrix} 2 & -2\\\\[0.3em] -1 & 5\\end{bmatrix}$.....(4) Multiplying equation (3) with (2), we get$\\\\$ $2(2X+3Y)=2\\begin{bmatrix} 2 & 3\\\\[0.3em] 4 & 0\\end{bmatrix}\\\\ \\implies 4X+6Y=\\begin{bmatrix} 4 & 6\\\\[0.3em] 8 & 0\\end{bmatrix}....(5)$$\\\\ Multiplying equation (4) with (3), we get\\\\ 3(3X+2Y)=3 \\begin{bmatrix} 2 & -2\\\\[0.3em] -1 & 5\\end{bmatrix}\\\\ \\implies 9X+6Y=\\begin{bmatrix} 6 & -6\\\\[0.3em] -3 & 15\\end{bmatrix}.....(6)$$\\\\$ From(5) and (6),we have$\\\\$ $(4X+6Y)-(9X+6Y) =\\begin{bmatrix}", "& 0 & 1 & 0 \\\\ - \\sinh \\, \\tanh^{-1} \\, \\frac{v}{c} & 0 & 0 & \\cosh \\, \\tanh^{-1} \\, \\frac{v}{c} \\end{pmatrix} \\begin{pmatrix} \\Delta x \\\\ \\Delta y \\\\ \\Delta z \\\\ c \\Delta t \\end{pmatrix}$. $\\begin{pmatrix}\\Delta x' \\\\ \\Delta y' \\\\ \\Delta z' \\\\ c \\Delta t' \\end{pmatrix} = \\begin{pmatrix} \\frac{1}{\\sqrt{1 - \\frac{v^2}{c^2}}} & 0 & 0 & \\frac{-v}{c \\sqrt{1 - \\frac{v^2}{c^2}}} \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ \\frac{-v}{c \\sqrt{1 - \\frac{v^2}{c^2}}} & 0 & 0 & \\frac{1}{\\sqrt{1 - \\frac{v^2}{c^2}}} \\end{pmatrix} \\begin{pmatrix} \\Delta x \\\\ \\Delta y \\\\ \\Delta z \\\\ c \\Delta t \\end{pmatrix}$. $\\begin{pmatrix}\\Delta x' \\\\ \\Delta y' \\\\ \\Delta z' \\\\ \\Delta t' \\end{pmatrix} = \\begin{pmatrix} \\cosh \\, \\tanh^{-1} \\, \\frac{v}{c} & 0 & 0 & - c \\, \\sinh \\, \\tanh^{-1} \\, \\frac{v}{c} \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ - \\frac{1}{c} \\, \\sinh \\, \\tanh^{-1} \\, \\frac{v}{c} & 0 & 0 & \\cosh \\, \\tanh^{-1} \\, \\frac{v}{c} \\end{pmatrix} \\begin{pmatrix} \\Delta x \\\\ \\Delta y \\\\ \\Delta z \\\\ \\Delta t \\end{pmatrix}$. $\\begin{pmatrix}\\Delta x' \\\\ \\Delta y' \\\\ \\Delta z' \\\\ \\Delta t' \\end{pmatrix} = \\begin{pmatrix} \\frac{1}{\\sqrt{1 - \\frac{v^2}{c^2}}} & 0 & 0 & \\frac{-v}{\\sqrt{1 - \\frac{v^2}{c^2}}} \\\\ 0 &", "\\theta^B_i \\end{bmatrix}\\sim MVN \\begin{pmatrix} \\begin{bmatrix} \\mu^{A} \\\\ \\mu^{B} \\end{bmatrix} , \\begin{bmatrix} \\Sigma^{A} & \\Sigma^{AB} \\\\ \\Sigma^{BA} & \\Sigma^{B} \\end{bmatrix} \\end{pmatrix}$$", "\\cos \\theta \\end{pmatrix}$ Assume that $A^k=\\begin{pmatrix} \\cos k \\theta & - \\sin k \\theta \\\\ \\sin k \\theta & \\cos k \\theta \\end{pmatrix}$ Then $A^{k+1}=A^kA=\\begin{pmatrix} \\cos k \\theta & - \\sin k \\theta \\\\ \\sin k \\theta & \\cos k \\theta \\end{pmatrix} \\begin{pmatrix} \\cos \\theta & - \\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{pmatrix}$ Now make the multiplication and use these formula : $\\cos(a+b)=\\cos(a)\\cos(b)-\\sin(a)\\sin(b)$ $\\sin(a+b)=\\sin(a)\\cos(b)+\\sin(b)\\cos(a)$ and you're done", "am just wondering how they derived at equation 21, ie $ \\displaystyle \\frac{d }{d \\theta}(r cos (\\theta) sin (\\phi)) = - sin (\\theta) $ I thought it would have been $- r sin (\\theta) sin (\\phi)$?? 4. You're correct. However, that's not quite what they're doing. For one thing, they're dividing by the length. So, if you first compute $\\displaystyle\\frac{d\\mathbf{r}}{d\\theta}=\\begin{bm atrix}-r\\sin(\\theta)\\sin(\\phi)\\\\ r\\cos(\\theta)\\sin(\\phi)\\\\ 0\\end{bmatrix},$ and then find its length, which is $\\displaystyle\\left|\\frac{d\\mathbf{r}}{d\\theta}\\rig ht|=r\\sqrt{\\sin^{2}(\\theta)\\sin^{2}(\\phi)+\\cos^{2} (\\theta)\\sin^{2}(\\phi)}=r|\\sin(\\phi)|=r\\sin(\\phi),$ on account of the range of $\\phi$ (the polar angle on this website), then you'll find that $\\hat{\\mathbf{\\theta}}=\\dfrac{\\dfrac{d\\mathbf{r}}{d \\theta}}{\\left|\\dfrac{d\\mathbf{r}}{d\\theta}\\right| }=\\dfrac{1}{r\\sin(\\phi)}\\begin{bmatrix}-r\\sin(\\theta)\\sin(\\phi)\\\\ r\\cos(\\theta)\\sin(\\phi)\\\\ 0\\end{bmatrix}=\\begin{bmatrix}-\\sin(\\theta)\\\\ \\cos(\\theta)\\\\ 0\\end{bmatrix},$ as they have. Does that make it clearer, perhaps? 5. Originally Posted by Ackbeet You're correct. However, that's not quite what they're doing. For one thing, they're dividing by the length. So, if you first compute $\\displaystyle\\frac{d\\mathbf{r}}{d\\theta}=\\begin{bm atrix}-r\\sin(\\theta)\\sin(\\phi)\\\\ r\\cos(\\theta)\\sin(\\phi)\\\\ 0\\end{bmatrix},$ and then find its length, which is $\\displaystyle\\left|\\frac{d\\mathbf{r}}{d\\theta}\\rig ht|=r\\sqrt{\\sin^{2}(\\theta)\\sin^{2}(\\phi)+\\cos^{2} (\\theta)\\sin^{2}(\\phi)}=r|\\sin(\\phi)|=r\\sin(\\phi),$ on account of the range of $\\phi$ (the polar angle on this website), then you'll find that $\\hat{\\mathbf{\\theta}}=\\dfrac{\\dfrac{d\\mathbf{r}}{d \\theta}}{\\left|\\dfrac{d\\mathbf{r}}{d\\theta}\\right| }=\\dfrac{1}{r\\sin(\\phi)}\\begin{bmatrix}-r\\sin(\\theta)\\sin(\\phi)\\\\ r\\cos(\\theta)\\sin(\\phi)\\\\ 0\\end{bmatrix}=\\begin{bmatrix}-\\sin(\\theta)\\\\ \\cos(\\theta)\\\\ 0\\end{bmatrix},$ as they have. Does that make it clearer, perhaps? Yes, thats almost making sense. Just couldnt find any reference in my books regarding the magnitude of a 3*1 matrix, ie your second line. Im guessing its an extended form of pythagoras, ie SRSS's... Thanks 6. Originally Posted", "+ i\\sin(3t) - 3i\\cos(3t) + 3\\sin(3t)\\end{pmatrix} =\\begin{pmatrix} -2\\cos(3t) \\\\ \\cos(3t) + 3\\sin(3t)\\end{pmatrix} + i \\begin{pmatrix} -2\\sin(3t) \\\\ \\sin(3t) - 3\\cos(3t)\\end{pmatrix}$$ Thus, the general solution of this system is $$\\textbf{x}(t) = c_1\\begin{pmatrix} -2\\cos(3t) \\\\ \\cos(3t) + 3\\sin(3t)\\end{pmatrix} + c_2 \\begin{pmatrix} -2\\sin(3t) \\\\ \\sin(3t) - 3\\cos(3t)\\end{pmatrix}$$ b) As $t\\to\\infty$, solution circulates in clockwise direction in elliptical shapes. See attached image.", "\\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 2. $\\begin{pmatrix} -3 \\\\ 3 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 3. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 4. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k+\\begin{pmatrix} 0 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 5. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 14 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 6. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 6 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ Problem 2 Produce two descriptions of this set that are different than this one. $\\{\\begin{pmatrix} 2 \\\\ -5 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 3 Show that the three descriptions given at the start of this subsection all describe the same set. This exercise is recommended for all readers. Problem 4 Show that these sets are equal $\\{\\begin{pmatrix} 1 \\\\ 4 \\\\ 1 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}z\\,\\big|\\, z\\in\\mathbb{R} \\} \\quad\\text{and}\\quad \\{\\begin{pmatrix} 0 \\\\ 4 \\\\ 2 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R} \\},$ and that both describe the solution set of this system. $\\begin{array}{*{4}{rc}r} x &- &y &+ &z &+", "& \\frac{u}{c} \\frac{1}{\\sqrt{1 - \\frac{u^2}{c^2}}} & 0 & 0 \\\\ \\frac{u}{c} \\frac{1}{\\sqrt{1 - \\frac{u^2}{c^2}}} & \\frac{1}{\\sqrt{1 - \\frac{u^2}{c^2}}} & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix} = \\begin{pmatrix} \\cosh \\, \\rho & \\sinh \\, \\rho & 0 & 0 \\\\ \\sinh \\, \\rho & \\cosh \\, \\rho & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\end{pmatrix} = \\textrm{exp} \\, \\begin{pmatrix} 0 & \\rho & 0 & 0 \\\\ \\rho & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\\\ 0 & 0 & 0 & 0 \\end{pmatrix}$ which is nice because $\\Lambda^T = \\Lambda$, we may also use the more general form: $\\Lambda = \\textrm{exp} \\, \\begin{pmatrix} 0 & \\rho_x & \\rho_y & \\rho_z \\\\ \\rho_x & 0 & \\theta_z & - \\theta_y \\\\ \\rho_y & -\\theta_z & 0 & \\theta_x \\\\ \\rho_z & \\theta_y & -\\theta_x & 0 \\end{pmatrix}$ where for any square matrix, M, we have $\\textrm{exp} \\, M = \\sum_{k=0}^{\\infty} \\frac{1}{k!} M^k = I + \\sum_{k=1}^{\\infty} \\frac{1}{k!} M^k$ from which it follows that $\\left( \\textrm{exp} \\, M \\right)^T = \\textrm{exp} \\, M^T$. In this more general Lorentz transform, $\\Lambda^T = \\Lambda$ is not always true.", "\\\\ \\mathbf{K}_r^s \\mathbf{L}_s \\mathbf{K}_s^r &= \\begin{bmatrix} \\cos(\\theta_r) & -\\sin(\\theta_r) \\\\ +\\sin(\\theta_r) & \\cos(\\theta_r) \\end{bmatrix} \\begin{bmatrix} 4 \\cos(\\theta_r) & 4\\sin(\\theta) \\\\ -2 \\sin(\\theta_r) & 2 \\cos(\\theta_r) \\end{bmatrix} \\\\ \\mathbf{K}_r^s \\mathbf{L}_s \\mathbf{K}_s^r &= \\begin{bmatrix} 4 \\cos^2(\\theta_r) + 2 \\sin^2(\\theta_r) & 4 \\sin(\\theta_r)\\cos(\\theta_r) - 2 \\sin(\\theta_r)\\cos(\\theta_r) \\\\ 4 \\sin(\\theta_r)\\cos(\\theta_r) - 2 \\sin(\\theta_r)\\cos(\\theta_r) & 4 \\sin^2(\\theta_r) + 2 \\cos^2(\\theta_r) \\end{bmatrix} \\\\ \\mathbf{K}_r^s \\mathbf{L}_s \\mathbf{K}_s^r &= \\begin{bmatrix} 2 + 2 \\cos(2\\theta_r) + 1 - \\cos(2\\theta_r) & \\sin(2\\theta_r) \\\\ \\sin(2\\theta_r) & 2 - 2\\cos(2\\theta_r) + 1 + \\cos(2\\theta_r) \\end{bmatrix} \\\\ \\mathbf{K}_r^s \\mathbf{L}_s \\mathbf{K}_s^r &= \\begin{bmatrix} 3 + \\cos(2\\theta_r) & \\sin(2\\theta_r) \\\\ \\sin(2\\theta_r) & 3 - \\cos(2\\theta_r) \\end{bmatrix} \\end{align} The fourth line follows from the third using the Power Reduction Identity (given for sine and cosine) and the Double Angle Identity (given for sine). The vector flux linkage equation is finished. $$$\\boxed{\\vec{\\lambda}_{abs} = \\begin{bmatrix} 3 + \\cos(2\\theta_r) & \\sin(2\\theta_r) \\\\ \\sin(2\\theta_r) & 3 - \\cos(2\\theta_r) \\end{bmatrix} \\vec{i}_{abs}}$$$ ## Alumni Liaison Basic linear algebra uncovers and clarifies very important geometry and algebra. Dr. Paul Garrett", "\\sin \\theta + \\sin \\beta \\cos \\theta) & -1 \\\\ \\frac {1} {\\rho} (\\cos \\beta \\sin \\theta + \\sin \\beta \\cos \\theta) & 0 \\end{pmatrix} \\begin{pmatrix} \\nu \\\\ \\gamma \\end{pmatrix}$$ Simplifying and then substituting $$\\theta = -(\\alpha + \\beta)$$ gives: $$= \\begin{pmatrix} - \\cos (\\beta + \\theta) & 0 \\\\ - \\frac {1} {\\rho} \\sin (\\beta + \\theta) & -1 \\\\ \\frac {1} {\\rho} \\sin (\\beta + \\theta) & 0 \\end{pmatrix} \\begin{pmatrix} \\nu \\\\ \\gamma \\end{pmatrix} = \\begin{pmatrix} - \\cos \\alpha &0 \\\\ {\\frac {\\sin \\alpha} {\\rho}} & -1 \\\\ - {\\frac {\\sin \\alpha} {\\rho}} & 0 \\end{pmatrix} \\begin{pmatrix} \\nu \\\\ \\gamma \\end{pmatrix}$$ So, we can write: $$\\begin{pmatrix} \\dot \\rho \\\\ \\dot \\alpha \\\\ \\dot \\beta \\end{pmatrix} = \\begin{pmatrix} -cos \\alpha & 0 \\\\ \\frac {sin \\alpha} {\\rho} & -1 \\\\ -\\frac {sin \\alpha} {\\rho} & 0 \\end{pmatrix} \\begin{pmatrix} \\nu \\\\ \\gamma \\end{pmatrix}, \\text { if } \\alpha \\in \\begin{pmatrix} -\\frac {\\pi} {2} , \\frac {\\pi} {2} \\end{pmatrix}$$ The equations $$\\nu = k _{\\rho} \\rho$$ $$\\gamma = k _{\\alpha} \\alpha + k _{\\beta} \\beta$$ are the linear control law which, when applied to the equation we just derived gives us: $$\\begin{pmatrix} \\dot \\rho \\\\ \\dot \\alpha \\\\ \\dot \\beta \\end{pmatrix} = \\begin{pmatrix} - k _{\\rho} \\rho \\cos \\alpha \\\\ k _{\\rho} \\rho \\sin \\alpha - k _{\\alpha} \\alpha", "$O_{\\theta}=\\begin{pmatrix} \\cos(2\\theta) & \\sin(2\\theta) \\\\ \\sin(2\\theta) & -\\cos(2\\theta) \\end{pmatrix}$ Another way is to write $O_{\\theta}=\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$ and to consider that $|v_{\\theta}\\rangle$ and $|v^{\\bot}_{\\theta}\\rangle$ are eigenvectors with +1 and -1 as the eigenvalues, respectively, i.e. $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}.\\begin{pmatrix} \\cos(\\theta) \\\\ \\sin(\\theta) \\end{pmatrix}=\\begin{pmatrix} \\cos(\\theta) \\\\ \\sin(\\theta) \\end{pmatrix}$ and $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}.\\begin{pmatrix} -\\sin(\\theta) \\\\ \\cos(\\theta) \\end{pmatrix}=-\\begin{pmatrix} -\\sin(\\theta) \\\\ \\cos(\\theta) \\end{pmatrix}$ So we have 4 equations with 4 unknowns. This can be solved to obtain again the previous result. by , 29 Sep 2015 11:42 02 Jun 2014 03:47 in discussion Hidden / Per page discussions » Ex5 4 To show the decomposition of $I$, note that $K(0) = T(0) = R(0) = I$. Thus, (1) $$K(0)T(0)R(0)T(0) = IIII = I .$$ To show the decomposition of $X$, note that (2) \\begin{align} T(\\pi/2) = \\left(\\begin{array}{cc} \\mathbf{i} & 0 \\\\ 0 & -\\mathbf{i} \\end{array}\\right) \\quad \\text{and} \\quad R(\\pi/2) = \\left(\\begin{array}{cc} 0 & 1 \\\\ -1 & 0 \\end{array}\\right). \\end{align} Thus, (3) \\begin{align} -\\mathbf{i}T(\\pi/2)R(\\pi/2)T(0) & = -\\mathbf{i}\\left(\\begin{array}{cc} \\mathbf{i} & 0 \\\\ 0 & -\\mathbf{i} \\end{array}\\right) \\left(\\begin{array}{cc} 0 & 1 \\\\ -1 & 0 \\end{array}\\right) I \\\\ \\\\ & = -\\mathbf{i}\\left(\\begin{array}{cc} 0 & \\mathbf{i} \\\\ \\mathbf{i} & 0 \\end{array}\\right) = \\left(\\begin{array}{cc} 0 & 1 \\\\ 1", "Question 42: Let A = $\\begin{pmatrix} x+2 & 3x\\\\ 3 & x+2 \\end{pmatrix}$ , B = $\\begin{pmatrix} x &0 \\\\ 5 & x+2 \\end{pmatrix}$ . Then all solutions of the equation det (AB) = 0 is 1. a. 1, –1, 0, 2 2. b. 1, 4, 0, –2 3. c. 1, –1, 4, 3 4. d. –1, 4, 0, 3 Solution: Given A = $\\begin{pmatrix} x+2 & 3x\\\\ 3 & x+2 \\end{pmatrix}$ , B = $\\begin{pmatrix} x &0 \\\\ 5 & x+2 \\end{pmatrix}$ det |AB| = $\\begin{vmatrix} x^{2}+17x &3x^{2}+6x \\\\ 8x+10& (x+2)^{2} \\end{vmatrix}$ ⇒ x(x + 2)(x-4)(x–1) = 0 ⇒ x = 0, -2, 1, 4 Question 43: The value of det A, where A = $\\begin{pmatrix} 1 & \\cos \\theta & 0 \\\\ -\\cos \\theta & 1 & \\cos \\theta \\\\ -1& -\\cos \\theta & 1 \\end{pmatrix}$ lies 1. a. in the closed interval [1, 2] 2. b. in the closed interval [0, 1] 3. c. in the open interval (0, 1) 4. d. in the open interval (1, 2) Solution: Given A = $\\begin{pmatrix} 1 & \\cos \\theta & 0 \\\\ -\\cos \\theta & 1 & \\cos \\theta \\\\ -1& -\\cos \\theta & 1 \\end{pmatrix}$ R1 -> R1+R3 det A ⇒ $\\begin{vmatrix} 0 & 0 &1 \\\\ -\\cos \\theta & 1 &\\cos \\theta \\\\ -1 &", "\\\\ 0 & 1 & -\\frac{r}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}} & -\\frac{r}{\\sqrt{2}} & 1 \\end{pmatrix}, \\;\\;\\; \\Sigma_{3e}(r)= \\begin{pmatrix} 1 & 0 & 0 \\\\ 0 & 1 & r \\\\ 0 & r & 1 \\end{pmatrix}, \\;\\;\\; \\Sigma_{4a}(r)= \\begin{pmatrix} 1 & 0 & 0 & \\frac{r}{\\sqrt{2}} \\\\ 0 & 1 & -r & \\frac{r}{\\sqrt{2}} \\\\ 0 & -r & 1 & -\\frac{1}{\\sqrt{2}} \\\\ \\frac{r}{\\sqrt{2}} & \\frac{r}{\\sqrt{2}} & -\\frac{1}{\\sqrt{2}} & 1 \\end{pmatrix},$ $\\Sigma_{4b}(r)= \\begin{pmatrix} 1 & 0 & r & \\frac{r}{\\sqrt{2}} \\\\ 0 & 1 & 0 & \\frac{r}{\\sqrt{2}} \\\\ r & 0 & 1 & \\frac{1}{\\sqrt{2}} \\\\ \\frac{r}{\\sqrt{2}} & \\frac{r}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & 1 \\end{pmatrix}, \\;\\;\\; \\Sigma_{4c}(r)= \\begin{pmatrix} 1 & 0 & \\frac{1}{\\sqrt{2}} & -\\frac{r}{\\sqrt{2}} \\\\ 0 & 1 & -\\frac{r}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & -\\frac{r}{\\sqrt{2}} & 1 & -r \\\\ -\\frac{r}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & -r & 1 \\end{pmatrix}\\;\\;\\text{and}\\;\\; \\Sigma_{4d}(r)= \\begin{pmatrix} 1 & r & \\frac{1}{\\sqrt{2}} & \\frac{r}{\\sqrt{2}} \\\\ r & 1 & \\frac{r}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} & \\frac{r}{\\sqrt{2}} & 1 & r \\\\ \\frac{r}{\\sqrt{2}} & \\frac{1}{\\sqrt{2}} & r & 1 \\end{pmatrix}.$ # Estimation methods Given the form of bridge function $$F$$, obtaining a moment-based estimation $$\\widehat \\sigma_{jk}$$ requires inversion of $$F$$. latentcor implements two methods for calculation of the inversion: Both methods calculate inverse bridge function applied to each element of sample Kendall’s $$\\tau$$ matrix. Because the", "$$\\mathbf{r}_{PB} = -\\mathbf{r}_{BA} + \\mathbf{r}_{PA}$$, we have $$-r_B \\cos\\theta_B \\mathbf{E}_1 + r_B \\sin\\theta_B \\mathbf{E}_2 = -b \\mathbf{E}_1 - r_A \\cos\\theta_A \\mathbf{E}_1 + r_A \\sin\\theta_A \\mathbf{E}_2$$ or \\begin{align} -r_B \\cos\\theta_B & = -b - r_A \\cos\\theta_A \\\\ r_B \\sin\\theta_B & = r_A \\sin\\theta_A \\end{align} In matrix form $$\\begin{bmatrix} \\cos\\theta_A & -\\cos\\theta_B \\\\ \\sin\\theta_A & -\\sin\\theta_B \\end{bmatrix} \\begin{bmatrix} r_A \\\\ r_B \\end{bmatrix} = \\begin{bmatrix} - b \\\\ 0 \\end{bmatrix}$$ The determinant of the $$2 \\times 2$$ matrix is $$D := -\\cos\\theta_A \\sin\\theta_B + \\sin\\theta_A \\cos\\theta_B = \\sin(\\theta_A - \\theta_B)$$ Therefore, inverting the matrix, we have $$\\begin{bmatrix} r_A \\\\ r_B \\end{bmatrix} = \\frac{1}{\\sin(\\theta_A -\\theta_B)} \\begin{bmatrix} -\\sin\\theta_B & \\cos\\theta_B \\\\ -\\sin\\theta_A & \\cos\\theta_A \\end{bmatrix}\\begin{bmatrix} - b \\\\ 0 \\end{bmatrix}$$ or \\boxed{ \\begin{align} r_A &= \\frac{b\\sin\\theta_B}{\\sin(\\theta_A -\\theta_B)} \\\\ r_B &= \\frac{b\\sin\\theta_A}{\\sin(\\theta_A -\\theta_B)} \\end{align} } Plugging in $$b = 300$$ mm, $$\\theta_A = 60^\\circ$$, $$\\theta_B = 20^\\circ$$, we have $$r_A = 159.6$$ mm and $$r_B = 404.2$$ mm. Now that we know the geometry, we can do the kinematics. Choose a rotating frame that has origin at point A, and the coordinate axes \\begin{align} \\mathbf{e}_1 &= \\sin\\theta_A \\mathbf{E}_1 + \\cos\\theta_A \\mathbf{E}_2 \\\\ \\mathbf{e}_2 &= -\\cos\\theta_A \\mathbf{E}_1 + \\sin\\theta_A \\mathbf{E}_2 = \\mathbf{r}_{PA}/||\\mathbf{r}_{PA}||\\\\ \\mathbf{e_3} &= \\mathbf{E}_3. \\end{align} or, inverting the relation, \\begin{align} \\mathbf{E}_1 &= \\sin\\theta_A \\mathbf{e}_1 - \\cos\\theta_A \\mathbf{e}_2 \\\\ \\mathbf{E}_2 &= \\cos\\theta_A \\mathbf{e}_1 + \\sin\\theta_A" ]

[ "1. The dilation S is minimal, i.e. \\begin{align}H=\\overline{span\\left \\{ R_{\\pm i}(S)h|h\\in \\mathfrak{H},n \\in \\left \\{ 0 \\right \\}\\cup \\mathbb{N} \\right \\}}\\end{align} if the spaces $E_{+}=\\overline{\\Psi \\mathfrak{H}}, E_{-}= \\overline{\\Phi^{*} \\mathfrak{H}}$ are separable. The following expressions was used for the proof: \\begin{align}R_{-i}^{n}(S)\\begin{pmatrix} 0\\\\ h_{0}\\\\ 0 \\end{pmatrix}=\\begin{pmatrix} 0\\\\ a_{n}\\\\ b_{n} \\end{pmatrix} \\end{align} where $n\\in \\mathbb{N},a_{n}=R^{n}h_{0}$, \\begin{align}b_{n}=e^{-t}\\sum_{k=1}^{n}\\frac{t^{n-k}}{(n-k)!i^{n-k-1}}\\Psi R^{k-1}h_{0}.\\\\ R_{i}^{n}(S)\\begin{pmatrix} 0 \\\\ h_{0} \\\\ 0 \\end{pmatrix}=\\begin{pmatrix} c_{n} \\\\ d_{n} \\\\ 0 \\end{pmatrix},\\end{align} where $d_{n}=R^{*n}h_{0}$, \\begin{align}c_{n}=e^{t}\\sum_{k=1}^{n}\\frac{t^{n-k}}{(n-k)!(-i)^{n-k-1}}\\Phi ^{*}R^{*k-1}h_{0},\\end{align} where $h_{0}\\in \\mathfrak{H}$. Keywords:dilation, self-adjoint operator, unbounded dissipative operator, minimality, operator knot UDC: 517.432", "clockwise by an angle of $\\theta$ about the origin are given by $\\begin{pmatrix} x’ \\\\ y’ \\end{pmatrix} = \\begin{pmatrix} \\cos \\theta &-\\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix}$ For $(x, y) = (2,-3)$ and rotation of $\\theta = 90^{\\circ}$ about the origin, we have \\begin{aligned} \\begin{pmatrix} x’ \\\\ y’ \\end{pmatrix} & = \\begin{pmatrix} \\cos 90^{\\circ} &-\\sin 90^{\\circ} \\\\ \\sin 90^{\\circ} & \\cos 90^{\\circ} \\end{pmatrix} \\begin{pmatrix} 2 \\\\-3 \\end{pmatrix} \\\\ & = \\begin{pmatrix} 0 &-1 \\\\ 1 & 0 \\end{pmatrix} \\begin{pmatrix} 2 \\\\-3 \\end{pmatrix} \\\\ & = \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} \\end{aligned} Thus, the coordinates of the image of point $P$ are $\\boxed{P'(3,2)}$ [collapse] Problem Number 7 Given a point $P(-8,12)$. The dilation $[P,1]$ maps point $(-4,8)$ to point $\\cdots \\cdot$ A. $(-4,8)$ D. $(4,-16)$ B. $(-4,16)$ E. $(4,-8)$ C. $(-4,-8)$ Solution The dilation rule is given as follows. If a point $(x,y)$ is dilated around center point $(a,b)$ by a scale factor of $k$, then the image point lies in $(k(x-a)+a, k(y-b)+b)$. This means that the coordinates of the image of the point $(-4, 8)$ after being dilated around the center point $(-8, 12)$ by a scale factor of $1$ are $(1(-4-(-8))+(-8), 1(8-12)+12)$ $= (-4, 8)$ Dilation $[P, 1]$ maps point $(-4,8)$ to $\\boxed{(-4, 8)}$ [collapse] Problem Number 8 The", "= \\begin{pmatrix} 0\\\\ 1 \\end{pmatrix} \\text{ lands on } \\begin{pmatrix} a.x+b.y\\\\ c.x+d.y \\end{pmatrix} = \\begin{pmatrix} - \\sin \\theta \\\\ \\cos \\theta \\end{pmatrix}$$ From which we can deduce: $$\\begin{pmatrix} 1.a+0.b\\\\ 1.c+0.d \\end{pmatrix} = \\begin{pmatrix} \\cos \\theta \\\\ \\sin \\theta \\end{pmatrix}$$ $$\\begin{pmatrix} 0.a+1.b\\\\ 0.c+1.d \\end{pmatrix} = \\begin{pmatrix} - \\sin \\theta \\\\ \\cos \\theta \\end{pmatrix}$$ Easy to deduce $a = \\cos \\left( \\theta \\right)$, $b = - \\sin \\left( \\theta \\right)$, $c = \\sin \\left( \\theta \\right)$ and $d = \\cos \\left( \\theta \\right)$. Which gives us our desired rotation matrix: $$\\begin{pmatrix} a & b\\\\c & d \\end{pmatrix} = \\begin{pmatrix} \\cos \\theta & -\\sin \\theta\\\\ \\sin \\theta & \\cos \\theta \\end{pmatrix}$$ Congratulations! You know of to define scaling, reflexion, shearing and rotation transformation matrices. So what is missing? ## 3x3 transformation matrices If you’re still with me at this point, maybe you’re wondering why any of this is useful. If it’s the case, you missed the point of this article, which is to understand affine transformations in order to apply them in code :mortar_board: . This is useful because at this point you know what a transformation matrix looks like, and you know how to compute one given a few position vectors, and it is also a great accomplishment by itself. But here’s the thing: $2 \\times 2$ matrices are limited", "$$\\mathbf{r}_{PB} = -\\mathbf{r}_{BA} + \\mathbf{r}_{PA}$$, we have $$-r_B \\cos\\theta_B \\mathbf{E}_1 + r_B \\sin\\theta_B \\mathbf{E}_2 = -b \\mathbf{E}_1 - r_A \\cos\\theta_A \\mathbf{E}_1 + r_A \\sin\\theta_A \\mathbf{E}_2$$ or \\begin{align} -r_B \\cos\\theta_B & = -b - r_A \\cos\\theta_A \\\\ r_B \\sin\\theta_B & = r_A \\sin\\theta_A \\end{align} In matrix form $$\\begin{bmatrix} \\cos\\theta_A & -\\cos\\theta_B \\\\ \\sin\\theta_A & -\\sin\\theta_B \\end{bmatrix} \\begin{bmatrix} r_A \\\\ r_B \\end{bmatrix} = \\begin{bmatrix} - b \\\\ 0 \\end{bmatrix}$$ The determinant of the $$2 \\times 2$$ matrix is $$D := -\\cos\\theta_A \\sin\\theta_B + \\sin\\theta_A \\cos\\theta_B = \\sin(\\theta_A - \\theta_B)$$ Therefore, inverting the matrix, we have $$\\begin{bmatrix} r_A \\\\ r_B \\end{bmatrix} = \\frac{1}{\\sin(\\theta_A -\\theta_B)} \\begin{bmatrix} -\\sin\\theta_B & \\cos\\theta_B \\\\ -\\sin\\theta_A & \\cos\\theta_A \\end{bmatrix}\\begin{bmatrix} - b \\\\ 0 \\end{bmatrix}$$ or \\boxed{ \\begin{align} r_A &= \\frac{b\\sin\\theta_B}{\\sin(\\theta_A -\\theta_B)} \\\\ r_B &= \\frac{b\\sin\\theta_A}{\\sin(\\theta_A -\\theta_B)} \\end{align} } Plugging in $$b = 300$$ mm, $$\\theta_A = 60^\\circ$$, $$\\theta_B = 20^\\circ$$, we have $$r_A = 159.6$$ mm and $$r_B = 404.2$$ mm. Now that we know the geometry, we can do the kinematics. Choose a rotating frame that has origin at point A, and the coordinate axes \\begin{align} \\mathbf{e}_1 &= \\sin\\theta_A \\mathbf{E}_1 + \\cos\\theta_A \\mathbf{E}_2 \\\\ \\mathbf{e}_2 &= -\\cos\\theta_A \\mathbf{E}_1 + \\sin\\theta_A \\mathbf{E}_2 = \\mathbf{r}_{PA}/||\\mathbf{r}_{PA}||\\\\ \\mathbf{e_3} &= \\mathbf{E}_3. \\end{align} or, inverting the relation, \\begin{align} \\mathbf{E}_1 &= \\sin\\theta_A \\mathbf{e}_1 - \\cos\\theta_A \\mathbf{e}_2 \\\\ \\mathbf{E}_2 &= \\cos\\theta_A \\mathbf{e}_1 + \\sin\\theta_A", "irrelevant as we are about to take moments about $A$. So, on we go: ${\\textstyle\\frac12}\\mathbf p\\times\\mathbf W+\\mathbf p\\times\\mathbf F+\\mathbf p\\times\\mathbf N=\\mathbf 0$, which we can write as ${\\textstyle\\frac12}\\mathbf p\\times(\\mathbf W+2\\mathbf F+2\\mathbf N)=\\mathbf 0$. This gives $\\begin{pmatrix}l\\sin\\alpha\\\\l\\cos\\alpha\\,\\sin\\thet a\\\\l\\cos\\alpha\\,\\cos\\theta\\end{pmatrix}\\times\\begi n{pmatrix}-2N\\\\-2F\\cos\\theta\\\\2F\\sin\\theta-W\\end{pmatrix}=\\begin{pmatrix}0\\\\0\\\\0\\end{pmatrix}$. Expanding the vector product on the left after cancelling $l$, $\\begin{pmatrix}2F\\cos\\alpha-W\\sin\\theta\\,\\cos\\alpha\\\\W\\sin\\alpha-2F\\sin\\theta\\,\\sin\\alpha-2N\\cos\\theta\\,\\cos\\alpha\\\\2N\\sin\\theta\\,\\cos\\alpha-2F\\cos\\theta\\,\\sin\\alpha\\end{pmatrix}=\\begin{pmatr ix}0\\\\0\\\\0\\end{pmatrix}$. The first equation gives $F={\\textstyle\\frac12}W\\sin\\theta$. The third equation is $F=N\\frac{\\tan\\theta}{\\tan\\alpha}$, and since $F\\leq\\mu N$ we see that $\\tan\\theta\\leq\\mu\\tan\\alpha$. So the maximum value of $\\theta=\\arctan(\\mu\\tan\\alpha)$ which leads by symmetry to the given answer. Eliminating $F$ we see that $N={\\textstyle\\frac12}W\\tan\\alpha\\cos\\theta$, and at the extreme positions $\\cos\\theta=\\frac1{\\sqrt{1+\\tan^2\\theta}}=\\frac1{\\s qrt{1+\\mu^2\\tan^2\\alpha}}$. This gives your other answer, my friend.", "plt.show() Dilating the plane by a factor of 0.5: $f \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}0.5 & 0\\\\0 & 0.5\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix},$ in Numpy: a = .5*np.eye(2) dst = linear_transformation(src, a) plt.imshow(dst) plt.show() Scaling the plane in the $y$-axis by a factor of 0.5: $f \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}1 & 0\\\\0 & 0.5\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix},$ in Numpy: a = np.array([[1, 0], [0, .5]]) dst = linear_transformation(src, a) plt.imshow(dst) plt.show() Shearing about the $y$-axis with a vertical displacement of $+x/2$: $f \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}1 & 0\\\\{1 \\over 2} & 0\\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix},$ in Numpy: a = np.array([[1, 0], [.5, 1]]) dst = linear_transformation(src, a) plt.imshow(dst) plt.show() Rotation through $45^{\\circ}$ about the origin: $f \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix} \\cos{\\pi \\over 4} & -\\sin{\\pi \\over 4} \\\\ \\sin{\\pi \\over 4} & \\cos{\\pi \\over 4} \\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix},$ in Numpy: alpha = np.pi/4 a = np.array([[np.cos(alpha), -np.sin(alpha)], [np.sin(alpha), np.cos(alpha)]]) dst = linear_transformation(src, a) plt.imshow(dst) plt.show() Reflexion in a line with inclination of $45^{\\circ}$ through the origin: $f \\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix} \\cos{\\pi \\over 2} & \\sin{\\pi \\over 2} \\\\ \\sin{\\pi \\over 2} & -\\cos{\\pi \\over 2} \\end{pmatrix} \\begin{pmatrix}x\\\\y\\end{pmatrix},$ in Numpy: alpha = np.pi/4 a = np.array([[np.cos(2*alpha), np.sin(2*alpha)], [np.sin(2*alpha), -np.cos(2*alpha)]]) dst = linear_transformation(src, a) plt.imshow(dst) plt.show()", "\\\\ \\sin \\theta & \\cos \\theta \\end{bmatrix}$ or $\\begin{bmatrix} a & -b \\\\ b & a \\end{bmatrix}$, where $a^2 + b^2 = 1$ Rotation through angle $\\theta$ combined with scaling by $r$ $\\begin{bmatrix} a & -b \\\\ b & a \\end{bmatrix} = r \\begin{bmatrix} \\cos \\theta & - \\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{bmatrix}$ Horizontal shear $\\begin{bmatrix} 1 & k \\\\ 0 & 1 \\end{bmatrix}$ Vertical shear $\\begin{bmatrix} 1 & 0 \\ k & 1 \\end{bmatrix}$ # 2.3 Matrix Products Rotation combined with scaling. Suppose • $T_1 \\mathbb{R}^2 \\to \\mathbb{R}^2$ gives rotation counter clockwise by angle $\\theta$ • $T_2 \\mathbb{R}^2 \\to \\mathbb{R}^2$ Scales by $k > 0$ This is in the form $T_2 (T_1(\\vec{x}))$ $T_2 T_1 : \\mathbb{R}^2 \\to \\mathbb{R}^2 \\text{ function composition}$ $(T_2 T_1)(\\vec{x}) = k \\begin{bmatrix} \\cos \\theta & - \\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{bmatrix} \\vec{x}$ What is the matrix? $\\begin{bmatrix} k\\cos \\theta & -k \\sin \\theta \\\\ k \\sin \\theta & k \\cos \\theta \\end{bmatrix} = \\begin{bmatrix} k & 0 \\\\ 0 & k \\end{bmatrix} \\begin{bmatrix} \\cos \\theta & -\\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{bmatrix}$ $\\text{Composition of Transformations} \\leftrightarrow \\text{Matrix Product}$ The matrix product BA: Suppose $B$ is an $n\\times p$ matrix and $A$ is a $p \\times m$ matrix. Size of $BA$: $[n \\times", "\\frac{0}{1} \\right) \\vec{e}_2 - \\vec{e}_1 = \\begin{bmatrix} -1 \\\\ 0 \\end{bmatrix}$ $A = \\begin{bmatrix} -1 & 0 \\\\ 0 & 1 \\end{bmatrix}$ ### Scaling For $k > 0,\\ T(\\vec{x}) = k \\vec{x}$. $\\begin{bmatrix} k & 0 \\\\ 0 & k \\end{bmatrix}$ • $k > 1$ : Dilation • $0 < k < 1$ : Contraction Question: Can we interpret the transformation $T(\\vec{x}) = \\begin{bmatrix} 0 & -1 \\\\ 1 & 0 \\end{bmatrix} \\vec{x}$ geometrically? Answer: Rotation counterclockwise by $\\frac{\\pi}{2}$ or 90 degrees. ### Rotation Counterclockwise by angle $\\theta$. $T\\left( \\vec{e}_1 \\right) = \\begin{bmatrix} \\cos \\theta \\\\ \\sin \\theta \\end{bmatrix}$ $T\\left( \\vec{e}_2 \\right) = \\begin{bmatrix} \\cos \\left( \\theta + \\frac{\\pi}{2} \\right) \\\\ \\sin \\left( \\theta + \\frac{\\pi}{2} \\right) \\end{bmatrix} = \\begin{bmatrix} -\\sin \\theta \\\\ \\cos \\theta \\end{bmatrix}$ $\\therefore A = \\begin{bmatrix} \\cos \\theta & - \\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{bmatrix}$ ## Transformation Recap: Transformation Matrix Scaling (by $k$) $kI_2 = \\begin{bmatrix} k & 0 \\\\ 0 & k \\end{bmatrix}$ Orthogonal projection onto line $L$ $\\begin{bmatrix} u_1^2 & u_1u_2 \\\\ u_1u_2 & u_2^2 \\end{bmatrix}$, where $\\begin{bmatrix} u_1 \\\\ u_2 \\end{bmatrix}$ is a unit vector parallel to $L$ Reflection about a line $\\begin{bmatrix} a & b \\\\ b & -a \\end{bmatrix}$, where $a^2 + b^2 = 1$ Rotation through angle $\\theta$ $\\begin{bmatrix} \\cos \\theta & - \\sin", "- \\sin^2 \\phi \\end{bmatrix} = \\begin{bmatrix} \\cos^2 \\phi & \\sin \\phi \\cos \\phi \\\\ \\sin \\phi \\cos \\phi & \\cos^2 \\phi \\end{bmatrix},$ which is the matrix for projection onto $L$. 2. When $k=-1$ we have $\\begin{bmatrix} 1 - 2 \\sin^2 \\phi & 2 \\sin \\phi \\cos \\phi \\\\ 2 \\sin \\phi \\cos \\phi & 1 - 2 \\cos^2 \\phi \\end{bmatrix} = \\begin{bmatrix} \\cos(2\\phi) & \\sin (2\\phi) \\\\ \\sin (2\\phi) & - \\cos (2\\phi) \\end{bmatrix}$ which is the matrix for reflection in $L$. Exercise 31 1. A reflection is a dilation by factor $-1$, so applying it twice gives a dilation with factor $(-1)^2 = 1$, i.e. the identity. 2. A projection is a dilation by factor $0$, so applying it twice gives a dilation with factor $0^2 = 0$, i.e. the same projection. Exercise 32 1. A projection matrix clearly has the desired properties. For the converse, suppose $M$ is symmetric ($b=c$), $ad = b^2$ and $a+d=1$. Since $a$ and $d$ have sum and product $\\geq 0$, both are $\\geq 0$; since they sum to $1$, both lie in $[0,1]$. Hence we have $a = \\cos^2 \\phi$ and $b = \\sin^2 \\phi$ for some $\\phi$. Then $b^2 = \\cos^2 \\phi \\sin^2 \\phi$ so $b = \\sin \\phi \\cos \\phi$ or $- \\sin \\phi \\cos \\phi$. In the first", "combined by matrix multiplication. Here’s a scaling of $$p, q, r$$ units followed by a rotation of $$\\theta$$ radians around the third axis followed by a rotation of $$\\phi$$ radians around the second axis: $\\begin{split}\\begin{bmatrix} x\\\\ y\\\\ z\\\\ \\end{bmatrix} = \\begin{bmatrix} \\cos(\\phi) & 0 & \\sin(\\phi) \\\\ 0 & 1 & 0 \\\\ -\\sin(\\phi) & 0 & \\cos(\\phi) \\\\ \\end{bmatrix} \\begin{bmatrix} \\cos(\\theta) & -\\sin(\\theta) & 0 \\\\ \\sin(\\theta) & \\cos(\\theta) & 0 \\\\ 0 & 0 & 1 \\\\ \\end{bmatrix} \\begin{bmatrix} p & 0 & 0 \\\\ 0 & q & 0 \\\\ 0 & 0 & r \\\\ \\end{bmatrix} \\begin{bmatrix} i\\\\ j\\\\ k\\\\ \\end{bmatrix}\\end{split}$ This can also be written: \\begin{align}\\begin{aligned}\\begin{split}M = \\begin{bmatrix} \\cos(\\phi) & 0 & \\sin(\\phi) \\\\ 0 & 1 & 0 \\\\ -\\sin(\\phi) & 0 & \\cos(\\phi) \\\\ \\end{bmatrix} \\begin{bmatrix} \\cos(\\theta) & -\\sin(\\theta) & 0 \\\\ \\sin(\\theta) & \\cos(\\theta) & 0 \\\\ 0 & 0 & 1 \\\\ \\end{bmatrix} \\begin{bmatrix} p & 0 & 0 \\\\ 0 & q & 0 \\\\ 0 & 0 & r \\\\ \\end{bmatrix}\\end{split}\\\\\\begin{split}\\begin{bmatrix} x\\\\ y\\\\ z\\\\ \\end{bmatrix} = M \\begin{bmatrix} i\\\\ j\\\\ k\\\\ \\end{bmatrix}\\end{split}\\end{aligned}\\end{align} This might be obvious because the matrix multiplication is the result of applying each transformation in turn on the coordinates output from the previous transformation. Combining the transformations into a single matrix $$M$$ works because", "all $\\mathbf{q},\\mathbf{k},m$. This allows us to decompose $\\Theta_f$ as $\\Theta_f(\\mathbf{x}, y) = \\Theta(\\mathbf{x}) + \\varphi(y)$. Examining the case of $m = n + 1$ reveals that $$\\varphi(m) - \\varphi(m-1) = \\Theta_g(\\mathbf{q}, \\mathbf{k}, 1) + \\Theta(\\mathbf{q}) - \\Theta(\\mathbf{k})$$ Since the right-hand side does not depend on $m$, the left hand side must not either and so $\\varphi$ is an arithmetic progression. Setting the initial values $\\varphi(0)=0$ and $\\varphi(1)=\\theta$, we have $\\varphi(m)=m\\theta$. Putting all of these pieces together, we get the final formula for the rotary positional embedding: $$f(\\mathbf{q}, m) = R_f(\\mathbf{q}, m)e^{i\\Theta_f(\\mathbf{q}, m)}=\\mathbf{q}e^{i(\\Theta(\\mathbf{q})+m\\mathbf{\\theta})} = \\sum_{j=1}^{d/2} q_je^{im\\theta_j} \\vec{e_j}$$ and likewise for $\\mathbf{k}$. Since computers tend to like real numbers and matrices more than complex numbers, its convenient to convert this expression into the matrix equation f(\\mathbf{q}, m) = \\begin{pmatrix} M_1 & & & \\\\ & M_2 & & \\\\ & & \\ddots & \\\\ & & & M_{d/2} \\end{pmatrix} \\begin{pmatrix} q_1\\\\ q_2\\\\ \\vdots\\\\ q_d \\end{pmatrix} = \\mathbf{\\Theta_m Q_m} = \\mathbf{\\Theta_m W_q X_m} where $M_j=\\begin{pmatrix}\\cos m\\theta_j & -\\sin m\\theta_j \\\\sin m\\theta_j & \\cos m\\theta_j\\end{pmatrix}$, $\\mathbf{\\Theta_m}$ is the block diagonal rotation matrix, $\\mathbf{W_q}$ is the learned query weights, and $\\mathbf{X_m}$ is the embedding of the $m^{th}$ token. Again, we also have the corresponding equation for $\\mathbf{k}$. ### Extension to multiple dimensions With relative ease RoPE can be extended into the multidimensional case.", "\\right) \\vec{e}_2 - \\vec{e}_1 = \\begin{bmatrix} -1 \\\\ 0 \\end{bmatrix}$ $A = \\begin{bmatrix} -1 & 0 \\\\ 0 & 1 \\end{bmatrix}$ ### Scaling For $k > 0,\\ T(\\vec{x}) = k \\vec{x}$. $\\begin{bmatrix} k & 0 \\\\ 0 & k \\end{bmatrix}$ • $k > 1$ : Dilation • $0 < k < 1$ : Contraction Question: Can we interpret the transformation $T(\\vec{x}) = \\begin{bmatrix} 0 & -1 \\\\ 1 & 0 \\end{bmatrix} \\vec{x}$ geometrically? Answer: Rotation counterclockwise by $\\frac{\\pi}{2}$ or 90 degrees. ### Rotation Counterclockwise by angle $\\theta$. $T\\left( \\vec{e}_1 \\right) = \\begin{bmatrix} \\cos \\theta \\\\ \\sin \\theta \\end{bmatrix}$ $T\\left( \\vec{e}_2 \\right) = \\begin{bmatrix} \\cos \\left( \\theta + \\frac{\\pi}{2} \\right) \\\\ \\sin \\left( \\theta + \\frac{\\pi}{2} \\right) \\end{bmatrix} = \\begin{bmatrix} -\\sin \\theta \\\\ \\cos \\theta \\end{bmatrix}$ $\\therefore A = \\begin{bmatrix} \\cos \\theta & - \\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{bmatrix}$ ## Transformation Recap: Transformation Matrix Scaling (by $k$) $kI_2 = \\begin{bmatrix} k & 0 \\\\ 0 & k \\end{bmatrix}$ Orthogonal projection onto line $L$ $\\begin{bmatrix} u_1^2 & u_1u_2 \\\\ u_1u_2 & u_2^2 \\end{bmatrix}$, where $\\begin{bmatrix} u_1 \\\\ u_2 \\end{bmatrix}$ is a unit vector parallel to $L$ Reflection about a line $\\begin{bmatrix} a & b \\\\ b & -a \\end{bmatrix}$, where $a^2 + b^2 = 1$ Rotation through angle $\\theta$ $\\begin{bmatrix} \\cos \\theta & - \\sin \\theta", "0$ makes an angle of θ with the x-axis, then $\\tan\\theta = f'(0)$. To rotate this an angle of $-\\theta$, we use the transformation $\\begin{pmatrix} u \\\\ v\\end{pmatrix} = \\begin{pmatrix} \\cos \\theta & \\sin\\theta \\\\ -\\sin\\theta & \\cos\\theta \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\implies \\begin{pmatrix} x \\\\ y\\end{pmatrix} = \\begin{pmatrix} \\cos\\theta & -\\sin\\theta \\\\ \\sin\\theta & \\cos\\theta\\end{pmatrix} \\begin{pmatrix} u \\\\ v \\end{pmatrix}.$ Substituting into $y = f(x)$ and partial differentiating with respect to u (noting that θ is constant) gives: $\\sin\\theta + \\cos\\theta \\frac{dv}{du} = f'(\\cos\\theta \\cdot u - \\sin\\theta\\cdot v) (\\cos\\theta - \\sin\\theta \\frac{dv}{du}).$ [ Side remark: note that $\\frac{dv}{du} = 0$ at $(u,v) = (0,0)$ since $\\tan\\theta = f'(0)$, which is what we want. ] Now, differentiating again and putting u=0, v=0, we get: $\\cos\\theta \\frac{d^2 v}{du^2} = f'(0)(-\\sin\\theta \\frac{d^2 v}{du^2}) + f''(0)(\\cos\\theta - \\sin\\theta \\frac{dv}{du})^2.$ Simplifying with $\\frac{dv}{du}=0$ and $\\tan\\theta = f'(0)$ gives the curvature: $\\frac{d^2 v}{du^2} = f''(0) \\cos^3\\theta = \\frac{f''(0)}{(1 + f'(0)^2)^{3/2}},$ which is typically what you see in the definition of curvature. So in short, to define the curvature of the graph of $y = f(x)$ at the point $(x_0, f(x_0))$, we use the formula $k = \\frac{f''(x_0)}{(1 + f'(x_0)^2)^{3/2}}.$ This is also known as the signed curvature. The unsigned curvature is usually denoted by $\\kappa = |k|$. Exercises. 1. Compute", "transformations are defined by the first translation matrix $\\mathbf{C}$, the rotation matrix $\\mathbf{B}$, and the last translation matrix $\\mathbf{A}$. $$\\mathbf{C} = \\begin{pmatrix} 1 & 0 & -t_{x} \\\\ 0 & 1 & -t_{y} \\\\ 0 & 0 & 1 \\end{pmatrix} \\text{ , } \\mathbf{B} = \\begin{pmatrix} \\cos \\theta & -\\sin \\theta & 0 \\\\ \\sin \\theta & \\cos \\theta & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\text{ and } \\mathbf{A} = \\begin{pmatrix} 1 & 0 & t_{x} \\\\ 0 & 1 & t_{y} \\\\ 0 & 0 & 1 \\end{pmatrix}$$ The composition of these transformations gives us the following product: \\begin{aligned} \\mathbf{A} . \\mathbf{B} . \\mathbf{C} &= \\begin{pmatrix} 1 & 0 & t_{x} \\\\ 0 & 1 & t_{y} \\\\ 0 & 0 & 1 \\end{pmatrix} . \\begin{pmatrix} \\cos \\theta & -\\sin \\theta & 0 \\\\ \\sin \\theta & \\cos \\theta & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} . \\begin{pmatrix} 1 & 0 & -t_{x} \\\\ 0 & 1 & -t_{y} \\\\ 0 & 0 & 1 \\end{pmatrix}\\\\\\\\ &= \\begin{pmatrix} 1 & 0 & t_{x} \\\\ 0 & 1 & t_{y} \\\\ 0 & 0 & 1 \\end{pmatrix} . \\begin{pmatrix} \\cos \\theta & -\\sin \\theta & -\\cos \\theta.t_{x} +\\sin \\theta.t_{y} \\\\ \\sin \\theta & \\cos \\theta & -\\sin \\theta.t_{x} -\\cos \\theta.t_{y} \\\\ 0 &", "Dec 21, 2013 at 21:37 • So I just tried to implement the code. It turns out it doesn't work because many of the matrices I have are singular... Dec 22, 2013 at 2:59 • @bigTree: I think that this should work even for singular or non-diagonalizable matrices. But I'm not sure, and asked this question for details of matrices raised to fractional powers. – MvG Dec 22, 2013 at 14:42 If you scale, shear, and rotate (in that order) then you'll have: $$\\begin{bmatrix} A_{11} & A_{12} \\ \\\\ A_{21} & A_{22} \\end{bmatrix} = \\begin{bmatrix} \\cos \\theta & -\\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{bmatrix} \\begin{bmatrix} 1 & m \\\\ 0 & 1 \\end{bmatrix} \\begin{bmatrix} s & 0 \\\\ 0 & s \\end{bmatrix}.$$ Then \\begin{align*} A_{11} &= s \\cos \\theta\\\\ A_{12} &= s m \\cos \\theta - s \\sin \\theta\\\\ A_{21} &= s \\sin \\theta\\\\ A_{22} &= s m \\sin \\theta + s \\cos \\theta\\\\ \\end{align*} From the first and third equations, $$s = \\sqrt{A_{11}^2 + A_{21}^2},$$ and $$\\theta = \\tan^{-1}\\left( \\frac{A_{21}}{A_{11}} \\right).$$ Finally, substituting the first and third into the second and fourth, $$m = \\frac{A_{12} + A_{21}}{A_{11}} = \\frac{A_{22} - A_{11}}{A_{21}}.$$ • I solved the equations and found the same results as you, but when I tried to test the equality for the matrix", "\\\\ 1 & 0 \\end{bmatrix},$ i.e., the imaginary unit $$i$$ corresponds to the matrix $$\\begin{bmatrix} 0 & -1 \\\\ 1 & 0 \\end{bmatrix}$$ and $$i^2 = -1$$ becomes $\\begin{bmatrix} 0 & -1 \\\\ 1 & 0 \\end{bmatrix} \\begin{bmatrix} 0 & -1 \\\\ 1 & 0 \\end{bmatrix} = -\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}.$ ## Polar Form (Decomposition) Writing $$z = re^{i \\theta} = r(\\cos \\theta + i \\sin \\theta)$$, we find \\begin{align*} Z &= r \\begin{bmatrix} \\cos \\theta & -\\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{bmatrix} \\\\[4pt] &= \\begin{bmatrix} r & 0 \\\\ 0 & r \\end{bmatrix} \\begin{bmatrix} \\cos \\theta & -\\sin \\theta \\\\ \\sin \\theta & \\cos \\theta \\end{bmatrix} \\end{align*} corresponding to a stretch factor $$r$$ multiplied by a 2D rotation matrix. In particular, multiplication by $$i$$ corresponds to the rotation with angle $$\\theta = \\pi /2$$ and $$r = 1$$. We will not make a lot of use of the matrix representation of complex numbers, but later it will help us remember certain formulas and facts.", "\\\\ \\sin \\alpha & = & {{\\tan \\alpha} \\over \\sqrt{1 + \\tan^2 \\alpha}} \\end{align} the rotation matrix becomes: $R_i = {1 \\over \\sqrt{1 + \\tan^2 \\gamma_i}} \\begin{bmatrix} 1 & -\\tan \\gamma_i \\\\ \\tan \\gamma_i & 1 \\end{bmatrix}$ The expression for the rotated vector $v_i = R_i v_{i-1}$ then becomes: $v_i = {1 \\over \\sqrt{1 + \\tan^2 \\gamma_i}} \\begin{bmatrix} 1 & -\\tan \\gamma_i \\\\ \\tan \\gamma_i & 1 \\end{bmatrix} \\begin{bmatrix} x_{i-1} \\\\ y_{i-1} \\end{bmatrix}$ where $x_{i-1}$ and $y_{i-1}$ are the components of $v_{i-1}$. Restricting the angles $\\gamma_{i}$ so that $\\tan \\gamma_{i}$ takes on the values $\\pm 2^{-i}$, the multiplication with the tangent can be replaced by a division by a power of two, which is efficiently done in digital computer hardware using a bit shift. The expression then becomes: $v_i = K_i \\begin{bmatrix} 1 & -\\sigma_i 2^{-i} \\\\ \\sigma_i 2^{-i} & 1 \\end{bmatrix} \\begin{bmatrix} x_{i-1} \\\\ y_{i-1} \\end{bmatrix}$ where $K_i = {1 \\over \\sqrt{1 + 2^{-2i}}}$ and $\\sigma_i$ can have the values of −1 or 1, and is used to determine the direction of the rotation; if the angle $\\beta_i$ is positive then $\\sigma_i$ is +1, otherwise it is −1. $K_i$ can be ignored in the iterative process and then applied afterward with a scaling factor: $K(n) = \\prod_{i=0}^{n-1} K_i = \\prod_{i=0}^{n-1} 1/\\sqrt{1 + 2^{-2i}}$ which is calculated in", "$\\begin{pmatrix} q_1 \\\\ q_2 \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & \\phi_1 \\\\ 0 & 1 & \\phi_2 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} p_1 \\\\ p_2 \\\\ 1 \\end{pmatrix}$ ### 2D Rotation To rotate a point about the origin by $\\theta$ $\\begin{pmatrix} q_1 \\\\ q_2 \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} cos(\\theta) & -sin(\\theta) & 0 \\\\ sin(\\theta) & cos(\\theta) & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} p_1 \\\\ p_2 \\\\ 1 \\end{pmatrix}$ Likewise to rotate a vector by $\\theta$ $\\begin{pmatrix} v_1 \\\\ v_2 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix} cos(\\theta) & -sin(\\theta) & 0 \\\\ sin(\\theta) & cos(\\theta) & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} u_1 \\\\ u_2 \\\\ 0 \\end{pmatrix}$ ### 2D Scale To scale a point by factors $(s_x, s_y)$ about the origin $\\begin{pmatrix} s_x p_1 \\\\ s_y p_2 \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} s_x & 0 & 0 \\\\ 0 & s_y & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} p_1 \\\\ p_2 \\\\ 1 \\end{pmatrix}$ Likewise to scale a vector by $(s_x, s_y)$ $\\begin{pmatrix} s_x v_1 \\\\ s_y v_2 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix} s_x & 0 & 0 \\\\ 0 & s_y & 0 \\\\ 0 & 0 & 1 \\end{pmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\\\ 0 \\end{pmatrix}$ ### 2D Shear", "\\end{align} respectively. In order to satisfy the boundary conditions, one calculates that \\begin{align} \\begin{pmatrix} -\\cos(\\nu) & \\cosh(\\nu) \\\\ \\sin(\\nu) & \\sinh(\\nu) \\end{pmatrix} \\begin{pmatrix} 1 \\\\ a \\end{pmatrix} = 0, \\quad \\begin{pmatrix} -\\sin(\\nu) & \\sinh(\\nu) \\\\ -\\cos(\\nu) & \\cosh(\\nu) \\end{pmatrix} \\begin{pmatrix} 1 \\\\ b \\end{pmatrix} = 0. \\end{align} The points at which either matrix becomes singular, are the roots $r_c$ and $r_s$ of \\begin{align} t_c := \\cos(\\nu)\\sinh(\\nu)+\\cosh(\\nu)\\sin(\\nu), \\quad t_s := \\cos(\\nu)\\sinh(\\nu)-\\cosh(\\nu)\\sin(\\nu), \\end{align} respectively. Both have infinitely many (single) roots which can be approximated by $-\\pi/4+i\\pi$ and $\\pi/4+i\\pi$, respectively, since for $k > 7$, these equal the true $i$-th roots $r_c^{(i)}$ and $r_s^{(i)}$ up to $20$ decimal places. Earlier values are easily found numerically. We therefor define $\\nu_{1,2} = 0, \\ \\nu_{1+2i} = r^{(i)}_c$ and $\\nu_{2+2i} = r^{(i)}_s$. After some calculus, we obtain \\begin{align} \\sqrt{1+\\frac{\\cos(\\nu_{1+2i})^2}{\\cosh(\\nu_{1+2i})^2}} f_{1+2i} & = \\cos(\\nu_{1+2i}x)+\\frac{\\cos(\\nu_{1+2i})}{\\cosh(\\nu_{1+2i})} \\cosh(\\nu_{1+2i}x) \\\\ \\sqrt{1-\\frac{\\sin(\\nu_{2+2i})^2}{\\sinh(\\nu_{2+2i})^2}} f_{2+2i} & = \\sin(\\nu_{2+2i}x)+\\frac{\\sin(\\nu_{2+2i})}{\\sinh(\\nu_{2+2i})} \\sinh(\\nu_{2+2i}x), \\quad i \\in \\mathbb{N} \\end{align} All these roots are distinct. The orthogonality conditions are easy to verify, considering \\begin{align} \\int_{-1}^1 f_i f_j = \\frac{1}{{\\nu_i}^4} \\int_{-1}^1 f_i^{(4)} f_j = -\\frac{1}{{\\nu_i}^4} \\int_{-1}^1 f_i f_j^{(4)} = \\frac{\\nu_j^4}{{\\nu_i}^4} \\int_{-1}^1 f_i f_j \\end{align} since either $\\nu_j \\neq \\nu_i$ or $\\nu_j = 0$, $j = 1,2$, for $j < i$. One can further calculate \\begin{align} \\int_{-1}^1 f_i'' f_j'' = \\delta_{ij} \\nu_i^4. \\end{align} The $f_k$ are hence in correct order", "0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ \\end{bmatrix}$ y $R (\\mathbf{e}_y,\\theta) = \\begin{bmatrix} 1 & 0 & 0 & 0 \\\\ 0 & \\cos \\theta & 0 & \\sin \\theta \\\\ 0 & 0 & 1 & 0 \\\\ 0 & -\\sin \\theta & 0 & \\cos \\theta \\\\ \\end{bmatrix}$ $B (\\mathbf{e}_y,\\phi) = \\begin{bmatrix} \\cosh\\phi & 0 & -\\sinh\\phi & 0 \\\\ 0 & 1 & 0 & 0 \\\\ -\\sinh\\phi & 0 & \\cosh\\phi & 0 \\\\ 0 & 0 & 0 & 1 \\\\ \\end{bmatrix}$ z $R (\\mathbf{e}_z,\\theta) = \\begin{bmatrix} 1 & 0 & 0 & 0 \\\\ 0 & \\cos \\theta & -\\sin \\theta & 0 \\\\ 0 & \\sin \\theta & \\cos \\theta & 0\\\\ 0 & 0 & 0 & 1\\\\ \\end{bmatrix}$ $B (\\mathbf{e}_z,\\phi) = \\begin{bmatrix} \\cosh\\phi & 0 & 0 & -\\sinh\\phi \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 1 & 0 \\\\ -\\sinh\\phi & 0 & 0 & \\cosh\\phi \\\\ \\end{bmatrix}$ The general rotation matrix is (in block form for compactness) $R(\\mathbf{a},\\theta) = \\begin{bmatrix} 1 & 0 \\\\ 0 & \\mathbf{I} + \\mathbf{A}^2(1-\\cos\\theta) + \\mathbf{A}\\sin\\theta \\end{bmatrix} \\,,$ and general boost matrix is $B(\\mathbf{n},\\phi) = \\begin{bmatrix} \\cosh\\phi & - \\mathbf{n}^\\mathrm{T} \\sinh\\phi \\\\ -\\mathbf{n}\\sinh\\phi" ]

[ "+0 # help 0 123 1 For a certain value of k, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 3y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of k? Feb 22, 2022 edited by Alan Feb 22, 2022 #1 +33151 +1 As follows: Feb 22, 2022", "+0 0 237 2 HALPPPP! For a certain value of $k,$ the system \\begin{align*} 3a + 4b &= 7,\\\\ 6a + 4b &= k- 4b \\end{align*}has infinitely many solutions $(a,b).$ What is $k$? I had this question on my math test. Got my results, but they didn't tell me correct answer. Plz tell me correct answer with steps so i know how i got it wrong. Jul 22, 2021 #1 +124524 +2 3a + 4b = 7 multiply through by 2 6a + 8b = 14 And note that the second equation simplifies to 6a + 8b = k Which....if we have infinite solutions implies that k = 14 Jul 22, 2021 #2 0 Thank you. I understand now. Jul 22, 2021", "+0 +2 198 3 +142 For a certain value of k, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 2y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of k? Aug 3, 2019 #1 +6046 +2 when k=6 the solutions for x,y,z all have 0 in the denominator which is a nono Aug 3, 2019 #2 +8810 +2 For a certain value of k, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 2y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of k? $$2x+2y+6z=20\\\\ \\underline{-4x+2y+5z=7}\\\\ 6x{\\color{blue}{+z}}=13$$ Thanks for the help heureka! $$\\underline{kx+z=3}\\\\ 6x-kx=10\\\\ x=\\frac{10}{6-k}\\\\ \\color{blue}k=6$$ For the value of k = 6, x is indefinite and the functions have no solution. ! Aug 3, 2019 edited by asinus Aug 3, 2019 #3 +23850 +2 For a certain value of k, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 2y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of $$k$$? $$\\begin{array}{|rcll|} \\hline \\begin{vmatrix} 1&1&3\\\\ -4 & 2 & 5\\\\ k & 0 &1 \\\\ \\end{vmatrix} &=& 0 \\\\\\\\ k\\begin{vmatrix} 1&3\\\\ 2 & 5\\\\ \\end{vmatrix}", "+0 # Help! 0 25 1 What negative value of k, gives one solution for this system of equations, I know that this means the discriminant needs to be 0, but I am stuck on the rest. Can someone help? Thanks! \\begin{align*} y &= 2x^2 + kx + 6 \\\\ y &= -x + 4? \\end{align*} Apr 19, 2022 #1 +1 Nevermind, I figured it out. Sry about that. Thanks though! Apr 19, 2022", "+0 # System 0 65 1 find the value of k for which the system of equations has a unique solution 2x+ky-9=0 , kx+3y-13=0. ​ Jun 15, 2022 #1 +9458 +1 The system is equivalent to $$\\begin{cases} 2x + ky &=& 9\\\\ kx + 3y &=& 13 \\end{cases}$$. As long as the lines represented by the equations are not parallel, the system would have a unique solution. $$-\\dfrac2k \\neq -\\dfrac k3\\\\ k^2 \\neq 6\\\\ k \\neq \\pm \\sqrt 6$$ For $$k \\neq \\sqrt 6$$ and $$k \\neq -\\sqrt 6$$, the system has a unique solution. Jun 16, 2022", "Comment Share Q) # If the system of equations :$x-ky-z=0,kx-y-z=0,x+y-z=0$ has a non zero solutions then the possible value of k are (A) -1,2 (B) 1,2 (C) 0,1 (D) -1,1 (i.e) $\\begin{vmatrix}1 &-k &-1\\\\k&-1&-1\\\\1 &1&-1\\end{vmatrix}$ $\\Rightarrow 1(1+1)+k(-k+1)-1(k+1)=0$ $\\Rightarrow 2-k^2+k-k-1=0$ $k^2=1$ $k=\\pm 1$", "# The number of values of $k$ for which the linear equations I was helping my comrade answering some questions taken from review classes when I stumbled upon this question. It looks like this: The number of values of $k$ for which the linear equations \\begin{align}4x + ky + 2z &= 0 \\\\kx + 4y + z &= 0 \\\\ 2x + 2y + kz &= 0\\end{align} possess a nonzero solution is..... My work I honestly don't know how to answer this question. I don't understand what the statement \"possess a nonzero solution is.....\" means. I thought that it would mean $x$ and $y$ and $z$ musn't equal to zero. Moreover, I'm dealing in $4$ unknowns and $3$ equations, making matters worse. How to answer the above question? • It means for which values of $k$ do the equations above, have at least one solution in $(x,y,z)$ which is not $(0,0,0)$ ? – zwim Sep 27 '17 at 8:32 • Let's link to this important justification of the determinant method : math.stackexchange.com/questions/1455226/… – zwim Sep 27 '17 at 10:02 We have the matrix:$$\\begin{bmatrix} 4&k&2\\\\k&4&1\\\\2&2&k\\\\\\end{bmatrix}$$ Since the determinant of the matrix equals $0$, \\begin{align}4(4k-2)-k(k^2-2)+2(2k-8)&=0 \\\\ 16k-8-k^3+2k+4k-16&=0 \\\\ -k^3+22k-24&=0 \\\\ (k-4)(k^2+4k-6)&=0 \\end{align} So either $k^2+4k-6=0$ or $k-4=0$ If the former applies, then \\begin{align}k^2+4k&=6\\\\k^2+4k+4&=10\\\\(k+2)^2&=10 \\\\k+2&=\\pm\\sqrt{10}\\end{align} so $k=\\sqrt{10}-2,$ or $-\\sqrt{10}-2$. If the", "+0 # System 0 83 1 Find all values of k for which the system x + ky - z = 0 kx - y - z = 0 7x + 3y - 2kz = 0 has a non-trivial solution. (In other words, find all values of k for which the system has a solution other than (x,y,z) = (0,0,0).) May 14, 2022 #1 +9459 0 For a homogeneous system $$A \\vec x = \\vec 0$$ to have non-trivial solutions, we must have $$\\det A = 0$$. Then $$\\det \\begin{pmatrix}1&k&-1\\\\k&-1&-1\\\\7&3&-2k\\end{pmatrix}= 0$$. Please simplify the determinant and solve the resulting cubic equation. May 14, 2022 edited by MaxWong May 14, 2022", "+0 # Find all real constants such that the system 0 111 1 +180 Find all real constants $$k$$ such that the system \\begin{align*} x + 3y &= kx, \\\\ 3x + y &= ky. \\end{align*} has a solution other than $$(x,y) = (0,0).$$ Jun 27, 2020 ### 1+0 Answers #1 0 For k = 4, all x = y is a solution, so this is the answer. Jun 27, 2020", "+0 # help +1 87 1 Find all real constants such that the system \\begin{align*} x + 3y &= kx, \\\\ 3x + y &= ky. \\end{align*} has a solution other than (x,y) = (0,0).\\$ Jul 31, 2020", "Question # The value of k for which the set of equations $$3x+ky- 2z=0, x + ky + 3z = 0$$ and $$2x+3y- 4z=0$$ has a non-trivial solution is- A 15 B 16 C 31/2 D 33/2 Solution ## The correct option is D $$33/2$$Given system has a non-trivial solution.$$\\begin{vmatrix} 3 & k & -2 \\\\ 1 & k & 3 \\\\ 2 & 3 & -4 \\end{vmatrix}=0$$$$\\Rightarrow 3(-4k-9)+10k-2(3-2k)=0$$$$\\Rightarrow 2k=33$$$$\\therefore k=33/2$$Hence, option D.Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More", "# System of three linear equations with unknown constant I have a linear system with three equations: \\begin{align} x + y + z &= 3a\\\\ x + 2y + (a+2)z &= a\\\\ x - (a+1)y - z &= 0 \\end{align} I want to find values for a where the system is inconsistent in addition to the values of $$x, y, z$$ and $$a$$ for where the system is consistent. How should one approach a problem like this? With $$z=x-(a+1)y$$ we get $$2x+y-(a+1)y=3a$$ and $$2x-ay=3a$$ eliminating $$x$$ we obtain $$-\\frac{3a(a+3)}{2}y=-\\frac{3a(a+3)}{2}$$ Now you can do the rest!", "$x - ky - z = 0$ $kx - y - z = 0$ $x + y + z = 0$ has a UNIQUE solution (0,0,0), if and only if $k^2 - 1 \\neq 0$, and infinitely many otherwise. If we are looking for a non-zero solution, it follows that $k^2 - 1 = 0$ (so that either $k = 1$ , or $k = -1$), and we must choose a non-zero value for $y$.", "Thanks Let $l = 1/k$. \\begin{aligned} \\Delta &= (4m)^2\\left(k-\\frac 1k\\right)^2 - 4(4m^2)\\left(k^2 + \\frac{1}{k^2}\\right) \\\\ &= 16m^2\\left(k^2 - 2 + \\frac{1}{k^2} - k^2 - \\frac{1}{k^2}\\right) \\\\ &= 16m^2(-2) \\\\ &= -32m^2\\end{aligned} which is negative for all nonzero real $m$.", "# For what values of k does this system of equations have a unique / infinite / no solutions? My system of equations is: $$\\begin{cases} x + 5y- 6z = 2 \\\\ kx + y - z = 3 \\\\ 5x - ky + 3z = 7 \\end{cases}$$ So the augmented matrix is: $$\\left[ \\begin{array}{ccc|c} 1&5&-6&2\\\\ k&1&-1&3\\\\ 5&-k&3&7 \\end{array} \\right]$$ I reduced it to this (with the shown steps), but I'm not sure if i've done so correctly and i'm not sure if i need to do Gauss-Jordan elimination or just Gauss elimination: \\begin{align} r_2 -> r_2 - 2r_1 \\end{align} \\begin{align} r_3 -> r+3 - 5r_1 \\end{align} \\begin{align} r_2 -> r_2 * \\frac{1}{-9} \\end{align} \\begin{align} r_1 -> r_1 - 5r_2 \\end{align} \\begin{align} r_3 -> r_3 +27r_2 \\end{align} $$\\left[ \\begin{array}{ccc|c} 1-5(\\frac{k-2}{-9})&0&\\frac{1}{9}&\\frac{13}{9}\\\\ \\frac{k-2}{-9}&1&\\frac{-11}{9}&\\frac{1}{9}\\\\ 27(\\frac{k-2}{-9})&-k+2&0&0 \\end{array} \\right]$$ From this reduced from that i got, i noticed that there is an infinite solution for k=2. However, I believe that i need to get one of the values in the 4th column of the matrix in terms of k so that i can find all the solutions i need, but i am unsure on how to do this. Could someone please help me and show full working, including the steps taken to reduce the matrix? • Hi and welcome to the site! Since", "+0 # question pls help ty +1 120 2 +25 For a certain value of $k$, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 3y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of $k$? Jun 21, 2021 #1 +2252 0 3x + 3y + 9z = 30 -4x + 3y + 5z = 7 7x + 4z = 23 4kx + 4z = 3 If k = 7/4. 4(7/4)x + 4z = 3 7x + 4z = 3, but 7x + 4z = 23?? That is impossible so it doesn't have any solutions. =^._.^= Jun 21, 2021 #2 +26213 +2 For a certain value of $$k$$, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 3y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of $$k$$? $$\\begin{array}{|rcll|} \\hline \\begin{vmatrix} 1 & 1 & 3 \\\\ -4 & 3 & 5 \\\\ k & 0 & 1 \\\\ \\end{vmatrix} &=& 0 \\\\\\\\ k* \\begin{vmatrix} 1 & 3 \\\\ 3 & 5 \\\\ \\end{vmatrix} +1* \\begin{vmatrix} 1 & 1 \\\\ -4 & 3 \\\\ \\end{vmatrix} &=&0 \\\\\\\\ k*(1*5-3*3) + 1*(1*3-(-4)*1) &=& 0 \\\\\\\\ k*(-4) + 1*7 &=& 0 \\\\\\\\ k*(-4)", "\\\\ &= \\frac{1}{n^{2}+n+1}\\sum_{k=1}^{n}k. \\end{align}", "+0 +1 60 6 For a certain value of $k$, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 3y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*}has no solutions. What is this value of $k$? Apr 6, 2021 #1 +485 +2 Hint: one way to define no solutions means that when you simplify everything(without possibly dividing by 0, such as 3x=5x, x=0 is a solution), you are left with two unequal numbers equaling each other, such as $3=5$. Apr 6, 2021 #2 +285 +3 So, just solve x an z. WillBillDillPickle Apr 6, 2021 #3 +118666 +3 Multiply the first equation by -3 -3x - 3y - 9z = -30 Add this to the second equation and we get that -7x -4z = -22 Mutltiply the last equation by -4 -4kx - 4z = -12 The system will have no solutions when -4k = -7 k = -7/-4 k = 7/4 Apr 6, 2021 edited by CPhill Apr 6, 2021 #4 +485 +2 Great job @CPhill! RiemannIntegralzzz Apr 6, 2021 #5 +118666 +1 THX!!!! CPhill Apr 6, 2021 #6 +1 Thanks! Its correct!! :D Apr 6, 2021", "+0 HALPPPP! ANSWER NEEDED NOW!! Ps. The answer is not 6. 0 55 1 For a certain value of $k$, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 3y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*}has no solutions. What is this value of $k$? Aug 9, 2021 #1 0 By Cramer's Rule, |1 1 3| |-4 3 5| = 0 |0 k 1| Then 7 - 17k = 0, so k = 7/17. Hope this helps! Aug 9, 2021", "k\\in\\mathbb{Z} ,\\\\ \\theta &=\\frac{2\\pi k}{n+1} . \\end{align} Thus, for $$n>1$$ we have non-trivial solutions of the form \\begin{align} z&= \\cos\\left(\\frac{2\\pi k}{n+1}\\right) + i\\cdot\\sin\\left(\\frac{2\\pi k}{n+1}\\right) ,\\quad k\\in\\mathbb{Z} . \\end{align}" ]

[ "− 3z = 5 We need to compute D, Dx, Dy, and Dz. We have: $D=\\begin{vmatrix} 1 & -5 & 2\\\\ 4 & 10 & 1 \\\\ -2 & 0 & -3\\end{vmatrix} = -40$ $D_x=\\begin{vmatrix} -7 & -5 & 2\\\\ 11 & 10 & 1 \\\\ 5& 0 & -3\\end{vmatrix} = -80$ $D_y=\\begin{vmatrix} 1 & -7 & 2\\\\ 4 & 11 & 1 \\\\ -2 &5 & -3\\end{vmatrix} = -24$ $D_z=\\begin{vmatrix} 1 & -5 & -7\\\\ 4 & 10 & 11 \\\\ -2 & 0 & 5\\end{vmatrix} = 120$ Then $\\displaystyle x = \\frac{D_x}{D} = \\frac{-80}{-40} = 2$ , $y = \\frac{D_y}{D} = \\frac{-24}{-40}= \\frac{3}{5}$ , and $z = \\frac{D_z}{D}=\\frac{120}{-40}=-3$. Our solution is $\\left(2, \\frac{3}{5}, -3\\right)$. Try to solve the following systems using Cramer's Rule on your own. Check you answer by mousing on the black rectangle. If you need additional help, look at the pencasts. Check yourself! System Solution 6x − 3y = 7 12x + 9y = 4 (56,-23) 9x − 4y = 13 3x + 10y = 10 (53,12) 4x − 3y + 8z = 33 6y + 10z = 31 -8x + 6y − 4z = -24 (34,-23,72) ## 6 When D = 0 Recall that the solution to a system of linear equations is either no solution, a unique solution, or infinitely many", "-3 \\end{vmatrix}$$ $$224$$ 20) $$\\begin{vmatrix} 6 & -1 & 2\\\\ -4 & -3 & 5\\\\ 1 & 9 & -1 \\end{vmatrix}$$ 21) $$\\begin{vmatrix} 5 & 1 & -1\\\\ 2 & 3 & 1\\\\ 3 & -6 & -3 \\end{vmatrix}$$ $$15$$ 22) $$\\begin{vmatrix} 1.1 & 2 & -1\\\\ -4 & 0 & 0\\\\ 4.1 & -0.4 & 2.5 \\end{vmatrix}$$ 23) $$\\begin{vmatrix} 2 & -1.6 & 3.1\\\\ 1.1 & 3 & -8\\\\ -9.3 & 0 & 2 \\end{vmatrix}$$ $$-17.03$$ 24) $$\\begin{vmatrix} -\\frac{1}{2} & \\frac{1}{3} & \\frac{1}{4}\\\\ \\frac{1}{5} & -\\frac{1}{6} & \\frac{1}{7}\\\\ 0 & 0 & \\frac{1}{8} \\end{vmatrix}$$ For the exercises 25-34, solve the system of linear equations using Cramer’s Rule. 25) \\begin{align*} 2x-3y &= -1\\\\ 4x+5y &= 9 \\end{align*} $$(1,1)$$ 26) \\begin{align*} 5x-4y &= 2\\\\ -4x+7y &= 6 \\end{align*} 27) \\begin{align*} 6x-3y &= 2\\\\ -8x+9y &= -1 \\end{align*} $$\\left (\\dfrac{1}{2}, \\dfrac{1}{3} \\right )$$ 28) \\begin{align*} 2x+6y &= 12\\\\ 5x-2y &= 13 \\end{align*} 29) \\begin{align*} 4x+3y &= 23\\\\ 2x-y &= -1 \\end{align*} $$(2,5)$$ 30) \\begin{align*} 10x-6y &= 2\\\\ -5x+8y &= -1 \\end{align*} 31) \\begin{align*} 4x-3y &= -3\\\\ 2x+6y &= -4 \\end{align*} $$\\left (-1, -\\dfrac{1}{3} \\right )$$ 32) \\begin{align*} 4x-5y &= 7\\\\ -3x+9y &= 0 \\end{align*} 33) \\begin{align*} 4x+10y &= 180\\\\ -3x-5y &= -105 \\end{align*} $$(15,12)$$ 34) \\begin{align*} 8x-2y &= -3\\\\ -4x+6y &= 4 \\end{align*} For the exercises 35-44, solve the system", "& = x_3\\\\ x_2' & = x_4\\\\ x_3' & = -2 \\frac{k}{m_1} x_1 + \\frac{k}{m_1} x_2\\\\ x_4' & = \\frac{k}{m_2} x_1 - 2\\frac{k}{m_2} x_2 \\end{align*} We can represent this system in the matrix form ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ where \\begin{equation*} A = \\begin{pmatrix} 0 & 0 & 1 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ -2k/m_1 & k/m_1 & 0 & 0 \\\\ k / m_2 & - 2k/m_2 & 0 & 0 \\end{pmatrix}. \\end{equation*} We will learn how to analyze and solve such systems in the next two sections. Subsection3.8.1Higher-Order Linear Systems We can write the system \\begin{align*} x_1' & = a_{11} x_1 + \\cdots a_{1n} x_n\\\\ x_2' & = a_{21} x_1 + \\cdots a_{2n} x_n\\\\ & \\vdots\\\\ x_n' & = a_{n1} x_1 + \\cdots a_{nn} x_n \\end{align*} in matrix form ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ where \\begin{equation*} A = \\begin{pmatrix} a_{11} & a_{12} & \\cdots & a_{1n} \\\\ a_{21} & a_{22} & \\cdots & a_{2n} \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ a_{n1} & a_{n2} & \\cdots & a_{nn} \\end{pmatrix} \\qquad\\text{and}\\qquad {\\mathbf x} = \\begin{pmatrix} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n \\end{pmatrix}. \\end{equation*} The strategy for finding solutions to the system ${\\mathbf x}' = A {\\mathbf x}$ is the same as for systems of two equations.", "system $\\Rightarrow z=-4$ So, the solution is $x=1,y=3,z=-4$ B) Cramer's rule Let $A=\\begin{pmatrix}1 & 2 & 3\\\\2 & 1 & 1\\\\1 & 1 & -1\\end{pmatrix}$ be the matrix of system. We have $d=\\det A=7$. $d_x=\\begin{vmatrix}-5 & 2 & 3\\\\1 & 1 & 1\\\\8 & 1 & -1\\end{vmatrix}=7$, $d_y=\\begin{vmatrix}1 & -5 & 3\\\\2 & 1 & 1\\\\1 & 8 & -1\\end{vmatrix}=21$, $d_z=\\begin{vmatrix}1 & 2 & -5\\\\2 & 1 & 1\\\\1 & 1 & 8\\end{vmatrix}=-28$ Now, $\\displaystyle x=\\frac{d_x}{d}=\\frac{7}{7}=1$, $\\displaystyle y=\\frac{d_y}{d}=\\frac{21}{7}=3$, $\\displaystyle z=\\frac{d_z}{d}=-\\frac{28}{7}=-4$ C) Matrix method The system can be written as $A\\cdot X=B$, where $A$ is the matrix of the system (see B)), $X=\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}$, $B=\\begin{pmatrix}-5\\\\1\\\\8\\end{pmatrix}$ $\\det A=7\\Rightarrow A$ is invertible. Multiplying both members of the matrix equation, to the left, with $A^{-1}$ we have $X=A^{-1}\\cdot B=\\begin{pmatrix}1\\\\3\\\\-4\\end{pmatrix}$ (I hope you know how the find $A^{-1}$). • Jul 19th 2007, 12:24 AM red_dog Quote: 6. Solve the linear inequality 3(2x-1) - 2(x-4) > 7 + 2(3+4x) $3(2x-1)-2(x-4)>7+2(3+4x)\\Leftrightarrow 6x-3-2x+8>7+6+8x\\Leftrightarrow$ $\\Leftrightarrow -4x>8\\Leftrightarrow x<-2\\Leftrightarrow x\\in(-\\infty ,-2)$ Quote: 7. Solve each compound inequality A) 2x-5 > -1 or 3x < 3 B) 5x + 3 < 18 and 2x - 7 < 5 A) $2x-5>-1\\Rightarrow x>2\\Rightarrow x\\in (2,\\infty )$ $3x<3\\Rightarrow x<1\\Rightarrow x\\in (-\\infty ,1)$ Then $x\\in (-\\infty ,1)\\cup (2,\\infty )$ B) $5x+3<18\\Rightarrow x<3\\Rightarrow x\\in (-\\infty ,3)$ $2x-7<5\\Rightarrow x<6\\Rightarrow x\\in (-\\infty ,6)$ Then $x\\in (-\\infty, 3)\\cap (-\\infty", "out with using the Cramer’s Rule to solve this system later on. • Following the $2 \\times 2$ determinant formula, we can write the determinant matrix as: $D = \\begin{vmatrix} 2 & 3 \\\\ { – 3 } & 4 \\end{vmatrix}$ We labeled it “D.” • Putting the constant coefficients from the system in the first column (in place of $x$s), we can write another matrix: $D_{ x } = \\begin{vmatrix} 7 & 3 \\\\ { 15 } & 4 \\end{vmatrix}$ We labeled it “$D_{ x }$” and call it the x-matrix. • Similarly, putting the constants from the system in the second column (in place of $y$s), we can write another matrix: $D_{ y } = \\begin{vmatrix} 2 & 7 \\\\ { – 3 } & 15 \\end{vmatrix}$ We labeled it “$D_{ y }$” and call it the y-matrix. Now, the formula of Cramer’s Rule to solve for the variables $x$ and $y$ are shown below: $x = \\frac{ D_{ x } }{ D } = \\frac{ \\begin{vmatrix} 7 & 3 \\\\ { 15 } & 4 \\end{vmatrix} }{ \\begin{vmatrix} 2 & 3 \\\\ { – 3 } & 4 \\end{vmatrix} }$ $y = \\frac{ D_{ y } }{ D } = \\frac{ \\begin{vmatrix} 2 & 7 \\\\ { – 3 } & 15 \\end{vmatrix} }{ \\begin{vmatrix}", "20 } \\end{vmatrix} $We will use the formula to calculate the determinants of$ 2 \\times 2 $matrices to calculate the matrices$ D $,$ D_{ x } $, and$ D_{ y } $.$ D = \\begin{vmatrix} 5 & 2 \\\\ { – 1 } & { 4 } \\end{vmatrix} = ( 5 )( 4 ) – ( 2 )( -1 ) = 20 + 2 = 22 D_{ x } = \\begin{vmatrix} 10 & 2 \\\\ { 20 } & { 4 } \\end{vmatrix} = ( 10 )( 4 ) – ( 2 )( 20 ) = 40 – 40 = 0 D_{ y } = \\begin{vmatrix} 5 & { 10 } \\\\ { – 1 } & { 20 } \\end{vmatrix} = ( 5 )( 20 ) – ( 10 )( -1 ) = 100 + 10 = 110 $Now, we use the formulas learned in Cramer’s Rule to find the values of the variables:$ x = \\frac{ D_{ x } }{ D } = \\frac{ 0 }{ 22 } = 0 y = \\frac{D_{ y } }{ D } = \\frac{ 110 }{ 22 } = 5 $The solution set of the system is$ (0, 5) $. 2. The first step is to write the determinants of this system of equations, determinant ($ D $),$ x –", "x\\\\y\\\\z \\end{pmatrix} = \\begin{pmatrix}3k\\\\0\\\\k\\end{pmatrix}. \\end{equation*} Now use Gaussian elimination on the augmented matrix (I've skipped the details of how to do this): \\begin{equation*} \\left(\\begin{array}{ccc|c} 6 & -6 & 1 & 3k \\\\ 3 & 6 & 1 & 0 \\\\ 4 & 2 & 1 & k \\end{array}\\right) \\to \\left(\\begin{array}{ccc|c} 1 & 0 & \\frac{2}{9} & \\frac{k}{3}\\\\ 0 & 1 & \\frac{1}{18} & \\frac{-k}{6} \\\\ 0 & 0 & 0 & 0\\end{array}\\right)\\end{equation*} This matrix is equivalent to the system \\begin{eqnarray*} x +\\frac{2}{9}z&=& \\frac{k}{3} \\\\ y + \\frac{1}{18}z &=& \\frac{-k}{6}\\end{eqnarray*} Now we see that given any value of $z$, we find a solution by setting \\begin{eqnarray*} x &=& \\frac{k}{3} - \\frac{2}{9}z\\\\ y &=& \\frac{-k}{6} - \\frac{1}{18}z \\end{eqnarray*} Therefore, for any $k$, the system has infinitely many solutions. The solution set is a line in $\\mathbb{R}^3$, with parametrization \\begin{equation*} \\left(\\frac{k}{3},\\frac{-k}{6}, 0\\right) + t\\left(\\frac{-2}{9}, \\frac{-1}{18}, 1\\right).\\end{equation*} - I think you've made an error. Note that the determinant of the matrix is zero, so the system is degenerate. There is another solution: $(k/3,-k/6,0)$, and another: $(0,-k/4,3k/2)$. In fact, I believe the entire plane containing your solution and these 2 other points solves the system. – Rick Goldstein May 15 '14 at 22:23 Rick, thanks for the correction. All three solutions are collinear, so the solution space is a line. As the matrix", "& 1 & -2\\\\{\\color{red}21} & -3 & 2\\end{vmatrix}}{-16}=\\frac{-32}{-16}=2 \\qquad y = \\frac{\\begin{vmatrix}2 & {\\color{red}-7} & -1\\\\5 & {\\color{red}-3} & -2\\\\1 & {\\color{red}21} & 2\\end{vmatrix}}{-16}=\\frac{48}{-16}=-3 \\qquad z = \\frac{\\begin{vmatrix}2 & 2 & {\\color{red}-7}\\\\5 & 1 & {\\color{red}-3}\\\\1 & -3 & {\\color{red}21}\\end{vmatrix}}{-16}=\\frac{-80}{-16}=5$ So the solution is (2, -3, 5). ### Guided Practice Use Cramer’s Rule to solve the systems below. 1. $2x+5y &=7\\\\x+3y &=2$ 2. $4x-y &=6\\\\-8x+2y &=10$ 3. $x+2y+3z &=8\\\\2x-y+4z &=3\\\\-x-4y+3z &=14$ 1. Find the $det \\ A = \\begin{vmatrix}2 & 5\\\\1 & 3\\end{vmatrix} = 1.$ Now find $x$ and $y$ as follows: $x= \\frac{\\begin{vmatrix}7 & 5\\\\2 & 3 \\end{vmatrix}}{1} = \\frac{11}{1}=11 \\qquad y= \\frac{\\begin{vmatrix}2 & 7\\\\1 & 2 \\end{vmatrix}}{1} = \\frac{-3}{1}=-3, \\quad \\text{solution:} \\ (11,-3)$ 2. Find the $det \\ A= \\begin{vmatrix}4 & -1 \\\\-8 & 2 \\end{vmatrix} = 8-8 = 0.$ Therefore, there is no unique solution. We must use linear combination or the substitution method to determine whether there are an infinite number of solutions or no solutions. Using linear combinations we can multiply the first equation by 2 and get the following: $& \\quad 8x-2y=12 \\\\& \\ \\underline{-8x+2y=10 \\; \\;} \\ \\Rightarrow \\quad \\text{Therefore, there is no solution}.\\\\& \\qquad \\quad \\ \\ 0=22$ 3. Find the $det \\ A=\\begin{vmatrix} 1 & 2 & 3\\\\2 & -1 & 4\\\\-1 & -4 & 3 \\end{vmatrix}=-34.$ Now find", "& -3 & 5 \\\\ 1 & -2 & 6 \\end{vmatrix} }{ \\begin{vmatrix} 2 & 1 & -2 \\\\ 3 & -3 & -1 \\\\ 1 & -2 & 3 \\end{vmatrix} } = \\frac{-26}{-26} = 1$The solution set of the system is$ (1, – 1,1) $. Practice Questions 1. Solve the system of equations shown below using Cramer’s Rule:$ \\begin{align*} { 5x } + 2y &= \\, { 10 } \\\\ { – x } + 4y &= { 20 } \\end{align*} $2. Solve the system of equations shown below using Cramer’s Rule:$ \\begin{align*} 3x – 4y + z &= \\, -5 \\\\ x – y – z &= – 10 \\\\ 6x – 8y + 2z = 10 \\end{align*} $### Answers 1. The first step is to write the determinants of this system of equations, determinant ($ D $),$ x – $determinant ($ D_{ x } $), and the$ y – $determinant ($ D_{ y } $). Let’s use the formula we learned and write them up:$ D = \\begin{vmatrix} 5 & 2 \\\\ { – 1 } & { 4 } \\end{vmatrix} D_{ x } = \\begin{vmatrix} 10 & 2 \\\\ { 20 } & { 4 } \\end{vmatrix} D_{ y } = \\begin{vmatrix} 5 & { 10 } \\\\ { – 1 } & {", "0 & 1 \\end{bmatrix} \\begin{bmatrix} v_1\\\\ v_2\\\\ v_3 \\end{bmatrix} = \\begin{bmatrix} 2v_1\\\\ 2v_2\\\\ 2v_3 \\end{bmatrix} \\end{align} From this matrix equation above we have the following system of equations: (4) \\begin{align} \\quad 3v_1 + v_2 + v_3 = 2v_1 \\\\ \\quad 2v_2 = 2v_2 \\\\ \\quad v_3 = 2v_3 \\end{align} The third equation above implies that $v_3 = 0$. Furthermore, the second equation implies that $v_2$ can be any number, say $v_2 = b \\in \\mathbb{R}$. The first equation implies that $3v_1 + b = 2v_1$ so $v_1 = -b$. Therefore, the corresponding eigenvectors to the eigenvalue $\\lambda_2 = 2$ are $\\begin{bmatrix} -b \\\\ b\\\\ 0 \\end{bmatrix}$. Lastly, consider the the eigenvalue $\\lambda_3 = 3$ and consider the following equation: (5) \\begin{align} Av = \\quad \\begin{bmatrix} 3 & 1 & 1\\\\ 0 & 2 & 0\\\\ 0 & 0 & 1 \\end{bmatrix} \\begin{bmatrix} v_1\\\\ v_2\\\\ v_3 \\end{bmatrix} = \\begin{bmatrix} 3v_1\\\\ 3v_2\\\\ 3v_3 \\end{bmatrix} \\end{align} From this matrix equation above we have the following system of equation: (6) \\begin{align} \\quad 3v_1 + v_2 + v_3 = 3v_1 \\\\ \\quad 2v_2 = 3v_2 \\\\ \\quad v_3 = 3v_3 \\end{align} The second and third equations imply that $v_2 = 0$ and $v_3 = 0$. Plugging this into the first equation and we get that $3v_1 = 3v_1$, so $v_1$ can be any", "equation. \\frac{dP}{dt}=kP\\left(1-\\frac{P}{K}\\right) \\int\\frac{dP}{P\\left(1-\\frac{P}{K}\\right)}=\\int k\\,dt To evaluate the integral on the left side, we simplify the fraction \\frac{1}{P\\left(1-\\frac{P}{K}\\right)}=\\frac{K}{P\\left(K-P\\right)} and then, we decompose the fraction into partial fractions \\frac{K}{P\\left(K-P\\right)}=\\frac{1}{P}+\\frac{1}{K-P} Thus we have \\int\\left(\\frac{1}{P}+\\frac{1}{K-P}\\right)\\,dP=\\int k\\,dt \\ln\\begin{vmatrix}P\\end{vmatrix}-\\ln\\begin{vmatrix}K-P\\end{vmatrix}=kt+C \\ln\\begin{vmatrix}K-P\\end{vmatrix}-\\ln\\begin{vmatrix}P\\end{vmatrix}=-kt-C \\ln\\begin{vmatrix}\\cfrac{K-P}{P}\\end{vmatrix}=-kt-C \\begin{vmatrix}\\cfrac{K-P}{P}\\end{vmatrix}=e^{-kt-C} \\begin{vmatrix}\\cfrac{K-P}{P}\\end{vmatrix}=e^{-C}e^{-kt} \\frac{K-P}{P}=\\pm e^{-C}e^{-kt} Let A=\\pm e^{-C}. \\frac{K-P}{P}=Ae^{-kt} \\frac{K}{P}-1=Ae^{-kt} \\frac{K}{P}=1+Ae^{-kt} \\frac{P}{K}=\\frac{1}{1+Ae^{-kt}} P=\\frac{K}{1+Ae^{-kt}} Therefore, the solution to the logistic equation is P\\left(t\\right)=\\frac{K}{1+Ae^{-kt}} To find A, let t=0 and P\\left(0\\right)=P_0. Then we have P_0=\\frac{K}{1+Ae^0} Noting that e^0=1, and solving for A we get A=\\frac{K-P_0}{P_0} ## Partial differential equations The method of separation of variables is also used to solve a wide range of linear partial differential equations with boundary and initial conditions, such as heat equation, wave equation, Laplace equation and Helmholtz equation. ### Homogeneous case Consider the one-dimensional heat equation.The equation is \\frac{\\partial u}{\\partial t}-\\alpha\\frac{\\partial^{2}u}{\\partial x^{2}}=0 (1) The boundary condition is homogeneous, that is u\\big|_{x=0}=u\\big|_{x=L}=0 (2) Let us attempt to find a solution which is not identically zero satisfying the boundary conditions but with the following property: u is a product in which the dependence of u on x, t is separated, that is: u(x,t) = X(x) T(t). (3) Substituting u back into equation (1) and using the product rule, \\frac{T'(t)}{\\alpha T(t)} = \\frac{X''(x)}{X(x)}. (4) Since the right hand side depends only on x and the left hand side only on t, both sides", "\\end {pmatrix}$$ Taking the corresponding part of the equal matrix: $$\\frac 1x$$ = $$\\frac 12$$ ∴ x = 2 ∴ y = 3 ∴ x = 2 and y = 3 Ans Here, $$\\frac 4x$$ + 3y = 11.................................(1) 3xy - 8 -5x = 0 or, $$\\frac {3xy}{x}$$ - $$\\frac 8x$$ - $$\\frac {5x}{x}$$ = 0 or, -$$\\frac 8x$$ + 3y = 5...................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 11\\\\ 5\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -8 & 3\\\\ \\end {vmatrix}$$ = 4 × 3 - (-8) × 3 = 12 + 24 = 36 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{36}$$$$\\begin {pmatrix} 3 & -3\\\\ 8 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{36} & \\frac {-3}{36}\\\\ \\frac {8}{36} & \\frac {4}{36}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{12} & \\frac {-1}{12}\\\\ \\frac {2}{9} & \\frac {1}{9}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix}", "+0 # question pls help ty +1 120 2 +25 For a certain value of $k$, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 3y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of $k$? Jun 21, 2021 #1 +2252 0 3x + 3y + 9z = 30 -4x + 3y + 5z = 7 7x + 4z = 23 4kx + 4z = 3 If k = 7/4. 4(7/4)x + 4z = 3 7x + 4z = 3, but 7x + 4z = 23?? That is impossible so it doesn't have any solutions. =^._.^= Jun 21, 2021 #2 +26213 +2 For a certain value of $$k$$, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 3y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of $$k$$? $$\\begin{array}{|rcll|} \\hline \\begin{vmatrix} 1 & 1 & 3 \\\\ -4 & 3 & 5 \\\\ k & 0 & 1 \\\\ \\end{vmatrix} &=& 0 \\\\\\\\ k* \\begin{vmatrix} 1 & 3 \\\\ 3 & 5 \\\\ \\end{vmatrix} +1* \\begin{vmatrix} 1 & 1 \\\\ -4 & 3 \\\\ \\end{vmatrix} &=&0 \\\\\\\\ k*(1*5-3*3) + 1*(1*3-(-4)*1) &=& 0 \\\\\\\\ k*(-4) + 1*7 &=& 0 \\\\\\\\ k*(-4)", "= -F_3 \\\\ G_1 \\frac{\\partial x}{\\partial z} + G_2 \\frac{\\partial y}{\\partial z} = - G_3 \\end{matrix}\\right. \\end{align} This system is linear (thinking of $\\frac{\\partial x}{\\partial z}$ and $\\frac{\\partial y}{\\partial z}$ as the \"variables\"). Now if you have taken a course on Linear Algebra, you might have heard of Cramer's Rule. We can apply Cramer's rule to find the partial derivative $\\left ( \\frac{\\partial x}{\\partial z} \\right)_w$ (5) \\begin{align} \\left ( \\frac{\\partial x}{\\partial z} \\right )_w = \\frac{\\det \\begin{vmatrix} -F_3 & F_2\\\\ -G_3 & G_2 \\end{vmatrix}}{\\det \\begin{vmatrix} F_1 & F_2\\\\ G_1 & G_2 \\end{vmatrix}} = \\frac{-F_3G_2 + F_2G_3}{F_1G_2 - F_2G_1} = -\\frac{F_3G_2 - F_2G_3}{F_1G_2 - F_2G_1} \\end{align} Similarly we can obtain $\\left ( \\frac{\\partial x}{\\partial z} \\right)_y$ as: (6) \\begin{align} \\left ( \\frac{\\partial x}{\\partial z} \\right )_y = \\frac{\\det \\begin{vmatrix} F_1 & -F_3\\\\ G_1 & -G_3 \\end{vmatrix}}{\\det \\begin{vmatrix} F_1 & F_2\\\\ G_1 & G_2 \\end{vmatrix}} = \\frac{-F_1G_3 +F_3G_1}{F_1G_2 - F_2G_1} = -\\frac{F_1G_3 - F_3G_1}{F_1G_2 - F_2G_1} \\end{align} We will see the significance of these partial derivatives soon. Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License", "# Solve this PDE $(y+z)u_x+(z+x)u_y+(x+y)u_z=0$ Solve $$(y+z)u_x+(z+x)u_y+(x+y)u_z=0$$ My method is to use characteristics. $$dx/(y+z)=dy/(z+x)=dz/(x+y).$$ Solve this?? $d(x+y)/(x+y+2z)=dz/(x+y)$ which implies $$(x+y-2z)^2(x+y+z)=C_1.$$ And by symmetry, another independent first integral is $$(y+z-2x)^2(x+y+z)=C_2$$ and thus the solution of the PDE is $$u=\\phi((x+y-2z)^2(x+y+z),(y+z-2x)^2(x+y+z)).$$Is this right? Since you have $x'=y+z$, $y'=z+x$, and $z'=x+y$, you have the system $$\\frac{d}{dt} \\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix} = \\begin{pmatrix}0&1&1\\\\1&0&1\\\\1&1&0\\end{pmatrix}\\begin{bmatrix} x \\\\ y \\\\ z \\end{bmatrix}$$ In general, $\\vec{v}'=A\\vec{v} \\implies \\vec{v}=\\mathrm{e}^{At}\\vec{v}_0$. Supposing you can find $\\mathrm{e}^{At}$ (I bet $A$ in this case is easily diagonalizable), then you have $\\frac{du}{dt}=0\\implies u(x(t;\\vec{v}_0),y(t;\\vec{v}_0),z(t;\\vec{v}_0)) = c$. It's difficult to say more without some side conditions, but at this point it's now a standard exercise in characteristics.", "\\dfrac{\\partial y_1}{\\partial x_3} \\\\[3pt] \\dfrac{\\partial y_2}{\\partial x_1} & \\dfrac{\\partial y_2}{\\partial x_2} & \\dfrac{\\partial y_2}{\\partial x_3} \\\\[3pt] \\dfrac{\\partial y_3}{\\partial x_1} & \\dfrac{\\partial y_3}{\\partial x_2} & \\dfrac{\\partial y_3}{\\partial x_3} \\\\[3pt] \\dfrac{\\partial y_4}{\\partial x_1} & \\dfrac{\\partial y_4}{\\partial x_2} & \\dfrac{\\partial y_4}{\\partial x_3} \\\\ \\end{bmatrix}=\\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 0 & 5 \\\\ 0 & 8x_2 & -2 \\\\ x_3\\cos(x_1) & 0 & \\sin(x_1) \\end{bmatrix}.$ This example shows that the Jacobian need not be a square matrix. Example 4. The Jacobian determinant of the function F: R3R3 with components \\begin{align} y_1 &= 5x_2 \\\\ y_2 &= 4x_1^2 - 2 \\sin (x_2x_3) \\\\ y_3 &= x_2 x_3 \\end{align} is $\\begin{vmatrix} 0 & 5 & 0 \\\\ 8 x_1 & -2 x_3 \\cos(x_2 x_3) & -2x_2\\cos(x_2 x_3) \\\\ 0 & x_3 & x_2 \\end{vmatrix} = -8 x_1 \\cdot \\begin{vmatrix} 5 & 0 \\\\ x_3 & x_2 \\end{vmatrix} = -40 x_1 x_2.$ From this we see that F reverses orientation near those points where x1 and x2 have the same sign; the function is locally invertible everywhere except near points where x1 = 0 or x2 = 0. Intuitively, if one starts with a tiny object around the point (1,2,3) and apply F to that object, one will get a resulting object with approximately 40×1×2 = 80 times the volume of", "will make this matrix have a zero determinant. \\begin{align*} \\detname{A} &= \\begin{vmatrix} 2 & 4 \\\\ 3 & k \\end{vmatrix} = 2k - 12 \\end{align*} so $\\detname{A} = 0$ only when $k = 6\\text{.}$ ###### M11. Find a value of $k$ so that the matrix \\begin{equation*} A = \\begin{bmatrix} 1 & 2 & 1\\\\ 2 & 0 & 1 \\\\ 2 & 3 & k \\end{bmatrix} \\end{equation*} has $\\det(A) = 0\\text{,}$ or explain why it is not possible. Solution We find \\begin{align*} \\detname{A} &= \\begin{vmatrix} 1 & 2 & 1 \\\\ 2 & 0 & 1 \\\\ 2& 3 & k \\end{vmatrix} = 7 - 4k\\text{.} \\end{align*} Thus, $\\detname{A} = 0$ only when $k = \\frac{7}{4}\\text{.}$ ###### M15. Given the matrix \\begin{equation*} B=\\begin{bmatrix} 2 - x & 1 \\\\ 4 & 2 - x \\end{bmatrix}\\text{,} \\end{equation*} find all values of $x$ that are solutions of $\\det(B) = 0\\text{.}$ Solution Using the formula for the determinant of a $2\\times 2$ matrix given in Theorem DMST, we have \\begin{align*} \\det{(B)} &= \\begin{vmatrix} 2 - x & 1\\\\ 4 & 2-x \\end{vmatrix} = (2 - x)(2 - x) - 4 = x^2 - 4x = x(x-4) \\end{align*} and thus $\\det{(B)} = 0$ only when $x=0$ or $x=4\\text{.}$ ###### M16. Given the matrix \\begin{equation*} B =\\begin{bmatrix} 4 - x & -4 &", "## Section3.2Planar Systems A first-order linear system of $$n$$ equations and $$n$$ variables is any system that can be written in the form \\begin{align*} \\frac{dx_1}{dt} & = a_{11}(t) x_1(t) + \\cdots + a_{1n}(t) x_n(t) + f_1(t),\\\\ \\frac{dx_2}{dt} & = a_{21}(t) x_1(t) + \\cdots + a_{2n}(t) x_n(t) + f_2(t),\\\\ & \\vdots\\\\ \\frac{dx_n}{dt} & = a_{n1}(t) x_1(t) + \\cdots + a_{nn}(t) x_n(t) + f_n(t). \\end{align*} If each of the coefficients is constant and the functions $$f_i$$ vanish, then we have a homogeneous first-order linear system with constant coefficients, \\begin{align*} \\frac{dx_1}{dt} & = a_{11} x_1 + \\cdots + a_{1n} x_n\\\\ \\frac{dx_2}{dt} & = a_{21} x_1 + \\cdots + a_{2n} x_n,\\\\ & \\vdots\\\\ \\frac{dx_n}{dt} & = a_{n1} x_1 + \\cdots + a_{nn} x_n. \\end{align*} We will concentrate on $$2 \\times 2$$ homogeneous first-order linear systems or planar systems for the time being, \\begin{align} \\frac{dx}{dt} & = a x + b y,\\tag{3.2.1}\\\\ \\frac{dy}{dt} & = c x + d y.\\tag{3.2.2} \\end{align} ### Subsection3.2.1Planar Systems and $$2 \\times 2$$ Matrices We will use linear systems of differential equations to illustrate how we can use systems of differential equations to model how subtances flow back and forth between two or more compartments. Suppose that we have two tanks ($$A$$ and $$B$$) between which a mixture of brine flows (Figure 3.2.1). Tank $$A$$ contains 300 liters", "I find myself getting a totally different answer from the first one I worked out.) In order to find out what the values of x, y and z are, I divided the values of their determinants by D: $x=\\frac{D_x}{D}\\rightarrow\\frac{21}{7}=3$ $y=\\frac{D_y}{D}\\rightarrow\\frac{-7}{7}=-1$ $z=\\frac{D_z}{D}\\rightarrow\\frac{7}{7}=1$ Threrefore: x=3, y=-1, z=1. I know that the solution for y is wrong but I can't see what I did wrong. Thank you for your help so far! 8. Originally Posted by quikwerk Alrighty! Here is the working I used in its unabridged form: I started my working by defining system of equations and Determinant $D$: $2x-y=4$ $3x+y-z=10$ $y+z=3$ $ \\begin{vmatrix} 2 & -1 & 0\\\\ 3 & 1 & -1\\\\ 0 & 1 & 1 \\end{vmatrix} $ Then I defined determinants $D_x$, $D_y$ and $D_z$: $D_x$: $ \\begin{vmatrix} 4 & -1 & 0\\\\ 10 & 1 & -1\\\\ 3 & 1 & 1 \\end{vmatrix} $ $D_y$: $ \\begin{vmatrix} 2 & 4 & {\\color{red}-1}\\\\ 3 & 10 & {\\color{red}1}\\\\ 0 & 3 & 1 \\end{vmatrix} $ $D_z$: $ \\begin{vmatrix} 2 & -1 & 4\\\\ 3 & 1 & 10\\\\ 0 & 1 & 3 \\end{vmatrix} $ I expanded the determinants by their minor elements, in this instance I expanded them all about their first row elements. I used the following template to help me remember the signs", "by some scalar $k$. • matrix $\\boldsymbol{B}_2$ is obtained by multiplying the second row of $\\boldsymbol{A}$ by some scalar $k$. Our goal is to show the following: \\begin{align*} \\mathrm{det}(\\boldsymbol{B}_1)&=k\\cdot\\mathrm{det}(\\boldsymbol{A})\\\\ \\mathrm{det}(\\boldsymbol{B}_2)&=k\\cdot\\mathrm{det}(\\boldsymbol{A}) \\end{align*} The reason why we consider the two cases above is that we are going to compute the determinant by performing cofactor expansion along the first row. Therefore, we must consider the case when we modify the first row as well as the case when we modify any other row. Firstly, the determinant of $\\boldsymbol{B}_1$ is: \\begin{align*} \\det(\\boldsymbol{B}_1)&= ka_{11}\\begin{vmatrix} a_{22}&a_{23}\\\\a_{32}&a_{33}\\\\ \\end{vmatrix}- ka_{12}\\begin{vmatrix} a_{21}&a_{23}\\\\a_{31}&a_{33}\\\\ \\end{vmatrix}+ ka_{13}\\begin{vmatrix} a_{21}&a_{22}\\\\a_{31}&a_{32}\\\\ \\end{vmatrix}\\\\ &=k\\left(a_{11}\\begin{vmatrix} a_{22}&a_{23}\\\\a_{32}&a_{33}\\\\ \\end{vmatrix}- a_{12}\\begin{vmatrix} a_{21}&a_{23}\\\\a_{31}&a_{33}\\\\ \\end{vmatrix}+ a_{13}\\begin{vmatrix} a_{21}&a_{22}\\\\a_{31}&a_{32}\\\\ \\end{vmatrix}\\right)\\\\ &=k\\cdot\\det(\\boldsymbol{A}) \\end{align*} Next, the determinant of $\\boldsymbol{B}_2$ is: $$\\det(\\boldsymbol{B}_2)= a_{11}\\begin{vmatrix} ka_{22}&ka_{23}\\\\a_{32}&a_{33}\\\\ \\end{vmatrix}- a_{12}\\begin{vmatrix} ka_{21}&ka_{23}\\\\a_{31}&a_{33}\\\\ \\end{vmatrix}+ a_{13}\\begin{vmatrix} ka_{21}&ka_{22}\\\\a_{31}&a_{32}\\\\ \\end{vmatrix}\\\\$$ We then use our inductive assumption that $\\mathrm{det}(\\boldsymbol{B})=k\\cdot\\mathrm{det}(\\boldsymbol{A})$ for $2\\times2$ case: \\begin{align*} \\det(\\boldsymbol{B}_2) &=ka_{11}\\begin{vmatrix} a_{22}&a_{23}\\\\a_{32}&a_{33}\\\\ \\end{vmatrix}- ka_{12}\\begin{vmatrix} a_{21}&a_{23}\\\\a_{31}&a_{33}\\\\ \\end{vmatrix}+ ka_{13}\\begin{vmatrix} a_{21}&k_{22}\\\\a_{31}&a_{32}\\\\ \\end{vmatrix}\\\\ &=k\\left(a_{11}\\begin{vmatrix} a_{22}&a_{23}\\\\a_{32}&a_{33}\\\\ \\end{vmatrix}- a_{12}\\begin{vmatrix} a_{21}&a_{23}\\\\a_{31}&a_{33}\\\\ \\end{vmatrix}+ a_{13}\\begin{vmatrix} a_{21}&k_{22}\\\\a_{31}&a_{32}\\\\ \\end{vmatrix}\\right)\\\\ &=k\\cdot{\\det(\\boldsymbol{A})} \\end{align*} This completes the proof. Theorem. # Effect of interchanging two rows on the determinant Let $\\boldsymbol{A}$ be a square matrix. If we swap one row with another row to produce matrix $\\boldsymbol{B}$, then: $$\\mathrm{det}(\\boldsymbol{B}) =-\\mathrm{det}(\\boldsymbol{A})$$ Proof. We prove this by induction. We will consider the two cases below: • case when we interchange the first and second rows. • case when" ]

[ "+0 # question pls help ty +1 120 2 +25 For a certain value of $k$, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 3y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of $k$? Jun 21, 2021 #1 +2252 0 3x + 3y + 9z = 30 -4x + 3y + 5z = 7 7x + 4z = 23 4kx + 4z = 3 If k = 7/4. 4(7/4)x + 4z = 3 7x + 4z = 3, but 7x + 4z = 23?? That is impossible so it doesn't have any solutions. =^._.^= Jun 21, 2021 #2 +26213 +2 For a certain value of $$k$$, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 3y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of $$k$$? $$\\begin{array}{|rcll|} \\hline \\begin{vmatrix} 1 & 1 & 3 \\\\ -4 & 3 & 5 \\\\ k & 0 & 1 \\\\ \\end{vmatrix} &=& 0 \\\\\\\\ k* \\begin{vmatrix} 1 & 3 \\\\ 3 & 5 \\\\ \\end{vmatrix} +1* \\begin{vmatrix} 1 & 1 \\\\ -4 & 3 \\\\ \\end{vmatrix} &=&0 \\\\\\\\ k*(1*5-3*3) + 1*(1*3-(-4)*1) &=& 0 \\\\\\\\ k*(-4) + 1*7 &=& 0 \\\\\\\\ k*(-4)", "+0 +2 198 3 +142 For a certain value of k, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 2y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of k? Aug 3, 2019 #1 +6046 +2 when k=6 the solutions for x,y,z all have 0 in the denominator which is a nono Aug 3, 2019 #2 +8810 +2 For a certain value of k, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 2y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of k? $$2x+2y+6z=20\\\\ \\underline{-4x+2y+5z=7}\\\\ 6x{\\color{blue}{+z}}=13$$ Thanks for the help heureka! $$\\underline{kx+z=3}\\\\ 6x-kx=10\\\\ x=\\frac{10}{6-k}\\\\ \\color{blue}k=6$$ For the value of k = 6, x is indefinite and the functions have no solution. ! Aug 3, 2019 edited by asinus Aug 3, 2019 #3 +23850 +2 For a certain value of k, the system \\begin{align*} x + y + 3z &= 10, \\\\ -4x + 2y + 5z &= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of $$k$$? $$\\begin{array}{|rcll|} \\hline \\begin{vmatrix} 1&1&3\\\\ -4 & 2 & 5\\\\ k & 0 &1 \\\\ \\end{vmatrix} &=& 0 \\\\\\\\ k\\begin{vmatrix} 1&3\\\\ 2 & 5\\\\ \\end{vmatrix}", "$$7 \\leq k\\equiv 3\\pmod 4$$ such that $$X^2-mkX+m\\frac{(kp+1)}{4} = 0$$ has integer roots. Suppose we can always find a solution of the form $$n(4m-p)-1 = dp \\mid mp$$ for some positive integers $$(m,n,d)$$. Then $$d$$ divides $$m$$, so we may replace $$m$$ with $$dm$$. i.e. solving $$n(4dm-p)-1 = dp \\mid (dm)p$$ Let $$n=k-d$$ and $$w=k-2d$$, then this becomes \\begin{align*} n(4md-p)-1 &= dp\\\\ (k-d)(4md-p) &= dp+1\\\\ 4md(k-d) &= kp+1\\\\ m(2d)(2k-2d) &= kp+1\\\\ m(k-w)(k+w) &= kp+1\\\\ mk^2-kp-1 &= mw^2\\\\ m^2k^2-mkp-m &= (mw)^2 \\end{align*} Now since $$m^2k^2-mkp-m$$ is the discriminant $$D$$ of the equation $$X^2-mkX+m\\frac{kq+1}{4}=0$$ This shows that the equation has solutions $$\\frac{mk\\pm \\sqrt{D}}{2} = \\frac{mk \\pm mw}{2}$$ which must be integral (either $$m$$ even or $$k$$ and $$w$$ can be shown to be odd). We also require $$k\\equiv 3\\pmod 4$$, which can be shown as follows: \\begin{align*} n(4m-p)-1 &= dp\\\\ (k-d)(4m-p)-1 &= dp\\\\ (k-d)(-1)-1 &\\equiv d \\pmod 4\\\\ d-k-1 &\\equiv d \\pmod 4\\\\ k &\\equiv 3 \\pmod 4 \\end{align*} Since $$k=n+d$$ it is always positive, but there is a small problem that it might be equal to $$3$$.", "will make this matrix have a zero determinant. \\begin{align*} \\detname{A} &= \\begin{vmatrix} 2 & 4 \\\\ 3 & k \\end{vmatrix} = 2k - 12 \\end{align*} so $\\detname{A} = 0$ only when $k = 6\\text{.}$ ###### M11. Find a value of $k$ so that the matrix \\begin{equation*} A = \\begin{bmatrix} 1 & 2 & 1\\\\ 2 & 0 & 1 \\\\ 2 & 3 & k \\end{bmatrix} \\end{equation*} has $\\det(A) = 0\\text{,}$ or explain why it is not possible. Solution We find \\begin{align*} \\detname{A} &= \\begin{vmatrix} 1 & 2 & 1 \\\\ 2 & 0 & 1 \\\\ 2& 3 & k \\end{vmatrix} = 7 - 4k\\text{.} \\end{align*} Thus, $\\detname{A} = 0$ only when $k = \\frac{7}{4}\\text{.}$ ###### M15. Given the matrix \\begin{equation*} B=\\begin{bmatrix} 2 - x & 1 \\\\ 4 & 2 - x \\end{bmatrix}\\text{,} \\end{equation*} find all values of $x$ that are solutions of $\\det(B) = 0\\text{.}$ Solution Using the formula for the determinant of a $2\\times 2$ matrix given in Theorem DMST, we have \\begin{align*} \\det{(B)} &= \\begin{vmatrix} 2 - x & 1\\\\ 4 & 2-x \\end{vmatrix} = (2 - x)(2 - x) - 4 = x^2 - 4x = x(x-4) \\end{align*} and thus $\\det{(B)} = 0$ only when $x=0$ or $x=4\\text{.}$ ###### M16. Given the matrix \\begin{equation*} B =\\begin{bmatrix} 4 - x & -4 &", "-3 \\end{vmatrix}$$ $$224$$ 20) $$\\begin{vmatrix} 6 & -1 & 2\\\\ -4 & -3 & 5\\\\ 1 & 9 & -1 \\end{vmatrix}$$ 21) $$\\begin{vmatrix} 5 & 1 & -1\\\\ 2 & 3 & 1\\\\ 3 & -6 & -3 \\end{vmatrix}$$ $$15$$ 22) $$\\begin{vmatrix} 1.1 & 2 & -1\\\\ -4 & 0 & 0\\\\ 4.1 & -0.4 & 2.5 \\end{vmatrix}$$ 23) $$\\begin{vmatrix} 2 & -1.6 & 3.1\\\\ 1.1 & 3 & -8\\\\ -9.3 & 0 & 2 \\end{vmatrix}$$ $$-17.03$$ 24) $$\\begin{vmatrix} -\\frac{1}{2} & \\frac{1}{3} & \\frac{1}{4}\\\\ \\frac{1}{5} & -\\frac{1}{6} & \\frac{1}{7}\\\\ 0 & 0 & \\frac{1}{8} \\end{vmatrix}$$ For the exercises 25-34, solve the system of linear equations using Cramer’s Rule. 25) \\begin{align*} 2x-3y &= -1\\\\ 4x+5y &= 9 \\end{align*} $$(1,1)$$ 26) \\begin{align*} 5x-4y &= 2\\\\ -4x+7y &= 6 \\end{align*} 27) \\begin{align*} 6x-3y &= 2\\\\ -8x+9y &= -1 \\end{align*} $$\\left (\\dfrac{1}{2}, \\dfrac{1}{3} \\right )$$ 28) \\begin{align*} 2x+6y &= 12\\\\ 5x-2y &= 13 \\end{align*} 29) \\begin{align*} 4x+3y &= 23\\\\ 2x-y &= -1 \\end{align*} $$(2,5)$$ 30) \\begin{align*} 10x-6y &= 2\\\\ -5x+8y &= -1 \\end{align*} 31) \\begin{align*} 4x-3y &= -3\\\\ 2x+6y &= -4 \\end{align*} $$\\left (-1, -\\dfrac{1}{3} \\right )$$ 32) \\begin{align*} 4x-5y &= 7\\\\ -3x+9y &= 0 \\end{align*} 33) \\begin{align*} 4x+10y &= 180\\\\ -3x-5y &= -105 \\end{align*} $$(15,12)$$ 34) \\begin{align*} 8x-2y &= -3\\\\ -4x+6y &= 4 \\end{align*} For the exercises 35-44, solve the system", "a {\\mathbf x}_{\\text{Re}}(t) + b{\\mathbf x}_{\\text{Im}}(t). \\end{align*} Notice that the solutions \\begin{equation*} {\\mathbf x}(t) = c_1 \\begin{pmatrix} \\cos \\beta t \\\\ - \\sin \\beta t \\end{pmatrix} + c_2 \\begin{pmatrix} \\sin \\beta t\\\\ \\cos \\beta t \\end{pmatrix} \\end{equation*} are periodic with period $2 \\pi / \\beta\\text{.}$ This type of system is called a center. ##### Example3.26 Consider the initial value problem \\begin{align*} \\frac{dx}{dt} \\amp = 2y\\\\ \\frac{dy}{dt} \\amp = -2x\\\\ x(0) \\amp = 1\\\\ y(0) \\amp = 2. \\end{align*} The eigenvalues of this system are $\\lambda = \\pm 2i\\text{.}$ Therefore, the general solution to the system is \\begin{align*} x(t) \\amp = c_1 \\cos 2t + c_2 \\sin 2t\\\\ y(t) \\amp = - c_1 \\sin 2t + c_2 \\cos 2t. \\end{align*} Using the initial conditions to solve for $c_1$ and $c_2\\text{,}$ the solution to our initial value problem is \\begin{align*} x(t) \\amp = \\cos 2t + 2 \\sin 2t\\\\ y(t) \\amp = - \\sin 2t + 2 \\cos 2t. \\end{align*} The phase portrait is a circle of radius 2 about the origin (Figure 3.27). # Subsection3.4.2Spiral Sinks and Sources¶ permalink Now let us consider the system ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ where \\begin{equation*} A = \\begin{pmatrix} \\alpha & \\beta \\\\ - \\beta & \\alpha \\end{pmatrix} \\end{equation*} and $\\alpha$ and $\\beta$ are nonzero real numbers. The characteristic equation of $A$", "If you want to align the formulas you can use the align environment. The align environment is a mathematical environment itself, so it doesn’t have to be put inside $..$ or $..$. LaTeX aligns the lines at & and a line is ended with \\\\. If you use align* the equation won’t be numbered. \\begin{align} f(x) &= x^2 + 1 \\\\ g(x) &= x^3 - 3 \\end{align} \\begin{align} f(x) &= x^2 + 1 \\\\ g(x) &= x^3 - 3 \\end{align} ## Matrices Matrices can be made with the pmatrix environment. For different brackets you can use bmatrix or vmatrix. $\\begin{pmatrix} a & b & c \\\\ d & e & f \\\\ g & h & i \\end{pmatrix} \\qquad \\begin{bmatrix} a & b & c \\\\ d & e & f \\\\ g & h & i \\end{bmatrix} \\qquad \\begin{vmatrix} a & b & c \\\\ d & e & f \\\\ g & h & i \\end{vmatrix}$ \\begin{pmatrix} a & b & c \\\\ d & e & f \\\\ g & h & i \\end{pmatrix} \\begin{bmatrix} a & b & c \\\\ d & e & f \\\\ g & h & i \\end{bmatrix} \\begin{vmatrix} a & b & c \\\\ d & e & f \\\\ g & h & i \\end{vmatrix} In generic matrices", "# KKT stationarity condition in simple SDP This question is part of my continuing efforts to answer my original question in A puzzling KKT for LMI vs. scalar constraint. Consider an SDP \\begin{align} &\\min - \\text{Tr}(P)\\\\ &\\ \\ \\text{s.t.} \\begin{pmatrix} A^TP + PA +C^TPC + Q & PB + C^TPD \\\\ B^T P + D^TPC& R+D^TPD \\end{pmatrix}\\succeq0, P\\succeq0 \\end{align} In a paper, they define a dual variable as \\begin{align} Z= \\begin{pmatrix} S&U^T&0\\\\ U&T&0\\\\ 0&0&W \\end{pmatrix}, \\end{align} and claim that \\begin{align} AS+SA^T + CSC^T + BU + DUC^T + U^TB^T + CU^TD^T + D^TTD + W + I=0. \\end{align} By computing the KKT stationarity condition, we get that the Trace of this equation is zero rather than its argument. What am I missing? 1. How will the equation change if we change to objective to $$-\\text{Tr}(R+D^TPD)$$. 2. Is it a standard SDP? I learned that the objective should be expressed as $$c^Tx$$. You can write the primal as: \\begin{align} &\\min - \\text{Tr}(P)\\\\ &\\ \\ \\text{s.t.} \\begin{pmatrix} A^TP + PA +C^TPC + Q & PB + C^TPD & 0 \\\\ B^T P + D^TPC& R+D^TPD & 0 \\\\ 0 & 0 & P \\end{pmatrix}\\succeq0 \\end{align} The Lagrangian is: \\begin{align} L(P,Z) = - \\text{Tr}(P) - \\langle \\begin{pmatrix} A^TP + PA +C^TPC + Q & PB + C^TPD & 0 \\\\ B^T", "+ b \\begin{pmatrix} \\sin \\beta t \\\\ \\cos \\beta t \\end{pmatrix}\\\\ & = a {\\mathbf x}_{\\text{Re}}(t) + b{\\mathbf x}_{\\text{Im}}(t). \\end{align*} Notice that the solutions \\begin{equation*} {\\mathbf x}(t) = c_1 \\begin{pmatrix} \\cos \\beta t \\\\ - \\sin \\beta t \\end{pmatrix} + c_2 \\begin{pmatrix} \\sin \\beta t\\\\ \\cos \\beta t \\end{pmatrix} \\end{equation*} are periodic with period $$2 \\pi / \\beta\\text{.}$$ This type of system is called a center. Consider the initial value problem \\begin{align*} \\frac{dx}{dt} \\amp = 2y\\\\ \\frac{dy}{dt} \\amp = -2x\\\\ x(0) \\amp = 1\\\\ y(0) \\amp = 2. \\end{align*} The eigenvalues of this system are $$\\lambda = \\pm 2i\\text{.}$$ Therefore, the general solution to the system is \\begin{align*} x(t) \\amp = c_1 \\cos 2t + c_2 \\sin 2t\\\\ y(t) \\amp = - c_1 \\sin 2t + c_2 \\cos 2t. \\end{align*} Using the initial conditions to solve for $$c_1$$ and $$c_2\\text{,}$$ the solution to our initial value problem is \\begin{align*} x(t) \\amp = \\cos 2t + 2 \\sin 2t\\\\ y(t) \\amp = - \\sin 2t + 2 \\cos 2t. \\end{align*} The phase portrait is a circle of radius 2 about the origin (Figure 3.4.3). ### Subsection3.4.2Spiral Sinks and Sources Now let us consider the system $${\\mathbf x}' = A {\\mathbf x}\\text{,}$$ where \\begin{equation*} A = \\begin{pmatrix} \\alpha & \\beta \\\\ - \\beta & \\alpha \\end{pmatrix} \\end{equation*} and $$\\alpha$$ and", "= \\frac{-x}{3}+\\frac{4}{3} && \\frac{-y}{-1}=\\frac{-5x}{-1}+\\frac{4}{-1}\\\\ & \\boxed{y = \\frac{-1}{3}x+\\frac{4}{3}} && \\boxed{y=5x-4}\\end{align*} There is one point of intersection (1, 1). The system is consistent and independent. Summary In this lesson you have learned that a \\begin{align*}2 \\times 2\\end{align*} system of linear equations consists of two equations with two variables. You learned to solve these systems by graphing and you learned that the resulting lines may a) intersect at one point, b) coincide or c) be parallel. You also learned that a system that resulted in one point of intersection is consistent and independent. A system that resulting in lines that coincided is consistent and dependent. A system that resulted in two parallel lines is inconsistent. You also learned that a system of equations that was consistent had equivalent systems that could be determined by adding the original equations of the system. The final concept that you learned was how to solve the systems using technology. Problem Set Without graphing, determine whether the system is consistent and independent, consistent and dependent, or inconsistent. 1. . \\begin{align*}\\begin{Bmatrix} 2x+3y=6 \\\\ 6x+9y=18 \\end{Bmatrix}\\end{align*} 1. . \\begin{align*}\\begin{Bmatrix} 2x-y=-14 \\\\ 12x-6y=-11 \\end{Bmatrix}\\end{align*} 1. . \\begin{align*}\\begin{Bmatrix} 3x+2y=14 \\\\ 5x-y=6 \\end{Bmatrix}\\end{align*} 1. . \\begin{align*}\\begin{Bmatrix} 2x+3y=-12 \\\\ 3x-y=3 \\end{Bmatrix}\\end{align*} 1. . \\begin{align*}\\begin{Bmatrix} 20x+15y=-30 \\\\ 4x+3y=18 \\end{Bmatrix}\\end{align*} For each of the following linear systems that are consistent, write equivalent systems:", "assume that no backward reactions occure and that the final products $D,E$ are removed immediatley after producion, this means $k_{-2}=0$ and $k_{-4}=0$ and symlifying the system further by assuming $k_{-1}=k_{-3}=0$ the dynamical system reads \\begin{align} \\frac{dX}{dt}&=1-(b+1)X+aX^2Y\\\\ \\frac{dY}{dt}&=bX-aX^2Y \\tag{1} \\end{align} We obtain the fixed point by setting the system of simultaneous nonlinear differential equations (1) to zero. \\begin{align} 1-(b+1)X+aX^2Y&=0\\\\ bX-aX^2Y&=0 \\tag{2} \\end{align} Adding the equations of (2) yields $x=1$. Plugging this result into one of the equations of (2) gives $y=b/a$. So there is only one singular point \\begin{align} (x^*,y^*)=\\left(1,\\frac{b}{a}\\right) \\end{align} What about the stability of this fixed point? To answer this we compute the Jacobian \\begin{align} Df(x,y)=\\begin{pmatrix}1-b + 2aXY & aX^2\\\\ b-2aXY &aX^2 \\end{pmatrix} \\end{align} and plug in the values for the fixed point. \\begin{align} Df(x,y)=\\begin{pmatrix}1-b & a\\\\ -b &a \\end{pmatrix} \\end{align} To determine the solution of the charakteristic polynomial we compute the trace $\\tau$ and the determinant $\\Delta$ of the Jacobian \\begin{align} \\tau &=trace(Df(1,b/a))=1-b+a\\\\ \\Delta&=det(Df(1,b/a))=a \\end{align} Now notice, for $\\tau >0$ the fixed point is unstable, while for $\\tau <0$ it is stable, changing the stability at $\\tau =0$. So we find the critical parameter values by the equation \\begin{align} \\tau &=1-b+a=0\\\\ \\end{align} Parameterspace for the Brusselator If you plug in the point $(a,b)=(0,0)$ of the parameterspace, then $\\tau$ becomes positive and is therefore unstable at this", "points would be when $x'=0$ and $y'=0$: \\begin{align}0=(1+x)\\sin(y)\\end{align} \\begin{align}0=1-x-\\cos(y)\\end{align} From $(1)$, $\\sin(y)$ would be $0$ when $y=n\\pi$ where $n=0, 1, 2, 3,...$. Since there is a $\\cos(y)$ in $(2)$, the previous equation can be split into two cases: $y=2n\\pi$ where $n=0, 1, 2, 3,....$ and $y=n\\pi$ where $n=1, 3, 5,....$. Then in case 1 $\\cos(y)=1$ and in case 2 $cos(y)=-1$. Plugging both of these into $(2)$ gives the critical points: \\begin{align}(0, 2n\\pi) \\quad where \\quad n=0, 1, 2, 3,...\\end{align} \\begin{align}(2, n\\pi) \\quad where \\quad n=1, 3, 5,...\\end{align} The Jacobian matrix would be: \\begin{align}J=\\begin{pmatrix}F_{x}&F_{y}\\\\G_{x}&G_{y}\\end{pmatrix}=\\begin{pmatrix}\\sin(y)&-\\cos(y)(1+x)\\\\-1&\\sin(y)\\end{pmatrix}\\end{align} At $(0, 2n\\pi)$: \\begin{align}J=\\begin{pmatrix}0&1\\\\-1&0\\end{pmatrix}\\end{align} This gives eigenvalues $r_{1}=i$ and $r_{2}=-i$. Since $r_{1}, r_{2}=\\lambda\\pm i\\mu$ where $\\lambda=0$ the nonlinear system would be an indeterminate center or spiral point. At $(2, n\\pi)$: \\begin{align}J=\\begin{pmatrix}0&-3\\\\-1&0\\end{pmatrix}\\end{align} The eigenvalues would be $r_{1}=\\sqrt{3}$ and $r_{2}=-\\sqrt{3}$. Since $r_{2}<0<r_{1}$ the nonlinear system would be an unstable saddle point. 5 Term Test 2 / Re: TT2--P3M « on: March 28, 2018, 10:43:24 AM » I found my mistake. I think you're correct. 6 Term Test 2 / Re: TT2--P3M « on: March 28, 2018, 01:26:54 AM » I got something different for part (b). I got that $u_{2}'$ should be: \\begin{align}u_{2}'=\\frac{-6e^{3t}}{e^t+1}\\end{align} Therefore $u_{2}$ would be: \\begin{align}u_{2}(t) = -3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2}\\end{align} Then the general solution would be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these", "0 & 1 \\end{bmatrix} \\begin{bmatrix} v_1\\\\ v_2\\\\ v_3 \\end{bmatrix} = \\begin{bmatrix} 2v_1\\\\ 2v_2\\\\ 2v_3 \\end{bmatrix} \\end{align} From this matrix equation above we have the following system of equations: (4) \\begin{align} \\quad 3v_1 + v_2 + v_3 = 2v_1 \\\\ \\quad 2v_2 = 2v_2 \\\\ \\quad v_3 = 2v_3 \\end{align} The third equation above implies that $v_3 = 0$. Furthermore, the second equation implies that $v_2$ can be any number, say $v_2 = b \\in \\mathbb{R}$. The first equation implies that $3v_1 + b = 2v_1$ so $v_1 = -b$. Therefore, the corresponding eigenvectors to the eigenvalue $\\lambda_2 = 2$ are $\\begin{bmatrix} -b \\\\ b\\\\ 0 \\end{bmatrix}$. Lastly, consider the the eigenvalue $\\lambda_3 = 3$ and consider the following equation: (5) \\begin{align} Av = \\quad \\begin{bmatrix} 3 & 1 & 1\\\\ 0 & 2 & 0\\\\ 0 & 0 & 1 \\end{bmatrix} \\begin{bmatrix} v_1\\\\ v_2\\\\ v_3 \\end{bmatrix} = \\begin{bmatrix} 3v_1\\\\ 3v_2\\\\ 3v_3 \\end{bmatrix} \\end{align} From this matrix equation above we have the following system of equation: (6) \\begin{align} \\quad 3v_1 + v_2 + v_3 = 3v_1 \\\\ \\quad 2v_2 = 3v_2 \\\\ \\quad v_3 = 3v_3 \\end{align} The second and third equations imply that $v_2 = 0$ and $v_3 = 0$. Plugging this into the first equation and we get that $3v_1 = 3v_1$, so $v_1$ can be any", "&= 7, \\\\ kx + z &= 3 \\end{align*} has no solutions. What is this value of $$k$$? $$\\begin{array}{|rcll|} \\hline \\begin{vmatrix} 1&1&3\\\\ -4 & 2 & 5\\\\ k & 0 &1 \\\\ \\end{vmatrix} &=& 0 \\\\\\\\ k\\begin{vmatrix} 1&3\\\\ 2 & 5\\\\ \\end{vmatrix} + \\begin{vmatrix} 1&1\\\\ -4 & 2\\\\ \\end{vmatrix} &=& 0 \\\\\\\\ k*(5-6)+2+4 &=& 0\\\\ -k &=& -6 \\\\ \\mathbf{k} &=& \\mathbf{6} \\\\ \\hline \\end{array}$$ Aug 3, 2019", "coordinate system, and $$oxyz$$ is the camera coordinate system attached to the camera. $$O$$ and $$o$$ represent the origin of each. The vector is expressed as being independent of the coordinate system. I will write $$\\vec{Oo}=\\vec{t}$$. Then $$P$$ can be expressed as \\begin{align*} \\vec{OP}=\\vec{Oo}+\\vec{oP}=\\vec{t}+\\vec{oP}=\\vec{t}+x\\vec{e_1}+y\\vec{e_2}+z\\vec{e_3} \\end{align*} as a point in space. If $$\\vec{e_1}$$, $$\\vec{e_2}$$, and $$\\vec{e_3}$$ are the basis vectors of the coordinate system $$oxyz$$, the coordinates of the $$P$$ point in the coordinate system $$oxyz$$ are $$(x, y, z)$$. On the other hand, the coordinates of the $$P$$ point in the coordinate system $$OXYZ$$ are assumed to be $$(X, Y, Z)$$. Here, if all the vectors are displayed as components using the coordinate system $$OXYZ$$, the result is \\begin{align*} \\vec{OP}=\\begin{pmatrix} X \\\\ Y \\\\ Z \\end{pmatrix} .\\end{align*} Also assume that it is \\begin{align*} \\vec{t}=\\begin{pmatrix} t_1 \\\\ t_2 \\\\ t_3 \\end{pmatrix}, \\vec{e_1}=\\begin{pmatrix} e_{11} \\\\ e_{12} \\\\ e_{13} \\end{pmatrix}, \\vec{e_2}=\\begin{pmatrix} e_{21} \\\\ e_{22} \\\\ e_{23} \\end{pmatrix}, \\vec{e_3}=\\begin{pmatrix} e_{31} \\\\ e_{32} \\\\ e_{33} \\end{pmatrix} .\\end{align*} Then \\begin{align*} \\vec{OP}=\\vec{t}+x\\vec{e_1}+y\\vec{e_2}+z\\vec{e_3} \\end{align*} becomes \\begin{align*} \\begin{pmatrix} X \\\\ Y \\\\ Z \\end{pmatrix}= \\begin{pmatrix} t_1 \\\\ t_2 \\\\ t_3 \\end{pmatrix}+ \\begin{pmatrix} e_{11} \\\\ e_{12} \\\\ e_{13} \\end{pmatrix}x+ \\begin{pmatrix} e_{21} \\\\ e_{22} \\\\ e_{23} \\end{pmatrix}y+ \\begin{pmatrix} e_{31} \\\\ e_{32} \\\\ e_{33} \\end{pmatrix}z= \\begin{pmatrix} e_{11} & e_{21} & e_{31} \\\\ e_{12} & e_{22} & e_{32} \\\\ e_{13} &", "x \\\\ y \\end{pmatrix}. \\end{equation*} If we let $I$ denote the $2 \\times 2$ identity matrix, \\begin{equation*} I = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix}, \\end{equation*} we can rewrite this equation in the form $$(A - \\lambda I) \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix}. \\label{equation-linear01-characteristic}\\tag{3.1.2}$$ We know that $A - \\lambda I$ is a $2 \\times 2$ matrix and that this system will only have nonzero solutions if $\\det(A - \\lambda I) = 0\\text{.}$ In our example, \\begin{align*} \\det(A - \\lambda I) & = \\det\\begin{pmatrix} 1 - \\lambda & 2 \\\\ 4 & 3 - \\lambda \\end{pmatrix} \\\\ & = (1 - \\lambda) (3 - \\lambda ) - 8\\\\ & = \\lambda^2 - 4\\lambda - 5\\\\ & = (\\lambda - 5)(\\lambda +1 ). \\end{align*} Thus, $\\lambda = 5$ or $-1\\text{.}$ To see this from a different perspective, we will rewrite equation (3.1.2) as \\begin{align*} x + 2 y & = \\lambda x\\\\ 4 x + 3 y & = \\lambda y. \\end{align*} This system is equivalent to \\begin{align*} (1 - \\lambda) x + 2 y & = 0\\\\ 4 x + (3 - \\lambda) y & = 0 \\end{align*} which can be reduced to \\begin{align*} (1 - \\lambda) x + 2 y & = 0\\\\ (\\lambda^2 - 4\\lambda - 5)", "## Change of variables, from dxdy to dudv Suppose we have a function $f(x, y)$ and we rewrite it as $f(u(x, y), v(x, y))$, i.e. perform a change of variables, then the caculation of area using double integral can be carried out with these new variables $u, v$. But we have to figure out the relation between $dudv$ and $dxdy$. Here is the thing, with the variables $x, y$, $dxdy$ is the area of the parallelogram between the vectors $(dx, 0)$ and $(0, dy)$, under the new variables, these vectors are transformed into two new vectors $(u_x dx, v_x dx)$, $(u_y dy, v_y dy)$, and $dudv$ is the area of the parallelogram between these two new vectors. $\\begin{array}{cc} (dx, 0) & (u_x dx, v_x dx) \\\\ (0, dy) & (u_y dy, v_y dy) \\end{array}$ To get the area of this parallelogram, we need the determinant of this matrix: \\begin{align*} M &= \\begin{pmatrix} u_x dx & u_y dy \\\\ v_x dx & v_y dy \\end{pmatrix} \\\\ &= \\begin{pmatrix} u_x & u_y \\\\ v_x & v_y \\end{pmatrix} \\begin{pmatrix} dx & 0 \\\\ 0 & dy \\end{pmatrix} \\\\ &= J \\begin{pmatrix} dx & 0 \\\\ 0 & dy \\end{pmatrix}, \\end{align*} where $J$ stands for the Jacobian matrix of $f$. So the determinant is \\begin{align*} \\det(M) &= \\det(J) dxdy \\end{align*} Remember in", "to $x_2$, and then to $x_3$, and so on, until $x_n$, and then back to $x_1$. For example, for the unit cube in 3 dimensions, the curl is counterclockwise (when viewed opposite the origin). However, applying a matrix with a single negative rescaling and thus a determinant of $-1$, one of the sides of the unit cube is flipped in the opposite direction. This causes the curl to reverse its orientation from counterclockwise to clockwise. However, applying a matrix with two negative rescalings and thus a determinant of $1$, two of the sides of the unit cube are flipped in the opposite direction, and the curl maintains its counterclockwise orientation. Practice Problems Decompose the following matrices into products of rescalings and shears. (You can view the solution by clicking on the problem.) \\begin{align*} 1) \\hspace{.5cm} \\begin{pmatrix} 2 & 1 \\\\ 0 & 1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} & \\mbox{answers may vary; one correct answer is} \\\\ & \\begin{pmatrix} 1 & 1 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 2 & 0 \\\\ 0 & 1 \\end{pmatrix} \\end{align*} \\begin{align*} 2) \\hspace{.5cm} \\begin{pmatrix} 3 & 0 \\\\ 1 & 2 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} & \\mbox{answers may vary; one correct answer is} \\\\ & \\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{3} & 1 \\end{pmatrix} \\begin{pmatrix} 3 & 0 \\\\ 0 & 2", "$$3$$. It is easy to calculate the first few values of this sum. Namely, \\begin{align} f(0) &= 0 \\\\ f(1) &= 1 \\\\ f(2) &= 5 \\\\ f(3) &= 14 \\end{align} I claim that these four points are sufficient to uniquely determine $$f$$. Since $$\\deg f = 3$$, we have in general that $$f(x)=\\sum\\limits_{k=0}^3 c_kx^k$$ which when be combined with the four computed values above results in the following system of equations: $$\\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 1 & 1 & 1 & 1 \\\\ 1 & 2 & 4 & 8 \\\\ 1 & 3 & 9 & 27 \\end{pmatrix} \\begin{pmatrix} c_0 \\\\ c_1 \\\\ c_2 \\\\ c_3 \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 1 \\\\ 5 \\\\ 14 \\end{pmatrix}$$ Note that $$c_0 = 0$$ trivially, and so we can help ourselves by writing the reduced matrix equation for the other three coefficients as $$\\begin{pmatrix} 1 & 1 & 1 \\\\ 2 & 4 & 8 \\\\ 3 & 9 & 27 \\end{pmatrix} \\begin{pmatrix} c_1 \\\\ c_2 \\\\ c_3 \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ 5 \\\\ 14 \\end{pmatrix}$$ Call the above matrix $$V$$. This matrix is a Vandermonde matrix which has a well-known determinant \\begin{align} \\det(V) &= 1\\cdot 2\\cdot 3\\cdot(2-1)(3-1)(3-2) \\\\ &\\neq 0 \\end{align} Because its determinant is nonzero, the matrix is invertible, and", "and then conjugating by $M_B$ is a roundabout way of doing that right-multiplication. – Marc van Leeuwen Nov 9 '12 at 13:49 The answer is $\\begin{bmatrix} T(e_1) & T(e_2) \\end{bmatrix}$ where $e_1 = \\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}$ and $e_2 = \\begin{bmatrix} 0 \\\\ 1 \\end{bmatrix}$. We just need to figure out what $T(e_1)$ and $T(e_2)$ are. Let $v_1 = \\begin{bmatrix} -1 \\\\ -2 \\end{bmatrix}$ and $v_2 = \\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix}$. Notice that $$e_1 = 3 v_1 + 2 v_2$$ and $$e_2 = -2 v_1 - v_2.$$ These equations allow us to compute $T(e_1)$ and $T(e_2)$. - Let's write $$\\begin{bmatrix} a & b \\\\ c & d\\end{bmatrix}$$ for the matrix we want to find. Then the two given equations read \\begin{align*} -a - 2b &= 17\\\\ -c - 2d &= 11\\\\[3mm] 2a + 3b &= -30\\\\ 2c + 3d &= -17 \\end{align*} We have to solve this system, so let's first look at the equations for $a$ and $b$ \\begin{align*} -a - 2b &= 17\\\\ 2a + 3b &= -30 \\end{align*} Adding the first two times to the second gives \\begin{align*} -a - 2b &= 17\\\\ -b &= 4 \\end{align*} Hence $b = -4$ and $a = -9$. The second system reads \\begin{align*} -c - 2d &= 11\\\\ 2c + 3d &= -17 \\end{align*} Adding again," ]

[ "# Thread: no of solutions to w+x+y+z=8? 1. ## no of solutions to w+x+y+z=8? How many solutions in non-negative integers are there to the equation w +x+y +z = 8? and How many solutions in positive integers? I think for the first part i am just looking at the number of 8 element multisubsets of a 4 element set e.g (8+4-1) choose 8? For the second part I am a bit confused, thanks for any help 2. For the second part think this way. Put a one in each variable. That leaves four to put into four places. $\\dbinom{4+4-1}{4}$. 3. Ah thank you that makes sense 4. Hello, hmmmm! How many solutions in non-negative integers are there to the equation: . $w +x+y +z \\:=\\: 8$ ? Place 8 objects in a row, leaving a space before, after and between them. . . $ \\_\\;\\circ\\;\\_\\;\\circ\\;\\_\\;\\circ\\;\\_\\;\\circ\\;\\_\\;\\c irc\\;\\_\\; \\circ\\;\\_\\;\\circ\\;\\_\\; \\circ\\;\\_$ Place 3 \"dividers\" in any of the 9 spaces. . . $\\circ \\circ | \\circ \\circ \\circ| \\circ \\circ |\\circ\\;\\text{ represents }\\,2 + 3 + 2 + 1$ . . $\\circ \\circ || \\circ \\circ \\circ \\circ \\circ |\\circ\\;\\text{ represents }\\,2 + 0 + 5 + 1$ . . $|\\circ \\circ \\circ | \\circ \\circ \\circ \\circ \\circ|\\;\\text{ represents }\\,0 + 3 + 5 + 0$ . . $\\circ \\circ", "360^\\circ = 20^\\circ$ The pie chart of the above data is shown below: 1 Comment • Anonymous having good solution", ",\\quad 3x-6$$ 14. 14 Find x so that the following are three terms of an arithmetic sequence. $$x-3,\\quad 2x+7, \\quad 5x-3$$ Good test question! $$e=mc^2$$", "$$\\displaystyle{s}_{{1}}=$$ $${s}_{{2}}=$$ $${s}_{{3}}=$$ b) Classify the sequence as arithmetic, geometric, both, or neither. arithmeticgeometric bothneither If arithmetic, give d, if geometric, give r, if both, give d followed by r. (If both, enter your answers as a comma-separated list. If neither, enter NONE.) Polynomial arithmetic", "### Home > CCAA8 > Chapter cca10 > Lesson cca10.2.3 > Problem10-56 10-56. How many terms are in the arithmetic sequence shown below? Homework Help ✎ $15,\\ 7,\\ −1,\\ −9,...,\\ −225$ Write an equation for the arithmetic sequence using $a_n−m\\left(n-1\\right)+a_1$ Let $a$ sub $n=−225$ and solve for $n$.", "it out on your calculator for $a=1^\\circ$, then for fractions of a degree such as $a=.1^\\circ$, $a=.01^\\circ$, and so on.", "you stop reading right now and pick up a pencil and try to do it yourself. Hint: start with $\\cos {72}^{\\circ}$. As I write this section I get the message: This answer is getting pretty long. Consider making it more concise so students aren't intimidated. Please don't be intimidated, this part is all optional. It turns out all angles that are multiples of three degrees are constructible with straightedge and compass, and thus they all have a trig functions that are made out of the usual operations and square roots. It's not too hard to solve for those. Here's our plan. We're going to find $\\cos {72}^{\\circ}$ and $\\cos {144}^{\\circ}$ by solving $\\cos \\left(2 x\\right) = \\cos \\left(3 x\\right)$. Then we note $\\cos {36}^{\\circ} = - \\cos {144}^{\\circ}$ and ${24}^{\\circ} = {144}^{\\circ} - {120}^{\\circ} .$ Reminder: $\\cos x = \\cos a$ has solutions $x = \\setminus \\pm a + {360}^{\\circ} k ,$ integer $k$. Solve $\\cos \\left(2 x\\right) = \\cos \\left(3 x\\right)$ for $x$ $2 x = \\setminus \\pm 3 x + {360}^{\\circ} k$ $2 x \\setminus \\pm 3 x = {360}^{\\circ} k$ $5 x = {360}^{\\circ} k \\mathmr{and} x = - {360}^{\\circ} k$ $x = {72}^{\\circ} k \\quad$ This includes $- {360}^{\\circ} k$ so we don't need to write that separately. So we see our equation gives", "+0 # Algebraic structures 0 150 2 +336 Find the solution: $$τ=(1,2,5,7)(3,4,6,7)(1,4,8,9,3)(4,7,2)$$ The answer is $$(1,7)(2,8,9,4,3,6)(5) => (2,8,9,4,3,6)(1,7)$$ An other ex. is this:$$σ=(1,3,5)(2,7)(11,1,4)(5,6,2,8,9)(7,3)(4,5,2,1)$$ The answer is $$(1,11,3,2,4,6,7,5,8,9)(10) =>(1,11,3,2,4,6,7,5,8,9)$$ I don't know/understand the steps to find this answer. Thank you! Jun 2, 2019 #1 +23137 +2 2.) An other ex. is this: $$\\sigma=(1,3,5)(2,7)(11,1,4)(5,6,2,8,9)(7,3)(4,5,2,1)$$ The answer is $$(1,11,3,2,4,6,7,5,8,9)(10) \\Rightarrow (1,11,3,2,4,6,7,5,8,9)$$ Start with $$x = 1 \\text{ in } k(x)$$ $$\\small{ \\begin{array}{|ccccccccccc|c|cccccc|} \\hline f &\\circ& g &\\circ& h &\\circ& i &\\circ& j &\\circ& k & x & k(x) & j(k(x)) & i(j(k(x))) & \\tiny{h(i(j(k(x))))} & \\tiny{g(h(i(j(k(x)))))} & \\tiny{f(g(h(i(j(k(x))))))} \\\\ \\hline (1,3,5)&\\circ&(2,7)&\\circ&(11,1,4)&\\circ&(5,6,2,8,9)&\\circ&(7,3)&\\circ&(4,5,2,1) &\\color{red} 1 & 4 &4 & 4 & 11 &11 & 11 \\\\ & & & & & & & & & & & 11 & 11 &11 & 11 & 1 &1 &3 \\\\ & & & & & & & & & & & 3 & 3 &7 & 7 & 7 &2 &2 \\\\ & & & & & & & & & & & 2 & 1 &1 & 1 & 4 &4 &4 \\\\ & & & & & & & & & & & 4 & 5 &5 & 6 & 6 &6 &6 \\\\ & & & & & & & & & & & 6 & 6 &6 & 2 & 2", "3} x + 1 = 3x^4 + 5x^3 + 4x^2 + 3x + 1 is an identity, (b) To calculate: The lopynomial P(x) = 6x^5 - 3x^4 + 9x^3 + 6x^2 -8x + 12 without powers of x as in patr (a). Consider the following sequence. $$s_{n} = 2n − 1$$ (a) Find the first three terms of the sequence whose nth term is given. $$s_{1} =$$ $$s_{2} =$$ $$s_{3} =$$ (b) Classify the sequence as arithmetic, geometric, both, or neither. arithmetic, geometric bothneither If arithmetic, give d, if geometric, give r, if both, give d followed by r. (If both, enter your answers as a comma-separated list. If neither, enter NONE.) 1. Explain with numerical examples what Real Numbers and Algebraic Expressions are. 2. Explain with numerical examples Factoring and finding LCMs (least common multiples). Explain factoring of a larger number. 3. Explain with numerical examples arithmetical operations (addition, subtraction, multiplication, division) with fractions 4, Explain with numerical examples arithmetical operations (addition, subtraction, multiplication, division) with percentages 5. Explain with numerical examples exponential notation 6. Explain with numerical examples order (precedence) of arithmetic operations 7. Explain with numerical examples the concept and how to find perimeter, area, volume, and circumference (use related formulas) Given the following function: $$\\displaystyle{f{{\\left({x}\\right)}}}={1.01}{e}^{{{4}{x}}}-{4.62}{e}^{{{3}{x}}}-{3.11}{e}^{{{2}{x}}}+{12.2}{e}^{{{x}}}-{1.99}$$ a)Use three-digit rounding frithmetic, the assumption that $$\\displaystyle{e}^{{{1.53}}}={4.62}$$, and", "## anonymous 3 years ago Solve the equation: a.) 2 cos^2x-cos x=1 1. anonymous let's define $t:= \\cos(x)$ we have the quadratic equation: $2t^2-t=1$ can you find it's solutions? 2. anonymous Would it be cos=1? Which means cos=0 degrees or 360 degrees? 3. anonymous there are two solutions, $\\cos(x)=1$ and $\\cos(x)=-\\frac{1}{2}$ so x can be either $360^\\circ{} k$ (for every integer k) or $120^\\circ{}+ 360^{\\circ}k$ or $240^\\circ{}+ 360^{\\circ}k$", "Write all solutions on the same line, separated by commas. ##### Question 3 Solve for the unknown: $-8x+x^2=-6-x-x^2$8x+x2=6xx2 1. Write all solutions on the same line, separated by commas. ##### Question 4 Solve the following equation: $x-\\frac{45}{x}=4$x45x=4 1. Write all solutions on the same line, separated by commas.", "the special in which $L = \\mathbb{N}$, and instead of $L^*$ I will use comma-separated sequences. Here is my enumeration: $$\\epsilon; 1; 2; 1,1; 3; 1,2; 2,1; 1,1,1; 4; \\ldots$$ and so on. See if you can figure out a pattern (or if you don't like my pattern, come up with a different one).", "### Home > CCAA8 > Chapter 10 Unit 11 > Lesson CCA: 10.2.3 > Problem10-56 10-56. How many terms are in the arithmetic sequence shown below? $15,\\ 7,\\ −1,\\ −9,...,\\ −225$ Write an equation for the arithmetic sequence using $a_n−m\\left(n-1\\right)+a_1$ Let $a$ sub $n=−225$ and solve for $n$.", "## Introductory Algebra for College Students (7th Edition) 44$^\\circ$ Find the solution by adding the temperature variation to the low temperature. -56+100=44$^\\circ$", "Find the solutions of equation 2^ x equiv x^ 5 rounded to one decitrial place. Enter your answers as a comma-separated .", "as $[a_1, b_1]$. Repeating this method produces a sequence of nested intervals $[a_0, b_0] \\supset [a_1, b_1] \\supset [a_2, b_2] \\supset \\cdots$, each of which contains a solution, and each of which is one tenth the size of the previous interval. If $[a_0, b_0] = [0,1]$, then each successive $[a_n, b_n]$ determines a successive decimal digit of the solution.", "terms of each arithmetic sequence. $$a_{1}=300, d=50$$... ##### In Exercises $35-37$ , use Theorem I to verify the formula. $$\\frac{d}{d x} \\csc x=-\\csc x \\cot x$$ In Exercises $35-37$ , use Theorem I to verify the formula. $$\\frac{d}{d x} \\csc x=-\\csc x \\cot x$$...", "(a) Find the range of (i) $${f_1} \\circ {g_1}$$ ; (ii) $${f_3} \\circ {g_2}$$ . (b) Find all the solutions of $${f_1} \\circ {g_2}(k) = {f_2} \\circ {g_1}(k)$$ . (c) Find all the solutions of $${f_3}(m,{\\text{ }}n) = p$$ in each of the cases p =1 and p = 2 . ## Markscheme (a) (i) $${f_1} \\circ {g_1}(k) = k + 4$$ M1 Range $$({f_1} \\circ {g_1}) = \\mathbb{Z}$$ A1 (ii) $${f_3} \\circ {g_2}(k) = 0$$ M1 Range $$({f_3} \\circ {g_2}) = \\{ 0\\}$$ A1 [4 marks] (b) the equation to solve is $$k – \\left| k \\right| + 4 = \\left| {2k} \\right|$$ M1A1 the positive solution is k = 2 A1 the negative solution is k = –1 A1 [4 marks] (c) the equation factorizes: $$(m + n)(m – n) = p$$ (M1) for p = 1 , the possible factors over $$\\mathbb{Z}$$ are $$m + n = \\pm 1,{\\text{ }}m – n = \\pm 1$$ (M1)(A1) with solutions (1, 0) and (–1, 0) A1 for p = 2 , the possible factors over $$\\mathbb{Z}$$ are $$m + n = \\pm 1,{\\text{ }} \\pm 2;{\\text{ }}m – n = \\pm 2,{\\text{ }} \\pm 1$$ M1A1 there are no solutions over $$\\mathbb{Z} \\times \\mathbb{Z}$$ A1 [7 marks] Total [15 marks] ## Examiners report The majority of candidates", "## A quick description of your question... Find all solutions of the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places where appropriate.) 3 csc^2 ? ? 4 = 0 ? =", "the given interval. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x) = √x , [0, 25] 1. ### Calculus 1 Consider the following. B(x) = 3x^(2/3) − x (a) Find the interval of increase.(Enter your answer using interval notation.) Find the interval of decrease. (Enter your answer using interval notation.) (b) Find the local maximum 2. ### Pre-Calc sin(2θ)=1/2 Find all solutions in the interval [0, 2π). (Enter your answers as a comma-separated list.) k = any integer What I have so far: pi/12 + kpi, 5pi/12 + kpi However, I do not know how to solve for the interval. Please 3. ### Math An equation is given. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.) Find the solutions in the interval: 4. ### mathematics Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places where appropriate.) sin(x) = √3/2 x = ?rad 1. ### math Find two quadratic equations having the given solutions. (There are many correct answers. Use x for the variable. Enter your answers as a comma-separated list of equations.) 4+5√(5), 4-5√(5) 2. ### calculus" ]

[ "## A community for students. Sign up today! Here's the question you clicked on: ## haleyking345 one year ago Solve the equation: a.) 2 cos^2x-cos x=1 • This Question is Closed 1. Marx let's define $t:= \\cos(x)$ we have the quadratic equation: $2t^2-t=1$ can you find it's solutions? 2. haleyking345 Would it be cos=1? Which means cos=0 degrees or 360 degrees? 3. Marx there are two solutions, $\\cos(x)=1$ and $\\cos(x)=-\\frac{1}{2}$ so x can be either $360^\\circ{} k$ (for every integer k) or $120^\\circ{}+ 360^{\\circ}k$ or $240^\\circ{}+ 360^{\\circ}k$ #### Ask your own question Ask a Question Find more explanations on OpenStudy", "335.9^\\circ+i \\sin 335.9^\\circ)]^7 &= (\\sqrt{6})^7(\\cos (7 \\cdot 335.9^\\circ)+i \\sin (7 \\cdot 335.9^\\circ))\\\\&= 216 \\sqrt{6}(\\cos 2351.3^\\circ+i \\sin 2351.3^\\circ)\\\\&= 216 \\sqrt{6}(-0.981-0.196i)\\\\&= -519.04-103.7i$ ### Concept Problem Solution The original problem was: $[5(\\cos 135^\\circ + i \\sin 135^\\circ)]^4$ In this problem, $r = 5, \\theta = \\frac{3\\pi}{4}$ Using $z^n = [r(\\cos \\theta + i \\sin \\theta)]^n = r^n(\\cos n\\theta + i \\sin n\\theta)$ , you can substitute to get: $[5(\\cos \\frac{3\\pi}{4} + i \\sin \\frac{3\\pi}{4})]^4\\\\ = 5^4(\\cos 4\\frac{3\\pi}{4} + i \\sin 4\\frac{3\\pi}{4})\\\\= 625(\\cos(3\\pi) + i\\sin(3\\pi)\\\\= 625(-1 + i0)\\\\= -625\\\\$ ### Practice Use DeMoivre's Theorem to evaluate each expression. Write your answer in standard form. 1. $(\\cos\\frac{\\pi}{4}+i\\sin\\frac{\\pi}{4})^3$ 2. $[2(\\cos\\frac{\\pi}{6}+i\\sin\\frac{\\pi}{6})]^2$ 3. $[3(\\cos\\frac{3\\pi}{2}+i\\sin\\frac{\\pi}{2})]^5$ 4. $(1+i)^5$ 5. $(1-\\sqrt{3}i)^3$ 6. $(1+2i)^6$ 7. $(\\sqrt{3}-i)^5$ 8. $(\\frac{1}{2}+\\frac{i\\sqrt{3}}{2})^4$ 9. $[3(\\cos\\frac{\\pi}{4}+i\\sin\\frac{\\pi}{4})]^5$ 10. $(2-\\sqrt{5}i)^5$ 11. $(\\sqrt{2}+\\sqrt{2}i)^4$ 12. $[2(\\cos\\frac{\\pi}{12}+i\\sin\\frac{\\pi}{12})]^8$ 13. $(-1+\\sqrt{2}i)^6$ 14. $[5(\\cos\\frac{5\\pi}{3}+i\\sin\\frac{5\\pi}{3})]^3$ 15. $(3-4i)^6$", "could do $cos(66.5^{\\circ})$ it would would work the same way alright so $cos^2(22.5^{\\circ}) = \\frac{1 + cos2\\cdot22.5^{\\circ}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{1 + cos45^{\\circ}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{1 + \\frac{1}{\\sqrt{2}}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{\\sqrt{2} + 1}{2\\sqrt{2}} = \\sqrt{\\frac{\\sqrt{2} + 1}{2\\sqrt{2}}} \\Rightarrow \\sqrt{\\frac{\\sqrt{2} + 2}{4}} = cos(22.5^{\\circ})$ in exact form now do $cos(66.5^{\\circ})$ in similar fasion and multiply them together :)", "us all the multiples of ${72}^{\\circ}$. Let's go at it a different way, using the double and triple angle formulas: Solve $\\cos \\left(2 x\\right) = \\cos \\left(3 x\\right)$ for $\\cos \\left(x\\right)$ $2 {\\cos}^{2} x - 1 = 4 {\\cos}^{3} x - 3 \\cos x$ Let's abbreviate $c = \\cos x$ $2 {c}^{2} - 1 = 4 {c}^{3} - 3 c$ $4 {c}^{3} - 2 {c}^{2} - 3 c + 1 = 0$ That's a cubic equation, but one of the easy ones. All the cosines of multiples of ${72}^{\\circ}$ will be roots of this equation. That includes $\\cos 0 = 1$. So $c - 1$ is a factor, and with long division we get $\\left(c - 1\\right) \\left(4 {c}^{2} + 2 c - 1\\right) = 0$ $\\cos {0}^{\\circ} ,$ $\\cos {72}^{\\circ} ,$ and $\\cos {144}^{\\circ}$ are all different, so they're the three roots to this cubic equation. All other multiples of ${72}^{\\circ}$ share one of these cosines. $c = \\frac{1}{4} \\left(- 1 \\setminus \\pm \\sqrt{5}\\right)$ Clearly $\\cos {72}^{\\circ}$ is the positive one and $\\cos {144}^{\\circ}$ is the negative one: $\\cos {72}^{\\circ} = \\frac{1}{4} \\left(\\setminus \\sqrt{5} - 1\\right)$ $\\cos {144}^{\\circ} = \\frac{1}{4} \\left(- \\setminus \\sqrt{5} - 1\\right)$ $\\cos {36}^{\\circ} = - \\setminus \\cos \\left({180}^{\\circ} - {36}^{\\circ}\\right) = - \\cos {144}^{\\circ} = \\frac{1}{4} \\left(1 + \\setminus \\sqrt{5}\\right)$ That's half the", "# How do you solve a^2=3^2+5^2-2(3)(5)cos85? May 1, 2018 $a = 31.4$ units #### Explanation: ${a}^{2} = {3}^{2} + {5}^{2} - \\left(2 \\cdot 3 \\cdot 5\\right) \\cos {85}^{\\circ}$. it seems best to first simplify the expression on the right-hand side. ${3}^{2} + {5}^{2} = 9 + 25 = 34$ $2 \\cdot 3 \\cdot 5 = 30 \\rightarrow - \\left(2 \\cdot 3 \\cdot 5\\right) \\cos {85}^{\\circ} = - 30 \\cos {85}^{\\circ}$/ ${3}^{2} + {5}^{2} - \\left(2 \\cdot 3 \\cdot 5\\right) \\cos {85}^{\\circ} = 34 - 30 \\cos {85}^{\\circ}$. hence, ${a}^{2} = 34 - 30 \\cos {85}^{\\circ}$. since side length cannot be negative, $a$ is the positive square root of $34 - 30 \\cos {85}^{\\circ}$. putting $\\sqrt{34 - 30 \\cos {85}^{\\circ}}$ into a calculator gives $31.38532771756$. this can be rounded to $3$ significant figures to give $31.4$. therefore $a = 31.4$ units in length.", "# Difference between revisions of \"1996 USAMO Problems/Problem 1\" ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos 5^\\circ)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot", "# Difference between revisions of \"1996 USAMO Problems/Problem 1\" ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos5)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$,", "20:50 Understandable - I just found this to be a little less messy. – user17794 Aug 27 '13 at 20:52 Hint: \\begin{align} 0&=\\cos x+\\sin x\\\\ \\sin x &= -\\cos x \\\\ \\text{Divide}&\\text{ both sides by }\\cos x\\\\ \\tan x &= -1 \\end{align} Now you can solve for $x$ using inverse trigonometric functions. - if you use the Double angle formulae $$\\sin^2(x)\\cos^2(x)=(1-\\cos2x)/2 \\cdot (1+\\cos2x)/2 = (1-\\cos^2(2x))/4 = \\sin^2(2x)/4 = (1-\\cos4x)/8$$ so minimum is $1/8$ at $x=45\\circ$ and maximum is 1/4 at $x=90\\circ$ remember $\\sin^2(x)=(1-\\cos2x)/2$ and $\\cos^2(x)=(1+\\cos2x)/2$ - The equation $\\cos x+\\sin x=0$ is really the same as $\\cos x-\\sin x=0$ once you make the transformation $x\\mapsto x+\\frac{\\pi}2$, because $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and $\\sin (x+\\frac{\\pi}2)=\\cos(x)$. The latter can be solved by resorting to $\\cos ^2+\\sin ^2=1$. Edit: We can derive $\\cos (x+\\frac{\\pi}2)=-\\sin x$ and similar results (cf. Daniel Fischer's comment) from the well known formulas $\\cos (x+y)=\\cos x\\cos y-\\sin x\\sin y$ and $\\sin (x+y)=\\cos x\\sin y+\\cos y\\sin x$. These can either be proved geometrically, or by resorting to some definition of $\\sin$ and $\\cos$, for example $\\sin x=(e^{ix}-e^{-ix})/2i, \\cos x=(e^{ix}-e^{-ix})/2$ -", "B = \\frac{c^2 + a^2 − b^2}{2ca}$$ $$⇒ cos B = \\frac{4^2 + 2^2 − 3^2}{ 2(4)(2)} = 0.6875$$ $$\\boxed{⇒ B = 46.6^\\circ}$$ Now we use the formula for $$c^{2}$$ to find C: $$c^2 = a^2 + b^2 − 2ab\\;cos C ⇒ cos C = \\frac{a^2 + b^2 − c^2}{ 2ab}$$ $$⇒ cos C = \\frac{2^2 + 3^2 − 4^2}{ 2(2)(3)} = −0.25$$ $$\\boxed{⇒ C = 104.5^\\circ}$$ Thus, A = 180º −BC = 180º −46.6º −104.5º⇒ $$\\boxed{A = 28.9º}$$ . It may seem that there is always a solution in Case 4 (given all three sides), but that is not true, as the following example shows. Example 4 Case 4: Three sides. Solve the triangle ∆ABC given a = 2, b = 3, and c = 6. Solution: If we blindly try to use the Law of Cosines to find A, we get $$a^2 = b^2 + c^2 − 2bc cos A ⇒ cos A = \\frac{b^2 + c^2 − a^2}{2bc} = \\frac{3^2 + 6^2 − 2^2}{2(3)(6)} = 1.139 ,$$ which is impossible since |cos A| ≤ 1. Thus, there is $$\\boxed{\\text{no solution}}$$ . We could have saved ourselves some effort by recognizing that the length of one of the sides (c = 6) is greater than the sums of the lengths of the remaining sides (a =", "# How to find the exact value of $\\cos(36^\\circ)$? [duplicate] Noting that $t=\\frac{\\pi}{5}$ satisfies $3t=\\pi-2t$, find the exact value of $$\\cos(36^\\circ)$$ it says that you may find useful the following identities: $$\\cos^2 t+\\sin^2 t = 1,\\\\ \\sin 2t = 2\\sin t\\cos t,\\\\ \\sin 3t = 3\\sin t - 4\\sin^3 t.$$ Do I have to do a system of linear equations in function of ..what? $t$? $\\cos$? ## marked as duplicate by Xander Henderson, José Carlos Santos, Cesareo, YiFan, Eevee TrainerMar 30 at 1:15 Let $t=\\frac{\\pi}{5}$ (so $t$ is $36^\\circ$). Since $108=180-72$, we have $3t=\\pi-2t$ and therefore $$\\sin(3t)=\\sin(\\pi-2t).$$ But $\\sin(\\pi-2t)=\\sin(2t)=2\\sin t\\cos t$. Also, by the identity you were given, $\\sin(3t)=3\\sin t-4\\sin^3 t$. Thus $$3\\sin t-4\\sin^3 t=2\\sin t\\cos t.$$ But $\\sin t\\ne 0$, so we can cancel a $\\sin t$ and obtain $$3-4\\sin^2 t=2\\cos t.$$ Substitute $1-\\cos^2 t$ for $\\sin^2 t$ and simplify a bit. We get $$4\\cos^2 t-2\\cos t-1=0.$$ Use the Quadratic Formula to solve this quadratic equation for $\\cos t$, rejecting the negative root. We get $$\\cos t=\\frac{2+\\sqrt{20}}{8}.$$ We can simplify this to $\\dfrac{1+\\sqrt{5}}{4}.$ • Very well explained, thanks very much :) deeply appreciated – Maximilian1988 Feb 11 '13 at 8:31 To find $\\cos{\\pi/5}$, note that $$\\sin{(3 \\pi/5)} = \\sin{(2 \\pi/5)}$$ and $$\\sin{(3 \\pi/5)} = 3 \\sin{(\\pi/5)} - 4 \\sin^3{(\\pi/5)} = 2 \\sin{(\\pi/5)} \\cos{(\\pi/5)}$$ Thus $$2 \\cos{(\\pi/5)} =", "\\text{cos}(44 ^ \\circ ) \\\\ \\red x^2 = 14^2 + 10^2 -2 \\cdot 14 \\cdot 10 \\text{cos}(44 ^ \\circ ) \\\\ \\red x^2 = 296 -280 \\text{cos}(44 ^ \\circ) \\\\ \\red x^2 = 94.5848559051777 \\\\ \\red x = \\sqrt{ 94.5848559051777} \\\\ \\red x = 9.725474585087234$$ ##### Example 2 Given : 3 sides $$a^2 = b^2 + c^2 - 2bc\\cdot \\text{cos}(\\red A) \\\\ 25^2 = 32^2 + 37^2 -2 \\cdot 32 \\cdot 37 \\cdot \\text{cos}(\\red A) \\\\ 625 =2393 - 2368\\cdot \\text{cos}(\\red A) \\\\ \\frac{625-2393}{ - 2368}= cos(\\red A) \\\\ 0.7466216216216216 = cos(\\red A ) \\\\ \\red A = cos^{-1} (0.7466216216216216 ) \\\\ \\red A = 41.70142633732469 ^ \\circ$$ ### Practice Problem The problems below are ones that ask you to apply the formula to solve straight forward questions. If they start to seem too easy, try our more challenging problems. ##### Problem 1 $$c^2 = a^2 + b^2 - 2ab\\cdot \\text{cos}( 66 ^\\circ) \\\\ c^2 = 20^2 + 13^2 - 2\\cdot20\\cdot 13 \\cdot \\text{cos}( 66 ^\\circ) \\\\ c^2 = 20^2 + 13^2 - 2\\cdot20\\cdot 13 \\cdot \\text{cos}( 66 ^\\circ) \\\\ c^2 =357.4969456005839 \\\\ c = \\sqrt{357.4969456005839} \\\\ c = 18.907589629579544$$ ##### Problem 2 $$a^2 = b^2 + c^2 - 2bc\\cdot \\text{cos}(A) \\\\ x^2 = y^2 + z^2 - 2yz\\cdot \\text{cos}(X ) \\\\ 14^2 = 20^2 + 12^2 -", "## anonymous 3 years ago Solve the equation: a.) 2 cos^2x-cos x=1 1. anonymous let's define $t:= \\cos(x)$ we have the quadratic equation: $2t^2-t=1$ can you find it's solutions? 2. anonymous Would it be cos=1? Which means cos=0 degrees or 360 degrees? 3. anonymous there are two solutions, $\\cos(x)=1$ and $\\cos(x)=-\\frac{1}{2}$ so x can be either $360^\\circ{} k$ (for every integer k) or $120^\\circ{}+ 360^{\\circ}k$ or $240^\\circ{}+ 360^{\\circ}k$", "(4, 30o) 2. (2.5, $$\\ \\pi$$) 3. $$\\ \\left(-1, \\frac{\\pi}{3}\\right)$$ 4. $$\\ \\left(3, \\frac{5 \\pi}{6}\\right)$$ 5. (−2, 300o) Solution Example 3 Use a graphing calculator or plotting program to plot the following equations. 1. r=1+3sinθ 2. r=1+2cosθ Solution 1. Review the steps above under graphing using technology if you are having trouble. Example 4 Recall that $$\\ \\pi r a d=180^{\\circ} \\text { and } 1 \\mathrm{rad}=\\frac{180}{\\pi} \\approx 57.3^{\\circ}$$ 1. $$\\ \\frac{\\pi}{2}$$ 2. 5.17 3. $$\\ \\frac{3 \\pi}{2}$$ Solution 1. If $$\\ \\pi r a d=180^{\\circ}$$ then $$\\ \\frac{\\pi}{2} r a d=90^{\\circ}$$ 2. If $$\\ 1 rad \\approx 57.3^{\\circ}$$ then $$\\ 5.17 rad \\approx 296^{\\circ}$$ 3. If $$\\ \\pi r a d=180^{\\circ}$$ then $$\\ \\frac{3 \\pi}{2} r a d=270^{\\circ}$$ Example 5 Recall that $$\\ \\frac{180^{\\circ}}{\\pi}=57.3^{\\circ} \\approx 1 \\mathrm{rad}$$. 1. 251o 2. 360o 3. 327o Solution 1. If $$\\ 57.3^{\\circ} \\approx 1rad$$ then $$\\ 251^{\\circ} \\approx 4.38 \\mathrm{rad} \\approx 1.4 \\pi \\mathrm{rad}$$ 2. If $$\\ 57.3^{\\circ} \\approx 1 rad$$ then $$\\ 360^{\\circ} \\approx 6.28 rad$$ 3. If $$\\ 57.3^{\\circ} \\approx 1 rad$$ then $$\\ \\frac{327^{\\circ}}{57.3^{\\circ}} \\approx 5.71 rad$$ Example 6 1. 90o 2. 270o 3. 45o Solution 1. If $$\\ \\pi r a d=180^{\\circ}$$ then $$\\ \\frac{\\pi}{2} r a d=90^{\\circ}$$ 2. If $$\\ \\pi r a d=180^{\\circ} \\text { and } \\frac{\\pi}{2} r a d=90^{\\circ}$$ then $$\\ 1 \\frac{1}{2}", "1996 USAMO Problems/Problem 1 Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. Solution Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos 5^\\circ)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$, as desired. $\\Box$ 1996 USAMO (Problems • Resources)", "Brian Bi ## Section 15.5. Constructions with Ruler and Compass Exercise 15.5.1 \\begin{align*} \\cos 15^\\circ &= \\cos(45^\\circ - 30^\\circ) \\\\ &= \\cos 45^\\circ \\cos 30^\\circ + \\sin 45^\\circ \\sin 30^\\circ \\\\ &= \\frac{\\sqrt{2}}{2} \\frac{\\sqrt{3}}{2} + \\frac{\\sqrt{2}}{2} \\frac{1}{2} \\\\ &= \\frac{\\sqrt{6} + \\sqrt{2}}{4} \\end{align*} Exercise 15.5.2 1. Observe that $$\\cos 72^\\circ = \\cos 288^\\circ = 2\\cos^2 144^\\circ - 1 = 2(2 \\cos^2 72^\\circ - 1)^2 - 1$$. So, if we put $$x = \\cos 72^\\circ$$ then we have $$x = 2(2x^2 - 1)^2 - 1$$ which can be expanded to yield a quartic equation for $$x$$. However, two of the roots $$x_1, x_2$$ of this quartic are obviously just the roots of $$x = 2x^2 - 1$$, which lie in a quadratic extension $$\\Q(\\sqrt{d})$$ of the rationals. Dividing out $$(x - x_1)(x - x_2)$$ from the quartic yields a quadratic polynomial with coefficients in $$\\Q(\\sqrt{d})$$, so the remaining two roots lie in some quadratic extension of this field. One of these four roots is $$\\cos 72^\\circ$$, so by Theorem 15.5.6, the latter is constructible. This implies that the regular pentagon is constructible. 2. It is possible to devise the construction by writing out the explicit form of $$\\cos 72^\\circ$$, which is $$\\frac{\\sqrt{5} - 1}{4}$$, and applying Lemma 15.5.11, constructing first 5, then $$\\sqrt{5}$$, then $$\\sqrt{5} - 1$$, and finally", "Thread: Is this simultaneous equation possible? 1. Is this simultaneous equation possible? I'm working on a question and I've come up with this: $a^2 = 400+c^2-40c*\\cos A$ $400=a^2+c^2 - 2ac * \\cos 120$ $c^2=a^2+400-40a*\\cos C$ Is this possible to do? 2. I'll take a swing! For starters, cos120 = -0.5 Assuming that a and A are the same value, we get the following: a^2 = 400 + c^2 - 40c * cos(a) 400 = a^2 + c^2 + ac c^2 = a^2 + 400 - 40a * cos(c) a^2 = 400 + a^2 + 400 - 40a * cos(c) <=> 0 = 800 - 40a * cos(c) <=> 0 = 800 - 40c * cos(a) <=> 800 - 40a * cos(c) = 800 - 40c * cos(a) <=> 40a * cos(c) = 40c * cos(a) <=> a * cos(c) = c * cos(a) <=> a = c * cos(a) / cos(c) <=> c = a * cos(c) / cos(a) And that's as far as I got I'm probably forgetting some trig rules. Do you know anything else about these values? 3. Originally Posted by jgv115 I'm working on a question and I've come up with this: $a^2 = 400+c^2-40c*\\cos A$ $400=a^2+c^2 - 2ac * \\cos 120$ $c^2=a^2+400-40a*\\cos C$ Is this possible to do? All you have done is", "$$x(x-8)$$ gives $$4(x-8)=12(x)$$, distributing the 4 over the values within the parentheses yields $$x-8=3x$$ or $$x=-4$$. Therefore, the value of $$\\frac{x}{2}=\\frac{-4}{2}=-2$$. 12- D Given the two equations, substitute the numerical value of a into the second equation to solve for $$x$$. $$a=\\sqrt{3}, 4a=\\sqrt{4x}$$ Substituting the numerical value for a into the equation with $$x$$ is as follows. $$4(\\sqrt{3})=\\sqrt{4x}$$, From here, distribute the $$4$$. $$4\\sqrt{3}=\\sqrt{4x}$$ Now square both side of the equation. $$(4\\sqrt{3})^2=(\\sqrt{4x})^2$$ Remember to square both terms within the parentheses. Also, recall that squaring a square root sign cancels them out. $$4^2 \\sqrt{3}^2=4x, 16(3)=4x, 48=4x, x=12$$ 13- D In the figure angle A is labeled $$(3x-2)$$ and it measures $$37$$. Thus, $$3x-2=37$$ and $$3x=39$$ or $$x=13$$. That means that angle B, which is labeled $$(5x)$$, must measure $$5×13=65$$. Since the three angles of a triangle must add up to $$180, 37+65+y-8=180$$, then: $$y+94=180→y=180-94=86$$ 14- C Let $$x$$ be all expenses, then $$\\frac{22}{100} x=660 →x=\\frac{100×660}{22}=3,000$$ Mr. Jones spent for his rent: $$\\frac{ 27}{100}×3,000=810$$ 15- A The amount of money for $$x$$ bookshelf is: $$200x$$, Then, the total cost of all bookshelves is equal to: $$200x+600$$, The total cost, in dollar, per bookshelf is: $$\\frac{Total \\ cost}{number \\ of \\ items}=\\frac{200x+600}{x}$$ 16- B It is given that g(5)=4. Therefore, to find the value of f(g(5)), then f(g(5))=f(4)=6 17- A Area", "have that $\\sin18^\\circ=\\frac{-1+\\sqrt{5}}{4}.$ Find all solutions between $0^\\circ$ and $360^\\circ$ of the equation $4\\tan 2x=\\cot x.$ We play a similar game here to that of the first part of the question. Firstly, we can write $\\tan x=\\frac{s}{c}, \\quad \\cot x=\\frac{1}{\\tan x}=\\frac{c}{s},$ so the equation we need to solve becomes $4\\frac{\\sin(2x)}{\\cos(2x)}=\\frac{c}{s} \\implies 4\\sin(2x)s-\\cos(2x)c=0 \\implies 8s^2c-(1-2s^2)c=0 \\implies (10s^2-1)c=0.$ So we have that $\\cos x=0$ solves this, which gives us the values $x=90^\\circ, 270^\\circ$ between $0^\\circ$ and $360^\\circ$. We also have that $\\sin x=\\pm\\frac{1}{\\sqrt{10}}$ is a solution, so that $x=\\sin^{-1}(\\pm1/\\sqrt{10})$ are also solutions. To work out how many solutions between $0^\\circ$ and $360^\\circ$, consider We see that $\\sin x=\\pm1/\\sqrt{10}$ four times between $0^\\circ$ and $360^\\circ$. Using a calculator, one finds that $\\sin^{-1}(1/\\sqrt{10})=18.4^\\circ$ to one decimal place. From the symmetries of sine, we see that the other angles are then $180^\\circ-18.4^\\circ=161.6^\\circ$, $180^\\circ+18.4^\\circ=198.4^\\circ$ and $360^\\circ-18.4^\\circ=341.6^\\circ$, giving us six solutions in total.", "125}$ Now using Law of Cosines on $\\triangle AQP$ we get: $625 = a^2 + c^2 - 2ac\\cos{A}$ Notice $\\cos{A} = \\frac{c}{a+b}$ Substituting and Simplifying: $625 = a^2 + c^2 - 2ac\\frac{c}{a+b}$ $625 = a^2 + c^2 - 2ac\\frac{c}{a+\\frac{400}{a}}$ $625 = c^2 + 125 + c^2 - 2\\frac{(ac)^2}{a^2+400}$ $625 = c^2 + 125 + c^2 - 2\\frac{c^2(c^2+125)}{c^2+125+400}$ Now we solve for $c$ using regular algebra which actually turns out to be very easy. We get $c = 5\\sqrt{35}$ and from the above relations between the variables we quickly determine $d = 3\\sqrt{35}$, $a = 10\\sqrt{10}$ and $b = 4\\sqrt{10}$ Therefore $AB\\cdot AC = (a+b)\\cdot(c+d) = 560\\sqrt{14}$ So the answer is $560 + 14 = \\boxed{574}$ By asr41", "# Prove that $2\\cdot \\cos \\frac{72}{2}\\cdot \\cos \\frac{24}{2}+2\\cdot \\sin \\frac{96}{2}\\cdot \\sin \\frac{72}{2}=0.5$. Additional data added To solve it I have tried some options, where in one of them I applied product to sum formulas, which seemed to be very helpful, but didn't get the answer. Used these formulae: $$\\cos(a+b)+ \\cos(a-b)=2\\cdot \\cos(a)\\cdot \\cos(b)$$ $$−\\cos(a+b)+\\cos(a−b)=2\\sin(a)\\sin(b)$$ And now, I am stuck at this: $$\\cos(48)+\\cos(24)-\\cos(84)+\\cos(12)=0.5$$ Then tried to make the arguments similar using a double angle and a half angle formulae $$2\\cos^2(24)-1+\\cos(24)+\\sqrt{\\frac{1+\\cos(24)}{2}}+\\cos(84)=0.5$$ Anyway, I can't show that expression above really is 0.5. • I don’t understand. Why do you write $\\frac{72}2$ instead of $36$, for instance? – Lubin Jun 20 '17 at 20:13 • So was written on that sheet of paper, but I know that it is 36 :). – user36339 Jun 20 '17 at 20:15 • Do you know these formulea ? \\begin{eqnarray*} \\cos(a+b)+\\cos(a-b) = 2 \\cos(a) \\cos(b) \\\\ -\\cos(a+b)+\\cos(a-b) = 2 \\sin(a) \\sin(b) \\end{eqnarray*} – Donald Splutterwit Jun 20 '17 at 20:15 • No, I didn't.Thanks! – user36339 Jun 20 '17 at 20:50 • Note that if these are degrees, then the degree sign is obligatory; otherwise the unit is taken to be radians. For instance, $\\sin 90^\\circ = 1$, while $\\sin 90 = 0.89399\\ldots$. The degree sign can be typeset using ^\\circ. – Théophile Jun 20 '17 at" ]

[ "## A community for students. Sign up today! Here's the question you clicked on: ## haleyking345 one year ago Solve the equation: a.) 2 cos^2x-cos x=1 • This Question is Closed 1. Marx let's define $t:= \\cos(x)$ we have the quadratic equation: $2t^2-t=1$ can you find it's solutions? 2. haleyking345 Would it be cos=1? Which means cos=0 degrees or 360 degrees? 3. Marx there are two solutions, $\\cos(x)=1$ and $\\cos(x)=-\\frac{1}{2}$ so x can be either $360^\\circ{} k$ (for every integer k) or $120^\\circ{}+ 360^{\\circ}k$ or $240^\\circ{}+ 360^{\\circ}k$ #### Ask your own question Ask a Question Find more explanations on OpenStudy", "335.9^\\circ+i \\sin 335.9^\\circ)]^7 &= (\\sqrt{6})^7(\\cos (7 \\cdot 335.9^\\circ)+i \\sin (7 \\cdot 335.9^\\circ))\\\\&= 216 \\sqrt{6}(\\cos 2351.3^\\circ+i \\sin 2351.3^\\circ)\\\\&= 216 \\sqrt{6}(-0.981-0.196i)\\\\&= -519.04-103.7i$ ### Concept Problem Solution The original problem was: $[5(\\cos 135^\\circ + i \\sin 135^\\circ)]^4$ In this problem, $r = 5, \\theta = \\frac{3\\pi}{4}$ Using $z^n = [r(\\cos \\theta + i \\sin \\theta)]^n = r^n(\\cos n\\theta + i \\sin n\\theta)$ , you can substitute to get: $[5(\\cos \\frac{3\\pi}{4} + i \\sin \\frac{3\\pi}{4})]^4\\\\ = 5^4(\\cos 4\\frac{3\\pi}{4} + i \\sin 4\\frac{3\\pi}{4})\\\\= 625(\\cos(3\\pi) + i\\sin(3\\pi)\\\\= 625(-1 + i0)\\\\= -625\\\\$ ### Practice Use DeMoivre's Theorem to evaluate each expression. Write your answer in standard form. 1. $(\\cos\\frac{\\pi}{4}+i\\sin\\frac{\\pi}{4})^3$ 2. $[2(\\cos\\frac{\\pi}{6}+i\\sin\\frac{\\pi}{6})]^2$ 3. $[3(\\cos\\frac{3\\pi}{2}+i\\sin\\frac{\\pi}{2})]^5$ 4. $(1+i)^5$ 5. $(1-\\sqrt{3}i)^3$ 6. $(1+2i)^6$ 7. $(\\sqrt{3}-i)^5$ 8. $(\\frac{1}{2}+\\frac{i\\sqrt{3}}{2})^4$ 9. $[3(\\cos\\frac{\\pi}{4}+i\\sin\\frac{\\pi}{4})]^5$ 10. $(2-\\sqrt{5}i)^5$ 11. $(\\sqrt{2}+\\sqrt{2}i)^4$ 12. $[2(\\cos\\frac{\\pi}{12}+i\\sin\\frac{\\pi}{12})]^8$ 13. $(-1+\\sqrt{2}i)^6$ 14. $[5(\\cos\\frac{5\\pi}{3}+i\\sin\\frac{5\\pi}{3})]^3$ 15. $(3-4i)^6$", "us all the multiples of ${72}^{\\circ}$. Let's go at it a different way, using the double and triple angle formulas: Solve $\\cos \\left(2 x\\right) = \\cos \\left(3 x\\right)$ for $\\cos \\left(x\\right)$ $2 {\\cos}^{2} x - 1 = 4 {\\cos}^{3} x - 3 \\cos x$ Let's abbreviate $c = \\cos x$ $2 {c}^{2} - 1 = 4 {c}^{3} - 3 c$ $4 {c}^{3} - 2 {c}^{2} - 3 c + 1 = 0$ That's a cubic equation, but one of the easy ones. All the cosines of multiples of ${72}^{\\circ}$ will be roots of this equation. That includes $\\cos 0 = 1$. So $c - 1$ is a factor, and with long division we get $\\left(c - 1\\right) \\left(4 {c}^{2} + 2 c - 1\\right) = 0$ $\\cos {0}^{\\circ} ,$ $\\cos {72}^{\\circ} ,$ and $\\cos {144}^{\\circ}$ are all different, so they're the three roots to this cubic equation. All other multiples of ${72}^{\\circ}$ share one of these cosines. $c = \\frac{1}{4} \\left(- 1 \\setminus \\pm \\sqrt{5}\\right)$ Clearly $\\cos {72}^{\\circ}$ is the positive one and $\\cos {144}^{\\circ}$ is the negative one: $\\cos {72}^{\\circ} = \\frac{1}{4} \\left(\\setminus \\sqrt{5} - 1\\right)$ $\\cos {144}^{\\circ} = \\frac{1}{4} \\left(- \\setminus \\sqrt{5} - 1\\right)$ $\\cos {36}^{\\circ} = - \\setminus \\cos \\left({180}^{\\circ} - {36}^{\\circ}\\right) = - \\cos {144}^{\\circ} = \\frac{1}{4} \\left(1 + \\setminus \\sqrt{5}\\right)$ That's half the", "# Using the R method for finding all solutions to $\\sin(2a) - \\cos(2a) = \\frac{\\sqrt{6}}{2}$. My solution differs from official answer. How many solutions does $$\\sin(2a) - \\cos(2a) = \\frac{\\sqrt{6}}{2}$$ have between $$-90^\\circ$$ and $$90^\\circ$$? I used the R method and got $$2a-45^\\circ = \\arcsin\\left(\\frac{\\sqrt{3}}{2}\\right).$$ Since $$a$$ is between $$-90^\\circ$$ and $$90^\\circ$$, then $$2a$$ is between $$-180^\\circ$$ and $$180^\\circ$$. The RHS can be $$60^\\circ$$, $$120^\\circ$$, $$-240^\\circ$$, and $$-300^\\circ$$. Only $$60^\\circ$$ and $$120^\\circ$$ fit the criteria, but the answer is 4 solutions. Where did I go wrong? You are right. Here is the image of the function using google. • Thanks, but i want to know whats wrong with this solution that got 4 as the answer. Sin 2a - cos 2a = root 6 /2, then he squared it, getting 1-2sin2acos2a= 3/2 which is sin 4a = -1/2 which does have 4 solutions. – SuperMage1 Jan 14 at 2:54 • Since I don't see how they got 4 solutions, there is no way to explain what they did wrong. Do you know which solutions they got? – Andrei Jan 14 at 2:56 • No solutions were given – SuperMage1 Jan 14 at 2:58 • @SuperMage1 when we square we add solutions $x=2$ square $x^2 = 4$ two solutions -/+2. If you square you need to return to the", "could do $cos(66.5^{\\circ})$ it would would work the same way alright so $cos^2(22.5^{\\circ}) = \\frac{1 + cos2\\cdot22.5^{\\circ}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{1 + cos45^{\\circ}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{1 + \\frac{1}{\\sqrt{2}}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{\\sqrt{2} + 1}{2\\sqrt{2}} = \\sqrt{\\frac{\\sqrt{2} + 1}{2\\sqrt{2}}} \\Rightarrow \\sqrt{\\frac{\\sqrt{2} + 2}{4}} = cos(22.5^{\\circ})$ in exact form now do $cos(66.5^{\\circ})$ in similar fasion and multiply them together :)", "# Difference between revisions of \"1996 USAMO Problems/Problem 1\" ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos5)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$,", "Brian Bi ## Section 15.5. Constructions with Ruler and Compass Exercise 15.5.1 \\begin{align*} \\cos 15^\\circ &= \\cos(45^\\circ - 30^\\circ) \\\\ &= \\cos 45^\\circ \\cos 30^\\circ + \\sin 45^\\circ \\sin 30^\\circ \\\\ &= \\frac{\\sqrt{2}}{2} \\frac{\\sqrt{3}}{2} + \\frac{\\sqrt{2}}{2} \\frac{1}{2} \\\\ &= \\frac{\\sqrt{6} + \\sqrt{2}}{4} \\end{align*} Exercise 15.5.2 1. Observe that $$\\cos 72^\\circ = \\cos 288^\\circ = 2\\cos^2 144^\\circ - 1 = 2(2 \\cos^2 72^\\circ - 1)^2 - 1$$. So, if we put $$x = \\cos 72^\\circ$$ then we have $$x = 2(2x^2 - 1)^2 - 1$$ which can be expanded to yield a quartic equation for $$x$$. However, two of the roots $$x_1, x_2$$ of this quartic are obviously just the roots of $$x = 2x^2 - 1$$, which lie in a quadratic extension $$\\Q(\\sqrt{d})$$ of the rationals. Dividing out $$(x - x_1)(x - x_2)$$ from the quartic yields a quadratic polynomial with coefficients in $$\\Q(\\sqrt{d})$$, so the remaining two roots lie in some quadratic extension of this field. One of these four roots is $$\\cos 72^\\circ$$, so by Theorem 15.5.6, the latter is constructible. This implies that the regular pentagon is constructible. 2. It is possible to devise the construction by writing out the explicit form of $$\\cos 72^\\circ$$, which is $$\\frac{\\sqrt{5} - 1}{4}$$, and applying Lemma 15.5.11, constructing first 5, then $$\\sqrt{5}$$, then $$\\sqrt{5} - 1$$, and finally", "# Difference between revisions of \"1996 USAMO Problems/Problem 1\" ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos 5^\\circ)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot", "(4, 30o) 2. (2.5, $$\\ \\pi$$) 3. $$\\ \\left(-1, \\frac{\\pi}{3}\\right)$$ 4. $$\\ \\left(3, \\frac{5 \\pi}{6}\\right)$$ 5. (−2, 300o) Solution Example 3 Use a graphing calculator or plotting program to plot the following equations. 1. r=1+3sinθ 2. r=1+2cosθ Solution 1. Review the steps above under graphing using technology if you are having trouble. Example 4 Recall that $$\\ \\pi r a d=180^{\\circ} \\text { and } 1 \\mathrm{rad}=\\frac{180}{\\pi} \\approx 57.3^{\\circ}$$ 1. $$\\ \\frac{\\pi}{2}$$ 2. 5.17 3. $$\\ \\frac{3 \\pi}{2}$$ Solution 1. If $$\\ \\pi r a d=180^{\\circ}$$ then $$\\ \\frac{\\pi}{2} r a d=90^{\\circ}$$ 2. If $$\\ 1 rad \\approx 57.3^{\\circ}$$ then $$\\ 5.17 rad \\approx 296^{\\circ}$$ 3. If $$\\ \\pi r a d=180^{\\circ}$$ then $$\\ \\frac{3 \\pi}{2} r a d=270^{\\circ}$$ Example 5 Recall that $$\\ \\frac{180^{\\circ}}{\\pi}=57.3^{\\circ} \\approx 1 \\mathrm{rad}$$. 1. 251o 2. 360o 3. 327o Solution 1. If $$\\ 57.3^{\\circ} \\approx 1rad$$ then $$\\ 251^{\\circ} \\approx 4.38 \\mathrm{rad} \\approx 1.4 \\pi \\mathrm{rad}$$ 2. If $$\\ 57.3^{\\circ} \\approx 1 rad$$ then $$\\ 360^{\\circ} \\approx 6.28 rad$$ 3. If $$\\ 57.3^{\\circ} \\approx 1 rad$$ then $$\\ \\frac{327^{\\circ}}{57.3^{\\circ}} \\approx 5.71 rad$$ Example 6 1. 90o 2. 270o 3. 45o Solution 1. If $$\\ \\pi r a d=180^{\\circ}$$ then $$\\ \\frac{\\pi}{2} r a d=90^{\\circ}$$ 2. If $$\\ \\pi r a d=180^{\\circ} \\text { and } \\frac{\\pi}{2} r a d=90^{\\circ}$$ then $$\\ 1 \\frac{1}{2}", "# How do you solve a^2=3^2+5^2-2(3)(5)cos85? May 1, 2018 $a = 31.4$ units #### Explanation: ${a}^{2} = {3}^{2} + {5}^{2} - \\left(2 \\cdot 3 \\cdot 5\\right) \\cos {85}^{\\circ}$. it seems best to first simplify the expression on the right-hand side. ${3}^{2} + {5}^{2} = 9 + 25 = 34$ $2 \\cdot 3 \\cdot 5 = 30 \\rightarrow - \\left(2 \\cdot 3 \\cdot 5\\right) \\cos {85}^{\\circ} = - 30 \\cos {85}^{\\circ}$/ ${3}^{2} + {5}^{2} - \\left(2 \\cdot 3 \\cdot 5\\right) \\cos {85}^{\\circ} = 34 - 30 \\cos {85}^{\\circ}$. hence, ${a}^{2} = 34 - 30 \\cos {85}^{\\circ}$. since side length cannot be negative, $a$ is the positive square root of $34 - 30 \\cos {85}^{\\circ}$. putting $\\sqrt{34 - 30 \\cos {85}^{\\circ}}$ into a calculator gives $31.38532771756$. this can be rounded to $3$ significant figures to give $31.4$. therefore $a = 31.4$ units in length.", "# How to find the exact value of $\\cos(36^\\circ)$? [duplicate] Noting that $t=\\frac{\\pi}{5}$ satisfies $3t=\\pi-2t$, find the exact value of $$\\cos(36^\\circ)$$ it says that you may find useful the following identities: $$\\cos^2 t+\\sin^2 t = 1,\\\\ \\sin 2t = 2\\sin t\\cos t,\\\\ \\sin 3t = 3\\sin t - 4\\sin^3 t.$$ Do I have to do a system of linear equations in function of ..what? $t$? $\\cos$? ## marked as duplicate by Xander Henderson, José Carlos Santos, Cesareo, YiFan, Eevee TrainerMar 30 at 1:15 Let $t=\\frac{\\pi}{5}$ (so $t$ is $36^\\circ$). Since $108=180-72$, we have $3t=\\pi-2t$ and therefore $$\\sin(3t)=\\sin(\\pi-2t).$$ But $\\sin(\\pi-2t)=\\sin(2t)=2\\sin t\\cos t$. Also, by the identity you were given, $\\sin(3t)=3\\sin t-4\\sin^3 t$. Thus $$3\\sin t-4\\sin^3 t=2\\sin t\\cos t.$$ But $\\sin t\\ne 0$, so we can cancel a $\\sin t$ and obtain $$3-4\\sin^2 t=2\\cos t.$$ Substitute $1-\\cos^2 t$ for $\\sin^2 t$ and simplify a bit. We get $$4\\cos^2 t-2\\cos t-1=0.$$ Use the Quadratic Formula to solve this quadratic equation for $\\cos t$, rejecting the negative root. We get $$\\cos t=\\frac{2+\\sqrt{20}}{8}.$$ We can simplify this to $\\dfrac{1+\\sqrt{5}}{4}.$ • Very well explained, thanks very much :) deeply appreciated – Maximilian1988 Feb 11 '13 at 8:31 To find $\\cos{\\pi/5}$, note that $$\\sin{(3 \\pi/5)} = \\sin{(2 \\pi/5)}$$ and $$\\sin{(3 \\pi/5)} = 3 \\sin{(\\pi/5)} - 4 \\sin^3{(\\pi/5)} = 2 \\sin{(\\pi/5)} \\cos{(\\pi/5)}$$ Thus $$2 \\cos{(\\pi/5)} =", "$0^\\circ$ and $360^\\circ$. Using the $\\cos^{-1}$ function on a calculator gives $2x=38.94^\\circ$ (to four significant figures). Thus the general solution is $2x=-38.94 + 360n$, or $2x=38.94+360n$. This gives the four solutions (all to three significant figures) for $x$ between $0^\\circ$ and $360^\\circ$: $x=19.5^\\circ, 161^\\circ, 199^\\circ, 341^\\circ.$ Using a similar method to solve $\\cos 2y=\\dfrac{1}{3}$ gives (all to three significant figures): $y=35.3^\\circ, 145^\\circ, 215^\\circ, 325^\\circ.$ We now need to pair the $x$ values with the corresponding $y$-values. The first equation is equivalent to $3\\cos 2x+2\\cos 2y=3$, and we know the values of $\\cos 2x$ and $\\cos 2y$, so this will not help us further. However, we squared the second equation (thus introducing extra roots), so if we return to the original here, it should tell us more. So we have $\\sin 2y=\\dfrac{3}{2}\\sin 2x$. When $x=19.5$, $\\sin 2y=\\dfrac{3}{2}\\sin 2x=0.943$, so $2y=70.5$ or $109$ or $431$ or $469$, giving $y=35.3$, $54.7$, $215$ or $235$. Comparing these with our list of possible values of $y$ from $\\cos 2y=\\dfrac{1}{3}$ shows that $y=35.3$ or $215$ in this case. Likewise, when $x=161^\\circ$, $\\sin 2y=\\dfrac{3}{2}\\sin 2x=-0.943$, so $2y=251$ or $289$ or $611$ or $649$, giving $y=125$ or $145$ or $305$ or $325$. Thus, comparing with the possible $y$ values gives $y=145$ or $325$ in this case. In the third case, when $x=199$, we have $\\sin 2y=0.943$", "have that $\\sin18^\\circ=\\frac{-1+\\sqrt{5}}{4}.$ Find all solutions between $0^\\circ$ and $360^\\circ$ of the equation $4\\tan 2x=\\cot x.$ We play a similar game here to that of the first part of the question. Firstly, we can write $\\tan x=\\frac{s}{c}, \\quad \\cot x=\\frac{1}{\\tan x}=\\frac{c}{s},$ so the equation we need to solve becomes $4\\frac{\\sin(2x)}{\\cos(2x)}=\\frac{c}{s} \\implies 4\\sin(2x)s-\\cos(2x)c=0 \\implies 8s^2c-(1-2s^2)c=0 \\implies (10s^2-1)c=0.$ So we have that $\\cos x=0$ solves this, which gives us the values $x=90^\\circ, 270^\\circ$ between $0^\\circ$ and $360^\\circ$. We also have that $\\sin x=\\pm\\frac{1}{\\sqrt{10}}$ is a solution, so that $x=\\sin^{-1}(\\pm1/\\sqrt{10})$ are also solutions. To work out how many solutions between $0^\\circ$ and $360^\\circ$, consider We see that $\\sin x=\\pm1/\\sqrt{10}$ four times between $0^\\circ$ and $360^\\circ$. Using a calculator, one finds that $\\sin^{-1}(1/\\sqrt{10})=18.4^\\circ$ to one decimal place. From the symmetries of sine, we see that the other angles are then $180^\\circ-18.4^\\circ=161.6^\\circ$, $180^\\circ+18.4^\\circ=198.4^\\circ$ and $360^\\circ-18.4^\\circ=341.6^\\circ$, giving us six solutions in total.", "1996 USAMO Problems/Problem 1 Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. Solution Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos 5^\\circ)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$, as desired. $\\Box$ 1996 USAMO (Problems • Resources)", "+ 315^\\circ) + i \\sin(56.3^\\circ + 315^\\circ)\\\\ & = 5.09(\\cos 371.3^\\circ + i \\sin 371.3^\\circ)\\\\ & = 5.09(\\cos 11.3^\\circ + i \\sin 11.3^\\circ)$ $^*$ Note: Angles are expressed $0^\\circ \\le \\theta \\le 360^\\circ$ unless otherwise stated. #### Example B Find the product of $5 \\left(\\cos \\frac{3\\pi}{4}+i \\sin \\frac{3\\pi}{4} \\right ) \\cdot \\sqrt{3} \\left (\\cos \\frac{\\pi}{2}+i \\sin \\frac{\\pi}{2} \\right )$ Solution: First, calculate $r_1r_2=5 \\cdot \\sqrt{3}=5\\sqrt{3}$ and $\\theta =\\theta_1+\\theta_2=\\frac{3\\pi}{4}+\\frac{\\pi}{2}=\\frac{5\\pi}{4}$ $5\\sqrt{3} \\left (\\cos \\frac{5\\pi}{4}+i \\sin \\frac{5\\pi}{4} \\right )$ #### Example C Find the product of the numbers $r_1 = 1 + i$ and $r_2 = \\sqrt{3} - i$ by first converting them to trigonometric form. Solution: First, convert $1 + i$ to polar form: $r_1 = \\sqrt{1^2 + 1^2} = \\sqrt{2}\\\\\\theta_1 = \\arctan \\frac{1}{1}= \\frac{\\pi}{4}\\\\$ And now do the same with $\\sqrt{3} - i$ : $r_2 = \\sqrt{\\sqrt{3}^2 + (-1)^2} = 2\\\\\\theta_2 = \\arctan -\\frac{1}{\\sqrt{3}} = -\\frac{\\pi}{6}\\\\$ And now substituting these values into the product theorem: $r_1 \\times r_2 =\\\\ (2)(\\sqrt{2}) \\left( \\cos \\left( -\\frac{\\pi}{6} + \\frac{\\pi}{4} \\right) + i\\sin \\left( -\\frac{\\pi}{6} + \\frac{\\pi}{4} \\right) \\right)\\\\(2)(\\sqrt{2}) \\left( \\cos \\left( \\frac{\\pi}{12} \\right) + i\\sin \\left( \\frac{\\pi}{12} \\right) \\right)$ ### Vocabulary Product Theorem: The product theorem is a theorem showing a simplified way to multiply complex numbers. ### Guided Practice 1. Multiply together the following complex numbers. If they are not in", "B = \\frac{c^2 + a^2 − b^2}{2ca}$$ $$⇒ cos B = \\frac{4^2 + 2^2 − 3^2}{ 2(4)(2)} = 0.6875$$ $$\\boxed{⇒ B = 46.6^\\circ}$$ Now we use the formula for $$c^{2}$$ to find C: $$c^2 = a^2 + b^2 − 2ab\\;cos C ⇒ cos C = \\frac{a^2 + b^2 − c^2}{ 2ab}$$ $$⇒ cos C = \\frac{2^2 + 3^2 − 4^2}{ 2(2)(3)} = −0.25$$ $$\\boxed{⇒ C = 104.5^\\circ}$$ Thus, A = 180º −BC = 180º −46.6º −104.5º⇒ $$\\boxed{A = 28.9º}$$ . It may seem that there is always a solution in Case 4 (given all three sides), but that is not true, as the following example shows. Example 4 Case 4: Three sides. Solve the triangle ∆ABC given a = 2, b = 3, and c = 6. Solution: If we blindly try to use the Law of Cosines to find A, we get $$a^2 = b^2 + c^2 − 2bc cos A ⇒ cos A = \\frac{b^2 + c^2 − a^2}{2bc} = \\frac{3^2 + 6^2 − 2^2}{2(3)(6)} = 1.139 ,$$ which is impossible since |cos A| ≤ 1. Thus, there is $$\\boxed{\\text{no solution}}$$ . We could have saved ourselves some effort by recognizing that the length of one of the sides (c = 6) is greater than the sums of the lengths of the remaining sides (a =", "# Thread: Need help on this tricky question 1. ## Need help on this tricky question Find the sum: Cos^2 (0) + Cos^2 (2) + Cos^2 (4) + Cos^2 (356) + Cos^2 (358) + Cos^2 (360) 2. Originally Posted by JohnLord123 Find the sum: Cos^2 (0) + Cos^2 (2) + Cos^2 (4) + Cos^2 (356) + Cos^2 (358) + Cos^2 (360) 1. I assume that the argument of the trigonometric functions is in degree (?). 2. You can simplify the given term a little bit: Since $\\cos^2(0^\\circ) = \\cos^2(360^\\circ) = 1$ $\\cos^2(2^\\circ) = \\cos^2(358^\\circ)$ $\\cos^2(4^\\circ) = \\cos^2(356^\\circ)$ you'll get: $\\begin{array}{l} \\cos^2 (0^\\circ) + \\cos^2 (2^\\circ) + \\cos^2 (4^\\circ) + \\cos^2 (356^\\circ) + \\cos^2 (358^\\circ) + \\cos^2 (360^\\circ)=\\\\ 2 + 2\\cos^2(2^\\circ)+2\\cos^2(4^\\circ)\\end{array}$ Since $2\\cos^2(2^\\circ) = \\cos(4^\\circ)+1$ the term simplifies to: $2+1+ \\cos(4^\\circ)+2\\cos^2(4^\\circ) = \\boxed{3+ \\cos(4^\\circ)(1+2\\cos(4^\\circ))}$", "down the four roots of $-8 + 8\\sqrt{3}i$. • $2(\\cos 30^{\\circ} + i \\sin 30^{\\circ})$ • $2(\\cos 120^{\\circ} + i \\sin 120^{\\circ})$ • $2(\\cos 210^{\\circ} + i \\sin 210^{\\circ})$ • $2(\\cos 300^{\\circ} + i \\sin 300^{\\circ})$ We can even convert the roots to rectangular form as shown by evaluating the cosine and sine values then distributing $2$ each time. Polar Form Rectangular Form $2(\\cos 30^{\\circ} + i \\sin 30^{\\circ})$ \\begin{aligned} 2(\\cos 30^{\\circ} + i \\sin 30^{\\circ}) &= 2\\left(\\dfrac{\\sqrt{3}}{2}+ \\dfrac{1}{2}i\\right) \\\\&= 2 \\cdot \\dfrac{\\sqrt{3}}{2}+ 2\\cdot \\dfrac{1}{2}i \\\\&=\\sqrt{3} + i \\end{aligned} $2(\\cos 120^{\\circ} + i \\sin 120^{\\circ})$ \\begin{aligned} 2(\\cos 120^{\\circ} + i \\sin 120^{\\circ}) &= 2\\left(-\\dfrac{1}{2}+ \\dfrac{\\sqrt{3}}{2}i\\right) \\\\&= 2 \\cdot -\\dfrac{1}{2}+ 2\\cdot \\dfrac{\\sqrt{3}}{2} i \\\\&=-1 + \\sqrt{3}i \\end{aligned} $2(\\cos 210^{\\circ} + i \\sin 210^{\\circ})$ \\begin{aligned} 2(\\cos 210^{\\circ} + i \\sin 210^{\\circ}) &= 2\\left(-\\dfrac{\\sqrt{3}}{2}- \\dfrac{1}{2}i\\right) \\\\&= 2 \\cdot -\\dfrac{\\sqrt{3}}{2}- 2\\cdot \\dfrac{1}{2} i \\\\&=-\\sqrt{3} – i \\end{aligned} $2(\\cos 300^{\\circ} + i \\sin 300^{\\circ})$ \\begin{aligned} 2(\\cos 300^{\\circ} + i \\sin 300^{\\circ}) &= 2\\left(\\dfrac{1}{2}- \\dfrac{\\sqrt{3}}{2}i\\right) \\\\&= 2 \\cdot \\dfrac{1}{2}- 2\\cdot \\dfrac{\\sqrt{3}}{2} i \\\\&=1 – \\sqrt{3}i \\end{aligned} Hence, we’ve just shown that we can find the remaining roots geometrically and even convert the result in rectangular form. ### Practice Questions 1. Which of the following shows the complex fourth roots of $16\\left(\\cos \\dfrac{4\\pi}{3} + i\\sin \\dfrac{4\\pi}{3}\\right)$? 2. Which of the following shows the complex fourth", "Here's the question you clicked on: ## haleyking345 Group Title Solve the equation: a.) 2 cos^2x-cos x=1 one year ago one year ago • This Question is Closed 1. Marx Group Title let's define $t:= \\cos(x)$ we have the quadratic equation: $2t^2-t=1$ can you find it's solutions? 2. haleyking345 Group Title Would it be cos=1? Which means cos=0 degrees or 360 degrees? 3. Marx Group Title there are two solutions, $\\cos(x)=1$ and $\\cos(x)=-\\frac{1}{2}$ so x can be either $360^\\circ{} k$ (for every integer k) or $120^\\circ{}+ 360^{\\circ}k$ or $240^\\circ{}+ 360^{\\circ}k$", "# 1996 USAMO Problems/Problem 1 ## Problem Prove that the average of the numbers $n\\sin n^{\\circ}\\; (n = 2,4,6,\\ldots,180)$ is $\\cot 1^\\circ$. ## Solution ### Solution 1 First, as $180\\sin{180^\\circ}=0,$ we omit that term. Now, we multiply by $\\sin 1^\\circ$ to get, after using product to sum, $(\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos5)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)$. This simplifies to $\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5^\\circ+\\cos 7^\\circ+...+\\cos 177^\\circ-89\\cos 179^\\circ$. Since $\\cos x=-\\cos(180-x),$ this simplifies to $90\\cos 1^\\circ$. We multiplied by $\\sin 1^\\circ$ in the beginning, so we must divide by it now, and thus the sum is just $90\\cot 1^\\circ$, so the average is $\\cot 1^\\circ$, as desired. $\\Box$ ### Solution 2 Notice that for every $n\\sin n^\\circ$ there exists a corresponding pair term $(180^\\circ - n)\\sin{180^\\circ - n} = (180^\\circ - n)\\sin n^\\circ$, for $n$ not $90^\\circ$. Pairing gives the sum of all $n\\sin n^\\circ$ terms to be $90(\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ)$, and thus the average is $$S = (\\sin 2^\\circ + \\sin 4^\\circ + ... + \\sin 178^\\circ). (*)$$ We need to show that $S = \\cot 1^\\circ$. Multiplying (*) by $2\\sin 1^\\circ$ and using sum-to-product and telescoping gives $2\\sin 1^\\circ S = \\cos 1^\\circ - \\cos 179^\\circ = 2\\cos 1^\\circ$. Thus, $S = \\frac{\\cos 1^\\circ}{\\sin 1^\\circ} = \\cot 1^\\circ$, as desired. $\\Box$ ###" ]

[ "+0 # let x and y be vectors such that and compute the ratio ​ 0 61 1 let x and y be vectors such that $$\\operatorname{proj}_{\\mathbf{x}}(\\mathbf{y})=\\dbinom{5}{3}$$and $$\\operatorname{proj}_{\\mathbf{y}}(\\mathbf{x})=\\dbinom{2}{-2}$$ compute the ratio $$\\dfrac{\\|\\mathbf{x}\\|}{\\|\\mathbf{y}\\|}$$ Feb 13, 2022", "+0 # Let $\\mathbf{v}$ and $\\mathbf{w}$ be vectors such that \\[\\operatorname{proj}_{\\bold{w}}( \\bold{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.\\ 0 113 1 Let v and w be vectors such that $$\\operatorname{proj}_{{w}}({v})= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.$$ Find $$\\operatorname{proj}_{-2{w}} (3{v})$$. Oct 9, 2019", "= 0\\text{,}$$ the variable $$y$$ can assume any value and $$x = - (b/a)y\\text{.}$$ Thus, the solutions to our system lie along a line through the origin. In fact, we will always get a line of solutions through the origin as long as at least one entry in our matrix is nonzero. 1 Subsection3.1.2Linear Independence We say that two vectors $${\\mathbf x}$$ and $${\\mathbf y}$$ in $${\\mathbb R}^2$$ are linearly independent if they do not lie on the same line through the origin. If, on the other hand, they do lie on the same line, then the vectors are linearly dependent. Equivalently, two vectors are linearly dependent if one vector is a multiple of the other. We leave the proof of the following theorem as an exercise. If we have a pair of linearly independent vectors in $${\\mathbb R}^2\\text{,}$$ then we can write any vector in $${\\mathbb R}^2$$ as a unique linear combination of the two vectors. That is, given two linearly independent vectors $${\\mathbf x} = (x_1, x_2)$$ and $${\\mathbf y} = (y_1, y_2)\\text{,}$$ we can write $${\\mathbf z} = (z_1, z_2)$$ as \\begin{equation*} \\begin{pmatrix} z_1 \\\\ z_2 \\end{pmatrix} = \\alpha \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} + \\beta \\begin{pmatrix} y_1 \\\\ y_2 \\end{pmatrix}, \\end{equation*} where $$\\alpha$$ and $$\\beta$$ are unique. To see why this is true, we", "# Vectors# Author(s): Translation: ## Definition# The term vector has several interpretations: mathematical, geometric, physical, etc. The exact meaning of the term depends on the context. Formally, a vector is defined as an element of vector space – a set with addition and multiplication of a vector by a number (scalar) operations, which satisfy 8 axioms. For simplicity, let us consider the rectangular (Cartesian) coordinate system in the plane, familiar from school days – two perpendicular to each other axes $$x$$ and $$y$$, selected unit vectors (orthogonal) $$\\mathbf{e}_1$$, $$\\mathbf{e}_2$$ on them and the origin of coordinates. The vector $$\\mathbf{a}$$ in this coordinate system can be written as follows: $$\\mathbf{a} = a_1 \\mathbf{e}_1 + a_2 \\mathbf{e}_2 = \\begin{pmatrix} a_1 \\\\ a_2 \\end{pmatrix}$$ Note 1: Vector coordinates do not uniquely define its position in the plane, but only the position of the vector’s end relative to its start. For example, a vector $$\\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix}$$ can be drawn either from the origin $$\\begin{pmatrix} 0 , 0 \\end{pmatrix}$$ with the end at $$\\begin{pmatrix} 3 , 4 \\end{pmatrix}$$, or from an arbitrary point, e.g. $$\\begin{pmatrix} 1 , 1 \\end{pmatrix}$$ with the end at $$\\begin{pmatrix} 4 , 5 \\end{pmatrix}$$. Both of these vectors correspond to 3 orthogonal unite vectors on the $$x$$-axis and 4 on the $$y$$-axis. Normally, unless otherwise stated,", "of the problem. Assume that two images are given, where $\\mx{P}_L$ is the camera matrix corresponding to the first image and $\\mx{P}_R$ is the camera matrix for the second image. Assume that one 3D point with homogeneous coordinate vector $\\vc{u}$ is seen in both images. In the first image, the homogeneous coordinate vector is $\\vc{v}_L$ and in the other image the vector is $\\vc{v}_R$. The camera matrix equations for these two images are $$\\lambda_L \\vc{v}_L = \\mx{P}_L \\vc{u}\\ \\ \\ \\ \\mathrm{and} \\ \\ \\ \\ \\lambda_R \\vc{v}_R = \\mx{P}_R \\vc{u}.$$ (7.92) These two matrix equations can be written as a joint matrix equation, i.e., $$\\begin{pmatrix} \\mx{P}_L & \\vc{v}_L & \\vc{0} \\\\ \\mx{P}_R & \\vc{0} & \\vc{v}_R \\end{pmatrix} \\begin{pmatrix} \\vc{u}\\\\ -\\lambda_L \\\\ -\\lambda_R \\end{pmatrix} = \\begin{pmatrix} \\vc{0}\\\\ \\vc{0} \\end{pmatrix} .$$ (7.93) Notice that the matrices and vectors are made out of blocks of other matrices or vectors. One often talks of block matrices. The matrix equation is of type $\\mx{A} \\vc{x} = \\vc{0}$. Here $\\mx{A}$ is a square matrix of size $6 \\times 6$. Notice that the matrix equation has a solution $\\begin{pmatrix}\\vc{u}\\\\-\\lambda_L \\\\-\\lambda_R \\end{pmatrix}$ that is not the zero vector. According to Theorem 7.10, this means that $\\det \\mx{A} = 0$. If we expand the determinant first along the fifth column and then along the sixth column", "# Find a set of vector B Find a set of vector B such that $$\\operatorname{span}(B)=\\{(x,y,z,w)^T\\mid 2x+y+w=0, y+2z=0\\}$$ in $\\mathbb{R}^4$. This is what I have so far I don't know if it is correct. $$x\\begin{pmatrix} a_1 \\\\ b_1 \\\\ c_1 \\\\ d_1 \\end{pmatrix}+ y\\begin{pmatrix} a_2 \\\\ b_2 \\\\ c_2 \\\\ d_2 \\end{pmatrix}+ z\\begin{pmatrix} a_3 \\\\ b_3 \\\\ c_3 \\\\ d_3 \\end{pmatrix}+w \\begin{pmatrix} a_4 \\\\ b_4 \\\\ c_4 \\\\ d_4 \\end{pmatrix}$$ where $2x+y+w=0$ and $y+2z=0$. Then $w=-2x-y$, $y=2z$, $w=-2x+2z$. Then $$x\\begin{pmatrix} b_1 \\\\ b_2 \\\\ b_3 \\\\ b_4 \\end{pmatrix}+ y\\begin{pmatrix} α_1 \\\\ α_2 \\\\ α_3 \\\\ α_4 \\end{pmatrix}+ z\\begin{pmatrix} -2c_1 \\\\ -2c_2 \\\\ -2c_3 \\\\ -2c_4 \\end{pmatrix} +w\\begin{pmatrix} -2b-α_1 \\\\ -2b-α_2 \\\\ -2b-α_3 \\\\ -2b-α_4 \\end{pmatrix}$$ Thus $$2b+α_1+-2b_1-α_1=0$$ and $$α_1-2c=0$$ where $$c_1=1/2α_1$$ Not sure where you deduced $w=-2x-y$ and $y+2z=0$. What you need to find is the solution set of $$\\begin{cases} 2x+y+w=0 \\\\ y+2z=0 \\end{cases}$$ so $w=-2x-y$ and $y=-2z$; hence $w=-2x+2z$, so the vectors are of the form $$\\begin{pmatrix} a \\\\ -2b \\\\ b \\\\ -2a-2b \\end{pmatrix}$$ where $a$ and $b$ are arbitrary. Since $$\\begin{pmatrix} a \\\\ -2b \\\\ b \\\\ -2a-2b \\end{pmatrix} = a\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\\\ -2 \\end{pmatrix} + b\\begin{pmatrix} 0 \\\\ -2 \\\\ 1 \\\\ -2 \\end{pmatrix}$$ you are done. • Thanks, it helped me a lot. – marie Feb 15 '18 at", "no solution $(a,b)$, or explain why no such vector exists. 1. Give an example of a vector $w = \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}$ so that the equation \\begin{equation*}a \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} + b \\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\end{equation*} has no solution $(a,b)$, or explain why no such vector exists. 2. Give an example of a vector $w = \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}$ so that the equation \\begin{equation*} a \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} + b \\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\end{equation*} has exactly one solution $(a,b)$, or explain why no such vector exists. 1. Give an example of a vector $X = \\begin{pmatrix} x \\\\ y \\\\ z\\end{pmatrix}$ such that the equation \\begin{equation*} a \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} + b \\begin{pmatrix} 0 \\\\ 0 \\\\ 1\\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}x \\\\ y \\\\ z \\end{pmatrix} \\end{equation*} has no solution $(a,b,c)$, or explain why no such vector exists. 2. Give an example of a vector $X = \\begin{pmatrix} x \\\\ y \\\\ z\\end{pmatrix}$ such that the equation \\begin{equation*} a \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} + b \\begin{pmatrix}", "to the basis $$\\{ {\\mathbf v}_1, {\\mathbf v}_2 \\}\\text{.}$$ The vectors $$(1, 1)$$ and $$(-1, -1)$$ do not form a basis for $${\\mathbb R}^2$$ since these two vectors lie along the same line. If $$2 \\times 2$$ matrices and the rest of what we have described above make you nervous, you should work through the exercises at the end of this section. Activity3.1.1.Matrix Operations. Given the matrices and vectors \\begin{equation*} A = \\begin{pmatrix} 5 & 3 \\\\ -6 & 4 \\end{pmatrix}, B = \\begin{pmatrix} 2 & 3 \\\\ 1 & 2 \\end{pmatrix}, \\mathbf x = \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} \\mathbf y = \\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix} \\end{equation*} compute each of the following expressions. (a) $$AB\\text{,}$$ $$BA$$ (b) $$A^{-1}\\text{,}$$ $$B^{-1}\\text{,}$$ $$(AB)^{-1}\\text{,}$$ $$B^{-1}A^{-1}$$ (c) $$\\det(A), \\det(B), \\det(AB), \\det(A^{-1})$$ (d) $$A \\mathbf x\\text{,}$$ $$A \\mathbf y\\text{,}$$ $$\\mathbf y^T \\mathbf x\\text{,}$$ $$\\mathbf x \\mathbf y^T\\text{,}$$ where $$\\mathbf y^T = (2, -1)$$ Subsection3.1.3Finding Eigenvalues and Eigenvectors A nonzero vector $${\\mathbf v}$$ is an eigenvector of $$A$$ if $$A {\\mathbf v} = \\lambda {\\mathbf v}$$ for some $$\\lambda \\in {\\mathbf R}\\text{.}$$ The constant $$\\lambda$$ is called an eigenvalue of $$A\\text{.}$$ Letting \\begin{equation*} A = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\quad \\text{and} \\quad \\mathbf v = \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\neq \\mathbf 0, \\end{equation*} we have $$A \\mathbf", "7 \\end{pmatrix}$$ The answer-vector is $\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$ away from $\\begin{pmatrix} 12 \\\\ 6 \\end{pmatrix}$ and $\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$ is a third of the sum of the x-vector and the y-vector. So if we do $6\\frac{1}{3}$ of the x-vector and $\\frac{1}{3}$ of the y-vector we'll get the right answer. That is, $x = 6\\frac{1}{3} = \\frac{19}{3}$ and $y = \\frac{1}{3}$. As you can see, there's no \"method\" per se, but thinking of it as a vector equation can help. -", "must solve the equations \\begin{align*} z_1 & = \\alpha x_1 + \\beta y_1\\\\ z_2 & = \\alpha x_2 + \\beta y_2 \\end{align*} for $$\\alpha$$ and $$\\beta\\text{.}$$ However, this system has a unique solution since \\begin{equation*} \\det \\begin{pmatrix} x_1 & y_1 \\\\ x_2 & y_2 \\end{pmatrix} \\neq 0. \\end{equation*} Two vectors are said to be a basis for $${\\mathbb R}^2$$ if we can write any vector in $${\\mathbb R}^2$$ as a linear combination of these two vectors. By our arguments above, any two linearly independent vectors will form a basis for $${\\mathbb R}^2\\text{.}$$ The vectors $${\\mathbf e}_1 = (1, 0)$$ and $${\\mathbf e}_2 = (0, 1)$$ form a basis for $${\\mathbb R}^2\\text{.}$$ Indeed, if $${\\mathbf z} = (z_1, z_2)\\text{,}$$ then we can write \\begin{equation*} {\\mathbf z} = z_1 {\\mathbf e}_1 + z_2 {\\mathbf e}_2. \\end{equation*} The vectors $${\\mathbf e}_1$$ and $${\\mathbf e}_2$$ are called the standard basis for $${\\mathbb R}^2\\text{.}$$ Let $${\\mathbf v}_1 = (2,1)$$ and $${\\mathbf v}_2 = (3, 2)\\text{.}$$ Since \\begin{equation*} \\det \\begin{pmatrix} 2 & 3 \\\\ 1 & 2 \\end{pmatrix} \\neq 0, \\end{equation*} these vectors form a basis for $${\\mathbb R}^2\\text{.}$$ If $${\\mathbf z} = (-5, -4)\\text{,}$$ then we can write \\begin{equation*} {\\mathbf z} = 2 {\\mathbf v}_1 - 3 {\\mathbf v}_2. \\end{equation*} We say that the coordinates of $${\\mathbf z}$$ are $$(2, -3)$$ with respect", "# Thread: Spans and vectors 1. ## Spans and vectors If a question asks, Is the vector $\\mathbf{b}=\\begin{pmatrix}-2\\\\ -6\\\\ -4\\end{pmatrix}\\in \\mbox{span}(\\mathbf{v_1, v_2, v_3, v_4})$ where: $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 3\\\\ 0\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}2\\\\ 2\\\\ 1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ -1\\end{pmatrix}$, $\\mathbf{v_4}=\\begin{pmatrix}1\\\\ -2\\\\ 1\\end{pmatrix}$ Do we just make it a matrix and solve the matrix? And how would we solve it if we have 4 variables but only 3 equations? 2) Is the set of vectors $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 2\\\\ 3\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}1\\\\ 1\\\\ -1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ 5\\end{pmatrix}$ a spanning set of $\\mathbb{R}^3$? Do we just make this a matrix with say $\\mathbf{x}=\\begin{pmatrix}x_1\\\\ x_2\\\\ x_3\\end{pmatrix}$ as our solution and see if we can solve the matrix? 2. So your first question is asking whether there exist scalars $a_{1},a_{2},a_{3},a_{4}$ such that $\\mathbf{b}=a_{1}\\mathbf{v}_{1}+a_{2}\\mathbf{v}_{2} +a_{3}\\mathbf{v}_{3}+a_{4}\\mathbf{v}_{4}.$ This equation you can translate into a normal matrix problem of the form $A\\mathbf{x}=\\mathbf{b}$. Row reduce that problem. If you find any solutions (one or infinitely many), then the answer is yes. If you find that there are no solutions, then the answer is no. Your second question, again, is asking whether there exist scalars $c_{1}, c_{2}, c_{3}$ such that $c_{1}\\mathbf{v}_{1}+c_{2}\\mathbf{v}_{2}+c_{3}\\math bf{v}_{3}=\\mathbf{x}$ for your arbitrary $\\mathbf{x}.$ This, again, can be translated into a standard matrix problem. Since this problem will result in a square matrix, you can use the theory of determinants to help you out.", "is defined as the sum of the products of the components. Symbolically we write $\\begin{pmatrix} a_1 \\\\ a_2 \\end{pmatrix}\\cdot\\begin{pmatrix}b_1 \\\\ b_2 \\end{pmatrix} = a_1b_1 + a_2b_2$ For example, $\\begin{pmatrix} 3 \\\\ 5 \\end{pmatrix}\\cdot\\begin{pmatrix} 1 \\\\ -2 \\end{pmatrix}=3-10=-7$ The dot product of two vectors has an alternate form: $\\mathbf{a}\\cdot\\mathbf{b}=|\\mathbf{a}||\\mathbf{b}|\\cos{\\theta}$ The angle θ then is important, as it shows that the dot product of two vectors is related to the angle between them. More specifically, we can calculate the dot product of two vectors - if the dot product is zero we can then say that the two vectors are perpendicular. For example, consider simply $\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}\\cdot\\begin{pmatrix}1 \\\\ -1\\end{pmatrix}=1-1=0$ Plot these vectors on the plane and verify for yourself that these vectors are perpendicular. ## Vectors of 3D-lines Cartesian equation: $\\tfrac{x-a}{l}=\\tfrac{y-b}{m}=\\tfrac{z-c}{n}$. where a, b, and c are the coordinates on the vector line. Vector equation: r = $\\begin{pmatrix} a \\\\ b \\\\ c \\end{pmatrix}$", "of the smallest such set S which exists. ## Preliminaries It is useful to view our mxn matrix A as a collection of m points located in an n-dimensional space (i.e. each row of A represents a point in $$\\mathbb{R}^{n}$$). For example, the matrix below gives us the location of 4 points in 3D space: $A = \\begin{pmatrix} 1 & 1 & 1 \\\\ 2 & 3 & 2 \\\\ 1 & 2 & 3 \\\\ 4 & 2 & 1 \\\\ \\end{pmatrix}$ If we use the Cartesian Coordinate System, we can describe each point by three numbers, one for each of the x, y, and z axes. In other words, we can describe the point (1, 2, 3) as a linear combination of three vectors, each of which represents one of x, y, and z axes: $\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\\\ \\end{pmatrix} = 1 \\times \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\\\ \\end{pmatrix} + 2 \\times \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\\\ \\end{pmatrix} + 3 \\times \\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\\\ \\end{pmatrix}$ Note that we can represent any point in $$\\mathbb{R}^3$$ as a linear combination of these three vectors - these three vectors are thus said to Span the whole of $$\\mathbb{R}^{3}$$ and are collectively said to form a Basis for $$\\mathbb{R}^{3}$$.", "+0 # Help with vector geometry +1 88 1 +38 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w} \\text{ and }\\mathbf{x}$$ are linearly independent, answer with 0. If they aren't, find coefficients a,b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a+b}{c}$$ Jun 5, 2019", "be $$c\\begin{pmatrix}x\\\\y\\end{pmatrix} = \\begin{pmatrix}cx\\\\cy\\end{pmatrix}$$ Of course, vectors need not be two-dimensional. They can have 3, 4, or any positive integer number of dimensions. As an example, below is an image of the three-dimensional vector $\\displaystyle{\\begin{pmatrix}2\\\\3\\\\5\\end{pmatrix}}$ The addition and scalar multiplication of these higher-dimensional vectors is defined in a perfectly analogous way. Consider these operations defined three-dimensional vectors: $$\\begin{pmatrix}x_1\\\\y_1\\\\z_1\\end{pmatrix} + \\begin{pmatrix}x_2\\\\y_2\\\\z_2\\end{pmatrix} = \\begin{pmatrix}x_1+x_2\\\\y_1+y_2\\\\z_1+z_2\\end{pmatrix} \\quad \\textrm{ and } \\quad c\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} = \\begin{pmatrix}cx\\\\cy\\\\cz\\end{pmatrix}$$", "&1 &\\sqrt{2}\\\\ 0 &0 &\\sqrt{2}+1\\end{bmatrix}. $ The equations these computations represent are as follows: $\\begin{bmatrix}1 &1 &\\sqrt{2}\\\\ 0 &0 &\\sqrt{2}+1\\end{bmatrix}\\begin{bmatrix}x\\\\y\\\\z\\end {bmatrix}=\\begin{bmatrix}0\\\\0\\end{bmatrix},$ and hence we see immediately that $z=0.$ Plugging this back in, we see that $x+y=0,$ or $x=-y.$ Setting $t=y$ as our parameter, we can represent the equation of the line as $\\mathbf{r}(t)=t\\begin{bmatrix}-1\\\\1\\\\0\\end{bmatrix}.$ Hence, a vector parallel to the line generated by the intersection of these two planes is given by $\\mathbf{r}(1)-\\mathbf{r}(0)=\\begin{bmatrix}-1\\\\1\\\\0\\end{bmatrix}.$ Perform this same procedure with the intersection of the tilted wall and the ground, and you'll get another vector. This constitutes step 1. Does all this make sense? 3. Most of it, but where does $\\mathbf{r}(t)=t\\begin{bmatrix}-1\\\\1\\\\0\\end{bmatrix}.$ come from? 4. It comes from the general equation of a line in 3D. You have $\\mathbf{r}(t)=t\\mathbf{v}+\\mathbf{b},$ where $\\mathbf{v}$ is a vector in the direction of the line, and $\\mathbf{b}$ is a point on the line. It's essentially the same thing as your $y=mx+b$. In this case, the $\\mathbf{b}=0,$ since the line goes through the origin (plug in $t=0$, and you'll see that). Make sense? 5. the procedure yes i think so...but will take a bit of going over. 6. Ok. Let me know if you have more questions. 7. So once I have the two vectors, they go into $\\displaystyle\\alpha=\\cos^{-1}\\left(\\frac{\\mathbf{x}\\cdot\\mathbf{y}}{|\\mathbf{ x}|\\,|\\mathbf{y}|}\\right)$ (step 3 is just a reordered step 2?)", "(-5, -4)\\text{,}$ then we can write \\begin{equation*} {\\mathbf z} = 2 {\\mathbf v}_1 - 3 {\\mathbf v}_2. \\end{equation*} We say that the coordinates of ${\\mathbf z}$ are $(2, -3)$ with respect to the basis $\\{ {\\mathbf v}_1, {\\mathbf v}_2 \\}\\text{.}$ ###### Example3.1.4 The vectors $(1, 1)$ and $(-1, -1)$ do not form a basis for ${\\mathbb R}^2$ since these two vectors lie along the same line. If $2 \\times 2$ matrices and the rest of what we have described above make you nervous, you should work through the exercises at the end of this section. ### Subsection3.1.3Finding Eigenvalues and Eigenvectors A nonzero vector ${\\mathbf v}$ is an eigenvector of $A$ if $A {\\mathbf v} = \\lambda {\\mathbf v}$ for some $\\lambda \\in {\\mathbf R}\\text{.}$ The constant $\\lambda$ is called an eigenvalue of $A\\text{.}$ Letting \\begin{equation*} A = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\quad \\text{and} \\quad \\mathbf v = \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\neq \\mathbf 0, \\end{equation*} we have $A \\mathbf x = \\lambda \\mathbf x$ or $A \\mathbf x - \\lambda \\mathbf x = \\mathbf 0\\text{.}$ In matrix form this is \\begin{align*} \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\end{pmatrix} - \\lambda \\begin{pmatrix} x \\\\ y \\end{pmatrix} \\amp = \\begin{pmatrix} a & b \\\\ c & d", "Vectors Satisfy $A\\mathbf{x}=0$ then Find Another Solution Suppose that the vectors \\[\\mathbf{v}_1=\\begin{bmatrix} -2 \\\\ 1 \\\\ 0 \\\\ 0 \\\\ 0 \\end{bmatrix}, \\qquad \\mathbf{v}_2=\\begin{bmatrix} -4 \\\\ 0... Close", "of geometric object does $$N\\cdot X= N\\cdot P$$ describe? 7. Let $$u=\\begin{pmatrix}1\\\\1\\\\1\\\\ \\vdots \\\\ 1\\end{pmatrix} {\\rm ~and~} v= \\begin{pmatrix}1\\\\2\\\\3\\\\ \\vdots\\\\ \\! 101\\!\\end{pmatrix}$$ Find the projection of $$v$$ onto $$u$$ and the projection of $$u$$ onto $$v$$. ($$\\textit{Hint:}$$ Remember that two vectors $$u$$ and $$v$$ define a plane, so first work out how to project one vector onto another in a plane. The picture from Section 14.4 could help.) 8. If the solution set to the equation $$A(x)=b$$ is the set of vectors whose tips lie on the paraboloid $$z=x^{2}+y^{2}$$, then what can you say about the function $$A$$? 9. Find a system of equations whose solution set is $$\\left\\{ \\begin{pmatrix}1\\\\1\\\\2\\\\0\\end{pmatrix} +c_1 \\begin{pmatrix}-1\\\\-1\\\\0\\\\1\\end{pmatrix} +c_2 \\begin{pmatrix}0\\\\0\\\\-1\\\\-3\\end{pmatrix} \\middle| \\,c_1,c_2\\in \\Re \\right\\}.$$ Give a general procedure for going from a parametric description of a hyperplane to a system of equations with that hyperplane as a solution set. 10. If $$A$$ is a linear operator and both $$x=v$$ and $$x=cv$$ (for any real number $$c$$) are solutions to $$Ax=b$$, then what can you say about $$b$$?", "(iii)$ The row vectors of $\\mx{B}$ constitute an orthonormal basis. $\\spc (iv)$ $\\mx{B}^{-1} = \\mx{B}^{\\T}$ $(i)$ and $(ii)$ are simply the definition, and so need not be proved. Next, we show that $(iv)$ and $(ii)$ are equivalent. Assume that we have a orthonormal basis (Definition 3.5) consisting of the following set of vectors, $\\{\\vc{b}_1,\\vc{b}_2,\\vc{b}_3,\\dots,\\vc{b}_n\\}$. Let us put them as column vectors in a matrix, $\\mx{B}$, i.e., \\begin{align} \\mx{B} &= \\begin{pmatrix} | & | & \\dots & | \\\\ \\vc{b}_{1} & \\vc{b}_{2} & \\dots & \\vc{b}_{n} \\\\ | & | & \\dots & | \\\\ \\end{pmatrix}, \\end{align} (6.87) and now the transpose of $\\mx{B}$ multiplied by itself then becomes \\begin{align} \\mx{B}^{\\T} \\mx{B} &= \\begin{pmatrix} -\\,\\,\\, \\vc{b}_{1}^\\T - \\\\ -\\,\\,\\, \\vc{b}_{2}^\\T - \\\\ -\\,\\,\\, \\vc{b}_{3}^\\T - \\\\ \\vdots \\\\ -\\,\\,\\, \\vc{b}_{n}^\\T - \\end{pmatrix} \\begin{pmatrix} | & | & | & \\dots & | \\\\ \\vc{b}_{1} & \\vc{b}_{2} & \\vc{b}_{3} & \\dots & \\vc{b}_{n} \\\\ | & | & | & \\dots & | \\\\ \\end{pmatrix}\\\\ &= \\begin{pmatrix} \\vc{b}_{1}^\\T \\vc{b}_{1} & \\vc{b}_{1}^\\T \\vc{b}_{2} & \\vc{b}_{1}^\\T \\vc{b}_{3} & \\dots & \\vc{b}_{1}^\\T \\vc{b}_{n} \\\\ \\vc{b}_{2}^\\T \\vc{b}_{1} & \\vc{b}_{2}^\\T \\vc{b}_{2} & \\vc{b}_{2}^\\T \\vc{b}_{3} & \\dots & \\vc{b}_{2}^\\T \\vc{b}_{n} \\\\ \\vc{b}_{3}^\\T \\vc{b}_{1} & \\vc{b}_{3}^\\T \\vc{b}_{2} & \\vc{b}_{3}^\\T \\vc{b}_{3} & \\dots & \\vc{b}_{3}^\\T \\vc{b}_{n} \\\\ \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\ \\vc{b}_{n}^\\T" ]

[ "\\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} - \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} = \\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ On the other hand, the reflection of $\\mathbf{x}$ in the plane $$L^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\mathbf{v} \\cdot \\mathbf{y} = 0\\}$$ with normal vector $\\mathbf{v}$ is given by $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = 2\\operatorname{Proj}_{L^\\perp}(\\mathbf{x}) - \\mathbf{x} = 2\\left(\\mathbf{x} - \\operatorname{Proj}_L(\\mathbf{x})\\right) - \\mathbf{x} = \\mathbf{x} - 2\\operatorname{Proj}_L(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v},$$ which in your case yields $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} - 2 \\cdot \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ One last cultural note: the reflection $\\operatorname{Refl}_{L^\\perp}(\\mathbf{x})$ of $\\mathbf{x}$ in the plane $L^\\perp$ with normal vector $\\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image $$H_{\\mathbf{v}}(\\mathbf{x}) := \\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}$$ of $\\mathbf{x}$ under the Householder transformation $H_\\mathbf{v}$ corresponding to $\\mathbf{v}$. So, in your case, as we just saw, $$H_{\\mathbf{v}}(\\mathbf{x}) = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ Suppose you mean a vector ($\\mathbf{v}$) \"specular\" to the given one ($\\mathbf{u}$) with respect to the line $L$. Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin. If you do the cross product $\\mathbf{w} = \\mathbf{u} \\times \\mathbf{L}$ you get a vector which is orthogonal either to the line and to $\\mathbf{u}$, i.e. it is normal to the plane containing the line and $\\mathbf{u}$. Then $\\mathbf{v} \\times \\mathbf{L}$ shall be equal to $-\\mathbf{w}$", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "the representation {\\displaystyle {\\begin{aligned}{\\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}}&=a\\cdot {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}+b\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}+c\\cdot {\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}\\\\[0.5em]&=a\\cdot \\left({\\frac {1}{4}}\\cdot {\\begin{pmatrix}2\\\\1\\\\0\\end{pmatrix}}+{\\frac {1}{4}}\\cdot {\\begin{pmatrix}2\\\\-1\\\\0\\end{pmatrix}}\\right)+b\\cdot \\left({\\frac {1}{2}}\\cdot {\\begin{pmatrix}2\\\\1\\\\0\\end{pmatrix}}-{\\frac {1}{2}}\\cdot {\\begin{pmatrix}2\\\\-1\\\\0\\end{pmatrix}}\\right)+c\\cdot {\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}\\\\[0.5em]&={\\frac {a+2b}{4}}\\cdot {\\begin{pmatrix}2\\\\1\\\\0\\end{pmatrix}}+{\\frac {a-2b}{4}}\\cdot {\\begin{pmatrix}2\\\\-1\\\\0\\end{pmatrix}}+c\\cdot {\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}\\end{aligned}}} Thus ${\\displaystyle (a,b,c)^{T}\\in \\operatorname {span} (\\{v_{1},v_{2},v_{3}\\})}$. Since this vector was arbitrarily chosen, every vector from ${\\displaystyle \\mathbb {R} ^{3}}$ is representable as a linear combination of the linearly independent vectors ${\\displaystyle v_{1},v_{2}}$ and ${\\displaystyle v_{3}}$. Thus ${\\displaystyle \\{v_{1},v_{2},v_{3}\\}}$ is a generator of ${\\displaystyle \\mathbb {R} ^{3}}$. Therefore, we cannot add another vector to ${\\displaystyle v_{1},v_{2}}$ and ${\\displaystyle v_{3}}$, so the system remains linearly independent, since every other vector from ${\\displaystyle \\mathbb {R} ^{3}}$ can be represented as a linear combination of ${\\displaystyle v_{1},v_{2}}$ and ${\\displaystyle v_{3}}$. In other words, ${\\displaystyle v_{1},v_{2}}$ and ${\\displaystyle v_{3}}$ form a maximal system of linearly independent vectors. In summary, we proceeded as follows to find a maximal system of linearly independent vectors: We start with a vector that is not the zero vector. That means we should not consider the null vector space here. Then we proceed step by step: Once we have found linearly independent vectors ${\\displaystyle v_{1},...,v_{n}}$, we form the span of these vectors. If this is the entire vector space, we have found a generator and are done. We choose a vector ${\\displaystyle v_{n+1}}$ that is not in ${\\displaystyle \\operatorname {span} (\\{v_{1},...,v_{n}\\})}$. This contributes a new direction", "E}$${\\ displaystyle {\\ vec {a}} = {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}}}$ ${\\ displaystyle {\\ vec {OF}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {\\ left [{\\ begin {pmatrix} 8 \\\\ 8 \\ \\ -1 \\ end {pmatrix}} - {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}} \\ right] \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}} {{\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {27} {9}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}}}$. The distance of the point from the plane thus satisfies ${\\ displaystyle d (P; E)}$${\\ displaystyle P}$${\\ displaystyle E}$ ${\\ displaystyle d (P; E) = d (P; F) \\; {\\ widehat {=}} \\; | {\\ vec {PF}} | = \\ left | {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}} -", "\\end{pmatrix} - \\begin{pmatrix} \\hat \\phi \\\\ 1 \\end{pmatrix} }\\right)$ $\\displaystyle$ $=$ $\\displaystyle A^n \\begin{pmatrix} \\frac \\phi {\\sqrt 5} \\\\ \\frac 1 {\\sqrt 5} \\end{pmatrix} - A^n \\begin{pmatrix} \\frac {\\hat \\phi} {\\sqrt 5} \\\\ \\frac 1 {\\sqrt 5} \\end{pmatrix}$ $\\displaystyle$ $=$ $\\displaystyle \\phi^n \\begin{pmatrix} \\frac \\phi {\\sqrt 5} \\\\ \\frac 1 {\\sqrt 5} \\end{pmatrix} - \\hat \\phi^n \\begin{pmatrix} \\frac {\\hat \\phi} {\\sqrt 5} \\\\ \\frac 1 {\\sqrt 5} \\end{pmatrix}$ $\\displaystyle$ $=$ $\\displaystyle \\frac 1 {\\sqrt 5} \\begin{pmatrix} \\phi^n \\cdot \\phi - \\hat \\phi^n \\cdot \\hat \\phi \\\\ \\phi^n - \\hat \\phi^n \\end{pmatrix}$ Hence the result. $\\blacksquare$ ## Source of Name This entry was named for Jacques Philippe Marie Binet and Leonhard Paul Euler. ## Also known as The Euler-Binet Formula is also known as Binet's formula.", "1 & 0 \\\\ 1 & 1 \\end{pmatrix}, N_3 = \\begin{pmatrix} 1 & 1 \\\\ 0 & 1 \\end{pmatrix}, N_4 = \\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix}$ Does $\\{ N_1, N_2, N_3, N_4 \\}$ generate V? $M_1 = -\\frac23 N_1 + \\frac13(N_2 + N_3 + N_4)$ $M_2 = -\\frac23 N_2 + \\frac13(N_1 + N_3 + N_4)$ $M_3 = -\\frac23 N_3 + \\frac13(N_1 + N_2 + N_4)$ $M_4 = -\\frac23 N_4 + \\frac13(N_1 + N_2 + N_3)$ Then $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = aM_1 + bM_2 + cM_3 + dM_4 =$ $a \\cdot \\left( -\\frac23 N_1 + \\frac13(N_2 + N_3 + N_4) \\right) + b \\cdot \\left( -\\frac23 N_2 + \\frac13(N_1 + N_3 + N_4) \\right) + \\ldots$ Theorem: If $S \\in \\mathbf V$, then $\\operatorname{span}(S) =$ {all l.c. of elements of S} is a subspace of V.", "( 7,5,-1 ) \\\\ &= \\left( \\frac{437}{75}, -\\frac{4}{15}, -\\frac{266}{75} \\right) \\\\ &\\approx \\left( 5.82667, -0.266665, -3.54667 \\right), \\end{align*} and $t$ must equal $$t=\\frac{41}{75} \\approx \\boxed{0.546667}.$$ $\\textbf{Remark}$. It is worth noting that the distance between points $a$ and $v(0.546667)$ is \\begin{align*} ||\\text{ortho}_{u}w || &= || w-\\text{proj}_u w || \\\\ &= \\Bigg|\\Bigg| \\langle 2,7,8\\rangle -\\frac{41}{75} \\langle 7,5,-1 \\rangle \\Bigg|\\Bigg| \\\\ &= \\Bigg|\\Bigg| \\left( -\\frac{137}{75} , \\frac{64}{15}, \\frac{641}{75} \\right) \\Bigg|\\Bigg| \\\\ &= \\sqrt{\\frac{7094}{75}} \\\\ &\\approx 9.72557. \\end{align*} $$v-a = \\begin{pmatrix} 2 \\\\ -3 \\\\ -3 \\end{pmatrix} + \\begin{pmatrix} 7 \\\\ 5 \\\\ -1 \\end{pmatrix} t -\\begin{pmatrix} 4 \\\\ 4 \\\\ 5 \\end{pmatrix} = \\begin{pmatrix} 2+ 7t-4 \\\\-3+ 5t -4\\\\ -3-1t-5 \\end{pmatrix} = \\begin{pmatrix} -2+ 7t \\\\-7+ 5t\\\\ -8-1t \\end{pmatrix}$$ The magnitude of $v-a$ is the square root of $(-2+ 7t)^2 + (-7+ 5t)^2 + (-8-1t)^2$. Do you know how to minimise this expression? Write the vector form of the difference between the two vectors. Then find an expression in $t$ representing the square of the magnitude of the difference vector. If you minimize the square of the distance, you also minimize the distance. Find the value of $t$ which minimizes the square of the distance using the first derivative of the expression. Vector $v$ can be written as $$\\begin{pmatrix} 2 +7t \\\\ -3 +5t\\\\ -3 -1t\\end{pmatrix}$$ Then the distance between $a$", "=\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\left [\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}\\end{pmatrix} -\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\right ]+\\begin{pmatrix}1 \\\\ 5\\end{pmatrix} \\\\ & =\\begin{pmatrix}-1 & 0 \\\\ 0 & -1\\end{pmatrix}\\begin{pmatrix}\\frac{3x}{5}+\\frac{4y}{5}-\\frac{6}{5}-1\\\\ -\\frac{4x}{5}+\\frac{3y}{5}-\\frac{2}{5}-2\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}\\end{pmatrix} +\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{11}{5}+1\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{12}{5}+5\\end{pmatrix}\\\\ & =\\begin{pmatrix}-\\frac{3x}{5}-\\frac{4y}{5}+\\frac{16}{5}\\\\ \\frac{4x}{5}-\\frac{3y}{5}+\\frac{37}{5}\\end{pmatrix} \\end{align*} Is that correct so far? How could we continue? Klaas van Aarsen MHB Seeker Staff member The matrix for $\\delta$ is correct. But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. mathmari Well-known member MHB Site Helper But how did you get that $\\gamma$ has matrix $-I$? Since $\\gamma(B)=A$ and since $C$ is not halfway between $A$ and $B$ that cannot be correct. It is not a point reflection. Instead we need to show that $(\\gamma\\circ\\delta)$ is a point reflection. Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? Klaas van Aarsen MHB Seeker Staff member Is the point $C=\\begin{pmatrix}1 \\\\ 5\\end{pmatrix}$ correct? That point is in between $A$ and $B$. But $C$ is on the circumcircle of triangle $ABZ$, which implies that it cannot be between $A$ and $B$. Klaas van Aarsen MHB Seeker Staff member I've drawn a picture now: \\begin{tikzpicture} \\usetikzlibrary", "\\\\ 0\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (1 \\cdot 0 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 0 + 1 \\cdot 0) \\\\ 1^2 + 0^2 - 0^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\1 \\end{pmatrix} \\\\ \\phi(|1\\rangle) = \\phi\\left(\\begin{pmatrix}0 \\\\ 1\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (0 \\cdot 1 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 1 + 0 \\cdot 0) \\\\ 0^2 + 0^2 - 1^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\-1 \\end{pmatrix} \\\\ \\phi(|+\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}} + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\frac{1}{\\sqrt{2}} + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|-\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(-\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} -1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|{+i}\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{i}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot 0 + 0 \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ 2 \\cdot \\left(-0 \\cdot 0 + \\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - 0^2 - \\left(\\frac{1}{\\sqrt{2}}\\right)^2 \\end{pmatrix} = \\begin{pmatrix}0\\\\1\\\\0 \\end{pmatrix} \\\\ \\phi(|{-i}\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{i}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot 0 + 0 \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right)\\right) \\\\ 2 \\cdot \\left(-0 \\cdot 0", "& \\frac{1}{2} \\end{pmatrix} \\cdot \\begin{pmatrix} 2 \\\\ 2 \\end{pmatrix} = \\begin{pmatrix} 1-\\sqrt{3} \\\\ 1+\\sqrt{3} \\end{pmatrix}$$", "0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -5 & 0\\\\ 0 & 0 & 0\\end{array}\\right) \\mbox{ and } \\left(\\begin{array}{cc|c} 5 & -5 & 0\\\\ 1 & -1 & 0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -1 & 0\\\\ 0 & 0 & 0\\end{array}\\right)\\, ,$$ respectively, so are $$v_{2}=\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}$$ and $$v_{-2} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$. Hence $$M^{12} v_{2} = 2^{12} v_{2}$$ and $$M^{12}v_{-2} = (-2)^{12} v_{-2}$$. Now, $$\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\frac{x-y}{4}\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix} -\\frac{x-5y}{4} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$ (this was obtained by solving the linear system $$a v_{2} + b v_{-2} =$$ for $$a$$ and $$b$$). Thus $$M \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\frac{x-y}{4} M v_{2} -\\frac{x-5y}{4} M v_{-2}$$ $$= 2^{12} \\Big(\\frac{x-y}{4} v_{2} -\\frac{x-5y}{4} v_{-2}\\Big) = 2^{12} \\begin{pmatrix} x \\\\ y \\end{pmatrix}\\, .$$ Thus $$M^{12}=\\begin{pmatrix} 4096 & 0 \\\\ 0 & 4096\\end{pmatrix}\\, .$$ $$\\textit{If you understand the above explanation, then you have a good understanding of diagonalization. A quicker route}$$ $$\\textit{is simply to observe that}$$ $$M^{2} = \\begin{pmatrix}4 & 0 \\\\ 0 & 4\\end{pmatrix}$$. 8. $$P_{M}(\\lambda) = (-1)^{2} {\\rm det}\\begin{pmatrix} a-\\lambda & b \\\\ c &d-\\lambda\\end{pmatrix} =(\\lambda-a)(\\lambda-d) - bc\\, .$$ Thus $$P_{M}(M)=(M-a I )(M- d I) - bc I$$ $$= \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}a&0\\\\0&a\\end{pmatrix}\\right) \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}d&0\\\\0&d\\end{pmatrix}\\right)-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}$$ $$=\\begin{pmatrix}0& b\\\\c&d-a\\end{pmatrix}\\begin{pmatrix}a-d&b\\\\ c&0\\end{pmatrix}-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}=0\\, .$$ Observe that any $$2\\times 2$$ matrix is a zero of its own characteristic polynomial ($$\\textit{in fact this holds", "the complete solution is% \\begin{align*} \\bar{x} & =\\bar{x}_{h}+\\bar{x}_{p}\\\\ & =c_{1}% \\begin{pmatrix} 1\\\\ -2 \\end{pmatrix} e^{-2t}+c_{2}% \\begin{pmatrix} 2\\\\ 1 \\end{pmatrix} e^{5t}+% \\begin{pmatrix} \\frac{1}{490}e^{-2t}\\left( -1127t^{2}+728t+104\\right) \\\\ \\frac{1}{245}e^{-2t}\\left( 1127t^{2}+182t+26\\right) \\end{pmatrix} \\end{align*} To find the constants, we apply initial conditions. Att=0% \\begin{align*}% \\begin{pmatrix} 7\\\\ 3 \\end{pmatrix} & =c_{1}% \\begin{pmatrix} 1\\\\ -2 \\end{pmatrix} +c_{2}% \\begin{pmatrix} 2\\\\ 1 \\end{pmatrix} +% \\begin{pmatrix} \\frac{52}{245}\\\\ \\frac{26}{245}% \\end{pmatrix} \\\\ c_{1}% \\begin{pmatrix} 1\\\\ -2 \\end{pmatrix} +c_{2}% \\begin{pmatrix} 2\\\\ 1 \\end{pmatrix} & =% \\begin{pmatrix} 7\\\\ 3 \\end{pmatrix} -% \\begin{pmatrix} \\frac{52}{245}\\\\ \\frac{26}{245}% \\end{pmatrix} \\\\% \\begin{pmatrix} 1 & 2\\\\ -2 & 1 \\end{pmatrix}% \\begin{pmatrix} c_{1}\\\\ c_{2}% \\end{pmatrix} & =% \\begin{pmatrix} \\frac{1663}{245}\\\\ \\frac{709}{245}% \\end{pmatrix} \\\\% \\begin{pmatrix} 1 & 2\\\\ 0 & 5 \\end{pmatrix}% \\begin{pmatrix} c_{1}\\\\ c_{2}% \\end{pmatrix} & =% \\begin{pmatrix} \\frac{1663}{245}\\\\ \\frac{807}{49}% \\end{pmatrix} \\end{align*} Hence5c_{2}=\\frac{807}{49}$or$c_{2}=\\frac{807}{245}$and$c_{1}% +2c_{2}=\\frac{1663}{245}$, hence$c_{1}=\\frac{1663}{245}-2\\left( \\frac {807}{245}\\right) =\\frac{1}{5}\\$. Therefore the solution becomes% $\\bar{x}=\\frac{1}{5}% \\begin{pmatrix} 1\\\\ -2 \\end{pmatrix} e^{-2t}+\\frac{807}{245}% \\begin{pmatrix} 2\\\\ 1 \\end{pmatrix} e^{5t}+% \\begin{pmatrix} \\frac{1}{490}e^{-2t}\\left( -1127t^{2}+728t+104\\right) \\\\ \\frac{1}{245}e^{-2t}\\left( 1127t^{2}+182t+26\\right) \\end{pmatrix}$ \\end{document} This is how the HTML looks like generated by tex4ht using the above command How does this file look in latexml HTML? lwarp HTML? This is the PDF file generated by lualatex if needed. • arxiv-vanity.com uses LaTeXML. – ShreevatsaR Feb 25 '18 at 23:15 • Do you have a sample document to test? – michal.h21 Feb 25 '18 at 23:26 • @Nasser here is your sample file with different", "\\frac{1-\\sqrt{5}}{2} \\end{pmatrix}^{-1}\\begin{pmatrix} 1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2} \\end{pmatrix} \\begin{pmatrix}C_{1}\\\\ C_{2}\\end{pmatrix} = \\begin{pmatrix} 1 & 1 \\\\ \\frac{1+\\sqrt{5}}{2} & \\frac{1-\\sqrt{5}}{2} \\end{pmatrix}^{-1} \\begin{pmatrix}1\\\\ \\frac{-5}{4}\\end{pmatrix} \\begin{pmatrix}C_{1}\\\\ C_{2}\\end{pmatrix} = \\frac{-1}{\\sqrt{5}}\\begin{pmatrix} \\frac{1-\\sqrt{5}}{2} & -1 \\\\ \\frac{-1-\\sqrt{5}}{2} & 1 \\end{pmatrix} \\begin{pmatrix}1\\\\ \\frac{-5}{4}\\end{pmatrix} which gives us C_{1}=\\frac{-2(1-\\sqrt{5})-5}{4\\sqrt{5}}\\ (7) C_{2}=\\frac{2(1+\\sqrt{5})+5}{4\\sqrt{5}}\\ (8) Finally, by plugging (7) and (8) into (3) we get F_{n}=-4\\left (\\left ( \\frac{-2(1-\\sqrt{5})-5}{4\\sqrt{5}} \\right )\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{n}+\\left ( \\frac{2(1+\\sqrt{5})+5}{4\\sqrt{5}} \\right )\\left ( \\frac{1-\\sqrt{5}}{2} \\right )^{n}\\right ) (9) We can now “easily” compute F_{386} which is given by F_{386}=-4\\left (\\left ( \\frac{-2(1-\\sqrt{5})-5}{4\\sqrt{5}} \\right )\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{386}+\\left ( \\frac{2(1+\\sqrt{5})+5}{4\\sqrt{5}} \\right )\\left ( \\frac{1-\\sqrt{5}}{2} \\right )^{386}\\right ) Google returns F_{386} \\cong 5.2783459\\times 10^{80} which is just an approximation, but now we have a formula for the precise answer. If you wish to check for yourself that out formula (9) is indeed correct, click on the terms below to see the computation by Google: F_{0}=-4 F_{1}=5 F_{2}=1 F_{3}=6 F_{4}=7 F_{5}=13 On a side note – Raising a number to the power of 386 may seem like a long computation but it really requires only 10 multiplications! Don’t believe me? Consider this: \\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{386}=\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{256}\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{128}\\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{2} To calculate \\left ( \\frac{1+\\sqrt{5}}{2} \\right )^{256} we use the fact that 1.)\\ \\left", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "\\\\ 2 & 3 \\end{pmatrix} x = \\begin{pmatrix} 5 \\\\ 6 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{5} \\begin{pmatrix} 9 \\\\ 4 \\end{pmatrix} \\end{align*} \\begin{align*} 12) \\hspace{.5cm} \\begin{pmatrix} −1 & 5 \\\\ 1 & 1 \\end{pmatrix} x = \\begin{pmatrix} 2 \\\\ −1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{6} \\begin{pmatrix} −7 \\\\ 1 \\end{pmatrix} \\end{align*} \\begin{align*} 13) \\hspace{.5cm} \\begin{pmatrix} 3 & −7 \\\\ −9 & 21 \\end{pmatrix} x = \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\text{no inverse} \\end{align*} \\begin{align*} 14) \\hspace{.5cm} \\begin{pmatrix} 2 & 7 \\\\ 5 & −1 \\end{pmatrix} x = \\begin{pmatrix} 3 \\\\ 4 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\frac{1}{37} \\begin{pmatrix} 31 \\\\ 7 \\end{pmatrix} \\end{align*} \\begin{align*} 15) \\hspace{.5cm} \\begin{pmatrix} 1 & 0 & 1 \\\\ 0 & −1 & 1 \\\\ 1 & 2 & 2 \\end{pmatrix} x = \\begin{pmatrix} 1 \\\\ −2 \\\\ −1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} x = \\begin{pmatrix} 3 \\\\ 0 \\\\ −2 \\end{pmatrix}\\end{align*} \\begin{align*} 16) \\hspace{.5cm} \\begin{pmatrix} 7 & 3 & 4 \\\\ 1 & 2 & 3 \\\\ 4 & −3 & −5 \\end{pmatrix} x = \\begin{pmatrix} 3 \\\\ 2 \\\\ 1 \\end{pmatrix} \\end{align*} Solution: \\begin{align*} \\text{no inverse} \\end{align*} \\begin{align*} 17) \\hspace{.5cm} \\begin{pmatrix} 3 & 4 & 1 \\\\ −2 & 3 & 1 \\\\ 0 & 3 & 2 \\end{pmatrix} x = \\begin{pmatrix} 5 \\\\", "$\\begin{pmatrix}x\\\\y\\end{pmatrix}_p=t\\begin{pmatrix}-\\frac{1}{3}\\\\\\frac{1}{6}\\end{pmatrix} + \\begin{pmatrix}-\\frac{5}{54}\\\\\\frac{7}{216}\\end{pmatrix}$ So, my final solution, as before, is $\\begin{cases} x(t)= -4C_1e^{4t}+C_2e^{9t}-\\frac{1}{3}t-\\frac{5}{54}\\\\ y(t)= C_1e^{4t}+C_2e^{9t}+\\frac{1}{6}t+\\frac{7}{216} \\end{cases}$", "instructive to consider a few special cases. Let us denote the map with $$\\phi: \\mathcal{H} \\to \\mathbb{R}^3$$ where $$\\mathcal{H}$$ is the Hilbert space of a qubit. We calculate that $$\\phi(|0\\rangle) = \\phi\\left(\\begin{pmatrix}1 \\\\ 0\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (1 \\cdot 0 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 0 + 1 \\cdot 0) \\\\ 1^2 + 0^2 - 0^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\1 \\end{pmatrix} \\\\ \\phi(|1\\rangle) = \\phi\\left(\\begin{pmatrix}0 \\\\ 1\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot (0 \\cdot 1 + 0 \\cdot 0) \\\\ 2 \\cdot (-0 \\cdot 1 + 0 \\cdot 0) \\\\ 0^2 + 0^2 - 1^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 0\\\\0\\\\-1 \\end{pmatrix} \\\\ \\phi(|+\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}} + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\frac{1}{\\sqrt{2}} + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} 1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|-\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ -\\frac{1}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + 0 \\cdot 0\\right) \\\\ 2 \\cdot \\left(-0 \\cdot \\left(-\\frac{1}{\\sqrt{2}}\\right) + \\frac{1}{\\sqrt{2}} \\cdot 0\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 + 0^2 - \\left(-\\frac{1}{\\sqrt{2}}\\right)^2 - 0^2 \\end{pmatrix} = \\begin{pmatrix} -1\\\\0\\\\0 \\end{pmatrix} \\\\ \\phi(|{+i}\\rangle) = \\phi\\left(\\begin{pmatrix}\\frac{1}{\\sqrt{2}} \\\\ \\frac{i}{\\sqrt{2}}\\end{pmatrix}\\right) = \\begin{pmatrix} 2 \\cdot \\left(\\frac{1}{\\sqrt{2}} \\cdot 0 + 0 \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ 2 \\cdot \\left(-0 \\cdot 0 + \\frac{1}{\\sqrt{2}} \\cdot \\frac{1}{\\sqrt{2}}\\right) \\\\ \\left(\\frac{1}{\\sqrt{2}}\\right)^2 +", "be: \\begin{align}\\textbf{x}(t)=(18\\ln(e^t+1)+c_{1})e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+(-3e^{2t}+6e^t-6\\ln(e^t+1)+c_{2})e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}\\end{align} Subbing in the initial condition $\\textbf{x}(0)=\\begin{pmatrix}1-3\\ln2\\\\-3-3\\ln2\\end{pmatrix}$: \\begin{align}1-3\\ln2=c_{1}+3+c_{2}\\end{align} \\begin{align}-3-3\\ln2=-6-2c_{2}\\end{align} Solving these equations for the constants gives: \\begin{align}c_{1}=\\frac{-19-9\\ln2}{2}\\end{align} \\begin{align}c_{2}=\\frac{-3+3\\ln2}{2}\\end{align} Therefore the general solution for the IVP is: \\begin{align}\\textbf{x}(t)=\\frac{-19-9\\ln2}{2}e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}+\\frac{-3+3\\ln2}{2}e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+18\\ln(e^t+1)e^{-t}\\begin{pmatrix}1\\\\0\\end{pmatrix}-3e^{-t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}+6e^{-2t}\\begin{pmatrix}1\\\\-2\\end{pmatrix}-6\\ln(e^t+1)e^{-3t}\\begin{pmatrix}1\\\\-2\\end{pmatrix} \\end{align}", "\\end{pmatrix} \\tag{1}$$ ## Multiplying on the left Step by step (so we can compare with multiplying on the right shortly): $$\\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 5 \\\\ -2 & 3 \\end{pmatrix} = \\begin{pmatrix} 4 & 5 \\\\ 0 & \\frac{11}{2} \\end{pmatrix} \\tag{2}$$ Rescaling, $$\\begin{pmatrix} 1 & 0 \\\\ 0 & \\frac{2}{11} \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 5 \\\\ -2 & 3 \\end{pmatrix} = \\begin{pmatrix} 4 & 5 \\\\ 0 & 1 \\end{pmatrix} \\tag{3}$$ Then $$\\begin{pmatrix} 1 & -5 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & \\frac{2}{11} \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 5 \\\\ -2 & 3 \\end{pmatrix} = \\begin{pmatrix} 4 & 0 \\\\ 0 & 1 \\end{pmatrix} \\tag{4}$$ and finally $$\\begin{pmatrix} \\frac{1}{4} & 0 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & -5 \\\\ 0 & 1 \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ 0 & \\frac{2}{11} \\end{pmatrix} \\begin{pmatrix} 1 & 0 \\\\ \\frac{1}{2} & 1 \\end{pmatrix} \\begin{pmatrix} 4 & 5 \\\\ -2 & 3 \\end{pmatrix} = \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} \\tag{5}$$ Let's then work this the opposite way to see what the row operations above are doing when we multiply on the right instead. ##", "$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$ $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 5 & -2\\\\ \\end {vmatrix}$$ = -4 - 15 = -19≠ 0 It has a unique solution. X = A-1B A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$ X = A-1B or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -2 & -3\\\\ -5 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\frac {1}{-19}$$$$\\begin {pmatrix} -10 - 9\\\\ -25 + 6\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-19}{-19}\\\\ \\frac {-19}{-19}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, 3x - 5y = 3..........................(1) 4x + 3y = 4.........................(2) Given equation in matrix form; $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 3\\\\ 4\\\\ \\end {pmatrix}$$ or, AX = B where, A = $$\\begin {pmatrix} 3 & -5\\\\ 4 & 3\\\\ \\end {pmatrix}$$," ]

[ "\\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} - \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} = \\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ On the other hand, the reflection of $\\mathbf{x}$ in the plane $$L^\\perp = \\{\\mathbf{y} \\in \\mathbb{R}^3 \\mid \\mathbf{v} \\cdot \\mathbf{y} = 0\\}$$ with normal vector $\\mathbf{v}$ is given by $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = 2\\operatorname{Proj}_{L^\\perp}(\\mathbf{x}) - \\mathbf{x} = 2\\left(\\mathbf{x} - \\operatorname{Proj}_L(\\mathbf{x})\\right) - \\mathbf{x} = \\mathbf{x} - 2\\operatorname{Proj}_L(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v},$$ which in your case yields $$\\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix} - 2 \\cdot \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix} = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ One last cultural note: the reflection $\\operatorname{Refl}_{L^\\perp}(\\mathbf{x})$ of $\\mathbf{x}$ in the plane $L^\\perp$ with normal vector $\\mathbf{v}$ is better known in more advanced contexts by another name, namely as the image $$H_{\\mathbf{v}}(\\mathbf{x}) := \\operatorname{Refl}_{L^\\perp}(\\mathbf{x}) = \\mathbf{x} - 2\\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}$$ of $\\mathbf{x}$ under the Householder transformation $H_\\mathbf{v}$ corresponding to $\\mathbf{v}$. So, in your case, as we just saw, $$H_{\\mathbf{v}}(\\mathbf{x}) = -\\frac{1}{9}\\begin{pmatrix}23\\\\-20\\\\23\\end{pmatrix}.$$ Suppose you mean a vector ($\\mathbf{v}$) \"specular\" to the given one ($\\mathbf{u}$) with respect to the line $L$. Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin. If you do the cross product $\\mathbf{w} = \\mathbf{u} \\times \\mathbf{L}$ you get a vector which is orthogonal either to the line and to $\\mathbf{u}$, i.e. it is normal to the plane containing the line and $\\mathbf{u}$. Then $\\mathbf{v} \\times \\mathbf{L}$ shall be equal to $-\\mathbf{w}$", "Review question # Which point $P$ is on both this line and this plane? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R6006 ## Solution The line $l$ has the equation $\\mathbf{r}= \\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}+ \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}, \\hspace{1cm} \\lambda \\in \\mathbb{R}.$ 1. Show that $l$ lies in the plane whose equation is $\\mathbf{r}.\\begin{pmatrix}-1\\\\ 2\\\\ 0\\end{pmatrix}=-5.$ We need to show that every vector $\\mathbf{r}$ that satisfies the line equation also satisfies the plane equation. Substituting in, we find $\\left[\\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}+ \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix} \\right].\\begin{pmatrix}-1\\\\ 2\\\\ 0\\end{pmatrix}=\\begin{pmatrix}5+2\\lambda\\\\ \\lambda\\\\ 5\\end{pmatrix} .\\begin{pmatrix}-1\\\\ 2\\\\ 0\\end{pmatrix}= -5 -2\\lambda+2\\lambda=-5.$ This is true for every $\\lambda$, so we’ve shown that the line does indeed lie in the given plane. 1. Find the position vector of $L$, the foot of the perpendicular from the origin $O$ to $l$. Let $\\overrightarrow{OL} = \\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}+ \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}.$ We need $\\left[\\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}+ \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix} \\right].\\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}=0$ which is true if and only if $10+4\\lambda+\\lambda =0,$ or $\\lambda = -2.$ Thus $L$ is the point $\\begin{pmatrix}1\\\\ -2\\\\ 5\\end{pmatrix}$. 1. Find an equation of the plane containing $O$ and $l$. We know that $\\overrightarrow{OL}$ is in the plane, and putting $\\lambda = 0$, we know also that", "shown below) $$r·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$(3×1)+(2×[-5])+(7×8)=49$$ Therefore this means that the distanve of the vector d = 49 To complete the equation we need to write down the value of the whole equation in cartesian form $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$x-5y+8z=49$$ x - 5y + 8z - 49 = 0 #### Question 2 Write the equation of the plane through (2,1,3), given that the vector (4,5,6) is perpendicular $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 1\\\\\\ 3\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$(2×4)+(1×5)+(3×6)=31$$ Therefore the distance of the vector is d = 31 $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$4x+5y+6z=31$$ 4x + 5y + 6z - 31 = 0 #### Intersection of a line and a plane Find the point of intersection between the line The normal (perpendicular) to a plane is $$r=\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ with plane $$5x+y-z=1$$ First we need to replace r with the matrix (x,y,z) in order to find the values of x,y and z $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ We can change this equation into 3 different equations by seperating the matrices into: x = 2 + λ y = 3 + 2λ z = 4 - λ And we know that 5x+y-z=1 Replacing the values of x,y and z with these new values we get", "1 & 0 \\\\ 1 & 1 \\end{pmatrix}, N_3 = \\begin{pmatrix} 1 & 1 \\\\ 0 & 1 \\end{pmatrix}, N_4 = \\begin{pmatrix} 1 & 1 \\\\ 1 & 0 \\end{pmatrix}$ Does $\\{ N_1, N_2, N_3, N_4 \\}$ generate V? $M_1 = -\\frac23 N_1 + \\frac13(N_2 + N_3 + N_4)$ $M_2 = -\\frac23 N_2 + \\frac13(N_1 + N_3 + N_4)$ $M_3 = -\\frac23 N_3 + \\frac13(N_1 + N_2 + N_4)$ $M_4 = -\\frac23 N_4 + \\frac13(N_1 + N_2 + N_3)$ Then $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = aM_1 + bM_2 + cM_3 + dM_4 =$ $a \\cdot \\left( -\\frac23 N_1 + \\frac13(N_2 + N_3 + N_4) \\right) + b \\cdot \\left( -\\frac23 N_2 + \\frac13(N_1 + N_3 + N_4) \\right) + \\ldots$ Theorem: If $S \\in \\mathbf V$, then $\\operatorname{span}(S) =$ {all l.c. of elements of S} is a subspace of V.", "we begin with the point $(\\frac{1}{2},\\frac{1}{3})$ and keep applying the matrix $A$ over to the result? $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{1}{3})=(\\frac{1}{3},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{6})$ (remember we are reducing modulo 1) x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{5},{1}{6})=(\\frac{1}{6},0)$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},0)=(0,\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (0,\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{2})=(\\frac{1}{2},\\frac{5}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{2},\\frac{5}{6})=(\\frac{5}{6},\\frac{1}{3})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{5}{6},\\frac{1}{3})=(\\frac{1}{3},\\frac{1}{6})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{3},\\frac{1}{6})=(\\frac{1}{6},\\frac{1}{2})$ x $\\begin{pmatrix} 0 & 1 \\\\1 & 1 \\end{pmatrix}\\cdot (\\frac{1}{6},\\frac{1}{2})=(\\frac{1}{2},\\frac{1}{3})$ x and we are back to where we started, after 13 steps. The collection 12 points we collected along the way: $(\\frac{1}{3},\\frac{5}{6}), (\\frac{5}{6},\\frac{1}{6}), \\ldots$ is called the orbit of the initial point $(\\frac{1}{3},\\frac{5}{6})$. Any of the points in the orbit of $(\\frac{1}{3},\\frac{5}{6})$ has exactly the same 12 points as its orbit. This is because the matrix $A$ is an invertible matrix with inverse $A^{-1}=\\begin{pmatrix} -1 & 1 \\\\1 & 0 \\end{pmatrix}$ Just as for the point we", "0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -5 & 0\\\\ 0 & 0 & 0\\end{array}\\right) \\mbox{ and } \\left(\\begin{array}{cc|c} 5 & -5 & 0\\\\ 1 & -1 & 0\\end{array}\\right)\\sim \\left(\\begin{array}{cc|c} 1 & -1 & 0\\\\ 0 & 0 & 0\\end{array}\\right)\\, ,$$ respectively, so are $$v_{2}=\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix}$$ and $$v_{-2} = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$. Hence $$M^{12} v_{2} = 2^{12} v_{2}$$ and $$M^{12}v_{-2} = (-2)^{12} v_{-2}$$. Now, $$\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\frac{x-y}{4}\\begin{pmatrix} 5 \\\\ 1 \\end{pmatrix} -\\frac{x-5y}{4} \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}$$ (this was obtained by solving the linear system $$a v_{2} + b v_{-2} =$$ for $$a$$ and $$b$$). Thus $$M \\begin{pmatrix} x \\\\ y \\end{pmatrix} = \\frac{x-y}{4} M v_{2} -\\frac{x-5y}{4} M v_{-2}$$ $$= 2^{12} \\Big(\\frac{x-y}{4} v_{2} -\\frac{x-5y}{4} v_{-2}\\Big) = 2^{12} \\begin{pmatrix} x \\\\ y \\end{pmatrix}\\, .$$ Thus $$M^{12}=\\begin{pmatrix} 4096 & 0 \\\\ 0 & 4096\\end{pmatrix}\\, .$$ $$\\textit{If you understand the above explanation, then you have a good understanding of diagonalization. A quicker route}$$ $$\\textit{is simply to observe that}$$ $$M^{2} = \\begin{pmatrix}4 & 0 \\\\ 0 & 4\\end{pmatrix}$$. 8. $$P_{M}(\\lambda) = (-1)^{2} {\\rm det}\\begin{pmatrix} a-\\lambda & b \\\\ c &d-\\lambda\\end{pmatrix} =(\\lambda-a)(\\lambda-d) - bc\\, .$$ Thus $$P_{M}(M)=(M-a I )(M- d I) - bc I$$ $$= \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}a&0\\\\0&a\\end{pmatrix}\\right) \\left(\\begin{pmatrix}a&b\\\\c&d\\end{pmatrix}-\\begin{pmatrix}d&0\\\\0&d\\end{pmatrix}\\right)-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}$$ $$=\\begin{pmatrix}0& b\\\\c&d-a\\end{pmatrix}\\begin{pmatrix}a-d&b\\\\ c&0\\end{pmatrix}-\\begin{pmatrix}bc&0\\\\0&bc\\end{pmatrix}=0\\, .$$ Observe that any $$2\\times 2$$ matrix is a zero of its own characteristic polynomial ($$\\textit{in fact this holds", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "Since $(21,26)=1$, there exists $21^{-1}\\in\\mathbb{Z}/26\\mathbb{Z}$. Remember that finding $21^{-1}\\in\\mathbb{Z}/26\\mathbb{Z}$ is equivalent to solving the diophantine equation $21x+26y=1$. \\begin{align*} 26&=21\\cdot 1+5\\\\ 21&=5\\cdot 4+1 \\end{align*} and \\begin{align*} 1&=21-5\\cdot 4\\\\ &=21-(26-21)4\\\\ &=5\\cdot 21+26(-4) \\end{align*} Thus, we find $21^{-1}\\equiv 5\\mod 26$. \\begin{align*} A^{-1}&=\\begin{pmatrix} 5\\cdot 8 & -5\\cdot 3\\\\ -5\\cdot 7 & 5\\cdot 2 \\end{pmatrix}\\\\ &=\\begin{pmatrix} 40 & -15\\\\ -35 & 10 \\end{pmatrix}\\\\ &\\equiv\\begin{pmatrix} 14 & 11\\\\ 17 & 10 \\end{pmatrix}\\mod 26 \\end{align*} Let $A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$. Then $$\\begin{pmatrix} x’\\\\ y’ \\end{pmatrix}=A\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\begin{pmatrix} x\\\\ y \\end{pmatrix}=\\begin{pmatrix} ax+by\\\\ cx+dy \\end{pmatrix}$$ This gives a linear map or a linear transformation from vectors to vectors. We introduce the following theorem without proof. Theorem. Let $A=\\begin{pmatrix} a & b\\\\ c & d \\end{pmatrix}\\in M_2(\\mathbb{Z}/N\\mathbb{Z})$ and set $D=ad-bc$ as before. Then the following are equivalent: 1. $(D,N)=1$. 2. $A$ has an inverse matrix. 3. if $x$ and $y$ are not both $0$ in $\\mathbb{Z}/N\\mathbb{Z}$, then $A\\begin{pmatrix}x\\\\y \\end{pmatrix}\\ne\\begin{pmatrix}0\\\\0\\end{pmatrix}$. 4. $A$ gives a one-to-one correspondence of $(\\mathbb{Z}/N\\mathbb{Z})^2$ with itself. Example. Solve the following systems of simultaneous congruences. 1. \\begin{align*}2x+3y&\\equiv 1\\mod 26\\\\ 7x+8y&\\equiv 2\\mod 26 \\end{align*} 2. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 2\\mod 26 \\end{align*} 3. \\begin{align*} x+3y&\\equiv 1\\mod 26\\\\ 7x+9y&\\equiv 1\\mod 26 \\end{align*} Solution. 1. The system of simultaneous congruences can be written as the", "end up with a form from which you can read the center coordinates. \\begin{align*} x'^2 + 18\\,x' &= \\phantom{1.00}(x' + 9)^2 - 81 & x_c' &= -9 \\\\ 6.25\\,y'^2 - 25\\,y' &= 6.25(y' - 2)^2 - 25 & y_c' &= \\phantom+2 \\\\ \\end{align*} Note that the center coordinates $x_c'$ and $y_c'$ describe the center in the already rotated coordinate system. To obtain the original center you'd multiply again with the first matrix from $(2)$: $$\\begin{pmatrix}x_c\\\\y_c\\end{pmatrix}= \\left(\\begin{array}{rr}0.6&-0.8\\\\0.8&0.6\\end{array}\\right)\\cdot \\left(\\begin{array}{r}-9\\\\2\\end{array}\\right)= \\begin{pmatrix}-7\\\\-6\\end{pmatrix}$$ You can also do a second change of coodinate system, this time into one where the ellipse is not only axis-aligned but also centered at the origin. # Scaling The completed squares above contributed some more terms to the constant factor $F$, since the equation by now has become $$(x' + 9)^2 - 81 + 6.25(y' - 2)^2 - 25 + 81 = 0$$ The combined constant term here is $81 - 81 - 25 = -25$. Now you move this to the right side of the equation, then divide the whole equation by this number so that you get the $=1$ from the desired form. Then you can deduce the radii $a$ and $b$. \\begin{align*} (x' + 9)^2 + 6.25(y' - 2)^2 &= 25 \\\\ \\frac{(x' + 9)^2}{25} + \\frac{6.25(y' - 2)^2}{25} &= 1 \\end{align*} \\begin{align*} \\frac1{a^2}&=\\frac{1}{25} & a", "# Shortest Distance between two lines in the 3D plane Find the shortest distance between the line $x$ $=$ $2$ $-$ $t$ $y$ $=$ $-1$ $+$ $2t$ and $z$ $=$ $-1$ $+$ $t$ and the line $x$ $=$ $5$ $+$ $3s$ $y$ $=$ $0$ and $z$ $=$ $2$ $-$ $s$. Find the points on these two lines that give this shortest distance. Using the cross product, I got the normal vector to be: \\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix} The I selected two points on the lines with $t$ $=$ $-1$: \\begin{pmatrix}3\\\\ -3\\\\ -2\\end{pmatrix} and with $s$ $=$ $1$: \\begin{pmatrix}8\\\\ 0\\\\ 1\\end{pmatrix} Therefore the distance between the these two points gave the vector: \\begin{pmatrix}-5\\\\ \\:-3\\\\ \\:-3\\end{pmatrix} So then I projected this vector: $\\left(\\frac{\\left(-5,\\:-3,\\:-3\\right)\\cdot \\left(-5,\\:-3,\\:-3\\right)}{\\left(-2,\\:-2,\\:-6\\right)\\cdot \\left(-2,\\:-2,\\:-6\\right)}\\right)\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ This left me with: $\\frac{43}{\\:44}\\cdot$ $\\begin{pmatrix}-2\\\\ -2\\\\ -6\\end{pmatrix}$ Which left me with a final distance of: $\\frac{43}{44}\\sqrt{44}$ But this answer isn't correct. I think I have made a mistake in my projection calculation but I am not sure where. Also, how would I find the two points that are closest together. I am also a little confused about that. Any help would be highly appreciated! • $(-1,2,1)\\times(3,0,-1)\\ne(-2,-2,-6)$. – amd Apr 6 '18 at 21:50 • Thank you! I guess I missed the sign. So it should be $(-2, 2, -6)$? But this will", "# Finding the orthogonal projection of a vector I can't figure this out. How the book explains to do this problem doesn't make sense to me. I'm hoping someone can simplify this so I can understand how to do this step by step. Let $L$ be the line in $\\mathbb{R}^3$ that consists of all scalar multiples of $$\\left[ \\begin{array}{ccc} 2\\\\ 1\\\\ 2\\\\ \\end{array} \\right]$$ Find the reflection of the vector $$\\left[ \\begin{array}{ccc} 1\\\\ 4\\\\ 1\\\\ \\end{array} \\right]$$ onto $L$ • Are you asking for reflection (as in the body of the question) or projection (as in the subject)? – Thomas Jul 11 '16 at 17:38 Let $L$ be the line spanned by some non-zero vector $\\mathbf{v} \\in \\mathbb{R}^3$, so that $L = \\{a\\mathbf{v} \\mid a \\in \\mathbb{R}\\}$ is the space of all scalar multiples of $\\mathbf{v}$. Then the orthogonal projection of a vector $\\mathbf{x} \\in \\mathbb{R}^3$ onto the line $L$ can be computed as $$\\operatorname{Proj}_L(\\mathbf{x}) = \\frac{\\mathbf{v} \\cdot \\mathbf{x}}{\\mathbf{v}\\cdot\\mathbf{v}} \\mathbf{v}.$$ So, in this case, we have $$\\mathbf{v} = \\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}, \\quad \\mathbf{x} = \\begin{pmatrix}1\\\\4\\\\1\\end{pmatrix},$$ so that $$\\mathbf{v} \\cdot \\mathbf{x} = 2 \\cdot 1 + 1 \\cdot 4 + 2 \\cdot 1 = 8, \\quad \\mathbf{v} \\cdot \\mathbf{v} = 2^2 + 1^2 + 2^2 = 9,$$ and hence $$\\operatorname{Proj}_L(\\mathbf{x}) = \\frac{8}{9}\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}.$$ Now, you probably wanted to compute the orthogonal projection", "# Decide a number $a, b, c$ (expressed by $\\lambda$) so that $a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$ See the following vectors in $$\\mathbb{R}^3$$: $$\\mathbf{u}_1= \\begin{pmatrix} 1 \\\\ -1\\\\ 1 \\\\ 1 \\end{pmatrix}$$, $$\\mathbf{u}_2 = \\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\\\ 2 \\end{pmatrix}$$, $$\\mathbf{u}_3 = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\end{pmatrix}$$ and $$\\mathbf{v}_1= \\begin{pmatrix} 5 \\\\ -3\\\\ -4 \\\\ -1 \\end{pmatrix}$$, $$\\mathbf{v}_2 = \\begin{pmatrix} 6 \\\\ 4 \\\\ 1 \\\\ 8 \\end{pmatrix}$$, $$\\mathbf{v}_3 = \\begin{pmatrix} 7 \\\\ 2 \\\\ -1 \\\\ 6 \\end{pmatrix}$$ We are told that $$\\mathcal{B} = (\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3)$$ and $$\\mathcal{C} = (\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3)$$ both are bases for the same subspace $$\\mathcal{U}$$. Let $$\\mathcal{\\lambda}$$ be an unknown number og see the vector $$\\mathbf{x} = \\mathbf{v}_1 + \\lambda\\mathbf{v}_2 - \\mathbf{v}_3$$. Decide a number $$a, b, c$$ (expressed by $$\\lambda$$) so that $$a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$$ 1. Find some $$\\alpha_i$$ such that $$v_1=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3$$ 2. Find some $$\\beta_i$$ such that $$v_2=\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3$$ 3. Find some $$\\gamma_i$$ such that $$v_3=\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3$$ 4. Simplify the expression $$x=v_1+\\lambda v_2-v_3=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3+\\lambda(\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3)-(\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3)$$ • Thanks 5xum! When I reduce it, I get $\\begin{pmatrix} 8 \\\\ 3\\\\ -9 \\\\ 1 \\end{pmatrix} + a\\begin{pmatrix} 2 \\\\ -2\\\\ 1 \\\\ 1 \\end{pmatrix} + b\\begin{pmatrix} 3 \\\\ 6\\\\ 0 \\\\ 6 \\end{pmatrix} + c\\begin{pmatrix} 1 \\\\", "+0 # dumb precalculus 0 45 1 For $$\\bold{v} = \\begin{pmatrix} \\phantom -2 \\\\ \\phantom -3 \\\\ -1 \\end{pmatrix} and \\bold{w} = \\begin{pmatrix} \\phantom -2 \\\\ -1 \\\\ \\phantom -0 \\end{pmatrix}$$ compute the projection of $\\mathbf v$ onto $\\mathbf w.$ i literally dont know how to solve this Apr 18, 2022 #1 +9369 +3 We have the formula: $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac{\\langle\\mathbf v, \\mathbf w\\rangle}{\\|\\mathbf w\\|^2}\\mathbf{w}$$ Now, we compute $$\\langle\\mathbf v, \\mathbf w\\rangle$$ and $$\\|\\mathbf w\\|^2$$. $$\\begin{array}{cl} &\\langle\\mathbf v, \\mathbf w\\rangle\\\\ =& 2(2) + 3(-1) + (-1)(0)\\\\ =& 1\\\\ &\\|\\mathbf w\\|^2\\\\ =& 2^2 + (-1)^2 + 0^2\\\\ =&5 \\end{array}$$ Therefore, substituting back into the formula, we have $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac15\\mathbf{w} = \\begin{pmatrix}\\frac25\\\\-\\frac15\\\\0\\end{pmatrix}$$ Apr 18, 2022 #1 +9369 +3 We have the formula: $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac{\\langle\\mathbf v, \\mathbf w\\rangle}{\\|\\mathbf w\\|^2}\\mathbf{w}$$ Now, we compute $$\\langle\\mathbf v, \\mathbf w\\rangle$$ and $$\\|\\mathbf w\\|^2$$. $$\\begin{array}{cl} &\\langle\\mathbf v, \\mathbf w\\rangle\\\\ =& 2(2) + 3(-1) + (-1)(0)\\\\ =& 1\\\\ &\\|\\mathbf w\\|^2\\\\ =& 2^2 + (-1)^2 + 0^2\\\\ =&5 \\end{array}$$ Therefore, substituting back into the formula, we have $$\\operatorname{Proj}_{\\mathbf{w}} (\\mathbf v) = \\dfrac15\\mathbf{w} = \\begin{pmatrix}\\frac25\\\\-\\frac15\\\\0\\end{pmatrix}$$ MaxWong Apr 18, 2022", "y(0) & = -3 \\end{align*} ##### 6. \\begin{align*} x' & = 5x + 4y\\\\ y' & = -9x - 7y\\\\ x(0) & = 2\\\\ y(0) & = 1 \\end{align*} ##### 7. \\begin{align*} x' & = -x + y\\\\ y' & = -x - 3y\\\\ x(0) & = 2\\\\ y(0) & = 2 \\end{align*} ##### 8. \\begin{align*} x' & = 2x + y\\\\ y' & = -x\\\\ x(0) & = 0\\\\ y(0) & = -5 \\end{align*} #### 9. Consider the linear system $$d \\mathbf x/dt = A \\mathbf x\\text{,}$$ where \\begin{equation*} A = \\begin{pmatrix} 3 \\amp 1 \\\\ -4 \\amp -1 \\end{pmatrix}. \\end{equation*} Suppose the initial conditions for the solution curve are $$x(0) = -2$$ and $$y(0) = 5\\text{.}$$ We can use the following Sage code to plot the phase portrait of this system, including a solution curve and the straight-line solution. x, y, t = var('x y t') #declare the variables F = [3*x + y, -4*x - y] #declare the system # normalize the vector fields so that all of the arrows are the same length n = sqrt(F[0]^2 + F[1]^2) # plot the vector field p = plot_vector_field((F[0]/n, F[1]/n), (x, -20, 20), (y, -20, 20), aspect_ratio = 1) # solve the system for the initial condition t = 0, x = -2, y = 5 P1", "the representation {\\displaystyle {\\begin{aligned}{\\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}}&=a\\cdot {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}+b\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}+c\\cdot {\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}\\\\[0.5em]&=a\\cdot \\left({\\frac {1}{4}}\\cdot {\\begin{pmatrix}2\\\\1\\\\0\\end{pmatrix}}+{\\frac {1}{4}}\\cdot {\\begin{pmatrix}2\\\\-1\\\\0\\end{pmatrix}}\\right)+b\\cdot \\left({\\frac {1}{2}}\\cdot {\\begin{pmatrix}2\\\\1\\\\0\\end{pmatrix}}-{\\frac {1}{2}}\\cdot {\\begin{pmatrix}2\\\\-1\\\\0\\end{pmatrix}}\\right)+c\\cdot {\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}\\\\[0.5em]&={\\frac {a+2b}{4}}\\cdot {\\begin{pmatrix}2\\\\1\\\\0\\end{pmatrix}}+{\\frac {a-2b}{4}}\\cdot {\\begin{pmatrix}2\\\\-1\\\\0\\end{pmatrix}}+c\\cdot {\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}\\end{aligned}}} Thus ${\\displaystyle (a,b,c)^{T}\\in \\operatorname {span} (\\{v_{1},v_{2},v_{3}\\})}$. Since this vector was arbitrarily chosen, every vector from ${\\displaystyle \\mathbb {R} ^{3}}$ is representable as a linear combination of the linearly independent vectors ${\\displaystyle v_{1},v_{2}}$ and ${\\displaystyle v_{3}}$. Thus ${\\displaystyle \\{v_{1},v_{2},v_{3}\\}}$ is a generator of ${\\displaystyle \\mathbb {R} ^{3}}$. Therefore, we cannot add another vector to ${\\displaystyle v_{1},v_{2}}$ and ${\\displaystyle v_{3}}$, so the system remains linearly independent, since every other vector from ${\\displaystyle \\mathbb {R} ^{3}}$ can be represented as a linear combination of ${\\displaystyle v_{1},v_{2}}$ and ${\\displaystyle v_{3}}$. In other words, ${\\displaystyle v_{1},v_{2}}$ and ${\\displaystyle v_{3}}$ form a maximal system of linearly independent vectors. In summary, we proceeded as follows to find a maximal system of linearly independent vectors: We start with a vector that is not the zero vector. That means we should not consider the null vector space here. Then we proceed step by step: Once we have found linearly independent vectors ${\\displaystyle v_{1},...,v_{n}}$, we form the span of these vectors. If this is the entire vector space, we have found a generator and are done. We choose a vector ${\\displaystyle v_{n+1}}$ that is not in ${\\displaystyle \\operatorname {span} (\\{v_{1},...,v_{n}\\})}$. This contributes a new direction", "+0 # help vectors 0 46 2 Let v and w be vectors such that $$\\text{proj}_{{w}} {v} = \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ Find $$\\text{proj}_{-2 {w}} (3 {v})$$. Feb 23, 2020 #1 0 Projection is linear, so the answer is 2*3*(4,-7) = (24,-42). Feb 24, 2020 #2 +24366 +1 Let $$\\mathbf{v} \\text{ and } \\mathbf{w} \\text{ be vectors such that } \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$. Find $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}})$$. $$\\begin{array}{|rcll|} \\hline \\vec{p} &=& \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} ) \\\\\\\\ &=& \\dbinom{4}{-7} \\\\\\\\ &=& \\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\ \\hline \\vec{P} &=& \\operatorname{proj}_{\\mathbf{-2w}}( \\mathbf{3v} ) \\\\\\\\ &=& \\dfrac{(-2\\vec{w})(3\\vec{v})}{|-2w|^2}(-2\\vec{w}) \\\\\\\\ &=& 6\\dfrac{\\vec{w}\\vec{v}}{2w^2}\\vec{w} \\\\\\\\ &=& 3\\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\\\\\ &=& 3\\vec{p} \\\\\\\\ &=& 3\\dbinom{4}{-7} \\\\\\\\ &=& \\dbinom{12}{-21} \\\\ \\hline \\end{array}$$ $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}}) = \\dbinom{12}{-21}.$$ Feb 24, 2020", "# Compute the orthogonal projection $w$ of $u$ on a line $L$ A vector $u$ and a line $L$ in $\\mathbb{R}^2$ are given. Compute the orthogonal projection $w$ of $u$ on $L$, and use it to compute the distance $d$ from the endpoint of $u$ to $L$ $u = \\begin{pmatrix} 5 \\\\ 0 \\end{pmatrix}$ $y = 0$ I know to that $w = \\frac{u \\cdot v}{||v||^2}v$ and that $d = u - w$ The answer in the book says the answer is $w = \\begin{pmatrix}5 \\\\0 \\end{pmatrix}$ and $d = 0$ How do I find $v$ to solve the problem? • What is your line $L$? Is it the line given by the equation $y = 0$? – Eli Rose Dec 14 '15 at 4:16 • Yes, the book only gives $u$ and the equation of Line $L$ – user273323 Dec 14 '15 at 4:25 The vector $v$ can be any vector in the direction of the line $L$. For a line $ax + by = c$, the vector $\\begin{pmatrix}b\\\\-a\\end{pmatrix}$, or any non-zero multiple of it, lies in the direction of the line. So for $L: y=0$, alias $0x+1y=0$, the vector $v = \\begin{pmatrix}1\\\\0\\end{pmatrix}$ will do. Put this into your equation gives $w=\\begin{pmatrix}5\\\\0\\end{pmatrix}$ as expected. Or, more simply, you can observe that $u$ is a vector parallel to", "is: $$cosθ = \\frac{10}{6\\sqrt{26}}$$ $$θ = cos^{-1}(\\frac{10}{6\\sqrt{26}}) = 71.44°$$ #### 3-D Vector Equation of a line $$\\underline{r} = \\begin{pmatrix}3\\\\\\ 4\\\\\\ 1\\end{pmatrix} + λ\\begin{pmatrix}2\\\\\\ 1\\\\\\ -1\\end{pmatrix}$$ has position $$\\begin{pmatrix}3\\\\\\ 4\\\\\\ 1\\end{pmatrix}$$ and direction $$\\begin{pmatrix}2\\\\\\ 1\\\\\\ -1\\end{pmatrix}$$ Find the equation of the line In order to find the equation of the line we need to find the values of x,y and z in terms of λ and then we can set them equal to eachother To do this we do the same as we did with 2-D vectors and take the x,y and z terms from the equation seperately like so: First we find the equation for x: $$x=3+2λ$$ $$x-3=2λ$$ $$λ=\\frac{x-3}{2}$$ Now we find the equation for y: $$y=4+λ$$ $$λ=\\frac{y-4}{1}$$ Finally we find the equation for z: $$z=1-λ$$ $$λ=\\frac{z-1}{-1}$$ Setting all these equations as equal to eachother gives us the vector equation of the line: $$\\frac{x-3}{2}=\\frac{y-4}{1}=\\frac{z-1}{-1}$$ This can be re-written as: $$\\frac{x-3}{2}=y-4=-z+1$$", "\\frac{-3}{2}, \\frac{1}{3},240\\right)=\\left(0, \\frac{-5}{3},247\\right)$ $7\\left( \\frac{1}{5},\\frac{1}{7},\\frac{1}{9}\\right)=\\left( \\frac{7}{5},1,\\frac{7}{9}\\right)$ Ex.2. $V=M_{m\\times n}(F)=\\left\\lbrace\\begin{pmatrix} a_{11} & \\cdots & a_{1n} \\\\ \\vdots & & \\vdots \\\\ a_{m1} & \\cdots & a_{mn}\\end{pmatrix}: a_{ij} \\in F \\right\\rbrace$ $M_{3\\times 2}( \\mathbb{R})\\ni \\begin{pmatrix} 7 & -7 \\\\ \\pi & \\mathit{e} \\\\ -5 & 2 \\end{pmatrix}$ $\\mbox{Addition by adding entry by entry:}$ $M_{2\\times 2}\\ \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix}+\\begin{pmatrix} b_{11} & b_{12} \\\\ b_{21} & b_{22} \\end{pmatrix}=\\begin{pmatrix} {a_{11}+b_{11}} & {a_{12}+b_{12}} \\\\ {a_{21}+b_{21}} & {a_{22}+b_{22}} \\end{pmatrix}$ $\\mbox{Multiplication by multiplying scalar c to all entries by M.}$ $c\\cdot M_{2\\times 2}\\ \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix}=\\begin{pmatrix} c\\cdot a_{11} & c\\cdot a_{12} \\\\ c\\cdot a_{21} & c\\cdot a_{22} \\end{pmatrix}$ $\\mbox{Zero matrix has all entries = 0:}$ $0_{M_{m\\times n}}=\\begin{pmatrix} 0 & \\cdots & 0 \\\\ \\vdots & & \\vdots \\\\ 0 & \\cdots & 0\\end{pmatrix}$ Ex.3. $\\mathbb{C}$ form a vector space over $\\mathbb{R}$. Ex.4. $\\mbox{F is a vector space over itself.}$ Ex.5. $\\mathbb{R}$ is a vector space over $\\mathbb{Q}$. Ex.6. $\\mbox{Let S be a set. Let}$ $\\mathcal{F}(S,\\mathbb{R})=\\big\\{f:S\\to \\mathbb{R} \\big\\}$ $f,g \\in \\mathcal{F}(S,\\mathbb{R})$ $(f+g)(t)=f(t)+g(t)\\ for\\ any\\ t\\in S$ $(af)(t)=a\\cdot f(t)\\$" ]

