Convergence of Product of Sines

Date: 10/17/2003 at 06:54:51
From: Vitor
Subject: Imaginary/Complex numbers

How can we prove that, for any n >= 2,

 (sin(pi/n))*(sin(2pi/n))*...*(sin((n-1)pi/n)) = n/(2^(n-1))?

Date: 10/17/2003 at 09:46:52
From: Doctor Anthony
Subject: Re: Imaginary/Complex numbers

Hello, Vitor. 

Consider factors of x^(2n) - 2x^n*cos(na) + 1 = 0  which is a 
quadratic in x^n with roots cos(na) +/- i*sin(na).

The 2n values of x are  

  x = cos(a+2k*pi/n) +/- i*sin(a+2k*pi/n)

with k ranging from 0 to n-1.

With the product of the factors with the particular value k=r we get 

  x - [cos(a+2r*pi/n) + i*sin(a+2r*pi/n)]  times

  x - [cos(a+2r*pi/n) - i*sin(a+2r*pi/n)]  which leads to

  x^2 - 2x*cos(a+2r*pi/n) + 1

It follows that 

  x^(2n) - 2x^n*cos(na) + 1 

            = Product(k=0 to n-1)[x^2 - 2x*cos(a+2k*pi/n) + 1]

and putting x=1, a = 2b in this we get
 
  sin(nb) = 2^(n-1) * Product(0 to n-1)[sin(b+k*pi/n]

              [This is a standard formula that can be used 
              in many such problems.]

  sin(nb) =   2^(n-1)
            * sin(b)
            * sin(b+pi/n)
            * sin(b+2pi/n) 
            * ... 
            * sin(b+(n-1)pi/n)


  sin(nb)
  ------- =   2^(n-1)
  sin(b)    * sin(b+pi/n)
            * sin(b+2pi/n)
            * ...
            * sin(b+(n-1)pi/n)
  
As b->0 we obtain

  sin(nb)
  ------- =   2^(n-1)
  sin(b)    * sin(pi/n)
            * sin(2pi/n)
            * ...
            * sin((n-1)pi/n)

We must evaluate sin(nb)/sin(b) as b -> 0.  Using l'Hospital's rule, 
we obtain 

    n*cos(nb)/cos(b) = n   as b-> 0

and so the product of the sines

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


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