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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[3]/2, we then see that

     Cos[Pi/9] = (e^(I*Pi/9) + e^(-I*Pi/9))/2
          = ((1 + I Sqrt[3])^(1/3) + (1 - I Sqrt[3])^(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[17] - Sqrt[2(17-Sqrt[17])] + 
                Sqrt[2(34+6Sqrt[17]+Sqrt[2(17-Sqrt[17])] - 
                Sqrt[34(17-Sqrt[17])] + 8Sqrt[2(17+Sqrt[17])])])/2]/4.


- Doctor Pete, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Imaginary/Complex Numbers
High School Trigonometry

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