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$\arcsin(x)$ is the inverse function of $\sin(x)$, this means that: $$\arcsin(\sin(x))=x$$ If I want to solve the equation: $$\sin(x)=\sin(y)$$ using this link, $x=\pi - y$ for instance. but, also we can say that: $$x=\arcsin(\sin x)=\arcsin(\sin y)=y$$ which is for me a contradiction. Where is the missing point in my writings?

Remember
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    The identity $\arcsin(\sin x)=x$ holds only when $x\in[-\pi/2,\pi/2]$. If $x$ is in that range, then $y=\pi-x$ will not be. Neither will $z=x+2\pi$ even though also $\sin z=\sin x$. In other words $\sin x$ has no global inverse (that is one that would hold for all $x$). Periodicity alone would make that impossible. – Jyrki Lahtonen Jul 26 '22 at 08:36
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    $\arcsin$ is only the inverse of $\sin$ in a suitably restricted domain. Read the section "Principal values" in the following Wikipedia article. – projectilemotion Jul 26 '22 at 08:38
  • And, if $x=3,$ you can't just write that $x=\sin(\arcsin x),$ either. – ryang Jul 26 '22 at 08:50
  • $\sin x$ is not one-to-one fuction, is many-to-one, then there are no inverse function for whole domain of $\sin x$. That's why inverse function covers only part of domain of $\sin x$, exactly $[-\pi/2;\pi/2]$. That's why $x=\arcsin \sin(x)$ is correct only for this part of domain of $\sin(x)$. – Ivan Kaznacheyeu Jul 26 '22 at 09:59

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WHat's wrong is your very first sentence. The function $\sin:\Bbb{R}\to\Bbb{R}$ DOES NOT have an inverse. It is a periodic function, so not injective (one-to-one) so definitely cannot be inverted.

On the other hand, if you restrict the domain and target, you get $\sin|_{[-\pi/2,\pi/2]}:[-\pi/2,\pi/2]\to [-1,1]$. This restricted function (which is different from the usual $\sin$, since we changed the domain and target) is invertible, and it's inverse is called $\arcsin:[-1,1]\to [-\pi/2,\pi/2]$. So, \begin{align} \arcsin:= \left(\sin|_{[-\pi/2,\pi/2]}\right)^{-1}. \end{align}

Some remarks: we can restrict the $\sin$ function to other intervals and still get invertible functions. FOr example, $\sin|_{[5\pi/2,7\pi/2]}:[5\pi/2,7\pi/2]\to[-1,1]$ is also invertible, but its inverse function is NOT $\arcsin$, it is $\arcsin +2\pi$. There are actually infinitely many other intervals we can restrict to and get an invertible function, and it turns out the inverses differ by multiples of $\pi$

The name $\arcsin$, rather than $\sin^{-1}$ is thus appropriate. The function $\sin$ is not invertible, and $\arcsin$ refers to the inverse of a very specific restriction of $\sin$. Also, it is purely convention that we define $\arcsin$ by restricting $\sin$ to $[-\pi/2,\pi/2]$, the so-called principal branch.

peek-a-boo
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