Q.101.

Question 101.
Using de Moivre's theorem, find the constants a, b and c given that
sin6θ/sinθ = a cos^5(θ) + b cos^3(θ) + c cos(θ), where sinθ  ≠ 0.

Answer 101.
By de Moivre's theorem,
(cos θ + i sin θ)^6 = cos 6θ + i sin 6θ

Expanding LHS using Binomial Theorem,
LHS
= cos^6(θ) + 6i cos^5(θ)*sinθ - 15cos^4(θ) sin^2(θ) - 20i cos^3(θ) sin^3(θ) + 15cos^2(θ)   sin^2(θ) + 20i cosθ sin^5(θ) - sin^6(θ)

Comparing imaginary terms,
sin 6θ = 6cos^5(θ) sinθ - 20cos^3(θ) sin^3(θ) + 6cosθ sin^5(θ)

=> (sin 6θ) / sin θ
= 6 cos^5(θ) - 20 cos^3(θ) sin^2(θ) + 6 cosθ sin^4(θ)
= 6cos^5(θ) - 20 cos^3(θ)*[1 - cos^2(θ)] + 6 cosθ *[1 - cos^2(θ)]^2
= 6 cos^5(θ) - 20 cos^3(θ) + 20 cos^5(θ) + 6cosθ - 12 cos^3(θ) + 6 cos^5(θ)
= 32 cos^5(θ) - 32 cos^3θ + 6cosθ

=> a = 32, b = - 32, c = 6.

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