# $f(x)=\cos\left(\frac{\arctan x }3\right)$

Consider the function $$f(x)=\cos\left(\frac{\arctan x }3\right)$$ from domain $$\mathbb R$$.

1. Determine the range of $$f$$.

Regarding the first question, I tried to express $$f$$ without using trigonometric functions, but it got complicated, so I thought of the following: As $$-1\le\cos x\le1$$ we have that Im$$f\subseteq[-1,1]$$. And then how $$f$$ takes the values of $$-1$$ and $$1$$, for $$x=\tan(3\pi+6k\pi),k\in\mathbb Z$$ and $$x=\tan(6k\pi),k\in\mathbb Z$$, respectively. And by the T.V.I the image is proven.

1. Prove whether $$f$$ is periodic.

I think $$f$$ is not periodic because $$\arctan x$$ is injective, so there is no $$p$$ such that $$f(x+p)=f(x)$$.

• $\arctan$ is usually chosen to have image between $-\pi/2$ and $\pi/2$. You can’t choose a different branch to get the $x$ you want on the first part. Aug 19, 2022 at 9:37
• Also, $\arctan$ is injective, not bijective. This actually makes the argument easier. Just because $\arctan$ isn’t periodic, you can’t conclude its composition with $\cos$ isn’t periodic. Aug 19, 2022 at 9:39
• Your argument for the second part is also not valid. Aug 19, 2022 at 9:39

The problem with your reasoning is that $$\tan$$ is invertible only when it is defined as a function $$\tan: \left(-\frac{\pi}{2},\frac{\pi}{2} \right) \to \mathbb{R}$$. Of course, it is invertible over other choice of domain too but this is the natural domain which is used. So, it follows that $$\arctan: \mathbb{R} \to \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$ is the inverse that you want. So, we see that: $$\forall x \in \mathbb{R}: -\frac{\pi}{6} < \frac{\arctan(x)}{3} < \frac{\pi}{6}$$ Observe that $$\cos$$ is an even function. So, we can just consider the cases where $$\frac{\arctan(x)}{3} \geq 0$$. But now, notice that in the interval $$\left[0,\frac{\pi}{6} \right)$$, you have that $$\cos$$ is decreasing. Since $$\cos$$ is continuous, it follows that: $$\text{Im}(f) = \left(\frac{\sqrt{3}}{2},1 \right]$$
As for your second argument, you have a decent idea but you need to elaborate further. Assume that $$f$$ is periodic with some period $$p > 0$$. Then, it follows that: $$\forall x \in \mathbb{R}: f(x+p) = f(x)$$ But now, notice that: $$f(0) = f(p) \implies 1 = \cos \left(\frac{\arctan(p)}{3} \right)$$ Using the information we have derived above about the range of $$\arctan$$, this implies that $$\arctan(p) = 0$$. But this means that $$p = 0$$ and that is a contradiction. It follows that $$f$$ is not periodic.