Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

APPENDIX D - Trigonometry - D Exercises - Page A 33: 57



Work Step by Step

We need to prove the identity $sin3\theta+sin\theta=2sin2\theta$ $cos\theta$ $sin3\theta$ can be written as: $sin3\theta=sin(2\theta+\theta)$ Use the 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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