## Precalculus (6th Edition) Blitzer

$[-\frac{\pi}{2}, \frac{\pi}{2}]$, $sin^{-1}(x)$.
To have an inverse, a function needs to pass the horizontal line test. For $y=sin(x)$, we need to restrict the domain so that an inverse can be found, and the restricted domain is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. We use a special symbol $f^{-1}$ to denote an inverse, and the inverse of $y=sin(x)$ is $y=sin^{-1}(x)$.