# Cos/Sin as roots of rationals

1. Apr 6, 2006

### gonzo

This is indirectly related to issues with cyclotomic polynomials and glaois groups.

Is there some easy way to know if you are dealing with a cos or sin that is expressable in terms of roots of rationals? Like $\pi/3$ for example? If so, is there any straightforward way of figuring it out?

I'm asking because I'm trying to do a problem involving the galois group for the 7th roots of unity. The fact that there is a order 2 and order 3 subgroup of the galois group makes me think that Q(i) must be a fixed field and then some cube root another one for the subgroups. This would further imply that $cos(4\pi/7)$ is expressable as a ratioal function plus a cube root (unless I'm way off in space, which is possible).

Thanks

2. Apr 7, 2006

### leach

A very interesting problem. I'm not very fluent with Galois theory these days, so please take my words with a grain of salt.

For the 7th roots of unity we have the polynomial

$$(x-1)^7 \quad = \quad (x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$$

So the interesting polynomial for these roots is:

$$p \quad = \quad x^6+x^5+x^4+x^3+x^2+x+1$$

Now, it is obvious that this polynomial has as Galois group $$\mathbb{Z}_7$$. If we call $$\omega$$ some seventh root of unity (not 1), the other roots of $$p$$ are the powers $$\omega^k,\, k=2,\ldots, 6$$. Hence the field for this polynomial is $$\mathbb{Q}(\omega)$$, extension of dimension 7 over $$\mathbb{Q}$$, and its automorphisms are obviously the powers of the roots.

The group $$\mathbb{Z}_7$$ has no nontrivial subgroups. I think you are erroneously taking $$\mathbb{Z}_6$$ as the Galois group of the polynomial p.

Anyway, we come to the conclusion that any root of our polynomial p has the form $$a_0 + a_1\omega + a_2\omega^2 + a_3\omega^3 + a_4\omega^4 + a_5\omega^5 + a_6\omega^6$$, where the $$a_i$$ are rationals. Since $$\omega$$ is a radical (a seventh root of unity), this trivially means that all the roots of p are radicals.

However, I think that in order to get better information on the structure of the roots, the coefficients $$a_i$$, etc., you could try reading modern elaborations of Gauss' works on cyclotomic groups. I hope this link helps : http://en.wikipedia.org/wiki/Gaussian_period

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If you take sixth roots of unity, you have $$\mathbb{Z}_6 \cong \mathbb{Z}_3\times\mathbb{Z}_2$$, so the Galois group should be generated by two radicals $$\mathbb{Q}(i,\xi)$$, with $$\xi^3 = 1$$.

This means that the sixth roots of unity should be writeable in the form: $$a_0 + a_1 i + a_2 \xi + a_3 i\xi + a_4\xi^2 + a_5i\xi^2$$. Again, more finese in the form of these roots should be achieved turning to Gauss' works.