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Solve cos((1/3)*arccos(x))

  1. Sep 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Find an albebraic expression for cos((1/3)*arccos(x)), in order to get rid of the trigonometric operands.

    2. Relevant equations

    3. The attempt at a solution
    cos(x)= 4cos(x/3)^3 - 3cos(x/3), I can reduce cos(cos(x)/3)
    by noting that cos(arccos(x))=x. Then I solve the cubic:
    1 = 4y^3 - 3y
    This was a tips I have found, but, it gave me nothing :P
    Need help!
  2. jcsd
  3. Sep 10, 2009 #2


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    The equation you want to solve is x=4y^3-3y, isn't it? You want to solve for y in terms of x. Don't ask me about the algebra of solving the cubic. I really don't like solving cubics.
  4. Sep 10, 2009 #3
    Do you really need too use cubic to solve cos((1/3)*arccos(x))?
    However, is it posible to get an albebraic expression for cos((1/3)*arccos(x)) without the trigonometric operands?

    I can solve the cubic if you get it on the form ax^3+bx^2+cx+d=0 for me ;)
  5. Sep 10, 2009 #4
    This is the answer to the problem. It is exactly solving a cubic.
  6. Sep 10, 2009 #5


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    I don't know your teacher -- but "an algebraic expression for foo" does not necessarily mean that "foo" has to be isolated. It might be enough to simply have a purely algebraic equation involving foo.
  7. Sep 10, 2009 #6


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    If you can solve cubics, then the problem is solve a*y^3+b*y^2+c*y+d=0, where a=4, b=0, c=-3 and d=(-x). (-x) is just a constant. I think Zetison may actually be expected to solve the cubic. There is already one post detailing an explicit cubic solution. I gave up half way through. I couldn't handle it. I did an explicit cubic solution once in my life. Because I REALLY wanted an expression for the real root. I hope never to have to do it again.
    Last edited: Sep 11, 2009
  8. Sep 11, 2009 #7


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    Just for amusement I plugged that into Maple 10 and got this answer:

    [tex] y = \frac{(x + \sqrt{-1+x^2})^{1/3}}{2} + \frac 1 {2(x+\sqrt{-1+x^2})^{1/3}} [/tex]
  9. Sep 11, 2009 #8
    Makes no sense to me, because x is defined only in [-1,1]. While your equation is defined in (-infinite,-1]U[1,infinite)
    Last edited: Sep 11, 2009
  10. Sep 11, 2009 #9

    Ben Niehoff

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    No, the expression is actually defined for all x. You get imaginary numbers, but the first term is the conjugate of the second term, so the imaginary parts cancel. When cos and arccos are analytically continued to the complex plane, the expression

    [tex]\cos \arccos x = x[/tex]

    is true for all complex x (and in particular, for all real x).
  11. Sep 11, 2009 #10
    So with other words:

    http://folk.ntnu.no/jonvegar/3 [Broken]
    for all x and none irrational numbers?
    Last edited by a moderator: May 4, 2017
  12. May 18, 2010 #11
    Maple says:
    [PLAIN]http://folk.ntnu.no/jonvegar/images/capture.png [Broken]

    It did not work. I still need help! :smile:
    Last edited by a moderator: May 4, 2017
  13. May 18, 2010 #12


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    That's because you're equating the wrong things. The correct solution is given in post #7 by LCKurtz.

    What you've done in post #11 is the following

    cos(1/3 \arccos(x))= 4\cos(x/3)^3 - 3\cos(x/3)

    which is wrong. Just read your own post again (post #1).
  14. May 18, 2010 #13
    No solution (without cheating like by using Maple) after 8 months? :confused:
  15. May 18, 2010 #14
    You can solve this problem easily as follows.

    Using the identity:

    cos(x) = [exp(i x) + exp(-i x)]/2

    express arccos(x) in terms of logarithms.

    Then compute cos[1/3 arccos(x)] using the obtained expression for arccos(x) and the above identity for cos(x).
  16. May 18, 2010 #15
    Hmmm, it shall be possible to write the equation without any logarithms.
    If it is so easy, can you just write down the formula for me? :smile: I need it actually just to solv an other problem.
  17. May 19, 2010 #16


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    You've been given two methods. Either you solve the cubic equation or you use Count Iblis' method which is a lot faster. Perhaps it's time you show some work?
  18. May 19, 2010 #17
    Belive me, I have tried. But I don't understand the answer i get if I use the 2. method:

    [PLAIN]http://folk.ntnu.no/jonvegar/images/imaginary.jpg [Broken]

    I want a answer without any imaginary parts. My x is defined only for [0,9], but when I cancel the imaginary part I do not get the right answer as demonstrated above...
    Last edited by a moderator: May 4, 2017
  19. May 19, 2010 #18


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    Maybe you want to explain what you're doing instead of spitting out some maple math with some random numbers plugged in. I suggest once again that you use Count Iblis' method to derive the result instead of using some computer program.
    Last edited: May 19, 2010
  20. May 19, 2010 #19
    Can you define Count Iblis' method for me?
  21. May 19, 2010 #20
    You obtain an expression for y = arccos(x) by solving the equation:

    [exp(i y) + exp(-iy)]/2 = x

    Putting z = exp(iy), yields:

    z + 1/z = 2 x

    Can you take it from here?
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