# What is wrong with my trivial solution to finding a cubic polynomial with roots $\cos{2\pi/7}$, $\cos{4\pi/7}$, $\cos{6\pi/7}$?

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?

• Nothing wrong. You will get the same answer, but you will need to do trigonometric transformations to get numbers. – Sonal_sqrt May 23 '18 at 3:23
• 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. – fleablood May 23 '18 at 3:48
• 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. – fleablood May 23 '18 at 3:50
• math.stackexchange.com/questions/638874/… – lab bhattacharjee May 23 '18 at 7:19

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.
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 $8c^3 + 4 c^2 - 4 c - 1 = 0.$