The value of frac{(cos theta + i sin theta)^4 | vedclass

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Question

(D) Given expression: $E = \frac{(\cos 	heta + i \sin 	heta)^4}{(\sin 	heta + i \cos 	heta)^5}$ Using De Moivre's Theorem,the numerator is $(\cos

Vedclass · Accepted Answer

$\sin 9 	heta - i \cos 9 	heta$

Vedclass · Answer

(D) Given expression: $E = \frac{(\cos 	heta + i \sin 	heta)^4}{(\sin 	heta + i \cos 	heta)^5}$ Using De Moivre's Theorem,the numerator is $(\cos 	heta + i \sin 	heta)^4 = \cos 4 	heta + i \sin 4 	heta$. For the denominator,rewrite $\sin 	heta + i \cos 	heta$ as $i(\cos 	heta - i \sin 	heta) = i(\cos(-	heta) + i \sin(-	heta))$. So,$(\sin 	heta + i \cos 	heta)^5 = i^5 (\cos(-	heta) + i \sin(-	heta))^5 = i (\cos(-5 	heta) + i \sin(-5 	heta)) = i(\cos 5 	heta - i \sin 5 	heta)$. Thus,$E = \frac{\cos 4 	heta + i \sin 4 	heta}{i(\cos 5 	heta - i \sin 5 	heta)} = \frac{1}{i} \cdot \frac{\cos 4 	heta + i \sin 4 	heta}{\cos 5 	heta - i \sin 5 	heta}$. Using the property $\frac{\cos \alpha + i \sin \alpha}{\cos \beta + i \sin \beta} = \cos(\alpha - \beta) + i \sin(\alpha - \beta)$,we have: $E = -i \cdot \frac{\cos 4 	heta + i \sin 4 	heta}{\cos(-5 	heta) + i \sin(-5 	heta)} = -i (\cos(4 	heta - (-5 	heta)) + i \sin(4 	heta - (-5 	heta)))$ $E = -i (\cos 9 	heta + i \sin 9 	heta) = -i \cos 9 	heta - i^2 \sin 9 	heta = \sin 9 	heta - i \cos 9 	heta$.

The value of $\frac{(\cos \theta + i \sin \theta)^4}{(\sin \theta + i \cos \theta)^5}$ is equal to,where $i = \sqrt{-1}$:

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