Solution 10400: Calculating the Inverse Sine of 2 Using the TI-85.
Why does the TI-85 return an answer for the inverse sine of 2?
For complex numbers x+iy, the sine function is defined as sin(x+iy) = sin(x)cosh(y)+icos(x)sinh(y).
cosh(y) = (ey + e-y)/2
sinh(y) = (ey - e-y)/2
When y=0, cosh(y)=1 and sinh(y)=0, so sin(x+i*0) = sin(x), the function commonly used. However, sin(p/2+1.316i) = sin(p/2)cosh(1.316)+i*cos(p/2)sinh(1.316) = cosh(1.316)+i*0=~ 2.
Since this functionality is built into the TI-85, arcsin(2) returns p/2+1.316...i, which the TI-85 displays as (1.57... , 1.316...).
NOTE: The above information is based on the identity ei*y = cos(y)+i*sin(y).