What is $\cos\left(\frac{\arctan(x)}{3}\right)$ and $\sin\left(\frac{\arctan(x)}{3}\right)$?

Question

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?

Answer 1

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.

Answer 2

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.