How do you verify #(2cos^2x-1)^2/(cos^4x-sin^4x)#?

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Answer 1 · 2018-04-30T03:13:08

#cos2x#

Knowing that:
#cos2x= cos^2x-sin^2x= 2cos^2x-1#
and #sin^2x+cos^2x=1#

Start:
#(2cos^2x-1)^2/(cos^4x-sin^4x)=#

#(cos2x)^2/((cos^2x+sin^2x)(cos^2x-sin^2x))=#

#(cos2x)^2/((1)(cos^2x-sin^2x))=#

#(cos2x)^2/(cos^2x-sin^2x)=#

#(cos2x)^2/(cos2x)=#

#cos2x#

End