[ "Find the inverse of P where $P = \\begin{pmatrix} 1&-2\\\\ -1&3\\\\ \\end{pmatrix}$ c) Verify that $A = P^{-1} \\begin{pmatrix} 1&0\\\\ 0&2\\\\ \\end{pmatrix} P$ d) Compute A5 by using part (b) and (c). e) Compute A100", "## Inverse and a power Let $$\\mathbf B$$ be the inverse of $$\\mathbf A^2$$. Prove that $$\\mathbf A$$ is regular and determine its inverse. From $$\\mathbf A^2\\mathbf B = \\mathbf A\\mathbf A\\mathbf B = \\mathbf I_n$$ we get that $$\\mathbf A^{-1}=\\mathbf A\\mathbf B$$.", "$$\\mathbf{C}$$ is an inverse of $$\\mathbf{A}$$, then $$\\mathbf{A}$$ is an inverse of $$\\mathbf{C}$$. This is immediate from the definition. • If $$\\mathbf{C}$$ and $$\\mathbf{D}$$ are both inverses of $$\\mathbf{A}$$, then $$\\mathbf{C} = \\mathbf{D}$$. Proof: If $$\\mathbf{C}$$ and $$\\mathbf{D}$$ are inverses of $$\\mathbf{A}$$, then we have \\begin{aligned} \\mathbf{A} \\mathbf{C} = \\mathbf{I} &\\Leftrightarrow \\mathbf{D} (\\mathbf{A} \\mathbf{C}) = \\mathbf{D} \\mathbf{I} \\\\ &\\Leftrightarrow (\\mathbf{D} \\mathbf{A}) \\mathbf{C} = \\mathbf{D} \\\\ &\\Leftrightarrow \\mathbf{I} \\mathbf{C} = \\mathbf{D} \\\\ &\\Leftrightarrow \\mathbf{C} = \\mathbf{D}. \\end{aligned} As a consequence of this property, we write the inverse of $$\\mathbf{A}$$, when it exists, as $$\\mathbf{A}^{-1}$$. • The inverse of the inverse of $$\\mathbf{A}$$ is $$\\mathbf{A}$$: $$(\\mathbf{A}^{-1})^{-1} = \\mathbf{A}$$. • If the inverse of $$\\mathbf{A}$$ exists, then the inverse of its transpose is the transpose of the inverse: $$(\\mathbf{A}^\\top)^{-1} = (\\mathbf{A}^{-1})^\\top$$. • Matrix inversion inverts scalar multiplication: if $$c \\neq 0$$, then $$(c \\mathbf{A})^{-1} = (1/c) \\mathbf{A}^{-1}$$. • The identity matrix is its own inverse: $$\\mathbf{I}_n^{-1} = \\mathbf{I}_n$$. Some matrices are not invertible; i.e., their inverse does not exist. As a simple example, think of $\\mathbf{A} = \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix}.$ It’s easy to see that, for any $$2 \\times 2$$ matrix $$\\mathbf{B}$$, we have $\\mathbf{A} \\mathbf{B} = \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix} \\neq \\mathbf{I}_2.$ Therefore, $$\\mathbf{A}$$ does not have an inverse.", "## Matrix inverses Show that $\\mathbf{R_1} = \\begin{pmatrix}-1 & 1 \\\\ 6& -4 \\\\ 4& -3\\end{pmatrix}$ and $\\mathbf{R_2} = \\begin{pmatrix}1 & -1 \\\\ -4& 6 \\\\ -4& 5\\end{pmatrix}$ are both right-inverses of the matrix $\\mathbf{A} = \\begin{pmatrix}1 &1 &-1 \\\\ 4&0 &1 \\end{pmatrix}$. Use the right-inverses $\\mathbf{R_1}$ and $\\mathbf{R_2}$ to find two solutions $\\mathbf{x_1}$ and $\\mathbf{x_2}$ of the equation $\\mathbf{Ax = b}$, where $\\mathbf{b} =\\begin{pmatrix}0\\\\ 8\\end{pmatrix}$. By what I understand, the only way to solve Ax = b is with an inverse: $\\mathbf{A^{-1}Ax = A^{-1}b}$ $\\mathbf{x = A^{-1}b}$ and matrix $\\mathbf{A}$ doesnt have an inverse but the question asks to use the right-inverse and this is what I dont understand", "given $\\mathbf{A}$ and $\\mathbf{b}$. If we can find $\\mathbf{A}^{-1}$ (the inverse of $\\mathbf{A}$), then our solution $\\mathbf{x}$ is given by $\\mathbf{A}^{-1}\\mathbf{b}$, since $$\\mathbf{A}^{-1}\\,\\mathbf{b} = \\mathbf{A}^{-1}(\\mathbf{Ax}) = (\\mathbf{A}^{-1}\\mathbf{A})\\mathbf{x} = \\mathbf{x}.$$ How can we compute $\\mathbf{A}^{-1}$? While there are formulas for inverting small matrices by hand, doing this is not practical for larger matrices. Instead, we will use the Python programming language as a calculator: In [7]: \"\"\" Python code \"\"\" import numpy as np A = np.array([[ 2, 4, 2], [-2, -3, 1], [ 2, 2, -3]]) A_inv = np.linalg.inv(A) print(A_inv) [[ 3.5 8. 5. ] [-2. -5. -3. ] [ 1. 2. 1. ]] Don't worry about the code for now. All you need to know is that the inverse of $\\mathbf{A}$ has been computed (A_inv in the code above), and is given by: $$\\mathbf{A}^{-1} = \\begin{pmatrix} 3.5 & 8 & 5 \\\\ -2 & -5 & -3 \\\\ 1 & 2 & 1 \\end{pmatrix}$$ Substituting this into our equation for $\\mathbf{x}$, we get $$\\mathbf{x} = \\mathbf{A}^{-1}\\, \\mathbf{b} = \\begin{pmatrix} 3.5 & 8 & 5 \\\\ -2 & -5 & -3 \\\\ 1 & 2 & 1 \\end{pmatrix} \\begin{pmatrix} 16 \\\\ -5 \\\\ -3 \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}.$$ Thus, $x=1, y=2, z=3$ solves our system of equations. (Verify this for yourself!) ## Reviewing", "do you confirm that f and g are inverses by showing that f(g(x))=x and g(f(x))=x ? Problem: f(x)=3x-2 and g(x)= x+2 --- 3 Please help, don't know where to begin? 2. ### Math Find the $2 \\times 2$ matrix $\\bold{A}$ such that $\\bold{A} \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix}$ and \\[\\bold{A} \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} -7 \\\\ 4", "= e \\left(= 7\\right) .$ Therefore, the Inverse of $2$, denoted by, ${2}^{-} 1$, under the operation $\\ast$ is $12 \\in \\mathbb{Z} .$ N.B.: In fact, it is easy to see that, $\\forall a \\in \\mathbb{Z} , {a}^{-} 1 = 14 - a .$ Enjoy Maths.!", "• You're supposed to multiply the expression for the partitioned inverse with $\\begin{pmatrix}\\bf \\tt{wa}\\\\\\mathbf 0\\end{pmatrix}$, thus resulting in expressions for $\\begin{pmatrix}\\mathbf v\\\\\\bf \\tt{bb}\\end{pmatrix}$. – J. M.'s ennui Dec 1 '15 at 2:24 • @J.M. Thank you. I think I got it now. – Jack LaVigne Dec 6 '15 at 15:32 • Looks right to me. :) – J. M.'s ennui Dec 9 '15 at 2:49", "# If $\\begin{bmatrix} 1 & -1 & x \\\\ 1 & x & 1 \\\\ x & -1 &1 \\end {bmatrix}$ has no inverse, then the real value of x is : $\\begin {array} {1 1} (1)\\;2 & \\quad (2)\\;3 \\\\ (3)\\;0 & \\quad (4)\\;1 \\end {array}$", "+0 # Matrices 0 69 1 Given $$\\mathbf{A}^{-1} = \\begin{pmatrix} 2 & 3 \\\\ -1 & 0 \\end{pmatrix} \\quad \\text{and} \\quad \\mathbf{B}^{-1} = \\begin{pmatrix} 0 & 4 \\\\ 1 & 5 \\end{pmatrix} ,$$ find the inverse of AB. Thanks! Apr 2, 2021 #1 +22020 +2 1) Find the inverse of A-1; this will be A. 2) Find the inverse of B-1; this will be B. 3) Find the product of A and B; this will be AB. 4) Lastly, find the inverse of AB (your answer to step 3). Apr 2, 2021", "can find the inverse just by turning the matrix upside down: $$\\sigma^{-1} = \\begin{pmatrix} 3 &4 &1& 2 \\\\ 1& 2& 3&4 \\end{pmatrix}.$$ Of course the ordering is different, but you can still read off that ,for example $\\sigma^{-1}(3)=1, \\sigma^{-1}(4)=2$, and so on.", "B is the target, then $|f(A)| = |B|$, where $f(A)$ is the image of A under f. True False Does not make sense I don't know yet 6. Is the following statement true, false, or neither? Given a bijective function $f:A \\rightarrow B$, then $|A| = |B|$. True False Does not make sense I don't know yet 7. Find the inverse of the following function, if it exists. $$\\begin{pmatrix} -3&-2&-1&0&1&2&3 \\\\ -1&3&1&-2&0&2&-3 \\end{pmatrix}$$ 8. Find the inverse of the following function, if it exists. $$\\begin{pmatrix} 0&1 \\\\ 1&0 \\end{pmatrix}$$ 9. Find the inverse of the following function, if it exists. $$\\begin{pmatrix} 0&1&2&3&4 \\\\ 4&3&1&0&1 \\end{pmatrix}$$ 10. Find the inverse of the following function, if it exists. $$f : \\{ x : x^2 < 3 \\} \\rightarrow \\{\\alpha, \\beta, \\gamma, \\delta\\}$$ 11. Given $$f := \\begin{pmatrix} 0&1&2&3&4 \\\\ 0&4&1&2&3 \\end{pmatrix},$$ $$g := \\begin{pmatrix} 0&1&2&3&4 \\\\ 4&1&3&0&2 \\end{pmatrix},$$ find the composition f@g. 12. Given $$f := \\begin{pmatrix} 0&1&2&3&4 \\\\ 0&4&1&2&3 \\end{pmatrix},$$ $$g := \\begin{pmatrix} 0&1&2&3&4 \\\\ 4&1&3&0&2 \\end{pmatrix},$$ find the composition g@f. 13. Given $$f := \\begin{pmatrix} 0&1&2&3&4 \\\\ 1&0&3&2&4 \\end{pmatrix},$$ $$g := \\begin{pmatrix} 0&1&2&3&4 \\\\ 3&2&1&4&0 \\end{pmatrix},$$ find the composition f@g. 14. Given $$f := \\begin{pmatrix} -2&-1&0&1&2 \\\\ 0&-2&1&2&-1 \\end{pmatrix},$$ $$g := \\begin{pmatrix} 0&1&2&3&4 \\\\ 0&2&3&1&4 \\end{pmatrix},$$ find the composition f@g. 15. Given $$f := \\begin{pmatrix} -2&-1&0&1&2 \\\\ 2&0&1&4&3", "# The inverse of the following statement If $x=3,$ then $x^2=9$ is $\\begin{array}{1 1} If \\;x^2=9,then \\;x=3 \\\\If \\;x \\neq 3,then \\;x^2 \\neq 3 \\\\ If \\;x^2=9,then \\;x \\neq 3 \\\\ none\\;of\\; these \\end{array}$", "D \\end{pmatrix}$ and being nonsingular this has an inverse $Z$: $\\begin{pmatrix} A & B \\\\ C & D \\end{pmatrix}YZX =I$ If $\\overline{A}$ is small, its inverse would not be computationally complex, and likewise D is easy to invert, and the rest is row/column operations. -", "0 & 1 & 2 \\\\ 0 & 0 & 1\\end{pmatrix}, \\quad \\mathbf N = \\begin{pmatrix} 0 & 0 & 1 \\\\ 0 & 1 & 3 \\\\ 1 & 3 & 6\\end{pmatrix}.$ This entry was posted in Uncategorized and tagged , , , , , . Bookmark the permalink.", "find the inverse of $a$ in $\\mathbb{Z}_n=\\mathbb{Z}/n\\mathbb{Z}$.", "is just $a$ and $a$ has an inverse in $\\mathbb Z[i]$ iff $a=\\pm 1$. This is because, the multiplicative inverse is $\\frac{1}{a}$, which will be in $\\mathbb Z$ iff $a=\\pm 1$. - If $z,w\\in\\mathbb{Z}[i]$ are such that $zw=1$ (i.e. $z$ is a unit and $w$ its inverse), then $|z|^2|w|^2=|zw|^2=1$, or $$(a^2+b^2)(c^2+d^2)=1, \\quad z=a+bi,\\; w=c+di.$$ Now $a,b,c,d$ are all integers, so $a^2+b^2$ and $c^2+d^2$ must both be nonnegative integers, which must both equal exactly $1$ and no greater in order to multiply to $1$ in the integers. And if $a^2+b^2=1$, we have $a^2$ and $b^2\\le1$. Check by hand the only solutions here correspond to $(a,b)=(\\pm1,0)$ or $(0,\\pm1)$. - Equivalently, by taking conjugates we conclude that $\\bar{z}\\bar{w}=1$ and therefore $(z\\bar z)(w\\bar{w})=1$ and therefore $z\\bar{z}=1$. – André Nicolas Feb 11 '12 at 8:14", "\\end{pmatrix}$ Using the definition of the inner product $(\\mathbf u, \\mathbf v) = \\mathbf v^* \\mathbf u$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (5-i)(1+3i) + (6-4i)(2+4i) = 36 + 30i$ , but using $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\begin{pmatrix} 1 \\\\ 2 \\\\ \\end{pmatrix} , \\begin{pmatrix} 5 \\\\ 6 \\\\ \\end{pmatrix}) + i(\\begin{pmatrix} 3 \\\\ 4 \\\\ \\end{pmatrix} , \\begin{pmatrix} 1 \\\\ 4 \\\\ \\end{pmatrix}) = 5 + 12 + i(3 + 16) = 17 + 19i$. What went wrong with my interpretation of the formula , $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$ ? Shouldn't both computations be mathematically sound and give the same result? Thanks in advance for any help!", "\\\\ 2 & - 3 & 4 \\\\ 0 & - 1 & 1\\end{bmatrix}$ Q 10 | Page 34 Find the inverse $\\begin{bmatrix}1 & 2 & 0 \\\\ 2 & 3 & - 1 \\\\ 1 & - 1 & 3\\end{bmatrix}$ Q 11 | Page 34 Find the inverse $\\begin{bmatrix}2 & - 1 & 3 \\\\ 1 & 2 & 4 \\\\ 3 & 1 & 1\\end{bmatrix}$ Q 12 | Page 34 Find the inverse $\\begin{bmatrix}1 & 1 & 2 \\\\ 3 & 1 & 1 \\\\ 2 & 3 & 1\\end{bmatrix}$ Q 13 | Page 34 Find the inverse $\\begin{bmatrix}2 & - 1 & 4 \\\\ 4 & 0 & 7 \\\\ 3 & - 2 & 7\\end{bmatrix}$ Q 14 | Page 34 Find the inverse $\\begin{bmatrix}3 & 0 & - 1 \\\\ 2 & 3 & 0 \\\\ 0 & 4 & 1\\end{bmatrix}$ Q 15 | Page 34 Find the inverse $\\begin{bmatrix}1 & 3 & - 2 \\\\ - 3 & 0 & - 1 \\\\ 2 & 1 & 0\\end{bmatrix}$ Q 16 | Page 34 Find the inverse of each of the following matrices by using elementary row transformations: $\\begin{bmatrix}- 1 & 1 & 2 \\\\ 1 & 2 & 3 \\\\ 3 & 1 & 1\\end{bmatrix}$ #### Chapter 7: Adjoint and Inverse of a", "\\\\ 0 & \\frac{1}{cosh{y}}\\end{pmatrix} *j$ = $\\begin{pmatrix} \\frac{1}{cosh^2{y}} & 0 \\\\ 0 & (\\frac{1}{cosh{y}})^4 \\end{pmatrix}$ there was a mistake, it was the transpose not the inverse. 5. Jun 30, 2015 ### Simone Furcas I solved it. I solved the exercise." ]

[ "\\end{pmatrix}$ Using the definition of the inner product $(\\mathbf u, \\mathbf v) = \\mathbf v^* \\mathbf u$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (5-i)(1+3i) + (6-4i)(2+4i) = 36 + 30i$ , but using $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\begin{pmatrix} 1 \\\\ 2 \\\\ \\end{pmatrix} , \\begin{pmatrix} 5 \\\\ 6 \\\\ \\end{pmatrix}) + i(\\begin{pmatrix} 3 \\\\ 4 \\\\ \\end{pmatrix} , \\begin{pmatrix} 1 \\\\ 4 \\\\ \\end{pmatrix}) = 5 + 12 + i(3 + 16) = 17 + 19i$. What went wrong with my interpretation of the formula , $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$ ? Shouldn't both computations be mathematically sound and give the same result? Thanks in advance for any help!", "given $\\mathbf{A}$ and $\\mathbf{b}$. If we can find $\\mathbf{A}^{-1}$ (the inverse of $\\mathbf{A}$), then our solution $\\mathbf{x}$ is given by $\\mathbf{A}^{-1}\\mathbf{b}$, since $$\\mathbf{A}^{-1}\\,\\mathbf{b} = \\mathbf{A}^{-1}(\\mathbf{Ax}) = (\\mathbf{A}^{-1}\\mathbf{A})\\mathbf{x} = \\mathbf{x}.$$ How can we compute $\\mathbf{A}^{-1}$? While there are formulas for inverting small matrices by hand, doing this is not practical for larger matrices. Instead, we will use the Python programming language as a calculator: In [7]: \"\"\" Python code \"\"\" import numpy as np A = np.array([[ 2, 4, 2], [-2, -3, 1], [ 2, 2, -3]]) A_inv = np.linalg.inv(A) print(A_inv) [[ 3.5 8. 5. ] [-2. -5. -3. ] [ 1. 2. 1. ]] Don't worry about the code for now. All you need to know is that the inverse of $\\mathbf{A}$ has been computed (A_inv in the code above), and is given by: $$\\mathbf{A}^{-1} = \\begin{pmatrix} 3.5 & 8 & 5 \\\\ -2 & -5 & -3 \\\\ 1 & 2 & 1 \\end{pmatrix}$$ Substituting this into our equation for $\\mathbf{x}$, we get $$\\mathbf{x} = \\mathbf{A}^{-1}\\, \\mathbf{b} = \\begin{pmatrix} 3.5 & 8 & 5 \\\\ -2 & -5 & -3 \\\\ 1 & 2 & 1 \\end{pmatrix} \\begin{pmatrix} 16 \\\\ -5 \\\\ -3 \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}.$$ Thus, $x=1, y=2, z=3$ solves our system of equations. (Verify this for yourself!) ## Reviewing", "$$\\mathbf{C}$$ is an inverse of $$\\mathbf{A}$$, then $$\\mathbf{A}$$ is an inverse of $$\\mathbf{C}$$. This is immediate from the definition. • If $$\\mathbf{C}$$ and $$\\mathbf{D}$$ are both inverses of $$\\mathbf{A}$$, then $$\\mathbf{C} = \\mathbf{D}$$. Proof: If $$\\mathbf{C}$$ and $$\\mathbf{D}$$ are inverses of $$\\mathbf{A}$$, then we have \\begin{aligned} \\mathbf{A} \\mathbf{C} = \\mathbf{I} &\\Leftrightarrow \\mathbf{D} (\\mathbf{A} \\mathbf{C}) = \\mathbf{D} \\mathbf{I} \\\\ &\\Leftrightarrow (\\mathbf{D} \\mathbf{A}) \\mathbf{C} = \\mathbf{D} \\\\ &\\Leftrightarrow \\mathbf{I} \\mathbf{C} = \\mathbf{D} \\\\ &\\Leftrightarrow \\mathbf{C} = \\mathbf{D}. \\end{aligned} As a consequence of this property, we write the inverse of $$\\mathbf{A}$$, when it exists, as $$\\mathbf{A}^{-1}$$. • The inverse of the inverse of $$\\mathbf{A}$$ is $$\\mathbf{A}$$: $$(\\mathbf{A}^{-1})^{-1} = \\mathbf{A}$$. • If the inverse of $$\\mathbf{A}$$ exists, then the inverse of its transpose is the transpose of the inverse: $$(\\mathbf{A}^\\top)^{-1} = (\\mathbf{A}^{-1})^\\top$$. • Matrix inversion inverts scalar multiplication: if $$c \\neq 0$$, then $$(c \\mathbf{A})^{-1} = (1/c) \\mathbf{A}^{-1}$$. • The identity matrix is its own inverse: $$\\mathbf{I}_n^{-1} = \\mathbf{I}_n$$. Some matrices are not invertible; i.e., their inverse does not exist. As a simple example, think of $\\mathbf{A} = \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix}.$ It’s easy to see that, for any $$2 \\times 2$$ matrix $$\\mathbf{B}$$, we have $\\mathbf{A} \\mathbf{B} = \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix} \\neq \\mathbf{I}_2.$ Therefore, $$\\mathbf{A}$$ does not have an inverse.", "\\ \\ \\ \\ \\ \\ \\$ $-6$ $-20.5$ $-5.5$ $-11.5$ $-5$ $-5.5$ $-4.5$ $-3.5$ $-4$ $-1.5$ 1. -5 2. -11.5 3. -3.5 4. -5.5 3. The following table gives some of the values of $f(x)$, a function which has an inverse. Using this information, find $f^{-1}(-4)$. $\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ mathbf{x} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\$ $\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\mathbf{f(x)} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\$ $-6$ $8$ $-5$ $4$ $-4$ $0$ $-3$ $-4$ $-2$ $-8$ 1. -4 2. 0 3. -3 4. -5 4. What is the value of $f^{-1}(6)$, given the following table? Assume that $f(x)$ has an inverse. $\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ mathbf{x} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\$ $\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\mathbf{f(x)} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\$ $1/36$ $6$ $1/6$ $10$ $1$ $23$ $6$ $54$ $36$ $103$ 1. $10$ 2. $1/36$ 3. $54$ 4. It cannot be determined with the information given. 5. Determine the value of $f^{-1}(0.4)$, given the following table. Assume that $f(x)$ has an inverse. $\\ \\ \\ \\ \\ \\ \\", "is just $a$ and $a$ has an inverse in $\\mathbb Z[i]$ iff $a=\\pm 1$. This is because, the multiplicative inverse is $\\frac{1}{a}$, which will be in $\\mathbb Z$ iff $a=\\pm 1$. - If $z,w\\in\\mathbb{Z}[i]$ are such that $zw=1$ (i.e. $z$ is a unit and $w$ its inverse), then $|z|^2|w|^2=|zw|^2=1$, or $$(a^2+b^2)(c^2+d^2)=1, \\quad z=a+bi,\\; w=c+di.$$ Now $a,b,c,d$ are all integers, so $a^2+b^2$ and $c^2+d^2$ must both be nonnegative integers, which must both equal exactly $1$ and no greater in order to multiply to $1$ in the integers. And if $a^2+b^2=1$, we have $a^2$ and $b^2\\le1$. Check by hand the only solutions here correspond to $(a,b)=(\\pm1,0)$ or $(0,\\pm1)$. - Equivalently, by taking conjugates we conclude that $\\bar{z}\\bar{w}=1$ and therefore $(z\\bar z)(w\\bar{w})=1$ and therefore $z\\bar{z}=1$. – André Nicolas Feb 11 '12 at 8:14", "0 & 1 & 2 \\\\ 0 & 0 & 1\\end{pmatrix}, \\quad \\mathbf N = \\begin{pmatrix} 0 & 0 & 1 \\\\ 0 & 1 & 3 \\\\ 1 & 3 & 6\\end{pmatrix}.$ This entry was posted in Uncategorized and tagged , , , , , . Bookmark the permalink.", "= e \\left(= 7\\right) .$ Therefore, the Inverse of $2$, denoted by, ${2}^{-} 1$, under the operation $\\ast$ is $12 \\in \\mathbb{Z} .$ N.B.: In fact, it is easy to see that, $\\forall a \\in \\mathbb{Z} , {a}^{-} 1 = 14 - a .$ Enjoy Maths.!", "f)$$, which means $$\\forall x\\in({{\\mathbf{1}}}+{{\\mathcal A}}_1).~ f(a_1(x)) = a_2(({\\mathsf{id}}_{{{\\mathbf{1}}}} + f)(x))$$. Now, if $$x\\in({{\\mathbf{1}}}+{{\\mathcal A}}_1)$$ then either $$x\\in{{\\mathbf{1}}}$$ or $$x\\in{{\\mathcal A}}_1$$, so there are two cases to consider. 1. If $$x\\in{{\\mathbf{1}}}$$ then $$x{{\\,{=}\\,}}{\\ast}$$ and $$a_1(x){{\\,{=}\\,}}z_1$$ so $$f(a_1(x)){{\\,{=}\\,}}f(z_1)$$. If $$x{{\\,{=}\\,}}{\\ast}$$ then $$({\\mathsf{id}}_{{{\\mathbf{1}}}} + f)(x){{\\,{=}\\,}}{\\mathsf{id}}_{{{\\mathbf{1}}}}(x){{\\,{=}\\,}}x{{\\,{=}\\,}}{\\ast}$$, so $$a_2(({\\mathsf{id}}_{{{\\mathbf{1}}}} + f)(x)){{\\,{=}\\,}}a_2({\\ast}){{\\,{=}\\,}}z_2$$. Thus the equation $$f(a_1(x)) = a_2(({\\mathsf{id}}_{{{\\mathbf{1}}}} + f)(x))$$ reduces to $$f(z_1){{\\,{=}\\,}}z_2$$. 2. If $$x\\in{{\\mathcal A}}_1$$ then $$a_1(x){{\\,{=}\\,}}s_1(x)$$, so $$f(a_1(x)){{\\,{=}\\,}}f(s_1(x))$$. If $$x\\in{{\\mathcal A}}_1$$ then $$({\\mathsf{id}}_{{{\\mathbf{1}}}} + f)(x){{\\,{=}\\,}}f(x)$$, so $$a_2(({\\mathsf{id}}_{{{\\mathbf{1}}}} + f)(x)){{\\,{=}\\,}}a_2(f(x)){{\\,{=}\\,}}s_2(f(x))$$. Thus the equation $$f(a_1(x)) = a_2(({\\mathsf{id}}_{{{\\mathbf{1}}}} + f)(x))$$ reduces to $$f(s_1(x)){{\\,{=}\\,}}s_2(f(x))$$. A morphism from the $${{{\\Bbb{F}}}_{{{\\scriptsize{{{\\Bbb N}}}}}}}$$-algebra corresponding to the natural numbers $$({{\\Bbb N}},0,{{\\small\\mathsf{S}}})$$ to an $${{{\\Bbb{F}}}_{{{\\scriptsize{{{\\Bbb N}}}}}}}$$-algebra corresponding to $$({{\\mathcal A}},z,s)$$ is a function $$f:{{\\Bbb N}}{{\\;{\\to}\\;}}{{\\mathcal A}}$$ such that $$f(0){\\,{=}\\,}z$$ and $$\\forall n\\in{{\\Bbb N}}.\\,f({{\\small\\mathsf{S}}}(n)){\\,{=}\\,}s(f(n))$$, i.e.: $$f(n)={{{\\mathtt{if}}}~n{=}0~{{\\mathtt{then}}}~z~{{\\mathtt{else}}}~s(f(n{-}1))}$$. This is the recursive equation characterising the natural numbers. A function $$g:{{\\mathcal C}}_1{{\\;{\\to}\\;}}{{\\mathcal C}}_2$$ is a morphism from an $${{\\Bbb{F}}}$$-coalgebra $$({{\\mathcal C}}_1, c_1)$$ to an $${{\\Bbb{F}}}$$-coalgebra $$({{\\mathcal C}}_2, c_2)$$ if and only if $$c_2 \\circ g {{\\,{=}\\,}} {{\\Bbb{F}}}(g) \\circ c_1$$. To see what morphisms between coalgebras are, consider $${{{\\Bbb{F}}}_{{{\\scriptsize{{{\\Bbb N}}}}}}}$$-coalgebras. Recall: $${{{\\Bbb{F}}}_{{{\\scriptsize{{{\\Bbb N}}}}}}}(X) {{\\,{=}\\,}} {{\\mathbf{1}}}+ X$$. Suppose $$g:{{\\mathcal C}}_1{{\\;{\\to}\\;}}{{\\mathcal C}}_2$$ is a morphism from an $${{{\\Bbb{F}}}_{{{\\scriptsize{{{\\Bbb N}}}}}}}$$-coalgebra $$c_1:{{\\mathcal C}}_1{{\\;{\\to}\\;}}{{{\\Bbb{F}}}_{{{\\scriptsize{{{\\Bbb N}}}}}}}({{\\mathcal C}}_1)$$ corresponding to a Peano coalgebra $$({{\\mathcal C}}_1,isz_1,p_1)$$ to an $${{{\\Bbb{F}}}_{{{\\scriptsize{{{\\Bbb N}}}}}}}$$-coalgebra $$c_2:{{\\mathcal C}}_2{{\\;{\\to}\\;}}{{{\\Bbb{F}}}_{{{\\scriptsize{{{\\Bbb N}}}}}}}({{\\mathcal C}}_2)$$ corresponding to a Peano coalgebra $$({{\\mathcal C}}_2,isz_2,p_2)$$. If", "• You're supposed to multiply the expression for the partitioned inverse with $\\begin{pmatrix}\\bf \\tt{wa}\\\\\\mathbf 0\\end{pmatrix}$, thus resulting in expressions for $\\begin{pmatrix}\\mathbf v\\\\\\bf \\tt{bb}\\end{pmatrix}$. – J. M.'s ennui Dec 1 '15 at 2:24 • @J.M. Thank you. I think I got it now. – Jack LaVigne Dec 6 '15 at 15:32 • Looks right to me. :) – J. M.'s ennui Dec 9 '15 at 2:49", "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", "why $x^2+y^2=(-y)^2+x^2$ (?!), which makes for a rather poor question, I am sorry to say. Note also that every sequence $(\\mathbf x_n)$ defined by $\\mathbf x_{n+1}=L(\\mathbf x_n)$ has period $4$, as an elementary computation shows. I know that to solve the exercise I need to show that F is constant along the orbits of L. That is, I need to show that $F \\text{ o } L(x) = L(x)$. No. That $F$ is an invariant of the dynamical system associated to $L$ means that, if $$\\dot{\\mathbf x}(t)=L(\\mathbf x(t)),$$ then the quantity $F(\\mathbf x(t))$ does not depend on $t$. This is usually done showing that $u'(t)=0$ for every $t$, where $$u(t)=F(\\mathbf x(t)),$$ which is implied by the condition that, for every $\\mathbf x$, $$\\nabla F(\\mathbf x)\\cdot L(\\mathbf x)=0.$$ In your case, $$\\nabla F(\\mathbf x)=2\\mathbf x,$$ hence one is left with $$\\nabla F(\\mathbf x)\\cdot L(\\mathbf x)=2\\mathbf x\\cdot L(\\mathbf x)=2\\begin{pmatrix}x\\\\y\\end{pmatrix}\\cdot\\begin{pmatrix}y\\\\-x\\end{pmatrix}=0.$$ Here is a rough sketch of some solutions $(\\mathbf x(t))$: $\\hspace2in$", "this: $(\\mathbf A \\mathbf B)^T = \\mathbf B^T \\mathbf A^T$ needs to be like so: $(\\mathbf A \\mathbf B)^T = \\mathbf B^T \\mathbf A^T$ and so on. It seems that MathJax has no problem with it, but nbconvert needs this format.", "3(x^{2} + x + 1) ]^{2} < 0$, Or, $(x^{2} – 3x – 1 + 3\\times 2 + 3x + 3) (x^{2} – 3x – 1 – 3\\times 2 – 3x -3) < 0$ Or, $(4 \\times 2 + 2) (- 2 \\times 2 – 6x – 4) < 0,$ Or, (2 \\times 2 + 1) (x^{2} + 3x + 2) > 0, Or, (2 \\times 2 + 1) (x + 1) (x + 2) > 0, Therefore, $\\mathbf{x\\in \\;(-\\infty,\\; -\\;2)\\;(-\\;1, \\infty ) }$", "\\mathbf{A}_{2,1} & \\mathbf{A}_{2,2} \\end{bmatrix} \\mbox { , } \\mathbf{B} = \\begin{bmatrix} \\mathbf{B}_{1,1} & \\mathbf{B}_{1,2} \\\\ \\mathbf{B}_{2,1} & \\mathbf{B}_{2,2} \\end{bmatrix} \\mbox { , } \\mathbf{C} = \\begin{bmatrix} \\mathbf{C}_{1,1} & \\mathbf{C}_{1,2} \\\\ \\mathbf{C}_{2,1} & \\mathbf{C}_{2,2} \\end{bmatrix}$ with $\\mathbf{A}_{i,j}, \\mathbf{B}_{i,j}, \\mathbf{C}_{i,j} \\in R^{2^{n-1} \\times 2^{n-1}}$ then $\\mathbf{C}_{1,1} = \\mathbf{A}_{1,1} \\mathbf{B}_{1,1} + \\mathbf{A}_{1,2} \\mathbf{B}_{2,1}$ $\\mathbf{C}_{1,2} = \\mathbf{A}_{1,1} \\mathbf{B}_{1,2} + \\mathbf{A}_{1,2} \\mathbf{B}_{2,2}$ $\\mathbf{C}_{2,1} = \\mathbf{A}_{2,1} \\mathbf{B}_{1,1} + \\mathbf{A}_{2,2} \\mathbf{B}_{2,1}$ $\\mathbf{C}_{2,2} = \\mathbf{A}_{2,1} \\mathbf{B}_{1,2} + \\mathbf{A}_{2,2} \\mathbf{B}_{2,2}$ With this construction we have not reduced the number of multiplications. We still need 8 multiplications to calculate the Ci,j matrices, the same number of multiplications we need when using standard matrix multiplication. Now comes the important part. We define new matrices $\\mathbf{M}_{1} := (\\mathbf{A}_{1,1} + \\mathbf{A}_{2,2}) (\\mathbf{B}_{1,1} + \\mathbf{B}_{2,2})$ $\\mathbf{M}_{2} := (\\mathbf{A}_{2,1} + \\mathbf{A}_{2,2}) \\mathbf{B}_{1,1}$ $\\mathbf{M}_{3} := \\mathbf{A}_{1,1} (\\mathbf{B}_{1,2} - \\mathbf{B}_{2,2})$ $\\mathbf{M}_{4} := \\mathbf{A}_{2,2} (\\mathbf{B}_{2,1} - \\mathbf{B}_{1,1})$ $\\mathbf{M}_{5} := (\\mathbf{A}_{1,1} + \\mathbf{A}_{1,2}) \\mathbf{B}_{2,2}$ $\\mathbf{M}_{6} := (\\mathbf{A}_{2,1} - \\mathbf{A}_{1,1}) (\\mathbf{B}_{1,1} + \\mathbf{B}_{1,2})$ $\\mathbf{M}_{7} := (\\mathbf{A}_{1,2} - \\mathbf{A}_{2,2}) (\\mathbf{B}_{2,1} + \\mathbf{B}_{2,2})$ which are then used to express the Ci,j in terms of Mk. Because of our definition of the Mk we can eliminate one matrix multiplication and reduce the number of multiplications to 7 (one multiplication for each Mk) and express the Ci,j as $\\mathbf{C}_{1,1} = \\mathbf{M}_{1} + \\mathbf{M}_{4} - \\mathbf{M}_{5} + \\mathbf{M}_{7}$ $\\mathbf{C}_{1,2} = \\mathbf{M}_{3} + \\mathbf{M}_{5}$", "these (e) None of these Explanation: $({{a}^{2}}+ab+{{b}^{2}})({{a}^{2}}-ab-{{b}^{2}})$ $={{a}^{2}}({{a}^{2}}-ab-{{b}^{2}})+ab({{a}^{2}}-ab-{{b}^{2}})+{{b}^{2}}({{a}^{2}}-ab-{{b}^{2}})$ $={{a}^{4}}-{{a}^{3}}^{~}b-{{a}^{2}}{{b}^{2}}+{{a}^{3}}b-{{a}^{2}}{{b}^{2}}-a{{b}^{3}}+{{a}^{2}}{{b}^{2}}-a{{b}^{3}}-{{b}^{4}}$ $={{a}^{4}}-{{a}^{2}}^{~}{{b}^{2}}-a{{b}^{3}}-a{{b}^{3}}-{{b}^{4}}$ $={{a}^{4}}-{{a}^{2}}^{~}{{b}^{2}}-2a{{b}^{3}}-{{b}^{4}}$ Divide: $\\mathbf{2}{{\\mathbf{x}}^{\\mathbf{2}}}+\\mathbf{3x}+\\mathbf{1}\\text{ }\\mathbf{by}\\text{ }\\left( \\mathbf{x}+\\mathbf{1} \\right)?$ (a) $3x+2$ (b) $2x+1$ (c) $5x-3$ (d) All of these (e) None of these Explanation: Quotient of the division$=2x+1$is the solution of the expression. Simplify the following, ${{\\mathbf{x}}^{0}}-\\mathbf{xy}-\\mathbf{8}{{\\mathbf{y}}^{0}}.$ (a) $\\left( xy+7 \\right)$ (b) $\\left( xy-7 \\right)$ (c) $-\\left( xy+7 \\right)$ (d) All of these (e) None of these Explanation: ${{x}^{0}}-xy-8{{y}^{0}}=1-xy-8=-xy-7=-\\left( xy+7 \\right).$ Express the ratio 45 :108 in its simplest form: (a) 4 : 12 (b) 5 : 12 (c) 5 : 13 (d) 3 : 12 (e) None of these Explanation: In order to express the given ratio in its simplest form we divide its first and second terms by their HCF. We have, $45=3\\times 3\\times 5$ and $108=2\\times 2\\times 3\\times 3\\times 3$ So, HCF of 45 and 108 is $3\\times 3=9.$ $\\therefore 45:108=\\frac{45}{108}=\\frac{45\\div 9}{108\\div 9}=\\frac{5}{12}=5:12$ Divide Rs. 1250 between Raman and Rahul in the ratio 2 : 3. (a) Rs. 750 (b) Rs. 745 (c) Rs. 760 (d) Rs. 755 (e) None of these Explanation: We have, sum of the terms of the ratio $=\\left( 2+3 \\right)=5$ $\\therefore$ Raman's share $=Rs.\\left( \\frac{2}{5}\\times 1250 \\right)=Rs.\\left( 2\\times 250 \\right)=Rs.500$ Rahul’s share $=Rs.\\left( \\frac{3}{5}\\times 1250 \\right)=Rs.\\left( 3\\times 250 \\right)=Rs.750$ #### Other Topics ##### 30 20 You need to login to perform this", "it (with an appropriate multiplier) to make all the numbers in the second column, zero, that is: $\\begin{bmatrix} \\mathbf{0} && \\mathbf{0} && ... && \\mathbf{0}\\\\ \\mathbf{0} && a'_{22} && ... && a'_{2N}\\\\ \\mathbf{0} && a'_{32} && ... && a'_{3N}\\\\ ... && ... && .. && ...\\\\ \\mathbf{0} && a'_{N2} && ... && a'_{NN}\\\\ \\end{bmatrix}\\\\ = \\ell_1u_1+ \\begin{bmatrix} 0\\\\ 1\\\\ \\ell_{23}\\\\ .\\\\ .\\\\ \\ell_{2N}\\\\ \\end{bmatrix} \\begin{bmatrix} 0 && a'_{22} && a'_{23} && ... && a_{1N}\\\\ \\end{bmatrix} + \\begin{bmatrix} \\mathbf{0} && \\mathbf{0} && ... && \\mathbf{0}\\\\ \\mathbf{0} && \\mathbf{0} && ... && \\mathbf{0}\\\\ \\mathbf{0} && \\mathbf{0} && ... && a''_{3N}\\\\ ... && ... && .. && ...\\\\ \\mathbf{0} && \\mathbf{0} && ... && a''_{NN}\\\\ \\end{bmatrix}\\\\$ ### Step 4 Call the first term on the LHS as $$\\ell_2u_2$$, so that: $\\begin{bmatrix} \\mathbf{0} && \\mathbf{0} && ... && \\mathbf{0}\\\\ \\mathbf{0} && a'_{22} && ... && a'_{2N}\\\\ \\mathbf{0} && a'_{32} && ... && a'_{3N}\\\\ ... && ... && .. && ...\\\\ \\mathbf{0} && a'_{N2} && ... && a'_{NN}\\\\ \\end{bmatrix}\\\\ = \\ell_1u_1+ \\ell_2u_2+ \\begin{bmatrix} \\mathbf{0} && \\mathbf{0} && ... && \\mathbf{0}\\\\ \\mathbf{0} && \\mathbf{0} && ... && \\mathbf{0}\\\\ \\mathbf{0} && \\mathbf{0} && ... && a''_{3N}\\\\ ... && ... && .. && ...\\\\ \\mathbf{0} && \\mathbf{0} && ... && a''_{NN}\\\\ \\end{bmatrix}\\\\$ ### Step 5 I hope you can see the pattern, we are gradually aiming", "true. So $A^n =\\begin{pmatrix}a^n & 0\\\\0 & b^n\\end{pmatrix}, \\forall n \\in \\mathbb{N}$", "+0 # Let $n$ be the inverse of $2\\pmod{17}$. 0 24 1 +122 Let $n$ be the inverse of $2\\pmod{17}$. That is, let $n$ be the integer $0\\leq n < 17$ for which $2n \\equiv 1 \\pmod{17}$. What is $\\left(2^n\\right)^2 - 2 \\pmod{17}$? Express your answer as an integer from $0$ to $16$, inclusive. Sort: #1 +19376 +1 Let $n$ be the inverse of $2\\pmod{17}$. That is, let $n$ be the integer $0\\leq n < 17$ for which $2n \\equiv 1 \\pmod{17}$. What is $\\left(2^n\\right)^2 - 2 \\pmod{17}$? Express your answer as an integer from $0$ to $16$, inclusive. $$\\begin{array}{|rcll|} \\hline 2n &\\equiv& 1 \\pmod {17} \\quad & | \\quad : 2 \\\\\\\\ n &\\equiv& \\frac12 \\pmod {17} \\\\ \\mathbf{n} &\\mathbf{\\equiv}& \\mathbf{ 2^{-1} \\pmod {17}} \\\\ \\hline \\end{array}$$ $$\\text{ Solve 2^{-1} \\pmod {17} }$$ $$\\text{Because gcd(2,17)=1 we can continue.}$$ $$\\begin{array}{|rcll|} \\hline 2^{-1} \\pmod {17} &\\equiv& 2^{\\varphi(17)-1} \\pmod {17} \\quad & | \\quad \\varphi(p) = p - 1 \\quad \\text{p is a prime number} \\\\\\\\ && \\varphi(17) = 16 \\\\\\\\ &\\equiv& 2^{16-1} \\pmod {17} \\\\ &\\equiv& 2^{15} \\pmod {17} \\\\ &\\equiv& 32768 \\pmod {17} \\\\ 2^{-1} \\pmod {17}&\\equiv& 9 \\pmod {17} \\\\ \\mathbf{n} &\\mathbf{=}& \\mathbf{9} \\\\ \\hline \\end{array}$$ $$Proof:\\\\ 2\\cdot 9 \\equiv 1 \\pmod{17} \\ \\checkmark$$ $$\\begin{array}{|rcll|} \\hline && (2^n)^2 - 2 \\pmod{17} \\quad & | \\quad n =", "value of $(x^2+y^2)^2$ $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=(6-2xy)^2+4x^2y^2=2(2xy-3)^2+18\\ge18$$ The equality occurs if $2xy=3$ You can rephrase $$y_{\\min} \\leq Ax \\leq y_{\\max},$$ as $$\\begin{pmatrix}A \\\\ -A\\end{pmatrix}x \\leq \\begin{pmatrix}y_{\\max} \\\\ -y_{\\min}\\end{pmatrix},$$ which is in the format of what quadprog accepts....", "B) \\times (\\mathbf A + \\mathbf B)$$ Now the cross product $\\times$ is linear, but you also get terms $\\mathbf A \\times \\mathbf B$ and $\\mathbf B \\times \\mathbf A$, which cancel by a property of the cross product. So in the end you get $$\\mathbf C \\times \\mathbf C = \\mathbf A \\times \\mathbf A + \\mathbf B \\times \\mathbf B$$ Which just says that $0=0+0$. Another way to see that your conclusion does not hold, is simply to see that $\\mathbf A \\times \\mathbf B$ is non-zero in general, so by taking the magnitude, we find that $|\\mathbf A \\circ \\mathbf B|$ is non-zero in general, and hence $\\mathbf A \\circ \\mathbf B$ is non-zero in general. 5. Aug 4, 2016 ### Devil Moo Now it is defined as the angle between $\\mathbf A$ and $\\mathbf B$ where $\\theta$ is smaller than or equal to $\\pi$ and it is commutative. How do you find out that it is not distributive? Um... How are these two examples related to the derivation? 6. Aug 4, 2016 ### Staff: Mentor By looking at an example. $$A = \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}, B = \\begin{pmatrix} -1 \\\\ 0 \\\\ 0 \\end{pmatrix}, C = \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}$$ $$(A+B) \\circ C = 0 \\neq A" ]

[ "given $\\mathbf{A}$ and $\\mathbf{b}$. If we can find $\\mathbf{A}^{-1}$ (the inverse of $\\mathbf{A}$), then our solution $\\mathbf{x}$ is given by $\\mathbf{A}^{-1}\\mathbf{b}$, since $$\\mathbf{A}^{-1}\\,\\mathbf{b} = \\mathbf{A}^{-1}(\\mathbf{Ax}) = (\\mathbf{A}^{-1}\\mathbf{A})\\mathbf{x} = \\mathbf{x}.$$ How can we compute $\\mathbf{A}^{-1}$? While there are formulas for inverting small matrices by hand, doing this is not practical for larger matrices. Instead, we will use the Python programming language as a calculator: In [7]: \"\"\" Python code \"\"\" import numpy as np A = np.array([[ 2, 4, 2], [-2, -3, 1], [ 2, 2, -3]]) A_inv = np.linalg.inv(A) print(A_inv) [[ 3.5 8. 5. ] [-2. -5. -3. ] [ 1. 2. 1. ]] Don't worry about the code for now. All you need to know is that the inverse of $\\mathbf{A}$ has been computed (A_inv in the code above), and is given by: $$\\mathbf{A}^{-1} = \\begin{pmatrix} 3.5 & 8 & 5 \\\\ -2 & -5 & -3 \\\\ 1 & 2 & 1 \\end{pmatrix}$$ Substituting this into our equation for $\\mathbf{x}$, we get $$\\mathbf{x} = \\mathbf{A}^{-1}\\, \\mathbf{b} = \\begin{pmatrix} 3.5 & 8 & 5 \\\\ -2 & -5 & -3 \\\\ 1 & 2 & 1 \\end{pmatrix} \\begin{pmatrix} 16 \\\\ -5 \\\\ -3 \\end{pmatrix} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}.$$ Thus, $x=1, y=2, z=3$ solves our system of equations. (Verify this for yourself!) ## Reviewing", "## Matrix inverses Show that $\\mathbf{R_1} = \\begin{pmatrix}-1 & 1 \\\\ 6& -4 \\\\ 4& -3\\end{pmatrix}$ and $\\mathbf{R_2} = \\begin{pmatrix}1 & -1 \\\\ -4& 6 \\\\ -4& 5\\end{pmatrix}$ are both right-inverses of the matrix $\\mathbf{A} = \\begin{pmatrix}1 &1 &-1 \\\\ 4&0 &1 \\end{pmatrix}$. Use the right-inverses $\\mathbf{R_1}$ and $\\mathbf{R_2}$ to find two solutions $\\mathbf{x_1}$ and $\\mathbf{x_2}$ of the equation $\\mathbf{Ax = b}$, where $\\mathbf{b} =\\begin{pmatrix}0\\\\ 8\\end{pmatrix}$. By what I understand, the only way to solve Ax = b is with an inverse: $\\mathbf{A^{-1}Ax = A^{-1}b}$ $\\mathbf{x = A^{-1}b}$ and matrix $\\mathbf{A}$ doesnt have an inverse but the question asks to use the right-inverse and this is what I dont understand", "## Inverse and a power Let $$\\mathbf B$$ be the inverse of $$\\mathbf A^2$$. Prove that $$\\mathbf A$$ is regular and determine its inverse. From $$\\mathbf A^2\\mathbf B = \\mathbf A\\mathbf A\\mathbf B = \\mathbf I_n$$ we get that $$\\mathbf A^{-1}=\\mathbf A\\mathbf B$$.", "$$\\mathbf{C}$$ is an inverse of $$\\mathbf{A}$$, then $$\\mathbf{A}$$ is an inverse of $$\\mathbf{C}$$. This is immediate from the definition. • If $$\\mathbf{C}$$ and $$\\mathbf{D}$$ are both inverses of $$\\mathbf{A}$$, then $$\\mathbf{C} = \\mathbf{D}$$. Proof: If $$\\mathbf{C}$$ and $$\\mathbf{D}$$ are inverses of $$\\mathbf{A}$$, then we have \\begin{aligned} \\mathbf{A} \\mathbf{C} = \\mathbf{I} &\\Leftrightarrow \\mathbf{D} (\\mathbf{A} \\mathbf{C}) = \\mathbf{D} \\mathbf{I} \\\\ &\\Leftrightarrow (\\mathbf{D} \\mathbf{A}) \\mathbf{C} = \\mathbf{D} \\\\ &\\Leftrightarrow \\mathbf{I} \\mathbf{C} = \\mathbf{D} \\\\ &\\Leftrightarrow \\mathbf{C} = \\mathbf{D}. \\end{aligned} As a consequence of this property, we write the inverse of $$\\mathbf{A}$$, when it exists, as $$\\mathbf{A}^{-1}$$. • The inverse of the inverse of $$\\mathbf{A}$$ is $$\\mathbf{A}$$: $$(\\mathbf{A}^{-1})^{-1} = \\mathbf{A}$$. • If the inverse of $$\\mathbf{A}$$ exists, then the inverse of its transpose is the transpose of the inverse: $$(\\mathbf{A}^\\top)^{-1} = (\\mathbf{A}^{-1})^\\top$$. • Matrix inversion inverts scalar multiplication: if $$c \\neq 0$$, then $$(c \\mathbf{A})^{-1} = (1/c) \\mathbf{A}^{-1}$$. • The identity matrix is its own inverse: $$\\mathbf{I}_n^{-1} = \\mathbf{I}_n$$. Some matrices are not invertible; i.e., their inverse does not exist. As a simple example, think of $\\mathbf{A} = \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix}.$ It’s easy to see that, for any $$2 \\times 2$$ matrix $$\\mathbf{B}$$, we have $\\mathbf{A} \\mathbf{B} = \\begin{bmatrix} 0 & 0 \\\\ 0 & 0 \\end{bmatrix} \\neq \\mathbf{I}_2.$ Therefore, $$\\mathbf{A}$$ does not have an inverse.", "# Decide a number $a, b, c$ (expressed by $\\lambda$) so that $a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$ See the following vectors in $$\\mathbb{R}^3$$: $$\\mathbf{u}_1= \\begin{pmatrix} 1 \\\\ -1\\\\ 1 \\\\ 1 \\end{pmatrix}$$, $$\\mathbf{u}_2 = \\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\\\ 2 \\end{pmatrix}$$, $$\\mathbf{u}_3 = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\end{pmatrix}$$ and $$\\mathbf{v}_1= \\begin{pmatrix} 5 \\\\ -3\\\\ -4 \\\\ -1 \\end{pmatrix}$$, $$\\mathbf{v}_2 = \\begin{pmatrix} 6 \\\\ 4 \\\\ 1 \\\\ 8 \\end{pmatrix}$$, $$\\mathbf{v}_3 = \\begin{pmatrix} 7 \\\\ 2 \\\\ -1 \\\\ 6 \\end{pmatrix}$$ We are told that $$\\mathcal{B} = (\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3)$$ and $$\\mathcal{C} = (\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3)$$ both are bases for the same subspace $$\\mathcal{U}$$. Let $$\\mathcal{\\lambda}$$ be an unknown number og see the vector $$\\mathbf{x} = \\mathbf{v}_1 + \\lambda\\mathbf{v}_2 - \\mathbf{v}_3$$. Decide a number $$a, b, c$$ (expressed by $$\\lambda$$) so that $$a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$$ 1. Find some $$\\alpha_i$$ such that $$v_1=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3$$ 2. Find some $$\\beta_i$$ such that $$v_2=\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3$$ 3. Find some $$\\gamma_i$$ such that $$v_3=\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3$$ 4. Simplify the expression $$x=v_1+\\lambda v_2-v_3=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3+\\lambda(\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3)-(\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3)$$ • Thanks 5xum! When I reduce it, I get $\\begin{pmatrix} 8 \\\\ 3\\\\ -9 \\\\ 1 \\end{pmatrix} + a\\begin{pmatrix} 2 \\\\ -2\\\\ 1 \\\\ 1 \\end{pmatrix} + b\\begin{pmatrix} 3 \\\\ 6\\\\ 0 \\\\ 6 \\end{pmatrix} + c\\begin{pmatrix} 1 \\\\", "right inverse: $\\mathbf{E^{-1}} \\mathbf{E} = \\mathbf{E} \\mathbf{E^{-1}} = \\mathbf{I}$ It is a bit out of our way to prove that a matrix with a left inverse must also have a right inverse. If you accept that on faith for now, it is easy to prove that, if there is a right inverse, it must be the same as the left inverse. Call the left inverse $$\\mathbf{L}$$ and the right inverse $$\\mathbf{R}$$: $\\begin{split} \\mathbf{LA} = \\mathbf{I}\\\\ \\mathbf{AR} = \\mathbf{I}\\\\ \\end{split}$ then: $\\begin{split} \\mathbf{LAR} = \\mathbf{LAR}\\\\ \\mathbf{L(AR)} = \\mathbf{(LA)R}\\\\ \\mathbf{L} = \\mathbf{R} \\end{split}$ So, in our case, where we want to find the transformations following the first affine, we can do this: $\\begin{split} \\mathbf{F} \\triangleq \\mathbf{C} \\mathbf{B} \\\\ \\mathbf{D} = \\mathbf{F} \\mathbf{A} \\\\ \\mathbf{D} \\mathbf{A^{-1}} = \\mathbf{F} \\mathbf{A} \\mathbf{A^{-1}} \\\\ \\mathbf{D} \\mathbf{A^{-1}} = \\mathbf{F} \\end{split}$ For our actual affines: third_with_second = combined @ npl.inv(first_affine) third_with_second array([[ 0.9211, -0. , 0.3894, 10. ], [ 0. , 1. , 0. , 20. ], [-0.3894, -0. , 0.9211, 30. ], [ 0. , 0. , 0. , 1. ]]) # This is the same as F = third_affine @ second_affine F array([[ 0.9211, 0. , 0.3894, 10. ], [ 0. , 1. , 0. , 20. ], [-0.3894, 0. , 0.9211, 30. ], [ 0. , 0. , 0. , 1. ]]) assert", "# Solving modular equations of two variables $$7x + 9y \\equiv 0 \\pmod {31}$$ $$2x - 5y \\equiv 2 \\pmod {31}$$ I'm supposed to find the x, but I'm stuck in the middle. First I tried to get rid of y. $$53x \\equiv 18 \\pmod {31}$$ Here now I don't know what to do with this. Can anyone help me out?Thanks. You need to find the inverse of $53\\equiv 22$ modulo $31$. To do it, you may use the Euclidean algorithm. ote you can also look at this as a $$\\begin{pmatrix}7&9\\\\2&-5\\end{pmatrix}\\begin{pmatrix}x\\\\y\\end{pmatrix}=\\begin{pmatrix}0\\\\2\\end{pmatrix}$$ Note that $$\\det\\begin{pmatrix}7&9\\\\2&-5\\end{pmatrix}=-53\\equiv 9$$ But $9\\times 7=63\\equiv 1$ so $$\\begin{pmatrix}x\\\\y\\end{pmatrix}=7\\begin{pmatrix}-5&-9\\\\-2&7\\end{pmatrix}\\begin{pmatrix}0\\\\2\\end{pmatrix}=\\begin{pmatrix} 29\\\\23\\end{pmatrix}$$ That is, one can work as usual using Cramer's rule as long as $\\det A\\not \\equiv 0$ modulo what you're working in. • How so? Why am I finding the inverse of 53 = 22 modulo 31? Oct 2 '13 at 22:50 • When you have an equation of the form $ax=b$; how do you \"find\" $x$? You divide by $a$, right? Here, we can also \"divide\" since $31$ is prime, meaning every nonzero number has an inverse, that is, if $31\\not\\mid x$ then we can find $y$ such that $xy\\equiv 1\\mod 31$. If you find this number, you can find what $x$ is. Oct 2 '13 at 22:53 • In this case, and almost always", "Free Version Easy # Find 3x3 Inverse - 3 LINALG-TGB9X1 Find the inverse of the matrix: $$\\begin{pmatrix}1&1&3\\\\\\ 4&1&6\\\\\\ 7&1&9\\end{pmatrix}$$ ...if it exists. A $\\begin{pmatrix}1&2&1\\\\\\ 26&4&-6\\\\\\ 4&2&-2\\end{pmatrix}$ B $\\begin{pmatrix}1&2&1\\\\\\ 24&4&-6\\\\\\ 4&2&-2\\end{pmatrix}$ C $\\begin{pmatrix}5&-2&3\\\\\\ -2&-4&2\\\\\\ -4&2&-\\cfrac{5}{2}\\end{pmatrix}$ D $\\begin{pmatrix}5&-2&-3\\\\\\ -2&-4&-2\\\\\\ -4&2&\\cfrac{5}{2}\\end{pmatrix}$ E This matrix is not invertible", "\\end{pmatrix}$ Using the definition of the inner product $(\\mathbf u, \\mathbf v) = \\mathbf v^* \\mathbf u$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (5-i)(1+3i) + (6-4i)(2+4i) = 36 + 30i$ , but using $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\begin{pmatrix} 1 \\\\ 2 \\\\ \\end{pmatrix} , \\begin{pmatrix} 5 \\\\ 6 \\\\ \\end{pmatrix}) + i(\\begin{pmatrix} 3 \\\\ 4 \\\\ \\end{pmatrix} , \\begin{pmatrix} 1 \\\\ 4 \\\\ \\end{pmatrix}) = 5 + 12 + i(3 + 16) = 17 + 19i$. What went wrong with my interpretation of the formula , $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$ ? Shouldn't both computations be mathematically sound and give the same result? Thanks in advance for any help!", "is just $a$ and $a$ has an inverse in $\\mathbb Z[i]$ iff $a=\\pm 1$. This is because, the multiplicative inverse is $\\frac{1}{a}$, which will be in $\\mathbb Z$ iff $a=\\pm 1$. - If $z,w\\in\\mathbb{Z}[i]$ are such that $zw=1$ (i.e. $z$ is a unit and $w$ its inverse), then $|z|^2|w|^2=|zw|^2=1$, or $$(a^2+b^2)(c^2+d^2)=1, \\quad z=a+bi,\\; w=c+di.$$ Now $a,b,c,d$ are all integers, so $a^2+b^2$ and $c^2+d^2$ must both be nonnegative integers, which must both equal exactly $1$ and no greater in order to multiply to $1$ in the integers. And if $a^2+b^2=1$, we have $a^2$ and $b^2\\le1$. Check by hand the only solutions here correspond to $(a,b)=(\\pm1,0)$ or $(0,\\pm1)$. - Equivalently, by taking conjugates we conclude that $\\bar{z}\\bar{w}=1$ and therefore $(z\\bar z)(w\\bar{w})=1$ and therefore $z\\bar{z}=1$. – André Nicolas Feb 11 '12 at 8:14", "Find $A^2=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}\\times \\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}$ So, $B=\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}^2-3\\begin{pmatrix}2 & -5 \\\\3 & 1\\end{pmatrix}+17\\begin{pmatrix}1 & 0 \\\\0 & 1\\end{pmatrix}$", "the additive inverse $-v$ is unique. (e) $0v=\\mathbf{0}$ for every $v\\in V$, where $0\\in\\R$ is the zero scalar. (f) $a\\mathbf{0}=\\mathbf{0}$ for every scalar $a$. (g) If $av=\\mathbf{0}$, then $a=0$ or $v=\\mathbf{0}$. (h) $(-1)v=-v$. The first two properties are called the cancellation law. Problem 710 Find a basis for $\\Span(S)$ where $S= \\left\\{ \\begin{bmatrix} 1 \\\\ 2 \\\\ 1 \\end{bmatrix} , \\begin{bmatrix} -1 \\\\ -2 \\\\ -1 \\end{bmatrix} , \\begin{bmatrix} 2 \\\\ 6 \\\\ -2 \\end{bmatrix} , \\begin{bmatrix} 1 \\\\ 1 \\\\ 3 \\end{bmatrix} \\right\\}$. Problem 709 Let $S=\\{\\mathbf{v}_{1},\\mathbf{v}_{2},\\mathbf{v}_{3},\\mathbf{v}_{4},\\mathbf{v}_{5}\\}$ where $\\mathbf{v}_{1}= \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\\\ -1 \\end{bmatrix} ,\\;\\mathbf{v}_{2}= \\begin{bmatrix} 1 \\\\ 3 \\\\ 1 \\\\ 1 \\end{bmatrix} ,\\;\\mathbf{v}_{3}= \\begin{bmatrix} 1 \\\\ 5 \\\\ -1 \\\\ 5 \\end{bmatrix} ,\\;\\mathbf{v}_{4}= \\begin{bmatrix} 1 \\\\ 1 \\\\ 4 \\\\ -1 \\end{bmatrix} ,\\;\\mathbf{v}_{5}= \\begin{bmatrix} 2 \\\\ 7 \\\\ 0 \\\\ 2 \\end{bmatrix} .$ Find a basis for the span $\\Span(S)$. Problem 708 Let $A=\\begin{bmatrix} 2 & 4 & 6 & 8 \\\\ 1 &3 & 0 & 5 \\\\ 1 & 1 & 6 & 3 \\end{bmatrix}$. (a) Find a basis for the nullspace of $A$. (b) Find a basis for the row space of $A$. (c) Find a basis for the range of $A$ that consists of column vectors of $A$. (d) For each column vector which is not a", "+0 # Let $n$ be the inverse of $2\\pmod{17}$. 0 24 1 +122 Let $n$ be the inverse of $2\\pmod{17}$. That is, let $n$ be the integer $0\\leq n < 17$ for which $2n \\equiv 1 \\pmod{17}$. What is $\\left(2^n\\right)^2 - 2 \\pmod{17}$? Express your answer as an integer from $0$ to $16$, inclusive. Sort: #1 +19376 +1 Let $n$ be the inverse of $2\\pmod{17}$. That is, let $n$ be the integer $0\\leq n < 17$ for which $2n \\equiv 1 \\pmod{17}$. What is $\\left(2^n\\right)^2 - 2 \\pmod{17}$? Express your answer as an integer from $0$ to $16$, inclusive. $$\\begin{array}{|rcll|} \\hline 2n &\\equiv& 1 \\pmod {17} \\quad & | \\quad : 2 \\\\\\\\ n &\\equiv& \\frac12 \\pmod {17} \\\\ \\mathbf{n} &\\mathbf{\\equiv}& \\mathbf{ 2^{-1} \\pmod {17}} \\\\ \\hline \\end{array}$$ $$\\text{ Solve 2^{-1} \\pmod {17} }$$ $$\\text{Because gcd(2,17)=1 we can continue.}$$ $$\\begin{array}{|rcll|} \\hline 2^{-1} \\pmod {17} &\\equiv& 2^{\\varphi(17)-1} \\pmod {17} \\quad & | \\quad \\varphi(p) = p - 1 \\quad \\text{p is a prime number} \\\\\\\\ && \\varphi(17) = 16 \\\\\\\\ &\\equiv& 2^{16-1} \\pmod {17} \\\\ &\\equiv& 2^{15} \\pmod {17} \\\\ &\\equiv& 32768 \\pmod {17} \\\\ 2^{-1} \\pmod {17}&\\equiv& 9 \\pmod {17} \\\\ \\mathbf{n} &\\mathbf{=}& \\mathbf{9} \\\\ \\hline \\end{array}$$ $$Proof:\\\\ 2\\cdot 9 \\equiv 1 \\pmod{17} \\ \\checkmark$$ $$\\begin{array}{|rcll|} \\hline && (2^n)^2 - 2 \\pmod{17} \\quad & | \\quad n =", "\\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 2. $\\begin{pmatrix} -3 \\\\ 3 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 3. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ 4. $\\begin{pmatrix} -3 \\\\ 3 \\\\ 4 \\end{pmatrix}$, $\\{\\begin{pmatrix} 1 \\\\ -1 \\\\ 2 \\end{pmatrix}k+\\begin{pmatrix} 0 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 5. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 14 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ 6. $\\begin{pmatrix} 1 \\\\ 4 \\\\ 6 \\end{pmatrix}$, $\\{\\begin{pmatrix} 2 \\\\ 2 \\\\ 5 \\end{pmatrix}k+\\begin{pmatrix} -1 \\\\ 0 \\\\ 2 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ Problem 2 Produce two descriptions of this set that are different than this one. $\\{\\begin{pmatrix} 2 \\\\ -5 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 3 Show that the three descriptions given at the start of this subsection all describe the same set. This exercise is recommended for all readers. Problem 4 Show that these sets are equal $\\{\\begin{pmatrix} 1 \\\\ 4 \\\\ 1 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}z\\,\\big|\\, z\\in\\mathbb{R} \\} \\quad\\text{and}\\quad \\{\\begin{pmatrix} 0 \\\\ 4 \\\\ 2 \\\\ 1 \\end{pmatrix} +\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix}k\\,\\big|\\, k\\in\\mathbb{R} \\},$ and that both describe the solution set of this system. $\\begin{array}{*{4}{rc}r} x &- &y &+ &z &+", "of $(\\mathbb{Q}_p-\\{0\\},Mul)$ inherits from that of $(\\mathbb{Z}[\\frac{1}{p}]_{\\geq0},Mul)$. And we want to find an inverse for each element in it. For example, an non-zero element writes $a=p^ux$ where $x_0\\in F_p-\\{0\\}$. So if we can find an inverse for $x$, say $y$ such that $xy=1$, then it is wasy to see that $p^{-u}y$ is the inverse of $a=p^ux$. So we can just suppose that $a\\in\\mathbb{Z}_p(a_0\\neq0)$.The method is to go back to the $\\mathbb{Z}_{\\geq0}$ world, and then take limits. Specifically, suppose that $a\\in\\mathbb{Z}_p$ with $s(a,0)=0$, then using again the approximation sequence $A_M=\\sum_{0\\leq n\\leq M}a_np^n$. If we can find the inverse $B_M$of $A_M$ for each $M$, then since $A_M$ is a Cauchy sequence, so is $B_M$(because $d(A_M,0)d(B_M,0)=d(A_MB_M,0)=d(1,0)=1$, and $d(A_M,0)$ converge to a non-zero number). So things are reduced to find these $B_M$, or more generally, find the inverse of any positive integer $A$ prime to $p$. Suppose the inverse of $A$ is $b=\\sum_{n\\geq0}b_np^n$(if it exists). Then consider the approximating sequence $B_N$ to $b$. In order that $d(AB_N,0)\\rightarrow d(Ab,0)=1$.For $N=0$, since $gcd(A,p)=1$, there is an integer $b_0\\in F_p$ such that $Ab_0=1(\\text{mod}p)$. So, we have that $d(Ab_0-1,0)\\leq e^{-1}$. For $N=2$, since again $gcd(A,p^2)=1$, there is $b_1\\in F_p$ such that $A(b_0+b_1p)=1(\\text{mod}p^2)$. Similarly, we have that $d(A(b_0+b_1p)-1,0)\\leq e^{-2}$. In fact, $(b_0,b_1)$ is the only element in $F_p^2$ such that $Ab_0=1(\\text{mod}p),A(b_0+b_1p)=1(\\text{mod}p^2)$. We can continue this process to", "B is the target, then $|f(A)| = |B|$, where $f(A)$ is the image of A under f. True False Does not make sense I don't know yet 6. Is the following statement true, false, or neither? Given a bijective function $f:A \\rightarrow B$, then $|A| = |B|$. True False Does not make sense I don't know yet 7. Find the inverse of the following function, if it exists. $$\\begin{pmatrix} -3&-2&-1&0&1&2&3 \\\\ -1&3&1&-2&0&2&-3 \\end{pmatrix}$$ 8. Find the inverse of the following function, if it exists. $$\\begin{pmatrix} 0&1 \\\\ 1&0 \\end{pmatrix}$$ 9. Find the inverse of the following function, if it exists. $$\\begin{pmatrix} 0&1&2&3&4 \\\\ 4&3&1&0&1 \\end{pmatrix}$$ 10. Find the inverse of the following function, if it exists. $$f : \\{ x : x^2 < 3 \\} \\rightarrow \\{\\alpha, \\beta, \\gamma, \\delta\\}$$ 11. Given $$f := \\begin{pmatrix} 0&1&2&3&4 \\\\ 0&4&1&2&3 \\end{pmatrix},$$ $$g := \\begin{pmatrix} 0&1&2&3&4 \\\\ 4&1&3&0&2 \\end{pmatrix},$$ find the composition f@g. 12. Given $$f := \\begin{pmatrix} 0&1&2&3&4 \\\\ 0&4&1&2&3 \\end{pmatrix},$$ $$g := \\begin{pmatrix} 0&1&2&3&4 \\\\ 4&1&3&0&2 \\end{pmatrix},$$ find the composition g@f. 13. Given $$f := \\begin{pmatrix} 0&1&2&3&4 \\\\ 1&0&3&2&4 \\end{pmatrix},$$ $$g := \\begin{pmatrix} 0&1&2&3&4 \\\\ 3&2&1&4&0 \\end{pmatrix},$$ find the composition f@g. 14. Given $$f := \\begin{pmatrix} -2&-1&0&1&2 \\\\ 0&-2&1&2&-1 \\end{pmatrix},$$ $$g := \\begin{pmatrix} 0&1&2&3&4 \\\\ 0&2&3&1&4 \\end{pmatrix},$$ find the composition f@g. 15. Given $$f := \\begin{pmatrix} -2&-1&0&1&2 \\\\ 2&0&1&4&3", "+0 # Matrices 0 69 1 Given $$\\mathbf{A}^{-1} = \\begin{pmatrix} 2 & 3 \\\\ -1 & 0 \\end{pmatrix} \\quad \\text{and} \\quad \\mathbf{B}^{-1} = \\begin{pmatrix} 0 & 4 \\\\ 1 & 5 \\end{pmatrix} ,$$ find the inverse of AB. Thanks! Apr 2, 2021 #1 +22020 +2 1) Find the inverse of A-1; this will be A. 2) Find the inverse of B-1; this will be B. 3) Find the product of A and B; this will be AB. 4) Lastly, find the inverse of AB (your answer to step 3). Apr 2, 2021", "+ 3 x_{22}\\\\ \\end{pmatrix} \\end{align*} So our equation is $$\\begin{pmatrix} 9 & -3 \\\\ 5 & -5 \\end{pmatrix} - \\begin{pmatrix} -9 x_{11} +8 x_{12} & -2x_{11}+ 3 x_{12}\\\\ -9 x_{21} + 8 x_{22} & -2 x_{21} + 3 x_{22}\\\\ \\end{pmatrix}= \\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\\\ \\end{pmatrix}$$ Now we write it as a system of equations: \\begin{align*} 1&=9- (-9 x_{11} +8x_{12})\\\\ 0&= -3-(-2x_{11} +3x_{12}) \\\\ 0&=5-(-9x_{21} +8 x_{21}) \\\\ 1&=-5 - (-2x_{21}+3x_{22}) \\\\ \\end{align*} If you need help solving this system tell me. - Thank you very much for your excellent answer! – Lukas Arvidsson Mar 15 at 9:27 You are welcome :) – Dominic Michaelis Mar 15 at 9:28 First of all rearrange to get X[-9 -2, 8 5] = [8 -3, 5 -6]. Then find the inverse of [-9 -2, 8 5], and multiply both sides of the equation by this inverse on the right, which will leave X = [8 -3, 5 -6][-9 -2, 8 5]^-1 as your solution. - Denote $B=\\begin{pmatrix}-9&-2\\\\8&5\\end{pmatrix}$. Then we have: $$\\begin{pmatrix}9&-3\\\\5&-5\\end{pmatrix}-E=XB$$ Observe that $\\det B=-45+16\\neq0$ and hence $B$ is invertible. Multiplying by $B^{-1}$ we have: $$X=\\left(\\begin{pmatrix}9&-3\\\\5&-5\\end{pmatrix}-E\\right)B^{-1}$$ Do you know how to find $B^{-1}$? - Thanks for your answer! I get the final result to be: $\\left[ \\begin{array}{cc} \\frac{-64}{29} & \\frac{-43}{29}\\\\ \\frac{-48}{29} & \\frac{-54}{29}\\end{array} \\right]$ Is that correct?", "\\vec{w} \\\\\\\\ & &=& \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} \\times \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} \\\\\\\\ & &=& \\begin{vmatrix}1&1&1 \\\\ -4&0&2 \\\\ 0&-3&-1 \\end{vmatrix} \\\\\\\\ & &=& \\begin{pmatrix} 6 \\\\ -4 \\\\ 12 \\end{pmatrix} \\\\ \\hline n_1+n_2+n_3: & 6-4+12 &=& 14 \\quad | \\quad \\cdot \\dfrac{7}{14} \\\\\\\\ & 6\\cdot\\dfrac{7}{14}-4\\cdot\\dfrac{7}{14}+12\\cdot\\dfrac{7}{14} &=& 14\\cdot\\dfrac{7}{14} \\\\\\\\ &\\mathbf{ 3-2+6 }&=& \\mathbf{7} \\\\\\\\ & \\vec{n} &=& \\begin{pmatrix}\\mathbf{3} \\\\ \\mathbf{-2} \\\\ \\mathbf{6} \\end{pmatrix} \\\\ \\hline \\end{array} }$$ 2. Vector dot product $$\\begin{array}{|lrcll|} \\hline & \\vec{n}\\cdot \\vec{v} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} &=& 0 \\\\ (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ \\hline & \\vec{n}\\cdot \\vec{w} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} &=& 0 \\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ \\hline (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ \\hline \\hline \\end{array}$$ $$\\begin{array}{|lrcll|} \\hline (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ &2n_3&=&4n_1 \\quad | \\quad :2 \\\\ & \\mathbf{n_3} &=& \\mathbf{2n_1} \\\\\\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ & \\mathbf{n_3} &=& \\mathbf{-3n_2} \\\\\\\\ & n_3= 2n_1 &=& -3n_2 \\\\ & 2n_1 &=& -3n_2 \\quad | \\quad :2 \\\\ & \\mathbf{n_1} &=& \\mathbf{-\\dfrac{3}{2}n_2} \\\\\\\\ (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ & -\\dfrac{3}{2}n_2 + n_2 + -3n_2 &=& 7 \\\\ & -\\dfrac{3}{2}n_2 -2n_2 &=& 7 \\quad | \\quad", "• You're supposed to multiply the expression for the partitioned inverse with $\\begin{pmatrix}\\bf \\tt{wa}\\\\\\mathbf 0\\end{pmatrix}$, thus resulting in expressions for $\\begin{pmatrix}\\mathbf v\\\\\\bf \\tt{bb}\\end{pmatrix}$. – J. M.'s ennui Dec 1 '15 at 2:24 • @J.M. Thank you. I think I got it now. – Jack LaVigne Dec 6 '15 at 15:32 • Looks right to me. :) – J. M.'s ennui Dec 9 '15 at 2:49" ]

[ "## Q). In the figure chord AB and CD when produced meet at point P. If , then prove that $\\angle APC=\\frac{\\left(\\theta -\\alpha \\right)}{2}$. Subject: Math, asked 2 days, 18 hours ago", "respectively. Then since the areas of triangles $ABP$, $APC$ are proportional to bases $BP, PC$ [Euclid VI,1], and are also proportional to heights $AE,CD$, since they share base $AP$, we have $$\\frac{BP}{PC}=\\frac{AE}{CD}$$And since $\\angle ABE = \\angle BAP = \\angle CAD$, therefore $\\triangle ABE$ and $\\triangle CAD$ are similar, making $$\\frac{AE}{CD}=\\frac{AB}{AC}$$Therefore,$$\\frac{BP}{PC}=\\frac{AB}{AC}$$", "when Two Lines Intersect each Other? Answer: Let AB and CD be two lines intersecting at a single point, say, P. APC and BPD, and APD and BOC form two pairs of vertically opposite angles. Additionally, ray AP is perpendicular on line CD. It implies the sum of APC and APD is equal to 180o. Now, the sum of angles APD and BPD is equal to 180o. By using these two equations, it’s clear that the sum of angles APC and APD is equal to the sum of APD and BPD angles. It implies angle APC is equal to BPD. Similarly, it can be shown that angle APD is equal to angle BPC.", "The coordinates of the points $A$, $B$, $C$ are $(-2,1)$, $(2,7)$, $(5,5)$ respectively. Prove that these points form three corners of a rectangle and that $AB=2BC$. If $D$ is the fourth corner of the rectangle, calculate the distance of $C$ from the diagonal $BD$.", "Find $S_{PCD}$ given $S_{ABCD},S_{ABP},S_{BPC}$ In the following figure,$ABCD$ is a paralleogram in which point $P$ lies in triangle $ABD$ and the areas of $ABCD,ABP,BPC$ are known to be $s_1,s_2,s_3$ respectively. Find $S_{PCD}$ I think the number of unknowns in this problem is more than the number of equations that can be formed!! $S_{ABP}+S_{PCD}=\\frac{1}{2}PE\\cdot AB+\\frac{1}{2}PF\\cdot CD=\\frac{1}{2}(PE+PF)\\cdot AB=\\frac{1}{2}EF\\cdot AB=\\frac{1}{2}S_{ABCD}$ • So what is the usage of $S_{BPC}$? – Hamid Reza Ebrahimi Aug 18 '17 at 5:22 • @HamidRezaEbrahimi $S_{BPC}$ is not used or needed, unless you want to determine $S_{APD}$ too. – dxiv Aug 18 '17 at 5:24", "+0 0 44 1 A segment with endpoints at $A(2, -2)$ and $B(14, 4)$ is extended through $B$ to point $C$. If $BC = \\frac{1}{3} \\cdot AB$, what are the coordinates for point $C$? Express your answer as an ordered pair. Mar 14, 2021", "is located at $(24, 24). \\, \\blacksquare$ The distance $DE$ is equal to the distance between $(12, 12)$ and $(24, 24)$, which is $\\boxed{\\textbf{(D)} ~12\\sqrt{2}}$ 2、Using Stewart's Theorem we find $BP = 8\\sqrt{2}$. From the similar triangles $BPA\\sim DPC$ and $BPC\\sim EPA$ we have$$DP = BP\\cdot\\frac{PC}{PA} = 2BP$$$$EP = BP\\cdot\\frac{PA}{PC} = \\frac12 BP$$So$$DE = \\frac{3}{2}BP = \\boxed{\\textbf{(D) }12\\sqrt2}$$ 3、Let $x$ be the length $PE$. From the similar triangles $BPA\\sim DPC$ and $BPC\\sim EPA$ we have$$BP = \\frac{PA}{PC}x = \\frac12 x$$$$PD = \\frac{PA}{PC}BP = \\frac14 x$$Therefore $BD = DE = \\frac{3}{4}x$. Now extend line $CD$ to the point $Z$ on $AE$, forming parallelogram $ZABC$. As $BD = DE$ we also have $EZ = ZC = 13$ so $EC = 26$. We now use the Law of Cosines to find $x$ (the length of $PE$):$$x^2 = EC^2 + PC^2 - 2(EC)(PC)\\cos{(PCE)} = 26^2 + 10^2 - 2\\cdot 26\\cdot 10\\cos(\\angle PCE)$$As $\\angle PCE = \\angle BAC$, we have (by Law of Cosines on triangle $BAC$)$$\\cos(\\angle PCE) = \\frac{13^2 + 15^2 - 14^2}{2\\cdot 13\\cdot 15}.$$Therefore\\begin{align*} x^2 &= 26^2 + 10^2 - 2\\cdot 26\\cdot 10\\cdot\\frac{198}{2\\cdot 13\\cdot 15}\\\\ &= 776 - 264\\\\ &= 512 \\end{align*}And $x = 16\\sqrt2$. The answer is then $\\frac34x = \\boxed{\\textbf{(D) }12\\sqrt2}$ 4、Let the brackets denote areas. By Heron's Formula, we have\\begin{align*} [ABC]&=\\sqrt{\\frac{13+14+15}{2}\\left(\\frac{13+14+15}{2}-13\\right)\\left(\\frac{13+14+15}{2}-14\\right)\\left(\\frac{13+14+15}{2}-15\\right)} \\\\ &=\\sqrt{21\\left(21-13\\right)\\left(21-14\\right)\\left(21-15\\right)} \\\\ &=\\sqrt{21\\left(8\\right)\\left(7\\right)\\left(6\\right)} \\\\ &=\\sqrt{\\left(3\\cdot7\\right)\\left(2^3\\right)\\left(7\\right)\\left(2\\cdot3\\right)} \\\\", "# 5. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD. Direction ratios of AB are $(4-1),(5-2),(7-3) = 3,3,4$ and Direction ratios of CD are $(2-(-4)), (9-3), (2-(-6)) = 6,6,8$ So, it can be noticed that, $\\frac{a_{1}}{a_{2}} = \\frac{b_{1}}{b_{2}} =\\frac{c_{1}}{c_{2}} = \\frac{1}{2}$ Therefore, AB is parallel to CD. Thus, we can easily say the angle between AB and CD which is either $0^{\\circ}\\ or\\ 180^{\\circ}$. Exams Articles Questions", "C'A &= AP \\cdot \\frac{\\sin \\angle APC'}{\\sin \\angle AC'P}. \\end{align*} Hence, \\begin{align*} &\\frac{AB'}{B'C} \\cdot \\frac{CA'}{A'B} \\cdot \\frac{BC'}{C'A} \\\\ &= \\frac{\\sin \\angle APB'}{\\sin \\angle AB'P} \\cdot \\frac{\\sin \\angle CB'P}{\\sin \\angle CPB'} \\cdot \\frac{\\sin \\angle CPA'}{\\sin \\angle CA'P} \\cdot \\frac{\\sin \\angle BA'P}{\\sin \\angle BPA'} \\cdot \\frac{\\sin \\angle BPC'}{\\sin \\angle BC'P} \\cdot \\frac{\\sin \\angle AC'P}{\\sin \\angle APC'}. \\end{align*} Since angles $\\angle AB'P$ and $\\angle CB'P$ are supplementary or equal, depending on the position of $B'$ on $AC$, $$\\sin \\angle AB'P = \\sin \\angle CB'P.$$ Similarly, \\begin{align*} \\sin \\angle CA'P &= \\sin \\angle BA'P, \\\\ \\sin \\angle BC'P &= \\sin \\angle AC'P. \\end{align*} By the reflective property, $\\angle APB'$ and $\\angle BPA'$ are supplementary or equal, so $$\\sin \\angle APB' = \\sin \\angle BPA'.$$ Similarly, \\begin{align*} \\sin \\angle CPA' &= \\sin \\angle APC', \\\\ \\sin \\angle BPC' &= \\sin \\angle CPB'. \\end{align*} Therefore, $$\\frac{AB'}{B'C} \\cdot \\frac{CA'}{A'B} \\cdot \\frac{BC'}{C'A} = 1,$$ so by Menelaus's theorem, $A'$, $B'$, and $C'$ are collinear.", "= 2r_2 \\cdot \\sin(\\angle BAP)$ Conclusion, since $r_1 = r_2$, $BP$ has to be the same value as $BQ$ and $\\Delta BPQ$ has to be an isosceles triangle.", "as desired. Equality occurs when $x=y$ which is when P is the midpoint of BC. In this case, it is easy to show that triangle ABC is right angled at A. Derivation of condition for existence of at least one such P Let N be the midpoint of AC. Then triangle ABC is an enlargement of triangle AMN, as AB = 2AM and AC = 2AN, so ∠MNA = ∠BCA. The set of points P such that ∠MPA = ∠BCA = ∠MNA is the set of points located on the circumcircle of triangle AMN, on the same side of AM as N; this is due to the same segment theorem. Hence, for there to be such a P located on BC, the circumcircle of AMN must touch BC at least once. Clearly this does not always happen: Sometimes there are two such P: And sometimes, in the case we will now examine, there is exactly one such P. In this case we can apply secant-tangent theorem: $BP^{2}=BM\\cdot BA=\\frac{c^{2}}{2}$ $CP^{2}=CN\\cdot CA=\\frac{b^{2}}{2}$ $BP^{2}\\cdot CP^{2}=\\frac{b^{2}\\cdot c^{2}}{4}$ $BP\\cdot CP=\\frac{bc}{2}$ Also $BP+PC=a$ $BP^{2}+PC^{2}+2BP\\cdot PC=a^{2}$ $\\frac{b^{2}+c^{2}+2bc}{2}=a^{2}$ $(b+c)^{2}=2a^{2}$ $b+c=a\\sqrt{2}$ Fix the side BC and consider for which points A there exists exactly one P. a√2 is constant so the locus of A is an ellipse, since the sum of the distances to each of two", "to $\\overline{AB}$ passing through $C$. The slope of this line is the same as the slope of $\\overline{AB}$, or $\\tfrac{12}{5}$, and has equation $y=\\tfrac{12}{5}x-\\tfrac{168}{5}$. The intersection of this line with $y=x$ is $(24, 24)$. Therefore point $E$ is located at $(24, 24). \\, \\blacksquare$ The distance $DE$ is equal to the distance between $(12, 12)$ and $(24, 24)$, which is $\\boxed{\\textbf{(D)} ~12\\sqrt{2}}$ ## Solution 2 Using Stewart's Theorem we find $BP = 8\\sqrt{2}$. From the similar triangles $BPA\\sim DPC$ and $BPC\\sim EPA$ we have $$DP = BP\\cdot\\frac{PC}{PA} = 2BP$$ $$EP = BP\\cdot\\frac{PA}{PC} = \\frac12 BP$$ So $$DE = \\frac{3}{2}BP = \\boxed{\\textbf{(D) }12\\sqrt2}$$ ## Solution 3 Let $x$ be the length $PE$. From the similar triangles $BPA\\sim DPC$ and $BPC\\sim EPA$ we have $$BP = \\frac{PA}{PC}x = \\frac12 x$$ $$PD = \\frac{PA}{PC}BP = \\frac14 x$$ Therefore $BD = DE = \\frac{3}{4}x$. Now extend line $CD$ to the point $Z$ on $AE$, forming parallelogram $ZABC$. As $BD = DE$ we also have $EZ = ZC = 13$ so $EC = 26$. We now use the Law of Cosines to find $x$ (the length of $PE$): $$x^2 = EC^2 + PC^2 - 2(EC)(PC)\\cos{(PCE)} = 26^2 + 10^2 - 2\\cdot 26\\cdot 10\\cos(\\angle PCE)$$ As $\\angle PCE = \\angle BAC$, we have (by Law of Cosines on triangle $BAC$) $$\\cos(\\angle PCE) = \\frac{13^2", "\\parallel CD$. ## Solution 2 First note that $\\overline{Q_1 Q_2} \\parallel \\overline{AB}$ if and only if the altitudes from $Q_1$ and $Q_2$ to $\\overline{AB}$ are the same, or $|Q_1B|\\sin \\angle ABQ_1 =|Q_2A|\\sin \\angle BAQ_2$. Similarly $\\overline{Q_1 Q_2} \\parallel \\overline{CD}$ iff $|Q_1C|\\sin \\angle DCQ_1 =|Q_2D|\\sin \\angle CDQ_2$. If we define $S =\\frac{|Q_1B|\\sin \\angle ABQ_1}{|Q_2A|\\sin \\angle BAQ_2} \\times \\frac{|Q_2D|\\sin \\angle CDQ_2}{|Q_1C|\\sin \\angle DCQ_1}$, then we are done if we can show that S=1. By the law of sines, $\\frac{|Q_1B|}{|Q_1C|}=\\frac{\\sin\\angle Q_1CB}{\\sin\\angle Q_1BC}$ and $\\frac{|Q_2D|}{|Q_2A|}=\\frac{\\sin\\angle Q_2AD}{\\sin\\angle Q_2DA}$. So, $S=\\frac{\\sin \\angle ABQ_1}{\\sin \\angle BAQ_2}\\cdot\\frac{\\sin \\angle CDQ_2}{\\sin \\angle DCQ_1}\\cdot\\frac{\\sin \\angle BCQ_1}{\\sin \\angle CBQ_1}\\cdot\\frac{\\sin \\angle DAQ_2}{\\sin \\angle ADQ_2}$ By the terms of the problem, $S=\\frac{\\sin \\angle PBC}{\\sin \\angle PAD}\\cdot\\frac{\\sin \\angle PDA}{\\sin \\angle PCB}\\cdot\\frac{\\sin \\angle PCD}{\\sin \\angle PBA}\\cdot\\frac{\\sin \\angle PAB}{\\sin \\angle PDC}$. (If two subangles of an angle of the quadrilateral are equal, then their complements at that quadrilateral angle are equal as well.) Rearranging yields $S= \\frac{\\sin \\angle PBC}{\\sin \\angle PCB}\\cdot\\frac{\\sin \\angle PDA}{\\sin \\angle PAD}\\cdot\\frac{\\sin \\angle PCD}{\\sin \\angle PDC}\\cdot\\frac{\\sin \\angle PAB}{\\sin \\angle PBA}$. Applying the law of sines to the triangles with vertices at P yields $S=\\frac{|PC|}{|PB|}\\frac{|PA|}{|PD|}\\frac{|PD|}{|PC|}\\frac{|PB|}{|PA|}=1$.", "& \\implies \\frac{PC}{PB}+\\frac{PB}{PC}-\\frac{196}{PC\\cdot PB}=-\\sqrt{3} \\\\ & \\implies PC\\cdot PB = 196\\left(\\frac{1}{\\frac{PC}{PB}+\\frac{PB}{PC}+\\sqrt{3}}\\right). \\end{align*} By AM-GM, $\\frac{PC}{PB}+\\frac{PB}{PC}\\geq{2}$, so $$PB\\cdot PC\\leq 196\\left(\\frac{1}{2+\\sqrt{3}}\\right)=196(2-\\sqrt{3}).$$Finally, $$[\\triangle{BPC}]=\\frac12 \\cdot PB\\cdot PC\\cdot\\sin{150^\\circ}=\\frac14 \\cdot PB\\cdot PC,$$and the maximum area would be $49(2-\\sqrt{3})=98-49\\sqrt{3},$ so the answer is $\\boxed{150}$. ## Solution 3 Proceed as in Solution 2 until you find $\\angle CPB = 150$. The locus of points $P$ that give $\\angle CPB = 150$ is a fixed arc from $B$ to $C$ ($P$ will move along this arc as $D$ moves along $BC$) and we want to maximise the area of [$\\triangle BPC$]. This means we want $P$ to be farthest distance away from $BC$ as possible, so we put $P$ in the middle of the arc (making $\\triangle BPC$ isosceles). We know that $BC=14$ and $\\angle CPB = 150$, so $\\angle PBC = \\angle PCB = 15$. Let $O$ be the foot of the perpendicular from $P$ to line $BC$. Then the area of [$\\triangle BPC$] is the same as $7OP$ because base $BC$ has length $14$. We can split $\\triangle BPC$ into two $15-75-90$ triangles $BOP$ and $COP$, with $BO=CO=7$ and $OP=7 \\tan 15=7(2-\\sqrt{3})=14-7\\sqrt3$. Then, the area of [$\\triangle BPC$] is equal to $7 \\cdot OP=98-49\\sqrt{3}$, and so the answer is $98+49+3=\\boxed{150}$.", "+0 # pls help 0 42 1 In trapezoid ABCD segments AB and CD are parallel. Point P is the intersection of diagonals AC and BD. The area of APAB is 16 square units, and the area of ▲PCD is 25 square units. What is the area of trapezoid ABCD? Aug 30, 2022 #1 0 Triangles $PAB$ and $PCD$ are similar, so $\\frac{[PAB]}{[PCD]} = \\left( \\frac{AB}{CD} \\right)^2 = \\frac{x^2}{y^2}.$ But we are given that $[PAB] = 8$ and $[PCD] = 18$, so $\\frac{x^2}{y^2} = \\frac{8}{18} = \\frac{4}{9},$ which means $x/y = 2/3$. Then $AP/PC = AB/CD = 2/3$. Triangles $ABP$ and $BCP$ have the same height with respect to base $\\overline{AC}$, so $\\frac{[BCP]}{[ABP]} = \\frac{CP}{AP} = \\frac{3}{2},$ which means $[BCP] = 3/2 \\cdot [ABP] = 3/2 \\cdot 8 = 12$. Also, $BP/PD = AB/CD = 2/3$. Triangles $ABP$ and $ADP$ have the same height with respect to base $\\overline{BD}$, so $\\frac{[ADP]}{[ABP]} = \\frac{DP}{BP} = \\frac{3}{2},$ which means $[ADP] = 3/2 \\cdot [ABP] = 3/2 \\cdot 8 = 12$. Therefore, the area of trapezoid $ABCD$ is \\$[ABP] + [BCP] + [CDP] + [DAP] = 8 + 12 + 18 + 12 = 50 Aug 30, 2022", "Hence $\\frac{AC'}{C'B}=\\frac{AC''}{C''B}$, meaning that $C'$ and $C''$ are in fact the same point. Thus the point $P$ lies on the segment $CC'$ as well. \\end{proof} \\begin{remark} Ceva's theorem has analogues both in spherical and hyperbolic geometries. Instead of quotients of lengths of sides, quotients of certain functions of these lengths are used (sines for spherical geometry and hyperbolic sines for hyperbolic geometry). \\end{remark} We can also find the ratio in which the point $P$ divides the cevians $AA',BB'$ and $CC'$ in terms of the ratios in which the points $A',B'$ and $C'$ divide the sides of the triangle. Indeed, it follows from the lemma that $\\frac{AP}{PA'}\\cdot\\area(A'PC)=\\area(PCA)$ and similarly that $\\frac{AP}{PA'}\\cdot\\area(BA'P)=\\area(BPA)$. By adding these two equalities we get that $\\frac{AP}{PA'}\\cdot\\area(BCP)=\\area(PCA)+\\area(BPA)$, or $\\frac{AP}{PA'}=\\frac{\\area(PCA)}{\\area(BCP)}+\\frac{\\area(BPA)}{\\area(BCP)}$. But from the corrollary we know that $\\frac{\\area(PCA)}{\\area(BCP)}=\\frac{AC'}{C'B}$ and $\\frac{\\area(BPA)}{\\area(BCP)}=\\frac{AB'}{B'C}$. Thus $$\\frac{AP}{PA'}=\\frac{AC'}{C'B}+\\frac{AB'}{B'C}$$. Now we will use Ceva's theorem to prove a couple of classical results. Recall that a median in a triangle is a line connecting a vertex with the midpoint of the opposite side. An angle bisector is a line dividing an internal angle of a triangle at a vertex into two equal parts. An altitude is a perpendicular dropped from a vertex to the opposite side of the triangle. We will prove that the three medians of any triangle are concurrent. The same holds for", "BPD = \\angle BOH$. • And if one angle is $\\dfrac\\pi8$ , then the three lines quadrisect the right angle into four equal pieces. Oct 15, 2015 at 23:58 $\\angle ABM=\\angle CAB$ (isocele triangle), and $\\angle HBC=\\angle CAB$ as well ($HBC$ $HAB$ similar triangles). $\\angle ABD=\\angle DBC$ by definition, thus you have $\\angle DBM=\\angle ABD-\\angle ABM=\\angle DBC-\\angle HBC=\\angle HBD$, which is what you wanted to prove.", "formatting. 2 \\times 3 $$2 \\times 3$$ 2^{34} $$2^{34}$$ a_{i-1} $$a_{i-1}$$ \\frac{2}{3} $$\\frac{2}{3}$$ \\sqrt{2} $$\\sqrt{2}$$ \\sum_{i=1}^3 $$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$ Sort by: b=c square - 3 years, 9 months ago Solution (SPOILER) : We let $$AB = c, BC = a, CA = b$$ for simplicity. Note that $$AE \\cdot AQ = (a + c) \\cdot \\frac{bc}{a + c} = bc$$, by Angle Bisector Theorem. Also, $$AP \\cdot AM = AP \\cdot (AP + PM) = AP^2 + AP \\cdot PM = AP^2 + BP \\cdot CP$$, by Power of Point. Since by Stewart's Theorem, $$AP^2 + BP \\cdot CP = \\frac{AB^2 \\cdot CP + AC^2 \\cdot BP}{BC} = \\frac{c^2(\\frac{ab}{b + c}) + b^2(\\frac{ac}{b + c})}{a} = \\frac{bc(b + c)}{b + c} = bc$$, we have $$AP \\cdot AM = bc$$. Consider the following transformation (commonly known as $$\\sqrt{bc}$$-inversion.) : 1) Reflect every point $$X$$ across the angle bisector of $$\\angle BAC$$. 2) Invert about point $$A$$ with radius $$\\sqrt{bc}$$. So, it follows immediately that $$E$$ and $$Q$$ map to each other. Similarly, $$P$$ and $$M$$ map to each other. Thus, the circumcircle of $$BME$$ is mapped to the circumcircle of $$CPQ$$, with center $$O$$. Therefore, $$AO$$ is isogonal to $$AO_1$$ in $$\\angle BAC$$, since the reflection of $$O_1$$ across the angle bisector of $$\\angle", "common side $$CD$$ and the fact $$ADB$$ is a straight line, so $$\\angle ADC$$ and $$\\angle BDC$$ are supplementary. Hence $$CD$$, $$\\angle ADC$$ and $$\\angle BDC$$ must appear in our equations if those facts are to be brought to bear. Moreover, we want $$AD = m$$ and $$BD = n$$ to appear too, since they were indicated as of particular interest (the ratio $$\\triangle ABC$$ was divided in for the textbook; the lengths marked on the wall in the OP's question). Needing to represent two sides and an included angle in each of the smaller triangles rules out the law of sines; we could apply the law of cosines if we wanted $$AC$$ and $$BC$$ to appear in our equations also. Alternatively we can apply $$(4)$$ if we prefer two sides and two angles in each equation. At this stage we have a choice. Treating $$AD$$ as included by $$A$$ and $$\\pi - \\theta$$, and $$BD$$ as included by $$B$$ and $$\\theta$$, \\begin{align} m &= CD \\sin (\\pi - \\theta) \\left(\\cot (\\pi - \\theta) + \\cot A \\right) = CD \\sin \\theta \\left( -\\cot \\theta + \\cot A \\right) \\\\ n &= CD \\sin \\theta \\left( \\cot \\theta + \\cot B \\right) \\\\ \\implies \\frac{m}{n} &= \\frac{-\\cot \\theta + \\cot A }{\\cot \\theta + \\cot B} \\tag{7} \\end{align}", "so $AP=PP'$. Thus, we get that $AP+BP+CP=BP+PP'+C'P'$ as $CP=C'P'$. Now, WLOG let $\\angle B\\leq\\angle C$. We have that $\\angle BCC'\\leq 180^{\\circ}$. We note that $AP+BP+CP=BP+PP'+C'P'\\geq BC'$ with equality if and only if $P,P'\\in BC'$. This means that $\\angle APC'=60^{\\circ}$ (as then $P'\\in CP$). Thus, we see that $APCC'$ is cyclic, so thus as $P$ lies on $BC'$, we see that $P$ is the intersection of the circumcircle of $ACC'$ and $BC'$ (not $C'$). Thus, note that as $\\angle APC'=\\angle CPC'=60^{\\circ}$, $\\angle APC=120^{\\circ}$. Similarly, $\\angle CPB=180^{\\circ}-\\angle CPC'=120^{\\circ}$. Thus, we have found the Fermat Point. ### Vector Proof (Due to Titu Andreescu and Oleg Mushkarov) We will let our origin be the point $P$ with $\\angle APB=\\angle BPC=\\angle CPA=120^{\\circ}$. Consider the point in the plane $X$. Let $a=\\vec{A},b=\\vec{B},c=\\vec{C},x=\\vec{X}$ and $i,j,k$ the unit vectors along $a,b,c$. Then, $|a|=a\\cdot i=(a-x)\\cdot i+x\\cdot i\\leq |a-x|+x\\cdot i$. Similarly, $|b|=b\\cdot j=(b-x)\\cdot j+x\\cdot j\\leq |b-x|+x\\cdot j$ and $|c|=c\\cdot k=(c-x)\\cdot k+x\\cdot k\\leq |c-x|+x\\cdot k$. Noting that $i+j+k=\\vec{0}$ and adding, we see that $|a|+|b|+|c|\\leq |a-x|+|b-x|+|c-x|$, or $AP+BP+CP\\leq AX+BX+CX$. Thus, the origin or point $P$ is the desired point. ## Generalizations There are two main generalizations: ### Weighted Generalization The problem goes as following: which point $P$ minimizes $m\\cdot AP+n\\cdot BP+p\\cdot CP$, where $m,n,p$ are positive reals? ### Polygon Generalization The problem goes as following: for the polygon $A_1A_2\\cdots A_n$," ]

[ "= OP(CR(OA,rA), c); pair OB = (O+B)/2; real rB = length(B-O)/2; pair Bp = OP(CR(OB,rB), c); X=extension(A,Ap,B,Bp); draw(A--B--C--A, s); draw(C--D^^B--O--A^^Ap--O--X, gray+0.25); draw(c^^A--X--B); dot(\"A\", A, N); dot(\"B\", B, SE); dot(\"C\", C, SW); dot(\"D\", D, 0.2*(D-C)); dot(\"I\", X, 0.5*(X-C)); dot(\"P\", Ap, 0.3*(Ap-O)); dot(\"Q\", Bp, 0.3*(Bp-O)); dot(\"O\", O, W); label(\"\\beta\",B,10*dir(157)); label(\"\\alpha\",A,5*dir(-55)); label(\"\\theta\",X,5*dir(55)); [/asy]$ Let $AP=AD=x$, let $BQ=BD=y$, and let $IP=IQ=z$. Let $OD=r$. We find $AB=37$. Let $\\alpha$, $\\beta$, and $\\theta$ be the angles $OAD$, $OBD$, and $OPI$ respectively. Then $\\alpha + \\beta + \\theta = 90^\\circ$, so $$\\theta = 90^\\circ - (\\alpha+\\beta).$$ The perimeter of $\\triangle ABI$ is $2(x+y+z)=2(37+z)$. The desired ratio is then $$\\rho = 2\\left(1+\\frac z{37}\\right)$$ We need to find $z$. In $\\triangle OPI$, $z=r\\cot\\theta = r\\tan (\\alpha+\\beta)$. We get $$\\tan\\alpha = \\frac{OD}{AD} = \\frac 12 \\frac{CD}{AD} = \\frac 12 \\tan A = \\frac 12 \\frac{BC}{AC} = \\frac{35}{24}.$$ Similarly, $\\tan\\beta = \\tfrac 6{35}$. Then $$z = r\\cdot \\tan (\\alpha+\\beta) = r\\cdot \\frac{\\tan\\alpha + \\tan\\beta}{1-\\tan\\alpha\\tan\\beta}= \\frac{37^2\\cdot r}{18\\cdot 35}$$ Computing $[ABC]$ in two ways we get $CD = \\tfrac{12\\cdot 35}{37}$, so $r=\\tfrac{6\\cdot 35}{37}$. Using this value of $r$ we get $z=\\tfrac {37}3$. Thus $$\\rho = 2\\left(1+\\frac 1{3}\\right) = \\frac 8{3},$$ and $8+3=\\boxed{011}$. ## Solution 5 This solution is not a real solution and is solving the problem with a ruler and compass. Draw $AC = 4.8, BC = 14, AB =", "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of", "\\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7}.$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) ### Solution 1.2 (Version 2) Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$ and $\\cos \\theta$,", "draw(A--F); draw(X--F); draw(E--F); draw(circumcircle(A,B,C)); draw(circumcircle(M,F,E)); dot(D); dot(F); dot(A); dot(B); dot(C); dot(E); dot(X); dot(R); dot(I); label(\"A\",A,dir(220)); label(\"B\",B,dir(110)); label(\"C\",C,dir(250)); label(\"D\",D,dir(60)); label(\"E\",E,dir(0)); label(\"F\",F,dir(315)); label(\"X\",X,dir(180)); [/asy]$ First of all, use the Angle Bisector Theorem to find that $BD=35/8$ and $CD=21/8$, and use Stewart's Theorem to find that $AD=15/8$. Then use Power of a Point to find that $DE=49/8$. Then use the circumradius of a triangle formula to find that the length of the circumradius of $\\triangle ABC$ is $\\frac{7\\sqrt{3}}{3}$. Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of the circumcircle of $\\triangle ABC$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\frac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=a$, and $DF=b$. Then, by the Pythagorean Theorem, $$x^2+b^2=\\left(\\frac{49}{8}\\right)^2=\\frac{2401}{64}$$ and $$x^2+(a+b)^2=\\left(\\frac{14\\sqrt{3}}{3}\\right)^2=\\frac{196}{3}.$$ Subtracting the first equation from the second, the $x^2$ term cancels out and we obtain: $$(a+b)^2-b^2=\\frac{196}{3}-\\frac{2401}{64}$$ $$a^2+2ab = \\frac{5341}{192}.$$ By Power of a Point, $ab=BD \\cdot DC=735/64=2205/192$, so $$a^2+2 \\cdot \\frac{2205}{192}=\\frac{5341}{192}$$ $$a^2=\\frac{931}{192}.$$ Since $a=XD$, $XD=\\frac{7\\sqrt{19}}{8\\sqrt{3}}$. Because $\\angle EXF$ and $\\angle EAF$ intercept the same arc in circle $\\omega$ and the same goes for $\\angle XFA$ and $\\angle XEA$, $\\angle EXF\\cong\\angle EAF$ and $\\angle XFA\\cong\\angle XEA$. Therefore, $\\triangle XDE\\sim\\triangle ADF$ by AA Similarity. Since side lengths in similar triangles are proportional, $$\\frac{AF}{\\frac{15}{8}}=\\frac{\\frac{14\\sqrt{3}}{3}}{\\frac{7\\sqrt{19}}{8\\sqrt{3}}}$$ $$\\frac{AF}{\\frac{15}{8}}=\\frac{16}{\\sqrt{19}}$$ $$AF \\cdot \\sqrt{19} = 30$$ $$AF = \\frac{30}{\\sqrt{19}}.$$", "$\\frac{{\\rm Eq} \\ (1)}{{\\rm Eq} \\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7} .$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) Solution 2 Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$", "# Difference between revisions of \"2010 USAMO Problems/Problem 1\" ## Problem Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\\angle XOZ$, where $O$ is the midpoint of segment $AB$. ## Solution 1 Let $\\alpha = \\angle BAZ$, $\\beta = \\angle ABX$. Since $XY$ is a chord of the circle with diameter $AB$, $\\angle XAY = \\angle XBY = \\gamma$. From the chord $YZ$, we conclude $\\angle YAZ = \\angle YBZ = \\delta$. $[asy] import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label(\"O\", O, plain.S); pair A = r * plain.W; dot(A); label(\"A\", A, unit(A)); pair B = r * plain.E; dot(B); label(\"B\", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label(\"X\", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label(\"Y\", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label(\"Z\", Z, unit(Z)); // Feet of perpendiculars", "# 2010 USAMO Problems/Problem 1 ## Problem Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\\angle XOZ$, where $O$ is the midpoint of segment $AB$. ## Solution 1 Let $\\alpha = \\angle BAZ$, $\\beta = \\angle ABX$. Since $XY$ is a chord of the circle with diameter $AB$, $\\angle XAY = \\angle XBY = \\gamma$. From the chord $YZ$, we conclude $\\angle YAZ = \\angle YBZ = \\delta$. $[asy] import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label(\"O\", O, plain.S); pair A = r * plain.W; dot(A); label(\"A\", A, unit(A)); pair B = r * plain.E; dot(B); label(\"B\", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label(\"X\", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label(\"Y\", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label(\"Z\", Z, unit(Z)); // Feet of perpendiculars from Y pair P", "# Difference between revisions of \"2010 USAMO Problems/Problem 1\" ## Problem Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\\angle XOZ$, where $O$ is the midpoint of segment $AB$. ## Solution Let $\\alpha = \\angle BAZ$, $\\beta = \\angle ABX$. Since $XY$ is a chord of the circle with diameter $AB$, $\\angle XAY = \\angle XBY = \\gamma$. From the chord $YZ$, we conclude $\\angle YAZ = \\angle YBZ = \\delta$. $[asy] import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label(\"O\", O, plain.S); pair A = r * plain.W; dot(A); label(\"A\", A, unit(A)); pair B = r * plain.E; dot(B); label(\"B\", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label(\"X\", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label(\"Y\", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label(\"Z\", Z, unit(Z)); // Feet of perpendiculars from", "have $AB = 3$, $AC = 1$, $BC = 2$, $AP = 2 + r$, $BP = 1 + r$, and $CP = 3 - r$, where $r$ is the radius of the circle centered at $P$ that we seek. Thus: $$3 \\cdot 1 \\cdot 2 + 3 {\\left(3-r\\right)}^2 = 1 {\\left(1+r\\right)}^2 + 2 {\\left(2+r\\right)}^2$$ $$6 + 3\\left(9 - 6r + r^2\\right) = \\left(1 + 2r + r^2\\right) + 2\\left(4 + 4r + r^2\\right)$$ $$33 - 18r + 3r^2 = 9 + 10r + 3r^2$$ $$28r = 24$$ $$r = \\boxed{\\frac{6}{7}} \\to \\boxed{\\textbf{(B)}}$$ NOTICE to proficient editors: please label the points on the diagrams, thanks! ## Solution 2 $[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); [/asy]$ Like the solution above, connecting the centers of the circles results in triangle $\\Delta ABP$ with cevian $PC$. The two triangles $\\Delta APC$ and $\\Delta ABP$ share angle $A$, which means we can use Law of Cosines to set up a system of 2 equations that solve for $r$ respectively: $(2 + r)^2 + 1^2 - 2(2 + r)(1)cosA = (3 - r)^2$ (notice that the diameter of the largest semicircle is 6, so its radius is 3 and $PC$ is 3 - r) $(2 + r)^2 + 3^2 - 2(2", "have $AB = 3$, $AC = 1$, $BC = 2$, $AP = 2 + r$, $BP = 1 + r$, and $CP = 3 - r$, where $r$ is the radius of the circle centered at $P$ that we seek. Thus: $$3 \\cdot 1 \\cdot 2 + 3 {\\left(3-r\\right)}^2 = 1 {\\left(1+r\\right)}^2 + 2 {\\left(2+r\\right)}^2$$ $$6 + 3\\left(9 - 6r + r^2\\right) = \\left(1 + 2r + r^2\\right) + 2\\left(4 + 4r + r^2\\right)$$ $$33 - 18r + 3r^2 = 9 + 10r + 3r^2$$ $$28r = 24$$ $$r = \\boxed{\\frac{6}{7}} \\to \\boxed{\\textbf{(B)}}$$ NOTICE to proficient editors: please label the points on the diagrams, thanks! ## Solution 2 $[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); [/asy]$ Like the solution above, connecting the centers of the circles results in triangle $\\Delta ABP$ with cevian $PC$. The two triangles $\\Delta APC$ and $\\Delta ABP$ share angle $A$, which means we can use Law of Cosines to set up a system of 2 equations that solve for $r$ respectively: $(2 + r)^2 + 1^2 - 2(2 + r)(1)cosA = (3 - r)^2$ (notice that the diameter of the largest semicircle is 6, so its radius is 3 and $PC$ is 3 - r) $(2 + r)^2 + 3^2 - 2(2", "+ r)(3)cosA = (r+1)^2$ We can eliminate the extra variable of angle $A$ by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find $r$: $2(r^2 + 4r + 4) - 6 = 2r^2 - 20r + 26$ $8r + 2 = -20r + 26$ $28r = 24$, so $r$ = $6/7$ $\\boxed{(B)}$ ## Solution 3 $[asy] size(7.5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((-3,12/7)--P); draw((3,12/7)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label(\"A\",A,SW); label(\"C\",C,S); label(\"B\",B,SE); label(\"P\",P,N); [/asy]$ Let $C$ be the center of the largest semicircle. We know that $AC = 1$, $CB = 2$, $AP = r + 2$, $BP = r + 1$, and $CP = 3 - r$. Notice that $\\Delta ACP$ and $\\Delta CBP$ are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of $\\Delta CBP$ must be twice that of $\\Delta ACP$, since the area of a triangle is $\\frac{1}{2} Base \\cdot Height$. Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles. $[asy] size(5cm); pair A", "# Difference between revisions of \"Mock AIME 1 2006-2007 Problems/Problem 8\" ## Problem Let $ABCDE$ be a convex pentagon with $\\frac{AB}{\\sqrt{2}}=BC=CD=DE$, $\\angle ABC=150^\\circ$, $\\angle BCD=75^\\circ$, and $\\angle CDE=165^\\circ$. If $\\angle ABE=\\frac{m}{n}^\\circ$ where $m$ and $n$ are relatively prime positive integers, find $m+n$. ## Solution $[asy]defaultpen(fontsize(8)); pair A=expi(pi*5/12)+expi(0)+expi(pi/2), B=expi(pi*5/12), C=(0,0), D=expi(0), E=expi(0)+expi(pi/12), P=expi(pi*5/12)+expi(0); draw(A--B--C--D--E--A);draw(B--P--E--B);draw(D--P--A); label(\"A\",A,(1,0));label(\"B\",B,(-1,0));label(\"C\",C,(-1,0));label(\"D\",D,(1,-1)); label(\"E\",E,(1,0));label(\"P\",P,(1,0)); dot(A^^B^^C^^D^^E^^P);[/asy]$ Let $P$ be a point in $ABCDE$ such that $\\angle ABP=\\angle BAP=45^\\circ$. We see that $\\angle CBP=115^\\circ$ and thus $BP||CD$. Since $BP=BC=CD$, we have that $BCDP$ is a rhombus. Therefore $\\angle CDP=115^\\circ$ so $\\angle PDE=60^\\circ$. Since $PD=CD=DE$ we have that $\\triangle PDE$ is equilateral. $\\angle BPE=75+60^\\circ=180^\\circ-2\\angle PBE\\implies \\angle PBE=\\frac{45}{2}$. $\\angle ABE=\\angle ABP+\\angle PBE=45+\\frac{45}{2}^\\circ=\\frac{135}{2}^\\circ\\implies m+n=\\boxed{137}$.", "Extended Law of Sine to find that the length of the circumradius of $\\triangle ABC$ is $\\tfrac{7\\sqrt{3}}{3}$. $[asy] size(175); defaultpen(fontsize(9pt)); pair A,B,C,D,E,F,W; B=MP(\"B\",origin,dir(180)); C=MP(\"C\",(7,0),dir(0)); A=MP(\"A\",IP(CR(B,5),CR(C,3)),N); D=MP(\"D\",extension(B,C,A,bisectorpoint(C,A,B)),dir(220)); path omega=circumcircle(A,B,C); E=MP(\"E\",OP(omega,A--(A+20*(D-A))),S); path gamma=CR(midpoint(D--E),length(D-E)/2); F=MP(\"F\",OP(omega,gamma),SE); pair X=MP(\"X\",IP(omega,F--(F+2*(D-F))),N); draw(omega^^A--B--C--cycle^^gamma); draw(A--E--F--cycle, gray); draw(E--X--F, royalblue); dot(\"W\",circumcenter(A,B,C),dir(180)); dot(circumcenter(D,E,F)); label(\"\\gamma\",gamma,dir(180)); label(\"u\",X--D,dir(60)); label(\"v\",D--F,dir(70)); [/asy]$ Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of $\\omega$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\tfrac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=u$, and $DF=v$. Then $XE^2-XF^2=EF^2=DE^2-DF^2$, so we get $$(u+v)^2-v^2=\\frac{196}{3}-\\frac{2401}{64}$$ which simplifies to $$u^2+2uv = \\frac{5341}{192}.$$ By Power of Point $D$, $uv=BD \\cdot DC=735/64$. Combining with above, we get $$XD^2=u^2=\\frac{931}{192}.$$ Note that $\\triangle XDE\\sim \\triangle ADF$ and the ratio of similarity is $\\rho = AD : XD = \\tfrac{15}{8}:u$. Then $AF=\\rho\\cdot XE = \\tfrac{15}{8u}\\cdot R$ and $$AF^2 = \\frac{225}{64}\\cdot \\frac{R^2}{u^2} = \\frac{900}{19}.$$ The answer is $900+19=\\boxed{919}$. -Solution by TheBoomBox77 ## Solution 4 Use Law of Cosines in $\\triangle ABC$ to get $\\angle BAC=120^\\circ$. Because $AE$ bisects $\\angle A$, $E$ is the midpoint of major arc $BC$ so $BE=CE,$ and $\\angle BEC=60^\\circ.$ Thus $\\triangle BEC$ is equilateral. Notice now that $\\angle BFC=\\angle BFE= 60^\\circ.$ But $\\angle DFE=90^\\circ$ so $FD$ bisects $\\angle BFC.$ Thus, $$\\frac{BF}{CF}=\\frac{BD}{CD}=\\frac{BA}{CA}=\\frac{5}{3}.$$ Let $BF=5k, CF=3k.$ Use Law of Cosines on $\\triangle BFC$ to get$$25k^2+9k^2-15k^2 =", "to get the point $D^\\prime$. $\\angle PCD$ is congruent to $\\angle D^\\prime CD$, since $P$, $D^\\prime$, and $C$ are collinear. Therefore, we can just find $\\cos(\\angle D^\\prime CD)$. Note that $\\triangle CD^\\prime D$ is a right triangle with $\\angle CD^\\prime D$ as a right angle. $[asy] size(150); import patterns; import olympiad; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux,R; add(\"hatch\",hatch()); //AA=new A and etc. draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; draw(BB--DD); P=midpoint(AA--BB); Q=midpoint(CC--DD); R=midpoint(AA--CC); pair X=intersectionpoints(P--CC,BB--R)[0]; draw(AA--CC,dashed); draw(DD--X,dashed); draw(X--CC,dashed); draw(rightanglemark(CC,X,DD)); dot(P); dot(Q); dot(X); label(\"A\",AA,W); label(\"B\",BB,S); label(\"C\",CC,E); label(\"D\",DD,N); label(\"P\",P,S); label(\"Q\",Q,E); label(\"D^\\prime\",X,W); //Credit to TheMaskedMagician for the diagram //Changes made by Treetor10145[/asy]$ As given by the problem, $CD=1$. Note that $D^\\prime$ is the centroid of equilateral $\\triangle ABC$. Additionally, since $\\triangle ABC$ is equilateral, $D^\\prime$ is also the orthocenter. Due to this, the distance from $C$ to $D^\\prime$ is $\\frac{2}{3}$ of the altitude of $\\triangle ABC$. Therefore, $CD^\\prime=\\frac{\\sqrt{3}}{3}$. Since $\\cos(\\angle D^\\prime CD)=\\cos(\\angle PCQ)=\\frac{CD^\\prime}{CD}$, $\\cos(\\angle PCQ)=\\frac{\\frac{\\sqrt{3}}{3}}{1}=\\frac{\\sqrt{3}}{3}$ $$PQ^2=CP^2+CQ^2-2(CP)(CQ)\\cos(\\angle PCQ)$$ $$PQ^2=\\frac{3}{4}+\\frac{1}{4}-2\\left(\\frac{\\sqrt{3}}{4}\\right)\\left(\\frac{1}{2}\\right)\\left(\\frac{\\sqrt{3}}{3}\\right)$$ Simplifying, $PQ^2=\\frac{1}{2}$. Therefore, $PQ=\\frac{\\sqrt{2}}{2}\\Rightarrow$ $\\boxed{\\textbf{C}}$ Solution by treetor10145 ## Solution 2 (less overkill) Notice, like above said, that $P$ is the midpoint of $AB$ and $Q$ is the midpoint of $CD$. To find the length of $PQ$, first draw in lines $CP$ and $DP$. Notice that $DP$ is an altitude of $\\triangle ADP$. We find that $\\angle{DAP} = 60 ^{\\circ}$ (since $\\triangle ABD$ is equilateral), and $AD=\\frac{1}{2}$. Use the", "(2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0, 3.6), S); [/asy]$ We know that $AU = UM$, $AV = VC$, so $UV = \\frac{1}{2} MC$. As $\\angle UPM = \\angle VPC = 90$, we can see that $\\triangle UPM \\cong \\triangle VPC$ and $\\triangle UVP \\sim \\triangle MPC$ with a side ratio of $1 : 2$. So $UP = VP = 4$, $MP = PC = 8$. With that, we can see that $S\\triangle UPM = 16$, and the area of trapezoid $MUVC$ is 72. As said in solution 1, $S\\triangle AMC = 72 / \\frac{3}{4} = \\boxed{\\textbf{(C) } 96}$. ## Solution 5 (Medians, Height, Pythagorean Theorem, Lots of symmetry, A little extra work) [asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); draw((-2, 6), (2, 6)) label(\"K\", (0, 6), NE); label(\"M\", (-4,0), W); label(\"C\", (4,0), E); label(\"A\", (0, 12), N); label(\"V\", (2, 6), NE); label(\"U\", (-2, 6), NW); label(\"P\", (0.5, 4), E); label(\"B\", (0, 0), S); [/asy] It is well known that medians divide each other into segments of $2:1$ ratio. From this, we have $PC=MP=8$ and $UP=UV=4$. From right triangle $\\triangle{MPC}$, $MC^2=MP^2+MC^2=8^2+8^2=128$, which implies $MC=\\sqrt{128}=8\\sqrt{2}$. Then the area of $\\triangle{AMC}$ is $\\dfrac{8\\sqrt{2} \\cdot AB}{2}$, so our goal is to find $AB$. Note that $AB=AP+BP$. Since $\\triangle{AMC}$ is isosceles, by symmetry $MB=MC$, because $AB$ is the altitude. Knowing this, $BP$", "$$\\frac{PC}{PB}=\\frac{PB}{PD}, \\ \\angle CPB=\\angle BPD \\implies \\triangle CPB\\sim \\triangle BPD$$ These triangles have the same angles ($$\\alpha,\\beta,\\gamma$$) as shown. $$\\frac{PC}{PA}=\\frac{PA}{PD}, \\ \\angle CPA=\\angle APD \\implies \\triangle CPA\\sim \\triangle APD$$ These triangles have the same angles ($$\\delta,\\varphi,\\varepsilon$$) as shown. In quadrilateral $$ABCD:$$ $$\\angle C= \\varphi+\\gamma$$ $$\\angle D= \\beta+\\varepsilon$$ $$\\angle C +\\angle D=\\varphi+\\gamma + \\beta+\\varepsilon=180^\\circ$$ The last is obvious from triangle $$ABC$$. Sum of opposite angles is $$180^\\circ$$ and therefore the quadrilateral is concyclic.", "circle at $G$. Draw line $BG$. $A[\\Delta ABD] = A[\\Delta ABG] = 120$ (Since $\\square ABGD$ is cyclic) But $A[\\Delta ABE]$ is common in both with an area of 60. So, $A[\\Delta AED] = A[\\Delta BEG]$. \\therefore $A[\\Delta AED] \\cong A[\\Delta BEG]$ (SAS Congruency Theorem). In $\\Delta AED$, let $DI$ be the median of $\\Delta AED$. Which means $A[\\Delta AID] = 30 = A[\\Delta EID]$ Rotate $\\Delta DEA$ to meet $D$ at $B$ and $A$ at $G$. $DE$ will fit exactly in $BE$ (both are radii of the circle). From the above solutions, $\\frac{AE}{EF} = 2:1$. $AE$ is a radius and $EF$ is half of it implies $EF$ = $\\frac{radius}{2}$. Which means $A[\\Delta BEF] \\cong A[\\Delta DEI]$ Thus $A[\\Delta BEF] = \\boxed{\\textbf{(B) }30}$ ~phoenixfire & flamewavelight ## Solution 8 $[asy] import geometry; unitsize(2cm); pair A,B,C,DD,EE,FF, M; B = (0,0); C = (3,0); M = (1.45,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); draw(EE--M,StickIntervalMarker(1,1)); label(\"M\",M,S); draw(A--DD,invisible,StickIntervalMarker(1,1)); dot((DD+C)/2); draw(DD--C,invisible,StickIntervalMarker(2,1)); [/asy]$ Using the ratio of $\\overline{AD}$ and $\\overline{CD}$, we find the area of $\\triangle ADB$ is $120$ and the area of $\\triangle BDC$ is $240$. Also using the fact that $E$ is the midpoint of $\\overline{BD}$, we know $\\triangle ADE = \\triangle", "axis for beta Dim ptz2() As Double 'pt to define new x axis for gamma Dim A() As Double, B() As Double, C() As Double Dim dir_cos() As Double Dim dir_ang() As Double x = -3: y = -4: z = 5 A = pt(x, y, z) 'draw the autocad line vec_draw1 A ptX = pt(x, 0, 0) ptx2 = pt(x + 1, 0, 0) ucs ptX, ptx2, A, \"ucs_alpha\" ptX = pt(Abs(x), 0, 0) dim_ang pt0, ptX, A, midpt2(ptX, A) label_pt A, \"A\" ptY = pt(0, y, 0) pty2 = pt(0, y + 1, 0) ucs ptY, pty2, A, \"ucs_beta\" ptY = pt(0, Abs(y), 0) dim_ang pt0, ptY, A, midpt2(ptY, A) ptZ = pt(0, 0, z) ptz2 = pt(0, 0, z + 1) ucs ptZ, ptz2, A, \"ucs_gamma\" ptZ = pt(0, 0, Abs(z)) dim_ang pt0, ptZ, A, midpt2(ptZ, A) 'the box is a visual aid not always required 'Dim pt1() As Double 'pt1 = pt(x / 2, y / 2, z / 2) 'solidbox pt1, Abs(x), Abs(y), Abs(z) 'here are the calculations dir_cos = dir_cos1(A) Debug.Print \"< \" & x; \", \" & y & \", \" & z & \" >\" Debug.Print \"alpha_cos \" & dir_cos(0) Debug.Print \"beta_cos \" & dir_cos(1) Debug.Print \"gamma_cos \" & dir_cos(2) Debug.Print 'direction cosines are the xyz values where the line", "\\frac M6 = L = \\frac{a^5+b^5 + c^5}5$ Yes. Long and tedious and little insight. But methodical. The theory for this type of questions are Newton's Identities. For a cubic polynomial \\begin{align}ax^3+bx^2+cx+d\\end{align}\\\\ Vieta's formulas for the roots $\\alpha_1, \\alpha_2, \\alpha_3$ are: \\begin{align}a(\\alpha_1 + \\alpha_2 + \\alpha_3) + b=0\\tag{V}\\\\ a(\\alpha_1\\alpha_2 + \\alpha_2\\alpha_3 + \\alpha_3\\alpha_1) - c=0\\\\ a\\alpha_1\\alpha_2\\alpha_3 + d=0\\end{align} Newton's Identities for the power sums $P_i:=\\alpha_1^i+\\alpha_2^i+\\alpha_3^i$ are: \\begin{align} aP_1 ~+ ~&b&=0\\tag{N1}\\\\ aP_2 ~+ ~&bP_1 +~ 2c&=0\\tag{N2}\\\\ aP_3 ~+ ~&bP_2 +~ cP_1 +~ 3d&=0\\tag{N3}\\\\ aP_4 ~+ ~&bP_3 +~ cP_2 +~ dP_1&=0\\tag{N4}\\\\ aP_5 ~+ ~&bP_4 +~ cP_3 +~ dP_2&=0\\tag{N5}\\\\ \\end{align} If $\\alpha_1 + \\alpha_2 + \\alpha_3=0$ then from $(V)$ it follows that $b=0$. Using $(N5)$ \\begin{align} aP_5 +~ cP_3 +~ dP_2&=0\\\\ \\end{align} multiply by $6$ \\begin{align} 6aP_5 +~ 3\\cdot2cP_3 +~ 2\\cdot3 dP_2&=0\\\\ \\end{align} using $(N2)$ and $(N3)$ \\begin{align} 6aP_5 -~ 3\\cdot(aP_2+bP_1)P_3 -~ 2\\cdot(aP_3+bP_2+cP_1)P_2=\\end{align} \\begin{align} 6aP_5-3aP_2P_3 -~ 2aP_3P_2=0\\\\ \\end{align} If $a\\neq 0$ then \\begin{align} 6P_5=5P_2P_3\\\\ \\end{align}", "$D\\Theta.\\,$ Its extension cuts $AB\\,$ in point $H.$ In $\\Delta BEA,\\,$ from the Angle Bisector Property, $\\displaystyle \\frac{E\\Theta}{\\Theta A} = \\frac{BE}{BA}=\\frac{a-c}{2c} = \\frac{ED}{DC}.$ Thus, by converse of Thales Theorem we take $D\\Theta\\parallel AC,\\,$ therefore $\\Delta H\\,$ is altitude in $\\Delta ADB.$ P.S. Symbols different from those picture: \\displaystyle \\begin{align} &\\Gamma= C,\\\\ &\\Delta = D,\\\\ &\\gamma = c [= (AB)]\\\\ &\\alpha= a [= (BC)]\\\\ &\\beta = b [= (AC)]. \\end{align} ### Solution 5 Let $AZ\\,$ be an angle bisector and $DE\\,$ the altitude in $\\Delta BAD\\,$ with $K\\,$ their intersection point. $AK\\,$ cuts $BD\\,$ in point $\\Theta.\\,$ We shall prove that $\\Theta\\,$ is the midpoint of $BD.$ Let $KM=KD\\,(1).$ Since $ED\\parallel AC,\\,$ $\\displaystyle\\frac{EB}{EA}=\\frac{BD}{CD}\\,(2)\\,$ and from $\\displaystyle \\frac{BD}{BE}=\\frac{BC}{BA},\\,$ $\\displaystyle \\frac{BD}{DE}=\\frac{BC}{DC}\\,(3).\\,$ In $\\Delta BED,\\,$ $AK\\,$ is an angle bisector. and from the related property we have $\\displaystyle \\frac{KD}{KE}=\\frac{BD}{BE},\\,$ implying (via (1),(3)) that \\displaystyle\\begin{align}&\\frac{KM}{KE}=\\frac{BC}{DC}&\\Longrightarrow\\\\ &\\frac{KM-KE}{KE}=\\frac{BD-DC}{DC}&\\Longrightarrow\\\\ &\\frac{ME}{KE}=\\frac{DB}{DC}\\,\\text{(due to (2))}&\\Longrightarrow\\\\ &\\frac{ME}{KE}=\\frac{AB}{EA}&\\Longrightarrow\\\\ &MB\\parallel AK. \\end{align} Thus, in $\\Delta MBD\\,$ $K\\,$ is the midpoint of side $MD\\,$and $K\\Theta\\parallel MB.\\,$ Therefore, $\\Theta\\,$ is the midpoint if $BD.$ Symbols different from those in the diagram: \\begin{align} \\Gamma = C,\\\\ \\Delta = D. \\end{align} ### Acknowledgment The problem has been posted at the Οι Ρομαντικοι της Γεωμετριας (Romantics of Geometry) facebook group. Takis Chronopoulos has generously volunteered to translate the posted proofs into English. Solution 2 is by Stathis" ]

[ "draw(A--F); draw(X--F); draw(E--F); draw(circumcircle(A,B,C)); draw(circumcircle(M,F,E)); dot(D); dot(F); dot(A); dot(B); dot(C); dot(E); dot(X); dot(R); dot(I); label(\"A\",A,dir(220)); label(\"B\",B,dir(110)); label(\"C\",C,dir(250)); label(\"D\",D,dir(60)); label(\"E\",E,dir(0)); label(\"F\",F,dir(315)); label(\"X\",X,dir(180)); [/asy]$ First of all, use the Angle Bisector Theorem to find that $BD=35/8$ and $CD=21/8$, and use Stewart's Theorem to find that $AD=15/8$. Then use Power of a Point to find that $DE=49/8$. Then use the circumradius of a triangle formula to find that the length of the circumradius of $\\triangle ABC$ is $\\frac{7\\sqrt{3}}{3}$. Since $DE$ is the diameter of circle $\\gamma$, $\\angle DFE$ is $90^\\circ$. Extending $DF$ to intersect circle $\\omega$ at $X$, we find that $XE$ is the diameter of the circumcircle of $\\triangle ABC$ (since $\\angle DFE$ is $90^\\circ$). Therefore, $XE=\\frac{14\\sqrt{3}}{3}$. Let $EF=x$, $XD=a$, and $DF=b$. Then, by the Pythagorean Theorem, $$x^2+b^2=\\left(\\frac{49}{8}\\right)^2=\\frac{2401}{64}$$ and $$x^2+(a+b)^2=\\left(\\frac{14\\sqrt{3}}{3}\\right)^2=\\frac{196}{3}.$$ Subtracting the first equation from the second, the $x^2$ term cancels out and we obtain: $$(a+b)^2-b^2=\\frac{196}{3}-\\frac{2401}{64}$$ $$a^2+2ab = \\frac{5341}{192}.$$ By Power of a Point, $ab=BD \\cdot DC=735/64=2205/192$, so $$a^2+2 \\cdot \\frac{2205}{192}=\\frac{5341}{192}$$ $$a^2=\\frac{931}{192}.$$ Since $a=XD$, $XD=\\frac{7\\sqrt{19}}{8\\sqrt{3}}$. Because $\\angle EXF$ and $\\angle EAF$ intercept the same arc in circle $\\omega$ and the same goes for $\\angle XFA$ and $\\angle XEA$, $\\angle EXF\\cong\\angle EAF$ and $\\angle XFA\\cong\\angle XEA$. Therefore, $\\triangle XDE\\sim\\triangle ADF$ by AA Similarity. Since side lengths in similar triangles are proportional, $$\\frac{AF}{\\frac{15}{8}}=\\frac{\\frac{14\\sqrt{3}}{3}}{\\frac{7\\sqrt{19}}{8\\sqrt{3}}}$$ $$\\frac{AF}{\\frac{15}{8}}=\\frac{16}{\\sqrt{19}}$$ $$AF \\cdot \\sqrt{19} = 30$$ $$AF = \\frac{30}{\\sqrt{19}}.$$", "$$\\frac{PC}{PB}=\\frac{PB}{PD}, \\ \\angle CPB=\\angle BPD \\implies \\triangle CPB\\sim \\triangle BPD$$ These triangles have the same angles ($$\\alpha,\\beta,\\gamma$$) as shown. $$\\frac{PC}{PA}=\\frac{PA}{PD}, \\ \\angle CPA=\\angle APD \\implies \\triangle CPA\\sim \\triangle APD$$ These triangles have the same angles ($$\\delta,\\varphi,\\varepsilon$$) as shown. In quadrilateral $$ABCD:$$ $$\\angle C= \\varphi+\\gamma$$ $$\\angle D= \\beta+\\varepsilon$$ $$\\angle C +\\angle D=\\varphi+\\gamma + \\beta+\\varepsilon=180^\\circ$$ The last is obvious from triangle $$ABC$$. Sum of opposite angles is $$180^\\circ$$ and therefore the quadrilateral is concyclic.", "# Difference between revisions of \"Mock AIME 1 2006-2007 Problems/Problem 8\" ## Problem Let $ABCDE$ be a convex pentagon with $\\frac{AB}{\\sqrt{2}}=BC=CD=DE$, $\\angle ABC=150^\\circ$, $\\angle BCD=75^\\circ$, and $\\angle CDE=165^\\circ$. If $\\angle ABE=\\frac{m}{n}^\\circ$ where $m$ and $n$ are relatively prime positive integers, find $m+n$. ## Solution $[asy]defaultpen(fontsize(8)); pair A=expi(pi*5/12)+expi(0)+expi(pi/2), B=expi(pi*5/12), C=(0,0), D=expi(0), E=expi(0)+expi(pi/12), P=expi(pi*5/12)+expi(0); draw(A--B--C--D--E--A);draw(B--P--E--B);draw(D--P--A); label(\"A\",A,(1,0));label(\"B\",B,(-1,0));label(\"C\",C,(-1,0));label(\"D\",D,(1,-1)); label(\"E\",E,(1,0));label(\"P\",P,(1,0)); dot(A^^B^^C^^D^^E^^P);[/asy]$ Let $P$ be a point in $ABCDE$ such that $\\angle ABP=\\angle BAP=45^\\circ$. We see that $\\angle CBP=115^\\circ$ and thus $BP||CD$. Since $BP=BC=CD$, we have that $BCDP$ is a rhombus. Therefore $\\angle CDP=115^\\circ$ so $\\angle PDE=60^\\circ$. Since $PD=CD=DE$ we have that $\\triangle PDE$ is equilateral. $\\angle BPE=75+60^\\circ=180^\\circ-2\\angle PBE\\implies \\angle PBE=\\frac{45}{2}$. $\\angle ABE=\\angle ABP+\\angle PBE=45+\\frac{45}{2}^\\circ=\\frac{135}{2}^\\circ\\implies m+n=\\boxed{137}$.", "have $AB = 3$, $AC = 1$, $BC = 2$, $AP = 2 + r$, $BP = 1 + r$, and $CP = 3 - r$, where $r$ is the radius of the circle centered at $P$ that we seek. Thus: $$3 \\cdot 1 \\cdot 2 + 3 {\\left(3-r\\right)}^2 = 1 {\\left(1+r\\right)}^2 + 2 {\\left(2+r\\right)}^2$$ $$6 + 3\\left(9 - 6r + r^2\\right) = \\left(1 + 2r + r^2\\right) + 2\\left(4 + 4r + r^2\\right)$$ $$33 - 18r + 3r^2 = 9 + 10r + 3r^2$$ $$28r = 24$$ $$r = \\boxed{\\frac{6}{7}} \\to \\boxed{\\textbf{(B)}}$$ NOTICE to proficient editors: please label the points on the diagrams, thanks! ## Solution 2 $[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); [/asy]$ Like the solution above, connecting the centers of the circles results in triangle $\\Delta ABP$ with cevian $PC$. The two triangles $\\Delta APC$ and $\\Delta ABP$ share angle $A$, which means we can use Law of Cosines to set up a system of 2 equations that solve for $r$ respectively: $(2 + r)^2 + 1^2 - 2(2 + r)(1)cosA = (3 - r)^2$ (notice that the diameter of the largest semicircle is 6, so its radius is 3 and $PC$ is 3 - r) $(2 + r)^2 + 3^2 - 2(2", "have $AB = 3$, $AC = 1$, $BC = 2$, $AP = 2 + r$, $BP = 1 + r$, and $CP = 3 - r$, where $r$ is the radius of the circle centered at $P$ that we seek. Thus: $$3 \\cdot 1 \\cdot 2 + 3 {\\left(3-r\\right)}^2 = 1 {\\left(1+r\\right)}^2 + 2 {\\left(2+r\\right)}^2$$ $$6 + 3\\left(9 - 6r + r^2\\right) = \\left(1 + 2r + r^2\\right) + 2\\left(4 + 4r + r^2\\right)$$ $$33 - 18r + 3r^2 = 9 + 10r + 3r^2$$ $$28r = 24$$ $$r = \\boxed{\\frac{6}{7}} \\to \\boxed{\\textbf{(B)}}$$ NOTICE to proficient editors: please label the points on the diagrams, thanks! ## Solution 2 $[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); [/asy]$ Like the solution above, connecting the centers of the circles results in triangle $\\Delta ABP$ with cevian $PC$. The two triangles $\\Delta APC$ and $\\Delta ABP$ share angle $A$, which means we can use Law of Cosines to set up a system of 2 equations that solve for $r$ respectively: $(2 + r)^2 + 1^2 - 2(2 + r)(1)cosA = (3 - r)^2$ (notice that the diameter of the largest semicircle is 6, so its radius is 3 and $PC$ is 3 - r) $(2 + r)^2 + 3^2 - 2(2", "+ r)(3)cosA = (r+1)^2$ We can eliminate the extra variable of angle $A$ by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find $r$: $2(r^2 + 4r + 4) - 6 = 2r^2 - 20r + 26$ $8r + 2 = -20r + 26$ $28r = 24$, so $r$ = $6/7$ $\\boxed{(B)}$ ## Solution 3 $[asy] size(7.5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((-3,12/7)--P); draw((3,12/7)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label(\"A\",A,SW); label(\"C\",C,S); label(\"B\",B,SE); label(\"P\",P,N); [/asy]$ Let $C$ be the center of the largest semicircle. We know that $AC = 1$, $CB = 2$, $AP = r + 2$, $BP = r + 1$, and $CP = 3 - r$. Notice that $\\Delta ACP$ and $\\Delta CBP$ are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of $\\Delta CBP$ must be twice that of $\\Delta ACP$, since the area of a triangle is $\\frac{1}{2} Base \\cdot Height$. Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles. $[asy] size(5cm); pair A", "So $\\dfrac{BP}{BC}=\\dfrac{BQ}{BA}$, plugging in the given values we get $\\dfrac{BP}{4}=\\dfrac{4.5}{5}$, so $BP=\\dfrac{18}{5}$, and $m+n=\\boxed{023}$. ### Solution 3 Let $r=BO$. Drawing perpendiculars, $BM=MC=2$ and $BN=NA=2.5$. From there, $OM=\\sqrt{r^2-4}$. Thus, $OQ=\\frac{\\sqrt{4r^2+9}}{2}$. Using $\\triangle{BOQ}$, we get $r=3$. Now let's find $NP$. After some calculations with $\\triangle{BON}$ ~ $\\triangle{OPN}$, ${NP=11/10}$. Therefore, $BP=\\frac{5}{2}+\\frac{11}{10}=18/5$. $18+5=\\boxed{023}$. ## Solution 4 Let $\\angle{BQO}=\\alpha$. Extend $OB$ to touch the circumcircle at a point $K$. Then, note that $\\angle{KAC}=\\angle{CBK}=\\angle{QBO}=90^\\circ-\\alpha$. But since $BK$ is a diameter, $\\angle{KAB}=90^\\circ$, implying $\\angle{CAB}=\\alpha$. It follows that $APCQ$ is a cyclic quadrilateral. Let $BP=x$. By Power of a Point, $$5x=4\\cdot\\frac 9 2\\implies x=\\frac{18}{5}.$$The answer is $18+5=\\boxed{023}$. ## Solution 5 Note: This is not a very good solution, but it is relatively natural and requires next to no thinking. Denote the circumradius of $ABC$ to be $R$, the circumcircle of $ABC$ to be $O$, and the shortest distance from $Q$ to circle $O$ to be $x$. Using Power of a Point on $Q$ relative to circle $O$, we get that $x(x+2r) = 0.5 \\cdot 4.5 = \\frac{9}{4}$. Using Pythagorean Theorem on triangle $QOB$ to get $(x + r)^2 + r^2 = \\frac{81}{4}$. Subtracting the first equation from the second, we get that $2r^2 = 18$ and therefore $r = 3$. Now, set $\\cos{ABC} = y$. Using law of cosines on $ABC$ to find $AC$ in terms", "formatting. 2 \\times 3 $$2 \\times 3$$ 2^{34} $$2^{34}$$ a_{i-1} $$a_{i-1}$$ \\frac{2}{3} $$\\frac{2}{3}$$ \\sqrt{2} $$\\sqrt{2}$$ \\sum_{i=1}^3 $$\\sum_{i=1}^3$$ \\sin \\theta $$\\sin \\theta$$ \\boxed{123} $$\\boxed{123}$$ Sort by: b=c square - 3 years, 9 months ago Solution (SPOILER) : We let $$AB = c, BC = a, CA = b$$ for simplicity. Note that $$AE \\cdot AQ = (a + c) \\cdot \\frac{bc}{a + c} = bc$$, by Angle Bisector Theorem. Also, $$AP \\cdot AM = AP \\cdot (AP + PM) = AP^2 + AP \\cdot PM = AP^2 + BP \\cdot CP$$, by Power of Point. Since by Stewart's Theorem, $$AP^2 + BP \\cdot CP = \\frac{AB^2 \\cdot CP + AC^2 \\cdot BP}{BC} = \\frac{c^2(\\frac{ab}{b + c}) + b^2(\\frac{ac}{b + c})}{a} = \\frac{bc(b + c)}{b + c} = bc$$, we have $$AP \\cdot AM = bc$$. Consider the following transformation (commonly known as $$\\sqrt{bc}$$-inversion.) : 1) Reflect every point $$X$$ across the angle bisector of $$\\angle BAC$$. 2) Invert about point $$A$$ with radius $$\\sqrt{bc}$$. So, it follows immediately that $$E$$ and $$Q$$ map to each other. Similarly, $$P$$ and $$M$$ map to each other. Thus, the circumcircle of $$BME$$ is mapped to the circumcircle of $$CPQ$$, with center $$O$$. Therefore, $$AO$$ is isogonal to $$AO_1$$ in $$\\angle BAC$$, since the reflection of $$O_1$$ across the angle bisector of $$\\angle", "$\\frac{{\\rm Eq} \\ (1)}{{\\rm Eq} \\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7} .$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) Solution 2 Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$", "# Tangent to the Incircle ### Problem Let $ABC$ be an isosceles triangle, with the base angle $\\alpha$ and incircle $\\Gamma$. $D$ and $E$ are points on the sides $AB$ and $AC$ such that $DE$ is tangent to $\\Gamma.$ Show that $\\displaystyle\\frac{AD}{DB}+\\frac{AE}{EC}=\\frac{1-\\cos\\alpha}{\\cos\\alpha}$ ### Lemma $PA,PB,PC$ are three chords in a circle, $\\angle APB=\\angle APC=\\phi.$ Then $\\displaystyle\\frac{PB+PC}{PA}=2\\cos\\phi.$ Proof of Lemma Let $O$ be the center of the circle, $AD$ a diameter. Introduce $\\beta=\\angle DCP=\\angle DAP.$ In right $\\Delta ADP,$ $AP=AD\\cdot\\cos\\beta=2R\\cos\\beta,$ where R is the radius of circle $(O).$ In $\\Delta BCP,$ $\\angle BCP=90^{\\circ}-\\phi +\\beta,$ $\\angle CBP=90^{\\circ}-\\phi -\\beta,$ (or the other way round). By the Law of Sines, $\\displaystyle\\frac{PB}{\\sin(90^{\\circ}-\\phi +\\beta)}=\\frac{PC}{\\sin(90^{\\circ}-\\phi -\\beta)}=2R.$ Since $\\sin(90^{\\circ}-\\phi \\pm\\beta)=\\cos\\phi\\cdot\\cos\\beta\\mp\\sin\\phi\\cdot\\sin\\beta,$ we have $PB+PC=4R\\cdot\\cos\\phi\\cdot\\cos\\beta,$ Dividing by $AP=2R\\cos\\beta$ proves the lemma. ### Proof 1 Joint $O$ to $D$ and $E$ and let $K,L,M$ be the points of tangency of $(O),$ as shown. $\\angle KAM=180^{\\circ}-2\\alpha$ so that $\\angle KOM=2\\alpha.$ Further, $OD$ bisects $\\angle KOL,$ $OE$ bisects $\\angle LOM,$ implying that $\\angle DOE=\\alpha .$ Draw $AF\\parallel OE$ and $AG\\parallel OD,$ where $F$ is on $CO,$ $G$ on $BO,$ both produced. First of all, note that, by the construction, $\\angle FAG=\\angle DOE=\\alpha.$ Second of all, $\\angle FOG=\\angle BOC=180^{\\circ}-\\alpha.$ It follows that $\\angle FAG+\\angle FOG=180^{\\circ},$ making quadrilateral $AFOG$ cyclic. Finally note that $OA$ bisects $\\angle FOG$ because $O$ is the incenter of", "49 \\qquad \\Longrightarrow\\qquad k=\\frac 7{\\sqrt{19}}$$ Use Ptolemy's Theorem on $BFCA$, to get $$15k+15k=7\\cdot AF, \\qquad \\Longrightarrow\\qquad AF= \\frac{30}{\\sqrt{19}},$$ so $AF^2=\\frac{900}{19}$ and the answer is $900+19=919$ ~abacadaea ## Solution 5 Denote $AB = c, BC = a, AC = b, \\angle A = 2 \\alpha.$ Let M be midpoint BC. Let $\\theta$ be the circle centered at $A$ with radius $\\sqrt{AB \\cdot AC} =\\sqrt{bc}.$ We calculate the length of some segments. The median $AM = \\sqrt{\\frac {b^2}{2} + \\frac {c^2}{2} - \\frac {a^2}{4}}.$ The bisector $AD = \\frac {2 b c \\cos \\alpha}{b+c}.$ One can use Stewart's Theorem in both cases. $AD$ is bisector of $\\angle A \\implies BD = \\frac {a c}{b + c}, CD = \\frac {a b}{b + c} \\implies$ $$BD \\cdot CD = \\frac {a^2 bc }{(b+c)^2}.$$ We use Power of Point $D$ and get $AD \\cdot DE = BD \\cdot CD.$ $$AE = AD + DE = AD + \\frac {BD \\cdot CD}{AD},$$ $$AE =\\frac {2 b c \\cos \\alpha}{b+c} + \\frac {a^2 bc \\cdot (b+c) }{(b+c)^2 \\cdot 2 b c \\cos \\alpha} =$$ $$= \\frac {b c \\cos^2 \\alpha + a^2}{2(b+c)\\cos \\alpha} =\\frac {4bc \\cos^2 \\alpha + b^2 +c^2 -2 b c \\cos 2\\alpha}{2(b+c) \\cos \\alpha} = \\frac {b+c}{2} \\implies AD \\cdot AE = 2 bc \\cos \\alpha.$$ We consider the inversion with", "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of", "# 2010 USAMO Problems/Problem 1 ## Problem Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\\angle XOZ$, where $O$ is the midpoint of segment $AB$. ## Solution 1 Let $\\alpha = \\angle BAZ$, $\\beta = \\angle ABX$. Since $XY$ is a chord of the circle with diameter $AB$, $\\angle XAY = \\angle XBY = \\gamma$. From the chord $YZ$, we conclude $\\angle YAZ = \\angle YBZ = \\delta$. $[asy] import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label(\"O\", O, plain.S); pair A = r * plain.W; dot(A); label(\"A\", A, unit(A)); pair B = r * plain.E; dot(B); label(\"B\", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label(\"X\", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label(\"Y\", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label(\"Z\", Z, unit(Z)); // Feet of perpendiculars from Y pair P", "that $$\\angle BPE=60°.$$ However, I emphasise that (as OOP confirmed under the comments on OP) the assumed condition $$a=b$$ is not part of the original problem. Hence my claim above that the problem has no unique solution. However, we may get a parametric solution as follows. Fix any $$0 then the position of the cevian $$CE$$ is fixed. Also, as is clear from the above argument, $$m,$$ and therefore the other cevian $$BD,$$ is fixed too. Thus, the angle $$BPE$$ sought is also now determined. Therefore, let $$\\angle BPE=\\angle DPC=\\theta.$$ Let us make the following abbreviations, letting $$\\angle PEB=\\omega,\\angle PBC=\\phi,\\angle PCB=\\phi',\\angle CDP=\\omega'.$$ Now, applying the sine rule in $$\\triangle EBC,$$ we obtain $$\\frac {1}{\\sin\\omega}=\\frac{b}{\\sin\\phi'},$$ so that $$\\sin\\phi'=b\\sin\\omega.$$ But in the same triangle, we have $$\\omega+60°+\\phi'=180°,$$ upon substitution of which in the previous equation and some other transformations, gives $$\\cos(30°-\\omega)=(1-a)\\sin\\omega.$$ Performing similar operations in the triangle $$DBC$$ gives the equation $$\\sin\\phi=m\\cos(30°-\\phi).$$ Thus, the angles $$\\phi$$ and $$\\omega$$ are now explicitly known. Finally, in $$\\triangle EBP,$$ we have $$\\omega+(60°-\\phi)+\\theta=180°,$$ which gives the angle sought $$\\theta=120°+\\phi-\\omega.$$ This is only determined uniquely once we select any real number $$a$$ such that $$0 PS. It should be noted that the procedure above will apply mutatis mutandis to an arbitrary triangle with given sides, since then the angles are also determined by the cosine", "\\ (6)}$, we get $$\\tan \\theta = \\frac{60}{21/2} = \\frac{40}{7}.$$ Therefore, by writing this answer in the form of $\\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \\boxed{047}$. ~ Steven Chen (www.professorchenedu.com) ### Solution 1.2 (Version 2) Since we are asked to find $\\tan \\theta$, we can find $\\sin \\theta$ and $\\cos \\theta$ separately and use their values to get $\\tan \\theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$. $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label(\"A\",A,NW); label(\"B\",B,NE); label(\"C\",C,SE); label(\"D\",D,SW); dot(X); label(\"X\",X,S); label(\"5\",(A+B)/2,N); label(\"6\",(B+C)/2,E); label(\"9\",(C+D)/2,S); label(\"7\",(D+A)/2,W); label(\"\\theta\",X,2.5E); label(\"a\",(A+X)/2,NE); label(\"b\",(B+X)/2,NW); label(\"c\",(C+X)/2,SW); label(\"d\",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\\sin \\theta$ and $\\cos \\theta$,", "dir(D-F)*I, deepmagenta); label(\"4\\sqrt{2}\", (D+E)/2, dir(E-D)*I, deepmagenta); label(\"a\", (B+X)/2, dir(B-X)*I, gray(0.3)); label(\"a\\sqrt{2}\", (D+X)/2, dir(D-X)*I, gray(0.3)); [/asy]$ Notice that $$\\angle DFE=\\angle CFE-\\angle CFD=\\angle CBE-\\angle CAD=180-B-A=C.$$By the Law of Cosines, $$\\cos C=\\frac{AC^2+BC^2-AB^2}{2\\cdot AC\\cdot BC}=\\frac34.$$Then, $$DE^2=DF^2+EF^2-2\\cdot DF\\cdot EF\\cos C=32\\implies DE=4\\sqrt2.$$Let $X=\\overline{AB}\\cap\\overline{CF}$, $a=XB$, and $b=XD$. Then, $$XA\\cdot XD=XC\\cdot XF=XB\\cdot XE\\implies b(a+4)=a(b+4\\sqrt2)\\implies b=a\\sqrt2.$$However, since $\\triangle XFD\\sim\\triangle XAC$, $XF=\\tfrac{4+a}3$, but since $\\triangle XFE\\sim\\triangle XBC$, $$\\frac75=\\frac{4+a}{3a}\\implies a=\\frac54\\implies BE=a+a\\sqrt2+4\\sqrt2=\\frac{5+21\\sqrt2}4,$$and the requested sum is $5+21+2+4=\\boxed{032}$. (Solution by TheUltimate123) ## Solution 2 Define $\\omega_1$ to be the circumcircle of $\\triangle ACD$ and $\\omega_2$ to be the circumcircle of $\\triangle EBC$. Because of exterior angles, $\\angle ACB = \\angle CBE - \\angle CAD$ But $\\angle CBE = \\angle CFE$ because $CBFE$ is cyclic. In addition, $\\angle CAD = \\angle CFD$ because $CAFD$ is cyclic. Therefore, $\\angle ACB = \\angle CFE - \\angle CFD$. But $\\angle CFE - \\angle CFD = \\angle DFE$, so $\\angle ACB = \\angle DFE$. Using Law of Cosines on $\\triangle ABC$, we can figure out that $\\cos(\\angle ACB) = \\frac{3}{4}$. Since $\\angle ACB = \\angle DFE$, $\\cos(\\angle DFE) = \\frac{3}{4}$. We are given that $DF = 2$ and $FE = 7$, so we can use Law of Cosines on $\\triangle DEF$ to find that $DE = 4\\sqrt{2}$. Let $G$ be the intersection of segment $\\overline{AE}$ and $\\overline{CF}$. Using Power of a Point with respect to $G$", "know that $\\overline{PC}\\parallel\\overline{OM}.$ We construct $D$ on $\\overline{PC}$ such that $\\overline{XD}\\perp\\overline{PC}.$ Clearly, $MXDC$ is a rectangle. Since $\\angle XPD=\\angle OXP$ by alternate interior angles, we have $\\triangle XPD\\sim\\triangle OXP$ by the AA Similarity, with ratio of similitude $\\frac{XP}{OX}=\\frac 35.$ Therefore, we get that $PD=\\frac 95$ and $PC=PD+DC=PD+MX=\\frac 95 + \\frac 32 = \\frac{33}{10}.$ The area of $\\triangle PQR$ is$$\\frac12(RQ)(PC)=\\frac12\\left(3\\sqrt3\\right)\\left(\\frac{33}{10}\\right)=\\frac{99\\sqrt3}{20},$$and the answer is $99+3+20=\\boxed{\\textbf{(D) } 122}.$ ~MRENTHUSIASM Solution 2 Diagram $[asy] draw(circle((7,0),7)); pair A = (0, 0); pair B = (14, 0); draw(A--B); draw(circle((11,3),3)); label(\"C\", (7, 0), S); label(\"O\", (11, 3), E); label(\"P\", (11, 0), S); pair C = (7, 0); pair O = (11, 3); pair P = (11, 0); pair Q = intersectionpoints(circle(C, 7), circle(O, 3))[1]; pair R = intersectionpoints(circle(C, 7), circle(O, 3))[0]; draw(C--O); draw(C--Q); draw(C--R); draw(Q--R); draw(O--P); draw(O--Q); draw(O--R); draw(P--Q); draw(P--R); label(\"Q\", Q, N); label(\"R\", R, E); [/asy]$ Solution Suppose we label the points as shown in the diagram above, where $C$ is the center of the semicircle and $O$ is the center of the circle tangent to $\\overline{AB}$. Since $\\angle QPR = 60^{\\circ}$, we have $\\angle QOR = 2\\cdot 60^{\\circ}=120^{\\circ}$ and $\\triangle QOR$ is a $30-30-120$ triangle, which can be split into two $30-60-90$ triangles by the altitude from $O$. Since $QR=3\\sqrt{3},$ we know $OQ=OR=\\tfrac{3\\sqrt{3}}{\\sqrt{3}}=3$ by $30-60-90$ triangles. The area of this part of $\\triangle", "A + \\cot B + \\cot C$$ : $\\cot A + \\cot B + \\cot C = \\frac{{({a^2} + {b^2} + {c^2})}}{{4\\Delta }}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\, \\ldots \\left( 5 \\right)$ From (4) and (5), the required sum is zero. Example - 49 Consider a triangle $$\\Delta ABC$$ . Let $$\\Psi$$ be the circumcircle of $$\\Delta ABC$$ . At each of the three points A, B, C, tangents are drawn to $$\\Psi$$ to form another triangle, $$\\Delta PQR$$ . Find the sides of $$\\Delta PQR$$ . Solution: Let us evaluate the side PQ. Note that $\\angle PCB = \\angle PBC = \\angle A\\qquad\\qquad(why?)$ so that $\\angle P = \\pi - 2\\angle A$ Applying the sine rule in $$\\Delta PBC$$ , we have \\begin{align} & \\qquad \\frac{{PB}}{{\\sin A}} = \\frac{{PC}}{{\\sin A}} = \\frac{{BC}}{{\\sin 2A}} \\\\ &\\Rightarrow \\quad PC = \\frac{{BC\\sin A}}{{\\sin 2A}} = \\frac{a}{{2\\cos A}} \\\\ \\end{align} Similarly, \\begin{align}QC = \\frac{b}{{2\\cos B}}\\end{align} . Thus, \\begin{align}&PQ = PC + CQ = \\frac{a}{{2\\cos A}} + \\frac{b}{{2\\cos B}} = \\frac{1}{2}\\frac{{(a\\cos B + b\\cos A)}}{{\\cos A\\cos B}} \\\\ &\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\qquad\\;= \\frac{{\\cos C}}{{2\\cos A\\cos B}} \\\\ \\end{align} Similar expressions will hold for QR and RP.", "lie on the circle, and $\\angle ADC=\\angle CDQ$. $\\angle EAD=\\angle ECD$ because $A$,$C$,$D$ and $E$ all lie on the circle. $\\angle DAC=\\angle ECF$ because $CD=EF$, and $A$,$C$,$D$,$E$ and $F$ all lie on the circle. Then, $\\angle EAC=\\angle EAD+\\angle DAC=\\angle ECD+\\angle ECF=\\angle DCQ$ Therefore, $\\triangle AEC$ and $\\triangle CDQ$ are similar, so $AC/CE=CQ/QD$. Lastly, let's show that $\\triangle CPQ$ and $\\triangle EPD$ are similar. $[asy] pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=IntersectionPoint(A--D,C--F); draw(A--B); draw(B--C); draw(C--D); draw(D--E,green); draw(E--F); draw(F--A); draw(A--C); draw(A--Q); draw(A--E); draw(B--E); draw(P--E,green); draw(Q--F); draw(C--P,red); draw(Q--P,red); draw(C--Q,red); draw(D--P,green); draw(circle((0,0),1)); label(\"$$A$$\",A,W); label(\"$$B$$\",B,N); label(\"$$C$$\",C,N); label(\"$$D$$\",D,NE); label(\"$$E$$\",E,SE); label(\"$$F$$\",F,S); label(\"$$P$$\",(0.3,0.8),S); label(\"$$Q$$\",(-0.15,0.4),S); [/asy]$ Because $CD=EF$ and $C$,$D$,$E$ and $F$ all lie on the circle, $CF$ is parallel to $DE$. So, $\\triangle CPQ$ and $\\triangle EPD$ are similar, and $CQ/DE=CP/PE$. Putting it all together, $\\frac{CP}{PE}=\\frac{CQ}{DE}=\\frac{AC}{CE}\\cdot \\frac{QD}{DE}=(\\frac{AC}{CE})^2$. Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol943.html", "and $CPD$, \\begin{align}\\tag{2}x^2+z^2=180\\end{align}. It follows that \\begin{align}\\tag{3}z^2-y^2=32\\end{align}. Since $\\sin\\alpha=\\sqrt{1-\\cos^2\\alpha}$, $$\\sqrt{1-\\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\\sqrt{1-\\frac{(x^2+z^2-260)^2}{4x^2z^2}}$$ which simplifies to $$\\frac{48^2}{y^2}=\\frac{80^2}{z^2} \\qquad \\Longrightarrow \\qquad 5y=3z.$$ Plugging this back to equations $(1)$, $(2)$, and $(3)$, it can be solved that $x=\\sqrt{130},y=3\\sqrt{2},z=5\\sqrt{2}$. Then, the area of the quadrilateral is $$x(y+z)\\sin\\alpha=\\sqrt{130}\\cdot8\\sqrt{2}\\cdot\\frac{14}{\\sqrt{260}}=\\boxed{112}$$ --Solution by MicGu ## Solution 5 As in all other solutions, we can first find that either $AP=CP$ or $BP=DP$, but it's an AIME problem, we can take $AP=CP$, and assume the other choice will lead to the same result (which is true). From $AP=CP$, we have $[DAP]=[DCP]$, and $[BAP]=[BCP] \\implies [ABD] = [CBD]$, therefore, \\begin{align} \\nonumber \\frac 12 \\cdot AB\\cdot AD\\sin A &= \\frac 12 \\cdot BC\\cdot CD\\sin C \\\\ \\Longrightarrow \\hspace{1in} 7\\sin C &= \\sqrt{65}\\sin A \\end{align} By Law of Cosines, \\begin{align} \\nonumber 10^2+14^2-2\\cdot 10\\cdot 14\\cos C &= 10^2+4\\cdot 65-2\\cdot 10\\cdot 2\\sqrt{65}\\cos A \\\\ \\Longrightarrow \\hspace{1in} -\\frac 85 - 7\\cos C &= \\sqrt{65}\\cos A \\tag{2} \\end{align} Square $(1)$ and $(2)$, and add them, to get $$\\left(\\frac 85\\right)^2 + 2\\cdot \\frac 85 \\cdot 7\\cos C + 7^2 = 65$$ Solve, $\\cos C = 3/5 \\implies \\sin C = 4/5$, $$[ABCD] = 2[BCD] = BC\\cdot CD\\cdot \\sin C = 14\\cdot 10\\cdot \\frac 45 = \\boxed{112}$$ -Mathdummy ## Solution 6 Either $PA=PC$ or $PD=PB$. Let $PD=PB=s$. Applying Stewart's Theorem on $\\triangle ABD$ and $\\triangle BCD$, dividing by $2s$" ]

[ "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "# Routh's Theorem In triangle $ABC$, $D$, $E$ and $F$ are points on sides $BC$, $AC$, and $AB$, respectively. Let $r=\\frac{AF}{FB}$, $s=\\frac{BD}{DC}$, and $t=\\frac{CE}{AE}$. Let $G$ be the intersection of $AD$ and $BC$, $H$ be the intersection of $BE$ and $CF$, and $I$ be the intersection of $CF$ and $AD$. Then, Routh's Theorem states that $$[GHI]=\\dfrac{(rst-1)^2}{(rs+r+1)(st+s+1)(tr+t+1)}[ABC]$$ $[asy] unitsize(5); defaultpen(fontsize(10)); pair A,B,C,D,E,F,G,H,I; A=(10,20); B=(0,0); C=(30,0); D=(20,0); E=(16.66,13.33); F=(5,10); G=(14.585,11.6298); H=(9.998,8); I=(17.5,5); draw(A--B); draw(B--C); draw(C--A); draw(A--D); draw(B--E); draw(C--F); label(\"A\",A,N); label(\"B\",B,SW); label(\"C\",C,SE); label(\"D\",D,S); label(\"E\",E,NE); label(\"F\",F,NW); label(\"G\",G,N); label(\"H\",H,N); label(\"I\",I,SW);[/asy]$ ## Proof Assume triangle$ABC$'s area to be 1. We can then use Menelaus's Theorem on triangle $ABD$ and line $FHC$. $\\frac{AF}{FB}\\times\\frac{BC}{CD}\\times\\frac{DG}{GA}= 1$ This means $\\frac{DG}{GA}= \\frac{BF}{FA}\\times\\frac{DC}{CB} = \\frac{rs}{s+1}$", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle CBD = 5^{\\circ}$, and $\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw $E'$ such that $EE'B = 60^{\\circ}$ and so that $E'$ is on $\\overline{AE}$, and draw $E''$ such that $\\angle EE''C = 60^{\\circ}$ and $E''$ is on $\\overline{DE}$. It follows that $\\triangle BEE'$ and $\\triangle CEE''$ are both equilateral. Also, it is easy to see that $\\triangle BEC \\cong \\triangle DE''C$ and $\\triangle BCE \\cong \\triangle BAE'$ by construction, so that $DE'' = BE = EE'$ and $EE'' = CE = E'A$. Thus, $AE = AE' + E'E = EE'' + DE'' = DE$, so $\\triangle ADE$ is isosceles. Since $\\angle AED = 120^{\\circ}$, then $\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and $\\angle BAD = 30 + 55 = 85^{\\circ}$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf);", "D=C+a*dir(120); draw(A--B--C--D--cycle); pair P1=B+3*a*dir(145), P2=C+3*a*dir(205); pair P=intersectionpoint(B--P1,C--P2); draw(B--P--C); label(\"A\",A,SW); label(\"B\",B,SE); label(\"C\",C,NE); label(\"D\",D,N); label(\"P\",P,W); label(\"35^\\circ\",B + dir(180-17.5)); label(\"35^\\circ\",B + dir(180-35-17.5)); label(\"85^\\circ\",C + .5*dir(120+42.5)); label(\"85^\\circ\",C + .5*dir(120+85+42.5)); [/asy]$ ### Solution 2 Draw the diagonals $\\overline{BD}$ and $\\overline{AC}$, and suppose that they intersect at $E$. Then, $\\triangle ABC$ and $\\triangle BCD$ are both isosceles, so by angle-chasing, we find that $\\angle BAC = 55^{\\circ}$, $\\angle$CBD = 5^{\\circ}$, and$\\angle BEA = 180 - \\angle EBA - \\angle BAE = 60^{\\circ}$. Draw$E'$such that$EE'B = 60^{\\circ}$and so that$E'$is on$\\overline{AE}$, and draw$E$such that$\\angle EEC = 60^{\\circ}$and$E$is on$\\overline{DE}$. It follows that$\\triangle BEE'$and$\\triangle CEE$are both equilateral. Also, it is easy to see that$\\triangle BEC \\cong \\triangle DEC$and$\\triangle BCE \\cong \\triangle BAE'$by construction, so that$DE = BE = EE'$and$EE = CE = E'A$. Thus,$AE = AE' + E'E = EE + DE = DE$, so$\\triangle ADE$is isosceles. Since$\\angle AED = 120^{\\circ}$, then$\\angle DAC = \\frac{180 - 120}{2} = 30^{\\circ}$, and$\\angle BAD = 30 + 55 = 85^{\\circ}\\$. $[asy] import graph; size(6.73cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.237,xmax=2.492,ymin=-0.16,ymax=1.947; pen evefev=rgb(0.898,0.937,0.898), qqwuqq=rgb(0,0.392,0); filldraw(arc((1,0),0.141,115,175)--(1,0)--cycle,evefev,qqwuqq); filldraw(arc((0.658,0.94),0.051,175,235)--(0.658,0.94)--cycle,evefev,qqwuqq); draw((0,0)--(1,0)); draw((1,0)--(0.658,0.94)); draw((0.658,0.94)--(0.158,1.806)); draw((0.158,1.806)--(0,0)); draw((0,0)--(0.658,0.94)); draw((0.158,1.806)--(1,0)); draw((0.058,0.082)--(1,0)); draw((0.558,0.948)--(0.658,0.94)); dot((0,0),ds); label(\"A\",(-0.096,0.005),NE*lsf); dot((1,0),ds); label(\"B\",(1.117,0.028),NE*lsf); dot((0.658,0.94),ds); label(\"C\",(0.727,0.996),NE*lsf); dot((0.158,1.806),ds); label(\"D\",(0.187,1.914),NE*lsf); dot((0.6,0.857),ds); label(\"E\",(0.479,0.825),NE*lsf); dot((0.058,0.082),ds); label(\"E'\",(0.1,0.23),NE*lsf); label(\"60^\\circ\",(0.767,0.091),NE*lsf,qqwuqq); dot((0.558,0.948),ds); label(\"E''\",(0.423,0.957),NE*lsf); label(\"60^\\circ\",(0.761,0.886),NE*lsf,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$", "similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ### Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ### Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. ## Pythagorean", "the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 $[asy] unitsize(0.5cm); pair A,B,C,D,E; B=(0,0); C=(14,0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"A\",A,N); label(\"B\",B,S); label(\"C\",C,S); label(\"E\",E,NW); label(\"D\",D,NE); label(\"13\",A--B,NW); label(\"15\",A--C,NE); label(\"14\",B--C,S); label(\"6.5\",B--E,N); label(\"7\",C--D,N); [/asy]$ Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, then $BD=CD=7$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "(1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ We draw line $FD$ so that we can define a variable $x$ for the area of $\\triangle BEF = \\triangle DEF$. Knowing that $\\triangle ABE$ and $\\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$. Since we have a rule where 2 triangles, ($\\triangle A$ which has base $a$ and vertex $c$), and ($\\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\\frac{ \\triangle ABF}{\\triangle ACF} = \\frac{\\triangle DBF}{\\triangle DCF}$. Substituting $x$ into the equation we get: $$\\frac{x+60}{300-x} = \\frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\\triangle BEF=30.$ ~$\\bold{\\color{blue}{onionheadjr}}$ ## Video Solutions Associated video - https://www.youtube.com/watch?v=DMNbExrK2oo https://youtu.be/Ns34Jiq9ofc —DSA_Catachu https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution) ## Solution 15 (Straightforward Solution) $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D", "problem at hand. $[asy] size(170); pair C = (1, 3), A = (0,0), B = (1.7,0); real a = 0.5, b= 0.25, c = 0.25; pair P = a*A + b*B + c*C; pair D = b/(b+c)*B + c/(b+c)*C; pair EE = c/(c+a)*C + a/(c+a)*A; pair F = a/(a+b)*A + b/(a+b)*B; draw(A--B--C--cycle); draw(A--P); draw(B--P--C); draw(P--D, dotted); draw(EE--P--F, dotted); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"D\", D, NE); label(\"E\", EE, NW); label(\"F\", F, S); label(\"P\", P, E); [/asy]$ Since $AP = PD = 6$, this means that $[APB] + [APC] = [BPC]$; thus $\\triangle BPC$ has half the area of $\\triangle ABC$. And since $PE = 3 = \\dfrac{1}{3}BP$, we can conclude that $\\triangle APC$ has one third of the combined areas of triangle $BPC$ and $APB$, and thus $\\dfrac{1}{4}$ of the area of $\\triangle ABC$. This means that $\\triangle APB$ is left with $\\dfrac{1}{4}$ of the area of triangle $ABC$: $$[BPC]: [APC]: [APB] = 2:1:1.$$ Since $[APC] = [APB]$, and since $\\dfrac{[APC]}{[APB]} = \\dfrac{CD}{DB}$, this means that $D$ is the midpoint of $BC$. Furthermore, we know that $\\dfrac{CP}{PF} = \\dfrac{[APC] + [BPC]}{[APB]} = 3$, so $CP = \\dfrac{3}{4} \\cdot CF = 15$. We now apply Stewart's theorem to segment $PD$ in $\\triangle BPC$—or rather, the simplified version for a median. This tells us that $$2", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "+0 0 40 1 The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label(\"$W$\", A, N); label(\"$X$\", B, SW); label(\"$Y$\", C, SE); label(\"$Z$\", D, SW); [/asy] Oct 9, 2022", "# 2009 AIME II Problems/Problem 3 ## Problem In rectangle $ABCD$, $AB=100$. Let $E$ be the midpoint of $\\overline{AD}$. Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$. ## Solution ### Solution 1 $[asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10), Q=(0,5); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); pair F=(6.7,6.7); label(\"$$E$$\",E,N); label(\"$$A$$\",A,NW); label(\"$$B$$\",B,SW); label(\"$$C$$\",C,SE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,W); label(\"$$100$$\",Q,W); [/asy]$ From the problem, $AB=100$ and triangle $FBA$ is a right triangle. As $ABCD$ is a rectangle, triangles $BCA$, and $ABE$ are also right triangles. By $AA$, $\\triangle FBA \\sim \\triangle BCA$, and $\\triangle FBA \\sim \\triangle ABE$, so $\\triangle ABE \\sim \\triangle BCA$. This gives $\\frac {AE}{AB}= \\frac {AB}{BC}$. $AE=\\frac{AD}{2}$ and $BC=AD$, so $\\frac {AD}{2AB}= \\frac {AB}{AD}$, or $(AD)^2=2(AB)^2$, so $AD=AB \\sqrt{2}$, or $100 \\sqrt{2}$, so the answer is $\\boxed{141}$. ### Solution 2 Let $x$ be the ratio of $BC$ to $AB$. On the coordinate plane, plot $A=(0,0)$, $B=(100,0)$, $C=(100,100x)$, and $D=(0,100x)$. Then $E=(0,50x)$. Furthermore, the slope of $\\overline{AC}$ is $x$ and the slope of $\\overline{BE}$ is $-x/2$. They are perpendicular, so they multiply to $-1$, that is, $$x\\cdot-\\frac{x}{2}=-1,$$ which implies that $-x^2=-2$ or $x=\\sqrt 2$. Therefore $AD=100\\sqrt 2\\approx 141.42$ so $\\lfloor AD\\rfloor=\\boxed{141}$. ### Solution 3 Similarly to Solution 2, let the positive x-axis be in the direction of ray $BC$ and let", "ABE = 60$. Let $M$ be a point such $\\overline{EM}$ is parellel to $\\overline{CD}$. We immediatley know that $\\triangle BEM \\sim BDC$ by $2$. Using that we can conclude $EM$ has ratio $1$. Using $\\triangle EFM \\sim \\triangle AFC$, we get $EF:AE = 1:2$. Therefore using the fact that $\\triangle EBF$ is in $\\triangle ABF$, the area has ratio $\\triangle BEF : \\triangle ABE=1:2$ and we know $\\triangle ABE$ has area $60$ so $\\triangle BEF$ is $\\boxed{\\textbf{(B) }30}$. - fath2012 ## Solution 9 (Menelaus's Theorem) $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label(\"A\",A,N); dot(B); label(\"B\", B,SW);dot(C); label(\"C\",C,SE); dot(DD); label(\"D\",DD,NE); dot(EE); label(\"E\",EE,NW); dot(FF); label(\"F\",FF,S); [/asy]$ By Menelaus's Theorem on triangle $BCD$, we have $$\\dfrac{BF}{FC} \\cdot \\dfrac{CA}{DA} \\cdot \\dfrac{DE}{BE} = 3\\dfrac{BF}{FC} = 1 \\implies \\dfrac{BF}{FC} = \\dfrac13 \\implies \\dfrac{BF}{BC} = \\dfrac14.$$ Therefore, $$[EBF] = \\dfrac{BE}{BD}\\cdot\\dfrac{BF}{BC}\\cdot [BCD] = \\dfrac12 \\cdot \\dfrac 14 \\cdot \\left( \\dfrac23 \\cdot [ABC]\\right) = \\boxed{\\textbf{(B) }30}.$$ ## Solution 10 (Graph Paper) $[asy] unitsize(2cm); pair A,B,C,D,E,F,a,b,c,d,e,f; A = (2,3); B = (0,2); C = (2,0); D = (2/3)*A+(1/3)*C; E = (B+D)/2; F = intersectionpoint(B--C,A--A+2*(E-A)); a = (0,0); b = (1,0); c = (2,1); d = (1,3); e = (0,3); f = (0,1); draw(a--C,dashed); draw(f--c,dashed); draw(e--A,dashed); draw(a--e,dashed); draw(b--d,dashed); draw(A--B--C--cycle); draw(A--F); draw(B--D);", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "P, dir(20)); dot(\"Q\", Q, dir(150)); pair W, X, Y, Z; W = extension(A, P, D, C); X = extension(B, Q, C, D); Y = extension(C, Q, A, B); Z = extension(D, P, A, B); label(\"W\", W, dir(100)); label(\"X\", X, dir(60)); label(\"Y\", Y, dir(50)); label(\"Z\", Z, dir(140)); draw(A--W--X--B--Y--C--D--Z--B--C--Y--A--D); [/asy]$ Let $W, X, Y, Z = \\overline{AP} \\cap \\overline{CD}, \\overline{BQ} \\cap \\overline{CD}, \\overline{CQ} \\cap \\overline{AB}, \\overline{DP} \\cap \\overline{AB}$ respectively. Since $\\angle{DCQ} = \\angle{BCQ}, \\angle{CBQ} = \\angle{ABQ}$ we have $\\angle{QCB} + \\angle{CBQ} = 90 \\iff \\overline{BX} \\perp \\overline{CY};$ similarly we get $\\overline{AW} \\perp \\overline{DZ}.$ Thus, $\\overline{BQ}$ is both an angle bisector and altitude of $\\triangle{CBY}$ so $BC = BY.$ Using the same logic on $\\triangle{BCX}$ gives $BC = BX \\iff BYXC$ is a rhombus; similarly, $ADWZ$ is a rhombus. Then, $[ABQCDP] = [ABCD] - \\frac{1}{4}\\left([BYXC] + [ADWZ]\\right) = 15h - \\frac{1}{4}(7h + 12h) = 12h$ where $h$ is the height of trapezoid $ABCD.$ Finding $h$ is the same as finding the altitude to the side of length $8$ in a $5-7-8$ triangle, and using Heron's, the area of such a triangle is $\\sqrt{10(5)(3)(2)} = 10 \\sqrt{3} = 4h \\iff h = \\frac{5\\sqrt{3}}{2}.$ Multiply to get our answer is $\\boxed{\\mathrm{B}}.$ ## Solution 5 Like above solutions, find out the height of $ABCD$ is $\\frac{5 \\sqrt{3}}{2}.$ Let $M$ be the intersection of the", "and the area of $\\triangle BDC$ is $240$. Also using the fact that $E$ is the midpoint of $\\overline{BD}$, we know $\\triangle ADE = \\triangle ABE = 60$. Let $M$ be a point such $\\overline{EM}$ is parellel to $\\overline{CD}$. We immediatley know that $\\triangle BEM \\sim BDC$ by $2$. Using that we can conclude $EM$ has ratio $1$. Using $\\triangle EFM \\sim \\triangle AFC$, we get $EF:AE = 1:2$. Therefore using the fact that $\\triangle EBF$ is in $\\triangle ABF$, the area has ratio $\\triangle BEF : \\triangle ABE=1:2$ and we know $\\triangle ABE$ has area $60$ so $\\triangle BEF$ is $\\boxed{\\textbf{B} \\, 30}$. - fath2012 ## Solution 9 $[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(F--D); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"x\", (D+E+F)/3); label(\"x\", (B+E+F)/3); label(\"120+2x\", (D+F+C)/3); [/asy]$ Labeling the areas in the diagram, we have: $[DBC]=240=[BFE]+[FED]+[FDC]=x+x+120+2x=120+4x$ so $240=120+4x, 120=4x, 30=x$. So our answer is $\\boxed{\\textbf{(B)} 30}$. ~~RWhite ## Solution 10 (Menelaus's Theorem) $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF", "dot(A); label(\"A\",A,NE); dot(B); label(\"B\",B,dir(180)); dot(C); label(\"C\",C,SE); dot(D); label(\"D\",D,dir(0)); dot(E); label(\"E\",E,SE); dot(F); label(\"F\",F,SW); [/asy]$ Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper. As triangle $ABC$ is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles. As point $D$ splits line segment $\\overline{AC}$ in a $1:2$ ratio, we draw $\\overline{AC}$ as a vertical line segment $3$ units long. Point $D$ is thus $1$ unit below point $A$ and $2$ units above point $C$. By definition, Point $E$ splits line segment $\\overline{BD}$ in a $1:1$ ratio, so we draw $\\overline{BD}$ $2$ units long directly left of $D$ and draw $E$ directly between $B$ and $D$, $1$ unit away from both. We then draw line segments $\\overline{AB}$ and $\\overline{BC}$. We can easily tell that triangle $ABC$ occupies $3$ square units of space. Constructing line $AE$ and drawing $F$ at the intersection of $AE$ and $BC$, we can easily see that triangle $EBF$ forms a right triangle occupying $\\frac{1}{4}$ of a square unit of space. The ratio of the areas of triangle $EBF$ and triangle $ABC$ is thus", "of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2,", "## anonymous one year ago In square $ABCD$, $E$ is the midpoint of $\\overline{BC}$, and $F$ is the midpoint of $\\overline{CD}$. Let $G$ be the intersection of $\\overline{AE}$ and $\\overline{BF}$. Prove that $DG = AB$. • This Question is Open 1. anonymous Asymptote code below [asy] unitsize(2.5 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (0,1); C = (1,1); D = (1,0); E = (B + C)/2; F = (C + D)/2; G = extension(A,E,B,F); draw(A--B--C--D--cycle); draw(A--E); draw(B--F); draw(D--G); label(\"$A$\", A, SW); label(\"$B$\", B, NW); label(\"$C$\", C, NE); label(\"$D$\", D, SE); label(\"$E$\", E, N); label(\"$F$\", F, dir(0)); label(\"$G$\", G, WSW); [/asy]" ]

[ "# Help reducing the following expression involving Heron's formula for the area of a triangle. Let $(ABC) = \\sqrt{s \\cdot (s - a) \\cdot (s - b) \\cdot (s - c).}$ This is Heron's formula for $(ABC),$ the area of $\\Delta ABC,$ with sides $a, b, c,$ using $s = \\frac{a + b + c}{2},$ the semiperimeter. I have the following step that I don't understand in a problem I am working using the above notation. $$(ABC) \\cdot (\\frac{1}{s - a} + \\frac{1}{s - b} + \\frac{1}{s - c} - \\frac{1}{s}) =$$ $$\\frac{(ABC) \\cdot abc}{s \\cdot (s - a) \\cdot (s - b) \\cdot (s - c)}$$ I realize it's probably \"just\" algebra and suspect that the reduction uses the fact that $s - b = \\frac{a - b + c}{2},$ and similarly for $s - a$ and $s - c.$ But all my attempts to show the reduction (as given in a solution) have gotten complicated and I haven't been able to derive the result. For context, the problem is 3.2.4 in Coxeter and Greitzer's book Geometry Revisited: In the notation of Section 1.4, $$r_a + r_b + r_c - r = 4R,$$ where $r_a, r_b, r_c$ are the excircle radii of $\\Delta ABC,$ and $r, R$ are the incircle and circumcircle radii of $\\Delta ABC,$ respectively.", "# In triangle ABC, A=31.4, B=53.7, <C=61.3°, how do you find the area? May 28, 2015 First you can use The Law of Cosines to evaluate the length of the third side $C$: ${C}^{2} = {A}^{2} + {B}^{2} - 2 A B \\cos \\gamma$ ${C}^{2} = {31.4}^{2} + {53.7}^{2} - 2 \\cdot 31.4 \\cdot 53.7 \\cdot \\cos {61.3}^{\\circ}$ ${C}^{2} = 985.96 + 2883.69 - 3372.36 \\cdot 0.4802$ ${C}^{2} = 2250.2427$ $C = 47.44$ Now, for the area of the triangle we can use the Heron's formula: $P = \\sqrt{p \\left(p - a\\right) \\left(p - b\\right) \\left(p - c\\right)}$ where $p = \\frac{a + b + c}{2}$ is half of the circumference. $a = 31.4 , b = 53.7 , c = 47.44 , p = 66.27$ $P = \\sqrt{66.27 \\cdot 34.87 \\cdot 12.57 \\cdot 18.83} = \\sqrt{546958.6761}$ $P = 739.57$", "ABC$ with cevian $\\overline{AD}$, we have $$(3m)^2\\cdot 20 + 20\\cdot10\\cdot10 = 10^2\\cdot10 + (30 - 2c)^2\\cdot 10.$$ Our earlier result from Power of a Point was that $2m^2 = (10 - c)^2$, so we combine these two results to solve for $c$ and we get $$9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \\quad \\Longrightarrow \\quad c^2 - 12c + 20 = 0.$$ Thus $c = 2$ or $= 10$. We discard the value $c = 10$ as extraneous (it gives us an equilateral triangle) and are left with $c = 2$, so our triangle has area $\\sqrt{28 \\cdot 18 \\cdot 8 \\cdot 2} = 24\\sqrt{14}$ and so the answer is $24 + 14 = \\boxed{038}$.", "\\cdot CX \\cdot \\cos(150) = 3^2+4^2-24 \\cdot \\cos(150) = 25+12\\sqrt{3}$$ and thus the area of $ABC$ equals $$BC^2\\frac{\\sqrt{3}}{4} = 25\\frac{\\sqrt{3}}{4}+9.$$ so our final answer is $3+4+25+9 = \\boxed{041.}$", "\\cdot 2$ $z^2+(r-3)^2=r^2 \\longrightarrow z^2=(2r-3) \\cdot 3$ [The relationship between the half-sides, the extension of the radius beyond the triangle, and the radius is another pretty pattern.] That gives 3 equations in 4 variables. I needed one more to solve…. The area of $\\Delta ABC$ can be expressed two ways: as the sum of the areas of the isosceles triangles, and using Heron’s formula. From the areas of the isosceles triangles, $Area( \\Delta ABC) = \\frac{1}{2}(2x)(r-1) + \\frac{1}{2}(2y)(r-2) + \\frac{1}{2}(2z)(r-3)$ $Area( \\Delta ABC) = x \\cdot (r-1) + y \\cdot (r-2) + z \\cdot (r-3)$ The sides of $\\Delta ABC$ are 2x, 2y, and 2z, so its semiperimeter is x+y+z and Heron’s formula gives its area as $Area( \\Delta ABC) = \\sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)}$. The area of a given triangle is unique, so the two different area expressions are equivalent, giving a fourth equation. SOLVING A SYSTEM & ANSWERING THE QUESTION With four equations in four variables, I had a system of equations. The algebra was messy, so I invoked my CAS to crunch it for me. The question asked for the area of the triangle, so I just substituted my values back into the area formulas. And 17.186… is clearly not one of the choices in the original problem. A PLEA… Recognizing the perpendicular bisectors, seeing all the right", "# Equilateral triangle ABC has side length of 1, and squares ABDE, BCHI, CAFG lie outside the triangle. What is the area of hexagon DEFGHI? Dec 17, 2017 $\\left(\\sqrt{3} + 3\\right) \\text{ sq.unit}$. #### Explanation: We will use, to find the Area of $\\Delta A B C$, the Formula, $\\text{Area of \"DeltaABC=} \\frac{1}{2} \\cdot A B \\cdot A C \\cdot \\sin \\angle B A C .$ Observe that ${\\Delta}^{s} A F E , B D I , \\mathmr{and} C H G$ are all congruent, so, they all have the same Area, $= \\frac{1}{2} \\cdot A F \\cdot A E \\cdot \\sin \\angle E A F ,$ $= \\frac{1}{2} \\cdot 1 \\cdot 1 \\cdot \\sin \\left\\{{360}^{\\circ} - \\left({90}^{\\circ} + {90}^{\\circ} + {60}^{\\circ}\\right)\\right\\} ,$ $= \\frac{1}{2} \\cdot \\sin {120}^{\\circ}$, $= \\frac{1}{2} \\cdot \\sin \\left({180}^{\\circ} - {60}^{\\circ}\\right)$, $= \\frac{1}{2} \\cdot \\sin {60}^{\\circ}$, $= \\frac{1}{2} \\cdot \\frac{\\sqrt{3}}{2} ,$ $= \\frac{\\sqrt{3}}{4.}$ Also, Area of the equilateral $\\Delta A B C = \\frac{1}{2} \\cdot 1 \\cdot 1 \\cdot \\sin {60}^{\\circ} = \\frac{\\sqrt{3}}{4}$. $\\text{Area of each square=1}$. Hence, The Area of the Hexagon $D E F G H I$, $= 3 \\times \\frac{\\sqrt{3}}{4} + \\frac{\\sqrt{3}}{4} + 3 \\times 1$, $= \\left(\\sqrt{3} + 3\\right) \\text{ sq.unit}$. Dec 17, 2017 Area of hexagon$=$Area of $\\Delta A B C +$Area of three squares$+$Area of three triangles $E A", "# A Flock Of Herons Geometry Level 5 Triangles $\\Delta ABC$ and $\\Delta XYZ$ are Heronian triangles, which are triangles that have integer sides with integer areas. Points $A,B,C$ are centers of circles, the radii of which are Radius of circle $A=1092=2^2\\cdot 3\\cdot 7\\cdot 13$ Radius of circle $B=608=2^5\\cdot 19$ Radius of circle $C=1140=2^2\\cdot 3\\cdot 5\\cdot 19$ The sides of triangle $\\Delta ABC$ are $AB=1700=2^2\\cdot 5^2\\cdot 17$ $BC=5491=17^2\\cdot 19$ $CA=4437=3^2\\cdot 17\\cdot 29$ Triangle $\\Delta XYZ$ has the maximum area of any triangle (Heronian or not) that has one vertex on the circumference of each of the circles $A,B,C$ Let $\\dfrac { \\Delta XYZ }{ \\Delta ABC } =\\dfrac { p }{ q }$ be the ratio of the areas of triangles $\\Delta XYZ$ and $\\displaystyle \\Delta ABC$ where $p,q$ are $4$ digit co-prime integers. The difference $p-q$ is a $4$ digit prime number. Find that prime number. For your convenience, area of triangle $\\displaystyle \\Delta ABC$ works out to $3261654=2\\cdot 3^3\\cdot 11\\cdot 17^2\\cdot 19$ Note: Triangle $\\Delta XYZ$ as drawn in graphic does not have the maximum area Also, as you solve this, you'll come across more Herons. A helpful hint, I hope. ×", "perimeter was 27, and I could use Heron’s Formula to define its area: $Area(\\Delta EFC) = \\sqrt{13.5(1.5)(13.5-x)(x-1.5)}$ But I didn’t know the exact area, so that was a dead end. Since $\\Delta ABC$ is equilateral, $m \\angle C=60^{\\circ}$ , I then thought about expressing the area using trigonometry. With trig, the area of a triangle is half the product of any two sides multiplied by the sine of the contained angle. That meant $Area(\\Delta EFC) = \\frac{1}{2} \\cdot 12 \\cdot (15-x) \\cdot sin(60^{\\circ}) = 3(15-x) \\sqrt3$. Now I had two expressions for the same area, so I could solve for x. $3\\sqrt{3}(15-x) = \\sqrt{13.5(1.5)(13.5-x)(x-1.5)}$ Squaring both sides revealed a quadratic in x. I could do this algebra, if necessary, but this was clearly a CAS moment. I had two solutions, but this felt WAY too complicated. Also, Five Triangles problems are generally accessible to middle school students. The trigonometric form of a triangle’s area is not standard middle school fare. There had to be an easier way. A Quicker Ending Thinking trig opened me up to angle measures. If I let $m \\angle CEF = \\theta$, then $m \\angle EFC = 120^{\\circ}-\\theta$, making $m \\angle DFB = \\theta$, and I suddenly had my simple breakthrough! Because their angles were congruent, I knew $\\Delta CEF \\sim \\Delta BFD$. Because", "will be easy. The area of the triangle ABC can be calculated by Heron's formula: So, the semi-perimeter, where $a = 3 cm,\\ b =4cm\\ and\\ c = 5cm.$ $s = \\frac{a+b+c}{2} = \\frac{3+4+5}{2} = 6cm$ Heron's Formula for calculating the area: $A = \\sqrt{s(s-a)(s-b)(s-c)}$ $= \\sqrt{6(6-3)(6-4)(6-5)}= \\sqrt{6(3)(2)(1)} = \\sqrt{36} = 6\\ cm^2$ And the sides of the triangle ACD are $a' =4cm,\\ b' = 5cm\\ and\\ c' = 5cm.$ So, the semi-perimeter of the triangle: $s' = \\frac{a'+b'+c'}{2} = \\frac{4+5+5}{2} = \\frac{14}{2} = 7cm$ Therefore, the area will be given by, Heron's formula $A = \\sqrt{s'(s'-a')(s'-b')(s'-c')}$. $= \\sqrt{7(7-4)(7-5)(7-5)}$ $= \\sqrt{7(3)(2)(2)} = 2\\sqrt{21} = 9.2\\ cm^2\\ \\ \\ \\ (Approx.)$ Then, the total area of the quadrilateral will be: $= Area\\ of\\ triangle\\ ABC + Area\\ of\\ triangle\\ ACD$ $\\Rightarrow$ $Total\\ area\\ of\\ quadrilateral\\ ABCD = 6+9.2 = 15.2\\ cm^2$ Hence, the area of the quadrilateral ABCD is $15.2 \\ cm^2.$ The total area of the paper used will be the sum of the area of the sections I, II, III, IV, and V. i.e., $Total\\ area = I +II+III+IV+V$ For section I: Here, the sides are $a = 1cm\\ and\\ b =c = 5cm.$ So, the Semi-perimeter will be: $s = \\frac{a+b+c}{2} = \\frac{5+5+1}{2} = 5.5\\ cm$ Therefore, the area of section I will be given by", "area of ABC can be found from: $3A=\\frac{1}{2}bc\\sin(A)+\\frac{1}{2}ac\\sin(B)+\\frac {1}{2}ab\\sin(C)$ and so: $A=\\frac{1}{6}\\left(bc\\sin(A)+ac\\sin(B)+ab\\sin(C) \\right)$ For the 3 triangles outside of RST, I used the same formula: $A_O=\\frac{1}{2}\\cdot\\frac{2}{3}b\\cdot\\frac{1}{3}c \\sin(A)+\\frac{1}{2}\\cdot\\frac{1}{3}a\\cdot\\frac{2}{ 3}c\\sin(B)+\\frac{1}{2}\\cdot\\frac{2}{3}a\\cdot\\frac{ 1}{3}b\\sin(C)$ $A_O=\\frac{1}{9}\\left(bc\\sin(A)+ac\\sin(B)+ab\\sin(C) \\right)$", "# How would I find the area of a triangle given three sides and using either the sine/cosine laws? Triangle ABC has sides $8.5m$ (a), $7.1$ (b), and $9$ (c). I have been asked to find the area of the triangle using trigonometry. - Hint: Heron's Formula. You can also use the Sine Formula if you want sines and cosines. Regards – Amzoti Jan 21 '13 at 2:55 If you must use Trigonometry, we can use this formula: $$K = \\frac{1}{2} ab \\sin C$$ In order to find the $\\sin C$, we can use the law of cosines: $$c^2 = a^2 + b^2 - 2ab \\cos C$$ $$9^2 = 8.5^2 + 7.1^2 - 2 \\cdot 8.5 \\cdot 7.1 \\cos C$$ $$-41.66 = -120.7 \\cos C$$ $$0.34515327257 = \\cos C$$ $$C \\approx 69.80^{\\circ}$$ Now, we have: $$K = \\frac{1}{2} \\cdot 8.5 \\cdot 7.1 \\sin 69.8$$ $$K \\approx 30.175 \\sin 69.8$$ $$\\color{green}{K \\approx 28.32}$$ If you only have a scientific calculator, you can also avoid the inverse cosine by using: $$\\sin x = \\sqrt{1 - \\cos^2 x}$$ So: $$K = \\frac{1}{2} \\cdot 8.5 \\cdot 7.1 \\sqrt{1 - 0.34515327257^2}$$ $$K = \\frac{1}{2} \\cdot 8.5 \\cdot 7.1 \\cdot 0.9385463325985667$$ $$K = 28.3206355862$$ Applying Heron's Formula verifies this result. We have $s = 12.3$. $$\\sqrt{12.3(12.3 - 8.5)(12.3 - 7.1)(12.3 - 9)}$$ $$\\sqrt{802.058} \\approx", "\\triangle BDF}$ The area of $$\\triangle ADE$$ is $$\\frac{1}{2} \\cdot y \\cdot (x\\cdot \\sin(A)) = \\frac{1}{2} \\cdot x\\cdot y \\cdot \\sin(A)$$. The area of $$\\triangle BDF$$ is $\\frac{1}{2} \\cdot (2 y) \\cdot (z\\cdot \\sin(B)) = \\frac{1}{2} \\cdot y \\cdot (2z)\\cdot \\sin(B) = \\frac{1}{2}\\cdot y \\cdot (x\\cdot \\sin(A)) = \\underbrace{\\frac{1}{2}\\cdot x \\cdot y \\cdot \\sin(A)}_{\\text{area of } \\triangle ADE} .$ Also, $\\frac{3y}{\\sin(C)} = \\frac{2x}{\\sin(B)} = \\frac{4z}{\\sin(A)} \\Longrightarrow \\sin(C) = \\frac{3y\\sin(B)}{2x} = \\frac{3y\\sin(A)}{4z}$ The area of $$\\triangle CEF$$ is $\\frac{1}{2} \\cdot x \\cdot [ (3z)\\cdot \\sin(180^\\circ - C)] = \\frac{1}{2} \\cdot x \\cdot [ (3z)\\cdot \\sin(C) ] = \\frac{1}{2} \\cdot x \\cdot 3z \\cdot \\frac{3y}{4z}\\cdot \\sin(A) = \\underbrace{\\frac{1}{2}\\cdot x \\cdot y \\cdot \\sin(A)}_{\\text{area of } \\triangle ADE} \\cdot \\frac{9}{4}.$ So the area of $$\\triangle CEF$$ is $$\\frac{9}{4}$$-th the area of $$\\triangle ADE$$. Lastly, the area of $$\\triangle ABC$$ gives us the following identity: $24 = \\frac{1}{2} \\cdot (3y) \\cdot [ 2x\\cdot \\sin(A) ]$ which implies the area of $$\\triangle ADE$$ is $\\frac{1}{2}\\cdot x\\cdot y \\cdot \\sin(A) = 4$ It should be fairly straightforward to deduce that the area of $$\\triangle DEF$$ is $24 - (4+4+9) = 7$ This analytic solution is really the same as the geometric solution I posted earlier ( http://www.hpmuseum.org/forum/thread-795...l#pid71463 ) Graph 3D | QPI | SolveSys 04-08-2017, 03:37 AM (This post was last modified: 04-08-2017 03:37", "# How do you use Heron's formula to determine the area of a triangle with sides of that are 35, 28, and 41 units in length? Apr 17, 2018 $483$ units squared #### Explanation: Heron's formula states that, $A = \\sqrt{s \\left(s - a\\right) \\left(s - b\\right) \\left(s - c\\right)}$ • $s$ is the semiperimeter of the triangle, given by $s = \\frac{a + b + c}{2}$. • $a , b , c$ are the sides of the triangle Let $a = 35 , b = 28 , c = 41$ $\\therefore s = \\frac{35 + 28 + 41}{2} = \\frac{104}{2} = 52$ So, the area of this triangle will be: $A = \\sqrt{52 \\left(52 - 35\\right) \\left(52 - 28\\right) \\left(52 - 41\\right)}$ $= \\sqrt{52 \\cdot 17 \\cdot 24 \\cdot 11}$ $= \\sqrt{233376}$ $\\approx 483$ So, the area of the triangle will be $483$ units squared.", "into four triangles. The area of $\\triangle ABC$ is therefore $12 - (\\frac{4 \\cdot 2}{2}+\\frac{3 \\cdot 1}{2}+\\frac{3 \\cdot 1}{2}) = 12 - (4 + \\frac{3}{2} + \\frac{3}{2}) = 12 - 7 = 5$. That means that $\\triangle ABC$ takes up $\\frac{5}{30} = \\boxed{\\textbf{(A)}~\\frac{1}{6}}$ of the grid. ### Solution 5 Using Pick's Theorem, the area of the triangle is $4 + \\dfrac{4}{2} - 1=5$. Therefore, the triangle takes up $\\dfrac{5}{30}=\\boxed{\\textbf{(A)}~\\frac{1}{6}}$ of the grid. ### Solution 6 (Heron's Formula, Not Recommended) We can find the lengths of the sides by using the Pythagorean Theorem. Then, we apply Heron's Formula to find the area. $$\\sqrt{\\left(\\frac{\\sqrt{10}+\\sqrt{10}+2\\sqrt{5}}{2}\\right)\\left(\\frac{\\sqrt{10}+\\sqrt{10}+2\\sqrt{5}}{2}-\\sqrt{10}\\right)\\left(\\frac{\\sqrt{10}+\\sqrt{10}+2\\sqrt{5}}{2}-\\sqrt{10}\\right)\\left(\\frac{\\sqrt{10}+\\sqrt{10}+2\\sqrt{5}}{2}-2\\sqrt{5}\\right)}.$$ This simplifies to $$\\sqrt{\\left(\\sqrt{10}+\\sqrt{5}\\right)\\left(\\sqrt{10}+\\sqrt{5}-\\sqrt{10}\\right)\\left(\\sqrt{10}+\\sqrt{5}-\\sqrt{10}\\right)\\left(\\sqrt{10}+\\sqrt{5}-2\\sqrt{5}\\right)}.$$ Again, we simplify to get $$\\sqrt{\\left(\\sqrt{10}+\\sqrt{5}\\right)\\left(\\sqrt{5}\\right)\\left(\\sqrt{5}\\right)\\left(\\sqrt{10}-\\sqrt{5}\\right)}.$$ The middle two terms inside the square root multiply to $5$, and the first and last terms inside the square root multiply to $\\sqrt{10}^2-\\sqrt{5}^2=10-5=5.$ This means that the area of the triangle is $$\\sqrt{5\\cdot 5}=5.$$ The area of the grid is $6\\cdot 5=30.$ Thus, the answer is $\\frac{5}{30}=\\boxed{\\textbf{(A) }\\frac{1}{6}}$. -BorealBear ~savannahsolver", "the sum of the areas is $3\\cdot \\frac{\\sqrt{3}}{32}=\\frac{3\\sqrt{3}}{32}$ Therefore, the area of the convex hexagon is $\\frac{9\\sqrt{3}}{16}-\\frac{3\\sqrt{3}}{32}=\\frac{18\\sqrt{3}}{32}-\\frac{3\\sqrt{3}}{32}=\\boxed{\\frac{15\\sqrt{3}}{32}}\\implies \\boxed{C}$ ## Solution 5 Dividing $\\triangle MNO$ into two right triangles congruent to $\\triangle PMN$, we see that $[MPNQOR]=\\dfrac{5}{8}[ACE]$. Because $[ACE] = \\dfrac{1}{2}[ABCDEF]$, we have $[MPNQOR]=\\dfrac{5}{16}[ABCDEF]$. From here, you should be able to tell that the answer will have a factor of $5$, and $\\boxed{\\textbf{(C)} \\frac {15}{32}\\sqrt{3}}$ is the only answer that has a factor of $5$. However, if you want to actually calculate the area, you would calculate $[ABCDEF]$ to be $6 \\cdot \\dfrac{\\sqrt{3}}{2 \\cdot 2} = \\dfrac{3\\sqrt{3}}{2}$, so $[MPNQOR] = \\dfrac{5}{16} \\cdot \\dfrac{3\\sqrt{3}}{2} = \\boxed{\\textbf{(C)} \\frac {15}{32}\\sqrt{3}}$.", "- \\angle EOA = \\angle DOE$ $60^\\circ - 30^\\circ = \\angle DOE = 30^\\circ$ So, the area of sector $DOE$ is $\\frac{30}{360} \\cdot \\pi \\cdot 2^2 = \\frac{\\pi}{3}$. The area of triangle $DOE$ is $\\frac{1}{2}\\cdot 2 \\cdot 2 \\cdot \\sin 30^\\circ = 1$. Since $AB=CB=1$ , $DB=ED=(\\sqrt{3}-1)$. So, the area of triangle $DBE$ is $\\frac{1}{2} \\cdot (\\sqrt{3}-1)^2 = 2-\\sqrt{3}$. Therefore, the shaded area is $(\\frac{\\pi}{3}) - (1) + (2-\\sqrt{3}) = \\frac{\\pi}{3}+1-\\sqrt{3} \\Longrightarrow \\boxed{\\mathrm{(A)}}$ Non-Trig Approach $\\triangle{ODA}$ has the same height as $\\triangle{OBD}$ which is $1.$ We already know that $BD = \\sqrt{3} - 1.$ Therefore the area is $(\\sqrt{3}-1) \\cdot 1 \\cdot \\frac{1}{2} = \\frac{\\sqrt{3}-1}{2}.$ Since $\\triangle{ODB} = \\triangle{OBE} = \\frac{\\sqrt{3}-1}{2}.$ Therefore the sum of the areas is $2 \\cdot \\frac{\\sqrt{3}-1}{2} = \\sqrt{3}-1.$ Then the area of the shaded area becomes $\\frac{\\pi}{3} - (\\sqrt{3} - 1) = \\boxed{\\textbf{(A)}\\frac{\\pi}{3} - \\sqrt{3} + 1}.$ ~mathboy282 Solution 2 From the pythagorean theorem, we can see that $DA$ is $\\sqrt{3}$. Then, $DB = DA - BA = \\sqrt{3} - 1$. The area of the shaded element is the area of sector $DOE$ minus the areas of triangle $DBO$ and triangle $EBO$ combined. Below is an image to help. $[asy] defaultpen(linewidth(0.8)); pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B; fill(Arc(O, 2, 0, 90)--O--cycle, mediumgray); clip(B--Arc(O, 2, 30, 60)--cycle); draw(Circle(origin, 2)); draw((-2,0)--(2,0)^^(0,-2)--(0,2));", "have, $${BD}^2 = {BC}^2 + {CD}^2$$ ...(By Pythagoras theorem) i.e., = $${12}^2 + {5}^2 = 144 + 25 = 169$$ i.e., $${BD}^2 = {13}^2$$ i.e., BD = 13m Also, In $$\\triangle{ABD}$$, we have, AB = 9m, BD = 13m, DA = 8m Now, we know that, $$s = \\frac{a + b + c}{2}$$ Therefore, $$s = \\frac{9 + 13 + 8}{2} = \\frac{30}{2} = 15m$$ Now, Area of triangle = $$\\sqrt{15(15 - 9)(15 - 13)(15 - 8)}$$ ...(Since, Heron's formula [area = $$\\sqrt{s(s - a)(s - b)(s - c)}$$]) = $$\\sqrt{15 × 6 × 2 × 7}$$ = $$\\sqrt{3 × 5 × 3 × 2 × 2 × 7}$$ = $$3 × 2\\sqrt{5 × 7} {m}^2$$ = $$6\\sqrt{35} {m}^2$$ = 6 × 5.9$${m}^2$$ = 35.4$${m}^2$$(approx) ...(i) Since, $$\\triangle{BCD}$$ is an right angled triangle, Area of $$\\triangle{BCD}$$ = $$frac{1}{2}$$ × BC × CD ...(Since, Area of triangle = $$frac{1}{2}$$ × Base × Height) = $$frac{1}{2} × 12 × 5 = 30{m}^2$$ ...(ii) Area of quadrilateral ABCD = Area of $$\\triangle{ABD}$$ + Area of $$\\triangle{BCD}$$ Hence from (i) and (ii), Area of quadrilateral ABCD = $$35.4{m}^2 + 30{m}^2 = 65.4{m}^2$$ 2. Find the area of a quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 4cm, DA = 5cm and AC = 5cm. Open in new tab", "$ABC$. Note that the area of $\\triangle ADE$ is equal to $\\frac{1}{2} \\cdot 25 \\cdot 5 \\cdot \\sin{A}$ and the area of triangle $ABC$ is equal to $\\frac{1}{2} \\cdot 50 \\cdot 10 \\cdot \\sin A$. The ratio $\\frac{[ADE]}{[ABC]}$ is thus equal to $\\frac{1}{4}$ and the area of triangle $ADE$ is $\\frac{1}{4} \\cdot 120 = 30$. Let side $BC$ be equal to $6x$, then $BG = 5x, GC = x$ by the angle bisector theorem. Similarly, we find the area of triangle $AGC$ to be $\\frac{1}{2} \\cdot 10 \\cdot x \\cdot \\sin C$ and the area of triangle $ABC$ to be $\\frac{1}{2} \\cdot 6x \\cdot 10 \\cdot \\sin C$. A ratio between these two triangles yields $\\frac{[ACG]}{[ABC]} = \\frac{x}{6x} = \\frac{1}{6}$, so $[AGC] = 20$. Now we just need to find the area of triangle $AFE$ and subtract it from the combined areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are", "## College Algebra 7th Edition $a.\\quad$ Please see \"step-by-step\" $b.\\quad$ Area = $10$ a. Reading the ccoordinates from the graph, we have $A=(2,2), B=(3, -1)$, and $C=(-3, -3)$. Calculate the sides' lengths: $d(A, B)=\\sqrt{(3-2)^{2}+(-1-2)^{2}}=\\sqrt{1^{2}+(-3)^{2}}=\\sqrt{1+9}=\\sqrt{10}.$ $d(C, B)=\\sqrt{(3-(-3))^{2}+(-1-(-3))^{2}}=\\sqrt{6^{2}+2^{2}}=\\sqrt{36+4}\\\\=\\sqrt{40}=2\\sqrt{10}.$ $d(A, C)=\\sqrt{(-3-2)^{2}+(-3-2)^{2}}=\\sqrt{(-5)^{2}+(-5)^{2}}\\\\=\\sqrt{25+25}=\\sqrt{50}=5\\sqrt{2}$. Note that $[d(A, B)]^{2}+[d(C, B)]^{2}=10+4\\cdot 10=50$ $[d(A, C)]^{2}=25\\cdot 2=50,$ so, $[d(A, B)]^{2}+[d(C, B)]^{2}=[d(A, C)]^{2}$. By the converse of the Pythagorean Theorem, we conclude that the triangle is a right triangle. b. The area of the triangle is $\\displaystyle \\frac{1}{2}\\cdot d(C, B)\\cdot d(A, B)=\\frac{1}{2}\\cdot\\sqrt{10}\\cdot 2\\sqrt{10}=10$.", "+0 # Help pls 0 236 2 Find the inradius of triangle ABC if it's sides lengths are 29, 29, 40. Jan 29, 2022 #1 +1224 +2 The inradius can be found using the formula A = rs, where A is the triangle's area, r is the inradius, and s is the triangle's semiperimeter. $$s= \\frac{29+29+40}{2} = 49$$ $$A = \\sqrt{s(s-a)(s-b)(s-c)} = \\sqrt{(49)(20)(20)(9)} = 420$$ $$\\rightarrow r = \\frac{A}{s} = \\frac{420}{49} = \\boxed{\\frac{60}{7}}$$ Jan 29, 2022 #2 +13890 +2 Find the inradius of triangle ABC if it's sides lengths are 29, 29, 40. Hello Guest! a = b = 29 c = 40 $$cos(\\alpha)=\\frac{c}{2\\cdot b}=\\frac{40}{2\\cdot 29 }\\\\ \\alpha =46.397^o\\\\ f(x)=tan( \\frac{\\alpha}{2})x\\\\ x=\\frac{c}{2}=20\\\\ r=tan( \\frac{\\alpha}{2})\\cdot x=tan(\\frac{46.397^o}{2})\\cdot 20$$ $$r=8.\\overline{571428}=\\large \\frac{60}{7}$$ ! Jan 30, 2022" ]

[ "the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 $[asy] unitsize(0.5cm); pair A,B,C,D,E; B=(0,0); C=(14,0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"A\",A,N); label(\"B\",B,S); label(\"C\",C,S); label(\"E\",E,NW); label(\"D\",D,NE); label(\"13\",A--B,NW); label(\"15\",A--C,NE); label(\"14\",B--C,S); label(\"6.5\",B--E,N); label(\"7\",C--D,N); [/asy]$ Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, then $BD=CD=7$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "area of a triangle equals $\\frac{1}{2} \\cdot ab \\cdot \\sin(C)$. If angle $DAE = Z$, we have that $$\\frac{[ADE]}{[ABC]} = \\frac{\\frac{1}{2} \\cdot 19 \\cdot 14 \\cdot \\sin(Z)}{\\frac{1}{2} \\cdot 25 \\cdot 42 \\cdot \\sin(Z)} = \\frac{19 \\cdot 14}{25 \\cdot 42} = \\frac{ab}{cd}$$. - SuperJJ ## Video Solution CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00 ## Solution 2 (no trig) $[asy] unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label(\"A\", A, N); label(\"B\", B, SW); label(\"C\", C, SE); label(\"D\", D, W); label(\"E\", E, NE); label(\"19\", (A + D)/2, W); label(\"6\", (B + D)/2, W); label(\"14\", (A + E)/2, NE); label(\"28\", (C + E)/2, NE); [/asy]$ We can let $[ADE]=x$. Since $EC=2 \\cdot EA$, $[DEC]=2x$. So, $[ADC]=3x$. This means that $[BDC]=\\frac{6}{19}\\cdot3x=\\frac{18x}{19}$. Thus, $\\frac{[ADE]}{[BCED]} = \\frac{x}{\\frac{18x}{19}+2x}= \\boxed{\\textbf{(D) }\\frac{19}{56}}.$ -Conantwiz2023 ## Solution 3 (trig) The area of a triangle is $\\frac{1}{2} bc\\sin A$. Using this formula: $[ADE]=\\frac{1}{2}\\cdot19\\cdot14\\cdot\\sin A = 133\\sin A$ $[ABC]=\\frac{1}{2}\\cdot25\\cdot42\\cdot\\sin A = 525\\sin A$ Since the area of $BCED$ is equal to the area of $ABC$ minus the area of $ADE$, $[BCED] = 525\\sin A - 133\\sin A = 392\\sin A$. Therefore, the desired ratio is $\\frac{133\\sin A}{392\\sin A}=\\boxed{\\textbf{(D) }\\frac{19}{56}}$ Note: $BC=39$ was not used in this problem. ## Solution 4 Let", "get $$b^2+9a^2-2\\cdot 3a\\cdot b\\cdot \\cos(\\angle ADC)=7\\qquad (5)$$ $$b^2+a^2+2\\cdot a\\cdot b\\cdot \\cos(\\angle ADC)=9.\\qquad (6)$$ $(5)+(6)$, and according to $(4)$, we can get $$37a^2+2b^2=48. \\qquad (7)$$ Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\\frac{\\sqrt{22}}{2}$. Finally, we use Law of Cosines for $\\triangle ADB$, $$4(\\frac{\\sqrt{22}}{2})^2+1+2\\cdot2(\\frac{\\sqrt{22}}{2})\\cdot \\cos(ADC)=AB^2$$ then $AB=2\\sqrt{7}$, so the height of this $\\triangle ABC$ is $\\sqrt{4^2-(\\sqrt{7})^2}=3$. Then the area of $\\triangle ABC$ is $3\\sqrt{7}$, so the answer is $\\boxed{010}$. ### Solution 3 Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. $[asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label(\"A\",A,SW);label(\"B\",B,SE);label(\"C\",C,N);label(\"D\",D,NE);label(\"L\",L,NW);label(\"M\",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label(\"X\",X,S);label(\"E\",EE,NW);label(\"Y\",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy]$ Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$. Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\\frac{b}{2}$, we know that $$MX=\\frac{b}{2}-\\frac{3b}{7}=\\frac{b}{14}.$$ Also, as $CE:EM=7:1$, we know that $EM=\\frac{1}{\\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\\triangle {XCM}$, we know that $$\\frac{b^2}{196}+h^2=\\left(\\sqrt{7}+\\frac{1}{\\sqrt{7}}\\right)^2=\\frac{64}{7}.$$ Also, as $LE:BE=5:3$, we know that $BL=\\frac{8}{5}\\cdot 3=\\frac{24}{5}$.", "the length of the segment perpendicular to the line with one endpoint on the line and the other on the point. This means the altitude from $P$ to $\\overline{AD}$ is $x$, so the area of $\\triangle ADP$ is equal to $\\frac12\\cdot AD\\cdot x=\\frac72x$. Similarly, the area of $\\triangle BCQ$ is $\\frac12\\cdot BC\\cdot x=\\frac52x$. The altitude of the trapezoid is $2x$, because it is the sum of the distances from either $P$ or $Q$ to $\\overline{AB}$ and $\\overline{CD}$. This means the area of trapezoid $ABCD$ is $\\frac12(AB+CD)\\cdot2x=\\frac12(11+19)\\cdot2x=30x$. Now, the area of hexagon $ABQCDP$ is the area of trapezoid $ABCD$, minus the areas of triangles $ADP$ and $BCQ$. This is $30x-\\frac72x-\\frac52x=24x$. Now it remains to find $x$. $[asy] unitsize(0.6cm); import olympiad; pair A,B,C,D,R,S; A=(11/2,5sqrt(3)/2); B=(33/2,5sqrt(3)/2); C=(19,0); D=(0,0); R=(11/2,0); S=(33/2,0); draw(A--B--C--D--cycle); draw(A--R); draw(B--S); label(\"A\",A,dir(135)); label(\"B\",B,dir(45)); label(\"C\",C,dir(315)); label(\"D\",D,dir(225)); label(\"R\",R,dir(270)); label(\"S\",S,dir(270)); label(\"11\",midpoint(A--B),dir(90)); label(\"5\",midpoint(B--C),dir(45)); label(\"11\",midpoint(R--S),dir(270)); label(\"7\",midpoint(D--A),dir(135)); label(\"r\",midpoint(R--D),dir(270)); label(\"s\",midpoint(C--S),dir(270)); label(\"19\",midpoint(C--D),5*dir(270)); label(\"2x\",midpoint(A--R),dir(0)); label(\"2x\",midpoint(B--S),dir(180)); draw(rightanglemark(A,R,D,15)); draw(rightanglemark(B,S,C,15)); [/asy]$ We let $R$ and $S$ be the feet of the altitudes of $A$ and $B$, respectively, to $\\overline{CD}$. We define $r=RD$ and $s=SC$. We know that $AB=RS$, so $RS=11$ and $r+s=19-11=8$. By the Pythagorean Theorem on $\\triangle ADR$ and $\\triangle BCS$, we get $r^2+(2x)^2=7^2$ and $s^2+(2x)^2=5^2$, respectively. Subtracting the second equation from the first gives us $r^2-s^2=49-25=24$. The left hand side of this equation is a difference of squares and", "areas of $[ADE]$ and $[ACG]$, since we count it twice. Note that the angle bisector theorem also applies for $\\triangle ADE$ and $\\frac{AE}{AD} = \\frac{1}{5}$, so thus $\\frac{EF}{ED} = \\frac{1}{6}$ and we find $[AFE] = \\frac{1}{6} \\cdot 30 = 5$, and the area outside $FDBG$ must be $[ADE] + [AGC] - [AFE] = 30 + 20 - 5 = 45$, and we finally find $[FDBG] = [ABC] - 45 = 120 -45 = \\boxed{75}$, and we are done. ## Solution 6: Areas $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"3Y\",(2.5,0.75),N); label(\"Y\",(1,0.2),N); label(\"X\",(0.5,0.5),N); label(\"3X\",(1.25,1.75),N); [/asy]$ Give triangle $AEF$ area X. Then, by similarity, since $\\frac{AC}{AE} = \\frac{2}{1}$, $ACG$ has area 4X. Thus, $FGCE$ has area 3X. Doing the same for triangle $AGB$, we get that triangle $AFD$ has area Y and quadrilateral $GFDB$ has area 3Y. Since $AEF$ has the same height as $AFD$, the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, $\\frac{CG}{GB} = \\frac{1}{5}$. So, $\\frac{[AEF]}{[AFD]} = \\frac{1}{5}$. Since $AEF$ has area X, we can write the equation 5X = Y and substitute 5X for Y. $[asy] draw((0,0)--(1,3)--(5,0)--cycle); draw((0,0)--(2,2.25)); draw((0.5,1.5)--(2.5,0)); label(\"A\",(0,0),SW); label(\"B\",(5,0),SE); label(\"C\",(1,3),N); label(\"G\",(2,2.25),NE); label(\"D\",(2.5,0),S); label(\"E\",(0.5,1.5),NW); label(\"\",(2.5,0.75),N); label(\"\",(1,0.2),N); label(\"F\", (1, 1.5), N); label(\"\",(0.5,1.5),N); label(\"\",(1.25,1.75),N); [/asy]$ Now we can solve for X by adding up", "sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B =", "of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{(B) 30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2,", "the sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\\frac{BF}{FC}$ is equal to $\\frac{60}{180}$=$\\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\\boxed{\\textbf{(B) }30}$. ~~SmileKat32 ## Solution 4 (Similar Triangles) Extend $\\overline{BD}$ to $G$ such that $\\overline{AG} \\parallel \\overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SE); label(\"F\", F, S); label(\"G\", G, ENE); [/asy]$ Then $\\triangle ADG \\sim \\triangle CDB$ and $\\triangle AEG \\sim \\triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$. Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \\boxed{\\textbf{(B) }30}$. $[asy] size(8cm); pair A, B, C, D, E, F, G; B", "$\\frac{1}{4}\\div3=\\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\\frac{1}{12}\\times360=\\boxed{\\textbf{(B) }30}$. ~emerald_block Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90 ## Solution 11 $[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label(\"A\",A,N); label(\"B\", B,SW); label(\"C\",C,SE); label(\"D\",DD,NE); label(\"E\",EE,NW); label(\"F\",FF,S); label(\"G\",G,S); [/asy]$ We know that $AD = \\dfrac{1}{3} AC$, so $[ABD] = \\dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \\dfrac{1}{2} BD$, $[ABE] = \\dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\\overline{BC}$ such that $\\overline{DG}$ is parallel to $\\overline{AF}$ and create segment $DG$. We then observe that $\\triangle AFC \\sim \\triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\\triangle DBG \\sim \\triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \\dfrac{1}{4} BC$. Thus, $[BFA] = \\dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\\triangle EBF$ is $[BFA] - [ABE] = \\boxed{\\textbf{(B) }30}$. ~i_equal_tan_90 ## Solution", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 (Coordinate Geometry) We will put triangle ACE on a xy-coordinate plane with C being the origin. The area of triangle ACE is 96. To", "= \\frac 14$, and therefore $\\frac{BC}{AC} = \\frac{DE}{CE} = \\frac{FA}{EA} = \\frac 34$. Draw the height from $F$ onto $AB$ as in the picture below: $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",0.5*C + 0.5*B,4*W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label(\"G\",G,W); [/asy]$ Now consider the area of $\\triangle ABF$. Clearly the triangles $\\triangle AFG$ and $\\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\\frac {AF}{AE} = \\frac 34$, hence $FG = \\frac 34 \\cdot CE$. Now the area $S_{ABF}$ of $\\triangle ABF$ can be computed as $S_{ABF} = \\frac 12 \\cdot AB \\cdot FG$ = $\\frac 12 \\cdot \\left( \\frac 14 \\cdot AC \\right) \\cdot \\left( \\frac 34 \\cdot EC \\right) = \\frac 14 \\cdot \\frac 34 \\cdot S_{ACE}$. Similarly we can find that $S_{BCD} = S_{DEF} = \\frac 3{16}\\cdot S_{ACE}$ as well. Hence $S_{BDF} = S_{ACE} - 3\\cdot\\left( \\frac 3{16} \\cdot S_{ACE} \\right) = \\frac 7{16} \\cdot S_{ACE}$, and the answer is $\\frac{S_{BDF}}{S_{ACE}} = \\boxed{\\frac 7{16}}$. ## Solution 3 The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9),", "\\frac {1}{2} \\cdot 15 \\cdot 36 = \\boxed{\\mathrm{(C)}\\ 210}$$ ### Solution 2 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$. We get the following triangle: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0); draw(A--B--C--cycle); draw(C--F,dashed); label(\"$$A'$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,N); label(\"$$C'$$\",F,SE); label(\"5\",(A+C)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ The length of $A'B$ in this triangle is equal to the length of the original $AB$, minus the length of $CD$. Thus $A'B = 52 - 39 = 13$. Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\\frac{A'C\\cdot BC}2 = \\frac{5\\cdot 12}2 = 30$, and therefore its altitude $CC'$ is $\\frac{[A'BC]}{A'B} = \\frac{60}{13}$. Now the area of the original trapezoid is $\\frac{(AB+CD)\\cdot CC'}2 = \\frac{91 \\cdot 60}{13 \\cdot 2} = 7\\cdot 30 = \\boxed{\\mathrm{(C)}\\ 210}$ ### Solution 3 Draw altitudes from points $C$ and $D$: $[asy] unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,S); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,N); label(\"$$D'$$\",F,SSE); label(\"$$C'$$\",E,S); label(\"39\",(C+D)/2,N); label(\"52\",(A+B)/2,S); label(\"5\",(A+D)/2,W); label(\"12\",(B+C)/2,ENE); [/asy]$ Call the length of $AD'$ to be $y$, the length of $BC'$ to be $z$, and the height of the trapezoid to be $x$. By", "triangles of area 1 be $x$, then the base length of $\\Delta ADE=4x$. Notice, $\\big(\\frac{DE}{BC}\\big)^2=\\frac{1}{40}\\implies \\frac{x}{BC}=\\frac{1}{\\sqrt{40}}$, then $4x=\\frac{4BC}{\\sqrt{40}}\\implies \\big[ADE\\big]=\\big(\\frac{4}{\\sqrt{40}}\\big)^2\\cdot \\big[ABC\\big]=\\frac{2}{5}\\big[ABC\\big]$ Thus, $\\big[DBCE\\big]=\\big[ABC\\big]-\\big[ADE\\big]=\\big[ABC\\big]\\big(1-\\frac{2}{5}\\big)=\\frac{3}{5}\\cdot 40=\\boxed{24}.$ $[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label(\"B\",(0,0),SW); label(\"C\",(60,0),SE); label(\"E\",(50,10),E); label(\"D\",(10,10),W); label(\"A\",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]$ Solution by ktong ## Solution 4 The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$. Therefore the area of the trapezoid is $40-16=\\boxed{24}$. ## Solution 5 You can see that we can create a \"stack\" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be $7+5+3+1=16$, so to find the area of such trapezoid $BCED$, we just take $40-16=\\boxed{24}$, like so. ∎ --anna0kear ## Solution 6 The combined area of the small triangles is $7$, and from the fact that each small triangle has an area of $1$, we can deduce that the larger triangle above has an area of $9$ (as the sides of the triangles are in a proportion of $\\frac{1}{3}$, so will their areas have a proportion that is the square of the", "and the areas of similar triangles are proportional to the squares of corresponding side lengths, $\\frac{[ABC]}{AB^2} = \\frac{[CBH]}{CB^2} = \\frac{[ACH]}{AC^2}$. But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. Proof 2 Consider a circle $\\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\\omega$. Let the line $AB$ meet $\\omega$ at $Y$ and $X$, as shown in the diagram: Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\\omega$, we see $AC^2 = AY \\cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. Proof 3 $ABCD$ and $EFGH$ are squares. $[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label(\"A\", A, NNW); label(\"B\", B, ENE); label(\"C\", C, ESE); label(\"D\", D, SSW); draw(E--F--G--H--cycle); label(\"E\", E, N); label(\"F\", F,SE); label(\"G\", G, S); label(\"H\", H, W); label(\"a\", A--B,N); label(\"a\", B--F,SE); label(\"a\", C--G,S); label(\"a\", H--D,W); label(\"b\", E--B,N); label(\"b\", F--C,SE); label(\"b\", G--D,S); label(\"b\", A--H,W); label(\"c\", E--H,NW); label(\"c\", E--F); label(\"c\", F--G,SE); label(\"c\", G--H,SW); [/asy]$ $(a+b)^2=c^2+4\\left(\\frac{1}{2}ab\\right)\\implies a^2+2ab+b^2=c^2+2ab\\implies a^2 + b^2=c^2$. Common Pythagorean Triples A Pythagorean Triple is", "$BD=7$. Thus we get the base of triangle $ADE, DE$, to be $\\frac{1}{2}$ units long. We can now use Heron's Formula on $ABC$. $$s=\\frac{AB+BC+AC}{2}=21$$ $$[ABC]=\\sqrt{(s)(s-AB)(s-BC)(s-AC)}=\\sqrt{(21)(8)(7)(6)}=84$$ $$\\frac{DE}{BC}[ABC]=\\frac{\\frac{1}{2}}{14}\\cdot 84=3$$ Therefore, the answer is $\\mathrm{C}$. ## Solution 3 pair A,B,C,D,E; B=(0,0); C=(14;0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label(\"$A$\",A,N); label(\"$B$\",B,S); label(\"$C$\",C,S); label(\"$E$\",E,S); label(\"$D$\",D,S); label(\"$13$\",A--B,NW); label(\"$15$\",A--C,NE); label(\"$14$\",B--C,S); (Error compiling LaTeX. 32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy: 7.6: syntax error error: could not load module '32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy') Let's find the area of $\\Delta ABC$ by Heron, $s=\\frac{a+b+c}{2}\\\\\\\\s=\\frac{14+15+13}{2}\\to\\boxed{s=21}$ Then, $A^2=s(s-a)(s-b)(s-c)\\\\\\\\A^2=21(21-14)(21-15)(21-13)\\\\\\\\\\boxed{A=84}$ Knowing that D is the midpoint of BC, thus the area of $\\Delta ADC=\\Delta ABD$, which is half the area of $\\Delta ABC$, so $\\Delta ADC=\\Delta ABD=42$. By Angle Bisector Theorem we know that: $\\frac{13}{BE}=\\frac{15}{CE}$ $\\boxed{CE=14-BE}$ $\\frac{13}{BE}=\\frac{15}{14-BE}$ $\\boxed{BE=6.5}\\Rightarrow CE=7.5$ Also, we know that: $\\frac{A_{\\Delta ADE}}{A_{\\Delta ABC}}=\\frac{ED}{BC}$ And, we can easily see that $DE=0.5$, so, $\\frac{A_{\\Delta ADE}}{84}=\\frac{0.5}{14}$ $\\boxed{A_{\\Delta ADE}=3}$", "= (-1,0); pair C = (0,0); pair B = (2,0); draw((-1,0)--(2,0)); pair P = (9/7,12/7); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--P); label(\"A\",A,W); label(\"C\",C,S); label(\"B\",B,E); label(\"P\",P,N); [/asy]$ Let $A_1$ equal to the area of $\\Delta ACP$ and $A_2$ equal to the area of $\\Delta CBP$. Heron's Formula states that the area of an triangle with sides $a$ $b$ and $c$ is $$\\sqrt{s(s-a)(s-b)(s-c)}$$ where $s$, or the semiperimeter, is half of the triangle's perimeter. The semiperimeter $s_1$ of $\\Delta ACP$ is $$\\frac{[(r + 2) + (3 - r) + 1]}{2} = \\frac{6}{2} = 3$$ Use Heron's Formula to obtain $$A_1 = \\sqrt{3(2)(3-2-r)(3-3+r)} = \\sqrt{6r(1-r)} = \\sqrt{6r-6r^2}$$ Using Heron's Formula again, find the area of $\\Delta CBP$ with sides $r+1$, $2$, and $3-r$. $$s_2 = \\frac{(r + 1) + 2 + (3 - r)}{2} = 3$$ $$A_2 = \\sqrt{3(3-2)(3-1-r)(3-3+r)} = \\sqrt{3(2r-r^2)} = \\sqrt{6r-3r^2}$$ Now, $$2 \\cdot A_1 = A_2$$ $$2\\sqrt{6r-6r^2} = \\sqrt{6r-3r^2}$$ $$4(6r-6r^2) = 6r-3r^2$$ $$24r-24r^2 = 6r-3r^2$$ $$18r = 21r^2$$ $$r = \\frac{18}{21} = \\boxed{\\frac{6}{7}} \\to \\boxed{\\textbf{(B)}}$$ ## See Also 2017 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 •", "$\\triangle AGH$ is the altitude of $\\triangle ABC$, or $\\frac{3\\sqrt{7}}{2}$. However, it's not too hard to see that $GB = HC = 1$, and therefore $[AGH] = [ABC]$. From here, we get that the area of $\\triangle ABC$ is $\\frac{15\\sqrt{7}}{14} \\implies \\boxed{036}$, by similarity. ~awang11 ## Solution 2 Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that $$AB\\cdot M_AC + AC\\cdot M_AB = BC\\cdot M_AA \\implies M_AA = 2M_AI.$$ In particular, we have $AI = IM_A$. $[asy] defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot(\"A\", A, dir(120)); dot(\"B\", B, dir(220)); dot(\"C\", C, dir(320)); dot(\"D\", D, dir(230)); dot(\"E\", E, dir(330)); dot(\"F\", F, dir(250)); dot(\"I\", I, dir(80)); dot(\"M_A\", MA,", "$\\triangle ABC$ is$$s = \\frac{AB + BC + CA}2 = \\frac{15}2,$$and Heron's Formula gives the area of $\\triangle ABC$ as$$\\sqrt{s(s-AB)(s-BC)(s-CA)} = \\frac{15\\sqrt7}4.$$This area is also $rs$ implying that $r = \\frac{\\sqrt7}2$. Stewart's Theorem gives $AD =3\\sqrt2$. Because the ratio of the areas of $\\triangle IBC$ and $\\triangle ABC$ is $\\frac{ID}{AD},$ it follows that$$ID = AD\\cdot\\frac{\\frac{r\\cdot BC}2}{\\frac{15\\sqrt7}4} = \\sqrt2.$$Thus $IM = MD - ID = \\frac{\\sqrt2}2$. $[asy] unitsize(0.8 cm); pair A, B, C, D, E, F, I, M; real angleC = aCos(3/4); C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); M = (A + D)/2; draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5)); dot(\"A\", A, SW); dot(\"B\", B, dir(0)); dot(\"C\", C, N); dot(\"D\", D, dir(0)); dot(\"E\", E, N); dot(\"F\", F, dir(0)); dot(\"I\", I, dir(120)); dot(\"M\", M, W); [/asy]$ Note that $\\angle EFI = 90^{\\circ} - \\angle FIA = 90^{\\circ} - \\angle CID = 90^{\\circ} - \\frac{\\angle A}{2} - \\frac{\\angle C}{2} = \\frac{\\angle B}{2} = \\angle IBC$. Thus $\\triangle IBC \\sim \\triangle IFE$. The height of $\\triangle IBC$ to $I$ is $r=\\frac{\\sqrt7}2$, and the height of $\\triangle IFE$ to $I$", "= (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label(\"A\", A, N); label(\"B\", B, WSW); label(\"C\", C, ESE); label(\"D\", D, dir(0)*1.5); label(\"E\", E, SSE); label(\"F\", F, S); label(\"60\", (A+E+D)/3); label(\"60\", (A+E+B)/3); label(\"120\", (D+E+C)/3); label(\"x\", (B+E+F)/3); label(\"120-x\", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\\triangle ABD=\\frac{1}{3} \\cdot 360 = 120.$ Similarly, $\\triangle DBC = \\frac{2}{3} \\cdot 360 = 240.$ Now, since $E$ is a midpoint of $BD$, $\\triangle ABE = \\triangle AED = 120 \\div 2 = 60.$ We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\\triangle BEC$ and $\\triangle DEC$ share 2 sides. We know that $\\triangle BEC=\\triangle DEC=240 \\div 2 = 120$ since $E$ is a midpoint of $BD.$ Let's label $\\triangle BEF$ $x$. We know that $\\triangle EFC$ is $120-x$ since $\\triangle BEC = 120.$ Note that with this information now, we can deduct more things that are needed to finish the solution. Note that $\\frac{EF}{AE} = \\frac{120-x}{180} = \\frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$ We want to find $x.$ This is a simple equation, and solving we get $x=\\boxed{\\textbf{(B)}30}.$ ~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea. ## Note This question is extremely similar to 1971 AHSME Problem 26.", "# Problem #2136 2136 In the right triangle $\\triangle ACE$, we have $AC=12$, $CE=16$, and $EA=20$. Points $B$, $D$, and $F$ are located on $AC$, $CE$, and $EA$, respectively, so that $AB=3$, $CD=4$, and $EF=5$. What is the ratio of the area of $\\triangle DBF$ to that of $\\triangle ACE$? $\\mathrm{(A) \\ } \\frac{1}{4} \\qquad \\mathrm{(B) \\ } \\frac{9}{25} \\qquad \\mathrm{(C) \\ } \\frac{3}{8} \\qquad \\mathrm{(D) \\ } \\frac{11}{25} \\qquad \\mathrm{(E) \\ } \\frac{7}{16}$ $[asy] unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label(\"A\",A,N); label(\"B\",B,W); label(\"C\",C,SW); label(\"D\",D,S); label(\"E\",E,SE); label(\"F\",F,NE); label(\"3\",A--B,W); label(\"9\",C--B,W); label(\"4\",C--D,S); label(\"12\",D--E,S); label(\"5\",E--F,NE); label(\"15\",F--A,NE); [/asy]$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved." ]

[ "# finding a direction vector , perpendicular to two lines $(x+1)/3=(y+1)/8=(z+1)/12$ and $(1-2x)/-4=(3y+1)/9=z/6$ .", "# Perpendicular lines and vectors My textbook is confusing me a little. Here is a worked example from my textbook: Line $$l$$ has the equation $$\\begin{pmatrix}3\\\\ -1\\\\ 0\\end{pmatrix}+\\lambda \\begin{pmatrix}1\\\\ -1\\\\ 1\\end{pmatrix}$$ and point $$A$$ has co-ordinates $$(3, 9, -2)$$. Find the coordinates of point $$B$$ on $$l$$ so that $$AB$$ is perpendicular to $$l$$. $$\\vec{AB\\cdot }\\begin{pmatrix}1\\\\ -1\\\\ 1\\end{pmatrix}=0$$ $$\\vec{OB}=r=\\begin{pmatrix}3+\\lambda \\\\ -1-\\lambda \\\\ \\lambda \\end{pmatrix}$$ $$\\vec{AB}=\\begin{pmatrix}\\lambda \\\\ -10-\\lambda \\\\ \\lambda +2\\end{pmatrix}$$ $$\\begin{pmatrix}\\lambda \\\\ -10-\\lambda \\\\ \\lambda +2\\end{pmatrix}\\cdot \\begin{pmatrix}1\\\\ -1\\\\ 1\\end{pmatrix}=0$$ $$3\\lambda= -12, \\lambda = -4$$ Coordinates of $$B$$: $$(-1, 3, -4)$$ The thing I don't understand is why they found the dot product of the line AB and the direction vector of line l. My textbook does mention that to check whether two vectors are perpendicular, $$a\\cdot b = 0$$ and for lines, the dot product of their direction vectors = 0. So why did they mix both here? Didn't they use the entire line $$AB$$ and then just the direction vector of line l? Or am I missing something as usual? There is nothing special here the dot product $AB\\cdot v$ gives the condition of orthogonality between the vector from $A$ to $B$ and the direction vector of the given line. Note also that vector $AB$ is a direction vector for the line orthogonal to the given line", "values in the above equation, we have (ii) If the line are perpendicular, then $\\theta = {90^ \\circ }$, so putting this values in the above equation, we have This is the condition for two lines to perpendicular to each other.", "point. - • For a vector $(a,b)$ the perpendicular direction is $(-b,a)$ (this transformation turns left, multiply by $-1$ if you want to go the other way). • If you want to have a vector along direction $(a,b)$ of length $d$, then calculate $\\frac{d}{|(a,b)|}(a,b)$ where $|v|$ is the length of $v$. @Walkerneo You are, $(a,b) = P_2-P_1$, $M = \\frac{P_1+P_2}{2}$, $X = M + \\alpha \\cdot (-b,a)$ for some $\\alpha \\in \\mathbb{R}$; try calculating $\\alpha$ on your own. – dtldarek Mar 14 '13 at 7:21", "perpendicular to the line $x=1+t, y=1-t, z=2t$ The direction vector you need is $<1,1,1>\\times<1,-1,2>$.", "& 2 & 5 & 4 \\end{array} \\right |$$ Which will give you a vector perpendicular to the n-1 you already have.", "if they are perpendicular to each other. Part 2 You can find the direction vector of $\\mathbb{r}$ by cross product. The direction vector of $\\mathbb{s}$ is $2\\vec{i}+3\\vec{j}-6\\vec{k}$. Find the dot product of the two direction vectors.", "point, there is exactly one. I assume you want the equation for that line. Since the line passes through both the given point and the given line, a vector in the direction of this perpendicular line would be the vector $\\begin{pmatrix}3-(-1) \\\\ 1-(-2+t) \\\\ -2-(-1+t)\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 3-t \\\\ -1-t\\end{pmatrix}$. Let's find a vector in the direction of the given line by finding two points along the line (we can choose any two points, so let's say $t=0, t=1$). So, a vector in the direction of the given line is $\\begin{pmatrix}-1-(-1) \\\\ -2+1 - (-2+0) \\\\ -1+1 - (-1+0)\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Since we want these vectors to be perpendicular, we want the dot product to be zero. Hence: $(4\\cdot 0) + (1\\cdot (3-t)) + (1\\cdot (-1-t)) = 2-2t = 0$ implies $t=1$. Hence, a vector in the direction of the perpendicular line is $\\begin{pmatrix}4 \\\\ 3-1 \\\\ -1-1\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 2 \\\\ -2\\end{pmatrix}$. So, the perpendicular line would have parametric equation $x=3+4t, y=1+2t, z=-2-2t$. 4. ## Re: equation of line through point orthogonal to line so the point (does this mean you are treating the point as a bounded vector) minus the line gives a direction vector along the perpendicular line then I take the given direction vector of the line which", "and a direction ~v that is parallel(1) to the line. Two visualize, place two cubes side-by-side.", "the two vectors as sides. 6. ## Re: Linear Algebra It is also the vector that is normal (or perpendicular) to the plane containing the two vectors. The vector $m+n$ is in the same plane as both $m$ and $n$, so the same vector that is perpendicular to both $m$ and $n$ will also be perpendicular with $m$ and $m+n$ and $n$ and $m+n$.", "of the line. it describes the direction the line goes in according to the vector here we were given a line. the normal vector to the plane perpendicular to it is the same as its direction vector WARNING: by taking the cross product here, you are finding a vector in the plane, not perpendicular to it", "# Thread: finding a direction vector , perpendicular to two lines 1. ## finding a direction vector , perpendicular to two lines Find a direction vector of the line that is perpendicular to both: $\\displaystyle (x+1)/3=(y+1)/8=(z+1)/12$ and $\\displaystyle (1-2x)/-4=(3y+1)/9=z/6$ . 2. Find a direction vector for each line and then take the cross-product of them both.", "is in fact the direction vector - going along the line. In the case of the plane, the normal vector is perpendicular to the plane, so is at $90^\\circ$ to the plane. This means that the two problems are rotations of each other: one is sin and the other is cos.", "Comment Share Q) # The direction cosine of a line which is $\\perp$ to two perpendicular lines whose direction cosines are $(l_1,m_1,n_1)\\:\\:and\\:\\:(l_2,m_2,n_2)$ is ? $\\begin{array}{1 1} (a)\\:(m_1n_2-m_2n_1,\\:n_1l_2-n_2l_1\\:,l_1,m_2-m_1l_2) \\\\ (b)\\:(l_1+l_2,\\:m_1+m_2,\\:n_1+n_2) \\\\ (c)\\:(l_1+l_2,\\:m_1+m_2,\\:n_1+n_2) \\\\ (d)\\:None\\:of\\:these. \\end{array}$", "be easily calculated using vectors. $d=|\\vec{PF}|$ $=\\left| \\begin{pmatrix} -1-(-1) \\\\ 0-0 \\\\ 1-3 \\end{pmatrix}\\right|$ $=\\left| \\begin{pmatrix} 0 \\\\ 0 \\\\ -2 \\end{pmatrix}\\right|$ $=\\sqrt{(-2)^2}$ $=2$ The distance between the two lines $g$ and $h$ is 2 LU.", "A line is perpendicular to two lines having direction ratios Question: A line is perpendicular to two lines having direction ratios $1,-2,-2$ and $0,2,1$. The direction cosines of the line are A. $\\frac{-2}{3}, \\frac{1}{3}, \\frac{2}{3}$ B. $\\frac{2}{3}, \\frac{1}{3}, \\frac{-1}{3}$ C. $\\frac{2}{3}, \\frac{-1}{3}, \\frac{2}{3}$ D. none of these Solution:", "to vector form $$y = \\frac{x}{3} + 2$$ We want this in the form $$\\underline{r} =$$ 1st point + direction Using the cartesian equation we can find the direction as we know that the gradient of a line is: $$m = \\frac{\\text{Change in y}}{\\text{Change in x}}$$ We can see that for every time y is increased by 1 x is increases by 1/3 as x is divided by 3 giving us: $$m = \\frac{1}{(\\frac{1}{3})}=\\frac{3}{1}$$ This means we can write the direction as: $$\\underline{r} =$$ 1st point + $$λ\\begin{pmatrix}3\\\\\\ 1\\end{pmatrix}$$ Now we juust need to find the 1st point to do this we simply set the value of λ=0 as this is where the line starts From the equation $$y = \\frac{x}{3} + 2$$ We can see that when x=0 y=2 so this is where our line starts Therefore the vector equation can be written as: $$\\underline{r} = \\begin{pmatrix}0\\\\\\ 2\\end{pmatrix} + λ\\begin{pmatrix}3\\\\\\ 1\\end{pmatrix}$$ #### Intersection of two lines What about if we need to find the exact point that two lines intersect knowing only the vector equations of these two lines? $$\\underline{r_1} = \\begin{pmatrix}2\\\\\\ 3\\end{pmatrix} + λ\\begin{pmatrix}1\\\\\\ 2\\end{pmatrix}$$ $$\\underline{r_2} = \\begin{pmatrix}6\\\\\\ 1\\end{pmatrix} + μ\\begin{pmatrix}1\\\\\\ -3\\end{pmatrix}$$ For the two lines to intercept there must be a point when the value of the two equations are the same this means we", "and if the flag is between the two lines. We just need to check the result for a single point, because then every point will rotate by the same amount and so the flag will remain intact (and will have rotated by the same amount). The diagram says it all, really: the line segment joining the point to the origin has rotated by an angle of $\\alpha+\\alpha+(\\theta-\\alpha)+(\\theta -\\alpha)=2\\theta$ in the direction of $\\theta$.", "by the above relation actually passes throught the point $\\;A:=\\begin{pmatrix}a\\\\a\\\\a\\end{pmatrix}\\,$ (very easy) ,and that the vector $\\;A-u\\;$ is perpendicular to the given line there (very easy, too). Can you explain where the $A-u$ vector comes from? I don't understand why showing $A-u$ to be perpendicular to the line $x=y=z$ is helpful. – Xoque55 Nov 25 '13 at 4:33", "## Elementary Geometry for College Students (6th Edition) When $n=1$, line 1 passes through the point $(1,2,-3)$. Line 2 also passes through this point when $r=0$. Therefore, the lines have at least one point in common. A direction vector for line 1 is $(1,2,-3)$ and a direction vector for line 2 is $(-1,-2,3)$. When $n=-1$, the product of $n$ and the direction vector of line 1 is equal to the direction vector of line 2. Therefore, the lines are parallel. Since the lines share at least one point and the lines are parallel, the lines are coincident." ]

[ "## Directional and orthogonal form of a line Directional and orthogonal form of a line Directional form: \\begin{align*} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} &= \\begin{pmatrix} a_1 \\\\ a_2 \\end{pmatrix} + \\alpha \\begin{pmatrix} d_1 \\\\ d_2 \\end{pmatrix} \\end{align*} or, concisely \\begin{align} x = a + \\alpha d \\end{align} Let $v$ be a vector perpendicular to $d$. Transpose both sides and multiply by $v$: \\begin{align} x'v &= a'v + \\alpha d'v = a'v = c \\end{align} Where c is a constant scalar. This is called the orthogonal form. If you see it as $f(x) = x'v$, then $v$ is the gradient of $f$.", "vectors form an orthogonal set. Solution. To show that vectors $\\boldsymbol{v}$ and $\\boldsymbol{w}$ form an orthogonal set, we must check that they are perpendicular to each other. We know from theoremlink that if the dot product of two vectors is zero, then they are perpendicular. So let's compute the dot product between them: \\begin{align*} \\begin{pmatrix} 1\\\\-1 \\end{pmatrix}\\cdot \\begin{pmatrix} 2\\\\2 \\end{pmatrix}&= (1)(2)+(-1)(2)\\\\&=0 \\end{align*} Because $\\boldsymbol{v}$ and $\\boldsymbol{w}$ are orthogonal to each other, we conclude that they form an orthogonal set. Let's also visually confirm this: Indeed, they are perpendicular! Definition. # Orthonormal set An orthonormal set is an orthogonal set of unit vectors. In other words, a set of vectors is orthonormal if and only if: • every pair of vectors is perpendicular. • every vector has a length of one. Example. ## Showing that a set is orthonormal Consider the standard basis vectors for $\\mathbb{R}^2$ below: $$\\boldsymbol{e}_1= \\begin{pmatrix} 1\\\\0 \\end{pmatrix},\\;\\;\\;\\;\\; \\boldsymbol{e}_2= \\begin{pmatrix}0\\\\1 \\end{pmatrix}$$ Show that these vectors form an orthonormal set. Solution. Let's first check that they are perpendicular to each other: \\begin{align*} \\boldsymbol{e}_1\\cdot\\boldsymbol{e}_2&= \\begin{pmatrix} 1\\\\0 \\end{pmatrix}\\cdot \\begin{pmatrix} 0\\\\1 \\end{pmatrix}\\\\ &=(1)(0)+(0)(1)\\\\ &=0 \\end{align*} Because their dot product is zero, they are perpendicular. We can also easily tell that these vectors have both length one, and so $\\boldsymbol{e}_1$ and $\\boldsymbol{e}_2$ form an orthonormal set by definition! We can visually", "# Another Line Equation Case At Line $L_{1}$ has equation $r = \\begin{pmatrix} -5\\\\ -3\\\\ 2 \\end{pmatrix} + \\lambda \\begin{pmatrix} -1\\\\ 2\\\\ 2 \\end{pmatrix}$ A line $L_{2}$ passing through the origin intersects $L_{1}$ and perpendicular to $L_{1}$. Find the vector equation of $L_{2}$! I find the vector for $L_{2}$ using cross product $\\begin{pmatrix} i & j & k\\\\ -5 & -3 & 2\\\\ -1 & 2 & 2 \\end{pmatrix}$, and I don't have any idea to do with that vector.... • You need to solve $r\\cdot\\begin{pmatrix}-1\\\\ 2\\\\ 2 \\end{pmatrix}=0$ for $\\lambda$, it gives the foot of the perpendicular. Jun 3, 2015 at 1:51 • What is $r$ on that dot product? I know that $\\begin{pmatrix}-1\\\\ 2\\\\ 2 \\end{pmatrix}$ is the direction vector. Jun 3, 2015 at 2:12 • It's $r$ for $L_1$ from above Jun 3, 2015 at 2:13 • So $r\\cdot\\begin{pmatrix}-1\\\\ 2\\\\ 2 \\end{pmatrix}=0$ is for searching the lambda. After do the dot product, I get : $\\begin{pmatrix} a\\\\ b\\\\ c \\end{pmatrix}\\cdot \\begin{pmatrix} -1\\\\ 2\\\\ 2 \\end{pmatrix}=0$ and become : $-a+2b+2c=0$ Jun 3, 2015 at 2:18 You had the right idea that the cross product will give you an orthogonal vector, but taking $(-5,-3,2)\\times(-1,2,2)$ will give you a vector orthogonal to the direction of the line and a point on the line, not necessarily the direction perpendicular", "point, there is exactly one. I assume you want the equation for that line. Since the line passes through both the given point and the given line, a vector in the direction of this perpendicular line would be the vector $\\begin{pmatrix}3-(-1) \\\\ 1-(-2+t) \\\\ -2-(-1+t)\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 3-t \\\\ -1-t\\end{pmatrix}$. Let's find a vector in the direction of the given line by finding two points along the line (we can choose any two points, so let's say $t=0, t=1$). So, a vector in the direction of the given line is $\\begin{pmatrix}-1-(-1) \\\\ -2+1 - (-2+0) \\\\ -1+1 - (-1+0)\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Since we want these vectors to be perpendicular, we want the dot product to be zero. Hence: $(4\\cdot 0) + (1\\cdot (3-t)) + (1\\cdot (-1-t)) = 2-2t = 0$ implies $t=1$. Hence, a vector in the direction of the perpendicular line is $\\begin{pmatrix}4 \\\\ 3-1 \\\\ -1-1\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 2 \\\\ -2\\end{pmatrix}$. So, the perpendicular line would have parametric equation $x=3+4t, y=1+2t, z=-2-2t$. 4. ## Re: equation of line through point orthogonal to line so the point (does this mean you are treating the point as a bounded vector) minus the line gives a direction vector along the perpendicular line then I take the given direction vector of the line which", "Orthogonal Vectors Two vectors, x and y, are orthogonal if their dot product is zero. For example $e \\cdot f = \\begin{pmatrix} 2 & 5 & 4 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 5 \\end{pmatrix} = 2*4 + (5)*(-2) + 4*5 = 8-10+20 = 18$ Vectors e and f are not orthogonal. $g \\cdot h = \\begin{pmatrix} 2 & 3 & -2 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 1 \\end{pmatrix} = 2*4 + (3)*(-2) + (-2)*1 = 8-6-2 = 0$ However, vectors g and h are orthogonal. Orthogonal can be thought of as an expansion of perpendicular for higher dimensions. Let $x_1, x_2, \\ldots , x_n$ be m-dimensional vectors. Then a linear combination of $x_1, x_2, \\ldots , x_n$ is any m-dimensional vector that can be expressed as $c_1 x_1 + c_2 x_2 + \\ldots + c_n x_n$ where $c_1, \\ldots, c_n$ are all scalars. For example: $x_1 =\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix}$ $y =\\begin{pmatrix} -5 \\\\ 12 \\\\ -8 \\end{pmatrix} = 1*\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix} + (-2)* \\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix} = 1*x_1 + (-2)*x_2$ So y is a linear combination of $x_1$ and $x_2$. The set of all linear combinations of $x_1, x_2, \\ldots , x_n$ is called", "# Perpendicular lines and vectors My textbook is confusing me a little. Here is a worked example from my textbook: Line $$l$$ has the equation $$\\begin{pmatrix}3\\\\ -1\\\\ 0\\end{pmatrix}+\\lambda \\begin{pmatrix}1\\\\ -1\\\\ 1\\end{pmatrix}$$ and point $$A$$ has co-ordinates $$(3, 9, -2)$$. Find the coordinates of point $$B$$ on $$l$$ so that $$AB$$ is perpendicular to $$l$$. $$\\vec{AB\\cdot }\\begin{pmatrix}1\\\\ -1\\\\ 1\\end{pmatrix}=0$$ $$\\vec{OB}=r=\\begin{pmatrix}3+\\lambda \\\\ -1-\\lambda \\\\ \\lambda \\end{pmatrix}$$ $$\\vec{AB}=\\begin{pmatrix}\\lambda \\\\ -10-\\lambda \\\\ \\lambda +2\\end{pmatrix}$$ $$\\begin{pmatrix}\\lambda \\\\ -10-\\lambda \\\\ \\lambda +2\\end{pmatrix}\\cdot \\begin{pmatrix}1\\\\ -1\\\\ 1\\end{pmatrix}=0$$ $$3\\lambda= -12, \\lambda = -4$$ Coordinates of $$B$$: $$(-1, 3, -4)$$ The thing I don't understand is why they found the dot product of the line AB and the direction vector of line l. My textbook does mention that to check whether two vectors are perpendicular, $$a\\cdot b = 0$$ and for lines, the dot product of their direction vectors = 0. So why did they mix both here? Didn't they use the entire line $$AB$$ and then just the direction vector of line l? Or am I missing something as usual? There is nothing special here the dot product $AB\\cdot v$ gives the condition of orthogonality between the vector from $A$ to $B$ and the direction vector of the given line. Note also that vector $AB$ is a direction vector for the line orthogonal to the given line", "$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$ and the second has normal vector $\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $v_1,v_2$ is $v_1\\wedge v_2$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\wedge \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $y=0$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $x=0, z=1$, thus the line of intersection is $\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$ -", "the two points $A$ and $C$. $M_{AB}(\\frac{5+1}{2}|\\frac{2+4}{2})$ $M_{AB}(3|3)$ 2. #### Normal vector on two sides A vertical vector is now determined for each of the two sides selected. This serves as the direction vector for the line, so that it lies vertically on the side (prerequisite for a perpendicular bisector). Direction vector for $g_{AB}$ A vector must be found with which the scalar product results in zero (= vectors perpendicular). $\\vec{AB}\\cdot\\vec{n_{AB}}=0$ $\\begin{pmatrix} -4 \\\\ 0 \\end{pmatrix}\\cdot\\color{blue}{\\begin{pmatrix} \\, \\\\ \\, \\end{pmatrix}} = 0$ It is particularly easy if you swap the two coordinates and change exactly one sign. $\\begin{pmatrix} \\color{red}{-4} \\\\ \\color{red}{0} \\end{pmatrix}\\cdot\\begin{pmatrix} \\color{blue}{0} \\\\ \\color{blue}{+4} \\end{pmatrix} = 0$ $\\vec{n_{AB}}=\\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$ Direction vector for $g_{AC}$ $\\vec{AC}\\cdot\\vec{n_{AC}}=0$ $\\begin{pmatrix} -4 \\\\ 2 \\end{pmatrix}\\cdot\\color{blue}{\\begin{pmatrix} \\, \\\\ \\, \\end{pmatrix}} = 0$ $\\begin{pmatrix} \\color{red}{-4} \\\\ \\color{red}{2} \\end{pmatrix}\\cdot\\begin{pmatrix} \\color{blue}{2} \\\\ \\color{blue}{+4} \\end{pmatrix} = 0$ $\\vec{n_{AC}}=\\begin{pmatrix} 2 \\\\ 4 \\end{pmatrix}$ 3. #### Equation of a line Set up equation of a line with the support vector and the direction vector. Equation of a line for $g_{AB}$ $g_{AB}: \\vec{x} = \\vec{OM_{AB}} + r \\cdot \\vec{n_{AB}}$ $g_{AB}: \\vec{x} = \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$ Equation of a line for $g_{AC}$ $g_{AC}: \\vec{x} = \\vec{OM_{AC}} + s \\cdot \\vec{n_{AC}}$ $g_{AC}: \\vec{x} = \\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix} +", "Math Relative position of lines Determining point of intersection # Intersecting lines Two lines can intersect and then have a point of intersection. ! ### Requirements 1. Direction vectors are not collinear 2. When equating the linear equations, we receive a distinct result ## Calculating point of intersection To calculate the point of intersection, we follow the scheme above. The result from step 2 for $r$ or $s$ is then inserted into the corresponding linear equation. ### Example $\\text{g: } \\vec{x} = \\begin{pmatrix} 10 \\\\ 9 \\\\ 7 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}$ $\\text{h: } \\vec{x} = \\begin{pmatrix} 4 \\\\ 4 \\\\ 6 \\end{pmatrix} + s \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ 1. #### Direction vectors not collinear? First, we examine if the direction vectors are multiples of each other (=collinear). $\\vec{a}=t\\cdot\\vec{b}$ $\\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}=t\\cdot\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ Calculate $t$ for every row 1. $9=t\\cdot1$ 2. $8=t\\cdot1$ 3. $7=t\\cdot2$ 1. $t=9$ 2. $t=8$ 3. $t=3.5$ $t$ does not have the same value everywhere. Thus the direction vectors are not collinear. Therefore the lines are skew or have a point of intersection. 2. #### Equate $g$ and $h$ Now equate the linear equations and calculate $r$ and $s$. $\\begin{pmatrix} 10 \\\\ 9 \\\\ 7", "# Find the value of $a$ for which two lines are perpendicular Two vectors are perpendicular when their dot product is zero. But how to get the dot product of two lines? $r_1= (8,-3,1)+s(12,-5,0)$ $r_2= (1,14,3)+t(5,a,b)$ The question asks for the value of $a$ given that the two lines are perpendicular. • The 2 lines are perpendicular if $(12,-5,0).(5,a,b)=0$ which gives $a=12$. – Paracosmiste Aug 23 '14 at 14:44 The vectors $(12,-5,0)$ and $(5,a,b)$ are the direction vectors of the lines. If the dot product of those two vectors is zero, then the lines are perpendicular. Hence you have to calculate: $$(12,-5,0)\\cdot(5,a,b)=0$$ This gives: $12\\cdot5-5\\cdot a=0 \\Longleftrightarrow 60-5a=0 \\Longrightarrow \\boxed{a=12}$", "if the direction vectors are multiples of each other (=collinear). $\\vec{a}=t\\cdot\\vec{b}$ $\\begin{pmatrix} -3 \\\\ 5 \\\\ 2 \\end{pmatrix}=t\\cdot\\begin{pmatrix} -3 \\\\ 5 \\\\ 2 \\end{pmatrix}$ Calculate $t$ for every row $-3=t\\cdot(-3)$ $5=t\\cdot5$ $2=t\\cdot2$ If $t$ equals the same in each row (here: $t=1$) then the vectors are collinear. 2. #### Insert the vector of $h$ into $g$ Now the position vector of one line (here: $h$) is inserted into the other. $\\begin{pmatrix} -3 \\\\ 14 \\\\ 10 \\end{pmatrix} = \\begin{pmatrix} 3 \\\\ 4 \\\\ 6 \\end{pmatrix} + r \\cdot \\begin{pmatrix} -3 \\\\ 5 \\\\ 2 \\end{pmatrix}$ Now we set up the equation system and solve it. If $r$ equals the same in every equation, then the terminal point of the position vector is on both lines and thus they are coincident. 1. $-3=3-3r$ 2. $14=4+5r$ 3. $10=6+2r$ 1. $r=2$ 2. $r=2$ 3. $r=2$ => The lines are coincident.", "& \\frac{-5}{14} \\\\ \\frac{2}{14} & \\frac{4}{14} \\end{pmatrix}$ For finding the matrix inverse in general, you can use Gauss-Jordan Algorithm. However, this is a rather complicated algorithm, so usually one relies upon the computer or calculator to find the matrix inverse. Orthogonal Vectors Two vectors, x and y, are orthogonal if their dot product is zero. For example $e \\cdot f = \\begin{pmatrix} 2 & 5 & 4 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 5 \\end{pmatrix} = 2*4 + (5)*(-2) + 4*5 = 8-10+20 = 18$ Vectors e and f are not orthogonal. $g \\cdot h = \\begin{pmatrix} 2 & 3 & -2 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 1 \\end{pmatrix} = 2*4 + (3)*(-2) + (-2)*1 = 8-6-2 = 0$ However, vectors g and h are orthogonal. Orthogonal can be thought of as an expansion of perpendicular for higher dimensions. Let $$x_1, x_2, \\ldots , x_n$$ be m-dimensional vectors. Then a linear combination of $$x_1, x_2, \\ldots , x_n$$ is any m-dimensional vector that can be expressed as $c_1 x_1 + c_2 x_2 + \\ldots + c_n x_n$ where $$c_1, \\ldots, c_n$$ are all scalars. For example: $x_1 =\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix}$ $y =\\begin{pmatrix} -5 \\\\ 12 \\\\ -8 \\end{pmatrix} = 1*\\begin{pmatrix} 3 \\\\ 8", "really treating the possible triangle sides as vectors. And, with perpendicular vectors, we can see that we can get something like one of these two pairs (though perpendicular vectors don’t have to look like this): $\\mathtt{\\begin{bmatrix} x_1\\\\x_2 \\end{bmatrix} \\textrm{and} \\begin{bmatrix} -x_2\\\\ \\,\\,\\,\\,x_1 \\end{bmatrix} \\textrm{or} \\begin{bmatrix} x_1\\\\x_2 \\end{bmatrix} \\textrm{and} \\begin{bmatrix} \\,\\,\\,\\,x_2\\\\-x_1 \\end{bmatrix}}$ The dot product is defined as the sum of the element-wise products of the vector components. In the case of perpendicular vectors, the dot product is 0. Here is the dot product of our vectors: $\\mathtt{(x_1 \\cdot -x_2) + (x_2 \\cdot x_1) = 0}$ Some Programming One reason why this way of defining perpendicularity (with a single value) is helpful is that we avoid nasty zero denominators and, therefore, undefined slopes. With the two vectors at the right, we get $\\mathtt{\\begin{bmatrix}0\\\\5\\end{bmatrix} \\cdot \\begin{bmatrix}4\\\\0\\end{bmatrix} = (0)(4) + (5)(0) = 0}$ We can take all of the points and run them through a program to find all the connected perpendicular vectors. The result ((0, 1), (1, 0)), ((0, 1), (2, 3)) below means that the vector connecting (0, 1) and (1, 0) and the vector connecting (0, 1) and (2, 3) are perpendicular. This gives us all the perpendicular vector pairs, though it doesn’t filter out those vectors with unequal magnitudes, which we wanted in order to identify the", "Well i didn't know how to put $\\begin{pmatrix}t\\\\\\!\\!\\!-t\\\\3t\\end{pmatrix}$ in. I am in engineering i learned about \"dot product\" before inner product so i associate Euclidean inner product with dot product. Since no vector space is mentioned in the question it is probably ok to assume it's euclidean. That being said i don't really know much about linear algebra, it is by far my weakest course. I hope i cleared everything up for you. 4. Okay, \"letting t= 1\" in that gives $\\begin{pmatrix}1 \\\\ -1\\\\ 3\\end{pmatrix}$. A vector, $\\begin{pmatrix}a \\\\ b\\\\ c\\end{pmatrix}$, will be orthogonal to that if and only if $\\begin{pmatrix}a \\\\ b\\\\ c\\end{pmatrix}\\cdot\\begin{pmatrix}1 \\\\ -1 \\\\ 3\\end{pmatrix}= a- b+ 3c= 0$ From a- b+ 3c= 0, b= a+ 3c so $\\begin{pmatrix}a \\\\ b\\\\ c\\end{pmatrix}= \\begin{pmatrix}a \\\\ a+ 3c \\\\ c\\end{pmatrix}= a\\begin{pmatrix}1 \\\\ 1 \\\\ 0\\end{pmatrix}+ c\\begin{pmatrix}0 \\\\ 3 \\\\ 1\\end{pmatrix}$. 5. Thank you halls, i was heading along the right track, just couldn't think straight. Thank very much!", "# How to find the position vector for the point of intersection of a line and the perpendicular line through a point C How could I find the exact coordinates of the point N for example which is the point of intersection of the line $$L=\\begin{pmatrix} 0 \\\\ -1 \\\\ 2 \\end{pmatrix} +t\\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}$$ and a perpendicular line that passes through a point C(1,2,3) not on the line. What I have done so far is find the value of point N in terms of $$t$$ to find the vector $$\\overrightarrow{CN}$$: $$\\longrightarrow N\\begin{pmatrix} -3t \\\\ -2t-1 \\\\ -3t+2 \\end{pmatrix}$$ $$\\longrightarrow \\overrightarrow{CN}=\\begin{pmatrix} -3t \\\\ -2t-1 \\\\ -3t+2 \\end{pmatrix}-\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ $$=\\begin{pmatrix} -3t-1 \\\\ -2t-3 \\\\ -3t-1 \\end{pmatrix}$$ After this I made the scalar product eual zero: $$\\overrightarrow{CN} \\cdot \\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}=0$$ With this I find that, $$t=\\frac{-6}{11}$$ But my textbook says that $$t=\\frac{5}{11}$$ Can anyone confirm which answer for $$t$$ is right and if it is $$t=\\frac{5}{11}$$ then what did I do wrong? • For me, your answer it correct. Jul 30 '20 at 12:32 • WA agrees. Maybe there's the line $L$ points swapped in the solution? Jul 30 '20 at 12:42 • Thank you for the confirmation Jul 30 '20 at 12:42 Your derivation seems", "any other vector is defined to be a right angle). Thus vectors from $\\mathbb{R}^n$ are orthogonal (or perpendicular) if and only if their dot product is zero. Example 2.8 These vectors are orthogonal. $\\begin{pmatrix} 1 \\\\ -1 \\end{pmatrix}\\cdot\\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}=0$ The arrows are shown away from canonical position but nevertheless the vectors are orthogonal. Example 2.9 The $\\mathbb{R}^3$ angle formula given at the start of this subsection is a special case of the definition. Between these two the angle is $\\arccos(\\frac{(1)(0)+(1)(3)+(0)(2)}{\\sqrt{1^2+1^2+0^2}\\sqrt{0^2+3^2+2^2}}) =\\arccos(\\frac{3}{\\sqrt{2}\\sqrt{13}})$ approximately $0.94 \\text{radians}$. Notice that these vectors are not orthogonal. Although the $yz$-plane may appear to be perpendicular to the $xy$-plane, in fact the two planes are that way only in the weak sense that there are vectors in each orthogonal to all vectors in the other. Not every vector in each is orthogonal to all vectors in the other. ## Exercises This exercise is recommended for all readers. Problem 1 Find the length of each vector. 1. $\\begin{pmatrix} 3 \\\\ 1 \\end{pmatrix}$ 2. $\\begin{pmatrix} -1 \\\\ 2 \\end{pmatrix}$ 3. $\\begin{pmatrix} 4 \\\\ 1 \\\\ 1 \\end{pmatrix}$ 4. $\\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$ 5. $\\begin{pmatrix} 1 \\\\ -1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 2 Find the angle between each two, if it is defined.", "From the list below, select the vector that is NOT orthogonal (perpendicular) to: $\\begin{bmatrix} \\cos\\theta \\\\ \\sin\\theta \\\\ 1 \\end{bmatrix}$", "is the same you got setting t values <0,1,1> then set up the dot product and solve for t and input t into the direction vector I got from taking the point minus the line, then write the new parametric equations based off my original point and the direction vector I'll do that way but for sake of completeness (1) the book suggests find the intersection (2) can I use projection to find it as well?? 5. ## Re: equation of line through point orthogonal to line I got two points along the given line by plugging in $t=0$ and $t=1$. When $t=0$, I get the point $(-1,-2+0,-1+0) = (-1,-2,-1)$. When $t=1$, I get the point $(-1,-2+1,-1+1) = (-1,-1,0)$. So, the vector from the first point to the second point is $\\begin{pmatrix}-1-(-1) \\\\ -1-(-2) \\\\ 0-(-1)\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. That is a vector in the direction of the given line (Essentially, this a vector containing some multiple of the coefficients of $t$ in the parametric equation of the given line). You can rewrite the line as $\\begin{pmatrix}-1 \\\\ -2 \\\\ -1\\end{pmatrix} + t\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. This immediately gives a vector in the direction of the given line to be $\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Essentially, what I did was I found all possible", "it from? But, for this question, I got the same answer as Cullen_Im. aleksanderjanczewski: June 1, 2015, 11:25 p.m. I got different vectors than you, I guess you shoud try to derive identity matrix for change of basis and then multiply vector <a,b> and <c,d> by this matrix. This will give you these vectors with respect to your friend's basis. TLindberg: June 2, 2015, 4:39 p.m. A more concise way to state:$\\langle \\begin{pmatrix} a \\\\ b \\end{pmatrix} , \\begin{pmatrix} c \\\\ d \\end{pmatrix}\\rangle$is 2ac + 4bc + 4ad + 10bd.$\\\\$You can do this by computing the dot product of the the two vectors$\\begin{pmatrix} a+ 3b \\\\ a + b \\end{pmatrix} \\bullet \\begin{pmatrix} c + 3d \\\\ c + d \\end{pmatrix}\\\\\\$ I'm not sure if this is better, but this might be a safer answer on the test, because it shows that you more thoroughly understand the concept. Cullen_Im: June 3, 2015, 12:27 a.m. @tedallen: I don't believe there is a an absolute value around (v dot product w) because you're trying to find the angle between vectors and not between lines. The absolute value makes the angle be less than 90 degrees, but the angle between vectors can be greater than 90 degrees so the absolute value is not needed.", "$AB$ and $CD$ have the same direction, then $r\\cdot AB=CD$ for some real number $r\\geq 0$ (If there is such $r$, but it is negative, they are exactly in opposite directions! Sometimes when people say \"same direction\", they include the case $r<0$ which can be quite confusing...). In your case this means $r\\cdot\\begin{pmatrix}2\\\\1\\\\3\\end{pmatrix}=\\begin{pmatrix}4\\\\3\\\\6\\end{pmatrix}$ which means $\\begin{pmatrix}r\\cdot 2\\\\r\\cdot 1\\\\r\\cdot 3\\end{pmatrix}=\\begin{pmatrix}4\\\\3\\\\6\\end{pmatrix}$, so there must be a $r\\geq 0$ such that \\begin{align*}2r&=4 \\\\r&=3\\\\3r&=6\\end{align*} hold at the same time. Well, that's not possible, because if $r=3$, then $2r=6\\neq 4$. Just for illustration why this is the correct way (only 2D, works for 3D the same): So if we want to know if $AB$ and $CD$ have the same direction, we fix the starting point of $AB$ and $CD$ to be in the same place (usually $(0,0)$, so the end point of the vectors is the same as in the vector representation, but it doesn't really matter). The vectors (and their direction) don't change when we \"move\" them around like that which is an important point in the concept of vectors. Then, we clearly see that in this example they're not in the same direction. (And we observe that there's no $r$ such that $r\\cdot AB=CD$, which you can proove the same way as I did with your example above.) Take another example:" ]

[ "## Directional and orthogonal form of a line Directional and orthogonal form of a line Directional form: \\begin{align*} \\begin{pmatrix} x_1 \\\\ x_2 \\end{pmatrix} &= \\begin{pmatrix} a_1 \\\\ a_2 \\end{pmatrix} + \\alpha \\begin{pmatrix} d_1 \\\\ d_2 \\end{pmatrix} \\end{align*} or, concisely \\begin{align} x = a + \\alpha d \\end{align} Let $v$ be a vector perpendicular to $d$. Transpose both sides and multiply by $v$: \\begin{align} x'v &= a'v + \\alpha d'v = a'v = c \\end{align} Where c is a constant scalar. This is called the orthogonal form. If you see it as $f(x) = x'v$, then $v$ is the gradient of $f$.", "# Perpendicular lines and vectors My textbook is confusing me a little. Here is a worked example from my textbook: Line $$l$$ has the equation $$\\begin{pmatrix}3\\\\ -1\\\\ 0\\end{pmatrix}+\\lambda \\begin{pmatrix}1\\\\ -1\\\\ 1\\end{pmatrix}$$ and point $$A$$ has co-ordinates $$(3, 9, -2)$$. Find the coordinates of point $$B$$ on $$l$$ so that $$AB$$ is perpendicular to $$l$$. $$\\vec{AB\\cdot }\\begin{pmatrix}1\\\\ -1\\\\ 1\\end{pmatrix}=0$$ $$\\vec{OB}=r=\\begin{pmatrix}3+\\lambda \\\\ -1-\\lambda \\\\ \\lambda \\end{pmatrix}$$ $$\\vec{AB}=\\begin{pmatrix}\\lambda \\\\ -10-\\lambda \\\\ \\lambda +2\\end{pmatrix}$$ $$\\begin{pmatrix}\\lambda \\\\ -10-\\lambda \\\\ \\lambda +2\\end{pmatrix}\\cdot \\begin{pmatrix}1\\\\ -1\\\\ 1\\end{pmatrix}=0$$ $$3\\lambda= -12, \\lambda = -4$$ Coordinates of $$B$$: $$(-1, 3, -4)$$ The thing I don't understand is why they found the dot product of the line AB and the direction vector of line l. My textbook does mention that to check whether two vectors are perpendicular, $$a\\cdot b = 0$$ and for lines, the dot product of their direction vectors = 0. So why did they mix both here? Didn't they use the entire line $$AB$$ and then just the direction vector of line l? Or am I missing something as usual? There is nothing special here the dot product $AB\\cdot v$ gives the condition of orthogonality between the vector from $A$ to $B$ and the direction vector of the given line. Note also that vector $AB$ is a direction vector for the line orthogonal to the given line", "# Another Line Equation Case At Line $L_{1}$ has equation $r = \\begin{pmatrix} -5\\\\ -3\\\\ 2 \\end{pmatrix} + \\lambda \\begin{pmatrix} -1\\\\ 2\\\\ 2 \\end{pmatrix}$ A line $L_{2}$ passing through the origin intersects $L_{1}$ and perpendicular to $L_{1}$. Find the vector equation of $L_{2}$! I find the vector for $L_{2}$ using cross product $\\begin{pmatrix} i & j & k\\\\ -5 & -3 & 2\\\\ -1 & 2 & 2 \\end{pmatrix}$, and I don't have any idea to do with that vector.... • You need to solve $r\\cdot\\begin{pmatrix}-1\\\\ 2\\\\ 2 \\end{pmatrix}=0$ for $\\lambda$, it gives the foot of the perpendicular. Jun 3, 2015 at 1:51 • What is $r$ on that dot product? I know that $\\begin{pmatrix}-1\\\\ 2\\\\ 2 \\end{pmatrix}$ is the direction vector. Jun 3, 2015 at 2:12 • It's $r$ for $L_1$ from above Jun 3, 2015 at 2:13 • So $r\\cdot\\begin{pmatrix}-1\\\\ 2\\\\ 2 \\end{pmatrix}=0$ is for searching the lambda. After do the dot product, I get : $\\begin{pmatrix} a\\\\ b\\\\ c \\end{pmatrix}\\cdot \\begin{pmatrix} -1\\\\ 2\\\\ 2 \\end{pmatrix}=0$ and become : $-a+2b+2c=0$ Jun 3, 2015 at 2:18 You had the right idea that the cross product will give you an orthogonal vector, but taking $(-5,-3,2)\\times(-1,2,2)$ will give you a vector orthogonal to the direction of the line and a point on the line, not necessarily the direction perpendicular", "point, there is exactly one. I assume you want the equation for that line. Since the line passes through both the given point and the given line, a vector in the direction of this perpendicular line would be the vector $\\begin{pmatrix}3-(-1) \\\\ 1-(-2+t) \\\\ -2-(-1+t)\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 3-t \\\\ -1-t\\end{pmatrix}$. Let's find a vector in the direction of the given line by finding two points along the line (we can choose any two points, so let's say $t=0, t=1$). So, a vector in the direction of the given line is $\\begin{pmatrix}-1-(-1) \\\\ -2+1 - (-2+0) \\\\ -1+1 - (-1+0)\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Since we want these vectors to be perpendicular, we want the dot product to be zero. Hence: $(4\\cdot 0) + (1\\cdot (3-t)) + (1\\cdot (-1-t)) = 2-2t = 0$ implies $t=1$. Hence, a vector in the direction of the perpendicular line is $\\begin{pmatrix}4 \\\\ 3-1 \\\\ -1-1\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 2 \\\\ -2\\end{pmatrix}$. So, the perpendicular line would have parametric equation $x=3+4t, y=1+2t, z=-2-2t$. 4. ## Re: equation of line through point orthogonal to line so the point (does this mean you are treating the point as a bounded vector) minus the line gives a direction vector along the perpendicular line then I take the given direction vector of the line which", "$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$ and the second has normal vector $\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $v_1,v_2$ is $v_1\\wedge v_2$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\wedge \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. We can try $y=0$ and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $x=0, z=1$, thus the line of intersection is $\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb R\\rbrace}$ -", "vectors form an orthogonal set. Solution. To show that vectors $\\boldsymbol{v}$ and $\\boldsymbol{w}$ form an orthogonal set, we must check that they are perpendicular to each other. We know from theoremlink that if the dot product of two vectors is zero, then they are perpendicular. So let's compute the dot product between them: \\begin{align*} \\begin{pmatrix} 1\\\\-1 \\end{pmatrix}\\cdot \\begin{pmatrix} 2\\\\2 \\end{pmatrix}&= (1)(2)+(-1)(2)\\\\&=0 \\end{align*} Because $\\boldsymbol{v}$ and $\\boldsymbol{w}$ are orthogonal to each other, we conclude that they form an orthogonal set. Let's also visually confirm this: Indeed, they are perpendicular! Definition. # Orthonormal set An orthonormal set is an orthogonal set of unit vectors. In other words, a set of vectors is orthonormal if and only if: • every pair of vectors is perpendicular. • every vector has a length of one. Example. ## Showing that a set is orthonormal Consider the standard basis vectors for $\\mathbb{R}^2$ below: $$\\boldsymbol{e}_1= \\begin{pmatrix} 1\\\\0 \\end{pmatrix},\\;\\;\\;\\;\\; \\boldsymbol{e}_2= \\begin{pmatrix}0\\\\1 \\end{pmatrix}$$ Show that these vectors form an orthonormal set. Solution. Let's first check that they are perpendicular to each other: \\begin{align*} \\boldsymbol{e}_1\\cdot\\boldsymbol{e}_2&= \\begin{pmatrix} 1\\\\0 \\end{pmatrix}\\cdot \\begin{pmatrix} 0\\\\1 \\end{pmatrix}\\\\ &=(1)(0)+(0)(1)\\\\ &=0 \\end{align*} Because their dot product is zero, they are perpendicular. We can also easily tell that these vectors have both length one, and so $\\boldsymbol{e}_1$ and $\\boldsymbol{e}_2$ form an orthonormal set by definition! We can visually", "Math Relative position of lines Determining point of intersection # Intersecting lines Two lines can intersect and then have a point of intersection. ! ### Requirements 1. Direction vectors are not collinear 2. When equating the linear equations, we receive a distinct result ## Calculating point of intersection To calculate the point of intersection, we follow the scheme above. The result from step 2 for $r$ or $s$ is then inserted into the corresponding linear equation. ### Example $\\text{g: } \\vec{x} = \\begin{pmatrix} 10 \\\\ 9 \\\\ 7 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}$ $\\text{h: } \\vec{x} = \\begin{pmatrix} 4 \\\\ 4 \\\\ 6 \\end{pmatrix} + s \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ 1. #### Direction vectors not collinear? First, we examine if the direction vectors are multiples of each other (=collinear). $\\vec{a}=t\\cdot\\vec{b}$ $\\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}=t\\cdot\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ Calculate $t$ for every row 1. $9=t\\cdot1$ 2. $8=t\\cdot1$ 3. $7=t\\cdot2$ 1. $t=9$ 2. $t=8$ 3. $t=3.5$ $t$ does not have the same value everywhere. Thus the direction vectors are not collinear. Therefore the lines are skew or have a point of intersection. 2. #### Equate $g$ and $h$ Now equate the linear equations and calculate $r$ and $s$. $\\begin{pmatrix} 10 \\\\ 9 \\\\ 7", "# How to find the position vector for the point of intersection of a line and the perpendicular line through a point C How could I find the exact coordinates of the point N for example which is the point of intersection of the line $$L=\\begin{pmatrix} 0 \\\\ -1 \\\\ 2 \\end{pmatrix} +t\\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}$$ and a perpendicular line that passes through a point C(1,2,3) not on the line. What I have done so far is find the value of point N in terms of $$t$$ to find the vector $$\\overrightarrow{CN}$$: $$\\longrightarrow N\\begin{pmatrix} -3t \\\\ -2t-1 \\\\ -3t+2 \\end{pmatrix}$$ $$\\longrightarrow \\overrightarrow{CN}=\\begin{pmatrix} -3t \\\\ -2t-1 \\\\ -3t+2 \\end{pmatrix}-\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ $$=\\begin{pmatrix} -3t-1 \\\\ -2t-3 \\\\ -3t-1 \\end{pmatrix}$$ After this I made the scalar product eual zero: $$\\overrightarrow{CN} \\cdot \\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}=0$$ With this I find that, $$t=\\frac{-6}{11}$$ But my textbook says that $$t=\\frac{5}{11}$$ Can anyone confirm which answer for $$t$$ is right and if it is $$t=\\frac{5}{11}$$ then what did I do wrong? • For me, your answer it correct. Jul 30 '20 at 12:32 • WA agrees. Maybe there's the line $L$ points swapped in the solution? Jul 30 '20 at 12:42 • Thank you for the confirmation Jul 30 '20 at 12:42 Your derivation seems", "if the direction vectors are multiples of each other (=collinear). $\\vec{a}=t\\cdot\\vec{b}$ $\\begin{pmatrix} -3 \\\\ 5 \\\\ 2 \\end{pmatrix}=t\\cdot\\begin{pmatrix} -3 \\\\ 5 \\\\ 2 \\end{pmatrix}$ Calculate $t$ for every row $-3=t\\cdot(-3)$ $5=t\\cdot5$ $2=t\\cdot2$ If $t$ equals the same in each row (here: $t=1$) then the vectors are collinear. 2. #### Insert the vector of $h$ into $g$ Now the position vector of one line (here: $h$) is inserted into the other. $\\begin{pmatrix} -3 \\\\ 14 \\\\ 10 \\end{pmatrix} = \\begin{pmatrix} 3 \\\\ 4 \\\\ 6 \\end{pmatrix} + r \\cdot \\begin{pmatrix} -3 \\\\ 5 \\\\ 2 \\end{pmatrix}$ Now we set up the equation system and solve it. If $r$ equals the same in every equation, then the terminal point of the position vector is on both lines and thus they are coincident. 1. $-3=3-3r$ 2. $14=4+5r$ 3. $10=6+2r$ 1. $r=2$ 2. $r=2$ 3. $r=2$ => The lines are coincident.", "Orthogonal Vectors Two vectors, x and y, are orthogonal if their dot product is zero. For example $e \\cdot f = \\begin{pmatrix} 2 & 5 & 4 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 5 \\end{pmatrix} = 2*4 + (5)*(-2) + 4*5 = 8-10+20 = 18$ Vectors e and f are not orthogonal. $g \\cdot h = \\begin{pmatrix} 2 & 3 & -2 \\end{pmatrix} * \\begin{pmatrix} 4 \\\\ -2 \\\\ 1 \\end{pmatrix} = 2*4 + (3)*(-2) + (-2)*1 = 8-6-2 = 0$ However, vectors g and h are orthogonal. Orthogonal can be thought of as an expansion of perpendicular for higher dimensions. Let $x_1, x_2, \\ldots , x_n$ be m-dimensional vectors. Then a linear combination of $x_1, x_2, \\ldots , x_n$ is any m-dimensional vector that can be expressed as $c_1 x_1 + c_2 x_2 + \\ldots + c_n x_n$ where $c_1, \\ldots, c_n$ are all scalars. For example: $x_1 =\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix}$ $y =\\begin{pmatrix} -5 \\\\ 12 \\\\ -8 \\end{pmatrix} = 1*\\begin{pmatrix} 3 \\\\ 8 \\\\ -2 \\end{pmatrix} + (-2)* \\begin{pmatrix} 4 \\\\ -2 \\\\ 3 \\end{pmatrix} = 1*x_1 + (-2)*x_2$ So y is a linear combination of $x_1$ and $x_2$. The set of all linear combinations of $x_1, x_2, \\ldots , x_n$ is called", "the two points $A$ and $C$. $M_{AB}(\\frac{5+1}{2}|\\frac{2+4}{2})$ $M_{AB}(3|3)$ 2. #### Normal vector on two sides A vertical vector is now determined for each of the two sides selected. This serves as the direction vector for the line, so that it lies vertically on the side (prerequisite for a perpendicular bisector). Direction vector for $g_{AB}$ A vector must be found with which the scalar product results in zero (= vectors perpendicular). $\\vec{AB}\\cdot\\vec{n_{AB}}=0$ $\\begin{pmatrix} -4 \\\\ 0 \\end{pmatrix}\\cdot\\color{blue}{\\begin{pmatrix} \\, \\\\ \\, \\end{pmatrix}} = 0$ It is particularly easy if you swap the two coordinates and change exactly one sign. $\\begin{pmatrix} \\color{red}{-4} \\\\ \\color{red}{0} \\end{pmatrix}\\cdot\\begin{pmatrix} \\color{blue}{0} \\\\ \\color{blue}{+4} \\end{pmatrix} = 0$ $\\vec{n_{AB}}=\\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$ Direction vector for $g_{AC}$ $\\vec{AC}\\cdot\\vec{n_{AC}}=0$ $\\begin{pmatrix} -4 \\\\ 2 \\end{pmatrix}\\cdot\\color{blue}{\\begin{pmatrix} \\, \\\\ \\, \\end{pmatrix}} = 0$ $\\begin{pmatrix} \\color{red}{-4} \\\\ \\color{red}{2} \\end{pmatrix}\\cdot\\begin{pmatrix} \\color{blue}{2} \\\\ \\color{blue}{+4} \\end{pmatrix} = 0$ $\\vec{n_{AC}}=\\begin{pmatrix} 2 \\\\ 4 \\end{pmatrix}$ 3. #### Equation of a line Set up equation of a line with the support vector and the direction vector. Equation of a line for $g_{AB}$ $g_{AB}: \\vec{x} = \\vec{OM_{AB}} + r \\cdot \\vec{n_{AB}}$ $g_{AB}: \\vec{x} = \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$ Equation of a line for $g_{AC}$ $g_{AC}: \\vec{x} = \\vec{OM_{AC}} + s \\cdot \\vec{n_{AC}}$ $g_{AC}: \\vec{x} = \\begin{pmatrix} 3 \\\\ 3 \\end{pmatrix} +", "lines have the same $x$-intercept. • These lines both go through the point $(1, 5)$. • These lines … Can you sort them into the correct pairs and complete the final description? While sketching the lines could be helpful, 12 might be too many on one graph. Are there any pairs we can match up by just looking at the equations? $\\mathbf{r}= \\begin{pmatrix} 0 \\\\ -4 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} 4 \\\\ 6 \\end{pmatrix}$ $\\mathbf{r}= \\begin{pmatrix} 0 \\\\ -4 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} 1 \\\\ 6 \\end{pmatrix}$ Equations 7 and 10 have the same position vector which also happens to be their $y$-interept. Do you think there are particular features of lines that are easier to spot when the equation is written in vector form compared to Cartesian form? $\\mathbf{r}= \\begin{pmatrix} 2 \\\\ 9.5 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} -2 \\\\ -8 \\end{pmatrix}$ $\\mathbf{r}= \\begin{pmatrix} 0 \\\\ 4 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} 1 \\\\ 4 \\end{pmatrix}$ Equations 3 and 6 are parallel lines, but it looks like they have different direction vectors. How do you know these lines are parallel? What is the same about the two direction vectors? $\\mathbf{r}= \\begin{pmatrix} 0 \\\\ -6 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} -1 \\\\ 4 \\end{pmatrix}$ $\\mathbf{r}= \\begin{pmatrix} 1 \\\\ 1 \\\\ \\end{pmatrix} + \\mu \\begin{pmatrix} 4 \\\\", "\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ 2. Calculate perpendicular point The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$ $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ 3. Calculate the distance between the two points The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can be easily calculated using vectors. $d=|\\vec{PF}|$ $=\\left| \\begin{pmatrix} -1-(-1) \\\\ 0-0 \\\\ 1-3 \\end{pmatrix}\\right|$ $=\\left| \\begin{pmatrix} 0 \\\\ 0 \\\\ -2 \\end{pmatrix}\\right|$ $=\\sqrt{(-2)^2}$ $=2$ The distance from the line $g$ to the point $P$ is 2 LU.", "to the other line The distance can now be calculated as described under distance point and line. First set up an auxiliary plane with $P$ as the support point and direction vector as the normal vector. $\\text{H: } (\\vec{x} - \\vec{a}) \\cdot \\vec{n}=0$ $\\text{H: } (\\vec{x} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}) \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$. $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can", "$$z$$ term. Hence, the direction vector for the first line is: $\\begin{bmatrix} 2 \\\\ 3\\\\ 1 \\end{bmatrix}$ and the direction vector the the second line is: $\\begin{bmatrix} 6 \\\\9\\\\3 \\end{bmatrix}.$ You can see that if you multiply the first direction vector by 3, you will get the second direction vector. This means that the two lines must be parallel, since the direction vectors are pointing in the same direction. This next example will give the lines in vector form. Determine whether the following lines intersect. If they intersect, find the point of intersection. If they do not intersect, determine whether they are skewed or parallel. \\begin{align} \\vec{r}_1 & = \\begin{bmatrix} 2 \\\\ -4 \\\\ 3 \\end{bmatrix} + t \\begin{bmatrix} 3 \\\\ 11 \\\\ -2 \\end{bmatrix} \\\\ \\vec{r}_2 & = \\begin{bmatrix} 1 \\\\ 6 \\\\ 3 \\end{bmatrix} + s \\begin{bmatrix} 4 \\\\ 1\\\\ -2 \\end{bmatrix}. \\end{align} Solution Firstly, you can tell the lines are not parallel since the direction vectors are not multiples of one another. This means that they must either intersect or be skewed. To determine this, first set the equations equal to one another: $\\begin{bmatrix} 2 \\\\ -4 \\\\ 3 \\end{bmatrix} + t \\begin{bmatrix} 3 \\\\ 11 \\\\ -2 \\end{bmatrix} = \\begin{bmatrix} 1 \\\\ 6 \\\\ 3 \\end{bmatrix} + s \\begin{bmatrix}4 \\\\ 1\\\\ -2 \\end{bmatrix}.$ This", "# Find the value of $a$ for which two lines are perpendicular Two vectors are perpendicular when their dot product is zero. But how to get the dot product of two lines? $r_1= (8,-3,1)+s(12,-5,0)$ $r_2= (1,14,3)+t(5,a,b)$ The question asks for the value of $a$ given that the two lines are perpendicular. • The 2 lines are perpendicular if $(12,-5,0).(5,a,b)=0$ which gives $a=12$. – Paracosmiste Aug 23 '14 at 14:44 The vectors $(12,-5,0)$ and $(5,a,b)$ are the direction vectors of the lines. If the dot product of those two vectors is zero, then the lines are perpendicular. Hence you have to calculate: $$(12,-5,0)\\cdot(5,a,b)=0$$ This gives: $12\\cdot5-5\\cdot a=0 \\Longleftrightarrow 60-5a=0 \\Longrightarrow \\boxed{a=12}$", "Math Relative position of lines Skew lines Skew lines Lines that are neither parallel nor intersect, are called skew. ! Requirements 1. Direction vectors are not collinear 2. A false statement results when we equate the linear equations Example $\\text{g: } \\vec{x} = \\begin{pmatrix} 2 \\\\ 4 \\\\ 6 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 0 \\\\ 4 \\\\ 0 \\end{pmatrix}$ $\\text{h: } \\vec{x} = \\begin{pmatrix} 6 \\\\ 4 \\\\ 8 \\end{pmatrix} + s \\cdot \\begin{pmatrix} 0 \\\\ 0 \\\\ 8 \\end{pmatrix}$ 1. Direction vectors not collinear? First, examine if the direction vectors are multiples of each other (=collinear). $\\vec{a}=t\\cdot\\vec{b}$ $\\begin{pmatrix} 0 \\\\ 4 \\\\ 0\\end{pmatrix}=t\\cdot\\begin{pmatrix} 0 \\\\ 0 \\\\ 8 \\end{pmatrix}$ Determine $t$ for every row 1. $0=t\\cdot0$ 2. $4=t\\cdot0$ 3. $0=t\\cdot8$ 1. $0=0$ 2. $4=0$ 3. $t=0$ A contradiction results in the second row. Thus the direction vectors are not collinear. 2. Equate $g$ and $h$ By equating the linear equations we try to find a point of intersection. If a contradiction occurs the lines are skew. $\\begin{pmatrix} 2 \\\\ 4 \\\\ 6 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 0 \\\\ 4 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 6 \\\\ 4 \\\\ 8 \\end{pmatrix} + s \\cdot \\begin{pmatrix} 0 \\\\ 0 \\\\ 8 \\end{pmatrix}$ Now we set up an equation system and solve it. 1. $2+r\\cdot0=6+s\\cdot0$ 2. $4+r\\cdot4=4+s\\cdot0$ 3.", "1. $\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix}, \\begin{pmatrix} 1 \\\\ 4 \\end{pmatrix}$ 2. $\\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 4 \\\\ 1 \\end{pmatrix}$ 3. $\\begin{pmatrix} 1 \\\\ 2 \\end{pmatrix}, \\begin{pmatrix} 1 \\\\ 4 \\\\ -1 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 3 During maneuvers preceding the Battle of Jutland, the British battle cruiser Lion moved as follows (in nautical miles): $1.2$ miles north, $6.1$ miles $38$ degrees east of south, $4.0$ miles at $89$ degrees east of north, and $6.5$ miles at $31$ degrees east of north. Find the distance between starting and ending positions (O'Hanian 1985). Problem 4 Find $k$ so that these two vectors are perpendicular. $\\begin{pmatrix} k \\\\ 1 \\end{pmatrix} \\qquad \\begin{pmatrix} 4 \\\\ 3 \\end{pmatrix}$ Problem 5 Describe the set of vectors in $\\mathbb{R}^3$ orthogonal to this one. $\\begin{pmatrix} 1 \\\\ 3 \\\\ -1 \\end{pmatrix}$ This exercise is recommended for all readers. Problem 6 1. Find the angle between the diagonal of the unit square in $\\mathbb{R}^2$ and one of the axes. 2. Find the angle between the diagonal of the unit cube in $\\mathbb{R}^3$ and one of the axes. 3. Find the angle between the diagonal of the unit cube in $\\mathbb{R}^n$ and one of the axes. 4. What is the limit, as $n$ goes to $\\infty$,", "is the same you got setting t values <0,1,1> then set up the dot product and solve for t and input t into the direction vector I got from taking the point minus the line, then write the new parametric equations based off my original point and the direction vector I'll do that way but for sake of completeness (1) the book suggests find the intersection (2) can I use projection to find it as well?? 5. ## Re: equation of line through point orthogonal to line I got two points along the given line by plugging in $t=0$ and $t=1$. When $t=0$, I get the point $(-1,-2+0,-1+0) = (-1,-2,-1)$. When $t=1$, I get the point $(-1,-2+1,-1+1) = (-1,-1,0)$. So, the vector from the first point to the second point is $\\begin{pmatrix}-1-(-1) \\\\ -1-(-2) \\\\ 0-(-1)\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. That is a vector in the direction of the given line (Essentially, this a vector containing some multiple of the coefficients of $t$ in the parametric equation of the given line). You can rewrite the line as $\\begin{pmatrix}-1 \\\\ -2 \\\\ -1\\end{pmatrix} + t\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. This immediately gives a vector in the direction of the given line to be $\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Essentially, what I did was I found all possible", "with the parameter $t$ has its whole body in the line— it is a direction vector for the line. Note that points on the line to the left of $x=1$ are described using negative values of $t$. In $\\mathbb{R}^3$, the line through $(1,2,1)$ and $(2,3,2)$ is the set of (endpoints of) vectors of this form and lines in even higher-dimensional spaces work in the same way. If a line uses one parameter, so that there is freedom to move back and forth in one dimension, then a plane must involve two. For example, the plane through the points $(1,0,5)$, $(2,1,-3)$, and $(-2,4,0.5)$ consists of (endpoints of) the vectors in $\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix} +t\\cdot\\begin{pmatrix} 1 \\\\ 1 \\\\ -8 \\end{pmatrix} +s\\cdot\\begin{pmatrix} -3 \\\\ 4 \\\\ -4.5 \\end{pmatrix} \\,\\big|\\, t,s\\in\\mathbb{R} \\}$ (the column vectors associated with the parameters $\\begin{pmatrix} 1 \\\\ 1 \\\\ -8 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 1 \\\\ -3 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix} \\qquad \\begin{pmatrix} -3 \\\\ 4 \\\\ -4.5 \\end{pmatrix} = \\begin{pmatrix} -2 \\\\ 4 \\\\ 0.5 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix}$ are two vectors whose whole bodies lie in the plane). As with the line, note that some points in this plane are described with negative $t$'s or negative $s$'s or" ]

