hcju/proofwikiqa huggingface dataset:
Let the radius of convergence of $\map f x$ be $R$. Suppose $x \in I$ such that $x$ is not an endpoint of $I$. Then there exists $x_0 \in I$ such that $x$ lies between $x_0$ and $\xi$. Thus: :$\size {x - \xi} < \size {x_0 - \xi} < R$ Consider the series: :$\displaystyle \sum_{n \mathop = 2}^\infty \size {\frac {n \paren {n - 1} } 2 a_n X_0^{n - 2} }$ where $X_0 = x_0 - \xi$ From Radius of Convergence from Limit of Sequence, we have: :$\displaystyle \frac 1 R = \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = \limsup_{n \mathop \to \infty} \paren {\size {\frac {n \paren {n - 1} } 2} a_n}^{1/n}$ Thus we may deduce that: :$\displaystyle \sum_{n \mathop = 2}^\infty \size {\frac {n \paren {n - 1} } 2 a_n X_0^{n - 2} }$ converges. Put $\delta = \size {x - x_0}$. For values of $y$ such that $0 < \size {x - y} < \delta$, we consider: :$\displaystyle \Delta = \frac {\map f y - \map f x} {y - x} - \sum_{n \mathop = 1}^\infty n a_n \paren {x - \xi}^{n - 1} = \sum_{n \mathop = 1}^\infty a_n \paren {\frac {Y^n - X^n} {Y - X} - n X^{n - 1} }$ where $X = x - \xi, Y = y - \xi$. We want to show that $\Delta \to 0$ as $Y \to X$. From Difference of Two Powers, we have: {{begin-eqn}} {{eqn | l = \frac {Y^n - X^n} {Y - X} - n X^{n-1}       | r = \paren {Y^{n - 1} + Y^{n - 2} X + \cdots + Y X^{n - 2} + X^{n - 1} } - n X^{n - 1}       | c =  }} {{eqn | r = \paren {Y^{n - 1} - X^{n - 1} } + X \paren {Y^{n - 2} - X^{n - 2} } + \cdots + x^{n - 2} \paren {Y - X}       | c =  }} {{end-eqn}} It follows that from all of these terms we can extract $\left({Y - X}\right)$ as a factor. So the {{RHS}} reduces to a product of $\paren {Y - X}$ with the sum of $n \paren {n - 1} / 2$ terms of the form $X^r Y^s$ where $r + s = n - 2$. The number $n \paren {n - 1} / 2$ arises from Closed Form for Triangular Numbers: :$1 + 2 + \cdots + \paren {n - 1} = \dfrac {n \paren {n - 1} } 2$ Since: :$\size X = \size {x - \xi} < \size {x_0 - \xi} = \size {X_0}$ :$\size Y = \size {y - \xi} < \size {x - \xi} + \delta = \size {X_0}$ we have: :$\size {\dfrac {Y^n - X^n} {Y - X} - n X^{n - 1} } < n \paren {n - 1} / 2 \size {X_0}^{n - 2} \size {Y - X}$ So: :$\displaystyle \size \Delta \le \size {Y - X} \sum_{n \mathop = 2}^\infty \frac {n \paren {n - 1} } 2 \size {a_n X_0^{n - 2} } \to 0$ as $Y \to X$ So $f$ is differentiable at all $x \in I$ which is not an endpoint, and that: :$\displaystyle \map {D_x} {\map f x} = \sum_{n \mathop = 1}^\infty n a_n \paren {x - \xi}^{n - 1}$ Now we need to investigate the question of left hand continuity and right hand continuity at the endpoints of $I$. Abel's Theorem tells us that if $\displaystyle \sum_{k \mathop = 0}^\infty a_k$ is convergent, then $\displaystyle \lim_{x \mathop \to 1^-} \paren {\sum_{k \mathop = 0}^\infty a_k x^k} = \sum_{k \mathop = 0}^\infty a_k$. {{proof wanted}}
