Proving $2 ( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} )$ is a root of$ \sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$ How can one show that the number
$2 \left( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} \right)$
is a root of the equation
$\sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$?
 A: This solution is basically the stupidest possible way to solve the problem. I'm just posting it in case the direct calculation here sheds any light on the problem, it probably doesn't though. Especially frustrating is this does not seem to show any connection between the discriminant $19^2$ and the cyclotomic field involved.
Define $e(q) = e^{2 i \pi q}$ so that $2 \cos(2 \pi q) = e(q) + e(-q)$ and $x = 2 \left( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} \right) = e(2/19) + e(3/19) + e(5/19) + \text{mi}$. "mi" means mirror image, I'm going to omit these terms because there are a lot of terms involved. Since $x$ is real its square and cube will be too, that means (by algebraic independence of roots of unity) we must have the mirror image there too.. this save a bit of work.
Squaring it is a simple process it just means multiplying out the bracket and using the identity $e(q)e(r) = e(q+r)$. We get
x^2 = 6 + 2 e(1/19) + 2 e(2/19) + 2 e(3/19)
        + e(4/19) + 2 e(5/19) + e(6/19)
        + 2 e(7/19) + 2 e(8/19) + e(9/19)    + mi

Cubing is slightly more work but in the end we get
x^3 = 12 + 9 e(1/19) + 15 e(2/19) + 15 e(3/19)
         + 10 e(4/19) + 15 e(5/19) + 10 e(6/19)
         + 9 e(7/19) + 9 e(8/19) + 10 e(9/19)    + mi

Note that I have also reduced terms, e.g. $e(11/19) = e((11-19)/19) = e(-8)$. There is another reduction we can make too, since the terms contain the complete sum $e(1/19)+e(2/19)+...+e(9/19) + \text{mi} = -1$ (which can be seen consider the points evenly spaced around the unit circle) we apply this reduction to get:
x^2 = 4 - e(4/19) - e(6/19) - e(9/19) + mi

x^3 = 3 + 6 e(2/19) + 6 e(3/19) + e(4/19)
        + 6 e(5/19) + e(6/19) + e(9/19) + mi

It is now very easy to see that this satisfies the polynomial $x^3+x^2-6x-7$ and in fact you could have figured out that polynomial from this method.
What I would really like to do is start with the polynomial and produce an exponential series solving it..
A: Let $$f(x) = 2(\cos \frac{4\pi x}{19} + \cos \frac{6\pi x}{19} + \cos \frac{10\pi x}{19})$$
Then we claim that the following are true:
$$f(1) + f(4)^2 = 4$$
$$f(4) + f(16)^2 = 4$$
$$f(16) + f(64)^2 = 4$$
Noting that $\displaystyle f(4) \lt 0$, $\displaystyle f(16) \lt 0$ and $\displaystyle f(64) = f(1) \gt 0$, we can show the identity about $\displaystyle f(1)$.
In order to prove the above three identities,
We use the fact that if 
$\displaystyle P(z) = z^4 + z^{-4} + z^6 + z^{-6} + z^{10} + z^{-10}$
and if $\displaystyle w$ is a root of $\displaystyle z^{19} = -1$ then
$$P(w) + P^2(w^4) = 4$$
After some (slightly tedious, but easy) algebraic manipulation, the above identity involving $\displaystyle P(w)$ basically boils down to proving that
$$w^{18} - w^{17} + w^{16} - w^{15} + \dots + w^{2} - w = -1$$
which follows easily from the fact that
$$ 0 = w^{19} + 1 = (w+1)(w^{18} - w^{17} + w^{16} - w^{15} + \dots + w^{2} - w + 1)$$
Now since $\displaystyle P(z) = P(-z)$, the identity is also true of $\displaystyle w$ is a root of $\displaystyle z^{19} = 1$
Thus applying the identity with $\displaystyle P(w)$ three times to $\displaystyle c, c^4, c^{16}$, where $\displaystyle c = \cos \frac{\pi}{19} + i \sin \frac{\pi}{19}$ we get the above three identities involving $\displaystyle f(1), f(4), f(16)$ and thus we are done.
