《2025年小题狂做高中数学必修第二册人教版巅峰版》

22. [2023 新高考Ⅰ卷,17]已知在△ABC 中,$A + B = 3C$,$2\sin(A - C) = \sin B$.
(1) 求$\sin A$;
(2) 若$AB = 5$,求边$AB$上的高.

答案： 22.
(1) 因为$A + B = 3C$，且$A + B + C = \pi$，所以$4C = \pi$，即$C = \frac{\pi}{4}$。因为$2\sin(A - C) = \sin B$，所以$2\sin(A - \frac{\pi}{4}) = \sin(\frac{3\pi}{4}-A) = \sin(\pi-\frac{3\pi}{4} + A) = \sin(\frac{\pi}{4} + A)$。展开得$2(\sin A\cos\frac{\pi}{4}-\cos A\sin\frac{\pi}{4}) = \sin A\cos\frac{\pi}{4} + \cos A\sin\frac{\pi}{4}$，即$\sqrt{2}\sin A-\sqrt{2}\cos A = \frac{\sqrt{2}}{2}\sin A + \frac{\sqrt{2}}{2}\cos A$，整理得$\frac{\sqrt{2}}{2}\sin A = \frac{3\sqrt{2}}{2}\cos A$，所以$\tan A = 3$。因为$A\in(0,\pi)$，所以$\sin A = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$。
(2) 由
(1)知$\sin A = \frac{3\sqrt{10}}{10}$，$\cos A = \frac{\sqrt{10}}{10}$，$C = \frac{\pi}{4}$，所以$B = \pi-A-C = \pi-\arcsin\frac{3\sqrt{10}}{10}-\frac{\pi}{4}$。由正弦定理$\frac{BC}{\sin A} = \frac{AB}{\sin C}$，得$BC = \frac{AB\sin A}{\sin C} = \frac{5×\frac{3\sqrt{10}}{10}}{\frac{\sqrt{2}}{2}} = 3\sqrt{5}$。设边$AB$上的高为$h$，则$S_{\triangle ABC} = \frac{1}{2}AB· h = \frac{1}{2}BC· AC·\sin C$。又由正弦定理$\frac{AC}{\sin B} = \frac{AB}{\sin C}$，且$\sin B = \sin(A + C) = \sin A\cos C + \cos A\sin C = \frac{3\sqrt{10}}{10}×\frac{\sqrt{2}}{2} + \frac{\sqrt{10}}{10}×\frac{\sqrt{2}}{2} = \frac{4\sqrt{20}}{20} = \frac{2\sqrt{5}}{5}$，所以$AC = \frac{AB\sin B}{\sin C} = \frac{5×\frac{2\sqrt{5}}{5}}{\frac{\sqrt{2}}{2}} = 2\sqrt{10}$。所以$\frac{1}{2}×5× h = \frac{1}{2}×3\sqrt{5}×2\sqrt{10}×\frac{\sqrt{2}}{2}$，即$5h = 3\sqrt{5}×2\sqrt{10}×\frac{\sqrt{2}}{2} = 3\sqrt{5}×\sqrt{20} = 3\sqrt{5}×2\sqrt{5} = 30$，所以$h = 6$。故边$AB$上的高为$6$。

23. [2025 北京卷,16]在△ABC 中,$\cos A = -\frac{1}{3}$,$a\sin C = 4\sqrt{2}$.
(1) 求$c$.
(2) 在以下三个条件中选择一个作为已知,使得△ABC 存在,求边$BC$上的高.
①$a = 6$;②$b\sin C = \frac{10\sqrt{2}}{3}$;③△ABC 的面积为$10\sqrt{2}$.

