Compute the minimal polynomial of $u$ over $\Bbb{Q}$ without using cyclotomic polynomials Let $k\in \Bbb Z\setminus 7\Bbb Z$ and $a_k=\frac{2k\pi}{7}$. Compute the minimal polynomial of $u=$ over $\Bbb{Q}$
The "natural" (in my opinion) way to solve this problem, requires the use of $\Phi_7(t)=t^6+t^5+t^4+t^3+t^2+t+1$, the seventh cyclotomic polynomial, since it has the complex number $e^{2k\pi/7}$ as root. Then, noticing that 
$$u=e^{2k\pi i/7}+e^{-2k\pi i/7}$$
and dividing $\Phi_7(t)$ by $t^3$, doing some calculations result that a polynomial which has $u$ as one of its roots is 
$$P(t)=t^3+t^2-2t-1$$
But $P(t)$ is monic and has no roots in $\Bbb{Z}$, hence it is irreducible, so it has to be the minimal polynomial of $u$ over $\Bbb{Q}$.
However, I'm not entirely happy with this answer. It feels that there is a simpler way to get to the same result, or at least to find the degree of the required polynomial. I'm thinking that somehow, it might be related to the fact that the $n$-th roots of unity form under multiplication a cyclic group of order $n$. So my question would be the following:

Is there another way to solve this problem (if not, a way to calculate the degree of the minimal polynomial) without using cyclotomic polynomials?

As always, any answer would be highly appreciate, and thank you in advance!

Note that $u$ is the real part of $e^{i(2\pi/7)}$. Let $w=2\pi/7$ and note that $$1+\cos{w}+\cos2w+\dotsb+\cos6w=0.$$ because of the identity
$$\sum\limits_{k=0}^n \cos(kx+y)=\frac{\cos(\frac{n}{2}x+y)\sin(\frac{n+1}{2}x)}{\sin(\frac{x}{2})}$$
Then $\cos{w}=\cos6w$, while $\cos{2w}=\cos{5w}$, and $\cos{3w}=\cos{4w}$.
It follows that $$1+2\cos{w}+2\cos{2w}+2\cos{3w}=0.$$ or equivalently
$$-1-4\cos{w}+4\cos^2{w}+8\cos^3{w}=0.$$ 
This gives a polynomial of which $\cos{w}$ is a root.
