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

Question

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?

Answer 1

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.

Answer 2

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