Compute $\sin 510^\circ$.
I know that the sine function is periodic with a period of $360^\circ$, so I can subtract multiples of $360^\circ$ from the angle until I get an equivalent angle in the standard range of $0^\circ$ to $360^\circ$. So, $\sin 510^\circ = \sin (510^\circ - 360^\circ) = \sin 150^\circ$. Now, I recall that $\sin 150^\circ$ is one of the special angles that I can find using the unit circle and the fact that $\sin \theta$ is the y-coordinate of the point on the unit circle that makes an angle of $\theta$ with the positive x-axis. I also remember that $\sin 150^\circ = \sin (90^\circ + 60^\circ)$, and I can use the sine addition formula to write $\sin 150^\circ = \sin 90^\circ \cos 60^\circ + \cos 90^\circ \sin 60^\circ$. I know that $\sin 90^\circ = 1$, $\cos 90^\circ = 0$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$, and $\cos 60^\circ = \frac{1}{2}$, so I can plug these values into the formula and simplify. Therefore, $\sin 150^\circ = 1 \cdot \frac{1}{2} + 0 \cdot \frac{\sqrt{3}}{2} = \frac{1}{2}$. 