Proving that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ I am trying to solve the cubic equation $x^3-15x-4=0$ using Cardano's formula.  I already know that the solutions are $x=4$, $x= \sqrt{3}-2$ and $x= -\sqrt{3}-2$ and that using the formula in this problem requires finding the cube roots of $2+11i$ and $2-11i$, which are $2+i$ and $2-i$.  But when I try to use the formula on my calculator, a TI-89 Titanium, I get $2\sqrt 5 \sin \left( \frac{\arctan(\frac{2}{11})}{3}+\pi/3 \right)$ instead of $4$.  For some reason, the fact that $(2+i)^3 = 2 +11i$ and $x = 4$ is a zero of $x^3-15x-4$ feels like a byproduct of something else. So I have tried for more than a month to prove that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ without using either of these results.  

Let $\theta =\arctan(11/2)$.  Then, $\cos(\theta)=\frac{2}{5\sqrt 5}$.  
Now, let  $\alpha$ be given by $\alpha=3\arccos(2/\sqrt 5)$.  Using the triple angle formula
$$\cos(3x)=\cos^3(x)-3\sin^2(x)\cos(x)$$
we find that 
$$\begin{align}
\cos(\alpha)&=\cos(3\arccos(2/\sqrt 5))\\\\
&=\cos^3(\arccos(2/\sqrt 5))-3\sin^2(\arccos(2/\sqrt 5))\cos(\arccos(2/\sqrt 5))\\\\
&=\frac{8}{5\sqrt 5}-3\left(\frac15\right)\left(\frac{2}{\sqrt 5}\right)\\\\
&=\frac{2}{5\sqrt 5}
\end{align}$$
Therefore, $\alpha = \theta$ and we are done!
