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Homework Help: Equality math problem

  1. Jun 4, 2010 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    I found this equality in the thread https://www.physicsforums.com/showthread.php?t=407130"
    [tex]
    cos\left(\frac{1}{3}arccos(z)\right) = \frac{\left(z + \sqrt{z^2-1}\right)^{1/3}}{2} + \frac 1 {2\left(z+\sqrt{z^2-1}\right)^{1/3}}
    [/tex]
    and I'd like to know how it works.

    Note: I haven't studied this, but I do know about complex numbers and I got some hints from that thread on what to do.


    3. The attempt at a solution
    First of all, I assumed it was true.

    For the left side of the equality, let [tex]y=cos\left(\frac{1}{3}arccos(z)\right)[/tex]

    so [tex]z=cos\left(3arccos(y)\right)[/tex]

    by trig identities, [tex]cos\left(3arccos(y)\right)=4y^3-3y[/tex]

    So the solutions to [tex]y=cos\left(\frac{1}{3}arccos(z)\right)[/tex] are the solutions (not exactly sure which of the 3) to the cubic [tex]4y^3-3y-z=0[/tex]

    and now taking the right side, [tex]y=\frac{\left(z + \sqrt{z^2-1}\right)^{1/3}}{2} + \frac 1 {2\left(z+\sqrt{z^2-1}\right)^{1/3}} [/tex]

    [tex]z+\sqrt{z^2-1}=cos(z)+isin(z)[/tex]

    simplifying this gives [tex]y=cos(z/3)[/tex]

    So hence for this equality to be true, [tex]4\left(cos(z/3)\right)^3-3\left(cos(z/3)\right)-z=0[/tex] for all z, but this isn't the case.

    Please help me understand this more :smile:
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Jun 4, 2010 #2

    tiny-tim

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    Hi Mentallic! :smile:
    You mean [itex]4\left(cos(z/3)\right)^3-3\left(cos(z/3)\right)-cos(z)=0[/itex], which is true …

    just expand cos(z/3 + 2z/3). :wink:
     
  4. Jun 4, 2010 #3

    Mentallic

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    Re: Equality?

    But [tex]z=cos\left(3arccos(y)\right)[/tex]

    and [tex]cos\left(3arccos(y)\right)=4y^3-3y[/tex] where [tex]y=cos\left(\frac{1}{3}arccos(z)\right)[/tex]

    So shouldn't it then be [tex]4y^3-3y=z[/tex] ???

    Yeah if I change it to cos(z) it works, but I still don't see where I've gone wrong.
     
  5. Jun 4, 2010 #4

    tiny-tim

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    You're interchangeably using z as an angle and as a cosine.

    In particular, your …
    isn't true. :wink:

    (unless cosz = z)
     
  6. Jun 4, 2010 #5

    Mentallic

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    Re: Equality?

    Oh of course... :blushing:
     
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