<s>[INST] 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.  
 [/INST] The fact you want to prove is equivalent to
$$
\frac{\arctan\left(\frac{11}{2}\right)}{3} = \arccos\left(\frac{2}{\sqrt 5}\right),
$$
that is,
$$
\arctan\left(\frac{11}{2}\right) = 3 \arccos\left(\frac{2}{\sqrt 5}\right),
$$
that is,
$$
\frac{11}{2} = \tan\left(3 \arccos\left(\frac{2}{\sqrt 5}\right)\right).
$$
The angle $\arccos\left(\frac{2}{\sqrt 5}\right)$ is the angle
opposite the shorter leg in a right triangle with legs $1$ and $2$,
so 
$\arccos\left(\frac{2}{\sqrt 5}\right) = \arctan\left(\frac12\right)$,
and the fact you want to prove is therefore equivalent to
$$
\tan\left(3 \arctan\left(\frac12\right)\right) = \frac{11}{2}.
$$
Using the triple-angle formula
$$
\tan(3x) = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x}
$$
with $x = \arctan\left(\frac12\right)$, so $\tan x = \frac12$ and
\begin{align}
\tan\left(3 \arctan\left(\frac12\right)\right)
& = \frac{3 \tan x - \tan^3 x}{1 - 3 \tan^2 x} \\
& = \frac{3 \left(\frac12\right) - \left(\frac12\right)^3}
          {1 - 3 \left(\frac12\right)^2} \\
& = \frac{11}{2}
\end{align}
which is what you needed to show.
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