Calculus: Early Transcendentals 8th Edition

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

APPENDIX D - Trigonometry - D Exercises: 46



Work Step by Step

We need to prove the identity $(sinx+cosx)^{2}=1+sin2x$ We have: $(sinx+cosx)^{2}=sin^{2}x+cos^{2}x+2sinx cosx$ Also, $2sinx cosx= sin2x$, and $sin^{2}x+cos^{2}x=1$ Hence, $(sinx+cosx)^{2}=1+sin2x$
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