Answer
$\sin^{-1}0=0$
Work Step by Step
$y=\sin^{-1}x$
Domain: $[-1, 1] $, Range:$ [-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$
meaning:
"$y$ is the number from $[-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ such that $\sin y=x$"
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$\sin 0=0$, and $0\displaystyle \in[-\frac{\pi}{2}, \displaystyle \frac{\pi}{2}]$ so
$\sin^{-1}0=0$
