Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises: 54

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