Alternative form of ArcSin[Sin[x]]

Clash Royale CLAN TAG#URR8PPP

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Is it possible to get the following result in Mathematica by using only built-in functions:

$$arcsin( sin(x)) = x $$ if $x in [-pi /2 , pi/2]$

$$arcsin ( sin(x)) = pi - x $$ if $x in [pi /2 , 3pi/2]$

simplifying-expressions trigonometry education

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edited 11 hours ago

David G. Stork

22k21747

asked 11 hours ago

Gennaro Arguzzi

315210

add a comment | 

up vote
2
down vote

favorite

1

Is it possible to get the following result in Mathematica by using only built-in functions:

$$arcsin( sin(x)) = x $$ if $x in [-pi /2 , pi/2]$

$$arcsin ( sin(x)) = pi - x $$ if $x in [pi /2 , 3pi/2]$

simplifying-expressions trigonometry education

share|improve this question

edited 11 hours ago

David G. Stork

22k21747

asked 11 hours ago

Gennaro Arguzzi

315210

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

Is it possible to get the following result in Mathematica by using only built-in functions:

$$arcsin( sin(x)) = x $$ if $x in [-pi /2 , pi/2]$

$$arcsin ( sin(x)) = pi - x $$ if $x in [pi /2 , 3pi/2]$

simplifying-expressions trigonometry education

share|improve this question

edited 11 hours ago

David G. Stork

22k21747

asked 11 hours ago

Gennaro Arguzzi

315210

Is it possible to get the following result in Mathematica by using only built-in functions:

$$arcsin( sin(x)) = x $$ if $x in [-pi /2 , pi/2]$

$$arcsin ( sin(x)) = pi - x $$ if $x in [pi /2 , 3pi/2]$

simplifying-expressions trigonometry education

simplifying-expressions trigonometry education

share|improve this question

edited 11 hours ago

David G. Stork

22k21747

asked 11 hours ago

Gennaro Arguzzi

315210

share|improve this question

edited 11 hours ago

David G. Stork

22k21747

asked 11 hours ago

Gennaro Arguzzi

315210

share|improve this question

share|improve this question

edited 11 hours ago

David G. Stork

22k21747

edited 11 hours ago

David G. Stork

22k21747

edited 11 hours ago

David G. Stork

22k21747

22k21747

asked 11 hours ago

Gennaro Arguzzi

315210

asked 11 hours ago

Gennaro Arguzzi

315210

asked 11 hours ago

Gennaro Arguzzi

315210

315210

add a comment | 

add a comment | 

5 Answers
5

active

oldest

votes

up vote
4
down vote

We can take the Floor and Ceiling from Carl's answer and expand them out:

PiecewiseExpand[
 PowerExpand[ArcSin[Sin[x]], Assumptions -> x ∈ Reals], 
 -π/2 < x < 3π/2
]

share|improve this answer

answered 6 hours ago

Chip Hurst

19.5k15585

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @ChipHurst, can you tell me why in the second line of the output there is True instead of "-pi/2<=x<=pi/2" please?
  – Gennaro Arguzzi
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

@kglr was on the right track with PowerExpand. With the default option Assumptions->Automatic, Mathematica may return a result that is not valid. On the other hand, if you give PowerExpand a non-default assumption, then it will return a result valid given those assumptions. So, for your example:

Assuming[
 x ∈ Reals,
 Simplify @ PowerExpand[ArcSin[Sin[x]], Assumptions -> x ∈ Reals]
]

1/2 (-1)^(Ceiling[1/2 + x/π] + Floor[-(1/2) + x/π] +
Floor[1/2 + x/π]) (π + (-1)^(
Ceiling[1/2 + x/π] + Floor[1/2 + x/π]) π + 2 x -
2 π Floor[1/2 + x/π])

share|improve this answer

edited 5 hours ago

answered 8 hours ago

Carl Woll

64.2k283167

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @CarlWoll, I have some difficult to interpret the solution. Can you explain me how it is related to the solution which I expected (the one in the question) please?
  – Gennaro Arguzzi
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Plot[ArcSin[Sin[x]], π/2 TriangleWave[x/(2 π)], Abs[Mod[x - π/2, 2 π] - π] - π/2 + 
 0, -5, 5 // Evaluate,x, -10 π, 10 π]

