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Re: How to simplify ArcSin formula

• To: mathgroup at smc.vnet.net
• Subject: [mg123363] Re: How to simplify ArcSin formula
• From: Dana DeLouis <dana01 at me.com>
• Date: Sun, 4 Dec 2011 02:49:57 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com

>...I am interested in finding out, in general, how to manipulate
>   formulas of this type.

Hi.  I may be wrong, but I believe the problem with this equation is that
ArcSin[Sin[ t ] ]  in general does not really equal t.

As t gets larger, The inner Sin function caps the value between -1 &1.
ArcSin returns a number between -Pi/2 and Pi/2.  Hence, you can't get back your larger input.

r=Interval[{-10,10}];

Sin[r]
Interval[{-1,1}]

ArcSin[%]
Interval[{ -Pi/2, Pi/2} ]

- - - - - - -
Here's just an idea.  Here's your equation:

 xxx= t + ArcSin[aa] - ArcSin[aa Cos[t] + Sqrt[1 - aa^2] Sin[t]  ]

Inside your second ArcSin is:
   aa Cos[t] + Sqrt[1 - aa^2] Sin[t]

This is really just 1 wave form, just phase shifted.

aa Cos[t]+Sqrt[1-aa^2] Sin[t];

Let's set aa to a random number, say 1/2.
aa=1/2;

The magnitude of this one signal is simply 1 from observation, but...

mag=Simplify[Norm[ {aa,Sqrt[1-aa^2]}],-1<aa<1]
1

I'll use the Sin function, so the phase shift is:

SinPhase=ArcSin[aa/mag]
Pi / 6

Hense, the new equation could be:

f2 = t + ArcSin[aa] - ArcSin[mag*Sin[t + SinPhase]]    /. aa -> 1/2

Pi/6 + t - ArcSin[Sin[Pi/6 + t]]

FullSimplify[f2 == xxx]
True

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>>


= = = = = = = = = = = = = = = =
HTH  :>)
Dana DeLouis
Mac, Ver #8
= = = = = = = = = = = = = = = =



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

• Follow-Ups:
  • Re: How to simplify ArcSin formula
    • From: Alexei Boulbitch <Alexei.Boulbitch@iee.lu>
  • Re: How to simplify ArcSin formula
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>