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 $m$

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 $e^{2imx}-1=\Pi_{k=0,..m-1} {(e^{2ix}-e^{-\frac{2\pi ik}{m}})}$)

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.