872

在凸四边形 $A B C D$ 中，分别用 $A, B, C, D$ 表示四个内角，如果有 $\sin A = \cos B$ ，求 $S = 3 \sin A + 4 \sin B + 5 \sqrt{2} \sin C + 5 \sqrt{2} \sin D$ 的取值范围.

解钱莘芪

由 $\sin A = \cos B = \sin (\frac{π}{2} - B)$ ，得 $A = \frac{π}{2} - B$ 或 $A = B + \frac{π}{2}$ .

对 $A = \frac{π}{2} - B$ 的情形，有 $C + D = \frac{3 π}{2}$ .则 $\begin{array}{r} S = 3 \cos B + 4 \sin B + 5 \sqrt{2} (\sin C - \cos C) = 5 \sin (B + φ) + 10 \sin (C - \frac{π}{4}), φ = \arctan \frac{3}{4} . \end{array}$
由于 $0 < B < \frac{π}{2} < C < \frac{3 π}{2}$ ,故 $5 \sin (B + φ) \in (3, 5], 10 \sin (C - \frac{π}{4}) \in (5 \sqrt{2}, 10]$ ，故 $S \in (3 + 5 \sqrt{2}, 15]$ .
对 $A = B + \frac{π}{2}$ 的情形，有 $C + D = \frac{3 π}{2} - 2 B$ . $A < π \Rightarrow B < \frac{π}{2}$ .一方面， $\begin{aligned} S & = 3 \cos B + 4 \sin B + 5 \sqrt{2} (1 + \sin 2 B) \sin C - 5 \sqrt{2} \cos 2 B \cos C \\ \leq 3 \cos B + 4 \sin B + 5 \sqrt{2} \sqrt{2 + 2 \sin 2 B} \\ = \frac{\sqrt{2}}{2} [7 \cos (B - \frac{π}{4}) + 20 | \cos (B - \frac{π}{4}) | + \sin (B - \frac{π}{4})] \leq \frac{\sqrt{2}}{2} \cdot \sqrt{27^{2} + 1} = \sqrt{365}, \end{aligned}$
另一方面 $\sin C + \sin D = 2 \sin \frac{C + D}{2} \cos \frac{C - D}{2} = 2 \sin (\frac{3 π}{4} - B) \cos \frac{C - D}{2}$ .不妨设 $C \geq D$ ，则 $\frac{C - D}{2} \in [0, \frac{3 π}{4} - B) \subset [0, π]$ .而 $\cos x$ 在 $[0, π]$ 上单减，且 $\sin (\frac{3 π}{4} - B) > 0$ ,故 $\begin{array}{r} \sin C + \sin D > 2 \sin (\frac{3 π}{4} - B) \cos (\frac{3 π}{4} - B) = - \cos 2 B . \end{array}$
则 $\begin{array}{r} S > 3 \cos B + 4 \sin B - 5 \sqrt{2} \cos 2 B = g (B), \end{array}$
而 $g^{'} (B) = 10 \sqrt{2} \sin 2 B + 3 (\cos B - \sin B) + \cos B > 0$ ，故 $g (B) > g (\frac{π}{4}) = \frac{7 \sqrt{2}}{2}$ .
再由值域的连续性得 $S \in (\frac{7 \sqrt{2}}{2}, \sqrt{365}]$ .

将两个范围取并集，得 $S \in (\frac{7 \sqrt{2}}{2}, \sqrt{365}]$ .