Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Mid-Chapter Check Point - Page 684: 22


$-\frac{\sqrt {26}}{26}$

Work Step by Step

Step 1. Given $sin\alpha=\frac{3}{5}, \frac{\pi}{2}\lt\alpha\lt\pi$, we know $\alpha$ is in quadrant II; thus $cos\alpha=-\sqrt {1-sin^2\alpha}=-\frac{4}{5}$ and $tan\alpha=-\frac{3}{4}$ Step 2. Given $cos\beta=-\frac{12}{13}, \pi\lt\beta\lt \frac{3\pi}{2}$, we know $\beta$ is in quadrant III; thus $sin\beta=-\sqrt {1-cos^2\beta}=-\frac{5}{13}$ and $tan\beta=\frac{5}{12}$ Step 3. Use the Half-Angle Formula, with $\frac{\pi}{2}\lt \frac{\beta}{2} \lt \frac{3\pi}{4}$, we have $cos( \frac{\beta}{2})=-\sqrt {\frac{1+cos\beta}{2}}=-\sqrt {\frac{1+(-\frac{12}{13})}{2}}=-\frac{\sqrt {26}}{26}$
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