《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. 已知 $\sin \theta = \frac{4}{5}$，且 $\frac{5\pi}{2} ＜ \theta ＜ 3\pi$，求 $\sin \frac{\theta}{2}$, $\cos \frac{\theta}{2}$, $\tan \frac{\theta}{2}$.
【分析】 已知条件中的角 $\theta$ 与所求角中的 $\frac{\theta}{2}$ 成二倍关系，从而选择半角公式求值.

答案： 例1:
∵$\sin\theta = \frac{4}{5}, \frac{5\pi}{2} ＜ \theta ＜ 3\pi$,
∴$\cos\theta = - \sqrt{1 - \sin^{2}\theta} = - \frac{3}{5}$.
$\frac{5\pi}{4} ＜ \frac{\theta}{2} ＜ \frac{3\pi}{2}$,
∴$\sin\frac{\theta}{2} = - \sqrt{\frac{1 - \cos\theta}{2}} = - \frac{2\sqrt{5}}{5}$
$\cos\frac{\theta}{2} = - \sqrt{\frac{1 + \cos\theta}{2}} = - \sqrt{\frac{\sqrt{5}}{5}, \tan\frac{\theta}{2}} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = 2$.

对点训练1
设 $\pi ＜ \theta ＜ 2\pi$, $\cos \frac{\theta}{2} = -\frac{3}{5}$，求：
(1) $\sin \theta$ 的值;
(2) $\cos \theta$ 的值;
(3) $\sin^2 \frac{\theta}{4}$ 的值.

答案： 对点训练1:
(1)$\because \pi ＜ \theta ＜ 2\pi$,$\therefore \frac{\pi}{2} ＜ \frac{\theta}{2} ＜ \pi$,
又$\cos\frac{\theta}{2} = - \frac{3}{5}$,$\therefore \sin\frac{\theta}{2} = \sqrt{1 - \cos^{2}\frac{\theta}{2}} =$
$\sqrt{1 - \left( - \frac{3}{5} \right)^{2}} = \frac{4}{5}$,
$\therefore \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2 × \left( - \frac{3}{5} \right) × \frac{4}{5} = - \frac{24}{25}$.
(2)$\cos\theta = 2\cos^{2}\frac{\theta}{2} - 1 = 2 × \left( - \frac{3}{5} \right)^{2} - 1 = - \frac{7}{25}$.
(3)$\sin^{2}\frac{\theta}{4} = \frac{1 - \cos\frac{\theta}{2}}{2} = \frac{1 - \left( - \frac{3}{5} \right)}{2} = \frac{4}{5}$.

例2. 化简：$\frac{(1 - \sin \alpha - \cos \alpha)(\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2})}{\sqrt{2 - 2\cos \alpha}} (-\pi ＜ \alpha ＜ 0)$.

答案： 例2:原式$= \frac{\left( \frac{2\sin^{2}\frac{\alpha}{2} - 2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2} \right)\left( \sin\frac{\alpha}{2} + \cos\frac{\alpha}{2} \right)}{\sqrt{\frac{2 × 2\sin^{2}\frac{\alpha}{2}}{2}}}$
$= \frac{2\sin\frac{\alpha}{2}\left( \sin\frac{\alpha}{2} - \cos\frac{\alpha}{2} \right)\left( \sin\frac{\alpha}{2} + \cos\frac{\alpha}{2} \right)}{2^{\left| \sin\frac{\alpha}{2} \right|}}$
$= \frac{\frac{\sin\frac{\alpha}{2}\left( \sin^{2}\frac{\alpha}{2} - \cos^{2}\frac{\alpha}{2} \right)}{\sin\frac{\alpha}{2}}}{\sin\frac{\alpha}{2}}$
因为$ - \pi ＜ \alpha ＜ 0$,所以$- \frac{\pi}{2} ＜ \frac{\alpha}{2} ＜ 0$,
所以$\sin\frac{\alpha}{2} ＜ 0$,所以原式$= \frac{- \sin\frac{\alpha}{2}\cos\alpha}{- \sin\frac{\alpha}{2}} = \cos\alpha$.