《2025年金版新学案高三总复习数学北师大版》

典例 1 (1) 数学家华罗庚倡导的“0.618 优选法”在各领域都应用广泛，0.618 就是黄金分割比 $ m = \frac{\sqrt{5} - 1}{2} $ 的近似值，黄金分割比还可以表示成 $ 2\sin 18^{\circ} $，则 $\frac{2m\sqrt{4 - m^2}}{2\cos^2 27^{\circ} - 1} =$(A)

A.$4$
B.$\sqrt{5} + 1$
C.$2$
D.$\sqrt{5} - 1$

答案：
(1)A 根据题意,可得$m = 2\sin18^{\circ} = 2\cos72^{\circ}$,则$\frac{2m\sqrt{4 - m^{2}}}{2\cos^{2}27^{\circ} - 1} = \frac{4\cos72^{\circ}\sqrt{4 - 4\cos^{2}72^{\circ}}}{\cos54^{\circ}} = \frac{4\sin144^{\circ}}{\cos54^{\circ}} = \frac{4\sin(90^{\circ} + 54^{\circ})}{\cos54^{\circ}} = \frac{4\cos54^{\circ}}{\cos54^{\circ}} = 4$. 故选A.

(2) (2025·河南焦作模拟)$\frac{\tan 80^{\circ} - \tan 20^{\circ}}{1 + \frac{1}{2\cos 20^{\circ}}}$ 的值为$2\sqrt{3}$.

答案：
(2)$2\sqrt{3}$ $\tan80^{\circ} - \tan20^{\circ} = \tan(80^{\circ} - 20^{\circ})(1 + \tan80^{\circ}\tan20^{\circ}) = \sqrt{3}(1 + \frac{\sin80^{\circ}\sin20^{\circ}}{\cos80^{\circ}\cos20^{\circ}}) = \sqrt{3}(1 + \frac{2\cos^{2}10^{\circ}}{\cos20^{\circ}}) = \sqrt{3}(1 + \frac{1 + \cos20^{\circ}}{\cos20^{\circ}}) = \sqrt{3}(2 + \frac{1}{\cos20^{\circ}})$,所以$\frac{\tan80^{\circ} - \tan20^{\circ}}{1 + \frac{1}{2\cos20^{\circ}}} = 2\sqrt{3}$.

典例 2 (1) (2025·浙江杭州模拟)已知 $\alpha \in (\frac{\pi}{2}, \pi)$，$\beta \in (0, \frac{\pi}{2})$，若 $\sin (\alpha + \beta) = \frac{1}{3}$，$\cos \beta = \frac{\sqrt{3}}{3}$，则 $\cos 2\alpha =$(D)

A.$\frac{1}{3}$
B.$-\frac{1}{3}$
C.$\frac{23}{27}$
D.$-\frac{23}{27}$

答案：
(1)D 由于$\alpha\in(\frac{\pi}{2},\pi),\beta\in(0,\frac{\pi}{2})$,则$\alpha + \beta\in(\frac{\pi}{2},\frac{3\pi}{2})$,而$\sin(\alpha + \beta) = \frac{1}{3}$,故$\alpha + \beta\in(\frac{\pi}{2},\pi)$,所以$\cos(\alpha + \beta) = - \sqrt{1 - \sin^{2}(\alpha + \beta)} = - \frac{2\sqrt{2}}{3}$,由$\cos\beta = \frac{\sqrt{3}}{3},\beta\in(0,\frac{\pi}{2})$,可得$\sin\beta = \frac{\sqrt{6}}{3}$,则$\cos\alpha = \cos[(\alpha + \beta) - \beta] = \cos(\alpha + \beta)\cos\beta + \sin(\alpha + \beta)\sin\beta = - \frac{2\sqrt{2}}{3}×\frac{\sqrt{3}}{3} + \frac{1}{3}×\frac{\sqrt{6}}{3} = - \frac{\sqrt{6}}{9}$,故$\cos2\alpha = 2\cos^{2}\alpha - 1 = 2×( - \frac{\sqrt{6}}{9})^{2} - 1 = - \frac{23}{27}$. 故选D.

(2) (2024·九省适应性测试)已知 $\theta \in (\frac{3\pi}{4}, \pi)$，$\tan 2\theta = -4\tan (\theta + \frac{\pi}{4})$，则 $\frac{1 + \sin 2\theta}{2\cos^2 \theta + \sin 2\theta} =$(A)

A.$\frac{1}{4}$
B.$\frac{3}{4}$
C.$1$
D.$\frac{3}{2}$

答案：
(2)A 由$\tan2\theta = - 4\tan(\theta + \frac{\pi}{4})$得,$\frac{2\tan\theta}{1 - \tan^{2}\theta} = \frac{- 4(\tan\theta + 1)}{1 - \tan\theta} \Rightarrow - 4(\tan\theta + 1)^{2} = 2\tan\theta$,则$2\tan^{2}\theta + 5\tan\theta + 2 = 0$,得$\tan\theta = - \frac{1}{2}$或$- 2$,因为$\theta\in(\frac{3\pi}{4},\pi)$,所以$\tan\theta\in( - 1,0)$,所以$\tan\theta = - \frac{1}{2}$.

