The real part of (1 - cos theta + 2isin theta | vedclass

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(D) Let $z = (1 - \cos 	heta ) + i(2\sin 	heta )$. We want the real part of $z^{-1} = \frac{1}{z}$.Using trigonometric identities,$1 - \cos 	heta =

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$\frac{1}{5 + 3\cos 	heta }$

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(D) Let $z = (1 - \cos 	heta ) + i(2\sin 	heta )$. We want the real part of $z^{-1} = \frac{1}{z}$.Using trigonometric identities,$1 - \cos 	heta = 2\sin^2(\frac{	heta}{2})$ and $2\sin 	heta = 4\sin(\frac{	heta}{2})\cos(\frac{	heta}{2})$.So,$z = 2\sin(\frac{	heta}{2}) [\sin(\frac{	heta}{2}) + 2i\cos(\frac{	heta}{2})]$.Then,$z^{-1} = \frac{1}{2\sin(\frac{	heta}{2})} \cdot \frac{1}{\sin(\frac{	heta}{2}) + 2i\cos(\frac{	heta}{2})}$.Multiply numerator and denominator by the conjugate $\sin(\frac{	heta}{2}) - 2i\cos(\frac{	heta}{2})$:$z^{-1} = \frac{1}{2\sin(\frac{	heta}{2})} \cdot \frac{\sin(\frac{	heta}{2}) - 2i\cos(\frac{	heta}{2})}{\sin^2(\frac{	heta}{2}) + 4\cos^2(\frac{	heta}{2})}$.The real part is $\frac{\sin(\frac{	heta}{2})}{2\sin(\frac{	heta}{2}) [\sin^2(\frac{	heta}{2}) + 4\cos^2(\frac{	heta}{2})]} = \frac{1}{2[\sin^2(\frac{	heta}{2}) + 4\cos^2(\frac{	heta}{2})]}$.Using $\sin^2(\frac{	heta}{2}) = \frac{1 - \cos 	heta}{2}$ and $\cos^2(\frac{	heta}{2}) = \frac{1 + \cos 	heta}{2}$:Real part $= \frac{1}{2[\frac{1 - \cos 	heta}{2} + 4(\frac{1 + \cos 	heta}{2})]} = \frac{1}{1 - \cos 	heta + 4 + 4\cos 	heta} = \frac{1}{5 + 3\cos 	heta }$.

The real part of $(1 - \cos \theta + 2i\sin \theta )^{-1}$ is

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