《2025年金版新学案高三总复习数学北师大版》

（2025·江西南昌模拟）在△ABC 中，角 A，B，C 所对的边分别记为 a，b，c，且 tan A = $\frac{\cos B - \sin C}{\cos C + \sin B}$.
（1）若 B = $\frac{\pi}{6}$，求 C 的大小；
（2）若 a = 2，求 b + c 的取值范围.

答案： （1）因为$\tan A = \frac{\cos B - \sin C}{\cos C + \sin B}$，所以$\frac{\sin A}{\cos A} = \frac{\cos B - \sin C}{\cos C + \sin B}$，即$\sin A\cos C + \sin A\sin B = \cos A\cos B - \cos A\sin C$，即$\sin A\cos C + \cos A\sin C = \cos A\cos B - \sin A\sin B$，所以$\sin(A + C) = \cos(A + B)$，即$\sin B = \cos(A + B)$，而$A,B \in (0,\pi)$，所以$B + A + B = \frac{\pi}{2}$或$B - (A + B) = \frac{\pi}{2}$，所以$A + 2B = \frac{\pi}{2}$或$A = -\frac{\pi}{2}$（舍去），又因为$B = \frac{\pi}{6}$，所以$A = \frac{\pi}{6}$，所以$C = \frac{2\pi}{3}$。
(2)由
(1)得$A + 2B = \frac{\pi}{2}$，因为$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$，所以$b = \frac{a\sin B}{\sin A} = \frac{2\sin B}{\sin A} = \frac{2\sin B}{\cos2B}$，$c = \frac{a\sin C}{\sin A} = \frac{2\sin C}{\sin A} = \frac{2\cos B}{\cos2B}$，则$b + c = \frac{2(\sin B + \cos B)}{\cos 2B} = \frac{2(\sin B + \cos B)}{\cos^{2}B - \sin^{2}B} = \frac{2}{\cos B - \sin B} = \frac{\sqrt{2}}{\cos(B + \frac{\pi}{4})}$，由$0 ＜ B ＜ \frac{\pi}{2}$，$0 ＜ \frac{\pi}{2} - 2B ＜ \pi$，得$0 ＜ B ＜ \frac{\pi}{4}$，所以$\frac{\pi}{4} ＜ B + \frac{\pi}{4} ＜ \frac{\pi}{2}$，所以$0 ＜ \cos(B + \frac{\pi}{4}) ＜ \frac{\sqrt{2}}{2}$，所以$b + c \in (2, +\infty)$。

(1)求sin A；

答案： 因为$A + B = 3C$，所以$\pi - C = 3C$，即$C = \frac{\pi}{4}$.
又$2\sin(A - C) = \sin B = \sin(A + C)$，
所以$2\sin A\cos C - 2\cos A\sin C = \sin A\cos C + \cos A\sin C$，
所以$\sin A\cos C = 3\cos A\sin C$，所以$\sin A = 3\cos A$，
即$\tan A = 3$，所以$0 ＜ A ＜ \frac{\pi}{2}$，所以$\sin A = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$.

(2)设AB=5，求AB边上的高.

答案： 由
(1)知，$\cos A = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$，由$\sin B = \sin(A + C) = \sin A\cos C + \cos A\sin C = \frac{\sqrt{2}}{2}(\frac{3\sqrt{10}}{10} + \frac{\sqrt{10}}{10}) = \frac{2\sqrt{5}}{5}$.
由正弦定理，$\frac{AB}{\sin C} = \frac{AC}{\sin B}$，可得$AC = \frac{5 × \frac{2\sqrt{5}}{5}}{\frac{\sqrt{2}}{2}} = 2\sqrt{10}$.
所以$h = AC · \sin A = 2\sqrt{10} × \frac{3\sqrt{10}}{10} = 6$.

(1)求角B；

答案： 因为$2b\cos(A - \frac{\pi}{3}) = a + c$，
由正弦定理可得$2\sin B(\frac{1}{2}\cos A + \frac{\sqrt{3}}{2}\sin A) = \sin A + \sin C$，
即$\sin B\cos A + \sqrt{3}\sin A\sin B = \sin A + \sin(A + B)$，
即$\sin B\cos A + \sqrt{3}\sin A\sin B = \sin A + \sin A\cos B + \cos A\sin B$，
所以$\sqrt{3}\sin B\sin A = \sin A + \sin A\cos B$，
在三角形中，$\sin A ＞ 0$，所以$\sqrt{3}\sin B - \cos B = 1$，
即$\sin(B - \frac{\pi}{6}) = \frac{1}{2}$，因为$B \in (0,\pi)$，则$B - \frac{\pi}{6} \in (-\frac{\pi}{6},\frac{5\pi}{6})$，可得$B - \frac{\pi}{6} = \frac{\pi}{6}$，则$B = \frac{\pi}{3}$.

(2)若AC边上的高h=$\frac{\sqrt{3}}{4}$b，求cos A cos C.

