《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》

对点训练3
已知$\cos \alpha=\frac{1}{7}, \cos (\alpha+\beta)=-\frac{11}{14}$, 且$\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, 求$\beta$的值.

答案： 对点训练3:$\alpha, \beta \in (0, \frac{\pi}{2})$,$\therefore 0 ＜ \alpha + \beta ＜ \pi$,
$\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - (-\frac{11}{14})^2} = \frac{5\sqrt{3}}{14}$,$\sin(\alpha + \beta) = \sqrt{1 - (-\frac{11}{14})^2} = \frac{5\sqrt{3}}{14}$,
$\therefore \cos \beta = \cos[(\alpha + \beta) - \alpha] = \cos(\alpha + \beta)\cos \alpha + \sin(\alpha + \beta)\sin \alpha$
=$\frac{11}{14} × \frac{1}{7} + \frac{5\sqrt{3}}{14} × \frac{4\sqrt{3}}{7} - \frac{1}{2}$,
$\therefore \beta = \frac{\pi}{3}$.

1. 计算$\sin 43^{\circ} \cos 13^{\circ}-\cos 43^{\circ} \sin 13^{\circ}$的结果等于(A)

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

答案： 1 A $\because \sin 43° \cos 13° - \cos 43° \sin 13° = \sin(43° - 13°) =$
$\sin 30° = \frac{1}{2}$.故选A.

2. 已知$\cos \alpha=\frac{12}{13}, \alpha \in\left(\frac{3 \pi}{2}, 2 \pi\right)$, 则$\cos \left(\alpha-\frac{\pi}{4}\right)=$(D)

A.$\frac{5 \sqrt{2}}{13}$
B.$\frac{7 \sqrt{2}}{13}$
C.$\frac{17 \sqrt{2}}{26}$
D.$\frac{7 \sqrt{2}}{26}$

答案： 2 D 因为 $\cos \alpha = \frac{12}{13}$,$\alpha \in (\frac{3\pi}{2}, 2\pi)$,可得 $\sin \alpha = -\sqrt{1 - \cos^2 \alpha}$
=$-\frac{5}{13}$,则 $\cos(\alpha - \frac{\pi}{4}) = \frac{\sqrt{2}}{2} \cos \alpha + \frac{\sqrt{2}}{2} \sin \alpha = \frac{\sqrt{2}}{2} × \frac{12}{13} + \frac{\sqrt{2}}{2} ×$
($-\frac{5}{13}) = \frac{7\sqrt{2}}{26}$.故选D.

3. 若$\tan \alpha=2, \tan \beta=\frac{1}{3}$, 则$\tan (\alpha+\beta)$等于(C)

A.$1$
B.$-1$
C.$7$
D.$-7$

答案： 3 C $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha · \tan \beta} = \frac{2 + \frac{1}{3}}{1 - 2 × \frac{1}{3}} = 7$,故
选C.

4. 已知$0＜\beta＜\alpha＜\frac{\pi}{2}$, 且$\cos (\alpha-\beta)=\frac{12}{13}, \cos 2 \beta=\frac{3}{5}$, 则$\sin (\alpha+\beta)=$(A)

A.$\frac{63}{65}$
B.$\frac{33}{65}$
C.$\frac{48}{65}$
D.$\frac{16}{65}$

答案： 4 A 因为$0 ＜ \beta ＜ \alpha ＜ \frac{\pi}{2}$ 所以$0 ＜ \alpha - \beta ＜ \frac{\pi}{2}$,又 $\cos(\alpha - \beta) =$
$\frac{12}{13}$,所以 $\sin(\alpha - \beta) = \sqrt{1 - \cos^2(\alpha - \beta)} = \sqrt{1 - (\frac{12}{13})^2} = \frac{5}{13}$,
因为$0 ＜ \beta ＜ \frac{\pi}{2}$,所以$0 ＜ 2\beta ＜ \pi$,因为 $\cos 2\beta = \frac{3}{5}$,所以 $\sin 2\beta$
=$\sqrt{1 - \cos^2 2\beta} = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$,所以 $\sin(\alpha + \beta) = \sin[(\alpha$
$- \beta) + 2\beta] = \sin(\alpha - \beta)\cos 2\beta + \cos(\alpha - \beta)\sin 2\beta = \frac{5}{13} × \frac{3}{5} +$
$\frac{12}{13} × \frac{4}{5} = \frac{63}{65}$.故选A.

5. 如图, 在平面直角坐标系$x O y$中, 以$O x$轴为始边作两个锐角$\alpha 、 \beta$, 它们的终边分别与单位圆交于$A 、 B$两点. 已知$A 、 B$的横坐标分别为$\frac{\sqrt{2}}{10} 、 \frac{2 \sqrt{5}}{5}$.
(1) 求$\tan (\alpha+\beta)$的值;
(2) 求$\alpha+2 \beta$的值.

答案： 5
(1)由三角函数的定义可知 $\cos \alpha = \frac{\sqrt{2}}{10}$,$\cos \beta = \frac{2\sqrt{5}}{5}$;所以 $\sin \alpha$
=$\frac{7\sqrt{2}}{10}$,$\sin \beta = \frac{\sqrt{5}}{5}$,
所以 $\tan \alpha = 7$,$\tan \beta = \frac{1}{2}$,
于是 $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = -3$.
(2)$\tan(\alpha + 2\beta) = \tan[(\alpha + \beta) + \beta]$
=$\frac{\tan(\alpha + \beta) + \tan \beta}{1 - \tan(\alpha + \beta)\tan \beta} = \frac{-3 + \frac{1}{2}}{1 - (-3) × \frac{1}{2}} = -1$.
又$0 ＜ \alpha ＜ \frac{\pi}{2}$,所以$0 ＜ \alpha + 2\beta ＜ \frac{3\pi}{2}$,所以 $\alpha + 2\beta = \frac{3\pi}{4}\pi$.