1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equality math problem

  1. Jun 4, 2010 #1

    Mentallic

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper

    Re: Equality?

    Oh of course... :blushing:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Equality math problem
  1. A Math Problem (Replies: 14)

Loading...