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Summation of a trigonometric function

  1. May 23, 2015 #1
    1. The problem statement, all variables and given/known data
    By considering ∑z2n-1, where z=e, show that Σcos(2n-1)θ=sin(2Nθ)/2sinθ. (Σ means summation from 1 to N)


    2. Relevant equations
    Just a guess. S=a(1-r^n)/(1-r)

    3. The attempt at a solution
    I was thinking this but it doesn't seem to work very well. Σz2n-1=Σ(cosθ+isinθ)2n-1=Σ(cos(2n-1)θ+isin(2n-1)θ). Rearranging,
    Σcos(2n-1)θ=Σ(cosθ+isinθ)2n-1-Σsin(2n-1)θ
    =(cosθ+isinθ)(1-(cosθ+isinθ)2N/(1-(cosθ+isinθ)2) - isinθ(1-(isinθ)2N/(1-(isinθ)^2)
     
  2. jcsd
  3. May 23, 2015 #2

    Orodruin

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    I suggest not rewriting in terms of sines and cosines until after doing the summation.
     
  4. May 23, 2015 #3
    Okay, so I got:
    Σcos(2n-1)θ = ∑z2n-1 - Σsin(2n-1)θ
    =z(1-z2N)/(1-z2) - isinθ(1-(isinθ)2)/(1-(isinθ)2)

    =(z-z2N+1)/(1-z2) - (isinθ - isin(2N+1)θ)/(1 - isin2θ)

    =(z - izsin2θ - z2N+1 + iz2N+1sin2θ - isinθ + isin(2N+1)θ + iz2sinθ - iz2sin(2N+1)θ)/(1 - z2 - isin2θ + iz2sin2θ)

    =(cosθ + isinθ - icosθsin2θ + sinθsin2θ -cos(2N+1)θ -isin(2N+1)θ + icos(2N+1)θsin2θ - sin(2N+1)sin2θ - isinθ + isin(2N+1)θ + icos2θsinθ - sin2θsinθ - icos2θsin(2N+1)θ + sin2θsin(2N+1)θ)/(1 - cos2θ - isin2θ - isin2θ + icos2θsin2θ - sin2θsin2θ)

    =(cosθ - icosθsin2θ - cos(2N+1)θ + icos(2N+1)θsin2θ)/(1 - cos2θ - 2isin2θ + icos2θsin2θ - sin2θsin2θ) and I'm stuck :(
     
  5. May 24, 2015 #4

    Ray Vickson

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    If ##z = e^{i \theta}## then ##z^{2n-1} = e^{(2n-1) i \theta} = \cos((2n-1) \theta) + i \sin((2n-1) \theta)##.
     
  6. May 24, 2015 #5
    I found my mistake. I wrongly assumed that (sinθ)^n=sin(nθ). Here's my solution:
    Σcos(2n-1)θ=Re(Σz2n-1)
    =Re(e(1-e2Niθ)/(1-e2iθ))
    =Re((1-e2Niθ)/(e-iθ-e))
    =Re((1-cos(2Nθ)-isin(2Nθ)/(-2isinθ))
    =Re((i-icos(2Nθ)+sin(2Nθ)/(2sinθ))
    =sin(2Nθ)/2sinθ
     
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