The question is given in the chap. 1.4 of the book titled: Calculus Problems for the new century, by Robert Fraga.
What is the domain of $\operatorname{arcsec}(\sin x)$?
I feel confused as $\operatorname{arcsec}( x ) = \cos( x )$, as $\sec( x) =\arccos( x )$. I hope that am not be making error in terms of domain values here, as domain of $\cos( x) $ is entire real number line; while that of $\sec( x )$ is in the interval $\{-1,\cdots 1\}$.
So, the question reduces to finding domain of $\cos(\sin x)$.
The domain of $\cos(x)$ is the range of $\sin(x)$, so domain of input to $\cos(x)$, is $-1\le x\le 1$.
But, the book states answer as below, in an unexpected way:
The $\operatorname{arcsec}(x)$ is defined only on $\{x|\,|x|\ge 1\}$. However, $|\sin\, x\,|\ge 1$ only if $x$ is an odd multiple of $\frac{\pi}2$, so the domain consists of odd multiples of $\frac{\pi}2$.
Best Answer
$$\operatorname{arcsec}{(x)}\ \text{or}\ \sec^{-1}{(x)} \ne \cos{(x)}$$
That statement is wrong. When we write something like $\sec^{-1}{x}$, it does not mean $\frac{1}{\sec{x}}$. It means "inverse secant function". It's just a notation thing. I know it's confusion, but that's just how things are in modern mathematics.
As for the question itself, the range of the function $\sin{x}$ is $[-1,1]$. The inverse secant function, $\operatorname{arcsec}(x)$ (many times written as $\sec^{-1}{x}$), is only defined on $(-\infty,-1]\cup[1,+\infty)$. The intersection of those two sets consists of the following two elements: $\{-1,1\}$. So, the domain of $f(x)=\sec^{-1}{(\sin{x})}$ must be all values of $x$ where $\sin{x}$ equals either $-1$ or $1$:
$$\bigg\{\frac{\pi}{2}+k\pi,\ k\in\mathbb{Z}\bigg\}.$$