《2025年高考总复习优化设计高中数学北师大版》

1. 思考辨析(判断下列结论是否正确，正确的画“√”，错误的画“×”).
(1) 两角和与差的正弦、余弦、正切公式中的角$\alpha$，$\beta$是任意的. (×)
(2) 公式$a\sin x + b\cos x=\sqrt{a^{2}+b^{2}}\sin(x+\varphi)$中$\varphi$的取值与$a$，$b$的值无关. (×)
(3) $\cos\theta=2\cos^{2}\frac{\theta}{2}-1=1 - 2\sin^{2}\frac{\theta}{2}$. (√)
(4) $\frac{1 - \tan\theta}{1 + \tan\theta}=\tan\left(\frac{\pi}{4}+\theta\right)$. (×)

答案： 1.
(1)×
(2)×
(3)√
(4)×

2. 在$\triangle ABC$中，$\cos A=\frac{4}{5}$，$\tan B = 2$，则$\tan(2A + 2B)=$$\frac{44}{117}$.

答案： 2.$\frac{44}{117}$ 解析（方法一）在$\triangle ABC$中，由$\cos A = \frac{4}{5}$，$0 ＜ A ＜ \pi$，得$\sin A = \sqrt{1 - \cos ^{2} A} = \sqrt{1 - (\frac{4}{5})^{2}} = \frac{3}{5}$，所以$\tan A = \frac{\sin A}{\cos A} = \frac{3}{5} × \frac{5}{4} = \frac{3}{4}$，$\tan 2A = \frac{2\tan A}{1 - \tan ^{2} A} = \frac{2 × \frac{3}{4}}{1 - (\frac{3}{4})^{2}} = \frac{\frac{24}{4}}{ \frac{7}{16}} = \frac{24}{7}$。又$\tan B = 2$，所以$\tan 2B = \frac{2\tan B}{1 - \tan ^{2} B} = \frac{2 × 2}{1 - 2^{2}} = \frac{4}{3}$。于是$\tan(2A + 2B) = \frac{\tan 2A + \tan 2B}{1 - \tan 2A \tan 2B} = \frac{\frac{24}{7} + \frac{4}{3}}{1 - \frac{24}{7} × \frac{4}{3}} = \frac{\frac{72 + 28}{21}}{1 - \frac{96}{21}} = \frac{\frac{100}{21}}{-\frac{75}{21}} = -\frac{44}{117}$。
（方法二）在$\triangle ABC$中，由$\cos A = \frac{4}{5}$，$0 ＜ A ＜ \pi$，得$\sin A = \sqrt{1 - \cos ^{2} A} = \sqrt{1 - (\frac{4}{5})^{2}} = \frac{3}{5}$，所以$\tan A = \frac{\sin A}{\cos A} = \frac{3}{5} × \frac{5}{4} = \frac{3}{4}$。又$\tan B = 2$，所以$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\frac{3}{4} + 2}{1 - \frac{3}{4} × 2} = \frac{\frac{11}{4}}{1 - \frac{3}{2}} = -\frac{11}{2}$。所以$\tan(2A + 2B) = \tan[2(A + B)] = \frac{2\tan(A + B)}{1 - \tan ^{2}(A + B)} = \frac{2 × (-\frac{11}{2})}{1 - (-\frac{11}{2})^{2}} = \frac{-11}{1 - \frac{121}{4}} = \frac{-11}{-\frac{117}{4}} = \frac{44}{117}$。

3. 已知$\sin(\alpha-\beta)\cos\alpha-\cos(\beta-\alpha)\sin\alpha=\frac{3}{5}$，$\beta$是第三象限角，则$\sin\left(\beta+\frac{5\pi}{4}\right)=$$-\frac{7\sqrt{2}}{10}$.

答案： 3.$\frac{7\sqrt{2}}{10}$ 解析由已知得$\sin(\alpha - \beta) \cos \alpha - \cos(\alpha - \beta) \sin \alpha = \frac{3}{5}$，即$\sin[(\alpha - \beta) - \alpha] = \frac{3}{5}$，所以$\sin(-\beta) = \frac{3}{5}$，则$\sin \beta = -\frac{3}{5}$，又$\beta$是第三象限角，所以$\cos \beta = -\frac{4}{5}$，故$\sin(\beta + \frac{5\pi}{4}) = \sin \beta \cos \frac{5\pi}{4} + \cos \beta \cdot \sin \frac{5\pi}{4} = \frac{\sqrt{2}}{2}(\sin \beta + \cos \beta) = \frac{\sqrt{2}}{2}(-\frac{3}{5} - \frac{4}{5}) = -\frac{7\sqrt{2}}{10}$。

4. (2021·全国甲，理9，文11)若$\alpha\in\left(0,\frac{\pi}{2}\right)$，$\tan2\alpha=\frac{\cos\alpha}{2 - \sin\alpha}$，则$\tan\alpha=$(A)

A.$\frac{\sqrt{15}}{15}$
B.$\frac{\sqrt{5}}{5}$
C.$\frac{\sqrt{5}}{3}$
D.$\frac{\sqrt{15}}{3}$

答案： 4.A 解析$\because \tan 2\alpha = \frac{\cos \alpha}{2 - \sin \alpha}$，$\therefore \tan 2\alpha = \frac{\sin 2\alpha}{2\sin \alpha \cos \alpha} ÷ \frac{\cos \alpha}{2 - \sin \alpha} = \frac{\cos \alpha}{2 - \sin \alpha}$，$\because \alpha \in (0,\frac{\pi}{2})$，$\therefore \cos \alpha \neq 0$，$\therefore 2\sin \alpha = \frac{1}{2 - \sin \alpha}$，解得$\sin \alpha = \frac{1}{4}$，$\therefore \cos \alpha = \sqrt{1 - \sin ^{2} \alpha} = \frac{\sqrt{15}}{4}$，$\therefore \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\sqrt{15}}{15}$。

5. (2022·浙江，13)若$3\sin\alpha-\sin\beta=\sqrt{10}$，$\alpha+\beta=\frac{\pi}{2}$，则$\sin\alpha=$$\frac{3\sqrt{10}}{10}$，$\cos2\beta=$$\frac{4}{5}$.

答案： 5.$\frac{3\sqrt{10}}{10}$ $\frac{4}{5}$ 解析（方法一 辅助角公式法）$\because \alpha + \beta = \frac{\pi}{2}$，$\therefore \sin \beta = \cos \alpha$，$\cos \beta = \sin \alpha$，$\therefore 3\sin \alpha - \cos \alpha = \sqrt{10}$，即$\sqrt{10}(\frac{3\sqrt{10}}{10} \sin \alpha - \frac{\sqrt{10}}{10} \cos \alpha) = \sqrt{10}$，令$\sin \theta = \frac{\sqrt{10}}{10}$，$\cos \theta = \frac{3\sqrt{10}}{10}$，则$\sqrt{10} \sin(\alpha - \theta) = \sqrt{10}$，$\therefore \alpha - \theta = \frac{\pi}{2} + 2k\pi$，$k \in Z$，即$\alpha = \theta + \frac{\pi}{2} + 2k\pi$，$\therefore \sin \alpha = \sin(\theta + \frac{\pi}{2} + 2k\pi) = \cos \theta = \frac{3\sqrt{10}}{10}$，则$\cos 2\beta = 2\cos ^{2} \beta - 1 = 2\sin ^{2} \alpha - 1 = \frac{4}{5}$。
（方法二 同角三角函数关系法）$\because \alpha + \beta = \frac{\pi}{2}$，$\therefore \sin \beta = \cos \alpha$，$\cos \beta = \sin \alpha$，$\therefore 3\sin \alpha - \cos \alpha = \sqrt{10}$，又$\sin ^{2} \alpha + \cos ^{2} \alpha = 1$，将$\cos \alpha = 3\sin \alpha - \sqrt{10}$代入得$10\sin ^{2} \alpha - 6\sqrt{10} \sin \alpha + 9 = 0$，解得$\sin \alpha = \frac{3\sqrt{10}}{10}$，则$\cos 2\beta = 2\cos ^{2} \beta - 1 = 2\sin ^{2} \alpha - 1 = \frac{4}{5}$。

例1 (1)(2023·新高考Ⅰ，8)已知$\sin(\alpha-\beta)=\frac{1}{3}$，$\cos\alpha\sin\beta=\frac{1}{6}$，则$\cos(2\alpha + 2\beta)=$(B)

A.$\frac{7}{9}$
B.$\frac{1}{9}$
C.$-\frac{1}{9}$
D.$-\frac{7}{9}$

答案： 例1
(1)B 解析（1）由题意，$\because \sin(\alpha - \beta) = \frac{1}{3}$，$\cos \alpha \sin \beta = \frac{1}{6}$，$\therefore \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta = \sin \alpha \cos \beta - \frac{1}{6} = \frac{1}{3}$，解得$\sin \alpha \cos \beta = \frac{1}{2}$。$\because \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$，$\therefore \cos(2\alpha + 2\beta) = \cos[2(\alpha + \beta)] = 1 - 2\sin ^{2}(\alpha + \beta) = 1 - 2 × (\frac{2}{3})^{2} = \frac{1}{9}$。故B选。

