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sin(x + pi/2) = cos(x) = cosh(ix)

Therefore, sin (pi/2 + i arcosh(2)) = cosh(-arcosh(2)) = 2

where arcosh(y) is the familiar inverse of the hyperbolic cosine, i.e. log(y + sqrt(y^2 - 1)), so we find sin(pi/2 + i log(2 + sqrt(3))) = 2.




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