《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 2. 求 $y = \sin \left( \frac{\pi}{3} - 2\theta \right) + \cos \left( \frac{\pi}{3} + 2\theta \right)$ 的最大值和周期.
【分析】由函数的解析式化为 $y = A \sin (\omega \theta + \varphi)$ 的形式，然后求其最大值和周期.

答案： 例2:$y = \sin(\frac{\pi}{3} - 2\theta) + \cos(\frac{\pi}{3} + 2\theta)$
=$\sin \frac{\pi}{3}\cos 2\theta - \cos \frac{\pi}{3}\sin 2\theta + \cos \frac{\pi}{3}\cos 2\theta - \sin \frac{\pi}{3}\sin 2\theta$
=$\frac{1 + \sqrt{3}}{2}(\cos 2\theta - \sin 2\theta) = - \frac{\sqrt{2} + \sqrt{6}}{2}\sin(2\theta - \frac{\pi}{4})$,
当$2\theta - \frac{\pi}{4} = 2k\pi - \frac{\pi}{2}$,即$\theta = k\pi - \frac{\pi}{8}(k \in \mathbf{Z})$时,
$y_{\max} = \frac{\sqrt{2} + \sqrt{6}}{2},T = \frac{2\pi}{2} = \pi$,
$\therefore$函数的最大值是$\frac{\sqrt{2} + \sqrt{6}}{2}$,周期为$\pi$.

对点训练 2
已知函数 $f(x) = \sqrt{3} \sin x \sin \left( x + \frac{\pi}{2} \right) - \sin^2 x$.
(1) 求 $f(x)$ 的最小正周期;
(2) 若 $x \in \left( 0, \frac{\pi}{4} \right)$, 求函数 $f(x)$ 的值域.

答案： 对点训练2:
(1)$\because f(x) = \sqrt{3}\sin x\sin(x + \frac{\pi}{2}) - \sin^2x = \sqrt{3}\sin x\cos x - \frac{1 - \cos 2x}{2} = \frac{\sqrt{3}}{2}\sin 2x + \frac{1}{2}\cos 2x - \frac{1}{2} = \sin(2x + \frac{\pi}{6}) - \frac{1}{2}$,
$\therefore f(x)$的最小正周期为$T = \frac{2\pi}{2} = \pi$.
(2)$\because x \in (0, \frac{\pi}{4})$,$\therefore 2x + \frac{\pi}{6} \in (\frac{\pi}{6}, \frac{2\pi}{3})$,
$\therefore \sin(2x + \frac{\pi}{6}) \in (\frac{1}{2}, 1]$,$\therefore f(x) \in (0, \frac{1}{2}]$.
$\therefore f(x)$的值域为$(0, \frac{1}{2}]$.

例 3. 已知三个电流瞬时值的函数式分别是 $I_1 = \sqrt{2} \sin \omega t$, $I_2 = 2 \sin \left( \omega t + \frac{\pi}{4} \right)$, $I_3 = 4 \sin \left( \omega t - \frac{\pi}{4} \right)$, 其中 $\omega$ 为常数, $t$ 为线圈旋转的时间. 求它们合成后的电流瞬时值的函数式, 并求出这个函数的振幅.

答案： 例3:$I = I_1 + I_2 + I_3 = \sqrt{2}\sin \omega t + 2\sin(\omega t + \frac{\pi}{4}) + 4\sin(\omega t - \frac{\pi}{4}) = \sqrt{2}\sin \omega t + 2(\sin \omega t\cos \frac{\pi}{4} + \cos \omega t\sin \frac{\pi}{4}) + 4(\sin \omega t\cos \frac{\pi}{4} - \cos \omega t\sin \frac{\pi}{4})$
=$4\sqrt{2}\sin \omega t - \sqrt{2}\cos \omega t$
=$\sqrt{34}(\frac{4}{\sqrt{17}}\sin \omega t - \frac{1}{\sqrt{17}}\cos \omega t)$
=$\sqrt{34}(\sin \omega t\cos \theta - \cos \omega t\sin \theta) = \sqrt{34}\sin(\omega t - \theta)$.
其中$\sin \theta = \frac{1}{\sqrt{17}},\cos \theta = \frac{4}{\sqrt{17}}$,
$\therefore I = \sqrt{34}\sin(\omega t - \theta)$,且它的振幅是$\sqrt{34}$.