物理

[TINY NSD]三角系列（续）：三角形中的关系

${（以下内容中，A，B，C为三角形三个内角，D为BC边上一点，a，b，c表示对应角所对的边，默认所有式子均有意义）}$

${1.三角恒等式}$

${1.1~\sin A+\sin B+\sin C=4\cos\frac{A}{2}\cos \frac{B}{2} \cos \frac{C}{2}}$

${[证明] \sin A+\sin B+\sin C=\sin A+\sin B+\sin (A+B)}$

${=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}+2\sin\frac{A+B}{2}\cos\frac{A+B}{2}}$

${=2\sin\frac{A+B}{2}(\cos\frac{A+B}{2}+\cos\frac{A-B}{2})}$

${=2\cos\frac{C}{2}\cdot(2\cos\frac{A}{2}\cos\frac{B}{2})}$

${=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}}$

${1.2~\cos A+\cos B+\cos C=4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}+1}$

${[证明]\cos A+\cos B+\cos C=\cos A+\cos B-\cos (A+B)}$

${=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}-(2\cos^2\frac{A+B}{2}-1)}$

${=2\cos\frac{A+B}{2}(\cos\frac{A-B}{2}-\cos\frac{A+B}{2})+1}$

${=2\sin \frac{C}{2}\cdot(2\sin\frac{A}{2}\sin \frac{B}{2})+1}$

${=4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}+1}$

${1.3~\tan A+\tan B+\tan C=\tan A\tan B\tan C}$

${[证明]\tan A+\tan B+\tan C=\tan (A+B)(1-\tan A \tan B)-\tan (A+B)}$

${=-\tan (A+B) \tan A \tan B}$

${=\tan A \tan B \tan C}$

${1.4~\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C}$

${[证明]\sin2A+\sin2B+\sin2C=2\sin(A+B)\cos(A-B)+2\sin C\cos C}$

${=2\sin C\cos (A-B)+2\sin C\cos C}$

${=2\sin C [\cos (A-B)-\cos (A+B)]}$

${=2\sin C \cdot (2 \sin A \sin B)}$

${=4 \sin A\sin B\sin C}$

${1.5~\cos 2A+\cos 2B+\cos 2C=-1-4\cos A\cos B\cos C}$

${[证明]\cos2A+\cos2B+\cos2C=2\cos (A+B)\cos (A-B)+2\cos^2 C-1}$

${=-1-2\cos C \cos (A+B)-2\cos (A-B) \cos C}$

${=-1-2\cos C [\cos(A-B)+\cos (A+B)]}$

${=-1-2\cos C \cdot (2\cos A \cos B)}$

${=-1-4\cos A \cos B \cos C}$

${1.6~\cos^2 A+\cos^2 B+\cos^2 C=1-2\cos A\cos B \cos C}$

${[证明]\cos^2A+\cos^2B+\cos^2C=\frac{1+\cos 2A}{2}+\frac{1+\cos 2B}{2}+\frac{1+\cos2C}{2}}$

${=\frac{3}{2}+\frac{\cos2A+\cos2B+\cos2C}{2}}$

${=\frac{3}{2}+\frac{-1-4\cos A \cos B \cos C}{2}（由1.5）}$

${=1-2\cos A \cos B \cos C}$

${1.7~\sin^2 A+\sin^2 B+\sin^2 C=2+2\cos A\cos B \cos C}$

${[证明]\sin^2A+\sin^2B+\sin^2C=\frac{1-\cos 2A}{2}+\frac{1-\cos 2B}{2}+\frac{1-\cos2C}{2}}$

${=\frac{3}{2}-\frac{\cos2A+\cos2B+\cos2C}{2}}$

${=\frac{3}{2}-\frac{-1-4\cos A \cos B \cos C}{2}（由1.5）}$

${=2+2\cos A \cos B \cos C}$

${1.8~\tan \frac{A}{2} \tan\frac{B}{2}+\tan\frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan\frac{A}{2}=1}$

${[证明]\tan\frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2}\tan\frac{A}{2}=\tan \frac{B}{2} (\tan \frac{A}{2}+\tan\frac{C}{2})+\tan\frac{A}{2} \tan \frac{C}{2}}$

${=\tan \frac{B}{2} \tan\frac{A+C}{2}(1-\tan \frac{A}{2} \tan \frac{C}{2})+\tan \frac{A}{2} \tan\frac{C}{2}}$

