Euler in the Product of a Regular Polygon's Diagonal LengthsDate: 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/ |
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