Question

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

Answer 1

`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`