s$in^{-1} e^{i\theta} \;=\; sin^{-1} ⁡(cos⁡θ+i sin⁡θ ) \\ \; \\ \; \\ Let, \; sin^{-1} ⁡(cos⁡θ+i sin⁡θ ) \;=\; x+iy \\ \; \\ \;\\ \therefore sin⁡(x+iy)=cos⁡θ+i sin⁡θ \\ \; \\ \; \\ \therefore cos⁡θ+i sin⁡θ \; = \; sin⁡x cos⁡(iy)+cos⁡x sin⁡(iy) \;= \; sin⁡x cosh⁡y+i cos⁡x sinh⁡y \\ \; \\ \; \\ $

Comparing Real and Imaginary parts,

$ \therefore cos⁡θ=sin⁡x cosh⁡y \; \; and \; \; sin⁡θ=cos⁡x sinh⁡y \\ \; \\ \; \\ We \; know \; that \; cos^2 x+sin^2 x=1 \\ \; \\ \; \\ \therefore \dfrac{sin^2 θ}{sinh^2 y}+ \dfrac{cos^2 θ}{cosh^2 y}=1 \\ \; \\ \; \\ \therefore sin^2 θcosh^2 y+sinh^2 ycos^2 θ=sinh^2 ycosh^2 y=sinh^2 y(1+sinh^2 y) \\ \; \\ \; \\ \therefore sin^2 θ(1+sinh^2 y)+sinh^2 y(1-sin^2 θ)=sinh^2 y(1+sinh^2 y) \\ \; \\ \; \\ \therefore sin^2 θ+sin^2 θsinh^2 y+sinh^2 y-sinh^2 ysin^2 θ=sinh^2 y+sinh^4 y \\ \; \\ \; \\ \therefore sinh^4 y=sin^2 θ \\ \; \\ \; \\ \therefore sinh y= \sqrt{sin θ} \\ \; \\ \; \\ \mathbf{\therefore y \;=\; sinh^{-1} \sqrt{sin θ} \;=\; log[ \sqrt{sin θ} + \sqrt{sin θ+1} ] } \\ \; \\ \; \\ Consider\; cosh^2 y-sinh^2 y=1 \\ \; \\ \; \\ \therefore \dfrac{cos^2 θ}{sin^2 x}- \dfrac{sin^2 θ}{cos^2 x}=1 \\ \; \\ \; \\ \therefore cos^2 θcos^2 x-sin^2 xsin^2 θ=sin^2 xcos^2 x \\ \; \\ \; \\ \therefore cos^2 θcos^2 x-(1-cos^2 x)(1-cos^2 θ)=(1-cos^2 x) cos^2 x \\ \; \\ \; \\ \therefore cos^2 θcos^2 x-1+cos^2 θ+cos^2 x-cos^2 xcos^2 θ=cos^2 x-cos^4 x \\ \; \\ \; \\ \therefore cos^4 x=1-cos^2 θ=sin^2 θ \\ \; \\ \; \\ \therefore cos \; x \;=\; \sqrt{sin θ} \; \; \; \; \mathbf{\therefore x\;=\; cos^{-1}\sqrt{sin θ}} $