1

Currently I'm learning to use Cardano's Method to solve cubic equations and one of the interesting things is that, even for an equation with simple solutions, you may still have to generate some difficult expression to arrive at those solutions. I've come across several difficult expressions that equate to a simple rational number, but I don't have the mathematical tools to evaluate them.

  1. For example, the following expression: $$\frac{2\sqrt{19}}{3}\cos\left(\frac13\tan^{-1}\left(\frac{135}{28\sqrt3}\right)\right)$$ is equal to $\frac83$ (according to my calculator) and this checks out when used in subsequent calculations. Can you evaluate this expression analytically? I could if the $\frac13$ wasn't between the $\cos$ and $\tan^{-1}$, but I'm lost otherwise.
  2. This expression: $$\sqrt[3]{\frac{440}{27}+\frac{56\sqrt2}{3}}+\sqrt[3]{\frac{440}{27}-\frac{56\sqrt2}{3}}$$ is equal to $\frac43$. Same as above, but how do you even begin with this one?

Can you evaluate these expressions? Or is it just a case of proving they converge to the answer? Does my calculator "know" the answer, or is it just guessing?

Stephen Douglas Allen
  • 158
  • 1
    For the second example, there are algorithms for denesting radicals, but it is a big mess. A classic paper about this is here: https://www.sciencedirect.com/science/article/pii/S0747717185800146 – Cheerful Parsnip Aug 03 '24 at 04:50
  • Mathematica confirms the analytic value is $\frac{8}{3}$, but does not give the steps of simplification. There seems to be a typo in your second term. – David G. Stork Aug 03 '24 at 04:50
  • 1
    I wouldn't consider the first example as a difficult expression which is equal to rational number as it is $\cos\left(\frac13\tan^{-1}\left(\frac{135}{28\sqrt3}\right)\right) = \dfrac4{\sqrt{19}}$ in disguise. – User Aug 03 '24 at 04:59
  • 1
    For the 2nd question, I favor using Quanto's answer to this MathSE denesting question – user2661923 Aug 03 '24 at 05:54
  • 1
    For the second expression which is in the form of $\sqrt[3]{a+b}+\sqrt[3]{a-b}$, I could find similar questions. For example, see https://math.stackexchange.com/questions/825741/proving-an-expression-is-4?noredirect=1 https://math.stackexchange.com/questions/1097558/show-that-sqrt33-sqrt21-8-sqrt33-sqrt21-8-1?noredirect=1&lq=1 The second one has several related links too. – User Aug 03 '24 at 06:03

1 Answers1

1

For the first expression, we need to write $\tan^{-1}x$ in the form of $\cos^{-1}x$, then: $$ \tan^{-1}\left(\frac{135}{28\sqrt3}\right)=\cos^{-1}\left(\frac{28}{19\sqrt{19}}\right) $$ Moreover, apply the formula for $\cos3\theta$, which is: $$\cos3\theta=4\cos^3\theta-3\cos\theta$$ $$\theta=\frac13(\cos^{-1}(4\cos^3\theta-3\cos\theta))$$ Consider $\cos\theta$ to be $x$, then: $$x=\frac13(\cos^{-1}(4x^3-3x))$$ We have $4x^3-3x=\frac{28}{19\sqrt{19}}$, and solving this depressed cubic equation analytically gives us $x=\frac{4}{\sqrt{19}}$.

Therefore, $$\frac{2\sqrt{19}}{3}\cos\left(\cos^{-1}\frac{4}{\sqrt{19}}\right)=\frac{8}{3}$$

For the second expression, we need to find the $(a+b)^3$ form here: $$\frac{440}{27}+\frac{56\sqrt2}{3}=\frac{1}{27}(440+504\sqrt2)$$

Here, since there is a radical term $\sqrt{2}$, one of the terms in the expression $(a+b)^3$, either $a$ or $b$ must have a $\sqrt2$ term in it. We can use coefficient comparison to find so, considering $b=c\sqrt2$: $$ (a+c\sqrt2)^3=a^3+2c^3\sqrt2+3a^2c\sqrt2+6ac^2 $$ Here, $$2c^3+3a^2c=504$$$$a^3+6ac^2=440$$ You can go through rigorous mathematical process and solve this analytically. But, you can also guess. Since $a^3$ and $c^3$ are perfect cube, start taking numbers around $440$ and $168$, that are perfect cubes, surely you will set a solution.

Here, I found $a=2$ and $b=6$. Therefore, $$\sqrt[3]{\left(\frac{440}{27}+\frac{56\sqrt2}{3}\right)}$$ $$ =\sqrt[3]{\frac{1}{27}\left(440+504\sqrt2\right)} $$ $$=\frac13(2+6\sqrt2)$$ Similarly, the other term evaluates to $$=\frac13(2-6\sqrt2)$$, and their sum is $\frac{4}{3}$

M.Riyan
  • 1,450