Convergence of Product of SinesDate: 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/ |
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