Example 1 Given that in the right-angled $\triangle ABC$, the two legs are $AC=2, CB=3$, and $CP$ is the angle bisector of $\angle ACB$ (point $P$ is on the hypotenuse $AB$). Fold the right-angled triangle along $CP$ to form a dihedral angle $A-CP-B$. When $AB=2\sqrt{2}$, the size of the dihedral angle $A-CP-B$ is $\qquad$
Solution: As shown in Figure 2, we know
$$
\begin{array}{l}
\alpha=\beta=45^{\circ}, \\
\cos \gamma \\
=\cos \angle A C B \\
=\frac{2^{2}+3^{2}-(2 \sqrt{2})^{2}}{2 \times 3 \times 2} \\
=\frac{5}{12} .
\end{array}
$$

By the formula for the cosine of the dihedral angle, we get
$$
\cos \theta=\frac{\frac{5}{12}-\frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2}}=-\frac{1}{6} \text {. }
$$

Thus, $\theta=\arccos \left(-\frac{1}{6}\right)$.