## Precalculus (6th Edition)

Published by Pearson

# Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 693: 20

#### Answer

$\displaystyle \cos\theta=-\frac{\sqrt{30}}{6}$ $\displaystyle \sin\theta=\frac{\sqrt{6}}{6}$

#### Work Step by Step

$\theta$ terminates in Q.II, where its cosine is negative, its sine is positive. Double-Angle Identity: $\cos 2\theta=2\cos^{2}\theta- \mathrm{l}$ $\displaystyle \frac{2}{3}+1=2\cos^{2}\theta$ $\displaystyle \frac{5}{3}=2\cos^{2}\theta$ $\displaystyle \cos^{2}\theta=\frac{5}{6}$ ... $\cos\theta$ is negative, $\displaystyle \cos\theta=-\sqrt{\frac{5}{6}}=-\frac{\sqrt{5}}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=-\frac{\sqrt{30}}{6}$ Pythagorean Identity ($\sin\theta$ is positive): $\sin\theta=+\sqrt{1-\cos^{2}\theta}$ $=\displaystyle \sqrt{1-\frac{5}{6}}= \sqrt{\frac{1}{6}}=\frac{1}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{6}}{6}$

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