## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 27

#### Answer

$-\frac{1}{2}\sqrt{2-\sqrt{2}}$

#### Work Step by Step

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. $\sin \frac{9\pi}{8}$ $=\sin \frac{\frac{9\pi}{4}}{2}$ $=-\sqrt{\frac{1-\cos \frac{9\pi}{4}}{2}}$ $=-\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}$ $=-\sqrt{\frac{(1-\frac{\sqrt{2}}{2})*2}{2*2}}$ $=-\sqrt{\frac{2-\sqrt{2}}{4}}$ $=-\frac{\sqrt{2-\sqrt{2}}}{\sqrt{4}}$ $=-\frac{1}{2}\sqrt{2-\sqrt{2}}$

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