How do you prove #cos4x=1-2sin^2(2x)#?

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Answer 1 · 2018-04-13T12:46:48

Please see the proof below

#"Reminder"#

#cos(A+B)=cosAcosB-sinAsinB#

#cos^2theta+sin^2theta=1#

Therefore,

#LHS=cos4x=cos(2x+2x)#

#=cos2xcos2x-sin2xsin2x#

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

#=1-sin^2 2x-sin^2 2x#

#=1-2sin^2 2x#

#=RHS#

#QED#