9
$\begingroup$

So, I was minding my own business and I thought I had defined inverse cotangent in the natural fashion. In particular, we define inverse tangent as the inverse of tangent restricted to $(-\pi/2, \pi/2)$. We all know this. So, I thought, $y = \cot(x)$ has vertical asymptotes at $x = n \pi$ for any integer $n$ so surely the natural domain to restrict cotangent for inversion would be $(0, \pi)$. Then $0 < \cot^{-1}(x) < \pi$ and $\lim_{ x\rightarrow \infty} \cot^{-1}(x) = 0$ whereas $\lim_{x \rightarrow -\infty} \cot^{-1}(x) = \pi$.

Then I made the mistake of checking my work. I found that I have not followed the apparent proper path and inverted cotangent across its asymptote at $x=0$. Apparently, $cot^{-1}$ is understood as the inverse for cotangent restricted to $(-\pi/2,0) \cup (0, \pi/2]$ which leads to the graph of inverse cotangent having a hideous discontinuity as it jumps from $-\pi/2$ to $\pi/2$ at $x=0$.

See Mathword at Wolfram where the offensive (to my taste) definition is discussed at length. I guess this goes back to Euler. However, I also noticed I am not alone as the website above says:

A different but common convention (e.g., Zwillinger 1995, p. 466; Bronshtein and Semendyayev, 1997, p. 70; Jeffrey 2000, p. 125) defines the range of $\cot^{-1}x$ as $(0,\pi)$, thus giving a function that is continuous on the real line R. Extreme care should be taken where examining identities involving inverse trigonometric functions, since their range of applicability or precise form may differ depending on the convention being used.

My question is simply this:

Which definition of inverse cotangent do you favor in your teaching ?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ I discussed this issue here($\operatorname{arccot}(x)\ne\arctan \left(\frac{1}{x}\right)$ for negative $x\,?$) and addressed your closing question in the addendum. Typo correction: you mean $\left(-\frac{\pi}2,0\right)\cup\left(0,\frac{\pi}2\right]$ instead of $\left(-\frac{\pi}2,0\right)\cup\left(0,\frac{\pi}2\right).$ $\endgroup$
    – ryang
    Feb 20 at 10:03
  • 1
    $\begingroup$ @ryang thanks for the correction and the pointer to that answer. Glad to hear Desmos graphs it correctly :) $\endgroup$
    – James S. Cook
    Feb 20 at 19:34

3 Answers 3

Reset to default
7
$\begingroup$

For teaching, I think this situation gives a good excuse to talk about how in mathematics, sometimes we get to embrace ambiguity.

One thing that differentiates the two definitions is that they satisfy different "functional equations." For one, we have

$$\hbox{arccot}(x)=\displaystyle{\frac{\pi}{2}}-\arctan(x)$$ while for the other, we have $$\hbox{arccot}(x)=\arctan\left(\frac{1}{x}\right)$$

Both are attractive. The second is an odd function, the first is not.

The second is not continuous at the origin, which many people find disturbing ("offensive" in the OP), while the first is not continuous at infinity.

One nice thing about the second one is that the branch cut for $f(z)=\hbox{arccot}(z)$ is compact and confined: it goes along the $y$-axis from $-i$ to $i$. The branch cut for the first wanders out to infinity. If I were to be offended by one or the other, I would be offended by the ugly branch cut for the first. I think that the compact branch cut is more aesthetic, but we have to trade this for continuity at the origin, caused by the fact that the origin is right in the middle of the compact branch cut.

When teaching this in calculus, I refer to the "debate" and suggest to students that this is one of the things that makes mathematics interesting. I mention briefly how it connects to complex numbers. I use different computation engines to evaluate $\hbox{arccot}(-1)$. Some will give $3\pi/4$ and others $-\pi/4$. You can show how both make sense by looking at the unit circle. When teaching complex variables, much more time can be spent on this issue, and it's really fun!

