 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2022/9/b/solution

Use the imaginary part of the finite complex sum:

\sum_{n = 0}^{N} (n N) sin (n θ + ϕ) = Im [e^{i ϕ} \sum_{n = 0}^{N} (n N) e^{in θ}] = Im [e^{i ϕ} (1 + e^{i θ})^{N}] .

The identity

1 + e^{i θ} = 2 e^{i θ /2} cos (θ /2)

then gives

n = 0 \sum N (n N) sin (n θ + ϕ) = 2^{N} cos^{N} (θ /2) sin (Nθ /2 + ϕ) .

Here

N

is a nonnegative integer. Keeping the signed cosine raised to the integer power makes the result valid even when

cos (θ /2) < 0

; inserting an absolute value would lose the sign for odd

N

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