From Higher Derivatives of Exponential Function, we have: :$\forall n \in \N: \map {f^{\paren n} } {\exp x} = \exp x$ Since $\exp 0 = 1$, the Taylor series expansion for $\exp x$ about $0$ is given by: :$\displaystyle \exp x = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ From Radius of Convergence of Power Series over Factorial, we know that this power series expansion converges for all $x \in \R$. From Taylor's Theorem, we know that :$\displaystyle \exp x = 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} + \frac {x^n} {n!} \map \exp \eta$ where $0 \le \eta \le x$. Hence: {{begin-eqn}} {{eqn | l = \size {\exp x - \paren {1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} } }       | r = \size {\frac {x^n} {n!} \map \exp \eta}       | c =  }} {{eqn | o = \le       | r = \frac {\size {x^n} } {n!} \map \exp {\size x}       | c = Exponential is Strictly Increasing }} {{eqn | o = \to       | r = 0       | rr= \text { as } n \to \infty       | c = Series of Power over Factorial Converges }} {{end-eqn}} So the partial sums of the power series converge to $\exp x$. The result follows. {{qed}}
From the definition of the cosine function, we have: :$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$ Then: {{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\cos x}       | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n 2 n \frac {x^{2 n - 1} } {\paren {2 n}!}       | c = Power Series is Differentiable on Interval of Convergence }} {{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n - 1} } {\paren {2 n - 1}!} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^{n + 1} \frac {x^{2 n + 1} } {\paren {2 n + 1}!}       | c = changing summation index }} {{eqn | r = -\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} }} {{end-eqn}} The result follows from the definition of the sine function. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\cos x}       | r = \lim_{h \mathop \to 0} \frac {\map \cos {x + h} - \cos x} h       | c = {{Defof|Derivative of Real Function at Point}} }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\cos x \cos h - \sin x \sin h - \cos x} h       | c = Cosine of Sum }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\cos x \cos h - \cos x} h + \lim_{h \mathop \to 0} \frac {-\sin x \sin h} h       | c = Sum Rule for Limits of Functions }} {{eqn | r = \cos x \lim_{h \mathop \to 0} \frac {\cos h - 1} h - \sin x \lim_{h \mathop \to 0} \frac {\sin h} h       | c = Multiple Rule for Limits of Functions }} {{eqn | r = \cos x \times 0 - \sin x \times 1       | c = Limit of (Cosine (X) - 1) over X and Limit of Sine of X over X }} {{eqn | r = -\sin x }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac \d {\d x} \cos x       | r = \frac \d {\d x} \map \sin {\frac \pi 2 - x}       | c = Sine of Complement equals Cosine }} {{eqn | r = -\map \cos {\frac \pi 2 - x}       | c = Derivative of Sine Function and Chain Rule for Derivatives  }} {{eqn | r = -\sin x       | c = Cosine of Complement equals Sine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\frac \d {\d x} } {\cos x}       | r = \lim_{h \mathop \to 0} \frac {\map \cos {x + h} - \cos x} h       | c = {{Defof|Derivative of Real Function at Point}} }} {{eqn | r = \lim_{h \mathop \to 0} \frac {\map \cos {\paren {x + \frac h 2} + \frac h 2} - \map \cos {\paren {x + \frac h 2} - \frac h 2} } h }} {{eqn | r = \lim_{h \mathop \to 0} \frac {-2 \map \sin {x + \frac h 2} \map \sin {\frac h 2} } h       | c = Simpson's Formula for Sine by Sine }} {{eqn | r = -\lim_{h \mathop \to 0} \map \sin {x + \frac h 2} \lim_{h \mathop \to 0} \frac {\map \sin {\frac h 2} } {\frac h 2}       | c = Multiple Rule for Limits of Functions and Product Rule for Limits of Functions }} {{eqn | r = -\sin x \times 1       | c = Continuity of Sine