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Homework Help: Cosine function

  1. Dec 22, 2008 #1
    1. The problem statement, all variables and given/known data

    what is cos(2n*pi)

    2. Relevant equations



    3. The attempt at a solution

    I understand that cos(npi)=(-1)^n
    so is cos(2n*pi)=2(-1)^n ??
     
  2. jcsd
  3. Dec 22, 2008 #2

    cristo

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    Make the substitution m=2n in cos(mπ)=(-1)^m
     
  4. Dec 22, 2008 #3
    so you mean cos(2n^2)=(-1)^2n
    ??
     
  5. Dec 22, 2008 #4

    cristo

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    No, I mean cos(2n pi)=(-1)^{2n}
     
  6. Dec 22, 2008 #5

    HallsofIvy

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    Do you know what cos(0) is? Do you know that cosine is periodic with period 2 pi?
     
  7. Dec 22, 2008 #6
    yep cos0=1
    yep, cos is periodic with period 2pi
     
  8. Dec 22, 2008 #7

    HallsofIvy

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    So cos(2n pi)= cos(0+ n(2pi))= ?
     
  9. Dec 23, 2008 #8
    oh right,
    so =(-1)^(n+1)
    is that right?
     
  10. Dec 23, 2008 #9

    Mark44

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    No. Look at posts 4 and 7.
     
  11. Dec 23, 2008 #10

    HallsofIvy

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    Okay, what does "periodic" mean????
     
  12. Dec 23, 2008 #11
    ...I think it is necessary to know the graph of cos(x), which may help a lot. so, find one.

    edit (:shy: trying not to be ambiguous)
    ...I think it is necessary for one to know the graph of cos(x), which may also help a lot. (regardless of this particular problem)...
    "periodic" is really the key:approve:
     
    Last edited: Dec 24, 2008
  13. Dec 24, 2008 #12

    HallsofIvy

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    It might help. It is not necessary. All that is necessary is to know what "periodic" means. No computation is required.
     
  14. Dec 24, 2008 #13
    I understand that periodic means that cosine function repeats after multiples of 2 pi. but how would that have anything to do with writing cos(2n*pi) ?
    cos (npi)=(-1)^n because as long as n is an integer, the value will alternate from -1 and 1 (clearly form the graph)
     
  15. Dec 24, 2008 #14

    HallsofIvy

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    Would it be easier if it were written n*(2pi) rather than 2n*pi? This is about multiples of 2pi!

    cos(2pi)= cos(0+ 2pi)= cos(0)= 1

    cos(4pi)= cos(2pi+ 2pi)= cos(2pi)= 1

    cos(6pi)= cos(4pi+ 2pi)= cos(4pi)= 1

     
  16. Dec 24, 2008 #15
    oh right!!! so cos(n2pi) has to always be 1...i feel very stupid, i should have known that. for all n, cos(2npi) must be 1 as long as n is an integer.

    thank you very much
     
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