Re: How to simplify ArcSin formula
- To: mathgroup at smc.vnet.net
- Subject: [mg123388] Re: How to simplify ArcSin formula
- From: Dana DeLouis <dana01 at me.com>
- Date: Mon, 5 Dec 2011 05:15:18 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
...with -1 < aa < 1 ...The answer I know is xxx = 0 Hi. Given a value of aa, the return value is 0 only when t falls within a certain range. RangeOfT[aa_] := Interval[{-Pi + ArcCos[aa], ArcCos[aa]}] So, if aa were 1/2. RangeOfT[1/2] Interval[{-((2*Pi)/3), Pi/3}] RangeOfT[1/2] //N Interval[{-2.094, 1.047}] If given aa as 1/2, then the equation is zero only when t is between -2.094 and 1.047 Chop[xxx /. {aa -> 1/2, t -> 1.04}] 0 Chop[xxx /. {aa -> 1/2, t -> 1.05}] 0.0056048976068160394 = = = = = = HTH Dana DeLouis = = = = = = On Nov 29, 7:06 am, David Sagan <david.sa... at gmail.com> wrote: > I am trying to discover how to simplify xxx where xxx is defined to > be: > xxx= t + ArcSin[aa] - ArcSin[aa Cos[t] + Sqrt[1 - aa^2] Sin[t]] > with > -1 < aa < 1 > The answer I know is xxx = 0 but the reason I am posing the question > is that I am interested in finding out, in general, how to manipulate > formulas of this type. I tried: > FullSimplify[xxx, -1<a<1] > but that did not work. Can anyone tell me how to do this? > > -- Thanks, David