I know that the cosine function is periodic with period $360^\circ$, so $\cos 300^\circ$ is the same as $\cos (300 - 360)^\circ = \cos (-60)^\circ$. I also know that the cosine function is an even function, which means that $\cos (-x) = \cos x$ for any $x$. Therefore, $\cos 300^\circ = \cos (-60)^\circ = \cos 60^\circ$. I remember that there is a special right triangle with angles $30^\circ, 60^\circ, 90^\circ$ and side lengths in the ratio $1:\sqrt{3}:2$. If I draw this triangle and label the sides and angles, I can see that $\cos 60^\circ$ is the ratio of the adjacent side to the hypotenuse, which is $\frac{1}{2}$. Therefore, $\cos 300^\circ = \cos 60^\circ = \frac{1}{2}$.


Compute $\cos 300^\circ$.