Example 20 (9th "Hope Cup" Competition Training Question) In the tetrahedron $P-ABC$, $\angle APB=\angle BPC=\angle CPA=90^{\circ}$, $D$ is a point within the base $ABC$, $\angle APD=45^{\circ}$, $\angle BPD=60^{\circ}$. Find the cosine value of $\angle CPD$.
Construct a rectangular cuboid as shown in Figure 4-1. Line $P D_{1}$ is the diagonal of the cuboid, so we have $\cos ^{2} \angle A P D+\cos ^{2} \angle B P D+\cos ^{2} \angle C P D=\frac{a^{2}+b^{2}+c^{2}}{a^{2}+b^{2}+c^{2}}=1$, thus $\cos ^{2} \angle C P D=\frac{1}{4}$, which means the cosine value of $\angle C P D$ is $\frac{1}{2}$.