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Perhaps this question will not be considered appropriate for MO - so be it. But hear me out before you dismiss it as completely elementary.

As the question suggests, I would like to know when $\sin(p\pi/q)$ can be expressed in radicals (in the way that $\sin(\pi/4) = \sqrt{2}/2$ and $\sin(\pi/3) = \sqrt{3}/2$ can). Let $\alpha = \sin(x)$, and consider the field extension $\mathbb{Q}[\alpha]$. Using $(\cos(x) + i\sin(x))^k = \cos(kx) + i \sin(kx)$ together with the binomial formula and the Pythagorean identity relating sine and cosine, we can see that $\sin(kx)$ lies in a solvable extension of $\mathbb{Q}[\alpha]$. Thus $\sin(p\pi/q )$ is expressible in radicals if $\sin(\pi/q)$ is.

To handle $\sin(\pi/q)$, we start by using the same trick (which most people also learn in an elementary trig class). Write $-1 = (\cos(\pi/q) + i\sin(\pi/q))^q$, use the binomial theorem to expand, compare imaginary parts, and express the right-hand-side in terms of sine using the Pythagorean identity. This gives an explicit equation for any $q$ one of whose solutions is $\sin(\pi/q)$. This equation is not a polynomial in $\sin(\pi/q)$ since it involves terms of the form $\sqrt{1 - \sin^2(\pi/q)}$, but it is enough to prove that $\sin(\pi/q)$ is algebraic.

So I am curious about the number theoretic properties of this equation. What can be said about the Galois group of its "splitting field" over $\mathbb{Q}$? Can we at least determine when it is solvable? Note that if the prime factors of $q$ are $p_1, \ldots p_k$ and we can express each $\sin(\pi/p_j)$ in radicals, then the same is true for $\sin(\pi/q)$. So it suffices to consider the case where $q$ is prime. That's about all the progress I have made.

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Isn't the Galois group always abelian, as your extension lives in $\mathbb{Q}(\omega,i)$, where $\omega$ is a pth root of unity? –  Steve D Aug 21 '10 at 13:17
    
Sorry, I mean qth root of unity, when q is prime, as in your last paragraph. –  Steve D Aug 21 '10 at 13:18
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2 Answers

up vote 13 down vote accepted

As $\cos x=\pm\sqrt{1-\sin^2 x}$ and $e^{ix}=\cos x +i\sin x$, and $\sin x=(e^{ix}-1/e^{ix})/2i$ then $\sin x$ is in a radical extension of $\mathbb{Q}$ iff $e^{ix}$ is. For rational $r$ with denominator $d$, $e^{2\pi i r}$ is a primitive $d$-th roots of unity. The extension of $\mathbb{Q}$ generated by a root of unity is a cyclotomic field. Every cyclotomic field is an abelian extension of $\mathbb{Q}$. (By the Kronecker-Weber theorem any abelian extension is contained in a cyclotomic field.) The Galois group of the $n$-th cyclotomic field is isomorphic to the multplicative group $(\mathbb{Z}/n\mathbb{Z})^*$

So we can obtain any $\sin r\pi$ for $r$ rational in terms of radicals, both in a trivial way if you allow an $n$-th root of unity as $1^{1/n}$, and also in a stricter sense, if you insist that you ascend through a chain of fields by adjoining at each stage a root of $x^m-a$ where this polynomial is irreducible over the previous field.

However if you insist on this more exacting definition, you will need radicals of non-real numbers unless $n$ is a product of distinct Fermat primes. All this is well-known.

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Wow, I don't think there is anything left to say. One might have thought I would have gotten a clue when I wrote $-1 = (\cos(\pi/q) + i \sin(\pi/q))^q$, but I guess that isn't always how the brain works. Thanks for your answer! –  Paul Siegel Aug 23 '10 at 21:10
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Your question has been thoroughly answered, so I don't really have much to add, except that there is a paper where you can see examples of these ideas in action with a very elementary presentation (i.e. suitable for undergraduates who have a basic knowledge of Galois theory).

Skip Garibaldi
Somewhat more than governors need to know about trigonometry
Mathematics Magazine 81 (2008) #3, 191-200.

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I've been wondering about this myself, so I wish to thank you, good sir, for pointing out this excellent paper. :) –  J. M. Aug 22 '10 at 1:42
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