《2025年小题狂做高中数学必修第二册人教版巅峰版》

26. [2023 新高考Ⅱ卷,17]记△ABC 的内角$A,B,C$的对边分别为$a,b,c$,已知△ABC 的面积为$\sqrt{3}$,$D$为$BC$的中点,且$AD = 1$.
(1) 若$\angle ADC = \frac{\pi}{3}$,求$\tan B$;
(2) 若$b^2 + c^2 = 8$,求$b,c$.

答案： 26.
(1) 如图，因为$D$为$BC$中点，所以$S_{\triangle ABC} = 2S_{\triangle ADC}$。又$S_{\triangle ABC} = \sqrt{3}$，所以$S_{\triangle ADC} = \frac{\sqrt{3}}{2}$。在$\triangle ADC$中，$S_{\triangle ADC} = \frac{1}{2}· AD· DC·\sin\angle ADC = \frac{1}{2}×1× DC×\sin\frac{\pi}{3} = \frac{\sqrt{3}}{4}DC = \frac{\sqrt{3}}{2}$，所以$DC = 2$，故$BC = 2DC = 4$，即$a = 4$。在$\triangle ADC$中，由余弦定理得$AC^2 = AD^2 + DC^2-2AD· DC\cos\angle ADC = 1 + 4-2×1×2×\frac{1}{2} = 3$，所以$AC = \sqrt{3}$，即$b = \sqrt{3}$。在$\triangle ABC$中，由余弦定理得$\cos B = \frac{AB^2 + BC^2-AC^2}{2AB· BC} = \frac{c^2 + 16-3}{2c×4} = \frac{c^2 + 13}{8c}$。又$S_{\triangle ABC} = \frac{1}{2}AB· BC\sin B = \frac{1}{2}c×4\sin B = 2c\sin B = \sqrt{3}$，所以$c\sin B = \frac{\sqrt{3}}{2}$。由$\sin^2 B + \cos^2 B = 1$，得$\left(\frac{\sqrt{3}}{2c}\right)^2 + \left(\frac{c^2 + 13}{8c}\right)^2 = 1$，即$\frac{3}{4c^2} + \frac{(c^2 + 13)^2}{64c^2} = 1$，两边乘以$64c^2$得$48 + (c^2 + 13)^2 = 64c^2$，即$c^4 + 26c^2 + 169 + 48 = 64c^2$，$c^4-38c^2 + 217 = 0$，解得$c^2 = 19\pm6\sqrt{3}$。因为$c ＞ 0$，且由$c\sin B = \frac{\sqrt{3}}{2}$知$c ＞ \frac{\sqrt{3}}{2}$，检验后$c^2 = 19 + 6\sqrt{3} = (\sqrt{3} + 3\sqrt{2})^2?$ 实际上$19 + 6\sqrt{3} = (3+\sqrt{3})^2$，$19-6\sqrt{3} = (3-\sqrt{3})^2$，所以$c = 3+\sqrt{3}$或$c = 3-\sqrt{3}$。当$c = 3+\sqrt{3}$时，$\sin B = \frac{\sqrt{3}}{2c} = \frac{\sqrt{3}}{2(3+\sqrt{3})} = \frac{\sqrt{3}(3-\sqrt{3})}{2(9-3)} = \frac{3\sqrt{3}-3}{12} = \frac{\sqrt{3}-1}{4}$，$\cos B = \frac{c^2 + 13}{8c} = \frac{(3+\sqrt{3})^2 + 13}{8(3+\sqrt{3})} = \frac{9+6\sqrt{3}+3+13}{8(3+\sqrt{3})} = \frac{25+6\sqrt{3}}{8(3+\sqrt{3})} = \frac{(25+6\sqrt{3})(3-\sqrt{3})}{8(9-3)} = \frac{75-25\sqrt{3}+18\sqrt{3}-18}{48} = \frac{57-7\sqrt{3}}{48}$，$\tan B = \frac{\sin B}{\cos B} = \frac{\frac{\sqrt{3}-1}{4}}{\frac{57-7\sqrt{3}}{48}} = \frac{12(\sqrt{3}-1)}{57-7\sqrt{3}}$。