典例 3 (1) 若 $\alpha, \beta \in (\frac{\pi}{2}, \pi)$，满足 $(1 - \sin 2\alpha) \sin \beta = \cos \beta \cos 2\alpha$，下列结论正确的是(A)

A.$\beta - \alpha = \frac{\pi}{4}$
B.$\alpha - \beta = \frac{\pi}{4}$
C.$\alpha + \beta = \frac{\pi}{4}$
D.$\alpha = \beta$

答案：
(1)A 因为$\alpha,\beta\in(\frac{\pi}{2},\pi)$,所以$\begin{cases}2\alpha\in(\pi,2\pi)\\\cos\beta ＜ 0\\\sin\beta ＞ 0\\\sin\alpha ＞ 0\end{cases}$
$\begin{cases}1 - \sin2\alpha ＞ 0\\\cos2\alpha ＜ 0\end{cases}$,所以$2\alpha\in(\pi,\frac{3}{2}\pi)\Rightarrow\alpha\in(\frac{\pi}{2},\frac{3}{4}\pi),\tan\alpha ＜ - 1$,所以$(1 - \sin2\alpha)\sin\beta = \cos\beta\cos2\alpha\Rightarrow\tan\beta = \frac{\cos2\alpha}{1 - \sin2\alpha} = \frac{\cos^{2}\alpha - \sin^{2}\alpha}{\cos^{2}\alpha + \sin^{2}\alpha - 2\sin\alpha\cos\alpha} = \frac{1 - \tan^{2}\alpha}{(\tan\alpha - 1)^{2}} = \frac{1 + \tan\alpha}{1 - \tan\alpha} = \tan(\frac{\pi}{4} + \alpha)$. 所以$\beta = \frac{\pi}{4} + \alpha$,即$\beta - \alpha = \frac{\pi}{4}$. 故选A.

(2) (2025·黑龙江双鸭山模拟)已知 $\alpha, \beta \in (0, \frac{\pi}{4})$，$\cos^2 \alpha - \sin^2 \alpha = \frac{1}{7}$，且 $3\sin \beta = \sin (2\alpha + \beta)$，则 $\alpha + \beta$ 的值为(D)

A.$\frac{\pi}{12}$
B.$\frac{\pi}{6}$
C.$\frac{\pi}{4}$
D.$\frac{\pi}{3}$

答案：
(2)D 因为$\cos^{2}\alpha - \sin^{2}\alpha = \frac{4}{7},\cos^{2}\alpha + \sin^{2}\alpha = 1$,所以$\cos^{2}\alpha = \frac{4 + 3}{14} = \frac{4}{7},\sin^{2}\alpha = \frac{3}{7}$,因为$\alpha\in(0,\frac{\pi}{4})$,所以$\cos\alpha = \frac{2}{\sqrt{7}},\sin\alpha = \frac{\sqrt{3}}{\sqrt{7}}$,所以$\tan\alpha = \frac{\sqrt{3}}{2}$. 由$3\sin\beta = \sin(2\alpha + \beta)$,得$3\sin[(\alpha + \beta) - \alpha] = \sin[(\alpha + \beta) + \alpha]$,即$3\sin(\alpha + \beta)\cos\alpha - 3\cos(\alpha + \beta)\sin\alpha = \sin(\alpha + \beta)\cos\alpha + \cos(\alpha + \beta)\sin\alpha$,所以$\sin(\alpha + \beta)\cos\alpha = 2\cos(\alpha + \beta)\sin\alpha$,所以$\tan(\alpha + \beta) = 2\tan\alpha = \sqrt{3}$. 又$0 ＜ \alpha + \beta ＜ \frac{\pi}{2}$,所以$\alpha + \beta = \frac{\pi}{3}$. 故选D.

(1) (2025·福建南平模拟)已知 $\tan (\alpha + \frac{\pi}{6}) = \frac{1}{2}$，则 $\cos (2\alpha - \frac{2\pi}{3}) =$(A)
A. $-\frac{3}{5}$
B. $\frac{3}{4}$
C. $-\frac{4}{5}$
D. $\frac{4}{5}$
(2) (多选题)(2025·江苏盐城模拟)下列各式正确的是(BD)
A. $\cos^2 15^{\circ} - \sin 15^{\circ} \cos 15^{\circ} = \sqrt{3}$
B. $(1 + \tan 1^{\circ})(1 + \tan 44^{\circ}) = 2$
C. $\frac{1}{\sin 10^{\circ}} - \frac{\sqrt{3}}{\cos 10^{\circ}} = 2$
D. $\frac{3 - \sin 70^{\circ}}{2 - \cos^2 10^{\circ}} = 2$
(3) 若 $\cos (\alpha - \beta) = \frac{\sqrt{5}}{5}$，$\cos 2\alpha = \frac{\sqrt{10}}{10}$，且 $\alpha, \beta$ 均为锐角，$\alpha ＜ \beta$，则 $\alpha + \beta =$$\frac{3\pi}{4}$.

