 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/41d/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let

C_{j} (ξ)

denote the

j

ξ_{n} = (- 1)^{n} x_{n}

. For

0 \leq k < N

C_{N - 1 - k} (ξ) = n = 0 \sum N - 1 (- 1)^{n} x_{n} cos (π (n + \frac{1}{2}) - \frac{π}{N} (n + \frac{1}{2}) (k + 1)) = n = 0 \sum N - 1 x_{n} sin (\frac{π}{N} (n + \frac{1}{2}) (k + 1)) = Z_{k} .

The identity uses

cos (π (n + 1/2) - u) = (- 1)^{n} sin u

. Sign-modulating the input and reversing the

N

cosine-transform outputs each cost

O (N)

operations, so part (c) gives the sine transform in

O (N lo g N)

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