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Re: How to simplify ArcSin formula

  • To: mathgroup at smc.vnet.net
  • Subject: [mg123415] Re: How to simplify ArcSin formula
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 6 Dec 2011 03:13:38 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201112040749.CAA21486@smc.vnet.net> <3A8A16B2-5407-470C-9413-F2EC033B0AFE@mimuw.edu.pl> <8872BEF1-58A9-48E5-84B5-0514955655B2@mimuw.edu.pl> <7BB0E96E72E84E41B24A1FF8ED13F5E61568B8C80C@IE2RD2XVS581.red002.local>

The point was to show that under the assumption Cos[t + \[Theta]] >= 0 
we always get zero but under the assumption Cos[t + \[Theta]] < 0 we 
don't. (There is also the case Cos[t + \[Theta]] < 0 and 
Cos[\[Theta]]<0, which gives 0, but this does not fall into the region 
-Pi/2<=\[Theta]<=Pi/2 so I should not have included it). In any case 
this is all you need to show that with the assumption 
-Pi/2<=\[Theta]<=Pi/2,  Cos[t + \[Theta]] >= 0 is necessary and 
sufficient for xxx==0.
Your suggested approach also works, of course.

Andrzej


On 5 Dec 2011, at 09:36, Alexei Boulbitch wrote:

> 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: Re: 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
>
>
>




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