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Viete's relations problem

  1. May 21, 2008 #1
    [SOLVED] viete's relations problem

    1. The problem statement, all variables and given/known data
    The zeros of the polynomial P(x) = x^3 -10x+11 are u,v,and w. Determine the value of arctan u +arctan v+ arctan w.


    2. Relevant equations
    http://en.wikipedia.org/wiki/Vi├Ęte's_formulas

    3. The attempt at a solution
    I must admit I have no idea how to do this problem. I usually try to show some work but I have no clue how Vietes relations allow you to connect u,v,and w to the inverse trig functions. Sorry. Someone please just give me a nudge in the right direction.
     
  2. jcsd
  3. May 21, 2008 #2

    rock.freak667

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    Homework Helper

    Let [itex]P(x)=x^3 -10x+11=0[/itex]

    what is u+v+w,uv+uw+vw and uvw ?

    and what you want to get is [itex]tan^{-1}(u)+tan^{-1}(v)+tan^{-1}(w)[/itex]

    if we let [itex]A=tan^{-1}(u);B=tan^{-1}(v);C=tan^{-1}(w)[/itex]

    thus making tanA=u;tanB=v and tanC=w, are able with some trig identity to help you?
     
  4. May 21, 2008 #3
    u+v+w = 0

    uv+uw+vw = -10

    uvw = -11

    The only relevant identity I can think of is

    [tex]\tan(a\pm b) = \frac{\tan a \pm \tan b}{1\mp \tan a \tan b} [/tex]

    which does not seem very useful here...

    We can of course obtain that

    [tex]\tan^3 A +\tan^3 B + \tan ^3 C = 33 [/tex]
     
    Last edited: May 21, 2008
  5. May 21, 2008 #4

    rock.freak667

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    Homework Helper

    Good.

    Now onto a new part.

    It is relevant but what you need to do is extend it to three things inside the brackets.

    i.e. tan(A+B+C)=tan(A+(B+C))
     
  6. May 21, 2008 #5
    Wow that worked out really really nicely.

    I get tan(A+B+C) = 1 which implies that the desired sum is [itex]\frac{\pi}{4}[/itex]. Thanks.
     
  7. May 21, 2008 #6
    On second thought, do we know that the sum is not 5 pi/4 or -3 pi/4?
     
  8. May 21, 2008 #7
    Never mind, I got it.
     
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