《2026年非常学案高中数学选择性必修第一册人教版》

[典例讲评] 2. 已知 $\tan \alpha = 3$，求下列各式的值：
(1) $\frac{4 \sin \alpha - \cos \alpha}{3 \sin \alpha + 5 \cos \alpha}$；
(2) $\frac{\sin^2 \alpha - 2 \sin \alpha · \cos \alpha - \cos^2 \alpha}{4 \cos^2 \alpha - 3 \sin^2 \alpha}$；
(3) $\frac{3}{4} \sin^2 \alpha + \frac{1}{2} \cos^2 \alpha$.

答案： 典例讲评2.解:
(1)原式$=\frac{4\tan\alpha-1}{3\tan\alpha+5}=\frac{4×3-1}{3×3+5}=\frac{11}{14}$.
(2)原式$=\frac{\tan^2\alpha-2\tan\alpha-1}{4-3\tan^2\alpha}=\frac{9-2×3-1}{4-3×9}=-\frac{2}{23}$.
(3)原式$=\frac{3 - \frac{4}{5}\sin^2\alpha+\frac{1}{2}\cos^2\alpha}{\sin^2\alpha+\cos^2\alpha}$
$=\frac{\frac{3}{4}\tan^2\alpha+\frac{1}{2}}{\tan^2\alpha+1}=\frac{\frac{3}{4}×9+\frac{1}{2}}{9+1}=\frac{29}{40}$.

[典例讲评] 3. 已知 $\sin \alpha + \cos \alpha = -\frac{1}{3}, 0 ＜ \alpha ＜ \pi$.
(1) 求 $\sin \alpha \cos \alpha$ 的值；
(2) 求 $\sin \alpha - \cos \alpha$ 的值.

答案： 典例讲评3.解:
(1)由$\sin\alpha+\cos\alpha=-\frac{1}{3}$，得$(\sin\alpha+\cos\alpha)^2=\frac{1}{9}$，即$\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha=\frac{1}{9}$，
所以$\sin\alpha\cos\alpha=-\frac{4}{9}$.
(2)因为$0＜\alpha＜\pi$，
所以$\sin\alpha＞0,\cos\alpha＜0$，所以$\sin\alpha-\cos\alpha＞0$.
所以$\sin\alpha-\cos\alpha=\sqrt{(\sin\alpha-\cos\alpha)^2}$
$=\sqrt{1-2\sin\alpha\cos\alpha}=\frac{\sqrt{17}}{3}$.

2. (1) 已知 $\alpha \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right], \sin \alpha + \cos \alpha = -\frac{1}{5}$，则 $\tan \alpha = \underline{\hspace{2cm}}$

A.$-\frac{4}{3}$
B.$-\frac{3}{4}$
C.$\frac{3}{4}$
D.$\frac{4}{3}$

答案： 学以致用　2.
(1)A
(2)$\frac{1}{3}$或$-\frac{11}{3}$ $\left[(1)\right.$由题意，$\alpha\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$，
$\sin\alpha+\cos\alpha=-\frac{1}{5}$，$\therefore\cos\alpha＞0,\sin\alpha＜0$，
$(\sin\alpha+\cos\alpha)^2=\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha=1+2\sin\alpha\cos\alpha=\frac{1}{25}$，
解得$2\sin\alpha\cos\alpha=-\frac{24}{25}$，
$\therefore\sin\alpha-\cos\alpha=-\sqrt{\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha}$
$=-\sqrt{1-\left(-\frac{24}{25}\right)}=-\frac{7}{5}$，
$\therefore\begin{cases}\sin\alpha=-\frac{4}{5},\\\cos\alpha=\frac{3}{5},\end{cases}$
$\therefore\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=-\frac{4}{3}$，故选A.
(2)由题中等式易知$\cos\alpha\neq0$，
则$2\cos^2\alpha-3\sin\alpha\cos\alpha=\frac{2\cos^2\alpha-3\sin\alpha\cos\alpha}{\sin^2\alpha+\cos^2\alpha}=\frac{2-3\tan\alpha}{\tan^2\alpha+1}=\frac{9}{10}$，
整理得$9\tan^2\alpha+30\tan\alpha-11=0$，
即$(3\tan\alpha-1)(3\tan\alpha+11)=0$，
解得$\tan\alpha=\frac{1}{3}$或$\tan\alpha=-\frac{11}{3}$.]

(2) 已知 $2 \cos^2 \alpha - 3 \sin \alpha \cos \alpha = \frac{9}{10}$，则 $\tan \alpha = \underline{\hspace{3cm}}$.$\frac{1}{3}$或$-\frac{11}{3}$

答案：
(2)$\frac{1}{3}$或$-\frac{11}{3}$

4. (1) 化简 $\frac{2 \sin^2 \alpha - 1}{1 - 2 \cos^2 \alpha}$.
(2) 求证：$\frac{\sin \alpha - \cos \alpha + 1}{\sin \alpha + \cos \alpha - 1} =$$\frac{1 + \sin \alpha}{\cos \alpha}$.

答案： 典例讲评　4.
(1)1　[原式$=\frac{2\sin^2\alpha-1}{1-2(1-\sin^2\alpha)}=\frac{2\sin^2\alpha-1}{2\sin^2\alpha-1}=1$.]
(2)证明:法一:左边$=\frac{(\sin\alpha-\cos\alpha+1)(\sin\alpha+\cos\alpha+1)}{(\sin\alpha+\cos\alpha-1)(\sin\alpha+\cos\alpha+1)}=\frac{(\sin\alpha+1)^2-\cos^2\alpha}{(\sin\alpha+\cos\alpha)^2-1}$
$=\frac{\sin^2\alpha+2\sin\alpha+1-\cos^2\alpha}{2\sin\alpha\cos\alpha}=\frac{2\sin^2\alpha+2\sin\alpha}{2\sin\alpha\cos\alpha}=\frac{\sin\alpha+1}{\cos\alpha}=$右边.
所以原等式成立.
法二:因为$(\sin\alpha-\cos\alpha+1)\cos\alpha$
$=\sin\alpha\cos\alpha-\cos^2\alpha+\cos\alpha$
$=\sin\alpha\cos\alpha+\cos\alpha-(1-\sin^2\alpha)$
$=\cos\alpha(\sin\alpha+1)-(1+\sin\alpha)(1-\sin\alpha)$
$=(1+\sin\alpha)(\sin\alpha+\cos\alpha-1)$，
且$\sin\alpha+\cos\alpha-1\neq0,\cos\alpha\neq0$，
所以$\frac{\sin\alpha-\cos\alpha+1}{\sin\alpha+\cos\alpha-1}=\frac{1+\sin\alpha}{\cos\alpha}$.