Proving $ z^n + \frac{1}{z^n} = 2\cos(n\theta) $ for $z = \cos (\theta) + i\sin(\theta)$ 

Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then
$$ z^n +  {1\over z^n} = 2\cos(n\theta) $$



What I have attempted
If $$  z = \cos (\theta) + i\sin(\theta) $$
then $$ z^n = \cos (n\theta) + i\sin(n\theta) $$
$$ z^n +  {1\over z^n} $$
$$ \cos (n\theta) + i\sin(n\theta) +  {1\over \cos (n\theta) + i\sin(n\theta)} $$
$$ (\cos (n\theta) + i\sin(n\theta))\cdot(\cos (n\theta) + i\sin(n\theta)) + 1 $$
$$ \left[ {(\cos (n\theta) + i\sin(n\theta))\cdot(\cos (n\theta) + i\sin(n\theta)) + 1\over \cos (n\theta) + i\sin(n\theta)}  \right]  $$
$$ \left[ {\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta) - \sin^2(n\theta) + 1\over \cos (n\theta) + i\sin(n\theta)}  \right]  $$
Now this is where I am stuck.. I tried to use a double angle identity but I can't eliminate the imaginary part..
 A: 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, 
$$=\cos(n\theta)+i\sin(n\theta)+\cos(-n\theta)+i\sin(-n\theta)$$
$$=\cos(n\theta)+i\sin(n\theta)+\cos(n\theta)-i\sin(n\theta)$$
$$=2\cos(n\theta)$$
A: Hint. If $$z=\cos (\theta) + i\sin(\theta)= e^{i\theta}$$ then
$$
\frac1{z^n}=z^{-n}=e^{-in\theta}=\cos (n\theta) - i\sin(n\theta).
$$
A: If you want to continue in the way you started, you can simply rewrite $1=\cos^2(n\theta)+\sin^2(n\theta)$ and then simplify:
$$
\frac{\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta) - \sin^2(n\theta) + 1}{ \cos (n\theta) + i\sin(n\theta)}=
\frac{\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta) - \sin^2(n\theta) + \cos^2(n\theta)+\sin^2(n\theta)}{ \cos (n\theta) + i\sin(n\theta)}=
\frac{2\cos^2(n\theta) + 2i\sin(n\theta)\cos (n\theta)}{ \cos (n\theta) + i\sin(n\theta)}=
\frac{2\cos(n\theta) (\cos(n\theta)+i\sin(n\theta))}{ \cos (n\theta) + i\sin(n\theta)}= \underline{\underline{2\cos(n\theta)}}
$$ 
Which shows that you were almost there. (But it is still useful that you posted your question here - you have seen other approaches.)
A: Hint: To simplify $\dfrac{1}{\cos(n\theta)+i\sin(n\theta)}$, multiply top and bottom by $\cos(n\theta)-i\sin(n\theta)$.
A: Using geometry of complex:

By construction, point $C$ corresponds to $z^n$ an $C^\prime$ to $z^{-n}$. i.e. $\beta=-n\theta$ (the symmetry of the image is due to the fact that $|z|=1$, and thus $z\overline{z}=1$). Then $z^n+z^{-n}$ is, interpreting as a sum of vectors, the vector $\vec{OD}$. Then, by means of Cosine Law's, it is easy to conclude.  
