The real part of sin^{-1}(e^{itheta}) is | vedclass

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(A) Let $\sin^{-1}(e^{i	heta}) = x + iy$. Then,$e^{i	heta} = \sin(x + iy)$. Using the identity $\sin(x + iy) = \sin x \cosh y + i \cos x \sinh y$,we

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$\cos^{-1}(\sqrt{\sin	heta})$

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(A) Let $\sin^{-1}(e^{i	heta}) = x + iy$. Then,$e^{i	heta} = \sin(x + iy)$. Using the identity $\sin(x + iy) = \sin x \cosh y + i \cos x \sinh y$,we have: $\sin x \cosh y + i \cos x \sinh y = \cos 	heta + i \sin 	heta$. Comparing real and imaginary parts: $\sin x \cosh y = \cos 	heta$ and $\cos x \sinh y = \sin 	heta$. Thus,$\cosh y = \frac{\cos 	heta}{\sin x}$ and $\sinh y = \frac{\sin 	heta}{\cos x}$. Using the identity $\cosh^2 y - \sinh^2 y = 1$: $\frac{\cos^2 	heta}{\sin^2 x} - \frac{\sin^2 	heta}{\cos^2 x} = 1$. $\cos^2 	heta \cos^2 x - \sin^2 	heta \sin^2 x = \sin^2 x \cos^2 x$. $\cos^2 	heta \cos^2 x - \sin^2 	heta (1 - \cos^2 x) = (1 - \cos^2 x) \cos^2 x$. $\cos^2 	heta \cos^2 x - \sin^2 	heta + \sin^2 	heta \cos^2 x = \cos^2 x - \cos^4 x$. $\cos^2 x (\cos^2 	heta + \sin^2 	heta) - \sin^2 	heta = \cos^2 x - \cos^4 x$. $\cos^2 x - \sin^2 	heta = \cos^2 x - \cos^4 x$. $\cos^4 x = \sin^2 	heta$. Taking the square root,$\cos^2 x = \sin 	heta$,so $x = \cos^{-1}(\sqrt{\sin 	heta})$.

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

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