How do you prove #(sinx + cosx)^2=1 + sin2x#?

1 Answer
Apr 15, 2015

It depends on how far back you need to go in your proof.
If you can use:
#sin^2(x)+cos^2(x)=1# (which can be derived from the Pythagorean Theorem)
and #sin(2x) =2sin(x)*cos(x)# (the Double Angle Formula for sin, which is considerably more complicate to prove)
then the requested proof is fairly simple:

#(sin(x)+cos(x))^2#

#= sin^2(x) + 2sin(x)*cos(x) +cos^2(x)#

#= [sin^2(x)+cos^2(x)] + 2 sin(x)*cos(x)#

#= 1 + sin(2x)#