Finding the value of Inverse Trigonometric functions beyond their Real Domain I wanted to ask how can we calculate the values of the inverse of trigonometric functions beyond their domain of definition, for example $\arcsin{2}$ beyond its domain of $-\frac{\pi}{2}<x<\frac{\pi}{2}$.
I tried to use the Euler form but did not get much. Thanks.

@Neelesh Vij, he makes an interesting point.  I will do my best to explain.
We have as follows:$$\sin(z)=x, z=\arcsin(x)$$$$e^{iz}=\cos(z)+i\sin(z)$$$$e^{-iz} =\cos(-z)+i\sin(-z)=\cos(z)-i\sin(z)$$And we are trying to solve for an unknown z, given x.
Made clear, the last line has the equation $e^{-iz}=\cos(z)-i\sin(z)$.  This was found with trigonometric identities.
Now subtract the last two equations to get the following:$$e^{iz}-e^{-iz}=2i\sin(z)$$$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$Recall from the very top that $\sin(z)=x$.$$x=\frac{e^{iz}-e^{-iz}}{2i}=\frac1{2i}e^{iz}-\frac1{2i}e^{-iz}$$Multiply both sides by $e^{iz}$ to get a solvable quadratic.$$xe^{iz}=\frac1{2i}e^{2iz}-\frac1{2i}\to0=\frac1{2i}e^{2iz}-xe^{iz}-\frac1{2i}$$We can now use quadratic formula to solve for $e^{iz}$.  If you are confused, try substituting $e^{iz}=y$.$$e^{iz}=\frac{x\pm\sqrt{x^2-1}}{\frac1i}=i(x\pm\sqrt{x^2-1})$$Now, solve for z.$$iz=\ln[i(x\pm\sqrt{x^2-1})]=\ln(i)+\ln[(x\pm\sqrt{x^2-1})]=\pi(\frac12+2in)+\ln[x\pm\sqrt{x^2-1}],n=\pm(0,1,2,3,4,\ldots)$$
We get a grand and wondrous solution:$$z=\frac{\pi(\frac12+2in)+\ln[x\pm\sqrt{x^2-1}]}i,n=\pm(0,1,2,3,4,\ldots)$$So for $x=2$, we can find $z=\arcsin(2)$.$$\arcsin(2)=\frac{\pi(\frac12+2in)+\ln[2\pm\sqrt{2^2-1}]} i,n=\pm(0,1,2,3,4,\ldots)$$
Lastly, take note that $\arcsin(x)=\arccos(x)-\frac{\pi}2\pm2\pi m, m=0,1,2,3,4,\ldots$ so that we have:$$\arccos(x)=\frac{\pi(\frac12+2in)+\ln[x\pm\sqrt{x^2-1}]}i+\frac{\pi}2\pm2\pi m, m=0,1,2,3,4,\ldots,n=\pm(0,1,2,3,4,\ldots)$$
