《2025年学霸高考黑题数学人教版》

典型例题4. (2022·新高考全国Ⅰ)记△ABC的内角A,B,C的对边分别为a,b,c,已知cos A/(1+sin A)=sin 2B/(1+cos 2B).
(1)若C=2π/3,求B;
(2)求(a²+b²)/c²的最小值.

答案： 典型例题4. 解：
(1)$\frac{\cos A}{1 + \sin A} = \frac{\sin 2B}{1 + \cos 2B}$,$\frac{\cos \frac{A}{2} - \sin \frac{A}{2}}{\cos \frac{A}{2} + \sin \frac{A}{2}} + \sin^{2} \frac{A}{2} = \frac{\cos^{2} \frac{A}{2} - \sin^{2} \frac{A}{2}}{\cos^{2} \frac{A}{2} + \sin^{2} \frac{A}{2} + 2 \sin \frac{A}{2} \cos \frac{A}{2}} = \frac{1 - \tan \frac{A}{2}}{1 + \tan \frac{A}{2}} = \tan B$,$\frac{1 - \tan \frac{A}{2}}{1 + \tan \frac{A}{2}} = \tan B$,结合正切函数的单调性,$\frac{\pi}{4} - \frac{A}{2} = B$,则
$A + 2B = \frac{\pi}{2}$,结合$A + B = \frac{\pi}{3}$,解得$B = \frac{\pi}{6}$.
(2)由
(1)知,$\sin B = - \cos C ＞ 0$,所以$\frac{\pi}{2} ＜ C ＜ \pi$,$0 ＜ B ＜ \frac{\pi}{2}$,而$\sin B =\cos C = \sin (C - \frac{\pi}{2})$,所以$C = \frac{\pi}{2} + B$,即有$A = \frac{\pi}{2} - 2B$,所以$B \in(0,\frac{\pi}{4})$,$C \in (\frac{\pi}{2},\frac{3\pi}{4})$,所
以$\frac{a^{2} + b^{2}}{c^{2}} = \frac{\sin^{2}A + \sin^{2}B}{\sin^{2}C} =\frac{\cos^{2}2B + 1 - \cos^{2}B}{\cos^{2}B} = \frac{(2\cos^{2}B - 1)^{2} + 1 - \cos^{2}B}{\cos^{2}B} = 4\cos^{2}B + \frac{2}{\cos^{2}B} - 5 \geq 2\sqrt{4\cos^{2}B × \frac{2}{\cos^{2}B}} - 5 = 4\sqrt{2} - 5$.当且仅当$\cos^{2}B = \frac{\sqrt{2}}{2}$
时取等号,所以$\frac{a^{2} + b^{2}}{c^{2}}$的最小值为$4\sqrt{2} - 5$.

变式训练7. (2025·四川泸州一模)设△ABC的内角A,B,C的对边分别为a,b,c,且cos A/a=(cos B+cos C)/(b+c).
(1)求A;
(2)若2b²+c²的最大值为6+2√3,求a的值.

