Sine SeriesDate: 09/09/98 at 15:12:12 From: Daniel Meyer Subject: Sine series We were given a sine series: sin1*sin2*sin3***sin179. These are in degrees. We are to determine what this equals. The only progress I have made is recognizing that sin1 = sin179, sin2 = sin178, etc. Thank you for your help. Daniel Date: 05/21/2006 at 17:33:32 From: Doctor Pete Subject: Re: Sine series The general product is P[n] = Product[Sin[k Pi/n], {k,1,n-1}] = n/2^(n-1), and the special case is P[180] = 180*2^(-179) To prove this, let v = Exp[Pi i/n] be a primitve (2n)th root of unity. Then we have Sin[k Pi/n] = (v^k - v^-k)/(2i) = (1 - v^(-2k))(v^k/(2i)). Therefore P[n] = Product[1 - v^(-2k), {k,1,n-1}] Product[v^k/(2i), {k,1,n-1}]. But the first product is simply the factorization of (z^(2n) - 1)/(z^2 - 1) = z^(2n-2) + z^(2n-4) + ... + 1 with z = 1. Consequently, this first product has value n. The second product is found by summing the exponents, giving P[n] = n v^Sum[k, {k,1,n-1}]/(2i)^(n-1) = n v^((n-1)n/2) /(2i)^(n-1) = n Exp[Pi i (n-1)/2]/(2 Exp[Pi i/2])^(n-1) = n/2^(n-1) Exp[Pi i (n-1)/2 - Pi i(n-1)/2] = n/2^(n-1), and the claim is proven. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/ |
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