and Limit of Sine of X over X }} {{eqn | r = -\sin x }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = 1       | r = \cos^2 x + \sin^2 x       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = \map {D_x} 1       | r = \map {D_x} {\cos^2 x + \sin^2 x}       | c = differentiating both sides }} {{eqn | ll= \leadsto       | l = 0       | r = 2 \map {D_x} {\cos x} \cos x  + 2 \sin x \cos x       | c = Product Rule for Derivatives or Chain Rule for Derivatives and Derivative of Sine Function  }} {{eqn | ll= \leadsto       | l = \map {D_x} {\cos x}       | r = -\sin x       | c = provided $\cos x \ne 0$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sin x       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}       | c = {{Defof|Real Sine Function}} }} {{eqn | r = \left({-1}\right)^0 \frac{x^{2 \cdot 0 + 1} } { \left({2 \cdot 0 + 1}\right)!} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} }} {{eqn | r = x + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \lim_{x \mathop \to 0} \frac {\sin x} x       | r = \lim_{x \mathop \to 0} \frac {x + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} } x }} {{eqn | r = \lim_{x \mathop \to 0} \frac x x + \lim_{x \mathop \to 0} \frac{\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} } x }} {{eqn | r = 1 + \lim_{x \mathop \to 0} \frac {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} } 1       | c = Power Series is Differentiable on Interval of Convergence and L'Hôpital's Rule }} {{eqn | r = 1 + \lim_{x \mathop \to 0} \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} }} {{eqn | r = 1 + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {0^{2 n} } {\paren {2 n}!}       | c = Real Polynomial Function is Continuous }} {{eqn | r = 1 }} {{end-eqn}} {{qed}}
From Sine of Zero is Zero: :$\sin 0 = 0$ From Derivative of Sine Function: :$D_x \left({\sin x}\right) = \cos x$ Then by Cosine of Zero is One: :$\cos 0 = 1$ From Derivative of Identity Function: :$D_x \left({x}\right) = 1$ Thus L'Hôpital's Rule applies and so: :$\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x = \lim_{x \mathop \to 0} \frac {D_x \left({\sin x}\right)} {D_x \left({x}\right)} = \lim_{x \mathop \to 0} \frac {\cos x} 1 = \frac 1 1 = 1$ {{qed}}
{{begin-eqn}} {{eqn | l = 1       | r = \cos 0       | c = Cosine of Zero is One }} {{eqn | r = \map \cos {x - x}       | c =  }} {{eqn | r = \cos x \, \map \cos {-x} - \sin x \, \map \sin {-x}       | c = Cosine of Sum }} {{eqn | r = \cos x \cos x - \paren {-\sin x \sin x}       | c = Cosine Function is Even and Sine Function is Odd }} {{eqn | r = \cos^2 x + \sin^2 x       | c =  }} {{end-eqn}} {{qed}} === Notes === Note that we need to start from the algebraic definitions of sine and cosine: :$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$ :$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$ and then use the proofs of the Cosine of Sum that derive directly from these. Otherwise these proofs are circular.
From the trigonometric definitions of sine and cosine: {{begin-eqn}} {{eqn | l = \sin x       | r = \frac{\text{opposite} } {\text{hypotenuse} } }} {{eqn | l = \cos x       | r = \frac{\text{adjacent} } {\text{hypotenuse} } }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \sin^2 x + \cos^2 x       | r = \frac{\text{opposite}^2 + \text{adjacent}^2} {\text{hypotenuse}^2}       | c squaring both sides and adding }} {{eqn | r = \frac{\text{hypotenuse}^2} {\text{hypotenuse}^2}       | c = Pythagoras's Theorem }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}}
