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    Hey, am currently attempting a question off the old P6 papers to do with de Moivre's. The question is:

    a) Use de Moivre’s theorem to show that tan3theta = (3tantheta-tan^3theta) / (1-3tan^2theta)
    b) Hence show that t = tan pi/9 is a solution of the equation:
    t^3 - (3rt3)t^2 - 3t + rt3 = 0

    I'm fine with part a, but I can't see how the equation in part b has any resemblance to the bit in part a, it seems that they've multiplied the denominator by rt 3 but that's all I can tell... if anyone could help I'd be very grateful!
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      (Original post by Jemeter)
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      Well, if we write x = \tan\theta then what is the equation in part (a) in terms of x? (Leave \tan 3\theta as it is for now)
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      (Original post by Kolya)
      Well, if we write x = \tan\theta then what is the equation in part (a) in terms of x? (Leave \tan 3\theta as it is for now)
      Okay, I have tan3theta = (3x - x^3)/(1-3x^2), still unsure as where to go next..
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        (Original post by Jemeter)
        Okay, I have tan3theta = (3x - x^3)/(1-3x^2), still unsure as where to go next..
        How about trying to rearrange what you have to get it in the form of a cubic equation, like you are given in part (b)? If you try to get the two into similar forms, you might be able to spot the similarities between the two.
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        (Original post by Kolya)
        How about trying to rearrange what you have to get it in the form of a cubic equation, like you are given in part (b)? If you try to get the two into similar forms, you might be able to spot the similarities between the two.
        Aha, I think I've cracked, thank you!
       
       
       
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