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: 53

Answer

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

Work Step by Step

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