## Precalculus (6th Edition)

$-\displaystyle \frac{\sqrt{3}}{2}$
Letting $A=2\displaystyle \theta, \quad \cos A=\frac{1}{2}$ we express $\displaystyle \theta=\frac{A}{2},\quad\cos\theta=\cos\frac{A}{2}$ ($\cos\theta$ is negative as $\theta$ terminates in Q.II.) Half-Angle identity: $\cos\theta$=$\displaystyle \cos\frac{A}{2}=-\sqrt{\frac{1+\cos A}{2}}=-\sqrt{\frac{1+\frac{1}{2}}{2}}$ $=-\displaystyle \sqrt{\frac{3}{4}}=-\frac{\sqrt{3}}{2}$