三角形中的几个恒等式

在 \(\Delta ABC\) 中，以下等式成立

正余弦平方和

\[ \sin A+\sin B+\sin C=4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\ \tag{1} \]

证：

\[ \begin{aligned} LHS &= 2\,\sin \frac{A+B}{2}\,\cos \frac{A-B}{2} +\sin C \\ &= 2 \,\cos \frac{C}{2}\,\cos \frac{A-B}{2}+2\,\sin \frac{C}{2}\,\cos \frac{C}{2}\\ &=2\,\cos \frac{C}{2}\,(\cos \frac{A-B}{2}+\cos \frac{A+B}{2})\\ &=4\,\cos \frac{A}{2}\,\cos \frac{B}{2}\,\cos \frac{C}{2} \\ \end{aligned} \]

\[ \cos A+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\\ \tag{2} \]

证：

\[ \begin{aligned} LHS &= 2\,\cos \frac{A+B}{2}\,\cos \frac{A-B}{2} + 1-2\,\sin ^2\frac{C}{2} \\ &= 1-2\,\sin \frac{C}{2}\,(\sin \frac{C}{2}-\cos \frac{A-B}{2}) \\ &=1-2\,\sin \frac{C}{2}\,(\cos \frac{A+B}{2}-\cos \frac{A-B}{2}) \\ &=1-2\,\sin \frac{C}{2}\,(-2\,\sin \frac{A}{2}\,\sin \frac{B}{2})\\ &=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \end{aligned} \]

正余半角平方和

\[ \sin ^2\frac{A}{2}+\sin ^2\frac{B}{2}+\sin ^2\frac{C}{2}= 1-2\,\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\tag{3} \]

证：

\[ \begin{aligned} RHS &= 1+\sin \frac{C}{2}\,(\cos \frac{A+B}{2}-\cos \frac{A-B}{2}) \\ &= 1+\sin ^2\frac{C}{2}- \cos \frac{A+B}{2}\cos \frac{A-B}{2}\\ &= 1+\sin ^2\frac{C}{2}-\frac{\cos A+\cos B}{2} \\ &= 1+\sin ^2\frac{C}{2}-\frac{2-2\sin ^2\frac{A}{2}-2\sin ^2\frac{B}{2}}{2} \\ &= \sin ^2\frac{A}{2}+\sin ^2\frac{B}{2}+\sin ^2\frac{C}{2} \end{aligned} \]

同理易得：

\[ \cos ^2\frac{A}{2}+\cos ^2\frac{B}{2}+\cos ^2\frac{C}{2}= 2+2\,\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\tag{4} \]

正割余割和

\[ \tan A+\tan B+\tan C=\tan A\;\tan B\;\tan C \tag{5} \]

证：

\[ \begin{gather*} \text{和角公式：}\tan C = -\frac{\tan A+\tan B}{1-\tan A\;\tan B}\\ \tan C-\tan A\;\tan B\;\tan C = -(\tan A+\tan B)\\ \text{移项后得到结论} \end{gather*} \]

推论：

\[ \cot A\,\cot B+\cot A\,\cot C+\cot B\,\cot C =1\tag{6} \]

证：

\[ \begin{aligned} \text{由 (5)得到: }\frac{1}{\tan A\,\tan B}&=\frac{\tan C}{\tan A+\tan B+\tan C}\\ \frac{1}{\tan A\,\tan C}&=\frac{\tan B}{\tan A+\tan B+\tan C}\\ \frac{1}{\tan B\,\tan C}&=\frac{\tan A}{\tan A+\tan B+\tan C}\\ \end{aligned} \]

三个式子相加即可

正余割半角

\[ \cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2} = \cot \frac{A}{2}\,\cot \frac{B}{2}\,\cot \frac{C}{2} \tag{7} \]

证： 受到 (5) 启发，只需证明

\[ cot \frac{C}{2} = -\frac{\cot \frac{A}{2}+\cot \frac{B}{2}}{1-\cot \frac{A}{2}\,\cot \frac{B}{2}} \]

即证：

\[ \begin{gather*} \tan\frac{C}{2} =\frac{\cot \frac{A}{2}\,\cot \frac{B}{2}-1}{\cot \frac{A}{2}+\cot \frac{B}{2}}\\ \Leftarrow \tan\frac{C}{2} = \frac{1-\tan\frac{A}{2}\tan\frac{B}{2}}{\tan\frac{A}{2}+\tan\frac{B}{2}} \end{gather*} \]

推论：

\[ \tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{A}{2}\tan\frac{C}{2} = 1 \tag{8} \]

由 (7) 移项得到

\[ \begin{gather*} \tan\frac{A}{2}\tan\frac{B}{2}=\frac{\tan\frac{A}{2}\,\tan\frac{B}{2}\,\tan\frac{B}{2}\,\cot\frac{C}{2}}{\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}}\\ \tan\frac{A}{2}\tan\frac{C}{2}=\frac{\tan\frac{A}{2}\,\tan\frac{B}{2}\,\tan\frac{B}{2}\,\cot\frac{B}{2}}{\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}}\\ \tan\frac{B}{2}\tan\frac{C}{2}=\frac{\tan\frac{A}{2}\,\tan\frac{B}{2}\,\tan\frac{B}{2}\,\cot\frac{A}{2}}{\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}}\\ \end{gather*} \\ \]

三个式子相加即可