#### Answer

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

#### Work Step by Step

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