Evaluate: frac{(cos theta + isin theta)^4}{(s | vedclass

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Question

(D) Given expression: $\frac{(\cos 	heta + i\sin 	heta)^4}{(\sin 	heta + i\cos 	heta)^5}$ We know that $\sin 	heta + i\cos 	heta = i(\cos 	heta

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

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(D) Given expression: $\frac{(\cos 	heta + i\sin 	heta)^4}{(\sin 	heta + i\cos 	heta)^5}$ We know that $\sin 	heta + i\cos 	heta = i(\cos 	heta - i\sin 	heta) = i(\cos 	heta + i\sin 	heta)^{-1}$. Substituting this into the denominator: $\frac{(\cos 	heta + i\sin 	heta)^4}{[i(\cos 	heta + i\sin 	heta)^{-1}]^5} = \frac{(\cos 	heta + i\sin 	heta)^4}{i^5(\cos 	heta + i\sin 	heta)^{-5}}$ Since $i^5 = i$,we have: $\frac{(\cos 	heta + i\sin 	heta)^4}{i(\cos 	heta + i\sin 	heta)^{-5}} = \frac{1}{i} (\cos 	heta + i\sin 	heta)^{4 - (-5)} = \frac{1}{i} (\cos 	heta + i\sin 	heta)^9$ Using De Moivre's Theorem: $\frac{1}{i} (\cos 9	heta + i\sin 9	heta) = -i(\cos 9	heta + i\sin 9	heta) = -i\cos 9	heta - i^2\sin 9	heta = \sin 9	heta - i\cos 9	heta$.

Evaluate: $\frac{(\cos \theta + i\sin \theta)^4}{(\sin \theta + i\cos \theta)^5}$

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