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Product of Exponential Form (easy)

  1. Sep 2, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]e^{i\theta_1}e^{i\theta_2} = e^{i(\theta_1 + \theta_2)}}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    For some reason every I multiply [tex](cos\theta_1 + isin\theta_1)(cos\theta_2 + isin\theta_2)[/tex] I am getting

    [tex](cos\theta_1 cos\theta_2 + sin\theta_1 sin\theta_2) + i(sin\theta_1 cos\theta_2 + cos\theta_1 sin\theta_2)[/tex]

    according to my book the first part should be [tex](cos\theta_1 cos\theta_2 - sin\theta_1 sin\theta_2)[/tex]

    what am I missing here? Is it some basic fundamental from calculus I have forgotten?

    What I am doing is [tex]cos\theta_1 cos\theta_2 - (isin\theta_1)(isin\theta_2) = cos\theta_1 cos\theta_2 - i^2 sin\theta_1 sin\theta_2[/tex] since [tex]i^2 = -1[/tex] that makes it positive. But this can't be right because both the book and my notes from class cannot be wrong. So please enlighten me =]
     
  2. jcsd
  3. Sep 2, 2009 #2
    [tex]e^{ix} = \cos x + i\sin x,\,\![/tex]

    Substitute x=[itex]\theta_1 + \theta_2[/itex] and then use:

    [tex]\sin(\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \,[/tex]

    [tex]\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta\,[/tex]
     
  4. Sep 2, 2009 #3

    Mark44

    Staff: Mentor

    For the real part you should be getting cos(th1)cos(th2) + i^2*sin(th1)sin(th2). I think you omitted the i^2 factor.
     
  5. Sep 2, 2009 #4
    Why are you subtracting? Once you answer that, you should be all set.
     
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