Chapter 3 - Section 3.2 - Radians and Degrees - 3.2 Problem Set - Page 133: 80

$\frac{-8+3\sqrt 3}{8}$

$-1+\frac{3}{4}\cos(2\times\frac{\pi}{6}-\frac{\pi}{2})=-1+\frac{3}{4}\cos(\frac{\pi}{3}-\frac{\pi}{2})$ $=-1+\frac{3}{4}\cos(-\frac{\pi}{6})=-1+\frac{3}{4}\cos\frac{\pi}{6}=-1+\frac{3}{4}\times\frac{\sqrt 3}{2}$ $=-1+\frac{3\sqrt 3}{8}=\frac{-8+3\sqrt 3}{8}$