What is the range of the function $f(x) = sin\lfloor{x}\rfloor$ given that it's domain is $(\dfrac{-\pi}{4}, \dfrac {\pi}{4})$

Question

This is how I did it, let me know where I went wrong

$Dom(f) = (\dfrac{-\pi}{4}, \dfrac {\pi}{4})$
$\therefore \dfrac{-\pi}{4} < x < \dfrac{\pi}{4}$
$\therefore$ Approximately, $-0.785 < x < 0.785$
$\therefore \lfloor{x}\rfloor$ can have values : $-1$ and $0$ only
So, $sin\lfloor{x}\rfloor$ can only have two values : $sin(-1)$ and $sin(0)$
$\therefore$ In my opinion, $Range(f) = ${$sin(-1),sin(0)$} which can be written as {$-sin(1),sin(0)$}

But as per my textbook, the range is {$-sin(1),sin(0),sin(1)$}, which means that according to the book, $\lfloor{x}\rfloor$ can be equal to $1$, but for it to happen, the value of x should be greater than or equal to $1$, which is not in the domain.

Am I wrong, or is the book wrong?

Let me know, thanks...