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: $(sin~x+cos~x)^2=sin^2x+2~sin~x~cos~x+cos^2x=(sin^2x+cos^2x)+2~sin~x~cos~x=1+sin~2x$
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