题目

( 2025年12月T8高三年级检测训练第$14$题 )

已知 $\alpha, \beta \in \mathbb{R}$,$(\sin \alpha - |\sin \beta|) \cdot (\cos \beta - |\cos \alpha|) = 0$,则 $\sin \alpha + \cos \beta - 2$ 的最小值为_____.

解析

由$(\sin \alpha - |\sin \beta|) \cdot (\cos \beta - |\cos \alpha|) = 0$得
$$\sin \alpha = |\sin \beta| \quad \text{或} \quad \cos \beta = |\cos \alpha|.$$

若 $\sin \alpha = |\sin \beta|$，则
$$\begin{aligned}\sin \alpha + \cos \beta - 2 = &|\sin \beta|+ \cos \beta - 2\\\\\geqslant &0 + (-1) - 2\\\\
= &-3.\end{aligned}$$当 $|\sin \beta| = 0$ 且 $\cos \beta = -1$即$\beta =2k\pi+ \pi(k\in\mathbb{Z})$时取等号.
若 $\cos \beta = |\cos \alpha|$，则
$$\begin{aligned}\sin \alpha + \cos \beta - 2 = &\sin \alpha + |\cos \alpha| - 2\\\ \geqslant & -1 + 0 - 2\\\\
= &-3.\end{aligned}$$
当 $\sin \alpha = -1$ 且 $|\cos \alpha| = 0$即$\alpha =2k\pi -\frac{\pi}{2}(k\in\mathbb{Z})$时取等号.

综上，$\sin \alpha + \cos \beta - 2$ 的最小值为 $\boxed{-3}$.