## Precalculus (6th Edition)

$-\displaystyle \frac{\pi}{2}$
$y=\sin^{-1}x$ Domain: $[-1, 1]$Range: $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ $y$ is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ such that $\sin y=-1$ ---------------- $\displaystyle \sin(-\frac{\pi}{2})=-1$, and $-\displaystyle \frac{\pi}{2}\in[-\frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ so $\displaystyle \sin^{-1}(-1)=-\frac{\pi}{2}$