Cosine function

  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. cristo

    cristo 8,407
    Staff Emeritus
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    Make the substitution m=2n in cos(mπ)=(-1)^m
     
  4. so you mean cos(2n^2)=(-1)^2n
    ??
     
  5. cristo

    cristo 8,407
    Staff Emeritus
    Science Advisor

    No, I mean cos(2n pi)=(-1)^{2n}
     
  6. HallsofIvy

    HallsofIvy 40,382
    Staff Emeritus
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    Do you know what cos(0) is? Do you know that cosine is periodic with period 2 pi?
     
  7. yep cos0=1
    yep, cos is periodic with period 2pi
     
  8. HallsofIvy

    HallsofIvy 40,382
    Staff Emeritus
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    So cos(2n pi)= cos(0+ n(2pi))= ?
     
  9. oh right,
    so =(-1)^(n+1)
    is that right?
     
  10. Mark44

    Staff: Mentor

    No. Look at posts 4 and 7.
     
  11. HallsofIvy

    HallsofIvy 40,382
    Staff Emeritus
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    Okay, what does "periodic" mean????
     
  12. ...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. HallsofIvy

    HallsofIvy 40,382
    Staff Emeritus
    Science Advisor

    It might help. It is not necessary. All that is necessary is to know what "periodic" means. No computation is required.
     
  14. 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. HallsofIvy

    HallsofIvy 40,382
    Staff Emeritus
    Science Advisor

    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. 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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