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DeMoivre's theorem: cos3θ in powers of cosθ

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Use de Moivre's Theorem to express cos3θ in powers of cosθ


    2. Relevant equations

    z^n = [r(cosθ + isinθ)]^n = r^n (cos(nθ) + i sin(nθ))

    3. The attempt at a solution

    cos3θ = Re(cos3θ +isin3θ) = Re[(cosθ +isinθ)^3]

    I've then expanded the brackets using binomial theorem and got;

    (cosθ)^3 + 3[(cosθ)^2][isinθ] + 3(cosθ)[(isinθ)^2] + (isinθ)^3

    So (cosθ)^3 is the real part and 3[(cosθ)^2][isinθ] + 3(cosθ)[(isinθ)^2] + (isinθ)^3 the imaginary part.

    If anyone has any suggestions...

    Thank you
     
    Last edited: Oct 30, 2011
  2. jcsd
  3. Oct 30, 2011 #2

    lurflurf

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    use i^2=-1 to write (cosθ)^3 + 3[(cosθ)^2][isinθ] + 3(cosθ)[(isinθ)^2] + (isinθ)^3 in a+bi form
     
  4. Oct 30, 2011 #3
    Okay, so doing that I'm getting:

    (cosθ)^3 - 3cosθ(sinθ)^2 + 3i(cosθ)^2 (sinθ) - i(sinθ)^3

    So the real part is (cosθ)^3 - 3cosθ(sinθ)^2

    the imaginary part 3i(cosθ)^2 (sinθ) - i(sinθ)^3
     
  5. Oct 30, 2011 #4

    lurflurf

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    Yes, though the imaginary part is usually defined to not have the i
    also the question asked for an answer in cosθ so eliminate sinθ
     
  6. Oct 30, 2011 #5
    Okay, thanks. I've got the answer.
     
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