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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?

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  • $\begingroup$ $\sin (1)$ shouldn't be there. $\endgroup$
    – user261263
    Feb 11, 2017 at 9:41
  • $\begingroup$ try to write in math mode $\endgroup$
    – Arun
    Feb 11, 2017 at 9:45
  • $\begingroup$ yeah $\sin(1)$ shouldn't be there. $\endgroup$
    – Arun
    Feb 11, 2017 at 9:51
  • $\begingroup$ 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 $\sin(-1), \sin(0), \sin(1)$? $\endgroup$
    – mavavilj
    May 10, 2018 at 7:17

2 Answers 2

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The floor function of a float/real returns the integer right below the number.

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 $\{\sin(−1),\sin(0),\sin(1)\}$.

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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 $x \ge 0$, $f(x) = \sin[x] = \sin(0) = 0$.

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  • $\begingroup$ Expect when $1<= x <= \frac{\pi}{4}$ and then $floor(x)=1$? $\endgroup$
    – mavavilj
    May 10, 2018 at 7:22

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