TITLE: Simplify Product of sines
QUESTION [6 upvotes]: Is there a way simplify this product?
$$
\sin\left({n} \frac{\pi}{2}\right)  \sin\left({n} \frac{\pi}{3}\right) \sin\left({n} \frac{\pi}{4}\right) ...\sin\left({n} \frac{\pi}{n-1}\right) 
$$
And, is this the correct way to write it?
$$
 \prod_{m=2}^{n-1} \sin\left(n \frac{\pi}{m}\right)
$$
I'm not a professional so I'd appreciate a simple explanation.

REPLY [2 votes]: I don't think so.  Using de Moivre's formula $e^{ix} = \cos x + i \sin x$ your product is related to all the finite series:
$$ n\pi  \times \left( \pm \frac{1}{2} \pm \frac{1}{3} \dots \pm \frac{1}{n-1} \right)  \mod 1$$
Your product is a weighted sum of $e^{[\dots]}$ for all series of this kind, behaving like the Boltzmann partition function.  In fact the numbers in question wrap around the number like $[0,1]$ in a somewhat random fashion.

It's unlikely there is any simplification unless we try to estimate this series.
$$ \sum e^{n \pi i \cdot \left(  \pm \frac{1}{2} \pm \frac{1}{3} \dots \pm \frac{1}{n-1}  \right)}$$
That average is very likely to be close to $0$ since these numbers are equidistributed on the unit circle.

A histogram shows a little bit of variance, but basically the same idea:

It's very difficult to say how close to $0$ this result is.  However there may be classical results for estimating random-ish sums of this type.