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Deriving arccos from arcsin

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data

    I want to show that:

    [tex] \arccos \,z= \left( -1 \right) ^{n+1} \arcsin \,z+\pi/2\, +n\pi [/tex]

    2. Relevant equations

    There is a trigonometric identity that says
    [tex] \arccos \,z= \pi/2\, -\arcsin \,z [/tex]

    3. The attempt at a solution

    So far, I have come up to this
    [tex] \arccos\,z=\pi/2\, -{\it i}\log\ \left( -iz+\sqrt {1-{z}^{2}} \right) [/tex]

    What is left to show is that

    [tex] \arcsin\, \left( -z \right) = \arcsin\,z [/tex]

    My plan is to add the periodicity (is that the term?) later on since
    [tex] -\cos\,z= \cos\ \left( z+\pi \right) [/tex]
     
  2. jcsd
  3. Mar 8, 2009 #2

    gabbagabbahey

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    Why would you want to show this?...It isn't true!

    For example, for n=2 you have [tex] \left( -1 \right) ^{n+1} \arcsin \,z+\pi/2\, +n\pi =- \arcsin \,z+\pi/2\ +2\pi=\arccos \,z +2\pi \neq \arccos \,z[/tex]....

    Not sure what you did to get to this point, but you have a sign error; you should have

    [tex] \arccos\,z=\pi/2\, +{\it i}\log\ \left( -iz+\sqrt {1-{z}^{2}} \right) [/tex]

    Again, why would you want to show this? The fact that [itex]\sin(-x)=-\sin(x)[/itex] should tell you that it's not true.

    Is this part of a larger problem? What is the original question?
     
  4. Mar 8, 2009 #3
    This problem is not part of a bigger problem:

    [tex] \arccos \,z= \left( -1 \right) ^{n+1} \arcsin \,z+\pi/2\, +n\pi [/tex]

    I have seen a similar computation http://math.fullerton.edu/mathews/c2003/ComplexFunTrigInverseMod.html" [Broken].

    I am not sure though how [tex] \arcsin\, \left( -z \right) = \arcsin\,z [/tex] was derived.
     
    Last edited by a moderator: May 4, 2017
  5. Mar 8, 2009 #4

    gabbagabbahey

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    That link contains a typo (at least one): [itex]\arcsin(-z)=-\arcsin(z)\neq\arcsin(z)[/itex].....plug in some numbers into your calculator if you don't believe me.


    While your at it, plug some numbers into the equation your trying to prove....you'll find that it isn't true (except for n=0 of course)
     
  6. Mar 8, 2009 #5
    Let's say you're right. How then do I resolve this? Do I provide a counterexample instead of proving it? The problem was explicit in saying that I should prove it.

    This problem is given by my professor in Complex Variables which I am taking under an MS Math degree program.
     
  7. Mar 9, 2009 #6

    gabbagabbahey

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    Provide a counter example...say z=0 and n=1 for example....

    Are you sure you weren't asked to show

    [tex]\cos\left(\left( -1 \right) ^{n+1} \arcsin \,z+\pi/2\, +n\pi \right)=z[/tex]

    Because that actually is true....the reason that you can't just take the arccos of both sides is that arcos always returns the principle value, so arcos(z) is always between 0 and 2pi, while [itex]\left( -1 \right) ^{n+1} \arcsin \,z+\pi/2\, +n\pi[/itex] need not be....this is why what you say you are trying to prove, is not true.
     
  8. Mar 9, 2009 #7
    I'm sure that the problem I posted was the original problem.

    Thanks for your advice. I'll ask my professor about it.
     
  9. Mar 9, 2009 #8
    I just thought, isn't it that since that arccos is expressed in terms of log, then it should be multiple-valued since log is?
     
  10. Mar 9, 2009 #9

    gabbagabbahey

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    Since when is Log multiple valued? :confused:

    ArcCos and Log are both defined in a way that makes them single valued.
     
  11. Mar 9, 2009 #10
    This is our definition:

    [tex] \log \,z= \ln \, \left| z \right| +i Arg \left( z+2\,n\pi \right) [/tex]

    as opposed to

    [tex] Log \,z= \ln \, \left| z \right| +i Arg z [/tex]

    So log here becomes multiple valued, whereas Log is not.
     
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