### Instruction: Compute $\sin(-60^\circ)$.### Response: I know the sine function is periodic with period $360^\circ$, so $\sin(-60^\circ) = \sin(-60^\circ + 360^\circ) = \sin(300^\circ)$.
I also know that the sine function is symmetrical about the origin, so $\sin(-x) = -\sin(x)$ for any angle $x$.
Therefore, $\sin(-60^\circ) = -\sin(60^\circ)$.
To find $\sin(60^\circ)$, I can use the special right triangle with angles $30^\circ$, $60^\circ$, and $90^\circ$ and sides in the ratio $1:\sqrt{3}:2$.
The sine of an angle is the ratio of the opposite side to the hypotenuse, so $\sin(60^\circ) = \frac{\sqrt{3}}{2}$.
Therefore, $\sin(-60^\circ) = -\frac{\sqrt{3}}{2}$.