[ "find the smallest positive integer $m$, and $0$ is not a positive integer. – MJD Nov 9 '14 at 2:32 $$\\left(\\sqrt{3}+i\\right)^m=\\left(\\sqrt{3}-i\\right)^m$$ $$2^m\\left(\\frac{\\sqrt{3}}{2}+\\frac{i}{2}\\right)^m=2^m\\left(\\frac{\\sqrt{3}}{2}-\\frac{i}{2}\\right)^m$$ $$e^{i\\theta}=\\cos\\theta+i\\sin\\theta$$ Using above formula we can write $$2^m(e^{i\\pi/6})^m=2^m(e^{-i\\pi/6})^m$$ $$e^{mi\\pi/6}=e^{-mi\\pi/6}$$ $$e^{2mi\\pi/6}=e^{mi\\pi/3}=1$$ $$e^{mi\\pi/3}=e^{2ik\\pi}$$ $$\\frac{mi\\pi}{3}=2ik\\pi$$ $$\\frac{m}{3}=2k$$ $$m=6k$$ where $k\\in\\mathbb Z$", "# Math Help - Find the smallest integer 1. ## Find the smallest integer Find the smallest integer n for which (1+i/1-i)^n=1 2. Hello, Originally Posted by varunnayudu Find the smallest integer n for which (1+i/1-i)^n=1 $1+i=\\sqrt{2} e^{i \\frac{\\pi}{4}}$ $1-i=\\sqrt{2} e^{-i\\frac{\\pi}{4}}$ So $\\frac{1+i}{1-i}=e^{i \\frac{\\pi}{2}}=i$ Hence $\\left(\\frac{1+i}{1-i}\\right)^n=i^n$ n=0 or n=4 etc... 3. ## Where to find a complete explanation for this Thanks for ur reply but please can u tell me from where have u gottten this from, i need it as a reference. If u could please recommend me a book from which i can refer this from. 4. Hmm I don't know any reference in English. I used this formula : $a+ib=\\sqrt{a^2+b^2} \\cdot e^{i \\arctan \\frac ab}$ then use the properties of i : $i^2=-1$ then $i^4=1$ etc.. http://en.wikipedia.org/wiki/Complex...the_polar_form", "# Find the smallest positive integer $n$ The smallest positive integer $$n$$ for which the difference $$\\sqrt {n} - \\sqrt {n-1}$$ becomes less than $$0.01$$ I wanted to check the error: I made using differentials, which gives an approximation to the root The answer $$2501$$ is right, but check out this method. 1) $$(\\sqrt{n}-\\sqrt{n-1})(\\sqrt{n}+\\sqrt{n-1})=n-(n-1)=1$$ by difference of squares. 2) You can't have $$\\sqrt{n}+\\sqrt{n-1}>100$$ until you have $$\\sqrt{n}>50$$. On the other hand you surely have $$\\sqrt{n}+\\sqrt{n-1}>100$$ when $$\\sqrt{n-1}\\ge 50$$. So from (2) the minimum $$n$$ for $$\\sqrt{n}+\\sqrt{n-1}>100$$ is $$50^2+1=2501$$. And then from (1) that is also the minimum for $$\\sqrt{n}-\\sqrt{n-1}<0.01$$.", "# Michael's trigonometric expression Algebra Level 3 Find the smallest positive integer $$x$$ such that $\\tan{x^{\\circ}}=\\dfrac{\\cos{2013^{\\circ}}+\\sin{2013^{\\circ}}}{\\cos{2013^{\\circ}}-\\sin{2013^{\\circ}}}.$ This problem is shared by Michael T. from AIME. ×", "jump to navigation ## Problem of the Day #9: A Simple Number Theory ProblemMarch 28, 2011 Posted by Seungln in : potd , trackback Let $V(x)$ be the possible number of positive integers $n$ such that $n^{-1}$ in modulus $x$ exists – that is, there exists an integer r such that $n \\equiv \\frac{1}{r} \\pmod{x}$. Find $p$ such that $V(1998)\\equiv p \\pmod{330}$ ## Comments» no comments yet - be the first?", "Closure. However: $m$ was the smallest (strictly) positive integer such that $a^m \\in H$ and: $0 \\le r < m$ Therefore it follows that: $r = 0$ Therefore: $n = q m$ and: $b = a^n = \\paren {a^m}^q$. We conclude that any arbitrary element $b = a^n$ of $H$ is a power of $a^m$. So, by definition, $H = \\gen {a^m}$ is cyclic. $\\blacksquare$", "# smallest positive number How to find the smallest positive number $x$ such that $2011x^{2}+1$ is a square number $$2011x^{2}+1=y^{2}$$ $x,y$ are positive numbers - Try using continued fractions. – dinoboy Jan 24 '13 at 19:35 According to Maple, $510732021116138713675018566232201605320997$ – Robert Israel Jan 24 '13 at 19:37 It seems the question is not What is the smallest positive number etc. but How to find it. – Did Jan 24 '13 at 20:02 $X_0 = 0\\\\ Y_0 = 1$ $X_{n+1} = P X_n + Q Y_n \\\\ Y_{n+1} = R X_n + S Y_n \\\\ \\\\ P = 22 903355 954053 525066 202335 319378 237605 968890 (44 \\text{ digits})\\\\ Q = 510732 021116 138713 675018 566232 201605 320997 (42 \\text{ digits})\\\\ R = 1027 082094 464554 953200 462336 692957 428300 524967 (46 \\text{ digits})\\\\ S = 22 903355 954053 525066 202335 319378 237605 968890 (44 \\text{ digits})\\\\$ It will give you a step by step solution - a=0;b=1;c=19742928095;d=441354968;e=883151290968;f=19742928095 ;l=10;z=vector(l);y=vector(l);z[1]=a;y[1]=b;for(n=2,l,z[n]=c*z[n-1]+d*y[n-1];y[n]=e*z [n-1]+f*y[n-1];print([z[n],y[n]])) (Pari gp) [441354968, 19742928095] [17427278795190051920, 779566419528680658049] [688131024289910853831312029832, 30781847532062692044137637045215] [27171442664988816842285870050168151808160, 1215447604913533872118296302514247336972801] [1072887677544578769792061644645677816761262916480568, 47992989334095735859126116621354243081754352590442975] [42363888543548329618109290861429128960866113212881520825707760, 1895044274974308089824464492379048791500336482990271031908792449] [1672774410679737895614308190979727225255931422523452909015625403410553832, 74827445715318345179514414300506375814316454587342868891991725485181466335] [66050929818412130693077441046821016092265334803401759338667273028766982617779712320, 2954625760580091857825190852826899891492369204494107266499709578 694843084725227571201] [2608077516025604206696668091297297960193250473178344064814349517650835355756863832363680706968, 11666592787753487807500058472831110573381478047673851 4851627220646522719069184329815980490117855] So minimal x= 441354968 Sorry i mistook and solved $2001x^{2}-y^{2}+1 =0$ - Alpha shows this is not correct. – Ross Millikan Jan 26 '13 at 23:30", "$$1^5+2^5+...+n^5$$ Find the smallest positive integer $$n>1$$, such that $\\sum_{k=1}^n k^5$ is a perfect square. ×", "+0 # Help 0 133 1 Let $a_1,$ $a_2,$ $\\dots$ be a sequence of real numbers such that for all positive integers $n,$ $\\sum_{k = 1}^n a_k \\left( \\frac{k}{n} \\right)^2 = 1.$Find the smallest $n$ such that $a_n < \\frac{1}{2018}.$ Aug 25, 2021", "+0 # Help 0 89 1 Let $a_1,$ $a_2,$ $\\dots$ be a sequence of real numbers such that for all positive integers $n,$ $\\sum_{k = 1}^n a_k \\left( \\frac{k}{n} \\right)^2 = 1.$Find the smallest $n$ such that $a_n < \\frac{1}{2018}.$ Aug 25, 2021", "# $a_n$ is the smallest positive integer number such that $\\sqrt{a_n+\\sqrt{a_{n-1}+…+\\sqrt{a_1}}}$ is positive integer An infinite sequence of pairwise distinct numbers $a_1, a_2, a_3, ...$ is defined thus: $a_n$ is the smallest positive integer number such that $\\sqrt{a_n+\\sqrt{a_{n-1}+...+\\sqrt{a_1}}}$ is positive integer. Prove that the sequence $a_1, a_2, a_3, ...$ contains all positive integers numbers. ### My work: Let $a_1=1$. Then $\\sqrt{a_2+1}$ is positive integer and $a_2$ is the smallest positive integer then $a_2=3$. Then $\\sqrt{a_3+2}$ is positive integer and $a_3$ is the smallest positive integer then $a_3=2$. Then $\\sqrt{a_4+\\sqrt{a_3+\\sqrt{a_{2}+\\sqrt{a_1}}}}=\\sqrt{a_4+2}$ is positive integer and $a_4$ is the smallest positive integer and $a_4\\not=a_1,a_2,a_3$ then $a_4=7$. • Interesting: the sequence, which (if I've computed it correctly) starts out $1,3,2,7,6,13,5,\\ldots$, does not (yet) appear in the OEIS. – Barry Cipra Oct 12 '17 at 13:11 • Someone should add it. – Ivan Neretin Oct 12 '17 at 13:40 • @IvanNeretin I would do it but I don't want to claim credit for the sequence, so I'd only it Roman83 is ok with it. Also I feel like we should feel around a bit more for references. – orlp Oct 12 '17 at 13:44 • Where does this problem come from? – mrf Oct 12 '17 at 19:31 • So, should I create an OEIS submission? – orlp Oct 13 '17 at 9:08", "# Can we find the smallest positive integer $n$ , such that $x^2+n$ is composite for $0\\le x\\le 10^4$? Let $n$ be a positive integer. Define $f(n)$ to be the smallest non-negative number $x$, such that $x^2+n$ is prime (assuming that $x^2+n$ always is prime for some $x$, which is probably not proven yet, but a consequence of the Bunyakovsky-conjecture). In particalar, if $n$ is prime, we have $f(n)=0$ For example $f(54168539)=1008$ because $x^2+54168539$ is composite for $0\\le x\\le 1007$ and prime for $x=1008$. Moreover, $54168539$ is the smallest positive integer $n$ satisfying $f(n)>1000$ • Given a number $k$, can we efficiently (without brute force) determine the smallest positive integer $n$ with $f(n)>k$ ? For example, what is the smallest $n$ with $f(n)>10^4$ ? • Is $f(n)$ surjective , in other words, does for every $k\\ge 0$ exist a number $n>0$ with $f(n)=k$ ? The existence of an $n$ with $f(n)>k$ for every $k$ can be shown with the chinese remainder theorem : We choose $n$ , such that $n\\equiv 0\\mod 2$ , $n\\equiv -1\\mod 3$ , $n\\equiv -4\\mod 5$ , $\\cdots$, $n\\equiv -(k-1)^2\\mod p_{k}$ where $p_k$ is the $k$-th prime. Then, the numbers $x^2+n$ with $0\\le x\\le k-1$ all have a prime factor not exceeding $p_k$. If we choose $n>p_k$, those numbers must be composite. • You", "+0 # HELP MATRIX 0 25 1 Let $\\mathbf{M} = \\begin{pmatrix} 2 & 1 \\\\ -1 & 4 \\end{pmatrix},$ and let $\\mathbf{M}^{2016} = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}.$Find $d.$ Express your answer in the form $m \\cdot p^n,$ where $m,$ $n,$ $p$ are positive integers, $p$ is prime, and $m$ is not divisible by $p.$ Oct 5, 2020", "168 views Let $A$ be the $2 \\times 2$ matrix $\\begin{pmatrix} \\sin\\frac{\\pi}{18}&-\\sin \\frac{4\\pi}{9} \\\\ \\sin \\frac{4\\pi}{9}&\\sin \\frac {\\pi}{18} \\end{pmatrix}$. Then the smallest number $n \\in \\mathbb{N}$ such that $A^{n}=1$ is. 1. $3$ 2. $9$ 3. $18$ 4. $27$ | 168 views +1 vote By solving Characteristic Equation $\\lambda = sin(\\pi/18) +\\mathbf{i} sin(4\\pi/9 )$ $\\lambda = sin(\\pi/18) -\\mathbf{i} sin(4\\pi/9 )$ By cayley hamilton theorem $\\lambda ^n=1$ and solving for n(minimum) get n= 9 so Option B is correct by (223 points) +1 vote", "Smallest positive integer for equation I am having trouble identify the smallest positive integer $n$ such that $(\\frac{1+i}{1-i})^n = 1$ Can someone please throw on approach? (Also, please correct the equation in the form of Tex/Latex format). Thanks. - LaTeX is not introduced, did I interpreted your question appropriately? – Quixotic Dec 13 '11 at 13:27 @MaX: $\\left( \\frac{1+n}{1-n} \\right)^n = 1$ has no positive integer solutions – Daniel Freedman Dec 13 '11 at 13:28 I would also like to know that many users find questions posted in the imperative (\"Show that\", \"Prove\", \"Do\", \"Find\") unpleasant and somewhat rude. – Quixotic Dec 13 '11 at 13:29 I am sorry Max, I would just update the post. – Nikhil Mulley Dec 13 '11 at 13:30 This is real-analysis?! – Quixotic Dec 13 '11 at 13:34 If your questions means \"Find the smallest positive integer $n$ such that $\\left(\\frac{1+i}{1-i}\\right)^n = 1$, where $i$ is the imaginary unit\", then I would first make the denominator of your fraction real (by multiplying top and bottom by $1+i$). After simplification, we find that we are looking for the smallest $n$ such that $i^n = 1$. Alternatively, as GEdgar suggests above, we could convert to polar coordinates. What moduli do $1+i$ and $1-i$ have? What are their arguments? So what is the modulus and", "$x$ is the smallest positive number. Clearly $x\\le 1$ and also $x^2\\gt 0$. Since $x$ is the smallest positive number, $x^2$ can't be smaller then $x$, so we must have $x^2\\geq x$. We can divide both sides of this by the positive number $x$ to get $x \\geq 1$. Since $x$ is both less than or equal to $1$ and greater than or equal to $1$, $x$ must equal $1$. Thus the smallest positive number is $1$. Clare wrote: There is no smallest positive number. If $0\\lt x\\lt1,$ then we know $0\\lt x^2 \\lt x\\times1 = x.$ Therefore for any positive number $x$ below $1$ that you might think is the smallest positive number, there is always a smaller positive number $x^2.$ $\\frac x2$ will also be a positve number less than $x.$ So we can never find the smallest positive number - it doesn't exist. This means that proof P2 begins with an invalid assumption, so the proof is invalid. In fact, you could consider P2 to be a proof by contradiction that there is no smallest positive number. It begins by assuming there is a smallest positive number, $x,$ and leads to nonsense. Hans found the problem with P3: ${\\mathbf{1=-1}}$. Click to see the proof. Proof: Clearly, $-1=-1$ and $\\frac{1}{1} = \\frac{-1}{-1}$ Therefore, $-1\\times \\frac{1}{1}=-1\\times \\frac{-1}{-1}$", "+0 # help +1 84 1 Let $$z = 3[\\cos(85^{\\circ})+i\\sin(85^{\\circ})]$$. Find the smallest positive value of $$r$$ such that $$r[\\cos(45^{\\circ})+i\\sin(45^{\\circ})] = z^n$$ for some positive integer $$n$$. Dec 20, 2019", "# Perfect Square in 987654321... Find the smallest positive integer $N$ such that $N^2$ starts with 987654321. Note: If you use a calculator for this problem, it can be solved in under a minute. There is no need to do extensive trial and error. You could push a maximum of 100 buttons (inclusive of each digit). ×", "say \"The smallest integer whose square ends in 269,696 is: \" j output: The smallest integer whose square ends in 269,696 is: 25264 ### examine remainder after dividing by one million /*REXX program finds the lowest (positive) integer whose square ends in 269,696. */ do j=2 by 2 /*start J at two, increment by two. */ if ((j * j) // 1000000) == 269696 then leave /*is square mod one million our target?*/ end /*◄── signifies the end of the DO loop.*/ /* [↑] // is division remainder.*/ say \"The smallest integer whose square ends in 269,696 is: \" j output is identical to the 1st REXX version. ### examine only numbers ending in 4 or 6 /*REXX program finds the lowest (positive) integer whose square ends in 269,696. */ /*─────────────────── we will only examine integers that are ending in four or six. */ do j=4 by 10 /*start J at four, increment by ten.*/ k = j /*set K to J's value. */ if right(k * k, 6) == 269696 then leave /*examine right-most 6 decimal digits. */ /* == means exactly equal to. */ k = j+2 /*set K to J+2 value. */ if right(k * k, 6) == 269696 then leave /*examine right-most 6 decimal digits. */ end /*◄── signifies the end of the DO", "# Perfect how? Divide $$297$$ by the smallest possible positive integer $$a$$ so that the result is the square of a positive integer $$b$$. What is $$a+b$$? ×" ]

[ "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of", "# Getting $x,y$ position on an image based on given value This should be simple but my math skills are really bad ... I have an image of 36 images (6 by 6 matrix). These small images are 36 instances of a direction arrow (like from Google maps GPS), each image rotated by $10^\\circ$ from $0^\\circ$ to $350^\\circ$. I don't know how to get image $X$ and $Y$ position of each small image when I have the rotation amount. Images are $50\\times50\\,\\mathrm{px}$, so... • for $10^\\circ$ image is at $50\\times0\\,\\mathrm{px}$, • for $20^\\circ$ image is at $100\\times0\\,\\mathrm{px}$ • for $90^\\circ$ image is at $150\\times50\\,\\mathrm{px}$ (because it is on row 2) • ... • for $350^\\circ$ image is at $250\\times250\\,\\mathrm{px}$ What is the formula to get this $X-Y$ value? - Given that the angle of the $n$th image is $n\\cdot10^\\circ$ for $n=0,1,\\dots,35$, you can have the $y$-index $j$ to be $\\lfloor{n/6}\\rfloor$ (int part of $n/6$) and the $x$-index $i$ to be $n-6j$. You then multiply $i$ and $j$ for the respective $\\Delta x$ and $\\Delta j$ which in your problem is $50\\,\\mathrm{px}$ each. \\begin{align} j &= \\lfloor{n/6}\\rfloor \\\\ i &= n-6j \\\\ x &=i\\cdot50\\,\\mathrm{px} \\\\ y &=j\\cdot50\\,\\mathrm{px} \\end{align} Programaticaly (this will be pseudocode, but you might be able to adapt to your language of choice): for n = 0", "= \\angle XBW = 90^{\\circ} = \\angle AMH = \\angle ATH$$ so $H$ lies on this line as well and we may conclude $\\square$ --mathguy623 03:10, 12 August 2016 (EDT) ## Solution 3 (Complex Bash) $[asy] unitsize(0.8cm); draw((0,0)--(14,0)--cycle); draw((0,0)--(5,12)--cycle); draw((5,12)--(14,0)--cycle); draw((2.071,4.97)--(14,0)--cycle); draw((8.96,6.72)--(0,0)--cycle); draw((2.071,4.97)--(9,0)--cycle); draw((8.96,6.72)--(9,0)--cycle); draw((0,0)--(0,2.083)--cycle); draw((14,0)--(14,6.75)--cycle); draw((0,2.083)--(14,6.75)--cycle); draw((0,2.083)--(2.071,4.97)--cycle); draw((0,2.083)--(9,0)--cycle); draw((8.96,6.72)--(14,6.75)--cycle); draw((14,6.75)--(9,0)--cycle); draw(circle((4.5,1.0415),4.61895)); draw(circle((11.5,3.375),4.20007)); label((0,0),\"C\",SW); label((14,0),\"B\",SE); label((5,12),\"A\",N); label((8.96,7),\"N\",N); label((2.071,4.97),\"M\",NW); label((8.25,0),\"W\",S); label((5,4),\"H\",N); label((0,2.083),\"Y\",NW); label((14,6.75),\"X\",NE); [/asy]$ For any point $P,$ let $p$ denote the complex number for point $P.$ First off, let $C$ be the origin. Now, since $WY$ is the diameter of the circumcircle of triangle $CMW$, we must have $\\angle{YMW}=\\angle{YCH}=90^{\\circ}.$ Since angles inscribed in the same arc are congruent, $\\angle{MYW}=\\angle{MCW}=\\angle{C}.$ This means that $\\frac{|w-m|}{|y-m|}=\\tan{\\angle{C}}.$ Combining this with the fact that $\\angle{YMW}$ is right, we find that $$i(y-m)\\tan{\\angle{C}}=(w-m).$$ Solving, we find that $$y=\\frac{w+(i\\tan{\\angle{C}}-1)m}{i\\tan{\\angle{C}}}.$$ We wish to simplify $(i\\tan{\\angle{C}}-1)m$ first. Note that $$m=\\frac{|CM|}{|CA|}\\cdot(a)=\\frac{(|BC|)\\cos{\\angle{C}}}{|CA|}\\cdot(|CA|)(\\cos{\\angle{C}}+i\\sin{\\angle{C}})=(|BC|)\\cos{\\angle{C}}(\\cos{\\angle{C}}+i\\sin{\\angle{C}}).$$ This means that \\begin{align*} (i\\tan{\\angle{C}}-1)m&=(i\\tan{\\angle{C}}-1)((|BC|)\\cos{\\angle{C}}(\\cos{\\angle{C}}+i\\sin{\\angle{C}}))\\\\ &=\\left(\\frac{i\\sin{\\angle{C}}}{\\cos{\\angle{C}}}-1\\right)((|BC|)\\cos{\\angle{C}}(\\cos{\\angle{C}}+i\\sin{\\angle{C}}))\\\\ &=(i\\sin{\\angle{C}}-\\cos{\\angle{C}})(|BC|)(\\cos{\\angle{C}}+i\\sin{\\angle{C}})\\\\ &=-|BC| \\end{align*} This means that $$y=\\frac{w-|BC|}{i\\tan{\\angle{C}}}=\\frac{|BC|-w}{\\tan{\\angle{C}}}i.$$ Now, since $WX$ is a diameter of the circumcircle of triangle $NBW,$ we must have $\\angle{WNX}=\\angle{WBX}=90^{\\circ}.$ Since angles inscribed in the same arc are congruent, $\\angle{NXW}=\\angle{NBW}=\\angle{B}.$ This means that $\\frac{|x-n|}{|w-n|}=\\cot{\\angle{B}}.$ Combining this with the fact that $\\angle{WNX}$ is right, we find that $$(x-n)=i\\cot{\\angle{B}}(w-n).$$ Solving, we find that $$x=wi\\cot{\\angle{B}}+n(1-i\\cot{\\angle{B}}).$$ We wish to simplify $n(1-i\\cot{\\angle{B}})$ first. Note that $$n=e^{(90-\\angle{B})i}|CN|=(\\cos{(90-\\angle{B})}+i\\sin{(90-\\angle{B})})(|BC|)\\sin{\\angle{B}}=(|BC|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}})).$$ This means that \\begin{align*} n(1-i\\cot{\\angle{B}})&=(|BC|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}}))(1-i\\cot{\\angle{B}})\\\\ &=(|BC|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}}))\\left(1-\\frac{\\cos{\\angle{B}}}{\\sin{\\angle{B}}}i\\right)\\\\ &=(|BC|)(\\sin{\\angle{B}}+i\\cos{\\angle{B}})(\\sin{\\angle{B}}-i\\cos{\\angle{B}})\\\\ &=(|BC|)(1)", "811 calculations, way too tedious for humans. But there is a better way, and we have touched on it quite a few times! We notice that we could make the jump just like before into rational mathematics: ${\\displaystyle {\\begin{matrix}216a&=&1+811b\\\\0&\\equiv &1+163b&{\\pmod {216}}\\\\\\end{matrix}}}$ We jump into rational maths again ${\\displaystyle {\\begin{matrix}216c&=&1+163b\\\\53c&\\equiv &1&{\\pmod {163}}\\\\\\end{matrix}}}$ We jump once more ${\\displaystyle {\\begin{matrix}53c&=&1+163d\\\\0&\\equiv &1+4d&{\\pmod {53}}\\\\\\end{matrix}}}$ Now the pattern is clear, we shall start from the beginning so that the process is not broken: ${\\displaystyle {\\begin{matrix}216a&=&1+811b\\\\216c&=&1+163b\\\\53c&=&1+163d\\\\53e&=&1+4d\\\\e&=&1+4f\\\\\\end{matrix}}}$ Now all we have to do is choose a value for f and substitute it back to find a! Remember a is the inverse of 216 mod 811. We choose f = 0, therefore e = 1, d = 13, c = 40, b = 53 and finally a = 199! If f is chosen to be 1 we will get a different value for a. The very perceptive reader should have noticed that this is just Euclid's gcd algorithm in reverse. Here are a few more examples of this ingenious method in action: Example 1 Find the smallest positive value of a: ${\\displaystyle {\\begin{matrix}33a&\\equiv &1\\ \\ {\\mbox{(mod 101)}}\\\\33a&=&1+101b\\\\33c&=&1+2b\\\\c&=&1+2d\\\\\\end{matrix}}}$ Choose d = 0, therefore a = 49. Example 2 Find the smallest positive value of a: ${\\displaystyle {\\begin{matrix}27a&\\equiv &1\\ \\ {\\mbox{(mod 821)}}\\\\27a&=&1+821b\\\\27c&=&1+11b\\\\5c&=&1+11d\\\\5e&=&1+d\\\\\\end{matrix}}}$ Choose e = 0, therefore a =", "y2=\"300\" /> <rect x=\"0\" y=\"0\" width=\"200\" height=\"100\" fill=\"red\" opacity=\"0.9\" transform=\"matrix(0.866025, 0.5, -0.5, 0.866025, 50.0, 13.39746)\"></rect> </svg> </div> ## ⑧ transform: skewX(a) skewX means x For the inclination in the axis direction, the trigonometric function will also be used here. Similarly, there is the following in radian system: $$a'=\\frac{\\pi}{180}*a$$ Because tilting only occurs in x Axial direction, from which: $$\\begin{cases} newX = \\Delta x + oldX = tan(a')*oldY + oldX\\\\ newY = oldY \\end{cases}$$ Therefore, the transformation function is matrix(1, 0, tan(a'), 1, 0, 0) \\begin{aligned} \\begin{pmatrix} newX \\\\ newY \\\\ 1 \\\\ \\end{pmatrix} & = \\begin{pmatrix} 1 & tan(a') & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ \\end{pmatrix} \\begin{pmatrix} oldX \\\\ oldY \\\\ 1 \\\\ \\end{pmatrix} \\\\ & = \\begin{pmatrix} 1*oldX+tan(a')*oldY+0 \\\\ 0*oldX+1*oldY+0 \\\\ 1 \\\\ \\end{pmatrix}\\\\ &= \\begin{pmatrix} tan(a')*oldY + oldX\\\\ oldY \\\\ 1 \\\\ \\end{pmatrix} \\end{aligned} <div> <h1>transform: skewX(a)</h1> <svg x=\"0px\" y=\"0px\" width=\"600px\" height=\"300px\"> <line label=\"axisX\" fill=\"none\" stroke=\"black\" stroke-width=\"10\" x1=\"0\" y1=\"0\" x2=\"600\" y2=\"0\" /> <line label=\"axisY\" fill=\"none\" stroke=\"black\" stroke-width=\"10\" x1=\"0\" y1=\"0\" x2=\"0\" y2=\"300\" /> <rect x=\"0\" y=\"0\" width=\"200\" height=\"100\" fill=\"red\" opacity=\"0.9\" transform=\"skewX(30)\"></rect> </svg> <h1>transform: matrix(1, 0, tan(a'), 1, 0, 0)</h1> <svg x=\"0px\" y=\"0px\" width=\"600px\" height=\"300px\"> <line label=\"axisX\" fill=\"none\" stroke=\"black\" stroke-width=\"10\" x1=\"0\" y1=\"0\" x2=\"600\" y2=\"0\" /> <line label=\"axisY\" fill=\"none\" stroke=\"black\" stroke-width=\"10\" x1=\"0\" y1=\"0\" x2=\"0\" y2=\"300\" /> <rect x=\"0\" y=\"0\" width=\"200\"", "+0 # Confusing 0 116 4 +282 i hate this lizagame Feb 28, 2017 Sort: #1 0 each quadrant is 90 degrees, so you will turn it 3 times, because 90 * 3 = 270 Guest Feb 28, 2017 #2 +77001 0 Rotating the figure 270° counter-clockwise is the same as rotating the figure 90° clockwise... And each point will swap coordinates and both will have positive signs....so (-2,1) → (1, 2) (-4,2) → (2, 4) (-3,5) → (5, 3) C is correct CPhill Feb 28, 2017 #3 +90586 0 Try watching this. There are a lot of resourced on the web to help with mathematical concepts. Melody Feb 28, 2017 #4 +18624 +5 Matrix Rotation counterclockwise: $$\\begin{array}{|rcll|} \\hline \\begin{pmatrix} \\cos(\\varphi) & \\sin (\\varphi) \\\\ -\\sin(\\varphi) & \\cos (\\varphi) \\\\ \\end{pmatrix} \\stackrel{\\varphi=270^{\\circ}} \\rightarrow \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} \\\\ \\hline \\end{array}$$ The point P becomes to P': $$\\begin{array}{|rcll|} \\hline \\binom{x}{y}\\cdot \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} = \\binom{y}{-x} \\\\ \\hline \\end{array}$$ $$\\text{Let}\\ X =\\binom{-2}{1} \\\\ \\text{Let}\\ Y =\\binom{-4}{2} \\\\ \\text{Let}\\ Z =\\binom{-3}{5} \\\\$$ $$\\begin{array}{|rcll|} \\hline X'=\\binom{-2}{1}\\cdot \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} = \\binom{1}{2} \\\\ Y'=\\binom{-4}{2}\\cdot \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} = \\binom{2}{4} \\\\ Z'=\\binom{-3}{5}\\cdot \\begin{pmatrix} 0 & -1 \\\\", "= \\text{angle} \\left ( \\sum_{j=1}^n w \\cdot \\exp(i \\cdot \\alpha_j) \\right )$ Missing values in angles are omitted from the calculations. References Examples 1. Circular mean of a 1-D array of angles, in radians >>> import pingouin as pg >>> angles = [0.785, 1.570, 3.141, 0.839, 5.934] >>> round(pg.circ_mean(angles), 4) 1.013 Compare with SciPy: >>> from scipy.stats import circmean >>> import numpy as np >>> round(circmean(angles, low=0, high=2*np.pi), 4) 1.013 1. Using a 2-D array of angles in degrees >>> np.random.seed(123) >>> deg = np.random.randint(low=0, high=360, size=(3, 5)) >>> deg array([[322, 98, 230, 17, 83], [106, 123, 57, 214, 225], [ 96, 113, 126, 47, 73]]) We first need to convert from degrees to radians: >>> rad = np.round(pg.convert_angles(deg, low=0, high=360), 4) array([[-0.6632, 1.7104, -2.2689, 0.2967, 1.4486], [ 1.85 , 2.1468, 0.9948, -2.5482, -2.3562], [ 1.6755, 1.9722, 2.1991, 0.8203, 1.2741]]) >>> pg.circ_mean(rad) # On the first axis (default) array([1.27532162, 1.94336576, 2.23195927, 0.52110503, 1.80240563]) >>> pg.circ_mean(rad, axis=-1) # On the last axis (default) array([0.68920819, 2.49334852, 1.5954149 ]) >>> round(pg.circ_mean(rad, axis=None), 4) # Across the entire array 1.6954 Missing values are omitted from the calculations: >>> rad[0, 0] = np.nan array([1.76275 , 1.94336576, 2.23195927, 0.52110503, 1.80240563]) 1. Using binned angles >>> np.random.seed(123) >>> nbins = 18 # Number of bins to divide the unit circle >>> angles_bins = np.linspace(0, 2", "an even number of them must be reflections, we either reflect $T$ $0$ times or $2$ times. Case 1: 0 reflections on T In this case, we must use $3$ rotations to return $T$ to its original position. Notice that our set of rotations, $\\{90^\\circ,180^\\circ,270^\\circ\\}$, contains every multiple of $90^\\circ$ except for $0^\\circ$. We can start with any two rotations $a,b$ in $\\{90^\\circ,180^\\circ,270^\\circ\\}$ and there must be exactly one $c \\equiv -a - b \\pmod{360^\\circ}$ such that we can use the three rotations $(a,b,c)$ which ensures that $a + b + c \\equiv 0^\\circ \\pmod{360^\\circ}$. That way, the composition of rotations $a,b,c$ yields a full rotation. For example, if $a = b = 90^\\circ$, then $c \\equiv -90^\\circ - 90^\\circ = -180^\\circ \\pmod{360^\\circ}$, so $c = 180^\\circ$ and the rotations $(90^\\circ,90^\\circ,180^\\circ)$ yields a full rotation. The only case in which this fails is when $c$ would have to equal $0^\\circ$. This happens when $(a,b)$ is already a full rotation, namely, $(a,b) = (90^\\circ,270^\\circ),(180^\\circ,180^\\circ),$ or $(270^\\circ,90^\\circ)$. However, we can simply subtract these three cases from the total. Selecting $(a,b)$ from $\\{90^\\circ,180^\\circ,270^\\circ\\}$ yields $3 \\cdot 3 = 9$ choices, and with $3$ that fail, we are left with $6$ combinations for case 1. Case 2: 2 reflections on T In this case, we first eliminate the possibility of having two", "4. Recalling that $i^2=-1$, this is quickly done: . $1$ $i$ $-i$ $-1$ $1$ $1$ $i$ $-i$ $-1$ $i$ $i$ $-1$ $1$ $-i$ $-i$ $-i$ $1$ $-1$ $i$ $-1$ $-1$ $-i$ $i$ $1$ 5. The transformations we use must be limited to rotations where our pentagon ultimately ends up in the same area it originally occupied after the rotation. Given that a pentagon has 5 sides, the smallest such rotation (besides rotating by $0^{\\circ}$) is a rotation of $360^{\\circ} / 5=72^{\\circ}$. One could rotate the pentagon by any multiple of $72^{\\circ}$ and occupy its original area, but notice $5 \\cdot 72^{\\circ} = 360^{\\circ}$ is equivalent to no rotation at all. As such, we really only have 5 distinct transformations: rotating by $0^{\\circ}$ (doing nothing), $72^{\\circ}$, $144^{\\circ}$, $216^{\\circ}$, or $288^{\\circ}$. Calling these rotations $R_{0}, R_{72}, R_{144}, R_{216}$, and $R_{288},$ we can flesh out the table for their combinations: $R_{0}$ $R_{72}$ $R_{144}$ $R_{216}$ $R_{288}$ $R_{0}$ $R_{0}$ $R_{72}$ $R_{144}$ $R_{216}$ $R_{288}$ $R_{72}$ $R_{72}$ $R_{144}$ $R_{216}$ $R_{288}$ $R_{0}$ $R_{144}$ $R_{144}$ $R_{216}$ $R_{288}$ $R_{0}$ $R_{72}$ $R_{216}$ $R_{216}$ $R_{288}$ $R_{0}$ $R_{72}$ $R_{144}$ $R_{288}$ $R_{288}$ $R_{0}$ $R_{72}$ $R_{144}$ $R_{216}$ 6. In both cases, we simply do the normal operation (addition or multiplication) and then find the remainder upon division by 5 of the result. This remainder is our answer. So for example: $$4 \\cdot 3", "that $w = \\operatorname{cis}(30^{\\circ})$ and $z = \\operatorname{cis}(120^{\\circ})$ because the cosine and sine sums of those angles give the values of $w$ and $z$, respectively. By Demoivre's theorem, $\\operatorname{cis}(\\theta)^n = \\operatorname{cis}(n\\theta)$. When you multiply by $i$, we can think of that as rotating the complex number $90^{\\circ}$ counterclockwise in the complex plane. Therefore, by the equation we know that $30r + 90$ and $120s$ land on the same angle. This means that$$30r + 90 \\equiv 120s \\pmod{360},$$which we can simplify to$$r+3 \\equiv 4s \\pmod{12}.$$Notice that this means that $r$ cycles by $12$ for every value of $s$. This is because once $r$ hits $12$, we get an angle of $360^{\\circ}$ and the angle laps onto itself again. By a similar reasoning, $s$ laps itself every $3$ times, which is much easier to count. By listing the possible values out, we get the pairs $(r,s)$:$$\\begin{array}{cccccccc} (1,1) & (5,2) & (9,3) & (13,1) & (17,2) & (21,3) & \\ldots & (97,1) \\\\ (1,4) & (5,5) & (9,6) & (13,4) & (17,5) & (21,6) & \\ldots & (97,4) \\\\ (1,7) & (5,8) & (9,9) & (13,7) & (17,8) & (21,9) & \\ldots & (97,7) \\\\ [-1ex] \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots & \\vdots \\\\ (1,100) & (5,98) & (9,99) & (13,100) & (17,98)", "terminal side of $\\gamma$ lies on the negative $x$-axis, $\\gamma$ is a quadrantal angle. All angles coterminal with $\\gamma$ are of the form $\\theta=540^{\\circ}+360^{\\circ} \\cdot k$, where $k$ is an integer. Working through the arithmetic, we find three such angles: $900^{\\circ}, 180^{\\circ}$ and $-180^{\\circ}$. 4. The Greek letter $\\varphi$ is pronounced 'fee' or 'fie'9 and since $\\varphi=-750 \\quad$ is negative, we begin our rotation clockwise from the positive $x$-axis. Two full rotations account for $720^{\\circ}$, with just $30^{\\circ}$ or $\\frac{1}{12}$ of a revolution to go. We find that $\\varphi$ is a Quadrant IV angle. To find coterminal angles, we compute $\\theta=-750^{\\circ}+360^{\\circ} \\cdot k$ for a few integers $k$ and obtain $-390^{\\circ},-30^{\\circ}$ and $330^{\\circ} .$ by Platinum (132,156 points) 0 like 0 dislike 1 like 0 dislike 0 like 0 dislike 0 like 0 dislike 0 like 0 dislike 0 like 0 dislike 0 like 0 dislike 0 like 0 dislike 0 like 0 dislike", "of rotation, and in our case we’re starting from angle 0, we expect that for any parameter $t$ we’ll get the half-angle $t\\arccos d$. This lets us define the absolute angular error: $e = 2|t\\arccos d - \\arccos \\frac{(1-t')+dt'}{\\sqrt{(1-d^2){t'}^2 + (1-t'+dt')^2}}|$ We can now measure the maximum error for $nlerp$ and all three representations and get: $e_{nlerp} = 1.42229 * 10^{-1} = 8.15^{\\circ}$ $K_0: e_0 = 6.96632 * 10^{-3} = 0.40^{\\circ}$ $K_1: e_1 = 1.09562 * 10^{-3} = 0.06^{\\circ}$ $K_2: e_2 = 7.76255 * 10^{-4} = 0.04^{\\circ}$ This is pretty good - remember, this is the maximum absolute error across our entire range! We can clearly see that our efforts to make the approximation more precise paid off - using a more involved approximation for $K$ together as well as carefully fitting the coefficients reduced the error by an order of magnitude. Also note that all errors reach their maximum value for quaternions that are at almost $180^{\\circ}$ rotation from each other. If we reduce the interval so that initial quaternions are at most $90^{\\circ}$ from each other, we get: $e_{nlerp} = 1.60363 * 10^{-2} = 0.91^{\\circ}$ $K_0: e_0 = 1.12533 * 10^{-4} = 0.006^{\\circ}$ $K_1: e_1 = 1.24728 * 10^{-4} = 0.007^{\\circ}$ $K_2: e_2 = 7.22881 * 10^{-5} = 0.004^{\\circ}$ It’s interesting that while our approximations do", "+0 # Confusing 0 133 4 +282 i hate this lizagame Feb 28, 2017 Sort: #1 0 each quadrant is 90 degrees, so you will turn it 3 times, because 90 * 3 = 270 Guest Feb 28, 2017 #2 +79894 0 Rotating the figure 270° counter-clockwise is the same as rotating the figure 90° clockwise... And each point will swap coordinates and both will have positive signs....so (-2,1) → (1, 2) (-4,2) → (2, 4) (-3,5) → (5, 3) C is correct CPhill Feb 28, 2017 #3 +91254 0 Try watching this. There are a lot of resourced on the web to help with mathematical concepts. If you want us to help you find them then ask specifically. :) Melody Feb 28, 2017 #4 +18777 +5 Matrix Rotation counterclockwise: $$\\begin{array}{|rcll|} \\hline \\begin{pmatrix} \\cos(\\varphi) & \\sin (\\varphi) \\\\ -\\sin(\\varphi) & \\cos (\\varphi) \\\\ \\end{pmatrix} \\stackrel{\\varphi=270^{\\circ}} \\rightarrow \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} \\\\ \\hline \\end{array}$$ The point P becomes to P': $$\\begin{array}{|rcll|} \\hline \\binom{x}{y}\\cdot \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} = \\binom{y}{-x} \\\\ \\hline \\end{array}$$ $$\\text{Let}\\ X =\\binom{-2}{1} \\\\ \\text{Let}\\ Y =\\binom{-4}{2} \\\\ \\text{Let}\\ Z =\\binom{-3}{5} \\\\$$ $$\\begin{array}{|rcll|} \\hline X'=\\binom{-2}{1}\\cdot \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} = \\binom{1}{2} \\\\ Y'=\\binom{-4}{2}\\cdot \\begin{pmatrix} 0 & -1 \\\\ 1", "# OEDrawCurvedArrow¶ void OEDrawCurvedArrow(OEImageBase& image, const OE2DPoint& center, double bgnAngle, double endAngle, double radius, bool clockwise, const OEPen& pen); Draws a curved arrow. See example in Figure: Example of drawing curved arrows. Example of drawing curved arrows image The image in which the arrow is drawn. center The center of the arc. bgnAngle, endAngle The two endpoints of the arc (in degrees). Both angles are in degrees and their values have to be in a range of $$[0.0^{\\circ}, 360.0^{\\circ}]$$. Angles are interpreted such that $$0.0^{\\circ}$$ and $$360.0^{\\circ}$$ degrees are at the 12 o’clock position, $$90.0^{\\circ}$$ degrees corresponds to 3 o’clock, etc.", "# Dodecagon (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff) A dodecagon is a 12-sided polygon. The sum of its internal angles is $1800^{\\circ}$. Each of its exterior angles has measure $30^{\\circ}$. A regular dodecagon can be seen below: $[asy] for(int i = 0; i <= 11; ++i) { draw(dir(360/12*i)--dir(360/12*(i + 1))); } pair A,B,C,D,E,F,G,H,I,J,K,L; A=dir(360/12*0); B=dir(360/12*1); C=dir(360/12*2); D=dir(360/12*3); E=dir(360/12*4); F=dir(360/12*5); G=dir(360/12*6); H=dir(360/12*7); I=dir(360/12*8); J=dir(360/12*9); K=dir(360/12*10); L=dir(360/12*11); label(\"A\",A,dir(0)); label(\"B\",B,dir(30)); label(\"C\",C,dir(60)); label(\"D\",D,dir(90)); label(\"E\",E,dir(120)); label(\"F\",F,dir(150)); label(\"G\",G,dir(180)); label(\"H\",H,dir(210)); label(\"I\",I,dir(240)); label(\"J\",J,dir(270)); label(\"K\",K,dir(300)); label(\"L\",L,dir(330)); draw(dir(360/12*0)--dir(360/12*6)); dot((dir(360/12*0)+dir(360/12*6))/2); pair O = (dir(360/12*0)+dir(360/12*6))/2; label(\"O\",O,S); draw(A--O); draw(Circle(O,1)); [/asy]$ The area of a regular dodecagon can be calculated by the formula $3R^2$, where $R$ is the circumradius of the dodecagon. In this case, $R$ would be $OA$. Also, each small triangle ($AOB$, $BOC$, etc.) is congruent, so $\\angle AOB=\\angle BOC=\\angle COD$ (etc) $=30^{\\circ}$.", "there we have it. The inverse of ${V}$ does exist (although the matrix has numerical rank ${1}$) and it is >> inv(V) warning: matrix singular to machine precision, rcond = 1.11022e-16 ans = 1.0000e+00 4.5036e+15 0.0000e+00 4.5036e+15 and note that $\\texttt{4.5036e+15}$ is indeed just the inverse of the machine precision, so this inverse is actually 100% accurate. Recombining gives >> inv(V)*D*V warning: matrix singular to machine precision, rcond = 1.11022e-16 ans = 1 0 0 1 which is not even close to our original ${A}$. How is that? Here is a solution. The matrix $\\displaystyle \\tilde A = \\begin{pmatrix} 1 & 1\\\\ 0 & 1+\\epsilon^{-2} \\end{pmatrix}$ is indistinguishable from ${A}$ numerically. However, it has two distinct eigenvalues, so it is diagonalizable. Indeed, a basis of eigenvectors is $\\displaystyle \\tilde V = \\begin{pmatrix} 1 & -1\\\\ 0 & \\epsilon^{-2} \\end{pmatrix}$ which is indistinguishable from ${V}$ above and it holds that $\\displaystyle \\tilde V^{-1}\\tilde A\\tilde V = \\begin{pmatrix} 1 & 0\\\\ 0 & 1+\\epsilon^{-2}. \\end{pmatrix}$ which is indistinguishable from $D$. Im Wintersemster 2018/19 habe ich die Vorlesung “Lineare Algebra 1” gehalten. Hier die lecture notes dazu: This is a short note to self: Let $A$ be a symmetric positive semidefinite matrix with one-dimensional kernel spanned by $v$. How to solve $Ax=b$ (if you know that $b$ is in the range", "arrow pointing in one of the eight cardinal directions, subject to the compatibility requirement. We even allow the arrow in a corner square to point in the eighth forbidden direction. Define a function $N$, from the set of all pairs of adjacent elements of $S$ to the set $\\{ - 45^{\\circ},0^{\\circ},45^{\\circ} \\}$, as follows. If $A,B \\in S$ are adjacent, define $N(A,B)$ as the angle $\\theta \\in \\{ \\pm 45^{\\circ},0^{\\circ} \\}$ that we must turn the arrow in $A$ to obtain the arrow in $B$. (Counterclockwise rotations are considered positive.) Clearly, $N(B,A) = - N(A,B)$ for any pair $(A,B)$ of adjacent elements of $S$. Next, define a function $W: V \\to \\mathbb{R}$, which we call the winding-number function, as follows: • For any vertex $v$ that is not a corner of the cube, let $$W(v) \\stackrel{\\text{df}}{=} N(A,B) + N(B,C) + N(C,D) + N(D,A),$$ where $(A,B,C,D)$ is any counterclockwise ordering of the four squares surrounding $v$; from the compatibility requirement, we have $W(v) = 0^{\\circ}$. • For any vertex $v$ that is a corner of the cube, let $$W(v) \\stackrel{\\text{df}}{=} N(A,B) + N(B,C) + N(C,A),$$ where $(A,B,C)$ is any counterclockwise ordering of the three squares surrounding $v$; it is not difficult to check that $W(v) = - 90^{\\circ}$. Now, $\\displaystyle \\sum_{v \\in V} W(v) = 0^{\\circ}$ as any two", "$x(3x +2y -30)=0$, $y(2y-x-6)=0$ we get cases \\begin{align*} &x=0,\\ y=0 &&\\implies A_1=(0,0),\\\\ &x=0,\\ 2y-x-6=0 &&\\implies A_2=(0,3)\\\\ &y=0,\\ 3x+2y-30=0 &&\\implies A_3=(10,0),\\\\ &3x+2y-30=0,\\ 2y-x-6=0 &&\\implies A_4=(6,6). \\end{align*} (b) Linearizations at these points have matrices \\begin{align*} &A_1&&A_2&&A_3&&A_4\\\\ & \\begin{pmatrix} -30 &\\ \\ 0\\\\ 0 &-6 \\end{pmatrix} && \\begin{pmatrix} -24 &\\ \\ 0\\\\ -3 &6 \\end{pmatrix} && \\begin{pmatrix} 30 &20\\\\ 0 &-16 \\end{pmatrix} && \\begin{pmatrix} 18 &12\\\\ -6 &12 \\end{pmatrix}\\\\ \\text{with eigenvalues}\\\\ &\\{-30,-6\\} &&\\{-24,6\\} && \\{30,-16\\} && \\{15-\\sqrt{63}i,15-\\sqrt{63}i \\} \\end{align*} correspondingly. Therefore $A_1$ is a stable node, $A_2$ and $A_3$ are saddles, $A_4$ is an unstable focal point and since left bottom number is $-6<0$ it is clockwise oriented. Directions are: $A_1$: $\\mathbf{f}_1=(1,0)^T$, $\\mathbf{f}_2=(0,1)^T$ ; Since $-30=\\lambda_1 <\\lambda_2=-6$, all trajectories have an entry directions $\\pm \\mathbf{f}_2$ (except two, which have entry directions $\\pm \\mathbf{f}_1$). $A_2$: $\\mathbf{f}_1=(10,1)^T$ ($\\lambda_1=-24$)--stable, $\\mathbf{f}_2=(0,1)^T$ ($\\lambda_2=6$)--unstable. $A_3$: $\\mathbf{f}_1=(1,0)^T$ ($\\lambda_1=30$)--unstable, $\\mathbf{f}_2=(1,-2.3)^T$ ($\\lambda_2=-16$)--stable. (c)--(d) One should observe that either $x=0$ in every point of the trajectory, or in no point; and that $y=0$ in every point of the trajectory, or in no point. It allows us to make a skeleton'' of the phase portrait (thick black lines) on the figure; red lines are very approximate Remark: $A_4$ may not look as a focal point on the computer-generated plot but it is. The reason for this disparity is that for one rotation", "45^\\circ = 360^\\circ$, so in total it's as if we didn't rotate at all. Therefore $(1 + i)^8 = 16$. Alternatively, the way to see this with algebra is \\begin{aligned} &\\phantom{=}(1 + i)^8 \\\\\\\\ &= \\left(\\sqrt{2}\\cdot(\\cos(45^\\circ) + i \\sin(45^\\circ) \\right)^8 \\\\ &= (\\sqrt{2})^8 \\cdot \\left( \\cos(\\underbrace{45^\\circ + \\cdots + 45^\\circ}_{\\text{8 times}}) + i\\sin(\\underbrace{45^\\circ + \\cdots + 45^\\circ}_{\\text{8 times}}) \\right) \\\\\\\\ &= 16 \\left(\\cos(360^\\circ) + i\\sin(360^\\circ) \\right) \\\\\\\\ &= 16 \\end{aligned} ### Example 3: $z^5 = 1$ Now let's start asking the reverse question: Is there a number $z$ such that after multiplying everything in the plane by $z$ five successive times, things are back to where they started? In other words, can we solve the equation $z^5 = 1$? One simple answer is $z = 1$, but let's see if we can find any others. First off, the magnitude of such a number would have to be $1$, since if it were more than $1$, the plane would keep stretching, and if it were less than $1$, it would keep shrinking. Rotation is a different animal, though, since you can get back to where you started after repeating certain rotations. In particular, if you rotate $\\dfrac{1}{5}$ of the way around, like this then doing this $5$ successive times will bring you back to where you started. The number", "D_4$. Since $D_4$ has only $8$ elements, this calculation is fairly easy. I shall illustrate the case $n=4, k=6$. We can view the $4\\times 4$ matrix as a block matrix $$\\begin{pmatrix} A & B \\\\ C & D\\end{pmatrix}$$ where each block is a $2\\times 2$ matrix. Clearly the identity fixes all $\\binom{16}{6}=8008$ matrices. Rotation by $90^\\circ$ counterclockwise permutes this to give $$\\begin{pmatrix} B & D \\\\ A & C\\end{pmatrix}$$ so the matrix is fixed iff $A=B=D=C$. But this means the number of ones in the matrix is a multiple of $4$, yet $k$ is not divisible by $4$, thus no matrices are fixed by rotation by $90^\\circ$. It is easy to see that the result is the same for rotation by $270^\\circ$. Rotation by $180^\\circ$ permutes the original matrix to $$\\begin{pmatrix} D & C \\\\ B & A\\end{pmatrix}$$ which is the same as the original iff $A=D$ and $B=C$. Thus since $k=6$ we can either have $3$ ones in $A$, $2$ in $A$ and $1$ in $B$, $1$ in $A$ and $2$ in $B$, or $3$ in $B$. Since choosing $A$ and $B$ completely determines the matrix, the number of matrices fixed by rotation by $180^\\circ$ is $$\\binom43\\binom40+\\binom42\\binom41+\\binom41\\binom42+\\binom40\\binom43=56.$$ Now for reflection. Vertical or horizontal reflection is easy; a matrix is fixed iff its left (upper) half is the" ]

[ "# Math Help - Find the smallest positive 1. ## Find the smallest positive Find the smallest positive integer solution to $\\tan{19x^{\\circ}}=\\dfrac{\\cos{96^{\\circ}}+\\sin{96^ {\\circ}}}{\\cos{96^{\\circ}}-\\sin{96^{\\circ}}}$. 2. $\\color{white}.\\quad.$ $\\frac{\\cos{96^\\circ}+\\sin{96^\\circ}}{\\cos{96^\\circ }-\\sin{96^\\circ}}$ $=\\ \\frac{\\sin{45^\\circ}\\cos{96^\\circ}+\\cos{45^\\circ}\\ sin{96^\\circ}}{\\cos{45^\\circ}\\cos{96^\\circ}-\\sin{45^\\circ}\\sin{96^\\circ}}$ $=\\ \\frac{\\sin{(96^\\circ+45^\\circ)}}{\\cos{(96^\\circ+45 ^\\circ)}}$ $=\\ \\tan{141^\\circ}$ $=\\ \\tan{(141^\\circ+180n^\\circ)}\\quad n\\in\\mathbb{Z}$ So find the smallest positive integer x such that $19x\\equiv141\\!\\!\\!\\!\\pmod{180}$. The answer should be $x = 159$. (I worked it out using Excel. ) 3. Originally Posted by JaneBennet So find the smallest positive integer x such that $19x\\equiv141\\!\\!\\!\\!\\pmod{180}$. The answer should be $x = 159$. (I worked it out using Excel. )[/SIZE][/FONT] Here is a non-excel solution. First $180 = 2^2\\cdot 3^2\\cdot 5$. Thus, this is equivalent to, $19x\\equiv 141(\\bmod 4)$ $19x\\equiv 141(\\bmod 9)$ $19x\\equiv 141(\\bmod 5)$ Which gets simplified to, $-x\\equiv 1(\\bmod 4)$ $x\\equiv 6(\\bmod 9)$ $-x\\equiv 1(\\bmod 5)$ The first and third congruences combine since $\\gcd(4,5)=1$. $x\\equiv -1(\\bmod 20)$ $x\\equiv 6(\\bmod 9)$ In first congruence subtract 20 once, in second congruence subtract 9 three times. To get*, $x\\equiv - 21(\\bmod 20)$ $x\\equiv -21(\\bmod 9)$ Thus, $x\\equiv -21(\\bmod 180) \\implies x\\equiv 159(\\bmod 180)$. *)You can use Chinese Remainder Theorem here, but I do not like to use that theorem, it is too computational. What you see here is my approach to combining modolus.", "101b\\\\ 33c &=& 1 + 2b\\\\ c &=& 1 + 2d\\\\ \\end{matrix}$ Choose d = 0, therefore a = 49. Example 2 Find the smallest positive value of a: $\\begin{matrix} 27a & \\equiv & 1 \\ \\ \\mbox{(mod 821)}\\\\ 27a &=& 1 + 821b\\\\ 27c &=& 1 + 11b\\\\ 5c &=& 1 + 11d\\\\ 5e &=& 1 + d\\\\ \\end{matrix}$ Choose e = 0, therefore a = -152 = 669 Example 3 Find the smallest positive value of a: $\\begin{matrix} 34a & \\equiv & 1 \\ \\ \\mbox{(mod 55)}\\\\ 34a &=& 1 + 55b\\\\ 34c &=& 1+ 21b\\\\ 13c& =& 1 + 21d\\\\ 13e& =& 1 + 8d\\\\ 5e &=& 1 + 8f\\\\ 5g& = &1 + 3f\\\\ 2g& =& 1 + 3h\\\\ 2i& = &1 + h\\\\ \\end{matrix}$ Set i = 0, then a = -21 = 34. Why is this so slow for two numbers that are so small? What can you say about the coefficients? Example 4 Find the smallest positive value of a: $\\begin{matrix} 21a & \\equiv & 1 \\ \\ \\mbox{(mod 102)}\\\\ 21a &=& 1 + 102b\\\\ 21c &=& 1 + 18b\\\\ 3c &=& 1 + 18d\\\\ 3d &=& 1\\\\ \\end{matrix}$ Now d is not an integer, therefore 21 does not have an inverse mod 102. What we have discussed so far is", "+0 0 511 1 What is the smallest integer $n$, greater than $1$, such that $n^{-1}\\pmod{1320}$ is defined? May 21, 2019 #1 +25228 +3 What is the smallest integer n, greater than 1, such that $$n^{-1}\\pmod{1320}$$is defined? According to Euler's theorem, if a is coprime to m, that is, gcd(a, m) = 1, then $${\\displaystyle a^{\\phi (m)}\\equiv 1{\\pmod {m}},}$$ where $${\\displaystyle \\phi }$$ is Euler's totient function. A modular multiplicative inverse can be found directly: $${\\displaystyle a^{\\phi (m)-1}\\equiv a^{-1}{\\pmod {m}}.}$$ n must be coprime to 1320 or gcd(n,1320) = 1, Factorization of 1320: $$1320 = 2^3×3×5×11$$ (6 prime factors, 4 distinct) The smallest integer n > 1 coprime to 1320 is 7. $$\\begin{array}{|rcll|} \\hline && 7^{-1}{\\pmod {1320}} \\\\ &\\equiv& 7^{\\phi (1320)-1}{\\pmod {1320}} \\quad | \\quad \\phi (1320) = 320 \\\\ &\\equiv& 7^{320-1}{\\pmod {1320}} \\\\ &\\equiv& 7^{319}{\\pmod {1320}} \\\\ &\\equiv& \\mathbf{943} {\\pmod {1320}} \\\\ \\hline \\end{array}$$ May 21, 2019 edited by heureka May 22, 2019", "+0 # Please help, ASAP! 0 125 1 I usually don't do this but I'm really sick and missed class. Any help is appreciated! 1) Trapezoid $EFGH$ is inscribed in a circle, with $EF \\parallel GH$. If arc $GH$ is $70$ degrees, arc $EH$ is $x^2 - 2x$ degrees, and arc $FG$ is $56 - 3x$ degrees, where $x > 0,$ find arc $EPF$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E; A = dir(170); B = dir(30); D = dir(130); C = dir(70); E = dir(280); draw(Circle((0,0),1)); draw(A--B--C--D--cycle); label(\"$E$\", A, W); label(\"$F$\", B, dir(0)); label(\"$G$\", C, NE); label(\"$H$\", D, NW); dot(\"$P$\", E, S); [/asy] 2) In cyclic quadrilateral $PQRS,$ $\\frac{\\angle P}{2} = \\frac{\\angle Q}{3} = \\frac{\\angle R}{4}.$Find the largest angle of quadrilateral $PQRS,$ in degrees. 3) In the diagram, $\\angle U = 30^\\circ$, arc $XY$ is $170^\\circ$, and arc $VW$ is $110^\\circ$. Find arc $WY$, in degrees. [asy] unitsize(2 cm); pair A, B, C, D, E, F; D = dir(160); B = dir(200); E = dir(0); C = dir(270); A = extension(B,C,D,E); F = extension(B,E,C,D); draw(Circle((0,0),1)); draw(C--A--E); draw(C--D--B--E); label(\"$U$\", A, W); label(\"$V$\", B, SW); label(\"$W$\", C, S); label(\"$X$\", D, NW); label(\"$Y$\", E, dir(0)); //label(\"$F$\", F, dir(60)); [/asy] 4) In rectangle $EFGH$, $EH = 3$ and $EF = 4$. Let $M$ be the midpoint of", "# Getting $x,y$ position on an image based on given value This should be simple but my math skills are really bad ... I have an image of 36 images (6 by 6 matrix). These small images are 36 instances of a direction arrow (like from Google maps GPS), each image rotated by $10^\\circ$ from $0^\\circ$ to $350^\\circ$. I don't know how to get image $X$ and $Y$ position of each small image when I have the rotation amount. Images are $50\\times50\\,\\mathrm{px}$, so... • for $10^\\circ$ image is at $50\\times0\\,\\mathrm{px}$, • for $20^\\circ$ image is at $100\\times0\\,\\mathrm{px}$ • for $90^\\circ$ image is at $150\\times50\\,\\mathrm{px}$ (because it is on row 2) • ... • for $350^\\circ$ image is at $250\\times250\\,\\mathrm{px}$ What is the formula to get this $X-Y$ value? - Given that the angle of the $n$th image is $n\\cdot10^\\circ$ for $n=0,1,\\dots,35$, you can have the $y$-index $j$ to be $\\lfloor{n/6}\\rfloor$ (int part of $n/6$) and the $x$-index $i$ to be $n-6j$. You then multiply $i$ and $j$ for the respective $\\Delta x$ and $\\Delta j$ which in your problem is $50\\,\\mathrm{px}$ each. \\begin{align} j &= \\lfloor{n/6}\\rfloor \\\\ i &= n-6j \\\\ x &=i\\cdot50\\,\\mathrm{px} \\\\ y &=j\\cdot50\\,\\mathrm{px} \\end{align} Programaticaly (this will be pseudocode, but you might be able to adapt to your language of choice): for n = 0", "+0 # gveme 0 175 3 +1167 Find the least positive integer x that satisfies $$x+4609 \\equiv 2104 \\pmod{12}$$. tertre Mar 23, 2017 Sort: #1 0 The smallest positive integer =3 General formula: 3 + 12n, for n = 0, 1, 2,.....etc. Guest Mar 23, 2017 #2 +6810 +1 if $$x+4609 \\equiv 2104 \\pmod{12}$$ , x + 2505 is divisible by 12 x must be an odd number in order to make x + 2505 divisible by 12. Trial and error: Try x = 1, 1 + 2505 = 2506 <-- not divisible by 12. Try x = 3, 3 + 2505 = 2508 <-- divisible by 12(2508 = 2 x 2 x 3 x 11 x 19) Therefore the least positive integer x is 3. MaxWong Mar 23, 2017 #3 +18625 +1 Find the least positive integer x that satisfies $$x+4609 \\equiv 2104 \\pmod{12}$$. $$\\begin{array}{|lrcll|} \\hline & x+4609 &\\equiv& 2104 \\pmod{12} \\\\ & x+4609-2104 &=& n \\cdot 12 \\quad & | \\quad n \\in \\mathbb{N} \\\\ & x+2505 &=& n \\cdot 12 \\\\ & x&=& n \\cdot 12 -2505 \\\\\\\\ \\text{so } & n \\cdot 12 &=& 2505 \\\\ & n &=& \\uparrow \\frac{2505}{12} \\uparrow \\\\ & n &= & \\uparrow 208.75 \\uparrow \\\\ & n &= & 209 \\\\\\\\ & x&=& 209 \\cdot 12 -2505 \\\\", "+0 # Confusing 0 133 4 +282 i hate this lizagame Feb 28, 2017 Sort: #1 0 each quadrant is 90 degrees, so you will turn it 3 times, because 90 * 3 = 270 Guest Feb 28, 2017 #2 +79894 0 Rotating the figure 270° counter-clockwise is the same as rotating the figure 90° clockwise... And each point will swap coordinates and both will have positive signs....so (-2,1) → (1, 2) (-4,2) → (2, 4) (-3,5) → (5, 3) C is correct CPhill Feb 28, 2017 #3 +91254 0 Try watching this. There are a lot of resourced on the web to help with mathematical concepts. If you want us to help you find them then ask specifically. :) Melody Feb 28, 2017 #4 +18777 +5 Matrix Rotation counterclockwise: $$\\begin{array}{|rcll|} \\hline \\begin{pmatrix} \\cos(\\varphi) & \\sin (\\varphi) \\\\ -\\sin(\\varphi) & \\cos (\\varphi) \\\\ \\end{pmatrix} \\stackrel{\\varphi=270^{\\circ}} \\rightarrow \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} \\\\ \\hline \\end{array}$$ The point P becomes to P': $$\\begin{array}{|rcll|} \\hline \\binom{x}{y}\\cdot \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} = \\binom{y}{-x} \\\\ \\hline \\end{array}$$ $$\\text{Let}\\ X =\\binom{-2}{1} \\\\ \\text{Let}\\ Y =\\binom{-4}{2} \\\\ \\text{Let}\\ Z =\\binom{-3}{5} \\\\$$ $$\\begin{array}{|rcll|} \\hline X'=\\binom{-2}{1}\\cdot \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} = \\binom{1}{2} \\\\ Y'=\\binom{-4}{2}\\cdot \\begin{pmatrix} 0 & -1 \\\\ 1", "+0 What is the smallest positive integer $n$ for which $(12{,}500{,}000)\\cdot n$ leaves a remainder of $111$ when divided by $999{,}999{,}999$? +1 802 7 +474 That is the smallest positive integer N for which $$(12{,}500{,}000)\\cdot n$$ leaves a remainder of 111 when divided by 999,999,999? Jan 29, 2018 #3 +22307 +2 What is the smallest positive integer N for which $$(12{,}500{,}000)\\cdot n$$ leaves a remainder of 111 when divided by 999,999,999? 1. $$\\begin{array}{|rcll|} \\hline (12~500~000)\\cdot n & \\equiv & 111 \\pmod{999~999~999} \\quad & | \\quad :(12~500~000) \\\\ n & \\equiv & 111\\cdot \\dfrac{1}{12~500~000} \\pmod{999~999~999} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|lcll|} \\hline \\text{According to Euler's theorem,} \\\\ \\text{if a is coprime to m, that is, gcd(a, m) = 1, then }\\\\ {\\displaystyle a^{\\phi (m)}\\equiv 1{\\pmod {m}},}\\\\ \\text{where {\\displaystyle \\phi } is Euler's totient function.}\\\\ \\text{Therefore, a modular multiplicative inverse can be found directly:} \\\\ {\\displaystyle a^{\\phi (m)-1}\\equiv a^{-1}{\\pmod {m}}.} \\\\ \\hline \\end{array}$$ 2. $$\\text{greatest common divisor gcd(12~500~000,999~999~999) = 1 }$$ $$\\begin{array}{|rcll|} \\hline && \\dfrac{1}{12~500~000} \\pmod{999~999~999} \\\\\\\\ & \\mathbf{\\equiv} & \\mathbf{12~500~000^{-1} \\pmod{999~999~999}} \\\\\\\\ & \\equiv & 12~500~000^{\\phi (999~999~999)-1} \\pmod{999~999~999} \\\\\\\\ & \\equiv & 12~500~000^{648~646~704-1} \\pmod{999~999~999} \\\\\\\\ & \\equiv & 12~500~000^{648~646~703} \\pmod{999~999~999} \\\\\\\\ & \\mathbf{\\equiv} & \\mathbf{80 \\pmod{999~999~999}} \\\\ \\hline \\end{array}$$ 3. n = ? $$\\begin{array}{|rcll|} \\hline n & \\equiv & 111\\cdot \\dfrac{1}{12~500~000} \\pmod{999~999~999} \\\\\\\\ & \\equiv & 111\\cdot \\left( 12~500~000^{-1} \\pmod{999~999~999} \\right)", "+0 # Confusing 0 116 4 +282 i hate this lizagame Feb 28, 2017 Sort: #1 0 each quadrant is 90 degrees, so you will turn it 3 times, because 90 * 3 = 270 Guest Feb 28, 2017 #2 +77001 0 Rotating the figure 270° counter-clockwise is the same as rotating the figure 90° clockwise... And each point will swap coordinates and both will have positive signs....so (-2,1) → (1, 2) (-4,2) → (2, 4) (-3,5) → (5, 3) C is correct CPhill Feb 28, 2017 #3 +90586 0 Try watching this. There are a lot of resourced on the web to help with mathematical concepts. Melody Feb 28, 2017 #4 +18624 +5 Matrix Rotation counterclockwise: $$\\begin{array}{|rcll|} \\hline \\begin{pmatrix} \\cos(\\varphi) & \\sin (\\varphi) \\\\ -\\sin(\\varphi) & \\cos (\\varphi) \\\\ \\end{pmatrix} \\stackrel{\\varphi=270^{\\circ}} \\rightarrow \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} \\\\ \\hline \\end{array}$$ The point P becomes to P': $$\\begin{array}{|rcll|} \\hline \\binom{x}{y}\\cdot \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} = \\binom{y}{-x} \\\\ \\hline \\end{array}$$ $$\\text{Let}\\ X =\\binom{-2}{1} \\\\ \\text{Let}\\ Y =\\binom{-4}{2} \\\\ \\text{Let}\\ Z =\\binom{-3}{5} \\\\$$ $$\\begin{array}{|rcll|} \\hline X'=\\binom{-2}{1}\\cdot \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} = \\binom{1}{2} \\\\ Y'=\\binom{-4}{2}\\cdot \\begin{pmatrix} 0 & -1 \\\\ 1 & 0 \\\\ \\end{pmatrix} = \\binom{2}{4} \\\\ Z'=\\binom{-3}{5}\\cdot \\begin{pmatrix} 0 & -1 \\\\", "+0 # help 0 237 1 What is the smallest positive integer $n$ for which $9n-2$ and $7n + 3$ share a common factor greater than $1$? May 12, 2020 #1 +25565 +1 What is the smallest positive integer $$n$$ for which $$9n-2$$ and $$7n + 3$$ share a common factor greater than $$1$$? My attempt: $$\\gcd(9n-2,\\ 7n+3) \\ne 1$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{\\gcd(9n-2,\\ 7n+3)} &=& \\gcd(9n-2-( 7n+3),\\ 7n+3) \\\\ &=& \\gcd(2n-5,\\ 7n+3) \\\\ &=& \\gcd(7n+3,\\ 2n-5) \\\\ &=& \\gcd(7n+3-3(2n-5),\\ 2n-5) \\\\ &=& \\gcd(n+18,\\ 2n-5) \\\\ &=& \\gcd(2n-5,\\ n+18) \\\\ &=& \\gcd(2n-5-2(n+18),\\ n+18) \\\\ &=& \\mathbf{\\gcd(-41,\\ n+18)} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline \\mathbf{\\gcd(-41,\\ 41)} &=& 41 \\\\ n+18 &=& 41 \\\\ n &=& 41-18 \\\\ \\mathbf{n} &=& \\mathbf{23} \\quad | \\quad \\text{smallest positive integer n }\\\\ \\hline \\end{array}$$ In general: $$\\begin{array}{|rcll|} \\hline n+18 &\\equiv& 0 \\pmod{41} \\quad | \\quad -18\\\\ n &\\equiv& -18 \\pmod{41} \\\\ n &\\equiv& -18 +41 \\pmod{41} \\\\ n &\\equiv& 23 \\pmod{41} \\\\ n-23 &=& z\\cdot 41 \\quad | \\quad z\\in \\mathbb{Z} \\\\ \\mathbf{n} &=& \\mathbf{23+z\\cdot 41} \\\\ \\hline \\end{array}$$ May 12, 2020", "811 calculations, way too tedious for humans. But there is a better way, and we have touched on it quite a few times! We notice that we could make the jump just like before into rational mathematics: ${\\displaystyle {\\begin{matrix}216a&=&1+811b\\\\0&\\equiv &1+163b&{\\pmod {216}}\\\\\\end{matrix}}}$ We jump into rational maths again ${\\displaystyle {\\begin{matrix}216c&=&1+163b\\\\53c&\\equiv &1&{\\pmod {163}}\\\\\\end{matrix}}}$ We jump once more ${\\displaystyle {\\begin{matrix}53c&=&1+163d\\\\0&\\equiv &1+4d&{\\pmod {53}}\\\\\\end{matrix}}}$ Now the pattern is clear, we shall start from the beginning so that the process is not broken: ${\\displaystyle {\\begin{matrix}216a&=&1+811b\\\\216c&=&1+163b\\\\53c&=&1+163d\\\\53e&=&1+4d\\\\e&=&1+4f\\\\\\end{matrix}}}$ Now all we have to do is choose a value for f and substitute it back to find a! Remember a is the inverse of 216 mod 811. We choose f = 0, therefore e = 1, d = 13, c = 40, b = 53 and finally a = 199! If f is chosen to be 1 we will get a different value for a. The very perceptive reader should have noticed that this is just Euclid's gcd algorithm in reverse. Here are a few more examples of this ingenious method in action: Example 1 Find the smallest positive value of a: ${\\displaystyle {\\begin{matrix}33a&\\equiv &1\\ \\ {\\mbox{(mod 101)}}\\\\33a&=&1+101b\\\\33c&=&1+2b\\\\c&=&1+2d\\\\\\end{matrix}}}$ Choose d = 0, therefore a = 49. Example 2 Find the smallest positive value of a: ${\\displaystyle {\\begin{matrix}27a&\\equiv &1\\ \\ {\\mbox{(mod 821)}}\\\\27a&=&1+821b\\\\27c&=&1+11b\\\\5c&=&1+11d\\\\5e&=&1+d\\\\\\end{matrix}}}$ Choose e = 0, therefore a =", "# To find: The smallest the smallest positive integer such that if we divide it by two, three and four, the remainder is one but seven divides the number evenly. Congruence To find: The smallest the smallest positive integer such that if we divide it by two, three and four, the remainder is one but seven divides the number evenly. 2021-04-15 Suppose the required smallest positive integer is x, then from the given information, three congruence equations are $$\\displaystyle{x}\\equiv{1}\\pm{o}{d}{2},{x}\\equiv{1}\\pm{o}{d}{3},{x}\\equiv{1}\\pm{o}{d}{4}$$. The congruence $$\\displaystyle{x}\\equiv{1}\\pm{o}{d}{2}$$ means if x divided by 2 the remainder is 1. So the number x is one of the number from the following list: 1,3,5,7,9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39,41, 43, 45, 47, 49,51... Similarly The congruence $$\\displaystyle{x}\\equiv{1}\\pm{o}{d}{3}$$ means if x divided by 3 the remainder is 1. So the number x is one of the number from the following list: 1,4,7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49... The congruence $$\\displaystyle{x}\\equiv{1}\\pm{o}{d}{4}$$ means if x divided by 4 the remainder is 1. So the number x is one of the number from the following list: 1,5, 9, 13, 17,21, 25, 29, 33, 37,41, 45, 49, 53, 57,... Let a and b be the two numbers, a divides b evenly means", "+0 # i) Binomial mathematics 0 945 2 +105 i) Show that for all positivew integers n, $$x[(1+x)^{n-1}+(1+x)^{n-2}+\\dots (1+x)+1]=(1+x)^n-1\\\\$$ I can do this first bit no worries but then part 2 is Hence show that for $$1\\le k\\le n$$ $$\\begin{pmatrix}n-1 \\\\k-1\\end{pmatrix}+ \\begin{pmatrix}n-2 \\\\k-1\\end{pmatrix}+ \\begin{pmatrix}n-3 \\\\k-1\\end{pmatrix}+\\dots + \\begin{pmatrix}k-1 \\\\k-1\\end{pmatrix}= \\begin{pmatrix}n \\\\k\\end{pmatrix}$$ I don't know how to do this second bit. Can someone help me please? Sep 7, 2016 #1 0 Give n and k numerical value, which might be easier to visualize: i. e. n=11, k=3 (11 -1)C(3 - 1)=45 + (11 - 2)C(3 - 1)=36+ (11 - 3)C(3 - 1)=28+(11 - 4)C(3-1)=21........=11C3=165 45 + 36 + 28 + 21........+15+ 10+6+3+1 =165 Sep 7, 2016 #2 +118117 +5 i) Show that for all positiveintegers n, $$x[(1+x)^{n-1}+(1+x)^{n-2}+\\dots (1+x)+1]=(1+x)^n-1\\qquad (1)\\\\$$ I can do this first bit no worries but then part 2 is Hence show that for $$1\\le k\\le n$$ $$\\begin{pmatrix}n-1 \\\\k-1\\end{pmatrix}+ \\begin{pmatrix}n-2 \\\\k-1\\end{pmatrix}+ \\begin{pmatrix}n-3 \\\\k-1\\end{pmatrix}+\\dots + \\begin{pmatrix}k-1 \\\\k-1\\end{pmatrix}= \\begin{pmatrix}n \\\\k\\end{pmatrix}$$ I don't know how to do this second bit. Can someone help me please? Ok what you need to see is that is the coefficient of x^k in the right hand side of equation 1 is nCk So I need to find the coefficient of x^k in the left hand side. Those two coefficients mucs be equal Looking at equation", "= \\angle XBW = 90^{\\circ} = \\angle AMH = \\angle ATH$$ so $H$ lies on this line as well and we may conclude $\\square$ --mathguy623 03:10, 12 August 2016 (EDT) ## Solution 3 (Complex Bash) $[asy] unitsize(0.8cm); draw((0,0)--(14,0)--cycle); draw((0,0)--(5,12)--cycle); draw((5,12)--(14,0)--cycle); draw((2.071,4.97)--(14,0)--cycle); draw((8.96,6.72)--(0,0)--cycle); draw((2.071,4.97)--(9,0)--cycle); draw((8.96,6.72)--(9,0)--cycle); draw((0,0)--(0,2.083)--cycle); draw((14,0)--(14,6.75)--cycle); draw((0,2.083)--(14,6.75)--cycle); draw((0,2.083)--(2.071,4.97)--cycle); draw((0,2.083)--(9,0)--cycle); draw((8.96,6.72)--(14,6.75)--cycle); draw((14,6.75)--(9,0)--cycle); draw(circle((4.5,1.0415),4.61895)); draw(circle((11.5,3.375),4.20007)); label((0,0),\"C\",SW); label((14,0),\"B\",SE); label((5,12),\"A\",N); label((8.96,7),\"N\",N); label((2.071,4.97),\"M\",NW); label((8.25,0),\"W\",S); label((5,4),\"H\",N); label((0,2.083),\"Y\",NW); label((14,6.75),\"X\",NE); [/asy]$ For any point $P,$ let $p$ denote the complex number for point $P.$ First off, let $C$ be the origin. Now, since $WY$ is the diameter of the circumcircle of triangle $CMW$, we must have $\\angle{YMW}=\\angle{YCH}=90^{\\circ}.$ Since angles inscribed in the same arc are congruent, $\\angle{MYW}=\\angle{MCW}=\\angle{C}.$ This means that $\\frac{|w-m|}{|y-m|}=\\tan{\\angle{C}}.$ Combining this with the fact that $\\angle{YMW}$ is right, we find that $$i(y-m)\\tan{\\angle{C}}=(w-m).$$ Solving, we find that $$y=\\frac{w+(i\\tan{\\angle{C}}-1)m}{i\\tan{\\angle{C}}}.$$ We wish to simplify $(i\\tan{\\angle{C}}-1)m$ first. Note that $$m=\\frac{|CM|}{|CA|}\\cdot(a)=\\frac{(|BC|)\\cos{\\angle{C}}}{|CA|}\\cdot(|CA|)(\\cos{\\angle{C}}+i\\sin{\\angle{C}})=(|BC|)\\cos{\\angle{C}}(\\cos{\\angle{C}}+i\\sin{\\angle{C}}).$$ This means that \\begin{align*} (i\\tan{\\angle{C}}-1)m&=(i\\tan{\\angle{C}}-1)((|BC|)\\cos{\\angle{C}}(\\cos{\\angle{C}}+i\\sin{\\angle{C}}))\\\\ &=\\left(\\frac{i\\sin{\\angle{C}}}{\\cos{\\angle{C}}}-1\\right)((|BC|)\\cos{\\angle{C}}(\\cos{\\angle{C}}+i\\sin{\\angle{C}}))\\\\ &=(i\\sin{\\angle{C}}-\\cos{\\angle{C}})(|BC|)(\\cos{\\angle{C}}+i\\sin{\\angle{C}})\\\\ &=-|BC| \\end{align*} This means that $$y=\\frac{w-|BC|}{i\\tan{\\angle{C}}}=\\frac{|BC|-w}{\\tan{\\angle{C}}}i.$$ Now, since $WX$ is a diameter of the circumcircle of triangle $NBW,$ we must have $\\angle{WNX}=\\angle{WBX}=90^{\\circ}.$ Since angles inscribed in the same arc are congruent, $\\angle{NXW}=\\angle{NBW}=\\angle{B}.$ This means that $\\frac{|x-n|}{|w-n|}=\\cot{\\angle{B}}.$ Combining this with the fact that $\\angle{WNX}$ is right, we find that $$(x-n)=i\\cot{\\angle{B}}(w-n).$$ Solving, we find that $$x=wi\\cot{\\angle{B}}+n(1-i\\cot{\\angle{B}}).$$ We wish to simplify $n(1-i\\cot{\\angle{B}})$ first. Note that $$n=e^{(90-\\angle{B})i}|CN|=(\\cos{(90-\\angle{B})}+i\\sin{(90-\\angle{B})})(|BC|)\\sin{\\angle{B}}=(|BC|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}})).$$ This means that \\begin{align*} n(1-i\\cot{\\angle{B}})&=(|BC|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}}))(1-i\\cot{\\angle{B}})\\\\ &=(|BC|)\\sin{\\angle{B}}(\\sin{\\angle{B}}+i\\cos{\\angle{B}}))\\left(1-\\frac{\\cos{\\angle{B}}}{\\sin{\\angle{B}}}i\\right)\\\\ &=(|BC|)(\\sin{\\angle{B}}+i\\cos{\\angle{B}})(\\sin{\\angle{B}}-i\\cos{\\angle{B}})\\\\ &=(|BC|)(1)", "28 at 12:01 • @PeterTaylor This was an assumption that I thought should be true, since (255,255,255) is the room diagonal, I thought it should have 45° angles for teta and phi. For phi that is correct but as it turns out teta has a 54,7° angle = cos^-1(1/sqrt(3)). – Lucas Raphael Pianegonda Feb 28 at 12:54 So how would one generate RGB vectors with a constant norm that are still valid RGB values? ($$0 \\le r,g,b \\le 255$$) There is a simpler algorithm. Let $$L$$ be the given norm. • Verify that $$0\\le L\\le 255\\sqrt3$$. • Let r be a random number between $$\\sqrt{L^2-\\min(2\\cdot255^2, L^2)}$$ and $$\\min(255, L)$$. • Let $$M=\\sqrt{L^2-r^2}$$. Let g be a random number between $$\\sqrt{M^2-\\min(255^2, M^2)}$$ and $$\\min(255,M)$$. • Let $$b =\\sqrt{L^2-r^2-g^2}$$. Some minor adjustment is needed if $$r,g,b$$ should be integers, in which case only approximate solutions can be found in general. Here is the right computation if we want to find the range of $$\\theta$$ and $$\\phi$$. • If $$0\\le L\\le 255$$, then $$0\\le\\theta,\\phi\\le\\pi/2$$ • If $$255\\lt L\\le 255\\sqrt3$$, let $$M=\\sqrt{\\max(0, L^2-2\\cdot255^2})$$. we have \\begin{align} &\\quad\\quad\\begin{cases}b = L \\sin(\\theta) \\cos(\\phi)\\\\ r = L \\sin(\\theta) \\sin(\\phi)\\\\ g = L \\cos(\\theta)\\\\ 0\\le b,r,g\\le 255\\\\ 0\\le\\theta,\\phi\\le\\pi/2 \\end{cases}\\\\ &\\Longleftrightarrow \\begin{cases}b = L \\sin(\\theta) \\cos(\\phi)\\\\ r = L \\sin(\\theta) \\sin(\\phi)\\\\ g = L \\cos(\\theta)\\\\ \\sqrt{\\max\\left(0,(L\\sin(\\theta))^2-255^2\\right)}\\le b,r\\le", "- 40\\sqrt{10}$. Through power of a point, we can find out that $(CE)(DE)=20\\sqrt{10} - 20$, so $CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\\sqrt{10}) + 2(20\\sqrt{10} - 20) = \\boxed{\\textbf{(E) } 100}$. ~Math_Wiz_3.14 ## Solution 4 (Reflections) $[asy] draw(circle((0,0),7.07)); dot((-7.07,0)); label(\"A\",(-7.07,0),W); dot((7.07,0)); label(\"B\",(7.07,0),E); dot((0,0)); label(\"O\",(0,0),N); dot((-2.24,-6.71)); label(\"C\",(-2.24,-6.71),SSW); dot((6.71,2.24)); label(\"D\",(6.71,2.24),NE); draw((-2.24,-6.71)--(6.71,2.24)); dot((6.71,-2.24)); label(\"D'\",(6.71,-2.24),SE); draw((4.51,0)--(6.71,-2.24)); dot((4.51,0)); label(\"E\",(4.51,0),NNW); draw((-7.07,0)--(7.07,0)); draw((0,0)--(-2.24,-6.71)); draw((0,0)--(6.71,-2.24)); draw((-2.24,-6.71)--(6.71,-2.24)); [/asy]$ Let $O$ be the center of the circle. By reflecting $D$ across the line $AB$ to produce $D'$, we have that $\\angle BED'=45$. Since $\\angle AEC=45$, $\\angle CED'=90$. Since $DE=ED'$, by the Pythagorean Theorem, our desired solution is just $CD'^2$. Looking next to circle arcs, we know that $\\angle AEC=\\frac{\\overarc{AC}+\\overarc{BD}}{2}=45$, so $\\overarc{AC}+\\overarc{BD}=90$. Since $\\overarc{BD'}=\\overarc{BD}$, and $\\overarc{AC}+\\overarc{BD'}+\\overarc{CD'}=180$, $\\overarc{CD'}=90$. Thus, $\\angle COD'=90$. Since $OC=OD'=5\\sqrt{2}$, by the Pythagorean Theorem, the desired $CD'^2= \\boxed{\\textbf{(E) } 100}$. ~sofas103 ## See Also 2020 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's", "a: $\\begin{matrix} 33a & \\equiv & 1 \\ \\ \\mbox{(mod 101)}\\\\ 33a &=& 1 + 101b\\\\ 33c &=& 1 + 2b\\\\ c &=& 1 + 2d\\\\ \\end{matrix}$ Choose d = 0, therefore a = 49. Example 2 Find the smallest positive value of a: $\\begin{matrix} 27a & \\equiv & 1 \\ \\ \\mbox{(mod 821)}\\\\ 27a &=& 1 + 821b\\\\ 27c &=& 1 + 11b\\\\ 5c &=& 1 + 11d\\\\ 5e &=& 1 + d\\\\ \\end{matrix}$ Choose e = 0, therefore a = -152 = 669 Example 3 Find the smallest positive value of a: $\\begin{matrix} 34a & \\equiv & 1 \\ \\ \\mbox{(mod 55)}\\\\ 34a &=& 1 + 55b\\\\ 34c &=& 1+ 21b\\\\ 13c& =& 1 + 21d\\\\ 13e& =& 1 + 8d\\\\ 5e &=& 1 + 8f\\\\ 5g& = &1 + 3f\\\\ 2g& =& 1 + 3h\\\\ 2i& = &1 + h\\\\ \\end{matrix}$ Set i = 0, then a = -21 = 34. Why is this so slow for two numbers that are so small? What can you say about the coefficients? Example 4 Find the smallest positive value of a: $\\begin{matrix} 21a & \\equiv & 1 \\ \\ \\mbox{(mod 102)}\\\\ 21a &=& 1 + 102b\\\\ 21c &=& 1 + 18b\\\\ 3c &=& 1 + 18d\\\\ 3d &=& 1\\\\ \\end{matrix}$ Now d is not an integer, therefore", "3 111 and 369 are not coprime 4. smaller larger 2021 4032 2011 2021 10 2011 1 10 0 1 2021 and 4032 are coprime 2.We first calculate the gcd for all combinations smaller larger 15 510 0 15 smaller larger 15 375 0 15 smaller larger 375 510 135 375 105 135 30 105 15 30 0 15 The gcd for any combination of the numbers is 15 so the gcd is 15 for the three numbers. Diophantine equation Exercises 1. ${\\displaystyle {\\begin{matrix}216x&=&1+816b\\\\216c&=&1+168b\\\\48c&=&1+168d\\\\48e&=&1+24d\\\\24e&=&1+24f\\\\\\end{matrix}}}$ There is no solution, because can never become an integer. 2. ${\\displaystyle {\\begin{matrix}42x&=&7+217b\\\\42c&=&7+7b\\\\7c&=&0+7d\\\\\\end{matrix}}}$ We choose d=1, then x=26. 3. (a) smaller larger PQ 33 101 3 2 33 16 1 2 2 0 1 3 16 2 0 1 3 49 101 1 0 1 16 33 4. (a) smaller larger PQ 17 317 18 11 17 1 6 11 1 5 6 1 1 5 5 0 1 18 1 1 1 5 0 1 18 19 37 56 317 1 0 1 1 2 3 17 Chinese remainder theorem exercises 1. ${\\displaystyle {\\begin{matrix}3x&\\equiv &5{\\pmod {14}}\\\\x&\\equiv &11{\\pmod {14}}\\\\x&=&11+14a\\\\2x&=&2(11+14a)&\\equiv &-3{\\pmod {17}}\\\\&&22+28a&\\equiv &-3{\\pmod {17}}\\\\&&11a&\\equiv &-8{\\pmod {17}}\\\\&&a&=&7+17b\\\\x&=&11+14(7+17b)&\\equiv &6{\\pmod {15}}\\\\&=&109+238b&\\equiv &6{\\pmod {15}}\\\\&=&4+13b&\\equiv &6{\\pmod {15}}\\\\&=&13b&\\equiv &2{\\pmod {15}}\\\\&&b&\\equiv &14{\\pmod {15}}\\\\&&b&=&14+15c\\\\x&=&109+238(14+15c)\\\\x&=&3441+3570c\\end{matrix}}}$ Question 1 Show that the divisible-by-3 theorem works for any 3 digits numbers (Hint: Express a 3 digit number", "+0 # This is too confsing -1 106 5 +172 What is the smallest positive integer n such that $$3n \\equiv 1356 \\pmod{22}?$$ Jul 13, 2020 #1 +744 +1 only try to post a few or a single questions at a time, please Jul 13, 2020 #3 +279 0 yeah forumofweb2.0cal do your own homework #4 +2313 +1 I think this is Alcumus.... CalTheGreat Jul 13, 2020 #2 0 I think it is 12. Jul 13, 2020 #5 +25543 +1 What is the smallest positive integer n such that $$3n \\equiv 1356 \\pmod{22}$$ $$\\begin{array}{|rcll|} \\hline 3n &\\equiv& 1356 \\pmod{22} \\\\ 3n &\\equiv& 1356-61*22 \\pmod{22} \\\\ 3n &\\equiv& 14 \\pmod{22} \\\\ n &\\equiv& \\dfrac{14}{3} \\pmod{22} \\\\ n &\\equiv& 14*3^{-1} \\pmod{22} \\\\ && \\boxed{ 3^{-1} \\pmod{22} = 3^{\\phi(22)-1} \\pmod{22} \\quad | \\quad \\phi(22)=22*\\left( 1-\\dfrac{1}{2}\\right)*\\left( 1-\\dfrac{1}{11}\\right)=10 \\\\ =3^{10-1} \\pmod{22} \\\\ = 3^9 \\pmod{22} \\\\ = 19683 \\pmod{22} \\\\ = 19683-894*22 \\pmod{22} \\\\ = 15 \\pmod{22} \\\\\\mathbf{3^{-1} \\pmod{22}=15 \\pmod{22}} } \\\\ n &\\equiv& 14*15 \\pmod{22} \\\\ n &\\equiv& 210 \\pmod{22} \\\\ n &\\equiv& 210-9*22 \\pmod{22} \\\\ \\mathbf{n} &\\equiv& \\mathbf{12 \\pmod{22}} \\\\ \\hline \\end{array}$$ The smallest positive integer n is $$\\mathbf{12}$$ Jul 14, 2020 edited by heureka Jul 14, 2020", "angle $-\\arctan(A/C')$ to cancel $A$ $$(A,0,C')\\to(0,0,C'').$$ • Assuming I need to execute the transformations one after another given only (A,B,C) ? without calculating the angle arctan(A/C') as this is something I can't do without calculating the first rotation? May 6 '15 at 17:35 • $C'=\\sqrt{B^2+C^2}$ and $C''=\\sqrt{A^2+B^2+C^2}$, obviously. – user65203 May 6 '15 at 17:42 • Assuming the (A,B,C) = (1,0,0), we divide by zero. What do you say? May 6 '15 at 17:45 • No problem. For the first rotation the angle is arbitrary and the second, $\\pi/2$. The equations are coherent. – user65203 May 6 '15 at 17:47 What you want to do is rotate your vector $(A,B,C)^T$ three times, around each axis, and find the appropriate angles of rotation. You can write: $$\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}=R_X\\cdot R_Y\\cdot R_Z\\cdot\\begin{pmatrix}A\\\\B\\\\C\\end{pmatrix}$$ $$\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}=\\begin{pmatrix}1&0&0\\\\0&\\cos X&-\\sin X\\\\0&\\sin X&\\cos X\\end{pmatrix}\\begin{pmatrix}\\cos Y&0&\\sin Y\\\\0&1&0\\\\-\\sin Y&0&\\cos Y\\end{pmatrix}\\begin{pmatrix}\\cos Z&-\\sin Z&0\\\\\\sin Z&\\cos Z&0\\\\0&0&1\\end{pmatrix}\\begin{pmatrix}A\\\\B\\\\C\\end{pmatrix}$$ Now you can multiply these matrices, and you should end up with a system of three equations with three unknowns, namely the angles $X$, $Y$ and $Z$. • I don't think this is the best option as this is like \"brute forcing\" the angles. There must be a way to calculate the angels like @Yves Daoust suggested, just without the need of calculating the second angle after calculating the first transformation. May 6 '15 at 17:43 For" ]

[ "Find curve in $x-y$ plane which passes through$(1,1)$ . Intersects line $y=x$ at right angle at that point and satisfies $xy''+2y''=0$ $(a)\\;y=\\frac{1}{x^2} \\\\ (b)\\;y = \\frac{1}{x} \\\\ (c)\\;y=0 \\\\ (d)\\;y=x^2$", "and $$run=1$$. So, to find another point on the line, we add $$3$$ to the y-coordinate of $$(0,2)$$ and add $$1$$ to the x-coordinate of $$(0,2)$$ to get the point $$(1,5)$$ 3. Plot the y-intercept and the point we found in the previous step on the xy-plane: 4. Finally, we draw the line that goes through both the points: chat loading...", "• 5. Determine whether the line given by $<5-t, -5, 1+2t>$ and the $xz$-plane will intersect. If so, find the intersection point.", "## Elementary Geometry for College Students (7th Edition) Clone The line $l$ intersects plane $X$ at $P$", "# Where Line Meets Plane Geometry Level 4 Consider a plane passing through the point $$(1,2,3)$$ with normal vector $$(4,5,6)$$. Suppose we start at the point $$(-1,0,5)$$ and travel in the direction of the vector $$(-1,-1,2)$$ until we intersect the plane. If the intersection point is $$(a,b,c)$$, what is $$a+b+c$$? × Problem Loading... Note Loading... Set Loading...", "passes through points (4, -5) and (4, 4). $$\\textrm{line passing through the points }( 4, -(5) )\\textrm{ and }( 4, 4 )$$", "Changing It Up Geometry Level 3 Given that a straight line in the $xy$-plane passes through the point $(3,48)$ in the first quadrant and intersects positive $x$- and $y$-axes at points $P$ and $Q$ respectively. Let $O$ designate the origin. Find the minimum value of $OP+OQ$. ×", "## In $$xy$$-plane, line $$K$$ passes through the points $$A(6, -7)$$ and $$B(4, 5).$$ Does line $$K$$ also pass through po ##### This topic has expert replies Moderator Posts: 1178 Joined: 29 Oct 2017 Thanked: 1 times Followed by:5 members ### In $$xy$$-plane, line $$K$$ passes through the points $$A(6, -7)$$ and $$B(4, 5).$$ Does line $$K$$ also pass through po by M7MBA » Tue Sep 22, 2020 11:59 pm 00:00 A B C D E ## Global Stats In $$xy$$-plane, line $$K$$ passes through the points $$A(6, -7)$$ and $$B(4, 5).$$ Does line $$K$$ also pass through point $$C?$$ (1) Coordinates of Point $$C$$ are $$( 5, -1).$$ (2) Point $$C$$ is equidistant from Point $$A$$ and Point $$B.$$", "of the quadrilateral, $\\overline{XY}$ , would have to intersect the line $\\overleftrightarrow{ZW}$ by the plane separation theorem. If $\\overline{XY}$ intersects $\\overline{ZW}$ , this is a contradiction which proves that X amd Y lie in the same half-plane which proves that W and Z lie on the same side of the line $\\overleftrightarrow{XY}$ . If $\\overline{XY}$ intersects the line $\\overleftrightarrow{ZW}$ outside the segment $\\overline{ZW}$ , then Z and W lie in the same half-plane determined by the line $\\overleftrightarrow{XY}$ .", "the point $(1,1)$. 3. Draw a line connecting these two points.", "Find the coordinates of the point where the line through (3, - 4, -5) and (2, -3, 1) crosses the plane 2x+y+z=7. ### Question Asked by a Student from EXXAMM.com Team Q 2677556486. Find the coordinates of the point where the line through (3, - 4, -5) and (2, -3, 1) crosses the plane 2x+y+z=7. #### HINT (Provided By a Student and Checked/Corrected by EXXAMM.com Team) #### Access free resources including • 100% free video lectures with detailed notes and examples • Previous Year Papers • Mock Tests • Practices question categorized in topics and 4 levels with detailed solutions • Syllabus & Pattern Analysis", "# Go straight ! Geometry Level 4 The number of lines in the $$xy$$- plane ,whose distance from $$(-1,2)$$ is $$2$$ and from $$(2,6)$$ is $$3$$ , is : ###### This problem is a part of the set advanced is basic × Problem Loading... Note Loading... Set Loading...", "found as the intersection between the line $$BN$$ and the line $$N_1N_2$$, as in Figure 4.", "## Algebra 1 The line goes through the points $(-2,0)$ and $(0,-3)$. Since the line goes through the point $(0,-3)$, the line crosses the y-axis at $-3$. (A line with all x-coordinates 0 is the x-axis).", "Go straight! Geometry Level 4 The number of lines in the $$xy$$- plane ,whose distance from $$(-1,2)$$ is $$2$$ and from $$(2,6)$$ is $$3$$ , is : ×", "the point $$(\\bar{x}, \\bar{y})$$ lies on the line.", "# 1, 2, 3, 4, 5 Geometry Level 1 The point $$(1 , 2)$$ and the line $$3x - 4y - 5 = 0$$ are on the $$xy$$-plane. What is the shortest distance between the line and the point? ×", "to $y=\\dfrac{3}{4}x$ and the line $k$ is parallel to $l$. If $k$ passes through the point $(-2, -3)$, which of the following shows the graphs of $l$ and $k$?", "$Q$. The third point that it passes through is the point $P$, and we draw the vertical line through $P$ to find the point $P'$, which is equal to $2Q$. Now we discuss the case described by the third picture, where the line going through the two points $P$ and $Q$ which we want to “add” is a vertical line. To explain what happens, we need the notion of a “point at infinity” (see Projective Geometry). We write the point at infinity as $0$, expressing the idea that it is the identity element of our group. We cannot find this point at infinity in the $x$$y$ plane, but we can think of it as the third point that the vertical line passes through aside from $P$ and $Q$. In this case, of course, there is no need to draw another vertical line – we simply write $P+Q=0$. Finally we come to the case described by the fourth picture; this is simply a combination of the earlier cases we have described above. The vertical line is tangent to the curve at the point $P$, so we can think of it as passing through $P$ twice, and the third point is passes through is the point at infinity $0$, so we can write $2P=0$. We will not prove explicitly that", "# Line passes through a point parallet to parametric equations 1. Sep 15, 2009 ### megr_ftw 1. The problem statement, all variables and given/known data Consider the line which passes through the point P(-1, 4, 3), and which is parallel to the line x = 1 + 3t, y = 2 + 4t, z = 3 + 5t Find the point of intersection of this new line with each of the coordinate planes: xy-plane: ( , , 0 ) xz-plane: ( , 0 , ) yz-plane: ( 0 , , ) 2. Relevant equations 3. The attempt at a solution I can figure out the whole intersection thing, I just don't understand how I'm suppose to make it parallel to a line at the same time. 2. Sep 15, 2009 ### Staff: Mentor I think you're approaching this in the wrong order. First, find the parameteric equations for the line through (-1, 4, 3), then figure out where this line intersects the coordinate planes. 3. Sep 15, 2009 ### tiny-tim Hi megr_ftw! I don't understand what you're saying here an intersection is a point, it can't be parallel to a line (or to anything). You're asked what is the line through P parallel to the given line. 4. Sep 15, 2009 ### megr_ftw so how do i go" ]

[ "4t=6+2(-1)}$ substituted u back into equation 1 ${\\displaystyle t=1}$ solved for t Now to check if it works in the final equation, substitute the values for u and t into the equation that you didn't use. ${\\displaystyle 2t=-1-3u}$ ${\\displaystyle 2(1)=-1-3(-1)}$ ${\\displaystyle 2=2}$ If you get a value expression (a constant equals itself) then this means that they intersect, else they don't. This is known as a skew. ### The closest point to another point on a line equation and the distance between them So let's say there's a vector equation ${\\displaystyle {\\textbf {r}}={\\begin{pmatrix}2\\\\2\\\\2\\end{pmatrix}}+t{\\begin{pmatrix}1\\\\-2\\\\3\\end{pmatrix}}}$ and there's a point P with a position vector: ${\\displaystyle {\\overrightarrow {OP}}={\\begin{pmatrix}1\\\\1\\\\2\\end{pmatrix}}}$ . We want to find both the coordinate of the point on the line that is closest to P and the distance between that point and P. We'll refer to this point on the line as X. First we need to use our dot product rule, because the line from P to X will be perpendicular to the line itself. So to do this we'll need to find the direction of the line and the vector from P to X. The direction of the line is just the direction vector of the line (which we'll call A): ${\\displaystyle {\\textbf {A}}={\\begin{pmatrix}1\\\\-2\\\\3\\end{pmatrix}}}$ As for the direction of P to X, we'll need to find X", "x=-5+\\frac{3}{7}z$ , and thus the intersection line is given by $\\left\\{\\begin{pmatrix}-5+\\frac{3}{7}t\\\\{}\\\\-1-\\frac{2}{7}t\\\\{}\\\\t\\end{pmatrix}\\,,\\,t\\in\\mathbb{R }\\right\\}$ . The above is already a parametrization, but your teacher may want you to write it as $u+tv\\,,\\,u,v$ vectors on the line, $t=$ the real parameter. Tonio", "Math Relative position of lines Determining point of intersection # Intersecting lines Two lines can intersect and then have a point of intersection. ! ### Requirements 1. Direction vectors are not collinear 2. When equating the linear equations, we receive a distinct result ## Calculating point of intersection To calculate the point of intersection, we follow the scheme above. The result from step 2 for $r$ or $s$ is then inserted into the corresponding linear equation. ### Example $\\text{g: } \\vec{x} = \\begin{pmatrix} 10 \\\\ 9 \\\\ 7 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}$ $\\text{h: } \\vec{x} = \\begin{pmatrix} 4 \\\\ 4 \\\\ 6 \\end{pmatrix} + s \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ 1. #### Direction vectors not collinear? First, we examine if the direction vectors are multiples of each other (=collinear). $\\vec{a}=t\\cdot\\vec{b}$ $\\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}=t\\cdot\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ Calculate $t$ for every row 1. $9=t\\cdot1$ 2. $8=t\\cdot1$ 3. $7=t\\cdot2$ 1. $t=9$ 2. $t=8$ 3. $t=3.5$ $t$ does not have the same value everywhere. Thus the direction vectors are not collinear. Therefore the lines are skew or have a point of intersection. 2. #### Equate $g$ and $h$ Now equate the linear equations and calculate $r$ and $s$. $\\begin{pmatrix} 10 \\\\ 9 \\\\ 7", "line segment $$PQ$$ has an end-point $$P\\begin{pmatrix}-3,4\\end{pmatrix}$$ and its mid-point has coordinates $$M\\begin{pmatrix}2,-1\\end{pmatrix}$$. Find the coordinates of the segment's other end-point $$Q$$. 5. A line segment $$AB$$ has an end-point $$A\\begin{pmatrix}1,-3\\end{pmatrix}$$ and its mid-point has coordinates $$M\\begin{pmatrix}5,2\\end{pmatrix}$$. Find the coordinates of the segment's other end-point $$B$$. Note: this exercise can be downloaded as a worksheet to practice with: 1. We find $$B\\begin{pmatrix}4,13\\end{pmatrix}$$. 2. We find $$Q\\begin{pmatrix}-5,-1\\end{pmatrix}$$. 3. We find $$T\\begin{pmatrix}6,-2\\end{pmatrix}$$. 4. We find $$Q\\begin{pmatrix}7,-6\\end{pmatrix}$$. 5. We find $$B\\begin{pmatrix}9,7\\end{pmatrix}$$.", "starting with two vectors: one vector $\\vec{r}_0$ drawn from the origin to a chosen point of the line (the so-called position vector of this point) and another vector (the so-called {\\em direction vector}) $\\vec{u}$ determining the direction of the line. Then there exists a real number $t$ such that the position vector $\\vec{r}$ of an arbitrary point of the line can be expressed as \\begin{align} \\vec{r} \\;=\\; \\vec{r}_0\\!+\\!t\\vec{u}. \\end{align} This is the {\\em vector-formed equation of the line}. Since every vector can be given by three coordinates of the end-point, we may write $$\\vec{r} \\;=\\; \\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix},\\quad \\vec{r}_0 \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix},\\quad \\vec{u} \\;=\\; \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!.$$ $$\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix} \\;=\\; \\begin{pmatrix} x_0\\\\y_0\\\\z_0 \\end{pmatrix}+t \\begin{pmatrix} a\\\\b\\\\c \\end{pmatrix}\\!,$$ and we obtain three scalar equations \\begin{align} \\begin{cases} x \\;=\\; x_0\\!+\\!ta\\\\ y \\;=\\; y_0\\!+\\!tb\\\\ z \\;=\\; z_0\\!+\\!tc. \\end{cases} \\end{align} These are the {\\em parametric equations of the line} (3), with the parameter $t$ taking all real values. The contents of (4) is often written with proportions: $$\\frac{x\\!-\\!x_0}{a} \\;=\\; \\frac{y\\!-\\!y_0}{b} \\;=\\; \\frac{z\\!-\\!z_0}{c}\\;\\;\\; (= t)$$ \\begin{thebibliography}{8} \\bibitem{LL}{\\sc L. Lindel\\\"of}: {\\em Analyyttisen geometrian oppikirja}.\\, Kolmas painos.\\, Suomalaisen Kirjallisuuden Seura, Helsinki (1924). \\end{thebibliography} %%%%% %%%%% \u001bnd{document}", "shown below) $$r·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$(3×1)+(2×[-5])+(7×8)=49$$ Therefore this means that the distanve of the vector d = 49 To complete the equation we need to write down the value of the whole equation in cartesian form $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$x-5y+8z=49$$ x - 5y + 8z - 49 = 0 #### Question 2 Write the equation of the plane through (2,1,3), given that the vector (4,5,6) is perpendicular $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 1\\\\\\ 3\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$(2×4)+(1×5)+(3×6)=31$$ Therefore the distance of the vector is d = 31 $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$4x+5y+6z=31$$ 4x + 5y + 6z - 31 = 0 #### Intersection of a line and a plane Find the point of intersection between the line The normal (perpendicular) to a plane is $$r=\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ with plane $$5x+y-z=1$$ First we need to replace r with the matrix (x,y,z) in order to find the values of x,y and z $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ We can change this equation into 3 different equations by seperating the matrices into: x = 2 + λ y = 3 + 2λ z = 4 - λ And we know that 5x+y-z=1 Replacing the values of x,y and z with these new values we get", "Given points A(3 , -4) , B(-2 , 1) $$\\therefore$$ $$\\overrightarrow {AB}$$ = $$\\begin {pmatrix} -5 \\\\ 5 \\end{pmatrix}$$ Ans. Magnitude of $$\\overrightarrow {AB}$$ = $$\\sqrt { (-5) ^{2} + 5^{2}}$$ = $$\\sqrt{25 + 25}$$ = 5$$\\sqrt{2}$$ units. Direction of $$\\overrightarrow {AB}$$ , tan $$\\theta$$ = $$\\frac{5}{-5}$$ = -1 = tan135o. Ans. Given points A(1 , 1) , B(2 , 4) $$\\therefore$$ Vector $$\\overrightarrow {AB}$$ = $$\\begin{pmatrix}1 \\\\ 3 \\end{pmatrix}$$ Magnitude of $$\\overrightarrow {AB}$$ = $$\\sqrt{1 + 3^{2}}$$ = $$\\sqrt{ 1+9}$$ = $$\\sqrt{10}$$ units. Ans. Direction of $$\\overrightarrow {AB}$$ is tan $$\\theta$$ = $$\\frac{3}{1}$$ = tan(tan-13) $$\\theta$$ = tan-13 Ans. Given points A( -1 , 3) , B(1 , -3) $$\\therefore$$ Vector $$\\overrightarrow {AB}$$ = $$\\begin{pmatrix} 2 \\\\ -6 \\end {pmatrix}$$ Ans. $$\\therefore$$ Magnitude of $$\\overrightarrow {AB}$$ = $$\\sqrt{2 ^{2} + 6 ^{2}}$$ = $$\\sqrt{4 + 36}$$ = $$\\sqrt{4 + 36}$$ = $$\\sqrt{40}$$ = 2$$\\sqrt{10}$$units. Ans. Direction of $$\\overrightarrow {AB}$$ , tan$$\\theta$$ = $$\\frac{-6}{2}$$ = -3 = tan(tan-1(-3) $$\\therefore$$ $$\\theta$$ =tan-1(-3) . Given points A(-2 , 0) , B(-3 , -2) $$\\therefore$$ Vector $$\\overrightarrow {AB}$$ = $$\\begin {pmatrix} - 1 \\\\ -2 \\end{pmatrix}$$ Ans. Magnitude of $$\\overrightarrow {AB}$$ = $$\\sqrt { -1^{2} + -2 ^{2}}$$ = $$\\sqrt{1 +4}$$ = $$\\sqrt{5}$$ Ans. Direction of $$\\overrightarrow {AB}$$ , tan$$\\theta$$ = $$\\frac{-2}{-1}$$ = tan(tan-12) $$\\theta$$ = (tan-12) Here , A(x1 y1)", "point, there is exactly one. I assume you want the equation for that line. Since the line passes through both the given point and the given line, a vector in the direction of this perpendicular line would be the vector $\\begin{pmatrix}3-(-1) \\\\ 1-(-2+t) \\\\ -2-(-1+t)\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 3-t \\\\ -1-t\\end{pmatrix}$. Let's find a vector in the direction of the given line by finding two points along the line (we can choose any two points, so let's say $t=0, t=1$). So, a vector in the direction of the given line is $\\begin{pmatrix}-1-(-1) \\\\ -2+1 - (-2+0) \\\\ -1+1 - (-1+0)\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Since we want these vectors to be perpendicular, we want the dot product to be zero. Hence: $(4\\cdot 0) + (1\\cdot (3-t)) + (1\\cdot (-1-t)) = 2-2t = 0$ implies $t=1$. Hence, a vector in the direction of the perpendicular line is $\\begin{pmatrix}4 \\\\ 3-1 \\\\ -1-1\\end{pmatrix} = \\begin{pmatrix}4 \\\\ 2 \\\\ -2\\end{pmatrix}$. So, the perpendicular line would have parametric equation $x=3+4t, y=1+2t, z=-2-2t$. 4. ## Re: equation of line through point orthogonal to line so the point (does this mean you are treating the point as a bounded vector) minus the line gives a direction vector along the perpendicular line then I take the given direction vector of the line which", "& 0 \\\\ 1 & -1 & 0 \\\\ 0 & 0 & 0\\end{pmatrix} \\ \\longrightarrow \\begin{pmatrix}-1 & -1 & 0 \\\\ 0 & -2 & 0 \\\\ 0 & 0 & 0\\end{pmatrix}\\end{equation*} We get the equations \\begin{align*}-v_1-v_2=&0 \\\\ -2v_2=&0\\end{align*} The set of fixed vectors is therefore \\begin{equation*}\\text{Fix}=\\left \\{v_3\\begin{pmatrix}0 \\\\ 0 \\\\ 1\\end{pmatrix}\\mid v_3\\in \\mathbb{R}\\right \\}\\end{equation*} Is this a fixpoint vector along the axis, and that the lengths and angles of vectors are preserved and so we have a rotation? 5. Originally Posted by mathmari Hey!! Which is the geometric interpretation of the following maps? $$v\\mapsto \\begin{pmatrix} 0&-1&0\\\\ 1&0&0\\\\ 0&0&-1\\end{pmatrix}v$$ and (1, 0, 0) is mapped to (0, 1, 0), (0, 1, 0) is mapped to (-1, 0, 0), and (0, 0, -1). Geometrically, a three dimensional figure is rotated 90 degrees in the xy-plane and the z-direction is reversed. Quote: $$v\\mapsto \\begin{pmatrix} 1& 0&0\\\\0&\\frac{1}{2} &-\\frac{\\sqrt{3}}{2}\\\\ 0&\\frac{\\sqrt{3}}{2}&\\frac{1}{2}\\end{pmatrix}v$$ $cos(-60)= \\frac{1}{2}$ and $sin(-60)= -\\frac{\\sqrt{3}}{2}$ so this is a rotation by -60 degrees (60 degrees counter-clockwise) in the yz-plane. 6. Originally Posted by mathmari For the kernel we have: \\begin{equation*}\\begin{pmatrix}0 & -1 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 1\\end{pmatrix} \\ \\overset{Z_1\\leftrightarrow Z_2}{\\longrightarrow } \\ \\begin{pmatrix}1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1\\end{pmatrix}\\end{equation*} Shouldn't the matrix", "The right matrix (prior to being inverted) will map the $x$ direction to the direction of the line of intersection, and it will map the $y$ direction to the direction perpendicular to that line within the $z=0$ plane. The $z$ axis is preserved, so this is a rotation around the $z$ axis. The inverse of that matrix will map the direction of the line of intersection to the $x$ direction, from where the left matrix will map it back to the line of intersection. You have to make sure that the second column of the right matrix has the correct sign, otherwise you'll end up describing a reflection (in the angle bisector plane) instead of a rotation. I did this at first, accidentially. To get the offset right, simply plug in a point on the line of intersection (i.e. one which satisfies $x+3y+1=0$ and $z=0$), and compute the difference between the transformed and the original point. You want to subtract that difference, so that the point remains fixed. In the end you obtain $$\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}\\mapsto \\frac1{110}\\begin{pmatrix} -\\sqrt{11} + 99 & -3 \\, \\sqrt{11} - 33 & 10 \\, \\sqrt{11} \\\\ -3 \\, \\sqrt{11} - 33 & -9 \\, \\sqrt{11} + 11 & 30 \\, \\sqrt{11} \\\\ -10 \\, \\sqrt{11} & -30 \\, \\sqrt{11} & -10 \\, \\sqrt{11} \\end{pmatrix}\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}- \\frac1{110}\\begin{pmatrix}", "to the other line The distance can now be calculated as described under distance point and line. First set up an auxiliary plane with $P$ as the support point and direction vector as the normal vector. $\\text{H: } (\\vec{x} - \\vec{a}) \\cdot \\vec{n}=0$ $\\text{H: } (\\vec{x} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}) \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$. $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can", "position vector of their point of intersection $A$ is then $\\begin{pmatrix}6\\\\-3\\\\-2\\end{pmatrix} + 3\\begin{pmatrix}-1\\\\2\\\\3\\end{pmatrix} = \\begin{pmatrix}3\\\\3\\\\7\\end{pmatrix}$ or equivalently $\\begin{pmatrix}-5\\\\15\\\\3\\end{pmatrix} + 4\\begin{pmatrix}2\\\\-3\\\\1\\end{pmatrix} = \\begin{pmatrix}3\\\\3\\\\7\\end{pmatrix}$ For part (b) we find the dot product of the direction vectors of the two lines. We get $(-1, 2, 3) \\cdot (2, -3, 1) = -5$ but also $(-1, 2, 3) \\cdot (2, -3, 1) = |(-1, 2, 3)||(2, -3, 1)|\\cos\\theta = 14\\cos\\theta$ Equating the two gives $\\cos\\theta = \\frac{-5}{14}$ so $\\theta = \\arccos(\\frac{-5}{14}) = 110.9248324^{\\circ}$, and therefore the acute angle must be $180^{\\circ} - \\theta = 69.1^{\\circ}$ (to nearest $0.1^{\\circ}$). For part (c) we observe that if $B$ lies on $l_1$, there is a $\\lambda$ such that $\\begin{pmatrix}5\\\\-1\\\\1\\end{pmatrix} = \\begin{pmatrix}6-\\lambda\\\\-3+2\\lambda\\\\-2+3\\lambda\\end{pmatrix}$ By inspection, the required value is $\\lambda = 1$. Finally, for part (d) it is helpful to draw the following sketch of the situation: We know that point $A$ has coordinates $(3, 3, 7)$ and point $B$ has coordinates $(5, -1, 1)$, and therefore the length of the line from $B$ to $A$ is $|\\overrightarrow{BA}| = |(3, 3, 7) - (5, -1, 1)| = |(-2, 4, 6)| = \\sqrt{(-2)^2 + 4^2 + 6^2} = \\sqrt{56}$ The shortest distance from $B$ to the line $l_2$ is the length of the perpendicular from $B$ which is shown in the sketch as intersecting line $l_2$ at a", "## Dividing a Line Segment in a Given Ratio Suppose the line between $A(2,3)$ and $B((9,20)$ is to be divided in the ratio $2:3$ . Ket the point that divides the line segment in this ratio be $P$ . The vector from $A$ to $B$ is $\\vec{AB}=\\vec{OB}=\\vec{OA}= \\begin{pmatrix}9\\\\20\\end{pmatrix}-\\begin{pmatrix}2\\end{pmatrix}=\\begin{pmatrix}7\\\\17\\end{pmatrix}$ . The vector $\\vec{AP}=\\frac{2}{5} \\vec{AB}=\\frac{2}{5}\\begin{pmatrix}7\\\\17\\end{pmatrix}=\\begin{pmatrix}14/5\\\\34/5\\end{pmatrix}$ . Then $\\vec{OP}=\\vec{OA}+ \\vec{AP}=\\begin{pmatrix}2\\\\3\\end{pmatrix}+\\begin{pmatrix}14/5\\\\34/5\\end{pmatrix}=\\begin{pmatrix}24/5\\\\49/5\\end{pmatrix}$ . The point $P$ is $(24/5,49/5)$ .", "to vector form $$y = \\frac{x}{3} + 2$$ We want this in the form $$\\underline{r} =$$ 1st point + direction Using the cartesian equation we can find the direction as we know that the gradient of a line is: $$m = \\frac{\\text{Change in y}}{\\text{Change in x}}$$ We can see that for every time y is increased by 1 x is increases by 1/3 as x is divided by 3 giving us: $$m = \\frac{1}{(\\frac{1}{3})}=\\frac{3}{1}$$ This means we can write the direction as: $$\\underline{r} =$$ 1st point + $$λ\\begin{pmatrix}3\\\\\\ 1\\end{pmatrix}$$ Now we juust need to find the 1st point to do this we simply set the value of λ=0 as this is where the line starts From the equation $$y = \\frac{x}{3} + 2$$ We can see that when x=0 y=2 so this is where our line starts Therefore the vector equation can be written as: $$\\underline{r} = \\begin{pmatrix}0\\\\\\ 2\\end{pmatrix} + λ\\begin{pmatrix}3\\\\\\ 1\\end{pmatrix}$$ #### Intersection of two lines What about if we need to find the exact point that two lines intersect knowing only the vector equations of these two lines? $$\\underline{r_1} = \\begin{pmatrix}2\\\\\\ 3\\end{pmatrix} + λ\\begin{pmatrix}1\\\\\\ 2\\end{pmatrix}$$ $$\\underline{r_2} = \\begin{pmatrix}6\\\\\\ 1\\end{pmatrix} + μ\\begin{pmatrix}1\\\\\\ -3\\end{pmatrix}$$ For the two lines to intercept there must be a point when the value of the two equations are the same this means we", "# Find distance from point to line I am asked to find the distance between a point ( 5,1,1 ) and a line $\\displaystyle \\left\\{\\begin{matrix} x + y + z = 0\\\\ x - 2y + z = 0 \\end{matrix}\\right.$ What ive done so far is to simplify by gauss elimination and I get to: $\\displaystyle \\begin{bmatrix} 1 &1 &1 &0 \\\\ 0 &1 &0 &0 \\end{bmatrix}$ In turn I put this back in the form of an equation $\\left\\{\\begin{matrix} x + y + z = 0\\\\ y = 0 \\end{matrix}\\right.$ What I need to find the distance between the point and line is any point on the line and a directional vector, right? Am i right in assuming that we have $\\begin{pmatrix} x,y,z \\end{pmatrix} = \\begin{pmatrix} t,0,-t \\end{pmatrix}$ So a point on this line could be (1,0,-1) or (12,0,-12)? If this is correct, how would I go on about finding the directional vector so I can put this all on the form of $\\begin{Vmatrix} \\overrightarrow{PQ} \\times \\overrightarrow{v} \\end{Vmatrix} \\div \\begin{Vmatrix} \\overrightarrow{v} \\end{Vmatrix}$ So to sum up, Q is given, the line is given in a system of equations, how do I extract a point P on the line and the vector v? For $t=1$, a possible directional vector for your line is $(1,0,-1)$. The projection of $(5,1,1)$", "# Compute the orthogonal projection $w$ of $u$ on a line $L$ A vector $u$ and a line $L$ in $\\mathbb{R}^2$ are given. Compute the orthogonal projection $w$ of $u$ on $L$, and use it to compute the distance $d$ from the endpoint of $u$ to $L$ $u = \\begin{pmatrix} 5 \\\\ 0 \\end{pmatrix}$ $y = 0$ I know to that $w = \\frac{u \\cdot v}{||v||^2}v$ and that $d = u - w$ The answer in the book says the answer is $w = \\begin{pmatrix}5 \\\\0 \\end{pmatrix}$ and $d = 0$ How do I find $v$ to solve the problem? • What is your line $L$? Is it the line given by the equation $y = 0$? – Eli Rose Dec 14 '15 at 4:16 • Yes, the book only gives $u$ and the equation of Line $L$ – user273323 Dec 14 '15 at 4:25 The vector $v$ can be any vector in the direction of the line $L$. For a line $ax + by = c$, the vector $\\begin{pmatrix}b\\\\-a\\end{pmatrix}$, or any non-zero multiple of it, lies in the direction of the line. So for $L: y=0$, alias $0x+1y=0$, the vector $v = \\begin{pmatrix}1\\\\0\\end{pmatrix}$ will do. Put this into your equation gives $w=\\begin{pmatrix}5\\\\0\\end{pmatrix}$ as expected. Or, more simply, you can observe that $u$ is a vector parallel to", "# How to find the position vector for the point of intersection of a line and the perpendicular line through a point C How could I find the exact coordinates of the point N for example which is the point of intersection of the line $$L=\\begin{pmatrix} 0 \\\\ -1 \\\\ 2 \\end{pmatrix} +t\\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}$$ and a perpendicular line that passes through a point C(1,2,3) not on the line. What I have done so far is find the value of point N in terms of $$t$$ to find the vector $$\\overrightarrow{CN}$$: $$\\longrightarrow N\\begin{pmatrix} -3t \\\\ -2t-1 \\\\ -3t+2 \\end{pmatrix}$$ $$\\longrightarrow \\overrightarrow{CN}=\\begin{pmatrix} -3t \\\\ -2t-1 \\\\ -3t+2 \\end{pmatrix}-\\begin{pmatrix} 1 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ $$=\\begin{pmatrix} -3t-1 \\\\ -2t-3 \\\\ -3t-1 \\end{pmatrix}$$ After this I made the scalar product eual zero: $$\\overrightarrow{CN} \\cdot \\begin{pmatrix} -3 \\\\ -2 \\\\ -3 \\end{pmatrix}=0$$ With this I find that, $$t=\\frac{-6}{11}$$ But my textbook says that $$t=\\frac{5}{11}$$ Can anyone confirm which answer for $$t$$ is right and if it is $$t=\\frac{5}{11}$$ then what did I do wrong? • For me, your answer it correct. Jul 30 '20 at 12:32 • WA agrees. Maybe there's the line $L$ points swapped in the solution? Jul 30 '20 at 12:42 • Thank you for the confirmation Jul 30 '20 at 12:42 Your derivation seems", "+0 # Help! 0 49 1 +11 Consider the line parametrized by \\begin{align*} x&= 2t + 1,\\\\ y& = -3t +2. \\end{align*} Find a vector $\\begin{pmatrix}a\\\\b\\end{pmatrix}$ that's parallel to this line and satisfies $a+b = 3$. Jul 27, 2020 edited by FearlessIsland3 Jul 27, 2020", "\\frac{-v_3}{v_1} & 0 & 1\\\\ \\end{pmatrix}$$ and thus a vector $\\vec x$ lies in the cylinder iff $\\|P(\\vec x-\\vec p)\\|\\leq r$. Note that the projection of the box is a convex set and is equal to the convex hull of the images of its vertices, and that either the image of center of the cylinder is contained in the image of the box or the boundary of the image of the box intersects the image of the cylinder. In the first case, we have that the line $\\vec p + t\\vec v$ intersects one of the three planes $x=x_{min},y=y_{min},z=z_{min}$ within the boundary of the box. The points of intersection are given by $$\\begin{pmatrix} x_{min}\\\\ p_2+\\frac{x_{min}-p_1}{v_1}v_2\\\\ p_3+\\frac{x_{min}-p_1}{v_1}v_3\\\\ \\end{pmatrix}, \\begin{pmatrix} p_1+\\frac{y_{min}-p_2}{v_2}v_1\\\\ y_{min}\\\\ p_3+\\frac{y_{min}-p_2}{v_2}v_3\\\\ \\end{pmatrix}, \\begin{pmatrix} p_1+\\frac{z_{min}-p_3}{v_3}v_1\\\\ p_2+\\frac{z_{min}-p_3}{v_3}v_2\\\\ z_{min}\\\\ \\end{pmatrix}$$ and are nonexistant if $v_1,v_2$ or $v_3$ respectively are $0$. It is easy to check whether these are contained in the box. If any of them are, the intersection is nonempty. In the second case, let $\\vec x_1+t\\vec y_1,\\ldots,\\vec x_{12}+t\\vec y_{12}$ be the lines which form the edges of the box (and are exactly the edges when $0\\leq t\\leq 1$), which are easy to determine from the vertices. We want to know whether there is a solution to $\\|P(\\vec x_i+t\\vec y_i-\\vec p)\\|\\leq r$ for any $i$ with $0\\leq t\\leq 1$. Observe", "16:58 • The point $(2,2)$ does not go to $(2,0)$. Since rotation preserves distance, $(2,2)$ gets sent to the point north of $(1,1)$ at a distance of $\\sqrt{2}$. – John Douma Jul 5 at 17:50 First, translate the plane so that $$(1,1)$$ goes to the origin. This gives $$\\begin{pmatrix} x-1 \\\\ y-1 \\end{pmatrix}$$ Then rotate by $$\\frac{\\pi}{4}$$ to get $$\\begin{pmatrix}\\frac{1}{\\sqrt2}\\frac{-1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\begin{pmatrix} x-1 \\\\ y-1 \\end{pmatrix}=\\begin{pmatrix} \\frac{x-y}{\\sqrt{2}} \\\\ \\frac{x+y-2}{\\sqrt{2}}\\end{pmatrix}$$ Finally, do the inverse translation, i.e. add $$1$$ to each coordinate to get $$\\begin{pmatrix} \\frac{x-y+\\sqrt{2}}{\\sqrt{2}} \\\\ \\frac{x+y+\\sqrt{2}-2}{\\sqrt{2}}\\end{pmatrix}$$" ]

[ "x=-5+\\frac{3}{7}z$ , and thus the intersection line is given by $\\left\\{\\begin{pmatrix}-5+\\frac{3}{7}t\\\\{}\\\\-1-\\frac{2}{7}t\\\\{}\\\\t\\end{pmatrix}\\,,\\,t\\in\\mathbb{R }\\right\\}$ . The above is already a parametrization, but your teacher may want you to write it as $u+tv\\,,\\,u,v$ vectors on the line, $t=$ the real parameter. Tonio", "16:58 • The point $(2,2)$ does not go to $(2,0)$. Since rotation preserves distance, $(2,2)$ gets sent to the point north of $(1,1)$ at a distance of $\\sqrt{2}$. – John Douma Jul 5 at 17:50 First, translate the plane so that $$(1,1)$$ goes to the origin. This gives $$\\begin{pmatrix} x-1 \\\\ y-1 \\end{pmatrix}$$ Then rotate by $$\\frac{\\pi}{4}$$ to get $$\\begin{pmatrix}\\frac{1}{\\sqrt2}\\frac{-1}{\\sqrt{2}} \\\\ \\frac{1}{\\sqrt{2}} \\frac{1}{\\sqrt{2}}\\end{pmatrix}\\begin{pmatrix} x-1 \\\\ y-1 \\end{pmatrix}=\\begin{pmatrix} \\frac{x-y}{\\sqrt{2}} \\\\ \\frac{x+y-2}{\\sqrt{2}}\\end{pmatrix}$$ Finally, do the inverse translation, i.e. add $$1$$ to each coordinate to get $$\\begin{pmatrix} \\frac{x-y+\\sqrt{2}}{\\sqrt{2}} \\\\ \\frac{x+y+\\sqrt{2}-2}{\\sqrt{2}}\\end{pmatrix}$$", "4t=6+2(-1)}$ substituted u back into equation 1 ${\\displaystyle t=1}$ solved for t Now to check if it works in the final equation, substitute the values for u and t into the equation that you didn't use. ${\\displaystyle 2t=-1-3u}$ ${\\displaystyle 2(1)=-1-3(-1)}$ ${\\displaystyle 2=2}$ If you get a value expression (a constant equals itself) then this means that they intersect, else they don't. This is known as a skew. ### The closest point to another point on a line equation and the distance between them So let's say there's a vector equation ${\\displaystyle {\\textbf {r}}={\\begin{pmatrix}2\\\\2\\\\2\\end{pmatrix}}+t{\\begin{pmatrix}1\\\\-2\\\\3\\end{pmatrix}}}$ and there's a point P with a position vector: ${\\displaystyle {\\overrightarrow {OP}}={\\begin{pmatrix}1\\\\1\\\\2\\end{pmatrix}}}$ . We want to find both the coordinate of the point on the line that is closest to P and the distance between that point and P. We'll refer to this point on the line as X. First we need to use our dot product rule, because the line from P to X will be perpendicular to the line itself. So to do this we'll need to find the direction of the line and the vector from P to X. The direction of the line is just the direction vector of the line (which we'll call A): ${\\displaystyle {\\textbf {A}}={\\begin{pmatrix}1\\\\-2\\\\3\\end{pmatrix}}}$ As for the direction of P to X, we'll need to find X", "\\frac{-v_3}{v_1} & 0 & 1\\\\ \\end{pmatrix}$$ and thus a vector $\\vec x$ lies in the cylinder iff $\\|P(\\vec x-\\vec p)\\|\\leq r$. Note that the projection of the box is a convex set and is equal to the convex hull of the images of its vertices, and that either the image of center of the cylinder is contained in the image of the box or the boundary of the image of the box intersects the image of the cylinder. In the first case, we have that the line $\\vec p + t\\vec v$ intersects one of the three planes $x=x_{min},y=y_{min},z=z_{min}$ within the boundary of the box. The points of intersection are given by $$\\begin{pmatrix} x_{min}\\\\ p_2+\\frac{x_{min}-p_1}{v_1}v_2\\\\ p_3+\\frac{x_{min}-p_1}{v_1}v_3\\\\ \\end{pmatrix}, \\begin{pmatrix} p_1+\\frac{y_{min}-p_2}{v_2}v_1\\\\ y_{min}\\\\ p_3+\\frac{y_{min}-p_2}{v_2}v_3\\\\ \\end{pmatrix}, \\begin{pmatrix} p_1+\\frac{z_{min}-p_3}{v_3}v_1\\\\ p_2+\\frac{z_{min}-p_3}{v_3}v_2\\\\ z_{min}\\\\ \\end{pmatrix}$$ and are nonexistant if $v_1,v_2$ or $v_3$ respectively are $0$. It is easy to check whether these are contained in the box. If any of them are, the intersection is nonempty. In the second case, let $\\vec x_1+t\\vec y_1,\\ldots,\\vec x_{12}+t\\vec y_{12}$ be the lines which form the edges of the box (and are exactly the edges when $0\\leq t\\leq 1$), which are easy to determine from the vertices. We want to know whether there is a solution to $\\|P(\\vec x_i+t\\vec y_i-\\vec p)\\|\\leq r$ for any $i$ with $0\\leq t\\leq 1$. Observe", "shown below) $$r·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$(3×1)+(2×[-5])+(7×8)=49$$ Therefore this means that the distanve of the vector d = 49 To complete the equation we need to write down the value of the whole equation in cartesian form $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$x-5y+8z=49$$ x - 5y + 8z - 49 = 0 #### Question 2 Write the equation of the plane through (2,1,3), given that the vector (4,5,6) is perpendicular $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 1\\\\\\ 3\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$(2×4)+(1×5)+(3×6)=31$$ Therefore the distance of the vector is d = 31 $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$4x+5y+6z=31$$ 4x + 5y + 6z - 31 = 0 #### Intersection of a line and a plane Find the point of intersection between the line The normal (perpendicular) to a plane is $$r=\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ with plane $$5x+y-z=1$$ First we need to replace r with the matrix (x,y,z) in order to find the values of x,y and z $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ We can change this equation into 3 different equations by seperating the matrices into: x = 2 + λ y = 3 + 2λ z = 4 - λ And we know that 5x+y-z=1 Replacing the values of x,y and z with these new values we get", "Math Relative position of lines Determining point of intersection # Intersecting lines Two lines can intersect and then have a point of intersection. ! ### Requirements 1. Direction vectors are not collinear 2. When equating the linear equations, we receive a distinct result ## Calculating point of intersection To calculate the point of intersection, we follow the scheme above. The result from step 2 for $r$ or $s$ is then inserted into the corresponding linear equation. ### Example $\\text{g: } \\vec{x} = \\begin{pmatrix} 10 \\\\ 9 \\\\ 7 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}$ $\\text{h: } \\vec{x} = \\begin{pmatrix} 4 \\\\ 4 \\\\ 6 \\end{pmatrix} + s \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ 1. #### Direction vectors not collinear? First, we examine if the direction vectors are multiples of each other (=collinear). $\\vec{a}=t\\cdot\\vec{b}$ $\\begin{pmatrix} 9 \\\\ 8 \\\\ 7 \\end{pmatrix}=t\\cdot\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}$ Calculate $t$ for every row 1. $9=t\\cdot1$ 2. $8=t\\cdot1$ 3. $7=t\\cdot2$ 1. $t=9$ 2. $t=8$ 3. $t=3.5$ $t$ does not have the same value everywhere. Thus the direction vectors are not collinear. Therefore the lines are skew or have a point of intersection. 2. #### Equate $g$ and $h$ Now equate the linear equations and calculate $r$ and $s$. $\\begin{pmatrix} 10 \\\\ 9 \\\\ 7", "The right matrix (prior to being inverted) will map the $x$ direction to the direction of the line of intersection, and it will map the $y$ direction to the direction perpendicular to that line within the $z=0$ plane. The $z$ axis is preserved, so this is a rotation around the $z$ axis. The inverse of that matrix will map the direction of the line of intersection to the $x$ direction, from where the left matrix will map it back to the line of intersection. You have to make sure that the second column of the right matrix has the correct sign, otherwise you'll end up describing a reflection (in the angle bisector plane) instead of a rotation. I did this at first, accidentially. To get the offset right, simply plug in a point on the line of intersection (i.e. one which satisfies $x+3y+1=0$ and $z=0$), and compute the difference between the transformed and the original point. You want to subtract that difference, so that the point remains fixed. In the end you obtain $$\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}\\mapsto \\frac1{110}\\begin{pmatrix} -\\sqrt{11} + 99 & -3 \\, \\sqrt{11} - 33 & 10 \\, \\sqrt{11} \\\\ -3 \\, \\sqrt{11} - 33 & -9 \\, \\sqrt{11} + 11 & 30 \\, \\sqrt{11} \\\\ -10 \\, \\sqrt{11} & -30 \\, \\sqrt{11} & -10 \\, \\sqrt{11} \\end{pmatrix}\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}- \\frac1{110}\\begin{pmatrix}", "line segment $$PQ$$ has an end-point $$P\\begin{pmatrix}-3,4\\end{pmatrix}$$ and its mid-point has coordinates $$M\\begin{pmatrix}2,-1\\end{pmatrix}$$. Find the coordinates of the segment's other end-point $$Q$$. 5. A line segment $$AB$$ has an end-point $$A\\begin{pmatrix}1,-3\\end{pmatrix}$$ and its mid-point has coordinates $$M\\begin{pmatrix}5,2\\end{pmatrix}$$. Find the coordinates of the segment's other end-point $$B$$. Note: this exercise can be downloaded as a worksheet to practice with: 1. We find $$B\\begin{pmatrix}4,13\\end{pmatrix}$$. 2. We find $$Q\\begin{pmatrix}-5,-1\\end{pmatrix}$$. 3. We find $$T\\begin{pmatrix}6,-2\\end{pmatrix}$$. 4. We find $$Q\\begin{pmatrix}7,-6\\end{pmatrix}$$. 5. We find $$B\\begin{pmatrix}9,7\\end{pmatrix}$$.", "diagonal, length 3, of the enclosed square. Now that we have the position vectors of vertices A and B we could find the length of the line AB using Pythagoras. However, you might notice that the area of the square will be given by $AB^2$ and we can write this down directly, $AB^2 = 2 \\times \\left(\\frac{3}{2}\\right)^2 = \\frac{9}{2}.$ You could also have noticed that the right-angled triangle joining points A and B is isosceles. This tells us that the square could be divided up into four of these triangles. We can now show that the area of the square is again equal to $4 \\times \\frac{1}{2}\\left(\\frac{3}{2}\\right)^2 = \\frac{9}{2}.$ What can we now say about the positions of the other vertices and the diagonals of the square? Find a vector equation for each of the two diagonals. In order to write down a vector equation of a line we need to know the position of a point on that line and a direction vector. For the vertical diagonal of the square we know that point B, with position vector $\\begin{pmatrix}2\\\\4\\end{pmatrix}$ lies on the line and that the direction vector is $\\begin{pmatrix}0\\\\1\\end{pmatrix}$. A vector equation of the vertical diagonal of the square is therefore $\\mathbf{r}=\\begin{pmatrix}2\\\\4\\end{pmatrix} + p\\begin{pmatrix}0\\\\1\\end{pmatrix}.$ We could also have written $\\mathbf{r}=\\begin{pmatrix}2\\\\1\\end{pmatrix} + p\\begin{pmatrix}0\\\\1\\end{pmatrix},$ or any vector equation", "2z + 2 = 0.$$ Notice that, these are two equations in three variables, so you have a free variable say $z=t$, then we have $$x + 2y = 1-t \\\\ 2x + 3y = 2t-2.$$ Solving the last system gives $$\\left\\{ x=-7+7\\,t,y=4-4\\,t \\right\\} .$$ Then the parametrized equation of the line is given by $$(x,y,z)= (-7+7t, 4-4t,t)=(-7,4,0)+(7,-4,1)t .$$ • Can you explain how you got the parameterized result? Sep 23, 2021 at 4:46 From the coefficients of x, y and z of the general form equations, the first plane has normal vector $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}$$ and the second has normal vector $$\\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}$$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $$v_1,v_2$$ is $$v_1\\times v_2$$, so the direction vector of the line of intersection is $$\\begin{pmatrix}1\\\\2\\\\1\\end{pmatrix}\\times \\begin{pmatrix}2\\\\3\\\\-2\\end{pmatrix}=\\begin{pmatrix}-7\\\\4\\\\-1\\end{pmatrix}$$Next, we need to find a particular point on the line. Here, since a line that isn't parallel to any coordinate plane passes through all three, you can check if it is parallel to one by using the directional vector. Since here, the line passes through all three planes, we can try $$y=0$$(since it passes through the $$x-z$$ plane) and solve the resulting system of linear equations:\\begin{align}x+z-1&=&0\\\\2x-2z+2&=&0\\end{align} giving $$x=0, z=1$$, thus the line of intersection is $$\\lbrace{\\begin{pmatrix}-7t\\\\4t\\\\1-t\\end{pmatrix}:t\\in \\Bbb", "any point in this space, we need two vectors. For example, to get to the point $\\begin{pmatrix}7\\\\-4\\end{pmatrix}$ we need to go along the first vector $7$ times, and along the second vector, backwards 4 times. Having only one vector in our set we can only move along the direction of that vector. Having any more would again be redundant, since these two are enough. $$\\mathbb{R}^3 = \\text{span}\\left\\{\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix},\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix},\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}\\right\\}$$ This is usually called $3-$space. We can think of this space as actually being the space we live in. We have up and down, left and right, forward and backward. So three vectors will be needed to describe coordinates here. To get to any point in this space, we need three vectors. To get to the point $\\begin{pmatrix}1\\\\-2\\\\1\\end{pmatrix}$ we need to go along the direction of the first vector $1$ time, and along the second vector backwards 2 times, and along the third vector once. Having only one or two vectors in our set we can only move along the plane spanned by them. For example, I couldn't get to $\\begin{pmatrix}0\\\\0\\\\2\\end{pmatrix}$ if the third vector was deleted from our set. The bases I gave above are called the Standard Bases for the respective vector spaces. Most you encounter will have a defined 'standard' basis. It's just the simplest to use since,", "point $$\\begin{pmatrix}a,b \\end{pmatrix}$$ by making $$y$$ the subject in: $y - b = m \\begin{pmatrix}x - a \\end{pmatrix}$ Since the point we're working with is $$\\begin{pmatrix}3,5 \\end{pmatrix}$$ and the gradient, we found in step 1, is $$m = 6$$, we this becomes: $y - 5 = 6 \\begin{pmatrix}x - 3 \\end{pmatrix}$ Distributing the $$6$$ on the right hand side: $y - 5 = 6x - 18$ Adding $$5$$ to each side: $y = 6x - 18 +5$ Finally we find the equation of the tangent to the curve at the point $$\\begin{pmatrix}3,5 \\end{pmatrix}$$: $y = 6x - 13$ ## Exercise 1 1. Given the function defined by: $f(x) = 2x^2 - 2$ 1. Find $$f'(x)$$. 2. Calculate $$f'(2)$$. 3. Find the equation of the tangent to this function's graph $$y=2x^2-2$$ at the point along its length with coordinates $$\\begin{pmatrix}2,6 \\end{pmatrix}$$ 2. Consider the function defined by: $f(x) = 5x+\\frac{3}{x}$ 1. Calculate the $$y$$-coordinate of the curve $$y=f(x)$$ when $$x = 1$$. State the coordinates of the point on the curve with $$x$$-coordinate $$x = 1$$. 2. Find an expression for $$f'(x)$$. 3. Find the equation of the tangent to the curve $$y=f(x)$$ at the point along its length with $$x$$-coordinate $$x = 1$$. 1. Given the function defined by: $f(x) = 2x^2 - 2$ 1. For $$f'(x)$$, we", "to vector form $$y = \\frac{x}{3} + 2$$ We want this in the form $$\\underline{r} =$$ 1st point + direction Using the cartesian equation we can find the direction as we know that the gradient of a line is: $$m = \\frac{\\text{Change in y}}{\\text{Change in x}}$$ We can see that for every time y is increased by 1 x is increases by 1/3 as x is divided by 3 giving us: $$m = \\frac{1}{(\\frac{1}{3})}=\\frac{3}{1}$$ This means we can write the direction as: $$\\underline{r} =$$ 1st point + $$λ\\begin{pmatrix}3\\\\\\ 1\\end{pmatrix}$$ Now we juust need to find the 1st point to do this we simply set the value of λ=0 as this is where the line starts From the equation $$y = \\frac{x}{3} + 2$$ We can see that when x=0 y=2 so this is where our line starts Therefore the vector equation can be written as: $$\\underline{r} = \\begin{pmatrix}0\\\\\\ 2\\end{pmatrix} + λ\\begin{pmatrix}3\\\\\\ 1\\end{pmatrix}$$ #### Intersection of two lines What about if we need to find the exact point that two lines intersect knowing only the vector equations of these two lines? $$\\underline{r_1} = \\begin{pmatrix}2\\\\\\ 3\\end{pmatrix} + λ\\begin{pmatrix}1\\\\\\ 2\\end{pmatrix}$$ $$\\underline{r_2} = \\begin{pmatrix}6\\\\\\ 1\\end{pmatrix} + μ\\begin{pmatrix}1\\\\\\ -3\\end{pmatrix}$$ For the two lines to intercept there must be a point when the value of the two equations are the same this means we", "line, then the parametric equations are given by ${x = x_1 + r \\ cos (\\theta) \\ and \\ y = y_1 + r \\ sin (\\theta),}$ where ${r}$ is the distance of ${(x,y)}$ from ${(x_1, y_1)}$ • ### General Equation of 1st Order in 2 Variables The general equation of 1st order in 2 variables is of the form ${ax + by + c = 0}$ It represents a straight line. The necessary condition is both ${a}$ and ${b}$ cannot be simultaneously zero. • ### Angle Between Any 2 Lines Let ${\\theta}$ be the angle between 2 lines having slopes ${m_1}$ and ${m_2}$. Then, ${tan (\\theta) = \\begin {vmatrix} \\frac {m_1 - m_2}{1+ m_1 m_2} \\end {vmatrix}}$ • ### Point of Intersection of 2 Lines We know that 2 non-parallel lines intersect each other at only 1 point. Let the equations of lines be ${a_1 x + b_1 y = c_1 \\ and \\ a_2 x + b_2 y = c_2}$ The coordinates of point of intersection are ${\\begin {pmatrix}\\frac {\\begin {vmatrix} c_1 & c_2 \\\\ b_1 & b_2 \\end {vmatrix}}{\\begin {vmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end {vmatrix}} , \\frac {\\begin {vmatrix} a_1 & a_2 \\\\ c_1 & c_2 \\end {vmatrix}}{\\begin {vmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end {vmatrix}} \\end {pmatrix}}$ The", "repeat what we did for the last quation setting the x and y values equal to eachother as this will be the intersection point Equating the y values gives us (Equation 1): 6-2λ=1+μ Equating the x values gives us (Equation 2): 1+λ=1+2μ 2+2λ=2+4μ Equation 1 + Equation 2 8=3+5μ 5=5μ μ=1 Now we can find the position vector of the intersection point by replacing μ=-2 in the original equation as we know this is when the two lines meet $$\\underline{r_2} = \\begin{pmatrix}1\\\\\\ 1\\end{pmatrix} + 1\\begin{pmatrix}2\\\\\\ 1\\end{pmatrix} = \\begin{pmatrix}3\\\\\\ 2\\end{pmatrix}= 3\\underline{i}+2\\underline{j}$$ #### Angles between two vectors The angle between two vectors can be found using this equation: $$cosθ = \\frac{\\underline{a} · \\underline{b}}{|a||b|}=\\frac{\\text{Scalar Product}}{\\text{Multiply Vector Magnitudes}}$$ Scalar Product: Scalar Product of $$\\begin{pmatrix}4\\\\\\ 3\\end{pmatrix} \\text{ and } \\begin{pmatrix}3\\\\\\ 2\\end{pmatrix} = (x_1×x_2)+(y_1×y_2) = (4×3)+(3×2) = 24+6 = 30$$ What happens when the angle on the two vecors is 90°? Using the Equation to find the angle we can see $$cos90° = \\frac{\\underline{a} · \\underline{b}}{|a||b|}=\\frac{\\text{Scalar Product}}{\\text{Multiply Vector Magnitudes}}$$ However we know that cos90°=0 this means that the entire equation is equal to 0: $$0 = \\frac{\\underline{a} · \\underline{b}}{|a||b|}=\\frac{\\text{Scalar Product}}{\\text{Multiply Vector Magnitudes}}$$ The only way this can be true is if the scalar product also equals 0 This means we can find out easily if two vectors are perpendicular as there scalar product will", "& 0 \\\\ 1 & -1 & 0 \\\\ 0 & 0 & 0\\end{pmatrix} \\ \\longrightarrow \\begin{pmatrix}-1 & -1 & 0 \\\\ 0 & -2 & 0 \\\\ 0 & 0 & 0\\end{pmatrix}\\end{equation*} We get the equations \\begin{align*}-v_1-v_2=&0 \\\\ -2v_2=&0\\end{align*} The set of fixed vectors is therefore \\begin{equation*}\\text{Fix}=\\left \\{v_3\\begin{pmatrix}0 \\\\ 0 \\\\ 1\\end{pmatrix}\\mid v_3\\in \\mathbb{R}\\right \\}\\end{equation*} Is this a fixpoint vector along the axis, and that the lengths and angles of vectors are preserved and so we have a rotation? 5. Originally Posted by mathmari Hey!! Which is the geometric interpretation of the following maps? $$v\\mapsto \\begin{pmatrix} 0&-1&0\\\\ 1&0&0\\\\ 0&0&-1\\end{pmatrix}v$$ and (1, 0, 0) is mapped to (0, 1, 0), (0, 1, 0) is mapped to (-1, 0, 0), and (0, 0, -1). Geometrically, a three dimensional figure is rotated 90 degrees in the xy-plane and the z-direction is reversed. Quote: $$v\\mapsto \\begin{pmatrix} 1& 0&0\\\\0&\\frac{1}{2} &-\\frac{\\sqrt{3}}{2}\\\\ 0&\\frac{\\sqrt{3}}{2}&\\frac{1}{2}\\end{pmatrix}v$$ $cos(-60)= \\frac{1}{2}$ and $sin(-60)= -\\frac{\\sqrt{3}}{2}$ so this is a rotation by -60 degrees (60 degrees counter-clockwise) in the yz-plane. 6. Originally Posted by mathmari For the kernel we have: \\begin{equation*}\\begin{pmatrix}0 & -1 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 1\\end{pmatrix} \\ \\overset{Z_1\\leftrightarrow Z_2}{\\longrightarrow } \\ \\begin{pmatrix}1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 1\\end{pmatrix}\\end{equation*} Shouldn't the matrix", "$\\,\\dim(W_1\\cap W_2)=1\\,$ (why?) , we get that the above condition completely determines the elements of the intersection, and thus $$W_1\\cap W_2=\\left\\{\\,\\begin{pmatrix}a+2b\\\\2a+b\\\\3a+b\\end{pmatrix}\\;;\\;2a+b=0\\,\\right\\}=\\left\\{\\,\\begin{pmatrix}-3a\\\\0\\\\a\\end{pmatrix}\\;;\\;a\\in\\Bbb R\\,\\right\\}=\\operatorname{Span}\\left\\{\\,\\begin{pmatrix}-3\\\\0\\\\1\\end{pmatrix}\\,\\right\\}$$ - The two sub spaces are not the same, because $W_2$ has no extent in the second axis while $W_1$ does. The intersection is then a line in the $xz$ plane, which $W_2$ spans. If you can find a vector in $W_1$ in that plane that is your basis -", "+0 # Help! 0 49 1 +11 Consider the line parametrized by \\begin{align*} x&= 2t + 1,\\\\ y& = -3t +2. \\end{align*} Find a vector $\\begin{pmatrix}a\\\\b\\end{pmatrix}$ that's parallel to this line and satisfies $a+b = 3$. Jul 27, 2020 edited by FearlessIsland3 Jul 27, 2020", "with the parameter $t$ has its whole body in the line— it is a direction vector for the line. Note that points on the line to the left of $x=1$ are described using negative values of $t$. In $\\mathbb{R}^3$, the line through $(1,2,1)$ and $(2,3,2)$ is the set of (endpoints of) vectors of this form and lines in even higher-dimensional spaces work in the same way. If a line uses one parameter, so that there is freedom to move back and forth in one dimension, then a plane must involve two. For example, the plane through the points $(1,0,5)$, $(2,1,-3)$, and $(-2,4,0.5)$ consists of (endpoints of) the vectors in $\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix} +t\\cdot\\begin{pmatrix} 1 \\\\ 1 \\\\ -8 \\end{pmatrix} +s\\cdot\\begin{pmatrix} -3 \\\\ 4 \\\\ -4.5 \\end{pmatrix} \\,\\big|\\, t,s\\in\\mathbb{R} \\}$ (the column vectors associated with the parameters $\\begin{pmatrix} 1 \\\\ 1 \\\\ -8 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ 1 \\\\ -3 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix} \\qquad \\begin{pmatrix} -3 \\\\ 4 \\\\ -4.5 \\end{pmatrix} = \\begin{pmatrix} -2 \\\\ 4 \\\\ 0.5 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 0 \\\\ 5 \\end{pmatrix}$ are two vectors whose whole bodies lie in the plane). As with the line, note that some points in this plane are described with negative $t$'s or negative $s$'s or", "to the other line The distance can now be calculated as described under distance point and line. First set up an auxiliary plane with $P$ as the support point and direction vector as the normal vector. $\\text{H: } (\\vec{x} - \\vec{a}) \\cdot \\vec{n}=0$ $\\text{H: } (\\vec{x} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}) \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$. $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can" ]

[ "+0 # Pls help me +1 32 1 +57 The vectors $\\mathbf{a},$ $\\mathbf{b},$ and $\\mathbf{c}$ satisfy $\\|\\mathbf{a}\\| = \\|\\mathbf{b}\\| = 1,$ $\\|\\mathbf{c}\\| = 2,$ and $\\mathbf{a} \\times (\\mathbf{a} \\times \\mathbf{c}) + \\mathbf{b} = \\mathbf{0}.$If $\\theta$ is the angle between $\\mathbf{a}$ and $\\mathbf{c},$ then find all possible values of $\\theta,$ in degrees. Apr 25, 2021", "construction, $\\mathbf{e}_2$ is a unit vector orthogonal to $\\mathbf{e}_1,$ so that in particular $\\{\\mathbf{e}_1,\\mathbf{e}_2\\}$ is a linearly independent set in $\\mathbf{V}.$ Moreover, we have that $\\lambda_1 \\leq \\lambda_2,$ since $S(\\mathbf{V}_2)$ is a subset of $S(\\mathbf{V}_1).$", "$\\mathbf{u}^T\\mathbf{v} = \\Vert\\mathbf{u}\\Vert\\Vert\\mathbf{v}\\Vert\\cos\\theta$ This is a very important connection between the notion of inner product and trigonometry. As a quick check, note that if $$\\mathbf{u}$$ and $$\\mathbf{v}$$ are nonzero, and $$\\mathbf{u}^T\\mathbf{v} = 0$$, then $$\\cos\\theta = 0.$$ In other words, the angle between $$\\mathbf{u}$$ and $$\\mathbf{v}$$ is 90 degrees (or 270 degrees). So this agrees with our definition of orthogonality. One implication in particular concerns unit vectors. $\\mathbf{u}^T\\mathbf{v} = \\Vert\\mathbf{u}\\Vert\\Vert\\mathbf{v}\\Vert\\cos\\theta$ So $\\frac{\\mathbf{u}^T\\mathbf{v}}{\\Vert\\mathbf{u}\\Vert\\Vert\\mathbf{v}\\Vert} = \\cos\\theta$ $\\frac{\\mathbf{u}^T}{\\Vert\\mathbf{u}\\Vert}\\frac{\\mathbf{v}}{\\Vert\\mathbf{v}\\Vert} = \\cos\\theta$ Note that $$\\frac{\\mathbf{u}}{\\Vert\\mathbf{u}\\Vert}$$ and $$\\frac{\\mathbf{v}}{\\Vert\\mathbf{v}\\Vert}$$ are unit vectors. So we have the very simple rule, that for two unit vectors, their inner product is the cosine of the angle between them! u = np.array([0.95, np.sin(np.arccos(0.95))]) theta = (np.pi/3)+np.arccos(0.95) v = np.array([np.cos(theta), np.sin(theta)]) ax = ut.plotSetup(-1.3,1.3,-1.3,1.3,(12,6)) ut.centerAxes(ax) plt.axis('equal') angles = 2.0*np.pi * np.array(range(101))/100.0 plt.plot(np.cos(angles),np.sin(angles),'r-') ax.annotate('', xy=(0.2*v[0], 0.2*v[1]), xycoords='data', xytext=(0.3*u[0], 0.3*u[1]), textcoords='data', size=15, #bbox=dict(boxstyle=\"round\", fc=\"0.8\"), arrowprops={'arrowstyle': 'simple', 'fc': '0', 'ec': 'none', mid = 0.4*(u+v)/2.0 ax.text(mid[0],mid[1],r'$\\theta$',size=20) ax.text(u[0]+.1,u[1]+.1,r'$(%0.2f,%0.2f)$'% (u[0],u[1]),size=16) ax.text(v[0]+.1,v[1]+.1,r'$(%0.2f,%0.2f)$'% (v[0],v[1]),size=16); Here $$\\mathbf{u} = \\begin{bmatrix}0.95\\\\0.31\\end{bmatrix},$$ and $$\\mathbf{v} = \\begin{bmatrix}0.20\\\\0.98\\end{bmatrix}.$$ So $$\\mathbf{u}^T\\mathbf{v} = (0.95\\cdot 0.20) + (0.31 \\cdot 0.98) = 0.5$$ So $$\\cos\\theta = 0.5.$$ So $$\\theta = 60$$ degrees. Example. Find the angle formed by the vectors: $\\begin{split}\\mathbf{u} = \\begin{bmatrix}1\\\\3\\\\-7\\\\-2\\end{bmatrix} \\;\\;\\mbox{and}\\;\\; \\mathbf{v} = \\begin{bmatrix}8\\\\-2\\\\4\\\\6\\end{bmatrix}\\end{split}$ Solution. First normalize the vectors: $\\Vert\\mathbf{u}\\Vert = \\sqrt{1^2 + 3^2 + (-7)^2 + (-2)^2} = 7.93$ $\\Vert\\mathbf{v}\\Vert = \\sqrt{8^2 + (-2)^2", "equal or negative of the other. You could even define a plane in $\\Bbb R^3$ to be a set of vectors, $P = \\{\\mathbf x \\in \\Bbb R^3 \\ | \\ \\mathbf a , \\mathbf b \\in P \\ \\Rightarrow \\ \\frac{\\mathbf x \\times \\mathbf a}{|| \\mathbf x \\times \\mathbf a ||} = \\frac{\\mathbf x \\times \\mathbf b}{|| \\mathbf x \\times \\mathbf b ||} \\lor \\frac{\\mathbf x \\times \\mathbf a}{|| \\mathbf x \\times \\mathbf a ||} = -1\\frac{\\mathbf x \\times \\mathbf b}{|| \\mathbf x \\times \\mathbf b ||} \\}$. Then three vectors $u,v,w$ are coplanar iff $u,v,w \\in P$. That is they are contained in some plane $P$ defined above.", "+0 # help 0 64 1 Let $$\\mathbf{a}, \\mathbf{b}, \\mathbf{c}$$ be three vectors that lie on the same line. Find $$\\mathbf{a} \\times \\mathbf{b} + \\mathbf{b} \\times \\mathbf{c} + \\mathbf{c} \\times \\mathbf{a}$$ May 1, 2020", "is the unit vector in direction of$a \\times b$if$a=(2,1,1)$and$b=(-1,2,2) ?$0 answers 4 views Which of the following is the unit vector in direction of$a \\times b$if$a=(2,1,1)$and$b=(-1,2,2) ?$Which of the following is the unit vector in direction of$a \\times b$if$a=(2,1,1)$and$b=(-1,2,2) ?$&nbsp; A.$\\left(-\\frac{1}{2}, \\frac{1}{2}, ...", "planes to coincide. In higher dimension, a dihedral angle represents the angle between two hyperplanes.[1] ## Alternative definitions Since a plane can be defined in several ways (e.g., by vectors or points in them, or by their normal vectors), there are several equivalent definitions of a dihedral angle. Any plane can be defined by two non-collinear vectors lying in that plane; taking their cross product and normalizing yields the normal unit vector to the plane. Thus, a dihedral angle can be defined by four, pairwise non-collinear vectors. We may also define the dihedral angle of three non-collinear vectors $\\mathbf{b}_1$, $\\mathbf{b}_2$ and $\\mathbf{b}_3$ (red, green and blue, respectively, in the diagram). The vectors $\\mathbf{b}_1$ and $\\mathbf{b}_2$ define the first plane, whereas $\\mathbf{b}_2$ and $\\mathbf{b}_3$ define the second plane. The dihedral angle corresponds to an exterior spherical angle, given by $\\varphi = \\operatorname{atan2} \\left( \\left([\\mathbf{b}_1 \\times \\mathbf{b}_2]\\times [\\mathbf{b}_2 \\times \\mathbf{b}_3]\\right) \\cdot \\frac{\\mathbf{b}_2}{|\\mathbf{b}_2|}, [\\mathbf{b}_1 \\times \\mathbf{b}_2] \\cdot [\\mathbf{b}_2 \\times \\mathbf{b}_3] \\right),$ as derived in.[2] ## Dihedral angles in polyhedra Every polyhedron, regular and irregular, convex and concave, has a dihedral angle at every edge. A dihedral angle (also called the face angle) is the internal angle at which two adjacent faces meet. An angle of zero degrees means the face normal vectors are antiparallel and the faces overlap each other (Implying", "Vectors $\\mathbf{a}$ and $\\mathbf{b}$ must be going in the same direction.", "# Math Help - scalar products 1. ## scalar products The position vectors of three points A, B, and C with respect to a fixed origin O are 2i - 2j + k , 4i + 2j + k and i + j + 3k respectively. Find the unit vectors in the directions of CA and CB. < Calculate the ange ACB in degrees. 2. Originally Posted by Oasis1993 The position vectors of three points A, B, and C with respect to a fixed origin O are 2i - 2j + k , 4i + 2j + k and i + j + 3k respectively. Find the unit vectors in the directions of CA and CB. < Calculate the ange ACB in degrees. $CA = CO + OA$ $= -OC + OA$ $= OA - OC$ $= 2\\mathbf{i} - 2\\mathbf{j} + \\mathbf{k} - (4\\mathbf{i} + 2\\mathbf{j} + \\mathbf{k})$ $= -2\\mathbf{i} - 4\\mathbf{j} + 0\\mathbf{k}$. To find the unit vector in this direction, divide it by its length. $|CA| = \\sqrt{(-2)^2 + (-4)^2 + 0^2}$ $= \\sqrt{4 + 16 + 0}$ $= \\sqrt{20}$ $= 2\\sqrt{5}$. So $\\frac{OC}{|OC|} = \\frac{-2\\mathbf{i} - 4\\mathbf{j} + 0\\mathbf{k}}{2\\sqrt{5}}$ $= \\frac{-\\mathbf{i} - 2\\mathbf{j} + 0\\mathbf{k}}{\\sqrt{5}}$ $= \\frac{-\\sqrt{5}\\mathbf{i} - 2\\sqrt{5}\\mathbf{j} + 0\\mathbf{k}}{5}$ $= -\\frac{\\sqrt{5}}{5}\\mathbf{i} - \\frac{2\\sqrt{5}}{5}\\mathbf{j} + 0\\mathbf{k}$. Can you do the same to find the", "• # question_answer Direction: Let a, b and c are three non-coplanar vectors, i.e., $[a\\,b\\,c]\\,\\ne 0$. The three new vectors $a',\\,\\,b'$ and c' defined by the equation $a'=\\frac{b\\times c}{[a\\,\\,b\\,\\,c]},\\,\\,b'=\\frac{c\\times a}{[a\\,\\,b\\,\\,c]}$and$c'=\\frac{a\\times b}{[a\\,\\,b\\,\\,c]}$ are called reciprocal system to the vectors a, b and c. If a, b, c and a', b', c' are reciprocal system of vectors, then the value of$a\\times a'+b\\times b'+c\\times c'$ is A) $a\\times a'+b\\times b'+c\\times c'$ B) $2\\,(a'\\times b'\\times c')$ C) $\\frac{[a\\,\\,b\\,\\,c]}{2}$ D) 0 $\\therefore \\,\\,a\\times a'+b\\times b'+c+c'$$=\\frac{1}{[a\\,b\\,c]}[a\\times (b\\times c)+b\\times (c\\times a)+c\\times (a\\times b)]$(for cyclic order) $=\\frac{1}{[a\\,b\\,c]}[0]=0$", "# If $\\overrightarrow{a}$ and $\\overrightarrow{b}$ are unit vectors and $|\\overrightarrow{a} \\times \\overrightarrow{b}|=1$, then the angle between $\\overrightarrow{a}$ and $\\overrightarrow{b}$ is : ( A ) $\\frac{\\pi}{2}$ ( B ) $\\pi$ ( C ) $\\frac{\\pi}{3}$ ( D ) $\\frac{\\pi}{4}$", "and $C$ have position vectors $\\mathbf{a}, \\mathbf{b}$ and $\\mathbf{c}$ respectively.", "independent. $k_1\\neq 0$ since $S$ has no parabolic point, and $k_1\\neq k_2$ since $S$ has no umbilical points. Hence, $\\mathbf{y}_u\\times \\mathbf{y}_v\\neq\\mathbf{0}$. Similarly, we have $$\\mathbf{z}_u=\\mathbf{x}_u-\\frac{(k_2)_u}{k_2^2}N+\\frac{1}{k_2}N_u=(1-\\frac{k_1}{k_2})\\mathbf{x}_u-\\frac{(k_2)_u}{k_2^2}N,$$ $$\\mathbf{z}_v=\\mathbf{x}_v-\\frac{(k_2)_v}{k_2^2}N+\\frac{1}{k_2}N_v=-\\frac{(k_2)_v}{k_2^2}N,$$ which implies that $$\\mathbf{z}_u\\times \\mathbf{z}_v=-\\frac{(k_2)_v}{k_2^2}(1-\\frac{k_1}{k_2})\\mathbf{x}_u\\times N.$$ Again, $\\mathbf{x}_u\\times N\\neq\\mathbf{0}$ becasue they are linearly independent. $k_2\\neq 0$ since $S$ has no parabolic point, and $k_1\\neq k_2$ since $S$ has no umbilical points. Hence, $\\mathbf{z}_u\\times \\mathbf{z}_v\\neq\\mathbf{0}$.", "# Solve this following Question: If $\\hat{\\mathrm{a}}$ and $\\hat{\\mathrm{b}}$ are unit vectors such that $(\\hat{\\mathrm{a}}+\\mathrm{b})$ is a unit vector, what is the angle between $\\hat{\\mathrm{a}}$ and $\\hat{\\mathrm{b}}$ ? Solution:", "+0 # help vectors 0 64 1 Let $$\\mathbf{u}$$ and $$\\mathbf{v}$$ be linearly independent unit vectors. Find the set of all possible values of $$\\mathbf{u}\\cdot\\mathbf{v}$$. Give your answer in interval notation. Feb 7, 2020", "# Math Help - Vector geometry question 1. ## Orthogonal vectors Let $\\mathbf{a,b,c}$ be three vectors in $\\mathbb{R}^3$ which satisfy the relations $\\mathbf{b=c \\times a}\\ and\\ \\mathbf{c=a \\times b}$ 1) Show that $\\mathbf{a, b\\ and\\ c}$ are a set of mutually orthogonal vectors 2) Show that $\\mathbf{b\\ and\\ c}$ are of equal length and that if $\\mathbf{b \\neq 0}$, then $\\mathbf{a}$ is a unit vector. 2. Originally Posted by acevipa Let $\\mathbf{a,b,c}$ be three vectors in $\\mathbb{R}^3$ which satisfy the relations $\\mathbf{b=c \\times a}\\ and\\ \\mathbf{c=a \\times b}$ 1) Show that $\\mathbf{a, b\\ and\\ c}$ are a set of mutually orthogonal vectors 2) Show that $\\mathbf{b\\ and\\ c}$ are of equal length and that if $\\mathbf{b \\neq 0}$, then $\\mathbf{a}$ is a unit vector. This is pretty straight forward isn't it? $\\mathbf{b= c\\times a}$ is, by definition of \"cross product\", orthogonal to both $\\mathbf{a}$ and $mathbf{c}$ and $\\mathbf{c= a\\times b}$ is orthogonal to a. Further, since $\\mathbf{a}$ is orthogonal to $\\mathbf{c}$, $|\\mathbf{b}|= |\\mathbf{a}\\times\\mathbf{c}|= |\\mathbf{a}||\\mathbf{c}\\sin(\\pi/2)$ so that $1= (|\\mathbf{a}|)(1)(1)$.", "degree, the angles that a diagonal of a box with dimensions $10 \\mathrm{~cm} \\times 11 \\mathrm{~cm} \\times 25 \\mathrm{~cm}$ makes ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is orthogonal. Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is orthogonal.Let $S=\\{\\mathbf{i}, \\mathbf{j}, \\mathbf{k}\\}$ be the set of standard unit vectors in $R^{3} .$ Show that each ordered pair of vectors in $S$ is ortho ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Find a unit vector that is orthogonal to both $\\mathbf{u}=(1,0,1)$ and $\\mathbf{v}=(0,1,1)$ Find a unit vector that is orthogonal to both $\\mathbf{u}=(1,0,1)$ and $\\mathbf{v}=(0,1,1)$Find a unit vector that is orthogonal to both $\\mathbf{u}=(1,0,1)$ and $\\mathbf{v}=(0,1,1)$ ... close Show that if $\\mathbf{v}$ is orthogonal to both $\\mathbf{w}_{1}$ and $\\mathbf{w}_{2}$, then $\\mathbf{v}$ is orthogonal to $k_{1} \\mathbf{w}_{1}+k_{2} \\mathbf{w}_{2}$ for all scalars $k_{1}$ and $k_{2}$.Show that if $\\mathbf{v}$ is orthogonal to both $\\mathbf{w}_{1}$ and $\\mathbf{w}_{2}$, then $\\mathbf{v}$ is orthogonal to $k_{1} \\mathbf{w}_{1}+k_{2} ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Which of the following is the unit vector in direction of$a \\times", "+ 3k, $\\vec{B}$ = i – 2j + 2k Solution: Let $\\vec{C}$ be the unit vector. Cross product of A and B, $\\vec{A} \\times \\vec{B}$ = 4i – j – 3k Now unit vector in the direction of $\\vec{C}$ is $\\vec{C}/|\\vec{C}|$. $|\\vec{C}|$ = √26 Therefore desired unit vector is (1/√26) (4i – j – 3k) Example 6: If $\\mathbf{a}+\\mathbf{b}+\\mathbf{c}=\\mathbf{0},$ then which relation is correct $A)\\ \\mathbf{a}=\\mathbf{b}=\\mathbf{c}=\\mathbf{0}\\\\ B)\\ \\mathbf{a}\\,.\\,\\mathbf{b}=\\mathbf{b}\\,.\\,\\mathbf{c}=\\mathbf{c}\\,.\\,\\mathbf{a}\\\\ C)\\ \\mathbf{a}\\times \\mathbf{b}=\\mathbf{b}\\times \\mathbf{c}=\\mathbf{c}\\times \\mathbf{a}\\\\ D) \\ None \\ of \\ these$ Solution: Since, $\\mathbf{a}+\\mathbf{b}+\\mathbf{c}=0 \\\\\\Rightarrow \\mathbf{a}\\times (\\mathbf{a}+\\mathbf{b}+\\mathbf{c})=\\mathbf{0}\\\\\\Rightarrow \\mathbf{a}\\times \\mathbf{a}+\\mathbf{a}\\times \\mathbf{b}+\\mathbf{a}\\times \\mathbf{c}=\\mathbf{0} \\\\\\Rightarrow \\mathbf{a}\\times \\mathbf{b}=-\\mathbf{a}\\times \\mathbf{c}=\\mathbf{c}\\times \\mathbf{a}…..(i)\\\\$ $Similarly, \\mathbf{b}\\times (\\mathbf{a}+\\mathbf{b}+\\mathbf{c})=\\mathbf{0}\\\\\\Rightarrow \\mathbf{a}\\times \\mathbf{b}=\\mathbf{b}\\times \\mathbf{c}…..(ii)\\\\$ $\\ By \\ (i) \\ and \\ (ii), \\ we \\ get \\ \\mathbf{a}\\times \\mathbf{b}=\\mathbf{b}\\times \\mathbf{c}=\\mathbf{c}\\times \\mathbf{a}.$ Example 7: What is the number of vectors of unit length perpendicular to vectors a=(1,1,0) and b=(0,1,1)? Solution: The vector perpendicular to $\\mathbf{a} \\text \\ and \\ \\mathbf{b}$ is $\\mathbf{a}\\times \\mathbf{b}=\\left| \\begin{matrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ 1 & 1 & 0 \\\\ 0 & 1 & 1 \\\\ \\end{matrix} \\right|=\\mathbf{i}-\\mathbf{j}+\\mathbf{k}\\\\$ Since the length of this vector is $\\sqrt{3},\\ the \\ unit \\ vector \\ perpendicular \\ to \\ \\mathbf{a} \\ and \\ \\mathbf{b} \\ is \\pm \\frac{\\mathbf{a}\\times \\mathbf{b}}{|\\mathbf{a}\\times \\mathbf{b}|}\\\\=\\pm \\frac{1}{\\sqrt{3}}(\\mathbf{i}-\\mathbf{j}+\\mathbf{k})\\\\$ Hence there are two such vectors.", "independent vectors (i.e. they are not parallel), we can equate the coefficients of $\\mathbf{a}$ and $\\mathbf{b}$ on either side of this equation. Thus $h = \\frac{k}{6} \\;\\text{ and }\\; 0 = 1-\\frac{2k}{3} \\quad\\implies\\quad k = \\frac{3}{2} \\;\\text{ and }\\; h = \\frac{1}{4}.$ So $\\overrightarrow{OZ} = \\dfrac{1}{4}\\mathbf{a}$, and we have the position vector of $Z$. What would happen if $\\mathbf{a}$ and $\\mathbf{b}$ were parallel or anti-parallel?", "included angle between vectors is 90°. See figure. Problem : If a and b are two non-collinear unit vectors and | a + b | = √3, then find the value of expression : $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)\\end{array}$ Solution : The given expression is scalar product of two vector sums. Using distributive property we can expand the expression, which will comprise of scalar product of two vectors a and b . $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2\\mathbf{a}\\mathbf{.}\\mathbf{a}+\\mathbf{a}\\mathbf{.}\\mathbf{b}-\\mathbf{b}\\mathbf{.}2\\mathbf{a}+\\left(-\\mathbf{b}\\right)\\mathbf{.}\\left(-\\mathbf{b}\\right)=2{a}^{2}-\\mathbf{a}\\mathbf{.}\\mathbf{b}-{b}^{2}\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta \\end{array}$ We can evaluate this scalar product, if we know the angle between them as magnitudes of unit vectors are each 1. In order to find the angle between the vectors, we use the identity, $\\begin{array}{l}\\mathbf{A}\\mathbf{.}\\mathbf{A}={A}^{2}\\end{array}$ Now, $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}=\\left(\\mathbf{a}+\\mathbf{b}\\right)\\mathbf{.}\\left(\\mathbf{a}+\\mathbf{b}\\right)={a}^{2}+{b}^{2}+2ab\\mathrm{cos}\\theta =1+1+2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}\\theta \\end{array}$ $\\begin{array}{l}⇒{|\\mathbf{a}+\\mathbf{b}|}^{2}=2+2\\mathrm{cos}\\theta \\end{array}$ It is given that : $\\begin{array}{l}{|\\mathbf{a}+\\mathbf{b}|}^{2}={\\left(\\surd 3\\right)}^{2}=3\\end{array}$ Putting this value, $\\begin{array}{l}⇒2\\mathrm{cos}\\theta ={|\\mathbf{a}+\\mathbf{b}|}^{2}-2=3-2=1\\end{array}$ $\\begin{array}{l}⇒\\mathrm{cos}\\theta =\\frac{1}{2}\\\\ ⇒\\theta =60°\\end{array}$ Using this value, we now proceed to find the value of given identity, $\\begin{array}{l}\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=2{a}^{2}-{b}^{2}-ab\\mathrm{cos}\\theta =2\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}{1}^{2}-{1}^{2}-1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}1\\phantom{\\rule{2pt}{0ex}}x\\phantom{\\rule{2pt}{0ex}}\\mathrm{cos}60°\\end{array}$ $\\begin{array}{l}⇒\\left(\\mathbf{a}-\\mathbf{b}\\right)\\mathbf{.}\\left(2\\mathbf{a}+\\mathbf{b}\\right)=\\frac{1}{2}\\end{array}$ ## Evaluation of dot product Problem : In an experiment of light reflection, if a , b and c are the unit vectors in the direction of incident ray, reflected ray and normal to the reflecting surface, then prove that : $\\begin{array}{l}⇒\\mathbf{b}=\\mathbf{a}-2\\left(\\mathbf{a}\\mathbf{.}\\mathbf{c}\\right)\\mathbf{c}\\end{array}$ Solution : Let us consider vectors in a coordinate system in which “x” and “y” axes of the coordinate system are in the direction of reflecting surface and normal to the reflecting" ]

[ "f_1(\\mathbf{v}) + f_2(\\mathbf{v})$ then $\\frac{\\partial f}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial f_1}{\\partial \\mathbf{v}} + \\frac{\\partial f_2}{\\partial \\mathbf{v}}\\right)\\cdot\\mathbf{u}$ 2. If $f(\\mathbf{v}) = f_1(\\mathbf{v})~ f_2(\\mathbf{v})$ then $\\frac{\\partial f}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial f_1}{\\partial \\mathbf{v}}\\cdot\\mathbf{u}\\right)~f_2(\\mathbf{v}) + f_1(\\mathbf{v})~\\left(\\frac{\\partial f_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} \\right)$ 3. If $f(\\mathbf{v}) = f_1(f_2(\\mathbf{v}))$ then $\\frac{\\partial f}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\frac{\\partial f_1}{\\partial f_2}~\\frac{\\partial f_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u}$ ### Derivatives of vector valued functions of vectors Let $\\mathbf{f}(\\mathbf{v})$ be a vector valued function of the vector $\\mathbf{v}$. Then the derivative of $\\mathbf{f}(\\mathbf{v})$ with respect to $\\mathbf{v}$ (or at $\\mathbf{v}$) in the direction $\\mathbf{u}$ is the second-order tensor defined as $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = D\\mathbf{f}(\\mathbf{v})[\\mathbf{u}] = \\left[\\frac{d }{d \\alpha}~\\mathbf{f}(\\mathbf{v} + \\alpha~\\mathbf{u})\\right]_{\\alpha = 0}$ for all vectors $\\mathbf{u}$. Properties: 1. If $\\mathbf{f}(\\mathbf{v}) = \\mathbf{f}_1(\\mathbf{v}) + \\mathbf{f}_2(\\mathbf{v})$ then $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial \\mathbf{f}_1}{\\partial \\mathbf{v}} + \\frac{\\partial \\mathbf{f}_2}{\\partial \\mathbf{v}}\\right)\\cdot\\mathbf{u}$ 2. If $\\mathbf{f}(\\mathbf{v}) = \\mathbf{f}_1(\\mathbf{v})\\times\\mathbf{f}_2(\\mathbf{v})$ then $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial \\mathbf{f}_1}{\\partial \\mathbf{v}}\\cdot\\mathbf{u}\\right)\\times\\mathbf{f}_2(\\mathbf{v}) + \\mathbf{f}_1(\\mathbf{v})\\times\\left(\\frac{\\partial \\mathbf{f}_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} \\right)$ 3. If $\\mathbf{f}(\\mathbf{v}) = \\mathbf{f}_1(\\mathbf{f}_2(\\mathbf{v}))$ then $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\frac{\\partial \\mathbf{f}_1}{\\partial \\mathbf{f}_2}\\cdot\\left(\\frac{\\partial \\mathbf{f}_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} \\right)$ ### Derivatives of scalar valued functions of second-order tensors Let $f(\\boldsymbol{S})$ be a real valued function of the second order tensor $\\boldsymbol{S}$. Then the derivative of $f(\\boldsymbol{S})$ with respect to $\\boldsymbol{S}$ (or at $\\boldsymbol{S}$) in the direction $\\boldsymbol{T}$ is the second order tensor defined as $\\frac{\\partial f}{\\partial \\boldsymbol{S}}:\\boldsymbol{T} = Df(\\boldsymbol{S})[\\boldsymbol{T}] = \\left[\\frac{d }{d \\alpha}~f(\\boldsymbol{S} + \\alpha\\boldsymbol{T})\\right]_{\\alpha =", "{1}{\\sqrt {\\det g}}}{\\frac {\\partial \\left({\\sqrt {\\frac {\\det g}{g_{ii}}}}\\,{\\hat {F}}^{i}\\right)}{\\partial x^{i}}}.}$ See § In curvilinear coordinates for further discussion. Properties The following properties can all be derived from the ordinary differentiation rules of calculus. Most importantly, the divergence is a linear operator, i.e., ${\\displaystyle \\operatorname {div} (a\\mathbf {F} +b\\mathbf {G} )=a\\operatorname {div} \\mathbf {F} +b\\operatorname {div} \\mathbf {G} }$ for all vector fields F and G and all real numbers a and b. There is a product rule of the following type: if φ is a scalar-valued function and F is a vector field, then ${\\displaystyle \\operatorname {div} (\\varphi \\mathbf {F} )=\\operatorname {grad} \\varphi \\cdot \\mathbf {F} +\\varphi \\operatorname {div} \\mathbf {F} ,}$ or in more suggestive notation ${\\displaystyle \\nabla \\cdot (\\varphi \\mathbf {F} )=(\\nabla \\varphi )\\cdot \\mathbf {F} +\\varphi (\\nabla \\cdot \\mathbf {F} ).}$ Another product rule for the cross product of two vector fields F and G in three dimensions involves the curl and reads as follows: ${\\displaystyle \\operatorname {div} (\\mathbf {F} \\times \\mathbf {G} )=\\operatorname {curl} \\mathbf {F} \\cdot \\mathbf {G} -\\mathbf {F} \\cdot \\operatorname {curl} \\mathbf {G} ,}$ or ${\\displaystyle \\nabla \\cdot (\\mathbf {F} \\times \\mathbf {G} )=(\\nabla \\times \\mathbf {F} )\\cdot \\mathbf {G} -\\mathbf {F} \\cdot (\\nabla \\times \\mathbf {G} ).}$ The Laplacian of a scalar field is the divergence of the field's gradient:", "\\sqrt{\\left({\\operatorname{d}\\!r\\over\\operatorname{d}\\!t}\\right)^2 + r^2\\left({\\operatorname{d}\\!\\theta\\over\\operatorname{d}\\!t}\\right)^2 } \\operatorname{d}\\!t = \\int_{\\theta(t_1)}^{\\theta(t_2)} \\sqrt{\\left({\\operatorname{d}\\!r\\over\\operatorname{d}\\!\\theta}\\right)^2 + r^2 } \\operatorname{d}\\!\\theta.$ Now let $\\mathbf{C}(t) = (r(t), \\theta(t), \\phi(t))$ be a curve expressed in spherical coordinates where $\\theta$ is the polar angle measured from the positive $z$-axis and $\\phi$ is the azimuthal angle. The mapping that transforms from spherical coordinates to rectangular coordinates is $\\mathbf{x}(r,\\theta,\\phi) = (r\\sin\\theta\\cos\\phi, r\\sin\\theta\\sin\\phi, r\\cos\\theta).$ Using the chain rule again shows that $D(\\mathbf{x}\\circ\\mathbf{C}) = \\mathbf{x}_r r' + \\mathbf{x}_{\\theta}\\theta' + \\mathbf{x}_{\\phi}\\phi'$. All dot products $\\mathbf{x}_i \\cdot \\mathbf{x}_j$ where $i$ and $j$ differ are zero, so the squared norm of this vector is $(\\mathbf{x}_r\\cdot \\mathbf{x}_r )(r'^2) + (\\mathbf{x}_{\\theta} \\cdot \\mathbf{x}_{\\theta})(\\theta')^2 + (\\mathbf{x}_{\\phi}\\cdot \\mathbf{x}_{\\phi})(\\phi')^2 = (r')^2 + r^2(\\theta')^2 + r^2 \\sin^2\\theta (\\phi')^2.$ So for a curve expressed in spherical coordinates, the arc length is $\\int_{t_1}^{t_2} \\sqrt{\\left({\\operatorname{d}\\!r\\over\\operatorname{d}\\!t}\\right)^2 + r^2\\left({\\operatorname{d}\\!\\theta\\over\\operatorname{d}\\!t}\\right)^2 + r^2\\sin^2\\theta \\left({\\operatorname{d}\\!\\phi\\over\\operatorname{d}\\!t}\\right)^2 } \\operatorname{d}\\!t.$ A very similar calculation shows that the arc length of a curve expressed in cylindrical coordinates is $\\int_{t_1}^{t_2} \\sqrt{\\left({\\operatorname{d}\\!r\\over\\operatorname{d}\\!t}\\right)^2 + r^2\\left({\\operatorname{d}\\!\\theta\\over\\operatorname{d}\\!t}\\right)^2 + \\left({\\operatorname{d}\\!z\\over\\operatorname{d}\\!t}\\right)^2 } \\operatorname{d}\\!t.$ ## Simple cases ### Arcs of circles Arc lengths are denoted by s, since the Latin word for length (or size) is spatium. In the following lines, $r$ represents the radius of a circle, $d$ is its diameter, $C$ is its circumference, $s$ is the length of an arc of the circle, and $\\theta$ is the angle which the", "integral. You can then 'cancel' the dot product. Setting $d\\mathbf{S} = \\mathbf{n}\\,dS$ where $\\mathbf{n}$ is the outward unit normal and $d\\mathbf{l} = \\mathbf{t}\\,d\\ell$ where $\\mathbf{t}$ is the unit tangent and taking advantage of the cyclic nature of the scalar triple product is also useful. For example, to prove the first identity: Let $\\mathbf{c}$ be an arbitrary constant vector. Then $$\\nabla \\cdot (\\mathbf{f} \\times \\mathbf{c}) = \\mathbf{c} \\cdot (\\nabla \\times \\mathbf{f}).$$ Now apply the divergence theorem: $$\\mathbf{c} \\cdot \\int_V \\nabla \\times \\mathbf{f}\\,dV = \\int_{\\partial V} (\\mathbf{f} \\times \\mathbf{c}) \\cdot \\mathbf{n}\\,dS = \\mathbf{c} \\cdot \\int_{\\partial V} \\mathbf{n} \\times \\mathbf{f}\\,dS.$$ Hence $$\\mathbf{c} \\cdot \\left(\\int_V \\nabla \\times \\mathbf{f}\\,dV - \\int_{\\partial V} \\mathbf{n} \\times \\mathbf{f}\\,dS\\right) = 0$$ and since this holds for all $\\mathbf{c}$ it holds in particular for each of the cartesian standard basis vectors, and we must have $$\\int_V \\nabla \\times \\mathbf{f}\\,dV = \\int_{\\partial V} \\mathbf{n} \\times \\mathbf{f}\\,dS.$$ vanhees71", "reciprocal frame vectors are \\begin{aligned}\\mathbf{x}^a &= \\mathbf{x}_b \\cdot \\frac{1}{{ \\mathbf{x}_a \\wedge \\mathbf{x}_b }} = - \\frac{\\gamma_1 \\exp\\left( { \\gamma_1 \\gamma_0 b } \\right)}{\\sinh(b)} \\\\ \\mathbf{x}^b &= \\mathbf{x}_a \\cdot \\frac{1}{{ \\mathbf{x}_b \\wedge \\mathbf{x}_a }} = - \\frac{\\gamma_1}{\\sinh(b)},\\end{aligned} which satisfy $\\mathbf{x}^a \\cdot \\mathbf{x}_a = \\mathbf{x}^b \\cdot \\mathbf{x}_b = 1$, and $\\mathbf{x}^a \\cdot \\mathbf{x}_b = \\mathbf{x}^b \\cdot \\mathbf{x}_a = 0$, which can be shown explicitly with relative ease using scalar selection operations. The vector derivative is $$\\partial = - \\frac{\\gamma_1}{\\sinh(b)} \\left( { \\frac{\\partial {}}{\\partial {b}} + \\exp\\left( { \\gamma_1 \\gamma_0 b } \\right) \\frac{\\partial {}}{\\partial {a}} } \\right).$$ This is enough to explicitly express the left and right hand sides of the fundamental theorem identity $$\\int d^2 \\mathbf{x} \\partial F=\\gamma_0 \\int da db \\left( { \\frac{\\partial {}}{\\partial {b}} + \\exp\\left( { \\gamma_1 \\gamma_0 b } \\right) \\frac{\\partial {}}{\\partial {a}} } \\right) F,$$ and $$-\\int d\\mathbf{x}_b {\\left.{{F}}\\right\\vert}_{{\\Delta a}}+\\int d\\mathbf{x}_a {\\left.{{F}}\\right\\vert}_{{\\Delta b}}=\\gamma_0 \\int db \\exp\\left( { \\gamma_1 \\gamma_0 b } \\right) {\\left.{{F}}\\right\\vert}_{{\\Delta a}}+ \\gamma_0 \\int da {\\left.{{F}}\\right\\vert}_{{\\Delta b}}.$$ As expected by the theorem, these are clearly identical. According to the book Geometric Algebra for Physicists, by Doran and Lasenby, \"the [fundamental] theorem is still valid in a Lorentzian space\" (page 200, in chapter 6). The next section, \"Embedded surfaces and vector manifolds\" (6.5) might also be of interest to you. Doesn't directly express", "1 - 1\\cdot (-1)] - \\mathbf{j}[2 \\cdot 1 - 1 \\cdot 3] + \\mathbf{k}[2 \\cdot (-1) - 2 \\cdot 3]$ $= 3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}$. I agree with the answer you have given for $\\mathbf{v} \\times \\mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect. Anyway, to find the unit vector in the direction of $\\mathbf{v} \\times \\mathbf{w}$, divide it by its length. $|\\mathbf{v}\\times\\mathbf{w}| = \\sqrt{3^2 + (-1)^2 + (-8)^2}$ $= \\sqrt{9 + 1 + 64}$ $= \\sqrt{74}$. Therefore: $\\frac{\\mathbf{v}\\times\\mathbf{w}}{|\\mathbf{v}\\time s\\mathbf{w}|} = \\frac{3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}}{\\sqrt{74}}$ $= \\frac{3}{\\sqrt{74}}\\mathbf{i} - \\frac{1}{\\sqrt{74}}\\mathbf{j} - \\frac{8}{\\sqrt{74}}\\mathbf{k}$ $= \\frac{3\\sqrt{74}}{74}\\mathbf{i} - \\frac{\\sqrt{74}}{74}\\mathbf{j} - \\frac{4\\sqrt{74}}{37}\\mathbf{k}$. 3. Hello, Paymemoney! You've done nothing wrong . . . Their answer is the negative of yours. There are two vectors perpendicular to both $\\vec v\\text{ and }\\vec w,$ . . one \"up\" and one \"down\". They picked one, you picked the other. 4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k", "dual covariant basis defined as $\\mathbf{e}_i= \\frac{\\partial\\mathbf{x}}{\\partial{u^i}} \\to \\mathbf{e}_i\\cdot\\mathbf{e}^j = \\delta_{i}^{j}~,$ and therefore relates to the contravariant vectors as $\\mathbf{e}_i = \\frac{\\mathbf{e}^j\\times\\mathbf{e}^k}{\\mathbf{e}^i\\cdot\\mathbf{e}^j\\times\\mathbf{e}^k} = \\sqrt{g}\\;\\mathbf{e}^j\\times\\mathbf{e}^k ~,$ where $(i,j,k)$ are cyclic permutations of $(1,2,3)$ and we have used the notation $(u^1, u^2, u^3) = (\\psi,\\theta,\\phi)$. The Jacobian $\\sqrt{g}$ is defined below. Similarly $\\mathbf{e}^i = \\frac{\\mathbf{e}_j\\times\\mathbf{e}_k}{\\sqrt{g}} ~.$ Any vector field $\\mathbf{B}$ can be represented as $\\mathbf{B} = (\\mathbf{B}\\cdot\\mathbf{e}^i)\\mathbf{e}_i = B^i\\mathbf{e}_i$ or $\\mathbf{B} = (\\mathbf{B}\\cdot\\mathbf{e}_i)\\mathbf{e}^i = B_i\\mathbf{e}^i ~.$ In particular any basis vector $\\mathbf{e}_i = (\\mathbf{e}_i\\cdot\\mathbf{e}_j)\\mathbf{e}^j$. The metric tensor is defined as $g_{ij} = \\mathbf{e}_i\\cdot\\mathbf{e}_j \\; ; \\; g^{ij} = \\mathbf{e}^i\\cdot\\mathbf{e}^j \\; ; \\; g^j_i = \\mathbf{e}_i\\cdot\\mathbf{e}^j = \\delta_i^j ~.$ The metric tensors can be used to raise or lower indices. Take $\\mathbf{B} = B_i\\mathbf{e}^i = B_i g^{ij}\\mathbf{e}_j = B^j\\mathbf{e}_j~,$ so that $B^j = g^{ij} B_i~.$ ### Jacobian The Jacobian of the coordinate transformation $\\mathbf{x}(\\psi, \\theta, \\phi)$ is defined as $J = \\det\\left(\\frac{\\partial(x,y,z)}{\\partial(\\psi,\\theta,\\phi)}\\right) = \\frac{\\partial\\mathbf{x}}{\\partial{\\psi}}\\cdot\\frac{\\partial\\mathbf{x}}{\\partial{\\theta}} \\times \\frac{\\partial\\mathbf{x}}{\\partial{\\phi}}$ and that of the inverse transformation $J^{-1} = \\det\\left(\\frac{\\partial(\\psi,\\theta,\\phi)}{\\partial(x,y,z)}\\right) = \\nabla{\\psi}\\cdot\\nabla{\\theta} \\times \\nabla{\\phi}$ It can be seen that [1] $g \\equiv \\det(g_{ij}) = J^2 \\Rightarrow J = \\sqrt{g}$ ### Some surface elements Consider a surface defined by a constant value of $\\phi$. Then, the surface element is $d{\\mathbf S}_\\phi = \\mathbf{e}_\\psi\\times\\mathbf{e}_\\theta d\\psi d\\theta = \\sqrt{g}\\, \\nabla\\phi d\\psi d\\theta .$ As for a surface defined by", "$$\\mathbf{F} = \\phi\\mathbf{c} \\times \\mathbf{r}$$ in the Divergence Theorem as you quoted above, where $$\\mathbf{c}$$ is an arbitrary yet constant vector field. By the divergence product rule for a scalar field and a vector field: $$\\nabla \\cdot (\\phi\\mathbf{c} \\times \\mathbf{r}) = \\nabla \\phi \\cdot (\\mathbf{c} \\times \\mathbf{r}) + \\phi \\nabla \\cdot (\\mathbf{c} \\times \\mathbf{r}) = \\mathbf{c} \\cdot (\\mathbf{r} \\times \\nabla \\phi)$$, by the cyclic properties of the triple vector product and the fact that $$\\nabla \\cdot (\\mathbf{c} \\times \\mathbf{r}) = \\mathbf{0}$$ (check this directly). Now note from the Divergence Theorem for the vector field $$\\mathbf{F}$$: $$\\iiint_R \\mathbf{c} \\cdot (\\mathbf{r} \\times \\nabla \\phi) \\,dV = \\iint_{\\partial R} (\\phi\\mathbf{c} \\times \\mathbf{r}) \\cdot \\mathbf{n} \\,dS$$ $$\\implies \\mathbf{c} \\cdot \\iiint_R (\\mathbf{r} \\times \\nabla \\phi) \\,dV = \\mathbf{c} \\cdot \\iint_{\\partial R} \\mathbf{r} \\times \\phi \\mathbf{n}\\,dS$$ where the right hand side of the last expression follows again from the cyclic properties of the triple product. Recalling $$\\mathbf{c}$$ is arbitrary, we're done.", "# On the integral $\\iint_{\\partial B_1(0)}\\frac{d\\Omega}{\\left(1+a\\cdot r\\right)\\left(1+b\\cdot r\\right)}$ Let $a, b$ be arbitrary vectors in $\\mathbb{R}^3$ and define $\\partial B_1(0):=\\{r\\in\\mathbb{R}^3:|r|=1\\}$ as the boundary of the three-dimensional unit sphere. I am trying to evaluate the following integral: $$I = \\iint_{\\partial B_1(0)}\\frac{d\\Omega}{\\left(1+a\\cdot r\\right)\\left(1+b\\cdot r\\right)}$$ So far, I've said that we can, given a vector $a$ and a fixed vector $b$, write the following: $$a = c b + x$$ where $c = \\frac{a\\cdot b}{|b|}$ and $x\\in\\mathbb{R}^3$ such that $x$ orthogonal to $b$. Then, if we choose our coordinate system so that the positive $z$-axis runs parallel to the vector $b$, then, considering spherical coordinates, the angle between $r$ and $b$ is given by $\\theta$, and we have: \\begin{align} I = \\int_0^\\pi d\\theta\\sin\\theta\\int_0^{2\\pi}d\\phi\\frac{1}{\\left(1+|a|\\cos\\theta\\right)\\left(1+c\\cos\\theta + x\\cdot r\\right)} \\end{align} Now, I know that, since $x\\cdot b = 0$, $x$ must lie in the $x-y-$plane, so we can express it as $(\\rho\\cos\\zeta, \\rho\\sin\\zeta, 0)$ for some modulus $\\rho$ and some polar angle $\\zeta$. Thus: $$x\\cdot r = \\rho\\sin\\theta\\left(\\cos\\zeta\\cos\\phi+\\sin\\zeta\\sin\\phi\\right) = \\rho\\sin\\theta\\cos\\left(\\phi-\\zeta\\right)$$ Letting $\\phi-\\zeta=:\\xi$, we have $d\\xi = d\\phi$ and thus, since all periods of the cosine function are equal, we have: \\begin{align} I&=\\int_0^\\pi d\\theta\\frac{\\sin\\theta}{1+|a|\\cos\\theta}\\int_{-\\zeta}^{2\\pi-\\zeta}\\frac{d\\xi}{1+c\\cos\\theta+\\rho\\sin\\theta\\cos\\xi}\\\\ &=\\int_{-1}^1 \\frac{du}{1+|a|u}\\int_{0}^{2\\pi}\\frac{d\\xi}{1+cu+\\rho\\sqrt{1-u^2}\\cos\\xi} \\end{align} And this is where I'm stumped. I've been thinking of maybe converting the second integral into a contour integral over the unit circle, which reduces to evaluating $$\\oint_{|z|=1}\\frac{dz}{z^2+\\alpha", "a)_{ij} = \\frac{\\partial a_i}{\\partial x_j} \\tag{16}$$ where $(\\nabla \\mathbf a)_{ij}$ is the component of $\\nabla \\mathbf a$ associated with the outer product $\\mathbf e_i \\otimes \\mathbf e_j$. ## A Note About Antisymmetric Second-Order Tensors If a second-order tensor $\\mathbf F$ is antisymmetric (i.e. $F_{ji} = -F_{ij}$), then there exists some scalars $f_k$ such that, $$F_{ij} = \\epsilon_{ijk} f_k \\tag{17}$$ or, equivalently, $$f_k = \\frac{1}{2}\\epsilon_{ijk}F_{ij} \\tag{18}$$ and the vector $\\mathbf f$ for which $f_k$ is the $k$-th component is called the axial vector of $\\mathbf F$. Note that the definition for the vector product in (14) also involves the components of the alternating tensor. This is no accident, as the vector product in $\\mathbb R^3$ between two vectors $\\mathbf a$ and $\\mathbf f$ is actually the result of an inner product between an antisymmetric tensor $\\mathbf F$ and a vector $\\mathbf a$, $$F_{ij}a_j = \\epsilon_{ijk} a_j f_k \\tag{19}$$ ## The Lorentz Force Now we are in a position to consider the equation you have written for $d \\vec p/dt$ using clearer notation, \\begin{align} \\frac{d\\mathbf p}{dt} & = e\\left[\\left(-\\frac{\\partial \\mathbf A}{\\partial t} - \\nabla \\phi \\right) + \\mathbf v \\cdot \\nabla \\mathbf A - \\nabla \\mathbf A \\cdot \\mathbf v\\right] \\\\ & = e\\left(\\mathbf E + \\mathbf v \\cdot \\nabla \\mathbf A - \\nabla \\mathbf A \\cdot \\mathbf v\\right) \\end{align}", "A) $\\frac{1}{2|b|}\\sqrt{{{d}^{2}}-{{a}^{2}}}$ B) $\\frac{1}{2|a|}\\sqrt{{{d}^{2}}-{{a}^{2}}}$ C) $\\frac{1}{2|d|}\\sqrt{{{d}^{2}}-{{a}^{2}}}$ D) None of these • question_answer139) The value of $'a'$ for which the function $f(x)=(4a-3)(x+\\log 5)+2(a-7)$ $\\cot \\frac{x}{2}\\cdot {{\\sin }^{2}}\\frac{x}{2}$does not possess critical points is A) $(-\\infty ,\\,\\,2)$ B) $(-\\infty ,\\,\\,-1)$ C) $[1,\\,\\,\\infty )$ D) $\\left( -\\infty ,\\,\\,\\frac{-4}{3} \\right)\\cup (2,\\,\\,\\infty )$ • question_answer140) $\\mathbf{a}$and $\\mathbf{c}$ are unit vectors and$|\\mathbf{b}|\\,\\,=4$. If angle between $\\mathbf{b}$ and $\\mathbf{c}$ is and ${{\\cos }^{-1}}\\left( \\frac{1}{4} \\right)$$\\text{and}$$\\mathbf{a}\\times \\mathbf{b}=2\\mathbf{a}\\times \\mathbf{c}$, then$\\mathbf{b}=\\lambda \\mathbf{a}+2\\mathbf{c}$, where$\\lambda$is equal to A) $\\pm \\frac{1}{4}$ B) $\\pm \\frac{1}{2}$ C) $\\pm \\,\\,1$ D) None of the above • question_answer141) If the planes$x-cy-bz=0$, $cx-y+az=0$and$bx+ay-z=0$ pass through a straight line, then the value of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2abc$is A) $-1$ B) $2$ C) $1$ D) $0$ • question_answer142) The plane $ax+by=0$ is rotated through an angle $\\alpha$ about its line of intersection with the plane$z=0$, then the equation to the plane in new position is A) $ax-by\\pm z\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot \\cot \\alpha =0$ B) $ax-by\\pm z\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot \\tan \\alpha =0$ C) $ax+by\\pm z\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot \\cot \\alpha =0$ D) $ax+by\\pm z\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot \\tan \\alpha =0$ • question_answer143) If$y=a\\cos (\\log x)-b\\sin (\\log x)$, then the value of${{x}^{2}}\\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\\frac{dy}{dx}+y$is A) $0$ B) $1$ C) $2$ D) $3$ • question_answer144) The inverse of the function$f(x)=\\log ({{x}^{2}}+3x+1),\\,\\,x\\in [1,\\,\\,3]$, assuming it to be an onto function, is A) $\\frac{-3+\\sqrt{5+4{{e}^{x}}}}{2}$ B) $\\frac{-3\\pm \\sqrt{5+4{{e}^{x}}}}{2}$ C) $\\frac{-3-\\sqrt{5+4{{e}^{x}}}}{2}$ D) None of the above • question_answer145) $\\underset{x\\to 0}{\\mathop{\\lim }}\\,\\frac{\\sqrt{1-\\cos", "## Thomas' Calculus 13th Edition The diagonal is ${\\bf u}+{\\bf v}$ The cosine of the angle $\\theta_{1}$, between ${\\bf u}+{\\bf v}$ and ${\\bf u}$ is $\\displaystyle \\cos\\theta_{1}=\\frac{({\\bf u+v})\\cdot{\\bf u}}{|{\\bf u+v}||{\\bf u}|}$ The cosine of the angle $\\theta_{2}$, between ${\\bf u}+{\\bf v}$ and ${\\bf v}$ is $\\displaystyle \\cos\\theta_{2}=\\frac{({\\bf u+v})\\cdot{\\bf v}}{|{\\bf u+v}||{\\bf v}|}$ The denominators in these expressions are equal, because $|{\\bf u}|=|{\\bf v}|$. If we prove that the numerators are equal, because cosine is one-to-one on $[0,\\pi]$, it would follow that $\\theta_{1}=\\theta_{2}.$ We will apply Properties of the Dot Product (boxed on p. 709) $({\\bf u+v})\\cdot{\\bf u}={\\bf u}\\cdot({\\bf u+v}) \\quad$ ... property 1 =${\\bf u}\\cdot{\\bf u} +{\\bf u}\\cdot{\\bf v}\\quad$ ... property 3 = $|{\\bf u}|^{2} +{\\bf u}\\cdot{\\bf v}\\quad$ ... property 4 = $|{\\bf v}|^{2} +{\\bf u}\\cdot{\\bf v}\\quad$ ... given in the problem =${\\bf v}\\cdot{\\bf v} +{\\bf u}\\cdot{\\bf v}\\quad$ ... property $4$ $={\\bf v}\\cdot({\\bf u+v}) \\quad$ ... property $3$ $=({\\bf u+v})\\cdot{\\bf v} \\quad$ ... property 1 The numerators are equal as well. Thus, $\\theta_{1}=\\theta_{2}.$", "B$ is defined by $\\vec A \\cdot \\vec B = \\left | \\vec A \\right | \\left | \\vec B \\right | \\cos \\theta$ This formula can be arranged to determine the angle between two non-zero vectors. $\\vec A \\cdot \\vec B = \\left | \\vec A \\right | \\left | \\vec B \\right | \\cos \\theta$ $\\frac { \\vec A \\cdot \\vec B} {\\left | \\vec A \\right | \\left | \\vec B \\right |} = \\cos \\theta$ $\\arccos \\left ( \\frac {\\vec A \\cdot \\vec B} {\\left | \\vec A \\right | \\left | \\vec B \\right |} \\right ) = \\theta$ ## Geometric Properties of the Dot Product Here are the geometric properties of the dot product. We first put restriction on $\\theta$. We take $0 \\le \\theta \\le \\pi$ • If $\\mathbf{a}$ and $\\mathbf{b}$ are nonzero vectors then $\\mathbf{a} \\cdot \\mathbf{b} = 0$ if and only if $\\cos \\theta = 0$. $\\cos \\theta = 0$ when $\\theta = \\frac{\\pi} {2}$. It makes sense for to call $\\mathbf{a}$ and $\\mathbf{b}$ perpendicular (or orthogonal). This applies to two-dimensional and three-dimensional anything higher is not a property but a definition of perpendicularity. Notice that if $\\mathbf{a}$ or $\\mathbf{b}$ is the zero vector then we can say the zero vector is perpendicular to every vector. • The dot", "\\cos\\theta_v\\,\\id + \\sin\\theta_v\\,V/\\|V\\|$: the “scalar” matrix $\\id$ has superposition weight $\\cos\\theta_v=\\mathbf{b}\\cdot\\mathbf{c}$ whereas the skew-Hermitian “vector” matrix $\\sin\\theta_v\\,V/\\|V\\|$ is equal to $(\\mathbf{b}\\times\\mathbf{c})_x\\,\\hat{s}_x+(\\mathbf{b}\\times\\mathbf{c})_y\\,\\hat{s}_y+(\\mathbf{b}\\times\\mathbf{c})_z\\,\\hat{s}_z$. So now we work out what the product $(\\mathbf{a}\\cdot\\mathbf{b},\\,\\mathbf{a}\\times\\mathbf{b})\\,(\\mathbf{b}\\cdot\\mathbf{c},\\,\\mathbf{b}\\times\\mathbf{c})$ must be if it is to correspond to the $SU(2)$ element $e^V\\,e^U$ by calling on $\\eqref{SU2Example_13}$ and $\\eqref{SU2Example_14}$ to work out the Hermitian and skew-Hermitian parts of the product: the skew-Hermitian part, from $\\eqref{SU2Example_13}$, corresponds to $(\\mathbf{b}\\cdot \\mathbf{c})\\,\\mathbf{a}\\times \\mathbf{b}+(\\mathbf{a}\\cdot \\mathbf{b})\\,\\mathbf{b}\\times \\mathbf{c} + (\\mathbf{a}\\times \\mathbf{b})\\times(\\mathbf{c}\\times \\mathbf{d})$, which, after simplification through standard vector identities, reduces to $\\mathbf{c}\\times\\mathbf{a}$; we must take heed that $\\mathbf{a},\\,\\mathbf{b},\\,\\mathbf{c}$ are all unit vectors. The Hermitian part of the product, from $\\eqref{SU2Example_14}$, corresponds to $(\\mathbf{a}\\cdot \\mathbf{b})\\,(\\mathbf{b}\\cdot \\mathbf{c}) + (\\mathbf{a}\\times \\mathbf{b})\\cdot (\\mathbf{b}\\times \\mathbf{c})$, which simplifies to $\\mathbf{c}\\cdot\\mathbf{a}$. Written out in full, the product formation is: $$\\label{SU2Example_16}\\begin{array}{lcl}\\mathbf{u}\\,\\mathbf{v} &=& \\left(\\mathbf{a}\\cdot \\mathbf{b},\\,\\mathbf{a}\\times\\mathbf{b}\\right)\\,\\left(\\mathbf{b}\\cdot \\mathbf{c},\\,\\mathbf{b}\\times\\mathbf{c}\\right)\\\\&=& \\left((\\mathbf{a}\\cdot \\mathbf{b})\\,(\\mathbf{b}\\cdot \\mathbf{c}) + (\\mathbf{a}\\times \\mathbf{b})\\cdot (\\mathbf{b}\\times \\mathbf{c}),\\,(\\mathbf{b}\\cdot \\mathbf{c})\\,\\mathbf{a}\\times \\mathbf{b}+(\\mathbf{a}\\cdot \\mathbf{b})\\,\\mathbf{b}\\times \\mathbf{c} + (\\mathbf{a}\\times \\mathbf{b})\\times(\\mathbf{c}\\times \\mathbf{d})\\right)\\\\&=&\\left(\\mathbf{c}\\cdot \\mathbf{a},\\,\\mathbf{c}\\times\\mathbf{a}\\right)\\\\&=&\\mathbf{w}\\end{array}$$ but of course this is nothing more than the quaternion pairing $(\\mathbf{c}\\cdot\\mathbf{a},\\, \\mathbf{c}\\times\\mathbf{a})$ that describes the arc $\\mathbf{w}$ in Figure 1.1. So the unit quaternion ($SU(2)$) product $\\mathbf{u}\\,\\mathbf{v}$ is indeed defined by the pair of vectors $\\mathbf{c},\\,\\mathbf{a}$ and by the arc $\\mathbf{w}$ on the unit sphere which closes the spherical triangle formed by $\\mathbf{a},\\,\\mathbf{b},\\,\\mathbf{c}$ and unit quaternion multiplication can be interpreted as a spherical arc addition rule. Unit quaternions, i.e. members", "\\right) \\right]d\\mathbf{x}, \\end{align} since the terms in $\\mathbf{E}\\cdot \\nabla E_i$ and $\\mathbf{e}_i\\cdot\\left( (\\mathbf{E}\\cdot\\nabla)\\mathbf{E}\\right)$ cancel. This means we can write the whole integrand as the divergence of some vector field, and use the divergence theorem: \\begin{align}{} \\int_V \\left[ \\mathbf{E}(\\nabla\\cdot \\mathbf{E})-\\mathbf{E} \\times(\\nabla \\times \\mathbf{E}) \\right]d\\mathbf{x} &= \\sum_i \\mathbf{e}_i \\int_V \\nabla\\cdot\\left[ E_i \\mathbf{E} - \\frac12 \\mathbf{e}_i E^2 \\right]d\\mathbf{x} \\\\ & = \\sum_i \\mathbf{e}_i \\oint_S\\left[ E_i \\mathbf{E} - \\frac12 \\mathbf{e}_i E^2 \\right]\\cdot d\\mathbf{a}. \\tag 1 \\end{align} In terms of conservation law structure, we're essentially done, as we've reduced the rate of change of momentum to a surface term. However, it is crying out for some simplification. In particular, this expression is basis-dependent, but it is so close to being basis independent that it's worth a closer look. The first term, for instance, is simply crying out for a simplification that would look something like $$\\sum_i \\mathbf{e}_i \\oint_S E_i \\mathbf{E}\\cdot d\\mathbf{a} = \\oint_S \\mathbf{E}\\, \\mathbf{E}\\cdot d\\mathbf{a}$$ if we could only make sense of an object like $\\mathbf{E}\\, \\mathbf{E}$. Even better, if we could make sense of such a combination, then it turns out that the seemingly basis-dependent combination that would come up in the second term, $\\sum_i \\mathbf{e}_i\\,\\mathbf{e}_i$, turns out to be basis independent: one can prove that for any two orthonormal bases $\\{\\mathbf{e}_1, \\mathbf{e}_2, \\mathbf{e}_3\\}$ and $\\{\\mathbf{e}_1', \\mathbf{e}_2', \\mathbf{e}_3'\\}$, those combinations are", "that $$\\mathbf{J}_{o}$$ is formulated (as far as I am aware) to represent the axis-angle change in orientation with respect to joint velocities. Thus, the pseudo-inverse of $$\\mathbf{J}_{o}$$ will return proper joint velocities when multiplied with the axis-angle representation of angular change. To accomplish this given $$\\mathbf{e}_{q}$$, one could apply the following: $$\\text{let} \\ r_{i} \\triangleq q_{i}\\text{sin}(\\frac{\\psi}{2}), \\ \\text{for} \\ i \\in \\{1,2,3\\} \\\\ ||\\mathbf{e}_q|| = \\sqrt{r_{1}^{2} + r_{2}^{2} + r_{3}^{2}} \\\\$$ Then, as $$||\\mathbf{q}|| = \\sqrt{q_{1}^{2} + q_{2}^{2} + q_{3}^{2}} = 1$$ by the definition of a quaternion (versor), we can find the rotation angle from the following result: $$||\\mathbf{e}_{q}|| = ||\\mathbf{q} \\cdot \\text{sin}(\\frac{\\psi}{2})|| = | \\text{sin}(\\frac{\\psi}{2})| \\cdot ||\\mathbf{q}|| = | \\text{sin}(\\frac{\\psi}{2}) | \\cdot 1$$ Where we can pull the $$| \\sin({\\frac{\\psi}{2}}) |$$ out of the norm by the definition of the Euclidian norm. Now, we need to deal with the absolute value portion of the $$| \\sin(\\frac{\\psi}{2}) |$$ term. So, choose an arbitrary $$r_{j}, \\text{where} \\ 1 \\le j \\le 3$$, and we can now define: $$\\psi \\triangleq 2 \\sin^{-1}(||\\mathbf{e}_{q}||) \\cdot \\text{sgn}(r_{j})$$ In other words, we assume $$q_{j}$$ is always positive and determine the correct sign of our angle with the direction of $$r_{j}$$. Note that this should not cause any issues as we will be multiplying $$q_j$$ and $$\\psi$$ resulting in the same original sign (we", "(in direction n) $t' = \\gamma \\left(t - \\frac{\\mathbf{r} \\cdot v\\mathbf{n}}{c^2} \\right) \\,,$ $\\mathbf{r}' = \\mathbf{r} + (\\gamma-1)(\\mathbf{r}\\cdot\\mathbf{n})\\mathbf{n} - \\gamma t v\\mathbf{n} \\,.$ In the position transformation, the position vector can be neatly factorized using the dot and dyadic products of vectors. Although dyadic tensors are an archaic formalism not often used in this context, the notation is well-suited to the equations here, which are $t' = \\gamma t - \\gamma\\frac{v\\mathbf{n} \\cdot \\mathbf{r}}{c^2} \\,,$ $\\mathbf{r}' = - \\gamma t v\\mathbf{n} + ( \\mathbf{I} + (\\gamma-1)\\mathbf{n}\\mathbf{n})\\cdot\\mathbf{r} \\,.$ where I is the unit dyadic and nn is the dyadic product of n with itself.[nb 2] The identities I · r = r and nn · r = n(n · r) follow from the definitions of the dot and dyadic products. The full set of equations are $ct' = \\gamma c t - \\gamma \\frac{v n_x}{c}x - \\gamma \\frac{v n_y}{c}y - \\gamma \\frac{v n_z}{c} z \\,.$ $x' = - \\gamma \\frac{v n_x}{c} c t + (1 + (\\gamma-1) n_x^2 )x + (\\gamma-1) n_x n_y y + (\\gamma-1) n_x n_z z \\,,$ $y' = - \\gamma \\frac{v n_y}{c} ct + (\\gamma-1) n_y n_x x + (1 + (\\gamma-1) n_y^2 ) y + (\\gamma-1) n_y n_z z \\,,$ $z' = - \\gamma \\frac{v n_z}{c} ct + (\\gamma-1) n_z n_x x + (\\gamma-1) n_z", "\\cdot \\mathbf{v}&=(1,1,2) \\cdot (-1,3,0) \\\\ &=1*(-1)+1*3+2*0 \\\\ &=2. \\end{align*} The angle between Recall the we had defined the dot product so the that angle between the vectors, $$\\theta$$ can be found as \\begin{align*} \\cos(\\theta)=\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| * || \\mathbf{v}||}. \\end{align*} We found most of the required information, but we still need to find $$||\\mathbf{v}||$$ is we want to calculate this angle. Therefore, we find that \\begin{align*} ||\\mathbf{v}||&=\\sqrt{\\mathbf{v} \\cdot \\mathbf{v}} \\\\ &=\\sqrt{(-1,3,0) \\cdot (-1,3,0)} \\\\ &=\\sqrt{(-1)(-1)+3*3+0*0} \\\\ &=\\sqrt{10}. \\end{align*} Using the information we already have, we then find that \\begin{align*} \\cos(\\theta)&=\\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| * || \\mathbf{v}||} \\\\ &=\\frac{2}{\\sqrt{6}*\\sqrt{10}} \\\\ &=\\frac{2}{\\sqrt{60}} \\\\ &=\\frac{1}{\\sqrt{15}}. \\end{align*} Hence, we find that the angle between the vectors is $$\\cos^{-1}(\\frac{1}{\\sqrt{15}})$$. Vectors in $$C[0,1]$$. Now that we have worked through an example in $$\\mathbf{R}^{2}$$ we will look work with an example using continuous function on the interval $$[0,1]$$. We know that $$C[0,1]$$ is a vector space under the usual addition and scalar multiplication, but we still need an inner product. As such, we will define our inner product for $$C[0,1]$$ as, for $$f,g \\in P_{1}$$, \\begin{align*} \\langle f, g \\rangle =\\int_{0}^{1}f(x)g(x)dx. \\end{align*} Before continuing, we should do a quick check that this inner product does satisfy the definition of inner product. For $$f(x),g(x),h(x) \\in C[0,1]$$ and $$c \\in \\mathbb{R}$$, we note that 1. $$\\int_{0}^{1}f(x)g(x)dx=\\int_{0}^{1}g(x)f(x)dx$$, that is", "can produce a linearly independant set of equations by choosing linearly independant tangent vectors. Choosing vectors that will break $\\tau_b$ into components $\\tau_{bx}$ and $\\tau_{by}$: $\\mathbf{t_x} = \\begin{pmatrix}1 \\\\ 0 \\\\ \\frac{\\partial b}{\\partial x}\\end{pmatrix}$ $\\mathbf{t_y} = \\begin{pmatrix}0 \\\\ 1 \\\\ \\frac{\\partial b}{\\partial y}\\end{pmatrix}$ This gives us the system of equations: $\\beta^2 \\frac{\\mathbf{t_x}}{||\\mathbf{t_x}||} \\cdot \\mathbf{v} = \\frac{\\mathbf{t_x}}{||\\mathbf{t_x}||} \\cdot (\\mathbf{T} \\frac{\\mathbf{n_b}}{||\\mathbf{n_b}||}) = \\tau_bx$ $\\beta^2 \\frac{\\mathbf{t_y}}{||\\mathbf{t_y}||} \\cdot \\mathbf{v} = \\frac{\\mathbf{t_y}}{||\\mathbf{t_y}||} \\cdot (\\mathbf{T} \\frac{\\mathbf{n_b}}{||\\mathbf{n_b}||}) = \\tau_by$ Expanding the vector and tensor products: $\\beta^2 ||\\mathbf{n_b}|| (u + w\\frac{\\partial b}{\\partial x}) = T_{xx}\\frac{\\partial b}{\\partial x} + T_{xy}\\frac{\\partial b}{\\partial y} - T_{xz} - T_{zz}\\frac{\\partial b}{\\partial x}$ $\\beta^2 ||\\mathbf{n_b}|| (u + w\\frac{\\partial b}{\\partial y}) = T_{yx}\\frac{\\partial b}{\\partial x} + T_{yy}\\frac{\\partial b}{\\partial y} - T_{yz} - T_{zz}\\frac{\\partial b}{\\partial y}$ We make the assumption here that the vertical velocity component at the base is much larger than the horizontal component, which is valid as long as bedrock gradients are not too large (??) and melt rates are negligable. Hence, we drop the w term in the right-hand-side and substitute deviatoric stresses for total stresses, observing that the pressure at the base is $\\rho_i g H$: $\\beta^2 u ||\\mathbf{n_b}|| = (2T'_{xx} + T'_{yy} - \\rho_i g H)\\frac{\\partial b}{\\partial x} + T'_{xy}\\frac{\\partial b}{\\partial y} - T'_{xz} + \\rho_i g H \\frac{\\partial b}{\\partial x}$ $\\beta^2 v ||\\mathbf{n_b}|| = (2T'_{yy} + T'_{xx}", "a fleshing out of the hint by @GerryMyerson. We assume that ${\\bf a}\\ne {\\bf 0}$. Let ${\\bf d} = {\\bf b}-{\\bf c}$. Then ${\\bf a}\\times{\\bf d} = {\\bf 0}$ and ${\\bf a}\\cdot{\\bf d} = 0$, and so $\\|{\\bf a}\\| \\|{\\bf d}\\|\\sin\\theta = 0$ and $\\|{\\bf a}\\| \\|{\\bf d}\\|\\cos\\theta = 0$, where $\\theta$ is the angle between ${\\bf a}$ and ${\\bf d}$. Since there is no $\\theta$ for which $\\sin\\theta = \\cos\\theta = 0$, we can conclude that ${\\bf d} = {\\bf 0}$, and so ${\\bf b} = {\\bf c}$. - is $a \\times (b-c) = a \\times b - a \\times c$ ?? I thought the equation is correct for addition... i guess you can write b+(-c) – ASROMA Oct 14 '12 at 5:23 Hint: Consider the identity $$\\|\\mathbf{x} \\times \\mathbf{y}\\|^2 + (\\mathbf{x}\\cdot \\mathbf{y})^2 = 2\\|\\mathbf{x}\\|^2\\|\\mathbf{y}\\|^2$$ In particular, this is a special case of the angle identities that Gerry was mentioning. - but how does this help? i have to prove that vectors are equal, not their scalars – ASROMA Oct 14 '12 at 4:50 Can you think of some way to apply this identity to your problem? Try proving that $\\mathbf{b - c} = \\mathbf{0}$. – EuYu Oct 14 '12 at 4:51 there is a gray area when it comes to angles, i have to use different" ]

[ "a \\to b, b \\to c, c \\to a$$ This gives $${\\bf A} = \\frac{{\\bf b}\\times {\\bf c}}{{\\bf a}\\cdot({\\bf b}\\times {\\bf c})} \\quad {\\bf B} = \\frac{{\\bf c}\\times {\\bf a}}{{\\bf b}\\cdot({\\bf c}\\times {\\bf a})} \\quad {\\bf C} = \\frac{{\\bf a}\\times {\\bf b}}{{\\bf c}\\cdot({\\bf a}\\times {\\bf b})}$$ and $${\\bf a} = \\frac{{\\bf B}\\times {\\bf C}}{{\\bf A}\\cdot({\\bf B}\\times {\\bf C})} \\quad {\\bf b} = \\frac{{\\bf C}\\times {\\bf A}}{{\\bf B}\\cdot({\\bf C}\\times {\\bf A})} \\quad {\\bf c} = \\frac{{\\bf A}\\times {\\bf B}}{{\\bf C}\\cdot({\\bf A}\\times {\\bf B})}$$ [..] the \"crystallographer's\" definition, comes from defining the reciprocal lattice to be $e^{2 \\pi i\\mathbf{K}\\cdot\\mathbf{R}}=1$ which changes the definitions of the reciprocal lattice vectors to be $\\mathbf{b_{1}}=\\frac{\\mathbf{a_{2}} \\times \\mathbf{a_{3}}}{\\mathbf{a_{1}} \\cdot (\\mathbf{a_{2}} \\times \\mathbf{a_{3}})}$ and so on for the other vectors. The crystallographer's definition has the advantage that the definition of $\\mathbf{b_{1}}$ is just the reciprocal magnitude of $\\mathbf{a_{1}}$ in the direction of $\\mathbf{a_{2}} \\times \\mathbf{a_{3}}$, dropping the factor of $2 \\pi$. Checking the magnitude: $$\\lVert{\\bf A}\\rVert = \\frac{\\lVert {\\bf b} \\times{\\bf c}\\rVert}{\\lVert{\\bf a}\\rVert\\lVert{\\bf b}\\times {\\bf c}\\rVert\\cos\\angle({\\bf a}, {\\bf b} \\times{\\bf c})} = \\frac{1}{\\lVert{\\bf a}\\rVert ({\\bf e_a} \\cdot {\\bf e}_{{\\bf b} \\times{\\bf c}})}$$ Solving the question: Using the \"bac-cab\" rule ${\\bf a}\\times({\\bf b}\\times{\\bf c}) = {\\bf b}({\\bf a}\\cdot {\\bf c}) - {\\bf c}({\\bf a}\\cdot {\\bf b})$ we go for the nominator of ${\\bf B}\\times{\\bf C}$: $$({\\bf c}\\times", "# Orthogonal Nonzero Vectors Are Linearly Independent ## Problem 591 Let $S=\\{\\mathbf{v}_1, \\mathbf{v}_2, \\dots, \\mathbf{v}_k\\}$ be a set of nonzero vectors in $\\R^n$. Suppose that $S$ is an orthogonal set. (a) Show that $S$ is linearly independent. (b) If $k=n$, then prove that $S$ is a basis for $\\R^n$. ## Proof. ### (a) Show that $S$ is linearly independent. Consider the linear combination $c_1\\mathbf{v}_1+c_2\\mathbf{v}_2+\\cdots +c_k \\mathbf{v}_k=\\mathbf{0}.$ Our goal is to show that $c_1=c_2=\\cdots=c_k=0$. We compute the dot product of $\\mathbf{v}_i$ and the above linear combination for each $i=1, 2, \\dots, k$: \\begin{align*} 0&=\\mathbf{v}_i\\cdot \\mathbf{0}\\\\ &=\\mathbf{v}_i \\cdot (c_1\\mathbf{v}_1+c_2\\mathbf{v}_2+\\cdots +c_k \\mathbf{v}_k)\\\\ &=c_1\\mathbf{v}_i \\cdot \\mathbf{v}_1+c_2\\mathbf{v}_i \\cdot \\mathbf{v}_2+\\cdots +c_k \\mathbf{v}_i \\cdot\\mathbf{v}_k. \\end{align*} As $S$ is an orthogonal set, we have $\\mathbf{v}_i\\cdot \\mathbf{v}_j=0$ if $i\\neq j$. Hence all terms but the $i$-th one are zero, and thus we have $0=c_i\\mathbf{v}_i\\cdot \\mathbf{v}_i=c_i \\|\\mathbf{v}_i\\|^2.$ Since $\\mathbf{v}_i$ is a nonzero vector, its length $\\|\\mathbf{v}_i\\|$ is nonzero. It follows that $c_i=0$. As this computation holds for every $i=1, 2, \\dots, k$, we conclude that $c_1=c_2=\\cdots=c_k=0$. Hence the set $S$ is linearly independent. ### (b) If $k=n$, then prove that $S$ is a basis for $\\R^n$. Suppose that $k=n$. Then by part (a), the set $S$ consists of $n$ linearly independent vectors in the dimension $n$ vector space $\\R^n$. Thus, $S$ is also a spanning set of $\\R^n$,", "f_1(\\mathbf{v}) + f_2(\\mathbf{v})$ then $\\frac{\\partial f}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial f_1}{\\partial \\mathbf{v}} + \\frac{\\partial f_2}{\\partial \\mathbf{v}}\\right)\\cdot\\mathbf{u}$ 2. If $f(\\mathbf{v}) = f_1(\\mathbf{v})~ f_2(\\mathbf{v})$ then $\\frac{\\partial f}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial f_1}{\\partial \\mathbf{v}}\\cdot\\mathbf{u}\\right)~f_2(\\mathbf{v}) + f_1(\\mathbf{v})~\\left(\\frac{\\partial f_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} \\right)$ 3. If $f(\\mathbf{v}) = f_1(f_2(\\mathbf{v}))$ then $\\frac{\\partial f}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\frac{\\partial f_1}{\\partial f_2}~\\frac{\\partial f_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u}$ ### Derivatives of vector valued functions of vectors Let $\\mathbf{f}(\\mathbf{v})$ be a vector valued function of the vector $\\mathbf{v}$. Then the derivative of $\\mathbf{f}(\\mathbf{v})$ with respect to $\\mathbf{v}$ (or at $\\mathbf{v}$) in the direction $\\mathbf{u}$ is the second-order tensor defined as $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = D\\mathbf{f}(\\mathbf{v})[\\mathbf{u}] = \\left[\\frac{d }{d \\alpha}~\\mathbf{f}(\\mathbf{v} + \\alpha~\\mathbf{u})\\right]_{\\alpha = 0}$ for all vectors $\\mathbf{u}$. Properties: 1. If $\\mathbf{f}(\\mathbf{v}) = \\mathbf{f}_1(\\mathbf{v}) + \\mathbf{f}_2(\\mathbf{v})$ then $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial \\mathbf{f}_1}{\\partial \\mathbf{v}} + \\frac{\\partial \\mathbf{f}_2}{\\partial \\mathbf{v}}\\right)\\cdot\\mathbf{u}$ 2. If $\\mathbf{f}(\\mathbf{v}) = \\mathbf{f}_1(\\mathbf{v})\\times\\mathbf{f}_2(\\mathbf{v})$ then $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\left(\\frac{\\partial \\mathbf{f}_1}{\\partial \\mathbf{v}}\\cdot\\mathbf{u}\\right)\\times\\mathbf{f}_2(\\mathbf{v}) + \\mathbf{f}_1(\\mathbf{v})\\times\\left(\\frac{\\partial \\mathbf{f}_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} \\right)$ 3. If $\\mathbf{f}(\\mathbf{v}) = \\mathbf{f}_1(\\mathbf{f}_2(\\mathbf{v}))$ then $\\frac{\\partial \\mathbf{f}}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} = \\frac{\\partial \\mathbf{f}_1}{\\partial \\mathbf{f}_2}\\cdot\\left(\\frac{\\partial \\mathbf{f}_2}{\\partial \\mathbf{v}}\\cdot\\mathbf{u} \\right)$ ### Derivatives of scalar valued functions of second-order tensors Let $f(\\boldsymbol{S})$ be a real valued function of the second order tensor $\\boldsymbol{S}$. Then the derivative of $f(\\boldsymbol{S})$ with respect to $\\boldsymbol{S}$ (or at $\\boldsymbol{S}$) in the direction $\\boldsymbol{T}$ is the second order tensor defined as $\\frac{\\partial f}{\\partial \\boldsymbol{S}}:\\boldsymbol{T} = Df(\\boldsymbol{S})[\\boldsymbol{T}] = \\left[\\frac{d }{d \\alpha}~f(\\boldsymbol{S} + \\alpha\\boldsymbol{T})\\right]_{\\alpha =", "# Multivariable Calculus: Problem Set 6 ## 1. Equality of the mixed derivatives Let $f(x, y) = \\frac{xy(x^2-y^2)}{x^2 + y^2},$ away from $(0,0)$ and zero otherwise. Show that $\\pd{^2 f}{y\\partial x} \\neq \\pd{^2 f}{x\\partial y}$ at the origin. Spoiler ## 2. Vector identities ### Part a (sum proof) For vector fields $\\mathbf{F}$, $\\mathbf{G}$, show that $\\nabla \\cdot (\\mathbf{F} \\times \\mathbf{G}) = \\mathbf{G} \\cdot (\\nabla \\times \\mathbf{F}) - \\mathbf{F}\\cdot (\\nabla \\times \\mathbf{G}).$ Spoiler ### Part a (index notation) Prove the same identity using index notation. Spoiler ### Part b Show also that $\\nabla \\times (\\mathbf{F} \\times \\mathbf{G}) = \\mathbf{F}(\\nabla \\cdot \\mathbf{G}) - \\mathbf{G}(\\nabla \\cdot \\mathbf{F}) + (\\mathbf{G} \\cdot \\nabla)\\mathbf{F} - (\\mathbf{F} \\cdot \\nabla)\\mathbf{G}.$ Spoiler ### Part b (index notation) Prove the same identity using index notation. Spoiler ## 3. Change of coordinates Let $\\phi(x,y,z) = y^2 - xz$ and $\\mathbf{f}(x,y,z) = (z^2, x^2, y^2)$. a) Find $\\nabla \\phi$, $\\nabla \\times \\mathbf{f}$, and $\\nabla \\cdot \\mathbf{f}$. Spoiler b) For the orthogonal basis $\\mathbf{e}_1 = [0, -1, 0]$, $\\mathbf{e}_2 = [1, 0, -1]/\\sqrt{2}$, $\\mb{e}_3 = [1,0,1]/\\sqrt{2}$, create new coordinates $X$, $Y$, $Z$, such that $X \\mb{e}_1 + Y\\mb{e}_2 + Z\\mb{e}_3 = x\\mb{i} + y\\mb{j} + z\\mb{k}.$ Spoiler c) Determine $x$, $y$, $z$ in terms of $X$, $Y$, $Z$. Find also $\\Phi$, $F_1$, $F_2$, and $F_3$ such that $\\Phi(X,Y,Z) = \\phi(x,y,z)$ and", "a_k\\mathbf v_k \\wedge a_1, a_2, \\dots, a_k \\in F\\}$ • This set of vectors is a linearly independent set if the only linear combination of these vectors that produces the zero vector is the trivial one. I.e. $a_1\\mathbf v_1 + a_2\\mathbf v_2 + \\cdots + a_k\\mathbf v_k = 0 \\implies a_1=a_2=\\cdots=a_k=0$. • A set of vectors is a linearly dependent set if it is not a linearly independent set. Now let's show that a linearly dependent set has at least one vector which is a linear combination of the others. Let $\\mathbf a,\\mathbf b,\\mathbf c \\in \\Bbb R^3$ be a set of linearly dependent vectors. Then by definition, the equation $$x\\mathbf a+y\\mathbf b+z\\mathbf c=\\mathbf 0$$ for scalars $x,y,z$ has more than one solution ($x=y=z=0$ is definitely a solution, but it's not the only one). Thus at least one of the scalars $x,y,z$ are nonzero. WLOG let's say it's $x$. Then we can rearrange this equation by subtracting all of the vectors except $x\\mathbf a$ on both sides. $$x\\mathbf a=-y\\mathbf b-z\\mathbf c$$ Now because $x \\ne 0$, we can divide it on both sides to get $$\\mathbf a=-\\frac yx\\mathbf b -\\frac zx\\mathbf c$$ Thus $\\mathbf a$ is a linear combination of $\\mathbf b$ and $\\mathbf c$. You can see why this would fail in the linearly independent case --", "all linear combinations of linear combinations of $$\\mathbf{v}_{1},\\cdots,\\mathbf{v}_{k}$$ is called the linear span of S and is denoted by $$L(S)$$, namely $L(S)=\\{c_{1}\\mathbf{v}_{1}+\\cdots+c_{k}\\mathbf{v}_{k}|\\ c_{1}\\in\\mathbb{R},\\cdots,c_{k}\\in\\mathbb{R}\\}.$ Let $$S=\\{\\mathbf{v}\\}$$, and $$\\mathbf{v}\\neq O$$ is a vector in $$\\mathbb{R}^{3}$$. Then $$L(S)=\\{c\\mathbf{v}|\\ c\\in\\mathbb{R}\\}$$ is the line passing through the origin in the direction of $$\\mathbf{v}.$$ Let u and v be two non-parallel vectors in space. Then $$L(S)=\\{c_{1}\\mathbf{u}+c_{2}\\mathbf{v}|\\ c_{1}\\in\\mathbb{R},\\ c_{2}\\in\\mathbb{R}\\}$$ is the plane passing through the origin generated by u and v. Every set $$S=\\{\\mathbf{v}_{1},\\cdots,\\mathbf{v}_{k}\\}$$ spans the zero vector. Or $$\\mathbf{O}\\in L(S)$$. Def: A set S = {A,, . . . ,Ak} of vectors in , is said to span X uniquely if it spans X and if Theorem: A set S spans every vector in L(S) uniquely if and only if S spans the zero vector uniquely. ## Linear Independence A set $$S=\\{\\mathbf{v}_{1},\\cdots,\\mathbf{v}_{k}\\}$$ which spans the zero vector uniquely is said to be a linearly independent set of vectors. Otherwise, S is called linearly dependent. ## Change of Basis Use of the coordinate matrix of a vector $$\\mathbf{v}$$ relative to an ordered basis $$\\mathbb{\\mathcal{B}}$$ $\\mathbf{v}=\\left[\\begin{array}{c} x_{1}\\\\ \\vdots\\\\ x_{n} \\end{array}\\right],$ is often more convenient than writing it as an n–tuple $$(x_{1},\\cdots,x_{n})$$ of coordinates. We often write the coordinate matrix as $\\left[\\mathbf{v}\\right]_{\\mathbb{\\mathcal{B}}}$ to indicate its dependence on the basis. If $$\\mathbb{\\mathcal{B}}$$ is the standard basis, then we", "derivation the book gave is ambiguous. Kanos's answer provides one alternative, but the vector potential is not needed. What you really need is to express the relationship using notation that is a little more detailed, but unambiguous. We split the $$r$$ from the Biot-Savart law to get two independent variables - one that is an integration variable, the other that is not. $$\\mathbf{B}(\\mathbf{r}) = \\frac{\\mu_0}{4\\pi} \\int \\mathbf{J}(\\mathbf{r}')\\times \\frac{\\mathbf{r}-\\mathbf{r}'}{\\left|\\mathbf{r}-\\mathbf{r}'\\right|^3} \\operatorname{d}^3 r'.$$ Notice: $$\\mathbf{J}(\\mathbf{r}')$$ is not a function of $$\\mathbf{r}$$, so when you try to take any derivatives with respect to any $$\\mathbf{r}$$ coordinate you'll get zero. So, we get: \\begin{align} \\nabla\\cdot \\mathbf{B}(\\mathbf{r}) & = \\frac{\\mu_0}{4\\pi} \\int \\nabla\\cdot \\left[\\mathbf{J}(\\mathbf{r}')\\times \\frac{\\mathbf{r}-\\mathbf{r}'}{\\left|\\mathbf{r}-\\mathbf{r}'\\right|^3} \\right] \\operatorname{d}^3 r'. \\end{align} Now, apply the vector calculus cross product identity, $$\\nabla\\cdot(\\mathbf{A}\\times\\mathbf{B}) = (\\nabla\\times \\mathbf{A})\\cdot \\mathbf{B} - \\mathbf{A}\\cdot(\\nabla\\times\\mathbf{B})$$ with $$\\mathbf{A}=\\mathbf{J}$$ and $$\\mathbf{B}=\\frac{\\mathbf{r}-\\mathbf{r}'}{\\left|\\mathbf{r}-\\mathbf{r}'\\right|^3}$$ to get \\begin{align} \\nabla\\cdot \\mathbf{B}(\\mathbf{r}) & = \\frac{\\mu_0}{4\\pi} \\int \\left[(\\nabla\\times\\mathbf{J}(\\mathbf{r}'))\\cdot \\frac{\\mathbf{r}-\\mathbf{r}'}{\\left|\\mathbf{r}-\\mathbf{r}'\\right|^3} - \\mathbf{J}(\\mathbf{r}')\\cdot \\left(\\nabla\\times\\frac{\\mathbf{r}-\\mathbf{r}'}{\\left|\\mathbf{r}-\\mathbf{r}'\\right|^3} \\right)\\right] \\operatorname{d}^3 r'. \\end{align} The first cross product vanishes because $$\\mathbf{J}(\\mathbf{r}')$$ is not a function of $$\\mathbf{r}$$ and $$\\nabla$$ is a derivative in the $$\\mathbf{r}$$ coordinates. That the second cross product vanishes can be shown using a little algebra, or some tricks discussed in other answers. The curl of a vector field $$\\vec v$$ $$\\nabla \\times \\vec v$$ measures the rotational motion of the vector field. If the thumb is the model for", "y[1], \\cdots )$ to the shifted-one-place-to-the-right version of it, namely $( \\cdots, y[-2], y[-1], y[0], \\cdots )$. In linear algebra, we know that a matrix can be used to map one vector to another vector provided the operation is linear, and in fact, shifting a vector is indeed a linear operation. In abstract terms, there exists a linear operator $S \\colon \\mathbb{R}^\\infty \\rightarrow \\mathbb{R}^\\infty$ that shifts a sequence one place to the right: $S(\\ ( \\cdots, y[-1], y[0], y[1], \\cdots ) \\ ) = ( \\cdots, y[-2], y[-1], y[0], \\cdots )$. Letting $\\mathbf{x}$ and $\\mathbf{y}$ denote the vectors $( \\cdots, x[-1], x[0], x[1], \\cdots )$ and $( \\cdots, y[-1], y[0], y[1], \\cdots )$ respectively, we can rigorously write our system as $\\mathbf{y} = \\frac12 S \\mathbf{y} + \\mathbf{x}$. This is precisely the same as what we did when we said the first row of the table is equal to the second row of the table plus the third row of the table. Our natural instinct is to collect like terms: if $I$ denotes the identity operator (think of a matrix with ones on the diagonal), so that $I\\mathbf{y} = \\mathbf{y}$, then $\\mathbf{y} = \\frac12 S \\mathbf{y} + \\mathbf{x}$ is equivalent to $\\left( I - \\frac12 S \\right) \\mathbf{y} = \\mathbf{x}$, that is, $\\mathbf{y} = \\left( I - \\frac12", "## Proof That the Volume Integral of the Curl of a Vector Normal to a Surface Everwhere is Zero Theorem If $\\mathbf{F}$ is a vector always normal to a surface $S$ enclosing a volume $V$ then $\\int \\int \\int_V \\mathbf{\\nabla} \\times \\mathbf{F} \\: dV=0$ . Proof Let $\\mathbf{a}$ be a constant vector. The Divergence Theorem for the field $\\mathbf{F} \\times \\mathbf{a}$ states $\\int \\int \\int_V \\mathbf{\\nabla} \\cdot ( \\mathbf{F} \\times \\mathbf{a}) dV = \\int \\int_S ( \\mathbf{F} \\times \\mathbf{a}) \\cdot \\mathbf{n} \\: dS$ Since $\\mathbf{a}$ is a constant vector, $\\mathbf{\\nabla} \\cdot ( \\mathbf{F} \\times \\mathbf{a}) = \\mathbf{a} \\cdot ( \\mathbf{\\nabla} \\times \\mathbf{f}) - \\mathbf{F} \\cdot ( \\mathbf{\\nabla} \\times \\mathbf{a}) = \\mathbf{a} \\cdot ( \\mathbf{\\nabla} \\times \\mathbf{f})$ and $(\\mathbf{F} \\times \\mathbf{a}) \\cdot \\mathbf{n} = \\mathbf{F} \\cdot (\\mathbf{a} \\times \\mathbf{n} ) =(\\mathbf{a} \\times \\mathbf{n} ) \\cdot \\mathbf{F} = \\mathbf{a} \\cdot ( \\mathbf{n} \\times \\mathbf{F} )$ Hence $\\int \\int \\int_V \\mathbf{a} \\cdot ( \\mathbf{\\nabla} \\times \\mathbf{F}) \\: dV = \\int \\int_S \\mathbf{a} \\cdot ( \\mathbf{n} \\times \\mathbf{F}) \\: dS$ or $\\mathbf{a} \\cdot \\int \\int \\int_V ( \\mathbf{\\nabla} \\times \\mathbf{F}) \\: dV = \\mathbf{a} \\cdot \\int \\int_S ( \\mathbf{n} \\times \\mathbf{F}) \\: dS$ Hence $\\int \\int \\int_V ( \\mathbf{\\nabla} \\times \\mathbf{F}) \\: dV = \\int \\int_S ( \\mathbf{n} \\times \\mathbf{F}) \\: dS$ $\\mathbf{F}$ is normal to $S$ so $\\mathbf{F} \\times \\mathbf{n} =0$", "do. It seems that part (b) in the question is answered by part (c) in the solution and that part (b) of the solution is completely extraneous. It seems to answer a different question altogether. Addendum 2 in response to OP's questions The fact that $\\mathbf{u}\\notin \\mathrm{span}(\\mathbf{v},\\ \\mathbf{w})$ comes simply from the fact that $\\{\\mathbf{u},\\ \\mathbf{v},\\ \\mathbf{w}\\}$ is a linearly independent set. Certainly you can recast the argument so that contradiction is not used, and here is one example of a proof using the contrapositive instead of contradiction. This proof makes it clear that the result follows essentially from the definition of linear independence. Theorem: Suppose that $\\{\\mathbf{v}_1,\\ \\cdots,\\ \\mathbf{v}_n\\}$ is a linearly independent set of vectors. Then $\\mathbf{v}_1 \\notin \\mathrm{span}(\\mathbf{v}_2,\\ \\cdots,\\ \\mathbf{v}_n)$. Proof: The contrapositive of the statement is: Suppose that $\\mathbf{v}_1 \\in \\mathrm{span}(\\mathbf{v}_2,\\ \\cdots,\\ \\mathbf{v}_n)$. Then $\\{\\mathbf{v}_1,\\ \\cdots,\\ \\mathbf{v}_n\\}$ is linearly dependent. But this is essentially trivial. The fact that $\\mathbf{v}_1 \\in \\mathrm{span}(\\mathbf{v}_2,\\ \\cdots,\\ \\mathbf{v}_n)$ means there exists some linear combination of $\\{\\mathbf{v}_2,\\ \\cdots,\\ \\mathbf{v}_n\\}$ adding up to $\\mathbf{v}_1$, i.e. there exists scalars $\\{c_i\\}$ such that $$\\mathbf{v}_1 = c_2\\mathbf{v}_2 + \\cdots + c_n\\mathbf{v}_n$$ But then we also have $$-\\mathbf{v}_1 + c_2\\mathbf{v}_2 + \\cdots + c_n\\mathbf{v}_n = \\mathbf{0}$$ which is a non-trivial linear combination of $\\{\\mathbf{v}_1,\\ \\cdots,\\ \\mathbf{v}_n\\}$ adding up to the zero vector. By definition, this", "# Prove that any non-zero orthogonal vectors in $\\mathbb R^{n}$ form a basis for $\\mathbb R^{n}$ Proposition. Any $$n$$ non-zero orthogonal vectors in $$\\mathbb R^{n}$$ form a basis for $$\\mathbb R^{n}$$. My attempt: Suppose we have non-zero orthogonal vectors $$\\bf (v_{1},v_{2},\\cdots,v_{n})$$ We know that $$n$$ linearly independent vectors in $$\\mathbb R^{n}$$ span $$\\mathbb R^{n}$$. or in other words $$n$$ linearly independent vectors in $$\\mathbb R^{n}$$ form a basis. Therefore, if we show that $$\\bf (v_{1},v_{2},\\cdots,v_{n})$$ are linearly independent, we automatically show that they form a basis. We will prove linear independence by contradiction. Suppose $$\\bf (v_{1},v_{2},\\cdots,v_{n})$$ are linearly dependent. We have $$k_1\\mathbf{v_{1}} + k_2\\mathbf{v_{2}}+ \\cdots + k_j\\mathbf{v_{j}} + \\cdots + k_n\\mathbf{v_{n}} = \\mathbf O$$ Where at least one scalar, call it $$k_{j}$$, is not zero. Premultiply both sides by $$\\bf v_{j}^{T}$$ $$\\mathbf{v_{j}}^{T}\\bigl(k_1\\mathbf{v_{1}} + k_2\\mathbf{v_{2}}+ \\cdots + k_j\\mathbf{v_{j}} + \\cdots + k_n\\mathbf{v_{n}}\\bigr) = \\mathbf {v_{j}}^{T}\\mathbf O \\implies$$ $$k_1\\mathbf{v_{j}}^{T}\\mathbf{v_{1}} + k_2\\mathbf{v_{j}}^{T}\\mathbf{v_{2}}+ \\cdots + k_j\\mathbf{v_{j}}^{T}\\mathbf{v_{j}} + \\cdots + k_n\\mathbf{v_{j}}^{T}\\mathbf{v_{n}} = 0\\implies$$ Because our vectors are orthogonal, then $$a≠b \\implies \\mathbf {v_{a}}^{T}\\mathbf {v_{b}} = 0$$ Thus, we have $$0 + 0 + \\cdots k_{j}||\\mathbf{v_{j}}||^{2} \\cdots 0 = k_{j}||\\mathbf{v_{j}}||^{2} = 0$$ We know that $$||\\mathbf{v_{j}}||^{2} > 0$$ (it doesn't equal to zero because task specifies our vectors must be non-zero) We also know that $$k ≠ 0$$ In this case, $$k_{j}||\\mathbf{v_{j}}||^{2}$$ cannot equal", "reciprocal frame vectors are \\begin{aligned}\\mathbf{x}^a &= \\mathbf{x}_b \\cdot \\frac{1}{{ \\mathbf{x}_a \\wedge \\mathbf{x}_b }} = - \\frac{\\gamma_1 \\exp\\left( { \\gamma_1 \\gamma_0 b } \\right)}{\\sinh(b)} \\\\ \\mathbf{x}^b &= \\mathbf{x}_a \\cdot \\frac{1}{{ \\mathbf{x}_b \\wedge \\mathbf{x}_a }} = - \\frac{\\gamma_1}{\\sinh(b)},\\end{aligned} which satisfy $\\mathbf{x}^a \\cdot \\mathbf{x}_a = \\mathbf{x}^b \\cdot \\mathbf{x}_b = 1$, and $\\mathbf{x}^a \\cdot \\mathbf{x}_b = \\mathbf{x}^b \\cdot \\mathbf{x}_a = 0$, which can be shown explicitly with relative ease using scalar selection operations. The vector derivative is $$\\partial = - \\frac{\\gamma_1}{\\sinh(b)} \\left( { \\frac{\\partial {}}{\\partial {b}} + \\exp\\left( { \\gamma_1 \\gamma_0 b } \\right) \\frac{\\partial {}}{\\partial {a}} } \\right).$$ This is enough to explicitly express the left and right hand sides of the fundamental theorem identity $$\\int d^2 \\mathbf{x} \\partial F=\\gamma_0 \\int da db \\left( { \\frac{\\partial {}}{\\partial {b}} + \\exp\\left( { \\gamma_1 \\gamma_0 b } \\right) \\frac{\\partial {}}{\\partial {a}} } \\right) F,$$ and $$-\\int d\\mathbf{x}_b {\\left.{{F}}\\right\\vert}_{{\\Delta a}}+\\int d\\mathbf{x}_a {\\left.{{F}}\\right\\vert}_{{\\Delta b}}=\\gamma_0 \\int db \\exp\\left( { \\gamma_1 \\gamma_0 b } \\right) {\\left.{{F}}\\right\\vert}_{{\\Delta a}}+ \\gamma_0 \\int da {\\left.{{F}}\\right\\vert}_{{\\Delta b}}.$$ As expected by the theorem, these are clearly identical. According to the book Geometric Algebra for Physicists, by Doran and Lasenby, \"the [fundamental] theorem is still valid in a Lorentzian space\" (page 200, in chapter 6). The next section, \"Embedded surfaces and vector manifolds\" (6.5) might also be of interest to you. Doesn't directly express", "The set $\\left{\\mathbf{v}_1, \\ldots, \\mathbf{v}_p\\right}$ is said to be linearly dependent if (2.2) has a solution where not all of $c_1, \\ldots, c_p$ are zero (a nontrivial solution). In such a case, (2.2) with at least one $c_i$ nonzero gives a linear dependence relation among $\\left{\\mathbf{v}_1, \\ldots, \\mathbf{v}_p\\right}$. An arbitrary set $S \\subseteq V$ is said to be linearly independent if everry finitè subsét of $S$ is linearly indēpendent. Thẻ set $S$ is linearly dependent, if it is not linearly independent. We will see in this section that any n-dimensional vector space over $\\mathbb{F}$ “works the same” as $\\mathbb{F}^n$, which simplifies the study of such vector spaces tremendously. To make this idea more precise, we have to discuss coordinate systems. We start with the following result. Theorem 2.5.1 Let $\\mathcal{B}=\\left{\\mathbf{v}_1, \\ldots, \\mathbf{v}_n\\right}$ be a basis for a vector space $V$ over $\\mathbb{F}$. Then for each $\\mathbf{v} \\in V$ there exists unique $c_1, \\ldots, c_n \\in \\mathbb{F}$ so that $$\\mathbf{v}=c_1 \\mathbf{v}_1+\\cdots+c_n \\mathbf{v}_n .$$ Proof. Let $\\mathbf{v} \\in V$. As $\\operatorname{Span} \\mathcal{B}=V$, we have that $\\mathbf{v}=c_1 \\mathbf{v}_1+\\cdots+c_n \\mathbf{v}_n$ for some $c_1, \\ldots, c_n \\in \\mathbb{F}$. Suppose that we also have $\\mathbf{v}=d_1 \\mathbf{v}_1+\\cdots+d_n \\mathbf{v}_n$ for some $d_1, \\ldots, d_n \\in \\mathbb{F}$. Then $$\\mathbf{0}=\\mathbf{v}-\\mathbf{v}=\\sum_{j=1}^n c_j \\mathbf{v}j-\\sum{j=1}^n d_j \\mathbf{v}_j=\\left(c_1-d_1\\right) \\mathbf{v}_1+\\cdots+\\left(c_n-d_n\\right) \\mathbf{v}_n .$$ As $\\left{\\mathbf{v}_1, \\ldots, \\mathbf{v}_n\\right}$ is linearly independent, we must have $c_1-d_1=0,", "as an axial vector. Intuitively, what this means is that the orientation of $\\mathbf{B}$ isn't symmetry with regard to mirror reflections. For example if I reverse orientation (i.e. Line Element: $d\\ell \\mapsto -d\\ell$, Displacement Vector : $\\hat{\\mathbf{r}} \\mapsto -\\hat{\\mathbf{r}}$), then I actually end up with the same field, as per the Biot-Savart Law. Mathematically, this says that $\\mathbf{B}$ doesn't transform under $\\mathsf{O}(n)$ as a \"standard\" vector does. In fact, this was probably the first physical realization of W. Clifford's forgotten but extremely useful concept of a Clifford Algebra. This odd behavior of the magnetic field is formally characterized by saying that the magnetic field is actually a two-form, i.e. $\\mathbf{B} \\in \\bigwedge^{2}{\\mathbb{R}^3}$. Geometrically, this fact acknowledges the arbitrarily choice of a right-handed system, $\\mathbf{a},\\mathbf{b},\\mathbf{a}\\times\\mathbf{b}$ — the cross product could just as easily give a resultant that yields left-handed coordinates, $\\mathbf{a},\\mathbf{b},-\\mathbf{a}\\times\\mathbf{b}$. With this key concept in hand, we can now meaningfully write $\\mathbf{B} = B_{ij}\\, dx^{i}\\wedge dx^{j}$. From Gauss's equation on magnetic monopoles, $\\nabla\\cdot\\mathbf{B} = 0$ and the fact that $d \\mathbf{B} = \\frac{\\partial B_{ij}}{\\partial x^{k}} \\, dx^{k} \\wedge dx^{i} \\wedge dx^{j} = \\text{div}\\mathbf{B}$, we know that $\\mathbf{B}$ must be a closed 2-form. One thing that has always bothered me is the rather hand-waving proof one receives in E&M classes about the existence of the Magnetic Vector potential, $\\mathbf{A}$,", "to the boundary and therefore any cross product involving $$\\hat{\\bf n}$$ will be perpendicular to $$\\hat{\\bf n}$$. The corresponding component of the current density is $$\\hat{\\bf n} \\times \\left( \\hat{\\bf n} \\times {\\bf J}_s \\right)$$, so Equation \\ref{m0023_eBC1} may be equivalently written as follows: $\\hat{\\bf n} \\times \\left( {\\bf H}_2 -{\\bf H}_1 \\right) = \\hat{\\bf n} \\times \\hat{\\bf n} \\times {\\bf J}_s \\label{m0023_eBC2}$ Applying a vector identity ($$\\mathbf{A} \\times(\\mathbf{B} \\times \\mathbf{C})=\\mathbf{B}(\\mathbf{A} \\cdot \\mathbf{C})-\\mathbf{C}(\\mathbf{A} \\cdot \\mathbf{B})$$ of Section B3) to the right side of Equation \\ref{m0023_eBC2} we obtain: \\begin{aligned} \\hat{\\bf n} \\times \\hat{\\bf n} \\times {\\bf J}_s &= \\hat{\\bf n} \\left(\\hat{\\bf n}\\cdot{\\bf J}_s\\right) - {\\bf J}_s\\left(\\hat{\\bf n}\\cdot\\hat{\\bf n}\\right) \\nonumber \\\\ &= \\hat{\\bf n} \\left(0\\right) - {\\bf J}_s\\left(1\\right) \\nonumber \\\\ &= -{\\bf J}_s\\end{aligned} Therefore: $\\hat{\\bf n} \\times \\left( {\\bf H}_2 -{\\bf H}_1 \\right) = -{\\bf J}_s$ The minus sign on the right can be eliminated by swapping $${\\bf H}_2$$ and $${\\bf H}_1$$ on the left, yielding $\\boxed{ \\hat{\\bf n} \\times \\left( {\\bf H}_1 -{\\bf H}_2 \\right) = {\\bf J}_s }$ This is the form in which the boundary condition is most commonly expressed. It is worth noting what this means for the magnetic field intensity $${\\bf B}$$. Since $${\\bf B}=\\mu{\\bf H}$$: In the absence of surface current, the tangential component of $${\\bf B}$$ across the boundary between two material regions", "a_n {\\bf u}_n = {\\bf 0} ,$$ and so ${\\bf v} = a_1^{-1} \\left( -a_2 {\\bf u}_2 - \\cdots - a_n {\\bf u}_n \\right) = - \\left( a_1^{-1} a_2 \\right) {\\bf u}_2 - \\cdots - \\left( a_1^{-1} a_n \\right) {\\bf u}_n .$ Since v is a linear combination of u1, ... , un, which are elements of S, we conclude that v belongs to the span of S. Conversely, let $${\\bf v} \\in \\mbox{span}(S) .$$ Then there exist vectors v1, ... , vm in S and scalars b1, b2, ... , bm such that $${\\bf v} = b_1 {\\bf v}_1 + b_2 {\\bf v}_2 + \\cdots + b_m {\\bf v}_m .$$ Hence, ${\\bf 0} = b_1 {\\bf v}_1 + b_2 {\\bf v}_2 + \\cdots + b_m {\\bf v}_m + (-1)\\,{\\bf v} .$ Since $${\\bf v} \\ne {\\bf v}_i$$ for $$i=1,2,\\ldots , m ,$$ the coefficient of v in this linear combination is nonzero, and so the set $$\\left\\{ {\\bf v} , {\\bf v}_1 , {\\bf v}_2 , \\ldots , {\\bf v}_m \\right\\}$$ is linearly dependent. Therefore, $$S \\cup \\{ {\\bf v} \\}$$ is linearly dependent. Theorem: If a vector space V is generated by a finite set S, then some subset of S is a basis for V If $$S = \\varnothing \\ \\mbox{or } \\ S =", "d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I} \\right) \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.118) On the left our integral can be rewritten as", "# How can I see this equation describes advection? My notes say that this is the equation for the \"advection of a quantity A at speed v\": $$\\frac{\\partial \\mathbf{A}}{\\partial t} = \\nabla\\times(\\mathbf{v}\\times\\mathbf{A}) .$$ Is this true? How can I see (mathematically) that this equation corresponds to advection? - Is there any special property that $\\mathbf{A}$ has to satisfy? – Mr. Gentleman Apr 9 '14 at 15:05 no, $\\mathbf{A}$ is a general vector property. In my example it is the magnetic field $\\mathbf{B}$ – SuperCiocia Apr 9 '14 at 15:11 Well, B does have special properties, such as always-zero divergence. – Kvothe Apr 9 '14 at 15:13 This term looks like Faraday's law that is used in Ideal magnetohydrodynamics (MHD). So yes, it is true. In order to see the mathematical advection, you'll need to apply some vector calculus: $$\\nabla\\times\\mathbf a\\times\\mathbf b = \\mathbf a\\left(\\nabla\\cdot\\mathbf b\\right) - \\mathbf b\\left(\\nabla\\cdot\\mathbf a\\right) + \\left(\\mathbf b\\cdot\\nabla\\right)\\mathbf a - \\left(\\mathbf a\\cdot\\nabla\\right)\\mathbf b$$ Then grouping the $\\mathbf a$ and $\\mathbf b$ terms, $$\\nabla\\times\\mathbf a\\times\\mathbf b = \\left(\\nabla\\cdot\\mathbf b+ \\mathbf b\\cdot\\nabla\\right)\\mathbf a - \\left(\\nabla\\cdot\\mathbf a+\\mathbf a\\cdot\\nabla\\right)\\mathbf b$$ which then reduces to give the advection equation in the form $$\\frac{\\partial\\mathbf A}{\\partial t}=\\nabla\\times\\mathbf v\\times\\mathbf A=\\nabla\\cdot\\left(\\mathbf A\\mathbf v^T - \\mathbf v\\mathbf A^T\\right)\\tag{1}$$ This now gives us a time-derivative of vector $\\mathbf A$ equal to the gradient of $\\mathbf", "is constant. Also $\\nabla \\cdot \\mathbf{r}=3$ and $(\\mathbf{c}\\cdot\\nabla)\\mathbf{r}=\\mathbf{c}$ so the whole expression reduces to $3\\mathbf{c}-\\mathbf{c}=2\\mathbf{c}$ which is why I though that the answer should then be parallel to $\\mathbf{c}$. However, I think the point of the question was to justify this intuitively without explicitly doing the calculation above. And that is where I'm unsure. Thanks 4. May 4, 2017 BvU I see. 'Parallel' in the sense of 'linearly dependent'. I was under the impresssion you worked out the components of $\\vec c \\times\\vec r$ and then applied the $\\vec \\nabla \\times$ to the result. That, to me, is an explicit calculation. I tried it and I think it yields $2\\,\\vec c$ as you found. So you are fine. However, with the triple product expansion expression in the link I gave, I managed to confuse myself: the Lagrange formula reads $${\\bf a}\\times\\left ( {\\bf b} \\times {\\bf c} \\right ) = {\\bf b} \\left ( {\\bf a} \\cdot {\\bf c} \\right ) - {\\bf c} \\left ( {\\bf a} \\cdot {\\bf b} \\right )$$so that $$\\nabla \\times(\\mathbf{c}\\times\\mathbf{r}) = \\mathbf{c} (\\nabla \\cdot \\mathbf{r}) - \\mathbf{r} (\\nabla\\cdot \\mathbf{c}\\ ) \\ ,$$ only two terms, and yielding $3\\bf c$.... Perhaps some math expert can put me right ?", "# If the Nullity of a Linear Transformation is Zero, then Linearly Independent Vectors are Mapped to Linearly Independent Vectors ## Problem 722 Let $T: \\R^n \\to \\R^m$ be a linear transformation. Suppose that the nullity of $T$ is zero. If $\\{\\mathbf{x}_1, \\mathbf{x}_2,\\dots, \\mathbf{x}_k\\}$ is a linearly independent subset of $\\R^n$, then show that $\\{T(\\mathbf{x}_1), T(\\mathbf{x}_2), \\dots, T(\\mathbf{x}_k) \\}$ is a linearly independent subset of $\\R^m$. ## Proof. Suppose that we have a linear combination $c_1T(\\mathbf{x}_1)+c_2T(\\mathbf{x}_2)+\\cdots+c_k T(\\mathbf{x}_k)=\\mathbf{0}_m,$ where $\\mathbf{0}_m$ is the $m$ dimensional zero vector in $\\R^m$. To show that the set $\\{T(\\mathbf{x}_1), T(\\mathbf{x}_2), \\dots, T(\\mathbf{x}_k) \\}$ is linearly independent, we need to show that $c_1=c_2=\\cdots=c_k=0$. Using the linearity of $T$, we have $T(c_1\\mathbf{x}_1+c_2\\mathbf{x}_2+\\cdots+c_k \\mathbf{x}_k)=\\mathbf{0}_m.$ Then the vector $c_1\\mathbf{x}_1+c_2\\mathbf{x}_2+\\cdots+c_k \\mathbf{x}_k$ is in the nullspace $\\calN(T)$ of $T$. Since the nullity, which is the dimension of the nullspace, is zero, we have $\\calN(T)=\\{\\mathbf{0}_n\\}$. This yields $c_1\\mathbf{x}_1+c_2\\mathbf{x}_2+\\cdots+c_k \\mathbf{x}_k=\\mathbf{0}_n.$ Since the vectors $\\mathbf{x}_1, \\mathbf{x}_2,\\dots, \\mathbf{x}_k$ are linearly independent, we must have $c_1=c_2=\\dots=c_k=0$ as required. Thus we conclude that $\\{T(\\mathbf{x}_1), T(\\mathbf{x}_2), \\dots, T(\\mathbf{x}_k) \\}$ is a linearly independent subset of $\\R^m$. ### More from my site • Linear Transformation to 1-Dimensional Vector Space and Its Kernel Let $n$ be a positive integer. Let $T:\\R^n \\to \\R$ be a non-zero linear transformation. Prove the followings. (a) The nullity of $T$ is $n-1$. That" ]

[ "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "+0 # Help with vector geometry +1 88 1 +38 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w} \\text{ and }\\mathbf{x}$$ are linearly independent, answer with 0. If they aren't, find coefficients a,b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a+b}{c}$$ Jun 5, 2019", "+0 # Help fast! 0 86 1 Find the value of $x$ such that $$$\\mathbf{v} = \\begin{pmatrix} 1 \\\\ 2 \\\\ x \\end{pmatrix}$$$is a vector parallel to the plane through the points $A = (0,1,1), B = (1,1,0)$ and $C = (1,0,3).$ Apr 10, 2019", "vector $\\mathbf{P}=\\begin{pmatrix}0 \\\\ -1 \\end{pmatrix}$ So then the general solution is $\\mathbf{X}=c_1\\begin{pmatrix} 1 \\\\ 3\\end{pmatrix}+c_2\\left[t\\begin{pmatrix} 1 \\\\ 3\\end{pmatrix}+ \\begin{pmatrix}0 \\\\ -1 \\end{pmatrix}\\right]$ thanks for that! i had a major brain fart!", "or $\\mathbf W_2 \\subset \\mathbf W_1$. (See HW2 pp. 20-21, #19.) ## Linear combinations Definition: A vector u is a \"linear combination\" (l.c.) of vectors u1, ..., un if there exists scalars a1, ..., an such that $u = a_1 u_1 + a_2 u_2 + \\ldots + a_n u_n$ Example: $\\mathbb P_n(F) = \\{ \\mbox{Polynomials of degree at most } n \\mbox{ with coefficients in } F \\}$ $= \\left\\{ \\sum_{i=0}^n a_i x^i : a_i \\in F \\right\\}$ Definition: A subset $S \\subset \\mathbf V$ \"generates\" or \"spans\" V iff the set of linear combinations of elements of S is all of V. Example: Let $\\mathbf V = \\mathbb M_{n \\times n}(\\real)$ Let $M_1 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix}, M_2 = \\begin{pmatrix} 0 & 1 \\\\ 0 & 0 \\end{pmatrix}, M_3 = \\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \\end{pmatrix}, M_4 = \\begin{pmatrix} 0 & 0 \\\\ 0 & 1 \\end{pmatrix}$ Then $S = \\{ M_1, M_2, M_3, M_4 \\}$ generates V. Proof: Given $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} \\in \\mathbb M_{2 \\times 2}(\\real)$, write $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix} = aM_1 + bM_2 + cM_3 + dM_4.$ Example: Let $N_1 = \\begin{pmatrix} 0 & 1 \\\\ 1 & 1 \\end{pmatrix}, N_2 = \\begin{pmatrix}", "+0 # How do i find (a-b)/c? 0 56 1 +39 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w}$$, and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$and answer with $$\\dfrac{a-b}{c}$$. I have tried this multiple times and I got a=0, b=0, and c=0 but it was incorrect. I tried distributing the vectors and solving for a, b, and c with the 3 equations. Can someone please tell me if I am doing anything wrong? Thanks! Apr 30, 2020 #1 +24983 +3 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}$$, $$\\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}$$. If the vectors $$\\mathbf{v}$$, $$\\mathbf{w}$$ and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a-b}{c}$$. $$\\begin{array}{|rcll|} \\hline \\begin{array}{|rcll|} \\hline a\\begin{pmatrix}", "vector from $(3,3)$ to $(2,5)$ in $\\mathbb{R}^2$ 3. the vector from $(1,0,6)$ to $(5,0,3)$ in $\\mathbb{R}^3$ 4. the vector from $(6,8,8)$ to $(6,8,8)$ in $\\mathbb{R}^3$ This exercise is recommended for all readers. Problem 2 Decide if the two vectors are equal. 1. the vector from $(5,3)$ to $(6,2)$ and the vector from $(1,-2)$ to $(1,1)$ 2. the vector from $(2,1,1)$ to $(3,0,4)$ and the vector from $(5,1,4)$ to $(6,0,7)$ This exercise is recommended for all readers. Problem 3 Does $(1,0,2,1)$ lie on the line through $(-2,1,1,0)$ and $(5,10,-1,4)$? This exercise is recommended for all readers. Problem 4 1. Describe the plane through $(1,1,5,-1)$, $(2,2,2,0)$, and $(3,1,0,4)$. 2. Is the origin in that plane? Problem 5 Describe the plane that contains this point and line. $\\begin{pmatrix} 2 \\\\ 0 \\\\ 3 \\end{pmatrix} \\qquad \\{\\begin{pmatrix} -1 \\\\ 0 \\\\ -4 \\end{pmatrix} +\\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix}t \\,\\big|\\, t\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 6 Intersect these planes. $\\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\end{pmatrix}t+ \\begin{pmatrix} 0 \\\\ 1 \\\\ 3 \\end{pmatrix}s \\,\\big|\\, t,s\\in\\mathbb{R}\\} \\qquad \\{\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} +\\begin{pmatrix} 0 \\\\ 3 \\\\ 0 \\end{pmatrix}k+ \\begin{pmatrix} 2 \\\\ 0 \\\\ 4 \\end{pmatrix}m \\,\\big|\\, k,m\\in\\mathbb{R}\\}$ This exercise is recommended for all readers. Problem 7 Intersect each pair, if possible. 1. $\\{\\begin{pmatrix} 1 \\\\", "\\quad \\mathbf{b} = \\begin{pmatrix} x_b \\\\y_b\\end{pmatrix}$$ If these are to reach exactly the same set of points we must have, for any integers $N$ and $M$ that $$N\\begin{pmatrix} x_a \\\\ y_a\\end{pmatrix} +M\\begin{pmatrix} x_b\\\\ y_b\\end{pmatrix}\\mathbf{b} = \\begin{pmatrix} 2p \\\\ q\\end{pmatrix}$$ Clearly to reach these, and only these points, I would need both $a$ and $b$ to have even $x$ component. Furthermore, if I can reach the points $(2, 0)$ and $(0, 1)$ with combinations of my base vectors then I would be able to reach all of the points $(2n, m)$ by adding together $n$ and $m$ lots of these combinations. Algebra quickly becomes very complicated, as there are too many variable to give me anything to 'solve'. So, look at some more particular cases. Look at a base vector with an $x$-component of $4$. $$\\begin{pmatrix} 4 \\\\ 0\\end{pmatrix}$$ Clearly, to be able to reach the point $(2, 0)$ the other base vector would need to be $\\begin{pmatrix} \\pm 2 \\\\ 0\\end{pmatrix}$, but this would not allow us to reach the point $(0, 1)$. Thus, any other possible pairs of base vectors would have to have non-zero $x$ and $y$ components. A suitable pair would be $$\\mathbf{a}= \\begin{pmatrix} 4 \\\\ 1\\end{pmatrix}\\,, \\mathbf{b} =\\begin{pmatrix}2 \\\\ 1\\end{pmatrix}$$ To see why, note that the combinations $\\mathbf{a}-\\mathbf{b}$ and $2\\mathbf{b}-\\mathbf{a}$ take us to the", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "vector: $$\\boldsymbol{v}=\\begin{pmatrix} 1\\\\2\\\\ \\end{pmatrix}$$ Show that this vector does not form a basis for $\\mathbb{R}^2$. Solution. Since we only have one vector, this vector must be linearly independent. However, $\\boldsymbol{v}$ clearly does not span $\\mathbb{R}^2$ because some vectors in $\\mathbb{R}^2$ cannot be expressed as a scalar multiple of $\\boldsymbol{v}$. Therefore, $\\boldsymbol{v}$ does not form a basis for $\\mathbb{R}^2$. We have already covered the standard basis of $\\mathbb{R}^2$ in the earlier sectionlink. We can also generalize the idea to $\\mathbb{R}^n$. Definition. # Standard basis of Rn The standard basis vectors of $\\mathbb{R}^n$ is defined as a set containing the following $n$ vectors: $$\\left\\{ \\begin{pmatrix} 1\\\\0\\\\\\vdots\\\\0 \\end{pmatrix}, \\begin{pmatrix} 0\\\\1\\\\\\vdots\\\\0 \\end{pmatrix},\\;\\cdots\\;, \\begin{pmatrix} 0\\\\0\\\\\\vdots\\\\1 \\end{pmatrix} \\right\\}$$ Edited by 0 others thumb_up thumb_down Comment Citation Ask a question or leave a feedback... thumb_up 0 thumb_down 0 chat_bubble_outline 0 settings Enjoy our search Hit / to insta-search docs and recipes!", "+0 0 108 3 +226 Let a,b,c be vectors such that $$\\mathbf{a} \\times \\mathbf{b} = \\begin{pmatrix} -1 \\\\- 1 \\\\ -1 \\end{pmatrix}, \\mathbf{a} \\times \\mathbf{c} = \\begin{pmatrix} -1 \\\\ 2 \\\\ 1 \\end{pmatrix}, \\text{ and } \\mathbf{b} \\times \\mathbf{c} = \\begin{pmatrix} 0 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ Then evaluate $$(\\mathbf{b} + \\mathbf{c})\\times \\mathbf{b}, \\mathbf{a}\\times(\\mathbf{b} + 4 \\mathbf{a}), (\\mathbf{a} + \\mathbf{b} + \\mathbf{c})\\times \\mathbf{a}$$ Oct 30, 2020 #2 +31698 +3 For cross products we have the following a x a = 0 (simiarly for b x b etc) a x b = - b x a etc. (b + c) x a = b x+ c x a = -x b - a x c etc. Can you take it from here? Oct 30, 2020 #3 +112499 +1 Thanks Alan, that sure made it easy Oct 31, 2020", "it is assumed that the vector is deferred from the origin. Note 2: A vector can be represented either as a column vector $$\\begin{pmatrix} a_1 \\ a_2 \\end{pmatrix}$$ or as a row vector $$\\begin{pmatrix} a_1 \\ a_2 \\end{pmatrix}$$. Hereinafter, by a vector we mean a column vector, unless otherwise stated. The notion of a vector in the plane can be generalized to 3-dimensional space, and, in general, to $$n$$-dimensional space (which can no longer be visualized): $\\begin{split} \\mathbf{a} = a_1 \\mathbf{e}_1 + a_2 \\mathbf{e}_2 + ... + a_n \\mathbf{e}_n = \\begin{pmatrix} a_1 \\ a_2 \\\\ ... \\ a_n \\end{pmatrix} \\end{split}$ ## Vector operations# As stated earlier, there are two basic operations on vectors in the formal definition: $\\begin{split} \\mathbf{a} + \\mathbf{b} = (a_1 + b_1) \\mathbf{e}_1 + (a_2 + b_2) \\mathbf{e}_2 + ... + (a_n + b_n) \\mathbf{e}_n = \\ = \\begin{pmatrix} a_1 \\ a_2 \\\\ ... \\ a_n \\end{pmatrix} + \\begin{pmatrix} b_1 \\\\ b_2 \\\\ ... \\ b_n \\end{pmatrix} = \\begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\\\ ... \\ a_n + b_n \\end{pmatrix} \\end{split}$ • multiplication of a vector by a number (scalar): $\\begin{split} \\lambda \\mathbf{a} = \\lambda a_1 \\mathbf{e}_1 + \\lambda a_2 \\mathbf{e}_2 + ... + \\lambda a_n \\mathbf{e}_n = \\lambda \\begin{pmatrix} a_1 \\\\ a_2 \\\\ ... \\\\ a_n \\end{pmatrix} = \\begin{pmatrix} \\lambda a_1", "formula: if $D$ is invertible, $$\\det \\pmatrix{A & B\\cr C & D\\cr} = \\det(D) \\det(A-BD^{-1} C)$$ Take $D=I$. Curiously enough, this question can be answered quite simply from first principles, without reference to determinants; we merely need to show that any vector mapped to $0$ by the matrix $\\mathscr M = \\begin{bmatrix} A & B \\\\ C & I \\end{bmatrix} \\tag 1$ is itself $0$ ; so let $\\mathscr X$ be a vector such that $\\mathscr M \\mathscr X = \\begin{bmatrix} A & B \\\\ C & I \\end{bmatrix}\\mathscr X = 0. \\tag 2$ If the size of $A$ is $p$ and the size of $I$ is $q$, then the size of $\\mathscr M$ is $p + q$ whilst $B$ is $p \\times q$ and $C$ is $q \\times p$. We can then write $\\mathscr X$ in terms of two vectors $\\mathbf x$ and $\\mathbf y$, where $\\dim \\mathbf x = p$ and $\\dim \\mathbf y = q$, thusly: $\\mathscr X = \\begin{pmatrix} \\mathbf x \\\\ \\mathbf y \\end{pmatrix}; \\tag 3$ we have: $\\mathscr M \\mathscr X = \\begin{bmatrix} A & B \\\\ C & I \\end{bmatrix}\\begin{pmatrix} \\mathbf x \\\\ \\mathbf y \\end{pmatrix} = \\begin{pmatrix} A\\mathbf x + B\\mathbf y \\\\ C\\mathbf x + I\\mathbf y \\end{pmatrix} = \\begin{pmatrix} A\\mathbf x + B\\mathbf y \\\\ C\\mathbf x +", "\\mathbb{F}$ such that $a = \\lambda \\cdot b$. Now consider your example. Let $a = \\begin{pmatrix} 1 - \\mathrm{i} \\\\ 2 - 2\\mathrm{i}\\end{pmatrix} \\in V$ and $b = \\begin{pmatrix} \\mathrm{i} - 1 \\\\ 2\\mathrm{i} - 2 \\end{pmatrix} \\in V$. The vectors would be linearly dependent, if $a = -1 \\cdot b$ and $-1 \\in \\mathbb{F}$. This is true for the field $\\mathbb{F} := \\mathbb{C}$, but also for $\\mathbb{F} := \\mathbb{R}$ or any other field of which $-1$ is an element.", "= {2\\mathbf x^T A \\cdot \\mathbf x^T\\mathbf x - \\mathbf x^T A \\mathbf x \\cdot 2\\mathbf x^T \\over (\\mathbf x^T\\mathbf x)^2}(^1_0)$$ In other words, if $A=A^T$: $$Df(\\mathbf x) = {2\\mathbf x^T A \\cdot \\mathbf x^T\\mathbf x - \\mathbf x^T A \\mathbf x \\cdot 2\\mathbf x^T \\over (\\mathbf x^T\\mathbf x)^2}$$ Last edited: Jan 24, 2013 3. Jan 24, 2013 ### libragirl79 Since I forgot a lot of my matrix algebra, when you say A (x y) it looks like x choose y, but i am assuming that's not what you meant... 4. Jan 24, 2013 ### I like Serena Since you are talking about x transpose, that means that x is a vector. A vector is commonly written as: $$\\mathbf x = \\begin{pmatrix}x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n\\end{pmatrix}$$ I have simplified it to: $$\\mathbf x = \\begin{pmatrix}x \\\\ y\\end{pmatrix}$$ So indeed, this does not mean x choose y. ;) Furthermore: $$\\mathbf x^T = \\begin{pmatrix}x_1 & x_2 & \\dots & x_n\\end{pmatrix}$$ and $$\\mathbf x^T \\mathbf x = \\begin{pmatrix}x_1 & x_2 & \\dots & x_n\\end{pmatrix}\\begin{pmatrix}x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n\\end{pmatrix} = x_1^2 + x_2^2 + \\dots + x_n^2$$ 5. Jan 24, 2013 ### libragirl79 Ok, I see! Thanks!! :)", "that $\\mathbf{v}\\mathbf{w}=0$ for a Fixed Vector $\\mathbf{v}$ Let $\\mathbf{v} = \\begin{bmatrix} 2 & -5 & -1 \\end{bmatrix}$. Find all $3 \\times 1$ column vectors $\\mathbf{w}$ such that...", "\\mathbf{b} = \\begin{vmatrix} \\hat{\\imath} & \\hat{\\jmath} & \\hat{k} \\\\ a_x & a_y & 0 \\\\ b_x & b_y & 0 \\end{vmatrix}$$ For part b, find the new displacement vector from the point to the point defined by r. gvcalamike Thanks! I got it.", "0 & 1 & 2 \\\\ 0 & 0 & 1\\end{pmatrix}, \\quad \\mathbf N = \\begin{pmatrix} 0 & 0 & 1 \\\\ 0 & 1 & 3 \\\\ 1 & 3 & 6\\end{pmatrix}.$ This entry was posted in Uncategorized and tagged , , , , , . Bookmark the permalink.", "0 \\ 0 \\ \\vdots \\ 0 \\end{pmatrix} = \\begin{pmatrix} u_1 \\ u_2 \\ \\vdots \\ u_n \\end{pmatrix}$$ Similarly, the vector $\\mathbf{0} + \\mathbf{u}$ is given by: $$\\mathbf{0} + \\mathbf{u} = \\begin{pmatrix} 0 \\ 0 \\ \\vdots \\ 0 \\end{pmatrix} + \\begin{pmatrix} u_1 \\ u_2 \\ \\vdots \\ u_n \\end{pmatrix} = \\begin{pmatrix} u_1 \\ u_2 \\ \\vdots \\ u_n \\end{pmatrix}$$ Since $\\mathbf{u} + \\mathbf{0} = \\mathbf{0} + \\mathbf{u}$, it follows that $\\mathbf{u} + \\mathbf{0} = \\mathbf{0} + \\mathbf{u} = \\mathbf{u}$. (d) To prove this statement, we can again write out the components of the vector $\\mathbf{u}$: $$\\mathbf{u} = \\begin{pmatrix} u_1 \\ u_2 \\ \\vdots \\ u_n \\end{pmatrix}$$ Then, the vector $\\mathbf{u} + (-\\mathbf{u})$ is given by: $$\\mathbf{u} + (-\\mathbf{u}) = \\begin{pmatrix} u_1 \\ u_2 \\ \\vdots \\ u_n \\end{pmatrix} + \\begin{pmatrix} -u_1 \\ -u_2 \\ \\vdots \\ -u_n \\end{pmatrix} = \\begin{pmatrix} u_1 - u_1 \\ u_2 - u_2 \\ \\vdots \\ u_n - u_n \\end{pmatrix} = \\begin{pmatrix} 0 \\ 0 \\ \\vdots \\ 0 \\end{pmatrix} = \\mathbf{0}$$ Thus, $\\mathbf{u} + (-\\mathbf{u}) = \\mathbf{0}$. (e) To prove this statement, we can again write out the components of the vectors $\\mathbf{u}$ and $\\mathbf{v}$: $$\\mathbf{u} = \\begin{pmatrix} u_1 \\ u_2 \\ \\vdots \\ u_n \\end{pmatrix} \\quad \\mathbf{v} = \\begin{pmatrix} v_1 \\ v_2 \\ \\vdots \\ v_n \\end{pmatrix}$$ Then, the vector $k(\\mathbf{u}", "# Thread: Spans and vectors 1. ## Spans and vectors If a question asks, Is the vector $\\mathbf{b}=\\begin{pmatrix}-2\\\\ -6\\\\ -4\\end{pmatrix}\\in \\mbox{span}(\\mathbf{v_1, v_2, v_3, v_4})$ where: $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 3\\\\ 0\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}2\\\\ 2\\\\ 1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ -1\\end{pmatrix}$, $\\mathbf{v_4}=\\begin{pmatrix}1\\\\ -2\\\\ 1\\end{pmatrix}$ Do we just make it a matrix and solve the matrix? And how would we solve it if we have 4 variables but only 3 equations? 2) Is the set of vectors $\\mathbf{v_1}=\\begin{pmatrix}1\\\\ 2\\\\ 3\\end{pmatrix}$, $\\mathbf{v_2}=\\begin{pmatrix}1\\\\ 1\\\\ -1\\end{pmatrix}$, $\\mathbf{v_3}=\\begin{pmatrix}-1\\\\ 0\\\\ 5\\end{pmatrix}$ a spanning set of $\\mathbb{R}^3$? Do we just make this a matrix with say $\\mathbf{x}=\\begin{pmatrix}x_1\\\\ x_2\\\\ x_3\\end{pmatrix}$ as our solution and see if we can solve the matrix? 2. So your first question is asking whether there exist scalars $a_{1},a_{2},a_{3},a_{4}$ such that $\\mathbf{b}=a_{1}\\mathbf{v}_{1}+a_{2}\\mathbf{v}_{2} +a_{3}\\mathbf{v}_{3}+a_{4}\\mathbf{v}_{4}.$ This equation you can translate into a normal matrix problem of the form $A\\mathbf{x}=\\mathbf{b}$. Row reduce that problem. If you find any solutions (one or infinitely many), then the answer is yes. If you find that there are no solutions, then the answer is no. Your second question, again, is asking whether there exist scalars $c_{1}, c_{2}, c_{3}$ such that $c_{1}\\mathbf{v}_{1}+c_{2}\\mathbf{v}_{2}+c_{3}\\math bf{v}_{3}=\\mathbf{x}$ for your arbitrary $\\mathbf{x}.$ This, again, can be translated into a standard matrix problem. Since this problem will result in a square matrix, you can use the theory of determinants to help you out." ]

