25~26高三上·T8联考·第14题

$-3$

令$x = \sin\alpha,y = \cos\beta$，
则$\left(x – \sqrt{1 – y^2}\right)\cdot\left(y – \sqrt{1 – x^2}\right) = 0$，
$\therefore x = \sqrt{1 – y^2}$，或$y = \sqrt{1 – x^2}$，
$\therefore x^2 + y^2 = 1(x \geq 0)$，或$x^2 + y^2 = 1(y \geq 0)$．

$\therefore$点$(\sin\alpha,\cos\beta)$在圆$x^2 + y^2 = 1$位于第一、二、四象限（包括坐标轴）的部分上．

$\because$点$(\sin\alpha,\cos\beta)$到直线$x + y – 2 = 0$距离为$\frac{|\sin\alpha + \cos\beta – 2|}{\sqrt{2}} = d$，
又$\sin\alpha + \cos\beta – 2 \leq 0$，$\therefore \sin\alpha + \cos\beta – 2 = -\sqrt{2}d$．

下求$d$的最大值．如图，$d$的最大值为点$(-1,0)$到直线$x + y – 2 = 0$的距离，$\therefore d_{\text{max}} = \frac{|-1 + 0 – 2|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$，
$\therefore (\sin\alpha + \cos\beta – 2)_{\text{min}} = -\sqrt{2} \times \frac{3}{\sqrt{2}} = -3$．