答案： 23.
(1) 由正弦定理$\frac{a}{\sin A} = \frac{c}{\sin C}$，得$a\sin C = c\sin A$。已知$a\sin C = 4\sqrt{2}$，所以$c\sin A = 4\sqrt{2}$。因为$\cos A = -\frac{1}{3}$，所以$A$为钝角，$\sin A = \sqrt{1-\cos^2 A} = \sqrt{1-\frac{1}{9}} = \frac{2\sqrt{2}}{3}$。代入得$c×\frac{2\sqrt{2}}{3} = 4\sqrt{2}$，解得$c = 6$。
(2) 由
(1)知$c = 6$，$\sin A = \frac{2\sqrt{2}}{3}$，$\cos A = -\frac{1}{3}$。
选择条件①：$a = 6$。由正弦定理$\frac{a}{\sin A} = \frac{c}{\sin C}$，得$\frac{6}{\frac{2\sqrt{2}}{3}} = \frac{6}{\sin C}$，所以$\sin C = \frac{2\sqrt{2}}{3}$。因为$c = 6$，$a = 6$，所以$a = c$，则$A = C$，又$\cos A = -\frac{1}{3}$，所以$C$为钝角，$\cos C = -\frac{1}{3}$。所以$B = \pi-A-C = \pi-2A$，$\sin B = \sin(\pi-2A) = \sin2A = 2\sin A\cos A = 2×\frac{2\sqrt{2}}{3}×(-\frac{1}{3}) = -\frac{4\sqrt{2}}{9}$（舍去，因为$\sin B ＞ 0$）或利用$\cos2A = 2\cos^2 A-1 = 2×\frac{1}{9}-1 = -\frac{7}{9}$，$\sin B = \sin(\pi-2A) = \sin2A = \sqrt{1-\cos^2 2A} = \sqrt{1-\frac{49}{81}} = \frac{4\sqrt{2}}{9}$。由正弦定理$\frac{b}{\sin B} = \frac{a}{\sin A}$，得$b = \frac{a\sin B}{\sin A} = \frac{6×\frac{4\sqrt{2}}{9}}{\frac{2\sqrt{2}}{3}} = 4$。设边$BC$上的高为$h_a$，则$S_{\triangle ABC} = \frac{1}{2}ah_a = \frac{1}{2}bc\sin A$，所以$h_a = \frac{bc\sin A}{a} = \frac{4×6×\frac{2\sqrt{2}}{3}}{6} = \frac{8\sqrt{2}}{3}$。
选择条件②：$b\sin C = \frac{10\sqrt{2}}{3}$。由正弦定理$\frac{b}{\sin B} = \frac{c}{\sin C}$，得$b\sin C = c\sin B = 6\sin B = \frac{10\sqrt{2}}{3}$，所以$\sin B = \frac{5\sqrt{2}}{9}$。因为$\sin A = \frac{2\sqrt{2}}{3} \approx 0.9428$，$\sin B = \frac{5\sqrt{2}}{9} \approx 0.7857$，且$A$为钝角，$B$为锐角，则$\cos B = \sqrt{1-\sin^2 B} = \sqrt{1-\frac{50}{81}} = \frac{\sqrt{31}}{9}$。所以$\sin C = \sin(A + B) = \sin A\cos B + \cos A\sin B = \frac{2\sqrt{2}}{3}×\frac{\sqrt{31}}{9} + (-\frac{1}{3})×\frac{5\sqrt{2}}{9} = \frac{2\sqrt{62}}{27}-\frac{5\sqrt{2}}{27} = \frac{\sqrt{2}(2\sqrt{31}-5)}{27}$。由正弦定理$\frac{a}{\sin A} = \frac{c}{\sin C}$，得$a = \frac{c\sin A}{\sin C} = \frac{6×\frac{2\sqrt{2}}{3}}{\frac{\sqrt{2}(2\sqrt{31}-5)}{27}} = \frac{4}{\frac{2\sqrt{31}-5}{27}} = \frac{108}{2\sqrt{31}-5}$。设边$BC$上的高为$h_a$，则$S_{\triangle ABC} = \frac{1}{2}ah_a = \frac{1}{2}bc\sin A$，所以$h_a = \frac{bc\sin A}{a} = \frac{b×6×\frac{2\sqrt{2}}{3}}{a} = \frac{4\sqrt{2}b}{a}$。又由正弦定理$\frac{b}{\sin B} = \frac{c}{\sin C}$，得$b = \frac{c\sin B}{\sin C} = \frac{6×\frac{5\sqrt{2}}{9}}{\frac{\sqrt{2}(2\sqrt{31}-5)}{27}} = \frac{\frac{10\sqrt{2}}{3}}{\frac{\sqrt{2}(2\sqrt{31}-5)}{27}} = \frac{10}{3}×\frac{27}{2\sqrt{31}-5} = \frac{90}{2\sqrt{31}-5}$。所以$h_a = \frac{4\sqrt{2}×\frac{90}{2\sqrt{31}-5}}{\frac{108}{2\sqrt{31}-5}} = \frac{4\sqrt{2}×90}{108} = \frac{10\sqrt{2}}{3}$。
选择条件③：△ABC 的面积为$10\sqrt{2}$。$S_{\triangle ABC} = \frac{1}{2}bc\sin A = \frac{1}{2}× b×6×\frac{2\sqrt{2}}{3} = 2\sqrt{2}b = 10\sqrt{2}$，所以$b = 5$。由余弦定理$a^2 = b^2 + c^2-2bc\cos A = 25 + 36-2×5×6×(-\frac{1}{3}) = 61 + 20 = 81$，所以$a = 9$。设边$BC$上的高为$h_a$，则$S_{\triangle ABC} = \frac{1}{2}ah_a = 10\sqrt{2}$，所以$h_a = \frac{20\sqrt{2}}{a} = \frac{20\sqrt{2}}{9}$。

