Prove that the LHS equals the RHS: (cosx - cosy) = -2sin[(x+y)/2] * sin [(x-y)/2] thanks?

Answer 1

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Expand the RHS -2[sin(x/2)cos(y/2)+sin(y/2)cos(x/2)][sin(x/2)cos(y/2)-sin(y/2)cos(x/2)]=-2[sin^2(x/2cos^2(y/2)-sin^2(y/2)cos^2(x/2)] using difference of two squares

change the cos^2(y/2)=1-sin^2(y/2) and cos^2(x/2)=(1-sin^2x/2) to obtain

-2(sin^2(x/2)-sin^2(y/2)) convert the sin^2(x/2) and sin^2(y/2) to the double angle

-2((1-cosx)/2-(1-cosy)/2=-(1-cosx-1+cosy)=cosx-cosy