Recently, I found this identities in a sheet of paper I was given as studying material: $$\prod^n_{k=1}\sin\left(\frac{k\pi}{2n+1}\right)=\frac{\sqrt{2n+1}}{2^n}\tag1$$ $$\prod^n_{k=1}\cos\left(\frac{k\pi}{2n+1}\right)=\frac{1}{2^n}\tag2$$ $$\prod^n_{k=1}\tan\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}\tag3$$

They were there as auxiliary identites, meaning that we could use them if we needed to. But I found them really interesting, so I tried to prove them. After some time struggling with equations, I managed to prove $(2)$ for $n=1,2,3$, with increasing difficulty using the sum of cosines formula, the product of cosines, and the formula for the arithmetic progression of cosines. I couldn't finish the proof for $n=4$, because I needed that: $$\left(2\cos\frac{\pi}{9}\right)\left(2\cos\frac{2\pi}{9}-1\right)=1\tag{4}$$ Equivalently, showing that $\cos\frac{\pi}{9}$ is a solution of $$8x^3-2x-1=0$$

And no idea with $(1)$ and $(3)$.The best I could think of is that since the square root can't appear out of the blue, considering the square and doing a clever trick that results in a sum that yields $2n+1$ in some way would do it. But I have had no luck until now. Any ideas?

Edit: Actually, the proof for $(4)$ wasn't that hard: $$2*2\cos\frac{\pi}{9}\cos\frac{2\pi}{9}-2\cos\frac{\pi}{9}=1$$ $$\iff 2*\left(\cos\frac{\pi}{3}+\cos\frac{\pi}{9}\right)-2\cos\frac{\pi}{9}=1$$ $$\iff 2*\left(\frac{1}{2}+\cos\frac{\pi}{9}\right)-2\cos\frac{\pi}{9}=1$$ The last one is clearly true.

  • $\begingroup$ Did you try writing $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$? That'll give you the $2^n$ in the denominator, and working on the numerator should become be a bit easier than on the $\sin$'s. $\endgroup$ – Clement C. Jan 24 '14 at 21:49
  • $\begingroup$ @ClementC. I don't know complex numbers (only the basics), but I am not really sure if this has an elementary proof since this wasn't an exercise. So if that results in a proof, it would be great too! $\endgroup$ – chubakueno Jan 24 '14 at 21:55
  • $\begingroup$ Look at the accepted answer of math.stackexchange.com/questions/55120/… $\endgroup$ – lab bhattacharjee Jan 27 '14 at 14:15


$$\prod_{k=1}^{2n} \sin \left(\frac{k\pi}{2n+1}\right).$$

By symmetry, it's the square of your first product, and expanding it with Euler's formula, we find

$$\begin{align} \prod_{k=1}^{2n} \sin \left(\frac{k\pi}{2n+1}\right) &= \frac{1}{(2i)^{2n}}\prod_{k=1}^{2n} \left(\exp \frac{k\pi i}{2n+1} - \exp \frac{-k\pi i}{2n+1}\right)\\ &= \frac{1}{(2i)^{2n}} \exp\left(\frac{\pi i (2n)(2n+1)}{2(2n+1)}\right) \prod_{k=1}^{2n}\left(1-\exp \frac{-2k\pi i}{2n+1}\right)\\ &= \frac{1}{2^{2n}}\prod_{k=1}^{2n}\left(1-\exp \frac{-2k\pi i}{2n+1}\right). \end{align}$$

Now consider the polynomial

$$P(X) = \prod_{k=1}^{2n} \left(X - \exp \frac{-2k\pi i}{2n+1}\right).$$

All its roots are $2n+1$-th roots of unity other than $1$, and they are all distinct, since

$$\frac{-k}{2n+1} - \frac{-j}{2n+1} \notin \mathbb{Z}$$

for $k\neq j$ and $1 \leqslant k,j \leqslant 2n$. So we have

$$P(X) = \frac{X^{2n+1}-1}{X-1} = X^{2n} + X^{2n-1} + \dotsc + X + 1,$$

and we see that $P(1) = 2n+1$, which proves the first equation.

