• $\mathrm{Given}:\sin^{-1}(e^{i\theta})$
• $Let \sin^{-1}(e^{i\theta}) = x + iy$

$e^{i\theta}= \sin (x +iy)$

$\cos\theta + i\sin\theta = \sin x\ \cos\ hy + i\cos x\ \sin\ hy$

• Comparing real and imaginary parts,

$\cos \theta = \sin x\ \cos\ hy\cdots \mathrm{Equation \ 1}$

$\sin x= \dfrac{\cos\theta}{\cos h y}$

$\sin \theta = \cos x\ \sin hy\cdots \mathrm{Equation \ 2}$

$\cos x= \dfrac{\sin\theta}{\sin h y}$

• $But\ \sin^2x + \cos^2x = 1$

$ \dfrac{\cos^2\theta}{\cos h^2 y} + \dfrac{\sin^2\theta}{\sin h^2 y}=1$

$ \cos^2\theta\ \sin\ h^2 y\ + \sin^2\theta\ \cos h^2 y =\cos h^2 y \times \sin h^2 y$

$(1-\sin^2\theta)\ \sin h^2 y\ + \sin^2\theta\ (1+\sin h^2 y) =(1+\sin h^2 y) \times \sin h^2 y$$

$\sin h^2 y\ - \sin^2\theta\ \sin h^2 y\ + \sin^2\theta\ +\sin^2\theta\ \sin h^2 y =\sin h^2 y +\sin h^4 y$

$sin^2 \theta = sin h^4y$

$sin \theta = sin h^2y\cdots \mathrm{Equation \ 3}$

$\sqrt{sin \theta} = sinhy$

$y = sinh^{-1} \sqrt{sin \theta}$

• $ Squaring\ equation\ 2, \sin^2 \theta = \cos^2 x\ \sinh^2y$

$\sin^2 \theta = \cos^2 x\ \sin \theta\cdots \mathrm{From\ Equation \ 3}$

$\cos^2 x = \sin\theta$

$\cos x = \sqrt {\sin\theta}$

$ x = \cos^{-1}(\sqrt {\sin\theta})$

• $Thus, \sin^{-1}(e^{i\theta}) = x + iy = \cos^{-1}(\sqrt {\sin\theta}) + i \sinh^{-1} \sqrt{\sin \theta}$
• $Hence, Real\ part\ is\ x = \cos^{-1}(\sqrt {\sin\theta}) and\ Imaginary\ part\ is\ y = \sinh^{-1} \sqrt{\sin \theta}$