Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 58


a. $ -\frac{4}{5}$ b. $ -\frac{3}{5}$ c. $ \frac{3}{4}$

Work Step by Step

Given $sin\alpha=\frac{4}{5}$ with $\alpha$ in quadrant I, we have $cos\alpha=\sqrt {1-sin^2\alpha}=\frac{3}{5}$ and $tan\alpha=\frac{sin\alpha}{cos\alpha}=\frac{4}{3}$. Similarly, given $sin\beta=\frac{7}{25}$ with $\beta$ in quadrant II, we have $cos\beta=-\sqrt {1-sin^2\beta}=-\frac{24}{25}$ and $tan\beta=\frac{sin\beta}{cos\beta}=-\frac{7}{24}$. Using the Addition Formulas, we have: a. $cos(\alpha+\beta)=cos(\alpha) cos(\beta)-sin(\alpha) sin(\beta)=(\frac{3}{5})(-\frac{24}{25})-(\frac{4}{5})(\frac{7}{25})=-\frac{100}{125}=-\frac{4}{5}$ b. $sin(\alpha+\beta)=sin(\alpha) cos(\beta)+cos(\alpha) sin(\beta)=(\frac{4}{5})(-\frac{24}{25})+(\frac{3}{5})(\frac{7}{25})=-\frac{75}{125}=-\frac{3}{5}$ c. $tan( \alpha+\beta)=\frac{tan(\alpha)+tan(\beta)}{1-tan(\alpha) tan(\beta)}=\frac{(\frac{4}{3})+(-\frac{7}{24})}{1-(\frac{4}{3})(-\frac{7}{24})}=\frac{3}{4}$
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