Re: How to simplify ArcSin formula
- To: mathgroup at smc.vnet.net
- Subject: [mg123399] Re: How to simplify ArcSin formula
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Mon, 5 Dec 2011 05:17:20 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201112040749.CAA21486@smc.vnet.net>
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