With[x=RandomReal[-100,100,1000],
 ArcSin[Sin[x]] - # //Chop//Union]& /@ π/2 TriangleWave[x/(2 π)], Abs[Mod[x-π/2,2 π]-π]-π/2

0, 0

share|improve this answer

edited 30 mins ago

answered 1 hour ago

chyanog

6,78921545

add a comment | 

up vote
0
down vote

Answer to the first question

Simplify[x == ArcSin[Sin[x]], -Pi/2 <= x <= Pi/2]
(* True *)

share|improve this answer

answered 9 hours ago

Ulrich Neumann

5,617413

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @UlrichNeumann, if possible i'd like to avoid to specify the range -Pi/2 <= x <= Pi/2. I prefer a more general statement because the above one is valid also for instance when Simplify[x == ArcSin[Sin[x]], 0 <= x <= Pi/2]
  – Gennaro Arguzzi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Gennaro Arguzzi You want to prove ArcSin[Sin[x]]=="triangle function"?
  – Ulrich Neumann
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm looking for built-in functions that allow me to "simplify" arcsin(sin(x)).
  – Gennaro Arguzzi
  9 hours ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

An easy way is to define a new function sin which will work as intended:

sin /: (ArcSin[sin[x :(_Real | _Integer| _Rational)]]
 /;-(Pi/2)<=Mod[x, 2 Pi, -Pi/2] <= Pi/2) := x;
sin /: (ArcSin[sin[x :(_Real | _Integer| _Rational)]]
 /; Pi/2<Mod[x, 2Pi,-Pi/2] < (3 Pi)/2) := Pi - x;
sin[x_] := Sin[x];

I intentionally restricted to real numbers (why I used heads Real, Integer, and Rational) so that I can use Mod.

Instead of trying to convert each Sin to sin, we can automate it by

$Pre = Function[# /. Sin -> sin];

Now, we do not have to do anything: The code works as expected. For example

Plot[ArcSin[Sin[x]], x, -Pi, 3 Pi]

share|improve this answer

answered 5 hours ago

Soner

77349

add a comment | 

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

We can take the Floor and Ceiling from Carl's answer and expand them out:

PiecewiseExpand[
 PowerExpand[ArcSin[Sin[x]], Assumptions -> x ∈ Reals], 
 -π/2 < x < 3π/2
]

share|improve this answer

answered 6 hours ago

Chip Hurst

19.5k15585

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @ChipHurst, can you tell me why in the second line of the output there is True instead of "-pi/2<=x<=pi/2" please?
  – Gennaro Arguzzi
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

We can take the Floor and Ceiling from Carl's answer and expand them out:

PiecewiseExpand[
 PowerExpand[ArcSin[Sin[x]], Assumptions -> x ∈ Reals], 
 -π/2 < x < 3π/2
]

share|improve this answer

answered 6 hours ago

Chip Hurst

19.5k15585

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @ChipHurst, can you tell me why in the second line of the output there is True instead of "-pi/2<=x<=pi/2" please?
  – Gennaro Arguzzi
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

We can take the Floor and Ceiling from Carl's answer and expand them out:

PiecewiseExpand[
 PowerExpand[ArcSin[Sin[x]], Assumptions -> x ∈ Reals], 
 -π/2 < x < 3π/2
]

share|improve this answer

answered 6 hours ago

Chip Hurst

19.5k15585

We can take the Floor and Ceiling from Carl's answer and expand them out:

PiecewiseExpand[
 PowerExpand[ArcSin[Sin[x]], Assumptions -> x ∈ Reals], 
 -π/2 < x < 3π/2
]

share|improve this answer

answered 6 hours ago

Chip Hurst

19.5k15585

share|improve this answer

share|improve this answer

answered 6 hours ago

Chip Hurst

19.5k15585

answered 6 hours ago

Chip Hurst

19.5k15585

answered 6 hours ago

Chip Hurst

19.5k15585

19.5k15585

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @ChipHurst, can you tell me why in the second line of the output there is True instead of "-pi/2<=x<=pi/2" please?
  – Gennaro Arguzzi
  1 hour ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @ChipHurst, can you tell me why in the second line of the output there is True instead of "-pi/2<=x<=pi/2" please?
  – Gennaro Arguzzi
  1 hour ago
  

