By considering N∑n=1z2n−1, where z=eiθ, show that
N∑n=1cos(2n−1)θ=sin(2Nθ)2sinθ,
where sinθ≠0
I approached this question by starting with that the sum of real part of z2n−1 is cos(2n−1)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)θ−sinθ)/(−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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