当$c = 3-\sqrt{3}$时，$\sin B = \frac{\sqrt{3}}{2(3-\sqrt{3})} = \frac{\sqrt{3}(3+\sqrt{3})}{2(9-3)} = \frac{3\sqrt{3}+3}{12} = \frac{\sqrt{3}+1}{4}$，$\cos B = \frac{(3-\sqrt{3})^2 + 13}{8(3-\sqrt{3})} = \frac{9-6\sqrt{3}+3+13}{8(3-\sqrt{3})} = \frac{25-6\sqrt{3}}{8(3-\sqrt{3})} = \frac{(25-6\sqrt{3})(3+\sqrt{3})}{8(9-3)} = \frac{75+25\sqrt{3}-18\sqrt{3}-18}{48} = \frac{57+7\sqrt{3}}{48}$，$\tan B = \frac{\frac{\sqrt{3}+1}{4}}{\frac{57+7\sqrt{3}}{48}} = \frac{12(\sqrt{3}+1)}{57+7\sqrt{3}}$。但题目要求$\tan B$，可能只需一种情况。由$D$为中点及$\angle ADC = \frac{\pi}{3}$，可考虑向量法或几何性质。实际上，由$AD = 1$，$DC = 2$，$\angle ADC = \frac{\pi}{3}$，在$\triangle ADC$中可求$AC = \sqrt{3}$。在$\triangle ABD$中，$BD = 2$，$AD = 1$，$\angle ADB = \pi-\frac{\pi}{3} = \frac{2\pi}{3}$，由余弦定理得$AB^2 = AD^2 + BD^2-2AD· BD\cos\frac{2\pi}{3} = 1 + 4-2×1×2×(-\frac{1}{2}) = 7$，所以$c = \sqrt{7}$。然后在$\triangle ABC$中，$a = 4$，$b = \sqrt{3}$，$c = \sqrt{7}$，由余弦定理得$\cos B = \frac{a^2 + c^2-b^2}{2ac} = \frac{16 + 7-3}{2×4×\sqrt{7}} = \frac{20}{8\sqrt{7}} = \frac{5}{2\sqrt{7}}$，$\sin B = \sqrt{1-\frac{25}{28}} = \frac{\sqrt{3}}{2\sqrt{7}}$，所以$\tan B = \frac{\sin B}{\cos B} = \frac{\frac{\sqrt{3}}{2\sqrt{7}}}{\frac{5}{2\sqrt{7}}} = \frac{\sqrt{3}}{5}$。此结果与上面不同，应以几何关系直接计算为准。所以$\tan B = \frac{\sqrt{3}}{5}$。
(2) 因为$D$为$BC$中点，所以$\overrightarrow{AD} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC})$，则$|\overrightarrow{AD}|^2 = \frac{1}{4}(|\overrightarrow{AB}|^2 + |\overrightarrow{AC}|^2 + 2\overrightarrow{AB}·\overrightarrow{AC}) = \frac{1}{4}(c^2 + b^2 + 2bc\cos A)$。又$AD = 1$，所以$1 = \frac{1}{4}(b^2 + c^2 + 2bc\cos A)$，即$b^2 + c^2 + 2bc\cos A = 4$。已知$b^2 + c^2 = 8$，所以$8 + 2bc\cos A = 4$，即$2bc\cos A = -4$，$bc\cos A = -2$。又△ABC的面积$S = \frac{1}{2}bc\sin A = \sqrt{3}$，所以$bc\sin A = 2\sqrt{3}$。两式平方相加得$(bc)^2(\cos^2 A + \sin^2 A) = 4 + 12 = 16$，所以$(bc)^2 = 16$，因为$bc ＞ 0$，所以$bc = 4$。代入$bc\cos A = -2$得$\cos A = -\frac{1}{2}$，所以$A = \frac{2\pi}{3}$。由$b^2 + c^2 = 8$，$bc = 4$，得$(b + c)^2 = b^2 + c^2 + 2bc = 8 + 8 = 16$，所以$b + c = 4$。又$b^2 + c^2 = 8$，联立可解得$b = c = 2$。

27. [2022 新高考Ⅱ卷,18]记△ABC 的内角$A,B,C$的对边分别为$a,b,c$,分别以$a,b,c$为边长的三个正三角形的面积依次为$S_1,S_2,S_3$.已知$S_1 - S_2 + S_3 = \frac{\sqrt{3}}{2}$,$\sin B = \frac{1}{3}$.
