 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/12/c/solution

z = e^{i θ}

, so

2 i sin θ = z - z^{- 1}

(2 i sin θ)^{5} = (z^{5} - z^{- 5}) - 5 (z^{3} - z^{- 3}) + 10 (z - z^{- 1}) .

Use

z^{k} - z^{- k} = 2 i sin (k θ)

and

(2 i)^{5} = 32 i

to obtain

sin^{5} θ = \frac{10 sin θ - 5 sin 3 θ + sin 5 θ}{16} .

This is a finite Fourier sine series, with only odd harmonics.

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