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I want to find an expression for $cos(\frac{2\pi}{5})$ containing only rational numbers and square roots of rational numbers.

Sub questions of this exercise (which I have already solved) give me the following information:

$X^5 - 1 = (X-1)(X^4 + X^3 + X^2 + X + 1) := (X-1)\Phi$ where $\Phi$ is irreducible in $\mathbb{Q}[X]$. Define $$M := \mathbb{Q}[X]/\Phi, \,\,\, \zeta := X + (\Phi), \,\,\, \beta := X + X^4 + (\Phi) \in M, \,\,\, L := \mathbb{Q}[\beta] \subset M.$$

I also found that the minimal polynomial $f_\mathbb{Q}^\beta = X^2 + X - 1$, that $[M:L] = 2$, and that $f_\mathbb{Q}^\zeta = X^2 −\beta X + 1 $.

So now, what I want to do is to find an expression for $cos(\frac{2\pi}{5})$ containing only rational numbers and square roots of rational numbers. However, I am stuck on this part. I've tried to fill in some rational numbers into the equations obtained from the subquestions, but it feels like stupid work. I also don't really see how this question relates to the subquestions...

Help is appreciated! :)

hello
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2 Answers2

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Rewrite $x^4+x^3+x^2+x+1=0\iff (x^2+x^{-2})+(x+x^{-1})+1=0$

Let's call $a=\dfrac{2\pi}{5}$ then use above relation to calculate $\cos(a)+\cos(2a)=-\frac 12$

Use trigonometric formulas to find $\cos(a)\times\cos(2a)=-\frac 14$

Now solve $x^2-sx+p=0$ for sum and product.

zwim
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  • Thanks! I don't really understand why the first iff statement holds though.. Could you maybe elaborate on this? – hello Jun 13 '20 at 15:19
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    I just divided by $x^2\neq 0$ (since $\cos$ is not zero for this angle). – zwim Jun 13 '20 at 15:35
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Hint: for direct geometric solution to your question, use the fact that a regular pentagon can be constructed with straightedge and compass.

Ethan Bolker
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