Algebraic numbers expressible in terms of real-valued radicals Here's a question that I bumped into a couple of times over the years when I was still working in maths, but which somehow never yielded to my efforts. Already when I was a student, I heard a professor remark that, although all cubic equations can be solved in terms of radicals, not all of them can be solved in terms of radicals where only roots of real numbers are extracted. Sadly, this piece of information diminished the appeal of the "cubic formula" for me.


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*Is there a general characterization of algebraic numbers which can be expressed in terms of real-valued radicals? That is to say, all operations expressed by the formula should take place within the field of real algebraic numbers.

*As a more modest question, can it be shown that $\sin(2 \pi/7)$ can not be expressed in terms of real-valued radicals?
 A: I will expand on i-m-soloveichik's comment, because it really answers the question. In Solution of Polynomials by Real Radicals (Amer. Math. Monthly 92 (1985), 571-575), Isaacs prove the following theorem.
Theorem. If $x \in \overline {\mathbb Q}$ is an algebraic number having the following properties:


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*$x$ is totally real: it is real, and so are all of its conjugates [in other words: its minimal polynomial splits over $\mathbb R$];

*$x$ is a real radical element: it lies in a field $F_r \subset \mathbb R$ at the top of a tower of extensions
$$F_0 = \mathbb Q \subset F_1 \subset F_2 \subset \ldots \subset F_r$$
such that forall $j$, $F_j/F_{j-1}$ is an extension generated by an element $\alpha_j$ which has a power in $F_{j-1}$. [In other words, $x$ is expressible in termes of real-valued radicals];


Then $x$ is constructible.
I find this theorem truly remarkable: if I allow you to take all the real $n$th roots that you want, you couldn't construct a single totally real number more than the kid down the block with just his square root. 
Obviously, the situation is better for non totally real numbers, starting with $\sqrt[n]{2}$ and the like, but it makes you reconsider the definition of solvable by radicals, does it not?
For the second part of your question, it suffices to remark that $\kappa_n = \cos(2 \pi/n)$ is a totally real number. (Note that $\mathbb Q(\kappa_n)$ is a Galois extension of $\mathbb Q$, because it is a subextension of the $n$th cyclotomic field, whose Galois group is abelian, so every conjugate of $\kappa_n$ must lie in $\mathbb Q(\kappa_n) \subset \mathbb R$).
As such, one can apply your theorem to $\kappa_n$, which will be "expressible in terms of real-valued radicals" if and only if it is constructible, that is, if and only if the regular $n$-gon is straightedge-and-compass-constructible. It's famously not the case for $n = 7$. 
(One easily transfers the result from $\kappa_n = \cos(i 2 \pi/n)$ to $\sigma_n = \sin(i 2 \pi/n)$, since the relation $\kappa_n^2 + \sigma_n^2 = 1$ means that one is constructible (resp. expressible in termes of real radicals) if and only if the other is).
A: Following up on the earlier answer given by @PseudoNeo and my comment on it, here are some more results, quoted from the 1998 paper Real fields and repeated radical extensions (on the arXiv here) by Isaacs and his student Moulton (numbering as in the paper, notation slightly altered):
Theorem B. Suppose that $F$ is a real field and that $F \subset L$ is a repeated radical extension with $[L:F]$ odd. If $F \subset K \subset L$ is an intermediate field extension, then $K$ is a repeated radical extension of $F$. 
Here, a radical extension means an extension of the form $F[\alpha]$ with some power of $\alpha$ in $F$, and a repeated radical extension means what it appears to mean.
Theorem C. Suppose that $F$ is a real field and that $f \in F[X]$ is irreducible of odd degree. If $f$ has some root $\alpha$ in a real repeated radical extension of $F$, then $\alpha$ is the only real root of $f$.
In particular, this yields a large class of examples of non-"real-solvable" polynomials: all odd-degree irreducible polynomials with more than one real root! 
Theorem D. Suppose that $F \subset L$, where $L$ is a real repeated radical extension of $F$. If $[L:F]$ is a power of two and $F \subset K \subset L$ is an intermediate field extension, then $K$ is a repeated radical extension of $F$.
I haven't looked at the particulars of the proofs too closely, but the ideas seem to be both accessible and interesting, so I will definitely keep that on my to-do list.
