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 - Page 694: 62


$\displaystyle \frac{\sqrt{5}}{5}$

Work Step by Step

$90^{o} < \theta < 180^{o}\quad/\div 2$ $45^{o} < \displaystyle \frac{\theta}{2} < 90^{o}$ so, $\theta$ is quadrant II where cos$\theta$ and sec$\theta$ are negative, $\displaystyle \frac{\theta}{2} $is in quadrant I, $\displaystyle \tan\frac{\theta}{2}$ and $\displaystyle \cot\frac{\theta}{2}$ are positive. Pythagorean identity: $\displaystyle \sec^{2}\theta=\tan^{2}\theta+1=\frac{5}{4}+1=\frac{9}{4}$ $\displaystyle \sec\theta=-\frac{3}{2}\Rightarrow\cos\theta=-\frac{2}{3}$ $\displaystyle \tan\frac{\theta}{2}=+\sqrt{\frac{1-(-\frac{2}{3})}{1+(-\frac{2}{3})}}=\sqrt{\frac{\frac{5}{3}}{\frac{1}{3}}}=\sqrt{5}$ So, cot and tan being reciprocal, $\displaystyle \cot\frac{\theta}{2}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$
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