物理

[TINY NSD]三角系列（续）：三角形中的关系

${（有同学反映帖子有点卡，所以把不等式移到下一个帖子了）}$

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

${1.三角恒等式}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

${=1}$