Let f () sin (arctan) x x =. What is the range of f?

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