15. Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $1, E$ and $F$ are moving points on edges $A B$ and $B C$ respectively, and $A E=B F$. Find the minimum value of the angle formed by lines $A_{1} E$ and $C_{1} F$ (express the result using inverse trigonometric functions).
15. Solution 1: As shown in Figure 1, extend $DC$ to point $G$ such that $CG = AE$, and connect $C_1G$ and $FG$. From the problem, we know that $A_1E \parallel C_1G$, and the angle formed by $A_1E$ and $C_1F$ is equal to $\angle FC_1G$.
Let $AE = CG = x (0 \leqslant x \leqslant 1)$, then we have
$$
\begin{array}{l}
CF = 1 - x, \quad C_1G = \sqrt{1 + x^2}, \\
C_1F = \sqrt{1 + (1 - x)^2}, \quad FG = \sqrt{x^2 + (1 - x)^2}.
\end{array}
$$

In $\triangle C_1FG$, by the cosine rule, we get
$$
\begin{array}{l}
\cos \angle FC_1G = \frac{(C_1F)^2 + (C_1G)^2 - (FG)^2}{2(C_1F)(C_1G)} \\
= \frac{2}{2 \sqrt{1 + x^2} \cdot \sqrt{1 + (1 - x)^2}} \\
= \frac{1}{\sqrt{1 + x^2} \cdot \sqrt{1 + (1 - x)^2}}.
\end{array}
$$
The minimum value of $\angle FC_1G$ is when $\cos \angle FC_1G$ is maximized, which is when $S(x) = (1 + x^2)[1 + (1 - x)^2]$ is minimized.
Using the identity
$$
\begin{array}{l}
(1^2 + a^2)(1^2 + b^2) = (1 - ab)^2 + (a + b)^2, \\
\text{we get } S(x) = [1 - x(1 - x)]^2 + [x + (1 - x)]^2 \\
= (x^2 - x + 1)^2 + 1 = \left[\left(x - \frac{1}{2}\right)^2 + \frac{3}{4}\right]^2 + 1.
\end{array}
$$

Therefore, when $x = \frac{1}{2}$, $S(x)$ reaches its minimum value $\frac{25}{16}$.
Thus, $\cos \angle FC_1G = \frac{1}{\sqrt{S(x)}} = \frac{4}{5}$, which means
$$
\angle FC_1G = \arccos \frac{4}{5}.
$$

Hence, the minimum angle formed by $A_1E$ and $C_1F$ is $\arccos \frac{4}{5}$, at which point $E$ and $F$ are the midpoints of edges $AB$ and $BC$, respectively.