MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: x-ArcSin[Sin[x]]

  • To: mathgroup at
  • Subject: [mg50035] Re: [mg50007] x-ArcSin[Sin[x]]
  • From: DrBob <drbob at>
  • Date: Thu, 12 Aug 2004 05:43:58 -0400 (EDT)
  • References: <>
  • Reply-to: drbob at
  • Sender: owner-wri-mathgroup at

>> I would like to know if anything has been changed in the Mathematica implementation of
>> ArcSin[Sin[x]].

There is not, and cannot be, an implementation of ArcSin[Sin[x]]. Sin and ArcSin are separate functions, separately implemented.

Moreover, Sin is periodic on the real line, and hence it cannot have a left inverse for all x. Therefore, as explained at the link, ArcSin[Sin[x]]==x is false in general.

If you want it to be true for x in a certain interval, you may be able to define an appropriate branch of ArcSin FOR THAT INTERVAL. That is, it wouldn't be a function of x alone, but also a function of the interval.

As x varies from 0 to 2Pi, the quantity Pi/2-x sweeps out the following interval:

Pi/2 - Interval[{0, 2*Pi}]
Interval[{-((3*Pi)/2), Pi/2}]

Max[%] - Min[%]

Now, on any interval of that length, Sin takes on all values between -1 and 1 TWICE. So again, on an interval like that, Sin has no left inverse.

In summary: the value you give for that integral is wrong, no matter how many books agree with you.


On Wed, 11 Aug 2004 05:53:03 -0400 (EDT), Paul Muller < at> wrote:

> Dear all,
> I know this question was already discussed in 1995 (that's what I found in Google), but
> I would like to know if anything has been changed in the Mathematica implementation of
> ArcSin[Sin[x]].
> I am calculating the magnetic field in any point of the plane of a spire (or spiral
> inductor). When I apply this to the center of the spire, I get the following integral:
> Integrate[Sin[x+ArcSin[Sin[Pi/2-x]]], {x, 0, 2Pi}]
> According to textbook results for magnetic fields, the result should be 2Pi. Because of
> the definition of ArcSin[Sin[x]] in Mathematica, the integral results in Pi.
> Maybe I'm not careful enough when using the ArcSin function on the angles of
> my geometric problem (I'm an engineer, not a mathematician), but it would
> be helpful if this integral was solved "correctly".
> Any updates on this are welcome.
> Thanks in advance
> Paul Muller

DrBob at

  • References:
  • Prev by Date: Re: how to graphically fill the tails of a normaldistribution
  • Next by Date: RE: how to graphically fill the tails of a normal distribution
  • Previous by thread: x-ArcSin[Sin[x]]
  • Next by thread: Re: x-ArcSin[Sin[x]]