## Calculus: Early Transcendentals 8th Edition

$(sinx+cosx)^{2}=1+sin2x$
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, $(sinx+cosx)^{2}=1+sin2x$