Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix D - Trigonometry - D Exercises: 57


$sin3\theta+sin\theta=2sin2\theta$ $cos\theta$

Work Step by Step

Need to prove the identity $sin3\theta+sin\theta=2sin2\theta$ $cos\theta$ Since $sin3\theta$ can be written as: $sin3\theta=sin(2\theta+\theta)$ Use sum identity for sine. $sin3\theta=sin(2\theta+\theta)=sin2\theta cos\theta+cos2\theta sin\theta$ Thus, $sin3\theta+sin\theta=sin2\theta cos\theta+cos2\theta sin\theta+sin\theta$ $=sin2\theta cos\theta+(2cos^{2}\theta-1) sin\theta+sin\theta$ $=sin2\theta cos\theta+2cos^{2}\theta sin\theta$ $=sin2\theta cos\theta+(2 sin\theta cos\theta) cos \theta$ $ =sin2\theta cos\theta+ sin2\theta cos \theta $ Hence, $sin3\theta+sin\theta=2sin2\theta$ $cos\theta$
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