# 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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