Show that $z+\frac{1}{z} = 2\cos(\theta)$ and $z^n+\frac{1}{z^n} = 2\cos (n\theta)$. Hence express $\cos^6 (\theta)$ Show that $z+ \frac{1}{z} = 2\cos (\theta)$ and $z^n+\frac{1}{z^n} = 2\cos (n\theta)$
Hence express $\cos^6(\theta)$
So I have nailed down the proof:
Setting the value of $z=\cos(\theta) + i\sin(\theta)$, $$z^n+\frac{1}{z^n}$$$$=z^n+z^{-n}$$
$$=(\cos\theta+i\sin\theta)^n+(\cos\theta+i\sin\theta)^{-n}$$
Using d-Moivre's theorem, 
$$=\cos(n\theta)+i\sin(n\theta)+\cos(-n\theta)+i\sin(-n\theta)$$
$$=\cos(n\theta)+i\sin(n\theta)+\cos(n\theta)-i\sin(n\theta)$$
$$=2\cos(n\theta)$$
I'm stuck from here since, I assume we somehow need to use the theorem.
and equate $\cos^6$ to something based on $z^n+\frac{1}{z^n}$

I assume you want to express $\cos^6(x)$ in terms of a linear combination of terms from $\{\cos(kx)\}_{k=0}^6$, as this is a standard question. By the binomial theorem,
$$2^6\cos^6(x)=(z+z^{-1})^6=(z^6+z^{-6})+6(z^4+z^{-4})+\binom{6}{2}(z^2+z^{-2})+\binom{6}{3}$$
Which reduces to
$$2^5\cos^6(x)=\cos(6x)+6\cos(4x)+15\cos(2x)+10$$
The overarching point is that we can form a recursive relation between $\cos^nx$ and $\cos^{n-2}(x)$ to prove any power of cosine is expressible as a linear combination of $\{\cos(kx)\}_{k=0}^n$. The polynomial $T_n(x)$ for which $T_n(\cos(x))=\cos(nx)$ is known as the Chebyshev polynomial of the first kind.
