Mathematical syntax when defining domain and range What is the correct syntax when writing about the domain and range of a function?

For example, let's say: 

$f(y)=\cos{y}$ and $y(x)=\arcsin{x}$ 

In order to simplify the function $f(y(x))=\cos{\arcsin{x}}$ I want to write down the domains and ranges of all functions involved in a thorough way. 

So I would start with the domain of $\arcsin{x}$ which is $[-1,1]$ then the range, which is $[-\frac{\pi}{2},\frac{\pi}{2}]$, which is also the domain of $\cos{y}$, which implies that the range of $\cos{y}$ is always positive.

How to write the above using the right mathematical syntax? I mean, using the right symbols?

Further question: what does the "element of" $\epsilon$ symbol mean and how to use it? Is that symbol useful in this case?

Furthermore I also sometimes see symbols like: $\mathfrak{D}$ and $\mathfrak{R}$. How are they correctly used?
There are several mistakes which are probably just the result of sloppy writing, but can also lead to some wrong results:


The function $f(x)$ is not equal to $\cos \arcsin x$. If $f(y)=\cos y$, then $f(x)=\cos x$. I believe what you want to write down is $f(y(x))$.
Second of all, you write that the range of $\arcsin$ is $\left[-\frac\pi2, \frac\pi2\right]$ which is true, but then you say "which is also the domain of $\cos y$", and this is not true. The domain of $\cos y$ is $\mathbb R$.


Other than that, it's hard to give you a full answer because I don't understand your question. Do you want to write down the domain of the function $\cos\arcsin x$? Then that domain is equal to the domain of $\arcsin x$, which is $[-1,1]$. Or do you also want to calculate the range?
You have two functions.

First function:

$f(x)=\cos x$

Domain: $x \in \mathbb R$

Range: $-1 \le f(x) \le 1$

Second function:

$g(x)=\arcsin x$

Domain: $-1 \le x \le 1$

Range: $-\frac {\pi}2 \le g(x) \le \frac {\pi}2$

Now consider $h(x)=f(g(x))$

The domain of $g(x)$ is $-1 \le x \le 1$.

The range of $g(x))$ is $-\frac {\pi}2 \le g(x) \le \frac {\pi}2$, which is all suitable as a domain for $f(x)$.

However, since the domain of $f(x)$ has now been restricted to $-\frac {\pi}2 \le g(x) \le \frac {\pi}2$, the range of $f(x)$ is restricted to $0 \le h(x) \le 1$

In conclusion:

Domain of $h(x)$ is $-1 \le x \le 1$.

Range of $h(x)$ is $0 \le h(x) \le 1$