## Precalculus (10th Edition)

$\frac{\pi}{4}.$
The domain of $\sin^{-1}{x}$ is the range of $\sin{x}$, which is $[−1, 1]$. The range of $\sin^{-1}{x}$ is a specified domain of $\sin{x}$, i.e. $\left[− \frac{\pi}{2} , \frac{\pi}{2} \right]$. We know that if $\sin y=x$, then $\sin^{-1}(x)=y.$ We know that $\sin(\frac{\pi}{4})=\frac{\sqrt2}{2}$, hence $\sin^{-1}(\frac{\sqrt2}{2})=\frac{\pi}{4}.$