$\begin{array}{l} \text { Example } 4 . \text { Simplify: } \cos ^{2}(A-\alpha)+\cos ^{2}(B-\alpha) \\ -2 \cos (A-B) \cos (A-\alpha) \cos (B-\alpha) .\end{array}$
Solve the original expression $=\cos ^{2}(\mathbf{A}-\boldsymbol{\alpha})+1-\sin ^{2}(B-\boldsymbol{\alpha})$
$$
\begin{array}{l}
-2 \cos (\mathbf{A}-\mathbf{B}) \cos (\mathbf{A}-\boldsymbol{\alpha}) \cos (\mathbf{B}-\boldsymbol{\alpha}) \\
= \cos (\mathbf{A}+\mathrm{B}-2 \boldsymbol{\alpha}) \cos (\mathbf{A}-\mathbf{B}) \\
-2 \cos (\mathbf{A}-\mathbf{B}) \cos (\mathbf{A}-\boldsymbol{\alpha}) \\
\cos (\mathbf{B}-\boldsymbol{\alpha})+1 \\
= \cos (\mathbf{A}-\mathbf{B})\{\cos (\mathbf{A}+\mathbf{B}-2 \boldsymbol{a}) \\
-2 \cos (\mathbf{A}-\boldsymbol{\alpha}) \cos (\mathbf{B}-\boldsymbol{\alpha})\}+1 \\
= \cos (\mathbf{A}-\mathbf{B})\{\cos (\mathbf{A}+\mathbf{B}-2 \boldsymbol{\alpha}) \\
-[\cos (\mathbf{A}+\mathrm{B}-2 \boldsymbol{\alpha})+\cos (\mathbf{A}-\mathrm{B})]\} \\
+1 \\
=-\cos ^{2}(\mathbf{A}-\mathrm{B})+\mathbf{1}=\sin ^{2}(\mathbf{A}-\mathbf{B}) .
\end{array}
$$

Note: (1) In Example 3, the middle two terms are transformed into products using formula (1).
(2) Use the formula $\sin ^{2} x+\cos ^{2} x=1$ to first transform Example 4 into a form suitable for using formula (2). The simplified result of Example 4 explains the original expression.