So you have composed two functions,
h(x)=sin(x) and g(x)=arctan(x)
→f=h∘g
meaning
f(x)=h(g(x))
g:R→[−1;1]
h:R→[−π2;π2]
And since
[−1;1]∈R→f is defined ∀x∈R
And since arctan(x) is strictly increasing and continuous in [-1;1] ,
h(g(]−∞;∞[))=h([−1;1])=[arctan(−1);arctan(1)]
Meaning
f:R→[arctan(−1);arctan(1)]=[−π4;π4]
so there's your domain