Let $P = \tuple {x, y}$ be a point on the circumference of a unit circle whose center is at the origin of a cartesian plane. From Sine of Angle in Cartesian Plane and Cosine of Angle in Cartesian Plane: :$P = \tuple {\cos \theta, \sin \theta}$ The graph of the unit circle is the locus of: :$x^2 + y^2 = 1$ as given by Equation of Circle. Substituting $x = \cos \theta$ and $y = \sin \theta$ yields: :$\cos^2 \theta + \sin^2 \theta = 1$ {{qed}}
{{begin-eqn}} {{eqn | l = \cos^2 x + \sin^2 x       | r = \left({\cos x + i \, \sin x}\right) \, \left({\cos x - i \, \sin x}\right)       | c = factoring over the complex numbers }} {{eqn | r = \left({\cos x + i \, \sin x}\right) \, \left({\cos \left({-x}\right) + i \, \sin \left({-x}\right)}\right)       | c = Cosine Function is Even and Sine Function is Odd }} {{eqn | r = e^{ix} \, e^{-ix}       | c = Euler's Formula }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos^2 x + \sin^2 x       | r = \paren {\frac {e^{i x} + e^{-i x} } 2}^2 + \sin^2 x       | c = Cosine Exponential Formulation }} {{eqn | r = \paren {\frac {e^{i x} + e^{-i x} } 2}^2 + \paren {\frac {e^{i x} - e^{-i x} } {2 i} }^2       | c = Sine Exponential Formulation }} {{eqn | r = \frac {\paren {e^{i x} }^2 + 2 e^{-i x} e^{i x} + \paren {e^{-i x} }^2} 4 + \paren {\frac {e^{i x} - e^{-i x} } {2 i} }^2       | c = Square of Sum }} {{eqn | r = \frac {\paren {e^{i x} }^2 + 2 e^{-i x} e^{i x} + \paren {e^{-i x} }^2} 4 + \frac {\paren {e^{i x} }^2 - e^{-i x} e^{i x} + \paren {e^{-i x} }^2} {-4}        | c = Square of Difference and $i^2 = -1$ }} {{eqn | r = \frac {e^{2 i x} + 2 + e^{-2 i x} } 4 + \frac {e^{2 i x} - 2 + e^{-2 i x} } {-4}        | c = Exponential of Sum }} {{eqn | r = \frac {e^{2 i x} + 2 + e^{-2 i x} - e^{2 i x} + 2 - e^{-2 i x} } 4       | c = simplifying }} {{eqn | r = \frac 4 4       | c = simplifying }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \cos {a + b} + i \, \map \sin {a + b}       | r = e^{i \paren {a + b} }       | c = Euler's Formula }} {{eqn | r = e^{i a} e^{i b}       | c = Exponential of Sum }} {{eqn | r = \paren {\cos a + i \sin a} \paren {\cos b + i \sin b}       | c = Euler's Formula }} {{eqn | r = \paren {\cos a \cos b - \sin a \sin b} + i \paren {\sin a \cos b + \cos a \sin b}       | c = Complex Numbers form Field }} {{end-eqn}} The result follows by equating the real parts. {{qed}}
Recall the analytic definitions of sine and cosine: :$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1}} {\left({2 n + 1}\right)!}$ :$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n}} {\left({2 n}\right)!}$ Let: {{begin-eqn}} {{eqn | l = g \left({a}\right)       | r = \sin \left({a + b}\right) - \sin a \cos b - \cos a \sin b }} {{eqn | l = h \left({a}\right)       | r = \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b }} {{end-eqn}} Let us differentiate these with respect to $a$, keeping $b$ constant. Then from Derivative of Sine Function and Derivative of Cosine Function, we have: {{begin-eqn}} {{eqn | l = g' \left({a}\right)       | r = \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b = h \left({a}\right) }} {{eqn | l = h' \left({a}\right)       | r = - \sin \left({a + b}\right) + \sin a \cos b + \cos a \sin b = - g \left({a}\right) }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = D_a \left({\left({g \left({a}\right)}\right)^2 + \left({h \left({a}\right)}\right)^2}\right)       | r = 2 g \left({a}\right) g' \left({a}\right) + 2 h \left({a}\right) h' \left({a}\right) }} {{eqn | r = 0 }} {{end-eqn}} Thus from Derivative of Constant: : $\forall a \in \R: g \left({a}\right)^2 + h \left({a}\right)^2 = c$ In particular, it is true for $a = 0$, and so: :$g \left({0}\right)^2 + h \left({0}\right)^2 = 0$ So: :$g \left({a}\right)^2 + h \left({a}\right)^2 = 0$ But from Square of Real Number is Non-Negative: :$g \left({a}\right)^2 \ge 0$ and: :$h \left({a}\right)^2 \ge 0$ So it follows that: :$g \left({a}\right) = 0$ and: :$h \left({a}\right) = 0$ Hence the result. {{qed}}
:File:Tri1.PNG $AB$, $AC$, $AE$, and $AD$ are radii of the circle centered at $A$. Let $\angle BAC = a$ and $\angle DAC = \angle BAE = b$. By Euclid's First Postulate, we can construct line segments $BD$ and $CE$. By Euclid's second common notion, $\angle DAB = \angle CAE$. Thus by Triangle Side-Angle-Side Equality, $\triangle DAB \cong \triangle CAE$. Therefore, $DB = CE$.   We now assign Cartesian coordinates to the points $B$, $C$, $D$, and $E$: {{begin-eqn}} {{eqn | l = B       | r = \tuple {1, 0}       | c =  }} {{eqn | l = C       | r = \tuple {\cos a, \sin a}       | c =  }} {{eqn | l = D       | r = \tuple {\map \cos {a + b}, \map \sin {a + b} }       | c =  }} {{eqn | l = E       | r = \tuple {\cos b, -\sin b}       | c = Cosine Function is Even and Sine Function is Odd }} {{end-eqn}} We use the definition of the distance function on the Euclidean space $\struct {\R^2, d}$ as defined by the Euclidean metric: :$\forall x, y \in \R^2: \map d {x, y} = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$ where $x = \tuple {x_1, y_1}, y = \tuple {x_2, y_2}$. Thus: :$DB \cong CE \iff \map d {D, B} = \map d {C, E}$ So, plugging in the coordinates of $B, C, D, E$, we get: {{begin-eqn}} {{eqn | l = \paren {\map \cos {a + b} } - 1)^2 + \map {\sin^2} {a + b}       | r = \paren {\cos a - \cos b}^2 + \paren {\sin a + \sin b}^2 }} {{eqn | ll= \leadsto       | l = \cos^2 \left({a + b}\right) + \sin^2 \left({a + b}\right)       | o =        | c = multiplying out {{LHS}} }} {{eqn | l = {} - 2 \, \map \cos {a + b} + 1       | r = \paren {\cos a - \cos b}^2 + \paren {\sin a + \sin b}^2       | c =  }} {{eqn | ll= \leadsto       | l = 1 - 2 \, \map \cos {a + b} + 1       | r = \paren {\cos a - \cos b}^2 + \paren {\sin a + \sin b}^2       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = 2 - 2 \, \map \cos {a + b}       | r = \cos^2 a - 2 \cos a \cos b + \cos^2 b       | c = multiplying out {{RHS}} }} {{eqn | o =        | ro= +       | r = \sin^2 a + 2 \sin a \sin b + \sin^2 b }} {{eqn | ll= \leadsto       | l = 2 - 2 \, \map \cos {a + b}       | r = 2 - 2 \cos a \cos b + 2 \sin a \sin b       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = \map \cos {a + b}       | r = \cos a \cos b - \sin a \sin b }} {{end-eqn}} {{qed}}
