《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2
求证：$\frac{\cos^2 \alpha}{\frac{1}{1 - \tan \frac{\alpha}{2}}} = \frac{1}{4} \sin 2\alpha$.

答案： 对点训练2:【证明】 证法一:左边$= \frac{\cos^{2}\frac{\alpha}{2} - \sin^{2}\frac{\alpha}{2}}{\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}} =$
$\frac{\cos^{2}\frac{\alpha}{2}\sin\frac{\alpha}{2} - \cos\frac{\alpha}{2}\cos\frac{\alpha}{2}\sin\frac{\alpha}{2}}{\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}} =$
$\frac{\cos\frac{\alpha}{2}\sin\frac{\alpha}{2} - \cos\alpha}{\frac{1}{2}\sin\alpha\cos\alpha} = \frac{\frac{1}{2}\sin 2\alpha}{\frac{1}{2}\sin 2\alpha} =$右边.
$\therefore$原式成立.
证法二:左边$= \frac{\frac{\cos^{2}\alpha}{1 + \cos\alpha} - \frac{1 - \cos\alpha}{\sin\alpha}}{\sin\alpha} =$
$= \frac{\frac{1}{2}\sin\alpha\cos\alpha = \frac{1}{4}\sin 2\alpha} =右边$.
$\therefore$原式成立.
证法三:左边$= \frac{\frac{\cos^{2}\frac{\alpha}{2}\tan\frac{\alpha}{2}}{1 - \tan^{2}\frac{\alpha}{2}} = \frac{\frac{1}{2}\cos^{2}\alpha · \frac{2\tan\frac{\alpha}{2}}{1 - \tan^{2}\frac{\alpha}{2}}}{} =$
$\frac{\frac{1}{2}\cos^{2}\alpha · \tan\alpha = \frac{1}{2}\cos\alpha\sin\alpha = \frac{1}{4}\sin 2\alpha} =右边$.
$\therefore$原式成立.

例3. 已知函数 $f(x) = \sqrt{3} \sin(2x - \frac{\pi}{6}) + 2\sin^2(x - \frac{\pi}{12}) (x \in \mathbb{R})$.
(1) 求函数 $f(x)$ 的最小正周期;
(2) 求使函数 $f(x)$ 取得最大值的 $x$ 的集合.

答案： 例3:
(1)$\because f(x) = \sqrt{3}\sin\left( 2x - \frac{\pi}{6} \right) + 2\sin^{2}\left( x - \frac{\pi}{12} \right)$
$= \sqrt{3}\sin\left[ 2\left( x - \frac{\pi}{12} \right) \right] + 1 - \cos\left[ 2\left( x - \frac{\pi}{12} \right) \right]$
$= 2\left\{ \frac{\sqrt{3}}{2}\sin\left[ 2\left( x - \frac{\pi}{12} \right) \right] - \frac{1}{2}\cos\left[ 2\left( x - \frac{\pi}{12} \right) \right] \right\} + 1$
$= 2\sin\left[ 2\left( x - \frac{\pi}{12} \right) - \frac{\pi}{6} \right] + 1$
$= 2\sin\left( 2x - \frac{\pi}{3} \right) + 1$,
$\therefore f(x)$的最小正周期为$T = \frac{2\pi}{2} = \pi$.
(2)当$f(x)$取得最大时$,\sin\left( 2x - \frac{\pi}{3} \right) = 1$,
有$2x - \frac{\pi}{3} = 2k\pi + \frac{\pi}{2}(k \in \mathbf{Z})$,即$x = k\pi + \frac{5\pi}{12}(k \in \mathbf{Z})$,
$\therefore$所求$x$的集合为$\left\{ x \mid x = k\pi + \frac{5\pi}{12},k \in \mathbf{Z} \right\}$

对点训练3
(1) 函数 $f(x) = \sin(2x + \frac{\pi}{3}) - \sin(\frac{\pi}{6} - 2x) (0 \leq x \leq \frac{\pi}{2})$ 的值域为 (A)
A. $[\frac{1 - \sqrt{3}}{2}, \sqrt{2}]$
B. $[\frac{\sqrt{3} - 1}{2}, \sqrt{2}]$
C. $[\frac{1 - \sqrt{3}}{2}, \frac{\sqrt{3} - 1}{2}]$
D. $[\frac{1 - \sqrt{3}}{2}, 2]$
(2) 已知当 $x = x_0$ 时，函数 $f(x) = \sin x - 2\cos x$ 取得最大值，则 $\cos x_0 =$$- \frac{2\sqrt{5}}{5}$.

