《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 1. (1)若$\sin A=\frac{4}{5}$,则$\frac{5\sin A + 8}{15\cos A - 7}$的值为6或$-\frac{3}{4}$；
(2)已知$\theta\in(0,\pi)$,$\sin\theta+\cos\theta=\frac{\sqrt{3}-1}{2}$,则$\tan\theta$的值为$-\sqrt{3}$；
(3)化简$\frac{\sqrt{1 - 2\sin20^{\circ}·\cos20^{\circ}}}{\sin160^{\circ}-\sqrt{1-\sin^{2}20^{\circ}}}$.
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答案： 例1:
(1)6或$-\frac{3}{4}$
(2)$-\sqrt{3}$
(3)见解析
【解析】
(1)因为$\sin A=\frac{4}{5}＞0$，所以$A$是第一或第二象限角.
当$A$是第一象限角时，$\cos A=\sqrt{1-\sin^{2}A}=\frac{3}{5}$，所以
$\frac{5\sin A+8}{15\cos A-7}=\frac{5×\frac{4}{5}+8}{15×\frac{3}{5}-7}=6$；
当$A$是第二象限角时，$\cos A=-\sqrt{1-\sin^{2}A}=-\frac{3}{5}$，
所以$\frac{5\sin A+8}{15\cos A-7}=\frac{5×\frac{4}{5}+8}{15×(-\frac{3}{5})-7}=-\frac{3}{4}$.
(2)方法一:将$\sin\theta+\cos\theta=\frac{\sqrt{3}-1}{2}$两边平方，得$1+2\sin\theta\cos\theta$
$=1-\frac{\sqrt{3}}{2}$，即$\sin\theta\cos\theta=-\frac{\sqrt{3}}{4}$，易知$\theta\neq\frac{\pi}{2}$
故$\sin\theta\cos\theta=\frac{\sin\theta\cos\theta}{\sin^{2}\theta+\cos^{2}\theta}=\frac{\tan\theta}{\tan^{2}\theta+1}=-\frac{\sqrt{3}}{4}$，
解得$\tan\theta=-\sqrt{3}$或$\tan\theta=-\frac{\sqrt{3}}{3}$
$\because\theta\in(0,\pi),\sin\theta\cos\theta=-\frac{\sqrt{3}}{4}＜0,\therefore\theta\in(\frac{\pi}{2},\pi)$，由$\sin\theta$
$+\cos\theta=\frac{\sqrt{3}-1}{2}＞0$可知$\sin\theta＞-\cos\theta$，即$|\sin\theta|＞|\cos\theta|$，故$\theta$
$\in(\frac{\pi}{2},\frac{3\pi}{4})$，则$\tan\theta＜-1,\therefore\tan\theta=-\sqrt{3}$.
方法二:本题若利用$\sin\theta\pm\cos\theta$与$\sin\theta\cos\theta$之间的关系，则会得到更为简捷的解法.
由$\sin\theta+\cos\theta=\frac{\sqrt{3}-1}{2}$ ①，得$\sin\theta\cos\theta=-\frac{\sqrt{3}}{4}＜0$，
又$\theta\in(0,\pi)，\therefore\sin\theta＞0,\cos\theta＜0,\therefore\sin\theta-\cos\theta＞0$.
又$(\sin\theta-\cos\theta)^{2}=1-2\sin\theta\cos\theta=1+\frac{\sqrt{3}}{2}=\frac{(\sqrt{3}+1)^{2}}{4}$，
$\therefore\sin\theta-\cos\theta=\frac{\sqrt{3}+1}{2}$ ②.
联立①②解得$\sin\theta=\frac{\sqrt{3}}{2},\cos\theta=-\frac{1}{2},\therefore\tan\theta=-\sqrt{3}$.
(3)原式$=\frac{\sqrt{\cos^{2}20^{\circ}-2\sin20^{\circ}\cos20^{\circ}+\sin^{2}20^{\circ}}}{\sin(180^{\circ}-20^{\circ})-\cos20^{\circ}}$
$=\frac{|\cos20^{\circ}-\sin20^{\circ}|}{\sin20^{\circ}-\cos20^{\circ}}=\frac{\cos20^{\circ}-\sin20^{\circ}}{\sin20^{\circ}-\cos20^{\circ}}=-1$.

例 2. (1)已知角$\alpha$的顶点在坐标原点,始边与$x$轴的非负半轴重合,将角$\alpha$的终边按
顺时针方向旋转$\frac{\pi}{6}$后经过点$(-3,4)$,则$\cos\alpha=$
A.$\frac{3\sqrt{3}+4}{10}$
B.$\frac{4 - 3\sqrt{3}}{10}$
C.$\frac{3\sqrt{3}-4}{10}$
D.$-\frac{3\sqrt{3}+4}{10}$
(2)已知$0＜\alpha＜\frac{\pi}{2}$,$-\frac{\pi}{2}＜\beta＜0$,$\sin\alpha=\frac{5}{13}$,$\cos\beta=\frac{4}{5}$,则$\sin(\alpha+\beta)$的值是
A.$\frac{16}{65}$
B.$\frac{56}{65}$
C.$-\frac{16}{65}$
D.$-\frac{56}{65}$
(3)已知$\tan\alpha$和$\tan(\frac{\pi}{4}-\alpha)$是方程$ax^{2}+bx+c=0$的两个根,则$a,b,c$的关系
是
A.$b=a + c$
B.$2b=a + c$
C.$c=b + a$
D.$c=ab$
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答案： 例2:
(1)D
(2)C
(3)C
(1)由三角函数的定义可知
$\cos(\alpha-\frac{\pi}{6})=\frac{-3}{\sqrt{(-3)^{2}+4^{2}}}=-\frac{3}{5}$，
$\sin(\alpha-\frac{\pi}{6})=\frac{4}{\sqrt{(-3)^{2}+4^{2}}}=\frac{4}{5}$
$\therefore\cos\alpha=\cos[(\alpha-\frac{\pi}{6})+\frac{\pi}{6}]$
$=\cos(\alpha-\frac{\pi}{6})\cos\frac{\pi}{6}-\sin(\alpha-\frac{\pi}{6})·\sin\frac{\pi}{6}$
$=(-\frac{3}{5})×\frac{\sqrt{3}}{2}-\frac{4}{5}×\frac{1}{2}=-\frac{3\sqrt{3}+4}{10}$
(2)因为$0＜\alpha＜\frac{\pi}{2},-\frac{\pi}{2}＜\beta＜0,\sin\alpha=\frac{5}{13},\cos\beta=\frac{4}{5}$
$\therefore\cos\alpha=\sqrt{1-\sin^{2}\alpha}=\frac{12}{13},\sin\beta=-\sqrt{1-\cos^{2}\beta}=-\frac{3}{5}$
$\therefore\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$
$=\frac{5}{13}×\frac{4}{5}+\frac{12}{13}×(-\frac{3}{5})=-\frac{16}{65}$.
(3)$\tan\alpha+\tan(\frac{\pi}{4}-\alpha)=-\frac{b}{a},\tan\alpha\tan(\frac{\pi}{4}-\alpha)=\frac{c}{a}$
$\therefore\tan\frac{\pi}{4}=\tan[\alpha+(\frac{\pi}{4}-\alpha)]=\frac{\tan\alpha+\tan(\frac{\pi}{4}-\alpha)}{1-\tan\alpha\tan(\frac{\pi}{4}-\alpha)}$
$=\frac{-\frac{b}{a}}{1-\frac{c}{a}}=1$.
$\therefore-\frac{b}{a}=1-\frac{c}{a},\therefore-b=a-c$，即$c=a+b$.