《2025年金版新学案高三总复习数学北师大版》

典例2 (1)(2025·河南郑州模拟)函数$f(x) = \sin\left(x + \dfrac{\pi}{5}\right) + \sqrt{3}\cos\left(x + \dfrac{8\pi}{15}\right)$的最大值为(A)

A.$1$
B.$\sqrt{3}$
C.$\sqrt{5}$
D.$\sqrt{7}$

答案：
(1)A

(2)已知$3\sin x - 4\cos x = 5\sin(x + \varphi)$，则$\varphi$所在的象限为(D)

A.第一象限
B.第二象限
C.第三象限
D.第四象限

答案：
(2)D

(1)(2025·四川宜宾模拟)若$\cos\left(\alpha - \dfrac{\pi}{3}\right) + \cos\alpha = -1$，则$\cos\left(\alpha - \dfrac{\pi}{6}\right) =$(A)
A. $-\dfrac{\sqrt{3}}{3}$
B. $\dfrac{\sqrt{3}}{3}$
C. $\dfrac{2\sqrt{3}}{3}$
D. $-\dfrac{2\sqrt{3}}{3}$
(2)(2025·湖北武汉模拟)函数$f(x) = 3\cos x - 4\sin x$，当$f(x)$取得最大值时，$\sin x =$(B)
A. $\dfrac{4}{5}$
B. $-\dfrac{4}{5}$
C. $\dfrac{3}{5}$
D. $-\dfrac{3}{5}$

答案：
(1)A
(2)B

典例3 (1)$2\cos\left(2x + \dfrac{\pi}{3}\right)\sin\left(2x - \dfrac{\pi}{3}\right) =$(D)

A.$\dfrac{1}{2} + \cos 4x$
B.$\dfrac{1}{2} - \sin 4x$
C.$\dfrac{\sqrt{3}}{2} + \cos 4x$
D.$-\dfrac{\sqrt{3}}{2} + \sin 4x$

答案：
(1)D

(2)(2025·安徽淮北模拟)已知$\sin x\cos y + \cos x\sin y = \dfrac{1}{2}$，$\cos 2x - \cos 2y = \dfrac{1}{4}$，则$\sin(x - y) =$(D)

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

答案：
(2)D

(2025·浙江嘉兴模拟)已知$\alpha$，$\beta \in (0, \pi)$且满足$\sin\alpha + \sin\beta = \sqrt{3}(\cos\alpha + \cos\beta)$，则(B)

A.$\tan(\alpha + \beta) = \sqrt{3}$
B.$\tan(\alpha + \beta) = -\sqrt{3}$
C.$\cos(\alpha + \beta) = \dfrac{\sqrt{3}}{2}$
D.$\cos(\alpha + \beta) = -\dfrac{\sqrt{3}}{2}$

答案： B

[真题再现] (2024·新课标Ⅰ卷)已知$\cos(\alpha + \beta) = m$，$\tan\alpha\tan\beta = 2$，则$\cos(\alpha - \beta) =$(A)

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

答案： A

[教材呈现] （北师必修二P163B组T3）已知$\sin(\alpha + \beta) = \dfrac{1}{2}$，$\sin(\alpha - \beta) = \dfrac{1}{3}$，求$\dfrac{\tan(\alpha + \beta) - \tan\alpha - \tan\beta}{\tan^2\beta\tan(\alpha + \beta)}$的值.

答案： $5$