《2025年更高更妙的高中数学思想与方法高中数学必修第一册》

【例 6】已知$\alpha \in \left(0,\dfrac{\pi }{2}\right)$且满足$\tan \alpha = 3$，求下列各式的值：
(1)$\dfrac{\sin \alpha - 4\cos \alpha }{5\sin \alpha + 2\cos \alpha }$；
(2)$\sin ^{2}\alpha + 2\sin \alpha \cos \alpha - 3\cos ^{2}\alpha$。

答案： 解答
(1)
方法1
已知$\tan\alpha = 3$，$\alpha \in \left(0,\dfrac{\pi}{2}\right)$，由$\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = 3$及$\sin^2\alpha + \cos^2\alpha = 1$，得：
$\sin\alpha = \dfrac{3\sqrt{10}}{10},\quad \cos\alpha = \dfrac{\sqrt{10}}{10}$
代入原式：
$\dfrac{\sin\alpha - 4\cos\alpha}{5\sin\alpha + 2\cos\alpha} = \dfrac{\dfrac{3\sqrt{10}}{10} - 4×\dfrac{\sqrt{10}}{10}}{5×\dfrac{3\sqrt{10}}{10} + 2×\dfrac{\sqrt{10}}{10}} = \dfrac{-\dfrac{\sqrt{10}}{10}}{\dfrac{17\sqrt{10}}{10}} = -\dfrac{1}{17}$
方法2
分子分母同除以$\cos\alpha$（$\cos\alpha \neq 0$）：
$\dfrac{\sin\alpha - 4\cos\alpha}{5\sin\alpha + 2\cos\alpha} = \dfrac{\tan\alpha - 4}{5\tan\alpha + 2} = \dfrac{3 - 4}{5×3 + 2} = \dfrac{-1}{17} = -\dfrac{1}{17}$
(2)
方法1
由
(1)知$\sin\alpha = \dfrac{3\sqrt{10}}{10}$，$\cos\alpha = \dfrac{\sqrt{10}}{10}$，代入原式：
$\sin^2\alpha + 2\sin\alpha\cos\alpha - 3\cos^2\alpha = \left(\dfrac{3\sqrt{10}}{10}\right)^2 + 2×\dfrac{3\sqrt{10}}{10}×\dfrac{\sqrt{10}}{10} - 3\left(\dfrac{\sqrt{10}}{10}\right)^2$
$= \dfrac{90}{100} + \dfrac{60}{100} - \dfrac{30}{100} = \dfrac{120}{100} = \dfrac{6}{5}$
方法2
分子分母同除以$\cos^2\alpha$（$\cos\alpha \neq 0$），并利用$\sin^2\alpha + \cos^2\alpha = 1$：
$\sin^2\alpha + 2\sin\alpha\cos\alpha - 3\cos^2\alpha = \dfrac{\sin^2\alpha + 2\sin\alpha\cos\alpha - 3\cos^2\alpha}{\sin^2\alpha + \cos^2\alpha} = \dfrac{\tan^2\alpha + 2\tan\alpha - 3}{\tan^2\alpha + 1}$
代入$\tan\alpha = 3$：
$\dfrac{3^2 + 2×3 - 3}{3^2 + 1} = \dfrac{9 + 6 - 3}{10} = \dfrac{12}{10} = \dfrac{6}{5}$
答案
(1) $-\dfrac{1}{17}$
(2) $\dfrac{6}{5}$

【例 7】已知$\sin \alpha + \cos \alpha = \dfrac{\sqrt{2}}{2}$，$\alpha \in (0,\pi )$，求$\tan \alpha$的值。

答案： 已知$\sin\alpha + \cos\alpha = \dfrac{\sqrt{2}}{2}$，$\alpha \in (0,\pi)$。
1. 两边平方得：$(\sin\alpha + \cos\alpha)^2 = \left(\dfrac{\sqrt{2}}{2}\right)^2$，即$1 + 2\sin\alpha\cos\alpha = \dfrac{1}{2}$，解得$2\sin\alpha\cos\alpha = -\dfrac{1}{2}$。
2. 因为$\alpha \in (0,\pi)$，且$\sin\alpha\cos\alpha = -\dfrac{1}{4} ＜ 0$，所以$\sin\alpha ＞ 0$，$\cos\alpha ＜ 0$，则$\sin\alpha - \cos\alpha ＞ 0$。
3. $(\sin\alpha - \cos\alpha)^2 = 1 - 2\sin\alpha\cos\alpha = 1 - \left(-\dfrac{1}{2}\right) = \dfrac{3}{2}$，故$\sin\alpha - \cos\alpha = \dfrac{\sqrt{6}}{2}$。
4. 联立$\begin{cases}\sin\alpha + \cos\alpha = \dfrac{\sqrt{2}}{2} \\ \sin\alpha - \cos\alpha = \dfrac{\sqrt{6}}{2}\end{cases}$，解得$\sin\alpha = \dfrac{\sqrt{2} + \sqrt{6}}{4}$，$\cos\alpha = \dfrac{\sqrt{2} - \sqrt{6}}{4}$。
5. $\tan\alpha = \dfrac{\sin\alpha}{\cos\alpha} = \dfrac{\dfrac{\sqrt{2} + \sqrt{6}}{4}}{\dfrac{\sqrt{2} - \sqrt{6}}{4}} = \dfrac{\sqrt{2} + \sqrt{6}}{\sqrt{2} - \sqrt{6}} = -(2 + \sqrt{3})$。
结论：$\tan\alpha = -(2 + \sqrt{3})$。

【例 8】已知$\theta \in (0,\pi )$，且$\sin \theta$，$\cos \theta$是方程$25x^{2} - 5x - 12 = 0$的两个实根，求$\sin ^{3}\theta + \cos ^{3}\theta$和$\tan \theta - \dfrac{1}{\tan \theta}$的值。

答案： 由韦达定理得：$\sin\theta + \cos\theta = \frac{1}{5}$，$\sin\theta \cos\theta = -\frac{12}{25}$。
1. $\sin^3\theta + \cos^3\theta = (\sin\theta + \cos\theta)(\sin^2\theta - \sin\theta \cos\theta + \cos^2\theta) = \frac{1}{5}\left(1 - \left(-\frac{12}{25}\right)\right) = \frac{1}{5} × \frac{37}{25} = \frac{37}{125}$。
2. $\theta \in (0,\pi)$，$\sin\theta \cos\theta ＜ 0$，则$\sin\theta ＞ 0$，$\cos\theta ＜ 0$。$\sin\theta - \cos\theta = \sqrt{(\sin\theta - \cos\theta)^2} = \sqrt{1 - 2\sin\theta \cos\theta} = \sqrt{1 + \frac{24}{25}} = \frac{7}{5}$。
$\tan\theta - \frac{1}{\tan\theta} = \frac{\sin^2\theta - \cos^2\theta}{\sin\theta \cos\theta} = \frac{(\sin\theta + \cos\theta)(\sin\theta - \cos\theta)}{\sin\theta \cos\theta} = \frac{\frac{1}{5} × \frac{7}{5}}{-\frac{12}{25}} = -\frac{7}{12}$。
$\sin^3\theta + \cos^3\theta = \frac{37}{125}$；$\tan\theta - \frac{1}{\tan\theta} = -\frac{7}{12}$。