I know that the cosine function is even, meaning that $\cos (-x) = \cos x$ for any angle $x$. So, $\cos (-150^\circ) = \cos (150^\circ)$. I also know that 150 degrees is in the second quadrant, where cosine is negative. I can use the reference angle of 30 degrees, which is the angle formed by the terminal side of 150 degrees and the negative x-axis. I recall that $\cos (30^\circ) = \frac{\sqrt{3}}{2}$, which is a special value of cosine. So, $\cos (150^\circ) = -\cos (30^\circ) = -\frac{\sqrt{3}}{2}$. Therefore, $\cos (-150^\circ) = -\frac{\sqrt{3}}{2}$. 
Compute $\cos (-150^\circ)$.