(1) 求△ABC 的面积;
(2) 若$\sin A\sin C = \frac{\sqrt{2}}{3}$,求$b$.

答案： 27.
(1) 边长为$x$的正三角形的面积为$\frac{\sqrt{3}}{4}x^2$。所以$S_1 = \frac{\sqrt{3}}{4}a^2$，$S_2 = \frac{\sqrt{3}}{4}b^2$，$S_3 = \frac{\sqrt{3}}{4}c^2$。由$S_1-S_2 + S_3 = \frac{\sqrt{3}}{2}$得$\frac{\sqrt{3}}{4}(a^2-b^2 + c^2) = \frac{\sqrt{3}}{2}$，所以$a^2-b^2 + c^2 = 2$。由余弦定理$b^2 = a^2 + c^2-2ac\cos B$，所以$a^2 + c^2 = b^2 + 2ac\cos B$。代入$a^2-b^2 + c^2 = 2$得$b^2 + 2ac\cos B-b^2 = 2$，即$2ac\cos B = 2$，$ac\cos B = 1$。又$\sin B = \frac{1}{3}$，所以$\cos B = \sqrt{1-\sin^2 B} = \frac{2\sqrt{2}}{3}$（因为$B$为三角形内角，$\sin B = \frac{1}{3}$，故$\cos B$可取正）。所以$ac = \frac{1}{\cos B} = \frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{4}$。故△ABC的面积$S = \frac{1}{2}ac\sin B = \frac{1}{2}×\frac{3\sqrt{2}}{4}×\frac{1}{3} = \frac{\sqrt{2}}{8}$。
(2) 由正弦定理$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$，得$a = 2R\sin A$，$c = 2R\sin C$。由
(1)知$ac = \frac{3\sqrt{2}}{4}$，所以$4R^2\sin A\sin C = \frac{3\sqrt{2}}{4}$。又已知$\sin A\sin C = \frac{\sqrt{2}}{3}$，所以$4R^2×\frac{\sqrt{2}}{3} = \frac{3\sqrt{2}}{4}$，解得$4R^2 = \frac{9}{4}$，即$R^2 = \frac{9}{16}$，$R = \frac{3}{4}$（因为$R ＞ 0$）。所以$b = 2R\sin B = 2×\frac{3}{4}×\frac{1}{3} = \frac{1}{2}$。

28. [2022 新高考Ⅰ卷,18]记△ABC 的内角$A,B,C$的对边分别为$a,b,c$,已知$\frac{\cos A}{1 + \sin A} = \frac{\sin 2B}{1 + \cos 2B}$.
(1) 若$C = \frac{2\pi}{3}$,求$B$;
(2) 求$\frac{a^2 + b^2}{c^2}$的最小值.

答案： 28.
(1) 由已知$\frac{\cos A}{1 + \sin A} = \frac{\sin 2B}{1 + \cos 2B} = \frac{2\sin B\cos B}{2\cos^2 B} = \frac{\sin B}{\cos B} = \tan B$，所以$\cos A = \tan B(1 + \sin A)$。若$C = \frac{2\pi}{3}$，则$A + B = \pi-C = \frac{\pi}{3}$，所以$B = \frac{\pi}{3}-A$，代入得$\cos A = \tan(\frac{\pi}{3}-A)(1 + \sin A)$。即$\cos A = \frac{\sin(\frac{\pi}{3}-A)}{\cos(\frac{\pi}{3}-A)}(1 + \sin A)$。展开$\sin(\frac{\pi}{3}-A) = \sin\frac{\pi}{3}\cos A-\cos\frac{\pi}{3}\sin A = \frac{\sqrt{3}}{2}\cos A-\frac{1}{2}\sin A$，$\cos(\frac{\pi}{3}-A) = \cos\frac{\pi}{3}\cos A + \sin\frac{\pi}{3}\sin A = \frac{1}{2}\cos A + \frac{\sqrt{3}}{2}\sin A$。所以$\cos A = \frac{\frac{\sqrt{3}}{2}\cos A-\frac{1}{2}\sin A}{\frac{1}{2}\cos A + \frac{\sqrt{3}}{2}\sin A}(1 + \sin A)$。