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This is a homework problem I have. I don't remember learning the difference, and searching hasn't helped explain the difference between the capitalization.

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    $\begingroup$ Other than the way it appears, there is no difference. I guess it's the way people write them. In the older days with typewritters, people could not type $sin^{-1}(x)$ on a type writter. So they used $arcsin(x)$ instead. $\endgroup$ – KingDuken Jul 1 '16 at 14:47
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    $\begingroup$ Sometimes people use capitalization to refer to the principal value, but I haven't really seen that used myself. $\endgroup$ – Clement C. Jul 1 '16 at 14:49
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    $\begingroup$ @MaxLi That's interesting. I personally have never heard of capitalizing a trig function to have a different meaning. $\endgroup$ – KingDuken Jul 1 '16 at 14:52
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    $\begingroup$ In high school, they taught us that the upper case Sin$^{-1}$ returned the principle value (obviously they didn't put it in those words). I haven't seen anyone worry about the distinction since high school though. $\endgroup$ – Merkh Jul 1 '16 at 15:09
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    $\begingroup$ $\displaystyle \sin^{-1}\left(x\right)$ can be misleading to $\displaystyle {1 \over \sin\left(x\right)}$. It's always better to use $\displaystyle \arcsin\left(x\right)$. It even exists as a $\displaystyle\LaTeX$ command. $\endgroup$ – Felix Marin Mar 16 at 4:34
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While I'm almost certain the other answers are correct about $\arcsin$ being equivalent to $\sin^{-1}$, the majority of sources I can find contradict the answers given by MarianD and Polygon:

According to these sources $$\arcsin x =\sin^{-1}(x) = \{ \theta \in\Bbb{R} : \sin \theta = x\} ,$$ whereas $$\operatorname{Arcsin} x =\operatorname{Sin}^{-1}(x)= \theta,$$ where $\theta$ is the unique angle between $-\pi/2$ and $\pi/2$ (inclusive) such that $\sin\theta = x$, which is often called the principal value.

Side note

I'd like to comment that in practice, in most cases the lower case variants are used to denote the principal value, and the upper case variants are not used at all. However, when they are used, this does appear to be the correct answer regarding their meaning.

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  • $\begingroup$ MathOnWeb.com is an example of a source, which strictly contradict your answer. $\endgroup$ – MarianD Mar 16 at 5:59
  • $\begingroup$ @MarianD I've retracted my downvote on your answer in light of the new source, but imo the majority of sources appear to support my answer, including ones I have high confidence in, like Wolfram MathWorld. $\endgroup$ – jgon Mar 16 at 18:32
  • $\begingroup$ I agree with you, +1. $\endgroup$ – MarianD Mar 16 at 18:40
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We may symbolically write:

  • $\arcsin\ $ = $\sin^{-1}\, $
  • Arc$\sin\ $ = Sin$^{-1}\, $

because

  • $\arcsin\ $ and $\sin^{-1}\, $ are only two different notations for the same function,

  • Arc$\sin\ $ and Sin$^{-1}\, $are only two different notations for the same maping.

Definition: The Arcsine of $x$, denoted $\operatorname{Arcsin}(x)$, is defined as "the set of all angles whose sine is $x$". It may be interpreted as a one-to-many relation.

Definition: The arcsine of x, denoted $\arcsin(x)$, is defined as "the (only) angle from the closed interval $[-\pi/2, +\pi/2]$ whose sine is $x$". It may be interpreted as a one-to-one relation.


So, for example

$$\arcsin\left(\frac1 2\right)= \sin^{-1}\left(\frac1 2\right) = \frac{\pi}{6} $$

while

$$\operatorname{Arcsin}\left(\frac1 2\right)= \operatorname{Sin^{-1}}\left(\frac1 2\right) = \left.\left\{\frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi\ \right| \ k \in \mathbb Z\right\} $$

See also definitions and illustrations of these functions / mappings in The Algebra Help e-book on MathOnWeb.com site.


Edit:

It seems that definitions for capitalized and lowercase names are swapped. Several other sources - see the jgon's answer - defines them in the opposite manner.

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  • $\begingroup$ Downvote retracted in light of new source $\endgroup$ – jgon Mar 16 at 12:16
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Sin$^{-1}$ (with a capital S) returns every value you could put into the sine function to get your input. So $\operatorname{Sin}^{-1}(\sqrt2/2) = \{\cdots\frac{-7\pi}4,\frac{-5\pi}4,\frac\pi4,\frac{3\pi}4,\frac{9\pi}4,\frac{11\pi}4\cdots\}$.

But with a lowercase s, it only returns values between $\frac{-\pi}2$ and $\frac\pi2$.

So $\sin^{-1}(\frac{\sqrt{2}}2)$ is just $\frac\pi4$.

The notation with the capital S is rarely used because it is not a function; one input gives you infinitely many outputs.

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  • $\begingroup$ Is it then possible to give a statement right at the beginning, for eg, consider the domain ___< x < _____, then we use capital S, for inverse sine, such that we only have limited answers within { } instead of infinite answers? $\endgroup$ – NetUser5y62 Jul 1 '16 at 15:12
  • $\begingroup$ It says on my sheet that Arcsin refers to 2 quadrants while arcsin refers to 4 quadrants. Is it the reverse for $\sin^{-1}?$ $\endgroup$ – Max Li Jul 5 '16 at 20:29
  • $\begingroup$ That is a strange way of describing arcsin vs Arcsin, but it is true. If you look at the graph of arcsin(x), you will see that it is only in quadrants I and III, but Arcsin(x) is in all four. It is the same for $\sin^{-1}$. $\endgroup$ – Polygon Jul 6 '16 at 0:14
  • $\begingroup$ (-1) I can find no sources supporting this answer, see the sources linked in my answer. $\endgroup$ – jgon Mar 16 at 4:57
  • $\begingroup$ @jgon It looks like you're right. My answer was based off what I had learned in my high school math course, but it looks like I had written it down wrong. I just found two sources supporting what you said. $\endgroup$ – Polygon Mar 16 at 5:02

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