$$ 2^{n-1} \prod^{n-1}_{k=1} (\cos \theta-\cos \frac{kπ}{n})= ?? $$
In this question I thought that in some way or the other complex numbers should be involved along with roots of unity, but actually I could not resolve it into any equation. I thought of forming an equation like $\prod^{n-1}_{k=1} (x-\cos \frac{kπ}{n})$, but because only real part of the n th root of unity was involved I couldn't solve further. Help me please.
$\cos(\frac {k\pi} n)$ for $k=1...n-1$ are the roots of $U_{n-1}$, the Chebyshev polynomial of the second kind of degree $n-1$.
The leading coefficient of $U_{n-1}$ is $2^{n-1}$. Therefore that polynomial can be written as $$U_{n-1}(X)=2^{n-1} \prod^{n-1}_{k=1} \left (X-\cos \frac{kπ}{n}\right)$$ Evaluating at $X=\cos\theta$ shows that the expression you're looking for is $U_{n-1}(\cos \theta)$.
Using a famous identity for that polynomial leads to $$2^{n-1} \prod^{n-1}_{k=1} (\cos \theta-\cos \frac{kπ}{n})= \frac{\sin(n\theta)}{\sin \theta}$$