Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I found a lecture notes that claims the following. Is this standard?

The notation $\overline{\text{arc}}\text{ sin }x$ is the inverse function of $\sin x$ restricted to $\left [ -\frac{\pi}{2},\frac{\pi}{2}\right ]$ and $\text{arc sin }x $ mean all those $y$ satisfying $\sin x=y.$

share|cite|improve this question
$\arctan$, $\tan^{-1}$ are usually used. – Pedro Tamaroff Oct 26 '13 at 15:30
Why are you using \text{arc sin} rather than \arcsin? ($\arcsin$)? – Asaf Karagila Oct 26 '13 at 15:31
If I saw correctly, the notes has a space between the letters c and s. – forumreader Oct 26 '13 at 15:33
up vote 2 down vote accepted

The most common notation used is either $\,\arcsin x\,$ or $\,\sin^{-1}x$.

When the desired value of $\,f(x) = \arcsin x\,$ is restricted to those values lying in $[-\pi/2, \pi/2]$, this is usually stated explicitly. I presume the lecturer introduced $\overline{\text{arc}}\sin x$ to spare the need from restricting the range of solutions repeatedly.

share|cite|improve this answer
So is $\arcsin x$ the set of all solutions to $\sin y=x$ or is it just some real number satisfying the equation? – forumreader Oct 26 '13 at 15:40
@amWhy I disagree. $\arcsin x$ is a function in its own right, defined over $[-1,1]$ and taking values on $[-\pi/2,\pi/2]$. – Pedro Tamaroff Oct 26 '13 at 15:47
I don't think I've said anything to contradict that, @Pedro. – amWhy Oct 26 '13 at 15:52
Your last comment. – Pedro Tamaroff Oct 26 '13 at 15:52
forumreader: Yes, $y= \arcsin x$ is the set of solutions to $\sin y = x$, where $y \in [-\pi/2, \pi/2]$ – amWhy Oct 26 '13 at 16:01

Your Answer


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.