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Question: Is the value of $\sin({\frac{\pi}{n}})$ expressible by fractions, radicals and natural numbers for each given $n$? If not, for which $n$ can we prove this non-expressibility?

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Yes if N-th roots are allowed and no if only square roots are allowed. The latter is the classical problem of which regular polygons are ruler and compass constructible. The former is from the field of N-th roots of unity having commutative Galois group. –  zyx Nov 18 '13 at 18:54
I think there is no closed form for $n=7$ in terms of square roots –  Norbert Nov 18 '13 at 18:54
@zyx: Please give me more explanations and introduce some references in an answer. –  Saint Georg Nov 18 '13 at 18:57
There are two very different versions of this question: (easy one) where complex roots are allowed and (hard one) where only real roots are allowed (which is related to casus irreducibilis etc). Turns out cos(2pi/n) can be expressed in real radicals iff the regular n-gon is constructible –  Grigory M Nov 18 '13 at 20:01
See rational angles with sines expressible with radicals, including the comments. Incidentally, here's the Math Forum archived version of the sci.math post I gave a google-groups URL for (in case that URL requires you to sign in or something). –  Dave L. Renfro Nov 18 '13 at 21:45
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5 Answers 5

You'll find an excellent start here: http://www.efnet-math.org/Meta/sine1.htm

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As was alluded to in comments this relates to the nth roots of unity. First of all you distinguish between 'primitive' and non-primitive. Then you look if those primitive roots are expressible in radicals and so on. The case where they are not happens for example with the Casus irreducibilis, which sometimes (!) happens in trisecting the angle (but never in bisecting the angle). Long story short it turns out for example that for $0<=k<=2^n$ all $\sin(2\pi k/2^n)$ and $\cos(2\pi k/2^n)$ are expressible in radicals!

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I don't think it works for $n=90$ because that would essentially mean sin2° and that it as far as I know not expressible in terms of radical terms

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(Many)radicals are irrational too. Eg. $\sqrt{2}$ –  Priyatham Nov 18 '13 at 18:58
yes, I realized, see edit. I essentially concur with Norbert... –  imranfat Nov 18 '13 at 19:00
But it works for $ sin(\pi/60) $ , $$sin\frac{\pi}{60} = \frac{1}{16}[2(1-\sqrt3)\sqrt{5+\sqrt5}+\sqrt2(\sqrt5-1)(\sqrt3+1)]$$ –  Alan Nov 18 '13 at 20:52
If $ g(x) = 4x^3- 3x + 1/16 [(2 (1-\sqrt(3)) \sqrt(5+\sqrt(5))+\sqrt(2) (\sqrt(5)-1) (\sqrt(3)+1))] $ , then $ g(\sin(\pi/180) = 0 $ . The smallest real root is sin(1) (degree) and x ~ 0.01745240643728351281941897851631619247225272031 –  Alan Nov 18 '13 at 23:20
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Cyclotomic polynomials are solvable, so their imaginary parts are expressible in radicals. A separate issue is whether the values are constructible -- i.e., whether all the radical sign are square roots.

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If $f(x) = 64x^6 - 112x^4 + 56x^2 - 7$ then $f(\sin(\pi/7)) = 0$ , but I think this was known in some form to Kepler.

Which suggests that we look for the minimal polynomial which has $\sin(\pi/n)$ as a root.

Here is a reference:Scott Beslin and Valerio de Angelis, The minimal polynomials of sin(2 pi/p) and cos(2 pi/n), Math. Mag., 77 (2004), 146-149.

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