《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 3
已知两个电流瞬时值函数式分别是 $I_1 = 12 \sin \left( \omega t - \frac{\pi}{6} \right)$, $I_2 = 10 \sin \left( \omega t + \frac{\pi}{6} \right)$, 求合成后的电流 $I = I_1 + I_2$.

答案： 对点训练3:$I = I_1 + I_2 = 12\sin(\omega t - \frac{\pi}{6}) + 10\sin(\omega t + \frac{\pi}{6})$
=$12(\sin \omega t\cos \frac{\pi}{6} - \cos \omega t\sin \frac{\pi}{6}) + 10(\sin \omega t\cos \frac{\pi}{6} + \cos \omega t\sin \frac{\pi}{6})$
=$6\sqrt{3}\sin \omega t - 6\cos \omega t + 5\sqrt{3}\sin \omega t + 5\cos \omega t$
=$11\sqrt{3}\sin \omega t - \cos \omega t$
=$\sqrt{364}(\frac{11\sqrt{3}}{\sqrt{364}}\sin \omega t - \frac{1}{\sqrt{364}}\cos \omega t)$
=$2\sqrt{91}\sin(\omega t - \varphi)$,
其中$\cos \varphi = \frac{11\sqrt{3}}{2\sqrt{91}},\sin \varphi = \frac{1}{2\sqrt{91}}$,

1. 计算 $\sin 7° \cos 23° + \sin 83° \cos 67°$ 的值为 (B)

A.$-\frac{1}{2}$
B.$\frac{1}{2}$
C.$\frac{\sqrt{3}}{2}$
D.$-\frac{\sqrt{3}}{2}$

答案： 1.B 原式=$\cos 83^{\circ}\cos 23^{\circ} + \sin 83^{\circ}\sin 23^{\circ} = \cos(83^{\circ} - 23^{\circ}) = \cos 60^{\circ} = \frac{1}{2}$

2. $\cos \alpha - \sqrt{3} \sin \alpha$ 化简的结果可以是 (B)

A.$\frac{1}{2} \cos \left( \frac{\pi}{6} - \alpha \right)$
B.$2 \cos \left( \frac{\pi}{3} + \alpha \right)$
C.$\frac{1}{2} \cos \left( \frac{\pi}{3} - \alpha \right)$
D.$2 \cos \left( \frac{\pi}{6} - \alpha \right)$

答案： 2.B $\cos \alpha - \sqrt{3}\sin \alpha = 2(\frac{1}{2}\cos \alpha - \frac{\sqrt{3}}{2}\sin \alpha) = 2\cos(\frac{\pi}{3} + \alpha)$.

3. 求值: $\frac{1}{2} \cos \frac{\pi}{12} + \frac{\sqrt{3}}{2} \sin \frac{\pi}{12} = \underline{\hspace{5em}}$

答案： 3.$\frac{\sqrt{2}}{2}\frac{1}{2}\cos \frac{\pi}{12} + \frac{\sqrt{3}}{2}\sin \frac{\pi}{12} = \cos(\frac{\pi}{3} - \frac{\pi}{12}) = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$.

4. $f(x) = | \sin x - \cos x |$ 的最小正周期为$\pi$$\underline{\hspace{5em}}$.

答案： 4.$\pi$函数$f(x) = |\sin x - \cos x| = \sqrt{2}\sin(x - \frac{\pi}{4})$的最小正周期为$\frac{1}{2} · 2\pi = \pi$.

5. 已知 $f(x) = \cos x + \cos \left( x + \frac{\pi}{3} \right)$.
(1) 求 $f(x)$ 的最小正周期;
(2) 求 $f(x)$ 的单调递增区间.

答案： 5.$f(x) = \cos x + \cos x\cos \frac{\pi}{3} - \sin x\sin \frac{\pi}{3} = \frac{3}{2}\cos x - \frac{\sqrt{3}}{2}\sin x = \sqrt{3}(\frac{\sqrt{3}}{2}\cos x - \frac{1}{2}\sin x) = \sqrt{3}\cos(x + \frac{\pi}{6})$.
(1)$T = 2\pi$.
(2)由$-\pi + 2k\pi \leqslant x + \frac{\pi}{6} \leqslant 2k\pi(k \in \mathbf{Z})$得$-\frac{7\pi}{6} + 2k\pi \leqslant x \leqslant -\frac{\pi}{6} + 2k\pi(k \in \mathbf{Z})$,即$f(x)$的单调递增区间为$[-\frac{7\pi}{6} + 2k\pi, -\frac{\pi}{6} + 2k\pi](k \in \mathbf{Z})$.

知识点 积化和差、和差化积公式

答案： 由于题目未给出具体题目内容，无法给出具体答案选项。