Prove that $\sin(x)$ is strictly increasing in $(-\frac{\pi}{2},\frac{\pi}{2})$ Following this article:
Let $S$ and $C$ be two $\mathbb{R}\to\mathbb{R}$ functions such that for a given $p>0$ the following properties are met

*

*$S(p)=1$

*$S(x)\geq 0$ for all $x\in[0,p]$

*$C(x-y)=C(x)C(y)+S(x)S(y)$ for all $x$, $y\in\mathbb{R}$
It is known that $S$ and $C$ are uniquely determined by this properties. It is also known that:

*

*$S$ and $C$ are continuous

*$C$ is even and $S$ is odd.

*$C^2(x)+S^2(x)=1$.

*$S(x)$, $C(x)\in[0,1]$ for $x\in[0,p]$.

*$S(0)=C(p)=0$, and $C(0)=1$.

*$C(p-x)=S(x)$, $S(p-x)=S(x)$, $S(p+x)=C(x)$ and $C(p+x)=-S(x)$.

*$S(2p+x)=-S(x)$ and $C(2p+x)=-C(x)$.

*$S$ and $C$ are $4p$-periodic.

*$S$ is nondecreasing in $[0,p]$ and $C$ is nonincreasing in $[0,p]$

*The limit $\lim_{x\to 0}\frac{S(x)}{x}$ exists.

Let $L:=\lim_{x\to 0}\frac{S(x)}{x}$ for $p=1$ and define $\sin(x):=S(x)$ and $\cos(x):=C(x)$ with $p=L$. It is known that $L=\frac{\pi}{2}$.
It is easy to prove then that $\sin'(x)=\cos(x)$ and that $\cos'(x)=-\sin(x)$ and hence derive the usual power series representation of the sine and the cosine.
Then, how do I prove that $\sin(x)$ is strictly increasing in $(-\frac{\pi}{2},\frac{\pi}{2})$? I tried the obvious $\sin'(x)=\cos(x)\geq 0$, but I can't show it is guaranteed that $\cos(x)>0$
 A: So, $\frac{dy}{dx} = 0$, and there are stationary points at $x = {-\frac{\pi}{2}, \frac{\pi}{2}}$. Prove that $x = -\frac{\pi}{2}$ is a minimum and $x = \frac{\pi}{2}$ is maximum by showing $\frac{d^2y}{dx^2} > 0$ for $x = -\frac{\pi}{2}$ and $\frac{d^2y}{dx^2} < 0$ for $x = \frac{\pi}{2}$. Since these stationary points are consecutive and the minimum comes first, the function can only be increasing.
Alternatively, you could linearise its $\frac{dy}{dx}$ over the interval (avg. tangent gradient) to prove its $>0$ and that its strictly increasing for $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

A: We know that if $\sin'(x)>0$ por all $x\in(0,\frac{\pi}{2})$ then $\sin$ is strictly increasing in $(0,\frac{\pi}{2})$.
If $x_0\in(0,\frac{\pi}{2})$ is a number such that $\sin'(x_0)=\cos(x_0)=0$, then, for being $\cos$ non-incresing and non-negative in $(0,\frac{\pi}{2})$ we have that $\cos(y)=0$ for all $y\in[x_0,\frac{\pi}{2}]$, hence $\cos'(y)=0$ for all $y\in(x_0,\frac{\pi}{2})$.
That being said, $\sin$ must be constant in $(x_0,\frac{\pi}{2})$, but $\sin(\frac{\pi}{2})=1$ so $\sin(y)=1$ for all $y\in(x_0,\frac{\pi}{2})$, and $\cos'(y)=-\sin(y)=-1$ for all $y\in(x_0,\frac{\pi}{2})$, but this is a contradiction.
