0
$\begingroup$

The range of $\sin^{-1}(x)$ is (-$\pi/2, \pi/2)$, so is that of cosec$^−$$^1$(x) {except the 0 since 1/0 isn't defined}

The range of $\cos^{-1}(x)$ is (0, $\pi)$, so is that of $\sec^{-1}(x)$ {except the $\pi/2$ since 1/0 isn't defined}

Why isn't the same true for $\tan^{-1}(x)$ and $\cot^{-1}(x)$?

Why is the range of $\tan^{-1}(x)$ = (-$\pi/2, \pi/2)$ but that of $\cot^{-1}(x)$ = (0, $\pi)$

aren't they reciprocals?

enter image description here

from the graph, it's clear that that would be possible too (i.e. they can have the same range too, except cot$^−$$^1$(x) would not have the 0 for obvious reasons)

$\endgroup$
7
  • $\begingroup$ It is, in Wolfram Alpha or if you define $\tan^{-1}(\frac1x)=\cot^{-1}(x)$. There are different conventions $\endgroup$
    – Тyma Gaidash
    Commented Nov 5, 2023 at 14:02
  • $\begingroup$ because $\cot (\theta) = \frac{\cos (\theta)}{\sin (\theta)}$. And $\sin(\theta) = 0$ when $\theta = 0$ and $\theta = \pi$ $\endgroup$
    – Miranda
    Commented Nov 5, 2023 at 14:20
  • $\begingroup$ Extending the comment of @Miranda, note that the range of the arctan function, $~-\pi/2 < \theta < \pi/2,~$ matches the range of the arcsine function, except for its endpoints. So, you would expect the range of the arccot function to match the range of the arccos function, excepting its endpoints, which is $~0 < \theta < \pi.$ $\endgroup$
    – user2661923
    Commented Nov 5, 2023 at 16:23
  • $\begingroup$ and why would I expect that @user2661923? only because they are in the numerator? Well, arcsec matches the range of arccos, as I mentioned in the question. Wouldn't arccot match the range of arcsin by that logic? My only point of this argument was that it would be so convenient (and it works too) for the principal domain of the cot function to match that of the tan function, only because they are reciprocals, just like all other trig functions do. $\endgroup$
    – Maddy
    Commented Nov 6, 2023 at 12:29
  • $\begingroup$ Tangent is sine/cosine. So, arctan function's range matches the range of the inverse function, arcsine, that corresponds to the numerator of the tangent function. Exception is that endpoints excluded, because you can't have the denominator, cosine, equal to zero. Perfectly analogous logic focuses on cotan = cos/sine, so range of arccot function should match range of arccos function, with endpoints excluded, since (again), you can't have denominator (in this case the sine function) equal to zero. ...see my next comment $\endgroup$
    – user2661923
    Commented Nov 6, 2023 at 12:33

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .