《2025年创新设计高考总复习数学浙江专版》

考点一 公式的基本应用
例 1 (1)(2024·新高考 I 卷)已知 $ \cos(\alpha + \beta) = m $，$ \tan \alpha \tan \beta = 2 $，则 $ \cos(\alpha - \beta) = $(A)

A.$ -3m $
B.$ -\frac{m}{3} $
C.$ \frac{m}{3} $
D.$ 3m $

答案： 例1
(1)A[
(1)由$\cos(\alpha + \beta) = m$得$\cos \alpha \cos \beta - \sin \alpha \sin \beta = m.$由$\tan \alpha \tan \beta = 2$得$\frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = 2,$由①②得$\begin{cases} \cos \alpha \cos \beta = -m, \\ \sin \alpha \sin \beta = -2m, \end{cases}$所以$\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta = -3m,$故选A.]

(2)(2024·全国甲卷)已知 $ \frac{\cos \alpha}{\cos \alpha - \sin \alpha} = \sqrt{3} $，则 $ \tan(\alpha + \frac{\pi}{4}) = $(B)

A.$ 2\sqrt{3} + 1 $
B.$ 2\sqrt{3} - 1 $
C.$ \frac{\sqrt{3}}{2} $
D.$ 1 - \sqrt{3}$$$

答案： 例1
(2)B[
(2)根据题意有$\frac{\cos \alpha - \sin \alpha}{\cos \alpha} = \frac{\sqrt{3}}{3},$即$1 - \tan \alpha = \frac{\sqrt{3}}{3},$所以$\tan \alpha = 1 - \frac{\sqrt{3}}{3},$所以$\tan(\alpha + \frac{\pi}{4}) = \frac{\tan \alpha + 1}{1 - \tan \alpha} = \frac{2 - \frac{\sqrt{3}}{3}}{\frac{\sqrt{3}}{3}} = 2\sqrt{3} - 1,$故选B.]

训练 1 (1)(2025·石家庄质检)已知 $ \sin(\frac{\pi}{6} + \alpha) = \frac{\sqrt{2}}{3} $，则 $ \cos(2\alpha + \frac{4\pi}{3}) $ 的值为(B)

A.$ \frac{5}{9} $
B.$ -\frac{5}{9} $
C.$ \frac{1}{3} $
D.$ -\frac{1}{3} $

答案： 训练$1(1)B[(1)\cos(2\alpha + \frac{4\pi}{3})= \cos(2\alpha + \frac{\pi}{3} + \pi)= - \cos(2\alpha + \frac{\pi}{3})= 2\sin^{2}(\alpha + \frac{\pi}{6}) - 1= \frac{5}{9}.]$

(2)计算：$ \frac{\cos 55° + \sin 25° \cos 60°}{\cos 25°} = $(B)

A.$ -\frac{\sqrt{3}}{2} $
B.$ \frac{\sqrt{3}}{2} $
C.$ -\frac{1}{2} $
D.$ \frac{1}{2}$$$

答案： 训练$1(2)B[(2)\frac{\cos 55^{\circ} + \sin 25^{\circ} \cos 60^{\circ}}{\cos 25^{\circ}}= \frac{\cos(30^{\circ} + 25^{\circ}) + \frac{1}{2} \sin 25^{\circ}}{\cos 25^{\circ}}=]$

考点二 公式的逆用及变形
例 2 (1)(多选)(2025·合肥质检)下列代数式的值为 $ \frac{1}{4} $ 的是(BCD)

A.$ \cos^2 75° - \sin^2 75° $
B.\frac{\tan 15^{\circ}}{1 + \tan^{2}15^{\circ}}$ \frac{\tan 15°}{1 + \tan^2 15°} $
C.$ \cos 36° \cos 72° $
D.$ 2\cos 20° \cos 40° \cos 80° $

