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,

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.