《2025年创新设计高考总复习高三数学人教版A版》

(2024·新高考Ⅱ卷)记$\triangle ABC$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$A$;
(2)若$a = 2$,$\sqrt{2}b\sin C = c\sin 2B$,求$\triangle ABC$的周长.解 (1)由$\sin A+\sqrt{3}\cos A = 2$，得$\frac{1}{2}\sin A+\frac{\sqrt{3}}{2}\cos A = 1$，所以$\sin(A+\frac{\pi}{3}) = 1$.因为$0＜A＜\pi$，所以$\frac{\pi}{3}＜A+\frac{\pi}{3}＜\frac{4\pi}{3}$，所以$A+\frac{\pi}{3}=\frac{\pi}{2}$，故$A=\frac{\pi}{6}$.(2)由$\sqrt{2}b\sin C = c\sin 2B$，得$\sqrt{2}b\sin C = 2c\sin B\cos B$，由正弦定理，得$\sqrt{2}bc = 2cb\cos B$，所以$\cos B=\frac{\sqrt{2}}{2}$，因为$0＜B＜\pi$，所以$B=\frac{\pi}{4}$.$C=\pi-(A + B)=\frac{7\pi}{12}$，所以$\sin C=\sin\frac{7\pi}{12}=\sin(\frac{\pi}{3}+\frac{\pi}{4})$$=\sin\frac{\pi}{3}\cos\frac{\pi}{4}+\cos\frac{\pi}{3}\sin\frac{\pi}{4}$$=\frac{\sqrt{3}}{2}×\frac{\sqrt{2}}{2}+\frac{1}{2}×\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$.由正弦定理得$b=\frac{a\sin B}{\sin A}=\frac{2\sin\frac{\pi}{4}}{\sin\frac{\pi}{6}}=2\sqrt{2}$，$c=\frac{a\sin C}{\sin A}=\frac{2\sin\frac{7\pi}{12}}{\sin\frac{\pi}{6}}=\sqrt{6}+\sqrt{2}$，

答案： 训练3 解
(1)由$\sin A+\sqrt{3}\cos A = 2$，得$\frac{1}{2}\sin A+\frac{\sqrt{3}}{2}\cos A = 1$，所以$\sin(A+\frac{\pi}{3}) = 1$.因为$0＜A＜\pi$，所以$\frac{\pi}{3}＜A+\frac{\pi}{3}＜\frac{4\pi}{3}$，所以$A+\frac{\pi}{3}=\frac{\pi}{2}$，故$A=\frac{\pi}{6}$.
(2)由$\sqrt{2}b\sin C = c\sin 2B$，得$\sqrt{2}b\sin C = 2c\sin B\cos B$，由正弦定理，得$\sqrt{2}bc = 2cb\cos B$，所以$\cos B=\frac{\sqrt{2}}{2}$，因为$0＜B＜\pi$，所以$B=\frac{\pi}{4}$.$C=\pi-(A + B)=\frac{7\pi}{12}$，所以$\sin C=\sin\frac{7\pi}{12}=\sin(\frac{\pi}{3}+\frac{\pi}{4})=\sin\frac{\pi}{3}\cos\frac{\pi}{4}+\cos\frac{\pi}{3}\sin\frac{\pi}{4}=\frac{\sqrt{3}}{2}×\frac{\sqrt{2}}{2}+\frac{1}{2}×\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$.由正弦定理得$b=\frac{a\sin B}{\sin A}=\frac{2\sin\frac{\pi}{4}}{\sin\frac{\pi}{6}}=2\sqrt{2}$，$c=\frac{a\sin C}{\sin A}=\frac{2\sin\frac{7\pi}{12}}{\sin\frac{\pi}{6}}=\sqrt{6}+\sqrt{2}$，所以$\triangle ABC$的周长为$a + b + c = 2+\sqrt{5}+3\sqrt{2}$.

典例
在$\triangle ABC$中,内角$A$,$B$,$C$的对边分别为$a$,$b$,$c$,若$\triangle ABC$为锐角三角形,且满足$\sin B(1 + 2\cos C)=2\sin A\cos C+\cos A\sin C$,则下列等式成立的是 (A)

A.$a = 2b$
B.$b = 2a$
C.$A = 2B$
D.$B = 2A$

答案： 典例A [法一 因为$\sin B(1 + 2\cos C)=2\sin A\cos C+\cos A\sin C$,所以$\sin B + 2\sin B\cos C=\sin A\cos C+\sin(A + C)$,所以$\sin B + 2\sin B\cos C=\sin A\cos C+\sin B$,即$\cos C(2\sin B-\sin A)=0$,所以$\cos C = 0$或$2\sin B=\sin A$,即$C = 90^{\circ}$或$2b = a$,又$\triangle ABC$为锐角三角形,所以$0^{\circ}＜C＜90^{\circ}$,故$2b = a$. 法二 由正弦和余弦定理得$b(1+\frac{a^{2}+b^{2}-c^{2}}{ab})=2a×\frac{a^{2}+b^{2}-c^{2}}{2ab}+c×\frac{b^{2}+c^{2}-a^{2}}{2bc}$所以$2b^{2}(1+\frac{a^{2}+b^{2}-c^{2}}{ab})=a^{2}+3b^{2}-c^{2}$,即$\frac{2b}{a}(a^{2}+b^{2}-c^{2})=a^{2}+3b^{2}-c^{2}$,即$(a^{2}+b^{2}-c^{2})(\frac{2b}{a}-1)=0$,所以$a^{2}+b^{2}=c^{2}$或$2b = a$,又$\triangle ABC$为锐角三角形,所以$a^{2}+b^{2}＞c^{2}$,故$2b = a$. 法三 由正弦定理及射影定理,得$b + 2b\cos C = 2a\cos C + c\cos A=a\cos C+(a\cos C + c\cos A)=a\cos C + b$,即$2b\cos C = a\cos C$,又因为$\triangle ABC$为锐角三角形,所以$\cos C\neq0$,所以$2b = a$.]

训练
(2025·济南模拟)已知$a$,$b$,$c$分别为$\triangle ABC$三个内角$A$,$B$,$C$的对边,且$a\cos C+\sqrt{3}a\sin C = b$,则$A=$ (A)

A.$\frac{\pi}{6}$
B.$\frac{\pi}{4}$
C.$\frac{\pi}{3}$
D.$\frac{\pi}{2}$

答案： 训练A [法一 由$a\cos C+\sqrt{3}a\sin C = b$及正弦定理得$\sin A\cos C+\sqrt{3}\sin A\sin C=\sin B=\sin(A + C)=\sin A\cos C+\cos A\sin C$,所以$\sqrt{3}\sin A\sin C=\cos A\sin C$,因为$\sin C\neq0$,所以$\sqrt{3}\sin A=\cos A$,$\tan A=\frac{\sqrt{3}}{3}$,由$A\in(0,\pi)$,得$A=\frac{\pi}{6}$. 法二(射影定理) 由射影定理知$b = a\cos C + c\cos A$,所以$a\cos C+\sqrt{3}a\sin C = a\cos C + c\cos A$,即$\sqrt{3}a\sin C = c\cos A$,由正弦定理得$\sqrt{3}\sin A\sin C=\sin C\cos A$,又$\sin C\neq0$,所以$\tan A=\frac{\sqrt{3}}{3}$,由$A\in(0,\pi)$,得$A=\frac{\pi}{6}$.]