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Dear Student,
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),
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Plz Approve the answer...!!!!
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Aman Bansal
Askiitian Expert
The answer is ZERO!!
As the series is in increasing order... therefore, there is sin90° in between... which is 0 and multiplying with 0 is alway 0...
plz approve my answer!! :)
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