${=(1-\tan \frac{A}{2}\tan\frac{C}{2})+\tan \frac{A}{2} \tan \frac{C}{2}}$

${=1}$

${2.三角不等式（高考党慎入）}$

${2.1~\sin \frac{A}{2}+\sin\frac{B}{2}+\sin \frac{C}{2} \le \frac{3}{2}}$

${[证明]令f(x)=\sin x(0 \lt x \lt \frac{\pi}{2})，则f''(x)=-\sin x \lt 0}$

${又\frac{A}{2},\frac{B}{2}, \frac{C}{2} \in (0, \frac{\pi}{2})且\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}}$

${由Jensen不等式，\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2} \le 3\sin\frac{\pi}{6} =\frac{3}{2}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.2~\cos \frac{A}{2}+\cos\frac{B}{2}+\cos \frac{C}{2} \le \frac{3\sqrt{3}}{2}}$

${[证明]令f(x)=\cos x(0 \lt x \lt \frac{\pi}{2})，则f''(x)=-\cos x \lt 0}$

${又\frac{A}{2},\frac{B}{2}, \frac{C}{2} \in (0, \frac{\pi}{2})且\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}}$

${由Jensen不等式，\cos \frac{A}{2}+\cos \frac{B}{2}+\cos \frac{C}{2} \le 3\cos\frac{\pi}{6} =\frac{3\sqrt{3}}{2}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.3~\tan \frac{A}{2}+\tan\frac{B}{2}+\tan \frac{C}{2} \ge \sqrt{3}}$

${[证明]令f(x)=\tan x(0 \lt x \lt \frac{\pi}{2})，则f''(x)=\frac{2\sin x}{\cos^3 x} \gt 0}$

${又\frac{A}{2},\frac{B}{2}, \frac{C}{2} \in (0, \frac{\pi}{2})且\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}}$

${由Jensen不等式，\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2} \ge 3\tan\frac{\pi}{6} =\sqrt{3}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.4~\sin \frac{A}{2} \sin \frac{B}{2}\sin \frac{C}{2} \le \frac{1}{8}}$

${[证明]由AM-GM不等式与2.1，\sin\frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \le (\frac{\sin \frac{A}{2}+\sin\frac{B}{2}+\sin \frac{C}{2}}{3})^3 \le (\frac{1}{2})^3 =\frac{1}{8}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.5~\cos \frac{A}{2} \cos \frac{B}{2}\cos \frac{C}{2} \le \frac{3\sqrt{3}}{8}}$

${[证明]由AM-GM不等式与2.1，\cos\frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \le (\frac{\cos \frac{A}{2}+\cos\frac{B}{2}+\cos \frac{C}{2}}{3})^3 \le (\frac{\sqrt{3}}{2})^3=\frac{3\sqrt{3}}{8}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$                  

${2.6~\tan\frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2} \le \frac{\sqrt{3}}{9}}$

${[证明]由1.8与AM-GM不等式，\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2} =\sqrt{\tan^2 \frac{A}{2}\tan^2\frac{B}{2}\tan^2\frac{C}{2}}}$

${\le \sqrt{(\frac{\tan\frac{A}{2} \tan\frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2}\tan\frac{A}{2}}{3})^3}=(\frac{1}{3})^\frac{3}{2}=\frac{\sqrt{3}}{9}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.7~\sin A+\sin B+\sin C\le\frac{3\sqrt{3}}{2}}$

${[证法1]令f(x)=\sin x(0 \lt x \lt \pi),则f''(x)=-\sin x\lt 0}$

${而A, B, C \in(0,\pi)且A+B+C=\pi}$

${由Jensen不等式，\sin A+\sin B+\sin C \le 3\sin \frac{\pi}{3}=\frac{3\sqrt{3}}{2}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${[证法2]由2.5及1.1，\sin A+\sin B+\sin C=4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \le 4 \cdot\frac{3\sqrt{3}}{8}=\frac{3\sqrt{3}}{2}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.8~\cos A+\cos B+\cos C \le\frac{3}{2}}$

${[证法1]不妨设A \ge B \ge C，则B \lt \frac{\pi}{2}, C\le\frac{\pi}{3}, A \ge \frac{\pi}{3}}$

${令f(x)=\cos x+\cos(\theta-x), 其中\theta \in (\frac{\pi}{2}, \pi), x\in(0, \theta)}$

${则f'(x)=\sin(\theta-x)-\sin x=2\cos \frac{\theta}{2} \sin (\frac{\theta}{2}-x)}$

${由于\sin\frac{\theta}{2}\lt 0, \cos \frac{\theta}{2} \lt 0, f(x)在(0,\frac{\theta}{2})时单调递增，在(\frac{\theta}{2},\theta)单调递减}$