The product of the cosines is similar, we get

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


$$Q(X) = \prod_{k=1}^{2n}\left(X + \exp \frac{-2k\pi i}{2n+1}\right) = P(-X),$$

we find $Q(1) = 1$, so

$$\prod_{k=1}^{2n} \cos \frac{k\pi}{2n+1} = \frac{(-1)^n}{2^{2n}}.$$

Since the factor for $k = 2n+1-j$ is the negative of the factor for $j$, $1 \leqslant j \leqslant n$, we have

$$\prod_{k=1}^n \cos \frac{k\pi}{2n+1} = \frac{1}{2^n}.$$

The formula for the product of the tangents is a direct consequence.

  • $\begingroup$ I missed that simmetry! Although couldn't have done this argument(but I understood it) because I didn't know that $\sin x=\frac{e^{ix}−e^{−ix}}{2i}$ previous to Clement's comment. I will see if I can make something useful out of it to do a purely real proof :) $\endgroup$ – chubakueno Jan 24 '14 at 22:44
  • $\begingroup$ If you find a purely real proof, please ping me, I'd be interested to see one. $\endgroup$ – Daniel Fischer Jan 24 '14 at 22:46
  • $\begingroup$ Done! (Of course, assuming my proof isn't wrong) $\endgroup$ – chubakueno Jan 25 '14 at 7:26

Here is an elementary proof of $(2)$ that can be extended to $(1)$ with some work: First, since $\cos nx=2\cos x\cos((n-1)x)-\cos((n-2)x)$, by induction $\cos nx$ is always a polynomial of $y=\cos x$ with degree $n$ and the same parity as $n$. Then, we see that the coefficient of the highest term is twice the previous one, so by another induction the coefficient of $y^n$ in $\cos nx$ is $2^{n-1}$. Let's consider the zeroes of $f(x)=\cos((2n+1)x)-1$. Since we know that $\cos((2n+1)x)$ is an odd polynomial of $\cos x$, it does not have a constant term. So the constant term of $f$ is $-1$. So by Vieta

$$\prod^{2n}_{k=0} \cos\frac{2k\pi}{2n+1} = \frac{1}{2^{2n}}$$

Taking advantage of the simmetry $\cos x = -\cos(x-\pi)=-\cos(\pi-x)$ and forgetting $k=0$, we get $$(-1)^{n}\prod^{2n}_{k=1} \cos\frac{k\pi}{2n+1}=2^{-2n}$$ By the same symmetry, we have $$\left(\prod^{n}_{k=1}\cos\frac{k\pi}{2n+1}\right)^2=2^{-2n}$$

Every term inside the parenthesis is positive , hence $(2)$

  • $\begingroup$ Nice. I would think that also for polynomials in $\cos x$, the usual term is "constant term", not "free term". That's probably a direct translation from how it's called in your native language? $\endgroup$ – Daniel Fischer Jan 25 '14 at 10:14
  • $\begingroup$ @DanielFischer Kind of, yes. Thank you for the correction, I also fixed some typos I made. This was written without preview in an old small cellphone, so I would say it went better than expected. $\endgroup$ – chubakueno Jan 25 '14 at 15:24
  • 1
    $\begingroup$ Without preview from a phone? Awesome typing! $\endgroup$ – Daniel Fischer Jan 25 '14 at 15:29

Interesting fact used in the answer of Daniel Fischer: the complex numbers $\cos{k\pi\over n}+i\sin{k\pi\over n}$, $k=1,\cdots n$ are the complex roots of unit (roots of the polynomial $z^n-1$. Many similar formulas can be proved using this. Two examples: $$ \sin{\pi\over n}\sin{2\pi\over n}\sin{3\pi\over n}\cdots\sin{(n-1)\pi\over n} = {n\over 2^{n-1}}. $$

$$ 1+\cos\theta+\cos 2\theta+\cdots+\cos n\theta = {1-\cos\theta+\cos n\theta-\cos(n+1)\theta\over 2-2\cos\theta}. $$

  • $\begingroup$ Thank you, for the reference, I will try to prove them on my own. The second one is not that hard though, if you pass the denominator to multiply you get a nice telescopy :) $\endgroup$ – chubakueno Jan 26 '14 at 1:13

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