Prove trigonometric relation

Question

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.

Answer 1

Consider

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

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

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

Considering

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

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.

Answer 2

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 $(2)$

Answer 3

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: $$ \sin{\pi\over n}\sin{2\pi\over n}\sin{3\pi\over n}\cdots\sin{(n-1)\pi\over n} = {n\over 2^{n-1}}. $$

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