Expand binomially to prove trigonometric identity Prompt: By expanding $\left(z+\frac{1}{z}\right)^4$ show that $\cos^4\theta = \frac{1}{8}(\cos4\theta + 4\cos2\theta + 3).$
I did the expansion using binomial equation as follows 
$$\begin{align*}
\left(z+\frac{1}{z}\right)^4 &= z^4 + \binom{4}{1}z^3.\frac{1}{z} + \binom{4}{2}z^2.\frac{1}{z^2} + \binom{4}{3}z^3.\frac{1}{z}+\frac{1}{z^4}\\
&=z^4+4z^2+6+\frac{4}{z^2}+\frac{1}{z^4}\\
&=z^4+\frac{1}{z^4}+4\left(z^2+\frac{1}{z^2}\right) + 6.  -(eqn 1)
\end{align*}
$$
I'm not sure how to go on about rest of the problem.
[update]
Reading comments, I tried assuming $z = e^{i\theta}$
$2\cos\theta = e^{i\theta} + e^{-i\theta}$
$(2\cos\theta)^4 = (e^{i\theta} + e^{-i\theta})^4$
$=e^{4i\theta} + e^{-4i\theta} + 8(e^{2i\theta}+e^{-2i\theta})+6$  (from eqn 1)

Following your idea, you can suppose that $|z|=1$ and then $z=\text{cis}\theta$ and $z^{-1}=\text{cis}(-\theta)$
$$z+z^{-1}=2\cos\theta \to (z+z^{-1})^4=2^4\cos^4\theta$$
on the other hand:
$$z^2+z^{-2}=(z+z^{-1})^2-2=2^2\cos^2\theta-2=2\cos2\theta$$
$$z^4+z^{-4}=(z^2+z^{-2})^2-2=(2\cos2\theta)^2-2=2\cos4\theta$$
Replacing on your last equation you get:
$$2^4\cos^4\theta=2\cos4\theta+8\cos2\theta+6\\
\cos^4\theta=\frac{1}{8}(\cos4\theta+4\cos2\theta+3)$$
