Domain and range of a cyclometric function My textbooks tells me: 
$$f(x) = \arctan(x)$$
$D_f = \mathbb{R}$ and $B_f = \left\langle -\dfrac{1}{2} \pi, \dfrac{1}{2} \pi \right\rangle$
$$g(x) = \arcsin(x)$$
$D_g = [-1, 1]$ and $ B_g = [ -\dfrac{1}{2} \pi, \dfrac{1}{2} \pi]$ 
For $y=\tan(x)$ the domain is $ < -\dfrac{1}{2} \pi, \dfrac{1}{2} \pi >$, for $y=\sin(x) = [ -\dfrac{1}{2} \pi, \dfrac{1}{2} \pi ]$ and for $\cos(x)$ it's $[0, \pi]$.
I have the following problem. You can for example easily calculate $\sin(8\pi)$, eventhough it is outside of its domain technically. My idea is just that with domain in the case of the trig functions they just mean the 1 period that gets repeated to infinity.
However, if you try to do that for a cyclometric function, for example $\arcsin(8 \pi)$ it'll give you an error. Why is this? Why does this not work for cyclometric functions?
 A: In principle $\text{dom}(\sin)=\Bbb R$. But since $\sin$ is not injective in $\Bbb R$, it doesn't make sense to talk about its inverse.
However there are sets (and intervals) in which the restriction of $\sin$ to those sets is injective, in particular some intervals of length $\pi$, namely any interval equal to $[a,a+\pi]$, for all $a\in \Bbb R$ such that there exists $k\in \Bbb Z$ such that $\displaystyle a=k\pi+\frac{\pi}{2}$.
Now considering the restriction of $\sin$ to given interval, $\sin \lvert_{[a,a+\pi]}$, it is injective and therefore it has an inverse, (if you consider its codomain to be $\text{im}(\sin \lvert_{[a,a+\pi]})\color{grey}{=[-1,1\textbf{]}}$). The domain of its inverse will be $[-1,1\textbf{]}$ and the codomain of the inverse will be $[a,a+\pi\textbf{]}$. Under these restrictions we choose to name the inverse $(\sin \lvert_{[a,a+\pi]}) ^{-1}$ as $\arcsin _a$. Usually people will refrain from indicating the restriction and just write $\sin$ when they actually mean its restriction somewhere and also the subscript $a$ in $\arcsin _a$ is dropped.
In practice it's common to consider $a=-\pi/2$ or $a=0$. And of course here it doesn't make sense to talk about $\sin \lvert_{[-\pi-/2,\pi /2]}(8\pi)$. But $\sin (8\pi)$ does make sense.
