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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 $f(x+p)=f(x)$.

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    $\begingroup$ $\arctan$ 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. $\endgroup$ Aug 19, 2022 at 9:37
  • $\begingroup$ 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. $\endgroup$ Aug 19, 2022 at 9:39
  • $\begingroup$ Your argument for the second part is also not valid. $\endgroup$ Aug 19, 2022 at 9:39

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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.

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