## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.2 - Sum and Difference Formulas - Exercise Set - Page 669: 62

#### Answer

a. $-\frac{2\sqrt 2+\sqrt 3}{6}$ b. $\frac{2\sqrt 6-1}{6}$ c. $\frac{8\sqrt 2-9\sqrt 3}{5}$

#### Work Step by Step

Step 1. Given $cos\alpha=\frac{1}{2}$ and $\alpha$ in quadrant IV, we have $sin\alpha=-\sqrt {1-cos^2\alpha}=-\frac{\sqrt 3}{2}$ and $tan\alpha=\frac{sin\alpha}{cos\alpha}=-\sqrt 3$ Step 2. Given $sin\beta=-\frac{1}{3}$ and $\beta$ in quadrant III, we have $cos\beta=-\sqrt {1-sin^2\beta}=-\frac{2\sqrt 2}{3}$ and $tan\beta=\frac{sin\beta}{cos\beta}=\frac{\sqrt 2}{4}$ a. $cos(\alpha+\beta)=cos\alpha cos\beta - sin\alpha sin\beta =(\frac{1}{2})(-\frac{2\sqrt 2}{3})-(-\frac{\sqrt 3}{2})(-\frac{1}{3})=-\frac{2\sqrt 2+\sqrt 3}{6}$ b. $sin(\alpha+\beta)=sin\alpha cos\beta + cos\alpha sin\beta =(-\frac{\sqrt 3}{2})(-\frac{2\sqrt 2}{3})+(\frac{1}{2})(-\frac{1}{3})=\frac{2\sqrt 6-1}{6}$ c. $tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}=\frac{(-\sqrt 3)+(\frac{\sqrt 2}{4})}{1-(-\sqrt 3)(\frac{\sqrt 2}{4})}=\frac{\sqrt 2-4\sqrt 3}{4+\sqrt 6}=\frac{8\sqrt 2-9\sqrt 3}{5}$

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