答案： 因为$AC$边上的高$h = \frac{\sqrt{3}}{4}b$，
所以$S_{\triangle ABC} = \frac{1}{2}b · h = \frac{1}{2}b · \frac{\sqrt{3}}{4}b = \frac{\sqrt{3}}{8}b^{2}$①.
又$S_{\triangle ABC} = \frac{1}{2}ac\sin B = \frac{1}{2}ac × \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}ac$②.
由①②可得$b^{2} = 2ac$，由正弦定理可得$\sin^{2}B = 2\sin A\sin C$，
结合
(1)中$B = \frac{\pi}{3}$可得$\sin A\sin C = \frac{3}{8}$，
因为$\cos B = - \cos(A + C) = - \cos A\cos C + \sin A\sin C = \frac{1}{2}$，
所以$\cos A\cos C = \sin A\sin C - \frac{1}{2} = \frac{3}{8} - \frac{1}{2} = - \frac{1}{8}$.

(1)若∠ADC=$\frac{\pi}{3}$，求tan B；

答案：
在$\triangle ABC$中，因为$D$为$BC$的中点，$\angle ADC = \frac{\pi}{3}$，$AD = 1$，则$S_{\triangle ADC} = \frac{1}{2}AD · DC\sin\angle ADC = \frac{1}{2} × 1 × \frac{1}{2}a × \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}a = \frac{1}{2}S_{\triangle ABC} = \frac{\sqrt{3}}{2}$，解得$a = 4$.
在$\triangle ABD$中，$\angle ADB = \frac{2\pi}{3}$，由余弦定理得$c^{2} = BD^{2} + AD^{2} - 2BD · AD\cos\angle ADB$，
即$c^{2} = 4 + 1 - 2 × 2 × 1 × (-\frac{1}{2}) = 7$，解得$c = \sqrt{7}$，
则$\cos B = \frac{7 + 4 - 1}{2\sqrt{7} × 2} = \frac{5\sqrt{7}}{14}$，$\sin B = \sqrt{1 - \cos^{2}B} = \sqrt{1 - (\frac{5\sqrt{7}}{14})^{2}} = \frac{\sqrt{21}}{14}$，所以$\tan B = \frac{\sin B}{\cos B} = \frac{\sqrt{3}}{5}$.
法二：在$\triangle ABC$中，因为$D$为$BC$的中点，$\angle ADC = \frac{\pi}{3}$，$AD = 1$，
则$S_{\triangle ADC} = \frac{1}{2}AD · DC\sin\angle ADC = \frac{1}{2} × 1 × \frac{1}{2}a × \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}a$，解得$a = 4$.
在$\triangle ACD$中，由余弦定理得$b^{2} = CD^{2} + AD^{2} - 2CD · AD\cos\angle ADC$，
即$b^{2} = 4 + 1 - 2 × 2 × 1 × \frac{1}{2} = 3$，解得$b = \sqrt{3}$，有$AC^{2} + AD^{2} = 4 = CD^{2}$，
则$\angle CAD = \frac{\pi}{2}$，$C = \frac{\pi}{6}$，如图，过$A$作$AE \perp BC$于$E$，

于是$CE = AC\cos C = \frac{3}{2}$，$AE = AC\sin C = \frac{\sqrt{3}}{2}$，$BE = \frac{5}{2}$，所以$\tan B = \frac{AE}{BE} = \frac{\sqrt{3}}{5}$.

(2)若b²+c²=8，求b，c.

答案： 在$\triangle ABD$与$\triangle ACD$中，由余弦定理得$\begin{cases}c^{2} = \frac{1}{4}a^{2} + 1 - 2 × \frac{1}{2}a × 1 × \cos(\pi - \angle ADC) \\b^{2} = \frac{1}{4}a^{2} + 1 - 2 × \frac{1}{2}a × 1 × \cos\angle ADC\end{cases}$
整理得$\frac{1}{2}a^{2} + 2 = b^{2} + c^{2}$，而$b^{2} + c^{2} = 8$，则$a = 2\sqrt{3}$.
又$S_{\triangle ADC} = \frac{1}{2} × \sqrt{3} × 1 × \sin\angle ADC = \frac{\sqrt{3}}{2}$，解得$\sin\angle ADC = 1$，而$0 ＜ \angle ADC ＜ \pi$，于是$\angle ADC = \frac{\pi}{2}$，所以$b = c = \sqrt{AD^{2} + CD^{2}} = 2$.
法二：在$\triangle ABC$中，因为$D$为$BC$的中点，则$2\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{AC}$，
又$\overrightarrow{CB} = \overrightarrow{AB} - \overrightarrow{AC}$，
于是$4\overrightarrow{AD}^{2} + \overrightarrow{CB}^{2} = (\overrightarrow{AB} + \overrightarrow{AC})^{2} + (\overrightarrow{AB} - \overrightarrow{AC})^{2} = 2(b^{2} + c^{2}) = 16$，
即$4 + a^{2} = 16$，解得$a = 2\sqrt{3}$.
又$S_{\triangle ADC} = \frac{1}{2} × \sqrt{3} × 1 × \sin\angle ADC = \frac{\sqrt{3}}{2}$，解得$\sin\angle ADC = 1$，而$0 ＜ \angle ADC ＜ \pi$，于是$\angle ADC = \frac{\pi}{2}$，所以$b = c = \sqrt{AD^{2} + CD^{2}} = 2$.