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: 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}
A: The left hand side is a Chebyshev polynomial of the second kind in $\cos(x)$ (by definition) - the Wikipedia page tells you how to express it as a polynomial. You get the leading coefficient from the recurrence formula for the Chebyshev polynomials (also on the wikipedia page). Serre is trying to make you work, but it is really not necessary.
