<s>[INST] Proving Trig Identities (Complex Numbers) 

Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then
$$ z^n +  {1\over z^n} = 2\cos(n\theta) $$
Hence prove that $\cos^6(\theta)$ $=$ $\frac 1{32}$$(\cos(6\theta)$ + $6\cos(4\theta)$ + $15\cos(2\theta)$ + $10$)


I learnt to prove the first part in another post linked here.
The second part is where I am confused because there is a 'hence'
so I thought of taking 2 approaches:
either $$ z^6 + \frac 1{z^6} $$
or $$ z^6 $$ and equating real parts.
I will start with my first approach 
$$ z^6 + \frac 1{z^6} $$
$$ (\cos(x)+i\sin(x))^6 + \frac 1{(\cos(x)+i\sin(x))^6} $$
$$ (\cos(x)+i\sin(x))^6 + {(\cos(x)+i\sin(x))^{-6}} $$
$$ \cos(6x) + i\sin(6x) + \cos(-6x)+ i \sin(-6x) $$
$$ 2\cos(6x) $$ 
Which is no where near what I am suppose to prove..
So with my second approach (expanding and equation real parts)
$$ z^6 $$ 
$$ (\cos(x) + i \sin(x))^6 $$
Using pascals 
$$ \cos^6(x) + i*6\cos^5(x)\sin(x) + i^2*15\cos^4(x)\sin^2(x) + i^3*20\cos^3(x)\sin(x) + i^4*15\cos^2(x)\sin^4(x) + i^5*6\cos(x)\sin^5(x) + i^6 * \sin^6(x) $$
Simplifying
$$ \cos^6(x) - 15\cos^4(x)\sin^2(x)+ 15\cos^2(x)\sin^4(x) - \sin^6(x) +i(6\cos^5(x)\sin(x)-20\cos^3(x)\sin(x) + 6cos(x)\sin^5(x)$$
Now considering only real 
$$ \cos^6(x) - 15\cos^4(x)\sin^2(x)+ 15\cos^2(x)\sin^4(x) - \sin^6(x) $$
At this point I'm confused , am I on the right approach?
 [/INST] I don't think that the hint is really that useful. It would be better to do
$$
\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}
$$
that gives you
\begin{align}
64\cos^6\theta
&=(e^{i\theta}+e^{-i\theta})^6 \\[3px]
&=e^{6i\theta}+6e^{4i\theta}+15e^{2i\theta}
  +20+15e^{-2i\theta}+6e^{-4i\theta}+e^{-6i\theta} \\[3px]
&=(e^{6i\theta}+e^{-6i\theta})+
  6(e^{4i\theta}+e^{-4i\theta})+
  15(e^{2i\theta}+e^{-2i\theta})+
  20 \\[3px]
&=2\cos6\theta+12\cos4\theta+30\cos2\theta+20
\end{align}
This is no more than De Moivre’s formula in disguise, but much easier to manage.
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