If the real part of the complex number (1-cos | vedclass

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(D) Let $z = (1 - \cos 	heta + 2i \sin 	heta)^{-1} = \frac{1}{(1 - \cos 	heta) + 2i \sin 	heta}$.Multiplying numerator and denominator by the conj

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(D) Let $z = (1 - \cos 	heta + 2i \sin 	heta)^{-1} = \frac{1}{(1 - \cos 	heta) + 2i \sin 	heta}$.Multiplying numerator and denominator by the conjugate $((1 - \cos 	heta) - 2i \sin 	heta)$:$z = \frac{(1 - \cos 	heta) - 2i \sin 	heta}{(1 - \cos 	heta)^2 + 4 \sin^2 	heta}$.The real part is $	ext{Re}(z) = \frac{1 - \cos 	heta}{(1 - \cos 	heta)^2 + 4 \sin^2 	heta} = \frac{1}{5}$.Using $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$ and $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$:$	ext{Re}(z) = \frac{2 \sin^2 \frac{	heta}{2}}{4 \sin^4 \frac{	heta}{2} + 16 \sin^2 \frac{	heta}{2} \cos^2 \frac{	heta}{2}} = \frac{2 \sin^2 \frac{	heta}{2}}{4 \sin^2 \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})} = \frac{1}{5}$.Thus,$2(\sin^2 \frac{	heta}{2} + 4 \cos^2 \frac{	heta}{2}) = 5 \implies 2(1 - \cos^2 \frac{	heta}{2} + 4 \cos^2 \frac{	heta}{2}) = 5$.$2(1 + 3 \cos^2 \frac{	heta}{2}) = 5 \implies 2 + 6 \cos^2 \frac{	heta}{2} = 5 \implies 6 \cos^2 \frac{	heta}{2} = 3 \implies \cos^2 \frac{	heta}{2} = \frac{1}{2}$.Since $	heta \in (0, \pi)$,$\frac{	heta}{2} \in (0, \frac{\pi}{2})$,so $\cos \frac{	heta}{2} = \frac{1}{\sqrt{2}}$,which gives $\frac{	heta}{2} = \frac{\pi}{4} \implies 	heta = \frac{\pi}{2}$.Finally,$\int_{0}^{\frac{\pi}{2}} \sin x \,dx = [-\cos x]_{0}^{\frac{\pi}{2}} = -(\cos \frac{\pi}{2} - \cos 0) = -(0 - 1) = 1$.

If the real part of the complex number $(1-\cos \theta+2 i \sin \theta)^{-1}$ is $\frac{1}{5}$ for $\theta \in(0, \pi)$,then the value of the integral $\int_{0}^{\theta} \sin x \,dx$ is equal to:

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