Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.4 Trigonometric Functions - Exercises - Page 30: 24


$\sin 2\theta=\frac{2\sqrt 2}{3}$ and $\cos 2\theta=-\frac{1}{3}$

Work Step by Step

$\sec^{2}\theta=1+tan^{2}\theta=1+2=3$ $\implies \sec\theta= \sqrt 3$ ($\sec \theta$ is positive as $0\leq\theta\lt\frac{\pi}{2}$) $\cos\theta=\frac{1}{\sec\theta}=\frac{1}{\sqrt 3}$ $\sin^{2}\theta=1-\cos^{2}\theta=1-\frac{1}{3}=\frac{2}{3}$ $\implies \sin \theta=\frac{\sqrt 2}{\sqrt 3}$ ($\sin \theta$ is positive as $0\leq\theta\lt\frac{\pi}{2}$) Now, $\sin 2\theta=2\sin\theta\cos\theta=2\times\frac{\sqrt 2}{\sqrt 3}\times\frac{1}{\sqrt 3}=\frac{2\sqrt 2}{3}$ $\cos 2\theta= \cos^{2}\theta-\sin^{2}\theta=\frac{1}{3}-\frac{2}{3}=-\frac{1}{3}$
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