Show that cosx-cosy=-2sin(x+y)/2*sin(x-y)/2 I am stuck

= -2sin(x+y)/2*sin(x-y)/2

= -2(sin(x/2)cos(y/2)+cos(x/2)sin(y/2))( sin(x/2)cos(y/2)-cos(x/2)sin(y/2))

= -2(sin^2(x/2)cos^2(y/2)+cos(x/2)sin(y/2) sin(x/2)cos(y/2)-cos(x/2)sin(y/2) sin(x/2)cos(y/2)+cos^2(x/2)sin^2(y/2))

= -2(sin^2(x/2)cos^2(y/2)+cos^2(x/2) sin^2(y/2))

=???

What do I do next? Thank You.

I think it might be more helpful to use a product to sum identity. Call (x+y)/2 A and (x-y)/2 B. then you have -2 sin A sin B. when you are doing cos (A+B) you get cos A cos B -sin A sin B. When you do cos (A-B) you get cos A cos B +sin A sin B. if you took cos (A+B)-cos (A-B)
you would get -2 sin A sin B. Remember A is (x+y)/2 and B is (x-y)/2.
so you have cos ((x+y)/2 +(x-y)/2) -cos ((x+y)/2 -(x-y)/2) which when
you simplify is cos x -cos y. That's it.