Question

using demoivre's theorem simplify: (cos3a+ isin3a)4(cos4a+isin4a)2/ (cos2a+isina)5 (cos5a+isin5a)3​

Answer 1

Answer:

De Moivre's formula reads

(cosθ+isinθ)n=cos(nθ)+isin(nθ)

Of course this identity implies the real part should be also equality. That is

cos(nθ)=R{(cosθ+isinθ)n}

Hence we have

cos(3θ)=R{cos3θ+3icos2θsinθ−3cosθsin2θ−isin3θ}=cos3θ−3cosθsin2θ

Step-by-step explanation:

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