[ "\\end{pmatrix}$ Using the definition of the inner product $(\\mathbf u, \\mathbf v) = \\mathbf v^* \\mathbf u$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (5-i)(1+3i) + (6-4i)(2+4i) = 36 + 30i$ , but using $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\begin{pmatrix} 1 \\\\ 2 \\\\ \\end{pmatrix} , \\begin{pmatrix} 5 \\\\ 6 \\\\ \\end{pmatrix}) + i(\\begin{pmatrix} 3 \\\\ 4 \\\\ \\end{pmatrix} , \\begin{pmatrix} 1 \\\\ 4 \\\\ \\end{pmatrix}) = 5 + 12 + i(3 + 16) = 17 + 19i$. What went wrong with my interpretation of the formula , $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$ ? Shouldn't both computations be mathematically sound and give the same result? Thanks in advance for any help!", "\\mathbf {B} ={\\begin{pmatrix}B_{11}&B_{12}&\\cdots &B_{1p}\\\\B_{21}&B_{22}&\\cdots &B_{2p}\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\B_{m1}&B_{m2}&\\cdots &B_{mp}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\cdots &\\mathbf {b} _{p}\\end{pmatrix}}}$ where ${\\displaystyle {\\mathbf {a} }_{i}={\\begin{pmatrix}A_{i1}&A_{i2}&\\cdots &A_{im}\\end{pmatrix}}\\,,\\quad {\\mathbf {b} }_{i}={\\begin{pmatrix}B_{1i}\\\\B_{2i}\\\\\\vdots \\\\B_{mi}\\end{pmatrix}}}$ Then: ${\\displaystyle \\mathbf {AB} ={\\begin{pmatrix}\\mathbf {a} _{1}\\\\\\mathbf {a} _{2}\\\\\\vdots \\\\\\mathbf {a} _{n}\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {b} _{1}&\\mathbf {b} _{2}&\\dots &\\mathbf {b} _{p}\\end{pmatrix}}={\\begin{pmatrix}(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{1}\\cdot \\mathbf {b} _{p})\\\\(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{2}\\cdot \\mathbf {b} _{p})\\\\\\vdots &\\vdots &\\ddots &\\vdots \\\\(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{1})&(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{2})&\\dots &(\\mathbf {a} _{n}\\cdot \\mathbf {b} _{p})\\end{pmatrix}}}$ It is also possible to express a matrix product in terms of concatenations of products of matrices and row or column vectors: ${\\displaystyle {\\mathbf {AB} }={\\begin{pmatrix}{\\mathbf {A} }{\\mathbf {b} }_{1}&{\\mathbf {A} }{\\mathbf {b} }_{2}&\\dots &{\\mathbf {A} }{\\mathbf {b} }_{p}\\end{pmatrix}}={\\begin{pmatrix}{\\mathbf {a} }_{1}{\\mathbf {B} }\\\\{\\mathbf {a} }_{2}{\\mathbf {B} }\\\\\\vdots \\\\{\\mathbf {a} }_{n}{\\mathbf {B} }\\end{pmatrix}}}$ These decompositions are particularly useful for matrices that are envisioned as concatenations of particular types of row vectors or column vectors, e.g. orthogonal matrices (whose rows and columns are unit vectors orthogonal to each other) and Markov matrices (whose rows or columns sum to 1).{{ safesubst:#invoke:Unsubst||date=__DATE__ |B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }} Outer product The outer product (also known as the dyadic product or tensor product) of two", "0 & \\mathbf1\\\\ \\mathbf1 & 0 & 0 & 0 & 0 & 0 & \\mathbf1\\\\ \\mathbf1 & \\mathbf1 & 0 & 0 & 0 & \\mathbf1 & \\mathbf1\\\\ \\mathbf1 &\\mathbf1 &\\mathbf1 &\\mathbf1 &\\mathbf1 &\\mathbf1 &\\mathbf1 \\end{bmatrix}$$ The bold entries remain constant, while the rest will update as we improve our approximation. We then normally put in some non-zero initial guess for each point to be computed, just a very rough approximation. For instance, in this case we could put $$M = \\begin{bmatrix} \\mathbf1 &\\mathbf1 &\\mathbf1 &\\mathbf1 &\\mathbf1 &\\mathbf1 &\\mathbf1\\\\ \\mathbf1 & \\mathbf1 & 0.66 & 0.66 & 0.66 & \\mathbf1 & \\mathbf1\\\\ \\mathbf1 & 0.66 & 0.33 & 0.33 & 0.33 & 0.66 & \\mathbf1\\\\ \\mathbf1 & 0.66 & 0.33 & \\mathbf0 & 0.33 & 0.66 & \\mathbf1\\\\ \\mathbf1 & 0.66 & 0.33 & 0.33 & 0.33 & 0.66 & \\mathbf1\\\\ \\mathbf1 & \\mathbf1 & 0.66 & 0.66 & 0.66 & \\mathbf1 & \\mathbf1\\\\ \\mathbf1 &\\mathbf1 &\\mathbf1 &\\mathbf1 &\\mathbf1 &\\mathbf1 &\\mathbf1 \\end{bmatrix}$$ We are now ready to begin the iterative process ## Iteration: What we use now (and will justify later, maybe) is the fact that on a $2$D plane where $\\phi$ is a function satisfying $$\\nabla^2\\phi=\\frac{\\partial^2\\phi}{\\partial x^2}+\\frac{\\partial^2\\phi}{\\partial y^2}=0$$ the following is approximately true: Suppose that $\\phi(x_0,y_0)=\\phi_0$ Let $\\Delta x$ and $\\Delta y$ be \"sufficiently small\" and", "\\mathbf{U}^T\\mathbf{U} = \\mathbf{I}$$, $$\\mathbf{V}\\mathbf{V}^T = \\mathbf{V}^T\\mathbf{V} = \\mathbf{I}$$, and $$\\Sigma$$ containing the singular values of $$\\mathbf{A}$$, $$\\sigma_i$$, $$1 \\leq I \\leq \\min{(m, n)}$$, along the diagonal. Then, the factorization of $$\\mathbf{B}$$ in terms of $$\\mathbf{U}$$, $$\\mathbf{V}$$, and $$\\Sigma$$ is the following: $$\\mathbf{B} = \\begin{pmatrix} \\mathbf{0} & \\mathbf{A}^T \\\\ \\mathbf{A} & \\mathbf{0} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{0} & \\mathbf{V}\\Sigma^T\\mathbf{U}^T \\\\ \\mathbf{U}\\Sigma\\mathbf{V}^T & \\mathbf{0} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{V} & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{U} \\end{pmatrix} \\begin{pmatrix} \\mathbf{0} & \\Sigma^T \\\\ \\Sigma & \\mathbf{0} \\end{pmatrix} \\begin{pmatrix} \\mathbf{V}^T & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{U}^T \\end{pmatrix} = \\mathbf{Q}_1 \\mathbf{C} \\mathbf{Q}_1^T$$ where $$\\mathbf{Q}_1 = \\begin{pmatrix} \\mathbf{V} & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{U} \\end{pmatrix}$$ and $$\\mathbf{C} = \\begin{pmatrix} \\mathbf{0} & \\Sigma^T \\\\ \\Sigma & \\mathbf{0} \\end{pmatrix}$$. We can readily verify that $$\\mathbf{Q}_1$$ is orthogonal since $$\\mathbf{Q}_1\\mathbf{Q}_1^T = \\mathbf{Q}_1^T\\mathbf{Q}_1 = \\mathbf{I}$$. However, $$\\mathbf{C}$$ has not the expected diagonal structure. Instead, the singular values, $$\\sigma_i$$, of $$\\mathbf{A}$$ are distributed in the following way: $$\\mathbf{C} = \\begin{pmatrix} \\mathbf{0} & \\Sigma^T \\\\ \\Sigma & \\mathbf{0} \\\\ \\end{pmatrix} = \\left( \\begin{array}{ccc|ccc} & & & \\sigma_1 & & \\\\ & \\mathbf{0} & & & \\sigma_2 & \\\\ & & & & & \\ddots \\\\ \\hline \\sigma_1 & & & & & \\\\ & \\sigma_2 & & & \\mathbf{0} & \\\\ & & \\ddots & & & \\end{array} \\right)$$ That is, $$\\mathbf{C}$$ contains", "Therefore $\\mathbf{W^T}= \\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\end{bmatrix}$ and $\\mathbf{W} \\cdot ( \\mathbf{Kd-F} ) = 1[ \\sum^6_{j=1} \\mathbf{K}_{1j} \\mathbf{d}_j- \\mathbf{F}_1] + 0[ \\sum^6_{j=1} \\mathbf{K}_{2j} \\mathbf{d}_j- \\mathbf{F}_2] + 0 [ \\sum^6_{j=1} \\mathbf{K}_{3j} \\mathbf{d}_j- \\mathbf{F}_3] + 0 [ \\sum^6_{j=1} \\mathbf{K}_{4j} \\mathbf{d}_j- \\mathbf{F}_4] + 0 [ \\sum^6_{j=1} \\mathbf{K}_{5j} \\mathbf{d}_j- \\mathbf{F}_5] + 0 [ \\sum^6_{j=1} \\mathbf{K}_{6j} \\mathbf{d}_j- \\mathbf{F}_6]$ Thus, the result becomes $\\sum^6_{j=1} \\mathbf{K}_{1j} \\mathbf{d}_j= \\mathbf{F}_1$ Which is designated to be equation (a). 2nd Choice: Choose W such that W2=1, W1=W3=W4=W5=W6=0 so that: $\\mathbf{W^T}= \\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\end{bmatrix}$ Therefore: $\\mathbf{W} \\cdot ( \\mathbf{Kd-F} ) = 0[ \\sum^6_{j=1} \\mathbf{K}_{1j} \\mathbf{d}_j- \\mathbf{F}_1] + 1[ \\sum^6_{j=1} \\mathbf{K}_{2j} \\mathbf{d}_j- \\mathbf{F}_2] + 0 [ \\sum^6_{j=1} \\mathbf{K}_{3j} \\mathbf{d}_j- \\mathbf{F}_3] + 0 [ \\sum^6_{j=1} \\mathbf{K}_{4j} \\mathbf{d}_j- \\mathbf{F}_4] + 0 [ \\sum^6_{j=1} \\mathbf{K}_{5j} \\mathbf{d}_j- \\mathbf{F}_5] + 0 [ \\sum^6_{j=1} \\mathbf{K}_{6j} \\mathbf{d}_j- \\mathbf{F}_6]$ Thus, the result becomes $\\sum^6_{j=1} \\mathbf{K}_{2j} \\mathbf{d}_j= \\mathbf{F}_2$ Which is designated to be equation (b) 3rd Choice: C hoose W such that W3=1, W1=W2=W4=W5=W6=0 so that $\\mathbf{W^T}= \\begin{bmatrix} 0 & 0 & 1 & 0 & 0 & 0 \\end{bmatrix}$ $\\mathbf{W} \\cdot ( \\mathbf{Kd-F} ) = 0[ \\sum^6_{j=1} \\mathbf{K}_{1j} \\mathbf{d}_j- \\mathbf{F}_1] + 0[ \\sum^6_{j=1} \\mathbf{K}_{2j} \\mathbf{d}_j- \\mathbf{F}_2] + 1 [ \\sum^6_{j=1} \\mathbf{K}_{3j} \\mathbf{d}_j- \\mathbf{F}_3] + 0 [ \\sum^6_{j=1} \\mathbf{K}_{4j} \\mathbf{d}_j- \\mathbf{F}_4] +", "& -8\\\\ 2 & -3\\\\ \\end {vmatrix}$$ = 1 × (-3) - 2 × (-8) = -3 + 16 = 13 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$ We know that, AX =B X = A-1B or, X =$$\\frac {1}{13}$$$$\\begin {pmatrix} -3 & -8\\\\ -2 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 39\\\\ 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -117 + 104\\\\ -78 + 13\\\\ \\end {pmatrix}$$ or, X = $$\\frac {1}{13}$$$$\\begin {pmatrix} -13\\\\ -65\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-13}{13}\\\\ \\frac {-65}{13}\\\\ \\end {pmatrix}$$ ∴ X = $$\\begin {pmatrix} -1\\\\ -5\\\\ \\end {pmatrix}$$ ∴ x = -1 and y = -5 Ans Here, $$\\frac 2x$$ + $$\\frac 3y$$ = 2..............................................(1) $$\\frac 4x$$ - $$\\frac 9y$$ = -1.............................................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B = $$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & 3\\\\ 4 & -9\\\\ \\end {vmatrix}$$ = 2 ×", "\\mathbf{A}_{2,1} & \\mathbf{A}_{2,2} \\end{bmatrix} \\mbox { , } \\mathbf{B} = \\begin{bmatrix} \\mathbf{B}_{1,1} & \\mathbf{B}_{1,2} \\\\ \\mathbf{B}_{2,1} & \\mathbf{B}_{2,2} \\end{bmatrix} \\mbox { , } \\mathbf{C} = \\begin{bmatrix} \\mathbf{C}_{1,1} & \\mathbf{C}_{1,2} \\\\ \\mathbf{C}_{2,1} & \\mathbf{C}_{2,2} \\end{bmatrix}$ with $\\mathbf{A}_{i,j}, \\mathbf{B}_{i,j}, \\mathbf{C}_{i,j} \\in R^{2^{n-1} \\times 2^{n-1}}$ then $\\mathbf{C}_{1,1} = \\mathbf{A}_{1,1} \\mathbf{B}_{1,1} + \\mathbf{A}_{1,2} \\mathbf{B}_{2,1}$ $\\mathbf{C}_{1,2} = \\mathbf{A}_{1,1} \\mathbf{B}_{1,2} + \\mathbf{A}_{1,2} \\mathbf{B}_{2,2}$ $\\mathbf{C}_{2,1} = \\mathbf{A}_{2,1} \\mathbf{B}_{1,1} + \\mathbf{A}_{2,2} \\mathbf{B}_{2,1}$ $\\mathbf{C}_{2,2} = \\mathbf{A}_{2,1} \\mathbf{B}_{1,2} + \\mathbf{A}_{2,2} \\mathbf{B}_{2,2}$ With this construction we have not reduced the number of multiplications. We still need 8 multiplications to calculate the Ci,j matrices, the same number of multiplications we need when using standard matrix multiplication. Now comes the important part. We define new matrices $\\mathbf{M}_{1} := (\\mathbf{A}_{1,1} + \\mathbf{A}_{2,2}) (\\mathbf{B}_{1,1} + \\mathbf{B}_{2,2})$ $\\mathbf{M}_{2} := (\\mathbf{A}_{2,1} + \\mathbf{A}_{2,2}) \\mathbf{B}_{1,1}$ $\\mathbf{M}_{3} := \\mathbf{A}_{1,1} (\\mathbf{B}_{1,2} - \\mathbf{B}_{2,2})$ $\\mathbf{M}_{4} := \\mathbf{A}_{2,2} (\\mathbf{B}_{2,1} - \\mathbf{B}_{1,1})$ $\\mathbf{M}_{5} := (\\mathbf{A}_{1,1} + \\mathbf{A}_{1,2}) \\mathbf{B}_{2,2}$ $\\mathbf{M}_{6} := (\\mathbf{A}_{2,1} - \\mathbf{A}_{1,1}) (\\mathbf{B}_{1,1} + \\mathbf{B}_{1,2})$ $\\mathbf{M}_{7} := (\\mathbf{A}_{1,2} - \\mathbf{A}_{2,2}) (\\mathbf{B}_{2,1} + \\mathbf{B}_{2,2})$ which are then used to express the Ci,j in terms of Mk. Because of our definition of the Mk we can eliminate one matrix multiplication and reduce the number of multiplications to 7 (one multiplication for each Mk) and express the Ci,j as $\\mathbf{C}_{1,1} = \\mathbf{M}_{1} + \\mathbf{M}_{4} - \\mathbf{M}_{5} + \\mathbf{M}_{7}$ $\\mathbf{C}_{1,2} = \\mathbf{M}_{3} + \\mathbf{M}_{5}$", "3 & 1 & −2 \\\\ −2 & 0 & 5 \\end{pmatrix}$$ $$\\mathbf{F} = \\begin{pmatrix} 3 & 4 \\\\ 5 & 6 \\end{pmatrix}$$ $$\\mathbf{G} = \\begin{pmatrix} 2 & −4 \\\\ 7 & 1 \\\\ −1 & 1 \\end{pmatrix}$$ $$\\mathbf{H} = \\begin{pmatrix} −8 & −1 & −3 \\\\ −4 & 2 & −2 \\\\ −1 & 0 & 0 \\end{pmatrix}$$ Calculate each of the following 1. $$|\\mathbf{a} − (\\mathbf{b} + \\mathbf{3c})|$$ 7. $$(\\mathbf{EH})^T$$ 2. Component of $$\\mathbf{c}$$ along $$\\mathbf{a}$$ 8. $$|\\mathbf{HE}|$$ 3. Angle between $$\\mathbf{c}$$ and $$\\mathbf{d}$$ 9. $$\\mathbf{EHG}$$ 4. $$(\\mathbf{b} \\times \\mathbf{d}) \\cdot \\mathbf{a}$$ 10. $$\\mathbf{EG} − \\mathbf{HG}$$ 5. $$(\\mathbf{b} \\times \\mathbf{d}) \\times \\mathbf{a}$$ 11. $$\\mathbf{EH} − \\mathbf{H}^T \\mathbf{E}^T$$ 6. $$\\mathbf{b}\\times (\\mathbf{d} \\times \\mathbf{a})$$ 12. $$\\mathbf{F}^{−1}$$ 2. For what values of $$a$$ are the vectors $$\\mathbf{A} = 2a\\hat{i} − 2\\hat{j} + a\\hat{k}$$ and $$\\mathbf{B} = a\\hat{i} + 2a\\hat{j}+ 2\\hat{k}$$ perpendicular? 3. Show that the triple scalar product $$(A \\times B) \\cdot C$$ can be written as $(\\mathbf{A} \\times \\mathbf{B}) \\cdot \\mathbf{C} = \\begin{vmatrix} A_1 & A_2 & A_3 \\\\ B_1 & B_2 & B_3 \\\\ C_1 & C_2 & C_3 \\end{vmatrix} \\nonumber$ Show also that the product is unaffected by interchange of the scalar and vector product operations or by change in the order of $$A, B, C$$ as long as they are in cyclic order, that is", "0 & 1 & 2 \\\\ 0 & 0 & 1\\end{pmatrix}, \\quad \\mathbf N = \\begin{pmatrix} 0 & 0 & 1 \\\\ 0 & 1 & 3 \\\\ 1 & 3 & 6\\end{pmatrix}.$ This entry was posted in Uncategorized and tagged , , , , , . Bookmark the permalink.", "this to $\\mathbf{b}$, we must add the $x$ values and $y$ values separately. Doing so, we get the answer to be: $2\\mathbf{a}+\\mathbf{b}=\\begin{pmatrix}6\\\\16\\end{pmatrix}+\\begin{pmatrix}-7\\\\2\\end{pmatrix}=\\begin{pmatrix}-1\\\\18\\end{pmatrix}$ Firstly, to multiply $\\mathbf{a}$ by $3$, we must multiply both of its components: $3\\mathbf{a}=3\\times\\begin{pmatrix}2\\\\7\\end{pmatrix}=\\begin{pmatrix}6\\\\21\\end{pmatrix}$ Then, in order subtract $2\\mathbf{b}$, we must first multiply $\\mathbf{b}$ by $2$ $2\\mathbf{b}=2\\times\\begin{pmatrix}-5\\\\3\\end{pmatrix}=\\begin{pmatrix}-10\\\\6\\end{pmatrix}$ Thus the calculation is: $3\\mathbf{a}-2\\mathbf{b}=\\begin{pmatrix}6\\\\21\\end{pmatrix}-\\begin{pmatrix}-10\\\\6\\end{pmatrix}=\\begin{pmatrix}16\\\\15\\end{pmatrix}$ Firstly, to multiply $\\mathbf{b}$ by $2$, we must multiply both of its components: $2\\mathbf{b}=2\\times\\begin{pmatrix}5\\\\-3\\end{pmatrix}=\\begin{pmatrix}10\\\\-6\\end{pmatrix}$ Then, we can add $\\mathbf{a}$ and $2\\mathbf{b}$: $\\mathbf{a}+2\\mathbf{b}=\\begin{pmatrix}6\\\\2\\end{pmatrix}+\\begin{pmatrix}10\\\\-6\\end{pmatrix}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}$ The final calculation is to subtract $\\mathbf{c}$: $\\mathbf{a}+2\\mathbf{b}-\\mathbf{c}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}-\\begin{pmatrix}2\\\\1\\end{pmatrix} = \\begin{pmatrix}14\\\\-5\\end{pmatrix}$ Level 4-5 ### Learning resources you may be interested in We have a range of learning resources to compliment our website content perfectly. Check them out below.", "or, Q3= 3($$\\frac{N + 1}{4}$$)th item or, Q3= 3($$\\frac{7 + 1}{4}$$)th item or, Q3= 3($$\\frac{8}{4}$$)th item or, Q3= 3 $$\\times$$ 2th item or, Q3= 6th item i.e.Q3= x + 25 Then, or,Q3= x + 25 or, 60 = x + 25 or, x + 60 - 25 $$\\therefore$$ x = 35 Solution: Now, Arranging in ascending order 50, 52, 56, 61, 64, 68, 72 numbers = 7 We know that, or, Q1 = $$\\begin{pmatrix}\\frac{n + 1}{4} \\end{pmatrix}$$thitem or, Q1 = $$\\begin{pmatrix}\\frac{7 + 1}{4} \\end{pmatrix}$$thitem or, Q1 = $$\\begin{pmatrix}\\frac{8}{4} \\end{pmatrix}$$thitem or, Q1 = 2thitem or,Q1 = 52 Again, or, Q2= $$\\begin{pmatrix}\\frac{n + 1}{2} \\end{pmatrix}$$thitem or, Q2= $$\\begin{pmatrix}\\frac{7 + 1}{2} \\end{pmatrix}$$thitem or, Q2= $$\\begin{pmatrix}\\frac{8}{2} \\end{pmatrix}$$thitem or, Q2= 4thitem or, Q2= 61 Similarly, or, Q3= 3$$\\begin{pmatrix}\\frac{n + 1}{4} \\end{pmatrix}$$thitem or, Q3= 3$$\\begin{pmatrix}\\frac{7 + 1}{4} \\end{pmatrix}$$thitem or, Q3= 3$$\\begin{pmatrix}\\frac{8}{4} \\end{pmatrix}$$thitem or, Q3= 3 $$\\times$$2thitem or, Q3= 6thitem or, Q3 = 68 $$\\therefore$$ The value ofQ1, Q2 and Q3is 52, 61 and 68 respectively. Solution: Marks No. of students(f) cumulative frequency (c.f.) 10 2 2 20 4 6 30 6 12 40 12 24 50 8 32 60 6 38 70 1 39 N = 39 We have, Q1= value of $$\\begin{pmatrix}\\frac{n + 1}{4} \\end{pmatrix}$$th term = value of $$\\begin{pmatrix}\\frac{39 + 1}{4} \\end{pmatrix}$$th term = value of 10th term In c.f. column just", "$[\\mathbf{A},\\mathbf{B}]=\\mathbf{A}\\mathbf{B}-\\mathbf{B}\\mathbf{A} \\nonumber$ ##### Example $$\\PageIndex{3}$$ Given $$\\mathbf{A}=\\begin{pmatrix} 3&1 \\\\ 2&0 \\end{pmatrix}$$ and $$\\mathbf{B}=\\begin{pmatrix} 1&0 \\\\ -1&2 \\end{pmatrix}$$ Calculate the commutator $$[\\mathbf{A},\\mathbf{B}]$$ ###### Solution $[\\mathbf{A},\\mathbf{B}]=\\mathbf{A}\\mathbf{B}-\\mathbf{B}\\mathbf{A}\\nonumber$ $\\mathbf{A}\\mathbf{B}=\\begin{pmatrix} 3&1 \\\\ 2&0 \\end{pmatrix}\\begin{pmatrix} 1&0 \\\\ -1&2 \\end{pmatrix}=\\begin{pmatrix} 3\\times 1+1\\times (-1)&3\\times 0 +1\\times 2 \\\\ 2\\times 1+0\\times (-1)&2\\times 0+ 0\\times 2 \\end{pmatrix}=\\begin{pmatrix} 2&2 \\\\ 2&0 \\end{pmatrix}\\nonumber$ $\\mathbf{B}\\mathbf{A}=\\begin{pmatrix} 1&0 \\\\ -1&2 \\end{pmatrix}\\begin{pmatrix} 3&1 \\\\ 2&0 \\end{pmatrix}=\\begin{pmatrix} 1\\times 3+0\\times 2&1\\times 1 +0\\times 0 \\\\ -1\\times 3+2\\times 2&-1\\times 1+2\\times 0 \\end{pmatrix}=\\begin{pmatrix} 3&1 \\\\ 1&-1 \\end{pmatrix}\\nonumber$ $[\\mathbf{A},\\mathbf{B}]=\\mathbf{A}\\mathbf{B}-\\mathbf{B}\\mathbf{A}=\\begin{pmatrix} 2&2 \\\\ 2&0 \\end{pmatrix}-\\begin{pmatrix} 3&1 \\\\ 1&-1 \\end{pmatrix}=\\begin{pmatrix} -1&1 \\\\ 1&1 \\end{pmatrix}\\nonumber$ $\\displaystyle{\\color{Maroon}[\\mathbf{A},\\mathbf{B}]=\\begin{pmatrix} -1&1 \\\\ 1&1 \\end{pmatrix}}\\nonumber$ ## Multiplication of a vector by a scalar The multiplication of a vector $$\\vec{v_1}$$ by a scalar $$n$$ produces another vector of the same dimensions that lies in the same direction as $$\\vec{v_1}$$; $n\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\begin{pmatrix} nx \\\\ ny \\end{pmatrix} \\nonumber$ The scalar can stretch or compress the length of the vector, but cannot rotate it (figure [fig:vector_by_scalar]). ## Multiplication of a square matrix by a vector The multiplication of a vector $$\\vec{v_1}$$ by a square matrix produces another vector of the same dimensions of $$\\vec{v_1}$$. For example, we can multiply a $$2\\times 2$$ matrix and a 2-dimensional vector: $\\begin{pmatrix} a&b \\\\ c&d \\end{pmatrix}\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\begin{pmatrix} ax+by \\\\ cx+dy \\end{pmatrix} \\nonumber$ For example, consider the matrix $\\mathbf{A}=\\begin{pmatrix} -2 &0", "\\in \\mathbb{R}^5$, and $\\mathbf{c} \\in \\mathbb{R}^3$. Written out fully we would have: $$\\begin{pmatrix} a_1^{(t)}\\\\ a_2^{(t)}\\\\ a_3^{(t)}\\\\ a_4^{(t)}\\\\ a_5^{(t)} \\end{pmatrix} = \\begin{pmatrix} b_1^{(t)}\\\\ b_2^{(t)}\\\\ b_3^{(t)}\\\\ b_4^{(t)}\\\\ b_5^{(t)} \\end{pmatrix} + \\begin{pmatrix} w_{11} & w_{12} & w_{13} & w_{14} & w_{15}\\\\ w_{21} & w_{22} & w_{23} & w_{24} & w_{25}\\\\ w_{31} & w_{32} & w_{33} & w_{34} & w_{35}\\\\ w_{41} & w_{42} & w_{43} & w_{44} & w_{45}\\\\ w_{51} & w_{52} & w_{53} & w_{54} & w_{55} \\end{pmatrix} \\begin{pmatrix} h_1^{(t-1)}\\\\ h_2^{(t-1)}\\\\ h_3^{(t-1)}\\\\ h_4^{(t-1)}\\\\ h_5^{(t-1)} \\end{pmatrix} + \\begin{pmatrix} u_{11} & u_{12} & u_{13} & u_{14}\\\\ u_{21} & u_{22} & u_{23} & u_{24}\\\\ u_{31} & u_{32} & u_{33} & u_{34}\\\\ u_{41} & u_{42} & u_{43} & u_{44}\\\\ u_{51} & u_{52} & u_{53} & u_{54} \\end{pmatrix} \\begin{pmatrix} x_1^{(t)}\\\\ x_2^{(t)}\\\\ x_3^{(t)}\\\\ x_4^{(t)} \\end{pmatrix}$$ Which we can write a little more compactly as \\begin{align} a_1^{(t)} &= b_1^{(t)} + \\sum\\limits_{i=1}^5 w_{1i}h_i^{(t-1)} + \\sum\\limits_{j=1}^4 u_{1j}x_j^{(t)}\\\\ a_2^{(t)} &= b_2^{(t)} + \\sum\\limits_{i=1}^5 w_{2i}h_i^{(t-1)} + \\sum\\limits_{j=1}^4 u_{2j}x_j^{(t)}\\nonumber\\\\ a_3^{(t)} &= b_3^{(t)} + \\sum\\limits_{i=1}^5 w_{3i}h_i^{(t-1)} + \\sum\\limits_{j=1}^4 u_{3j}x_j^{(t)}\\nonumber\\\\ a_4^{(t)} &= b_4^{(t)} + \\sum\\limits_{i=1}^5 w_{4i}h_i^{(t-1)} + \\sum\\limits_{j=1}^4 u_{4j}x_j^{(t)}\\nonumber\\\\ a_5^{(t)} &= b_5^{(t)} + \\sum\\limits_{i=1}^5 w_{5i}h_i^{(t-1)} + \\sum\\limits_{j=1}^4 u_{5j}x_j^{(t)}\\nonumber \\end{align} We could do the same for the output so that we arrive at $$\\label{o_vector} \\begin{pmatrix} o_1^{(t)}\\\\ o_2^{(t)}\\\\ o_3^{(t)} \\end{pmatrix} = \\begin{pmatrix} c_1^{(t)}\\\\ c_2^{(t)}\\\\ c_3^{(t)}\\end{pmatrix} + \\begin{pmatrix} v_{11} & v_{12} & v_{13} & v_{14} & v_{15}\\\\ v_{21}", "multiply both of its components by $2$: $2\\mathbf{a}=2\\times\\begin{pmatrix}3\\\\8\\end{pmatrix}=\\begin{pmatrix}6\\\\16\\end{pmatrix}$ Then, to add this to $\\mathbf{b}$, we must add the $x$ values and $y$ values separately. Doing so, we get the answer to be: $2\\mathbf{a}+\\mathbf{b}=\\begin{pmatrix}6\\\\16\\end{pmatrix}+\\begin{pmatrix}-7\\\\2\\end{pmatrix}=\\begin{pmatrix}-1\\\\18\\end{pmatrix}$ Firstly, to multiply $\\mathbf{a}$ by $3$, we must multiply both of its components: $3\\mathbf{a}=3\\times\\begin{pmatrix}2\\\\7\\end{pmatrix}=\\begin{pmatrix}6\\\\21\\end{pmatrix}$ Then, in order subtract $2\\mathbf{b}$, we must first multiply $\\mathbf{b}$ by $2$. $2\\mathbf{b}=2\\times\\begin{pmatrix}-5\\\\3\\end{pmatrix}=\\begin{pmatrix}-10\\\\6\\end{pmatrix}$ Thus the calculation is: $3\\mathbf{a}-2\\mathbf{b}=\\begin{pmatrix}6\\\\21\\end{pmatrix}-\\begin{pmatrix}-10\\\\6\\end{pmatrix}=\\begin{pmatrix}16\\\\15\\end{pmatrix}$ Firstly, to multiply $\\mathbf{b}$ by $2$, we must multiply both of its components: $2\\mathbf{b}=2\\times\\begin{pmatrix}5\\\\-3\\end{pmatrix}=\\begin{pmatrix}10\\\\-6\\end{pmatrix}$ Then, we can add $\\mathbf{a}$ and $2\\mathbf{b}$: $\\mathbf{a}+2\\mathbf{b}=\\begin{pmatrix}6\\\\2\\end{pmatrix}+\\begin{pmatrix}10\\\\-6\\end{pmatrix}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}$ The final calculation is to subtract $\\mathbf{c}$: $\\mathbf{a}+2\\mathbf{b}-\\mathbf{c}=\\begin{pmatrix}16\\\\-4\\end{pmatrix}-\\begin{pmatrix}2\\\\1\\end{pmatrix} = \\begin{pmatrix}14\\\\-5\\end{pmatrix}$ ## Related Topics #### Coordinates and Ratios Level 4-5Level 6-7KS3 #### Direct and Inverse Proportion Level 4-5Level 6-7KS3 ## Worksheet and Example Questions ### (NEW) Column Vectors Exam Style Questions - MMe Level 4-5 GCSENewOfficial MME ## You May Also Like... ### GCSE Maths Revision Cards Revise for your GCSE maths exam using the most comprehensive maths revision cards available. These GCSE Maths revision cards are relevant for all major exam boards including AQA, OCR, Edexcel and WJEC. £8.99 ### GCSE Maths Revision Guide The MME GCSE maths revision guide covers the entire GCSE maths course with easy to understand examples, explanations and plenty of exam style questions. We also provide a separate answer book to make checking your answers easier!", "\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}, \\quad \\bfb = \\begin{pmatrix}1\\\\3\\\\2\\end{pmatrix}.$$ Recall that Hence, $$\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix} = \\begin{pmatrix} 1/3 & 1 & -1/3 \\\\ -1/2 & 1/2 & 0 \\\\ 1/6 & -1/2 & 1/3 \\end{pmatrix} \\begin{pmatrix}1\\\\3\\\\2 \\end{pmatrix} = \\begin{pmatrix}8/3\\\\1\\\\-2/3\\end{pmatrix}$$ Hence, the polynomial through the points $(-1, 1), (1, 3),$ and $(2,2)$ is $$y = \\frac{8}{3} + x - \\frac{2}{3} x^2$$", "\\\\ {6} & {1} & {9} \\\\ {4} & {-3} & {2} \\end{bmatrix} \\quad \\mathbf{ B} = \\begin{bmatrix} {2} & {-1} & {3} \\\\ {1} & {2} & {5} \\\\ {-3} & {-5} & {2} \\end{bmatrix}\\nonumber$ determine $$\\mathbf{ A}+\\mathbf{ B}$$ and $$\\mathbf{ A}-3 \\mathbf{ B}$$. 23. Given the following $$1\\times 4$$ row matrices $\\mathbf{ a} = \\begin{bmatrix} {3} & {1} & {5} & {6} \\end{bmatrix}, \\quad \\mathbf{ b} = \\begin{bmatrix} {6} & {-2} & {0} & {4} \\end{bmatrix}\\nonumber$ determine $$2\\mathbf{ a}+\\mathbf{ b}$$. 24.$$\\ddagger$$ Compute the $$4\\times 4$$ matrix defined by $$\\mathbf{ A}-\\mathbf{ B}+\\mathbf{ C}$$ where $$\\mathbf{A}$$, $$\\mathbf{B}$$ and $$\\mathbf{C}$$ are given by $\\mathbf{ A} = \\begin{bmatrix} {0.856} & {0.328} & {0.0663} & {0.908} \\\\ {0.529} & {0.426} & {0.364} & {0.434} \\\\ {0.506} & {0.812} & {0.400} & {0.137} \\\\ {0.652} & {0.402} & {0.0276} & {0.995} \\end{bmatrix} \\quad \\mathbf{ B} = \\begin{bmatrix} {1.142} & {0.438} & {0.0884} & {1.211} \\\\ {0.705} & {0.568} & {0.485} & {0.579} \\\\ {0.675} & {1.083} & {0.534} & {0.183} \\\\ {0.869} & {0.535} & {0.368} & {1.326} \\end{bmatrix}\\nonumber$ $\\mathbf{ C} = \\begin{bmatrix} {0.285} & {0.109} & {0.0221} & {0.303} \\\\ {0.176} & {0.142} & {0.121} & {0.145} \\\\ {0.169} & {0.271} & {0.133} & {0.0457} \\\\ {0.217} & {0.134} & {0.0092} & {0.332} \\end{bmatrix}\\nonumber$ 25.$$\\ddagger$$ Consider a project in which all", "_{n+1}}$ (18) substituting back into Eqn(17) and using the definitions in Eqns(8)–(10), we recover the relation ${\\displaystyle \\left(1-\\alpha \\right)p_{n+1}=\\left(\\mathbf {G} _{f}-\\alpha \\mathbf {G} \\right)\\mathbf {y} _{n+1}}$ (19) On the other hand the structure is advanced in time following the time integration rule of Eqn(14). Substituting Eqn(19) into Eqn(14) and collecting terms we obtain ${\\displaystyle {\\begin{pmatrix}\\mathbf {I} &\\mathbf {-L_{n+1}} \\\\\\alpha \\mathbf {G} -\\mathbf {G} _{f}&1-\\alpha \\end{pmatrix}}{\\begin{pmatrix}\\mathbf {y} _{n+1}\\\\p_{n+1}\\end{pmatrix}}={\\begin{pmatrix}\\mathbf {A} &\\mathbf {L} _{n}\\\\0&0\\end{pmatrix}}{\\begin{pmatrix}\\mathbf {y} _{n}\\\\p_{n}\\end{pmatrix}}}$ (20) which allows us to write synthetically ${\\displaystyle \\mathbf {z} _{n+1}=\\mathbf {A} _{ex}\\mathbf {z} _{n}}$ (21) where ${\\displaystyle \\mathbf {A} _{ex}:={\\begin{pmatrix}\\mathbf {I} &\\mathbf {-L} _{n+1}\\\\\\alpha \\mathbf {G} -\\mathbf {G} _{f}&1-\\alpha \\end{pmatrix}}^{-1}{\\begin{pmatrix}\\mathbf {A} &\\mathbf {L} _{n}\\\\0&0\\end{pmatrix}}}$ (22) and ${\\displaystyle \\mathbf {z} _{n+1}:={\\begin{pmatrix}\\mathbf {y} _{n+1}\\\\p_{n+1}\\end{pmatrix}}}$ (23) Eqn(21) yields the amplification form of the exact solution for the coupled problem. 2.3 A fractional step-like algorithm In the previous section we expressed the exact solution in an amplification form, relating pressures, velocities and displacements at two consecutive time steps. Unfortunately in real life not all the information needed is available at the same time. Therefore an approximate coupling strategy needs to be devised. In this section we propose and analyse a fractional step-like partitioned approach including the following steps: Predictive step: • Compute the fluid forces acting on the structure, • Advance in time the structure under to the", "\\\\ y\\end{pmatrix} \\cdot d\\mathbf r + \\hat {\\mathbf y} \\oint \\begin{pmatrix}z \\\\ 0 \\\\ -x\\end{pmatrix} \\cdot d\\mathbf r + \\hat {\\mathbf z} \\oint \\begin{pmatrix}-y \\\\ x \\\\ 0\\end{pmatrix} \\cdot d\\mathbf r$$ Applying the Kelvin-Stokes theorem we get: $$\\oint \\mathbf r \\times d\\mathbf r = \\hat {\\mathbf x} \\iint (\\nabla \\times \\begin{pmatrix}0 \\\\ -z \\\\ y\\end{pmatrix}) \\cdot d\\mathbf S + \\hat {\\mathbf y} \\iint (\\nabla \\times \\begin{pmatrix}z \\\\ 0 \\\\ -x\\end{pmatrix}) \\cdot d\\mathbf S + \\hat {\\mathbf z} \\iint (\\nabla \\times \\begin{pmatrix}-y \\\\ x \\\\ 0\\end{pmatrix}) \\cdot d\\mathbf S$$ $$\\oint \\mathbf r \\times d\\mathbf r = \\hat {\\mathbf x} \\iint 2 \\hat {\\mathbf x} \\cdot d\\mathbf S + \\hat {\\mathbf y} \\iint 2 \\hat {\\mathbf y} \\cdot d\\mathbf S + \\hat {\\mathbf z} \\iint 2 \\hat {\\mathbf z} \\cdot d\\mathbf S$$ $$\\oint \\mathbf r \\times d\\mathbf r = 2 \\iint d\\mathbf S = 2 \\iint dS \\hat {\\mathbf n}$$ However, this is a factor 2 different from the equation you proposed. Last edited: Aug 28, 2011 5. Aug 29, 2011 daudaudaudau Sorry, I was missing a factor of two in the equation i posted, so your proof is correct :) 6. Sep 1, 2011 Anthony Alternatively, for arbitrary fixed vector $\\mathbf{a} \\in \\mathbf{R}^3$ $$\\mathbf{a} \\cdot \\oint \\mathbf{r} \\wedge \\mathrm{d}\\mathbf{r} = \\oint \\mathbf{a} \\cdot (\\mathbf{r} \\wedge \\mathrm{d}\\mathbf{r}) = \\oint (\\mathbf{a}", "## Solving a 3 x 3 System of Simultaneous Equation Using Matrix Techniques Suppose we have a system of equations $a_{11} x + a_{12}y + a_{13} z= \\alpha$ $a_{21} x + a_{22}y + a_{23} z= \\beta$ $a_{31} x + a_{32}y + a_{33} z= \\gamma$ We can write this in matrix form as $\\left( \\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\\\ a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \\end{array} \\right) \\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} = \\begin{pmatrix} \\alpha \\\\ \\beta \\\\ \\gamma \\end{pmatrix}$ Let $\\mathbf{a} = \\begin{pmatrix}a_{11}\\\\a_{21}\\\\a_{31}\\end{pmatrix}, \\: \\mathbf{b} = \\begin{pmatrix}a_{12}\\\\a_{22}\\\\a_{32}\\end{pmatrix} , \\: \\mathbf{c} = \\begin{pmatrix}a_{13}\\\\a_{23}\\\\a_{33}\\end{pmatrix}$ Then $A^{-1} = \\frac{1}{det (A)} ( \\mathbf{b} \\times \\mathbf{c} \\: \\mathbf{c} \\times \\mathbf{a} \\; \\mathbf{a} \\times \\mathbf{b} )$ . Hence $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} =\\frac{1}{det (A)} ( \\mathbf{b} \\times \\mathbf{c} \\: \\mathbf{c} \\times \\mathbf{a} \\; \\mathbf{a} \\times \\mathbf{b} ) \\begin{pmatrix} \\alpha \\\\ \\beta \\\\ \\gamma \\end{pmatrix} = \\begin{pmatrix} \\frac{( \\mathbf{b} \\times \\mathbf{c} ) \\cdot (\\alpha \\mathbf{i} + \\beta \\mathbf{j} + \\gamma \\mathbf{j})}{det(A)} \\\\ \\frac{( \\mathbf{c} \\times \\mathbf{a} ) \\cdot (\\alpha \\mathbf{i} + \\beta \\mathbf{j} + \\gamma \\mathbf{j})}{det(A)} \\\\ \\frac{( \\mathbf{a} \\times \\mathbf{b} ) \\cdot (\\alpha \\mathbf{i} + \\beta \\mathbf{j} + \\gamma \\mathbf{j})}{det(A)} \\end{pmatrix}$ . In matrix form $x= \\frac{\\left| \\begin{array}{ccc} \\alpha & a_{12} & a_{13} \\\\ \\beta & a_{22} & a_{23} \\\\ \\gamma & a_{32} & a_{33} \\end{array} \\right|}{\\left| \\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\\\", "= \\begin{pmatrix} \\begin{pmatrix} \\mathbb{A}_{\\overline{K}} & \\mathcal{O}\\\\ \\mathcal{O} & \\mathbb{A}_{L} \\end{pmatrix}^T \\mathbb{Q} + \\mathbb{Q} \\begin{pmatrix} \\mathbb{A}_{\\overline{K}} & \\mathcal{O}\\\\ \\mathcal{O} & \\mathbb{A}_{L} \\end{pmatrix} + \\mathbb{I}_{2n} & \\mathbb{Q} \\begin{pmatrix} D_{11} & D_{12}\\\\ D_{21} & D_{22} \\end{pmatrix}\\\\ \\begin{pmatrix} D_{11} & D_{12}\\\\ D_{21} & D_{22} \\end{pmatrix}^T \\mathbb{Q}^T & -\\mu \\mathbb{I}_{q} \\end{pmatrix} + \\begin{pmatrix} \\mathbb{Z}^{-1} \\overline{B} \\overline{K}\\\\ \\mathcal{O}\\\\ \\mathcal{O} \\end{pmatrix} \\begin{pmatrix} \\mathcal{O} & \\tilde{B} & \\mathcal{O} \\end{pmatrix} + \\begin{pmatrix} \\mathcal{O}\\\\ \\tilde{B}^T\\\\ \\mathcal{O} \\end{pmatrix} \\begin{pmatrix} \\left(\\overline{B} \\overline{K}\\right)^T \\mathbb{Z}^{-1} & \\mathcal{O} & \\mathcal{O} \\end{pmatrix}$$ If you expand further, you'll see that the following needs to be true for them to be equal: 1. $$\\mathbb{Q} \\begin{pmatrix} \\mathbb{A}_{\\overline{K}} & \\mathcal{O}\\\\ \\mathcal{O} & \\mathbb{A}_{L} \\end{pmatrix} + \\mathbb{Q} \\begin{pmatrix} \\mathbb{A}_{\\overline{K}} & \\mathcal{O}\\\\ \\mathcal{O} & \\mathbb{A}_{L} \\end{pmatrix}$$ is the same as $$\\begin{pmatrix} \\mathbb{A}_{\\overline{K}} & \\mathcal{O}\\\\ \\mathcal{O} & \\mathbb{A}_{L} \\end{pmatrix}^T \\mathbb{Q} + \\mathbb{Q} \\begin{pmatrix} \\mathbb{A}_{\\overline{K}} & \\mathcal{O}\\\\ \\mathcal{O} & \\mathbb{A}_{L} \\end{pmatrix}$$ 2. Given that $$\\mathbb{Q} = \\begin{pmatrix} \\mathbb{Z}^{-1} & \\mathcal{O}\\\\ \\mathcal{O} & \\mathbb{P} \\end{pmatrix}$$, $$\\mathbb{P}$$ is a positive-definite matrix with matrices $$\\mathbb{Z}$$ and $$\\mathbb{Q}$$ being symmetric, $$2 \\hspace{0.1cm} \\mathbb{Q} \\hspace{0.1cm} \\begin{pmatrix} \\mathcal{O} & \\overline{B} \\overline{K} \\tilde{B}\\\\ \\mathcal{O} & \\mathcal{O} \\end{pmatrix}$$ is the same as $$\\begin{pmatrix} \\mathcal{O} & \\mathbb{Z}^{-1} \\hspace{1mm} \\overline{B} \\overline{K} \\hspace{1mm} \\tilde{B}\\\\ \\tilde{B} \\hspace{1mm} \\left(\\overline{B} \\overline{K}\\right)^T \\mathbb{Z}^{-1} & \\mathcal{O} \\end{pmatrix}$$ 3. $$2 \\hspace{0.1cm} \\xi^T \\hspace{0.1cm} \\mathbb{Q} \\hspace{0.1cm} \\begin{pmatrix} D_{11} & D_{12}\\\\ D_{21} & D_{22} \\end{pmatrix} \\eta$$ is the same as $$\\eta" ]

[ "# Decide a number $a, b, c$ (expressed by $\\lambda$) so that $a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$ See the following vectors in $$\\mathbb{R}^3$$: $$\\mathbf{u}_1= \\begin{pmatrix} 1 \\\\ -1\\\\ 1 \\\\ 1 \\end{pmatrix}$$, $$\\mathbf{u}_2 = \\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\\\ 2 \\end{pmatrix}$$, $$\\mathbf{u}_3 = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\end{pmatrix}$$ and $$\\mathbf{v}_1= \\begin{pmatrix} 5 \\\\ -3\\\\ -4 \\\\ -1 \\end{pmatrix}$$, $$\\mathbf{v}_2 = \\begin{pmatrix} 6 \\\\ 4 \\\\ 1 \\\\ 8 \\end{pmatrix}$$, $$\\mathbf{v}_3 = \\begin{pmatrix} 7 \\\\ 2 \\\\ -1 \\\\ 6 \\end{pmatrix}$$ We are told that $$\\mathcal{B} = (\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3)$$ and $$\\mathcal{C} = (\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3)$$ both are bases for the same subspace $$\\mathcal{U}$$. Let $$\\mathcal{\\lambda}$$ be an unknown number og see the vector $$\\mathbf{x} = \\mathbf{v}_1 + \\lambda\\mathbf{v}_2 - \\mathbf{v}_3$$. Decide a number $$a, b, c$$ (expressed by $$\\lambda$$) so that $$a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$$ 1. Find some $$\\alpha_i$$ such that $$v_1=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3$$ 2. Find some $$\\beta_i$$ such that $$v_2=\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3$$ 3. Find some $$\\gamma_i$$ such that $$v_3=\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3$$ 4. Simplify the expression $$x=v_1+\\lambda v_2-v_3=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3+\\lambda(\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3)-(\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3)$$ • Thanks 5xum! When I reduce it, I get $\\begin{pmatrix} 8 \\\\ 3\\\\ -9 \\\\ 1 \\end{pmatrix} + a\\begin{pmatrix} 2 \\\\ -2\\\\ 1 \\\\ 1 \\end{pmatrix} + b\\begin{pmatrix} 3 \\\\ 6\\\\ 0 \\\\ 6 \\end{pmatrix} + c\\begin{pmatrix} 1 \\\\", "\\vec{w} \\\\\\\\ & &=& \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} \\times \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} \\\\\\\\ & &=& \\begin{vmatrix}1&1&1 \\\\ -4&0&2 \\\\ 0&-3&-1 \\end{vmatrix} \\\\\\\\ & &=& \\begin{pmatrix} 6 \\\\ -4 \\\\ 12 \\end{pmatrix} \\\\ \\hline n_1+n_2+n_3: & 6-4+12 &=& 14 \\quad | \\quad \\cdot \\dfrac{7}{14} \\\\\\\\ & 6\\cdot\\dfrac{7}{14}-4\\cdot\\dfrac{7}{14}+12\\cdot\\dfrac{7}{14} &=& 14\\cdot\\dfrac{7}{14} \\\\\\\\ &\\mathbf{ 3-2+6 }&=& \\mathbf{7} \\\\\\\\ & \\vec{n} &=& \\begin{pmatrix}\\mathbf{3} \\\\ \\mathbf{-2} \\\\ \\mathbf{6} \\end{pmatrix} \\\\ \\hline \\end{array} }$$ 2. Vector dot product $$\\begin{array}{|lrcll|} \\hline & \\vec{n}\\cdot \\vec{v} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{-4} \\\\ \\mathbf{0} \\\\ \\mathbf{2} \\end{pmatrix} &=& 0 \\\\ (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ \\hline & \\vec{n}\\cdot \\vec{w} &=& 0 \\\\ & \\begin{pmatrix}n_1 \\\\ n_2 \\\\ n_3 \\end{pmatrix} \\cdot \\begin{pmatrix}\\mathbf{0} \\\\ \\mathbf{-3} \\\\ \\mathbf{-1} \\end{pmatrix} &=& 0 \\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ \\hline (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ \\hline \\hline \\end{array}$$ $$\\begin{array}{|lrcll|} \\hline (1)& \\mathbf{-4n_1+2n_3} &=& \\mathbf{0} \\\\ &2n_3&=&4n_1 \\quad | \\quad :2 \\\\ & \\mathbf{n_3} &=& \\mathbf{2n_1} \\\\\\\\ (2) & \\mathbf{-3n_2-n_3} &=& \\mathbf{0} \\\\ & \\mathbf{n_3} &=& \\mathbf{-3n_2} \\\\\\\\ & n_3= 2n_1 &=& -3n_2 \\\\ & 2n_1 &=& -3n_2 \\quad | \\quad :2 \\\\ & \\mathbf{n_1} &=& \\mathbf{-\\dfrac{3}{2}n_2} \\\\\\\\ (3) & \\mathbf{n_1+n_2+n_3} &=& \\mathbf{7} \\\\ & -\\dfrac{3}{2}n_2 + n_2 + -3n_2 &=& 7 \\\\ & -\\dfrac{3}{2}n_2 -2n_2 &=& 7 \\quad | \\quad", "\\end{pmatrix}$ Using the definition of the inner product $(\\mathbf u, \\mathbf v) = \\mathbf v^* \\mathbf u$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (5-i)(1+3i) + (6-4i)(2+4i) = 36 + 30i$ , but using $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$, I get $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\begin{pmatrix} 1 \\\\ 2 \\\\ \\end{pmatrix} , \\begin{pmatrix} 5 \\\\ 6 \\\\ \\end{pmatrix}) + i(\\begin{pmatrix} 3 \\\\ 4 \\\\ \\end{pmatrix} , \\begin{pmatrix} 1 \\\\ 4 \\\\ \\end{pmatrix}) = 5 + 12 + i(3 + 16) = 17 + 19i$. What went wrong with my interpretation of the formula , $([\\mathbf x_1 , \\mathbf x_2] , [\\mathbf y_1 , \\mathbf y_2])_{X_C} = (\\mathbf x_1, \\mathbf y_1)_X + i(\\mathbf x_2, \\mathbf y_2)_X$ ? Shouldn't both computations be mathematically sound and give the same result? Thanks in advance for any help!", "3 & 1 & −2 \\\\ −2 & 0 & 5 \\end{pmatrix}$$ $$\\mathbf{F} = \\begin{pmatrix} 3 & 4 \\\\ 5 & 6 \\end{pmatrix}$$ $$\\mathbf{G} = \\begin{pmatrix} 2 & −4 \\\\ 7 & 1 \\\\ −1 & 1 \\end{pmatrix}$$ $$\\mathbf{H} = \\begin{pmatrix} −8 & −1 & −3 \\\\ −4 & 2 & −2 \\\\ −1 & 0 & 0 \\end{pmatrix}$$ Calculate each of the following 1. $$|\\mathbf{a} − (\\mathbf{b} + \\mathbf{3c})|$$ 7. $$(\\mathbf{EH})^T$$ 2. Component of $$\\mathbf{c}$$ along $$\\mathbf{a}$$ 8. $$|\\mathbf{HE}|$$ 3. Angle between $$\\mathbf{c}$$ and $$\\mathbf{d}$$ 9. $$\\mathbf{EHG}$$ 4. $$(\\mathbf{b} \\times \\mathbf{d}) \\cdot \\mathbf{a}$$ 10. $$\\mathbf{EG} − \\mathbf{HG}$$ 5. $$(\\mathbf{b} \\times \\mathbf{d}) \\times \\mathbf{a}$$ 11. $$\\mathbf{EH} − \\mathbf{H}^T \\mathbf{E}^T$$ 6. $$\\mathbf{b}\\times (\\mathbf{d} \\times \\mathbf{a})$$ 12. $$\\mathbf{F}^{−1}$$ 2. For what values of $$a$$ are the vectors $$\\mathbf{A} = 2a\\hat{i} − 2\\hat{j} + a\\hat{k}$$ and $$\\mathbf{B} = a\\hat{i} + 2a\\hat{j}+ 2\\hat{k}$$ perpendicular? 3. Show that the triple scalar product $$(A \\times B) \\cdot C$$ can be written as $(\\mathbf{A} \\times \\mathbf{B}) \\cdot \\mathbf{C} = \\begin{vmatrix} A_1 & A_2 & A_3 \\\\ B_1 & B_2 & B_3 \\\\ C_1 & C_2 & C_3 \\end{vmatrix} \\nonumber$ Show also that the product is unaffected by interchange of the scalar and vector product operations or by change in the order of $$A, B, C$$ as long as they are in cyclic order, that is", "\\\\ 0\\end{pmatrix} + \\beta e^{\\lambda_2 t} T \\begin{pmatrix} 0 \\\\ 1 \\end{pmatrix}\\\\ & = \\alpha e^{\\lambda_2 t} \\mathbf v_1 + \\beta e^{\\lambda_2 t} \\mathbf v_2. \\end{align*} Suppose $$d{\\mathbf x}/dt = A {\\mathbf x}\\text{,}$$ where \\begin{equation*} A = \\begin{pmatrix} 1 & 2 \\\\ 4 & 3 \\end{pmatrix}. \\end{equation*} The eigenvalues of $$A$$ are $$\\lambda_1 = 5$$ and $$\\lambda_2 = -1$$ and the associated eigenvectors are $$(1, 2)$$ and $$(1, -1)\\text{,}$$ respectively. In this case, our matrix $$T$$ is \\begin{equation*} \\begin{pmatrix} 1 & 1 \\\\ 2 & -1 \\end{pmatrix}. \\end{equation*} If $${\\mathbf e}_1 = (1, 0)$$ and $${\\mathbf e}_2 = (0, 1)\\text{,}$$ then $$T {\\mathbf e}_i = {\\mathbf v}_i$$ for $$i = 1, 2\\text{.}$$ Consequently, $$T^{-1} {\\mathbf v}_i = {\\mathbf e}_i$$ for $$i = 1, 2\\text{,}$$ where \\begin{equation*} T^{-1} = \\begin{pmatrix} 1/3 & 1/3 \\\\ 2/3 & -1/3 \\end{pmatrix}. \\end{equation*} Thus, \\begin{equation*} T^{-1} A T = \\begin{pmatrix} 1/3 & 1/3 \\\\ 2/3 & -1/3 \\end{pmatrix} \\begin{pmatrix} 1 & 2 \\\\ 4 & 3 \\end{pmatrix} \\begin{pmatrix} 1 & 1 \\\\ 2 & -1 \\end{pmatrix} = \\begin{pmatrix} 5 & 0 \\\\ 0 & -1 \\end{pmatrix}. \\end{equation*} The eigenvalues of the matrix \\begin{equation*} \\begin{pmatrix} 5 & 0 \\\\ 0 & -1 \\end{pmatrix} \\end{equation*} are $$\\lambda_1 = 5$$ and $$\\lambda_2 = -1$$ with eigenvectors $$(1, 0)$$ and $$(0, 1)\\text{,}$$ respectively. Thus, the general solution", "A \\mathbf x\\text{,}$$ where \\begin{equation*} A = \\begin{pmatrix} 5 & 1 \\\\ -4 & 1 \\end{pmatrix}. \\end{equation*} The characteristic polynomial of $$A$$ is $$\\lambda^2 - 6 \\lambda + 9 = (\\lambda - 3)^2\\text{,}$$ we have only a single eigenvalue $$\\lambda = 3$$ with eigenvector $$\\mathbf v = (1, -2)\\text{.}$$ Any other eigenvector for $$\\lambda$$ is a multiple of $$\\mathbf v\\text{.}$$ If we choose $$\\mathbf w = (1, 0)\\text{,}$$ then $$\\mathbf v$$ and $$\\mathbf w$$ are linearly independent. Furthermore, \\begin{equation*} A \\mathbf w = \\begin{pmatrix} 5 \\\\ - 4 \\end{pmatrix} = 2 \\begin{pmatrix} 1 \\\\ -2 \\end{pmatrix} + \\lambda \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = 2 \\begin{pmatrix} 1 \\\\ -2 \\end{pmatrix} + 3 \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix}. \\end{equation*} So we can let $$\\mathbf u = (1/2) \\mathbf w = (1/2, 0)\\text{.}$$ Therefore, the matrix that we seek is \\begin{equation*} T = \\begin{pmatrix} 1 \\amp 1/2 \\\\ -2 \\amp 0 \\end{pmatrix}, \\end{equation*} and \\begin{equation*} T^{-1} A T = \\begin{pmatrix} -1/2 & 2 \\\\ 1 & 1 \\end{pmatrix} \\begin{pmatrix} 5 & 1 \\\\ -4 & 1 \\end{pmatrix} \\begin{pmatrix} 1 \\amp 1/2 \\\\ -2 \\amp 0\\end{pmatrix} = \\begin{pmatrix} 3 & 1 \\\\ 0 & 3 \\end{pmatrix}. \\end{equation*} From Section 3.3, we know that the general solution to the system \\begin{equation*} \\begin{pmatrix} dx/dt \\\\ dy/dt \\end{pmatrix} = \\begin{pmatrix} 3 & 1 \\\\", "= {2\\mathbf x^T A \\cdot \\mathbf x^T\\mathbf x - \\mathbf x^T A \\mathbf x \\cdot 2\\mathbf x^T \\over (\\mathbf x^T\\mathbf x)^2}(^1_0)$$ In other words, if $A=A^T$: $$Df(\\mathbf x) = {2\\mathbf x^T A \\cdot \\mathbf x^T\\mathbf x - \\mathbf x^T A \\mathbf x \\cdot 2\\mathbf x^T \\over (\\mathbf x^T\\mathbf x)^2}$$ Last edited: Jan 24, 2013 3. Jan 24, 2013 ### libragirl79 Since I forgot a lot of my matrix algebra, when you say A (x y) it looks like x choose y, but i am assuming that's not what you meant... 4. Jan 24, 2013 ### I like Serena Since you are talking about x transpose, that means that x is a vector. A vector is commonly written as: $$\\mathbf x = \\begin{pmatrix}x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n\\end{pmatrix}$$ I have simplified it to: $$\\mathbf x = \\begin{pmatrix}x \\\\ y\\end{pmatrix}$$ So indeed, this does not mean x choose y. ;) Furthermore: $$\\mathbf x^T = \\begin{pmatrix}x_1 & x_2 & \\dots & x_n\\end{pmatrix}$$ and $$\\mathbf x^T \\mathbf x = \\begin{pmatrix}x_1 & x_2 & \\dots & x_n\\end{pmatrix}\\begin{pmatrix}x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n\\end{pmatrix} = x_1^2 + x_2^2 + \\dots + x_n^2$$ 5. Jan 24, 2013 ### libragirl79 Ok, I see! Thanks!! :)", "$$\\mathbf A= \\begin{pmatrix} 0&1&0&0&-1\\\\ 0&0&0&0&0\\\\ 0&0&0&1&-1\\\\ 0&0&0&0&1\\\\ 0&0&0&0&0 \\end{pmatrix}$$ • #### Solution It has a single eigenvalue $$\\lambda=0$$ of algebraic multiplicity 5. The rank of $$\\mathbf A$$ is 3, so $$\\dim(\\ker A)=2$$ and we have to linearly independent eigenvectors. When we would like for the eigenvector $$\\mathbf x_1$$ calculate a generalized eigenvector $$\\mathbf x_2$$, it would mean to solve $$(\\mathbf A-\\lambda \\mathbf I)\\mathbf x_2=\\mathbf x_1$$, what is equivalent to solving $$(\\mathbf A-\\lambda \\mathbf I)^2\\mathbf x_2 = (\\mathbf A-\\lambda \\mathbf I)\\mathbf x_1 = \\mathbf 0$$. In other words, $$\\mathbf x_2 \\in \\ker((\\mathbf A-\\lambda \\mathbf I)^2) \\setminus \\ker(\\mathbf A-\\lambda \\mathbf I)$$ etc. for the other generalized eigenvectors in the chain. In our case with $$\\lambda=0$$ we calculate $\\mathbf A^2= \\begin{pmatrix} 0&0&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&0&1\\\\ 0&0&0&0&0\\\\ 0&0&0&0&0 \\end{pmatrix}$ a $\\mathbf A^3= \\begin{pmatrix} 0&0&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&0&0\\\\ 0&0&0&0&0 \\end{pmatrix}$ Clearly $$\\dim(\\ker \\mathbf A^2)=4$$ and $$\\dim(\\ker \\mathbf A^3)=5$$, so there is only one linearly independent vector in $$\\ker \\mathbf A^3 \\setminus \\ker \\mathbf A^2$$, e.g. $$x_3=(0{,}0,0{,}0,1)^T$$. This shall satisfy $$\\mathbf A\\mathbf x_3=\\mathbf x_2$$, so we claculate $$\\mathbf x_2=(-1{,}0,-1{,}1,0)^T$$ and $$\\mathbf x_1=\\mathbf A\\mathbf x_2=(0{,}0,1{,}0,0)^T$$ (this is an eigenvector of the matrix $$\\mathbf A$$). This yields a single Jordan cell of size 3. The other cell will be of size 2 and again we will start at the end of the chain, i.e. by the generalized", "0 \\\\ 4 \\end{pmatrix}$$. 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$. 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$. 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$. Note: this exercise can be downloaded as a worksheet to practice with: ## Solution Without Working 1. For vector $$\\vec{a} = \\begin{pmatrix} 8 \\\\ 6 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{a} \\end{vmatrix} = 10$ 2. For vector $$\\vec{b} = \\begin{pmatrix} 2 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{b} \\end{vmatrix} = \\sqrt{29} \\quad \\begin{pmatrix}5.39 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{c} = \\begin{pmatrix} -2 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{c} \\end{vmatrix} = \\sqrt{13} \\quad \\begin{pmatrix}3.61 \\ \\text{2 d.p} \\end{pmatrix}$ 4. For vector $$\\vec{d} = \\begin{pmatrix} -4 \\\\ 5 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{d} \\end{vmatrix} = \\sqrt{41} \\quad \\begin{pmatrix}6.40 \\ \\text{2 d.p} \\end{pmatrix}$ 1. For vector $$\\vec{e} = \\begin{pmatrix} 0 \\\\ 4 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{e} \\end{vmatrix} = 4$ 2. For vector $$\\vec{f} = \\begin{pmatrix} 2 \\\\ -2 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{f} \\end{vmatrix} = \\sqrt{8} = 2\\sqrt{2} \\quad \\begin{pmatrix}2.83 \\ \\text{2 d.p} \\end{pmatrix}$ 3. For vector $$\\vec{g} = \\begin{pmatrix} 5 \\\\ -12 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{g} \\end{vmatrix} = 13$ 4. For vector $$\\vec{h} = \\begin{pmatrix} -4 \\\\ 3 \\end{pmatrix}$$, we find: $\\begin{vmatrix} \\vec{h} \\end{vmatrix} = 5$", "2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right)$ . The 1's in the second diagonal are an approximation to $5^{1/4}$ , the 2's an approximation to $5^{2/4}$ , the 4's an approximation to $5^{3/4}$ . Let $\\mathbf{v}= \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix}$ . Then $A \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}1\\\\1\\\\1\\\\1\\end{pmatrix} = \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix}$ . $A^2 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}24\\\\16\\\\12\\\\8\\end{pmatrix} = \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix}$ . $A^3 \\mathbf{v}= \\left( \\begin{array}{cccc} 4 & 5 & 5 & 10 \\\\ 2 & 4 & 5 & 5 \\\\ 1 & 2 & 4 & 5 \\\\ 1 & 1 & 2 & 4 \\end{array} \\right) \\begin{pmatrix}316\\\\212\\\\144\\\\96\\end{pmatrix} = \\begin{pmatrix}4004\\\\2680\\\\1796\\\\1200\\end{pmatrix}$ . Hence $5^{3/4} \\simeq \\frac{4004}{1200}=3.336666667$ , $5^{2/4} \\simeq \\frac{2680}{1200}=2.233333333$ and $5^{1/4} \\simeq \\frac{1796}{1200}=2.1.496666667$ . Compare these with the true values $5^{3/4} = 3.343701525$ , $5^{2/4} = 2.236067977$ , and $5^{1/4} = 1.495348781$ , all to 10 significant figures.", "\\mathbf{U}^T\\mathbf{U} = \\mathbf{I}$$, $$\\mathbf{V}\\mathbf{V}^T = \\mathbf{V}^T\\mathbf{V} = \\mathbf{I}$$, and $$\\Sigma$$ containing the singular values of $$\\mathbf{A}$$, $$\\sigma_i$$, $$1 \\leq I \\leq \\min{(m, n)}$$, along the diagonal. Then, the factorization of $$\\mathbf{B}$$ in terms of $$\\mathbf{U}$$, $$\\mathbf{V}$$, and $$\\Sigma$$ is the following: $$\\mathbf{B} = \\begin{pmatrix} \\mathbf{0} & \\mathbf{A}^T \\\\ \\mathbf{A} & \\mathbf{0} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{0} & \\mathbf{V}\\Sigma^T\\mathbf{U}^T \\\\ \\mathbf{U}\\Sigma\\mathbf{V}^T & \\mathbf{0} \\end{pmatrix} = \\begin{pmatrix} \\mathbf{V} & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{U} \\end{pmatrix} \\begin{pmatrix} \\mathbf{0} & \\Sigma^T \\\\ \\Sigma & \\mathbf{0} \\end{pmatrix} \\begin{pmatrix} \\mathbf{V}^T & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{U}^T \\end{pmatrix} = \\mathbf{Q}_1 \\mathbf{C} \\mathbf{Q}_1^T$$ where $$\\mathbf{Q}_1 = \\begin{pmatrix} \\mathbf{V} & \\mathbf{0} \\\\ \\mathbf{0} & \\mathbf{U} \\end{pmatrix}$$ and $$\\mathbf{C} = \\begin{pmatrix} \\mathbf{0} & \\Sigma^T \\\\ \\Sigma & \\mathbf{0} \\end{pmatrix}$$. We can readily verify that $$\\mathbf{Q}_1$$ is orthogonal since $$\\mathbf{Q}_1\\mathbf{Q}_1^T = \\mathbf{Q}_1^T\\mathbf{Q}_1 = \\mathbf{I}$$. However, $$\\mathbf{C}$$ has not the expected diagonal structure. Instead, the singular values, $$\\sigma_i$$, of $$\\mathbf{A}$$ are distributed in the following way: $$\\mathbf{C} = \\begin{pmatrix} \\mathbf{0} & \\Sigma^T \\\\ \\Sigma & \\mathbf{0} \\\\ \\end{pmatrix} = \\left( \\begin{array}{ccc|ccc} & & & \\sigma_1 & & \\\\ & \\mathbf{0} & & & \\sigma_2 & \\\\ & & & & & \\ddots \\\\ \\hline \\sigma_1 & & & & & \\\\ & \\sigma_2 & & & \\mathbf{0} & \\\\ & & \\ddots & & & \\end{array} \\right)$$ That is, $$\\mathbf{C}$$ contains", "b_{21} & b_{22} \\end{pmatrix} \\begin{pmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\end{pmatrix} \\Rightarrow \\\\ \\begin{cases}\\require{cancel}\\cancel{a_{11}b_{11}}+a_{12}b_{21}=\\cancel{b_{11}a_{11}}+b_{12}a_{21}\\\\ a_{11}b_{12}+a_{12}b_{22}=b_{11}a_{12}+b_{12}a_{22}\\\\ a_{21}b_{11}+a_{22}b_{21}=b_{21}a_{11}+b_{22}a_{21}\\\\ a_{21}b_{12}+\\cancel{a_{22}b_{22}}=b_{21}a_{12}+\\cancel{b_{22}a_{22}}\\end{cases}$$ From$(1)$, since$b_{12}$and$b_{21}$can be any number, in particular,$b_{12}=0$and$b_{21}\\ne 0$, we get:$a_{12}=0$. Similarly, for$b_{12}\\ne 0$and$b_{21}=0$, we get$a_{21}=0$. From$(2)$, since$a_{12}=0$and$b_{12}$is an arbitrary number, we get$a_{11}b_{12}=b_{12}a_{22} \\Rightarrow a_{11}=a_{22}\\$.", "1 \\\\ -4 & -2 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix} 2 \\\\ -4 \\end{pmatrix} = 2 {\\mathbf v}_1. \\end{equation*} Thus, we can take ${\\mathbf v}_2 = (1/2)\\mathbf w = (1/2, 0)\\text{,}$ and our second solution is \\begin{equation*} {\\mathbf x}_2 = e^{\\lambda t} ({\\mathbf v}_2 + t {\\mathbf v}_1) = e^{3t} \\begin{pmatrix} 1/2 + t \\\\ -2t \\end{pmatrix} \\end{equation*} Thus, our general solution is \\begin{equation*} {\\mathbf x} = c_1 {\\mathbf x}_1 + c_2 {\\mathbf x}_2 = c_1 e^{3t} \\begin{pmatrix} 1 \\\\ -2 \\end{pmatrix} + c_2 e^{3t} \\begin{pmatrix} 1/2 + t \\\\ -2t \\end{pmatrix}. \\end{equation*} If the eigenvalue is positive, we will have a nodal source. If it is negative, we will have a nodal sink. Notice that we have only given a recipe for finding a solution to $\\mathbf x' = A \\mathbf x\\text{,}$ where $A$ has a repeated eigenvalue and any two eigenvectors are linearly dependent. We will justify our procedure in the next section (Section 3.6). ### Subsection3.5.3Important Lessons • If \\begin{equation*} A = \\begin{pmatrix} \\lambda & 1 \\\\ 0 & \\lambda \\end{pmatrix}, \\end{equation*} then $A$ has one repeated real eigenvalue. The general solution to the system ${\\mathbf x}' = A {\\mathbf x}$ is \\begin{equation*} {\\mathbf x}(t) = \\alpha e^{\\lambda t} \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} + \\beta e^{\\lambda t} \\begin{pmatrix} t \\\\ 1", "\\mathbf O_{n-1} \\\\ 0 & \\end{pmatrix}=\\begin{pmatrix} x_{0} & 0 & \\ldots & 0 \\\\ 0 & & & \\\\ \\vdots & & \\mathbf O_{n-1} \\\\ 0 & & & \\end{pmatrix}\\end{align}\\\\ \\begin{align}&\\Big(I_n-\\frac{X X^T}{\\Vert X \\Vert^{2}} \\Big) \\begin{pmatrix} \\ast & \\ast & \\ldots & \\ast \\\\ \\ast & & \\\\ \\vdots & & S \\\\ \\ast & & & \\end{pmatrix} \\Big(I_n-\\frac{X X^\\top}{\\Vert X \\Vert^{2}} \\Big)^T\\\\ =&\\mathcal R_X\\begin{pmatrix} \\ast & \\ast & \\ldots & \\ast \\\\ \\ast & & \\\\ \\vdots & & S \\\\ \\ast & & & \\end{pmatrix}\\mathcal R_X^T =\\begin{pmatrix} 0 & 0 & \\ldots & 0 \\\\ \\ast & & \\\\ \\vdots & & S \\\\ \\ast & & & \\end{pmatrix}\\mathcal R_X^T\\\\ =&\\begin{pmatrix} 0 & 0 & \\ldots & 0 \\\\ 0 & & \\\\ \\vdots & & S \\\\ 0 & & & \\end{pmatrix}\\end{align} Hence $$\\Gamma=x_0\\mathcal P_XI_n+\\mathcal R_X\\tilde S\\mathcal R_X^T$$ Thanks a lot ! I can't figure out where I was wrong. Was it my expression for $\\mathrm{I}_{n} - \\frac{X X^\\top}{\\Vert X \\Vert^{2}}$ ? – jibounet Nov 22 '13 at 15:26", "$[\\mathbf{A},\\mathbf{B}]=\\mathbf{A}\\mathbf{B}-\\mathbf{B}\\mathbf{A} \\nonumber$ ##### Example $$\\PageIndex{3}$$ Given $$\\mathbf{A}=\\begin{pmatrix} 3&1 \\\\ 2&0 \\end{pmatrix}$$ and $$\\mathbf{B}=\\begin{pmatrix} 1&0 \\\\ -1&2 \\end{pmatrix}$$ Calculate the commutator $$[\\mathbf{A},\\mathbf{B}]$$ ###### Solution $[\\mathbf{A},\\mathbf{B}]=\\mathbf{A}\\mathbf{B}-\\mathbf{B}\\mathbf{A}\\nonumber$ $\\mathbf{A}\\mathbf{B}=\\begin{pmatrix} 3&1 \\\\ 2&0 \\end{pmatrix}\\begin{pmatrix} 1&0 \\\\ -1&2 \\end{pmatrix}=\\begin{pmatrix} 3\\times 1+1\\times (-1)&3\\times 0 +1\\times 2 \\\\ 2\\times 1+0\\times (-1)&2\\times 0+ 0\\times 2 \\end{pmatrix}=\\begin{pmatrix} 2&2 \\\\ 2&0 \\end{pmatrix}\\nonumber$ $\\mathbf{B}\\mathbf{A}=\\begin{pmatrix} 1&0 \\\\ -1&2 \\end{pmatrix}\\begin{pmatrix} 3&1 \\\\ 2&0 \\end{pmatrix}=\\begin{pmatrix} 1\\times 3+0\\times 2&1\\times 1 +0\\times 0 \\\\ -1\\times 3+2\\times 2&-1\\times 1+2\\times 0 \\end{pmatrix}=\\begin{pmatrix} 3&1 \\\\ 1&-1 \\end{pmatrix}\\nonumber$ $[\\mathbf{A},\\mathbf{B}]=\\mathbf{A}\\mathbf{B}-\\mathbf{B}\\mathbf{A}=\\begin{pmatrix} 2&2 \\\\ 2&0 \\end{pmatrix}-\\begin{pmatrix} 3&1 \\\\ 1&-1 \\end{pmatrix}=\\begin{pmatrix} -1&1 \\\\ 1&1 \\end{pmatrix}\\nonumber$ $\\displaystyle{\\color{Maroon}[\\mathbf{A},\\mathbf{B}]=\\begin{pmatrix} -1&1 \\\\ 1&1 \\end{pmatrix}}\\nonumber$ ## Multiplication of a vector by a scalar The multiplication of a vector $$\\vec{v_1}$$ by a scalar $$n$$ produces another vector of the same dimensions that lies in the same direction as $$\\vec{v_1}$$; $n\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\begin{pmatrix} nx \\\\ ny \\end{pmatrix} \\nonumber$ The scalar can stretch or compress the length of the vector, but cannot rotate it (figure [fig:vector_by_scalar]). ## Multiplication of a square matrix by a vector The multiplication of a vector $$\\vec{v_1}$$ by a square matrix produces another vector of the same dimensions of $$\\vec{v_1}$$. For example, we can multiply a $$2\\times 2$$ matrix and a 2-dimensional vector: $\\begin{pmatrix} a&b \\\\ c&d \\end{pmatrix}\\begin{pmatrix} x \\\\ y \\end{pmatrix}=\\begin{pmatrix} ax+by \\\\ cx+dy \\end{pmatrix} \\nonumber$ For example, consider the matrix $\\mathbf{A}=\\begin{pmatrix} -2 &0", "Review question # Can we find the equation of the plane containing $l_1$ and $PQ$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R5089 ## Solution The lines $l_1$ and $l_2$ have equations $\\mathbf{r} = \\mathbf{a} + s\\mathbf{b}$ and $\\mathbf{r} = \\mathbf{c} + t\\mathbf{d}$, where $s$ and $t$ are real numbers and $\\mathbf{a} = 5\\mathbf{i}+ 6\\mathbf{j}- \\mathbf{k}, \\quad\\mathbf{b} = \\mathbf{i} +\\mathbf{j}+ \\mathbf{k}, \\quad\\mathbf{c} = \\mathbf{i}+ 13\\mathbf{j} +7\\mathbf{k}, \\quad\\mathbf{d} = \\mathbf{i} + 6\\mathbf{j} + 3\\mathbf{k}.$ Also, the point $P$ on $l_1$ and the point $Q$ on $l_2$ are such that $PQ$ is perpendicular to both $l_1$ and $l_2.$ In any order: 1. show that $PQ = \\sqrt{38},$ 2. find the position vectors of $P$ and $Q,$ We can write the position vector of $P$ as $\\mathbf{p} = \\begin{pmatrix}5+s \\\\ 6+s \\\\ -1+s\\end{pmatrix}$, and the position vector of $Q$ as $\\mathbf{q} = \\begin{pmatrix}1+t \\\\ 13+6t\\\\ 7+3t\\end{pmatrix}.$ Thus $\\mathbf{q} - \\mathbf{p} = \\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}$. This vector has to be perpendicular to both $l_1$ and $l_2$, and so using the scalar product, $\\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}.\\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=0 \\quad\\text{and}\\quad \\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}.\\begin{pmatrix}1\\\\ 6 \\\\ 3\\end{pmatrix}=0.$ This gives us the two equations $10t-3s+11=0$ and $46t-10s+62=0$, which solve to give $s = -3$, $t", "we have $\\mathbf{i} + \\mathbf{j} + \\mathbf{k} + a_i \\mathbf{i} + a_j \\mathbf{j} + a_k \\mathbf{k} = 2\\mathbf{i} - 3\\mathbf{j} - \\mathbf{k}$ Comparing co-efficients, we have $\\begin{cases} 1 + a_i = 2 \\\\ 1 + a_j = -3 \\\\ 1 + a_k = -1 \\end{cases}$ Thus $$a_i = 1$$, $$a_j = -4$$, $$a_k = -2$$, and therefore $$\\mathbf{x} = \\mathbf{i} - 4 \\mathbf{j} - 2 \\mathbf{k}$$ Exercise 10 $$\\mathbf{0} = \\begin{pmatrix}0 \\\\ 0 \\\\ 0\\end{pmatrix}\\quad \\mathbf{i} = \\begin{pmatrix}1 \\\\ 0 \\\\ 0\\end{pmatrix} \\quad \\mathbf{j} = \\begin{pmatrix}0 \\\\ 1 \\\\ 0\\end{pmatrix} \\quad \\mathbf{k} = \\begin{pmatrix}0 \\\\ 0 \\\\ 1\\end{pmatrix}$$ ### Vector Equation of a Straight Line If $$A$$ and $$B$$ are any two distinct points on a straight line in space with position vectors $$\\mathbf{a}$$ and $$\\mathbf{b}$$ respectively, then the vector equation of a straight line is $\\mathbf{r} = (1 - s)\\mathbf{a} + s \\mathbf{b}$ where $$\\mathbf{r}$$ represents the position vector of any point on the line. In Cartesian coordinate form, with $\\mathbf{a} = a_1 \\mathbf{i} + a_2 \\mathbf{j} + a_3 \\mathbf{k}$ $\\mathbf{b} = b_1 \\mathbf{i} + b_2 \\mathbf{j} + b_3 \\mathbf{k}$ $\\mathbf{r} = x \\mathbf{i} + y \\mathbf{j} + z \\mathbf{k}$ the equation of the line is $x = a_1 + s (b_1 - a_1)\\quad,\\quad y = a_2 + s (b_2 - a_2)\\quad,\\quad z = a_3 + s", "it (with an appropriate multiplier) to make all the numbers in the second column, zero, that is: $\\begin{bmatrix} \\mathbf{0} && \\mathbf{0} && ... && \\mathbf{0}\\\\ \\mathbf{0} && a'_{22} && ... && a'_{2N}\\\\ \\mathbf{0} && a'_{32} && ... && a'_{3N}\\\\ ... && ... && .. && ...\\\\ \\mathbf{0} && a'_{N2} && ... && a'_{NN}\\\\ \\end{bmatrix}\\\\ = \\ell_1u_1+ \\begin{bmatrix} 0\\\\ 1\\\\ \\ell_{23}\\\\ .\\\\ .\\\\ \\ell_{2N}\\\\ \\end{bmatrix} \\begin{bmatrix} 0 && a'_{22} && a'_{23} && ... && a_{1N}\\\\ \\end{bmatrix} + \\begin{bmatrix} \\mathbf{0} && \\mathbf{0} && ... && \\mathbf{0}\\\\ \\mathbf{0} && \\mathbf{0} && ... && \\mathbf{0}\\\\ \\mathbf{0} && \\mathbf{0} && ... && a''_{3N}\\\\ ... && ... && .. && ...\\\\ \\mathbf{0} && \\mathbf{0} && ... && a''_{NN}\\\\ \\end{bmatrix}\\\\$ ### Step 4 Call the first term on the LHS as $$\\ell_2u_2$$, so that: $\\begin{bmatrix} \\mathbf{0} && \\mathbf{0} && ... && \\mathbf{0}\\\\ \\mathbf{0} && a'_{22} && ... && a'_{2N}\\\\ \\mathbf{0} && a'_{32} && ... && a'_{3N}\\\\ ... && ... && .. && ...\\\\ \\mathbf{0} && a'_{N2} && ... && a'_{NN}\\\\ \\end{bmatrix}\\\\ = \\ell_1u_1+ \\ell_2u_2+ \\begin{bmatrix} \\mathbf{0} && \\mathbf{0} && ... && \\mathbf{0}\\\\ \\mathbf{0} && \\mathbf{0} && ... && \\mathbf{0}\\\\ \\mathbf{0} && \\mathbf{0} && ... && a''_{3N}\\\\ ... && ... && .. && ...\\\\ \\mathbf{0} && \\mathbf{0} && ... && a''_{NN}\\\\ \\end{bmatrix}\\\\$ ### Step 5 I hope you can see the pattern, we are gradually aiming", "Express the vector $\\mathbf{u}$ as a linear combination of $\\mathbf{v}$ and $\\mathbf{w}$. (b) Compute $A^5\\mathbf{v}$. (c) Compute $A^5\\mathbf{w}$. (d) Compute $A^5\\mathbf{u}$. ## Problem 691 In this problem, we use the following vectors in $\\R^2$. $\\mathbf{a}=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}, \\mathbf{b}=\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}, \\mathbf{c}=\\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix}, \\mathbf{d}=\\begin{bmatrix} 3 \\\\ 2 \\end{bmatrix}, \\mathbf{e}=\\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}, \\mathbf{f}=\\begin{bmatrix} 5 \\\\ 6 \\end{bmatrix}.$ For each set $S$, determine whether $\\Span(S)=\\R^2$. If $\\Span(S)\\neq \\R^2$, then give algebraic description for $\\Span(S)$ and explain the geometric shape of $\\Span(S)$. (a) $S=\\{\\mathbf{a}, \\mathbf{b}\\}$ (b) $S=\\{\\mathbf{a}, \\mathbf{c}\\}$ (c) $S=\\{\\mathbf{c}, \\mathbf{d}\\}$ (d) $S=\\{\\mathbf{a}, \\mathbf{f}\\}$ (e) $S=\\{\\mathbf{e}, \\mathbf{f}\\}$ (f) $S=\\{\\mathbf{a}, \\mathbf{b}, \\mathbf{c}\\}$ (g) $S=\\{\\mathbf{e}\\}$ ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 672 For an integer $n > 0$, let $\\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \\{ 1 , x , x^2 , \\cdots , x^n \\}$ is a basis of $\\mathrm{P}_n$, called the standard basis. Let $T : \\mathrm{P}_n \\rightarrow \\mathrm{P}_{n+1}$ be the map defined by, for $f \\in \\mathrm{P}_n$, $T (f) (x) =", "will use the standard basis vectors \\begin{equation*} {\\mathbf e}_1 = \\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\end{pmatrix}, {\\mathbf e}_2 = \\begin{pmatrix} 0 \\\\ 1 \\\\ 0 \\end{pmatrix}, {\\mathbf e}_3 = \\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}. \\end{equation*} Thus, the general solution to our system is \\begin{equation*} {\\mathbf x}(t) = c_1 e^{-t} \\begin{pmatrix} 1 - 18t + 5t^2/2 \\\\ 8t - 5t^2/2 \\\\ -7t + 10t^2 \\end{pmatrix} + c_2 e^{-t} \\begin{pmatrix} -46t +13t^2/2 \\\\ 1 + 20t -13t^2/2 \\\\ -15t + 26t^2 \\end{pmatrix} + c_3 e^{-t} \\begin{pmatrix} - 7t + t^2 \\\\ 3t - t^2 \\\\ 1 - 2t + 4t^2 \\end{pmatrix}. \\end{equation*} Subsection3.9.2Generalized Eigenvalues Example3.9.10 Consider the system ${\\mathbf x}' = A {\\mathbf x}\\text{,}$ where \\begin{equation*} A = \\begin{pmatrix} 7 \\amp -1 \\amp 2 \\\\ 18 \\amp -2 \\amp 6 \\\\ -9 \\amp 2 \\amp -1 \\end{pmatrix}\\text{.} \\end{equation*} The characteristic polynomial of $A$ is \\begin{equation*} \\det(A - \\lambda I) = \\lambda^3 - 4\\lambda^2 + 5 \\lambda - 2 = (\\lambda - 1)^2 (\\lambda - 2). \\end{equation*} The eigenvalue $\\lambda_1 = 1$ has eigenvector $\\mathbf v_1 = (2, 6, -3)$ and the eigenvalue $\\lambda_2 = 2$ has eigenvector $\\mathbf v_2 = (1, 3, -1)\\text{.}$ Thus, we can find two linearly independent solutions in this case \\begin{equation*} {\\mathbf x}_1(t) = e^{t} \\begin{pmatrix} 2 \\\\ 6 \\\\ -3 \\end{pmatrix} \\qquad\\text{and}\\qquad" ]

[ "^ 3\\right) \\tan \\left(\\frac{1}{x} ^ 3\\right)$", "\\left(t\\right)}=\\frac{1-\\tan t}{1+\\tan t}$", "+ \\tan x) \\right]_0^{\\pi/4}$$", "on the final step, but your evaluation is correct.$$-\\tan\\left(-\\frac\\pi3\\right)=\\sqrt 3$$", "\\pi + {\\tan}^{-} 1 \\left(\\frac{- 1}{4}\\right) + 2 \\pi\\right) + i \\sin \\left({\\tan}^{-} 1 \\left(6\\right) + \\pi + {\\tan}^{-} 1 \\left(\\frac{- 1}{4}\\right) + 2 \\pi\\right)\\right)$", "$\\mathrm{tan}\\left(\\alpha -\\beta \\right)=\\left(1-n\\right)\\mathrm{tan}\\alpha$ What are you looking for?", "= tan^{-1} \\left ( \\frac{1}{2.3} \\right )$ -Dan", "1 \\left(\\frac{72.0881}{-} 0.3893\\right)$ ${\\theta}_{a e} = \\pi - {\\tan}^{-} 1 \\left(\\frac{72.0881}{0.3893}\\right)$", "\\left(\\tan ^{-1} \\left(4x\\right)\\right)$$ Section 6.4 Solving Trig Equations 439", "\\frac{\\tan 2 \\theta}{2 \\theta} \\times 2 = 2 \\left(1\\right) \\times 2 = 4$", "\\left(\\theta\\right) < \\frac{5 \\pi}{4}$", "\\left(\\frac{1}{7}\\right)\\right) + i \\sin \\left({\\tan}^{-} 1 \\left(\\frac{2}{5}\\right) - {\\tan}^{-} 1 \\left(\\frac{1}{7}\\right)\\right)\\right)$", "# Prove that $\\tan \\left(\\frac{3\\pi}{11}\\right)+4\\sin \\left(\\frac{2\\pi}{11}\\right)=\\sqrt {11}$ [duplicate] Prove that $$\\tan \\left(\\frac{3\\pi}{11}\\right)+4\\sin \\left(\\frac{2\\pi}{11}\\right)=\\sqrt {11}$$", "$\\tag{12} {µ_0}=\\frac{\\sin{\\left( \\alpha \\right) }}{\\cos{\\left( \\alpha \\right) }}$ or converted to $\\tag{13} {µ_0}=\\tan{\\left( \\alpha \\right) }$", "\\frac{16 \\sqrt{3}}{9} {\\tan}^{-} 1 \\left(\\frac{\\sqrt{3}}{9} \\left(2 x + 2\\right)\\right) + C$", "can say that, $\\frac{{dy}}{{dx}} = \\; - \\tan \\theta$ ⇒ ${\\left[ {\\frac{{dy}}{{dx}}} \\right]_{\\left( {\\theta = \\frac{\\pi }{4}} \\right)}} = \\; - \\tan \\frac{\\pi }{4} = \\; - 1$ Hence, option A is the correct answer.", "\\right) \\tan \\left( \\frac{3\\pi}{4} \\right) = - 4 < 0$ $\\text{ So,} x = \\frac{3\\pi}{4}\\text { is a point of local maximum }.$ $\\text { The local maximum value is given by }$ $f\\left( \\frac{3\\pi}{4} \\right) = \\tan \\left( \\frac{3\\pi}{4} \\right) - 2 \\times \\frac{3\\pi}{4} = - 1 - \\frac{3\\pi}{2}$ Is there an error in this question or solution?", "$\\tan \\left(\\frac{3 \\pi}{8}\\right) = \\csc \\left(\\frac{3 \\pi}{4}\\right) - \\cot \\left(\\frac{3 \\pi}{4}\\right) = \\sqrt{2} - \\left(- 1\\right) = \\sqrt{2} + 1$", "$$\\tan\\left(\\frac{\\theta_{\\varphi=\\pi/2}}{2}\\right)\\sin(\\varphi)=\\tan\\left(\\frac{\\theta}{2}\\right)$$ $\\hspace{3.2cm}$", "{2 - \\frac{1}{2}} \\right) = m\\left( {1 - 0} \\right)$$ ⇒ m = 3 / 2 ⇒ tan θ = 3 / 2 $$\\theta = {\\tan ^{ - 1}}\\left( {\\frac{3}{2}} \\right)$$" ]

[ "obtain 15° by subtracting 45° - 30°. We will now convert back to radians. $$45°=\\frac{π}{180 °}=\\frac{π}{4}$$ $$30°=\\frac{π}{180 °}=\\frac{π}{6}$$ For reference, let's obtain the tangent of each: $$\\text{tan}\\frac{π}{4}=1$$ $$\\text{tan}\\frac{π}{6}=\\frac{\\sqrt{3}}{3}$$ Now, we return to our problem. Let's use our difference identity for tangent: $$\\text{tan}\\frac{π}{12}$$ $$=\\text{tan}\\left(\\frac{π}{4}- \\frac{π}{6}\\right)$$ $$=\\frac{1 - \\large{\\frac{\\sqrt{3}}{3}}}{1 + \\large{\\frac{\\sqrt{3}}{3}}}$$ Simplify: $$=\\frac{1 - \\large{\\frac{\\sqrt{3}}{3}}}{1 + \\large{\\frac{\\sqrt{3}}{3}}}\\cdot \\frac{1 - \\large{\\frac{\\sqrt{3}}{3}}}{1 - \\large{\\frac{\\sqrt{3}}{3}}}$$ $$=\\frac{1 - \\large{\\frac{2\\sqrt{3}}{3}}+ \\large{\\frac{1}{3}}}{1 - \\large{\\frac{1}{3}}}$$ $$=\\frac{\\Large{\\frac{4 - 2 \\sqrt{3}}{3}}}{\\Large{\\frac{2}{3}}}$$ $$=\\require{cancel}\\frac{4 - 2\\sqrt{3}}{\\cancel{3}}\\cdot \\frac{\\cancel{3}}{2}$$ $$=\\frac{\\cancel{2}(2 - \\sqrt{3})}{\\cancel{2}}$$ $$=2 - \\sqrt{3}$$ We will also see problems that ask us to simplify. Let's look at an example. Example #3: Simplify. $$\\text{sin}\\hspace{.1em}3 θ \\hspace{.1em}\\text{cos}({-}2θ) - \\text{cos}\\hspace{.1em}3θ \\hspace{.1em}\\text{sin}(-2 θ)$$ To simplify, we use our difference identity for sine: $$\\text{sin}(3θ - (-2θ))=\\text{sin}\\hspace{.1em}5 θ$$ ### Finding Function Values and the Quadrant of A + B Another type of problem involves finding trigonometric function values using the sum and difference identities. Let's look at some examples. Example #4: Find sin(θ + β), tan(θ + β), and the quadrant of A + B $$\\text{sin}\\hspace{.1em}θ=\\frac{4}{5}$$ θ is in quadrant II. $$\\text{cos}\\hspace{.1em}β=-\\frac{5}{13}$$ β is in quadrant III. To find the sine of (θ + β), we need to obtain the sine of β and the cosine of θ: $$\\text{cos}^2 β + \\text{sin}^2 β=1$$ $$\\left(-\\frac{5}{13}\\right)^2 + \\text{sin}^2 β=1$$ $$\\frac{25}{169}+ \\text{sin}^2 β=1$$ $$\\text{sin}^2 β=\\frac{169}{169}- \\frac{25}{169}$$ $$\\text{sin}^2 β=\\frac{144}{169}$$ $$\\text{sin}\\hspace{.1em}β=\\pm \\sqrt{\\frac{144}{169}}$$ Since", "even $\\cos(-x) = \\cos x$, we can simplify the above equation to: $$\\tan(-x) = -\\left(\\frac{\\sin x }{\\cos x}\\right) = -\\tan x$$ which means that the tangent function is an odd function.", "again? The period of the tangent function is $\\pi$. 6. so is this including tan=1 as well, coz is that how you got $\\frac{\\pi}{4}=\\frac{\\pi}{24}$ 7. $\\frac{\\pi}{4}\\neq \\frac{\\pi}{24}$. I went from $\\frac{25\\pi}{12}$ to $\\frac{\\pi}{12}$ because you have gone around the unit circle once. Though I may have made a mistake in doing that too early... Will edit.", "\\leq \\pi$$, or $$0^{\\circ} \\leq x \\leq 180^{\\circ}$$ in degree mode. Note that we could also have taken $$-\\pi \\leq x \\leq 0$$, but for convenience we take $$x$$ to be a positive angle. Lastly, for $$\\tan(x)$$ we can take a full period around the origin, so $$-\\frac{\\pi}{2} \\leq x \\leq \\frac{\\pi}{2}$$, or $$-90^{\\circ} \\leq x \\leq 90^{\\circ}$$ in degree mode.", "+ m1m2 Which the goal is to have an equal 180 degrees or 179.9 as possible. #### jamescv31 ##### New member Could I manage to have an exact 180 degrees with the formula of tangent m2-m1/1+m1m2 ? - - - Updated - - - Thank you for the time. #### MarkFL Staff member Yes I'm referring on the interior problem, since our teacher provided this formula only for the discussion the Tangent = m2+m1/ 1 + m1m2 Which the goal is to have an equal 180 degrees or 179.9 as possible. Okay, well using the formula you originally gave, we find: $$\\displaystyle A=\\tan^{-1}\\left(\\frac{m_b-m_c}{1+m_am_b} \\right)=\\tan^{-1}\\left(\\frac{3-\\frac{7}{4}}{1+3\\cdot\\frac{7}{4}} \\right)=\\tan^{-1}\\left(\\frac{1}{5} \\right)\\approx0.197395559849881$$ $$\\displaystyle B=\\tan^{-1}\\left(\\frac{m_c-m_a}{1+m_am_c} \\right)=\\tan^{-1}\\left(\\frac{\\frac{1}{2}-\\frac{7}{4}}{1+\\frac{1}{2}\\cdot\\frac{7}{4}} \\right)=\\tan^{-1}\\left(\\frac{2}{3} \\right)\\approx0.588002603547568$$ $$\\displaystyle C=\\pi+\\tan^{-1}\\left(\\frac{m_a-m_b}{1+m_am_b} \\right)=\\pi+\\tan^{-1}\\left(\\frac{\\frac{1}{2}-3}{1+\\frac{1}{2}\\cdot3} \\right)=\\pi+\\tan^{-1}\\left(-1 \\right)=\\pi-\\frac{\\pi}{4}=\\frac{3\\pi}{4}$$ Note: For angle $C$ it was necessary to add $\\pi$ to get the angle in the correct quadrant. If we convert the angles to degrees, we find: $$\\displaystyle A\\approx11.31^{\\circ}$$ $$\\displaystyle B\\approx33.69^{\\circ}$$ $$\\displaystyle C=135^{\\circ}$$ We see then that: $$\\displaystyle A+B+C\\approx180^{\\circ}$$", "# Tangent of 165 Degrees ## Theorem $\\tan 165^\\circ = \\tan \\dfrac {11 \\pi} {12} = - \\left({2 - \\sqrt 3}\\right)$ where $\\tan$ denotes tangent. ## Proof $\\displaystyle \\tan 165^\\circ$ $=$ $\\displaystyle \\tan \\left({90^\\circ + 75^\\circ}\\right)$ $\\displaystyle$ $=$ $\\displaystyle - \\cot 75^\\circ$ Tangent of Angle plus Right Angle $\\displaystyle$ $=$ $\\displaystyle - \\left({2 - \\sqrt 3}\\right)$ Cotangent of $75^\\circ$ $\\blacksquare$", "# Tangent of a complex number Calculator for the tangent of a complex number ## Tangent calculator This function returns the tangent to an angle specified as a complex number. To perform the calculation, enter the complex number. Then click on the 'Calculate' button. Complex number tangent calculator Input Complex number + i Decimal places 0 1 2 3 4 6 8 10 Result Cosine ## Formulas for tangent of complex number In the following description, $$z$$ stands for the complex number. $$x$$ stands for the real value $$Re$$ and $$y$$ for the imaginary value $$Im$$. $$\\displaystyle tan(z) = \\frac{sin(2x)}{cos(2x)+cosh(2y)} + \\frac{sinh(2y)}{cos(2x)+cosh(2y)}i$$ Example $$tan(z) = sinh(0.3+0.5i)$$ $$\\displaystyle tan(z) = \\frac{sin(2\\cdot0.3)}{cos(2\\cdot0.3)+cosh(2\\cdot 0.5)} + \\frac{sinh(2\\cdot0.5)}{cos(2\\cdot0.3)+cosh(2\\cdot 0.5)}i$$ $$\\displaystyle tan(z) = \\frac{sin(0.6)}{cos(0.6)+cosh(1)} + \\frac{sinh(1)}{cos(0.6)+cosh(1)}i$$ $$\\displaystyle tan(z) = \\frac{0.565}{2.368} + \\frac{1.175}{2.368}i$$ $$\\displaystyle tan(z) = tan(0.3+0.5i) = 0.238+0.496i$$", "# Tangent of 225 Degrees ## Theorem $\\tan 225^\\circ = \\tan \\dfrac {5 \\pi} 4 = 1$ where $\\tan$ denotes tangent. ## Proof $$\\displaystyle \\tan 225^\\circ$$ $$=$$ $$\\displaystyle \\tan \\left({360^\\circ - 135^\\circ}\\right)$$ $$\\displaystyle$$ $$=$$ $$\\displaystyle -\\tan 135^\\circ$$ Tangent of Conjugate Angle $$\\displaystyle$$ $$=$$ $$\\displaystyle 1$$ Tangent of 135 Degrees $\\blacksquare$", "$0 < x < 180$ as was stated in your original post. 7. Jun 15, 2015 ### Tangeton Ahh okay I forgot about that bit.. I guess I would have to remember to use the CAST diagram for my answer. Thanks for all help, everyone. 8. Jun 15, 2015 ### ehild You know that the tangent function is periodic with period of 180°. And you need to find solution in the range 0<θ<180°. Adding 180° to your result is also solution, just in the desired range.", "# Tangent of 225 Degrees Jump to: navigation, search ## Theorem $\\tan 225^\\circ = \\tan \\dfrac {5 \\pi} 4 = 1$ where $\\tan$ denotes tangent. ## Proof $$\\displaystyle \\tan 225^\\circ$$ $$=$$ $$\\displaystyle \\tan \\left({360^\\circ - 135^\\circ}\\right)$$ $$\\displaystyle$$ $$=$$ $$\\displaystyle -\\tan 135^\\circ$$ Tangent of Conjugate Angle $$\\displaystyle$$ $$=$$ $$\\displaystyle 1$$ Tangent of 135 Degrees $\\blacksquare$", "# How do you find the period of y = 2 tan 3x? $\\frac{\\pi}{3}$ The period of the tangent function is $\\pi$. Therefore, the period of $\\tan \\left(3 x\\right)$ is $\\frac{\\pi}{3}$ ($3 x$ will increase by $\\pi$ units anytime $x$ increases by $\\frac{\\pi}{3}$ units). This means the period of $y = 2 \\tan \\left(3 x\\right)$ is $\\frac{\\pi}{3}$ as well. The \"2\" is irrelevant.", "The period of sin is 2$pi$ . The range of the function is [-1,1]. #### The cosine function cos : R -> R The period of sin is 2$pi$ . The range of the function is [-1,1]. #### The tangent function tan : R -> R The range of the function is R . Now, the period is $pi$ and the function is undefined at x = ($pi$/2) + k$pi$, k=0,1,2. The graph of the tangent function on the interval 0 - $pi$ #### The cotangent function cot : R -> R The range of the function is R . The period is $pi$ and that the function is undefined at x = k$pi$, k=0,1,2. #### The values of sin, cos, tan, cot at the angles of 0°, 30°, 60°, 90°, 120°, 135°, 150°, 180°, 210°, 225°, 240°, 270°, 300°, 315°, 330°, 360° $A^o$ ^o30^o45^o60^o90^o120^o135^o150^o180^o210^o225^o240^o270^o300^o315^o330^o360^oA rad $frac<6>$ $frac<4>$ $frac<3>$ $frac<2>$ $frac<2pi><3>$ $frac<3pi><4>$ $frac<5pi><6>$ $pi$ $frac<7pi><6>$ $frac<5pi><4>$ $frac<4pi><3>$ $frac<3pi><2>$ $frac<5pi><3>$ $frac<7pi><4>$ $frac<11pi><6>$ $2pi$ $sin A$ frac<1><2>frac><2>frac><2>1frac><2>frac><2>frac<1><2> $-frac<1><2>$ $-frac><2>$ $-frac><2>$ $-1$ $-frac><2>$ $-frac><2>$ $-frac<1><2>$ cos A1frac><2>frac><2>frac<1><2> $-frac<1><2>$ $-frac><2>$ $-frac><2>$ $-1$ $-frac><2>$ $-frac><2>$ $-frac<1><2>$ frac<1><2>frac><2>frac><2>1 an A $frac><3>$ $1$ $sqrt<3>$ $-$ $-sqrt<3>$ $-1$ $-frac><3>$ frac><3>1sqrt<3>--sqrt<3>-1-frac><3> $cot A$ $-$ $sqrt<3>$ $1$ $frac><3>$ -frac><3>-1-sqrt<3>-sqrt<3>1frac><3> $-frac><3>$ $-1$ $-sqrt<3>$ $-$ The easiest way to remember the basic values of sin and cos at the angles", "NZ Level 7 (NZC) Level 2 (NCEA) Period changes for tangents Lesson We saw, in another chapter, how the period of the sine and cosine functions is affected by the coefficient $k$k that multiplies $x$x in $\\sin kx$sinkx and $\\cos kx$coskx. The fact that the sine and cosine functions both have period $360^\\circ$360° is expressed by the statements $\\sin(x+360^\\circ n)=\\sin x$sin(x+360°n)=sinx and $\\cos(x+360^\\circ n)=\\cos x$cos(x+360°n)=cosx for all integers $n$n. So it must be true that since $\\tan x=\\frac{\\sin x}{\\cos x}$tanx=sinxcosx, then $\\tan(x+360^\\circ n)=\\tan x$tan(x+360°n)=tanx. However, $360^\\circ$360° is not the smallest interval at which the tangent function repeats. A glance at the graph of the tangent function, shown below, should convince you that this function has a period of $180^\\circ$180°. That is, $\\tan(x+180^\\circ n)=\\tan x$tan(x+180°n)=tanx for all integers $n$n. The fact that the tangent function repeats at intervals of $180^\\circ$180° can be verified by considering the unit circle diagram. If $180^\\circ$180° is added to an angle $\\alpha$α, then the diagram shows that $\\sin(\\alpha+180^\\circ)$sin(α+180°) has the same magnitude as $\\sin\\alpha$sinα but opposite sign. The same relation holds between $\\cos(\\alpha+180^\\circ)$cos(α+180°) and $\\cos\\alpha$cosα. We make use of the definition: $\\tan\\alpha=\\frac{\\sin\\alpha}{\\cos\\alpha}$tanα=sinαcosα $\\tan(\\alpha+180^\\circ)$tan(α+180°) $=$= $\\frac{\\sin(\\alpha+180^\\circ)}{\\cos(\\alpha+180^\\circ)}$sin(α+180°)cos(α+180°)​ $=$= $\\frac{-\\sin\\alpha}{-\\cos\\alpha}$−sinα−cosα​ $=$= $\\tan\\alpha$tanα We are now in a position to determine the period of the function $\\tan kx$tankx where $k$k is any number. We can define a new", "# Tangent Function The tangent function is a periodic function which is very important in trigonometry. The simplest way to understand the tangent function is to use the unit circle. For a given angle measure $\\theta$ draw a unit circle on the coordinate plane and draw the angle centered at the origin, with one side as the positive $x$ -axis. The $x$ -coordinate of the point where the other side of the angle intersects the circle is $\\mathrm{cos}\\left(\\theta \\right)$ and the $y$ -coordinate is $\\mathrm{sin}\\left(\\theta \\right)$ . There are a few sine and cosine values that should be memorized, based on $30°-60°-90°$ triangles and $45°-45°-90°$ triangles. Based on these, you can work out the related values for tangent. $\\mathrm{sin}\\left(\\theta \\right)$ $\\mathrm{cos}\\left(\\theta \\right)$ $\\mathrm{tan}\\left(\\theta \\right)$ $\\mathrm{sin}\\left(0°\\right)=0$ $\\mathrm{cos}\\left(0°\\right)=1$ $\\mathrm{tan}\\left(0°\\right)=\\frac{0}{1}=0$ $\\mathrm{sin}\\left(30°\\right)=\\frac{1}{2}$ $\\mathrm{cos}\\left(30°\\right)=\\frac{\\sqrt{3}}{2}$ $\\mathrm{tan}\\left(30°\\right)=\\frac{1}{2}\\cdot \\frac{2}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}$ $\\mathrm{sin}\\left(45°\\right)=\\frac{\\sqrt{2}}{2}$ $\\mathrm{cos}\\left(45°\\right)=\\frac{\\sqrt{2}}{2}$ $\\mathrm{tan}\\left(45°\\right)=\\frac{\\sqrt{2}}{2}\\cdot \\frac{2}{\\sqrt{2}}=1$ $\\mathrm{sin}\\left(60°\\right)=\\frac{\\sqrt{3}}{2}$ $\\mathrm{cos}\\left(60°\\right)=\\frac{1}{2}$ $\\mathrm{tan}\\left(60°\\right)=\\frac{\\sqrt{3}}{2}\\cdot \\frac{2}{1}=\\sqrt{3}$ $\\mathrm{sin}\\left(90°\\right)=1$ $\\mathrm{cos}\\left(90°\\right)=0$ $\\mathrm{tan}\\left(90°\\right)=\\frac{1}{0}=\\text{undef}.$ Note that: • for angles with their terminal arm in Quadrant II, since sine is positive and cosine is negative, tangent is negative. • for angles with their terminal arm in Quadrant III, since sine is negative and cosine is negative, tangent is positive. • for angles with their terminal arm in Quadrant IV, since sine is negative and cosine is positive, tangent is negative. You can plot these points on a coordinate plane to show", "\\left(\\frac{\\pi}{2}\\right) = \\frac{1}{0}$$, which is not defined. Whenever $$\\cos(\\theta) = 0$$, the tangent is not defined and has a vertical asymptote. Like the sine and cosine functions, the tangent function is periodic. This makes sense because it is defined using sine and cosine. The period of tangent is only $$\\pi$$ while the period of sine and cosine is $$2\\pi$$.", "range of inverse tangent is $\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)$ ... $(-90^\\circ,90^\\circ)$ , quads I and IV 6. ## Thank you! Thanks so much!", "tangent is $\\pi$-periodic (or $180°$-periodic if you prefer), $\\tan 225° = \\tan 45°=1$, so the previous equality holds and your sum is actually $1$. -", "# How do you convert (15pi)/4 into degrees? $\\frac{15 \\pi}{4}$ has the same trig functions as$\\left(\\frac{7 \\pi}{4} + 2 \\pi\\right)$ $\\pi \\to 180$ deg $\\frac{7 \\pi}{4}$--> x deg? $\\left(\\frac{180}{\\pi}\\right) . \\frac{7 \\pi}{4}$ = 315 deg", "your calculator. Round to 4 decimal places. a) $\\sin 78^\\circ$ b) $\\cos 60^\\circ$ c) $\\tan 15^\\circ$ Solution: Depending on your calculator, you enter the degree and then press the trig button or the other way around. Also, make sure the mode of your calculator is in DEGREES. a) $\\sin 78^\\circ = 0.9781$ b) $\\cos 60^\\circ = 0.5$ c) $\\tan 15^\\circ = 0.2679$ ## Finding the Sides of a Triangle using Trig Ratios One application of the trigonometric ratios is to use them to find the missing sides of a right triangle. Example 5: Find the value of each variable. Round your answer to the nearest tenth. Solution: We are given the hypotenuse. Use sine to find $b$, and cosine to find $a$. $\\sin 22^\\circ &= \\frac{b}{30} && \\quad \\ \\ \\cos 22^\\circ = \\frac{a}{30}\\\\30 \\cdot \\sin 22^\\circ &= b && 30 \\cdot \\cos 22^\\circ = a\\\\b & \\approx 11.2 && \\qquad \\quad \\ \\ \\ a \\approx 27.8$ Example 6: Find the value of each variable. Round your answer to the nearest tenth. Solution: We are given the adjacent leg to $42^\\circ$. To find $c$, use cosine and tangent to find $d$. $\\cos 42^\\circ &= \\frac{adjacent}{hypotenuse} = \\frac{9}{c}&& \\quad \\tan 42^\\circ = \\frac{opposite}{adjacent} = \\frac{d}{9}\\\\c \\cdot \\cos 42^\\circ &= 9 && 9 \\cdot \\tan 42^\\circ = d\\\\c &=", "# Prove the trigonometric identity (cos(45)tan(45)+sin(45))/(tan(45))=2cos(45) ## \\frac{\\cos\\left(45\\right)\\cdot \\tan\\left(45\\right)+\\sin\\left(45\\right)}{\\tan\\left(45\\right)}=2\\cos\\left(45\\right) Go! 1 2 3 4 5 6 7 8 9 0 x y (◻) ◻/◻ 2 e π ln log lim d/dx d/dx > < >= <= sin cos tan cot sec csc asin acos atan acot asec acsc sinh cosh tanh coth sech csch asinh acosh atanh acoth asech acsch $true$ ## Step by step solution Problem $\\frac{\\cos\\left(45\\right)\\cdot \\tan\\left(45\\right)+\\sin\\left(45\\right)}{\\tan\\left(45\\right)}=2\\cos\\left(45\\right)$ 1 Calculating the sine of $45$ degrees $\\frac{0.8509+\\tan\\left(45\\right)\\cdot \\cos\\left(45\\right)}{\\tan\\left(45\\right)}=\\cos\\left(45\\right)\\cdot 2$ 2 Calculating the cosine of $45$ degrees $\\frac{0.8509+\\tan\\left(45\\right)\\cdot 0.5253}{\\tan\\left(45\\right)}=0.5253\\cdot 2$ 3 Multiply $2$ times $\\frac{83}{158}$ $\\frac{0.8509+\\tan\\left(45\\right)\\cdot 0.5253}{\\tan\\left(45\\right)}=1.0506$ 4 Calculating the tangent of $45$ degrees $\\frac{0.8509+1.6198\\cdot 0.5253}{1.6198}=1.0506$ 5 Multiply $\\frac{83}{158}$ times $\\sqrt[4]{47}$ $\\frac{0.8509+0.8509}{1.6198}=1.0506$ 6 Add the values $\\frac{\\sqrt[3]{8}}{2}$ and $\\frac{\\sqrt[3]{8}}{2}$ $\\frac{1.7018}{1.6198}=1.0506$ 7 Divide $\\sqrt[3]{8}$ by $\\sqrt[4]{47}$ $1.0506=1.0506$ 8 $1.0506$ equal to $1.0506$ $true$ $true$ ### Struggling with math? Access detailed step by step solutions to millions of problems, growing every day! ### Main topic: Trigonometric identities 0.2 seconds 148" ]

[ "# Tangent of a complex number Calculator for the tangent of a complex number ## Tangent calculator This function returns the tangent to an angle specified as a complex number. To perform the calculation, enter the complex number. Then click on the 'Calculate' button. Complex number tangent calculator Input Complex number + i Decimal places 0 1 2 3 4 6 8 10 Result Cosine ## Formulas for tangent of complex number In the following description, $$z$$ stands for the complex number. $$x$$ stands for the real value $$Re$$ and $$y$$ for the imaginary value $$Im$$. $$\\displaystyle tan(z) = \\frac{sin(2x)}{cos(2x)+cosh(2y)} + \\frac{sinh(2y)}{cos(2x)+cosh(2y)}i$$ Example $$tan(z) = sinh(0.3+0.5i)$$ $$\\displaystyle tan(z) = \\frac{sin(2\\cdot0.3)}{cos(2\\cdot0.3)+cosh(2\\cdot 0.5)} + \\frac{sinh(2\\cdot0.5)}{cos(2\\cdot0.3)+cosh(2\\cdot 0.5)}i$$ $$\\displaystyle tan(z) = \\frac{sin(0.6)}{cos(0.6)+cosh(1)} + \\frac{sinh(1)}{cos(0.6)+cosh(1)}i$$ $$\\displaystyle tan(z) = \\frac{0.565}{2.368} + \\frac{1.175}{2.368}i$$ $$\\displaystyle tan(z) = tan(0.3+0.5i) = 0.238+0.496i$$", "your calculator. Round to 4 decimal places. a) $\\sin 78^\\circ$ b) $\\cos 60^\\circ$ c) $\\tan 15^\\circ$ Solution: Depending on your calculator, you enter the degree and then press the trig button or the other way around. Also, make sure the mode of your calculator is in DEGREES. a) $\\sin 78^\\circ = 0.9781$ b) $\\cos 60^\\circ = 0.5$ c) $\\tan 15^\\circ = 0.2679$ ## Finding the Sides of a Triangle using Trig Ratios One application of the trigonometric ratios is to use them to find the missing sides of a right triangle. Example 5: Find the value of each variable. Round your answer to the nearest tenth. Solution: We are given the hypotenuse. Use sine to find $b$, and cosine to find $a$. $\\sin 22^\\circ &= \\frac{b}{30} && \\quad \\ \\ \\cos 22^\\circ = \\frac{a}{30}\\\\30 \\cdot \\sin 22^\\circ &= b && 30 \\cdot \\cos 22^\\circ = a\\\\b & \\approx 11.2 && \\qquad \\quad \\ \\ \\ a \\approx 27.8$ Example 6: Find the value of each variable. Round your answer to the nearest tenth. Solution: We are given the adjacent leg to $42^\\circ$. To find $c$, use cosine and tangent to find $d$. $\\cos 42^\\circ &= \\frac{adjacent}{hypotenuse} = \\frac{9}{c}&& \\quad \\tan 42^\\circ = \\frac{opposite}{adjacent} = \\frac{d}{9}\\\\c \\cdot \\cos 42^\\circ &= 9 && 9 \\cdot \\tan 42^\\circ = d\\\\c &=", "obtain 15° by subtracting 45° - 30°. We will now convert back to radians. $$45°=\\frac{π}{180 °}=\\frac{π}{4}$$ $$30°=\\frac{π}{180 °}=\\frac{π}{6}$$ For reference, let's obtain the tangent of each: $$\\text{tan}\\frac{π}{4}=1$$ $$\\text{tan}\\frac{π}{6}=\\frac{\\sqrt{3}}{3}$$ Now, we return to our problem. Let's use our difference identity for tangent: $$\\text{tan}\\frac{π}{12}$$ $$=\\text{tan}\\left(\\frac{π}{4}- \\frac{π}{6}\\right)$$ $$=\\frac{1 - \\large{\\frac{\\sqrt{3}}{3}}}{1 + \\large{\\frac{\\sqrt{3}}{3}}}$$ Simplify: $$=\\frac{1 - \\large{\\frac{\\sqrt{3}}{3}}}{1 + \\large{\\frac{\\sqrt{3}}{3}}}\\cdot \\frac{1 - \\large{\\frac{\\sqrt{3}}{3}}}{1 - \\large{\\frac{\\sqrt{3}}{3}}}$$ $$=\\frac{1 - \\large{\\frac{2\\sqrt{3}}{3}}+ \\large{\\frac{1}{3}}}{1 - \\large{\\frac{1}{3}}}$$ $$=\\frac{\\Large{\\frac{4 - 2 \\sqrt{3}}{3}}}{\\Large{\\frac{2}{3}}}$$ $$=\\require{cancel}\\frac{4 - 2\\sqrt{3}}{\\cancel{3}}\\cdot \\frac{\\cancel{3}}{2}$$ $$=\\frac{\\cancel{2}(2 - \\sqrt{3})}{\\cancel{2}}$$ $$=2 - \\sqrt{3}$$ We will also see problems that ask us to simplify. Let's look at an example. Example #3: Simplify. $$\\text{sin}\\hspace{.1em}3 θ \\hspace{.1em}\\text{cos}({-}2θ) - \\text{cos}\\hspace{.1em}3θ \\hspace{.1em}\\text{sin}(-2 θ)$$ To simplify, we use our difference identity for sine: $$\\text{sin}(3θ - (-2θ))=\\text{sin}\\hspace{.1em}5 θ$$ ### Finding Function Values and the Quadrant of A + B Another type of problem involves finding trigonometric function values using the sum and difference identities. Let's look at some examples. Example #4: Find sin(θ + β), tan(θ + β), and the quadrant of A + B $$\\text{sin}\\hspace{.1em}θ=\\frac{4}{5}$$ θ is in quadrant II. $$\\text{cos}\\hspace{.1em}β=-\\frac{5}{13}$$ β is in quadrant III. To find the sine of (θ + β), we need to obtain the sine of β and the cosine of θ: $$\\text{cos}^2 β + \\text{sin}^2 β=1$$ $$\\left(-\\frac{5}{13}\\right)^2 + \\text{sin}^2 β=1$$ $$\\frac{25}{169}+ \\text{sin}^2 β=1$$ $$\\text{sin}^2 β=\\frac{169}{169}- \\frac{25}{169}$$ $$\\text{sin}^2 β=\\frac{144}{169}$$ $$\\text{sin}\\hspace{.1em}β=\\pm \\sqrt{\\frac{144}{169}}$$ Since", "the function $y=0.5\\tan(\\frac{\\pi}{2}x)\\\\$. Solution First, we identify A and B. Figure 2 Because $A=0.5$ and $B=\\frac{\\pi}{2}\\\\$, we can find the stretching/compressing factor and period. The period is $\\frac{\\pi}{\\frac{\\pi}{2}}=2\\\\$, so the asymptotes are at $x=\\pm 1$. At a quarter period from the origin, we have $\\begin{array}f(0.5)=0.5\\tan(\\frac{0.5\\pi}{2}) \\hfill& \\\\ =0.5\\tan(\\frac{\\pi}{4}) \\hfill& \\\\ =0.5 \\end{array}\\\\$ This means the curve must pass through the points(0.5,0.5),(0,0),and(−0.5,−0.5).The only inflection point is at the origin. Figure shows the graph of one period of the function. Figure 3 Try It 1 Sketch a graph of $f(x)=3\\tan\\left(\\frac{\\pi}{6}x\\right)\\\\$. Solution Graphing One Period of a Shifted Tangent Function Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we add C and D to the general form of the tangent function. $f(x)=A\\tan(Bx−C)+D\\\\$ The graph of a transformed tangent function is different from the basic tangent function tan x in several ways: A General Note: Features of the Graph of y = Atan(Bx−C)+D • The stretching factor is |A|. • The period is $\\frac{\\pi}{|B|}\\\\$. • The domain is $x\\ne\\frac{C}{B}+\\frac{\\pi}{|B|}k\\\\$, where k is an integer. • The range is (−∞,−|A|] ∪ [|A|, ∞). • The vertical asymptotes occur at $x=\\frac{C}{B}+\\frac{\\pi}{2|B|}k\\\\$, where k is an odd integer. • There is no amplitude. • $y=A\\tan(Bx)$ is an", "again? The period of the tangent function is $\\pi$. 6. so is this including tan=1 as well, coz is that how you got $\\frac{\\pi}{4}=\\frac{\\pi}{24}$ 7. $\\frac{\\pi}{4}\\neq \\frac{\\pi}{24}$. I went from $\\frac{25\\pi}{12}$ to $\\frac{\\pi}{12}$ because you have gone around the unit circle once. Though I may have made a mistake in doing that too early... Will edit.", "even $\\cos(-x) = \\cos x$, we can simplify the above equation to: $$\\tan(-x) = -\\left(\\frac{\\sin x }{\\cos x}\\right) = -\\tan x$$ which means that the tangent function is an odd function.", "# Prove the trigonometric identity (cos(45)tan(45)+sin(45))/(tan(45))=2cos(45) ## \\frac{\\cos\\left(45\\right)\\cdot \\tan\\left(45\\right)+\\sin\\left(45\\right)}{\\tan\\left(45\\right)}=2\\cos\\left(45\\right) Go! 1 2 3 4 5 6 7 8 9 0 x y (◻) ◻/◻ 2 e π ln log lim d/dx d/dx > < >= <= sin cos tan cot sec csc asin acos atan acot asec acsc sinh cosh tanh coth sech csch asinh acosh atanh acoth asech acsch $true$ ## Step by step solution Problem $\\frac{\\cos\\left(45\\right)\\cdot \\tan\\left(45\\right)+\\sin\\left(45\\right)}{\\tan\\left(45\\right)}=2\\cos\\left(45\\right)$ 1 Calculating the sine of $45$ degrees $\\frac{0.8509+\\tan\\left(45\\right)\\cdot \\cos\\left(45\\right)}{\\tan\\left(45\\right)}=\\cos\\left(45\\right)\\cdot 2$ 2 Calculating the cosine of $45$ degrees $\\frac{0.8509+\\tan\\left(45\\right)\\cdot 0.5253}{\\tan\\left(45\\right)}=0.5253\\cdot 2$ 3 Multiply $2$ times $\\frac{83}{158}$ $\\frac{0.8509+\\tan\\left(45\\right)\\cdot 0.5253}{\\tan\\left(45\\right)}=1.0506$ 4 Calculating the tangent of $45$ degrees $\\frac{0.8509+1.6198\\cdot 0.5253}{1.6198}=1.0506$ 5 Multiply $\\frac{83}{158}$ times $\\sqrt[4]{47}$ $\\frac{0.8509+0.8509}{1.6198}=1.0506$ 6 Add the values $\\frac{\\sqrt[3]{8}}{2}$ and $\\frac{\\sqrt[3]{8}}{2}$ $\\frac{1.7018}{1.6198}=1.0506$ 7 Divide $\\sqrt[3]{8}$ by $\\sqrt[4]{47}$ $1.0506=1.0506$ 8 $1.0506$ equal to $1.0506$ $true$ $true$ ### Struggling with math? Access detailed step by step solutions to millions of problems, growing every day! ### Main topic: Trigonometric identities 0.2 seconds 148", "# Tangent of 165 Degrees ## Theorem $\\tan 165^\\circ = \\tan \\dfrac {11 \\pi} {12} = - \\left({2 - \\sqrt 3}\\right)$ where $\\tan$ denotes tangent. ## Proof $\\displaystyle \\tan 165^\\circ$ $=$ $\\displaystyle \\tan \\left({90^\\circ + 75^\\circ}\\right)$ $\\displaystyle$ $=$ $\\displaystyle - \\cot 75^\\circ$ Tangent of Angle plus Right Angle $\\displaystyle$ $=$ $\\displaystyle - \\left({2 - \\sqrt 3}\\right)$ Cotangent of $75^\\circ$ $\\blacksquare$", "# Tangent Function The tangent function is a periodic function which is very important in trigonometry. The simplest way to understand the tangent function is to use the unit circle. For a given angle measure $\\theta$ draw a unit circle on the coordinate plane and draw the angle centered at the origin, with one side as the positive $x$ -axis. The $x$ -coordinate of the point where the other side of the angle intersects the circle is $\\mathrm{cos}\\left(\\theta \\right)$ and the $y$ -coordinate is $\\mathrm{sin}\\left(\\theta \\right)$ . There are a few sine and cosine values that should be memorized, based on $30°-60°-90°$ triangles and $45°-45°-90°$ triangles. Based on these, you can work out the related values for tangent. $\\mathrm{sin}\\left(\\theta \\right)$ $\\mathrm{cos}\\left(\\theta \\right)$ $\\mathrm{tan}\\left(\\theta \\right)$ $\\mathrm{sin}\\left(0°\\right)=0$ $\\mathrm{cos}\\left(0°\\right)=1$ $\\mathrm{tan}\\left(0°\\right)=\\frac{0}{1}=0$ $\\mathrm{sin}\\left(30°\\right)=\\frac{1}{2}$ $\\mathrm{cos}\\left(30°\\right)=\\frac{\\sqrt{3}}{2}$ $\\mathrm{tan}\\left(30°\\right)=\\frac{1}{2}\\cdot \\frac{2}{\\sqrt{3}}=\\frac{\\sqrt{3}}{3}$ $\\mathrm{sin}\\left(45°\\right)=\\frac{\\sqrt{2}}{2}$ $\\mathrm{cos}\\left(45°\\right)=\\frac{\\sqrt{2}}{2}$ $\\mathrm{tan}\\left(45°\\right)=\\frac{\\sqrt{2}}{2}\\cdot \\frac{2}{\\sqrt{2}}=1$ $\\mathrm{sin}\\left(60°\\right)=\\frac{\\sqrt{3}}{2}$ $\\mathrm{cos}\\left(60°\\right)=\\frac{1}{2}$ $\\mathrm{tan}\\left(60°\\right)=\\frac{\\sqrt{3}}{2}\\cdot \\frac{2}{1}=\\sqrt{3}$ $\\mathrm{sin}\\left(90°\\right)=1$ $\\mathrm{cos}\\left(90°\\right)=0$ $\\mathrm{tan}\\left(90°\\right)=\\frac{1}{0}=\\text{undef}.$ Note that: • for angles with their terminal arm in Quadrant II, since sine is positive and cosine is negative, tangent is negative. • for angles with their terminal arm in Quadrant III, since sine is negative and cosine is negative, tangent is positive. • for angles with their terminal arm in Quadrant IV, since sine is negative and cosine is positive, tangent is negative. You can plot these points on a coordinate plane to show", "in DEGREES. a) $\\sin 78^\\circ = 0.97815$ b) $\\cos 60^\\circ = 0.5$ c) $\\tan 15^\\circ = 0.26795$ Example B Find the value of each variable. Round your answer to the nearest tenth. We are given the hypotenuse. Use sine to find $b$ , and cosine to find $a$ . Use your calculator to evaluate the sine and cosine of the angles. $\\sin 22^\\circ &= \\frac{b}{30} && \\quad \\ \\ \\cos 22^\\circ = \\frac{a}{30}\\\\30 \\cdot \\sin 22^\\circ &= b && 30 \\cdot \\cos 22^\\circ = a\\\\b & \\approx 11.2 && \\qquad \\quad \\ \\ \\ a \\approx 27.8$ Example C Find the value of each variable. Round your answer to the nearest tenth. We are given the adjacent leg to $42^\\circ$ . To find $c$ , use cosine and use tangent to find $d$ . $\\cos 42^\\circ &= \\frac{adjacent}{hypotenuse} = \\frac{9}{c}&& \\quad \\tan 42^\\circ = \\frac{opposite}{adjacent} = \\frac{d}{9}\\\\c \\cdot \\cos 42^\\circ &= 9 && 9 \\cdot \\tan 42^\\circ = d\\\\c &= \\frac{9}{\\cos 42^\\circ} \\approx 12.1 && \\qquad \\quad \\ \\ d \\approx 27.0$ Any time you use trigonometric ratios, use only the information that you are given. This will result in the most accurate answers. Guided Practice 1. What is $\\tan 45^\\circ$ ? 2. Find the length of the missing sides and round your answers to the nearest tenth:", "# Make the internet a better place to learn ## How do you convert (5pi)/7 from radians to degree? Sonnhard Featured 1 year ago It is ${900}^{\\setminus \\circ} / 7$ #### Explanation: From ${\\alpha}^{\\setminus \\circ} / \\alpha = {360}^{\\setminus \\circ} / \\left(2 \\cdot \\pi\\right)$ we get ${\\alpha}^{\\setminus \\circ} = {180}^{\\setminus \\circ} / \\pi \\cdot \\frac{5}{7} \\cdot \\pi$ canceling the $\\pi$ we get ${\\alpha}^{\\setminus \\circ} = {900}^{\\setminus \\circ} / 7$ ## How do you simplify tan(x+y) to trigonometric functions of x and y? mason m 3 years ago $\\tan \\left(x + y\\right) = \\frac{\\tan x + \\tan y}{1 - \\tan x \\tan y}$ #### Explanation: This can be expanded through the tangent angle addition formula: $\\tan \\left(\\alpha + \\beta\\right) = \\frac{\\tan \\alpha + \\tan \\beta}{1 - \\tan \\alpha \\tan \\beta}$ Thus, $\\tan \\left(x + y\\right) = \\frac{\\tan x + \\tan y}{1 - \\tan x \\tan y}$ The tangent addition formula can be found using the sine and cosine angle addition formulas. $\\sin \\left(\\alpha + \\beta\\right) = \\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta$ $\\cos \\left(\\alpha + \\beta\\right) = \\cos \\alpha \\cos \\beta - \\sin \\alpha \\sin \\beta$ Since $\\tan x = \\sin \\frac{x}{\\cos} x$, $\\tan \\left(\\alpha + \\beta\\right) = \\sin \\frac{\\alpha + \\beta}{\\cos} \\left(\\alpha + \\beta\\right) = \\frac{\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta}{\\cos \\alpha", "# Chapter 4 - Review Exercises - Page 184: 47 $\\color{blue}{y=2 \\tan{(\\frac{\\pi}{2})}}$ #### Work Step by Step The graph looks like that of the basic tangent function $y=\\tan{x}$. However, the period of the given function is $2\\pi$, while that of the basic tangent function is $\\pi$. RECALL: The period of the tangent function $y=\\tan{(bx)}$ is $\\frac{\\pi}{b},b\\gt 0$. Thus, $2\\pi=\\frac{\\pi}{b} \\\\2b\\pi=\\pi \\\\\\frac{2b\\pi}{2\\pi}=\\frac{\\pi}{2\\pi} \\\\b=2$ Thus, the tentative equation of the function whose graph is given is $y=a\\cdot \\tan{(\\frac{\\pi}{2})}$. The given graph contains the point $(\\frac{\\pi}{2}, 2)$. Substituting the x and y values of this point into the tentative equation above gives: $y=a \\cdot \\tan{(\\frac{\\pi}{2})} \\\\2=a \\cdot \\tan{(\\frac{\\frac{\\pi}{2}}{2})} \\\\2=a \\cdot \\tan{\\frac{\\pi}{4}} \\\\2=a \\cdot 1 \\\\2=a$ Therefore, the equation of the the function whose graph is shown is $\\color{blue}{y=2 \\tan{(\\frac{\\pi}{2})}}$. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "= \\frac {2}{3}\\\\ \\tan\\theta = \\frac {2}{3}\\\\ \\theta = \\arctan (\\frac {2}{3})$ This is not one of your special angles, so you will need a calculator. $\\theta = 33.69^\\circ$ $\\tan \\theta$ is not a 1-1 function. The inverse function maps back to a value in $[0,180^\\circ)$. There is second answer in the interval $[0,360^\\circ)$. $\\tan \\theta = \\tan (\\theta + 180)$ $213.69^\\circ$ is also in the solution set. The next one might even be a little bit trickier. $\\cos 3\\theta = -0.5\\\\ 3\\theta = \\arccos (-0.5)\\\\ 3\\theta = 120^\\circ$ but again, this isn't the only answer. $\\theta = \\arccos (-0.5)$ has two answers in the 360 degree interval, and $3\\theta = \\arccos (-0.5)$ will have $6.$ $\\cos \\theta = \\cos (-\\theta)$ and $\\cos \\theta = \\cos (360+\\theta)$ using these identities we find other values of theta. $3\\theta = 240^\\circ, 480^\\circ, 600^\\circ, 840^\\circ, 960^\\circ$ are all viable, as well. $\\theta \\in \\{40, 80, 160, 200, 280, 320\\}$ a) Reduce to $$\\tan\\theta=\\frac23.$$ Then as the tangent function has period $180°$ and is monotonic in a period, $$\\tan\\theta=\\tan(\\theta+k180°)$$ and all solutions are of the form $$\\theta+k180°=\\arctan\\frac23.$$ Find all $k$ such that $0\\le\\theta\\le360°$. b) The cosine has period $360°$ and is an even function monotonic in $[0°,180°]$ so that $$\\cos3\\theta=\\cos(k360°\\pm3\\theta).$$ Then all solutions are of the form $$k120°\\pm\\theta=\\frac13\\arccos(-0.5).$$ Again, find all $k$", "$\\cos 60^\\circ$ c) $\\tan 15^\\circ$ Depending on your calculator, you enter the degree and then press the trig button or the other way around. Also, make sure the mode of your calculator is in DEGREES. a) $\\sin 78^\\circ = 0.97815$ b) $\\cos 60^\\circ = 0.5$ c) $\\tan 15^\\circ = 0.26795$ #### Example B Find the value of each variable. Round your answer to the nearest tenth. We are given the hypotenuse. Use sine to find $b$ , and cosine to find $a$ . Use your calculator to evaluate the sine and cosine of the angles. $\\sin 22^\\circ &= \\frac{b}{30} && \\quad \\ \\ \\cos 22^\\circ = \\frac{a}{30}\\\\30 \\cdot \\sin 22^\\circ &= b && 30 \\cdot \\cos 22^\\circ = a\\\\b & \\approx 11.2 && \\qquad \\quad \\ \\ \\ a \\approx 27.8$ #### Example C Find the value of each variable. Round your answer to the nearest tenth. We are given the adjacent leg to $42^\\circ$ . To find $c$ , use cosine and use tangent to find $d$ . $\\cos 42^\\circ &= \\frac{adjacent}{hypotenuse} = \\frac{9}{c}&& \\quad \\tan 42^\\circ = \\frac{opposite}{adjacent} = \\frac{d}{9}\\\\c \\cdot \\cos 42^\\circ &= 9 && 9 \\cdot \\tan 42^\\circ = d\\\\c &= \\frac{9}{\\cos 42^\\circ} \\approx 12.1 && \\qquad \\quad \\ \\ d \\approx 27.0$ Any time you use trigonometric ratios, use only the information that", "the mode of your calculator is in DEGREES. a) $\\sin 78^\\circ = 0.97815$ b) $\\cos 60^\\circ = 0.5$ c) $\\tan 15^\\circ = 0.26795$ Example B Find the value of each variable. Round your answer to the nearest tenth. We are given the hypotenuse. Use sine to find $b$ , and cosine to find $a$ . Use your calculator to evaluate the sine and cosine of the angles. $\\sin 22^\\circ &= \\frac{b}{30} && \\quad \\ \\ \\cos 22^\\circ = \\frac{a}{30}\\\\30 \\cdot \\sin 22^\\circ &= b && 30 \\cdot \\cos 22^\\circ = a\\\\b & \\approx 11.2 && \\qquad \\quad \\ \\ \\ a \\approx 27.8$ Example C Find the value of each variable. Round your answer to the nearest tenth. We are given the adjacent leg to $42^\\circ$ . To find $c$ , use cosine and use tangent to find $d$ . $\\cos 42^\\circ &= \\frac{adjacent}{hypotenuse} = \\frac{9}{c}&& \\quad \\tan 42^\\circ = \\frac{opposite}{adjacent} = \\frac{d}{9}\\\\c \\cdot \\cos 42^\\circ &= 9 && 9 \\cdot \\tan 42^\\circ = d\\\\c &= \\frac{9}{\\cos 42^\\circ} \\approx 12.1 && \\qquad \\quad \\ \\ d \\approx 27.0$ Any time you use trigonometric ratios, use only the information that you are given. This will result in the most accurate answers. Guided Practice 1. What is $\\tan 45^\\circ$ ? 2. Find the length of the missing sides and round", "tangent function has period $$\\pi .$$ Proof The result follows immediately from our definitions. $$\\quad$$ Q.E.D. Proposition $$\\PageIndex{3}$$ For any $$x, y \\in D$$ with $$x+y \\in D$$, $\\tan (x+y)=\\frac{\\tan (x)+\\tan (y)}{1-\\tan (x) \\tan (y)}.$ Proof Suppose $$y_{1}, y_{2} \\in\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right)$$ with $$y_{1}+y_{2} \\in\\left(-\\frac{\\pi}{2}, \\frac{\\pi}{2}\\right) .$$ Let $$x_{1}=\\tan \\left(y_{1}\\right)$$ and $$x_{2}=\\tan \\left(y_{2}\\right) .$$ Note that if $$x_{1}>0,$$ then $$x_{1} x_{2} \\geq 1$$ would imply that $x_{2} \\geq \\frac{1}{x_{1}},$ which in turn implies that \\begin{aligned} y_{1}+y_{2} &=\\arctan \\left(x_{1}\\right)+\\arctan \\left(x_{2}\\right) \\\\ & \\geq \\arctan \\left(x_{1}\\right)+\\arctan \\left(\\frac{1}{x_{1}}\\right) \\\\ &=\\frac{\\pi}{2}, \\end{aligned} contrary to our assumptions. Similarly, if $$x_{1}<0,$$ then $$x_{1} x_{2} \\geq 1$$ would imply that $x_{2} \\leq \\frac{1}{x_{1}},$ which in turn implies that \\begin{aligned} y_{1}+y_{2} &=\\arctan \\left(x_{1}\\right)+\\arctan \\left(x_{2}\\right) \\\\ & \\leq \\arctan \\left(x_{1}\\right)+\\arctan \\left(\\frac{1}{x_{1}}\\right) \\\\ &=-\\frac{\\pi}{2}, \\end{aligned} contrary to our assumptions. Thus we must have $$x_{1} x_{2}<1 .$$ Moreover, suppose $$u$$ is a number between $$-x_{1}$$ and $$x_{2} .$$ If $$x_{1}>0,$$ then $x_{2}<\\frac{1}{x_{1}},$ and so $u<\\frac{1}{x_{1}}.$ If $$x_{1}<0,$$ then $x_{2}>\\frac{1}{x_{1}},$ and so $u>\\frac{1}{x_{1}}.$ Now let $x=\\frac{x_{1}+x_{2}}{1-x_{1} x_{2}}.$ We want to show that $\\arctan (x)=\\arctan \\left(x_{1}\\right)+\\arctan \\left(x_{2}\\right),$ which will imply that $\\frac{\\tan \\left(y_{1}\\right)+\\tan \\left(y_{2}\\right)}{1-\\tan \\left(y_{1}\\right) \\tan \\left(y_{2}\\right)}=\\tan \\left(y_{1}+y_{2}\\right).$ We need to compute $\\arctan (x)=\\arctan \\left(\\frac{x_{1}+x_{2}}{1-x_{1} x_{2}}\\right)=\\int_{0}^{\\frac{x_{1}+x_{2}}{1-x_{1} x_{2}}} \\frac{1}{1+t^{2}} d t.$ Let $t=\\varphi(u)=\\frac{x_{1}+u}{1-x_{1} u},$ where $$u$$ varies between $$-x_{1},$$ where $$t=0,$$ and $$x_{2},$$ where $$t=x .$$ Now $\\varphi^{\\prime}(u)=\\frac{\\left(1-x_{1} u\\right)-\\left(x_{1}+u\\right)\\left(-x_{1}\\right)}{\\left(1-x_{1} u\\right)^{2}}=\\frac{1+x_{1}^{2}}{\\left(1-x_{1} u\\right)^{2}},$ which is", "3 }}{{P - 3}}} \\right)} \\right) = - 180$$ $$\\Rightarrow 180 - 60^\\circ + {\\tan ^{ - 1}}\\left( {\\frac{{3\\sqrt 3 }}{{P - 3}}} \\right) = 180^\\circ$$ $$\\Rightarrow {\\tan ^{ - 1}}\\left( {\\frac{{3\\sqrt 3 }}{{P - 3}}} \\right) = 60^\\circ$$ $$\\Rightarrow \\frac{{3\\sqrt 3 }}{{P - 3}} = \\sqrt 3$$ ⇒ P – 3 = 3 ⇒ P = 6", "+ m1m2 Which the goal is to have an equal 180 degrees or 179.9 as possible. #### jamescv31 ##### New member Could I manage to have an exact 180 degrees with the formula of tangent m2-m1/1+m1m2 ? - - - Updated - - - Thank you for the time. #### MarkFL Staff member Yes I'm referring on the interior problem, since our teacher provided this formula only for the discussion the Tangent = m2+m1/ 1 + m1m2 Which the goal is to have an equal 180 degrees or 179.9 as possible. Okay, well using the formula you originally gave, we find: $$\\displaystyle A=\\tan^{-1}\\left(\\frac{m_b-m_c}{1+m_am_b} \\right)=\\tan^{-1}\\left(\\frac{3-\\frac{7}{4}}{1+3\\cdot\\frac{7}{4}} \\right)=\\tan^{-1}\\left(\\frac{1}{5} \\right)\\approx0.197395559849881$$ $$\\displaystyle B=\\tan^{-1}\\left(\\frac{m_c-m_a}{1+m_am_c} \\right)=\\tan^{-1}\\left(\\frac{\\frac{1}{2}-\\frac{7}{4}}{1+\\frac{1}{2}\\cdot\\frac{7}{4}} \\right)=\\tan^{-1}\\left(\\frac{2}{3} \\right)\\approx0.588002603547568$$ $$\\displaystyle C=\\pi+\\tan^{-1}\\left(\\frac{m_a-m_b}{1+m_am_b} \\right)=\\pi+\\tan^{-1}\\left(\\frac{\\frac{1}{2}-3}{1+\\frac{1}{2}\\cdot3} \\right)=\\pi+\\tan^{-1}\\left(-1 \\right)=\\pi-\\frac{\\pi}{4}=\\frac{3\\pi}{4}$$ Note: For angle $C$ it was necessary to add $\\pi$ to get the angle in the correct quadrant. If we convert the angles to degrees, we find: $$\\displaystyle A\\approx11.31^{\\circ}$$ $$\\displaystyle B\\approx33.69^{\\circ}$$ $$\\displaystyle C=135^{\\circ}$$ We see then that: $$\\displaystyle A+B+C\\approx180^{\\circ}$$", "# Thread: Equation For The Line Tangent To The Graph 1. ## Equation For The Line Tangent To The Graph Hey everyone, I have this question, “Find an equation for the line tangent to the graph y = sec x at (pi/3 , 2)” So far I have that, f(x) = sec x f '(x) = sec x . tan x I then subbed in x = (pi/3) into f '(x) to get the gradient which I found was 0, is this right? I wanted to make I was approaching this question correctly and if I have that just means I sub m into the equation, y - y1 = m(x - x1) and finish it off. Thanks! 2. Yep sounds right if all your calculations are right 3. Originally Posted by 11rdc11 Yep sounds right if all you calculations are right Awesome, thanks ^^ 5. Originally Posted by 11rdc11 Really? What did I do wrong. 6. $f'\\left(\\frac{\\pi}{3}\\right) = \\sec \\left(\\frac{\\pi}{3}\\right) \\cdot \\tan \\left(\\frac{\\pi}{3}\\right) = 2 \\cdot \\sqrt{3}$ 7. $sec\\bigg(\\frac{\\pi}{3}\\bigg) = 2$ $tan\\bigg(\\frac{\\pi}{3}\\bigg) = \\sqrt{3}$ 8. Originally Posted by o_O $f'\\left(\\frac{\\pi}{3}\\right) = \\sec \\left(\\frac{\\pi}{3}\\right) \\cdot \\tan \\left(\\frac{\\pi}{3}\\right) = 2 \\cdot \\sqrt{3}$ Originally Posted by 11rdc11 $sec\\bigg(\\frac{\\pi}{3}\\bigg) = 2$ $tan\\bigg(\\frac{\\pi}{3}\\bigg) = \\sqrt{3}$ Ohhh I understand now for some reason I had that Tan pi/3 = 0 Thanks for the", "the equation for the hill is $$Y = X \\cdot tan(20^\\circ)$$ Substituting for Y in the parabola equation (equation A) gives us $$X \\cdot tan(20^\\circ) = X - 4.9 \\cdot \\left( \\frac{X^2}{V_{ix}^2} \\right)$$ $$tan(20^\\circ) = 1 - 4.9 \\cdot \\left( \\frac{X}{V_{ix}^2} \\right)$$ $$\\left(\\frac{4.9}{V_{ix}^2}\\right) \\cdot X = 1 - tan(20^\\circ)$$ $$X = \\frac{V_{ix}^2(1 - tan(20^\\circ))}{4.9} = \\frac{(50 \\cdot cos(45^\\circ))^2 \\cdot (1 - tan(20^\\circ))}{4.9}$$ $$X = \\frac{35.355^2 \\cdot 0.636}{4.9} = \\frac{1250 \\cdot 0.636}{4.9} = \\frac{795.037}{4.9} = 162.252$$ Plug this value back into equation A, above, and you get Y = 59.055 The distance up the his is found using the Pythagorean Theorem: $$D^2 = X^2 + Y^2 = 162.252^2 + 59.055^2 = 29813.373$$ and D = 172.665 m (note: I only rounded off in the text, not in my actual calculations) Last edited: Jun 12, 2010" ]

[ "6\\end{pmatrix}) = -3$ • $C_{2,3} = (-1)^{5} * det(M_{2,3}) = -det(M_{2,3}) = \\textbf{-6}$ $det(M_{2,3})=det(\\begin{pmatrix} -1 & 4 \\\\ -3 & 6 \\end{pmatrix}) = 6$ • $C_{3,3} = (-1)^{6} * det(M_{3,3}) = det(M_{3,3}) = \\textbf{-3}$ $det(M_{2,3})=det(\\begin{pmatrix} -1 & 4 \\\\ 2 & -5\\end{pmatrix}) = -3$ Hence, the result is $\\begin{split} det(M) = -7 * C_{1,3} + 8 * C_{2,3} + -9 * C_{3,3} \\\\ = -7 * -3 + 8 * -6 + -9 * -3 \\\\ = 0 \\\\ \\end{split}$", "a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf b_4)] \\\\ &\\ \\ \\ \\ + 5[6\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf a_4) + 2\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf b_4)] \\\\ &= -6\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf a_4) - 2\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf b_4) \\\\ &\\ \\ \\ \\ + 30\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf a_4) + 10\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf b_4)\\end{align} Antisymmetry The fact that $\\det$ is antisymmetric means that if you swap any two arguments of $\\det$, you'll negate the sign. Thus $$\\det(\\mathbf a_1, \\mathbf a_2) = -\\det(\\mathbf a_2, \\mathbf a_1)$$ Likewise \\begin{align}\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf a_4) &= -\\det(\\mathbf a_1, \\mathbf a_4, \\mathbf a_3, \\mathbf a_2) \\\\ &= \\det(\\mathbf a_4, \\mathbf a_1, \\mathbf a_3, \\mathbf a_2)\\end{align} So swapping an even number of columns doesn't change the sign, but swapping an odd number will produce a negative sign. One consequence of this antisymmetry is that if two columns are scalar multiples of each other, then the determinant must be zero. This is because after pulling the constant multiple out front the two columns will be the same. Then if you swap those two columns, the determinant will look the same -- but it'll have a minus sign out front. And thus the determinant will", "prove that $$\\det(B)\\ge\\det(A)$$ (so that $$\\det(B)>0$$). Let $$A=\\pmatrix{p&s_1a&s_2b\\\\ s_1a&q&s_3c\\\\ s_2b&s_3c&r} \\ \\text{ and }\\ B=\\pmatrix{p&a&b\\\\ a&q&c\\\\ b&c&r}$$ where $$p,q,r>0,\\,a,b,c\\ge0$$ and $$s_1,s_2,s_3\\in\\{1,-1\\}$$. Then \\begin{aligned} \\det(A) &=pqr+2s_1s_2s_3abc-(pc^2+qb^2+ra^2)\\\\ &\\le pqr+2abc-(pc^2+qb^2+ra^2)\\\\ &=\\det(B). \\end{aligned}", "\\det(\\mathbf{A})^{-1}\\, \\mathrm{adj}(\\mathbf{A}).$ All the best.", "$\\mod (\\det A)=n^{n/2}$ . 4. ## Re: Find mod ( det A ) Originally Posted by FernandoRevilla Modulus of the complex number $\\det (A)$ . Right, $A$ is a Vandermonde matrix. Being this the Math Challenge Problem forum, I know the solution (and its resolution) . Another version is: prove that $\\mod (\\det A)=n^{n/2}$ . so the question is to find $|\\det(A)|.$ so far i got $|\\det(A)| = 2^{\\binom{n}{2}}a^n,$ where $a = \\prod_{k=1}^m \\sin(k \\theta)$ and $\\theta = \\frac{\\pi}{n}$ and $m = \\lfloor \\frac{n-1}{2} \\rfloor.$ so the problem now is to evaluate $a.$ 5. ## Re: Find mod ( det A ) A better approach is to find previously $A^2$ .", "the Math Challenge Problem forum, I know the solution (and its resolution) :) . Another version is: prove that $\\mod (\\det A)=n^{n/2}$ . • Dec 18th 2011, 02:17 AM NonCommAlg Re: Find mod ( det A ) Quote: Originally Posted by FernandoRevilla Modulus of the complex number $\\det (A)$ . Right, $A$ is a Vandermonde matrix. Being this the Math Challenge Problem forum, I know the solution (and its resolution) :) . Another version is: prove that $\\mod (\\det A)=n^{n/2}$ . so the question is to find $|\\det(A)|.$ so far i got $|\\det(A)| = 2^{\\binom{n}{2}}a^n,$ where $a = \\prod_{k=1}^m \\sin(k \\theta)$ and $\\theta = \\frac{\\pi}{n}$ and $m = \\lfloor \\frac{n-1}{2} \\rfloor.$ so the problem now is to evaluate $a.$ • Dec 18th 2011, 05:28 AM FernandoRevilla Re: Find mod ( det A ) A better approach is to find previously $A^2$ .", "Math Help - Find determinant 1. Find determinant $det(2A^8 B^{-1})$ I tried to break it up but I'm not sure where the 2 goes. $det(2A^8) det(B^{-1})$ ? Any help is appreciated! 2. $ \\det(2A^{8}B^{-1})= 2^{3}\\det(A^{8})\\det(B^{-1})= 8\\frac{\\det(A)^{8}}{\\det(B)}= 8\\frac{ (-1)^{8}}{-2} = -4 $", "# If A and B are 4 times 4 matrices, det(A) = 1, det(B) = 4, then det(AB) = ? det(3A) = ? det(A^T) = ? det(B^{-1}) = ? det(B^4) = ? If A and B are $4×4$ matrices, det(A) = 1, det(B) = 4, then det(AB) = ? det(3A) = ? $det\\left({A}^{T}\\right)=?$ $det\\left({B}^{-1}\\right)=?$ $det\\left({B}^{4}\\right)=?$ You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Corben Pittman Given that A and B are $4×4$ matrices: det(A)=1 : det(B)=4 Then (1) $det\\left(AB\\right)=det\\left(A\\right)\\cdot det\\left(B\\right)$ $=1\\cdot 4$ det(AB)=4 (2) $det\\left(3A\\right)=3\\cdot det\\left(A\\right)$ $=3\\cdot \\left(1\\right)$ det(3A)=3 (3) $det\\left({A}^{T}\\right)=det\\left(A\\right)$ $det\\left({A}^{T}\\right)=1$ (4)$det\\left({B}^{-1}\\right)=\\frac{1}{|B|}$ $det\\left({B}^{-1}\\right)=\\frac{1}{4}$ (5) $det\\left({B}^{4}\\right)=det\\left(B\\cdot B\\cdot B\\cdot B\\right)$ $=det\\left(B\\right)\\cdot det\\left(B\\right)\\cdot det\\left(B\\right)\\cdot det\\left(B\\right)$ $=4\\cdot 4\\cdot 4\\cdot 4$ =256 Jeffrey Jordon", "all $\\mathbf {N}$ lines are multiplied by $\\mathbf {\\beta}$, we have : $\\mathbf {Det (A * B) = Det (A * \\beta I) = \\beta^N Det (A * I) = \\beta^ N Det (A)}$ $\\mathbf {Det (A) * Det (B) = Det (A)\\beta^N Det (I) = \\beta^N Det (A)}$ **c) **Any N-dimensional shape represented by a matrix can be broken down into an amount $\\mathbf {s}$ of small cube matrices $\\mathbf {B_k}$ ($\\mathbf {k}$ is an index from 1 to s) of dimension $\\mathbf {\\beta}$ (diagonal matrix with only $\\mathbf {\\beta}$value), where $\\mathbf {\\beta}$ is arbitrarily small and therefore $\\mathbf {s}$ arbitrarily large. Look at the volume below the colorful surface, above the plane $\\mathbf {z=0}$ (vertical axis), below the plane $\\mathbf {y=0}$ (negative y) to down and left, right to the place $\\mathbf {x=0}$ (positive x). Imagine put inside this volume thousands and thousands ($\\mathbf {s}$) of tiny cubes. $\\mathbf {Det(B) = Det(B_1) + ... + Det(B_s)}$ $\\mathbf { Det (A) * Det (B) = Det (A) * (Det (B_1) + ... + Det (B_s)) = s Det (A) \\beta^N \\qquad}$ (c1) $\\mathbf { Det (A * B) = Det (A * (B1 + ... + B_s)) = Det (A * B_1) + ... + Det (A * B_s)}$ By (b) above $\\mathbf {= \\beta^N Det(A) +", "$$\\mathrm{det}\\left(A + uv^T\\right) = \\mathrm{det}(A) \\mathrm{det}\\left( I + A^{-1} u v^T \\right)$$ but I don't think it's very useful.", "# What can be said if $A^2+B^2+2AB=0$ for some real $2\\times2$ matrices $A$ and $B$? Let $A,B\\in M_2(\\mathbb{R})$ be such that $A^2+B^2+2AB=0$ and $\\det A= \\det B$. Our goal is to compute $\\det(A^2 – B^2)$. According to the chain of comments on Art of Problem Solving, the following statements are true: 1. $\\det(A^2+B^2)+\\det(A^2-B^2)=2(\\det A^2+\\det B^2)$. (Is this well known?) 2. (1) $\\implies \\det(A^2-B^2)=0$. 3. If $A,B\\in M_2(\\mathbb{C})$ satisfy $A^2+B^2+2AB=0$, then $AB=BA$. 4. $(A+B)^2=0 \\implies \\det(A^2-B^2)=0$. Can someone help me with justifying these statements? Edit: Doug M provided an explanation for (1) in the answers. Here is an explanation for (2): $A^2+B^2+2AB=O_2 \\implies A^2+B^2=-2AB$. So $\\det(A^2+B^2)=4\\det(AB)$. Now using (1), $\\det(A^2-B^2)= 2(\\det(A^2)-2\\det(AB)+\\det(B^2)) = 2((\\det(A)^2-\\det(B)^2) = 0$. #### Solutions Collecting From Web of \"What can be said if $A^2+B^2+2AB=0$ for some real $2\\times2$ matrices $A$ and $B$?\" Of course, the OP did not understand much about the proposed exercise. In particular, in $M_2(\\mathbb{C})$, $A^2 +B^2+2AB=0$ implies $\\det(A)=\\det(B)$. In fact, the interesting result is Proposition. Let $A,B\\in M_2(\\mathbb{C})$ be such that $(1)$ $A^2+B^2+2AB=0$; then $AB=BA$. Note that the result works only in dimension $2$. Proof. If $H=A+B$, then $H^2=HA-AH$ and $trace(H^2)=0$. Moreover $H^3=H^2A-HAH=HAH-AH^2$ implies that $trace(H^3)=0$. We deduce that the eigenvalues of $H$ are $0$, $H$ is nilpotent, $H^2=0$, $AH=HA$, $AB=BA$. $\\square$ Remark. According to Proposition, $(1)$ implies that $(A+B)^2=0$, $\\det(A+B)=0$, $\\det(A^2-B^2)=0$;", "$1 = \\det(\\mathbf I_n) = \\det(\\mathbf{A} \\mathbf{A}^{-1}) = \\det(\\mathbf{A}) \\det(\\mathbf{A}^{-1}),$ and if det(A) is a unit then (*) above shows that $\\mathbf{A}^{-1} = \\det(\\mathbf{A})^{-1}\\, \\mathrm{adj}(\\mathbf{A}).$ ## Properties The adjugate has the properties $\\mathrm{adj}(\\mathbf{I}) = \\mathbf{I}\\,$ $\\mathrm{adj}(\\mathbf{AB}) = \\mathrm{adj}(\\mathbf{B})\\,\\mathrm{adj}(\\mathbf{A})\\,$ for all n×n matrices A and B. The adjugate preserves transposition: $\\mathrm{adj}(\\mathbf{A}^T) = \\mathrm{adj}(\\mathbf{A})^T\\,$. Furthermore $\\det\\big(\\mathrm{adj}(\\mathbf{A})\\big) = \\det(\\mathbf{A})^{n-1}.\\,$ If p(t) = det(A − tI) is the characteristic polynomial of A and we define the polynomial q(t) = (p(0) − p(t))/t, then $\\mathrm{adj}(\\mathbf{A}) = q(\\mathbf{A}) = -(p_1 \\mathbf{I} + p_2 \\mathbf{A} + p_3 \\mathbf{A}^2 + \\cdots + p_{n} \\mathbf{A}^{n-1}),$ where pj are the coefficients of p(t), $p(t) = p_0 + p_1 t + p_2 t^2 + \\cdots + p_{n} t^{n}.$ The adjugate also appears in the formula of the derivative of the determinant. ## References • Strang, Gilbert (1988). \"Section 4.4: Applications of determinants\". Linear Algebra and its Applications (3rd ed.). Harcourt Brace Jovanovich. pp. 231–232. ISBN 0-15-551005-3. Wikimedia Foundation. 2010. ### Look at other dictionaries: • Matrix - получить на Академике действующий промокод Pharmacosmetica или выгодно matrix купить со скидкой на распродаже в Pharmacosmetica • Matrix (mathematics) — Specific elements of a matrix are often denoted by a variable with two subscripts. For instance, a2,1 represents the element at the second row and first column of a matrix A. In", "# Find $\\det A$ and $Tr A$ if $\\det(A+\\sqrt[n]{3}I_n)=0$ $$A \\in M_{n}(\\mathbb{C})$$ and I have to find $$\\det A$$ and $$Tr A$$ if $$\\det(A+\\sqrt[n]{3}I_n)=0$$. I observed that $$\\sqrt[n]{3}$$ is an eigenvalue of $$A$$,but I don’t know how to continue.", "= -2ae - 3cd - 2bf + 2ce + 2bd + 3af = -3 det(V) = 3ae + 5cd + bf - ce - 3bd - 5af You can see that det(V) = -$\\frac{1}{2}$ det(P) - det(U). So we get det(V) = -5 + 3 = -2", "$\\det\\pmatrix{A+xI&B\\\\ C&D}=\\det\\left((A+xI)D-CB\\right)$. Put $x=0$, the assertion follows. The proofs for the other three cases are similar.", "can establish the recurrence $$p_k=p_{k-1}\\cos\\frac{\\pi}{n}-\\frac{\\cos^2 \\theta_{k-1}}{4}p_{k-2}$$ and then $\\det\\mathbf T=p_n$. There is also an explicit formula for the first and last columns of the inverse of a tridiagonal matrix (the result of $\\mathbf T^{-1}\\mathbf U$) which I will have to look up (unless somebody beats me to it), and proving the nonnegativity of $\\det\\mathbf A$ boils down to proving that either $p_n$ and $\\det(\\mathbf I+\\mathbf V^T\\mathbf T^{-1}\\mathbf U)$ have the same sign, or that one of the two is zero. - I guess your $\\det\\mathbf A=(\\det\\mathbf T)\\det(\\mathbf I+\\mathbf V^T\\mathbf A^{-1}\\mathbf U)$ should be $\\det\\mathbf A=(\\det\\mathbf T)\\det(\\mathbf I+\\mathbf V^T\\mathbf T^{-1}\\mathbf U)$. the Sherman-Morrison-Woodbury formula for determinants is usually for $U,V$ of rank one, nd.edu/~jeff/Teaching/CBE60478/ShermanMorrison.pdf . Can you show me the reference for your formula $\\det\\mathbf A=(\\det\\mathbf T)\\det(\\mathbf I+\\mathbf V^T\\mathbf T^{-1}\\mathbf U)$. – Sunni May 18 '11 at 0:18 Yes @Sunni; I'll correct that typo. Faddev/Faddeva or Aitken's old books should have it, but if you're knowledgable with Schur complements, it shouldn't be too hard to reckon it yourself. – Ｊ. Ｍ. Jun 2 '11 at 4:01 This is not an aswer, but was too long for a comment. I suspect that for $N>3$ the matrix is (semi)positive definite, a stronger property than what is asked. As I pointed in a comment, to prove this is equivalent to prove the", "of $\\mathbb{R}^{n+m}$ then $\\det D(x,y)f=\\det(e_{n+1},...,e_{n+m},e_1,...,e_n)=(-1)^{mn}\\det(e_1,...,e_n,e_{n+1},...,e_{n+m})=(-1)^{mn}$. So $\\mathrm{deg}\\,f=(-1)^{mn}$. -", "# Does $k[B]^{B'}=k[\\det(b),\\det(b)^{-1}]$? Let $k=\\overline{k}$. Suppose $GL_2(k)$ acts on $GL_2(k)$ by left multiplication. Then $k[GL_2]^{SL_2} = k[\\det(g),\\det(g)^{-1}]$. Now for $B=\\left\\{ \\left( \\begin{array}{cc} b_{11} & b_{12} \\\\ 0 & b_{22} \\\\ \\end{array}\\right) : b_{ii}\\not=0, b_{12}\\in k \\right\\}$, let $B'=\\{ b\\in B: \\det(B)=1 \\}$. Suppose $B$ acts on $B$ by left multiplication. Then does $k[B]^{B'}=k[b_{11}, b_{22},b_{11}^{-1},b_{22}^{-1}]$ or does $k[B]^{B'}=k[\\det(b),\\det(b)^{-1}]$? -", "can arrive at correct result. – mez Apr 9 '13 at 16:00 There are some mistakes in your answer though. $\\det A = n^{m-1}$ and that $B-JAJ^{T} = B-\\frac{1}{n}JJ^{T}$, but idea in general is correct. – mez Apr 14 '13 at 10:36", "You can use $\\det(A \\cdot B)= \\det(A) \\cdot \\det(B)$. And consider that $$(\\mathbf v+9\\mathbf w, \\mathbf w+3\\mathbf u, \\mathbf u+3\\mathbf v) = (\\mathbf u, \\mathbf v, \\mathbf w) \\cdot \\begin{bmatrix}0&3&1\\\\ 1&0&3\\\\ 9&1&1\\end{bmatrix}$$ Ignore this you did edit ur post op so #### Petrus ##### Well-known member progress so far: $\\det(v+9w, w+3u, u+3v) =9\\det(v, w+3u, u+3v) + 9\\det(w, w+3u, u+3v) =$ $27\\det(v, w, u+3v) + 27\\det(w, u, u+3v) + 27\\det(w, w, u+3v) + 27\\det(w, u, u+3v)$ $= 81\\det(v, w, u) + 81\\det(v, w, v) + 81\\det(w, u, u) + 81\\det(w, u, v)+$ $81/det(w, w, u) + 81\\det(w, w, v) + 81\\det(w, u, u) + 81\\det(w, u, v)$ The next step im kinda unsure how to do it, I guess ima swap road or something and try get exemple $\\det(w, w, w)$ cause that is equal to zero? I am correct? How do i swap row or column? #### Klaas van Aarsen ##### MHB Seeker Staff member progress so far: $\\det(v+9w, w+3u, u+3v) =9\\det(v, w+3u, u+3v) + 9\\det(w, w+3u, u+3v) =$ You have a 9 instead of a 1. It should be: $\\det(1v+9w, w+3u, u+3v) =1\\det(v, w+3u, u+3v) + 9\\det(w, w+3u, u+3v) =$ $27\\det(v, w, u+3v) + 27\\det(w, u, u+3v) + 27\\det(w, w, u+3v) + 27\\det(w, u, u+3v)$ You have repeated the same mistake here. $= 81\\det(v, w, u) +" ]

[ "x= \\boxed{\\mathbf{\\pm 4}}$$", "= \\boxed{\\textbf{(B)}\\ 24}.$", "is $\\boxed{\\textbf{(D)}~\\dfrac{7}{16}}$.", "6\\end{pmatrix}) = -3$ • $C_{2,3} = (-1)^{5} * det(M_{2,3}) = -det(M_{2,3}) = \\textbf{-6}$ $det(M_{2,3})=det(\\begin{pmatrix} -1 & 4 \\\\ -3 & 6 \\end{pmatrix}) = 6$ • $C_{3,3} = (-1)^{6} * det(M_{3,3}) = det(M_{3,3}) = \\textbf{-3}$ $det(M_{2,3})=det(\\begin{pmatrix} -1 & 4 \\\\ 2 & -5\\end{pmatrix}) = -3$ Hence, the result is $\\begin{split} det(M) = -7 * C_{1,3} + 8 * C_{2,3} + -9 * C_{3,3} \\\\ = -7 * -3 + 8 * -6 + -9 * -3 \\\\ = 0 \\\\ \\end{split}$", "1\\} =$ 3. $\\{0, 1\\} \\cap \\{1, 3, 4\\} =$ 4. $\\{-2, 4, 6\\} \\cap \\mathbf{N} =$ #### Solution 1. $\\{0\\} \\cap \\varnothing = \\boxed{\\varnothing}$ 2. $\\{0, 1\\} \\cap \\{0, 1\\} = \\boxed{\\{0, 1\\}}$ 3. $\\{0, 1\\} \\cap \\{1, 3, 4\\} = \\boxed{\\{1\\}}$ 4. $\\{-2, 4, 6\\} \\cap \\mathbf{N} = \\boxed{\\{4, 6\\}}$", "C}}{3.21 \\mathrm{~\\mu F} }= \\boxed{28 \\,\\text{V}}$$", "a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf b_4)] \\\\ &\\ \\ \\ \\ + 5[6\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf a_4) + 2\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf b_4)] \\\\ &= -6\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf a_4) - 2\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf b_4) \\\\ &\\ \\ \\ \\ + 30\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf a_4) + 10\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf b_4)\\end{align} Antisymmetry The fact that $\\det$ is antisymmetric means that if you swap any two arguments of $\\det$, you'll negate the sign. Thus $$\\det(\\mathbf a_1, \\mathbf a_2) = -\\det(\\mathbf a_2, \\mathbf a_1)$$ Likewise \\begin{align}\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf a_4) &= -\\det(\\mathbf a_1, \\mathbf a_4, \\mathbf a_3, \\mathbf a_2) \\\\ &= \\det(\\mathbf a_4, \\mathbf a_1, \\mathbf a_3, \\mathbf a_2)\\end{align} So swapping an even number of columns doesn't change the sign, but swapping an odd number will produce a negative sign. One consequence of this antisymmetry is that if two columns are scalar multiples of each other, then the determinant must be zero. This is because after pulling the constant multiple out front the two columns will be the same. Then if you swap those two columns, the determinant will look the same -- but it'll have a minus sign out front. And thus the determinant will", "\\det(\\mathbf{A})^{-1}\\, \\mathrm{adj}(\\mathbf{A}).$ All the best.", "then $$16^{3/4}=(16^{\\frac{1}{4}})^{3}=(2)^3=\\boxed{8}$$ - Have you read the other answers? – Servaes May 18 '14 at 16:09", "for $1 \\le n \\le 31$), and so our answer is $\\boxed{\\textbf{(B) } 31}$", "be divisible by $4$, and this first occurs when $n = \\boxed{ 127 }$ (note that if $4m'-n > 1$, then $n > 250$). Indeed, this gives $m' = 32$ and the fraction $\\frac {32}{127}\\approx 0.25196 \\ldots$).", "not divisible by $3$ We try $M=7$ this does work giving $N=16,~M=7$ and thus $3\\times 16=\\boxed{\\textbf{(B)},~48}$ games in their division.", "the sequence. $\\boxed{\\textbf{(B)}\\ 21}$.", "6 is $\\boxed{\\textbf{(D) } 13}$. - mathleticguyyy", "$m+n=\\boxed{831}$.", "$\\det(A-I)=-4(1-\\cos \\theta). \\ \\Box$", "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", "is $\\boxed{\\textbf{C) }13+4\\sqrt{2}}$. ~IceMatrix", "$\\boxed{\\textbf{(D) }2\\le S<6}$. ~Baolan ~ pi_is_3.14", "is $|x| = 8$. Therefore, the x can be $8$ or $-8$. The product is then $-8 \\times 8 = \\boxed{\\textbf{(A)} -64}$." ]

[ "6\\end{pmatrix}) = -3$ • $C_{2,3} = (-1)^{5} * det(M_{2,3}) = -det(M_{2,3}) = \\textbf{-6}$ $det(M_{2,3})=det(\\begin{pmatrix} -1 & 4 \\\\ -3 & 6 \\end{pmatrix}) = 6$ • $C_{3,3} = (-1)^{6} * det(M_{3,3}) = det(M_{3,3}) = \\textbf{-3}$ $det(M_{2,3})=det(\\begin{pmatrix} -1 & 4 \\\\ 2 & -5\\end{pmatrix}) = -3$ Hence, the result is $\\begin{split} det(M) = -7 * C_{1,3} + 8 * C_{2,3} + -9 * C_{3,3} \\\\ = -7 * -3 + 8 * -6 + -9 * -3 \\\\ = 0 \\\\ \\end{split}$", "a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf b_4)] \\\\ &\\ \\ \\ \\ + 5[6\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf a_4) + 2\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf b_4)] \\\\ &= -6\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf a_4) - 2\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf b_4) \\\\ &\\ \\ \\ \\ + 30\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf a_4) + 10\\det(\\mathbf a_1, \\mathbf b_2, \\mathbf a_3, \\mathbf b_4)\\end{align} Antisymmetry The fact that $\\det$ is antisymmetric means that if you swap any two arguments of $\\det$, you'll negate the sign. Thus $$\\det(\\mathbf a_1, \\mathbf a_2) = -\\det(\\mathbf a_2, \\mathbf a_1)$$ Likewise \\begin{align}\\det(\\mathbf a_1, \\mathbf a_2, \\mathbf a_3, \\mathbf a_4) &= -\\det(\\mathbf a_1, \\mathbf a_4, \\mathbf a_3, \\mathbf a_2) \\\\ &= \\det(\\mathbf a_4, \\mathbf a_1, \\mathbf a_3, \\mathbf a_2)\\end{align} So swapping an even number of columns doesn't change the sign, but swapping an odd number will produce a negative sign. One consequence of this antisymmetry is that if two columns are scalar multiples of each other, then the determinant must be zero. This is because after pulling the constant multiple out front the two columns will be the same. Then if you swap those two columns, the determinant will look the same -- but it'll have a minus sign out front. And thus the determinant will", "x= \\boxed{\\mathbf{\\pm 4}}$$", "1\\} =$ 3. $\\{0, 1\\} \\cap \\{1, 3, 4\\} =$ 4. $\\{-2, 4, 6\\} \\cap \\mathbf{N} =$ #### Solution 1. $\\{0\\} \\cap \\varnothing = \\boxed{\\varnothing}$ 2. $\\{0, 1\\} \\cap \\{0, 1\\} = \\boxed{\\{0, 1\\}}$ 3. $\\{0, 1\\} \\cap \\{1, 3, 4\\} = \\boxed{\\{1\\}}$ 4. $\\{-2, 4, 6\\} \\cap \\mathbf{N} = \\boxed{\\{4, 6\\}}$", "# If A and B are 4 times 4 matrices, det(A) = 1, det(B) = 4, then det(AB) = ? det(3A) = ? det(A^T) = ? det(B^{-1}) = ? det(B^4) = ? If A and B are $4×4$ matrices, det(A) = 1, det(B) = 4, then det(AB) = ? det(3A) = ? $det\\left({A}^{T}\\right)=?$ $det\\left({B}^{-1}\\right)=?$ $det\\left({B}^{4}\\right)=?$ You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it Corben Pittman Given that A and B are $4×4$ matrices: det(A)=1 : det(B)=4 Then (1) $det\\left(AB\\right)=det\\left(A\\right)\\cdot det\\left(B\\right)$ $=1\\cdot 4$ det(AB)=4 (2) $det\\left(3A\\right)=3\\cdot det\\left(A\\right)$ $=3\\cdot \\left(1\\right)$ det(3A)=3 (3) $det\\left({A}^{T}\\right)=det\\left(A\\right)$ $det\\left({A}^{T}\\right)=1$ (4)$det\\left({B}^{-1}\\right)=\\frac{1}{|B|}$ $det\\left({B}^{-1}\\right)=\\frac{1}{4}$ (5) $det\\left({B}^{4}\\right)=det\\left(B\\cdot B\\cdot B\\cdot B\\right)$ $=det\\left(B\\right)\\cdot det\\left(B\\right)\\cdot det\\left(B\\right)\\cdot det\\left(B\\right)$ $=4\\cdot 4\\cdot 4\\cdot 4$ =256 Jeffrey Jordon", "prove that $$\\det(B)\\ge\\det(A)$$ (so that $$\\det(B)>0$$). Let $$A=\\pmatrix{p&s_1a&s_2b\\\\ s_1a&q&s_3c\\\\ s_2b&s_3c&r} \\ \\text{ and }\\ B=\\pmatrix{p&a&b\\\\ a&q&c\\\\ b&c&r}$$ where $$p,q,r>0,\\,a,b,c\\ge0$$ and $$s_1,s_2,s_3\\in\\{1,-1\\}$$. Then \\begin{aligned} \\det(A) &=pqr+2s_1s_2s_3abc-(pc^2+qb^2+ra^2)\\\\ &\\le pqr+2abc-(pc^2+qb^2+ra^2)\\\\ &=\\det(B). \\end{aligned}", "\\det(\\mathbf{A})^{-1}\\, \\mathrm{adj}(\\mathbf{A}).$ All the best.", "the Math Challenge Problem forum, I know the solution (and its resolution) :) . Another version is: prove that $\\mod (\\det A)=n^{n/2}$ . • Dec 18th 2011, 02:17 AM NonCommAlg Re: Find mod ( det A ) Quote: Originally Posted by FernandoRevilla Modulus of the complex number $\\det (A)$ . Right, $A$ is a Vandermonde matrix. Being this the Math Challenge Problem forum, I know the solution (and its resolution) :) . Another version is: prove that $\\mod (\\det A)=n^{n/2}$ . so the question is to find $|\\det(A)|.$ so far i got $|\\det(A)| = 2^{\\binom{n}{2}}a^n,$ where $a = \\prod_{k=1}^m \\sin(k \\theta)$ and $\\theta = \\frac{\\pi}{n}$ and $m = \\lfloor \\frac{n-1}{2} \\rfloor.$ so the problem now is to evaluate $a.$ • Dec 18th 2011, 05:28 AM FernandoRevilla Re: Find mod ( det A ) A better approach is to find previously $A^2$ .", "$\\mod (\\det A)=n^{n/2}$ . 4. ## Re: Find mod ( det A ) Originally Posted by FernandoRevilla Modulus of the complex number $\\det (A)$ . Right, $A$ is a Vandermonde matrix. Being this the Math Challenge Problem forum, I know the solution (and its resolution) . Another version is: prove that $\\mod (\\det A)=n^{n/2}$ . so the question is to find $|\\det(A)|.$ so far i got $|\\det(A)| = 2^{\\binom{n}{2}}a^n,$ where $a = \\prod_{k=1}^m \\sin(k \\theta)$ and $\\theta = \\frac{\\pi}{n}$ and $m = \\lfloor \\frac{n-1}{2} \\rfloor.$ so the problem now is to evaluate $a.$ 5. ## Re: Find mod ( det A ) A better approach is to find previously $A^2$ .", "$y = 11$. This does work. We found one valid solution so the answer is $\\boxed{\\textbf{(B) }1}$.", "You can use $\\det(A \\cdot B)= \\det(A) \\cdot \\det(B)$. And consider that $$(\\mathbf v+9\\mathbf w, \\mathbf w+3\\mathbf u, \\mathbf u+3\\mathbf v) = (\\mathbf u, \\mathbf v, \\mathbf w) \\cdot \\begin{bmatrix}0&3&1\\\\ 1&0&3\\\\ 9&1&1\\end{bmatrix}$$ Ignore this you did edit ur post op so #### Petrus ##### Well-known member progress so far: $\\det(v+9w, w+3u, u+3v) =9\\det(v, w+3u, u+3v) + 9\\det(w, w+3u, u+3v) =$ $27\\det(v, w, u+3v) + 27\\det(w, u, u+3v) + 27\\det(w, w, u+3v) + 27\\det(w, u, u+3v)$ $= 81\\det(v, w, u) + 81\\det(v, w, v) + 81\\det(w, u, u) + 81\\det(w, u, v)+$ $81/det(w, w, u) + 81\\det(w, w, v) + 81\\det(w, u, u) + 81\\det(w, u, v)$ The next step im kinda unsure how to do it, I guess ima swap road or something and try get exemple $\\det(w, w, w)$ cause that is equal to zero? I am correct? How do i swap row or column? #### Klaas van Aarsen ##### MHB Seeker Staff member progress so far: $\\det(v+9w, w+3u, u+3v) =9\\det(v, w+3u, u+3v) + 9\\det(w, w+3u, u+3v) =$ You have a 9 instead of a 1. It should be: $\\det(1v+9w, w+3u, u+3v) =1\\det(v, w+3u, u+3v) + 9\\det(w, w+3u, u+3v) =$ $27\\det(v, w, u+3v) + 27\\det(w, u, u+3v) + 27\\det(w, w, u+3v) + 27\\det(w, u, u+3v)$ You have repeated the same mistake here. $= 81\\det(v, w, u) +", "# What can be said if $A^2+B^2+2AB=0$ for some real $2\\times2$ matrices $A$ and $B$? Let $A,B\\in M_2(\\mathbb{R})$ be such that $A^2+B^2+2AB=0$ and $\\det A= \\det B$. Our goal is to compute $\\det(A^2 – B^2)$. According to the chain of comments on Art of Problem Solving, the following statements are true: 1. $\\det(A^2+B^2)+\\det(A^2-B^2)=2(\\det A^2+\\det B^2)$. (Is this well known?) 2. (1) $\\implies \\det(A^2-B^2)=0$. 3. If $A,B\\in M_2(\\mathbb{C})$ satisfy $A^2+B^2+2AB=0$, then $AB=BA$. 4. $(A+B)^2=0 \\implies \\det(A^2-B^2)=0$. Can someone help me with justifying these statements? Edit: Doug M provided an explanation for (1) in the answers. Here is an explanation for (2): $A^2+B^2+2AB=O_2 \\implies A^2+B^2=-2AB$. So $\\det(A^2+B^2)=4\\det(AB)$. Now using (1), $\\det(A^2-B^2)= 2(\\det(A^2)-2\\det(AB)+\\det(B^2)) = 2((\\det(A)^2-\\det(B)^2) = 0$. #### Solutions Collecting From Web of \"What can be said if $A^2+B^2+2AB=0$ for some real $2\\times2$ matrices $A$ and $B$?\" Of course, the OP did not understand much about the proposed exercise. In particular, in $M_2(\\mathbb{C})$, $A^2 +B^2+2AB=0$ implies $\\det(A)=\\det(B)$. In fact, the interesting result is Proposition. Let $A,B\\in M_2(\\mathbb{C})$ be such that $(1)$ $A^2+B^2+2AB=0$; then $AB=BA$. Note that the result works only in dimension $2$. Proof. If $H=A+B$, then $H^2=HA-AH$ and $trace(H^2)=0$. Moreover $H^3=H^2A-HAH=HAH-AH^2$ implies that $trace(H^3)=0$. We deduce that the eigenvalues of $H$ are $0$, $H$ is nilpotent, $H^2=0$, $AH=HA$, $AB=BA$. $\\square$ Remark. According to Proposition, $(1)$ implies that $(A+B)^2=0$, $\\det(A+B)=0$, $\\det(A^2-B^2)=0$;", "6 is $\\boxed{\\textbf{(D) } 13}$. - mathleticguyyy", "# Does $k[B]^{B'}=k[\\det(b),\\det(b)^{-1}]$? Let $k=\\overline{k}$. Suppose $GL_2(k)$ acts on $GL_2(k)$ by left multiplication. Then $k[GL_2]^{SL_2} = k[\\det(g),\\det(g)^{-1}]$. Now for $B=\\left\\{ \\left( \\begin{array}{cc} b_{11} & b_{12} \\\\ 0 & b_{22} \\\\ \\end{array}\\right) : b_{ii}\\not=0, b_{12}\\in k \\right\\}$, let $B'=\\{ b\\in B: \\det(B)=1 \\}$. Suppose $B$ acts on $B$ by left multiplication. Then does $k[B]^{B'}=k[b_{11}, b_{22},b_{11}^{-1},b_{22}^{-1}]$ or does $k[B]^{B'}=k[\\det(b),\\det(b)^{-1}]$? -", "all $\\mathbf {N}$ lines are multiplied by $\\mathbf {\\beta}$, we have : $\\mathbf {Det (A * B) = Det (A * \\beta I) = \\beta^N Det (A * I) = \\beta^ N Det (A)}$ $\\mathbf {Det (A) * Det (B) = Det (A)\\beta^N Det (I) = \\beta^N Det (A)}$ **c) **Any N-dimensional shape represented by a matrix can be broken down into an amount $\\mathbf {s}$ of small cube matrices $\\mathbf {B_k}$ ($\\mathbf {k}$ is an index from 1 to s) of dimension $\\mathbf {\\beta}$ (diagonal matrix with only $\\mathbf {\\beta}$value), where $\\mathbf {\\beta}$ is arbitrarily small and therefore $\\mathbf {s}$ arbitrarily large. Look at the volume below the colorful surface, above the plane $\\mathbf {z=0}$ (vertical axis), below the plane $\\mathbf {y=0}$ (negative y) to down and left, right to the place $\\mathbf {x=0}$ (positive x). Imagine put inside this volume thousands and thousands ($\\mathbf {s}$) of tiny cubes. $\\mathbf {Det(B) = Det(B_1) + ... + Det(B_s)}$ $\\mathbf { Det (A) * Det (B) = Det (A) * (Det (B_1) + ... + Det (B_s)) = s Det (A) \\beta^N \\qquad}$ (c1) $\\mathbf { Det (A * B) = Det (A * (B1 + ... + B_s)) = Det (A * B_1) + ... + Det (A * B_s)}$ By (b) above $\\mathbf {= \\beta^N Det(A) +", "which is precisely the given answer. Alternatively, use the properties: \\begin{align}\\text{adj}(cA)&=c^{n-1}\\text{adj}(A) \\quad (1)\\\\ \\det(cA)&=c^n\\det(A) \\quad (2)\\\\ \\det(\\text{adj}(A))&=(\\det(A))^{n-1} \\quad (3)\\\\ \\det(A^{-1})&=(\\det(A))^{-1} \\quad (4) \\end{align} to get: $$\\det(3\\,\\text{adj}(2A^{-1}))\\stackrel{(1)}{=}\\\\ \\det(3\\cdot 2^2\\,\\text{adj}(A^{-1}))\\stackrel{(2)}{=}\\\\ 12^3\\det(\\text{adj}(A^{-1}))\\stackrel{(3)}{=}\\\\ 12^3(\\det(A^{-1}))^2\\stackrel{(4)}{=}\\\\ 12^3((\\det(A))^{-1})^2=\\\\ 12^3\\cdot 4^{-2}=\\\\ 108.$$", "can establish the recurrence $$p_k=p_{k-1}\\cos\\frac{\\pi}{n}-\\frac{\\cos^2 \\theta_{k-1}}{4}p_{k-2}$$ and then $\\det\\mathbf T=p_n$. There is also an explicit formula for the first and last columns of the inverse of a tridiagonal matrix (the result of $\\mathbf T^{-1}\\mathbf U$) which I will have to look up (unless somebody beats me to it), and proving the nonnegativity of $\\det\\mathbf A$ boils down to proving that either $p_n$ and $\\det(\\mathbf I+\\mathbf V^T\\mathbf T^{-1}\\mathbf U)$ have the same sign, or that one of the two is zero. - I guess your $\\det\\mathbf A=(\\det\\mathbf T)\\det(\\mathbf I+\\mathbf V^T\\mathbf A^{-1}\\mathbf U)$ should be $\\det\\mathbf A=(\\det\\mathbf T)\\det(\\mathbf I+\\mathbf V^T\\mathbf T^{-1}\\mathbf U)$. the Sherman-Morrison-Woodbury formula for determinants is usually for $U,V$ of rank one, nd.edu/~jeff/Teaching/CBE60478/ShermanMorrison.pdf . Can you show me the reference for your formula $\\det\\mathbf A=(\\det\\mathbf T)\\det(\\mathbf I+\\mathbf V^T\\mathbf T^{-1}\\mathbf U)$. – Sunni May 18 '11 at 0:18 Yes @Sunni; I'll correct that typo. Faddev/Faddeva or Aitken's old books should have it, but if you're knowledgable with Schur complements, it shouldn't be too hard to reckon it yourself. – Ｊ. Ｍ. Jun 2 '11 at 4:01 This is not an aswer, but was too long for a comment. I suspect that for $N>3$ the matrix is (semi)positive definite, a stronger property than what is asked. As I pointed in a comment, to prove this is equivalent to prove the", "is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = 3 \\pm i$ and $|c| = \\boxed{\\textbf{(E) } \\sqrt{10}}$.", "we see that $f(m,n)=1$ for all $m\\ge 5$. In particular, $f(2015,2) = \\boxed{\\textbf{(B)} \\; 1}$.", "$m+n=2+89=\\boxed{091}$." ]

[ "of ordered pairs $$\\{(a,1), (b,3), (c,2)\\}.$$", "# Number of ordered pairs Find the number of ordered pairs of whole numbers $$(x,y)$$ such that $$(xy-7)^2 = x^2 +y^2$$. ×", "+0 0 78 2 One ordered pair $$(a,b)$$ satisfies the two equations $$ab^4 = 384$$ and $$a^2 b^5 = 4608$$. What is the value of $$a$$ in this ordered pair? May 9, 2020 #1 0 a = 2 \\sqrt[3]{4} (and b = 4 \\sqrt[3]{6}) May 9, 2020 #2 0 umm its wrong, thats all i know Guest May 9, 2020", "+0 0 186 3 +51 Find the number of ordered pairs $(a,b)$ of integers such that $\\frac{a + 2}{a + 5} = \\frac{b}{3}.$ May 10, 2020 #3 +25451 +4 Find the number of ordered pairs $$(a,b)$$ of integers such that $$\\dfrac{a + 2}{a + 5} = \\dfrac{b}{3}$$.", "+0 # HELP ASAP PLSS 0 37 1 Find the number of ordered pairs $(a,b)$ of integers such that $\\frac{a + 2}{a + 5} = \\frac{b}{3}.$ May 4, 2020", "+0 # PLS HALPAROOOO 0 65 1 Find the number of ordered pairs $(a,b)$ of integers such that $\\frac{a + 2}{a + 5} = \\frac{b}{3}.$ Jun 26, 2022", "# Ordered Pairs Discrete Mathematics Level 3 Find the number of ordered pairs of integers $$(m,n)$$ that satisfy $\\frac{m}{12}=\\frac{12}{n}.$ ×", "Determine the number of ordered pairs $$(a,b)$$ of integers such that $$\\log_a b + 6\\log_b a=5, 2 \\leq a \\leq 2005,$$ and $$2 \\leq b \\leq 2005.$$ (第二十三届AIME2 2005 第5题)", "# How do you find the ordered pair that satisfies x - 2y = 0 and 2x - 5y = -4? (2) 2x - 5y = - 4 --> 4y - 5y = - 4 -> $y = 4$ -> $x = 2 y = 8$", "possible ordered pairs for $(a,b)$ are $(504, 503)$, $(505, 504)$, ... , $(1004, 1003)$, so $a$ has $\\boxed{501}$ possible values.", "# Arshdeep's ordered pairs Number Theory Level 5 Find the total number of ordered pairs of integers $$(a,b)$$ satisfying $a^2 (b^2 + 1) + b^2(a^2 + 16) = 448.$ This problem is shared by Arshdeep D, who came across it in Mathematical Mayham. ×", "Solve this (Inspired by Mehul Arora's comment) If $$abc\\neq{0}$$ and $$a,b,c\\in { Z }^{ + }$$. Find the no. of ordered pair for $$(a,b,c)$$ for: $a^{2015}+b^{2015}=c^{2015}$ ×", "# Arshdeep's ordered pairs Find the total number of ordered pairs of integers $$(a,b)$$ satisfying $a^2 (b^2 + 1) + b^2(a^2 + 16) = 448.$ This problem is shared by Arshdeep D, who came across it in Mathematical Mayham. ×", "× # ordered pair help plzzz How many real order pairs of $$(x,p)$$ satisfy $$\\frac{100x^3}{3}-(150+\\frac{p}{8})x^2 + (125 +\\frac{5p}{8})x - (\\frac{59p}{96} - \\frac{75}{2}) = 0$$ such that $$0\\leq x\\leq2.5$$ ? Note by Ashok Kumar 3 years, 2 months ago There are no comments in this discussion. ×", "+0 Help! 0 133 1 For certain ordered pairs (a,b) of real numbers, the system of equations [\\begin{aligned} ax+by&=1 \\\\ x^2 + y^2 &= 50 \\end{aligned} has at least one solution, and each solution is an ordered pair (x,y) of integers. How many such ordered pairs (a,b) are there? Dec 16, 2018 $$\\text{there are only so many pairs of integers that satisfy the second equation}\\\\ \\text{They are }(\\pm 1, \\pm 7),(\\pm 7,\\pm 1),(\\pm 5,\\pm 5)\\\\ \\text{each of these produces 4 ordered pairs }(a,b) \\text{ thus there are 12 total}$$", "going to look at the intersection of A with BxC, and both B and C have integers > 0 and < 5 let's restrict the pairs in A to between 0 and 5. Thus we have ordered pairs from {0,1,2,3,4,5} which have a difference of <=1. {(0,0),(0,1), (1,0),(1,1),(1,2), (2,1),(2,2),(2,3), (3,2),(3,3),(3,4), (4,3),(4,4),(4,5), (5,4),(5,5)} Now you can determine BxC to get another set of ordered pairs, then find the intersection. Hope that helps.", "+0 0 242 1 +606 If $1 \\le a \\le 10$ and $1 \\le b \\le 36$, for how many ordered pairs of integers $(a, b)$ is $\\sqrt{a + \\sqrt{b}}$ an integer? Apr 21, 2018 #1 +98044 +1 $$1 \\le a \\le 10 and 1 \\le b \\le 36$$ $$\\sqrt{a + \\sqrt{b}}$$ b is only an integer when b = 1, 4, 9, 16, 25 , 36 The possible pairs of (a, b) are (1, 9) (2, 4) (3, 36) (4, 25) (5, 16) (6, 9) (7, 4) (8, 1) (10, 36) So.....nine possible ordered pairs Apr 21, 2018", "## Basic College Mathematics (10th Edition) a) $-2,-2$ b) $-8,-8$ c) $10,10$ The answers to each pair are the same because are the same operations, just changing the order. a) $$-5 + 3 = -2$$ $$3 + (-5) = 3-5 = -2$$ b) $$-2 + (-6) = -2 - 6 = -8$$ $$-6 + (-2) = -6-2 = -8$$ c) $$17 + (-7) = 17 - 7 = 10$$ $$-7 + 17 = 10$$ The answers to each pair are the same because are the same operations, just changing the order.", "Number Theory Level 4 How many ordered pairs of integers $$a,b,c$$ exist such that $$a^{2}+b^{2}-8c=6$$?", "+0 # Help! 0 39 1 For certain ordered pairs (a,b) of real numbers, the system of equations [\\begin{aligned} ax+by&=1 \\\\ x^2 + y^2 &= 50 \\end{aligned} has at least one solution, and each solution is an ordered pair (x,y) of integers. How many such ordered pairs (a,b) are there? Dec 16, 2018 #1 +3563 +2 $$\\text{there are only so many pairs of integers that satisfy the second equation}\\\\ \\text{They are }(\\pm 1, \\pm 7),(\\pm 7,\\pm 1),(\\pm 5,\\pm 5)\\\\ \\text{each of these produces 4 ordered pairs }(a,b) \\text{ thus there are 12 total}$$ . Dec 16, 2018" ]

[ "(Did you ever notice there was a suspicious gap there between $$60^\\circ$$ and $$90^\\circ$$?) Since $$75^\\circ=30^\\circ+45^\\circ$$, we can use $$\\alpha=30^\\circ$$ and $$\\beta=45^\\circ$$ in our formulas to find the sine and cosine: \\begin{align*} \\sin(75^\\circ) &= \\sin(30^\\circ)\\cos(45^\\circ)+\\cos(30^\\circ)\\sin(45^\\circ)\\\\ &= \\dfrac12\\cdot\\dfrac{\\sqrt2}2+\\dfrac{\\sqrt3}2\\cdot\\dfrac{\\sqrt2}2\\\\ &= \\dfrac{1\\cdot\\sqrt2}4+\\dfrac{\\sqrt3\\cdot\\sqrt2}4\\\\ &= \\dfrac{\\sqrt2+\\sqrt6}4 \\end{align*}\\\\ \\begin{align*} \\cos(75^\\circ) &= \\cos(30^\\circ)\\cos(45^\\circ)+\\sin(30^\\circ)\\sin(45^\\circ)\\\\ &= \\dfrac{\\sqrt3}2\\cdot\\dfrac{\\sqrt2}2-\\dfrac{1}2\\cdot\\dfrac{\\sqrt2}2\\\\ &= \\dfrac{\\sqrt3\\cdot\\sqrt2}4-\\dfrac{1\\cdot\\sqrt2}4\\\\ &= \\dfrac{\\sqrt6-\\sqrt2}4 \\end{align*} So the $$75^\\circ$$ lands at the point $$\\left(\\dfrac{\\sqrt6-\\sqrt2}{4},\\dfrac{\\sqrt6+\\sqrt2}{4}\\right)$$, which is about $$(0.966,0.259)$$. ## Other derived formulas We have similar formulas for how to deal with the difference of two angles: \\begin{align*} \\sin(\\alpha-\\beta) &= \\sin(\\alpha)\\cos(\\beta)-\\cos(\\alpha)\\sin(\\beta)\\\\ \\cos(\\alpha-\\beta) &= \\cos(\\alpha)\\cos(\\beta)+\\sin(\\alpha)\\sin(\\beta)\\\\ \\end{align*} These formulas are so similar to the sum formulas above that we often just write them together: \\begin{align*} \\sin(\\alpha\\pm\\beta) &= \\sin(\\alpha)\\cos(\\beta)\\pm\\cos(\\alpha)\\sin(\\beta)\\\\ \\cos(\\alpha\\pm\\beta) &= \\cos(\\alpha)\\cos(\\beta)\\mp\\sin(\\alpha)\\sin(\\beta)\\\\ \\end{align*} Note the \"$$\\mp$$\" in the last formula! This means that the sign used in this part of the formula is the opposite of the one at the beginning. You might be thinking, if we've got formulas for sine and cosine, what about for tangent? You're in luck — there are formulas for that too! \\begin{align*} \\tan(\\alpha+\\beta) &= \\dfrac{\\tan(\\alpha)+\\tan(\\beta)}{1-\\tan(\\alpha)\\tan(\\beta)}\\\\ \\tan(\\alpha-\\beta) &= \\dfrac{\\tan(\\alpha)-\\tan(\\beta)}{1+\\tan(\\alpha)\\tan(\\beta)} \\end{align*} Or again we can write these as one formula if we'd like: $\\tan(\\alpha\\pm\\beta) = \\dfrac{\\tan(\\alpha)\\pm\\tan(\\beta)}{1\\mp\\tan(\\alpha)\\tan(\\beta)}$ In the questions below, you'll show where most of these formulas come from. # Preview Activity 15 Answer these questions and submit your answers as", "the problem gives us: We don’t seem to know much about the side $PR$, whereas parts of $PQ$ and $QR$ are known. Hence, in calculating the area of $\\Delta PQR$, it seems like using the formula $\\text{Area of } \\Delta PQR = \\displaystyle\\frac{1}{2} |PQ||QR|\\sin PQR = \\frac{1}{2} |PQ||QR| \\sin 60^\\circ$ could work. Let’s label the diagram further to give more info on $PQ$ and $QR$: We have 2 unknown quantities we would like to find: $|PQ|$ and $|QR|$. We can use the sine rule to reduce that to just finding $|PQ|$: \\begin{aligned} \\displaystyle\\frac{|PQ|}{\\sin 45^\\circ} &= \\frac{|QR|}{\\sin 75^\\circ}, \\\\ |QR| &= \\frac{|PQ|}{\\sin 45^\\circ} \\sin 75^\\circ = \\sqrt{2}|PQ| \\sin 75^\\circ. \\end{aligned} From the diagram, we know that $|PQ| = 2 + |AP|$. We can use the sine rule again to find the value of $|AP|$: \\begin{aligned} \\displaystyle\\frac{|AP|}{\\sin 45^\\circ} &= \\frac{|AF|}{\\sin 75^\\circ}, \\\\ |AP| &= \\frac{1}{\\sin 75^\\circ} \\sin 45^\\circ = \\frac{1}{\\sqrt{2}\\sin 75^\\circ}. \\end{aligned} Hence, we have \\begin{aligned} \\text{Area} &= \\displaystyle\\frac{1}{2}|PQ| \\cdot \\sqrt{2} |PQ| \\sin 75^\\circ \\cdot \\sin 60^\\circ \\\\ &= \\frac{\\sqrt{2}}{2}\\frac{\\sqrt{3}}{2} \\sin 75^\\circ |PQ|^2 \\\\ &= \\frac{\\sqrt{6}}{4} \\sin 75^\\circ \\left( 2 + \\frac{1}{\\sqrt{2}\\sin 75^\\circ} \\right)^2 \\\\ &= \\frac{\\sqrt{6}}{4}\\sin 75^\\circ \\left( 4 + \\frac{4}{\\sqrt{2}\\sin 75^\\circ} + \\frac{1}{2\\sin^2 75^\\circ} \\right) \\\\ &= \\sqrt{6}\\sin 75^\\circ + \\sqrt{3} + \\frac{\\sqrt{6}}{8 \\sin 75^\\circ}. \\end{aligned} It remains to get rid of the $\\sin 75^\\circ$ term. $\\sin 75^\\circ", "the circumscribed circle. Since $\\alpha = 45^\\circ$, \\begin{align*} a & = 2R\\sin\\alpha\\\\ & = 2 \\cdot 2 \\cdot \\sin(45^\\circ)\\\\ & = 4 \\cdot \\frac{\\sqrt{2}}{2}\\\\ & = 2\\sqrt{2} \\end{align*} Since $\\gamma = 60^\\circ$, \\begin{align*} c & = 2R\\sin\\gamma\\\\ & = 2 \\cdot 2 \\cdot \\sin(60^\\circ)\\\\ & = 4 \\cdot \\frac{\\sqrt{3}}{2}\\\\ & = 2\\sqrt{3} \\end{align*} The area of the triangle is $$A = \\frac{1}{2}ac\\sin\\beta$$ since $c\\sin\\beta$ is the length of the altitude to side $\\overline{BC}$, which has length $a$. By the Angle Sum Theorem for Triangles, \\begin{align*} \\beta & = 180^\\circ - 60^\\circ - 45^\\circ\\\\ & = 75^\\circ \\end{align*} To find the exact value of $\\sin(75^\\circ)$, we use the Sum of Angles Formula $$\\sin(\\theta + \\varphi) = \\sin\\theta\\cos\\varphi + \\cos\\theta\\sin\\varphi$$ with $\\theta = 30^\\circ$ and $\\varphi = 45^\\circ$, which yields \\begin{align*} \\sin(75^\\circ) & = \\sin(30^\\circ)\\cos(45^\\circ) + \\cos(30^\\circ)\\sin(45^\\circ)\\\\ & = \\frac{1}{2} \\cdot \\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{2}}{2}\\\\ & = \\frac{\\sqrt{2} + \\sqrt{6}}{4} \\end{align*} Hence, the area of triangle $ABC$ is \\begin{align*} A & = \\frac{1}{2}ac\\sin\\beta\\\\ & = \\frac{1}{2} \\cdot 2\\sqrt{2} \\cdot 2\\sqrt{3} \\cdot \\frac{\\sqrt{2} + \\sqrt{6}}{4}\\\\ & = \\frac{1}{2} \\cdot \\sqrt{6}(\\sqrt{2} + \\sqrt{6})\\\\ & = \\frac{1}{2}(\\sqrt{12} + 6)\\\\ & = \\frac{1}{2}(2\\sqrt{3} + 6)\\\\ & = \\sqrt{3} + 3 \\end{align*} square units. If $O$ is the circumcenter of $ABC$, we have: $$\\Delta = [OAB]+[OAC]+[OBC] = \\frac{R^2}{2}\\left(\\sin(2A)+\\sin(2B)+\\sin(2C)\\right).\\tag{1}$$ In our case, we have $A=45^\\circ,C=60^\\circ$,", "4. . \\scriptsize \\begin{align*}\\cos {{50}^\\circ} & =\\cos ({{25}^\\circ}+{{25}^\\circ})\\\\=2{{\\cos }^{2}}{{25}^\\circ}-1\\\\\\therefore 2{{\\cos }^{2}}{{25}^\\circ} & =\\cos {{50}^\\circ}+1\\\\\\therefore {{\\cos }^{2}}{{25}^\\circ} & =\\displaystyle \\frac{{\\cos {{{50}}^\\circ}+1}}{2}\\\\\\therefore \\cos {{25}^\\circ} & =\\sqrt{{\\displaystyle \\frac{{\\cos {{{50}}^\\circ}+1}}{2}}}\\quad \\quad \\cos {{25}^\\circ} \\gt 0\\therefore {{\\cos }^{2}}{{25}^\\circ} \\gt 0\\\\&=\\sqrt{{\\displaystyle \\frac{{\\cos ({{{90}}^\\circ}-{{{40}}^\\circ})+1}}{2}}}\\\\&=\\sqrt{{\\displaystyle \\frac{{\\sin {{{40}}^\\circ}+1}}{2}}}\\\\&=\\sqrt{{\\displaystyle \\frac{{b+1}}{2}}}\\end{align*} 5. . \\scriptsize \\begin{align*}\\cos {{80}^\\circ}&=\\cos ({{40}^\\circ}+{{40}^\\circ})\\\\&=1-2{{\\sin }^{2}}{{40}^\\circ}\\\\&=1-2{{b}^{2}}\\end{align*} 6. . \\scriptsize \\begin{align*}\\cos (-{{800}^\\circ})&=\\cos (-{{80}^\\circ}-2\\cdot {{360}^\\circ})\\\\&=\\cos (-{{80}^\\circ})\\\\&=\\cos {{80}^\\circ}\\\\&=1-2{{b}^{2}}\\quad \\quad \\text{From e}\\text{.}\\end{align*} 2. . \\scriptsize \\begin{align*}\\text{LHS}&=\\displaystyle \\frac{1}{{\\sin 2A}}-\\displaystyle \\frac{1}{{\\tan 2A}}\\\\&=\\displaystyle \\frac{1}{{\\sin 2A}}-\\displaystyle \\frac{{\\cos 2A}}{{\\sin 2A}}\\\\&=\\displaystyle \\frac{{1-\\cos 2A}}{{\\sin 2A}}\\\\&=\\displaystyle \\frac{{1-(1-2{{{\\sin }}^{2}}A)}}{{2\\sin A\\cos A}}\\\\&=\\displaystyle \\frac{{2{{{\\sin }}^{2}}A}}{{2\\sin A\\cos A}}\\\\&=\\displaystyle \\frac{{\\sin A}}{{\\cos A}}\\\\&=\\tan A=\\text{RHS}\\end{align*} 3. $\\scriptsize 13\\cos \\theta =-5$ and $\\scriptsize {{180}^\\circ}\\le \\theta \\le {{360}^\\circ}$ 1. . \\scriptsize \\begin{align*}\\cos \\theta & =-\\displaystyle \\frac{5}{{13}}\\\\\\cos 2\\theta & =2{{\\cos }^{2}}\\theta -1\\\\&=2{{\\left( {-\\displaystyle \\frac{5}{{13}}} \\right)}^{2}}-1\\\\&=2\\left( {\\displaystyle \\frac{{25}}{{169}}} \\right)-1\\\\&=\\displaystyle \\frac{{50}}{{169}}-1\\\\&=\\displaystyle \\frac{{50-169}}{{169}}\\\\&=-\\displaystyle \\frac{{119}}{{169}}\\end{align*} 2. . \\scriptsize \\begin{align*}\\sin 2\\theta &=2\\sin \\theta \\cos \\theta \\\\&=2\\times \\left( {-\\displaystyle \\frac{{12}}{{13}}\\times -\\displaystyle \\frac{5}{{13}}} \\right)\\\\&=2\\times \\left( {\\displaystyle \\frac{{60}}{{169}}} \\right)\\\\&=\\displaystyle \\frac{{120}}{{169}}\\end{align*} 3. . \\scriptsize \\begin{align*}\\tan 2\\theta &=\\displaystyle \\frac{{\\sin 2\\theta }}{{\\cos 2\\theta }}\\\\&=\\displaystyle \\frac{{\\displaystyle \\frac{{120}}{{169}}}}{{-\\displaystyle \\frac{{119}}{{169}}}}\\\\&=-\\displaystyle \\frac{{120}}{{169}}\\times \\displaystyle \\frac{{169}}{{119}}\\\\&=-\\displaystyle \\frac{{120}}{{119}}\\end{align*} 4. . \\scriptsize \\begin{align*}\\sin ({{180}^\\circ}-2\\theta )&=\\sin 2\\theta \\\\&=\\displaystyle \\frac{{120}}{{169}}\\end{align*} Back to Exercise 1.2 ### Unit 1: Assessment 1. . \\scriptsize \\displaystyle \\begin{align*}\\text{LHS}&=\\sin {{75}^\\circ}+\\sin {{15}^\\circ}\\\\&=\\sin ({{45}^\\circ}+{{30}^\\circ})+\\sin ({{45}^\\circ}-{{30}^\\circ})\\\\&=\\sin {{45}^\\circ}\\cos {{30}^\\circ}+\\cos {{45}^\\circ}\\sin {{30}^\\circ}+\\sin {{45}^\\circ}\\cos {{30}^\\circ}-\\cos {{45}^\\circ}\\sin {{30}^\\circ}\\\\&=2\\sin {{45}^\\circ}\\cos {{30}^\\circ}\\\\&=2\\times \\displaystyle \\frac{1}{{\\sqrt{2}}}\\times \\displaystyle \\frac{{\\sqrt{3}}}{2}\\\\&=\\displaystyle \\frac{{\\sqrt{3}}}{{\\sqrt{2}}}\\\\&=\\displaystyle \\frac{{\\sqrt{6}}}{2}=\\text{RHS}\\end{align*} 2. . \\scriptsize \\displaystyle \\begin{align*}\\text{LHS}&=\\displaystyle \\frac{{\\sin 2\\theta }}{{\\sin \\theta }}-\\displaystyle \\frac{{\\cos 2\\theta }}{{\\cos \\theta }}\\\\&=\\displaystyle \\frac{{2\\sin \\theta", "& = 180^\\circ\\\\ m \\angle A & = 57^\\circ\\end{align*} Now, use the Law of Sines to set up ratios for \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*}. \\begin{align*}\\frac{\\sin 57^\\circ}{a} = \\frac{\\sin 85^\\circ}{b} = \\frac{\\sin 38^\\circ}{12}\\end{align*} \\begin{align*}\\frac{\\sin 57^\\circ}{a} & = \\frac{\\sin 38^\\circ}{12} && \\quad \\frac{\\sin 85^\\circ}{b} = \\frac{\\sin 38^\\circ}{12}\\\\ a \\cdot \\sin 38^\\circ & = 12 \\cdot \\sin 57^\\circ && b \\cdot \\sin 38^\\circ = 12 \\cdot \\sin 85^\\circ\\\\ a & = \\frac{12 \\cdot \\sin 57^\\circ}{\\sin 38^\\circ} \\approx 16.4 && \\qquad \\quad \\ \\ b = \\frac{12 \\cdot \\sin 85^\\circ}{\\sin 38^\\circ} \\approx 19.4\\end{align*} Example 2: Solve the triangle using the Law of Sines. Round decimal answers to the nearest tenth. Solution: Set up the ratio for \\begin{align*}\\angle B\\end{align*} using Law of Sines. \\begin{align*}\\frac{\\sin 95^\\circ}{27} & = \\frac{\\sin B}{16}\\\\ 27 \\cdot \\sin B & = 16 \\cdot \\sin 95^\\circ\\\\ \\sin B & = \\frac{16 \\cdot \\sin 95^\\circ}{27} \\rightarrow \\sin^{-1} \\left ( \\frac{16 \\cdot \\sin 95^\\circ}{27} \\right ) = 36.2^\\circ\\end{align*} To find \\begin{align*}m \\angle C\\end{align*} use the Triangle Sum Theorem. \\begin{align*}m \\angle C + 95 ^\\circ + 36.2^\\circ = 180^\\circ \\rightarrow m \\angle C = 48.8^\\circ\\end{align*} To find \\begin{align*}c\\end{align*}, use the Law of Sines again. \\begin{align*}\\frac{\\sin 95^\\circ}{27} = \\frac{\\sin 48.8^\\circ}{c}\\end{align*} \\begin{align*}c \\cdot \\sin 95^\\circ & = 27 \\cdot \\sin 48.8^\\circ\\\\ c & = \\frac{27 \\cdot \\sin 48.8^\\circ}{\\sin 95^\\circ} \\approx 20.4\\end{align*} ## Law of Cosines Law", "of length $\\,3,\\, 4,\\, 6\\,$. • Note that $\\,6 < 3 + 4\\,$. By SSS, the triangle is unique. • Use the Law of Cosines with the longest side: $$\\begin{gather} \\cssId{sb36}{6^2 = 3^2 + 4^2 - 2\\cdot 3\\cdot 4\\cdot\\cos F}\\cr\\cr \\cssId{sb37}{\\cos F \\approx -0.4583} \\cr\\cr \\cssId{sb38}{F = \\arccos(\\cos F) \\approx 117.3^\\circ} \\end{gather}$$ • Use the Law of Sines with the shortest side: $$\\begin{gather} \\cssId{sb40}{\\frac{\\sin 117.3^\\circ}{6} = \\frac{\\sin D}{3}}\\cr\\cr \\cssId{sb41}{\\sin D = \\frac{3\\sin 117.3^\\circ}{6} \\approx 0.4443}\\cr\\cr \\cssId{sb42}{D = \\arcsin(0.4443) \\approx 26.38^\\circ} \\end{gather}$$ • $\\,E = 180^\\circ - 117.3^\\circ - 26.38^\\circ = 36.32^\\circ$ • Summary: ANGLES OPPOSITE SIDES $D = 26.38^\\circ$ $d = 3$ $E = 36.32^\\circ$ $e = 4$ $F = 117.3^\\circ$ $f = 6$ This is the correct triangle! Note: At left, the corrections would be: $$\\begin{gather} \\cssId{sb56}{F = 180^\\circ - \\arcsin(0.8886) \\approx 117.3^\\circ}\\cr\\cr \\cssId{sb57}{E = 180^\\circ - 26.38^\\circ - 117.3^\\circ = 36.32^\\circ} \\end{gather}$$ ## Congruence Theorems CONGRUENCE THEOREMS A unique triangle is determined by each of these configurations: SAS side-angle-side (two sides and an included angle) SSS side-side-side (three sides) AAS/SAA angle-angle-side (or, side-angle-angle) (two angles and a non-included side) ASA angle-side-angle (two angles and an included side) If $$\\begin{gather} \\cssId{sb65}{e > 0}\\cr \\cssId{sb66}{f > 0}\\cr \\cssId{sb67}{0^\\circ < D < 180^\\circ} \\end{gather}$$ then there exists a unique triangle having the two sides and included angle. Name the", "\\pm 40^\\circ : n \\in \\mathbb Z \\} \\\\ \\end{align} Solution set: $$x \\in (\\{60^\\circ n \\pm 40^\\circ : n \\in \\mathbb Z \\} \\cup \\{180^\\circ n : n \\in \\mathbb Z\\}) \\cap [0^\\circ, 360^\\circ]$$ $$x \\in \\left\\{ \\begin{array}{rrrrr} 0^\\circ, & 20^\\circ, & 40^\\circ, & 80^\\circ, & 100^\\circ, \\\\ 140^\\circ, & 160^\\circ, & 180^\\circ, & 200^\\circ, & 220^\\circ, \\\\ 260^\\circ, & 280^\\circ, & 320^\\circ, & 340^\\circ, & 360^\\circ \\\\ \\end{array} \\right\\}$$ Using $\\sin3B=3\\sin B-4\\sin^3B$ $$32\\sin^3x\\cdot\\cos^2x\\cdot\\cos2x=\\sin x(3-4\\sin^2x)$$ If $\\sin x=0,x=180^\\circ n$ where $n$ is any integer Else using $\\cos2A=2\\cos^2A-1=1-2\\sin^2A,$ $$32\\cdot\\dfrac{1-\\cos2x}2\\cdot\\dfrac{1+\\cos2x}2\\cdot\\cos2x=3-2(1-\\cos2x)$$ $$\\iff8(c-c^3)=1+2c\\iff4c^3-3c=-\\dfrac12$$ $$\\implies\\cos6x=-\\dfrac12,6x=360^\\circ m\\pm120^\\circ\\iff x=60^\\circ m\\pm20^\\circ$$ where $m$ is any integer • @Théophile, One set of solution is $$0\\le180n\\le360\\iff0\\le n\\le2$$ Nov 2 '16 at 17:57", "\\ sin (15 ^ {\\ circ}) = \\ cos (75 ^ {\\ circ}) = {\\ frac {1} {4}} \\ cdot {\\ sqrt {2}} \\ cdot \\ left ({ \\ sqrt {3}} - 1 \\ right)}$ • ${\\ displaystyle \\ sin (18 ^ {\\ circ}) = \\ cos (72 ^ {\\ circ}) = {\\ frac {1} {4}} \\ cdot \\ left ({\\ sqrt {5}} - 1 \\ right)}$ • ${\\ displaystyle \\ sin (21 ^ {\\ circ}) = \\ cos (69 ^ {\\ circ}) = {\\ frac {1} {16}} \\ cdot \\ left (- {\\ sqrt {2}} \\ cdot \\ left ({\\ sqrt {3}} - 1 \\ right) \\ cdot \\ left ({\\ sqrt {5}} + 1 \\ right) +2 \\ cdot \\ left ({\\ sqrt {3}} + 1 \\ right ) \\ cdot {\\ sqrt {5 - {\\ sqrt {5}}}} \\ right)}$ • ${\\ displaystyle \\ sin (24 ^ {\\ circ}) = \\ cos (66 ^ {\\ circ}) = {\\ frac {1} {8}} \\ cdot \\ left (- {\\ sqrt {2}} \\ cdot {\\ sqrt {5 - {\\ sqrt {5}}}} + {\\ sqrt {3}} \\ cdot \\ left ({\\ sqrt {5}} + 1 \\ right) \\ right)}$ • ${\\ displaystyle \\ sin (27 ^ {\\ circ}) = \\ cos (63 ^ {\\ circ}) = {\\ frac {1} {8}}", "inverse tangent of that you get \\begin{align} x_1 &= \\tan^{-1} y_1 =\\tan^{-1} 1.0000 = 45^\\circ\\\\ x_2 &= \\tan^{-1} y_2 =\\tan^{-1} 0.2679= 15^\\circ \\end{align} • I think you missed multiplying 1+tanx on the R.H.S – NoLand'sMan May 27 at 8:24 • right .. ill edit – Ahmad Bazzi May 27 at 8:25 $$\\frac{\\sqrt{3} +1}{2\\sqrt{2}}=\\frac{\\sqrt{3}\\sqrt{2} +\\sqrt{2}}{4}=\\frac{\\sqrt{2}}{2}\\cdot\\frac{\\sqrt{3}}{2}+\\frac{\\sqrt{2}}{2}\\cdot\\frac{1}{2}=\\sin{45^\\circ}\\cos{30^\\circ}+\\cos{45^\\circ}\\sin{30^\\circ}=\\sin{(45^\\circ+30^\\circ)}=\\sin{75^\\circ}$$", "{2} - \\dfrac{2\\cdot 2\\sin(60^\\circ)} {2} \\right) \\\\\\\\ A &=& 2\\left( \\dfrac{10\\cdot 5\\sin(60^\\circ)} {2} - \\dfrac{2\\cdot 2\\sin(60^\\circ)} {2} \\right) \\\\\\\\ A &=& \\dfrac{2\\sin(60^\\circ)}{2} \\left( 10\\cdot 5 -2\\cdot 2 \\right) \\\\\\\\ A &=& \\dfrac{2\\sin(60^\\circ)}{2} \\cdot 46 \\\\\\\\ A &=& 2\\sin(60^\\circ) \\cdot 23 \\quad |\\quad \\sin(60^\\circ)=\\dfrac{\\sqrt{3} }{2} \\\\\\\\ A &=& 2\\cdot \\dfrac{\\sqrt{3} }{2} \\cdot 23 \\\\\\\\ \\mathbf{A} &=& \\mathbf{23\\sqrt{3}} \\\\ \\mathbf{A} &=& \\mathbf{39.8371685741~ u^2} \\\\ \\hline \\end{array}$$ The area of hexagon IJKLMN is $$\\mathbf{\\approx 40~u^2}$$ May 12, 2020", "## Precalculus (6th Edition) Blitzer Published by Pearson # Chapter 6 - Review Exercises - Page 798: 20 #### Answer The two roads measure $404\\text{ and }551\\text{ feet}$ respectively. #### Work Step by Step For the given triangle ABC, $A=55{}^\\circ,B=46{}^\\circ,a=460$ Using the angle sum property, compute the measure of angle C as follows: \\begin{align} & 55{}^\\circ +46{}^\\circ +C=180{}^\\circ \\\\ & C=180{}^\\circ -101{}^\\circ \\\\ & C=79{}^\\circ \\end{align} Now, using the law of sines we will obtain the value of both b and c: \\begin{align} & \\frac{a}{\\sin A}=\\frac{c}{\\sin C} \\\\ & \\frac{460}{\\sin 55{}^\\circ }=\\frac{c}{\\sin 79{}^\\circ } \\\\ & c\\cdot \\sin 55{}^\\circ =460\\cdot \\sin 79{}^\\circ \\\\ & c=\\frac{460\\cdot \\sin 79{}^\\circ }{\\sin 55{}^\\circ } \\end{align} This gives $c\\approx 551$ to the nearest tenth. Also, \\begin{align} & \\frac{a}{\\sin A}=\\frac{b}{\\sin B} \\\\ & \\frac{460}{\\sin 55{}^\\circ }=\\frac{b}{\\sin 46{}^\\circ } \\\\ & b\\cdot \\sin 55{}^\\circ =460\\cdot \\sin 46{}^\\circ \\\\ & b=\\frac{460\\cdot \\sin 46{}^\\circ }{\\sin 55{}^\\circ } \\end{align} This gives $b\\approx 404$ to the nearest tenth. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "= b \\sin A \\rightarrow \\frac{a}{b} = \\sin A \\rightarrow \\frac{22}{31} = \\sin A \\rightarrow A = 45.2^\\circ$ 3. $a > b \\sin A \\rightarrow \\frac{a}{b} > \\sin A \\rightarrow \\frac{22}{31} > \\sin A \\rightarrow A < 45.2^\\circ$ 4. This problem can be done entirely with right triangle trig, but there are several different ways to solve this particular problem. $BD & = \\sqrt{13.7^2 - 9.8^2} = 9.6 && \\tan 42.6^\\circ = \\frac{9.8}{DC} \\rightarrow DC = 10.7 \\\\\\sin 42.6^\\circ & = \\frac{9.8}{AC} \\rightarrow AC = 14.5 && BC = 9.6 + 10.7 = 20.3 \\\\\\sin B & = \\frac{9.8}{13.7} \\rightarrow \\angle{B} = 45.7^\\circ && \\angle{A} = 180^\\circ - 45.7^\\circ - 42.6^\\circ = 91.7^\\circ$ 1. $\\angle{EBD} \\Rightarrow 7.6^2 = 9.9^2 + 10.2^2 - 2 \\cdot 9.9 \\cdot 10.2 \\cos EBD \\Rightarrow 44.4^\\circ$ 2. $\\angle{BDE} \\Rightarrow \\frac{\\sin BDE}{9.9} = \\frac{\\sin 44.4^\\circ}{7.6} \\Rightarrow 65.7^\\circ$ 3. $\\angle{DEB} \\Rightarrow 180^\\circ - 65.7^\\circ - 44.4^\\circ \\Rightarrow 69.9^\\circ$ 4. $\\angle{BDC} \\Rightarrow 180^\\circ - 65.7^\\circ \\Rightarrow 114.3^\\circ$ 5. $\\angle{BEA} \\Rightarrow 180^\\circ - 69.9^\\circ \\Rightarrow 110.1^\\circ$ 6. $\\angle{DBC} \\Rightarrow 180^\\circ - 114.3^\\circ - 21.8^\\circ \\Rightarrow 43.9^\\circ$ 7. $\\angle{ABE} \\Rightarrow 109.6^\\circ - 43.9^\\circ - 44.4^\\circ \\Rightarrow 21.3^\\circ$ 8. $\\angle{BAE} \\Rightarrow 180^\\circ - 21.3^\\circ - 110.1^\\circ \\Rightarrow 48.6^\\circ$ 9. $BC \\Rightarrow \\frac{\\sin 114.3^\\circ}{BC} = \\frac{\\sin 21.8^\\circ}{10.2} \\Rightarrow 25.0$ 10. $AB \\Rightarrow \\frac{\\sin 110.1^\\circ}{AB} = \\frac{\\sin 48.6^\\circ}{9.9} \\Rightarrow 12.4$ 11.", "and since $7.55 < 10 < 12$ we have $h < b < a$ so we are looking for one solution. Hence, ${b}^{2} = {a}^{2} + {c}^{2} - 2 a c \\cos B$ ${12}^{2} = {a}^{2} + {10}^{2} - 2 \\left(10\\right) \\left(a\\right) \\cos {49}^{\\circ}$ $144 = {a}^{2} + 100 - \\left(20 \\cos {49}^{\\circ}\\right) a$ $0 = {a}^{2} - \\left(20 \\cos {49}^{\\circ}\\right) a - 44$ $a = \\frac{\\left(20 \\cos {49}^{\\circ}\\right) \\pm \\sqrt{{\\left(20 \\cos {49}^{\\circ}\\right)}^{2} - 4 \\left(1\\right) \\left(- 44\\right)}}{2}$ $a = \\frac{\\left(20 \\cos {49}^{\\circ}\\right) + \\sqrt{{\\left(20 \\cos {49}^{\\circ}\\right)}^{2} + 176}}{2}$ or $a = \\frac{\\left(20 \\cos {49}^{\\circ}\\right) - \\sqrt{{\\left(20 \\cos {49}^{\\circ}\\right)}^{2} + 176}}{2}$ a~~15.89 or cancel(a~~-2.77 $\\therefore a \\approx 15.89$ To find the measure of angle A we use the law of cosine and solve for A. That is, $A = {\\cos}^{-} 1 \\left[\\frac{{12}^{2} + {10}^{2} - {a}^{2}}{2 \\cdot 12 \\cdot 10}\\right] = {92}^{\\circ} 02 '$ and therefore $C = {180}^{\\circ} - {49}^{\\circ} - {92}^{\\circ} 02 ' = {38}^{\\circ} 58 '$", "\\cdot \\sin A \\left(\\sin A + 2 \\cos A\\right) \\end{align*} after angle addition identity. Similarly, $\\angle{EAC} = 90 - \\angle A,$ and by Law of Sines $CE = 8 \\sin{\\angle{EAC}} = 8 \\cos A.$ Note that $\\angle{E} = \\angle{ACB}$ from inscribed angles, so \\begin{align*} [ACE] &= \\dfrac{1}{2} AC \\cdot CE \\sin{\\angle C} \\\\ &= \\dfrac{1}{2} \\cdot \\dfrac{8 \\sqrt{5}}{5} \\cdot 8 \\cos A \\sin{\\left[180 - \\left(90 - \\angle A\\right) - \\angle E\\right]} \\\\ &= \\dfrac{32 \\sqrt{5}}{5} \\cdot \\cos A \\sin{\\left[\\left(90 - \\angle A\\right) + \\angle{ACB}\\right]} \\\\ &= \\dfrac{32 \\sqrt{5}}{5} \\cdot \\cos A \\left(\\dfrac{2 \\sqrt{5} \\cos A}{5} + \\dfrac{\\sqrt{5} \\sin A}{5}\\right) \\\\ &= \\dfrac{32}{5} \\cdot \\cos A \\left(\\sin A + 2 \\cos A\\right) \\end{align*} after angle addition identity. Setting the two areas equal, we get $$\\tan A = \\frac{\\sin A}{\\cos A} = 8 \\iff \\sin A = \\frac{8}{\\sqrt{65}}, \\cos A = \\frac{1}{\\sqrt{65}}$$ after Pythagorean Identity. Now plug back in and the common area is $\\frac{64}{65} \\iff \\boxed{129}.$ ## Solution 2 $[asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label(\"A\",A,NE);label(\"O_1\",O_1,NE);label(\"O_2\",O_2,NE);label(\"B\",B,SW);label(\"C\",C,SW);label(\"D\",D,NE);label(\"E\",E,NE);label(\"N\",N,W);label(\"K\",(-24/15,0.2));label(\"L\",(24/15,0.2));label(\"n\",(-0.8,-0.12));label(\"p\",((29/15,-48/15)));label(\"\\mathcal{P}\",(-1.6,1.1));label(\"\\mathcal{Q}\",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); [/asy]$ Call $O_1$ and $O_2$ the centers of circles $\\mathcal{P}$ and $\\mathcal{Q}$, respectively, and extend $CB$ and $O_2O_1$ to meet at point $N$. Call $K$ and $L$ the feet of the altitudes from $B$ to $O_1N$ and $C$ to $O_2N$, respectively. Using the fact that $\\triangle{O_1BN} \\sim", "then simplify the answer. \\begin{align*} \\tan{30^\\circ} - \\sin{60^\\circ} & = \\frac{1}{\\sqrt{3}} - \\frac{\\sqrt{3}}{2} \\\\ & = \\frac{2 - \\left(\\sqrt{3}\\right)\\left(\\sqrt{3}\\right)}{(2)\\left(\\sqrt{3}\\right)} \\\\ & = \\frac{2 - 3}{2\\sqrt{3}} \\\\ & = \\frac{-1}{2\\sqrt{3}} \\end{align*} $$\\sin{30^\\circ} - \\tan{45^\\circ} - \\sin{30^\\circ}$$ $\\begin{array}{c c c c c} -\\frac{\\sqrt{3}}{2} & -1 &-\\frac{1}{\\sqrt{3}} &-\\frac{\\sqrt{3}}{1} &-\\frac{7}{2\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\sin{30^\\circ} - \\tan{45^\\circ} - \\sin{30^\\circ} & = \\frac{1}{2} - \\frac{1}{1} - \\frac{1}{2} \\\\ & = \\frac{1 - 2 - 1}{2} \\\\ & = -1 \\end{align*} $$\\tan{30^\\circ} \\div \\tan{30^\\circ} \\div \\sin{45^\\circ}$$ $\\begin{array}{c c c c c} \\frac{\\sqrt{3}}{1} &\\frac{2\\sqrt{3}}{\\sqrt{2}} &\\frac{2}{\\sqrt{3}} &\\frac{\\sqrt{2}}{1} &\\frac{2\\sqrt{2}}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\tan{30^\\circ} \\div \\tan{30^\\circ} \\div \\sin{45^\\circ} & = \\frac{1}{\\sqrt{3}} \\div \\frac{1}{\\sqrt{3}} \\div \\frac{1}{\\sqrt{2}} \\\\ & = \\left(\\frac{1}{\\sqrt{3}} \\times \\frac{\\sqrt{3}}{1}\\right) \\div \\frac{1}{\\sqrt{2}} \\\\ & = 1 \\times \\frac{\\sqrt{2}}{1} \\\\ & = \\frac{\\sqrt{2}}{1} \\end{align*} $$\\sin{45^\\circ} \\div \\sin{30^\\circ} \\div \\cos{45^\\circ}$$ $\\begin{array}{c c c c c} \\frac{\\sqrt{2}}{\\sqrt{3}} &\\frac{1}{\\sqrt{2}} &\\frac{4}{\\sqrt{3}} &2 &\\frac{2\\sqrt{2}}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\sin{45^\\circ} \\div \\sin{30^\\circ} \\div \\cos{45^\\circ}", "(90^{\\circ} - \\alpha )$. And, of course, $\\sin (72^{\\circ})=\\sqrt{2(5+\\sqrt{5})}/4$ because $\\sin (72^{\\circ}) = \\cos (18^{\\circ})$. Finally, let's compute $\\sin 54^{\\circ}:\\;$ \\displaystyle\\begin{align} \\sin 54^{\\circ} &= \\sin 36^{\\circ}\\cos 18^{\\circ}+\\cos 36^{\\circ}\\sin 18^{\\circ}\\\\ &= \\displaystyle\\frac{\\sqrt{2(5-\\sqrt{5})}}{4}\\cdot \\displaystyle\\frac{\\sqrt{2(5+\\sqrt{5})}}{4} + \\displaystyle\\frac{1 + \\sqrt{5}}{4}\\cdot\\displaystyle\\frac{\\sqrt{5}-1}{4}\\\\ &=\\displaystyle\\frac{\\sqrt{5}+1}{4}, \\end{align} as expected since $\\sin 54^{\\circ}=\\cos 36^{\\circ}.$ [an error occurred while processing this directive] [an error occurred while processing this directive]", "+ \\ left ({\\ sqrt {5}} - 1 \\ right) \\ right)}$ • ${\\ displaystyle \\ sin (81 ^ {\\ circ}) = \\ cos (9 ^ {\\ circ}) = {\\ frac {1} {8}} \\ cdot \\ left ({\\ sqrt {2}} \\ cdot \\ left ({\\ sqrt {5}} + 1 \\ right) +2 \\ cdot {\\ sqrt {5 - {\\ sqrt {5}}}} \\ right)}$ • ${\\ displaystyle \\ sin (84 ^ {\\ circ}) = \\ cos (6 ^ {\\ circ}) = {\\ frac {1} {8}} \\ cdot \\ left ({\\ sqrt {2}} \\ cdot { \\ sqrt {5 - {\\ sqrt {5}}}} + {\\ sqrt {3}} \\ cdot \\ left ({\\ sqrt {5}} + 1 \\ right) \\ right)}$ • ${\\ displaystyle \\ sin (87 ^ {\\ circ}) = \\ cos (3 ^ {\\ circ}) = {\\ frac {1} {16}} \\ cdot \\ left ({\\ sqrt {2}} \\ cdot \\ left ({\\ sqrt {3}} - 1 \\ right) \\ cdot \\ left ({\\ sqrt {5}} - 1 \\ right) +2 \\ cdot \\ left ({\\ sqrt {3}} + 1 \\ right) \\ cdot {\\ sqrt {5 + {\\ sqrt {5}}}} \\ right)}$ • ${\\ displaystyle \\ sin (90 ^ {\\ circ}) = \\ cos (0 ^ {\\ circ}) = 1}$ The halving of the angle can be achieved by", "n\\theta\\right) \\end{align} ...which has no imaginary part! EDIT: Since you can gleen from the initial problem that: \\begin{align} \\frac{\\sqrt{3}}{2} + \\frac{i}{2} =& \\cos\\left(60^\\circ\\right) + i\\sin\\left(60^\\circ\\right) = e^{i60^\\circ} \\\\ \\frac{\\sqrt{3}}{2} - \\frac{i}{2} =& \\cos\\left(-60^\\circ\\right) + i\\sin\\left(-60^\\circ\\right) = e^{-i60^\\circ} \\end{align} We can conclude with $\\theta = 60^\\circ$ that the final result is: \\begin{align} 2\\cdot1^{107}\\cdot\\cos\\left(107\\cdot60\\right) =& 2\\cos\\left(17\\cdot360^\\circ + 300^\\circ\\right) \\\\ =& 2\\cdot\\cos\\left(300^\\circ\\right) \\\\ =& 2\\cdot\\cos\\left(-60^\\circ\\right) = 2\\cdot\\cos\\left(60^\\circ \\right) \\\\ =& 2 \\cdot\\frac{\\sqrt{3}}{2} = \\sqrt{3} \\end{align} or \\begin{align} 2\\cdot1^{107}\\cdot\\cos\\left(107\\cdot-60\\right) =& 2\\cos\\left(-18\\cdot360^\\circ +60^\\circ\\right) \\\\ =& 2\\cdot\\cos\\left(60^\\circ\\right) \\\\ =& 2 \\cdot\\frac{\\sqrt{3}}{2} = \\sqrt{3} \\end{align} EDIT: This post is false. Sorry. In general: $$f(a+bi)+f(a-bi)$$always has a zero imaginary part, no matter what $f$ is. Reason: The two terms are conjugates of each other.* Adding two conjugates to each other gives you a real number.** *This is true, because conjugation distributes over everything: $$\\overline{f(z)}=f(\\overline z)$$Basically, what conjugation does is it replaces $i$ with $-i$. If you think about it, there's no difference between the two! They both square to $-1$, and they both aren't real. ($-i$ isn't negative, because imaginary numbers don't have a concept of \"being less than zero\" - there is no concept of \"less than\" at all!) **$(a+bi)+(a-bi)=2a$ • I don't see how this is possible. Certainly it's true of the identity function, but I seriously doubt it's true of all functions. –", "75^{\\circ }={\\frac {\\sqrt {2}}{8}},}$ ${\\displaystyle \\cos 15^{\\circ }\\cdot \\cos 75^{\\circ }={\\frac {1}{4}}.}$ Similarly, ${\\displaystyle \\tan 50^{\\circ }\\cdot \\tan 60^{\\circ }\\cdot \\tan 70^{\\circ }=\\tan 80^{\\circ },}$ ${\\displaystyle \\tan 40^{\\circ }\\cdot \\tan 30^{\\circ }\\cdot \\tan 20^{\\circ }=\\tan 10^{\\circ }.}$ The following is perhaps not as readily generalized to an identity containing variables (but see explanation below): ${\\displaystyle \\cos 24^{\\circ }+\\cos 48^{\\circ }+\\cos 96^{\\circ }+\\cos 168^{\\circ }={\\frac {1}{2}}.}$ Degree measure ceases to be more felicitous than radian measure when we consider this identity with 21 in the denominators: {\\displaystyle {\\begin{aligned}&\\cos {\\frac {2\\pi }{21}}+\\cos \\left(2\\cdot {\\frac {2\\pi }{21}}\\right)+\\cos \\left(4\\cdot {\\frac {2\\pi }{21}}\\right)\\\\[10pt]&{}\\qquad {}+\\cos \\left(5\\cdot {\\frac {2\\pi }{21}}\\right)+\\cos \\left(8\\cdot {\\frac {2\\pi }{21}}\\right)+\\cos \\left(10\\cdot {\\frac {2\\pi }{21}}\\right)={\\frac {1}{2}}.\\end{aligned}}} The factors 1, 2, 4, 5, 8, 10 may start to make the pattern clear: they are those integers less than 21/2 that are relatively prime to (or have no prime factors in common with) 21. The last several examples are corollaries of a basic fact about the irreducible cyclotomic polynomials: the cosines are the real parts of the zeroes of those polynomials; the sum of the zeroes is the Möbius function evaluated at (in the very last case above) 21; only half of the zeroes are present above. The two identities preceding this last one arise in the same fashion with 21 replaced by 10", "$\\sin 90^\\circ= 1$, $c \\left( \\frac{1}{2} \\right) = (10) (1)$ $c = 20$ If the above problem asked to find the radius of the circumcircle of $\\vartriangle ABC$, the law of sines could help to find the diameter. $\\frac{a}{\\sin A} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C} = 2r$ $\\frac{b}{\\sin B} = 2r$ Substituting the values for $b, B$, $\\frac{10}{\\sin 30^\\circ} = 2r$ Since $\\sin 30^\\circ= \\frac{1}{2}$, $\\cfrac{10}{\\left( \\frac{1}{2} \\right)} = 2r$ $20 = 2r$ $10 = r$ # References All images were made by the page's author using Adobe Photoshop and Cinderella2." ]

[ "the identity. This is the opposite of sine of a sum, where a positive sign is used on the other side of the identity. ### Examples Example 1 Earlier, you were asked to evaluate $$\\sin 15^{\\circ}$$ and $$\\sin 75^{\\circ}$$ exactly without a calculator. To do this you need to use the sine of a difference and sine of a sum. $$\\begin{array}{l} \\sin \\left(45^{\\circ}-30^{\\circ}\\right)=\\sin 45^{\\circ} \\cos 30^{\\circ}-\\cos 45^{\\circ} \\sin 30^{\\circ}=\\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{3}}{2}-\\frac{\\sqrt{2}}{2} \\cdot \\frac{1}{2}=\\frac{\\sqrt{6}-\\sqrt{2}}{4} \\\\ \\sin \\left(45^{\\circ}+30^{\\circ}\\right)=\\sin 45^{\\circ} \\cos 30^{\\circ}+\\cos 45^{\\circ} \\sin 30^{\\circ}=\\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{3}}{2}+\\frac{\\sqrt{2}}{2} \\cdot \\frac{1}{2}=\\frac{\\sqrt{6}+\\sqrt{2}}{4} \\end{array}$$ Example 2 Find the exact value of $$\\tan 15^{\\circ}$$ without using a calculator. $$\\tan 15^{\\circ}=\\tan \\left(45^{\\circ}-30^{\\circ}\\right)=\\frac{\\tan 45^{\\circ}-\\tan 30^{\\circ}}{1+\\tan 45^{\\circ} \\tan 30^{\\circ}}=\\frac{1-\\frac{\\sqrt{3}}{3}}{1+1 \\cdot \\frac{\\sqrt{3}}{3}}=\\frac{3-\\sqrt{3}}{3+\\sqrt{3}}$$ A final solution will not have a radical in the denominator. In this case multiplying through by the conjugate of the denominator will eliminate the radical. $$=\\frac{(3-\\sqrt{3}) \\cdot(3-\\sqrt{3})}{(3+\\sqrt{3}) \\cdot(3-\\sqrt{3})}$$ $$=\\frac{(3-\\sqrt{3})^{2}}{9-3}$$ $$=\\frac{(3-\\sqrt{3})^{2}}{6}$$ Example 3 Evaluate the expression exactly without using a calculator. $$\\cos 50^{\\circ} \\cos 5^{\\circ}+\\sin 50^{\\circ} \\sin 5^{\\circ}$$ Once you know the general form of the sum and difference identities then you will recognize this as cosine of a difference. $$\\cos 50^{\\circ} \\cos 5^{\\circ}+\\sin 50^{\\circ} \\sin 5^{\\circ}=\\cos \\left(50^{\\circ}-5^{\\circ}\\right)=\\cos 45^{\\circ}=\\frac{\\sqrt{2}}{2}$$ Example 4 Use a sum or difference identity to find an exact value of $$\\cot \\left(\\frac{5 \\pi}{12}\\right)$$. Start with the definition of cotangent as the inverse of tangent. \\begin{aligned}", "\\pm 40^\\circ : n \\in \\mathbb Z \\} \\\\ \\end{align} Solution set: $$x \\in (\\{60^\\circ n \\pm 40^\\circ : n \\in \\mathbb Z \\} \\cup \\{180^\\circ n : n \\in \\mathbb Z\\}) \\cap [0^\\circ, 360^\\circ]$$ $$x \\in \\left\\{ \\begin{array}{rrrrr} 0^\\circ, & 20^\\circ, & 40^\\circ, & 80^\\circ, & 100^\\circ, \\\\ 140^\\circ, & 160^\\circ, & 180^\\circ, & 200^\\circ, & 220^\\circ, \\\\ 260^\\circ, & 280^\\circ, & 320^\\circ, & 340^\\circ, & 360^\\circ \\\\ \\end{array} \\right\\}$$ Using $\\sin3B=3\\sin B-4\\sin^3B$ $$32\\sin^3x\\cdot\\cos^2x\\cdot\\cos2x=\\sin x(3-4\\sin^2x)$$ If $\\sin x=0,x=180^\\circ n$ where $n$ is any integer Else using $\\cos2A=2\\cos^2A-1=1-2\\sin^2A,$ $$32\\cdot\\dfrac{1-\\cos2x}2\\cdot\\dfrac{1+\\cos2x}2\\cdot\\cos2x=3-2(1-\\cos2x)$$ $$\\iff8(c-c^3)=1+2c\\iff4c^3-3c=-\\dfrac12$$ $$\\implies\\cos6x=-\\dfrac12,6x=360^\\circ m\\pm120^\\circ\\iff x=60^\\circ m\\pm20^\\circ$$ where $m$ is any integer • @Théophile, One set of solution is $$0\\le180n\\le360\\iff0\\le n\\le2$$ Nov 2 '16 at 17:57", "the circumscribed circle. Since $\\alpha = 45^\\circ$, \\begin{align*} a & = 2R\\sin\\alpha\\\\ & = 2 \\cdot 2 \\cdot \\sin(45^\\circ)\\\\ & = 4 \\cdot \\frac{\\sqrt{2}}{2}\\\\ & = 2\\sqrt{2} \\end{align*} Since $\\gamma = 60^\\circ$, \\begin{align*} c & = 2R\\sin\\gamma\\\\ & = 2 \\cdot 2 \\cdot \\sin(60^\\circ)\\\\ & = 4 \\cdot \\frac{\\sqrt{3}}{2}\\\\ & = 2\\sqrt{3} \\end{align*} The area of the triangle is $$A = \\frac{1}{2}ac\\sin\\beta$$ since $c\\sin\\beta$ is the length of the altitude to side $\\overline{BC}$, which has length $a$. By the Angle Sum Theorem for Triangles, \\begin{align*} \\beta & = 180^\\circ - 60^\\circ - 45^\\circ\\\\ & = 75^\\circ \\end{align*} To find the exact value of $\\sin(75^\\circ)$, we use the Sum of Angles Formula $$\\sin(\\theta + \\varphi) = \\sin\\theta\\cos\\varphi + \\cos\\theta\\sin\\varphi$$ with $\\theta = 30^\\circ$ and $\\varphi = 45^\\circ$, which yields \\begin{align*} \\sin(75^\\circ) & = \\sin(30^\\circ)\\cos(45^\\circ) + \\cos(30^\\circ)\\sin(45^\\circ)\\\\ & = \\frac{1}{2} \\cdot \\frac{\\sqrt{2}}{2} + \\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{2}}{2}\\\\ & = \\frac{\\sqrt{2} + \\sqrt{6}}{4} \\end{align*} Hence, the area of triangle $ABC$ is \\begin{align*} A & = \\frac{1}{2}ac\\sin\\beta\\\\ & = \\frac{1}{2} \\cdot 2\\sqrt{2} \\cdot 2\\sqrt{3} \\cdot \\frac{\\sqrt{2} + \\sqrt{6}}{4}\\\\ & = \\frac{1}{2} \\cdot \\sqrt{6}(\\sqrt{2} + \\sqrt{6})\\\\ & = \\frac{1}{2}(\\sqrt{12} + 6)\\\\ & = \\frac{1}{2}(2\\sqrt{3} + 6)\\\\ & = \\sqrt{3} + 3 \\end{align*} square units. If $O$ is the circumcenter of $ABC$, we have: $$\\Delta = [OAB]+[OAC]+[OBC] = \\frac{R^2}{2}\\left(\\sin(2A)+\\sin(2B)+\\sin(2C)\\right).\\tag{1}$$ In our case, we have $A=45^\\circ,C=60^\\circ$,", "properties of the sine function, we can simplify this to $$R = 4 \\cdot \\sum_{k = 1}^{5} \\sin \\left( k\\cdot \\frac{360^\\circ}{12} \\right)$$ The five-element sum is just $\\sin 30^\\circ + \\sin 60^\\circ + \\sin 90^\\circ + \\sin 120^\\circ + \\sin 150^\\circ$. We know that $\\sin 30^\\circ = \\sin 150^\\circ = \\frac 12$, $\\sin 60^\\circ = \\sin 120^\\circ = \\frac{\\sqrt 3}2$, and $\\sin 90^\\circ = 1$. Hence our sum evaluates to: $$R = 4 \\cdot \\left( 2\\cdot \\frac 12 + 2\\cdot \\frac{\\sqrt 3}2 + 1 \\right) = 8 + 4\\sqrt 3$$ Therefore the answer is $8+4+3 = \\boxed{015}$. ## Solution 2 Note that $x^k + x^{k-1} + \\dots + x + 1 = \\frac{x^{k+1} - 1}{x - 1}$. Our sum can be reformed as $$\\frac{x(x^{47} - 1) + x^2(x^{45} - 1) + \\dots + x^{24}(x - 1)}{x-1}$$ So $$\\frac{x^{48} + x^{47} + x^{46} + \\dots + x^{25} - x^{24} - x^{23} - \\dots - x}{x-1} = 0$$ $x(x^{47} + x^{46} + \\dots - x - 1) = 0$ $x^{47} + x^{46} + \\dots - x - 1 = 0$ $x^{47} + x^{46} + \\dots + x + 1 = 2(x^{23} + x^{22} + \\dots + x + 1)$ $\\frac{x^{48} - 1}{x - 1} = 2\\frac{x^{24} - 1}{x - 1}$ $x^{48} - 1 - 2x^{24} + 2 = 0$ $(x^{24}", "then simplify the answer. \\begin{align*} \\tan{30^\\circ} - \\sin{60^\\circ} & = \\frac{1}{\\sqrt{3}} - \\frac{\\sqrt{3}}{2} \\\\ & = \\frac{2 - \\left(\\sqrt{3}\\right)\\left(\\sqrt{3}\\right)}{(2)\\left(\\sqrt{3}\\right)} \\\\ & = \\frac{2 - 3}{2\\sqrt{3}} \\\\ & = \\frac{-1}{2\\sqrt{3}} \\end{align*} $$\\sin{30^\\circ} - \\tan{45^\\circ} - \\sin{30^\\circ}$$ $\\begin{array}{c c c c c} -\\frac{\\sqrt{3}}{2} & -1 &-\\frac{1}{\\sqrt{3}} &-\\frac{\\sqrt{3}}{1} &-\\frac{7}{2\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\sin{30^\\circ} - \\tan{45^\\circ} - \\sin{30^\\circ} & = \\frac{1}{2} - \\frac{1}{1} - \\frac{1}{2} \\\\ & = \\frac{1 - 2 - 1}{2} \\\\ & = -1 \\end{align*} $$\\tan{30^\\circ} \\div \\tan{30^\\circ} \\div \\sin{45^\\circ}$$ $\\begin{array}{c c c c c} \\frac{\\sqrt{3}}{1} &\\frac{2\\sqrt{3}}{\\sqrt{2}} &\\frac{2}{\\sqrt{3}} &\\frac{\\sqrt{2}}{1} &\\frac{2\\sqrt{2}}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\tan{30^\\circ} \\div \\tan{30^\\circ} \\div \\sin{45^\\circ} & = \\frac{1}{\\sqrt{3}} \\div \\frac{1}{\\sqrt{3}} \\div \\frac{1}{\\sqrt{2}} \\\\ & = \\left(\\frac{1}{\\sqrt{3}} \\times \\frac{\\sqrt{3}}{1}\\right) \\div \\frac{1}{\\sqrt{2}} \\\\ & = 1 \\times \\frac{\\sqrt{2}}{1} \\\\ & = \\frac{\\sqrt{2}}{1} \\end{align*} $$\\sin{45^\\circ} \\div \\sin{30^\\circ} \\div \\cos{45^\\circ}$$ $\\begin{array}{c c c c c} \\frac{\\sqrt{2}}{\\sqrt{3}} &\\frac{1}{\\sqrt{2}} &\\frac{4}{\\sqrt{3}} &2 &\\frac{2\\sqrt{2}}{\\sqrt{3}} \\end{array}$ We need to use special angles to help us solve this problem. First write down each ratio using special angles and then simplify the answer. \\begin{align*} \\sin{45^\\circ} \\div \\sin{30^\\circ} \\div \\cos{45^\\circ}", "be hard so I didn't mention it. Constructing such $F$ and then deriving $AC||DF$ is quite easy, but when you construct $AC||DF,\\,F\\in AB$ then proving $\\angle FCB=30^\\circ$ requires non-obvious steps, you can try it. – Alexey Burdin Jun 10 '20 at 0:03 • I felt that the parallel line approach was easier.. Noting CDFA is an isosceles trapezoid gives triangles CDA and AFC congruent, and also triangle ADF is congruent to CFD. We then get DF = CF and CF=FB. Now DFB is isosceles with F=100, so we are done. – Isomorphism Jun 10 '20 at 5:08 Let $$AD\\cap BC=\\{E\\}$$. Thus, since $$\\frac{AE}{CE}\\cdot\\frac{CE}{DE}\\cdot\\frac{DE}{BE}\\cdot\\frac{BE}{AE}=1$$ we obtain: $$\\frac{\\sin50^{\\circ}}{\\sin20^{\\circ}}\\cdot\\frac{\\sin60^{\\circ}}{\\sin50^{\\circ}}\\cdot\\frac{\\sin(70^{\\circ}-x)}{\\sin{x}}\\cdot\\frac{\\sin80^{\\circ}}{\\sin30^{\\circ}}=1$$ or $$\\frac{\\sin(70^{\\circ}-x)}{\\sin{x}}=\\frac{\\sin20^{\\circ}\\sin30^{\\circ}}{\\sin60^{\\circ}\\sin80^{\\circ}}.$$ But it's obvious that $$x$$ is an acute angle and $$\\frac{\\sin(70^{\\circ}-x)}{\\sin{x}}$$ decreases. Id est, it's enough to prove that: $$\\frac{\\sin10^{\\circ}}{\\sin60^{\\circ}}=\\frac{\\sin20^{\\circ}\\sin30^{\\circ}}{\\sin60^{\\circ}\\sin80^{\\circ}}$$ or $$2\\sin10^{\\circ}\\sin80^{\\circ}=\\sin20^{\\circ},$$ which is obvious. • Thank you for a wonderful solution. Sorry that I could not produce a meaningful attempt. – Isomorphism Jun 8 '20 at 15:03 We draw first the red rhombus $$ABFE$$ with angles $$100^\\circ$$ and $$80^\\circ$$. Then we draw the lines $$g$$, $$h$$ through $$A$$ and $$F$$ which make an angle of $$20^\\circ$$ with two sides of the rhombus. Due to symmetry these lines intersect at a point $$D$$ on the symmetry axis $$BE$$ of the rhombus. All black noted angle sizes in the figure can", "may notice I used the $\\sin$/$\\cos$ addition formulas twice, but if you multiply your expression by $\\cos(10^\\circ)$ and \"couple\" the right terms in a slick way, you get the answer with less computations. However, my approach or Gelfand's intended one are more or less the same. – Jack D'Aurizio May 17 '16 at 0:36 • @SirJony: I think the formula has a different name depending on your location: Briggs, Werner or prostapheresis (en.wikipedia.org/wiki/Prosthaphaeresis). I just call it an instance of the addition formulas. – Jack D'Aurizio May 17 '16 at 0:38 • This is a nice solution but I don't think it's the one the problem-setter had in mind, since it doesn't make use of the hint given. – Jack Kinsella Jun 24 at 12:58 From the triple-angle formula $\\sin (3\\theta) = - 4\\sin^3\\theta + 3\\sin\\theta$ when $\\sin (3\\theta) = 1/2$, we get that $\\sin(10^\\circ)$, $\\sin(50^\\circ)$, $\\sin(-70^\\circ)$ are the roots of $8x^3-6x+1$. Therefore $$\\sin(10^\\circ) \\sin(50^\\circ) \\sin(70^\\circ) =-\\sin(10^\\circ) \\sin(50^\\circ) \\sin(-70^\\circ) =-(-\\frac18) =\\frac18.$$ • – lhf May 17 '16 at 0:59 We can prove $$4\\sin(60^\\circ-x)\\sin x\\sin(60^\\circ+x)=\\sin3x$$ Proof $\\#1:$ $$\\sin(60^\\circ-x)\\sin(60^\\circ+x)=\\sin^260^\\circ-\\sin^2x?$$ and $\\sin3x=3\\sin x-4\\sin^3x$ Proof$\\#2:$ If $\\sin3x=\\sin3A, 3x=180^\\circ n+(-1)^n3A$ where $n$ is any integer $\\implies x=60^\\circ n+(-1)^nA$ where $n\\equiv-1,0,1\\pmod3$ $\\implies x=-(60^\\circ+A), A,60^\\circ-A$ As $\\sin3x=3\\sin x-4\\sin^3x,3\\sin x-4\\sin^3x=\\sin3A\\iff4\\sin^3x-3\\sin x-\\sin3A=0$ Using Vieta's formula, $$\\prod_{n=-1}^1\\sin\\left(60^\\circ n+(-1)^nA\\right)=\\dfrac{\\sin3A}4$$ Use $\\sin\\left\\{-(60^\\circ+A)\\right\\}=-\\sin(60^\\circ+A)$ • I don't understand how this (and the", "75^{\\circ }={\\frac {\\sqrt {2}}{8}},}$ ${\\displaystyle \\cos 15^{\\circ }\\cdot \\cos 75^{\\circ }={\\frac {1}{4}}.}$ Similarly, ${\\displaystyle \\tan 50^{\\circ }\\cdot \\tan 60^{\\circ }\\cdot \\tan 70^{\\circ }=\\tan 80^{\\circ },}$ ${\\displaystyle \\tan 40^{\\circ }\\cdot \\tan 30^{\\circ }\\cdot \\tan 20^{\\circ }=\\tan 10^{\\circ }.}$ The following is perhaps not as readily generalized to an identity containing variables (but see explanation below): ${\\displaystyle \\cos 24^{\\circ }+\\cos 48^{\\circ }+\\cos 96^{\\circ }+\\cos 168^{\\circ }={\\frac {1}{2}}.}$ Degree measure ceases to be more felicitous than radian measure when we consider this identity with 21 in the denominators: {\\displaystyle {\\begin{aligned}&\\cos {\\frac {2\\pi }{21}}+\\cos \\left(2\\cdot {\\frac {2\\pi }{21}}\\right)+\\cos \\left(4\\cdot {\\frac {2\\pi }{21}}\\right)\\\\[10pt]&{}\\qquad {}+\\cos \\left(5\\cdot {\\frac {2\\pi }{21}}\\right)+\\cos \\left(8\\cdot {\\frac {2\\pi }{21}}\\right)+\\cos \\left(10\\cdot {\\frac {2\\pi }{21}}\\right)={\\frac {1}{2}}.\\end{aligned}}} The factors 1, 2, 4, 5, 8, 10 may start to make the pattern clear: they are those integers less than 21/2 that are relatively prime to (or have no prime factors in common with) 21. The last several examples are corollaries of a basic fact about the irreducible cyclotomic polynomials: the cosines are the real parts of the zeroes of those polynomials; the sum of the zeroes is the Möbius function evaluated at (in the very last case above) 21; only half of the zeroes are present above. The two identities preceding this last one arise in the same fashion with 21 replaced by 10", "& = 180^\\circ\\\\ m \\angle A & = 57^\\circ\\end{align*} Now, use the Law of Sines to set up ratios for \\begin{align*}a\\end{align*} and \\begin{align*}b\\end{align*}. \\begin{align*}\\frac{\\sin 57^\\circ}{a} = \\frac{\\sin 85^\\circ}{b} = \\frac{\\sin 38^\\circ}{12}\\end{align*} \\begin{align*}\\frac{\\sin 57^\\circ}{a} & = \\frac{\\sin 38^\\circ}{12} && \\quad \\frac{\\sin 85^\\circ}{b} = \\frac{\\sin 38^\\circ}{12}\\\\ a \\cdot \\sin 38^\\circ & = 12 \\cdot \\sin 57^\\circ && b \\cdot \\sin 38^\\circ = 12 \\cdot \\sin 85^\\circ\\\\ a & = \\frac{12 \\cdot \\sin 57^\\circ}{\\sin 38^\\circ} \\approx 16.4 && \\qquad \\quad \\ \\ b = \\frac{12 \\cdot \\sin 85^\\circ}{\\sin 38^\\circ} \\approx 19.4\\end{align*} Example 2: Solve the triangle using the Law of Sines. Round decimal answers to the nearest tenth. Solution: Set up the ratio for \\begin{align*}\\angle B\\end{align*} using Law of Sines. \\begin{align*}\\frac{\\sin 95^\\circ}{27} & = \\frac{\\sin B}{16}\\\\ 27 \\cdot \\sin B & = 16 \\cdot \\sin 95^\\circ\\\\ \\sin B & = \\frac{16 \\cdot \\sin 95^\\circ}{27} \\rightarrow \\sin^{-1} \\left ( \\frac{16 \\cdot \\sin 95^\\circ}{27} \\right ) = 36.2^\\circ\\end{align*} To find \\begin{align*}m \\angle C\\end{align*} use the Triangle Sum Theorem. \\begin{align*}m \\angle C + 95 ^\\circ + 36.2^\\circ = 180^\\circ \\rightarrow m \\angle C = 48.8^\\circ\\end{align*} To find \\begin{align*}c\\end{align*}, use the Law of Sines again. \\begin{align*}\\frac{\\sin 95^\\circ}{27} = \\frac{\\sin 48.8^\\circ}{c}\\end{align*} \\begin{align*}c \\cdot \\sin 95^\\circ & = 27 \\cdot \\sin 48.8^\\circ\\\\ c & = \\frac{27 \\cdot \\sin 48.8^\\circ}{\\sin 95^\\circ} \\approx 20.4\\end{align*} ## Law of Cosines Law", "How can I prove that $\\sin (10^\\circ), \\sin(1^\\circ), \\sin(2^\\circ), \\sin(3^\\circ), \\tan(10^\\circ)$ are irrational Prove that $\\sin (10^\\circ)$, $\\sin(1^\\circ)$, $\\sin(2^\\circ)$, $\\sin(3^\\circ)$, and $\\tan(10^\\circ)$ are irrational. My Attempt: Let $x = 10^\\circ$. Then \\begin{align} x &= 10^\\circ\\\\ 3x &= 30^\\circ\\\\ \\sin (3x) &= \\sin (30^\\circ)\\\\ 3\\sin (10^\\circ)-4\\sin^3(0^\\circ) &= \\frac{1}{2} \\end{align} Now let $y = \\sin (10^\\circ)$. Then \\begin{align} 3y-4y^3 &= \\frac{1}{2}\\\\ 6y-8y^3 &= 1\\\\ \\tag18y^3-6y+1 &= 0 \\end{align} How can I calculate the roots of $(1)$? - You don't need to find the roots, you just need to show that they aren't rational. – Git Gud May 19 '13 at 16:00 I'm assuming you mean, e.g. $\\sin(10^\\circ)$ and not $\\sin(10^0)$? – amWhy May 19 '13 at 16:01 Thanks Git Gud but how can i prove that they are not rational – juantheron May 19 '13 at 16:02 -", "How to change the appearence of the correct answer of $\\cos55^\\circ\\cdot\\cos65^\\circ\\cdot\\cos175^\\circ$ I represented the problem in the following view and solved it: \\begin{align}-\\sin35^\\circ\\cdot\\sin25^\\circ\\cdot\\sin85^\\circ\\cdot\\sin45^\\circ&=A\\cdot\\sin45^\\circ\\\\ -\\frac{1}{2}(\\cos20^\\circ-\\cos70^\\circ)\\cdot\\frac{1}{2}(\\cos50^\\circ-\\cos120^\\circ)&=A\\cdot\\sin45^\\circ\\\\ \\cos20^\\circ\\cdot\\cos50^\\circ-\\cos50^\\circ\\cdot\\cos70^\\circ+\\frac{\\cos20^\\circ}{2}-\\frac{\\cos70^\\circ}{2}&=A\\cdot(-4)\\cdot \\sin45^\\circ\\\\ \\frac{1}{2}(\\cos30^\\circ+\\cos70^\\circ)-\\frac{1}{2}(\\cos20^\\circ+\\cos120^\\circ)&=A\\cdot(-4)\\cdot \\sin45^\\circ\\\\ \\frac{\\sqrt{3}}{4}+\\frac{\\cos70^\\circ}{2}-\\frac{\\cos20^\\circ}{2}+\\frac{1}{4}+\\frac{\\cos20^\\circ}{2}-\\frac{\\cos70^\\circ}{2}&=-2\\sqrt{2}A\\\\ A&=-\\frac{\\sqrt{6}+\\sqrt{2}}{16} \\end{align} I believe that the above answer is true. But that didn't match a variant below: A) $$-\\frac{1}{8}$$ B)$$-\\frac{\\sqrt{3}}{8}$$ C) $$\\frac{\\sqrt{3}}{8}$$ D) $$-\\frac{1}{8}\\sqrt{2-\\sqrt{3}}$$ E) $$-\\frac{1}{8}\\sqrt{2+\\sqrt{3}}$$ I did the problem over again. After getting the same result, I thought that the apperance of my answer could be changed to match one above, so I tried to implement one of formulae involving radical numbers: all to no avail. How to change that? As $\\cos175^\\circ=\\cos(180^\\circ-5^\\circ)=-\\cos5^\\circ,$ $$4\\cos(60^\\circ-5^\\circ)\\cos5^\\circ\\cos(60^\\circ+5^\\circ)=\\cos(3\\cdot5^\\circ)$$ Now use $15=60-45$ or $=45-30$", "an angle in terms of simpler angles. Example $$\\PageIndex{1}$$ Add text here. For the automatic number to work, you need to add the \"AutoNum\" template (preferably at the end) to the page. 1 Find the exact value of $$\\cos (75{}^\\circ )$$. Since $$75{}^\\circ =30{}^\\circ +45{}^\\circ$$, we can evaluate $$\\cos (75{}^\\circ )$$ as $$\\cos (75{}^\\circ )=\\cos (30{}^\\circ +45{}^\\circ )$$ Apply the cosine sum of angles identity $$=\\cos (30{}^\\circ )\\cos (45{}^\\circ )-\\sin (30{}^\\circ )\\sin (45{}^\\circ )$$ Evaluate $$=\\frac{\\sqrt{3} }{2} \\cdot \\frac{\\sqrt{2} }{2} -\\frac{1}{2} \\cdot \\frac{\\sqrt{2} }{2}$$ Simply $=\\frac{\\sqrt{6} -\\sqrt{2} }{4}$ Try it Now 1. Find the exact value of $$\\sin \\left(\\frac{\\pi }{12} \\right)$$. Example $$\\PageIndex{1}$$ Add text here. For the automatic number to work, you need to add the \"AutoNum\" template (preferably at the end) to the page. 2 Rewrite $$\\sin \\left(x-\\frac{\\pi }{4} \\right)$$ in terms of sin(x) and cos(x). $$\\sin \\left(x-\\frac{\\pi }{4} \\right)$$ Use the difference of angles identity for sine = $$\\sin \\left(x\\right)\\cos \\left(\\frac{\\pi }{4} \\right)-\\cos \\left(x\\right)\\sin \\left(\\frac{\\pi }{4} \\right)$$ Evaluate the cosine and sine and rearrange $=\\frac{\\sqrt{2} }{2} \\sin \\left(x\\right)-\\frac{\\sqrt{2} }{2} \\cos \\left(x\\right)$ Additionally, these identities can be used to simplify expressions or prove new identities Example $$\\PageIndex{1}$$ Add text here. For the automatic number to work, you need to add the \"AutoNum\" template (preferably at the end) to the page. 3 Prove $$\\frac{\\sin (a+b)}{\\sin (a-b)} =\\frac{\\tan", "$\\sin 15^\\circ$ and $\\sin 75^\\circ$ exactly without a calculator, you need to use the sine of a difference and sine of a sum. $\\sin (45^\\circ-30^\\circ) &= \\sin 45^\\circ \\cos 30^\\circ-\\cos 45^\\circ \\sin 30^\\circ=\\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{3}}{2} - \\frac{\\sqrt{2}}{2} \\cdot \\frac{1}{2}=\\frac{\\sqrt{6}-\\sqrt{2}}{4}\\\\\\sin (45^\\circ+30^\\circ) &= \\sin 45^\\circ \\cos 30^\\circ + \\cos 45^\\circ \\sin 30^\\circ=\\frac{\\sqrt{2}}{2} \\cdot \\frac{\\sqrt{3}}{2}+\\frac{\\sqrt{2}}{2} \\cdot \\frac{1}{2}=\\frac{\\sqrt{6}+\\sqrt{2}}{4}$ #### Vocabulary The Greek letters used in this concept refer to unknown angles. They are $\\alpha$ - alpha , $\\beta$ - beta , $\\theta$ - theta , $\\gamma$ - gamma The symbol $\\pm$ is short hand for “ plus or minus .” The symbol $\\mp$ is shorthand for “ minus or plus .” The order is important because for cosine of a sum, the negative sign is used on the other side of the identity. This is the opposite of sine of a sum, where a positive sign is used on the other side of the identity. #### Guided Practice 1. Prove the sine of a difference identity. 2. Use a sum or difference identity to find an exact value of $\\cot \\left(\\frac{5 \\pi}{12}\\right)$ . 3. Prove the following identity: $\\frac{\\sin (x-y)}{\\sin (x+y)}=\\frac{\\tan x-\\tan y}{\\tan x+\\tan y}$ 1. Start with the cofunction identity and then distribute and work out the cosine of a sum and cofunction identities. $\\sin (\\alpha-\\beta) &= \\cos \\left(\\frac{\\pi}{2}-(\\alpha - \\beta)\\right)\\\\&=", "But, like others, I have a really hard time to digest your proof and I must look at the follow-up post by Opalg to fully decipher the reasoning behind your proof. Still, I am struggling to see why there must be true that the line $BGH$ is a tangent to the circle at the point $G$. I am sorry... I want to also thank to Opalg, for your willingness to answer to johng's posts, because I always think it's the OP's responsibility to answer to anyone who has found the proof confounding... My solution: Ceva's theorem tells us the product of the ratios of the pairs of segments formed on each side of the triangle by the intersection point is equal to 1. Hence we have $\\dfrac{\\sin 13^{\\circ}}{\\sin 13^{\\circ}} \\cdot \\dfrac{\\sin 73^{\\circ}}{\\sin 30^{\\circ}} \\cdot \\dfrac{\\sin x}{\\sin (51^{\\circ}-x)}=1$ $\\sin 73^{\\circ}\\cdot \\sin x=\\sin 30^{\\circ} \\cdot \\sin (51^{\\circ}-x)$ $\\sin 73^{\\circ}\\cdot \\sin x=\\sin 30^{\\circ} \\cdot (\\sin 51^{\\circ} \\cos x-\\cos 51^{\\circ} \\sin x)$ $\\sin x(\\sin 73^{\\circ}+\\sin 30^{\\circ}\\cos 51^{\\circ})=\\sin 30^{\\circ} \\sin 51^{\\circ} \\cos x$ \\begin{align*}\\tan x&=\\dfrac{\\sin 30^{\\circ} \\sin 51^{\\circ}}{\\sin 73^{\\circ}+\\sin 30^{\\circ}\\cos 51^{\\circ}}\\\\&=\\dfrac{\\dfrac{1}{2} \\sin 51^{\\circ}}{\\sin 73^{\\circ}+\\dfrac{1}{2}\\cos 51^{\\circ}}\\\\&=\\dfrac{\\sin 51^{\\circ}}{2\\sin 73^{\\circ}+\\cos 51^{\\circ}}\\\\&=\\dfrac{\\sin 3(17)^{\\circ}}{2\\sin (90-17)^{\\circ}+\\cos 3(17)^{\\circ}}\\\\&=\\dfrac{\\sin 17^{\\circ}(3-4\\sin^2 17^{\\circ})}{2\\cos 17^{\\circ}+\\cos 17^{\\circ}(4\\cos^2 17^{\\circ}-3)}\\\\&=\\dfrac{\\sin 17^{\\circ}(3-4\\sin^2 17^{\\circ})}{\\cos 17^{\\circ}(2+4\\cos^2 17^{\\circ}-3)}\\\\&=\\dfrac{\\sin 17^{\\circ}(3-4\\sin^2 17^{\\circ})}{\\cos 17^{\\circ}(4(1-\\sin^2 17^{\\circ})-1)}\\\\&=\\tan 17^{\\circ}\\end{align*} $\\therefore x=17^{\\circ}$ #### Opalg ##### MHB Oldtimer Staff member Albert, Opalg, someone please help. I still don't get", "u_{1}-u_{2}=2 \\\\ & a-a r=2 \\\\ & a(1-r)=2 &\\cdots (1)\\\\ & S=50\\\\ &\\frac{a}{1-r}=50 & \\cdots (2)\\\\ (1)\\times (2):& a^2=100\\\\& a=\\pm 10\\end{array}$$ If $a=-10, 1-r=\\frac{2}{-10}$. Thus $r=\\frac 65$, which is impossible because $|r| < 1.$ Thus $a=10$. ## Solution 4 Given: $O$ is the centre of the circle To find: $\\angle R Q T$ \\begin{aligned} &\\text { Solution: } \\beta=90^{\\circ}(\\because \\text { radius } \\perp \\text { tangent }) \\\\ &\\qquad \\begin{aligned} \\delta &=\\beta+34^{\\circ}=124^{\\circ} \\\\ \\alpha &=\\frac{\\delta}{2}=\\frac{124^{\\circ}}{2}=62^{\\circ} \\\\ \\therefore & \\angle R Q T=\\alpha=62^{\\circ} \\end{aligned} \\end{aligned} ## Solution 5 \\begin{aligned} \\sin 2 x(1+2 \\cos x) &=\\sin 2 x+2 \\sin 2 x \\cos x\\\\ &=\\sin 2x+\\left(\\sin(2x+x)+\\sin(2x-x)\\right)\\\\ &=\\sin 2x+\\sin 3x+\\sin x\\\\ &=\\sin x+\\sin 2 x+\\sin 3 x \\end{aligned} ## Solution 6 \\begin{aligned} &\\lim _{x \\rightarrow 2} \\frac{x^{3}-8}{\\sqrt{x+2}-2}\\\\ &=\\lim _{x \\rightarrow 2} \\frac{(x-2)\\left(x^{2}+2 x+4\\right)}{\\sqrt{x+2}-2} \\times \\frac{\\sqrt{x+2}+2}{\\sqrt{x+2}+2}\\\\ &=\\lim _{x \\rightarrow 2} \\frac{(x-2)\\left(x^{2}+2 x+4\\right)(\\sqrt{x+2}+2)}{x+2-4}\\\\ &=\\lim _{x \\rightarrow 2} \\frac{(x-2)\\left(x^{2}+2 x+4\\right)(\\sqrt{x+2}+2)}{x-2}\\\\ &=\\lim _{x \\rightarrow 2}\\left[\\left(x^{2}+2 x+4\\right)(\\sqrt{x+2}+2)\\right]\\\\ &=\\left(2^{2}+2(2)+4\\right)(\\sqrt{2+2}+2)\\\\ &=12 \\times 4\\\\ &=48 \\end{aligned} \\begin{aligned} &\\lim _{x \\rightarrow 0} \\frac{\\cos x-1}{\\sin ^{2} x} \\\\ =& \\lim _{x \\rightarrow 0} \\frac{\\cos x-1}{1-\\cos ^{2} x} \\\\ =& \\lim _{x \\rightarrow 0} \\frac{\\cos x-1}{(1+\\cos x)(1-\\cos x)} \\\\ =& \\lim _{x \\rightarrow 0} \\frac{-(1-\\cos x)}{(1+\\cos x)(1-\\cos x)} \\\\ =& \\lim _{x \\rightarrow 0} \\frac{-1}{1+\\cos x} \\\\ =& \\frac{-1}{1+\\cos 0} \\\\ =& \\frac{-1}{1+1} \\\\ =&-\\frac{1}{2} \\end{aligned} ## Solution 7(a) \\begin{aligned} f(x) &=\\frac{x}{x-3},", "The number of ordered pairs Question: The number of ordered pairs $(r, k)$ for which 6. ${ }^{35} C_{r}=\\left(k^{2}-3\\right) \\cdot{ }^{36} C_{r+1}$, where $k$ is an integer, is: 1. (1) 3 2. (2) 2 3. (3) 6 4. (4) 4 Correct Option: , 4 Solution: $\\frac{36}{r+1} \\times{ }^{35} \\mathrm{C}_{r}\\left(k^{2}-3\\right)={ }^{35} \\mathrm{C}_{r} \\cdot 6$ $\\Rightarrow \\quad k^{2}-3=\\frac{r+1}{6}$ $\\Rightarrow \\quad k^{2}=3+\\frac{r+1}{6}$ $r$ can be 5,35 for $k \\in I$ $r=5, k=\\pm 2$ $r=35, k=\\pm 3$ Hence, number of ordered pairs $=4$.", "make sure I got it: Prove of $$\\frac{2\\cdot\\sin2^{\\circ}+4\\cdot\\sin4^{\\circ}+\\cdots+178\\cdot\\sin178^{\\circ}+180\\cdot\\sin180^{\\circ}}{90} = \\cot 1^{\\circ}$$: L.H.S.= $$\\frac{2\\cdot\\sin2^{\\circ}+4\\cdot\\sin4^{\\circ}+\\cdots+178\\cdot\\sin178^{\\circ}+180\\cdot\\sin180^{\\circ}}{90} = \\cot 1^{\\circ}$$ We can use the identity $$\\sin x = \\sin (180^{\\circ} - x)$$ to get: L.H.S.= $$\\frac{2\\cdot\\sin2^{\\circ}+178\\cdot\\sin178^{\\circ}+180\\cdot\\sin180^{\\circ}+4\\cdot\\sin4^{\\circ}+\\cdots}{90} = \\cot 1^{\\circ}$$ We can simplify to get: L.H.S.= $$2\\cdot(\\sin 2^{\\circ}+\\sin 4^{\\circ}+\\cdots+\\sin 88^{\\circ})+\\sin 90^{\\circ}$$ We can multiply by $$\\frac{\\sin1^{\\circ}}{\\sin1^{\\circ}}$$ to get: L.H.S.= $$\\frac{2\\cdot(\\sin 2^{\\circ}\\cdot\\sin 1^{\\circ}+\\sin 4^{\\circ}\\cdot\\sin 1^{\\circ}+\\cdots+\\sin 88^{\\circ}\\cdot\\sin 1^{\\circ})+\\sin 90^{\\circ}\\cdot\\sin 1^{\\circ}}{\\sin1^{\\circ}}$$ Then, we can use the sum-to-product (or Prosthaphaeresis :)) identities to get: L.H.S.= $$\\frac{\\cos1^{\\circ}-\\cos3^{\\circ}+\\cos3^{\\circ}-\\cos5^{\\circ}+\\cdots+\\cos87^{\\circ}-\\cos89^{\\circ}+\\sin90^{\\circ}\\cdot\\sin1^{\\circ}}{\\sin1^{\\circ}}$$ We can cancel the terms to get: L.H.S.= $$\\frac{\\cos1^{\\circ}-\\cos89^{\\circ}+\\frac{\\cos89^{\\circ}-\\cos91^{\\circ}}{2}}{\\sin1^{\\circ}}$$ We can simplify to get: L.H.S.= $$\\frac{\\cos1^{\\circ}-\\frac{\\cos89^{\\circ}+\\cos91^{\\circ}}{2}}{\\sin1^{\\circ}}$$ We can use the identity $$\\cos{x}=\\cos{(180-x)}$$ to get: L.H.S.= $$\\frac{\\cos1^{\\circ}-\\frac{\\cos89^{\\circ}-\\cos89^{\\circ}}{2}}{\\sin1^{\\circ}}$$ L.H.S.= $$\\frac{\\cos1^{\\circ}}{\\sin1^{\\circ}}=\\cot1^{\\circ}$$ R.H.S.= $$\\cot1^{\\circ}$$ $$L.H.S.=R.H.S.$$ If $$\\frac{2\\cdot\\sin2^{\\circ}+4\\cdot\\sin4^{\\circ}+\\cdots+178\\cdot\\sin178^{\\circ}+180\\cdot\\sin180^{\\circ}}{90} = \\cot 1^{\\circ}$$, that means the average of $$n \\sin n^\\circ$$($$n = 2, 4, 6, \\ldots, 180$$) is $$\\cot 1^\\circ$$. Therefore, the average of $$n \\sin n^\\circ$$($$n = 2, 4, 6, \\ldots, 180$$)is $$\\cot 1^\\circ$$. Jul 17, 2019", "- \\sin x = \\cos x - \\cos(90^\\circ - x)$$ And with the identity: $$\\cos p - \\cos q = -2\\sin\\frac{p+q}2\\sin\\frac{p-q}2$$ It follows that: $$\\cos x - \\sin x = \\cos x - \\cos(90^\\circ - x) = -2 \\sin\\frac{x+(90^\\circ - x)}2\\sin\\frac{x-(90^\\circ - x)}2 = -2 \\cdot \\frac 12\\sqrt{2} \\cdot \\sin(x-45^\\circ)$$ If $\\cos y < 0$, then $90^\\circ < y < 270^\\circ$ plus a possible multiple of $360^\\circ$. Starting from: $$\\cos(x+45^\\circ)<0$$ we find: \\begin{array}{lcccl}90^\\circ &<& x + 45^\\circ &<& 270^\\circ \\\\ 45^\\circ &<& x &<& 225^\\circ \\end{array} All plus a possible multiple of $360^\\circ$. Well of course, how wonderful! Thank you so much I like Serena! $$\\displaystyle \\cos(x)=\\cos\\left(0^{\\circ}-(-x) \\right)=\\cos\\left(0^{\\circ} \\right)\\cos(-x)+\\sin\\left(0^{\\circ} \\right)\\sin(-x)=\\cos(-x)$$ $$\\displaystyle \\sin(x)=\\cos\\left(90^{\\circ}-x \\right)=\\cos\\left(360^{\\circ}-\\left(90^{\\circ}-x \\right) \\right)=\\cos\\left(270^{\\circ}+x \\right)$$ Oh ok, thank you! I understand now. However, how would you solve the previously mentioned inequality? #### MarkFL Staff member ... Oh ok, thank you! I understand now. However, how would you solve the previously mentioned inequality? How did you solve the first version? #### Prove It ##### Well-known member MHB Math Helper Forum, do you have any idea how to solve the trigonometric inequality $$\\displaystyle \\cos (x) < \\sin (x)$$ strictly algebraically? The conventional(?) approach is to first solve $$\\displaystyle \\cos(x) = \\sin(x)$$ and then draw the graphs for each function in order to find the correct interval. However, I would", "# Tag Info 7 $\\sin(x)$ is an odd function. $\\sin(-x) = -\\sin(x)$ for all $x$. It is $\\cos$ that is even. 6 Your form, $\\sqrt{\\frac{2-\\sqrt3}{4}}$, is already close: \\begin{align*} \\sin15^\\circ &=\\sqrt{\\frac{2-\\sqrt3}{4}}\\\\ &= \\sqrt{\\frac{4-2\\sqrt3}{8}}\\\\ &= \\sqrt{\\frac{1-2\\sqrt3+3}{8}}\\\\ &= \\sqrt{\\frac{(1-\\sqrt3)^2}{8}}\\\\ &= \\frac{\\sqrt3 -1 }{2\\sqrt2} \\end{align*} 6 By definition, $\\sin tA=\\frac1{2i}(e^{itA}-e^{-itA})$. So you need only to show that $$t\\mapsto e^{itA},\\quad t\\in\\Bbb R$$ is continuous. (For the $e^{-itA}$ part, just note that it is the inverse of $e^{itA}$ and that taking inverse is a continuous map in the invertible matrix space. [Or rather, we don't even need to bother with taking inverse, thanks to ... 5 Let me show the first identity. By the double-angle identity: \\begin{align} & \\sin^2(10^\\circ) - \\sin^2(20^\\circ) - \\sin^2(40^\\circ) \\\\ = & \\frac{1 - \\cos(20^\\circ)}{2} - \\frac{1 - \\cos(40^\\circ)}{2} - \\frac{1 - \\cos(80^\\circ)}{2} \\\\ = & -\\frac{1}{2} - \\frac{1}{2}(\\cos(20^\\circ) - \\cos(40^\\circ) - \\cos(80^\\circ)).\\\\ \\end{align} So to show the ... 5 Looking at derivatives of the functions $\\sin(\\alpha t) e^{-\\lambda t}$ and $\\cos(\\alpha t) e^{-\\lambda t}$, it's reasonable to try for an antiderivative of the form $$F(t) = (A \\sin(\\alpha t) + B \\cos(\\alpha t)) e^{-\\lambda t}$$ where $A$ and $B$ are constants. Differentiating this, $$F'(t) = ((-A \\lambda - B \\alpha) \\sin(\\alpha t) + (A \\alpha - B ... 5 Recall that$$\\cos 3x=4\\cos^3 x- 3\\cos x\\implies 2(4\\cos^3 12^{\\circ}-3\\cos 12^{\\circ})=2\\cos36^\\circ$$Using$$\\cos", "(\\beta )+\\sin (\\alpha )(-\\sin (\\beta ))} \\\\ {\\cos (\\alpha )\\cos (\\beta )-\\sin (\\alpha )\\sin (\\beta )} \\end{array}$$ The sum and difference of angles identities are often used to rewrite expressions in other forms, or to rewrite an angle in terms of simpler angles. EXAMPLE 1 Find the exact value of $$\\cos (75{}^\\circ )$$. Solution Since $$75{}^\\circ =30{}^\\circ +45{}^\\circ$$, we can evaluate $$\\cos (75{}^\\circ )$$ as $$\\cos (75{}^\\circ )=\\cos (30{}^\\circ +45{}^\\circ )$$ Apply the cosine sum of angles identity $$=\\cos (30{}^\\circ )\\cos (45{}^\\circ )-\\sin (30{}^\\circ )\\sin (45{}^\\circ )$$ Evaluate $$=\\frac{\\sqrt{3} }{2} \\cdot \\frac{\\sqrt{2} }{2} -\\frac{1}{2} \\cdot \\frac{\\sqrt{2} }{2}$$ Simply $$=\\frac{\\sqrt{6} -\\sqrt{2} }{4}$$ Exercise Find the exact value of $$\\sin \\left(\\frac{\\pi }{12} \\right)$$. $$\\sin \\left(\\frac{\\pi }{12} \\right)=\\sin \\left(\\frac{\\pi }{3} -\\frac{\\pi }{4} \\right)=\\sin \\left(\\frac{\\pi }{3} \\right)\\cos \\left(\\frac{\\pi }{4} \\right)-\\cos \\left(\\frac{\\pi }{3} \\right)\\sin \\left(\\frac{\\pi }{4} \\right)$$ $$=\\frac{\\sqrt{3} }{2} \\frac{\\sqrt{2} }{2} -\\frac{1}{2} \\frac{\\sqrt{2} }{2}$$ $$\\frac{\\sqrt{6} -\\sqrt{2} }{4}$$ EXAMPLE 2 Rewrite $$\\sin \\left(x-\\frac{\\pi }{4} \\right)$$ in terms of sin($$x$$) and cos($$x$$). Solution $$\\sin \\left(x-\\frac{\\pi }{4} \\right)$$ Use the difference of angles identity for sine = $$\\sin \\left(x\\right)\\cos \\left(\\frac{\\pi }{4} \\right)-\\cos \\left(x\\right)\\sin \\left(\\frac{\\pi }{4} \\right)$$ Evaluate the cosine and sine and rearrange $$=\\frac{\\sqrt{2} }{2} \\sin \\left(x\\right)-\\frac{\\sqrt{2} }{2} \\cos \\left(x\\right)$$ Additionally, these identities can be used to simplify expressions or prove new identities EXAMPLE 3 Prove $$\\frac{\\sin (a+b)}{\\sin (a-b)} =\\frac{\\tan (a)+\\tan (b)}{\\tan (a)-\\tan (b)}$$. Solution As" ]

[ "Comment Share Q) # Using properties of determinants, evaluate $A=\\begin{vmatrix}43 & 1 & 6\\\\35 & 7 & 4\\\\17 & 3 &2\\end{vmatrix}$ $\\begin{array}{1 1} 7 \\\\ 0 \\\\ -7 \\\\ 1 \\end{array}$", "& c\\end{vmatrix}$ Ex. 6.2 | Q 1.4 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}1 & - 3 & 2 \\\\ 4 & - 1 & 2 \\\\ 3 & 5 & 2\\end{vmatrix}$ Ex. 6.2 | Q 1.5 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}1 & 4 & 9 \\\\ 4 & 9 & 16 \\\\ 9 & 16 & 25\\end{vmatrix}$ Ex. 6.2 | Q 1.6 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}6 & - 3 & 2 \\\\ 2 & - 1 & 2 \\\\ - 10 & 5 & 2\\end{vmatrix}$ Ex. 6.2 | Q 1.7 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}1 & 3 & 9 & 27 \\\\ 3 & 9 & 27 & 1 \\\\ 9 & 27 & 1 & 3 \\\\ 27 & 1 & 3 & 9\\end{vmatrix}$ Ex. 6.2 | Q 2.01 | Page 57 Without expanding, show that the value of the following determinant is zero: $\\begin{vmatrix}8 & 2 & 7 \\\\ 12 & 3 & 5 \\\\ 16 & 4 & 3\\end{vmatrix}$ Ex. 6.2 | Q 2.02 | Page 57 Without expanding, show that the value of the following determinant is zero: $\\begin{vmatrix}6 & - 3 & 2 \\\\ 2 & - 1 & 2 \\\\ - 10 & 5 & 2\\end{vmatrix}$ Ex. 6.2 | Q", "# Without expanding evaluate the determinent $\\begin{vmatrix} (a^x+a^{-x})^2 & (a^x-a^{-x})^2 & 1 \\\\ (a^y+a^{-y})^2 & (a^y-a^{-y})^2 & 1 \\\\ (a^z+a^{-z})^2 & (a^z-a^{-z})^2 & 1 \\end{vmatrix}$ where $a>0$ ,and $x,y,z \\in R$ $\\begin{array}{1 1} 0 \\\\ 1 \\\\ -1 \\\\ \\pm 1 \\end{array}$", "& y_2 & z_2 \\\\ x_3 & y_3 & z_3 \\end{vmatrix}$ References", "x_3 & y_3 & z_3 \\end{vmatrix}$$ give the second and the third component of the question vector, equal to $$0$$.", "1 \\\\ z & x\\end{vmatrix} \\right\\} \\left[ \\text{ Expanding along } C_1 \\right]$ $= \\left( x + y + z \\right) \\left( x - z \\right)^2$ Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 12 Maths Chapter 6 Determinants Exercise 6.2 | Q 43 | Page 61", "y_2 & 1 \\\\ x_3 & y_3 & 1 \\\\ \\end{vmatrix}}~ \\right)$$", "\\\\ x_2 & y_2 & z_2 & 1 \\\\ x_3 & y_3 & z_3 & 1 \\\\ \\end{vmatrix}$$", "Browse Questions # Using properties of determinants, solve the following for x : $\\begin{vmatrix} x+a & x & x \\\\ x & x+a & x \\\\ x & x & x+a \\end{vmatrix} = 0$ This question has appeared in textbook.", "the way, the expression above evaluates to 1.", "\\\\ 3 - x & 8\\end{bmatrix}$ Q 12.2 | Page 11 For what value of x the matrix A is singular? $A = \\begin{bmatrix}x - 1 & 1 & 1 \\\\ 1 & x - 1 & 1 \\\\ 1 & 1 & x - 1\\end{bmatrix}$ #### Pages 57 - 62 Q 1.1 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}1 & 3 & 5 \\\\ 2 & 6 & 10 \\\\ 31 & 11 & 38\\end{vmatrix}$ Q 1.2 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}67 & 19 & 21 \\\\ 39 & 13 & 14 \\\\ 81 & 24 & 26\\end{vmatrix}$ Q 1.3 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}a & h & g \\\\ h & b & f \\\\ g & f & c\\end{vmatrix}$ Q 1.4 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}1 & - 3 & 2 \\\\ 4 & - 1 & 2 \\\\ 3 & 5 & 2\\end{vmatrix}$ Q 1.5 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}1 & 4 & 9 \\\\ 4 & 9 & 16 \\\\ 9 & 16 & 25\\end{vmatrix}$ Q 1.6 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}6 & - 3 & 2 \\\\ 2 & - 1 & 2 \\\\ - 10 & 5 & 2\\end{vmatrix}$ Q 1.7 | Page 57 Evaluate", "\\end{vmatrix}\\:\\:\\:$ is $2$ $2e$ $2 \\pi$ $2i$", "# Evaluate :$\\begin{vmatrix}y+z&z&y\\\\z&z+x&x\\\\y&x&x+y\\end{vmatrix}$ $\\begin{array}{1 1} 2xy \\\\ 4xy \\\\ 4xyz \\\\ None\\;of\\;these \\end{array}$ Apply $C_1\\rightarrow C_1-(C_2+C_3)$ $\\begin{vmatrix}(y+z)-(z+y)&z&y\\\\z-(z+2x)&z+x&x\\\\y-(2x+y)&x&x+y\\end{vmatrix}$ $\\begin{vmatrix}0 &z &y\\\\-2x&z+x&x\\\\-2x&x&x+y\\end{vmatrix}$ $-2x\\begin{vmatrix}0 &z&y\\\\1 &z+x&x\\\\1&x&x+y\\end{vmatrix}$ Apply $R_2\\rightarrow R_2-R_3$ $\\Rightarrow -2x\\begin{vmatrix}0 &z&y\\\\0&z&-y\\\\1 &x&x+y\\end{vmatrix}$ $\\Rightarrow -2x(1)(-yz-yz)$ $\\Rightarrow -2x(-2yz)$ $\\Rightarrow 4xyz$ Hence (c) is the correct answer.", "& 17\\\\22 & 3 & 19\\\\26 & 5 & 21\\end{vmatrix}$$ 23. Evaluate $$\\begin{vmatrix}4 & 9 & 7\\\\3 & 5 & 7\\\\5 & 4 & 5\\end{vmatrix}$$ 24. Evaluate $$\\begin{vmatrix}1^2 & 2^2 & 3^2\\\\2^2 & 3^2 & 4^2\\\\3^2 & 4^2 & 5^2\\end{vmatrix}$$ 25. Evaluate $$\\begin{vmatrix}1 & 1 & 1\\\\1 & 1 + x & 1\\\\1 & 1 & 1 + y\\end{vmatrix}$$ 26. Evaluate $$\\begin{vmatrix}1 & 1 & 1\\\\a & b & c\\\\a^3 & b^3 & c^3\\end{vmatrix}$$ 27. Evaluate $$\\begin{vmatrix}1 & b + c & b^2 + c^2\\\\1 & c + a & c^2 + a^2\\\\ 1 & a + b & a^2 + b^2\\end{vmatrix}$$ 28. Evaluate $$\\begin{vmatrix}1 & a & a^2 - bc\\\\1 & b & b^2 - ca \\\\ 1 & c & c^2 - ab\\end{vmatrix}$$ 29. Evaluate $$\\begin{vmatrix}1 & bc & bc(b + c)\\\\1 & ca & ca(c + a)\\\\1 & ab & ab(a + b)\\end{vmatrix}$$ 30. Prove that $$\\begin{vmatrix}1 & a & b + c\\\\1 & b & c + a\\\\1 & c & a + b\\end{vmatrix} = 0$$ 31. If $$a, b, c$$ be the $$p\\text{th}, q\\text{th}, r\\text{th}$$ terms respectively of an H.P., show that $$\\begin{vmatrix}bc & p & 1\\\\ ca & q & 1\\\\ ab & r & 1\\end{vmatrix} = 0$$ 32. If $$\\begin{vmatrix}x^2 + 3x & x - 1 & x + 3\\\\x +1", "# The least value of the product $xyz$ for which the determinant $\\begin{vmatrix} x& 1& 1 \\\\ 1 & y & 1 \\\\ 1 & 1 & z \\end{vmatrix}$ is non negative, is: ( A ) $-8$ ( B ) $-16 \\sqrt{2}$ ( C ) $-1$ ( D ) $-2\\sqrt{2}$", "Browse Questions # Find the value of determinant $\\begin{vmatrix}1&x&y+z\\\\1&y&z+x\\\\1&z&x+y\\end{vmatrix}$ $(a)\\;0\\qquad(b)\\;x+y+z\\qquad(c)\\;1+x+y+z\\qquad(d)\\;(x-y)(y-z)(z-x)$ Can you answer this question? $\\begin{vmatrix}1&x&y+z\\\\1&y&z+x\\\\1&z&x+y\\end{vmatrix}$ Apply $C_3\\rightarrow C_3+C_2$ $\\begin{vmatrix}1&x&y+z+x\\\\1&y&z+x+y\\\\1&z&x+y+z\\end{vmatrix}$ Taking out (x+y+z) we have Since $C_1 and \\;C_3$ are identical. $\\Rightarrow \\Delta=0$ Hence (a) is the correct answer. answered Nov 26, 2013 edited Mar 20, 2014", "Browse Questions # Evaluate the determinants: $\\begin{vmatrix} 3&-4&5 \\\\ 1&1&-2 \\\\2&3&1 \\end{vmatrix}$ Note: This is part 2 of a 4 part question, split as 4 separate questions here. Toolbox: • To evaluate a matrix of order $3\\times 3$ • $\\mid A\\mid=\\begin{vmatrix}a_{11} & a_{12} & a_{13}\\\\a_{21} & a_{22} & a_{23}\\\\a_{31} & a_{32} & a_{33}\\end{vmatrix}$ • Therefore $\\mid A\\mid=a_{11}(a_{22}\\times a_{33}-a_{23}\\times a_{32})-a_{12}(a_{21}\\times a_{32}-a_{23}\\times a_{31})+a_{13}(a_{21}\\times a_{32}-a_{22}\\times a_{31})$ Given: $Evaluate:\\begin{vmatrix}3 & -4 & 5\\\\1& 1 &-2\\\\2 & 3 & 1\\end{vmatrix}$ We know to evaluate the value of the determinant of order $3\\times 3$ Therefore $\\mid A\\mid=a_{11}(a_{22}\\times a_{33}-a_{23}\\times a_{32})-a_{12}(a_{21}\\times a_{32}-a_{23}\\times a_{31})+a_{13}(a_{21}\\times a_{32}- a_{22}\\times a_{31})$ $\\mid A\\mid=3[(1\\times 1-(3\\times -2)]-(-4)[(1\\times 1)-(-2\\times 2]+5[(1\\times 3)-(1\\times 2)]$ $\\qquad=3(1+6)+4(1+4)+52(3-2)$ $\\qquad=21+20+5$ $\\qquad=46$ edited Feb 18, 2014", "&i \\end{vmatrix}$$ $$=2\\begin{vmatrix}a & b &c \\\\ d&e&f\\\\g&h&i \\end{vmatrix}$$ Take the coomon term from first row outside of determinant. $$= 2 \\times 4 \\dotsc \\text{ Subtitute } \\begin{vmatrix}a & b &c \\\\ d&e&f\\\\g&h&i \\end{vmatrix}=4$$ $$= 8$$ Hence, the value of the determinant asked in the question is equal to $$\\begin{vmatrix}2a & 2b &2c \\\\ d&e&f\\\\g&h&i \\end{vmatrix} = 8$$", "Browse Questions # Evaluate : $\\begin{vmatrix}0 &a-b&b-c\\\\b-a&0&b-c\\\\c-a&c-b&0\\end{vmatrix}$ (A) 1 (B) 2 (C) 3 (D) 0 Can you answer this question? Applying $R_1 \\rightarrow R_1-R_2$ $R_2\\rightarrow R_2-R_3$ We get, $\\begin{vmatrix}a-b&a-b&a-b\\\\b-c&b-c&b-c\\\\c-a&c-b&0\\end{vmatrix}$ $\\Rightarrow (a-b)(b-c)\\begin{vmatrix}1&1&1\\\\1&1&1\\\\c-a&c-b&0\\end{vmatrix}$ $\\Rightarrow 0$ Hence (d) is the correct answer. answered Nov 19, 2013", "&7 \\\\ 3& 8& 8\\\\ 5& 9&9 \\end{vmatrix}$ [$\\dpi{100} \\because$ Here two columns are identical ] $\\dpi{100} = 0$ Therefore we have the value of determinant = 0. Exams Articles Questions" ]

