The real part of (1-cos theta+i sin theta)^{- | vedclass

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Question

(A) Given the expression $(1-\cos 	heta+i \sin 	heta)^{-1} = \frac{1}{1-\cos 	heta+i \sin 	heta}$.Using trigonometric identities $1-\cos 	heta =

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$1/2$

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(A) Given the expression $(1-\cos 	heta+i \sin 	heta)^{-1} = \frac{1}{1-\cos 	heta+i \sin 	heta}$.Using trigonometric identities $1-\cos 	heta = 2 \sin^2(	heta/2)$ and $\sin 	heta = 2 \sin(	heta/2) \cos(	heta/2)$:$\frac{1}{2 \sin^2(	heta/2) + i(2 \sin(	heta/2) \cos(	heta/2))} = \frac{1}{2 \sin(	heta/2) [\sin(	heta/2) + i \cos(	heta/2)]}$.Multiplying the numerator and denominator by the conjugate $[\sin(	heta/2) - i \cos(	heta/2)]$:$= \frac{\sin(	heta/2) - i \cos(	heta/2)}{2 \sin(	heta/2) [\sin^2(	heta/2) + \cos^2(	heta/2)]} = \frac{\sin(	heta/2) - i \cos(	heta/2)}{2 \sin(	heta/2)}$.$= \frac{\sin(	heta/2)}{2 \sin(	heta/2)} - i \frac{\cos(	heta/2)}{2 \sin(	heta/2)} = \frac{1}{2} - i \frac{1}{2} \cot(	heta/2)$.The real part is $1/2$.

The real part of $(1-\cos \theta+i \sin \theta)^{-1}$ is

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