Improve this answer
$\endgroup$
1
  • $\begingroup$ Excellent answer. Exactly the sort of insight I hoped for here. $\endgroup$
    – James S. Cook
    Feb 22 at 5:53
2
$\begingroup$

enter image description here

This is from Algebra and Trigonometry by Paul Foerster. His definition has Cot following Cos, which to me is perfectly logical, and what we use in my HS.

Respectfully, Wolfram is no more an authority than Foerster. It seems we agree, and if a student approached me regarding this, I'll stick with "apparently, there are multiple conventions. you need to use what your teacher has stated in their class." Add this to the list. (Deprecating septagon and Pluto were the first two on my long list of changes between college and working in a HS nearly 30 years later.)

Edit: when I engaged in a related conversation, I was led to This Document which offers -

enter image description here

And at least was written by a body with some authority. If they disagreed with us, I'd have to rethink my strongly held belief. ISO is a well recognized organization offering agree upon standards.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ I see, well, you can disrespect Wolfram all you like, no offense here. I merely used mathworld as an example. It happened the first few things I found in my initial search on the question also lined-up with mathworld. Anyway, thanks for the references. $\endgroup$
    – James S. Cook
    Feb 22 at 1:09
  • $\begingroup$ Understood, my wording may have been poor. In the end, I appreciate your question wasn't phrased "which is correct?" but rather, "which do you use?" I'm still wondering if there's a higher source. Wolfram has a PHD in particle physics, Foerster, an MS in Mathematics. Would educated mathematicians consider ISO a legitimate standard for the final say in mathematical notation? $\endgroup$
    – JTP - Apologise to Monica
    Feb 22 at 2:30
-3
$\begingroup$

Not 100% sure if I'm following you, but if I am, the former. All it is is the cot function rotated 90 degrees. Graph it and you'll see. Just the arccot has horizontal asymptotes and the cot has vertical ones.

And I'm fine with allowing the definition to include multiples of 2pi for either of them. So what if it's a relation instead of a function?

Don't have to restrict it to the primary interval. If you do, sure, have a discontinuity. Who cares... I mean regular cot and tan do funky things when sin or cos become zero. So what if their inverses do also.

Just graph it manually and I think you'll feel better about it. Never bothered me. And this is how both books I have show the graphs. Lyman Kells et al Plane and Spherical Trig. And Frank Ayres Schaum's Outline for First Year College Math.

Improve this answer
$\endgroup$
6
  • $\begingroup$ I agree the discontinuous version is popular and logical, it's a matter of taste. $\endgroup$
    – James S. Cook
    Feb 20 at 19:36
  • 5
    $\begingroup$ This answer (and in particular the sentence "All it is is the cot function rotated 90 degrees") misses the point entirely. For one thing, it's not a rotation, but a reflection of the graph across the line $y = x$. More significantly, the cotangent function is not one-to-one, hence not invertible, so in order to define an "inverse cotangent" one must restrict to some interval on which cot is monotone. The graph of the inverse cotangent function is of necessity only a reflection of part of the graph of the cotangent function... 1/ $\endgroup$
    – mweiss
    Feb 20 at 19:45
  • 1
    $\begingroup$ ... The question "Which part?" is important because it determines the range of the function, which is otherwise ambiguous. Furthermore, the inverse tangent function does not have a discontinuity, precisely because the universal convention is to avoid it by restricting the domain of the tangent function to $(-\pi/2, \pi/2)$. $\endgroup$
    – mweiss
    Feb 20 at 19:45
  • 2
    $\begingroup$ This is just wrong; -1. If you rotate the cotangent function you don't get the inverse, and any simple answer like this results in something that is not a function. $\endgroup$
    – Chris Cunningham ♦
    Feb 21 at 17:41
  • $\begingroup$ You are correct, that it's a reflection across the dihedral angle, not a rotation. Of course, this doesn't change the operation on the vertical asymptotes. Or the conversion of a function to a relation. But yes, thanks for the precision, curvatures are different. But of course, it was the asymptote that was the concern. So really the precision is not addressing the issue. $\endgroup$
    – guest
    Mar 3 at 14:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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