[ "& -3 & 5 \\\\ 1 & -2 & 6 \\end{vmatrix} }{ \\begin{vmatrix} 2 & 1 & -2 \\\\ 3 & -3 & -1 \\\\ 1 & -2 & 3 \\end{vmatrix} } = \\frac{-26}{-26} = 1$The solution set of the system is$ (1, – 1,1) $. Practice Questions 1. Solve the system of equations shown below using Cramer’s Rule:$ \\begin{align*} { 5x } + 2y &= \\, { 10 } \\\\ { – x } + 4y &= { 20 } \\end{align*} $2. Solve the system of equations shown below using Cramer’s Rule:$ \\begin{align*} 3x – 4y + z &= \\, -5 \\\\ x – y – z &= – 10 \\\\ 6x – 8y + 2z = 10 \\end{align*} $### Answers 1. The first step is to write the determinants of this system of equations, determinant ($ D $),$ x – $determinant ($ D_{ x } $), and the$ y – $determinant ($ D_{ y } $). Let’s use the formula we learned and write them up:$ D = \\begin{vmatrix} 5 & 2 \\\\ { – 1 } & { 4 } \\end{vmatrix} D_{ x } = \\begin{vmatrix} 10 & 2 \\\\ { 20 } & { 4 } \\end{vmatrix} D_{ y } = \\begin{vmatrix} 5 & { 10 } \\\\ { – 1 } & {", "Comment Share Q) # The determinant $\\begin{vmatrix}xp+y&x&y\\\\yp+z&y&z\\\\0 &xp+y&yp+z\\end{vmatrix}=0$ if (A) x,y,z are in A.P (B) x,y,z are in G.P (C) x,y,z are in H.P (D) xy,yz,zx are in A.P $\\begin{vmatrix}xp+y&x&y\\\\yp+z&y&z\\\\0 &xp+y&yp+z\\end{vmatrix}=0$ operating $C_1=C_1-PC_2-C_3$ $\\begin{vmatrix}0&x&y\\\\0&y&z\\\\-(xp^2+2py+z) &xp+y&yp+z\\end{vmatrix}=0$ $\\Rightarrow (xz-y^2)(xp^2+2py+z)=0$ $\\Rightarrow xz-y^2=0$ $\\Rightarrow D=0$ for other quad.factor in p which shows second factor is a perfect square. $\\Rightarrow y^2=xz$ $\\Rightarrow x,y,z$ are in G.P.", "F_x & F_y & F_z \\end{vmatrix}$$ Then you evaluate the determinant. • Thank you - this gives me a starting point. I think I can see how to proceed. – Proioxis Nov 5 '14 at 1:43 • Not a problem. You can now equate the i, j, and k pieces and work from there. – dustin Nov 5 '14 at 1:44", "of determinants. Let $\\displaystyle A = \\begin{bmatrix}a&b&c \\\\ d&e&f \\\\ g&h&i \\end{bmatrix}$ Given that $\\displaystyle |A| = \\alpha\\text{ and }\\alpha \\ne 0$, express each of the following terms of $\\displaystyle \\alpha:$ $\\displaystyle (a)\\;|2A|$ Answer: .$\\displaystyle 8\\alpha$ $\\displaystyle (b)\\;|A^{-1}|$ Answer: .$\\displaystyle \\frac{1}{\\alpha}$ $\\displaystyle (c)\\;\\begin{vmatrix}2a&2b&2c \\\\ d&e&f \\\\ g&h&i\\end{vmatrix}\\right|$ Answer: .$\\displaystyle 2\\alpha$ $\\displaystyle (d)\\;\\begin{vmatrix}a&b&c \\\\ d&e&f \\\\ g&h&i\\end{vmatrix}$ Answer: .$\\displaystyle \\alpha$ $\\displaystyle (e)\\;\\begin{vmatrix}a&b&c \\\\ d-3a & e-3b & f-3c \\\\ g&h&i\\end{vmatrix}$ Answer: .$\\displaystyle \\alpha$ 7. ## Re: Determinants of this matrix - pls help Thank you very much for your clear answer. How did you type it up like that? 8. ## Re: Determinants of this matrix - pls help Im a little stuck with this one, how do I work out this determinant? can anyone help with this? Thanks $\\displaystyle \\begin{vmatrix}a &b &c \\\\d-a&e-b &f-c \\\\g-2a&h-2b &i-2c\\end{vmatrix}$ 9. ## Re: Determinants of this matrix - pls help Use the property det(AB) = det(A).det(B)... AB = $\\displaystyle \\begin{vmatrix}a &b &c \\\\d-a&e-b &f-c \\\\g-2a&h-2b &i-2c\\end{vmatrix}$ Let B = $\\displaystyle \\begin{vmatrix}a &b &c \\\\d&e &f \\\\g&h &i\\end{vmatrix}$ we already know det(B) = alpha Now find a matrix A multiplied by B,that will give you AB... I will leave this to you to do, if you have trouble finding sucha matrix then post again. Then Find det(A) multiply by det(B)... I have not", "1. Determinant 2 $|A|=\\begin{vmatrix}1&1&1&\\dots &1\\\\1& 1-x& 1&\\dots &1\\\\1&1& 2-x &\\dots &1\\\\\\vdots&&&&\\\\1&1&1&\\dots &n-x\\end{vmatrix}$ For which $x$ does $|A|=0$? 2. Substract the first row from the others: $|A|=\\begin{vmatrix}1 & 1 & 1 & \\dots & 1\\\\ 0 & -x & 0 & \\ldots & 0\\\\ 0 & 0 & 1-x & \\ldots & 0\\\\ \\vdots\\\\ 0 & 0 & 0 & \\ldots & n-1-x\\end{vmatrix}= -x(1-x)(2-x)\\ldots (n-1-x)$ $|A|=0\\Rightarrow x_1=0, \\ x_2=1, \\ldots,x_n=n-1$", "Q # By using properties of determinants show that determinant x x^2 yz y y^2 zx z z^2 xy =(x-y)(y-z)(z-x)(xy++yz+zx) Ex 4.2 Q : 9 Q : 9 By using properties of determinants, show that: $\\dpi{100} \\begin{vmatrix} x & x^2 & yz\\\\ y & y^2 &zx \\\\ z & z^2 & xy \\end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)$ Views We have the determinant: $\\triangle = \\begin{vmatrix} x & x^2 & yz\\\\ y & y^2 &zx \\\\ z & z^2 & xy \\end{vmatrix}$ Applying the row transformations $R_{1} \\rightarrow R_{1}- R_{3}$ and then $R_{2} \\rightarrow R_{2}- R_{3}$, we have; $\\triangle = \\begin{vmatrix} x-z & x^2-z^2 & yz-xy\\\\ y-z & y^2-z^2 &zx-xy \\\\ z & z^2 & xy \\end{vmatrix}$ $= \\begin{vmatrix} x-z & (x-z)(x+z) & y(z-x)\\\\ y-z & (y-z)(y+z) &x(z-y) \\\\ z & z^2 & xy \\end{vmatrix}$ $= (x-z)(y-z)\\begin{vmatrix} 1 & (x+z) & -y\\\\ 1 & (y+z) &-x \\\\ z & z^2 & xy \\end{vmatrix}$ Now, applying $R_{1} \\rightarrow R_{1} - R_{2}$; we have $= (x-z)(y-z)\\begin{vmatrix} 0 & (x-y) & (x-y)\\\\ 1 & (y+z) &-x \\\\ z & z^2 & xy \\end{vmatrix}$ $= (x-z)(y-z)(x-y)\\begin{vmatrix} 0 & 1 & 1\\\\ 1 & (y+z) &-x \\\\ z & z^2 & xy \\end{vmatrix}$ Now, expanding the remaining determinant; $= (x-z)(y-z)(x-y) \\left [ (xy+zx) + (z^2 - zy-z^2) \\right]$ $= -(x-z)(y-z)(x-y) \\left [ xy+zx + zy \\right]$ $= (x-y)(y-z)(z-x)", "matrix: (1) \\begin{vmatrix} \\vec{i} & \\vec{j} & \\vec{k}\\\\ u_1 & u_2 & 0\\\\ v_1 & v_2 & 0 \\end{vmatrix} • If we evaluate this determinant by cofactor expansion along the first row, we obtain that: (2) \\begin{align} \\vec{u} \\times \\vec{v} = \\vec{i} \\begin{vmatrix} u_2 & 0 \\\\ v_2 & 0 \\end{vmatrix} - \\vec{j} \\begin{vmatrix} u_1 & 0 \\\\ v_1 & 0 \\end{vmatrix} + \\vec{k} \\begin{vmatrix} u_1 & u_2 \\\\ v_1 & v_2 \\end{vmatrix} \\end{align} • However, we note that the determinants $\\vec{i} \\begin{vmatrix} u_2 & 0 \\\\ v_2 & 0 \\end{vmatrix} = 0$ and $- \\vec{j} \\begin{vmatrix} u_1 & 0 \\\\ v_1 & 0 \\end{vmatrix} = 0$. Thus it follows that: (3) \\begin{align} \\vec{u} \\times \\vec{v}= \\vec{k} \\begin{vmatrix} u_1 & u_2 \\\\ v_1 & v_2 \\end{vmatrix} \\end{align} • Now let's take the norm of both sides to obtain that: (4) \\begin{align} \\| \\vec{u} \\times \\vec{v} \\| = \\| \\vec{k} \\| \\mathrm{abs} \\begin{vmatrix} u_1 & u_2 \\\\ v_1 & v_2 \\end{vmatrix} \\end{align} • However, $\\vec{k}$ is a standard unit vector so $\\| \\vec{k} \\| = 1$ and thus $\\| \\vec{u} \\times \\vec{v} \\| = \\mathrm{abs} \\begin{vmatrix}u_1 & u_2 \\\\ v_1 & v_2 \\end{vmatrix}$, and of course, this is equal to the area of the parallelogram. $\\blacksquare$", "0 & 0 & i \\end{vmatrix} =\\begin{vmatrix} a & 0 & 0\\cr d & e & 0\\cr g & h & i \\end{vmatrix}$ = a × e × i. There you go!! These are the basic properties of determinants and we are sure you want to learn more. To know more about different concepts like matrices and determinants across different subjects download BYJU’S-The Learning App. #### Practise This Question In the below figure, ___ angles are present and they are ___", "using Equation \\ref{eq:determinants1}. Example $$\\PageIndex{1}$$ Find $$x$$ in the following system of equations: \\begin{align*}2x+3y+8z &=0 \\\\[4pt] x-\\frac{1}{2}y-3z &=\\frac{1}{2} \\\\ -x-y-z &=\\frac{1}{2}\\end{align*} Solution We can calculate the $$x$$ as; $x=\\frac{D_1}{D} \\nonumber$ where $D=\\begin{vmatrix} 2 &3 & 8 \\\\ 1 &-1/2 &-3 \\\\ -1 &-1 &-1 \\end{vmatrix} \\nonumber$ and $D_1=\\begin{vmatrix} 0 &3 & 8 \\\\ 1/2 &-1/2 &-3 \\\\ 1/2 &-1 &-1 \\end{vmatrix} \\nonumber$ $$D$$ is a 3x3 determinant $$D$$ and can be expanded using quation \\ref{eq:determinants1}: \\begin{align*} D &= 2\\begin{vmatrix} -1/2&-3\\\\ -1&-1 \\end{vmatrix}- 3\\begin{vmatrix} 1&-3\\\\ -1&-1 \\end{vmatrix}+ 8\\begin{vmatrix} 1&-1/2\\\\ -1&-1 \\end{vmatrix} \\\\[4pt] &=2\\times(-5/2)-(3)\\times(-4)+8\\times(-3/2)=-5 \\end{align*} The determinant $$D_1$$ is similarly expanded: \\begin{align*} D_1 &= 0\\begin{vmatrix} -1/2&-3\\\\ -1&-1 \\end{vmatrix}- 3\\begin{vmatrix} 1/2&-3\\\\ 1/2&-1 \\end{vmatrix} + 8\\begin{vmatrix} 1/2&-1/2\\\\ 1/2&-1 \\end{vmatrix} \\\\[4pt] &=0\\times(-5/2)-(3)\\times(1)+8\\times(-1/4)=-5 \\end{align*} So, $\\displaystyle{\\color{Maroon}x=\\frac{D_1}{D}=1} \\nonumber$ To practice, finish this problem and obtain $$y$$ and $$z$$ (Problem 13.1). Example $$\\PageIndex{2}$$ Show that a $$3\\times 3$$ determinant that contains zeros below the principal diagonal (top left to bottom right) is the product of the diagonal elements. Solution We are asked to prove that $D=\\begin{vmatrix} a &b&c \\\\ 0&d &e \\\\ 0& 0 &f \\end{vmatrix}=adf \\nonumber$ Because we have two zeros in the first column, it makes more sense to calculate the determinant by expanding along the first column instead of the first row. Yet, if you feel uncomfortable doing this at this point we", "Browse Questions # Using the property of determinants, evalute: $\\begin{vmatrix}x& x+y & x+2y\\\\x+2y & x & x+y\\\\x+y & x+2y & x\\end{vmatrix}$ $\\begin{array}{1 1} 9y^2(x+y) \\\\ 9x^2(x+y) \\\\ y^2(x+y) \\\\ x^2(x+y) \\end{array}$ $\\Delta=\\begin{vmatrix}x& x+y & x+2y\\\\x+2y & x & x+y\\\\x+y & x+2y & x\\end{vmatrix}$ Step 1: Apply $R_1\\rightarrow R_1+R_2+R_3$ $\\Delta=\\begin{vmatrix}3x+3y& 3x+3y & 3x+3y\\\\x+2y & x & x+y\\\\x+y & x+2y & x\\end{vmatrix}$", "1. ## Determinant 2 $|A|=\\begin{vmatrix}1&1&1&\\dots &1\\\\1& 1-x& 1&\\dots &1\\\\1&1& 2-x &\\dots &1\\\\\\vdots&&&&\\\\1&1&1&\\dots &n-x\\end{vmatrix}$ For which $x$ does $|A|=0$? 2. Substract the first row from the others: $|A|=\\begin{vmatrix}1 & 1 & 1 & \\dots & 1\\\\ 0 & -x & 0 & \\ldots & 0\\\\ 0 & 0 & 1-x & \\ldots & 0\\\\ \\vdots\\\\ 0 & 0 & 0 & \\ldots & n-1-x\\end{vmatrix}= -x(1-x)(2-x)\\ldots (n-1-x)$ $|A|=0\\Rightarrow x_1=0, \\ x_2=1, \\ldots,x_n=n-1$", "# Determinant of a particular matrix. What is the best way to find determinant of the following matrix? $$A=\\left(\\begin{matrix} 1&ax&a^2+x^2\\\\1&ay&a^2+y^2\\\\ 1&az&a^2+z^2 \\end{matrix}\\right)$$ I thought it looks like a Vandermonde matrix, but not exactly. I can't use $$|A+B|=|A|+|B|$$ to form a Vandermonde matrix. Please suggest. Thanks. • Did you mean $\\vert A + B \\vert = \\vert A \\vert + \\vert B \\vert$? – Bladewood Feb 15 at 22:20 \\begin{align} &|A|\\\\ &=\\det\\left(\\begin{matrix} 1&ax&a^2+x^2\\\\1&ay&a^2+y^2\\\\ 1&az&a^2+z^2 \\end{matrix}\\right) \\\\ &=\\begin{vmatrix} 1&ax&a^2\\\\1&ay&a^2\\\\ 1&az&a^2 \\end{vmatrix}+\\begin{vmatrix} 1&ax&x^2\\\\1&ay&y^2\\\\ 1&az&z^2 \\end{vmatrix} \\tag{multilinearity on 3rd column} \\\\ &=0+a\\begin{vmatrix} 1&x&x^2\\\\1&y&y^2\\\\ 1&z&z^2 \\end{vmatrix} \\\\ &= a (x-y)(y-z)(z-x) \\end{align} Note that $$\\det\\left(\\begin{matrix} 1&ax&a^2+x^2\\\\1&ay&a^2+y^2\\\\ 1&az&a^2+z^2 \\end{matrix}\\right) =\\det \\left(\\begin{matrix} 1&ax&x^2\\\\1&ay&y^2\\\\ 1&az&z^2 \\end{matrix}\\right)=a\\cdot\\det \\left(\\begin{matrix} 1&x&x^2\\\\1&y&y^2\\\\ 1&z&z^2 \\end{matrix}\\right)$$ which boils down to Vandermonde determinant $$a(x-y)(y-z)(z-x).$$ I think the best way it's the following: $$\\Delta=\\sum_{cyc}(ay(a^2+z^2)-ay(a^2+x^2))=a\\sum_{cyc}(x^2z-x^2y)=a(x-y)(y-z)(z-x).$$", "allow us to stop writing $$\\det$$ over and over again: $\\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} := \\det\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$ Using the additivity of the bilinearity of the determinant on the first vector we can split this into: $\\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} = \\begin{vmatrix} a & b \\\\ 0 & d \\end{vmatrix} + \\begin{vmatrix} 0 & b \\\\ c & d \\end{vmatrix}$ And on the second vector...: $$\\label{additivity} ... = \\begin{vmatrix} a & b \\\\ 0 & 0 \\end{vmatrix} + \\begin{vmatrix} a & 0 \\\\ 0 & d \\end{vmatrix} + \\begin{vmatrix} 0 & b \\\\ c & 0 \\end{vmatrix} + \\begin{vmatrix} 0 & 0 \\\\ c & d \\end{vmatrix}$$ Then we can use homogeneity to extract the scalars: $$\\label{homogeneity} ... = ab\\begin{vmatrix} 1 & 1 \\\\ 0 & 0 \\end{vmatrix} + ad\\begin{vmatrix} 1 & 0 \\\\ 0 & 1 \\end{vmatrix} + bc\\begin{vmatrix} 0 & 1 \\\\ 1 & 0 \\end{vmatrix} + cd\\begin{vmatrix} 0 & 0 \\\\ 1 & 1 \\end{vmatrix}$$ Lastly we use that the determinant is alternating together with the determinant of an identity matrix being 1: $$\\label{final} ... = ab \\cdot 0 + ad\\cdot 1 + bc\\cdot -1 + cd\\cdot 0$$ And we conclude that the determinant of a general $$2\\times2$$ matrix", "with the pivot operations of adding ${\\displaystyle v_{1}{\\vec {\\rho }}_{1}}$ to ${\\displaystyle v_{i}{\\vec {\\beta }}}$ and ${\\displaystyle w_{1}{\\vec {\\rho }}_{1}}$ to ${\\displaystyle w_{i}{\\vec {\\rho }}_{1}}$, etc. Multilinearity allows us to expand a determinant into a sum of determinants, each of which involves a simple matrix. Example 3.4 We can use multilinearity to split this determinant into two, first breaking up the first row ${\\displaystyle {\\begin{vmatrix}2&1\\\\4&3\\end{vmatrix}}={\\begin{vmatrix}2&0\\\\4&3\\end{vmatrix}}+{\\begin{vmatrix}0&1\\\\4&3\\end{vmatrix}}}$ and then separating each of those two, breaking along the second rows. ${\\displaystyle ={\\begin{vmatrix}2&0\\\\4&0\\end{vmatrix}}+{\\begin{vmatrix}2&0\\\\0&3\\end{vmatrix}}+{\\begin{vmatrix}0&1\\\\4&0\\end{vmatrix}}+{\\begin{vmatrix}0&1\\\\0&3\\end{vmatrix}}}$ We are left with four determinants, such that in each row of each matrix there is a single entry from the original matrix. Example 3.5 In the same way, a ${\\displaystyle 3\\!\\times \\!3}$ determinant separates into a sum of many simpler determinants. We start by splitting along the first row, producing three determinants (the zero in the ${\\displaystyle 1,3}$ position is underlined to set it off visually from the zeroes that appear in the splitting). ${\\displaystyle {\\begin{vmatrix}2&1&-1\\\\4&3&{\\underline {0}}\\\\2&1&5\\end{vmatrix}}={\\begin{vmatrix}2&0&0\\\\4&3&{\\underline {0}}\\\\2&1&5\\end{vmatrix}}+{\\begin{vmatrix}0&1&0\\\\4&3&{\\underline {0}}\\\\2&1&5\\end{vmatrix}}+{\\begin{vmatrix}0&0&-1\\\\4&3&{\\underline {0}}\\\\2&1&5\\end{vmatrix}}}$ Each of these three will itself split in three along the second row. Each of the resulting nine splits in three along the third row, resulting in twenty seven determinants ${\\displaystyle ={\\begin{vmatrix}2&0&0\\\\4&0&0\\\\2&0&0\\end{vmatrix}}+{\\begin{vmatrix}2&0&0\\\\4&0&0\\\\0&1&0\\end{vmatrix}}+{\\begin{vmatrix}2&0&0\\\\4&0&0\\\\0&0&5\\end{vmatrix}}+{\\begin{vmatrix}2&0&0\\\\0&3&0\\\\2&0&0\\end{vmatrix}}+\\dots +{\\begin{vmatrix}0&0&-1\\\\0&0&{\\underline {0}}\\\\0&0&5\\end{vmatrix}}}$ such that each row contains a single entry from the starting matrix. So an ${\\displaystyle n\\!\\times \\!n}$ determinant expands into a sum of ${\\displaystyle", "the determinant $\\begin {vmatrix} a & b & c\\\\ a + 2x & b + 2y & c + 2z\\\\ x & y & z\\\\ \\end{vmatrix}$. Applying $\\text{R}_1 \\rightarrow \\text{R}_1 + 2 \\text{R}_3$ $\\begin {vmatrix} a & b & c\\\\ a + 2x & b + 2y & c + 2z\\\\ x & y & z\\\\ \\end{vmatrix} = \\begin {vmatrix} a + 2x & b + 2y & c + 2z\\\\ a + 2x & b + 2y & c + 2z\\\\ x & y & z\\\\ \\end{vmatrix} = 0$ (Elements of rows $\\text{R}_1$ and $\\text{R}_2$ are identical). Example 2: Show that $\\begin {vmatrix} x & p & q\\\\ p & x & q\\\\n q & q & x\\\\ \\end{vmatrix} = (x – p) \\left(x^2 + px – 2q^2 \\right)$ Applying $\\text{C}_1 \\rightarrow \\text{C}_1 – \\text{C}_2$ $\\begin {vmatrix} x & p & q\\\\ p & x & q\\\\ q & q & x\\\\ \\end{vmatrix} = \\begin {vmatrix} x – p & p & q\\\\ p – x & x & q\\\\ q – q & q & x\\\\ \\end{vmatrix} = \\begin {vmatrix} x – p & p & q\\\\ p – x& x & q\\\\ 0 & q & x\\\\ \\end{vmatrix} = \\begin {vmatrix} x – p & p & q\\\\ -(x – p)& x & q\\\\ 0", "\\end{vmatrix} \\end{align*} For the third element in the first row, we use the same strategy from the second element, where we use our fingers to block out all of the elements in the same row as the corresponding coefficient. That coefficient is now $4$, so we see that this $2$ by $2$ matrix will contain the elements $7$, $24$, $2$, and $6$. This gives us the matrix below. \\begin{align*} \\begin{vmatrix} 7 & 24 \\\\ 2 & 6 \\end{vmatrix} \\end{align*} Plugging in the $2$ by $2$ matrices, their corresponding coefficients, and the coefficients’ new signs, we have \\begin{align*} \\begin{vmatrix} 0 & 2 & 4 \\\\ 7 & 24 & 1 \\\\ 2 & 6 & 8 \\end{vmatrix} = 0 - 2 \\times \\begin{vmatrix} 7 & 1 \\\\ 2 & 8 \\end{vmatrix} + 4 \\times \\begin{vmatrix} 7 & 24 \\\\ 2 & 6 \\end{vmatrix} \\end{align*} We can now solve each $2$ by $2$ matrix as usual, multiply those determinants by the coefficients, and add them up to find the determinant of the $3$ by $3$ matrix. Doing this with our example, we have \\begin{align*} \\begin{vmatrix} 0 & 2 & 4 \\\\ 7 & 24 & 1 \\\\ 2 & 6 & 8 \\end{vmatrix} &= 0 - 2 \\times \\begin{vmatrix} 7 & 1 \\\\ 2 & 8 \\end{vmatrix} + 4 \\times", "Browse Questions # Using properties of determinants,prove the following $\\begin{vmatrix}x& x+y & x+2y\\\\x+2y & x & x+y\\\\x+y & x+2y & x\\end{vmatrix}=9y^2(x+y)$ This question appeared in both 65-1 and 65-2 versions of the paper in 2013. $\\Delta=\\begin{vmatrix}x& x+y & x+2y\\\\x+2y & x & x+y\\\\x+y & x+2y & x\\end{vmatrix}$ Step 1: Apply $R_1\\rightarrow R_1+R_2+R_3$ $\\Delta=\\begin{vmatrix}3x+3y& 3x+3y & 3x+3y\\\\x+2y & x & x+y\\\\x+y & x+2y & x\\end{vmatrix}$ Step 2:Apply $C_2\\rightarrow C_2-C_3$ and $C_3\\rightarrow C_3-C_1$ $\\Delta=3(x+y)\\begin{vmatrix}1& 0 & 0\\\\x+2y & -y & -y\\\\x+y & 2y & -y\\end{vmatrix}$ Therefore $\\Delta=3(x+y)(3y^2)$ $\\qquad\\qquad\\;\\;=9y^2(x+y)$ Hence proved. edited Mar 23, 2013 by meena.p", "0 & 0 \\\\ -2 & 1 & 0 \\\\ -2 & 0 & 1 \\end{vmatrix}}_{=1} \\cdot \\begin{vmatrix} 1 & 4 & 1 \\\\ 2 & 6 & 3 \\\\ 2 & 10 & 3 \\end{vmatrix} = \\begin{vmatrix} 1 & 4 & 1 \\\\ 0 & -2 & 1 \\\\ 0 & 2 & 1 \\end{vmatrix} = \\begin{vmatrix} -2 & 1 \\\\ 2 & 1 \\end{vmatrix} =-4 \\end{align*} \\end{document} Somehow, it would make more sense if you tried to demonstrate the column manipulation, because then the triangle unit-determinant matrix would come on the right side. But even this is completely clear IMHO. Here's another style I made. If you aren't doing determinants, replace the \"vmatrix\" with \"bmatrix\". I don't know how to post a picture of what it looks like. \\paragraph{Example} Evaluate the determinant \\begin{align*} \\begin{vmatrix*}[r] 1 & -12 & 5 & -3 \\\\ 2 & -21 & 22 & -7 \\\\ 2 & -9 & 41 & -10 \\\\ 0 & 3 & 7 & 3 \\end{vmatrix*}. \\end{align*} \\paragraph{Solution} \\begin{align*} \\begin{vmatrix*}[r] 1 & -12 & 5 & -3 \\\\ 2 & -21 & 22 & -7 \\\\ 2 & -9 & 41 & -10 \\\\ 0 & 3 & 7 & 3 \\end{vmatrix*} &= \\begin{vmatrix*}[r] 1 & -12 & 5 & -3 \\\\ 0 & 3 &", "& h\\\\i & j & k & l\\\\m & n & o & p \\end{vmatrix}$ $=a\\begin{vmatrix} f & g & h\\\\j & k & l\\\\n & o & p \\end{vmatrix}$ $-b\\begin{vmatrix} e & g & h\\\\i & k & l\\\\m & o & p \\end{vmatrix}$$+c\\begin{vmatrix} e & f & h\\\\i & j & l\\\\m & n & p \\end{vmatrix}$$-d\\begin{vmatrix} e & f & g\\\\i & j & k\\\\m & n & o \\end{vmatrix}.$ Another way to define the determinant is expressed in terms of the columns of the matrix. If we write an n × n matrix A in terms of its column vectors $A = \\begin{bmatrix} a_1, & a_2, & \\ldots, & a_n \\end{bmatrix}$ where the $a_j$ are vectors of size n, then the determinant of A is defined so that \\begin{align} & \\det \\begin{bmatrix} a_1, & \\ldots, & b a_j + c v, & \\ldots, a_n \\end{bmatrix} = b \\det(A) + c \\det \\begin{bmatrix} a_1, & \\ldots, & v, & \\ldots, a_n \\end{bmatrix} \\\\ & \\det \\begin{bmatrix} a_1, & \\ldots, & a_j, & a_{j+1}, & \\ldots, a_n \\end{bmatrix} = - \\det \\begin{bmatrix} a_1, & \\ldots, & a_{j+1}, & a_j, & \\ldots, a_n \\end{bmatrix} \\\\ & \\det(I) = 1 \\end{align} where b and c are scalars, v is any vector of size n and I is the", "& j & k & l\\\\m & n & o & p \\end{vmatrix} =a\\begin{vmatrix} f & g & h\\\\j & k & l\\\\n & o & p \\end{vmatrix} -b\\begin{vmatrix} e & g & h\\\\i & k & l\\\\m & o & p \\end{vmatrix}+c\\begin{vmatrix} e & f & h\\\\i & j & l\\\\m & n & p \\end{vmatrix}-d\\begin{vmatrix} e & f & g\\\\i & j & k\\\\m & n & o \\end{vmatrix}. Another way to define the determinant is expressed in terms of the columns of the matrix. If we write an n × n matrix A in terms of its column vectors \\( A = \\begin{bmatrix} a_1, & a_2, & \\ldots, & a_n \\end{bmatrix} where the \\(a_j are vectors of size n, then the determinant of A is defined so that \\( \\begin{align} & \\det \\begin{bmatrix} a_1, & \\ldots, & b a_j + c v, & \\ldots, a_n \\end{bmatrix} = b \\det(A) + c \\det \\begin{bmatrix} a_1, & \\ldots, & v, & \\ldots, a_n \\end{bmatrix} \\\\ & \\det \\begin{bmatrix} a_1, & \\ldots, & a_j, & a_{j+1}, & \\ldots, a_n \\end{bmatrix} = - \\det \\begin{bmatrix} a_1, & \\ldots, & a_{j+1}, & a_j, & \\ldots, a_n \\end{bmatrix} \\\\ & \\det(I) = 1 \\end{align} where b and c are scalars, v is any vector of size n and I is the" ]

[ "Comment Share Q) # The determinant $\\begin{vmatrix}xp+y&x&y\\\\yp+z&y&z\\\\0 &xp+y&yp+z\\end{vmatrix}=0$ if (A) x,y,z are in A.P (B) x,y,z are in G.P (C) x,y,z are in H.P (D) xy,yz,zx are in A.P $\\begin{vmatrix}xp+y&x&y\\\\yp+z&y&z\\\\0 &xp+y&yp+z\\end{vmatrix}=0$ operating $C_1=C_1-PC_2-C_3$ $\\begin{vmatrix}0&x&y\\\\0&y&z\\\\-(xp^2+2py+z) &xp+y&yp+z\\end{vmatrix}=0$ $\\Rightarrow (xz-y^2)(xp^2+2py+z)=0$ $\\Rightarrow xz-y^2=0$ $\\Rightarrow D=0$ for other quad.factor in p which shows second factor is a perfect square. $\\Rightarrow y^2=xz$ $\\Rightarrow x,y,z$ are in G.P.", "Browse Questions # Using the property of determinants, evalute: $\\begin{vmatrix}x& x+y & x+2y\\\\x+2y & x & x+y\\\\x+y & x+2y & x\\end{vmatrix}$ $\\begin{array}{1 1} 9y^2(x+y) \\\\ 9x^2(x+y) \\\\ y^2(x+y) \\\\ x^2(x+y) \\end{array}$ $\\Delta=\\begin{vmatrix}x& x+y & x+2y\\\\x+2y & x & x+y\\\\x+y & x+2y & x\\end{vmatrix}$ Step 1: Apply $R_1\\rightarrow R_1+R_2+R_3$ $\\Delta=\\begin{vmatrix}3x+3y& 3x+3y & 3x+3y\\\\x+2y & x & x+y\\\\x+y & x+2y & x\\end{vmatrix}$", "& -3 & 5 \\\\ 1 & -2 & 6 \\end{vmatrix} }{ \\begin{vmatrix} 2 & 1 & -2 \\\\ 3 & -3 & -1 \\\\ 1 & -2 & 3 \\end{vmatrix} } = \\frac{-26}{-26} = 1$The solution set of the system is$ (1, – 1,1) $. Practice Questions 1. Solve the system of equations shown below using Cramer’s Rule:$ \\begin{align*} { 5x } + 2y &= \\, { 10 } \\\\ { – x } + 4y &= { 20 } \\end{align*} $2. Solve the system of equations shown below using Cramer’s Rule:$ \\begin{align*} 3x – 4y + z &= \\, -5 \\\\ x – y – z &= – 10 \\\\ 6x – 8y + 2z = 10 \\end{align*} $### Answers 1. The first step is to write the determinants of this system of equations, determinant ($ D $),$ x – $determinant ($ D_{ x } $), and the$ y – $determinant ($ D_{ y } $). Let’s use the formula we learned and write them up:$ D = \\begin{vmatrix} 5 & 2 \\\\ { – 1 } & { 4 } \\end{vmatrix} D_{ x } = \\begin{vmatrix} 10 & 2 \\\\ { 20 } & { 4 } \\end{vmatrix} D_{ y } = \\begin{vmatrix} 5 & { 10 } \\\\ { – 1 } & {", "& c\\end{vmatrix}$ Ex. 6.2 | Q 1.4 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}1 & - 3 & 2 \\\\ 4 & - 1 & 2 \\\\ 3 & 5 & 2\\end{vmatrix}$ Ex. 6.2 | Q 1.5 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}1 & 4 & 9 \\\\ 4 & 9 & 16 \\\\ 9 & 16 & 25\\end{vmatrix}$ Ex. 6.2 | Q 1.6 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}6 & - 3 & 2 \\\\ 2 & - 1 & 2 \\\\ - 10 & 5 & 2\\end{vmatrix}$ Ex. 6.2 | Q 1.7 | Page 57 Evaluate the following determinant: $\\begin{vmatrix}1 & 3 & 9 & 27 \\\\ 3 & 9 & 27 & 1 \\\\ 9 & 27 & 1 & 3 \\\\ 27 & 1 & 3 & 9\\end{vmatrix}$ Ex. 6.2 | Q 2.01 | Page 57 Without expanding, show that the value of the following determinant is zero: $\\begin{vmatrix}8 & 2 & 7 \\\\ 12 & 3 & 5 \\\\ 16 & 4 & 3\\end{vmatrix}$ Ex. 6.2 | Q 2.02 | Page 57 Without expanding, show that the value of the following determinant is zero: $\\begin{vmatrix}6 & - 3 & 2 \\\\ 2 & - 1 & 2 \\\\ - 10 & 5 & 2\\end{vmatrix}$ Ex. 6.2 | Q", "1. Determinant 2 $|A|=\\begin{vmatrix}1&1&1&\\dots &1\\\\1& 1-x& 1&\\dots &1\\\\1&1& 2-x &\\dots &1\\\\\\vdots&&&&\\\\1&1&1&\\dots &n-x\\end{vmatrix}$ For which $x$ does $|A|=0$? 2. Substract the first row from the others: $|A|=\\begin{vmatrix}1 & 1 & 1 & \\dots & 1\\\\ 0 & -x & 0 & \\ldots & 0\\\\ 0 & 0 & 1-x & \\ldots & 0\\\\ \\vdots\\\\ 0 & 0 & 0 & \\ldots & n-1-x\\end{vmatrix}= -x(1-x)(2-x)\\ldots (n-1-x)$ $|A|=0\\Rightarrow x_1=0, \\ x_2=1, \\ldots,x_n=n-1$", "# Q : 9 Evaluate $\\dpi{100} \\begin{vmatrix} x & y &x+y \\\\ y & x+y &x \\\\ x+y & x & y \\end{vmatrix}$ We have determinant $\\triangle = \\begin{vmatrix} x & y &x+y \\\\ y & x+y &x \\\\ x+y & x & y \\end{vmatrix}$ Applying row transformations; $R_{1} \\rightarrow R_{1}+R_{2}+R_{3}$ , we have then; $\\triangle = \\begin{vmatrix} 2(x+y) & 2(x+y) &2(x+y) \\\\ y & x+y &x \\\\ x+y & x & y \\end{vmatrix}$ Taking out the common factor 2(x+y) from the row first. $= 2(x+y)\\begin{vmatrix} 1 & 1 &1 \\\\ y & x+y &x \\\\ x+y & x & y \\end{vmatrix}$ Now, applying the column transformation; $C_{1} \\rightarrow C_{1} - C_{2}$ and $C_{2} \\rightarrow C_{2} - C_{1}$ we have ; $= 2(x+y)\\begin{vmatrix} 0 & 0 &1 \\\\ -x & y &x \\\\ y & x-y & y \\end{vmatrix}$ Expanding the remaining determinant; $= 2(x+y)(-x(x-y)-y^2) = 2(x+y)[-x^2+xy-y^2]$ $= -2(x+y)[x^2-xy+y^2] = -2(x^3+y^3)$. Exams Articles Questions", "Browse Questions # Using properties of determinants,prove the following $\\begin{vmatrix}x& x+y & x+2y\\\\x+2y & x & x+y\\\\x+y & x+2y & x\\end{vmatrix}=9y^2(x+y)$ This question appeared in both 65-1 and 65-2 versions of the paper in 2013. $\\Delta=\\begin{vmatrix}x& x+y & x+2y\\\\x+2y & x & x+y\\\\x+y & x+2y & x\\end{vmatrix}$ Step 1: Apply $R_1\\rightarrow R_1+R_2+R_3$ $\\Delta=\\begin{vmatrix}3x+3y& 3x+3y & 3x+3y\\\\x+2y & x & x+y\\\\x+y & x+2y & x\\end{vmatrix}$ Step 2:Apply $C_2\\rightarrow C_2-C_3$ and $C_3\\rightarrow C_3-C_1$ $\\Delta=3(x+y)\\begin{vmatrix}1& 0 & 0\\\\x+2y & -y & -y\\\\x+y & 2y & -y\\end{vmatrix}$ Therefore $\\Delta=3(x+y)(3y^2)$ $\\qquad\\qquad\\;\\;=9y^2(x+y)$ Hence proved. edited Mar 23, 2013 by meena.p", "Browse Questions # Using the properties of determinants, evaluate $\\begin{vmatrix} a+x & y & z \\\\ x & a+y & z \\\\ x & y & a+z \\end{vmatrix}$ Toolbox: • If each element of a row(or a column) of A is expressed as the sum of two or more terms,then the determinant of A can be expressed as the sum of determinant of two or more matrices. • If each element of a row (or column) of a determinant is multiplied by a constant 'k' then |A|=k|A|. Let $\\Delta=\\begin{vmatrix}a+x & y & z\\\\x & a+y & z\\\\x & y & a+z\\end{vmatrix}$ Apply $C_1+C_2+C_3$ $\\Delta=\\begin{vmatrix}a+x+y+z & y & z\\\\x+a+y+z & a+y & z\\\\x+y+a+z & y & a+z\\end{vmatrix}$ Let us take (a+x+y+z) as a common factor from $C_1$. $\\Delta=(a+x+y+z)\\begin{vmatrix}1 & y & z\\\\1 & a+y & z\\\\1 & y & a+z\\end{vmatrix}$ Apply $R_2\\rightarrow R_2-R_3$ and $R_3\\rightarrow R_3-R_1$ $\\Delta=(a+x+y+z)\\begin{vmatrix}1 & y & z\\\\0 & a & -a\\\\0 & 0 & a\\end{vmatrix}$ On expanding along $R_1$ $\\Delta=(a+x+y+z)[1(a^2)-0+0]$ $\\quad=(a+x+y+z).a^2$ Hence $\\Delta=a^2(a+x+y+z)$ edited Mar 14, 2013", "(v_y w_z - v_z w_y)\\hat{i} -$$ $$(v_x w_z - w_x v_z)\\hat{j} +$$ $$(v_x w_y - w_x v_y)\\hat{k} ] =$$ $$u_x(v_y w_z - v_z w_y) -$$ $$u_y(v_x w_z - w_x v_z) +$$ $$u_z(v_x w_y - w_x v_y)$$ We could certainly multiply the components of vector $$\\vec{u}$$ through each factor, but for reasons you will see later, we will leave the equation as it is. To sum up, we have calculated $$\\vec{u} \\cdot (\\vec{v} \\times \\vec{w}) =$$ $$u_x(v_y w_z - v_z w_y) -$$ $$u_y(v_x w_z - w_x v_z) +$$ $$u_z(v_x w_y - w_x v_y) ~~~~~ (1)$$ Okay, now let's calculate the determinant $$\\begin{vmatrix} u_x & u_y & u_z \\\\ v_x & v_y & v_z \\\\ w_x & w_y & w_z \\end{vmatrix}$$ We will go across the top row (although going down the first column will give us the same result, going across the top row makes the algebra come out the way want it to). $$\\begin{vmatrix} u_x & u_y & u_z \\\\ v_x & v_y & v_z \\\\ w_x & w_y & w_z \\end{vmatrix} =$$ $$u_x(v_y w_z - v_z w_y) -$$ $$u_y(v_x w_z - w_x v_z) +$$ $$u_z(v_x w_y - w_x v_y)$$ Notice this last equation is the same as equation (1) above. This is easier to see since we did not multiply out all factors in Equation (1). So,", "Q # By using properties of determinants show that determinant x x^2 yz y y^2 zx z z^2 xy =(x-y)(y-z)(z-x)(xy++yz+zx) Ex 4.2 Q : 9 Q : 9 By using properties of determinants, show that: $\\dpi{100} \\begin{vmatrix} x & x^2 & yz\\\\ y & y^2 &zx \\\\ z & z^2 & xy \\end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)$ Views We have the determinant: $\\triangle = \\begin{vmatrix} x & x^2 & yz\\\\ y & y^2 &zx \\\\ z & z^2 & xy \\end{vmatrix}$ Applying the row transformations $R_{1} \\rightarrow R_{1}- R_{3}$ and then $R_{2} \\rightarrow R_{2}- R_{3}$, we have; $\\triangle = \\begin{vmatrix} x-z & x^2-z^2 & yz-xy\\\\ y-z & y^2-z^2 &zx-xy \\\\ z & z^2 & xy \\end{vmatrix}$ $= \\begin{vmatrix} x-z & (x-z)(x+z) & y(z-x)\\\\ y-z & (y-z)(y+z) &x(z-y) \\\\ z & z^2 & xy \\end{vmatrix}$ $= (x-z)(y-z)\\begin{vmatrix} 1 & (x+z) & -y\\\\ 1 & (y+z) &-x \\\\ z & z^2 & xy \\end{vmatrix}$ Now, applying $R_{1} \\rightarrow R_{1} - R_{2}$; we have $= (x-z)(y-z)\\begin{vmatrix} 0 & (x-y) & (x-y)\\\\ 1 & (y+z) &-x \\\\ z & z^2 & xy \\end{vmatrix}$ $= (x-z)(y-z)(x-y)\\begin{vmatrix} 0 & 1 & 1\\\\ 1 & (y+z) &-x \\\\ z & z^2 & xy \\end{vmatrix}$ Now, expanding the remaining determinant; $= (x-z)(y-z)(x-y) \\left [ (xy+zx) + (z^2 - zy-z^2) \\right]$ $= -(x-z)(y-z)(x-y) \\left [ xy+zx + zy \\right]$ $= (x-y)(y-z)(z-x)", "& 1 \\\\ x_1 & y_1 & z_1 &1 \\\\ x_2 & y_2 & z_2 & 1 \\\\ x_3 & y_3 & z_3 & 1 \\end{vmatrix} = 0\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\left(15 \\right )$$Expanding this determinant using the cofactors [12] of the elements $$x$$, $$y$$, $$z$$, and 1 in the first row of the determinant yields$$x\\:\\begin{vmatrix} y_1 & z_1 & 1\\\\ y_2 & z_2 & 1\\\\ y_3 & z_3 & 1 \\end{vmatrix} - y\\:\\begin{vmatrix} x_1 & z_1 & 1\\\\ x_2 & z_2 & 1\\\\ x_3 & z_3 & 1 \\end{vmatrix} + z\\:\\begin{vmatrix} x_1 & y_1 & 1\\\\ x_2 & y_2 & 1\\\\ x_3 & y_3 & 1 \\end{vmatrix} - \\begin{vmatrix} x_1 & y_1 & z_1\\\\ x_2 & y_2 & z_2\\\\ x_3 & y_3 & z_3 \\end{vmatrix} = 0\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\left(16 \\right)$$Equation 16 may be rewritten in more compact form as$$Ax + By+Cz+D = 0\\;\\;\\;\\;\\;\\;\\;\\;\\;\\;\\left(17 \\right )$$where $$A$$, $$B$$, $$C$$, and $$D$$ represent the determinants in Equation 16. The determinants $$A$$, $$B$$, and $$C$$ may be expanded using the cofactors of the elements in their third columns as follows$$\\begin{matrix} A= \\begin{vmatrix} y_1 & z_1 & 1\\\\ y_2 & z_2 & 1\\\\ y_3 & z_3 & 1 \\end{vmatrix}= \\begin{vmatrix} y_2 & z_2\\\\ y_3 & z_3 \\end{vmatrix}-\\begin{vmatrix} y_1 & z_1\\\\ y_3 & z_3 \\end{vmatrix}+\\begin{vmatrix} y_1 & z_1\\\\ y_2 & z_2 \\end{vmatrix} & \\left(18a \\right ) \\\\ B=", "matrix: (1) \\begin{vmatrix} \\vec{i} & \\vec{j} & \\vec{k}\\\\ u_1 & u_2 & 0\\\\ v_1 & v_2 & 0 \\end{vmatrix} • If we evaluate this determinant by cofactor expansion along the first row, we obtain that: (2) \\begin{align} \\vec{u} \\times \\vec{v} = \\vec{i} \\begin{vmatrix} u_2 & 0 \\\\ v_2 & 0 \\end{vmatrix} - \\vec{j} \\begin{vmatrix} u_1 & 0 \\\\ v_1 & 0 \\end{vmatrix} + \\vec{k} \\begin{vmatrix} u_1 & u_2 \\\\ v_1 & v_2 \\end{vmatrix} \\end{align} • However, we note that the determinants $\\vec{i} \\begin{vmatrix} u_2 & 0 \\\\ v_2 & 0 \\end{vmatrix} = 0$ and $- \\vec{j} \\begin{vmatrix} u_1 & 0 \\\\ v_1 & 0 \\end{vmatrix} = 0$. Thus it follows that: (3) \\begin{align} \\vec{u} \\times \\vec{v}= \\vec{k} \\begin{vmatrix} u_1 & u_2 \\\\ v_1 & v_2 \\end{vmatrix} \\end{align} • Now let's take the norm of both sides to obtain that: (4) \\begin{align} \\| \\vec{u} \\times \\vec{v} \\| = \\| \\vec{k} \\| \\mathrm{abs} \\begin{vmatrix} u_1 & u_2 \\\\ v_1 & v_2 \\end{vmatrix} \\end{align} • However, $\\vec{k}$ is a standard unit vector so $\\| \\vec{k} \\| = 1$ and thus $\\| \\vec{u} \\times \\vec{v} \\| = \\mathrm{abs} \\begin{vmatrix}u_1 & u_2 \\\\ v_1 & v_2 \\end{vmatrix}$, and of course, this is equal to the area of the parallelogram. $\\blacksquare$", "using Equation \\ref{eq:determinants1}. Example $$\\PageIndex{1}$$ Find $$x$$ in the following system of equations: \\begin{align*}2x+3y+8z &=0 \\\\[4pt] x-\\frac{1}{2}y-3z &=\\frac{1}{2} \\\\ -x-y-z &=\\frac{1}{2}\\end{align*} Solution We can calculate the $$x$$ as; $x=\\frac{D_1}{D} \\nonumber$ where $D=\\begin{vmatrix} 2 &3 & 8 \\\\ 1 &-1/2 &-3 \\\\ -1 &-1 &-1 \\end{vmatrix} \\nonumber$ and $D_1=\\begin{vmatrix} 0 &3 & 8 \\\\ 1/2 &-1/2 &-3 \\\\ 1/2 &-1 &-1 \\end{vmatrix} \\nonumber$ $$D$$ is a 3x3 determinant $$D$$ and can be expanded using quation \\ref{eq:determinants1}: \\begin{align*} D &= 2\\begin{vmatrix} -1/2&-3\\\\ -1&-1 \\end{vmatrix}- 3\\begin{vmatrix} 1&-3\\\\ -1&-1 \\end{vmatrix}+ 8\\begin{vmatrix} 1&-1/2\\\\ -1&-1 \\end{vmatrix} \\\\[4pt] &=2\\times(-5/2)-(3)\\times(-4)+8\\times(-3/2)=-5 \\end{align*} The determinant $$D_1$$ is similarly expanded: \\begin{align*} D_1 &= 0\\begin{vmatrix} -1/2&-3\\\\ -1&-1 \\end{vmatrix}- 3\\begin{vmatrix} 1/2&-3\\\\ 1/2&-1 \\end{vmatrix} + 8\\begin{vmatrix} 1/2&-1/2\\\\ 1/2&-1 \\end{vmatrix} \\\\[4pt] &=0\\times(-5/2)-(3)\\times(1)+8\\times(-1/4)=-5 \\end{align*} So, $\\displaystyle{\\color{Maroon}x=\\frac{D_1}{D}=1} \\nonumber$ To practice, finish this problem and obtain $$y$$ and $$z$$ (Problem 13.1). Example $$\\PageIndex{2}$$ Show that a $$3\\times 3$$ determinant that contains zeros below the principal diagonal (top left to bottom right) is the product of the diagonal elements. Solution We are asked to prove that $D=\\begin{vmatrix} a &b&c \\\\ 0&d &e \\\\ 0& 0 &f \\end{vmatrix}=adf \\nonumber$ Because we have two zeros in the first column, it makes more sense to calculate the determinant by expanding along the first column instead of the first row. Yet, if you feel uncomfortable doing this at this point we", "& (x-z)(x+z) & y(z-x)\\\\ y-z & (y-z)(y+z) &x(z-y) \\\\ z & z^2 & xy \\end{vmatrix}$ $= (x-z)(y-z)\\begin{vmatrix} 1 & (x+z) & -y\\\\ 1 & (y+z) &-x \\\\ z & z^2 & xy \\end{vmatrix}$ Now, applying $R_{1} \\rightarrow R_{1} - R_{2}$; we have $= (x-z)(y-z)\\begin{vmatrix} 0 & (x-y) & (x-y)\\\\ 1 & (y+z) &-x \\\\ z & z^2 & xy \\end{vmatrix}$ $= (x-z)(y-z)(x-y)\\begin{vmatrix} 0 & 1 & 1\\\\ 1 & (y+z) &-x \\\\ z & z^2 & xy \\end{vmatrix}$ Now, expanding the remaining determinant; $= (x-z)(y-z)(x-y) \\left [ (xy+zx) + (z^2 - zy-z^2) \\right]$ $= -(x-z)(y-z)(x-y) \\left [ xy+zx + zy \\right]$ $= (x-y)(y-z)(z-x) \\left [ xy+zx + zy \\right]$ Hence proved. Question:10(i) By using properties of determinants, show that: $\\dpi{100} \\begin{vmatrix} x+4 &2x &2x \\\\ 2x & x+4 & 2x\\\\ 2x & 2x & x+4 \\end{vmatrix}=(5x+4)(4-x)$ Given determinant: $\\begin{vmatrix} x+4 &2x &2x \\\\ 2x & x+4 & 2x\\\\ 2x & 2x & x+4 \\end{vmatrix}$ Applying row transformation: $R_{1} \\rightarrow R_{1} + R_{2} + R_{3}$ then we have; $\\triangle = \\begin{vmatrix} 5x+4 &5x+4 &5x+4 \\\\ 2x & x+4 & 2x\\\\ 2x & 2x & x+4 \\end{vmatrix}$ Taking a common factor: 5x+4 $= (5x+4)\\begin{vmatrix} 1 &1 &1 \\\\ 2x & x+4 & 2x\\\\ 2x & 2x & x+4 \\end{vmatrix}$ Now, applying column transformations $C_{1} \\rightarrow C_{1}- C_{2}$ and $C_{2} \\rightarrow", "# Using Cofactors of elements of third column, evaluate $\\Delta = \\begin{vmatrix} 1&x&yz \\\\ 1&y&zx \\\\ 1&z&xy \\end{vmatrix}$ $\\begin{array}{1 1} (x+y)(y+z)(z+x) \\\\ (x-y(z-y)(x-z) \\\\ (x-y)(y-z)(z-x) \\\\ (y-z)(y-x)(z-x) \\end{array}$ Toolbox: • $\\bigtriangleup$=Sum of the product of any row (or column) with their corresponding cofactors. • Given: • Using cofactors of elements of third column evaluate $\\bigtriangleup=\\begin{vmatrix}1 & x & yz\\\\1 & y & zx\\\\1 & z & xy\\end{vmatrix}$ we know that expanding the determinant $\\bigtriangleup$ along $C_3$ we have, $\\bigtriangleup=a_{13}A_{13}+a_{23}A_{23}+a_{33}A_{33}$ Minor of$M_{13}=\\begin{vmatrix}1 & y\\\\1 & z\\end{vmatrix}=(z-y)$ Minor of$M_{23}=\\begin{vmatrix}1 & x\\\\1 & z\\end{vmatrix}=(z-x)$ Minor of$M_{33}=\\begin{vmatrix}1 & x\\\\1 & y\\end{vmatrix}=(y-x)$ $A_{13}=(-1)^{1+3}(z-y)=z-y.$ $A_{23}=(-1)^{2+3}(z-x)=-(z-x).$ $A_{33}=(-1)^{3+3}(y-x)=y-x.$ Hence $a_{13}A_{13}+a_{23}A_{23}+a_{33}A_{33}=yz(z-y)+zx(-)(z-x)+xy(y-x)$ $\\;\\;\\;=(z^2y-y^2z)-(zx^2-x^2z)+(xy^2-x^2y)$ $\\bigtriangleup=(x^2z-y^2z)+(yz^2-xz^2)+(xy^2-x^2y)$ Taking the common factor $\\bigtriangleup=z(x^2-y^2)+z^2(y-x)+xy(y-x)$ We know $(a^2-b^2)=(a+b)(a-b)$ Therefore $\\bigtriangleup=z(x+y)(x-y)+z^2(y-x)+z^2(y-x)+xy(y-x)$ Taking (x-y)as a common factor, $\\bigtriangleup=(x-y)[z(x+y)-z^2-xy]$ $\\quad=(x-y)[zx+zy-z^2-xy]$ $\\quad=(x-y)[z(x-z)+y(z-x)]$ Taking the common factor x(z-x) $\\bigtriangleup=(x-y)[(z-x)(y-z)]$ $\\bigtriangleup=(x-y)(z-x)(y-z)$ Hence $\\bigtriangleup=(x-y)(y-z)(z-x)$", "Browse Questions # Find the value of determinant $\\begin{vmatrix}1&x&y+z\\\\1&y&z+x\\\\1&z&x+y\\end{vmatrix}$ $(a)\\;0\\qquad(b)\\;x+y+z\\qquad(c)\\;1+x+y+z\\qquad(d)\\;(x-y)(y-z)(z-x)$ Can you answer this question? $\\begin{vmatrix}1&x&y+z\\\\1&y&z+x\\\\1&z&x+y\\end{vmatrix}$ Apply $C_3\\rightarrow C_3+C_2$ $\\begin{vmatrix}1&x&y+z+x\\\\1&y&z+x+y\\\\1&z&x+y+z\\end{vmatrix}$ Taking out (x+y+z) we have Since $C_1 and \\;C_3$ are identical. $\\Rightarrow \\Delta=0$ Hence (a) is the correct answer. answered Nov 26, 2013 edited Mar 20, 2014", "explicitly spell out everything. – achille hui Oct 29 '13 at 11:29 \\begin{align}x^3+y^3+z^3-3xyz & = x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2 \\\\ & =(x+y)^3+z^3-3xy(x+y+z)\\\\ &= (x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z) \\\\ & =(x+y+z)(x^2+2xy+y^2+z^2-xy-xz-3xy) \\\\ & =(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\\end{align} Note that (can be easily seen with rule of Sarrus)$$\\begin{vmatrix} x & y & z \\\\ z & x & y \\\\ y & z & x \\\\ \\end{vmatrix}=x^3+y^3+z^3-3xyz$$ On the other hand, it is equal to (if we add to the first row 2 other rows) $$\\begin{vmatrix} x+y+z & x+y+z & x+y+z \\\\ z & x & y \\\\ y & z & x \\\\ \\end{vmatrix}=(x+y+z)\\begin{vmatrix} 1 & 1 & 1 \\\\ z & x & y \\\\ y & z & x \\\\ \\end{vmatrix}=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$$ just as we wanted. The last equality follows from the expansion of the determinant by first row. A polynomial from $\\mathbb{Q}[x,y,z]$ is a polynomial from $\\mathbb{Q}[x,y][z]$, so it can be viewed as a polynomial in $z$ with coefficients from the integral domain $\\mathbb{Q}[x,y]$. $$p(z)=z^3-3xy \\cdot z +x^3+y^3$$ So we can try our methods to factor a polynomial of degree 3 over an integral domain: If it can be factored then there is a factor of degree $1$, we call it $z-u(x,y)$ and $u(x,y)$ divides the constant term of $p(z)$ which is $x^3+y^3$. The latter is can be factored to $(x+y)(x^2-xy+y^2)$ We check each of the", "of determinants. Let $\\displaystyle A = \\begin{bmatrix}a&b&c \\\\ d&e&f \\\\ g&h&i \\end{bmatrix}$ Given that $\\displaystyle |A| = \\alpha\\text{ and }\\alpha \\ne 0$, express each of the following terms of $\\displaystyle \\alpha:$ $\\displaystyle (a)\\;|2A|$ Answer: .$\\displaystyle 8\\alpha$ $\\displaystyle (b)\\;|A^{-1}|$ Answer: .$\\displaystyle \\frac{1}{\\alpha}$ $\\displaystyle (c)\\;\\begin{vmatrix}2a&2b&2c \\\\ d&e&f \\\\ g&h&i\\end{vmatrix}\\right|$ Answer: .$\\displaystyle 2\\alpha$ $\\displaystyle (d)\\;\\begin{vmatrix}a&b&c \\\\ d&e&f \\\\ g&h&i\\end{vmatrix}$ Answer: .$\\displaystyle \\alpha$ $\\displaystyle (e)\\;\\begin{vmatrix}a&b&c \\\\ d-3a & e-3b & f-3c \\\\ g&h&i\\end{vmatrix}$ Answer: .$\\displaystyle \\alpha$ 7. ## Re: Determinants of this matrix - pls help Thank you very much for your clear answer. How did you type it up like that? 8. ## Re: Determinants of this matrix - pls help Im a little stuck with this one, how do I work out this determinant? can anyone help with this? Thanks $\\displaystyle \\begin{vmatrix}a &b &c \\\\d-a&e-b &f-c \\\\g-2a&h-2b &i-2c\\end{vmatrix}$ 9. ## Re: Determinants of this matrix - pls help Use the property det(AB) = det(A).det(B)... AB = $\\displaystyle \\begin{vmatrix}a &b &c \\\\d-a&e-b &f-c \\\\g-2a&h-2b &i-2c\\end{vmatrix}$ Let B = $\\displaystyle \\begin{vmatrix}a &b &c \\\\d&e &f \\\\g&h &i\\end{vmatrix}$ we already know det(B) = alpha Now find a matrix A multiplied by B,that will give you AB... I will leave this to you to do, if you have trouble finding sucha matrix then post again. Then Find det(A) multiply by det(B)... I have not", "Find the value of determinant $\\begin{vmatrix}1&x&y+z\\\\1&y&z+x\\\\1&z&x+y\\end{vmatrix}$ $(a)\\;0\\qquad(b)\\;x+y+z\\qquad(c)\\;1+x+y+z\\qquad(d)\\;(x-y)(y-z)(z-x)$ $\\begin{vmatrix}1&x&y+z\\\\1&y&z+x\\\\1&z&x+y\\end{vmatrix}$ Apply $C_3\\rightarrow C_3+C_2$ $\\begin{vmatrix}1&x&y+z+x\\\\1&y&z+x+y\\\\1&z&x+y+z\\end{vmatrix}$ Taking out (x+y+z) we have Since $C_1 and \\;C_3$ are identical. $\\Rightarrow \\Delta=0$ Hence (a) is the correct answer. edited Mar 20, 2014", "the given result is proved. Q9: By using the properties of determinants, show that: $\\begin{vmatrix} x & x^{2} & yz \\\\ y & y^{2} & zx \\\\ z & z^{2} xy & \\end{vmatrix} = (x – y)(y-z)(z-x)(xy+yz+zx)$ Sol: Let $\\Delta = \\begin{vmatrix} x & x^{2} & yz \\\\ y & y^{2} & zx \\\\ z & z^{2} xy & \\end{vmatrix}$ Applying $R_{2}\\rightarrow R_{2}-R_{1} \\;\\; and R_{3}\\rightarrow R_{3}-R_{1}$ we have: $\\\\\\boldsymbol{\\Rightarrow }$ $\\Delta = \\begin{vmatrix} x & x^{2} & yz \\\\ y-x & y^{2}-x^{2} & zx-yz \\\\ z-x & z^{2}-x^{2} xy-yz & -y(z-x) \\end{vmatrix}\\\\$ = $\\\\\\begin{vmatrix} x & x^{2} & yz \\\\ -(x-y) & -(x-y)(x+y) & z(x-y) \\\\ z-x & (z-x)(z+x) & -y(z-x) \\end{vmatrix}\\\\$ = $\\\\(x-y)(z-x)\\begin{vmatrix} x & x^{2} & yz \\\\ -1 & -x-y & z \\\\ 1 & (z+x) & -y \\end{vmatrix}\\\\$ Applying $R_{3}\\rightarrow R_{3}+R_{2}$ we have: $\\boldsymbol{\\Rightarrow }$ $\\Delta = (x-y)(z-x)\\begin{vmatrix} x & x^{2} & yz \\\\ -1 & -x-y & z \\\\ 0 & z-y & z-y \\end{vmatrix}\\\\$ = $\\\\(x-y)(z-x)(z-y)\\begin{vmatrix} x & x^{2} & yz \\\\ -1 & -x-y & z \\\\ 0 & 1 & 1 \\end{vmatrix}\\\\$ Expanding along $R_{3}$ we have: $\\boldsymbol{\\Rightarrow }$ $\\Delta = \\left [ (x-y)(z-x)(z-y) \\right ] \\left [ (-1) \\begin{vmatrix} x & yz \\\\ -1 & z \\end{vmatrix} + 1 \\begin{vmatrix} x & x^{2} \\\\ -1 & -x-y \\end{vmatrix}" ]

[ "= 50^{\\circ}$?", "\\circ$.", "{45}^{\\circ}$.", "$^\\circ$C.", "6^\\circ$ -", "^{\\circ}$", "\\circ h' =g$.", "^ {\\ circ}}$", "\\circ m(f)$", "(37^\\circ+21^\\circ)=180^\\circ-58^\\circ=122^\\circ:$", "to this problem is $$\\{114.3^\\circ,335.7^\\circ\\}$$", "- 273.15) + 32 = 71.33 ^{\\circ}F$$", "\\cos^{-1} (\\frac{1}{3}) = 70.53^\\circ$", "subtract the two from $$180^{\\circ}$$.", "H - N - H = {109.5}^{\\circ}$.", "$73.7^{\\circ}$ (e) $x=2$ 37. (a) $p=4, q=3$ (b) $60^{\\circ}$(c) $16.5^{\\circ}$(d) $19.1^{\\circ}$ 38. (a) $141.3^{\\circ}$ (b) $A D =6.24$ (c) $A=7.81$", "calculate by hand now, although $\\cos 30^\\circ$ is quite close and gives one a sense of it. – Dylan Moreland Feb 5 '12 at 20:57", "T\\circ \\phi^{-1}.$", "K}}$ or $- {39.43}^{\\circ} \\text{C}$.", "Find $$\\hat{N}$$. $\\hat{N} = 83^{\\circ}$" ]

[ "us all the multiples of ${72}^{\\circ}$. Let's go at it a different way, using the double and triple angle formulas: Solve $\\cos \\left(2 x\\right) = \\cos \\left(3 x\\right)$ for $\\cos \\left(x\\right)$ $2 {\\cos}^{2} x - 1 = 4 {\\cos}^{3} x - 3 \\cos x$ Let's abbreviate $c = \\cos x$ $2 {c}^{2} - 1 = 4 {c}^{3} - 3 c$ $4 {c}^{3} - 2 {c}^{2} - 3 c + 1 = 0$ That's a cubic equation, but one of the easy ones. All the cosines of multiples of ${72}^{\\circ}$ will be roots of this equation. That includes $\\cos 0 = 1$. So $c - 1$ is a factor, and with long division we get $\\left(c - 1\\right) \\left(4 {c}^{2} + 2 c - 1\\right) = 0$ $\\cos {0}^{\\circ} ,$ $\\cos {72}^{\\circ} ,$ and $\\cos {144}^{\\circ}$ are all different, so they're the three roots to this cubic equation. All other multiples of ${72}^{\\circ}$ share one of these cosines. $c = \\frac{1}{4} \\left(- 1 \\setminus \\pm \\sqrt{5}\\right)$ Clearly $\\cos {72}^{\\circ}$ is the positive one and $\\cos {144}^{\\circ}$ is the negative one: $\\cos {72}^{\\circ} = \\frac{1}{4} \\left(\\setminus \\sqrt{5} - 1\\right)$ $\\cos {144}^{\\circ} = \\frac{1}{4} \\left(- \\setminus \\sqrt{5} - 1\\right)$ $\\cos {36}^{\\circ} = - \\setminus \\cos \\left({180}^{\\circ} - {36}^{\\circ}\\right) = - \\cos {144}^{\\circ} = \\frac{1}{4} \\left(1 + \\setminus \\sqrt{5}\\right)$ That's half the", "58: Line 92: - $c^{2} = a^{2} + b^{2} - 2ab \\cos C$ + ::$c^{2} = a^{2} + b^{2} - 2ab \\cos C$ [[Image:SAS triangle.jpg]] [[Image:SAS triangle.jpg]] Line 67: Line 101: - $c^{2} = 6^{2} + (6 \\sqrt{2})^{2} -2 (6) (6 \\sqrt{2}) \\cos 45^\\circ$ + :$c^{2} = 6^{2} + (6 \\sqrt{2})^{2} -2 (6) (6 \\sqrt{2}) \\cos 45^\\circ$ Simplify for Simplify for - $c^{2} =36 + 36 (2) - 72 \\sqrt{2}) \\cos 45^\\circ$ + ::$c^{2} =36 + 36 (2) - 72 \\sqrt{2}) \\cos 45^\\circ$ Since $\\cos 45^\\circ = \\frac{1}{\\sqrt{2}}$, substitution gives us Since $\\cos 45^\\circ = \\frac{1}{\\sqrt{2}}$, substitution gives us - $c^{2} =36 + 36 (2) - 72 \\sqrt{2} (\\frac{1}{\\sqrt{2}})$ + ::$c^{2} =36 + 36 (2) - 72 \\sqrt{2} (\\frac{1}{\\sqrt{2}})$ Simplify for Simplify for - $c^{2} =36 + 72 - 72$ + ::$c^{2} =36 + 72 - 72$ - $c^{2} =36$ + ::$c^{2} =36$ Taking the square root of both sides gives us Taking the square root of both sides gives us - $c =6$ + ::$c =6$ ---- ---- Line 93: Line 127: - $b^{2} = a^{2} + c^{2} - 2ab \\cos B$ + ::$b^{2} = a^{2} + c^{2} - 2ab \\cos B$ Substituting in the appropriate side lengths gives us Substituting in the appropriate side lengths gives us - $(6 \\sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6)", "$C=\\frac{-1 \\pm \\sqrt{5}}{2}$; we know $\\cos(72º)>0$, so only the positive branch makes sense. Therefore $\\cos(72º) = \\frac{\\sqrt{5}-1}{2}$. I like this method a lot - it drops out super-neatly. However, it feels a bit like the triangles have appeared by magic; it's easy to prove it once you have the scaffolding in place, but I wouldn't have come up with the scaffolding on my own. ## Colin Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove. 1. I have not yet read the book, but think highly of Mark and trust that it's a good one. [] ### 3 comments on “$\\cos(72º)$, revisited: a geometric method” • ##### Barney Maunder-Taylor The other evening, I was trying to work out what shape triangles I needed to make a rectified pentagonal prism (as you do), and as part of the calculations needed cos(36deg). Being too lazy to get out of bed and find a calculator, I observed that if a=36deg then cos(3a)=cos(180-2a)since both are cos(108deg). Expanding both sides down to single angle gave a cubic equation which easily factorised into a linear and a quadratic. Solving the quadratic gave cos(36deg)=(1+sqrt5)/4, whence it’s easy to get cos72", "$$\\frac{2}{3}$$ we rewrite the ratio 3a = 2b cross-multiply $$\\frac{3 a}{3}$$ = $$\\frac{2 b}{3}$$ Divide both sides by 3 a = $$\\frac{2 b}{3}$$ m∠RQS + m∠SQP = 90° ∠RQS and ∠SQP are complementary b + a = 90 b + $$\\frac{2 b}{3}$$ = 90 substitute $$\\frac{3 b+2 b}{3}$$ = 90 Determine b $$\\frac{5 b}{3}$$ = 90 5b = 3 ∙ 90 5b = 270 $$\\frac{5 b}{5}$$ = $$\\frac{270}{5}$$ b = 54 a = $$\\frac{2 \\cdot 54}{3}$$ = 36 Determine a a = 36 b = 54 Question 32. $$\\overleftrightarrow{P R}$$ is a straight line. a : b = 2 : 3 we are given $$\\frac{a}{b}$$ = $$\\frac{2}{3}$$ we rewrite the ratio 3a = 2b cross-multiply $$\\frac{3 a}{3}$$ = $$\\frac{2 b}{3}$$ Divide bothsides by 3 a = $$\\frac{2 b}{3}$$ m∠RQS + m∠SQP = 180° ∠RQS and ∠SQP are supplementary b + a = 180 substitute b + $$\\frac{2 b}{3}$$ = 180 $$\\frac{3 b+2 b}{3}$$ = 180 $$\\frac{5 b}{3}$$ = 180 5b = 3 ∙ 180 5b = 540 $$\\frac{5 b}{5}$$ = $$\\frac{540}{5}$$ b = 108 a = $$\\frac{2 \\cdot 108}{3}$$ = 72 a = 72 b = 108 Solve. Question 33. The diagram shows the pattern on a stained glass window. $$\\overleftrightarrow{A C}$$ is a straight line. ∠EBD and ∠DBA are complementary angles and m∠DBA = 30°. Find the", "which satisfied $A + B + C = 180^\\circ$ 20. $\\cos C = \\frac{1}{2} = \\frac{a^2 + b^2 - c^2}{2ab} \\Rightarrow (1 + \\sqrt{3}^2) + 4 - c^2 = 2(1 + \\sqrt{3})$ $\\Rightarrow c^2 = 6 \\Rightarrow c = \\sqrt{6}$ $\\cos A = \\frac{b^2 + c^2 - a^2}{2bc} = \\frac{4 + 6 - (1 + \\sqrt{3})^2}{4\\sqrt{6}}$ $= \\frac{6 + 2\\sqrt{3}}{\\sqrt{6}} = \\sqrt{6} + \\sqrt{2} \\Rightarrow A = 75^\\circ$ Thus, $\\angle B = 45^\\circ$ 21. Greatest angle will be opposite to greatest side i.e. $\\sqrt{x^2 + xy + y^2}$ $\\cos\\theta = \\frac{x^2 + y^2 - x^2 - xy - y^2}{2.x.y} = -\\frac{1}{2} \\Rightarrow \\theta = 120^\\circ$ 22. Given sides are $2x + 3, x^2 + 3x + 3$ and $x^2 + 2x.$ Since lengths of sides is a positive quantity, therefore $x^2 + 2x > 0 \\Rightarrow x > 0$ This leads to the fact that $x^2 + 3x + 3$ will be greatest side. $\\cos\\theta = \\frac{(2x + 3)^2 + (x^2 + 2x)^2 - (x^2 + 3x + 3)^2}{2(2x + 3)(x^2 + 2x)}$ $= \\frac{4x^2 + 12x + 9 + x^4 + 4x^3 + 4x^2 - x^4 - 9x^2 - 9 - 6x^3 - 6x^2 - 18x}{4x^3 + 14x^2 + 12x}$ $= \\frac{-2x^3 -7x^2 - 6x}{2(2x^3 + 7x^2 + 6x)} = -\\frac{1}{2} = \\cos120^\\circ$ 23. Given, $3a =", "# $4 \\sin 72^\\circ \\sin 36^\\circ = \\sqrt 5$ How do I establish this and similar values of trigonometric functions? $$4 \\sin 72^\\circ \\sin 36^\\circ = \\sqrt 5$$ A key observation is $36 + 72 + 72 = 180$. This means $$\\sin 72 = \\sin (36 + 72) = \\sin 36 \\cos 72 + \\cos 36 \\sin 72$$ Using the identity $\\sin 72 = 2\\sin 36 \\cos 36$, we get $$2\\sin 36 \\cos 36 = \\sin 36 \\cos 72 + 2 \\sin 36 \\cos^2 36$$ Now notice $\\sin 36$ appears everywhere, so we can divide by it... $$2\\cos 36 = \\cos 72 + 2\\cos^2 36$$ Use the identity $\\cos 72 = 2 \\cos^2 36 - 1$, we get $$2\\cos 36 = 4\\cos^2 36 - 1$$ So $\\cos36$ is one of the solutions to $2x = 4x^2 - 1$. Using the quadratic formula, you see that the solutions are $$\\frac {2 \\pm \\sqrt{20}}4 = \\frac {1 \\pm \\sqrt 5}2$$ But you know that $\\cos 36$ must be positive, so you know the correct value. Now you can get $\\sin 72$ from any identity you wish, and multiply. How to get such values \"in general\": This is hard. The reason $36$ and $72$ are easy values is that a small integer multiple of them gives you a multiple of $180$.", "using the double and triple angle formulas $\\cos 3 \\theta = \\cos 2 \\theta$ becomes $4 {x}^{3} - 3 x = 2 {x}^{2} - 1$ $4 {x}^{3} - 2 {x}^{2} - 3 x + 1 = 0$ Since $\\theta = {72}^{\\circ} k$ we know $\\theta = 0$ works, so $x = \\cos 0 = 1$ is a root of this polynomial. We factor: $\\left(x - 1\\right) \\left(4 {x}^{2} + 2 x - 1\\right) = 0$ The quadratic has solutions $x = \\frac{1}{4} \\left(- 1 \\pm \\sqrt{5}\\right)$ $\\cos {144}^{\\circ}$ is the only negative root of the original cubic. $\\cos {0}^{\\circ}$ and $\\cos {72}^{\\circ}$ are the other two roots, both positive. $\\cos {144}^{\\circ} = - \\frac{1}{4} \\left(1 + \\sqrt{5}\\right)$ Like I said, negative one half the Golden Ratio $\\phi = \\frac{1 + \\sqrt{5}}{2.}$ $\\sin {144}^{\\circ}$ is positive. Unfortunately it has a nested square root; we live with it. $\\sin {144}^{\\circ} = \\sqrt{1 - {\\cos}^{2} {144}^{\\circ}} = \\sqrt{1 - \\frac{1}{16} \\left(6 + 2 \\sqrt{5}\\right)} = \\setminus \\sqrt{\\frac{1}{8} \\left(5 - \\sqrt{5}\\right)}$ Now we can finally apply the difference angle formula: $\\sin {24}^{\\circ} = \\sin \\left({144}^{\\circ} - {120}^{\\circ}\\right)$ $= \\sin {144}^{\\circ} \\cos {120}^{\\circ} - \\cos {144}^{\\circ} \\sin {120}^{\\circ}$ $= \\sqrt{\\frac{1}{8} \\left(5 - \\sqrt{5}\\right)} \\left(- \\frac{1}{2}\\right) - \\left(- \\frac{1}{4} \\left(1 + \\sqrt{5}\\right)\\right) \\left(\\frac{\\sqrt{3}}{2}\\right)$ = -1/8 sqrt{16/8(5-sqrt{5}) } + 1/8(sqrt{3} + sqrt{15})) $\\sin {24}^{\\circ} =", "B = \\frac{c^2 + a^2 − b^2}{2ca}$$ $$⇒ cos B = \\frac{4^2 + 2^2 − 3^2}{ 2(4)(2)} = 0.6875$$ $$\\boxed{⇒ B = 46.6^\\circ}$$ Now we use the formula for $$c^{2}$$ to find C: $$c^2 = a^2 + b^2 − 2ab\\;cos C ⇒ cos C = \\frac{a^2 + b^2 − c^2}{ 2ab}$$ $$⇒ cos C = \\frac{2^2 + 3^2 − 4^2}{ 2(2)(3)} = −0.25$$ $$\\boxed{⇒ C = 104.5^\\circ}$$ Thus, A = 180º −BC = 180º −46.6º −104.5º⇒ $$\\boxed{A = 28.9º}$$ . It may seem that there is always a solution in Case 4 (given all three sides), but that is not true, as the following example shows. Example 4 Case 4: Three sides. Solve the triangle ∆ABC given a = 2, b = 3, and c = 6. Solution: If we blindly try to use the Law of Cosines to find A, we get $$a^2 = b^2 + c^2 − 2bc cos A ⇒ cos A = \\frac{b^2 + c^2 − a^2}{2bc} = \\frac{3^2 + 6^2 − 2^2}{2(3)(6)} = 1.139 ,$$ which is impossible since |cos A| ≤ 1. Thus, there is $$\\boxed{\\text{no solution}}$$ . We could have saved ourselves some effort by recognizing that the length of one of the sides (c = 6) is greater than the sums of the lengths of the remaining sides (a =", "# Prove that $2\\cdot \\cos \\frac{72}{2}\\cdot \\cos \\frac{24}{2}+2\\cdot \\sin \\frac{96}{2}\\cdot \\sin \\frac{72}{2}=0.5$. Additional data added To solve it I have tried some options, where in one of them I applied product to sum formulas, which seemed to be very helpful, but didn't get the answer. Used these formulae: $$\\cos(a+b)+ \\cos(a-b)=2\\cdot \\cos(a)\\cdot \\cos(b)$$ $$−\\cos(a+b)+\\cos(a−b)=2\\sin(a)\\sin(b)$$ And now, I am stuck at this: $$\\cos(48)+\\cos(24)-\\cos(84)+\\cos(12)=0.5$$ Then tried to make the arguments similar using a double angle and a half angle formulae $$2\\cos^2(24)-1+\\cos(24)+\\sqrt{\\frac{1+\\cos(24)}{2}}+\\cos(84)=0.5$$ Anyway, I can't show that expression above really is 0.5. • I don’t understand. Why do you write $\\frac{72}2$ instead of $36$, for instance? – Lubin Jun 20 '17 at 20:13 • So was written on that sheet of paper, but I know that it is 36 :). – user36339 Jun 20 '17 at 20:15 • Do you know these formulea ? \\begin{eqnarray*} \\cos(a+b)+\\cos(a-b) = 2 \\cos(a) \\cos(b) \\\\ -\\cos(a+b)+\\cos(a-b) = 2 \\sin(a) \\sin(b) \\end{eqnarray*} – Donald Splutterwit Jun 20 '17 at 20:15 • No, I didn't.Thanks! – user36339 Jun 20 '17 at 20:50 • Note that if these are degrees, then the degree sign is obligatory; otherwise the unit is taken to be radians. For instance, $\\sin 90^\\circ = 1$, while $\\sin 90 = 0.89399\\ldots$. The degree sign can be typeset using ^\\circ. – Théophile Jun 20 '17 at", "= 3.5$$ weeks 22- Choice D is correct The difference of the file added, and the file deleted is: $$487,236 – 397,756 = 89,480, 987,345 + 89,480 = 1,076,825$$ 23- Choice C is correct 40 students did not have to go to summer school. $$50 – 10 = 40, \\frac{40}{50} = \\frac{4}{5}$$ 24- Choice B is correct Diameter $$= 2r ⇒4 = 2r⇒r = 2$$, Circumference $$= 2πr ⇒ C = 2π (2) ⇒ C = 4π$$ 25- Choice D is correct Write a proportion and solve. $$\\frac{\\frac{1}{4}}{1}= \\frac{60}{x}$$ $$x = \\frac{60}{\\frac{1}{4}} = 240$$ 26- Choice A is correct All angles in a triable add up to 180 degrees. $$90^\\circ + 60^\\circ = 150^\\circ ⇒ x = 180^\\circ − 150^\\circ = 30^\\circ$$ 27- Choice C is correct $$\\sqrt{81} = 9, \\sqrt{49} = 7, 9 × 7 = 63$$ 28- Choice D is correct If $$a = 5$$ then: $$b = \\frac{a^3}{5} -2 ⇒ b = \\frac{5^3}{5} -2 = 25 -2 = 23$$ 29- Choice B is correct An obtuse angle is an angle of greater than 90 degrees and less than 180 degrees. Only choice A is an obtuse angle. 30- Choice D is correct Only choice D is not equal to $$3^2$$ 31- Choice B is correct $$\\sqrt[4]{625} = 5, (5^4=5×5×5×5=625)$$ 32- Choice B is correct ($$r", "3 \\\\ 2ab = 36 \\\\ ab^2 + c = 42 \\end {cases} \\end{align} We can solve this by first substituting $$a=3$$ in the second equation, i.e. we have that $$2\\cdot 3 \\cdot b = 36$$ From which we have that $$b = \\frac {36}{6} = 6$$. We can then substitute the value $$b=64$$ into $$b^2 + c = 42$$, which gives that $$3\\cdot 36 + c = 42$$ And thus that $$c=42-108=-66$$. Therefore, $$3x^2 + 36x + 42 = 3(x+3)^2 - 66$$ • Example 4.3.6 Solve the equation $$3x^2 + 36x + 42 = 0.$$ We use the completed-square form of the quadratic (which we found in the previous example). This means that the values of $$x$$ we are interested in are those which satisfy $$3(x+3)^2 - 66 = 0.$$ We then add $$66$$ to both sides, which gets us the equivalent result $$3(x+3)^2 = 66.$$ Then we divide by $$3$$, which gives \\begin{align} (x+3)^2 &= \\frac {66}{3} \\\\ &= 22 \\end{align} Therefore, \\begin{align} x + 3 &= \\pm \\sqrt {22}. \\end{align} • Common mistake 4.3.1 Don’t forget the $$\\pm$$! The expression $$\\sqrt {x}$$ refers only to the positive square root of $$x$$! However, the negative one solves the equation just as well. Thus the values of $$x$$ which solve the equation are $$x = -3 \\pm", "could do $cos(66.5^{\\circ})$ it would would work the same way alright so $cos^2(22.5^{\\circ}) = \\frac{1 + cos2\\cdot22.5^{\\circ}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{1 + cos45^{\\circ}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{1 + \\frac{1}{\\sqrt{2}}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{\\sqrt{2} + 1}{2\\sqrt{2}} = \\sqrt{\\frac{\\sqrt{2} + 1}{2\\sqrt{2}}} \\Rightarrow \\sqrt{\\frac{\\sqrt{2} + 2}{4}} = cos(22.5^{\\circ})$ in exact form now do $cos(66.5^{\\circ})$ in similar fasion and multiply them together :)", "}\\qquad k\\in\\Bbb{Z}$$ - why is $4sin^{2}(\\frac{x}{2}+10^{\\circ})=\\frac{4}{cos^{2}(\\frac{x}{2}+10^{\\circ})}$ – aric May 11 at 18:55 @Foga- also you got: $1=4$ so i think something is wrong in your solution. – aric May 11 at 19:08 Nothing! You obtain $1=4$ by dividing by $0$... – Foga May 11 at 19:39 @Foga, But $x=180^\\circ n-20^\\circ$ will make the LHS $\\dfrac00$ – lab bhattacharjee May 12 at 6:55 I've edited the answer to make as clear as possible... – Foga May 12 at 11:09 $4sin^2(\\frac{x + 20^\\circ}2) = 4\\cdot1/2\\cdot(1-\\cos(x+20^\\circ)) = 2-2\\cos(x+20^\\circ)$ Applying componendo and dividendo on $\\frac{tg(x+20^{\\circ })-sin(x+20^{\\circ })}{tg(x+20^{\\circ })+sin(x+20^{\\circ })}=4sin^{2}(\\frac{x}{2}+10^{\\circ })$ , $-\\sec(x+20^\\circ) = \\frac{2-2\\cos(x+20^\\circ)+1}{2-2\\cos(x+20^\\circ) -1}$, $cos(x+20^\\circ)\\ne1/2$ $-\\sec(x+20^\\circ)\\cdot(1-{2\\cos(x+20^\\circ)}) = {3-2\\cos(x+20^\\circ)}$ $-\\sec(x+20^\\circ)+2=3-2\\cos(x+20^\\circ)$ $2cos(x+20^\\circ)-\\sec(x+20^\\circ)-1=0$ $2cos^2(x+20^\\circ)-cos(x+20^\\circ)-1=0$ Can you solve it further? -", "Prove that $\\dfrac{1}{2}\\sqrt{4\\sin^2(36^{\\circ})-1} = \\cos(72^{\\circ})$ Prove that $\\dfrac{1}{2}\\sqrt{4\\sin^2(36^{\\circ})-1} = \\cos(72^{\\circ})$ This question seemed pretty simple so I first started out by turning the left-hand side into terms of $\\cos(x)$. We have $\\dfrac{1}{2}\\sqrt{4(1-\\cos^2(36^{\\circ}))-1} = \\dfrac{1}{2}\\sqrt{3-4\\cos^2(36^{\\circ})}$. Then I would use $\\cos(72^{\\circ}) = 2\\cos^2(36^{\\circ})-1$. Thus, $\\dfrac{1}{2}\\sqrt{3-4\\cos^2(36^{\\circ})} = \\dfrac{1}{2}\\sqrt{3-2(\\cos(72^{\\circ})+1)} = \\dfrac{1}{2}\\sqrt{1-2\\cos(72^{\\circ})}$. I get stuck here. • multiply the equality by $2$, square both sides and use your formula for $\\cos(72^{\\circ})$. – Surb Jan 4 '16 at 21:10 • I still don't see how that proves it. – Jacob Willis Jan 4 '16 at 21:13 Note that we have the system of equations $$\\begin{cases}\\cos 72^\\circ&=2\\cos^2 36^\\circ-1,\\\\\\cos 36^\\circ&=1-2\\cos^2 72^\\circ.\\end{cases}$$ Adding the two equations together yields $$\\cos 72^\\circ+\\cos 36^\\circ=2(\\cos^236^\\circ-\\cos^272^\\circ)\\implies\\cos 36^\\circ-\\cos 72^\\circ=\\frac12.$$ Now subtracting the first equation from the second gives $$\\cos 36^\\circ-\\cos 72^\\circ=2-2(\\cos^236^\\circ+\\cos^272^\\circ)\\implies\\cos^236^\\circ+\\cos^272^\\circ=\\frac34.$$ To finish, note that this equality rearranges to $$1-\\cos^2 36^\\circ-\\cos^272^\\circ=\\sin^236^\\circ-\\cos^272^\\circ=\\frac14.$$ Hence $4\\sin^236^\\circ-1=4\\cos^272^\\circ$, and taking the square root of both sides and dividing by $2$ gives the desired result. • As a sidenote, the main idea behind these manipulations (computing $\\cos 36^\\circ-\\cos 72^\\circ$) is actually AHSME 1975 #30 - remembering the solution to that problem was very helpful in solving this one. – David Altizio Jan 4 '16 at 21:30 We need $$4\\sin^236^\\circ-4\\cos^272^\\circ=1$$ Using $\\cos2A=1-2\\sin^2A=2\\cos^2A-1,$ $$\\iff2(1-\\cos72^\\circ)-2(1+\\cos144^\\circ)=1$$ As $\\cos(180^\\circ-B)=-\\cos B,$ $$\\iff\\cos36^\\circ-\\cos72^\\circ=\\dfrac12$$", "subject of this formula: $$a^2-b^2-c^2=-2bc\\cdot\\cos{A}$$ $$\\cos{A}=\\frac{a^2-b^2-c^2}{-2bc}$$ $$\\cos{A}=\\frac{b^2+c^2-a^2}{2bc}$$ now you just need to plug in the values, but notice that this formula is finding $A$ while you are looking for the angle B, so you could swap the a's for the b's (because the letters in these formulae are arbitrary, all you need to keep consistent is the lowercase letter side length with its uppercase letter opposite angle). So $$\\cos{B} = \\frac{a^2+c^2-b^2}{2ac}$$ 3. Oct 28, 2013 ### Staff: Mentor mathopressor,", "$kx$ and, dividing these by $k$, at last the values of the angle $x$. Example 1. $\\sin{x}\\cos{x}+\\frac{1}{4}=0$ $2\\sin{x}\\cos{x}=-\\frac{1}{2}$ $\\sin{2x}=-\\frac{1}{2}$ $\\sin{2x}=\\sin{210^{\\circ}}$ $2x=210^{\\circ}+n\\cdot 360^{\\circ}\\quad\\lor\\quad 2x=180^{\\circ}-210^{\\circ}+n% \\cdot 360^{\\circ}$ $x=105^{\\circ}+n\\cdot 180^{\\circ}\\quad\\lor\\quad x=-15^{\\circ}+n\\cdot 180^{\\circ}$ On the third line one used the double angle formula of sine. It may happen that the form (1) cannot be attained, but instead e.g. the form $\\displaystyle f(kx)=f(\\alpha),$ (2) where $x$ can be in $\\alpha$. Example 2. $\\sin 2x=\\cos 3x$ $\\cos(90^{\\circ}-2x)=\\cos 3x$ $90^{\\circ}-2x=3x+n\\cdot 360^{\\circ}\\quad\\lor\\quad 90^{\\circ}-2x=-3x+n\\cdot 360% ^{\\circ}$ $-2x-3x=-90^{\\circ}+n\\cdot 360^{\\circ}\\quad\\lor\\quad-2x+3x=-90^{\\circ}+n\\cdot 3% 60^{\\circ}$ $x=18^{\\circ}+n\\cdot 72^{\\circ}\\quad\\lor\\quad x=-90^{\\circ}+n\\cdot 360^{\\circ}$ On the second line one of the complement formulas was utilized. Example 3. $\\sin{2x}=-\\sin{3x}$ $\\sin{2x}=\\sin(-3x)$ $2x=-3x+n\\cdot 360^{\\circ}\\quad\\lor\\quad 2x=180^{\\circ}-(-3x)+n\\cdot 360^{\\circ}$ $x=n\\cdot 72^{\\circ}\\quad\\lor\\quad x=180^{\\circ}+n\\cdot 360^{\\circ}$ On the second line the opposite angle formula (http://planetmath.org/GoniometricFormulae) of sine was utilized. Example 4. $\\cos{2x}=-\\cos{3x}$ $\\cos{2x}=\\cos(180^{\\circ}-3x)$ $2x=\\pm(180^{\\circ}-3x)+n\\cdot 360^{\\circ}$ $x=36^{\\circ}+n\\cdot 72^{\\circ}\\quad\\lor\\quad x=180^{\\circ}+n\\cdot 360^{\\circ}$ On the second line the supplement formula (http://planetmath.org/GoniometricFormulae) of sine was utilized. Title trigonometric equations TrigonometricEquations 2013-03-22 17:46:25 2013-03-22 17:46:25 pahio (2872) pahio (2872) 12 pahio (2872) Definition msc 33B10 msc 26A09 msc 00A35 Equation trigonometric equation", "Brian Bi ## Section 15.5. Constructions with Ruler and Compass Exercise 15.5.1 \\begin{align*} \\cos 15^\\circ &= \\cos(45^\\circ - 30^\\circ) \\\\ &= \\cos 45^\\circ \\cos 30^\\circ + \\sin 45^\\circ \\sin 30^\\circ \\\\ &= \\frac{\\sqrt{2}}{2} \\frac{\\sqrt{3}}{2} + \\frac{\\sqrt{2}}{2} \\frac{1}{2} \\\\ &= \\frac{\\sqrt{6} + \\sqrt{2}}{4} \\end{align*} Exercise 15.5.2 1. Observe that $$\\cos 72^\\circ = \\cos 288^\\circ = 2\\cos^2 144^\\circ - 1 = 2(2 \\cos^2 72^\\circ - 1)^2 - 1$$. So, if we put $$x = \\cos 72^\\circ$$ then we have $$x = 2(2x^2 - 1)^2 - 1$$ which can be expanded to yield a quartic equation for $$x$$. However, two of the roots $$x_1, x_2$$ of this quartic are obviously just the roots of $$x = 2x^2 - 1$$, which lie in a quadratic extension $$\\Q(\\sqrt{d})$$ of the rationals. Dividing out $$(x - x_1)(x - x_2)$$ from the quartic yields a quadratic polynomial with coefficients in $$\\Q(\\sqrt{d})$$, so the remaining two roots lie in some quadratic extension of this field. One of these four roots is $$\\cos 72^\\circ$$, so by Theorem 15.5.6, the latter is constructible. This implies that the regular pentagon is constructible. 2. It is possible to devise the construction by writing out the explicit form of $$\\cos 72^\\circ$$, which is $$\\frac{\\sqrt{5} - 1}{4}$$, and applying Lemma 15.5.11, constructing first 5, then $$\\sqrt{5}$$, then $$\\sqrt{5} - 1$$, and finally", "# The angle $168^\\circ$ is constructible Prove that the angle $\\theta=168^\\circ$ is constructible using a straightedge and a compass. It is enough to show that the number $\\cos\\theta$ is constructible, and WolframAlpha gave $\\cos\\theta=\\frac18 (1-\\sqrt5)-\\frac14 \\sqrt{{\\frac32} (5+\\sqrt5)}$. I think after that it is immediate because we first add $\\sqrt5$ to $\\mathbb{Q}$ and then add $\\sqrt{{\\frac32} (5+\\sqrt5)}$, so both extensions are of degree $2$. But then how can we find that representation of $\\cos\\theta$? And is there a nicer way? EDIT I found online a cool criterion: can we just state that the order of $168^\\circ$ is $15$, since $168*15$ is divisible by 360, and then since $15=3\\times5$ is a product of Fermat primes we are done? And if so, Do we have to check all numbers $1,...,15$ to see $15$ is indeed the order? • You know(?) that you can construct a regular hexagon ($60°$), and a regular pentagon ($72°$). Then you can also construct $12° = 72° - 60°$. Constructing $12°$ is clearly equivalent to constructing $168° = 180° - 12°$. – Daniel Fischer Feb 9 '14 at 21:51 The canonical pentagon is indeed constructible, as mentioned by the OP, and so is its angle $$\\varphi=\\frac{3\\pi}{5}=\\frac{540^\\circ}{5}=108^\\circ$$ Also $60^\\circ$ is the angle of equilateral triangle, and so it is another constructible angle. Hence $\\vartheta=108^\\circ+60^\\circ=168^\\circ$ is also constructible. The", "## A community for students. Sign up today! Here's the question you clicked on: ## haleyking345 one year ago Solve the equation: a.) 2 cos^2x-cos x=1 • This Question is Closed 1. Marx let's define $t:= \\cos(x)$ we have the quadratic equation: $2t^2-t=1$ can you find it's solutions? 2. haleyking345 Would it be cos=1? Which means cos=0 degrees or 360 degrees? 3. Marx there are two solutions, $\\cos(x)=1$ and $\\cos(x)=-\\frac{1}{2}$ so x can be either $360^\\circ{} k$ (for every integer k) or $120^\\circ{}+ 360^{\\circ}k$ or $240^\\circ{}+ 360^{\\circ}k$ #### Ask your own question Ask a Question Find more explanations on OpenStudy", "draw(B--C--(1,0)--cycle,blue+1bp); pair P = 0.73*C+0.27*B; draw(O--P); dot(P); label(\"1\",(C+P)/2,NE); label(\"r\",(B+P)/2,NE); [/asy]$ then using the Pythagorean theorem we have: $(r+1)^2=(2s)^2+(r-1)^2$ which is equivalent to: $r^2+2r+1=4s^2+r^2-2r+1$ subtracting $r^2-2r+1$ from both sides $4r=4s^2$ solving for s we get: $$s=\\sqrt{r}$$ next we can find the area of the frustum and of the sphere and we know $V_{\\text{frustum}}=2V_{\\text{sphere}}$ so we can solve for $s$ using $V_{\\text{frustum}}=\\frac{\\pi*h}{3}(R^2+r^2+Rr)$ we get: $$V_{\\text{frustum}}=\\frac{\\pi*2\\sqrt{r}}{3}(r^2+r+1)$$ using $V_{\\text{sphere}}=\\dfrac{4r^{3}\\pi}{3}$ we get $$V_{\\text{sphere}}=\\dfrac{4(\\sqrt{r})^{3}\\pi}{3}$$ so we have: $$\\frac{\\pi*2\\sqrt{r}}{3}(r^2+r+1)=2*\\dfrac{4(\\sqrt{r})^{3}\\pi}{3}$$ dividing by $\\frac{2\\pi*\\sqrt{r}}{3}$ we get $$r^2+r+1=4r$$ which is equivalent to $$r^2-3r+1=0$$ $r=\\dfrac{3\\pm\\sqrt{(-3)^2-4*1*1}}{2*1}$ so $$r=\\dfrac{3+\\sqrt{5}}{2}\\longrightarrow \\boxed{E}$$ Let's once again look at the cross section of the frustum. Let the angle from the center of the sphere to a point on the circumference of the bottom circle be $\\theta.$ This implies that the angle from the center of the sphere to a point on the circumference of the top circle is $90 - \\theta.$ Hence the bottom radius is $r\\tan{\\theta}$ and the top radius is $\\frac {r}{\\tan {\\theta}}.$ This means that the radio between the bottom radius and top radius is $(\\tan {\\theta})^2.$ Using the frustum volume formula, we find that the are of this figure is $\\frac{2\\pi r}{3}(r^2(\\tan {\\theta})^2 + r^2 + \\frac {r^2} {(\\tan {\\theta})^2}).$ We can equate this to $\\frac {8\\pi*r^3} 3.$ Simplifying, we are left with a quadratic conveniently in $(\\tan {\\theta})^2.$ The quadratic" ]

[ "us all the multiples of ${72}^{\\circ}$. Let's go at it a different way, using the double and triple angle formulas: Solve $\\cos \\left(2 x\\right) = \\cos \\left(3 x\\right)$ for $\\cos \\left(x\\right)$ $2 {\\cos}^{2} x - 1 = 4 {\\cos}^{3} x - 3 \\cos x$ Let's abbreviate $c = \\cos x$ $2 {c}^{2} - 1 = 4 {c}^{3} - 3 c$ $4 {c}^{3} - 2 {c}^{2} - 3 c + 1 = 0$ That's a cubic equation, but one of the easy ones. All the cosines of multiples of ${72}^{\\circ}$ will be roots of this equation. That includes $\\cos 0 = 1$. So $c - 1$ is a factor, and with long division we get $\\left(c - 1\\right) \\left(4 {c}^{2} + 2 c - 1\\right) = 0$ $\\cos {0}^{\\circ} ,$ $\\cos {72}^{\\circ} ,$ and $\\cos {144}^{\\circ}$ are all different, so they're the three roots to this cubic equation. All other multiples of ${72}^{\\circ}$ share one of these cosines. $c = \\frac{1}{4} \\left(- 1 \\setminus \\pm \\sqrt{5}\\right)$ Clearly $\\cos {72}^{\\circ}$ is the positive one and $\\cos {144}^{\\circ}$ is the negative one: $\\cos {72}^{\\circ} = \\frac{1}{4} \\left(\\setminus \\sqrt{5} - 1\\right)$ $\\cos {144}^{\\circ} = \\frac{1}{4} \\left(- \\setminus \\sqrt{5} - 1\\right)$ $\\cos {36}^{\\circ} = - \\setminus \\cos \\left({180}^{\\circ} - {36}^{\\circ}\\right) = - \\cos {144}^{\\circ} = \\frac{1}{4} \\left(1 + \\setminus \\sqrt{5}\\right)$ That's half the", "58: Line 92: - $c^{2} = a^{2} + b^{2} - 2ab \\cos C$ + ::$c^{2} = a^{2} + b^{2} - 2ab \\cos C$ [[Image:SAS triangle.jpg]] [[Image:SAS triangle.jpg]] Line 67: Line 101: - $c^{2} = 6^{2} + (6 \\sqrt{2})^{2} -2 (6) (6 \\sqrt{2}) \\cos 45^\\circ$ + :$c^{2} = 6^{2} + (6 \\sqrt{2})^{2} -2 (6) (6 \\sqrt{2}) \\cos 45^\\circ$ Simplify for Simplify for - $c^{2} =36 + 36 (2) - 72 \\sqrt{2}) \\cos 45^\\circ$ + ::$c^{2} =36 + 36 (2) - 72 \\sqrt{2}) \\cos 45^\\circ$ Since $\\cos 45^\\circ = \\frac{1}{\\sqrt{2}}$, substitution gives us Since $\\cos 45^\\circ = \\frac{1}{\\sqrt{2}}$, substitution gives us - $c^{2} =36 + 36 (2) - 72 \\sqrt{2} (\\frac{1}{\\sqrt{2}})$ + ::$c^{2} =36 + 36 (2) - 72 \\sqrt{2} (\\frac{1}{\\sqrt{2}})$ Simplify for Simplify for - $c^{2} =36 + 72 - 72$ + ::$c^{2} =36 + 72 - 72$ - $c^{2} =36$ + ::$c^{2} =36$ Taking the square root of both sides gives us Taking the square root of both sides gives us - $c =6$ + ::$c =6$ ---- ---- Line 93: Line 127: - $b^{2} = a^{2} + c^{2} - 2ab \\cos B$ + ::$b^{2} = a^{2} + c^{2} - 2ab \\cos B$ Substituting in the appropriate side lengths gives us Substituting in the appropriate side lengths gives us - $(6 \\sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6)", "# $4 \\sin 72^\\circ \\sin 36^\\circ = \\sqrt 5$ How do I establish this and similar values of trigonometric functions? $$4 \\sin 72^\\circ \\sin 36^\\circ = \\sqrt 5$$ A key observation is $36 + 72 + 72 = 180$. This means $$\\sin 72 = \\sin (36 + 72) = \\sin 36 \\cos 72 + \\cos 36 \\sin 72$$ Using the identity $\\sin 72 = 2\\sin 36 \\cos 36$, we get $$2\\sin 36 \\cos 36 = \\sin 36 \\cos 72 + 2 \\sin 36 \\cos^2 36$$ Now notice $\\sin 36$ appears everywhere, so we can divide by it... $$2\\cos 36 = \\cos 72 + 2\\cos^2 36$$ Use the identity $\\cos 72 = 2 \\cos^2 36 - 1$, we get $$2\\cos 36 = 4\\cos^2 36 - 1$$ So $\\cos36$ is one of the solutions to $2x = 4x^2 - 1$. Using the quadratic formula, you see that the solutions are $$\\frac {2 \\pm \\sqrt{20}}4 = \\frac {1 \\pm \\sqrt 5}2$$ But you know that $\\cos 36$ must be positive, so you know the correct value. Now you can get $\\sin 72$ from any identity you wish, and multiply. How to get such values \"in general\": This is hard. The reason $36$ and $72$ are easy values is that a small integer multiple of them gives you a multiple of $180$.", "could do $cos(66.5^{\\circ})$ it would would work the same way alright so $cos^2(22.5^{\\circ}) = \\frac{1 + cos2\\cdot22.5^{\\circ}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{1 + cos45^{\\circ}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{1 + \\frac{1}{\\sqrt{2}}}{2} \\Rightarrow cos^2(22.5^{\\circ}) = \\frac{\\sqrt{2} + 1}{2\\sqrt{2}} = \\sqrt{\\frac{\\sqrt{2} + 1}{2\\sqrt{2}}} \\Rightarrow \\sqrt{\\frac{\\sqrt{2} + 2}{4}} = cos(22.5^{\\circ})$ in exact form now do $cos(66.5^{\\circ})$ in similar fasion and multiply them together :)", "Prove that $\\dfrac{1}{2}\\sqrt{4\\sin^2(36^{\\circ})-1} = \\cos(72^{\\circ})$ Prove that $\\dfrac{1}{2}\\sqrt{4\\sin^2(36^{\\circ})-1} = \\cos(72^{\\circ})$ This question seemed pretty simple so I first started out by turning the left-hand side into terms of $\\cos(x)$. We have $\\dfrac{1}{2}\\sqrt{4(1-\\cos^2(36^{\\circ}))-1} = \\dfrac{1}{2}\\sqrt{3-4\\cos^2(36^{\\circ})}$. Then I would use $\\cos(72^{\\circ}) = 2\\cos^2(36^{\\circ})-1$. Thus, $\\dfrac{1}{2}\\sqrt{3-4\\cos^2(36^{\\circ})} = \\dfrac{1}{2}\\sqrt{3-2(\\cos(72^{\\circ})+1)} = \\dfrac{1}{2}\\sqrt{1-2\\cos(72^{\\circ})}$. I get stuck here. • multiply the equality by $2$, square both sides and use your formula for $\\cos(72^{\\circ})$. – Surb Jan 4 '16 at 21:10 • I still don't see how that proves it. – Jacob Willis Jan 4 '16 at 21:13 Note that we have the system of equations $$\\begin{cases}\\cos 72^\\circ&=2\\cos^2 36^\\circ-1,\\\\\\cos 36^\\circ&=1-2\\cos^2 72^\\circ.\\end{cases}$$ Adding the two equations together yields $$\\cos 72^\\circ+\\cos 36^\\circ=2(\\cos^236^\\circ-\\cos^272^\\circ)\\implies\\cos 36^\\circ-\\cos 72^\\circ=\\frac12.$$ Now subtracting the first equation from the second gives $$\\cos 36^\\circ-\\cos 72^\\circ=2-2(\\cos^236^\\circ+\\cos^272^\\circ)\\implies\\cos^236^\\circ+\\cos^272^\\circ=\\frac34.$$ To finish, note that this equality rearranges to $$1-\\cos^2 36^\\circ-\\cos^272^\\circ=\\sin^236^\\circ-\\cos^272^\\circ=\\frac14.$$ Hence $4\\sin^236^\\circ-1=4\\cos^272^\\circ$, and taking the square root of both sides and dividing by $2$ gives the desired result. • As a sidenote, the main idea behind these manipulations (computing $\\cos 36^\\circ-\\cos 72^\\circ$) is actually AHSME 1975 #30 - remembering the solution to that problem was very helpful in solving this one. – David Altizio Jan 4 '16 at 21:30 We need $$4\\sin^236^\\circ-4\\cos^272^\\circ=1$$ Using $\\cos2A=1-2\\sin^2A=2\\cos^2A-1,$ $$\\iff2(1-\\cos72^\\circ)-2(1+\\cos144^\\circ)=1$$ As $\\cos(180^\\circ-B)=-\\cos B,$ $$\\iff\\cos36^\\circ-\\cos72^\\circ=\\dfrac12$$", "Brian Bi ## Section 15.5. Constructions with Ruler and Compass Exercise 15.5.1 \\begin{align*} \\cos 15^\\circ &= \\cos(45^\\circ - 30^\\circ) \\\\ &= \\cos 45^\\circ \\cos 30^\\circ + \\sin 45^\\circ \\sin 30^\\circ \\\\ &= \\frac{\\sqrt{2}}{2} \\frac{\\sqrt{3}}{2} + \\frac{\\sqrt{2}}{2} \\frac{1}{2} \\\\ &= \\frac{\\sqrt{6} + \\sqrt{2}}{4} \\end{align*} Exercise 15.5.2 1. Observe that $$\\cos 72^\\circ = \\cos 288^\\circ = 2\\cos^2 144^\\circ - 1 = 2(2 \\cos^2 72^\\circ - 1)^2 - 1$$. So, if we put $$x = \\cos 72^\\circ$$ then we have $$x = 2(2x^2 - 1)^2 - 1$$ which can be expanded to yield a quartic equation for $$x$$. However, two of the roots $$x_1, x_2$$ of this quartic are obviously just the roots of $$x = 2x^2 - 1$$, which lie in a quadratic extension $$\\Q(\\sqrt{d})$$ of the rationals. Dividing out $$(x - x_1)(x - x_2)$$ from the quartic yields a quadratic polynomial with coefficients in $$\\Q(\\sqrt{d})$$, so the remaining two roots lie in some quadratic extension of this field. One of these four roots is $$\\cos 72^\\circ$$, so by Theorem 15.5.6, the latter is constructible. This implies that the regular pentagon is constructible. 2. It is possible to devise the construction by writing out the explicit form of $$\\cos 72^\\circ$$, which is $$\\frac{\\sqrt{5} - 1}{4}$$, and applying Lemma 15.5.11, constructing first 5, then $$\\sqrt{5}$$, then $$\\sqrt{5} - 1$$, and finally", "using the double and triple angle formulas $\\cos 3 \\theta = \\cos 2 \\theta$ becomes $4 {x}^{3} - 3 x = 2 {x}^{2} - 1$ $4 {x}^{3} - 2 {x}^{2} - 3 x + 1 = 0$ Since $\\theta = {72}^{\\circ} k$ we know $\\theta = 0$ works, so $x = \\cos 0 = 1$ is a root of this polynomial. We factor: $\\left(x - 1\\right) \\left(4 {x}^{2} + 2 x - 1\\right) = 0$ The quadratic has solutions $x = \\frac{1}{4} \\left(- 1 \\pm \\sqrt{5}\\right)$ $\\cos {144}^{\\circ}$ is the only negative root of the original cubic. $\\cos {0}^{\\circ}$ and $\\cos {72}^{\\circ}$ are the other two roots, both positive. $\\cos {144}^{\\circ} = - \\frac{1}{4} \\left(1 + \\sqrt{5}\\right)$ Like I said, negative one half the Golden Ratio $\\phi = \\frac{1 + \\sqrt{5}}{2.}$ $\\sin {144}^{\\circ}$ is positive. Unfortunately it has a nested square root; we live with it. $\\sin {144}^{\\circ} = \\sqrt{1 - {\\cos}^{2} {144}^{\\circ}} = \\sqrt{1 - \\frac{1}{16} \\left(6 + 2 \\sqrt{5}\\right)} = \\setminus \\sqrt{\\frac{1}{8} \\left(5 - \\sqrt{5}\\right)}$ Now we can finally apply the difference angle formula: $\\sin {24}^{\\circ} = \\sin \\left({144}^{\\circ} - {120}^{\\circ}\\right)$ $= \\sin {144}^{\\circ} \\cos {120}^{\\circ} - \\cos {144}^{\\circ} \\sin {120}^{\\circ}$ $= \\sqrt{\\frac{1}{8} \\left(5 - \\sqrt{5}\\right)} \\left(- \\frac{1}{2}\\right) - \\left(- \\frac{1}{4} \\left(1 + \\sqrt{5}\\right)\\right) \\left(\\frac{\\sqrt{3}}{2}\\right)$ = -1/8 sqrt{16/8(5-sqrt{5}) } + 1/8(sqrt{3} + sqrt{15})) $\\sin {24}^{\\circ} =", "which satisfied $A + B + C = 180^\\circ$ 20. $\\cos C = \\frac{1}{2} = \\frac{a^2 + b^2 - c^2}{2ab} \\Rightarrow (1 + \\sqrt{3}^2) + 4 - c^2 = 2(1 + \\sqrt{3})$ $\\Rightarrow c^2 = 6 \\Rightarrow c = \\sqrt{6}$ $\\cos A = \\frac{b^2 + c^2 - a^2}{2bc} = \\frac{4 + 6 - (1 + \\sqrt{3})^2}{4\\sqrt{6}}$ $= \\frac{6 + 2\\sqrt{3}}{\\sqrt{6}} = \\sqrt{6} + \\sqrt{2} \\Rightarrow A = 75^\\circ$ Thus, $\\angle B = 45^\\circ$ 21. Greatest angle will be opposite to greatest side i.e. $\\sqrt{x^2 + xy + y^2}$ $\\cos\\theta = \\frac{x^2 + y^2 - x^2 - xy - y^2}{2.x.y} = -\\frac{1}{2} \\Rightarrow \\theta = 120^\\circ$ 22. Given sides are $2x + 3, x^2 + 3x + 3$ and $x^2 + 2x.$ Since lengths of sides is a positive quantity, therefore $x^2 + 2x > 0 \\Rightarrow x > 0$ This leads to the fact that $x^2 + 3x + 3$ will be greatest side. $\\cos\\theta = \\frac{(2x + 3)^2 + (x^2 + 2x)^2 - (x^2 + 3x + 3)^2}{2(2x + 3)(x^2 + 2x)}$ $= \\frac{4x^2 + 12x + 9 + x^4 + 4x^3 + 4x^2 - x^4 - 9x^2 - 9 - 6x^3 - 6x^2 - 18x}{4x^3 + 14x^2 + 12x}$ $= \\frac{-2x^3 -7x^2 - 6x}{2(2x^3 + 7x^2 + 6x)} = -\\frac{1}{2} = \\cos120^\\circ$ 23. Given, $3a =", "}\\qquad k\\in\\Bbb{Z}$$ - why is $4sin^{2}(\\frac{x}{2}+10^{\\circ})=\\frac{4}{cos^{2}(\\frac{x}{2}+10^{\\circ})}$ – aric May 11 at 18:55 @Foga- also you got: $1=4$ so i think something is wrong in your solution. – aric May 11 at 19:08 Nothing! You obtain $1=4$ by dividing by $0$... – Foga May 11 at 19:39 @Foga, But $x=180^\\circ n-20^\\circ$ will make the LHS $\\dfrac00$ – lab bhattacharjee May 12 at 6:55 I've edited the answer to make as clear as possible... – Foga May 12 at 11:09 $4sin^2(\\frac{x + 20^\\circ}2) = 4\\cdot1/2\\cdot(1-\\cos(x+20^\\circ)) = 2-2\\cos(x+20^\\circ)$ Applying componendo and dividendo on $\\frac{tg(x+20^{\\circ })-sin(x+20^{\\circ })}{tg(x+20^{\\circ })+sin(x+20^{\\circ })}=4sin^{2}(\\frac{x}{2}+10^{\\circ })$ , $-\\sec(x+20^\\circ) = \\frac{2-2\\cos(x+20^\\circ)+1}{2-2\\cos(x+20^\\circ) -1}$, $cos(x+20^\\circ)\\ne1/2$ $-\\sec(x+20^\\circ)\\cdot(1-{2\\cos(x+20^\\circ)}) = {3-2\\cos(x+20^\\circ)}$ $-\\sec(x+20^\\circ)+2=3-2\\cos(x+20^\\circ)$ $2cos(x+20^\\circ)-\\sec(x+20^\\circ)-1=0$ $2cos^2(x+20^\\circ)-cos(x+20^\\circ)-1=0$ Can you solve it further? -", "\\pm 1}{4}$ where we can choose the correct value of the four possible values by noting that, because 72° is between 45° and 90°, $$\\cos 72^\\circ$$ must lie between $$1 / \\sqrt{2}$$ and 0. Because $$\\cos 72^\\circ$$ is positive, we choose the \"+\" before the fraction, and because $$\\cos 72^\\circ$$ is less than $$1 / \\sqrt{2}$$, which in turn is less than $$\\frac{\\sqrt{5}+1}{4}$$, we choose the \"-\" in the numerator: $\\cos 72^\\circ = \\frac{\\sqrt{5}-1}{4}$ Constructing a regular pentagon So we can construct $$\\cos 72^\\circ$$. For example, the diagonal of a 1-by-2 rectangle is $$\\sqrt{5}$$. We could cut off one unit from a segment of length $$\\sqrt{5}$$, then divide the segment of length $$\\sqrt{5}-1$$ into four pieces of length $$\\frac{\\sqrt{5} -1} {4}$$. (Or we could construct the appropriate solution to the equation $$4x^2 +2x -1 = 0$$. See my post on Solving quadratic equations via geometric construction.) Construct a unit circle centered at O, and construct a radius $$\\overline{OA}$$. Construct the point B on $$\\overline{OA}$$ so that $$\\overline{OB}$$ has length $$\\cos 72^\\circ$$. If C is a point on the circle so that $$\\overline{BC}$$ is perpendicular to $$\\overline{OA}$$, then $$\\angle COA$$ is a 72° angle, and both A and C are vertices of a regular pentagon inscribed in the circle.", "$$\\frac{2}{3}$$ we rewrite the ratio 3a = 2b cross-multiply $$\\frac{3 a}{3}$$ = $$\\frac{2 b}{3}$$ Divide both sides by 3 a = $$\\frac{2 b}{3}$$ m∠RQS + m∠SQP = 90° ∠RQS and ∠SQP are complementary b + a = 90 b + $$\\frac{2 b}{3}$$ = 90 substitute $$\\frac{3 b+2 b}{3}$$ = 90 Determine b $$\\frac{5 b}{3}$$ = 90 5b = 3 ∙ 90 5b = 270 $$\\frac{5 b}{5}$$ = $$\\frac{270}{5}$$ b = 54 a = $$\\frac{2 \\cdot 54}{3}$$ = 36 Determine a a = 36 b = 54 Question 32. $$\\overleftrightarrow{P R}$$ is a straight line. a : b = 2 : 3 we are given $$\\frac{a}{b}$$ = $$\\frac{2}{3}$$ we rewrite the ratio 3a = 2b cross-multiply $$\\frac{3 a}{3}$$ = $$\\frac{2 b}{3}$$ Divide bothsides by 3 a = $$\\frac{2 b}{3}$$ m∠RQS + m∠SQP = 180° ∠RQS and ∠SQP are supplementary b + a = 180 substitute b + $$\\frac{2 b}{3}$$ = 180 $$\\frac{3 b+2 b}{3}$$ = 180 $$\\frac{5 b}{3}$$ = 180 5b = 3 ∙ 180 5b = 540 $$\\frac{5 b}{5}$$ = $$\\frac{540}{5}$$ b = 108 a = $$\\frac{2 \\cdot 108}{3}$$ = 72 a = 72 b = 108 Solve. Question 33. The diagram shows the pattern on a stained glass window. $$\\overleftrightarrow{A C}$$ is a straight line. ∠EBD and ∠DBA are complementary angles and m∠DBA = 30°. Find the", "# Prove that $2\\cdot \\cos \\frac{72}{2}\\cdot \\cos \\frac{24}{2}+2\\cdot \\sin \\frac{96}{2}\\cdot \\sin \\frac{72}{2}=0.5$. Additional data added To solve it I have tried some options, where in one of them I applied product to sum formulas, which seemed to be very helpful, but didn't get the answer. Used these formulae: $$\\cos(a+b)+ \\cos(a-b)=2\\cdot \\cos(a)\\cdot \\cos(b)$$ $$−\\cos(a+b)+\\cos(a−b)=2\\sin(a)\\sin(b)$$ And now, I am stuck at this: $$\\cos(48)+\\cos(24)-\\cos(84)+\\cos(12)=0.5$$ Then tried to make the arguments similar using a double angle and a half angle formulae $$2\\cos^2(24)-1+\\cos(24)+\\sqrt{\\frac{1+\\cos(24)}{2}}+\\cos(84)=0.5$$ Anyway, I can't show that expression above really is 0.5. • I don’t understand. Why do you write $\\frac{72}2$ instead of $36$, for instance? – Lubin Jun 20 '17 at 20:13 • So was written on that sheet of paper, but I know that it is 36 :). – user36339 Jun 20 '17 at 20:15 • Do you know these formulea ? \\begin{eqnarray*} \\cos(a+b)+\\cos(a-b) = 2 \\cos(a) \\cos(b) \\\\ -\\cos(a+b)+\\cos(a-b) = 2 \\sin(a) \\sin(b) \\end{eqnarray*} – Donald Splutterwit Jun 20 '17 at 20:15 • No, I didn't.Thanks! – user36339 Jun 20 '17 at 20:50 • Note that if these are degrees, then the degree sign is obligatory; otherwise the unit is taken to be radians. For instance, $\\sin 90^\\circ = 1$, while $\\sin 90 = 0.89399\\ldots$. The degree sign can be typeset using ^\\circ. – Théophile Jun 20 '17 at", "= a^{2} + b^{2} - 2ab \\cos C$ ### Solution To find the side length $c$, $c^{2} = 6^{2} + (6 \\sqrt{2})^{2} -2 (6) (6 \\sqrt{2}) \\cos 45^\\circ$ Simplify for $c^{2} =36 + 36 (2) - 72 \\sqrt{2}) \\cos 45^\\circ$ Since $\\cos 45^\\circ = \\frac{1}{\\sqrt{2}}$, substitution gives us $c^{2} =36 + 36 (2) - 72 \\sqrt{2} (\\frac{1}{\\sqrt{2}})$ Simplify for $c^{2} =36 + 72 - 72$ $c^{2} =36$ Taking the square root of both sides gives us $c =6$ Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure $B$, $b^{2} = a^{2} + c^{2} - 2ab \\cos B$ Substituting in the appropriate side lengths gives us $(6 \\sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \\cos B$ Simplify for $36 (2) = 36 + 36 - 72 \\cos B$ $72 = 72 - 72 \\cos B$ Subtracting $72$ from both sides gives us $0 = - 72 \\cos B$ Dividing both sides by $-72$ gives us $0 = \\cos B$ Using inverse trigonometry, we know that $B = 90^\\circ$ And we can find the last angle measure $A$ by subtracting the other two measures from $180^\\circ$ $180^\\circ - 90^\\circ - 45^\\circ = 45^\\circ$ $A=45^\\circ$", "= 3.5$$ weeks 22- Choice D is correct The difference of the file added, and the file deleted is: $$487,236 – 397,756 = 89,480, 987,345 + 89,480 = 1,076,825$$ 23- Choice C is correct 40 students did not have to go to summer school. $$50 – 10 = 40, \\frac{40}{50} = \\frac{4}{5}$$ 24- Choice B is correct Diameter $$= 2r ⇒4 = 2r⇒r = 2$$, Circumference $$= 2πr ⇒ C = 2π (2) ⇒ C = 4π$$ 25- Choice D is correct Write a proportion and solve. $$\\frac{\\frac{1}{4}}{1}= \\frac{60}{x}$$ $$x = \\frac{60}{\\frac{1}{4}} = 240$$ 26- Choice A is correct All angles in a triable add up to 180 degrees. $$90^\\circ + 60^\\circ = 150^\\circ ⇒ x = 180^\\circ − 150^\\circ = 30^\\circ$$ 27- Choice C is correct $$\\sqrt{81} = 9, \\sqrt{49} = 7, 9 × 7 = 63$$ 28- Choice D is correct If $$a = 5$$ then: $$b = \\frac{a^3}{5} -2 ⇒ b = \\frac{5^3}{5} -2 = 25 -2 = 23$$ 29- Choice B is correct An obtuse angle is an angle of greater than 90 degrees and less than 180 degrees. Only choice A is an obtuse angle. 30- Choice D is correct Only choice D is not equal to $$3^2$$ 31- Choice B is correct $$\\sqrt[4]{625} = 5, (5^4=5×5×5×5=625)$$ 32- Choice B is correct ($$r", "B = \\frac{c^2 + a^2 − b^2}{2ca}$$ $$⇒ cos B = \\frac{4^2 + 2^2 − 3^2}{ 2(4)(2)} = 0.6875$$ $$\\boxed{⇒ B = 46.6^\\circ}$$ Now we use the formula for $$c^{2}$$ to find C: $$c^2 = a^2 + b^2 − 2ab\\;cos C ⇒ cos C = \\frac{a^2 + b^2 − c^2}{ 2ab}$$ $$⇒ cos C = \\frac{2^2 + 3^2 − 4^2}{ 2(2)(3)} = −0.25$$ $$\\boxed{⇒ C = 104.5^\\circ}$$ Thus, A = 180º −BC = 180º −46.6º −104.5º⇒ $$\\boxed{A = 28.9º}$$ . It may seem that there is always a solution in Case 4 (given all three sides), but that is not true, as the following example shows. Example 4 Case 4: Three sides. Solve the triangle ∆ABC given a = 2, b = 3, and c = 6. Solution: If we blindly try to use the Law of Cosines to find A, we get $$a^2 = b^2 + c^2 − 2bc cos A ⇒ cos A = \\frac{b^2 + c^2 − a^2}{2bc} = \\frac{3^2 + 6^2 − 2^2}{2(3)(6)} = 1.139 ,$$ which is impossible since |cos A| ≤ 1. Thus, there is $$\\boxed{\\text{no solution}}$$ . We could have saved ourselves some effort by recognizing that the length of one of the sides (c = 6) is greater than the sums of the lengths of the remaining sides (a =", "# [Solutions] Baltic Way Mathematical Competition 2014 1. Show that $\\cos(56^{\\circ}) \\cdot \\cos(2 \\cdot 56^{\\circ}) \\cdot \\cos(2^2\\cdot 56^{\\circ})\\cdot . . . \\cdot \\cos(2^{23}\\cdot 56^{\\circ}) = \\frac{1}{2^{24}} .$ 2. Let $a_0, a_1, . . . , a_N$ be real numbers satisfying $a_0 = a_N = 0$ and $a_{i+1} - 2a_i + a_{i-1} = a^2_i$ for $i = 1, 2, . . . , N - 1.$ Prove that $a_i\\leq 0$ for $i = 1, 2, . . . , N- 1.$ 3. Positive real numbers $a, b, c$ satisfy $\\frac{1}{a} +\\frac{1}{b} +\\frac{1}{c} = 3.$ Prove the inequality $\\frac{1}{\\sqrt{a^3+ b}}+\\frac{1}{\\sqrt{b^3 + c}}+\\frac{1}{\\sqrt{c^3 + a}}\\leq \\frac{3}{\\sqrt{2}}.$ 4. Find all functions $f$ defined on all real numbers and taking real values such that $f(f(y)) + f(x - y) = f(xf(y) - x),$ for all real numbers $x, y.$ 5. Given positive real numbers $a, b, c, d$ that satisfy equalities $a^2 + d^2 - ad = b^2 + c^2 + bc, \\ \\ a^2 + b^2 = c^2 + d^2$ find all possible values of the expression $$\\frac{ab+cd}{ad+bc}.$$ 6. In how many ways can we paint $16$ seats in a row, each red or green, in such a way that the number of consecutive seats painted in the same colour is always odd? 7. Let $p_1, p_2, . . . , p_{30}$", "draw(B--C--(1,0)--cycle,blue+1bp); pair P = 0.73*C+0.27*B; draw(O--P); dot(P); label(\"1\",(C+P)/2,NE); label(\"r\",(B+P)/2,NE); [/asy]$ then using the Pythagorean theorem we have: $(r+1)^2=(2s)^2+(r-1)^2$ which is equivalent to: $r^2+2r+1=4s^2+r^2-2r+1$ subtracting $r^2-2r+1$ from both sides $4r=4s^2$ solving for s we get: $$s=\\sqrt{r}$$ next we can find the area of the frustum and of the sphere and we know $V_{\\text{frustum}}=2V_{\\text{sphere}}$ so we can solve for $s$ using $V_{\\text{frustum}}=\\frac{\\pi*h}{3}(R^2+r^2+Rr)$ we get: $$V_{\\text{frustum}}=\\frac{\\pi*2\\sqrt{r}}{3}(r^2+r+1)$$ using $V_{\\text{sphere}}=\\dfrac{4r^{3}\\pi}{3}$ we get $$V_{\\text{sphere}}=\\dfrac{4(\\sqrt{r})^{3}\\pi}{3}$$ so we have: $$\\frac{\\pi*2\\sqrt{r}}{3}(r^2+r+1)=2*\\dfrac{4(\\sqrt{r})^{3}\\pi}{3}$$ dividing by $\\frac{2\\pi*\\sqrt{r}}{3}$ we get $$r^2+r+1=4r$$ which is equivalent to $$r^2-3r+1=0$$ $r=\\dfrac{3\\pm\\sqrt{(-3)^2-4*1*1}}{2*1}$ so $$r=\\dfrac{3+\\sqrt{5}}{2}\\longrightarrow \\boxed{E}$$ Let's once again look at the cross section of the frustum. Let the angle from the center of the sphere to a point on the circumference of the bottom circle be $\\theta.$ This implies that the angle from the center of the sphere to a point on the circumference of the top circle is $90 - \\theta.$ Hence the bottom radius is $r\\tan{\\theta}$ and the top radius is $\\frac {r}{\\tan {\\theta}}.$ This means that the radio between the bottom radius and top radius is $(\\tan {\\theta})^2.$ Using the frustum volume formula, we find that the are of this figure is $\\frac{2\\pi r}{3}(r^2(\\tan {\\theta})^2 + r^2 + \\frac {r^2} {(\\tan {\\theta})^2}).$ We can equate this to $\\frac {8\\pi*r^3} 3.$ Simplifying, we are left with a quadratic conveniently in $(\\tan {\\theta})^2.$ The quadratic", "(c) EH= $7.07$ (d) $70.4$ (e) $114.3^{\\circ}$ 12. (a) Show (b) $x=3$ (c) $70.9^{\\circ}$ (d) $20 \\sqrt{3}$ 13. (a) Show (b) $x=2$ 14. (a) $50.5^{\\circ}$ (b) $x=4.55$ 15. (a)(i) 12 (ii) $5 \\sqrt{3}$ (b) $40.3^{\\circ}$ (c) $63.6^{\\circ}$ (d) $76.1$ (e) $V=168$ 16. (a) $A=62.964$, $B=38.573^{\\circ}, C=78.4637$ (b) $14.6$ (c) 13 17. (a) Show (b) $V F=9.43$ (c) $A=38.7$ (d) $\\theta=33.69$ 18. (a)(i) $v=6 t-2 t^{2}$ (ii) $s=3 t^{2}-\\frac{2 t^{3}}{3}+5$ (b) (i) $t=3$ (ii) 14 19. (a) $A E=10.7$ (b) $\\theta=58^{\\circ}$ 20. (a) $\\theta=71.8^{\\circ}$ (b) $A=34.2$ 21. (a) $4\\sqrt{3}$ (b) 120.6 (c) 40.6 22. (a) Show (c) $B G=15.7$ (d) $67 \\cdot 6^{\\circ}$ (e) $34.7^{\\circ}$ 23. $\\left(\\right.$ a) $66.7^{\\circ}, 113.3^{\\circ}$ (b) $5.54$ 24. (a) CH=8.54 (b) $A H=12.1$. (c) $F N=9.55$ (d) $70.7^{\\circ}$ (e) $18.3^{\\circ}$ 25. (a) $x=4.86$ (b) $30.3$ 26. (a) Show (b) $x=8$ (c) $58.8^{\\circ}$ (d) 47.9 27. (a) Show (b) $43.9^{\\circ}$ (c) Show (d) $98.2^{\\circ}$ 28. (a) $10x$ (b) $EA=10\\sqrt{2}x$ (c) $59.0^{\\circ}$ (d) $73.7^\\circ$ (e) $x=1.804$ 29. (a) $\\angle B=74.9^{\\circ}, 21.1^{\\circ}$ (b) $A=19$ 30. $\\angle A=40.8^{\\circ}$ 31. $(a) 39.4^{\\circ}$ (b) $33.9$ 32. (a) Show (b) $VA=16$ (c) $DX=8 \\sqrt{3}$ (d) $54.7^{\\circ}$ (e) $109.5$ 33. $x=2 \\cdot 15$ 34. (a) Show (b) $x=\\frac 54$ 35. (a) Show (b) $EN=14.8$ (c) $19.8^{\\circ}$ (d) $42.8^{\\circ}$ 36. (a) $\\quad V N=5 x \\sqrt{3}$ (b) $V A=10 x$ (c) $65.2^{\\circ}$ (d)", "# Using the R method for finding all solutions to $\\sin(2a) - \\cos(2a) = \\frac{\\sqrt{6}}{2}$. My solution differs from official answer. How many solutions does $$\\sin(2a) - \\cos(2a) = \\frac{\\sqrt{6}}{2}$$ have between $$-90^\\circ$$ and $$90^\\circ$$? I used the R method and got $$2a-45^\\circ = \\arcsin\\left(\\frac{\\sqrt{3}}{2}\\right).$$ Since $$a$$ is between $$-90^\\circ$$ and $$90^\\circ$$, then $$2a$$ is between $$-180^\\circ$$ and $$180^\\circ$$. The RHS can be $$60^\\circ$$, $$120^\\circ$$, $$-240^\\circ$$, and $$-300^\\circ$$. Only $$60^\\circ$$ and $$120^\\circ$$ fit the criteria, but the answer is 4 solutions. Where did I go wrong? You are right. Here is the image of the function using google. • Thanks, but i want to know whats wrong with this solution that got 4 as the answer. Sin 2a - cos 2a = root 6 /2, then he squared it, getting 1-2sin2acos2a= 3/2 which is sin 4a = -1/2 which does have 4 solutions. – SuperMage1 Jan 14 at 2:54 • Since I don't see how they got 4 solutions, there is no way to explain what they did wrong. Do you know which solutions they got? – Andrei Jan 14 at 2:56 • No solutions were given – SuperMage1 Jan 14 at 2:58 • @SuperMage1 when we square we add solutions $x=2$ square $x^2 = 4$ two solutions -/+2. If you square you need to return to the", "6^{2} + (6 \\sqrt{2})^{2} -2 (6) (6 \\sqrt{2}) \\cos 45^\\circ$ Simplify for $c^{2} =36 + 36 (2) - 72 \\sqrt{2}) \\cos 45^\\circ$ Since $\\cos 45^\\circ = \\frac{1}{\\sqrt{2}}$, substitution gives us $c^{2} =36 + 36 (2) - 72 \\sqrt{2} (\\frac{1}{\\sqrt{2}})$ Simplify for $c^{2} =36 + 72 - 72$ $c^{2} =36$ Taking the square root of both sides gives us $c =6$ Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure $B$, $b^{2} = a^{2} + c^{2} - 2ab \\cos B$ Substituting in the appropriate side lengths gives us $(6 \\sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \\cos B$ Simplify for $36 (2) = 36 + 36 - 72 \\cos B$ $72 = 72 - 72 \\cos B$ Subtracting $72$ from both sides gives us $0 = - 72 \\cos B$ Dividing both sides by $-72$ gives us $0 = \\cos B$ Using inverse trigonometry, we know that $B = 90^\\circ$ And we can find the last angle measure $A$ by subtracting the other two measures from $180^\\circ$ $180^\\circ - 90^\\circ - 45^\\circ = 45^\\circ$ $A=45^\\circ$ # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. Have questions about the image or the explanations on this page? [[Category:]]" ]

[ "+0 # 3d matrix stuff 0 28 1 Let $\\mathbf{a},$ $\\mathbf{b},$ $\\mathbf{c}$ be vectors, and let $D$ be the determinant of the matrix whose column vectors are $\\mathbf{a},$ $\\mathbf{b},$ and $\\mathbf{c}.$ Then the determinant of the matrix whose column vectors are $\\mathbf{a} \\times \\mathbf{b},$ $\\mathbf{b} \\times \\mathbf{c},$ and $\\mathbf{c} \\times \\mathbf{a}$ is equal to $k \\cdot D^n.$Enter the ordered pair $(k,n).$ Oct 5, 2020 #1 0 (k,n) = (2,4). Oct 5, 2020", "{\\begin{pmatrix} 1 & 2\\\\ 3 & 4\\\\ 5 & 6 \\end{pmatrix}}= \\begin{pmatrix} 22& 28 \\\\ 49& 64\\\\ \\end{pmatrix} \\end{align} Notice that the order of the product is reduced to 2 x 2. In general if an m x n order matrix is multiplied by an n x k order matrix the result will be an m x k matrix.", "# Sufficient conditions for Loewner ordering of two matrices? Le $$\\mathbf{A}$$ and $$\\mathbf{B}$$ two psd matrices of size $$n$$. Additionally we assume that the entries are real and non-negative. Does the following hold: $$\\forall (i,j) \\in [n], \\mathbf{A}_{ij} \\leq \\mathbf{B}_{ij} \\implies \\mathbf{A}^2 \\preccurlyeq \\mathbf{B}^2$$ where $$\\preccurlyeq$$ is the Loewner partial order on the cone of psd matrices. No. E.g. $$A=\\pmatrix{1&0\\\\ 0&0}\\le \\pmatrix{1&1\\\\ 1&1}=B$$ entrywise, but $$B^2-A^2=\\pmatrix{1&2\\\\ 2&2}$$ is indefinite.", "## Matrix od order n $$\\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & … & n \\\\ -1 & 0 & 2 & 3 & 4 & … & n-1 \\\\ -1 & -2 & 0 & 3 & 4 & … & n-1 \\\\ -1 & -2 & -3 & 0 & 4 & … & n-1 \\\\ -1 & -2 & -3 & -4 & 0 & … & n-1 \\\\ \\vdots & \\vdots & \\vdots & \\vdots & & \\ddots & \\\\ -1 & -2 & -3 & -4 & … & 1-n & 0 \\\\ \\end{pmatrix}$$. The determinant of the matrix is $$n!$$.", "+0 # Long solution matrix part 2 0 237 1 Let A be a $$2 \\times 2$$ matrix. Suppose that for every two-dimensional vector v, there exists a two-dimensional vector w such that $$\\mathbf{A} \\mathbf{w} = \\mathbf{v}$$ Show that we can find a matrix B such that $$\\mathbf{A} \\mathbf{B} = \\mathbf{I}$$. Mar 13, 2019 $$\\exists w_1 \\ni A w_1 = \\begin{pmatrix}1\\\\0\\end{pmatrix}\\\\ \\exists w_2 \\ni A w_2 = \\begin{pmatrix}0\\\\1\\end{pmatrix}\\\\ B = \\begin{pmatrix}w_1 &w_2\\end{pmatrix}$$", "where $\\mathbf{i}, \\mathbf{j}, \\mathbf{k}$ are unit vectors along the coordinate axes. From (4) $\\mathbf{a.b} = (a_1\\mathbf{i} + a_2\\mathbf{j} + a_3\\mathbf{k}) .\\mathbf{b} = a_1\\mathbf{i}.\\mathbf{b} + a_2\\mathbf{j}.\\mathbf{b} + a_3\\mathbf{k}.\\mathbf{b}$ but $\\mathbf{b} = b_1\\mathbf{i} + b_2\\mathbf{j} + b_3\\mathbf{k}$ so: $\\mathbf{a.b} = b_1a_1\\mathbf{i .i} + b_2a_1\\mathbf{i .j} + b_3a_1\\mathbf{i .k} + b_1a_2\\mathbf{j.i} + b_2a_2\\mathbf{j.j} + b_3a_2\\mathbf{j.k} + b_1a_3\\mathbf{k.i} + b_2a_3\\mathbf{k.j} + b_3a_3\\mathbf{k.k}$ $\\mathbf{i .j, i .k, j .k,}$ etc. are all zero because the vectors are orthogonal, also $\\mathbf{i .i, j.j}$ and $\\mathbf{k.k}$ are all one (these are unit vectors defined to be 1 unit in length). Using these results: $\\mathbf{a.b} = a_1b_1 + a_2b_2 + a_3b_3$ ## Matrices Matrices are sets of numbers arranged in a rectangular array. They are especially important in linear algebra because they can be used to represent the elements of linear equations. 11a + 2b = c 5a + 7b = d The constants in the equation above can be represented as a matrix: $\\mathbf{A} = \\begin{bmatrix} 11 & 2 \\\\ 5 & 7 \\\\ \\end{bmatrix}$ The elements of matrices are usually denoted symbolically using lower case letters: $\\mathbf{A} = \\begin{bmatrix} a_{11} & a_{12} \\\\ a_{21} & a_{22} \\\\ \\end{bmatrix}$ Matrices are said to be equal if all of the corresponding elements are equal. Eg: if $a_{ij}= b_{ij}$ Then $\\mathbf{A} = \\mathbf{B}$ Matrices are added by", "# Matrix Entrywise Addition/Examples/Real 2 x 2 Let $\\mathbf A = \\begin {pmatrix} p & q \\\\ r & s \\end {pmatrix}$ and $\\mathbf B = \\begin {pmatrix} w & x \\\\ y & z \\end {pmatrix}$ be order $2$ square matrices over the real numbers. Then the matrix sum of $\\mathbf A$ and $\\mathbf B$ is given by: $\\mathbf A + \\mathbf B = \\begin {pmatrix} p + w & q + x \\\\ r + y & s + z \\end {pmatrix}$", "gives $q=-5p$. We now know that $\\mathbf{e}= t\\begin{pmatrix}1\\\\-5 \\\\ -2\\end{pmatrix}$. We also know $\\big \\vert\\mathbf{e}\\big \\vert = \\sqrt{30}$ and $p$ is negative, hence $t=-1$ and we have $\\mathbf{e}= \\begin{pmatrix}-1\\\\5 \\\\ 2\\end{pmatrix}$ and $\\mathbf{f}= \\begin{pmatrix}5\\\\2 \\\\-1\\end{pmatrix}$. It might be tempting to proceed with vectors perpendicular to $\\pi$ and $\\pi'$ being $\\mathbf{b} \\times \\mathbf{c}$ and $\\mathbf{e} \\times \\mathbf{f}$. $\\mathbf{b}\\times \\mathbf{c} = \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ 2&-1&-1 \\\\ -1&2&1\\end{matrix}\\right \\vert = \\mathbf{i}-\\mathbf{j}+3\\mathbf{k}.$ $\\mathbf{e}\\times \\mathbf{f}= \\left \\vert \\begin{matrix}\\mathbf{i}&\\mathbf{j}&\\mathbf{k}\\\\ p&q&r \\\\ q&r&p\\end{matrix}\\right \\vert = (qp-r^2)\\mathbf{i}+(rq-p^2)\\mathbf{j}+(pr-q^2)\\mathbf{k}.$ However, the resulting set of simultaneous equations is very difficult to solve. By adding them we can then simplify using $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ in other words $p^2+q^2+r^2=30$. We can think about $(p+q+r)^2$ to continue the manipulation. Try it if you like! Show that $\\pi'$ is the reflection of $\\pi$ in the origin. Using the form of the plane equation $\\mathbf{r}.\\mathbf{n}=\\mathbf{a}.\\mathbf{n}$ with $\\mathbf{n}$ as the normal vector we found earlier, we have the equation of $\\pi$ as $x - y + 3z = -2.$ Similarly using the point $\\mathbf{d}$, the plane $\\pi'$ is $x - y + 3z = 2.$ If $(x_1,y_1,z_1)$ is a point on $\\pi$ then $x_1 - y_1 + 3z_1 = -2$. Multiplying by $-1$ gives $(-x_1) -(-y_1) + 3(-z_1) = 2$. So the point $(-x_1,-y_1,-z_1)$ lies on $\\pi'$. Since every point on", "### Solving Matrix Equations #### Definition Given matrices $\\mathbf{A}$ and $\\mathbf{B}$, we can solve the equation $\\mathbf{A} \\mathbf{X} = \\mathbf{B}$ for the unknown matrix $\\mathbf{X}$ when the matrix inverse of $\\mathbf{A}$ exists. We do this by multiplying both sides of the equation on the left by $\\mathbf{A}^{-1}$, providing it exists, to give \\begin{align} \\mathbf{A} \\mathbf{X} &= \\mathbf{B} \\\\ \\mathbf{A}^{-1} \\mathbf{A} \\mathbf{X} &= \\mathbf{A}^{-1} \\mathbf{B} \\\\ \\mathbf{X} &= \\mathbf{A}^{-1} \\mathbf{B} \\end{align} as $\\mathbf{A}^{-1} \\mathbf{A} = \\mathbf{I}$ and multiplying any matrix by the identity matrix leaves it unaltered. It is important that both sides of the equation are multiplied on the left by $\\mathbf{A}^{-1}$ because matrix multiplication does not commute - multiplying on the right could give a different result. #### Worked Example ###### Example 1 Let \\begin{align} \\mathbf{A} &= \\begin{pmatrix}1&2\\\\3&-5\\end{pmatrix}, \\\\ \\mathbf{B} &= \\begin{pmatrix}4\\\\1\\end{pmatrix}. \\end{align} Solve the equation $\\mathbf{A} \\mathbf{X} = \\mathbf{B}$. ###### Solution We have the equation $\\begin{pmatrix}1&2\\\\3&-5\\end{pmatrix}\\mathbf{X} = \\begin{pmatrix}4\\\\1\\end{pmatrix}$ To get $\\mathbf{X}$ on its own, multiply both sides by the inverse of $\\mathbf{A}$. Note that $\\lvert \\mathbf{A} \\rvert = -5-6 = -11$. Therefore, $\\mathbf{A}^{-1} = -\\dfrac{1}{11}\\begin{pmatrix} -5&-2\\\\-3&1\\end{pmatrix}$ So $\\mathbf{X} = \\mathbf{A}^{-1} \\mathbf{B} = -\\dfrac{1}{11}\\begin{pmatrix} -5&-2\\\\-3&1\\end{pmatrix}\\begin{pmatrix}4\\\\1\\end{pmatrix}$ Now, use matrix multiplication to find $\\mathbf{X}$. \\begin{align} \\mathbf{X} &= -\\frac{1}{11}\\begin{pmatrix} -5&-2\\\\-3&1\\end{pmatrix}\\begin{pmatrix}4\\\\1\\end{pmatrix}\\\\ &= -\\frac{1}{11}\\begin{pmatrix}(-20)+(-2)\\\\(-12)+1\\end{pmatrix}\\\\ &= -\\frac{1}{11}\\begin{pmatrix}-22\\\\-11\\end{pmatrix}\\\\ &= \\begin{pmatrix}2\\\\1\\end{pmatrix} \\end{align} #### Simultaneous Equations We can apply this method to solving a system of", "that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is the projection matrix $P=\\frac{1}{\\mathbf{a}^T\\mathbf{a}}\\mathbf{a}\\mathbf{a}^T,$ which is an $n\\times n$ matrix. To this end, Prof. Strang writes the projection vector as $\\mathbf{p}=\\mathbf{a}\\,x$. Now, you should see $\\mathbf{p}$ and $\\mathbf{a}$ as $n\\times1$ matrices and $x$ as a one dimensional vector or a $1\\times1$ matrix, so that $\\mathbf{a}$ and $x$ can be multiplied. Thus, we have $x\\,\\mathbf{a}=\\mathbf{a}\\,x.$ This identity does not mean that $\\mathbf{a}$ and $x$ are two matrices that commute. Just read the left and right sides as stated above. It is true that the notation may be a bit misleading. Perhaps it would be better to write $(x)$ instead of $x$ if $x$ should be seen as a $1\\times 1$ matrix (parentheses are used to delimit matrices). For example, $4\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix} =\\begin{pmatrix} 4 \\\\ 8 \\\\ 12\\end{pmatrix} =\\begin{pmatrix} 1 \\\\ 2 \\\\ 3\\end{pmatrix}(4).$ The advantage of expressing $\\mathbf{p}$ as $\\mathbf{a}x$ is that generalization is then possible. If $S$ is spanned by the linearly independent vectors $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, the orthogonal projection $\\mathbf{p}$ of a vector $\\mathbf{b}$ onto $S$ should be a linear combination of $\\mathbf{a}_1,\\ldots,\\mathbf{a}_k$, that is, $\\mathbf{p}=x_1\\mathbf{a}_1+\\cdots+x_k\\mathbf{a}_k=AX$ with $A=\\left(\\mathbf{a}_1 \\vert\\ldots \\vert\\mathbf{a}_k\\right)$ and $X=\\begin{pmatrix}x_1 \\\\ \\vdots \\\\ x_k\\end{pmatrix}$ The reasoning in Prof. Strang’s lecture would then show that $\\mathbf{p}=P\\mathbf{b}$, where $P$ is now the projection matrix $P=A(A^TA)^{-1}A^T.$ Returning to your question, you", "matrix $\\mathbf B$ which results from the row operation $\\mathbf R$ is: $\\mathbf B = \\begin {pmatrix} 1 & 0 & 1 & 1 \\\\ 0 & 0 & 3 & 2 \\\\ 0 & 1 & 6 & 10 \\end {pmatrix}$ ### Arbitrary Matrix $3$ Let $\\mathbf A$ be the matrix: $\\mathbf A = \\begin {pmatrix} 0 & 4 & 7 & 10 \\\\ 2 & 5 & 8 & 11 \\\\ 3 & 6 & 9 & 12 \\end {pmatrix}$ The matrix $\\mathbf R$ corresponding to the row operation to clear the first column is: $\\mathbf R = \\begin {pmatrix} 0 & \\dfrac 1 2 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & -\\dfrac 3 2 & 1 \\end {pmatrix}$ and the matrix $\\mathbf B$ which results from the row operation $\\mathbf R$ is: $\\mathbf B = \\begin {pmatrix} 1 & \\dfrac 5 2 & 4 & \\dfrac {11} 2 \\\\ 0 & 4 & 7 & 10 \\\\ 0 & -\\dfrac 3 2 & -3 & -\\dfrac 9 2 \\end {pmatrix}$ ## Also presented as Some sources are not concerned about making $b_{1 1}$ equal to $1$. Hence they do not use step $3$ of the algorithm.", "The rank of the following matrix is $\\\\ \\begin{pmatrix} 1 & 1 & 0 & -2 \\\\ 2 & 0 & 2 & 2 \\\\ 4 & 1 & 3 & 1 \\end{pmatrix}$ 1. $1$ 2. $2$ 3. $3$ 4. $4$ Answer:", "## Symmetric matrix of a form Let $$\\mathbf A=\\begin{pmatrix} 1 & -2 & 0 \\\\ 2 & 0 & -1 \\\\ -2 & 2 & 0 \\\\ \\end{pmatrix}$$ be the matrix of a bilinear form on $$\\mathbb K^3$$. Determine the analytic expression of this form and also for the associated quadratic form. Find a symmetric matrix for this quadratic form. (All tasks w.r.t. the same basis.) • #### Variant 1 For $$\\mathbb K=\\mathbb R$$. • #### Variant 2 For $$\\mathbb K=\\mathbb Z_2$$ (the number 2 in $$\\mathbb Z_2$$ interpret as 0). • #### Variant 3 For $$\\mathbb K=\\mathbb Z_3$$.", "+0 # Help Quick +1 38 2 Let $\\mathbf{A}$ be a matrix and $\\mathbf{x}$ be a non-zero vector such that $\\mathbf{A} \\mathbf{x} = 2 \\mathbf{x}.$Then we have that $\\mathbf{A}^7\\mathbf{x} = a \\mathbf{x}$some value of $a$. What is $a$ equal to? Nov 12, 2020 ### 1+0 Answers #2 +31296 +2 Think about $$\\mathbf{A^7}\\mathbf{x}= \\mathbf{A^6}(\\mathbf{A}\\mathbf{x})$$ Nov 13, 2020", "\\qquad \\mathbf=\\begin -2 \\ 1 \\ 0 \\end, \\qquad \\mathbf=\\begin 1 \\ 1 \\end.$ For each of the vectors $\\mathbf, \\mathbf, \\mathbf$, determine whether the vector is in the null space $\\cal N(A)$. (b) Find a basis of the null space of the matrix $B=\\begin 1 & 1 & 2 \\ -2 &-2 &-4 \\end$. (e) If $A$ or $B$ is nonsingular, then $AB$ is similar to $BA$. Read solution Let $V$ be a real vector space of all real sequences $(a_i)_^=(a_1, a_2, \\dots).$ Let $U$ be the subspace of $V$ consisting of all real sequences that satisfy the linear recurrence relation $a_-5a_ 3a_=0$ for $k=1, 2, \\dots$. • ###### Determinants of Matrices Problems in Mathematics Problems of Determinants of Matrices. From introductory exercise problems to linear algebra exam problems from various universities. Basic to advanced level.… • ###### CHAPTER 8 MATRICES and DETERMINANTS SECTION 8.1 MATRICES and SYSTEMS OF EQUATIONS PART A MATRICES A matrix is basically an organized box or “array” of numbers or other expressions. In this chapter, we will typically assume that our matrices contain only numbers. Example Here is a matrix of size 2 3 “2 by 3”, because it has 2 rows and 3 columns 10 2 015… • ###### Exercises and Problems in Linear Algebra - edu Exercises and Problems in", "element $x_{i,0}$ which always equals to $1$. So our required function becomes $w_0 x_{i,0} + w_1 x_{i,1} + \\dots w_d x_{i,d}$. We denote it as $h_{\\mathbf{w}}(\\mathbf{x}_i)$. Using matrices, we can write $h_{\\mathbf{w}}(\\mathbf{x}_i)$ in a much more compact form. Conventionally, we use column matrices to represent vectors. Thus we have $\\mathbf{w} = \\begin{pmatrix} w_0\\\\ w_1\\\\ w_2\\\\ \\vdots\\\\ w_d \\end{pmatrix}, \\hspace{1cm}\\mathbf{x}_i = \\begin{pmatrix} x_{i,0}\\\\ x_{i,1}\\\\ x_{i,2}\\\\ \\vdots\\\\ x_{i,d} \\end{pmatrix}$ Our function $h_{\\mathbf{w}}(\\mathbf{x}_i)$ thus can be written as $\\mathbf{w}^\\intercal\\mathbf{x}_i$, or equivalently, as $\\mathbf{x}_i^\\intercal\\mathbf{w}$. If there are $n$ data points $\\mathbf{x}_1, \\mathbf{x}_2, \\dots, \\mathbf{x}_n$, the outputs corresponding to all these can be kept together in a column matrix as follows: $\\begin{pmatrix} \\mathbf{x}_1^\\intercal\\mathbf{w}\\\\ \\mathbf{x}_2^\\intercal\\mathbf{w}\\\\ \\vdots\\\\ \\mathbf{x}_n^\\intercal\\mathbf{w} \\end{pmatrix}$ Note that this column matrix can be decomposed into the following product $\\begin{pmatrix} \\mathbf{x}_1^\\intercal\\mathbf{w}\\\\ \\mathbf{x}_2^\\intercal\\mathbf{w}\\\\ \\vdots\\\\ \\mathbf{x}_n^\\intercal\\mathbf{w} \\end{pmatrix} = \\begin{pmatrix} —\\mathbf{x}_1^\\intercal— \\\\ —\\mathbf{x}_2^\\intercal—\\\\ \\vdots\\\\ —\\mathbf{x}_n^\\intercal— \\end{pmatrix} \\times \\mathbf{w}$ It is important to keep in mind that even though the last product looks like a scalar multiplication, $\\mathbf{w}$ is in fact a matrix, and that the LHS and RHS are equal nevertheless. Understandably, we call the first matrix in the decomposition $\\mathbf{X}$. It is an $n\\times (d+1)$ matrix, each of whose rows represents a data point. Thus we can compute the outputs $h_\\mathbf{w}(\\mathbf{x}_i)$ corresponding to all the data points $\\mathbf{x}_i$ using the single matrix product $\\mathbf{X}\\mathbf{w}$. ##", "of a row matrix with a column matrix; that is, \\begin{align*} {\\mathbf x} \\cdot {\\mathbf y} & = {\\mathbf x}^{\\rm t} {\\mathbf y}\\\\ & = \\begin{pmatrix} x_1 & x_2 & \\cdots & x_n \\end{pmatrix} \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix}\\\\ & = x_{1}y_{1} + x_{2}y_{2} + \\cdots + x_{n}y_{n}. \\end{align*} ##### Example8.19 Suppose that the words to be encoded consist of all binary 3-tuples and that our encoding scheme is even-parity. To encode an arbitrary 3-tuple, we add a fourth bit to obtain an even number of 1s. Notice that an arbitrary $n$-tuple ${\\mathbf x} = (x_1, x_2, \\ldots, x_n)^{\\rm t}$ has an even number of 1s exactly when $x_1 + x_2 + \\cdots + x_n = 0$; hence, a 4-tuple ${\\mathbf x} = (x_1, x_2, x_3, x_4)^{\\rm t}$ has an even number of 1s if $x_1+ x_2+ x_3+ x_4 = 0$, or \\begin{equation*}{\\mathbf x} \\cdot {\\mathbf 1} = {\\mathbf x}^{\\rm t} {\\mathbf 1} = \\begin{pmatrix} x_1 & x_2 & x_3 & x_4 \\end{pmatrix} \\begin{pmatrix} 1 \\\\ 1 \\\\ 1 \\\\ 1 \\end{pmatrix} = 0.\\end{equation*} This example leads us to hope that there is a connection between matrices and coding theory. Let ${\\mathbb M}_{m \\times n}({\\mathbb Z}_2)$ denote the set of all $m \\times n$ matrices with entries in ${\\mathbb Z}_2$. We do matrix operations", "The matrix $\\begin{pmatrix}0&0&1&0\\cr0&0&0&1\\cr-1&0&0&0\\cr0&-1&0&0\\end{pmatrix}$ generates a cyclic group of order $4$ under matrix multiplication.", "Matrix $M_1 = M(f', \\mathcal{B}, \\mathcal{B '})$ of $f$ $f: \\mathbb{R}^5 \\rightarrow \\mathcal{M}_{2}(\\mathbb{R})$ $(x_1,x_2,x_3,x_4,x_5) \\rightarrow \\begin{pmatrix} x_1 & x_2 \\\\ x_1 + x_3 & x_5 \\end{pmatrix}$ 1. Determine the basis of $Kerf$ 2. Determine a basis of $Imf$ 3. Determine the matrix $M_1 = M(f', \\mathcal{B}, \\mathcal{B '})$ of $f$ 1 . $x = (x_1,x_2,x_3,x_4,x_5) \\in Kerf \\iff f(x_1,x_2,x_3,x_4,x_5) = \\begin{pmatrix} x_1 & x_2 \\\\ x_1 + x_3 & x_5 \\end{pmatrix} =\\begin{pmatrix} 0 & 0 \\\\ 0 & 0\\end{pmatrix}$ I got $(x_1,x_2,x_3,x_4,x_5) = (0,0,0,x_4,0)$. A base of $Kerf$ is $(0,0,0,1,0)$. 1. $Imf^= Vect <f(e_1), f(e_2), f(e_3), f(e_4), f(e_5) > = <E_{11}, E_{12}, E_{21}, E_{22}>$ So the family $(E_{11}, E_{12}, E_{21}, E_{22})$ is basis. with $E_{ij}$ is the elementary matrix. The coordinate $x_4$ is not on the matrix. I need help with question 3, too. Thank you. • $f$ should be a map from $\\mathbb{R}^5$. – tattwamasi amrutam Jul 11 '17 at 21:29 • For starters: you have ignored $x_3$ in the kernel. So it should be $(0,0,x_3,x_4,0) \\in K$. This mean kernel has dimension $2$. Likewise the image also needs fixing. Because the image can only have dimension $3$. – Anurag A Jul 11 '17 at 21:31 • @AnuragA I just made a mistake typing. I have just corrected it. Thank you for noticing. – Zouhair El Yaagoubi", "+0 Let $\\mathbf{A}$ be a matrix, and let $\\mathbf{x}$ and $\\mathbf{y}$ be linearly independent vectors such that \\[\\mathbf{A} \\mathbf{x} = \\mat +1 273 1 Let A be a matrix, and let x and y be linearly independent vectors such that $$\\mathbf{A} \\mathbf{x} = \\mathbf{y}, \\mathbf{A} \\mathbf{y} = \\mathbf{x} + 2\\mathbf{y}$$ Then we have that $$\\mathbf{A}^{-1} \\mathbf{x} = a \\mathbf{x} + b\\mathbf{y}$$ for some scalars a and b. Find the ordered pair (a,b). Mar 1, 2019 #1 +23071 +2 Let $$\\mathbf{A}$$ be a matrix, and let $$\\mathbf{x}$$ and $$\\mathbf{y}$$ be linearly independent vectors such that $$\\mathbf{A} \\mathbf{x} = \\mathbf{y}, \\mathbf{A} \\mathbf{y} = \\mathbf{x} + 2\\mathbf{y}$$ Then we have that $$\\mathbf{A}^{-1} \\mathbf{x} = a \\mathbf{x} + b\\mathbf{y}$$ or some scalars a and b. Find the ordered pair (a,b). $$\\begin{array}{|rcll|} \\hline Ax &=& y \\quad | \\quad \\cdot A^{-1} \\\\ A^{-1}Ax &=& A^{-1}y \\quad | \\quad A^{-1}A = I \\\\ x &=& A^{-1}y \\\\ \\mathbf{A^{-1}y} &\\mathbf{=}&\\mathbf{x} \\\\ \\hline \\end{array}$$ $$\\begin{array}{|rcll|} \\hline Ay &=& x + 2y \\quad | \\quad \\cdot A^{-1} \\\\ A^{-1}Ay &=& A^{-1}x + 2A^{-1}y \\quad | \\quad A^{-1}A = I \\\\ y &=& A^{-1}x + 2A^{-1}y \\quad | \\quad \\mathbf{A^{-1}y=x} \\\\ y &=& A^{-1}x + 2x \\\\ \\mathbf{A^{-1}x} &\\mathbf{=}&\\mathbf{-2x+y } \\\\ \\hline \\end{array}$$ $$\\mathbf{(a,b) = (-2,1)}$$ Mar 1, 2019" ]

[ "\\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {vmatrix}$$ = 3 × (-2) - 9 × 4 = -6 - 36 = -42 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{-42}$$$$\\begin {pmatrix} -2 & -4\\\\ -9 & 3\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {-2}{-42} & \\frac {-4}{-42}\\\\ \\frac {-9}{-42} & \\frac {3}{-42}\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$ We know that: AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{21} & \\frac {2}{21}\\\\ \\frac {9}{42} & \\frac {3}{42}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{21} + \\frac {5}{21}\\\\ \\frac {9}{21} - \\frac {15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 5}{21}\\\\ \\frac {36 - 15}{84}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {7}{21}\\\\ \\frac {21}{84}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ i.e. $$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{3}\\\\ \\frac {1}{4}\\\\ \\end {pmatrix}$$ Taking the corresponding parts of the equal matrix: $$\\frac 1x$$ = $$\\frac 13$$ ∴ x = 3 $$\\frac 1y$$ = $$\\frac 14$$", "48x + 54 y &= 27+8=35. \\\\ \\end{align*} Write the two equations in matrix form: $$\\begin{pmatrix} 24 & -18 \\\\ 48 & 54 \\\\ \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ \\end{pmatrix} = \\begin{pmatrix} 5 \\\\ 35 \\\\ \\end{pmatrix}.$$ Write this as an augmented matrix: $$\\begin{pmatrix} 24 & -18 & 5\\\\ 48 & 54 & 35 \\\\ \\end{pmatrix}$$ and then row-reduce (multiply the first row by $-2$ and then add the corresponding numbers to the second row): $$\\begin{pmatrix} 24 & -18 & 5\\\\ 0 & 90 & 25 \\\\ \\end{pmatrix}.$$ Divide the second row by $5$ to get: $$\\begin{pmatrix} 24 & -18 & 5\\\\ 0 & 18 & 5 \\\\ \\end{pmatrix}.$$ Add the second row to the first row: $$\\begin{pmatrix} 24 & 0 & 10\\\\ 0 & 18 & 5 \\\\ \\end{pmatrix}.$$ Divide the first row by $24$ and divide the second row by $18$: $$\\begin{pmatrix} 1 & 0 & \\frac{5}{12}\\\\ 0 & 1 & \\frac{5}{18} \\\\ \\end{pmatrix}.$$ So $$x= \\frac{5}{12} \\mbox{ and } y= \\frac{5}{18},$$ and we conclude $$\\boxed{x+y = \\frac{5}{12} + \\frac{5}{18} = \\frac{25}{36}}.$$ Hint: let $s=x+y\\,$, then substituting $x=s-y$ back into the equations gives: \\begin{cases} \\begin{align} 2\\left((s-y)-\\frac{1}{3} \\right) - \\frac{3}{2}\\left(y-\\frac{1}{6}\\right) &= 0 \\quad\\iff\\quad 2s - \\frac{7}{2}y = \\frac{5}{12}\\\\ \\frac{8}{3}\\left((s-y)-\\frac{1}{6}\\right) + 3\\left(y-\\frac{1}{2}\\right) &= 0 \\quad\\iff\\quad \\frac{8}{3}s + \\frac{1}{3}y=\\frac{35}{18} \\end{align} \\end{cases} Eliminate $y$ between the latter equations (for", "$$7 \\leq k\\equiv 3\\pmod 4$$ such that $$X^2-mkX+m\\frac{(kp+1)}{4} = 0$$ has integer roots. Suppose we can always find a solution of the form $$n(4m-p)-1 = dp \\mid mp$$ for some positive integers $$(m,n,d)$$. Then $$d$$ divides $$m$$, so we may replace $$m$$ with $$dm$$. i.e. solving $$n(4dm-p)-1 = dp \\mid (dm)p$$ Let $$n=k-d$$ and $$w=k-2d$$, then this becomes \\begin{align*} n(4md-p)-1 &= dp\\\\ (k-d)(4md-p) &= dp+1\\\\ 4md(k-d) &= kp+1\\\\ m(2d)(2k-2d) &= kp+1\\\\ m(k-w)(k+w) &= kp+1\\\\ mk^2-kp-1 &= mw^2\\\\ m^2k^2-mkp-m &= (mw)^2 \\end{align*} Now since $$m^2k^2-mkp-m$$ is the discriminant $$D$$ of the equation $$X^2-mkX+m\\frac{kq+1}{4}=0$$ This shows that the equation has solutions $$\\frac{mk\\pm \\sqrt{D}}{2} = \\frac{mk \\pm mw}{2}$$ which must be integral (either $$m$$ even or $$k$$ and $$w$$ can be shown to be odd). We also require $$k\\equiv 3\\pmod 4$$, which can be shown as follows: \\begin{align*} n(4m-p)-1 &= dp\\\\ (k-d)(4m-p)-1 &= dp\\\\ (k-d)(-1)-1 &\\equiv d \\pmod 4\\\\ d-k-1 &\\equiv d \\pmod 4\\\\ k &\\equiv 3 \\pmod 4 \\end{align*} Since $$k=n+d$$ it is always positive, but there is a small problem that it might be equal to $$3$$.", "(-5) × 3 = 8 + 15 = 23 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{23}$$ $$\\begin {pmatrix} 4 & -3\\\\ 5 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} -4\\\\ -13\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-16}{23} + \\frac {39}{23}\\\\ \\frac {-20}{23} - \\frac {26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {-16 + 39}{23}\\\\ \\frac {-20 - 26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {23}{23}\\\\ \\frac {-46}{23}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ -2\\ \\end {pmatrix}$$ ∴ x = 1 and y = -2Ans Here, 4x + 3y = 5.........................................(1) y - 3x = -7...........................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 &3\\\\ -3 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -3 & 1\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -3 & 1\\\\", "$x'+x = -3$, $xx'=1$. These numbers are the roots of the quadratic equation $$\\lambda^2 +3\\lambda +1 = 0.$$ Thus, we have $x = \\frac{-3+\\sqrt 5}2$ and $x'= \\frac{-3-\\sqrt 5}2$. With these, we have \\begin{align} (A+xB)(A+x'B) &= A^2 -3AB + B^2 + x(BA-AB) \\\\ &= BA-AB + x(BA-AB) = (BA-AB)(1+x).\\end{align} We now have that $$\\det(A+xB)(A+x'B) =q \\in \\mathbb{Q},$$ and $$q=(1+x)^n \\det(BA-AB)$$ Then it follows by $A, B\\in M_n(\\mathbb{Q})$ and $(1+x)^n\\in\\mathbb{R} \\backslash \\mathbb{Q}$ that $\\det(BA-AB)=0$. An example with $AB\\neq BA$ For the example that @Jonas Meyer requested, here it is: $$A=\\begin{pmatrix} 1 & 1 \\\\ 0 & \\frac{-3+\\sqrt 5}2\\end{pmatrix}, \\ \\ B=\\begin{pmatrix} \\frac{3+\\sqrt 5}2 & 1 \\\\ 0 & -1\\end{pmatrix}.$$ Then $$A-B= \\begin{pmatrix} \\frac{-1-\\sqrt 5}2 & 0 \\\\ 0 & \\frac{-1+\\sqrt 5}2\\end{pmatrix}$$, $$(A-B)^2 = \\begin{pmatrix} \\frac{3+\\sqrt 5}2 & 0 \\\\ 0 & \\frac{3-\\sqrt 5}2\\end{pmatrix} = AB.$$ But, $$BA =\\begin{pmatrix} \\frac{3+\\sqrt 5}2 & \\sqrt 5 \\\\ 0 & \\frac{3-\\sqrt 5}2\\end{pmatrix} \\neq AB.$$ • Could you clarify how you conclude that $q\\in\\mathbb Q$? – stewbasic Apr 19 '17 at 2:23 • Its conjugate inside $\\mathbb{Q}(\\sqrt 5)$ is the same with itself. – i707107 Apr 19 '17 at 2:25 • Ah right, thanks. So is there a counterexample over $\\mathbb C$ then? – stewbasic Apr 19 '17 at 2:26 • @SangchulLee Arin Chaudhuri's comment is not a counterexample to this because it gives", "= $$\\frac {1}{-13}$$$$\\begin {pmatrix} -3 & -1\\\\ -4 & 3\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 51\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153}{13} & \\frac {3}{13}\\\\ \\frac {204}{13} & \\frac {-9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153 + 3}{13}\\\\ \\frac {204 - 9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {156}{13}\\\\ \\frac {195}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 12\\\\ 5\\\\ \\end {pmatrix}$$ ∴ x = 12 and y = 5 Ans Here, 2x - 5y = 1.................................(1) 7x + 3y = 24.............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {vmatrix}$$ = 2 × 3 - 7 × (-5) = 6 + 35 = 41 ≠ 0 It has a unique solution. A-1= $$\\frac", "-2c_2 + 8c_3 &= 0,\\\\ -2c_1 + 2c_2 -2c_3 &= 0 \\end{align} Solving the above system of equations shows that the only possible solution is $$c_1 = c_2 = c_3 = 0$$. Thus $$\\{ x_1 , x_2 , x_3 \\}$$ is linearly independent. One way to solve the system of equations is shown below. First, subtract (4/3) times the 1st equation from the 2nd equation. $-\\frac{4}{3}(3c_1 + 4c_2 +6c_3) + (4c_1 -2c_2 + 8c_3) = -\\frac{22}{3}c_2 = -\\frac{4}{3}0 + 0 = 0 \\Rightarrow c_2 = 0$ Then add the 1st and 3 times the 3rd equations together, and substitute in $$c_2 = 0$$. $(3c_1 + 4c_2 +6c_3) + 3*(-2c_1 + 2c_2 -2c_3) = -3c_1 + 10 c_2 = -3c_1 + 10*0 = 0 + 3*0 = 0 \\Rightarrow c_1 = 0$ Now, substituting both $$c_1 = 0$$ and $$c_2 = 0$$ into equation 2 gives. $4c_1 -2c_2 + 8c_3 = 4*0 -2*0 + 8c_3 = 0 \\Rightarrow c_3 = 0$ So $$c_1 = c_2 = c_3 = 0$$, and $$\\{ x_1 , x_2 , x_3 \\}$$ are linearly independent. ### Example M.5.2 $x_1 =\\begin{pmatrix} 1 \\\\ -8 \\\\ 8 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 2 \\end{pmatrix}, x_3 =\\begin{pmatrix} 1 \\\\ 3 \\\\ -2 \\end{pmatrix}$ In this case $$\\{ x_1 , x_2 , x_3 \\}$$are linearly dependent,", "0$$. That's: $$\\begin{pmatrix} x - 5 \\end{pmatrix}\\begin{pmatrix} x + 5 \\end{pmatrix} = 0$$ if and only if: $$x - 5 = 0$$ or $$x+5 = 0$$, which leads to: $$x = 5$$ or $$x = -5$$. Finally we can state that this quadratic equation has two solutions: $$x = 5$$ and $$x = -5$$. 4. We factor and solve $$4x^2 - 9=0$$ in two steps: • Step 1: we notice that $$4x^2 - 9$$ is the difference of two squares. Indeed $$x^2 - 49 = \\begin{pmatrix}2x\\end{pmatrix}^2 - 3^2$$. So we can write it in factored form: $x^2 - 49 = \\begin{pmatrix} 2x - 3 \\end{pmatrix}\\begin{pmatrix} 2x + 3 \\end{pmatrix}$ • Step 2: we use the null product law to solve the factored equation $$\\begin{pmatrix} 2x - 3 \\end{pmatrix}\\begin{pmatrix} 3x + 3 \\end{pmatrix} = 0$$. That's: $$\\begin{pmatrix} 2x - 3 \\end{pmatrix}\\begin{pmatrix} 2x + 3 \\end{pmatrix} = 0$$ if and only if: $$2x - 3 = 0$$ or $$2x+3 = 0$$, which leads to: $$2x = 3$$ or $$2x = -3$$. Which, in turn, leads to: $$x = \\frac{3}{2}$$ or $$x = -\\frac{3}{2}$$. Finally we can state that this quadratic equation has two solutions: $$x = \\frac{3}{2}$$ and $$x = -\\frac{3}{2}$$. 5. We factor and solve $$16x^2 - 81=0$$ in two steps: • Step 1: we notice that $$16x^2 -", "0\\\\0 & \\frac12 & 0\\\\0 & 0 & 1\\end{pmatrix}.$$$$ The problem of calculating $$A^n$$ is thus reduced to calculating $$J^n$$. Let $$a_{ij}^{(n)}$$ denote the entry of $$J^n$$ in the $$i$$-th row and $$j$$-th column. The product of an arbitrary $$3\\times3$$-matrix with $$J$$ is given by: $$$$\\begin{pmatrix} a&b&c\\\\d&e&f\\\\g&h&i \\end{pmatrix} J = \\begin{pmatrix} \\frac a2&a+\\frac b2&c\\\\\\frac d2&d+\\frac e2&f\\\\\\frac g2&g+\\frac h2&i \\end{pmatrix}.$$$$ We can deduce that, for all $$n\\in\\Bbb N$$: \\begin{align} a_{11}^{(n)}&=a_{22}^{(n)}=\\frac1{2^n}, \\\\a_{21}^{(n)}&=a_{31}^{(n)}=0,\\\\ a_{13}^{(n)}&=a_{23}^{(n)}=a_{32}^{(n)}=0, \\\\ a_{33}^{(n)}&=1,\\\\ a_{12}^{(n+1)}&=a_{11}^{(n)}+\\frac{a_{12}^{(n)}}2=\\frac1{2^n}+\\frac{a_{12}^{(n)}}2. \\end{align} Thus, all $$a_{ij}^{(n)}$$ are explicitly known except for $$a_{12}^{(n)}$$. Note that, by the last equation, $$$$a_{12}^{(n+1)}=2^{-n}+\\frac{a_{12}^{(n)}}2 = 2^{-n}+2^{-n}+\\frac{a_{12}^{(n-1)}}4 = \\dots = (n+1)\\cdot2^{-n}.$$$$ Thus, $$$$J^n=\\begin{pmatrix}2^{-n}&n\\cdot 2^{1-n} & 0\\\\0 & 2^{-n} & 0\\\\0 & 0 & 1\\end{pmatrix}.$$$$ And by some calculations, we find that $$$$A^n=\\big(v_1,v_2,v_3\\big)J^n\\big(v_1,v_2,v_3\\big)^{-1}= \\begin{pmatrix} 2^{-n} & n\\cdot 2^{-n-1} - 2^{-n-1} + \\frac12 & {1-\\frac{n+1}{2^n}\\over2}\\\\ 0 & {2^{-n}+1\\over2} & {1-2^{-n}\\over2} \\\\ 0 & {1-2^{-n}\\over2} & {2^{-n}+1\\over2} \\end{pmatrix}.$$$$ • I don't understand very well the \"your\" in the first sentence of this answer \"Note that your matrix...\" ; in fact, as far I have understood, you are answering your own question ... – Jean Marie May 25 at 17:39 • @JeanMarie indeed I am answering my own question. On the questions on this forum that I saw that were answered by the OP, the OP was always talking about himself in second person; so I", "\\end {vmatrix}$$ = 4 × 1 - 3 × -3 = 4 + 9 = 13 ≠ 0 It has unique solution. A-1=$$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{13}$$$$\\begin {pmatrix} 1 & -3\\\\ 3 & 4\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{13} & \\frac {-3}{13}\\\\ \\frac {3}{13} & \\frac {4}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{13} & \\frac {-3}{13}\\\\ \\frac {3}{13} & \\frac {4}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{13} + \\frac {21}{13}\\\\ \\frac {15}{13} - \\frac {28}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5 + 21}{13}\\\\ \\frac {15 - 28}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {26}{13}\\\\ \\frac {-13}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 2\\\\ -1\\\\ \\end {pmatrix}$$ ∴ x = 2 and y = -1 Ans Here, 2x - y = 1...............................(1) x + 2y = 3..............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -1\\\\ 1 & 2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -1\\\\ 1 & 2\\\\ \\end {pmatrix}$$, X = $$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2", "value problem, \\begin{align*} \\frac{dx}{dt} & = - 3 x + \\frac{4}{3} y,\\\\ \\frac{dy}{dt} & = 3 x - 3 y,\\\\ x(0) & = 100,\\\\ y(0) & = 0. \\end{align*} If we write our system in matrix form, $${\\mathbf x}' = A {\\mathbf x}\\text{,}$$ then \\begin{equation*} A = \\begin{pmatrix} -3 & 4/3 \\\\ 3 & -3 \\end{pmatrix}. \\end{equation*} It is easy to check that we have eigenvalues $$\\lambda = -1$$ and $$\\mu = -5$$ with eigenvectors $${\\mathbf u} = (2, 3)$$ and $${\\mathbf v} = (-2, 3)\\text{,}$$ respectively. Thus, we have two solutions to our system, \\begin{align*} {\\mathbf x}_1(t) & = e^{-t} {\\mathbf u},\\\\ {\\mathbf x}_2(t) & = e^{-5t} {\\mathbf v}. \\end{align*} Since any linear combination of solutions is also a solution, \\begin{equation*} {\\mathbf x}(t) = c_1 \\begin{pmatrix} 2e^{-t} \\\\ 3e^{-t} \\end{pmatrix} + c_2 \\begin{pmatrix} -2e^{-5t} \\\\ 3e^{-5t} \\end{pmatrix} \\end{equation*} is a solution to our system. Using the initial values $$x(0) = 100$$ and $$y(0) = 0\\text{,}$$ we can determine that $$c_1 = 25$$ and $$c_2 = -25\\text{.}$$ We now have the solution that we seek, \\begin{align*} x(t) & = 50 e^{-t} + 50 e^{-5t}\\\\ y(t) & = 75 e^{-t} - 75 e^{-5t}. \\end{align*} ### Subsection3.2.3Solving Linear Systems Our goal is to prove the following theorem. If $${\\mathbf v}_1$$ and $${\\mathbf v}_2$$ are linearly dependent, then there exists $$\\alpha", "case 2 If $\\frac{1}{2}k^{2}-\\frac{9}{2}k+10=0$, then $z=t$, a free parameter, which can be any value. Second row in (1) now gives \\begin{align*} \\frac{4}{3}y+\\left( k-\\frac{7}{3}\\right) t & =0\\\\ y & =-\\frac{3}{4}t\\left( k-\\frac{7}{3}\\right) \\end{align*} And first row gives \\begin{align*} 3x+2y+z & =0\\\\ 3x & =-2\\left( -\\frac{3}{4}t\\left( k-\\frac{7}{3}\\right) \\right) -t\\\\ x & =\\frac{1}{2}t\\left( k-3\\right) \\end{align*} Hence in this case, the NullSpace is $$\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} =t \\begin{pmatrix} \\frac{1}{2}\\left( k-3\\right) \\\\ -\\frac{3}{4}\\left( k-\\frac{7}{3}\\right) \\\\ 1 \\end{pmatrix}$$ Since $\\frac{1}{2}k^{2}-\\frac{9}{2}k+10=0$, the solution is $k=5,k=4$. Substituting these values into the above gives the NullSpace as $$\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} = t\\left\\{ \\begin{pmatrix} 1\\\\ -2\\\\ 1 \\end{pmatrix} , \\begin{pmatrix} \\frac{1}{2}\\\\ -\\frac{5}{4}\\\\ 1 \\end{pmatrix} \\right\\}$$ For any $t$. For only $t=0$, this gives the result gives by Mathematica. So Wolfram Mathematica only considered the case 1 above, and did not consider case 2. Wolfram Alpha, which has additional A.I. engine, seems to have looked more into this. • Here's how to use Mathematica for constructing the reduced form: UpperTriangularize[First[LUDecomposition[{{3, 2, 1}, {7, 6, k}, {k, 4, 3}}]]]. Mar 27 '17 at 4:50 • @J.M. Thank you. Mar 27 '17 at 5:04 • @J.M. btw, this only works for square matrices. For non-square matrices, I had to write an M function to do this. Mar 27 '17 at 5:20 • Indeed, it is unfortunate that Mathematica", "of (2*3)*5 A: 36 B: 11 C: 25 D: 55 E: No option 24: If p and q are two non zero numbers and 18(p+q) = (18+p)q, which of the following must be true? A: q = 18 B: p = 18 C: p<1 D: q<1 E: No option 25: If $\\begin{vmatrix} x & 3\\\\ 2 & 7 \\end{vmatrix} = 15$ , find the value of x A: 3 B: 5 C: 4 D: 2 E: No option 26: Evaluate $\\begin{vmatrix} 2 & 0 & 5\\\\ 4 & 6 & 3\\\\ 8 & 9 & 1 \\end{vmatrix}$ A: -42 B: 102 C: 18 D: -102 E: No option 27: If $P = \\begin{pmatrix} 2 & -3\\\\ 1 & 1 \\end{pmatrix}$ , what is P-1 A: $\\begin{pmatrix} \\frac{1}{5} & \\frac{3}{5}\\\\ \\frac{1}{5} & \\frac{2}{5} \\end{pmatrix}$ B: $\\begin{pmatrix} \\frac{1}{5} & \\frac{3}{5}\\\\ -\\frac{1}{5} & \\frac{2}{5} \\end{pmatrix}$ C: $\\begin{pmatrix} -\\frac{1}{5} & \\frac{3}{5}\\\\ -\\frac{1}{5} & \\frac{2}{5} \\end{pmatrix}$ D: $\\begin{pmatrix} -\\frac{1}{5} & -\\frac{3}{5}\\\\ -\\frac{1}{5} & -\\frac{2}{5} \\end{pmatrix}$ E: No option 28: From the diagram below, find x A: 65o B: 50o C: 55o D: 75o E: No option 29: The interior angles of a quadrilateral are (x + 15)o, (2x - 45)o, (x - 30)o and (x + 10)o. Find the value of the least interior angle A: 102o B: 52o C: 82o D: 112o", "& 3(z^{2}-1) \\\\ -2(z^{2}-1) & -1+4z-z^{2} \\end{pmatrix}\\!.$$ From Step 1 we see that $$k=\\dim \\mathscr{E}_{\\mathbf{B}}=1$$. We compute $$\\overline{\\mathbf{A}}=\\mathcal{I}_{1}\\overline{\\mathbf{B}}$$ by computing its symbol. $$\\overline{\\mathbf{A}}^{\\ast}(z)=\\frac{1}{16}\\begin{pmatrix} 4z^{-1}(1+z)^{3} & -3z^{-1}(z+1) \\\\ 2z^{-2}(z^{2}-1)^{2} & -1+4z-z^{2} \\end{pmatrix}\\!.$$ In this step we transform back to the original basis $$\\mathbf{A}=\\overline{R}\\,\\overline{\\mathbf{A}}\\,\\overline{R}^{-1}$$, by deriving A*(z). \\begin{align} \\mathbf{A}^{\\ast}(z)=\\frac{1}{32}z^{-2}\\begin{pmatrix} z^{4}+16z^{3}+18z^{2}+7z-2 & \\enspace 3z^{4}+8z^{3}+14z^{2}+z-2 \\\\[0.3cm] 7z^{4}+8z^{3}+12z^{2}+7z+2 & \\enspace 5z^{4}+16z^{3}+4z^{2}+z+2 \\end{pmatrix}\\!. \\end{align} (4.22) It follows from the analysis preceeding Procedure 4.16 that SA is C2. To verify Remark 4.17 we show that $$\\mathscr{E}_{\\mathbf{A}}$$ has the same basis as $$\\mathscr{E}_{\\mathbf{B}}$$. We compute $$\\mathbf{A}^{\\ast}(1)=\\frac{1}{8}\\Big(\\begin{array}{r r} 10 & 6 \\\\[0.1cm] 9 & 7 \\end{array}\\Big), \\quad \\mathbf{A}^{\\ast}(-1)=\\frac{1}{16}\\Big(\\begin{array}{r r} -3 & 3 \\\\ 3 & -3 \\end{array}\\Big)$$ and their eigenvalues and eigenvectors: \\begin{align} &\\textrm{For } \\mathbf{A}^{\\ast}(1): \\quad \\textrm{eigenvalues}: 2, \\tfrac{1}{8},\\quad \\text{eigenvectors: }\\Big(\\begin{array}{r r} 1 \\\\ 1 \\end{array}\\Big), \\Big(\\begin{array}{r r} -2 \\\\ 3 \\end{array}\\Big), \\text{ resp. }\\\\ \\nonumber &\\textrm{For } \\mathbf{A}^{\\ast}(-1): \\quad \\textrm{eigenvalues}: 0, -\\tfrac{3}{8}, \\quad \\text{eigenvectors: }\\Big(\\begin{array}{r r} 1 \\\\ 1 \\end{array}\\Big), \\Big(\\begin{array}{r r} -1 \\\\ 1 \\end{array}\\Big),\\text{ resp. } \\end{align} (4.23) Therefore by (4.21), (4.23) and Lemma 4.6, $$\\mathscr{E}_{\\mathbf{A}}$$ and $$\\mathscr{E}_{\\mathbf{B}}$$ are spanned by $${1 \\choose 1}$$. Note that the eigenvectors corresponding to the eigenvalues which have modulus less than 1, of MA and MB are different. Thus in order to generate a C3 scheme from SA, a new canonical transformation has to be computed. Also,", "& -1\\\\ 1 & 2\\\\ \\end {vmatrix}$$ = 2 × 2 - 1 × -1 = 4 + 1 = 5 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{5}$$$$\\begin {pmatrix} 2 & 1\\\\ -1 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {-2}{5} & \\frac {1}{5}\\\\ \\frac {-1}{5} & \\frac {2}{5}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 1\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2}{5} + \\frac {3}{5}\\\\ \\frac {-1}{5} + \\frac {6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {2 + 3}{5}\\\\ \\frac {-1 + 6}{5}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {5}{5}\\\\ \\frac {5}{5}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ 1\\\\ \\end {pmatrix}$$ ∴ x = 1 and y = 1 Ans Here, x + 2y = 8........................................(1) 2x + 3y = 11..................................(2) Given equation in matrix form: $$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ , AX = B where, A =$$\\begin {pmatrix} 1 & 2\\\\ 2 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 8\\\\ 11\\\\ \\end {pmatrix}$$ $$\\begin", "the inverse of $M$ exists, and it's not hard to see that: (3) \\begin{align} \\quad M^{-1} = \\begin{bmatrix} -\\frac{1}{2} & \\frac{7}{12} & \\frac{1}{3}\\\\ 0 & -\\frac{1}{6} & \\frac{1}{3}\\\\ \\frac{1}{2} & -\\frac{1}{12} & -\\frac{1}{3} \\end{bmatrix} \\end{align} We saw earlier that for $\\mathbf{x} = [2, 3, 1]$ that $\\phi_M(\\mathbf{x}) = [9, 7, 13]$ and so: (4) \\begin{align} \\quad \\phi_M^{-1}(\\phi_M(\\mathbf{x})) = \\phi_M(\\mathbf{x})M^{-1} = \\begin{bmatrix} 9 & 7 & 13 \\end{bmatrix} \\begin{bmatrix} -\\frac{1}{2} & \\frac{7}{12} & \\frac{1}{3}\\\\ 0 & -\\frac{1}{6} & \\frac{1}{3}\\\\ \\frac{1}{2} & -\\frac{1}{12} & -\\frac{1}{3} \\end{bmatrix} = \\begin{bmatrix} 2 & 3 & 1 \\end{bmatrix} \\end{align}", "$$M=\\begin{pmatrix} \\frac{4- \\cos(\\rho)}{3} & \\frac{2- 2\\cos(\\rho)}{3} & 0 & -\\frac{\\sin(\\rho)}{\\sqrt{3}} \\\\ \\frac{2\\cos(\\rho) - 2}{3} & \\frac{4- \\cos(\\rho)}{3} & 0 & \\frac{2\\sin(\\rho)}{\\sqrt{3}}\\\\ 0 & 0 & 1 & 0 \\\\ -\\frac{\\sin(\\rho)}{\\sqrt{3}} & -\\frac{2\\sin(\\rho)}{\\sqrt{3}} & 0 & \\cos(\\rho) \\\\ \\end{pmatrix}$$ whose logarithm you are invited to take. I suspect that entry to be something like $(4\\cos \\rho -1)/3$ instead---see below. Your time appears to be in the 4th component, unlike the first one in the conventional notation. In any case, observe $M=\\mathbb{1}$ as $\\rho\\to 0$, so to find its logarithm, we expand in the first two powers of ρ, $$M=\\mathbb{1} -\\frac{\\rho}{\\sqrt{3}}\\begin{pmatrix} 0 & 0 & 0 & 1\\\\ 0 & 0 & 0 & -2\\\\ 0 & 0 & 0 & 0 \\\\ 1 & 2& 0 & 0 \\\\ \\end{pmatrix}+ \\frac{\\rho^2}{6} \\begin{pmatrix} 1 & 2 & 0 & 0 \\\\ - 2 & 1& 0 & 0\\\\ 0 & 0 & 0 & 0 \\\\ 0 &0 & 0 & -3 \\\\ \\end{pmatrix} +Ο(\\rho^3).$$ Let us call the first big matrix A and the second one B. Note $$A^2=\\begin{pmatrix} 1 & 2 & 0 & 0 \\\\ - 2 & -4& 0 & 0\\\\ 0 & 0 & 0 & 0 \\\\ 0 &0 & 0 & -3 \\\\ \\end{pmatrix} .$$ Now, if the 22 entry of B were", "designate your choice for the first and the second point. Designating the points will reduce the risk of entering the values in the wrong location of the formula. i) $& \\begin{pmatrix} x_1, & y_1 \\\\ -3, & -8 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ 5, & 8 \\end{pmatrix}\\\\& \\text{Substitute the values into the formula} && m=\\frac{y_2-y_1}{x_2-x_1}\\\\& && m=\\frac{8--8}{5--3}\\\\& \\text{Simplify} && m=\\frac{8+8}{5+3}\\\\& \\text{Calculate} && m=\\frac{16}{8}\\\\& \\text{Simplify} && m=\\frac{2}{1}$ ii) $& \\begin{pmatrix} x_1, & y_1 \\\\ 9, & 5 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ -1, & 6 \\end{pmatrix}\\\\& m =\\frac{y_2-y_1}{x_2-x_1}\\\\& m =\\frac{6-5}{-1-9}\\\\& m =\\frac{1}{-10}$ There should never be a negative sign in the denominator of a fraction. $m =-\\frac{1}{10}$ iii) $& \\begin{pmatrix} x_1, & y_1 \\\\ -2, & 7 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ -3, & -1 \\end{pmatrix}\\\\& m =\\frac{y_2-y_1}{x_2-x_1}\\\\& m =\\frac{-1-7}{-3--2}\\\\& m =\\frac{-1-7}{-3+2}\\\\& m =\\frac{-8}{-1}\\\\& m =\\frac{8}{1}$ iv) $& \\begin{pmatrix} x_1, & y_1 \\\\ 4, & 0 \\end{pmatrix} \\quad \\begin{pmatrix} x_2, & y_2 \\\\ 0, & -3 \\end{pmatrix} && \\text{Express the} \\ x- \\text{and} \\ y- \\text{intercepts as coordinates of a point.}\\\\& m =\\frac{y_2-y_1}{x_2-x_1}\\\\& m =\\frac{-3-0}{0-4}\\\\& m =\\frac{-3}{-4}\\\\& m =\\frac{3}{4}$ Example C If you look at the following graphs, a slope or gradient is not present. (a) Two points on this line are (–5, 5) and (4, 5). (b) Two points on this line", "solutions: $$x = 5$$ and $$x = -5$$. 4. We factor and solve $$4x^2 - 9=0$$ in two steps: • Step 1: we notice that $$4x^2 - 9$$ is the difference of two squares. Indeed $$x^2 - 49 = \\begin{pmatrix}2x\\end{pmatrix}^2 - 3^2$$. So we can write it in factored form: $x^2 - 49 = \\begin{pmatrix} 2x - 3 \\end{pmatrix}\\begin{pmatrix} 2x + 3 \\end{pmatrix}$ • Step 2: we use the null product law to solve the factored equation $$\\begin{pmatrix} 2x - 3 \\end{pmatrix}\\begin{pmatrix} 3x + 3 \\end{pmatrix} = 0$$. That's: $$\\begin{pmatrix} 2x - 3 \\end{pmatrix}\\begin{pmatrix} 2x + 3 \\end{pmatrix} = 0$$ if and only if: $$2x - 3 = 0$$ or $$2x+3 = 0$$, which leads to: $$2x = 3$$ or $$2x = -3$$. Which, in turn, leads to: $$x = \\frac{3}{2}$$ or $$x = -\\frac{3}{2}$$. Finally we can state that this quadratic equation has two solutions: $$x = \\frac{3}{2}$$ and $$x = -\\frac{3}{2}$$. 5. We factor and solve $$16x^2 - 81=0$$ in two steps: • Step 1: we notice that $$16x^2 - 81$$ is the difference of two squares. Indeed $$16x^2 - 81 = \\begin{pmatrix}4x\\end{pmatrix}^2 - 9^2$$. So we can write it in factored form: $16x^2 - 81 = \\begin{pmatrix} 4x - 9 \\end{pmatrix}\\begin{pmatrix} 4x + 9 \\end{pmatrix}$ • Step 2: we use the null product law to", "2c_2 -2c_3 &= 0 \\end{align} Solving the above system of equations shows that the only possible solution is $c_1 = c_2 = c_3 = 0$. Thus $\\{ x_1 , x_2 , x_3 \\}$ is linearly independent. One way to solve the system of equations is shown below. First, subtract (4/3) times the 1st equation from the 2nd equation. $-\\frac{4}{3}(3c_1 + 4c_2 +6c_3) + (4c_1 -2c_2 + 8c_3) = -\\frac{22}{3}c_2 = -\\frac{4}{3}0 + 0 = 0 \\Rightarrow c_2 = 0$ Then add the 1st and 3 times the 3rd equations together, and substitute in $c_2 = 0$. $(3c_1 + 4c_2 +6c_3) + 3*(-2c_1 + 2c_2 -2c_3) = -3c_1 + 10 c_2 = -3c_1 + 10*0 = 0 + 3*0 = 0 \\Rightarrow c_1 = 0$ Now, substituting both $c_1 = 0$ and $c_2 = 0$ into equation 2 gives. $4c_1 -2c_2 + 8c_3 = 4*0 -2*0 + 8c_3 = 0 \\Rightarrow c_3 = 0$ So $c_1 = c_2 = c_3 = 0$, and $\\{ x_1 , x_2 , x_3 \\}$ are linearly independent. ### Example M.5.2 $x_1 =\\begin{pmatrix} 1 \\\\ -8 \\\\ 8 \\end{pmatrix}, x_2 =\\begin{pmatrix} 4 \\\\ -2 \\\\ 2 \\end{pmatrix}, x_3 =\\begin{pmatrix} 1 \\\\ 3 \\\\ -2 \\end{pmatrix}$ In this case $\\{ x_1 , x_2 , x_3 \\}$are linearly dependent, because if $c = (-1, 1, -2)$," ]

[ "-abc 6. If A is a square matrix of order 3, then |kA| is (a) k|A| (b) -k|A| (c) k3|A| (d) -k3|A| 7. adj (AB) is equal to (a) (b) (c) (d) 8. The inverse matrix of $\\begin{pmatrix} \\frac { 1 }{ 5 } & \\frac { 5 }{ 25 } \\\\ \\frac { 2 }{ 5 } & \\frac { 1 }{ 2 } \\end{pmatrix}$ is (a) ${{7}\\over{30}}\\begin{pmatrix} \\frac { 1 }{ 2 } & \\frac { 5 }{ 12 } \\\\ \\frac { 2 }{ 5 } & \\frac { 4 }{ 5 } \\end{pmatrix}$ (b) ${{7}\\over{30}}\\begin{pmatrix} \\frac { 1 }{ 2 } & \\frac { -5 }{ 12 } \\\\ \\frac { -2 }{ 5 } & \\frac { 1 }{ 5 } \\end{pmatrix}$ (c) ${{30}\\over{7}}\\begin{pmatrix} \\frac { 1 }{ 2 } & \\frac { 5 }{ 12 } \\\\ \\frac { 2 }{ 5 } & \\frac { 4 }{ 5 } \\end{pmatrix}$ (d) ${{30}\\over{7}}\\begin{pmatrix} \\frac { 1 }{ 2 } & \\frac { -5 }{ 12 } \\\\ \\frac { -2 }{ 5 } & \\frac { 4 }{ 5 } \\end{pmatrix}$ 9. If A = $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$such that ad - bc $\\neq$ 0 then A-1 is (a) ${{1}\\over{ad-bc}}\\begin{pmatrix} d & b \\\\-c & a\\end{pmatrix}$ (b)", "$dy = a\\,dx$ $\\frac{dy}{dx} = a$ $dy = \\mathbf{a}\\,d\\mathbf{x}$ $\\frac{dy}{d\\mathbf{x}} = \\mathbf{a}$ $dy = {\\rm tr}(\\mathbf{A}\\,d\\mathbf{X})$ $\\frac{dy}{d\\mathbf{X}} = \\mathbf{A}$ $d\\mathbf{y} = \\mathbf{a}\\,dx$ $\\frac{d\\mathbf{y}}{dx} = \\mathbf{a}$ $d\\mathbf{y} = \\mathbf{A}\\,d\\mathbf{x}$ $\\frac{d\\mathbf{y}}{d\\mathbf{x}} = \\mathbf{A}$ $d\\mathbf{Y} = \\mathbf{A}\\,dx$ $\\frac{d\\mathbf{Y}}{dx} = \\mathbf{A}$ 4. ^ a b c Here, $\\mathbf{0}$ refers to a column vector of all 0's, of size n, where n is the length of x. 5. ^ a b Here, $\\mathbf{0}$ refers to a matrix of all 0's, of the same shape as X. 6. Petersen, Kaare Brandt and Michael Syskind Pedersen. The Matrix Cookbook. November 14, 2008. http://matrixcookbook.com. [3] This book uses a mixed layout, i.e. by Y in $\\frac{\\partial \\mathbf{Y}}{\\partial x},$ by X in $\\frac{\\partial y}{\\partial \\mathbf{X}}.$ $\\frac{d\\,\\ln au}{dx} = \\frac{1}{au}\\frac{d(au)}{dx} = \\frac{1}{au}a\\frac{du}{dx} = \\frac{1}{u}\\frac{du}{dx} = \\frac{d\\,\\ln u}{dx}.$", "= $$\\frac {1}{-13}$$$$\\begin {pmatrix} -3 & -1\\\\ -4 & 3\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {3}{13} & \\frac {1}{13}\\\\ \\frac {4}{13} & \\frac {-3}{13}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 51\\\\ 3\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153}{13} & \\frac {3}{13}\\\\ \\frac {204}{13} & \\frac {-9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {153 + 3}{13}\\\\ \\frac {204 - 9}{13}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {156}{13}\\\\ \\frac {195}{13}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 12\\\\ 5\\\\ \\end {pmatrix}$$ ∴ x = 12 and y = 5 Ans Here, 2x - 5y = 1.................................(1) 7x + 3y = 24.............................(2) Given equation in matrix form: $$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 1\\\\ 24\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 2 & -5\\\\ 7 & 3\\\\ \\end {vmatrix}$$ = 2 × 3 - 7 × (-5) = 6 + 35 = 41 ≠ 0 It has a unique solution. A-1= $$\\frac", "0s. We instead get the matrix \\begin{align*} \\begin{bmatrix} 3 & 1 & 9 & 4 & x \\\\ 5 & 7 & 3 & 2 & y \\\\ 2 & 8 & 3 & 7 & z \\end{bmatrix} \\end{align*} which we can row reduce to the matrix \\begin{align*} \\begin{bmatrix} 1 & 0 & 0 & \\frac{-49}{36} & \\frac{1}{72}(-x+23y-20z) \\\\ 0 & 1 & 0 & \\frac{11}{12} & \\frac{1}{24}(-x-y+4z) \\\\ 0 & 0 & 1 & \\frac{43}{54} & \\frac{1}{108}(13x-11y+8z) \\end{bmatrix}. \\end{align*} Since we only need one solution, it will be easiest to let $$c_{4}=0$$, then $$c_{1}=\\frac{1}{72}(-x+23y-20z), c_{2}=\\frac{1}{24}(-x-y+4z),$$ and $$c_{3}=\\frac{1}{108}(13x-11y+8z)$$. We are now ready for the proof. #### Proof Let $$S=\\left\\{(3,5,2),(1,7,8),(9,3,3),(4,2,7)\\right\\}$$ and $$(x,y,z) \\in \\mathbb{R}^{3}$$. Then note that \\begin{align*} &\\frac{1}{72}(-x+23y-20z)(3,5,2)+\\frac{1}{24}(-x-y+4z)(1,7,8) \\\\ &+\\frac{1}{24}(-x-y+4z)(9,3,3)+0(4,2,7)=(x,y,z). \\end{align*} Hence, $$S$$ spans $$\\mathbb{R}^{3}$$. ### Basis Here, we can just look at the work we have already done. Since $$S$$ is not linearly independent, we know that $$S$$ is not a basis of $$\\mathbb{R}^{3}$$. ## Conclusion Here we had a set that was spanning but not linearly independent. The set was therefore not a basis. In the next post, we will look at other examples of vector spaces to see which properties it has. I hope this helped you with determining whether or not a set spanned a given vector space or was linearly independent and how", "48x + 54 y &= 27+8=35. \\\\ \\end{align*} Write the two equations in matrix form: $$\\begin{pmatrix} 24 & -18 \\\\ 48 & 54 \\\\ \\end{pmatrix} \\begin{pmatrix} x \\\\ y \\\\ \\end{pmatrix} = \\begin{pmatrix} 5 \\\\ 35 \\\\ \\end{pmatrix}.$$ Write this as an augmented matrix: $$\\begin{pmatrix} 24 & -18 & 5\\\\ 48 & 54 & 35 \\\\ \\end{pmatrix}$$ and then row-reduce (multiply the first row by $-2$ and then add the corresponding numbers to the second row): $$\\begin{pmatrix} 24 & -18 & 5\\\\ 0 & 90 & 25 \\\\ \\end{pmatrix}.$$ Divide the second row by $5$ to get: $$\\begin{pmatrix} 24 & -18 & 5\\\\ 0 & 18 & 5 \\\\ \\end{pmatrix}.$$ Add the second row to the first row: $$\\begin{pmatrix} 24 & 0 & 10\\\\ 0 & 18 & 5 \\\\ \\end{pmatrix}.$$ Divide the first row by $24$ and divide the second row by $18$: $$\\begin{pmatrix} 1 & 0 & \\frac{5}{12}\\\\ 0 & 1 & \\frac{5}{18} \\\\ \\end{pmatrix}.$$ So $$x= \\frac{5}{12} \\mbox{ and } y= \\frac{5}{18},$$ and we conclude $$\\boxed{x+y = \\frac{5}{12} + \\frac{5}{18} = \\frac{25}{36}}.$$ Hint: let $s=x+y\\,$, then substituting $x=s-y$ back into the equations gives: \\begin{cases} \\begin{align} 2\\left((s-y)-\\frac{1}{3} \\right) - \\frac{3}{2}\\left(y-\\frac{1}{6}\\right) &= 0 \\quad\\iff\\quad 2s - \\frac{7}{2}y = \\frac{5}{12}\\\\ \\frac{8}{3}\\left((s-y)-\\frac{1}{6}\\right) + 3\\left(y-\\frac{1}{2}\\right) &= 0 \\quad\\iff\\quad \\frac{8}{3}s + \\frac{1}{3}y=\\frac{35}{18} \\end{align} \\end{cases} Eliminate $y$ between the latter equations (for", "- \\mathbf{\\mu}_{i})^{T}$ and $\\mathbf{\\mu}_{i} = \\frac{\\sum_{j: y_{j}=i}\\mathbf{x}_{j}}{n_{i}}$. However, the between class covariance matrix $\\mathbf{S}_{B}$ is not easy to obtain. One of the simplifications is that we may assume that the total covariance $\\mathbf{S}_{T}$ of the data is constant, since $\\mathbf{S}_{W}$ is easy to compute, we can get $\\mathbf{S}_{B}$ using the following relationship: \\begin{align} \\mathbf{S}_{B} = \\mathbf{S}_{T} - \\mathbf{S}_{W} \\end{align} Actually, there is another generation for $\\mathbf{S}_{B}$. Denote a total mean vector $\\mathbf{\\mu}$ by \\begin{align} \\mathbf{\\mu} = \\frac{1}{n}\\sum_{i}\\mathbf{x_{i}} = \\frac{1}{n}\\sum_{j=1}^{k}n_{j}\\mathbf{\\mu}_{j} \\end{align} Thus the total covariance matrix $\\mathbf{S}_{T}$ is \\begin{align} \\mathbf{S}_{T} = \\sum_{i}(\\mathbf{x_{i}-\\mu})(\\mathbf{x_{i}-\\mu})^{T} \\end{align} Thus we obtain \\begin{align} & \\mathbf{S}_{T} = \\sum_{i=1}^{k}\\sum_{j: y_{j}=i}(\\mathbf{x}_{j} - \\mathbf{\\mu}_{i} + \\mathbf{\\mu}_{i} - \\mathbf{\\mu})(\\mathbf{x}_{j} - \\mathbf{\\mu}_{i} + \\mathbf{\\mu}_{i} - \\mathbf{\\mu})^{T} \\\\& = \\sum_{i=1}^{k}\\sum_{j: y_{j}=i}(\\mathbf{x}_{j}-\\mathbf{\\mu}_{i})(\\mathbf{x}_{j}-\\mathbf{\\mu}_{i})^{T}+ \\sum_{i=1}^{k}\\sum_{j: y_{j}=i}(\\mathbf{\\mu}_{i}-\\mathbf{\\mu})(\\mathbf{\\mu}_{i}-\\mathbf{\\mu})^{T} \\\\& = \\mathbf{S}_{W} + \\sum_{i=1}^{k} n_{i}(\\mathbf{\\mu}_{i}-\\mathbf{\\mu})(\\mathbf{\\mu}_{i}-\\mathbf{\\mu})^{T} \\end{align} Since the total covariance $\\mathbf{S}_{T}$ is the sum of the within class covariance $\\mathbf{S}_{W}$ and the between class covariance $\\mathbf{S}_{B}$, we can denote the second term as the general between class covariance matrix $\\mathbf{S}_{B}$, thus we obtain \\begin{align} \\mathbf{S}_{B} = \\sum_{i=1}^{k} n_{i}(\\mathbf{\\mu}_{i}-\\mathbf{\\mu})(\\mathbf{\\mu}_{i}-\\mathbf{\\mu})^{T} \\end{align} Therefore, \\begin{align} \\mathbf{S}_{T} = \\mathbf{S}_{W} + \\mathbf{S}_{B} \\end{align} Recall that in the two class case problem, we have \\begin{align} & \\mathbf{S}_{B^{\\ast}} = (\\mathbf{\\mu}_{1}-\\mathbf{\\mu}_{2})(\\mathbf{\\mu}_{1}-\\mathbf{\\mu}_{2})^{T} \\\\ & = (\\mathbf{\\mu}_{1}-\\mathbf{\\mu}+\\mathbf{\\mu}-\\mathbf{\\mu}_{2})(\\mathbf{\\mu}_{1}-\\mathbf{\\mu}+\\mathbf{\\mu}-\\mathbf{\\mu}_{2})^{T} \\\\ & = ((\\mathbf{\\mu}_{1}-\\mathbf{\\mu})-(\\mathbf{\\mu}_{2}-\\mathbf{\\mu}))((\\mathbf{\\mu}_{1}-\\mathbf{\\mu})-(\\mathbf{\\mu}_{2}-\\mathbf{\\mu}))^{T} \\\\ & = (\\mathbf{\\mu}_{1}-\\mathbf{\\mu})(\\mathbf{\\mu}_{1}-\\mathbf{\\mu})^{T}+(\\mathbf{\\mu}_{2}-\\mathbf{\\mu})(\\mathbf{\\mu}_{2}-\\mathbf{\\mu})^{T} \\end{align} From the general form, \\begin{align} & \\mathbf{S}_{B} = n_{1}(\\mathbf{\\mu}_{1}-\\mathbf{\\mu})(\\mathbf{\\mu}_{1}-\\mathbf{\\mu})^{T} + n_{2}(\\mathbf{\\mu}_{2}-\\mathbf{\\mu})(\\mathbf{\\mu}_{2}-\\mathbf{\\mu})^{T} \\end{align} Apparently,", "0\\\\0 & \\frac12 & 0\\\\0 & 0 & 1\\end{pmatrix}.$$$$ The problem of calculating $$A^n$$ is thus reduced to calculating $$J^n$$. Let $$a_{ij}^{(n)}$$ denote the entry of $$J^n$$ in the $$i$$-th row and $$j$$-th column. The product of an arbitrary $$3\\times3$$-matrix with $$J$$ is given by: $$$$\\begin{pmatrix} a&b&c\\\\d&e&f\\\\g&h&i \\end{pmatrix} J = \\begin{pmatrix} \\frac a2&a+\\frac b2&c\\\\\\frac d2&d+\\frac e2&f\\\\\\frac g2&g+\\frac h2&i \\end{pmatrix}.$$$$ We can deduce that, for all $$n\\in\\Bbb N$$: \\begin{align} a_{11}^{(n)}&=a_{22}^{(n)}=\\frac1{2^n}, \\\\a_{21}^{(n)}&=a_{31}^{(n)}=0,\\\\ a_{13}^{(n)}&=a_{23}^{(n)}=a_{32}^{(n)}=0, \\\\ a_{33}^{(n)}&=1,\\\\ a_{12}^{(n+1)}&=a_{11}^{(n)}+\\frac{a_{12}^{(n)}}2=\\frac1{2^n}+\\frac{a_{12}^{(n)}}2. \\end{align} Thus, all $$a_{ij}^{(n)}$$ are explicitly known except for $$a_{12}^{(n)}$$. Note that, by the last equation, $$$$a_{12}^{(n+1)}=2^{-n}+\\frac{a_{12}^{(n)}}2 = 2^{-n}+2^{-n}+\\frac{a_{12}^{(n-1)}}4 = \\dots = (n+1)\\cdot2^{-n}.$$$$ Thus, $$$$J^n=\\begin{pmatrix}2^{-n}&n\\cdot 2^{1-n} & 0\\\\0 & 2^{-n} & 0\\\\0 & 0 & 1\\end{pmatrix}.$$$$ And by some calculations, we find that $$$$A^n=\\big(v_1,v_2,v_3\\big)J^n\\big(v_1,v_2,v_3\\big)^{-1}= \\begin{pmatrix} 2^{-n} & n\\cdot 2^{-n-1} - 2^{-n-1} + \\frac12 & {1-\\frac{n+1}{2^n}\\over2}\\\\ 0 & {2^{-n}+1\\over2} & {1-2^{-n}\\over2} \\\\ 0 & {1-2^{-n}\\over2} & {2^{-n}+1\\over2} \\end{pmatrix}.$$$$ • I don't understand very well the \"your\" in the first sentence of this answer \"Note that your matrix...\" ; in fact, as far I have understood, you are answering your own question ... – Jean Marie May 25 at 17:39 • @JeanMarie indeed I am answering my own question. On the questions on this forum that I saw that were answered by the OP, the OP was always talking about himself in second person; so I", "value problem, \\begin{align*} \\frac{dx}{dt} & = - 3 x + \\frac{4}{3} y,\\\\ \\frac{dy}{dt} & = 3 x - 3 y,\\\\ x(0) & = 100,\\\\ y(0) & = 0. \\end{align*} If we write our system in matrix form, $${\\mathbf x}' = A {\\mathbf x}\\text{,}$$ then \\begin{equation*} A = \\begin{pmatrix} -3 & 4/3 \\\\ 3 & -3 \\end{pmatrix}. \\end{equation*} It is easy to check that we have eigenvalues $$\\lambda = -1$$ and $$\\mu = -5$$ with eigenvectors $${\\mathbf u} = (2, 3)$$ and $${\\mathbf v} = (-2, 3)\\text{,}$$ respectively. Thus, we have two solutions to our system, \\begin{align*} {\\mathbf x}_1(t) & = e^{-t} {\\mathbf u},\\\\ {\\mathbf x}_2(t) & = e^{-5t} {\\mathbf v}. \\end{align*} Since any linear combination of solutions is also a solution, \\begin{equation*} {\\mathbf x}(t) = c_1 \\begin{pmatrix} 2e^{-t} \\\\ 3e^{-t} \\end{pmatrix} + c_2 \\begin{pmatrix} -2e^{-5t} \\\\ 3e^{-5t} \\end{pmatrix} \\end{equation*} is a solution to our system. Using the initial values $$x(0) = 100$$ and $$y(0) = 0\\text{,}$$ we can determine that $$c_1 = 25$$ and $$c_2 = -25\\text{.}$$ We now have the solution that we seek, \\begin{align*} x(t) & = 50 e^{-t} + 50 e^{-5t}\\\\ y(t) & = 75 e^{-t} - 75 e^{-5t}. \\end{align*} ### Subsection3.2.3Solving Linear Systems Our goal is to prove the following theorem. If $${\\mathbf v}_1$$ and $${\\mathbf v}_2$$ are linearly dependent, then there exists $$\\alpha", "} & a_{ 22 } \\end{bmatrix}$$ Then, a11 = $$\\frac{ 1 }{ 1 } = 1$$ a12 = $$\\frac{ 1 }{ 2 }$$ a21 = $$\\frac{ 2 }{ 1 } = \\frac{ 2 }{ 1 }$$ a22 = $$\\frac{ 2 }{ 2 } = 1$$ Hence, the required matrix is A = $$\\begin{bmatrix} 1 & \\frac{ 1 }{ 2 } \\\\ 2 & 1 \\end{bmatrix}$$ (c) As per the data given in the question: mij = $$\\frac{ \\left ( i + 2j \\right )^{ 2 } }{ 2 }$$ A ( 2 × 2 ) matrix is usually given by, A = $$\\begin{bmatrix} a_{ 11 } & a_{ 12 } \\\\ a_{ 21 } & a_{ 22 } \\end{bmatrix}$$ Then, a11 = $$\\frac{ \\left ( 1 + 2 \\times 1 \\right )^{ 2 } }{ 2 } = \\frac{ 9 }{ 2 }$$ a12 = $$\\frac{ \\left ( 1 + 2 \\times 2 \\right )^{ 2 } }{ 2 } = \\frac{ 25 }{ 2 }$$ a21 = $$\\frac{ \\left ( 2 + 2 \\times 1 \\right )^{ 2 } }{ 2 } = \\frac{ 16 }{ 2 } = 8$$ a22 = $$\\frac{ \\left ( 2 + 2 \\times 2 \\right )^{ 2 } }{ 2 } = \\frac{ 36 }{ 2 } = 18$$ Hence, the", "# Matrix Equation, Solving for Variables. I'm going through my exercises, and came across a problem that wasn't covered in our lectures. Here's the question: \\begin{align} \\begin{bmatrix} a-b & b+c\\\\ 3d+c & 2a-4d \\end{bmatrix} \\end{align} = \\begin{align} \\begin{bmatrix} 8 & 1\\\\ 7 & 6 \\end{bmatrix} \\end{align} What I have done so far is: \\begin{align} \\begin{bmatrix} a-b-8 & b+c-1\\\\ 3d+c-7 & 2a-4d-6 \\end{bmatrix} \\end{align} = \\begin{align} \\begin{bmatrix} 0 & 0\\\\ 0 & 0 \\end{bmatrix} \\end{align} And the solving for variables, $$a-b-8 = 0$$ $$a-b = 8$$ $$a = \\frac{8}{-b}$$ $$b+c-1=0$$ $$b+c=1$$ $$b=\\frac{1}{c}$$ $$3d+c-7=0$$ $$3d+c=7$$ $$3d=\\frac{7}{c}$$ $$d=\\frac{7}{3c}$$ $$2a-4d-6=0$$ $$2a-4d=6$$ $$\\frac{16}{-2b}-\\frac{28}{12c}=6$$ - Whoa. Yes, quite a few very basic mistakes here. In fact, every single one of your subproblems is wrong; you are dividing when you should be subtracting. I hope you will not take offense but I would strongly recommend a conference with your instructor. – Michael Grant May 22 '13 at 2:48 +1 for showing your work. – response May 22 '13 at 2:51 @MichaelC.Grant No none taken, we haven't done anything like this in our labs or during the lectures, and I was just guessing as to what I should be doing. – Matt Brzezinski May 22 '13 at 2:55 Cool. It does seem rather unfortunate that they would foist this multiequation problem on you before you had", "so we have $$\\begin{pmatrix} c_1 \\\\ c_2 \\\\ c_3 \\end{pmatrix} = V^{-1}\\cdot\\begin{pmatrix} 1 \\\\ 5 \\\\ 14 \\end{pmatrix}$$ from which $$f(x)$$ can be determined directly. However, if you're like most people, inverting matrices doesn't exactly tickle your fancy! Luckily, now that we see that the interpolating cubic is unique, we could find it through the described matrix multiplication, but we would get to the same result if we proceeded a different route as well. This is where Lagrange polynomials come to the rescue. Using the general formula, we have immediately that \\begin{align} f(x) &= 0\\cdot(\\dots)+1\\cdot\\frac{x(x-2)(x-3)}{1(1-2)(1-3)}+5\\cdot\\frac{x(x-1)(x-3)}{2(2-1)(2-3)}+14\\cdot\\frac{x(x-1)(x-2)}{3(3-1)(3-2)} \\\\ &= \\frac{1}{2}\\left(x^3-5x^2+6x\\right)-\\frac{5}{2}\\left(x^3-4x^2+3x\\right)+\\frac{14}{6}\\left(x^3-3x^2+2x\\right) \\\\ &= \\frac{1}{6}\\left(2x^3+3x^2+x\\right) \\\\ &= \\frac{1}{6}x\\left(x+1\\right)\\left(2x+1\\right) \\end{align} You can generalize this approach to find expressions for $$\\sum n^p\\quad\\forall p\\in\\mathbb{N}$$. Or, you know, there's always Faulhaber's formula. Here, we present an approach that uses only the sum of the arithmetic progression $\\sum_{k=1}^nk=\\frac12n(n+1)$. Here, we note that $\\sum_{j=1}^k(1)=k$. Then, we can write \\begin{align} \\sum_{k=1}^nk^2&=\\sum_{k=1}^nk\\sum_{j=1}^k(1)\\\\\\\\ &=\\sum_{j=1}^n\\sum_{k=j}^n\\,k\\\\\\\\ &=\\frac12\\sum_{j=1}^n(n+1-j)(j+n)\\\\\\\\ &=\\frac12\\sum_{j=1}^n\\left(n(n+1)+j-j^2\\right)\\\\\\\\ &=\\frac12n^2(n+1)+\\frac14n(n+1)-\\frac12\\sum_{j=1}^nj^2\\\\\\\\ \\frac32\\sum_{k=1}^nk^2&=\\frac{(2n+1)n(n+1)}{4}\\\\\\\\ \\sum_{k=1}^nk^2&=\\frac{(2n+1)n(n+1)}{6} \\end{align} Late to the party, but I hope I am contributing something new to the answers already posted here. We seek $$\\sum_{i=1}^Ni^2=1^2+2^2+\\cdots+N^2$$ which, geometrically speaking, is equivalent to adding the area of $$N$$ squares with side lengths $$1,2,...,N$$. A visual depiction when $$N=2$$ is By adding the missing tile on the top-left corner, we may form a rectangle as follows, with area", "\\end{bmatrix}$$ Then, a11 = $$\\frac{ \\left ( 1 + 1 \\right )^{ 2 } }{ 2 } = \\frac{ 4 }{ 2 } = 2$$ a12 = $$\\frac{ \\left ( 1 + 2 \\right )^{ 2 } }{ 2 } = \\frac{ 9 }{ 2 }$$ a21 = $$\\frac{ \\left ( 2 + 1 \\right )^{ 2 } }{ 2 } = \\frac{ 9 }{ 2 }$$ a22 = $$\\frac{ \\left ( 2 + 2 \\right )^{ 2 } }{ 2 } = \\frac{ 16 }{ 2 } = 8$$ Hence, the required matrix is A = $$\\begin{bmatrix} 2 & \\frac{ 9 }{ 2 } \\\\ \\frac{ 9 }{ 2 } & 8 \\end{bmatrix}$$. (b) As per the data : mij = $$\\frac{ i }{ j }$$ A ( 2 × 2 ) matrix is given by: A = $$\\begin{bmatrix} a_{ 11 } & a_{ 12 } \\\\ a_{ 21 } & a_{ 22 } \\end{bmatrix}$$ Then, a11 = $$\\frac{ 1 }{ 1 } = 1$$ a12 = $$\\frac{ 1 }{ 2 }$$ a21 = $$\\frac{ 2 }{ 1 } = \\frac{ 2 }{ 1 }$$ a22 = $$\\frac{ 2 }{ 2 } = 1$$ Hence Based on the formula Matrices, the required matrix is A = $$\\begin{bmatrix} 1 & \\frac{ 1 }{ 2 } \\\\ 2 &", "is simply (10.6.8)$(\\mathbf{K}+\\mathbf{M})\\mathbf{w} = \\mathbf{f}.$ The matrix $$\\mathbf{K}$$ is called the stiffness matrix and $$\\mathbf{M}$$ is called the mass matrix. By their definitions, they are symmetric. The last piece of the puzzle is to make some selection of $$\\phi_1,\\ldots,\\phi_m$$ and obtain a fully specified algorithm. Example 10.6.2 Suppose we are given $$-u''+4u=x$$, with $$u(0)=u(\\pi)=0$$. We could choose the basis functions $$\\phi_k=\\sin(kx)$$ for $$k=1,2,3$$. Then \\begin{align*} M_{ij} & = 4 \\int_0^\\pi \\sin(ix) \\sin(jx)\\, dx, \\\\ K_{ij} & = ij \\int_0^\\pi \\cos(ix) \\cos(jx)\\, dx, \\\\ f_i &= \\int_0^\\pi x \\sin(ix)\\, dx. \\end{align*} With some calculation (or computer algebra), we find $\\begin{split} \\mathbf{M} = 2\\pi \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 1 & 0 \\\\ 0 & 0 & 1 \\end{bmatrix}, \\qquad \\mathbf{K} = \\frac{\\pi}{2} \\begin{bmatrix} 1 & 0 & 0 \\\\ 0 & 4 & 0 \\\\ 0 & 0 & 9 \\end{bmatrix}, \\qquad \\mathbf{f} = \\pi \\begin{bmatrix} 1 \\\\ -1/2 \\\\ 1/3 \\end{bmatrix}. \\end{split}$ Upon solving the resulting diagonal linear system, the approximate solution is $\\frac{2}{5}\\sin(x) - \\frac{1}{8} \\sin(2x) + \\frac{2}{39}\\sin(3x).$ Finite elements¶ One useful and general choice for the $$\\phi_i$$ are the piecewise linear hat functions constructed in Section 5.2. As usual, we select nodes $$a=t_0 < t_1 < \\cdots < t_n=b$$. Also define $h_i = t_i - t_{i-1}, \\qquad i=1,\\ldots,n.$ Then we set $$m=n-1$$,", "# Matrix Equation, Solving for Variables. I'm going through my exercises, and came across a problem that wasn't covered in our lectures. Here's the question: \\begin{align} \\begin{bmatrix} a-b & b+c\\\\ 3d+c & 2a-4d \\end{bmatrix} \\end{align} = \\begin{align} \\begin{bmatrix} 8 & 1\\\\ 7 & 6 \\end{bmatrix} \\end{align} What I have done so far is: \\begin{align} \\begin{bmatrix} a-b-8 & b+c-1\\\\ 3d+c-7 & 2a-4d-6 \\end{bmatrix} \\end{align} = \\begin{align} \\begin{bmatrix} 0 & 0\\\\ 0 & 0 \\end{bmatrix} \\end{align} And the solving for variables, $$a-b-8 = 0$$ $$a-b = 8$$ $$a = \\frac{8}{-b}$$ $$b+c-1=0$$ $$b+c=1$$ $$b=\\frac{1}{c}$$ $$3d+c-7=0$$ $$3d+c=7$$ $$3d=\\frac{7}{c}$$ $$d=\\frac{7}{3c}$$ $$2a-4d-6=0$$ $$2a-4d=6$$ $$\\frac{16}{-2b}-\\frac{28}{12c}=6$$ - Whoa. Yes, quite a few very basic mistakes here. In fact, every single one of your subproblems is wrong; you are dividing when you should be subtracting. I hope you will not take offense but I would strongly recommend a conference with your instructor. – Michael C. Grant May 22 '13 at 2:48 +1 for showing your work. – response May 22 '13 at 2:51 @MichaelC.Grant No none taken, we haven't done anything like this in our labs or during the lectures, and I was just guessing as to what I should be doing. – user1327636 May 22 '13 at 2:55 Cool. It does seem rather unfortunate that they would foist this multiequation problem on you before you had", "calculate such an expression is by diagonalizing $M$ (or, more generally, by computing the Jordan form of $M$). The eigenvalues of $M$ are easily calculated to be $\\frac{1 \\pm \\sqrt{5}}{2}$, and so for some matrix $P$, \\begin{align*} f(n) &= \\begin{pmatrix} 1 & 1 \\end{pmatrix} P \\begin{pmatrix} \\left(\\frac{1+\\sqrt{5}}{2}\\right)^n & 0 \\\\ 0 & \\left(\\frac{1-\\sqrt{5}}{2}\\right)^n \\end{pmatrix} P^{-1} \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} \\\\ &= A \\left(\\frac{1+\\sqrt{5}}{2}\\right)^n + B \\left(\\frac{1-\\sqrt{5}}{2}\\right)^n, \\end{align*} for some coefficients $A,B$. To determine $A,B$ we can either diagnoalize $M$ explicitly, or just set up a linear system using known values of $f$. Since $f(0) = 1$ and $f(1) = 2$, the latter approach shows that $$A + B = 1 \\\\ \\frac{1+\\sqrt{5}}{2} A + \\frac{1-\\sqrt{5}}{2} B = 2$$ Solving this system (e.g. using Gaussian elimination), we find out that $A = \\frac{5+3\\sqrt{5}}{10}$ and $B = \\frac{5-3\\sqrt{5}}{10}$. Therefore \\begin{align*} f(n) &= \\frac{5+3\\sqrt{5}}{10} \\left(\\frac{1+\\sqrt{5}}{2}\\right)^n + \\frac{5-3\\sqrt{5}}{10} \\left(\\frac{1-\\sqrt{5}}{2}\\right)^n \\\\ &= \\Theta(\\lambda_{\\max}^n), \\text{ where } \\lambda_{\\max} = \\frac{1+\\sqrt{5}}{2}. \\end{align*} The same approach works for every regular language. • This is a fun approach! I always love seeing linear algebraic reductions for path related problems :) – Lee Nov 16 '16 at 17:42 @Lee Gao's is too complex (I haven't even read the whole thing), here is a simplistic approach: Let f(n) be all desired strings out of which let a(n) be", "(-5) × 3 = 8 + 15 = 23 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj. A = $$\\frac {1}{23}$$ $$\\begin {pmatrix} 4 & -3\\\\ 5 & 2\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$ We know that, AX = B X = A-1B or, X =$$\\begin {pmatrix} \\frac {4}{23} & \\frac {-3}{23}\\\\ \\frac {5}{23} & \\frac {2}{23}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} -4\\\\ -13\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {-16}{23} + \\frac {39}{23}\\\\ \\frac {-20}{23} - \\frac {26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {-16 + 39}{23}\\\\ \\frac {-20 - 26}{23}\\\\ \\end {pmatrix}$$ or, X = $$\\begin {pmatrix} \\frac {23}{23}\\\\ \\frac {-46}{23}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} 1\\\\ -2\\ \\end {pmatrix}$$ ∴ x = 1 and y = -2Ans Here, 4x + 3y = 5.........................................(1) y - 3x = -7...........................................(2) Given equation in matrix form: $$\\begin {pmatrix} 4 &3\\\\ -3 & 1\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$, AX =B where, A =$$\\begin {pmatrix} 4 & 3\\\\ -3 & 1\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix} x\\\\ y\\\\ \\end {pmatrix}$$ and B =$$\\begin {pmatrix} 5\\\\ -7\\\\ \\end {pmatrix}$$ $$\\begin {vmatrix} A\\\\ \\end {vmatrix}$$ =$$\\begin {vmatrix} 4 & 3\\\\ -3 & 1\\\\", "$$\\frac{d}{d\\mathbf{V}}$$, which is the same shape of $$\\mathbf{V}$$, which is $$r \\times m$$. So somehow we need to take a $$n \\times m$$ matrix and a $$n \\times r$$ matrix, multiply them together (perhaps with some transposes) to get a $$r \\times m$$. We can do this by multiplying $$U^\\top$$ by $$(\\mathbf{X} - \\mathbf{U}\\mathbf{V})$$. Thus, we can guess the solution to (17.4.49) is (17.4.52)$\\frac{d}{d\\mathbf{V}} \\|\\mathbf{X} - \\mathbf{U}\\mathbf{V}\\|_2^{2}= 2\\mathbf{U}^\\top(\\mathbf{X} - \\mathbf{U}\\mathbf{V}).$ To show we that this works, we would be remiss to not provide a detailed computation. If we already believe that this rule-of-thumb works, feel free to skip past this derivation. To compute (17.4.53)$\\frac{d}{d\\mathbf{V}} \\|\\mathbf{X} - \\mathbf{U}\\mathbf{V}\\|_2^2,$ we must find for every $$a$$, and $$b$$ (17.4.54)$\\frac{d}{dv_{ab}} \\|\\mathbf{X} - \\mathbf{U}\\mathbf{V}\\|_2^{2}= \\frac{d}{dv_{ab}} \\sum_{i, j}\\left(x_{ij} - \\sum_k u_{ik}v_{kj}\\right)^2.$ Recalling that all entries of $$\\mathbf{X}$$ and $$\\mathbf{U}$$ are constants as far as $$\\frac{d}{dv_{ab}}$$ is concerned, we may push the derivative inside the sum, and apply the chain rule to the square to get (17.4.55)$\\frac{d}{dv_{ab}} \\|\\mathbf{X} - \\mathbf{U}\\mathbf{V}\\|_2^{2}= \\sum_{i, j}2\\left(x_{ij} - \\sum_k u_{ik}v_{kj}\\right)\\left(\\sum_k u_{ik}\\frac{dv_{kj}}{dv_{ab}} \\right).$ As in the previous derivation, we may note that $$\\frac{dv_{kj}}{dv_{ab}}$$ is only non-zero if the $$k=a$$ and $$j=b$$. If either of those conditions do not hold, the term in the sum is zero, and we may freely discard it. We see that (17.4.56)$\\frac{d}{dv_{ab}} \\|\\mathbf{X} - \\mathbf{U}\\mathbf{V}\\|_2^{2}= \\sum_{i}2\\left(x_{ib} -", "$$7 \\leq k\\equiv 3\\pmod 4$$ such that $$X^2-mkX+m\\frac{(kp+1)}{4} = 0$$ has integer roots. Suppose we can always find a solution of the form $$n(4m-p)-1 = dp \\mid mp$$ for some positive integers $$(m,n,d)$$. Then $$d$$ divides $$m$$, so we may replace $$m$$ with $$dm$$. i.e. solving $$n(4dm-p)-1 = dp \\mid (dm)p$$ Let $$n=k-d$$ and $$w=k-2d$$, then this becomes \\begin{align*} n(4md-p)-1 &= dp\\\\ (k-d)(4md-p) &= dp+1\\\\ 4md(k-d) &= kp+1\\\\ m(2d)(2k-2d) &= kp+1\\\\ m(k-w)(k+w) &= kp+1\\\\ mk^2-kp-1 &= mw^2\\\\ m^2k^2-mkp-m &= (mw)^2 \\end{align*} Now since $$m^2k^2-mkp-m$$ is the discriminant $$D$$ of the equation $$X^2-mkX+m\\frac{kq+1}{4}=0$$ This shows that the equation has solutions $$\\frac{mk\\pm \\sqrt{D}}{2} = \\frac{mk \\pm mw}{2}$$ which must be integral (either $$m$$ even or $$k$$ and $$w$$ can be shown to be odd). We also require $$k\\equiv 3\\pmod 4$$, which can be shown as follows: \\begin{align*} n(4m-p)-1 &= dp\\\\ (k-d)(4m-p)-1 &= dp\\\\ (k-d)(-1)-1 &\\equiv d \\pmod 4\\\\ d-k-1 &\\equiv d \\pmod 4\\\\ k &\\equiv 3 \\pmod 4 \\end{align*} Since $$k=n+d$$ it is always positive, but there is a small problem that it might be equal to $$3$$.", "\\times x_{i+1}\\right)^\\frac12 = \\left(2 \\times s \\times \\left(s - x_{i+1}\\right)\\right)^\\frac12$ To avoid cancellation error in this and $$\\hat{a}_{i+1}$$ it's prudent to give $$s$$ the opposite sign to $$x_{i+1}$$ so that $s = \\begin{cases} +\\left( \\sum_{j=i+1}^{n-1} \\; x_j^2\\right)^\\frac12 & x_{i+1} \\leqslant 0\\\\ -\\left( \\sum_{j=i+1}^{n-1} \\; x_j^2\\right)^\\frac12 & x_{i+1} > 0 \\end{cases}$ For example, given the three by three symmetric matrix $\\mathbf{M} = \\begin{pmatrix} 1 & -4 & 3\\\\ -4 & 2 & -1\\\\ 3 & -1 & 2 \\end{pmatrix}$ we have $s = +\\left(\\left(-4\\right)^2 + 3^2\\right)^\\frac12 = 5$ and consequently \\begin{align*} 2 \\times \\hat{\\mathbf{a}} \\times \\mathbf{x} &= \\left(2 \\times 5 \\times \\left(5 + 4\\right)\\right)^\\frac12 = 90^\\frac12\\\\ \\hat{\\mathbf{a}} &= \\begin{pmatrix} 0\\\\ \\frac{-4-5}{90^\\frac12}\\\\ \\frac{3}{90^\\frac12} \\end{pmatrix} = \\begin{pmatrix} 0\\\\ -\\frac{9}{90^\\frac12}\\\\ \\frac{3}{90^\\frac12} \\end{pmatrix} = \\begin{pmatrix} 0\\\\ -\\frac{3}{10^\\frac12}\\\\ \\frac{1}{10^\\frac12} \\end{pmatrix} \\end{align*} and an outer product of $\\hat{\\mathbf{a}} \\otimes \\hat{\\mathbf{a}} = \\begin{pmatrix} 0 & 0 & 0\\\\ 0 & \\frac{9}{10} & -\\frac{3}{10}\\\\ 0 & -\\frac{3}{10} & \\frac{1}{10} \\end{pmatrix}$ This gives us the Householder transformation matrix $\\mathbf{H}_0 = \\mathbf{I} - 2 \\times \\hat{\\mathbf{a}} \\otimes \\hat{\\mathbf{a}} = \\begin{pmatrix} 1 & 0 & 0\\\\ 0 & 1-\\frac{9}{5} & -\\frac{3}{5}\\\\ 0 & -\\frac{3}{5} & 1-\\frac{1}{5} \\end{pmatrix} = \\begin{pmatrix} 1 & 0 & 0\\\\ 0 & -\\frac{4}{5} & \\frac{3}{5}\\\\ 0 & \\frac{3}{5} & \\frac{4}{5} \\end{pmatrix}$ Pre multiplying $$\\mathbf{M}$$ by it first yields $\\mathbf{H}_0 \\times \\mathbf{M} = \\begin{pmatrix} 1 & 0", "&= \\frac{1}{3}(2a - b) \\end{align} Since $$n,m$$ are nonnegative integers, we have \\begin{align} 0 &\\leq \\frac{1}{3}(2b - a) \\\\ 0 &\\leq \\frac{1}{3}(2a - b) \\end{align} The first of these inequalities implies $$b \\geq \\frac{a}{2}$$, and the second implies $$b \\leq 2a$$, i.e. $$2a \\geq b \\geq \\frac{a}{2}$$. Conversely, suppose that conditions 1 and 2 hold for some point $$(a,b)$$. By condition 1, there exists an integer $$k$$ such that $$a + b = 3k$$. Take $$n = b - k$$ and $$m = a - k$$. Note that $$k = \\frac{a + b}{3}$$. We compute \\begin{align} n &= b - k = b - \\frac{a + b}{3} = \\frac{2b - a}{3} \\geq \\frac{2\\frac{a}{2} - a}{3} = \\frac{a - a}{3} = 0 \\\\ m &= a - k = a - \\frac{a + b}{3} = \\frac{2a - b}{3} \\geq \\frac{2a - 2a}{3} = 0 \\end{align} Thus, $$n$$ and $$m$$ are nonnegative integers. Moreover \\begin{align} (n + 2m, 2n + m) &= ((b - k) + 2(a - k) , 2(b - k) + (a - k)) = (b + 2a - 3k, 2b + a - 3k) \\\\&= (b + a + a - 3k, b + b + a - 3k) = (3k + a - 3k, b + 3k - 3k) = (a,b) \\end{align} By our original" ]

[ "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "+0 # Consider the projection matrix P that satisfies 0 169 1 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}.$$ Use the picture below to find 3 vectors with integer components and magnitude no greater than 8 that satisfy Find $$\\mathbf{P}^{-1} \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$$$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}.$$ Aug 6, 2019", "? Free Version Moderate # Projection Onto Column Space LINALG-WLW1LH The matrix: $$A = \\begin{pmatrix} 2/3 & -2/3 \\\\\\ 1/3 & 2/3 \\\\\\ 2/3 & 1/3 \\end{pmatrix}$$ ...is orthogonal. Compute the projection of $\\mathbf{v}=(1,1,0)$ onto the column space of $A$. A $(2/3,1/3,2/3)$ B $(1,0,0)$ C $(1/3,2/3,-2/3)$ D $(2/3,1/3,-1/3)$ E $(0,1,1)$", "+0 # Matrices 0 40 1 Let w = $$\\begin{pmatrix} 2 \\\\ -1 \\end{pmatrix}$$. There exists a $$2 \\times 2$$ matrix $${P}$$ such that $$\\text{proj}_{{w}} {v} = {P} {v}$$ for all 2-dimensional vectors $${v}$$. Find $${P}$$. I'm not really sure where to start for this problem. Should I plug w into the projection formula? Mar 6, 2021 ### 1+0 Answers #1 0 The matrix is $$\\begin{pmatrix} 2 & -1 \\\\ -1 & 0 \\end{pmatrix}$$ Mar 7, 2021", "How do you do this vector projection? For $\\textbf{v} = \\begin{pmatrix} 2 \\\\ 3 \\\\ -1 \\end{pmatrix}$ and $\\textbf{w} = \\begin{pmatrix} 2 \\\\ -1 \\\\ 0 \\end{pmatrix}$, compute $\\text{proj}_{\\textbf{w}} \\textbf{v}$. I don't know how to project vectors of this form... I can only do it in 2D. Is there a formula for this? • You do it the same way. There are just more components to compute. – amd Mar 19 '17 at 1:01", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "# Orthogonal projection onto span of vectors using weighted inner product • Jul 14th 2008, 07:11 PM JCS007 Orthogonal projection onto span of vectors using weighted inner product Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ • Jul 14th 2008, 09:43 PM CaptainBlack Quote: Originally Posted by JCS007 Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ Let: $b_1=\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and: $b_2=\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ and: $ u_1=\\frac{b_1}{||b_1||} $ $ u_2=\\frac{b_2-\\langle b_2,u_1 \\rangle u_1}{||b_2-\\langle b_2,u_1 \\rangle u_1||} $ Then the orthogonal projection is: $p=\\langle v,u_1\\rangle u_1 + \\langle v,u_2 \\rangle u_2$ Where $||x||=\\langle x,x \\rangle^{\\frac{1}{2}}$ RonL", "to an arbitrary vector $(x,y,z)$, and see if the resulting vector is orthogonal to the normal vector of that given plane, thus implying that the vector is projected successfully to the plane? This is necessary but not sufficient. If $n$ is normal to the plane and $\\langle n, Tx\\rangle = 0$ for all $x$, then the image of $T$ is contained in the plane, but that doesn't necessarily mean that $T$ is a projection onto the plane. For example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$ This matrix has the property that $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. So the image lies in the plane whose normal vector is $\\begin{pmatrix}1 \\\\ 0 \\\\ 0\\end{pmatrix}$. But it is not a projection onto that plane because the image only has dimension 1. For another example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 2 \\end{pmatrix}$$ Once again, we have $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. The dimension of the image is correct (2), but this", "+0 0 61 2 Consider the projection matrix P that satisfies Pv = Projection of v onto $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$. Find 3 vectors with integer components and magnitude no greater than 8 that satisfy Pv = $$\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$. Use this to help you find $$P^{-1}\\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ . Dec 30, 2019 #1 0 You can take the vectors [3,4], [4,1], and [5,-2]. Dec 30, 2019 #2 0 I tries this answer and it is incorrect. Thank you though. Guest Dec 30, 2019", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "# Linear Algebra: Projection Maps I would like to check if my understanding of projection maps is correct. I have been given the following subset of $\\mathbb{R}^3$: $$A=\\left\\{\\begin{pmatrix} x \\\\ y \\\\ -x+2y \\end{pmatrix} \\middle| x,y,z\\in\\mathbb{R}\\right\\}$$ A basis for this subset is $\\mathscr{B}=\\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\right\\}$, and to extend this basis to one for the vector space $\\mathbb{R^3}$ we simply add to the basis the vector: $$\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}$$ To obtain $\\mathscr{C} = \\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right\\}$, a basis for $\\mathbb{R}^3$. We can call $B = Span\\left\\{\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\\right\\}$, and then we can say $\\mathbb{R}^3=A\\bigoplus B$. What I want to know is if I am correct in interpreting the definition of projection map. Let $P:\\mathbb{R}\\to\\mathbb{R}$ be the projection map onto A. The question asks me to calculate $P(e_1)$, $P(e_2)$ and $P(e_3)$ then write down the matrix of $P$ with respect to the standard basis of $\\mathbb{R}^3$. Without explicitly giving my answer (I want to check my method, not my answers), this is my method: 1. Write each vector $e_1$, $e_2$ and $e_3$ as a linear combination of the vectors in $\\mathscr{C}$, so,", "similarly for the nullspace of $1-P$. 1. since $P$ is a projection matrix $P^2=P$. Using this it is easy to show that (I-P)(I-P) = I-P. Hence I-P is also Projection (symmetry property also preserves). 2. Since the eigenvalues of a projection matrix are either ones or zeros you can use eigendecomposition: $P = U \\begin{pmatrix} I_M & \\bf{0}\\\\ \\bf{0} &\\bf{0} \\end{pmatrix} U^T$ where $M$ is the dimension of the subspace to which the projectios is carried it's easy to show that $I - U \\begin{pmatrix} I_M & \\bf{0}\\\\ \\bf{0} &\\bf{0} \\end{pmatrix} U^T = U \\begin{pmatrix} \\bf{0} & \\bf{0}\\\\ \\bf{0} & I_{N-M} \\end{pmatrix} U^T$. From here it is easy to extract the 4 sub-spaces of each projection operator. The above analysis is correct for orthogonal projectors. For oblique projectors a similar calculation can be done using singular value decomposition. • Oblique projectors don't diagonalise like this – Jazzmaniac Mar 27 '16 at 12:09 • Right. Obviously the problem consider orthogonal projectors. – Nir Regev Mar 27 '16 at 12:13 • I don't see any reference to orthogonal projection in the question – Jazzmaniac Mar 27 '16 at 12:14 • @Jazzmaniac you're right. I updated the answer – Nir Regev Mar 27 '16 at 12:29", "maps between vector spaces of different dimension An example of a linear map between two vector spaces with different dimensions is the following projection of the space ${\\displaystyle \\mathbb {R} ^{3}}$ onto the plane ${\\displaystyle \\mathbb {R} ^{2}}$: ${\\displaystyle f\\colon \\mathbb {R} ^{3}\\to \\mathbb {R} ^{2};\\quad {\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}\\mapsto {\\begin{pmatrix}x\\\\y\\end{pmatrix}}}$ We now check whether the vector addition is preserved. That means, for vectors ${\\displaystyle a,b\\in \\mathbb {R} ^{3}}$ we need that ${\\displaystyle f(a+b)=f(a)+f(b)}$ This can be verified directly: {\\displaystyle {\\begin{aligned}f(a+b)&=f\\left({\\begin{pmatrix}a_{x}\\\\a_{y}\\\\a_{z}\\end{pmatrix}}+{\\begin{pmatrix}b_{x}\\\\b_{y}\\\\b_{z}\\end{pmatrix}}\\right)=f\\left({\\begin{pmatrix}a_{x}+b_{x}\\\\a_{y}+b_{y}\\\\a_{z}+b_{z}\\end{pmatrix}}\\right)\\\\[0.5em]&={\\begin{pmatrix}a_{x}+b_{x}\\\\a_{y}+b_{y}\\end{pmatrix}}={\\begin{pmatrix}a_{x}\\\\a_{y}\\end{pmatrix}}+{\\begin{pmatrix}b_{x}\\\\b_{y}\\end{pmatrix}}\\\\[0.5em]&=f\\left({\\begin{pmatrix}a_{x}\\\\a_{y}\\\\a_{z}\\end{pmatrix}}\\right)+f\\left({\\begin{pmatrix}b_{x}\\\\b_{y}\\\\b_{z}\\end{pmatrix}}\\right)=f(a)+f(b).\\end{aligned}}} Now we check homogeneity. For all ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle a\\in \\mathbb {R} ^{2}}$ we need: ${\\displaystyle f(\\lambda \\cdot a)=\\lambda \\cdot f(a).}$ We have that {\\displaystyle {\\begin{aligned}f(\\lambda \\cdot a)&=f\\left(\\lambda \\cdot {\\begin{pmatrix}a_{x}\\\\a_{y}\\\\a_{z}\\end{pmatrix}}\\right)=f\\left({\\begin{pmatrix}\\lambda a_{x}\\\\\\lambda a_{y}\\\\\\lambda a_{z}\\end{pmatrix}}\\right)\\\\[0.5em]&={\\begin{pmatrix}\\lambda a_{x}\\\\\\lambda a_{y}\\end{pmatrix}}=\\lambda \\cdot {\\begin{pmatrix}a_{x}\\\\a_{y}\\end{pmatrix}}=\\lambda \\cdot f\\left({\\begin{pmatrix}a_{x}\\\\a_{y}\\\\a_{z}\\end{pmatrix}}\\right)\\\\[0.5em]&=\\lambda \\cdot f(a).\\end{aligned}}} So the projection ${\\displaystyle f}$ is a linear map. ### A non-linear map Next, we investigate some examples for non-linear maps. It is easy to come up with such maps: basically any function on ${\\displaystyle \\mathbb {R} }$ whose graph is not a line is a non-linear map. So \"most maps are non-linear\". Of course, there are also examples for non-linear maps on ${\\displaystyle \\mathbb {R} ^{2}}$. For instance, consider the norm mapping on the plane which assigns the length to every vector: ${\\displaystyle \\|\\cdot \\|_{2}\\,\\colon \\,\\mathbb {R} ^{2}\\to \\mathbb {R} ;\\quad {\\begin{pmatrix}x\\\\y\\end{pmatrix}}\\mapsto {\\sqrt", "# Math Help - Orthogonal projection onto span of vectors using weighted inner product 1. ## Orthogonal projection onto span of vectors using weighted inner product Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ 2. Originally Posted by JCS007 Find the orthogonal projection of $v=\\begin{pmatrix}1 \\\\ 2 \\\\ -1 \\\\ 2 \\end{pmatrix}$ onto the span of $\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and $\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ using the weighted inner product $=4v_1w_1+3v_2w_2+2v_3w_3+v_4w_4$ Let: $b_1=\\begin{pmatrix}1 \\\\ -1 \\\\ 2 \\\\ 5 \\end{pmatrix}$ and: $b_2=\\begin{pmatrix}2 \\\\ 1 \\\\ 0 \\\\ -1 \\end {pmatrix}$ and: $ u_1=\\frac{b_1}{||b_1||} $ $ u_2=\\frac{b_2-\\langle b_2,u_1 \\rangle u_1}{||b_2-\\langle b_2,u_1 \\rangle u_1||} $ Then the orthogonal projection is: $p=\\langle v,u_1\\rangle u_1 + \\langle v,u_2 \\rangle u_2$ Where $||x||=\\langle x,x \\rangle^{\\frac{1}{2}}$ RonL", "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "The orthogonal projection $\\mbox{proj}_{M}({\\vec{v}\\,})$ of a vector is its projection onto $M$ along $M^\\perp$. Example 3.5 In $\\mathbb{R}^3$, to find the orthogonal complement of the plane $P=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, 3x+2y-z=0 \\}$ we start with a basis for $P$. $B=\\langle \\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\rangle$ Any $\\vec{v}$ perpendicular to every vector in $B$ is perpendicular to every vector in the span of $B$ (the proof of this assertion is Problem 10). Therefore, the subspace $P^\\perp$ consists of the vectors that satisfy these two conditions. $\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0 \\qquad \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0$ We can express those conditions more compactly as a linear system. $P^\\perp=\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\begin{pmatrix} 1 &0 &3 \\\\ 0 &1 &2 \\end{pmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix} =\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix} \\}$ We are thus left with finding the nullspace of the map represented by the matrix, that is, with calculating the solution set of a homogeneous linear system. $P^\\perp =\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\;\\begin{array}{*{3}{rc}r} v_1 & & &+ &3v_3 &= &0 \\\\ & &v_2 &+ &2v_3 &= &0 \\end{array}\\; \\} =\\{k\\begin{pmatrix} -3", "& 0 \\\\ 0 & 0 & -1 & 0 \\\\ 0 & 0 & 0 & -1 \\end{pmatrix}$ and $B=\\begin{pmatrix} 0 & -1 & -1 & 1 \\\\ 1 & 0 & -1 & -1 \\\\ -1 & -1 & 0 & 1 \\\\ 1 & -1 & -1 & 0 \\end{pmatrix}$, the aforementioned operator has matrix $\\begin{pmatrix} 0 & 1 \\\\ -1 & 0 \\end{pmatrix}$ which has no real eigenvalue. - Oh wow! Yeah, it is a complex vector space, sorry! – Jon Beardsley Apr 6 '11 at 19:03 In that case, my previous answer gives a solution: every endomorphism of a complex vector space has an eigenvector. – Plop Apr 6 '11 at 19:12 Okay, thankyou. I'm not entirely clear on how to think of these projections on to something with respect to something else. So I will have to suss all that out. Or even why it is the case that such a vector space must be the sum of two such spaces. – Jon Beardsley Apr 6 '11 at 19:27 Sorry about that, English is not my native language. According to Wikipedia, I should have written \"projection along $E^B_{-j}$ onto $E^B_j$\" – Plop Apr 6 '11 at 19:34 Is it obvious that such a projection always exists, and in this case is an", "+0 # Help hard quick 0 68 0 Consider the projection matrix P that satisfies $$\\mathbf{P} \\mathbf{v} = \\text{Projection of \\mathbf{v} onto } \\begin{pmatrix} 4 \\\\ 1 \\end{pmatrix}$$ Use the picture below to find vectors with integer components and magnitude no greater than 8 that satisfy $$\\mathbf{P} \\mathbf{v} = \\begin{pmatrix} 4 \\\\1 \\end{pmatrix}$$" ]

[ "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "{\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}+{\\begin{pmatrix}x_{2}\\\\y_{2}\\\\z_{2}\\end{pmatrix}}={\\begin{pmatrix}x_{1}+x_{2}-1\\\\y_{1}+y_{2}\\\\z_{1}+z_{2}\\end{pmatrix}}\\qquad r{\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}={\\begin{pmatrix}rx-r+1\\\\ry\\\\rz\\end{pmatrix}}}$ 1. Show that it is not a subspace of ${\\displaystyle \\mathbb {R} ^{3}}$ . (Hint. See Example 2.5). 2. Show that it is a vector space. Note that by the prior item, Lemma 2.9 can not apply. 3. Show that any subspace of ${\\displaystyle \\mathbb {R} ^{3}}$ must pass through the origin, and so any subspace of ${\\displaystyle \\mathbb {R} ^{3}}$ must involve zero in its description. Does the converse hold? Does any subset of ${\\displaystyle \\mathbb {R} ^{3}}$ that contains the origin become a subspace when given the inherited operations? 1. It is not a subspace because these are not the inherited operations. For one thing, in this space, ${\\displaystyle 0\\cdot {\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}}$ while this does not, of course, hold in ${\\displaystyle \\mathbb {R} ^{3}}$ . 2. We can combine the argument showing closure under addition with the argument showing closure under scalar multiplication into one single argument showing closure under linear combinations of two vectors. If ${\\displaystyle r_{1},r_{2},x_{1},x_{2},y_{1},y_{2},z_{1},z_{2}}$ are in ${\\displaystyle \\mathbb {R} }$ then ${\\displaystyle r_{1}{\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}+r_{2}{\\begin{pmatrix}x_{2}\\\\y_{2}\\\\z_{2}\\end{pmatrix}}={\\begin{pmatrix}r_{1}x_{1}-r_{1}+1\\\\r_{1}y_{1}\\\\r_{1}z_{1}\\end{pmatrix}}+{\\begin{pmatrix}r_{2}x_{2}-r_{2}+1\\\\r_{2}y_{2}\\\\r_{2}z_{2}\\end{pmatrix}}={\\begin{pmatrix}r_{1}x_{1}-r_{1}+r_{2}x_{2}-r_{2}+1\\\\r_{1}y_{1}+r_{2}y_{2}\\\\r_{1}z_{1}+r_{2}z_{2}\\end{pmatrix}}}$ (note that the definition of addition in this space is that the first components combine as ${\\displaystyle (r_{1}x_{1}-r_{1}+1)+(r_{2}x_{2}-r_{2}+1)-1}$ , so the first component of the last vector does not say \"${\\displaystyle +2}$ \"). Adding the three components of the last vector gives ${\\displaystyle r_{1}(x_{1}-1+y_{1}+z_{1})+r_{2}(x_{2}-1+y_{2}+z_{2})+1=r_{1}\\cdot 0+r_{2}\\cdot 0+1=1}$ . Most", "${\\displaystyle ={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}{\\begin{pmatrix}1/2\\end{pmatrix}}{\\begin{pmatrix}-1&0&1\\end{pmatrix}}={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}{\\begin{pmatrix}-1/2&0&1/2\\end{pmatrix}}={\\begin{pmatrix}1/2&0&-1/2\\\\0&0&0\\\\-1/2&0&1/2\\end{pmatrix}}}$ followed by matrix-vector multiplication ${\\displaystyle {\\mbox{proj}}_{M}({\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix}}){\\begin{pmatrix}1/2&0&-1/2\\\\0&0&0\\\\-1/2&0&1/2\\end{pmatrix}}{\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}}$ Problem 5 Check that the operation of Definition 3.1 is well-defined. That is, in Example 3.2 and 3.3, doesn't the answer depend on the choice of bases? No, a decomposition of vectors ${\\displaystyle {\\vec {v}}={\\vec {m}}+{\\vec {n}}}$ into ${\\displaystyle {\\vec {m}}\\in M}$ and ${\\displaystyle {\\vec {n}}\\in N}$ does not depend on the bases chosen for the subspaces— this was shown in the Direct Sum subsection. Problem 6 What is the orthogonal projection onto the trivial subspace? The orthogonal projection of a vector onto a subspace is a member of that subspace. Since a trivial subspace has only one member, ${\\displaystyle {\\vec {0}}}$, the projection of any vector must equal ${\\displaystyle {\\vec {0}}}$. Problem 7 What is the projection of ${\\displaystyle {\\vec {v}}}$ onto ${\\displaystyle M}$ along ${\\displaystyle N}$ if ${\\displaystyle {\\vec {v}}\\in M}$? The projection onto ${\\displaystyle M}$ along ${\\displaystyle N}$ of a ${\\displaystyle {\\vec {v}}\\in M}$ is ${\\displaystyle {\\vec {v}}}$. Decomposing ${\\displaystyle {\\vec {v}}={\\vec {m}}+{\\vec {n}}}$ gives ${\\displaystyle {\\vec {m}}={\\vec {v}}}$ and ${\\displaystyle {\\vec {n}}={\\vec {0}}}$, and dropping the ${\\displaystyle N}$ part but retaining the ${\\displaystyle M}$ part results in a projection of ${\\displaystyle {\\vec {m}}={\\vec {v}}}$. Problem 8 Show that if ${\\displaystyle M\\subseteq \\mathbb {R} ^{n}}$ is a subspace with orthonormal basis ${\\displaystyle \\langle {\\vec {\\kappa }}_{1},\\ldots ,{\\vec", "does not hold in general. See the counterexample: ${\\displaystyle {\\Biggl (}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}{\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}{\\Biggr )}^{+}={\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}={\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}\\quad \\neq \\quad {\\begin{pmatrix}{\\tfrac {1}{4}}&0\\\\{\\tfrac {1}{4}}&0\\end{pmatrix}}={\\begin{pmatrix}0&{\\tfrac {1}{2}}\\\\0&{\\tfrac {1}{2}}\\end{pmatrix}}{\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}={\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}^{+}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}}$ ### Projectors ${\\displaystyle P=AA^{+}}$ and ${\\displaystyle Q=A^{+}A}$ are orthogonal projection operators, that is, they are Hermitian (${\\displaystyle P=P^{*}}$, ${\\displaystyle Q=Q^{*}}$) and idempotent (${\\displaystyle P^{2}=P}$ and ${\\displaystyle Q^{2}=Q}$). The following hold: • ${\\displaystyle PA=AQ=A}$ and ${\\displaystyle A^{+}P=QA^{+}=A^{+}}$ • ${\\displaystyle P}$ is the orthogonal projector onto the range of ${\\displaystyle A}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A^{*}}$). • ${\\displaystyle Q}$ is the orthogonal projector onto the range of ${\\displaystyle A^{*}}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A}$). • ${\\displaystyle (I-Q)=\\left(I-A^{+}A\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A}$. • ${\\displaystyle (I-P)=\\left(I-AA^{+}\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A^{*}}$.[8] The last two properties imply the following identities: • ${\\displaystyle A\\,\\ \\left(I-A^{+}A\\right)=\\left(I-AA^{+}\\right)A\\ \\ =0}$ • ${\\displaystyle A^{*}\\left(I-AA^{+}\\right)=\\left(I-A^{+}A\\right)A^{*}=0}$ Another property is the following: if ${\\displaystyle A\\in \\mathbb {K} ^{n\\times n}}$ is Hermitian and idempotent (true if and only if it represents an orthogonal projection), then, for any matrix ${\\displaystyle B\\in \\mathbb {K} ^{m\\times n}}$ the following equation holds:[10] ${\\displaystyle A(BA)^{+}=(BA)^{+}}$ This can be proven by defining matrices ${\\displaystyle C=BA}$, ${\\displaystyle D=A(BA)^{+}}$, and checking that ${\\displaystyle D}$ is indeed a pseudoinverse for ${\\displaystyle C}$ by verifying that", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "The orthogonal projection $\\mbox{proj}_{M}({\\vec{v}\\,})$ of a vector is its projection onto $M$ along $M^\\perp$. Example 3.5 In $\\mathbb{R}^3$, to find the orthogonal complement of the plane $P=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, 3x+2y-z=0 \\}$ we start with a basis for $P$. $B=\\langle \\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\rangle$ Any $\\vec{v}$ perpendicular to every vector in $B$ is perpendicular to every vector in the span of $B$ (the proof of this assertion is Problem 10). Therefore, the subspace $P^\\perp$ consists of the vectors that satisfy these two conditions. $\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0 \\qquad \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0$ We can express those conditions more compactly as a linear system. $P^\\perp=\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\begin{pmatrix} 1 &0 &3 \\\\ 0 &1 &2 \\end{pmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix} =\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix} \\}$ We are thus left with finding the nullspace of the map represented by the matrix, that is, with calculating the solution set of a homogeneous linear system. $P^\\perp =\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\;\\begin{array}{*{3}{rc}r} v_1 & & &+ &3v_3 &= &0 \\\\ & &v_2 &+ &2v_3 &= &0 \\end{array}\\; \\} =\\{k\\begin{pmatrix} -3", "vector ${\\displaystyle (-1,0,-1)^{T}}$ is a multiple of ${\\displaystyle (1,0,1)^{T}}$. That is, the direction represented by ${\\displaystyle (-1,0,-1)^{T}}$ is the same as the direction represented by ${\\displaystyle (1,0,1)^{T}}$. Hence, we can remove this vector from ${\\displaystyle E}$ and obtain a new generator ${\\displaystyle E_{1}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}},{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Can we reduce the size of this generator, as well? Yes, since ${\\displaystyle {\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}+{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}}$ So ${\\displaystyle (2,1,1)^{T}}$ adds no new direction that is not already spanned by ${\\displaystyle (1,0,1)^{T}}$ and ${\\displaystyle (1,1,0)^{T}}$. We thus obtain a smaller generator ${\\displaystyle E_{2}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Now we cannot reduce the generator any further without losing the property of it being a generator. For if we remove any of the three vectors in ${\\displaystyle E_{2}}$, it is no longer in the span of the remaining two vectors. For ${\\displaystyle (1,0,1)^{T}}$, for example, we see this as follows: Suppose we had some ${\\displaystyle \\lambda ,\\mu \\in \\mathbb {R} }$, so that ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=\\lambda {\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+\\mu {\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}}$ Then ${\\displaystyle \\lambda =-\\mu }$ would have to hold because the second component must be zero on both sides. Because of the third component, ${\\displaystyle \\mu ={\\frac {1}{2}}}$ must be true. Thus we obtain the contradiction ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=-{\\frac {1}{2}}{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+{\\frac {1}{2}}{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}0.5\\\\0\\\\1\\end{pmatrix}}}$ So ${\\displaystyle (1,0,1)^{T}}$ is not in ${\\displaystyle \\operatorname {span} ((1,1,0)^{T},(2,1,2)^{T})}$. So we have found a generator containing linearly independent vectors (a minimal generator). In summary, we proceeded", "# Linear Algebra: Projection Maps I would like to check if my understanding of projection maps is correct. I have been given the following subset of $\\mathbb{R}^3$: $$A=\\left\\{\\begin{pmatrix} x \\\\ y \\\\ -x+2y \\end{pmatrix} \\middle| x,y,z\\in\\mathbb{R}\\right\\}$$ A basis for this subset is $\\mathscr{B}=\\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\right\\}$, and to extend this basis to one for the vector space $\\mathbb{R^3}$ we simply add to the basis the vector: $$\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}$$ To obtain $\\mathscr{C} = \\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right\\}$, a basis for $\\mathbb{R}^3$. We can call $B = Span\\left\\{\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\\right\\}$, and then we can say $\\mathbb{R}^3=A\\bigoplus B$. What I want to know is if I am correct in interpreting the definition of projection map. Let $P:\\mathbb{R}\\to\\mathbb{R}$ be the projection map onto A. The question asks me to calculate $P(e_1)$, $P(e_2)$ and $P(e_3)$ then write down the matrix of $P$ with respect to the standard basis of $\\mathbb{R}^3$. Without explicitly giving my answer (I want to check my method, not my answers), this is my method: 1. Write each vector $e_1$, $e_2$ and $e_3$ as a linear combination of the vectors in $\\mathscr{C}$, so,", "f}$ is an epimorphism. Solution By construction, ${\\displaystyle f}$ is a linear map. We want to show that ${\\displaystyle \\{f(e_{1}),f(e_{2}),f(e_{3})\\}=\\{(1,1),(1,-1)\\}}$ is a generator of ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have with the above theorem that ${\\displaystyle f}$ is an epimorphism. So, we must represent every ${\\displaystyle (x,y)\\in \\mathbb {R} ^{2}}$ as a linear combination of vectors ${\\displaystyle (1,1),(1,-1)}$. Accordingly, we search linear combination scalars ${\\displaystyle \\lambda ,\\mu \\in \\mathbb {R} }$ such that {\\displaystyle {\\begin{aligned}{\\begin{pmatrix}x\\\\y\\end{pmatrix}}=\\lambda {\\begin{pmatrix}1\\\\1\\end{pmatrix}}+\\mu {\\begin{pmatrix}1\\\\-1\\end{pmatrix}}.\\end{aligned}}} From this we get the linear system of equations {\\displaystyle {\\begin{aligned}{\\begin{aligned}x&=\\lambda +\\mu \\\\y&=\\lambda -\\mu ,\\end{aligned}}\\end{aligned}}} which is solved by ${\\displaystyle \\lambda ={\\frac {x+y}{2}}}$ and ${\\displaystyle \\mu ={\\frac {x-y}{2}}}$. So we have ${\\displaystyle {\\begin{pmatrix}x\\\\y\\end{pmatrix}}={\\frac {x+y}{2}}{\\begin{pmatrix}1\\\\1\\end{pmatrix}}+{\\frac {x-y}{2}}{\\begin{pmatrix}1\\\\-1\\end{pmatrix}}}$ for all ${\\displaystyle x,y\\in \\mathbb {R} }$. Thus ${\\displaystyle \\{(1,1),(1,-1)\\}}$ is a generator. This proves that ${\\displaystyle f}$ is an epimorphism. Exercise Consider the function space ${\\displaystyle \\operatorname {Fun} ([0,1],\\mathbb {R} )}$ of all function s mapping ${\\displaystyle \\mathbb {R} }$ to ${\\displaystyle \\mathbb {R} }$, as well as the map {\\displaystyle {\\begin{aligned}\\phi \\colon \\operatorname {Fun} ([0,1],\\mathbb {R} )&\\to \\mathbb {R} ,\\\\f&\\mapsto f(0).\\end{aligned}}} Show that ${\\displaystyle \\phi }$ is an epimorphism. Solution The operations on the function space are defined element-wise in each case. That is: for ${\\displaystyle f,g\\in \\operatorname {Fun} ([0,1],\\mathbb {R} )}$, ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle x\\in [0,1]}$ we", "|}\\,q\\in \\mathbb {C} \\}}$, and ${\\displaystyle {\\mathcal {N}}(t_{0}^{2})}$ is the space of quadratic polynomials with no ${\\displaystyle x^{2}}$ term ${\\displaystyle \\{p+qx\\,{\\big |}\\,p,q\\in \\mathbb {C} \\}}$, and this is the end ${\\displaystyle {\\mathcal {N}}_{\\infty }(t_{0})={\\mathcal {N}}(t_{0}^{2})}$. 2. The second power ${\\displaystyle {\\begin{pmatrix}a\\\\b\\end{pmatrix}}{\\stackrel {t_{1}}{\\longmapsto }}{\\begin{pmatrix}0\\\\a\\end{pmatrix}}{\\stackrel {t_{1}}{\\longmapsto }}{\\begin{pmatrix}0\\\\0\\end{pmatrix}}}$ is the zero map. Consequently, the chain of rangespaces ${\\displaystyle \\mathbb {R} ^{2}\\supset \\{{\\begin{pmatrix}0\\\\p\\end{pmatrix}}\\,{\\big |}\\,p\\in \\mathbb {C} \\}\\supset \\{{\\vec {0}}\\,\\}=\\cdots }$ and the chain of nullspaces ${\\displaystyle \\{{\\vec {0}}\\,\\}\\subset \\{{\\begin{pmatrix}q\\\\0\\end{pmatrix}}\\,{\\big |}\\,q\\in \\mathbb {C} \\}\\subset \\mathbb {R} ^{2}=\\cdots }$ each has length two. The generalized rangespace is the trivial subspace and the generalized nullspace is the entire space. 3. Iterates of this map cycle around ${\\displaystyle a+bx+cx^{2}{\\stackrel {t_{2}}{\\longmapsto }}b+cx+ax^{2}{\\stackrel {t_{2}}{\\longmapsto }}c+ax+bx^{2}{\\stackrel {t_{2}}{\\longmapsto }}a+bx+cx^{2}\\;\\cdots }$ and the chains of rangespaces and nullspaces are trivial. ${\\displaystyle {\\mathcal {P}}_{2}={\\mathcal {P}}_{2}=\\cdots \\qquad \\{{\\vec {0}}\\,\\}=\\{{\\vec {0}}\\,\\}=\\cdots }$ Thus, obviously, generalized spaces are ${\\displaystyle {\\mathcal {R}}_{\\infty }(t_{2})={\\mathcal {P}}_{2}}$ and ${\\displaystyle {\\mathcal {N}}_{\\infty }(t_{2})=\\{{\\vec {0}}\\,\\}}$. 4. We have ${\\displaystyle {\\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}}\\mapsto {\\begin{pmatrix}a\\\\a\\\\b\\end{pmatrix}}\\mapsto {\\begin{pmatrix}a\\\\a\\\\a\\end{pmatrix}}\\mapsto {\\begin{pmatrix}a\\\\a\\\\a\\end{pmatrix}}\\mapsto \\cdots }$ and so the chain of rangespaces ${\\displaystyle \\mathbb {R} ^{3}\\supset \\{{\\begin{pmatrix}p\\\\p\\\\r\\end{pmatrix}}\\,{\\big |}\\,p,r\\in \\mathbb {C} \\}\\supset \\{{\\begin{pmatrix}p\\\\p\\\\p\\end{pmatrix}}\\,{\\big |}\\,p\\in \\mathbb {C} \\}=\\cdots }$ and the chain of nullspaces ${\\displaystyle \\{{\\vec {0}}\\,\\}\\subset \\{{\\begin{pmatrix}0\\\\0\\\\r\\end{pmatrix}}\\,{\\big |}\\,r\\in \\mathbb {C} \\}\\subset \\{{\\begin{pmatrix}0\\\\q\\\\r\\end{pmatrix}}\\,{\\big |}\\,q,r\\in \\mathbb {C} \\}=\\cdots }$ each has length two. The generalized spaces are the final ones shown above in each chain.", "{\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}\\mapsto 1v_{1}+2v_{2}+3v_{3}}$ with respect to the standard bases and show that ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}}$ is a member of the perp of the nullspace. Prove that ${\\displaystyle {\\mathcal {N}}(f)^{\\perp }}$ is equal to the span of this vector. 2. Generalize that to apply to any ${\\displaystyle f:\\mathbb {R} ^{n}\\to \\mathbb {R} }$. 3. Represent ${\\displaystyle f:\\mathbb {R} ^{3}\\to \\mathbb {R} ^{2}}$ ${\\displaystyle {\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}\\mapsto {\\begin{pmatrix}1v_{1}+2v_{2}+3v_{3}\\\\4v_{1}+5v_{2}+6v_{3}\\end{pmatrix}}}$ with respect to the standard bases and show that ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}},\\;{\\begin{pmatrix}4\\\\5\\\\6\\end{pmatrix}}}$ are both members of the perp of the nullspace. Prove that ${\\displaystyle {\\mathcal {N}}(f)^{\\perp }}$ is the span of these two. (Hint. See the third item of Problem 14.) 4. Generalize that to apply to any ${\\displaystyle f:\\mathbb {R} ^{n}\\to \\mathbb {R} ^{m}}$. This, and related results, is called the Fundamental Theorem of Linear Algebra in (Strang 1993). 1. The representation of ${\\displaystyle {\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}{\\stackrel {f}{\\longmapsto }}1v_{1}+2v_{2}+3v_{3}}$ is this. ${\\displaystyle {\\rm {Rep}}_{{\\mathcal {E}}_{3},{\\mathcal {E}}_{1}}(f)={\\begin{pmatrix}1&2&3\\end{pmatrix}}}$ By the definition of ${\\displaystyle f}$ ${\\displaystyle {\\mathcal {N}}(f)=\\{{\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}\\,{\\big |}\\,1v_{1}+2v_{2}+3v_{3}=0\\}=\\{{\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}\\,{\\big |}\\,{\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}=0\\}}$ and this second description exactly says this. ${\\displaystyle {\\mathcal {N}}(f)^{\\perp }=[\\{{\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}\\}]}$ 2. The generalization is that for any ${\\displaystyle f:\\mathbb {R} ^{n}\\to \\mathbb {R} }$ there is a vector ${\\displaystyle {\\vec {h}}}$ so that ${\\displaystyle {\\begin{pmatrix}v_{1}\\\\\\vdots \\\\v_{n}\\end{pmatrix}}{\\stackrel {f}{\\longmapsto }}h_{1}v_{1}+\\dots +h_{n}v_{n}}$ and ${\\displaystyle {\\vec {h}}\\in {\\mathcal {N}}(f)^{\\perp }}$. We can prove this by, as in the prior item, representing ${\\displaystyle f}$", "}$-vector space. In addition, the map is well-defined. Let ${\\displaystyle (x_{1},x_{2})^{T}}$ and ${\\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have: {\\displaystyle {\\begin{aligned}f\\left({\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\\\(x_{1}+y_{1})-5\\cdot (x_{2}+y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{separate vectors}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}x_{1}+x_{2}\\\\x_{1}-5x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}+y_{2}\\\\y_{1}-5y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+f{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\end{aligned}}} Proof step: homogeneity Let ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle (x_{1},x_{2})^{T}\\in \\mathbb {R} ^{2}}$. Then: {\\displaystyle {\\begin{aligned}f\\left(\\lambda \\cdot {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}\\lambda \\cdot x_{1}\\\\\\lambda \\cdot x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda x_{1}+\\lambda x_{2}\\\\\\lambda x_{1}-5\\lambda x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda (x_{1}+x_{2})\\\\\\lambda (x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{scalar multiplication}}\\right.}\\\\[0.3em]&=\\lambda \\cdot {\\begin{pmatrix}(x_{1}+x_{2})\\\\(x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=\\lambda \\cdot f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\\\[0.3em]\\end{aligned}}} Thus the map is linear. ## A linear map in the vector space of sequences Next, we consider the space of all sequences of real numbers. This space is infinite-dimensional, because there are not finitely many sequences generating this sequence space. But it is a vector space, as we have shown in the chapter about sequence spaces. Exercise (Sequence space) Let ${\\displaystyle V}$ be the ${\\displaystyle \\mathbb {R} }$-vector space of all real-valued sequences. Show that the map {\\displaystyle {\\begin{aligned}f:V&\\to V\\\\(a_{0},a_{1},a_{2},\\ldots )&\\mapsto (a_{1},a_{2},a_{3},\\ldots )\\end{aligned}}} is linear. How to get to the proof? (Sequence space) To show linearity, two properties need to be checked: 1. ${\\displaystyle f}$ is additive: ${\\displaystyle f(v+w)=f(v)+f(w)}$ for all ${\\displaystyle v,w\\in V}$ 2. ${\\displaystyle", "projection in $$A \\subseteq M$$ which we can decompose in the way you want. $$A$$ can be thought of as the infinite tensor product of $$M_2$$. So $$A = \\otimes_1^{\\infty}M_2$$, which is really the norm closure of $$\\cup_n (M_2^{\\otimes n}\\otimes 1\\otimes\\cdots)$$. Let $$p = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes 1 \\otimes\\cdots.$$ We can write this projection in the following way. Let $$p_1 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes 1 \\otimes\\cdots$$ and $$p_2 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 0 & 0 \\\\ 0 & 1 \\end{pmatrix} \\otimes 1 \\otimes\\cdots.$$ Now its clear that $$p_1,p_2 < p$$ and that $$p = p_1 + p_2$$. Edit: Note that you can keep on doing this, so these projections can't be minimal! This is a \"less naive\" way of seeing the chain $$M_2(\\mathbb C)\\subset M_4(\\mathbb C)\\subset\\cdots\\subset B(H)$$ for infinite-dimensional separable $$H$$ (and it is was was done in PStheman's answer, just a bit more explicit here). You see $$M_2(\\mathbb C)$$ as $$\\begin{bmatrix} a&b\\\\ c&d\\\\ &&a&b\\\\ &&c&d\\\\ &&&&a&b\\\\ &&&&c&d\\\\ &&&&&&a&b\\\\ &&&&&&c&d\\\\ &&&&&&&&\\ddots \\end{bmatrix},$$ then $$M_4(\\mathbb C)$$ as $$\\begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\\\ a_{21}&a_{22}&a_{23}&a_{24}\\\\ a_{31}&a_{32}&a_{33}&a_{34}\\\\ a_{41}&a_{42}&a_{43}&a_{44}\\\\ &&&&a_{11}&a_{12}&a_{13}&a_{14}\\\\ &&&&a_{21}&a_{22}&a_{23}&a_{24}\\\\ &&&&a_{31}&a_{32}&a_{33}&a_{34}\\\\ &&&&a_{41}&a_{42}&a_{43}&a_{44}\\\\ &&&&&&&&\\ddots \\end{bmatrix}.$$ So for example take $$E_{11}^{(2)}\\in M_2(\\mathbb C)$$, and let's", "## Linear Algebra and Its Applications, Exercise 3.4.2 Exercise 3.4.2. Given two orthonormal vectors $a_1 = (\\frac{2}{3}, \\frac{2}{3}, -\\frac{1}{3})$ and $a_2 = (-\\frac{1}{3}, \\frac{2}{3}, \\frac{2}{3})$ and the vector $b = (0, 3, 0)$, project $b$ onto $a_1$ and $a_2$. Also find the projection $p$ of $b$ onto the plane formed by $a_1$ and $a_2$. Answer: Since $a_1$ is orthonormal, the projection of $b$ onto $a_1$ is given by $a_1^Tba_1 = \\begin{bmatrix} \\frac{2}{3}&\\frac{2}{3}&-\\frac{1}{3} \\end{bmatrix} \\begin{bmatrix} 0 \\\\ 3 \\\\ 0 \\end{bmatrix} \\begin{bmatrix} \\frac{2}{3} \\\\ \\frac{2}{3} \\\\ -\\frac{1}{3} \\end{bmatrix} = 2 \\cdot \\begin{bmatrix} \\frac{2}{3} \\\\ \\frac{2}{3} \\\\ -\\frac{1}{3} \\end{bmatrix} = \\begin{bmatrix} \\frac{4}{3} \\\\ \\frac{4}{3} \\\\ -\\frac{2}{3} \\end{bmatrix}$ Similarly the projection of $b$ onto $a_2$ is given by $a_2^Tba_2 = \\begin{bmatrix} -\\frac{1}{3}&\\frac{2}{3}&\\frac{2}{3} \\end{bmatrix} \\begin{bmatrix} 0 \\\\ 3 \\\\ 0 \\end{bmatrix} \\begin{bmatrix} -\\frac{1}{3} \\\\ \\frac{2}{3} \\\\ \\frac{2}{3} \\end{bmatrix} = 2 \\cdot \\begin{bmatrix} -\\frac{1}{3} \\\\ \\frac{2}{3} \\\\ \\frac{2}{3} \\end{bmatrix} = \\begin{bmatrix} -\\frac{2}{3} \\\\ \\frac{4}{3} \\\\ \\frac{4}{3} \\end{bmatrix}$ The projection $p$ of $b$ onto the plane formed by $a_1$ and $a_2$ is simply the sum of the projections above: $p = a_1^Tba_1 + a_2^Tba_2 = \\begin{bmatrix} \\frac{4}{3} \\\\ \\frac{4}{3} \\\\ -\\frac{2}{3} \\end{bmatrix} + \\begin{bmatrix} -\\frac{2}{3} \\\\ \\frac{4}{3} \\\\ \\frac{4}{3} \\end{bmatrix} = \\begin{bmatrix} \\frac{2}{3} \\\\ \\frac{8}{3} \\\\ \\frac{2}{3} \\end{bmatrix}$ NOTE: This continues a series of posts containing worked out exercises from the (out of print)", "This exercise is recommended for all readers. Problem 5 Decide if ${\\displaystyle 1+2x}$ is in the range of the map from ${\\displaystyle \\mathbb {R} ^{3}}$ to ${\\displaystyle {\\mathcal {P}}_{2}}$ represented with respect to ${\\displaystyle {\\mathcal {E}}_{3}}$ and ${\\displaystyle \\langle 1,1+x^{2},x\\rangle }$ by this matrix. ${\\displaystyle {\\begin{pmatrix}1&3&0\\\\0&1&0\\\\1&0&1\\end{pmatrix}}}$ Denote the given basis of ${\\displaystyle {\\mathcal {P}}_{2}}$ by ${\\displaystyle B}$. Then application of the linear map is represented by matrix-vector addition. Thus, the first vector in ${\\displaystyle {\\mathcal {E}}_{3}}$ is mapped to the element of ${\\displaystyle {\\mathcal {P}}_{2}}$ represented with respect to ${\\displaystyle B}$ by ${\\displaystyle {\\begin{pmatrix}1&3&0\\\\0&1&0\\\\1&0&1\\end{pmatrix}}{\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}}$ and that element is ${\\displaystyle 1+x}$. The other two images of basis vectors are calculated similarly. ${\\displaystyle {\\begin{pmatrix}1&3&0\\\\0&1&0\\\\1&0&1\\end{pmatrix}}{\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}={\\begin{pmatrix}3\\\\1\\\\0\\end{pmatrix}}={\\rm {Rep}}_{B}(4+x^{2})\\quad {\\begin{pmatrix}1&3&0\\\\0&1&0\\\\1&0&1\\end{pmatrix}}{\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}={\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}={\\rm {Rep}}_{B}(x)}$ We can thus decide if ${\\displaystyle 1+2x}$ is in the range of the map by looking for scalars ${\\displaystyle c_{1}}$, ${\\displaystyle c_{2}}$, and ${\\displaystyle c_{3}}$ such that ${\\displaystyle c_{1}\\cdot (1)+c_{2}\\cdot (1+x^{2})+c_{3}\\cdot (x)=1+2x}$ and obviously ${\\displaystyle c_{1}=1}$, ${\\displaystyle c_{2}=0}$, and ${\\displaystyle c_{3}=1}$ suffice. Thus it is in the range, and in fact it is the image of this vector. ${\\displaystyle 1\\cdot {\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}+0\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}+1\\cdot {\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}}}$ Problem 6 Example 2.8 gives a matrix that is nonsingular, and is therefore associated with maps that are nonsingular. 1. Find the set of column vectors representing the members of the nullspace of any map represented by this matrix. 2.", "Linear Algebra/Orthogonal Projection Onto a Line/Solutions Solutions This exercise is recommended for all readers. Problem 1 Project the first vector orthogonally onto the line spanned by the second vector. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}$ 4. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}$ Each is a straightforward application of the formula from Definition 1.1. 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\-2\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}={\\frac {4}{13}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}={\\begin{pmatrix}12/13\\\\-8/13\\end{pmatrix}}}$ 2. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\0\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}={\\frac {2}{3}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}={\\begin{pmatrix}2\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}={\\frac {-1}{6}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}={\\begin{pmatrix}-1/6\\\\-1/3\\\\1/6\\end{pmatrix}}}$ 4. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}={\\frac {1}{3}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}={\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$ This exercise is recommended for all readers. Problem 2 Project the vector orthogonally onto the line. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}},\\{c{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ 2. ${\\displaystyle {\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}}$, the line ${\\displaystyle y=3x}$ 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}}{{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}={\\frac {-19}{19}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}={\\begin{pmatrix}3\\\\-1\\\\3\\end{pmatrix}}}$ 2. Writing the line as ${\\displaystyle \\{c\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ gives this projection. ${\\displaystyle {\\frac {{\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\frac {-4}{10}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\begin{pmatrix}-2/5\\\\-6/5\\end{pmatrix}}}$ Problem 3 Although the development of Definition 1.1 is guided by the pictures, we are not restricted to spaces that we can draw. In ${\\displaystyle \\mathbb {R} ^{4}}$ project this vector onto this line. ${\\displaystyle {\\vec {v}}={\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\qquad \\ell =\\{c\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}}{{\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}={\\frac {3}{4}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}={\\begin{pmatrix}-3/4\\\\3/4\\\\-3/4\\\\3/4\\end{pmatrix}}}$ This exercise", "E}$${\\ displaystyle {\\ vec {a}} = {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}}}$ ${\\ displaystyle {\\ vec {OF}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {\\ left [{\\ begin {pmatrix} 8 \\\\ 8 \\ \\ -1 \\ end {pmatrix}} - {\\ begin {pmatrix} 1 \\\\ 3 \\\\ 4 \\ end {pmatrix}} \\ right] \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}} {{\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}}}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 8 \\\\ 8 \\\\ - 1 \\ end {pmatrix}} - {\\ frac {27} {9}} \\ cdot {\\ begin {pmatrix} 1 \\\\ 2 \\\\ - 2 \\ end {pmatrix}} = {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}}}$. The distance of the point from the plane thus satisfies ${\\ displaystyle d (P; E)}$${\\ displaystyle P}$${\\ displaystyle E}$ ${\\ displaystyle d (P; E) = d (P; F) \\; {\\ widehat {=}} \\; | {\\ vec {PF}} | = \\ left | {\\ begin {pmatrix} 5 \\\\ 2 \\\\ 5 \\ end {pmatrix}} -", "^{2}}$ is an ${\\displaystyle \\mathbb {R} }$-vector space. In addition, the map is well-defined. Let ${\\displaystyle (x_{1},x_{2})^{T}}$ and ${\\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have: {\\displaystyle {\\begin{aligned}f\\left({\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\\\(x_{1}+y_{1})-5\\cdot (x_{2}+y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{separate vectors}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}x_{1}+x_{2}\\\\x_{1}-5x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}+y_{2}\\\\y_{1}-5y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+f{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\end{aligned}}} Proof step: homogeneity Let ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle (x_{1},x_{2})^{T}\\in \\mathbb {R} ^{2}}$. Then: {\\displaystyle {\\begin{aligned}f\\left(\\lambda \\cdot {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}\\lambda \\cdot x_{1}\\\\\\lambda \\cdot x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda x_{1}+\\lambda x_{2}\\\\\\lambda x_{1}-5\\lambda x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda (x_{1}+x_{2})\\\\\\lambda (x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{scalar multiplication}}\\right.}\\\\[0.3em]&=\\lambda \\cdot {\\begin{pmatrix}(x_{1}+x_{2})\\\\(x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=\\lambda \\cdot f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\\\[0.3em]\\end{aligned}}} Thus the map is linear. ### Important special cases Exercise (The identity is a linear map) Let ${\\displaystyle V}$ be a ${\\displaystyle K}$-vector space. Prove that the identity ${\\displaystyle \\operatorname {id} :V\\to V}$ with ${\\displaystyle \\operatorname {id} (v)=v}$ is a linear map. Proof (The identity is a linear map) The identity is additive: Let ${\\displaystyle v,w\\in V}$, then. ${\\displaystyle \\operatorname {id} (v+w)=v+w=\\operatorname {id} (v)+\\operatorname {id} (w)}$ The identity is homogeneous: Let ${\\displaystyle \\lambda \\in K}$ and ${\\displaystyle v\\in V}$, then ${\\displaystyle \\operatorname {id} (\\lambda \\cdot v)=\\lambda \\cdot v=\\lambda \\cdot \\operatorname {id} (v)}$ Exercise (The map to zero is a linear map) Let ${\\displaystyle V,W}$ be two ${\\displaystyle K}$-vector spaces. Show that", "^{+}\\}}$ where ${\\displaystyle \\mathbb {R} ^{+}}$ is the set of reals greater than zero 5. Under the inherited operations, ${\\displaystyle \\{{\\begin{pmatrix}x\\\\y\\end{pmatrix}}\\in \\mathbb {R} ^{2}\\,{\\big |}\\,x+3y=4{\\text{ and }}2x-y=3{\\text{ and }}6x+4y=10\\}}$ Problem 5 Define addition and scalar multiplication operations to make the complex numbers a vector space over ${\\displaystyle \\mathbb {R} }$. This exercise is recommended for all readers. Problem 6 Is the set of rational numbers a vector space over ${\\displaystyle \\mathbb {R} }$ under the usual addition and scalar multiplication operations? Problem 7 Show that the set of linear combinations of the variables ${\\displaystyle x,y,z}$ is a vector space under the natural addition and scalar multiplication operations. Problem 8 Prove that this is not a vector space: the set of two-tall column vectors with real entries subject to these operations. ${\\displaystyle {\\begin{pmatrix}x_{1}\\\\y_{1}\\end{pmatrix}}+{\\begin{pmatrix}x_{2}\\\\y_{2}\\end{pmatrix}}={\\begin{pmatrix}x_{1}-x_{2}\\\\y_{1}-y_{2}\\end{pmatrix}}\\qquad r\\cdot {\\begin{pmatrix}x\\\\y\\end{pmatrix}}={\\begin{pmatrix}rx\\\\ry\\end{pmatrix}}}$ Problem 9 Prove or disprove that ${\\displaystyle \\mathbb {R} ^{3}}$ is a vector space under these operations. 1. ${\\displaystyle {\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}+{\\begin{pmatrix}x_{2}\\\\y_{2}\\\\z_{2}\\end{pmatrix}}={\\begin{pmatrix}0\\\\0\\\\0\\end{pmatrix}}\\quad {\\text{and}}\\quad r{\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}={\\begin{pmatrix}rx\\\\ry\\\\rz\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}+{\\begin{pmatrix}x_{2}\\\\y_{2}\\\\z_{2}\\end{pmatrix}}={\\begin{pmatrix}0\\\\0\\\\0\\end{pmatrix}}\\quad {\\text{and}}\\quad r{\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}={\\begin{pmatrix}0\\\\0\\\\0\\end{pmatrix}}}$ This exercise is recommended for all readers. Problem 10 For each, decide if it is a vector space; the intended operations are the natural ones. 1. The diagonal ${\\displaystyle 2\\!\\times \\!2}$ matrices ${\\displaystyle \\{{\\begin{pmatrix}a&0\\\\0&b\\end{pmatrix}}\\,{\\big |}\\,a,b\\in \\mathbb {R} \\}}$ 2. This set of ${\\displaystyle 2\\!\\times \\!2}$ matrices ${\\displaystyle \\{{\\begin{pmatrix}x&x+y\\\\x+y&y\\end{pmatrix}}\\,{\\big |}\\,x,y\\in \\mathbb {R} \\}}$ 3. This set ${\\displaystyle \\{{\\begin{pmatrix}x\\\\y\\\\z\\\\w\\end{pmatrix}}\\in" ]

[ "${\\displaystyle {\\vec {p}}={\\frac {{\\begin{pmatrix}0\\\\-2\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\begin{pmatrix}-3/5\\\\-9/5\\end{pmatrix}}}$ The distance between ${\\displaystyle {\\vec {v}}}$ and ${\\displaystyle {\\vec {p}}}$ is approximately ${\\displaystyle 0.63}$ miles and so the sub must move to get in range. This subsection has developed a natural projection map: orthogonal projection onto a line. As suggested by the examples, it is often called for in applications. The next subsection shows how the definition of orthogonal projection onto a line gives us a way to calculate especially convienent bases for vector spaces, again something that is common in applications. The final subsection completely generalizes projection, orthogonal or not, onto any subspace at all. Exercises This exercise is recommended for all readers. Problem 1 Project the first vector orthogonally onto the line spanned by the second vector. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}$ 4. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}$ This exercise is recommended for all readers. Problem 2 Project the vector orthogonally onto the line. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}},\\{c{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ 2. ${\\displaystyle {\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}}$, the line ${\\displaystyle y=3x}$ Problem 3 Although the development of Definition 1.1 is guided by the pictures, we are not restricted to spaces that we can draw. In ${\\displaystyle \\mathbb {R} ^{4}}$ project this vector onto this line. ${\\displaystyle {\\vec {v}}={\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\qquad \\ell =\\{c\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector $\\mathbf{x}$ onto a vector $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "+0 # help 0 39 1 Find the projection of $$\\mathbf{a}$$ onto $$\\mathbf{b} = \\begin{pmatrix} 2 \\\\ 6 \\\\ 3 \\end{pmatrix}$$ if $$\\mathbf{a} \\cdot \\mathbf{b} = 8.$$ Nov 22, 2019", "## Projection of a Vector Onto Another Vector in Higher Dimensions We can find the projection of a vector onto another vector in any dimension, not just in two and three dimensional space. The general formula for the projection of a vector &nbsp $\\mathbf{x}$ &nbsp onto a vector &nbsp $\\mathbf{y}$ is $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{\\mathbf{x} \\cdot \\mathbf{y}}{| \\mathbf{y} |} \\mathbf{y}$ Example: Take &nbsp $\\mathbf{x} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\: \\mathbf{y} = \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}$ $\\mathbf{x} \\cdot \\mathbf{y} = \\begin{pmatrix}1\\\\0\\\\2\\\\3\\end{pmatrix} \\cdot \\begin{pmatrix}-3\\\\2\\\\1\\\\0\\end{pmatrix}=-3+0+2+0=-1$ $| \\mathbf{y} | = \\sqrt{(-3)^2+2^2+1^2+0^2} = \\sqrt{14}$ Then $Proj_{\\mathbf{y}} \\mathbf{x} = \\frac{-1}{\\sqrt{14}} = - \\frac{\\sqrt{14}}{14}$", "${\\displaystyle ={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}{\\begin{pmatrix}1/2\\end{pmatrix}}{\\begin{pmatrix}-1&0&1\\end{pmatrix}}={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}{\\begin{pmatrix}-1/2&0&1/2\\end{pmatrix}}={\\begin{pmatrix}1/2&0&-1/2\\\\0&0&0\\\\-1/2&0&1/2\\end{pmatrix}}}$ followed by matrix-vector multiplication ${\\displaystyle {\\mbox{proj}}_{M}({\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix}}){\\begin{pmatrix}1/2&0&-1/2\\\\0&0&0\\\\-1/2&0&1/2\\end{pmatrix}}{\\begin{pmatrix}0\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}-1\\\\0\\\\1\\end{pmatrix}}}$ Problem 5 Check that the operation of Definition 3.1 is well-defined. That is, in Example 3.2 and 3.3, doesn't the answer depend on the choice of bases? No, a decomposition of vectors ${\\displaystyle {\\vec {v}}={\\vec {m}}+{\\vec {n}}}$ into ${\\displaystyle {\\vec {m}}\\in M}$ and ${\\displaystyle {\\vec {n}}\\in N}$ does not depend on the bases chosen for the subspaces— this was shown in the Direct Sum subsection. Problem 6 What is the orthogonal projection onto the trivial subspace? The orthogonal projection of a vector onto a subspace is a member of that subspace. Since a trivial subspace has only one member, ${\\displaystyle {\\vec {0}}}$, the projection of any vector must equal ${\\displaystyle {\\vec {0}}}$. Problem 7 What is the projection of ${\\displaystyle {\\vec {v}}}$ onto ${\\displaystyle M}$ along ${\\displaystyle N}$ if ${\\displaystyle {\\vec {v}}\\in M}$? The projection onto ${\\displaystyle M}$ along ${\\displaystyle N}$ of a ${\\displaystyle {\\vec {v}}\\in M}$ is ${\\displaystyle {\\vec {v}}}$. Decomposing ${\\displaystyle {\\vec {v}}={\\vec {m}}+{\\vec {n}}}$ gives ${\\displaystyle {\\vec {m}}={\\vec {v}}}$ and ${\\displaystyle {\\vec {n}}={\\vec {0}}}$, and dropping the ${\\displaystyle N}$ part but retaining the ${\\displaystyle M}$ part results in a projection of ${\\displaystyle {\\vec {m}}={\\vec {v}}}$. Problem 8 Show that if ${\\displaystyle M\\subseteq \\mathbb {R} ^{n}}$ is a subspace with orthonormal basis ${\\displaystyle \\langle {\\vec {\\kappa }}_{1},\\ldots ,{\\vec", "# Thread: Projection of point onto line 1. ## Projection of point onto line Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line 2. Originally Posted by acevipa Find the projection of $\\begin{pmatrix}-2\\\\2\\\\7\\end{pmatrix}$ on the direction of the line $\\mathbf{x}=\\begin{pmatrix}1\\\\0\\\\2\\end{pmatrix}+\\la mbda\\begin{pmatrix}-1\\\\1\\\\2\\end{pmatrix}$ I'm used to finding the projection of a vector onto another vector. But this has confused me a little as it is the projection of the vector onto a line It's exactly the same thing- the only use you make of the vector is to determine the line. Here the direction of the line is given by the \"coefficient\" vector $\\begin{pmatrix}-1 \\\\ 1 \\\\ 2\\end{pmatrix}$. Project onto that vector.", "The orthogonal projection $\\mbox{proj}_{M}({\\vec{v}\\,})$ of a vector is its projection onto $M$ along $M^\\perp$. Example 3.5 In $\\mathbb{R}^3$, to find the orthogonal complement of the plane $P=\\{\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix}\\,\\big|\\, 3x+2y-z=0 \\}$ we start with a basis for $P$. $B=\\langle \\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\rangle$ Any $\\vec{v}$ perpendicular to every vector in $B$ is perpendicular to every vector in the span of $B$ (the proof of this assertion is Problem 10). Therefore, the subspace $P^\\perp$ consists of the vectors that satisfy these two conditions. $\\begin{pmatrix} 1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0 \\qquad \\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix}\\cdot\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}=0$ We can express those conditions more compactly as a linear system. $P^\\perp=\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\begin{pmatrix} 1 &0 &3 \\\\ 0 &1 &2 \\end{pmatrix} \\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix} =\\begin{pmatrix} 0 \\\\ 0 \\end{pmatrix} \\}$ We are thus left with finding the nullspace of the map represented by the matrix, that is, with calculating the solution set of a homogeneous linear system. $P^\\perp =\\{\\begin{pmatrix} v_1 \\\\ v_2 \\\\ v_3 \\end{pmatrix}\\,\\big|\\, \\;\\begin{array}{*{3}{rc}r} v_1 & & &+ &3v_3 &= &0 \\\\ & &v_2 &+ &2v_3 &= &0 \\end{array}\\; \\} =\\{k\\begin{pmatrix} -3", "it extends to the endpoint ${\\displaystyle (b_{1}-a_{1},b_{2}-a_{2})}$. We typically just refer to \"the point ${\\displaystyle {\\begin{pmatrix}1\\\\2\\end{pmatrix}}}$\" rather than \"the endpoint of the canonical position of\" that vector. Thus, we will call both of these sets ${\\displaystyle \\mathbb {R} ^{2}}$. ${\\displaystyle \\{(x_{1},x_{2})\\,{\\big |}\\,x_{1},x_{2}\\in \\mathbb {R} \\}\\qquad \\{{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\,{\\big |}\\,x_{1},x_{2}\\in \\mathbb {R} \\}}$ In the prior section we defined vectors and vector operations with an algebraic motivation; ${\\displaystyle r\\cdot {\\begin{pmatrix}v_{1}\\\\v_{2}\\end{pmatrix}}={\\begin{pmatrix}rv_{1}\\\\rv_{2}\\end{pmatrix}}\\qquad {\\begin{pmatrix}v_{1}\\\\v_{2}\\end{pmatrix}}+{\\begin{pmatrix}w_{1}\\\\w_{2}\\end{pmatrix}}={\\begin{pmatrix}v_{1}+w_{1}\\\\v_{2}+w_{2}\\end{pmatrix}}}$ we can now interpret those operations geometrically. For instance, if ${\\displaystyle {\\vec {v}}}$ represents a displacement then ${\\displaystyle 3{\\vec {v}}\\,}$ represents a displacement in the same direction but three times as far, and ${\\displaystyle -1{\\vec {v}}\\,}$ represents a displacement of the same distance as ${\\displaystyle {\\vec {v}}\\,}$ but in the opposite direction. And, where ${\\displaystyle {\\vec {v}}}$ and ${\\displaystyle {\\vec {w}}}$ represent displacements, ${\\displaystyle {\\vec {v}}+{\\vec {w}}}$ represents those displacements combined. The long arrow is the combined displacement in this sense: if, in one minute, a ship's motion gives it the displacement relative to the earth of ${\\displaystyle {\\vec {v}}}$ and a passenger's motion gives a displacement relative to the ship's deck of ${\\displaystyle {\\vec {w}}}$, then ${\\displaystyle {\\vec {v}}+{\\vec {w}}}$ is the displacement of the passenger relative to the earth. Another way to understand the vector sum is with the parallelogram rule. Draw the parallelogram formed by the vectors", "line spanned by a nonzero ${\\displaystyle {\\vec {s}}\\,}$ is this vector. ${\\displaystyle {\\mbox{proj}}_{[{\\vec {s}}\\,]}({\\vec {v}})={\\frac {{\\vec {v}}\\cdot {\\vec {s}}}{{\\vec {s}}\\cdot {\\vec {s}}}}{\\vec {s}}}$ Problem 13 checks that the outcome of the calculation depends only on the line and not on which vector ${\\displaystyle {\\vec {s}}\\,}$ happens to be used to describe that line. Remark 1.2 The wording of that definition says \"spanned by ${\\displaystyle {\\vec {s}}\\,}$\" instead the more formal \"the span of the set ${\\displaystyle \\{{\\vec {s}}\\,\\}}$\". This casual first phrase is common. Example 1.3 To orthogonally project the vector ${\\displaystyle {\\binom {2}{3}}}$ onto the line ${\\displaystyle y=2x}$, we first pick a direction vector for the line. For instance, ${\\displaystyle {\\vec {s}}={\\begin{pmatrix}1\\\\2\\end{pmatrix}}}$ will do. Then the calculation is routine. ${\\displaystyle {\\frac {{\\begin{pmatrix}2\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\2\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}={\\displaystyle {\\frac {8}{5}}}\\cdot {\\begin{pmatrix}1\\\\2\\end{pmatrix}}={\\begin{pmatrix}8/5\\\\16/5\\end{pmatrix}}}$ Example 1.4 In ${\\displaystyle \\mathbb {R} ^{3}}$, the orthogonal projection of a general vector ${\\displaystyle {\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}}$ onto the ${\\displaystyle y}$-axis is ${\\displaystyle {\\frac {{\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}}{{\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix}}={\\begin{pmatrix}0\\\\y\\\\0\\end{pmatrix}}}$ which matches our intuitive expectation. The picture above with the stick figure walking out on the line until ${\\displaystyle {\\vec {v}}}$'s tip is overhead is one way to think of the orthogonal projection of a vector onto a line. We finish this subsection with two other ways. Example 1.5 A railroad car left on an east-west track without its brake is pushed", "{\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}+{\\begin{pmatrix}x_{2}\\\\y_{2}\\\\z_{2}\\end{pmatrix}}={\\begin{pmatrix}x_{1}+x_{2}-1\\\\y_{1}+y_{2}\\\\z_{1}+z_{2}\\end{pmatrix}}\\qquad r{\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}={\\begin{pmatrix}rx-r+1\\\\ry\\\\rz\\end{pmatrix}}}$ 1. Show that it is not a subspace of ${\\displaystyle \\mathbb {R} ^{3}}$ . (Hint. See Example 2.5). 2. Show that it is a vector space. Note that by the prior item, Lemma 2.9 can not apply. 3. Show that any subspace of ${\\displaystyle \\mathbb {R} ^{3}}$ must pass through the origin, and so any subspace of ${\\displaystyle \\mathbb {R} ^{3}}$ must involve zero in its description. Does the converse hold? Does any subset of ${\\displaystyle \\mathbb {R} ^{3}}$ that contains the origin become a subspace when given the inherited operations? 1. It is not a subspace because these are not the inherited operations. For one thing, in this space, ${\\displaystyle 0\\cdot {\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}}}$ while this does not, of course, hold in ${\\displaystyle \\mathbb {R} ^{3}}$ . 2. We can combine the argument showing closure under addition with the argument showing closure under scalar multiplication into one single argument showing closure under linear combinations of two vectors. If ${\\displaystyle r_{1},r_{2},x_{1},x_{2},y_{1},y_{2},z_{1},z_{2}}$ are in ${\\displaystyle \\mathbb {R} }$ then ${\\displaystyle r_{1}{\\begin{pmatrix}x_{1}\\\\y_{1}\\\\z_{1}\\end{pmatrix}}+r_{2}{\\begin{pmatrix}x_{2}\\\\y_{2}\\\\z_{2}\\end{pmatrix}}={\\begin{pmatrix}r_{1}x_{1}-r_{1}+1\\\\r_{1}y_{1}\\\\r_{1}z_{1}\\end{pmatrix}}+{\\begin{pmatrix}r_{2}x_{2}-r_{2}+1\\\\r_{2}y_{2}\\\\r_{2}z_{2}\\end{pmatrix}}={\\begin{pmatrix}r_{1}x_{1}-r_{1}+r_{2}x_{2}-r_{2}+1\\\\r_{1}y_{1}+r_{2}y_{2}\\\\r_{1}z_{1}+r_{2}z_{2}\\end{pmatrix}}}$ (note that the definition of addition in this space is that the first components combine as ${\\displaystyle (r_{1}x_{1}-r_{1}+1)+(r_{2}x_{2}-r_{2}+1)-1}$ , so the first component of the last vector does not say \"${\\displaystyle +2}$ \"). Adding the three components of the last vector gives ${\\displaystyle r_{1}(x_{1}-1+y_{1}+z_{1})+r_{2}(x_{2}-1+y_{2}+z_{2})+1=r_{1}\\cdot 0+r_{2}\\cdot 0+1=1}$ . Most", "vector ${\\displaystyle (-1,0,-1)^{T}}$ is a multiple of ${\\displaystyle (1,0,1)^{T}}$. That is, the direction represented by ${\\displaystyle (-1,0,-1)^{T}}$ is the same as the direction represented by ${\\displaystyle (1,0,1)^{T}}$. Hence, we can remove this vector from ${\\displaystyle E}$ and obtain a new generator ${\\displaystyle E_{1}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}},{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Can we reduce the size of this generator, as well? Yes, since ${\\displaystyle {\\begin{pmatrix}2\\\\1\\\\1\\end{pmatrix}}={\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}+{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}}$ So ${\\displaystyle (2,1,1)^{T}}$ adds no new direction that is not already spanned by ${\\displaystyle (1,0,1)^{T}}$ and ${\\displaystyle (1,1,0)^{T}}$. We thus obtain a smaller generator ${\\displaystyle E_{2}=\\left\\{{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}},{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}},{\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}\\right\\}}$ Now we cannot reduce the generator any further without losing the property of it being a generator. For if we remove any of the three vectors in ${\\displaystyle E_{2}}$, it is no longer in the span of the remaining two vectors. For ${\\displaystyle (1,0,1)^{T}}$, for example, we see this as follows: Suppose we had some ${\\displaystyle \\lambda ,\\mu \\in \\mathbb {R} }$, so that ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=\\lambda {\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+\\mu {\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}}$ Then ${\\displaystyle \\lambda =-\\mu }$ would have to hold because the second component must be zero on both sides. Because of the third component, ${\\displaystyle \\mu ={\\frac {1}{2}}}$ must be true. Thus we obtain the contradiction ${\\displaystyle {\\begin{pmatrix}1\\\\0\\\\1\\end{pmatrix}}=-{\\frac {1}{2}}{\\begin{pmatrix}1\\\\1\\\\0\\end{pmatrix}}+{\\frac {1}{2}}{\\begin{pmatrix}2\\\\1\\\\2\\end{pmatrix}}={\\begin{pmatrix}0.5\\\\0\\\\1\\end{pmatrix}}}$ So ${\\displaystyle (1,0,1)^{T}}$ is not in ${\\displaystyle \\operatorname {span} ((1,1,0)^{T},(2,1,2)^{T})}$. So we have found a generator containing linearly independent vectors (a minimal generator). In summary, we proceeded", "# Linear Algebra: Projection Maps I would like to check if my understanding of projection maps is correct. I have been given the following subset of $\\mathbb{R}^3$: $$A=\\left\\{\\begin{pmatrix} x \\\\ y \\\\ -x+2y \\end{pmatrix} \\middle| x,y,z\\in\\mathbb{R}\\right\\}$$ A basis for this subset is $\\mathscr{B}=\\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix} \\right\\}$, and to extend this basis to one for the vector space $\\mathbb{R^3}$ we simply add to the basis the vector: $$\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}$$ To obtain $\\mathscr{C} = \\left\\{ \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 1 \\\\ 2 \\end{pmatrix},\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix} \\right\\}$, a basis for $\\mathbb{R}^3$. We can call $B = Span\\left\\{\\begin{pmatrix} 0 \\\\ 0 \\\\ 1 \\end{pmatrix}\\right\\}$, and then we can say $\\mathbb{R}^3=A\\bigoplus B$. What I want to know is if I am correct in interpreting the definition of projection map. Let $P:\\mathbb{R}\\to\\mathbb{R}$ be the projection map onto A. The question asks me to calculate $P(e_1)$, $P(e_2)$ and $P(e_3)$ then write down the matrix of $P$ with respect to the standard basis of $\\mathbb{R}^3$. Without explicitly giving my answer (I want to check my method, not my answers), this is my method: 1. Write each vector $e_1$, $e_2$ and $e_3$ as a linear combination of the vectors in $\\mathscr{C}$, so,", "}$-vector space. In addition, the map is well-defined. Let ${\\displaystyle (x_{1},x_{2})^{T}}$ and ${\\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have: {\\displaystyle {\\begin{aligned}f\\left({\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\\\(x_{1}+y_{1})-5\\cdot (x_{2}+y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{separate vectors}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}x_{1}+x_{2}\\\\x_{1}-5x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}+y_{2}\\\\y_{1}-5y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+f{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\end{aligned}}} Proof step: homogeneity Let ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle (x_{1},x_{2})^{T}\\in \\mathbb {R} ^{2}}$. Then: {\\displaystyle {\\begin{aligned}f\\left(\\lambda \\cdot {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}\\lambda \\cdot x_{1}\\\\\\lambda \\cdot x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda x_{1}+\\lambda x_{2}\\\\\\lambda x_{1}-5\\lambda x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda (x_{1}+x_{2})\\\\\\lambda (x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{scalar multiplication}}\\right.}\\\\[0.3em]&=\\lambda \\cdot {\\begin{pmatrix}(x_{1}+x_{2})\\\\(x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=\\lambda \\cdot f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\\\[0.3em]\\end{aligned}}} Thus the map is linear. ## A linear map in the vector space of sequences Next, we consider the space of all sequences of real numbers. This space is infinite-dimensional, because there are not finitely many sequences generating this sequence space. But it is a vector space, as we have shown in the chapter about sequence spaces. Exercise (Sequence space) Let ${\\displaystyle V}$ be the ${\\displaystyle \\mathbb {R} }$-vector space of all real-valued sequences. Show that the map {\\displaystyle {\\begin{aligned}f:V&\\to V\\\\(a_{0},a_{1},a_{2},\\ldots )&\\mapsto (a_{1},a_{2},a_{3},\\ldots )\\end{aligned}}} is linear. How to get to the proof? (Sequence space) To show linearity, two properties need to be checked: 1. ${\\displaystyle f}$ is additive: ${\\displaystyle f(v+w)=f(v)+f(w)}$ for all ${\\displaystyle v,w\\in V}$ 2. ${\\displaystyle", "does not hold in general. See the counterexample: ${\\displaystyle {\\Biggl (}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}{\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}{\\Biggr )}^{+}={\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}={\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}\\quad \\neq \\quad {\\begin{pmatrix}{\\tfrac {1}{4}}&0\\\\{\\tfrac {1}{4}}&0\\end{pmatrix}}={\\begin{pmatrix}0&{\\tfrac {1}{2}}\\\\0&{\\tfrac {1}{2}}\\end{pmatrix}}{\\begin{pmatrix}{\\tfrac {1}{2}}&0\\\\{\\tfrac {1}{2}}&0\\end{pmatrix}}={\\begin{pmatrix}0&0\\\\1&1\\end{pmatrix}}^{+}{\\begin{pmatrix}1&1\\\\0&0\\end{pmatrix}}^{+}}$ ### Projectors ${\\displaystyle P=AA^{+}}$ and ${\\displaystyle Q=A^{+}A}$ are orthogonal projection operators, that is, they are Hermitian (${\\displaystyle P=P^{*}}$, ${\\displaystyle Q=Q^{*}}$) and idempotent (${\\displaystyle P^{2}=P}$ and ${\\displaystyle Q^{2}=Q}$). The following hold: • ${\\displaystyle PA=AQ=A}$ and ${\\displaystyle A^{+}P=QA^{+}=A^{+}}$ • ${\\displaystyle P}$ is the orthogonal projector onto the range of ${\\displaystyle A}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A^{*}}$). • ${\\displaystyle Q}$ is the orthogonal projector onto the range of ${\\displaystyle A^{*}}$ (which equals the orthogonal complement of the kernel of ${\\displaystyle A}$). • ${\\displaystyle (I-Q)=\\left(I-A^{+}A\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A}$. • ${\\displaystyle (I-P)=\\left(I-AA^{+}\\right)}$ is the orthogonal projector onto the kernel of ${\\displaystyle A^{*}}$.[8] The last two properties imply the following identities: • ${\\displaystyle A\\,\\ \\left(I-A^{+}A\\right)=\\left(I-AA^{+}\\right)A\\ \\ =0}$ • ${\\displaystyle A^{*}\\left(I-AA^{+}\\right)=\\left(I-A^{+}A\\right)A^{*}=0}$ Another property is the following: if ${\\displaystyle A\\in \\mathbb {K} ^{n\\times n}}$ is Hermitian and idempotent (true if and only if it represents an orthogonal projection), then, for any matrix ${\\displaystyle B\\in \\mathbb {K} ^{m\\times n}}$ the following equation holds:[10] ${\\displaystyle A(BA)^{+}=(BA)^{+}}$ This can be proven by defining matrices ${\\displaystyle C=BA}$, ${\\displaystyle D=A(BA)^{+}}$, and checking that ${\\displaystyle D}$ is indeed a pseudoinverse for ${\\displaystyle C}$ by verifying that", "projection in $$A \\subseteq M$$ which we can decompose in the way you want. $$A$$ can be thought of as the infinite tensor product of $$M_2$$. So $$A = \\otimes_1^{\\infty}M_2$$, which is really the norm closure of $$\\cup_n (M_2^{\\otimes n}\\otimes 1\\otimes\\cdots)$$. Let $$p = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes 1 \\otimes\\cdots.$$ We can write this projection in the following way. Let $$p_1 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes 1 \\otimes\\cdots$$ and $$p_2 = \\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \\end{pmatrix} \\otimes \\begin{pmatrix} 0 & 0 \\\\ 0 & 1 \\end{pmatrix} \\otimes 1 \\otimes\\cdots.$$ Now its clear that $$p_1,p_2 < p$$ and that $$p = p_1 + p_2$$. Edit: Note that you can keep on doing this, so these projections can't be minimal! This is a \"less naive\" way of seeing the chain $$M_2(\\mathbb C)\\subset M_4(\\mathbb C)\\subset\\cdots\\subset B(H)$$ for infinite-dimensional separable $$H$$ (and it is was was done in PStheman's answer, just a bit more explicit here). You see $$M_2(\\mathbb C)$$ as $$\\begin{bmatrix} a&b\\\\ c&d\\\\ &&a&b\\\\ &&c&d\\\\ &&&&a&b\\\\ &&&&c&d\\\\ &&&&&&a&b\\\\ &&&&&&c&d\\\\ &&&&&&&&\\ddots \\end{bmatrix},$$ then $$M_4(\\mathbb C)$$ as $$\\begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\\\ a_{21}&a_{22}&a_{23}&a_{24}\\\\ a_{31}&a_{32}&a_{33}&a_{34}\\\\ a_{41}&a_{42}&a_{43}&a_{44}\\\\ &&&&a_{11}&a_{12}&a_{13}&a_{14}\\\\ &&&&a_{21}&a_{22}&a_{23}&a_{24}\\\\ &&&&a_{31}&a_{32}&a_{33}&a_{34}\\\\ &&&&a_{41}&a_{42}&a_{43}&a_{44}\\\\ &&&&&&&&\\ddots \\end{bmatrix}.$$ So for example take $$E_{11}^{(2)}\\in M_2(\\mathbb C)$$, and let's", "{\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}\\mapsto 1v_{1}+2v_{2}+3v_{3}}$ with respect to the standard bases and show that ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}}$ is a member of the perp of the nullspace. Prove that ${\\displaystyle {\\mathcal {N}}(f)^{\\perp }}$ is equal to the span of this vector. 2. Generalize that to apply to any ${\\displaystyle f:\\mathbb {R} ^{n}\\to \\mathbb {R} }$. 3. Represent ${\\displaystyle f:\\mathbb {R} ^{3}\\to \\mathbb {R} ^{2}}$ ${\\displaystyle {\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}\\mapsto {\\begin{pmatrix}1v_{1}+2v_{2}+3v_{3}\\\\4v_{1}+5v_{2}+6v_{3}\\end{pmatrix}}}$ with respect to the standard bases and show that ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}},\\;{\\begin{pmatrix}4\\\\5\\\\6\\end{pmatrix}}}$ are both members of the perp of the nullspace. Prove that ${\\displaystyle {\\mathcal {N}}(f)^{\\perp }}$ is the span of these two. (Hint. See the third item of Problem 14.) 4. Generalize that to apply to any ${\\displaystyle f:\\mathbb {R} ^{n}\\to \\mathbb {R} ^{m}}$. This, and related results, is called the Fundamental Theorem of Linear Algebra in (Strang 1993). 1. The representation of ${\\displaystyle {\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}{\\stackrel {f}{\\longmapsto }}1v_{1}+2v_{2}+3v_{3}}$ is this. ${\\displaystyle {\\rm {Rep}}_{{\\mathcal {E}}_{3},{\\mathcal {E}}_{1}}(f)={\\begin{pmatrix}1&2&3\\end{pmatrix}}}$ By the definition of ${\\displaystyle f}$ ${\\displaystyle {\\mathcal {N}}(f)=\\{{\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}\\,{\\big |}\\,1v_{1}+2v_{2}+3v_{3}=0\\}=\\{{\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}\\,{\\big |}\\,{\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}v_{1}\\\\v_{2}\\\\v_{3}\\end{pmatrix}}=0\\}}$ and this second description exactly says this. ${\\displaystyle {\\mathcal {N}}(f)^{\\perp }=[\\{{\\begin{pmatrix}1\\\\2\\\\3\\end{pmatrix}}\\}]}$ 2. The generalization is that for any ${\\displaystyle f:\\mathbb {R} ^{n}\\to \\mathbb {R} }$ there is a vector ${\\displaystyle {\\vec {h}}}$ so that ${\\displaystyle {\\begin{pmatrix}v_{1}\\\\\\vdots \\\\v_{n}\\end{pmatrix}}{\\stackrel {f}{\\longmapsto }}h_{1}v_{1}+\\dots +h_{n}v_{n}}$ and ${\\displaystyle {\\vec {h}}\\in {\\mathcal {N}}(f)^{\\perp }}$. We can prove this by, as in the prior item, representing ${\\displaystyle f}$", "^{2}}$ is an ${\\displaystyle \\mathbb {R} }$-vector space. In addition, the map is well-defined. Let ${\\displaystyle (x_{1},x_{2})^{T}}$ and ${\\displaystyle (y_{1},y_{2})^{T}}$ be any vectors from the plane ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have: {\\displaystyle {\\begin{aligned}f\\left({\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}x_{1}+y_{1}\\\\x_{2}+y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+y_{1})+(x_{2}+y_{2})\\\\(x_{1}+y_{1})-5\\cdot (x_{2}+y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}(x_{1}+x_{2})+(y_{1}+y_{2})\\\\(x_{1}-5x_{2})+(y_{1}-5y_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{separate vectors}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}x_{1}+x_{2}\\\\x_{1}-5x_{2}\\end{pmatrix}}+{\\begin{pmatrix}y_{1}+y_{2}\\\\y_{1}-5y_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}+f{\\begin{pmatrix}y_{1}\\\\y_{2}\\end{pmatrix}}\\end{aligned}}} Proof step: homogeneity Let ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle (x_{1},x_{2})^{T}\\in \\mathbb {R} ^{2}}$. Then: {\\displaystyle {\\begin{aligned}f\\left(\\lambda \\cdot {\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\right)&=f{\\begin{pmatrix}\\lambda \\cdot x_{1}\\\\\\lambda \\cdot x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda x_{1}+\\lambda x_{2}\\\\\\lambda x_{1}-5\\lambda x_{2}\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{distributive law}}\\right.}\\\\[0.3em]&={\\begin{pmatrix}\\lambda (x_{1}+x_{2})\\\\\\lambda (x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{scalar multiplication}}\\right.}\\\\[0.3em]&=\\lambda \\cdot {\\begin{pmatrix}(x_{1}+x_{2})\\\\(x_{1}-5x_{2})\\end{pmatrix}}\\\\[0.3em]&{\\color {OliveGreen}\\left\\downarrow \\ {\\text{definition of }}f\\right.}\\\\[0.3em]&=\\lambda \\cdot f{\\begin{pmatrix}x_{1}\\\\x_{2}\\end{pmatrix}}\\\\[0.3em]\\end{aligned}}} Thus the map is linear. ### Important special cases Exercise (The identity is a linear map) Let ${\\displaystyle V}$ be a ${\\displaystyle K}$-vector space. Prove that the identity ${\\displaystyle \\operatorname {id} :V\\to V}$ with ${\\displaystyle \\operatorname {id} (v)=v}$ is a linear map. Proof (The identity is a linear map) The identity is additive: Let ${\\displaystyle v,w\\in V}$, then. ${\\displaystyle \\operatorname {id} (v+w)=v+w=\\operatorname {id} (v)+\\operatorname {id} (w)}$ The identity is homogeneous: Let ${\\displaystyle \\lambda \\in K}$ and ${\\displaystyle v\\in V}$, then ${\\displaystyle \\operatorname {id} (\\lambda \\cdot v)=\\lambda \\cdot v=\\lambda \\cdot \\operatorname {id} (v)}$ Exercise (The map to zero is a linear map) Let ${\\displaystyle V,W}$ be two ${\\displaystyle K}$-vector spaces. Show that", "Linear Algebra/Orthogonal Projection Onto a Line/Solutions Solutions This exercise is recommended for all readers. Problem 1 Project the first vector orthogonally onto the line spanned by the second vector. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}$ 2. ${\\displaystyle {\\begin{pmatrix}2\\\\1\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}$ 4. ${\\displaystyle {\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$, ${\\displaystyle {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}$ Each is a straightforward application of the formula from Definition 1.1. 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\-2\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}={\\frac {4}{13}}\\cdot {\\begin{pmatrix}3\\\\-2\\end{pmatrix}}={\\begin{pmatrix}12/13\\\\-8/13\\end{pmatrix}}}$ 2. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\0\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}={\\frac {2}{3}}\\cdot {\\begin{pmatrix}3\\\\0\\end{pmatrix}}={\\begin{pmatrix}2\\\\0\\end{pmatrix}}}$ 3. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}={\\frac {-1}{6}}\\cdot {\\begin{pmatrix}1\\\\2\\\\-1\\end{pmatrix}}={\\begin{pmatrix}-1/6\\\\-1/3\\\\1/6\\end{pmatrix}}}$ 4. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}{{\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}={\\frac {1}{3}}\\cdot {\\begin{pmatrix}3\\\\3\\\\12\\end{pmatrix}}={\\begin{pmatrix}1\\\\1\\\\4\\end{pmatrix}}}$ This exercise is recommended for all readers. Problem 2 Project the vector orthogonally onto the line. 1. ${\\displaystyle {\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}},\\{c{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ 2. ${\\displaystyle {\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}}$, the line ${\\displaystyle y=3x}$ 1. ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}2\\\\-1\\\\4\\end{pmatrix}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}}{{\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}={\\frac {-19}{19}}\\cdot {\\begin{pmatrix}-3\\\\1\\\\-3\\end{pmatrix}}={\\begin{pmatrix}3\\\\-1\\\\3\\end{pmatrix}}}$ 2. Writing the line as ${\\displaystyle \\{c\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ gives this projection. ${\\displaystyle {\\frac {{\\begin{pmatrix}-1\\\\-1\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}{{\\begin{pmatrix}1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\frac {-4}{10}}\\cdot {\\begin{pmatrix}1\\\\3\\end{pmatrix}}={\\begin{pmatrix}-2/5\\\\-6/5\\end{pmatrix}}}$ Problem 3 Although the development of Definition 1.1 is guided by the pictures, we are not restricted to spaces that we can draw. In ${\\displaystyle \\mathbb {R} ^{4}}$ project this vector onto this line. ${\\displaystyle {\\vec {v}}={\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\qquad \\ell =\\{c\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\,{\\big |}\\,c\\in \\mathbb {R} \\}}$ ${\\displaystyle \\displaystyle {\\frac {{\\begin{pmatrix}1\\\\2\\\\1\\\\3\\end{pmatrix}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}}{{\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}={\\frac {3}{4}}\\cdot {\\begin{pmatrix}-1\\\\1\\\\-1\\\\1\\end{pmatrix}}={\\begin{pmatrix}-3/4\\\\3/4\\\\-3/4\\\\3/4\\end{pmatrix}}}$ This exercise", "f}$ is an epimorphism. Solution By construction, ${\\displaystyle f}$ is a linear map. We want to show that ${\\displaystyle \\{f(e_{1}),f(e_{2}),f(e_{3})\\}=\\{(1,1),(1,-1)\\}}$ is a generator of ${\\displaystyle \\mathbb {R} ^{2}}$. Then, we have with the above theorem that ${\\displaystyle f}$ is an epimorphism. So, we must represent every ${\\displaystyle (x,y)\\in \\mathbb {R} ^{2}}$ as a linear combination of vectors ${\\displaystyle (1,1),(1,-1)}$. Accordingly, we search linear combination scalars ${\\displaystyle \\lambda ,\\mu \\in \\mathbb {R} }$ such that {\\displaystyle {\\begin{aligned}{\\begin{pmatrix}x\\\\y\\end{pmatrix}}=\\lambda {\\begin{pmatrix}1\\\\1\\end{pmatrix}}+\\mu {\\begin{pmatrix}1\\\\-1\\end{pmatrix}}.\\end{aligned}}} From this we get the linear system of equations {\\displaystyle {\\begin{aligned}{\\begin{aligned}x&=\\lambda +\\mu \\\\y&=\\lambda -\\mu ,\\end{aligned}}\\end{aligned}}} which is solved by ${\\displaystyle \\lambda ={\\frac {x+y}{2}}}$ and ${\\displaystyle \\mu ={\\frac {x-y}{2}}}$. So we have ${\\displaystyle {\\begin{pmatrix}x\\\\y\\end{pmatrix}}={\\frac {x+y}{2}}{\\begin{pmatrix}1\\\\1\\end{pmatrix}}+{\\frac {x-y}{2}}{\\begin{pmatrix}1\\\\-1\\end{pmatrix}}}$ for all ${\\displaystyle x,y\\in \\mathbb {R} }$. Thus ${\\displaystyle \\{(1,1),(1,-1)\\}}$ is a generator. This proves that ${\\displaystyle f}$ is an epimorphism. Exercise Consider the function space ${\\displaystyle \\operatorname {Fun} ([0,1],\\mathbb {R} )}$ of all function s mapping ${\\displaystyle \\mathbb {R} }$ to ${\\displaystyle \\mathbb {R} }$, as well as the map {\\displaystyle {\\begin{aligned}\\phi \\colon \\operatorname {Fun} ([0,1],\\mathbb {R} )&\\to \\mathbb {R} ,\\\\f&\\mapsto f(0).\\end{aligned}}} Show that ${\\displaystyle \\phi }$ is an epimorphism. Solution The operations on the function space are defined element-wise in each case. That is: for ${\\displaystyle f,g\\in \\operatorname {Fun} ([0,1],\\mathbb {R} )}$, ${\\displaystyle \\lambda \\in \\mathbb {R} }$ and ${\\displaystyle x\\in [0,1]}$ we", "|}\\,q\\in \\mathbb {C} \\}}$, and ${\\displaystyle {\\mathcal {N}}(t_{0}^{2})}$ is the space of quadratic polynomials with no ${\\displaystyle x^{2}}$ term ${\\displaystyle \\{p+qx\\,{\\big |}\\,p,q\\in \\mathbb {C} \\}}$, and this is the end ${\\displaystyle {\\mathcal {N}}_{\\infty }(t_{0})={\\mathcal {N}}(t_{0}^{2})}$. 2. The second power ${\\displaystyle {\\begin{pmatrix}a\\\\b\\end{pmatrix}}{\\stackrel {t_{1}}{\\longmapsto }}{\\begin{pmatrix}0\\\\a\\end{pmatrix}}{\\stackrel {t_{1}}{\\longmapsto }}{\\begin{pmatrix}0\\\\0\\end{pmatrix}}}$ is the zero map. Consequently, the chain of rangespaces ${\\displaystyle \\mathbb {R} ^{2}\\supset \\{{\\begin{pmatrix}0\\\\p\\end{pmatrix}}\\,{\\big |}\\,p\\in \\mathbb {C} \\}\\supset \\{{\\vec {0}}\\,\\}=\\cdots }$ and the chain of nullspaces ${\\displaystyle \\{{\\vec {0}}\\,\\}\\subset \\{{\\begin{pmatrix}q\\\\0\\end{pmatrix}}\\,{\\big |}\\,q\\in \\mathbb {C} \\}\\subset \\mathbb {R} ^{2}=\\cdots }$ each has length two. The generalized rangespace is the trivial subspace and the generalized nullspace is the entire space. 3. Iterates of this map cycle around ${\\displaystyle a+bx+cx^{2}{\\stackrel {t_{2}}{\\longmapsto }}b+cx+ax^{2}{\\stackrel {t_{2}}{\\longmapsto }}c+ax+bx^{2}{\\stackrel {t_{2}}{\\longmapsto }}a+bx+cx^{2}\\;\\cdots }$ and the chains of rangespaces and nullspaces are trivial. ${\\displaystyle {\\mathcal {P}}_{2}={\\mathcal {P}}_{2}=\\cdots \\qquad \\{{\\vec {0}}\\,\\}=\\{{\\vec {0}}\\,\\}=\\cdots }$ Thus, obviously, generalized spaces are ${\\displaystyle {\\mathcal {R}}_{\\infty }(t_{2})={\\mathcal {P}}_{2}}$ and ${\\displaystyle {\\mathcal {N}}_{\\infty }(t_{2})=\\{{\\vec {0}}\\,\\}}$. 4. We have ${\\displaystyle {\\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}}\\mapsto {\\begin{pmatrix}a\\\\a\\\\b\\end{pmatrix}}\\mapsto {\\begin{pmatrix}a\\\\a\\\\a\\end{pmatrix}}\\mapsto {\\begin{pmatrix}a\\\\a\\\\a\\end{pmatrix}}\\mapsto \\cdots }$ and so the chain of rangespaces ${\\displaystyle \\mathbb {R} ^{3}\\supset \\{{\\begin{pmatrix}p\\\\p\\\\r\\end{pmatrix}}\\,{\\big |}\\,p,r\\in \\mathbb {C} \\}\\supset \\{{\\begin{pmatrix}p\\\\p\\\\p\\end{pmatrix}}\\,{\\big |}\\,p\\in \\mathbb {C} \\}=\\cdots }$ and the chain of nullspaces ${\\displaystyle \\{{\\vec {0}}\\,\\}\\subset \\{{\\begin{pmatrix}0\\\\0\\\\r\\end{pmatrix}}\\,{\\big |}\\,r\\in \\mathbb {C} \\}\\subset \\{{\\begin{pmatrix}0\\\\q\\\\r\\end{pmatrix}}\\,{\\big |}\\,q,r\\in \\mathbb {C} \\}=\\cdots }$ each has length two. The generalized spaces are the final ones shown above in each chain." ]

