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