Proving that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$

Question

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.

Answer 1

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.

Answer 2

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!

Answer 3

Let $\cos\theta=\frac2{\sqrt5}$. Then $$\cos3\theta=4\cos^3\theta-3\cos\theta=\frac2{5\sqrt5}$$ and so $$\tan3\theta=\frac{\sqrt{(5\sqrt5)^2-2^2}}2=\frac{11}2\ .$$

Answer 4

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= $2\over \sqrt{5}$

Answer 5

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!