  
  
  
  

Hello @ChipHurst, can you tell me why in the second line of the output there is True instead of "-pi/2<=x<=pi/2" please?
– Gennaro Arguzzi
1 hour ago

Hello @ChipHurst, can you tell me why in the second line of the output there is True instead of "-pi/2<=x<=pi/2" please?
– Gennaro Arguzzi
1 hour ago

add a comment | 

up vote
3
down vote

@kglr was on the right track with PowerExpand. With the default option Assumptions->Automatic, Mathematica may return a result that is not valid. On the other hand, if you give PowerExpand a non-default assumption, then it will return a result valid given those assumptions. So, for your example:

Assuming[
 x ∈ Reals,
 Simplify @ PowerExpand[ArcSin[Sin[x]], Assumptions -> x ∈ Reals]
]

1/2 (-1)^(Ceiling[1/2 + x/π] + Floor[-(1/2) + x/π] +
Floor[1/2 + x/π]) (π + (-1)^(
Ceiling[1/2 + x/π] + Floor[1/2 + x/π]) π + 2 x -
2 π Floor[1/2 + x/π])

share|improve this answer

edited 5 hours ago

answered 8 hours ago

Carl Woll

64.2k283167

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @CarlWoll, I have some difficult to interpret the solution. Can you explain me how it is related to the solution which I expected (the one in the question) please?
  – Gennaro Arguzzi
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

@kglr was on the right track with PowerExpand. With the default option Assumptions->Automatic, Mathematica may return a result that is not valid. On the other hand, if you give PowerExpand a non-default assumption, then it will return a result valid given those assumptions. So, for your example:

Assuming[
 x ∈ Reals,
 Simplify @ PowerExpand[ArcSin[Sin[x]], Assumptions -> x ∈ Reals]
]

1/2 (-1)^(Ceiling[1/2 + x/π] + Floor[-(1/2) + x/π] +
Floor[1/2 + x/π]) (π + (-1)^(
Ceiling[1/2 + x/π] + Floor[1/2 + x/π]) π + 2 x -
2 π Floor[1/2 + x/π])

share|improve this answer

edited 5 hours ago

answered 8 hours ago

Carl Woll

64.2k283167

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @CarlWoll, I have some difficult to interpret the solution. Can you explain me how it is related to the solution which I expected (the one in the question) please?
  – Gennaro Arguzzi
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

@kglr was on the right track with PowerExpand. With the default option Assumptions->Automatic, Mathematica may return a result that is not valid. On the other hand, if you give PowerExpand a non-default assumption, then it will return a result valid given those assumptions. So, for your example:

Assuming[
 x ∈ Reals,
 Simplify @ PowerExpand[ArcSin[Sin[x]], Assumptions -> x ∈ Reals]
]

1/2 (-1)^(Ceiling[1/2 + x/π] + Floor[-(1/2) + x/π] +
Floor[1/2 + x/π]) (π + (-1)^(
Ceiling[1/2 + x/π] + Floor[1/2 + x/π]) π + 2 x -
2 π Floor[1/2 + x/π])

share|improve this answer

edited 5 hours ago

answered 8 hours ago

Carl Woll

64.2k283167

@kglr was on the right track with PowerExpand. With the default option Assumptions->Automatic, Mathematica may return a result that is not valid. On the other hand, if you give PowerExpand a non-default assumption, then it will return a result valid given those assumptions. So, for your example:

Assuming[
 x ∈ Reals,
 Simplify @ PowerExpand[ArcSin[Sin[x]], Assumptions -> x ∈ Reals]
]

