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

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

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

Prove that

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

• Incidentally, the proof given in fiktor's answer below can be modified to show that $\sin nx=2^{n-1}\prod_{k=0}^{n-1} \sin\left( x + \frac{k\pi}{n} \right)$, a very pretty multiple-angle identity which is not as widely know as it deserves to be. Dividing by $\sin x$ and letting $x\to 0$ reduces that identity to the one in the question. – Hans Lundmark Oct 31 '10 at 14:38
• And here's a kill-a-mosquito-with-a-cannon proof of the identity in my previous comment: combine Gauss's multiplication formula for the gamma function, $\Gamma(nx) = \frac{n^{nx-1/2}}{(2\pi)^{(n-1)/2}} \prod_{k=0}^{n-1} \Gamma(x+\frac{k}{n})$, with Euler's reflection formula $\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$. – Hans Lundmark Oct 31 '10 at 16:43
• And another comment... I just ran into this on Wikipedia: en.wikipedia.org/wiki/Morrie%27s_law – Hans Lundmark Nov 4 '10 at 8:55
• – lhf May 4 '18 at 16:33
• One can also construct a simple three-diagonal matrix with the eigenvalues $(2\sin\frac{\pi k}{n})^2$ and express the product in terms of determinants of its minors. – DVD Jul 20 '18 at 21:38

$$P=\prod_{k=1}^{n-1}\sin(k\pi/n)=(2i)^{1-n}\prod_{k=1}^{n-1}(e^{ik\pi/n}-e^{-ik\pi/n})=(2i)^{1-n}e^{-i\pi n(n-1)/(2n)}\prod_{k=1}^{n-1}(e^{2ik\pi/n}-1)=$$ $$(-2)^{1-n}\prod_{k=1}^{n-1}(\xi^k-1)=2^{1-n}\prod_{k=1}^{n-1}(1-\xi^k),$$ where $\xi=e^{2i\pi/n}$. Now note, that $x^n-1=(x-1)\sum_{k=0}^{n-1}x^k$ and $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$, thus 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 $P=n2^{1-n}$.

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 of L.H.S. and R.H.S. we can find, that $Q(x)$ has degree $0$. Comparing highest coefficients we can 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 $P=n2^{1-n}$.

• Can you explain a little bit more why $x^n-1=\prod_{k=0}^n (x-\xi^k)$? – Robert Smith Oct 30 '10 at 21:09
• @Robert: well, the $\xi^k$ are the nth roots of unity... – J. M. is a poor mathematician Oct 30 '10 at 22:50
• @J.M. Yes, I know. But I wanted to know about the validity of the equality. – Robert Smith Oct 30 '10 at 23:36
• It's just an expansion of $x^n-1$ in terms of its zeroes, e.g. $x^2-1=(x+1)(x-1)$. – J. M. is a poor mathematician Oct 30 '10 at 23:43
• @J.M: I think it should be $\prod_{k=0}^n (x-\xi^k)=(x-1)(x^{n}-1)$ instead of $\prod_{k=0}^n (x-\xi^k)=x^n-1$ – Robert Smith Oct 31 '10 at 0:55

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.

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

## protected by user99914 Feb 10 '18 at 7:06

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