《2026年南方凤凰台5A新考案高中数学二轮基础版》

变式 2
(1) (2025·漳州二模)已知 $\alpha \in \left(0, \dfrac{\pi}{2}\right)$，若 $\sin\left(\dfrac{3\pi}{2} + 2\alpha\right) + \cos\left(\alpha - \dfrac{\pi}{4}\right) = 0$，则 $\alpha =$(A)
A.$\dfrac{\pi}{12}$
B.$\dfrac{\pi}{6}$
C.$\dfrac{5\pi}{12}$
D.$\dfrac{\pi}{3}$

答案： 变式2
(1)A 【解析】因为$\sin \left( \frac{3 \pi}{2} + 2 \alpha \right) = - \cos 2 \alpha$，所以$- \cos 2 \alpha + \cos \left( \alpha - \frac{\pi}{4} \right) = 0$，即$\cos 2 \alpha = \cos \left( \alpha - \frac{\pi}{4} \right)$，即$\cos^{2} \alpha - \sin^{2} \alpha = \frac{\sqrt{2}}{2} (\cos \alpha + \sin \alpha)$。又因为$\alpha \in \left( 0, \frac{\pi}{2} \right)$，所以$\cos \alpha + \sin \alpha ＞ 0$，所以$\cos \alpha - \sin \alpha = \frac{\sqrt{2}}{2}$，即$\cos \left( \alpha + \frac{\pi}{4} \right) = \frac{1}{2}$。又$\alpha \in \left( 0, \frac{\pi}{2} \right)$，所以$\alpha + \frac{\pi}{4} \in \left( \frac{\pi}{4}, \frac{3 \pi}{4} \right)$，所以$\alpha + \frac{\pi}{4} = \frac{\pi}{3}$，所以$\alpha = \frac{\pi}{12}$。

(2) (2025·珠海模拟)设 $\alpha \in \left(0, \dfrac{\pi}{2}\right), \beta \in \left(0, \dfrac{\pi}{2}\right)$，且 $\tan \alpha + \tan \beta = \dfrac{1}{\cos \beta}$，则(D)

A.$3\alpha - \beta = \dfrac{\pi}{2}$
B.$2\alpha - \beta = \dfrac{\pi}{2}$
C.$3\alpha + \beta = \dfrac{\pi}{2}$
D.$2\alpha + \beta = \dfrac{\pi}{2}$

答案：
(2)D 【解析】由题意知$\frac{\sin \alpha}{\cos \alpha} + \frac{\sin \beta}{\cos \beta} = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta} = \frac{\sin (\alpha + \beta)}{\cos \alpha \cos \beta} = \frac{1}{\cos \beta}$，所以$\sin (\alpha + \beta) = \cos \alpha$。因为$\alpha \in \left( 0, \frac{\pi}{2} \right)$，$\beta \in \left( 0, \frac{\pi}{2} \right)$，所以$\alpha + \beta \in (0, \pi)$。又$\sin (\alpha + \beta) = \cos \alpha = \sin \left( \frac{\pi}{2} \pm \alpha \right)$，所以$\alpha + \beta = \frac{\pi}{2} \pm \alpha$或$\alpha + \beta + \frac{\pi}{2} \pm \alpha = \pi$，即$2 \alpha + \beta = \frac{\pi}{2}$或$\beta = \frac{\pi}{2}$（舍去），故$2 \alpha + \beta = \frac{\pi}{2}$。

例 3 (1) (2025·沈阳三模)函数 $f(x) = \cos 2x + 6\cos\left(\dfrac{\pi}{2} - x\right)$ 的最小值为$- 7$。

答案： 例3
(1)$ - 7$ 【解析】$f (x) = \cos 2x + 6 \cos \left( \frac{\pi}{2} - x \right) = 1 - 2 \sin^{2} x + 6 \sin x$，令$t = \sin x \in [-1, 1]$，则$y = - 2t^{2} + 6t + 1$，且该二次函数的对称轴为直线$t = \frac{3}{2}$，故函数$y = - 2t^{2} + 6t + 1$在$[-1, 1]$上单调递增，所以$y_{\min} = - 2 × (-1)^{2} - 6 + 1 = - 7$，即函数$f (x)$的最小值为$- 7$。

