hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \int \frac {\sin a x \ \mathrm d x} {\sin \left({a x + \phi}\right)}       | r = \frac 1 {\cos \phi} \int \frac {\sin a x \cos \phi \ \mathrm d x} {\sin \left({a x + \phi}\right)}       | c = multiplying top and bottom by $\cos \phi$ }} {{eqn | r = \frac 1 {\cos \phi} \int \frac {\left({\sin a x \cos \phi + \cos a x \sin \phi - \cos a x \sin \phi}\right) \ \mathrm d x} {\sin \left({a x + \phi}\right)}       | c =  }} {{eqn | r = \frac 1 {\cos \phi} \int \frac {\left({\sin \left({a x + \phi}\right) - \cos a x \sin \phi}\right) \ \mathrm d x} {\sin \left({a x + \phi}\right)}       | c = Sine of Sum }} {{eqn | r = \frac 1 {\cos \phi} \int \frac {\left({\sin \left({a x + \phi}\right) - \cos a x \sin \phi}\right) \ \mathrm d x} {\sin \left({a x + \phi}\right)}       | c = Sine of Sum }} {{eqn | r = \frac 1 {\cos \phi} \int \mathrm d x  - \frac {\sin \phi} {\cos \phi} \int \frac {\cos a x \ \mathrm d x} {\sin \left({a x + \phi}\right)}       | c = Linear Combination of Integrals }} {{eqn | r = \frac x {\cos \phi} - \tan \phi \int \frac {\cos a x \ \mathrm d x} {\sin \left({a x + \phi}\right)} + C       | c = Tangent is Sine divided by Cosine }} {{end-eqn}} {{qed}} Category:Primitives involving Sine Function Category:Primitives involving Cosine Function 8u5q1cmoonorym0tk2lhyqd4ysmepto
{{begin-eqn}} {{eqn | l = \cosh x       | r = 2 \cosh^2 \frac x 2 - 1       | c = Hyperbolic Cosine Double Angle Formula }} {{eqn | ll= \iff       | l = 1 + \cosh x       | r = 2 \cosh^2 \frac x 2       | c = adding $1$ to both sides }} {{eqn | ll= \iff       | l = \frac 1 {1 + \cosh x}       | r = \frac 1 2 \frac 1 {\cosh^2 \frac x 2}       | c = taking the reciprocal of both sides }} {{eqn | r = \frac 1 2 \operatorname{sech}^2 \frac x 2       | c = Definition of Hyperbolic Secant }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cosh x       | r = 1 + 2 \sinh^2 \frac x 2       | c = Hyperbolic Cosine Double Angle Formula }} {{eqn | ll= \leadstoandfrom       | l = \cosh x - 1       | r = 2 \sinh^2 \frac x 2       | c = subtracting $1$ from both sides }} {{eqn | ll= \leadstoandfrom       | l = \frac 1 {\cosh x - 1}       | r = \frac 1 2 \frac 1 {\sinh^2 \frac x 2}       | c = taking the reciprocal of both sides }} {{eqn | r = \frac 1 2 \csch^2 \frac x 2       | c = {{Defof|Hyperbolic Cosecant|index = 2}} }} {{end-eqn}} {{qed}}
Let $a$, $b$, and $c$ are the sides opposite $A$, $B$ and $C$ respectively. By Length of Angle Bisector, $\omega_\alpha, \omega_\beta$ are given by: :$\omega_\alpha^2 = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}$ :$\omega_\beta^2 = \dfrac {b a} {\paren {b + a}^2} \paren {\paren {b + a}^2 - c^2}$ Equating the two equations,  {{begin-eqn}} {{eqn | l = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}       | r = \dfrac {b a} {\paren {b + a}^2} \paren {\paren {b + a}^2 - c^2} }} {{eqn | ll = \leadsto       | l = b c \paren {b + a}^2 \paren {\paren {b + c}^2 - a^2}       | r = b a \paren {b + c}^2 \paren {\paren {b + a}^2 - c^2} }} {{eqn | ll = \leadsto       | l = b c \paren {b + a}^2 \paren {b + c + a} \paren {b + c - a}       | r = b a \paren {b + c}^2 \paren {b + a + c} \paren {b + a - c} }} {{eqn | ll = \leadsto       | l = c \paren {b + a}^2 \paren {b + c - a}       | r = a \paren {b + c}^2 \paren {b + a - c} \quad \quad a + b + c > 0 \text { as each } a, b, c > 0 }} {{eqn | ll = \leadsto       | l = a^2 b c + a^2 c^2 - a^3 c + 2 a b^2 c + 2 a b c^2 - 2 a^2 b c + b^3 c + b^2 c^2 - a b^2 c       | r = a b^3 + a^2 b^2 - a b^2 c + 2 a b^2 c + 2 a^2 b c - 2 a b c^2 + a b c^2 + a^2 c^2 - a c^3 }} {{eqn | ll = \leadsto       | l = - a^3 c + 2 a b c^2 - a^2 b c + b^3 c + b^2 c^2       | r = a b^3 + a^2 b^2 + 2 a^2 b c - a b c^2 - a c^3 }} {{eqn | ll = \leadsto       | l = a c^3 - a^3 c + 3 a b c^2 - 3 a^2 b c + b^3 c - a b^3 + b^2 c^2 - a^2 b^2        | r = 0 }} {{eqn | ll = \leadsto       | l = \paren {c - a} \paren {a c \paren {c + a} + 3 a b c + b^3 + b^2 \paren {c + a} }       | r = 0 }} {{eqn | ll = \leadsto       | l = c - a       | r = 0 \quad \quad a c \paren {c + a} + 3 a b c + b^3 + b^2 \paren {c + a} > 0 \text { as each } a, b, c > 0 }} {{eqn | ll = \leadsto       | l = c       | r = a }} {{end-eqn}} Therefore $ABC$ is an isosceles triangle. {{qed}}
Let $\triangle ABC$ and $\triangle DEF$ be two triangles having sides $AB = DE$ and $AC = DF$, and with $\angle ABC = \angle DEF = 90^\circ$. By Pythagoras' Theorem: : $BC = \sqrt {AB^2 + AC^2}$ and: : $EF = \sqrt {DE^2 + DF^2}$ :$\therefore BC = \sqrt {AB^2 + AC^2} = \sqrt {DE^2 + DF^2} = EF$ The part that the remaining two angles are equal to their respective remaining angles follows from Triangle Side-Side-Side Equality. {{qed}}
From Sum of Internal Angles of Polygon, we have that the sum $S$ of all internal angles of a $n$-gon is: :$S = \paren {n - 2} 180 \degrees$ From the definition of a regular polygon, all the internal angles of a regular polygon are equal. Therefore, the size $A$ of each internal angle of a regular polygon with $n$ sides is: :$A = \dfrac {\paren {n - 2} 180 \degrees} n$ {{qed}}
For convex polygons, name a vertex as $A_1$, go clockwise and name the vertices as $A_2, A_3, \ldots, A_n$. By joining $A_1$ to every vertex except $A_2$ and $A_n$, one can form $\paren {n - 2}$ triangles. From Sum of Angles of Triangle equals Two Right Angles, the sum of the internal angles of a  triangle is $180 \degrees$. Therefore, the sum of internal angles of a polygon with $n$ sides is $\paren {n - 2} 180 \degrees$. {{qed}} {{proof wanted|concave polygons?}}
Consider a convex regular polyhedron $P$. Let $m$ be the number of sides of each of the regular polygons that form the faces of $P$. Let $n$ be the number of those polygons which meet at each vertex of $P$. From Internal Angles of Regular Polygon, the internal angles of each face of $P$ measure $180^\circ - \dfrac {360^\circ} m$. The sum of the internal angles must be less than $360^\circ$. So: {{begin-eqn}} {{eqn | l = n \paren {180^\circ - \dfrac {360^\circ} m}       | o = <       | r = 360^\circ       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {1 - \dfrac 2 m}       | o = <       | r = 2       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {m - 2}       | o = <       | r = 2m       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {m - 2} - 2 \paren {m - 2}       | o = <       | r = 2m - 2 \paren {m - 2}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {m - 2} \paren {n - 2}       | o = <       | r = 4       | c =  }} {{end-eqn}} But $m$ and $n$ are both greater than $2$. So: :if $m = 3$, $n$ can only be $3$, $4$ or $5$ :if $m = 4$, $n$ can only be $3$ :if $m = 5$, $n$ can only be $3$ and $m$ cannot be greater than $3$. There are $5$ possibilities in all. Therefore all platonic solids have been accounted for. {{qed}}
