In this case arcsin() is the unrestricted sin inverse function. I know that it is either undefined or has the value of 4. It could be undefined because arcsin() has only a doman of -1...1 and 4 is out of the domain. On the other hand, it could be that since they are inverses the intermediary result does not matter and they will cancel to get back 4. Which one is it?
$\endgroup$3 Answers
$\begingroup$If you are working with real numbers, $\arcsin(4)$ doesn't exist.
Note that the formula
$$\sin(\arcsin(x))=x$$ only holds over the domain of $\arcsin(x)$ which is $x \in [-1,1]$.
Things are different when you work with complex numbers though....
$\endgroup$ $\begingroup$If you are willing to allow complex numbers, $\arcsin(4)$ does exist: it is $\pi/2 - i \ln(\sqrt{15}+4)$ (using the principal branch of arcsin). If not, as others have said, the expression is undefined.
$\endgroup$ $\begingroup$Complex values aside, this expression cannot be evaluated since the $arcsin$ can only be taken from valus between -1 and 1 (inclusive) so the cancelation property of the inverses cannot be applied here. Cancelation property of inverses can be used for values that are in the respective domains of the functions. Think about $\sqrt(-2)^2=-2$ where I cancelled the squareroot against the square.
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