Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$

Question

Using $\text{n}^{\text{th}}$ root of unity

$$\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$$

Prove that

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

Answer 1

$$ \begin{align*} P & = \prod_{k=1}^{n-1}\sin(k\pi/n) \\ & = (2i)^{1-n}\prod_{k=1}^{n-1}(e^{ik\pi/n}-e^{-ik\pi/n}) \\ & = (2i)^{1-n} e^{-i \frac{n(n-1)}{2}\frac{\pi}{n}} \prod_{k=1}^{n-1}(e^{2ik\pi/n}-1) \\ & = (-2)^{1-n}\prod_{k=1}^{n-1}(\xi^k-1) \\ & = 2^{1-n}\prod_{k=1}^{n-1}(1-\xi^k) \\ \end{align*} $$ where $\xi=e^{2i\pi/n}$.

Now note that $x^n-1=(x-1)\sum_{k=0}^{n-1}x^k$ and $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$.

Cancelling $(x-1)$ we have $\prod_{k=1}^{n-1} (x-\xi^k) =\sum_{k=0}^{n-1}x^k$. Substituting $x=1$ we have $\prod_{k=1}^{n-1} (1-\xi^k)=n$. $$ \therefore \boxed{P=n2^{1-n}}$$

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Edit:

In order to note that $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$, note that $1,\xi,\dots,\xi^{n-1}$ are roots of $x^n-1$. Therefore by polynomial reminder theorem we have $x^n-1=Q(x) \prod_{k=0}^{n-1} (x-\xi^k)$. Comparing degrees we find $Q(x)$ has degree $0$. Comparing highest coefficients we conclude $Q(x)=1$.

Edit:

We may instead use the identity $\left\lvert 1 - e^{2ik\pi/n} \right\rvert = 2\sin(k\pi/n), k = 1, ..., n - 1,$ to establish immediately that $P \equiv \prod_{k=1}^{n-1}\sin(k\pi/n)= 2^{1-n}\prod_{k=1}^{n-1}\left\lvert 1 - e^{2ik\pi/n} \right\rvert = 2^{1 - n}\left\lvert \prod_{k=1}^{n-1}(1 - e^{2ik\pi/n}) \right\rvert$, and continue by applying the foregoing logic to the product to obtain $P=n2^{1-n}$.

Answer 2

Consider $z^n=1$, each root is $$\xi_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n} = e^{i\frac{2k\pi}{n}}, k=0,1,2,...,n-1 $$ So, we have $$ z^n -1 = \prod_{k=0}^{n-1}(z-\xi_k)$$ $$\Longrightarrow (z-1)(z^{n-1}+...+z^2+z+1) = (z-\xi_0)\prod_{k=1}^{n-1}(z-\xi_k)$$ $$\Longrightarrow (z-1)(z^{n-1}+...+z^2+z+1) = (z-1)\prod_{k=1}^{n-1}(z-\xi_k)$$ $$\Longrightarrow z^{n-1}+...+z^2+z+1 = \prod_{k=1}^{n-1}(z-\xi_k)$$ By substituting z=1, $$\Longrightarrow n = \prod_{k=1}^{n-1}(1-\xi_k) $$

Next, take the modulus on both sides, $$ |n| = n = |\prod_{k=1}^{n-1}(1-\xi_k)| = \prod_{k=1}^{n-1}|(1-\xi_k)|$$ $$ 1 - \xi_k = 1-(\cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}) = 2\sin\frac{k\pi}{n}(\sin\frac{k\pi}{n} -i\cos\frac{k\pi}{n})$$ $$ |1 - \xi_k| = 2\sin\frac{k\pi}{n} $$ So, $$ n = 2^{n-1}\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}$$ $$\prod_{k=1}^{n-1}\sin\frac{k\pi}{n} = \frac{n}{2^{n-1}} $$

Answer 3

Here is a more "1st principles" pf. I use a hint in Marsden's book.

