frac{(cos alpha + isin alpha )^4}{(sin beta + | vedclass

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Question

(C) Given expression: $\frac{(\cos \alpha + i\sin \alpha )^4}{(\sin \beta + i\cos \beta )^5}$ Using De Moivre's theorem,the numerator is $\cos 4\alpha

Vedclass · Accepted Answer

$\sin (4\alpha + 5\beta ) - i\cos (4\alpha + 5\beta )$

Vedclass · Answer

(C) Given expression: $\frac{(\cos \alpha + i\sin \alpha )^4}{(\sin \beta + i\cos \beta )^5}$ Using De Moivre's theorem,the numerator is $\cos 4\alpha + i\sin 4\alpha$. For the denominator,we write $\sin \beta + i\cos \beta = i(\cos \beta - i\sin \beta)$. So,the denominator becomes $i^5(\cos \beta - i\sin \beta)^5 = i(\cos 5\beta - i\sin 5\beta)$. Thus,the expression is $\frac{\cos 4\alpha + i\sin 4\alpha}{i(\cos 5\beta - i\sin 5\beta)} = \frac{1}{i}(\cos 4\alpha + i\sin 4\alpha)(\cos 5\beta + i\sin 5\beta)$. Since $\frac{1}{i} = -i$,we have $-i[\cos (4\alpha + 5\beta) + i\sin (4\alpha + 5\beta)]$. $= -i\cos (4\alpha + 5\beta) - i^2\sin (4\alpha + 5\beta) = \sin (4\alpha + 5\beta) - i\cos (4\alpha + 5\beta)$.

$\frac{(\cos \alpha + i\sin \alpha )^4}{(\sin \beta + i\cos \beta )^5} = $

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