Compute $\cos 210^\circ$.
I know that $\cos 210^\circ$ is the same as $\cos (180^\circ + 30^\circ)$, by adding $180^\circ$ to the angle. I also know that $\cos (180^\circ + \theta) = -\cos \theta$, by using the evenness property of cosine. So, $\cos 210^\circ = -\cos 30^\circ$. Now, I recall that $\cos 30^\circ = \frac{\sqrt{3}}{2}$, by using the special right triangle with angles $30^\circ, 60^\circ, 90^\circ$. Therefore, $\cos 210^\circ = -\frac{\sqrt{3}}{2}$. 