Services & ResourcesWolfram ForumsMathGroup Archive

[Date Index] [Thread Index] [Author Index]

Re: How to simplify ArcSin formula

• To: mathgroup at smc.vnet.net
• Subject: [mg123416] Re: How to simplify ArcSin formula
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Tue, 6 Dec 2011 03:13:49 -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> <EA8288A8-CE6C-4659-B74B-4A028ED5C40F@mimuw.edu.pl>

Sorry, there was a small mistake in my written argument (although not in 
the actual computation). The substitution should have been

FF[t_, \[Theta]_] = xxx /. aa -> Sin[\[Theta]];

since -1<=aa<=1 can be replaced by -Pi/2<=\[Theta]<Pi/2.

In this case everything works fine and you can see that the partial 
derivatives vanish only when Cos[t + \[Theta]] > 0.


Andrzej



On 5 Dec 2011, at 10:49, Andrzej Kozlowski wrote:

> 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
>>
>>
>>
>

• References:
  • Re: How to simplify ArcSin formula
    • From: Dana DeLouis <dana01@me.com>