Question

Why does arcsin have restrictions at (domain) [-1, 1] and (range) [(-pi/2),(pi/2)]?

Answer 1

Because the sine function is not bijective, specifically beacuse it's not injective.

The sine function, in fact, spans the values between -1 and 1 infinite times, as it waves between them infinite times.

The arcsin should be the inverse function of the sine function, so it should work like that: pick any number, and it returns an angle whose sine is the value you picked.

Two problems arises: first of all, you can't really pick any number: if you choose a number greater than 1 or smaller than -1, you can't possibly find an angle whose sine is the value you picked. For example, you can't ask "the sine of which angle equals 2?" This is why the domain is [-1,1]: you couldn't answer the question the arcsin function is posing otherwise.

Now, assume your question actually makes sense. For example, you want to define arcsin(0), i.e., you want to answer the question "the sine of which angle equals 0?" Now the answer is well-posed, but it has too many answers!. For example, you may pick the angles
`0, \pi, 2\pi, 3\pi, 42356\pi,\ldots`. But since the arcsine function is...a function (!), you need to associate only one output for every input. So, we got rid of the "too many" answer by deliberately choosing a portion of all possible angles, i.e.
`[-(\pi)/(2),(\pi)/(2)]`

Recap of all I said: the choice for the domain was made because otherwise you couldn't find an answer to the question, the choice for the range was made because otherwise there would have been too many answers to your question. Instead, you want an invertible function to be a bijection, i.e. a one-to-one relationship, so every input must have one and one ouput only.