3. (i) (Grade 11) Let $f(x)=\frac{1+x}{1-x}$, and denote $f_{1}(x)=f(x)$. If $f_{n+1}(x)=f\left(f_{n}(x)\right)$, then $f_{2006}(x)=$ ( ).
(A) $x$
(B) $-\frac{1}{x}$
(C) $\frac{1+x}{1-x}$
(D) $\frac{x-1}{x+1}$
(ii) (Grade 12) The base $ABCD$ of the pyramid $P-ABCD$ is a unit square $(A, B, C, D$ are arranged counterclockwise), and the side edge $PB$ is perpendicular to the base, with $PB=\sqrt{3}$. Let $\angle APD=\theta$. Then $\sin \theta=(\quad)$.
(A) $\frac{\sqrt{2}}{2}$
(B) $\frac{\sqrt{3}}{3}$
(C) $\frac{\sqrt{5}}{5}$
(D) $\frac{\sqrt{6}}{6}$
3. (i) B.

It is easy to find that $f_{4 n+1}(x)=\frac{1+x}{1-x}, f_{4 n+2}(x)=-\frac{1}{x}$,
$$
f_{4 n+3}(x)=\frac{x-1}{x+1}, f_{4 n+4}(x)=x \text {. }
$$

Therefore, $f_{2000}(x)=-\frac{1}{x}$.
(ii) C.
$$
\begin{array}{l}
P A=\sqrt{1^{2}+(\sqrt{3})^{2}}=2, \\
P D=\sqrt{B D^{2}+P B^{2}}=\sqrt{2+3}=\sqrt{5} .
\end{array}
$$

Since $P A \perp A D$, then $\sin \theta=\frac{A D}{P D}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$.