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 693: 56


$\displaystyle \frac{\sqrt{13}}{4}$

Work Step by Step

$\displaystyle \frac{\pi}{2} < x < \pi\qquad/\div 2$ $\displaystyle \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$, so, $\displaystyle \frac{x}{2}$ is in quadrant I, where the cosine is positive. Half-Angle Identity: $\displaystyle \sin\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{2}}=+\sqrt{\frac{1-(-\frac{5}{8})}{2}}$ $=\sqrt{\dfrac{\frac{13}{8}}{2} }=\sqrt{\dfrac{13}{16}} $ $=\displaystyle \frac{\sqrt{13}}{4}$
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