Ciara Mcdaniel

2022-07-10

I would like to show that ${\mathrm{sin}}^{-1}$ is an increasing function. In other words, I want to show that
${x}_{1}\le {x}_{2}\to {\mathrm{sin}}^{-1}\left({x}_{1}\right)\le {\mathrm{sin}}^{-1}\left({x}_{2}\right).$
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.

Elias Flores

Expert

Hint
If f is increasing, then ${f}^{-1}$ is increasing.
Indeed, let $u\le v$ and x,y such that $u=f\left(x\right)$ and $v=f\left(y\right)$. Then, since f is increasing,
$u\le v\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}f\left(x\right)\le f\left(y\right)\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x\le y\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{f}^{-1}\left(u\right)\le {f}^{-1}\left(v\right).$
PS: Use $\mathrm{arcsin}$ instead of ${\mathrm{sin}}^{-1}$

Janet Forbes

Expert

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

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