Precalculus (6th Edition) Blitzer

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

Chapter 5 - Test - Page 709: 2



Work Step by Step

Step 1. Given $sin\alpha=\frac{4}{5}, \frac{\pi}{2}\lt\alpha\lt\pi$, we can identify $\alpha$ to be in quadrant II, which means $cos\alpha=-\sqrt {1-sin^2\alpha}=-\frac{3}{5}$ and $tan\alpha=-\frac{4}{3}$ Step 2. Given $cos\beta=\frac{5}{13}, 0\lt\beta\lt \frac{\pi}{2}$, we can identify $\beta$ to be in quadrant I, which means $sin\beta=\sqrt {1-cos^2\beta}=\frac{12}{13}$ and $tan\beta=\frac{12}{5}$ Step 3. Use the the subtraction formula, we have $tan(\alpha-\beta)=\frac{tan\alpha-tan\beta}{1+tan\alpha tan\beta}=\frac{(-\frac{4}{3})-(\frac{12}{5})}{1+(-\frac{4}{3})(\frac{12}{5})}=\frac{56}{33}$
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