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
