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Derivation Trouble

  1. Sep 6, 2004 #1
    Derivation Trouble!!!!!

    Please read the attachment with this posting: Here's the problem, I have been attempting to derive this for a couple of days now... However, it seems that whatever I do all that I end up deriving is itself again; meaning I get back to where I started. Can anyone give me a few pointers because I'm flat out of luck here. Here I use Euler's Relationships...but no help....hmm..?

    e^(i*theta)-1=2i*sin(theta/2)*e^(i*(theta/2))

    I have the pretty version in the attachment. :grumpy: :rofl: :grumpy:

    Crazed Hannah
     

    Attached Files:

  2. jcsd
  3. Sep 6, 2004 #2
    liz this is how one posts
     
  4. Sep 6, 2004 #3

    Tide

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    Note that [itex]1=e^{2\pi i}[/itex] and [itex]a^2-b^2 = (a-b)(a+b)[/itex]
     
  5. Sep 6, 2004 #4

    mathman

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    I will use x for theta.

    eix-1=cosx+isinx-1
    =cos2x/2-sin2x/2+2icosx/2sinx/2-1
    =-2sin2x/2+2icosx/2sinx/2
    =2isinx/2(cosx/2+isinx/2)
    =2isinx/2eix/2
     
  6. Sep 6, 2004 #5
    I'm still a little confused on how you got the half angle in there? I seemed to have missed a step

    Hannah
     
  7. Sep 6, 2004 #6
    Because I know that e^(2ix) = cos2x/2-sin2x/2+2icosx/2sinx/2. Where'd you get e^(2ix), because all that I see is e^(ix)? Or is it emplied that 1=e^(2ix)? If that is true then why does your answer still contain a -1? Looking like this: cos2x/2-sin2x/2+2icosx/2sinx/2 -1?

    Hannah
     
  8. Sep 6, 2004 #7

    Tide

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    Even simpler:

    [tex]e^{i \theta} - 1 = e^{i \frac{\theta}{2}} \left(e^{i \frac{\theta}{2}} - e^{-i \frac{\theta}{2}}\right)[/tex]

    It's just factoring.
     
  9. Sep 7, 2004 #8
    Thanx so much....
     
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