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