# On the roots of $t^4-6\sqrt3t^3+8t^2+2\sqrt3t-1=0$

The Problem Prove that $\tan \frac{\pi}{15}$ is a root of the equation $t^4-6\sqrt3t^3+8t^2+2\sqrt3t-1=0$, and find the other roots.

Source: Question 17 from the complex numbers chapter from Bostock's Further Pure Mathematics

Thoughts Using the identity $\tan 5\theta =\frac{5\tan\theta-10\tan^3\theta-\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$, I can show that $\tan \frac{\pi}{15}$ is a root of $t^5+5\sqrt3t^4+10t^3-10\sqrt3t^2-5t+\sqrt3=0$, but it isn't clear how to proceed. Alternatively, I can derive and use a formula for $\tan 4\theta$ to obtain a quartic equation as desired, but that will require me to find $\tan \frac{4\pi}{15}$ which I cannot.

I suspect this is not difficult though I can't see how to proceed. Hints would be appreciated.

-
I believe that the coefficient of $\tan^5 \theta$ should be +1. – Calvin Lin Sep 21 '13 at 16:18
Hint: Note that $t = \tan(-\pi/3)$ will be also a solution of the polynomial. Remove the undesired factor. – Macavity Sep 21 '13 at 16:25

First of all, $\tan5\theta=\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$

Set $\tan5\theta=\sqrt3=\tan\frac\pi3$

$\implies 5\theta=n\pi+\frac\pi3=\frac{(3n+1)\pi}5$ where $n$ is any integer

So, the values of $\theta$ can be set from $n=0,1,2,3,4$

Observe that for $n=3,\tan\theta=\tan\frac{2\pi}3=\tan\left(\pi-\frac\pi3\right)=-\tan\frac\pi3=-\sqrt3$

So rearrange $\displaystyle\sqrt3=\frac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$ and divide by $\tan\theta -(-\sqrt3)=\tan\theta+\sqrt3$ to get the equation with roots $\displaystyle\tan\frac{(3n+1)\pi}{15}$ where $n=0,1,2,4$

-
Excellent, thanks. For the last sentence, you mean $n=0,1,2,4$. – Sarastro Sep 21 '13 at 16:48
@Sarastro, thanks for your observation. Hope I could explain the idea – lab bhattacharjee Sep 21 '13 at 16:50