Re: Re: x-ArcSin[Sin[x]]

*To*: mathgroup at smc.vnet.net*Subject*: [mg50091] Re: [mg50064] Re: x-ArcSin[Sin[x]]*From*: DrBob <drbob at bigfoot.com>*Date*: Sat, 14 Aug 2004 01:51:04 -0400 (EDT)*References*: <200408110953.FAA04108@smc.vnet.net> <cfffgn$ilf$1@smc.vnet.net> <200408130956.FAA03696@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

One typical situation is that you know Sin[theta] AND Cos[theta]. In that case, ArcTan[Cos[theta],Sin[theta]] is more likely to come out in the proper quadrant. I don't know if that will help in your problem. Bobby On Fri, 13 Aug 2004 05:56:38 -0400 (EDT), Paul Muller <paul.muller-at-epfl.ch at sicinfo2.epfl.ch> wrote: > Thanks for your answer. Just to make clear: no book stated that ArcSin[Sin[x]]=x. The > books give the magnetic field in the center of the spire, and that's where my integral > gives a wrong value. Now this is probably due to my too wide interpretation of the > ArcSin at the place it gets into the calculation, messing up the final result. > > Unfortunately, at some point in my calculation, I have sin(theta)= something and I need > to extract theta from this. It seemed obvious to me to use ArcSin, but according to the > whole discussion, this is what I'm not supposed to assume... > > Thanks again > > Paul > > DrBob wrote: >> >> 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[%] >> 2*Pi >> >> 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. >> >> Bobby >> > > > -- DrBob at bigfoot.com www.eclecticdreams.net

**References**:**x-ArcSin[Sin[x]]***From:*Paul Muller <paul.muller-at-epfl.ch@sicinfo2.epfl.ch>

**Re: x-ArcSin[Sin[x]]***From:*Paul Muller <paul.muller-at-epfl.ch@sicinfo2.epfl.ch>