- #1

Supernova123

- 12

- 0

## Homework Statement

By considering ∑z

^{2n-1}, where z=e

^{iθ}, show that Σcos(2n-1)θ=sin(2Nθ)/2sinθ. (Σ means summation from 1 to N)

## Homework Equations

Just a guess. S=a(1-r^

^{n})/(1-r)

## The Attempt at a Solution

I was thinking this but it doesn't seem to work very well. Σz

^{2n-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)