Let $T = \left({\R, \tau}\right)$ be the set of real numbers $\R$ with the countable complement topology. From Countable Complement Space is not $T_2$, $T$ is not a Hausdorff space. Suppose $\left\langle{x_n}\right\rangle$ is a sequence in $\R$ which converges to $x$. Then $C = \left\{{x_n: x_n \ne x}\right\}$ is closed in $T$ because it is countable. So $X \setminus C$ is a neighborhood of $x$. This means there is some $N \in \N$ such that: :$\forall n > N: x_n \in X \setminus C$ That is, $x_n = x$ for large $n$. This means that if $x_n \to y$ then $y = x$, proving limit points in $T$ are unique. {{qed}} Category:Hausdorff Spaces Category:Countable Complement Topology 7v3aad2g7ywuewc89zbk8mykmrsgxtu
Let $G_n = \sequence {a_0, a_1, \ldots, a_n}$ be natural numbers in geometric sequence such that $a_0 \perp a_n$. {{AimForCont}}, suppose there were to exist another set of natural numbers in geometric sequence: :$G\,'_n = \sequence {b_0, b_1, \cdots, b_n}$ with the same common ratio where: :$\forall k \in \N_{\le n}: a_k > b_k$ From {{EuclidPropLink|book = VII|prop = 14|title = Proportion of Numbers is Transitive}}: :$a_0 : a_n = b_0 : b_n$ But by hypothesis: :$a_0 \perp a_n$ and so from: :{{EuclidPropLink|prop = 21|title = Coprime Numbers form Fraction in Lowest Terms}} and: :{{EuclidPropLink|prop = 20|title = Ratios of Fractions in Lowest Terms}} it follows that: :$a_0 \divides b_0$ However, this contradicts the assumption that $b_0 < a_0$. Therefore $a_0, a_1, \cdots, a_n$ are the least of those with the same common ratio. {{qed}}
Let $G_n = \sequence {a_0, a_1, \ldots, a_n}$ be natural numbers in geometric sequence such that $a_0 \perp a_n$. {{AimForCont}} $G\,'_n = \sequence {b_0, b_1, \cdots, b_n}$ be another set of natural numbers in geometric sequence with the same common ratio where: :$\forall k \in \N_{\le n}: a_k > b_k$ By definition of geometric sequence: :$a_0 = r^n a_n$ :$b_0 = r^n b_n$ Hence: :$\dfrac {a_0} {a_n} = \dfrac {b_0} {b_n}$ By hypothesis: : $a_0 \perp a_n$ Thus $\dfrac {a_0} {a_n}$ is in canonical form. From Canonical Form of Rational Number is Unique it follows that $a_0$ and $a_n$ are the only two integers fulfilling these conditions. Thus: :$\forall p, q: \dfrac p q = \dfrac {a_0} {a_n}: a_0 \le p, a_n \le q$ But it was supposed that: :$b_0 < a_0$ :$b_n < a_n$ From this contradiction it follows that there can be no such $b_0, \ldots, b_n$. Hence the result. {{qed}}
From $2 \pi$ radians, a full rotation is defined to be $360^\circ$ or $2 \pi$ radians. Since lines are straight, it therefore follows that from any point on a line, the angle between one side of the line and the other is one half of a full rotation. Therefore, the measurement of a straight angle is: :$\dfrac{360^\circ} 2 = 180^\circ$ or: :$\dfrac {2 \pi} 2 = \pi$ {{qed}}
A right angle is equal to one half of a straight angle. From Measurement of Straight Angle it follows that the measurement of a right angle is $\dfrac {180 \degrees} 2 = 90 \degrees$ or $\dfrac \pi 2$. {{qed}}
From Full Rotation measures $2 \pi$ Radians, a full rotation is $2 \pi$ radians. By definition of degree of angle, a full rotation is $360^\circ$. Therefore, a full angle is $360^\circ$ or $2 \pi$. {{qed}} Category:Angles 3ohjiezeamys6ibjby32gvefryqjk7w
An acute angle is defined to be an angle whose measure is between that of a zero angle and a right angle. A zero angle measures $0$ by definition, and a right angle measures $90 \degrees$ or $\dfrac \pi 2$. Hence the result. {{qed}} Category:Angles 6mq2g6a4rxr0okwgk49ktafsz00xpnl
