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 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.

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    $\begingroup$ The first line is $\sin 72^{\circ}$. $\endgroup$ Commented Aug 9, 2018 at 0:52

2 Answers 2


Note that $\cos \left(\frac{2\pi}{5} \right) = \frac 12 (\zeta_5 + \zeta_5^{-1})$, where $\zeta_5 = e^{2\pi i/5}$. Thus we have that $\cos \left(\frac{2\pi}{5} \right)\in \mathbb{Q}(\zeta_5 + \zeta_5^{-1})$, which is a quadratic extension.

On the other side note that as the Galois group of the cyclotomic extension $\mathbb{Q}(\zeta_5)$ is $\mathbb{Z}/4\mathbb{Z}$, a cyclic group, there is a unique quadratic field. Now we have that the discriminant of the field extension $\mathbb{Q}(\zeta_5)$ is $5^3$. Also it's well-known that the square root of the discriminant is in the extension (this holds for normal extensions, which is the case here). Thus $5\sqrt{5} \in \mathbb{Q}(\zeta_5)$ and so $\mathbb{Q}(\sqrt{5}) \subset \mathbb{Q}(\zeta_5)$.

From here we conclude that the $\cos \left(\frac{2\pi}{5} \right) \in \mathbb{Q}(\sqrt{5})$, as it's the unique quadratic subfield. Now as $\cos \left(\frac{2\pi}{5} \right) \not \in \mathbb{Q}$ and $\mathbb{Q}(\sqrt{5})$ is a quadratic extension we get that $\mathbb{Q}\left(\cos \left(\frac{2\pi}{5} \right)\right) = \mathbb{Q}(\sqrt{5})$. Finally we observe that $\sqrt{5} \in \mathbb{Q}\left(\cos \left(\frac{2\pi}{5} \right)\right)$ and moreover it's the only square root contained in the field.

In your question you have the value of $\sin \left(\frac{2\pi}{5} \right)$, instead of $\cos \left(\frac{2\pi}{5} \right)$. Nevertheless we can consider that case too.

Squaring the expression you can get that $\sin \left(\frac{2\pi}{5} \right)$ satisfies $16x^4 - 20x^2 + 5$, which is irreducible polynomial by Eisenstein Criterion. Thus $\left[\mathbb{Q}\left(\sin \left(\frac{2\pi}{5} \right)\right): \mathbb{Q}\right] = 4$. With a bit of effort you can prove that the Galois group of the extension is $\mathbb{Z}/4\mathbb{Z}$. This can be done by noticing that the other roots are $\sin \left(\frac{4\pi}{5} \right),\sin \left(\frac{6\pi}{5} \right),\sin \left(\frac{8\pi}{5} \right)$ and the automorphism $\sin \left(\frac{2\pi}{5} \right) \to \sin \left(\frac{4\pi}{5} \right)$ generates the Galois group.

So again there is a unique quadratic subfield and as you've noticed it is $\mathbb{Q}(\sqrt{5})$. Thus $\sqrt{5}$ is the only square root contained in $\mathbb{Q}\left(\sin \left(\frac{2\pi}{5} \right)\right)$


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 $\frac {2\pi}5$.


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