{{begin-eqn}}  {{eqn | l = \cos a \cos b - \sin a \sin b       | r = \paren {\frac {e^{i a} + e^{-i a} } 2} \paren {\frac {e^{i b} + e^{-i b} } 2} - \sin a \sin b       | c = Cosine Exponential Formulation }} {{eqn | r = \paren {\frac {e^{i a} + e^{-i a} } 2} \paren {\frac {e^{i b} + e^{-i b} } 2}           - \paren {\frac {e^{i a} - e^{-i a} } {2 i} } \paren {\frac {e^{i b} - e^{-i b} } {2 i} }       | c = Sine Exponential Formulation }} {{eqn | r = \frac {e^{i a} e^{i b} + e^{i a} e^{-i b} + e^{-i a} e^{i b} + e^{-i a} e^{-i b} } 4 }} {{eqn | o =        | ro= -       | r = \frac {e^{i a} e^{i b} - e^{i a} e^{-i b} - e^{-i a} e^{i b} + e^{-i a} e^{-i b} } {4 i^2} }} {{eqn | r = \frac {e^{i a} e^{i b} + e^{i a} e^{-i b} + e^{-i a} e^{i b} + e^{-i a} e^{-i b} } 4 }} {{eqn | o =        | ro= +       | r = \frac {e^{i a} e^{i b} - e^{i a} e^{-i b} - e^{-i a} e^{i b} + e^{-i a} e^{-i b} } 4       | c = as $i^2 = -1$ }} {{eqn | r = \frac {e^{i a} e^{i b} + e^{-ia} e^{-ib} } 2 }} {{eqn | r = \frac {e^{i \paren {a + b} } + e^{-i \paren {a + b} } } 2       | c = Exponential of Sum  }} {{eqn | r = \map \cos {a + b}       | c = Cosine Exponential Formulation }} {{end-eqn}} {{qed}}
Let $\map s x: \R \to \R$, $\map c x: \R \to \R$ be two functions that satisfy: :$(1): \quad \map {s'} x = \map c x$ :$(2): \quad \map {c'} x = -\map s x$ :$(3): \quad \map s 0 = 0$ :$(4): \quad \map c 0 = 1$ :$(5): \quad \forall x: \map {s^2} x + \map {c^2} x = 1$ where $s'$ denotes the derivative {{WRT|Differentiation}} $x$. Let $\map f x: \R \to \R$, $\map g x: \R \to \R$ also be two functions that satisfy: :$(1): \quad \map {f'} x = \map g x$ :$(2): \quad \map {g'} x = -\map f x$ :$(3): \quad \map f 0 = 0$ :$(4): \quad \map g 0 = 1$ :$(5): \quad \forall x: \map {f^2} x + \map {g^2} x = 1$ It will be shown that: :$\map f x = \map s x$ and: :$\map g x = \map c x$ Define: :$\map h x = \paren {\map c x - \map g x}^2 + \paren {\map s x - \map f x}^2$ Notice that: :$\paren {\forall x: \map h x = 0} \iff \paren {\forall x: \map c x = \map g x, \map s x = \map f x}$ Then: {{begin-eqn}} {{eqn | l = \map h x       | r = \map {c^2} x - 2 \map c x \map g x + \map {g^2} x + \map {s^2} x - 2 \map s x \map f x + \map {f^2} x       | c = }} {{eqn | r = 2 - 2 \paren {\map c x \map g x + \map s x \map f x}       | c = Property $(5)$ }} {{end-eqn}} By taking $\map {h'} x$: {{begin-eqn}} {{eqn | l = \map {h'} x       | r = -2 \paren {\map c x \paren {-\map f x} + \map g x \paren {-\map s x} + \map s x \map g x + \map c x \map f x}       | c = Properties $(1)$ and $(2)$ and Product Rule }} {{eqn | r = 0       | c =   }} {{end-eqn}} By Zero Derivative implies Constant Function, $\map h x$ is a constant function: :$\map h x = k$ Also: {{begin-eqn}} {{eqn | l = \map h 0       | r = \paren {1 - 1}^2 + \paren {0 - 0}^2       | c = Properties $(3)$ and $(4)$ }} {{eqn | r = 0       | c =  }} {{end-eqn}} Since $\map h x$ is constant, then: : $\forall x: \map h x = 0$ Then: :$\map c x = \map g x$ and: :$\map s x = \map f x$ By: :Derivative of Sine Function :Derivative of Cosine Function :Sine of Zero is Zero :Cosine of Zero is One :Sum of Squares of Sine and Cosine both definitions satisfy all these properties. Therefore they must be the same. {{qed}} Category:Sine Function Category:Cosine Function Category:Definition Equivalences da4v7o00c8r9abiknxjy95m7utal6f4
