Please, I need full explanation on the cancellation property applied to the composition of functions below $$\;\arcsin{\big(\sin(x)\big)}=x,\; \forall\; x\in \Big[-\frac{\pi}{2},\frac{\pi}{2}\Big] ,\;\;\;\sin{\big(\arcsin(x)\big)}=x,\; \forall\; x\in [-1,1].$$ and why is $$\sin{\big(\arcsin(x)\big)}\neq x,\; \forall\; x\in \Bbb{R}?$$
I need some explanation on the kind of maps, how the values are calculated and perhaps, some drawings for clarity!
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This is all fairly simple.
By convention $\arcsin x$ is a function that takes value between -1 and 1 (inclusive) and returns angle between $-\pi/2$ and $+\pi/2$. That's it. $\arcsin$ will never return an angle outside of this segment.
But you can calculate sine function for any angle, outside of $[-\pi/2,+\pi/2]$ range. For such angles inverse sine function will never return the original angle.
For example:
$$\arcsin(\sin\frac{\pi}{6})=\frac{\pi}{6}$$
$$\arcsin(\sin\frac{5\pi}{6})=\frac{\pi}{6}$$
In other words, identity:
$$\arcsin(\sin x)=x$$
...is valid only for $x\in [-\pi/2, +\pi/2]$
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For example $$\sin(\dfrac{\pi}{4})={\sqrt2\over 2}$$but so does $\sin(\dfrac{3\pi}{4})$ , $\sin(\dfrac{9\pi}{4})$ , $\sin(\dfrac{11\pi}{4})$ , ... so what is the value of $\arcsin(x)$? There to avoid ambiguity, $\arcsin$ function has bee defines such that for each argument gives a result in $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$ since $\sin$ function is bijective at this interval.$$\arcsin(x)=\{\theta\in[-\dfrac{\pi}{2},\dfrac{\pi}{2}]\quad,\quad\sin\theta=x\}$$
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What is the answer to this question: for what $x$ the equation $\sin(x)=0$?
You can clearly see that you could say $x=0$ or $x=\pi$ as well $x=2\pi,3\pi,4\pi\dots$ and so on. All this answers are clearly different from each other, so if I don't restrict myself on a certain domain the question $$\arcsin(\sin(x))=x$$ has more than one answer. For example let us choose $x={\pi\over4}$ and $y={3\pi\over4}$ and compute $$\arcsin(\sin(y))=\arcsin(\frac{\sqrt{2}}{2}) = {\pi\over4} = x$$ but clearly $x\neq y$. So in general this equalities hold only on a smaller domain: if you want to invert the function $\sin(x)$ you have to do it in $[-{\pi\over 2},{\pi\over 2}]$ and in this case: $$\arcsin(x): [-1,1]\rightarrow [-{\pi\over 2},{\pi\over 2}]$$ and as well $$\sin(x):[-{\pi\over 2},{\pi\over 2}]\rightarrow[-1,1]$$ in this cases every element in the domain gets mapped to one element of the codomain, so the answer to that equations is unique
It's all about Domain and Range.
$$ {\rm sin}(x) \;: \;\mathbb{R} \rightarrow [-1, 1]$$
$$ {\rm arcsin}(x) \;: \;[-1, 1] \rightarrow \Big[-\frac{\pi}{2}, \frac{\pi}{2}\Big]$$
$$ \Downarrow $$
$$ {\rm sin(arcsin(}(x)) \;: \;[-1, 1] \xrightarrow[id]{} [-1, 1]$$
$$ {\rm arcsin(sin(}(x)) \;: \;\mathbb{R} \rightarrow \Big[-\frac{\pi}{2}, \frac{\pi}{2}\Big]$$
The last map is not an identity, but its restriction on $[-\frac{\pi}{2}, \frac{\pi}{2}] \subset \mathbb{R}$ is:
$$ {\rm arcsin(sin(}(x))_{[-\frac{\pi}{2}, \frac{\pi}{2}]} \;: \;\Big[-\frac{\pi}{2}, \frac{\pi}{2}\Big] \xrightarrow[id]{} \Big[-\frac{\pi}{2}, \frac{\pi}{2}\Big].$$