## Precalculus (6th Edition) Blitzer

a.)$\frac{\sqrt 3}{2}$ b.)$\frac{\sqrt 3}{2}$
a.)cos $\frac{\pi}{6}$ corresponds to x value. cos $\frac{\pi}{6}$ = $\frac{\sqrt 3}{2}$. b.)Cosine is an even function, so cos(-t) = cos t. cos-$\frac{\pi}{6}$ = cos$\frac{\pi}{6}$ = $\frac{\sqrt 3}{2}$