整理得$\cos A(\frac{1}{2}\cos A + \frac{\sqrt{3}}{2}\sin A) = (\frac{\sqrt{3}}{2}\cos A-\frac{1}{2}\sin A)(1 + \sin A)$。两边乘以2：$\cos A(\cos A + \sqrt{3}\sin A) = (\sqrt{3}\cos A-\sin A)(1 + \sin A)$。即$\cos^2 A + \sqrt{3}\sin A\cos A = \sqrt{3}\cos A + \sqrt{3}\cos A\sin A-\sin A-\sin^2 A$。所以$\cos^2 A = \sqrt{3}\cos A-\sin A-\sin^2 A$。因为$\cos^2 A + \sin^2 A = 1$，所以$1 = \sqrt{3}\cos A-\sin A + 1$，即$\sqrt{3}\cos A-\sin A = 0$，$\tan A = \sqrt{3}$。因为$A\in(0,\pi)$，且$A + B = \frac{\pi}{3}$，所以$A = \frac{\pi}{3}$，则$B = \frac{\pi}{3}-A = 0$，不合题意。故需重新检查。由$\cos A = \tan B(1 + \sin A)$，且$A + B = \frac{\pi}{3}$，所以$\cos A = \tan(\frac{\pi}{3}-A)(1 + \sin A)$。令$t = \tan\frac{A}{2}$，则$\sin A = \frac{2t}{1 + t^2}$，$\cos A = \frac{1-t^2}{1 + t^2}$，$\tan(\frac{\pi}{3}-A) = \frac{\tan\frac{\pi}{3}-\tan A}{1 + \tan\frac{\pi}{3}\tan A} = \frac{\sqrt{3}-\frac{2t}{1-t^2}}{1 + \sqrt{3}·\frac{2t}{1-t^2}}$（这里$\tan A = \frac{2t}{1-t^2}$）。代入求解较繁。考虑几何意义或利用恒等式。由$\frac{\cos A}{1 + \sin A} = \tan B$，且$A + B = \frac{\pi}{3}$，所以$\tan B = \tan(\frac{\pi}{3}-A) = \frac{\tan\frac{\pi}{3}-\tan A}{1 + \tan\frac{\pi}{3}\tan A} = \frac{\sqrt{3}-\tan A}{1 + \sqrt{3}\tan A}$。所以$\frac{\cos A}{1 + \sin A} = \frac{\sqrt{3}-\tan A}{1 + \sqrt{3}\tan A}$。令$t = \tan A$，则$\frac{\cos A}{1 + \sin A} = \frac{1-t^2}{1 + t^2}÷(1 + \frac{2t}{1 + t^2}) = \frac{1-t^2}{1 + t^2 + 2t} = \frac{1-t^2}{(1 + t)^2} = \frac{1-t}{1 + t}$。所以方程化为$\frac{1-t}{1 + t} = \frac{\sqrt{3}-t}{1 + \sqrt{3}t}$。交叉相乘得$(1-t)(1 + \sqrt{3}t) = (1 + t)(\sqrt{3}-t)$，即$1 + \sqrt{3}t-t-\sqrt{3}t^2 = \sqrt{3}-t + \sqrt{3}t-t^2$，整理得$1 + (\sqrt{3}-1)t-\sqrt{3}t^2 = \sqrt{3} + (\sqrt{3}-1)t-t^2$，所以$1-\sqrt{3}t^2 = \sqrt{3}-t^2$，即$1-\sqrt{3} = \sqrt{3}t^2-t^2 = t^2(\sqrt{3}-1)$，所以$t^2 = \frac{1-\sqrt{3}}{\sqrt{3}-1} = -1$，无解。这说明在$C = \frac{2\pi}{3}$的条件下，无解？但题目要求求$B$，可能$B$有特定值。我们回到最初等式：$\frac{\cos A}{1 + \sin A} = \tan B$。由$A + B + C = \pi$，$C = \frac{2\pi}{3}$，则$A + B = \frac{\pi}{3}$，所以$B = \frac{\pi}{3}-A$，代入得$\frac{\cos A}{1 + \sin A} = \tan(\frac{\pi}{3}-A)$。利用两角差的正切公式，$\tan(\frac{\pi}{3}-A) = \frac{\tan\frac{\pi}{3}-\tan A}{1 + \tan\frac{\pi}{3}\tan A} = \frac{\sqrt{3}-\tan A}{1 + \sqrt{3}\tan A}$。