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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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