$23 \cdot 35$ The base $ABCD$ of the pyramid $P-ABCD$ is a square, and the vertex $P$ is equidistant from $A, B, C, D$. If $AB=1, \angle APB=2\theta$, then the volume of the pyramid is
(A) $\frac{\sin \theta}{6}$.
(B) $\frac{\cot \theta}{6}$.
(C) $\frac{1}{6 \sin \theta}$.
(D) $\frac{1-\sin 2\theta}{6}$.
(E) $\frac{\sqrt{\cos 2\theta}}{6 \sin \theta}$.
(41st American High School Mathematics Examination, 1990)
[Solution] Draw the slant height $PH$ of the side $\triangle PAB$, and let $O$ be the center of the square base. Connect $HO$.
Given $\angle APB = 2\theta$, then $PH = \frac{1}{2} \operatorname{ctg} \theta$.
Also, $PO \perp OH$, in the right triangle $\triangle PHO$,
$$
\begin{array}{l} 
PO = \sqrt{PH^2 - HO^2} \\
= \sqrt{\left(\frac{1}{2} \operatorname{ctg} \theta\right)^2 - \left(\frac{1}{2}\right)^2} \\
= \frac{1}{2} \sqrt{\operatorname{ctg}^2 \theta - 1} \\
= \frac{1}{2} \sqrt{\frac{\cos^2 \theta - \sin^2 \theta}{\sin^2 \theta}} = \frac{1}{2} \frac{\sqrt{\cos 2\theta}}{\sin \theta}.
\end{array}
$$

Thus, $V_{P-ABCD} = \frac{1}{3} \cdot S_{ABCD} \cdot PO = \frac{1}{3} \cdot 1 \cdot \frac{1}{2} \cdot \frac{\sqrt{\cos 2\theta}}{\sin \theta} = \frac{\sqrt{\cos 2\theta}}{6 \sin \theta}$. Therefore, the answer is $(E)$.