Investigating the product $$\prod_{k=1}^{m}\sin(kt)$$ And became to the following enough trivial identity, but have a feeling that I am missing something here. Any hint about the following is appreciated.
\begin{align} \sum _{k=1}^m \log \sin (k t) &= \sum _{k=1}^m \left(-\sum _{n=1}^{\infty } \frac{\cos (2 k n t)}{n}-\log 2\right)=\\ &=-\sum _{n=1}^{\infty } \sum _{k=1}^m \frac{\cos (2 k n t)}{n}-m \log 2=\\ &=-\sum _{n=1}^{\infty } \frac{\csc (n t) \sin (m n t) \cos ((m+1) n t)}{n}-m \log 2 \end{align}
So the final result would be: $$\prod_{k=1}^{m}\sin(kt)=2^{-m} \prod _{k=1}^{\infty } \exp\left(-\frac{\csc (k t) \sin (k m t) \cos (k (m+1) t)}{k}\right)$$
Your final answer is true for some intervals of the real line, but not for all $t \in \mathbb{R}$. The problem is that $\ln(\sin(kt)) \not= -\sum _{n=1}^{\infty } \frac{\cos (2 k n t)}{n}-\log 2$ when $\sin(kt) < 0$. What is always true is $\ln\left|\sin(kt)\right| = -\sum _{n=1}^{\infty } \frac{\cos (2 k n t)}{n}-\log 2$. You can also see that the LHS of the product would be negative for some $t$, while the RHS would always be positive.
The identity would be right if it was instead $$\prod_{k=1}^{m}\left|\sin(kt)\right|=\left|\prod_{k=1}^{m}\sin(kt)\right|=2^{-m} \prod _{k=1}^{\infty } \exp\left(-\frac{\csc (k t) \sin (k m t) \cos (k (m+1) t)}{k}\right)$$