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Prove using de moivre's theorem of I=cos^2n theta

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

    Use de Moivre's Theorem and the result that integral (with limits pi, 0) of cos[nx]dx = 0 for all non-zero integers of n. when n does equal 0 the answer is 1.

    [this i proved in the first section of the qu using logic and limits(x->0).

    so we need to prove that

    integral[with limits pi, 0] [(cos(x))2n]dx = [pi / (2)2n] * [(2n)! / (n!)2]


    2. Relevant equations

    de moivre -> (cos(nx) + i*sin(nx)) = (cos(x) + i*sin(x))n

    eulers -> (cos(x) + i*sin(x)) = eix

    3. The attempt at a solution

    okay ive tried not using de moivres and trying to find a pattern by doing the integral a few times, and ive also thought that using eulers formula might help, but i seriously just cant get started...i know that i need to use the real part of the formula but i dont know what to do first.


    this is a past exam paper by the way and my uni doesnt issue mark schemes, plus it being bank hols weekend no one will be there until wednesday

    i appreciate any help or even just ideas on how to get started
    thanks
     
  2. jcsd
  3. May 2, 2009 #2
    cos(x) = (exp(ix) + exp(-ix))/2

    Take both sides to the power 2n

    Use the binomial theorem for the right-hand side

    Integrate
     
  4. May 2, 2009 #3
    well ive just done that between posting the qu and reading your reply and this is what ive done

    cos(x)2n = 1 / 2^(2n) * (values of cos + (2n)! / ((n!)2))

    the values of cos were found with setting n to various numbers and expanding
    ie-
    if n=2 :-values of cos were cos4x + 4cos2x
    if n=3 :-values of cos were cos6x + 6cos4x + 15cos2x

    but i dont know if i can do this without expanding and leaving n in there.

    so with this equ integrated the cos becomes sin and then fall to 0 because integar numbers of pi is 0, and (2n)! / (n!)2 becomes (2n)! / ((n!)2) *x and the times this by the 1 / 2^(2n) gives the answer.

    but once again i dont know whether i should know how to represent the 'values of cos' relating it to n [the main prob is the coefficients], cause i wont have this kind of time in the exam
     
    Last edited: May 2, 2009
  5. May 5, 2009 #4

    tiny-tim

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    Homework Helper

    Hi indie452! :smile:

    (have a pi: π :wink:)
    Use nealjking's :smile: two hints: cos(x) = (eix + e-ix)/2,

    and (a + b)2n = ∑akb2n-k 2nCk

    which coefficient(s) are the only ones with nonzero integral? :wink:
     
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