1/2 (-1)^(Ceiling[1/2 + x/π] + Floor[-(1/2) + x/π] +
Floor[1/2 + x/π]) (π + (-1)^(
Ceiling[1/2 + x/π] + Floor[1/2 + x/π]) π + 2 x -
2 π Floor[1/2 + x/π])

share|improve this answer

edited 5 hours ago

answered 8 hours ago

Carl Woll

64.2k283167

share|improve this answer

share|improve this answer

edited 5 hours ago

edited 5 hours ago

edited 5 hours ago

answered 8 hours ago

Carl Woll

64.2k283167

answered 8 hours ago

Carl Woll

64.2k283167

answered 8 hours ago

Carl Woll

64.2k283167

64.2k283167

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @CarlWoll, I have some difficult to interpret the solution. Can you explain me how it is related to the solution which I expected (the one in the question) please?
  – Gennaro Arguzzi
  1 hour ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @CarlWoll, I have some difficult to interpret the solution. Can you explain me how it is related to the solution which I expected (the one in the question) please?
  – Gennaro Arguzzi
  1 hour ago
  

  
  
  
  

Hello @CarlWoll, I have some difficult to interpret the solution. Can you explain me how it is related to the solution which I expected (the one in the question) please?
– Gennaro Arguzzi
1 hour ago

Hello @CarlWoll, I have some difficult to interpret the solution. Can you explain me how it is related to the solution which I expected (the one in the question) please?
– Gennaro Arguzzi
1 hour ago

add a comment | 

up vote
1
down vote

Plot[ArcSin[Sin[x]], π/2 TriangleWave[x/(2 π)], Abs[Mod[x - π/2, 2 π] - π] - π/2 + 
 0, -5, 5 // Evaluate,x, -10 π, 10 π]

With[x=RandomReal[-100,100,1000],
 ArcSin[Sin[x]] - # //Chop//Union]& /@ π/2 TriangleWave[x/(2 π)], Abs[Mod[x-π/2,2 π]-π]-π/2

0, 0

share|improve this answer

edited 30 mins ago

answered 1 hour ago

chyanog

6,78921545

add a comment | 

up vote
1
down vote

Plot[ArcSin[Sin[x]], π/2 TriangleWave[x/(2 π)], Abs[Mod[x - π/2, 2 π] - π] - π/2 + 
 0, -5, 5 // Evaluate,x, -10 π, 10 π]

With[x=RandomReal[-100,100,1000],
 ArcSin[Sin[x]] - # //Chop//Union]& /@ π/2 TriangleWave[x/(2 π)], Abs[Mod[x-π/2,2 π]-π]-π/2

0, 0

share|improve this answer

edited 30 mins ago

answered 1 hour ago

chyanog

6,78921545

add a comment | 

up vote
1
down vote

up vote
1
down vote

Plot[ArcSin[Sin[x]], π/2 TriangleWave[x/(2 π)], Abs[Mod[x - π/2, 2 π] - π] - π/2 + 
 0, -5, 5 // Evaluate,x, -10 π, 10 π]

With[x=RandomReal[-100,100,1000],
 ArcSin[Sin[x]] - # //Chop//Union]& /@ π/2 TriangleWave[x/(2 π)], Abs[Mod[x-π/2,2 π]-π]-π/2

0, 0

share|improve this answer

edited 30 mins ago

answered 1 hour ago

chyanog

6,78921545

Plot[ArcSin[Sin[x]], π/2 TriangleWave[x/(2 π)], Abs[Mod[x - π/2, 2 π] - π] - π/2 + 
 0, -5, 5 // Evaluate,x, -10 π, 10 π]

With[x=RandomReal[-100,100,1000],
 ArcSin[Sin[x]] - # //Chop//Union]& /@ π/2 TriangleWave[x/(2 π)], Abs[Mod[x-π/2,2 π]-π]-π/2

0, 0

share|improve this answer

edited 30 mins ago

answered 1 hour ago

chyanog

6,78921545

share|improve this answer

share|improve this answer

edited 30 mins ago

edited 30 mins ago

edited 30 mins ago

answered 1 hour ago

chyanog

6,78921545

answered 1 hour ago

chyanog

6,78921545

answered 1 hour ago

chyanog

6,78921545

6,78921545

add a comment | 

add a comment | 

up vote
0
down vote

Answer to the first question

Simplify[x == ArcSin[Sin[x]], -Pi/2 <= x <= Pi/2]
(* True *)

share|improve this answer

answered 9 hours ago

Ulrich Neumann

5,617413

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @UlrichNeumann, if possible i'd like to avoid to specify the range -Pi/2 <= x <= Pi/2. I prefer a more general statement because the above one is valid also for instance when Simplify[x == ArcSin[Sin[x]], 0 <= x <= Pi/2]
  – Gennaro Arguzzi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Gennaro Arguzzi You want to prove ArcSin[Sin[x]]=="triangle function"?
  – Ulrich Neumann
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm looking for built-in functions that allow me to "simplify" arcsin(sin(x)).
  – Gennaro Arguzzi
  9 hours ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

