Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I happened to find that the graphs of $\cos(\arctan(\tan x))$ and $\cos(x)$ are different. Why?

Is there something wrong with $\arctan(\tan x)=x$?

Thanks!

share|improve this question
    
Since $\tan x$ is periodic, $\arctan(\tan x)$ only holds over one period of $\tan x$. You can choose whichever period you want, but the standard choice is the interval $(-\pi/2,\pi/2)$. –  Alex Becker Sep 20 '12 at 19:06
    
The range of $\arctan$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$, so at best the formula can be valid for $|x| < \frac{\pi}{2}$. –  copper.hat Sep 20 '12 at 19:12

1 Answer 1

up vote 3 down vote accepted

The function $x \mapsto \tan x$ is $\pi$-periodic whereas $x \mapsto \cos x$ is $2\pi$-periodic. Now, $\arctan$ will land you somewhere in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, even if your original $\theta$ does not lie in this interval. If $\cos$ were $\pi$-period then it would be fine, but since it isn't, it means that we have things like, for instance, $$\cos(\arctan(\tan \pi)) = \cos(\arctan(0)) = \cos(0) = 1 \ne -1 = \cos \pi$$ More precisely, if $x \in \left(\frac{(2n-1)\pi}{2}, \frac{(2n+1)\pi}{2} \right)$, then $\cos(\arctan(\tan x)) = x$ if and only if $n$ is even, or $-x$ if $n$ is odd, since $\cos(x+(2k+1)\pi)=-\cos x$.


Generally $\arctan(\tan x)=x+n\pi$ for some $n\in \mathbb{Z}$, and $n=0$ if and only if $x \in (-\frac{\pi}{2},\frac{\pi}{2})$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.