I want to find the answer to the following problem: Construct a cubic polynomial with integer coefficients, whose roots - $\cos{\frac{2 \pi}{7}}$, $\cos{\frac{4 \pi}{7}}$ and $\cos{\frac{6 \pi}{7}}$.
I have the following idea. These trigonometric roots are the extremum points of the Chebyshev polynomial $T_7 (x) = T_7 (\cos{t}) = \cos{7t}$ and $T_7(x_k) = 1$, where $x_k$ required the roots of a cubic function (see problem). Then the 7 degree polynomial $T_n(x)-1$ will have roots $x_k$. But the problem is that we need a polynomial of degree 3. I'm sure these trigonometric roots will be associated with the Chebyshev polynomials.What do you think about this ? Can you suggest your ideas to solve?
How about the polynomial $$8\left(x-\cos\tfrac{2\pi}{7}\right)\left(x-\cos\tfrac{4\pi}{7}\right)\left(x-\cos\tfrac{6\pi}{7}\right)=8x^3+4x^2-4x-1?$$