Re: How to simplify ArcSin formula
- To: mathgroup at smc.vnet.net
- Subject: [mg123416] Re: How to simplify ArcSin formula
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Tue, 6 Dec 2011 03:13:49 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201112040749.CAA21486@smc.vnet.net> <3A8A16B2-5407-470C-9413-F2EC033B0AFE@mimuw.edu.pl> <8872BEF1-58A9-48E5-84B5-0514955655B2@mimuw.edu.pl> <7BB0E96E72E84E41B24A1FF8ED13F5E61568B8C80C@IE2RD2XVS581.red002.local> <EA8288A8-CE6C-4659-B74B-4A028ED5C40F@mimuw.edu.pl>
Sorry, there was a small mistake in my written argument (although not in the actual computation). The substitution should have been FF[t_, \[Theta]_] = xxx /. aa -> Sin[\[Theta]]; since -1<=aa<=1 can be replaced by -Pi/2<=\[Theta]<Pi/2. In this case everything works fine and you can see that the partial derivatives vanish only when Cos[t + \[Theta]] > 0. Andrzej On 5 Dec 2011, at 10:49, Andrzej Kozlowski wrote: > The point was to show that under the assumption Cos[t + \[Theta]] >= 0 we always get zero but under the assumption Cos[t + \[Theta]] < 0 we don't. (There is also the case Cos[t + \[Theta]] < 0 and Cos[\[Theta]]<0, which gives 0, but this does not fall into the region -Pi/2<=\[Theta]<=Pi/2 so I should not have included it). In any case this is all you need to show that with the assumption -Pi/2<=\[Theta]<=Pi/2, Cos[t + \[Theta]] >= 0 is necessary and sufficient for xxx==0. > Your suggested approach also works, of course. > > Andrzej > > > On 5 Dec 2011, at 09:36, Alexei Boulbitch wrote: > >> It is a nice solution. There is a minor miswriting as much as I see: the FullSimplify operations listed below give equal results. Indeed: >> >> xxx = t + ArcSin[aa] - ArcSin[aa*Cos[t] + Sqrt[1 - aa^2]*Sin[t]]; >> FF[t_, \[Theta]_] = xxx /. aa -> Cos[\[Theta]]; >> >> a = FullSimplify[ >> D[FF[t, \[Theta]], \[Theta]], {Cos[\[Theta]] > 0, >> Cos[t + \[Theta]] > 0}]; >> >> b = FullSimplify[ >> D[FF[t, \[Theta]], \[Theta]], {Cos[\[Theta]] < 0, >> Cos[t + \[Theta]] < 0}]; >> >> c = FullSimplify[ >> D[FF[t, \[Theta]], \[Theta]], {Cos[\[Theta]] < 0, >> Cos[t + \[Theta]] > 0}]; >> >> d = FullSimplify[ >> D[FF[t, \[Theta]], \[Theta]], {Cos[\[Theta]] > 0, >> Cos[t + \[Theta]] < 0}]; >> >> a == b >> a == c >> a == d >> >> >> True >> >> True >> >> True >> >> >> If one, instead, writes conditions on sin, rather than on cos, everything works: >> >> Simplify[D[FF[t, \[Theta]], \[Theta]], {Sin[\[Theta]] > 0, >> Sin[t - \[Theta]] > 0}] >> Simplify[D[FF[t, \[Theta]], \[Theta]], {Sin[\[Theta]] > 0, >> Sin[t - \[Theta]] < 0}] >> Simplify[D[FF[t, \[Theta]], \[Theta]], {Sin[\[Theta]] < 0, >> Sin[t + \[Theta]] > 0}] >> Simplify[D[FF[t, \[Theta]], \[Theta]], {Sin[\[Theta]] < 0, >> Sin[t + \[Theta]] < 0}] >> >> -2 >> >> 0 >> >> 2 >> >> 0 >> >> >> Best, Alexei >> >> >> Alexei BOULBITCH, Dr., habil. >> IEE S.A. >> ZAE Weiergewan, >> 11, rue Edmond Reuter, >> L-5326 Contern, LUXEMBOURG >> >> Office phone : +352-2454-2566 >> Office fax: +352-2454-3566 >> mobile phone: +49 151 52 40 66 44 >> >> e-mail: alexei.boulbitch at iee.lu >> >> >> -----Original Message----- >> From: Andrzej Kozlowski [mailto:akoz at mimuw.edu.pl] >> Sent: Sonntag, 4. Dezember 2011 23:36 >> To: mathgroup at smc.vnet.net Steve >> Cc: Dana DeLouis; Alexei Boulbitch >> Subject: Re: Re: How to simplify ArcSin formula >> >> Here is a way to obtain, I think, the complete solution to the original problem. >> >> xxx = t + ArcSin[aa] - ArcSin[aa*Cos[t] + Sqrt[1 - aa^2]*Sin[t]] >> >> Assuming that -1<=a<=1 we can make the substitution (as in Alexei Boulbitch's post) and define a function of two variables. >> >> FF[t_, \[Theta]_] = xxx /. aa -> Cos[\[Theta]]; >> >> Now we look for the conditions that make the partial derivatives vanish: >> >> FullSimplify[D[FF[t, \[Theta]], \[Theta]], {Cos[\[Theta]] > 0, Cos[t + \[Theta]] > 0}] >> >> FullSimplify[D[FF[t, \[Theta]], \[Theta]], {Cos[\[Theta]] > 0, Cos[t + \[Theta]] > 0}] >> >> FullSimplify[D[FF[t, \[Theta]], \[Theta]], {Cos[\[Theta]] < 0, Cos[t + \[Theta]] < 0}] >> >> FullSimplify[D[FF[t, \[Theta]], \[Theta]], {Cos[\[Theta]] < 0, Cos[t + \[Theta]] > 0}] >> >> FullSimplify[D[FF[t, \[Theta]], \[Theta]], {Cos[\[Theta]] > 0, Cos[t + \[Theta]] < 0}] >> >> which shows that the necessary condition is Cos[\[Theta] + t] >= 0. It is easy to show that it is also sufficient for the vanishing of FF[t, \[Theta]] and, hence, zz. >> >> We can now use Reduce: >> >> cond = >> DeleteCases[ >> Reduce[aa == Sin[\[Theta]] && -1 <= t <= 1 && >> Cos[\[Theta] + t] >= 0 && >> Inequality[-Pi/2, LessEqual, \[Theta], Less, Pi/2], t], >> \[Theta] == _, Infinity] >> >> (aa == -1 && >> 0 <= t <= 1) || (-1 < aa < >> Sin[(2 - Pi)/2] && (1/2)*(-Pi - 2*ArcSin[aa]) <= t <= 1) || >> (Sin[(2 - Pi)/2] <= aa <= Sin[(1/2)*(-2 + Pi)] && -1 <= t <= >> 1) || (Sin[(1/2)*(-2 + Pi)] < aa < 1 && >> -1 <= t <= (1/2)*(Pi - 2*ArcSin[aa])) >> >> I believe that give the subregion of the square -1<=aa<=1, -1<=t<=1 in which xxx is zero. The following Plot3D confirms this (notwithstanding some singular-like behaviour near the boundary) >> >> Plot3D[xxx, {aa, -1, 1}, {t, -1, 1}, >> RegionFunction -> Function[{aa, t}, cond], WorkingPrecision -> 30] >> >> >> Andrzej >> >> On 4 Dec 2011, at 15:09, Andrzej Kozlowski wrote: >> >>> >>> On 4 Dec 2011, at 08:49, Dana DeLouis wrote: >>> >>>>> >>>> >>>> This I find strange, in that Mathematica really does recognize this form. >>>> >>>> Anyway, at small values of t, >>>> ArcSin[Sin[t + Pi/6]] returns t+Pi/6. >>>> Which cancels out the other terms, and returns zero. >>>> >>>> However, as t gets large, >>>> ArcSin[Sin[t + Pi/6]] does NOT return t+Pi/6. >>>> >>>> And hence the return value is NOT 0. >>>> >>>> If you plot with say aa -> 1/2... >>>> >>>> Plot[xxx /. aa -> 1/2, {t, -5, 5}] >>>> >>>> Then the zero line is broken based on the phase shift when t gets to both >>>> >>>> NSolve[Pi/6 + t == Pi/2] >>>> {{t -> 1.047197}} >>>> >>>> NSolve[Pi/6 + t == -Pi/2] >>>> {{t -> -2.094395}} >>>> >>>> For your question, you should add a constraint for t also. >>>> However, it doesn't seem to work here for this equation either: :>( >>>> >>>> FullSimplify[xxx, Assumptions -> {-1 < aa < 1, -1 < t < 1}] >>>> <<unevaluated>> >>>> >>> >>> I don't really what it is that "doesn't work" here. As far as I can tell, what you are doing is: >>> >>> xxx = t + ArcSin[aa] - ArcSin[aa Cos[t] + Sqrt[1 - aa^2] Sin[t]] >>> >>> >>> FullSimplify[xxx, Assumptions -> {-1 < aa < 1, -1 < t < 1}] >>> >>> What do you expect to get here? Look at: >>> >>> Plot3D[xxx, {aa, -1, 1}, {t, -1, 1}, WorkingPrecision -> 30] >>> >>> Are you claiming the graph is wrong or what? (You obviously need to take a smaller interval for aa. ) >>> >>> Andrzej Kozlowski >> >> >> >
- References:
- Re: How to simplify ArcSin formula
- From: Dana DeLouis <dana01@me.com>
- Re: How to simplify ArcSin formula