Cosine 20 Degrees
Date: 03/24/2002 at 15:41:35 From: Matthew Gallaher Subject: The exact value for cosine 20 degrees I have tried to find the exact value of cosine 20 degrees but have not been able to come up with a solution. I tried to take a third of the exact value of Cosine 60 degrees but that did not work. I then tried trial and error to find the exact value, but of course concluded fruitlessly. I come to you now completely stumped. I am a pre-calculus student with a humble request: show me how to derive the exact value of cosine 20 degrees. This of course means without a calculator and in no approximate decimal form. Thank you very much for your time.
Date: 03/25/2002 at 07:20:47 From: Doctor Pete Subject: Re: The exact value for cosine 20 degrees Hi Matthew, If you don't mind imaginary numbers, then we can appeal to DeMoivre's formula and the polar representation of complex numbers. Recall that e^(It) = Cos[t] + I Sin[t], where I is the imaginary unit (I^2 = -1). Thus, e^(-It) = Cos[-t] + I Sin[-t] = Cos[t] - I Sin[t], and hence Cos[t] = (e^(It) + e^(-It))/2. Now consider DeMoivre's formula, which states that e^(Int) = (Cos[nt] + I Sin[nt]) = (Cos[t] + I Sin[t])^n. Thus let t = Pi/3, n = 1/3 to find e^(I*Pi/9) = (Cos[Pi/3] + I Sin[Pi/3])^(1/3), e^(-I*Pi/9) = (Cos[Pi/3] - I Sin[Pi/3])^(1/3), and since Cos[Pi/3] = 1/2, Sin[Pi/3] = Sqrt/2, we then see that Cos[Pi/9] = (e^(I*Pi/9) + e^(-I*Pi/9))/2 = ((1 + I Sqrt)^(1/3) + (1 - I Sqrt)^(1/3))/(2^(4/3)). I would understand if this answer leaves you unsatisfied, but it is actually the best that we can do (to understand why would involve a lengthy discussion on abstract algebra and Galois theory). Alternatively, we can write Cos[3x] = Cos[x]Cos[2x] - Sin[x]Sin[2x] = Cos[x](Cos[x]^2 - Sin[x]^2) - 2Sin[x]^2 Cos[x] = Cos[x](2Cos[x]^2 - 1) - 2(1-Cos[x]^2)Cos[x] = 2Cos[x]^3 - Cos[x] - 2Cos[x] + 2Cos[x]^3 = 4Cos[x]^3 - 3Cos[x], and thus choosing x = Pi/9, we find Cos[Pi/3] = 4(Cos[Pi/9])^3 - 3 Cos[Pi/9]. Let y = Cos[Pi/9]. Then y is a solution to the degree-3 polynomial 8y^3 - 6y - 1 = 0, and in particular, it is the only positive real root of this polynomial. So, to make you feel a little better about the situation, I will mention that many trigonometric functions of angles don't have "nice" exact expressions. However, there are some very unusual values which do work out exactly, for instance, Cos[2*Pi/17] = Sqrt[(15 + Sqrt - Sqrt[2(17-Sqrt)] + Sqrt[2(34+6Sqrt+Sqrt[2(17-Sqrt)] - Sqrt[34(17-Sqrt)] + 8Sqrt[2(17+Sqrt)])])/2]/4. - Doctor Pete, The Math Forum http://mathforum.org/dr.math/
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