1. ## Can I express cos(pi/7) with radicals?

I'm trying to express cos(pi/7) with radicals - that is, basic operations (addition, division, etc) and square and cube roots (or even higher roots if necessary).

I got this far:

If $s = 84\sqrt{3} \cdot i - 28$ and $c = \sqrt[3]{s}$, then $x = \frac{2+c}{12} + \frac{7}{3c} = \cos(\pi/7)$

Although this seems a complex expression, the imaginary part of x is actually zero, so x is in fact a real number (obviously, since cos(pi/7) is real). But I can't extract the real part as a "really real" radical expression (without i, that is).

2. Originally Posted by Maxim
I'm trying to express cos(pi/7) with radicals - that is, basic operations (addition, division, etc) and square and cube roots (or even higher roots if necessary).

I got this far:

If $s = 84\sqrt{3} \cdot i - 28$ and $c = \sqrt[3]{s}$, then $x = \frac{2+c}{12} + \frac{7}{3c} = \cos(\pi/7)$

Although this seems a complex expression, the imaginary part of x is actually zero, so x is in fact a real number (obviously, since cos(pi/7) is real). But I can't extract the real part as a "really real" radical expression (without i, that is).
Read this: Trigonometry Angles--Pi/7 -- from Wolfram MathWorld

3. Thanks. Still a bit confused though, the page says:
Originally Posted by Wolfram Mathworld
The angles $\pi\frac{m}{n}$ (with m,n integers) for which the trigonometric functions may be expressed in terms of finite root extraction of real numbers are limited to values of m which are precisely those which produce constructible polygons.
Which is also what I read elsewhere (except for the typo - that last condition should be "values of n")

But further down, it says:
In general, any trigonometric function can be expressed in radicals for arguments of the form $\pi\frac{m}{n}$ (...)
From the example, I figured that $\cos(\pi/7) = -\frac{1}{2} \cdot (-1)^{6/7} \cdot \left(1+(-1)^{2/7}\right)$, doesn't that fit the description "in terms of finite root extraction of real numbers"?