答案： 对点训练3:
(1)A
(2)$- \frac{2\sqrt{5}}{5}$
(1)$f(x) = \sin\left( 2x + \frac{\pi}{3} \right) -$
$\sin\left( \frac{\pi}{6} - 2x \right) = \sin\left( 2x + \frac{\pi}{3} \right) - \cos\left( 2x + \frac{\pi}{3} \right) = \sqrt{2}\sin\left( 2x + \frac{\pi}{12} \right)$.
又$0 \leq x \leq \frac{\pi}{2}$,所以$\frac{\pi}{12} \leq 2x + \frac{\pi}{12} \leq \frac{13\pi}{12}$,所以$\frac{\sqrt{2} - \sqrt{6}}{4} \leq$
$\sin\left( 2x + \frac{\pi}{12} \right) \leq 1$,所以$\frac{1 - \sqrt{3}}{2} \leq \sqrt{2}\sin\left( 2x + \frac{\pi}{12} \right) \leq \sqrt{2}$,所以函数
$f(x) = \sqrt{2}\sin\left( 2x + \frac{\pi}{12} \right)$的值域为$\left[ \frac{1 - \sqrt{3}}{2},\sqrt{2} \right]$.
(2)由辅助角公式,得$f(x) = \sin x - 2\cos x =$
$\sqrt{5}\left( \frac{\sqrt{5}}{5}\sin x - \frac{2\sqrt{5}}{5}\cos x \right) = \sqrt{5}\sin(x - \varphi)$,其中$\sin\varphi = \frac{2\sqrt{5}}{5},\cos\varphi = \frac{\sqrt{5}}{5}$.
当$x = x_{0}$时,函数$f(x)$取得最大值,即$\sin\left( x_{0} - \varphi \right) = 1$,则
$x_{0} - \varphi = 2k\pi + \frac{\pi}{2}(k \in \mathbf{Z})$,即$x_{0} = 2k\pi + \frac{\pi}{2} + \varphi(k \in \mathbf{Z})$,所以
$\cos x_{0} = \cos\left( 2k\pi + \frac{\pi}{2} + \varphi \right) = - \sin\varphi = - \frac{2\sqrt{5}}{5}$.

1. 若 $\cos \alpha = -\frac{4}{5}$, $\alpha$ 是第三象限角, 则 $\frac{1 + \tan \frac{\alpha}{2}}{1 - \tan \frac{\alpha}{2}} =$ (A)

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

答案： 1.A $\because \alpha$是第三象限角$,\cos\alpha = - \frac{4}{5}$,
$\therefore \sin\alpha = - \frac{3}{5}$.
$\frac{1 + \tan\frac{\alpha}{2}}{1 - \tan\frac{\alpha}{2}} = \frac{\frac{\cos\frac{\alpha}{2} + \sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}}{\frac{\cos\frac{\alpha}{2} - \sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}} = \frac{\cos\frac{\alpha}{2} + \sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2} - \sin\frac{\alpha}{2}} =$
$\frac{\frac{\cos\frac{\alpha}{2} + \sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}}{\frac{\cos\frac{\alpha}{2} + \sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}}} = \frac{1 + \sin\alpha}{\cos\alpha} = \frac{1 - \frac{3}{5}}{- \frac{4}{5}} = - \frac{1}{2}$.故选A.

