Answer
$-\frac{3\sqrt {10}}{10}$.
Work Step by Step
1. Let $tan^{-1}(-3)=t$, we have $tan(t)=-3$ and $t$ in quadrant IV.
2. Let $x=1, y=-3$, we have $r=\sqrt {1^2+(-3)^2}=\sqrt {10}$.
3. Thus $sin(tan^{-1}(-3))=sin(t)=\frac{y}{r}=-\frac{3\sqrt {10}}{10}$.
