## Precalculus: Mathematics for Calculus, 7th Edition

$-\frac{1}{2}\sqrt{2+\sqrt{3}}$
Use the half-angle formula, $\cos\frac{u}{2}=\pm\sqrt{\frac{1+\cos u}{2}}$. Note that $165^\circ$ is in Quadrant II, where cosine is negative, so we take the negative square root. $\cos 165^\circ$ $=\cos \frac{330^\circ}{2}$ $=-\sqrt{\frac{1+\cos 330^\circ}{2}}$ $=-\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}$ $=-\sqrt{\frac{(1+\frac{\sqrt{3}}{2})*2}{2*2}}$ $=-\sqrt{\frac{2+\sqrt{3}}{4}}$ $=-\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}$ $=-\frac{1}{2}\sqrt{2+\sqrt{3}}$