Summation of a trigonometric function

[Supernova123]

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)

 

[Orodruin]

I suggest not rewriting in terms of sines and cosines until after doing the summation.

 

[Supernova123]

Okay, so I got:
Σcos(2n-1)θ = ∑z^2n-1 - Σsin(2n-1)θ
=z(1-z^2N)/(1-z²) - isinθ(1-(isinθ)²)/(1-(isinθ)²)

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

=(z - izsin2θ - z^2N+1 + iz^2N+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 :(

 

[Ray Vickson]

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)

If $z = e^{i \theta}$ then $z^{2n-1} = e^{(2n-1) i \theta} = \cos((2n-1) \theta) + i \sin((2n-1) \theta)$.

 

[Supernova123]

I found my mistake. I wrongly assumed that (sinθ)^n=sin(nθ). Here's my solution:
Σcos(2n-1)θ=Re(Σz^2n-1)
=Re(e^iθ(1-e^2Niθ)/(1-e^2iθ))
=Re((1-e^2Niθ)/(e^-iθ-e^iθ))
=Re((1-cos(2Nθ)-isin(2Nθ)/(-2isinθ))
=Re((i-icos(2Nθ)+sin(2Nθ)/(2sinθ))
=sin(2Nθ)/2sinθ