Re: How to simplify ArcSin formula
- To: mathgroup at smc.vnet.net
- Subject: [mg123415] Re: How to simplify ArcSin formula
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Tue, 6 Dec 2011 03:13:38 -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>
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