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From this question Evaluation of a product of sines I allready know that: $$\prod_{k=1}^{n-1} \sin(\frac{k\pi}{n}) = \frac{n}{2^{n-1}}.$$

I am interested in a closed form for the following product: $$\prod_{k=2}^{n-1} \sin(\frac{n\pi}{k})$$ where $n$ is a fixed integer. Does one exist? My math abilities are sadly not sufficient to derive it from the linked question. The reason why I am interested in a closed form is that this product evaluates to 0 at composite numbers, as allready pointed out by @orion in the comments.

For illustration purposes: this is a graph of $f(x) = \prod_{k=2}^{n} \sin(\frac{\pi x}{k})$ with increasing $n$:

Illustration of formula

Judging by its looks, I doubt that a closed form exists. What we can say is that $f$ is periodic with period $n!$.

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SmallestUncomputableNumber
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  • i know nothing about this product,sorry – Dr. Sonnhard Graubner Oct 25 '18 at 16:43
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    Try: Using the fact that $\frac{e^{ix} - e^{-x}}{2i} = \sin(x)$, expand product, rewrite as a sum of $\sin$s – wlad Oct 25 '18 at 16:48
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    Also this: https://en.wikipedia.org/wiki/Orthogonal_functions may limit your ability to simplify the resulting expression beyond a point – wlad Oct 25 '18 at 16:52
  • Thanks for the interesting read. I'll try to grasp it – SmallestUncomputableNumber Oct 25 '18 at 16:55
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    If anything, it might relate to divisor functions and modular forms. For instance - it will be 0 for any nonprime $n$. Logarithms might help convert it into a sum (ugly sum, but it might be solvable with other standard tools - generating functions, Euler-Maclaurin and similar) – orion Oct 25 '18 at 22:14

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