I know that if I have a function $\sin(arcsin x)$, it's going to be defined on closed interval from $-1$ to $1$. But this is not working when I apply the same method for finding domain of a function for $\arcsin(\sin x)$ and it goes like this: First I look at the function $\sin x$ - I see that it is defined for every real number. Then I look at $\arcsin$, for which I know it is defined from on closed interval from $-1$ to $1$. And when I see the intersection of their domains it gives me closed interval from $-1$ to $1$. But the right answer is domain of function $\arcsin(\sin x)$ is defined for every real number. I truly don't know if something is wrong with my technique or I don't understand trigonometric functions at all. Thank you for your help! Appreciate it :)
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$\begingroup$You have to look at domain of $\sin$ as well as the the range of $\sin$, since the outputs of $\sin$ become the inputs for $\arcsin$.
As you said, $\sin(x)$ is defined for every real number, so this does not impose any restriction on the domain of $\arcsin(\sin(x))$.
The range of $\sin(x)$ is $[-1,1]$, so we need to determine whether those values are all valid inputs to $\arcsin$. Indeed, $\arcsin$ is defined for every input in $[-1,1]$, so we conclude that $\arcsin(\sin(x))$ is defined for every $x$.
$\endgroup$ $\begingroup$When you look at the domain of compound functions, you should not look at the intersection of the domains of the two, you should look at what values you can put into the first that give a result out the other end. As you say, the domain of $\sin x$ is all real numbers. The range of $\sin x$ is $[-1,1]$, which is the value you will try to take the $\arcsin$ of. The $\arcsin$ is happy doing that, so the domain of $\arcsin(\sin(x))$ is all the real line. The compound function will not be the identity over the whole line, but it gives a sensible answer with any real input.
$\endgroup$ 1 $\begingroup$Note that for every $x\in{\bf{R}}$, $\sin x$ is still in between (including) $-1$ and $1$, so we still can take $\arcsin$, so no problem.
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