答案： 例2
(1)BCD[
(1)对于$A,\cos^{2}75^{\circ} - \sin^{2}75^{\circ} = \cos 150^{\circ} = \cos(180^{\circ} - 30^{\circ}) = - \cos 30^{\circ} = \frac{\sqrt{3}}{2};$对于$B,\frac{\tan 15^{\circ}}{1 + \tan^{2}15^{\circ}} = \frac{\sin 15^{\circ} \cos 15^{\circ}}{\cos^{2}15^{\circ} + \sin^{2}15^{\circ}} = \frac{\sin 15^{\circ} \cos 15^{\circ}}{1} = \frac{1}{2} \sin 30^{\circ} = \frac{1}{4};$对于$C,\cos 36^{\circ} \cos 72^{\circ}= \frac{\sin 36^{\circ} \cos 36^{\circ} \cos 72^{\circ}}{\sin 36^{\circ}} = \frac{\frac{1}{2} \sin 72^{\circ} \cos 72^{\circ}}{\sin 36^{\circ}} = \frac{\frac{1}{4} \sin 144^{\circ}}{\sin 36^{\circ}} = \frac{1}{4};$对于$D,2\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}= 2\cos 20^{\circ} \sin 40^{\circ} \cos 80^{\circ}= \frac{\sin 20^{\circ}}{\sin 20^{\circ}} \cdot \sin 40^{\circ} \cos 40^{\circ} \cos 80^{\circ}= \frac{\frac{1}{2} \sin 80^{\circ} \cos 80^{\circ}}{\sin 20^{\circ}} = \frac{\frac{1}{4} \sin 160^{\circ}}{\sin 20^{\circ}} = \frac{1}{4}.]$

(2)$ \tan 70° + \tan 50° - \sqrt{3} \tan 70° \tan 50° = $____.

答案： 例$2(2)- \sqrt{3}[(2)$因为$\tan(70^{\circ} + 50^{\circ}) = \frac{\tan 70^{\circ} + \tan 50^{\circ}}{1 - \tan 70^{\circ} \tan 50^{\circ}}= \tan 120^{\circ} = - \sqrt{3},$所以$\tan 70^{\circ} + \tan 50^{\circ} = - \sqrt{3} + \sqrt{3} \tan 70^{\circ} \tan 50^{\circ},$所以$\tan 70^{\circ} + \tan 50^{\circ} - \sqrt{3} \tan 70^{\circ} \tan 50^{\circ} = - \sqrt{3}.]$

训练 2 (1)$ \sin^2(\alpha - \frac{\pi}{6}) + \sin^2(\alpha + \frac{\pi}{6}) - \sin^2 \alpha $ 等于(C)
A. $ -\frac{1}{2} $
B. $ -\frac{\sqrt{3}}{2} $
C. $ \frac{1}{2} $
D. $ \frac{\sqrt{3}}{2} $

答案： 训练2
(1)C[
(1)原式$=\frac{1 - \cos(2\alpha - \frac{\pi}{3})}{2} + \frac{1 - \cos(2\alpha + \frac{\pi}{3})}{2} - \sin^{2} \alpha= 1 - \frac{1}{2} \cdot [\cos(2\alpha - \frac{\pi}{3}) + \cos(2\alpha + \frac{\pi}{3})] - \sin^{2} \alpha= 1 - \cos 2\alpha \cos \frac{\pi}{3} - \sin^{2} \alpha= 1 - \frac{\cos 2\alpha}{2} - \frac{1 - \cos 2\alpha}{2} = \frac{1}{2}.]$

(2)化简：$ \cos 40°(1 + \sqrt{3} \tan 10°) = $____.

答案： 训练$2(2)1[(2)\cos 40^{\circ}(1 + \sqrt{3} \tan 10^{\circ})= \cos 40^{\circ}(1 + \frac{\sqrt{3} \sin 10^{\circ}}{\cos 10^{\circ}})= \cos 40^{\circ} \cdot \frac{\cos 10^{\circ} + \sqrt{3} \sin 10^{\circ}}{\cos 10^{\circ}}= 2\cos 40^{\circ} \cdot (\frac{1}{2} \cos 10^{\circ} + \frac{\sqrt{3}}{2} \sin 10^{\circ})= \frac{2\cos 40^{\circ} \sin 40^{\circ}}{\cos 10^{\circ}} = \frac{\sin 80^{\circ}}{\cos 10^{\circ}} = 1.]$