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,

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

$=cos2xcos2x-sin2xsin2x$

$=cos^2 2x- sin^2 2x$

$=1-sin^2 2x-sin^2 2x$

$=1-2sin^2 2x$

$=RHS$

$QED$

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