Prove that if \phi is not equal to 2k\pi for

Cobie Sadler

Cobie Sadler

Answered question

2022-02-28

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)}

Answer & Explanation

ImpudgeIntemnect

ImpudgeIntemnect

Beginner2022-03-01Added 6 answers

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[eiθt=0neitϕ]
=I[eiθ1ei(n+1)ϕ1eiϕ]
=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.

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