I am trying to find anything about the product $$\sin(t)\sin(2t)...\sin(nt)$$
The few things I have discovered
But I was unable to discover any details about the count and nature of those coefficients.
I need an efficient way to calculate this product. Any hint is appreciated.
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.
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 $n\to\infty$.
Here is a way for finding the power-series expansion of the expression to whatever order you want, though it gets more and more tedious the higher the order (and don't see why one would bother). This doesn't give a closed form for the coefficients for arbitrary large coefficients (which would probably be some horrible long and complicated expression), but we do find how fast the coefficients will grow with $n$.
Close to $x=0$ your product can be written as
$$f_n(x) = n! x^n\exp\left(-g_n(x)\right)\,\,\text{ where }\,\, g_n(x) \equiv -\sum_{i=1}^n \log\text{sinc}(ix)$$
Using the series expansion $$\log\text{sinc}(ix)= \sum_{m\geq 1}\frac{(-1)^{m+1}4^{m}B_{2m}}{2m(2m)!}i^{2m}x^{2m}$$ where $B_{2m}$ are the Bernoulli numbers. This gives us $$g_n(x) = \sum_{m\geq 1}\frac{4^{m}|B_{2k}|}{2m(2m)!}x^{2m}S_{2m}(n)\\ = \frac{1}{6}S_2(n)x^2 + \frac{1}{180}S_4(n)x^4 + \frac{1}{2835}S_6(n) x^6 + \ldots$$ where $S_k(n) = 1 + 2^k + 3^k + \ldots + n^k$ is the sum of the first $n$ powers. From this it's not hard to derive an expression for the first few terms in the power-series of $f_n(x)$. We only need to use the first few terms in the power-series of the exponential and the first few terms in $g_n(x)$ to get the first few terms in the power-series of $f_n(x)$:
$$f_n(x) = n! x^n - \frac{n!}{6}S_2(n)x^{2+n} + n!\left[\frac{1}{72}S_2^2(n) - \frac{1}{180}S_4(n)\right]x^{4+n} \\+ n!\left[\frac{1}{1080}S_2(n)S_4(n)- \frac{1}{2835}S_6(n)-\frac{1}{6^4}S_2^3(n)\right]x^{6+n} + \ldots$$
We can use $S_2(n) = \frac{n(n+1)(2n+1)}{6}$, $S_4(n) = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ and so on to simplify this. This gives for the first few coefficients
$$[x^n] = n!$$ $$[x^{n+2}] = -n!\frac{n(n+1)(2n+1)}{36}$$ $$[x^{n+4}] = n!\frac{n(n+1)(2n+1)(50 n^3 + 39 n^2 - 11 n + 12)}{64800}$$
Note that these are not always integers. It also follows that the coefficients grows with $n$ as $[x^{n+2m}] \sim \frac{S_2^m(n)}{m!6^m}\sim \frac{n! n^{3m}}{m!18^m}$.
