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  4. The domain of sin-1[x] is given by

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Q. The domain of $sin^{-1}[x]$ is given by

Inverse Trigonometric Functions

Solution:

Domain of $sin^{-1} x$ is $[-1, 1]$
and Domain of $sin^{-1}[x]$ is $\{x : -1 \le [x] \le 1\}$
$= \{x : [x] = -1, 0, 1\}$
But $[x] = \begin{cases} -1 & \forall \,-1 \le x < 0 \\ 0 & \forall \,0\le x < 1 \\ 1 & \forall \,1\le x < 2 \end{cases} $
$\therefore x \in [-1, 2) $ i.e., $-1 \le x < 2$