而$\frac{\cos A}{1 + \sin A} = \frac{1-\tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}}÷(1 + \frac{2\tan\frac{A}{2}}{1 + \tan^2\frac{A}{2}}) = \frac{1-\tan^2\frac{A}{2}}{(1 + \tan\frac{A}{2})^2} = \frac{1-\tan\frac{A}{2}}{1 + \tan\frac{A}{2}}$。令$u = \tan\frac{A}{2}$，则方程化为$\frac{1-u}{1 + u} = \frac{\sqrt{3}-\frac{2u}{1-u^2}}{1 + \sqrt{3}·\frac{2u}{1-u^2}} = \frac{\sqrt{3}(1-u^2)-2u}{(1-u^2) + 2\sqrt{3}u}$。即$\frac{1-u}{1 + u} = \frac{\sqrt{3}-\sqrt{3}u^2-2u}{1-u^2 + 2\sqrt{3}u}$。交叉相乘：$(1-u)(1-u^2 + 2\sqrt{3}u) = (1 + u)(\sqrt{3}-\sqrt{3}u^2-2u)$。左边=$(1-u)[(1-u^2) + 2\sqrt{3}u] = (1-u)(1-u^2) + 2\sqrt{3}u(1-u) = (1-u^2)(1-u) + 2\sqrt{3}u(1-u) = (1-u)[1-u^2 + 2\sqrt{3}u]$。右边=$(1 + u)[\sqrt{3}(1-u^2)-2u] = \sqrt{3}(1 + u)(1-u^2)-2u(1 + u) = \sqrt{3}(1-u^2)(1 + u)-2u(1 + u) = (1 + u)[\sqrt{3}(1-u^2)-2u]$。化简得$(1-u)(1-u^2 + 2\sqrt{3}u) = (1 + u)(\sqrt{3}(1-u^2)-2u)$。展开左边：$(1-u)(1-u^2) + 2\sqrt{3}u(1-u) = (1-u^2-u + u^3) + 2\sqrt{3}u-2\sqrt{3}u^2 = 1-u-u^2 + u^3 + 2\sqrt{3}u-2\sqrt{3}u^2 = 1 + (-1+2\sqrt{3})u + (-1-2\sqrt{3})u^2 + u^3$。右边：$(1 + u)\sqrt{3}(1-u^2)-2u(1 + u) = \sqrt{3}(1 + u)(1-u^2)-2u-2u^2 = \sqrt{3}(1-u^2 + u-u^3)-2u-2u^2 = \sqrt{3}(1 + u-u^2-u^3)-2u-2u^2 = \sqrt{3} + \sqrt{3}u-\sqrt{3}u^2-\sqrt{3}u^3-2u-2u^2 = \sqrt{3} + (\sqrt{3}-2)u + (-\sqrt{3}-2)u^2-\sqrt{3}u^3$。所以方程化为：$1 + (-1+2\sqrt{3})u + (-1-2\sqrt{3})u^2 + u^3 = \sqrt{3} + (\sqrt{3}-2)u + (-\sqrt{3}-2)u^2-\sqrt{3}u^3$。移项：$1-\sqrt{3} + [(-1+2\sqrt{3})-(\sqrt{3}-2)]u + [(-1-2\sqrt{3})-(-\sqrt{3}-2)]u^2 + (1 + \sqrt{3})u^3 = 0$。计算常数项：$1-\sqrt{3}$。$u$的系数：$-1+2\sqrt{3}-\sqrt{3}+2 = 1 + \sqrt{3}$。$u^2$的系数：$-1-2\sqrt{3} + \sqrt{3}+2 = 1-\sqrt{3}$。$u^3$的系数：$1 + \sqrt{3}$。所以方程为：$(1-\sqrt{3}) + (1 + \sqrt{3})u + (1-\sqrt{3})u^2 + (1 + \sqrt{3})u^3 = 0$。分组：$(1 + \sqrt{3})u^3 + (1-\sqrt{3})u^2 + (1 + \sqrt{3})u + (1-\sqrt{3}) = 0$。观察到$u = -1$是一个根？代入$u = -1$：$-(1 + \sqrt{3}) + (1-\sqrt{3})-(1 + \sqrt{3}) + (1-\sqrt{3}) = -1-\sqrt{3}+1-\sqrt{3}-1-\sqrt{3}+1-\sqrt{3} = (-1+1-1+1) + (-\sqrt{3}-\sqrt{3}-\sqrt{3}-\sqrt{3}) = -4\sqrt{3} \neq 0$。$u = 1$？代入：$(1 + \sqrt{3}) + (1-\sqrt{3}) + (1 + \sqrt{3}) + (1-\sqrt{3}) = 4$，不为0。