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How to simplify ArcSin formula
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David Sagan
Nov 29, 2011, 7:06:49 AM11/29/11
to
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
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
Dana DeLouis
Nov 30, 2011, 3:26:55 AM11/30/11
to
> The answer I know is xxx = 0...
I don't believe it's true for all cases of t.
aa=1/2;
xxx = t+ArcSin[aa]-ArcSin[aa Cos[t]+Sqrt[1-aa^2] Sin[t]];
Maybe for a small range...
xxx/. t->1. //Chop
0
However...
xxx/. t->2.
1.9056
xxx/. t->3.
3.9056
xxx/. t->4.
5.9056
= = = = = = = = = =
HTH
Dana DeLouis
= = = = = = = = = =
I don't believe it's true for all cases of t.
aa=1/2;
xxx = t+ArcSin[aa]-ArcSin[aa Cos[t]+Sqrt[1-aa^2] Sin[t]];
Maybe for a small range...
xxx/. t->1. //Chop
0
However...
xxx/. t->2.
1.9056
xxx/. t->3.
3.9056
xxx/. t->4.
5.9056
= = = = = = = = = =
HTH
Dana DeLouis
= = = = = = = = = =
Ray Koopman
Nov 30, 2011, 3:27:26 AM11/30/11
to
On Nov 29, 4:06 am, David Sagan <david.sa...@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 [...]
> 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
It is ?
xxx = t + ArcSin[aa] - ArcSin[aa Cos[t] + Sqrt[1 - aa^2] Sin[t]];
xxx /. t -> Pi/2
% /. aa -> 1/2
Pi/2 + ArcSin[aa] - ArcSin[Sqrt[1 - aa^2]]
Pi/3
Alexei Boulbitch
Nov 30, 2011, 7:06:35 AM11/30/11
to
Hi, David,
You should help Mathematica understanding what you would like to get. I would do it like follows:
xxx = t + ArcSin[aa] - ArcSin[aa Cos[t] + Sqrt[1 - aa^2] Sin[t]];
xxx1 = xxx /. aa -> Sin[\[CurlyPhi]] /.
Sqrt[1 - Sin[\[Alpha]_]^2] -> Cos[\[Alpha]]
This gives you
t + ArcSin[Sin[\[CurlyPhi]]] -
ArcSin[Cos[\[CurlyPhi]] Sin[t] + Cos[t] Sin[\[CurlyPhi]]]
Then make
Simplify[xxx1] /. ArcSin[Sin[\[Alpha]_]] -> \[Alpha]
Giving you 0.
Have fun, Alexei
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
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.b...@iee.lu<mailto:alexei.b...@iee.lu>
You should help Mathematica understanding what you would like to get. I would do it like follows:
xxx = t + ArcSin[aa] - ArcSin[aa Cos[t] + Sqrt[1 - aa^2] Sin[t]];
Sqrt[1 - Sin[\[Alpha]_]^2] -> Cos[\[Alpha]]
This gives you
t + ArcSin[Sin[\[CurlyPhi]]] -
ArcSin[Cos[\[CurlyPhi]] Sin[t] + Cos[t] Sin[\[CurlyPhi]]]
Then make
Simplify[xxx1] /. ArcSin[Sin[\[Alpha]_]] -> \[Alpha]
Giving you 0.
Have fun, Alexei
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
Alexei BOULBITCH, Dr., habil.
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
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.b...@iee.lu<mailto:alexei.b...@iee.lu>
Andrzej Kozlowski
Dec 1, 2011, 5:50:48 AM12/1/11
to
You are making several assumptions which hide the actual complexity of
the situation. The expression is only zero for a certain range of
parameter values, as can be clearly seen from this graph:
Plot3D[Chop[
Pi/2}, {aa, -1, 1}, AxesLabel -> {"t", "aa", "x"}]
Andrzej Kozlowski
the situation. The expression is only zero for a certain range of
parameter values, as can be clearly seen from this graph:
Plot3D[Chop[
t + ArcSin[aa] -
ArcSin[aa Cos[t] + Sqrt[1 - aa^2] Sin[t]]], {t, -Pi/2,
Pi/2}, {aa, -1, 1}, AxesLabel -> {"t", "aa", "x"}]
Andrzej Kozlowski
Alexei Boulbitch
Dec 1, 2011, 7:25:25 AM12/1/11
to
Of course, Andrzej, you are right. And it is quite clear, where these assumptions are hidden.
However, looking for a complete analysis of this (rather simple) problem is not the content of question asked, and it is not my task. The question is: "I am interested in finding out, in general, how to manipulate formulas of this type". My answer shows one way of such a manipulation. I am sure one may point out a couple of other ways.
Once I have already formulated here my attitude, but repeat it in the context of the present case.
I believe that in such a situation the complete analysis of the expression from the mathematical point of view is the "homework" of the one asking the question and (presumably) personally interested in the result. Indeed, if it is a student, the problem may be a part of his University task, and making his job you participate in falsification. If the question is asked by a scientist or an engineer, it is most probably a part of his/her job. Then he gains money for the solution that you do for him.
I believe that my role here may only be in helping one by showing him a specific Mathematica trick (of course, if I know one) that does not belong to general mathematics, (physics, biology and so on) and may take too much time to find it out. This is already a great help. At least, for me it is. All the rest is the business of the one asking a question.
Best regards, 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.b...@iee.lu
However, looking for a complete analysis of this (rather simple) problem is not the content of question asked, and it is not my task. The question is: "I am interested in finding out, in general, how to manipulate formulas of this type". My answer shows one way of such a manipulation. I am sure one may point out a couple of other ways.
Once I have already formulated here my attitude, but repeat it in the context of the present case.
I believe that in such a situation the complete analysis of the expression from the mathematical point of view is the "homework" of the one asking the question and (presumably) personally interested in the result. Indeed, if it is a student, the problem may be a part of his University task, and making his job you participate in falsification. If the question is asked by a scientist or an engineer, it is most probably a part of his/her job. Then he gains money for the solution that you do for him.
I believe that my role here may only be in helping one by showing him a specific Mathematica trick (of course, if I know one) that does not belong to general mathematics, (physics, biology and so on) and may take too much time to find it out. This is already a great help. At least, for me it is. All the rest is the business of the one asking a question.
Best regards, 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.b...@iee.lu
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