《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》

对点训练 1
(1) 已知$\alpha$是第三象限的角，$\cos\alpha = -\frac{12}{13}$, 则$\sin\alpha =$
A. $\frac{5}{13}$
B. $-\frac{5}{13}$
C. $\frac{5}{12}$
D. $-\frac{5}{12}$

答案：
(1)B

(2) 已知$\tan\alpha = 2\left(0 ＜ \alpha ＜ \frac{\pi}{2}\right)$, 则$\sin\alpha =$

A.$\frac{\sqrt{5}}{5}$
B.$-\frac{2\sqrt{5}}{5}$
C.$-\frac{\sqrt{5}}{5}$
D.$\frac{2\sqrt{5}}{5}$

答案：
(2)D

例 2. 已知$\frac{\tan\alpha}{\tan\alpha - 1} = 2$, 求下列各式的值：
(1) $\frac{2\sin\alpha - 3\cos\alpha}{4\sin\alpha - 9\cos\alpha}$;
(2) $\frac{2\sin^{2}\alpha - 3\cos^{2}\alpha}{4\sin^{2}\alpha - 9\cos^{2}\alpha}$;
(3) $4\sin^{2}\alpha - 3\sin\alpha\cos\alpha - 5\cos^{2}\alpha$.
【分析】所求式子都是关于$\sin\alpha$、$\cos\alpha$的分式齐次式（或可化为分式齐次式），将其分子、分母同除以$\cos\alpha$的整数次幂，就把所求式子用$\tan\alpha$表示，因此可先由已知条件求$\tan\alpha$的值，再求各式的值.

答案： 例2:由$\frac{\tan \alpha }{\tan \alpha -1 } =2,$得$\tan \alpha =2.$
$(1)\frac{2\sin \alpha -3\cos \alpha }{4\sin \alpha -9\cos \alpha } =\frac{2\tan \alpha -3}{4\tan \alpha -9 } $
$\because \tan \alpha =2,\therefore $原式$=\frac{2× 2-3}{4× 2-9} =-1.$
$(2)\frac{2\sin^{2} \alpha -3\cos^{2} \alpha }{4\sin^{2} \alpha -9\cos^{2} \alpha } =\frac{2\tan^{2} \alpha -3}{4\tan^{2} \alpha -9 } $
$\because \tan \alpha =2,\therefore $原式$=\frac{2× 4-3}{4× 4-9} =\frac{5}{7} .$
$(3)4\sin^{2} \alpha -3\sin \alpha \cos \alpha -5\cos^{2} \alpha $
$=\frac{4\sin^{2} \alpha -3\sin \alpha \cos \alpha -5\cos^{2} \alpha }{\sin^{2} \alpha +\cos^{2} \alpha } $
$=\frac{4\tan^{2} \alpha -3\tan \alpha -5}{\tan^{2} \alpha +1 } $
$\because \tan \alpha =2,\therefore $原式$=\frac{4× 4-3× 2-5}{4+1} =1.$

对点训练 2
已知$\tan\alpha = -\frac{1}{2}$, 求下列各式的值：
(1) $\sin\alpha + 2\cos\alpha$;
(2) $\frac{\cos\alpha - 5\sin\alpha}{3\cos\alpha + \sin\alpha}$;
(3) $\frac{\sin^{2}\alpha - \sin\alpha\cos\alpha - 3\cos^{2}\alpha}{5\sin\alpha\cos\alpha + \sin^{2}\alpha + 1}$;
(4) $2\sin^{2}\alpha - \sin\alpha\cos\alpha + \cos^{2}\alpha$.

答案： 对点训练$2:(1)\tan \alpha =\frac{\sin \alpha }{\cos \alpha } =-\frac{1}{2} ,$
$\therefore \cos \alpha =-2\sin \alpha .$
又$\sin^{2} \alpha +\cos^{2} \alpha =1,\therefore \sin^{2} \alpha +4\sin^{2} \alpha =1,$
$\therefore \sin^{2} \alpha =\frac{1}{5} ,\therefore \sin \alpha =\pm \frac{\sqrt{5} }{5} $
$\because \tan \alpha =-\frac{1}{2} \lt 0,\therefore \alpha$为第二,四象限角
当$\alpha$为第二象限角时$,\sin \alpha =\frac{\sqrt{5} }{5} ,\cos \alpha =-\frac{2\sqrt{5} }{5} ,$
$\sin \alpha +2\cos \alpha =-\frac{3\sqrt{5} }{5} ,$
当$\alpha$为第四象限角时$,\sin \alpha =-\frac{\sqrt{5} }{5} ,\cos \alpha =\frac{2\sqrt{5} }{5} ,$
$\sin \alpha +2\cos \alpha =\frac{3\sqrt{5} }{5} .$
$(2)\frac{\cos \alpha -5\sin \alpha }{3\cos \alpha +\sin \alpha } =\frac{1-5\tan \alpha }{3+\tan \alpha } =\frac{1-5× (-\frac{1}{2} )}{3-\frac{1}{2} } =\frac{7}{5} $
$(3)\frac{\sin^{2} \alpha -\sin \alpha \cos \alpha -3\cos^{2} \alpha }{5\sin \alpha \cos \alpha +\sin^{2} \alpha +1 } $
$=\frac{\sin^{2} \alpha -\sin \alpha \cos \alpha -3\cos^{2} \alpha }{5\sin \alpha \cos \alpha +2\sin^{2} \alpha +\cos^{2} \alpha } $
$=\frac{\tan^{2} \alpha -\tan \alpha -3}{2\tan^{2} \alpha +5\tan \alpha +1 } $
$=\frac{(-\frac{1}{2} )^{2} -(-\frac{1}{2} )-3}{2× (-\frac{1}{2} )^{2} +5× (-\frac{1}{2} )+1} $
$=\frac{\frac{1}{4} +\frac{1}{2} -3}{\frac{1}{2} -\frac{5}{2} +1} $
$=\frac{-\frac{9}{4} }{ -1} $
$=\frac{9}{4} $
$(4)2\sin^{2} \alpha -\sin \alpha \cos \alpha +\cos^{2} \alpha $
$=\frac{2\sin^{2} \alpha -\sin \alpha \cos \alpha +\cos^{2} \alpha }{\sin^{2} \alpha +\cos^{2} \alpha } $
$=\frac{2\tan^{2} \alpha -\tan \alpha +1}{\tan^{2} \alpha +1 } $
$=\frac{2× \frac{1}{4} -(-\frac{1}{2} )+1}{\frac{1}{4} +1} $
$=\frac{\frac{1}{2} +\frac{1}{2} +1}{\frac{5}{4} } $
$=\frac{8}{5}$