TITLE: Proving that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$
QUESTION [9 upvotes]: 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.

REPLY [3 votes]: 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!