Question

How do you solve the inverse trig function `sin(sin^-1 (1/3))`?

Answer 1

`sin^-1(1/3)` or `arcsin(1/3)` is the angle `theta` for which `sin(theta) = 1/3`

Therefore `sin(sin^-1(1/3))`
`= sin(theta)` for the value of `theta` for which `sin(theta) = 1/3`

That is
`sin(sin^-1(1/3)) = 1/3`