# Prove $\prod_{i=1}^{n-1} \sin(i\pi/n) = 2^{1-n} n$ without complex functions.

Note that $i$ here refers to indexing variable, not $\sqrt{-1}$.

$$\prod_{i=1}^{n-1} \sin\left(\frac{i \pi}{n}\right) = 2^{1-n} n$$ This formula was used here to give an 'elementary' proof of product of diagonals = N. Mathworld is the only place I can find the formula and that website cites a personal communication with 'T. Drane'

I have tried to prove it on my own but this is the only progress I've made.

Letting $n=2k+1$, I can show that

$$\prod_{i=1}^{2k} \sin\left(\frac{i \pi}{2k+1}\right) = \prod_{i=1}^{k} \cos^2\left(\frac{(2i-1) \pi}{4k+2}\right)$$

I used the sine product formula $\quad\sin(a)\sin(b)=\dfrac{\cos(a-b)-\cos(a+b)}{2}\quad$ to pair of the first/last terms, etc. Then used the cosine double-angle formula $\quad \cos(2a)+1 = 2\cos^2(a)\quad$ on the resulting product.

I don't know where to go on from on here and I was just investigating the special case of odd $n$. Any ideas on what to do next? Simpler approach that would also apply to $n=2k$?

PS: It would be preferred if the proof avoided complex functions in order to complete the answer to the diagonals question reference earlier.

• Both answers use complex functions. – genepeer Jul 2 '14 at 12:29
• @amWhy how has this been answered?! I specifically asked for no complex functions. – genepeer Jul 2 '14 at 17:04
• Have you looked at Chebyshev polynomials? They might help you. The idea I'm suggesting (for odd $n$) is to consider the polynomial $R_{n-1}(u)$ of degree $n-1$ defined by the equation $$\sin nx=\sin x R_{n-1}(\sin u).$$ For example, when $n=3$ we have the formula $$\sin 3x=3\sin x-4\sin^3x,$$ so $R_2(u)=3-4u^2$. Anyway, the idea is that the zeros of $R_{n-1}(u)$ are clearly $u_k=\sin(k\pi/n)$ for $k=1,2,\ldots,n-1$. Therefore, if $$R_{n-1}(u)=a_0+\cdots+a_{n-1}u^2,$$ we have $$u_1u_2\cdots u_{n-1}=(-1)^{n}a_0/a_{n-1}.$$ – Jyrki Lahtonen Jul 2 '14 at 18:12
• (cont'd) Known facts about Chebyshev polynomials probably imply that $a_0=n$ and $a_{n-1}=-2^{n-1}$ giving your claim. I'm afraid I'm too rusty on Chebyshev stuff to dig all that out. The formulas about Chebyshev polynomials can be proven without resorting to complex exponential. But I don't see right away how to handle the even case... – Jyrki Lahtonen Jul 2 '14 at 18:17
• Also, $u_1 u_2 \cdots u_{n-1} = (-1)^{n-1} a_0/a_{n-1}$. But I understood what you meant :) – genepeer Jul 2 '14 at 19:01

Thanks to Lahtonen's suggestion to use Chebyshev's polynomials, I was able to answer the question. I used this paper as a reference on the properties of the polynomials.

I'll prove $\frac{\sin((n+1)\theta)}{\sin(\theta)} = U_n(\cos(\theta))$ using induction (the paper used complex numbers to prove it).

\frac{\sin(\theta)}{\sin(\theta)} = 1 = U_0(\cos(\theta)), \qquad \frac{\sin(2\theta)}{\sin(\theta)} = 2\cos(\theta) = U_1(\cos(\theta)), \\ {~}\\ {~}\\ \begin{aligned} \frac{\sin((n+1)\theta)}{\sin(\theta)} &= 2\cos(\theta)\frac{\sin(n\theta)}{\sin(\theta)} - \frac{\sin((n-1)\theta}{\sin(\theta)} \\ &= 2\cos(\theta)U_{n-1}(\cos(\theta))-U_{n-2}(\cos(\theta)) \\ &= U_n(\cos(\theta)) \end{aligned}

Using Identities 7/9 in the linked paper, it followed that $$\sin((2m+1)\theta) = (-1)^mT_{2m+1}(\sin(\theta))$$

Since $T_{2m+1}$ is an odd polynomial, I can factor it as such: $$\sin((2m+1)\theta) = \sin(\theta)R_{2m}(\sin(\theta))$$

where $R_{2m}$ is an even polynomial of degree $2m$ with the following properties:

• unique roots are $u_k = \sin\left(\frac{k\pi}{2m+1}\right)$ for $k =\pm 1, \ldots, \pm m$.
• constant term is $2m+1$
• leading coefficient is $(-1)^m 2^{2m}$

Therefore, $$\prod_{k=1}^m u_k u_{-k} = (-1)^m \frac{2m+1}{2^{2m}}$$

By symmetry properties of the sine function, $$\prod_{k=1}^{2m} \sin\left(\frac{k\pi}{2m+1}\right) = \frac{2m+1}{2^{2m}} \qquad \square$$

This proves the identity for when $n$ is an odd integer. I'll introduce an abbreviation for simplicity's sake: let $F(n) = \prod_{k=1}^{n-1} \sin(k\pi/n)$. We have shown that $F(2m+1) = 2^{-2m}(2m+1)$ Also note that $$F(2m+1) = \prod_{k=1}^{m} \sin^2\left(\frac{k\pi}{2m+1}\right) = \prod_{k=1}^{m} \cos^2\left(\frac{(2m+1-2k)\pi}{4m+2}\right)$$ This will be useful soon. Now, let $n=2m$. $$F(2m) = \prod_{k=1}^{m-1} \sin^2\left(\frac{k\pi}{2m}\right) = \prod_{k=1}^{m-1} \cos^2\left(\frac{(m-k)\pi}{2m}\right) = \prod_{k=1}^{m-1} \cos^2\left(\frac{k\pi}{2m}\right)$$

I'll need to split this in two cases again: $n=4m$ and $n=4m+2$.

\begin{aligned} F(4m+2) &= \prod_{k=1}^{2m} \cos^2\left(\frac{k\pi}{4m+2}\right) \\ &= \prod_{k=1}^m \left(\cos\left(\frac{k\pi}{4m+2}\right)\cos^2\left(\frac{(2m+1-k)\pi}{4m+2}\right) \right)^2 \\ &= \prod_{k=1}^m \frac{1}{4} \cos^2 \left(\frac{(2m+1-2k)\pi}{4m+2}\right) \\ &= \frac{1}{2^{2m}}F(2m+1) \\ &= \frac{1}{2^{2m}}\cdot\frac{2m+1}{2^{2m}} &\square \end{aligned}

and finally, assume that $F(n)=\frac{n}{2^{n-1}}$ for all $n<4m$,

\begin{aligned} F(4m) &= \prod_{k=1}^{2m-1} \cos^2\left(\frac{k\pi}{4m}\right) \\ &= \frac{1}{2}\prod_{k=1}^{m-1} \left(\cos\left(\frac{k\pi}{4m}\right)\cos\left(\frac{(2m-k)\pi}{4m}\right) \right)^2 \\ &= \frac{1}{2}\prod_{k=1}^{m-1} \frac{1}{4}\cos^2\left(\frac{(m-k)\pi}{2m}\right) \\ &= \frac{1}{2^{2m-1}} F(2m) \\ &= \frac{1}{2^{2m-1}}\cdot \frac{2m}{2^{2m-1}} & \blacksquare \end{aligned}