# Proving A Trigonometric Identity- Double Angles

$(\cos(2x)-\sin(2x))(\sin(2x)+\cos(2x)) = \cos(4x)$ I'm trying to prove that the left side equals the right side. I'm just stuck on which double angle formula of cosine to use.

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$$(\cos(2x)-\sin(2x))(\sin(2x)+\cos(2x))=(\cos^2(2x)-\sin^2(2x)) = \cos(4x)$$ From $$\cos(a+b)=\cos a \cos b-\sin a\sin b$$ if $a=2x,b=2x$ then $$\cos(4x)=\cos^22x-\sin^22x$$
$cos(4x) = cos(2(2x)) = cos^2(2x)-sin^2(2x) = (cos(2x)-sin(2x))(cos(2x)+sin(2x))$