Answer
$-45^{\circ}$
Work Step by Step
Because $\sin 225^{\circ}=\sin(180^{\circ}+45^{\circ})=-\sin45^{\circ}=-\frac{\sqrt 2}{2}$, $\sin^{-1}(\sin225^{\circ})=\sin^{-1}(-\frac{\sqrt 2}{2})$ will be the angle $y$, $-90^{\circ}\leq y\leq90^{\circ}$, for which $\sin y=-\frac{\sqrt 2}{2}$.
As $\sin (-x)=-\sin x$ and $\sin 45^{\circ}=\frac{\sqrt 2}{2}$, we can write
$\sin (-45^{\circ})=-\frac{\sqrt 2}{2}$.
Therefore, $\sin^{-1}(\sin225^{\circ})=\sin^{-1}(-\frac{\sqrt 2}{2})=-45^{\circ}$
