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?

  • $\begingroup$ I'm confused... you want a cubic equation with specific roots. There is only $one$ such equation... so it either has integer coefficients or it doesn't, there's no constructing... $\endgroup$ – Asier Calbet Sep 23 '15 at 22:58
  • $\begingroup$ @Assaultous2 Only one, up to multiplication by non-zero scalars :) $\endgroup$ – Servaes Sep 23 '15 at 23:09
  • $\begingroup$ You are right. This will be a polynomial $(x-\cos{\frac{2 \pi}{7}})(x-\cos{\frac{4 \pi}{7}})(x-\cos{\frac{6 \pi}{7}})$, and the problem reduces to the difficult trigonometric calculations. I think this problem can be solved easier, using Chebyshev polynomials. $\endgroup$ – Victor Sep 23 '15 at 23:15
  • $\begingroup$ @ Servaes of course $\endgroup$ – Asier Calbet Sep 24 '15 at 10:11

How about the polynomial $$8\left(x-\cos\tfrac{2\pi}{7}\right)\left(x-\cos\tfrac{4\pi}{7}\right)\left(x-\cos\tfrac{6\pi}{7}\right)=8x^3+4x^2-4x-1?$$

  • $\begingroup$ Sorry, I made a mistake in the description of the problem. There should be 3 different roots $\cos{\frac{2 \pi}{7}}$, $\cos{\frac{4 \pi}{7}}$ and $\cos{\frac{6 \pi}{7}}$. $\endgroup$ – Victor Sep 23 '15 at 23:09
  • $\begingroup$ @Victor Thank you, I just noticed that one of the roots was indeed off. It is now correct. $\endgroup$ – Servaes Sep 23 '15 at 23:10
  • $\begingroup$ And how you got this look? You primnarily and revealed just the brackets? $\endgroup$ – Victor Sep 24 '15 at 11:27
  • $\begingroup$ @Victor I don't understand what you are asking. All I did was work out the brackets. $\endgroup$ – Servaes Sep 24 '15 at 12:29

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