Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 11 / a

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- Solution a

Solution

Put

c = cos θ

and

s = sin θ

. Taking the real part of

(c + i s)^{6}

in De Moivre's theorem gives

cos 6 θ = c^{6} - 15 c^{4} s^{2} + 15 c^{2} s^{4} - s^{6} .

Taking the imaginary parts of the second and fourth powers gives

sin 2 θ = 2 cs

and

sin 4 θ = 4 cs (c^{2} - s^{2})

. Hence

8 cos 6 θ + 15 sin 4 θ sin 2 θ = 8 (c^{6} - 15 c^{4} s^{2} + 15 c^{2} s^{4} - s^{6}) + 120 c^{2} s^{2} (c^{2} - s^{2}) = 8 (cos^{6} θ - sin^{6} θ) .

The mixed powers cancel exactly.

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