2. 若 $\theta \in [\frac{\pi}{4}, \frac{\pi}{2}]$, 且 $\sin 2\theta = \frac{3\sqrt{7}}{8}$，则 $\sin \theta =$ (D)

A.$\frac{3}{5}$
B.$\frac{4}{5}$
C.$\frac{\sqrt{7}}{4}$
D.$\frac{3}{4}$

答案： 2.D 本题主要考查简单的三角恒等变换、倍角公式及同角三角函数关系式$\theta \in \left[ \frac{\pi}{4},\frac{\pi}{2} \right]$,$\therefore 2\theta \in \left[ \frac{\pi}{2},\pi \right]$,$\therefore \sin\theta ＞ 0$,
$\cos 2\theta ＜ 0$,$\therefore \cos 2\theta = - \sqrt{1 - \sin^{2}2\theta} = - \frac{1}{8}$,又$\sin^{2}\theta =$
$\frac{1 - \cos 2\theta}{2}$,$\therefore \sin^{2}\theta = \frac{9}{16}$,$\therefore \sin\theta = \frac{3}{4}$,故选D.

3. 设 $-3\pi ＜ \alpha ＜ -\frac{5\pi}{2}$，则化简 $\sqrt{\frac{1 - \cos(\alpha - \pi)}{2}}$ 的结果是 (C)

A.$\sin \frac{\alpha}{2}$
B.$\cos \frac{\alpha}{2}$
C.$-\cos \frac{\alpha}{2}$
D.$-\sin \frac{\alpha}{2}$

答案： 3.C $\because - 3\pi ＜ \alpha ＜ - \frac{5\pi}{2}$,$\therefore - \frac{3\pi}{2} ＜ \frac{\alpha}{2} ＜ - \frac{5\pi}{4}$,$\therefore \cos\frac{\alpha}{2} ＜ 0$,
$\therefore$原式$= \sqrt{\frac{1 + \cos\alpha}{2}} = \cos\frac{\alpha}{2} = - \cos\frac{\alpha}{2}$

4. 设 $a = \frac{1}{2} \cos 6° - \frac{\sqrt{3}}{2} \sin 6°$, $b = 2 \sin 13° \cos 13°$, $c = \sqrt{\frac{1 - \cos 50°}{2}}$，则有 (C)

A.$c ＜ b ＜ a$
B.$a ＜ b ＜ c$
C.$a ＜ c ＜ b$
D.$b ＜ c ＜ a$

答案： 4.C $a = \sin 30^{\circ}\cos 6^{\circ} - \cos 30^{\circ}\sin 6^{\circ} = \sin\left( 30^{\circ} - 6^{\circ} \right) = \sin 24^{\circ}$,
$b = \sin 26^{\circ},c = \sqrt{\frac{2\sin^{2}25^{\circ}}{2}} = \sin 25^{\circ}$,$\therefore b ＞ c ＞ a$.故选C.

5. 在平面直角坐标系 $xOy$ 中，以 $x$ 轴的非负半轴为始边作两个锐角 $\alpha$, $\beta$，它们的终边分别与单位圆相交于 $A$, $B$ 两点，已知 $A$, $B$ 的横坐标分别为 $\frac{1}{3}$, $\frac{2}{3}$，求 $\cos \frac{\alpha}{2} + \sin \frac{\beta}{2} + \tan \frac{\alpha}{2}$ 的值.

答案： 5.依题意,得$\cos\alpha = \frac{1}{3},\cos\beta = \frac{2}{3}$
因为$\alpha,\beta$为锐角,
所以$\cos\frac{\alpha}{2} + \sin\frac{\beta}{2} + \tan\frac{\alpha}{2} =$
$\sqrt{\frac{1 + \cos\alpha}{2}} + \sqrt{\frac{1 - \cos\beta}{2}} + \sqrt{\frac{1 - \cos\alpha}{1 + \cos\alpha}} =$
$\sqrt{\frac{1}{2} + \frac{1}{2}} + \sqrt{\frac{1}{2} - \frac{1}{2}} + \sqrt{\frac{1}{1 + \frac{1}{3}}} = \frac{\sqrt{2} + \sqrt{6}}{2}$