${令\theta=A+C, 则\frac{\pi}{2} \lt \theta \lt \pi}$

${(1)若\frac{\pi}{2}\lt\theta \le \frac{2\pi}{3}，则A \ge \frac{\pi}{3}\ge\frac{\theta}{2}}$

${故f(A) \le f(\frac{\pi}{3})，即\cos A+\cos C \le \cos\frac{\pi}{3}+\cos (\frac{2\pi}{3}-B)}$

${令\theta=\frac{2\pi}{3}，则f(B) \le f(\frac{\pi}{3})，即\cos B+\cos(\frac{2\pi}{3}-B) \le 2\cos \frac{\pi}{3}}$

${故\cos A+\cos B+\cos C \le 3\cos \frac{\pi}{3} =\frac{3}{2}}$

${(2)若\frac{2\pi}{3}\lt\theta \lt \pi，则C \le \frac{\pi}{3}\lt\frac{\theta}{2}}$

${故f(C) \lt f(\frac{\pi}{3})，即\cos A+\cos C \lt \cos\frac{\pi}{3}+\cos (\frac{2\pi}{3}-B)}$

${令\theta=\frac{2\pi}{3}，则f(B) \le f(\frac{\pi}{3})，即\cos B+\cos(\frac{2\pi}{3}-B) \le 2\cos \frac{\pi}{3}}$

${故\cos A+\cos B+\cos C \lt 3\cos \frac{\pi}{3} =\frac{3}{2}}$

${综上，总有\cos A+\cos B+\cos C \le \frac{3}{2}，取等当且仅当A=B=C=\frac{\pi}{3}}$

${[证法2]不妨设A \ge B \ge C}$

${则0\lt\frac{A+B}{2} \lt \frac{\pi}{2}, \cos \frac{A+B}{2} \gt 0}$

${\cos A+\cos B=2\cos \frac{A+B}{2}\cos\frac{A-B}{2} \le 2\cos \frac{A+B}{2}}$

${令f(x)=\cos x(0\lt x \lt \frac{\pi}{2})，则f''(x)=-\cos x \lt 0}$

${而\frac{A+B}{2},C\in (0, \frac{\pi}{2})}$

${故由Jensen不等式，2\cos \frac{A+B}{2} +\cos C \le 3\cos \frac{\pi}{3}=\frac{3}{2}}$

${从而\cos A+\cos B+\cos C \le 2\cos \frac{A+B}{2}+\cos C \le \frac{3}{2}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${[证法3]由2.4及1.2，\cos A+\cos B+\cos C=4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}+1 \le 4 \cdot\frac{1}{8}+1=\frac{3}{2}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.9~\tan A+\tan B+\tan C \ge 3\sqrt{3} 或 \lt 0}$

${[证明]若\tan A+\tan B+\tan C \ge 0}$

${则由AM-GM不等式，\tan A+\tan B+\tan C \ge 3\sqrt[3]{\tan A\tan B\tan C}=3\sqrt[3]{\tan A+\tan B+\tan C}（由1.3）}$

${解得\tan A+\tan B+\tan C \ge 3\sqrt{3}}$

${故\tan A+\tan B+\tan C \lt 0 或 \ge 3\sqrt{3}，取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.10~\sin A \sin B \sin C \le\frac{3\sqrt{3}}{8}}$

${[证明]由2.7和AM-GM不等式，\sin A\sin B \sin C \le (\frac{\sin A+\sin B+\sin C}{3})^3 \le \frac{3\sqrt{3}}{8}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.11~\tan A \tan B \tan C \ge 3\sqrt{3} 或\lt 0}$

${[证明]由1.5和2.9，\tan A \tan B \tan C =\tan A+\tan B+\tan C \ge 3\sqrt{3} 或 \lt 0}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.12~-1 \lt \cos A\cos B\cos C \le\frac{1}{8}}$

${[证法1]若A,B,C \in (0,\frac{\pi}{2})，由2.8和AM-GM不等式，\cos A \cos B \cos C \le(\frac{\cos A+\cos B+\cos C}{3})^3 \le \frac{1}{8}}$

${否则，不妨设A \ge B \ge C，由于B \lt \frac{\pi}{2}，必有A \ge\frac{\pi}{2}，从而\cos A \cos B \cos C \le 0 \lt\frac{1}{8}}$

${而由于A,B,C \in(0,\pi),有|\cos A|, |\cos B|, |\cos C| \lt 1}$

${故\cos A \cos B\cos C \ge -|\cos A \cos B \cos C| \gt -1}$

${综上，-1 \lt \cos A\cos B\cos C \le \frac{1}{8}，右侧取等当且仅当A=B=C=\frac{\pi}{3}}$

${[证法2]由2.10及2.11：}$

${(1)若\tan A \tan B\tan C \lt 0，则由于\sin A \sin B \sin C \gt 0，有\cos A \cos B \cos C=\frac{\sin A \sin B \sin C}{\tan A \tan B \tan C} \lt 0 \lt \frac{1}{8}}$

