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Re: How to simplify ArcSin formula

• To: mathgroup at smc.vnet.net
• Subject: [mg123395] Re: How to simplify ArcSin formula
• From: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>
• Date: Mon, 5 Dec 2011 05:16:34 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201112040749.CAA21486@smc.vnet.net>

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: [mg123395] 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>