An obtuse angle is defined to be an angle whose measure is between that of a right angle and a straight angle. A right angle measures $90^\circ$ or $\dfrac \pi 2$ and a straight angle measures $180^\circ$ or $\pi$. Hence the result. {{qed}} Category:Angles fd1dnjje8x8b0xlq1mrfiz637oxciba
A reflex angle is defined to be an angle whose measure is between that of a straight angle and a full angle. A straight angle measures $180^\circ$ or $\pi$ and a full angle measures $360^\circ$ or $2 \pi$. Hence the result. {{qed}} Category:Angles 6mvlsmzuva8jr6i0ng49gra5jvx9cg3
:200px Let $AB$ and $CD$ be infinite straight lines. Let $EF$ be a transversal that cuts them. Let at least one pair of interior angles on the same side of the transversal be supplementary. {{WLOG}}, let those interior angles be $\angle BGH$ and $\angle DHG$. So, by definition, $\angle DHG + \angle BGH$ equals two right angles. Also, from Two Angles on Straight Line make Two Right Angles, $\angle AGH + \angle BGH$ equals two right angles. Then from Euclid's first and third common notion and Euclid's fourth postulate: :$\angle AGH = \angle DHG$ Finally, by Equal Alternate Angles implies Parallel Lines: :$AB \parallel CD$ {{qed}} {{Euclid Note|28|I|{{EuclidNoteConverse|prop = 29|title = Parallelism implies Supplementary Interior Angles|part = third}}|part = second}}
:200px Let $AB$ and $CD$ be infinite straight lines. Let $EF$ be a transversal that cuts them. Let at least one pair of corresponding angles be equal. {{WLOG}}, let $\angle EGB = \angle GHD$. By the Vertical Angle Theorem: :$\angle GHD = \angle EGB = \angle AGH$ Thus by Equal Alternate Angles implies Parallel Lines: :$AB \parallel CD$ {{qed}} {{Euclid Note|28|I|{{EuclidNoteConverse|prop = 29|title = Parallelism implies Equal Corresponding Angles|part = second}}|part = first}}
:200px Let $AB$ and $CD$ be parallel infinite straight lines. Let $EF$ be a transversal that cuts them. Aiming for a contradiction, suppose the alternate angles are not equal. Then one of the pair $\angle AGH$ and $\angle GHD$ must be greater. {{WLOG}}, let $\angle AGH$ be greater. From Two Angles on Straight Line make Two Right Angles, $\angle AGH + \angle BGH$ equal two right angles. So $\angle GHD + \angle BGH$ is less than two right angles. By Euclid's fifth postulate, lines extended infinitely from angles less than two right angles must meet. But the lines are parallel. So by definition the lines do not intersect. From this contradiction it follows that the alternate angles are be equal. {{qed}} {{Euclid Note|29|I|{{EuclidNoteConverse|prop = 27|title = Equal Alternate Angles implies Parallel Lines}}<br><br>This is the first proposition of {{BookLink|The Elements|Euclid}} to make use of Euclid's fifth postulate.|part = first}}
:200px Let $AB$ and $CD$ be parallel infinite straight lines. Let $EF$ be a transversal that cuts them. From Parallelism implies Equal Alternate Angles: :$\angle AGH = \angle DHG$ By the Vertical Angle Theorem: :$\angle EGB = \angle AGH = \angle DHG$ {{qed}} {{Euclid Note|29|I|{{EuclidNoteConverse|prop = 28|title = Equal Corresponding Angles implies Parallel Lines|part = first}}|part = second}}
Let $x$ be arbitrary. Then $x = 1 \times x$. The result follows from the definition of reflexivity and proportion. {{qed}} Category:Proportion Category:Reflexive Relations ihv11jn67cuzil8n3v37u8hpbhzun6m
