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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?

StubbornAtom
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JackLametta
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  • 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? – Jyrki Lahtonen May 06 '19 at 07:41
  • 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. – Jyrki Lahtonen May 06 '19 at 07:45
  • 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

     – JackLametta May 06 '19 at 07:48
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    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. – Jyrki Lahtonen May 06 '19 at 07:53
  • @JyrkiLahtonen do you think if would be better to leave trigonometry? – JackLametta May 06 '19 at 07:54
  • IDK. Depends on what you want to do with this. Cardano's formula calls for you to calculate complex cubic roots. Guess what! When you calculate those you will run into the problem of calculating $\sin(\alpha/3)$ and $\cos(\alpha)$ for a "known" $\alpha$! It all amounts to the same! You cannot usually find the roots a cubic with real radicals alone. Complex roots are required, and then you end up with these trig problems :-( – Jyrki Lahtonen May 06 '19 at 07:57
  • @JyrkiLahtonen I'm using a real symmetric matrix, I'm about to calculate eigenvalues so I need to find 3 couples ${u_{i},v_{k}}$ that makes the roots as real. (Does it help and make sense?) – JackLametta May 06 '19 at 08:00
  • I'm afraid it does not help. You know that all the eigenvalues will be real. This automatically means that your cubic falls into casus irreducibilis, and cannot be solved with real-valued radicals alone. – Jyrki Lahtonen May 06 '19 at 08:06
  • I guess there's nothing I can do :/ – JackLametta May 06 '19 at 08:08
  • If the entries of your matrix are random reals, then you are more or less forced to approximate them (and the roots) anyway (and none of this matters). If they are integers or rationals, then there is the small chance that one of the roots is rational. You can find if that is the case using the rational root test. Then the casus irreducibilis machinery doesn't come into play at all, and you can find the other roots with the quadratic formula. – Jyrki Lahtonen May 06 '19 at 08:11
  • Let us continue this discussion in chat. – JackLametta May 06 '19 at 08:12

2 Answers2

4

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.

0

Triple cosine formula is more well-known: $\cos(3\theta)=4\cos^3\theta-3\cos\theta$. If we let $3\theta=\arctan x$ and $y=\cos\theta$, then $4y^3-3y=\cos3\theta=\cos\arctan x=\frac{1}{\sqrt{x^2+1}}$. Hence, we have to solve the equation $$y^3-\frac34y-\frac{1}{4\sqrt{x^2+1}}=0.$$ Its Lagrange resolvent is $z^2--\frac{1}{4\sqrt{x^2+1}}+\frac{1}{4}=0$ with solution $z_{1,2}=\frac{1\pm xi}{8\sqrt{x^2+1}}.$ Hence the real solution of the equation is $$y=\frac{(1+xi)^{1/3}+(1-xi)^{1/3}}{2(x^2+1)^{1/6}}.$$ For $x=1$ guess what? $y=\cos 15^{\circ}$... We don't need WA for this particular example but when $x=2$ we need.

Bob Dobbs
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