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Help me derive Lagrange's Trigonometric Identity

  1. Jun 29, 2004 #1
    Using the fact that z=e^ibeta

    and the identity 1 + z + z^2...+z^n=1-z^(n+1)/1-z

    Help me derive

    1 + cosbeta +cos2beta + ... +cosnbeta= 1/2 + sin((2n+1)beta/2)/2sin(beta/2) where beta is between 0 and 2pi
     
  2. jcsd
  3. Jun 29, 2004 #2

    HallsofIvy

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    z= e= cos(β)+ i sin(β)
    z2= e2iβ= cos(2β)+ i sin(2β)
    .
    .
    .
    zn= eniβ= cos(nβ)+ i sin(nβ

    so the left hand side is the real part of 1+ z+ z2+...+zn.


    (By the way: the right hand side of the first sum should be (1- z^(n+1))/(1-z).
    The parentheses are necessary!)
     
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