Precalculus (6th Edition) Blitzer

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

Chapter 5 - Review Exercises - Page 708: 36


a. $-\frac{63}{65}$ b. $-\frac{56}{65}$ c. $\frac{63}{16}$ d. $\frac{24}{25}$ e. $\frac{5\sqrt {26}}{26}$

Work Step by Step

Given $tan\alpha=\frac{4}{3}, \pi\lt \alpha \lt \frac{3\pi}{2}$ and $tan\beta=\frac{5}{12}, 0\lt \beta \lt\frac{\pi}{2} $ For angle $\alpha$, form a triangle of sides $4,3,5$; we have: $sin\alpha=-\frac{4}{5}, cos\alpha=-\frac{3}{5}$; For angle $\beta$, form a triangle of sides $5,12,13$; we have $sin\beta=\frac{5}{13}, cos\beta=\frac{12}{13}$. a. Use the Addition Formulas, $sin(\alpha+\beta)=sin\alpha cos\beta+ cos\alpha sin\beta =(-\frac{4}{5})(\frac{12}{13})+(-\frac{3}{5})(\frac{5}{13})=-\frac{63}{65}$ b. Use the Subtraction Formulas, $cos(\alpha-\beta)=cos\alpha cos\beta+ sin\alpha sin\beta =(-\frac{3}{5})(\frac{12}{13})+(-\frac{4}{5})(\frac{5}{13})=-\frac{56}{65}$ c. Use the Addition Formulas, $tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}=\frac{(\frac{4}{3})+(\frac{5}{12})}{1-(\frac{4}{3})(-\frac{5}{12})}=\frac{63}{16}$ d. Use the Double-Angle Formula, $sin2\alpha = 2 sin\alpha cos\alpha = 2(-\frac{4}{5})(-\frac{3}{5})=\frac{24}{25}$ e. Use the Half-Angle Formula, $0\lt \frac{\beta}{2} \lt\frac{\pi}{4} $ and $cos\frac{\beta}{2}=\sqrt {\frac{1+cos\beta}{2}}=\sqrt {\frac{1+(\frac{12}{13})}{2}}=\frac{5\sqrt {26}}{26}$
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