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I came across a problem in a book recently that asked to find a cubic polynomial with roots $\cos{2\pi/7}, \cos{4\pi/7}, \cos{6\pi/7}$. There were no extra conditions on the problem. It just asks you to find a cubic polynomial with those roots. It was marked as one of the harder problems, so I was kind of confused because it seems obvious that a polynomial like $$\left(x-\cos\frac{2\pi}{7}\right)\left(x-\cos\frac{4\pi}{7}\right)\left(x-\cos\frac{6\pi}{7}\right)$$ should work.

But when I looked up the solution in the solutions manual, it turns out that you can use an obscure trig identity for $\cos{7\theta}$ to eventually construct the polynomial $$8x^3+4x^2-4x-1$$

I'm really lost. What's wrong with my trivial example?

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  • $\begingroup$ Nothing wrong. You will get the same answer, but you will need to do trigonometric transformations to get numbers. $\endgroup$ – Sonal_sqrt May 23 '18 at 3:23
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    $\begingroup$ The question was assuming, without stating it, that the polynomial will have integer coefficients. It will be the exact 8 times your polynomial but you have to show that the coefficients turn out to be the same. $\endgroup$ – fleablood May 23 '18 at 3:48
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    $\begingroup$ But as N8tron's answer says, the text made the mistake of not specifying the cubic need integer (or at least rational) coefficients. This is entire the texts fault. Not yours. $\endgroup$ – fleablood May 23 '18 at 3:50
  • $\begingroup$ math.stackexchange.com/questions/638874/… $\endgroup$ – lab bhattacharjee May 23 '18 at 7:19
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Well if they didn't specify anything else your solution is correct. But typically people specify things like find a polynomial with rational (or equivalently integer) coefficients meeting the specific property.

My favorite example of when this happens is when people say $\pi$ is transcendental because it's not solution of a polynomial equation. I usually point out

$$x-\pi$$ is a polynomial and then preach the importance of correctly qualifying expressions.

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easiest is to take $\omega$ as a primitive seventh root of unity, any one of $$ e^{2 \pi i / 7} \; , \; \; e^{4 \pi i / 7} \; , \; \;e^{6 \pi i / 7} \; , \; \; $$ so that $\omega^7 = 1$ but $\omega \neq 1,$ and $$ \omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1 = 0. $$ Next, for any of the three, take $$ x = \omega + \frac{1}{\omega} $$ First, $$ x^3 = \omega^3 + 3 \omega + \frac{3}{\omega} + \frac{1}{\omega^3} \; \; , \; \; $$ $$ x^2 = \omega^2 + 2 + \frac{1}{\omega^2} \; $$ $$ -2 x = -2 \omega - \frac{2}{\omega} $$ $$ -1 = -1 \; \; . $$ So $$ x^3 + x^2 - 2 x - 1 = \; \; \frac{\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1}{\omega^3}\; \; = \; 0 $$

In each case, we have $x = 2 \cos (2 k \pi i / 7),$ so taking $x = 2c$ we find $8c^3 + 4 c^2 - 4 c - 1 = 0.$

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