${(2)若\tan A \tan B\tan C \ge 3\sqrt{3}，则由于0 \lt \sin A \sin B \sin C \le\frac{3\sqrt{3}}{8}}$

${有\cos A \cos B\cos C=\frac{\sin A \sin B \sin C}{\tan A \tan B \tan C} \le\frac{\frac{3\sqrt{3}}{8}}{3\sqrt{3}}=\frac{1}{8}}$

${(3)若\tan A \tan B\tan C无意义，则三角形为直角三角形，显然\cos A \cos B \cos C =0 \lt\frac{1}{8}}$

${而由于A,B,C \in(0,\pi),有|\cos A|, |\cos B|, |\cos C| \lt 1}$

${故\cos A \cos B\cos C \ge -|\cos A \cos B \cos C| \gt -1}$

${综上，-1 \lt \cos A\cos B\cos C \le \frac{1}{8}，右侧取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.13~\sin 2A+\sin 2B+\sin 2C \le\frac{3\sqrt{3}}{2}}$

${[证明]由1.4及2.10，\sin2A+\sin 2B+\sin 2C =4\sin A \sin B \sin C \le \frac{3\sqrt{3}}{2}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${2.14~\cos 2A+\cos 2B+\cos 2C \ge-\frac{3}{2}}$

${[证法1]由1.5及2.12，\cos2A+\cos 2B+\cos 2C=-1-4\cos A \cos B \cos C \ge -1-4\cdot \frac{1}{8}=-\frac{3}{2}}$

${取等当且仅当A=B=C=\frac{\pi}{3}}$

${[证法2]不妨设A \ge B \ge C, 则B \lt \frac{\pi}{2}, 2(A+C)\in (\pi, 2\pi),A \ge \frac{\pi}{3} \ge C}$

${令f(x)=\cos x+\cos (\theta-x),其中\theta \in (\pi, 2\pi), x \in (0,\theta)}$

${则f'(x)=\sin(\theta-x)-\sin x=2\cos \frac{\theta}{2} \sin(\frac{\theta}{2}-x)}$

${而\cos\frac{\theta}{2} \lt 0，故f(x)在(0,\frac{\theta}{2})单调递减，(\frac{\theta}{2}, \theta)单调递增}$

${令\theta=2(A+C)}$

${(1)若\pi \lt\theta \le \frac{4\pi}{3}，则2A \ge \frac{2\pi}{3} \ge\frac{\theta}{2}}$

${故\cos 2A+\cos 2C=f(2A) \ge f(\frac{2\pi}{3}) =\cos \frac{2\pi}{3}+\cos (\theta-\frac{2\pi}{3})=\cos\frac{2\pi}{3}+\cos (\frac{4\pi}{3}-2B)}$

${又令\theta=\frac{4\pi}{3}，则f(2B) \le f(\frac{2\pi}{3}),\cos(\frac{4\pi}{3}-2B)+ \cos 2B \ge2\cos \frac{2\pi}{3}}$

${故\cos 2A+\cos2B+\cos 2C \ge \cos \frac{2\pi}{3}+\cos (\frac{4\pi}{3}-2B) +\cos 2B \ge 3\cos\frac{2\pi}{3}=-\frac{3}{2}}$

${(2)若\frac{4\pi}{3}\lt \theta \lt \pi，则2C \lt \frac{2\pi}{3} \le\frac{\theta}{2}}$

${故\cos 2A+\cos 2C=f(2C) \gt f(\frac{2\pi}{3}) =\cos \frac{2\pi}{3}+\cos (\theta-\frac{2\pi}{3})=\cos\frac{2\pi}{3}+\cos (\frac{4\pi}{3}-2B)}$

${又令\theta=\frac{4\pi}{3}，则f(2B) \le f(\frac{2\pi}{3}),\cos(\frac{4\pi}{3}-2B)+ \cos 2B \ge2\cos \frac{2\pi}{3}}$

${故\cos 2A+\cos2B+\cos 2C \gt \cos \frac{2\pi}{3}+\cos (\frac{4\pi}{3}-2B) +\cos 2B \ge 3\cos\frac{2\pi}{3}=-\frac{3}{2}}$

${综上，总有\cos 2A+\cos2B+\cos 2C \ge -\frac{3}{2}，取等当且仅当A=B=C=\frac{\pi}{3}}$

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