《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》

例3. (1)已知$\cos(\pi + \alpha) = -\frac{1}{2}$,则$\cos(\pi - \alpha)= $$-\frac{1}{2}$$$.
(2)已知$\sin(\frac{\pi}{6} - \alpha) = \frac{2\sqrt{2}}{3}, \cos(\frac{\pi}{6} - \alpha) = \frac{1}{3}$,求$\cos(\frac{5\pi}{6} + \alpha) · \sin(\frac{19\pi}{6} - \alpha) = $$\frac{2\sqrt{2}}{9}$$$.

答案： 例3:
(1)$-\frac{1}{2}$
(2)$\frac{2\sqrt{2}}{9}$
(1)方法一:因为$\cos(\pi + \alpha)=-\cos\alpha = -\frac{1}{2}$,所以$\cos\alpha = \frac{1}{2}$,则$\cos(\pi - \alpha)=-\cos\alpha = -\frac{1}{2}$.
方法二:记$\pi + \alpha = x$,$\pi - \alpha = y$,问题变为已知$\cos x$求$\cos y$.
显然$x + y = 2\pi$(目的是消去$\alpha$),
所以$\cos y=\cos(2\pi - x)=\cos(-x)=\cos x = -\frac{1}{2}$.
(2)$\cos(\frac{5\pi}{6} + \alpha)·\sin(\frac{19}{6}\pi - \alpha)=\cos[\pi - (\frac{\pi}{6} - \alpha)]·\sin(3\pi + \frac{\pi}{6} - \alpha)=-\cos(\frac{\pi}{6} - \alpha)·\sin[\pi + (\frac{\pi}{6} - \alpha)]=-\cos(\frac{\pi}{6} - \alpha)·[-\sin(\frac{\pi}{6} - \alpha)]=\cos(\frac{\pi}{6} - \alpha)·\sin(\frac{\pi}{6} - \alpha)=-\frac{1}{2}[-\frac{2\sqrt{2}}{3} × \frac{1}{3}]= \frac{2\sqrt{2}}{9}$.

对点训练3
(1)已知$\cos(53° - \alpha) = \frac{1}{5}$,则$\cos(127° + \alpha) =$ (D)
A. $\pm \frac{1}{5}$
B. $\frac{2\sqrt{6}}{5}$
C. $\frac{1}{5}$
D. $-\frac{1}{5}$
(2)已知$\cos(\frac{\pi}{6} - \alpha) = \frac{\sqrt{3}}{3}$,求$\cos(\frac{7\pi}{6} - \alpha) - \sin^2(\alpha - \frac{13\pi}{6})$的值(注: $\sin^2(\frac{\pi}{6} - \alpha) + \cos^2(\frac{\pi}{6} - \alpha) = 1, \alpha$为任意角).

答案： 对点训练3:
(1)D
(2)见解析
【解析】
(1)因为$\cos(53° - \alpha)=\frac{1}{5}$,所以$\cos(127° + \alpha)=\cos[180° - (53° - \alpha)]=-\cos(53° - \alpha)=-\frac{1}{5}$. 故选D.
(2)$\cos(\frac{7\pi}{6} - \alpha) - \sin^2(\alpha - \frac{13\pi}{6})=\cos[\pi + (\frac{\pi}{6} - \alpha)] - \sin^2[(\alpha - \frac{\pi}{6}) - 2\pi]$
$=-\cos(\frac{\pi}{6} - \alpha) - \sin^2(\frac{\pi}{6} - \alpha)=-\frac{\sqrt{3}}{3} - \frac{2}{3}=-\frac{\sqrt{3} + 2}{3}$.

1. $\sin \ 1215° =$ (A)

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

答案： 1 A $\sin 1215°=\sin(3 × 360° + 135°)=\sin 135°=\sin(180° - 45°)=\sin 45°=\frac{\sqrt{2}}{2}$

2. 若$\cos \ \alpha = m$,则$\cos(-\alpha) =$ (A)

A.$m$
B.$-m$
C.$|m|$
D.$m^2$

答案： 2 A $\cos(-\alpha)=\cos\alpha = m$.

3. 在平面直角坐标系$xOy$中,角$\alpha$与角$\beta$均以$Ox$为始边,它们的终边关于$y$轴对称. 若$\sin \ \alpha = \frac{1}{3}$,则$\sin \ \beta = $$\frac{1}{3}$$$.

答案： 3 $\frac{1}{3}$ 由题意可知$\alpha + \beta = \pi + 2k\pi$,$k \in \mathbf{Z}$.
$\because \sin\alpha = \frac{1}{3}$,
$\therefore \sin\beta = \sin(\pi + 2k\pi - \alpha)=\sin\alpha = \frac{1}{3}$

4. 化简: $\frac{\cos(180° + \alpha)\sin(\alpha + 360°)}{\sin(-\alpha - 180°)\cos(-180° - \alpha)}$.

答案： 4 原式$=\frac{(-\cos\alpha)\sin\alpha}{-\sin(\alpha + 180°)\cos(180° + \alpha)}$
$=\frac{\sin\alpha\cos\alpha}{\sin(\alpha + 180°)\cos(180° + \alpha)}$
$=\frac{\sin\alpha\cos\alpha}{(-\sin\alpha)(-\cos\alpha)} = 1$.