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_{(-(\pi)/(2))}=\sin (-(\pi)/(2))=-1 \\ g_{((\pi)/(2))}=\sin ((\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.