# How do you evaluate sin^-1(0)?

Jan 3, 2017

0

#### Explanation:

${\sin}^{-} 1 \left(0\right) = x$

Take sin to the left and to the right.

$\sin {\sin}^{-} 1 \left(0\right) = \sin x = 0$

$x = 0 + k \pi$, because $\sin k \pi = 0$

But what is the domain of $f \left(x\right) = {\sin}^{-} 1 x$ ?

Response: $f : \left[- 1 , 1\right] \rightarrow \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$