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: 37


a. $1$ b. $-\frac{3}{5}$ c. undefined d. $-\frac{3}{5}$ e. $-\frac{\sqrt {10+3\sqrt {10}}}{2\sqrt 5}$

Work Step by Step

Given $tan\alpha=-3,\frac{\pi}{2}\lt \alpha \lt\pi$ and $cot\beta=-3, \frac{3\pi}{2}\lt \beta \lt 2\pi $, For angle $\alpha$, form a right triangle of sides $3,1,\sqrt {10}$; we have $sin\alpha=\frac{3\sqrt {10}}{10}, cos\alpha=-\frac{\sqrt {10}}{10}$; For angle $\beta$, form a right triangle of sides $1,3,\sqrt {10}$; we have $sin\beta=-\frac{\sqrt {10}}{10}, cos\beta=\frac{3\sqrt {10}}{10}, tan\beta=-\frac{1}{3}$. a. Use the Addition Formulas, $sin(\alpha+\beta)=sin\alpha cos\beta+ cos\alpha sin\beta =(\frac{3\sqrt {10}}{10})(\frac{3\sqrt {10}}{10})+(-\frac{\sqrt {10}}{10})(-\frac{\sqrt {10}}{10})=1$ b. Use the Subtraction Formulas, $cos(\alpha-\beta)=cos\alpha cos\beta+ sin\alpha sin\beta =(-\frac{\sqrt {10}}{10})(\frac{3\sqrt {10}}{10})+(\frac{3\sqrt {10}}{10})(-\frac{\sqrt {10}}{10})=-\frac{3}{5}$ c. Use the Addition Formulas, $tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}=\frac{(-3)+(-\frac{1}{3})}{1-(-3)(-\frac{1}{3})}=-\infty$ undefined d. Use the Double-Angle Formula, $sin2\alpha = 2 sin\alpha cos\alpha = 2(\frac{3\sqrt {10}}{10})(-\frac{\sqrt {10}}{10})=-\frac{3}{5}$ e. Use the Half-Angle Formula, $ \frac{3\pi}{4}\lt \frac{\beta}{2} \lt \pi $ and $cos\frac{\beta}{2}=-\sqrt {\frac{1+cos\beta}{2}}=-\sqrt {\frac{1+\frac{3\sqrt {10}}{10}}{2}}$=$-\frac{\sqrt {10+3\sqrt {10}}}{2\sqrt 5}$
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