Find $2^{n-1}\prod_{k = 1}^{n-1}{\left(\cos\theta - \cos\left( \frac{k\pi}{n} \right)\right)}$ Evaluate:
$$2^{n-1} \left(\cos\theta - \cos\left( \frac{\pi}{n} \right)\right)\left(\cos\theta - \cos\left( \frac{\pi}{n} \right)\right)...\left(\cos\theta - \cos\left( \frac{(n-1)\pi}{n} \right)\right)$$
The answer is

 \begin{align} \frac{\sin(n\theta)}{sin\theta} \end{align}

 A: Let $u=e^{i\theta}$. Then $\cos(\theta)=\frac12\left(u+\frac1u\right)$ and
$$
\begin{align}
2^{n-1}\prod_{k=1}^{n-1}\left(\cos(\theta)-\cos\left(\frac{k\pi}n\right)\right)
&=\prod_{k=1}^{n-1}\left(u+\frac1u-e^{ik\pi/n}-e^{-ik\pi/n}\right)\tag{1}\\
&=\frac1{u^{n-1}}\prod_{k=1}^{n-1}\left(u-e^{ik\pi/n}\right)\prod_{k=1}^{n-1}\left(u-e^{-ik\pi/n}\right)\tag{2}\\
&=\frac1{u^{n-1}}\frac{u^{2n}-1}{u^2-1}\tag{3}\\[3pt]
&=\frac{u^n-u^{-n}}{u-u^{-1}}\tag{4}\\[6pt]
&=\frac{\sin(n\theta)}{\sin(\theta)}\tag{5}
\end{align}
$$
Explanation:
$(1)$: distribute $2$ over each term in the product$\vphantom{\prod\limits_{k=1}^{2n}}$
$(2)$: $u+\frac1u-e^{ik\pi/n}-e^{-ik\pi/n}=\frac1u\left(u-e^{ik\pi/n}\right)\left(u-e^{-ik\pi/n}\right)$
$(3)$: $\prod\limits_{k=1}^{2n}\left(u-e^{ik\pi/n}\right)=u^{2n}-1$, then divide out the $k=n$ and $k=2n$ terms
$(4)$: algebra
$(5)$: $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$\vphantom{\prod\limits_{k=1}^{2n}}$
A: I will use multiple angle formula to solve the problem: 
$$\sin(mx) = 2^{m-1}\prod_{k=0}^{m-1} \sin\left(x+\frac{k\pi}{m}\right)$$
Now, to solve our problem,
\begin{align}
&2^{n-1} \prod_{k=1}^{n-1} \left( \cos \theta - \cos \left( \frac{k\pi}{n}\right) \right)\\ & = 2^{n-1} \prod_{k=1}^{n-1} (-2) \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) \sin \left( \frac{\theta}{2}- \frac{k\pi}{2n}\right)  \label{1}\tag{1}\\ 
&= 2^{2n-2}\prod_{k=1}^{n-1} (-1) \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) \sin \left( \frac{\theta}{2}- \frac{k\pi}{2n}\right) \label{2}\tag{2}\\
&= 2^{2n-2}\prod_{k=1}^{n-1}  \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) \left( -\sin \left( \frac{\theta}{2}- \frac{k\pi}{2n}\right)\right) \label{3}\tag{3}\\
&= 2^{2n-2}\prod_{k=1}^{n-1}  \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) \sin \left( \frac{\theta}{2}- \frac{k\pi}{2n}+\pi\right) \label{4}\tag{4}\\
&= 2^{2n-2}\prod_{k=1}^{n-1}  \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) \sin \left( \frac{\theta}{2}+ \frac{(2n-k)\pi}{2n}\right) \label{5}\tag{5}\\
&= \frac{2^{2n-2}\sin \left(\frac{\theta}{2} \right) \sin \left( \frac{\theta}{2}+ \frac{\pi}{2}\right)\prod_{k=1}^{n-1}  \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) \sin \left( \frac{\theta}{2}+ \frac{(2n-k)\pi}{2n}\right)}{\sin \left(\frac{\theta}{2} \right) \sin \left( \frac{\theta}{2}+ \frac{\pi}{2}\right)} \label{6}\tag{6}\\
&= \frac{2^{2n-2}\prod_{k=0}^{2n-1}  \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) }{\sin \left(\frac{\theta}{2} \right) \sin \left( \frac{\theta}{2}+ \frac{\pi}{2}\right)} \label{7}\tag{7}\\
&= \frac{2^{2n-1}\prod_{k=0}^{2n-1}  \sin \left(\frac{\theta}{2}+\frac{k\pi}{2n} \right) }{2\sin \left(\frac{\theta}{2} \right) \sin \left( \frac{\theta}{2}+ \frac{\pi}{2}\right)} \label{8}\tag{8}\\
&= \frac{\sin((2n)\frac{\theta}{2})}{2\sin \left(\frac{\theta}{2} \right) \sin \left( \frac{\theta}{2}+ \frac{\pi}{2}\right)} \label{9}\tag{9}\\
&= \frac{\sin((n\theta)}{2\sin \left(\frac{\theta}{2} \right) \cos\left( \frac{\theta}{2}\right)} \label{10}\tag{10}\\
&= \frac{\sin((n\theta)}{\sin\theta}  \label{11}\tag{11}\\
\end{align}
where in $(\ref{1})$, we use $\cos A - \cos B = -2 \sin \left( \frac{A+B}{2}\right) \sin \left( \frac{A-B}{2}\right)$
in $(\ref{2})$, I just take out the $2$ from the product.
in $(\ref{4})$, we use $\sin(A+\frac{\pi}{2})=\cos(A)$
in $(\ref{6})$, I just multiply top and bottom by the same term.
in $(\ref{7})$, I just rearrange the product terms.
in $(\ref{9})$, I use multiple angle formula.
and finally, I use double angle formula in the denominator.
