(sin theta - i cos theta)^3 is equal to | vedclass

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(D) Given expression is $(\sin 	heta - i \cos 	heta)^3$. Factor out $-i$ from the expression: $(\sin 	heta - i \cos 	heta) = -i (\cos 	heta + i \

Vedclass · Accepted Answer

$(-i)^3(\cos 3 	heta + i \sin 3 	heta)$

Vedclass · Answer

(D) Given expression is $(\sin 	heta - i \cos 	heta)^3$. Factor out $-i$ from the expression: $(\sin 	heta - i \cos 	heta) = -i (\cos 	heta + i \sin 	heta)$. Now,raise this to the power of $3$: $[-i (\cos 	heta + i \sin 	heta)]^3 = (-i)^3 (\cos 	heta + i \sin 	heta)^3$. Using De Moivre's Theorem,$(\cos 	heta + i \sin 	heta)^n = \cos n 	heta + i \sin n 	heta$: $= (-i)^3 (\cos 3 	heta + i \sin 3 	heta)$. Since $(-i)^3 = -i^3 = -(-i) = i$,the expression simplifies to $i(\cos 3 	heta + i \sin 3 	heta) = i \cos 3 	heta - \sin 3 	heta$.

$(\sin \theta - i \cos \theta)^3$ is equal to

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