Complex Trigonometry

Assume $\theta$ is a real number such that $\pi/2\leq \theta \leq \pi$ . ( My method seems to only be valid for that range of angles, including addition by any multiple of $2\pi$ . I shortcutted the half angle formulas. I'll leave it for another ambitious soul to work out the proper details for considering all the angles.) Setting up the equation, $\sin{w}=\frac{1}{2i}(e^{i w}-e^{-i w})=z,$ and solving it for $w$ leads to the well-known formula for the arcsine function of a complex variable: $\sin^{-1}{z}=\frac{1}{i}\ln{\Big(iz+(1-z^2)^{1/2}\Big)}$ (where the exponent $1/2$ refers to both the positive and negative square roots). The expression can also be written in the following way: $\sin^{-1}{z}=\frac{1}{i}\ln{\Big(iz+\sqrt{|1-z^2|}e^{\frac{i}{2}arg{(1-z^2)}}\Big)}.$ I wish to use the principal branch of the arcsine (often denoted by "Arcsin"). This is accomplished by using the principal branch of the natural logarithm (often denoted by "Ln") and the principal branch of the argument (often denoted by "Arg"). $Arcsin{(z)}=\frac{1}{i}Ln{\Big(iz+\sqrt{|1-z^2|}e^{\frac{i}{2}Arg{(1-z^2)}}\Big)}.$ I now substitute $z=e^{i\theta}$ and get to work: $Arcsin{(e^{i\theta})}=\frac{1}{i}Ln{\Big(ie^{i\theta}+\sqrt{|1-e^{2i\theta}|}e^{\frac{i}{2}Arg{(1-e^{2i\theta})}}\Big)}$ $=\frac{1}{i}Ln{\Big(i\cos{\theta}-\sin{\theta}+\sqrt{\sqrt{(1-e^{2i\theta})(1-e^{-2i\theta})}}e^{\frac{i}{2}Tan^{-1}{\frac{Im{(1-e^{2i\theta})}}{Re{(1-e^{2i\theta})}}}}\Big)}$ $=\frac{1}{i}Ln{\Big(i\cos{\theta}-\sin{\theta}+\sqrt{\sqrt{2-(e^{2i\theta}+e^{-2i\theta})}}e^{\frac{i}{2}Tan^{-1}{\frac{(1-e^{2i\theta}-(1-e^{-2i\theta}))/(2i)}{(1-e^{2i\theta}+1-e^{-2i\theta})/2}}}\Big)}$ $=\frac{1}{i}Ln{\Big(i\cos{\theta}-\sin{\theta}+\sqrt{\sqrt{2-2\cos{2\theta}}}(e^{\frac{i}{2}Tan^{-1}{\frac{\sin{2\theta}}{\cos{2\theta}-1}}})\Big)}$ $=\frac{1}{i}Ln{\Big(i\cos{\theta}-\sin{\theta}+\sqrt{\sqrt{2-2\cos{2\theta}}}\Big(\cos{\frac{Tan^{-1}{\frac{\sin{2\theta}}{\cos{2\theta}-1}}}{2}}+i\sin{\frac{Tan^{-1}{\frac{\sin{2\theta}}{\cos{2\theta}-1}}}{2}\Big)}\Big)}$ $=\frac{1}{i}Ln{\Big(i\cos{\theta}-\sin{\theta}+\sqrt{\sqrt{2-2\cos{2\theta}}}\Big(\sqrt{\frac{1+\cos{(Tan^{-1}{\frac{\sin{2\theta}}{\cos{2\theta}-1}})}}{2}}+i\sqrt{\frac{1-\cos{(Tan^{-1}{\frac{\sin{2\theta}}{\cos{2\theta}-1}})}}{2}}\Big)\Big)}$ $=\frac{1}{i}Ln{\Big(i\cos{\theta}-\sin{\theta}+\sqrt{\sqrt{\frac{1-\cos{2\theta}}{2}}}\Big(\sqrt{1+\cos{(Tan^{-1}{\frac{\sin{2\theta}}{\cos{2\theta}-1}})}}+i\sqrt{1-\cos{(Tan^{-1}{\frac{\sin{2\theta}}{\cos{2\theta}-1}})}}\Big)\Big)}$ $=\frac{1}{i}Ln{\Big(i\cos{\theta}-\sin{\theta}+\sqrt{\sin{\theta}}\Big(\sqrt{1+\cos{(Tan^{-1}{\frac{\sin{2\theta}}{\cos{2\theta}-1}})}}+i\sqrt{1-\cos{(Tan^{-1}{\frac{\sin{2\theta}}{\cos{2\theta}-1}})}}\Big)\Big)}$ $=\frac{1}{i}Ln{\Big(i\cos{\theta}-\sin{\theta}+\sqrt{\sin{\theta}}\Big(\sqrt{1+\frac{1}{\sqrt{1+\Big(\frac{\sin{2\theta}}{\cos{2\theta}-1}\Big)^2}}}+i\sqrt{1-\frac{1}{\sqrt{1+\Big(\frac{\sin{2\theta}}{\cos{2\theta}-1}\Big)^2}}}\Big)\Big)}$ $=\frac{1}{i}Ln{\Big(i\cos{\theta}-\sin{\theta}+\sqrt{\sin{\theta}}\Big(\sqrt{1+\frac{1}{\frac{\sqrt{2-2\cos{2\theta}}}{1-\cos{2\theta}}}}+i\sqrt{1-\frac{1}{\frac{\sqrt{2-2\cos{2\theta}}}{1-\cos{2\theta}}}}\Big)\Big)}$ $=\frac{1}{i}Ln{\Big(i\cos{\theta}-\sin{\theta}+\sqrt{\sin{\theta}}\Big(\sqrt{1+\frac{1-\cos{2\theta}}{2\sin{\theta}}}+i\sqrt{1-\frac{1-\cos{2\theta}}{2\sin{\theta}}}\Big)\Big)}$ $=\frac{1}{i}Ln{\Big(i\cos{\theta}-\sin{\theta}+\sqrt{\sin{\theta}}\Big(\sqrt{1+\frac{1-(1-2\sin^2{\theta})}{2\sin{\theta}}}+i\sqrt{1-\frac{1-(1-2\sin^2{\theta})}{2\sin{\theta}}}\Big)\Big)}$ $=\frac{1}{i}Ln{\Big(i\cos{\theta}-\sin{\theta}+\sqrt{\sin{\theta}}\Big(\sqrt{1+\sin{\theta}}+i\sqrt{1-\sin{\theta}}\Big)\Big)}$ $=\frac{1}{i}Ln{\Big(\sqrt{\sin{\theta}(1+\sin{\theta})}-\sin{\theta}+i\Big(\sqrt{\sin{\theta}(1-\sin{\theta})}+\cos{\theta}\Big)\Big)}$ $=\frac{1}{i}\Big[Ln{\Big|\sqrt{\sin{\theta}(1+\sin{\theta})}-\sin{\theta}+i\Big(\sqrt{\sin{\theta}(1-\sin{\theta})}+\cos{\theta}\Big)\Big|}+i Arg\Big(\sqrt{\sin{\theta}(1+\sin{\theta})}-\sin{\theta}+i\Big(\sqrt{\sin{\theta}(1-\sin{\theta})}+\cos{\theta}\Big)\Big)\Big]$ The real part of the final expression above is what we desire: $Arg\Big(\sqrt{\sin{\theta}(1+\sin{\theta})}-\sin{\theta}+i\Big(\sqrt{\sin{\theta}(1-\sin{\theta})}+\cos{\theta}\Big)\Big)$ The final step is to use the formula $Arg(x+iy) = \cos^{-1}{\frac{x}{\sqrt{x^2+y^2}}}$ . I'm still working on being more precise about this whole thing to make it work out nicely. I'm a bit lazy to do all the steps properly at the moment. I'll probably be back on it eventually, but anyone is welcome to comment if they'd like to finish it off.