0

So I was solving a math problem when I came upon this weird function with some interesting points:

$\begin{equation} f(x) = \tan^{-1}\biggl(\frac{ 8\sin^{3}(x)\cos(x)}{ 1-8\sin^{2}(x)\cos^2(x) }\biggr) \end{equation}$

I found that, for some reason, the points $(\frac{1}{12}\pi, \frac{1}{12}\pi)$, $(\frac{1}{9}\pi, \frac{1}{3}\pi)$, $(\frac{1}{6}\pi, -\frac{1}{3}\pi)$ all lie on the function.

Is there any way to simplify $f(x)$ to make sense of all of this? Why would such nice numbers come out of a function looking as nasty as that?

crxyz
  • 346
  • I'm not sure if there's anything interesting about those points. Plugging in $\pi/12$, $\pi/9$, and $\pi/6$ give you "nice" numbers because, well, the function just outputs those numbers. A simplification can be found here. – Accelerator Apr 26 '23 at 00:25

1 Answers1

2

Note that $1 - 8\sin^2(x)\cos^2(x) = 1 - 2(2\sin(x)\cos(x))^2 = 1 - 2\sin^2(2x) = \cos(4x)$. So,

\begin{align*} f(x) &= \tan^{-1}\left(\frac{4\sin^2(x)\sin(2x)}{\cos(4x)}\right) \\ &= \tan^{-1}\left(\frac{4\sin^2(x)\sin(2x)}{\sin(4x)}\cdot\tan(4x)\right) \\ &= \tan^{-1}\left(\frac{2\sin^2(x)}{\sin(2x)}\cdot\tan(2x)\tan(4x)\right) \\ &= \tan^{-1}\left(\tan(x)\tan(2x)\tan(4x)\right) \end{align*}

However, I am not sure if this has a nice form. I suspect it has though... Maybe someone can finish it off :D

In particular, $\tan\left(\frac{\pi}{9}\right)\tan\left(\frac{2\pi}{9}\right)\tan\left(\frac{4\pi}{9}\right) = \sqrt{3}$ follows from this answer (Look at the polynomial $x^6 - 33x^4 + 27x^2 - 3 = 0$ and apply Vieta's). Also, $\tan\left(\frac{\pi}{12}\right)\tan\left(\frac{\pi}{6}\right)\tan\left(\frac{\pi}{3}\right) = \tan\left(\frac{\pi}{12}\right)$ and the rest follows.

Gareth Ma
  • 3,725
  • ah thank you! f(1/9 pi) is the exact form in which my math problem was in (it was $\tan(20)\tan(40)\tan(80)=\tan(x)$). – crxyz Apr 26 '23 at 01:09