How to show that $\sin^{-1}(x)$ is an increasing function?

Question

I would like to show that $\sin^{-1}$ is an increasing function. In other words, I want to show that $$x_{1} \leq x_{2} \rightarrow \sin^{-1}(x_{1}) \leq \sin^{-1}(x_{2}).$$

I fail to see how to start this. Also, if possible, I would like to show this without using other functions (e.g. $\sin$) known to be increasing/decreasing.

Answer 1

Hint

If $f$ is increasing, then $f^{-1}$ is increasing.

Indeed, let $u\leq v$ and $x,y$ such that $u=f(x)$ and $v=f(y)$. Then, since $f$ is increasing,

$$u\leq v\implies f(x)\leq f(y)\implies x\leq y\implies f^{-1}(u)\leq f^{-1}(v).$$

PS: Use $\arcsin$ instead of $\sin^{-1}$ !

Answer 2

HINT: $$\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}>0$$$$\ \ \forall \ \ -1<x<1$$