Calculus: Early Transcendentals 8th Edition

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

APPENDIX D - Trigonometry - D Exercises: 53

Answer

$sinx$*$sin2x$ $+cosx$$cos2x$ $=cosx$

Work Step by Step

We need to prove the identity $sinx$*$sin2x$ $+cosx$$cos2x$ $=cosx$ $sin2x=2sinxcosx$ and $cos2x=cos^{2}x-sin^{2}x$ Thus, $sinx$*$sin2x$ $+cosx$$cos2x$ $=sinx(2sinxcosx)+cosx(cos^{2}x-sin^{2}x)$ $sinx$*$sin2x$ $+cosx$$cos2x$ $=2sin^{2}x cosx+cos^{3}x-cosx\times sin^{2}x$ $sinx$*$sin2x$ $+cosx$$cos2x$ $=sin^{2}x cosx+cos^{3}x$ $sinx$*$sin2x$ $+cosx$$cos2x$ $=cosx (sin^{2}x +cos^{2}x)$ $sinx$*$sin2x$ $+cosx$$cos2x$ $=cosx (1)$ Hence, $sinx$*$sin2x$ $+cosx$$cos2x$ $=cosx$
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