Notice that sin(x) is not a one-to-one function since
[tex]\sin x=\sin (x+2\pi)[/tex]
(The period of the sine function is 2pi)
On the other hand, a function needs to be one-to-one in order to be invertible.
Thus, the answer to the first gap is 'is not one-to-one'.
As for the second gap, notice that
[tex]\begin{gathered} \sin (0)=\sin (\pi)=0 \\ \sin (-\frac{\pi}{2})=-1,\sin (\frac{\pi}{2})=1 \end{gathered}[/tex]
The maximum interval at which f(x) is one to one is
[tex]\text{one}-to-\text{one}=\mleft\lbrace x\in\R|-\frac{\pi}{2}\le x\le\frac{\pi}{2}\mright\rbrace=\lbrack-\frac{\pi}{2},\frac{\pi}{2}\rbrack[/tex]
This is the maximum interval that verifies the horizontal-line test.
The answer to the second gap is [-pi/2,pi/2]
Notice that if the domain of the inverse function is [-1,1]; then the domain of such function is [-pi/2,pi/2]
