Wednesday, 4 June 2014

trigonometry - Summation of cos (2n-1) theta


By considering Nn=1z2n1, where z=eiθ, show that
Nn=1cos(2n1)θ=sin(2Nθ)2sinθ,
where sinθ0





I approached this question by starting with that the sum of real part of z2n1 is cos(2n1)θ, and then I stated that the first term is a=cosθ+isinθ, and r=(cosθ+isinθ)2, and therefore said it's a geometric sum, so I used the formula of a geometric series, and after simplification I removed the imaginary parts and I ended up with
(sin(2n+1)θsinθ)/(2sinθcosθ)



I can't take it to the form they have given :/ can anyone please tell me where I went wrong , or if how to simplify this.

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