(2)(2024·福建泉州模拟)已知$\cos\left(\frac{\pi}{3}-\alpha\right)=2\cos\left(\alpha+\frac{\pi}{6}\right)$，则$\sin\left(2\alpha+\frac{\pi}{3}\right)=$(D)

A.$-\frac{4}{5}$
B.$-\frac{3}{5}$
C.$\frac{3}{5}$
D.$\frac{4}{5}$

答案：
(2)D 解析（2）$\because \cos(\frac{\pi}{3} - \alpha) = \cos[\frac{\pi}{2} - (\alpha + \frac{\pi}{6})] = \sin(\alpha + \frac{\pi}{6}) = 2\cos(\alpha + \frac{\pi}{6})$，$\therefore \tan(\alpha + \frac{\pi}{6}) = 2$，$\sin(2\alpha + \frac{\pi}{3}) = 2\sin(\alpha + \frac{\pi}{6}) \cos(\alpha + \frac{\pi}{6}) = \frac{2\sin(\alpha + \frac{\pi}{6}) \cos(\alpha + \frac{\pi}{6})}{\sin ^{2}(\alpha + \frac{\pi}{6}) + \cos ^{2}(\alpha + \frac{\pi}{6})} = \frac{2\tan(\alpha + \frac{\pi}{6})}{\tan ^{2}(\alpha + \frac{\pi}{6}) + 1} = \frac{4}{5}$。

(3)(2024·广东广州模拟)若$\tan\beta=\frac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}$，则(A)

A.$\tan(\alpha-\beta)=1$
B.$\tan(\alpha-\beta)=-1$
C.$\tan(\alpha+\beta)=1$
D.$\tan(\alpha+\beta)=-1$

答案：
(3)A 解析（3）因为$\tan \beta = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$，所以$\tan \beta = \frac{\tan \alpha - 1}{\tan \alpha + 1}$，所以$\tan \alpha \tan \beta + \tan \beta = \tan \alpha - 1$，所以$1 + \tan \alpha \tan \beta = \tan \alpha - \tan \beta$，所以$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} = 1$。

例2 (1)(多选题)(2024·湖南长沙模拟)计算下列各式，结果为$\sqrt{3}$的是(AD)

A.$\sqrt{2}\sin15^{\circ}+\sqrt{2}\cos15^{\circ}$
B.$\cos^{2}15^{\circ}-\sin15^{\circ}\cos75^{\circ}$
C.$\frac{\tan30^{\circ}}{1 - \tan^{2}30^{\circ}}$
D.$\frac{1 + \tan15^{\circ}}{1 - \tan15^{\circ}}$

答案： 例2
(1)AD 解析（1）对于选项A，由辅助角公式得$\sqrt{2} \sin 15^{\circ} + \sqrt{2} \cos 15^{\circ} = 2\sin(15^{\circ} + 45^{\circ}) = 2\sin 60^{\circ} = \sqrt{3}$，故A正确；对于选项B，$\cos ^{2} 15^{\circ} - \sin 15^{\circ} \cos 75^{\circ} = \sin 75^{\circ} \cos 15^{\circ} - \sin 15^{\circ} \cos 75^{\circ} = \sin(75^{\circ} - 15^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$，故B错误；对于选项C，$\frac{\tan 30^{\circ}}{1 - \tan ^{2} 30^{\circ}} = \frac{\frac{\sqrt{3}}{3}}{1 - (\frac{\sqrt{3}}{3})^{2}} = \frac{\sqrt{3}}{2}$，故C错误；对于选项D，$\frac{1 + \tan 15^{\circ}}{1 - \tan 15^{\circ}} = \frac{\tan 45^{\circ} + \tan 15^{\circ}}{1 - \tan 15^{\circ} × 45^{\circ}} = \tan(45^{\circ} + 15^{\circ}) = \tan 60^{\circ} = \sqrt{3}$，故D正确。故选AD。

(2)(2020·北京，14)若函数$f(x)=\sin(x+\varphi)+\cos x$的最大值为$2$，则常数$\varphi$的一个取值为$\frac{\pi}{2}$.

答案：
(2)$\frac{\pi}{2}$（答案不唯一，只要符合$2k\pi + \frac{\pi}{2}$，$k \in Z$均可） 解析（2）因为$f(x) = \cos \varphi \sin x + (\sin \varphi + 1) \cos x = \sqrt{\cos ^{2} \varphi + (\sin \varphi + 1)^{2}} \sin(x + \theta)$，其中$\tan \theta = \frac{\sin \varphi + 1}{\cos \varphi}$，所以$\sqrt{\cos ^{2} \varphi + (\sin \varphi + 1)^{2}} = 2$，解得$\sin \varphi = 1$，故可取$\varphi = \frac{\pi}{2}$。