Establishing $\frac{ \sin mx}{\sin x}=(-4)^{(m-1)/2}\prod_{1\leq j\leq(m-1)/2}\left(\sin^2x-\sin^2\frac{2\pi j}{m}\right) $ for odd $m$ 
Let $m$ be an odd positive integer. Prove that
$$ \dfrac{ \sin (mx) }{\sin x } = (-4)^{\frac{m-1}{2}} \prod_{1 \leq j
 \leq \frac{(m-1)}{2} }  \left( \sin^2 x - \sin^2 \left( \dfrac{ 2 \pi
 j }{m } \right) \right) $$

Atempt to the proof
My idea is to use induction on $m$. The base case is $m=3$ and we obtain
$$ \dfrac{ \sin (3x) }{\sin x } = (-4) ( \sin^2 x - \sin^2 (2 \pi /3 ) ) $$
and this holds if one uses the well known $\sin (3x) = 3 \sin x - 4 \sin^3 x $ identity.
Now, if we assume the result is true for $m = 2k-1$, then we prove it holds for $m=2k+1$. We have
$$ \dfrac{ \sin (2k + 1) x }{\sin x } = \dfrac{ \sin [(2k-1 + 2 )x] }{\sin x } = \dfrac{ \sin[(2k-1)x ] \cos (2x) }{\sin x } + \dfrac{ \cos [(2k-1) x ] \sin 2x }{\sin x }  $$
And this is equivalent to
$$ cos(2x) \cdot (-4)^{k-1} \prod_{1 \leq j
 \leq k-1  }\left( \sin^2 x - \sin^2 \left( \dfrac{ 2 \pi
 j }{m } \right) \right) + 2 \cos [(2k-1) x ] \cos x $$
Here I dont see any way to simplify it further. Am I on the right track?
 A: Note that $\sin (x-\frac{2\pi j }{m})=-\sin(x+\frac{(m-2j)\pi}{m})$ and $m-2j$ goes through the odd numbers $1,...m-2$ when $ 1\le j \le \frac{m-1}{2}$
By the paralelogram rule for sine $\sin^2 x- \sin^2 y=\sin(x-y)\sin(x+y)$ so we get that the RHS product 
$P=\sin x  \prod_{1 \leq j
 \leq \frac{(m-1)}{2} }  \left( \sin^2 x - \sin^2 \left( \dfrac{ 2 \pi
 j }{m } \right) \right)=(-1)^{\frac{m-1}{2}}\prod_{0 \leq j
 \leq m-1}\sin (x+\frac{j\pi}{m})=$
$=(-1)^{\frac{m-1}{2}}2^{-(m-1)}\sin mx$ by the classic product formula, so we are done!
(the product formula is obtained by taking the imaginary part of both sides in $e^{2imx}-1=\Pi_{k=0,..m-1} {(e^{2ix}-e^{-\frac{2\pi ik}{m}})}$)
A: Too long for a comment: Look up Chebyshev polynomial of second kind. They are literally what you are dealing with:
$$U_n(\cos x) = \dfrac{\sin((n+1)x)}{\sin x}.$$
Your attempt at induction basically reduces it to an equivalent problem that uses the first kind of Chebyshev polynomials, so I would not be fixated on inductive approach. 
