First off, I know this is a duplicate of this question. I'm asking this because I still don't quite understand the answer given there. But first, some graphs!

  • $y=\sin(x)$

    y=sin x

  • $y=\arcsin(x)$

    y=arcsin x

  • $y=\sin(\arcsin(x))$

    y=sin(arcsin x)

  • $y=\arcsin(\sin (x))$

    y=arcsin(sin x)

What I don't understand is why $f(x) = \arcsin(\sin(x))$ looks like it does. As stated in the previously linked question, $f(x)=\arcsin(x)$ is only defined between $[-1,1]$. So, how does it effect the rest of the graph? Shouldn't the domain be only values of $\sin x$ within the domain of $\arcsin x$?


  • $\begingroup$ Relevant: Why does $\sin^{-1}(\sin(\pi))$ not equal $\pi$ $\endgroup$ – MJD Feb 19 '14 at 2:45
  • $\begingroup$ Yes, arcsin is only defined for values in [-1,1] (complex numbers notwithstanding). And sin only takes values in [-1,1]. So therefore any x value can be plugged into arcsin(sin x). $\endgroup$ – anon Feb 19 '14 at 3:04

For $\pi\ge x\ge \frac{1}{2}\pi$, we have $\sin[x]=\sin[\pi-x]$. Therefore we have $$ \arcsin(\sin[x])=\arcsin(\sin[\pi-x])=\pi-x $$ Similarly for $-\pi\le x\le -\frac{\pi}{2}$, we have $\sin[x]=\sin[-\pi-x]$. Therefore we have $$ \arcsin(\sin[x])=\arcsin(\sin[-\pi-x])=\arcsin(-\sin[\pi+x])=-\arcsin(\sin(\pi+x))=-\pi-x $$

I think the rest of the graph follows by the same argument. I attached a graph, I do not know if it helps.

enter image description here

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Since $\sin(x)$ is always in the interval $[-1,1]$ where $\arcsin$ is defined, $\arcsin(\sin(x))$ is defined for all real $x$.

$\arcsin(y)$ is defined as the number $x$ within the interval $[-\pi/2, \pi/2]$ where $\sin(x) = y$. So for $x$ in the interval $[-\pi/2, \pi/2]$, $\arcsin(\sin(x)) = x$ satisfies that definition. For $x$ in the interval $[\pi/2, 3 \pi/2]$, $\arcsin(\sin(x))$ can't be $x$, but it can be $\pi - x$ which is in the interval $[-\pi/2, \pi/2]$ (notice that $\sin(\pi - x) = \sin(x)$). So that takes care of this part of the graph:

enter image description here

Now notice that $\sin(x)$ is periodic with period $2\pi$, so the graph looks the same if you shift it left or right by $2\pi$. That takes care of the rest of it.

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  • $\begingroup$ So I understand the jist of it; but, I don't get why $sin(π−x)=sin(x)$. Did you mean that this only works for the interval $[−π/2,π/2]$? $\endgroup$ – evamvid Feb 19 '14 at 3:19
  • $\begingroup$ To understand why sin(π−x)=sin(x), we need to start from the extended definition of sine for angles greater than π/2. sin(x) is defined as y-ordinate to the radius of the circle in question. That circle is centered at O(0, 0) and radius = $sqrt(x^2 + y^2)$. $\endgroup$ – Mick Feb 19 '14 at 4:43

So let $f(x)=\sin(x)$ and let $g(x)=\arcsin(x)$. Let $\operatorname{Dom}(f)$ be the domain of $f$ and let $\operatorname{Rng}(f)$ be the range of $f$. Thus we have $$ \begin{align} \operatorname{Dom}(f)&=\mathbb{R}\\ \operatorname{Rng}(f)&=[-1,1]\\ \operatorname{Dom}(g)&=[-1,1]\\ \operatorname{Rng}(g)&=\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \end{align} $$

If we restrict the $\operatorname{Dom}(f)$ to $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, then $f$ and $g$ are inverses of one another and

$$ (f\circ g)(x)=(g\circ f)(x) = x $$

on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$.

Now let $\operatorname{Dom}(f)=\mathbb{R}$ and consider $(f\circ g)(x)$. This graph should remain unchanged since for any $x\not\in[-1,1]$, $(f\circ g)(x)$ is not defined and so no picture can be drawn on $\mathbb{R}\setminus[-1,1]$.

On the other hand, when we let $\operatorname{Dom}(f)=\mathbb{R}$ and consider $(g\circ f)(x)$, this composition is defined over all of $\mathbb{R}$, but it behaves in two different ways:

1.) On the intervals $\left[\frac{\pi(2k-1)}{2},\frac{\pi\cdot(2k+1)}{2}\right]$ when $k$ is odd: As $x$ goes from $\frac{\pi(2k-1)}{2}$ to $\frac{\pi(2k+1)}{2}$, $f(x)$ goes from $-1$ to $1$. Thus $(g\circ f)$ goes from $\frac{\pi(2k-1)}{2}$ to $\frac{\pi(2k+1)}{2}$, forming a positively sloped line.

2.) On the intervals $\left[\frac{\pi(2k-1)}{2},\frac{\pi\cdot(2k+1)}{2}\right]$ when $k$ is even: As $x$ goes from $\frac{\pi(2k-1)}{2}$ to $\frac{\pi(2k+1)}{2}$, $f(x)$ goes from $1$ to $-1$. Thus $(g\circ f)$ goes from $\frac{\pi(2k+1)}{2}$ to $\frac{\pi(2k-1)}{2}$, forming a negatively sloped line.

This is why you see the jigsaw pattern when you graph $(g\circ f)$ and don't see any change in $(f\circ g)$.

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  • $\begingroup$ Great. But where do the$/frac{[π(2k−1)}{2}$ and $/frac{π⋅(2k+1)}{2}]$ come from? Specifically the $2k-1$ and $2k+1$ parts in the numberators. $\endgroup$ – evamvid Feb 19 '14 at 13:01
  • $\begingroup$ It is just a way to index the domain in chunks where the function behaves the same way. We are just a shifting the original domain $[\frac{-\pi}{2},\frac{\pi}{2}]$. The first few values of $k=-2,-1,0,1,2$ give $[\frac{-5\pi}{2},\frac{-3\pi}{2}]$, $[\frac{-3\pi}{2},\frac{-\pi}{2}]$, $[\frac{-\pi}{2},\frac{\pi}{2}]$, $[\frac{\pi}{2},\frac{3\pi}{2}]$, $[\frac{3\pi}{2},\frac{5\pi}{2}]$ etc... $\endgroup$ – Laars Helenius Feb 19 '14 at 13:28

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