Trig in surd form

How would I go about getting something such as sin(pi/7) in surd form? Obviously, through the half angle formulae I can get cos and sin(pi/2^n), but how to get the other numbers? The obvious thing to do would be to solve the equation x^6+x^5+x^4+x^3+x^2+x+1=0, to find sin(pi/7), but that doesn't look to be very soluble to me...

Reply 1

Glutamic Acid

Not all trigonometric expressions can be expressed in surd form (I guess because the roots of the respective polynomials can't be expressed in radicals, shown by Galois) -- and I don't think this one can.

Reply 2

Positronized

Second, this article exploits only the first two of five known Fermat primes: 3 and 5; and the trigonometric functions of other angles, such as 2π/7, 2π/9 (= 40°), and 2π/13 (as well as the other constructible polygons, 2π/17, 2π/257, or 2π/65537) are soluble by radicals.

Taken from http://en.wikipedia.org/wiki/Exact_trigonometric_constants

I couldn't figure out how to go about doing this at the moment though :-/ Not even sure if I'll ever be able to figure this one out by myself without losing interest first :P

*EDIT: from http://mathworld.wolfram.com/TrigonometryAnglesPi7.html, Mathematica tells me that

\sin(\frac{\pi}{7})=\frac{(-1)^{\frac{5}{14}}\left(1-(-1)^{\frac{2}{7}}\right)}{2}

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