Let $x, y$ be arbitrary. Let $x$ be proportional to $y$: :$x \propto y$ Then by definition: {{begin-eqn}} {{eqn | ll= \exists k \ne 0:       | l = x       | r = k \times y       | c =  }} {{eqn | lll= \implies       | l = y       | r = k^{-1} \times x       | c =  }} {{end-eqn}} The result follows from the definition of symmetry and proportion. {{qed}} Category:Proportion Category:Symmetric Relations sjurctni0ip6boyngisddlghex73gno
Let $x, y, z$ be arbitrary. Let $x$ be proportional to $y$ and $y$ to $z$: :$x \propto y \land y \propto z$ Then by definition: :$\exists j, k \ne 0: x = j \times y \land y = k \times z$ Substituting $k \times z$ for $y$: :$x = \left({j \times k}\right) \times z$ so $j \times k$ is the desired constant of proportion. The result follows from the definition of transitivity and proportion. {{qed}} Category:Proportion Category:Transitive Relations mvflj2wl0kv6fcqafmywaghzdcg5d14
* Proportion is Reflexive: $\forall x \in \R: x \propto x$ * Proportion is Symmetric: $\forall x, y \in \R: x \propto y \implies y \propto x$ * Proportion is Transitive: $\forall x, y, z \in \R: x \propto y \land y \propto z \implies x \propto z$ The result follows from the definition of an equivalence relation. {{qed}} Category:Proportion Category:Equivalence Relations 3vin5kgub8il4kmryq48cs3281yporg
:200px Let $AB$ and $CD$ be parallel infinite straight lines. Let $EF$ be a transversal that cuts them. From Parallelism implies Equal Corresponding Angles and Euclid's second common notion: :$\angle EGB + \angle BGH = \angle DHG + \angle BGH$ From Two Angles on Straight Line make Two Right Angles, $\angle EGB + \angle BGH$ equals two right angles. So by definition, $\angle BGH$ and $\angle DHG$ are supplementary. {{qed}} {{Euclid Note|29|I|{{EuclidNoteConverse|prop = 28|title = Supplementary Interior Angles implies Parallel Lines|part = second}}|part = third}}
Let $x \propto y$. Then $\exists k \ne 0: x = k \times y$ by the definition of proportion. Raising both sides of this equation to the $n$th power: {{begin-eqn}} {{eqn | l = x^n       | r = \left({k \times y}\right)^n }} {{eqn | r = k^n \times y^n }} {{end-eqn}} so $k^n$ is the desired constant of proportion. The result follows from the definition of proportion. {{qed}} Category:Proportion qpxl6b1brrbs6ujti8w4beangbxaaqz
From Sine of $90 \degrees$: :$\sin \paren {5 \times 18 \degrees} = \sin 90 \degrees = 1$. Consider the equation: : $\sin 5x = 1$ where $x = 18 \degrees$ is one of the solutions. From Quintuple Angle Formula of Sine: :$16 \sin^5 \theta - 20 \sin^3 \theta + 5 \sin \theta = 1$ Let $s = \sin \theta$: :$16 s^5 - 20 s^3 + 5s - 1 = 0$ That is: :$\paren {s - 1} \paren {4 s^2 + 2 s - 1}^2 = 0$ Therefore, either: :$s = 1$ or by the Quadratic Formula: :$s = \dfrac 1 4 \paren {\pm \sqrt 5 - 1}$ Since $0 < \sin 18 \degrees < 1$: :$\sin 18^\circ = \dfrac {\sqrt 5 - 1} 4$ {{qed}} Category:Sine Function q0tt2du7kh1dffh46qb53sjjb92qbxo
