How do you verify the identity #(sintheta+costheta)^2-1=sin2theta#?

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Answer 1 · 2016-12-31T09:35:22

See proof below

We need

#sin^2theta+cos^2theta=1#

#(a+b)^2=a^2+2ab+b^2#

#2sinthetacostheta=sin2theta#

LHS is

#(sintheta+costheta)^2-1#

#=sin^2theta+2sinthetacostheta+sin^2theta-1#

#=1+2sinthetacostheta-1#

#=2sinthetacostheta#

#=sin2theta#

#=RHS#

#QED#