jmc

Rotating $60^{\circ }$ clockwise is the same as rotating $360^{\circ }-60^{\circ }=300^{\circ }$ counterclockwise, so $\sin (-60^{\circ })=\sin (360^{\circ }-60^{\circ })=\sin 300^{\circ }$.

Let $P$ be the point on the unit circle that is $300^{\circ }$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.



Triangle $POD$ is a 30-60-90 triangle, so $DO=\frac{1}{2}$ and $DP=\frac{\sqrt{3}}{2}$. Therefore, the coordinates of $P$ are $\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$, so $\sin (-60^{\circ })=\sin 300^{\circ }=\boxed{-\frac{\sqrt{3}}{2}}$.