{{begin-eqn}} {{eqn | l = \cos 72 \degrees       | r = \map \cos {90 \degrees - 18 \degrees} }} {{eqn | r = \sin 18 \degrees       | c = Cosine of Complement equals Sine }} {{eqn | r = \dfrac {\sqrt 5 - 1} 4       | c = Sine of $18 \degrees$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos 72 \degrees       | r = 2 \cos 36 \degrees - 1 }} {{eqn | r = 2 \paren {\dfrac \phi 2}^2 - 1       | c = Cosine of $36 \degrees$ }} {{eqn | r = \dfrac {\phi^2} 2 - 1       | c =  }} {{eqn | r = \dfrac {\phi + 1} 2 - 1       | c = Square of Golden Mean equals One plus Golden Mean }} {{eqn | r = \dfrac {\phi - 1} 2       | c =  }} {{eqn | r = -\dfrac {1 - \phi} 2       | c =  }} {{eqn | r = \dfrac {\phi^{-1} } 2       | c = Reciprocal Form of One Minus Golden Mean }} {{eqn | r = \dfrac {\sqrt 5 - 1} 4       | c = {{Defof|Golden Mean|index = 2}}, and algebra }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sin 72 \degrees       | r = \sqrt {1 - \cos^2 72 \degrees}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \sqrt {1 - \paren {\dfrac{\sqrt 5 - 1} 4}^2}       | c = Cosine of $72 \degrees$ }} {{eqn | r = \sqrt {1 - \dfrac {6 - 2 \sqrt 5} {16} } }} {{eqn | r = \dfrac {\sqrt {10 + 2 \sqrt 5} } 4 }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos 18^\circ       | r = \cos \left({90^\circ - 72^\circ}\right) }} {{eqn | r = \sin 72^\circ       | c = Cosine of Complement equals Sine }} {{eqn | r = \frac {\sqrt {10 + 2 \sqrt 5} } 4       | c = Sine of 72 Degrees }} {{end-eqn}} {{qed}} Category:Cosine Function qfdwbrom2u1wtnh06j1q6ipf4ebudx0
:300px Let $\triangle ABC$ be a triangle. Let $BC$ be extended to a point $D$. Construct $CE$ through the point $C$ parallel to the straight line $AB$. We have that $AB \parallel CE$ and $AC$ is a transversal that cuts them. From Parallelism implies Equal Alternate Angles: :$\angle BAC = \angle ACE$ Similarly, we have that $AB \parallel CE$ and $BD$ is a transversal that cuts them. From Parallelism implies Equal Corresponding Angles: :$\angle ECD = \angle ABC$ Thus by Euclid's Second Common Notion: : $\angle ACD = \angle ABC + \angle BAC$ {{qed}}
{{begin-eqn}} {{eqn | l = \sin 3^\circ       | r = \sin \left({75^\circ - 72^\circ}\right) }} {{eqn | r = \sin 75^\circ \cos 72^\circ - \cos 75^\circ \sin 72^\circ       | c = Sine of Difference }} {{eqn | r = \dfrac {\sqrt 6 + \sqrt 2} 4 \times \dfrac {\sqrt 5 - 1} 4 - \cos 75^\circ \sin 72^\circ       | c = Sine of $75^\circ$, Cosine of $72^\circ$ ... }} {{eqn | r = \dfrac {\sqrt 6 + \sqrt 2} 4 \times \dfrac {\sqrt 5 - 1} 4 - \dfrac {\sqrt 6 - \sqrt 2} 4 \times \dfrac {\sqrt {10 + 2 \sqrt 5} } 4       | c = ... Cosine of $75^\circ$, Sine of $72^\circ$ }} {{eqn | r = \dfrac 1 {16} \times \left({\left({\sqrt 6 + \sqrt 2}\right) \times \left({\sqrt 5 - 1}\right) - \left({\sqrt 6 - \sqrt 2}\right) \times \left({\sqrt {10 + 2 \sqrt 5} }\right) }\right) }} {{eqn | r = \dfrac 1 {16} \times \left({\left({\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2}\right) - \left({\sqrt {60 + 12 \sqrt 5} - \sqrt {20 + 4 \sqrt 5} }\right) }\right) }} {{eqn | r = \dfrac 1 {16} \times \left({\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - \sqrt {60 + 12 \sqrt5 } + \sqrt {20 + 4 \sqrt 5} }\right) }} {{eqn | r = \dfrac 1 {16} \times \left({\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2\sqrt {15 + 3 \sqrt5 } + 2 \sqrt {5 + \sqrt 5} }\right) }} {{eqn | r = \dfrac {\sqrt{30} + \sqrt{10} - \sqrt 6 - \sqrt 2 - 2 \sqrt {15 + 3 \sqrt 5} + 2\sqrt {5 + \sqrt 5} } {16} }} {{end-eqn}} {{qed}} Category:Sine Function kjgsyns60rpt2i9bn4jxxxy59rzkp3s
