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Question of 88

Q.Simplify ((cos⁡α+isin⁡α)4(sin⁡β+icos⁡β)5)\left(\dfrac{(\cos\alpha + i\sin\alpha)^4}{(\sin\beta + i\cos\beta)^5}\right).

Telangana TsbieTelangana Board of Intermediate Education 2024Subjective· 2mImportance★★★★★
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Apply De Moivre's theorem to the numerator, rewrite the denominator as i e−iβi\,e^{-i\beta}, and simplify the resulting exponential form.

By De Moivre's theorem, (cos⁡α+isin⁡α)4=cos⁡4α+isin⁡4α=ei4α(\cos\alpha+i\sin\alpha)^4 = \cos4\alpha+i\sin4\alpha = e^{i4\alpha}.

For the denominator, note sin⁡β+icos⁡β=i(cos⁡β−isin⁡β)=i e−iβ\sin\beta+i\cos\beta = i(\cos\beta - i\sin\beta) = i\,e^{-i\beta}.

So (sin⁡β+icos⁡β)5=i5e−i5β=i e−i5β(\sin\beta+i\cos\beta)^5 = i^5 e^{-i5\beta} = i\,e^{-i5\beta} (since i5=ii^5=i).

Therefore: …

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