Answer
$-30^{\circ}$
Work Step by Step
Because $\sin 330^{\circ}=-\sin(360^{\circ}-330^{\circ})=-\sin30^{\circ}=-\frac{1}{2}$,
$\sin^{-1}(\sin330^{\circ})=\sin^{-1}(-\frac{1}{2})$ will be the angle $y$, $-90^{\circ}\leq y\leq90^{\circ}$, for which $\sin y=-\frac{1}{2}$.
As $\sin (-x)=-\sin x$ and $\sin 30^{\circ}=\frac{1}{2}$, we can write
$\sin (-30^{\circ})=-\frac{1}{2}$.
Therefore, $\sin^{-1}(\sin330^{\circ})=\sin^{-1}(-\frac{1}{2})=-30^{\circ}$