[ "both. A description of planes that is often encountered in algebra and calculus uses a single equation as the condition that describes the relationship among the first, second, and third coordinates of points in a plane. The translation from such a description to the vector description that we favor in this book is to think of the condition as a one-equation linear system and parametrize $x=(1/2)(4-y-z)$. Generalizing from lines and planes, we define a $k$-dimensional linear surface (or $k$-flat) in $\\mathbb{R}^n$ to be $\\{\\vec{p}+t_1\\vec{v}_1+t_2\\vec{v}_2+\\cdots+t_k\\vec{v}_k \\,\\big|\\, t_1,\\ldots ,t_k\\in\\mathbb{R}\\}$ where $\\vec{v}_1,\\ldots,\\vec{v}_k\\in\\mathbb{R}^n$. For example, in $\\mathbb{R}^4$, $\\{\\begin{pmatrix} 2 \\\\ \\pi \\\\ 3 \\\\ -0.5 \\end{pmatrix} +t\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\\\ 0 \\end{pmatrix} \\,\\big|\\, t\\in\\mathbb{R}\\}$ is a line, $\\{ \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 0 \\end{pmatrix} +t\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\\\ -1 \\end{pmatrix} +s\\begin{pmatrix} 2 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix} \\,\\big|\\, t,s\\in\\mathbb{R}\\}$ is a plane, and $\\{ \\begin{pmatrix} 3 \\\\ 1 \\\\ -2 \\\\ 0.5 \\end{pmatrix} +r\\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\\\ -1 \\end{pmatrix} +s\\begin{pmatrix} 1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix} +t\\begin{pmatrix} 2 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix} \\,\\big|\\, r,s,t\\in\\mathbb{R}\\}$ is a three-dimensional linear surface. Again, the intuition is that a line permits motion in one direction, a plane permits motion in combinations of two directions, etc. $L=\\{ \\begin{pmatrix}", "we get $\\begin{pmatrix}i & j & k & l \\\\ 1 & -2 & 0 & 1 \\\\ 2 & 0 & -1 & 1 \\\\ 0 & -1 & 2 & 1 \\\\ \\end{pmatrix}=( 5 )i + ( -1 )j + ( 3 )k + ( -7 )l$ That is, the same coefficients as in the first case I would like to notice only one, three points that you will enter should not be on one straight line, since in this case, the equation of the plane cannot be calculated due to the ambiguity of its position in space. Good luck! Copyright © 2021 AbakBot-online calculators. All Right Reserved. Author by Dmitry Varlamov", "coordinate system on the plane, we get $$\\mathtt H = \\mathtt B \\left(\\mathbf C \\mathbf \\Pi^T-\\mathbf C^T \\mathbf \\Pi \\mathtt I_4 \\right) \\begin{bmatrix}\\mathtt M^{-1}\\\\\\mathbf 0^T\\end{bmatrix}.$$ With $\\mathbf\\Pi=(0,0,1,0)^T$—the $x$-$y$ plane—and $$\\mathtt B = \\begin{bmatrix}1&0&0&0\\\\0&1&0&0\\\\0&0&0&1\\end{bmatrix}$$ this produces the same homography matrix as above.", "plane and defined by the parametric equations: $\\begin{cases} x = x \\left({t}\\right) \\\\ y = y \\left({t}\\right) \\end{cases}$ The curvature $k$ of $C$ at a point $P = \\left({x, y}\\right)$ is given by: $k = \\dfrac {x' y'' - y' x''} {\\left({x'^2 + y'^2}\\right)^{3/2} }$ where: $x' = \\dfrac {\\mathrm d x} {\\mathrm d t}$ is the derivative of $x$ with respect to $t$ at $P$ $y' = \\dfrac {\\mathrm d y} {\\mathrm d t}$ is the derivative of $y$ with respect to $t$ at $P$ $x''$ and $y''$ are the second derivatives of $x$ and $y$ with respect to $t$ at $P$.", "Math Planes Parametric equation # Parametric equation of a plane Planes play a major role in analytical geometry. Similar to lines, there is also a parametric equation for planes. However, there is a position vector and two direction vectors. $\\text{E: } \\vec{x} = \\vec{a} + r \\cdot \\vec{u} + s \\cdot \\vec{v}$ • $\\vec{x}$ is the general vector of the plane • $\\vec{a}$ is the position vector • $\\vec{u}, \\vec{v}$ are the direction vectors • $r, s$ are parameters ! ### Remember A plane is clearly defined by three points. ## Parametric equation using 3 points If 3 points $A$, $B$, $C$ are given, you can easily set up a parametric equation of the plane. $\\text{E: } \\vec{x} = \\vec{OA} + r \\cdot \\vec{AB} + s \\cdot \\vec{AC}$ i ### Method 1. Position vector of a point as position vector of the plane 2. Direction vectors: any two connecting vectors of the given points 3. Insert position vector and direction vector ### Example Determine the parametric equation of plane $E$ using points $A(2|1|1)$, $B(3|2|1)$ and $C(3|6|3)$. 1. #### Position vector $\\vec{OA}=\\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}$ 2. #### Connecting vector $\\vec{AB}$ $=\\begin{pmatrix} 3-2 \\\\ 2-1 \\\\ 1-1 \\end{pmatrix}$ $=\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $\\vec{AC}$ $=\\begin{pmatrix} 3-2 \\\\ 6-1 \\\\ 3-1 \\end{pmatrix}$ $=\\begin{pmatrix} 1 \\\\ 5", "+0 # Consider the following sets of points: -1 34 1 +82 Consider the following sets of points: \\begin{align*} &S_1 \\text{ is the set of all points (x, y, z) such that x+y + 2z = 1}, \\\\ &S_2 \\text{ is the set of all points Q such that }\\overrightarrow{OQ} = \\begin{pmatrix}1 \\\\ 2\\\\ 3 \\end{pmatrix} + s \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix} + t \\begin{pmatrix} 2 \\\\ 2 \\\\ 4 \\end{pmatrix} \\text{for some real s and t},\\\\ &S_3 \\text{ is the set of all points (x, y, z) such that x = y = 2z}, \\\\ &S_4 \\text{ is the set of all points Q such that }\\overrightarrow{OQ} = \\begin{pmatrix}1 \\\\ 2\\\\ 3 \\end{pmatrix} + s \\begin{pmatrix} 1 \\\\ 1 \\\\ 2 \\end{pmatrix} + t \\begin{pmatrix} 2 \\\\ 2 \\\\ 3 \\end{pmatrix} \\text{for some real s and t}. \\end{align*} Each of these sets of points is either a line or a plane. For each set above, enter \"line\" if it's a line, and \"plane\" if it's a plane. Enter the answers in the order they appear in the list above. Could someone help me? I thought it was plane, plane, line, line, but it's wrong. Mar 2, 2020", "Review question # Can we show the direction of $P'Q'$ is independent of $P$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R8565 ## Question The plane $\\pi$ has equation $\\mathbf{r} = \\mathbf{a} + \\lambda \\mathbf{b}+ \\mu \\mathbf{c}$ and the plane $\\pi'$ has equation $\\mathbf{r} = \\mathbf{d}+ \\lambda'\\mathbf{e}+\\mu'\\mathbf{f}$, where $\\lambda$, $\\mu$, $\\lambda'$ and $\\mu'$ are scalar parameters. It is given that $\\mathbf{a}=\\begin{pmatrix}1 \\\\ 0 \\\\ -1\\end{pmatrix}, \\mathbf{b}=\\begin{pmatrix}2\\\\ -1 \\\\ -1\\end{pmatrix}, \\mathbf{c} = \\begin{pmatrix}-1 \\\\ 2 \\\\ 1\\end{pmatrix}, \\mathbf{d} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}, \\mathbf{e} = \\begin{pmatrix}p \\\\ q \\\\ r\\end{pmatrix}, \\mathbf{f} = \\begin{pmatrix}q\\\\ r\\\\ p\\end{pmatrix},$ $\\big \\vert\\mathbf{e}\\big \\vert=\\big \\vert\\mathbf{f}\\big \\vert = \\sqrt{30}$ and that $p$ is negative. The planes $\\pi$ and $\\pi'$ are parallel. Find $p$, $q$ and $r$. Show that $\\pi'$ is the reflection of $\\pi$ in the origin. The line $L$ has equation $\\dfrac{x+1}{9}=\\dfrac{y-1}{-6}=\\dfrac{z}{-5}$ and lies in $\\pi$. The point $P$ is in $\\pi$ with parameters $\\lambda = l$ and $\\mu = m$, and $P'$ is in $\\pi'$ with $\\lambda' = l$ and $\\mu' = m$. The point $Q'$ is the reflection of $P$ in the origin. Show that when $P$ lies on $L$ the direction of $P'Q'$ is independent of the position of $P$ on $L$.", "# 11c. Lines and planes Work in progress The following questions illustrate the key concepts needed in this topic. See if you can answer them. Click on the title for detailed discussion of how to tackle them. Click on or hover over the question to show the answer. Consider the points $A(3,-1,6), B(1,-1,-4)$ and $C(5,3,0)$. Consider the points $A(3,-1,6), C(5,3,0), D(5,-2,5)$ and the lines $l_1: \\mathbf{r} =\\begin{pmatrix} 3 \\\\ -1 \\\\ 6 \\end{pmatrix} + \\lambda \\begin{pmatrix} 1 \\\\ 2 \\\\ -3 \\end{pmatrix}, \\lambda \\in \\mathbb{R} \\\\ l_2: \\mathbf{r} =\\begin{pmatrix} 1 \\\\ -1 \\\\ -4 \\end{pmatrix} + \\mu \\begin{pmatrix} 2 \\\\ -1 \\\\ -1 \\end{pmatrix}, \\mu \\in \\mathbb{R}$. The lines $l$ and $l_3$ and the planes $\\pi_1$ and $\\pi_2$ have equations $l_1: \\mathbf{r} =\\begin{pmatrix} 3 \\\\ -1 \\\\ 6 \\end{pmatrix} + \\lambda \\begin{pmatrix} 1 \\\\ 2 \\\\ -3 \\end{pmatrix}, \\lambda \\in \\mathbb{R} \\\\ l_3: \\mathbf{r} = \\begin{pmatrix} 5 \\\\ 3 \\\\ 0 \\end{pmatrix} + \\nu \\begin{pmatrix} 1 \\\\ -2 \\\\ 1 \\end{pmatrix}, \\nu \\in \\mathbb{R} \\\\ \\pi_1: x+y+z = 8 \\\\ \\pi_2: 2x-3y+5z=-2$. Consider the lines $l$ and $m$ and the planes $\\pi_1$ and $\\pi_2$ with equations: $l: \\mathbf{r} =\\begin{pmatrix} 3 \\\\ -1 \\\\ 6 \\end{pmatrix} + \\lambda \\begin{pmatrix} 1 \\\\ 2 \\\\ -3 \\end{pmatrix}, \\lambda \\in \\mathbb{R} \\\\ m: \\mathbf{r} = \\begin{pmatrix} 5 \\\\ 4 \\\\ -1 \\end{pmatrix}", "$\\overrightarrow{OM}$ is in the plane, where $\\overrightarrow{OM}$ = $\\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}$. Any point in the plane can be expressed as a linear combination of these vectors. So an equation of the plane is $\\mathbf{r} = \\alpha \\begin{pmatrix} 5 \\\\ 0 \\\\ 5 \\end{pmatrix}+ \\beta \\begin{pmatrix} 1 \\\\ -2 \\\\ 5 \\end{pmatrix}.$ Alternatively, we could write the equation as $\\mathbf{r}.\\mathbf{n}=0$ where $\\mathbf{n}$ is the normal to the plane. Substituting in the two points $L$ and $M$ tells us that $\\mathbf{n}$ is a multiple of $\\begin{pmatrix}1\\\\ -2\\\\ -1\\end{pmatrix}$, so the plane equation can be written as $\\mathbf{r}.\\begin{pmatrix}1\\\\ -2\\\\ -1\\end{pmatrix}=0.$ 1. Find the position vector of the point $P$ where $l$ meets the plane $\\pi$ whose equation is $\\mathbf{r}.\\begin{pmatrix}1\\\\ 2\\\\ 2\\end{pmatrix} = 11.$ Let $\\overrightarrow{OP} = \\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix} + \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}.$ If $P$ is on the plane, then $\\overrightarrow{OP}.\\begin{pmatrix}1\\\\ 2\\\\ 2\\end{pmatrix} = 11,$ and so \\begin{align*} \\left[ \\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}+ \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix} \\right].\\begin{pmatrix}1\\\\ 2\\\\ 2\\end{pmatrix} = 11. \\end{align*} This implies \\begin{align*} 15 + 4\\lambda = 11, \\end{align*} from which we conclude $\\lambda=-1$. Thus $\\overrightarrow{OP}=\\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix} - \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}=\\begin{pmatrix}3\\\\ -1\\\\ 5\\end{pmatrix}.$", "# the osculating planes of a curve pass through a fixed point $\\rightarrow$ the curve is a plane curve. If the osculating planes of a curve pass through a fixed point, the curve is a plane curve. How to prove it? Parametrize the curve by arclength, as usual, by $\\alpha(s)$. Say the fixed point is the origin. Then there are functions $a(s)$ and $b(s)$ so that $$\\alpha(s) = a(s)\\mathbf T(s) + b(s)\\mathbf N(s).$$ Now differentiate and apply Frenet. • @Pthrp97 If all the planes pass through $\\mathbf a$, just subtract that vector from $\\alpha(s)$ and proceed. Mar 20 '21 at 3:38 yes upon derivation, you should tao go to 0. Note that there is no B term. so the N' term will have a $\\tau$ $\\vec{B}$ term but this will go to zero.", "# hyperbolic plane in quadratic spaces A non-singular (http://planetmath.org/NonDegenerateQuadraticForm) isotropic quadratic space $\\mathcal{H}$ of dimension 2 (over a field) is called a hyperbolic plane. In other words, $\\mathcal{H}$ is a 2-dimensional vector space over a field equipped with a quadratic form $Q$ such that there exists a non-zero vector $v$ with $Q(v)=0$. Examples. Fix the ground field to be $\\mathbb{R}$, and $\\mathbb{R}^{2}$ be the two-dimensional vector space over $\\mathbb{R}$ with the standard basis $(0,1)$ and $(1,0)$. 1. 1. Let $Q_{1}(x,y)=xy$. Then $Q_{1}(a,0)=Q_{1}(0,b)=0$ for all $a,b\\in\\mathbb{R}$. $(\\mathbb{R}^{2},Q_{1})$ is a hyperbolic plane. When $Q_{1}$ is written in matrix form, we have $Q_{1}(x,y)=\\begin{pmatrix}x&y\\end{pmatrix}\\begin{pmatrix}0&\\frac{1}{2}\\\\ \\frac{1}{2}&0\\end{pmatrix}\\begin{pmatrix}x\\\\ y\\end{pmatrix}=\\begin{pmatrix}x&y\\end{pmatrix}M(Q_{1})\\begin{pmatrix}x\\\\ y\\end{pmatrix}.$ 2. 2. Let $Q_{2}(r,s)=r^{2}-s^{2}$. Then $Q_{2}(a,a)=0$ for all $a\\in\\mathbb{R}$. $(\\mathbb{R}^{2},Q_{2})$ is a hyperbolic plane. As above, $Q_{2}$ can be written in matrix form: $Q_{1}(x,y)=\\begin{pmatrix}x&y\\end{pmatrix}\\begin{pmatrix}1&0\\\\ 0&-1\\end{pmatrix}\\begin{pmatrix}x\\\\ y\\end{pmatrix}=\\begin{pmatrix}x&y\\end{pmatrix}M(Q_{2})\\begin{pmatrix}x\\\\ y\\end{pmatrix}.$ From the above examples, we see that the name “hyperbolic plane” comes from the fact that the associated quadratic form resembles the equation of a hyperbola in a two-dimensional Euclidean plane. It’s not hard to see that the two examples above are equivalent quadratic forms. To transform from the first form to the second, for instance, follow the linear substitutions $x=r-s$ and $y=r+s$, or in matrix form: $\\begin{pmatrix}1&1\\\\ -1&1\\end{pmatrix}M(Q_{1})\\begin{pmatrix}1&-1\\\\ 1&1\\end{pmatrix}=\\begin{pmatrix}1&1\\\\ -1&1\\end{pmatrix}\\begin{pmatrix}0&\\frac{1}{2}\\\\ \\frac{1}{2}&0\\end{pmatrix}\\begin{pmatrix}1&-1\\\\ 1&1\\end{pmatrix}=\\begin{pmatrix}1&0\\\\ 0&-1\\end{pmatrix}=M(Q_{2}).$ In fact, we have the following . Any two hyperbolic planes over a field", "the plane equation as $ax+by+cz+d=0$ . This has the effect of flipping the sign of $d$ . Our comments in Section 9.2.2 explaining our preference to put $d$ on the left side of the equals sign also apply here: our experience is that this form results in fewer terms and minus signs and a more intuitive geometric interpretation for $d$ . The vector $\\mathbf{n}$ is called the plane normal because it is perpendicular (normal) to the plane. Although $\\mathbf{n}$ is often normalized to unit length, this is not strictly necessary. We use a hat ( $\\stackrel{^}{\\mathbf{n}}$ ) when we are assuming unit length. The normal determines the orientation of the plane; $d$ defines its position. More specifically, it determines the signed distance to the plane from the origin, measured in the direction of the normal. Increasing $d$ slides the plane forward, in the direction of the normal. If $d>0$ , the origin is on the back side of the plane, and if $d<0$ , the origin is on the front side. (This assumes we put $d$ on the right-hand side of the equals sign, as in Equation (9.10). The standard homogenous form with $d$ on the left has the opposite sign conventions.) Let's verify that $\\mathbf{n}$ is perpendicular to the plane. Assume $\\mathbf{p}$ and $\\mathbf{q}$ are arbitrary points", "the plane by an equation of the form ${\\displaystyle ax+by+cz+d=0}$, solve the following system of equations: ${\\displaystyle \\,ax_{1}+by_{1}+cz_{1}+d=0}$ ${\\displaystyle \\,ax_{2}+by_{2}+cz_{2}+d=0}$ ${\\displaystyle \\,ax_{3}+by_{3}+cz_{3}+d=0.}$ This system can be solved using Cramer's rule and basic matrix manipulations. Let ${\\displaystyle D={\\begin{vmatrix}x_{1}&y_{1}&z_{1}\\\\x_{2}&y_{2}&z_{2}\\\\x_{3}&y_{3}&z_{3}\\end{vmatrix}}}$. If D is non-zero (so for planes not through the origin) the values for a, b and c can be calculated as follows: ${\\displaystyle a={\\frac {-d}{D}}{\\begin{vmatrix}1&y_{1}&z_{1}\\\\1&y_{2}&z_{2}\\\\1&y_{3}&z_{3}\\end{vmatrix}}}$ ${\\displaystyle b={\\frac {-d}{D}}{\\begin{vmatrix}x_{1}&1&z_{1}\\\\x_{2}&1&z_{2}\\\\x_{3}&1&z_{3}\\end{vmatrix}}}$ ${\\displaystyle c={\\frac {-d}{D}}{\\begin{vmatrix}x_{1}&y_{1}&1\\\\x_{2}&y_{2}&1\\\\x_{3}&y_{3}&1\\end{vmatrix}}.}$ These equations are parametric in d. Setting d equal to any non-zero number and substituting it into these equations will yield one solution set. #### Method 3 This plane can also be described by the \"point and a normal vector\" prescription above. A suitable normal vector is given by the cross product ${\\displaystyle {\\boldsymbol {n}}=({\\boldsymbol {p}}_{2}-{\\boldsymbol {p}}_{1})\\times ({\\boldsymbol {p}}_{3}-{\\boldsymbol {p}}_{1}),}$ and the point r0 can be taken to be any of the given points p1, p2 or p3 [6] (or any other point in the plane). ## Operations ### Distance from a point to a plane For a plane ${\\displaystyle \\Pi :ax+by+cz+d=0}$ and a point ${\\displaystyle {\\boldsymbol {p}}_{1}=(x_{1},y_{1},z_{1})}$ not necessarily lying on the plane, the shortest distance from ${\\displaystyle {\\boldsymbol {p}}_{1}}$ to the plane is ${\\displaystyle D={\\frac {\\left|ax_{1}+by_{1}+cz_{1}+d\\right|}{\\sqrt {a^{2}+b^{2}+c^{2}}}}.}$ It follows that ${\\displaystyle {\\boldsymbol {p}}_{1}}$ lies in the plane if and only if D=0. If ${\\displaystyle {\\sqrt {a^{2}+b^{2}+c^{2}}}=1}$ meaning", "+0 # math problem 0 246 2 The equation of the plane through the points triple A = (1,-2,1), B = (-3,1,4), C = (1,1,2) can be written as ax + by + cz = 13. What is the ordered triple (a, b, c)? Aug 15, 2019 #1 +24388 +3 The equation of the plane through the points triple A = (1,-2,1), B = (-3,1,4), C = (1,1,2) can be written as ax + by + cz = 13. What is the ordered triple (a, b, c)? Formula: $$\\vec{x} = \\vec{C} + r(\\vec{A}-\\vec{C}) +s(\\vec{B}-\\vec{C})$$ $$\\begin{array}{|lrcll|} \\hline & \\mathbf{\\vec{x}} &=& \\mathbf{ \\vec{C} + r(\\vec{A}-\\vec{C}) +s(\\vec{B}-\\vec{C}) } \\\\ & \\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} &=&\\begin{pmatrix} 1\\\\ 1\\\\ 2 \\end{pmatrix} +r\\left(\\begin{pmatrix} 1\\\\ -2\\\\ 1 \\end{pmatrix}-\\begin{pmatrix} 1\\\\ 1\\\\ 2 \\end{pmatrix}\\right) +s\\left(\\begin{pmatrix} -2\\\\ 1\\\\ 4 \\end{pmatrix}-\\begin{pmatrix} 1\\\\ 1\\\\ 2 \\end{pmatrix}\\right) \\\\ & \\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} &=&\\begin{pmatrix} 1\\\\ 1\\\\ 2 \\end{pmatrix} +r \\begin{pmatrix} 0\\\\ -3\\\\ -1 \\end{pmatrix} +s \\begin{pmatrix} -4\\\\ 0\\\\ 2 \\end{pmatrix} \\\\ \\hline (1) & x &=& 1-4s \\\\ & 4s &=& 1-x \\\\ & \\mathbf{s} &=& \\mathbf{\\dfrac{1}{4}(1-x)} \\\\\\\\ (2) & y &=& 1-3r \\\\ & 3r &=& 1-y \\\\ & \\mathbf{r} &=& \\mathbf{\\dfrac{1}{3}(1-y)} \\\\\\\\ (3) & z &=& 2-r+2s \\\\ & z &=& 2-\\dfrac{1}{3}(1-y)+\\dfrac{2}{4}(1-x) \\\\ & z &=& 2-\\dfrac{1}{3}(1-y)+\\dfrac{1}{2}(1-x) \\quad | \\quad \\cdot 6 \\\\ & 6z &=& 12-\\dfrac{6}{3}(1-y)+\\dfrac{6}{2}(1-x)", "we apply the second line$g(2) = 2 + g(1)$. Now 1 is odd so we use the third line$g(1) = 1 + g(0)$. and Finally 0 is the first line,$g(0) = 1$. Putting it all together,$g(0) = 1$,$g(1) = 2$,$g(2) = 4$and$g(4) = 6$. Think you'd be able to try out for$g(7)$and$g(12)$? I think I'd discuss (ii) with you since it's quite weird. ## Question 11 Do you recall an idea about$\\mathbf{r}\\cdot \\hat{\\mathbf{n}}$for equation of planes? Let's take a look at the plane (a separate example) $$\\mathbf{r} \\cdot \\begin{pmatrix} -2 \\\\ 1 \\\\ 2 \\end{pmatrix}= 5$$ The number 5 on the RHS has not much meaning except that points on the plane have to satisfy the equation to get 5. But let's transform this equation to the$\\mathbf{r}\\cdot \\hat{\\mathbf{n}}$form. Since$| \\mathbf{n} | = 3$, we divide our equation of the plane by 3 on both sides to get $$\\mathbf{r} \\cdot \\frac{\\begin{pmatrix} -2 \\\\ 1 \\\\ 2 \\end{pmatrix}}{3}= \\frac{5}{3}$$ Now$\\frac{5}{3}$has a meaning: it is the distance between the plane and the origin is$\\frac{5}{3}$. If a plane has equation$\\mathbf{r}\\cdot \\hat{\\mathbf{n}} = k$, then$k$is the perpendicular distance between the plane and the origin. In your earlier working you should have found an equation of$p$similar to$\\mathbf{r} \\cdot \\begin{pmatrix} -2 \\\\ 1 \\\\ 2 \\end{pmatrix}= -1$. Converting to the form we discussed we get$\\mathbf{r}", "is intersected by the set of planes is equal to $\\pi S ^ {2} / 2 \\overline{H}\\;$, where $\\overline{H}\\;$ is the total mean curvature. The integral invariant for a pair of planes is equal to the product of the integral invariants of the sets of planes. The kinematic measure in $E ^ {3}$ is equal to the product of the measure of the set of distinct oriented planes and the elementary kinematic measure in the orienting plane. The integral invariant for the rotation of a spatial figure having one fixed point is equal to $$1 + l _ {1} ^ {2} + l _ {2} ^ {2} + l _ {3} ^ {2} ,$$ where $l _ {i} = \\alpha _ {i} \\mathop{\\rm tan} ( \\phi / 2 )$, $i = 1 , 2 , 3$, $\\alpha _ {i}$ are the direction cosines of the axis of rotation and $\\phi$ is the angle of rotation around this axis. The measure of a set of bodies having a common point and differing by a rotation in space is equal to $\\pi ^ {2}$. Integral geometry on a surface is constructed by the introduction of a measure on the set of geodesics as the integral of an exterior differential form on the surface over the whole set. Thus, the exterior", "got the parametric equations of my plane to be: $x = \\frac{-5}{2} + \\frac{5}{2}\\lambda - \\mu$ $y = \\lambda$ $z = \\mu$ where $\\mu, \\lambda\\varepsilon \\mathbb{R}$ but the correct answer was: $\\begin{pmatrix}x\\\\y \\\\z\\end{pmatrix} = \\begin{pmatrix}1\\\\3 \\\\4\\end{pmatrix} + \\lambda \\begin{pmatrix}5\\\\2 \\\\0\\end{pmatrix} + \\mu\\begin{pmatrix}1\\\\0 \\\\-1\\end{pmatrix}$ Im not sure what I did wrong The parametric equation of the plane and the answer to the problem are two different things. The image of your plane is the product of the given matrix and the vector with the parametric representation of the plane. The parametric equation of the plane and the answer to the problem are two different things. The image of your plane is the product of the given matrix and the vector with the parametric representation of the plane. I see, but where did I go wrong with the vector with the parametric representation of the plane, cause I cant see what I did wrong. Mark44 Mentor I don't see anything wrong with that. I got about the same as what you got. It's possible that you and the book just have different representations of the same thing. You can check yours by substituting a few values of lambda and mu to see if you actually get points that are on the plane. I tried checking the x values so I", "Review question # Can we find the equation of the plane containing $l_1$ and $PQ$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R5089 ## Question The lines $l_1$ and $l_2$ have equations $\\mathbf{r} = \\mathbf{a} + s\\mathbf{b}$ and $\\mathbf{r} = \\mathbf{c} + t\\mathbf{d}$, where $s$ and $t$ are real numbers and $\\mathbf{a} = 5\\mathbf{i}+ 6\\mathbf{j}- \\mathbf{k}, \\quad\\mathbf{b} = \\mathbf{i} +\\mathbf{j}+ \\mathbf{k}, \\quad\\mathbf{c} = \\mathbf{i}+ 13\\mathbf{j} +7\\mathbf{k}, \\quad\\mathbf{d} = \\mathbf{i} + 6\\mathbf{j} + 3\\mathbf{k}.$ Also, the point $P$ on $l_1$ and the point $Q$ on $l_2$ are such that $PQ$ is perpendicular to both $l_1$ and $l_2.$ In any order: 1. show that $PQ = \\sqrt{38},$ 2. find the position vectors of $P$ and $Q,$ 3. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$.", "of the form Ax+ By+ Cz= E and E can be determined by a single point in the plane. You ca get parametric equations for Ax+ By+ Cz= D by, for example, solving for z: $z= \\frac{D}{C}-\\frac{A}{C}x- \\frac{B}{C}z$ and then using x and y as parameters. Or if you prefer using other letters, say u and v, $x= u$ $y= v$ $z= \\frac{D}{C}-\\frac{A}{C}u- \\frac{B}{C}v$", "plane was a two-dimensional one. So, it was $\\mathbb{Z_2^2}.$ $\\mathbb{Z_2^3}$ would be a small $3d-$plane with only $2^3 = 8$ points, all coordinates taken modulo $2.$ So, what we've seen is that the xor-operation is equivalent to vector addition in a $\\mathbb{Z}_2^d$ vector space. See how unnecessarily intimidating this simple idea sounds when written in formal math! Anyways, the problem: ### Problem 1 (Division 2 — C) Find the number of non-empty subsets, modulo $10^9 + 7,$ of a given set of size $1 \\le n \\le 10^5$ with range of elements $1 \\le a_i \\le 70,$ such that the product of it's elements is a square number. If you'd like to solve the problem first, then kindly pause and try it before reading on further. Solution #### Solution It's obvious that our solution will build on the constraint on $a_i,$ which is just $70.$ For a number to be square, each of it's prime divisors must have an even exponent in the prime factorization of the number. There are only $19$ primes upto $70.$ So, we can assign a mask of $19$ bits to each array element, denoting if the $i$'th prime occurs odd or even number of times in it by the $i$'th bit of the mask. So, the problem just boils down to finding out" ]

[ "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "to an arbitrary vector $(x,y,z)$, and see if the resulting vector is orthogonal to the normal vector of that given plane, thus implying that the vector is projected successfully to the plane? This is necessary but not sufficient. If $n$ is normal to the plane and $\\langle n, Tx\\rangle = 0$ for all $x$, then the image of $T$ is contained in the plane, but that doesn't necessarily mean that $T$ is a projection onto the plane. For example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$ This matrix has the property that $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. So the image lies in the plane whose normal vector is $\\begin{pmatrix}1 \\\\ 0 \\\\ 0\\end{pmatrix}$. But it is not a projection onto that plane because the image only has dimension 1. For another example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 2 \\end{pmatrix}$$ Once again, we have $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. The dimension of the image is correct (2), but this", "Since the distance can only be calculated if there is parallelism, you have to check whether the planes are parallel. We look to see whether the normal vectors are parallel. In contrast to the line, there is no check for perpendicularity. $\\vec{n_1}=r\\cdot\\vec{n_2}$ $\\begin{pmatrix} -4 \\\\ 4 \\\\ -8 \\end{pmatrix}=r\\cdot\\begin{pmatrix}2 \\\\ -2 \\\\ 4 \\end{pmatrix}$ $\\Rightarrow r=-2$ There is an $r$: The vectors are multiples of one another and therefore parallel. 2. #### Select point You can take any point on the plane. However, since you can easily read off the support point, this is a good option. $P(1|2|1)$ 3. #### Set up Hesse normal form $|\\vec{n}|=\\sqrt{2^2+(-2)^2+4^2}$ $=\\sqrt{24}$ $\\vec{n_0}= \\frac{\\vec{n}}{|\\vec{n}|}$ $=\\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix}$ $\\text{E: } \\left(\\vec{x} - \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}\\right) \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix}=0$ 4. #### Insert point $\\vec{p}=\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix}$ $d=$ $\\left|\\left(\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} - \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}\\right) \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix} \\right|$ $=\\left|\\begin{pmatrix} -1 \\\\ 1 \\\\ 0 \\end{pmatrix} \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix} \\right|$ $=|-\\frac4{\\sqrt{24}}|$ $\\approx0.82$", "# 9.2: Building Subspaces Consider the set $U= \\left\\{ \\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}, \\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix} \\right\\} \\subset \\Re^{3}.$ Because $$U$$ consists of only two vectors, it clear that $$U$$ is $$\\textit{not}$$ a vector space, since any constant multiple of these vectors should also be in $$U$$. For example, the $$0$$-vector is not in $$U$$, nor is $$U$$ closed under vector addition. But we know that any two vectors define a plane: In this case, the vectors in $$U$$ define the $$xy$$-plane in $$\\Re^{3}$$. We can view the $$xy$$-plane as the set of all vectors that arise as a linear combination of the two vectors in $$U$$. We call this set of all linear combinations the $$\\textit{span}$$ of $$U$$: $span(U)=\\left\\{ x \\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}+y \\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix} \\middle| x,y\\in \\Re \\right\\}.$ Notice that any vector in the $$xy$$-plane is of the form $\\begin{pmatrix}x\\\\y\\\\0\\end{pmatrix} = x \\begin{pmatrix}1\\\\0\\\\0\\end{pmatrix}+y \\begin{pmatrix}0\\\\1\\\\0\\end{pmatrix} \\in span(U).$ Definition Let $$V$$ be a vector space and $$S=\\{ s_{1}, s_{2}, \\ldots \\} \\subset V$$ a subset of $$V$$. Then the $$\\textit{span of S}$$ is the set: $span(S):=\\{ r^{1}s_{1}+r^{2}s_{2}+\\cdots + r^{N}s_{N} | r^{i}\\in \\Re, N\\in \\mathbb{N} \\}.$ That is, the span of $$S$$ is the set of all finite linear combinations (usually our vector spaces are defined over $$\\mathbb{R}$$, but in general we can have vector spaces defined over different base fields such as $$\\mathbb{C}$$ or $$\\mathbb{Z}_{2}$$. The", "## Vector Equation of a Plane The vector equation of a plane can be written as $r.n=a.n$ where $$a$$ is a position vector of a point on the plane and $$n$$ is a normal to the plane. LESSON 1 Determine the equation of the plane, in vector form and Cartesian form, which contains the point $$(2,-3,1)$$ with normal $$i-2j+3k$$. SOLUTION The equation of a plane can be written in the form $r.n=a.n$ where $$n$$ is a vector perpendicular to the plane and $$a$$ is a position vector of a point on the plane. State the position vector $\\begin{pmatrix} 2\\\\-3 \\\\1 \\end{pmatrix}$ State the normal to the plane $n=\\begin{pmatrix} 1\\\\-2 \\\\3 \\end{pmatrix}$ Substitute the vectors $$a$$ and $$n$$ into the general form of the vector equation of the plane, $$r.n=a.n$$ $r.\\begin{pmatrix} 1\\\\-2 \\\\3 \\end{pmatrix}=\\begin{pmatrix} 2\\\\-3 \\\\1 \\end{pmatrix}.\\begin{pmatrix} 1\\\\-2 \\\\3 \\end{pmatrix}$ Determine the dot product $$a.n$$ $r.\\begin{pmatrix} 1\\\\-2\\\\3 \\end{pmatrix}=2(1)+(-3)(-2)+1(3)$ $r.\\begin{pmatrix} 1\\\\-2\\\\3 \\end{pmatrix}=11$ CARTESIAN FORM Replace $$r$$ with $$\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix}$$ in the vector equation of the plane $\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix}.\\begin{pmatrix} 1\\\\-2\\\\3 \\end{pmatrix}=11$ Determine the dot product, $$r.n$$ $x-2y+3z=11$ TRY THIS Complete Questions 13 and 14 from worksheet.", "to the other line The distance can now be calculated as described under distance point and line. First set up an auxiliary plane with $P$ as the support point and direction vector as the normal vector. $\\text{H: } (\\vec{x} - \\vec{a}) \\cdot \\vec{n}=0$ $\\text{H: } (\\vec{x} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}) \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$. $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can", "the plane with equation $\\displaystyle{ \\ n_x au + n_y bv + n_z cw = d. }$ Let muu + mvv + mww = δ be the Hesse normal form of the new plane and $\\displaystyle{ \\;\\mathbf m = \\begin{bmatrix} m_u \\\\ m_v \\\\ m_w \\end{bmatrix}\\; }$ its unit normal vector. Hence $\\displaystyle{ \\mathbf e_0 = \\delta \\mathbf m \\; }$ is the center of the intersection circle and $\\displaystyle{ \\;\\rho = \\sqrt{1 - \\delta^2}\\; }$ Where mw = ±1 (i.e. the plane is horizontal), let $\\displaystyle{ \\ \\mathbf e_1 = \\begin{bmatrix} \\rho \\\\ 0 \\\\ 0 \\end{bmatrix},\\qquad \\mathbf e_2 = \\begin{bmatrix} 0 \\\\ \\rho \\\\ 0 \\end{bmatrix}. }$ Where mw ≠ ±1, let $\\displaystyle{ \\mathbf e_1 = \\frac{\\rho}{\\sqrt{m_u^2 + m_v^2}}\\, \\begin{bmatrix} m_v \\\\ -m_u \\\\ 0 \\end{bmatrix}\\, ,\\qquad \\mathbf e_2 = \\mathbf m \\times \\mathbf e_1\\ . }$ In any case, the vectors e1, e2 are orthogonal, parallel to the intersection plane and have length ρ (radius of the circle). Hence the intersection circle can be described by the parametric equation $\\displaystyle{ \\;\\mathbf u = \\mathbf e_0 + \\mathbf e_1\\cos t + \\mathbf e_2\\sin t\\;. }$ The reverse scaling (see above) transforms the unit sphere back to the ellipsoid and the vectors e0, e1, e2 are mapped onto vectors f0, f1, f2, which were wanted for the", "shown below) $$r·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$(3×1)+(2×[-5])+(7×8)=49$$ Therefore this means that the distanve of the vector d = 49 To complete the equation we need to write down the value of the whole equation in cartesian form $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}1\\\\\\ -5\\\\\\ 8\\end{pmatrix}$$ = $$x-5y+8z=49$$ x - 5y + 8z - 49 = 0 #### Question 2 Write the equation of the plane through (2,1,3), given that the vector (4,5,6) is perpendicular $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 1\\\\\\ 3\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ $$r·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$(2×4)+(1×5)+(3×6)=31$$ Therefore the distance of the vector is d = 31 $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}·\\begin{pmatrix}4\\\\\\ 5\\\\\\ 6\\end{pmatrix}$$ = $$4x+5y+6z=31$$ 4x + 5y + 6z - 31 = 0 #### Intersection of a line and a plane Find the point of intersection between the line The normal (perpendicular) to a plane is $$r=\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ with plane $$5x+y-z=1$$ First we need to replace r with the matrix (x,y,z) in order to find the values of x,y and z $$\\begin{pmatrix}x\\\\\\ y\\\\\\ z\\end{pmatrix}$$ = $$\\begin{pmatrix}2\\\\\\ 3\\\\\\ 4\\end{pmatrix}+λ\\begin{pmatrix}1\\\\\\ 2\\\\\\ -1\\end{pmatrix}$$ We can change this equation into 3 different equations by seperating the matrices into: x = 2 + λ y = 3 + 2λ z = 4 - λ And we know that 5x+y-z=1 Replacing the values of x,y and z with these new values we get", "that the plane $\\mathcal{P}$ can be described as the set of all vectors $\\vec{v}$ solving the following equation: $b(\\hat{n},\\vec{v}) + d = 0,$ where $d=-b(\\hat{n},\\vec{p})$ can be interpreted as the minimum distance between the plane and the origin of the euclidean space. The following figure is once again taken from Wikipedia: ### Computing the normal vector of a plane If a plane is given by its parametric form, $\\mathcal{P} = \\vec{p} + s\\vec{u} + t\\vec{v}$, then the normal vector $\\hat{n}$ is the normalized outer product of the vectors $\\vec{u}$ and $\\vec{v}$, written as $\\vec{u} \\times \\vec{v}$, or $\\vec{u} \\wedge \\vec{v}$. As above, let $\\vec{e}_1, \\vec{e}_2$ and $\\vec{e}_3$ denote the standard basis of $\\mathbb{R}^3$, then the outer product can be computed as follows: \\begin{align*} \\tilde{n} &= \\vec{u} \\times \\vec{v} \\\\ &= \\left(\\sum\\limits_{i=1}^3(\\vec{u} \\times \\vec{e}_i) \\otimes \\vec{e}_i \\right) \\cdot \\vec{v} \\\\ &= \\begin{pmatrix}0 & -u_3 & u_2 \\\\ u_3 & 0 & u_1 \\\\ u_2 & -u_1 & 0\\end{pmatrix} \\cdot \\begin{pmatrix}v_1 \\\\ v_2 \\\\ v_3\\end{pmatrix} \\\\ &= \\begin{pmatrix} u_2v_3 - u_3v_2 \\\\ -u_1v_3 + u_3v_1 \\\\ u_1v_2 - u_2v_1 \\end{pmatrix}.\\end{align*}Or, if you don't like tensors, you can \"cheat\" a little bit: \\begin{align*} \\tilde{n} &= \\vec{u} \\times \\vec{v} \\\\ &=\\operatorname{det}\\begin{pmatrix}\\vec{e}_1&u_1&v_1\\\\ \\vec{e}_2&u_2&v_2\\\\ \\vec{e}_3&u_3&v_3\\end{pmatrix} \\\\ & = \\vec{e}_1 \\cdot \\operatorname{det}\\begin{pmatrix}u_2&v_2\\\\u_3&v_3\\end{pmatrix} - \\vec{e}_2 \\cdot \\operatorname{det}\\begin{pmatrix}u_1&v_1\\\\u_3&v_3\\end{pmatrix} + \\vec{e}_3 \\cdot \\operatorname{det}\\begin{pmatrix}u_1&v_1\\\\u_2&v_2\\end{pmatrix} \\\\ &= \\vec{e}_1 \\cdot (u_2v_3-u_3v_2)", "$\\overrightarrow{OM}$ is in the plane, where $\\overrightarrow{OM}$ = $\\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}$. Any point in the plane can be expressed as a linear combination of these vectors. So an equation of the plane is $\\mathbf{r} = \\alpha \\begin{pmatrix} 5 \\\\ 0 \\\\ 5 \\end{pmatrix}+ \\beta \\begin{pmatrix} 1 \\\\ -2 \\\\ 5 \\end{pmatrix}.$ Alternatively, we could write the equation as $\\mathbf{r}.\\mathbf{n}=0$ where $\\mathbf{n}$ is the normal to the plane. Substituting in the two points $L$ and $M$ tells us that $\\mathbf{n}$ is a multiple of $\\begin{pmatrix}1\\\\ -2\\\\ -1\\end{pmatrix}$, so the plane equation can be written as $\\mathbf{r}.\\begin{pmatrix}1\\\\ -2\\\\ -1\\end{pmatrix}=0.$ 1. Find the position vector of the point $P$ where $l$ meets the plane $\\pi$ whose equation is $\\mathbf{r}.\\begin{pmatrix}1\\\\ 2\\\\ 2\\end{pmatrix} = 11.$ Let $\\overrightarrow{OP} = \\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix} + \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}.$ If $P$ is on the plane, then $\\overrightarrow{OP}.\\begin{pmatrix}1\\\\ 2\\\\ 2\\end{pmatrix} = 11,$ and so \\begin{align*} \\left[ \\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}+ \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix} \\right].\\begin{pmatrix}1\\\\ 2\\\\ 2\\end{pmatrix} = 11. \\end{align*} This implies \\begin{align*} 15 + 4\\lambda = 11, \\end{align*} from which we conclude $\\lambda=-1$. Thus $\\overrightarrow{OP}=\\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix} - \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}=\\begin{pmatrix}3\\\\ -1\\\\ 5\\end{pmatrix}.$", "# Finding the equation of a plane. How do I find the equation of a plane given by the points (0,1,1), (1,0,1) and (1,1,0)? Graphing it, it's a triangle when you connect the points. Can I use this somehow? - What do you know about equation of a plane? There are several formats, which one do you know of? 2-vector 1 point, normal vector and parametric. Sound familiar? – Calvin Lin Jan 26 '13 at 23:29 By inspection $x+y+z=2$ is an equation of the plane. – André Nicolas Jan 26 '13 at 23:30 a(x-x0)+b(y-y0)+c(z-z0)=0. Andre got the answer right but not sure on how to get it. – user1766888 Jan 26 '13 at 23:31 Sounds like you want the normal vector form. To find a normal vector to the plane, we take the cross product of 2 (linearly independent) vectors of the plane. In this case, we take the vectors $\\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 0 \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} -1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ to obtain $$\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix} \\times \\begin{pmatrix} -1 \\\\ 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix} -1 \\\\ -1", "$\\mathbf{r} = \\begin{pmatrix}1 \\\\ 0 \\\\ 3\\end{pmatrix}+s\\begin{pmatrix}-2 \\\\ -4 \\\\ 4\\end{pmatrix}+t\\begin{pmatrix}2 \\\\ 3 \\\\ -6\\end{pmatrix}.$ We can simplify this to $\\mathbf{r} = \\begin{pmatrix}1 \\\\ 0 \\\\ 3\\end{pmatrix}+s\\begin{pmatrix}1 \\\\ 2 \\\\ -2\\end{pmatrix}+t\\begin{pmatrix}2 \\\\ 3 \\\\ -6\\end{pmatrix},$ or $\\mathbf{r} = \\begin{pmatrix}1 +s+2t\\\\ 2s+3t \\\\ 3-2s-6t\\end{pmatrix}.$ Is the vector form of the equation of the plane unique?", "\\\\ -1 \\end{pmatrix}$$ Then, the plane satisfies $$\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\begin{pmatrix} -1 \\\\ -1 \\\\ -1 \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix} \\begin{pmatrix} -1 \\\\ -1 \\\\ -1 \\end{pmatrix} = -2,$$ or that $- x - y - z = - 2$. - Take one of the points as origin and use the other two to obtain two (linearly independent? check!) vectors. Then write the vectorial equation for the plane as $$(x,y,z)=P+\\lambda \\vec u + \\mu \\vec v.$$ -", "Math Relative position of planes Angle of intersection between two planes # Angle of intersection between two planes The angle of intersection between two planes is calculated like that of two lines (or like angles between two vectors). There are two planes (ideally in normal- or cartesian form for reading the normal vector): $E_1: (\\vec{x} - \\vec{a}) \\cdot \\vec{n_1}=0$ $E_2: (\\vec{x} - \\vec{b}) \\cdot \\vec{n_2}=0$ The two normal vectors of the planes are now used to calculate the angle: $\\cos(\\gamma) = \\frac{|\\vec{n_1}\\cdot\\vec{n_2}|}{|\\vec{n_1}|\\cdot|\\vec{n_2}|}$ $\\gamma = \\cos^{-1}\\left(\\frac{|\\vec{n_1}\\cdot\\vec{n_2}|}{|\\vec{n_1}|\\cdot|\\vec{n_2}|}\\right)$ i ### Method 2. Calculate the scalar product of the normal vectors 3. Calculate normal vector absolute value 4. Insert results in the formula ### Example $\\text{E: } 9x + 8y + 7z = 0$ $\\text{F: } x+y+2z=0$ 1. #### Read off normal vectors With the cartesian form, the normal vector can always be read off by the factors before x, y and z. $\\vec{n_1}=\\begin{pmatrix}9 \\\\ 8 \\\\ 7 \\end{pmatrix}$ $\\vec{n_2}=\\begin{pmatrix}1 \\\\ 1 \\\\ 2\\end{pmatrix}$ 2. #### Calculate scalar product $\\vec{n_1}\\cdot\\vec{n_2}$ $=\\begin{pmatrix}9 \\\\ 8 \\\\ 7 \\end{pmatrix}\\cdot\\begin{pmatrix}1 \\\\ 1 \\\\ 2\\end{pmatrix}$ $=9\\cdot1+8\\cdot1+7\\cdot2$ $=31$ 3. #### Calculate normal vector absolute value $|\\vec{n_1}|=\\sqrt{9^2+8^2+7^2}$ $=\\sqrt{194}$ $|\\vec{n_2}|=\\sqrt{1^2+1^2+2^2}$ $=\\sqrt{6}$ 4. #### Insert results in the formula $\\cos(\\gamma) = \\frac{|\\vec{n_1}\\cdot\\vec{n_2}|}{|\\vec{n_1}|\\cdot|\\vec{n_2}|}$ $\\cos(\\gamma) = \\frac{31}{\\sqrt{194}\\cdot\\sqrt{6}}$ $\\gamma = \\cos^{-1}\\left(\\frac{31}{\\sqrt{194}\\cdot\\sqrt{6}}\\right)$ $\\approx24.68°$", "the vector $\\overrightarrow{QP}$, we subtract the coordinates of $P$ from the coordinates of $Q$: $$\\overrightarrow{QP} = \\begin{pmatrix}0-1 \\ 2-0 \\ 0-0\\end{pmatrix} = \\begin{pmatrix}-1 \\ 2 \\ 0\\end{pmatrix} = -1\\hat{\\boldsymbol{\\imath}} + 2\\hat{\\boldsymbol{\\jmath}}$$ To calculate the vector $\\overrightarrow{QR}$, we subtract the coordinates of $R$ from the coordinates of $Q$: $$\\overrightarrow{QR} = \\begin{pmatrix}0-0 \\ 2-0 \\ 3-0\\end{pmatrix} = \\begin{pmatrix}0 \\ 2 \\ 3\\end{pmatrix} = 2\\hat{\\boldsymbol{\\jmath}} + 3\\hat{\\boldsymbol{k}}$$ So, as you said, $\\overrightarrow{QP} = \\hat{\\boldsymbol{\\imath}} – 2 \\hat{\\boldsymbol{\\jmath}}$ and $\\overrightarrow{QR} = -2\\hat{\\boldsymbol{\\jmath}} + 3\\hat{\\boldsymbol{k}}$. $\\vec{N} \\cdot \\vec{r}(t)=6$, where $\\vec{N}=\\langle 4,-3,-2\\rangle$. The equation $\\vec{N} \\cdot \\vec{r}(t)=6$ represents a plane in three-dimensional space, where $\\vec{N}$ is the normal vector to the plane and $\\vec{r}(t)$ is a vector pointing to any point on the plane at time $t$. Given that $\\vec{N}=\\langle 4,-3,-2\\rangle$, we can write the equation of the plane as: $$4x – 3y – 2z = 6$$ where $x$, $y$, and $z$ are the coordinates of any point on the plane. Alternatively, we can express the equation in vector form as: $$\\langle 4,-3,-2\\rangle \\cdot \\langle x,y,z\\rangle = 6$$ or $$\\langle 4,-3,-2\\rangle \\cdot \\vec{r}(t) = 6$$ where $\\vec{r}(t)=\\langle x(t),y(t),z(t)\\rangle$ is a vector function that describes the position of any point on the plane at time $t$. Note that there are infinitely many points on the plane described by this equation, since the equation has three", "Finding orthonormal basis for a subspace $W$ of the Euclidean space $\\mathbb{R}^3$. Problem: Let $\\mathbb{R}^3$ be an Euclidean space. Find an orthonormal basis for the subspace $W$ defined as $x + 2y-z = 0$. Attempt at solution: So this is a plane in $\\mathbb{R}^3$, so I guess I would need two vectors to span this? I did the following. Pick two arbitrary vectors that lie in this plane, for example $w_1 = (1,1,3)$ and $w_2 = (1,0,1)$. Then should I apply the Gram-Schmidt process to these? That would get me: \\begin{align*} v_1 &= w_1 \\\\ v_2 &= w_2 - \\frac{\\langle w_2, v_1 \\rangle}{\\langle v_1, v_1} v_1 \\end{align*} Then we would have $v_1 = (1,1,3)$ and $v_2 = (7/11, -4/11, -1/11)$. Now I should normalize these and I'm done? • Yes, but the results will be simpler if you set $w_1=(1,0,1)$. – Bernard Jul 27 '15 at 18:18 A simple way of getting independent vectors in this plane is to note that if $x+ 2y- z= 0$, then $z= x- 2y$ so any vector in this plane is of the form $\\begin{pmatrix} x \\\\ y \\\\ z\\end{pmatrix}= \\begin{pmatrix}x \\\\ y \\\\ x- 2y\\end{pmatrix}$. Taking $x= 1, y= 0$, that is $\\begin{pmatrix}1 \\\\ 0 \\\\ 1\\end{pmatrix}$. Taking $x= 0, y= 1$, this is $\\begin{pmatrix} 0 \\\\ 1 \\\\ -2\\end{pmatrix}$.", "in 3d is using the cross product. So start with the normal vector $(1,3,-1)$ of the plane, and an arbitrarily chosen second vector $(0,0,1)$ Compute the cross products $$v_1=\\begin{pmatrix}1\\\\3\\\\-1\\end{pmatrix}\\times\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix} =\\begin{pmatrix}3\\\\-1\\\\0\\end{pmatrix} \\\\ v_2=\\begin{pmatrix}1\\\\3\\\\-1\\end{pmatrix}\\times\\begin{pmatrix}3\\\\-1\\\\0\\end{pmatrix} =\\begin{pmatrix}-1\\\\-3\\\\-10\\end{pmatrix}$$ To preserve distances, you'll have to scale them to unit length. Then you can plug them into the columns of a transformation matrix, and choose an offset such that the image of the origin lies within your plane. For example like this: $$\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix}\\mapsto\\begin{pmatrix} \\frac{3}{\\sqrt{10}} & -\\frac{1}{\\sqrt{110}} & \\frac{1}{\\sqrt{11}} \\\\ -\\frac{1}{\\sqrt{10}} & -\\frac{3}{\\sqrt{110}} & \\frac{3}{\\sqrt{11}} \\\\ 0 & -\\frac{10}{\\sqrt{110}} & -\\frac{1}{\\sqrt{11}} \\end{pmatrix}\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} +\\begin{pmatrix}0\\\\0\\\\1\\end{pmatrix}$$ If you really want to rotate around the axis where both your planes intersect, so that points on that axis remain fixed, you'll need a bit more work. First find the direction of the common line. It is perpendicular to both the normals of the two planes. Since I used $(0,0,1)$ as the arbitrary vector in the first step above, and that's the normal of the $z=0$ plane, you already have $v_1$ as the direction of that axis. You can get the transformation matrix like this: $$\\begin{pmatrix} \\frac{3}{\\sqrt{10}} & -\\frac{1}{\\sqrt{110}} & \\frac{1}{\\sqrt{11}} \\\\ -\\frac{1}{\\sqrt{10}} & -\\frac{3}{\\sqrt{110}} & \\frac{3}{\\sqrt{11}} \\\\ 0 & -\\frac{10}{\\sqrt{110}} & -\\frac{1}{\\sqrt{11}} \\end{pmatrix}\\begin{pmatrix} \\frac{3}{\\sqrt{10}} & \\frac{1}{\\sqrt{10}} & 0 \\\\ -\\frac{1}{\\sqrt{10}} & \\frac{3}{\\sqrt{10}} & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}^{-1}$$", "+0 # Help with vector geometry +1 88 1 +38 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w} \\text{ and }\\mathbf{x}$$ are linearly independent, answer with 0. If they aren't, find coefficients a,b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a+b}{c}$$ Jun 5, 2019", "# Find an equation of the tangent plane to the given parametric surface at the specified point. x =... ## Question: Find an equation of the tangent plane to the given parametric surface at the specified point. ## Tangent plane to a surface This example illustrates how to find the tangent plane to a surface at a given point. In this particular case, the equation of the surface is given in terms of two parameters, and we must use partial differentiation and cross products. Suppose we have a surface with parametric equations {eq}\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} = \\begin{pmatrix} x(u, v)\\\\ y(u, v)\\\\ z(u, v) \\end{pmatrix}. {/eq} Then the vector which is normal to the surface at any particular point is given by {eq}\\begin{pmatrix} \\dfrac{\\partial x}{\\partial u}\\\\ \\dfrac{\\partial y}{\\partial u}\\\\ \\dfrac{\\partial z}{\\partial u} \\end{pmatrix} \\times \\begin{pmatrix} \\dfrac{\\partial x}{\\partial v}\\\\ \\dfrac{\\partial y}{\\partial v}\\\\ \\dfrac{\\partial z}{\\partial v} \\end{pmatrix}. {/eq} In this example, {eq}\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} = \\begin{pmatrix} u^2 + 1\\\\ v^3 + 1\\\\ u + v \\end{pmatrix} {/eq} and so the normal vector to the surface is {eq}\\begin{pmatrix} \\dfrac{\\partial x}{\\partial u}\\\\ \\dfrac{\\partial y}{\\partial u}\\\\ \\dfrac{\\partial z}{\\partial u} \\end{pmatrix} \\times \\begin{pmatrix} \\dfrac{\\partial x}{\\partial v}\\\\ \\dfrac{\\partial y}{\\partial v}\\\\ \\dfrac{\\partial z}{\\partial v} \\end{pmatrix} = \\begin{pmatrix} 2u\\\\ 0\\\\ 1 \\end{pmatrix} \\times \\begin{pmatrix} 0\\\\ 3v^2\\\\ 1 \\end{pmatrix} (*) {/eq} We need to find" ]

[ "= -2$. Thus $\\mathbf{p} = \\begin{pmatrix}2 \\\\ 3 \\\\ -4\\end{pmatrix}$ and $\\mathbf{q} = \\begin{pmatrix}-1 \\\\ 1\\\\ 1\\end{pmatrix}$. We also have $\\mathbf{q}-\\mathbf{p} = \\begin{pmatrix}-3 \\\\ -2\\\\ 5\\end{pmatrix}$, and so $\\big\\vert\\mathbf{q}-\\mathbf{p} \\big\\vert = \\sqrt{(-3)^2+(-2)^2+5^2} = \\sqrt{38}$. 1. obtain the Cartesian equation of the plane containing $l_1$ and $PQ$. Using vector product we can find a vector perpendicular to two others, thus $(\\mathbf{q}- \\mathbf{p}) \\times \\mathbf{b}=\\mathbf{n}$ where $\\mathbf{n}$ is a vector perpendicular to both $\\mathbf{b}$ (the direction vector of $l_1$) and $PQ$. $\\begin{pmatrix}-3\\\\-2\\\\5\\end{pmatrix} \\times \\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=\\begin{pmatrix}-2-5\\\\ -(-3-5)\\\\-3--2\\end{pmatrix}=\\begin{pmatrix}-7 \\\\ 8\\\\-1\\end{pmatrix}$ A normal vector immediately yields the vector equation of the plane $\\mathbf{r}.\\mathbf{n}=d$ where $d$ can be calculated from any known point on the plane. So $\\mathbf{q}.\\mathbf{n}= 7+8-1=d=14$. Hence the vector equation of the plane is $\\begin{pmatrix}x\\\\y\\\\z\\end{pmatrix} . \\begin{pmatrix} -7 \\\\ 8\\\\-1\\end{pmatrix}=14$ which gives rise to the Cartesian equation $-7x+8y-z=14$ or $7x-8y+z+14=0$ Check; is $\\mathbf{a}$ on this plane? Substituting in the coordinates we have $7(5)-8(6)+(-1)+14=0$, so yes. As an alternative method, an equation of the plane containing both $PQ$ and $l_1$ is $\\mathbf{r} = \\begin{pmatrix}x \\\\ y\\\\ z\\end{pmatrix}= \\begin{pmatrix}2\\\\ 3 \\\\ -4\\end{pmatrix}+ \\lambda\\begin{pmatrix}1 \\\\ 1\\\\ 1\\end{pmatrix} + \\mu\\begin{pmatrix}3 \\\\ 2\\\\ -5\\end{pmatrix}.$ Thus $x = 2+\\lambda + 3\\mu$, $y = 3+\\lambda + 2\\mu$, $z = -4+\\lambda -5\\mu$. Subtracting the first two, $x-y=-1+\\mu$, and so $\\mu = x-y+1$. Subtracting the second two, $y-z= 7+7\\mu$,", "that the plane $\\mathcal{P}$ can be described as the set of all vectors $\\vec{v}$ solving the following equation: $b(\\hat{n},\\vec{v}) + d = 0,$ where $d=-b(\\hat{n},\\vec{p})$ can be interpreted as the minimum distance between the plane and the origin of the euclidean space. The following figure is once again taken from Wikipedia: ### Computing the normal vector of a plane If a plane is given by its parametric form, $\\mathcal{P} = \\vec{p} + s\\vec{u} + t\\vec{v}$, then the normal vector $\\hat{n}$ is the normalized outer product of the vectors $\\vec{u}$ and $\\vec{v}$, written as $\\vec{u} \\times \\vec{v}$, or $\\vec{u} \\wedge \\vec{v}$. As above, let $\\vec{e}_1, \\vec{e}_2$ and $\\vec{e}_3$ denote the standard basis of $\\mathbb{R}^3$, then the outer product can be computed as follows: \\begin{align*} \\tilde{n} &= \\vec{u} \\times \\vec{v} \\\\ &= \\left(\\sum\\limits_{i=1}^3(\\vec{u} \\times \\vec{e}_i) \\otimes \\vec{e}_i \\right) \\cdot \\vec{v} \\\\ &= \\begin{pmatrix}0 & -u_3 & u_2 \\\\ u_3 & 0 & u_1 \\\\ u_2 & -u_1 & 0\\end{pmatrix} \\cdot \\begin{pmatrix}v_1 \\\\ v_2 \\\\ v_3\\end{pmatrix} \\\\ &= \\begin{pmatrix} u_2v_3 - u_3v_2 \\\\ -u_1v_3 + u_3v_1 \\\\ u_1v_2 - u_2v_1 \\end{pmatrix}.\\end{align*}Or, if you don't like tensors, you can \"cheat\" a little bit: \\begin{align*} \\tilde{n} &= \\vec{u} \\times \\vec{v} \\\\ &=\\operatorname{det}\\begin{pmatrix}\\vec{e}_1&u_1&v_1\\\\ \\vec{e}_2&u_2&v_2\\\\ \\vec{e}_3&u_3&v_3\\end{pmatrix} \\\\ & = \\vec{e}_1 \\cdot \\operatorname{det}\\begin{pmatrix}u_2&v_2\\\\u_3&v_3\\end{pmatrix} - \\vec{e}_2 \\cdot \\operatorname{det}\\begin{pmatrix}u_1&v_1\\\\u_3&v_3\\end{pmatrix} + \\vec{e}_3 \\cdot \\operatorname{det}\\begin{pmatrix}u_1&v_1\\\\u_2&v_2\\end{pmatrix} \\\\ &= \\vec{e}_1 \\cdot (u_2v_3-u_3v_2)", "comment? So you need a normal vector to your plane. That is, a vector normal to both $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ and $\\begin{pmatrix}-1 \\\\ a \\\\ 1\\end{pmatrix}$. As earboth already showed, this last vector is actually $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$. A normal vector must have a dot product of zero with the original vector. So a vector normal to $\\begin{pmatrix}0 \\\\ 1 \\\\ 2\\end{pmatrix}$ must be of the form $\\mathbf n = \\begin{pmatrix}* \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) To also be normal to $\\begin{pmatrix}-1 \\\\ 1 \\\\ 1\\end{pmatrix}$, it must be $\\mathbf n = \\begin{pmatrix}-1 \\\\ -2 \\\\ 1\\end{pmatrix}$. (Why?) Next we want to find the distance. Since the line is parallel to the plane, every point on it has the same distance, so we can pick any point on the line and calculate its distance to the plane. The distance of a point to a plane, is given by the projection of any vector from the point to the plane, onto the normal unit vector. Such a vector is $\\begin{pmatrix}2 \\\\ 2 \\\\ 2\\end{pmatrix} - \\begin{pmatrix}2 \\\\ 1 \\\\ 1\\end{pmatrix} = \\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix}$. Take the projection on the normal unit vector that we will call $\\mathbf{\\hat n}$: $d(l,V)=|\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot \\mathbf{\\hat n}| = |\\begin{pmatrix}0 \\\\ 1 \\\\ 1\\end{pmatrix} \\cdot", "the plane with equation $\\displaystyle{ \\ n_x au + n_y bv + n_z cw = d. }$ Let muu + mvv + mww = δ be the Hesse normal form of the new plane and $\\displaystyle{ \\;\\mathbf m = \\begin{bmatrix} m_u \\\\ m_v \\\\ m_w \\end{bmatrix}\\; }$ its unit normal vector. Hence $\\displaystyle{ \\mathbf e_0 = \\delta \\mathbf m \\; }$ is the center of the intersection circle and $\\displaystyle{ \\;\\rho = \\sqrt{1 - \\delta^2}\\; }$ Where mw = ±1 (i.e. the plane is horizontal), let $\\displaystyle{ \\ \\mathbf e_1 = \\begin{bmatrix} \\rho \\\\ 0 \\\\ 0 \\end{bmatrix},\\qquad \\mathbf e_2 = \\begin{bmatrix} 0 \\\\ \\rho \\\\ 0 \\end{bmatrix}. }$ Where mw ≠ ±1, let $\\displaystyle{ \\mathbf e_1 = \\frac{\\rho}{\\sqrt{m_u^2 + m_v^2}}\\, \\begin{bmatrix} m_v \\\\ -m_u \\\\ 0 \\end{bmatrix}\\, ,\\qquad \\mathbf e_2 = \\mathbf m \\times \\mathbf e_1\\ . }$ In any case, the vectors e1, e2 are orthogonal, parallel to the intersection plane and have length ρ (radius of the circle). Hence the intersection circle can be described by the parametric equation $\\displaystyle{ \\;\\mathbf u = \\mathbf e_0 + \\mathbf e_1\\cos t + \\mathbf e_2\\sin t\\;. }$ The reverse scaling (see above) transforms the unit sphere back to the ellipsoid and the vectors e0, e1, e2 are mapped onto vectors f0, f1, f2, which were wanted for the", "$\\overrightarrow{OM}$ is in the plane, where $\\overrightarrow{OM}$ = $\\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}$. Any point in the plane can be expressed as a linear combination of these vectors. So an equation of the plane is $\\mathbf{r} = \\alpha \\begin{pmatrix} 5 \\\\ 0 \\\\ 5 \\end{pmatrix}+ \\beta \\begin{pmatrix} 1 \\\\ -2 \\\\ 5 \\end{pmatrix}.$ Alternatively, we could write the equation as $\\mathbf{r}.\\mathbf{n}=0$ where $\\mathbf{n}$ is the normal to the plane. Substituting in the two points $L$ and $M$ tells us that $\\mathbf{n}$ is a multiple of $\\begin{pmatrix}1\\\\ -2\\\\ -1\\end{pmatrix}$, so the plane equation can be written as $\\mathbf{r}.\\begin{pmatrix}1\\\\ -2\\\\ -1\\end{pmatrix}=0.$ 1. Find the position vector of the point $P$ where $l$ meets the plane $\\pi$ whose equation is $\\mathbf{r}.\\begin{pmatrix}1\\\\ 2\\\\ 2\\end{pmatrix} = 11.$ Let $\\overrightarrow{OP} = \\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix} + \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}.$ If $P$ is on the plane, then $\\overrightarrow{OP}.\\begin{pmatrix}1\\\\ 2\\\\ 2\\end{pmatrix} = 11,$ and so \\begin{align*} \\left[ \\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix}+ \\lambda \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix} \\right].\\begin{pmatrix}1\\\\ 2\\\\ 2\\end{pmatrix} = 11. \\end{align*} This implies \\begin{align*} 15 + 4\\lambda = 11, \\end{align*} from which we conclude $\\lambda=-1$. Thus $\\overrightarrow{OP}=\\begin{pmatrix}5\\\\ 0\\\\ 5\\end{pmatrix} - \\begin{pmatrix}2\\\\ 1\\\\ 0\\end{pmatrix}=\\begin{pmatrix}3\\\\ -1\\\\ 5\\end{pmatrix}.$", "to an arbitrary vector $(x,y,z)$, and see if the resulting vector is orthogonal to the normal vector of that given plane, thus implying that the vector is projected successfully to the plane? This is necessary but not sufficient. If $n$ is normal to the plane and $\\langle n, Tx\\rangle = 0$ for all $x$, then the image of $T$ is contained in the plane, but that doesn't necessarily mean that $T$ is a projection onto the plane. For example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 0 & 0 \\\\ 0 & 0 & 1 \\end{pmatrix}$$ This matrix has the property that $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. So the image lies in the plane whose normal vector is $\\begin{pmatrix}1 \\\\ 0 \\\\ 0\\end{pmatrix}$. But it is not a projection onto that plane because the image only has dimension 1. For another example, consider the matrix $$T = \\begin{pmatrix}0 & 0 & 0 \\\\ 0 & 2 & 0 \\\\ 0 & 0 & 2 \\end{pmatrix}$$ Once again, we have $\\langle n, Tx\\rangle = 0$, where $n = \\begin{pmatrix}1 \\\\ 0 \\\\ 0 \\end{pmatrix}$ and $x$ is any vector. The dimension of the image is correct (2), but this", "Since the distance can only be calculated if there is parallelism, you have to check whether the planes are parallel. We look to see whether the normal vectors are parallel. In contrast to the line, there is no check for perpendicularity. $\\vec{n_1}=r\\cdot\\vec{n_2}$ $\\begin{pmatrix} -4 \\\\ 4 \\\\ -8 \\end{pmatrix}=r\\cdot\\begin{pmatrix}2 \\\\ -2 \\\\ 4 \\end{pmatrix}$ $\\Rightarrow r=-2$ There is an $r$: The vectors are multiples of one another and therefore parallel. 2. #### Select point You can take any point on the plane. However, since you can easily read off the support point, this is a good option. $P(1|2|1)$ 3. #### Set up Hesse normal form $|\\vec{n}|=\\sqrt{2^2+(-2)^2+4^2}$ $=\\sqrt{24}$ $\\vec{n_0}= \\frac{\\vec{n}}{|\\vec{n}|}$ $=\\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix}$ $\\text{E: } \\left(\\vec{x} - \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}\\right) \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix}=0$ 4. #### Insert point $\\vec{p}=\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix}$ $d=$ $\\left|\\left(\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} - \\begin{pmatrix} 2 \\\\ 1 \\\\ 1 \\end{pmatrix}\\right) \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix} \\right|$ $=\\left|\\begin{pmatrix} -1 \\\\ 1 \\\\ 0 \\end{pmatrix} \\cdot \\begin{pmatrix} 2/\\sqrt{24} \\\\ -2/\\sqrt{24} \\\\ 4/\\sqrt{24} \\end{pmatrix} \\right|$ $=|-\\frac4{\\sqrt{24}}|$ $\\approx0.82$", "$\\mathbf{r} = \\begin{pmatrix}1 \\\\ 0 \\\\ 3\\end{pmatrix}+s\\begin{pmatrix}-2 \\\\ -4 \\\\ 4\\end{pmatrix}+t\\begin{pmatrix}2 \\\\ 3 \\\\ -6\\end{pmatrix}.$ We can simplify this to $\\mathbf{r} = \\begin{pmatrix}1 \\\\ 0 \\\\ 3\\end{pmatrix}+s\\begin{pmatrix}1 \\\\ 2 \\\\ -2\\end{pmatrix}+t\\begin{pmatrix}2 \\\\ 3 \\\\ -6\\end{pmatrix},$ or $\\mathbf{r} = \\begin{pmatrix}1 +s+2t\\\\ 2s+3t \\\\ 3-2s-6t\\end{pmatrix}.$ Is the vector form of the equation of the plane unique?", "to the other line The distance can now be calculated as described under distance point and line. First set up an auxiliary plane with $P$ as the support point and direction vector as the normal vector. $\\text{H: } (\\vec{x} - \\vec{a}) \\cdot \\vec{n}=0$ $\\text{H: } (\\vec{x} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}) \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ The perpendicular point is the intersection of the line $g$ and the auxiliary plane $H$. To calculate the intersection, the equation of a line for $\\vec{x}$ is used in the plane. $\\left(\\color{red}{\\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} + r \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}} - \\begin{pmatrix} -1 \\\\ 0 \\\\ 3 \\end{pmatrix}\\right)$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ $\\begin{pmatrix} 2+r \\\\ 2+r \\\\ -2 \\end{pmatrix}$ $\\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}=0$ Calculate scalar product $(2+r)\\cdot1+(2+r)\\cdot1$ $+(-2)\\cdot0=0$ $4+2r=0\\quad|-4$ $2r=-4\\quad|:2$ $r=-2$ Insert $r$ in $g$ to get perpendicular point $F$. $\\vec{OF} = \\begin{pmatrix} 1 \\\\ 2 \\\\ 1 \\end{pmatrix} \\color{red}{-2} \\cdot \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ $= \\begin{pmatrix} 1-2 \\\\ 2-2 \\\\ 1-0 \\end{pmatrix}$ $= \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}$ $F(-1|0|1)$ The distance from the perpendicular point to point $P$ is also the distance from the line to this point. The distance between two points can", "## Vector Equation of a Plane The vector equation of a plane can be written as $r.n=a.n$ where $$a$$ is a position vector of a point on the plane and $$n$$ is a normal to the plane. LESSON 1 Determine the equation of the plane, in vector form and Cartesian form, which contains the point $$(2,-3,1)$$ with normal $$i-2j+3k$$. SOLUTION The equation of a plane can be written in the form $r.n=a.n$ where $$n$$ is a vector perpendicular to the plane and $$a$$ is a position vector of a point on the plane. State the position vector $\\begin{pmatrix} 2\\\\-3 \\\\1 \\end{pmatrix}$ State the normal to the plane $n=\\begin{pmatrix} 1\\\\-2 \\\\3 \\end{pmatrix}$ Substitute the vectors $$a$$ and $$n$$ into the general form of the vector equation of the plane, $$r.n=a.n$$ $r.\\begin{pmatrix} 1\\\\-2 \\\\3 \\end{pmatrix}=\\begin{pmatrix} 2\\\\-3 \\\\1 \\end{pmatrix}.\\begin{pmatrix} 1\\\\-2 \\\\3 \\end{pmatrix}$ Determine the dot product $$a.n$$ $r.\\begin{pmatrix} 1\\\\-2\\\\3 \\end{pmatrix}=2(1)+(-3)(-2)+1(3)$ $r.\\begin{pmatrix} 1\\\\-2\\\\3 \\end{pmatrix}=11$ CARTESIAN FORM Replace $$r$$ with $$\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix}$$ in the vector equation of the plane $\\begin{pmatrix} x\\\\y\\\\z \\end{pmatrix}.\\begin{pmatrix} 1\\\\-2\\\\3 \\end{pmatrix}=11$ Determine the dot product, $$r.n$$ $x-2y+3z=11$ TRY THIS Complete Questions 13 and 14 from worksheet.", "# Finding the equation of a Plane • Oct 20th 2009, 10:47 AM craig Finding the equation of a Plane Hi, another vector question I'm afraid ;) I have the 4 lines: $BC = \\begin{pmatrix} -2 \\\\ -4 \\\\ 1 \\end{pmatrix} + \\lambda\\begin{pmatrix} 3 \\\\ 6 \\\\ -3 \\end{pmatrix}$, $AC = \\begin{pmatrix} 1 \\\\ 2 \\\\ -2 \\end{pmatrix} + \\sigma\\begin{pmatrix} -2 \\\\ 2 \\\\ -4 \\end{pmatrix}$. $l2 = \\begin{pmatrix} 3 \\\\ 0 \\\\ 2 \\end{pmatrix} + \\beta\\begin{pmatrix} -2 \\\\ 2 \\\\ -4 \\end{pmatrix}$, $AD = \\begin{pmatrix} 0 \\\\ 0 \\\\ -1 \\end{pmatrix} + \\alpha\\begin{pmatrix} -3 \\\\ 0 \\\\ -3 \\end{pmatrix}$. I need to show that all 4 lines lie in a single plane and fine the Cartesian Equation of the plane. The Cartesian equation of a plane is in the form $Ax+By+Cz=d$, so from this I can set up the following equations: $-2A-4B+C=d$ $3A+2C=d$ $A+2B-2C=d$ From these I can get all $A,B,C$ in terms of $d$. $A=d$, $B=C=-d$ Putting these values back into the original Cartesian Equation I got: $dx-dy-dz = d$, which can be simplified to $x-y-z=1$. Is this correct so far or have I gone horribly wrong? Thanks • Oct 21st 2009, 11:40 AM craig Over 20 views and none of you have any ideas ;) Anyone got any comments at all? • Oct 21st 2009,", "the vector $\\overrightarrow{QP}$, we subtract the coordinates of $P$ from the coordinates of $Q$: $$\\overrightarrow{QP} = \\begin{pmatrix}0-1 \\ 2-0 \\ 0-0\\end{pmatrix} = \\begin{pmatrix}-1 \\ 2 \\ 0\\end{pmatrix} = -1\\hat{\\boldsymbol{\\imath}} + 2\\hat{\\boldsymbol{\\jmath}}$$ To calculate the vector $\\overrightarrow{QR}$, we subtract the coordinates of $R$ from the coordinates of $Q$: $$\\overrightarrow{QR} = \\begin{pmatrix}0-0 \\ 2-0 \\ 3-0\\end{pmatrix} = \\begin{pmatrix}0 \\ 2 \\ 3\\end{pmatrix} = 2\\hat{\\boldsymbol{\\jmath}} + 3\\hat{\\boldsymbol{k}}$$ So, as you said, $\\overrightarrow{QP} = \\hat{\\boldsymbol{\\imath}} – 2 \\hat{\\boldsymbol{\\jmath}}$ and $\\overrightarrow{QR} = -2\\hat{\\boldsymbol{\\jmath}} + 3\\hat{\\boldsymbol{k}}$. $\\vec{N} \\cdot \\vec{r}(t)=6$, where $\\vec{N}=\\langle 4,-3,-2\\rangle$. The equation $\\vec{N} \\cdot \\vec{r}(t)=6$ represents a plane in three-dimensional space, where $\\vec{N}$ is the normal vector to the plane and $\\vec{r}(t)$ is a vector pointing to any point on the plane at time $t$. Given that $\\vec{N}=\\langle 4,-3,-2\\rangle$, we can write the equation of the plane as: $$4x – 3y – 2z = 6$$ where $x$, $y$, and $z$ are the coordinates of any point on the plane. Alternatively, we can express the equation in vector form as: $$\\langle 4,-3,-2\\rangle \\cdot \\langle x,y,z\\rangle = 6$$ or $$\\langle 4,-3,-2\\rangle \\cdot \\vec{r}(t) = 6$$ where $\\vec{r}(t)=\\langle x(t),y(t),z(t)\\rangle$ is a vector function that describes the position of any point on the plane at time $t$. Note that there are infinitely many points on the plane described by this equation, since the equation has three", "Finding orthonormal basis for a subspace $W$ of the Euclidean space $\\mathbb{R}^3$. Problem: Let $\\mathbb{R}^3$ be an Euclidean space. Find an orthonormal basis for the subspace $W$ defined as $x + 2y-z = 0$. Attempt at solution: So this is a plane in $\\mathbb{R}^3$, so I guess I would need two vectors to span this? I did the following. Pick two arbitrary vectors that lie in this plane, for example $w_1 = (1,1,3)$ and $w_2 = (1,0,1)$. Then should I apply the Gram-Schmidt process to these? That would get me: \\begin{align*} v_1 &= w_1 \\\\ v_2 &= w_2 - \\frac{\\langle w_2, v_1 \\rangle}{\\langle v_1, v_1} v_1 \\end{align*} Then we would have $v_1 = (1,1,3)$ and $v_2 = (7/11, -4/11, -1/11)$. Now I should normalize these and I'm done? • Yes, but the results will be simpler if you set $w_1=(1,0,1)$. – Bernard Jul 27 '15 at 18:18 A simple way of getting independent vectors in this plane is to note that if $x+ 2y- z= 0$, then $z= x- 2y$ so any vector in this plane is of the form $\\begin{pmatrix} x \\\\ y \\\\ z\\end{pmatrix}= \\begin{pmatrix}x \\\\ y \\\\ x- 2y\\end{pmatrix}$. Taking $x= 1, y= 0$, that is $\\begin{pmatrix}1 \\\\ 0 \\\\ 1\\end{pmatrix}$. Taking $x= 0, y= 1$, this is $\\begin{pmatrix} 0 \\\\ 1 \\\\ -2\\end{pmatrix}$.", "# Find an equation of the tangent plane to the given parametric surface at the specified point. x =... ## Question: Find an equation of the tangent plane to the given parametric surface at the specified point. ## Tangent plane to a surface This example illustrates how to find the tangent plane to a surface at a given point. In this particular case, the equation of the surface is given in terms of two parameters, and we must use partial differentiation and cross products. Suppose we have a surface with parametric equations {eq}\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} = \\begin{pmatrix} x(u, v)\\\\ y(u, v)\\\\ z(u, v) \\end{pmatrix}. {/eq} Then the vector which is normal to the surface at any particular point is given by {eq}\\begin{pmatrix} \\dfrac{\\partial x}{\\partial u}\\\\ \\dfrac{\\partial y}{\\partial u}\\\\ \\dfrac{\\partial z}{\\partial u} \\end{pmatrix} \\times \\begin{pmatrix} \\dfrac{\\partial x}{\\partial v}\\\\ \\dfrac{\\partial y}{\\partial v}\\\\ \\dfrac{\\partial z}{\\partial v} \\end{pmatrix}. {/eq} In this example, {eq}\\begin{pmatrix} x\\\\ y\\\\ z \\end{pmatrix} = \\begin{pmatrix} u^2 + 1\\\\ v^3 + 1\\\\ u + v \\end{pmatrix} {/eq} and so the normal vector to the surface is {eq}\\begin{pmatrix} \\dfrac{\\partial x}{\\partial u}\\\\ \\dfrac{\\partial y}{\\partial u}\\\\ \\dfrac{\\partial z}{\\partial u} \\end{pmatrix} \\times \\begin{pmatrix} \\dfrac{\\partial x}{\\partial v}\\\\ \\dfrac{\\partial y}{\\partial v}\\\\ \\dfrac{\\partial z}{\\partial v} \\end{pmatrix} = \\begin{pmatrix} 2u\\\\ 0\\\\ 1 \\end{pmatrix} \\times \\begin{pmatrix} 0\\\\ 3v^2\\\\ 1 \\end{pmatrix} (*) {/eq} We need to find", "# Finding the equation of a plane. How do I find the equation of a plane given by the points (0,1,1), (1,0,1) and (1,1,0)? Graphing it, it's a triangle when you connect the points. Can I use this somehow? - What do you know about equation of a plane? There are several formats, which one do you know of? 2-vector 1 point, normal vector and parametric. Sound familiar? – Calvin Lin Jan 26 '13 at 23:29 By inspection $x+y+z=2$ is an equation of the plane. – André Nicolas Jan 26 '13 at 23:30 a(x-x0)+b(y-y0)+c(z-z0)=0. Andre got the answer right but not sure on how to get it. – user1766888 Jan 26 '13 at 23:31 Sounds like you want the normal vector form. To find a normal vector to the plane, we take the cross product of 2 (linearly independent) vectors of the plane. In this case, we take the vectors $\\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix}, \\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix} - \\begin{pmatrix} 1 \\\\ 0 \\\\ 1 \\end{pmatrix} = \\begin{pmatrix} -1 \\\\ 1 \\\\ 0 \\end{pmatrix}$ to obtain $$\\begin{pmatrix} -1 \\\\ 0 \\\\ 1 \\end{pmatrix} \\times \\begin{pmatrix} -1 \\\\ 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix} -1 \\\\ -1", "\\\\ -1 \\end{pmatrix}$$ Then, the plane satisfies $$\\begin{pmatrix} x \\\\ y \\\\ z \\end{pmatrix} \\begin{pmatrix} -1 \\\\ -1 \\\\ -1 \\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 1 \\\\ 1 \\end{pmatrix} \\begin{pmatrix} -1 \\\\ -1 \\\\ -1 \\end{pmatrix} = -2,$$ or that $- x - y - z = - 2$. - Take one of the points as origin and use the other two to obtain two (linearly independent? check!) vectors. Then write the vectorial equation for the plane as $$(x,y,z)=P+\\lambda \\vec u + \\mu \\vec v.$$ -", "both. A description of planes that is often encountered in algebra and calculus uses a single equation as the condition that describes the relationship among the first, second, and third coordinates of points in a plane. The translation from such a description to the vector description that we favor in this book is to think of the condition as a one-equation linear system and parametrize $x=(1/2)(4-y-z)$. Generalizing from lines and planes, we define a $k$-dimensional linear surface (or $k$-flat) in $\\mathbb{R}^n$ to be $\\{\\vec{p}+t_1\\vec{v}_1+t_2\\vec{v}_2+\\cdots+t_k\\vec{v}_k \\,\\big|\\, t_1,\\ldots ,t_k\\in\\mathbb{R}\\}$ where $\\vec{v}_1,\\ldots,\\vec{v}_k\\in\\mathbb{R}^n$. For example, in $\\mathbb{R}^4$, $\\{\\begin{pmatrix} 2 \\\\ \\pi \\\\ 3 \\\\ -0.5 \\end{pmatrix} +t\\begin{pmatrix} 1 \\\\ 0 \\\\ 0 \\\\ 0 \\end{pmatrix} \\,\\big|\\, t\\in\\mathbb{R}\\}$ is a line, $\\{ \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 0 \\end{pmatrix} +t\\begin{pmatrix} 1 \\\\ 1 \\\\ 0 \\\\ -1 \\end{pmatrix} +s\\begin{pmatrix} 2 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix} \\,\\big|\\, t,s\\in\\mathbb{R}\\}$ is a plane, and $\\{ \\begin{pmatrix} 3 \\\\ 1 \\\\ -2 \\\\ 0.5 \\end{pmatrix} +r\\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\\\ -1 \\end{pmatrix} +s\\begin{pmatrix} 1 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix} +t\\begin{pmatrix} 2 \\\\ 0 \\\\ 1 \\\\ 0 \\end{pmatrix} \\,\\big|\\, r,s,t\\in\\mathbb{R}\\}$ is a three-dimensional linear surface. Again, the intuition is that a line permits motion in one direction, a plane permits motion in combinations of two directions, etc. $L=\\{ \\begin{pmatrix}", "the surface does not lie in a plane, $X_{st}$ cannot be a linear combination of $X_s$ and $X_t$ (otherwise, the plane spanned by $X_s$ and $X_t$ would be fixed, and the surface would lie in plane). This means that $X_s$, $X_t$, $X_{st}$ is a basis of $\\mathbb{R}^3$, and, as such, there are equations of the form $$\\begin{pmatrix} dX_s& dX_t & dX_{st} \\end{pmatrix} = \\begin{pmatrix} X_s& X_t & X_{st} \\end{pmatrix} \\begin{pmatrix} f\\ ds& 0 & f_t\\ ds\\\\\\\\ 0 & g\\ dt & g_s\\ dt\\\\\\\\ dt & ds & f\\ ds + g\\ dt \\end{pmatrix}$$ (The equations for $dX_{st}$ follow since $(X_{st})_s = (f X_s)_t = f_t\\ X_s + f\\ X_{st}$, etc.) By comparing partials, or by using the structure equations above (i.e., expanding out the consequences of $d(d(X_{st}))=0$, etc.), one sees that $d(f\\ ds + g\\ dt) = 0$, so that there must exist a function $h$ such that $f = 2h_s$ and $g = 2h_t$. (The coefficient of $2$ avoids some fractions later.) The equation now becomes $$\\begin{pmatrix} dX_s& dX_t & dX_{st} \\end{pmatrix} = \\begin{pmatrix} X_s& X_t & X_{st} \\end{pmatrix} \\begin{pmatrix} 2h_s\\ ds& 0 & 2h_{st}\\ ds\\\\\\\\ 0 & 2h_t\\ dt & 2h_{st}\\ dt\\\\\\\\ dt & ds & 2h_s\\ ds + 2h_t\\ dt \\end{pmatrix}$$ Moreover, the structure equations now imply that $d(e^{-2h}h_{st})=0$, so $h_{st} = C\\ e^{2h}$ for", "$\\mathbb{R}^2$ is a subspace of $\\mathbb{R}^3$. By definition, for $\\mathbb{R}^2$ to be a subspace of $\\mathbb{R}^3$, $\\mathbb{R}^2$ must first be a subset of $\\mathbb{R}^3$. This is not true because vectors in $\\mathbb{R}^2$ have only two elements whereas vectors in $\\mathbb{R}^3$ have three elements. This means that vectors in $\\mathbb{R}^2$ do not belong in $\\mathbb{R}^3$ - for instance: $$\\begin{pmatrix} 2\\\\3 \\end{pmatrix}\\notin\\mathbb{R}^3$$ Therefore, $\\mathbb{R}^2$ is not a subspace of $\\mathbb{R}^3$. Example. ## A plane passing through the origin Consider the following plane: $$2x+4y+3z=0$$ Show that this is a subspace of $\\mathbb{R}^3$. Solution. To show that the plane is a subspace of $\\mathbb{R}^3$, we must check that the plane is closed under addition and scalar multiplication. Suppose the following two generic vectors lie on the plane: $$\\boldsymbol{v}_1= \\begin{pmatrix} x_1\\\\y_1\\\\z_1 \\end{pmatrix},\\;\\;\\;\\;\\; \\boldsymbol{v}_2= \\begin{pmatrix} x_2\\\\y_2\\\\z_2 \\end{pmatrix}$$ Firstly, since these points are on the plane, we have that: \\label{eq:M1IcX56LXG5HLrinvnh} \\begin{aligned} 2x_1+4y_1+3z_1&=0\\\\ 2x_2+4y_2+3z_2&=0\\\\ \\end{aligned} Now, let's check whether or not the plane is closed under addition. The vector $\\boldsymbol{v}_1+\\boldsymbol{v}_2$ is: \\label{eq:xWDCFePt8ckIu5VeeW0} \\begin{aligned}[b] \\boldsymbol{v}_1+ \\boldsymbol{v}_2&= \\begin{pmatrix} x_1\\\\y_1\\\\z_1 \\end{pmatrix}+ \\begin{pmatrix} x_2\\\\y_2\\\\z_2 \\end{pmatrix}\\\\ &= \\begin{pmatrix} x_1+x_2\\\\y_1+y_2\\\\z_1+z_2 \\end{pmatrix} \\end{aligned} Let's substitute \\eqref{eq:xWDCFePt8ckIu5VeeW0} into the equation of the plane to check whether this point is on the plane: \\begin{align*} 2(x_1+x_2)+4(y_1+y_2)+3(z_1+z_2) &=2x_1+2x_2+4y_1+4y_2+3z_1+3z_2\\\\ &=(2x_1+4y_1++3z_1)+(2x_2+4y_2+3z_2)\\\\ &=0 \\end{align*} The last step used \\eqref{eq:M1IcX56LXG5HLrinvnh}. Therefore, this equation is closed under addition. Now, we check whether", "the plane. Dec 7, 2021 at 18:38 In this question, we can find any point that will lie on the line intersecting the two planes, suppose $$(a,b,0)$$. Then we can simultaneously solve the the two planes equation by putting this point in it. $$a+2b=1\\qquad2a+3b=-3$$ After solving these two, we will get $$a=-7$$ and $$b=4$$. Now, we can find the direction of the line we need to find by taking cross product of normal vectors of two given planes. $$(i,2j,k) × ( 2i,3j,-2k) = (-7i,4j,-k)$$ Thus equation of line is $$-7(x+7)+4(y-4)-1(z-0)=0$$. Hence $$7x-4y+z+65=0$$ Let $$x$$ be in the intersection of the planes $$\\begin{equation*} \\langle m, x-b \\rangle = 0,\\\\ \\langle n, x-c \\rangle = 0 \\end{equation*}$$ and let $$p = m \\times n$$. Then for some $$\\lambda$$, $$\\begin{equation*} \\begin{pmatrix} m_0 & m_1 & m_2 \\\\ n_0 & n_2 & n_3 \\\\ p_0 & p_1 & p_2 \\\\ \\end{pmatrix} \\begin{pmatrix} x_0 \\\\ x_1 \\\\ x_2 \\end{pmatrix} = \\begin{pmatrix} \\langle m, b \\rangle \\\\ \\langle n, c \\rangle \\\\ \\lambda \\end{pmatrix}. \\end{equation*}$$ Using the vector identities $$\\begin{equation*} \\begin{split} u \\times (v \\times w) &= \\langle u, w \\rangle v - \\langle u, v \\rangle w, \\\\ \\Vert u \\times v \\Vert^2 &= \\Vert u \\Vert^2 \\Vert v \\Vert^2 - \\langle u, v \\rangle^2, \\end{split} \\end{equation*}$$ check that $$\\begin{equation*} \\begin{pmatrix} m_0 &" ]

[ "$$1^5+2^5+...+n^5$$ Find the smallest positive integer $$n>1$$, such that $\\sum_{k=1}^n k^5$ is a perfect square. ×", "### K - K-th Common Divisor ##### Languages: C, C++, Java, Python, ... (details) You are given positive integers $A$ and $B$. Find the $K$-th largest positive integer that divides both $A$ and $B$. The input guarantees that there exists such a number. #### Input The input contains one line with three integers $A$, $B$, $K$ with $1 \\leq A, B \\leq 100$ and $K ≥ 1$. The $K$-th largest positive integer that divides both $A$ and $B$ exists. #### Output Print the $K$-th largest positive integer that divides both $A$ and $B$. Input 8 12 2 Output 2 Input 100 50 4 Output 5 Input 1 1 1 Output 1", "# Perfect how? Divide $$297$$ by the smallest possible positive integer $$a$$ so that the result is the square of a positive integer $$b$$. What is $$a+b$$? ×", "integer $k\\geq4, 2^k < k!$. Then multiplying through by 2, we obtain $2^k < 2k!$ $2^k < (1+1)k!<(k+1)\\cdot k!$ since the smallest $k$ in the parenthesis is greater than 4. Therefore, $2^k <<(k+1)!$ Consequently, for all positive integer $n \\geq 4, P \\wedge Q.$ 8. Originally Posted by novice You started your proof incorrectly where I marked in red. You are not to begin with the final results. Novice, there was nothing wrong with the method.", "97) · k + 366 · 97k ≡ 0 (mod 7) ⇒ 146097k ≡ 0 (mod 7) Since, 7 ∣ 146097, the smallest positive integer k for which the above equivalence holds true is k = 1. So, p = 400. Hence, the calendar repeats every 400 years.", "× # Proof Let $$n$$ be an integer greater than 1 and $$k$$ be any positive integer. Prove that $$n-1$$ divides $$n^k - 1$$. Note by أحمد الحلاق 2 weeks, 1 day ago", "k is a positive integer, then 1999 divides 1#(k*999). If true, this cannot be proven by brute force computation, since k can be any of infinitely many values. You can use the fact that 999 is the smallest n>0 such that 1999 divides 1#n, since that fact has already been established by ceptimus's computations. Part three of the puzzle: Is there a simple algorithm that will find the smallest n>0 such that 1999 divides 1#n, without using any kind of \"big integer\" data types, i.e. can be done using only 16 bit integers? No need to find the result of the division, just find the number n. To test your algorithm, also find the smallest n such that 1997 divides 1#n. slimshady2357 Posts: 698 Joined: Sat Jun 05, 2004 4:53 pm xouper wrote: Part two of this puzzle: Is the following true or false and can you prove it? If and only if k is a positive integer, then 1999 divides 1#(k*999). If true, this cannot be proven by brute force computation, since k can be any of infinitely many values. You can use the fact that 999 is the smallest n>0 such that 1999 divides 1#n, since that fact has already been established by ceptimus's computations. . I think I can do the easy half of this", "× # Proof Let $$n$$ be an integer greater than 1 and $$k$$ be any positive integer. Prove that $$n-1$$ divides $$n^k - 1$$. Note by أحمد الحلاق 11 months, 3 weeks ago", "The Smallest $k$ Let $f(x)=7x^{11}+11x^{7}+9kx$ $g(x)=7x^{13}+13x^{7}+10kx$ Find the minimum positive value of $k$ for which $77| f(x)$ and $91|g(x)$ for every integer $x$ Details and Assumptions • We use \"$\\mid$\" to mean \"divides evenly into\". For example, $3\\mid6$, and $12\\mid132$. ×", "the square). The smallest number of the form K^2+L^2 for positive integers K,L that is on the white list is 13 for K,L=2,3 or 3,2: numbers 2,5,8 are not on the white list. And 18 = 3^2+3^2 that would be just a little bit bigger is not on the white list, either. The template 3(K^2+L^2) is clearly even less useful to create small solutions. The smallest allowed values of this tripled sum for positive integers K,L, namely 6,15,24..., are not on the white list, either. So far, K^2+L^2=9+4=13 is the only nontrivial small solution we found. Finally, we must deal with the template 2(K^2+L^2+-KL); of course, the negative sign is better to produce smaller results. This can only match the even numbers on the white list, namely 4,12,16..., because it is even. K^2+L^2+-KL would have to be equal to 2,6,8... However, 2,6,8 can't be written in this form: 2,6 are equal to 2 modulo 4, but K^2+L^2+-KL can't be 2 mod 4. And 8 is also impossible for similar reasons. So the only new small solution we found was one based on the hinges whose distance is sqrt(13). Top intersecting solution The assumption of just 2 vertices (hinges) that are not combinations of t,u,v,w is severely violated in the state-of-the-art best solutions. At the top, there is bbzippo's", "of Zelda using only the wooden sword.) Let $k$ be the smallest positive integer for which $a \\mid bk$. Suppose $l$ is the integer for which $bk = al$ and $m$ is the integer for which $b^2 = a^2 m$. Then $a b l = b^2 k = a^2 k m$, so that $b l = a k m$. Use the division algorithm and write $l = kq + r$ with $0 \\leq r < k$. Then $br = a k m - b k q = a (k m - l q)$. Thus, $a \\mid br$. By our assumption that $k$ is the smallest positive integer for which $a \\mid bk$, we find that $r$ cannot be positive, so is $0$. Thus, $l = kq$, so $bk = al = akq$, so $b = aq$ and $a$ divides $b$. (Of course this implies also that if $a^{2^k} \\mid b^{2^k}$, then $a \\mid b$. I don't see immediately an extension that works to prove that if $a^n \\mid b^n$, then $a \\mid b$, but am happy to edit if someone else does.) -", "# Make It An Ideal Expression Find the smallest positive integer $x$ such that $\\large \\frac{2x^2+5x+3}{3}$ is a perfect square. ×", "# Project Euler Problem 5: Smallest multiple 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20? In [1]: from six.moves import range In [2]: all_divides = lambda m, *numbers: all(m % n == 0 for n in numbers) In [3]: all_divides(2520, *range(1, 10)) Out[3]: True The least common multiple of the numbers 1 to 10 is 2520. We are asked to find that of the numbers 1 to 20. #### Version 1 - Integer Factorization¶ We already implemented lcm in problem 3, and the easiest way would be for us to simply use that. However, as we mentioned, it is not very efficient. Let's instead look at other ways of implementing it. #### Version 2 - Simple algorithm¶ Given a list of $n$ integers $X = (x_1, x_2, \\dotsc, x_n)$, we can find the least common multiple of the integers in $X$ by the following: Let $X^{(0)} = (x_1^{(0)}, x_2^{(0)}, \\dotsc, x_n^{(0)}) = (x_1, x_2, \\dotsc, x_n) = X$ and $X^{(m+1)} = (x_1^{(m+1)}, x_2^{(m+1)}, \\dotsc, x_n^{(m+1)})$ where $$x_k^{(m+1)} = x_k^{(m)} + \\begin{cases} x_k^{(0)} & \\text{if } \\min(X^{(m)}) = x_k^{(m)} \\\\ 0 & \\text{otherwise} \\end{cases}$$ Once all $X$", "or you will be at a disadvantage. # If k is the smallest positive integer such that 2,940k Author Message TAGS: ### Hide Tags Intern Joined: 16 Feb 2013 Posts: 6 If k is the smallest positive integer such that 2,940k [#permalink] ### Show Tags 04 May 2013, 09:17 4 00:00 Difficulty: 35% (medium) Question Stats: 66% (01:26) correct 34% (01:54) wrong based on 170 sessions ### HideShow timer Statistics If k is the smallest positive integer such that 2,940k is the square of an integer, then k must be A. 3 B. 5 C. 6 D. 15 E. 21 Intern Joined: 23 Apr 2013 Posts: 21 Re: If k is the smallest positive integer such that 2,940k [#permalink] ### Show Tags 04 May 2013, 09:52 1 streamingline wrote: If k is the smallest positive integer such that 2,940k is the square of an integer, then k must be A)3 B)5 C)6 D)15 E)21 Spoiler: :: OA 15 $$2940 = 2^2 * 3 * 5 * 7^2$$ So the smallest positive integer to be multiplied to make it a perfect square is 3 * 5 = 15 Correct Option is D Intern Joined: 16 Feb 2013 Posts: 6 Re: If k is the smallest positive integer such that 2,940k [#permalink] ### Show Tags 04 May 2013, 18:35 Thanks,", "exist an integer, N, such that N> z- and the fact that the set of positive integers is \"well ordered\"- any non-empty set of positive integers contains a smallest integer. Certainly if x- y> 1, x- y is positive. Let z= x- y. Then there exist a positive integer N> z= x- y so the set of \"all integers N, greater than x- y\" is non-empty and so has a smallest integer, n. Since n is the smallest integer in the set, n- 1 is NOT in the set. Consider the two cases: 1) n- 1= x- y 2) n-1< x- y", "# Find the divisors 448 views 0 Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\\cdot5^5\\cdot7^7.$ Find the number of positive integer divisors of n 0 Lifting the Exponent shows that so thus, divides . It also shows that so thus, divides . Now, multiplying by , we see and since and then meaning that we have that by LTE, divides . Since and all divide , the smallest value of working is their LCM, also . Thus the number of divisors is .", "# A number theory problem by Kian Chua Consider positive integers solutions $$(x,y,z,u)$$ of the systems of equations $\\begin{cases} x+y=3(z+u), \\\\ x+z=4(y+u), \\\\ x+u=5(y+z). \\\\ \\end{cases}$ What is the smallest value that $$x$$ can have? ×", "• # question_answer The least positive integer $n$ which will reduce ${{\\left( \\frac{i-1}{i+1} \\right)}^{n}}$ to a real number, is [Roorkee 1998] A) 2 B) 3 C) 4 D) 5 ${{\\left( \\frac{i-1}{i+1}\\times \\frac{i-1}{i-1} \\right)}^{n}}={{\\left( \\frac{-2i}{-2} \\right)}^{n}}={{i}^{n}}$ Hence, to make the real number the least positive integer is 2.", "# Inspried by Satyajit Mohanty Let $$S$$ be the set of numbers less than $$5$$ billions and can be expressed in the form $$n^2+5n+23$$, where $$n$$ is a positive integer. Let $$N$$ be the smallest prime number that divides for some elements of $$S$$. Let $$p$$ be the number of elements of $$S$$ that are prime. Let $$q$$ be the number of elements of $$S$$ that are divisible by $$N$$ . Find $$N+p+q$$. ×", "+0 # Help 0 89 1 Let $a_1,$ $a_2,$ $\\dots$ be a sequence of real numbers such that for all positive integers $n,$ $\\sum_{k = 1}^n a_k \\left( \\frac{k}{n} \\right)^2 = 1.$Find the smallest $n$ such that $a_n < \\frac{1}{2018}.$ Aug 25, 2021" ]

[ "## What are $$n$$th roots? Let $$n \\in \\mathbb{N}$$. We say that $$b$$ is an $$n$$th root of $$a$$ if $$b^n = a$$. When $$n = 2$$, $$b$$ is called a square root of $$a$$. When $$n = 3$$, $$b$$ is called a cube root of $$a$$. It is well known that every positive real number has two square roots. For example, the square roots of $$4$$ are $$2$$ and $$-2$$ because $$2^2 = 4$$ and $$(-2)^2 = 4$$. We now turn to finding $$n$$th roots for complex numbers. ## Motivating example Consider the complex numbers $$z = 2\\operatorname{cis}\\frac{\\pi}{6}$$ and $$w = 2\\operatorname{cis}\\frac{5\\pi}{6}$$. By De Moivre's Formula, we get $z^3 = 2^3 \\operatorname{cis}\\left(3\\cdot \\frac{\\pi}{6}\\right) = 8 \\operatorname{cis}\\frac{\\pi}{2}$ and $w^3 = 2^3 \\operatorname{cis}\\left(3\\cdot \\frac{5\\pi}{6}\\right) = 8 \\operatorname{cis}\\frac{5\\pi}{2} = 8 \\operatorname{cis}\\left(2\\pi + \\frac{\\pi}{2}\\right) = 8 \\operatorname{cis}\\frac{\\pi}{2}.$ Even though $$z\\neq w$$, we have $$z^3 = w^3 = 8 \\operatorname{cis} \\frac{\\pi}{2}$$. Hence, $$z$$ and $$w$$ are distinct cube roots of $$8 \\operatorname{cis} \\frac{\\pi}{2}$$. (Check that $$2\\operatorname{cis}\\frac{3\\pi}{2}$$ is also a cube root of $$8 \\operatorname{cis} \\frac{\\pi}{2}$$.) Observe that the argument of $$w$$, which is $$\\frac{5\\pi}{6}$$, is $$\\frac{2\\pi}{3}$$ more than the argument of $$z$$, which is $$\\frac{\\pi}{6}$$. Even if we did not know the actual arguments of $$z$$ and $$w$$, we could tell from this observation that the argument of $$w^3$$ is $$2\\pi$$ more", "-1 prove that$ $\\: z^6+1 = (z^2+1) (z^2-\\sqrt{3}z+1) (z^2+\\sqrt{3}z+1)$ $By using your usual method to find roots of unity you would get$ $z = i, -i, cis$ $(\\frac{\\pi}{6})$ $cis$ $(\\frac{-\\pi}{6})$ $cis$ $(\\frac{5\\pi}{6})$ $cis$ $(-\\frac{5\\pi}{6})$ $Then by using your usual linear factorisation of polynomials via their roots you would get$ $\\: z^6+1 = (z-i) (z+i) (z- cis$ $(\\frac{\\pi}{6}))$ $(z-cis$ $(\\frac{-\\pi}{6}))$ $(z-cis$ $(\\frac{5\\pi}{6}))$ $(z-cis$ $(\\frac{-5\\pi}{6}))$ $Now by expanding the pairs and simplifying you would get$ $(z^2+1) (z^2 -2cos$ $(\\frac{\\pi}{6})$ $z +1)$ $(z^2 -2cos$ $(\\frac{5\\pi}{6})$ $z +1)$ $Notice how the last two brackets above are in the form of the formula you mentioned, that is$ $z^2 -Sum of Roots + Product of Roots.$ $I think when you add stuff like the cis and its conjugate you$ $get the double of the real part which is essentially the 2 cos$ $\\frac{\\pi}{6}$ $and when you multiply (product) the the conjugate of two cis's you get 1 provided that that there is no coefficient in front of it$ $Therefore, from the equation above you just evaluate it and you should get$ $\\: z^6+1 = (z^2+1) (z^2-\\sqrt{3}z+1) (z^2+\\sqrt{3}z+1)$ Idk if that made any sense but hope that helps. This was the first time I used LaTeX so the input is killing me. Let me know if you need more help. There are currently 1 users", "# Problem finding zeros of complex polynomial I'm trying to solve this problem $$z^2 + (\\sqrt{3} + i)|z| \\bar{z}^2 = 0$$ So, I know $|z^2| = |z|^2 = a^2 + b ^2$ and $\\operatorname{Arg}(z^2) = 2 \\operatorname{Arg} (z) - 2k \\pi = 2 \\arctan (\\frac{b}{a} ) - 2 k\\pi$ for a $k \\in \\mathbb{Z}$. Regarding the other term, I know $|(\\sqrt{3} + i)|z| \\bar{z}^2 | = |z|^3 |\\sqrt{3} + i| = 2 |z|^3 = 2(a^2 + b^2)^{3/2}$ and because of de Moivre's theorem, I have $\\operatorname{Arg} [(\\sqrt{3} + i ) |z|\\bar{z}^2] = \\frac{\\pi}{6} + 2 \\operatorname{Arg} (z) - 2Q\\pi$. Using all of this I can rewrite the equation as follows \\begin{align*} &|z|^2 \\Bigl[ \\cos (2 \\operatorname{Arg} (z) - 2k \\pi) + i \\sin (2 \\operatorname{Arg}(z) - 2k \\pi)\\Bigr]\\\\ &\\qquad \\mathop{+} 2|z|^3 \\Biggl[\\cos \\left(\\frac{\\pi}{6} + 2 \\operatorname{Arg} (z) -2Q\\pi\\right) + i \\sin \\left(\\frac{\\pi}{6} + 2 \\operatorname{Arg} (z) -2Q\\pi\\right)\\Biggr] = 0 \\end{align*} Which, assuming $z \\neq 0$, can be simplified as \\begin{align*} &\\cos (2 \\operatorname{Arg} (z) - 2k \\pi) + i \\sin (2 \\operatorname{Arg} (z) - 2k \\pi) \\\\ &\\qquad\\mathop{+} 2 |z|\\Biggl[\\cos \\left(\\frac{\\pi}{6} + 2 \\operatorname{Arg} (z) -2Q \\pi \\right) + i \\sin \\left(\\frac{\\pi}{6} + 2 \\operatorname{Arg} (z) -2Q\\pi\\right)\\Biggr] = 0 \\end{align*} Now, from this I'm not sure how to go on. I tried a few things that got me", "= x + i\\sqrt{1 - x^2}$? (Note in this case to get the polynomial in $z$ I found the stupidly huge closed form of $x$ and then constructed $z$ and found the minimum polynomial.) ${}^\\dagger$By rational angle I mean an angle of the form $\\theta_{\\mathrm{rational}} = \\frac{2\\pi m}{n} \\quad \\mathrm{where} \\quad m,n \\in \\mathbb{Z},\\ n \\neq 0,\\ \\mathrm{and}\\ \\operatorname{GCD}(m, n) = 1$ • Sorry if I misunderstand, but if $z = x+i\\sqrt{1-x^2}$ then $x = (1+z^2)/(2z)$, so \\begin{align} 8x^3 + 8x^2 - 2x - 1 &= 8\\left(\\frac{1+z^2}{2z}\\right)^3 + 8\\left(\\frac{1+z^2}{2z}\\right)^2 - 2\\left(\\frac{1+z^2}{2z}\\right) - 1 \\\\ &= \\frac{z^6 + 2 z^5 + 2 z^4 + 3 z^3 + 2 z^2 + 2 z + 1}{z^3}, \\end{align} and the numerator is the polynomial you want. – Antonio Vargas Nov 19 '16 at 9:01 • @AntonioVargas thank you for your reply. I see you solved for x in terms of z but when I substitute $$x = (1 + z^2)/(2z)$$ into $$8x^3 + 8x^2 - 2x - 1$$ I get $$z^9+3 z^7+2 z^6+3 z^5+4 z^4+2 z^2-z-1$$ which is irreducible. How did you know to leave the $z^3$ denominator in order to get my desired polynomial? The desired root $z \\approx 0.4 + 0.9i$ value isn't even a root of the order 9 polynomial I got. – Brandon Enright Nov 19 '16 at", "# Operation with Complex number We've the following 3 complex numbers $z_1=2\\text{cis}(\\frac{\\pi}{6})~~$, $z_2=2\\text{cis}(\\frac{5\\pi}{6})~~$, $z_3=2\\text{cis}(\\frac{-\\pi}{2})$ We are asked to show that $z_1^{3n}+z_2^{3n}=2z_3^{3n}$ I have calculated $z_1^{3n}=2^{3n}\\text{cis}(\\frac{n\\pi}{2})~~$, $z_2^{3n}=2^{3n}\\text{cis}(\\frac{5n\\pi}{2})~~$, $z_3^{3n}=2^{3n}\\text{cis}(\\frac{-3n\\pi}{2})$. I'm stuck on how to proceed further to prove the above equations. Any help really appreciated, Please assist . Thank you , Arif • What is \"cis\" ? – Jean Marie Mar 29 '17 at 6:07 • @JeanMarie This. – dxiv Mar 29 '17 at 6:21 • @dxiv Thanks. I understand now the pedagogical interest of this notation... – Jean Marie Mar 29 '17 at 6:32 • $z_1^3 = 2^3 \\operatorname{cis}\\left(3 \\cdot \\cfrac{\\pi}{6}\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ • $z_2^3 = 2^3 \\operatorname{cis}\\left(3 \\cdot \\cfrac{5 \\pi}{6}\\right)=8\\operatorname{cis}\\left(\\cfrac{5 \\pi}{2}\\right)=8\\operatorname{cis}\\left(2\\pi+\\cfrac{\\pi}{2}\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ • $z_3^3 = 2^3 \\operatorname{cis}\\left(-3 \\cdot \\cfrac{\\pi}{2}\\right)=8\\operatorname{cis}\\left(- 3 \\cdot \\cfrac{\\pi}{2}+ 2 \\pi\\right)=8\\operatorname{cis}\\left(\\cfrac{\\pi}{2}\\right)=8i$ Therefore $\\;z_1^3=z_2^3=z_3^3\\,$.", "# Chapter 8 Exercise C Exercise 1 Because $$4 = \\operatorname{dim} \\mathbb{C}^4 = \\operatorname{dim} G(3, T) + \\operatorname{dim} G(5, T) + \\operatorname{dim} G(8, T),$$ it follows that the multiplicities of the eigenvalues of $T$ are at most $2$. Thus $p(z) = (z-3)^2(z-5)^2(z-8)^2$ is a polynomial multiple of the characteristic polynomial. Therefore, we must have $p(T) = 0$. Exercise 2 We have $$1 \\le \\dim G(5, T), \\dim G(6, T) \\le n – 1.$$ Hence $p(z) = (z – 5I)^{n-1}(z – 6I)^{n-1}$ is a polynomial multiple of the characteristic polynomial. Therefore, $p(T) = 0$. Exercise 3 Define $T \\in \\mathcal{L}(\\mathbb{C}^4)$ by $$T(z_1, z_2, z_3, z_4) = (7z_1, 7z_2, 8z_3, 8z_4).$$ Since $E(7, T) \\subset G(7, T), E(8, T) \\subset G(8, T)$ and $\\mathbb{C}^4 = E(7, T) \\oplus E(8, T)$, it follows that $E(7, T) = G(7, T)$ and $E(8, T) = G(8, T)$. We have $\\dim E(7, T) = \\dim E(8, T) = 2$. Hence, the characteristic polynomial of $T$ is $$(z – 7)^2(z – 8)^2.$$ Exercise 4 $$A = \\begin{pmatrix} 0 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 0 \\end{pmatrix}$$ Let $$A = \\begin{pmatrix} 1 & 1 & 0 & 0\\\\ 0 & 5 & 1 & 0\\\\ 0", "be the solutions to the equation $z^2=2+2\\sqrt 3i.$ Clearly, $z_1$ and $z_2$ are opposite complex numbers, so are $z_3$ and $z_4.$ This solution refers to the results of De Moivre's Theorem in Solution 2. From Solution 2, let $z_1=4\\operatorname{cis}\\phi$ for some $0<\\phi<\\frac{\\pi}{4}.$ It follows that $z_2=4\\operatorname{cis}(\\phi+\\pi).$ On the other hand, we have $z_3=2\\operatorname{cis}\\frac{\\pi}{6}$ and $z_4=2\\operatorname{cis}\\frac{7\\pi}{6}$ without the loss of generality. Since $\\tan(2\\phi)>\\tan\\frac{\\pi}{3},$ we deduce that $2\\phi>\\frac{\\pi}{3},$ from which $\\phi>\\frac{\\pi}{6}.$ In the complex plane, the positions of $z_1,z_2,z_3,$ and $z_4$ are shown below: $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -5; int xMax = 5; int yMin = -5; int yMax = 5; int numRays = 24; //Draws a polar grid that goes out to a number of circles //equal to big, with numRays specifying the number of rays: void polarGrid(int big, int numRays) { for (int i = 1; i < big+1; ++i) { draw(Circle((0,0),i), gray+linewidth(0.4)); } for(int i=0;i Note that the diagonals of every parallelogram partition the shape into four triangles with equal areas. Therefore, to find the area of the parallelogram with vertices $z_1,z_2,z_3,$ and $z_4,$ we find the area of the triangle with vertices $0,z_1,$ and $z_3,$ then multiply by $4.$ Recall that $|z_1|=4, |z_2|=2, \\sin\\phi=\\frac{\\sqrt6}{4},$ and $\\cos\\phi=\\frac{\\sqrt{10}}{4}$ from Solution 2. The area of the parallelogram is \\begin{align*} 4\\cdot\\left[\\frac12\\cdot|z_1|\\cdot|z_3|\\cdot\\sin\\left(\\phi-\\frac{\\pi}{6}\\right)\\right] &= 16\\sin\\left(\\phi-\\frac{\\pi}{6}\\right) \\\\", "# Finding the n-th root of a complex number I am trying to solve $z^6 = 1$ where $z\\in\\mathbb{C}$. So What I have so far is : $$z^6 = 1 \\rightarrow r^6\\operatorname{cis}(6\\theta) = 1\\operatorname{cis}(0 + \\pi k)$$ $$r = 1,\\ \\theta = \\frac{\\pi k}{6}$$ $$k=0: z=\\operatorname{cis}(0)=1$$ $$k=1: z=\\operatorname{cis}\\left(\\frac{\\pi}{6}\\right)=\\frac{\\sqrt{3} + i}{2}$$ $$k=2: z=\\operatorname{cis}\\left(\\frac{\\pi}{3}\\right)=\\frac{1 + \\sqrt{3}i}{2}$$ $$k=3: z=\\operatorname{cis}\\left(\\frac{\\pi}{2}\\right)=i$$ According to my book I have a mistake since non of the roots starts with $\\frac{\\sqrt{3}}{2}$, also even if I continue to $k=6$ I get different (new) results, but I thought that there should be (by the fundamental theorem) only 6 roots. Can anyone please tell me where my mistake is? Thanks! - So what is written in your book? – draks ... Feb 6 '12 at 13:46 $cis=cos+i\\cdot sin$? – Rasmus Feb 6 '12 at 14:01 yes its is. why do i have to write 12 chars? – yotamoo Feb 6 '12 at 14:10 You want $\\theta=\\frac{2\\pi k}6 = \\frac{\\pi k}3$. – Thomas Andrews Feb 6 '12 at 14:12 What ever \"start with ...\", I think there is a 2 missing. Shouldn't it be ${\\rm cis}(x)=e^{ix}=e^{i(x+2k\\pi)}$? This would also drop the solution that \"starts with $\\sqrt{3}/2$\", which is a root of -1 anyhow. – draks ... Feb 6 '12 at 14:15 You have to write your equation into the form $$z^6=e^{2in\\pi}=1$$", "# Difference between revisions of \"2021 AMC 12A Problems/Problem 22\" ## Problem Suppose that the roots of the polynomial $P(x)=x^3+ax^2+bx+c$ are $\\cos \\frac{2\\pi}7,\\cos \\frac{4\\pi}7,$ and $\\cos \\frac{6\\pi}7$, where angles are in radians. What is $abc$? $\\textbf{(A) }{-}\\frac{3}{49} \\qquad \\textbf{(B) }{-}\\frac{1}{28} \\qquad \\textbf{(C) }\\frac{\\sqrt[3]7}{64} \\qquad \\textbf{(D) }\\frac{1}{32}\\qquad \\textbf{(E) }\\frac{1}{28}$ ## Solution 1 (Complex Numbers: Vieta's Formulas) Let $z=e^{\\frac{2\\pi i}{7}}.$ Since $z$ is a $7$th root of unity, we have $z^7=1.$ For all integers $k,$ note that $\\cos\\frac{2k\\pi}{7}=\\operatorname{Re}\\left(z^k\\right)=\\operatorname{Re}\\left(z^{-k}\\right)$ and $\\sin\\frac{2k\\pi}{7}=\\operatorname{Im}\\left(z^k\\right)=-\\operatorname{Im}\\left(z^{-k}\\right).$ It follows that \\begin{alignat*}{4} \\cos\\frac{2\\pi}{7} &= \\frac{z+z^{-1}}{2} &&= \\frac{z+z^6}{2}, \\\\ \\cos\\frac{4\\pi}{7} &= \\frac{z^2+z^{-2}}{2} &&= \\frac{z^2+z^5}{2}, \\\\ \\cos\\frac{6\\pi}{7} &= \\frac{z^3+z^{-3}}{2} &&= \\frac{z^3+z^4}{2}. \\end{alignat*} By geometric series, we conclude that $$\\sum_{k=0}^{6}z^k=\\frac{1-1}{1-z}=0.$$ Alternatively, recall that the $7$th roots of unity satisfy the equation $z^7-1=0.$ By Vieta's Formulas, the sum of these seven roots is $0.$ As a result, we get $$\\sum_{k=1}^{6}z^k=-1.$$ Let $(r,s,t)=\\left(\\cos{\\frac{2\\pi}{7}},\\cos{\\frac{4\\pi}{7}},\\cos{\\frac{6\\pi}{7}}\\right).$ By Vieta's Formulas, the answer is \\begin{align*} abc&=[-(r+s+t)](rs+st+tr)(-rst) \\\\ &=(r+s+t)(rs+st+tr)(rst) \\\\ &=\\left(\\frac{\\sum_{k=1}^{6}z^k}{2}\\right)\\left(\\frac{2\\sum_{k=1}^{6}z^k}{4}\\right)\\left(\\frac{1+\\sum_{k=0}^{6}z^k}{8}\\right) \\\\ &=\\frac{1}{32}\\left(\\sum_{k=1}^{6}z^k\\right)\\left(\\sum_{k=1}^{6}z^k\\right)\\left(1+\\sum_{k=0}^{6}z^k\\right) \\\\ &=\\frac{1}{32}(-1)(-1)(1) \\\\ &=\\boxed{\\textbf{(D) }\\frac{1}{32}}. \\end{align*} ~MRENTHUSIASM (inspired by Peeyush Pandaya et al) ## Solution 2 (Complex Numbers: Trigonometric Identities) Let $z=e^{\\frac{2\\pi i}{7}}.$ In Solution 1, we conclude that $\\sum_{k=1}^{6}z^k=-1,$ so $$\\sum_{k=1}^{6}\\operatorname{Re}\\left(z^k\\right)=\\sum_{k=1}^{6}\\cos\\frac{2k\\pi}{7}=-1.$$ Since $\\cos\\theta=\\cos(2\\pi-\\theta)$ holds for all $\\theta,$ this sum becomes \\begin{align*} 2\\left(\\cos\\frac{2\\pi}{7}+\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}\\right)&=-1\\\\ \\cos\\frac{2\\pi}{7}+\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}&=-\\frac12. \\end{align*} Note that $\\theta=\\frac{2\\pi}{7},\\frac{4\\pi}{7},\\frac{6\\pi}{7}$ are roots of $$\\cos\\theta+\\cos(2\\theta)+\\cos(3\\theta)=-\\frac12, \\hspace{15mm} (\\bigstar)$$ as they can be verified either algebraically (by the identity $\\cos\\theta=\\cos(-\\theta)=\\cos(2\\pi-\\theta)$) or", "form $$\\ z=2 \\operatorname{cis}\\left(\\frac{5 \\pi}{3}\\right)$$ in polar form $$\\ x=(1-\\sqrt{3} i)^{1 / 3}$$ $$\\ x=\\left[2 \\operatorname{cis}\\left(\\frac{5 \\pi}{3}\\right)\\right]^{1 / 3}$$ Use De Moivre’s theorem to find the first solution: $$\\ x_{1}=2^{1 / 3} \\operatorname{cis}\\left(\\frac{5 \\pi / 3}{3}\\right) \\text { or } 2^{1 / 3} \\operatorname{cis}\\left(\\frac{5 \\pi}{9}\\right)$$ Leave answer in cis form to find the remaining solutions: n = 3 which means that the 3 solutions are $$\\ \\frac{2 \\pi}{3}$$ radians apart or $$\\ x_{2}=2^{1 / 3} \\operatorname{cis}\\left(\\frac{5 \\pi}{9}+\\frac{2 \\pi}{3}\\right)$$ and $$\\ x_{3}=2^{1 / 3} \\operatorname{cis}\\left(\\frac{5 \\pi}{9}+\\frac{2 \\pi}{3}+\\frac{2 \\pi}{3}\\right)$$ Note It is not necessary to add $$\\ \\frac{2 \\pi}{3}$$ again. Adding $$\\ \\frac{2 \\pi}{3}$$ three times equals $$\\ 2 \\pi$$. That would result in rotating around a full circle and to start where it all began- that is the first solution. The three solutions are: $$\\ \\begin{array}{l} x_{1}=2^{1 / 3} \\operatorname{cis}\\left(\\frac{5 \\pi}{9}\\right) \\\\ x_{2}=2^{1 / 3} \\operatorname{cis}\\left(\\frac{11 \\pi}{9}\\right) \\\\ x_{3}=2^{1 / 3} \\operatorname{cis}\\left(\\frac{17 \\pi}{9}\\right) \\end{array}$$ Each of these solutions, when graphed will be $$\\ \\frac{2 \\pi}{3}$$ apart. Check any one of these solutions to see if the results are confirmed. Checking the second solution: $$\\ \\begin{array}{l} x_{2}=2^{1 / 3} \\operatorname{cis}\\left(\\frac{11 \\pi}{9}\\right) \\\\ =1.260\\left[\\cos \\left(\\frac{11 \\pi}{9}\\right)+i \\sin \\left(\\frac{11 \\pi}{9}\\right)\\right] \\\\ =1.260[-0.766-0.643 i] \\\\ =-0.965-0.810 i \\\\ \\text { Does }(-0.965-0.810 i)^{3} \\text { or }(-0.965-0.810 i)(-0.965-0.810 i)(-0.965-0.810 i) \\\\ =(1-\\sqrt{3} i)", "Complex Roots - Not sure I did this right 1. Mar 27, 2009 Atena Hello. I'm not sure whether I did this right or messed up somewhere, just need to confirm my results...thanks to anybody who bothers answering. 1. The problem statement, all variables and given/known data Find all the roots of $$z^{4}=1-i$$ 2. Relevant equations I guess I should state De Moivre's here... $$(r cis(\\vartheta))^{n}=r^{n} cis (n\\vartheta)$$ 3. The attempt at a solution Firstly I re-wrote $$z^{4}=1-i$$ as $$z^{4}=\\sqrt{2} cis (\\frac{-\\pi}{4})$$ Using De Moivre's, $$z=(2\\frac{1}{2})^{\\frac{1}{4}} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+k2\\pi))$$ $$z=2\\frac{1}{8} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+k2\\pi))$$ I found the four roots letting k=0,1,2,3 $$z=2^\\frac{1}{8} cis (\\frac{-\\pi}{16})$$ $$z=2^\\frac{1}{8} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+2\\pi))=2^\\frac{1}{8} cis (\\frac{7\\pi}{16})$$ $$z=2^\\frac{1}{8} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+4\\pi))=2^\\frac{1}{8} cis (\\frac{15\\pi}{16})$$ $$z=2^\\frac{1}{8} cis (\\frac{1}{4}(\\frac{-\\pi}{4}+6\\pi))=2^\\frac{1}{8} cis (\\frac{23\\pi}{16})$$ 2. Mar 27, 2009 yyat Hi Atena! Your answer is correct. You can check this by taking the fourth powers of the solutions you got (using deMoivre).", "# Tag Info since $$\\sqrt{(x^2-2)^2+(x-3)^2}-\\sqrt{(x^2-1)^2+(x-0)^2}$$ let $$P(x,x^2),A(3,2),B(0,1)$$ so $$|PA|-|PB|\\le |AB|=\\sqrt{10}$$ if and only is $A,P,B$ on a line. The answer is $x=\\operatorname{li}^{-1}(100) \\approx 488.871909852807532.\\;$ My implementation computes $\\operatorname{li}^{-1}(x)$ as the zero of the function $f(z)=\\operatorname{li}(z)-x$ using Halley iterations $$z_{n+1}=z_n-\\frac{f(z_n)}{f'(z_n)}\\left[1 - \\frac{f(z_n)}{f'(z_n)}\\frac{f''(z_n)}{2f'(z_n)}\\right]^{-1}\\\\ =z_n-\\Delta_n \\ln(z_n) \\left[1 + ... 7 A multiple root is a common root of both f(x) and f^{\\prime}(x).$$f^{\\prime}(x)=5x^4-1$$and if you are interested only in real roots (i get this impression that you are) then$$x=\\frac{1}{\\sqrt[4]{5}}$$and further we must have$$f\\left(\\frac{1}{\\sqrt[4]{5}}\\right)=0$$this gives$$\\alpha=\\frac{4}{5\\sqrt[4]{5}}$$6 If$$z = e^z$$then$$-ze^{-z} = -1$$so$$-z = W(-1)$$and thus$$z = - W(-1),$$where W is any branch of the Lambert W function. 5 The fundamental theorem of algebra states that an n^{th} degree polynomial has exactly n roots, provided you count them with their multiplicity. This is equivalent to saying that any polynomial has a root. Because if you divide a polynomial of the n^{th} degree by the monomial x-r (where r is a root), you get an (n-1)^{th} polynomial, which has ... 4 Yes, what you have done is a very good start for the full root calculation. You now need to solve \\alpha^3=1 and \\alpha^3=8. The roots of the second equation are twice the roots of the first. To solve \\alpha^3=1, note that \\alpha^3-1=(\\alpha-1)(\\alpha^2+\\alpha+1). 4 Hint : Rewrite your equation as (z+1)^6-z^6=0 and since z=0 isn't a solution,", "what you had not being an equation because you seem to have a problem with writing exactly what you mean- and it hurt you here. Those four numbers are the fourth roots- you have already solved the problem at this point: $z^4= cis(\\pi)$ and $z = cis(\\frac{\\pi}{4})$, $cis(\\frac{5\\pi}{4})$, $cis(\\frac{9\\pi}{4})$, $cis(\\frac{13\\pi}{4})$ $z^4 = cis(\\frac{\\pi}{16})$, $cis(\\frac{5\\pi}{16})$, $cis(\\frac{9\\pi}{16})$, $cis(\\frac{13\\pi}{16}$ P.S 5. Originally Posted by Unknown008 I would have used the difference of two squares twice: $z^4 - 1 = (z^2+1)(z^2 -1) = (z^2+1)(z+1)(z-1)$ almost: $z^4+ 1= (z^2+ i)(z^2- i)= (z-(\\frac{\\sqrt{2}}{2}+ i\\frac{\\sqrt{2}}{2})(z+(\\frac{\\sqrt{2}}{2}+ i\\frac{\\sqrt{2}}{2}))(z-(\\frac{\\sqrt{2}}{2}- i\\frac{\\sqrt{2}}{2})(z+(\\frac{\\sqrt{2}}{2}- i\\frac{\\sqrt{2}}{2})$ Then, z = 1 and -1. For the other two complex roots, we get: $z^2 = -1$ Which you should be able to solve. 6. Originally Posted by HallsofIvy almost $z^4+ 1= (z^2+ i)(z^2- i)= (z-(\\sqrt{2}{2}+ i\\sqrt{2}{2})(z+(\\sqrt{2}{2}+ i\\sqrt{2}{2})(z-(\\sqrt{2}{2}- i\\sqrt{2}{2})(z+(\\sqrt{2}{2}- i\\sqrt{2}{2})$ Um... I don't understand why you look for $z^4+1$ while it's $z^4-1$... EDIT: Ah, okay. I understand now. 7. yes my bad, its meant to be $z^4+1=0$ srry for the confusion", "$c_1$ and $c_2\\text{,}$ it seems plausible that we could get rid of the square roots and fractions. Because we have initial values for $t(n)\\text{,}$ we are able to solve for $c_1$ and $c_2$ here. Evaluating at $n=0$ and $n=1$ we get \\begin{align*} 3 \\amp = c_1 + c_2\\\\ 8 \\amp = c_1\\frac{3+\\sqrt{5}}{2} + c_2 \\frac{3-\\sqrt{5}}{2}. \\end{align*} A little bit of computation gives \\begin{equation*} c_1 = \\frac{7\\sqrt{5}}{10} + \\frac{3}{2} \\quad\\text{and} \\quad c_2 = -\\frac{7\\sqrt{5}}{10} +\\frac{3}{2} \\end{equation*} so that \\begin{equation*} t(n) = \\left(\\frac{7\\sqrt{5}}{10} + \\frac{3}{2}\\right) \\left(\\frac{3+\\sqrt{5}}{2}\\right)^n+ \\left(-\\frac{7\\sqrt{5}}{10} +\\frac{3}{2}\\right) \\left(\\frac{3-\\sqrt{5}}{2}\\right)^n. \\end{equation*} ###### Example9.11. Find the general solution to the advancement operator equation \\begin{equation*} (A+1)(A-6)(A+4)f = 0. \\end{equation*} Solution By now, you shouldn't be surprised that we immediately make use of the roots of $p(A)$ and have that the solution is \\begin{equation*} f(n) = c_1(-1)^n + c_2 6^n + c_3 (-4)^n. \\end{equation*} By now, you should be able to see most of the pattern for solving homogeneous advancement operator equations. However, the examples we've considered thus far have all had one thing in common: the roots of $p(A)$ were all distinct. Solving advancement operator equations in which this is not the case is not much harder than what we've done so far, but we do need to treat it as a distinct case. ###### Example9.12. Find the general solution of the advancement", "don't want it in italics you do \\operatorname{cis}. – user38268 Mar 18 '12 at 8:04 This way is certainly less direct than the previous solution, but more straightforward. \\begin{eqnarray} z-z^2+z^3-z^4&=&-\\sum_{k=1}^4(-1)^ke^{2ik\\pi/5}=-\\sum_{k=1}^4\\left(-e^{2i\\pi/5}\\right)^k=-\\frac{-e^{2i\\pi/5}-(-e^{2i\\pi/5})^5}{1-(-e^{2i\\pi/5})}\\\\ &=&\\frac{e^{2i\\pi/5}-1}{e^{2i\\pi/5}+1}=\\frac{(e^{2i\\pi/5}-1)(e^{-2i\\pi/5}+1)}{(1+\\cos2\\pi/5)^2+\\sin^22\\pi/5}=\\frac{e^{2i\\pi/5}-e^{-2i\\pi/5}}{2+2\\cos2\\pi/5}\\\\ &=&\\frac{i\\sin2\\pi/5}{1+\\cos2\\pi/5}=\\frac{2i\\sin\\pi/5\\,\\cos\\pi/5}{2\\cos^22\\pi/5}= \\frac{i\\sin\\pi/5}{\\cos\\pi/5}=i\\tan\\pi/5. \\end{eqnarray} So it remains to check that $\\tan\\pi/5=2(\\sin2\\pi/5-\\sin\\pi/5)$. After a simplification, this amounts to check that $$1=2\\cos\\pi/5\\;(2\\cos\\pi/5-1),$$ and this can be seen directly from the fact that $\\cos\\pi/5=(1+\\sqrt5)/4$. -", "###### Taylor Polynomials The $n$th order Taylor polynomial of $f$ centered at $x = a$ is given by \\begin{align*} P_n(x) =\\mathstrut \\amp f(a) + f'(a)(x-a) + \\frac{f''(a)}{2!}(x-a)^2 + \\cdots + \\frac{f^{(n)}(a)}{n!}(x-a)^n\\\\ =\\mathstrut \\amp \\sum_{k=0}^n \\frac{f^{(k)}(a)}{k!}(x-a)^k\\text{.} \\end{align*} This degree $n$ polynomial approximates $f(x)$ near $x=a$ and has the property that $P_n^{(k)}(a) = f^{(k)}(a)$ for $k = 0, 1, \\ldots, n\\text{.}$ ###### Example7.53 Determine the third order Taylor polynomial for $f(x) = e^x\\text{,}$ as well as the general $n$th order Taylor polynomial for $f$ centered at $x=0\\text{.}$ Solution We know that $f'(x) = e^x$ and so $f''(x) = e^x$ and $f'''(x) = e^x\\text{.}$ Thus, \\begin{equation*} f(0) = f'(0) = f''(0) = f'''(0) = 1\\text{.} \\end{equation*} So the third order Taylor polynomial of $f(x) = e^x$ centered at $x=0$ is \\begin{align*} P_3(x) \\amp = f(0) + f'(0)(x-0) + \\frac{f''(0)}{2!}(x-0)^2 + \\frac{f'''(0)}{3!}(x-0)^3\\\\ \\amp = 1 + x + \\frac{x^2}{2} + \\frac{x^3}{6}\\text{.} \\end{align*} In general, for the exponential function $f$ we have $f^{(k)}(x) = e^x$ for every positive integer $k\\text{.}$ Thus, the $k$th term in the $n$th order Taylor polynomial for $f(x)$ centered at $x=0$ is \\begin{equation*} \\frac{f^{(k)}(0)}{k!}(x-0)^k = \\frac{1}{k!}x^k\\text{.} \\end{equation*} Therefore, the $n$th order Taylor polynomial for $f(x) = e^x$ centered at $x=0$ is \\begin{equation*} P_n(x) = 1+x+\\frac{x^2}{2!} + \\cdots + \\frac{1}{n!}x^n = \\sum_{k=0}^n \\frac{x^k}{k!}\\text{.} \\end{equation*} ###### Example7.54 We have just seen", "6 '13 at 13:27 • For what you did to make sense, $\\displaystyle \\Big|\\frac{(1+z)^{3}}{8z^{2}} \\Big| < 1$. From that you can determine an appropriate value for $R$. – Random Variable Oct 8 '13 at 21:20 You have a cubic in the denominator: $$8 z^2-(1-z)^3 = z^3+5 z^2+3 z-1$$ You should be able to see that $z=-1$ is a root of this cubic by inspection. By synthetic division, you will see that $$z^3+5 z^2+3 z-1 = (z+1)(z^2+4 z-1)$$ So we see that the other poles are at $z_{\\pm}=-2\\pm\\sqrt{5}$. Now use the fact that $$\\operatorname*{Res}_{z=z_0} \\frac{8 z}{z^3+5 z^2+3 z-1} = \\frac{8 z_0}{3 z_0^2+10 z_0+3}$$ at pole $z=z_0$. At each pole: $$\\operatorname*{Res}_{z=-1} \\frac{8 z}{z^3+5 z^2+3 z-1} = 2$$ $$\\operatorname*{Res}_{z=\\sqrt{5}-2} \\frac{8 z}{z^3+5 z^2+3 z-1} = \\frac{3}{\\sqrt{5}}-1$$ $$\\operatorname*{Res}_{z=-\\sqrt{5}-2} \\frac{8 z}{z^3+5 z^2+3 z-1} = -\\frac{3}{\\sqrt{5}}-1$$ Whether these poles contribute to the integral, as you say, depends on $R$: $$0 < R < \\sqrt{5}-2 \\implies 0$$ $$\\sqrt{5}-2 < R < 1 \\implies \\frac{3}{\\sqrt{5}}-1$$ $$1 < R < \\sqrt{5}+2 \\implies \\frac{3}{\\sqrt{5}}-1 + 2 = \\frac{3}{\\sqrt{5}}+1$$ $$R > \\sqrt{5} + 2 \\implies \\frac{3}{\\sqrt{5}}-1 + 2 - \\frac{3}{\\sqrt{5}}-1 = 0$$ This is essentially what you say. • Thanks a lot Dear @RonGordon:)) Now, I have a time to study your answer. And I understand without any question on my mind:) – B11b Oct 6 '13 at 17:25", "x2 1 + x2, and B3(x1, x2, x3) = x3 1 + 3 x1x2 + x3. Using these to specify the coefficients ci of the characteristic polynomial of a 2 × 2 matrix yields {displaystyle {begin{aligned}c_{2}=B_{0}=1,\\[4pt]c_{1}={frac {-1}{1!}}B_{1}(s_{1})=-s_{1}=-operatorname {tr} (A),\\[4pt]c_{0}={frac {1}{2!}}B_{2}(s_{1},-1!s_{2})={frac {1}{2}}(s_{1}^{2}-s_{2})={frac {1}{2}}((operatorname {tr} (A))^{2}-operatorname {tr} (A^{2})).end{aligned}}} The coefficient c0 gives the determinant of the 2 × 2 matrix, c1 minus its trace, while its inverse is given by {displaystyle A^{-1}={frac {-1}{det A}}(A+c_{1}I_{2})={frac {-2(A-operatorname {tr} (A)I_{2})}{(operatorname {tr} (A))^{2}-operatorname {tr} (A^{2})}}.} It is apparent from the general formula for cn−k, expressed in terms of Bell polynomials, that the expressions {displaystyle -operatorname {tr} (A)quad {text{and}}quad {tfrac {1}{2}}(operatorname {tr} (A)^{2}-operatorname {tr} (A^{2}))} always give the coefficients cn−1 of λn−1 and cn−2 of λn−2 in the characteristic polynomial of any n × n matrix, respectively. So, for a 3 × 3 matrix A, the statement of the Cayley–Hamilton theorem can also be written as {displaystyle A^{3}-(operatorname {tr} A)A^{2}+{frac {1}{2}}left((operatorname {tr} A)^{2}-operatorname {tr} (A^{2})right)A-det(A)I_{3}=O,} where the right-hand side designates a 3 × 3 matrix with all entries reduced to zero. Likewise, this determinant in the n = 3 case, is now {displaystyle {begin{aligned}det(A)&={frac {1}{3!}}B_{3}(s_{1},-1!s_{2},2!s_{3})={frac {1}{6}}(s_{1}^{3}+3s_{1}(-s_{2})+2s_{3})\\[5pt]&={tfrac {1}{6}}left((operatorname {tr} A)^{3}-3operatorname {tr} (A^{2})(operatorname {tr} A)+2operatorname {tr} (A^{3})right).end{aligned}}} This expression gives the negative of coefficient cn−3 of λn−3 in the general case, as seen below. Similarly, one can", "as a product of two quadratic factors with real co-efficients. 13. ## Re: HSC 2018 MX2 Marathon Originally Posted by fan96 An interesting roots of unity question from the SGS notes. i) Find the fourth roots of $-1$ in the form $a+ib$. ii) Hence write $z^4+1$ as a product of two quadratic factors with real co-efficients. $Roots: cis\\frac{\\pi}{4} (\\frac{1}{\\sqrt2} + \\frac{i}{\\sqrt2}), cis\\frac{-\\pi}{4} (\\frac{1}{\\sqrt2} + \\frac{-i}{\\sqrt2}),$ $cis\\frac{3\\pi}{4} (\\frac{-1}{\\sqrt2} + \\frac{i}{\\sqrt2}), cis\\frac{-3\\pi}{4} (\\frac{-1}{\\sqrt2} + \\frac{-i}{\\sqrt2})$ $Now z^4 + 1 = (z - cis\\frac{\\pi}{4})(z - cis\\frac{-\\pi}{4})(z - cis\\frac{3\\pi}{4})(z - cis\\frac{-3\\pi}{4})$ $\\noindentUsing the fact that z + \\bar z = 2\\mathrm{Re} (z) and that z\\bar z = x^2 + y^2 (and the fact the roots are in conjugate pairs)$ $z^4 + 1 = (z^2 - 2(\\frac{1}{\\sqrt2})z + 1)(z^2 + 2(\\frac{1}{\\sqrt2})z + 1)$ $z^4 + 1 = (z^2 - \\sqrt2 z + 1)(z^2 + \\sqrt2 z + 1)$ 14. ## Re: HSC 2018 MX2 Marathon $\\noindent If P(x) = 2x^4 + 7x^3 + 2x^2 -7x + 2 , solve:$ $\\noindent P(x) = 0 over C without using long division, sum/product of roots etc or the polynomial remainder theorem$ 15. ## Re: HSC 2018 MX2 Marathon Originally Posted by altSwift $\\noindent If P(x) = 2x^4 + 7x^3 + 2x^2 -7x + 2 , solve:$ $\\noindent P(x) = 0 over C without using long division,", "\\left [1- \\frac{(8 z+1)^2}{(8 z+3)^2} – \\frac{(8 z+1)^2}{(8 z+5)^2}+\\frac{(8 z+1)^2}{(8 z+7)^2} \\right ] \\cot{(\\pi z)} \\right ) \\right ]_{z=-1/8} \\\\ &= -\\frac1{128} \\pi^2 \\csc^2{\\frac{\\pi}{8}}\\end{align} The factor of $1/8^2$ comes from the fact that the residue calculation has us multiply $f$ by $(z+1/8)^2$. Similarly, $$\\operatorname*{Res}_{z=-3/8} \\pi \\, \\cot{(\\pi z)} \\, f(z) = \\frac1{128} \\pi^2 \\csc^2{\\frac{3 \\pi}{8}}$$ $$\\operatorname*{Res}_{z=-5/8} \\pi \\, \\cot{(\\pi z)} \\, f(z) = \\frac1{128} \\pi^2 \\csc^2{\\frac{5 \\pi}{8}}$$ $$\\operatorname*{Res}_{z=-7/8} \\pi \\, \\cot{(\\pi z)} \\, f(z) = -\\frac1{128} \\pi^2 \\csc^2{\\frac{7 \\pi}{8}}$$ Using $$\\sin^2{\\frac{\\pi}{8}} = \\sin^2{\\frac{7\\pi}{8}} = \\frac{2-\\sqrt{2}}{4} = \\frac{1/2}{2+\\sqrt{2}}$$ $$\\sin^2{\\frac{3\\pi}{8}} = \\sin^2{\\frac{5\\pi}{8}} = \\frac{2+\\sqrt{2}}{4} = \\frac{1/2}{2-\\sqrt{2}}$$ Then the sum is finally $$\\frac{\\pi^2}{128} 2 \\left (\\frac{2+\\sqrt{2}}{1/2} – \\frac{2-\\sqrt{2}}{1/2} \\right ) = \\frac{\\sqrt{2} \\pi^2}{16}$$" ]

[ "# Problem finding zeros of complex polynomial I'm trying to solve this problem $$z^2 + (\\sqrt{3} + i)|z| \\bar{z}^2 = 0$$ So, I know $|z^2| = |z|^2 = a^2 + b ^2$ and $\\operatorname{Arg}(z^2) = 2 \\operatorname{Arg} (z) - 2k \\pi = 2 \\arctan (\\frac{b}{a} ) - 2 k\\pi$ for a $k \\in \\mathbb{Z}$. Regarding the other term, I know $|(\\sqrt{3} + i)|z| \\bar{z}^2 | = |z|^3 |\\sqrt{3} + i| = 2 |z|^3 = 2(a^2 + b^2)^{3/2}$ and because of de Moivre's theorem, I have $\\operatorname{Arg} [(\\sqrt{3} + i ) |z|\\bar{z}^2] = \\frac{\\pi}{6} + 2 \\operatorname{Arg} (z) - 2Q\\pi$. Using all of this I can rewrite the equation as follows \\begin{align*} &|z|^2 \\Bigl[ \\cos (2 \\operatorname{Arg} (z) - 2k \\pi) + i \\sin (2 \\operatorname{Arg}(z) - 2k \\pi)\\Bigr]\\\\ &\\qquad \\mathop{+} 2|z|^3 \\Biggl[\\cos \\left(\\frac{\\pi}{6} + 2 \\operatorname{Arg} (z) -2Q\\pi\\right) + i \\sin \\left(\\frac{\\pi}{6} + 2 \\operatorname{Arg} (z) -2Q\\pi\\right)\\Biggr] = 0 \\end{align*} Which, assuming $z \\neq 0$, can be simplified as \\begin{align*} &\\cos (2 \\operatorname{Arg} (z) - 2k \\pi) + i \\sin (2 \\operatorname{Arg} (z) - 2k \\pi) \\\\ &\\qquad\\mathop{+} 2 |z|\\Biggl[\\cos \\left(\\frac{\\pi}{6} + 2 \\operatorname{Arg} (z) -2Q \\pi \\right) + i \\sin \\left(\\frac{\\pi}{6} + 2 \\operatorname{Arg} (z) -2Q\\pi\\right)\\Biggr] = 0 \\end{align*} Now, from this I'm not sure how to go on. I tried a few things that got me", "= x + i\\sqrt{1 - x^2}$? (Note in this case to get the polynomial in $z$ I found the stupidly huge closed form of $x$ and then constructed $z$ and found the minimum polynomial.) ${}^\\dagger$By rational angle I mean an angle of the form $\\theta_{\\mathrm{rational}} = \\frac{2\\pi m}{n} \\quad \\mathrm{where} \\quad m,n \\in \\mathbb{Z},\\ n \\neq 0,\\ \\mathrm{and}\\ \\operatorname{GCD}(m, n) = 1$ • Sorry if I misunderstand, but if $z = x+i\\sqrt{1-x^2}$ then $x = (1+z^2)/(2z)$, so \\begin{align} 8x^3 + 8x^2 - 2x - 1 &= 8\\left(\\frac{1+z^2}{2z}\\right)^3 + 8\\left(\\frac{1+z^2}{2z}\\right)^2 - 2\\left(\\frac{1+z^2}{2z}\\right) - 1 \\\\ &= \\frac{z^6 + 2 z^5 + 2 z^4 + 3 z^3 + 2 z^2 + 2 z + 1}{z^3}, \\end{align} and the numerator is the polynomial you want. – Antonio Vargas Nov 19 '16 at 9:01 • @AntonioVargas thank you for your reply. I see you solved for x in terms of z but when I substitute $$x = (1 + z^2)/(2z)$$ into $$8x^3 + 8x^2 - 2x - 1$$ I get $$z^9+3 z^7+2 z^6+3 z^5+4 z^4+2 z^2-z-1$$ which is irreducible. How did you know to leave the $z^3$ denominator in order to get my desired polynomial? The desired root $z \\approx 0.4 + 0.9i$ value isn't even a root of the order 9 polynomial I got. – Brandon Enright Nov 19 '16 at", "## What are $$n$$th roots? Let $$n \\in \\mathbb{N}$$. We say that $$b$$ is an $$n$$th root of $$a$$ if $$b^n = a$$. When $$n = 2$$, $$b$$ is called a square root of $$a$$. When $$n = 3$$, $$b$$ is called a cube root of $$a$$. It is well known that every positive real number has two square roots. For example, the square roots of $$4$$ are $$2$$ and $$-2$$ because $$2^2 = 4$$ and $$(-2)^2 = 4$$. We now turn to finding $$n$$th roots for complex numbers. ## Motivating example Consider the complex numbers $$z = 2\\operatorname{cis}\\frac{\\pi}{6}$$ and $$w = 2\\operatorname{cis}\\frac{5\\pi}{6}$$. By De Moivre's Formula, we get $z^3 = 2^3 \\operatorname{cis}\\left(3\\cdot \\frac{\\pi}{6}\\right) = 8 \\operatorname{cis}\\frac{\\pi}{2}$ and $w^3 = 2^3 \\operatorname{cis}\\left(3\\cdot \\frac{5\\pi}{6}\\right) = 8 \\operatorname{cis}\\frac{5\\pi}{2} = 8 \\operatorname{cis}\\left(2\\pi + \\frac{\\pi}{2}\\right) = 8 \\operatorname{cis}\\frac{\\pi}{2}.$ Even though $$z\\neq w$$, we have $$z^3 = w^3 = 8 \\operatorname{cis} \\frac{\\pi}{2}$$. Hence, $$z$$ and $$w$$ are distinct cube roots of $$8 \\operatorname{cis} \\frac{\\pi}{2}$$. (Check that $$2\\operatorname{cis}\\frac{3\\pi}{2}$$ is also a cube root of $$8 \\operatorname{cis} \\frac{\\pi}{2}$$.) Observe that the argument of $$w$$, which is $$\\frac{5\\pi}{6}$$, is $$\\frac{2\\pi}{3}$$ more than the argument of $$z$$, which is $$\\frac{\\pi}{6}$$. Even if we did not know the actual arguments of $$z$$ and $$w$$, we could tell from this observation that the argument of $$w^3$$ is $$2\\pi$$ more", "# Difference between revisions of \"2021 AMC 12A Problems/Problem 22\" ## Problem Suppose that the roots of the polynomial $P(x)=x^3+ax^2+bx+c$ are $\\cos \\frac{2\\pi}7,\\cos \\frac{4\\pi}7,$ and $\\cos \\frac{6\\pi}7$, where angles are in radians. What is $abc$? $\\textbf{(A) }{-}\\frac{3}{49} \\qquad \\textbf{(B) }{-}\\frac{1}{28} \\qquad \\textbf{(C) }\\frac{\\sqrt[3]7}{64} \\qquad \\textbf{(D) }\\frac{1}{32}\\qquad \\textbf{(E) }\\frac{1}{28}$ ## Solution 1 (Complex Numbers: Vieta's Formulas) Let $z=e^{\\frac{2\\pi i}{7}}.$ Since $z$ is a $7$th root of unity, we have $z^7=1.$ For all integers $k,$ note that $\\cos\\frac{2k\\pi}{7}=\\operatorname{Re}\\left(z^k\\right)=\\operatorname{Re}\\left(z^{-k}\\right)$ and $\\sin\\frac{2k\\pi}{7}=\\operatorname{Im}\\left(z^k\\right)=-\\operatorname{Im}\\left(z^{-k}\\right).$ It follows that \\begin{alignat*}{4} \\cos\\frac{2\\pi}{7} &= \\frac{z+z^{-1}}{2} &&= \\frac{z+z^6}{2}, \\\\ \\cos\\frac{4\\pi}{7} &= \\frac{z^2+z^{-2}}{2} &&= \\frac{z^2+z^5}{2}, \\\\ \\cos\\frac{6\\pi}{7} &= \\frac{z^3+z^{-3}}{2} &&= \\frac{z^3+z^4}{2}. \\end{alignat*} By geometric series, we conclude that $$\\sum_{k=0}^{6}z^k=\\frac{1-1}{1-z}=0.$$ Alternatively, recall that the $7$th roots of unity satisfy the equation $z^7-1=0.$ By Vieta's Formulas, the sum of these seven roots is $0.$ As a result, we get $$\\sum_{k=1}^{6}z^k=-1.$$ Let $(r,s,t)=\\left(\\cos{\\frac{2\\pi}{7}},\\cos{\\frac{4\\pi}{7}},\\cos{\\frac{6\\pi}{7}}\\right).$ By Vieta's Formulas, the answer is \\begin{align*} abc&=[-(r+s+t)](rs+st+tr)(-rst) \\\\ &=(r+s+t)(rs+st+tr)(rst) \\\\ &=\\left(\\frac{\\sum_{k=1}^{6}z^k}{2}\\right)\\left(\\frac{2\\sum_{k=1}^{6}z^k}{4}\\right)\\left(\\frac{1+\\sum_{k=0}^{6}z^k}{8}\\right) \\\\ &=\\frac{1}{32}\\left(\\sum_{k=1}^{6}z^k\\right)\\left(\\sum_{k=1}^{6}z^k\\right)\\left(1+\\sum_{k=0}^{6}z^k\\right) \\\\ &=\\frac{1}{32}(-1)(-1)(1) \\\\ &=\\boxed{\\textbf{(D) }\\frac{1}{32}}. \\end{align*} ~MRENTHUSIASM (inspired by Peeyush Pandaya et al) ## Solution 2 (Complex Numbers: Trigonometric Identities) Let $z=e^{\\frac{2\\pi i}{7}}.$ In Solution 1, we conclude that $\\sum_{k=1}^{6}z^k=-1,$ so $$\\sum_{k=1}^{6}\\operatorname{Re}\\left(z^k\\right)=\\sum_{k=1}^{6}\\cos\\frac{2k\\pi}{7}=-1.$$ Since $\\cos\\theta=\\cos(2\\pi-\\theta)$ holds for all $\\theta,$ this sum becomes \\begin{align*} 2\\left(\\cos\\frac{2\\pi}{7}+\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}\\right)&=-1\\\\ \\cos\\frac{2\\pi}{7}+\\cos\\frac{4\\pi}{7}+\\cos\\frac{6\\pi}{7}&=-\\frac12. \\end{align*} Note that $\\theta=\\frac{2\\pi}{7},\\frac{4\\pi}{7},\\frac{6\\pi}{7}$ are roots of $$\\cos\\theta+\\cos(2\\theta)+\\cos(3\\theta)=-\\frac12, \\hspace{15mm} (\\bigstar)$$ as they can be verified either algebraically (by the identity $\\cos\\theta=\\cos(-\\theta)=\\cos(2\\pi-\\theta)$) or", "} )$ is the linking number of $K _ { 0 }$. In particular, $\\mathsf{P} _ { K } ( 1,0 ) = a _ { 2 }$ (the second coefficient of the Alexander–Conway polynomial and the first non-trivial Vassiliev–Gusarov invariant). More generally, consider $P _ { L } ( v , z )$ as an element of the ring $R$ that is the subring of $\\mathbf{Z} [ v ^ { \\pm 1 } , z ^ { \\pm 1 } ]$ generated by $v \\pm 1$, $z$ and $( v ^ { - 1 } - v ) / z$. Then $P _ { L } ( v , z ) - P _ { T _ { \\text{com} ( L ) }} ( v , z )$ is a multiple of $z ( ( ( v ^ { - 1 } - v ) / z ) ^ { 2 } - 1 )$. Furthermore, \\begin{equation*} \\frac { P _ { L } ( v , z ) - P _ { T_{\\operatorname{ com } ( L )}} ( v , z ) } { z \\left( \\left( \\frac { v ^ { - 1 } - v } { z } \\right) ^ { 2 } - 1 \\right) } \\equiv \\end{equation*} \\begin{equation*} \\equiv -", "$-4I$ is a fourth power, $-4I=\\begin{bmatrix}1 & -1 \\\\1 & 1\\end{bmatrix}^4$, but $-I$ is not. Nov 6, 2020 at 0:03 Here's a necessary condition. Write the symmetric polynomial $$(x+y)^k - x^k - y^k$$ as a polynomial in the elementary symmetric polynomials $$x+y$$ and $$xy$$, say $$(x+y)^k - x^k - y^k = F_k(x+y,xy).$$ Then a necessary condition for $$A\\in\\operatorname{SL}_2(\\mathbb Z)$$ to be a $$k$$th power in $$\\operatorname{SL}_2(\\mathbb Z)$$ is that the following two conditions hold: 1. $$\\det A$$ is the $$k$$'th power of an integer, say $$\\det A=D^k$$. 2. The polynomial $$T^k - F_k(T,D) - \\operatorname{Trace}(A)$$ has a root in $$\\mathbb Z$$. The proof is easy enough, since if $$B^k=A$$, then $$\\operatorname{Trace}(B)$$ is an integer root of the polynomial. In particular, for the case $$k=8$$, we have $$F_k(u,v) = 8u^6v-20u^4v^2+16u^2v^3-2v^4.$$ So every matrix satisfying $$B^8=A$$ has the property that $$\\operatorname{Trace}(B)$$ is a root of the following polynomial, where $$D^k=\\det(A)$$, $$T^8 - 8DT^6 + 20D^2T^4 -16 D^3T^2+2D^4 - \\operatorname{Trace}(A).$$ In particular, if $$B$$ is required to have integer entries, then this polynomal has an integer root. Here is the necessary and sufficient condition, in terms of $$\\det A$$ and $$\\text{tr}(A)$$, in order that a $$2\\times2$$ matrix $$A$$ be the $$k$$-th power of some matrix with integer coefficients. Edit.Edit, 11.11.2020 The proof.The proof, is essentially routine; yet I post everything for", "# Chapter 8 Exercise C Exercise 1 Because $$4 = \\operatorname{dim} \\mathbb{C}^4 = \\operatorname{dim} G(3, T) + \\operatorname{dim} G(5, T) + \\operatorname{dim} G(8, T),$$ it follows that the multiplicities of the eigenvalues of $T$ are at most $2$. Thus $p(z) = (z-3)^2(z-5)^2(z-8)^2$ is a polynomial multiple of the characteristic polynomial. Therefore, we must have $p(T) = 0$. Exercise 2 We have $$1 \\le \\dim G(5, T), \\dim G(6, T) \\le n – 1.$$ Hence $p(z) = (z – 5I)^{n-1}(z – 6I)^{n-1}$ is a polynomial multiple of the characteristic polynomial. Therefore, $p(T) = 0$. Exercise 3 Define $T \\in \\mathcal{L}(\\mathbb{C}^4)$ by $$T(z_1, z_2, z_3, z_4) = (7z_1, 7z_2, 8z_3, 8z_4).$$ Since $E(7, T) \\subset G(7, T), E(8, T) \\subset G(8, T)$ and $\\mathbb{C}^4 = E(7, T) \\oplus E(8, T)$, it follows that $E(7, T) = G(7, T)$ and $E(8, T) = G(8, T)$. We have $\\dim E(7, T) = \\dim E(8, T) = 2$. Hence, the characteristic polynomial of $T$ is $$(z – 7)^2(z – 8)^2.$$ Exercise 4 $$A = \\begin{pmatrix} 0 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 0\\\\ 0 & 0 & 0 & 0 \\end{pmatrix}$$ Let $$A = \\begin{pmatrix} 1 & 1 & 0 & 0\\\\ 0 & 5 & 1 & 0\\\\ 0", "6 '13 at 13:27 • For what you did to make sense, $\\displaystyle \\Big|\\frac{(1+z)^{3}}{8z^{2}} \\Big| < 1$. From that you can determine an appropriate value for $R$. – Random Variable Oct 8 '13 at 21:20 You have a cubic in the denominator: $$8 z^2-(1-z)^3 = z^3+5 z^2+3 z-1$$ You should be able to see that $z=-1$ is a root of this cubic by inspection. By synthetic division, you will see that $$z^3+5 z^2+3 z-1 = (z+1)(z^2+4 z-1)$$ So we see that the other poles are at $z_{\\pm}=-2\\pm\\sqrt{5}$. Now use the fact that $$\\operatorname*{Res}_{z=z_0} \\frac{8 z}{z^3+5 z^2+3 z-1} = \\frac{8 z_0}{3 z_0^2+10 z_0+3}$$ at pole $z=z_0$. At each pole: $$\\operatorname*{Res}_{z=-1} \\frac{8 z}{z^3+5 z^2+3 z-1} = 2$$ $$\\operatorname*{Res}_{z=\\sqrt{5}-2} \\frac{8 z}{z^3+5 z^2+3 z-1} = \\frac{3}{\\sqrt{5}}-1$$ $$\\operatorname*{Res}_{z=-\\sqrt{5}-2} \\frac{8 z}{z^3+5 z^2+3 z-1} = -\\frac{3}{\\sqrt{5}}-1$$ Whether these poles contribute to the integral, as you say, depends on $R$: $$0 < R < \\sqrt{5}-2 \\implies 0$$ $$\\sqrt{5}-2 < R < 1 \\implies \\frac{3}{\\sqrt{5}}-1$$ $$1 < R < \\sqrt{5}+2 \\implies \\frac{3}{\\sqrt{5}}-1 + 2 = \\frac{3}{\\sqrt{5}}+1$$ $$R > \\sqrt{5} + 2 \\implies \\frac{3}{\\sqrt{5}}-1 + 2 - \\frac{3}{\\sqrt{5}}-1 = 0$$ This is essentially what you say. • Thanks a lot Dear @RonGordon:)) Now, I have a time to study your answer. And I understand without any question on my mind:) – B11b Oct 6 '13 at 17:25", "represented by: \\begin{align*} D(z) &= z + z^2 + z^3 + z^4 + z^5 + z^6 \\\\ &= z (1 + z) (1 - z + z^2) (1 + z + z^2) \\end{align*} Throwing two dice gives the distribution: $$\\begin{equation*} D^2(z) = z^2 (1 + z)^2 (1 - z + z^2)^2 (1 + z + z^2)^2 \\end{equation*}$$ So we want two dice, call them $$D_1(z)$$ and $$D_2(z)$$, such that $$D^2(z) = D_1(z) \\cdot D_2(z)$$. The respective polynomials have to satisfy several conditions: • We want that all faces have some eyes, i.e., the constant term of the polynomial has to be zero. We need to assign a $$z$$ factor to each. • The coefficients have to be integers (number of faces with each number of eyes). It is a fun fact that if a polynomial with integer coefficients factors, the primitive factors have integer coefficients (from Gauss' lemma). So this isn't a restriction. • The number of faces of the dice has to be six. This is the sum of the coefficients in the polynomial, which is just $$D_i(1)$$. The respective factors at $$z = 1$$ are $$1$$, $$2$$, $$1$$ and $$3$$. We have to give a $$1 + z$$ and a $$1 + z + z^2$$ to each, we can shuffle the $$1 - z + z^2$$", "computed as $\\boxed{36}$. ### Solution 3 (bashing) We first calculate that $z^4 = -1$. After a bit of calculation for the other even powers of $z$, we realize that they cancel out add up to zero. Now we can simplify the expression to $(z^{1^2} + z^{3^2} + ... + z^{11^2})(\\frac{1}{z^{1^2}} + \\frac{1}{z^{3^2}} + ... + \\frac{1}{z^{11^2}})$. Then, we calculate the first few odd powers of $z$. We notice that $z^1 = z^9$, so the values cycle after every 8th power. Since all of the odd squares are a multiple of $8$ away from each other, $z^1 = z^9 = z^{25} = ... = z^{121}$, so $z^{1^2} + z^{3^2} + ... + z^{11^2} = 6z^{1^2}$, and $\\frac{1}{z^{1^2}} + \\frac{1}{z^{3^2}} + ... + \\frac{1}{z^{11^2}} = \\frac{6}{z^{1^2}}$. When multiplied together, we get $6 * 6 = \\boxed{\\textbf{(C) } 36}$ as our answer. ~ Baolan", "y^{-1/6}$. Then for large $y>0$ (i.e. small $z>0$) the roots of $p(X,y)$ are $$r_1, r_2 = \\pm z^{-3} - z^6 \\mp z^9 \\pm \\frac32 z^{15} + 5 z^{18} + O(z^{21}),$$ $$r_3 = z^{-4} + \\frac13 z^{-2} - \\frac1{81} z^2 + \\frac{82}{243} z^4 + \\frac23 z^6 + \\frac{5585}{6561} z^8 + O(z^{10}),$$ and complex conjugate roots $r_4,r_5$ obtained from $r_3$ by multiplying $z$ by a cube root of unity. Let's warm up by showing that for large $y$ there are no integral roots, i.e. none of $r_1,r_2,r_3$ can be an integer (we needn't worry about $r_4,r_5$ because they're not even real). For $x=r_1,r_2$ it is enough to observe that $$x^2 - y = \\mp 2 z^3 - 2 z^6 + 4 z^{12} \\pm 12 z^{15} + 12 z^{18} + O(z^{21}),$$ which for large $y$ is a small nonzero real number and thus cannot be an integer. For $x=r_3$ we use the $11$ monomials in $x,y$ that have a pole of order at most $9$ in $z^2$, and find the polynomial $$y^3 + (3-2x) y^2 + (-x^3-3x^2+4x-4) y + 3x^4-3x^3-x^2-7x+4$$ with power series $$4 z^4 - \\frac43 z^6 - \\frac{104}{9} z^8 - \\frac{980}{81} z^{10} - \\frac{32}{3} z^{12} + O(z^{14}),$$ which again is nonzero but small for large $y$. For a quadratic factor the candidates are $(X-r_1)(X-r_2)$, $(X-r_1)(X-r_3)$, $(X-r_2)(X-r_3)$, and $(X-r_4)(X-r_5)$. The", "# Simple formula for integer polynomial with $2\\sin(2\\pi/n)$ as a root? Is there a simple formula an integer polynomial that $2\\sin(2\\pi/n)$ satisfies? For $2\\cos(2\\pi/n)$ the answer is relatively nice. For any given $n$, we have $2\\cos(2\\pi/n)= z + z^{-1}$ where $z = e^{2 \\pi i/n}$ satisfies a cyclotomic polynomial of degree $\\varphi(n) = 2k$, $$0 = a_{2k}z^{2k} + a_{2k-1}z^{2k-1} + \\ldots + a_{1}z + a_0,$$ where $a_{i} = a_{2k-i}$. Dividing by $\\zeta^k$ gives $$0 = a_{2k}z^k + \\ldots + a_k + a_0z^{-k}$$ Using the symmetry of the coefficients lets us write this as $$0 = a_0(z^k+z^{-k}) + \\ldots + a_k.$$ Then $$(z+z^{-1})^2 = z^2 + z^{-2} + \\binom{2}{1}$$ $$(z+z^{-1})^3 = z^3 + z^{-3} + \\binom{3}{1}(z+z^{-1})$$ $$(z+z^{-1})^4 = z^4 + z^{-4} + \\binom{4}{1}(z^2+z^{-2}) + \\binom{4}{2}$$ and so on, and in the end we get something fairly nice. What happens with $2\\sin(2\\pi/n)$? EDIT: I am aware that $2\\sin(2\\pi/n) = -i(z - z^{-1})$. EDIT: Arturo's comment made me realize that dividing by $(-z)^k$ or $(iz)^k$ may be the way to go. - Doesn't a similar argument hold, using the fact that $2\\sin(\\pi/n) = iz^{-1}-iz$, with $z$ satisfying a cyclotomic polynomial? Use the fact that $\\zeta_n=e^{2\\pi i/n} = \\cos(2\\pi/n) + i\\sin(2\\pi i/n)$, where $\\zeta_n$ is a primitive $n$-th root of unity, so $\\zeta_n - \\zeta_n^{-1} = 2i\\sin(2\\pi i/n)$. – Arturo Magidin Apr", "4z + 3z^2)^2, \\end{align*} keeping in mind that you are looking for a perfect square. Alternatively, you can posit a form of the common MGF to $X$ and $Y$, observing from the powers of $e^t$ that it needs to be $$M_X(t) = M_Y(t) = az^{-1} + b + cz$$ for suitable constants $a, b, c$, then equating coefficients of $M_X (t)^2$. • To factor every polynomial $$p(z)=a+bz+cz^2+bz^3+az^4$$ use the change of variable $$u=z+\\frac1z$$ and note that $$p(z)=z^2\\,\\left(a(z^2+1/z^2)+b(z+1/z)+c\\right)=z^2\\,\\left(a(u^2-2)+bu+c\\right)$$ then one can factor the quadratics in $u$, leading to $$p(z)=az^2(u-\\alpha)(u-\\beta)=a(z^2-\\alpha z+1)(z^2-\\beta z+1)$$ and now one must simply factor the quadratics $z^2-\\alpha z+1$ and $z^2-\\beta z+1$. If $$(a,b,c)=(9,24,34)$$ then $$a(u^2-2)+bu+c=9u^2+24u+16=(3u+4)^2$$ hence, indeed, as stated above, $$p(z)=(3z^2+4z+3)^2$$ – Did Aug 27 '16 at 10:04", "much work to check by hand, so we’ll do something else instead: another nice property of cyclotomic polynomials is that they only have integer coefficients, so we only need to look for $c,d$ such that $4\\frac{c}{d}$ is integer. The only nontrivial cases are $c=1,d=2$ and $c=3,d=4$, which tells us that the only beautiful cosines that are square roots of rationals are $\\pm\\sqrt{1/2}$ and $\\pm\\sqrt{3/4}$. The angles that give these cosines are the well-known $\\pi/4,7\\pi/4,3\\pi/4,5\\pi/4$ and $\\pi/6,11\\pi/6,5\\pi/6,5\\pi/6$. To finalize let’s consider cosines of the form $\\frac{a}{b}+\\sqrt{\\frac{c}{d}}$ for $a \\neq 0$ and $c \\neq 0$. We need the minimal polynomial for the hideous thing $z = \\frac{a}{b}+\\sqrt{\\frac{c}{d}}+i\\sqrt{1-\\left(\\frac{a}{b}+\\sqrt{\\frac{c}{d}}\\right)^2}.$ The previous case does not work, but if you think about it, it is easy to see that with $w = \\frac{a}{b}-\\sqrt{\\frac{c}{d}}+i\\sqrt{1-\\left(\\frac{a}{b}-\\sqrt{\\frac{c}{d}}\\right)^2}$ the polynomial $(x-z)(x-z^*)(x-w)(x-w^*)$ will do the job. Working it out with Mathematica, because that’s already above what I can do by hand, we get $x^4 – 4\\frac{a}{b}x^3 + \\left(2+4\\frac{a^2}{b^2}-4\\frac{c}{d}\\right)x^2 – 4\\frac{a}{b}x + 1.$ At this point it’s better to just look up what are the cyclotomic polynomials for $n=5,8,10,12$, which as we said before are the only ones with degree 4. They are \\begin{align*} \\Phi_5(x) &= x^4 + x^3 + x^2 + x +1 \\\\ \\Phi_8(x) &= x^4 + 1 \\\\ \\Phi_{10}(x) &= x^4 – x^3 + x^2 – x +", "+ 1.$ Calculating which integers have exactly four coprimes, like before, we see that they are 5, 8, 10, and 12, which is way too much work to check by hand, so we’ll do something else instead: another nice property of cyclotomic polynomials is that they only have integer coefficients, so we only need to look for $c,d$ such that $4\\frac{c}{d}$ is integer. The only nontrivial cases are $c=1,d=2$ and $c=3,d=4$, which tells us that the only beautiful cosines that are square roots of rationals are $\\pm\\sqrt{1/2}$ and $\\pm\\sqrt{3/4}$. The angles that give these cosines are the well-known $\\pi/4,7\\pi/4,3\\pi/4,5\\pi/4$ and $\\pi/6,11\\pi/6,5\\pi/6,5\\pi/6$. To finalize let’s consider cosines of the form $\\frac{a}{b}+\\sqrt{\\frac{c}{d}}$ for $a \\neq 0$ and $c \\neq 0$. We need the minimal polynomial for the hideous thing $z = \\frac{a}{b}+\\sqrt{\\frac{c}{d}}+i\\sqrt{1-\\left(\\frac{a}{b}+\\sqrt{\\frac{c}{d}}\\right)^2}.$ The previous case does not work, but if you think about it, it is easy to see that with $w = \\frac{a}{b}-\\sqrt{\\frac{c}{d}}+i\\sqrt{1-\\left(\\frac{a}{b}-\\sqrt{\\frac{c}{d}}\\right)^2}$ the polynomial $(x-z)(x-z^*)(x-w)(x-w^*)$ will do the job. Working it out with Mathematica, because that’s already above what I can do by hand, we get $x^4 – 4\\frac{a}{b}x^3 + \\left(2+4\\frac{a^2}{b^2}-4\\frac{c}{d}\\right)x^2 – 4\\frac{a}{b}x + 1.$ At this point it’s better to just look up what are the cyclotomic polynomials for $n=5,8,10,12$, which as we said before are the only ones with degree 4. They are \\begin{align*} \\Phi_5(x) &=", "-1 prove that$ $\\: z^6+1 = (z^2+1) (z^2-\\sqrt{3}z+1) (z^2+\\sqrt{3}z+1)$ $By using your usual method to find roots of unity you would get$ $z = i, -i, cis$ $(\\frac{\\pi}{6})$ $cis$ $(\\frac{-\\pi}{6})$ $cis$ $(\\frac{5\\pi}{6})$ $cis$ $(-\\frac{5\\pi}{6})$ $Then by using your usual linear factorisation of polynomials via their roots you would get$ $\\: z^6+1 = (z-i) (z+i) (z- cis$ $(\\frac{\\pi}{6}))$ $(z-cis$ $(\\frac{-\\pi}{6}))$ $(z-cis$ $(\\frac{5\\pi}{6}))$ $(z-cis$ $(\\frac{-5\\pi}{6}))$ $Now by expanding the pairs and simplifying you would get$ $(z^2+1) (z^2 -2cos$ $(\\frac{\\pi}{6})$ $z +1)$ $(z^2 -2cos$ $(\\frac{5\\pi}{6})$ $z +1)$ $Notice how the last two brackets above are in the form of the formula you mentioned, that is$ $z^2 -Sum of Roots + Product of Roots.$ $I think when you add stuff like the cis and its conjugate you$ $get the double of the real part which is essentially the 2 cos$ $\\frac{\\pi}{6}$ $and when you multiply (product) the the conjugate of two cis's you get 1 provided that that there is no coefficient in front of it$ $Therefore, from the equation above you just evaluate it and you should get$ $\\: z^6+1 = (z^2+1) (z^2-\\sqrt{3}z+1) (z^2+\\sqrt{3}z+1)$ Idk if that made any sense but hope that helps. This was the first time I used LaTeX so the input is killing me. Let me know if you need more help. There are currently 1 users", "? \\end{array}$$ Example 5 What are the two square roots of i? Solution Let $$\\ z=\\sqrt{0+i}$$. $$\\ r=1, \\theta=\\pi / 2 \\text { or } z=\\left[1 \\times \\operatorname{cis} \\frac{\\pi}{2}\\right]^{1 / 2}$$ Utilizing De Moivre’s theorem: $$\\ z_{1}=\\left[1 \\times \\operatorname{cis} \\frac{\\pi}{4}\\right] \\text { or } z_{2}=\\left[1 \\times \\operatorname{cis} \\frac{5 \\pi}{4}\\right]$$ $$\\ z_{1}=1\\left(\\cos \\frac{\\pi}{4}+i \\sin \\frac{\\pi}{4}\\right) \\text { or } z_{2}=1\\left(\\cos \\frac{5 \\pi}{4}+i \\sin \\frac{5 \\pi}{4}\\right)$$ $$\\ z_{1}=0.707+0.707 i \\text { or } z_{2}=-0.707-0.707 i$$ Check for z1 solution: $$\\ (0.707+0.707 i)^{2}=i ?$$ $$\\ 0.500+0.500 i+0.500 i+0.500 i^{2}=0.500+i+0.500(-1) \\text { or } i$$ Example 6 Calculate $$\\ \\sqrt[4]{(1+0 i)}$$. What are the four fourth roots of 1? Solution Let $$\\ z=1 \\text { or } z=1+0 i$$. Then the problem becomes find $$\\ z^{1 / 4}=(1+0 i)^{1 / 4}$$. Since $$\\ r=1 \\theta=0, z^{1 / 4}=[1 \\times \\operatorname{cis} 0]^{1 / 4}$$ with $$\\ z_{1}=1^{1 / 4}\\left(\\cos \\frac{0}{4}+i \\sin \\frac{0}{4}\\right)$$ or $$\\ 1(1+0)$$ or $$\\ 1$$ That root is not a surprise. Now use De Moivre’s to find the other roots: $$\\ z_{2}=1^{1 / 4}\\left[\\cos \\left(0+\\frac{\\pi}{2}\\right)+i \\sin \\left(0+\\frac{\\pi}{2}\\right)\\right]$$ Since there are 4 roots, dividing $$\\ 2 \\pi$$ by 4 yields $$\\ 0.5 \\pi$$ or $$\\ 0+i$$ or just $$\\ i$$ $$\\ z_{3}=1^{1 / 4}\\left[\\cos \\left(0+\\frac{2 \\pi}{2}\\right)+i \\sin \\left(0+\\frac{2 \\pi}{2}\\right)\\right]$$ which yields $$\\ z_{3}=-1$$ Finally, $$\\ z_{4}=1^{1 / 4}\\left[\\cos \\left(0+\\frac{3 \\pi}{2}\\right)+i \\sin", "equation $P(z) = 0$ by $z^{2}$, and since I had my own alternative method for proving the roots aren't real, I wanted to deny the reader the option of solving P(z) = 0 so as to make the question more of a challenge. Your method is kinda sneaky in that it circumvents the restriction imposed by the question while simultaneously utilizing the property of reducing P(z) to a quadratic in (z+1/z). Also, thanks for giving me the correct terminology for a polynomial with symmetric coefficients. I know what a palindromic word and a palindromic number is, but for some reason I didn't realize the same name applies for a polynomial haha 20. ## Re: Polynomials question \\begin{align*} P(z) &= 2z^{4}+5z^{3}+7z^{2}+5z+2 \\\\ &= 2z^{4}+5z^{3}+\\frac{7}{2}z^{2}+\\frac{7}{2}z^{2}+5z +2 \\\\ &= z^{2}(2z^{2}+5z+\\frac{7}{2})+(\\frac{7}{2}z^{2}+5z+ 2) \\end{align*} \\linebreak \\text{Both quadratics enclosed in brackets have positive leading coefficient and discriminant equal to} \\\\ \\bigtriangleup = -3<0,\\text{ so both are positive definite and hence so is } P. $\\text{QED}$ There are currently 1 users browsing this thread. (0 members and 1 guests) #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •", "# prime ideal decomposition in quadratic extensions of $\\mathbb{Q}$ Let $K$ be a quadratic number field, i.e. $K=\\mathbb{Q}(\\sqrt{d})$ for some square-free integer $d$. The discriminant of the extension is $D_{K/\\mathbb{Q}}=\\begin{cases}d,&\\text{ if }d\\equiv 1\\ \\operatorname{mod}\\ 4,% \\\\ 4d,&\\text{ if }d\\equiv 2,3\\operatorname{mod}\\ 4.\\end{cases}$ Let $\\mathcal{O}_{K}$ denote the ring of integers of $K$. We have: $\\mathcal{O}_{K}\\cong\\begin{cases}\\mathbb{Z}\\oplus\\frac{1+\\sqrt{d}}{2}\\mathbb{Z% },&\\text{ if }d\\equiv 1\\ \\operatorname{mod}\\ 4,\\\\ \\mathbb{Z}\\oplus\\sqrt{d}\\mathbb{Z},&\\text{ if }d\\equiv 2,3\\operatorname{mod}\\ % 4.\\end{cases}$ Prime ideals of $\\mathbb{Z}$ decompose as follows in $\\mathcal{O}_{K}$: ###### Theorem 1. Let $p\\in\\mathbb{Z}$ be a prime. 1. 1. If $p\\mid d$ (divides), then $p\\mathcal{O}_{K}=(p,\\sqrt{d})^{2}$; 2. 2. If $d$ is odd, then $2\\mathcal{O}_{K}=\\begin{cases}(2,1+\\sqrt{d})^{2},&\\text{ if }d\\equiv 3\\ % \\operatorname{mod}\\ 4,\\\\ \\left(2,\\frac{1+\\sqrt{d}}{2}\\right)\\left(2,\\frac{1-\\sqrt{d}}{2}\\right),&\\text{% if }d\\equiv 1\\ \\operatorname{mod}\\ 8,\\\\ \\text{prime},&\\text{ if }d\\equiv 5\\ \\operatorname{mod}\\ 8.\\end{cases}$ 3. 3. If $p\\neq 2$, $p$ does not divide $d$, then $p\\mathcal{O}_{K}=\\begin{cases}(p,n+\\sqrt{d})(p,n-\\sqrt{d}),&\\text{ if }d\\equiv n% ^{2}\\ \\operatorname{mod}\\ p,\\\\ \\text{prime},&\\text{ if d is not a square }\\operatorname{mod}\\ p.\\end{cases}$ ## References Title prime ideal decomposition in quadratic extensions of $\\mathbb{Q}$ PrimeIdealDecompositionInQuadraticExtensionsOfmathbbQ 2013-03-22 13:53:46 2013-03-22 13:53:46 alozano (2414) alozano (2414) 7 alozano (2414) Theorem msc 11R11 CalculatingTheSplittingOfPrimes ExamplesOfPrimeIdealDecompositionInNumberFields PrimeIdealDecompositionInCyclotomicExtensionsOfMathbbQ NumberField SplittingAndRamificationInNumberFieldsAndGaloisExtensions", "n}-1 &=(z+1)(z-1) \\prod_{k=1}^{n-1}\\left(z^{2}-2 z \\cos \\left(\\frac{k}{n} \\pi\\right)+1\\right) \\\\ z^{2 n+1}-1 &=(z-1) \\prod_{k=1}^{n}\\left(z^{2}-2 z \\cos \\left(\\frac{2 k}{2 n+1} \\pi\\right)+1\\right) \\end{aligned} Using the third identity, decompose the polynomial $p(z)=\\sum_{k=0}^{n-1} z^{2 k}$ into irreducible factors and prove the identity $$\\prod_{k=1}^{n-1} \\sin \\left(\\frac{k \\pi}{2 n}\\right)=\\frac{\\sqrt{n}}{2^{n-1}}$$ Still using the third identity, prove (see [98, p. 44]) that $$\\frac{\\sin n t}{\\sin t}=2^{n-1} \\prod_{k=1}^{n-1}\\left(\\cos t-\\cos \\frac{k \\pi}{n}\\right) .$$ Proof . The idea behind the four factorizations is that the polynomials are real, and hence their non-real roots appear in pairs, which lead to second-degree real polynomials: $$\\left(z-z_{0}\\right)\\left(z-\\overline{z_{0}}\\right)=z^{2}-2\\left(\\operatorname{Re} z_{0}\\right) z+\\left|z_{0}\\right|^{2}$$ We focus on the first and third equalities, and leave to the reader the proofs of the other two. The roots of the polynomial $z^{2 n}+1$ are $z_{k}=e^{i \\theta_{k}},$ with $$\\theta_{k}=\\frac{\\pi}{2 n}+\\frac{k \\pi}{n}, \\quad k=0, \\ldots, 2 n-1$$ The roots corresponding to $k=0, \\ldots, n-1$ are not conjugate to each other; indeed a pair of indices $\\left(k, k^{\\prime}\\right)$ corresponds to conjugate roots if $$\\frac{\\pi}{2 n}+\\frac{k \\pi}{n}=-\\frac{\\pi}{2 n}-\\frac{k^{\\prime} \\pi}{n} \\quad(\\bmod 2 \\pi)$$ that is $$\\frac{1}{n}+\\frac{k+k^{\\prime}}{n}=0 \\quad(\\bmod 2)$$ which cannot hold if both $k$ and $k^{\\prime}$ are between 0 and $n-1$. Thus $$z^{2 n}+1=\\prod_{k=0}^{2 n-1}\\left(z-z_{k}\\right)=\\prod_{k=0}^{n-1}\\left(z-z_{k}\\right)\\left(z-\\overline{z_{k}}\\right)$$ But $$\\left(z-z_{k}\\right)\\left(z-\\overline{z_{k}}\\right)=z^{2}-2 z \\cos \\theta_{k}+1$$ which concludes the proof of the first equality since $$\\cos \\theta_{k}=\\cos \\left(\\frac{(2 k+1) \\pi}{2 n}\\right)$$ We now prove the third equality. The roots of order $2 n$ of" ]

[ "Cscθ= Secθ= Cotθ= 10. ### Math Question Given tan θ = -8/5 and sin θ < 0, find sin θ, cos θ, csc θ, sec θ, and cot θ. More Similar Questions", "x \\tan x+\\cot x$$ 13. $$\\dfrac{1}{1−\\sin x}−\\dfrac{1}{1+\\sin x}=2\\tan x \\sec x$$ 14. $$\\csc^4 x−\\cot^4 x=\\csc ^2 x+\\cot^2x$$ 15. $$(\\sin x−\\tan x)(\\cos x−\\cot x)=(\\sin x−1)(\\cos x−1)$$", "help us here... ${\\sin}^{2} x + {\\cos}^{2} x = 1$ ${\\tan}^{2} x + 1 = {\\sec}^{2} x$ $\\underline{1 + {\\cot}^{2} x = {\\csc}^{2} x}$ I've underlined that last one because it applies to our problem. To see how, simply subtract ${\\cot}^{2} x$ in that identity: $1 = {\\csc}^{2} x - {\\cot}^{2} x$ And if this equation is supposed to hold for all values of $x$, it should hold for $x = 55$. Therefore: $\\sec 35 \\csc 55 - \\tan 35 \\cot 55 = {\\csc}^{2} 55 - {\\cot}^{2} 55 = 1$", "$$\\sin t, \\sec t, \\csc t, \\tan t, \\cot t$$. Answer If $$\\sin t=−\\dfrac{2\\sqrt{2}}{3}, \\sec t=−3, \\csc t=−\\csc t=−\\dfrac{3\\sqrt{2}}{4},\\tan t=2\\sqrt{2}, \\cot t= \\dfrac{\\sqrt{2}}{4}$$ 40) If $$\\tan t=\\dfrac{12}{5},$$ and $$0≤t< \\dfrac{π}{2}$$, find $$\\sin t, \\cos t, \\sec t, \\csc t,$$ and $$\\cot t$$. 41) If $$\\sin t= \\dfrac{\\sqrt{3}}{2}$$ and $$\\cos t=\\dfrac{1}{2},$$ find $$\\sec t, \\csc t, \\tan t,$$ and $$\\cot t$$. Answer $$\\sec t=2, \\csc t=\\csc t=\\dfrac{2\\sqrt{3}}{3}, \\tan t= \\sqrt{3}, \\cot t= \\dfrac{\\sqrt{3}}{3}$$ 42) If $$\\sin 40°≈0.643 \\; \\cos 40°≈0.766 \\; \\sec 40°,\\csc 40°,\\tan 40°, \\text{ and } \\cot 40°$$. 43) If $$\\sin t= \\dfrac{\\sqrt{2}}{2},$$ what is the $$\\sin (−t)$$? Answer $$−\\dfrac{\\sqrt{2}}{2}$$ 44) If $$\\cos t= \\dfrac{1}{2},$$ what is the $$\\cos (−t)$$? 45) If $$\\sec t=3.1,$$ what is the $$\\sec (−t)$$? Answer $$3.1$$ 46) If $$\\csc t=0.34,$$ what is the $$\\csc (−t)$$? 47) If $$\\tan t=−1.4,$$ what is the $$\\tan (−t)$$? Answer $$1.4$$ 48) If $$\\cot t=9.23,$$ what is the $$\\cot (−t)$$? ### Graphical For the exercises 49-51, use the angle in the unit circle to find the value of the each of the six trigonometric functions. 49) Answer $$\\sin t= \\dfrac{\\sqrt{2}}{2}, \\cos t= \\dfrac{\\sqrt{2}}{2}, \\tan t=1,\\cot t=1,\\sec t= \\sqrt{2}, \\csc t= \\csc t= \\sqrt{2}$$ 50) 51) Answer $$\\sin t=−\\dfrac{\\sqrt{3}}{2}, \\cos t=−\\dfrac{1}{2}, \\tan t=\\sqrt{3}, \\cot t= \\dfrac{\\sqrt{3}}{3}, \\sec t=−2, \\csc t=−\\csc t=−\\dfrac{2\\sqrt{3}}{3}$$ ### Technology For the exercises 52-61, use", "# If tanx=-2/9 sinx>0, how do you find the sin, cos, sec, csc, and cot? Apr 16, 2015 If $\\tan \\left(x\\right)$ is negative and $\\sin \\left(x\\right)$ is positive, $x$ must be in Quadrant 2 and evaluation is only a matter of reading the graph. Assuming $x \\epsilon \\left[0 , 2 \\pi\\right]$ $\\sin \\left(x\\right) = \\frac{2}{\\sqrt{85}}$ #cos(x) = - 9/sqrt(85) $\\csc \\left(x\\right) = \\frac{\\sqrt{85}}{2}$ $\\sec \\left(x\\right) = - \\frac{\\sqrt{85}}{9}$ $\\cot \\left(x\\right) = - \\frac{9}{2}$", "Suppose that $$\\sec x+\\tan x=\\frac{22}7$$ and that $$\\csc x+\\cot x=\\frac mn,$$ where $$\\frac mn$$ is in lowest terms. Find $$m+n^{}_{}.$$ (第九届AIME1991 第9题)", "13 } +cos\\frac { 5\\pi }{ 13 }=0$ . 1. If cosec A + sec A = cosec B + sec B, prove that cot$\\left( \\frac { A+B }{ 2 } \\right)$=tanA tanB 2. If y = 500 e7x + 600 e-7x then show that y2 - 49y = 0.", "^{2} x-\\cot ^{2} x\\right\\} d x=-3 \\cos x-4 \\sin x+$ $6 \\tan x+7 \\cot x+c$", "x - sin x ∴ 1 + sin x = cos x, sin x = cos x - 1 ∴ sec x + tan x = cot x - csc x = 1 cot x - csc x + 2 csc x = sec x + tan x + 2 csc x 2 csc x =? -2 sec x csc x =? -sec x", "do know the answer to! Take $$\\cos ^2(x) + \\sin ^2(x) = 1$$, divide through by $$\\sin ^2(x)$$ and obtain that $$\\cot ^2(x) + 1 = \\csc ^2(x) \\implies \\cot ^2(x) = \\csc ^2(x) - 1$$. Then,", "- 1)/sin x sin x + sin²x = cos²x - cos x 1 = cos x - sin x ∴ 1 + sin x = cos x, sin x = cos x - 1 ∴ sec x + tan x = cot x - csc x = 1 cot x - csc x + 2 csc x = sec x + tan x + 2 csc x 2 csc x =? -2 sec x csc x =? -sec x", "\\tan x+7 \\cot x+c$ Thus, $\\int\\left\\{3 \\sin x-4 \\cos x+\\frac{5}{\\cos ^{2} x}-\\frac{6}{\\sin ^{2} x}+\\tan ^{2} x-\\cot ^{2} x\\right\\} d x=-3 \\cos x-4 \\sin x+$ $6 \\tan x+7 \\cot x+c$", "## Precalculus (10th Edition) $sin^2(\\theta)+cos^2(\\theta)=1$, $tan^2(\\theta)+1=sec^2(\\theta)$, $cot^2(\\theta)+1=csc^2(\\theta)$. The identities are the following: $sin^2(\\theta)+cos^2(\\theta)=1$, $tan^2(\\theta)+1=sec^2(\\theta)$, $cot^2(\\theta)+1=csc^2(\\theta)$.", "# How do you verify sin x + cos x times cot x = csc x? May 4, 2016 see below #### Explanation: Left Side:$= \\sin x + \\cos x \\times \\cos \\frac{x}{\\sin} x$ $= \\frac{{\\sin}^{2} x + {\\cos}^{2} x}{\\sin} x$ $= \\frac{1}{\\sin} x$ $= \\csc x$ $=$Right Side", "theta sec theta ~ sin theta 2.) Find an expression equivalent to cos theta/sin theta . tan theta cot theta ~ sec theta csc theta 3.) 4. ### Math Trig Let theta be an angle in quad III such that sec theta = -13/5. Find the exact values of cot theta and sin theta? Thank you!!!", "$\\cosh x$ $\\sinh x$ $\\tanh x$ $\\ln( \\cosh x)$ $\\coth x$ $\\ln( \\sinh x)$ $\\text{sech} \\; x$ $2 \\tan^{-1}(e^x)$ $\\text{csch} \\; x$ $-\\ln (\\coth x + \\text{csch}\\; x)$", "x=\\csc2y+\\cot2y=\\dfrac{1+\\cos2y}{\\sin2y}=\\cot y$ $$\\sin\\theta=\\cos2y=\\dfrac{1-\\tan^2y}{1+\\tan^2y}=\\dfrac{\\cot^2y-1}{\\cot^2y+1}=?$$", "= - 0.70710678 = - \\sqrt { 2 } / 2$$ $$\\tan ( 5 / 4 ^ { * } \\pi ) = 1$$ $$\\cot ( 5 / 4 ^ { * } \\pi ) = 1$$ $$\\sec ( 5 / 4 ^ { * } \\pi ) = - 1.41421356 = - \\sqrt { 2 }$$ $$\\csc ( 5 / 4 ^ { * } \\pi ) = - 1.41421356 = - \\sqrt { 2 }$$ c) $$\\cos ( 1 ^ { * } \\pi ) = - 1$$ $$\\sec ( 1 * \\pi ) = - 1$$ $$\\sin ( 1 ^ { * } \\pi ) = 0$$ $$\\tan ( 1 ^ { * } \\pi ) = 0$$ $$\\cot ( 1 ^ { * } \\pi ) =$$ Undefined $$C S ( 1 ^ { * } \\pi ) =$$ Undefined d) $$\\sin ( - 1 / 3 ^ { * } \\pi ) = - 0.8660254 = - \\sqrt { 3 } / 2$$ $$\\cos ( - 1 / 3 ^ { * } \\pi ) = 0.5 = 1 / 2$$ $$\\tan ( - 1 / 3 ^ { * } \\pi ) = - 1.73205081 = - \\sqrt { 3 }$$ $$\\cot ( - 1 / 3 ^ { * } \\pi", "Click on the graph to view a larger graph For the given angle $x$ in the triangle given in the graph $\\sin x=$ ; $\\cos x=$ ; $\\tan x=$ ; $\\cot x=$ ; $\\sec x=$ ; $\\csc x=$ ; Your overall score for this problem is", "# How do you find sin, cos, tan, sec, csc, and cot given (0,6)? Dec 2, 2016 Find values of trig functions #### Explanation: The point (0, 6) is located on the Oy axis. The related arc (angle) is $t = \\frac{\\pi}{2}$. sin t = 1 cos t = 0 tan t = inf. cot t = 0 $\\sec t = \\frac{1}{\\cos} =$ inf. $\\csc t = \\frac{1}{\\sin} = 1$" ]

[ "for noticing this. In fact, I did the right thing, don't know why I wrote 4/3. – Mart Jul 28 '15 at 20:37 • Thanks again. but how do you that sin 2θ = $\\sqrt{16-9x^2}$ – Mart Jul 28 '15 at 20:55 • @Mart I edited the post to address this, we use $\\sin2\\theta=2\\sin\\theta\\cos\\theta$ – user265675 Jul 28 '15 at 21:17 Hint So far so good! The standard technique for handling this integrand is to invoke the double angle identity $$\\cos 2 \\theta = 2 \\cos^2 \\theta - 1.$$ $$\\int\\sqrt{16-9x^2}dx=$$ (Substitute $x=\\frac{4\\sin(u)}{3}$ and $dx=\\frac{4\\cos(u)}{3}du$ then $\\sqrt{16-9x^2}=\\sqrt{16-16\\sin^2(u)}=4\\cos(u)$ and $u=\\sin^{-1}\\left(\\frac{3x}{4}\\right)$): $$\\frac{4}{3}\\int 4\\cos^2(u)du=$$ $$\\frac{16}{3}\\int \\cos^2(u)du=$$ $$\\frac{16}{3}\\int \\left(\\frac{1}{2}\\cos(2u)+\\frac{1}{2}\\right)du=$$ $$\\frac{8}{3}\\int \\cos(2u)du+\\frac{8}{3}\\int 1 du=$$ (Substitute $s=2u$ and $ds=2du$): $$\\frac{4}{3}\\int \\cos(s)ds+\\frac{8}{3}\\int 1 du=$$ $$\\frac{4\\sin(s)}{3}+\\frac{8}{3}\\int 1 du=$$ $$\\frac{4\\sin(s)}{3}+\\frac{8u}{3}+C=$$ $$\\frac{4\\sin(2u)}{3}+\\frac{8u}{3}+C=$$ $$\\frac{8u}{3}+\\frac{8}{3}\\sin(u)\\cos(u)+C=$$ $$\\frac{8u}{3}+\\frac{8}{3}\\sin(u)\\sqrt{1-\\sin^2(u)}+C=$$ $$\\frac{8\\left(\\sin^{-1}\\left(\\frac{3x}{4}\\right)\\right)}{3}+\\frac{8}{3}\\sin\\left(\\left(\\sin^{-1}\\left(\\frac{3x}{4}\\right)\\right)\\right)\\sqrt{1-\\sin^2\\left(\\left(\\sin^{-1}\\left(\\frac{3x}{4}\\right)\\right)\\right)}+C=$$ $$\\frac{1}{2}\\sqrt{16-9x^2}x+\\frac{8}{3}\\sin^{-1}\\left(\\frac{3x}{4}\\right)+C$$ So: $$\\int\\sqrt{16-9x^2}dx=\\frac{1}{2}\\sqrt{16-9x^2}x+\\frac{8}{3}\\sin^{-1}\\left(\\frac{3x}{4}\\right)+C$$ $\\frac{16}{3}\\int(cos^2\\theta)$ = $\\frac{16}{3}\\int\\frac{1-cos2\\theta}{2}$ = $\\frac{16}{6}\\int(1-cos2\\theta)$ = $\\frac{16}{6}(\\int(1) - \\int(cos2\\theta))$ = $\\frac{16}{6}\\theta$ - $\\frac{16}{6}\\int(cos2\\theta)$ = $\\frac{8}{3}\\theta$ - $\\frac{8}{3}\\int(cos2\\theta)$ = $\\frac{8}{3}\\theta$ - $\\frac{8}{3}(\\frac{1}{2})\\int(cos(v))$, substitute $v = 2\\theta$ = $\\frac{8}{3}\\theta$ + $\\frac{4}{3}sin(v)$ = $\\frac{8}{3}\\theta$ + $\\frac{4}{3}sin(2\\theta)$ = $\\frac{8}{3}\\theta$ + $\\frac{4}{3}(2) sin(\\theta) cos (\\theta)$", "2 sin x / 2) then I raised what is in the bracket to the power of 2. and from there I began to doubt what I was doing so you did the following: sin 2x + 2 sin^2(45 - x ) = 2 * sin(x) * cos(x) + 2 * $$\\displaystyle \\left[\\frac{\\sqrt{2}}{2} * cos(x) - \\frac{\\sqrt{2}}{2} * sin(x) \\right]^2$$ After that I, I would factor out the (√2)/2 and realize that [(√2)/2]2 = 1/2 and continue..... #### Kellyanne Smith ##### New member You stopped just where you claim you are having trouble! Please show us your work so we can find your error. Do you know what the right hand side reduces to? so you did the following: sin 2x + 2 sin^2(45 - x ) = 2 * sin(x) * cos(x) + 2 * $$\\displaystyle \\left[\\frac{\\sqrt{2}}{2} * cos(x) - \\frac{\\sqrt{2}}{2} * sin(x) \\right]^2$$ After that I, I would factor out the (√2)/2 and realize that [(√2)/2]2 = 1/2 and continue..... I have continued and now after factoring out the (√2)/2 and realize that [(√2)/2]2 = 1/2 I have 2 sinx cos x + 2 (1/2 cos^2x - 2sinxcosx + 1/2sin^2x) reduced to 2sinx cos x + cos^2 x - 2sinxcosx + sin^2x = sin^2x + cos^2x is this correct? #### Subhotosh Khan ##### Super Moderator Staff", "14. $$\\cos^2A + \\cos^2\\left(\\frac{\\pi}{3} + A\\right) + \\cos^2\\left(\\frac{\\pi}{3} - A\\right)= \\frac{3}{2}$$ 15. $$(1 + \\sec2A)(1 + \\sec2^2A)(1 + \\sec2^3A) \\ldots (1 + \\sec2^nA) = \\frac{\\tan2^nA}{\\tan A}$$ 16. $$\\frac{\\sin2^nA}{\\sin A} = 2^n\\cos A\\cos 2A\\cos 2^2A\\ldots\\cos2^{n - 1}A$$ 17. $$3(\\sin A - \\cos A)^4 + 6(\\sin A + \\cos A)^2 + 4(\\sin^6A + \\cos^6A) = 13$$ 18. $$2(\\sin^6A + \\cos^6A) - 3(\\sin^4A + \\cos^4A) + 1 = 0$$ 19. $$\\cos^2A + \\cos^2(A + B) -2\\cos A\\cos B\\cos(A + B)$$ if independent of $$A.$$ 20. $$\\cos^3A\\cos 3A + \\sin^3A\\sin 3A = \\cos^32A$$ 21. $$\\tan A\\tan(60^\\circ - A)\\tan(60^\\circ + A) = \\tan 3A$$ 22. $$\\sin^2A + \\sin^3\\left(\\frac{2\\pi}{3} + A\\right) + \\sin^3\\left(\\frac{4\\pi}{3} + A\\right) = -\\frac{3}{4}\\sin 3A$$ 23. $$4(\\cos^310^\\circ + \\sin^320^\\circ) = 3(\\cos 10\\circ + \\sin 20^\\circ)$$ 24. $$\\sin A\\cos^3A - \\cos A\\sin^3A = \\frac{1}{4}\\sin 4A$$ 25. $$\\cos^3A\\sin3A + \\sin^3A\\cos 3A = \\frac{3}{4}\\sin 4A$$ 26. $$\\sin A\\sin(60^\\circ + A)\\sin(A + 120^\\circ) = \\sin 3A$$ 27. $$\\cot A + \\cot(60^\\circ + A) + \\cot(120^\\circ + A) = 3\\cot 3A$$ 28. $$\\cos 5A = 16\\cos^5A - 20\\cos^3A + 5\\cos A$$ 29. $$\\sin 5A = 5\\sin A - 20\\sin^3A + 16\\sin^5A$$ 30. $$\\cos 4A - \\cos 4B = 8(\\cos A - \\cos B)(\\cos A + \\cos B)(\\cos A - \\sin B)(\\cos A + \\sin B)$$ 31. $$\\tan 4A = \\frac{4\\tan A - 4\\tan^3A}{1 - 6\\tan^2A +", "think it's worth commenting that the \"trick\" here to get to your right hand expression is to multiply the 2sin(x/2)cos(x/2)/2 term by cos(x/2)/cos(x/2). – Esteban Sep 30 '20 at 14:51 Fairly obvious: $$\\sin^2\\left(\\frac{x}{2}\\right)\\tan\\left(\\frac{x}{2}\\right)=(1-\\cos^2\\left(\\frac{x}{2}\\right))\\tan\\left(\\frac{x}{2}\\right)=\\tan\\left(\\frac{x}{2}\\right)-\\cos\\left(\\frac{x}{2}\\right)\\sin\\left(\\frac{x}{2}\\right)=\\tan\\left(\\frac{x}{2}\\right)-\\frac{\\sin x}{2}$$ • All the tried trig simplifiers beg to differ it was fairly obvious, unfortunately. I like the route you took. – Esteban Sep 30 '20 at 14:54 $$\\frac{\\sin x}{2} = \\sin\\frac{x}{2}\\cos\\frac{x}{2} = \\frac{\\sin\\frac{x}{2}}{\\cos\\frac{x}{2}} \\cos^2\\frac{x}{2} = \\tan\\frac{x}{2} \\left(1 - \\sin^2\\frac{x}{2}\\right)$$", "# Solve $\\cos{(7x)}+\\sin{(3x)}=0$ Solve $\\cos(7x)+\\sin(3x)=0$ I did $\\cos(7x)=\\cos(4x+3x)=\\cos(4x)\\cos(3x)-\\sin(4x)\\sin(3x)$ So the original equation becomes $\\cos(4x)\\cos(3x)-\\sin(4x)\\sin(3x)+\\sin(3x)=0$ But that became long and ugly. What can I do now? $$\\displaystyle \\cos (7x) = -\\sin (3x) = \\sin (-3x) = \\cos \\left(\\frac{\\pi}{2}+3x\\right)$$ So we get $$\\displaystyle \\cos (7x) = \\cos \\left(\\frac{\\pi}{2}+3x\\right)$$ Using $\\displaystyle \\cos x= \\cos \\alpha\\;,$ Then $\\displaystyle x=2n\\pi\\pm \\alpha\\;,$ Where $n\\in \\mathbb{Z}$ So We get $$\\displaystyle 7x=2n\\pi\\pm \\left(\\frac{\\pi}{2}+3x\\right)\\;,$$ Where $n\\in \\mathbb{Z}$ • I believe you mean $n$ instead of $x$ in the final line – Brenton Oct 2 '15 at 18:58 • Sorry Brenton I have edited it. – juantheron Oct 2 '15 at 18:59 • No need to apologize. It's a great solution! +1 – Brenton Oct 2 '15 at 19:00 You have $$\\cos (7x)=-\\sin (3x)=\\sin(-3x)=\\cos (90--3x)$$ Hence, $$7x=\\pm(90+3x)+n.360, n\\epsilon\\mathbb{Z}$$ Well, since no one else has, here's the prosthaphaeresis solution: we have $$\\cos{7x}+\\cos{(\\tfrac{1}{2}\\pi-3x)} = 0,$$ using $\\cos{x}=\\sin{(\\frac{1}{2}\\pi-x)}$. We then have the prosthaphaeresis formula $$\\cos{A}+\\cos{B} = 2\\cos{\\frac{A+B}{2}}\\cos{\\frac{A-B}{2}},$$ which we use to write the equation in the form $$0 = \\cos{\\left( \\frac{7x-3x+\\pi/2}{2} \\right)}\\cos{\\left( \\frac{7x+3x-\\pi/2}{2} \\right)} = \\cos{\\left( 2x+\\frac{\\pi}{4} \\right)}\\cos{\\left( 5x-\\frac{\\pi}{4} \\right)}$$ Therefore the equation is satisfied precisely when one of these factors is zero. $\\cos{A}$ has roots when $A=(n+1/2)\\pi$ for integer $n$, and I'm sure you can take it from here. • Prosthaphaeresis? Whoah! I just call them sum-to-prduct formula – G-man", "\\frac{45}{49} \\quad\\Rightarrow\\quad \\sin A \\:=\\:\\frac{\\sqrt{45}}{7}$ Substitute into [1]: . $\\sin B \\:=\\:\\frac{8\\left(\\frac{\\sqrt{45}}{7}\\right)}{9} \\:=\\:\\frac{8\\sqrt{45}}{63} \\:=\\:\\frac{\\sqrt{2880}}{63}$ (c] Express $\\sin C$ in the form $\\frac{\\sqrt{P}}{3}$ Law of Sines: . $\\frac{\\sin C}{c} \\:=\\:\\frac{\\sin B}{b} \\quad\\Rightarrow\\quad \\sin C \\:=\\:\\frac{c\\sin B}{b}\\;\\;{\\color{blue}[2]}$ We have: . $c = 7,\\;b = x = 8,\\;\\sin B = \\frac{8\\sqrt{5}}{21}$ Substitute into [2]: . $\\sin C \\:=\\:\\frac{7\\left(\\frac{8\\sqrt{5}}{21}\\right)}{8} \\:=\\;\\frac{\\sqrt{5}}{3}$", "Add v1*sin(45) to both sides of (2) and then divide both sides by sin(45) v1 = v2*sin(30)/sin(45) ...(3) Replace v1 in (1) by (3) 17 = v2*sin(30)/sin(45)*cos(45) + v2*cos(30) Factor the right-hand side 17 = v2*[sin(30)/sin(45)*cos(45) + cos(30)] Divide both sides by [sin(30)/sin(45)*cos(45) + cos(30)] to get v2 v2 = 17/[sin(30)/sin(45)*cos(45) + cos(30)] ...(4) Replace v2 in (3) by (4) to get v1 v1 = 17/[sin(30)/sin(45)*cos(45) + cos(30)]*sin(30)/sin(45) $${\\mathtt{v1}} = {\\frac{{\\frac{{\\mathtt{17}}}{\\left[{\\frac{\\underset{\\,\\,\\,\\,^{{360^\\circ}}}{{sin}}{\\left({\\mathtt{30}}^\\circ\\right)}}{\\underset{\\,\\,\\,\\,^{{360^\\circ}}}{{sin}}{\\left({\\mathtt{45}}^\\circ\\right)}}}{\\mathtt{\\,\\times\\,}}\\underset{\\,\\,\\,\\,^{{360^\\circ}}}{{cos}}{\\left({\\mathtt{45}}^\\circ\\right)}{\\mathtt{\\,\\small\\textbf+\\,}}\\underset{\\,\\,\\,\\,^{{360^\\circ}}}{{cos}}{\\left({\\mathtt{30}}^\\circ\\right)}\\right]}}{\\mathtt{\\,\\times\\,}}\\underset{\\,\\,\\,\\,^{{360^\\circ}}}{{sin}}{\\left({\\mathtt{30}}^\\circ\\right)}}{\\underset{\\,\\,\\,\\,^{{360^\\circ}}}{{sin}}{\\left({\\mathtt{45}}^\\circ\\right)}}} \\Rightarrow {\\mathtt{v1}} = {\\mathtt{8.799\\: \\!847\\: \\!533\\: \\!482\\: \\!900\\: \\!7}}$$ $${\\mathtt{v2}} = {\\frac{{\\mathtt{17}}}{\\left[{\\frac{\\underset{\\,\\,\\,\\,^{{360^\\circ}}}{{sin}}{\\left({\\mathtt{30}}^\\circ\\right)}}{\\underset{\\,\\,\\,\\,^{{360^\\circ}}}{{sin}}{\\left({\\mathtt{45}}^\\circ\\right)}}}{\\mathtt{\\,\\times\\,}}\\underset{\\,\\,\\,\\,^{{360^\\circ}}}{{cos}}{\\left({\\mathtt{45}}^\\circ\\right)}{\\mathtt{\\,\\small\\textbf+\\,}}\\underset{\\,\\,\\,\\,^{{360^\\circ}}}{{cos}}{\\left({\\mathtt{30}}^\\circ\\right)}\\right]}} \\Rightarrow {\\mathtt{v2}} = {\\mathtt{12.444\\: \\!863\\: \\!728\\: \\!674\\: \\!910\\: \\!2}}$$ Alan Jul 18, 2014 #8 +1314 0 Thanks. Now I shouldnt get lost. Stu Jul 21, 2014 4 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details", "next? from here $\\L \\left(\\frac{\\sin^2{(2x)}}{4}\\right) \\left(\\frac{1 - \\cos{(2x)}}{2}\\right) =$ to here $\\L \\frac{1}{8} \\left[\\sin^2{(2x)} - \\sin^2{(2x)}\\cos{(2x)}\\right] =$ no big mystery ... I factored out the two constants in the denominator and multiplied (distributed) the numerators ... sin²(2x)[1 - cos(2x)] = sin²(2x) - sin²(2x)cos(2x) .. and thank you.. i was able to figure out the 2nd one! 5. then you would use substitution, u = sin2x and du= 1/2cos2x dx would your du not cancel out the cos 2x in the equation and then elave you with u^2 - u^2?? = 0? 6. $\\L \\frac{1}{8}\\int \\left[\\sin^2{(2x)} - \\sin^2{(2x)}\\cos{(2x)}\\right] dx$ $\\L \\frac{1}{8}\\int \\sin^2{(2x)} dx - \\frac{1}{8}\\int \\sin^2{(2x)}\\cos{(2x)}dx$ $\\L \\frac{1}{8} \\int \\frac{1 - \\cos{(4x)}}{2} dx - \\frac{1}{16}\\int \\sin^2{(2x)} \\cdot 2\\cos{(2x)}dx$ $\\L \\frac{1}{16} \\int 1 - \\cos{(4x)} dx - \\frac{1}{16}\\int u^2 du$ $\\L \\frac{1}{16}\\left[x - \\frac{\\sin{(4x)}}{4} - \\frac{\\sin^3{(2x)}}{3}\\right] + C$ #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •", "so \\begin{align} \\cos^4x\\sin^4x &=\\frac{1}{2^8}(e^{2ix}-e^{-2ix})^4\\\\[6px] &=\\frac{1}{2^8}(e^{8ix}-4e^{4ix}+6-4e^{-4ix}+e^{8ix})\\\\[6px] &=\\frac{1}{128}\\cos8x-\\frac{1}{32}\\cos4x+\\frac{3}{128} \\end{align} we now that $$\\sin 2x=2\\cos x\\sin x$$ $$\\cos x\\sin x=\\frac{\\sin 2x}{2}$$ so... $$\\cos^4 x\\sin^4 x=\\frac{\\sin^4 2x}{16}$$ $$\\frac{\\sin^4 2x}{16}=\\frac{(\\frac{1-\\cos 4x}{2})^2}{16}=\\frac{1-2\\cos 4x+\\cos^24x}{64}=\\frac{1-2\\cos 4x+\\frac{1+\\cos 8x}{2}}{64}$$", "= 45o in (1) we get 45o – B = 30o$\\Rightarrow$ B = 15o Hence, A = 45o and B = 15o Question.26: If 3x = cosec $\\theta$ and $\\frac{3}{x}$ = cot $\\theta$, find the value of $3\\left ( x^{2}-\\frac{1}{x^{2}} \\right )$. Solution: $3(x^{2} – \\frac{1}{x^{2}})$ $= \\frac{9}{3}(x^{2} – \\frac{1}{x^{2}})$ $= \\frac{1}{3}(9x^{2} – \\frac{9}{x^{2}})$ $= \\frac{1}{3}\\left [ (3x)^{2} – \\left ( \\frac{3}{x} \\right )^{2} \\right ]$ $= \\frac{1}{3}\\left [ (cosec \\; \\theta)^{2} – \\left ( \\cot \\theta \\right )^{2} \\right ]$ $= \\frac{1}{3}\\left [ (cosec ^{2}\\; \\theta – \\cot ^{2} \\theta ) \\right ]$ $= \\frac{1}{3} (1)$ = $\\frac{1}{3}$ Question.27: If sin (A + B) = sin A cos B + cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of (i) sin 75o and (ii) cos 15o Solution: Let A = $45^{\\circ}$ and B = $30^{\\circ}$ (i) As, $\\sin (A + B) = \\sin A \\cos B + \\cos A \\sin B$ $\\Rightarrow \\sin (45^{\\circ} + 30 ^{\\circ}) = \\sin 45^{\\circ} \\cos 30 ^{\\circ} + \\cos 45^{\\circ} \\sin 30 ^{\\circ}$ $\\Rightarrow \\sin (75 ^{\\circ}) = \\frac{1}{\\sqrt{2}} \\times \\frac{\\sqrt{3}}{2} + \\frac{1}{\\sqrt{2}} \\times \\frac{1}{2}$ $\\Rightarrow \\sin (75 ^{\\circ}) = \\frac{\\sqrt{3}}{2\\sqrt{2}} + \\frac{1}{2 \\sqrt{2}}$ $Therefore, \\sin (75 ^{\\circ}) = \\frac{\\sqrt{3} + 1}{2\\sqrt{2}}$ (ii) As, $\\cos (A –", "find this : $$\\displaystyle \\frac{12sin(x)^2+\\sqrt{3}*4sin(x)+1}{3\\sqrt{3}}$$ Idk how to further develop. In the identity I factorized $$\\displaystyle (sin+\\frac{1}{\\sqrt{3}})^2$$, found in (a-b) sinx-sinx =0 so there’s $$\\displaystyle \\frac{1}{\\sqrt{3}}$$ left. any help is welcome! Btw people here are so quick to answer this website is bomb, thank you really! Appreciate it Last edited: ##### Well-known member Hello SP3. you do not need $\\sin\\,x$ or $\\cos\\,x$ you are given $\\cos\\,x - \\sin\\,x = \\frac{1}{\\sqrt{3}}$ $\\cos^3 x - \\sin ^3 x = (\\cos\\,x - \\sin\\,x)^3 + 3 (\\cos\\,x - \\sin\\,x)(\\cos\\,x \\sin\\,x)\\cdots(1)$ so you need to evaluate $\\cos\\,x \\sin\\,x$ $\\cos\\,x - \\sin\\,x = \\frac{1}{\\sqrt{3}}$ square both sides to get $1 - 2\\cos\\,x \\sin\\,x = \\frac{1}{3}$ or $2\\cos\\,x \\sin\\,x = \\frac{2}{3}$ or $\\cos\\,x \\sin\\,x = \\frac{1}{3}$ you can put the value of $\\cos\\,x \\sin\\,x = \\frac{1}{3}$ and $\\cos\\,x - \\sin\\,x = \\frac{1}{\\sqrt{3}}$ in (1) to get $\\cos^3 x - \\sin ^3 x = (\\cos\\,x - \\sin\\,x)^3 + 3 (\\cos\\,x - \\sin\\,x)(\\cos\\,x \\sin\\,x)$ $= \\frac{1}{3\\sqrt{3}} + 3 \\frac{1}{\\sqrt{3}} \\frac{1}{3}$ $=\\frac{1+3}{3\\sqrt{3}}$ $=\\frac{4}{3\\sqrt{3}}$", "Prove that $\\tan\\frac{\\pi}{9}+ 4\\sin\\frac{\\pi}{9}= \\sqrt{3}$ . Prove that $$\\tan\\frac{\\pi}{9}+ 4\\sin\\frac{\\pi}{9}= \\sqrt{3}$$ I think the best solution here is using right triangle . . . I have one too, but not pretty. Multiplying by $$\\cos(\\pi/9)$$, this is equivalent to $$\\sin\\frac\\pi 9+2\\sin\\frac{2\\pi}9=\\sqrt3\\cos\\frac\\pi 9.\\tag1$$ But $$\\sqrt3\\cos\\frac\\pi 9-\\sin\\frac\\pi9= 2\\left(\\sin\\frac\\pi3\\cos\\frac\\pi 9-\\cos\\frac\\pi3\\sin\\frac\\pi9\\right) =2\\sin\\left(\\frac\\pi3-\\frac\\pi9\\right)=2\\sin\\frac{2\\pi}9$$ which proves $$(1)$$. If $$\\cos2x\\ne0,$$ $$\\tan2x+4\\sin(30^\\circ-x)$$ $$=\\dfrac{\\sin2x+4\\sin(30^\\circ-x)\\cos(30^\\circ-x)}{\\cos2x}$$ $$=\\dfrac{\\sin2x+2\\sin(60^\\circ-2x)}{\\cos2x}$$ $$=\\dfrac{\\sin2x+\\sqrt3\\cos2x-\\sin2x}{\\cos2x}=?$$ Here $$x=10^\\circ$$ $$(\\tan x+ 4\\sin x)^{2}- 3= - \\frac{1}{4}(2\\cos x+ 1)(2\\cos 3x- 1)\\csc^{2}\\left ( \\frac{\\pi}{4}- \\frac{x}{2} \\right )\\csc^{2}\\left ( \\frac{x}{2}+ \\frac{\\pi}{4} \\right )$$ Therefore $$\\tan\\frac{\\pi}{9}+ 4\\sin\\frac{\\pi}{9}= \\sqrt{3}$$ • .solve: @Eynjuhl – user707745 Sep 26 '19 at 6:03", "19:01 To attack this integral, we will need to make use of the following facts: $$(\\sin{x} + \\cos{x})^2 = 1+\\sin{2x}$$ $$(\\sin{x} - \\cos{x})^2 = 1-\\sin{2x}$$ $$\\text{d}(\\sin{x}+\\cos{x}) = (\\cos{x}-\\sin{x})\\text{d}x$$ $$\\text{d}(\\sin{x}-\\cos{x}) = (\\cos{x}+\\sin{x})\\text{d}x$$ Now, consider the denominator. It can be rewritten in two different ways as hinted by the above information. $$9+16\\sin{2x} = 25 - 16(1-\\sin{2x}) = 16(1+\\sin{2x})-7$$ $$9+16\\sin{2x} = 25 - 16(\\sin{x}-\\cos{x})^2 = 16(\\sin{x}+\\cos{x})^2-7$$ Also note that $$\\text{d}(\\sin{x}-\\cos{x})-\\text{d}(\\sin{x}+\\cos{x}) = 2\\sin{x}\\text{d}x$$ By making the substitutions $$u = \\sin{x}+\\cos{x}, v = \\sin{x}-\\cos{x}$$ The integral is transformed into two separate integrals which can be evaluated independently. $$2I = \\int \\frac{\\text{d}v - \\text{d}u}{9+16\\sin{2x}} = \\int \\frac{\\text{d}v}{25-16v^2} + \\int \\frac{\\text{d}u}{7-16u^2}$$ The remainder of this evaluation is left as an exercise to the reader. HINT: $$\\int\\frac{\\sin(x)}{9+16\\sin(2x)}\\space\\text{d}x=$$ Use the double angle formula $\\sin(2x)=2\\sin(x)\\cos(x)$: $$\\int\\frac{\\sin(x)}{32\\sin(x)\\cos(x)+9}\\space\\text{d}x=$$ Subsitute $u=\\tan\\left(\\frac{x}{2}\\right)$ and $\\text{d}u=\\frac{\\sec^2\\left(\\frac{x}{2}\\right)}{2}\\space\\text{d}x$. Then transform the integrand using the substitutions: $\\sin(x)=\\frac{2u}{u^2+1},\\cos(x)=\\frac{1-u^2}{u^2+1}$ and $\\text{d}x=\\frac{2}{u^2+1}\\space\\text{d}u$: $$\\int\\frac{4u}{\\left(u^2+1\\right)^2\\left(\\frac{64u(1-u^2)}{(u^2+1)^2}+9\\right)}\\space\\text{d}u=$$ $$\\int\\frac{4u}{9 u^4-64 u^3+18 u^2+64 u+9}\\space\\text{d}u$$", "+3A) - \\sin (4A -3A) - [ \\sin ( 2A+A) - \\sin (2A -A) ] }{ \\cos ( 4A - A) - \\cos ( 4A + A) + \\cos ( 6A + A) + \\cos ( 6A - A) }$ $\\displaystyle = \\frac{ \\sin 7A - \\sin A - \\sin 3A + \\sin A }{ \\cos 3A - \\cos 5A + \\cos 7A + \\cos 5A }$ $\\displaystyle = \\frac{ \\sin 7A - \\sin 3A }{ \\cos 3A + \\cos 7A }$ $\\displaystyle = \\frac{2 \\sin \\Big( \\frac{7A-3A}{2} \\Big) \\cos \\Big( \\frac{7A+3A}{2} \\Big)}{2 \\cos \\Big( \\frac{7A+3A}{2} \\Big) \\cos \\Big( \\frac{7A-3A}{2} \\Big)}$ $\\displaystyle = \\frac{2\\sin 2A \\cos 5A}{2\\cos 5A \\cos 2A}$ $\\displaystyle = \\frac{\\sin 2A }{\\cos 2A} = \\tan 2A =$ RHS $\\displaystyle \\text{ix) } \\text{LHS } = \\frac{\\sin A \\sin 2A + \\sin 3A \\sin 6A}{\\sin A \\cos 2A + \\sin 3A \\cos 6A}$ $\\displaystyle = \\frac{2\\sin A \\sin 2A + 2\\sin 3A \\sin 6A}{2\\sin A \\cos 2A + 2\\sin 3A \\cos 6A}$ $\\displaystyle = \\frac{ \\cos ( 2A -A) - \\cos (2A +A) + \\cos ( 6A-3A) - \\cos (6A+3A) }{ \\sin ( 2A + A) - \\sin ( 2A - A) + \\sin ( 6A + 3A) - \\sin ( 6A - 3A) }$ $\\displaystyle = \\frac{ \\cos A - \\cos 3A + \\cos 3A -", "yes, you can rewrite it using $\\sin(2x) = 2\\sin(x)\\cos(x)$ – Berci Oct 15 '12 at 0:57 Sorry for the typo, will update it now – Alex Oct 15 '12 at 0:58 I am not sure how you got the 2 in there? – Jillian Johnson Oct 15 '12 at 1:00 $$(1 - \\sin(x)\\cos(x))' = \\left(1 - \\frac12 \\sin(2x)\\right)' = -\\cos(2x).$$ Implicitly, I used the chain rule here, letting $f(x) = \\sin(x)$ and $g(x) = 2x$. - Where did you get the -(1/2)sin(2x)?? – Jillian Johnson Oct 15 '12 at 1:13 Sorry, I dont know the chain rule. But thank you for helping. – Jillian Johnson Oct 15 '12 at 1:16 By using $\\sin(2x) = 2\\sin(x)\\cos(x).$ – Ian Mateus Oct 15 '12 at 1:29 Got it, thanks! – Jillian Johnson Oct 15 '12 at 1:49 Using the well known trigonometric identity, $$\\sin(2x)=2\\sin(x)\\cos(x)$$ We can say that, $$y=1-\\sin(x)\\cos(x)=1-\\frac{2\\sin(x)\\cos(x)}{2}=1-\\frac{\\sin(2x)}{2}$$ and the derivative would be: $$y'=0-\\frac{2\\cos(2x)}{2}=-\\cos(2x)$$ - Upvoted for greatness. However, it won't help him in the long run because if he's asking this he clearly doesn't know about the product rule. – Damieh Oct 15 '12 at 15:10", "or trigonometric expansions are required. That should be enough of a hint. - No substitution needed, L'Hospital works at $\\pi$ too. – 1015 Feb 4 '13 at 14:03 @julien: You're right, I'm not sure why I suggested that. Thanks for pointing it out. ;-) – Noldorin Feb 4 '13 at 17:53 $$\\lim_{x\\to\\pi}\\frac{\\sin 3x}{\\sin 2x}=\\lim_{y\\to0}\\frac{\\sin(3\\pi+3y)}{\\sin(2\\pi+2y)}=\\lim_{y\\to0}\\frac{-\\sin(3y)}{\\sin(2y)}=-\\lim_{y\\to0}\\frac{\\sin(3y)}{3y}\\frac{2y}{\\sin(2y)}.\\frac{3}{2}=-\\frac{3}{2}$$ If you know L. Hospital's rule, then it is very easy. - Sorry, I did not see your restriction on LH rule. – Tapu Feb 4 '13 at 13:47 $$\\frac{\\sin3x}{\\sin2x}=\\frac{3\\sin x-4\\sin^3x}{2\\sin x\\cos x}=\\frac{3-4\\sin^2x}{2\\cos x} \\text{ if } \\sin x\\ne0$$ As $x\\to\\pi,\\sin x\\to\\sin \\pi\\implies \\sin x\\to 0\\implies \\sin x\\ne0$ So, $$\\lim_{x\\to \\pi}\\frac{\\sin3x}{\\sin2x}=\\lim_{x\\to \\pi}\\frac{3\\sin x-4\\sin^3x}{2\\sin x\\cos x}=\\lim_{x\\to \\pi}\\frac{3-4\\sin^2x}{2\\cos x}=-\\frac32 \\text{ as }\\cos\\pi=-1,\\sin\\pi=0$$ - This was originally posted (and then deleted) by another user. I fine-tuned it slightly (since it was a decent answer) and posted as a community wiki answer (since it certainly wasn't mine). We have $\\sin (3x)=-\\sin(3x-3\\pi)$ and $\\sin(2x)=\\sin (2x-2\\pi)$. So $$\\frac{\\sin(3x)}{\\sin(2x)}=-\\frac{\\sin(3(x-\\pi))}{\\sin(2(x-\\pi))}=-\\frac{\\sin (3u)}{\\sin (2u)}$$ where $u=x-\\pi$ now tends to $0$. Now use the famous $$\\lim_{y\\rightarrow 0}\\frac{\\sin y}{y}=1$$ and the fact that $$-\\frac{\\sin (3u)}{\\sin (2u)}=-\\frac{3}{2}\\frac{\\sin (3u)}{3u}\\frac{2u}{\\sin(2u)}$$ to get your limit of $-3/2$. -", "+0 # Get rid of fraction 0 36 3 +992 Hi friends....please a last question for today: The sum goes like this: $${{\\sqrt{2} \\over 2}Cos x +Sinx} \\over Sinx$$ becomes $$Sin x +Cos x \\over \\sqrt{2}Sinx$$ I understand the sqrt of 2 going to the bottom, but what happenms to the denominator \"2\"??. Thanks a lot for your patience today. All the best. Untill next time!!.. Mar 3, 2023 #1 +118455 +1 $$\\frac{\\frac{\\sqrt2}{2}cos\\;x\\;+\\;sin\\;x}{sin\\;x}\\\\~\\\\ =\\frac{\\frac{\\sqrt2cosx+2sinx}{2}}{sinx}\\\\ =\\frac{\\sqrt2cosx+2sinx}{2}\\div sinx\\\\ =\\frac{\\sqrt2cosx+2sinx}{2sinx}\\\\ =\\frac{\\frac{\\sqrt2cosx+2sinx}{\\sqrt2}}{\\frac{2sinx}{\\sqrt2}}\\\\ =\\frac{\\frac{\\sqrt2cosx+2sinx}{\\sqrt2}}{\\sqrt2sinx}\\\\ =\\frac{cosx+\\sqrt2sinx}{\\sqrt2sinx}\\\\$$ BUT If I put brackets into you question, this is what I get. $$\\frac{\\frac{\\sqrt2}{2}[cosx+sinx]}{sinx}\\\\ =\\frac{\\sqrt2}{2}[cosx+sinx]\\times\\frac{1}{sinx} \\\\ =\\frac{\\sqrt2}{2}[cosx+sinx]\\times\\frac{1}{sinx} \\times \\frac{\\sqrt2}{\\sqrt2}\\\\ =\\frac{2}{2\\sqrt2}[cosx+sinx]\\times\\frac{1}{sinx} \\\\ =\\frac{1}{\\sqrt2}[cosx+sinx]\\times\\frac{1}{sinx} \\\\ =\\frac{cosx+sinx}{\\sqrt2sinx} \\\\$$ LaTex: \\frac{\\frac{\\sqrt2}{2}cos\\;x\\;+\\;sin\\;x}{sin\\;x}\\\\~\\\\ =\\frac{\\frac{\\sqrt2cosx+2sinx}{2}}{sinx}\\\\ =\\frac{\\sqrt2cosx+2sinx}{2}\\div sinx\\\\ =\\frac{\\sqrt2cosx+2sinx}{2sinx}\\\\ =\\frac{\\frac{\\sqrt2cosx+2sinx}{\\sqrt2}}{\\frac{2sinx}{\\sqrt2}}\\\\ =\\frac{\\frac{\\sqrt2cosx+2sinx}{\\sqrt2}}{\\sqrt2sinx}\\\\ =\\frac{cosx+\\sqrt2sinx}{\\sqrt2sinx}\\\\ Mar 4, 2023 #2 +992 +1 Melody, once again, ..Thank you. So it's all about rasionalizing the numerator. I could do it like this:? $${{\\sqrt2 \\over2}(SinxCosx)} \\over Sinx$$ $${{{\\sqrt2 \\over2}(SinxCosx)} \\over Sinx}*{ \\sqrt2 \\over \\sqrt2}$$ which really means: $${{{\\sqrt2* {\\sqrt2} \\over2}(SinxCosx)} \\over \\sqrt2Sinx}$$ which will then also produce the final correct answer? By the way, the brackets were needed...I have forgotten to add them in the original sum...oopsie..sorry.. juriemagic Mar 4, 2023 #3 +118455 +1 Yes that is correct and it is easier than what I did.", "Write the value of sin Question: Write the value of $\\sin \\frac{\\pi}{15} \\sin \\frac{4 \\pi}{15} \\sin \\frac{3 \\pi}{10}$ Solution: $\\frac{\\pi}{15}=12^{\\circ}, \\frac{4 \\pi}{15}=48^{\\circ}, \\frac{3 \\pi}{10}=54^{\\circ}$ $\\sin 12^{\\circ} \\sin 48^{\\circ} \\sin 54^{\\circ}$ $=\\frac{1}{2}\\left[2 \\sin 12^{\\circ} \\sin 48^{\\circ}\\right] \\sin 54^{\\circ}$ $=\\frac{1}{2}\\left[\\cos \\left(12^{\\circ}-48^{\\circ}\\right)-\\cos \\left(12^{\\circ}+48^{\\circ}\\right)\\right] \\sin 54^{\\circ}$ $=\\frac{1}{2}\\left[\\cos \\left(-36^{\\circ}\\right)-\\cos 60^{\\circ}\\right] \\sin 54^{\\circ}$ $=\\frac{1}{2} \\sin 54^{\\circ}\\left[\\cos 36^{\\circ}-\\frac{1}{2}\\right]$ $=\\frac{1}{2}\\left[\\sin \\left(90^{\\circ}-36^{\\circ}\\right) \\cos 36^{\\circ}\\right]-\\frac{1}{4} \\sin \\left(90^{\\circ}-36^{\\circ}\\right)$ $=\\frac{1}{2} \\cos ^{2} 36^{\\circ}-\\frac{1}{4} \\cos 36^{\\circ}$ $=\\frac{1}{2}\\left(\\frac{\\sqrt{5}+1}{4}\\right)^{2}-\\left(\\frac{\\sqrt{5}+1}{16}\\right) \\quad\\left[\\cos 36^{\\circ}=\\frac{\\sqrt{5}+1}{4}\\right]$ $=\\frac{1}{2}\\left(\\frac{5+1+2 \\sqrt{5}}{16}\\right)-\\left(\\frac{\\sqrt{5}+1}{16}\\right)$ $=\\frac{6+2 \\sqrt{5}}{32}-\\frac{\\sqrt{5}+1}{16}$ $=\\frac{6+2 \\sqrt{5}-2 \\sqrt{5}-2}{32}$ $=\\frac{4}{32}$ $=\\frac{1}{8}$", "# How to convert $3\\sin(x)\\cos(x)$ into expression involving only $\\sin (x)$ I need to convert $3\\sin(x)\\cos(x)$ to an expression involving only $\\sin (x)$, but I don't know how. Can you please point me to solutions. Thanks!!! EDIT: I can do something like this: $$3\\sin(x)\\sin(\\frac{\\pi}{2}-x)$$ but how can I convert this so that I have only $\\sin(x)$ ? - $\\sin(2x)=2\\sin x\\cos x$. – David Mitra Feb 8 at 13:19 $\\cos x=\\sqrt {1-\\sin^2 x}$ (care with signs required) – Mark Bennet Feb 8 at 13:20 @DavidMitra Thanks for the answer, but can you explain to me what have you written – depecheSoul Feb 8 at 13:26 OK, I got is $sin(2x)=sin(x+x)=...$. It is adding two – depecheSoul Feb 8 at 13:27 Use the double angle formula for $\\sin(2x)$: $$2\\sin x \\cos x = \\sin(2x)\\;\\text{ and }\\;\\sin x \\cos x = \\frac 12 \\sin(2x)$$ $$3\\sin x \\cos x = 2\\sin x \\cos x + \\sin x \\cos x = \\sin(2x) + \\frac 12 \\sin (2x) = \\dfrac 32 \\sin (2x)$$ I did it a little diffrently $3*\\sin(x)*\\cos(x)=3*(\\frac{2}{2})*\\sin(x)*\\cos(x)=(\\frac{3}{2})*\\sin(2x)$ – depecheSoul Feb 8 at 13:51", "have $u$. - Remark that $cos'(t)=-sin( t)$ and $\\frac{d}{dt}(\\frac{t^2}{18})=\\frac{t}{9}$. Then $$\\frac{d}{dt}(90+45cos(\\frac{t^2}{18}))=\\frac{d}{dt}(90)+\\frac{d}{dt}(45cos(\\frac{t^2}{18}))=0+45\\frac{d}{dt}(cos(\\frac{t^2}{18}))$$ But setting $s=s(t)=\\frac{t^2}{18}$, $$\\frac{d}{dt}(cos(\\frac{t^2}{18}))=\\frac{d}{ds}(cos(s))\\frac{d}{dt}(s(t))=-sin(s).s'(t)=-sin(\\frac{t^2}{18})\\frac{t}{9}.$$ Thus $$\\frac{d}{dt}(90+45cos(\\frac{t^2}{18}))=45(-sin(\\frac{t^2}{18})\\frac{t}{9})=-5t.sin(\\frac{t^2}{18})$$ - You need to see the nesting: $$f(t) = 90 + 45\\cos\\left(\\frac{t^2}{18}\\right) \\\\ g(x) = 90 + 45\\cos(x) \\rightarrow g\\left(x = \\frac{t^2}{18}\\right) = f(t) = 90 + 45\\cos\\left(\\frac{t^2}{18}\\right) \\\\ g'(x) = 90 - 45\\sin(x) \\rightarrow g'\\left(x = \\frac{t^2}{18}\\right) = 90 - 45\\sin\\left(\\frac{t^2}{18}\\right) \\\\ x'(t) = \\frac{d}{dt}\\left(\\frac{t^2}{18}\\right) = \\frac{t}{9} \\\\ \\left.\\left.\\frac{d}{dt}\\right(f(t)\\right) = \\left.\\left.\\frac{d}{dt}\\right (g\\left(x(t)\\right)\\right) = \\frac{dx}{dt}\\cdot\\left(\\frac{dg}{dx} \\circ x(t)\\right) \\\\ \\frac{df}{dt} = \\frac{t}{9}\\left(90 - 45\\sin\\left(\\frac{t^2}{18}\\right)\\right)$$ -" ]

[ "1- cosx 1 = sinx / sinx 1=1 i hope it'll help... • Nov 29th 2008, 08:51 AM Soroban Hello, Chouman! Quote: $1)\\;\\text{Prove: }\\;\\frac{\\cos x-\\sin x +1}{\\cos x+ \\sin x-1} \\:=\\:\\frac{\\sin x}{1-\\cos x}$ . . Ignore their hint Multiply top and bottom by $(\\cos x + \\sin x - 1)$ . . $\\frac{\\cos x - \\sin x + 1}{\\cos x + \\sin x - 1}\\cdot {\\color{blue}\\frac{\\cos x + \\sin x - 1}{\\cos x + \\sin x - 1}}$ Numerator: . $\\cos^2\\!x + {\\color{red}\\rlap{/////////}}\\sin x\\cos x - {\\color{green}\\rlap{/////}}\\cos x - {\\color{red}\\rlap{/////////}}\\sin x\\cos x - \\sin^2x + \\sin x + {\\color{green}\\rlap{/////}}\\cos x + \\sin x - 1$ . . . $= \\;2\\sin x - \\sin^2\\!x - (1 - \\cos^2\\!x)$ . . . $=\\;2\\sin x - \\sin^2\\!x - \\sin^2\\!x$ . . . $=\\;2\\sin x - 2\\sin^2\\!x$ . . . $= \\;2\\sin x(1 - \\sin x)$ Denominator: . $\\cos^2\\!x + \\sin x\\cos x - \\cos x + \\sin x\\cos x + \\sin^2\\!x - \\sin x - \\cos x - \\sin x + 1$ . . . $= \\;\\underbrace{\\cos^2\\!x + \\sin^2\\!x}_{\\text{This is 1}} + 1 - 2\\sin x - 2\\cos x + 2\\sin x\\cos x$ . . . $= \\;2 - 2\\sin x - 2\\cos x + 2\\sin x\\cos x$ . . . $= \\;2(1-\\sin x) - 2\\cos x(1 - \\sin x)$ .", "2 sin x / 2) then I raised what is in the bracket to the power of 2. and from there I began to doubt what I was doing so you did the following: sin 2x + 2 sin^2(45 - x ) = 2 * sin(x) * cos(x) + 2 * $$\\displaystyle \\left[\\frac{\\sqrt{2}}{2} * cos(x) - \\frac{\\sqrt{2}}{2} * sin(x) \\right]^2$$ After that I, I would factor out the (√2)/2 and realize that [(√2)/2]2 = 1/2 and continue..... #### Kellyanne Smith ##### New member You stopped just where you claim you are having trouble! Please show us your work so we can find your error. Do you know what the right hand side reduces to? so you did the following: sin 2x + 2 sin^2(45 - x ) = 2 * sin(x) * cos(x) + 2 * $$\\displaystyle \\left[\\frac{\\sqrt{2}}{2} * cos(x) - \\frac{\\sqrt{2}}{2} * sin(x) \\right]^2$$ After that I, I would factor out the (√2)/2 and realize that [(√2)/2]2 = 1/2 and continue..... I have continued and now after factoring out the (√2)/2 and realize that [(√2)/2]2 = 1/2 I have 2 sinx cos x + 2 (1/2 cos^2x - 2sinxcosx + 1/2sin^2x) reduced to 2sinx cos x + cos^2 x - 2sinxcosx + sin^2x = sin^2x + cos^2x is this correct? #### Subhotosh Khan ##### Super Moderator Staff", "# Integrate $\\int\\frac{1}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x}dx$ Solve the indefinite integral $$I=\\int\\frac{1}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x}\\;dx$$ My Attempt: \\begin{align} I&=\\int\\frac{1}{\\sin x+\\cos x+\\frac{1}{\\sin x \\cos x}+\\frac{\\sin x +\\cos x}{\\sin x\\cos x}}\\;dx\\\\ \\\\ &=\\int\\frac{\\sin x\\cos x}{\\left(\\sin x+\\cos x\\right)\\left(\\sin x\\cos x \\right)+1+\\left(\\sin x+\\cos x\\right)}\\;dx \\end{align} How can I complete the solution from here? • Do you have any reason to think it has a nice answer? – Greg Martin Sep 18 '13 at 5:14 • Yeah... Good luck – Ali Caglayan Dec 25 '13 at 23:45 You want \\begin{align} \\int\\frac{dx}{\\sin x+\\cos x+\\tan x+\\cot x+\\csc x+\\sec x} &= \\int\\frac{dx}{\\sin x+\\cos x+\\frac{\\sin x}{\\cos x} +\\frac{\\cos x}{\\sin x}+\\frac1{\\sin x}+\\frac1{\\cos x}}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{\\sin^2 x \\cos x+\\cos^2 x \\sin x+\\sin^2 x +\\cos^2 x+\\cos x+\\sin x}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{\\sin x \\cos x(\\sin x+ \\cos x)+1+\\cos x+\\sin x}\\\\ &= \\int\\frac{\\sin x \\cos x\\ dx}{(\\sin x \\cos x+1)(\\sin x+ \\cos x)+1}\\\\ \\end{align}. Applying substitutions $\\sin x = \\frac{2t}{1+t^2}, \\cos x = \\frac{1-t^2}{1+t^2}, dx = \\frac{2 dt}{1+t^2}$, this becomes \\begin{align} \\int\\frac{\\sin x \\cos x\\ dx}{(\\sin x \\cos x+1)(\\sin x+ \\cos x)+1} &=\\int\\frac{\\frac{(2t)(1-t^2)(2dt)}{(1+t^2)^3}} {(\\frac{(2t)(1-t^2)}{(1+t^2)^2}+1)(\\frac{2t}{1+t^2}+ \\frac{1-t^2}{1+t^2})+1}\\\\ &=\\int\\frac{(2t)(1-t^2)(2dt)} {((2t)(1-t^2)+(1+t^2)^2)(1+2t-t^2)+(1+t^2)^3}\\\\ &=\\int\\frac{4t(1-t^2)dt} {((2t-2t^3)+1+2t^2+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\\\ &=\\int\\frac{4t(1-t^2)dt} {(1+2t+2t^2-2t^3+t^4)(1+2t-t^2)+1+3t^2+3t^4+t^6}\\\\ &=\\int\\frac{4t(1-t^2)dt} {2 (t+1) (2 t^4-3 t^3+3 t^2+t+1)} \\quad \\text{ (according to Wolfram Alpha)}\\\\ &=\\int\\frac{2t(1-t)dt} {2 t^4-3 t^3+3 t^2+t+1}\\\\ \\end{align}. According to Wolfram Alpha, that quartic can be factored as", "for noticing this. In fact, I did the right thing, don't know why I wrote 4/3. – Mart Jul 28 '15 at 20:37 • Thanks again. but how do you that sin 2θ = $\\sqrt{16-9x^2}$ – Mart Jul 28 '15 at 20:55 • @Mart I edited the post to address this, we use $\\sin2\\theta=2\\sin\\theta\\cos\\theta$ – user265675 Jul 28 '15 at 21:17 Hint So far so good! The standard technique for handling this integrand is to invoke the double angle identity $$\\cos 2 \\theta = 2 \\cos^2 \\theta - 1.$$ $$\\int\\sqrt{16-9x^2}dx=$$ (Substitute $x=\\frac{4\\sin(u)}{3}$ and $dx=\\frac{4\\cos(u)}{3}du$ then $\\sqrt{16-9x^2}=\\sqrt{16-16\\sin^2(u)}=4\\cos(u)$ and $u=\\sin^{-1}\\left(\\frac{3x}{4}\\right)$): $$\\frac{4}{3}\\int 4\\cos^2(u)du=$$ $$\\frac{16}{3}\\int \\cos^2(u)du=$$ $$\\frac{16}{3}\\int \\left(\\frac{1}{2}\\cos(2u)+\\frac{1}{2}\\right)du=$$ $$\\frac{8}{3}\\int \\cos(2u)du+\\frac{8}{3}\\int 1 du=$$ (Substitute $s=2u$ and $ds=2du$): $$\\frac{4}{3}\\int \\cos(s)ds+\\frac{8}{3}\\int 1 du=$$ $$\\frac{4\\sin(s)}{3}+\\frac{8}{3}\\int 1 du=$$ $$\\frac{4\\sin(s)}{3}+\\frac{8u}{3}+C=$$ $$\\frac{4\\sin(2u)}{3}+\\frac{8u}{3}+C=$$ $$\\frac{8u}{3}+\\frac{8}{3}\\sin(u)\\cos(u)+C=$$ $$\\frac{8u}{3}+\\frac{8}{3}\\sin(u)\\sqrt{1-\\sin^2(u)}+C=$$ $$\\frac{8\\left(\\sin^{-1}\\left(\\frac{3x}{4}\\right)\\right)}{3}+\\frac{8}{3}\\sin\\left(\\left(\\sin^{-1}\\left(\\frac{3x}{4}\\right)\\right)\\right)\\sqrt{1-\\sin^2\\left(\\left(\\sin^{-1}\\left(\\frac{3x}{4}\\right)\\right)\\right)}+C=$$ $$\\frac{1}{2}\\sqrt{16-9x^2}x+\\frac{8}{3}\\sin^{-1}\\left(\\frac{3x}{4}\\right)+C$$ So: $$\\int\\sqrt{16-9x^2}dx=\\frac{1}{2}\\sqrt{16-9x^2}x+\\frac{8}{3}\\sin^{-1}\\left(\\frac{3x}{4}\\right)+C$$ $\\frac{16}{3}\\int(cos^2\\theta)$ = $\\frac{16}{3}\\int\\frac{1-cos2\\theta}{2}$ = $\\frac{16}{6}\\int(1-cos2\\theta)$ = $\\frac{16}{6}(\\int(1) - \\int(cos2\\theta))$ = $\\frac{16}{6}\\theta$ - $\\frac{16}{6}\\int(cos2\\theta)$ = $\\frac{8}{3}\\theta$ - $\\frac{8}{3}\\int(cos2\\theta)$ = $\\frac{8}{3}\\theta$ - $\\frac{8}{3}(\\frac{1}{2})\\int(cos(v))$, substitute $v = 2\\theta$ = $\\frac{8}{3}\\theta$ + $\\frac{4}{3}sin(v)$ = $\\frac{8}{3}\\theta$ + $\\frac{4}{3}sin(2\\theta)$ = $\\frac{8}{3}\\theta$ + $\\frac{4}{3}(2) sin(\\theta) cos (\\theta)$", "4.5 Prove Trigonometric Identities Chapter Chapter 4 Section 4.5 Solutions 24 Videos Prove that \\cos2x=2\\cos^2x-1. 0.39mins Q1 Prove that \\cos2x=1-2\\sin^2x. 0.43mins Q2 Prove that \\sin(x+\\pi)=-\\sin x. 0.25mins Q3 Prove that \\displaystyle{\\sin\\left(\\frac{3\\pi}{2}-x\\right)=-\\cos x}. 0.44mins Q4 Prove that \\cos(\\pi-x)=-\\cos x. 0.26mins Q5 Prove that \\displaystyle{\\cos\\left(\\frac{3\\pi}{2}+x\\right)=\\sin x}. 0.46mins Q6 Prove that \\cos x=\\sin x\\cot x. 0.23mins Q7 Prove that 1+\\sin x=\\sin x(1+\\csc x). 0.40mins Q8 Prove that 1-2\\cos^2x=\\sin x\\cos x(\\tan x-\\cot x). 1.27mins Q9a Prove that \\csc^2x=1+\\cot^2x. 0.50mins Q10a Prove that \\sec^2x=1+\\tan^2x. 0.48mins Q10b Prove that \\displaystyle{\\frac{1-\\sin^2x}{\\cos x}=\\frac{\\sin2x}{2\\sin x}}. 2.26mins Q11 Prove that \\displaystyle{\\frac{\\csc^2x-1}{\\csc^2x}=1-\\sin^2x} 0.40mins Q12 Prove that \\displaystyle{\\frac{\\csc x}{\\cos x}=\\tan x+\\cot x}. 1.30mins Q13 Prove that 2\\sin x\\sin y=\\cos(x-y)-\\cos(x+y). 0.57mins Q15 Prove that \\sin2x+\\sin2y=2\\sin(x+y)\\cos(x-y). 1.22mins Q16 Use an appropriate compound angle formula to determine to determine an expression for \\sin3x in terms of \\sin x and \\cos x (a) Write 3x as the sum of two terms involving x. (b) Substitute the expression from part (a) into sin 3x. 0.14mins Q19ab Use an appropriate compound angle formula to expand the expression for \\sin(x + 2x). 0.12mins Q19c Use an appropriate compound angle formula to determine to determine an expression for \\sin3x in terms of \\sin x and \\cos x 0.58mins Q19d (a) Use graphing technology to determine whether it is reasonable to conjecture that \\sin^6x+\\cos^6x=1-3\\sin^2x\\cos^2x is an identity. (b) If", "\\frac{\\pi}{4} \\right ) & = \\sin x \\cos \\frac{\\pi}{4} - \\cos x \\sin \\frac{\\pi}{4} = \\frac{\\sqrt{2}}{2} \\sin x - \\frac{\\sqrt{2}}{2} \\cos x$ Set the two sides equal to each other: $\\frac{\\sqrt{3}}{2} \\sin x + \\frac{1}{2} \\cos x & = \\frac{\\sqrt{2}}{2} \\sin x - \\frac{\\sqrt{2}}{2} \\cos x \\\\\\sqrt{3} \\sin x + \\cos x & = \\sqrt{2} \\sin x - \\sqrt{2} \\cos x \\\\\\sqrt{3} \\sin x - \\sqrt{2} \\sin x & = - \\cos x - \\sqrt{2} \\cos x \\\\\\sin x \\left (\\sqrt{3} - \\sqrt{2} \\right ) & = \\cos x \\left (-1 - \\sqrt{2} \\right ) \\\\\\frac{\\sin x}{\\cos x} & = \\frac{-1- \\sqrt{2}}{\\sqrt{3} - \\sqrt{2}} \\\\\\tan x & = \\frac{-1- \\sqrt{2}}{\\sqrt{3} - \\sqrt{2}} \\cdot \\frac{\\sqrt{3} + \\sqrt{2}}{\\sqrt{3} + \\sqrt{2}} \\\\& = \\frac{- \\sqrt{3} - \\sqrt{2} + \\sqrt{6} - 2}{3-2} \\\\& = -2 + \\sqrt{6} - \\sqrt{3} - \\sqrt{2}$ As a decimal, this is $-2.69677$ , so $\\tan^{-1}(-2.69677) = x, x = 290.35^\\circ$ and $110.35^\\circ$ . ### Concept Problem Solution To find $\\sin \\left( \\frac{3\\pi}{2} + \\frac{\\pi}{4} \\right)$ , use the sine sum formula: $\\sin (a + b) = \\sin(a)\\cos(b) + \\cos(a)\\sin(b)\\\\\\sin \\left( \\frac{3\\pi}{2} + \\frac{\\pi}{4} \\right) = \\sin \\left( \\frac{3\\pi}{2} \\right) \\times \\cos \\left( \\frac{\\pi}{4} \\right) + \\cos \\left( \\frac{3\\pi}{2} \\right) \\times \\sin \\left( \\frac{\\pi}{4} \\right)\\\\= (-1)\\left( \\frac{\\sqrt{2}}{2} \\right) + (0)\\left( \\frac{\\sqrt{2}}{2} \\right)\\\\= -\\frac{\\sqrt{2}}{2}\\\\$ ### Practice Prove each identity.", "think it's worth commenting that the \"trick\" here to get to your right hand expression is to multiply the 2sin(x/2)cos(x/2)/2 term by cos(x/2)/cos(x/2). – Esteban Sep 30 '20 at 14:51 Fairly obvious: $$\\sin^2\\left(\\frac{x}{2}\\right)\\tan\\left(\\frac{x}{2}\\right)=(1-\\cos^2\\left(\\frac{x}{2}\\right))\\tan\\left(\\frac{x}{2}\\right)=\\tan\\left(\\frac{x}{2}\\right)-\\cos\\left(\\frac{x}{2}\\right)\\sin\\left(\\frac{x}{2}\\right)=\\tan\\left(\\frac{x}{2}\\right)-\\frac{\\sin x}{2}$$ • All the tried trig simplifiers beg to differ it was fairly obvious, unfortunately. I like the route you took. – Esteban Sep 30 '20 at 14:54 $$\\frac{\\sin x}{2} = \\sin\\frac{x}{2}\\cos\\frac{x}{2} = \\frac{\\sin\\frac{x}{2}}{\\cos\\frac{x}{2}} \\cos^2\\frac{x}{2} = \\tan\\frac{x}{2} \\left(1 - \\sin^2\\frac{x}{2}\\right)$$", "1Siny$\\frac{1}{Sin\\; y}$ = 135=53$\\frac{1}{\\frac{3}{5}} = \\frac{5}{3}$ Since, sin2y+cos2y=1$sin^{2}\\;y \\; + \\; cos^{2}\\;y = 1$ $\\Rightarrow$ cos2y=1sin2y$cos^{2} y = 1 – sin^{2} y$ $\\Rightarrow$ cos2y=1(35)2$cos^{2} y = 1 – \\left ( \\frac{3}{5} \\right )^{2}$ $\\Rightarrow$ cos2y=1925$cos^{2} y = 1 – \\frac{9}{25}$ $\\Rightarrow$ cos2y=1625$cos^{2} y = \\frac{16}{25}$ $\\Rightarrow$ cos y = ±45$\\pm \\frac{4}{5}$ Since, y lies in the 2nd quadrant, the value of cos y will be negative, Therefore, cos y = – 45$\\frac{4}{5}$ $\\Rightarrow$ sec y = 1cosy=1(45)=54$\\frac{1}{cos \\; y} = \\frac{1}{\\left (- \\frac{4}{5} \\right )} =\\; – \\frac{5}{4}$ $\\Rightarrow$ tan y = sinycosy=(35)(45)=34$\\frac{sin \\; y}{cos \\; y} = \\frac{\\left (\\frac{3}{5} \\right ) }{\\left (-\\frac{4}{5} \\right ) } = \\;- \\frac{3}{4}$ $\\Rightarrow$ cot y = 1tany=43$\\frac{1}{tan \\; y} =\\;- \\frac{4}{3}$ Q.3: Find the values of other five trigonometric functions if coty=34$cot \\; y = \\frac{3}{4}$, where y lies in the third quadrant. Sol: cot y = 34$\\frac{3}{4}$ Since, tan y = 1coty=134=43$\\frac{1}{cot \\; y}=\\frac{1}{\\frac{3}{4}}=\\frac{4}{3}$ Since, 1+tan2y=sec2y$1 + tan^{2}y = sec^{2}y$ $\\Rightarrow$ 1+(43)2=sec2y$1 + \\left ( \\frac{4}{3} \\right )^{2} = sec^{2}y$ $\\Rightarrow$ 1+169=sec2y$1 + \\frac{16}{9} = sec^{2}y$ $\\Rightarrow$ 259=sec2y$\\frac{25}{9} = sec^{2}y$ $\\Rightarrow$ sec y = ±53$\\pm \\frac{5}{3}$ Since, y lies in the 3rd quadrant, the value of sec y will be negative. Therefore, sec y = 53$\\;- \\frac{5}{3}$ cos y = 1secy=1(53)=35$\\frac{1}{sec \\; y} = \\frac{1}{\\left (- \\frac{5}{3} \\right )}", ". . $= \\;2(1-\\sin x)(1 - \\cos x)$ And the fraction becomes: . $\\frac{{\\color{green}\\rlap{/}}2\\cdot\\sin x\\cdot({\\color{red}\\rlap{////////}}1 - \\sin x)}{{\\color{green}\\rlap{/}}2\\cdot({\\color{red}\\rlap{////////}}1-\\sin x)\\cdot(1 - \\cos x)} \\;=\\;\\frac{\\sin x}{1 - \\cos x} \\quad\\hdots$ ta-DAA! • Nov 29th 2008, 09:42 AM Soroban Hello again, Chouman! I thinik the second one is even worse ... (Is that possible?) Quote: $2)\\;\\;\\frac{\\cot x}{1-\\tan x} + \\frac{\\tan x}{1-\\cot x} - 1 \\;=\\; \\sec x\\csc x$ The left side is: . $\\frac{\\dfrac{\\cos x}{\\sin x}} {1 - \\dfrac{\\sin x}{\\cos x}} + \\frac{\\dfrac{\\sin x}{\\cos x}}{1-\\dfrac{\\cos x}{\\sin x}} - 1$ Multiply top and bottom by $\\sin x\\cos x$ ... both fractions. . . $\\frac{\\sin x\\cos x\\left(\\dfrac{\\cos x}{\\sin x}\\right)} {\\sin x\\cos x\\left(1 - \\dfrac{\\sin x}{\\cos x}\\right)} + \\frac{\\sin x\\cos x\\left(\\dfrac{\\sin x}{\\cos x}\\right)}{\\sin x\\cos x\\left(1-\\dfrac{\\cos x}{\\sin x}\\right)} \\;-\\; 1$ . . $=\\;\\frac{\\cos^2\\!x}{\\sin x\\cos x - \\sin^2\\!x} + \\frac{\\sin^2\\!x}{\\sin x\\cos x - \\cos^2\\!x} \\:- \\:1$ . . $= \\;\\frac{\\cos^2\\!x}{\\sin x(\\cos x - \\sin x)} + \\frac{\\sin^2\\!x}{\\cos x(\\sin x - \\cos x)} \\:-\\:1$ . . $= \\;\\frac{\\cos^2\\!x}{\\sin x(\\cos x - \\sin x)} \\:{\\color{red}-}\\: \\frac{\\sin^2\\!x}{\\cos x(\\cos x \\:{\\color{red}-}\\: \\sin x)} \\:=\\:1$ . . $= \\;{\\color{blue}\\frac{\\cos x}{\\cos x}}\\cdot\\frac{\\cos^2\\!x}{\\sin x(\\cos x - \\sin x)} - {\\color{blue}\\frac{\\sin x}{\\sin x}}\\cdot\\frac{\\sin^2\\!x}{\\cos x(\\cos x - \\sin x)} \\:-\\:1$ . . $= \\;\\frac{\\overbrace{\\cos^3\\!x - \\sin^3\\!x}^{\\text{diff. of cubes}}}{\\sin x\\cos x(\\cos x -\\sin x)} \\:-\\:1$ . . $= \\;\\frac{({\\color{red}\\rlap{///////////}}\\cos x - \\sin x)(\\cos^2\\!x + \\sin x\\cos", "4:33 Why do you think that you've gone wrong? Your answer is correct. Edit. To show why the solution is the same as the one in your answers, we just need to show that $\\tan x-\\cot x=-2\\cot 2x$. \\begin{align*} &\\tan x-\\cot x+2\\cot 2x \\\\ &=\\frac{\\sin x}{\\cos x}-\\frac{\\cos x}{\\sin x}+2\\frac{\\cos 2x}{\\sin 2x} \\\\ &=\\frac{2\\cos x\\cos 2x\\sin x-\\cos^2 x\\sin 2x+\\sin^2 x\\sin 2x}{\\cos x\\sin x\\sin 2x} \\\\ &=\\frac{2\\cos x(\\cos^2 x-\\sin^2 x)\\sin x-\\cos^2 x(2\\sin x\\cos x)+\\sin^2 x(2\\sin x\\cos x)}{\\cos x\\sin x\\sin 2x} \\\\ &=\\frac{2\\cos^3 x\\sin x-2\\cos x\\sin^3 x-2\\sin x\\cos^3 x+2\\cos x\\sin^3 x}{\\cos x\\sin x\\sin 2x} \\\\ &=0 \\end{align*} • The answer in the back of this book is: $- \\cot x - 4x + \\tan x + c$ – seeker May 18 '13 at 2:40 • @Assad They are the same solution since $\\tan x-\\cot x=-2\\cot 2x$. Have you tried to show that they are equal? – Warren Moore May 18 '13 at 2:41 • Ah I see, I did not realise this... Thanks – seeker May 18 '13 at 2:44 • @Assad I've edited my answer with a proof of the equivalence for you as well. :) – Warren Moore May 18 '13 at 2:51 A cleaner way could be $$\\int (\\cot x-\\tan x)^2dx=\\int(\\cot^2x+\\tan^2x-2\\cot x\\tan x)dx$$ $$=\\int (\\csc^2x-1+\\sec^2x-1-2)dx$$ $$=\\int \\csc^2x dx+\\int \\sec^2x dx-4\\int dx$$ $$=-(\\cot x-\\tan x)-4x+C$$ From another answer $\\cot", "\\cos^3\\!x}{1 - 2\\cos^2\\!x}\\;= \\;\\frac{\\sec x - \\sin x}{\\tan x - 1}$ The left side has a sum of cubes ... factor it. . . . $\\frac{(\\sin x + \\cos x)(\\sin^2\\!x - \\sin x\\cos x + \\cos^2\\!x)}{1 - 2 \\cos^2\\!x}$ . . $= \\;\\frac{(\\sin x + \\cos x)(\\overbrace{\\sin^2\\!x + \\cos^2\\!x}^{\\text{This is 1}} - \\sin x\\cos x)}{\\underbrace{(1 - \\cos^2\\!x)}_{\\text{This is }\\sin^2\\!x} - \\cos^2\\!x}$ . . $= \\;\\frac{(\\sin x + \\cos x)(1 - \\sin x\\cos x)}{\\sin^2\\!x - \\cos^2\\!x}$ . . $= \\;\\frac{{\\color{red}\\rlap{////////////}}(\\sin x + \\cos x)(1 - \\sin x\\cos x)}{{\\color{red}\\rlap{////////////}}(\\sin x + \\cos x)(\\sin x - \\cos x)}$ . . $= \\;\\frac{1 - \\sin x\\cos x}{\\sin x - \\cos x}$ Divide top and bottom by $\\cos x$ . . $\\frac{\\dfrac{1}{\\cos x} - \\dfrac{\\sin x{\\color{red}\\rlap{/////}}\\cos x}{{\\color{red}\\rlap{/////}}\\cos x}} {\\dfrac{\\sin x}{\\cos x} - \\dfrac{{\\color{red}\\rlap{/////}}\\cos x}{{\\color{red}\\rlap{/////}}\\cos x}}$ . . $= \\;\\frac{\\sec x - \\sin x}{\\tan x - 1}$ 5. ## Thanks... I thank you all, especially Jhevon and Soroban. 6. Originally Posted by magentarita I thank you all, especially Jhevon and Soroban. skeeter had a valid point as well. the secant is undefined for the specified values of x. so that may cause problems. after graphing the two functions, i saw that the graphs were the same and so i overlooked it. i guess the point here was to do the algebraic manipulation to show", "1. ## Solving $\\tan2x = 8 \\cos^2x - \\cot x [0,\\frac{\\pi}{2}]$ 2. Arrange it all on one side so it's $\\tan 2x + \\cot x - 8 \\cos^2 x = 0$. Now the strategy is to factor the left side to get smaller pieces that are forced to be equal to zero. Start by recalling $\\tan 2x = \\frac{\\sin 2x}{\\cos 2x}$ and $\\cot x = \\frac{\\cos x}{\\sin x}$, so multiply through by $\\cos 2x \\sin x$. You get: $\\sin 2x \\sin x + \\cos x \\cos 2x - 8 \\cos^2 x \\cos 2x \\sin x = 0$. Now remember $\\sin 2x = 2\\sin x \\cos x$. So: $2 \\sin^2 x \\cos x + \\cos x \\cos 2x - 8 \\cos^2 x \\cos 2x \\sin x = 0$. And now we can make our first factor, which is $\\cos x$. $\\cos x \\left(2 \\sin^2 x + \\cos 2x - 8 \\cos x \\cos 2x \\sin x \\right) = 0$. No we know $\\cos x = 0$ when $x = \\frac{\\pi}{2}$ (in your range), so that is one solution. Let's forget about that factor now and concentrate on the rest: $2 \\sin^2 x + \\cos 2x - 8 \\cos x \\cos 2x \\sin x = 0$. Recall that $\\cos 2x = \\cos^2 x - \\sin^2 x$, and apply this to", "simplified form in terms of the second expression. 16. $\\frac{\\tan x+\\cot x}{\\csc x};\\cos x$ 17. $\\frac{\\sec x+\\csc x}{1+\\tan x};\\sin x$ 18. $\\frac{\\cos x}{1+\\sin x}+\\tan x;\\cos x$ 19. $\\frac{1}{\\sin x\\cos x}-\\cot x;\\cot x$ 20. $\\frac{1}{1-\\cos x}-\\frac{\\cos x}{1+\\cos x};\\csc x$ 21. $\\left(\\sec x+\\csc x\\right)\\left(\\sin x+\\cos x\\right)-2-\\cot x;\\tan x$ 22. $\\frac{1}{\\csc x-\\sin x};\\sec x\\text{ and }\\tan x$ 23. $\\frac{1-\\sin x}{1+\\sin x}-\\frac{1+\\sin x}{1-\\sin x};\\sec x\\text{ and }\\tan x$ 24. $\\tan x;\\sec x$ 25. $\\sec x;\\cot x$ 26. $\\sec x;\\sin x$ 27. $\\cot x;\\sin x$ 28. $\\cot x;\\csc x$ For the following exercises, verify the identity. 29. $\\cos x-{\\cos }^{3}x=\\cos x{\\sin }^{2}x$ 30. $\\cos x\\left(\\tan x-\\sec \\left(-x\\right)\\right)=\\sin x - 1$ 31. $\\frac{1+{\\sin }^{2}x}{{\\cos }^{2}x}=\\frac{1}{{\\cos }^{2}x}+\\frac{{\\sin }^{2}x}{{\\cos }^{2}x}=1+2{\\tan }^{2}x$ 32. ${\\left(\\sin x+\\cos x\\right)}^{2}=1+2\\sin x\\cos x$ 33. ${\\cos }^{2}x-{\\tan }^{2}x=2-{\\sin }^{2}x-{\\sec }^{2}x$ For the following exercises, prove or disprove the identity. 34. $\\frac{1}{1+\\cos x}-\\frac{1}{1-\\cos \\left(-x\\right)}=-2\\cot x\\csc x$ 35. ${\\csc }^{2}x\\left(1+{\\sin }^{2}x\\right)={\\cot }^{2}x$ 36. $\\left(\\frac{{\\sec }^{2}\\left(-x\\right)-{\\tan }^{2}x}{\\tan x}\\right)\\left(\\frac{2+2\\tan x}{2+2\\cot x}\\right)-2{\\sin }^{2}x=\\cos 2x$ 37. $\\frac{\\tan x}{\\sec x}\\sin \\left(-x\\right)={\\cos }^{2}x$ 38. $\\frac{\\sec \\left(-x\\right)}{\\tan x+\\cot x}=-\\sin \\left(-x\\right)$ 39. $\\frac{1+\\sin x}{\\cos x}=\\frac{\\cos x}{1+\\sin \\left(-x\\right)}$ For the following exercises, determine whether the identity is true or false. If false, find an appropriate equivalent expression. 40. $\\frac{{\\cos }^{2}\\theta -{\\sin }^{2}\\theta }{1-{\\tan }^{2}\\theta }={\\sin }^{2}\\theta$ 41. $3{\\sin }^{2}\\theta +4{\\cos }^{2}\\theta =3+{\\cos }^{2}\\theta$ 42. $\\frac{\\sec \\theta +\\tan \\theta }{\\cot \\theta +\\cos \\theta }={\\sec }^{2}\\theta$", "# Proving $\\tan^2\\left(\\frac{\\pi}{4}+x\\right)=\\frac{1+\\sin(2x)}{1-\\sin(2x)}$ Here is a question I am stuck on Show that $$\\tan^2\\left(\\frac{\\pi}{4}+x\\right)=\\frac{1+\\sin(2x)}{1-\\sin(2x)}$$ So far, I have been able to simplify to $$\\frac{\\cos^2x+\\sin^2x}{\\cos^2x-\\sin^2x}$$ But I don't know where to go from here. • Which side did you simplify? – J.G. Jun 22 '19 at 13:06 • @J.G. Simplifyed LHS first – Alex Jun 22 '19 at 13:07 $$\\displaystyle \\tan^2(\\frac{\\pi}4+x) \\\\= (\\frac{\\tan \\frac{\\pi}4 + \\tan x}{1 - \\tan \\frac{\\pi}4\\cdot \\tan x})^2 \\\\ = \\frac{(1+\\tan x)^2}{(1-\\tan x)^2} \\\\ = \\frac {(1+\\frac{\\sin x}{\\cos x})^2}{(1-\\frac{\\sin x}{\\cos x})^2} \\\\ = \\frac{(\\cos x + \\sin x)^2}{(\\cos x - \\sin x)^2}\\\\ = \\frac{\\sin^2 x + \\cos^2x + 2\\sin x \\cos x}{\\sin^2 x + \\cos^2x - 2\\sin x \\cos x} \\\\ = \\frac{1+ 2\\sin x\\cos x}{1 - 2\\sin x\\cos x}\\\\= \\frac{1 + \\sin 2x}{1 - \\sin 2x}$$ $$1\\pm\\sin2x=\\left(\\cos x\\pm\\sin x\\right)^2$$ $$\\dfrac{\\cos x+\\sin x}{\\cos x+\\sin x}=\\dfrac{1+\\tan x}{1-\\tan x}=\\tan\\left(x+\\dfrac\\pi4\\right)$$ Hint: $$\\frac{1+\\sin 2x}{1-\\sin 2x} = \\frac{(\\sin x + \\cos x)^2}{(\\sin x - \\cos x)^2}$$ $$\\tan (A+B) = \\frac{\\tan A + \\tan B}{1- \\tan A \\tan B}$$ From RHS $$\\sin(2x) = -\\cos\\bigg(\\frac{\\pi}{2}+2x\\bigg) = -\\cos\\bigg(2\\big(\\frac{\\pi}{4}+x\\big)\\bigg)$$ and use $$\\frac{1-\\cos(2\\alpha)}{1+\\cos(2\\alpha)}=\\tan^2\\alpha$$ to get LHS Or from LHS directly use the above identity to get $$\\tan^2\\alpha = \\frac{1-\\cos(2\\alpha)}{1+\\cos(2\\alpha)}$$ Here $$\\alpha = x+\\frac{\\pi}{4}$$ and $$\\cos(2\\frac{\\pi}{4}+2x) = \\cos(\\frac{\\pi}{2}+2x) = -\\sin(2x)$$ \\begin{align*} \\tan^2\\left(\\frac\\pi4+x\\right) &= \\sec^2\\left(\\frac\\pi4+x\\right)-1\\\\ &= \\frac1{\\cos^2\\left(\\frac\\pi4+x\\right)}-1\\\\ &= \\frac2{2\\cos^2\\left(\\frac\\pi4+x\\right)}-1\\\\ &= \\frac2{1+2\\cos^2\\left(\\frac\\pi4+x\\right)-1}-1\\\\ &= \\frac2{1+\\cos2\\left(\\frac\\pi4+x\\right)}-1\\\\ &= \\frac2{1-\\sin2x}-1\\\\ &= \\frac{1+\\sin2x}{1-\\sin2x}\\\\", "= 45o in (1) we get 45o – B = 30o$\\Rightarrow$ B = 15o Hence, A = 45o and B = 15o Question.26: If 3x = cosec $\\theta$ and $\\frac{3}{x}$ = cot $\\theta$, find the value of $3\\left ( x^{2}-\\frac{1}{x^{2}} \\right )$. Solution: $3(x^{2} – \\frac{1}{x^{2}})$ $= \\frac{9}{3}(x^{2} – \\frac{1}{x^{2}})$ $= \\frac{1}{3}(9x^{2} – \\frac{9}{x^{2}})$ $= \\frac{1}{3}\\left [ (3x)^{2} – \\left ( \\frac{3}{x} \\right )^{2} \\right ]$ $= \\frac{1}{3}\\left [ (cosec \\; \\theta)^{2} – \\left ( \\cot \\theta \\right )^{2} \\right ]$ $= \\frac{1}{3}\\left [ (cosec ^{2}\\; \\theta – \\cot ^{2} \\theta ) \\right ]$ $= \\frac{1}{3} (1)$ = $\\frac{1}{3}$ Question.27: If sin (A + B) = sin A cos B + cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of (i) sin 75o and (ii) cos 15o Solution: Let A = $45^{\\circ}$ and B = $30^{\\circ}$ (i) As, $\\sin (A + B) = \\sin A \\cos B + \\cos A \\sin B$ $\\Rightarrow \\sin (45^{\\circ} + 30 ^{\\circ}) = \\sin 45^{\\circ} \\cos 30 ^{\\circ} + \\cos 45^{\\circ} \\sin 30 ^{\\circ}$ $\\Rightarrow \\sin (75 ^{\\circ}) = \\frac{1}{\\sqrt{2}} \\times \\frac{\\sqrt{3}}{2} + \\frac{1}{\\sqrt{2}} \\times \\frac{1}{2}$ $\\Rightarrow \\sin (75 ^{\\circ}) = \\frac{\\sqrt{3}}{2\\sqrt{2}} + \\frac{1}{2 \\sqrt{2}}$ $Therefore, \\sin (75 ^{\\circ}) = \\frac{\\sqrt{3} + 1}{2\\sqrt{2}}$ (ii) As, $\\cos (A –", "A = 1.5,$ find $\\sin A.$ 53. If $\\cot A + \\cosec A = 5,$ find $\\cos A.$ 54. If $3\\sec^4 A + 8 = 10\\sec^2A,$ find the value of $\\tan A.$ 55. If $\\tan^2A + \\sec A = 5,$ find $\\cos A.$ 56. If $\\tan A + \\cot A = 2,$ find $\\sin A.$ 57. If $\\sec^2A = 2 + 2\\tan A,$ find $\\tan A.$ 58. If $\\tan A = \\frac{2x(x + 1)}{2x + 1},$ find $\\sin A$ and $\\cos A.$ 59. If $3\\sin A + 5\\cos A = 5,$ show that $5\\sin A - 3\\cos A = \\pm 3$ 60. If $\\sec A + \\tan A = \\sec A - \\tan A$ prove that each side is $\\pm 1$ 61. If $\\frac{\\cos^4 A}{\\cos^2 B} + \\frac{\\sin^4 A}{\\sin^2 B} = 1,$ prove that 1. $\\sin^4A + \\sin^4B = 2\\sin^2A \\sin^2B$ 2. $\\frac{\\cos^4 B}{\\cos^2 A} + \\frac{\\sin^4 B}{\\sin^2 A} = 1,$ 62. If $\\cos A + \\sin A = \\sqrt{2}\\cos A,$ prove that $\\cos A - \\sin A = \\pm \\sqrt{2}\\sin A$ 63. If $a\\cos A - b\\sin A = c,$ prove that $a\\sin A + b\\cos A = \\sqrt{a^2 + b ^2 - c^2}$ 64. If $1 - \\sin A = 1 + \\sin A,$ then prove that value of each side is $\\pm \\cos A$ 65.", "so \\begin{align} \\cos^4x\\sin^4x &=\\frac{1}{2^8}(e^{2ix}-e^{-2ix})^4\\\\[6px] &=\\frac{1}{2^8}(e^{8ix}-4e^{4ix}+6-4e^{-4ix}+e^{8ix})\\\\[6px] &=\\frac{1}{128}\\cos8x-\\frac{1}{32}\\cos4x+\\frac{3}{128} \\end{align} we now that $$\\sin 2x=2\\cos x\\sin x$$ $$\\cos x\\sin x=\\frac{\\sin 2x}{2}$$ so... $$\\cos^4 x\\sin^4 x=\\frac{\\sin^4 2x}{16}$$ $$\\frac{\\sin^4 2x}{16}=\\frac{(\\frac{1-\\cos 4x}{2})^2}{16}=\\frac{1-2\\cos 4x+\\cos^24x}{64}=\\frac{1-2\\cos 4x+\\frac{1+\\cos 8x}{2}}{64}$$", "- C)$ $\\cos A.\\cos\\frac{B + C}{2}\\sin\\frac{C - B}{2} = -2\\cos A.\\sin\\frac{B - C}{2}.\\cos\\frac{B - C}{2}$ If $B = C$ then above it $0 = 0$ i.e. triangle is isosceles. If $A = 90^\\circ$ then above is $0 =0$ i.e. triangle is right angled. 13. $\\because \\tan\\frac{A}{2}, \\tan\\frac{B}{2}, \\tan\\frac{C}{2}$ are in A.P. $\\tan\\frac{A}{2} - \\tan\\frac{B}{2} = \\tan\\frac{B}{2} - \\tan \\frac{C}{2}$ $\\frac{\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)}{\\cos\\frac{A}{2}\\cos\\frac{B}{2}} = \\frac{\\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)}{\\cos\\frac{B}{2}\\cos\\frac{C}{2}}$ $\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)\\cos\\frac{C}{2} = \\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)\\cos\\frac{A}{2}$ $\\sin\\left(\\frac{A}{2} - \\frac{B}{2}\\right)\\sin\\left(\\frac{A}{2} + \\frac{B}{2}\\right) = \\sin\\left(\\frac{B}{2} - \\frac{C}{2}\\right)\\sin\\left(\\frac{B}{2} + \\frac{C}{2}\\right)$ $\\Rightarrow \\cos B - \\cos A = \\cos C - \\cos B$ Thus, $\\cos A, \\cos B, \\cos C$ are in A.P. 14. Given, $a\\cos^2\\frac{C}{2} + c\\cos^2\\frac{A}{2} = \\frac{3b}{2}$ $a.\\frac{s(s - c)}{ab} + c.\\frac{s(s - a)}{bc} = \\frac{3b}{2}$ $\\frac{s}{b}[2s - a - c] = \\frac{3b}{2}$ $2s = 3b \\Rightarrow a + c = 2b$ We have to prove that $\\cot\\frac{A}{2} + \\cot\\frac{C}{2} = 2\\cot\\frac{B}{2}$ L.H.S. $= \\frac{s(s - a)}{\\Delta} + \\frac{s(s - c)}{\\Delta}$ $= \\frac{s}{\\Delta}(2s - a - c) = \\frac{2s(s - b)}{\\Delta} = 2\\cot\\frac{B}{2} =$ R.H.S. 15. Given, $a^2, b^2, c^2$ are in A.P. $\\Rightarrow b^2 - a^2 = c^2 - b^2$ $\\Rightarrow \\sin^2B - \\sin^2A = \\sin^2C - \\sin^2B$ $\\Rightarrow \\sin(B + A)\\sin(B - A) = \\sin(C + B)\\sin(C - B)$ $\\Rightarrow \\sin C\\sin(B - A) = \\sin A\\sin(C - B)$ $\\Rightarrow", "by 1, and that should not cause any problems. – Shobhit Dec 2 '17 at 9:42 • @user175989 you can set as solved if you are ok – gimusi Dec 3 '17 at 10:00 All you need to do now is$$\\frac{\\sin y\\cos x+\\cos y\\sin x}{\\sin y\\cos x-\\cos y\\sin x}=\\frac{\\frac{\\sin y\\cos x+\\cos y\\sin x}{\\cos x\\cos y}}{\\frac{\\sin y\\cos x-\\cos y\\sin x}{\\cos x\\cos y}}=\\frac{\\tan x+\\tan y}{\\tan x-\\tan y}.$$ Consider fractions that are equivalent to $2/3$. $$\\frac{2}{3}\\cdot\\frac{2}{2} = \\frac{4}{6}$$ $$\\frac{2}{3}\\cdot\\frac{3}{3} = \\frac{6}{9}$$ $$etc.$$ Conclusion ... you're allowed to multiply by one. Similarly, with the expression $\\displaystyle\\frac{\\sin y\\cos x+\\cos y \\sin x}{\\sin y \\cos x−\\cos y \\sin x}$, you're allowed to multiply by $\\displaystyle\\frac{\\frac{1}{\\cos y \\cos x}}{\\frac{1}{\\cos y \\cos x}}$ because $$\\displaystyle\\frac{\\frac{1}{\\cos y \\cos x}}{\\frac{1}{\\cos y \\cos x}} = 1$$ And just to make things mentally more clear, I may convert the original expression to a fraction of fractions as follows $$\\displaystyle\\frac{\\frac{\\sin y\\cos x+\\cos y \\sin x}{1}}{\\frac{\\sin y \\cos x−\\cos y \\sin x}{1}}$$ before multiplying. • What does $\\cos y \\cos x$ do to the expression ? there's nothing that can be cancelled out due to the '+' sign in the expression – user175089 Dec 3 '17 at 9:19 • Try multiplying the two expressions together, as suggested, and think of the distributive property when you do. – John Joy Dec 4 '17", "Prove that $\\tan\\frac{\\pi}{9}+ 4\\sin\\frac{\\pi}{9}= \\sqrt{3}$ . Prove that $$\\tan\\frac{\\pi}{9}+ 4\\sin\\frac{\\pi}{9}= \\sqrt{3}$$ I think the best solution here is using right triangle . . . I have one too, but not pretty. Multiplying by $$\\cos(\\pi/9)$$, this is equivalent to $$\\sin\\frac\\pi 9+2\\sin\\frac{2\\pi}9=\\sqrt3\\cos\\frac\\pi 9.\\tag1$$ But $$\\sqrt3\\cos\\frac\\pi 9-\\sin\\frac\\pi9= 2\\left(\\sin\\frac\\pi3\\cos\\frac\\pi 9-\\cos\\frac\\pi3\\sin\\frac\\pi9\\right) =2\\sin\\left(\\frac\\pi3-\\frac\\pi9\\right)=2\\sin\\frac{2\\pi}9$$ which proves $$(1)$$. If $$\\cos2x\\ne0,$$ $$\\tan2x+4\\sin(30^\\circ-x)$$ $$=\\dfrac{\\sin2x+4\\sin(30^\\circ-x)\\cos(30^\\circ-x)}{\\cos2x}$$ $$=\\dfrac{\\sin2x+2\\sin(60^\\circ-2x)}{\\cos2x}$$ $$=\\dfrac{\\sin2x+\\sqrt3\\cos2x-\\sin2x}{\\cos2x}=?$$ Here $$x=10^\\circ$$ $$(\\tan x+ 4\\sin x)^{2}- 3= - \\frac{1}{4}(2\\cos x+ 1)(2\\cos 3x- 1)\\csc^{2}\\left ( \\frac{\\pi}{4}- \\frac{x}{2} \\right )\\csc^{2}\\left ( \\frac{x}{2}+ \\frac{\\pi}{4} \\right )$$ Therefore $$\\tan\\frac{\\pi}{9}+ 4\\sin\\frac{\\pi}{9}= \\sqrt{3}$$ • .solve: @Eynjuhl – user707745 Sep 26 '19 at 6:03" ]