Answer to the first question

Simplify[x == ArcSin[Sin[x]], -Pi/2 <= x <= Pi/2]
(* True *)

share|improve this answer

answered 9 hours ago

Ulrich Neumann

5,617413

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @UlrichNeumann, if possible i'd like to avoid to specify the range -Pi/2 <= x <= Pi/2. I prefer a more general statement because the above one is valid also for instance when Simplify[x == ArcSin[Sin[x]], 0 <= x <= Pi/2]
  – Gennaro Arguzzi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Gennaro Arguzzi You want to prove ArcSin[Sin[x]]=="triangle function"?
  – Ulrich Neumann
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm looking for built-in functions that allow me to "simplify" arcsin(sin(x)).
  – Gennaro Arguzzi
  9 hours ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Answer to the first question

Simplify[x == ArcSin[Sin[x]], -Pi/2 <= x <= Pi/2]
(* True *)

share|improve this answer

answered 9 hours ago

Ulrich Neumann

5,617413

Answer to the first question

Simplify[x == ArcSin[Sin[x]], -Pi/2 <= x <= Pi/2]
(* True *)

share|improve this answer

answered 9 hours ago

Ulrich Neumann

5,617413

share|improve this answer

share|improve this answer

answered 9 hours ago

Ulrich Neumann

5,617413

answered 9 hours ago

Ulrich Neumann

5,617413

answered 9 hours ago

Ulrich Neumann

5,617413

5,617413

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @UlrichNeumann, if possible i'd like to avoid to specify the range -Pi/2 <= x <= Pi/2. I prefer a more general statement because the above one is valid also for instance when Simplify[x == ArcSin[Sin[x]], 0 <= x <= Pi/2]
  – Gennaro Arguzzi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Gennaro Arguzzi You want to prove ArcSin[Sin[x]]=="triangle function"?
  – Ulrich Neumann
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm looking for built-in functions that allow me to "simplify" arcsin(sin(x)).
  – Gennaro Arguzzi
  9 hours ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hello @UlrichNeumann, if possible i'd like to avoid to specify the range -Pi/2 <= x <= Pi/2. I prefer a more general statement because the above one is valid also for instance when Simplify[x == ArcSin[Sin[x]], 0 <= x <= Pi/2]
  – Gennaro Arguzzi
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Gennaro Arguzzi You want to prove ArcSin[Sin[x]]=="triangle function"?
  – Ulrich Neumann
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm looking for built-in functions that allow me to "simplify" arcsin(sin(x)).
  – Gennaro Arguzzi
  9 hours ago
  
  

  
  
  
  

Hello @UlrichNeumann, if possible i'd like to avoid to specify the range -Pi/2 <= x <= Pi/2. I prefer a more general statement because the above one is valid also for instance when Simplify[x == ArcSin[Sin[x]], 0 <= x <= Pi/2]
– Gennaro Arguzzi
9 hours ago

Hello @UlrichNeumann, if possible i'd like to avoid to specify the range -Pi/2 <= x <= Pi/2. I prefer a more general statement because the above one is valid also for instance when Simplify[x == ArcSin[Sin[x]], 0 <= x <= Pi/2]
– Gennaro Arguzzi
9 hours ago

@Gennaro Arguzzi You want to prove ArcSin[Sin[x]]=="triangle function"?
– Ulrich Neumann
9 hours ago

@Gennaro Arguzzi You want to prove ArcSin[Sin[x]]=="triangle function"?
– Ulrich Neumann
9 hours ago

I'm looking for built-in functions that allow me to "simplify" arcsin(sin(x)).
– Gennaro Arguzzi
9 hours ago

