Question

if g(x) = sin x on the domain [-pi/2 , pi/2], the domain of g^-1 is:

Answer 1

So we have:

$g_{(x)}=\sin x$

The inverse of the sine is the arcsine:

$g^{-1}_{(x)}=\arcsin x$

Now let's evaluate the sine on the extremes of its domain:

$\begin{gathered} g_{(-\frac{\pi}{2})}=\sin (-\frac{\pi}{2})=-1 \\ g_{(\frac{\pi}{2})}=\sin (\frac{\pi}{2})=1 \end{gathered}$

We also know that sin(x) cannot be bigger than 1 and smaller than -1 so we can assure that the image of the sine is [-1,1]. The good part is that this is also the domain of its inverse function, the arcsine.