(sin theta + icos theta )^n is equal to | vedclass

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(C) We know that $\sin 	heta = \cos \left( \frac{\pi }{2} - 	heta ight)$ and $\cos 	heta = \sin \left( \frac{\pi }{2} - 	heta ight)$.Substitut

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$\cos n\left( \frac{\pi }{2} - 	heta ight) + i\sin n\left( \frac{\pi }{2} - 	heta ight)$

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(C) We know that $\sin 	heta = \cos \left( \frac{\pi }{2} - 	heta ight)$ and $\cos 	heta = \sin \left( \frac{\pi }{2} - 	heta ight)$.Substituting these into the expression,we get $(\sin 	heta + i\cos 	heta )^n = \left[ \cos \left( \frac{\pi }{2} - 	heta ight) + i\sin \left( \frac{\pi }{2} - 	heta ight) ight]^n$.By De Moivre's theorem,$(\cos \phi + i\sin \phi )^n = \cos(n\phi) + i\sin(n\phi)$.Applying this with $\phi = \frac{\pi }{2} - 	heta$,we obtain $\cos n\left( \frac{\pi }{2} - 	heta ight) + i\sin n\left( \frac{\pi }{2} - 	heta ight)$.

$(\sin \theta + i\cos \theta )^n$ is equal to

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