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

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Answer 1 · 2018-04-30T04:09:01

Cosine double angle identity:
#cos(2x)=cos^2(x)-sin^2(x)#

Next, substitute rearranged #sin^2(x)+cos^2(x)=1#

Substitute #sin^2(x) = 1-cos^2(x)#:
#2cos^2(x)-1#
Sub #cos^2(x)#:
#1-2sin^2(x)#

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

The denominator #sin^4(x)-cos^4(x)# is a difference of squares:
#(sin^2(x)-cos^2(x))(sin^2(x)+cos^2(x))#
#=(cos(2x))*1#

Therefore:

#cos^2(2x)/cos(2x) = cos(2x)#

Factors cancel, leaving:
#cos(2x) = cos(2x)#