答案：
(1)A 因为$\tan(\alpha + \frac{\pi}{6}) = \frac{1}{2}$,所以$\frac{\sin(\alpha + \frac{\pi}{6})}{\cos(\alpha + \frac{\pi}{6})} = \frac{1}{2}$,且$\sin^{2}(\alpha + \frac{\pi}{6}) + \cos^{2}(\alpha + \frac{\pi}{6}) = 1$,解得$\cos(\alpha + \frac{\pi}{6}) = \frac{2}{\sqrt{5}}$,$\sin^{2}(\alpha + \frac{\pi}{6}) = \frac{1}{5}$,$\cos(2\alpha - \frac{2\pi}{3}) = \cos[2(\alpha + \frac{\pi}{6}) - \pi] = - \cos[2(\alpha + \frac{\pi}{6})] = - [1 - 2\sin^{2}(\alpha + \frac{\pi}{6})] = - (1 - 2×\frac{1}{5}) = - \frac{3}{5}$. 故选A.
(2)BD 对于A,由$\cos^{2}15^{\circ} - \sin15^{\circ}\cos15^{\circ} = \frac{1 + \cos30^{\circ}}{2} - \frac{1}{2}\sin30^{\circ} = \frac{1 + \sqrt{3}}{4}$,故A错误;对于B,$(1 + \tan1^{\circ})(1 + \tan44^{\circ}) = \tan44^{\circ} + \tan1^{\circ} + \tan44^{\circ}·\tan1^{\circ} + 1 = \tan(44^{\circ} + 1^{\circ})(1 - \tan44^{\circ}·\tan1^{\circ}) + \tan44^{\circ}·\tan1^{\circ} + 1 = 2$,故B正确;对于C,$\frac{1}{\sin10^{\circ}} - \frac{\sqrt{3}}{\cos10^{\circ}} = \frac{\cos10^{\circ} - \sqrt{3}\sin10^{\circ}}{\sin10^{\circ}·\cos10^{\circ}} = \frac{2\cos(60^{\circ} + 10^{\circ})}{\frac{1}{2}\sin20^{\circ}} = \frac{4\cos70^{\circ}}{\sin20^{\circ}} = 4$,故C错误;对于D,$\frac{3 - \sin70^{\circ}}{2 - \cos^{2}10^{\circ}} = \frac{3 - \cos20^{\circ}}{2 - \cos^{2}10^{\circ}} = \frac{3 - (2\cos^{2}10^{\circ} - 1)}{2 - \cos^{2}10^{\circ}} = \frac{4 - 2\cos^{2}10^{\circ}}{2 - \cos^{2}10^{\circ}} = 2$,故D正确. 故选BD.
(3)$\frac{3\pi}{4}$ 由题意可知,$0 ＜ \alpha ＜ \frac{\pi}{2},0 ＜ \beta ＜ \frac{\pi}{2},\alpha ＜ \beta$,所以$- \frac{\pi}{2} ＜ \alpha - \beta ＜ 0$,$\cos(\alpha - \beta) = \frac{\sqrt{5}}{5}$,得$\sin(\alpha - \beta) = - \frac{2\sqrt{5}}{5}$,$0 ＜ 2\alpha ＜ \pi$,且$\cos2\alpha = \frac{\sqrt{10}}{10}$,得$0 ＜ \alpha ＜ \frac{\pi}{4}$,$\sin2\alpha = \frac{3\sqrt{10}}{10}$,所以$\cos(\alpha + \beta) = \cos[2\alpha - (\alpha - \beta)] = \cos2\alpha\cos(\alpha - \beta) + \sin2\alpha\sin(\alpha - \beta) = \frac{\sqrt{10}}{10}×\frac{\sqrt{5}}{5} + \frac{3\sqrt{10}}{10}×( - \frac{2\sqrt{5}}{5}) = - \frac{\sqrt{2}}{2}$,因为$0 ＜ \alpha + \beta ＜ \pi$,所以$\alpha + \beta = \frac{3\pi}{4}$.

(1) 求 $\sin 2\alpha + \cos 2\alpha$ 的值；

答案： 解:
(1)因为$3\sin\alpha = 2\sin^{2}\frac{\alpha}{2} - 1$,
所以$3\sin\alpha = - \cos\alpha$,所以$\tan\alpha = - \frac{1}{3}$,
所以$\sin2\alpha + \cos2\alpha = \frac{2\sin\alpha\cos\alpha + \cos^{2}\alpha - \sin^{2}\alpha}{\sin^{2}\alpha + \cos^{2}\alpha} = \frac{2\tan\alpha + 1 - \tan^{2}\alpha}{\tan^{2}\alpha + 1} = 2×( - \frac{1}{3}) + 1 - \frac{1}{9} = \frac{1}{5}$.