Trigonometry Without CalculatorsDate: 01/06/97 at 20:15:08 From: Sriram Gollapalli Subject: Cosine 40 Hi! I was wondering how to find the cosine of 40 degrees in rational form and without a calculator. Thanks, Sriram Date: 01/07/97 at 07:15:23 From: Doctor Pete Subject: Re: Cosine 40 Hi, I am not quite sure what you mean by "in rational form," because Cos[40] is not a rational number. If you wish to find an equivalent expression for it, for example, the way you can by saying that Cos[60] = 1/2, or Sin[45] = 1/Sqrt[2], then this is also not quite as simple as it may seem. In any case, consider the double-angle formula: Cos[2x] = Cos[x]^2 - Sin[x]^2 = Cos[x]^2 - (1-Cos[x]^2) = 2 Cos[x]^2 - 1. Then Cos[40] = 2 Cos[20]^2 - 1. To find Cos[20], we note: Cos[3x] = Cos[x+2x] = Cos[x]Cos[2x] - Sin[x]Sin[2x] = Cos[x](Cos[x]^2-Sin[x]^2) - Sin[x](2 Sin[x]Cos[x]) = Cos[x]^3 - 3 Cos[x]Sin[x]^2 = Cos[x]^3 - 3 Cos[x](1-Cos[x]^2) = Cos[x]^3 - 3 Cos[x] + 3 Cos[x]^3 = 4 Cos[x]^3 - 3 Cos[x]. Therefore, Cos[60] = 4 Cos[20]^3 - 3 Cos[20] 1/2 = 4 Cos[20]^3 - 3 Cos[20] Hence Cos[20] is the real root of the irreducible cubic x^3 - 3/4 x - 1/8 = 0 This has the real root: r = (a^(1/3) + b^(1/3))/2 where a = (1+Sqrt[-3])/2, b = (1-Sqrt[-3])/2 are complex conjugates. Hence r is the value of Cos[20], and: Cos[20]^2 = (a^(2/3) + b^(2/3))/4 + 1/2 so Cos[40] = 2*Cos[20]^2 - 1 = (a^(2/3) + b^(2/3))/2, where a and b are given as above. Note that this alternative expression for Cos[40] is indeed a real number, because a and b are complex conjugates, and if the appropriate cube root is taken, these will also be conjugates, and hence their imaginary parts will cancel. There is no way of avoiding the usage of complex numbers in solving for the roots of the associated cubic for Cos[20]. Note that this shows that the regular polygon (a 9-gon) with interior angles of size Cos[40] is not constructible by Euclidean methods of straightedge and compass, because Cos [40] (the size of the angles you would need to draw to construct the polygon) is the solution of an irreducible cubic. If you are wondering how I obtained the real root r (you may substitute it and easily verify that it is indeed a root), it was found using Mathematica, a powerful symbolic algebra computer program. However, it is just as possible to find the solution to the cubic using pencil-and-paper methods. One last note: It can be shown that Cos[72] = 1/(1+Sqrt[5]), hence the regular pentagon is constructible by straightedge and compass. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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