How do you prove $cos(2x + π) = cos^2 (x - π/2) + cos(x + π) sin(x + π/2)$ using the double angle identity?

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Answer 1 · 2016-07-08T02:34:36

$RHS=cos^2(x-pi/2)+cos(x+pi)sin(x+pi/2)$

$=sin^2x-cosxcosx$

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

$=-cos2x$

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

How do you prove cos(2x + π) = cos^2 (x - π/2) + cos(x + π) sin(x + π/2)cos(2x + π) = cos^2 (x - π/2) + cos(x + π) sin(x + π/2) using the double angle identity?