Express $\cos(\pi/5)$ in terms of a sum of powers of the principal $100{th}$ root of unity.
Using the formula, $w_n = \cos(2\pi/n) + i \sin(2\pi/n)$
I have calculated, $w_{100} = \cos(\pi/50) + i \sin(\pi/50)$
$w_{100}^{10} = \cos(\pi/5) + i \sin(\pi/5)$
But $\cos(\pi/5)^{10} \neq w_{100}^{10}$, so I'm confused how to progress this problem.
Regards.
$$\frac12\left(w_{100}^{10}+w_{100}^{90}\right).$$
This works because
$$\begin{align} w_{100}&=\cos(\pi/50)+i\sin(\pi/50) \\ \Rightarrow w_{100}^{10}&=\cos(\pi/5)+i\sin(\pi/5) \\ \Rightarrow w_{100}^{90}&=\cos(9\pi /5)+i\sin(9\pi/5) \\ &=\cos(-\pi/5)+i\sin(-\pi/5) \\ \Rightarrow \frac{1}{2}\left(w_{100}^{10}+w_{100}^{90}\right)&=\frac{1}{2}\left(\cos(\pi/5)+i\sin(\pi/5)+\cos(-\pi/5)+i\sin(-\pi/5)\right) \end{align}$$
Now what do you know about $\cos(-A)$ and $\sin(-A)$?