《2025年成才之路高中新课程学习指导高中数学必修第二册北师大版》

1. $\cos 20^{\circ}=$

A.$\cos 30^{\circ} \cos 10^{\circ}-\sin 30^{\circ} \sin 10^{\circ}$
B.$\cos 30^{\circ} \cos 10^{\circ}+\sin 30^{\circ} \sin 10^{\circ}$
C.$\sin 30^{\circ} \cos 10^{\circ}-\sin 10^{\circ} \cos 30^{\circ}$
D.$\cos 30^{\circ} \cos 10^{\circ}-\sin 30^{\circ} \cos 10^{\circ}$

答案： 1.B $\cos 20° = \cos(30° - 10°) = \cos 30° \cos 10° + \sin 30° \sin 10°$,
故选B.

2. $\cos (\alpha-85^{\circ}) \cos (35^{\circ}+\alpha)+\sin (\alpha-85^{\circ}) \sin (35^{\circ}+\alpha)$的值为(A)

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

答案： 2.A 原式$ = \cos[(\alpha - 85°) - (35° + \alpha)] = \cos(-120°) = \cos 120° = - \cos 60° = - \frac{1}{2}$,故选A.

3. (多选)下列说法中，正确的是(AD)

A.存在$\alpha,\beta$的值，使$\cos (\alpha+\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta$
B.不存在无穷多个$\alpha,\beta$的值，使$\cos (\alpha+\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta$
C.对于任意的$\alpha,\beta$,都有$\cos (\alpha+\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta$
D.不存在$\alpha,\beta$的值，使$\cos (\alpha+\beta) \neq \cos \alpha \cos \beta-\sin \alpha \sin \beta$

答案： 3.AD 令$\alpha = \beta = 0$,则$\cos(\alpha + \beta) = 1$,$\cos \alpha \cos \beta + \sin \alpha \sin \beta = 1$,此时$\cos(\alpha + \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$,故A正确;令$\alpha = \beta = 2k\pi(k \in \mathbf{Z})$,$\cos(\alpha + \beta) = 1$,$\cos \alpha \cos \beta + \sin \alpha \sin \beta = 1$,此时$\cos(\alpha + \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$,故B错误;由两角和的余弦公式可知,对于任意的$\alpha$和$\beta$,$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$,故C错误;不存在$\alpha,\beta$的值,使$\cos(\alpha + \beta) \neq \cos \alpha \cos \beta - \sin \alpha \sin \beta$,若存在$\alpha$和$\beta$,则与两角和的余弦公式矛盾,故D正确.故选AD.

4. $\sin (\alpha-\beta) \sin \alpha+\cos (\alpha-\beta) \cos \alpha=$$\cos \beta$.

答案： 4.$\cos \beta$ 原式$ = \cos[(\alpha - \beta) - \alpha] = \cos( - \beta) = \cos \beta$.

5. 已知$\sin (\frac{3 \pi}{2}+\alpha)=\frac{3}{5},\alpha \in(\pi,\frac{3 \pi}{2})$,求$\sin (\frac{\pi}{6}+\alpha)$的值.

答案： 5.$\because \sin(\frac{3\pi}{2} + \alpha) = - \cos \alpha = - \frac{3}{5}$,
$\therefore \cos \alpha = - \frac{3}{5}$
又$\alpha \in (\pi, \frac{3\pi}{2})$,$\therefore \sin \alpha = - \frac{4}{5}$,
$\therefore \sin(\frac{\pi}{6} + \alpha) = \cos[\frac{\pi}{2} - (\frac{\pi}{6} + \alpha)] = \cos(\frac{\pi}{3} - \alpha)$
$ = \cos \frac{\pi}{3} · \cos \alpha + \sin \frac{\pi}{3} · \sin \alpha$
$ = \frac{1}{2} × ( - \frac{3}{5}) + \frac{\sqrt{3}}{2} × ( - \frac{4}{5}) = - \frac{3 + 4\sqrt{3}}{10}$

知识点1 两角和与差的正弦公式

答案： 知识点1 $S_{\alpha + \beta}$ $\sin \alpha \cos \beta + \cos \alpha \sin \beta$ $S_{\alpha - \beta}$ $\sin \alpha \cos \beta$
$-$ $\cos \alpha \sin \beta$

知识点2 两角和与差的正切公式

答案： 知识点2 $\tan \alpha + \tan \beta$
$1 - \tan \alpha \tan \beta$