# Find the range of $f(x)=\sin[x]$ where $(-\pi/4)\le x \le (\pi/4)$ and [x] is the greatest integer function of x

Question:

Find the range of $$f(x)=\sin[x]$$ where $$(-\pi/4)\le x \le (\pi/4)$$ and $$[x]$$ denotes the greatest integer function of $$x$$.

Since $$\pi=3.141...$$, $$\pi/4 < 1$$. $$\therefore [π/4]=0$$. So, upper limit of $$f(x)$$ is $$0$$ also $$-\pi/4 > -1$$ so $$[-\pi/4]$$ is $$-1$$. Therefore, the range should be $$\{\sin(-1),0\}$$. Though I think I'm right, the solution in the answer key is given as $$\{\sin(-1),0,\sin(1)\}$$.

How is it possible to get $$\sin(1)$$ as a result?

• $\sin (1)$ shouldn't be there.
– user261263
Feb 11, 2017 at 9:41
• try to write in math mode
– Arun
Feb 11, 2017 at 9:45
• yeah $\sin(1)$ shouldn't be there.
– Arun
Feb 11, 2017 at 9:51
• What you mean it shouldn't be there? For $0 <= x < 1$, $floor(x)=0$ and for $1 <= x <= \frac{\pi}{4}$, $floor(x)=1$? So the domain consists of $x \in \{-1, 0, 1\}$ and the range thus becomes $\sin(-1), \sin(0), \sin(1)$? May 10, 2018 at 7:17

For $$0\le x <1$$, $$\lfloor x\rfloor=0$$ and for $$1\le x \le \frac{\pi}{4}$$, $$\lfloor x\rfloor=1$$. So the domain consists of $$x \in \{-1, 0, 1\}$$ and the range thus becomes $$\{\sin(−1),\sin(0),\sin(1)\}$$.
You have determined that $$[x]$$ takes on the values $$-1$$ (when $$x < 0$$) and $$0$$ (when $$x \ge 0$$).
The range is the set of output values of $$\sin[x]$$. So, when $$x < 0$$, $$f(x) = \sin[x] = \sin(-1)$$.
And when $$x \ge 0$$, $$f(x) = \sin[x] = \sin(0) = 0$$.
• Expect when $1<= x <= \frac{\pi}{4}$ and then $floor(x)=1$? May 10, 2018 at 7:22