2025年金版新学案高三总复习数学北师大版


注:目前有些书本章节名称可能整理的还不是很完善,但都是按照顺序排列的,请同学们按照顺序仔细查找。练习册 2025年金版新学案高三总复习数学北师大版 答案主要是用来给同学们做完题方便对答案用的,请勿直接抄袭。



《2025年金版新学案高三总复习数学北师大版》

第97页
1. (多选题)下列说法中正确的是(
AB
)

A.存在$\alpha$,$\beta$,使等式$\sin(\alpha + \beta) = \sin\alpha + \sin\beta$成立
B.$\cos x + \dfrac{\sqrt{3}}{2}$化为积的形式为$2\cos\left(\dfrac{x}{2} + \dfrac{\pi}{12}\right)\cos\left(\dfrac{x}{2} - \dfrac{\pi}{12}\right)$
C.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
D.公式$\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$可以变形为$\tan\alpha + \tan\beta = \tan(\alpha + \beta)(1 - \tan\alpha\tan\beta)$,且对任意角$\alpha$,$\beta$都成立
答案: 1.AB
2. (链接北师必修二P156T2,改编)$\sin 20°\cos 10° - \cos 160°\sin 10°$等于(
D
)

A.$-\dfrac{\sqrt{3}}{2}$
B.$\dfrac{\sqrt{3}}{2}$
C.$-\dfrac{1}{2}$
D.$\dfrac{1}{2}$
答案: 2.D
3. (链接北师必修二P154T2,改编)若$\cos\alpha = -\dfrac{4}{5}$,$\alpha$是第三象限角,则$\sin\left(\alpha + \dfrac{\pi}{4}\right)$等于(
C
)

A.$-\dfrac{\sqrt{2}}{10}$
B.$\dfrac{\sqrt{2}}{10}$
C.$-\dfrac{7\sqrt{2}}{10}$
D.$\dfrac{7\sqrt{2}}{10}$
答案: 3.C
4. (链接北师必修二P156例4,改编)设锐角$\alpha$与$\beta$,若$\tan\alpha = 2$,$\tan\beta = 3$,则$\alpha + \beta = $
$\frac{3\pi }{4}$
$$.
答案: 4.$\frac{3\pi }{4}$
1. (2020·全国Ⅲ卷)已知$2\tan\theta - \tan\left(\theta + \dfrac{\pi}{4}\right) = 7$,则$\tan\theta =$(
D
)

A.$-2$
B.$-1$
C.$1$
D.$2$
答案: 1.D
2. (2025·江苏南京期末)已知$\cos(\alpha - \beta) = -\dfrac{3}{5}$,$\cos(\alpha + \beta) = \dfrac{1}{5}$,则$\sin\alpha\sin\beta =$(
B
)

A.$-\dfrac{3}{5}$
B.$-\dfrac{2}{5}$
C.$\dfrac{2}{5}$
D.$\dfrac{3}{5}$
答案: 2.B
3. 在$\triangle ABC$中,已知$\sin A = \dfrac{3}{5}$,$\cos B = \dfrac{5}{13}$,则$\cos C$等于(
A
)

A.$\dfrac{16}{65}$
B.$-\dfrac{16}{65}$
C.$\dfrac{16}{65}$或$\dfrac{56}{65}$
D.$-\dfrac{63}{65}$
答案: 3.A
1 (1)(多选题)计算下列几个式子,结果为$\sqrt{3}$的是(
ABD
)

A.$\tan 25° + \tan 35° + \sqrt{3}\tan 25°\tan 35°$
B.$2(\sin 35°\cos 25° + \sin 55°\cos 65°)$
C.$\dfrac{\tan\dfrac{\pi}{6}}{1 - \tan^2\dfrac{\pi}{6}}$
D.$\dfrac{1 + \tan 15°}{1 - \tan 15°}$
答案:
(1)ABD
(2)(2025·四川成都模拟)已知$\alpha$,$\beta$满足$\sin(2\alpha - \beta) = \dfrac{5}{12}$,$\cos(\alpha - \beta)\sin\alpha = \dfrac{1}{3}$,则$\sin\beta$的值为(
C
)

A.$\dfrac{1}{12}$
B.$-\dfrac{1}{12}$
C.$\dfrac{1}{4}$
D.$-\dfrac{1}{4}$
答案:
(2)C

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