Answer
The identity is verified.
$(sin~x+cos~x)^2=1+sin~2x$
Work Step by Step
Use: $cos^2x+sin^2x=1$ and $sin~2x=2~sin~x~cos~x$
Start on the left side:
$(sin~x+cos~x)^2=sin^2x+2~sin~x~cos~x+cos^2x=(sin^2x+cos^2x)+2~sin~x~cos~x=1+sin~2x$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 7 - Review Exercises - Page 553: 67
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