## Precalculus: Mathematics for Calculus, 7th Edition

$-\frac{1}{2}\sqrt{2-\sqrt{2}}$
Use the half-angle formula, $\sin\frac{u}{2}=\pm\sqrt{\frac{1-\cos u}{2}}$. Note that $\frac{9\pi}{8}$ is in Quadrant III, where sine is negative, so we take the negative square root. $\sin \frac{9\pi}{8}$ $=\sin \frac{\frac{9\pi}{4}}{2}$ $=-\sqrt{\frac{1-\cos \frac{9\pi}{4}}{2}}$ $=-\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}$ $=-\sqrt{\frac{(1-\frac{\sqrt{2}}{2})*2}{2*2}}$ $=-\sqrt{\frac{2-\sqrt{2}}{4}}$ $=-\frac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}$ $=-\frac{1}{2}\sqrt{2-\sqrt{2}}$