jmc

Rotating $360^{\circ }$ is the same as doing nothing, so rotating $600^{\circ }$ is the same as rotating $600^{\circ }-360^{\circ }=240^{\circ }$. Therefore, we have $\sin 600^{\circ }=\sin (600^{\circ }-360^{\circ })=\sin 240^{\circ }$.

Let $P$ be the point on the unit circle that is $240^{\circ }$ 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 $\sin 600^{\circ }=\sin 240^{\circ }=\boxed{-\frac{\sqrt{3}}{2}}$.