How do you verify #cos^4x - sin^4x = cos^2x - sin^2x#?

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Answer 1 · 2015-05-22T06:48:45

#cos^4x-sin^4x#

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

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

#= (cos^2x-sin^2x)xx1 = (cos^2x-sin^2x)#

based on the identities:

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

#cos^2x+sin^2x = 1# from Pythagoras (right angled triangle with hypotenuse of length 1).