Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.4 Double-Angle and Half-Angle Identities - 7.4 Exercises - Page 693: 18


$\displaystyle \cos\theta=-\frac{\sqrt{70}}{10}$ $\displaystyle \sin\theta=-\frac{\sqrt{30}}{10}$

Work Step by Step

Double-Angle Identity: $\cos 2\theta=2\cos^{2}\theta- \mathrm{l}$ $\displaystyle \frac{3}{4}+1=2\cos^{2}\theta$ $\displaystyle \frac{7}{5}=2\cos^{2}\theta$ $\displaystyle \cos^{2}\theta=\frac{7}{10}$ Since $\theta$ terminates in Q.III, its cosine (and sine) are negative $\displaystyle \cos\theta=-\sqrt{\frac{7}{10}}=-\frac{\sqrt{7}}{\sqrt{10}}\cdot\frac{\sqrt{10}}{\sqrt{10}}=-\frac{\sqrt{70}}{10}$ Pythagorean Identity ($\sin\theta$ is negative): $\sin\theta=-\sqrt{1-\cos^{2}\theta}$ $=-\sqrt{1-\frac{7}{10}}=- \sqrt{\frac{3}{10}}$ $=-\displaystyle \frac{\sqrt{3}}{\sqrt{10}}\cdot\frac{\sqrt{10}}{\sqrt{10}}=-\frac{\sqrt{30}}{10}$
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