所以因式分解较困难。考虑到$C = \frac{2\pi}{3}$是特殊角，可能$A,B$也是特殊角。尝试$B = \frac{\pi}{6}$，则$A = \frac{\pi}{6}$，但$A + B = \frac{\pi}{3}$，$C = \frac{2\pi}{3}$。检验：$\frac{\cos A}{1 + \sin A} = \frac{\cos\frac{\pi}{6}}{1 + \sin\frac{\pi}{6}} = \frac{\frac{\sqrt{3}}{2}}{1 + \frac{1}{2}} = \frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}} = \frac{\sqrt{3}}{3}$，$\tan B = \tan\frac{\pi}{6} = \frac{\sqrt{3}}{3}$，相等。所以$B = \frac{\pi}{6}$满足。再检验$B = \frac{\pi}{3}$？则$A = 0$，不合理。所以$B = \frac{\pi}{6}$。
(2) 由
(1)中推导知$\frac{\cos A}{1 + \sin A} = \tan B$，所以$\cos A = \tan B(1 + \sin A)$，即$\cos A-\sin A\tan B = \tan B$，所以$\cos A\cos B-\sin A\sin B = \sin B$，即$\cos(A + B) = \sin B$。所以$\cos(\pi-C) = \sin B$，即$-\cos C = \sin B$，所以$\sin B + \cos C = 0$，即$\cos C = -\sin B$。因为$A,B,C$为三角形内角，所以$C = \frac{\pi}{2} + B$或$C = \frac{3\pi}{2}-B$（舍去，因为$C ＜ \pi$），所以$C = \frac{\pi}{2} + B$，则$A = \pi-B-C = \pi-B-(\frac{\pi}{2} + B) = \frac{\pi}{2}-2B$。因为$A ＞ 0$，所以$B ＜ \frac{\pi}{4}$。由正弦定理$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$，所以$\frac{a^2 + b^2}{c^2} = \frac{\sin^2 A + \sin^2 B}{\sin^2 C} = \frac{\sin^2(\frac{\pi}{2}-2B) + \sin^2 B}{\sin^2(\frac{\pi}{2} + B)} = \frac{\cos^2 2B + \sin^2 B}{\cos^2 B} = \frac{(1-2\sin^2 B)^2 + \sin^2 B}{\cos^2 B} = \frac{1-4\sin^2 B + 4\sin^4 B + \sin^2 B}{1-\sin^2 B} = \frac{1-3\sin^2 B + 4\sin^4 B}{1-\sin^2 B}$。令$t = \sin^2 B$，则$t\in(0,\frac{1}{2})$（因为$B ＜ \frac{\pi}{4}$，所以$\sin B ＜ \frac{\sqrt{2}}{2}$，$t ＜ \frac{1}{2}$）。则$\frac{a^2 + b^2}{c^2} = \frac{4t^2-3t + 1}{1-t}$。令$f(t) = \frac{4t^2-3t + 1}{1-t}$，$t\in(0,\frac{1}{2})$。求导$f'(t) = \frac{(8t-3)(1-t)-(4t^2-3t + 1)(-1)}{(1-t)^2} = \frac{(8t-3)(1-t) + (4t^2-3t + 1)}{(1-t)^2} = \frac{8t-3-8t^2+3t + 4t^2-3t + 1}{(1-t)^2} = \frac{-4t^2 + 8t-2}{(1-t)^2} = \frac{-2(2t^2-4t + 1)}{(1-t)^2}$。令$f'(t)=0$得$2t^2-4t + 1=0$，解得$t = \frac{4\pm\sqrt{16-8}}{4} = \frac{4\pm2\sqrt{2}}{4} = 1\pm\frac{\sqrt{2}}{2}$。因为$t\in(0,\frac{1}{2})$，所以$t = 1-\frac{\sqrt{2}}{2}$。当$t\in(0,1-\frac{\sqrt{2}}{2})$时，$f'(t) ＜ 0$；当$t\in(1-\frac{\sqrt{2}}{2},\frac{1}{2})$时，$f'(t) ＞ 0$。所以$f(t)$在$t = 1-\frac{\sqrt{2}}{2}$处取得极小值，也是最小值。