Why do we choose $[-\pi/2,\pi/2]$ as the domain restriction for $\sin(x)$ when determining $\arcsin(x)$, clearly other restrictions are possible, which also yield a 1-1 function (like $[\pi/2,3\pi/2]$). No pre-calculus book I have seen goes beyond saying that this choice is completely arbitrary. However, I have noticed that every domain restriction for inverse trig functions contains the whole first quadrant in its entirety (with the exception of x values where there are vertical asymptopes). I think these domain restrictions we choose are the only possible choices to get a 1-1 function on a domain that includes "effectively" the whole first quadrant. Is this the reason why we choose these domain restrictions? Are there other reasons why these domain restrictions are more natural than others that would yield a 1-1 function?

  • $\begingroup$ Well, the interval is symmetric. Also $[3\pi / 2, 5\pi / 2]$ yields the same arguments as $[-\pi/2, \pi/2]$. $\endgroup$
    – AlexR
    Commented Sep 5, 2013 at 22:47
  • $\begingroup$ I'm not sure what you mean Alex; is an interval being symmetric a standard term? I've never heard this word used in this way. You could also make similar types of statements when adding $2\pi$ to the endpoints of any interval. $\endgroup$ Commented Sep 5, 2013 at 23:32
  • $\begingroup$ The domain should include the "normal" angles between $0$ and $90^\circ$, and be connected. Doesn't leave much choice for $\arcsin$! $\endgroup$ Commented Sep 5, 2013 at 23:45
  • $\begingroup$ "Symmetric", when applied to an interval in $\mathbb{R}$, or more generally a region in $\mathbb{R}^n$, means that for every point $x$ in the region, the point $-x$ is also in the region. $\endgroup$ Commented Sep 5, 2013 at 23:58
  • $\begingroup$ Oh I see, I thought Alex's second sentence was some how related to his definition of symmetry now I see that these were two separate statements. Thanks for the clarification. $\endgroup$ Commented Sep 6, 2013 at 20:14

1 Answer 1


It's nice to have a domain that includes $0$. The domain of a trig function is the range of the corresponding inverse function, and it's nice to have $0$ in the range.

Think of applications. If I'm building something, and I want to set two pieces at some angle, I want an angle between $0$ and $\pi$, because that's what I can measure with my tools. That's one reason that it's nice to use intervals either centered on $0$, or with $0$ as an endpoint.

Also, in calculus, we like to represent functions as power series, and these things are a lot easier to write down if we can center them at $x=0$. Of course, this isn't an explanation that would have much place in one of those Precalculus books that you mention, but it would seem strange to define the ranges of inverse trig functions to be one thing in Precal and something else in Calculus.

There may be other reasons, but I think these two are compelling.

  • $\begingroup$ This is a great point, but for say cos(x) $[-\pi,0]$ would also satisfy your condition that 0 should be included, although it would be an odd choice to have a completely non positive interval. I just pulled out my old calculus text book and there is also no explanation in there for the domain restriction (besides the standard 1-1 statements). I would up-vote this post, but unfortunately I am a newbie who doesn't have enough rep to do that. $\endgroup$ Commented Sep 5, 2013 at 23:24
  • $\begingroup$ Yeah, values of $\cos^{-1}$ are angles, and insofar as angles are physical quantities, it's natural for them to be non-negative. I don't know why they don't explain this in more textbooks. $\endgroup$ Commented Sep 5, 2013 at 23:27
  • $\begingroup$ I have the same doubt about the domains of inverse trigonometric functions. Unfortunatelly I would like a more objective answer but it seems that it is a conveniente convention. $\endgroup$ Commented Jun 27, 2019 at 20:16

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