Use De Moivre's theorem to express cos 5θ and sin 5θ in terms of sin θ and cos θ.

1 answer:

7 0

From the de Moivre's we have,

(cosθ+isinθ)^n=cos(nθ)+isin(nθ)

Therefore,

R((cosθ+isinθ)^5)=cos(5θ)I((cosθ+isinθ)^5)=sin(5θ)

Simplifying,

cos^5(θ)−10(sin^2(θ))(cos^3(θ))+5(sin^4(θ))(cosθ)=cos(5θ)

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