# Talk:Inverse trigonometric functions

WikiProject Mathematics (Rated B-class, Top-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
 B Class
 Top Importance
Field: Analysis
One of the 500 most frequently viewed mathematics articles.

Archives:

## proofs / citations

Consider the section Relationships between trigonometric functions and inverse trigonometric functions, which states useful rules without giving any background.

Ideally the section should lead with an introductory sentence or two that also outlines how these rules are derived (with a citation to any reliable source, even a math professor's website, that performs this derivation in detail).

Even failing this, there still ought at least to be some kind of citation given (even if an explicit proof is inappropriate). Cesiumfrog (talk) 04:21, 25 April 2012 (UTC)

They are easy to derive. For example, to get
$\sin (\arctan x) = \frac{x}{\sqrt{1+x^2}} \,,$
one calculates as follows. Let
$\theta = \arctan x \,,$
then
$x = \tan \theta$
$x^2 = \tan^2 \theta$
$1 + x^2 = 1 + \tan^2 \theta = \sec^2 \theta$
$\sqrt {1 + x^2} = \vert \sec \theta \vert = \sec \theta$
because θ is in (-π/2, π/2) where secant is positive
$\frac{1}{\sqrt {1 + x^2}} = \cos \theta$
$\frac{x}{\sqrt {1 + x^2}} = \tan \theta \cdot \cos \theta = \sin \theta \,.$
Substituting for θ, we get the desired formula. JRSpriggs (talk) 07:11, 25 April 2012 (UTC)
Yes a citation to where they are derived is the way to go okay, preferably to something that can be accessed easily. Dmcq (talk) 13:24, 25 April 2012 (UTC)
How about we just draw a triangle for each of the identities? Then there is no need to prove them, and the section benefits from some badly needed visual aids. Sławomir Biały (talk) 01:36, 29 April 2012 (UTC)
Yes. In the case above, you could put: θ in a corner, 1 on the adjacent side, x on the opposite side, and √(1+x2) on the hypotenuse. JRSpriggs (talk) 02:10, 29 April 2012 (UTC)

This doesn't appear to be done yet. Here is a diagram that may answer the above concerns for the section Relationships between trigonometric functions and inverse trigonometric functions?

The angle θ is an inverse trig function of x, or 1 + x2, or a ratio of them.

Best, M∧Ŝc2ħεИτlk 21:06, 14 April 2013 (UTC)

## History of the sin-1(x) notation?

Does anyone known where that awful sin-1(x) notation originated? I know it goes back a long way... I have seen in in 19th century textbooks. Tfr000 (talk) 13:28, 13 June 2012 (UTC)

Good question. To do it justice, we'd also need to know the histories of the notation for exponentation and of the notation for inverses of arbitrary functions. Cesiumfrog (talk) 06:45, 14 June 2012 (UTC)
Found it... and added it to the article. Tfr000 (talk) 19:35, 26 May 2015 (UTC)

## Extension to complex plane

I reverted an edit by Syed Wamiq Ahmed Hashmi (talk · contribs) to the section Inverse trigonometric functions#Derivatives of inverse trigonometric functions in which he restricted the domains of the derivatives to certain subsets of the real numbers. The formulas he changed were intended to apply to extensions of the inverse functions to the complex plane (a fact which he overlooked, but is now aware of). This brings my attention to an issue — this article mentions the complex extensions in several places, but never defines them. The article was apparently written with only the real versions in mind and then later patched to include pieces of information about the complex extensions. I think that this calls for a redesign of the whole article to give at least equal treatment to the complex extensions. And as Wamiq noticed, we need to indicate where the cuts in the complex domain are located. JRSpriggs (talk) 15:38, 14 April 2013 (UTC)

The problem I am seeing now is how do we handle the fact that there are multiple sheets in the complex domain for the inverse trigonometric functions. And they interact with the fact that the square-root has two sheets. JRSpriggs (talk) 23:41, 14 April 2013 (UTC)

There are several possible ways of defining the extensions of inverse trigonometric functions to the complex plane. In addition to how the functions are computed, they may differ in where one puts the cuts between different sheets of the functions. I would like to suggest the following:

$\arctan z = \int_0^z \frac{d z}{1 + z^2} \,$

provided that the contour of integration does not cross the part of the imaginary axis which does not lie strictly between -i and +i;

$\arcsin z = \arctan \frac{z}{\sqrt{1 - z^2}} \,$

where the square-root function has its cut along the negative real axis;

$\arccos z = \frac{\pi}{2} - \arcsin z \,;$
$\arccot z = \frac{\pi}{2} - \arctan z \,;$
$\arcsec z = \arccos \frac{1}{z} \,;$
$\arccsc z = \arcsin \frac{1}{z} \,.$

If we adopt that suggestion, then it may affect the signs of the derivatives of the functions. To verify what those derivatives would be:

$\frac{d \arcsin z}{d z} = \frac{1}{1 + \left( \frac{z}{\sqrt{1 - z^2}} \right)^2} \cdot \frac{\sqrt{1 - z^2} - z \left( \frac{- 2 z}{2 \sqrt{1 - z^2}} \right)}{1 - z^2} = \frac{1}{\sqrt{1 - z^2}} \,$

which is as expected;

