Simplification of $\prod\limits_{i=1}^{n}{\sin(it)}$

In terms of the q-Pochhammer symbol, we have

\begin{align}\prod_{k=1}^n\sin(kt)&=\prod_{k=1}^n\frac{e^{ikt}-e^{-ikt}}{2i}\\&=\frac{e^{in(n+1)t/2}}{(2i)^n}\prod_{k=1}^n(1-e^{-2ikt})\\&=\frac{e^{in(n+1)t/2}}{(2i)^n}(e^{-2it};e^{-2it})_n\end{align}

from which one can produce various identities.

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As $n\to\infty$, one can show that the product tends to $0$. We can bound how fast it goes to zero by considering the following:

$$|\sin(x)\sin(x+t)|\le\max\{\sin^2(t/2),\cos^2(t/2)\}$$

which gives us:

\begin{align}\left|\frac{\sin((n+1)t)}{\sin(t)}\right|\prod_{k=1}^n\sin^2(kt)&=\prod_{k=1}^n|\sin(kt)\sin(kt+t)|\\&\le\prod_{k=1}^n\max\{\sin^2(t/2),\cos^2(t/2)\}\\&=\max\{\sin^{2n}(t/2),\cos^{2n}(t/2)\}\end{align}

and thus,

$$\prod_{k=1}^n|\sin(kt)|\le\alpha^n\cdot\sqrt{\left|\frac{\sin(t)}{\sin((n+1)t)}\right|}$$

where $\alpha=\max\{|\sin(t/2)|,|\cos(t/2)|\}$, provided that $\sin(kt)\ne0$ for any natural $k$. In the event that $\sin(t/2)$ or $\cos(t/2)$ are $1$, then the product trivially reduces down to $0$. Since the irrationality measure of $\pi$ has an upper bound of $7.6063$, we have

$$\prod_{k=1}^n|\sin(kt)|\in\mathcal O\left(n^{-3.3031}\alpha^n\right)$$

as $n\to\infty$.