 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-2/5/b/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

(cos θ + i sin θ)^{5} = cos 5 θ + i sin 5 θ

. With

s = sin θ

and

cos^{2} θ = 1 - s^{2}

sin 5 θ = 5 (1 - s^{2})^{2} s - 10 (1 - s^{2}) s^{3} + s^{5} = 16 s^{5} - 20 s^{3} + 5 s .

The required equation cancels the fifth powers and becomes

5 s (1 - 4 s^{2}) = 0

. Thus

sin θ = 0, \pm 1/2

, giving all real solutions

θ = nπ or θ = nπ \pm \frac{π}{6}, n \in Z .

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