4. As shown in Figure 2, the axial section of the cone is an isosceles right $\triangle S A B, O$ is the center of the base circle, $C$ is the trisection point of the arc $\overparen{A B}$ on the circumference of the base circle, $\overparen{A C}=$ $2 \overparen{C B}$. Then the angle between $S A$ and $O C$ is ( ).
(A) $45^{\circ}$
(B) $60^{\circ}$
(C) $\arccos \frac{\sqrt{2}}{4}$
(D) $\arccos \frac{\sqrt{3}}{4}$
4. (C).

As shown in Figure 7, take another trisection point $D$ on the arc $\overparen{A B}$, such that $\overparen{A D}=\overparen{D C}=\overparen{C B}$. By the inscribed angle theorem, we know
$$
\angle B O C=\angle B A D \text {, }
$$

thus, $O C \parallel A D$.
Therefore, the angle between $O C$ and $S A$ is the same as the angle between $A D$ and $S A$.
Since $\overparen{A D}$ is $\frac{1}{6}$ of the circumference,

we have $A D=O A=\frac{\sqrt{2}}{2} S A$.
In the isosceles $\triangle S A D$, we have
$$
\begin{aligned}
\cos \angle S A D & =\frac{\frac{1}{2} A D}{S A} \\
& =\frac{\sqrt{2}}{4} .
\end{aligned}
$$

Thus, $\angle S A D=\arccos \frac{\sqrt{2}}{4}$.
Therefore, the angle between the skew lines $S A$ and $O C$ is $\arccos \frac{\sqrt{2}}{4}$.