《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》

知识点 1 辅助角公式
$a\sin x + b\cos x =$$\sqrt{a^2 + b^2}\sin(x + \varphi)$$\underline{\hspace{5em}}$. 其中 $\tan \varphi =$$\frac{b}{a}$$\underline{\hspace{3em}}$, $\varphi$ 所在象限由 $a$ 和 $b$ 的符号确定，或者 $\sin \varphi = \frac{b}{\sqrt{a^2 + b^2}}$, $\cos \varphi = \frac{a}{\sqrt{a^2 + b^2}}$

答案： 知识点1 $\sqrt{a^2 + b^2}\sin(x + \varphi)$ $\frac{b}{a}$

知识点 2 常见辅助角结论
(1) $\sin x \pm \cos x = \sqrt{2} \sin \left( x \pm \frac{\pi}{4} \right)$;
(2) $\cos x \pm \sin x = \sqrt{2} \cos \left( x \mp \frac{\pi}{4} \right)$;
(3) $\sin x \pm \sqrt{3} \cos x = 2 \sin \left( x \pm \frac{\pi}{3} \right)$;
(4) $\cos x \pm \sqrt{3} \sin x = 2 \cos \left( x \mp \frac{\pi}{3} \right)$.

答案： 上述辅助角结论正确，可直接应用于解题。

例 1. 化简下列各式：
(1) $\frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x$;
(2) $\sqrt{3} \sin x + \cos x$;
(3) $\sqrt{2} (\sin x - \cos x)$.

答案： 例1:
(1)原式=$\cos \frac{\pi}{3}\cos x - \sin \frac{\pi}{3}\sin x = \cos(\frac{\pi}{3} + x)$.
(2)原式=$2(\frac{\sqrt{3}}{2}\sin x + \frac{1}{2}\cos x) = 2(\sin x\cos \frac{\pi}{6} + \cos x\sin \frac{\pi}{6})$
=$2\sin(x + \frac{\pi}{6})$.
(3)原式=$2(\frac{\sqrt{2}}{2}\sin x - \frac{\sqrt{2}}{2}\cos x) = 2(\sin x\cos \frac{\pi}{4} - \cos x\sin \frac{\pi}{4})$
=$2\sin(x - \frac{\pi}{4})$.

对点训练 1
已知 $\cos \left( x - \frac{\pi}{6} \right) = -\frac{\sqrt{3}}{3}$, 则 $\cos x + \cos \left( x - \frac{\pi}{3} \right) =$ (C)

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

答案： 对点训练1:C $\because \cos(x - \frac{\pi}{6}) = - \frac{\sqrt{3}}{3}$, $\therefore \cos x + \cos(x - \frac{\pi}{3}) = \cos x + \frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x = \frac{3}{2}\cos x + \frac{\sqrt{3}}{2}\sin x = \sqrt{3}(\frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x) = \sqrt{3}\cos(x - \frac{\pi}{6}) = -1$.