How to prove $\cos 4x-\cos 4y=8(\cos x-\cos y)(\cos x+\cos y)(\cos x-\sin y)(\cos x+\sin y)$ Prove that 
$$\cos 4x-\cos 4y=8(\cos x-\cos y)(\cos x+\cos y)(\cos x-\sin y)(\cos x+\sin y)$$
 A: For the Right hand side, use
$$2(\cos x+\cos y)(\cos x-\cos y)=2\cos^2x-2\cos^2y=1+\cos2x-(1+\cos2y)$$
$$2(\cos x+\sin y)(\cos x-\sin y)=2\cos^2x-2\sin^2y=1+\cos2x-(1-\cos2y)=\cos2x-\cos2y$$
$$2(\cos x+\cos y)(\cos x-\cos y)\cdot2(\cos x+\sin y)(\cos x-\sin y)=(\cos2x+\cos2y)(\cos2x-\cos2y)$$
Now, $$2(\cos2x+\cos2y)(\cos2x-\cos2y)=2\cos^22x-2\cos^22y=1+\cos4x-(1+\cos4y)=?$$
A: It can be viewed as a consequence of relationship
$$\cos(4x)=8 \cos(x)^4 - 8\cos(x)^2 + 1$$
Remark: Any $\cos(nx)$ can be expressed as a polynomial in $\cos(x)$ ; these polynomials are known as Chebyshev polynomials of the first kind $T_n(\cos(\theta))$.
Setting $X:=cos(x)$ and $Y:=\cos(y)$, it amounts to prove the now non-trigonometric identity:
$$8(X^4-Y^4)-8(X^2-Y^2)=8(X-Y)(X+Y)(X-s\sqrt{1-Y^2})(X+s\sqrt{1-Y^2})$$
(where $s=1$ if $y$ is in the first 2 quadrants: $0 \leq (y \ mod \ 2 \pi) \leq \pi) $, $-1$ otherwise ; this is necessary to take into account the sign of $\sin(y)$).
or
$$8(X^2-Y^2)(X^2+Y^2)-8(X^2-Y^2)=8(X^2-Y^2)(X^2-(1-Y^2))$$
which is immediately proven.
