50
$\begingroup$

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

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

Prove that

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

$\endgroup$
  • 18
    $\begingroup$ 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. $\endgroup$ – Hans Lundmark Oct 31 '10 at 14:38
  • 10
    $\begingroup$ 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 $\Gamma(x) \Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$. $\endgroup$ – Hans Lundmark Oct 31 '10 at 16:43
  • 3
    $\begingroup$ And another comment... I just ran into this on Wikipedia: en.wikipedia.org/wiki/Morrie%27s_law $\endgroup$ – Hans Lundmark Nov 4 '10 at 8:55
  • $\begingroup$ See also math.stackexchange.com/questions/2766630/… $\endgroup$ – lhf May 4 '18 at 16:33
  • $\begingroup$ 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. $\endgroup$ – DVD Jul 20 '18 at 21:38
57
$\begingroup$

$$P=\prod_{k=1}^{n-1}\sin(k\pi/n)=(2i)^{1-n}\prod_{k=1}^{n-1}(e^{ik\pi/n}-e^{-ik\pi/n})=(2i)^{1-n}e^{-i\pi n(n-1)/(2n)}\prod_{k=1}^{n-1}(e^{2ik\pi/n}-1)=$$ $$(-2)^{1-n}\prod_{k=1}^{n-1}(\xi^k-1)=2^{1-n}\prod_{k=1}^{n-1}(1-\xi^k),$$ 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 $P=n2^{1-n}$.

$\endgroup$
  • $\begingroup$ Can you explain a little bit more why $x^n-1=\prod_{k=0}^n (x-\xi^k)$? $\endgroup$ – Robert Smith Oct 30 '10 at 21:09
  • 1
    $\begingroup$ @Robert: well, the $\xi^k$ are the nth roots of unity... $\endgroup$ – J. M. is a poor mathematician Oct 30 '10 at 22:50
  • $\begingroup$ @J.M. Yes, I know. But I wanted to know about the validity of the equality. $\endgroup$ – Robert Smith Oct 30 '10 at 23:36
  • $\begingroup$ It's just an expansion of $x^n-1$ in terms of its zeroes, e.g. $x^2-1=(x+1)(x-1)$. $\endgroup$ – J. M. is a poor mathematician Oct 30 '10 at 23:43
  • $\begingroup$ @J.M: I think it should be $\prod_{k=0}^n (x-\xi^k)=(x-1)(x^{n}-1)$ instead of $\prod_{k=0}^n (x-\xi^k)=x^n-1$ $\endgroup$ – Robert Smith Oct 31 '10 at 0:55
8
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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

$\endgroup$

protected by user99914 Feb 10 '18 at 7:06

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.