Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$ 
Show $\cos(x+y)\cos(x-y) - \sin(x+y)\sin(x-y) = \cos^2x - \sin^2x$

I have got as far as showing that:
$\cos(x+y)\cos(x-y) = \cos^2x\cos^2y -\sin^2x\sin^2y$
and 
$\sin(x+y)\sin(x-y) = \sin^2x\cos^2y - \cos^2x\sin^2y$
I get stuck at showing:
$\cos^2x\cos^2y -\sin^2x\sin^2y - \sin^2x\cos^2y - \cos^2x\sin^2y = \cos^2x - \sin^2x$
I know that $\sin^2x + \cos^2x = 1$ and I have tried rearranging this identity in various ways, but this has not helped me so far.

Looking at
$\cos^2x\cos^2y -\sin^2x\sin^2y - \sin^2x\cos^2y + \cos^2x\sin^2y$  (The last is a negative times a negative that forms a positive I believe.
You could pull $\sin^2x$ from the middle two terms to get this expression:
$ -\sin^2x\sin^2y - \sin^2x\cos^2y
  = -\sin^2x(\sin^2y+\cos^2y) = -\sin^2x $
There is a similar reduction with the first and last terms around $\cos^2x$ that should make this appear easier.
