10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I would like to apologize in advance in case I come across as rambling and being unhelpful. $\endgroup$ – cpiegore Feb 11 '16 at 4:10
  • $\begingroup$ You might try other methods of finding the roots, such as using some online root finder. (This is because maybe you'll get some third answer that contains pieces of the other two). I assume you've looked for unusual trig identities that help. $\endgroup$ – Nate 8 Feb 11 '16 at 4:30
  • $\begingroup$ Are you allowed to use multiple angle formulae? $\endgroup$ – David Feb 11 '16 at 4:34
  • $\begingroup$ @Nate8 Finding the roots is not the point. As I stated in the post I already knew what they are. The point is prove the equation in the title without using the fact x = 4 is one of the roots or the fact that (2+i)^3 = 2+11i. I know that what I want follows from these things. However I have this urge to find a proof that avoids them because, for reasons I cannot articulate clearly, they feel like byproducts of something more fundamental. $\endgroup$ – cpiegore Feb 11 '16 at 4:36
  • $\begingroup$ @David You are allowed to use anything as long as it does depend on (2+i)^3 = 2+11i or x = 4 is a root of x^3-15x-4. More importantly, I would like to point out that this is not a homework problem; I am not taking any math classes at this moment. I am doing mathematics as a hobby. $\endgroup$ – cpiegore Feb 11 '16 at 4:46
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Man, I had almost this exact thing in the works... beat me to it! +1 $\endgroup$ – Brevan Ellefsen Feb 11 '16 at 4:37
  • $\begingroup$ Submit it anyway. If your work is independent, I see no problem. There are no patents on answers here. $\endgroup$ – marty cohen Feb 11 '16 at 6:15
3
$\begingroup$

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:

arctan2 triangle

$\tan(\dfrac{\arctan(\frac{11}{2})}{3}) = \frac{1}{2} = t ,$

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!

$\endgroup$
2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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= $2\over \sqrt{5}$

$\endgroup$
1
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.