24. [2022 全国乙卷理,17]记△ABC 的内角$A,B,C$的对边分别为$a,b,c$,已知$\sin C\sin(A - B) = \sin B\sin(C - A)$.
(1) 求证:$2a^2 = b^2 + c^2$;
(2) 若$a = 5$,$\cos A = \frac{25}{31}$,求△ABC 的周长.

答案： 24.
(1) 由$\sin C\sin(A - B) = \sin B\sin(C - A)$，得$\sin C(\sin A\cos B-\cos A\sin B) = \sin B(\sin C\cos A-\cos C\sin A)$，即$\sin A\sin C\cos B-\cos A\sin B\sin C = \sin B\sin C\cos A-\sin A\sin B\cos C$。因为$\sin B\sin C \neq 0$，所以$\sin A\cos B-\cos A\sin B = \sin B\cos A-\frac{\sin A\sin B\cos C}{\sin C}$。整理得$\sin A\cos B = 2\sin B\cos A-\frac{\sin A\sin B\cos C}{\sin C}$。由正弦定理和余弦定理，将角化为边：$\sin A = \frac{a}{2R}$，$\cos B = \frac{a^2 + c^2-b^2}{2ac}$，$\sin B = \frac{b}{2R}$，$\cos A = \frac{b^2 + c^2-a^2}{2bc}$，$\cos C = \frac{a^2 + b^2-c^2}{2ab}$，代入整理可得$2a^2 = b^2 + c^2$。
另证：由$\sin C\sin(A - B) = \sin B\sin(C - A)$，根据和差化积公式或直接展开，利用正弦定理化边，可得：$a(\cos B-\cos A) = b(\cos A-\cos C)$？更直接的方法：由已知得$\sin C\sin A\cos B-\sin C\cos A\sin B = \sin B\sin C\cos A-\sin B\cos C\sin A$，即$\sin A\sin C\cos B + \sin A\sin B\cos C = 2\sin B\sin C\cos A$。由正弦定理$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$，得$\sin A = \frac{a}{2R}$，$\sin B = \frac{b}{2R}$，$\sin C = \frac{c}{2R}$。代入得$\frac{a}{2R}·\frac{c}{2R}\cos B + \frac{a}{2R}·\frac{b}{2R}\cos C = 2·\frac{b}{2R}·\frac{c}{2R}\cos A$，即$ac\cos B + ab\cos C = 2bc\cos A$。由余弦定理：$ac·\frac{a^2 + c^2-b^2}{2ac} + ab·\frac{a^2 + b^2-c^2}{2ab} = 2bc·\frac{b^2 + c^2-a^2}{2bc}$，所以$\frac{a^2 + c^2-b^2}{2} + \frac{a^2 + b^2-c^2}{2} = b^2 + c^2-a^2$，即$a^2 = b^2 + c^2-a^2$，所以$2a^2 = b^2 + c^2$。
(2) 由
(1)知$2a^2 = b^2 + c^2$，又$a = 5$，所以$b^2 + c^2 = 2×25 = 50$。由余弦定理$\cos A = \frac{b^2 + c^2-a^2}{2bc} = \frac{50-25}{2bc} = \frac{25}{2bc} = \frac{25}{31}$，所以$bc = \frac{25}{2}×\frac{31}{25} = \frac{31}{2}$。所以$(b + c)^2 = b^2 + c^2 + 2bc = 50 + 31 = 81$，所以$b + c = 9$。故△ABC的周长为$a + b + c = 5 + 9 = 14$。

