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.  
 A: 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.
A: Starting with 
$\cos(\dfrac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$
this also means starting with the triangle of trig ratios drawn and Pythagoras theorem:

$\tan(\dfrac{\arctan(\frac{11}{2})}{3}) = \frac{1}{2} = t ,$
Now use the $ \tan 3 \theta  = \dfrac{3 t - t^3}{1-3 t^2} \rightarrow \dfrac{11}{2}  $ triple angle formula and simplify, done!
A: Let $\cos\theta=\frac2{\sqrt5}$.  Then
$$\cos3\theta=4\cos^3\theta-3\cos\theta=\frac2{5\sqrt5}$$
and so
$$\tan3\theta=\frac{\sqrt{(5\sqrt5)^2-2^2}}2=\frac{11}2\ .$$
A: If $\cos(\theta / {3} )=x;$ then $\cos(\theta) = 4x^3 - 3x$ 
If $\tan(\theta) = $$11\over 2$; $\cos(\theta) = $$2\over 5\sqrt{5}$
Now, solving$ 4x^3 - 3x =$$2\over 5\sqrt{5}$ , You get x= $2\over \sqrt{5}$
A: Let $\theta =\arctan(11/2)$.  Then, $\cos(\theta)=\frac{2}{5\sqrt 5}$.  
Now, let  $\alpha$ be given by $\alpha=3\arccos(2/\sqrt 5)$.  Using the triple angle formula
$$\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!
