### Instruction: Compute $\sin 240^\circ$.### Response: I know that the sine function is periodic with a period of $360^\circ$, so I can reduce the angle modulo $360^\circ$.
That is, $\sin 240^\circ = \sin (240 - 360)^\circ = \sin (-120)^\circ$.
Now I can use the fact that the sine function is odd, meaning that $\sin (-x) = - \sin x$ for any angle $x$.
So $\sin (-120)^\circ = - \sin 120^\circ$.
To find $\sin 120^\circ$, I can use the special angle of $60^\circ$ and the angle sum formula for sine.
That is, $\sin 120^\circ = \sin (60 + 60)^\circ = \sin 60^\circ \cos 60^\circ + \cos 60^\circ \sin 60^\circ$.
I remember that $\sin 60^\circ = \sqrt{3}/2$ and $\cos 60^\circ = 1/2$, so I can plug those values in.
Then I get $\sin 120^\circ = (\sqrt{3}/2)(1/2) + (1/2)(\sqrt{3}/2) = \sqrt{3}/4 + \sqrt{3}/4 = \sqrt{3}/2$.