Prove that (cos 9x - cos 5x)/(sin 17x - sin 3x) = -(sin 2x)/(cos 10x).

$\sin C + \sin D = 2\sin\dfrac{C + D}{2}\cos\dfrac{C - D}{2}$, $\cos C + \cos D = 2\cos\dfrac{C + D}{2}\cos\dfrac{C - D}{2}$, $\cos C - \cos D = -2\sin\dfrac{C + D}{2}\sin\dfrac{C - D}{2}$. Products into sums: $2\cos x \cos y = \cos(x + y) + \cos(x - y)$ and so on.