I know that $$\cos(\dfrac{\pi}{3} - \arctan(x))= \dfrac{1}{2\sqrt{(1+x^2)}} + \dfrac{\sqrt{3}x}{2\sqrt{(1+x^2)}}$$

$\cos\left(\dfrac{\pi}{3} - \dfrac{\arctan(x)}{3}\right)$ = ?

$\cos\left(\dfrac{\pi}{3} - \dfrac{\arctan(x)}{3}\right) = \cos\left(\dfrac{\pi}{3}\right)\cos\left(\dfrac{\arctan(x)}{3}\right) + \sin\left(\dfrac{\pi}{3}\right)\sin\left(\dfrac{\arctan(x)}{3}\right) = \dfrac{1}{2}\cos\left(\dfrac{\arctan(x)}{3}\right) + \dfrac{\sqrt{3}}{2}\sin\left(\dfrac{\arctan(x)}{3}\right)$

but I can't go further since I don't know how to solve $\sin\left(\dfrac{\arctan(x)}{3}\right)$ and $\cos\left(\dfrac{\arctan(x)}{3}\right)$.

Any suggestion?

  • $\begingroup$ It is slightly more complicated. One way of describing the reason is that $\arctan x$ is really only determined up to an integer multiple of $\pi$. In other words, if $\tan \alpha=x$ we also have $\tan(\alpha+\pi)=x$ and $\tan(\alpha+2\pi)=x$. Therefore $\sin(\dfrac13\arctan x)$ can refer to any of $s_1=\sin(\alpha/3)$, $s_2=\sin(\alpha/3+\pi/3)$ or $s_3=\sin(\alpha/3+2\pi/3)$. I suspect (haven't checked yet, sorry) that it is possible to write down a cubic polynomial with roots $s_1,s_2,s_3$. Not sure, whether that's what you would want? $\endgroup$ – Jyrki Lahtonen May 6 at 7:41
  • $\begingroup$ Basically you can use the triple angle formula for tangent to write $x=\tan\alpha$ in terms of $\tan(\alpha/3)$, solve for $\tan(\alpha/3)$, and then convert that to $\sin(\alpha/3)$ or $\cos(\alpha/3)$ as you see fit. The resulting cubic has two other roots, and my first comment explains how they are related. $\endgroup$ – Jyrki Lahtonen May 6 at 7:45
  • $\begingroup$ All of this comes from a depressed cubic function. I calculated the angle $\Theta$ that is $\pi - \arctan (x)$ and the first root is $2\sqrt[3]{R}\cos{\dfrac{\Theta}{3}}$. It gets worst anytime lol $\endgroup$ – JackLametta May 6 at 7:48
  • 1
    $\begingroup$ Oops! So together we were running in circles :-) Yup, that is a known drag of using trigonometry to solve cubics. With numerical values you will be fine with a pocket calculator. Otherwise you may need to go to Cardano's formula, a local link. $\endgroup$ – Jyrki Lahtonen May 6 at 7:53
  • $\begingroup$ @JyrkiLahtonen do you think if would be better to leave trigonometry? $\endgroup$ – JackLametta May 6 at 7:54

Let us care about

$$t:=\tan\left(\frac{\arctan(x)}3\right), $$

using the fact that


By the triple angle formula, this equation writes $$\frac{3t-t^3}{1-3t^2}=x$$



We depress it with $u:=t-x$, giving


Now the discriminant is given by


As it is negative, the final expression will involve cubic roots of complex numbers, which cannot be expressed without… trigonometry, and you are circling in rounds.

  • $\begingroup$ So, I'm pretty stucked $\endgroup$ – JackLametta May 6 at 8:17
  • $\begingroup$ @JackLametta: mathematicians have been stuck on this for decades, if not centuries. $\endgroup$ – Yves Daoust May 6 at 8:18
  • $\begingroup$ I do feel them, now :) $\endgroup$ – JackLametta May 6 at 8:18
  • $\begingroup$ @JackLametta: you can still use the formula with complex cubic roots. wolframalpha.com/input/?i=t%5E3-3xt%5E2-3t%2Bx%3D0,+solve+for+t $\endgroup$ – Yves Daoust May 6 at 8:19
  • $\begingroup$ There is something unclear about this, cubic polynomials have always one real root (by the intermediate value theorem). So what do I don't understand? $\endgroup$ – Shashi May 6 at 9:07

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