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(ppi/q) can be expressed in radicals (in the way that sin(pi/4)=sqrt2/2 and sin(pi/3)=sqrt3/2 can). Let alpha=sin(x), and consider the field extension mathbbQ[alpha]. Using (cos(x)+isin(x))k=cos(kx)+isin(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 mathbbQ[alpha]. Thus sin(ppi/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)+isin(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 sqrt1−sin2(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 mathbbQ? Can we at least determine when it is solvable? Note that if the prime factors of q are p1,ldotspk and we can express each sin(pi/pj) 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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