{{begin-eqn}} {{eqn | l = \cos 3^\circ       | r = \cos \left({75^\circ - 72^\circ}\right) }} {{eqn | r = \cos 75^\circ \cos 72^\circ + \sin 75^\circ \sin 72^\circ       | c = Cosine of Difference }} {{eqn | r = \dfrac {\sqrt 6 - \sqrt 2} 4 \times \dfrac {\sqrt 5 - 1} 4 + \dfrac {\sqrt 6 + \sqrt 2} 4 \times \dfrac {\sqrt {10 + 2 \sqrt5 } } 4       | c = Cosine of 75 Degrees, Cosine of 72 Degrees, Sine of 75 Degrees, Sine of 72 Degrees }} {{eqn | r = \dfrac 1 {16} \times \left[{\left({\sqrt 6 - \sqrt 2}\right) \times \left({\sqrt 5 - 1}\right) + \left({\sqrt 6 + \sqrt 2}\right) \times \left({\sqrt {10 + 2 \sqrt 5} }\right) }\right] }} {{eqn | r = \dfrac 1 {16} \times \left[{\left({\sqrt{30} - \sqrt{10} - \sqrt 6 + \sqrt 2}\right) + \left({\sqrt{60 + 12 \sqrt 5} + \sqrt{20 + 4 \sqrt 5} }\right) }\right] }} {{eqn | r = \dfrac 1 {16} \times \left({\sqrt{30} - \sqrt{10} - \sqrt 6 + \sqrt 2 + \sqrt{60 + 12 \sqrt 5} + \sqrt{20 + 4 \sqrt 5} }\right) }} {{eqn | r = \dfrac 1 {16} \times \left({\sqrt{30} - \sqrt{10} - \sqrt 6 + \sqrt 2 + 2 \sqrt{15 + 3 \sqrt 5} + 2 \sqrt{5 + \sqrt 5} }\right) }} {{eqn | r = \dfrac{\sqrt{30} - \sqrt{10} - \sqrt 6 + \sqrt 2 + 2 \sqrt{15 + 3 \sqrt 5} + 2 \sqrt{5 + \sqrt 5} }{16} }} {{end-eqn}} {{qed}} Category:Cosine Function fl2cb66is22nvq38tpw30zkxhq1vq57
Let $\sqrt {a + b}$ be expressed in the form $\sqrt c + \sqrt d$. From Square of Sum: : $a + b = c + d + 2 \sqrt {c d}$ We now need to solve the simultaneous equations: : $a = c + d$ : $b = 2 \sqrt {c d}$ First: {{begin-eqn}} {{eqn | l = a       | r = c + d }} {{eqn | n = 1       | ll= \leadsto       | l = d       | r = a - c       | c = subtracting $c$ from both sides }} {{end-eqn}} Solving for $c$: {{begin-eqn}} {{eqn | l = b       | r = 2 \sqrt {c d} }} {{eqn | ll= \leadsto       | l = b^2       | r = 4 c d       | c = squaring both sides }} {{eqn | r = 4 c \paren {a - c}       | c = substituting $d = a - c$ from $(1)$ }} {{eqn | r = 4 a c - 4 c^2       | c = Real Multiplication Distributes over Addition }} {{eqn | ll= \leadsto       | l = 4 c^2 - 4 a c + b^2       | r = 0       | c = adding $4 c^2 - 4 a c$ to both sides }} {{eqn | n = 2       | ll= \leadsto       | l = c       | r = \frac a 2 \pm \frac {\sqrt {a^2 - b^2} } 2       | c = Quadratic Formula }} {{end-eqn}} Solving for $d$: {{begin-eqn}} {{eqn | l = d       | r = a - c }} {{eqn | r = a - \frac a 2 \mp \frac {\sqrt {a^2 - b^2} } 2       | c = substituting $c = \dfrac a 2 \pm \dfrac {\sqrt {a^2 - b^2} } 2$ from $(2)$ }} {{eqn | r = \frac a 2 \mp \frac {\sqrt {a^2 - b^2} } 2 }} {{end-eqn}} From Real Addition is Commutative, the sign of the square root may be chosen arbitrarily, provided opposite signs are chosen for $c$ and $d$. {{qed}}
From Sum of Square Roots as Square Root of Sum: :$\sqrt p + \sqrt q = \sqrt {p + q + \sqrt {4pq}}$ Let :$p = \dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2$, :$q = \dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2$. Then {{begin-eqn}} {{eqn | l = p + q       | r = \frac a 2 + \frac {\sqrt {a^2 - b^2} } 2 + \frac a 2 - \frac {\sqrt {a^2 - b^2} } 2 }} {{eqn | r = a }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sqrt {4pq}       | r = \sqrt {4 \paren {\frac a 2 + \frac {\sqrt {a^2 - b^2} } 2} \paren {\frac a 2 - \frac {\sqrt {a^2 - b^2} } 2} } }} {{eqn | r = \sqrt {\paren {a + \sqrt {a^2 - b^2} } \paren {a - \sqrt {a^2 - b^2} } }       | c = Real Multiplication Distributes over Addition }} {{eqn | r = \sqrt {a^2 - \paren {a^2 - b^2} }       | c = Difference of Two Squares }} {{eqn | r = \sqrt {b^2} }} {{eqn | r = b }} {{end-eqn}} Therefore: :$\sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2} = \sqrt {a + b}$ {{qed}}