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De Moivre's Formula Question

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Use de Moivre's formula to derive the following trigonometric identites:

    [tex](a) cos3\theta = cos^3\theta - 3cos\theta sin^2\theta[/tex]

    [tex](b) sin3\theta = 3cos^2\theta sin\theta - sin^3\theta[/tex]


    2. Relevant equations



    3. The attempt at a solution
    The only way I have even figured out to solve this is by just doing
    [tex](cos\theta + isin\theta)^3 = (cos^3\theta - 3cos\theta sin^2\theta) + i(3cos^2\theta sin\theta - sin^3\theta) = cos3\theta + isin3\theta[/tex]

    but I fear that this is not what the problem is asking me to do. I think on (a) I should be factoring out [tex]cos^3\theta - 3cos\theta sin^2\theta = cos\theta(cos^2\theta - 3sin^2\theta)[/tex]

    should I then use the trig. formula that [tex]cos^2 - sin^2 = cos2\theta[/tex] but the 3 in front of sin is throwing me off. Anyone have a clue as how this problem is supposed to be done in the way the question is asking? Thank you
     
  2. jcsd
  3. Sep 3, 2009 #2

    rock.freak667

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    Homework Helper

    [tex](cos\theta + i sin \theta)^3 = cos 3 \theta+ i sin 3 \theta[/tex]


    Expand out the left side and notice that cos3θ is the real part.
     
  4. Sep 3, 2009 #3
    Yea rock that's what I did I just do not know if that's what the question is asking for because I do not think it is but I was wondering if anyone else had any ideas on what to do besides what you mentioned.
     
  5. Sep 3, 2009 #4

    zcd

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    All the terms with i would = sin3θ, while all the terms without i (reals) would = cos3θ
     
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