已知$\dfrac{\cos^3\alpha}{\cos\beta}+\dfrac{\sin^3\alpha}{\sin\beta}=1$，求证：$\left(\dfrac{\cos\beta}{\cos\alpha}-\dfrac{\sin\beta}{\sin\alpha}\right)\left(\dfrac{\cos\beta}{\cos\alpha}+\dfrac{\sin\beta}{\sin\alpha}+1\right)=0$．
分析与解 根据已知，有\[\begin{aligned}\dfrac{\cos^3\alpha}{\cos\beta}-\cos^2\alpha+\dfrac{\sin^3\alpha}{\sin\beta}-\sin^2\alpha=0,\\\dfrac{\cos^3\alpha}{\cos\beta}-\cos^2\beta+\dfrac{\sin^3\alpha}{\sin\beta}-\sin^2\beta=0,\end{aligned}\]即\[\begin{aligned}\dfrac{\cos^3\alpha-\cos^2\alpha\cos\beta}{\cos\beta}&=\dfrac{\sin^2\alpha\sin\beta-\sin^3\alpha}{\sin\beta},\\\dfrac{\cos^3\alpha-\cos^3\beta}{\cos\beta}&=\dfrac{\sin^3\beta-\sin^3\alpha}{\sin\beta}.\end{aligned}\]
不难证明$\sin\alpha,\cos\alpha\ne 0$，接下来分类讨论．
情形一 $\cos\alpha-\cos\beta=0$或$\sin\beta-\sin\alpha=0$．
此时容易推得$\cos\alpha=\cos\beta$且$\sin\alpha=\sin\beta$，因此有\[\dfrac{\cos\beta}{\cos\alpha}-\dfrac{\sin\beta}{\sin\alpha}=0,\]原命题成立．
情形二 $\cos\alpha-\cos\beta\ne 0$且$\sin\beta-\sin\alpha\ne 0$．
两式相比，有\[\dfrac{\cos^2\alpha+\cos\alpha\cos\beta+\cos^2\beta}{\cos^2\alpha}=\dfrac{\sin^2\alpha+\sin\alpha\sin\beta+\sin^2\beta}{\sin^2\alpha},\]于是有\[\dfrac{\cos\beta}{\cos\alpha}+\dfrac {\cos^2\beta}{\cos^2\alpha}=\dfrac {\sin\beta}{\sin\alpha}+\dfrac {\sin^2\beta}{\sin^2\alpha},\]移项整理即得．