Can we express cos(pi/11) using radicals?

[mathbalarka]

This is the question I found long ago in another math forum. I thought that it would be a good sweat for everyone in order to find the answer :

Find all the closed from of the roots of this solvable quintic

\(\displaystyle 32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6x - 1 = 0\)

 

[mathbalarka]

Here is my solution :

Spoiler

\(\displaystyle 32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6x - 1 = 0\)

\(\displaystyle \Rightarrow (x + 1)(32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6x - 1)^2 = 0\)

Expanding this gives

\(\displaystyle 1024 x^{11} - 2816 x^9 + 2816 x^7 - 1232 x^5 + 220 x^3 -11 x + 1 = 0\)

\(\displaystyle \Rightarrow [1024 x^{11} - 2816 x^9 + 2816 x^7 - 1232 x^5 + 220 x^3 -11 x] + 1 = 0\)

Looking at the expression closed with 3-rd bracket carefully, we see that it's one of the Chebyshev polynomials. Hence, letting \(\displaystyle x = \cos(\theta)\), we get

\(\displaystyle \cos(11 \theta) + 1 = 0\)

So, the roots of the equation above are \(\displaystyle \theta = \frac{(2n + 1) \pi}{11}\).

Hence, the (uncertified) roots of the quintic of interest are \(\displaystyle x = \cos \left ( \frac{(2n + 1) \pi}{11} \right )\).

By a quick check, we see that only n = 0, 1, 2, 3, 4 works.

Hence, the roots are \(\displaystyle \cos \left ( \frac{\pi}{11} \right ), \cos \left ( \frac{3 \pi}{11} \right ) , \cos \left ( \frac{5 \pi}{11} \right ) , \cos \left ( \frac{7 \pi}{11} \right ) , \cos \left ( \frac{9 \pi}{11} \right )\)

Balarka
.

 

[mathbalarka]

Interestingly enough, the polynomial of the general interest has a solvable galois group (i.e, cyclic group of order 5) implying that all the roots over there can be expressed by finite number of elementary functions.

So, my question is how can we express at least, say, \(\displaystyle \cos (\pi / 11)\) in terms of radicals?

Maybe this one is a challenge too and probably a harder one!