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MathGroup Archive 1993

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Re: trig functions

  • To: mathgroup at yoda.physics.unc.edu
  • Subject: Re: trig functions
  • From: roland at afthp001.tuwien.ac.at (Roland Bauer)
  • Date: Thu, 17 Jun 93 12:30:44 +0200

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    Date: Mon, 7 Jun 93 21:30:10 -0500
    From: bappadit at ecn.purdue.edu (Banerjee Bappaditya)
    To: mathgroup at yoda.physics.unc.edu
    Subject: trig functions
    
    Hi!
    I am trying to find the appropriate quadrant (from -Pi to Pi) 
    to which the inverses of trignometric functions evaluate.  I have figured out the following rules...
    
    if Sin[theta]>=0 && Cos[theta]>=0, angle = theta
    if Sin[theta]>=0 && Cos[theta]<=0, angle = ArcCos[Cos[theta]]
    
    if Sin[theta]<0 && Cos[theta]<0, angle = ArcSin[Sin[theta]] + Pi/2
    if Sin[theta]=0 && Cos[theta]=-1, angle = -Pi  
    if Sin[theta]=-1 && Cos[theta]=0,angle = -Pi/2
    
    if Sin[theta]<=0 && Cos[theta]>=0, angle = ArcSin[Sin[theta]]
    
    How can I do this in Mma 1.2  so that a function 
    
    Angle[expression evaluating to a number] gives the correct theta ? 
    
    Or in other words, reduces the "expression " to modulo Pi.
    
    I have (unsuccessfully!!!) tried
    
    
    Angle[theta_] := If[Sin[theta]>=0 && Cos[theta]>=0,theta,
                     If[Sin[theta]>=0 && Cos[theta]<=0,ArcCos[Cos[theta]] ,
                     If[Sin[theta]=-1 && Cos[theta]=0,-Pi/2 ,
                     If[Sin[theta]<0  && Cos[theta]<0,ArcSin[Sin[theta]] + Pi/2 ,
                     If[Sin[theta]=0  && Cos[theta]=-1,-Pi ,
                     If[Sin[theta]<=0 && Cos[theta]>=0,ArcSin[Sin[theta]] 
                      ]]]]]] 
    
    
    Angle[theta_] := If[Sin[theta]>=0 && Cos[theta]>=0,theta,
                     If[Sin[theta]>=0 && Cos[theta]<=0,ArcCos[Cos[theta]] ,
                     If[Sin[theta]<0 && Cos[theta]<0,ArcSin[Sin[theta]] + Pi/2 ,
                     If[Sin[theta]<=0 && Cos[theta]>=0,ArcSin[Sin[theta]] 
                      ]]]]
    
    
    It does not work in the 3rd Quadrant i.e. from -Pi/2 to -Pi.
    
    I would appreciate any help.
    
    
    thanks,
    
    bappa.
    
    
    Bappaditya Banerjee
    bappadit at mn.ecn.purdue.edu
    Ray W. Herrick Laboratories
    Purdue University
    West Lafayette, IN 47907
    work : (317) 494 2132
           (317) 494 2147
    fax  : (317) 494 0787
    home : (317) 743 3982
    
     
Hi Bappa,

if you want to have a saw-tooth with the period a, You can define the
following function:

y[x_,a_]:=a(Mod[(x/a)+0.5,1]-0.5)

Then plot it with e.g.

Plot[y[x,360],{x,-500,+500}]

The function y transforms every angle into the range from
-(a/2) to +(a/2).

Maybe this was Your Problem?

Bye, Roland.

    
    


--

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| Roland BAUER                         tel: +43 (222) 58-801 / 4861       |
| Abt. Foerdertechnik                  fax: +43 (222) 586-58-47           |
|                                      email: bauer at afthp001.tuwien.ac.at |
| Getreidemarkt 9                                                         |
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| Technische Universitaet Wien                                            |
| Institut fuer Allgemeine Maschinenlehre und Foerdertechnik              |     
| Abteilung Foerdertechnik (321/2)                                        |
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