Showing ${\sin mx\over\sin x}=(-4)^{(m-1)/2}\prod_{j=1}^{(m-1)/2}\left(\sin^2x-\sin^2{2\pi j\over m}\right)$ for odd $m>0$ (from Serre's "Arithmetic") On pages 9 and 10 of Serre's A Course in Arithmetic, there is a lemma (towards a proof of the quadratic reciprocity law) stating that
for any positive odd integer $m$,
$${\sin mx\over\sin x}=(-4)^{(m-1)/2}\prod_{j=1}^{(m-1)/2}\left(\sin^2x-\sin^2{2\pi j\over m}\right)$$
Serre claims that this result is "elementary", saying that one should start by proving that $\sin(mx)/\sin(x)$ is a polynomial of degree $(m-1)/2$ in $\sin^2 x$ with $(m-1)/2$ roots given by $\sin^2(2\pi j/m)$ for $1\le j\le m$, then compare coefficients of
$e^{i(m-1)x}$ on both sides to get the factor of $(-4)^{(m-1)/2}$.
I was not able to follow this outline, as I'm not exactly sure which sine identity to start with. I suspect Serre would like me to expand
$${\sin mx\over \sin x} = {e^{imx}-e^{-imx}\over e^{ix} - e^{-ix}},$$
since he mentions coefficients of $e^{i(m-1)x}$. So I was thinking of performing induction over all odd $m$ (the case $m=1$ is easy), but have not been able to get the computations right. If anyone would be able to point me in the right direction, I'd very much appreciate it!

A direct proof.$\renewcommand\Re{\operatorname{Re}}$
Let $z=e^{-2ix}$ and $w=e^{2\pi i/m}$.
First note that $z\bar z=w\bar w=1$ and
\begin{align}
\{w^k:1\leq k\leq m-1\}
&=\{w^k:1\leq k\leq(m-1)/2\}\cup\{w^{-k}:1\leq k\leq(m-1)/2\}\\
&=\{w^{2k}:1\leq k\leq(m-1)/2\}\cup\{w^{-2k}:1\leq k\leq(m-1)/2\}
\end{align}
Then
\begin{align}
{\sin mx\over \sin x}
&= {e^{imx}-e^{-imx}\over e^{ix} - e^{-ix}}\\
&=e^{i(m-1)x} {1-e^{-2imx}\over 1 - e^{-2ix}}\\
&=e^{i(m-1)x}\frac{z^m-1}{z-1}\\
&=e^{i(m-1)x}\prod_{k=1}^{m-1}(z-w^k)\\
&=e^{i(m-1)x}\prod_{k=1}^{(m-1)/2}(z-w^{2k})(z-\bar w^{2k})\\
&=e^{i(m-1)x}\prod_{k=1}^{(m-1)/2}(z^2-2z\Re(w^{2k})+1)\\
&=\prod_{k=1}^{(m-1)/2}\bar z(z^2-2z\Re(w^{2k})+1)\\
&=\prod_{k=1}^{(m-1)/2}(z-2\Re(w^{2k})+\bar z)\\
&=\prod_{k=1}^{(m-1)/2}(2\Re(z)-2\Re(w^{2k}))\\
&=\prod_{k=1}^{(m-1)/2}2(\cos(2x)-\cos(4\pi k/m))\\
&=\prod_{k=1}^{(m-1)/2}(-4)(\sin^2(x)-\sin^2(2\pi k/m))\\
&=(-4)^{(m-1)/2}\prod_{k=1}^{(m-1)/2}(\sin^2(x)-\sin^2(2\pi k/m))
\end{align}