25. [2024 新高考Ⅰ卷,15]记△ABC 的内角$A,B,C$的对边分别为$a,b,c$,已知$\sin C = \sqrt{2}\cos B$,$a^2 + b^2 - c^2 = \sqrt{2}ab$.
(1) 求角$B$的大小;
(2) 若△ABC 的面积为$3 + \sqrt{3}$,求$c$.

答案： 25.
(1) 因为$\sin C = \sqrt{2}\cos B$，由正弦定理得$c = \sqrt{2}a\cos B$。又由余弦定理$a^2 + b^2-c^2 = 2ab\cos C$，结合已知$a^2 + b^2-c^2 = \sqrt{2}ab$，所以$2ab\cos C = \sqrt{2}ab$，因为$ab \neq 0$，所以$\cos C = \frac{\sqrt{2}}{2}$，又$C\in(0,\pi)$，所以$C = \frac{\pi}{4}$。因为$\sin C = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$，代入$\sin C = \sqrt{2}\cos B$得$\frac{\sqrt{2}}{2} = \sqrt{2}\cos B$，所以$\cos B = \frac{1}{2}$，又$B\in(0,\pi)$，所以$B = \frac{\pi}{3}$。
(2) 由
(1)知$B = \frac{\pi}{3}$，$C = \frac{\pi}{4}$，所以$A = \pi-B-C = \pi-\frac{\pi}{3}-\frac{\pi}{4} = \frac{5\pi}{12}$。由正弦定理$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$，得$a = 2R\sin A$，$b = 2R\sin B$，$c = 2R\sin C$。△ABC的面积$S = \frac{1}{2}ab\sin C = \frac{1}{2}(2R\sin A)(2R\sin B)\sin C = 2R^2\sin A\sin B\sin C = 3 + \sqrt{3}$。因为$\sin A = \sin\frac{5\pi}{12} = \sin(\frac{\pi}{4} + \frac{\pi}{6}) = \sin\frac{\pi}{4}\cos\frac{\pi}{6} + \cos\frac{\pi}{4}\sin\frac{\pi}{6} = \frac{\sqrt{2}}{2}×\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}×\frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}$，$\sin B = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$，$\sin C = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$，所以$2R^2×\frac{\sqrt{6}+\sqrt{2}}{4}×\frac{\sqrt{3}}{2}×\frac{\sqrt{2}}{2} = 3 + \sqrt{3}$。计算得$\frac{\sqrt{6}+\sqrt{2}}{4}×\frac{\sqrt{3}}{2}×\frac{\sqrt{2}}{2} = \frac{(\sqrt{6}+\sqrt{2})\sqrt{6}}{16} = \frac{6+\sqrt{12}}{16} = \frac{6+2\sqrt{3}}{16} = \frac{3+\sqrt{3}}{8}$。所以$2R^2×\frac{3+\sqrt{3}}{8} = 3 + \sqrt{3}$，解得$2R^2 = 8$，即$R^2 = 4$，$R = 2$。所以$c = 2R\sin C = 4×\frac{\sqrt{2}}{2} = 2\sqrt{2}$。