最小值为$f(1-\frac{\sqrt{2}}{2}) = \frac{4(1-\frac{\sqrt{2}}{2})^2-3(1-\frac{\sqrt{2}}{2}) + 1}{1-(1-\frac{\sqrt{2}}{2})} = \frac{4(1-\sqrt{2}+\frac{1}{2})-3 + \frac{3\sqrt{2}}{2} + 1}{\frac{\sqrt{2}}{2}} = \frac{4(\frac{3}{2}-\sqrt{2})-2 + \frac{3\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{6-4\sqrt{2}-2 + \frac{3\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{4-\frac{5\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{8-5\sqrt{2}}{\sqrt{2}} = \frac{8}{\sqrt{2}}-5 = 4\sqrt{2}-5$。所以$\frac{a^2 + b^2}{c^2}$的最小值为$4\sqrt{2}-5$。

29. [2021 新高考全国Ⅰ卷,19]记△ABC 的内角$A,B,C$的对边分别为$a,b,c$,已知$b^2 = ac$,点$D$在边$AC$上,$BD\sin\angle ABC = a\sin C$.
(1) 求证:$BD = b$;
(2) 若$AD = 2DC$,求$\cos\angle ABC$的值.

答案： 29.
(1) 在△ABC中，由正弦定理得$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$。因为$BD\sin\angle ABC = a\sin C$，即$BD\sin B = a\sin C$，所以$BD = \frac{a\sin C}{\sin B}$。又因为$b^2 = ac$，所以$\frac{a}{b} = \frac{b}{c}$。由正弦定理$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$，得$\frac{a}{b} = \frac{\sin A}{\sin B}$，$\frac{b}{c} = \frac{\sin B}{\sin C}$，所以$\frac{\sin A}{\sin B} = \frac{\sin B}{\sin C}$，即$\sin A\sin C = \sin^2 B$。所以$a\sin C = 2R\sin A\sin C = 2R\sin^2 B = b\sin B$（因为$b = 2R\sin B$）。所以$BD = \frac{a\sin C}{\sin B} = \frac{b\sin B}{\sin B} = b$。故$BD = b$。
(2) 因为$AD = 2DC$，所以$\overrightarrow{BD} = \frac{1}{3}\overrightarrow{BA} + \frac{2}{3}\overrightarrow{BC}$（或由定比分点公式）。所以$|\overrightarrow{BD}|^2 = \frac{1}{9}|\overrightarrow{BA}|^2 + \frac{4}{9}|\overrightarrow{BC}|^2 + \frac{4}{9}\overrightarrow{BA}·\overrightarrow{BC}$。即$b^2 = \frac{1}{9}c^2 + \frac{4}{9}a^2 + \frac{4}{9}ac\cos B$。又$b^2 = ac$，所以$ac = \frac{1}{9}c^2 + \frac{4}{9}a^2 + \frac{4}{9}ac\cos B$。两边乘以9得$9ac = c^2 + 4a^2 + 4ac\cos B$，所以$4ac\cos B = 9ac-c^2-4a^2$，即$\cos B = \frac{9ac-c^2-4a^2}{4ac}$。又由余弦定理$\cos B = \frac{a^2 + c^2-b^2}{2ac} = \frac{a^2 + c^2-ac}{2ac}$。所以$\frac{a^2 + c^2-ac}{2ac} = \frac{9ac-c^2-4a^2}{4ac}$，两边乘以$4ac$得$2(a^2 + c^2-ac) = 9ac-c^2-4a^2$，即$2a^2 + 2c^2-2ac = 9ac-c^2-4a^2$，整理得$6a^2 + 3c^2-11ac = 0$，即$2a^2 + c^2-\frac{11}{3}ac = 0$。令$\frac{a}{c} = t$，则$2t^2 + 1-\frac{11}{3}t = 0$，即$6t^2-11t + 3 = 0$，解得$t = \frac{1}{3}$或$t = \frac{3}{2}$。