I'm looking for built-in functions that allow me to "simplify" arcsin(sin(x)).
– Gennaro Arguzzi
9 hours ago

add a comment | 

up vote
0
down vote

An easy way is to define a new function sin which will work as intended:

sin /: (ArcSin[sin[x :(_Real | _Integer| _Rational)]]
 /;-(Pi/2)<=Mod[x, 2 Pi, -Pi/2] <= Pi/2) := x;
sin /: (ArcSin[sin[x :(_Real | _Integer| _Rational)]]
 /; Pi/2<Mod[x, 2Pi,-Pi/2] < (3 Pi)/2) := Pi - x;
sin[x_] := Sin[x];

I intentionally restricted to real numbers (why I used heads Real, Integer, and Rational) so that I can use Mod.

Instead of trying to convert each Sin to sin, we can automate it by

$Pre = Function[# /. Sin -> sin];

Now, we do not have to do anything: The code works as expected. For example

Plot[ArcSin[Sin[x]], x, -Pi, 3 Pi]

share|improve this answer

answered 5 hours ago

Soner

77349

add a comment | 

up vote
0
down vote

An easy way is to define a new function sin which will work as intended:

sin /: (ArcSin[sin[x :(_Real | _Integer| _Rational)]]
 /;-(Pi/2)<=Mod[x, 2 Pi, -Pi/2] <= Pi/2) := x;
sin /: (ArcSin[sin[x :(_Real | _Integer| _Rational)]]
 /; Pi/2<Mod[x, 2Pi,-Pi/2] < (3 Pi)/2) := Pi - x;
sin[x_] := Sin[x];

I intentionally restricted to real numbers (why I used heads Real, Integer, and Rational) so that I can use Mod.

Instead of trying to convert each Sin to sin, we can automate it by

$Pre = Function[# /. Sin -> sin];

Now, we do not have to do anything: The code works as expected. For example

Plot[ArcSin[Sin[x]], x, -Pi, 3 Pi]

share|improve this answer

answered 5 hours ago

Soner

77349

add a comment | 

up vote
0
down vote

up vote
0
down vote

An easy way is to define a new function sin which will work as intended:

sin /: (ArcSin[sin[x :(_Real | _Integer| _Rational)]]
 /;-(Pi/2)<=Mod[x, 2 Pi, -Pi/2] <= Pi/2) := x;
sin /: (ArcSin[sin[x :(_Real | _Integer| _Rational)]]
 /; Pi/2<Mod[x, 2Pi,-Pi/2] < (3 Pi)/2) := Pi - x;
sin[x_] := Sin[x];

I intentionally restricted to real numbers (why I used heads Real, Integer, and Rational) so that I can use Mod.

Instead of trying to convert each Sin to sin, we can automate it by

$Pre = Function[# /. Sin -> sin];

Now, we do not have to do anything: The code works as expected. For example

Plot[ArcSin[Sin[x]], x, -Pi, 3 Pi]

share|improve this answer

answered 5 hours ago

Soner

77349

An easy way is to define a new function sin which will work as intended:

sin /: (ArcSin[sin[x :(_Real | _Integer| _Rational)]]
 /;-(Pi/2)<=Mod[x, 2 Pi, -Pi/2] <= Pi/2) := x;
sin /: (ArcSin[sin[x :(_Real | _Integer| _Rational)]]
 /; Pi/2<Mod[x, 2Pi,-Pi/2] < (3 Pi)/2) := Pi - x;
sin[x_] := Sin[x];

I intentionally restricted to real numbers (why I used heads Real, Integer, and Rational) so that I can use Mod.

Instead of trying to convert each Sin to sin, we can automate it by

$Pre = Function[# /. Sin -> sin];

Now, we do not have to do anything: The code works as expected. For example

Plot[ArcSin[Sin[x]], x, -Pi, 3 Pi]

share|improve this answer

answered 5 hours ago

Soner

77349

share|improve this answer

share|improve this answer

answered 5 hours ago

Soner

77349

answered 5 hours ago

Soner

77349

answered 5 hours ago

Soner

77349

77349

add a comment | 

add a comment | 

 

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