Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

Prove that

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

share|cite|improve this question
13  
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
8  
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
2  
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
up vote 36 down vote accepted

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

share|cite|improve this answer
    
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. 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. 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

$$\begin{align*}=\prod_{k=1}^{n−1}\sin\frac{kπ}n&=(2i)^{1−n}\cdot\prod_{k=1}^{n−1}\left(e^{^\tfrac{ikπ}n}−e^{^{−\tfrac{ikπ}n}}\right)\\&=(2i)^{1−n}\cdot e^{^{−\tfrac{iπn(n−1)}{2n}}}\cdot\prod_{k=1}^{n−1}\left(e^{^\tfrac{2ikπ}n}−1\right)\\&=(−2)^{1−n}\cdot\prod_{k=1}^{n−1}\left(ξ^k−1\right)=2^{1−n}\cdot\prod_{k=1}^{n−1}\left(1−ξ^k\right),\qquad\text{where}\qquad ξ=e^{^\tfrac{2iπ}n}\end{align*}$$

Now, note that $\quad x^n−1=(x−1)\displaystyle\sum^{n−1}_{k=0}x^k,\quad$ and $\quad x^n−1=\displaystyle\prod^{n−1}_{k=\color{blue}0}\left(x−ξ^k\right).\quad$

Thus, cancelling $x−1$, we have $\quad\displaystyle\prod^{n−1}_{k=\color{blue}1}\left(x−ξ^k\right)=\sum^{n−1}_{k=0}x^k.\quad$

By substituting $x=1$ we have $\quad\displaystyle\prod^{n−1}_{k=1}\left(1−ξ^k\right)=n$. Therefore, $P=n\cdot2^{1−n}$

share|cite|improve this answer
2  
How is this different from the answer posted 30 October 2010 by @Fiktor ????? – Dr. MV Jul 6 '15 at 18:28
    
This answer looks like a perfect copy of the first, accepted one and almost three years later. Even with the same symbols, language... – Joanpemo Apr 6 at 19:29

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.

share|cite|improve this answer
    
can I upvote to my own answer? – GA316 Feb 24 at 7:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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