答案： 变式训练7. 解：
(1)由题设及正弦边角关系,有$\frac{\cos A}{\sin A} = \frac{\cos B + \sin C}{\sin B + \sin C}$,所
以$\sin B \cos A + \sin C \cos A = \sin A \cos B + \sin A \sin C$,整理得
$\sin B \cos A - \sin A \cos B = \sin A \sin C - \sin C \cos A$,即$\sin (B - A) = \sin (A - C)$,显然$B - A + A - C = B - C = \pi$不合题设,则$B - A = A - C$,所以$A = \frac{B + C}{2}$,而$A +B + C = \pi$,
则$A = \frac{\pi}{3}$.
(2)由$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$,可得$b = \frac{2a \sin B}{\sqrt{3}}$,$c = \frac{2a \sin C}{\sqrt{3}}$,所以$2b^{2} + c^{2} =\frac{4a^{2}}{3} (2\sin^{2}B + \sin^{2}C)$,由
(1)知,$B + C = \frac{2\pi}{3}$,
则$2b^{2} + c^{2} = \frac{2a^{2}}{3} \left[3 - 2\cos 2B - \cos (\frac{4\pi}{3} - 2B)\right] =\frac{2a^{2}}{3} \left[3 - 2\cos 2B - \cos \frac{4\pi}{3} \cos 2B - \sin \frac{4\pi}{3} \sin 2B\right] = 2a^{2} \left[1 + \frac{\sqrt{3}}{3} \sin (2B - \frac{\pi}{3})\right]$,由$0 ＜B ＜ \frac{2\pi}{3}$,则$- \frac{\pi}{3} ＜ 2B - \frac{\pi}{3} ＜ \pi$,又$2b^{2} + c^{2}$的最大值为$6 + 2\sqrt{3}$,所以
$2a^{2} \left[1 + \frac{\sqrt{3}}{3}\right] = 6 + 2\sqrt{3}$,可得$a = \sqrt{3}$(负值舍),综上,$a = \sqrt{3}$.

变式训练8. (2025·山东济南期末)在三角形ABC中,内角A,B,C的对边分别为a,b,c,且c cos B+4/5 b=a.
(1)求cos C;
(2)若a=2,且b²+c²＜4,求1/5 b+1/9 c²的取值范围.

答案： 变式训练8. 解：
(1)根据正弦定理可知$\sin C \cos B + \frac{4}{5} \sin B = \sin A \Rightarrow\sin C \cos B + \frac{4}{5} \sin B = \sin (B + C) \Rightarrow \frac{4}{5} \sin B = \sin (B + C - C)$.因为$B \in (0,\pi)$,
所以$\sin B \neq 0$,所以$\cos C = \frac{4}{5}$.
(2)由余弦定理可知$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc} = \frac{b^{2} + c^{2} - 4}{2bc}$.因为$b^{2} + c^{2} ＜ 4$,所以
$\cos A ＜ 0$,$A \in (\frac{\pi}{2},\pi)$,$\tan A ＜ 0$.因为$\cos C = \frac{4}{5} \in (\frac{\sqrt{2}}{2},\frac{\sqrt{3}}{2})$,所以$\sin C =\frac{3}{5}$,$C \in (\frac{\pi}{6},\frac{\pi}{4})$,由正弦定理得$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \Rightarrow b = \frac{5 \sin B}{\sin C}$,$c = \frac{5 \sin C}{\sin A}$,
所以$\frac{1}{5} + \frac{1}{9} c^{2} = \frac{8}{25} + \frac{25 \tan A}{25 \tan^{2} A} + \frac{4}{25 \tan^{2} A} =\frac{2}{25} \left[ \frac{3}{2} ( \sin^{2} A + \cos^{2} A ) + \frac{2}{\tan^{2} A} \right] + \frac{2}{25} = \frac{2}{25} \left[ \frac{3}{2} - \frac{3}{2} \cos 2A + \frac{2 (1 + \cos 2A)}{1 - \cos 2A} \right] + \frac{2}{25} = \frac{4}{25} \left( \frac{1}{1 - \cos 2A} - \frac{3}{4} \right)^{2} - \frac{39}{100}$.
因为$\frac{\pi}{2} ＜ A ＜ \frac{3\pi}{4}$,所以$\tan A \leq -1$,
$0 ＞ \frac{1}{\tan A} \geq -1$,所以$\frac{39}{100} + \frac{4}{25} (\frac{1}{\tan A})^{2} + \frac{9}{100} × \frac{1}{\tan A} + \frac{39}{100} × \frac{12}{25} = \frac{39}{100} + \frac{4}{25} × 1 + \frac{9}{100} × (-1) + \frac{39}{100} × \frac{12}{25} = \frac{12}{25}$.
所以$\frac{1}{5} - \frac{1}{9} + \frac{1}{9} c^{2}$的取值范围是$[\frac{39}{100},\frac{12}{25}]$.