From the algebraic definition of the real cosine function: : $\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$ it follows that $\cos x$ is a real function. Thus $\cos^2 x \ge 0$. From Sum of Squares of Sine and Cosine, we have that $\cos^2 x + \sin^2 x = 1$. Thus it follows that: : $\cos^2 x = 1 - \sin^2 x \le 1$ From Ordering of Squares in Reals and the definition of absolute value, we have that: :$x^2 \le 1 \iff \size x \le 1$ The result follows. {{qed}}
From Derivative of Sine Function, we have: : $D_x \paren {\sin x} = \cos x$ So by the Mean Value Theorem, there exists $\xi \in \R$ between $0$ and $x$ such that: :$\dfrac {\sin x - \sin 0} {x - 0} = \cos \xi$ From Real Cosine Function is Bounded we have that: : $\size {\cos \xi} \le 1$ {{qed}}
From Cosine of Zero is One we have that $\cos 0 = 1$. From Cosine Function is Even we have that $\cos x = \map \cos {-x}$. As the Cosine Function is Continuous, it follows that: :$\exists \xi > 0: \forall x \in \openint {-\xi} \xi: \cos x > 0$ {{AimForCont}} $\cos x$ were positive everywhere on $\R$. From Derivative of Cosine Function: :$\map {D_{xx} } {\cos x} = \map {D_x} {-\sin x} = -\cos x$ Thus $-\cos x$ would always be negative. Thus from Second Derivative of Concave Real Function is Non-Positive, $\cos x$ would be concave everywhere on $\R$. But from Real Cosine Function is Bounded, $\cos x$ is bounded on $\R$. By Differentiable Bounded Concave Real Function is Constant, $\cos x$ would then be a constant function. This contradicts the fact that $\cos x$ is not a constant function. Thus by Proof by Contradiction $\cos x$ can not be positive everywhere on $\R$. Therefore, there must exist a smallest positive $\eta \in \R_{>0}$ such that $\cos \eta = 0$. By definition, $\cos \eta = \map \cos {-\eta} = 0$ and $\cos x > 0$ for $-\eta < x < \eta$. Now we show that $\sin \eta = 1$. From Sum of Squares of Sine and Cosine: :$\cos^2 x + \sin^2 x = 1$ Hence as $\cos \eta = 0$ it follows that $\sin^2 \eta = 1$. So either $\sin \eta = 1$ or $\sin \eta = -1$. But $\map {D_x} {\sin x} = \cos x$. On the interval $\closedint {-\eta} \eta$, it has been shown that $\cos x > 0$. Thus by Derivative of Monotone Function, $\sin x$ is increasing on $\closedint {-\eta} \eta$. Since $\sin 0 = 0$ it follows that $\sin \eta > 0$. So it must be that $\sin \eta = 1$. Now we apply Sine of Sum and Cosine of Sum: :$\map \sin {x + \eta} = \sin x \cos \eta + \cos x \sin \eta = \cos x$ :$\map \cos {x + \eta} = \cos x \cos \eta - \sin x \sin \eta = -\sin x$ Hence it follows, after some algebra, that: :$\map \sin {x + 4 \eta} = \sin x$ :$\map \cos {x + 4 \eta} = \cos x$ Thus $\sin$ and $\cos$ are periodic on $\R$ with period $4 \eta$. {{qed}}
Let $f$ be differentiable and bounded on $\R$. Let $f$ be convex on $\R$. Let $\xi \in \R$. Aiming for a contradiction, suppose $f' \left({\xi}\right) > 0$. Then by Mean Value of Convex Real Function it follows that: : $f \left({x}\right) \ge f \left({\xi}\right) + f' \left({\xi}\right) \left({x - \xi}\right) \to + \infty$ as $x \to +\infty$ and therefore is not bounded. Similarly, suppose $f' \left({\xi}\right) < 0$. Then by Mean Value of Convex Real Function it follows that: : $f \left({x}\right) \ge f \left({\xi}\right) + f' \left({\xi}\right) \left({x - \xi}\right) \to + \infty$ as $x \to -\infty$ and therefore is likewise not bounded. Hence $f' \left({\xi}\right) = 0$. From Zero Derivative implies Constant Function, it follows that $f$ is constant. {{qed}}