[ "How do you graph y = cos(x- pi/6)? It is graph of y= cos x with a phase shift of $\\frac{\\pi}{6}$ The graph of y= cos x with a horizontal shift to the right by $\\frac{\\pi}{6}$, called phase shift, would represent the graph of $y = \\cos \\left(x - \\frac{\\pi}{6}\\right)$", "angle) * amperage y = sin(phase angle) * amperage", "# How do you graph y = cos(x- pi/6)? Sep 18, 2015 #### Answer: It is graph of y= cos x with a phase shift of $\\frac{\\pi}{6}$ #### Explanation: The graph of y= cos x with a horizontal shift to the right by $\\frac{\\pi}{6}$, called phase shift, would represent the graph of $y = \\cos \\left(x - \\frac{\\pi}{6}\\right)$", "h=22.5 we have a phase shift of 22.5 left. (A phase shift of `pi/8~~.393` when using radians.) The graph of `y=cos(theta)` and the transformed graph:", "# How do you find the phase shift of y=sin(theta-3 pi/4)? Aug 2, 2016 $$ Horizontal shift =3pi/4 #### Explanation: $y = \\sin \\left(\\theta - 3 \\frac{\\pi}{4}\\right)$ we have $a = 1$ $b = 1$ $c = 3 \\frac{\\pi}{4}$ $$ A phase shift is nothing but horizontal shift. Horizontal shift =3pi/4", "either that the phase shift is $+ \\pi$, or that the phase shift is $- \\pi$, and the graph would look the same. $y = \\sin \\left(x\\right)$ $y = - \\frac{1}{2} \\cdot \\sin \\left(x\\right)$ • 12 minutes ago • 13 minutes ago • 16 minutes ago • 17 minutes ago • 43 seconds ago • 2 minutes ago • 4 minutes ago • 4 minutes ago • 4 minutes ago • 11 minutes ago • 12 minutes ago • 13 minutes ago • 16 minutes ago • 17 minutes ago", "phase shift – a full period ($2\\pi$) ...", "## anonymous 3 years ago graphing $y=sin(2x-\\pi)$ did I graph this correctly? 1. anonymous |dw:1360883013269:dw| 2. anonymous I guess I can get as far as doing the phase shift but I don't know what to do with the 2 in front of the x $y=sin(2(x-\\pi/2))$|dw:1360883459233:dw| 3. anonymous @TuringTest 4. anonymous the phase shift is pi/2 isn't it? 5. TuringTest $\\sin(-\\pi)=0$so the graph should start at x=0, y=0 6. anonymous what does the phase shift do then? 7. anonymous oh I see 8. anonymous no not really 9. anonymous 10. TuringTest a phase shift of -pi will shift the graph a half-period to the right, which for the sin is equivalent to turning it upside-down 11. anonymous Oh i see. so the phase shift is pi and not pi/2 12. TuringTest right, you could also see that by applying$\\sin(\\alpha+\\beta)=\\sin\\alpha\\cos\\beta+\\cos\\alpha\\sin\\beta$ 13. anonymous makes sense. Thanks! 14. TuringTest welcome :D", "to be subtracting pi/2 phase. If z is upwards, y is to the right, surely out θ=wt is measured from the y axis anticlockwise to be consistent with your first step of the anticlockwise circle. So then if I want things to start at (0,0,0), I need to shift everything back by pi/2, i.e subtract it. 8. Sep 8, 2014 ### BvU My thingy starts at R,R for t = 0. So I have to let it roll a quarter turn, $\\pi/2$, to let z=0 and then y is $\\pi R/2$. The transform needed is $\\omega t'= \\omega t - \\pi/2$ and I realize that I also need to shift y: $y'= y - \\pi R/2$.", "2 = pi. y=2sin(pi(x)) Now, to phase shift 1/6 of a unit I do y = 2sin(pi(x) - pi/6) ? 11. Dec 11, 2012 ### Dick That looks right. 12. Dec 11, 2012 ### Feodalherren Thanks a bunch. Much appreciated. I totally bombed my final but at least I walked away with some new knowledge.", "How do you graph and list the amplitude, period, phase shift for y=sin(x-pi)-1? Nov 27, 2017 We can rewrite as $y = \\sin x \\cos \\pi - \\cos x \\sin \\left(\\pi\\right) - 1$ $y = - \\sin x - 1$ This is the basic sine graph reflected in the x-axis and transformed 1 unit down. Here is the graph: The amplitude is $1$, the phase shift is $0$ and the period is $2 \\pi$. Hopefully this helps!", "How do you graph y=sin(x-pi/4)? It would be graph of y= sin x , but with a lateral shift of $\\frac{\\pi}{4}$ to the right. The graph of sin (x-$\\frac{\\pi}{4}$) would be just like the graph of sinx with a horizontal shift of $\\frac{\\pi}{4}$ to the right. Thus it would require to graph y= sin x and then give it a lateral shift of $\\frac{\\pi}{4}$ to the right.", "# Trying to read phase shift constant from the graphs I have an equation $y(x,y) = y_0 \\sin (\\omega t \\pm kx \\pm \\phi)$ and there are two graphs. One is $y(x,t=0)$ in units $mm(mm)$ and the other is $y(t,x=0)$ in units $mm(s)$. Now I have to find all of the constants inside of the equation by looking at graphs and it goes well until I am to determine phase shift $\\phi$. If I calculate it by looking at $y(x=0,t=0)$ it works out well and I get right solution, but when I try to read $\\phi$ from graph $y(x,t=0)$ or $y(t,x=0)$ I get two wrong solutions. The idea for reading $\\phi$ from graphs was to express some distance $x'$ or time $t'$ in radians but it looks like I can't get it right. The sign in front of $\\phi$ is not a problem - if my sinus (in red) is shifted right it is minus otherwise it is plus. Can anyone clear things out for me and explain why I get wrong results which also differ from one-another? • Your result using $y(x, t=0)$ appears correct, it agrees to within measurement error. Your result for $\\phi$ from $y(x=0, t)$ should be replaced with $\\pi - \\phi$, since there is a relative minus sign between $\\omega t$ and $kx$", "# Laplace transform of phase-shifted sinusoid ## Homework Statement Find the laplace transform of [TEX]f(t)=sin(4t+\\frac{\\pi}{3})u(t)[/TEX]. http://img822.imageshack.us/img822/6996/codecogseqna.gif [Broken] ## Homework Equations Definition of laplace transform, and properties: http://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems\" ## The Attempt at a Solution I have a feeling that I need to use the time-shift property of the transform, but I need just a bit more direction on the strategy. Last edited by a moderator: $$f(t) = \\left[ sin(4t)cos(\\frac{\\pi}{3}) + cos(4t)sin(\\frac{\\pi}{3}) \\right] u(t).$$", "think of the phase shift is that it's just a new start to the graph. You can treat the phase shift as a place to actually begin your graph. So.. with all this information, we can now actually graph the equation of $y = 5 \\cos \\left(x - \\pi\\right) + 1$ . Here are some visual cues. My art is pretty bad, sographing the graph on a website should be much more different. I hope this helps.", "graph in a vertical direction either side of the x-axis by a factor of $5$. 3. ah icic cool thxs. so for the phase shift would be -4pi/5 correct? 4. Originally Posted by S2Krazy ah icic cool thxs. so for the phase shift would be -4pi/5 correct? Correct!", "and list the amplitude, period, phase shift for y=sin(1/2)(x-pi)+2? How do you graph and list the amplitude, period, phase shift for y=sin(1/2)(x-pi)+2?...", "practice, and in every country on this or any other planet. – Gerry Myerson Oct 22 '12 at 5:13 There seem to be different conventions about phase shift. See this Regents link, where it says that if $$y=\\sin(2x-(\\pi/2))=\\sin(2(x-(\\pi/4)))$$ then physicists and engineers say the phase shift is $\\pi/2$ but some math textbooks say it's $\\pi/4$. Find out for certain which convention your class is using, and use it. If it's too late for that, then declare which convention you are using, and stick with it.", "The value of $$C$$ is the phase shift (horizontal shift) of the sinusoidal function. The graph is shifted to the right if $$C > 0$$ and shifted to the left if $$C < 0$$. • The value of $$D$$ is the vertical shift of the sinusoid. The horizontal line $$y = D$$ is the so-called center line for the graph of the sinusoidal function. • The important notes about sinusoids on page 103.", "= 0 yeilds 0 = ym sin(- ), which means that the phase constant may be taken to be zero. You should show that the intergral multiples of π yield the same final results. Hence for this wave y = ym sin(kx-wt), and with the values just found, we obtain as the equation for the wave y = (0.50cm) sin {(0.27cm-1 )x - (730s-1 )t} where the quantity in bracket is an angel in radius." ]

[ "1x = np.linspace(-2*np.pi, 2*np.pi, 100) # domain 2np.sin(x-np.pi/2) # shifted right 3np.sin(x) 4np.sin(x+np.pi/2) # shifted left • The second subplot displays the original graph of $\\sin x$ defined within the domain $-2\\pi \\leq x \\leq 2\\pi$. • The first subplot displays the graph of $\\sin (x-\\frac{\\pi}{2})$ which is the graph of $\\sin x$ being shifted by $\\frac{\\pi}{2}$ units to the right. • The second subplot displays the graph of $\\sin (x+\\frac{\\pi}{2})$ which is the graph of $\\sin x$ being shifted by $\\frac{\\pi}{2}$ units to the left. ### Shifting in the $y$-direction Note Given $k$ is a positive real number: • The graph of $f(x)+k$ is the graph of $f(x)$ shifted vertically up by $k$ units. • The graph of $f(x)-k$ is the graph of $f(x)$ shifted vertically down by $k$ units. Example Shifting in the $y$-direction. 1x = np.linspace(-2*np.pi, 2*np.pi, 100) # domain 2np.sin(x)+1 # shifted up 3np.sin(x) 4np.sin(x)-1 # shifted down • The line in red displays the original graph of $\\sin x$ defined within the domain $-2\\pi \\leq x \\leq 2\\pi$. • The line in blue displays the graph of $\\sin(x)+1$ which is the graph of $\\sin x$ being shifted vertically up by $1$ unit. • The line in green displays the graph of $\\sin(x)-1$ which is the graph of $\\sin x$ being shifted vertically down", "geometric figures. /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ The first line real f1 (real x) {return sin(x);} defines the sin() function to be drawn, and the next line draw(graph(f1,-6.21,8.03)); actually draws the graph to the diagram. The graph() command comes from the Asymptote graph package. The next few lines draw several line segments. The attribute zzttqq is a pen whose color is mapped from some rgb value (with conversions ... 8 = y, 9 = z, a = a, b = b, ... ). ## Options An outline of the different available options. (The first three options are related and automatically update with respect to eath other). X/Y Units (cm): Indicates how many units in the picture correspond to 1 centimeter. Picture width/height: Indicates what size the Asymptote output should be, in centimeters. xmin/xmax/ymin/ymax: Indicates the dimensions of the picture in the picture's coordinates. (The next options change other features). Font size: Changes the default font size. The value defaults at 10pt, but I personally recommend 7pt for any file with multiple text lines, which more closely approximates the size of the font", "what translation to apply, however, we first factorise the function into the form $y=-2\\sin\\left(3\\left(x+\\frac{\\pi}{12}\\right)\\right)$y=2sin(3(x+π12)). In this form, we can see that the $x$x-values are increased by $\\frac{\\pi}{12}$π12 inside the function. This means that to get a particular $y$y-value, we can put in an $x$x-value that is $\\frac{\\pi}{12}$π12 smaller than before. Graphically, this corresponds to shifting the entire function to the left by $\\frac{\\pi}{12}$π12 units. The graph of $y=-2\\sin\\left(3x+\\frac{\\pi}{4}\\right)$y=−2sin(3x+π4​) Lastly, we translate the graph of $y=-2\\sin\\left(3x+\\frac{\\pi}{4}\\right)$y=2sin(3x+π4) upwards by $2$2 units, to obtain the final graph of $y=-2\\sin\\left(3x+\\frac{\\pi}{4}\\right)+2$y=2sin(3x+π4)+2. The graph of $y=-2\\sin\\left(3x+\\frac{\\pi}{4}\\right)+2$y=−2sin(3x+π4​)+2 Careful! When we geometrically apply each transformation to the graph of $y=\\sin x$y=sinx, it's important to consider the order of operations. If we had wanted to vertically translate the graph before reflecting about the $x$x-axis, we would have needed to translate the graph downwards first. ### The general case In the example above we were transforming the graph of $y=\\sin x$y=sinx. The particular function $y=\\sin x$y=sinx was not important, however. We could have just as easily transformed the graph of $y=\\cos x$y=cosx, or even a non-trigonometric function, using the same method! Consider a function $y=f\\left(x\\right)$y=f(x). Then we can obtain the graph of $y=af\\left(b\\left(x-c\\right)\\right)+d$y=af(b(xc))+d, where $a,b,c,d$a,b,c,d are constants, by applying a series of transformations to the graph of $y=f\\left(x\\right)$y=f(x). These transformations are summarised below. Summary To obtain the graph", "Here, the graphing calculator format is: $Y_1 = \\sin(X)$ $Y_2 = \\sin(X+\\pi/4)$ Using window coordinates $-6.5 ≤ x ≤ 6.5$ $-1.25 ≤ y ≤ 1.25,$ we get the following: $Y_1 = \\sin(X)$ is shown in red, and $Y_2 = \\sin(X+\\pi/4)$ is in yellow. We notice that addition of $\\pi/4$ to the argument has shifted the graph to the left by $\\pi/4$ units.In general: Replacing $x$ by $x+c$ shifts the graph to the left $c$ units. How would we shift the graph to the right $\\pi/4$ units? (c) Here, the graphing calculator format is: $Y_1 = \\sin(X)$ $Y_2 = \\sin(3*X)$ Using the same window coordinates and color coding as in part (b), we obtain the following graph. Notice that the graph of $\\sin(3*x)$ oscillates three times as fast as the graph of $\\sin(x).$ Thus, every $2\\pi$ units, the graph of $\\sin(3*x)$ oscillates three times. In general: Multiplying $x$ by $b$ multiplies the rate of oscillation by $b.$ (d) $Y_1 = \\sin(\\pi*X)$ $Y_2 = 2\\sin(\\pi(X+3))$ The graphs are shown in the window $-3.2 ≤ x ≤ 3.2$ $-2.5 ≤ y ≤ 2.5$ Notice several things: • The red graph, $y = \\sin(\\pix),$ crosses the $x-$axis at integral values $0 ±1, ±2, ...$ For instance, when $x = 1,$ we get $\\sin(\\pix) = \\sin(\\pi) = 0.$ • The yellow graph, $y", "draw(Label(\"$\\pi$\",1),shift(-0.15*Z)*scale3(0.5)*g,blue,Arrow3); path3 h=arc(O,1,170,-120,150,-120,CCW); draw(Label(\"$\\sigma$\",1),shift(-0.5*cos(pi/6)*Y-0.5*sin(pi/6)*X+0.35*Z)*scale3(0.5)*h,red,Arrow3); draw(shift(-0.5*cos(pi/6)*Y-0.5*sin(pi/6)*X-0.15*Z)*scale3(0.1)*unitsphere,gray+opacity(.3)); and the result You can easily modify the xpt and ypt values to find the best shape. For periodic spline, do not forget to have periodic value. O.G. • Great! But may I ask where the lower bound in (0.1,2*pi) comes from? – user121799 Jan 4 '18 at 0:28 • @marmot I add a comment. In view of your attempt, I transform the box (a,b) (-0.67,0.1),(0,2.032*pi) to (-0.67,0),(0.1,2*pi) so that the x variable varies from -0.67 to 0.1 while the y variable varies from 0 to 2*pi. Perhaps that (-0.67,0),(0,2*pi) should be ok... – O.G. Jan 4 '18 at 9:48", "layers. In order for my approach to work as intended, the layer settings window must look like this: However, quite annoyingly, when you close and reopen this saved document, the layer settings window looks like this: So, it must be changed back each time to allow trouble-free math drawing. # Finding Reciprocal Functions SEC, CSC, COT on TI-30XA First, you may want to refresh your memory of the regular trig functions: Now, let’s find $\\csc \\frac{5\\pi }{4}$. If we look it up on WolframAlpha, we find that it equals $-\\sqrt{2}$, or, roughly, $-1.4142$. To get this answer on the TI-30XA, first remember that $\\csc$ is equal to $\\frac{1}{\\sin }$. So, what we’re looking for could be expressed as $\\frac{1}{\\sin \\frac{5\\pi }{4}}$. Once we know this, we can type $5\\cdot \\pi$ into the TI-30XA. This gives us ~15.7. (Make sure you’re in radian mode. If not, press the DRG button — located by the 2nd button, upper-left — until you see RAD appear in small letters on the screen.) Now, we can divide 15.7 by 4. This gives us 3.925. If we take the $\\sin$ of this (by pressing the “sin” button while 3.925 is displayed), we get $-0.7$. IF we divide 1 by $-0.7$, we get $-1.43$. As you can see, this is close to the correct value.", "yields $$\\Delta p(x,y) = - a^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right) - b^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right)$$ and $$Q = - \\Delta p(x,y) = a^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right) + b^2 \\sin\\left(ax - \\frac{\\pi}{2}\\right) \\sin\\left(by - \\frac{\\pi}{2}\\right).$$ ## Input files The project file is square_1e3_poisson_sin_x_sin_y.prj. It describes the process to be solved and the related process variable together with their initial, boundary conditions and source terms. The definition of the source term $Q$ is in a Python script. The script for setting the source terms is referenced in the project file as follows: <python_script>sin_x_sin_y_source_term.py</python_script> In the source term description <source_term> <type>Python</type> <source_term_object>sinx_siny_source_term</source_term_object> </source_term> the domain is specified by the mesh-tag. The function $Q$ is defined by the Python object sinx_sinx_source_term that is created in the last line of the Python script: import OpenGeoSys from math import pi, sin a = 2.0*pi b = 2.0*pi def solution(x, y): return - sin(a*x-pi/2.0) * sin(b*y-pi/2.0) # - laplace(solution) = source term def laplace_solution(x, y): return a*a * sin(a*x-pi/2.0) * sin(b*y-pi/2.0) + b*b * sin(a*x-pi/2.0) * sin(b*y-pi/2.0) # source term for the benchmark class SinXSinYSourceTerm(OpenGeoSys.SourceTerm): def getFlux(self, t, coords, primary_vars): x, y, z = coords value = laplace_solution(x,y) Jac = [ 0.0 ] return (value, Jac) # instantiate source term object referenced in OpenGeoSys' prj file sinx_siny_source_term", "x) {return sin(x);} filldraw(graph(f2,0,pi )--(pi ,0)--(0,0)--cycle, fueaev, zzttqq); Some commands have to be defined with respect to a function. $\\texttt{f2}$ returns the $\\sin(x)$ function, and the $\\texttt{graph}$ command draws $\\texttt{f2}$ from $x = 0$ to $x = \\pi$. Label laxis; laxis.p = fontsize(10); string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); These commands draw the axes. The $\\texttt{laxis}$ label specifies the label styles of the axes (e.g., the field $\\texttt{laxis.p}$ defines the desired font pen). The next line, the string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} are functions that define the labels for both axes. These are abbreviated as $\\texttt{xlbl, ylbl}$ in the concise mode. The first defines axes with labels with respect to $\\pi$, and it formats the labels so that the $0$ label simply outputs $0$ instead of $0\\pi$, and $\\pm \\pi$ instead of $\\pm 1\\pi$. Finally, xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); are the commands that actually construct the xaxis and the yaxis. For documentation, see here. The next block of code does the actual construction of the", "calculate that part and end up with $35.75$. We also see with the formulas we used with the plug in that when you increase by $0.2$ the $35.75$ part decreases by $0.25$. The answer is then $\\boxed{(D) 36.8}$. ## Solution 13 1、First, move terms to get $1+5cos3x=3sinx$. After graphing, we find that there are $\\boxed{6}$ solutions (two in each period of $5cos3x$). -dstanz5 2、We can graph two functions in this case: $5\\cos{3x}$ and $3\\sin{x} -1$.$$\\newline$$Using transformation of functions, we know that $5\\cos{3x}$ is just a cos function with amplitude 5 and period $\\frac{2\\pi}{3}$. Similarly, $3\\sin{x} -1$ is just a sin function with amplitude 3 and shifted 1 unit downwards. So:$[asy] import graph; size(400,200,IgnoreAspect); real Sin(real t) {return 3*sin(t) - 1;} real Cos(real t) {return 5*cos(3*t);} draw(graph(Sin,0, 2pi),red,\"3\\sin{x} -1 \"); draw(graph(Cos,0, 2pi),blue,\"5\\cos{3x}\"); xaxis(\"x\",BottomTop,LeftTicks); yaxis(\"y\",LeftRight,RightTicks(trailingzero)); add(legend(),point(E),20E,UnFill); [/asy]$We have $\\boxed{(A) 6}$ ## Solution 14 1、This question is just about pythagorean theorem$$a^2+(a+2)^2-b^2 = (a+4)^2$$$$2a^2+4a+4-b^2 = a^2+8a+16$$$$a^2-4a+4-b^2 = 16$$$$(a-2+b)(a-2-b) = 16$$$$a=3, b=7$$With these calculation, we find out answer to be $\\boxed{\\textbf{(A) }24\\sqrt5}$ ~Lopkiloinm 2、Let $\\overline{AD}$ be $b$$\\overline{CD}$ be $a$$\\overline{MD}$ be $x$$\\overline{MC}$$\\overline{MA}$$\\overline{MB}$ be $t$$t-2$$t+2$ respectively. We have three equations:$$a^2 + x^2 = t^2$$$$a^2 + b^2 + x^2 = t^2 + 4t + 4$$$$b^2 + x^2 = t^2 - 4t + 4$$ Subbing in the first and third equation into the second equation, we get:$$t^2", "figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ /* dots and labels */ label(\"f\",(-6.12,-0.12),NE * labelscalefactor); label(\"a = 2\",(1.46,0.24),NE * labelscalefactor,zzttqq); dot((-3.14,0),dotstyle); label(\"A\",(-3.06,0.12),NE * labelscalefactor); dot((2.2,2.04),dotstyle); dot((4.5,2.08),dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$ Here is the Asymptote code produced: (with some spacing added) /* Geogebra to Asymptote conversion, documentation at userscripts.org/scripts/show/72997 */ import graph; size(14.26cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.22,xmax = 8.04,ymin = -4.18,ymax = 4.76; /* image dimensions */ pen fueaev = rgb(0.96,0.92,0.9); pen zzttqq = rgb(0.6,0.2,0); real f2 (real x) {return sin(x);} filldraw(graph(f2,0,pi )--(pi ,0)--(0,0)--cycle, fueaev, zzttqq); filldraw((-3.14,0)--(0,0)--(0,3.14)--(-3.14,pi )--cycle, fueaev, zzttqq); Label laxis; laxis.p = fontsize(10); string xaxislabel (real x) { int n=round(x/pi); if(n==-1) return \"$-\\pi$\"; if(n==1) return \"$\\pi$\"; if(n==0) return \"$0$\"; return \"$\" + string(n) + \"\\pi$\";} string yaxislabel (real x) {return \"$\" + string(x) + \"\\,\\mathrm{}$\";} xaxis(\"$x$\",-6.22,8.04,Ticks(laxis,xaxislabel,Step=3.141592653589793,Size=2),Arrows(6),above=true); yaxis(-4.18,4.76,Ticks(laxis,yaxislabel,Step=1.0,Size=2),Arrows(6),above=true); /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real", "SYSTEM label(\"$O$\", (0, 0, 0), NW); Label Lx = Label(\"$x$\", EndPoint, W); Label Ly = Label(\"$y$\", EndPoint, S); Label Lz = Label(\"$z$\", EndPoint, W); draw(Lx, O--2X, Arrow3(emissive(black))); draw(Ly, O--2Y, Arrow3(emissive(black))); draw(Lz, O--2Z, Arrow3(emissive(black))); // SURFACE triple S(pair uv) { real x = cos(uv.x)*sin(uv.y); real y = sin(uv.x)*sin(uv.y); real z = cos(uv.y); return (x-(1-z**2-y**2), y-0.3*sin(z*pi), z); } // BOUNDARY triple dS(real u) { return S((u, pi/2)); } // TANGENT VECTORS real e = 0.001; // *this... is... infinitesimal!* triple U(pair t) { return (S((e,0)+t)-S(t))/e; } triple V(pair t) { return (S((0,e)+t)-S(t))/e; } // NORMAL VECTOR triple N(pair uv) { return cross(U(uv),V(uv))/length(cross(U(uv),V(uv))); } // GRAPHICAL ELEMENTS surface s = surface(S, (0, 0), (2pi, pi/2), 64, 64, Spline); path3 ds = graph(dS, 0, 2pi); pair t = (-pi/8, pi/8); path3 n = S(t) -- shift(-N(t))*S(t); label(\"$\\Sigma$\", shift(0,0,0.1)*shift(adj)*S((pi/2,pi/5)), NE); Label LdS = Label(\"$\\partial\\Sigma$\", Relative(0.95), S); Label Ln = Label(\"$\\mathbf n$\", EndPoint, NW); draw(s, surfacepen=material(diffusepen=gray(0.6),emissivepen=gray(0.3),specularpen=gray(0.1))); draw(LdS, ds, Arrow3(emissive(black))); draw(shift(4*unit(currentprojection.camera))*ds, dashed); draw(Ln, n, Arrow3(emissive(black))); REQUESTED APPENDIX: on dashed lines I wanted to draw the dashed border over the surface. Since this is an orthographic projection, the easiest way is translating the object along the camera axis towards it (instead of checking what segments are visible, possibly complicated): draw(shift(4*unit(currentprojection.camera))*ds, dashed); unit(currentprojection.camera) is a versor parallel to the camera axis and pointing towards it, so it", "Math Help - Order of Operations with Trig Function 1. Order of Operations with Trig Function I'm trying to make a table of values for the trigonometric function $y = 2sin3(x-30) +1$, though I don't know exactly how to punch this into my calculator (confused with the order of operations). All I know is that from the parent function, $y = Sinx$, the following transformations have been made: Phase angle shift: $30deg$ right. Amplitude: V.E. by $2$. Therefore, $2$. Period: H.C. by $3$. Therefore, $120deg$. Vertical shift: $1$ up. How would I solve for y if $x = 0deg$? 2. Do you mean $y=2\\sin(3(x-30))+1$, or $y=2\\sin(3)(x-30)+1$? Putting careful parentheses around all function arguments is a very good way to enhance clarity of expressions. What kind of calculator do you have? If it's HP with the reverse Polish, let some other (poor) fellow help you out. If it's a more normal calculator, especially if it allows parentheses, you should be able to enter it more or less as it shows up on the page. Also, I would highly recommend you double-check to see if the calculator is in degrees or radian mode. I'm assuming the 30 is degrees? 3. That's the problem here, I don't know where the parenthesis are supposed to be; that's how much teacher typed it.", "# Math Help - Moving a Trigonometric function 1. ## Moving a Trigonometric function y=d+sin(bx+c) So if I have the equation: sin(2x)=y; I know that the period will be pi. So the whole graph will occur between (0,0) and (pi,0) Start at (0,0), Maximum (pi/4, 1); intercept at (pi/2, 0); Minimum (3Pi/4, -1); Intercept (pi,0) Now if I have: y=sin(x-(pi/4)) the graph gets shifted to the right pi/4 radians. So key points are Start: (pi/4, 0); Maximum: (3pi/4, 1); Intercept: (5pi/4,0); minimum (7pi/4, -1); Complete (9pi/4,0) If I put them together as Sin(2x-(pi/4) Why does my horizontal shift get cut in half? Why does it not start at (pi/4)? Why does it start at (pi/8)? In my graphing calculator I see everything gets moved my (pi/8) when I have this equation. 2. Originally Posted by IDontunderstand y=d+sin(bx+c) So if I have the equation: sin(2x)=y; I know that the period will be pi. So the whole graph will occur between (0,0) and (pi,0) Start at (0,0), Maximum (pi/4, 1); intercept at (pi/2, 0); Minimum (3Pi/4, -1); Intercept (pi,0) Now if I have: y=sin(x-(pi/4)) the graph gets shifted to the right pi/4 radians. So key points are Start: (pi/4, 0); Maximum: (3pi/4, 1); Intercept: (5pi/4,0); minimum (7pi/4, -1); Complete (9pi/4,0) If I put them together as Sin(2x-(pi/4) Why does my", "# Graphing sin, finding phase shift, period and transformations I need to graph $$y=2 \\cos\\left(x - \\frac{\\pi}{3}\\right)$$ and I am having trouble figuring out what points will be on the graph. The book tells me to split it into four parts, which doesn't make sense since they split it into five parts anyways. I know that the points will be between $\\pi/3$ and $(\\pi/3 + 2\\pi)$ so that gives me $\\pi/3 < x < 7\\pi/3$ The only problem is I have no idea how to split that up into equal parts to make graphing easier. - Indeed, it is useful to split into four parts, as follows: $$\\bigg[\\frac{{2\\pi }}{6},\\frac{{5\\pi }}{6}\\bigg],\\bigg[\\frac{{5\\pi }}{6},\\frac{{8\\pi }}{6}\\bigg],\\bigg[\\frac{{8\\pi }}{6},\\frac{{11\\pi }}{6}\\bigg],\\bigg[\\frac{{11\\pi }}{6},\\frac{{14\\pi }}{6}\\bigg].$$ The endpoints are $\\pi/3$ and $7\\pi / 3$. The middle point is then given by $$\\bigg(\\frac{\\pi }{3} + \\frac{{7\\pi }}{3}\\bigg)\\bigg /2 = \\frac{{8\\pi }}{6}.$$ Next, the middle point between $\\pi/3$ and $8\\pi/6$ is given by $$\\bigg(\\frac{\\pi }{3} + \\frac{{8\\pi }}{6}\\bigg)\\bigg/2 = \\bigg(\\frac{{2\\pi }}{6} + \\frac{{8\\pi }}{6}\\bigg)\\bigg/2 = \\frac{{5\\pi }}{6},$$ and the middle point between $8\\pi/6$ and $7\\pi/3$ is given by $$\\bigg(\\frac{8\\pi }{6} + \\frac{{7\\pi }}{3}\\bigg)\\bigg/2 = \\bigg(\\frac{{8\\pi }}{6} + \\frac{{14\\pi }}{6}\\bigg)\\bigg/2 = \\frac{{11\\pi }}{6}.$$ Adam: $\\pi/3 = 2\\pi/6$, $7\\pi/3 = 14\\pi/6$..., the book just reduced the fractions (Shai used 6 in the denominator to help establish the four parts, you can simply", "Solved Example. Graph$\\,y=\\mathrm{arccos}\\,x\\,$and state the domain and range of the function. How to find the base point given public and private key and EC parameters except the base point. The inverse of a 2x2 is easy... compared to larger matrices (such as a 3x3, 4x4, etc). Please help me answering this problem! Post-multiply both sides of … If you want to share a math-problem you need to explain what you're arguments are. For multiplication, it's division. Show … We can check … Are there any Pokémon that lose overall base stats when they evolve? Given a “special” input value, evaluate an inverse trigonometric function. [/latex], If$\\,x\\text{ is in }\\left[0,\\pi \\right],\\,$then$\\,{\\mathrm{sin}}^{-1}\\left(\\mathrm{cos}\\,x\\right)=\\frac{\\pi }{2}-x. The Inverse Park Transform block converts the time-domain direct, quadrature, and zero components in a rotating reference frame to the components of a three-phase system in an abc reference frame. Figure 11. If y = f(x) = ab x, then we may solve for x in terms of y using logarithms: .$, If$\\,x\\text{ is not in}\\,\\left[-\\frac{\\pi }{2},\\frac{\\pi }{2}\\right],\\,$then find another angle$\\,y\\text{ in }\\left[-\\frac{\\pi }{2},\\frac{\\pi }{2}\\right]\\,$such that$\\,\\mathrm{sin}\\,y=\\mathrm{sin}\\,x.$. prove that $\\det(ABC) = \\det(A) \\det(B) \\det(C)$ [for any $n×n$ matrices $A$, $B$, and $C$]. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical", "let us consider $g(x) =-2\\sin(x)+3\\text{.}$ In this case, we are multiplying the outputs of $\\sin(x)$ by $-2\\text{,}$ and then adding 3. Graphically, this results in a vertical stretch by a factor of $2\\text{,}$ a reflection across the $x$-axis, and finally a vertical shift up by 3 units (note that subtracting 3 would result in a shift down). We graph this below. You should verify this by creating a table like the one above. ###### Note57 When considering the function $g(x) =-2\\sin(x)+3$ above, we multiplied the outputs of $\\sin(x)$ by -2, which resulted in a stretch and reflection, and then added 3, which resulted in a shift. This is following the order of operations (PEMDAS) and is the order that should always be used when applying vertical transformations. Everything that we have done so far has involved $\\sin(x)\\text{.}$ However, if we replace $\\sin( x)$ with any of the other trigonometric functions, we will see the same results. ###### Example58 Describe how the graph of $g(x) = -4\\cos(x)+2$ differs from the graph of $f(x) = \\cos(x) \\text{.}$ Solution The graph of $g(x)$ is the result of starting with the graph of $f(x)$ and vertically stretching it by a factor of 4, then reflecting it over the $x$-axis, and finally shifting up by 2 units. ###### Example59 Describe how the graph", "not outweigh the negative by the cos and tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first solution is a little bit after $\\pi$, which is around $3.14$. Hence the answer is $\\boxed{\\textbf{(D)}}$. (Not original author) Here is the graph: $[asy] Label f; f.p=fontsize(6); xaxis(-5,5,Ticks(f, 1.0)); yaxis(-8,8,Ticks(f, 2.0)); real f(real x) { return sin(x); } draw(graph(f, -5,5)); real g(real x) { return 2*cos(x); } draw(graph(g, -5,5)); real h(real x) { return 3*tan(x); } draw(graph(h, -1.2,1.2)); draw(graph(h, 1.94, 4.34)); draw(graph(h, -4.34, -1.94)); [/asy]$ ## Solution 3(More Detailed Answer of Solution 1) Denote $D_{f}$ as the domain of $f(x).$ Obviously $D_{f}=\\left \\{x|x\\neq \\frac{(2k+1)\\pi}{2} \\right \\} .$ $f^{'}(x)=cosx-2sinx+3sec^{2}x=\\sqrt{5}cos(x+\\phi)+3sec^{2}x > 0$ for all $x$ in sub-intervals of $D_{f}.$ When $x\\in (0,\\frac{\\pi}{2}), f(0)=2, \\lim_{x \\to \\frac{\\pi}{2}^{-} }f(x)=+\\infty .$ Hence $f(x)>0$ in this interval. When $x\\in (\\frac{\\pi}{2},\\pi),\\lim_{x \\to \\frac{\\pi}{2}^{+} }f(x)=-\\infty, f(\\pi)=-2.$ Hence $f(x)<0$ in this interval. When $x\\in (\\pi,\\frac{3\\pi}{2}),f(\\pi)=-2<0.$ Notice that $\\frac{5\\pi}{4}<\\frac{3.15*5}{4}<4<\\frac{3\\pi}{2}, f(\\frac{5\\pi}{4})=3-\\frac{3\\sqrt{2}}{2}>0 .$ Hence there must be a root located in the interval $(3,4).$ Choose $\\boxed{\\textbf{(D)}}$. ~PythZhou", "# Finding $f'''(x)$ of $f(x)=\\sin{\\pi}x$ Finding $f'''(x)$ of $f(x)=\\sin{\\pi}x$. First $$f'(x)=\\frac{d\\sin{\\pi}x}{d({\\pi}x)}\\frac{d({\\pi}x)}{dx} ={\\pi}\\cos{\\pi}x$$ Second $$f''(x)=-{\\pi^2}\\sin{\\pi}x$$ Third $$f'''(x)= -{\\pi^3}\\cos {\\pi}x$$ These are the steps the teacher has written down on the blackboard, I have trouble understanding \"How it is solved\". Would you please provide the steps? - LaTeX tip: There are preexisting operators \\sin and \\cos for $\\sin x$ and $\\cos x$. There is a quick reference guide here. – Aeolian Mar 6 '13 at 7:50 \\begin{align*} f(x) = \\sin\\pi x \\end{align*} Here, note that $\\pi$ is a constant. So treat it like a coefficient of $x$. Progressively, what I will do is to: • Isolate the coefficient since it is not involved in the differentiation • Apply chain rule by differentiating the trigonometric function, followed by the function \"inside\" the trigo function - the $\\pi x$. Note that $\\frac{d}{dx}\\begin{pmatrix}\\pi x\\end{pmatrix} = \\pi$. \\begin{align*} f(x) &= \\sin\\pi x \\\\ f'(x) &= (\\cos \\pi x) \\times \\pi \\space\\space\\tag{by chain rule}\\\\ &= \\pi \\cos \\pi x \\\\\\\\\\\\f''(x) &= \\pi (-\\sin \\pi x)(\\pi) \\space\\space\\tag{by chain rule}\\\\ &= -\\pi^2 \\sin \\pi x \\\\\\\\\\\\f'''(x) &= \\pi^2 (-\\sin \\pi x)(\\pi) \\space\\space\\tag{by chain rule}\\\\ &= -\\pi^3 \\cos \\pi x \\end{align*} And that is how you do it. First, understand that $\\cos$ is a trigonometric function. I would recommend you to write $f(x)$ and $f'(x)$ as a composite", "this Concept, it is possible to shift an entire graph to the left or the right by changing the argument of the graph. So in this case, you can keep your graph by changing the function to $y = \\cos(x - 3)$ ### Practice Graph each of the following functions. 1. $y=\\cos(x-\\frac{\\pi}{2})$ 2. $y=\\sin(x+\\frac{3\\pi}{2})$ 3. $y=\\cos(x+\\frac{\\pi}{4})$ 4. $y=\\cos(x-\\frac{3\\pi}{4})$ 5. $y=-1+\\cos(x-\\frac{\\pi}{4})$ 6. $y=1+\\sin(x+\\frac{\\pi}{2})$ 7. $y=-2+\\cos(x+\\frac{\\pi}{4})$ 8. $y=3+\\cos(x-\\frac{3\\pi}{2})$ 9. $y=-4+\\sec(x-\\frac{\\pi}{4})$ 10. $y=3+\\csc(x-\\frac{\\pi}{2})$ 11. $y=2+\\tan(x+\\frac{\\pi}{4})$ 12. $y=-3+\\cot(x-\\frac{3\\pi}{2})$ 13. $y=1+\\cos(x-\\frac{3\\pi}{4})$ 14. $y=5+\\sec(x+\\frac{\\pi}{2})$ 15. $y=-1+\\csc(x+\\frac{\\pi}{4})$ 16. $y=3+\\tan(x-\\frac{3\\pi}{2})$ At Grade Sep 26, 2012 ## Last Modified: Oct 28, 2014 Files can only be attached to the latest version of Modality # Reviews Please wait... Please wait... Image Detail Sizes: Medium | Original MAT.TRG.244.L.1", "If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, find $n((A\\cup B)\\times(A\\cap B)\\times(A \\triangle B))$ 9. If p(A) denotes the power set of A, then find $n(P(P(P(\\phi)))).$ 10. 3 x 3 = 9 11. Compare and contrast the graph y = x2 - 1, y = 4(x2 - 1) and y = (4x)2 = 1. 12. By using the same concept applied in previous example, graphs of y = sin x and y = sin 2x, and also their combined graphs are given figures (a), (b) and (c). The minimum and maximum values of sin x and sin 2x are the same. But they have different x-intercepts. The x-intercepts for y = sin x are $\\pm n\\pi$ and for y = sin 2x are $\\pm{1\\over 2}n\\pi,\\ n\\in Z.$ ​​​ 13. Let us now draw the graph of y = 2 sin ( x - 1 ) + 3. 14. 2 x 5 = 10 15. Let f, g: $R \\rightarrow R$ be defined as f (x) = 2x -|x| and g(x) = 2x + |x|. find fog. 16. If $f:R\\rightarrow R$ is defined by f(x) = 2x- 3, prove that f is a bijection and find its inverse" ]

[ "I think! So, I let Maxima cut through all the algebra of the RC phase shift network. Lo and behold, here we have the results, achieved almost painlessly! $f = \\frac{1}{2 \\pi \\sqrt{R_2R_3C_2C_3 + R_1R_3C_2C_3 + R_1R_3C_1C_3 + R_1R_2C_2C_3 + R_1R_2C_1C_3 + R_1R_2C_1C_2}}$ For the benefit of the reader, I have also attached the Maxima batch files which did all the awesome work.", "1x = np.linspace(-2*np.pi, 2*np.pi, 100) # domain 2np.sin(x-np.pi/2) # shifted right 3np.sin(x) 4np.sin(x+np.pi/2) # shifted left • The second subplot displays the original graph of $\\sin x$ defined within the domain $-2\\pi \\leq x \\leq 2\\pi$. • The first subplot displays the graph of $\\sin (x-\\frac{\\pi}{2})$ which is the graph of $\\sin x$ being shifted by $\\frac{\\pi}{2}$ units to the right. • The second subplot displays the graph of $\\sin (x+\\frac{\\pi}{2})$ which is the graph of $\\sin x$ being shifted by $\\frac{\\pi}{2}$ units to the left. ### Shifting in the $y$-direction Note Given $k$ is a positive real number: • The graph of $f(x)+k$ is the graph of $f(x)$ shifted vertically up by $k$ units. • The graph of $f(x)-k$ is the graph of $f(x)$ shifted vertically down by $k$ units. Example Shifting in the $y$-direction. 1x = np.linspace(-2*np.pi, 2*np.pi, 100) # domain 2np.sin(x)+1 # shifted up 3np.sin(x) 4np.sin(x)-1 # shifted down • The line in red displays the original graph of $\\sin x$ defined within the domain $-2\\pi \\leq x \\leq 2\\pi$. • The line in blue displays the graph of $\\sin(x)+1$ which is the graph of $\\sin x$ being shifted vertically up by $1$ unit. • The line in green displays the graph of $\\sin(x)-1$ which is the graph of $\\sin x$ being shifted vertically down", "draw(Label(\"$\\pi$\",1),shift(-0.15*Z)*scale3(0.5)*g,blue,Arrow3); path3 h=arc(O,1,170,-120,150,-120,CCW); draw(Label(\"$\\sigma$\",1),shift(-0.5*cos(pi/6)*Y-0.5*sin(pi/6)*X+0.35*Z)*scale3(0.5)*h,red,Arrow3); draw(shift(-0.5*cos(pi/6)*Y-0.5*sin(pi/6)*X-0.15*Z)*scale3(0.1)*unitsphere,gray+opacity(.3)); and the result You can easily modify the xpt and ypt values to find the best shape. For periodic spline, do not forget to have periodic value. O.G. • Great! But may I ask where the lower bound in (0.1,2*pi) comes from? – user121799 Jan 4 '18 at 0:28 • @marmot I add a comment. In view of your attempt, I transform the box (a,b) (-0.67,0.1),(0,2.032*pi) to (-0.67,0),(0.1,2*pi) so that the x variable varies from -0.67 to 0.1 while the y variable varies from 0 to 2*pi. Perhaps that (-0.67,0),(0,2*pi) should be ok... – O.G. Jan 4 '18 at 9:48", "New Zealand Level 8 - NCEA Level 3 # Phase shifts for tangents ## Interactive practice questions Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. A phase shift has the same effect as a horizontal translation. True A False B True A False B Easy Less than a minute Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. Consider the graph of $y=\\tan x$y=tanx and state whether the following sentence is true or false. We want to identify how the coordinates of key points on the graph of $f\\left(x\\right)=\\tan x$f(x)=tanx change as we apply a phase shift to produce the graph of $g\\left(x\\right)=\\tan\\left(x-\\frac{\\pi}{3}\\right)$g(x)=tan(xπ3). ### Outcomes #### M8-2 Display and interpret the graphs of functions with the graphs of their inverse and/or reciprocal functions", "cos((-2*pi*ant_spacings(3)/lambda)*sin(theta(i)*(pi/180) - 2*pi/3))]; % Make the steering vectors complex exp((1i*(-2*pi*ant_spacings(1)/lambda)*sin(theta(i)*(pi/180) - pi/3))); exp((1i*(-2*pi*ant_spacings(2)/lambda)*sin(theta(i)*(pi/180) - pi/2))); exp((1i*(-2*pi*ant_spacings(3)/lambda)*sin(theta(i)*(pi/180) - 2*pi/3)))]; spectrum_func(i) = 1/((steering_vec'*noise_vecs)*noise_vecs'*steering_vec); end % grab theta corresponding to highest peak, round to nearest half integer [mx,argmax] = max(spectrum_func); estimated_AoA = floor(2*theta(argmax)+.5)/2; fprintf(\"Estimated AoA: %.1f \\t Hard-coded AoA: %.1f.\\n\", estimated_AoA, AoA) plot(theta,20*log10(abs((spectrum_func)))) title('MUSIC Psuedospectrum') xlabel('Angle of Arrival') $$$$ • Ah, I thought this might have had something to do with it. Is there any way you could elaborate on the 'using a real sinusoid will introduce a phase shift term in the opposite angle direction' aspect? – OGBerglemir Nov 19 '19 at 20:00 • @OGBerglemir When you multiply two complex exponentials, you can add the general phase terms together. If you recall, a real sinusoid is composed of two complex exponentials, which rotate in different directions. Multiplying by a real sinusoid uses both components, so you will shift a signal's phase using both, and not one as intended. – Envidia Nov 19 '19 at 20:07 • Ah, understood. – OGBerglemir Nov 19 '19 at 21:59", "0 to 2*pi in steps of 0.01 x1 = t; y1 = amplitude * sin(frequency*t); % Produce plot plot (x1,y1); title 'original' grid % Define transform parameters scale = 5; % 5 times bigger shift = [ -15 -10 ]; % shift 15 units left, and 10 units down theta = pi/6; % 30 degrees rotation_matrix = [ cos(theta), -sin(theta) sin(theta), cos(theta) ]; % Transform points via scaling, shifting, and rotation % make empty arrays to store new points in x2 = zeros(size(t)); y2 = zeros(size(t)); % Transform points one at a time for i = 1:length(t) old_xy = [ x1(i) y1(i) ]; % calculate transformed x and y new_xy = ( rotation_matrix * old_xy * scale ) + shift; % Store calculated values to vector x2(i) = new_xy(1); y2(i) = new_xy(2); end % Make a pretty picture figure; plot(x2,y2); set(gca,'XLim',[-20,20],'YLim',[-20,20]) title 'transformed' grid The ingredients for what you want are here: you just have to figure out how to get the sine wave to start and stop at your desired endpoints. I leave this exercise in trigonometry up to you. (Note: If you can't figure out how to find the transformation that will give a sine wave starting a point A and ending at point B, it would be better to ask math.stackexchange.com instead of StackOverflow -", "geometric figures. /* draw figures */ real f1 (real x) {return sin(x);} draw(graph(f1,-6.21,8.03)); draw((-3.14,0)--(0,0), zzttqq); draw((0,0)--(0,3.14), zzttqq); draw((0,3.14)--(-3.14,pi ), zzttqq); draw((-3.14,pi )--(-3.14,0), zzttqq); draw(circle((0,3.14),1.14)); draw(shift((3.35,2.06))*rotate(1)*xscale(1.59)*yscale(1.1)*unitcircle); pair hyperbolaLeft1 (real t) {return (0.49*(1+t^2)/(1-t^2),1.04*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (0.49*(-1-t^2)/(1-t^2),1.04*(-2)*t/(1-t^2));} draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaLeft1,-0.99,0.99)); draw(shift((3.35,2.06))*rotate(1)*graph(hyperbolaRight1,-0.99,0.99)); /* hyperbola construction */ The first line real f1 (real x) {return sin(x);} defines the sin() function to be drawn, and the next line draw(graph(f1,-6.21,8.03)); actually draws the graph to the diagram. The graph() command comes from the Asymptote graph package. The next few lines draw several line segments. The attribute zzttqq is a pen whose color is mapped from some rgb value (with conversions ... 8 = y, 9 = z, a = a, b = b, ... ). ## Options An outline of the different available options. (The first three options are related and automatically update with respect to eath other). X/Y Units (cm): Indicates how many units in the picture correspond to 1 centimeter. Picture width/height: Indicates what size the Asymptote output should be, in centimeters. xmin/xmax/ymin/ymax: Indicates the dimensions of the picture in the picture's coordinates. (The next options change other features). Font size: Changes the default font size. The value defaults at 10pt, but I personally recommend 7pt for any file with multiple text lines, which more closely approximates the size of the font", "plot (contours only) filled.contour() Level (contour) plot, filled The chunks below show some basic plot for these types, which can be highly customized if needed. For some more information about plotting check out the Plotting chapter. #### Generic X-Y Plot (Matrix) First, let us generate a matrix with some data which we will use for plotting. The following two lines generate a matrix m of dimension $$200 \\times 3$$ where each column contains one full period of sine along the unit circle ($$0$$ to $$2 \\cdot \\pi$$) with different phase shifts ($$0$$, $$\\frac{1}{2} \\pi$$, and $$\\pi$$). # Sequence from 0 to 2 * pi x <- seq(0, 2 * pi, length.out = 200) # Calculate sin(x + shift) m <- cbind(\"shift: 0\" = sin(x), \"shift: 1/2 pi\" = sin(x + 1 / 2 * pi), \"shift: pi\" = sin(x + pi)) head(m, n = 3) ## shift: 0 shift: 1/2 pi shift: pi ## [1,] 0.00000000 1.0000000 1.224647e-16 ## [2,] 0.03156855 0.9995016 -3.156855e-02 ## [3,] 0.06310563 0.9980069 -6.310563e-02 When calling plot(m), R will automatically take the first two columns of the matrix and plot them against each other, creating a 2d scatter plot. plot(m, main = \"Generic X-Y Plot\") This is nothing else than plotting plot(m[, 1], m[, 2]). #### Matrix plot matplot() is plotting data columnwise. Each", "$t$ then the term $-vt$ is just some phase-shift and we can see the graph is just a horizontal shift of the sine wave. In particular we shift the $y = A \\sin(x)$ graph $vt$-units to right to form $y = A \\sin(x-vt)$. This is a right-moving wave if we assume $v>0$. To see it, I think setting $0 \\leq x-vt \\leq 2\\pi$ might be helpful, then $vt \\leq x \\leq 2\\pi+vt$ and if we animate $t$ we can actually see the wave travel. Other more complicated wave graphs could be studied. I happen to have this graph of a solution to the wave equation sitting around. Of course, adding graphs has interesting interpretations in terms of constructive and destructive interference. Or, if we add time I think you can get beats. There is much to explore here. Edit: as Dave Renfro pointed out, I am using forbidden trigonometry. To, give a less exciting, but very much applied example consider: $$y = y_o- \\frac{g}{2}t^2$$ If we start with $y=t^2$ then multiplying b $-g/2$ stretches the graph and flips it over. The significance is that $g$ is the acceleration due to gravity. Then, adding $y_o$ shifts the graph $y=-gt^2/2$ up by $y_o$ and in this discussion $y_o$ has the significance of being the initial condition. A more ambitious formula", "# Get slopes of integer shift of an image in radians I tried to reproduce the result from this paper, However I'm stuck at the Section III of the paper which involve integer shift. Two images which their phase correlation $\\theta(k_1,k_2)$ was calculated at pixel-accuracy(integer shift). I get their phase component as shown(from -$\\pi$ to $\\pi$). It can be represented by $\\theta(k_1,k_2)=\\theta_1(k_1,k_2)-\\theta_2(k_1,k_2)$. Since the two images shifted in parallel, it can be approximated as $\\theta(k_1,k_2) \\approx ak_1+bk_2$ where $a$ and $b$ is the slopes obtained through least square method, $k_1$ and $k_2$ are the value $MxN$ of the image. Plotting $\\theta(k_1,k_2)$: And to get phase component with only sub-pixel accuracy(decimal shift), the phase component is subtracted with integer shift. The equation is $\\theta''(k_1,k_2)=\\theta(k_1,k_2)-(a'k_1+b'k_2)$ However, as you can see from the second figure, the value of $\\theta''(k_1,k_2)$ ranges wider than -$\\pi$ to $\\pi$(as compared to Figure 4 in the paper), which I suspect the slope($a'$,$b'$) was obtained inaccurately. The paragraph above the Equation (14) mentioned that they used conventional POC to obtain the slopes. However I'm not sure did I understood it correctly. The code I implemented: function [ output_args ] = phasecorrlsa( refIm, shifIm ) F=(fft2(double(refIm))); G=(fft2(double(shifIm))); [m, n]=size(refIm); [M,N] = meshgrid(1:m,1:n); X = [M(:), N(:)]; R=(F.*conj(G))./abs(F.*conj(G)); r=(ifft2(R)); % surf(r) [apx, bpx, rhat]=lsa(angle(r)); [~,w] = max(r(:)); [del_hat2p, del_hat1p]", "sure so I experimented, I decided to use a sine graph with a wavelength of 20 y=sin nx the wavelength is 2pi/n = 20 so n = pi/10 so my experimenting graph is $$y=sin(\\frac{\\pi }{10}x)\\\\ y=sin(\\frac{\\pi }{10}x)=sin(\\frac{\\pi }{10}(x-20))\\\\ let \\;\\;\\frac{x}{5}\\;\\;be\\;\\;substituted\\;\\;for\\;\\;x\\\\ y=sin(\\frac{\\pi }{10}\\frac{x}{5})=sin(\\frac{\\pi }{10}(\\frac{x}{5}-20))=sin(\\frac{\\pi }{10}(\\frac{x-100}{5}))$$ So the smallest a that we know with certainty is 100. Here is the graph i was playing with. https://www.desmos.com/calculator/oonutt20vz ------------------- Let me see if I can do this more simply. $$y=f(x)=f(x-20)\\\\ let \\;\\;\\frac{x}{5}\\;\\;be\\;\\;substituted\\;\\;for\\;\\;x\\\\ y=f(\\frac{x}{5})=f(\\frac{x}{5}-20)=f(\\frac{x-100}{5})$$ There you go, that was simple. The right shift is 100. Mar 27, 2018 #5 +100025 +1 2. What are the coordinates of the points where the graphs of $f(x)=x^3-x^2+x+1$ and $g(x)=x^3+x^2+x-1$ intersect? Give your answer as a list of points separated by semicolons, with the points ordered such that their $x$-coordinates are in increasing order. (So \"(1,-3); (2,3); (5,-7)\" - without the quotes - is a valid answer format.) $$x^3-x^2+x+1 = x^3+x^2+x-1\\\\ -x^2+1 = x^2-1\\\\ 2=2x^2\\\\ x^2=1\\\\ x=\\pm1\\\\ f(1)=1-1+1+1=2\\\\ f(-1)=-1-1-1+1=-2\\\\$$ so the points of intersection are (-1,-2); (1,2) check: Mar 27, 2018", "in the sense that the sine Fourier transform is imaginary, so a phase offset should represent some non-zero amplitude distributed in real and imaginary parts, and they should cancel when combined to take the absolute value. After reading up on the Fourier transform of a sinusoid with frequency shift (Fourier transform of the Cosine function with Phase Shift?), I have seen that it is the Dirac multiplied by the sum of a real and imaginary exponential: $\\pi (e^{-j\\theta}\\delta(w-w_0) + e^{j\\theta}\\delta(w-w_0))$ So I tried to do the same taking into account the Gaussian envelope: X(w) = $\\sum_{k=1}^N \\frac{A_k}{2} (e^{\\phi i} + e^{- \\phi i}) \\sqrt{2\\pi\\sigma_k^2} e^{-\\pi^2 (f_k-w)^2 2\\sigma_k^2}$ However it does not work. The following Python program implements what I have tried: #!/usr/bin/env python3 from matplotlib import pyplot as plt import numpy as np def time_gabor(x, var, mean, f, phi, amp): return (1.0 # * (1.0/(var*np.sqrt(2*np.pi))) * np.exp(-(x-mean)**2 / (2*var**2)) * np.cos((x-mean)*2*np.pi*f + phi)) * amp def fourier_gabor(x, var, mean, f, phi, amp): a = 2*var**2 g = np.sqrt(np.pi*a) * np.exp(-np.pi**2 * (f-w)**2 * a) return (np.exp(1.0j*phi) + np.exp(-1.0j*phi)) * g*amp/2 #return (np.cos(phi) - np.sin(phi))*g*amp def fft(y, sample_rate): ft = np.abs(np.fft.fft(y)[:len(y)//2]) / (sample_rate/2) ws = np.arange(len(y)) / len(y) * sample_rate return ws, ft N = 100000 x = np.linspace(-10,10,N).reshape(-1,1) w = np.linspace(0,5,N).reshape(-1,1) centers = np.array([0, 0, 8]) freq =", "$$\\pi k$$ (the c’s (cot and csc) have the “easier” ones). To get the new asymptotes, set the trig argument to asymptotes, and solve for x. For example, if we have $$\\tan \\left( {3x+\\pi } \\right)\\,\\,(\\text{which would be}\\,\\,\\tan \\left( {3\\left( {x+\\frac{\\pi }{3}} \\right)} \\right))$$, we would solve: $$\\displaystyle 3x+\\pi =\\frac{\\pi }{2}+\\pi k;\\,\\,\\,\\,\\,\\,3x=\\left( {\\frac{\\pi }{2}-\\pi } \\right)+\\pi k;$$ $$\\displaystyle 3x=-\\frac{\\pi }{2}+\\pi k;\\,\\,\\,\\,x=\\frac{{-\\frac{\\pi }{2}}}{3}+\\frac{{\\pi k}}{3};\\,\\,\\,\\,x=-\\frac{\\pi }{6}+\\frac{{\\pi k}}{3}$$. 3. The new range will still be $$\\left( {-\\infty ,\\,\\infty } \\right)$$. 4. Draw asymptotes first; phase shift will take care of itself. (To get an asymptote starting point, you can set k = 0 in your new asymptote equation, k = –1 for one to the left, k = 1 for one to the right). Right in between the asymptotes (you can take average of the x’s), draw the middle of the graph (but make sure it is shifted up or down according to the vertical shift, or d). 5. New period will be $$\\frac{\\pi }{b}$$ (since tan and cot have periods of ). This should be the distance between asymptotes (the coefficient of “k” in the asymptote equations). 6. If trig function is negative, make sure you flip the graph. 7. New domain will be all x values except for the asymptotes. 8. To get two more points on either side", "and cosine inputs at a precise time. That is, the graph of f(x) = 3sin x is shifted π … Take any phase shift and add or subtract any multiple of the period and that will produce another valid phase shift. Figure %: Horizontal shift The graph of sine is shifted to the left by units. $\\begingroup$ One can assign many phase shifts to the same function, each of which will produce the same graph. Amplitude, Period, Frequency and Phase Shift . For example, the function f(x) = 3sin (x – π) has a phase shift of π. A phase-shift mask relies on the fact that light passing through a transparent media will undergo a phase change as a function of its optical thickness. Phase Shift. If F {\\displaystyle F} is a \"canonical\" function for a class of signals, like sin ⁡ ( t ) {\\displaystyle \\sin(t)} is … So if has a phase shift of any multiple of , then it will just overlay the original graph. Phase-shift keying (PSK) is a digital modulation scheme based on changing, or modulating, the initial phase of a carrier signal. They are separated only by a phase shift: . Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math", "# Phase shift in frequency domain Hello, I'using Matlab to simulate phase shift in frequency domain (FD). I have got real and imaginary parts of the signal after FFT. I'd like to use phase shift in FD. This works: Y=fft(y); YY=Y.exp(-i*2*pi*nk/N*samples_delay); result=ifft(YY); But in my DSP I can't use the formula above and I need to use real and imaginary parts from fft signal. I supposed exp(-i*2*pi*nk/N*samples_delay) can be transfered to this: Shifted real part = re*cos(2*pi*nk/N*samples_delay) Shifted imaginary part = im*sin(-2*pi*nk/N*samples_delay) Then I transfered it back to time domain but result is inccorect. Could anybody help me? Thank you Related Other Physics Topics News on Phys.org DrGreg Gold Member Think about what you are doing. Is it true that$$(a + ib) (x + iy) = ax + iby?$$ sophiecentaur blue_leaf77 Homework Helper Shifted real part = re*cos(2*pi*nk/N*samples_delay) Shifted imaginary part = im*sin(-2*pi*nk/N*samples_delay) From what I understand, you actually want to calculate the real and imaginary parts of YY, am I right? Then you have to calculate Re[Y.exp(-i*2*pi*nk/N*samples_delay)] and Im[Y.exp(-i*2*pi*nk/N*samples_delay)]. Consider this $$Ze^{i\\theta} = (Re[Z] + i Im[Z])(\\cos(\\theta)+i\\sin(\\theta) )$$ Calculate the last expression to see which are the right real and imaginary parts of ##Ze^{i\\theta}##.", "layers. In order for my approach to work as intended, the layer settings window must look like this: However, quite annoyingly, when you close and reopen this saved document, the layer settings window looks like this: So, it must be changed back each time to allow trouble-free math drawing. # Finding Reciprocal Functions SEC, CSC, COT on TI-30XA First, you may want to refresh your memory of the regular trig functions: Now, let’s find $\\csc \\frac{5\\pi }{4}$. If we look it up on WolframAlpha, we find that it equals $-\\sqrt{2}$, or, roughly, $-1.4142$. To get this answer on the TI-30XA, first remember that $\\csc$ is equal to $\\frac{1}{\\sin }$. So, what we’re looking for could be expressed as $\\frac{1}{\\sin \\frac{5\\pi }{4}}$. Once we know this, we can type $5\\cdot \\pi$ into the TI-30XA. This gives us ~15.7. (Make sure you’re in radian mode. If not, press the DRG button — located by the 2nd button, upper-left — until you see RAD appear in small letters on the screen.) Now, we can divide 15.7 by 4. This gives us 3.925. If we take the $\\sin$ of this (by pressing the “sin” button while 3.925 is displayed), we get $-0.7$. IF we divide 1 by $-0.7$, we get $-1.43$. As you can see, this is close to the correct value.", "import numpy as np sr = 22050 ts = 1.0/sr duration = 2 t = np.arange(0,duration,ts) t_change = 1 freq1 =4.20 phase1 = 2*np.pi*freq1* t[t<=t_change] freq2 = 6.66 phase2 = 2*np.pi*freq2*t[t>t_change] + 2*np.pi*(freq1-freq2)*t_change total_phase = np.concatenate((phase1, phase2)) out = np.cos(total_phase) Considering more generalized cases with multiple steps, the phase offset to add for any given segment is: $$\\theta_n = \\sum_{m=1}^{n-1}2\\pi f_m t_m - \\sum_{m=1}^{n-1}2\\pi f_{m+1} t_m \\tag{1} \\label{1}$$ With the resulting phase ramp proceeding starting at $$t_n$$ as: $$\\theta(t) = 2\\pi f_n t + \\theta_n$$ This is demonstrated with a 3 segment ramp with reference to the figure below, where we note that each next phase ramp will start at the completed phase from the previous frequency. For example, the second phase offset given as $$\\theta_2$$ below is similarly determined as above by solving for the intersection of linear line equations to be: $$\\theta_2 = \\theta(t_2)-2\\pi f_3 t_2$$ $$= 2\\pi f_2 t_2 + \\theta_1 - 2\\pi f_3 t_2$$ $$= 2\\pi f_2 t_2 + 2\\pi (f_1-f_2)t_1 - 2\\pi f_3 t_2$$ Which is consistent with the generalized form given in $$\\ref{1}$$ as: $$\\theta_2 = 2\\pi f_1 t_1 + 2\\pi f_2 t_2 -2\\pi f_2 t_1 - 2\\pi f_3 t_2$$ With this, the third segment plotted below is given as: $$\\theta(t) = 2\\pi f_3 t + \\theta_2$$ Summary: In summary I have", "## Trigonometry (11th Edition) Clone Published by Pearson # Chapter 4 - Graphs of the Circular Functions - Section 4.3 Graphs of the Tangent and Cotangent Functions - 4.3 Exercises - Page 171: 9 B #### Work Step by Step The function is $y=\\tan (x-\\frac{\\pi}{4})$. When we compare $y=\\tan (x-\\frac{\\pi}{4})$ to the general form $y=c+a\\tan (bx-d)$, we find that $d=\\frac{\\pi}{4}$. $d$ represents the phase shift and since $d$ is positive, the phase shift is $\\frac{\\pi}{4}$ units to the right. Also, since $a=1$, $b=1$ and $c=0$, we can ascertain that the graph of $y=\\tan (x-\\frac{\\pi}{4})$ will be the same as the graph of $y=\\tan x$ except that it will be shifted $\\frac{\\pi}{4}$ units to the right. This means that the two vertical asymptotes will now be $-\\frac{\\pi}{2}+\\frac{\\pi}{4}=-\\frac{\\pi}{4}$ and $\\frac{\\pi}{2}+\\frac{\\pi}{4}=\\frac{3\\pi}{4}$. This means that the correct graph for $y=\\tan (x-\\frac{\\pi}{4})$ is B. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.", "# Coefficients in Trigonometric Functions So, previously in my math classes I have been taught that The phase shift of $$\\sin(ax+c)$$ is c. But in my new class they say that $$\\sin(ax+c) = \\sin(a(x+c))$$ which I believe to be wrong. But I have an exam tommorrow and I do not wish to be marked down for having the wrong phase shift. - The phase shift is $c$. The equation $\\sin(ax+c) = \\sin(a(x+c))$ is false in general (it's true if $a=1$ of course). The $a$ is the frequency parameter, and doesn't affect the phase shift. You can think of phase shift as the angle you get at time zero, or in this case when $x=0$. – Alan Guo Oct 22 '12 at 4:59 So I was correct? And my professor is teaching it wrong? In the textbook it says: $$y=\\csc(2x-\\pi/4) = \\csc 2(x-\\pi/8)$$ – Matt Jensen Oct 22 '12 at 5:03 The equation you cite from your textbook is very different from the one you display in the body of the question. – Gerry Myerson Oct 22 '12 at 5:08 Not in theory. It is essentially the same idea. Isn't it? – Matt Jensen Oct 22 '12 at 5:09 No, no, no. $ax+c$ is not the same as $a(x+c)$, but $2x-(\\pi/4)$ is the same as $2(x-(\\pi/8))$, in theory, in", "We’re rewarding the question askers & reputations are being recalculated! Read more. # Tag Info 0 You are really close! Change your signal and steering vectors to be complex. Specifically for the steering vectors, these coefficients are meant to act as phase shifts. Using a real sinusoid will introduce a phase shift term in the opposite angle direction, which you don't want. Doing this alone you will see an improvement in your pseudospectrum. In regards ... 4 You're almost there; you just need to connect a few dots. Let your $\\frac{10}{5 + i 2 \\pi f} = G(f)$. Then $g(t) = 10 u(t) e^{-5 t}$. Now we get into that exponent part. Your $h(t) = g(t - t_0)$ is correct, but you're applying it incorrectly. You need to apply it to the part that I've labeled as $g(t)$: $h(t) = g(t - t_0) = 10 u(t - t_0) e^{-5 (t - t_0)}... 0 The principal idea behind my recommended approach is to identify characteristic points in your cycle. You can then make a mapping from time within the cycle (as it is occuring) and the location of that point within your representative cycle. Once you have a set of mapping points, create a best fit function. This will then give you a mapping function of ... 1 You" ]

[ "A sequence is defined over non-negative integral indexes in the following way: $$a_{0}=a_{1}=3$$, $$a_{n+1}a_{n-1}=a_{n}^{2}+2007$$. Find the greatest integer that does not exceed $$\\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}$$ (第二十五届AIME1 2007 第14题)", "### Home > PC > Chapter 2 > Lesson 2.2.1 > Problem2-43 2-43. Write the following expression using summation notation. $1^2+2^2+3^2+\\dots+9^2+10^2$ How many terms are being added? Start your index at $1$. Stop at the last term. (Alternatively, you could start at $0$ and end at $9$.) $\\displaystyle\\sum\\limits_{k=1}^{10}$ Find the argument that will result in the desired expansion. $\\displaystyle\\sum\\limits_{k=1}^{10}k^2$", "index: click on a letter A B C D E F G H I J K L M N O P Q R S T U V W X Y Z A to Z index index: subject areas numbers & symbols sets, logic, proofs geometry algebra trigonometry advanced algebra & pre-calculus calculus advanced topics probability & statistics real world applications multimedia entries Bounded Sequence A sequence with terms that have an upper bound and a lower bound. For example, the harmonic sequence is bounded since no term is greater than 1 or less than 0.", "+0 help 0 89 1 The sum of the first terms of a certain sequence is $$n(n + 1)(n + 2)$$. Find the tenth term of the sequence. Mar 5, 2019 $$\\text{let }s_n = \\sum \\limits_{k=1}^n~a_k = n(n+1)(n+2)\\\\ s_{10}=s_9+a_{10}\\\\ a_{10} = s_{10}-s_9 = (10)(11)(12) - (9)(10)(11) = 330$$", "# Section5.1An Overview of Calculus¶ permalink When we studied sequences, we learned that the behavior of the sequence could be described in terms of its increments. (Recall 3.4.) The following summarizes what we learned about sequences. ##### Behavior of a Sequence Using Increments Every sequence $(x_k)$ has a corresponding increment sequence, $\\Delta x_k = x_{k} - x_{k-1}$. The dynamic behavior of the sequence $x$ can be described using the attributes of the increments as summarized below. • If the increments are positive on a range, $\\Delta x_k \\gt 0$ for all $k=m,\\ldots,n$, then the sequence $x$ is increasing on the index range $k=m-1,\\ldots,n$. • If the increments are negative on a range, $\\Delta x_k \\lt 0$ for all $k=m,\\ldots,n$, then the sequence $x$ is decreasing on the index range $k=m-1,\\ldots,n$. • If the increments are increasing on a range, $\\Delta x_k \\gt 0$ for all $k=m,\\ldots,n$, then the sequence $x$ is concave up on the index range $k=m-1,\\ldots,n$. • If the increments are decreasing on a range, $\\Delta x_k \\lt 0$ for all $k=m,\\ldots,n$, then the sequence $x$ is concave down on the index range $k=m-1,\\ldots,n$. ##### Note5.1.1 The range of index values for the sequence always starts one index earlier than the corresponding range of index values describing the increments because each increment is computed as a", "When computing the increments, we had to make a choice whether to do the backward difference or the forward difference, \\begin{equation*} \\Delta x_k = x_{k+1} - x_k\\text{.} \\end{equation*} The values of the backward differences and the forward differences are exactly the same. They differ only in terms of index values. I have chosen to use backward differences in order for our later work with accumulation sequences and summation to work out more cleanly. ###### Example3.1.6 For the sequence $x = (x_k)_{k=3}^{10} = (1,2,4,7,11,16,22,29)\\text{,}$ find the increments defined by the backward difference. Solution We can find the values by writing the sequence as a row of values. Then below the gaps between the values, we can write the differences. When the increments are defined by the backward difference, the index starts one value later than the original sequence. Thus, the increments are the sequence \\begin{equation*} \\nabla x = (\\nabla x_k)_{k=4}^{10} = (1, 2, 3, 4, 5, 6, 7)\\text{.} \\end{equation*} Because all of the increments are positive, $x$ is increasing on the full interval $\\{3,\\ldots,10\\}\\text{.}$ We can therefore analyze monotonicity of a sequence if we know the signs of its increments. ### Subsection3.1.4Concavity We now turn our attention to the concavity of a sequence. Concavity is based on whether the sequence of increments is increasing or decreasing. When the increments", "the original sequence {an}. Is it OK to use n again for the sequence of supremums of the tails? Yes. It really doesn't matter. The only place where indices have to be handled carefully is in the defining expression lim[sup{an:n≥k}], where k -> ∞. The sequence itself can be labelled with any index you want.", "the number of the last event that begins before the $$k^{th}$$ interval begins, defined formally as $n_k = \\sum_{i=1}^J I\\left[\\sum_{j=0}^i \\left(A_j + B_j\\right) < (k - 1) L \\right]$ for $$k = 1,...,K$$. It follows that $$W_k$$ will be equal to one if and only if event $$n_k$$ ends after the active part of interval $$k$$: $W_k = I \\left[\\sum_{j=1}^{n_k} \\left(B_{j-1} + A_j\\right) \\geq (k - 1) L + c \\right],$ for $$k=1,...,K$$. The summary measurement $$Y^W$$ is then calculated as the proportion of intervals during which the behavior is observed at any point: $$\\displaystyle{Y^W = \\sum_{k=1}^K W_k / K}$$.", "### Home > PC > Chapter 2 > Lesson 2.2.1 > Problem2-43 2-43. Write the following expression using summation notation. Homework Help ✎ 12 + 22 + 32 + … + 92 + 102 How many terms are being added? Start your index at 1. Stop at the last term. (Alternatively, you could start at 0 and end at 9.) $\\displaystyle\\sum\\limits_{k=1}^{10}$ Find the argument that will result in the desired expansion. $\\displaystyle\\sum\\limits_{k=1}^{10}k^2$", "sequence with $n$th term $u_n=311-7n$. Find the 100th term of the sequence. Find the first negative term in the sequence. Find the number of terms required for the sum of the series with $n$th term $u_n=3+4(n-1)$ to exceed 190. Evaluate $\\sum_{k=3}^{42}(6k-5)$.", "Precalculus (10th Edition) $\\sum_{k=0}^{n-1} ar^k$ We are given the sum: $a+ar+ar^2+...+ar^{n-1}$ The general term of the sum is $ar^k$. The number of terms is $n$, so the index $k$ goes from 0 to $n-1$. We write the sum: $a+ar+ar^2+...+ar^{n-1}=\\sum_{k=0}^{n-1} ar^k$", "sequence defined by $$z_n=r+s\\exp(2\\pi ni/3)+t\\exp(4\\pi ni/3)$$ has period $3$. All you need to do is to fins $r$, $s$ and $t$ such that the first three terms are $a$, $b$, $c$.", "intervals of $N$ consecutive composites, we can break each interval up into $N/M$ disjoint intervals of $M$ consecutive composites, giving a total of $kN/M$ intervals, and so $C(M, kN/M) \\leq C(N, k)$. So, we have $C(N,k) \\leq C(kN, 1) \\leq (kN+1)!+kN+1$. However, these bounds give $15=C(2,2)\\leq C(4,1)=27\\leq 5!+5=125$, which doesn't seem very tight. Can anyone come up with better bounds or asymptotics?", "$k$ to every term of the sequence to obtain a positive, monotone, decreasing sequence that converges to $k$.", "ar^{n-1} = \\sum_{k=0}^{n-1} ar^k = a \\sum_{k=0}^{n-1} r^k$ As for $S_n= \\frac{a}{1-r}$, that is in fact the limiting sum of a geometric sequence for $|r|< 1$, so yes, you're correct in that statement. Hope that helped", "using BIT, where $id_i$ is the original index of the interval, and the set of the first situation equals to $sum(1e9)-sum(y_i)$. Then we use $2^n-1$ minus the two sets and we can get $S_{i,0}$. As the bounds may be large, we need to discrete it. Total time complexity: $O(nlogn)$.", "# Definition:Geometric Sequence/Term Let $\\sequence {x_n}$ be a geometric sequence in $\\R$ defined as: $x_n = a r^n$ for $n = 0, 1, 2, 3, \\ldots$ $x_n$ for $n = 0, 1, 2, 3, \\ldots$ are the terms of $\\sequence {x_n}$.", "$$-\\ln(2) < 0$$. Therefore the sequence $$\\sum\\limits_{k=1}^{n}\\frac{(-1)^k}{k}(-1)^n=(-1)^n u_n$$", "$n$ matches the position in the sequence. However in the sequence we are working with, the first term is 4, which is 2², with the index being 1 more than $n$. The terms can then be given as ${2}^{n + 1}$ The general term would be ${T}_{n} = {2}^{n + 1}$", "2020-13 An integral sequence Let $$a_n$$ be a sequence defined recursively by $$a_0 = a_1 = \\dots = a_5 = 1$$ and $a_n = \\frac{a_{n-1} a_{n-5} + a_{n-2} a_{n-4} + a_{n-3}^2}{a_{n-6}}$ for $$n \\geq 6$$. Prove or disprove that every $$a_n$$ is an integer. GD Star Rating" ]

[ "\\pi}{2n+1}}{1-\\exp \\frac{i \\pi}{2n+1}}$$ So $$\\sum_{k=1}^n \\cos \\frac{k \\pi}{2n+1} (1 + i \\tan \\frac{k \\pi}{2n+1} ) = \\frac{1-\\exp \\frac{i (n+1) \\pi}{2n+1}}{1-\\exp \\frac{i \\pi}{2n+1}}$$ Sorry to keep prolonging this but I still don't see where this is going. I mean how are we ever going a product of tangents from this! tiny-tim Homework Helper Hi ehrenfest! You're getting confused. You've used ∑ instead of ∏. Your: $$\\sum_{k=1}^n \\cos \\frac{k \\pi}{2n+1} (1 + i \\tan \\frac{k \\pi}{2n+1} ) = \\sum_{k=1}^n (\\cos \\frac{\\pi}{2n+1}+ i \\sin \\frac{\\pi}{2n+1})^k$$ should be: $$\\Pi_{k=1}^n \\cos \\frac{k \\pi}{2n+1} (1 + i \\tan \\frac{k \\pi}{2n+1} ) = \\Pi_{k=1}^n (\\cos \\frac{\\pi}{2n+1}+ i \\sin \\frac{\\pi}{2n+1})^k = (\\cos \\frac{\\pi}{2n+1}+ i \\sin \\frac{\\pi}{2n+1})^{\\sum{k=1}^{n} k}\\\\,=\\,....$$ hmm … the LaTeX hasn't come out right … but I hope you see what I mean! Hi ehrenfest! You're getting confused. You've used ∑ instead of ∏. Your: should be: $$\\Pi_{k=1}^n \\cos \\frac{k \\pi}{2n+1} (1 + i \\tan \\frac{k \\pi}{2n+1} ) = \\Pi_{k=1}^n (\\cos \\frac{\\pi}{2n+1}+ i \\sin \\frac{\\pi}{2n+1})^k = (\\cos \\frac{\\pi}{2n+1}+ i \\sin \\frac{\\pi}{2n+1})^{\\sum{k=1}^{n} k}\\\\,=\\,....$$ hmm … the LaTeX hasn't come out right … but I hope you see what I mean! Well you said sum in post #4. Sorry I still don't see where we are going with this. I agree with what you posted above and the sum of the first n integers is n(n+1)/2 but", "since $a_n \\to 0$ then $\\mathop {\\lim }\\limits_{n \\to \\infty } {S_n} = \\mathop {\\lim }\\limits_{n \\to \\infty } (a_1 - a_{n+1}) = a_1$. Applying this fact for the sequence $a_k:=\\sin (\\frac{1}{2k})$, we get that \\begin{align*} \\sum\\limits_{n = 1}^\\infty {\\left( {\\sin \\frac{1}{{2n}} - \\sin \\frac{1}{{2n + 1}}} \\right)} = \\sin \\left( {\\frac{1}{2}} \\right) \\end{align*} - But here we have $\\sum (a_{2k} - a_{2k+1})$, or, if you wish, $\\sum (a_k - a_{k+\\frac{1}{2}})$, it doesn't telescope. – Daniel Fischer Apr 28 '14 at 11:25 Ah yes. You are right. But how come $k+\\frac{1}{2}$ in $a_{k+\\frac{1}{2}}$ – mwomath May 6 '14 at 11:18 @user45105 $a_k=\\sin(1/(2k))\\implies a_{k+1/2}=\\sin(1/(2(k+1/2)))=\\sin(1/(2k+1))$ – oldrinb May 7 '14 at 22:19 If we substitute $m=2n$, then \\begin{align} \\sum\\limits_{n = 1}^\\infty {\\left( {\\sin \\frac{1}{{2n}} - \\sin \\frac{1}{{2n + 1}}} \\right)} &= \\sum\\limits_{m = 2}^\\infty {\\left( {\\sin \\frac{1}{m} - \\sin \\frac{1}{{m + 1}}} \\right)} \\\\ &= \\sum\\limits_{m = 1}^\\infty {\\left( {\\sin \\frac{1}{m} - \\sin \\frac{1}{{m + 1}}} \\right)} - \\left( {\\sin 1 - \\sin \\frac{1}{2}} \\right) \\\\ &= \\sin 1 - \\left( {\\sin 1 - \\sin \\frac{1}{2}} \\right)\\\\ &= \\sin \\frac{1}{2} \\end{align} where as above $\\sum\\limits_{m = 1}^\\infty {\\left( {\\sin \\frac{1}{m} - \\sin \\frac{1}{{m + 1}}} \\right)} = \\sin \\left( 1 \\right)$ – mwomath Jun 3 '14 at 10:59", "- \\sin^2{x})\\cos{x}}{1 - 2\\sin^2{x}}\\,dx}$ Make the substitution $\\displaystyle u = \\sin{x} \\implies du = \\cos{x}\\,dx$ and the integral becomes $\\displaystyle \\int{\\frac{u^3(2 - u^2)}{1 - 2u^2}\\,du}$ Can you go from here? 4. (nevermind, my arithmetic and latex don't mix). 5. Originally Posted by Prove It \\displaystyle \\begin{align*} \\frac{1 - \\cos^4{x}}{\\cot{x} - \\tan{x}} &= \\frac{1 - \\cos^4{x}}{\\frac{\\cos{x}}{\\sin{x}} - \\frac{\\sin{x}}{\\cos{x}}}\\\\ &= \\frac{1 - \\cos^4{x}}{\\frac{\\cos^2{x} - \\sin^2{x}}{\\sin{x}\\cos{x}}} \\\\ &= \\frac{(1 - \\cos^2{x})(1 + \\cos^2{x})}{\\frac{\\cos^2{x} - \\sin^2{x}}{\\sin{x}\\cos{x}}} \\\\ &= \\frac{\\sin{x}\\cos{x}(1 - \\cos^2{x})(1 + \\cos^2{x})}{\\cos^2{x} - \\sin^2{x}} \\\\ &= \\frac{\\sin^3{x}\\cos{x}(1 + \\cos^2{x})}{\\cos^2{x} - \\sin^2{x}} \\\\ &= \\frac{\\sin^3{x}\\cos{x}(2 - \\sin^2{x})}{1 - 2\\sin^2{x}}\\end{align*} So $\\displaystyle \\int{\\frac{1 - \\cos^4{x}}{\\cot{x} - \\tan{x}}\\,dx} = \\int{\\frac{\\sin^3{x}(2 - \\sin^2{x})\\cos{x}}{1 - 2\\sin^2{x}}\\,dx}$ Make the substitution $\\displaystyle u = \\sin{x} \\implies du = \\cos{x}\\,dx$ and the integral becomes $\\displaystyle \\int{\\frac{u^3(2 - u^2)}{1 - 2u^2}\\,du}$ Can you go from here? should i divide by u^2 6. No, you should expand the numerator, then apply polynomial long division. It's probably easiest if you write it as $\\displaystyle \\frac{u^5 - 2u^3}{2u^2 - 1}$. 7. what should i do with the remainder 8. Partial Fractions 9. Hm, sorry for interrupting if you didn't like my solution, but since this is my first post I'd still like for it to be helpful. As mentioned, using the fact that: $\\tan 2x=\\frac{2\\tan x}{1-\\tan^2 x}$ you can rewrite the denominator as:", "\\frac{1}{2}\\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) \\\\ &=1 + \\frac{1}{3} + \\frac{1}{5} + \\dots + \\frac{1}{2n-1} + \\frac{1}{2}\\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) - \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) \\\\ &=1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{2n} - \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{n}\\right) \\\\ &=\\frac{1}{n+1} + \\frac{1}{n+2} + \\dots +\\frac{1}{2n}\\phantom{=} \\end{align*} - @GreenBeans, clearly this is the answer you are looking for. – leo Mar 1 '12 at 3:36 @leo You spoiled it. GreenBeans, this are the droids you are looking for. – Pedro Tamaroff Mar 1 '12 at 4:14 This sequence has popped up many times in SE. You can write it as $$s_n = \\sum_{k=1}^{n} \\frac{1}{n+k}$$ Note that the partial sums of the series for $\\log 2$ are \\eqalign{ & {p_1} = 1 \\cr & {p_2} = 1 - \\frac{1}{2} = \\frac{1}{2} \\cr & {p_3} = \\frac{1}{2} + \\frac{1}{3} \\cr & {p_4} = \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} = \\frac{1}{3} + \\frac{1}{4} \\cr & {p_5} = \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} \\cr & {p_6} = \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} \\cr & {p_7} = \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} \\cr & {p_8} = \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} + \\frac{1}{8} \\cr} I guess that with Henning's answer and this, you can probably relate the $s_n$ with", "and since $a_n \\to 0$ then $\\mathop {\\lim }\\limits_{n \\to \\infty } {S_n} = \\mathop {\\lim }\\limits_{n \\to \\infty } (a_1 - a_{n+1}) = a_1$. Applying this fact for the sequence $a_k:=\\sin (\\frac{1}{2k})$, we get that \\begin{align*} \\sum\\limits_{n = 1}^\\infty {\\left( {\\sin \\frac{1}{{2n}} - \\sin \\frac{1}{{2n + 1}}} \\right)} = \\sin \\left( {\\frac{1}{2}} \\right) \\end{align*} - But here we have $\\sum (a_{2k} - a_{2k+1})$, or, if you wish, $\\sum (a_k - a_{k+\\frac{1}{2}})$, it doesn't telescope. – Daniel Fischer Apr 28 at 11:25 Ah yes. You are right. But how come $k+\\frac{1}{2}$ in $a_{k+\\frac{1}{2}}$ – mwomath May 6 at 11:18 @user45105 $a_k=\\sin(1/(2k))\\implies a_{k+1/2}=\\sin(1/(2(k+1/2)))=\\sin(1/(2k+1))$ – oldrinb May 7 at 22:19 If we substitute $m=2n$, then \\begin{align} \\sum\\limits_{n = 1}^\\infty {\\left( {\\sin \\frac{1}{{2n}} - \\sin \\frac{1}{{2n + 1}}} \\right)} &= \\sum\\limits_{m = 2}^\\infty {\\left( {\\sin \\frac{1}{m} - \\sin \\frac{1}{{m + 1}}} \\right)} \\\\ &= \\sum\\limits_{m = 1}^\\infty {\\left( {\\sin \\frac{1}{m} - \\sin \\frac{1}{{m + 1}}} \\right)} - \\left( {\\sin 1 - \\sin \\frac{1}{2}} \\right) \\\\ &= \\sin 1 - \\left( {\\sin 1 - \\sin \\frac{1}{2}} \\right)\\\\ &= \\sin \\frac{1}{2} \\end{align} where as above $\\sum\\limits_{m = 1}^\\infty {\\left( {\\sin \\frac{1}{m} - \\sin \\frac{1}{{m + 1}}} \\right)} = \\sin \\left( 1 \\right)$ – mwomath Jun 3 at 10:59 You can determine whether a series converges or not by evaluating $a_n$ when $n \\to", "# Convergence of $\\sum_{n=1}^{\\infty}\\left(\\cos\\frac{1}{n}\\right)^{k(n)}\\,\\,\\,\\,; k(n)=\\frac{1}{\\sin^a\\left(\\frac{1}{n}\\right)}$ The convergence of the following series as $a \\in \\mathbb{R}$ $$\\sum_{n=1}^{\\infty}\\left(\\cos\\frac{1}{n}\\right)^{k(n)}\\,\\,\\,\\,; k(n)=\\frac{1}{\\sin^a\\left(\\frac{1}{n}\\right)}$$ As $n \\to +\\infty$ we have that $\\cos\\frac{1}{n} \\sim 1-\\frac{1}{2n^2}$ and that $\\sin^a\\left(\\frac{1}{n}\\right) \\sim \\frac{1}{n^a}$ but I can't figure out how to handle these results. Better to use comparision test? • Yes. You could use those familiar Maclaurin formula to estimate the general term. For the $x^y$ type, you may find it useful to consider $\\exp(x \\log(y))$ instead. – xbh Aug 16 '18 at 7:46 • Tip on MathJax: \\; would give a thick space. If possible \\quad would give a thicker space. – xbh Aug 16 '18 at 7:56 First you can have: $\\left(\\cos\\frac{1}{n}\\right)^{\\frac{1}{\\sin^a\\left(\\frac{1}{n}\\right)}}=e^{\\frac{1}{\\sin^a\\left(\\frac{1}{n}\\right)}\\log\\cos\\frac{1}{n}}.$ And $\\frac{1}{\\sin^a\\left(\\frac{1}{n}\\right)}\\log\\cos\\frac{1}{n}=\\frac{1}{\\sin^a\\left(\\frac{1}{n}\\right)}\\log(1+\\cos\\frac{1}{n}-1)\\sim\\frac{\\cos\\frac{1}{n}-1}{\\frac{1}{n^a}}\\sim\\frac{\\frac{-1}{2n^2}}{\\frac{1}{n^a}}=\\frac{-1}{2n^{2-a}},$ as $n\\to\\infty.$ So $$\\left(\\cos\\frac{1}{n}\\right)^{\\frac{1}{\\sin^a\\left(\\frac{1}{n}\\right)}}\\sim e^{\\frac{-1}{2}n^{a-2}}.$$ Base on this, you can give the convergence. HINT We have that $$\\cos\\frac{1}{n} \\sim 1-\\frac{1}{2n^2}$$ $$\\sin^a\\left(\\frac{1}{n}\\right) \\sim \\frac{1}{n^a}$$ therefore $$\\left(\\cos\\frac{1}{n}\\right)^{k(n)}=e^{k(n)\\log \\left(\\cos\\frac{1}{n}\\right)}\\sim e^{-\\frac12 n^{a-2}}$$ and the given series seems to converge for $a>2$. Now you can try to make it more rigorous taking into account the remainders and referring to limit comparison test. • It was a copy and paste of the OP! – gimusi Aug 16 '18 at 7:55", "$x^{2}+y^{2}=1;$ $A=(0,2),$ $B=(-\\sqrt{3},-1),$ $C=(\\sqrt{3},-1);$ $P=(\\cos t,\\sin t),$ $t\\in (\\frac{\\pi}{6},\\frac{5\\pi}{6});$ $AB:\\,y-2=x\\sqrt{3};$ \\begin{align}\\displaystyle PD &= 1+\\sin t,\\\\ PE &= \\frac{2-\\sqrt{3}\\cos t-\\sin t}{2}=\\frac{2-2\\sin (t+\\frac{\\pi}{3})}{2},\\\\ PF &= \\frac{2+\\sqrt{3}\\cos t-\\sin t}{2}=\\frac{2+2\\sin (t+\\frac{2\\pi}{3})}{2}. \\end{align} From here, \\begin{align}\\displaystyle \\sqrt{PD} &= \\sin\\frac{t}{2}+\\cos\\frac{t}{2},\\\\ \\sqrt{PE} &= \\sin(\\frac{t}{2}+\\frac{\\pi}{6})-\\cos(\\frac{t}{2}+\\frac{\\pi}{6}),\\\\ \\sqrt{PF} &= \\sin(\\frac{t}{2}+\\frac{\\pi}{3})+\\cos(\\frac{t}{2}+\\frac{\\pi}{3}).\\\\ \\end{align} With a little more effort one gets $\\sqrt{PD}=\\sqrt{PE}+\\sqrt{PF}.$ Note that this identity is a special case of a more general one $\\displaystyle\\sqrt{PD}\\cos\\frac{A}{2}=\\sqrt{PE}\\cos\\frac{B}{2}+\\sqrt{PF}\\cos\\frac{C}{2}$ where, for an equilateral triangle $A=B=C=\\frac{\\pi}{3}.$ Given that, the required identity follows from $\\displaystyle 2(xy+yz+zx)-(x^{2}+y^{2}+z^{2})=(\\sqrt{x}+\\sqrt{y}+\\sqrt{z})\\prod_{cycl}(-\\sqrt{x}+\\sqrt{y}+\\sqrt{z})=0,$ the product is of three factors each of the form $(\\pm\\sqrt{x}\\pm\\sqrt{y}\\pm\\sqrt{z}),$ with only one minus sign in each. • Angle Trisectors on Circumcircle • Equilateral Triangles On Sides of a Parallelogram • Pompeiu's Theorem • Pairs of Areas in Equilateral Triangle • The Eutrigon Theorem • Equilateral Triangle in Equilateral Triangle • Seven Problems in Equilateral Triangle • Spiral Similarity Leads to Equilateral Triangle • Parallelogram and Four Equilateral Triangles • Miguel Ochoa's van Schooten Like Theorem • Two Conditions for a Triangle to Be Equilateral • Incircle in Equilateral Triangle • When Is Triangle Equilateral: Marian Dinca's Criterion • Barycenter of Cevian Triangle • Excircle in Equilateral Triangle • Converse Construction in Pompeiu's Theorem • Wonderful Trigonometry In Equilateral Triangle • 60o Angle And Importance of Being The Other End of a Diameter • One More Property of Equilateral", "$x^{2}+y^{2}=1;$ $A=(0,2),$ $B=(-\\sqrt{3},-1),$ $C=(\\sqrt{3},-1);$ $P=(\\cos t,\\sin t),$ $t\\in (\\frac{\\pi}{6},\\frac{5\\pi}{6});$ $AB:\\,y-2=x\\sqrt{3};$ \\begin{align}\\displaystyle PD &= 1+\\sin t,\\\\ PE &= \\frac{2-\\sqrt{3}\\cos t-\\sin t}{2}=\\frac{2-2\\sin (t+\\frac{\\pi}{3})}{2},\\\\ PF &= \\frac{2+\\sqrt{3}\\cos t-\\sin t}{2}=\\frac{2+2\\sin (t+\\frac{2\\pi}{3})}{2}. \\end{align} From here, \\begin{align}\\displaystyle \\sqrt{PD} &= \\sin\\frac{t}{2}+\\cos\\frac{t}{2},\\\\ \\sqrt{PE} &= \\sin(\\frac{t}{2}+\\frac{\\pi}{6})-\\cos(\\frac{t}{2}+\\frac{\\pi}{6}),\\\\ \\sqrt{PF} &= \\sin(\\frac{t}{2}+\\frac{\\pi}{3})+\\cos(\\frac{t}{2}+\\frac{\\pi}{3}).\\\\ \\end{align} With a little more effort one gets $\\sqrt{PD}=\\sqrt{PE}+\\sqrt{PF}.$ Note that this identity is a special case of a more general one $\\displaystyle\\sqrt{PD}\\cos\\frac{A}{2}=\\sqrt{PE}\\cos\\frac{B}{2}+\\sqrt{PF}\\cos\\frac{C}{2}$ where, for an equilateral triangle $A=B=C=\\frac{\\pi}{3}.$ Given that, the required identity follows from $\\displaystyle 2(xy+yz+zx)-(x^{2}+y^{2}+z^{2})=(\\sqrt{x}+\\sqrt{y}+\\sqrt{z})\\prod_{cycl}(-\\sqrt{x}+\\sqrt{y}+\\sqrt{z})=0,$ the product is of three factors each of the form $(\\pm\\sqrt{x}\\pm\\sqrt{y}\\pm\\sqrt{z}),$ with only one minus sign in each. [an error occurred while processing this directive]", "$(2)$. Now, \\begin{align} &\\sum_{k=1}^{n}\\frac{1}{2k-1} \\\\ = & \\ \\sum_{k=1}^n\\frac{1}{2k-1} + \\sum_{k=1}^n\\frac{1}{2k} - \\sum_{k=1}^n\\frac{1}{2k}\\\\ = & \\ \\sum_{k=1}^{2n} \\frac{1}{k}- \\frac{1}{2}\\sum_{k=1}^n\\frac{1}{k}\\\\ \\end{align} And then, \\begin{align} &\\sum\\limits_{k=1}^n\\frac 1{2k-1} -\\frac 12\\sum\\limits_{k=1}^n\\frac 1k \\\\ = & \\ \\sum_{k=1}^{2n} \\frac{1}{k}- \\frac{1}{2}\\sum_{k=1}^n\\frac{1}{k} -\\frac 12\\sum\\limits_{k=1}^n\\frac 1k\\\\ = & \\ \\sum_{k=1}^{2n} \\frac{1}{k}- \\sum_{k=1}^n\\frac{1}{k} \\\\ = & \\ \\sum_{k=n+1}^{2n}\\frac{1}{k} = \\sum_{k=1}^{n}\\frac{1}{n+k} \\end{align} This gives $(2)$ to $(3)$ All in one chain: $$\\color{blue}{\\frac 12\\sum\\limits_{k=1}^n\\frac 1{2k-1}+\\frac 12\\sum\\limits_{k=1}^n\\frac 1{2k+1}-\\frac 12\\sum\\limits_{k=1}^n\\frac 1k+\\frac n{2n+1}=}$$ $$\\frac 12\\sum\\limits_{k=1}^n\\frac 1{2k-1}+\\frac 12\\left(\\sum\\limits_{k=1}^n\\frac 1{2k-1}-1+\\frac{1}{2n+1}\\right)-\\frac 12\\sum\\limits_{k=1}^n\\frac 1k+\\frac n{2n+1}=$$ $$\\sum\\limits_{k=1}^n\\frac 1{2k-1}+\\frac 12\\left(-\\frac{2n}{2n+1}\\right)-\\frac 12\\sum\\limits_{k=1}^n\\frac 1k+\\frac n{2n+1}=$$ $$\\color{blue}{\\sum\\limits_{k=1}^n\\frac 1{2k-1}-\\frac 12\\sum\\limits_{k=1}^n\\frac 1k=}$$ $$\\left(\\sum\\limits_{k=1}^{2n}\\frac 1{k}-\\sum\\limits_{k=1}^n\\frac 1{2k}\\right)-\\frac 12\\sum\\limits_{k=1}^n\\frac 1k=$$ $$\\left(\\sum\\limits_{k=1}^{2n}\\frac 1{k}-\\frac12\\sum\\limits_{k=1}^n\\frac 1{k}\\right)-\\frac 12\\sum\\limits_{k=1}^n\\frac 1k=$$ $$\\color{blue}{\\sum\\limits_{k=1}^{2n}\\frac 1k-\\sum\\limits_{k=1}^n\\frac 1k=}$$ $$\\left(\\sum\\limits_{k=1}^{n}\\frac 1k+\\sum\\limits_{k=n+1}^{2n}\\frac 1k\\right)-\\sum\\limits_{k=1}^n\\frac 1k=$$ $$\\sum\\limits_{k=n+1}^{2n}\\frac 1k=$$ $$\\color{blue}{\\sum\\limits_{k=1}^n\\frac 1{k+n}}$$", "- \\frac {x}{2},\\\\f_n'(x) &= - \\mbox{Re}\\left(\\sum_{n = 1}^{n}z^n\\right) - \\frac12.\\end{align*} After some simplifications we get $$f_n'(x) = \\frac {( - 1)^{n + 1}}{2}((1 - \\cos(x))\\frac {\\sin((n + 1)x)}{\\sin(x)} + \\cos((n + 1)x))$$ and $$f_n''(x) = \\frac {( - 1)^{n}}{2}\\frac {(n + 1)\\sin(nx) + n\\sin((n + 1)x)}{1 + \\cos(x)}.$$ The formula for $f_n''$ shows that $( - 1)^n f$ is convex for $0 < x < \\frac {\\pi}{n + 1}$. Since $f_n(0) = 0$ and $f_n'(0) = \\frac {( - 1)^{n + 1}}{2}$.We are ready when we can show that $( - 1)^{n + 1}f_n(\\frac {\\pi}{n + 1}) > 0$. We have to distinct between two different, but very similar cases, namely $n$ is odd, and $n$ is even. Let's restrict to the case $n$ is even. We prove $f_{2n}(\\frac {\\pi}{2n + 1}) < 0$. \\begin{align*}f_{2n}\\left( \\frac{\\pi}{2n+1} \\right) &=\\sum_{k=1}^{2n}{\\left( -1 \\right)}^{k+1}\\frac{\\sin \\left( \\frac{k\\pi}{2n+1} \\right)}{k}-\\frac{\\pi}{2\\left( 2n+1 \\right)}\\\\&=\\frac{\\pi}{2n+1}\\left( \\sum_{k=1}^n{\\frac{\\sin \\left( \\frac{\\left( 2k-1 \\right) \\pi}{2n+1} \\right)}{\\frac{\\left( 2k-1 \\right) \\pi}{2n+1}}}-\\sum_{k=1}^n{\\frac{\\sin \\left( \\frac{2k\\pi}{2n+1} \\right)}{\\frac{2k\\pi}{2n+1}}} \\right) -\\frac{\\pi}{2\\left( 2n+1 \\right)}.\\end{align*} The function $x \\mapsto \\frac {\\sin(x)}{x}$ is descending on $[0,\\pi]$, thus both sums lay between $a$ and $a + \\frac {2\\pi}{2n + 1}$, where $a = \\int_0^{\\pi}\\frac {\\sin(x)}{x}\\,dx$. Thus $$f_{2n}\\left(\\frac {\\pi}{2n + 1}\\right) < \\frac {\\pi}{2n + 1}\\cdot\\frac {2\\pi}{2n + 1} - \\frac {\\pi}{2(2n + 1)} < 0.$$ ## Euler-Maclaurin求和公式估计梯形积分公式的误差 $\\mathop {\\lim }\\limits_{n \\to \\infty }", "I don't think so. – Robert Israel Dec 4 '14 at 18:14 If you're happy to accept integrals in your general formula, then a really simple expression (via T.A.E.) is \\begin{align} \\frac{d^{n}}{dx^{n}}\\mathrm{sinc}(x) =\\frac{d^{n}}{dx^{n}}\\int_{-1}^{1}e^{ikx}\\, \\mathrm{d}k = \\int_{-1}^{1}(ik)^{n}e^{ikx}\\, \\mathrm{d}k \\end{align} $$\\frac{d^n}{dx^n}\\frac{\\sin(x)}{x}=\\frac{(-1)^{n}i}{2}(\\Gamma(n+1,ix)-\\Gamma(n+1,-ix))x^{-n-1}$$ $$=\\frac{(-1)^{n}n!i}{2}\\sum_{k=0}^{n}\\frac{i^k}{k!}(e^{-ix}-(-1)^ke^{ix})x^{-n+k-1}$$ $$=\\frac{(-1)^{n}n!i}{2}\\sum_{k=0}^{n}\\frac{i^k}{k!}((1-(-1)^k)\\cos(x)-(1+(-1)^k)i\\sin(x))x^{-n+k-1}$$ $$=(-1)^{n}n!2i\\sum_{k=0}^{n}\\frac{i^k}{k!}(\\text{ frac}\\left(\\frac{k}{2}\\right)\\cos(x)-\\text{ frac}\\left(\\frac{k+1}{2}\\right)i\\sin(x))x^{-n+k-1}$$ $$=(-1)^{n}n!2\\sum_{k=0}^{n}\\frac{i^{k+1}}{k!}\\text{ frac}\\left(\\frac{k}{2}\\right)\\frac{\\cos(x)}{x^{n-k+1}}+(-1)^{n}n!2\\sum_{k=0}^{n}\\frac{i^k}{k!}\\text{ frac}\\left(\\frac{k+1}{2}\\right)\\frac{\\sin(x)}{x^{n-k+1}}$$ $$=(-1)^{n}n!\\sum_{k=0}^{\\frac{n-1}{2}}\\frac{(-1)i^{k+1}}{(2k+1)!}\\frac{\\cos(x)}{x^{n-2k}}+(-1)^{n}n!\\sum_{k=0}^{\\frac{n}{2}}\\frac{(-1)^k}{(2k)!}\\frac{\\sin(x)}{x^{n-2k+1}}$$", "&= \\lim_{k \\to 0} \\frac{ 2\\cos{(x+\\frac{k}{2})} \\sin{\\frac{k}{2}} }{k} \\\\ &= \\lim_{k \\to 0} \\cos{(x+\\frac{k}{2})} \\cancelto{1}{\\lim_{k \\to 0} \\frac{ \\sin{\\frac{k}{2}} }{\\frac{k}{2}}} \\\\ &= \\cos{x} \\\\ \\end{align*} And finally \\begin{align*} \\ddx{\\cos x} &= \\lim_{k \\to 0} \\frac{ \\cos{(x+k)} - \\cos{x} }{k} \\\\ &= \\lim_{k \\to 0} \\frac{ \\cos{\\left( (x+\\frac{k}{2})+\\frac{k}{2} \\right)} - \\cos{\\left( (x+\\frac{k}{2})-\\frac{k}{2} \\right)} }{k} \\\\ &= \\lim_{k \\to 0} \\frac{ -2\\sin{(x+\\frac{k}{2})} \\sin{\\frac{k}{2}} }{k} \\\\ &= \\lim_{k \\to 0} -\\sin{(x+\\frac{k}{2})} \\cancelto{1}{\\lim_{k \\to 0} \\frac{ \\sin{\\frac{k}{2}} }{\\frac{k}{2}}} \\\\ &= -\\sin{x} \\\\ \\end{align*} A Parting Shot One beautiful result I've recently come across is given below \\[\\lim_{x \\to 0} \\frac{1}{x^2}- \\frac{1}{\\sin^2 x} In order to get there smoothly, I first want to establish the triple angle formula for $\\sin$. \\begin{align*} \\sin{3x} &= \\sin{2x}\\cos{x} + \\cos{2x}\\sin{x} &= 2\\sin{x}\\cos^2{x} + (1-2\\sin^2{x})\\sin{x} \\\\ &= 2\\sin{x}(1-\\sin^2{x}) + (1-2\\sin^2{x})\\sin{x} &= 2\\sin{x} - 2\\sin^3{x} + \\sin{x} - 2\\sin^3{x} \\\\ &= 3\\sin{x} - 4\\sin^3{x} \\end{align*} Now let's proceed \\begin{align*} \\lim_{x \\to 0} \\left( \\frac{1}{x^2}- \\frac{1}{\\sin^2 x} \\right) &= \\lim_{x \\to 0} \\left( \\frac{1}{x} - \\frac{1}{\\sin x} \\right)\\left( \\frac{1}{x} + \\frac{1}{\\sin x} \\right) \\\\ &= \\lim_{x \\to 0} \\frac{1}{x}\\left( \\frac{1}{x} - \\frac{1}{\\sin x} \\right)\\left( 1 + \\frac{x}{\\sin x} \\right) \\\\ &= \\lim_{x \\to 0} \\left( \\frac{1}{x^2} - \\frac{1}{x\\sin x} \\right) \\cancelto{2}{\\lim_{x \\to 0} \\left( 1 + \\frac{x}{\\sin x} \\right)} \\\\ &= 2\\lim_{x \\to 0} \\frac{\\sin x - x}{x^2\\sin x} \\\\", "even, then $\\cos(\\frac{\\pi}{2})=0$ appears in the product (when $k=n/2$) and $\\sin(\\frac{n\\pi}{2})=0$. If $n$ is odd, then combining $$\\lim_{z=1}\\frac{z^n+1}{z+1}=1\\tag{2a}$$ $$\\frac{z^n+1}{z+1}=\\prod_{k=1}^{n-1}(z+e^{2\\pi ik/n})\\tag{2b}$$ $$1+e^{i2k\\pi/n}=2\\cos(k\\pi/n)e^{ik\\pi/n}\\tag{2c}$$ and noting that $\\displaystyle\\sum_{k=1}^{n-1}k=\\frac{n(n-1)}{2}$ so that $\\displaystyle\\prod_{k=1}^{n-1}e^{ik\\pi/n}=(-1)^{(n-1)/2}$ which matches the sign of $\\sin(\\pi n/2)$, yields $$2^{n-1}\\prod_{k=1}^{n-1}\\cos(k\\pi/n)=(-1)^{(n-1)/2}=\\sin(\\pi n/2)$$ - Denote $w = e^{i \\pi/n}$. We have $$\\prod_{k = 1}^{n-1} \\sin \\left(\\frac{k\\pi}{n}\\right)= \\prod_{k = 1}^{n-1} \\frac{w^k - w^{-k}}{2i} = \\frac{1}{2^{n-1}} \\prod_{k = 1}^{n-1} \\frac{w^k}{i} (1-w^{-2k})$$ Since we have $$\\sum_{k = 0}^{n-1} x^k = \\prod_{k = 1}^{n-1} (x-w^{2k})$$ Setting $x=1$ yields $$\\prod_{k = 1}^{n-1} (1-w^{2k}) = n$$ So we get $$\\prod_{k = 1}^{n-1} \\sin \\left(\\frac{k\\pi}{n}\\right)= \\frac{n}{2^{n-1}} \\frac{w^{n(n-1)/2}}{i^{n-1}} = \\frac{i^{n-1}}{i^{n-1}} \\frac{n}{2^{n-1}} = \\frac{n}{2^{n-1}}$$ I guess (but did not check) that the same kind of reasoning gives the one with $\\cos$. - similar reasoning for odd $n$ (except you have to watch the sign), but very different, however easy, reasoning for even $n$. – robjohn Oct 6 '11 at 4:03 The second purported identity is equivalent to asking for the constant term of $\\dfrac{U_{n-1}(x)}{2^{n-1}}$ (i.e., $\\dfrac{U_{n-1}(0)}{2^{n-1}}$), where $U_n(x)$ is the Chebyshev polynomial of the second kind. Since $$\\frac{U_{n-1}(x)}{2^{n-1}}=\\frac{\\sin(n \\arccos\\,x)}{2^{n-1}\\sqrt{1-x^2}}$$ letting $x=0$ gives your identity. - This is really clever. – Joel Cohen Oct 6 '11 at 3:13 Define $\\zeta_n = e^{2 \\pi i/n}$. Proposition For odd integer $n \\geq 1$, \\begin{align} \\prod_{k = 1}^{n-1}(\\zeta_n^{k} - \\zeta_n^{-k}) = n. \\end{align} and \\begin{align} \\prod_{k", "$$(3)$$, we use that $$\\sum_{k\\geq1}\\frac{(-1)^k}{\\sqrt{k}}=(\\sqrt2-1)\\zeta(\\frac12)$$ (Question 2059712) and that $$\\sin(2\\pi n)=0$$ for $$n\\in\\mathbb{N}$$. We may evaluate the series in blue with \\begin{aligned} \\sum_{n\\ \\text{not square}}\\frac{\\left(-1\\right)^{n}}{n} &=\\sum_{n\\ge1}\\frac{\\left(-1\\right)^{n}}{n}-\\sum_{n\\ \\text{square}}\\frac{\\left(-1\\right)^{n}}{n} \\\\ &=-\\ln2-\\sum_{k\\ge1}\\frac{\\left(-1\\right)^{k^2}}{k^2} \\\\ &=-\\ln2-\\frac{-\\pi^2}{12} \\end{aligned} by using $$\\sum_{k\\geq1}\\frac{(-1)^k}{k^2}=\\frac{-\\pi^2}{12}$$ and that $$k^2$$ has the same parity as $$k$$. ### Partial analytic solution to $$f_1(k)$$ The two sums over $$n$$ and $$k$$, in the series for $$d$$, can be swapped, which gives us $$\\displaystyle d=\\frac{-1}{\\pi}\\sum_{k\\ge1}\\frac{f_1(k)}{k}$$, where $$\\displaystyle f_1(k)=\\sum_{n\\geq1}\\frac{(-1)^n\\sin(2\\pi k\\sqrt{n})}{n}$$, which has the fairly simple Laplace transform $$\\frac{1}{2\\pi}F\\left(\\frac{x}{2\\pi}\\right)$$, where $$\\displaystyle F(x)=\\sum_{n\\ge 1}\\frac{\\left(-1\\right)^{n}}{\\sqrt{n}\\left(n+x^2\\right)}$$. Therefore, $$\\displaystyle d=-\\frac{1}{2\\pi^{2}}\\sum_{k=1}^{\\infty}\\frac{\\mathcal{L}^{-1}\\left\\{F\\left(\\frac{s}{2\\pi}\\right),k\\right\\}}{k}$$. $$F(x)$$ has a similar series to the zeta functions. Substituting $$x$$ for $$x^2$$ in $$F$$ and taking the $$k$$'th derivative gives us $$\\small\\displaystyle \\frac{\\mathrm{d}^k}{\\mathrm{d}x^k}\\sum_{n\\ge 1}\\frac{\\left(-1\\right)^{n}}{\\sqrt{n}\\left(n+x\\right)}= \\sum_{n\\ge 1}\\frac{(-1)^{k}k!\\left(-1\\right)^{n}}{\\sqrt{n}\\left(n+x\\right)^{k+1}}$$, so we have the interesting Maclaurin series, $$\\small \\displaystyle F(x)=\\sum_{k\\ge0}\\left(-1\\right)^{k+1}\\left[1-{2^{-\\frac{1}{2}-k}}\\right]\\zeta\\left(k+\\frac{3}{2}\\right)x^{2k}$$. Although this only converges within a small radius of convergence, it does hint at the possibility of a closed form or analytic extension of this series. The expression for $$s$$ converges slower than River Li's but could possibly be tractable. It is similar to the series evaluated in the following questions, which would suggest that it could be amenable to the Euler-Maclaurin, Abel-Plana or Poisson summation formulae. It is also possible that it could be expressed in terms of theta or Bessel functions, though I am yet to find whether this is", "\\Im\\{X[k]\\} \\sin(2 \\pi (k/N) f_\\text{s} t) \\\\ & = \\frac1N \\left( X[0] + X[N/2] \\cos(2 \\pi (f_\\text{s}/2) t) + A_x \\sin(2 \\pi (f_\\text{s}/2) t) \\right) \\\\ & \\quad + \\frac2N \\sum\\limits_{k=1}^{\\frac{N}{2}-1} \\Re\\{X[k]\\} \\cos(2 \\pi (k/N) f_\\text{s} t) - \\Im\\{X[k]\\} \\sin(2 \\pi (k/N) f_\\text{s} t) \\\\ \\end{align} where $A_x$ can be any number, because that term will be zero at the sampling instances, when $t = nT = n/f_\\text{s}$. looking at it from a magnitude/phase POV, for $N$ odd $$x(t) = \\frac{X[0]}{N} + \\frac2N \\sum\\limits_{k=1}^{\\frac{N-1}{2}} |X[k]| \\cos(2 \\pi (k/N) f_\\text{s} t + \\arg\\{X[k]\\})$$ and for $N$ even \\begin{align} x(t) & = \\frac{X[0]}{N} + \\frac{X[N/2]}{N \\cos(\\theta)} \\cos(2 \\pi (f_\\text{s}/2) t + \\theta) \\\\ & \\quad + \\frac2N \\sum\\limits_{k=1}^{\\frac{N}{2}-1} |X[k]| \\cos(2 \\pi (k/N) f_\\text{s} t + \\arg\\{X[k]\\}) \\end{align} where $$\\Re\\{X[k]\\} = |X[k]| \\cos(\\arg\\{X[k]\\})$$ $$\\Im\\{X[k]\\} = |X[k]| \\sin(\\arg\\{X[k]\\})$$ or $$|X[k]| = \\sqrt{\\Re\\{X[k]\\}^2 + \\Im\\{X[k]\\}^2}$$ $$\\arg\\{X[k]\\} = \\operatorname{atan2}(\\Im\\{X[k]\\}, \\Re\\{X[k]\\})$$ and $\\theta$ is the phase of the Nyquist component and can be any real value except $\\pm \\frac{\\pi}{2}$ or $\\frac{\\pi}{2}+ m \\pi$ for integer $m$ (because you cannot divide by zero). the samples $x[n] \\triangleq x(n/f_\\text{s})$ will come out the same regardless of the phase of the Nyquist component as long as the amplitude of the component is adjusted accordingly by dividing by $\\cos(\\theta)$. • For real-valued sequences with $N$ even, you know both", "of $$n$$, \\begin{align}\\sum_{k=0}^{m-1}\\cos\\frac{2\\pi nkh}{b-a}&=\\sum_{k=0}^{m-1}\\cos\\frac{2\\pi nk}m=\\sum_{k=0}^{m-1}\\frac{\\sin\\frac{2\\pi n}m(k+1/2)-\\sin\\frac{2\\pi n}m(k-1/2)}{2\\sin\\frac{\\pi n}m}\\\\ &=\\frac{\\sin\\frac{2\\pi n}m(m-1/2)-\\sin\\frac{2\\pi n}m(-1/2)}{2\\sin\\frac{\\pi n}m}=0\\end{align} If $$m$$ is a divisor of $$n$$, then $$\\sum_{k=0}^{m-1}\\cos\\frac{2\\pi nkh}{b-a}=\\sum_{k=0}^{m-1}\\cos\\frac{2\\pi nk}m=m$$ Since $$\\cos\\alpha\\left(k+\\frac12\\right)-\\cos\\alpha\\left(k-\\frac12\\right)=-2\\sin\\alpha k\\sin\\alpha/2$$, if $$m$$ is not a divisor of $$n$$, \\begin{align}\\sum_{k=0}^{m-1}\\sin\\frac{2\\pi nkh}{b-a}&=\\sum_{k=0}^{m-1}\\sin\\frac{2\\pi nk}m=-\\sum_{k=0}^{m-1}\\frac{\\cos\\frac{2\\pi n}m(k+1/2)-\\cos\\frac{2\\pi n}m(k-1/2)}{2\\sin\\frac{\\pi n}m}\\\\ &=-\\frac{\\cos\\frac{2\\pi n}m(m-1/2)-\\cos\\frac{2\\pi n}m(-1/2)}{2\\sin\\frac{\\pi n}m}=0\\end{align} And even if $$m$$ is a divisor of $$n$$n $$\\sum_{k=0}^{m-1}\\sin\\frac{2\\pi nkh}{b-a}=\\sum_{k=0}^{m-1}\\sin\\frac{2\\pi nk}m=0$$ So the trapezoidal rule produces $$\\int_a^bf(x)dx\\approx(b-a)\\left(\\frac{a_0}2+\\sum_{n=1}^{\\infty}a_{mn}\\right)$$ Since the exact answer is $$\\int_a^bf(x)dx=(b-a)a_0/2$$ we see that the effect of the trapezoidal rule in this case is to approximate the function $$f(x)$$ by its $$2n-1$$ 'lowest energy' eigenfunctions and integrate the approximation. This is pretty much what Gaussian integration does so it's amusing to compare the two for periodic and nonperiodic functions. The program that produces my data looks like this: module Gmod use ISO_FORTRAN_ENV, only: wp=>REAL64 implicit none real(wp), parameter :: pi = 4*atan(1.0_wp) contains subroutine eval_legendre(n,x,p,q) integer, intent(in) :: n real(wp), intent(in) :: x real(wp), intent(out) :: p, q integer m real(wp) r if(n == 0) then p = 1 q = 0 else p = x q = 1 do m = 2, n-1, 2 q = ((2*m-1)*x*p-(m-1)*q)/m p = ((2*m+1)*x*q-m*p)/(m+1) end do if(m == n) then r = ((2*m-1)*x*p-(m-1)*q)/m q = p p = r end if end if end subroutine eval_legendre subroutine formula(n,x,w) integer, intent(in) :: n real(wp), intent(out)", "each of these parts, either $\\sin$, or $\\arcsin$ will be small, and in the limiting value will be $\\frac{\\pi}{2}$: In[28]:= Table[ N[1/n Sum[Cos[1/k] ArcCos[1/(n - k + 1)], {k, 1, n}], 50], {n, {5000, 10000, 100000}}] // N Out[28]= {1.56861, 1.56963, 1.57066} More rigorously: $$\\frac{1}{n} \\sum_{k=1}^n \\left( 1 - 2 \\sin^2\\left(\\frac{1}{2 k}\\right) \\right) \\left( \\frac{\\pi}{2} - \\arcsin\\left( \\frac{1}{n+1-k} \\right) \\right) = \\frac{\\pi}{2} - \\frac{\\pi}{n} \\sum_{k=1}^n \\sin^2 \\frac{1}{k} - \\frac{1}{n} \\sum_{k=1}^n \\arcsin\\frac{1}{k} + \\frac{2}{n} \\sum_{k=1}^n \\sin^2\\left( \\frac{1}{k}\\right) \\arcsin\\left(\\frac{1}{n+1-k}\\right)$$ Now: $$0 \\leqslant \\lim_{n \\to \\infty} \\frac{1}{n} \\sum_{k=1}^n \\sin^2\\left(\\frac{1}{k}\\right) \\leqslant \\lim_{n \\to \\infty} \\frac{1}{n} \\sum_{k=1}^n \\frac{1}{k^2} = 0$$ $$0 \\leqslant \\lim_{n\\to \\infty} \\frac{1}{n} \\sum_{k=1}^n \\arcsin\\left(\\frac{1}{k}\\right) \\leqslant \\lim_{n \\to \\infty} \\frac{\\pi}{2n} \\sum_{k=1}^n \\frac{1}{k} = \\lim_{n \\to \\infty} \\frac{\\pi}{2 n} \\ln(n) = 0$$ $$0 \\leqslant \\lim_{n \\to \\infty} \\frac{2}{n} \\sum_{k=1}^n \\sin^2 \\left( \\frac{1}{k} \\right) \\arcsin\\left(\\frac{1}{n+1-k} \\right) \\leqslant \\lim_{n \\to \\infty} \\frac{2}{n} \\sum_{k=1}^n \\frac{1}{k^2} \\frac{1}{n+1-k} \\leqslant \\lim_{n \\to \\infty} \\frac{2}{n} \\sum_{k=1}^n \\frac{1}{k^2} = 0$$ - could you provide with some more details pls about that splitting. I'm surrounded by mist because maybe it's something i don't catch it yet. This is from my highschool notebook and i was thinking that there is an easier way to solve it. I'm struggling to understand your way, now. – Chris's sis the artist May 25 '12 at 19:12 thanks for your solution. It's a bit ugly but", "\\begin{align} \\text{Range}(g) &= \\left[ \\frac{-1 - \\sin \\left( \\frac{1}{2} \\right)}{2 \\sin \\left( \\frac{1}{2} \\right)},\\frac{1 - \\sin \\left( \\frac{1}{2} \\right)}{2 \\sin \\left( \\frac{1}{2} \\right)} \\right] \\\\ &= [-1.54291482 \\ldots,0.54291482 \\ldots] \\\\ &\\subseteq [-2,2]. \\end{align} Conclusion: $\\displaystyle \\left| \\sum_{k=1}^{n} \\sin(k) \\right| < 2$ and $\\displaystyle \\left| \\sum_{k=1}^{n} \\cos(k) \\right| < 2$ for all $n \\in \\mathbb{N}$. - You have the formula $\\frac{1-x^{n+1}}{1-x}=\\sum_{k=1}^{n}x^k$, which is easy enough to see by algebra. So you can replace the LHS with $\\frac{1-e^{i(n+1)\\theta}}{1-e^{i\\theta}}$. – minimalrho Feb 1 '13 at 22:03 +1 Very nice hint. This is not a series, @Iuli, but a finite sum and, in fact, a rather easy one: a geometric sequence sum. Above you can read what you get from it, which with absolute values gives at once the wanted upper bound $\\,2\\,$ – DonAntonio Feb 1 '13 at 22:15 @CalvinLin , yes...so what? The absolute value of the expression hinted in the answer above is less than or equal what you wrote (in fact, it is always less). Besides this, in the given formula you must take the imaginary part as that the part that corresponds to the sum of sines... – DonAntonio Feb 1 '13 at 23:05 @HaskellCurry Yes, with the trig version I could see how to do it. The exp version was less obvious, since you still had", "x$ for $0\\le x \\le \\pi/2$. And from this set of inequalities we have that \\begin{align}\\sin(x)&\\le x \\tag 3\\\\\\\\ \\cos(x)&\\ge \\sqrt{1-x^2} \\tag 4 \\end{align} for $0\\le x \\le \\pi/2$. Using $(3)$ and $(4)$ in $(2)$ reveals for $n>1$ $$0 \\le e^{1-\\cos(1/n)}-1\\le \\frac{1}{2n^2\\sqrt{1-(1/n)^2}}\\le \\frac{1}{n^2}$$ Since the series $\\sum_{n=1}^{\\infty}\\frac{1}{n^2}=\\pi^2/6$ converges, the series of interest does likewise. And we are done. You could have used the ratio test to $$u_n=e^{1-\\cos\\left(\\frac{1}n\\right)} -1$$ $$\\frac{u_{n+1}}{u_n}=\\frac{e^{1-\\cos \\left(\\frac{1}{n+1}\\right)}-1}{e^{1-\\cos \\left(\\frac{1}{n}\\right)}-1}$$ and using Taylor series for large values of $n$ (just as T. Bongers answered) to get $$\\frac{u_{n+1}}{u_n}=1-\\frac{2}{n}+O\\left(\\frac{1}{n^2}\\right)$$ Similarly, using Raabe's test $$n\\left( \\frac{u_n}{u_{n+1}}-1\\right)=n \\left(\\frac{e^{1-\\cos \\left(\\frac{1}{n}\\right)}-1}{e^{1-\\cos \\left(\\frac{1}{n+1}\\right)}-1}-1\\right)=2+\\frac{1}{n}+O\\left(\\frac{1}{n^2}\\right)$$", "\\begin{align} I&=-\\frac{\\pi}{2}\\ln2+\\sum_{k\\geq1}\\frac{(-1)^{k+1}}{k^2}\\sin\\frac{\\pi k}{2}-\\left(-\\frac\\pi2\\ln2-\\sum_{k\\geq1}\\frac1{k^2}\\sin\\frac{\\pi k}2\\right)\\\\ &=-\\sum_{k\\geq1}\\frac{(-1)^{k}}{k^2}\\sin\\frac{\\pi k}{2}+\\sum_{k\\geq1}\\frac1{k^2}\\sin\\frac{\\pi k}2\\\\ &=\\sum_{k\\geq1}\\frac{1+(-1)^{k+1}}{k^2}\\sin\\frac{\\pi k}{2}\\\\ &=2\\sum_{k\\geq0}\\frac{1}{(2k+1)^2}\\sin\\frac{\\pi (2k+1)}{2}\\\\ &=2\\sum_{k\\geq0}\\frac{(-1)^k}{(2k+1)^2}\\\\ &=2\\mathrm{G}. \\end{align} Just an interesting side note, we recall that for $$|x|<\\pi/2$$: $$\\sec(x)=\\sum_{n\\geq0}(-1)^n\\frac{E_{2n}}{(2n)!}x^{2n}\\ ,$$ where $$E_n$$ are the Euler numbers. So $$\\sec(x+y)+\\sec(x-y)=2\\sum_{n\\geq0}(-1)^n\\frac{E_{2n}}{(2n)!}\\sum_{k=0}^{n}{2n\\choose 2k}x^{2n-2k}y^{2k}\\ ,$$ from $$(x+y)^{2n}+(x-y)^{2n}=2\\sum_{k=0}^{n}{2n\\choose 2k}x^{2n-2k}y^{2k}.$$ So our integral is \\begin{align} 2\\mathrm{G}&=\\int_0^{\\pi/4}\\int_0^{\\pi/4}2\\sum_{n\\geq0}(-1)^n\\frac{E_{2n}}{(2n)!}\\sum_{k=0}^{n}{2n\\choose 2k}x^{2n-2k}y^{2k}dxdy\\\\ &=2\\sum_{n\\geq0}(-1)^n\\frac{E_{2n}}{(2n)!}\\sum_{k=0}^{n}{2n\\choose 2k}\\int_0^{\\pi/4}x^{2n-2k}dx\\int_0^{\\pi/4}y^{2k}dy\\\\ &=2\\sum_{n\\geq0}(-1)^n\\frac{E_{2n}}{(2n)!}\\sum_{k=0}^{n}{2n\\choose 2k}\\frac{\\pi^{2n-2k+1}}{4^{2n-2k+1}(2n-2k+1)}\\cdot\\frac{\\pi^{2k+1}}{4^{2k+1}(2k+1)}\\\\ &=2\\sum_{n\\geq0}(-1)^n\\frac{E_{2n}}{(2n)!}\\left(\\frac\\pi4\\right)^{2n+2}\\sum_{k=0}^{n}\\frac{{2n\\choose 2k}}{(2n-2k+1)(2k+1)}. \\end{align} So we finally have the beautiful identity $$\\frac1{16}\\sum_{n\\geq0}(-1)^n\\frac{E_{2n}}{(2n)!}\\left(\\frac\\pi4\\right)^{2n}\\sum_{k=0}^{n}\\frac{{2n\\choose 2k}}{(2n-2k+1)(2k+1)}=\\frac{\\mathrm{G}}{\\pi^2}$$ • Since I'm more familiar with the Clausen Functions it felt like ages ago the last time I really used those Fourier series in order to evaluate an integral like that ^^. Nevertheless it adds an interesting alternative to this post (+1) – mrtaurho Jun 7 '19 at 15:59 • @mrtaurho Thank you so much! :) As I'm sure you know, the Fourier series for $\\ln\\sin x$ is really just $\\mathrm{Cl}_1$ in disguise. – clathratus Jun 7 '19 at 16:02 • Exactly. That's why I prefer using the machinery of the Clausen Functions as it is not only an elegant way of evaluating such integrals (not to forget it's close relation to the Polylogarithms) but also reduces what one has to write down quite tremendously :) – mrtaurho Jun 7 '19 at 16:10 • @mrtaurho What are your thoughts on the identity with $\\mathrm G/\\pi^2$? Do you think it's correct? I have never seen this identity" ]

[ "$(2)$. Now, \\begin{align} &\\sum_{k=1}^{n}\\frac{1}{2k-1} \\\\ = & \\ \\sum_{k=1}^n\\frac{1}{2k-1} + \\sum_{k=1}^n\\frac{1}{2k} - \\sum_{k=1}^n\\frac{1}{2k}\\\\ = & \\ \\sum_{k=1}^{2n} \\frac{1}{k}- \\frac{1}{2}\\sum_{k=1}^n\\frac{1}{k}\\\\ \\end{align} And then, \\begin{align} &\\sum\\limits_{k=1}^n\\frac 1{2k-1} -\\frac 12\\sum\\limits_{k=1}^n\\frac 1k \\\\ = & \\ \\sum_{k=1}^{2n} \\frac{1}{k}- \\frac{1}{2}\\sum_{k=1}^n\\frac{1}{k} -\\frac 12\\sum\\limits_{k=1}^n\\frac 1k\\\\ = & \\ \\sum_{k=1}^{2n} \\frac{1}{k}- \\sum_{k=1}^n\\frac{1}{k} \\\\ = & \\ \\sum_{k=n+1}^{2n}\\frac{1}{k} = \\sum_{k=1}^{n}\\frac{1}{n+k} \\end{align} This gives $(2)$ to $(3)$ All in one chain: $$\\color{blue}{\\frac 12\\sum\\limits_{k=1}^n\\frac 1{2k-1}+\\frac 12\\sum\\limits_{k=1}^n\\frac 1{2k+1}-\\frac 12\\sum\\limits_{k=1}^n\\frac 1k+\\frac n{2n+1}=}$$ $$\\frac 12\\sum\\limits_{k=1}^n\\frac 1{2k-1}+\\frac 12\\left(\\sum\\limits_{k=1}^n\\frac 1{2k-1}-1+\\frac{1}{2n+1}\\right)-\\frac 12\\sum\\limits_{k=1}^n\\frac 1k+\\frac n{2n+1}=$$ $$\\sum\\limits_{k=1}^n\\frac 1{2k-1}+\\frac 12\\left(-\\frac{2n}{2n+1}\\right)-\\frac 12\\sum\\limits_{k=1}^n\\frac 1k+\\frac n{2n+1}=$$ $$\\color{blue}{\\sum\\limits_{k=1}^n\\frac 1{2k-1}-\\frac 12\\sum\\limits_{k=1}^n\\frac 1k=}$$ $$\\left(\\sum\\limits_{k=1}^{2n}\\frac 1{k}-\\sum\\limits_{k=1}^n\\frac 1{2k}\\right)-\\frac 12\\sum\\limits_{k=1}^n\\frac 1k=$$ $$\\left(\\sum\\limits_{k=1}^{2n}\\frac 1{k}-\\frac12\\sum\\limits_{k=1}^n\\frac 1{k}\\right)-\\frac 12\\sum\\limits_{k=1}^n\\frac 1k=$$ $$\\color{blue}{\\sum\\limits_{k=1}^{2n}\\frac 1k-\\sum\\limits_{k=1}^n\\frac 1k=}$$ $$\\left(\\sum\\limits_{k=1}^{n}\\frac 1k+\\sum\\limits_{k=n+1}^{2n}\\frac 1k\\right)-\\sum\\limits_{k=1}^n\\frac 1k=$$ $$\\sum\\limits_{k=n+1}^{2n}\\frac 1k=$$ $$\\color{blue}{\\sum\\limits_{k=1}^n\\frac 1{k+n}}$$", "since $a_n \\to 0$ then $\\mathop {\\lim }\\limits_{n \\to \\infty } {S_n} = \\mathop {\\lim }\\limits_{n \\to \\infty } (a_1 - a_{n+1}) = a_1$. Applying this fact for the sequence $a_k:=\\sin (\\frac{1}{2k})$, we get that \\begin{align*} \\sum\\limits_{n = 1}^\\infty {\\left( {\\sin \\frac{1}{{2n}} - \\sin \\frac{1}{{2n + 1}}} \\right)} = \\sin \\left( {\\frac{1}{2}} \\right) \\end{align*} - But here we have $\\sum (a_{2k} - a_{2k+1})$, or, if you wish, $\\sum (a_k - a_{k+\\frac{1}{2}})$, it doesn't telescope. – Daniel Fischer Apr 28 '14 at 11:25 Ah yes. You are right. But how come $k+\\frac{1}{2}$ in $a_{k+\\frac{1}{2}}$ – mwomath May 6 '14 at 11:18 @user45105 $a_k=\\sin(1/(2k))\\implies a_{k+1/2}=\\sin(1/(2(k+1/2)))=\\sin(1/(2k+1))$ – oldrinb May 7 '14 at 22:19 If we substitute $m=2n$, then \\begin{align} \\sum\\limits_{n = 1}^\\infty {\\left( {\\sin \\frac{1}{{2n}} - \\sin \\frac{1}{{2n + 1}}} \\right)} &= \\sum\\limits_{m = 2}^\\infty {\\left( {\\sin \\frac{1}{m} - \\sin \\frac{1}{{m + 1}}} \\right)} \\\\ &= \\sum\\limits_{m = 1}^\\infty {\\left( {\\sin \\frac{1}{m} - \\sin \\frac{1}{{m + 1}}} \\right)} - \\left( {\\sin 1 - \\sin \\frac{1}{2}} \\right) \\\\ &= \\sin 1 - \\left( {\\sin 1 - \\sin \\frac{1}{2}} \\right)\\\\ &= \\sin \\frac{1}{2} \\end{align} where as above $\\sum\\limits_{m = 1}^\\infty {\\left( {\\sin \\frac{1}{m} - \\sin \\frac{1}{{m + 1}}} \\right)} = \\sin \\left( 1 \\right)$ – mwomath Jun 3 '14 at 10:59", "all $x \\in \\mathbb{R}$ we can establish convergence. Let: (2) \\begin{align} \\quad a_n = \\sin nx \\quad \\mathrm{and} \\quad b_n = \\frac{\\left ( 1 + \\frac{1}{2} + \\frac{1}{3} + ... + \\frac{1}{n} \\right )}{n} \\end{align} Let $(A_n)_{n=1}^{\\infty}$ denote the sequence of partial sums for $(a_n)_{n=1}^{\\infty}$. Then for each $n \\in \\mathbb{N}$ we have that: (3) \\begin{align} \\quad A_n = \\sum_{k=1}^{n} \\sin kx = \\sin x + \\sin 2x + \\sin 3x + ... + \\sin nx \\end{align} It can be shown that: (4) \\begin{align} \\quad A_n = \\frac{\\cos \\left ( \\frac{x}{2} \\right ) - \\cos \\left (n + \\frac{1}{2} \\right )x }{2 \\sin \\left ( \\frac{x}{2} \\right )} \\leq \\frac{1 + x}{\\sin \\left ( \\frac{x}{2} \\right )} \\end{align} Notice that the right hand side of the inequality above is fixed for every $x \\in \\mathbb{R}$. This shows that the sequence of partial sums $(A_n)_{n=1}^{\\infty}$ is bounded. We now establish that the sequence $(b_n)_{n=1}^{\\infty}$ is decreasing. Consider the difference, $b_{n+1} - b_n$: (5) \\begin{align} \\quad b_{n+1} - b_n & = \\frac{\\left ( 1 + \\frac{1}{2} + \\frac{1}{3} + ... + \\frac{1}{n+1} \\right )}{n+1} - \\frac{\\left ( 1 + \\frac{1}{2} + \\frac{1}{3} + ... + \\frac{1}{n} \\right )}{n} \\\\ \\quad b_{n+1} - b_n & = \\frac{\\left ( n + \\frac{n}{2} + \\frac{n}{3} + ... + \\frac{n}{n+1} \\right )}{n(n+1)} - \\frac{\\left", "and since $a_n \\to 0$ then $\\mathop {\\lim }\\limits_{n \\to \\infty } {S_n} = \\mathop {\\lim }\\limits_{n \\to \\infty } (a_1 - a_{n+1}) = a_1$. Applying this fact for the sequence $a_k:=\\sin (\\frac{1}{2k})$, we get that \\begin{align*} \\sum\\limits_{n = 1}^\\infty {\\left( {\\sin \\frac{1}{{2n}} - \\sin \\frac{1}{{2n + 1}}} \\right)} = \\sin \\left( {\\frac{1}{2}} \\right) \\end{align*} - But here we have $\\sum (a_{2k} - a_{2k+1})$, or, if you wish, $\\sum (a_k - a_{k+\\frac{1}{2}})$, it doesn't telescope. – Daniel Fischer Apr 28 at 11:25 Ah yes. You are right. But how come $k+\\frac{1}{2}$ in $a_{k+\\frac{1}{2}}$ – mwomath May 6 at 11:18 @user45105 $a_k=\\sin(1/(2k))\\implies a_{k+1/2}=\\sin(1/(2(k+1/2)))=\\sin(1/(2k+1))$ – oldrinb May 7 at 22:19 If we substitute $m=2n$, then \\begin{align} \\sum\\limits_{n = 1}^\\infty {\\left( {\\sin \\frac{1}{{2n}} - \\sin \\frac{1}{{2n + 1}}} \\right)} &= \\sum\\limits_{m = 2}^\\infty {\\left( {\\sin \\frac{1}{m} - \\sin \\frac{1}{{m + 1}}} \\right)} \\\\ &= \\sum\\limits_{m = 1}^\\infty {\\left( {\\sin \\frac{1}{m} - \\sin \\frac{1}{{m + 1}}} \\right)} - \\left( {\\sin 1 - \\sin \\frac{1}{2}} \\right) \\\\ &= \\sin 1 - \\left( {\\sin 1 - \\sin \\frac{1}{2}} \\right)\\\\ &= \\sin \\frac{1}{2} \\end{align} where as above $\\sum\\limits_{m = 1}^\\infty {\\left( {\\sin \\frac{1}{m} - \\sin \\frac{1}{{m + 1}}} \\right)} = \\sin \\left( 1 \\right)$ – mwomath Jun 3 at 10:59 You can determine whether a series converges or not by evaluating $a_n$ when $n \\to", "\\limits _{n=n_s} ^{n_s+L-1} s_I[n] \\cos 2\\pi\\frac{k}{N}n \\nonumber \\\\ &= \\sum \\limits _{n=n_s} ^{n_s+L-1} \\cos 2\\pi\\frac{k}{N}n \\label{eqIntroductionDerivation1} \\\\ &= \\sum \\limits _{n=n_s} ^{n_s+L-1} \\cos 2\\pi\\frac{k}{N}n~ \\frac{\\sin \\pi\\frac{k}{N}}{\\sin \\pi\\frac{k}{N}} \\nonumber \\end{align} Let $\\theta = 2\\pi k/N$ and using the identity $\\cos(A)\\sin(B) = 0.5 \\{ \\sin(A+B) – \\sin(A-B) \\}$, we get \\begin{align*} S_I[k] &= \\sum \\limits _{n=n_s} ^{n_s+L-1} \\cos n\\theta~ \\frac{\\sin \\theta/2}{\\sin \\theta/2} \\\\ &= \\frac{1}{2\\sin \\theta/2} \\sum \\limits _{n=n_s} ^{n_s+L-1} \\left[ \\sin \\left( n+\\frac{1}{2}\\right)\\theta – \\sin \\left( n-\\frac{1}{2}\\right)\\theta \\right] \\\\ &= \\frac{1}{2\\sin \\theta/2} \\left[ \\sin \\left( n_s+L-\\frac{1}{2}\\right)\\theta – \\sin \\left( n_s-\\frac{1}{2}\\right)\\theta \\right] \\\\ \\end{align*} where all the other terms in the summation cancel out. Using the same identity as above \\begin{align*} S_I[k] &= \\frac{1}{\\sin \\theta/2} \\cos \\left(n_s + \\frac{L-1}{2}\\right)\\theta \\cdot \\sin\\left(\\frac{L}{2} \\right)\\theta \\\\ &= \\frac{\\sin L \\theta/2}{\\sin \\theta/2} \\cos \\left(n_s + \\frac{L-1}{2}\\right)\\theta \\nonumber \\end{align*} Plugging in the expression back for $\\theta$, \\begin{align} S_I[k] &= \\frac{\\sin \\pi L k/N}{\\sin \\pi k/N} \\cos \\left[ 2\\pi \\frac{ k}{N}\\left(n_s + \\frac{L-1}{2}\\right) \\right]\\label{eqIntroductionDFTrectangleI} \\end{align} Similarly, $Q$ component can be found from DFT definition, \\begin{align*} S_Q[k] &= \\sum \\limits _{n=n_s} ^{n_s+L-1} -s_I[n] \\sin 2\\pi\\frac{k}{N}n \\\\ &= \\sum \\limits _{n=n_s} ^{n_s+L-1} -\\sin 2\\pi\\frac{k}{N}n \\end{align*} Following the same procedure as $I$ component, and using the identity $\\sin(A)\\sin(B) = \\frac{1}{2}$ $\\left\\{ \\cos(A-B) -\\right.$ $\\left. \\cos(A+B) \\right\\}$, the $Q$ component of its DFT is given by \\begin{align} S_Q[k] &= -\\frac{\\sin", ") ]$ Splitting the R.H.S into two separate summations, we will get: $2T sin(\\frac{u}{ 2}) = \\Sigma_{k = 1}^ {n} [ sin( [\\frac{ (2k + 1)} { 2} ]u ) ] - \\Sigma_{k = 1}^ {n} sin( [ \\frac{(2k - 1)} { 2} ]u ) ]$ Or: $2T sin(\\frac{u}{ 2})) = \\Sigma _{k = 2}^ {n } [ sin( [ \\frac{(2k - 1)} { 2} ]u ) ] - \\Sigma_{k = 1}^ {n} [ sin( [\\frac{ (2k - 1)} { 2} ]u ) ] $ The above is telescopic series: $2T sin(\\frac{u}{ 2}) $ $= sin( [ \\frac{(2n + 1)} { 2} ]u ) - sin(\\frac{u}{ 2}) $ $= sin [\\frac{ (2n + 1)} { 2} ]u ) - sin(\\frac{u}{ 2}) $ $= 2 sin(\\frac{nu }{ 2}) cos( \\frac{[ n + 1 ]u} { 2} ) $ Hence: $T =\\frac{ sin(\\frac{nu}{ 2}) cos( \\frac{[ n + 1 ]u} { 2} ) }{sin(\\frac{u}{ 2})}$ 3. First of all, let's write this in a form that is readable... $1 + 2\\sum_{k = 1}^{n}\\cos{(ku)} = \\frac{\\sin{\\left[\\frac{(2n + 1)u}{2}\\right]}}{\\sin{\\left(\\frac{u}{2}\\right)}}$. Base step $n = 1$: $LHS = 1 + 2\\cos{u}$. $RHS = \\frac{\\sin{\\left(\\frac{3u}{2}\\right)}}{\\sin{\\left( \\frac{u}{2}\\right)}}$ $= \\frac{\\sqrt{\\frac{1 - \\cos{3u}}{2}}}{\\sqrt{\\frac{1 - \\cos{u}}{2}}}$ $= \\sqrt{\\frac{\\frac{1 - \\cos{3u}}{2}}{\\frac{1 - \\cos{u}}{2}}}$ $= \\sqrt{\\frac{1 - \\cos{3u}}{1 - \\cos{u}}}$ $= \\sqrt{\\frac{1 - (4\\cos^3{u} - 3\\cos{u})}{1 - \\cos{u}}}$ $= \\sqrt{\\frac{1 +", "has been proven for some $n\\geq 1$. Then, again by definition, \\begin{align*} f\\left(\\frac{x}{2^n}\\right)=\\frac{1}{2}f\\left(\\frac{x}{2^{n+1}}\\right)+\\frac{x}{2^{n+2}}, \\end{align*} and if we put this into our formula (induction!), we obtain \\begin{align*} f(x)= \\frac{1}{2^n}\\left(\\frac{1}{2}f\\left(\\frac{x}{2^{n+1}}\\right)+\\frac{x}{2^{n+2}}\\right)+\\frac{x}{4^n}\\sum_{k=0}^{n-1}4^k= \\frac{1}{2^{n+1}}f\\left(\\frac{x}{2^{n+1}}\\right)+\\frac{x}{4^{n+1}}+\\frac{x}{4^n}\\sum_{k=0}^{n-1}4^k= \\frac{1}{2^{n+1}}f\\left(\\frac{x}{2^{n+1}}\\right)+\\frac{x}{4^{n+1}}\\sum_{k=0}^{n}4^k, \\end{align*} which proves $(\\star)$. Now, if we fix $x\\in\\mathbb R$ and take the limit $n\\to\\infty$ in $(\\star)$ we obtain due to the limit condition that $f(x)=x/3$. Try with a polynomial of degree $n$ $$\\sum_{k=0}^{k=n}a_kx^k$$ We have $$\\sum_{k=0}^{k=n}a_kx^k+x=\\sum_{k=0}^{k=n}2^{k+1}a_kx^k$$ It follows $$\\begin {cases}a_k=2^{k+1}a_k;\\space k=2,3,.....,n\\\\a_1x+x=4a_1x\\\\a_0=2a_0\\end{cases}$$ Hence the only polynomial satisfying the property is $$f(x)=\\frac x3$$ Similarly with an homography $$\\frac{ax+b}{cx+d}$$ one gets at once $c=0$ and $d=3a$ which comes to $f(x)=\\frac x3$ It is doubtful that the property is worth for rational or transcendental function. I stop here. Exists an elegant and general proof to show that $$f(x)=\\frac{x}{3}$$ is the only one solution. ## Elegant proof Using complex variable, rewrite the functional equation as $$2f(2z)-f(z)=z$$. Since $$2z$$ and $$z$$ are complex-differentiable at $$z\\in\\mathbb{C}$$ it follows that $$f(z)$$ is necessarily complex-differentiable for all $$z\\in\\mathbb{C}$$ and hence necessarily continuous. Moreover, since $$2$$, $$2z$$, $$-1$$ and $$z$$ are finite-grade polynomials then $$f(z)$$ too and $$\\deg(f) = \\max( \\frac{\\deg(z) - \\deg(2)}{\\deg(2z)} , \\frac{\\deg(z) - \\deg(-1)}{\\deg(z)} ) = \\max(1,1) = 1.$$ As a consequence, the general solution is $$f(z) = c_0+zc_1$$ where $$c_0 = 0$$ and $$c_1 = \\frac{1}{3}$$ by replacing in the functional equation. Therefore,", "× # Help:Sequence and Series Find the sum of the series: $$1 + 2(1-a) + 3(1-a)(1-2a) + 4(1-a)(1-2a)(1-3a) + 5(1-a)(1-2a)(1-3a)(1-4a)+......$$ to $$n$$ terms. 9 months ago Sort by: $S_n = \\frac{1 - (1 - a)(1 - 2a)(1- 3a)\\ldots(1-na)}{a}$ Use induction to prove easily as, $S_k + (k + 1)(1 - a)(1 - 2a)(1- 3a)\\ldots(1-ka) = \\frac{1 - (1 - a)(1 - 2a)(1- 3a)\\ldots(1-ka)(1 - a(k + 1))}{a} = S_{k + 1}$ EDIT: $t_k = k(1 - a)(1 - 2a)(1- 3a)\\ldots(1-(k - 1)a)$ Write $$k$$ as the following: $$k = \\dfrac{1 - (1 - ka)}{a} = \\dfrac{1}{a} - \\dfrac{(1 - ka)}{a}$$ so, $t_k = \\dfrac{(1 - a)(1 - 2a)(1- 3a)\\ldots(1-(k - 1)a)}{a} - \\dfrac{(1 - a)(1 - 2a)(1- 3a)\\ldots(1 - ka)}{a}$ which when summed, telescopes to, $S_n = \\frac{1 - (1 - a)(1 - 2a)(1- 3a)\\ldots(1-na)}{a}$ as claimed. · 9 months ago I was initially amazed that the answer was so nice, certainly wasn't expecting that. A slightly better way to present it would be to show that $$S_{k+1}-S_k = t_k$$, which follows easily from the factorization. Staff · 9 months ago I was looking for more of an Algebraic proof · 9 months ago I've added the motivation for the sum. Check it out. · 9 months ago Can you also see the other solution that i posted", "$(p-1)!\\sum_{n=1}^\\infty\\frac{n!}{(p+n-1)!}$. Now, this should simplify a lot... For $p=2$, we have $$\\frac {n!}{p(p+1) ... (p+n-1)} = \\frac{1}{n+1}$$ and hence the series diverges like the harmonic series. For $0<p<2$, we can do a comparison with $p=2$ to conclude that the series also diverges. Now, suppose $p>2$ and write $p=2+\\epsilon$. We can write the term as \\begin{align}a_n=\\frac {1}{ \\frac{p}2\\, \\frac{p+1}3\\, \\dots \\frac{p+n-2}n\\, (p+n-1) } &= \\frac {1}{ \\left(1+\\frac{\\epsilon}2\\right) \\left(1+\\frac{\\epsilon}3\\right) \\dots \\left(1+\\frac{\\epsilon}n\\right)\\, (p+n-1) } \\\\&= \\frac{\\frac{1}{n+1}}{ \\left(1+\\frac{\\epsilon}2\\right) \\left(1+\\frac{\\epsilon}3\\right) \\dots \\left(1+\\frac{\\epsilon}n\\right)\\, \\left(1+\\frac{\\epsilon}{n+1}\\right)\\, }. \\end{align} Let $b_n = n\\left(\\frac{a_n}{a_{n+1}}-1\\right)$. We can check that \\begin{align} b_n &= \\frac{n}{n+1}\\, (1+\\epsilon). \\end{align} Then $b_n\\to1+\\epsilon >1$ as $n\\to\\infty$ and hence $\\sum a_n$ converges by Raabe's test.", "100 terms of $\\sum \\frac{1}{n^2 + 3n + 2}.$ Solution: First we can factor the denominator, which will allow us to decompose this rational function into two: $$\\frac{1}{n^2 + 3n + 2} = \\frac{1}{(n + 1)(n + 2)}$$ Now let's decompose: \\begin{align} \\frac{1}{(n + 1)(n + 2)} &= \\frac{A}{n + 1} + \\frac{B}{n + 2}\\\\[5pt] 1 &= A(n + 2) + B(n + 1)\\\\[5pt] \\text{Let }\\; n &= -1, \\; \\text{ then } \\; A = 1 \\\\[5pt] \\text{Let }\\; n &= -2, \\; \\text{ then } \\; B = -1, \\; \\text{ so} \\\\[5pt] \\frac{1}{(n + 1)(n + 2)} &= \\frac{1}{n + 1} - \\frac{1}{n + 2} \\end{align} Now we'll telescope the series: \\begin{align} S &= \\left( \\frac{1}{2} - \\frac{1}{3} \\right) + \\left( \\frac{1}{3} - \\frac{1}{4} \\right) + \\left( \\frac{1}{4} - \\frac{1}{5} \\right) + \\dots + \\left( \\frac{1}{101} - \\frac{1}{102} \\right)\\\\[5pt] &= \\frac{1}{2} + \\left( \\frac{1}{3} - \\frac{1}{3} \\right) + \\left( \\frac{1}{4} - \\frac{1}{4} \\right) + \\left( \\frac{1}{5} - \\frac{1}{5} \\right) + \\dots - \\frac{1}{102}\\\\[5pt] &= \\frac{1}{2} - \\frac{1}{102} = \\frac{51 - 1}{102} = \\frac{25}{51} \\approx \\frac{1}{2} \\end{align} It's easy to see that the sum of this series as n → ∞ is ½. xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2016, Jeff Cruzan. All text and images", "be as follows: assume that this property holds for all $k≤n$. Then your inductive step for $n+1$ is: $$\\frac{1}{3^2} +\\ldots +\\frac{1}{n^2}+\\frac{1}{(n+1)^2}<\\frac{1}{2^2}+\\frac{1}{3^2}+\\ldots +\\frac{1}{n^2}$$ This holds as you can verify for yourself that $\\frac{1}{2^2}>\\frac{1}{(n+1)^2}$ for any $n≥2$. Since $\\frac{1}{n}>\\frac{1}{n+1}$, and by our assumption, we now have that the equation is: $$\\frac{1}{3^2} +\\ldots +\\frac{1}{(n+1)^2}< \\frac{1}{2^2}+\\ldots +\\frac{1}{n^2}<1-\\frac{1}{n}<1-\\frac{1}{n+1}$$ And our inductive step is shown. While this does not answer the specific question in the OP, I thought it would be instructive to present an approach that develops the inequality directly. To that end, we proceed. Note that we have \\begin{align} \\frac{1}{k^2}& \\le \\frac{1}{k(k-1)}\\\\\\\\ &=\\frac{1}{k-1}-\\frac{1}{k} \\end{align} Therefore, we can write \\begin{align} \\sum_{k=2}^n\\frac{1}{k^2}& \\le \\sum_{k=2}^n\\left(\\frac{1}{k-1}-\\frac{1}{k}\\right) \\tag 1\\\\\\\\ &=1-\\frac{1}{n} \\tag 2 \\end{align} where we used the telescoping nature of the summand to go from $(1)$ to $(2)$. And we are done! • Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark – Mark Viola Aug 12 '16 at 1:47", "Note that this summation is not really appropriate because the K-index is quasi-logarithmic, however, this is a common simple measure of quiet and disturbed days. (These might be interesting days for you to look at.) In [ ]: dfk['Ksum'] = dfk.sum(axis=1) Ksort = dfk.sort_values('Ksum') print('Disturbed days: \\n\\n', Ksort.tail(10)) # 2. Note on the Fast Fourier Transform¶ In the examples above we computed Fourier coefficients in the 'traditional' way, so that if $F(t)$ is a Fourier series representation of $f(t)$, then, \\begin{align} F(t) &= A_o+\\sum_{n=1}^N A_n \\cos\\left(\\frac{2\\pi nt}{T}\\right)+B_n \\sin\\left(\\frac{2\\pi nt}{T}\\right) \\end{align} where $T$ is the fundamental period of $F(t)$. The $A_n$ and $B_n$ are estimated by \\begin{align} A_o&=\\frac{1}{T}\\int_0^T f(t) dt\\\\ A_n&=\\frac{2}{T}\\int_0^T f(t)\\cos\\left(\\frac{2\\pi nt}{T}\\right) dt\\\\ B_n&=\\frac{2}{T}\\int_0^T f(t)\\sin\\left(\\frac{2\\pi nt}{T}\\right) dt \\end{align} With $N$ samples of digital data, the integral for $A_n$ may be replaced by the summation \\begin{align} A_n&=\\frac{2}{T}\\sum_{j=0}^{N-1} f_j\\cos\\left(\\frac{2\\pi nj\\Delta t}{T}\\right) \\Delta t\\\\ &=\\frac{2}{N}\\sum_{j=0}^{N-1} f_j\\cos\\left(\\frac{2\\pi nj}{N}\\right) \\end{align} where the sampling interval $\\Delta t$ is given by $T = N \\Delta t$ and $f_j = f(j \\Delta t)$. A similar expression applies for the $B_n$, and these are the coefficients returned by the function fourier above. The fast Fourier transform (FFT) offers a computationally efficient means of finding the Fourier coefficients. The conventions for the FFT and its inverse (IFFT) vary from package to package. In the scipy.fftpack package, the FFT", "# Radon Nikodym derivative when changing numeraires I note from Wikipedia that if $$Q$$ and $$Q^N$$ are two measures corresponding to numeraires $$M$$ and $$N$$, then the Radon Nikodym derivative is given by: $$\\frac{dQ^N}{dQ} = \\frac{M(0)}{M(T)}\\frac{N(T)}{N(0)}.$$ However I do not understand how this formula comes from the traditional definition of a Radon-Nikodym derivative, which is a random variable such that the following holds for all RV $$Z$$: $$E_N(Z)=E_M\\left(\\frac{dQ^N}{dQ}Z\\right)$$ For any price process $$Z$$, given that $$N$$ and $$M$$ are numeraire processes, \\begin{align*} E_N\\left(\\frac{Z_T}{N_T} \\right) = E_M\\left(\\frac{Z_T}{M_T}\\frac{M_0}{N_0} \\right), \\end{align*} as they both equal to $$Z_0/N_0$$. Note also that \\begin{align*} E_N\\left(\\frac{Z_T}{N_T} \\right) = E_M\\left(\\frac{dQ_N}{dQ_M}\\frac{Z_T}{N_T} \\right). \\end{align*} Then \\begin{align*} E_M\\left(\\frac{dQ_N}{dQ_M}\\frac{Z_T}{N_T} \\right) &= E_M\\left(\\frac{Z_T}{M_T}\\frac{M_0}{N_0} \\right)\\\\ &=E_M\\left(\\frac{N_T}{M_T}\\frac{M_0}{N_0}\\frac{Z_T}{N_T}\\right). \\end{align*} Since $$\\frac{Z_T}{N_T}$$ can be arbitrary, we conclude that \\begin{align*} \\frac{dQ_N}{dQ_M} = \\frac{N_T}{M_T}\\frac{M_0}{N_0}. \\end{align*} Any nonnegative random variable $$Z$$ with expectation 1 is a Radon-Nikodym derivative: $$\\mathbb{E}^{\\mathbb{P}} \\left(Z\\right) = \\mathbb{E}^{\\mathbb{P}} \\left(\\frac{\\mathrm{d}\\mathbb{Q}}{\\mathrm{d}\\mathbb{P}}\\right) = \\mathbb{E}^{\\mathbb{Q}} \\left(1\\right) = \\int{\\mathrm{d}\\mathbb{Q}} = 1$$ $$\\mathbb{Q} \\left(A\\right) = \\mathbb{E}^\\mathbb{P} \\left(Z 1_A\\right) \\in \\left[0, 1\\right]$$ If $$Z$$ is positive, the probability measure $$\\mathbb{Q}$$ that it defines is equivalent to the original probability measure $$\\mathbb{P}$$. Now, by definition of a numeraire, under its associated probability measure, all asset prices expressed as units of the numeraire are martingales. For $$\\mathbb{Q}$$ with numeraire $$M$$ and $$N$$ a positive asset price process, $$\\mathbb{E}^{\\mathbb{Q}} \\left(\\frac{N_T}{M_T}\\right) = \\frac{N_0}{M_0} \\Rightarrow \\mathbb{E}^{\\mathbb{Q}}", "I don't think so. – Robert Israel Dec 4 '14 at 18:14 If you're happy to accept integrals in your general formula, then a really simple expression (via T.A.E.) is \\begin{align} \\frac{d^{n}}{dx^{n}}\\mathrm{sinc}(x) =\\frac{d^{n}}{dx^{n}}\\int_{-1}^{1}e^{ikx}\\, \\mathrm{d}k = \\int_{-1}^{1}(ik)^{n}e^{ikx}\\, \\mathrm{d}k \\end{align} $$\\frac{d^n}{dx^n}\\frac{\\sin(x)}{x}=\\frac{(-1)^{n}i}{2}(\\Gamma(n+1,ix)-\\Gamma(n+1,-ix))x^{-n-1}$$ $$=\\frac{(-1)^{n}n!i}{2}\\sum_{k=0}^{n}\\frac{i^k}{k!}(e^{-ix}-(-1)^ke^{ix})x^{-n+k-1}$$ $$=\\frac{(-1)^{n}n!i}{2}\\sum_{k=0}^{n}\\frac{i^k}{k!}((1-(-1)^k)\\cos(x)-(1+(-1)^k)i\\sin(x))x^{-n+k-1}$$ $$=(-1)^{n}n!2i\\sum_{k=0}^{n}\\frac{i^k}{k!}(\\text{ frac}\\left(\\frac{k}{2}\\right)\\cos(x)-\\text{ frac}\\left(\\frac{k+1}{2}\\right)i\\sin(x))x^{-n+k-1}$$ $$=(-1)^{n}n!2\\sum_{k=0}^{n}\\frac{i^{k+1}}{k!}\\text{ frac}\\left(\\frac{k}{2}\\right)\\frac{\\cos(x)}{x^{n-k+1}}+(-1)^{n}n!2\\sum_{k=0}^{n}\\frac{i^k}{k!}\\text{ frac}\\left(\\frac{k+1}{2}\\right)\\frac{\\sin(x)}{x^{n-k+1}}$$ $$=(-1)^{n}n!\\sum_{k=0}^{\\frac{n-1}{2}}\\frac{(-1)i^{k+1}}{(2k+1)!}\\frac{\\cos(x)}{x^{n-2k}}+(-1)^{n}n!\\sum_{k=0}^{\\frac{n}{2}}\\frac{(-1)^k}{(2k)!}\\frac{\\sin(x)}{x^{n-2k+1}}$$", "We leave this material with two examples that use this new result. Examples: 1. Find $$\\displaystyle \\lim_{x\\to 0}\\frac{\\sin^2 x}{x}$$. Again, we use the limit laws, $$\\lim_{x\\to 0} \\frac{\\sin^2 x}{x} = \\left(\\lim_{x\\to 0} \\frac{\\sin x}{x}\\right) (\\lim_{x\\to 0} \\sin x) = 1\\cdot 0 = 0.$$ 2. Find $$\\displaystyle \\lim_{x\\to 0}\\frac{1-\\cos x}{x^2}$$. For this one we need to multiply by the conjugate of the numerator, \\begin{align*} \\lim_{x\\to 0}\\frac{1-\\cos x}{x^2} &= \\lim_{x\\to 0}\\frac{1-\\cos x}{x^2} \\left(\\frac{1+\\cos x}{1+\\cos x}\\right) = \\lim_{x\\to 0}\\frac{1-\\cos^2 x}{x^2(1+\\cos x)}\\\\ &= \\lim_{x\\to 0}\\frac{\\sin^2 x}{x^2(1+\\cos x)} = \\left(\\lim_{x\\to 0}\\frac{\\sin x}{x}\\right)^2 \\lim_{x\\to 0}\\frac{1}{1+\\cos x} = 1^2 \\cdot \\frac{1}{2} = \\frac{1}{2}. \\end{align*}", "\\Im\\{X[k]\\} \\sin(2 \\pi (k/N) f_\\text{s} t) \\\\ & = \\frac1N \\left( X[0] + X[N/2] \\cos(2 \\pi (f_\\text{s}/2) t) + A_x \\sin(2 \\pi (f_\\text{s}/2) t) \\right) \\\\ & \\quad + \\frac2N \\sum\\limits_{k=1}^{\\frac{N}{2}-1} \\Re\\{X[k]\\} \\cos(2 \\pi (k/N) f_\\text{s} t) - \\Im\\{X[k]\\} \\sin(2 \\pi (k/N) f_\\text{s} t) \\\\ \\end{align} where $A_x$ can be any number, because that term will be zero at the sampling instances, when $t = nT = n/f_\\text{s}$. looking at it from a magnitude/phase POV, for $N$ odd $$x(t) = \\frac{X[0]}{N} + \\frac2N \\sum\\limits_{k=1}^{\\frac{N-1}{2}} |X[k]| \\cos(2 \\pi (k/N) f_\\text{s} t + \\arg\\{X[k]\\})$$ and for $N$ even \\begin{align} x(t) & = \\frac{X[0]}{N} + \\frac{X[N/2]}{N \\cos(\\theta)} \\cos(2 \\pi (f_\\text{s}/2) t + \\theta) \\\\ & \\quad + \\frac2N \\sum\\limits_{k=1}^{\\frac{N}{2}-1} |X[k]| \\cos(2 \\pi (k/N) f_\\text{s} t + \\arg\\{X[k]\\}) \\end{align} where $$\\Re\\{X[k]\\} = |X[k]| \\cos(\\arg\\{X[k]\\})$$ $$\\Im\\{X[k]\\} = |X[k]| \\sin(\\arg\\{X[k]\\})$$ or $$|X[k]| = \\sqrt{\\Re\\{X[k]\\}^2 + \\Im\\{X[k]\\}^2}$$ $$\\arg\\{X[k]\\} = \\operatorname{atan2}(\\Im\\{X[k]\\}, \\Re\\{X[k]\\})$$ and $\\theta$ is the phase of the Nyquist component and can be any real value except $\\pm \\frac{\\pi}{2}$ or $\\frac{\\pi}{2}+ m \\pi$ for integer $m$ (because you cannot divide by zero). the samples $x[n] \\triangleq x(n/f_\\text{s})$ will come out the same regardless of the phase of the Nyquist component as long as the amplitude of the component is adjusted accordingly by dividing by $\\cos(\\theta)$. • For real-valued sequences with $N$ even, you know both", "\\begin{align} \\text{Range}(g) &= \\left[ \\frac{-1 - \\sin \\left( \\frac{1}{2} \\right)}{2 \\sin \\left( \\frac{1}{2} \\right)},\\frac{1 - \\sin \\left( \\frac{1}{2} \\right)}{2 \\sin \\left( \\frac{1}{2} \\right)} \\right] \\\\ &= [-1.54291482 \\ldots,0.54291482 \\ldots] \\\\ &\\subseteq [-2,2]. \\end{align} Conclusion: $\\displaystyle \\left| \\sum_{k=1}^{n} \\sin(k) \\right| < 2$ and $\\displaystyle \\left| \\sum_{k=1}^{n} \\cos(k) \\right| < 2$ for all $n \\in \\mathbb{N}$. - You have the formula $\\frac{1-x^{n+1}}{1-x}=\\sum_{k=1}^{n}x^k$, which is easy enough to see by algebra. So you can replace the LHS with $\\frac{1-e^{i(n+1)\\theta}}{1-e^{i\\theta}}$. – minimalrho Feb 1 '13 at 22:03 +1 Very nice hint. This is not a series, @Iuli, but a finite sum and, in fact, a rather easy one: a geometric sequence sum. Above you can read what you get from it, which with absolute values gives at once the wanted upper bound $\\,2\\,$ – DonAntonio Feb 1 '13 at 22:15 @CalvinLin , yes...so what? The absolute value of the expression hinted in the answer above is less than or equal what you wrote (in fact, it is always less). Besides this, in the given formula you must take the imaginary part as that the part that corresponds to the sum of sines... – DonAntonio Feb 1 '13 at 23:05 @HaskellCurry Yes, with the trig version I could see how to do it. The exp version was less obvious, since you still had", "# Limit of sequence lim n (1-cos(2/n)) 1. Homework Statement 2. Homework Equations 3. The Attempt at a Solution above picture !! Related Calculus and Beyond Homework Help News on Phys.org jbunniii Homework Helper Gold Member You can write \\begin{align} 1 - \\cos\\left(\\frac{2}{n}\\right) = 2\\left[\\frac{1}{2} - \\frac{1}{2}\\cos\\left(\\frac{2}{n}\\right)\\right] &= 2\\sin^2\\left(\\frac{1}{n}\\right) \\\\ \\end{align} LCKurtz Homework Helper Gold Member $$\\frac 1 {\\frac 1 n}$$doesn't go to 0 as $n\\to\\infty$. Another method is to use L'Hospital's rule on$$\\frac{1 - \\cos{\\frac 1 n}}{\\frac 1 n}$$ You can write \\begin{align} 1 - \\cos\\left(\\frac{2}{n}\\right) = 2\\left[\\frac{1}{2} - \\frac{1}{2}\\cos\\left(\\frac{2}{n}\\right)\\right] &= 2\\sin^2\\left(\\frac{1}{n}\\right) \\\\ \\end{align} where is 'n' gone? $$\\frac 1 {\\frac 1 n}$$doesn't go to 0 as $n\\to\\infty$. Another method is to use L'Hospital's rule on$$\\frac{1 - \\cos{\\frac 1 n}}{\\frac 1 n}$$ thanks LCKurtz jbunniii Homework Helper Gold Member where is 'n' gone? It didn't go anywhere. I was just suggesting how to rewrite the hard part. Now if we substitute back, we get $$n\\left[1 - \\cos\\left(\\frac{2}{n}\\right)\\right] = 2n \\sin^2\\left(\\frac{1}{n}\\right)$$ Can you see what to do now? It didn't go anywhere. I was just suggesting how to rewrite the hard part. Now if we substitute back, we get $$n\\left[1 - \\cos\\left(\\frac{2}{n}\\right)\\right] = 2n \\sin^2\\left(\\frac{1}{n}\\right)$$ Can you see what to do now? Thanks jbunniii", "if $k$ is very small. $k=\\frac{V}{nc}$ with $V\\ll c$, so this is exactly what you want. Another way of presenting this is to telescope the expression to arbitrary precision generating the geometric series in the process $$\\frac{1}{1+\\frac{V}{nc}}=\\frac{1}{1+\\frac{V}{nc}}\\frac{1-\\frac{V}{nc}}{1-\\frac{V}{nc}}=\\frac{1-\\frac{V}{nc}}{1-\\left(\\frac{V}{nc}\\right)^2}$$ and the next iteration $$\\frac{1}{1+\\frac{V}{nc}}=\\frac{1-\\frac{V}{nc}}{1-\\left(\\frac{V}{nc}\\right)^2}\\frac {1+\\left(\\frac{V}{nc}\\right)^2} {1+\\left(\\frac{V}{nc}\\right)^2}=\\frac{1-\\frac{V}{nc}+ \\left(\\frac{V}{nc}\\right)^2-\\left(\\frac{V}{nc}\\right)^3} {1-\\left(\\frac{V}{nc}\\right)^4}$$ We can stop at any stage once we are happy to approximate the denominator as $1$. The inverse of the first, multiplied by the second, is $$1-\\frac{V^2}{n^2c^2}$$ very near from $1$.", "${g=v^{\\frac{4}{n-2}}g_0}$, the mean curvature of ${(M,g)}$ will be $\\displaystyle h_g=\\frac{2}{n-2}v^{-\\frac{n}{n-2}}\\left(\\frac{\\partial v}{\\partial \\eta}+\\frac{n-2}{2}h_0 v\\right)$ where ${\\eta}$ is the normal outer unit vector under ${g_0}$. Using the above principle, let ${v=2^{\\frac{n-2}{2}}u}$, ${g_0}$ be the Euclidean flat metric, then ${h_0=1}$. $\\displaystyle \\frac{\\partial v}{\\partial \\eta}=(2\\epsilon)^{\\frac{n-2}{2}}(\\epsilon^2+1)^{-\\frac{n}{2}}(2-n)$ $\\displaystyle \\frac{n-2}{2}h_0 v=\\frac{n-2}{2}(2\\epsilon)^{\\frac{n-2}{2}}(\\epsilon^2+1)^{-\\frac{n-2}{2}}$ $\\displaystyle h_g=\\frac{2}{n-2}\\frac{(2\\epsilon)^{\\frac{n-2}{2}}(\\epsilon^2+1)^{-\\frac{n}{2}}(2-n)+\\frac{n-2}{2}(2\\epsilon)^{\\frac{n-2}{2}}(\\epsilon^2+1)^{-\\frac{n-2}{2}}}{(2\\epsilon)^{-\\frac{n}{2}}(\\epsilon^2+1)^{-\\frac{n}{2}}}=\\frac{\\epsilon^2-1}{2\\epsilon}$ Remark: Escobar. Conformal Defromation of a Riemannnian metric to a constant scalar curvature metric with constant mean curvature on the boundary. Indiana University Mathematics Journal 1996. ### Conformally invariant Laplacian On a compact manifold ${(M^n,g)}$, ${n\\geq 3}$, $\\displaystyle L_g=-\\Delta_g+\\frac{n-2}{4(n-1)}R_g$ is called conformal laplacian operator. This follows from the following fact. Suppose ${\\tilde{g}=\\phi^{\\frac{4}{n-2}}g}$, then $\\displaystyle L_{\\tilde{g}}(f)=\\phi^{-\\frac{n+2}{n-2}}L_g(\\phi f), \\quad\\forall\\,f\\in C^\\infty(M)$ Proof: Firstly suppose ${f>0}$, define ${\\bar{g}=(\\phi f)^{\\frac{4}{n-2}}g}$, then $\\displaystyle L_{\\bar{g}}(\\phi f)=\\frac{n-2}{4(n-1)}R_{\\bar{g}}(\\phi f)^{\\frac{n+2}{n-2}}$ on the other hand ${\\bar{g}=f^{\\frac{4}{n-2}}\\tilde{g}}$ $\\displaystyle L_{\\tilde{g}}(f)=\\frac{n-2}{4(n-1)}R_{\\bar{g}}(f)^{\\frac{n+2}{n-2}}$ So we proved the equality. For general ${f\\in C^\\infty(M)}$, ${\\exists\\, C>0}$ such that ${f+C>0}$. By the special case $\\displaystyle L_{\\tilde{g}}(f+C)=\\phi^{-\\frac{n+2}{n-2}}L_g(\\phi (f+C))$ $\\displaystyle L_{\\tilde{g}}(C)=\\phi^{-\\frac{n+2}{n-2}}L_g(\\phi (C))$ Thus the general case is also true. $\\Box$" ]

[ ") - 2C) Find total number of real s... Question # ) - 2C) Find total number of real solutions to the equation 52 + 122 = 132 15(S+ JEE/Engineering Exams Maths Solution 97 4.0 (1 ratings) here, given ( 5^{2^{2}+12^{2}=13^{2^{2}}} ) [ begin{array}{l}text { We know, } 5^{2}+12^{2}=13^{21} therefore frac{z^{2}=2}{17=pm sqrt{21}} text { Ary } text { total no of real son = two(2) }end{array} ]", "# Math Help - find number of solutions of x 1. ## find number of solutions of x find number of solutions of x ..where x belongs to real number for which $64^{\\frac{1}{x}} + 48^{\\frac{1}{x}} = 80^{\\frac{1}{x}}$ 2. Try factoring each number as a multiple of 8. You might get some cancellation there.", "another point is that, there are infinite numbers of solutions to the equation: $$\\frac{-16}{25} = \\sin (2 \\theta)$$, not just one as Swapnil mentioned.", "# find number of solutions of x #### banku12 find number of solutions of x ..where x belongs to real number for which $$\\displaystyle 64^{\\frac{1}{x}} + 48^{\\frac{1}{x}} = 80^{\\frac{1}{x}}$$ #### Ackbeet MHF Hall of Honor Try factoring each number as a multiple of 8. You might get some cancellation there.", "# Sine vs. Line Algebra Level 3 How many real numbers $$x$$ satisfy $$\\sin x = \\frac{x}{100}?$$ ×", "# find number of solutions of x • July 8th 2010, 09:38 AM banku12 find number of solutions of x find number of solutions of x ..where x belongs to real number for which $64^{\\frac{1}{x}} + 48^{\\frac{1}{x}} = 80^{\\frac{1}{x}}$ • July 8th 2010, 12:13 PM Ackbeet Try factoring each number as a multiple of 8. You might get some cancellation there.", "The number of distinct solutions of the equation Question: The number of distinct solutions of the equation $\\log _{\\frac{1}{2}}|\\sin x|=2-\\log _{\\frac{1}{2}}|\\cos x|$ in the interval $[0,2 \\pi]$, is___________. Solution:", "# A geometry problem by Aditya Pratap Singh Geometry Level 3 How many REAL solutions are there for the equation: sin(X)=X/10 ?", "# 'If clear with functions' Part-2 Geometry Level 4 Determine the number of roots of the equation $\\sin x =\\frac { x }{ 100 } .$ ×", "+0 # help 0 82 1 +103 Find the number of real solutions to $$x^{\\sqrt{x}} = (\\sqrt{x})^x$$ Dec 15, 2019", "### Home > A2C > Chapter 9 > Lesson 9.2.3 > Problem9-110 9-110. $x=\\frac{-b\\pm\\sqrt{b^{2}-4(5)(20)}}{2(5)}$ The equation will have real solutions as long as the value under the radical is greater than or equal to zero.", "# Solution for boxes Find the number of real solutions of the equation $\\frac{1}{[x]}+\\frac{1}{[2x]}= x-[x]+\\frac{1}{3}.$ Note: 1.$$[x]$$ stands for the greatest integer function. 2. This problem is not original. ×", "# Get Real! Algebra Level 5 Find the number of real solutions to the equation $\\sum_{r=1}^{100} \\dfrac{r}{x-r}=x$ ×", "+0 help 0 41 1 +103 Find the number of real solutions to $$x^{\\sqrt{x}} = (\\sqrt{x})^x$$ Dec 15, 2019", "real, then find them. 50. Form the equation whose one root is $$\\frac{1}{2 + \\sqrt{3}}$$", "29 views What is the number of solutions of: $x=\\frac{x^{2}}{50}-\\cos \\frac{x}{2}+2$ in $[0,10]$ ? 1. $0$ 2. $1$ 3. $2$ 4. $\\infty$", "# Solutions to Problems 1161-1170 Q1161. Find all values of $x$ (real number) satisfying $$\\frac{x-1}{2004} + \\frac{x-3}{2002} + \\frac{x-5}{2000} + \\cdots + \\frac{x-2003}{2} = \\cdots$$", "this equation solve this equation sin 30 cos x + cos 30 sinx = 1 where x is real number", "1. ## Trigonometry Help! Question: Find the solution of $\\sin{x}-1=0$ if $0 \\leq x < \\frac{1}{2}\\pi$. Attempt: $\\sin{x}-1=0$ $\\sin{x} = 1$ $x = \\sin^{-1}{(1)}$ $x = 90^o$ $x = \\frac{1}{2}\\pi$ Either my answer is wrong or the question is wrong because it was mentioned in the question that it is not greater than $\\frac{1}{2}\\pi$. Am I correct? 2. You're correct, so the answer is that no real number solves this equation on the given interval.", "equation has 100 solutions (by the fundamental theorem of algebra) lying on the unit circle. Furthermore they are distinct since the solutions of $z^{150}=1$ are distinct and $z^{100}+z^{50}+1$ is a factor of $z^{150}-1$. Therefore the answer is 100." ]

[ "answer is then $\\boxed{\\dfrac{145}{147}}$. ## Solution 5 $[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); draw((0.3,0.3)--(3.6,0.3), dashed); draw((0.3,2.7)--(0.3,0.3), dashed); label(\"S\", (-0.05,-0.05), NE); draw((0.3,0.3)--(1.41,1.91)); draw((1.63,1.78)--(1.48,1.56)); draw((1.28,1.70)--(1.48,1.56)); label(\"4\", (2,0), S); label(\"3\", (0,1.5), W); label(\"\\frac{10}{3}\", (2,0.3), N); label(\"\\frac{5}{2}\", (0.3,1.5), E); label(\"2\", (1,1.2), E); draw((3.6,0)--(3.6,0.3), dashed); draw((0,2.7)--(0.3,2.7), dashed); label(\"\\small{l}\", (3.6,0.15), W); label(\"\\small{l}\", (0.15,2.7), S); label(\"\\small{l}\", (0.3,0.15), E); label(\"\\small{l}\", (0.15,0.3), N); [/asy]$ On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\\frac{5}{2}$ and $\\frac{10}{3}$. With $l$ being the side length of the square, we need to find an expression for $l$. Using the hypotenuse, we can see that $\\frac{3}{2}+\\frac{8}{3}+\\frac{5}{4}l+\\frac{5}{3}l=5$. Simplifying, $\\frac{35}{12}l=\\frac{5}{6}$, or $l=2/7$. A different calculation would yield $l+\\frac{3}{4}l+\\frac{5}{2}=3$, so $\\frac{7}{4}l=\\frac{1}{2}$. In other words, $l=\\frac{2}{7}$, while to check, $l+\\frac{4}{3}l+\\frac{10}{3}=4$. As such, $\\frac{7}{3}l=\\frac{2}{3}$, and $l=\\frac{2}{7}$. Finally, we get $A(\\Square S)=l^2=\\frac{4}{49}$, to finish. As a proportion of the triangle with area $6$, the answer would be $1-\\frac{4}{49\\cdot6}=1-\\frac{2}{147}=\\frac{145}{147}$, so $\\boxed{\\textit D}$ is correct. ## Solution 6: Also Coordinate Geo Let the right angle be at $(0,0)$, the point $(x,x)$ be the far edge of the unplanted square and the hypotenuse be the line $y=-\\frac{3}{4}x+3$. Since the line from $(x,x)$ to the hypotenuse is the shortest possible distance, we know this line, call it line $\\l$, is perpendicular to the hypotenuse and therefore has", "1996 AJHSME Problems/Problem 5 Problem 5 The letters $P$, $Q$, $R$, $S$, and $T$ represent numbers located on the number line as shown. $[asy] unitsize(36); draw((-4,0)--(4,0)); draw((-3.9,0.1)--(-4,0)--(-3.9,-0.1)); draw((3.9,0.1)--(4,0)--(3.9,-0.1)); for (int i = -3; i <= 3; ++i) { draw((i,-0.1)--(i,0)); } label(\"-3\",(-3,-0.1),S); label(\"-2\",(-2,-0.1),S); label(\"-1\",(-1,-0.1),S); label(\"0\",(0,-0.1),S); label(\"1\",(1,-0.1),S); label(\"2\",(2,-0.1),S); label(\"3\",(3,-0.1),S); draw((-3.7,0.1)--(-3.6,0)--(-3.5,0.1)); draw((-3.6,0)--(-3.6,0.25)); label(\"P\",(-3.6,0.25),N); draw((-1.3,0.1)--(-1.2,0)--(-1.1,0.1)); draw((-1.2,0)--(-1.2,0.25)); label(\"Q\",(-1.2,0.25),N); draw((0.1,0.1)--(0.2,0)--(0.3,0.1)); draw((0.2,0)--(0.2,0.25)); label(\"R\",(0.2,0.25),N); draw((0.8,0.1)--(0.9,0)--(1,0.1)); draw((0.9,0)--(0.9,0.25)); label(\"S\",(0.9,0.25),N); draw((1.4,0.1)--(1.5,0)--(1.6,0.1)); draw((1.5,0)--(1.5,0.25)); label(\"T\",(1.5,0.25),N); [/asy]$ Which of the following expressions represents a negative number? $\\text{(A)}\\ P-Q \\qquad \\text{(B)}\\ P\\cdot Q \\qquad \\text{(C)}\\ \\dfrac{S}{Q}\\cdot P \\qquad \\text{(D)}\\ \\dfrac{R}{P\\cdot Q} \\qquad \\text{(E)}\\ \\dfrac{S+T}{R}$ Solution 1 First, note that $P < Q < 0 < R < S < T$. Thus, $P$ and $Q$ are negative, while $R$, $S$, and $T$ are positive. Option $A$ is the difference of two numbers. If the first number is lower than the second number, the answer will be negative. In this case, $P < Q$, so $P - Q < 0$, and $P-Q$ is thus indeed a negative number. So option $\\boxed{A}$ is correct. Option $B$ is the product of two negatives, which is always positive. Option $C$ starts with the quotient of a positive and a negative. This is negative. But this negative quotient is then multiplied by a negative number, which gives a positive number. Option $D$ has the product of two negatives in", "$m=1/3$, and $m=-1/2$. $[asy] unitsize(1cm); defaultpen(0.8); real m=1.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),E); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=0.5; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=1/3; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-1)*(1,m)) -- (3*(1,m)), dashed ); label(\"A\",(0,0),NW); label(\"B\",(1,1),N); label(\"C\",(6*m,0),E); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ $[asy] unitsize(2cm); defaultpen(0.8); real m=-1/2; draw( (0,0)--(1,1)--(6*m,0)--cycle ); draw( ((-2)*(1,m)) -- (1*(1,m)), dashed ); label(\"A\",(0,0),S); label(\"B\",(1,1),NE); label(\"C\",(6*m,0),W); pair D = intersectionpoint( (1,1)--(6*m,0), ((-1)*(1,m)) -- (3*(1,m)) ); dot(D); label(\"D\",D,S); [/asy]$ ## Solution 2 The line must pass through the triangle's centroid, whose coordinates are found by averaging those of the vertices. The slope of the line from the origin through the centroid is thus $\\frac{\\frac{1}{3}}{\\frac{1+6m}{3}}$, which is equal to $m$. Solving this equation, we arrive at the answer: $\\boxed{-\\frac{1}{6}}$.", "Problem #2045 2045 Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? $[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label(\"Figure\",(0.5,-1),S); label(\"0\",(0.5,-2.5),S); label(\"Figure\",(9.5,-1),S); label(\"1\",(9.5,-2.5),S); label(\"Figure\",(19.5,-1),S); label(\"2\",(19.5,-2.5),S); label(\"Figure\",(32.5,-1),S); label(\"3\",(32.5,-2.5),S); [/asy]$ $\\textbf{(A)}\\ 10401 \\qquad\\textbf{(B)}\\ 19801 \\qquad\\textbf{(C)}\\ 20201 \\qquad\\textbf{(D)}\\ 39801 \\qquad\\textbf{(E)}\\ 40801$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "Solving for $t$, we find $t = 20$. Our answer, $AP$, is equivalent to $8t$. Thus, $AP = 8 \\cdot 20 = \\boxed{160}$. ## Solution 3 We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$. Then we can produce the following diagram: $[asy] draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); label(\"100\",(0,0)--(100,0),S); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),S); dot((80,139)); label(\"M\",(80,139),N); [/asy]$ Now, if we drop an altitude from point$M$, we get : $[asy] size(300); draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),SE); dot((80,139)); label(\"M\",(80,139),N); draw((80,139)--(80,0),dashed); label(\"4x\",(0,0)--(80,0),S); label(\"100-4x\",(80,0)--(100,0),S); label(\"4x \\sqrt{3}\",(80,139)--(80,0),W); label(\"60^{\\circ}\", (3,1), NE); [/asy]$ We know this from the $30-60-90$ triangle that is formed. From this we get that: $$(7x)^2 = (4 \\sqrt{3})^2 + (100-4x)^2$$ $$\\Longrightarrow 49x^2 - 48x^2 = x^2 = (100-4x)^2$$ $$\\Longrightarrow 0=(100-4x)^2 - x^2 = (100-3x)(100-5x)$$. Therefore, we get that $x = \\frac{100}{3}$ or $x = 20$. Since $20< \\frac{100}{3}$, we have that $x=20$ (since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be $8 \\cdot x = 8 \\cdot 20 = \\boxed{160}$ meters. ~qwertysri987", "$n$ is not divisible by $3$ (therefore $9$) since $n>11$ and $n$ is prime, it follows that $n|\\underbrace{111\\cdots 111}_{x\\text{ times}}$. $\\Box$ ## Picture 1 $[asy]draw(Circle((1,1),2)); draw(Circle((sqrt(2),sqrt(3)/2),1)); dot((8/5,2/5)); dot((1,1)); draw((1,1)--(8/5,2/5),linetype(\"8 8\")); label(\"a\",(6/5,7/10),SSW); draw((8/5,2/5)--(12/5,-2/5),linetype(\"8 8\")); label(\"b\",(2,0),SSW); [/asy]$ $$\\text{Find the probability that }b>a \\text{.}$$ $[asy]unitsize(2inch); import olympiad; path c2 = dir(90)-dir(120)..dir(90)-dir(150)..dir(90)-dir(180); path c1 = dir(90)-dir(0)..dir(90)-dir(30)..dir(90)-dir(60); path c3 = dir(120)..dir(150)..dir(180); path c4 = dir(0)..dir(30)..dir(60); draw(dir(0)..dir(30)..dir(60)..dir(90)..dir(120)..dir(150)..dir(180)--dir(0)); draw(dir(90)-dir(0)..dir(90)-dir(30)..dir(90)-dir(60)..dir(90)-dir(90)..dir(90)-dir(120)..dir(90)-dir(150)..dir(90)-dir(180)--dir(90)-dir(0)); draw((-1.2,0.8)--(1.2,0.8)); label(\"l_{-n}\", (1.2,0.8),dir(0)); draw((-1.2,0.5)--(1.2,0.5)); label(\"l_0\", (1.2,0.5),dir(0)); draw((-1.2,0.2)--(1.2,0.2)); label(\"l_n\", (1.2,0.2),dir(0)); label(\"\\vdots\", (1.2, 0.4), dir(0)); label(\"\\vdots\", (0, 0.4)); label(\"\\vdots\", (1.2, 0.65), dir(0)); label(\"\\vdots\", (0, 0.65)); label(\"A_{-n}\", intersectionpoint(c1, (-1.2, 0.8)--(1.2, 0.8)), dir(135)); label(\"C_{-n}\", intersectionpoint(c3, (-1.2, 0.8)--(1.2, 0.8)), dir(135)); label(\"D_{-n}\", intersectionpoint(c4, (-1.2, 0.8)--(1.2, 0.8)), dir(45)); label(\"B_{-n}\", intersectionpoint(c2, (-1.2, 0.8)--(1.2, 0.8)), dir(45)); label(\"X\", intersectionpoint(c1, (-1.2, 0.5)--(1.2, 0.5)), dir(150)); label(\"Y\", intersectionpoint(c2, (-1.2, 0.5)--(1.2, 0.5)), dir(30)); label(\"C_{n}\", intersectionpoint(c3, (-1.2, 0.2)--(1.2, 0.2)), dir(135)); label(\"A_{n}\", intersectionpoint(c1, (-1.2, 0.2)--(1.2, 0.2)), dir(45)); label(\"B_{n}\", intersectionpoint(c2, (-1.2, 0.2)--(1.2, 0.2)), dir(135)); label(\"D_{n}\", intersectionpoint(c4, (-1.2, 0.2)--(1.2, 0.2)), dir(45)); dot(intersectionpoint(c1, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c2, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c3, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c4, (-1.2, 0.8)--(1.2, 0.8))); dot(intersectionpoint(c1, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c2, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c3, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c4, (-1.2, 0.5)--(1.2, 0.5))); dot(intersectionpoint(c1, (-1.2, 0.2)--(1.2, 0.2))); dot(intersectionpoint(c2, (-1.2, 0.2)--(1.2, 0.2))); dot(intersectionpoint(c3, (-1.2, 0.2)--(1.2, 0.2))); dot(intersectionpoint(c4, (-1.2, 0.2)--(1.2, 0.2))); [/asy]$ Two half-circles are drawn as shown above, with a line $l_0$ throught the two intersections points, $X,Y$ of the half-circles. Lines $l_k$ for $k=-n\\to", "on both tests? $[asy] draw((2,0)--(7,0)--(7,5)--(2,5)--cycle); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((5,0)--(5,5)); draw((6,0)--(6,5)); draw((2,1)--(7,1)); draw((2,2)--(7,2)); draw((2,3)--(7,3)); draw((2,4)--(7,4)); draw((.2,6.8)--(1.8,5.2)); draw(circle((4.5,1.5),.5),linewidth(.6 mm)); label(\"0\",(2.5,.2),N); label(\"0\",(3.5,.2),N); label(\"2\",(4.5,.2),N); label(\"1\",(5.5,.2),N); label(\"0\",(6.5,.2),N); label(\"0\",(2.5,1.2),N); label(\"0\",(3.5,1.2),N); label(\"1\",(4.5,1.2),N); label(\"1\",(5.5,1.2),N); label(\"1\",(6.5,1.2),N); label(\"1\",(2.5,2.2),N); label(\"3\",(3.5,2.2),N); label(\"5\",(4.5,2.2),N); label(\"2\",(5.5,2.2),N); label(\"0\",(6.5,2.2),N); label(\"1\",(2.5,3.2),N); label(\"4\",(3.5,3.2),N); label(\"3\",(4.5,3.2),N); label(\"0\",(5.5,3.2),N); label(\"0\",(6.5,3.2),N); label(\"2\",(2.5,4.2),N); label(\"2\",(3.5,4.2),N); label(\"1\",(4.5,4.2),N); label(\"0\",(5.5,4.2),N); label(\"0\",(6.5,4.2),N); label(\"F\",(1.5,.2),N); label(\"D\",(1.5,1.2),N); label(\"C\",(1.5,2.2),N); label(\"B\",(1.5,3.2),N); label(\"A\",(1.5,4.2),N); label(\"A\",(2.5,5.2),N); label(\"B\",(3.5,5.2),N); label(\"C\",(4.5,5.2),N); label(\"D\",(5.5,5.2),N); label(\"F\",(6.5,5.2),N); label(\"Test 1\",(-.5,5.2),N); label(\"Test 2\",(2.6,6),N); [/asy]$ $\\text{(A)}\\ 12\\% \\qquad \\text{(B)}\\ 25\\% \\qquad \\text{(C)}\\ 33\\frac{1}{3}\\% \\qquad \\text{(D)}\\ 40\\% \\qquad \\text{(E)}\\ 50\\%$ ## Problem 13 The perimeter of the polygon shown is $[asy] draw((0,0)--(0,6)--(8,6)--(8,3)--(2.7,3)--(2.7,0)--cycle); label(\"6\",(0,3),W); label(\"8\",(4,6),N); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((0.5,6)--(0.5,5.5)--(0,5.5)); draw((7.5,6)--(7.5,5.5)--(8,5.5)); draw((7.5,3)--(7.5,3.5)--(8,3.5)); draw((2.2,0)--(2.2,0.5)--(2.7,0.5)); draw((2.7,2.5)--(3.2,2.5)--(3.2,3)); [/asy]$ $\\text{(A)}\\ 14 \\qquad \\text{(B)}\\ 20 \\qquad \\text{(C)}\\ 28 \\qquad \\text{(D)}\\ 48$ $\\text{(E)}\\ \\text{cannot be determined from the information given}$ ## Problem 14 If $200\\leq a \\leq 400$ and $600\\leq b\\leq 1200$, then the largest value of the quotient $\\frac{b}{a}$ is $\\text{(A)}\\ \\frac{3}{2} \\qquad \\text{(B)}\\ 3 \\qquad \\text{(C)}\\ 6 \\qquad \\text{(D)}\\ 300 \\qquad \\text{(E)}\\ 600$ ## Problem 15 Sale prices at the Ajax Outlet Store are $50\\%$ below original prices. On Saturdays an additional discount of $20\\%$ off the sale price is given. What is the Saturday price of a coat whose original price is <dollar/>$180$? $\\text{(A)}$ <dollar/>$54$ $\\text{(B)}$ <dollar/>$72$ $\\text{(C)}$ <dollar/>$90$ $\\text{(D)}$ <dollar/>$108$ $\\text{(D)}$ <dollar/>$110$ ## Problem 16 A bar graph shows the number of hamburgers sold by a fast food chain each season. However, the bar", "draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); draw((1,1.732)--(2,0)); label(\"X\",(7,13),S); label(\"Y\",(14,0),S); label(\"Z\",(0,0),S); label(\"A\",(6,10.392),W); label(\"B\",(8,10.392),E); label(\"C\",(12,3.464),E); label(\"D\",(10,0),S); label(\"E\",(2,0),S); label(\"F\",(1,1.732),W); label(\"1\",(7,12.124)--(8,10.392),NE); label(\"1\",(6,10.392)--(8,10.392),S); label(\"4\",(8,10.392)--(12,3.464),NE); label(\"2\",(10,0)--(12,3.464),NW); label(\"2\",(14,0)--(12,3.464),NE); label(\"1\",(1,1.732)--(2,0),NE); label(\"1\",(0,0)--(2,0),S); label(\"1\",(0,0)--(1,1.732),NW); label(\"1\",(6,10.392)--(7,12.124),NW); label(\"2\",(10,0)--(14,0),S); [/asy]$ An equiangular hexagon can be made by drawing an equilateral triangle and cutting out smaller triangles from the corners. Labeling the triangle $X Y$ and $Z$ and drawing $AB$ of length one will remove one equilateral triangle of side length $1$, and drawing $CD$ will take out another equilateral triangle of side length $2$.Labeling the other sides of the smaller equilateral triangles, we can find that $XY$, or the side length of the equilateral triangle is $7$. Now, because we know what the side length of the triangle is, what $DY$ is, and it is given that $DE$ is $4$, we can find the length of $EZ$, $7-4-2=1$. Now, to calculate the area of the hexagon we can simply subtract the area of the smaller equilateral triangles from the larger equilateral triangle. The areas of the smaller equilateral triangles are $\\frac{1^2\\sqrt{3}}{4}\\implies\\frac{1\\sqrt{3}}{4}$, and $\\frac{2^2\\sqrt{3}}{4}\\implies\\frac{4\\sqrt{3}}{4}\\implies\\sqrt{3}$ and the area of the large equilateral triangle is $\\frac{7^2\\sqrt{3}}{4}\\implies\\frac{49\\sqrt{3}}{4}$ so the area of the hexagon would be $\\frac{49\\sqrt{3}}{4}-\\frac{\\sqrt {3}}{4}-\\frac{\\sqrt{3}}{4}-\\sqrt{3}\\implies\\boxed{\\frac{43\\sqrt{3}}{4}}$ Solution 2 $[asy] unitsize(0.5cm); draw((0,0)--(7,12.124)--(14,0)--cycle); draw((6,10.392)--(8,10.392)); draw((10,0)--(12,3.464)); draw((1,1.732)--(2,0)); label(\"X\",(7,13),S); label(\"Y\",(14,0),S); label(\"Z\",(0,0),S); label(\"A\",(6,10.392),W); label(\"B\",(8,10.392),E); label(\"C\",(12,3.464),E); label(\"D\",(10,0),S); label(\"E\",(2,0),S); label(\"F\",(1,1.732),W); label(\"1\",(7,12.124)--(8,10.392),NE); label(\"1\",(6,10.392)--(8,10.392),S); label(\"4\",(8,10.392)--(12,3.464),NE); label(\"2\",(10,0)--(12,3.464),NW); label(\"2\",(14,0)--(12,3.464),NE); label(\"1\",(1,1.732)--(2,0),NE); label(\"1\",(0,0)--(2,0),S); label(\"1\",(0,0)--(1,1.732),NW); label(\"1\",(6,10.392)--(7,12.124),NW); label(\"2\",(10,0)--(14,0),S); [/asy]$ An equiangular hexagon can", "- 40\\sqrt{10}$. Through power of a point, we can find out that $(CE)(DE)=20\\sqrt{10} - 20$, so $CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\\sqrt{10}) + 2(20\\sqrt{10} - 20) = \\boxed{\\textbf{(E) } 100}$. ~Math_Wiz_3.14 ## Solution 4 (Reflections) $[asy] draw(circle((0,0),7.07)); dot((-7.07,0)); label(\"A\",(-7.07,0),W); dot((7.07,0)); label(\"B\",(7.07,0),E); dot((0,0)); label(\"O\",(0,0),N); dot((-2.24,-6.71)); label(\"C\",(-2.24,-6.71),SSW); dot((6.71,2.24)); label(\"D\",(6.71,2.24),NE); draw((-2.24,-6.71)--(6.71,2.24)); dot((6.71,-2.24)); label(\"D'\",(6.71,-2.24),SE); draw((4.51,0)--(6.71,-2.24)); dot((4.51,0)); label(\"E\",(4.51,0),NNW); draw((-7.07,0)--(7.07,0)); draw((0,0)--(-2.24,-6.71)); draw((0,0)--(6.71,-2.24)); draw((-2.24,-6.71)--(6.71,-2.24)); [/asy]$ Let $O$ be the center of the circle. By reflecting $D$ across the line $AB$ to produce $D'$, we have that $\\angle BED'=45$. Since $\\angle AEC=45$, $\\angle CED'=90$. Since $DE=ED'$, by the Pythagorean Theorem, our desired solution is just $CD'^2$. Looking next to circle arcs, we know that $\\angle AEC=\\frac{\\overarc{AC}+\\overarc{BD}}{2}=45$, so $\\overarc{AC}+\\overarc{BD}=90$. Since $\\overarc{BD'}=\\overarc{BD}$, and $\\overarc{AC}+\\overarc{BD'}+\\overarc{CD'}=180$, $\\overarc{CD'}=90$. Thus, $\\angle COD'=90$. Since $OC=OD'=5\\sqrt{2}$, by the Pythagorean Theorem, the desired $CD'^2= \\boxed{\\textbf{(E) } 100}$. ~sofas103 ## See Also 2020 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's", "label(\"O\",(5,-1)); label(\"*\",(0,-1)); [/asy]$ ## circles $[asy] draw(Circle((0,0),3.5)); draw((-3.5,0)--(3.5,0)); label(\"7\", (0,0), dir(90)); dot((0,0)); draw(Circle((-2,1.4),1)); draw((-2,1.4)--(-1,1.4)); label(\"1\", (-1.5,1.4),dir(90)); [/asy]$ ## more circles $[asy] draw(Circle((0,0),20)); draw(Circle((0,0),14)); dot((0,0)); dot(20*dir(60)); dot(14*dir(180)); dot((-17,29.445)); draw(20*dir(60)--14*dir(180)--(-17,29.445)--cycle); label(\"O\",(0,0),dir(0)); label(\"A\",20*dir(60),dir(60)); label(\"B\",(-14,0),dir(180)); label(\"P\",(-17,29.445),dir(180)); draw((-17,29.445)--(0,0),red); [/asy]$ ## checkerboasrd $[asy] for(int i = 0; i < 8; ++i){ for(int j = 0; j < 8; ++j){ filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle,gray((i%3+3*(j%3))/8)); } } [/asy]$ ## Fermat point $[asy] import math; draw((0,0)--(4,0)--(0,3)--cycle); draw((0,0)--3*dir(30)--4*dir(-60)--cycle); draw((4,0)--4*dir(60)--(4*dir(60)+3*dir(30))--cycle); draw((0,3)--3*dir(150)--(3*dir(150)+4*dir(-60))--cycle); draw((0,0)--(4*dir(60)+3*dir(30)),blue+linetype(\"8 8\")); draw((0,3)--4*dir(-60),blue+linetype(\"8 8\")); draw((4,0)--3*dir(150),blue+linetype(\"8 8\")); draw((0,0)--(3*dir(150)+4*dir(-60)),red+linetype(\"8 8 0 8\")); draw((0,3)--4*dir(60),red+linetype(\"8 8 0 8\")); draw((4,0)--3*dir(30),red+linetype(\"8 8 0 8\"));[/asy]$ ## cenn driagrma $[asy] draw(Circle((1,0),2)); draw(Circle((-1,0),2)); label(\"3\",(0,0)); label(\"2\", (2,0)); label(\"2\", (-2,0)); label(\"Spotted\",(-2,2),dir(90)); label(\"5 Legs\",(2,2),dir(90)); [/asy]$ ## cyclic square $[asy] draw(Circle((0,0),0.5)); draw(0.5*dir(15)--0.5*dir(105)--0.5*dir(195)--0.5*dir(285)--cycle); label(scale(6)*\"CS\",(0,0)); [/asy]$ $[asy] import olympiad; size(10cm); draw(Circle((-5,0),4)); fill((0,0)--(-10,-1)--(-10,-5)--(0,-5)--cycle,white); dot(4*dir(60)-5); dot(4*dir(30)-5); dot(4*dir(100)-5); dot(4*dir(150)-5); label(scale(5)*\"Cyclic Squares\",(0,0)); draw((-0.75,2)--(-0.25,-2)--(9.25,-0.9)--(9,1.1)); draw(rightanglemark((-0.75,2),(-0.25,-2),(9.25,-0.9))); draw(rightanglemark((-0.25,-2),(9.25,-0.9),(9,1.1))); [/asy]$ ## diagram $$\\text{Given that }\\theta\\le 90^{\\circ}\\text{, prove }a^2+b^2\\le D^2\\text{, where }D\\text{ is the diameter of the circle.}$$ $[asy] draw(Circle((0,0),1)); draw(dir(0)--dir(40)--dir(170)--dir(260)--dir(0)--dir(170)--dir(260)--dir(40)); label(\"\\theta\", extension(dir(0),dir(170),dir(40),dir(260))-0.05*dir(30),-dir(30)); label(\"a\",(dir(170)+dir(260))/2,dir(215)); label(\"b\",(dir(0)+dir(40))/2,-dir(20)); [/asy]$ ## Cyclic squares DOTS DTOS TDORS $[asy] draw(Circle((0,0),90)); draw(Circle((30,40),10)); dot((37,38)); dot((25,39)); dot((20,30),gray(0.5)); dot((22,54),gray(0.6)); dot((36,27),gray(0.5)); dot((38,50),gray(0.4)); dot((10,36),gray(0.8)); dot((50,40),gray(0.75)); dot((30,20),gray(0.7)); dot((0,54),gray(0.85)); dot((4,23),gray(0.85)); dot((60,25),gray(0.9)); dot((30,70),gray(0.9)); [/asy]$", "be constant. It follows that the second differences of $(1,12,123,S)$ must be constant, as shown below: [asy] /* Made by MRENTHUSIASM */ size(10cm); label(\"\\boldmath{1}\",(0,0)); label(\"\\boldmath{12}\",(1,0)); label(\"\\boldmath{123}\",(2,0)); label(\"\\boldmath{S}\",(3,0)); label(\"11\",(0.5,-0.75)); label(\"111\",(1.5,-0.75)); label(\"d_1\",(2.5,-0.75)); label(\"100\",(1,-1.5)); label(\"d_2\",(2,-1.5)); for (real i=1; i<=3; ++i) { draw((0.1+(i-1),-0.15)--(0.4+(i-1),-0.6),red); } for (real i=1; i<=2; ++i) { draw((0.6+(i-1),-0.9)--(0.9+(i-1),-1.35),red); } for (real i=1; i<=3; ++i) { draw((0.6+(i-1),-0.6)--(0.9+(i-1),-0.15),red); } for (real i=1; i<=2; ++i) { draw((0.1+i,-1.35)--(0.4+i,-0.9),red); } label(\"\\textbf{First Differences}\",(-0.75,-0.75),align=W); label(\"\\textbf{Second Differences}\",(-0.75,-1.5),align=W); [/asy] Finally, we have $d_2=100,$ from which \\begin{align*} S&=123+d_1 \\\\ &=123+(111+d_2) \\\\ &=\\boxed{334}. \\end{align*} ~MRENTHUSIASM ## Solution 4 (Finite Differences by Algebra) Notice that we may rewrite the equations in the more compact form as: \\begin{align*} \\sum_{i=1}^{7}i^2x_i&=c_1, \\\\ \\sum_{i=1}^{7}(i+1)^2x_i&=c_2, \\\\ \\sum_{i=1}^{7}(i+2)^2x_i&=c_3, \\\\ \\sum_{i=1}^{7}(i+3)^2x_i&=c_4, \\end{align*} where $c_1=1, c_2=12, c_3=123,$ and $c_4$ is what we are trying to find. Now consider the polynomial given by $f(z) = \\sum_{i=1}^7 (i+z)^2x_i$ (we are only treating the $x_i$ as coefficients). Notice that $f$ is in fact a quadratic. We are given $f(0)=c_1, f(1)=c_2, f(2)=c_3$ and are asked to find $f(3)=c_4$. Using the concept of finite differences (a prototype of differentiation) we find that the second differences of consecutive values is constant, so that by arithmetic operations we find $c_4=\\boxed{334}$. Alternatively, applying finite differences, one obtains $$c_4 = {3 \\choose 2}f(2) - {3 \\choose 1}f(1) + {3 \\choose 0}f(0) =334.$$ ## Solution 5 (Assumption) The idea", "positions originally occupied by $a,c,e$ has become a 0. ### Case 2 $a,c>0$ and $e=0$. Define Operation B as the sequence of moves from Step 1 to Step 3, shown below: $[asy] size(300); defaultpen(fontsize(9)); label(\"d\",expi(0),(0,0)); label(\"c\",expi(pi/3),(0,0),red); label(\"b\",expi(2*pi/3),(0,0)); label(\"a\",expi(pi),(0,0),red); label(\"f\",expi(4*pi/3),(0,0)); label(\"0\",expi(5*pi/3),(0,0),red); label(\"Step 1\",(0,-2),(0,0)); label(\"c\",(5,0)+expi(0),(0,0)); label(\"c\",(5,0)+expi(pi/3),(0,0),red); label(\"a-c\",(5,0)+expi(2*pi/3),(0,0)); label(\"a\",(5,0)+expi(pi),(0,0),red); label(\"a\",(5,0)+expi(4*pi/3),(0,0)); label(\"0\",(5,0)+expi(5*pi/3),(0,0),red); label(\"Step 2\",(5,-2),(0,0)); label(\"c\",(10,0)+expi(0),(0,0)); label(\"\\begin{array}{l}2c-a\\\\ a-2c\\end{array}\",(10,0)+expi(pi/3),(0,0),red); label(\"a-c\",(10,0)+expi(2*pi/3),(0,0)); label(\"c\",(10,0)+expi(pi),(0,0),red); label(\"a\",(10,0)+expi(4*pi/3),(0,0)); label(\"0\",(10,0)+expi(5*pi/3),(0,0),red); label(\"Step 3\",(10,-2),(0,0)); [/asy]$ where in Step 3, we take the nonnegative choice of $2c-a$ or $a-2c$. $a,c,0$ is changed to either $c,2c-a,0$ or $c,a-2c,0$. If we have $c,2c-a,0$, their sum is $3c-a$ and this is less than $a+c+0$ (the original sum) unless $a=c$, but $a\\ne c$ since the original sum $a+c+0$ is odd by assumption. If we have $c,a-2c,0$, their sum is $a-c$, which is less than $a+c$. Operation B applied repeatedly will cause either $a$ or $c$ to become $0$. ### Case 3 $a>0$ and $c=e=0$. Define Operation C as the sequence of moves from Step 1 to Step 4, shown below: $[asy] size(400); defaultpen(fontsize(9)); label(\"d\",expi(0),(0,0)); label(\"0\",expi(pi/3),(0,0),red); label(\"b\",expi(2*pi/3),(0,0)); label(\"a\",expi(pi),(0,0),red); label(\"f\",expi(4*pi/3),(0,0)); label(\"0\",expi(5*pi/3),(0,0),red); label(\"Step 1\",(0,-2),(0,0)); label(\"0\",(5,0)+expi(0),(0,0)); label(\"0\",(5,0)+expi(pi/3),(0,0),red); label(\"a\",(5,0)+expi(2*pi/3),(0,0)); label(\"a\",(5,0)+expi(pi),(0,0),red); label(\"a\",(5,0)+expi(4*pi/3),(0,0)); label(\"0\",(5,0)+expi(5*pi/3),(0,0),red); label(\"Step 2\",(5,-2),(0,0)); label(\"0\",(10,0)+expi(0),(0,0)); label(\"0\",(10,0)+expi(pi/3),(0,0),red); label(\"a\",(10,0)+expi(2*pi/3),(0,0)); label(\"0\",(10,0)+expi(pi),(0,0),red); label(\"a\",(10,0)+expi(4*pi/3),(0,0)); label(\"0\",(10,0)+expi(5*pi/3),(0,0),red); label(\"Step 3\",(10,-2),(0,0)); label(\"0\",(15,0)+expi(0),(0,0)); label(\"0\",(15,0)+expi(pi/3),(0,0),red); label(\"0\",(15,0)+expi(2*pi/3),(0,0)); label(\"0\",(15,0)+expi(pi),(0,0),red); label(\"0\",(15,0)+expi(4*pi/3),(0,0)); label(\"0\",(15,0)+expi(5*pi/3),(0,0),red); label(\"Step 4\",(15,-2),(0,0)); [/asy]$", "is equilateral, and thus the desired length is $x = OC \\implies C$. ## Solution 3 $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.95,3),E); label(\"$$A$$\",(-3.6,2.5513),E); label(\"$$C$$\",(0.05,3.20),E); label(\"$$E$$\",(0.40,-3.60),E); label(\"$$B$$\",(-0.75,4.15),E); label(\"$$D$$\",(-2.62,1.5),E); label(\"$$F$$\",(-2.64,-1.43),E); label(\"$$G$$\",(-0.2,-2.8),E); label(\"$$\\sqrt{3}x$$\",(-1.5,-0.5),E); label(\"$$M$$\",(-2,-0.9),E); label(\"$$O$$\",(0.00,-0.10),E); label(\"$$1$$\",(-2.7,2.3),S); label(\"$$1$$\",(0.1,-3.4),S); label(\"$$8$$\",(-0.3,0),S); draw((0,-3.103)--(-2.687,1.5513)); draw((0.5,-3.9686)--(-0.5,3.9686));[/asy]$ Looking at the diagram above, we know that $BE$ is a diameter of circle $O$ due to symmetry. Due to Thales' theorem, triangle $ABE$ is a right triangle with $A = 90 ^\\circ$. $AE$ lies on $AD$ and $GE$ because $BAD$ is also a right angle. To find the length of $DG$, notice that if we draw a line from $F$ to $M$, the midpoint of line $DG$, it creates two $30$ - $60$ - $90$ triangles. Therefore, $MD = MG = \\frac{\\sqrt{3}x}{2} \\Rightarrow DG = \\sqrt{3}x$. $AE = 2 + \\sqrt{3}x$ Use the Pythagorean theorem on triangle $ABE$, we get $$(2+\\sqrt{3}x)^2 + x^2 = 8^2 \\Rightarrow 4 + 3x^2 + 4\\sqrt{3}x + x^2 = 64 \\Rightarrow x^2 + \\sqrt{3}x - 15 = 0$$ Using the quadratic formula to solve, we get $$x = \\frac{-\\sqrt{3} \\pm \\sqrt{3 -4(1)(-15)}}{2} = \\frac{\\pm 3\\sqrt{7} - \\sqrt{3}}{2}$$ $x$ must be positive, therefore $$x = \\frac{3\\sqrt{7} - \\sqrt{3}}{2} \\Rightarrow C$$ ~Zeric Hang ## Soultion 4 (coordinate bashing) $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.55,2.1),E); label(\"$$1$$\",(-0.5,3.8),S);[/asy]$", "be the intersection of the segments emanating from $A$ and $C$. By Law of Cosines, $$BP' = \\sqrt{1 + 1 - 2\\cos{120^\\circ}} = \\sqrt{3}.$$ So, $P'$ actually satisfies the conditions of the problem, and we can obtain again by Law of Cosines $$s = \\sqrt{4 + 1 - 4\\cos{120^\\circ}} = \\boxed{\\textbf{(B)} \\sqrt{7}}.$$ ~ hnkevin42 ## Solution 3 (Answer Choices) $[asy] unitsize(0.4inch); pen p = fontsize(10pt); draw((0,0)--(4,5.65)--(8,0)--cycle); label(\"A\", (4,5.65), N, p); label(\"C\", (8,0), SE, p); label(\"B\", (0,0), SW, p); label(\"P\", (3.5,3.5), E, p); label(\"E\", (2.8191,3.982), NW, p); label(\"F\", (4.848,4.452), NE, p); label(\"G\", (3.5,0), down, p); draw((0,0)--(3.5,3.5)); label(\"\\sqrt{3}\",(0,0)--(3.5,3.5), SE,p); draw((8,0)--(3.5,3.5)); label(\"2\",(8,0)--(3.5,3.5), SW,p); draw((4,5.65)--(3.5,3.5)); label(\"1\",(4,5.65)--(3.5,3.5), E,p); draw((3.5,3.5)--(2.8191,3.982)); draw((3.5,3.5)--(4.848,4.452)); draw((3.5,3.5)--(3.5,0)); [/asy]$ We begin by dropping altitudes from point $P$ down to all three sides of the triangle as shown above. We can therefore make equations regarding the areas of triangles $\\triangle{APC}$, $\\triangle{APB}$, and $\\triangle{BPC}$. Let $s$ be the side of the equilateral triangle, we use the Heron's formula: $$\\triangle{APC} = \\frac{s\\cdot PF}{2} = \\sqrt{\\frac{s+3}{2}\\left(\\frac{s+3}{2}-s\\right)\\left(\\frac{s+3}{2}-1\\right)\\left(\\frac{s+3}{2}-2\\right)}$$ $$\\implies PF = \\frac{\\sqrt{10s^2-s^4-9}}{2s}$$ Similarly, we obtain: $$PE = \\frac{\\sqrt{8s^2-s^4-4}}{2s}$$ $$PG = \\frac{\\sqrt{14s^2-s^4-1}}{2s}$$ By Viviani's theorem, $$\\frac{\\sqrt{10s^2-s^4-9}}{2s}+\\frac{\\sqrt{8s^2-s^4-4}}{2s}+\\frac{\\sqrt{14s^2-s^4-1}}{2s} = \\frac{\\sqrt{3}}{2}s$$ $$\\sqrt{10s^2-s^4-9}+\\sqrt{8s^2-s^4-4}+\\sqrt{14s^2-s^4-1} = \\sqrt{3}s^2$$ Note that from now on, the algebra will get extremely ugly and almost impossible to do by hand within the time frame. However, we do see that it's extremely easy to check the answer choices with", "\\pi - 3\\sqrt 2 \\qquad \\mathrm{(E) \\ } \\frac{8}{3} \\pi - 2\\sqrt 3$ $[asy] unitsize(1cm); defaultpen(0.8); pair O=(0,0), A=(0,1), B=(0,-1); path bigc1 = Circle(A,2), bigc2 = Circle(B,2), smallc = Circle(O,1); pair[] P = intersectionpoints(bigc1, bigc2); filldraw( arc(A,P[0],P[1])--arc(B,P[1],P[0])--cycle, lightgray, black ); draw(bigc1); draw(bigc2); unfill(smallc); draw(smallc); dot(O); dot(A); dot(B); label(\"A\",A,N); label(\"B\",B,S); draw( O--dir(30) ); draw( A--(A+2*dir(30)) ); draw( B--(B+2*dir(210)) ); label(\"1\", O--dir(30), N ); label(\"2\", A--(A+2*dir(30)), N ); label(\"2\", B--(B+2*dir(210)), S ); [/asy]$", "have $x-y\\geq1$, $y-z\\geq1$. The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too. $[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3); // three - currentprojection, orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), dashed+green); draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green); draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((0,0,0)--(1,0,0)--(1,1,1), red); draw((1,1,1)--(1,1,0)--(1,0.9,0), red); draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle); draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed); draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed); [/asy]$ The volume of this region is $V_2=\\dfrac{(n-2)^3}{6}$. So the probability is $p=\\dfrac{V_2}{V_1}=\\dfrac{(n-2)^3}{n^3}$. Substituting $n$ with the values in the choices, we find that when $n=10$, $p=\\frac{512}{1000}>\\frac{1}{2}$, when $n=9$, $p=\\frac{343}{729}<\\frac{1}{2}$. So $n\\geq 10$. So the answer is $\\boxed{\\textbf{(D)}\\ 10}$. ## Solution 2 Because $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$, which means that $x$, $y$, and $z$ distributes uniformly and independently in the interval $[0,n]$. So the point $(x, y, z)$ distributes uniformly in the cubic $0\\leqslant x, y, z \\leqslant n$, as shown in the figure below. The volume of this cubic is $V_0=n^3$. As we want to find the probablity of the incident $A=\\big\\{ |x-y| \\geq 1, |y-z| \\geq1, |z-x| \\geq", "draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); dot((72/5, 96/5)); label(\"A\",(72/5, 96/5),NE); dot((168/5, 24/5)); label(\"B\",(168/5, 24/5),NE); dot((24,0)); label(\"C\",(24,0),NW); dot((40, 0)); label(\"D\",(40, 0),NE); dot((24, 12)); label(\"E\",(24, 12),NE); dot((24, -12)); label(\"F\",(24, -12),SE); dot((54/5, 72/5)); label(\"G\",(54/5, 72/5),NW); dot((0, 0)); label(\"O_1\",(0, 0),S); dot((30, 0)); label(\"O_2\",(30, 0),S); [/asy]$ $r_1 = O_1A = 24$$r_2 = O_2B = 6$$AG = BO_2 = r_2 = 6$$O_1G = r_1 - r_2 = 24 - 6 = 18$$O_1O_2 = r_1 + r_2 = 30$ $\\triangle O_2BD \\sim \\triangle O_1GO_2$$\\frac{O_2D}{O_1O_2} = \\frac{BO_2}{GO_1}$$\\frac{O_2D}{30} = \\frac{6}{18}$$O_2D = 10$ $CD = O_2D + r_1 = 10 + 6 = 16$, $EF = 2EC = EA + EB = AB = GO_2 = \\sqrt{(O_1O_2)^2-(O_1G)^2} = \\sqrt{30^2-18^2} = 24$ $DEF = \\frac12 \\cdot EF \\cdot CD = \\frac12 \\cdot 24 \\cdot 16 = \\boxed{\\textbf{192}}$ ~isabelchen ## Solution 2 Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$. Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$, respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$. Since $\\triangle{APC} \\sim", "- 2x + 27 = 0$; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(m)(27) \\ge 0 \\Longleftrightarrow m \\le \\frac{1}{27}$. Thus, $$BP^2 \\le 405 + 729 \\cdot \\frac{1}{27} = \\boxed{432}$$ Equality holds when $x = 27$. ### Solution 2 $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$\\theta$$\",C + (-1.7,-0.2), NW); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ From the diagram, we see that $BQ = \\omega T + B\\omega\\sin\\theta = 9 + 9\\sin\\theta = 9(1 + \\sin\\theta)$, and that $QP = BA\\cos\\theta = 18\\cos\\theta$. \\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \\sin\\theta)^2 + 18^2\\cos^2\\theta\\\\ &= 9^2[1 + 2\\sin\\theta + \\sin^2\\theta + 4(1 - \\sin^2\\theta)]\\\\ BP^2 &= 9^2[5 + 2\\sin\\theta - 3\\sin^2\\theta]\\end{align*} This is a quadratic equation, maximized when $\\sin\\theta = \\frac { - 2}{ - 6} = \\frac {1}{3}$. Thus, $m^2 = 9^2[5 + \\frac {2}{3} - \\frac {1}{3}] = \\boxed{432}$. ### Solution 3 (Calculus Bash) $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) We let $AC=x$. From", "*/ pen qqwuqq = rgb(0,0,0); draw(arc((12,0),2,60,180)--(12,0)--cycle, qqwuqq); /* draw figures */ draw((0,0)--(12,0)); draw((0,0)--(18,10)); draw((18,10)--(12,0)); label(\"12\",(6,-1),SE*labelscalefactor); label(\" 2\\sqrt{76}-6 \",(14,5),SE*labelscalefactor); label(\"20\",(8,7),SE*labelscalefactor); draw((12,0)--(12,10)); draw((12,10)--(18,10)); /* dots and labels */ dot((0,0),dotstyle); label(\"A\", (0,0), NE * labelscalefactor); dot((12,0),dotstyle); label(\"B\", (12,0), NE * labelscalefactor); dot((18,10),dotstyle); label(\"C\", (18,10), NE * labelscalefactor); label(\"120^\\circ\", (11,1), NE * labelscalefactor,qqwuqq); dot((12,10),dotstyle); label(\"H\", (12,10), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$ Now, note that we have to find the coordinates of $E,F$ and $D$. Assume WLOG that $A$ is $(0,0)$ and $B$ be $(12,0)$. $E$ is obviously $(3,0)$, $F$ is going to be $(12+\\frac{9+\\sqrt{73}-12}{4}, \\frac{\\sqrt{219}-3\\sqrt3}{4})$ and the coordinates of $D$ is going to be $(\\frac{9+\\sqrt{73}}{4}, \\frac{\\sqrt{219}-3\\sqrt3}{4})$. Since $D$ and $F$ have the same $y$ value, we can find $[EDF]$ by using $\\frac12*b*h$. The base is going to be equal to the distance from $D$ to $F$, which is the same thing as $12+\\frac{9+\\sqrt{73}-12}{4}-\\frac{9+\\sqrt{73}}{4}=9=b$, and the height is the change in the $y$ coordinate from $E$ to $D$. This is the same thing as $\\frac{\\sqrt{219}-3\\sqrt3}{4}-0=\\frac{\\sqrt{219}-3\\sqrt3}{4}=h$. Hence, plugging these into $[EDF]=\\frac12*b*h$ give us $\\frac12*9*\\frac{\\sqrt{219}-3\\sqrt3}{4}=\\frac{9\\sqrt{219}-27\\sqrt3}{8}$. The answer is thus $9+219+27+3+8=\\boxed{266}$. ## Solution 2 Let $S$ be the area of the large triangle and let $x$ be the area we are trying to find. We have that $\\Delta ADE$ is similar to $\\Delta ABC$ with a ratio of $1:4$ while $\\Delta CDF$ is similar to $\\Delta", "\\frac{3\\sqrt{7}-\\sqrt{3}}{2}$. As previous solutions noted, $\\triangle BOC$ is equilateral, and thus the desired length is $x = OC \\implies (C)$. ## Solution 3 $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label(\"$$x$$\",(-1.95,3),E); label(\"$$A$$\",(-3.6,2.5513),E); label(\"$$C$$\",(0.05,3.20),E); label(\"$$E$$\",(0.40,-3.60),E); label(\"$$B$$\",(-0.75,4.15),E); label(\"$$D$$\",(-2.62,1.5),E); label(\"$$F$$\",(-2.64,-1.43),E); label(\"$$G$$\",(-0.2,-2.8),E); label(\"$$\\sqrt{3}x$$\",(-1.5,-0.5),E); label(\"$$M$$\",(-2,-0.9),E); label(\"$$O$$\",(0.00,-0.10),E); label(\"$$1$$\",(-2.7,2.3),S); label(\"$$1$$\",(0.1,-3.4),S); label(\"$$8$$\",(-0.3,0),S); draw((0,-3.103)--(-2.687,1.5513)); draw((0.5,-3.9686)--(-0.5,3.9686));[/asy]$ Looking at the diagram above, we know that $BE$ is a diameter of circle $O$ due to symmetry. Due to Thales' theorem, triangle $ABE$ is a right triangle with $A = 90 ^\\circ$. $AE$ lies on $AD$ and $GE$ because $BAD$ is also a right angle. To find the length of $DG$, notice that if we draw a line from $F$ to $M$, the midpoint of line $DG$, it creates two $30$ - $60$ - $90$ triangles. Therefore, $MD = MG = \\frac{\\sqrt{3}x}{2} \\Rightarrow DG = \\sqrt{3}x$. $AE = 2 + \\sqrt{3}x$ Applying the Pythagorean theorem to triangle $ABE$, we get $$(2+\\sqrt{3}x)^2 + x^2 = 8^2 \\Rightarrow 4 + 3x^2 + 4\\sqrt{3}x + x^2 = 64 \\Rightarrow x^2 + \\sqrt{3}x - 15 = 0$$ Using the quadratic formula to solve, we get $$x = \\frac{-\\sqrt{3} \\pm \\sqrt{3 -4(1)(-15)}}{2} = \\frac{\\pm 3\\sqrt{7} - \\sqrt{3}}{2}$$ $x$ must be positive, therefore $$x = \\frac{3\\sqrt{7} - \\sqrt{3}}{2} \\Rightarrow (C)$$ ~Zeric Hang ## Soultion 4 (coordinate bashing) $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat);" ]

[ "that $a+c=2+8=\\boxed{10}$. ### Explanation of the last step This is a property all rectangles in the coordinate plane have. For a proof, note that for any rectangle $ABCD$ its center can be computed as $(A+C)/2$ and at the same time as $(B+D)/2$. In our case, we can compute that the center is $\\left(\\frac{2+8}2,\\frac{5+3}2\\right)=(5,4)$, therefore $\\frac{a+c}2=5$, and $a+c=10$. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); draw ( (4,7) -- (6,1) ); draw ( (2,5) -- (8,3) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ ### An alternate last step We can easily compute $a$ and $c$ using our picture. $[asy] unitsize(0.5cm); pair X=(2,5), Y=(8,3); draw ( (-1,2) -- (4,7) -- (10,1) ); draw ( (-1,8) -- (6,1) -- (10,5) ); label(\"(2,5)\",X,W*1.5); label(\"(8,3)\",Y,E*1.5); label(\"(a,b)\",(4,7),N); label(\"(c,d)\",(6,1),S); [/asy]$ Consider the first graph on the interval $[2,8]$. The graph starts at height $5$, then rises for $a-2$ steps to the height $b=5+(a-2)$, and then falls for $8-a$ steps to the height $3=5+(a-2)-(8-a)$. Solving for $a$ we get $a=4$. Similarly we compute $c=6$, therefore $a+c=10$.", "\\frac{6\\sqrt{2}\\cdot 3\\sqrt{2} \\cdot 6}{3} \\right)=144$. Thus, our answer is $432-144=\\boxed{288}$. ### Solution 2 $[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(30,-20,10); real s = 6 * 2^.5; triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6),G=(s/2,-s/2,-6),H=(s/2,3*s/2,-6); draw(A--B--C--D--A--E--D); draw(B--F--C); draw(E--F); draw(A--G--B,dashed);draw(G--H,dashed);draw(C--H--D,dashed); label(\"A\",A,(-1,-1,0)); label(\"B\",B,( 2,-1,0)); label(\"C\",C,( 1, 1,0)); label(\"D\",D,(-1, 1,0)); label(\"E\",E,(0,0,1)); label(\"F\",F,(0,0,1)); label(\"G\",G,(0,0,-1)); label(\"H\",H,(0,0,-1)); [/asy]$ Extend $EA$ and $FB$ to meet at $G$, and $ED$ and $FC$ to meet at $H$. Now, we have a regular tetrahedron $EFGH$, which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s = 12\\sqrt{2}$. Using the formula for the volume of a regular tetrahedron, which is $V = \\frac{\\sqrt{2}S^3}{12}$, where S is the side length of the tetrahedron, the volume of our original solid is: $V = \\frac{1}{2} \\cdot \\frac{\\sqrt{2} \\cdot (12\\sqrt{2})^3}{12} = \\boxed{288}$. ### Solution 3 We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$; thus, we will integrate with respect to height from $0$ to $6$, noting that each cross section of height $dh$ is a rectangle. The volume is then $\\int_0^h(wl) \\ \\text{d}h$, where $w$ is the width of the rectangle and $l$ is the length. We can express $w$ in terms of $h$", "1996 AJHSME Problems/Problem 5 Problem 5 The letters $P$, $Q$, $R$, $S$, and $T$ represent numbers located on the number line as shown. $[asy] unitsize(36); draw((-4,0)--(4,0)); draw((-3.9,0.1)--(-4,0)--(-3.9,-0.1)); draw((3.9,0.1)--(4,0)--(3.9,-0.1)); for (int i = -3; i <= 3; ++i) { draw((i,-0.1)--(i,0)); } label(\"-3\",(-3,-0.1),S); label(\"-2\",(-2,-0.1),S); label(\"-1\",(-1,-0.1),S); label(\"0\",(0,-0.1),S); label(\"1\",(1,-0.1),S); label(\"2\",(2,-0.1),S); label(\"3\",(3,-0.1),S); draw((-3.7,0.1)--(-3.6,0)--(-3.5,0.1)); draw((-3.6,0)--(-3.6,0.25)); label(\"P\",(-3.6,0.25),N); draw((-1.3,0.1)--(-1.2,0)--(-1.1,0.1)); draw((-1.2,0)--(-1.2,0.25)); label(\"Q\",(-1.2,0.25),N); draw((0.1,0.1)--(0.2,0)--(0.3,0.1)); draw((0.2,0)--(0.2,0.25)); label(\"R\",(0.2,0.25),N); draw((0.8,0.1)--(0.9,0)--(1,0.1)); draw((0.9,0)--(0.9,0.25)); label(\"S\",(0.9,0.25),N); draw((1.4,0.1)--(1.5,0)--(1.6,0.1)); draw((1.5,0)--(1.5,0.25)); label(\"T\",(1.5,0.25),N); [/asy]$ Which of the following expressions represents a negative number? $\\text{(A)}\\ P-Q \\qquad \\text{(B)}\\ P\\cdot Q \\qquad \\text{(C)}\\ \\dfrac{S}{Q}\\cdot P \\qquad \\text{(D)}\\ \\dfrac{R}{P\\cdot Q} \\qquad \\text{(E)}\\ \\dfrac{S+T}{R}$ Solution 1 First, note that $P < Q < 0 < R < S < T$. Thus, $P$ and $Q$ are negative, while $R$, $S$, and $T$ are positive. Option $A$ is the difference of two numbers. If the first number is lower than the second number, the answer will be negative. In this case, $P < Q$, so $P - Q < 0$, and $P-Q$ is thus indeed a negative number. So option $\\boxed{A}$ is correct. Option $B$ is the product of two negatives, which is always positive. Option $C$ starts with the quotient of a positive and a negative. This is negative. But this negative quotient is then multiplied by a negative number, which gives a positive number. Option $D$ has the product of two negatives in", "\\frac {AD}{2} = 1004, \\quad ME = MC = \\frac {BC}{2} = 500$$ Thus $MN = NE - ME = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(N+H)/2+(0,1),S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(M+H)/2,W); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (504 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {DG}{CG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$. ### Solution 3 If you drop perpendiculars from B and C to AD, and call the points if you drop perpendiculars from B and C to AD and call the points where they meet AD E and F respectively and call FD = x and EA = $1008-x$ , then you", "Problem #2045 2045 Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? $[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label(\"Figure\",(0.5,-1),S); label(\"0\",(0.5,-2.5),S); label(\"Figure\",(9.5,-1),S); label(\"1\",(9.5,-2.5),S); label(\"Figure\",(19.5,-1),S); label(\"2\",(19.5,-2.5),S); label(\"Figure\",(32.5,-1),S); label(\"3\",(32.5,-2.5),S); [/asy]$ $\\textbf{(A)}\\ 10401 \\qquad\\textbf{(B)}\\ 19801 \\qquad\\textbf{(C)}\\ 20201 \\qquad\\textbf{(D)}\\ 39801 \\qquad\\textbf{(E)}\\ 40801$ This problem is copyrighted by the American Mathematics Competitions. Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.", "Solving for $t$, we find $t = 20$. Our answer, $AP$, is equivalent to $8t$. Thus, $AP = 8 \\cdot 20 = \\boxed{160}$. ## Solution 3 We can define $x$ to be the time elapsed since both Allie and Billie moved away from points $A$ and $B$ respectfully. Also, set the point of intersection to be $M$. Then we can produce the following diagram: $[asy] draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); label(\"100\",(0,0)--(100,0),S); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),S); dot((80,139)); label(\"M\",(80,139),N); [/asy]$ Now, if we drop an altitude from point$M$, we get : $[asy] size(300); draw((0,0)--(100,0)--(80,139)--cycle); label(\"8x\",(0,0)--(80,139),NW); label(\"7x\",(100,0)--(80,139),NE); dot((0,0)); label(\"A\",(0,0),S); dot((100,0)); label(\"B\",(100,0),SE); dot((80,139)); label(\"M\",(80,139),N); draw((80,139)--(80,0),dashed); label(\"4x\",(0,0)--(80,0),S); label(\"100-4x\",(80,0)--(100,0),S); label(\"4x \\sqrt{3}\",(80,139)--(80,0),W); label(\"60^{\\circ}\", (3,1), NE); [/asy]$ We know this from the $30-60-90$ triangle that is formed. From this we get that: $$(7x)^2 = (4 \\sqrt{3})^2 + (100-4x)^2$$ $$\\Longrightarrow 49x^2 - 48x^2 = x^2 = (100-4x)^2$$ $$\\Longrightarrow 0=(100-4x)^2 - x^2 = (100-3x)(100-5x)$$. Therefore, we get that $x = \\frac{100}{3}$ or $x = 20$. Since $20< \\frac{100}{3}$, we have that $x=20$ (since the problem asks for the quickest possible meeting point), so the distance Allie travels before meeting Billie would be $8 \\cdot x = 8 \\cdot 20 = \\boxed{160}$ meters. ~qwertysri987", "- 40\\sqrt{10}$. Through power of a point, we can find out that $(CE)(DE)=20\\sqrt{10} - 20$, so $CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\\sqrt{10}) + 2(20\\sqrt{10} - 20) = \\boxed{\\textbf{(E) } 100}$. ~Math_Wiz_3.14 ## Solution 4 (Reflections) $[asy] draw(circle((0,0),7.07)); dot((-7.07,0)); label(\"A\",(-7.07,0),W); dot((7.07,0)); label(\"B\",(7.07,0),E); dot((0,0)); label(\"O\",(0,0),N); dot((-2.24,-6.71)); label(\"C\",(-2.24,-6.71),SSW); dot((6.71,2.24)); label(\"D\",(6.71,2.24),NE); draw((-2.24,-6.71)--(6.71,2.24)); dot((6.71,-2.24)); label(\"D'\",(6.71,-2.24),SE); draw((4.51,0)--(6.71,-2.24)); dot((4.51,0)); label(\"E\",(4.51,0),NNW); draw((-7.07,0)--(7.07,0)); draw((0,0)--(-2.24,-6.71)); draw((0,0)--(6.71,-2.24)); draw((-2.24,-6.71)--(6.71,-2.24)); [/asy]$ Let $O$ be the center of the circle. By reflecting $D$ across the line $AB$ to produce $D'$, we have that $\\angle BED'=45$. Since $\\angle AEC=45$, $\\angle CED'=90$. Since $DE=ED'$, by the Pythagorean Theorem, our desired solution is just $CD'^2$. Looking next to circle arcs, we know that $\\angle AEC=\\frac{\\overarc{AC}+\\overarc{BD}}{2}=45$, so $\\overarc{AC}+\\overarc{BD}=90$. Since $\\overarc{BD'}=\\overarc{BD}$, and $\\overarc{AC}+\\overarc{BD'}+\\overarc{CD'}=180$, $\\overarc{CD'}=90$. Thus, $\\angle COD'=90$. Since $OC=OD'=5\\sqrt{2}$, by the Pythagorean Theorem, the desired $CD'^2= \\boxed{\\textbf{(E) } 100}$. ~sofas103 ## See Also 2020 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's", "a slope of $\\frac{4}{3}$. Since we know $m=\\frac{4}{3}$ , we can see that the line rises by $\\frac{8}{5}$ and moves to the right by $\\frac{6}{5}$ to meet the hypotenuse. (Let $2 = 5x$ and the rise be $4x$ and the run be $3x$ and then solve.) Therefore, line $\\l$ intersects the hypotenuse at the point $(x+\\frac{6}{5}, x+\\frac{8}{5})$. Plugging into the equation for the hypotenuse we have $x=\\frac{2}{7}$ , and after a bit of computation we get $\\boxed{\\textbf{(D) } \\frac{145}{147}}$ ## Solution 7(Slightly different from first solution): Same drawing as before: $[asy] draw((0,0)--(4,0)--(0,3)--(0,0)); draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0)); fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray); label(\"4\", (2,0), S); label(\"3\", (0,1.5), W); label(\"2\", (.8,1), E); label(\"S\", (0,0), NE); draw((0.3,0.3)--(1.4,1.9), dashed); draw((0.3,0.3)--(4,0), dashed); draw((0.3,0.3)--(0,3), dashed); label(\"\\small{a}\", (0.15,0.3), N); label(\"\\small{a}\", (0.3,0.15), E); [/asy]$ Let's assign $a$ as the side length of box S. We then get each of the smaller triangle areas. The sum of all the triangular areas(not including the box) is equal to $\\frac{(3-a) \\cdot a}{2} + \\frac{(4-a) \\cdot a}{2} + \\frac{5 \\cdot 2}{2} = \\frac{3 \\cdot 4}{2} - a^2$ You can solve for $a=\\frac{2}{7}$ Then, the ratio would be $1-\\dfrac{{\\frac{2}{7}}^2}{6}$ which is equal to $\\boxed{\\textbf{(D) } \\frac{145}{147}}$ ~Starshooter11 ## Solution 8(Similar but cleaner than first solution): Instead of dividing the large triangle into three right triangles plus a square, simply draw one diagonal of the square (can someone", "= \\frac{125}{251} \\Longrightarrow MN = \\frac{125}{126}EM.$$ Substituting, $1004 = NE = EM + MN = \\frac{251}{126}MN$, and $MN = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(A+F)/2,S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$1008-x$$\",(G+D)/2,S); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = AF$, so $FG = 1000$ and $DG = 2008 - x - 1000 = 1008 - x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {CG}{DG} \\Longleftrightarrow \\frac{h}{x} = \\frac{1008-x}{h} \\Longrightarrow h^2 + x^2 - 1008x = 0$$. Since $AH = 500+x$ and $AN = 1004$, it follows that $HN = AN - AH = 504 - x$. By the Pythagorean Theorem on $\\triangle MON$, $$MN^{2} = (504 - x)^2 + h^2 = 504^2 + h^2 + x^2 - 1008x = 504^2$$ and the answer is $MN = 504$.", "# 2010 AIME I Problems/Problem 11 ## Problem Let $\\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \\le 10$ and $3y - x \\ge 15$. When $\\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$, the volume of the resulting solid is $\\frac {m\\pi}{n\\sqrt {p}}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m + n + p$. ## Solution $[asy]size(280); import graph; real min = 2, max = 12; pen dark = linewidth(1); real P(real x) { return x/3 + 5; } real Q(real x) { return 10 - abs(x - 8); } path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12)); pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(0.9,0.9,0.9)); draw(graph(P,min,max),dark); draw(graph(Q,min,max),dark); draw(Arc((8,7.67),A,G,CW),dark,EndArrow(8)); draw(B--C--G--cycle,linetype(\"4 4\")); label(\"y \\ge x/3 + 5\",(max,P(max)),E,fontsize(10)); label(\"y \\le 10 - |x-8|\",(max,Q(max)),E,fontsize(10)); label(\"\\mathcal{R}\",(6,Q(6)),NW); /* axes */ Label f; f.p=fontsize(8); xaxis(0, max, Ticks(f, 6, 1)); yaxis(0, 10, Ticks(f, 5, 1)); [/asy]$ The inequalities are equivalent to $y \\ge x/3 + 5, y \\le 10 - |x - 8|$. We can set them equal to find the two points of intersection, $x/3 + 5 = 10 -", "- 2x + 27 = 0$; this is a quadratic, and its discriminant must be nonnegative: $(-2)^2 - 4(m)(27) \\ge 0 \\Longleftrightarrow m \\le \\frac{1}{27}$. Thus, $$BP^2 \\le 405 + 729 \\cdot \\frac{1}{27} = \\boxed{432}$$ Equality holds when $x = 27$. ### Solution 2 $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$\\theta$$\",C + (-1.7,-0.2), NW); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ From the diagram, we see that $BQ = \\omega T + B\\omega\\sin\\theta = 9 + 9\\sin\\theta = 9(1 + \\sin\\theta)$, and that $QP = BA\\cos\\theta = 18\\cos\\theta$. \\begin{align*}BP^2 &= BQ^2 + QP^2 = 9^2(1 + \\sin\\theta)^2 + 18^2\\cos^2\\theta\\\\ &= 9^2[1 + 2\\sin\\theta + \\sin^2\\theta + 4(1 - \\sin^2\\theta)]\\\\ BP^2 &= 9^2[5 + 2\\sin\\theta - 3\\sin^2\\theta]\\end{align*} This is a quadratic equation, maximized when $\\sin\\theta = \\frac { - 2}{ - 6} = \\frac {1}{3}$. Thus, $m^2 = 9^2[5 + \\frac {2}{3} - \\frac {1}{3}] = \\boxed{432}$. ### Solution 3 (Calculus Bash) $[asy] unitsize(3mm); pair B=(0,13.5), C=(23.383,0); pair O=(7.794, 9), P=(2*7.794,0); pair T=(7.794,0), Q=(0,0); pair A=(2*7.794,4.5); draw(Q--B--C--Q); draw(O--T); draw(A--P); draw(Circle(O,9)); dot(A);dot(B);dot(C);dot(T);dot(P);dot(O);dot(Q); label(\"$$B$$\",B,NW); label(\"$$A$$\",A,NE); label(\"$$\\omega$$\",O,N); label(\"$$P$$\",P,S); label(\"$$T$$\",T,S); label(\"$$Q$$\",Q,S); label(\"$$C$$\",C,E); label(\"$$9$$\", (B+O)/2, N); label(\"$$9$$\", (O+A)/2, N); label(\"$$9$$\", (O+T)/2,W); [/asy]$ (Diagram credit goes to Solution 2) We let $AC=x$. From", "= 2 and y = 5, then what is the value of \\frac{x^4+2y^2}{6} ? Level 1 Algebra We have $\\frac{x^4 + 2y^2}{6} = \\frac{2^4 + 2(5^2)}{6} = \\frac{16+2(25)}{6} = \\frac{16+50}{6} = \\frac{66}{6} = \\boxed{11}.$ The sequence of integers in the row of squares and in each of the two columns of squares form three distinct arithmetic sequences. What is the value of N? [asy] unitsize(0.35inch); draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(1,1)); draw((2,0)--(2,1)); draw((3,0)--(3,1)); draw((4,0)--(4,1)); draw((5,0)--(5,1)); draw((6,0)--(6,1)); draw((6,2)--(7,2)--(7,-4)--(6,-4)--cycle); draw((6,-1)--(7,-1)); draw((6,-2)--(7,-2)); draw((6,-3)--(7,-3)); draw((3,0)--(4,0)--(4,-3)--(3,-3)--cycle); draw((3,-1)--(4,-1)); draw((3,-2)--(4,-2)); label(\"21\",(0.5,0.8),S); label(\"14\",(3.5,-1.2),S); label(\"18\",(3.5,-2.2),S); label(\"N\",(6.5,1.8),S); label(\"-17\",(6.5,-3.2),S); [/asy] Level 2 Algebra Since 18 - 14 = 4, the common difference in the first column of squares is 4, so the number above 14 is 14 - 4 = 10, and the number above 10 is 10 - 4 = 6. This is also the fourth number in the row, so the common difference in the row is (6 - 21)/3 = -5. Then the seventh (and last) number in the row is 21 - 5 \\cdot 6 = -9. In the second column, the common difference is [(-17) - (-9)]/4 = -2, so N = -9 - (-2) = \\boxed{-7}. Tim wants to invest some money in a bank which compounds quarterly with an annual interest rate of 7\\%. To the nearest dollar, how much money should he invest if he", "2} = \\frac{3(4\\sqrt{3} - 2)}{44} = \\frac{6\\sqrt{3} - 3}{22}.$ $DC = \\frac{1}{2} - x = \\frac{1}{2} - \\frac{6\\sqrt{3} - 3}{22} = \\frac{14 - 6\\sqrt{3}}{22}$, and $AE = \\frac{7 - \\sqrt{27}}{22}$, so the answer is $\\boxed{056}$. $[asy] unitsize(8cm); pair a, o, d, r, e, m, cm, c,p; o =(0,0); d = (0.5, 0); r = (0,sqrt(3)/2); e = (-sqrt(3)/2,0); m = midpoint(d--r); draw(e--m); cm = foot(r, e, m); draw(L(r, cm,1, 1)); c = IP(L(r, cm, 1, 1), e--d); clip(r--d--e--cycle); draw(r--d--e--cycle); draw(rightanglemark(e, cm, c, 1.5)); a = -(4sqrt(3)+9)/11+0.5; dot(a); draw(a--r, dashed); draw(a--c, dashed); pair[] PPAP = {a, o, d, r, e, m, c}; for(int i = 0; i<7; ++i) { dot(PPAP[i]); } label(\"A\", a, W); label(\"E\", e, SW); label(\"C\", c, S); label(\"O\", o, S); label(\"D\", d, SE); label(\"M\", m, NE); label(\"R\", r, N); p = foot(a, r, c); label(\"P\", p, NE); draw(p--m, dashed); draw(a--p, dashed); dot(p); [/asy]$ ## Solution 2 Let $MP = x.$ Meanwhile, since $\\triangle R PM$ is similar to $\\triangle RCD$ (angle, side, and side- $RP$ and $RC$ ratio), $CD$ must be 2$x$. Now, notice that $AE$ is $x$, because of the parallel segments $\\overline A\\overline E$ and $\\overline P\\overline M$. Now we just have to calculate $ED$. Using the Law of Sines, or perhaps using altitude $\\overline R\\overline O$, we get $ED = \\frac{\\sqrt{3}+1}{2}$. $CA=RA$, which", "draw(circle((30,0),6)); draw((72/5, 96/5) -- (40,0)); draw((72/5, -96/5) -- (40,0)); draw((24, 12) -- (24, -12)); draw((0, 0) -- (40, 0)); draw((72/5, 96/5) -- (0,0)); draw((168/5, 24/5) -- (30,0)); draw((54/5, 72/5) -- (30,0)); dot((72/5, 96/5)); label(\"A\",(72/5, 96/5),NE); dot((168/5, 24/5)); label(\"B\",(168/5, 24/5),NE); dot((24,0)); label(\"C\",(24,0),NW); dot((40, 0)); label(\"D\",(40, 0),NE); dot((24, 12)); label(\"E\",(24, 12),NE); dot((24, -12)); label(\"F\",(24, -12),SE); dot((54/5, 72/5)); label(\"G\",(54/5, 72/5),NW); dot((0, 0)); label(\"O_1\",(0, 0),S); dot((30, 0)); label(\"O_2\",(30, 0),S); [/asy]$ $r_1 = O_1A = 24$$r_2 = O_2B = 6$$AG = BO_2 = r_2 = 6$$O_1G = r_1 - r_2 = 24 - 6 = 18$$O_1O_2 = r_1 + r_2 = 30$ $\\triangle O_2BD \\sim \\triangle O_1GO_2$$\\frac{O_2D}{O_1O_2} = \\frac{BO_2}{GO_1}$$\\frac{O_2D}{30} = \\frac{6}{18}$$O_2D = 10$ $CD = O_2D + r_1 = 10 + 6 = 16$, $EF = 2EC = EA + EB = AB = GO_2 = \\sqrt{(O_1O_2)^2-(O_1G)^2} = \\sqrt{30^2-18^2} = 24$ $DEF = \\frac12 \\cdot EF \\cdot CD = \\frac12 \\cdot 24 \\cdot 16 = \\boxed{\\textbf{192}}$ ~isabelchen ## Solution 2 Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$. Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$, respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$. Since $\\triangle{APC} \\sim", "three real solutions for $x$, so we are done and the answer is $\\boxed{\\textbf{(E) }a>\\frac12}$. ~ ccx09 ## Solution 8 Substituting $y = x^2 - a$ gives $x^2 + (x^2 - a)^2 = a^2$, which simplifies to $x^2 + x^4 - 2x^2a + a^2 = a^2$. This further simplifies to $x^2(1 + x^2 - 2a) = 0$. Thus, either $x^2 = 0$, or $x^2 - 2a + 1 = 0$. Since we care about $a$, we consider the second case. Then, we solve in terms of $a$ giving $a = \\frac{x^2}{2} + \\frac{1}{2}$. We see that in order to find the range in which $a$ lies, we must find the vertex of the that equation, which turns out to be $(0, \\frac{1}{2})$, so we know that the minimum is $\\frac{1}{2}$, which further implies that $\\boxed{\\textbf{(E) }a > \\frac{1}{2}}$ ## Solution 9 Now, let's graph these two equations. We want the blue parabola to be inside this red circle. $[asy] import graph; size(6cm); draw((0,0)--(0,10),EndArrow); draw((0,0)--(0,-10),EndArrow); draw((0,0)--(10,0),EndArrow); draw((0,0)--(-10,0),EndArrow); Label f; f.p=fontsize(6); xaxis(-10,10); yaxis(-10,10); real f(real x) { return x^2-5; } draw(graph(f,-4,4),blue+linewidth(1)); draw(circle((0,0),5),red); dot(scale(.7)*\"a\",(0,5),NE); dot(scale(.7)*\"-a\",(0,-5),N); dot(scale(.7)*\"a\",(5,0),NE); dot(scale(.7)*\"-a\",(-5,0),SE); [/asy]$ Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$. Expanding, we get $x^4-2ax^2+x^2=0$. Factoring out the $x$, we get $x^2(x^2-2a+1)=0$. Then we find that $x=0$ or $x=\\pm\\sqrt{2a-1}$. Therefore, $2a-1>0$, which means", "\\frac {AD}{2} = 1004, \\quad ME = MC = \\frac {BC}{2} = 500$$ Thus $MN = NE - ME = \\boxed{504}$. ### Solution 2 size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(N+H)/2,N); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(M+N)/2,W); (Error compiling LaTeX. 228aa9a09ec6a18b4a0ab11366b7b8f2ec44b852.asy: 27.6: no matching function 'label(string, pair[], pair)') Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (504 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {CG}{DG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow h^2 = (504 + x)(504 - x)\\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$.", "\\quad ME = MC = \\frac {BC}{2} = 500$$ Thus $MN = NE - ME = \\boxed{504}$. ### Solution 2 $[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints(N--E,B--C); pair F = foot(B,A,D), G=foot(C,A,D), H=foot(M[0],A,D); draw(A--B--C--D--cycle); draw(M[0]--N); draw(B--F,dashed); draw(C--G,dashed); draw(M[0]--H,dashed); label(\"$$A$$\",A,SW); label(\"$$B$$\",B,NW); label(\"$$C$$\",C,NE); label(\"$$D$$\",D,NE); label(\"$$F$$\",F,S); label(\"$$G$$\",G,SW); label(\"$$M$$\",M[0],SW); label(\"$$N$$\",N,S); label(\"$$H$$\",H,S); label(\"$$x$$\",(N+H)/2,S); label(\"$$h$$\",(B+F)/2,W); label(\"$$h$$\",(C+G)/2,W); label(\"$$1000$$\",(B+C)/2,NE); label(\"$$504-x$$\",(G+D)/2,S); label(\"$$504+x$$\",(A+F)/2,S); label(\"$$h$$\",(N+H)/2,W); [/asy]$ Let $F,G,H$ be the feet of the perpendiculars from $B,C,M$ onto $\\overline{AD}$, respectively. Let $x = NH$, so $DG = 1004 - 500 - x = 504 - x$ and $AF = 1004 - (504 - x) = 504 + x$. Also, let $h = BF = CG = HM$. By AA~, we have that $\\triangle AFB \\sim \\triangle CGD$, and so $$\\frac{BF}{AF} = \\frac {CG}{DG} \\Longleftrightarrow \\frac{h}{504+x} = \\frac{504-x}{h} \\Longrightarrow h^2 = (504 + x)(504 - x)\\Longrightarrow x^2 + h^2 = 504^2.$$ By the Pythagorean Theorem on $\\triangle MHN$, $$MN^{2} = x^2 + h^2 = 504^2,$$ so $MN = \\boxed{504}$.", "B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); pair[] ps={A,B,C,D,E,F,G}; dot(ps); label(\"A\",A,N); label(\"B\",B,N); label(\"C\",C,W); label(\"D\",D,S); label(\"E\",E,S); label(\"F\",F,E); label(\"G\",G,N); [/asy]$ Let $\\text{E}$ be the origin. Then, $\\text{D}=(1, 0)$ $\\text{A}=(0, 2)$ $\\text{B}=(1, 2)$ $\\text{F}=(2, 1)$ ${EB}$ can be represented by the line $y=2x$ Also, ${AF}$ can be represented by the line $y=-\\frac{1}{2}x+2$ Subtracting the second equation from the first gives us $\\frac{5}{2}x-2=0$. Thus, $x=\\frac{4}{5}$. Plugging this into the first equation gives us $y=\\frac{8}{5}$. Since $\\text{C} (0.8, 1.6)$, $G$ is $(0.8, 2)$, ${AB}=1$ and ${CG}=0.4$. Thus, $[ABC]=\\frac{1}{2} \\cdot {AB} \\cdot {CG}=\\frac{1}{2} \\cdot 1 \\cdot 0.4=0.2=\\frac{1}{5}$. The answer is $\\boxed{\\textbf{(B)}\\ \\frac15}$. ## Solution 3 $[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); pair H=(0,-1), I=(0.5,-1); draw(A--(2,0)); draw((0,-1)--F); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--F); draw(A--F); draw(B--E); draw(C--G); draw(H--I); pair[] ps={A,B,C,D,E,F,G, H, I}; dot(ps); label(\"A\",A,N); label(\"B\",B,N); label(\"C\",C,W); label(\"D\",D,S); label(\"E\",E,S); label(\"F\",F,E); label(\"G\",G,N); label(\"H\",H,W); label(\"I\",I,E); [/asy]$ Triangle $EAB$ is similar to triangle $EHI$; line $HI = 1/2$ Triangle $ACB$ is similar to triangle $FCI$ and the ratio of line $AB$ to line $IF = 1 : \\frac{3}{2} = 2: 3$. Based on similarity the length of the height of $GC$ is thus $\\frac{2}{5}\\cdot1 = \\frac{2}{5}$. Thus, $[ABC]=\\frac{1}{2} \\cdot {AB} \\cdot {CG}=\\frac{1}{2} \\cdot 1 \\cdot \\frac{2}{5}=\\frac{1}{5}$. The answer is $\\boxed{\\textbf{(B)}\\", "Note we can just use mass points to get \\left( \\frac{15}{26} \\right)^2= \\frac{225}{676}$which is \\boxed{901}$. ## Solution 2 First draw $\\overline{CP}$ and extend it so that it meets with $\\overline{AB}$ at point $X$. $[asy] size(10cm); pair A,B,C,D,E,X,P,Q,R; A=(0,0); B=(8,0); C=(1.9375,3.4994); D=(3.6696,2.4996); E=(1.4531,2.6246); X=(4.3636,0); P=(2.9639,2.0189); Q=(1.8462,0); R=(6.4615,0); dot(A); dot(B); dot(C); dot(D); dot(E); dot(X); dot(P); dot(Q); dot(R); label(\"A\",A,WSW); label(\"B\",B,ESE); label(\"C\",C,NNW); label(\"D\",D,NE); label(\"E\",E,WNW); label(\"X\",X,SSE); label(\"P\",P,NNE); label(\"Q\",Q,SSW); label(\"R\",R,SE); draw(A--B--C--cycle); draw(P--Q--R--cycle); draw(A--D); draw(B--E); draw(C--X); [/asy]$ We have that $[ABC]=\\frac{1}{2}\\cdot AC \\cdot BC\\sin{C}=\\frac{1}{2}\\cdot 4\\cdot {7}\\sin{C}=14\\sin{C}$ By Ceva's, $$3\\cdot{\\frac{2}{5}}\\cdot{\\frac{BX}{AX}}=1\\implies BX=\\frac{5\\cdot AX}{6}$$ That means that $$\\frac{11\\cdot {AX}}{6}=8\\implies AX=\\frac{48}{11} \\ \\text{and} \\ BX=\\frac{40}{11}$$ Now we apply mass points. Assume WLOG that $W_{A}=1$. That means that $$W_{C}=3, W_{B}=\\frac{6}{5}, W_{X}=\\frac{11}{5}, W_{D}=\\frac{21}{5}, W_{E}=4, W_{P}=\\frac{26}{5}$$ Notice now that $\\triangle{PBQ}$ is similar to $\\triangle{EBA}$. Therefore, $$\\frac{PQ}{EA}=\\frac{PB}{EB}\\implies \\frac{PQ}{3}=\\frac{10}{13}\\implies PQ=\\frac{30}{13}$$ Also, $\\triangle{PRA}$ is similar to $\\triangle{DBA}$. Therefore, $$\\frac{PA}{DA}=\\frac{PR}{DB}\\implies \\frac{21}{26}=\\frac{PR}{5}\\implies PR=\\frac{105}{26}$$ Because $\\triangle{PQR}$ is similar to $\\triangle{CAB}$, $\\angle{C}=\\angle{P}$. As a result, $[PQR]=\\frac{1}{2}\\cdot PQ \\cdot PR \\sin{C}=\\frac{1}{2}\\cdot \\frac{30}{13}\\cdot \\frac{105}{26}\\sin{P}=\\frac{1575}{338}\\sin{C}$. Therefore, $$\\frac{[PQR]}{[ABC]}=\\frac{\\frac{1575}{338}\\sin{C}}{14\\sin{C}}=\\frac{225}{676}\\implies 225+676=\\boxed{901}$$ • Not the author writing here, but a note is that Ceva's Theorem was actually not necessary to solve this problem. The information was just nice to know :)", "not outweigh the negative by the cos and tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first solution is a little bit after $\\pi$, which is around $3.14$. Hence the answer is $\\boxed{\\textbf{(D)}}$. (Not original author) Here is the graph: $[asy] Label f; f.p=fontsize(6); xaxis(-5,5,Ticks(f, 1.0)); yaxis(-8,8,Ticks(f, 2.0)); real f(real x) { return sin(x); } draw(graph(f, -5,5)); real g(real x) { return 2*cos(x); } draw(graph(g, -5,5)); real h(real x) { return 3*tan(x); } draw(graph(h, -1.2,1.2)); draw(graph(h, 1.94, 4.34)); draw(graph(h, -4.34, -1.94)); [/asy]$ ## Solution 3(More Detailed Answer of Solution 1) Denote $D_{f}$ as the domain of $f(x).$ Obviously $D_{f}=\\left \\{x|x\\neq \\frac{(2k+1)\\pi}{2} \\right \\} .$ $f^{'}(x)=cosx-2sinx+3sec^{2}x=\\sqrt{5}cos(x+\\phi)+3sec^{2}x > 0$ for all $x$ in sub-intervals of $D_{f}.$ When $x\\in (0,\\frac{\\pi}{2}), f(0)=2, \\lim_{x \\to \\frac{\\pi}{2}^{-} }f(x)=+\\infty .$ Hence $f(x)>0$ in this interval. When $x\\in (\\frac{\\pi}{2},\\pi),\\lim_{x \\to \\frac{\\pi}{2}^{+} }f(x)=-\\infty, f(\\pi)=-2.$ Hence $f(x)<0$ in this interval. When $x\\in (\\pi,\\frac{3\\pi}{2}),f(\\pi)=-2<0.$ Notice that $\\frac{5\\pi}{4}<\\frac{3.15*5}{4}<4<\\frac{3\\pi}{2}, f(\\frac{5\\pi}{4})=3-\\frac{3\\sqrt{2}}{2}>0 .$ Hence there must be a root located in the interval $(3,4).$ Choose $\\boxed{\\textbf{(D)}}$. ~PythZhou" ]

[ "+0 # Let $A,$ $B,$ $C$ be the angles of a triangle. 0 54 1 +467 Let a,b,c be the angles of a triangle. Evaluate$$\\begin{vmatrix} \\sin^2 A & \\cot A & 1 \\\\ \\sin^2 B & \\cot B & 1 \\\\ \\sin^2 C & \\cot C & 1 \\end{vmatrix}.$$ Nov 19, 2020 #1 0 Expanding along the third column, we find that most terms cancel, and we are left with a determinant of 1. Nov 19, 2020", "angles of a triangle, show that $$\\begin{vmatrix}-1 & \\cos C & \\cos B \\\\ \\cos C & -1 & \\cos A \\\\ \\cos B & \\cos A & 1\\end{vmatrix} = 0$$ 50. If $$A, B, C$$ are the angles of an isosceles triangle, evaluate $$\\begin{vmatrix}1 & 1& 1 \\\\ 1 + \\sin A & 1 + \\sin B & 1 + \\sin C \\\\ \\sin A + \\sin^2A & \\sin B + \\sin^2 B & \\sin C + \\sin^2 C\\end{vmatrix}$$", "# If $A,B,C$ are angles of a triangle and $\\begin{vmatrix} 1&1&1\\\\1+\\sin A&1+\\sin B&1+\\sin C\\\\\\sin A+\\sin ^2A&\\sin B+\\sin^2B& \\sin C+\\sin^2C\\end{vmatrix}=0$ $\\begin{array}{1 1}(A)\\;\\text{Isosceles}&(B)\\;\\text{Equilateral}\\\\(C)\\;\\text{Right angled triangle}&(D)\\;\\text{None of these}\\end{array}$", "# If $A,B$ and $C$ are the angles of a $\\Delta$le then $\\begin{vmatrix}\\sin 2A&\\sin C&\\sin B\\\\\\sin C&\\sin 2B&\\sin A\\\\\\sin B&\\sin A&\\sin 2C\\end{vmatrix}$ = $\\lambda\\sin A\\sin B\\sin C$, then $\\lambda$ is $(A)\\;1\\qquad(B)\\;0\\qquad(C)\\;2\\qquad(D)\\;3$", "Let $$a_{}^{}, b_{}^{}, c_{}^{}$$ be the three sides of a triangle, and let $$\\alpha_{}^{}, \\beta_{}^{}, \\gamma_{}^{}$$, be the angles opposite them. If $$a^2+b^2=1989^{}_{}c^2$$, find $$\\frac{\\cot \\gamma}{\\cot \\alpha+\\cot \\beta}$$ (第七届AIME1989 第10题)", "# If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\\Delta$ I'll state the question from my book below: If $$A$$, $$B$$ and $$C$$ are the angles of a triangle, then find the determinant value of $$\\Delta = \\begin{vmatrix}\\sin^2A & \\cot A & 1 \\\\ \\sin^2B & \\cot B & 1 \\\\ \\sin^2C & \\cot C & 1\\end{vmatrix}.$$ Here's how I tried solving the problem: $$\\Delta = \\begin{vmatrix}\\sin^2A & \\cot A & 1 \\\\ \\sin^2B & \\cot B & 1 \\\\ \\sin^2C & \\cot C & 1\\end{vmatrix}$$ $$R_2 \\to R_2 - R_1$$ $$R_3 \\to R_3 -R_1$$ $$= \\begin{vmatrix}\\sin^2A & \\cot A & 1 \\\\ \\sin^2B-\\sin^2A & \\cot B-\\cot A & 0 \\\\ \\sin^2C-\\sin^2A & \\cot C-\\cot A & 0\\end{vmatrix}$$ Expanding the determinant along $$C_3$$ \\begin{align} &= (\\sin^2B-\\sin^2A)(\\cot C-\\cot A)-(\\cot B-\\cot A)(\\sin^2C-\\sin^2A) \\\\ &= \\sin(B+A) \\sin(B-A) \\left[\\frac {\\cos C}{\\sin C} - \\frac {\\cos A}{\\sin A}\\right] - \\left[\\frac {\\cos B}{\\sin B} - \\frac {\\cos A}{\\sin A}\\right]\\sin(C+A) \\sin(C-A) \\\\ &= \\frac {\\sin(B+A) \\sin(B-A) \\sin(A-C)} {\\cos A \\cos C} - \\frac {\\sin(A-B) \\sin(C+A) \\sin(C-A)} {\\cos A \\cos C} \\\\ &= \\frac {\\sin(B-A) \\sin (A-C)} {\\cos A} \\left[\\frac {\\sin(A+B)} {\\cos C} - \\frac {\\sin(A+C)} {\\cos B}\\right] \\\\ &= \\frac {\\sin(B-A) \\sin (A-C)} {\\cos A} \\left[\\frac {\\sin C} {\\cos C} - \\frac {\\sin B} {\\cos B}\\right] \\\\", "# If $A,B,C$ are the angles of a $\\Delta$le then the value of the determinant $\\small\\begin{vmatrix}-1+\\cos B&\\cos C+\\cos B&\\cos B\\\\\\cos C+\\cos A&-1+\\cos A&\\cos A\\\\-1+\\cos B&-1+\\cos B&-1\\end{vmatrix}$ is $(a)\\;0\\qquad(b)\\;1\\qquad(c)\\;-1\\qquad(d)\\;2$", "is obtuse angled triangle. 45. Given, $\\begin{split}Area = \\begin{vmatrix} a & c & e & a \\\\ d & e & f & b \\end{vmatrix}\\end{split}$ Substituting the values and evaluating the determinant will yield the desired result. 46. $$a^t = \\log_t a . \\log_b t = \\log_b a$$ $$t = \\log_a(\\log_b a)$$", "angle triangle sin$$\\theta$$ = $$p\\over h$$; cos$$\\theta$$ = $$b\\over h$$; tan$$\\theta$$ = $$p\\over b$$; cosec$$\\theta$$ = $$h\\over p$$; sec$$\\theta$$ = $$h\\over b$$ and cot$$\\theta$$ = $$b\\over p$$ where ‘p’ is perpendicular ; ‘b’ is base and …", "angles is equal to the quotient of the subtraction of one from the product of cotangents of both angles by the sum of the cotangents of both angles. $(1) \\,\\,\\,\\,$ $\\cot{(A+B)}$ $\\,=\\,$ $\\dfrac{\\cot{B}\\cot{A}-1}{\\cot{B}+\\cot{A}}$ $(2) \\,\\,\\,\\,$ $\\cot{(x+y)}$ $\\,=\\,$ $\\dfrac{\\cot{y}\\cot{x}-1}{\\cot{y}+\\cot{x}}$ $(3) \\,\\,\\,\\,$ $\\cot{(\\alpha+\\beta)}$ $\\,=\\,$ $\\dfrac{\\cot{\\beta}\\cot{\\alpha}-1}{\\cot{\\beta}+\\cot{\\alpha}}$ Email subscription Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more", "common side $$CD$$ and the fact $$ADB$$ is a straight line, so $$\\angle ADC$$ and $$\\angle BDC$$ are supplementary. Hence $$CD$$, $$\\angle ADC$$ and $$\\angle BDC$$ must appear in our equations if those facts are to be brought to bear. Moreover, we want $$AD = m$$ and $$BD = n$$ to appear too, since they were indicated as of particular interest (the ratio $$\\triangle ABC$$ was divided in for the textbook; the lengths marked on the wall in the OP's question). Needing to represent two sides and an included angle in each of the smaller triangles rules out the law of sines; we could apply the law of cosines if we wanted $$AC$$ and $$BC$$ to appear in our equations also. Alternatively we can apply $$(4)$$ if we prefer two sides and two angles in each equation. At this stage we have a choice. Treating $$AD$$ as included by $$A$$ and $$\\pi - \\theta$$, and $$BD$$ as included by $$B$$ and $$\\theta$$, \\begin{align} m &= CD \\sin (\\pi - \\theta) \\left(\\cot (\\pi - \\theta) + \\cot A \\right) = CD \\sin \\theta \\left( -\\cot \\theta + \\cot A \\right) \\\\ n &= CD \\sin \\theta \\left( \\cot \\theta + \\cot B \\right) \\\\ \\implies \\frac{m}{n} &= \\frac{-\\cot \\theta + \\cot A }{\\cot \\theta + \\cot B} \\tag{7} \\end{align}", "interior angle measures of a triangle is $180^\\circ$180°, find the value of $\\theta$θ. ##### Question 5 Consider the triangle below. 1. Determine the value of $c$c. 2. Determine the value of $b$b. ##### Question 6 Consider the triangle below. 1. Determine the value of $a$a, giving your answer with a rational denominator. 2. Determine the value of $c$c, giving your answer with a rational denominator. 3. Determine the value of $c$c, giving your answer with a rational denominator.", "? Free Version Moderate # Many Angles Related to a Triangle SATMAT-UWNMFO In the figure above, which of the following must be equal to d? (No Calculator) A $a+b$ B $f+a$ C $f+c$ D $e+b$", "### Home > CCG > Chapter 7 > Lesson 7.2.5 > Problem7-98 7-98. $∠a$, $∠b$, and $∠c$ are exterior angles of the triangle below. Find $m∠a$, $m∠b$, and $m∠c$. Then find $m∠a+m∠b+m∠c$. How do the angles within a triangle relate to each other? How do the exterior angles relate to those within the triangle?", "Cot cot everywhere Geometry Level 3 $( \\cot B+ \\cot C)(\\cot C + \\cot A) (\\cot A + \\cot B)$ If $$A,B$$ and $$C$$ are angles of a triangle with non-zero area, which of these expressions in the answer choices is equivalent to the trigonometric expression above? × Problem Loading... Note Loading... Set Loading...", "# If $A,B,C$ are angles of a triangle and $\\begin{vmatrix} 1&1&1\\\\1+\\sin A&1+\\sin B&1+\\sin C\\\\\\sin A+\\sin ^2A&\\sin B+\\sin^2B& \\sin C+\\sin^2C\\end{vmatrix}=0$ $\\begin{array}{1 1}(A)\\;\\text{Isosceles}&(B)\\;\\text{Equilateral}\\\\(C)\\;\\text{Right angled triangle}&(D)\\;\\text{None of these}\\end{array}$ Applying $C_2-C_1$ and $C_3-C_1$ on the given determinant we get $\\begin{vmatrix} 1& 0& 0\\\\1+\\sin A&\\sin V-\\sin A&\\sin C-\\sin A\\\\\\sin A+\\sin ^2A&\\sin B-\\sin A+\\sin ^2B-\\sin ^2A&\\sin C-\\sin A+\\sin ^2C-\\sin ^2A\\end{vmatrix}=0$ $\\Rightarrow (\\sin B-\\sin A)(\\sin C-\\sin A)\\times \\begin{vmatrix} 1&1\\\\1+\\sin B+\\sin A&1+\\sin C+\\sin A\\end{vmatrix}=0$ $\\Rightarrow (\\sin B-\\sin A)(\\sin C-\\sin A)(\\sin C-\\sin B)=0$ $\\Rightarrow \\sin A=\\sin B$ or $(\\sin C=\\sin A)$ or $\\sin C=\\sin B$ $\\Rightarrow A=B$ Or $B=C$ (or) $C=A$ $\\Rightarrow$ triangle is isosceles. Hence (A) is the correct answer.", "sides of a triangle when one side and the three angles are known. (If two angles are known, the third is easily found since the sum of the angles is 180º.) See Solving Triangles Given ASA. It can also be used to find an angle when two sides and the angle opposite one side are known. ## Area of a triangleEdit The area of a triangle may be found in various ways. If all three sides are known, use Heron's theorem. If two sides and the included angle are known, consider the second diagram above. Let the sides b and c, and the angle between them α be known. From triangle ACO, the altitude h = CO is bsin(α) so the area is 12cbsin(α). If two angles and the included side are known, again consider the second diagram above. Let the side c and the angles α and γ be known. Let AO =x. Then $\\displaystyle \\frac{x}{h} = \\cot(\\alpha) \\text {; } \\frac{c-x}{h} = \\cot(\\gamma) \\text {; adding these, } \\frac{c}{h} = \\cot(\\alpha) + \\cot(\\gamma)$ Thus $\\displaystyle h = \\frac{c}{\\cot(\\alpha) + \\cot(\\gamma)} \\text { so area } = \\frac{c^2}{2(\\cot(\\alpha) + \\cot(\\gamma))}$.", "Find the value of $\\cot \\theta$ , if in a right triangle $ABC$ ,$AB=6\\;cm$ and $AC=8\\;cm$ right angled at $B$. $\\begin{array}{1 1} (a)\\;0 \\\\ (b)\\;10 \\\\ (c)\\;15 \\\\(d)\\;28 \\end{array}$", "\\wedge }{ k }$ respectively are the vector of a right angled, triangle. Also, find the remaining angles of the triangle. () points on the plane", "Back ## Triangles in a Triangle The measure of the exterior angle $c$ is always the sum of the measures of $a$ and $b:$ To see how we know this and learn a couple of other useful facts, keep reading. Or, you can skip ahead to the challenge if you feel ready." ]

[ "A\\cot A+\\sin A\\cos ecA}{\\left( \\cot A+\\cos ecA \\right)\\left( \\cot A-\\cos ecA \\right)} \\\\ \\end{align} Using the identity: $\\sin A\\cos ecA=1\\text{ and }(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$ we get, $L.H.S=\\dfrac{\\sin A\\cot A-1}{\\left( {{\\cot }^{2}}A-\\cos e{{c}^{2}}A \\right)}-\\dfrac{\\sin A\\cot A+1}{\\left( {{\\cot }^{2}}A-\\cos e{{c}^{2}}A \\right)}$ Applying the identity given by, $\\cos e{{c}^{2}}A-{{\\cot }^{2}}A=1$, we get, \\begin{align} & L.H.S=\\left( \\dfrac{\\sin A\\cot A-1}{-1} \\right)-\\left( \\dfrac{\\sin A\\cot A+1}{-1} \\right) \\\\ & =(1-\\sin A\\cot A)+(\\sin A\\cot A+1) \\\\ & =2 \\\\ & =R.H.S \\\\ \\end{align} Therefore, we have proved that, $\\dfrac{\\sin A}{\\left( \\cot A+\\cos ecA \\right)}-\\dfrac{\\sin A}{\\left( \\cot A-\\cos ecA \\right)}=2$. Hence, again shifting the trigonometric term to its original position we get, $\\dfrac{\\sin A}{\\left( \\cot A+\\cos ecA \\right)}=2+\\dfrac{\\sin A}{\\left( \\cot A-\\cos ecA \\right)}$. Note: It is important to note that this type of approach can be very useful in proving such types of questions which are difficult to prove from one side. An alternate method can be applied. We can start simplifying the left hand side term and we will stop when it cannot be further simplified. Now, we will start simplifying the right hand side term and simplify it till we get it equal to the simplified left hand side term.", "us to express in terms of the sides of the triangle: d= -a2+b2+c2 2c . For the height of the triangle we have that . By replacing with the formula given above and applying the difference of squares identity we get \\begin{align} h2&= 2-\\left( -a2+b2+c2 2c b \\right)2\\\\ &= (2bc-a2+b2+c2)(2bc+a2-b2-c2) 4c2 \\\\ &= ((b+c)2-a2)(a2-(b-c)2) 4c2 \\\\ &= (b+c-a)(b+c+a)(a+b-c)(a-b+c) 4c2 \\\\ &= 2(s-a) ⋅ 2s ⋅ 2(s-c) ⋅ 2(s-b) 4c2 \\\\ &= 4s(s-a)(s-b)(s-c) c2 . \\end{align} We now apply this result to the formula that calculates the area of a triangle from its height: \\begin{align} A&= ch 2 \\\\ &=\\sqrt{ c2 4 4s(s-a)(s-b)(s-c) c2 } \\\\ &= \\sqrt.\\end ### Trigonometric proof using the law of cotangents From the first part of the law of cotangents proof,[10] we have that the triangle's area is both \\begin{align} A&=r((s-a)+(s-b)+(s-c))= 2\\left( s-a r r + s-b r + s-c r \\right)\\\\ &= 2\\left(\\cot{ \\alpha 2 r } + \\cot + \\cot\\right)\\endand, but, since the sum of the half-angles is /2, the triple cotangent identity applies, so the first of these is \\begin{align} A&= 2\\left(\\cot{ \\alpha 2 r } \\cot \\cot\\right) = r^2\\left(\\frac \\cdot \\frac \\cdot \\frac\\right) \\\\ &= \\frac.\\end Combining the two, we get A2=s(s-a)(s-b)(s-c), from which the result follows. ## Numerical stability Heron's formula as given above is numerically unstable for triangles", "common side $$CD$$ and the fact $$ADB$$ is a straight line, so $$\\angle ADC$$ and $$\\angle BDC$$ are supplementary. Hence $$CD$$, $$\\angle ADC$$ and $$\\angle BDC$$ must appear in our equations if those facts are to be brought to bear. Moreover, we want $$AD = m$$ and $$BD = n$$ to appear too, since they were indicated as of particular interest (the ratio $$\\triangle ABC$$ was divided in for the textbook; the lengths marked on the wall in the OP's question). Needing to represent two sides and an included angle in each of the smaller triangles rules out the law of sines; we could apply the law of cosines if we wanted $$AC$$ and $$BC$$ to appear in our equations also. Alternatively we can apply $$(4)$$ if we prefer two sides and two angles in each equation. At this stage we have a choice. Treating $$AD$$ as included by $$A$$ and $$\\pi - \\theta$$, and $$BD$$ as included by $$B$$ and $$\\theta$$, \\begin{align} m &= CD \\sin (\\pi - \\theta) \\left(\\cot (\\pi - \\theta) + \\cot A \\right) = CD \\sin \\theta \\left( -\\cot \\theta + \\cot A \\right) \\\\ n &= CD \\sin \\theta \\left( \\cot \\theta + \\cot B \\right) \\\\ \\implies \\frac{m}{n} &= \\frac{-\\cot \\theta + \\cot A }{\\cot \\theta + \\cot B} \\tag{7} \\end{align}", "\\alpha/2 ) \\cot( \\beta/2 ) +1}{\\cot( \\alpha/2 ) \\cot ( \\beta/2 ) -1} \\\\[6pt] = {} & \\frac {\\cot ( \\alpha/2 ) +\\cot \\left( \\beta/2 \\right) +2\\,\\cot ( \\gamma/2 ) }{\\cot ( \\alpha/2 ) +\\cot ( \\beta/2 ) } = \\frac {4\\,s-a-b-2\\,c}{2\\,s-a-b}. \\end{align} Here, an extra step is required to transform a product into a sum, according to the sum/product formula. This gives the result: $\\dfrac {b+a}{c} = \\dfrac{\\cos ( \\alpha/2-\\beta/2 ) }{\\sin ( \\gamma/2 ) }$ as required.", "x_1 \\cot x_2 - (\\cot x_2)^2 -2)^2}{4(\\cot x_1 + \\cot x_2)^4}\\\\ =\\ &0. \\end{align} 2) $$n = 5$$. Let $$x_1, x_2 \\in (0, \\frac{\\pi}{2})$$ satisfying \\begin{align} -4(\\cot x_2)^2(\\cot x_1)^2 + (-2(\\cot x_2)^3 + 6\\cot x_2)\\cot x_1 + (\\cot x_2)^4 + 4(\\cot x_2)^2 - 1 = 0. \\end{align} Let $$y_1 = \\cot x_1, \\ y_2 = \\cot x_2$$. We have $$-4y_2^2y_1^2 + (-2y_2^3 + 6y_2) y_1 +y_2^4+4y_2^2-1 = 0$$ which results in $$\\sqrt{5}y_2^2-4y_1y_2-y_2^2+\\sqrt{5}+3 = 0$$ since $$x_1, x_2 \\in (0, \\frac{\\pi}{2}).$$ Let $$x_3 = x_2, \\ x_4 = x_1, \\ x_5 = \\pi - x_1 - x_2 - x_3 - x_4.$$ We have $$x_1, x_2, x_3, x_4, x_5 > 0; \\ x_1 + x_2+x_3+x_4+x_5=\\pi.$$ Note that $$\\big(\\frac{\\sin \\frac{\\pi}{5}}{\\sin \\frac{2\\pi}{5}}\\big)^5 = \\frac{5\\sqrt{5}-11}{2}$$. We have \\begin{align} &\\frac{\\sin x_1 \\sin x_2 \\sin x_3 \\sin x_4 \\sin x_5} {\\sin (x_1+x_2) \\sin (x_2 + x_3) \\sin (x_3 + x_4) \\sin (x_4+x_5) \\sin (x_5+x_1)} - \\Big(\\frac{\\sin \\frac{\\pi}{5}}{\\sin \\frac{2\\pi}{5}}\\Big)^5\\\\ =\\ &\\frac{(\\sin x_1)^2(\\sin x_2)^2\\sin (2x_1 + 2x_2)}{(\\sin (x_1+x_2))^2\\sin 2x_2 (\\sin (x_1 + 2x_2))^2} - \\frac{5\\sqrt{5}-11}{2}\\\\ =\\ &\\frac{(y_1y_2-1)(y_2^2+1)^2}{y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2} - \\frac{5\\sqrt{5}-11}{2}\\\\ =\\ &-\\frac{5\\sqrt{5}-11}{16}\\frac{ (\\sqrt{5}y_2^2+2y_1y_2+3y_2^2+\\sqrt{5}+1)(\\sqrt{5}y_2^2-4y_1y_2-y_2^2+\\sqrt{5}+3)^2} {y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2}\\\\ =\\ &0. \\end{align} • This is interesting +1. Since the general Jacobian matrix I computed was singular, maybe there are multiple points where the inequalities are attained. – dezdichado Jun 19 '19 at 19:45 • When $n = 4,$ it must be that $2x_1+x_2", "# User:Peter Mercator/Draft for spherical triginometry ### The cotangent four-part formulae The six elements of a triangle may be written in an anti-clockwise order as (aCbAcB\\$. The cotangent, or four-part, formulae relate two sides and two angles forming four \\emph{consecutive} elements around the triangle, for example (aCbA) or (BaCb). In such a set there are inner and outer elements: for example in the set (BaCb) the inner angle is C, the inner side is a, the outer angle is B, the outer side is b. The cotangent rule may be written as $\\begin{array}{rcl} \\cos(\\text{inner side}).\\cos(\\text{inner angle}) &{=}&\\;\\;\\cot(\\text{outer side}).\\sin(\\text{inner side})\\\\ &&-\\cot(\\text{outer angle}).\\sin(\\text{inner angle}) \\end{array}$ and the six possible equations are (with the relevant set shown at right): $\\begin{array}{lll} (CT1)\\quad& \\cos b\\,\\cos C=\\cot a\\,\\sin b - \\cot A \\,\\sin C ,\\qquad&(aCbA)\\\\[0ex] (CT2)& \\cos b\\,\\cos A=\\cot c\\,\\sin b - \\cot C \\,\\sin A,&(CbAc)\\\\[0ex] (CT3)& \\cos c\\,\\cos A=\\cot b\\,\\sin c - \\cot B \\,\\sin A,&(bAcB)\\\\[0ex] (CT4)& \\cos c\\,\\cos B=\\cot a\\,\\sin c - \\cot A \\,\\sin B,&(AcBa)\\\\[0ex] (CT5)& \\cos a\\,\\cos B=\\cot c\\,\\sin a - \\cot C \\,\\sin B,&(cBaC)\\\\[0ex] (CT6)& \\cos a\\,\\cos C=\\cot b\\,\\sin a - \\cot B \\,\\sin C,&(BaCb). \\end{array}$ ### Half-angle and half-side formulae With 2s=(a+b+c) and 2S=(A+B+C) $\\displaystyle \\begin{array}{lll} \\sin\\frac{A}{2}=\\left[\\frac{\\sin(s{-}b)\\sin(s{-}c)}{\\sin b\\sin c}\\right]^{1/2} &\\qquad &\\sin\\frac{a}{2}=\\left[\\frac{-\\cos S\\cos (S{-}A)}{\\sin B\\sin C}\\right]^{1/2}\\\\[2ex] \\cos\\frac{A}{2}=\\left[\\frac{\\sin s\\sin(s{-}a)}{\\sin b\\sin c}\\right]^{1/2} &\\qquad &\\cos\\frac{a}{2}=\\left[\\frac{\\cos (S{-}B)\\cos (S{-}C)}{\\sin B\\sin C}\\right]^{1/2}\\\\[2ex] \\tan\\frac{A}{2}=\\left[\\frac{\\sin(s{-}b)\\sin(s{-}c)}{\\sin s\\sin(s{-}a)}\\right]^{1/2} &\\qquad", "the inner side is a, the outer angle is B, the outer side is b. The cotangent rule may be written as (Todhunter,[1] Art.44) ${\\displaystyle \\cos({\\text{inner side}})\\cos({\\text{inner angle}})=\\cot({\\text{outer side}})\\sin({\\text{inner side}})\\ -\\ \\cot({\\text{outer angle}})\\sin({\\text{inner angle}}),}$ and the six possible equations are (with the relevant set shown at right): ${\\displaystyle {\\begin{array}{lll}{\\text{(CT1)}}\\quad &\\cos b\\,\\cos C=\\cot a\\,\\sin b-\\cot A\\,\\sin C,\\qquad &(aCbA)\\\\[0ex]{\\text{(CT2)}}&\\cos b\\,\\cos A=\\cot c\\,\\sin b-\\cot C\\,\\sin A,&(CbAc)\\\\[0ex]{\\text{(CT3)}}&\\cos c\\,\\cos A=\\cot b\\,\\sin c-\\cot B\\,\\sin A,&(bAcB)\\\\[0ex]{\\text{(CT4)}}&\\cos c\\,\\cos B=\\cot a\\,\\sin c-\\cot A\\,\\sin B,&(AcBa)\\\\[0ex]{\\text{(CT5)}}&\\cos a\\,\\cos B=\\cot c\\,\\sin a-\\cot C\\,\\sin B,&(cBaC)\\\\[0ex]{\\text{(CT6)}}&\\cos a\\,\\cos C=\\cot b\\,\\sin a-\\cot B\\,\\sin C,&(BaCb).\\end{array}}}$ To prove the first formula start from the first cosine rule and on the right-hand side substitute for ${\\displaystyle \\cos c}$ from the third cosine rule: {\\displaystyle {\\begin{aligned}\\cos a&=\\cos b\\cos c+\\sin b\\sin c\\cos A\\\\&=\\cos b\\ (\\cos a\\cos b+\\sin a\\sin b\\cos C)+\\sin b\\sin C\\sin a\\cot A\\\\\\cos a\\sin ^{2}b&=\\cos b\\sin a\\sin b\\cos C+\\sin b\\sin C\\sin a\\cot A.\\end{aligned}}} The result follows on dividing by ${\\displaystyle \\sin a\\sin b}$. Similar techniques with the other two cosine rules give CT3 and CT5. The other three equations follow by applying rules 1, 3 and 5 to the polar triangle. ### Half-angle and half-side formulae With ${\\displaystyle 2s=(a+b+c)}$ and ${\\displaystyle 2S=(A+B+C)}$, {\\displaystyle {\\begin{aligned}&\\sin {\\textstyle {\\frac {1}{2}}}A=\\left[{\\frac {\\sin(s{-}b)\\sin(s{-}c)}{\\sin b\\sin c}}\\right]^{1/2}&\\qquad &\\sin {\\textstyle {\\frac {1}{2}}}a=\\left[{\\frac {-\\cos S\\cos(S{-}A)}{\\sin B\\sin C}}\\right]^{1/2}\\\\[2ex]&\\cos {\\textstyle {\\frac {1}{2}}}A=\\left[{\\frac {\\sin s\\sin(s{-}a)}{\\sin b\\sin c}}\\right]^{1/2}&\\qquad &\\cos {\\textstyle {\\frac {1}{2}}}a=\\left[{\\frac", "# 20. Properties of Triangles’ Solutions Part 3¶ 1. The diagram is given below: Let $BD = DE = EC = x.$ Also let, $\\angle BAD = \\alpha, \\angle DAE = \\beta, EAC = \\gamma, CEA = \\theta$ Given, $\\tan\\alpha = t_1, \\tan\\beta = t_2, \\tan\\gamma = t_3$ Applying $m:n$ rule in $\\triangle ABC,$ we get $(2x + x)\\cot\\theta = 2x\\cot(\\alpha + \\beta) - x\\cot\\gamma$ From $\\triangle ADC,$ we get $2x\\cot\\theta = x\\cot\\beta - x\\cot\\gamma$ $\\Rightarrow \\frac{3}{2} = \\frac{2(\\alpha + \\beta) - \\cot\\gamma}{\\cot\\beta - \\cot\\gamma}$ $\\Rightarrow 3\\cot\\beta - 3\\cot\\gamma = 4\\cot(\\alpha + \\beta) - 2\\cot\\gamma$ $3\\cot\\beta - \\cot\\gamma = \\frac{4(\\cot\\alpha\\cot\\beta - 1)}{\\cot\\alpha + \\cot\\beta}$ $3\\cot^2\\beta - \\cot\\beta\\cot\\gamma + 3\\cot\\alpha\\cot\\beta - \\cot\\alpha\\cot\\gamma = 4\\cot\\alpha\\cot\\beta - 4$ $4 + 4\\cot^2\\beta = \\cot^2\\beta + \\cot\\alpha\\cot\\beta + \\cot\\beta\\cot\\gamma + \\cot\\alpha\\cot\\gamma$ $4(1 + \\cot^2\\beta) = (\\cot\\beta + \\cot\\alpha)(\\cot\\beta + \\cot\\gamma)$ $4\\left(1 + \\frac{1}{t_2^2}\\right) = \\left(\\frac{1}{t_1} + \\frac{1}{t_2}\\right)\\left(\\frac{1}{t_2} + \\frac{1}{t_3}\\right)$ 1. The diagram is given below: Let the medians be $AD, BE$ and $CF$ meet at $O.$ From question, $\\angle BOC=\\alpha, \\angle COA = \\beta, \\angle AOB = \\gamma$ Let $AD = p_1, BE=p_2, CF = p_3$ $AO:OD = 2:1 \\Rightarrow AO = \\frac{2}{3}p_1$ Similalrly, $OB = \\frac{2}{3}p_2, OC = \\frac{2}{3}p_3$ Applying cosine rule in $\\triangle AOC,$ $\\cos\\beta = \\frac{OA^2 + OC^2 - AC^2}{2.OA.OC} = \\frac{\\frac{4}{9}p_1^2 + \\frac{4}{9}p_3^2 - b^2}{2.\\frac{2}{3}p_1\\frac{2}{3}p_3}$ $\\cos\\beta = \\frac{4p_1^2 +", "A+\\cot A}{\\csc A+\\cot A}\\\\ &=&\\frac{\\frac{\\cos A-\\sin A +1}{\\sin A}}{\\frac{\\cos A+\\sin A -1}{\\sin A}}\\cdot\\frac{\\csc A+\\cot A}{\\csc A+\\cot A}\\\\ &=&\\frac{\\cot A - 1 + \\csc A}{\\cot A+1 -\\csc A}\\cdot\\frac{\\csc A+\\cot A}{\\csc A+\\cot A}\\\\ &=&\\frac{\\cot A - 1 + \\csc A}{(\\cot A+1 -\\csc A)(\\csc A+\\cot A)}\\cdot(\\csc A+\\cot A)\\\\ \\end{array}$$ Multiplying out the denominator we have $$\\cot A\\csc A+\\csc A\\color{blue}{-\\csc^2A+\\cot^2A}+\\cot A-\\csc A\\cot A$$ But rearranging the identity $$\\begin{array}{lll} \\cot^2A+1&=&\\csc^2A\\\\ \\color{blue}{\\cot^2A-\\csc^2A}&=&\\color{blue}{-1}\\\\ \\end{array}$$ our denominator becomes $$\\color{green}{\\csc A \\color{blue}{-1}+\\cot A}$$ continuing our proof $$\\begin{array}{lll} &=&\\frac{\\cot A - 1 + \\csc A}{(\\cot A+1 -\\csc A)(\\csc A+\\cot A)}\\cdot(\\csc A+\\cot A)\\\\ &=&\\frac{\\cot A - 1 + \\csc A}{\\color{green}{\\csc A -1 +\\cot A}}\\cdot(\\csc A+\\cot A)\\\\ &=&1\\cdot(\\csc A+\\cot A)\\\\ &=&\\csc A+\\cot A \\end{array}$$", "# 1989 AIME Problems/Problem 10 ## Problem Let $a$, $b$, $c$ be the three sides of a triangle, and let $\\alpha$, $\\beta$, $\\gamma$, be the angles opposite them. If $a^2+b^2=1989c^2$, find $\\frac{\\cot \\gamma}{\\cot \\alpha+\\cot \\beta}$ ## Solution ### Solution 1 We draw the altitude $h$ to $c$, to get two right triangles. $[asy] size(170); pair A = (0,0), B = (3, 0), C = (1, 4); pair P = .5*(C + reflect(A,B)*C); draw(A--B--C--cycle); draw(C--P, dotted); draw(rightanglemark(C,P, B, 4)); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"a\", (B+C)/2, NE); label(\"b\", (A+C)/2, NW); label(\"c\", (A+B)/2, S); label(\"h\", (C+P)/2, E);[/asy]$ Then $\\cot{\\alpha}+\\cot{\\beta}=\\frac{c}{h}$, from the definition of the cotangent. Let $K$ be the area of $\\triangle ABC.$ Then $h=\\frac{2K}{c}$, so $\\cot{\\alpha}+\\cot{\\beta}=\\frac{c^2}{2K}$. By identical logic, we can find similar expressions for the sums of the other two cotangents: \\begin{align*} \\cot \\alpha + \\cot \\beta &= \\frac{c^2}{2K} \\\\ \\cot \\beta + \\cot \\gamma &= \\frac{a^2}{2K} \\\\ \\cot \\gamma + \\cot \\alpha &= \\frac{b^2}{2K}. \\end{align*} Adding the last two equations, subtracting the first, and dividing by 2, we get $$\\cot \\gamma = \\frac{a^2 + b^2 - c^2}{4K}.$$ Therefore \\begin{align*} \\frac{\\cot \\gamma}{\\cot \\alpha + \\cot \\beta} &= \\frac{(a^2 + b^2 - c^2)/(4K)}{c^2/(2K)} \\\\ &= \\frac{a^2 + b^2 - c^2}{2c^2} \\\\ &= \\frac{1989 c^2 - c^2}{2c^2} \\\\ &= \\frac{1988}{2} = 994. \\end{align*} ### Solution 2", "\\frac{ch}{2}\\\\ & = \\sqrt{\\frac{c^2}{4}\\cdot \\frac{4s(s-a)(s-b)(s-c)}{c^2}}\\\\ & = \\sqrt{s(s-a)(s-b)(s-c)} \\end{align} ### Trigonometric proof using the Law of cotangents Geometrical significance of s-a, s-b, and s-c. See the Law of cotangents for the reasoning behind this. From the first part of the Law of cotangents proof,[6] we have that the triangle's area is both \\begin{align} A &= r\\big((s-a) + (s-b) + (s-c)\\big) = r^2\\left(\\frac{s-a}{r} + \\frac{s-b}{r} + \\frac{s-c}{r}\\right) \\\\ &= r^2\\left(\\cot{\\frac{\\alpha}{2}} + \\cot{\\frac{\\beta}{2}} + \\cot{\\frac{\\gamma}{2}}\\right) \\\\ \\end{align} and $A = rs$ but, since the sum of the half-angles is $\\tfrac{\\pi}{2}$, the triple cotangent identity applies, so the first of these is \\begin{align} A &= r^2\\left(\\cot{\\frac{\\alpha}{2}} \\cot{\\frac{\\beta}{2}} \\cot{\\frac{\\gamma}{2}}\\right) = r^2\\left( \\frac{s-a}{r}\\cdot \\frac{s-b}{r}\\cdot \\frac{s-c}{r}\\right) \\\\ &= \\frac{(s-a)(s-b)(s-c)}{r} \\\\ \\end{align} Combining the two, we get $A^2 = s(s-a)(s-b)(s-c)$ from which the result follows. ## Numerical stability Heron's formula as given above is numerically unstable for triangles with a very small angle. A stable alternative [7] [8] involves arranging the lengths of the sides so that $a \\ge b \\ge c$ and computing $A = \\frac{1}{4}\\sqrt{(a+(b+c)) (c-(a-b)) (c+(a-b)) (a+(b-c))}.$ The brackets in the above formula are required in order to prevent numerical instability in the evaluation. ## Other area formulas resembling Heron's formula Three other area formulas have the same structure as Heron's formula but are expressed in terms of different variables. First, denoting", "During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made. Difference between revisions of \"2000 AIME II Problems/Problem 15\" Problem Find the least positive integer $n$ such that $\\frac 1{\\sin 45^\\circ\\sin 46^\\circ}+\\frac 1{\\sin 47^\\circ\\sin 48^\\circ}+\\cdots+\\frac 1{\\sin 133^\\circ\\sin 134^\\circ}=\\frac 1{\\sin n^\\circ}.$ Solution We apply the identity \\begin{align*} \\frac{1}{\\sin n \\sin (n+1)} &= \\frac{1}{\\sin 1} \\cdot \\frac{\\sin (n+1) \\cos n - \\sin n \\cos (n+1)}{\\sin n \\sin (n+1)} \\\\ &= \\frac{1}{\\sin 1} \\cdot \\left(\\frac{\\cos n}{\\sin n} - \\frac{\\cos (n+1)}{\\sin (n+1)}\\right) \\\\ &= \\frac{1}{\\sin 1} \\cdot \\left(\\cot n - \\cot (n+1)\\right). \\end{align*} The motivation for this identity arises from the need to decompose those fractions, possibly into telescoping series. Thus our summation becomes $$\\sum_{k=23}^{67} \\frac{1}{\\sin (2k-1) \\sin 2k} = \\frac{1}{\\sin 1} \\left(\\cot 45 - \\cot 46 + \\cot 47 - \\cdots + \\cot 133 - \\cot 134 \\right).$$ Since $\\cot (180 - x) = - \\cot x$, the summation simply reduces to $\\frac{1}{\\sin 1} \\cdot \\left( \\cot 45 - \\cot 90 \\right) = \\frac{1 - 0}{\\sin 1} = \\frac{1}{\\sin 1^{\\circ}}$. Therefore, the answer is $\\boxed{001}$. Solution 2(Not Rigorous) We can make an approximation by observing the following points: Each of the terms is greater than 1, so the overall sum will be greater than 1. The average term is around the 60's which gives 4/3. There", "}{\\sin^2 u} \\mathop{du} &= 2 \\cdot \\int \\frac{\\cos(u) }{k^2} \\cdot \\frac{1}{\\cos u} \\mathop{dk} \\\\ &= 2 \\cdot \\int \\frac{ 1}{k^2} \\mathop{dk} \\\\ &= 2 \\cdot \\int \\frac{ 1}{k^2} \\mathop{dk} \\\\ &= - 2 k^{-1} + C_1 \\end{aligned} \\end{equation*} The other integrals are standard, giving \\begin{equation*} \\begin{aligned} \\int \\frac{1}{1 + \\sec(x/2)} \\mathop{dx} &= 2 \\cdot \\int \\frac{\\cos(u) }{\\sin^2 u} \\mathop{du} + 2 \\cdot \\int \\mathop{du} + 2 \\cdot \\int - \\csc^2 u \\mathop{du} \\\\ &= -2k^{-1} + 2 u - 2 \\cdot \\int \\csc^2 u \\mathop{du} \\\\ &= -2k^{-1} + 2 u - 2 \\cot(u) + C_2 \\end{aligned} \\end{equation*} Then subbing back in for $k,u$ \\begin{equation*} \\begin{aligned} -2k^{-1} + 2 u - 2 \\cot(u) + C_2 &= -(\\sin u)^{-1} + 2 u - 2 \\cot(u) + C_2 \\\\ &= -(\\sin (x/2))^{-1} + 2 (x/2) - 2 \\cot((x/2)) + C_2 \\end{aligned} \\end{equation*} ## 3 Answers Here is a simple (as I think) solution:- $$\\int \\dfrac{1}{1 + \\sec\\left( \\frac{x}{2} \\right)} dx = \\int \\dfrac{\\cos \\left( \\frac{x}{2} \\right)}{1 + \\cos \\left( \\frac{x}{2} \\right)} dx$$ $$= \\int \\dfrac{1 + \\cos \\left( \\frac{x}{4} \\right) - 1}{1 + \\cos \\left( \\frac{x}{2} \\right)} dx$$ $$= \\int dx - \\int \\dfrac{1}{2 \\cos^2\\left( \\frac{x}{4} \\right)} dx$$ $$= x - 2\\tan \\left( \\dfrac{x}{4} \\right) + c$$ Now, if you differentiate it properly, you will get the question back! Write", "Set $a=x+y$ and $b=z$. Then $$\\begin{eqnarray*} \\cot \\left( x+y+z\\right) &=&\\cot (a+b)=\\dfrac{\\cot a\\cot b-1}{\\cot b+\\cot a} \\\\ &=&\\dfrac{\\cot \\left( x+y\\right) \\cot z-1}{\\cot z+\\cot \\left( x+y\\right) }= \\dfrac{\\dfrac{\\cot x\\cot y-1}{\\cot x+\\cot y}\\cot z-1}{\\cot z+\\dfrac{\\cot x\\cot y-1}{\\cot x+\\cot y}} \\\\ &=&\\dfrac{\\dfrac{\\cot x\\cot y\\cot z-\\cot z-\\cot x-\\cot y}{\\cot x+\\cot y} }{\\dfrac{\\cot x\\cot z+\\cot y\\cot z+\\cot x\\cot y-1}{\\cot x+\\cot y}} \\\\ &=&\\dfrac{\\cot x\\cot y\\cot z-\\cot x-\\cot y-\\cot z}{\\cot x\\cot y+\\cot x\\cot z+\\cot y\\cot z-1}. \\end{eqnarray*}$$ -", "# Where did I mistake to integrate $I=\\int\\sqrt{\\frac{\\sin(x-\\alpha)}{\\sin(x+\\alpha)}} \\; dx\\; ?$ It was given to integrate $$I=\\int\\sqrt{\\frac{\\sin(x-\\alpha)}{\\sin(x+\\alpha)}} \\; dx.$$ # Attempt: \\begin{align}I&= \\int\\sqrt{\\frac{\\sin((x+\\alpha)-2\\alpha)}{\\sin(x+\\alpha)}}\\; dx\\\\&= \\sqrt{\\sin 2\\alpha}\\int\\sqrt{\\cot 2\\alpha - \\cot (x+\\alpha)}\\;dx\\; .\\end{align} Then I took \\begin{align}\\cot 2\\alpha - \\cot (x+\\alpha)= z^2 \\\\ \\implies [1 + \\cot^2(x+\\alpha)]\\;dx= 2z\\;dz\\\\ \\implies [1+\\{\\cot 2\\alpha - z^2\\}^2]\\; dx= 2z\\;dz\\; .\\end{align} Now, \\begin{align}\\int\\sqrt{\\cot 2\\alpha - \\cot (x+\\alpha)}\\; dx\\\\ &= \\int\\frac{z\\cdot 2z}{1+\\{\\cot 2\\alpha - z^2\\}^2}\\;dz\\\\ &= \\int\\frac{2}{\\dfrac{(1+ \\cot^2 2\\alpha)}{z^2} + z^2 -2\\cot2\\alpha}\\; dz\\\\ &= \\int \\frac{\\left(1+ \\dfrac{\\sqrt{(1+ \\cot^2 2\\alpha)}}{z^2}\\right)+ \\left(1- \\dfrac{\\sqrt{(1+ \\cot^2 2\\alpha)}}{z^2}\\right)}{\\dfrac{(1+ \\cot^2 2\\alpha)}{z^2} + z^2 -2\\cot2\\alpha}\\; dz \\; .\\end{align} This makes me to break the integral to get \\begin{align}\\int\\sqrt{\\cot 2\\alpha - \\cot (x+\\alpha)}\\; dx\\\\ &= \\int \\frac{\\left(1+ \\dfrac{\\sqrt{(1+ \\cot^2 2\\alpha)}}{z^2}\\right)+ \\left(1- \\dfrac{\\sqrt{(1+ \\cot^2 2\\alpha)}}{z^2}\\right)}{\\dfrac{(1+ \\cot^2 2\\alpha)}{z^2} + z^2 -2\\cot2\\alpha}\\; dz\\\\ \\\\ &= \\int\\frac{\\left(1+ \\dfrac{\\sqrt{(1+ \\cot^2 2\\alpha)}}{z^2}\\right)}{\\left(z-\\dfrac{\\sqrt{(1+ \\cot^2 2\\alpha)}}{z}\\right) +\\text{const.}}\\; dz +\\int\\frac{\\left(1- \\dfrac{\\sqrt{(1+ \\cot^2 2\\alpha)}}{z^2}\\right)}{\\left(z+\\dfrac{\\sqrt{(1+ \\cot^2 2\\alpha)}}{z}\\right) +\\text{const.}}\\; dz \\\\ &=-\\frac{1}{z- \\dfrac{(1+ \\cot^2 2\\alpha)}{z}}-\\frac{1}{z+ \\dfrac{(1+ \\cot^2 2\\alpha)}{z}} .\\end{align} So, $$I= -\\sqrt{\\sin 2\\alpha}\\left(\\frac{1}{z- \\dfrac{(1+ \\cot^2 2\\alpha)}{z}}+\\frac{1}{z+ \\dfrac{(1+ \\cot^2 2\\alpha)}{z}}\\right)\\; .$$ Unfortunately, my book gave the answer as $$I= \\cos\\alpha \\cos^{-1} (\\cos x\\sec\\alpha) - \\sin\\alpha\\log\\left(\\sin x+ \\sqrt{\\sin^2 x - \\sin^2\\alpha}\\right)\\; .$$ This is not near to my solution; I've made somewhere a big plunder. So, could anyone please tell me where I did mistake? Is my attempted technique wrong? • your mistake is in this step: \\begin{align}I&=", "y\\cot z.}$ Hermite's cotangent identity Charles Hermite demonstrated the following identity.[35] Suppose a1, ..., an are complex numbers, no two of which differ by an integer multiple of π. Let ${\\displaystyle A_{n,k}=\\prod _{\\begin{smallmatrix}1\\leq j\\leq n\\\\j\\neq k\\end{smallmatrix}}\\cot(a_{k}-a_{j})}$ (in particular, A1,1, being an empty product, is 1). Then ${\\displaystyle \\cot(z-a_{1})\\cdots \\cot(z-a_{n})=\\cos {\\frac {n\\pi }{2}}+\\sum _{k=1}^{n}A_{n,k}\\cot(z-a_{k}).}$ The simplest non-trivial example is the case n = 2: ${\\displaystyle \\cot(z-a_{1})\\cot(z-a_{2})=-1+\\cot(a_{1}-a_{2})\\cot(z-a_{1})+\\cot(a_{2}-a_{1})\\cot(z-a_{2}).}$ Ptolemy's theorem Ptolemy's theorem can be expressed in the language of modern trigonometry as: If w + x + y + z = π, then: {\\displaystyle {\\begin{aligned}\\sin(w+x)\\sin(x+y)&=\\sin(x+y)\\sin(y+z)&{\\text{(trivial)}}\\\\&=\\sin(y+z)\\sin(z+w)&{\\text{(trivial)}}\\\\&=\\sin(z+w)\\sin(w+x)&{\\text{(trivial)}}\\\\&=\\sin w\\sin y+\\sin x\\sin z.&{\\text{(significant)}}\\end{aligned}}} (The first three equalities are trivial rearrangements; the fourth is the substance of this identity.) Finite products of trigonometric functions For coprime integers n, m ${\\displaystyle \\prod _{k=1}^{n}\\left(2a+2\\cos \\left({\\frac {2\\pi km}{n}}+x\\right)\\right)=2\\left(T_{n}(a)+{(-1)}^{n+m}\\cos(nx)\\right)}$ where Tn is the Chebyshev polynomial. The following relationship holds for the sine function ${\\displaystyle \\prod _{k=1}^{n-1}\\sin \\left({\\frac {k\\pi }{n}}\\right)={\\frac {n}{2^{n-1}}}.}$ More generally [36] ${\\displaystyle \\sin(nx)=2^{n-1}\\prod _{k=0}^{n-1}\\sin \\left(x+{\\frac {k\\pi }{n}}\\right).}$ Linear combinations For some purposes it is important to know that any linear combination of sine waves of the same period or frequency but different phase shifts is also a sine wave with the same period or frequency, but a different phase shift. This is useful in sinusoid data fitting, because the measured or observed data are linearly related to the", "x_2) &= 0, \\qquad (1)\\\\ \\cot x_2 + \\cot (x_1 + x_2 + x_3) - 2\\cot (x_1 + x_2) - 2\\cot (x_2 + x_3) &= 0,\\\\ \\cot x_3 + \\cot (x_1 + x_2 + x_3) - 2\\cot (x_2 + x_3) &= 0. \\end{align} By letting $$u_1 =\\cot x_1, \\ u_2 = \\cot x_2, \\ u_3 = \\cot x_3$$, we have $$(1) \\Longrightarrow u_1u_2 + u_2u_3 + u_3u_1 - u_2^2 - 2 = 0$$. On the other hand, $$g(x_1, x_2, x_3) - \\frac{1}{4} = \\frac{(u_1u_2+u_2u_3+u_3u_1-u_2^2 - 2)^2}{4(u_1+u_2)^2(u_2+u_3)^2} = 0.$$ 2) We can see this from another view. Case $$n=4$$: Let $$u_i = \\cot x_i, \\ i=1,2,3$$. We have (noting that $$x_4 = \\pi - x_1-x_2-x_3$$) \\begin{align} &\\frac{\\sin x_1 \\sin x_2 \\sin x_3 \\sin x_4}{\\sin(x_1+x_2)\\sin (x_2+x_3)\\sin (x_3+x_4)\\sin (x_4+x_1)}\\nonumber\\\\ =\\ & \\frac{\\sin x_1\\sin x_2 \\sin x_3 \\sin (x_1+x_2+x_3)}{(\\sin (x_1+x_2))^2 (\\sin (x_2+x_3))^2}\\nonumber\\\\ =\\ & \\frac{(u_1u_2 + u_2u_3+u_3u_1-1)(1+u_2^2)}{(u_1+u_2)^2(u_2+u_3)^2}\\nonumber\\\\ =\\ & \\frac{1}{4} - \\frac{(u_1u_2+u_2u_3+u_3u_1-u_2^2 - 2)^2}{4(u_1+u_2)^2(u_2+u_3)^2}. \\end{align} Case $$n=5$$: Let $$u_i = \\cot x_i, \\ i=1,2,3, 4$$. We have (noting that $$x_5 = \\pi - x_1-x_2-x_3-x_4$$) \\begin{align} &\\frac{\\sin x_1 \\sin x_2 \\sin x_3 \\sin x_4 \\sin x_5}{\\sin(x_1+x_2)\\sin (x_2+x_3)\\sin (x_3+x_4)\\sin (x_4+x_5)\\sin (x_5+x_1)}\\nonumber\\\\ =\\ & \\frac{\\sin x_1 \\sin x_2 \\sin x_3 \\sin x_4 \\sin (x_1+x_2+x_3+x_4)}{\\sin(x_1+x_2)\\sin (x_2+x_3)\\sin (x_3+x_4)\\sin (x_1+x_2+x_3)\\sin (x_2+x_3+x_4)}\\nonumber\\\\ =\\ & \\frac{(u_1u_2u_3 + u_1u_2u_4 + u_1u_3u_4 + u_2u_3u_4 - u_1-u_2-u_3-u_4)(1+u_2^2)(1+u_3^2)} {(u_1+u_2)(u_2+u_3)(u_3+u_4)(u_1u_2+u_2u_3+u_3u_1-1)(u_2u_3+u_3u_4+u_4u_2-1)}.\\quad (2) \\end{align} Denote (2) as", "\\frac {s-b}r \\frac {s-c}r = \\frac {s-a}r + \\frac {s-b}r +\\frac {s-c}r$ But the right member evals to $s/r$, giving the value of $r^2$ and proving the second assertion. ## Some proofs using the law of cotangents A number of other results can be derived from the law of cotangents. • Heron's formula. Note that the area of triangle $ABC$ is also divided into 6 smaller triangles, also in 3 pairs, with the triangles in each pair having the same area. For example, the two triangles near vertex A, being right triangles of width s−a and height r, each have an area of $\\tfrac{1}{2}r(s-a)$. So those two triangles together have an area of $r(s-a)\\,$, and the area $S$ of the whole triangle is therefore \\begin{align} S &= r(s-a) + r(s-b) + r(s-c) = r(s-a + s-b + s-c) \\\\[8pt] &= r(3s - (a+b+c)) = r(3s - 2s) = rs \\\\[8pt] \\end{align} This gives the result: $S=\\sqrt{s(s-a)(s-b)(s-c)}\\,$ as required. $\\frac {\\sin ( \\alpha/2-\\beta/2 ) }{\\sin ( \\alpha/2+\\beta/2 ) } = \\frac {\\cot ( \\beta/2 ) -\\cot ( \\alpha/2 ) }{\\cot ( \\beta/2 ) +\\cot ( \\alpha/2 ) }= \\frac {a-b}{2s-a-b}.$ This gives the result: $\\dfrac {a-b}{c}=\\dfrac {\\sin ( \\alpha/2-\\beta/2 )}{\\cos ( \\gamma/2 ) }$ as required. \\begin{align} & \\frac {\\cos( \\alpha/2-\\beta/2 ) }{\\cos( \\alpha/2+\\beta/2 ) } = \\frac {\\cot(", "# Difference between revisions of \"2000 AIME II Problems/Problem 15\" ## Problem Find the least positive integer $n$ such that $\\frac 1{\\sin 45^\\circ\\sin 46^\\circ}+\\frac 1{\\sin 47^\\circ\\sin 48^\\circ}+\\cdots+\\frac 1{\\sin 133^\\circ\\sin 134^\\circ}=\\frac 1{\\sin n^\\circ}.$ ## Solution We apply the identity \\begin{align*} \\frac{1}{\\sin n \\sin (n+1)} &= \\frac{1}{\\sin 1} \\cdot \\frac{\\sin (n+1) \\cos n - \\sin n \\cos (n+1)}{\\sin n \\sin (n+1)} \\\\ &= \\frac{1}{\\sin 1} \\cdot \\left(\\frac{\\cos n}{\\sin n} - \\frac{\\cos (n+1)}{\\sin (n+1)}\\right) \\\\ &= \\frac{1}{\\sin 1} \\cdot \\left(\\cot n - \\cot (n+1)\\right). \\end{align*} The motivation for this identity arises from the need to decompose those fractions, possibly into telescoping series. Thus our summation becomes $$\\sum_{k=23}^{67} \\frac{1}{\\sin (2k-1) \\sin 2k} = \\frac{1}{\\sin 1} \\left(\\cot 45 - \\cot 46 + \\cot 47 - \\cdots + \\cot 133 - \\cot 134 \\right).$$ Since $\\cot (180 - x) = - \\cot x$, the summation simply reduces to $\\frac{1}{\\sin 1} \\cdot \\left( \\cot 45 - \\cot 90 \\right) = \\frac{1 - 0}{\\sin 1} = \\frac{1}{\\sin 1^{\\circ}}$. Therefore, the answer is $\\boxed{001}$.", "\\cfrac{AX}{\\sin(\\alpha+\\beta)}$ The above gives $z$ in terms of $\\,h,\\alpha,\\beta\\,$: $$\\,z = \\cfrac{h\\,\\sin(\\beta)}{\\sin(\\alpha)\\,\\sin(\\alpha+\\beta)}\\,$$ $\\,z'=AB\\,$ can be calculated in symmetric fashion (replacing $\\alpha \\mapsto \\pi - \\alpha$): $$\\;z' = \\cfrac{h\\,\\sin(\\beta)}{\\sin(\\alpha)\\,\\sin(\\alpha-\\beta)}\\,$$ Substituting the above in the equality $\\,z+z'=w\\,$ gives an equation that can be solved for $\\,\\beta\\,$: \\require{cancel} \\begin{align} \\frac{h}{\\sin(\\alpha)}\\left(\\frac{\\sin(\\beta)}{\\sin{(\\alpha+\\beta)}}+\\frac{\\sin(\\beta)}{\\sin{(\\alpha-\\beta)}}\\right) & = w \\\\ \\frac{1}{\\sin(\\alpha) \\cot(\\beta) + \\cos(\\alpha)} + \\frac{1}{\\sin(\\alpha) \\cot(\\beta) - \\cos(\\alpha)} & = \\frac{w \\sin(\\alpha)}{h} \\\\[6pt] \\frac{2 \\cancel{\\sin(\\alpha)} \\cot(\\beta)}{\\sin^2(\\alpha) \\cot^2(\\beta) - \\cos^2(\\alpha)}& = \\frac{w \\cancel{\\sin(\\alpha)}}{h} \\\\[5pt] w \\sin^2(\\alpha) \\cdot \\cot^2(\\beta) - 2 h \\cdot \\cot(\\beta) - w \\cos^2(\\alpha) &= 0 \\end{align} The latter is a quadratic in $\\,\\cot(\\beta)\\,$. Its determinant $\\frac{1}{4} \\Delta = h^2 + w^2 \\sin^2(\\alpha) \\cos^2(\\alpha) \\gt 0$ is positive, so the quadratic has two real roots, and their product is $-\\cot^2(\\alpha) \\lt 0$ so the roots have opposite signs. The positive root corresponds to a $\\beta \\in (0, \\pi / 2)$ which is then the unique solution. • But you don't know β nor z, so you have 2 unknowns in one equation. Both β and z change in function of w. – Enedin Jan 15 '17 at 9:44 • Thanks for the info @dxiv, but I don't see how I can find out the value of β with what you've posted. – Enedin Jan 15 '17 at 16:17 • @Enedin My mistake, I misread that" ]

