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Math Help - Prove using Binomial and DeMoivre's Theorem?

  1. #1
    Super Member fardeen_gen's Avatar
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    Prove using Binomial and DeMoivre's Theorem?

    If Z = cos θ + i sin θ, (θ belongs to set of real numbers, i = √-1), then according to DeMoivre's theorem, Z^n = cos nθ + i sin nθ, (n belongs to set of natural numbers). Using this theorem alongwith binomial expansion, prove that:

    (i) cos 2nθ =( r = 0 to n) (2n C 2r) cos^(2(n - r)) θ sin^(2r) θ
    (ii) cos 2nθ = (r = 0 to n) (2n C (2r + 1))cos^(2(n-r) - 1) θ sin^(2r + 1) θ

    NOTE: 2n C 2r is 2n choose 2r.
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  2. #2
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    Hi

    \cos(2n\theta) = Re(\cos(2n\theta) + i \sin(2n\theta))

    \cos(2n\theta) = Re\left((\cos \theta + i \sin \theta)^{2n}\right)

    \cos(2n\theta) = Re\left(\sum_{k=0}^{2n} \binom{2n}{k} \cos^{2n-k} \theta +(i \sin \theta)^{k}\right)

    (i \sin \theta)^{k} = i^k \sin^k \theta is pure complex if k is odd and real if k is even

    Set k=2r
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  3. #3
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by running-gag View Post
    Hi

    \cos(2n\theta) = Re\left(\sum_{k=0}^{2n} \binom{2n}{k} \cos^{2n-k} \theta + (i \sin \theta)^{k}\right)
    Instead of +, shoudn't it be . ?

    So, the question appears to have a small mistake, doesn't it? As in the R.H.S, it says in the question that the summation is from r = 0 to n instead of r = 0 to 2n as you proved. Thanks for the help!
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  4. #4
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    Quote Originally Posted by fardeen_gen View Post
    Instead of +, shoudn't it be . ?
    Yes sorry

    Quote Originally Posted by fardeen_gen View Post
    So, the question appears to have a small mistake, doesn't it? As in the R.H.S, it says in the question that the summation is from r = 0 to n instead of r = 0 to 2n as you proved. Thanks for the help!
    No
    k is from 0 to 2n and k=2r therefore r is from 0 to n
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  5. #5
    Super Member fardeen_gen's Avatar
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    No
    k is from 0 to 2n and k=2r therefore r is from 0 to n
    Oops. My bad. Thanks once again!
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  6. #6
    Super Member fardeen_gen's Avatar
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    But then in (ii), the sin term will disappear entirely(setting k = 2r + 1?). And also, the summation would be from r = -1/2(possible?) to r = (2n - 1)/2.
    What am I missing?
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  7. #7
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    Quote Originally Posted by fardeen_gen View Post
    (i) cos 2nθ =( r = 0 to n) (2n C 2r) cos^(2(n - r)) θ sin^(2r) θ
    (ii) cos 2nθ = (r = 0 to n) (2n C (2r + 1))cos^(2(n-r) - 1) θ sin^(2r + 1) θ

    NOTE: 2n C 2r is 2n choose 2r.
    To be clear
    (i) is false since (i \sin \theta)^{2r} = i^{2r} \sin^{2r} \theta = (-1)^r \sin^{2r} \theta
    (ii) is false because
    - it is \sin(2n\theta)
    - r cannot go up to n because you would get \binom{2n}{2n+1}
    Use the imaginary part instead of the real part to get the second relation
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