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

If g(x) = sin x on the domain [-pi/2 , pi/2], the domain of g^-1 is:-example-1
User Nekresh
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1 Answer

3 votes

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.

User Pete Rossi
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