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: 57


a. $-\frac{63}{65}$ b. $-\frac{16}{65}$ c. $\frac{16}{63}$

Work Step by Step

Given $sin\alpha=\frac{3}{5}$ with $\alpha$ in quadrant I, we have $cos\alpha=\sqrt {1-sin^2\alpha}=\frac{4}{5}$ and $tan\alpha=\frac{sin\alpha}{cos\alpha}=\frac{3}{4}$. Similarly, given $sin\beta=\frac{5}{12}$ with $\beta$ in quadrant II, we have $cos\beta=-\sqrt {1-sin^2\beta}=-\frac{12}{13}$ and $tan\beta=\frac{sin\beta}{cos\beta}=-\frac{5}{12}$. Using the Addition Formulas, we have: a. $cos(\alpha+\beta)=cos(\alpha) cos(\beta)-sin(\alpha) sin(\beta)=(\frac{4}{5})(-\frac{12}{13})-(\frac{3}{5})(\frac{5}{13})=-\frac{63}{65}$ b. $sin(\alpha+\beta)=sin(\alpha) cos(\beta)+cos(\alpha) sin(\beta)=(\frac{3}{5})(-\frac{12}{13})+(\frac{4}{5})(\frac{5}{13})=-\frac{16}{65}$ c. $tan( \alpha+\beta)=\frac{tan(\alpha)+tan(\beta)}{1-tan(\alpha) tan(\beta)}=\frac{(\frac{3}{4})+(-\frac{5}{12})}{1-(\frac{3}{4})(-\frac{5}{12})}=\frac{16}{63}$
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