How do we prove that #1+cos^2 2x=2(cos^4x+sin^4x)#?

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Answer 1 · 2017-08-25T16:47:09

#LHS=1+cos^2 2x#

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

#=cos^4x+sin^4x+2cos^2xsin^2x+cos^4x+sin^4x-2cos^2xsin^2x#

#=2(cos^4x+sin^4x)=RHS#

Answer 2 · 2017-08-25T17:29:20

Refer to the Explanation.

Using,

#2(a^2+b^2)=(a+b)^2+(a-b)^2,# with, #a=cos^2x, b=sin^2x,#

we have,

#2(cos^4x+sin^4x)=(cos^2x+sin^2x)^2+(cos^2x-sin^2x)^2#

#=1+cos^2 2x,# as desired.