As the title says, I don't understand how to find the domain of $\arcsin(\sqrt{1-x^2})$. I kinda understand how it would equate to it would be -1 < x < 1 (inclusive of 1 and -1) by definition of a real root, but how does the inverse sine function affect this domain? I know that for any function $\arcsin(f(x))$, the domain is $-1<f(x)<1$, but how exactly does the root affect this domain also?
Thanks for the help.
On
The domain of the function $f(x)=\arcsin(x)$, by definition, is equal to $-1\le x\le 1$. Now, since the argument in your function is $\sqrt{1-x^2}$ then we have to solve the following inequality: $$-1\le \sqrt{1-x^2} \le 1.$$
Since the function $f(x)=\sqrt{x}$ is greater than $0$ for all $x$, then we note that $$0 \le \sqrt{1-x^2} \le 1 $$ $$\therefore 0\le1-x^2\le1$$ $$\therefore-1\le-x^2\le0$$ $$\therefore 0\le x^2 \le1$$ $$\therefore x\in[-1,1]$$.
BEcause arcsin operates on values between $-1$ and $1$, you need to have $$ -1 \le \sqrt{1 - x^2} \le 1 $$ That's equivalent (by squaring, since the square root is always nonnegative) to $$ {1 - x^2} \le 1 $$ But with the constraint that $1 - x^2 \ge 0$, so that the square root is still defined. So we have $$ 0 \le {1 - x^2} \le 1 $$
Take $1 - \text{each term}$ and swap the comparisons to get $$1 \ge x^2 \ge 0 $$ whose solution is $-1 \le x \le 1$.