Calculus 8th Edition

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

Appendix D - Trigonometry - D Exercises - Page A32: 53


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

Work Step by Step

Need to prove the identity $sinx$ $sin2x$ $+cosx$$cos2x$ $=cosx$ Let us solve left side of the given identity. Since, $sin2x=2sinxcosx$ and $cos2x=cos^{2}x-sin^{2}x$ Thus, $sinx$ $sin2x$ $+cosx$$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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