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De Moivre's Theorum and Double-Angle Formulas

  1. Dec 31, 2007 #1
    I hope this is in the right place.

    I'm in grade 12, and I've been given an assignment involving complex numbers.

    The question reads:

    Use De Moivre's Theorum to verify the identities:
    [tex]cos(2\theta) = cos^2\theta - sin^2\theta[/tex]

    [tex]sin(2\theta) = 2sin\theta cos\theta[/tex]

    I've tried something like this:
    [tex]
    cos(2\theta) + i \cdot sin(2\theta) = (cos\theta + i \cdot sin\theta)^2
    [/tex]

    [tex]
    cos(2\theta) + i \cdot sin(2\theta) = cos^2\theta + i \cdot 2cos\theta sin\theta - sin^2\theta
    [/tex]

    [tex]
    cos(2\theta) = cos^2\theta - sin^2\theta + i \cdot 2cos\theta sin\theta - i \cdot sin(2\theta)
    [/tex]

    But I don't understand where to go from there. Can I somehow "separate" them?
    Any help would be appreciated.
     
    Last edited: Dec 31, 2007
  2. jcsd
  3. Dec 31, 2007 #2
    When the two complex numbers

    [tex] a+i\,b, \quad c+i\,d[/tex]

    are equal?
     
  4. Dec 31, 2007 #3
    I'm sorry, I don't understand.
     
  5. Dec 31, 2007 #4
    The equation [tex] a+i\,b=c+i\,d[/tex] gives

    [tex] a=c,\, \quad b=d[/tex].

    Apply this to your formulas
     
  6. Dec 31, 2007 #5
    Well, I didn't know that.
    Thanks for the help. :)

    EDIT: I just got it: I'm an idiot. Thanks again.
     
    Last edited: Dec 31, 2007
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