#### Answer

$-\frac{\sqrt{2+\sqrt{3}}}{2}$

#### Work Step by Step

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.
$\cos 165^\circ$
$=\cos\frac{330^\circ}{2}$
$=-\sqrt{\frac{1+\cos 330^\circ}{2}}$
$=-\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}$
$=-\sqrt{\frac{(1+\frac{\sqrt{3}}{2})*2}{2*2}}$
$=-\sqrt{\frac{2+\sqrt{3}}{4}}$
$=-\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}$
$=-\frac{\sqrt{2+\sqrt{3}}}{2}$