I know that one full rotation around the unit circle is $360^\circ$, so $\sin 600^\circ$ is the same as $\sin (600^\circ - 360^\circ) = \sin 240^\circ$. Now, I can use the fact that $\sin (180^\circ + \theta) = -\sin \theta$ for any angle $\theta$. This means that $\sin 240^\circ = \sin (180^\circ + 60^\circ) = -\sin 60^\circ$. I recall that $\sin 60^\circ = \frac{\sqrt{3}}{2}$ by using the 30-60-90 right triangle. Therefore, $\sin 600^\circ = -\frac{\sqrt{3}}{2}$. 
Compute $\sin 600^\circ$.