## Precalculus: Mathematics for Calculus, 7th Edition

$\frac{1}{2}\sqrt{2-\sqrt{3}}$
Use the half-angle formula, $\sin\frac{u}{2}=\pm\sqrt{\frac{1-\cos u}{2}}$. Note that $15^\circ$ is in Quadrant I, where sine is positive, so we take the positive square root. $\sin 15^\circ$ $=\sin \frac{30^\circ}{2}$ $=\sqrt{\frac{1-\cos 30^\circ}{2}}$ $=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}$ $=\sqrt{\frac{(1-\frac{\sqrt{3}}{2})*2}{2*2}}$ $=\sqrt{\frac{2-\sqrt{3}}{4}}$ $=\frac{\sqrt{2-\sqrt{3}}}{\sqrt{4}}$ $=\frac{1}{2}\sqrt{2-\sqrt{3}}$