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: 60


a. $ -\frac{6+4\sqrt 5}{15}$ b. $ \frac{8-3\sqrt 5}{15}$ c. $ \frac{54-25\sqrt 5}{22}$

Work Step by Step

Given $tan\alpha=-\frac{4}{3}$ with $\alpha$ in quadrant II, build a right triangle with angle $\alpha$ and sides $4,3,5$; we have $sin\alpha=\frac{4}{5}$ and $cos\alpha=-\frac{3}{5}$. Similarly, given $cos\beta=\frac{2}{3}$ with $\beta$ in quadrant I, build a right triangle with angle $\beta$ and sides $\sqrt 5,2,3$; we have $sin\beta=\frac{\sqrt 5}{3}$ and $tan\beta=\frac{\sqrt 5}{2}$. Using the Addition Formulas, we have: a. $cos(\alpha+\beta)=cos(\alpha) cos(\beta)-sin(\alpha) sin(\beta)=(-\frac{3}{5})(\frac{2}{3})-(\frac{4}{5})(\frac{\sqrt 5}{3})=-\frac{6+4\sqrt 5}{15}$ b. $sin(\alpha+\beta)=sin(\alpha) cos(\beta)+cos(\alpha) sin(\beta)=(\frac{4}{5})(\frac{2}{3})+(-\frac{3}{5})(\frac{\sqrt 5}{3})=\frac{8-3\sqrt 5}{15}$ c. $tan( \alpha+\beta)=\frac{tan(\alpha)+tan(\beta)}{1-tan(\alpha) tan(\beta)}=\frac{(-\frac{4}{3})+(\frac{\sqrt 5}{2})}{1-(-\frac{4}{3})(\frac{\sqrt 5}{2})}=-\frac{8-3\sqrt 5}{6+4\sqrt 5}=\frac{54-25\sqrt 5}{22}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.