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

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Bdub Share
Mar 8, 2018

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See Below

Explanation:

#LHS:#
#sin^4 x -cos^4 x =(sin^2x-cos^2x) (sin^2x+cos^2x)#->factor

#=(sin^2x-cos^2x)*1#

#=(1-cos^2x)-cos^2x#->use the property #sin^2x+cos^2x=1#

#=1-2cos^2x#

#=RHS#

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