[ "# If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\\Delta$ I'll state the question from my book below: If $$A$$, $$B$$ and $$C$$ are the angles of a triangle, then find the determinant value of $$\\Delta = \\begin{vmatrix}\\sin^2A & \\cot A & 1 \\\\ \\sin^2B & \\cot B & 1 \\\\ \\sin^2C & \\cot C & 1\\end{vmatrix}.$$ Here's how I tried solving the problem: $$\\Delta = \\begin{vmatrix}\\sin^2A & \\cot A & 1 \\\\ \\sin^2B & \\cot B & 1 \\\\ \\sin^2C & \\cot C & 1\\end{vmatrix}$$ $$R_2 \\to R_2 - R_1$$ $$R_3 \\to R_3 -R_1$$ $$= \\begin{vmatrix}\\sin^2A & \\cot A & 1 \\\\ \\sin^2B-\\sin^2A & \\cot B-\\cot A & 0 \\\\ \\sin^2C-\\sin^2A & \\cot C-\\cot A & 0\\end{vmatrix}$$ Expanding the determinant along $$C_3$$ \\begin{align} &= (\\sin^2B-\\sin^2A)(\\cot C-\\cot A)-(\\cot B-\\cot A)(\\sin^2C-\\sin^2A) \\\\ &= \\sin(B+A) \\sin(B-A) \\left[\\frac {\\cos C}{\\sin C} - \\frac {\\cos A}{\\sin A}\\right] - \\left[\\frac {\\cos B}{\\sin B} - \\frac {\\cos A}{\\sin A}\\right]\\sin(C+A) \\sin(C-A) \\\\ &= \\frac {\\sin(B+A) \\sin(B-A) \\sin(A-C)} {\\cos A \\cos C} - \\frac {\\sin(A-B) \\sin(C+A) \\sin(C-A)} {\\cos A \\cos C} \\\\ &= \\frac {\\sin(B-A) \\sin (A-C)} {\\cos A} \\left[\\frac {\\sin(A+B)} {\\cos C} - \\frac {\\sin(A+C)} {\\cos B}\\right] \\\\ &= \\frac {\\sin(B-A) \\sin (A-C)} {\\cos A} \\left[\\frac {\\sin C} {\\cos C} - \\frac {\\sin B} {\\cos B}\\right] \\\\", "height $h_A$ touches the base $e_A$ at $A'$, similar for height $h_B$ and $h_C$. Notice $u(A)^2$'s coefficient times area $|T|$ is: $$\\frac{|e_A|^2}{4|T|} = \\frac{|e_A|}{2 h_A} = \\frac{|BA'|+|A'C|}{2 h_A} = \\frac{1}{2} (\\cot\\angle B + \\cot\\angle C) .$$ similar for other two. The $u(A)u(B)$'s coefficient times area $|T|$ is: $$\\frac{|e_A||e_B|}{4|T|} \\cos\\angle C = \\frac{|e_A||e_B|}{2 |e_A| h_A} \\frac{|A'C|}{|e_B|} = \\frac{|A'C|}{2h_A} =\\frac{1}{2} \\cot \\angle C$$ Therefore, (2) becomes: \\begin{aligned} |\\nabla u |^2 &= u(A)^2\\frac{1}{2|T|} (\\cot\\angle B + \\cot\\angle C)\\\\ &\\quad + u(B)^2\\frac{1}{2|T|} (\\cot\\angle C + \\cot\\angle A) \\\\ &\\quad + u(C)^2\\frac{1}{2|T|} (\\cot\\angle A + \\cot\\angle B) \\\\ &\\quad - 2u(A) u(B) \\frac{1}{2|T|} \\cot \\angle C \\\\ &\\quad - 2u(B) u(C) \\frac{1}{2|T|} \\cot \\angle A \\\\ &\\quad - 2u(C) u(A) \\frac{1}{2|T|} \\cot \\angle B.\\tag{3} \\end{aligned} Finally for we are integrating a constant number on this triangle $T$: $$\\int_{T} |\\nabla u|^2 = |\\nabla u|^2 |T|,$$ and (3) multiplying with the area $|T|$ is the formula you gave. For a $u(x,y,z) = ax + by +cz +d$, the integration of $\\nabla u$ on a 3-simplex $T = V_1 V_2 V_3 V_4$ (a tetrahedron), denote the barycentric coordinate of vertex $V_i$ as $\\lambda_i$, $\\lambda_i(V_j) = \\delta_{ij}$. Also let $u(V_i) = u_i$, then: $$\\nabla u = \\sum_{i=1}^4 u_i \\nabla \\lambda_i.$$ And $\\nabla \\lambda_i$ is pointed in the opposite direction of the unit outer normal vector $\\nu_i$ to", "the inner side is a, the outer angle is B, the outer side is b. The cotangent rule may be written as (Todhunter,[1] Art.44) ${\\displaystyle \\cos({\\text{inner side}})\\cos({\\text{inner angle}})=\\cot({\\text{outer side}})\\sin({\\text{inner side}})\\ -\\ \\cot({\\text{outer angle}})\\sin({\\text{inner angle}}),}$ and the six possible equations are (with the relevant set shown at right): ${\\displaystyle {\\begin{array}{lll}{\\text{(CT1)}}\\quad &\\cos b\\,\\cos C=\\cot a\\,\\sin b-\\cot A\\,\\sin C,\\qquad &(aCbA)\\\\[0ex]{\\text{(CT2)}}&\\cos b\\,\\cos A=\\cot c\\,\\sin b-\\cot C\\,\\sin A,&(CbAc)\\\\[0ex]{\\text{(CT3)}}&\\cos c\\,\\cos A=\\cot b\\,\\sin c-\\cot B\\,\\sin A,&(bAcB)\\\\[0ex]{\\text{(CT4)}}&\\cos c\\,\\cos B=\\cot a\\,\\sin c-\\cot A\\,\\sin B,&(AcBa)\\\\[0ex]{\\text{(CT5)}}&\\cos a\\,\\cos B=\\cot c\\,\\sin a-\\cot C\\,\\sin B,&(cBaC)\\\\[0ex]{\\text{(CT6)}}&\\cos a\\,\\cos C=\\cot b\\,\\sin a-\\cot B\\,\\sin C,&(BaCb).\\end{array}}}$ To prove the first formula start from the first cosine rule and on the right-hand side substitute for ${\\displaystyle \\cos c}$ from the third cosine rule: {\\displaystyle {\\begin{aligned}\\cos a&=\\cos b\\cos c+\\sin b\\sin c\\cos A\\\\&=\\cos b\\ (\\cos a\\cos b+\\sin a\\sin b\\cos C)+\\sin b\\sin C\\sin a\\cot A\\\\\\cos a\\sin ^{2}b&=\\cos b\\sin a\\sin b\\cos C+\\sin b\\sin C\\sin a\\cot A.\\end{aligned}}} The result follows on dividing by ${\\displaystyle \\sin a\\sin b}$. Similar techniques with the other two cosine rules give CT3 and CT5. The other three equations follow by applying rules 1, 3 and 5 to the polar triangle. ### Half-angle and half-side formulae With ${\\displaystyle 2s=(a+b+c)}$ and ${\\displaystyle 2S=(A+B+C)}$, {\\displaystyle {\\begin{aligned}&\\sin {\\textstyle {\\frac {1}{2}}}A=\\left[{\\frac {\\sin(s{-}b)\\sin(s{-}c)}{\\sin b\\sin c}}\\right]^{1/2}&\\qquad &\\sin {\\textstyle {\\frac {1}{2}}}a=\\left[{\\frac {-\\cos S\\cos(S{-}A)}{\\sin B\\sin C}}\\right]^{1/2}\\\\[2ex]&\\cos {\\textstyle {\\frac {1}{2}}}A=\\left[{\\frac {\\sin s\\sin(s{-}a)}{\\sin b\\sin c}}\\right]^{1/2}&\\qquad &\\cos {\\textstyle {\\frac {1}{2}}}a=\\left[{\\frac", "common side $$CD$$ and the fact $$ADB$$ is a straight line, so $$\\angle ADC$$ and $$\\angle BDC$$ are supplementary. Hence $$CD$$, $$\\angle ADC$$ and $$\\angle BDC$$ must appear in our equations if those facts are to be brought to bear. Moreover, we want $$AD = m$$ and $$BD = n$$ to appear too, since they were indicated as of particular interest (the ratio $$\\triangle ABC$$ was divided in for the textbook; the lengths marked on the wall in the OP's question). Needing to represent two sides and an included angle in each of the smaller triangles rules out the law of sines; we could apply the law of cosines if we wanted $$AC$$ and $$BC$$ to appear in our equations also. Alternatively we can apply $$(4)$$ if we prefer two sides and two angles in each equation. At this stage we have a choice. Treating $$AD$$ as included by $$A$$ and $$\\pi - \\theta$$, and $$BD$$ as included by $$B$$ and $$\\theta$$, \\begin{align} m &= CD \\sin (\\pi - \\theta) \\left(\\cot (\\pi - \\theta) + \\cot A \\right) = CD \\sin \\theta \\left( -\\cot \\theta + \\cot A \\right) \\\\ n &= CD \\sin \\theta \\left( \\cot \\theta + \\cot B \\right) \\\\ \\implies \\frac{m}{n} &= \\frac{-\\cot \\theta + \\cot A }{\\cot \\theta + \\cot B} \\tag{7} \\end{align}", "x_1 \\cot x_2 - (\\cot x_2)^2 -2)^2}{4(\\cot x_1 + \\cot x_2)^4}\\\\ =\\ &0. \\end{align} 2) $$n = 5$$. Let $$x_1, x_2 \\in (0, \\frac{\\pi}{2})$$ satisfying \\begin{align} -4(\\cot x_2)^2(\\cot x_1)^2 + (-2(\\cot x_2)^3 + 6\\cot x_2)\\cot x_1 + (\\cot x_2)^4 + 4(\\cot x_2)^2 - 1 = 0. \\end{align} Let $$y_1 = \\cot x_1, \\ y_2 = \\cot x_2$$. We have $$-4y_2^2y_1^2 + (-2y_2^3 + 6y_2) y_1 +y_2^4+4y_2^2-1 = 0$$ which results in $$\\sqrt{5}y_2^2-4y_1y_2-y_2^2+\\sqrt{5}+3 = 0$$ since $$x_1, x_2 \\in (0, \\frac{\\pi}{2}).$$ Let $$x_3 = x_2, \\ x_4 = x_1, \\ x_5 = \\pi - x_1 - x_2 - x_3 - x_4.$$ We have $$x_1, x_2, x_3, x_4, x_5 > 0; \\ x_1 + x_2+x_3+x_4+x_5=\\pi.$$ Note that $$\\big(\\frac{\\sin \\frac{\\pi}{5}}{\\sin \\frac{2\\pi}{5}}\\big)^5 = \\frac{5\\sqrt{5}-11}{2}$$. We have \\begin{align} &\\frac{\\sin x_1 \\sin x_2 \\sin x_3 \\sin x_4 \\sin x_5} {\\sin (x_1+x_2) \\sin (x_2 + x_3) \\sin (x_3 + x_4) \\sin (x_4+x_5) \\sin (x_5+x_1)} - \\Big(\\frac{\\sin \\frac{\\pi}{5}}{\\sin \\frac{2\\pi}{5}}\\Big)^5\\\\ =\\ &\\frac{(\\sin x_1)^2(\\sin x_2)^2\\sin (2x_1 + 2x_2)}{(\\sin (x_1+x_2))^2\\sin 2x_2 (\\sin (x_1 + 2x_2))^2} - \\frac{5\\sqrt{5}-11}{2}\\\\ =\\ &\\frac{(y_1y_2-1)(y_2^2+1)^2}{y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2} - \\frac{5\\sqrt{5}-11}{2}\\\\ =\\ &-\\frac{5\\sqrt{5}-11}{16}\\frac{ (\\sqrt{5}y_2^2+2y_1y_2+3y_2^2+\\sqrt{5}+1)(\\sqrt{5}y_2^2-4y_1y_2-y_2^2+\\sqrt{5}+3)^2} {y_2(y_1+y_2)(2y_1y_2+y_2^2-1)^2}\\\\ =\\ &0. \\end{align} • This is interesting +1. Since the general Jacobian matrix I computed was singular, maybe there are multiple points where the inequalities are attained. – dezdichado Jun 19 '19 at 19:45 • When $n = 4,$ it must be that $2x_1+x_2", "# 1989 AIME Problems/Problem 10 ## Problem Let $a$, $b$, $c$ be the three sides of a triangle, and let $\\alpha$, $\\beta$, $\\gamma$, be the angles opposite them. If $a^2+b^2=1989c^2$, find $\\frac{\\cot \\gamma}{\\cot \\alpha+\\cot \\beta}$ ## Solution ### Solution 1 We draw the altitude $h$ to $c$, to get two right triangles. $[asy] size(170); pair A = (0,0), B = (3, 0), C = (1, 4); pair P = .5*(C + reflect(A,B)*C); draw(A--B--C--cycle); draw(C--P, dotted); draw(rightanglemark(C,P, B, 4)); label(\"A\", A, S); label(\"B\", B, S); label(\"C\", C, N); label(\"a\", (B+C)/2, NE); label(\"b\", (A+C)/2, NW); label(\"c\", (A+B)/2, S); label(\"h\", (C+P)/2, E);[/asy]$ Then $\\cot{\\alpha}+\\cot{\\beta}=\\frac{c}{h}$, from the definition of the cotangent. Let $K$ be the area of $\\triangle ABC.$ Then $h=\\frac{2K}{c}$, so $\\cot{\\alpha}+\\cot{\\beta}=\\frac{c^2}{2K}$. By identical logic, we can find similar expressions for the sums of the other two cotangents: \\begin{align*} \\cot \\alpha + \\cot \\beta &= \\frac{c^2}{2K} \\\\ \\cot \\beta + \\cot \\gamma &= \\frac{a^2}{2K} \\\\ \\cot \\gamma + \\cot \\alpha &= \\frac{b^2}{2K}. \\end{align*} Adding the last two equations, subtracting the first, and dividing by 2, we get $$\\cot \\gamma = \\frac{a^2 + b^2 - c^2}{4K}.$$ Therefore \\begin{align*} \\frac{\\cot \\gamma}{\\cot \\alpha + \\cot \\beta} &= \\frac{(a^2 + b^2 - c^2)/(4K)}{c^2/(2K)} \\\\ &= \\frac{a^2 + b^2 - c^2}{2c^2} \\\\ &= \\frac{1989 c^2 - c^2}{2c^2} \\\\ &= \\frac{1988}{2} = 994. \\end{align*} ### Solution 2", "# 19. Properties of Triangles’ Solutions Part 2¶ 1. The diagram is given below: From figure, $\\angle ADC = 90^\\circ + B$ By applying $m:n$ rule in triangle $ABC,$ we get $(1 + 1)\\cot(90^\\circ + B) = 1.\\cot90^\\circ - 1.\\cot(A-90^\\circ)$ $-2\\tan B = 0 - \\cot[-(90^\\circ - A)]$ $-2\\tan B = \\tan A \\Rightarrow \\tan A + 2\\tan B = 0$ 1. Given, $\\cot A + \\cot B + \\cot C = \\sqrt{3}$ Squaring, $\\cot^2A + \\cot^2B + \\cot^2C + 2\\cot A\\cot B + 2\\cot B\\cot C + 2\\cot C\\cot A = 3$ Since $A + B + C = \\pi \\Rightarrow A + B = \\pi - C$ $\\cot(A + B) = \\cot(\\pi - C)$ $\\frac{\\cot A\\cot B - 1}{\\cot A + \\cot B} = -\\cot C$ $\\cot A\\cot B + \\cot B\\cot C + \\cot C\\cot A = 1$ $\\Rightarrow \\cot^2A + \\cot^2B + \\cot^2C + 2\\cot A\\cot B + 2\\cot B\\cot C + 2\\cot C\\cot A = 3(\\cot A\\cot B + \\cot B\\cot C + \\cot C\\cot A)$ $\\frac{1}{2}[(\\cot A - \\cot B)^2 + (\\cot B - \\cot C)^2 + (\\cot C - \\cot A)^2] = 0$ $\\Rightarrow \\cot A = \\cot B = \\cot C$ $\\Rightarrow A = B = C$ i.e. triangle is equilateral. 2. Given, $(a^2 + b^2)\\sin(A - B) = (a^2", "+0 # Let $A,$ $B,$ $C$ be the angles of a triangle. 0 54 1 +467 Let a,b,c be the angles of a triangle. Evaluate$$\\begin{vmatrix} \\sin^2 A & \\cot A & 1 \\\\ \\sin^2 B & \\cot B & 1 \\\\ \\sin^2 C & \\cot C & 1 \\end{vmatrix}.$$ Nov 19, 2020 #1 0 Expanding along the third column, we find that most terms cancel, and we are left with a determinant of 1. Nov 19, 2020", "four-part, formulae relate two sides and two angles forming four ''consecutive'' parts around the triangle, for example (''aCbA'') or (''BaCb''). In such a set there are inner and outer parts: for example in the set (''BaCb'') the inner angle is ''C'', the inner side is ''a'', the outer angle is ''B'', the outer side is ''b''. The cotangent rule may be written as (Todhunter, Art.44) :$\\cos\\left(\\text\\right) \\cos\\left(\\text\\right) = \\cot\\left(\\text\\right) \\sin\\left(\\text\\right)\\ -\\ \\cot\\left(\\text\\right) \\sin\\left(\\text\\right),$ and the six possible equations are (with the relevant set shown at right): :$\\begin \\text\\quad& \\cos b\\,\\cos C=\\cot a\\,\\sin b - \\cot A \\,\\sin C ,\\qquad&\\left(aCbA\\right)\\\\ex\\text& \\cos b\\,\\cos A=\\cot c\\,\\sin b - \\cot C \\,\\sin A,&\\left(CbAc\\right)\\\\ex\\text& \\cos c\\,\\cos A=\\cot b\\,\\sin c - \\cot B \\,\\sin A,&\\left(bAcB\\right)\\\\ex\\text& \\cos c\\,\\cos B=\\cot a\\,\\sin c - \\cot A \\,\\sin B,&\\left(AcBa\\right)\\\\ex\\text& \\cos a\\,\\cos B=\\cot c\\,\\sin a - \\cot C \\,\\sin B,&\\left(cBaC\\right)\\\\ex\\text& \\cos a\\,\\cos C=\\cot b\\,\\sin a - \\cot B \\,\\sin C,&\\left(BaCb\\right). \\end$ To prove the first formula start from the first cosine rule and on the right-hand side substitute for $\\cos c$ from the third cosine rule: :$\\begin \\cos a & = \\cos b \\cos c + \\sin b \\sin c \\cos A \\\\ & = \\cos b\\ \\left(\\cos a \\cos b + \\sin a \\sin b \\cos C \\right) + \\sin b \\sin C \\sin a \\cot A \\\\", "+ c^2 - 4)}}{2} = \\frac{-bc + \\sqrt{(4 - b^2)(4 - c^2)}}{2}.$$ This asks for the trigonometric substitution $b = 2\\sin u$ and $c = 2\\sin v$, where $0^\\circ < u,v < 90^\\circ$. Then $$a = 2(-\\sin u\\sin v + \\cos u\\cos v) = 2\\cos (u + v),$$ and if we set $u = B/2$ and $v = C/2$, then $a = 2\\sin (A/2)$, $b = 2\\sin (B/2)$, and $c = \\sin (C/2)$, where $A$, $B$, and $C$ are the angles of a triangle. We have \\begin{align*} ab &= 4\\sin\\frac{A}{2}\\sin\\frac{B}{2} \\\\ &= 2\\sqrt{\\sin A\\tan\\frac{A}{2}\\sin B\\tan\\frac{B}{2}} = 2\\sqrt{\\sin A\\tan\\frac{B}{2}\\sin B\\tan\\frac{A}{2}} \\\\ &\\leq \\sin A\\tan\\frac{B}{2} + \\sin B\\tan\\frac{A}{2} \\\\ &= \\sin A\\cot\\frac{A + C}{2} + \\sin B\\cot\\frac{B + C}{2}, \\end{align*} where the inequality step follows from AM-GM. Likewise, \\begin{align*} bc &\\leq \\sin B\\cot\\frac{B + A}{2} + \\sin C\\cot\\frac{C + A}{2}, \\\\ ca &\\leq \\sin A\\cot\\frac{A + B}{2} + \\sin C\\cot\\frac{C + B}{2}. \\end{align*} Therefore \\begin{align*} ab + bc + ca &\\leq (\\sin A + \\sin B)\\cot\\frac{A + B}{2} + (\\sin B + \\sin C)\\cot\\frac{B + C}{2} + (\\sin C + \\sin A)\\cot\\frac{C + A}{2} \\\\ &= 2\\left(\\cos\\frac{A - B}{2}\\cos\\frac{A + B}{2} + \\cos\\frac{B - C}{2}\\cos\\frac{B + C}{2} + \\cos\\frac{C - A}{2}\\cos\\frac{C + A}{2} \\right)\\\\ &= 2(\\cos A + \\cos B + \\cos C) \\\\ &= 6 - 4\\left(\\sin^2\\frac{A}{2} + \\sin^2\\frac{B}{2}", "Set $a=x+y$ and $b=z$. Then $$\\begin{eqnarray*} \\cot \\left( x+y+z\\right) &=&\\cot (a+b)=\\dfrac{\\cot a\\cot b-1}{\\cot b+\\cot a} \\\\ &=&\\dfrac{\\cot \\left( x+y\\right) \\cot z-1}{\\cot z+\\cot \\left( x+y\\right) }= \\dfrac{\\dfrac{\\cot x\\cot y-1}{\\cot x+\\cot y}\\cot z-1}{\\cot z+\\dfrac{\\cot x\\cot y-1}{\\cot x+\\cot y}} \\\\ &=&\\dfrac{\\dfrac{\\cot x\\cot y\\cot z-\\cot z-\\cot x-\\cot y}{\\cot x+\\cot y} }{\\dfrac{\\cot x\\cot z+\\cot y\\cot z+\\cot x\\cot y-1}{\\cot x+\\cot y}} \\\\ &=&\\dfrac{\\cot x\\cot y\\cot z-\\cot x-\\cot y-\\cot z}{\\cot x\\cot y+\\cot x\\cot z+\\cot y\\cot z-1}. \\end{eqnarray*}$$ -", "\\frac {s-b}r \\frac {s-c}r = \\frac {s-a}r + \\frac {s-b}r +\\frac {s-c}r$ But the right member evals to $s/r$, giving the value of $r^2$ and proving the second assertion. ## Some proofs using the law of cotangents A number of other results can be derived from the law of cotangents. • Heron's formula. Note that the area of triangle $ABC$ is also divided into 6 smaller triangles, also in 3 pairs, with the triangles in each pair having the same area. For example, the two triangles near vertex A, being right triangles of width s−a and height r, each have an area of $\\tfrac{1}{2}r(s-a)$. So those two triangles together have an area of $r(s-a)\\,$, and the area $S$ of the whole triangle is therefore \\begin{align} S &= r(s-a) + r(s-b) + r(s-c) = r(s-a + s-b + s-c) \\\\[8pt] &= r(3s - (a+b+c)) = r(3s - 2s) = rs \\\\[8pt] \\end{align} This gives the result: $S=\\sqrt{s(s-a)(s-b)(s-c)}\\,$ as required. $\\frac {\\sin ( \\alpha/2-\\beta/2 ) }{\\sin ( \\alpha/2+\\beta/2 ) } = \\frac {\\cot ( \\beta/2 ) -\\cot ( \\alpha/2 ) }{\\cot ( \\beta/2 ) +\\cot ( \\alpha/2 ) }= \\frac {a-b}{2s-a-b}.$ This gives the result: $\\dfrac {a-b}{c}=\\dfrac {\\sin ( \\alpha/2-\\beta/2 )}{\\cos ( \\gamma/2 ) }$ as required. \\begin{align} & \\frac {\\cos( \\alpha/2-\\beta/2 ) }{\\cos( \\alpha/2+\\beta/2 ) } = \\frac {\\cot(", "us to express in terms of the sides of the triangle: d= -a2+b2+c2 2c . For the height of the triangle we have that . By replacing with the formula given above and applying the difference of squares identity we get \\begin{align} h2&= 2-\\left( -a2+b2+c2 2c b \\right)2\\\\ &= (2bc-a2+b2+c2)(2bc+a2-b2-c2) 4c2 \\\\ &= ((b+c)2-a2)(a2-(b-c)2) 4c2 \\\\ &= (b+c-a)(b+c+a)(a+b-c)(a-b+c) 4c2 \\\\ &= 2(s-a) ⋅ 2s ⋅ 2(s-c) ⋅ 2(s-b) 4c2 \\\\ &= 4s(s-a)(s-b)(s-c) c2 . \\end{align} We now apply this result to the formula that calculates the area of a triangle from its height: \\begin{align} A&= ch 2 \\\\ &=\\sqrt{ c2 4 4s(s-a)(s-b)(s-c) c2 } \\\\ &= \\sqrt.\\end ### Trigonometric proof using the law of cotangents From the first part of the law of cotangents proof,[10] we have that the triangle's area is both \\begin{align} A&=r((s-a)+(s-b)+(s-c))= 2\\left( s-a r r + s-b r + s-c r \\right)\\\\ &= 2\\left(\\cot{ \\alpha 2 r } + \\cot + \\cot\\right)\\endand, but, since the sum of the half-angles is /2, the triple cotangent identity applies, so the first of these is \\begin{align} A&= 2\\left(\\cot{ \\alpha 2 r } \\cot \\cot\\right) = r^2\\left(\\frac \\cdot \\frac \\cdot \\frac\\right) \\\\ &= \\frac.\\end Combining the two, we get A2=s(s-a)(s-b)(s-c), from which the result follows. ## Numerical stability Heron's formula as given above is numerically unstable for triangles", "# 20. Properties of Triangles’ Solutions Part 3¶ 1. The diagram is given below: Let $BD = DE = EC = x.$ Also let, $\\angle BAD = \\alpha, \\angle DAE = \\beta, EAC = \\gamma, CEA = \\theta$ Given, $\\tan\\alpha = t_1, \\tan\\beta = t_2, \\tan\\gamma = t_3$ Applying $m:n$ rule in $\\triangle ABC,$ we get $(2x + x)\\cot\\theta = 2x\\cot(\\alpha + \\beta) - x\\cot\\gamma$ From $\\triangle ADC,$ we get $2x\\cot\\theta = x\\cot\\beta - x\\cot\\gamma$ $\\Rightarrow \\frac{3}{2} = \\frac{2(\\alpha + \\beta) - \\cot\\gamma}{\\cot\\beta - \\cot\\gamma}$ $\\Rightarrow 3\\cot\\beta - 3\\cot\\gamma = 4\\cot(\\alpha + \\beta) - 2\\cot\\gamma$ $3\\cot\\beta - \\cot\\gamma = \\frac{4(\\cot\\alpha\\cot\\beta - 1)}{\\cot\\alpha + \\cot\\beta}$ $3\\cot^2\\beta - \\cot\\beta\\cot\\gamma + 3\\cot\\alpha\\cot\\beta - \\cot\\alpha\\cot\\gamma = 4\\cot\\alpha\\cot\\beta - 4$ $4 + 4\\cot^2\\beta = \\cot^2\\beta + \\cot\\alpha\\cot\\beta + \\cot\\beta\\cot\\gamma + \\cot\\alpha\\cot\\gamma$ $4(1 + \\cot^2\\beta) = (\\cot\\beta + \\cot\\alpha)(\\cot\\beta + \\cot\\gamma)$ $4\\left(1 + \\frac{1}{t_2^2}\\right) = \\left(\\frac{1}{t_1} + \\frac{1}{t_2}\\right)\\left(\\frac{1}{t_2} + \\frac{1}{t_3}\\right)$ 1. The diagram is given below: Let the medians be $AD, BE$ and $CF$ meet at $O.$ From question, $\\angle BOC=\\alpha, \\angle COA = \\beta, \\angle AOB = \\gamma$ Let $AD = p_1, BE=p_2, CF = p_3$ $AO:OD = 2:1 \\Rightarrow AO = \\frac{2}{3}p_1$ Similalrly, $OB = \\frac{2}{3}p_2, OC = \\frac{2}{3}p_3$ Applying cosine rule in $\\triangle AOC,$ $\\cos\\beta = \\frac{OA^2 + OC^2 - AC^2}{2.OA.OC} = \\frac{\\frac{4}{9}p_1^2 + \\frac{4}{9}p_3^2 - b^2}{2.\\frac{2}{3}p_1\\frac{2}{3}p_3}$ $\\cos\\beta = \\frac{4p_1^2 +", "& \\cos(B) & \\cos(C) \\\\ \\frac{1}{\\sin A}&\\frac{1}{\\sin B}&\\frac{1}{\\sin C}\\\\ \\frac{1}{\\cos A}&\\frac{1}{\\cos B}&\\frac{1}{\\cos C}\\end{pmatrix}=0$$ and by multiplying the first column by $\\cos(A)$, the second column by $\\cos(B)$ and the third column by $\\cos(C)$ we get that $$\\det\\begin{pmatrix}\\cos^2(A) & \\cos^2(B) & \\cos^2(C) \\\\ \\cot(A)&\\cot(B)&\\cot(C)\\\\ 1&1&1\\end{pmatrix}=0$$ and by replacing the first row with the difference between the third row and the first row $$\\det\\begin{pmatrix}\\sin^2(A) & \\sin^2(B) & \\sin^2(C) \\\\ \\cot(A)&\\cot(B)&\\cot(C)\\\\ 1&1&1\\end{pmatrix}=0$$ readily follows. One may also notice that if $\\theta\\in\\{A,B,C\\}$ then $$8R^2\\cdot \\sin^2\\theta + 4[ABC]\\cdot\\cot\\theta -(a^2+b^2+c^2)\\cdot 1 = 0$$ hence the nullspace of the given matrix has dimension $\\geq 1$. A hint, as requested: What you've done looks pretty good; I haven't checked every bit of the algebra, but the symmetry of the result makes me think that you've probably done it right. What you have not done is use the important fact given at the start: that $A, B,$ and $C$ are the angles of a triangle (hence they sum to $\\pi$). I don't see right away how to use that fact, but it's clearly important. • Umm, John, isn’t $\\cot(\\pi/2)=0$? – Lubin Feb 25 '18 at 14:45 • Ooops. More coffee needed. I'll edit. :) – John Hughes Feb 25 '18 at 15:10 • I used that fact as well when I got $sin (A+B) = \\sin C$ and", "(2 \\sin x-1)(\\sin x+1)\\end{align*} Solving the first part set equal to zero within the interval yields: 012x=2sinx1=sinx=π6,5π6,11π6,7π6\\begin{align*}0 &= 2 \\sin x-1\\\\ \\frac{1}{2} &= \\sin x\\\\ x &= \\frac{\\pi}{6}, \\frac{5 \\pi}{6}, - \\frac{11 \\pi}{6}, -\\frac{7 \\pi}{6}\\end{align*} Solving the second part set equal to zero yields: 01x=sinx+1=sinx=π2,3π2\\begin{align*}0 &= \\sin x+1\\\\ -1 &= \\sin x\\\\ x &= -\\frac{\\pi}{2}, \\frac{3 \\pi}{2}\\end{align*} These are the six solutions that will appear as intersections of the two graphs f(x)=cos2x\\begin{align*}f(x)=\\cos 2x\\end{align*} and g(x)=sinx\\begin{align*}g(x)=\\sin x\\end{align*} Note that the terms “general solution,” “completely solve”, and“solve exactly” #### Example 3 Determine the general solution to the following equation. cotx1=0\\begin{align*}\\cot x-1=0\\end{align*} cotx1cotx=0=1\\begin{align*}\\cot x-1 &= 0\\\\ \\cot x &= 1\\end{align*} One solution is x=π4\\begin{align*}x=\\frac{\\pi}{4}\\end{align*}. However, since this question asks for the general solution, you need to find every possible solution. You have to know that cotangent has a period of π\\begin{align*}\\pi\\end{align*} which means if you add or subtract π\\begin{align*}\\pi\\end{align*} from π4\\begin{align*}\\frac{\\pi}{4}\\end{align*} then it will also yield a height of 1. To capture all these other possible x\\begin{align*}x\\end{align*} values you should use this notation. x=π4±nπ\\begin{align*}x=\\frac{\\pi}{4} \\pm n \\cdot \\pi\\end{align*} where n\\begin{align*}n\\end{align*} is a integer Notice that trigonometric equations may have an infinite number of solutions that repeat in a certain pattern because they are periodic functions. When you see these directions remember to find all the solutions by using notation like in", "## Thursday, December 27, 2018 ### Trigonometry (Selected Problems) $\\def\\D{\\displaystyle}$ Question (1): In $\\Delta ABC$, if $\\cot A+\\cot B+\\cot C=\\sqrt 3$, then $\\Delta ABC$ is equilateral. $\\D \\begin{array}{|rl|}\\hline \\cot(A+B)&=\\D\\frac{\\cot A\\cot B-1}{\\cot A+\\cot B}\\\\ \\hline \\end{array}$ Proof: $\\cot C=\\cot(180-(B+C))=-\\cot(A+B)=\\frac{1-\\cot A\\cot B}{\\cot A+\\cot B}.$ Let $\\cot A=x,\\cot B=y.$ Hence, \\begin{eqnarray*} x+y+\\frac{1-xy}{x+y}&=&\\sqrt 3\\\\ (x^2+2xy+y^2) +(1-xy)&=&\\sqrt 3x+\\sqrt 3y\\\\ \\end{eqnarray*} For real solutions, $b^2-4ac\\ge 0.$ Thus \\begin{eqnarray*} (x-\\sqrt 3)^2-4(1)(x^2-\\sqrt 3x+1)&\\ge 0\\\\ 3x^2-2\\sqrt 3x+1&\\le&0\\\\ (\\sqrt 3x-1)^2&\\le&0\\\\ \\sqrt 3x-1&=&0. \\end{eqnarray*} Thus $x=1/\\sqrt 3$. By (1), $y=1/\\sqrt 3$. Therefore $\\cot A=\\cot B=\\frac{1}{\\sqrt 3}\\Longrightarrow A=B=60^{\\circ}.$ Question (2): In $\\Delta ABC$, if $\\sin^2 A+\\sin^2 B+\\sin^2 C= 2$, then $\\Delta ABC$ is a right triangle. $\\D\\begin{array}{|rl|}\\hline \\sin^2A&=\\D \\frac{1-\\cos2A}{2}\\\\ \\cos A+\\cos B&=\\D2\\cos\\frac{A+B}{2}\\cos\\frac{A-B}{2}\\\\ \\sin^2A&=1-\\cos^2A\\\\ \\sin A&=\\sin(180^{\\circ}-A)\\\\ \\hline \\end{array}$ Proof: \\begin{eqnarray*} \\sin^2A+\\sin^2B&=&\\frac{1-\\cos2A}{2}+\\frac{1-\\cos2B}{2} \\\\ &=&1-\\frac{1}{2}(\\cos2A+\\cos2B)\\\\ &=&1-\\frac{1}{2}\\left(2\\cos \\frac{2A+2B}{2}\\cos \\frac{2A-2B}{2}\\right)\\\\ \\sin^2A+\\sin^2B &=&1-\\cos(A+B)\\cos(A-B)\\cdots (1)\\\\ \\sin^2C&=&\\sin^2(180^{\\circ}-C)=\\sin^2(A+B)\\\\ \\sin^2C&=&1-\\cos^2(A+B)\\cdots (2) \\end{eqnarray*} (1)+(2): $\\D\\sin^2A+\\sin^2B+\\sin^2C$\\begin{eqnarray*} &=&2-\\cos(A+B)(\\cos(A+B)+\\cos(A-B))\\\\ 2&=&2-\\cos(A+B)\\left(\\cos A\\cos B-\\sin A\\sin B \\right.\\\\ 0&=&-2\\cos(A+B)\\cos A\\cos B \\end{eqnarray*} Hence, $\\D \\cos(A+B)=0$ or $\\D \\cos A=0$ or $\\D\\cos B=0.$ Thus $\\D A+B=90^{\\circ}$, ie $\\D C=90^{\\circ}$ or $\\D A=90^{\\circ}$ or $\\D B=90^{\\circ}.$ ## Wednesday, December 26, 2018 ### AP GP Series (IB Standard Level) $\\def\\D{\\displaystyle}$ 1.) In an arithmetic sequence, $\\D u_1 = 2$ and $\\D u_3 = 8.$ (a) Find $\\D d.$ (b) Find $\\D u_{20}.$ (c) Find $\\D S_{20}.$ (Total 6 marks) 2.) In an arithmetic sequence $\\D u_1 = 7, u_{20} =", "# Proof of Cot double angle formula $\\cot{2x}$, $\\cot{2\\theta}$, $\\cot{2A}$, $\\cot{2\\alpha}$ and etc. are examples for cot of double angle function. They can be expanded as given below in terms of cot of angle. $(1) \\,\\,\\,\\,\\,\\,$ $\\cot{2\\theta} \\,=\\, \\dfrac{\\cot^2{\\theta}-1}{2\\cot{\\theta}}$ $(2) \\,\\,\\,\\,\\,\\,$ $\\cot{2x} \\,=\\, \\dfrac{\\cot^2{x}-1}{2\\cot{x}}$ $(3) \\,\\,\\,\\,\\,\\,$ $\\cot{2A} \\,=\\, \\dfrac{\\cot^2{A}-1}{2\\cot{A}}$ $(4) \\,\\,\\,\\,\\,\\,$ $\\cot{2\\alpha} \\,=\\, \\dfrac{\\cot^2{\\alpha}-1}{2\\cot{\\alpha}}$ Let us prove cot double angle identity in trigonometric mathematics before using it as a formula. ### Proof Cot double angle identity is actually derived in trigonometry by geometric mathematics as per the mathematical relations between angles and sides. #### Construction of a Triangle with double angle $\\Delta DCE$ is a right angled triangle. Draw a line from point $C$ to side $\\overline{DE}$ to bisect the angle and each angle is $\\theta$. Draw a perpendicular line to side $\\overline{CE}$ from point $F$ and intersects the side $\\overline{CE}$ at point $G$. After that, draw a perpendicular line to side $\\overline{CD}$ from point $G$. It intersects the side $\\overline{CF}$ at point $H$ and also intersects the side $\\overline{CD}$ at point $I$. Draw perpendicular lines to sides $\\overline{DE}$ and $\\overline{GI}$ from points $H$ and $F$ respectively and their associated intersecting points are $K$ and $J$. #### Expressing Cotangent double angle in terms of ratio of sides Firstly, express cotangent of double angle in terms of ratio of", "$\\cot \\alpha = 2\\cot A+\\cot B$ and similarly $\\cot \\beta = 2\\cot A+\\cot C$. For convenience, let us take $\\cot A=c_1$, $\\cot B=c_2$ and $\\cot C=c_3$. By the addition formula we have: ... 2 Here is a complete proof along the lines of my comments to the OP. Most of it is well-known stuff, but it probably doesn't hurt to build it from scratch. First of all, the problem follows from the following claim: Theorem 1. Let $A$, $B$, $C$ and $D$ be four points on a circle. Let $X=AB\\cap CD$ and $Y=BC\\cap DA$. (I use the notation $\\ell\\cap m$ to denote ... 2 By the angle bisector theorem, we have: $$\\frac{CE}{AF}=\\frac{CD\\cos C}{AD\\cos A}=\\frac{BC\\cos C}{AB\\cos A}=\\frac{CI}{AI}$$ As $BF=BE$, by the converse of Ceva's theorem, we have that $AE, CF$ and $BI$ are concurrent. 2 Let $M$ and $N$ be the midpoints of $\\overline{BC}$ and $\\overline{AB}$, respectively, noting that $N$ is also the intersection of the lateral edges of trapezoid $\\square BCGE$. Now, if $\\angle AEC\\cong \\angle DGC$, then $\\angle BEC\\cong\\angle BGC$. This implies that $\\square BCGE$ is cyclic; since it's a trapezoid, it's therefore necessarily an isosceles ... 2 By Holder $\\left(\\sum\\limits_{cyc}\\frac{\\sqrt{b+c-a}}{a}\\right)^2\\sum\\limits_{cyc}a^2(b+c-a)^2\\geq(a+b+c)^3$. Hence, it remains to prove that $abc(a+b+c)\\geq\\sum\\limits_{cyc}a^2(b+c-a)^2$, which is $\\sum\\limits_{cyc}(a-b)^2(a+c-b)(b+c-a)\\geq0$, which is obvious. 2 Where is it located? Unfortunately I couldn't find a closed formula for", "in such a way that $\\widehat{AOG}=75^\\circ$; let $B'=BC\\cap OG$ and take $A'\\in FA$ such that $OA'\\perp OB'$; Let $\\Gamma$ be the circle with centre $O$ through $A'$ and $M$ be the midpoint of $OA'$; let $N$ be the ... 3 Since the cotangent function is odd, we may assume $A,B,C\\geq 0$ without loss of generality. So $A,B,C$ are the angles of a triangle. Let we deal first with the case of an acute triangle. By the Cauchy-Schwarz inequality: $$\\cot(A)^2+\\cot(B)^2+\\cot(C)^2 \\geq 3\\left(\\frac{\\cot A+\\cot B+\\cot C}{3}\\right)^2 = \\frac{\\cot(\\omega)^2}{3}$$ where $\\omega$ is the ... 3 For another way, since $\\cot^2x$ is convex when $x \\in [0,\\pi]$ and we need to worry only of positive $x$, so by Jensen's inequality: $$\\sum_{cyc} \\cot^2 A \\geqslant 3 \\cot^2\\frac{A+B+C}3=1$$ As equality is possible when $A=B=C=\\frac{\\pi}3$, we have the minimum. 3 Nope, $AC$ need not equal to other $5$ edges and the tetrahedron need not be regular. As long as the remaining $5$ edges are equal in length, we have $BC = CD = DB \\implies \\triangle BCD$ is equilateral $\\implies K$ coincides with centroid of $\\triangle BCD$. $AB = BD = AD \\implies \\triangle ABD$ is equilateral $\\implies M$ coincides with centroid of $\\... 3 $$(a-c)(a+c)^2+bc(a+c)=ab^2$$ $$\\implies a^3-a b^2-a c^2+b c^2+a^2 c+a b c-c^3=0$$ $$\\implies a(a^2-b^2)-c^2(a-b)+ca(a+b)-c^3=0$$ $$\\implies a(a+b)(a-b)-c^2(a-b)+ca(a+b)-c^3=0$$ $$\\implies (a-b)(a(a+b)-c^2)+c(a(a+b)-c^2)=0$$ $$\\implies (a-b+c)(a(a+b)-c^2)=0$$ Dividing the both" ]

[ "AM/GM VectorMethod 希望杯 Intermediate 2011 Problem - 2959 Let $G$ be the centroid of $\\triangle{ABC}$. Points $M$ and $N$ are on side $AB$ and $AC$, respectively such that $\\overline{AM} = m\\cdot\\overline{AB}$ and $\\overline{AN} = n\\cdot\\overline{AC}$ where $m$ and $n$ are two positive real numbers. Find the minimal value of $mn$. The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99.", "+0 # centroid 0 141 2 G is the centroid of triangle ABC. $$G_1,G_2,$$ and $$G_3$$ are the centroids of triangle BCG, triangle CAG, and triangle ABG, respectively. What is $$\\dfrac{[G_1G_2G_3]}{[ABC]}$$? Nov 11, 2021 #1 -1 $[G_1 G_2 G_3]/[ABC] = \\frac{1}{12}$ Nov 11, 2021 #2 -1 [G1G2G3] / [ABC] = 1 / 9 civonamzuk Nov 12, 2021 edited by Guest Nov 12, 2021", "medial triangles of $\\triangle ABC$, with common centroid $G$. - 5 years, 2 months ago I've added a full description of the configuration. I've started to write the results I've derived as problems. Here are links to the first two: - 5 years, 2 months ago", "# Centroid G of ABC: Author Van Khea Let $G$ be the centroid of $\\Delta ABC$. Let $AG, BG, CG$ cuts $BC, CA, AB$ at $D, E, F$ and cuts circumcircle of $\\Delta ABC$ at $X, Y, Z$. Let $YZ, ZX, XY$ cuts $AG, BG, CG$ at $P, Q, R$. a) Prove that: $\\displaystyle \\frac{GD}{GX}+\\frac{GE}{GY}+\\frac{GF}{GZ}=\\frac{3}{2}$ b) Prove that: $\\displaystyle \\frac{GD}{GP}+\\frac{GE}{GQ}+\\frac{GF}{GR}=3$ Author: Van Khea", "+0 # triangles 0 74 2 The centroid of ABC is G. Find BG. Apr 29, 2022 #1 0 BG = 3*sqrt(7). Apr 29, 2022 #2 +2440 +1 Draw median $$BM$$. The length of $$BM$$ is $$4 \\sqrt{13}$$. By the properties of centroids, $$GM: BG = 1:2$$ This means that $$BG$$ is $$\\color{brown}\\boxed{8 \\sqrt{13} \\over 3}$$ Apr 29, 2022", "+0 # triangles 0 130 2 The centroid of ABC is G. Find BG. Apr 29, 2022 #1 0 BG = 3*sqrt(7). Apr 29, 2022 #2 +2558 0 Draw median $$BM$$. The length of $$BM$$ is $$4 \\sqrt{13}$$. By the properties of centroids, $$GM: BG = 1:2$$ This means that $$BG$$ is $$\\color{brown}\\boxed{8 \\sqrt{13} \\over 3}$$ Apr 29, 2022", "is contained in $C_k$, and each $C_q$ has a point $p_q$ at its center. Then for every even $2\\leq n$, we can place the points at the corners of a regular n-gon.", "+0 -1 151 2 +0 G is the centroid of ABC and D and E are the midpoints of AB and AC respectively. AG and DE intersect at M. Find GM/GA Nov 19, 2019 #1 +1 I drew a diagram, and got GM = 2 and GA = 5, so I think the answer is 2/5. Nov 19, 2019 #2 +24407 +2 G is the centroid of ABC and D and E are the midpoints of AB and AC respectively. AG and DE intersect at M. Find GM/GA \\(\\dfrac{GM}{GA} = \\dfrac{1}{4} \\) Nov 20, 2019 edited by heureka Nov 20, 2019", "that the centroid ($G$), circumcenter ($O$), and orthocenter ($H$) of a triangle are collinear and that $OG:GH = 1:2$.", "# For $G$ the centroid in $\\triangle ABC$, if $AB+GC=AC+GB$, then $\\triangle ABC$ is isosceles. (Likewise, for the incenter.) Let $$G$$ be the centroid of $$\\triangle ABC$$. Prove that if $$AB+GC=AC+GB$$ then the triangle is isosceles! Of course, the equality is true, when we have isosceles triangle, but the other way is not trivial for me. I have tried using vectors, even the length of the medians. Let $$D$$ be the midpoint of $$BC$$. Let $$G'$$ be the reflection of $$G$$ over $$D$$. As $$BD=DC$$ and $$GD=DG'$$, $$GBG'C$$ is a parallelogram. Therefore, $$GB=G'C$$ and $$GC=G'B$$, whence, $$AB+BG'=AC+CG'$$. Thus, $$B$$ and $$C$$ lie on an ellipse $$e$$ with foci $$A$$ and $$G'$$. Let $$e'$$ be the reflection of $$e$$ over $$D$$. As $$BD=DC$$, $$B$$ and $$C$$ also lie on $$e'$$, and thus on $$e\\cap e'$$. As $$e$$ and $$e'$$ are distinct ellipses symmetric about $$AD$$, $$BC\\perp AD$$, whence $$AB=AC$$. $$\\blacksquare$$ (Note that we didn't use the fact that $$G$$ is the centroid of $$ABC$$ in the proof, we just used the fact that $$G$$ lies on $$AD$$. So, the same proof proves the following more general statement: If $$G$$ is a point on $$AD$$ such that $$AB+GC=AC+GB$$, then, $$\\triangle ABC$$ is isosceles.) • @BijayanRay, as the point of reflection($D$) lies on the axis($AG'$) of the first ellipse, the axis", "is the centroid of $\\triangle ABC$, which means $ABC$ is an equilateral.", "+0 # Triangle Medians and Altitudes +6 294 9 +290 Let $$P$$ be a point inside triangle $$ABC$$. Let $${G}_{1}$$$${G}_{2}$$, and $${G}_{3}$$ be the centroids of triangles $$PBC$$, $$PCA$$,and $$PAB$$, respectively. If the area of triangle $$ABC$$ is 18, then find the area of triangle $${G}_{1}{G}_{2}{G}_{3}$$. TheMathCoder Apr 20, 2018 #1 +290 +4 TheMathCoder Apr 21, 2018 #2 +963 0 I'm working on it, but this is hard! GYanggg Apr 21, 2018 #3 +3276 0 Same here! I'll try my best! tertre Apr 21, 2018 #4 +290 +3 I am still struggling to find the values... TheMathCoder Apr 21, 2018 #5 +1 X^2: Did you see this question, which was posted 2-3 days ago and hasn't been answered? I'm just curious for myself to know how to solve this seemingly very difficult Geometry-Trig problem. I don't even know how to approach the problem since I don't fully understand the properties of \"centroids\" of triangles. Thanks. Guest Apr 23, 2018 #6 +290 +2 HEEEEELP! TheMathCoder Apr 24, 2018 #7 +1 Let the height of the triangle ABC (the vertical height of A above BC), be h, and the height of P be $$\\alpha h \\text{ where } 0<\\alpha<1 .$$ The centroid of the triangle BPC, $$G_{1},$$ lies on the median of that triangle from P, (that's the line from", "# CENTROIDS AND SOME CHARACTERIZATIONS OF PARALLELOGRAMS • Kim, Dong-Soo ; • Lee, Kwang Seuk ; • Lee, Kyung Bum ; • Lee, Yoon Il ; • Son, Seongjin ; • Yang, Jeong Ki ; • Yoon, Dae Won • Published : 2016.07.31 • 49 5 #### Abstract For a polygon P, we consider the centroid $G_0$ of the vertices of P, the centroid $G_1$ of the edges of P and the centroid $G_2$ of the interior of P, respectively. When P is a triangle, the centroid $G_0$ always coincides with the centroid $G_2$. For the centroid $G_1$ of a triangle, it was proved that the centroid $G_1$ of a triangle coincides with the centroid $G_2$ of the triangle if and only if the triangle is equilateral. In this paper, we study the relationships between the centroids $G_0$, $G_1$ and $G_2$ of a quadrangle P. As a result, we show that parallelograms are the only quadrangles which satisfy either $G_0=G_1$ or $G_0=G_2$. Furthermore, we establish a characterization theorem for convex quadrangles satisfying $G_1=G_2$, and give some examples (convex or concave) which are not parallelograms but satisfy $G_1=G_2$. #### References 1. R. A. Johnson, Modem Geometry, Houghton-Mifflin Co., New York, 1929. 2. M. J. Kaiser, The perimeter centroid of a convex polygon, Appl. Math. Lett. 6 (1993), no. 3,", "$P$ such that $PC=3,PA=4$ and $PB=5$. Find the perimeter of $\\triangle ABC$. The solution was simply amazing and gave me the incentives to solve this one বড় ভালবাসি তোমায়,মা", "# Fill in the Blanks. - Mathematics Sum Fill in the blanks. Point G is the centroid of ABC. (1) If l(RG) = 2.5 then l(GC) = _____ (2) If l(BG) = 6 then l(BQ) = _____ (3) If l(AP) = 6 then l(AG) = _____ and l(GP) = _____ #### Solution In ∆ABC, the medians AP, BQ and CR to the sides BC, CA and AB respectively intersect at G. Since, centroid of a triangle divides the medians in the ratio of 2 : 1, then AG : GP = BG : GQ = CG : GR = 2 : 1. (1) We have, CG : GR = 2 : 1 ⇒ (GC)/(RG) = 2/1 ⇒ (GC)/(2.5) = 2/1 ⇒ GC = 5 (2) We have, BG : GQ = 2 : 1 ⇒ (BG)/(GQ) = 2/1 ⇒ (BG)/(BQ-BG) = 2/1 ⇒ (6)/(BQ-6) = 2 ⇒ BQ - 6 = 3 ⇒ BQ = 9 (3) We have, AG : GP = 2 : 1 ⇒ (AG)/(GP) = 2/1 ⇒ (AG)/(AP-AG) = 2 ⇒ (AG)/(6-AG) = 2 ⇒ 2(6 - AG) = AG ⇒ 12 - 2AG = AG ⇒ 3AG = 12 ⇒ AG = 12/3 ⇒ AG = 4 Now, GP = AP − AG = 6 − 4 = 2. Hence, we have, (1)", "the centroid is $G = \\frac{A+B+C}{3}$. Thus, $O, G, H$ are collinear and $\\frac{OG}{HG} = \\frac{1}{2}$", "by $$G\\left(\\dfrac{X_1+X_2+X_3}{3},\\dfrac{Y_1+Y_2+Y_3}{3}\\right).$$ For calculating the centroid of any triangle using above centroid calculator, you have to just write the values in the given input boxes and press the calculate button, you will get the result.", "G is the centroid in the equilateral △ ABC. If AB = 10 cm then length of AG is : [A] $\\frac{20\\sqrt{3}}{3}cm$ [B] $10\\sqrt{3}cm$ [C] $\\frac{10\\sqrt{3}}{3}cm$ [D] $5\\sqrt{3}cm$", "# if G is the centroid of the triangle ABC show that cot GAB+cot GBC+cot GCA= 3cot omega= cot ABG+cotBCG+cotCAG where cot omega=cot A+cot B+cot C Dear Student, let M be the midpoint of BC. Then applying sine rule on $△ABM$, we get Similarly we get the relations for cot GBC and cot GCA Using the fact that if $\\omega$ is the Brocard Angle, we have", "# If G Be the Centroid of a Triangle Abc, Prove That: Ab2 + Bc2 + Ca2 = 3 (Ga2 + Gb2 + Gc2) - Mathematics Sum If G be the centroid of a triangle ABC, prove that: AB2 + BC2 + CA2 = 3 (GA2 + GB2 + GC2) #### Solution Let A(x1,y1); B(x2,y2); C(x3,y3) be the coordinates of the vertices of ΔABC.Let us assume that centroid of the ΔABC is at the origin G.So, the coordinates of G are G(0,0). Now,(x_1+x_2+x_3)/3 =0; (y_1+y_2+y_3)/3 =0 So, x_1+x_2+x_3=0 ...........(1) y_1+y_2+y_3=0 ..........(2) Squaring (1) and (2), we get x_1^2+x_2^2+x_3^2+2x_1x_2+2x_2x_3+2x_3x_1=0 ..........(3) y_1^2+y_2^2+y_3^2+2y_1y_2+2y_2y_3+2y_3y_1=0 ..........(4) LHS=AB^2+BC^2+CA^2 =[sqrt((x_2-x_1)^2 +(y_2-y_1)^2]]^2 +[sqrt((x_3-x_2)^2+(y_3-y_2)^2)]^2 +[sqrt((x_3-x_1)^2+(y_3-y_1)^2)]^2 =(x_2-x_1)^2 +(y_2-y_1)^2+(x_3-x_2)^2+(y_3-y_2)^2+(x_3-x_1)^2+(y_3-y_1)^2 =x_1^2x_2^2-2x_1^2+y_1^2+y_2^2-2y_1y_2+x_2^2+x_3^2-2x_2x_3+y_2^2+y_2^2+y_3^2-2y_2y_3+x_1^2+x_3^2-2x_1x_3+y_1^2+y_3^2-2y_1v_3 =2(x_1^2+x_2^2+x_3^2)+2(y_1^2+y_2^2+y_3^2)-(2x_1x_2+2x_2x_3+2x_3x_1)-(2y_1y_2+2y_2y_3+2y_3y_1) =2(x_1^2+x_2^2+x_3^2)+2(y_1^2+y_2^2+y_3^2)+(x_1^2+x_2^2+x_3^2)+(y_1^2+y_2^2+y_3^2) =3(x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2) RHS =3(GA^2+GB^2+GC^2) =[{sqrt((x_1-0)^2+(y_1-0)^2)}^2 +{sqrt((x_2-0)^2+(y_2-0)^2)}^2 +{sqrt((x_3-0)^2+(y_3-0)^2)}^2] =3[x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2] Hence, AB^2+BC^2+CA^2=3(GA^2+GB^2+GC^2) Concept: Coordinate Geometry Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 10 Maths Chapter 6 Co-Ordinate Geometry Exercise 6.4 | Q 9 | Page 37" ]

[ "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "as the function of $X$ so that it satisfy the above table: \\begin{align} P_0 &= \\frac{1}{12} (X-2)(X-6)\\\\ P_1 &= -\\frac{1}{8} X(X-6)\\\\ P_2 &= \\frac{1}{24} X(X-2) \\end{align} If we replace $X$ in the above equation with \\begin{align} X &= (\\mathbf{S_i} + \\mathbf{S_{i+1}})(\\mathbf{S_i} + \\mathbf{S_{i+1}})\\\\ &= \\mathbf{S_i^2} + \\mathbf{S_{i+1}^2} + 2\\mathbf{S_i} \\cdot \\mathbf{S_{i+1}} \\\\ &= 4 + 2\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} \\end{align} then we get $$P_2 = \\frac{1}{6} (\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}})^2 + \\frac{1}{2} \\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} + \\frac{1}{3}$$ • Thanks for this answer to your own question :) It's simple yet effective especially the explanation of the projection operators ! – gertian May 4 '17 at 16:18", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "$2 + 2\\cdot 3 + 2\\cdot 3^2 + 2\\cdot 3^3$, so you normally you would form the array $(\\mathbf{2'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'})$, which has length $(2+1+1+1)+(2+3+1+1)+(2+3+3+1)+(2+3+3+3) = 32$. However, writing the ternary form as $2 + 3\\cdot(2 + 3\\cdot (2 + 3\\cdot 2))$, the array would be $(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'[(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'(\\mathbf{2'}\\cdot'\\mathbf{1'} +' \\mathbf{3'}\\cdot'\\mathbf{2'})]$, which only has length $(2+1+1+1)+3+(2+1+1)+3+(2+1)+(3+2) = 23$. In general, the smaller the total numbers that appear on your representation, the better. Using this technique, we are now able to reduce the required length to at most $94$, and we've now solved the problem! The ideas above are enough to pass the problem already, but I'm sure with a couple more tricks you can reduce the length even more. # Time Complexity: $O(N)$ # AUTHOR'S AND TESTER'S SOLUTIONS: This question is marked \"community wiki\". 1.7k586142 accept rate: 11% 19.7k350498541 4 “I loved solving this question.” The constraint is very restrictive, just use numbers very efficiently. So, I tried to make a LIS’s of length equal to MSB of K (for K = 4, length = 3). Now you need exactly K such LIS’s, so now try to create LIS’s in powers of two starting from the least power of 2 present, if power of 2 p, then", "${\\displaystyle \\mathbf {x} _{+}}$, such as ${\\displaystyle \\mathbf {x} _{2}}$, due to commuting or shifting ${\\displaystyle \\mathbf {a} _{k}}$ right an odd number of times, in the dividend blade ${\\displaystyle \\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n}}$, into its Template:Mvar-th spot. In general, ${\\displaystyle \\mathbf {x} _{k}}$ is solved as {\\displaystyle {\\begin{aligned}\\mathbf {x} _{k}&=(\\mathbf {a} _{1}\\wedge \\cdots \\wedge (\\mathbf {C} )_{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})\\cdot ((-1)^{k-1}\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})^{-1}\\\\[6pt]&={\\frac {(\\mathbf {a} _{1}\\wedge \\cdots \\wedge (\\mathbf {C} )_{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})\\cdot ((-1)^{k-1}\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})}{(\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})^{2}}}\\\\[6pt]&={\\frac {(-1)^{k-1}(\\mathbf {a} _{1}\\wedge \\cdots \\wedge (\\mathbf {C} )_{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})\\cdot (\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})}{(-1)^{\\frac {n(n-1)}{2}}(\\mathbf {a} _{n}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{1})\\cdot (\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})}}\\\\[6pt]&={\\frac {(-1)^{k-1}(\\mathbf {a} _{1}\\wedge \\cdots \\wedge (\\mathbf {C} )_{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})\\cdot (\\mathbf {a} _{n}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{1})}{(\\mathbf {a} _{n}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{1})\\cdot (\\mathbf {a} _{1}\\wedge \\cdots \\wedge", "\\mathbf{k}) \\cdot \\mathbf{p} + \\mathbf{p}^2 \\right)=e^{i \\mathbf{k}\\cdot \\mathbf{r}} (\\hbar \\mathbf{k} + \\mathbf{p})^2\\end{aligned} So, finally, our first commutator is \\begin{aligned}\\left[{H},{ e^{i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]&= \\frac{1}{{2m}}e^{i \\mathbf{k}\\cdot \\mathbf{r}} \\left((\\hbar\\mathbf{k})^2 + 2 (\\hbar \\mathbf{k}) \\cdot \\mathbf{p} \\right)\\end{aligned} \\hspace{\\stretch{1}}(3.16) The double commutator is then \\begin{aligned}\\left[{\\left[{H},{e^{i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]},{e^{-i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]&=e^{i\\mathbf{k} \\cdot \\mathbf{r}} \\frac{\\hbar \\mathbf{k}}{m} \\cdot \\left( \\mathbf{p} e^{-i \\mathbf{k} \\cdot \\mathbf{r}} - e^{-i \\mathbf{k} \\cdot \\mathbf{r}} \\mathbf{p} \\right)\\end{aligned} To simplify this we want \\begin{aligned}\\mathbf{k} \\cdot \\boldsymbol{\\nabla} e^{-i \\mathbf{k} \\cdot \\mathbf{r}} \\Psi &=k_n \\partial_n e^{-i k_m x_m } \\Psi \\\\ &=e^{-i \\mathbf{k} \\cdot \\mathbf{r} }\\left(k_n (-i k_n) \\Psi + k_n \\partial_n \\Psi \\right) \\\\ &=e^{-i \\mathbf{k} \\cdot \\mathbf{r} } \\left( -i \\mathbf{k}^2 + \\mathbf{k} \\cdot \\boldsymbol{\\nabla} \\right) \\Psi\\end{aligned} The double commutator is then left with just \\begin{aligned}\\left[{\\left[{H},{e^{i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]},{e^{-i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]&=- \\frac{1}{{m}} (\\hbar \\mathbf{k})^2 \\end{aligned} \\hspace{\\stretch{1}}(3.17) Now, returning to the energy expression \\begin{aligned}\\sum_n (E_n - E_s) {\\left\\lvert{{\\langle {n} \\rvert} e^{i\\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {s} \\rangle}}\\right\\rvert}^2&=\\sum_n (E_n - E_s) {\\langle {s} \\rvert} e^{-i\\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {n} \\rangle} {\\langle {n} \\rvert} e^{i\\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {s} \\rangle} \\\\ &=\\sum_n {\\langle {s} \\rvert} e^{-i\\mathbf{k} \\cdot \\mathbf{r}} H {\\lvert {n} \\rangle} {\\langle {n} \\rvert} e^{i\\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {s} \\rangle} -{\\langle {s} \\rvert} e^{-i\\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {n} \\rangle} {\\langle {n} \\rvert} e^{i\\mathbf{k} \\cdot \\mathbf{r}} H {\\lvert {s} \\rangle} \\\\ &={\\langle {s} \\rvert}", "(2) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\sqrt{1 \\cdot 1 + 2 \\cdot 2 + 3 \\cdot 3 + 4 \\cdot 4} = \\sqrt{1 + 4 + 9 + 16} = \\sqrt{30} \\end{align} Let's now look at some important properties regarding the Euclidean norm of a point $\\mathbf{x}$. Theorem 1: If $\\mathbf{x} = (x_1, x_2, ..., x_n) \\in \\mathbb{R}^n$ and $a \\in \\mathbb{R}$ then: a) $\\| \\mathbf{x} \\| \\geq 0$. b) $\\| a \\mathbf{x} \\| = \\mid a \\mid \\| \\mathbf{x} \\|$. c) $\\| \\mathbf{x} \\|^2 = \\mathbf{x} \\cdot \\mathbf{x}$. • Proof of a) Let $\\mathbf{x} \\in \\mathbb{R}^n$. • We note that $\\mathbf{x} \\cdot \\mathbf{x} = \\sum_{i=1}^{n} x_ix_i = \\sum_{i=1}^{n} x_i^2$. Since $x_i^2 \\geq 0$ for all $i \\in \\{1, 2, ..., n \\}$ we have that $\\mathbf{x} \\cdot \\mathbf{x} \\geq 0$. So: (3) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} \\geq 0 \\end{align} • Proof of b) Let $\\mathbf{x} \\in \\mathbb{R}^n$ and let $a \\in \\mathbb{R}$. Then: (4) \\begin{align} \\quad \\| a \\mathbf{x} \\| = \\sqrt{ (a \\mathbf{x} \\cdot a \\mathbf{x} } = \\sqrt{a^2 (\\mathbf{x} \\cdot \\mathbf{x})} = \\mid a \\mid \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\mid a \\mid \\| \\mathbf{x} \\| \\quad \\blacksquare \\end{align} • Proof of c) Let $\\mathbf{x} \\in \\mathbb{R}^n$. Then: (5) \\begin{align} \\quad \\| \\mathbf{x} \\|^2 = \\left", "+ (p^2)^2 + (p^3)^2 \\\\ &= -(\\frac{E}{c})^2 + (p_x)^2 + (p_y)^2 + (p_z)^2 \\\\ \\end{aligned} } \\displaystyle{ \\begin{aligned} \\| p \\| &= \\sqrt{p \\cdot p} &= \\sqrt{- \\left(\\frac{E}{c}\\right)^2 + (p_x)^2 + (p_y)^2 + (p_z)^2} \\\\ \\end{aligned} } \\displaystyle{ \\begin{aligned} \\left(\\frac{E}{c}\\right)^2 &= (p_x)^2 + (p_y)^2 + (p_z)^2 - p \\cdot p \\\\ \\left(\\frac{E}{c}\\right)^2 &= |\\mathbf {p} |^{2} - p \\cdot p \\\\ \\end{aligned} } . \\displaystyle{ \\begin{aligned} p \\cdot p &= -m^{2}c^{2} \\\\ \\end{aligned} } \\displaystyle{ \\begin{aligned} \\frac{E^2}{c^2} &= \\mathbf p \\cdot \\mathbf p + m^2 c^2 \\\\ \\end{aligned}} . \\displaystyle{ \\begin{aligned} \\frac{E^2}{c^2} &= \\mathbf p \\cdot \\mathbf p + m^2 c^2 \\\\ \\\\ \\frac{E}{c} &= \\sqrt{|\\mathbf p |^2 + m^2 c^2} \\\\ \\end{aligned}} . \\displaystyle{ \\begin{aligned} \\frac{E^2}{c^2} &= \\mathbf p \\cdot \\mathbf p + m^2 c^2 \\\\ \\end{aligned}} \\displaystyle{ \\begin{aligned} \\frac{E}{c} &= \\dotsb + \\dotsb \\\\ \\end{aligned}} . \\displaystyle{ \\begin{aligned} \\frac{E^2}{c^2} &= \\mathbf p \\cdot \\mathbf p + m^2 c^2 \\\\ &= 0 + m^2 c^2 \\\\ \\end{aligned}} \\displaystyle{ \\begin{aligned} \\frac{E^2}{c^2} &= m^2 c^2 \\\\ \\\\ E &= m c^2 \\\\ \\end{aligned}} . \\displaystyle{ \\begin{aligned} E_r &= m c^2 \\\\ \\end{aligned}} \\displaystyle{ \\begin{aligned} \\frac{E}{c} &= \\sqrt{|\\mathbf p |^2 + m^2 c^2} \\\\ \\end{aligned}} \\displaystyle{ \\begin{aligned} \\frac{E}{c} &= \\sqrt{p^2 + m^2 c^2} \\\\ \\\\ E &= c \\sqrt{p^2 + m^2 c^2} \\\\ \\end{aligned}} \\displaystyle{ \\begin{aligned} E_k &= E -", "i\\mathbf{q}\\cdot\\mathbf{r}}\\right) = \\frac{\\hbar}{2m}e^{+ i\\mathbf{q}\\cdot\\mathbf{r}}\\mathbf{q}\\cdot\\left(2\\mathbf{p}+\\hbar\\mathbf{q}\\right)$$ and $$[\\mathcal{H},e^{- i\\mathbf{q}\\cdot\\mathbf{r}}]= -\\frac{\\hbar}{2m}\\mathbf{q}\\cdot\\left(e^{- i\\mathbf{q}\\cdot\\mathbf{r}}\\mathbf{p}+\\mathbf{p}e^{- i\\mathbf{q}\\cdot\\mathbf{r}}\\right) =- \\frac{\\hbar}{2m}\\mathbf{q}\\cdot\\left(2\\mathbf{p}+\\hbar\\mathbf{q}\\right)e^{- i\\mathbf{q}\\cdot\\mathbf{r}}.$$ Putting either of these into the corresponding formula above, you get $$\\sum_n (\\mathcal{E_n}-\\mathcal{E_s})|\\langle n|e^{i\\mathbf{q}\\cdot\\mathbf{r}}|s \\rangle|^2 = \\frac{\\hbar}{2m}\\langle s|\\mathbf{q}\\cdot\\left(2\\mathbf{p}+\\hbar\\mathbf{q}\\right)|s \\rangle = \\frac{\\hbar^2}{2m}\\mathbf{q}^2+\\frac{\\hbar}{m}\\langle s|\\mathbf{q}\\cdot\\mathbf{p}|s \\rangle.$$ You can then always force the mean value of $\\mathbf{p}$ to vanish. - I) Let us work in the Schrödinger picture. Assume that the Hamiltonian operator $$\\tag{1} \\hat{H}~=~T(\\hat{\\bf p})+ V(\\hat{\\bf r}), \\qquad T(\\hat{\\bf p})~:=~ \\frac{\\hat{\\bf p}^2}{2m},$$ does not depend explicit on time. In particular, the point spectrum of energy eigenvalues $E_n$ does not depend on time. Let $\\psi_n({\\bf r})$ be the corresponding energy eigen-functions in position space, which satisfies the TISE $$\\tag{2} \\hat{H}\\psi_n({\\bf r})~=~E_n \\psi_n({\\bf r}).$$ We may assume that $\\psi_n({\\bf r})\\in{\\mathbb{R}}$ is a real function, see also this Phys.SE post. Let us consider a fixed time $t_0$. Let us pick a basis of time-dependent solutions $$\\tag{3} \\langle n, t |{\\bf r} \\rangle~=~ \\Psi_n({\\bf r},t)~=~ \\psi_n({\\bf r})e^{-\\frac{i}{\\hbar}E_n (t-t_0)}$$ that becomes real at the time $t=t_0$. It follows that for any (possibly complex) function $F=F({\\bf r})$ of position ${\\bf r}$, that the matrix element $$\\tag{4} \\langle n, t_0 |F(\\hat{\\bf r}) | m, t_0 \\rangle ~=~ \\int \\! d^3r ~ \\psi^{*}_n({\\bf r}) ~F({\\bf r})~\\psi_m({\\bf r})~=~ n \\leftrightarrow m$$ is symmetric in $n\\leftrightarrow m$ at $t=t_0$. The time dependence of the matrix element is just", "of this is 2 + 2\\cdot 3 + 2\\cdot 3^2 + 2\\cdot 3^3, so you normally you would form the array (\\mathbf{2'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{1'}) +' (\\mathbf{2'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}\\cdot'\\mathbf{3'}), which has length (2+1+1+1)+(2+3+1+1)+(2+3+3+1)+(2+3+3+3) = 32. However, writing the ternary form as 2 + 3\\cdot(2 + 3\\cdot (2 + 3\\cdot 2)), the array would be (\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'[(\\mathbf{2'}\\cdot' \\mathbf{1'}\\cdot' \\mathbf{1'}) +' \\mathbf{3'}\\cdot'(\\mathbf{2'}\\cdot'\\mathbf{1'} +' \\mathbf{3'}\\cdot'\\mathbf{2'})], which only has length (2+1+1+1)+3+(2+1+1)+3+(2+1)+(3+2) = 23. In general, the smaller the total numbers that appear on your representation, the better. Using this technique, we are now able to reduce the required length to at most 94, and we’ve now solved the problem! The ideas above are enough to pass the problem already, but I’m sure with a couple more tricks you can reduce the length even more. ### Time Complexity: [O(N)](https://en.wikipedia.org/wiki/Output-sensitive _ algorithm) ### AUTHOR’S AND TESTER’S SOLUTIONS: #2 “I loved solving this question.” The constraint is very restrictive, just use numbers very efficiently. So, I tried to make a LIS’s of length equal to MSB of K (for K = 4, length = 3). Now you need exactly K such LIS’s, so now try to create LIS’s in powers of two starting from the least power of 2 present, if power of 2 p, then create p pairs of numbers so", "+ 2 \\hbar \\mathbf{k} \\cdot \\mathbf{p}).\\end{aligned} \\hspace{\\stretch{1}}(3.21) For the double commutator we have \\begin{aligned}2m \\left[{\\left[{H},{e^{i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]},{e^{-i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]&=e^{i \\mathbf{k} \\cdot \\mathbf{r}} ((\\hbar \\mathbf{k})^2 + 2 \\hbar \\mathbf{k} \\cdot \\mathbf{p}) e^{-i \\mathbf{k} \\cdot \\mathbf{r}}-((\\hbar \\mathbf{k})^2 + 2 \\hbar \\mathbf{k} \\cdot \\mathbf{p}) \\\\ &=e^{i \\mathbf{k} \\cdot \\mathbf{r}} 2 (\\hbar \\mathbf{k} \\cdot \\mathbf{p}) e^{-i \\mathbf{k} \\cdot \\mathbf{r}}-2 \\hbar \\mathbf{k} \\cdot \\mathbf{p} \\\\ &=2 \\hbar \\mathbf{k} \\cdot (\\mathbf{p} - \\hbar \\mathbf{k})-2 \\hbar \\mathbf{k} \\cdot \\mathbf{p},\\end{aligned} so for the double commutator we have just a scalar \\begin{aligned}\\left[{\\left[{H},{e^{i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]},{e^{-i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]&= -\\frac{(\\hbar \\mathbf{k})^2}{m}.\\end{aligned} \\hspace{\\stretch{1}}(3.22) Now consider the expectation of this double commutator, expanded with some unintuitive steps that have been motivated by working backwards \\begin{aligned}{\\langle {s} \\rvert} \\left[{\\left[{H},{e^{i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]},{e^{-i \\mathbf{k} \\cdot \\mathbf{r}}}\\right] {\\lvert {s} \\rangle}&={\\langle {s} \\rvert} 2 H - e^{i \\mathbf{k} \\cdot \\mathbf{r}} H e^{-i \\mathbf{k} \\cdot \\mathbf{r}} - e^{-i \\mathbf{k} \\cdot \\mathbf{r}} H e^{i \\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {s} \\rangle} \\\\ &={\\langle {s} \\rvert} 2 e^{-i \\mathbf{k} \\cdot \\mathbf{r}} e^{i \\mathbf{k} \\cdot \\mathbf{r}} H- 2 e^{-i \\mathbf{k} \\cdot \\mathbf{r}} H e^{i \\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {s} \\rangle} \\\\ &=2 \\sum_n {\\langle {s} \\rvert} e^{-i \\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {n} \\rangle} {\\langle {n} \\rvert} e^{i \\mathbf{k} \\cdot \\mathbf{r}} H {\\lvert {s} \\rangle}- {\\langle {s} \\rvert} e^{-i \\mathbf{k} \\cdot \\mathbf{r}} H {\\lvert {n} \\rangle} {\\langle {n}", "of $$X$$, $$f_{X}(x)$$, is then \\begin{align*} f_{X}(0) &= \\mathbf{P}(X = 0) \\\\ &= \\mathbf{P}(\\mathrm{Hit\\:E})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &= 81/625 \\\\ f_{X}(1) &= \\mathbf{P}(X = 1) \\\\ &= 2 \\cdot \\mathbf{P}(\\mathrm{Hit\\:D})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &= 126/625 \\\\ f_{X}(2) &= \\mathbf{P}(X = 2) \\\\ &= \\begin{align*}& 2 \\cdot \\mathbf{P}(\\mathrm{Hit\\:C})\\mathbf{P}(\\mathrm{Hit\\:E}) \\\\ &+ \\mathbf{P}(\\mathrm{Hit\\:D})\\mathbf{P}(\\mathrm{Hit\\:D}) \\end{align*} \\\\ &= 139/625 \\end{align*} and so on. From (2), the expected value of $$X$$ is $$\\mathbf{E}(X) = 0.81/625 + 1 \\cdot 126/625 + 2 \\cdot 139/625 + 3 \\cdot 124/625 + ... = 2.4$$. From (4), the variance of $$X$$ is $$\\mathrm{Var}(X) = 2.4^{2} \\cdot 81/625 + 1.4^{2} \\cdot 126/625 + 0.4^{2} \\cdot 139/625 + 0.6^{2} \\cdot 124/625 + ... = 2.72$$. The distribution function of $$X$$, $$F_{X}(x)$$, is then \\begin{align*} F_{X}(0) &= \\mathbf{P}(X \\le 0) \\\\ &= f_{X}(0) \\\\ F_{X}(1) &= \\mathbf{P}(X \\le 1) \\\\ &= f_{X}(1) + f_{X}(0) \\\\ F_{X}(2) &= \\mathbf{P}(X \\le 2) \\\\ &= f_{X}(2) + f_{X}(1) + f_{X}(0) \\end{align*} and so on. The distribution function $$F_{X}(x)$$ is shown in Figure 4. Figure 4: The distribution function $$F_{X}$$ of $$X$$ for the archery target.", "on both ${\\mathbf{r}}$ and ${\\mathbf{r}'}$. We begin by writing $\\displaystyle \\nabla\\cdot\\left(\\frac{\\mathbf{J}}{d}\\right)=\\frac{1}{d}\\nabla\\cdot\\mathbf{J}+\\mathbf{J}\\cdot\\nabla\\left(\\frac{1}{d}\\right) \\ \\ \\ \\ \\ (11)$ We can also use the derivative with respect to ${\\mathbf{r}'}$: $\\displaystyle \\nabla'\\cdot\\left(\\frac{\\mathbf{J}}{d}\\right)=\\frac{1}{d}\\nabla'\\cdot\\mathbf{J}+\\mathbf{J}\\cdot\\nabla'\\left(\\frac{1}{d}\\right) \\ \\ \\ \\ \\ (12)$ We have (you can work this out by using 5 if you don’t believe me): $\\displaystyle \\nabla\\left(\\frac{1}{d}\\right)=-\\frac{\\hat{\\mathbf{d}}}{d^{2}}=-\\nabla'\\left(\\frac{1}{d}\\right) \\ \\ \\ \\ \\ (13)$ Therefore $\\displaystyle \\nabla'\\cdot\\left(\\frac{\\mathbf{J}}{d}\\right)$ $\\displaystyle =$ $\\displaystyle \\frac{1}{d}\\nabla'\\cdot\\mathbf{J}-\\mathbf{J}\\cdot\\nabla\\left(\\frac{1}{d}\\right)\\ \\ \\ \\ \\ (14)$ $\\displaystyle \\mathbf{J}\\cdot\\nabla\\left(\\frac{1}{d}\\right)$ $\\displaystyle =$ $\\displaystyle \\frac{1}{d}\\nabla'\\cdot\\mathbf{J}-\\nabla'\\cdot\\left(\\frac{\\mathbf{J}}{d}\\right) \\ \\ \\ \\ \\ (15)$ Inserting this into 11 we get $\\displaystyle \\nabla\\cdot\\left(\\frac{\\mathbf{J}}{d}\\right)=\\frac{1}{d}\\nabla\\cdot\\mathbf{J}+\\frac{1}{d}\\nabla'\\cdot\\mathbf{J}-\\nabla'\\cdot\\left(\\frac{\\mathbf{J}}{d}\\right) \\ \\ \\ \\ \\ (16)$ Now we need to work out the two divergences ${\\nabla\\cdot\\mathbf{J}}$ and ${\\nabla\\cdot\\mathbf{J}'}$. To do this, we need to remember that ${\\mathbf{J}=\\mathbf{J}\\left(\\mathbf{r}',t_{r}\\right)}$, so it depends on ${\\mathbf{r}}$ only via ${t_{r}}$ but it depends on ${\\mathbf{r}'}$ both explicitly through its first argument and implicitly through ${t_{r}}$. Using the chain rule, we get for the contribution from ${x}$ $\\displaystyle \\left(\\nabla\\cdot\\mathbf{J}\\right)_{x}$ $\\displaystyle =$ $\\displaystyle \\frac{\\partial\\mathbf{J}_{x}}{\\partial t_{r}}\\frac{\\partial t_{r}}{\\partial x}\\ \\ \\ \\ \\ (17)$ $\\displaystyle$ $\\displaystyle =$ $\\displaystyle -\\frac{1}{c}\\dot{\\mathbf{J}}_{x}\\frac{\\partial d}{\\partial x}\\ \\ \\ \\ \\ (18)$ $\\displaystyle$ $\\displaystyle =$ $\\displaystyle -\\frac{1}{c}\\dot{\\mathbf{J}}_{x}\\left(\\nabla d\\right)_{x} \\ \\ \\ \\ \\ (19)$ where the dot over the ${\\mathbf{J}}$ is a derivative with respect to ${t}$, which is the same as a derivative with respect to ${t_{r}}$ since ${t_{r}=t-d/c}$ and ${d}$ doesn’t", "B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot \\partial_2 B \\right) \\\\ &\\qquad - \\mathbf{x}_2 \\left( \\mathbf{x}_{34} \\cdot \\partial_1 B \\right) - \\mathbf{x}_3 \\left( \\mathbf{x}_{42} \\cdot \\partial_1 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{23} \\cdot \\partial_1 B \\right)} \\Bigr) \\\\", "\\mathbf{x}_3 \\cdot B} \\right) } \\right) - \\partial_2 \\left( { \\mathbf{x}_1 \\cdot \\left( { \\mathbf{x}_3 \\cdot B} \\right) } \\right)} \\Bigr) \\\\ &=\\frac{1}{2} d^3 u\\Bigl( {\\partial_2 B_{13} - \\partial_3 B_{12}+\\partial_3 B_{21} - \\partial_1 B_{23}+\\partial_1 B_{32} - \\partial_2 B_{31}} \\Bigr) \\\\ &=d^3 u\\Bigl( {\\partial_2 B_{13}+\\partial_3 B_{21}+\\partial_1 B_{32}} \\Bigr) \\\\ &= - \\frac{1}{2} d^3 u \\epsilon^{abc} \\partial_a B_{bc}. \\qquad\\square\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.152) ### (e) Four parameter volume, curl of bivector To start, we require lemma 3. For convenience lets also write our wedge products as a single indexed quantity, as in $\\mathbf{x}_{abc}$ for $\\mathbf{x}_a \\wedge \\mathbf{x}_b \\wedge \\mathbf{x}_c$. The expansion is \\begin{aligned}\\begin{aligned}d^4 \\mathbf{x} \\cdot \\left( \\boldsymbol{\\partial} \\wedge B \\right) &= d^4 u \\left( \\mathbf{x}_{1234} \\cdot \\mathbf{x}^i \\right) \\cdot \\partial_i B \\\\ &= d^4 u\\left( \\mathbf{x}_{123} \\cdot \\partial_4 B - \\mathbf{x}_{124} \\cdot \\partial_3 B + \\mathbf{x}_{134} \\cdot \\partial_2 B - \\mathbf{x}_{234} \\cdot \\partial_1 B \\right) \\\\ &= d^4 u \\Bigl( \\mathbf{x}_1 \\left( \\mathbf{x}_{23} \\cdot \\partial_4 B \\right) + \\mathbf{x}_2 \\left( \\mathbf{x}_{32} \\cdot \\partial_4 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{12} \\cdot \\partial_4 B \\right) \\\\ &\\qquad - \\mathbf{x}_1 \\left( \\mathbf{x}_{24} \\cdot \\partial_3 B \\right) - \\mathbf{x}_2 \\left( \\mathbf{x}_{41} \\cdot \\partial_3 B \\right) - \\mathbf{x}_4 \\left( \\mathbf{x}_{12} \\cdot \\partial_3 B \\right) \\\\ &\\qquad + \\mathbf{x}_1 \\left( \\mathbf{x}_{34} \\cdot \\partial_2 B \\right) + \\mathbf{x}_3 \\left( \\mathbf{x}_{41} \\cdot \\partial_2 B \\right) + \\mathbf{x}_4 \\left( \\mathbf{x}_{13} \\cdot \\partial_2", "adjoint action instead. Computing this action here, one can show that for $\\alpha \\rtimes \\beta$, we have \\begin{align*} \\widehat{\\phi_{p^{2k}}}(\\mathbf{u}) &= \\alpha \\cdot \\widehat{\\phi_{p^{2k}}}(\\mathbf{u}) = \\widehat{\\phi_{p^{2k}}}(\\alpha^T \\cdot \\mathbf{u}) \\\\ \\widehat{\\phi_{p^{2k}}}(\\mathbf{u}) &= \\beta \\cdot \\widehat{\\phi_{p^{2k}}}(\\mathbf{u}) = \\widehat{\\phi_{p^{2k}}}(\\beta^T \\cdot \\mathbf{u}) \\exp \\left( \\frac{2\\pi i \\langle \\mathbf{b}(\\beta), \\mathbf{u} \\rangle}{p^{2k}} \\right), \\end{align*} where \\begin{align*} \\alpha^T \\cdot \\mathbf{u} &= (\\alpha u_1, \\ldots, \\alpha^n u_n ) \\\\ \\beta^T \\cdot \\mathbf{u} &= \\left(\\binom{n}{1} \\beta, \\ldots, \\binom{n}{n}\\beta^n \\right) \\end{align*} and \\begin{align*} \\mathbf{b}(\\beta) = \\bigg( &u_1 + \\binom{n-1}{1} \\beta u_2 + \\cdots + \\binom{n-1}{n-1} \\beta^{n-1}u_n, \\\\ &u_2 + \\binom{n-2}{1} \\beta u_3 + \\cdots + \\binom{n-2}{n-2} \\beta^{n-2}u_n, \\\\ &\\ldots, \\\\ &u_n \\bigg). \\end{align*} For $m$ as in the statement of the lemma, i.e. $m$ is the smallest index such that $u_m \\neq 0$ and $m \\geq 2$, then we study the effect of $(\\alpha, \\beta)^T$ solely on $u_m$ and $u_{m-1}$. A simpler computation shows that \\begin{equation*} (\\alpha, \\beta)^T \\cdot \\mathbf{u} = (\\ldots, \\alpha^{m-1} u_{m-1} + (n - m + 1) \\beta \\alpha^m u_m, \\alpha^m u_m, \\ldots). \\end{equation*} The valuation of the $m$th coordinate is preserved, and there are $\\gg p^{2k - v_p(u_m)}$ different values $u_m$ produced by these group translations. For each fixed $u_m$, we count choices of $u_{m-1}$. We note that keeping $u_m$ fixed doesn't necessarily fix $\\alpha$ — instead it forces $\\alpha \\equiv \\zeta_m \\bmod p^{2k -", "1 - 1\\cdot (-1)] - \\mathbf{j}[2 \\cdot 1 - 1 \\cdot 3] + \\mathbf{k}[2 \\cdot (-1) - 2 \\cdot 3]$ $= 3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}$. I agree with the answer you have given for $\\mathbf{v} \\times \\mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect. Anyway, to find the unit vector in the direction of $\\mathbf{v} \\times \\mathbf{w}$, divide it by its length. $|\\mathbf{v}\\times\\mathbf{w}| = \\sqrt{3^2 + (-1)^2 + (-8)^2}$ $= \\sqrt{9 + 1 + 64}$ $= \\sqrt{74}$. Therefore: $\\frac{\\mathbf{v}\\times\\mathbf{w}}{|\\mathbf{v}\\time s\\mathbf{w}|} = \\frac{3\\mathbf{i} - \\mathbf{j} - 8\\mathbf{k}}{\\sqrt{74}}$ $= \\frac{3}{\\sqrt{74}}\\mathbf{i} - \\frac{1}{\\sqrt{74}}\\mathbf{j} - \\frac{8}{\\sqrt{74}}\\mathbf{k}$ $= \\frac{3\\sqrt{74}}{74}\\mathbf{i} - \\frac{\\sqrt{74}}{74}\\mathbf{j} - \\frac{4\\sqrt{74}}{37}\\mathbf{k}$. 3. Hello, Paymemoney! You've done nothing wrong . . . Their answer is the negative of yours. There are two vectors perpendicular to both $\\vec v\\text{ and }\\vec w,$ . . one \"up\" and one \"down\". They picked one, you picked the other. 4. ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k", "\\begin{aligned}\\mathbf{a}^{*} \\cdot \\mathbf{a} = 1,\\end{aligned} \\hspace{\\stretch{1}}(1.174) and \\begin{aligned}\\mathbf{a}^{*} \\cdot \\mathbf{b} = 0.\\end{aligned} \\hspace{\\stretch{1}}(1.175) \\begin{aligned}\\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right)=\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}.\\end{aligned} \\hspace{\\stretch{1}}(1.176) Dotting with $\\mathbf{a}$ we have \\begin{aligned}\\mathbf{a} \\cdot \\left( { \\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right) } \\right)=\\mathbf{a} \\cdot \\left( {\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}} \\right)=\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right)^2 - \\mathbf{b}^2 \\mathbf{a}^2,\\end{aligned} \\hspace{\\stretch{1}}(1.177) but dotting with $\\mathbf{b}$ yields zero \\begin{aligned}\\mathbf{b} \\cdot \\left( { \\mathbf{b} \\cdot \\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right) } \\right) &= \\mathbf{b} \\cdot \\left( {\\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}-\\mathbf{b}^2 \\mathbf{a}} \\right) \\\\ &= \\left( { \\mathbf{b} \\cdot \\mathbf{a} } \\right) \\mathbf{b}^2 - \\mathbf{b}^2 \\left( { \\mathbf{a} \\cdot \\mathbf{b} } \\right) \\\\ &= 0.\\end{aligned} \\hspace{\\stretch{1}}(1.178) To complete the proof, we note that the product in eq. 1.177 is just the wedge squared \\begin{aligned}\\left( { \\mathbf{a} \\wedge \\mathbf{b}} \\right)^2 &= \\left\\langle{{\\left( { \\mathbf{a} \\wedge \\mathbf{b} } \\right)^2}}\\right\\rangle \\\\ &= \\left\\langle{{\\left( { \\mathbf{a} \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} } \\right)\\left( { \\mathbf{a} \\mathbf{b} - \\mathbf{a} \\cdot \\mathbf{b} } \\right)}}\\right\\rangle \\\\ &= \\left\\langle{{\\mathbf{a} \\mathbf{b} \\mathbf{a} \\mathbf{b} - 2 \\left( {\\mathbf{a} \\cdot \\mathbf{b}} \\right) \\mathbf{a} \\mathbf{b}}}\\right\\rangle+\\left( { \\mathbf{a} \\cdot \\mathbf{b} } \\right)^2 \\\\ &= \\left\\langle{{\\mathbf{a} \\mathbf{b} \\left( { -\\mathbf{b} \\mathbf{a} + 2 \\mathbf{a} \\cdot \\mathbf{b} }", "${\\displaystyle w=p\\mathbf {e} _{1}\\cdot \\mathbf {n} _{3}+q\\mathbf {e} _{2}\\cdot \\mathbf {n} _{3}+r\\mathbf {e} _{3}\\cdot \\mathbf {n} _{3}}$. ${\\displaystyle u=c_{11}p+c_{12}q+c_{13}r}$, ${\\displaystyle v=c_{21}p+c_{22}q+c_{23}r}$, ${\\displaystyle w=c_{31}p+c_{32}q+c_{33}r}$, ${\\displaystyle {\\begin{bmatrix}u\\\\v\\\\w\\\\\\end{bmatrix}}={\\begin{bmatrix}c_{11}&c_{12}&c_{13}\\\\c_{21}&c_{22}&c_{23}\\\\c_{31}&c_{32}&c_{33}\\end{bmatrix}}{\\begin{bmatrix}p\\\\q\\\\r\\end{bmatrix}}}$. ${\\displaystyle c_{11}=\\mathbf {n} _{1}\\cdot \\mathbf {e} _{1}}$ ${\\displaystyle c_{12}=\\mathbf {n} _{1}\\cdot \\mathbf {e} _{2}}$ ${\\displaystyle c_{13}=\\mathbf {n} _{1}\\cdot \\mathbf {e} _{3}}$ ${\\displaystyle c_{21}=\\mathbf {n} _{2}\\cdot \\mathbf {e} _{1}}$ ${\\displaystyle c_{22}=\\mathbf {n} _{2}\\cdot \\mathbf {e} _{2}}$ ${\\displaystyle c_{23}=\\mathbf {n} _{2}\\cdot \\mathbf {e} _{3}}$ ${\\displaystyle c_{31}=\\mathbf {n} _{3}\\cdot \\mathbf {e} _{1}}$ ${\\displaystyle c_{32}=\\mathbf {n} _{3}\\cdot \\mathbf {e} _{2}}$ ${\\displaystyle c_{33}=\\mathbf {n} _{3}\\cdot \\mathbf {e} _{3}}$ ${\\displaystyle {\\mathbf {x} }=x_{1}{\\mathbf {e} }_{1}+x_{2}{\\mathbf {e} }_{2}+x_{3}{\\mathbf {e} }_{3}.}$ ${\\displaystyle {\\mathbf {y} }-{\\mathbf {x} }=(y_{1}-x_{1}){\\mathbf {e} }_{1}+(y_{2}-x_{2}){\\mathbf {e} }_{2}+(y_{3}-x_{3}){\\mathbf {e} }_{3}.}$ ${\\displaystyle {\\mathbf {x} }_{t}=t{\\mathbf {v} }+{\\mathbf {x} }_{0},}$ ${\\displaystyle {\\mathbf {F} }=m{\\mathbf {a} }}$ ${\\displaystyle E={\\mathbf {F} }\\cdot ({\\mathbf {x} }_{2}-{\\mathbf {x} }_{1}).}$ ${\\displaystyle {\\frac {df}{d\\tau }}=\\sum _{\\alpha =1}^{n}{\\frac {dx^{\\alpha }}{d\\tau }}{\\frac {\\partial f}{\\partial x^{\\alpha }}}.}$ ${\\displaystyle t^{\\alpha }={\\frac {dx^{\\alpha }}{d\\tau }}.}$ ${\\displaystyle {\\frac {d}{d\\tau }}=\\sum _{\\alpha }t^{\\alpha }{\\frac {\\partial }{\\partial x^{\\alpha }}}.}$ ${\\displaystyle \\mathbf {a} \\equiv a^{\\alpha }{\\frac {\\partial }{\\partial x^{\\alpha }}}.}$ Wikipedia", "{\\displaystyle {\\begin{aligned}\\int {\\frac {d^{3}k}{(2\\pi )^{3}}}\\mathbf {\\hat {k}} \\mathbf {\\hat {k}} {\\frac {\\exp \\left(i\\mathbf {k} \\cdot \\mathbf {r} \\right)}{k^{2}+m^{2}}}&=\\int {\\frac {d^{3}k}{(2\\pi )^{3}}}\\left[\\left(\\mathbf {\\hat {k}} \\cdot \\mathbf {\\hat {r}} \\right)^{2}\\mathbf {\\hat {r}} \\mathbf {\\hat {r}} +\\left(\\mathbf {\\hat {k}} \\cdot \\mathbf {\\hat {\\theta }} \\right)^{2}\\mathbf {\\hat {\\theta }} \\mathbf {\\hat {\\theta }} +\\left(\\mathbf {\\hat {k}} \\cdot \\mathbf {\\hat {\\phi }} \\right)^{2}\\mathbf {\\hat {\\phi }} \\mathbf {\\hat {\\phi }} \\right]{\\frac {\\exp \\left(i\\mathbf {k} \\cdot \\mathbf {r} \\right)}{k^{2}+m^{2}}}\\\\&={\\frac {e^{-mr}}{4\\pi r}}\\left\\{1+{\\frac {2}{mr}}-{2 \\over (mr)^{2}}\\left(e^{mr}-1\\right)\\right\\}\\left\\{\\mathbf {1} -{1 \\over 2}\\left[\\mathbf {1} -\\mathbf {\\hat {r}} \\mathbf {\\hat {r}} \\right]\\right\\}+\\int _{0}^{\\infty }{\\frac {k^{2}dk}{(2\\pi )^{2}}}\\int _{-1}^{1}du{\\frac {e^{ikru}}{k^{2}+m^{2}}}{1 \\over 2}\\left[\\mathbf {1} -\\mathbf {\\hat {r}} \\mathbf {\\hat {r}} \\right]\\\\&={1 \\over 2}{\\frac {e^{-mr}}{4\\pi r}}\\left[\\mathbf {1} -\\mathbf {\\hat {r}} \\mathbf {\\hat {r}} \\right]+{e^{-mr} \\over 4\\pi r}\\left\\{1+{\\frac {2}{mr}}-{2 \\over (mr)^{2}}\\left(e^{mr}-1\\right)\\right\\}\\left\\{{1 \\over 2}\\left[\\mathbf {1} +\\mathbf {\\hat {r}} \\mathbf {\\hat {r}} \\right]\\right\\}\\\\&={1 \\over 2}{\\frac {e^{-mr}}{4\\pi r}}\\left(\\left[\\mathbf {1} -\\mathbf {\\hat {r}} \\mathbf {\\hat {r}} \\right]+\\left\\{1+{\\frac {2}{mr}}-{2 \\over (mr)^{2}}\\left(e^{mr}-1\\right)\\right\\}\\left[\\mathbf {1} +\\mathbf {\\hat {r}} \\mathbf {\\hat {r}} \\right]\\right)\\end{aligned}}} Note that in the small m limit the integral reduces to ${\\displaystyle {1 \\over 2}{1 \\over 4\\pi r}\\left[\\mathbf {1} -\\mathbf {\\hat {r}} \\mathbf {\\hat {r}} \\right].}$ #### Transverse potential with mass ${\\displaystyle \\int {\\frac {d^{3}k}{(2\\pi )^{3}}}\\left[\\mathbf {1} -\\mathbf {\\hat {k}} \\mathbf {\\hat {k}} \\right]{\\exp \\left(i\\mathbf {k} \\cdot \\mathbf {r} \\right) \\over k^{2}+m^{2}}={1 \\over 2}{e^{-mr} \\over 4\\pi r}\\left\\{{2 \\over (mr)^{2}}\\left(e^{mr}-1\\right)-{2 \\over mr}\\right\\}\\left[\\mathbf" ]

[ "${\\displaystyle \\mathbf {x} _{+}}$, such as ${\\displaystyle \\mathbf {x} _{2}}$, due to commuting or shifting ${\\displaystyle \\mathbf {a} _{k}}$ right an odd number of times, in the dividend blade ${\\displaystyle \\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n}}$, into its Template:Mvar-th spot. In general, ${\\displaystyle \\mathbf {x} _{k}}$ is solved as {\\displaystyle {\\begin{aligned}\\mathbf {x} _{k}&=(\\mathbf {a} _{1}\\wedge \\cdots \\wedge (\\mathbf {C} )_{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})\\cdot ((-1)^{k-1}\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})^{-1}\\\\[6pt]&={\\frac {(\\mathbf {a} _{1}\\wedge \\cdots \\wedge (\\mathbf {C} )_{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})\\cdot ((-1)^{k-1}\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})}{(\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})^{2}}}\\\\[6pt]&={\\frac {(-1)^{k-1}(\\mathbf {a} _{1}\\wedge \\cdots \\wedge (\\mathbf {C} )_{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})\\cdot (\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})}{(-1)^{\\frac {n(n-1)}{2}}(\\mathbf {a} _{n}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{1})\\cdot (\\mathbf {a} _{1}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})}}\\\\[6pt]&={\\frac {(-1)^{k-1}(\\mathbf {a} _{1}\\wedge \\cdots \\wedge (\\mathbf {C} )_{k}\\wedge \\cdots \\wedge \\mathbf {a} _{n})\\cdot (\\mathbf {a} _{n}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{1})}{(\\mathbf {a} _{n}\\wedge \\cdots \\wedge \\mathbf {a} _{k}\\wedge \\cdots \\wedge \\mathbf {a} _{1})\\cdot (\\mathbf {a} _{1}\\wedge \\cdots \\wedge", "\\end{align*} Since point $A = (0,0,0), PA = 16$, and since we scaled the answer is $16 \\cdot 12 = \\boxed{192}$. ~Aaryabhatta1 ## Solution 3 Let $E$ be the vertex of the cube such that $ABED$ is a square. By the British Flag Theorem, we can easily we can show that $$PA^2 + PE^2 = PB^2 + PD^2$$ and $$PA^2 + PG^2 = PC^2 + PE^2$$ Hence, adding the two equations together, we get $2PA^2 + PG^2 = PB^2 + PC^2 + PD^2$. Substituting in the values we know, we get $2PA^2 + 7\\cdot 36^2 =10\\cdot60^2 + 5\\cdot 60^2 + 2\\cdot 120^2$. Thus, we can solve for $PA$, which ends up being $\\boxed{192}$. (Lokman GÖKÇE) ## Solution 4 For all points $X$ in space, define the function $f:\\mathbb{R}^{3}\\rightarrow\\mathbb{R}$ by $f(X)=PX^{2}-GX^{2}$. Then $f$ is linear; let $O=\\frac{2A+G}{3}$ be the center of $\\triangle BCD$. Then since $f$ is linear, \\begin{align*} 3f(O)=f(B)+f(C)+f(D)&=2f(A)+f(G) \\\\ \\left(PB^{2}-GB^{2}\\right)+\\left(PC^{2}-GC^{2}\\right)+\\left(PD^{2}-GD^{2}\\right)&=2\\left(PA^{2}-GA^{2}\\right)+PG^{2} \\\\ \\left(60\\sqrt{10}\\right)^{2}-2x^{2}+\\left(60\\sqrt{5}\\right)^{2}-2x^{2}+\\left(120\\sqrt{2}\\right)^{2}-2x^{2}&=2PA^{2}-2\\cdot 3x^{2}+\\left(36\\sqrt{7}\\right)^{2}, \\end{align*} where $x$ denotes the side length of the cube. Thus \\begin{align*} 36\\text{,}000+18\\text{,}000+28\\text{,}800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\\\ 82\\text{,}800-6x^{2}&=2PA^{2}-6x^{2}+9072 \\\\ 73\\text{,}728&=2PA^{2} \\\\ 36\\text{,}864&=PA^{2} \\\\ PA&=\\boxed{192}. \\end{align*}", "\\wedge d\\mathbf{x}_2 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I} \\right) \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.118) On the left our", "\\wedge d\\mathbf{x}_2 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I} \\right) \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.118) On the left our", "&= I_{1 2 \\cdots (k-1)} I \\left\\lvert {d\\mathbf{x}_1 \\wedge d\\mathbf{x}_2 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I}", "we can write $$D=D^\\prime + |DD^\\prime| \\frac{( C-B )\\times ( B \\times C )}{|BC|\\,|B\\times C|} \\tag{1}$$ where $$D^\\prime := \\frac12(B+C)$$ is the midpoint of $$\\overline{BC}$$. Further, by the Inscribed Angle Theorem, $$\\angle BOC\\cong\\angle BDC$$, and we conclude that $$\\overline{DD^\\prime}$$ is the altitude of an isosceles triangle with vertex angle $$\\alpha$$ and base $$|BC|$$. Therefore, $$|DD^\\prime| = \\frac12|BC|\\,\\cot\\frac12\\alpha$$, so we have $$\\frac{|DD^\\prime|}{|BC|\\,|B\\times C|} = \\frac{\\cot\\frac12\\alpha}{2bc\\sin\\alpha} = \\frac{\\cos\\frac12\\alpha\\,/\\,\\sin\\frac12\\alpha}{4bc\\sin\\frac12\\alpha\\cos\\frac12\\alpha}=\\frac{1}{4bc\\sin^2\\frac12\\alpha} = \\frac{1}{2bc(1-\\cos\\alpha)} \\tag{2}$$ Further, via a cross product identity, \\begin{align} (C-B)\\times(B\\times C) &= \\phantom{-}B\\,((C-B)\\cdot B) - C\\,((C-B)\\cdot C) \\\\ &=\\phantom{-}B\\left(B\\cdot C - |B|^2\\right)-C\\left(|C|^2 - B\\cdot C\\right) \\\\ &=-B\\,b(b-c\\cos\\alpha) - C\\,c(c-b\\cos\\alpha) \\end{align}\\tag{3} Altogether, this gives us \\begin{align}D\\;2bc(1-\\cos\\alpha) &= (B+C)bc(1-\\cos\\alpha)-B\\,b(b-c\\cos\\alpha)-C\\,c(c-b\\cos\\alpha) \\\\ &=Bb(c-b)+Cc(b-c) \\\\ &=(c-b)(B b-C c) \\tag{4a} \\end{align} (Note that, if $$b=c$$, then $$D=0$$, as we would expect. This confirms that we got our cross-product vector directions correct in $$(1)$$.) Likewise, \\begin{align} E\\,2ca(1-\\cos\\beta) &= (a-c)(C c-A a) \\tag{4b} \\\\ F\\,2ab(1-\\cos\\gamma) &= (b-a)(A a-B b) \\tag{4c} \\end{align} Finally, observe that, for $$a$$, $$b$$, $$c$$ not all equal (the only case with which we need concern ourselves), $$(\\text{eq }4a)\\;(a-c)(b-a) + (\\text{eq }4b)\\;(c-b)(b-a)+(\\text{eq }4c)\\;(a-c)(c-b) \\tag{5}$$ gives a non-trivial linear combination of $$D$$, $$E$$, $$F$$ that vanishes. Consequently, $$D$$, $$E$$, $$F$$ are linearly dependent; that is, they lie on a common plane through $$O$$, and the result follows. $$\\square$$ • Pardon my ignorance but I'm confused as to", "d\\mathbf{x}_{k-1} } \\right\\rvert \\\\ &\\quad -I_{1 \\cdots (k-2) k} I \\left\\lvert {d\\mathbf{x}_1 \\wedge \\cdots \\wedge d\\mathbf{x}_{k-2} \\wedge d\\mathbf{x}_k} \\right\\rvert+ \\cdots\\end{aligned}\\end{aligned} \\hspace{\\stretch{1}}(1.0.113) We’ve seen in eq. 1.0.106 and lemma 7 that the dual of vector $\\mathbf{a}$ with respect to the unit pseudoscalar $I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}$ in a subspace spanned by $\\left\\{ {\\mathbf{a}, \\cdots \\mathbf{c}, \\mathbf{d}} \\right\\}$ is \\begin{aligned}\\widehat{\\mathbf{a}^{*}} = I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}} \\frac{1}{{ I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}} }},\\end{aligned} \\hspace{\\stretch{1}}(1.0.114) or \\begin{aligned}\\widehat{\\mathbf{a}^{*}} I_{\\mathbf{a} \\cdots \\mathbf{c} \\mathbf{d}}^2=I_{\\mathbf{b} \\cdots \\mathbf{c} \\mathbf{d}}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.115) This allows us to write \\begin{aligned}d^{k-1} \\mathbf{x} I= I^2 \\sum_i \\widehat{\\mathbf{x}^i} d{A'}_i\\end{aligned} \\hspace{\\stretch{1}}(1.0.116) where $d{A'}_i = \\pm dA_i$, and $dA_i$ is the area of the boundary area element normal to $\\mathbf{x}^i$. Note that the $I^2$ term will now cancel cleanly from both sides of the divergence equation, taking both the metric and the orientation specific dependencies with it. This leaves us with \\begin{aligned}\\int_V dV \\boldsymbol{\\nabla} \\cdot \\mathbf{f} = (-1)^{k+1} \\int_{\\partial V} d{A'}_i \\widehat{\\mathbf{x}^i} \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.0.117) To spell out the details, we have to be very careful with the signs. However, that is a job best left for specific examples. ## Example: 2D divergence theorem Let’s start back at \\begin{aligned}\\int_A \\left\\langle{{ d^2 \\mathbf{x} \\boldsymbol{\\nabla} I \\mathbf{f} }}\\right\\rangle = \\int_{\\partial A} \\left( { d^1 \\mathbf{x} I} \\right) \\cdot \\mathbf{f}.\\end{aligned} \\hspace{\\stretch{1}}(1.118) On the left our integral can be rewritten as", "for $\\mathbf b$, we get \\begin{align} \\mathbf a\\cdot\\mathbf n_1 &= s_1' - (\\mathbf n_2\\cdot\\mathbf n_1) s_2',\\\\ \\mathbf a\\cdot\\mathbf n_2 &= (\\mathbf n_1\\cdot\\mathbf n_2) s_1' - s_2' . \\end{align} Solving for the distances $s_1'$ and $s_2'$ yields our desired expressions: \\begin{align} s_1' &=\\frac{(\\mathbf a\\cdot\\mathbf n_1)-(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_2)}{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2},\\\\ s_2' &= \\frac{- (\\mathbf a\\cdot\\mathbf n_2)+(\\mathbf n_1\\cdot\\mathbf n_2)(\\mathbf a\\cdot\\mathbf n_1) }{1-(\\mathbf n_1\\cdot\\mathbf n_2)^2}. \\end{align} Another way to rewrite the expressions for $s_1'$ and $s_2'$ is to define two vectors $\\mathbf c_1$ and $\\mathbf c_2$: \\begin{align} \\mathbf c_1 &= \\mathbf n_1 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_2,\\\\ \\mathbf c_2 &= \\mathbf n_2 -(\\mathbf n_1\\cdot\\mathbf n_2)\\mathbf n_1. \\end{align} These two vectors have the same magnitude $c$: c = c_1 = c_2 = \\sqrt{1 - (\\mathbf n_1\\cdot\\mathbf n_2)^2}. Thus, $s_1'$ and $s_2'$ simplify to \\begin{align} s_1' &= \\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2},\\\\ s_2' &= \\frac{-\\mathbf a\\cdot\\mathbf c_2}{c^2}. \\end{align} Using these expressions for $s_1'$ and $s_2'$, the vector $\\mathbf b$ becomes \\mathbf b = \\mathbf a - \\left(\\frac{\\mathbf a\\cdot\\mathbf c_2}{c^2}\\right)\\mathbf n_2-\\left(\\frac{\\mathbf a\\cdot\\mathbf c_1}{c^2}\\right)\\mathbf n_1. The magnitude $b$ of $\\mathbf b$ is the shortest distance between the two lines whose initial points are connected by the vector $\\mathbf a$ and whose directions are given by $\\mathbf n_1$ and $\\mathbf n_2$, with $\\mathbf c_1$ being the component of $\\mathbf n_1$ perpendicular to $\\mathbf n_2$ and $\\mathbf", "i\\mathbf{q}\\cdot\\mathbf{r}}\\right) = \\frac{\\hbar}{2m}e^{+ i\\mathbf{q}\\cdot\\mathbf{r}}\\mathbf{q}\\cdot\\left(2\\mathbf{p}+\\hbar\\mathbf{q}\\right)$$ and $$[\\mathcal{H},e^{- i\\mathbf{q}\\cdot\\mathbf{r}}]= -\\frac{\\hbar}{2m}\\mathbf{q}\\cdot\\left(e^{- i\\mathbf{q}\\cdot\\mathbf{r}}\\mathbf{p}+\\mathbf{p}e^{- i\\mathbf{q}\\cdot\\mathbf{r}}\\right) =- \\frac{\\hbar}{2m}\\mathbf{q}\\cdot\\left(2\\mathbf{p}+\\hbar\\mathbf{q}\\right)e^{- i\\mathbf{q}\\cdot\\mathbf{r}}.$$ Putting either of these into the corresponding formula above, you get $$\\sum_n (\\mathcal{E_n}-\\mathcal{E_s})|\\langle n|e^{i\\mathbf{q}\\cdot\\mathbf{r}}|s \\rangle|^2 = \\frac{\\hbar}{2m}\\langle s|\\mathbf{q}\\cdot\\left(2\\mathbf{p}+\\hbar\\mathbf{q}\\right)|s \\rangle = \\frac{\\hbar^2}{2m}\\mathbf{q}^2+\\frac{\\hbar}{m}\\langle s|\\mathbf{q}\\cdot\\mathbf{p}|s \\rangle.$$ You can then always force the mean value of $\\mathbf{p}$ to vanish. - I) Let us work in the Schrödinger picture. Assume that the Hamiltonian operator $$\\tag{1} \\hat{H}~=~T(\\hat{\\bf p})+ V(\\hat{\\bf r}), \\qquad T(\\hat{\\bf p})~:=~ \\frac{\\hat{\\bf p}^2}{2m},$$ does not depend explicit on time. In particular, the point spectrum of energy eigenvalues $E_n$ does not depend on time. Let $\\psi_n({\\bf r})$ be the corresponding energy eigen-functions in position space, which satisfies the TISE $$\\tag{2} \\hat{H}\\psi_n({\\bf r})~=~E_n \\psi_n({\\bf r}).$$ We may assume that $\\psi_n({\\bf r})\\in{\\mathbb{R}}$ is a real function, see also this Phys.SE post. Let us consider a fixed time $t_0$. Let us pick a basis of time-dependent solutions $$\\tag{3} \\langle n, t |{\\bf r} \\rangle~=~ \\Psi_n({\\bf r},t)~=~ \\psi_n({\\bf r})e^{-\\frac{i}{\\hbar}E_n (t-t_0)}$$ that becomes real at the time $t=t_0$. It follows that for any (possibly complex) function $F=F({\\bf r})$ of position ${\\bf r}$, that the matrix element $$\\tag{4} \\langle n, t_0 |F(\\hat{\\bf r}) | m, t_0 \\rangle ~=~ \\int \\! d^3r ~ \\psi^{*}_n({\\bf r}) ~F({\\bf r})~\\psi_m({\\bf r})~=~ n \\leftrightarrow m$$ is symmetric in $n\\leftrightarrow m$ at $t=t_0$. The time dependence of the matrix element is just", "Inequality with Roots, Squares and the Area Solution 1 Rotate $\\Delta CBP\\;$ around $B\\;$ and away from $A\\;$ through $60^{\\circ}$ into position $C'BP'.\\;$ Observe that this creates aquilateral triangles $BCC'\\;$ and $BPP'.\\;$ This gives us $PB=PP',\\;$ and $PC=P'C'\\;$ so that (1) $PA+PB+PC=AP+PP'+P'C'\\ge AC'.$ The Law of Cosines in $\\Delta ABC'\\;$ gives (with $\\angle ABC=\\beta )$ \\displaystyle\\begin{align} AC'^2 &= AB^2+BC'^2-2\\cdot AB\\cdot BC'\\cos\\angle ABC'\\\\ &=c^2+a^2-2ac\\cos (\\beta+60^{\\circ})\\\\ &=c^2+a^2-2ac\\left(\\cos 60^{\\circ}\\cos\\beta-\\sin 60^{\\circ}\\sin\\beta\\right)\\\\ &=c^2+a^2-2ac\\left(\\frac{1}{2}\\cos\\beta-\\frac{\\sqrt{3}}{2}\\sin\\beta\\right)\\\\ &=c^2+a^2-\\left(ac\\cos\\beta+\\sqrt{3}ac\\sin\\beta\\right)\\\\ &=c^2+a^2-\\frac{a^2+c^2-b^2}{2}+\\sqrt{3}\\cdot 2S\\\\ &=\\frac{2a^2+2c^2-a^2-c^2+b^2}{2}+2\\sqrt{3}S\\\\ &=\\frac{a^2+c^2+b^2}{2}+2\\sqrt{3}S. \\end{align} Thus $AC'=\\displaystyle\\sqrt{\\frac{a^2+b^2+c^2}{2}+2\\sqrt{3}S}.\\;$ With (1), this implies $\\displaystyle PA+PB+PC \\ge \\displaystyle\\sqrt{\\frac{a^2+b^2+c^2}{2}+2\\sqrt{3}S},$ which is the same as the required $\\displaystyle \\sqrt{2}(PA+PB+PC)\\ge\\sqrt{a^2+b^2+c^2+4\\sqrt{3}S}.$ For equality, we need $P,P'\\in AC'.$ In such a case, $\\angle BP'C'=120^{\\circ},\\;$ for, it's complementary to $\\angle BP'P.\\;$ This makes $\\angle BPC=120^{\\circ}.\\;$ Also, $\\angle BPA=120^{\\circ},\\;$ as complementary to $\\angle BPP'.\\;$ Thus, $\\angle APC=120^{\\circ}\\;$ also, which makes $P\\;$ the Fermat-Torricelli point in $\\Delta ABC.$ Naturally, this argument does not work when $\\angle ABC\\gt 120^{\\circ}.$ Solution 2 We'll consider the case in which $A,B,C\\lt 120^{\\circ}.\\;$ Let $T\\;$ be the Fermat-Toricelli point. Denote $TA=x,\\;$ $TB=y\\;$ and $TC=z.\\;$ Choose $T=0,\\;$ $A=x,\\;$ $B=yu\\;$ and $C=zu^2,\\;$ $\\displaystyle u=-\\frac{1}{2}+i\\frac{\\sqrt{3}}{2}.\\;$ We have: \\displaystyle\\begin{align} a^2 &= y^2+yz+z^2,\\\\ b^2 &= z^2+zx+x^2,\\\\ a^2 &= x^2+xy+y^2,\\\\ 4S\\sqrt{3}&=3(xy+yz+zx), \\end{align} so that $a^2+b^2+c^2+4S\\sqrt{3}=2(x+y+z)^2.\\;$ Thus, our inequality reduces to $PA+PB+PC\\ge TA+TB+TC,\\;$ which is known. Let's prove it, though. Let $P=w.\\;$ We need to show that $|w-x|+|w-yu|+|w-zu^2|\\ge x+y+z$ which is equivalent", "$DOP$ is similar to triangle $AOB$, and triangle $DPC$ is similar to triangle $ABC$. If $DP$ = $15y$, then $CP$ = $36y$, $OP$ = $10y$, and $OD$ = $5y\\sqrt {13}$. Since $OP$ + $CP$ = $46y$ = $\\frac {26}{3}$, $y$ = $\\frac {13}{69}$, and $AD$ = $\\frac {60\\sqrt{13}}{23}$ (by the pythagorean theorem on triangle $ABO$ we sum $AO$ and $OD$). The answer is $60$ + $13$ + $23$ = $\\boxed{096}$. ## Solution 2 Extend $AB$ and $CD$ to intersect at $P$. Note that since $\\angle ACB=\\angle PCB$ and $\\angle ABC=\\angle PBC=90^{\\circ}$ by ASA congruency we have $\\triangle ABC\\cong \\triangle PBC$. Therefore $AC=PC=13$. By the angle bisector theorem, $PD=\\dfrac{130}{23}$ and $CD=\\dfrac{169}{23}$. Now we apply Stewart's theorem to find $AD$: \\begin{align*}13\\cdot \\dfrac{130}{23}\\cdot \\dfrac{169}{23}+13\\cdot AD^2&=13\\cdot 13\\cdot \\dfrac{130}{23}+10\\cdot 10\\cdot \\dfrac{169}{23}\\\\ 13\\cdot \\dfrac{130}{23}\\cdot \\dfrac{169}{23}+13\\cdot AD^2&=\\dfrac{169\\cdot 130+169\\cdot 100}{23}\\\\ 13\\cdot \\dfrac{130}{23}\\cdot \\dfrac{169}{23}+13\\cdot AD^2&=1690\\\\ AD^2&=130-\\dfrac{130\\cdot 169}{23^2}\\\\ AD^2&=\\dfrac{130\\cdot 23^2-130\\cdot 169}{23^2}\\\\ AD^2&=\\dfrac{130(23^2-169)}{23^2}\\\\ AD^2&=\\dfrac{130(360)}{23^2}\\\\ AD&=\\dfrac{60\\sqrt{13}}{23}\\\\ \\end{align*} and our final answer is $60+13+23=\\boxed{096}$. ## Solution 3 Notice that by extending $AB$ and $CD$ to meet at a point $E$, $\\triangle ACE$ is isosceles. Now we can do a straightforward coordinate bash. Let $C=(0,0)$, $B=(12,0)$, $E=(12,-5)$ and $A=(12,5)$, and the equation of line $CD$ is $y=-\\dfrac{5}{12}x$. Let F be the intersection point of $AD$ and $BC$, and by using the Angle Bisector Theorem: $\\dfrac{BF}{AB}=\\dfrac{FC}{AC}$ we have $FC=\\dfrac{26}{3}$. Then the equation of", "(2) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\sqrt{1 \\cdot 1 + 2 \\cdot 2 + 3 \\cdot 3 + 4 \\cdot 4} = \\sqrt{1 + 4 + 9 + 16} = \\sqrt{30} \\end{align} Let's now look at some important properties regarding the Euclidean norm of a point $\\mathbf{x}$. Theorem 1: If $\\mathbf{x} = (x_1, x_2, ..., x_n) \\in \\mathbb{R}^n$ and $a \\in \\mathbb{R}$ then: a) $\\| \\mathbf{x} \\| \\geq 0$. b) $\\| a \\mathbf{x} \\| = \\mid a \\mid \\| \\mathbf{x} \\|$. c) $\\| \\mathbf{x} \\|^2 = \\mathbf{x} \\cdot \\mathbf{x}$. • Proof of a) Let $\\mathbf{x} \\in \\mathbb{R}^n$. • We note that $\\mathbf{x} \\cdot \\mathbf{x} = \\sum_{i=1}^{n} x_ix_i = \\sum_{i=1}^{n} x_i^2$. Since $x_i^2 \\geq 0$ for all $i \\in \\{1, 2, ..., n \\}$ we have that $\\mathbf{x} \\cdot \\mathbf{x} \\geq 0$. So: (3) \\begin{align} \\quad \\| \\mathbf{x} \\| = \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} \\geq 0 \\end{align} • Proof of b) Let $\\mathbf{x} \\in \\mathbb{R}^n$ and let $a \\in \\mathbb{R}$. Then: (4) \\begin{align} \\quad \\| a \\mathbf{x} \\| = \\sqrt{ (a \\mathbf{x} \\cdot a \\mathbf{x} } = \\sqrt{a^2 (\\mathbf{x} \\cdot \\mathbf{x})} = \\mid a \\mid \\sqrt{\\mathbf{x} \\cdot \\mathbf{x}} = \\mid a \\mid \\| \\mathbf{x} \\| \\quad \\blacksquare \\end{align} • Proof of c) Let $\\mathbf{x} \\in \\mathbb{R}^n$. Then: (5) \\begin{align} \\quad \\| \\mathbf{x} \\|^2 = \\left", "as the function of $X$ so that it satisfy the above table: \\begin{align} P_0 &= \\frac{1}{12} (X-2)(X-6)\\\\ P_1 &= -\\frac{1}{8} X(X-6)\\\\ P_2 &= \\frac{1}{24} X(X-2) \\end{align} If we replace $X$ in the above equation with \\begin{align} X &= (\\mathbf{S_i} + \\mathbf{S_{i+1}})(\\mathbf{S_i} + \\mathbf{S_{i+1}})\\\\ &= \\mathbf{S_i^2} + \\mathbf{S_{i+1}^2} + 2\\mathbf{S_i} \\cdot \\mathbf{S_{i+1}} \\\\ &= 4 + 2\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} \\end{align} then we get $$P_2 = \\frac{1}{6} (\\mathbf{S_i} \\cdot\\mathbf{S_{i+1}})^2 + \\frac{1}{2} \\mathbf{S_i} \\cdot\\mathbf{S_{i+1}} + \\frac{1}{3}$$ • Thanks for this answer to your own question :) It's simple yet effective especially the explanation of the projection operators ! – gertian May 4 '17 at 16:18", "+ GB + GH + DH + HE \\geq CF.$$ 6. Let $A_1A_2A_3A_4$ be a tetrahedron, $G$ its centroid, and $A'_1, A'_2, A'_3,$ and $A'_4$ the points where the circumsphere of $A_1A_2A_3A_4$ intersects $GA_1,GA_2,GA_3,$ and $GA_4,$ respectively. Prove that $GA_1 \\cdot GA_2 \\cdot GA_3 \\cdot GA_ \\cdot4 \\leq GA'_1 \\cdot GA'_2 \\cdot GA'_3 \\cdot GA'_4$ and $\\frac{1}{GA'_1} + \\frac{1}{GA'_2} + \\frac{1}{GA'_3} + \\frac{1}{GA'_4} \\leq \\frac{1}{GA_1} + \\frac{1}{GA_2} + \\frac{1}{GA_3} + \\frac{1}{GA_4}.$ 7. Let ABCD be a convex quadrilateral and O a point inside it. Let the parallels to the lines BC, AB, DA, CD through the point O meet the sides AB, BC, CD, DA of the quadrilateral ABCD at the points E, F, G, H, respectively. Then, prove that $\\sqrt {\\left|AHOE\\right|} + \\sqrt {\\left|CFOG\\right|}\\leq\\sqrt {\\left|ABCD\\right|}$, where $\\left|P_1P_2...P_n\\right|$ is an abbreviation for the non-directed area of an arbitrary polygon $P_1P_2...P_n$. 8. Suppose that $ABCD$ is a cyclic quadrilateral. Let $E = AC\\cap BD$ and $F = AB\\cap CD$. Denote by $H_{1}$ and $H_{2}$ the orthocenters of triangles $EAD$ and $EBC$, respectively. Prove that the points $F$, $H_{1}$, $H_{2}$ are collinear. ### Number Theory, Combinatorics 1. Let $k$ be a positive integer. Show that there are infinitely many perfect squares of the form $n \\cdot 2^k - 7$ where $n$ is a positive integer. 2. Let $\\mathbb{Z}$ denote the", "+ 2 \\hbar \\mathbf{k} \\cdot \\mathbf{p}).\\end{aligned} \\hspace{\\stretch{1}}(3.21) For the double commutator we have \\begin{aligned}2m \\left[{\\left[{H},{e^{i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]},{e^{-i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]&=e^{i \\mathbf{k} \\cdot \\mathbf{r}} ((\\hbar \\mathbf{k})^2 + 2 \\hbar \\mathbf{k} \\cdot \\mathbf{p}) e^{-i \\mathbf{k} \\cdot \\mathbf{r}}-((\\hbar \\mathbf{k})^2 + 2 \\hbar \\mathbf{k} \\cdot \\mathbf{p}) \\\\ &=e^{i \\mathbf{k} \\cdot \\mathbf{r}} 2 (\\hbar \\mathbf{k} \\cdot \\mathbf{p}) e^{-i \\mathbf{k} \\cdot \\mathbf{r}}-2 \\hbar \\mathbf{k} \\cdot \\mathbf{p} \\\\ &=2 \\hbar \\mathbf{k} \\cdot (\\mathbf{p} - \\hbar \\mathbf{k})-2 \\hbar \\mathbf{k} \\cdot \\mathbf{p},\\end{aligned} so for the double commutator we have just a scalar \\begin{aligned}\\left[{\\left[{H},{e^{i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]},{e^{-i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]&= -\\frac{(\\hbar \\mathbf{k})^2}{m}.\\end{aligned} \\hspace{\\stretch{1}}(3.22) Now consider the expectation of this double commutator, expanded with some unintuitive steps that have been motivated by working backwards \\begin{aligned}{\\langle {s} \\rvert} \\left[{\\left[{H},{e^{i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]},{e^{-i \\mathbf{k} \\cdot \\mathbf{r}}}\\right] {\\lvert {s} \\rangle}&={\\langle {s} \\rvert} 2 H - e^{i \\mathbf{k} \\cdot \\mathbf{r}} H e^{-i \\mathbf{k} \\cdot \\mathbf{r}} - e^{-i \\mathbf{k} \\cdot \\mathbf{r}} H e^{i \\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {s} \\rangle} \\\\ &={\\langle {s} \\rvert} 2 e^{-i \\mathbf{k} \\cdot \\mathbf{r}} e^{i \\mathbf{k} \\cdot \\mathbf{r}} H- 2 e^{-i \\mathbf{k} \\cdot \\mathbf{r}} H e^{i \\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {s} \\rangle} \\\\ &=2 \\sum_n {\\langle {s} \\rvert} e^{-i \\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {n} \\rangle} {\\langle {n} \\rvert} e^{i \\mathbf{k} \\cdot \\mathbf{r}} H {\\lvert {s} \\rangle}- {\\langle {s} \\rvert} e^{-i \\mathbf{k} \\cdot \\mathbf{r}} H {\\lvert {n} \\rangle} {\\langle {n}", "have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ #### Raerin ##### Member Re: If a nd b are unit vecotrs... Note that \\begin{aligned} (2\\mathbf{a}-\\mathbf{b}) \\cdot (\\mathbf{a}+3\\mathbf{b}) &= 2\\mathbf{a}\\cdot\\mathbf{a} +6\\mathbf{a}\\cdot\\mathbf{b} - \\mathbf{a}\\cdot\\mathbf{b} -3\\mathbf{b}\\cdot\\mathbf{b} \\\\ &= 2\\|\\mathbf{a}\\|^2 +5\\mathbf{a}\\cdot\\mathbf{b} - 3\\|\\mathbf{b}\\|^2\\\\ &= 5\\mathbf{a}\\cdot\\mathbf{b} - 1\\quad\\text{since \\mathbf{a} and \\mathbf{b} are unit vectors}\\end{aligned} Since $\\|\\mathbf{a}+\\mathbf{b}\\| = \\sqrt{2}$, squaring both sides and expanding via dot product leaves you with $\\|\\mathbf{a}\\|^2+ 2\\mathbf{a}\\cdot\\mathbf{b} + \\|\\mathbf{b}\\|^2 = 2 \\implies 2\\mathbf{a}\\cdot\\mathbf{b} = 0\\implies \\mathbf{a}\\cdot\\mathbf{b} = 0$ Therefore, we now have that $(2\\mathbf{a}-\\mathbf{b})\\cdot (\\mathbf{a}+3\\mathbf{b}) = 5\\mathbf{a}\\cdot\\mathbf{b} - 1 = -1$ I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0? Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? #### Chris L T521 ##### Well-known member Staff member Re: If a nd b are unit vecotrs... I don't understand how 2a . b = 0 becomes a . b = 0. Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1? Since $\\mathbf{a}\\cdot\\mathbf{b}$ is a scalar, then by the zero product property $2\\mathbf{a}\\cdot \\mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\\mathbf{a}\\cdot\\mathbf{b}=0$ (which", "# Calculating the following integral . I have found the following integral to be zero, but i don't think its correct. $$C = \\oint_K d\\mathbf r\\cdot \\mathbf A$$ Where $\\mathbf A = \\frac 1 2 \\mathbf n \\times \\mathbf r$ and $\\mathbf n \\cdot \\mathbf n=1$. Taking $K$ as a circle with radius $R$ and $\\mathbf n$ is the normal to the plane where the circle lives. Any kind of help is appreciated . The problem is that when i found $A$ it came out to be zero. i found normal as $\\nabla f$ where $f=x^2+y^2-1$ . Progress : I applied stokes theorem, and also $\\nabla \\times \\mathbf A =2\\mathbf n$ , then i get $C=\\pi R^2$ . am i right ?? how do i do it using line integral ? - In the following I assume the path $K$ is counter-clockwise with respect to the orientation $\\mathbf n$ of the plane. Let $$K\\equiv \\gamma(t) := R(\\cos(t), \\sin(t), 0)\\quad t\\in[0, 2\\pi]$$ we have $$\\mathbf A(\\gamma(t)) = \\frac 1 2 \\mathbf n\\times \\gamma(t) = \\frac R 2 (0, 0, 1)\\times (\\cos(t), \\sin(t), 0) = \\frac R 2 (-\\sin(t), \\cos(t), 0)$$ and so \\begin{align} \\oint_K d\\mathbf r \\cdot \\mathbf A &= \\int_0^{2\\pi} \\mathbf A(\\gamma(t))\\cdot \\dot\\gamma(t)dt \\\\ &= \\frac {R^2} 2 \\int_0^{2\\pi} (\\sin(t)^2 + \\cos(t)^2)dt\\\\ &= \\pi R^2 \\end{align} First note", "\\mathbf{k}) \\cdot \\mathbf{p} + \\mathbf{p}^2 \\right)=e^{i \\mathbf{k}\\cdot \\mathbf{r}} (\\hbar \\mathbf{k} + \\mathbf{p})^2\\end{aligned} So, finally, our first commutator is \\begin{aligned}\\left[{H},{ e^{i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]&= \\frac{1}{{2m}}e^{i \\mathbf{k}\\cdot \\mathbf{r}} \\left((\\hbar\\mathbf{k})^2 + 2 (\\hbar \\mathbf{k}) \\cdot \\mathbf{p} \\right)\\end{aligned} \\hspace{\\stretch{1}}(3.16) The double commutator is then \\begin{aligned}\\left[{\\left[{H},{e^{i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]},{e^{-i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]&=e^{i\\mathbf{k} \\cdot \\mathbf{r}} \\frac{\\hbar \\mathbf{k}}{m} \\cdot \\left( \\mathbf{p} e^{-i \\mathbf{k} \\cdot \\mathbf{r}} - e^{-i \\mathbf{k} \\cdot \\mathbf{r}} \\mathbf{p} \\right)\\end{aligned} To simplify this we want \\begin{aligned}\\mathbf{k} \\cdot \\boldsymbol{\\nabla} e^{-i \\mathbf{k} \\cdot \\mathbf{r}} \\Psi &=k_n \\partial_n e^{-i k_m x_m } \\Psi \\\\ &=e^{-i \\mathbf{k} \\cdot \\mathbf{r} }\\left(k_n (-i k_n) \\Psi + k_n \\partial_n \\Psi \\right) \\\\ &=e^{-i \\mathbf{k} \\cdot \\mathbf{r} } \\left( -i \\mathbf{k}^2 + \\mathbf{k} \\cdot \\boldsymbol{\\nabla} \\right) \\Psi\\end{aligned} The double commutator is then left with just \\begin{aligned}\\left[{\\left[{H},{e^{i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]},{e^{-i \\mathbf{k} \\cdot \\mathbf{r}}}\\right]&=- \\frac{1}{{m}} (\\hbar \\mathbf{k})^2 \\end{aligned} \\hspace{\\stretch{1}}(3.17) Now, returning to the energy expression \\begin{aligned}\\sum_n (E_n - E_s) {\\left\\lvert{{\\langle {n} \\rvert} e^{i\\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {s} \\rangle}}\\right\\rvert}^2&=\\sum_n (E_n - E_s) {\\langle {s} \\rvert} e^{-i\\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {n} \\rangle} {\\langle {n} \\rvert} e^{i\\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {s} \\rangle} \\\\ &=\\sum_n {\\langle {s} \\rvert} e^{-i\\mathbf{k} \\cdot \\mathbf{r}} H {\\lvert {n} \\rangle} {\\langle {n} \\rvert} e^{i\\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {s} \\rangle} -{\\langle {s} \\rvert} e^{-i\\mathbf{k} \\cdot \\mathbf{r}} {\\lvert {n} \\rangle} {\\langle {n} \\rvert} e^{i\\mathbf{k} \\cdot \\mathbf{r}} H {\\lvert {s} \\rangle} \\\\ &={\\langle {s} \\rvert}", "intersects $$BC$$ at $$D$$ such that $$DB$$ $$=$$ $$3CD$$(see Fig. below). Prove that $$2A{{B}^{2}}=2A{{C}^{2}}+B{{C}^{2}}$$ . Diagram ### Solution Reasoning: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Steps: In $$\\Delta ABC$$ $$AD\\bot BC$$ and $$\\,BD=3CD$$ \\begin{align}BD + CD &= BC\\\\3CD + CD &= BC\\\\4CD &= BC\\\\CD &= \\frac{1}{4}BC \\ldots \\left( 1 \\right)\\\\\\\\&{\\rm{and\\qquad }}\\\\\\\\BD &= \\frac{3}{4}BC \\ldots \\left( 2 \\right)\\end{align} In $$\\Delta ADC$$, \\begin{align}A{C^2} &= AD{}^2 + C{D^2}\\\\ & \\quad \\left[ {\\angle ADC = {{90}^0}} \\right]\\\\A{D^2} & = A{C^2} - C{D^2} \\ldots .\\left( 3 \\right)\\end{align} In $$\\Delta ADB$$, \\begin{align}A{B^2} &= AD{}^2 + B{D^2} \\\\ & \\quad \\left[ {\\angle ADB = {{90}^0}} \\right]\\\\A{B^2} &= A{C^2} - C{D^2} + B{D^2} \\\\ & \\quad\\left[ {{\\rm{from (3)}}} \\right]\\\\A{B^2}\\,\\, &= \\!\\! A{C^2} \\! +\\! {\\left( \\! {\\frac{3}{4}BC} \\! \\right)^2} -\\! {\\left(\\! {\\frac{1}{4}BC}\\! \\right)^2} \\\\ & \\quad\\left[ {{\\rm{from (1) and (2)}}}\\right]\\\\ \\\\ A{B^2} &= A{C^2} + \\frac{{9B{C^2} - B{C^2}}}{{16}}\\\\A{B^2} &= A{C^2} + \\frac{{8B{C^2}}}{{16}}\\\\A{B^2} &= A{C^2} + \\frac{1}{2}B{C^2}\\\\2A{B^2} &= 2A{C^2} + B{C^2}\\end{align} ## Question 15 In an equilateral triangle $$ABC$$, $$D$$ is a point on side $$BC$$ such that $$BD$$ $$=$$\\begin{align}\\frac{1}{3}BC\\end{align} Prove that $$9A{{D}^{2}}=7AB{}^{2}$$ . Diagram ### Solution Reasoning: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Steps:", "\\dot{\\mathbf{x}}) = 0\\end{aligned} \\hspace{\\stretch{1}}(3.26) By evaluating this, we can eventually show that we can construct a four vector equation. Doing this we have \\begin{aligned}\\frac{d}{dt} (\\gamma \\mathbf{v}) &=\\frac{d}{dt} \\left( \\left(1 - \\mathbf{v}^2/c^2\\right)^{-1/2} \\mathbf{v} \\right) \\\\ &=-2 (-1/2) \\mathbf{v} (\\mathbf{v} \\cdot \\dot{\\mathbf{v}})/c^2 \\left(1 - \\mathbf{v}^2/c^2\\right)^{-3/2} + \\left(1 - \\mathbf{v}^2/c^2\\right)^{-1/2} \\dot{\\mathbf{v}} \\\\ &=\\gamma \\left( \\frac{\\mathbf{v} (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) }{ c^2 - \\mathbf{v}^2 } + \\dot{\\mathbf{v}} \\right)\\end{aligned} Or \\begin{aligned}\\frac{\\mathbf{v} (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) }{ c^2 - \\mathbf{v}^2 } + \\dot{\\mathbf{v}} = 0\\end{aligned} \\hspace{\\stretch{1}}(3.27) Clearly $\\dot{\\mathbf{v}} = 0$ is a solution, but is it the only solution? By dotting this with $\\mathbf{v}$ we have \\begin{aligned}0 &= \\frac{\\mathbf{v}^2 (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) }{ c^2 - \\mathbf{v}^2 } + \\dot{\\mathbf{v}} \\cdot \\mathbf{v} \\\\ &= (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) \\left( 1 + \\frac{\\mathbf{v}^2}{c^2 - \\mathbf{v}^2} \\right) \\\\ &= (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) \\frac{c^2}{c^2 - \\mathbf{v}^2} \\end{aligned} This implies that $\\dot{\\mathbf{v}} = 0$ (a contraction) or that $\\mathbf{v} \\cdot \\dot{\\mathbf{v}} = 0$. To examine the perpendicularity question, let’s take cross products. This gives \\begin{aligned}0 =\\frac{(\\mathbf{v} \\times \\mathbf{v}) (\\mathbf{v} \\cdot \\dot{\\mathbf{v}}) }{ c^2 - \\mathbf{v}^2 } + \\dot{\\mathbf{v}} \\times \\mathbf{v}\\end{aligned} \\hspace{\\stretch{1}}(3.28) We have found that $\\mathbf{v} \\cdot \\dot{\\mathbf{v}} = 0$ and $\\mathbf{v} \\times \\dot{\\mathbf{v}} = 0$. This can only mean that $\\dot{\\mathbf{v}} = 0$, contradicting the assumption that is non-zero. We conclude that $\\dot{\\mathbf{v}} = 0$ is the only solution to 3.27." ]

[ "## Trigonometry (11th Edition) Clone Since the measure of angle is 4 radians, it lies between $\\frac{3\\pi}{2}$ and $\\pi$ radians. This means it lies in the third quadrant.", "points is A, then $\\frac{\\sin(A)}{\\sin(r)} = \\frac{sin(b)}{sin(x)}.$ The angle A is also the difference between new and old longitudes. Good Hunting (fishing?) RD Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", "= - \\sqrt{\\frac{1 - \\cos \\alpha}{2}}$ if $\\frac{\\alpha}{2}$ is located in the third or fourth quadrant. $\\cos \\frac{\\alpha}{2} = \\sqrt{\\frac{1 + \\cos \\alpha}{2}}$ if $\\frac{\\alpha}{2}$ is located in either the first or fourth quadrant. $\\cos \\frac{\\alpha}{2} = - \\sqrt{\\frac{1 + \\cos \\alpha}{2}}$ if $\\frac{\\alpha}{2}$ is located in either the second or fourth quadrant. Tip: This all depends how the values of cosine and sine are negative dependig which quadrant cosine or sine lies on.", "# Question #70b87 May 13, 2017 If Quadrant I: $\\frac{4}{5}$ If Quadrant II: $- \\frac{4}{5}$ #### Explanation: Use Pythagorean's Identity: ${\\sin}^{2} x + {\\cos}^{2} x = 1$ By substitution, we get: ${\\left(\\frac{3}{5}\\right)}^{2} + {\\cos}^{2} a = 1$ Subtract ${\\left(\\frac{3}{5}\\right)}^{2}$ from both sides: ${\\cos}^{2} a = 1 - {\\left(\\frac{3}{5}\\right)}^{2}$ ${\\cos}^{2} a = 1 - \\frac{9}{25}$ ${\\cos}^{2} a = \\frac{16}{25}$ Now, we take the square root of both sides: $\\cos a = \\pm \\sqrt{\\frac{16}{25}}$ $\\cos a = \\pm \\frac{4}{5}$ Therefore, if $\\angle A$ is in Quadrant I, $\\cos a = \\frac{4}{5}$ if $\\angle A$ is in Quadrant II, $\\cos a = - \\frac{4}{5}$", "# If z=a+ib lies in third quadrant, then Question: If $z=a+i b$ lies in third quadrant, then $\\frac{\\bar{z}}{z}$ also lies in third quadrant if (a) $a>b>0$ (b) $a (c)$b (d) $b>a>0$ Solution: Since, $z=a+i b$ lies in third quadrant. $\\Rightarrow a<0$ and $b<0$ Now, $\\frac{\\bar{z}}{z}=\\frac{\\overline{a+i b}}{a+i b}$ $=\\frac{a-i b}{a+i b}$ $=\\frac{a-i b}{a+i b} \\times \\frac{a-i b}{a-i b}$ $=\\frac{a^{2}+i^{2} b^{2}-2 a b i}{a^{2}-i^{2} b^{2}}$ $=\\frac{a^{2}-b^{2}-2 a b i}{a^{2}+b^{2}}$ Since, $\\frac{\\bar{z}}{z}$ also lies in third quadrant. $\\Rightarrow a^{2}-b^{2}<0$ $\\Rightarrow(a-b)(a+b)<0$ $\\Rightarrow a-b>0$ and $a+b<0$ $\\Rightarrow a>b \\quad \\ldots(2)$ From (1) and (2), \\$b Hence, the correct option is (c).", "### Still have math questions? Trigonometry Question Angle $$\\alpha$$ is in quadrant III and angle $$\\beta$$ is in quadrant IV. If $$\\sin \\alpha = - \\frac { 5 } { 6 }$$ and $$\\cos \\beta = \\frac { 1 } { 2 }$$ , find $$\\cos ( \\alpha + \\beta )$$ . $$\\frac { \\sqrt { 11 } + 5 \\sqrt { 3 } } { 6 }$$ $$\\frac { \\sqrt { 11 } 5 \\sqrt { 3 } } { 6 }$$ $$\\frac { \\sqrt { 11 } 5 \\sqrt { 3 } } { 12 }$$ $$\\frac { \\sqrt { 11 } | 5 \\sqrt { 3 } } { 12 }$$", "$\\tan { 22{ \\frac { 1 }{ 2 } }^{ ° } }$ 24. Find all other trigonometrical ratios if $\\sin x=\\frac{-2\\sqrt6}{5}$ and x lies in III quadrant? 25. Part D 2 x 5 = 10 26. Prove that sin(n+1)x sin(n+2) x+cos(n+1)xcos(n+2)x=cosx. 27. Prove that cos22x-cos26x = sin 4x.sin 8x", "takes a(n) approach to knowledge acquisition Aaron trusts what he himself observes, not claims by others. Aaron takes a(n) approach to knowledge acquisition... Given that the sine of a certain angle is $\\frac{5}{13}$, and that the angle lies in quadrant 1 of the coordinate plane, what is the tangent of the angle? $\\frac{5}{13}$ $\\frac{12}{13}$$\\frac{5}{12}$$\\frac{12}{5}$...", "# How do you find the exact values of sin^-1(1/2)? ##### 1 Answer Oct 15, 2016 ${\\sin}^{-} 1 \\left(\\frac{1}{2}\\right) = \\frac{\\pi}{6}$ #### Explanation: Find ${\\sin}^{-} 1 \\left(\\frac{1}{2}\\right)$ This problem is asking for the ANGLE with a sine of $\\frac{1}{2}$. The range of ${\\sin}^{-} 1$ or $\\arcsin$ is between $\\frac{\\pi}{2}$ and $- \\frac{\\pi}{2}$. If you are finding ${\\sin}^{-} 1$ of a positive value, the answer will be between 0 and $\\frac{\\pi}{2}$, or the first quadrant in the unit circle. Do NOT use the second quadrant angle with a sine of $\\frac{1}{2}$, because it does not fall within the range of ${\\sin}^{-} 1$. Using the unit circle, the angle with a sine of $\\frac{1}{2}$ in the first quadrant is $\\frac{\\pi}{6}$.", "# Difference between revisions of \"University of South Carolina High School Math Contest/1993 Exam/Problem 28\" ## Problem Suppose $\\triangle ABC$ is a triangle with 3 acute angles $A, B,$ and $C$. Then the point $( \\cos B - \\sin A, \\sin B - \\cos A)$ (A) can be in the 1st quadrant and can be in the 2nd quadrant only (B) can be in the 3rd quadrant and can be in the 4th quadrant only (C) can be in the 2nd quadrant and can be in the 3rd quadrant only (D) can be in the 2nd quadrant only (E) can be in any of the 4 quadrants ## Solution First note that $\\cos{A}, \\cos{B}, \\sin{A}, \\sin{B}>0$ and $A+B>90$. Since $\\cos B+\\sin A>0$, $\\cos B-\\sin A$ and $\\cos^2B-\\sin^2A$ have the same sign. Similarly, $\\sin B-\\cos A$ and $\\sin^2B-\\cos^2A$ have the same sign. Notice that $\\cos^2B-\\sin^2A=-(\\sin^2B-\\cos^2A)$. Then $(\\cos B-\\sin A, \\sin B-\\cos A)$ can only lie in the 2nd and 4th quadrants.", "quadrants from the first. You’ll see this in the following examples. Find the exact value of each of the following. In other words, don’t use a calculator. Show All SolutionsHide All Solutions 1. $$\\sin \\left( {\\frac{{2\\pi }}{3}} \\right)$$ and $$\\sin \\left( { - \\frac{{2\\pi }}{3}} \\right)$$ Show Solution The first evaluation here uses the angle $$\\frac{{2\\pi }}{3}$$. Notice that $$\\frac{{2\\pi }}{3} = \\pi - \\frac{\\pi }{3}$$. So $$\\frac{{2\\pi }}{3}$$ is found by rotating up $$\\frac{\\pi }{3}$$ from the negative $$x$$-axis. This means that the line for $$\\frac{{2\\pi }}{3}$$ will be a mirror image of the line for $$\\frac{\\pi }{3}$$ only in the second quadrant. The coordinates for $$\\frac{{2\\pi }}{3}$$ will be the coordinates for $$\\frac{\\pi }{3}$$ except the $$x$$ coordinate will be negative. Likewise, for $$- \\frac{{2\\pi }}{3}$$we can notice that $$- \\frac{{2\\pi }}{3} = - \\pi + \\frac{\\pi }{3}$$, so this angle can be found by rotating down $$\\frac{\\pi }{3}$$ from the negative $$x$$-axis. This means that the line for $$- \\frac{{2\\pi }}{3}$$ will be a mirror image of the line for $$\\frac{\\pi }{3}$$ only in the third quadrant and the coordinates will be the same as the coordinates for $$\\frac{\\pi }{3}$$ except both will be negative. Both of these angles along with their coordinates are shown on the following unit circle. From this unit circle we", "## Trigonometry (11th Edition) Clone $\\sin{\\frac{3\\pi}{4}} = \\frac{\\sqrt2}{2}$ Recall: (1) An angle and its reference angle either have the same trigonometric function values or differ only in signs. (2) $\\frac{\\pi}{4}$ is a special angle whose sine value is known to be $\\frac{\\sqrt2}{2}$. Note that the reference angle of $\\frac{3\\pi}{4}$ is $\\frac{\\pi}{4}$. However, since $\\frac{3\\pi}{4}$ is in Quadrant II, then its sine value is also positive. Therefore, $\\sin{\\frac{3\\pi}{4}} = \\frac{\\sqrt2}{2}$", "### Home > INT3 > Chapter 11 > Lesson 11.2.2 > Problem11-72 11-72. The sine of obtuse angle A is . What is the tangent of angle A? Homework Help ✎ First use sin2A + cos2A = 1 to find cosA. Note that A is in the second quadrant. $\\text{tan}A=\\frac{-3}{\\sqrt{91}}$", "# Math Help - Can anyone help me with this problem? 1. ## Can anyone help me with this problem? If sin(x)=-4/5 and (x) is in Quadrant III, find tan(x/2) 2. ## Re: Can anyone help me with this problem? Originally Posted by derek1008 If sin(x)=-4/5 and (x) is in Quadrant III, find tan(x/2) if $x$ is a quad III angle ... (1) by sketching a reference triangle in quad III, you can calculate the value of $\\cos{x}$. (2) $\\pi < x < \\frac{3\\pi}{2} \\implies \\frac{\\pi}{2} < \\frac{x}{2} < \\frac{3\\pi}{4}$ ... which says $\\frac{x}{2}$ is a quad II angle. Therefore, $\\tan\\left(\\frac{x}{2}\\right) < 0$. now use your half angle identity for tangent.", "$$\\frac{7π}{4}$$ $$\\frac{π}{4}$$, Quadrant IV, $$\\sin (\\frac{7π}{4})=−\\frac{\\sqrt{2}}{2}$$, $$\\cos (\\frac{7π}{4})=\\frac{\\sqrt{2}}{2}$$ For the following exercises, find the requested value. If $$\\cos (t)=\\frac{1}{7}$$ and $$t$$ is in the 4th quadrant, find $$\\sin (t)$$. If $$\\cos (t)=\\frac{2}{9}$$ and $$t$$ is in the 1st quadrant, find $$\\sin (t).$$ $$\\frac{\\sqrt{77}}{9}$$ If $$\\sin (t)=\\frac{3}{8}$$ and $$t$$ is in the 2nd quadrant, find $$\\cos (t)$$. If $$\\sin (t)=−\\frac{1}{4}$$ and $$t$$ is in the 3rd quadrant, find $$\\cos (t).$$ $$−\\frac{\\sqrt{15}}{4}$$ Find the coordinates of the point on a circle with radius 15 corresponding to an angle of $$220°$$. Find the coordinates of the point on a circle with radius 20 corresponding to an angle of $$120°$$. $$(−10,10\\sqrt{3})$$ Find the coordinates of the point on a circle with radius 8 corresponding to an angle of $$\\frac{7π}{4}$$. Find the coordinates of the point on a circle with radius 16 corresponding to an angle of $$\\frac{5π}{9}$$. $$(–2.778,15.757)$$ State the domain of the sine and cosine functions. State the range of the sine and cosine functions. $$[–1,1]$$ Graphical For the following exercises, use the given point on the unit circle to find the value of the sine and cosine of $$t$$ . $$\\sin t=\\frac{1}{2}, \\cos t=−\\frac{\\sqrt{3}}{2}$$ $$\\sin t=− \\frac{\\sqrt{2}}{2}, \\cos t=−\\frac{\\sqrt{2}}{2}$$ $$\\sin t=\\frac{\\sqrt{3}}{2},\\cos t=−\\frac{1}{2}$$ $$\\sin t=− \\frac{\\sqrt{2}}{2}, \\cos t=\\frac{\\sqrt{2}}{2}$$ $$\\sin t=0, \\cos t=−1$$ $$\\sin t=−0.596, \\cos t=0.803$$ $$\\sin t=\\frac{1}{2}, \\cos t= \\frac{\\sqrt{3}{2}}$$", "## anonymous 3 years ago pls help urgent help need simple trig question just can't figure what the right answer will be? 1. anonymous 2. anonymous @UnkleRhaukus do you think u could help me solve this question 3. anonymous think about where $$\\sin(\\theta)=\\cos(\\theta)$$ there are two places on the unit circle where this is true 4. anonymous i am not sure @satellite73 5. anonymous look at the unit circle on the last page of the attached cheat sheet then look to see for which points on the unit circle the first coordinate is the same as the second coordinate 6. anonymous if you look at the circle, you will see a coordinate in the first quadrant labelled $(\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{2}}{2})$ 7. anonymous the first coordinate is cosine, the second coordinate is sine, and they are equal there the angle that gives this is $$\\frac{\\pi}{4}$$ 8. UnkleRhaukus what do you get when $$\\theta=0$$?", "# What mistake did I make finding $\\arg (\\sqrt{3}-i)$ What mistake did I make finding $\\arg (\\sqrt{3}-i)$? I figured it will be in the 4th quadrant and look like: So $\\arg{z} = - \\arctan{\\frac{\\sqrt{3}}{1}} = -\\frac{\\pi}{3}$ The right answer is supposed to be $-\\frac{\\pi}{6}$ - Tangent is opposite over adjacent, which is $y$ coordinate over $x$ coordinate. You have it the other way around.", "angle $$\\frac{{2\\pi }}{3}$$. That’s not on our unit circle above, however notice that $$\\frac{{2\\pi }}{3} = \\pi - \\frac{\\pi }{3}$$. So $$\\frac{{2\\pi }}{3}$$ is found by rotating up $$\\frac{\\pi }{3}$$ from the negative $$x$$-axis. This means that the line for $$\\frac{{2\\pi }}{3}$$ will be a mirror image of the line for $$\\frac{\\pi }{3}$$ only in the second quadrant. The coordinates for $$\\frac{{2\\pi }}{3}$$ will be the coordinates for $$\\frac{\\pi }{3}$$ except the $$x$$ coordinate will be negative. Likewise, for $$- \\frac{{2\\pi }}{3}$$we can notice that $$- \\frac{{2\\pi }}{3} = - \\pi + \\frac{\\pi }{3}$$, so this angle can be found by rotating down $$\\frac{\\pi }{3}$$ from the negative $$x$$-axis. This means that the line for $$- \\frac{{2\\pi }}{3}$$ will be a mirror image of the line for $$\\frac{\\pi }{3}$$ only in the third quadrant and the coordinates will be the same as the coordinates for $$\\frac{\\pi }{3}$$ except both will be negative. Both of these angles, along with the coordinates of the intersection points, are shown on the following unit circle. From this unit circle we can see that $$\\sin \\left( {\\frac{{2\\pi }}{3}} \\right) = \\frac{{\\sqrt 3 }}{2}$$and $$\\sin \\left( { - \\frac{{2\\pi }}{3}} \\right) = - \\frac{{\\sqrt 3 }}{2}$$. This leads to a nice fact about the sine function. The sine function is called an odd function", "# How do you evaluate tan(sin^-1(-1/6))? $- \\frac{1}{\\sqrt{35}}$. Let $a = {\\sin}^{- 1} \\left(- \\frac{1}{6}\\right)$. Then, $\\sin a = - \\frac{1}{6} < 0$. $a$ is in the 3rd quadrant or in the 4th. On the other hand, he \"principal branch\" of the inverse sine corresponds to an angle in the first or fourth quadrant, not the third. So we pick the fourth quadrant angle, and $\\cos a = + \\frac{\\sqrt{35}}{6}$. The given expression $= \\tan a = \\sin \\frac{a}{\\cos} a = - \\frac{1}{\\sqrt{35}}$.", "Algebra and Trigonometry 10th Edition (a)$-\\frac{\\pi}{5}$ lies in the Fourth Quadrant. (b) $\\frac{7\\pi}{5}$ lies in the Third Quadrant. If $θ$ lies in the: - First Quadrant, then $0\\ltθ\\lt\\frac{\\pi}{2}$ - Second Quadrant, then $\\frac{\\pi}{2}\\ltθ\\lt\\pi$ - Third Quadrant, then $\\pi\\ltθ\\lt\\frac{3\\pi}{2}$ - Fourth Quadrant, then $\\frac{3\\pi}{2}\\ltθ\\lt2\\pi$ (a) $-\\frac{\\pi}{5}+2\\pi=\\frac{9\\pi}{5}$ $\\frac{3\\pi}{2}\\lt\\frac{9\\pi}{5}\\lt2\\pi$ , so $-\\frac{\\pi}{5}$ lies in the Fourth Quadrant. (b) $\\pi\\lt\\frac{7\\pi}{5}\\lt\\frac{3\\pi}{2}$, so $\\frac{7\\pi}{5}$ lies in the Third Quadrant." ]

[ "may show by graphing the given system that point P will lie in the second quadrant in the new coordinate system. Therefore, the abscissa of $P$ is negative, and is therefore $-\\frac{123}{13}$; similarly, the ordinate of $P$ is positive and is therefore $\\frac{526}{13}$. Thus, we have that $\\alpha=-\\frac{123}{13}$ and that $\\beta=\\frac{526}{13}$. It follows that $\\alpha+\\beta=-\\frac{123}{13}+\\frac{526}{13}=\\frac{403}{13}=\\boxed{031}$.", "# Question #cca9c Apr 27, 2017 $x = {143}^{o}$ #### Explanation: First of all, it helps to know that Sin (${37}^{o}$) = $\\frac{3}{5}$ and Cos (${37}^{o}$) = $\\frac{4}{5}$. Now according to your question, Sin x is positive and cos x is negative. This is only possible in the second quadrant of the cartesian plane, where Sin x is positive, and Cos x & Tan x are negative. In the second quadrant, all angles lie between ${90}^{o} \\mathmr{and} {180}^{o}$. So we know that $S \\in \\left(x\\right) = S \\in \\left({180}^{o} - x\\right)$ and $- C o s \\left(x\\right) = C o s \\left({180}^{o} - x\\right)$ Putting x as ${37}^{o}$, we get $S \\in \\left({37}^{o}\\right) = S \\in \\left({143}^{o}\\right)$ $S \\in \\left({143}^{o}\\right) = \\frac{3}{5}$ Also, $- C o s \\left({37}^{o}\\right) = C o s \\left({143}^{o}\\right)$ $C o s \\left({143}^{o}\\right) = - \\frac{4}{5}$ As all conditions of the question are satisfied, $x = {143}^{o}$ Apr 27, 2017 $x \\approx {143.13}^{\\circ}$ or 2.498 radians #### Explanation: We know that x is in the second quadrant, because the sine is positive and cosine is negative. If we use the cosine function and take its inverse, the calculator will do all of the work to give us the correct angle. $\\cos \\left(x\\right) = - \\frac{4}{5}$ Use the inverse cosine function on both sides: ${\\cos}^{-}", "quadrants from the first. You’ll see this in the following examples. Find the exact value of each of the following. In other words, don’t use a calculator. Show All SolutionsHide All Solutions 1. $$\\sin \\left( {\\frac{{2\\pi }}{3}} \\right)$$ and $$\\sin \\left( { - \\frac{{2\\pi }}{3}} \\right)$$ Show Solution The first evaluation here uses the angle $$\\frac{{2\\pi }}{3}$$. Notice that $$\\frac{{2\\pi }}{3} = \\pi - \\frac{\\pi }{3}$$. So $$\\frac{{2\\pi }}{3}$$ is found by rotating up $$\\frac{\\pi }{3}$$ from the negative $$x$$-axis. This means that the line for $$\\frac{{2\\pi }}{3}$$ will be a mirror image of the line for $$\\frac{\\pi }{3}$$ only in the second quadrant. The coordinates for $$\\frac{{2\\pi }}{3}$$ will be the coordinates for $$\\frac{\\pi }{3}$$ except the $$x$$ coordinate will be negative. Likewise, for $$- \\frac{{2\\pi }}{3}$$we can notice that $$- \\frac{{2\\pi }}{3} = - \\pi + \\frac{\\pi }{3}$$, so this angle can be found by rotating down $$\\frac{\\pi }{3}$$ from the negative $$x$$-axis. This means that the line for $$- \\frac{{2\\pi }}{3}$$ will be a mirror image of the line for $$\\frac{\\pi }{3}$$ only in the third quadrant and the coordinates will be the same as the coordinates for $$\\frac{\\pi }{3}$$ except both will be negative. Both of these angles along with their coordinates are shown on the following unit circle. From this unit circle we", "# Solution If $\\sin t = 1/4$, and the terminal side of angle $t$ is in quadrant II, find $\\cos t$. Combining the Pythagorean Identity $\\cos^2 t + \\sin^2 t = 1$ and the given $\\sin t = 1/4$, we have $$\\cos^2 t + \\left( \\frac{1}{4} \\right)^2 = 1$$ But then if we isolate the $\\cos^2 t$ term, we have $$\\cos^2 t = 1 - \\frac{1}{16} = \\frac{15}{16}$$ Hence, $$\\cos t = \\pm \\sqrt{\\frac{15}{16}} = \\pm \\frac{\\sqrt{15}}{4}$$ Note that $\\cos t$ is negative in quadrant II, so only one of these two values makes any sense $$\\cos t = -\\frac{\\sqrt{15}}{4}$$", "## Algebra and Trigonometry 10th Edition False. $sin\\frac{u}{2}=+\\sqrt {\\frac{1-cos~u}{2}}$ If $u$ is in the second quadrant: $\\frac{\\pi}{2}\\lt u\\lt\\pi$. Then, $\\frac{\\pi}{4}\\lt \\frac{u}{2}\\lt\\frac{\\pi}{2}$. That is $u$ is in the first quadrant. So, $sin\\frac{u}{2}\\gt0$", "= - \\sqrt{\\frac{1 - \\cos \\alpha}{2}}$ if $\\frac{\\alpha}{2}$ is located in the third or fourth quadrant. $\\cos \\frac{\\alpha}{2} = \\sqrt{\\frac{1 + \\cos \\alpha}{2}}$ if $\\frac{\\alpha}{2}$ is located in either the first or fourth quadrant. $\\cos \\frac{\\alpha}{2} = - \\sqrt{\\frac{1 + \\cos \\alpha}{2}}$ if $\\frac{\\alpha}{2}$ is located in either the second or fourth quadrant. Tip: This all depends how the values of cosine and sine are negative dependig which quadrant cosine or sine lies on.", "# Math Help - double angle problem 1. ## double angle problem It says the sinx= $14/15$ and it is in quadrant 2 what quadrant is the sin2x in? (NO CALCULATOR)i figured it out with a calc but i cant w/out a calc i really need help fast thnx 2. Originally Posted by brianH It says the sinx= $14/15$ and it is in quadrant 2 what quadrant is the sin2x in? (NO CALCULATOR)i figured it out with a calc but i cant w/out a calc i really need help fast thnx $\\sin(2x)=2\\sin(x)\\cos(x)=2\\cdot \\frac{14}{15}\\cdot \\frac{\\sqrt{29}}{15}$ 3. i got that and it was $-28 \\frac{\\sqrt{29}}{225}$ but how do i fugure out what quad thats in 4. Originally Posted by brianH It says the sinx= $14/15$ and it is in quadrant 2 what quadrant is the sin2x in? (NO CALCULATOR)i figured it out with a calc but i cant w/out a calc i really need help fast thnx We will use the identity $sin(2x)=2sinxcosx$ to help us. Since $sinx=\\frac{14}{15}$, we can determine that $cosx=\\frac{-\\sqrt{29}}{15}$ (since sine was defined in the 2nd quadrant). Plugging this into the identity, we have: $sin(2x)=2{\\left(\\frac{14}{15}\\right)}{\\left(\\frac{-\\sqrt{29}}{15}\\right)}=\\frac{-28\\sqrt{29}}{225}$. I'm going to conclude that $sin2x$ falls in the 4th quadrant, since $sin\\theta=\\frac{opp}{adj}$. Since the opposite is negative, it will reside either in the 3rd or 4th quadrants. Since the adjacent", "# Math Help - Can someone show me how to use the half angle formula? 1. ## Can someone show me how to use the half angle formula? Thanks 2. Originally Posted by qbkr21 Thanks You have that, $\\sin^2 x=\\frac{1-\\cos 2x}{2}$ $\\cos^2 x=\\frac{1+\\cos 2x}{2}$ So to get your value for $\\sin x,\\cos x$ you take the square root. CAREFUL, since there are two roots it can be negative or positive. How do you know? Simple, determine the quadrant of the angle and where it lies.", "worked example, it should hopefully make a bit more sense: We want to find $x$ such that $\\cos\\frac x2=\\frac{\\sqrt2}2,x\\in[0,2\\pi]$ The cosine we were given is a positive value, $\\frac{\\sqrt{2}}2$, so it must lie either in the first or fourth quadrant (think about your unit circle: the cosine is the $x$-coordinate, so it will be positive in quadrants I and IV). This means that our angle must lie either in $\\left[0,\\frac\\pi2\\right]$ or in $\\left[\\frac{3\\pi}2,2\\pi\\right]$. inverse cosine((square-root 2)/2)=45degrees If you are familiar with your unit circle, you should be able to determine that $\\arccos\\frac{\\sqrt2}2 = \\frac\\pi4$ or $45^\\circ$. This is the angle in the first quadrant, and looking at the unit circle you should be able to find the fourth quadrant angle as $\\frac{7\\pi}4 = 315^\\circ$. Note, however, that our condition for $x$ means that $0\\leq\\frac x2\\leq\\pi$, so $45^\\circ$ is the only angle we need consider. x/2 = 45 We found that our angle was 45°, but our angle is $\\frac x2$, so we have $\\frac x2 = 45^\\circ = \\frac\\pi4$ $\\Rightarrow x = 90^\\circ = \\frac\\pi2$ 7. Note, however, that our condition for $x$ means that $0\\leq\\frac x2\\leq\\pi$, so $45^\\circ$ is the only angle we need consider Why is 7pi/4 not possible? Isn't it smaller than 2pi? 8. Originally Posted by Stevo_Evo_22 Note, however, that our condition for $x$", "Home > CALC > Chapter 1 > Lesson 1.4.2 > Problem1-158 1-158. Visualize the angle on the unit circle. What side ratio correspond with that angle? What quadrant is it in? (This will determine whether your answer will be positive or negative.) $\\frac{1}{2}$ See hint in part (a). $\\text{Recall that tan}(a)=\\frac{\\text{sin}(a)}{\\text{cos}(a)}$ $\\text{Recall that sec}(a)=\\frac{1}{\\text{cos}(a)}$", "# Question of inverse trigonometry. How do I evaluate $\\sin \\left(\\sec^{-1}\\left(\\frac{-7}{5}\\right)\\right)$? The answer given is $\\frac{2\\sqrt{6}}{7}$, but I have no idea how to get it. Hope someone can explain it for me. Thanks in advance. where the last line is due to Pythagoras Theorem. Also think of which quadrant is $\\cos^{-1}\\left( -\\frac57 \\right)$ located.", "If $$\\frac{A}{2}$$ lies in quadrant I or II • If $$\\frac{A}{2}$$ lies in quadrant III or IV $$\\cos (\\frac{A}{2}) = \\sqrt{ \\frac{±1+\\cos(A)}{2}}$$ • If $$\\frac{A}{2}$$ lies in quadrant I or IV • If $$\\frac{A}{2}$$ lies in quadrant II or III ### Double and Triple Angle Formulas • $$\\sin 2A = 2\\sin A \\cos A$$ • $$\\cos 2A = \\cos^2 A – \\sin^2 A = 2 \\cos^2 – 1 = 1- \\sin^2 A$$ • $$\\sin 3A = 3\\sin A – 4 \\sin ^3 A$$ • $$\\cos 3A = 4 \\cos^3 A – 3\\cos A$$ • $$\\sin^4 A = 4 \\cos^3 A \\sin A – 4\\cos A \\sin^3 A$$ • $$\\cos^4 A = \\cos^4 A – 6\\cos^2 A\\sin^2 A +\\sin^4 A$$ • $$\\sin^2 A = \\frac{1–\\cos(2A)}{2}$$ • $$\\cos^2 A = \\frac{1+\\cos(2A)}{2}$$ ### Sum and Difference of Angles • $$\\sin(A + B) = \\sin(A)\\cos(B) + \\cos(A)\\sin(B)$$ • $$\\sin(B)\\sin(A−B)=\\sin(A)\\cos(B)−\\cos(A)\\sin(B)$$ • $$\\cos(A+B)=\\cos(A)\\cos(B)−\\sin(A)\\sin(B)$$ • $$\\cos(A−B)=\\cos(A)\\cos(B)+\\sin(A)\\sin(B)$$ • $$\\sin(A+B+C)=\\sin A\\cos B\\cos C+\\cos A\\sin B\\cos C+\\cos A\\cos B\\sin C−\\sin A\\sin B\\sin C$$ • $$\\cos(A+B+C)=\\cos A\\cos B\\cos C−\\sin A\\sin B\\cos C−\\sin A\\cos B\\sin C−\\sin A\\cos B\\sin C−\\cos A\\sin B\\sin C$$ • $$\\sin A + \\sin B = 2\\sin \\frac{(A+B)}{2}\\cos \\frac{(A−B)}{2}$$ • $$\\sin A – \\sin B = 2\\sin \\frac{(A−B)}{2}\\cos \\frac{(A+B)}{2}$$ • $$\\cos A + \\cos B = 2\\cos\\frac{(A+B)}{2} (A+B)2\\cos\\frac{(A−B)}{2}$$ • $$\\cos A + \\cos B = -2\\sin \\frac{(A+B)}{2}\\sin", "to $\\frac{1}{\\sqrt{2}}$ , but we are in the 3rd quadrant which means that both of the values has a negative sign : let’s put these together: $\\mathrm{tan}\\left(225°\\right)\\frac{\\mathrm{sin}\\left(225°\\right)}{\\mathrm{cos}\\left(225°\\right)}=\\frac{-\\frac{1}{\\sqrt{2}}}{-\\frac{1}{\\sqrt{2}}}=1$", "3) $7(cos135i+sin135j)$ what is your problem with this vector? Find the vector v with the given magnitude and the same direction as u $|v| = 1, u =<3,-3>$ v = < 1/sqrt(2) , -1/sqrt(2) > ... • July 19th 2010, 04:27 PM nhunhu9 Number 3 is also find the direction angle but since the fisrt and second problems have to do with quadrant i already firgured it out. Thank you", "$\\tan { 22{ \\frac { 1 }{ 2 } }^{ ° } }$ 24. Find all other trigonometrical ratios if $\\sin x=\\frac{-2\\sqrt6}{5}$ and x lies in III quadrant? 25. Part D 2 x 5 = 10 26. Prove that sin(n+1)x sin(n+2) x+cos(n+1)xcos(n+2)x=cosx. 27. Prove that cos22x-cos26x = sin 4x.sin 8x", "# 6.1 Positive and Negative Angles 6.1 Positive and Negative Angles 1. Positive angles are angles measure in an anticlockwise rotate from the positive x-axis about the origin, O. 2. Negative angles are angles measured in a clockwise rotation from the positive x-axis about the origin O. 3. One complete revolution is 360° or 2π radians. Example: Show each of the following angles on a separate diagram and state the quadrant in which the angle is situated. (a) 410° (b) 890° (e) –60o (f) –500° Solution: (a) 410° = 360° + 50° Based on the above circular diagram, the positive angle of 410° is in the first quadrant. (b) 890° = 720° + 170° Based on the above circular diagram, the positive angle of 890° is in the second quadrant. (c) $\\frac{22}{9}\\pi \\text{rad}=\\left(2\\pi +\\frac{4}{9}\\pi \\right)\\text{rad}={360}^{o}+{80}^{o}$ Based on the above circular diagram, the positive angle of is in the first quadrant. (d) $\\frac{10}{3}\\pi \\text{rad}=\\left(3\\pi +\\frac{1}{3}\\pi \\right)\\text{rad}={540}^{o}+{60}^{o}$ Based on the above circular diagram, the positive angle of is in the third quadrant. (e) Based on the above circular diagram, the negative angle of –60° is in the fourth quadrant. (f) –500° = –360° – 140° Based on the above circular diagram, the negative angle of –500° is in the third quadrant. (g) $-3\\frac{1}{4}\\pi \\text{rad}=\\left(-3\\pi -\\frac{1}{4}\\pi \\right)\\text{rad}=-{540}^{o}-{45}^{o}$ Based on the above", "# Question #70b87 May 13, 2017 If Quadrant I: $\\frac{4}{5}$ If Quadrant II: $- \\frac{4}{5}$ #### Explanation: Use Pythagorean's Identity: ${\\sin}^{2} x + {\\cos}^{2} x = 1$ By substitution, we get: ${\\left(\\frac{3}{5}\\right)}^{2} + {\\cos}^{2} a = 1$ Subtract ${\\left(\\frac{3}{5}\\right)}^{2}$ from both sides: ${\\cos}^{2} a = 1 - {\\left(\\frac{3}{5}\\right)}^{2}$ ${\\cos}^{2} a = 1 - \\frac{9}{25}$ ${\\cos}^{2} a = \\frac{16}{25}$ Now, we take the square root of both sides: $\\cos a = \\pm \\sqrt{\\frac{16}{25}}$ $\\cos a = \\pm \\frac{4}{5}$ Therefore, if $\\angle A$ is in Quadrant I, $\\cos a = \\frac{4}{5}$ if $\\angle A$ is in Quadrant II, $\\cos a = - \\frac{4}{5}$", "positive in the fourth quadrant and because $$\\sin(-300) = -\\sin(300) = --\\frac{\\sqrt{3}}{2}=\\frac{\\sqrt{3}}{2}$$ Just do it. $$\\cos A = \\frac 12$$ So what is $$\\sin A$$? Well $$\\cos^2 A + \\sin^2 A = 1$$ and so $$(\\frac 12)^2 + \\sin^2 A = 1$$ so $$\\sin^2 A= 1 - \\frac 14$$ so $$\\sin A =\\pm \\frac{\\sqrt {3}}2$$. But we are told $$\\sin A$$ and $$\\cos A$$ are the same sign so $$\\sin A =\\frac {\\sqrt 3}2$$. So what is $$\\sin -A$$? Well, we have a rule that $$\\sin (-A) = -\\sin A$$ (and ,yes, that is true for all quadrants) so $$\\sin (-A) = -\\sin A = -\\frac {\\sqrt 3}2$$. I'm not entirely sure where you trouble lies. ===== Given that $$\\cos A=1/2$$ Okay, that means that $$\\cos A > 0$$ so $$A$$ is in the first or fourth quadrant. and $$\\cos A$$ and $$\\sin A$$ have the same sign Okay, that means that $$A$$ is in first or third. In combo with $$\\cos A > 0$$ that means $$\\sin A > 0$$ and $$A$$ is in the first. Is it because this rule only applies to positive angles? What rule? There's nothing wrong and off here. We have $$\\sin A > 0$$ and so $$\\sin (-A) = - \\sin A < 0$$. This is exactly what we'd expect. Somehow", "of $A$ is $5$. We now need to pick a positive $y$ coordinate for $A$ so that we'll have $\\angle AOB=30^\\circ$. By the Pythagorean theorem we have $AO^2 = AB^2 + BO^2 = AB^2 + 25$. By the definition of sine, we have $\\frac{AB}{AO} = \\sin AOB = \\sin 30^\\circ = \\frac 12$, hence $AO=2\\cdot AB$. Substituting into the previous equation, we get $AB^2 = \\frac{25}3$, hence $AB=\\frac{5\\sqrt 3}3$. This means that the coordinates of $A$ are $\\left(5,\\frac{5\\sqrt 3}3\\right)$. After we rotate $OA$ $90^\\circ$ counterclockwise about $O$, it will be in the second quadrant and have the coordinates $\\boxed{ \\left( -\\frac{5\\sqrt 3}3, 5\\right) }$. So the answer is $\\boxed{\\text{B}}$. 2008 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions", "# How do you evaluate cos^-1(cos ((9pi)/8))? Oct 12, 2016 ${\\cos}^{-} 1 \\left(\\cos \\left(\\frac{9 \\pi}{8}\\right)\\right) = {\\cos}^{-} 1 \\left(\\cos \\left(\\frac{7 \\pi}{8}\\right)\\right) = \\frac{7 \\pi}{8}$ #### Explanation: Since the restriction for ${\\cos}^{-} 1 x$ is $\\left[0 , \\pi\\right] , \\frac{9 \\pi}{8}$ is in quadrant three and cosine is negative in quadrant three. So we need to find the reference angle which is $\\frac{9 \\pi}{8} - \\pi = \\frac{\\pi}{8}$. Hence the argument x in quadrant II is$\\frac{7 \\pi}{8}$ since cosine is negative in quadrant two from the restriction. Lastly, use the property f^-1(f(x)=x so ${\\cos}^{-} 1 \\left(\\cos \\left(\\frac{9 \\pi}{8}\\right)\\right) = {\\cos}^{-} 1 \\left(\\cos \\left(\\frac{7 \\pi}{8}\\right)\\right) = \\frac{7 \\pi}{8}$" ]

[ "# Question #cca9c Apr 27, 2017 $x = {143}^{o}$ #### Explanation: First of all, it helps to know that Sin (${37}^{o}$) = $\\frac{3}{5}$ and Cos (${37}^{o}$) = $\\frac{4}{5}$. Now according to your question, Sin x is positive and cos x is negative. This is only possible in the second quadrant of the cartesian plane, where Sin x is positive, and Cos x & Tan x are negative. In the second quadrant, all angles lie between ${90}^{o} \\mathmr{and} {180}^{o}$. So we know that $S \\in \\left(x\\right) = S \\in \\left({180}^{o} - x\\right)$ and $- C o s \\left(x\\right) = C o s \\left({180}^{o} - x\\right)$ Putting x as ${37}^{o}$, we get $S \\in \\left({37}^{o}\\right) = S \\in \\left({143}^{o}\\right)$ $S \\in \\left({143}^{o}\\right) = \\frac{3}{5}$ Also, $- C o s \\left({37}^{o}\\right) = C o s \\left({143}^{o}\\right)$ $C o s \\left({143}^{o}\\right) = - \\frac{4}{5}$ As all conditions of the question are satisfied, $x = {143}^{o}$ Apr 27, 2017 $x \\approx {143.13}^{\\circ}$ or 2.498 radians #### Explanation: We know that x is in the second quadrant, because the sine is positive and cosine is negative. If we use the cosine function and take its inverse, the calculator will do all of the work to give us the correct angle. $\\cos \\left(x\\right) = - \\frac{4}{5}$ Use the inverse cosine function on both sides: ${\\cos}^{-}", "quadrants from the first. You’ll see this in the following examples. Find the exact value of each of the following. In other words, don’t use a calculator. Show All SolutionsHide All Solutions 1. $$\\sin \\left( {\\frac{{2\\pi }}{3}} \\right)$$ and $$\\sin \\left( { - \\frac{{2\\pi }}{3}} \\right)$$ Show Solution The first evaluation here uses the angle $$\\frac{{2\\pi }}{3}$$. Notice that $$\\frac{{2\\pi }}{3} = \\pi - \\frac{\\pi }{3}$$. So $$\\frac{{2\\pi }}{3}$$ is found by rotating up $$\\frac{\\pi }{3}$$ from the negative $$x$$-axis. This means that the line for $$\\frac{{2\\pi }}{3}$$ will be a mirror image of the line for $$\\frac{\\pi }{3}$$ only in the second quadrant. The coordinates for $$\\frac{{2\\pi }}{3}$$ will be the coordinates for $$\\frac{\\pi }{3}$$ except the $$x$$ coordinate will be negative. Likewise, for $$- \\frac{{2\\pi }}{3}$$we can notice that $$- \\frac{{2\\pi }}{3} = - \\pi + \\frac{\\pi }{3}$$, so this angle can be found by rotating down $$\\frac{\\pi }{3}$$ from the negative $$x$$-axis. This means that the line for $$- \\frac{{2\\pi }}{3}$$ will be a mirror image of the line for $$\\frac{\\pi }{3}$$ only in the third quadrant and the coordinates will be the same as the coordinates for $$\\frac{\\pi }{3}$$ except both will be negative. Both of these angles along with their coordinates are shown on the following unit circle. From this unit circle we", "# Math Help - double angle problem 1. ## double angle problem It says the sinx= $14/15$ and it is in quadrant 2 what quadrant is the sin2x in? (NO CALCULATOR)i figured it out with a calc but i cant w/out a calc i really need help fast thnx 2. Originally Posted by brianH It says the sinx= $14/15$ and it is in quadrant 2 what quadrant is the sin2x in? (NO CALCULATOR)i figured it out with a calc but i cant w/out a calc i really need help fast thnx $\\sin(2x)=2\\sin(x)\\cos(x)=2\\cdot \\frac{14}{15}\\cdot \\frac{\\sqrt{29}}{15}$ 3. i got that and it was $-28 \\frac{\\sqrt{29}}{225}$ but how do i fugure out what quad thats in 4. Originally Posted by brianH It says the sinx= $14/15$ and it is in quadrant 2 what quadrant is the sin2x in? (NO CALCULATOR)i figured it out with a calc but i cant w/out a calc i really need help fast thnx We will use the identity $sin(2x)=2sinxcosx$ to help us. Since $sinx=\\frac{14}{15}$, we can determine that $cosx=\\frac{-\\sqrt{29}}{15}$ (since sine was defined in the 2nd quadrant). Plugging this into the identity, we have: $sin(2x)=2{\\left(\\frac{14}{15}\\right)}{\\left(\\frac{-\\sqrt{29}}{15}\\right)}=\\frac{-28\\sqrt{29}}{225}$. I'm going to conclude that $sin2x$ falls in the 4th quadrant, since $sin\\theta=\\frac{opp}{adj}$. Since the opposite is negative, it will reside either in the 3rd or 4th quadrants. Since the adjacent", "# How do you find the exact values of sin^-1(1/2)? ##### 1 Answer Oct 15, 2016 ${\\sin}^{-} 1 \\left(\\frac{1}{2}\\right) = \\frac{\\pi}{6}$ #### Explanation: Find ${\\sin}^{-} 1 \\left(\\frac{1}{2}\\right)$ This problem is asking for the ANGLE with a sine of $\\frac{1}{2}$. The range of ${\\sin}^{-} 1$ or $\\arcsin$ is between $\\frac{\\pi}{2}$ and $- \\frac{\\pi}{2}$. If you are finding ${\\sin}^{-} 1$ of a positive value, the answer will be between 0 and $\\frac{\\pi}{2}$, or the first quadrant in the unit circle. Do NOT use the second quadrant angle with a sine of $\\frac{1}{2}$, because it does not fall within the range of ${\\sin}^{-} 1$. Using the unit circle, the angle with a sine of $\\frac{1}{2}$ in the first quadrant is $\\frac{\\pi}{6}$.", "If $$\\frac{A}{2}$$ lies in quadrant I or II • If $$\\frac{A}{2}$$ lies in quadrant III or IV $$\\cos (\\frac{A}{2}) = \\sqrt{ \\frac{±1+\\cos(A)}{2}}$$ • If $$\\frac{A}{2}$$ lies in quadrant I or IV • If $$\\frac{A}{2}$$ lies in quadrant II or III ### Double and Triple Angle Formulas • $$\\sin 2A = 2\\sin A \\cos A$$ • $$\\cos 2A = \\cos^2 A – \\sin^2 A = 2 \\cos^2 – 1 = 1- \\sin^2 A$$ • $$\\sin 3A = 3\\sin A – 4 \\sin ^3 A$$ • $$\\cos 3A = 4 \\cos^3 A – 3\\cos A$$ • $$\\sin^4 A = 4 \\cos^3 A \\sin A – 4\\cos A \\sin^3 A$$ • $$\\cos^4 A = \\cos^4 A – 6\\cos^2 A\\sin^2 A +\\sin^4 A$$ • $$\\sin^2 A = \\frac{1–\\cos(2A)}{2}$$ • $$\\cos^2 A = \\frac{1+\\cos(2A)}{2}$$ ### Sum and Difference of Angles • $$\\sin(A + B) = \\sin(A)\\cos(B) + \\cos(A)\\sin(B)$$ • $$\\sin(B)\\sin(A−B)=\\sin(A)\\cos(B)−\\cos(A)\\sin(B)$$ • $$\\cos(A+B)=\\cos(A)\\cos(B)−\\sin(A)\\sin(B)$$ • $$\\cos(A−B)=\\cos(A)\\cos(B)+\\sin(A)\\sin(B)$$ • $$\\sin(A+B+C)=\\sin A\\cos B\\cos C+\\cos A\\sin B\\cos C+\\cos A\\cos B\\sin C−\\sin A\\sin B\\sin C$$ • $$\\cos(A+B+C)=\\cos A\\cos B\\cos C−\\sin A\\sin B\\cos C−\\sin A\\cos B\\sin C−\\sin A\\cos B\\sin C−\\cos A\\sin B\\sin C$$ • $$\\sin A + \\sin B = 2\\sin \\frac{(A+B)}{2}\\cos \\frac{(A−B)}{2}$$ • $$\\sin A – \\sin B = 2\\sin \\frac{(A−B)}{2}\\cos \\frac{(A+B)}{2}$$ • $$\\cos A + \\cos B = 2\\cos\\frac{(A+B)}{2} (A+B)2\\cos\\frac{(A−B)}{2}$$ • $$\\cos A + \\cos B = -2\\sin \\frac{(A+B)}{2}\\sin", "# Solution If $\\sin t = 1/4$, and the terminal side of angle $t$ is in quadrant II, find $\\cos t$. Combining the Pythagorean Identity $\\cos^2 t + \\sin^2 t = 1$ and the given $\\sin t = 1/4$, we have $$\\cos^2 t + \\left( \\frac{1}{4} \\right)^2 = 1$$ But then if we isolate the $\\cos^2 t$ term, we have $$\\cos^2 t = 1 - \\frac{1}{16} = \\frac{15}{16}$$ Hence, $$\\cos t = \\pm \\sqrt{\\frac{15}{16}} = \\pm \\frac{\\sqrt{15}}{4}$$ Note that $\\cos t$ is negative in quadrant II, so only one of these two values makes any sense $$\\cos t = -\\frac{\\sqrt{15}}{4}$$", "$\\tan { 22{ \\frac { 1 }{ 2 } }^{ ° } }$ 24. Find all other trigonometrical ratios if $\\sin x=\\frac{-2\\sqrt6}{5}$ and x lies in III quadrant? 25. Part D 2 x 5 = 10 26. Prove that sin(n+1)x sin(n+2) x+cos(n+1)xcos(n+2)x=cosx. 27. Prove that cos22x-cos26x = sin 4x.sin 8x", "= - \\sqrt{\\frac{1 - \\cos \\alpha}{2}}$ if $\\frac{\\alpha}{2}$ is located in the third or fourth quadrant. $\\cos \\frac{\\alpha}{2} = \\sqrt{\\frac{1 + \\cos \\alpha}{2}}$ if $\\frac{\\alpha}{2}$ is located in either the first or fourth quadrant. $\\cos \\frac{\\alpha}{2} = - \\sqrt{\\frac{1 + \\cos \\alpha}{2}}$ if $\\frac{\\alpha}{2}$ is located in either the second or fourth quadrant. Tip: This all depends how the values of cosine and sine are negative dependig which quadrant cosine or sine lies on.", "## Algebra and Trigonometry 10th Edition False. $sin\\frac{u}{2}=+\\sqrt {\\frac{1-cos~u}{2}}$ If $u$ is in the second quadrant: $\\frac{\\pi}{2}\\lt u\\lt\\pi$. Then, $\\frac{\\pi}{4}\\lt \\frac{u}{2}\\lt\\frac{\\pi}{2}$. That is $u$ is in the first quadrant. So, $sin\\frac{u}{2}\\gt0$", "may show by graphing the given system that point P will lie in the second quadrant in the new coordinate system. Therefore, the abscissa of $P$ is negative, and is therefore $-\\frac{123}{13}$; similarly, the ordinate of $P$ is positive and is therefore $\\frac{526}{13}$. Thus, we have that $\\alpha=-\\frac{123}{13}$ and that $\\beta=\\frac{526}{13}$. It follows that $\\alpha+\\beta=-\\frac{123}{13}+\\frac{526}{13}=\\frac{403}{13}=\\boxed{031}$.", "# If sin A = 1/3 and 0< A < 90, then what is cos A? Jun 4, 2018 See below #### Explanation: If $\\sin A = \\frac{1}{3}$ and $0 < A < 90$ Then using ${\\sin}^{2} A + {\\cos}^{2} A = 1$ ${\\cos}^{2} A = 1 - {\\sin}^{2} A = 1 - \\frac{1}{9} = \\frac{8}{9}$ Then $\\cos A = \\pm \\sqrt{\\frac{8}{9}} = \\pm \\frac{2 \\sqrt{2}}{3}$ But if 0< A< 90, then the cosine is positive and $\\cos A = \\frac{2 \\sqrt{2}}{3}$ Jun 4, 2018 $\\cos A = \\frac{2 \\sqrt{2}}{3}$ #### Explanation: Since $0 < A < {90}^{o}$ we know that $\\angle A$ is in the first quadrant. Consider the right $\\triangle O A B$ where $O$ is the origin and side $A B$ is opposite $\\angle A$ We are told that $\\sin A = \\frac{1}{3}$ $\\therefore$ side $A B = 1$ and side $O A = 3$ Applying Pythagoras ${\\left(O A\\right)}^{2} = {\\left(A B\\right)}^{2} + {\\left(O B\\right)}^{2}$ $\\therefore {3}^{2} = {1}^{2} + O {B}^{2}$ $O B = \\sqrt{9 - 1}$ [Since $O B > 0$ because $\\angle A$ is in the first quadrant] $O B = \\sqrt{8} = 2 \\sqrt{2}$ Now, $\\cos A = \\frac{O B}{O A}$ $\\cos A = \\frac{2 \\sqrt{2}}{3}$", "wrong here? The second zero calculates to cosx=(-1/2), and cosx is negative in quadrants two and three, not one and four ... I don't see my error. 2. Hello, Jeev! Welcome aboard! $2\\cos^2\\!x+\\cos x\\:=\\:0$ Solve for the domain $(0,\\:2\\pi)$ to nearest hundredth of a radian. Factor: . $\\cos x(2\\cos x + 1) \\:=\\:0$ Then: . $\\begin{array}{ccccccccccc}\\cos x \\:=\\:0 & \\Rightarrow &\\boxed{x \\:=\\:\\dfrac{\\pi}{2},\\:\\dfrac{3\\pi}{2}} \\\\ \\\\[-3mm] 2\\cos x + 1 \\:=\\:0 & \\Rightarrow & \\cos x \\:=\\:\\text{-}\\frac{1}{2} & \\Rightarrow &\\boxed{ x \\:=\\:\\dfrac{2\\pi}{3},\\:\\dfrac{4\\pi}{3}} \\end{array}$ 3. Thanks for giving me the solutions, but I am not understanding how you get the last two. The CAST rule states that when cosx<0, the angle is in quadrants two and three, which would mean: (pi) - 2.09 (pi) + 2.09 All I need to understand is, for the second zero, if cosx=(-1/2), why are we calculating the angle in quadrants one and four? 4. Originally Posted by Jeev Thanks for giving me the solutions, but I am not understanding how you get the last two. The CAST rule states that when cosx<0, the angle is in quadrants two and three, which would mean: (pi) - 2.09 (pi) + 2.09 All I need to understand is, for the second zero, if cosx=(-1/2), why are we calculating the angle in quadrants one and four? When you put", "worked example, it should hopefully make a bit more sense: We want to find $x$ such that $\\cos\\frac x2=\\frac{\\sqrt2}2,x\\in[0,2\\pi]$ The cosine we were given is a positive value, $\\frac{\\sqrt{2}}2$, so it must lie either in the first or fourth quadrant (think about your unit circle: the cosine is the $x$-coordinate, so it will be positive in quadrants I and IV). This means that our angle must lie either in $\\left[0,\\frac\\pi2\\right]$ or in $\\left[\\frac{3\\pi}2,2\\pi\\right]$. inverse cosine((square-root 2)/2)=45degrees If you are familiar with your unit circle, you should be able to determine that $\\arccos\\frac{\\sqrt2}2 = \\frac\\pi4$ or $45^\\circ$. This is the angle in the first quadrant, and looking at the unit circle you should be able to find the fourth quadrant angle as $\\frac{7\\pi}4 = 315^\\circ$. Note, however, that our condition for $x$ means that $0\\leq\\frac x2\\leq\\pi$, so $45^\\circ$ is the only angle we need consider. x/2 = 45 We found that our angle was 45°, but our angle is $\\frac x2$, so we have $\\frac x2 = 45^\\circ = \\frac\\pi4$ $\\Rightarrow x = 90^\\circ = \\frac\\pi2$ 7. Note, however, that our condition for $x$ means that $0\\leq\\frac x2\\leq\\pi$, so $45^\\circ$ is the only angle we need consider Why is 7pi/4 not possible? Isn't it smaller than 2pi? 8. Originally Posted by Stevo_Evo_22 Note, however, that our condition for $x$", "1. ## [SOLVED] Trignometry and Radians Here's the question I have solved the first part and got the following equation, which is correct: $y=(6sin2x)+3$ Now for part (ii) I have solved till $sin2x=\\frac{-1}{2}$ And don't know how to get the correct answer: Spoiler: 2. Originally Posted by unstopabl3 Here's the question I have solved the first part and got the following equation, which is correct: $y=(6sin2x)+3$ Now for part (ii) I have solved till $sin2x=\\frac{-1}{2}$ And don't know how to get the correct answer: Spoiler: Hi unstopabl3, $sin(2x)=-\\frac{1}{2}$ the first time this occurs is in the 3rd quadrant, as sin(angle) points out the y co-ordinate of a point on the circle. Hence the angle 2x is ${\\pi}+sin^{-1}\\left(\\frac{1}{2}\\right)$ $={\\pi}+\\frac{{\\pi}}{6}=\\frac{7{\\pi}}{6}$ x is half that. 3. Yea, I figured it out myself as well! Haha Thanks for the quick and easy to understand reply!", "Home > CALC > Chapter 1 > Lesson 1.4.2 > Problem1-158 1-158. Visualize the angle on the unit circle. What side ratio correspond with that angle? What quadrant is it in? (This will determine whether your answer will be positive or negative.) $\\frac{1}{2}$ See hint in part (a). $\\text{Recall that tan}(a)=\\frac{\\text{sin}(a)}{\\text{cos}(a)}$ $\\text{Recall that sec}(a)=\\frac{1}{\\text{cos}(a)}$", "positive in the fourth quadrant and because $$\\sin(-300) = -\\sin(300) = --\\frac{\\sqrt{3}}{2}=\\frac{\\sqrt{3}}{2}$$ Just do it. $$\\cos A = \\frac 12$$ So what is $$\\sin A$$? Well $$\\cos^2 A + \\sin^2 A = 1$$ and so $$(\\frac 12)^2 + \\sin^2 A = 1$$ so $$\\sin^2 A= 1 - \\frac 14$$ so $$\\sin A =\\pm \\frac{\\sqrt {3}}2$$. But we are told $$\\sin A$$ and $$\\cos A$$ are the same sign so $$\\sin A =\\frac {\\sqrt 3}2$$. So what is $$\\sin -A$$? Well, we have a rule that $$\\sin (-A) = -\\sin A$$ (and ,yes, that is true for all quadrants) so $$\\sin (-A) = -\\sin A = -\\frac {\\sqrt 3}2$$. I'm not entirely sure where you trouble lies. ===== Given that $$\\cos A=1/2$$ Okay, that means that $$\\cos A > 0$$ so $$A$$ is in the first or fourth quadrant. and $$\\cos A$$ and $$\\sin A$$ have the same sign Okay, that means that $$A$$ is in first or third. In combo with $$\\cos A > 0$$ that means $$\\sin A > 0$$ and $$A$$ is in the first. Is it because this rule only applies to positive angles? What rule? There's nothing wrong and off here. We have $$\\sin A > 0$$ and so $$\\sin (-A) = - \\sin A < 0$$. This is exactly what we'd expect. Somehow", "\\frac{4\\pi}{5}$ Related angle is $\\pi-\\frac{4\\pi}{5}=\\frac{\\pi}{5}$ $\\sin \\frac{4\\pi}{5}$ is a QII angle. Sin is positive in the first and second quadrants. Therefore, $\\sin -\\frac{4\\pi}{5}=-\\sin \\frac{\\pi}{5}$", "# If tan x=1, how do you find sin x/2 , cos x/2 , and tan x/2? Not enough information is provided. If $\\tan x = 1$ then x is in either quadrant 1 or 3 with reference angle $\\frac{\\pi}{4}$. So ${\\cos}^{2} x = \\frac{1}{2}$. If $x = \\frac{\\pi}{4}$ then for $\\frac{x}{2}$ all trig values are positive. If $x = \\frac{- 7 \\pi}{4}$ then $\\frac{x}{2}$ has negative sine and cosine. For some other possibilities for $x$, the half angle has sine positive and cosine negative, for still others, it is opposite.", "# Math Help - Can someone show me how to use the half angle formula? 1. ## Can someone show me how to use the half angle formula? Thanks 2. Originally Posted by qbkr21 Thanks You have that, $\\sin^2 x=\\frac{1-\\cos 2x}{2}$ $\\cos^2 x=\\frac{1+\\cos 2x}{2}$ So to get your value for $\\sin x,\\cos x$ you take the square root. CAREFUL, since there are two roots it can be negative or positive. How do you know? Simple, determine the quadrant of the angle and where it lies.", "positive values in the sum angle identity? And should the compound angle identity be equal to something ? Since you know that sin(x) is positive, that means x could be in the first and second quadrants, which means cos(x) could be positive or negative. Since you know that cos(y) is negative, that means y could be in the second or third quadrants, which means sin(x) could be positive or negative. So yes, since there are four possibilities, you will need to substitute all four values to get four possible results 6. ## Re: Trigonometric Identities Hello, jessm001! We are expected to find those missing values. $\\text{If }\\sin x = \\tfrac{2}{3}\\text{ and }\\cos y = \\text{-}\\tfrac{2}{7}\\text{, find the possible values of }\\cos(x+y).$ $\\sin x \\:=\\:\\frac{2}{3} \\:=\\:\\frac{opp}{hyp} \\qquad x\\text{ is in Quadrant 1 or 2.}$ . . $adj \\:=\\:\\pm\\sqrt{5} \\quad\\Rightarrow\\quad \\cos x \\:=\\:\\pm\\frac{\\sqrt{5}}{3}$ $\\cos y \\:=\\:\\frac{\\text{-}2}{7} \\:=\\:\\frac{adj}{hyp}\\qquad y\\text{ is in Quadrant 2 or 3.}$ . . $opp \\:=\\:\\pm\\sqrt{45} \\:=\\:\\pm3\\sqrt{5} \\quad\\Rightarrow\\quad \\sin y \\:=\\:\\pm\\frac{3\\sqrt{5}}{7}$ $\\cos(x + y) \\;=\\;\\cos x\\cos y - \\sin x\\sin y$ . . . . . . . . $=\\;\\left(\\pm\\frac{\\sqrt{5}}{3}\\right)\\left(-\\frac{2}{7}\\right) - \\left(\\frac{2}{3}\\right)\\left(\\pm\\frac{3\\sqrt{5}}{ 7}\\right)$ There are four possible values: . $\\frac{8\\sqrt{5}}{21},\\;-\\frac{8\\sqrt{5}}{21},\\;\\frac{4\\sqrt{5}}{21},\\;-\\frac{4\\sqrt{5}}{21}$ 7. ## Re: Trigonometric Identities Thanks a lot !" ]

[ "For certain ordered pairs $$(a,b)\\,$$ of real numbers, the system of equations $$ax+by=1\\,$$ $$x^2+y^2=50\\,$$ has at least one solution, and each solution is an ordered pair $$(x,y)\\,$$ of integers. How many such ordered pairs $$(a,b)\\,$$ are there? (第十二届AIME1994 第7题)", "Find the number of ordered pairs of real numbers $$(a,b)$$ such that whenever $$\\gamma$$ is a root of $$x^2+ax+b=0$$, $$\\gamma^{2}-2$$ is also a root of the same equation.", "+0 Help! 0 133 1 For certain ordered pairs (a,b) of real numbers, the system of equations [\\begin{aligned} ax+by&=1 \\\\ x^2 + y^2 &= 50 \\end{aligned} has at least one solution, and each solution is an ordered pair (x,y) of integers. How many such ordered pairs (a,b) are there? Dec 16, 2018 $$\\text{there are only so many pairs of integers that satisfy the second equation}\\\\ \\text{They are }(\\pm 1, \\pm 7),(\\pm 7,\\pm 1),(\\pm 5,\\pm 5)\\\\ \\text{each of these produces 4 ordered pairs }(a,b) \\text{ thus there are 12 total}$$", "+0 # Help! 0 39 1 For certain ordered pairs (a,b) of real numbers, the system of equations [\\begin{aligned} ax+by&=1 \\\\ x^2 + y^2 &= 50 \\end{aligned} has at least one solution, and each solution is an ordered pair (x,y) of integers. How many such ordered pairs (a,b) are there? Dec 16, 2018 #1 +3563 +2 $$\\text{there are only so many pairs of integers that satisfy the second equation}\\\\ \\text{They are }(\\pm 1, \\pm 7),(\\pm 7,\\pm 1),(\\pm 5,\\pm 5)\\\\ \\text{each of these produces 4 ordered pairs }(a,b) \\text{ thus there are 12 total}$$ . Dec 16, 2018", "# Back to basic algebra Algebra Level 4 Find all ordered pairs of real numbers $$(a, b)$$ such that $${ (x-a) }^{ 2 }+{ (2x-b) }^{ 2 }={ (x-3) }^{ 2 }+{ (2x) }^{ 2 }$$ is true for all values of $$x$$. Let $$n$$ be the sum of all $$a$$'s in ordered pair $$(a, b)$$ and $$m$$ be the sum of all $$b$$'s in the ordered pair $$(a, b)$$. What is $$\\frac { n }{ m }$$? ×", "l + \\begin{bmatrix}4\\\\0\\\\0\\\\1\\end{bmatrix} m,$$ where $k,l,m$ are some real numbers. by Professor selected by", "in terms of real-value functions. b. Also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as $t\\to \\infty$. $$\\mathbf{x}' =\\begin{pmatrix} -1 &-4\\\\ 1 &-1 \\end{pmatrix}\\mathbf{x}$$ 5 ##### Quiz-6 / Q6--T0901 « on: March 16, 2018, 08:17:56 PM » a. Express the general solution of the given system of equations in terms of real-valued functions. b. Also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as $t\\to \\infty$. $$\\mathbf{x}' =\\begin{pmatrix} 4 &-3\\\\ 8 &-6 \\end{pmatrix}\\mathbf{x}$$ 6 ##### Quiz-6 / Q6--T0801 « on: March 16, 2018, 08:17:05 PM » a. Express the general solution of the given system of equations in terms of real-valued functions. b. Also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as $t\\to \\infty$. $$\\mathbf{x}' =\\begin{pmatrix} 1 &-1\\\\ 5 &-3 \\end{pmatrix}\\mathbf{x}$$ 7 ##### Quiz-6 / Q6--T0701 « on: March 16, 2018, 08:16:03 PM » a. Express the general solution of the given system of equations in terms of real-valued functions. b. Also draw a direction field, sketch a few of the trajectories, and describe the behavior of the solutions as $t\\to \\infty$. $$\\mathbf{x}' =\\begin{pmatrix} 1 &2\\\\ -5 &-1 \\end{pmatrix}\\mathbf{x}$$ 8 ##### Quiz-6 / Q6--T0601 « on: March 16, 2018,", "+0 0 78 2 One ordered pair $$(a,b)$$ satisfies the two equations $$ab^4 = 384$$ and $$a^2 b^5 = 4608$$. What is the value of $$a$$ in this ordered pair? May 9, 2020 #1 0 a = 2 \\sqrt[3]{4} (and b = 4 \\sqrt[3]{6}) May 9, 2020 #2 0 umm its wrong, thats all i know Guest May 9, 2020", "we already are accounting for the ordering between the students: we're giving the first pair of questions to the first student, not the second pair (and so on). the main point is that if you sepcify three students with $A$,$B$,$C$, which means that we distinguish these three student, then we can conclude the answer is $\\{\\begin{pmatrix} 3\\\\ 1 \\end{pmatrix}+\\begin{pmatrix} 2\\\\ 1 \\end{pmatrix}+\\begin{pmatrix} 1\\\\ 1 \\end{pmatrix} \\} \\begin{pmatrix} 8\\\\ 2 \\end{pmatrix}\\begin{pmatrix} 6\\\\ 2 \\end{pmatrix}\\begin{pmatrix} 4\\\\ 2 \\end{pmatrix}$, not $3!(\\ldots)$", "the moment that we're happy with the real numbers and their properties. (Of course, there's a separate question - \"how do we construct the reals?\" - but let's leave that aside for now.) There are now a number of ways to rigorously define the complex numbers. Here's probably the simplest (although not the most useful): we let $\\mathbb{C}$ be the set of ordered pairs $(a, b)$ where $a, b\\in\\mathbb{R}$, and we define the operations $+$ and $\\times$ on $\\mathbb{C}$ as follows (intuitively, the pair $(a, b)$ represents the number $a+bi$): • $(a, b)+(c, d)=(a+b, c+d)$. • $(a, b)\\times (c, d)=(a\\times c-b\\times d, a\\times d+b\\times c)$. Note that on the right hand side of each of those expressions, I'm using the notion of $+$ and $\\times$ (and $-$) for real numbers. Arguably, a better way to write this would be • $(a, b)+_\\mathbb{C}(c, d)=(a+_\\mathbb{R}b, c+_\\mathbb{R}d)$. • $(a, b)\\times_\\mathbb{C} (c, d)=(a\\times_\\mathbb{R} c-_\\mathbb{R}b\\times_\\mathbb{R} d, a\\times_\\mathbb{R} d+_\\mathbb{R}b\\times_\\mathbb{R} c)$. That is, I've already defined operations $+_\\mathbb{R}, \\times_\\mathbb{R}, -_\\mathbb{R}$ for the reals, and now I'm defining new operations $+_\\mathbb{C}, \\times_\\mathbb{C}$ for the complex numbers. In general, it will always be clear from context which is meant; but it's a good habit when learning this material to get into to clarify things with subscripts whenever we're using both types of operation. We can use this", "$U=\\mathbf{Q}-L$. Then the ordered pair $< L,U >$ is called a section or a cut of the set of rational numbers. This section of the set of rational numbers is called a real number. Notation: The set of real numbers $\\alpha, \\beta, \\gamma, \\ldots$ is denote by $\\mathbf{R}$. Let $\\alpha = \\langle L,U \\rangle$ then $L$ and $U$ are called Lower and Upper Class of $\\alpha$ respectively. These classes will be denoted by $L(\\alpha)$ and $U(\\alpha)$ respectively.", "+0 # a bit of help +1 291 1 +816 Suppose that a and b are nonzero real numbers, and that the equation $$x^2+ax+b=0$$ has solutions a and b. Find the ordered pair (a,b). Dec 25, 2018 $$\\text{as }a \\text{ and }b \\text{ are solutions to }x^2 + ax + b = 0\\\\ \\text{we can write}\\\\ x^2 + ax + b = (x-a)(x-b) = x^2 -(a+b)x + ab\\\\ a = -(a+b) \\text{ and } b= a b\\\\ b = -2a\\\\ b = a(-2a) = -2a^2\\\\ -2a = -2a^2 \\Rightarrow a = 1\\\\ b = -2a = -2\\\\ (a,b) = (1,-2)$$", "# Circle and a line! $\\begin{cases} ax+by=1 \\\\ x^2 + y^2 = 50 \\end{cases}$ For certain ordered pairs $$(a,b)$$ of real numbers, the system of equations above has at least one solution, and each solution is an ordered pair $$(x,y)$$ of integers. How many such ordered pairs $$(a,b)$$ are there? ×", "1& 0 \\\\ 0 &1\\\\ 2 & 2\\end{pmatrix}$. Then $AB=I_2$ For the other one, consider $A=\\begin{pmatrix} 1& 2 \\end{pmatrix}$ and $B=\\begin{pmatrix} 2\\\\ -1 \\end{pmatrix}$. Then $AB=0$ • Nice solution!! – BAYMAX Jul 2 '18 at 15:51", "# 2 Mathematicians creating Madness Algebra Level 5 How many ordered triples of real numbers $$(a, b, c)$$ are there which satisfy: $a^{2} + b^{2} +c^{2} = 66$ $a^{3} + b^{3} + c^{3} = 408$$a^{4} + b^{4} + c^{4} = 2658$ This is part of Ordered Disorder. × Problem Loading... Note Loading... Set Loading...", "# A pair of equations. Algebra Level 4 How many ordered pairs of real numbers $$(x,y)$$ satisfy: $(x-1)(y^{2}+6)= y(x^{2}+1),$ $(y-1)(x^{2}+6)=x(y^{2}+1)?$ ×", "real numbers and $d,e$ and $f$ complex numbers, respectively. \\item First thing: \\begin{aligned}[t] c^2&=a^2+b^2\\\\ d^2&=e^2+f^2 \\end{aligned}, where $a,b$ and $c$ are supposed to be real numbers and $d,e$ and $f$ complex numbers, respectively. \\item \\mbox{}\\\\[-10pt]\\raisebox{17pt}{First thing:} \\begin{aligned}[b] c^2&=a^2+b^2\\\\ d^2&=e^2+f^2 \\end{aligned}, where $a,b$ and $c$ are supposed to be real numbers and $d,e$ and $f$ complex numbers, respectively. \\end{enumerate} \\end{document} • not every document class fails to indent a new paragraph in a list. Mar 3, 2018 at 21:32 • @barbarabeeton true but one thing at a time:-) (might be a rare case that \\\\ is better than a blank line, I agree. Mar 3, 2018 at 21:38 • no; if you're using aligned (which is sensible), \\\\ doesn't yield the same vertical spacing. it would be \\noindent. Mar 3, 2018 at 22:12", "× # ordered pair help plzzz How many real order pairs of $$(x,p)$$ satisfy $$\\frac{100x^3}{3}-(150+\\frac{p}{8})x^2 + (125 +\\frac{5p}{8})x - (\\frac{59p}{96} - \\frac{75}{2}) = 0$$ such that $$0\\leq x\\leq2.5$$ ? Note by Ashok Kumar 2 years, 12 months ago ## Comments There are no comments in this discussion. × Problem Loading... Note Loading... Set Loading...", "are solutions to the given equation. To shown an ordered pair is a solution we need to evaluate and see if you get a true statement. for the first one x=1 and y=1 so $6(1)-2(1) \\underbrace{=}_{\\mbox{is this true}}4$ So we simplify and see $4=4$ so yes it is a solution. For the next one x=2 y=4 $6(2)-2(4) \\underbrace{=}_{\\mbox{is this true}}4$ So we simplify and see $4=4$ so this is also a solution. I hope this clears it up. Good luck B 5. Originally Posted by armygirltiff 6x - 2y = 4; (1,1) (2,4) Show that the two ordered pairs are solutions to the given equation.", "# Construction of The Real Numbers DEFINITION1: The set $\\mathbb{R}$ with at least two distinct elements and satisfying the following five axioms is called the set of real numbers and each element of $\\mathbb{R}$ is called a real number: The function $+:\\mathbb{R}\\times \\mathbb{R}\\to \\mathbb{R}$ defined as $\\left( x,y \\right)\\to x+y\\in \\mathbb{R}$ for each $\\left( x,y \\right)$ in $\\mathbb{R}\\times \\mathbb{R}$ satisfies the following properties: I${{}_{1}}.\\,\\forall a,b\\in \\mathbb{R},a+b=b+a$, I${{}_{2}}.\\,\\forall a,b,c\\in \\mathbb{R},a+(b+c)=(a+b)+c$," ]

[ "v = 0.$$ Now $$\\bf w \\perp v$$, and $$\\bf w\\times v$$ is orthogonal to both, so $$\\bf v\\times (w\\times v)$$ is parallel to $$\\bf w$$ and has the same direction (explanation below). Now you only need to get the magnitude of $$\\bf w$$ right. This is quite easy, as many of our vectors are pairwise orthogonal: $$\\begin{gather} \\|\\bf w\\times v \\| = \\|w \\| \\, \\|v\\| \\sin \\angle(w,v) = \\|w \\| \\, \\|v\\| \\\\ \\bf \\|w\\| = \\frac {\\|w \\times v\\|}{\\|v\\|} = \\frac {\\|u \\times v\\|}{\\|v\\|} = \\|u\\|\\sin\\angle(u,v) \\tag{!} \\\\ \\bf\\| v\\times (w\\times v)\\| = \\|v\\| \\, \\|w\\| \\, \\|v\\| \\\\ \\bf w = \\frac { v\\times (w\\times v)}{\\| v\\times (w\\times v)\\|} \\|w\\| = \\frac { v\\times (u\\times v)}{\\|v\\| \\, \\|w\\| \\, \\|v\\|} \\|w\\| \\\\ \\bf w = \\frac 1{\\|v\\|^2} v\\times (u\\times v) = \\frac v{\\|v\\|} \\times (u\\times \\frac v{\\|v\\|}) \\end{gather}$$ However, there's an easier way. Because the scalar product of orthogonal vectors is zero: $$\\mathbf u\\cdot \\mathbf v = \\mathbf p\\cdot \\mathbf v = a \\mathbf v\\cdot \\mathbf v = a \\|\\mathbf v\\|^2,$$ whence $$\\mathbf p = a\\bf v = \\frac {u\\cdot v}{\\| v\\|^2} v = \\bigg(u\\cdot \\frac{v}{\\| v\\|}\\bigg) \\frac v{\\| v\\|}$$ $$\\bf \\|p\\| = \\|u\\| \\cos\\angle(u,\\frac v {\\|v\\|}) = \\|u\\| \\cos\\angle(u,v). \\tag{!}$$ Finally $$\\bf u = \\frac 1{\\|v\\|^2} \\bigg( (u\\cdot v) \\,v + v\\times (u\\times", "+0 # How do i find (a-b)/c? 0 56 1 +39 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w}$$, and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$and answer with $$\\dfrac{a-b}{c}$$. I have tried this multiple times and I got a=0, b=0, and c=0 but it was incorrect. I tried distributing the vectors and solving for a, b, and c with the 3 equations. Can someone please tell me if I am doing anything wrong? Thanks! Apr 30, 2020 #1 +24983 +3 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}$$, $$\\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}$$. If the vectors $$\\mathbf{v}$$, $$\\mathbf{w}$$ and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a-b}{c}$$. $$\\begin{array}{|rcll|} \\hline \\begin{array}{|rcll|} \\hline a\\begin{pmatrix}", "# Finding a set of vectors that spans the solution Find a set of vectors {u, v} in $\\mathbb{R}^4$ that spans the solution set of the equations: $x - y + 2z - 2w = 0$ $2x + 2y -z + 3w = 0$ ($u$ and $v$ are both $4 \\times 1$) $u = ?$, $v = ?$ I put the matrix in RREF to get $\\begin{bmatrix}1&0&3/4&-1/4\\\\0&1&-5/4&7/4\\end{bmatrix} = \\begin{bmatrix}0\\\\0\\end{bmatrix}$ Then I got $x = -\\frac{3}{4} z + \\frac{1}{4} w$ and $y = \\frac{5}{4} z - \\frac{7}{4} w$ But I'm not sure how to present the answer as they want it. • HINT: (Assuming your numbers are correct) you've found that $$\\pmatrix{x \\\\ y \\\\ z \\\\ w} = \\pmatrix{-\\frac 34z+\\frac 14w \\\\ \\frac 54z-\\frac 74w \\\\ z \\\\ w} = z\\pmatrix{-\\frac 34 \\\\ \\frac 54 \\\\ 1 \\\\ 0} + w\\pmatrix{\\frac 14 \\\\ -\\frac 74 \\\\ 0 \\\\1}$$ – user137731 Jul 8 '16 at 3:03 You can write the solution set $S$ as follows: $$S=\\{(x,y,z,w)\\in\\mathbb{R}^4: x=-\\frac{3}{4}z+\\frac{1}{4}w,\\, y=\\frac{5}{4}z-\\frac{7}{4}w;\\,\\,\\ z,w\\in\\mathbb{R}\\}.$$", "+0 # Help appreciated 0 102 1 Consider the vectors v = < 1 , 3 > w = < 3 , 2 >, and x = < 1 , 0 >. If the vectors v, w and x are linearly independent, answer with 0. If they aren't, find coefficients a,b and c, not all 0, such that a < 1 , 3 > + b < 3 , 2 > + c < 1 , 0 > = < 0 , 0 > and answer with {a+b}/c. May 19, 2019 #1 +6046 0 3 vectors having two coordinates each cannot possibly be linearlly independent. $$\\text{we have the equation}\\\\ \\begin{pmatrix}1&3&1\\\\3&2&0\\end{pmatrix}\\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}=\\begin{pmatrix}0\\\\0\\end{pmatrix}$$ $$\\text{apply Gaussian elimination}\\\\ \\begin{pmatrix}1&3&1\\\\3&2&0\\end{pmatrix}\\\\ \\begin{pmatrix}1&3&1\\\\0&-7&-3\\end{pmatrix}\\\\ \\begin{pmatrix}1&3&1\\\\0&1&\\frac 3 7\\end{pmatrix}\\\\ \\begin{pmatrix}1&0&-\\frac 2 7\\\\0&1&\\frac 3 7\\end{pmatrix}\\\\$$ $$a-\\dfrac 2 7 c = 0,~c=\\dfrac 7 2 a\\\\ b + \\dfrac 3 7 c = 0,~c = -\\dfrac 7 3 b\\\\ b = -\\dfrac 3 2 a\\\\ \\left(a,b,c\\right)= \\left(1,-\\dfrac 3 2,\\dfrac 7 2\\right)$$ . May 19, 2019 edited by Rom May 20, 2019", "## Elementary Linear Algebra 7th Edition $\\mathbf{z}=(\\frac{7}{2},3,\\frac{5}{2})$ $\\mathbf{u}=(1,2,3)\\quad\\mathbf{v}=(2,2-1)\\quad\\mathbf{w}=(4,0,-4)$ Find $\\mathbf{z}$ if $2\\mathbf{z}-3\\mathbf{u}=\\mathbf{w}$ We can solve for $\\mathbf{z}$: $2\\mathbf{z}=\\mathbf{w}+3\\mathbf{u}$ $\\mathbf{z}=\\frac{1}{2}\\mathbf{w}+\\frac{3}{2}\\mathbf{u}$. All we need to do now is plug in the vectors $\\mathbf{w}$ and $\\mathbf{u}$: $\\mathbf{z}=\\frac{1}{2}(4,0,-4)+\\frac{3}{2}(1,2,3)$ $\\mathbf{z}=(2,0,-2)+(\\frac{3}{2},3,\\frac{9}{2})=(\\frac{7}{2},3,\\frac{5}{2})$", "\\vec{\\mathbf u}=⟨3,4⟩$$ $$\\vec{\\mathbf v}=⟨\\frac{21}{5},\\frac{28}{5}⟩$$ 26) $$‖\\vec{\\mathbf v}‖=3,\\quad \\vec{\\mathbf u}=⟨−2,5⟩$$ 27) $$‖\\vec{\\mathbf v}‖=7,\\quad \\vec{\\mathbf u}=⟨3,−5⟩$$ $$\\vec{\\mathbf v}=⟨\\frac{21\\sqrt{34}}{34},−\\frac{35\\sqrt{34}}{34}⟩$$ 28) $$‖\\vec{\\mathbf v}‖=10,\\quad \\vec{\\mathbf u}=⟨2,−1⟩$$ For exercises 29-34, find the component form of vector $$\\vec{\\mathbf u}$$, given its magnitude and the angle the vector makes with the positive x-axis. Give exact answers when possible. 29) $$‖\\vec{\\mathbf u}‖=2, θ=30°$$ $$\\vec{\\mathbf u}=⟨\\sqrt{3},1⟩$$ 30) $$‖\\vec{\\mathbf u}‖=6, θ=60°$$ 31) $$‖\\vec{\\mathbf u}‖=5, θ=\\frac{π}{2}$$ $$\\vec{\\mathbf u}=⟨0,5⟩$$ 32) $$‖\\vec{\\mathbf u}‖=8, θ=π$$ 33) $$‖\\vec{\\mathbf u}‖=10, θ=\\frac{5π}{6}$$ $$\\vec{\\mathbf u}=⟨−5\\sqrt{3},5⟩$$ 34) $$‖\\vec{\\mathbf u}‖=50, θ=\\frac{3π}{4}$$ For exercises 35 and 36, vector $$\\vec{\\mathbf u}$$ is given. Find the angle $$θ∈[0,2π)$$ that vector $$\\vec{\\mathbf u}$$ makes with the positive direction of the x-axis, in a counter-clockwise direction. 35) $$\\vec{\\mathbf u}=5\\sqrt{2}\\hat{\\mathbf i}−5\\sqrt{2}\\hat{\\mathbf j}$$ $$θ=\\frac{7π}{4}$$ 36) $$\\vec{\\mathbf u}=−\\sqrt{3}\\hat{\\mathbf i}−\\hat{\\mathbf j}$$ 37) Let $$\\vec{\\mathbf a}=⟨a_1,a_2⟩, \\vec{\\mathbf b}=⟨b_1,b_2⟩$$, and $$\\vec{\\mathbf c}=⟨c_1,c_2⟩$$ be three nonzero vectors. If $$a_1b_2−a_2b_1≠0$$, then show there are two scalars, $$α$$ and $$β$$, such that $$\\vec{\\mathbf c}=α\\vec{\\mathbf a}+β\\vec{\\mathbf b}.$$ 38) Consider vectors $$\\vec{\\mathbf a}=⟨2,−4⟩, \\vec{\\mathbf b}=⟨−1,2⟩,$$ and $$\\vec{\\mathbf c}=\\vec{\\mathbf 0}$$ Determine the scalars $$α$$ and $$β$$ such that $$\\vec{\\mathbf c}=α\\vec{\\mathbf a}+β\\vec{\\mathbf b}$$. 39) Let $$P(x_0,f(x_0))$$ be a fixed point on the graph of the differentiable function $$f$$ with a domain that is the set of real numbers. a. Determine the real number $$z_0$$ such that point $$Q(x_0+1,z_0)$$ is situated on the line tangent to the graph", "that $\\mathbf{v}\\mathbf{w}=0$ for a Fixed Vector $\\mathbf{v}$ Let $\\mathbf{v} = \\begin{bmatrix} 2 & -5 & -1 \\end{bmatrix}$. Find all $3 \\times 1$ column vectors $\\mathbf{w}$ such that...", "- \\vec v$$ twice to $$\\vec v$$, because the vector $$[P\\vec{w}]\\vec v - \\vec v$$ is the \"distance vector\" from $$\\vec v$$ to $$Span(\\vec w)$$. Thus, $$$$\\begin{split} F\\vec{w}(\\vec v) &= \\begin{pmatrix} x\\\\y \\end{pmatrix} + 2 \\times (\\begin{pmatrix} \\frac{a^2x+aby}{a^2+b^2} \\frac{abx+b^2y}{a^2+b^2} \\end{pmatrix}- \\begin{pmatrix} x\\\\y \\end{pmatrix})\\\\ &= 2 \\times (\\begin{pmatrix} \\frac{a^2x+aby}{a^2+b^2} \\\\ \\frac{abx+b^2y}{a^2+b^2} \\end{pmatrix})- \\begin{pmatrix} x\\\\y \\end{pmatrix}\\\\ &= \\begin{pmatrix} \\frac{2a^2x+2aby}{a^2+b^2} -x \\\\ \\frac{2abx+2b^2y}{a^2+b^2}-y \\end{pmatrix}\\\\ &= \\begin{pmatrix} \\frac{a^2x+2aby-b^2x}{a^2+b^2} \\\\ \\frac{2abx+b^2y-a^2y}{a^2+b^2}\\end{pmatrix}\\\\ &= \\begin{pmatrix} \\frac{a^2-b^2}{a^2+b^2}x+\\frac{2ab}{a^2+b^2}y \\\\ \\frac{2ab}{a^2+b^2}x +\\frac{b^2-a^2}{a^2+b^2}y\\end{pmatrix} \\end{split}$$$$ From the formula, we know that $$F\\vec w$$ is a linear transformation just as $$P\\vec w$$, and that $$$$[F \\vec w] = \\begin{pmatrix} \\frac{a^2-b^2}{a^2+b^2} & \\frac{2ab}{a^2+b^2} \\\\ \\frac{2ab}{a^2+b^2} & \\frac{b^2-a^2}{a^2+b^2} \\end{pmatrix}$$$$ ### Subject:Number Theory TutorMe Question: Find, with full explanation, the remainder when dividing $$29^{1761}$$ by 23. Inactive Youtong W. This is a typical question of Modular arithmetics that directly utilizes Fermat's Little Theorem. We will give a little review of what the theorem says be fore we proceed. Fermat's Little Theorem: If $$p \\in N^+$$ is prime, then: 1. For all $$a\\in Z$$ with $$p\\nmid a$$, we have $$a^{p-1} \\equiv 1$$ (mod p); 2. For all $$a\\in Z$$, we have $$a^p \\equiv a$$ (mod p). Now 29 is prime, so is 23. So by the first part of Fermat's Little Theorem, $$29^{23-1}=29^{22} \\equiv 1$$(mod 23). Thus, $$(29^{22})^k \\equiv 1^k$$ for all $$k \\in N$$,", "Also, let $\\mathbf y$ be the $m$-dimensionl vector with your target values: $$\\mathbf y=\\begin{pmatrix}y^{(1)}\\\\y^{(2)}\\\\\\vdots\\\\y^{(m)}\\end{pmatrix}.$$ Now, since $h_w(x^{(i)})=(x^{(i)})^T\\mathbf w$, we can verify $$\\begin{array}{rcl}X\\mathbf w -\\mathbf y&=&\\begin{pmatrix}(x^{(1)})^T\\mathbf w\\\\\\vdots\\\\(x^{(m)})^T\\mathbf w\\end{pmatrix}-\\begin{pmatrix}y^{(1)}\\\\\\vdots\\\\y^{(m)}\\end{pmatrix}\\\\&=&\\begin{pmatrix}(x^{(1)})^T\\mathbf w-y^{(1)}\\\\\\vdots\\\\(x^{(m)})^T\\mathbf w-y^{(m)}\\end{pmatrix}.\\end{array}$$ Thus, using $\\mathbf v^T\\mathbf v = \\sum_i\\mathbf v_i^2$ for a vector $\\mathbf v$, we obtain $$\\begin{array}{rcl}E(w)&=&\\frac{1}{2}\\sum_{i=1}^m(h_w(x^{(i)}) - y^{(i)})^2\\\\&=&\\frac{1}{2}(X\\mathbf w-\\mathbf y)^T(X\\mathbf w-\\mathbf y).\\end{array}$$ Summarizing. To obtain $$\\mathbf w=\\begin{pmatrix}w_0\\\\w_1\\end{pmatrix},$$ proceed: 1. Set your collections $x$ and $y$: $$x=\\begin{pmatrix}1\\\\3\\\\5\\\\7\\\\9\\end{pmatrix}\\qquad y=\\begin{pmatrix}1\\\\3\\\\4\\\\3\\\\5\\end{pmatrix}.$$ 2. Loop until you think that $w_j$ converge (try with $10$ iterations) { for $i=1$ to $m$ { $$w_j:=w_j+\\eta\\sum_{i=1}^m(y^{(i)}-h_w(x^{(i)}))x_j^{(i)}\\qquad(\\text{for each }j)$$ } } Insted you can use 1'. Set your vectors $X$ and $\\mathbf y$: $$X=\\begin{pmatrix}1&1\\\\1&3\\\\1&5\\\\1&7\\\\1&9\\end{pmatrix}\\qquad\\mathbf y=\\begin{pmatrix}1\\\\3\\\\4\\\\3\\\\5\\end{pmatrix}.$$ 2'. Loop until you think that $\\mathbf w$ converge (try with $10$ iterations) { $$\\mathbf w:=\\mathbf w-\\frac{\\eta}{m}\\left((X\\mathbf w-\\mathbf y)^TX\\right)$$ } Note. With the above it can prove a analytical formula to obtain the parameters, using the normal equations to computed $\\mathbf w$ by $$\\mathbf w=(X^TX)^{-1}X^T\\mathbf y.$$ Obviously you need to use a computer software (like python or octave) or a calculator device (with matrix-vector support) to calculate the values.", "+0 # Let $\\mathbf{v}$ and $\\mathbf{w}$ be vectors such that \\[\\operatorname{proj}_{\\bold{w}}( \\bold{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.\\ 0 113 1 Let v and w be vectors such that $$\\operatorname{proj}_{{w}}({v})= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.$$ Find $$\\operatorname{proj}_{-2{w}} (3{v})$$. Oct 9, 2019", "Express the vector $\\mathbf{u}$ as a linear combination of $\\mathbf{v}$ and $\\mathbf{w}$. (b) Compute $A^5\\mathbf{v}$. (c) Compute $A^5\\mathbf{w}$. (d) Compute $A^5\\mathbf{u}$. ## Problem 691 In this problem, we use the following vectors in $\\R^2$. $\\mathbf{a}=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}, \\mathbf{b}=\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}, \\mathbf{c}=\\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix}, \\mathbf{d}=\\begin{bmatrix} 3 \\\\ 2 \\end{bmatrix}, \\mathbf{e}=\\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}, \\mathbf{f}=\\begin{bmatrix} 5 \\\\ 6 \\end{bmatrix}.$ For each set $S$, determine whether $\\Span(S)=\\R^2$. If $\\Span(S)\\neq \\R^2$, then give algebraic description for $\\Span(S)$ and explain the geometric shape of $\\Span(S)$. (a) $S=\\{\\mathbf{a}, \\mathbf{b}\\}$ (b) $S=\\{\\mathbf{a}, \\mathbf{c}\\}$ (c) $S=\\{\\mathbf{c}, \\mathbf{d}\\}$ (d) $S=\\{\\mathbf{a}, \\mathbf{f}\\}$ (e) $S=\\{\\mathbf{e}, \\mathbf{f}\\}$ (f) $S=\\{\\mathbf{a}, \\mathbf{b}, \\mathbf{c}\\}$ (g) $S=\\{\\mathbf{e}\\}$ ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 672 For an integer $n > 0$, let $\\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \\{ 1 , x , x^2 , \\cdots , x^n \\}$ is a basis of $\\mathrm{P}_n$, called the standard basis. Let $T : \\mathrm{P}_n \\rightarrow \\mathrm{P}_{n+1}$ be the map defined by, for $f \\in \\mathrm{P}_n$, $T (f) (x) =", "then $v=w$. (c) The zero vector $\\mathbf{0}$ is unique. (d) For each $v\\in V$, the additive inverse $-v$ is unique. (e) $0v=\\mathbf{0}$ for every $v\\in V$, where $0\\in\\R$ is the zero scalar. (f) $a\\mathbf{0}=\\mathbf{0}$ for every scalar $a$. (g) If $av=\\mathbf{0}$, then $a=0$ or $v=\\mathbf{0}$. (h) $(-1)v=-v$. The first two properties are called the cancellation law. ## Problem 709 Let $S=\\{\\mathbf{v}_{1},\\mathbf{v}_{2},\\mathbf{v}_{3},\\mathbf{v}_{4},\\mathbf{v}_{5}\\}$ where $\\mathbf{v}_{1}= \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\\\ -1 \\end{bmatrix} ,\\;\\mathbf{v}_{2}= \\begin{bmatrix} 1 \\\\ 3 \\\\ 1 \\\\ 1 \\end{bmatrix} ,\\;\\mathbf{v}_{3}= \\begin{bmatrix} 1 \\\\ 5 \\\\ -1 \\\\ 5 \\end{bmatrix} ,\\;\\mathbf{v}_{4}= \\begin{bmatrix} 1 \\\\ 1 \\\\ 4 \\\\ -1 \\end{bmatrix} ,\\;\\mathbf{v}_{5}= \\begin{bmatrix} 2 \\\\ 7 \\\\ 0 \\\\ 2 \\end{bmatrix} .$ Find a basis for the span $\\Span(S)$. ## Problem 708 Let $A=\\begin{bmatrix} 2 & 4 & 6 & 8 \\\\ 1 &3 & 0 & 5 \\\\ 1 & 1 & 6 & 3 \\end{bmatrix}$. (a) Find a basis for the nullspace of $A$. (b) Find a basis for the row space of $A$. (c) Find a basis for the range of $A$ that consists of column vectors of $A$. (d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$. ## Problem 707", "represents $f(x)$. In this basis, a vector that would be $(5,6,7)$ in $B$ is instead $(5,1,1)$, as shown here: $$(5,1,1)_{B'} = 5(1,1,1)+1(0,1,1)+1(0,0,1) = (5,6,7)_B$$ This affects the transformation as well: $$\\begin{bmatrix} 3&0&0\\\\ 1&2&0\\\\ 1&1&1 \\end{bmatrix} \\begin{pmatrix} 5\\\\6\\\\7 \\end{pmatrix} =\\begin{pmatrix} 15+0+0\\\\ 5+12+0\\\\ 5+6+7 \\end{pmatrix} =(15,17,18)_B\\\\ \\begin{bmatrix} 3&0&0\\\\ 0&2&0\\\\ 0&0&1 \\end{bmatrix} \\begin{pmatrix} 5\\\\1\\\\1 \\end{pmatrix} =\\begin{pmatrix} 15+0+0\\\\ 0+2+0\\\\ 0+0+1 \\end{pmatrix} =(15,2,1)_{B'}\\\\ (15,2,1)_{B'}=15(1,1,1)+2(0,1,1)+1(0,0,1)=(15,17,18)_B$$ # Caveats Notice how when we were finding $v_1$ above we picked $(1,1,1)$ to simplify. Unfortunately $v_1=(7,7,7)$, $v_2=(0,-3,-3)$ (and infinitely many more) are also valid in this context. To make every solution uniform, we normalize each of the vectors: $$v_1=\\frac{(1,1,1)}{\\sqrt{1^2+1^2+1^2}}=(\\frac{1}{\\sqrt{3}},\\frac{1}{\\sqrt{3}},\\frac{1}{\\sqrt{3}})\\\\ v_2=\\frac{(0,1,1)}{\\sqrt{1^2+1^2}}=(0,\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}})\\\\ v_3=\\frac{(0,0,1)}{\\sqrt{1^2}}=(0,0,1)$$ So we get a basis $B'=\\{(\\frac{1}{\\sqrt{3}},\\frac{1}{\\sqrt{3}},\\frac{1}{\\sqrt{3}}),(0,\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}}),(0,0,1)\\}$, which corresponds to: $$(5,1,1)_{B'} = 5(\\frac{1}{\\sqrt{3}},\\frac{1}{\\sqrt{3}},\\frac{1}{\\sqrt{3}})+1(0,\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}})+1(0,0,1)\\\\ = (\\frac{5}{\\sqrt{3}},\\frac{1}{\\sqrt{2}}+\\frac{5}{\\sqrt{3}},1+\\frac{1}{\\sqrt{2}}+\\frac{5}{\\sqrt{3}})_B$$ Additionally, we also add a restriction so that the first non-zero element of each eigenvector must be positive, flipping the signs of the rest of the vector if it was initially negative. # The Challenge Your task is to write a function or program which takes a black-box function or program and either: 1. outputs a function or program which accepts arguments in a special basis $B'$ as described above, performs the transformation on them, and returns the result in the standard basis $B$ 2. outputs the result in $B$ directly and takes the secondary arguments in $B'$ itself # Input You are", "equal to A) $|\\mathbf{a}{{|}^{2}}$ B) $2|\\mathbf{a}{{|}^{2}}$ C) $3|\\mathbf{a}{{|}^{2}}$ D) None of these • question_answer137) If the unit vectors $a$ and $b$ are inclined at an angle $2\\theta$ such that $|a-b|\\,\\,<1$ and$0\\le \\theta \\le \\pi$, then $\\theta$ lies in the interval A) $\\left[ 0,\\,\\,\\frac{\\pi }{6} \\right)$ B) $\\left[ \\frac{5\\pi }{6},\\,\\,\\pi \\right]$ C) $\\left[ \\frac{\\pi }{6},\\,\\,\\frac{\\pi }{2} \\right]$ D) $\\left[ \\frac{\\pi }{2},\\,\\,\\frac{5\\pi }{6} \\right]$ • question_answer138) The vectors$\\mathbf{a}=x\\mathbf{i}+(x+1)\\mathbf{j}+(x+2)\\mathbf{k}$,$\\mathbf{b}=(x+3)\\mathbf{i}+(x+4)\\mathbf{j}+(x+5)\\mathbf{k}$ and$\\mathbf{c}=(x+6)\\mathbf{i}+(x+7)\\mathbf{j}+(x+8)\\mathbf{k}$are coplanar for A) all values of$x$ B) only$x=1$ C) only$x=0$ D) None of these • question_answer139) If$y={{\\tan }^{-1}}\\left( \\frac{\\sqrt{1+{{x}^{2}}}+\\sqrt{1-{{x}^{2}}}}{\\sqrt{1+{{x}^{2}}}-\\sqrt{1-{{x}^{2}}}} \\right)$, then$\\frac{dy}{dx}$ equals to A) $\\frac{1}{\\sqrt{1-{{x}^{4}}}}$ B) $\\frac{-1}{\\sqrt{1-{{x}^{4}}}}$ C) $\\frac{x}{\\sqrt{1-{{x}^{4}}}}$ D) $\\frac{-x}{\\sqrt{1-{{x}^{4}}}}$ • question_answer140) If the parametric equation of a curve given by$x={{e}^{t}}\\cos t,\\,\\,y={{e}^{t}}\\sin t$, then the tangent to the curve at the point $=\\frac{\\pi }{4}$ makes with axis of $x$, the angle is A) $0$ B) $\\frac{\\pi }{4}$ C) $\\frac{\\pi }{3}$ D) $\\frac{\\pi }{2}$ • question_answer141) The function $f(x)=\\frac{x}{\\log x}$ increases on the interval A) $(0,\\,\\,\\infty )$ B) $(0,\\,\\,e)$ C) $(e,\\,\\,\\infty )$ D) None of these • question_answer142) If$f(x)={{\\tan }^{-1}}\\left\\{ \\frac{\\log \\left( \\frac{e}{{{x}^{2}}} \\right)}{\\log (e{{x}^{2}})} \\right\\}+{{\\tan }^{-1}}\\left( \\frac{1+2\\log x}{1-2\\log x} \\right)$, then$\\frac{dy}{dx}$is A) ${{\\tan }^{-1}}(\\log x)$ B) $0$ C) $\\frac{1}{2}$ D) None of these • question_answer143) If${{I}_{m}}=\\int_{1}^{x}{{{(\\log x)}^{m}}dx}$satisfies the relation ${{I}_{m}}=K-l{{I}_{m-1}}$,then A) $K=e$ B) $l=m$ C) $K=\\frac{1}{e}$ D) None of these • question_answer144) Let$I=\\int_{0}^{1}{\\frac{{{e}^{x}}}{x+1}dx}$, then the value of the integral $\\int_{0}^{1}{\\frac{x{{e}^{{{x}^{2}}}}}{{{x}^{2}}+1}dx}$is", "+0 # Help with vector geometry +1 88 1 +38 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w} \\text{ and }\\mathbf{x}$$ are linearly independent, answer with 0. If they aren't, find coefficients a,b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a+b}{c}$$ Jun 5, 2019", "## Linear Algebra: A Modern Introduction $||v||=4.41$. \\begin{bmatrix} \\frac{1.12}{4.41} \\\\ \\frac{-3.25}{4.41} \\\\ \\frac{2.07}{4.41} \\\\ \\frac{-1.83}{4.41} \\\\ \\end{bmatrix} I know that for the vector $v=\\begin{bmatrix} v_{1} \\\\ v_{2} \\\\ \\vdots\\\\ v_{n} \\end{bmatrix}$ The unit vector in the direction of $v$ is $\\frac{v}{||v||}$. Hence: $||v||=\\sqrt{1.12^2+(-3.25)^2+2.07^2+(-1.83)^2}=\\sqrt{1.2544+10.5625+4.2849+3.3489}=\\sqrt{19.4507}\\approx4.41$. Thus, the unit vector in the direction of $v$ is: \\begin{bmatrix} \\frac{1.12}{4.41} \\\\ \\frac{-3.25}{4.41} \\\\ \\frac{2.07}{4.41} \\\\ \\frac{-1.83}{4.41} \\\\ \\end{bmatrix}", "+0 # help vectors 0 46 2 Let v and w be vectors such that $$\\text{proj}_{{w}} {v} = \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$ Find $$\\text{proj}_{-2 {w}} (3 {v})$$. Feb 23, 2020 #1 0 Projection is linear, so the answer is 2*3*(4,-7) = (24,-42). Feb 24, 2020 #2 +24366 +1 Let $$\\mathbf{v} \\text{ and } \\mathbf{w} \\text{ be vectors such that } \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}$$. Find $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}})$$. $$\\begin{array}{|rcll|} \\hline \\vec{p} &=& \\operatorname{proj}_{\\mathbf{w}}( \\mathbf{v} ) \\\\\\\\ &=& \\dbinom{4}{-7} \\\\\\\\ &=& \\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\ \\hline \\vec{P} &=& \\operatorname{proj}_{\\mathbf{-2w}}( \\mathbf{3v} ) \\\\\\\\ &=& \\dfrac{(-2\\vec{w})(3\\vec{v})}{|-2w|^2}(-2\\vec{w}) \\\\\\\\ &=& 6\\dfrac{\\vec{w}\\vec{v}}{2w^2}\\vec{w} \\\\\\\\ &=& 3\\dfrac{\\vec{w}\\vec{v}}{w^2}\\vec{w} \\\\\\\\ &=& 3\\vec{p} \\\\\\\\ &=& 3\\dbinom{4}{-7} \\\\\\\\ &=& \\dbinom{12}{-21} \\\\ \\hline \\end{array}$$ $$\\operatorname{proj}_{-2 {\\mathbf{w}}} (3 {\\mathbf{v}}) = \\dbinom{12}{-21}.$$ Feb 24, 2020", "\\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 686 In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not. (a) The matrix $A$ is a $3 \\times 3$", "to an integer (either $$0$$ or $$1$$): $$\\mathbf{1}_{[a = b]} = \\begin{cases}1 &\\text{ if } a = b, \\\\ 0 &\\text{ otherwise.} \\end{cases}$$ 1. Bolded text for (I believe) non-vectors I think $$\\mathbf{t}$$ denotes a two-dimensional vector (as explained at the beginning of §2) and the expression $$\\mathbf{t} = 0$$ should be read as $$\\mathbf{t} = [0, 0]$$.", "1. ## Vector problem: \"The vectors v = i + 2j - 3k, u = 2i + 2j + 3k, and w = i + (2 - t)j + (t + 1)k are given. Find the value of t such that the three vectors u, v and w are coplanar. \" I have no idea how to solve this one!Where do I start? 2. let's take $v=(1,2,-3),u=(2,2,3),w=(1,2-t,t+1)$ Using Gauss method on vectors,we should have $\\begin{pmatrix} 1\\\\ 2\\\\ -3 \\end{pmatrix}$ $,\\begin{pmatrix} 2\\\\ 2\\\\ 3 \\end{pmatrix}$ $,\\begin{pmatrix} 1\\\\ 2-t\\\\ t+1 \\end{pmatrix}$ $\\Rightarrow$ $\\begin{pmatrix} 1\\\\ 2\\\\ -3 \\end{pmatrix}$ $,\\begin{pmatrix} 0\\\\ -2\\\\ 9 \\end{pmatrix}$ $,\\begin{pmatrix} 0\\\\ 0\\\\ \\frac{-7t+8}{2} \\end{pmatrix}$ for those three victors to be coplanar we must have $rang(\\begin{pmatrix} 1\\\\ 2\\\\ -3 \\end{pmatrix}$ $,\\begin{pmatrix} 0\\\\ -2\\\\ 9 \\end{pmatrix}$ , $\\begin{pmatrix} 0\\\\ 0\\\\ \\frac{-7t+8}{2} \\end{pmatrix} )$ $=2$ that is $\\frac{-7t+8}{2}=0$ which means $t=\\frac{7}{8}$ hope that's right 3. Originally Posted by Raoh let's take $v=(1,2,-3),u=(2,2,3),w=(1,2-t,t+1)$ Using Gauss method on vectors,we should have $\\begin{pmatrix} 1\\\\ 2\\\\ -3 \\end{pmatrix}$ $,\\begin{pmatrix} 2\\\\ 2\\\\ 3 \\end{pmatrix}$ $,\\begin{pmatrix} 1\\\\ 2-t\\\\ t+1 \\end{pmatrix}$ $\\Rightarrow$ $\\begin{pmatrix} 1\\\\ 2\\\\ -3 \\end{pmatrix}$ $,\\begin{pmatrix} 0\\\\ -2\\\\ 9 \\end{pmatrix}$ $,\\begin{pmatrix} 0\\\\ 0\\\\ \\frac{-7t+8}{2} \\end{pmatrix}$ for those three victors to be coplanar we must have $rang(\\begin{pmatrix} 1\\\\ 2\\\\ -3 \\end{pmatrix}$ $,\\begin{pmatrix} 0\\\\ -2\\\\ 9 \\end{pmatrix}$ , $\\begin{pmatrix} 0\\\\ 0\\\\ \\frac{-7t+8}{2} \\end{pmatrix} )$ $=2$ that is" ]

[ "represents $f(x)$. In this basis, a vector that would be $(5,6,7)$ in $B$ is instead $(5,1,1)$, as shown here: $$(5,1,1)_{B'} = 5(1,1,1)+1(0,1,1)+1(0,0,1) = (5,6,7)_B$$ This affects the transformation as well: $$\\begin{bmatrix} 3&0&0\\\\ 1&2&0\\\\ 1&1&1 \\end{bmatrix} \\begin{pmatrix} 5\\\\6\\\\7 \\end{pmatrix} =\\begin{pmatrix} 15+0+0\\\\ 5+12+0\\\\ 5+6+7 \\end{pmatrix} =(15,17,18)_B\\\\ \\begin{bmatrix} 3&0&0\\\\ 0&2&0\\\\ 0&0&1 \\end{bmatrix} \\begin{pmatrix} 5\\\\1\\\\1 \\end{pmatrix} =\\begin{pmatrix} 15+0+0\\\\ 0+2+0\\\\ 0+0+1 \\end{pmatrix} =(15,2,1)_{B'}\\\\ (15,2,1)_{B'}=15(1,1,1)+2(0,1,1)+1(0,0,1)=(15,17,18)_B$$ # Caveats Notice how when we were finding $v_1$ above we picked $(1,1,1)$ to simplify. Unfortunately $v_1=(7,7,7)$, $v_2=(0,-3,-3)$ (and infinitely many more) are also valid in this context. To make every solution uniform, we normalize each of the vectors: $$v_1=\\frac{(1,1,1)}{\\sqrt{1^2+1^2+1^2}}=(\\frac{1}{\\sqrt{3}},\\frac{1}{\\sqrt{3}},\\frac{1}{\\sqrt{3}})\\\\ v_2=\\frac{(0,1,1)}{\\sqrt{1^2+1^2}}=(0,\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}})\\\\ v_3=\\frac{(0,0,1)}{\\sqrt{1^2}}=(0,0,1)$$ So we get a basis $B'=\\{(\\frac{1}{\\sqrt{3}},\\frac{1}{\\sqrt{3}},\\frac{1}{\\sqrt{3}}),(0,\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}}),(0,0,1)\\}$, which corresponds to: $$(5,1,1)_{B'} = 5(\\frac{1}{\\sqrt{3}},\\frac{1}{\\sqrt{3}},\\frac{1}{\\sqrt{3}})+1(0,\\frac{1}{\\sqrt{2}},\\frac{1}{\\sqrt{2}})+1(0,0,1)\\\\ = (\\frac{5}{\\sqrt{3}},\\frac{1}{\\sqrt{2}}+\\frac{5}{\\sqrt{3}},1+\\frac{1}{\\sqrt{2}}+\\frac{5}{\\sqrt{3}})_B$$ Additionally, we also add a restriction so that the first non-zero element of each eigenvector must be positive, flipping the signs of the rest of the vector if it was initially negative. # The Challenge Your task is to write a function or program which takes a black-box function or program and either: 1. outputs a function or program which accepts arguments in a special basis $B'$ as described above, performs the transformation on them, and returns the result in the standard basis $B$ 2. outputs the result in $B$ directly and takes the secondary arguments in $B'$ itself # Input You are", "# Decide a number $a, b, c$ (expressed by $\\lambda$) so that $a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$ See the following vectors in $$\\mathbb{R}^3$$: $$\\mathbf{u}_1= \\begin{pmatrix} 1 \\\\ -1\\\\ 1 \\\\ 1 \\end{pmatrix}$$, $$\\mathbf{u}_2 = \\begin{pmatrix} 1 \\\\ 2 \\\\ 0 \\\\ 2 \\end{pmatrix}$$, $$\\mathbf{u}_3 = \\begin{pmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\end{pmatrix}$$ and $$\\mathbf{v}_1= \\begin{pmatrix} 5 \\\\ -3\\\\ -4 \\\\ -1 \\end{pmatrix}$$, $$\\mathbf{v}_2 = \\begin{pmatrix} 6 \\\\ 4 \\\\ 1 \\\\ 8 \\end{pmatrix}$$, $$\\mathbf{v}_3 = \\begin{pmatrix} 7 \\\\ 2 \\\\ -1 \\\\ 6 \\end{pmatrix}$$ We are told that $$\\mathcal{B} = (\\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3)$$ and $$\\mathcal{C} = (\\mathbf{v}_1, \\mathbf{v}_2, \\mathbf{v}_3)$$ both are bases for the same subspace $$\\mathcal{U}$$. Let $$\\mathcal{\\lambda}$$ be an unknown number og see the vector $$\\mathbf{x} = \\mathbf{v}_1 + \\lambda\\mathbf{v}_2 - \\mathbf{v}_3$$. Decide a number $$a, b, c$$ (expressed by $$\\lambda$$) so that $$a\\mathbf{u}_1 + b\\mathbf{u}_2 + c\\mathbf{u}_3 = \\mathbf{x}$$ 1. Find some $$\\alpha_i$$ such that $$v_1=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3$$ 2. Find some $$\\beta_i$$ such that $$v_2=\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3$$ 3. Find some $$\\gamma_i$$ such that $$v_3=\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3$$ 4. Simplify the expression $$x=v_1+\\lambda v_2-v_3=\\alpha_1 u_1+\\alpha_2u_2+\\alpha_3u_3+\\lambda(\\beta_1 u_1+\\beta_2u_2+\\beta_3u_3)-(\\gamma_1 u_1+\\gamma_2u_2+\\gamma_3u_3)$$ • Thanks 5xum! When I reduce it, I get $\\begin{pmatrix} 8 \\\\ 3\\\\ -9 \\\\ 1 \\end{pmatrix} + a\\begin{pmatrix} 2 \\\\ -2\\\\ 1 \\\\ 1 \\end{pmatrix} + b\\begin{pmatrix} 3 \\\\ 6\\\\ 0 \\\\ 6 \\end{pmatrix} + c\\begin{pmatrix} 1 \\\\", "+0 # Help with vector geometry +1 88 1 +38 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\3 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 3\\\\2 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}1 \\\\ 0 \\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w} \\text{ and }\\mathbf{x}$$ are linearly independent, answer with 0. If they aren't, find coefficients a,b, and c, not all 0, such that $$a \\begin{pmatrix} 1 \\\\ 3 \\end{pmatrix} + b \\begin{pmatrix} 3 \\\\ 2 \\end{pmatrix} + c \\begin{pmatrix} 1 \\\\ 0 \\end{pmatrix} = \\begin{pmatrix}0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a+b}{c}$$ Jun 5, 2019", "+0 # How do i find (a-b)/c? 0 56 1 +39 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}, \\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}.$$ If the vectors $$\\mathbf{v}, \\mathbf{w}$$, and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$and answer with $$\\dfrac{a-b}{c}$$. I have tried this multiple times and I got a=0, b=0, and c=0 but it was incorrect. I tried distributing the vectors and solving for a, b, and c with the 3 equations. Can someone please tell me if I am doing anything wrong? Thanks! Apr 30, 2020 #1 +24983 +3 Consider the vectors $$\\mathbf{v} = \\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}$$, $$\\mathbf{w} = \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix}$$, and $$\\mathbf{x} = \\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix}$$. If the vectors $$\\mathbf{v}$$, $$\\mathbf{w}$$ and $$\\mathbf{x}$$ are linearly independent, answer with ? (a question mark.) If they aren't, find coefficients a,b and c, not all 0, such that $$a\\begin{pmatrix} 1\\\\2\\\\1 \\end{pmatrix}+b \\begin{pmatrix} 1\\\\4 \\\\5 \\end{pmatrix} + c\\begin{pmatrix}-1 \\\\ 6 \\\\ 15\\end{pmatrix} = \\begin{pmatrix} 0 \\\\ 0 \\\\ 0 \\end{pmatrix}$$ and answer with $$\\dfrac{a-b}{c}$$. $$\\begin{array}{|rcll|} \\hline \\begin{array}{|rcll|} \\hline a\\begin{pmatrix}", "\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}, \\quad \\bfb = \\begin{pmatrix}1\\\\3\\\\2\\end{pmatrix}.$$ Recall that Hence, $$\\bfx = \\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix} = \\begin{pmatrix} 1/3 & 1 & -1/3 \\\\ -1/2 & 1/2 & 0 \\\\ 1/6 & -1/2 & 1/3 \\end{pmatrix} \\begin{pmatrix}1\\\\3\\\\2 \\end{pmatrix} = \\begin{pmatrix}8/3\\\\1\\\\-2/3\\end{pmatrix}$$ Hence, the polynomial through the points $(-1, 1), (1, 3),$ and $(2,2)$ is $$y = \\frac{8}{3} + x - \\frac{2}{3} x^2$$", "b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1\\right) \\\\ &=& \\left(16,0,48\\right). \\end{eqnarray}$$ Wikipedia has a decent explanation of the cross product. - By definition, the product of vectors $a= (a_x, a_y, a_z)$ and $b = (b_x, b_y, b_z)$ equals $$a \\times b = \\det \\begin{pmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k}\\\\ a_x & a_y & a_z\\\\ b_x & b_y & b_z \\end{pmatrix},$$ where $\\mathbf{i}$, $\\mathbf{j}$, $\\mathbf{k}$ is the standard basis. So we have $$a \\times b = \\det \\begin{pmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k}\\\\ 6 & 0 & -2\\\\ 0 & 8 & 0 \\end{pmatrix} = 16\\, \\mathbf{i} + 0\\, \\mathbf{j} + 48\\, \\mathbf{k}.$$ -", "Review question # Can we find the equation of the plane containing $l_1$ and $PQ$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource Ref: R5089 ## Solution The lines $l_1$ and $l_2$ have equations $\\mathbf{r} = \\mathbf{a} + s\\mathbf{b}$ and $\\mathbf{r} = \\mathbf{c} + t\\mathbf{d}$, where $s$ and $t$ are real numbers and $\\mathbf{a} = 5\\mathbf{i}+ 6\\mathbf{j}- \\mathbf{k}, \\quad\\mathbf{b} = \\mathbf{i} +\\mathbf{j}+ \\mathbf{k}, \\quad\\mathbf{c} = \\mathbf{i}+ 13\\mathbf{j} +7\\mathbf{k}, \\quad\\mathbf{d} = \\mathbf{i} + 6\\mathbf{j} + 3\\mathbf{k}.$ Also, the point $P$ on $l_1$ and the point $Q$ on $l_2$ are such that $PQ$ is perpendicular to both $l_1$ and $l_2.$ In any order: 1. show that $PQ = \\sqrt{38},$ 2. find the position vectors of $P$ and $Q,$ We can write the position vector of $P$ as $\\mathbf{p} = \\begin{pmatrix}5+s \\\\ 6+s \\\\ -1+s\\end{pmatrix}$, and the position vector of $Q$ as $\\mathbf{q} = \\begin{pmatrix}1+t \\\\ 13+6t\\\\ 7+3t\\end{pmatrix}.$ Thus $\\mathbf{q} - \\mathbf{p} = \\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}$. This vector has to be perpendicular to both $l_1$ and $l_2$, and so using the scalar product, $\\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}.\\begin{pmatrix}1 \\\\ 1\\\\1\\end{pmatrix}=0 \\quad\\text{and}\\quad \\begin{pmatrix}t-s-4 \\\\ 6t-s+7 \\\\ 3t-s+8\\end{pmatrix}.\\begin{pmatrix}1\\\\ 6 \\\\ 3\\end{pmatrix}=0.$ This gives us the two equations $10t-3s+11=0$ and $46t-10s+62=0$, which solve to give $s = -3$, $t", "+0 0 108 3 +226 Let a,b,c be vectors such that $$\\mathbf{a} \\times \\mathbf{b} = \\begin{pmatrix} -1 \\\\- 1 \\\\ -1 \\end{pmatrix}, \\mathbf{a} \\times \\mathbf{c} = \\begin{pmatrix} -1 \\\\ 2 \\\\ 1 \\end{pmatrix}, \\text{ and } \\mathbf{b} \\times \\mathbf{c} = \\begin{pmatrix} 0 \\\\ 2 \\\\ 3 \\end{pmatrix}$$ Then evaluate $$(\\mathbf{b} + \\mathbf{c})\\times \\mathbf{b}, \\mathbf{a}\\times(\\mathbf{b} + 4 \\mathbf{a}), (\\mathbf{a} + \\mathbf{b} + \\mathbf{c})\\times \\mathbf{a}$$ Oct 30, 2020 #2 +31698 +3 For cross products we have the following a x a = 0 (simiarly for b x b etc) a x b = - b x a etc. (b + c) x a = b x+ c x a = -x b - a x c etc. Can you take it from here? Oct 30, 2020 #3 +112499 +1 Thanks Alan, that sure made it easy Oct 31, 2020", "$$\\mathbf{u}$$, so (10.5.3) is an $$(n+1)\\times 1$$ set of nonlinear equations, amenable to solution by the techniques of Chapter 4. Example 10.5.1 Given the BVP $u'' - \\sin(xu) + \\exp(xu')=0, \\quad u(0)=-2, \\; u'(3/2)=1,$ we compare to the standard form (10.5.1) and recognize $\\phi(x,u,u') = \\sin(xu)-\\exp(xu').$ Suppose $$n=3$$ for an equispaced grid, so that $$h=\\frac{1}{2}$$, $$x_0=0$$, $$x_1=\\frac{1}{2}$$, $$x_2=1$$, and $$x_3=\\frac{3}{2}$$. There are four unknowns. We compute $\\begin{gather*} \\mathbf{D}_{xx} = \\frac{1}{1/4} \\begin{bmatrix} 2 & -5 & 4 & -1 \\\\ 1 & -2 & 1 & 0 \\\\ 0 & 1 & -2 & 1 \\\\ -1 & 4 & -5 & 2 \\end{bmatrix}, \\quad \\mathbf{D}_x = \\frac{1}{1} \\begin{bmatrix} -3 & 4 & -1 & 0 \\\\ -1 & 0 & 1 & 0 \\\\ 0 & -1 & 0 & 1 \\\\ 0 & 1 & -4 & 3 \\end{bmatrix}, \\\\[2mm] \\mathbf{E} \\mathbf{r}(\\mathbf{u}) = \\begin{bmatrix} \\sin\\left(\\frac{u_1}{2}\\right) - \\exp\\left(\\frac{u_2-u_0}{2}\\right) \\\\[1mm] \\sin(u_2) - \\exp\\left( u_3-u_1 \\right) \\end{bmatrix}, \\\\[2mm] \\mathbf{f}(\\mathbf{u}) = \\begin{bmatrix} (4u_0 -8u_1 + 4u_2) - \\sin\\left(\\frac{u_1}{2}\\right) + \\exp\\left(\\frac{u_2-u_0}{2}\\right) \\\\[1mm] (4u_1 -8u_2 + 4u_3) - \\sin(u_2) + \\exp\\left( u_3-u_1 \\right) \\\\[1mm] u_0 + 2 \\\\[1mm] (u_1 - 4u_2 + 3u_3) - 1 \\end{bmatrix}. \\end{gather*}$ ## Implementation¶ Our implementation using second-order finite differences is Function 10.5.2. It’s surprisingly short, considering how general it is, because we have laid a lot of", "1\\\\ \\end {vmatrix}$$ = 20 × 1 - 12 × 1 = 20 - 12 = 8 ≠ 0 It has a unique solution. A-1= $$\\frac {1}{\\begin {vmatrix} A\\\\ \\end {vmatrix}}$$ Adj.A = $$\\frac {1}{8}$$$$\\begin {pmatrix} 1 & -1\\\\ -12 & 20\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-12}{8} & \\frac {20}{8}\\\\ \\end {pmatrix}$$ = $$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$ We know that, AX = B or, X = A-1B or, X =$$\\begin {pmatrix} \\frac {1}{8} & \\frac {-1}{8}\\\\ \\frac {-3}{2} & \\frac {5}{2}\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} 35\\\\ 20\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35}{8} - \\frac {20}{8}\\\\ \\frac {-105}{2} + \\frac {100}{2}\\\\ \\end {pmatrix}$$ or, X =$$\\begin {pmatrix} \\frac {35 - 20}{8}\\\\ \\frac {-105 + 100}{2}\\\\ \\end {pmatrix}$$ ∴ X =$$\\begin {pmatrix} \\frac {15}{8}\\\\ \\frac {-5}{2}\\\\ \\end {pmatrix}$$ ∴ x = $$\\frac {15}{2}$$ and y = $$\\frac {-5}{2}$$ Ans Here, $$\\frac 3x$$ + $$\\frac 4y$$ = 2..........................................(1) $$\\frac 9x$$ - $$\\frac 2y$$ = $$\\frac 52$$....................(2) Given equation in the form of matrix: $$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$$$\\begin {pmatrix} \\frac 1x\\\\ \\frac1y\\\\ \\end {pmatrix}$$ =$$\\begin {pmatrix} 2\\\\ \\frac52\\\\ \\end {pmatrix}$$, AX = B where, A =$$\\begin {pmatrix} 3 & 4\\\\ 9 & -2\\\\ \\end {pmatrix}$$, X =$$\\begin {pmatrix}", "v = 0.$$ Now $$\\bf w \\perp v$$, and $$\\bf w\\times v$$ is orthogonal to both, so $$\\bf v\\times (w\\times v)$$ is parallel to $$\\bf w$$ and has the same direction (explanation below). Now you only need to get the magnitude of $$\\bf w$$ right. This is quite easy, as many of our vectors are pairwise orthogonal: $$\\begin{gather} \\|\\bf w\\times v \\| = \\|w \\| \\, \\|v\\| \\sin \\angle(w,v) = \\|w \\| \\, \\|v\\| \\\\ \\bf \\|w\\| = \\frac {\\|w \\times v\\|}{\\|v\\|} = \\frac {\\|u \\times v\\|}{\\|v\\|} = \\|u\\|\\sin\\angle(u,v) \\tag{!} \\\\ \\bf\\| v\\times (w\\times v)\\| = \\|v\\| \\, \\|w\\| \\, \\|v\\| \\\\ \\bf w = \\frac { v\\times (w\\times v)}{\\| v\\times (w\\times v)\\|} \\|w\\| = \\frac { v\\times (u\\times v)}{\\|v\\| \\, \\|w\\| \\, \\|v\\|} \\|w\\| \\\\ \\bf w = \\frac 1{\\|v\\|^2} v\\times (u\\times v) = \\frac v{\\|v\\|} \\times (u\\times \\frac v{\\|v\\|}) \\end{gather}$$ However, there's an easier way. Because the scalar product of orthogonal vectors is zero: $$\\mathbf u\\cdot \\mathbf v = \\mathbf p\\cdot \\mathbf v = a \\mathbf v\\cdot \\mathbf v = a \\|\\mathbf v\\|^2,$$ whence $$\\mathbf p = a\\bf v = \\frac {u\\cdot v}{\\| v\\|^2} v = \\bigg(u\\cdot \\frac{v}{\\| v\\|}\\bigg) \\frac v{\\| v\\|}$$ $$\\bf \\|p\\| = \\|u\\| \\cos\\angle(u,\\frac v {\\|v\\|}) = \\|u\\| \\cos\\angle(u,v). \\tag{!}$$ Finally $$\\bf u = \\frac 1{\\|v\\|^2} \\bigg( (u\\cdot v) \\,v + v\\times (u\\times", "• # question_answer Direction: Let a, b and c are three non-coplanar vectors, i.e., $[a\\,b\\,c]\\,\\ne 0$. The three new vectors $a',\\,\\,b'$ and c' defined by the equation $a'=\\frac{b\\times c}{[a\\,\\,b\\,\\,c]},\\,\\,b'=\\frac{c\\times a}{[a\\,\\,b\\,\\,c]}$and$c'=\\frac{a\\times b}{[a\\,\\,b\\,\\,c]}$ are called reciprocal system to the vectors a, b and c. If a, b, c and a', b', c' are reciprocal system of vectors, then the value of$a\\times a'+b\\times b'+c\\times c'$ is A) $a\\times a'+b\\times b'+c\\times c'$ B) $2\\,(a'\\times b'\\times c')$ C) $\\frac{[a\\,\\,b\\,\\,c]}{2}$ D) 0 $\\therefore \\,\\,a\\times a'+b\\times b'+c+c'$$=\\frac{1}{[a\\,b\\,c]}[a\\times (b\\times c)+b\\times (c\\times a)+c\\times (a\\times b)]$(for cyclic order) $=\\frac{1}{[a\\,b\\,c]}[0]=0$", "+0 # Let $\\mathbf{v}$ and $\\mathbf{w}$ be vectors such that \\[\\operatorname{proj}_{\\bold{w}}( \\bold{v} )= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.\\ 0 113 1 Let v and w be vectors such that $$\\operatorname{proj}_{{w}}({v})= \\begin{pmatrix} 4 \\\\ -7 \\end{pmatrix}.$$ Find $$\\operatorname{proj}_{-2{w}} (3{v})$$. Oct 9, 2019", "# How do you find two unit vectors orthogonal to a=⟨−1,2,1⟩ and b=⟨−4,0,−5⟩? May 25, 2017 $\\pm \\left(- \\frac{10}{7 \\sqrt{5}} , - \\frac{9}{7 \\sqrt{5}} , \\frac{8}{7 \\sqrt{5}}\\right) .$ #### Explanation: We know that, $\\vec{x} \\times \\vec{y}$ is $\\bot$to $\\vec{x} \\mathmr{and} \\vec{y} .$ Further, for $\\vec{v} \\ne \\vec{0} , \\pm \\frac{\\vec{v}}{|} | \\vec{v} | |$ are the unit vectors along $\\vec{v} .$ Now, $\\vec{a} \\times \\vec{b} = | \\left(i , j , k\\right) , \\left(- 1 , 2 , 1\\right) , \\left(- 4 , 0 , - 5\\right) |$ =(-10-0)i-(5-(-4))j+(0-(-8)k, $\\therefore \\vec{a} \\times \\vec{b} = - 10 i - 9 j + 8 k = \\left(- 10 , - 9 , 8\\right) .$ $\\therefore | | \\vec{a} \\times \\vec{b} | | = \\sqrt{{\\left(- 10\\right)}^{2} + {\\left(- 9\\right)}^{2} + {8}^{2}} = \\sqrt{245} = 7 \\sqrt{5.}$ Therefore, $\\pm \\left(- \\frac{10}{7 \\sqrt{5}} , - \\frac{9}{7 \\sqrt{5}} , \\frac{8}{7 \\sqrt{5}}\\right)$ are the desiredunit vectors. May 25, 2017 $\\hat{c} = \\pm \\frac{\\sqrt{5}}{35} \\left(\\begin{matrix}- 10 \\\\ - 9 \\\\ 8\\end{matrix}\\right)$ #### Explanation: We need to find a vector, $\\vec{c}$, such that $\\vec{c} \\bot \\vec{a}$ and $\\vec{c} \\bot \\vec{b}$. As such $\\vec{c} \\cdot \\vec{a} = 0$ and $\\vec{c} \\cdot \\vec{b} = 0$ Let $\\vec{c} = \\left(\\begin{matrix}x \\\\ y \\\\ z\\end{matrix}\\right)$ Then $- x + 2 y + z = 0$ and $- 4 x - 5", "points $(2, 0)$ and $(0, 1)$. Another suitable pair would be $$\\mathbf{a}= \\begin{pmatrix}6 \\\\ 2\\end{pmatrix}\\,, \\mathbf{b} =\\begin{pmatrix}4 \\\\ 1\\end{pmatrix}$$ To see why, note that the combinations $2\\mathbf{b}-\\mathbf{a}$ and $2\\mathbf{a}-3\\mathbf{a}$ take us to the points $(2, 0)$ and $(0, 1)$. For base vectors which yield none of the points in the original lattice, we would need irrational components, because if I had a base vector with rational components $\\frac{n}{m}$ and $\\frac{p}{q}$ then taking $2mq$ lots of this vector would take us to the point $(2qn, 2pm)$, which is in our original lattice of points.", "then $v=w$. (c) The zero vector $\\mathbf{0}$ is unique. (d) For each $v\\in V$, the additive inverse $-v$ is unique. (e) $0v=\\mathbf{0}$ for every $v\\in V$, where $0\\in\\R$ is the zero scalar. (f) $a\\mathbf{0}=\\mathbf{0}$ for every scalar $a$. (g) If $av=\\mathbf{0}$, then $a=0$ or $v=\\mathbf{0}$. (h) $(-1)v=-v$. The first two properties are called the cancellation law. ## Problem 709 Let $S=\\{\\mathbf{v}_{1},\\mathbf{v}_{2},\\mathbf{v}_{3},\\mathbf{v}_{4},\\mathbf{v}_{5}\\}$ where $\\mathbf{v}_{1}= \\begin{bmatrix} 1 \\\\ 2 \\\\ 2 \\\\ -1 \\end{bmatrix} ,\\;\\mathbf{v}_{2}= \\begin{bmatrix} 1 \\\\ 3 \\\\ 1 \\\\ 1 \\end{bmatrix} ,\\;\\mathbf{v}_{3}= \\begin{bmatrix} 1 \\\\ 5 \\\\ -1 \\\\ 5 \\end{bmatrix} ,\\;\\mathbf{v}_{4}= \\begin{bmatrix} 1 \\\\ 1 \\\\ 4 \\\\ -1 \\end{bmatrix} ,\\;\\mathbf{v}_{5}= \\begin{bmatrix} 2 \\\\ 7 \\\\ 0 \\\\ 2 \\end{bmatrix} .$ Find a basis for the span $\\Span(S)$. ## Problem 708 Let $A=\\begin{bmatrix} 2 & 4 & 6 & 8 \\\\ 1 &3 & 0 & 5 \\\\ 1 & 1 & 6 & 3 \\end{bmatrix}$. (a) Find a basis for the nullspace of $A$. (b) Find a basis for the row space of $A$. (c) Find a basis for the range of $A$ that consists of column vectors of $A$. (d) For each column vector which is not a basis vector that you obtained in part (c), express it as a linear combination of the basis vectors for the range of $A$. ## Problem 707", "Express the vector $\\mathbf{u}$ as a linear combination of $\\mathbf{v}$ and $\\mathbf{w}$. (b) Compute $A^5\\mathbf{v}$. (c) Compute $A^5\\mathbf{w}$. (d) Compute $A^5\\mathbf{u}$. ## Problem 691 In this problem, we use the following vectors in $\\R^2$. $\\mathbf{a}=\\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix}, \\mathbf{b}=\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix}, \\mathbf{c}=\\begin{bmatrix} 2 \\\\ 3 \\end{bmatrix}, \\mathbf{d}=\\begin{bmatrix} 3 \\\\ 2 \\end{bmatrix}, \\mathbf{e}=\\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}, \\mathbf{f}=\\begin{bmatrix} 5 \\\\ 6 \\end{bmatrix}.$ For each set $S$, determine whether $\\Span(S)=\\R^2$. If $\\Span(S)\\neq \\R^2$, then give algebraic description for $\\Span(S)$ and explain the geometric shape of $\\Span(S)$. (a) $S=\\{\\mathbf{a}, \\mathbf{b}\\}$ (b) $S=\\{\\mathbf{a}, \\mathbf{c}\\}$ (c) $S=\\{\\mathbf{c}, \\mathbf{d}\\}$ (d) $S=\\{\\mathbf{a}, \\mathbf{f}\\}$ (e) $S=\\{\\mathbf{e}, \\mathbf{f}\\}$ (f) $S=\\{\\mathbf{a}, \\mathbf{b}, \\mathbf{c}\\}$ (g) $S=\\{\\mathbf{e}\\}$ ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 672 For an integer $n > 0$, let $\\mathrm{P}_n$ be the vector space of polynomials of degree at most $n$. The set $B = \\{ 1 , x , x^2 , \\cdots , x^n \\}$ is a basis of $\\mathrm{P}_n$, called the standard basis. Let $T : \\mathrm{P}_n \\rightarrow \\mathrm{P}_{n+1}$ be the map defined by, for $f \\in \\mathrm{P}_n$, $T (f) (x) =", "\\mathbf{w}_3 \\|$. (e) $\\| 3 \\mathbf{w}_4 \\|$. (f) What is the distance between $\\mathbf{w}_2$ and $\\mathbf{w}_3$? ## Problem 688 Let $A$ be a $3\\times 3$ matrix and let $\\mathbf{v}=\\begin{bmatrix} 1 \\\\ 2 \\\\ -1 \\end{bmatrix} \\text{ and } \\mathbf{w}=\\begin{bmatrix} 2 \\\\ -1 \\\\ 3 \\end{bmatrix}.$ Suppose that $A\\mathbf{v}=-\\mathbf{v}$ and $A\\mathbf{w}=2\\mathbf{w}$. Then find the vector $A^5\\begin{bmatrix} -1 \\\\ 8 \\\\ -9 \\end{bmatrix}.$ ## Problem 687 For this problem, use the real vectors $\\mathbf{v}_1 = \\begin{bmatrix} -1 \\\\ 0 \\\\ 2 \\end{bmatrix} , \\mathbf{v}_2 = \\begin{bmatrix} 0 \\\\ 2 \\\\ -3 \\end{bmatrix} , \\mathbf{v}_3 = \\begin{bmatrix} 2 \\\\ 2 \\\\ 3 \\end{bmatrix} .$ Suppose that $\\mathbf{v}_4$ is another vector which is orthogonal to $\\mathbf{v}_1$ and $\\mathbf{v}_3$, and satisfying $\\mathbf{v}_2 \\cdot \\mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\\mathbf{v}_1 \\cdot \\mathbf{v}_2$. (b) $\\mathbf{v}_3 \\cdot \\mathbf{v}_4$. (c) $( 2 \\mathbf{v}_1 + 3 \\mathbf{v}_2 – \\mathbf{v}_3 ) \\cdot \\mathbf{v}_4$. (d) $\\| \\mathbf{v}_1 \\| , \\, \\| \\mathbf{v}_2 \\| , \\, \\| \\mathbf{v}_3 \\|$. (e) What is the distance between $\\mathbf{v}_1$ and $\\mathbf{v}_2$? ## Problem 686 In each of the following cases, can we conclude that $A$ is invertible? If so, find an expression for $A^{-1}$ as a linear combination of positive powers of $A$. If $A$ is not invertible, explain why not. (a) The matrix $A$ is a $3 \\times 3$", "- \\vec v$$ twice to $$\\vec v$$, because the vector $$[P\\vec{w}]\\vec v - \\vec v$$ is the \"distance vector\" from $$\\vec v$$ to $$Span(\\vec w)$$. Thus, $$$$\\begin{split} F\\vec{w}(\\vec v) &= \\begin{pmatrix} x\\\\y \\end{pmatrix} + 2 \\times (\\begin{pmatrix} \\frac{a^2x+aby}{a^2+b^2} \\frac{abx+b^2y}{a^2+b^2} \\end{pmatrix}- \\begin{pmatrix} x\\\\y \\end{pmatrix})\\\\ &= 2 \\times (\\begin{pmatrix} \\frac{a^2x+aby}{a^2+b^2} \\\\ \\frac{abx+b^2y}{a^2+b^2} \\end{pmatrix})- \\begin{pmatrix} x\\\\y \\end{pmatrix}\\\\ &= \\begin{pmatrix} \\frac{2a^2x+2aby}{a^2+b^2} -x \\\\ \\frac{2abx+2b^2y}{a^2+b^2}-y \\end{pmatrix}\\\\ &= \\begin{pmatrix} \\frac{a^2x+2aby-b^2x}{a^2+b^2} \\\\ \\frac{2abx+b^2y-a^2y}{a^2+b^2}\\end{pmatrix}\\\\ &= \\begin{pmatrix} \\frac{a^2-b^2}{a^2+b^2}x+\\frac{2ab}{a^2+b^2}y \\\\ \\frac{2ab}{a^2+b^2}x +\\frac{b^2-a^2}{a^2+b^2}y\\end{pmatrix} \\end{split}$$$$ From the formula, we know that $$F\\vec w$$ is a linear transformation just as $$P\\vec w$$, and that $$$$[F \\vec w] = \\begin{pmatrix} \\frac{a^2-b^2}{a^2+b^2} & \\frac{2ab}{a^2+b^2} \\\\ \\frac{2ab}{a^2+b^2} & \\frac{b^2-a^2}{a^2+b^2} \\end{pmatrix}$$$$ ### Subject:Number Theory TutorMe Question: Find, with full explanation, the remainder when dividing $$29^{1761}$$ by 23. Inactive Youtong W. This is a typical question of Modular arithmetics that directly utilizes Fermat's Little Theorem. We will give a little review of what the theorem says be fore we proceed. Fermat's Little Theorem: If $$p \\in N^+$$ is prime, then: 1. For all $$a\\in Z$$ with $$p\\nmid a$$, we have $$a^{p-1} \\equiv 1$$ (mod p); 2. For all $$a\\in Z$$, we have $$a^p \\equiv a$$ (mod p). Now 29 is prime, so is 23. So by the first part of Fermat's Little Theorem, $$29^{23-1}=29^{22} \\equiv 1$$(mod 23). Thus, $$(29^{22})^k \\equiv 1^k$$ for all $$k \\in N$$,", "+0 # Help appreciated 0 102 1 Consider the vectors v = < 1 , 3 > w = < 3 , 2 >, and x = < 1 , 0 >. If the vectors v, w and x are linearly independent, answer with 0. If they aren't, find coefficients a,b and c, not all 0, such that a < 1 , 3 > + b < 3 , 2 > + c < 1 , 0 > = < 0 , 0 > and answer with {a+b}/c. May 19, 2019 #1 +6046 0 3 vectors having two coordinates each cannot possibly be linearlly independent. $$\\text{we have the equation}\\\\ \\begin{pmatrix}1&3&1\\\\3&2&0\\end{pmatrix}\\begin{pmatrix}a\\\\b\\\\c\\end{pmatrix}=\\begin{pmatrix}0\\\\0\\end{pmatrix}$$ $$\\text{apply Gaussian elimination}\\\\ \\begin{pmatrix}1&3&1\\\\3&2&0\\end{pmatrix}\\\\ \\begin{pmatrix}1&3&1\\\\0&-7&-3\\end{pmatrix}\\\\ \\begin{pmatrix}1&3&1\\\\0&1&\\frac 3 7\\end{pmatrix}\\\\ \\begin{pmatrix}1&0&-\\frac 2 7\\\\0&1&\\frac 3 7\\end{pmatrix}\\\\$$ $$a-\\dfrac 2 7 c = 0,~c=\\dfrac 7 2 a\\\\ b + \\dfrac 3 7 c = 0,~c = -\\dfrac 7 3 b\\\\ b = -\\dfrac 3 2 a\\\\ \\left(a,b,c\\right)= \\left(1,-\\dfrac 3 2,\\dfrac 7 2\\right)$$ . May 19, 2019 edited by Rom May 20, 2019" ]