But it's complex!

Algebra Level pending

Find the value of \(\large \sin^{-1} \left(\frac{e^{2ix}-1}{2ie^{ix} }\right) \) at \(x=\frac\pi2\).

If the answer is \(A\pi\) , submit your answer as \(A\). [Hint:Use the eulers formula for complex numbers That is \(e^{ix}=\cos{x}+i\sin{x}\) ]

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