TITLE: Is $ \sqrt{2}$ an element of $ \mathbb{Q} ( \cos 72^\circ ) $?
QUESTION [8 upvotes]: There's an exact value for this angle related to the regular pentagon (1, 2):
$$\cos 72^\circ = \cos \frac{2\pi}{5} = \frac{1}{4} \sqrt{2(5 + \sqrt{5})}$$ 
This got me wondering about which square roots of integers are elements of the this field?
$$ \sqrt{2}, \sqrt{5} \stackrel{?}{\in} \mathbb{Q} \big( \cos 72^\circ \big) $$
This is certainly related to the fifth root of unity, but this field is certainly not $\mathbb{Q}( e^{2\pi i/5})$.  This field is a quadratic extension of another field:
$$ \big[\mathbb{Q}\big(\cos 72^\circ \big): \mathbb{Q}(\sqrt{5}) \big] = 2 $$
And therefore the Galois Group is $\mathbb{Z}/2\mathbb{Z}$.   However, none of this helps me solve the original problem.

REPLY [7 votes]: Taking into account the misprint pointed out by @i707107, your question amounts to the determination of the quadratic subfields of $\mathbf Q$(sin $\frac {2\pi}5)$. It is perhaps clearer to tackle right away the general case of $\mathbf Q$ (sin $\frac {2\pi}p)$ , where $p$ is a given odd prime. For simplification, denote $\theta = \frac {2\pi}p$ and $\zeta =$ cos $\frac {2\pi}p +$ i sin $\frac {2\pi}p$. The heart of the problem lies in the interplay between the fields $\mathbf Q(\zeta)$, $\mathbf Q$(sin $ \theta)$ and $\mathbf Q$(cos $ \theta)$ .
1) It is well known that the cyclotomic extension  $\mathbf Q(\zeta)/\mathbf Q$ is cyclic of degree $p-1$, hence it contains a unique quadratic subextension which is $\mathbf Q(\sqrt {p^*})/\mathbf Q$, where $p^*=(-1)^{\frac {p-1}2}p $ (this comes from a discriminant computation and classically constitutes the first step of one of the numerous proofs of the quadratic reciprocity law). At the other extreme, the unique subextension of degree $\frac {p-1}2$ is the maximal totally real subfield $\mathbf Q$(cos $ \theta)$ of $\mathbf Q(\zeta)$. Note that $\mathbf Q(\zeta)=\mathbf Q$(cos $ \theta$, i sin $\theta$) because $\zeta+\zeta^{-1}=2$cos $\theta$ and $\zeta.\zeta^{-1}=1$, but $\mathbf Q$(sin $ \theta)$ is not contained in $\mathbf Q(\zeta)$. 
2) Where does $\mathbf Q$(sin $\theta)$ sit in this picture ? It can be shown that $\mathbf Q$(sin $ \theta)=\mathbf Q$(cos $ \theta$, sin $\theta)$, see e.g. https://math.stackexchange.com/a/136757 or https://math.stackexchange.com/a/237530, but we'll not use this property. Introduce the auxiliary field $\mathbf Q(i,\zeta)=\mathbf Q$(i, cos $ \theta$, sin $\theta)$, which is a quadratic extension of $\mathbf Q$(cos $ \theta$, sin $\theta)$. Because $i$ is a primitive $4$-th root of $1$ and $\zeta$ a primitive $p$-th root, we know that $\mathbf Q(i)$ and $\mathbf Q(\zeta)$ are linearly disjoint, hence $\mathbf Q(i,\zeta)/\mathbf Q$ is abelian with Galois group $\cong C_2 \times C_{p-1}$ (in multiplicative notation). It follows that the maximal 2-quotient (= maximal quotient which is a 2-group) of exponent 2 of Gal$(\mathbf Q(i,\zeta)/\mathbf Q)$ is $\cong C_2 \times C_2$, so $\mathbf Q(i,\zeta)$ contains exactly three quadratic subfields. Let us list these. It is straightforward  that $\mathbf Q(i,\zeta)$ contains $\mathbf Q(i)$ and $\mathbf Q(\sqrt {p^*}$) (see 1)), hence the third of our list will be $\mathbf Q(\sqrt {p^{**}})$, with $p^{**}=(-1)^{\frac {p+1}2}p$ . Among these three, one and only one is totally real. Conclusion : $\mathbf Q$(sin $\theta)$ admits a unique quadratic subfield, which is $\mathbf Q(\sqrt {p^*})$ or $\mathbf Q(\sqrt{p^{**}})$ depending on the congruence $p\equiv \pm  1$ mod $4$ ; for $p\ge 5,\mathbf Q$(sin $\theta)$ and $\mathbf Q$(cos $\theta)$ share the same unique quadratic subfield. In your question, $p=p^*=5$, so the unique quadratic subfield of $\mathbf Q$(sin $\frac {2\pi}5)$ and of $\mathbf Q$(cos $\frac {2\pi}5)$ is $\mathbf Q(\sqrt 5)$. Note that we didn't need to know the explicit value of sin $\frac {2\pi}5$.