4. If $\triangle A B C$ is an obtuse triangle, then the range of $\operatorname{arccos}(\sin A)+\operatorname{arccos}(\sin B)+\operatorname{arccos}(\sin C)$ is $\qquad$
4. $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$.

Assume $\angle A$ is an obtuse angle, then $0<\angle A-\frac{\pi}{2}<\frac{\pi}{2}$.
$\therefore$ The original expression $=\arccos \left[\cos \left(A-\frac{\pi}{2}\right)\right]+\operatorname{arccos}\left[\cos \left(\frac{\pi}{2}\right.$.
$$
\begin{array}{l}
-B)]+\arccos \left[\cos \left(\frac{\pi}{2}-C\right)\right] \\
=\left(\angle A-\frac{\pi}{2}\right)+\left(\frac{\pi}{2}-\angle B\right)+\left(\frac{\pi}{2}-\angle C\right) \\
= 2 \angle A-\frac{\pi}{2} \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) .
\end{array}
$$