Euler in the Product of a Regular Polygon's Diagonal Lengths

Date: 04/06/2010 at 06:42:07
From: Robin
Subject: Why is sinpi/n.sin2pi/n....sin(n-1)pi/n=n/2^(n-1)

An n-sided regular polygon A1, A2, ... An is drawn on a circle of radius
1.

Numerically, the product A1A2 * A1A3 * ... * A1An seems to equal n, e.g.,

  2sin20 * 2sin40 * 2sin60 * ... 
              ... * 2sin160 = 9 (with error 1.6E-13 by calculator)
  
And I can prove the product of these polygon diagonals equals n for n = 3,
4, 5, 6, 8.

But there do not seem to be exact values for, e.g., sin20 = sin(pi/9), so
how can I get to a proof for any n? Is this a famous theorem?

Date: 04/08/2010 at 00:03:18
From: Doctor Vogler
Subject: Re: Why is sinpi/n.sin2pi/n....sin(n-1)pi/n=n/2^(n-1)

Hi Robin,

Thanks for writing to Dr. Math.

Here is a proof using complex numbers and Euler's equation, which is
described below, in case it's new to you:

  Imaginary exponents and Euler's equation.
    http://mathforum.org/dr.math/faq/faq.euler.equation.html 

You are trying to compute the product over 1 <= k <= n-1

    n-1
  product  2 * sin(k*pi/n).
    k=1

First of all, set

  z = exp(i*pi/n), or e^(i*pi/n).

Then we have

  z^k = cos(k*pi/n) + i*sin(k*pi/n)

and

  z^k - z^(-k) = 2*i*sin(k*pi/n).

Now define the polynomial

  f(x) = 1 + x + x^2 + ... + x^(n - 1)
       = (x^n - 1)/(x - 1).

This function is the product of x - r over its (n - 1) roots. And

  f(x) = 0 if x^n = 1,
  
but x is not 1.

Well, when x = z^(2k),

  x^n = (z^(2k))^n
      = exp(n*2k*i*pi/n)
      = exp(k*2*pi*i) = 1,

but z^(2k) is not 1 when 1 <= k <= n - 1.

So that means that

           n-1
  f(x) = product  (x - z^(2k)).
           k=1

Therefore, we have

           n-1
  f(1) = product  (1 - z^(2k)).
           k=1

But

  f(1) = 1^0 + 1^1 + 1^2 + 1^3 + ... + 1^(n - 1)
       = n.

Therefore, we have the equation

        n-1
  n = product  (1 - z^(2k))
        k=1

And we're almost there!

Now multiply both sides of the equation by

    n-1
  product i*z^(-k) = i^(n - 1)*z^-(1 + 2 + 3 + ... + (n - 1))
    k=1            = i^(n - 1)*z^(-n(n - 1)/2)

Use the fact that z^n = -1 to prove that this number is always 1, and then
conclude that ...

        n-1
  n = product -i*(z^k - z^(-k))
        k=1

... and therefore

        n-1
  n = product 2 * sin(k*pi/n)
        k=1

... just like you wanted.

If you have any questions about this or need more help, please write back
and show me what you have been able to do, and I will try to offer further
suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 04/08/2010 at 14:41:20
From: Robin
Subject: Thank you (Why is sinpi/n.sin2pi/n....sin(n-1)pi/n=n/2^(n-1))

Delighted with your rapid response -- very elegant! I love the use of
f(x).

Now I can complete my proof that

  INT (0 to infinity) 1/(1 + x^p)
    = pi/p/sin(pi/p)

for all real p > 1.

Since retiring from teaching, I have had more time to find out things I
should have learnt at university but was too busy (!) -- and nearly all of
them come back to Euler.

Is this proof down to Euler?

Thank you very much, indeed.

Robin Murphy, D.Phil, MA (Oxon)

Date: 04/11/2010 at 22:56:31
From: Doctor Vogler
Subject: Re: Thank you (Why is sinpi/n.sin2pi/n....sin(n-1)pi/n=n/2^(n-1))

Hi Robin,

Uh, I couldn't say who was the first to come up with this proof. I proved
it myself rather than finding it in a book, but I'm sure that I'm not the
first to do so.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/