sin^-1(3/5)+sin^-1(5/13)=sin^-1(56/65)?

Put u = arcsin(3/5), v = arcsin(5/13). Then sin(u) = (3/5) & sin(v) = (5/13), Now cos(u) = [1 - (3/5)^2]^(1/2) = (4/5) & cos(v) = [1 - (5/13)^2]^(1/2) = (12/13). sin(u+v) = sin(u)cos(v) +
cos(u)sin(v) = (3/5)(12/13) + (4/5)(5/13) = (36+20)/65 = (56/65). Then arcsin(u) + acsin(v) =
(u+v) = arcsin(56/65) = 59.48976259 deg., approx. = 59.5 deg.

Hint:Write arcsin(x) = -i ln(ix + sqrt(1 - x²)) then apply the properties of logarithms.