(2) 若锐角 $\alpha, \beta$ 满足 $\sin 2\alpha \sin 2\beta = 3(1 + \cos 2\alpha)(1 - \cos 2\beta)$，则 $\tan(\alpha - \beta)$ 的最大值是(D)
A.$-\dfrac{\sqrt{3}}{3}$
B.$\dfrac{3}{4}$
C.$\dfrac{3}{5}$
D.$\dfrac{\sqrt{3}}{3}$

答案：
(2)D 【解析】因为$\sin 2 \alpha \sin 2 \beta = 3 (1 + \cos 2 \alpha) (1 - \cos 2 \beta)$，所以$4 \sin \alpha \cos \alpha \sin \beta \cos \beta = 12 \cos^{2} \alpha \sin^{2} \beta$。因为$\alpha, \beta$是锐角，所以$\sin \alpha \cos \beta = 3 \cos \alpha \sin \beta$，即$\tan \alpha = 3 \tan \beta$，所以$\tan (\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = \frac{2 \tan \beta}{1 + 3 \tan^{2} \beta} = \frac{2}{\frac{1}{\tan \beta} + 3 \tan \beta}$。因为$\tan \beta ＞ 0$，所以$\frac{1}{\tan \beta} + 3 \tan \beta \geqslant 2 \sqrt{3}$，当且仅当$\tan \beta = \frac{\sqrt{3}}{3}$时取等号，所以$\tan (\alpha - \beta)$的最大值是$\frac{\sqrt{3}}{3}$。

(3) 某中学开展劳动实习，学生需要制作一个矩形框架的工艺品。要求将一个长和宽分别为 $20$ cm 和 $10$ cm 的矩形零件的四个顶点分别焊接在矩形框架的四条边上，则矩形框架周长的最大值为(D)

A.$20\sqrt{2}$ cm
B.$30\sqrt{5}$ cm
C.$40\sqrt{5}$ cm
D.$60\sqrt{2}$ cm

答案：

(3)D 【解析】如图，由题意知$EF = 10$，$FG = 20$，令$\angle AEF = \alpha$，$\alpha \in \left[ 0, \frac{\pi}{2} \right]$，则$AF = 10 \sin \alpha$，$\angle AFE = \frac{\pi}{2} - \alpha$，则$\angle BFG = \alpha$，$BF = 20 \cos \alpha$，$BG = 20 \sin \alpha$，$\angle BGF = \frac{\pi}{2} - \alpha$，则$\angle CGH = \alpha$，$CG = 10 \cos \alpha$，所以矩形框架的周长为$2AB + 2BC = 2 (10 \sin \alpha + 20 \cos \alpha) + 2 (20 \sin \alpha + 10 \cos \alpha) = 60 \sin \alpha + 60 \cos \alpha = 60 \sqrt{2} \sin \left( \alpha + \frac{\pi}{4} \right) \leqslant 60 \sqrt{2}$，当且仅当$\alpha = \frac{\pi}{4}$时取等号。

变式 3
(2025·秦皇岛二模改编)已知 $\beta$ 为锐角，且 $\tan \beta ＞ 2, \sin \alpha = \cos(2\beta - \alpha)$，则 $\tan \alpha$ 的取值范围为$(-3, -1)$。

答案： 变式3 ($ - 3, - 1$) 【解析】由$\sin \alpha = \cos (2 \beta - \alpha)$，可得$\sin \alpha = \cos 2 \beta \cos \alpha + \sin 2 \beta \sin \alpha$，即$(1 - \sin 2 \beta) \sin \alpha = \cos 2 \beta \cos \alpha$，因此$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\cos 2 \beta}{1 - \sin 2 \beta} = \frac{\cos^{2} \beta - \sin^{2} \beta}{(\sin \beta - \cos \beta)^{2}} = \frac{\sin \beta + \cos \beta}{\sin \beta - \cos \beta}$，即$\tan \alpha = \frac{\tan \beta + 1}{\tan \beta - 1} = - 1 - \frac{2}{\tan \beta - 1}$。又因为$\tan \beta ＞ 2$，所以$\tan \alpha \in (-3, -1)$。