当$t = \frac{a}{c} = \frac{1}{3}$时，$\cos B = \frac{a^2 + c^2-ac}{2ac} = \frac{\frac{1}{9}c^2 + c^2-\frac{1}{3}c^2}{2·\frac{1}{3}c^2} = \frac{\frac{1}{9} + 1-\frac{1}{3}}{\frac{2}{3}} = \frac{\frac{1}{9} + \frac{9}{9}-\frac{3}{9}}{\frac{2}{3}} = \frac{\frac{7}{9}}{\frac{2}{3}} = \frac{7}{9}×\frac{3}{2} = \frac{7}{6} ＞ 1$，舍去。当$t = \frac{a}{c} = \frac{3}{2}$时，$\cos B = \frac{(\frac{3}{2}c)^2 + c^2-\frac{3}{2}c^2}{2·\frac{3}{2}c^2} = \frac{\frac{9}{4}c^2 + c^2-\frac{3}{2}c^2}{3c^2} = \frac{\frac{9}{4} + 1-\frac{3}{2}}{3} = \frac{\frac{9}{4} + \frac{4}{4}-\frac{6}{4}}{3} = \frac{\frac{7}{4}}{3} = \frac{7}{12}$。所以$\cos\angle ABC = \frac{7}{12}$。

30. [2024 新高考Ⅱ卷,15]记△ABC 的内角$A,B,C$的对边分别为$a,b,c$,已知$\sin A + \sqrt{3}\cos A = 2$.
(1) 求角$A$的大小;
(2) 若$a = 2$,$\sqrt{2}b\sin C = c\sin 2B$,求△ABC 的周长.

答案： 30.
(1) 由$\sin A + \sqrt{3}\cos A = 2$，得$2(\frac{1}{2}\sin A + \frac{\sqrt{3}}{2}\cos A) = 2$，即$\sin A\cos\frac{\pi}{3} + \cos A\sin\frac{\pi}{3} = 1$，所以$\sin(A + \frac{\pi}{3}) = 1$。因为$A\in(0,\pi)$，所以$A + \frac{\pi}{3}\in(\frac{\pi}{3},\frac{4\pi}{3})$，所以$A + \frac{\pi}{3} = \frac{\pi}{2}$，得$A = \frac{\pi}{6}$。
(2) 由$\sqrt{2}b\sin C = c\sin 2B$，及正弦定理$\frac{b}{\sin B} = \frac{c}{\sin C} = 2R$，得$\sqrt{2}·2R\sin B·\sin C = 2R\sin C·2\sin B\cos B$。因为$\sin C ＞ 0$，$\sin B ＞ 0$，约去$2R\sin B\sin C$得$\sqrt{2} = 2\cos B$，所以$\cos B = \frac{\sqrt{2}}{2}$。因为$B\in(0,\pi)$，所以$B = \frac{\pi}{4}$。则$C = \pi-A-B = \pi-\frac{\pi}{6}-\frac{\pi}{4} = \frac{7\pi}{12}$。由正弦定理$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$，且$a = 2$，$\sin A = \sin\frac{\pi}{6} = \frac{1}{2}$，所以$2R = \frac{a}{\sin A} = \frac{2}{\frac{1}{2}} = 4$。所以$b = 2R\sin B = 4\sin\frac{\pi}{4} = 4×\frac{\sqrt{2}}{2} = 2\sqrt{2}$，$c = 2R\sin C = 4\sin\frac{7\pi}{12} = 4\sin(\frac{\pi}{4} + \frac{\pi}{6}) = 4(\sin\frac{\pi}{4}\cos\frac{\pi}{6} + \cos\frac{\pi}{4}\sin\frac{\pi}{6}) = 4(\frac{\sqrt{2}}{2}×\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}×\frac{1}{2}) = 4×\frac{\sqrt{2}(\sqrt{3}+1)}{4} = \sqrt{2}(\sqrt{3}+1) = \sqrt{6}+\sqrt{2}$。所以△ABC的周长为$a + b + c = 2 + 2\sqrt{2} + \sqrt{6}+\sqrt{2} = 2 + 3\sqrt{2} + \sqrt{6}$。