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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?

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  • $\begingroup$ 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
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    $\begingroup$ 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
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Let us care about

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

using the fact that

$$\tan\left(3\frac{\arctan(x)}3\right)=x.$$

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

or

$$t^3-3xt^2-3t+x=0.$$

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

$$u^3-3(x^2+1)u-2x(x^2+1)=0.$$

Now the discriminant is given by

$$(x(x^2+1))^2-(x^2+1)^3=-(x^2+1)^2.$$

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.

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  • $\begingroup$ 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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