$\frac{\sin(nx)}{\sin(x)}=(-4)^{(n-1)/2} \prod_{1\leq j \leq (n-1)/2}(\, \sin^2(x)-\sin^2(\frac{2\pi j}{n})\,)$ In Serre's A Course in Arithmetic, it states

For $n$ odd and positive integer, proof that $\frac{\sin(nx)}{\sin(x)}=(-4)^{(n-1)/2} \prod_{1\leq j \leq (n-1)/2}(\,\sin^2(x)-\sin^2(\frac{2\pi j}{n})\,)$ 

It says this is elementary, and suggest first proving that $\frac{\sin(nx)}{\sin(x)}$ is a polynomial of degree $(n-1)/2$ in $\sin^2(x)$, then remark that it has $\sin^2(\frac{2\pi j}{n})$ as roots, then comparing coefficients of $e^{i(n-1)/2}$ to get $(-4)^{(n-1)/2}$.
But I can't find useful form of polynomial to get roots, help please, other proof strategies are welcome.
 A: On the left you have an expression of the form
$$
\frac{z^n-z^{-n}}{z-z^{-1}}=z^{n-1}+z^{n-3}+z^{n-5}+...+z^{1-n}
$$
which divides perfectly and has the unit roots of degree 2n except $\pm 1$ as its roots. Because of symmetry, for every root $λ$ one has also $\bar λ$, $-λ$, $-\bar λ$ as roots. Because of $n$ odd the special case $λ=i$ does not occur, so that all roots can be grouped into these quadrupels. 
Instead of pairing conjugate roots one can here also pair mirror roots $λ$ and $-\bar λ$ relative to the imaginary axis resulting in 
\begin{align}
\frac{z^n-z^{-n}}{z-z^{-1}}
&=\prod_{k=1}^{(n-1)/2}(z-e^{i\pi \,k/n})(1+z^{-1}e^{-i\pi\, k/n})\;(z-e^{-i\pi \,k/n})(1+z^{-1}e^{i\pi\, k/n})\\
&=\prod_{k=1}^{(n-1)/2}(z-z^{-1}-e^{i\pi \,k/n}+e^{-i\pi\, k/n})(z-z^{-1}+e^{i\pi \,k/n}-e^{-i\pi\, k/n})\\
&=\prod_{k=1}^{(n-1)/2}(z-z^{-1}-2i\sin(\pi \,k/n))(z-z^{-1}+2i\sin(\pi \,k/n))
\end{align}
Replacing $z=e^{ix}$, collecting exponentials into sine terms and applying a binomial formula gives a formula similar to the stated result.
Check that also using $λ=e^{i2\pi\,k/n}$ gives one representant per root quadruple, here it is also important that $n$ is odd. This will finally result in the stated formula.
