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My objective is to prove that $|mathrm{Gal}(mathbb{Q}(cos(2pi/7))/mathbb{Q})|=3$. So far, I've been able to show that the degree of the field extension $mathbb{Q}(cos(2pi/7))/mathbb{Q}$ is $3$ using that $mathbb{Q}(e^{2pi/7})/mathbb{Q}(cos(2pi/7))$ has degree 2 and $mathbb{Q}(e^{2pi/7})/mathbb{Q}$ has degree 6. I don't know how to proceed from here. The standard way I know is to find the minimal polynomial of $cos(2pi/7)$ over the rationals and presenting the Galois group of the extension as a group of permutations of the roots of such polynomial contained in $mathbb{Q}(cos(2pi/7))$. If I could somehow prove that all such roots are contained in $mathbb{Q}(cos(2pi/7))$ it would be enough but I would like to avoid having to calculate the minimal polynomial. Thank you in advance.

The answer is almost contained in the answers of this question.

Galois group of $x^3 + x^2 - 2x - 1$.^2-2x-1

The second answer shows that the minimal polynomial of $cos({{2pi}over 7})$ is $x^3-x^2-2x-1$ and the first and second answer shows that the Galois group of this polynomial has order $3$.