《2025年金版新学案高三总复习数学北师大版》

(1) 若 $\alpha, \beta \in (\frac{\pi}{2}, \pi)$，且 $(1 - \cos 2\alpha)(1 + \sin \beta) = \sin 2\alpha \cos \beta$，则下列结论正确的是(A)
A. $2\alpha + \beta = \frac{5\pi}{2}$
B. $2\alpha - \beta = \frac{3\pi}{4}$
C. $\alpha + \beta = \frac{7\pi}{4}$
D. $\alpha - \beta = \frac{\pi}{2}$
(2) 若 $\tan \theta = 3$，则 $\frac{\cos \theta (1 + \sin 2\theta)}{\sin \theta + \cos \theta} =$$\frac{2}{5}$.

答案：
(1)A 因为$\alpha,\beta\in(\frac{\pi}{2},\pi)$,所以$\sin\alpha\neq0$,因为$(1 - \cos2\alpha)(1 + \sin\beta) = \sin2\alpha\cos\beta$,所以$2\sin^{2}\alpha(1 + \sin\beta) = 2\sin\alpha\cos\alpha\cos\beta$,即$\sin\alpha(1 + \sin\beta) = \cos\alpha\cos\beta$. 所以$\sin\alpha = \cos\alpha\cos\beta - \sin\alpha\sin\beta = \cos(\alpha + \beta)$,因为$\alpha,\beta\in(\frac{\pi}{2},\pi)$,所以$\pi ＜ \alpha + \beta ＜ 2\pi$,且$- \frac{\pi}{2} ＜ \frac{\pi}{2} - \alpha ＜ 0$,所以$\alpha + \beta = \frac{\pi}{2} - \alpha + 2\pi$,解得$2\alpha + \beta = \frac{5\pi}{2}$. 故选A.
(2)$\frac{2}{5}$ $\frac{\cos\theta(1 + \sin2\theta)}{\sin\theta + \cos\theta} = \frac{\cos\theta(\sin\theta + \cos\theta)^{2}}{\sin\theta + \cos\theta} = \cos\theta(\sin\theta + \cos\theta) = \cos\theta\sin\theta + \cos^{2}\theta = \frac{\sin^{2}\theta + \cos^{2}\theta}{\sin^{2}\theta + \cos^{2}\theta}·\frac{\tan\theta + 1}{\tan^{2}\theta + 1} = \frac{4}{10} = \frac{2}{5}$.

(2023·新课标Ⅰ卷)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： B 因为$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta = \frac{1}{3}$,而$\cos\alpha\sin\beta = \frac{1}{6}$. 因此$\sin\alpha\cos\beta = \frac{1}{2}$,则$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta = \frac{2}{3}$,
所以$\cos(2\alpha + 2\beta) = \cos2(\alpha + \beta) = 1 - 2\sin^{2}(\alpha + \beta) = 1 - 2×(\frac{2}{3})^{2} = \frac{1}{9}$. 故选B.

（北师必修二 P163B 组 T3 和 P164 例 1）(1) 已知 $\sin (\alpha + \beta) = \frac{1}{2}$，$\sin (\alpha - \beta) = \frac{1}{3}$，求 $\frac{\tan (\alpha + \beta) - \tan \alpha - \tan \beta}{\tan^2 \beta \tan (\alpha + \beta)}$.
(2) 已知角 $\alpha$ 是第二象限角，$\cos \alpha = -\frac{3}{5}$，求 $\sin 2\alpha$，$\cos 2\alpha$ 和 $\tan 2\alpha$ 的值.

答案：
(1) $5$；
(2) $\sin2\alpha=-\frac{24}{25}$，$\cos2\alpha=-\frac{7}{25}$，$\tan2\alpha=\frac{24}{7}$。

1. 用五点(画图)法作正弦函数和余弦函数的简图
(1)在正弦函数 $ y = \sin x $，$ x \in [0, 2\pi] $ 的图象中，五个关键点是：$(0, 0)$，$\left(\dfrac{\pi}{2}, 1\right)$，，$\left(\dfrac{3\pi}{2}, -1\right)$，$(2\pi, 0)$.
(2)在余弦函数 $ y = \cos x $，$ x \in [0, 2\pi] $ 的图象中，五个关键点是：$(0, 1)$，$\left(\dfrac{\pi}{2}, 0\right)$，，$\left(\dfrac{3\pi}{2}, 0\right)$，$(2\pi, 1)$.

答案：
(1) $(\pi, 0)$
(2) $(\pi, -1)$