Finding the equation whose roots are $\pm \tan(\pi/7)$, $\pm\tan(2\pi/7)$ and $\pm\tan(3\pi/7)$ using De Moivre's Theorem

Question

Find the equation whose roots are $\pm \tan(\pi/7)$, $\pm\tan(2\pi/7)$ and $\pm\tan(3\pi/7)$ using De Moivre's Theorem.

I tried converting $y=\cos x + i\sin x$ in terms of $\tan$ but it didn't work out. Please HELP . PLEASE USE DE MOIVRE'S THEOREM.

Answer 1

To build a polynomial equation whose roots are $(r_1,\ldots,r_n)$, you can easily build the following equation: $\prod_k (x-r_k) = 0$. With your example, you get the following equation: $(x-\tan(\pi/7))\cdot(x+\tan(\pi/7))\cdot(x-\tan(2\pi/7))\cdot(x+\tan(2\pi/7))\cdot(x-\tan(3\pi/7))\cdot(x+\tan(3\pi/7))=0$ Using the identity following identity: $(a+b)\cdot(a-b) = a^2 - b^2$, the equation becomes:

$(x^2 - \tan^2(\pi/7))(x^2 - \tan^2(2\pi/7))(x^2 - \tan^2(3\pi/7)) = 0$

Unfortunately, $\tan(\pi/7)$ doesn't have a nice way to be written.

Answer 2

What about $$ f(x) = (x^2- \tan ^2 (\pi /7))(x^2- \tan ^2 (2\pi /7))(x^2- \tan ^2 (3\pi /7))$$