$\frac{d \arcsec z}{d z} = \frac{-1}{\sqrt{1 - z^{-2}}} \cdot \frac{-1}{z^2} = \frac{1}{z^2 \sqrt{1 - z^{-2}}} = \pm \frac{1}{z \sqrt{z^2 - 1}} \,$

which may have a different sign than the usual expression. Similarly for arccsc. JRSpriggs (talk) 08:28, 17 April 2013 (UTC)

## Notation

<< I copied this from my talk page. JRSpriggs (talk) 08:01, 15 April 2013 (UTC) >>

Thanks a lot! That section looks better now ☺. Well, I see an issue with the notation used on Wikipedia for the inverse trigonometric functions, i.e., the convention here is to denote all functions with minuscule letters but to add the word arc with the inverse ones (sin x, arcsin x, etc...) but what we (along with our textbooks) do, is to denote regular functions with minuscule letters, e.g., sin x, cos x, etc., and the inverse functions with the first letter majuscule and a −1 superscript, e.g., Sin−1 x, Cos−1 x, etc., which causes no confusion between the inverse function (Sin−1 x) and the multiplicative inverse (sin−1 x). This notation is nowhere to be found here. I personally find the arc notation a bit odd. Do you find this (capital) notation at least worth mentioning in the article (if the arc notation is popular and cannot be removed)? Hoping to get a reply in the affirmative... Regards,

— Syɛd Шαмiq Aнмɛd Hαsнмi (тαlк) 06:26, 15 April 2013 (UTC)
I would rather not change the notation that way. Superscript minus one could be misinterpreted as the multiplicative inverse rather than the inverse with respect to composition. Please see the archive, Talk:Inverse trigonometric functions/Archive 1, for more discussion of this issue.
When referring to one of our articles or talk pages, please use [[this method]] rather than [http://en.wikipedia.org/wiki/this_method this method] . JRSpriggs (talk) 08:01, 15 April 2013 (UTC)
O.K., Now, I’ve got the things you said. I’ll do them as such. As to the notation, the article Inverse function, too, uses f−1 for the inverse function of f which here, means the compositional inverse (the −1 doesn’t mean multiplicative inverse which would be denoted by (f)−1...) So, are you satisfied as to the use of −1? Moreover, as I have already said, this notation doesn’t clash with that for the multiplicative ones. I have seen the archive and I do not demand replacement now, but just a bare mention (like somthing in the beginning of the article, saying that these notations are also used, which don’t cause confusion; for other people like me who don’t use and are unfamiliar with the arc notation).
— Syɛd Шαмiq Aнмɛd Hαsнмi (тαlк) 10:12, 15 April 2013 (UTC)

## Explanation of "z" in "Expression as definite integrals" section?

The "Expression as definite integrals" section lists expressions of the functions of x with a z^2 and dz on the right hand side, but no explanation of where this "z" comes from. It's been a while since I've had trig in school so I'm guessing I need to be reminded of what it is, and it would be good to have a small explanation at the side or bottom explaining where Z comes from. Or, if it's a typo, we need to change the expressions to have x^2 and dx rather than z. But my money is that I'm the one that needs educating. Could a section be added to clear up what "z" is? 74.10.5.213 (talk) 22:32, 1 October 2013 (UTC)

See definite integral. This is not a matter of trigonometry, but rather of your failure to understand the integral notation. For example, in
$\arccos x = \int_x^1 \frac{1}{ \sqrt{1 - z^2} } \, dz ,\qquad |x| \leq 1$,
if x=0.5, then
$\arccos 0.5 = \int_{0.5}^1 \frac{1}{ \sqrt{1 - z^2} } \, dz = \lim_{n \to \infty} \sum_{k = 1}^{n} \frac{1}{ \sqrt{ 1 - \left( 0.5 + ( k - \frac12 ) \frac{ (1 - 0.5) }{n} \right)^2 } } \frac{1 - 0.5}{n}$.
In other words, z is replaced by $0.5 + ( k - \frac12 ) \frac{ (1 - 0.5) }{n}$ and dz is replaced by $\frac{ (1 - 0.5) }{n}$. OK? JRSpriggs (talk) 02:11, 2 October 2013 (UTC)

## Definition of arccot

Wikipedia defines arccot to be the inversion of cot on the interval

]0,π[.

Whereas NIST (http://dlmf.nist.gov/4.23), as well as wolframalpha (http://www.wolframalpha.com/input/?i=arccotangent) use

]-π/2,π/2[.

Yesterday I just needed to plug arccot in somewhere and got quite confused. I think a caveat on this article would be nice explaining the different choice.

Furthermore: the plot of arccot in the complex plane (https://en.wikipedia.org/wiki/File:Complex_ArcCot.jpg) uses the mathematica/nist definition contradicting the article. Very confusing! — Preceding unsigned comment added by 93.194.241.203 (talk) 14:26, 10 January 2014 (UTC)

File:Arctangent Arccotangent.svg shows the correct version of the arccotangent. One must choose between
$\arccot (x) = \frac{\pi}{2} - \arctan (x) \,$
and
$\arccot (x) = \arctan \left( \frac{1}{x} \right) \,.$
I choose the former because it gives a value to arccot (0) and is continuous there whereas the latter definition is discontinuous and undefined at zero. JRSpriggs (talk) 07:20, 11 January 2014 (UTC)