(complex variables) Express $\cos3\phi$, $\cos4\phi$, and $\cos5\phi$ in terms of $\cos\phi$ and $\sin\phi$. I'm not sure what the intuition is supposed to be behind this question. This is my attempt at $\cos3\phi$. Does this look agreeable?
We can use the identity $e^{iz}=\cos z + i \sin z$. Let $z = 3 \phi$. 
\begin{align*}
\cos3\phi &= Re\left( e^{i3\phi} \right) \\
&= Re\left( \left( e^{i\phi} \right)^3 \right) \\
&= Re \left( \cos^3\phi + 2i\cos^2\phi\sin\phi - \cos\phi\sin^2\phi + i \cos^2\phi \sin \phi - 2 \cos\phi\sin^2\phi -i\sin^3\phi \right) \\
&= \cos^3\phi - 3\cos\phi\sin^2\phi
\end{align*}
And I suppose a similar argument would follow for $\cos4\phi$ and $\cos5\phi$. But is there an easier way than manually factoring $(\cos3\phi + i\sin3\phi)^4$ and $(\cos3\phi + i\sin3\phi)^5$?

Using Prosthaphaeresis Formula,
we can define reduction formulae of $\cos(mx)$ as follows:
$$\cos(2n+1)x+\cos x=2\cos(n+1)x\cos nx$$
$$\cos(2n)x+\cos2x=2\cos(n+1)x\cos(n-1)x$$
and for $\sin(mx),$
$$\sin(2n+1)x-\sin x=2\sin nx\cos(n+1)x$$
$$\sin(2n)x-\sin2x=2\sin(n-1)x\cos(n+1)x\text{ or }=2\sin(nx)\cos(nx)$$
