Use De Moivre's Theorem to express $\cos(4θ)$ in terms of sums and differences of powers of $\sin(θ)$ and $\cos(θ)$ So far I have gotten this far using the binomial theorem. 
$(x + iy)^4 = x^4 + 4x^3y + 6i^2x^2y^2 + 4i^3xy^3 + i^4y^4$=
$x^4 - 6x^2y^2 + y^4 + i(4x^3y - 4xy^3)$
$i^2 = -1 $ let $x = cosθ$  and y = $sinθ$
$e^{i4θ} = cos4θ + isin4θ = (cosθ + i sinθ)^4 $
$cos4θ = cos^4θ - 6cos^2θsin^2θ + sin^4θ$
I am not quite sure where to go next from here? Have I expressed this correctly in terms of sinθ and cosθ? 
Thanks for taking the time to review. 

${(\cos{\theta}+i\sin{\theta})^4}\to$
${(c+is)^4}\to$  (for brevity)
$\begin{matrix}
1&4&6&4&1\\
c^4&c^3&c^2&c&1\\
1&is&-s^2&-is^3&s^4\\
\hline
c^4&4ic^3s&-6c^2s^2&-4ics^3&s^4
\end{matrix}$
The terms pertaining to $\cos(4\theta)$  are the real ones; the imaginaries, after dispensing with $i$, pertain to $\sin(4\theta)$ .
So, $\cos(4\theta)=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta$ .
And, taking advantage of the fundamental relation between sine and cosine, $\cos(4\theta)=8\cos^4-8\cos^2+1=8\sin^4\theta-8\sin^2\theta+1$
