Prove that:
{∑(r = 0 to n) nCr cos^(n - r) x sin ^(r) x}{∑(r = 0 to n) (-1)^(r) nCr cos^(n - r) x sin^(r) x} = cos ^(n) 2x
NOTE: nCr is n choose r.
Remember that $\displaystyle (x+y)^n = \sum_{r=0}^n {n\choose r}x^ry^{n-r}$.
Therefore, $\displaystyle \sum_{r=0}^n {n\choose r}\cos^{n-r}x\sin^r x = (\cos x + \sin x) \text{ and }$$\displaystyle \sum_{r=0}^n (-1)^r{n\choose r}\cos^{n-r}x\sin^r x = (\cos x - \sin x)^n$
If you multiply them you get $\displaystyle (\cos^2 x - \sin^2 x)^n = \cos^n (2x)$.