Evaluating $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n}$ without using complex numbers

Question

I am looking to evaluate

$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n}$$

without using complex numbers. I can show the result if $n$ is a power of $2$, but if $n$ is anything else I reach a point where I can no longer proceed.

My current method (for when $n$ is a power of $2$) is to note that all of the angle that sin is being evaluated on are evenly distributed between $0$ and $\pi$, but more importantly, one of the angles will always be $\frac{\pi}{2}$ and all angles which are greater than $\frac{\pi}{2}$ can be re-written as the cos of an angle that is less than $\frac{\pi}{2}$ (using the complementary angle). When this is done, we get

$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \prod_{k = 1}^{\lfloor\frac{n - 1}{2}\rfloor}\frac{1}{2}\sin\frac{2k\pi}{n} = \frac{1}{2^{\lfloor\frac{n - 1}{2}\rfloor}}\prod_{k = 1}^{\lfloor\frac{n - 1}{2}\rfloor}\sin\frac{k\pi}{\frac{n}{2}}$$

This process can then be repeated recursively until we have just one $\sin$ term left in the product (which will be $\sin\frac{\pi}{2}$), then what is left is equivalent to

$$\frac{n}{2^{n - 1}}\;.$$

Can anyone suggest a way of generalizing this to when $n$ is not a power of $2$?

Thanks in advance.

Answer 1

I believe answer to your question is here on pages 128-130: https://sites.math.washington.edu//~billey/classes/lie/bulletins/damianou.2006.pdf. The proof is based on finding spectrum (set of all eigenvalues) of tridiagonal matrix with entries $-1,2,1$ and its determinant, which equals $n+1$ when size of matrix is $n$ (this is easy to show by induction and Laplace expansion). Then you need to know that product of all eigenvalues of matrix equals its determinant. All of this could be done without using complex numbers since eigenvalues of that tridiagonal matrix are real.

Answer 2

For any $z\in(0,1)$ we have $\sin(\pi z)=\frac{\pi}{\Gamma(z)\,\Gamma(1-z)}$, hence your question boils down to:

Q.: How can we prove the multiplication formula for the $\Gamma$ function using real-analytic tools?

The answer is given on PlanethMath: we may prove it through the functional relation $\Gamma(z+1)=z\,\Gamma(z)$ and Stirling's approximation. For an elementary proof of Stirling's approximation, have a look at this answer of mine: it can be done through creative telescoping and the above reflection formula.