《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 4. (1) 化简:$2\sqrt{1 + \sin 8} + \sqrt{2 + 2\cos 8}$;
(2) 设$\alpha \in (\frac{3\pi}{2}, 2\pi)$, 化简:$\sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\frac{1}{2} + \frac{1}{2}\cos 2\alpha}}$.
【分析】 (1)$1 + \sin 8 = \sin^{2}4 + 2\sin 4\cos 4 + \cos^{2}4 = (\sin 4 + \cos 4)^{2}, 2(1 + \cos 8) = 4\cos^{2}4$.
(2) 连续运用公式:$1 + \cos 2\alpha = 2\cos^{2}\alpha$.
[归纳提升]

答案： 例4：
(1)原式$= 2\sqrt{1 + 2\sin 4\cos 4 + \sqrt{4\cos^2 4}} =$
$2|\sin 4 + \cos 4| + 2|\cos 4|$.
因为$4 \in (\pi, \frac{3\pi}{2})$,
所以$\sin 4 ＜ 0$,$\cos 4 ＜ 0$.
故原式$= -2(\sin 4 + \cos 4) - 2\cos 4 = -2\sin 4 - 4\cos 4$.
(2)因为$\alpha \in (\frac{3\pi}{2}, 2\pi)$,所以$\cos \alpha ＞ 0$,$\cos \frac{\alpha}{2} ＜ 0$.
故原式$= \sqrt{\frac{1}{2} + \frac{1}{2}\sqrt{\cos^2 \alpha}} = \sqrt{\frac{1}{2} + \frac{1}{2}\cos \alpha} =$
$\sqrt{\cos^2 \frac{\alpha}{2}} = |\cos \frac{\alpha}{2}| = -\cos \frac{\alpha}{2}$.

对点训练 4
(1) 化简:$\sqrt{1 - \sin 40^{\circ}} + \sqrt{\frac{1 - \cos 40^{\circ}}{2}}$.
(2) 求证:$\cos^{2}(A + B) - \sin^{2}(A - B) = \cos 2A\cos 2B$.

答案： 对点训练4：
(1)原式$= \sqrt{(\sin 20° - \cos 20°)^2} +$
$\sqrt{1 - (1 - 2\sin^2 20°)}$
$= |\sin 20° - \cos 20°| + \sqrt{\sin^2 20°}$
$= \cos 20° - \sin 20° + \sin 20°$
$= \cos 20°$.
(2)证明：左边$= \frac{1 + \cos(2A + 2B)}{2} - \frac{1 - \cos(2A - 2B)}{2} =$
$\frac{\cos(2A + 2B) + \cos(2A - 2B)}{2} = \frac{1}{2} · (\cos 2A\cos 2B -$
$\sin 2A\sin 2B + \cos 2A\cos 2B + \sin 2A\sin 2B) = \cos 2A\cos 2B =$右边,所以等式成立.