Proofs that the sine function is strictly increasing on $[-\pi/2,\pi/2]$. Here on the forum there are some proofs that the sine function is strictly increasing on $[-\pi/2,\pi/2]$, but they all use the fact that $\sin '(x)=\cos(x)$. Is there any rigorous proof without going through the derivative?
(Define $\sin$ like a ratio between sides of a right triangle)
Thanks in advance.

Let $x,y∈[-π/2,π/2]$ such that $x>y$.
Then,
sin$x -$sin$y = 2 $sin$\frac{x-y}{2}$cos$\frac{x+y}{2}>0$
Thus , sin$x>$sin$y$ whenever $x>y$ in the given domain and sin$x$ is increasing.
