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GuDViN [60]
3 years ago
8

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

Mathematics
1 answer:
Sonja [21]3 years ago
7 0
From the de Moivre's we have,
<span>
(cosθ+isinθ)^n=cos(nθ)+isin(nθ)
</span><span>
Therefore, 
</span><span>
R((cosθ+isinθ)^5)=cos(5θ)I((cosθ+isinθ)^5)=sin(5θ)
</span><span>
Simplifying,
</span><span>
cos^5(θ)−10(sin^2(θ))(cos^3(θ))+5(sin^4(θ))(cosθ)=cos(5θ) </span><span>
</span>
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