1st, $\cos(A-B)-\cos(A+B)=2\sin A \sin B$ (1), which follows by angle summation formulas.

Next, we use Marsden's hint to consider roots of $(1-z)^n-1$. These satisfy

$$(1-z)^n=1 \leftrightarrow (1-z) \in \left\{\cos \frac{2 \pi k}{n}+i \sin \frac{2\pi k}{n}:k=0,...,n-1 \right\}$$

(the set of nth roots of 1)

$$\leftrightarrow z \in \left\{z_k= 1-\cos \frac{2 \pi k}{n}-i \sin \frac{2\pi k}{n}:k=0,...,n-1\right\}\;\;\; (2)$$.
Since $z_0,....,z_{n-1}$ are the roots of $(1-z)^n-1$, we have by factorization that

$$(1-z)^n-1=\prod_{k=0}^{n-1}(z_k-z)=-z \prod_{k=1}^{n-1}(z_k-z) \;\;(3)$$ (since, by (2), $z_0=0$)

In (3), the LHS and RHS are polynomials in z. Equating the coeffs in front of z, we get

$$-n=-\prod_{k=1}^{n-1}z_k \leftrightarrow n=\prod_{k=1}^{n-1}z_k\,.$$

Note

$$\prod_{k=1}^{n-1} \bar{z}_k=\overline{\prod_{k=1}^{n-1}z_k}=n$$

(since $n\in \mathbb{R}$), so

$$\prod_{k=1}^{n-1}|z_k|^2=\prod_{k=1}^{n-1} z_k \bar{z}_k=\prod_{k=1}^{n-1} z_k \prod_{k=1}^{n-1} \bar{z}_k=n^2\;\; (4).$$

Next,

$$|z_k|^2=(1-\cos \frac{2 \pi k}{n})^2+ \sin^2 \frac{2\pi k}{n}=2(1-\cos \frac{2 \pi k}{n})$$

using this in (4) gives

$$2^{n-1} \prod_{k=1}^{n-1}(1-\cos \frac{2 \pi k}{n})=n^2\;\;(5)$$.
Next,

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

(where in the last 2 steps, we exploit that the order of taking a product doesn't matter)

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

(by (1))

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

(using $\cos (\pi -x)=-\cos x$)

$$=n^2 /2^{2(n-1)}\;.$$ Applying a sqrt to everything gives the desired result.

Answer 4

Although this is an old question, I share my answer since I found a very interesting combinatorial proof using Kirchhoff's theorem.

For a graph $G$ with $n$ vertices, Laplacian matrix is defined as $L=D-A$, where $D$ is a diagonal matrix with diagonal entries being degree of each vertex and $A$ is the adjacency matrix. Then, by definition, $(1,\ldots,1)^\top$ is an eigenvector of $L$ with eigenvalue $0$. For a connected graph $G$, it is well known that the other $n-1$ eigenvalues are all positive.

Kirchhoff's theorem says that the number of spanning trees of a connected graph $G$ with $n$ vertices is equal to the product of $n-1$ positive eigenvalues of the Laplacian of $G$ divided by $n$.

Now let's consider $G$ as the graph of $n$ vertices connected in a circular shape.

Then we have $L=\begin{pmatrix} 2 & -1 & 0 & \cdots & 0 &-1\\ -1 & 2 & -1 & \cdots &0 & 0\\ 0 & -1 & 2 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 2 & -1\\ -1 & 0 & 0 & \cdots & -1 & 2 \end{pmatrix}$.

Its eigenvectors $v_m (m=0, ..., n-1)$ have components $(v_m)_k=e^{\frac{2\pi i mk}{n}}$ and eigenvalues $\lambda_m=4\sin^2(\frac{m\pi}{n})$. Thus, if we apply Kirchhoff's theorem, we get $$\prod_{m=1}^{n-1} 4\sin^2(\frac{m\pi}{n}) = n^2.$$ By taking square root and dividing both sides by $2^{n-1}$, we get the desired formula.