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

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Answer 1 · 2017-01-22T07:47:40

See proof below

We need

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

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

#sin2theta=2sinthetacostheta#

Therefore,

#LHS=(costheta+sintheta)^2-1#

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

#=1+2sinthetacostheta-1#

#=2sinthetacostheta#

#=sin2theta#

#=RHS#

#QED#

Answer 2 · 2017-01-22T07:48:01

#cancel1+2 sinthetacostheta cancel-1=2sinthetacostheta#
and the identity is verified.

Since

#sin2theta=2sinthetacostheta#,

you can substitute it in the given expression and expand the square of the bynomial:

#color(red)(sin^2theta+cos^2theta+2sinthetacostheta)-1 =color(green)(2 sinthetacostheta)#

Then, since

#sin^2theta+cos^2theta=1#, you get:

#cancel1+2 sinthetacostheta cancel-1=2sinthetacostheta#

and the identity is verified.