《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1. 求下列各三角函数式的值:
(1)$\sin \ 1320°$; (2)$\cos(-\frac{31\pi}{6})$.

答案： 例1:
(1)方法一:$\sin 1320°=\sin(3 × 360° + 240°)$
$=\sin 240°=\sin(180° + 60°)=-\sin 60° = -\frac{\sqrt{3}}{2}$.
方法二:$\sin 1320°=\sin(4 × 360° - 120°)=\sin(-120°)$
$=-\sin(180° - 60°)=-\sin 60° = -\frac{\sqrt{3}}{2}$.
(2)方法一:$\cos(-\frac{31\pi}{6})=\cos\frac{31\pi}{6}=\cos(4\pi + \frac{7\pi}{6})$
$=\cos(\pi + \frac{\pi}{6})=-\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}$.
方法二:$\cos(-\frac{31\pi}{6})=\cos(-6\pi + \frac{5\pi}{6})$
$=\cos(\pi - \frac{\pi}{6})=-\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}$.

对点训练1
(1)求下列各式的值:
$\sin \ 750° = $$\frac{1}{2}$$; \cos(-2040°) = $$-\frac{1}{2}$$;$
(2)计算: $\sin(-\frac{31\pi}{6}) - \cos(-\frac{10\pi}{3}) = $$1$$$.

答案： 对点训练1:
(1)$\frac{1}{2}$ -$\frac{1}{2}$
(2)1
(1)$\sin 750°=\sin(2 × 360° + 30°)=\sin 30° = \frac{1}{2}$.
$\cos(-2040°)=\cos 2040°=\cos(5 × 360° + 240°)$
$=\cos 240°=\cos(180° + 60°)=-\cos 60° = -\frac{1}{2}$.
(2)原式$=-\sin\frac{31\pi}{6}-\cos\frac{10\pi}{3}$
$=-\sin(4\pi + \pi + \frac{\pi}{6})-\cos(2\pi + \pi + \frac{\pi}{3})$
$=\sin\frac{\pi}{6} + \cos\frac{\pi}{3}=\frac{1}{2} + \frac{1}{2}=1$.

例2. (1)计算: $\cos \frac{\pi}{7} + \cos \frac{2\pi}{7} + \cos \frac{3\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{5\pi}{7} + \cos \frac{6\pi}{7} = $$0$$;$
(2)化简: $\frac{\cos(\pi + \alpha)\cos(3\pi - \alpha)}{\sin(\pi + \alpha)\cos(-\alpha - \pi)}$.

答案： 例2:
(1)0
(2)见解析
【解析】
(1)原式$=\cos\frac{\pi}{7} + \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} + \cos(\pi - \frac{3\pi}{7}) + \cos(\pi - \frac{2\pi}{7}) + \cos(\pi - \frac{\pi}{7})=\cos\frac{\pi}{7} + \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} - \cos\frac{3\pi}{7} - \cos\frac{2\pi}{7} - \cos\frac{\pi}{7}=0$.
(2)原式$=\frac{(-\cos\alpha)·(-\cos\alpha)}{(-\sin\alpha)·(-\cos\alpha)}·\frac{\cos\alpha}{\sin\alpha}$

对点训练2
已知$f(\alpha) = \frac{\sin(\pi + \alpha)\cos(2\pi - \alpha)\cos(-\alpha)}{\cos(-\pi - \alpha)\sin(-\pi - \alpha)}$.
(1)化简$f(\alpha)$;
(2)若$\alpha = -\frac{31\pi}{3}$,求$f(\alpha)$的值.

答案： 对点训练2:
(1)$f(\alpha)=\frac{-\sin\alpha\cos\alpha\cos\alpha}{(-\cos\alpha)\sin\alpha}=\cos\alpha$.
(2)$\because -\frac{31\pi}{3}=-6 × 2\pi + \frac{5\pi}{3}$,
$\therefore f(-\frac{31\pi}{3})=\cos(-6 × 2\pi + \frac{5\pi}{3})=\cos\frac{5\pi}{3}=\cos\frac{\pi}{3}=\frac{1}{2}$.