Prove that if ϕ is not equal to 2kπ for any integer k, then∑t=0nsin⁡(θ+tϕ)=sin⁡(n+1)ϕ2)sin⁡(θ+nϕ2)}{sin⁡(ϕ2)}

Use the exponential representation of the sines and cosines:cos⁡(θ+tϕ)=12(ei(θ+tϕ)+e−(θ+tϕ))=R[ei(θ+tϕ)]sin⁡(θ+tϕ)=12i(ei(θ+tϕ)−e−(θ+tϕ))=I[ei(θ+tϕ)]Then use a geometric series to sum.Specifically, for the sine series, write∑t=0nsin⁡(θ+tϕ)=I[eiθ∑t=0neitϕ]=I[eiθ1−ei(n+1)ϕ1−eiϕ]=I[ei(θ+nϕ2)]sin⁡[(n+1)ϕ2]sin⁡(ϕ2)=sin⁡(θ+nϕ2)sin⁡[(n+1)ϕ2]sin⁡(ϕ2)What is different for the cosine series?For∑t=0ncos2(2tθ)write cos2(2tθ)=12+(12)cos⁡(4tθ) and see if the work you did for the cosine series applies. Similar for the sin2(2tθ) series.

"Prove that if ϕ is not equal to..." solved and explained

Cobie Sadler

Answered question

2022-02-28

Prove that if

ϕ

is not equal to

2 k π

for any integer k, then

\sum_{t = 0}^{n} \sin (θ + t ϕ) = \frac{\sin (n + 1) ϕ}{2}) \sin (θ + \frac{n ϕ}{2})} {\sin (\frac{ϕ}{2})}

Answer & Explanation

ImpudgeIntemnect

Beginner2022-03-01Added 6 answers

Use the exponential representation of the sines and cosines:

\cos (θ + t ϕ) = \frac{1}{2} (e^{i (θ + t ϕ)} + e^{- (θ + t ϕ)}) = R [e^{i (θ + t ϕ)}]

\sin (θ + t ϕ) = \frac{1}{2 i} (e^{i (θ + t ϕ)} - e^{- (θ + t ϕ)}) = I [e^{i (θ + t ϕ)}]

Then use a geometric series to sum.
Specifically, for the sine series, write

\sum_{t = 0}^{n} \sin (θ + t ϕ) = I [e^{i θ} \sum_{t = 0}^{n} e^{i t ϕ}]

= I [e^{i θ} \frac{1 - e^{i (n + 1) ϕ}}{1 - e^{i ϕ}}]

= I [e^{i (θ + n \frac{ϕ}{2})}] \frac{\sin [(n + 1) \frac{ϕ}{2}]}{\sin (\frac{ϕ}{2})}

= \sin (θ + n \frac{ϕ}{2}) \frac{\sin [(n + 1) \frac{ϕ}{2}]}{\sin (\frac{ϕ}{2})}

What is different for the cosine series?
For

\sum_{t = 0}^{n} \cos^{2} (2 t θ)

write

\cos^{2} (2 t θ) = \frac{1}{2} + (\frac{1}{2}) \cos (4 t θ)

and see if the work you did for the cosine series applies. Similar for the

\sin^{2} (2 t θ)

series.

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