Question

How do you find the exact value of `tan (sin^-1 (2/3))`?

Answer 1

Since `tan(x)=sin(x)/cos(x)`, we must find `sin(sin^{-1}(2/3))` and `cos(sin^{-1}(2/3))`. Clearly, `sin(sin^{-1}(2/3))=2/3` by the definition of an inverse function (in this case, to be more precise, `sin(sin^{-1}(x))=x` for all `-1\leq x\leq 1`). Also, the cosine of the "angle" `sin^{-1}(2/3)` is positive, therefore by the Pythagorean identity, `cos(sin^{-1}(2/3))=sqrt{1-sin^{2}(sin^{-1}(2/3))}`
`=sqrt{1-(2/3)^2}=sqrt{5/9}=sqrt{5}/3.`

(You could also draw a right triangle, label one of the angles `sin^{-1}(2/3)`, label the side lengths appropriately, use the Pythagorean theorem and SOH, CAH, TOA appropriately to do this last calculation.)

Therefore, `tan(sin^[-1}(2/3))=(2/3)/(sqrt{5}/3)=2/sqrt{5}=(2sqrt{5})/5`