Use de Moivre's theorem to obtain an expression for $\sin^6x$ as a sum of terms in the form $\cos ax$ I'm not exactly sure if I'm on the right lines but I've started with a binomial expansion:
$(\cos x+i\sin x)^6=\cos 6x +i \sin 6x= \cos^6 x + i(6\cos^5x \sin x)-15\cos^4x \sin^2x-i(20\cos^3x \sin^3 x)+15 \cos^2x \sin^4x+i(6\cos x \sin^5 x)-\sin^6x$
If I rearranged this for $\sin^6x$ I'd get $\cos^kx$ terms, when the question asks for $\cos ax$
What should I do?

Let $z = e^{ix}$. $\sin x = \frac1{2i} (z-z^{-1})$ and
$$\sin^6 x = -\frac1{64} (z - z^{-1})^6 = -\frac1{64} (z^6 - 6z^4 + 15z^2 - 20 + 15 z^{-2} - 6 z^{-4} + z^{-6})$$
Now remember that $\cos ax = \frac12 (z^a + z^{-a})$ to get
$$\sin^6 x = -\frac1{32} \cos 6x + \frac3{16} \cos 4x - \frac{15}{32} \cos 2x + \frac5{16}$$
