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My book's throwing equations out of nowhere

  1. Aug 29, 2005 #1

    VietDao29

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    My book's throwing equations out of nowhere!!!

    On proving one theorem, my book says:
    [tex]1 + 2\sum_{k = 1} ^ {n} \cos k \alpha = \frac{\sin \left( n + \frac{1}{2} \right) \alpha}{\sin \frac{\alpha}{2}}[/tex]
    I have no idea where that formula comes from :confused:, and I tried in vain proving it, but I failed. :cry: Can anybody gives me a hint?
    Thank you,
    Any help will be appreciated,
    Viet Dao,
     
    Last edited: Aug 29, 2005
  2. jcsd
  3. Aug 29, 2005 #2

    Hurkyl

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    How did you try? If I remember, it's pretty easy.
     
  4. Aug 29, 2005 #3

    Tide

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    Think de Moivre! :)
     
  5. Aug 29, 2005 #4

    lurflurf

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    [tex]2\sin\left(\frac{a}{2}\right)\cos(k a)=\sin\left(k a+\frac{a}{2}\right) \ - \ \sin\left(k a-\frac{a}{2}\right)[/tex]
    so
    [tex]\cos(k a)=\frac{\sin\left(k a+\frac{a}{2}\right) \ - \ \sin\left(k a-\frac{a}{2}\right)}{2\sin\left(\frac{a}{2}\right)}[/tex]
     
  6. Aug 29, 2005 #5

    VietDao29

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    Thanks, After hours of struggling, I realised that it could be proved by induction. Not very hard, though. :smile:
    And I also tried lurflurf's way. Thanks.
    Thanks a lot,
    But... I still don't know how to prove it using de Moivre's formula...
    [tex](\cos x + i\sin x) ^ n = \cos (nx) + i \sin (nx)[/tex]
    How can I do from there?
    Viet Dao,
     
    Last edited by a moderator: Aug 29, 2005
  7. Aug 29, 2005 #6

    Hurkyl

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    Note that cos(nz) = Re[ cis(z)^n ]

    ( cis(z) := cos z + i sin z = exp(z) )
     
  8. Aug 30, 2005 #7

    VietDao29

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    Thanks a lot, guys. :smile:
    Viet Dao,
     
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