三角形中的几个恒等式

在 $\mathrm{\Delta }ABC$ 中，以下等式成立

正余弦平方和

$$\begin{array}{}\text{(1)}& \sin A+\sin B+\sin C=4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\end{array}$$

证：

$$\begin{array}{rl}LHS& =2\sin \frac{A+B}{2}\cos \frac{A-B}{2}+\sin C\\ & =2\cos \frac{C}{2}\cos \frac{A-B}{2}+2\sin \frac{C}{2}\cos \frac{C}{2}\\ & =2\cos \frac{C}{2}(\cos \frac{A-B}{2}+\cos \frac{A+B}{2})\\ & =4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}\end{array}$$

$$\begin{gathered} \begin{array}{}\text{(2)}& \cos A+\cos B+\cos C=1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \\ \end{array} \end{gathered}$$

证：

$$\begin{array}{rl}LHS& =2\cos \frac{A+B}{2}\cos \frac{A-B}{2}+1-2{\sin }^2\frac{C}{2}\\ & =1-2\sin \frac{C}{2}(\sin \frac{C}{2}-\cos \frac{A-B}{2})\\ & =1-2\sin \frac{C}{2}(\cos \frac{A+B}{2}-\cos \frac{A-B}{2})\\ & =1-2\sin \frac{C}{2}(-2\sin \frac{A}{2}\sin \frac{B}{2})\\ & =1+4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\end{array}$$

正余半角平方和

$$\begin{array}{}\text{(3)}& {\sin }^2\frac{A}{2}+{\sin }^2\frac{B}{2}+{\sin }^2\frac{C}{2}=1-2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\end{array}$$

证：

$$\begin{array}{rl}RHS& =1+\sin \frac{C}{2}(\cos \frac{A+B}{2}-\cos \frac{A-B}{2})\\ & =1+{\sin }^2\frac{C}{2}-\cos \frac{A+B}{2}\cos \frac{A-B}{2}\\ & =1+{\sin }^2\frac{C}{2}-\frac{\cos A+\cos B}{2}\\ & =1+{\sin }^2\frac{C}{2}-\frac{2-2{\sin }^2\frac{A}{2}-2{\sin }^2\frac{B}{2}}{2}\\ & ={\sin }^2\frac{A}{2}+{\sin }^2\frac{B}{2}+{\sin }^2\frac{C}{2}\end{array}$$

同理易得：

$$\begin{array}{}\text{(4)}& {\cos }^2\frac{A}{2}+{\cos }^2\frac{B}{2}+{\cos }^2\frac{C}{2}=2+2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\end{array}$$

正割余割和

$$\begin{array}{}\text{(5)}& \tan A+\tan B+\tan C=\tan A\tan B\tan C\end{array}$$

证：

$$\begin{array}{c}\text{和角公式：}\tan C=-\frac{\tan A+\tan B}{1-\tan A\tan B}\\ \tan C-\tan A\tan B\tan C=-(\tan A+\tan B)\\ \text{移项后得到结论}\end{array}$$

推论：

$$\begin{array}{}\text{(6)}& \cot A\cot B+\cot A\cot C+\cot B\cot C=1\end{array}$$

证：

$$\begin{array}{rl}\text{由 (5)得到: }\frac{1}{\tan A\tan B}& =\frac{\tan C}{\tan A+\tan B+\tan C}\\ \frac{1}{\tan A\tan C}& =\frac{\tan B}{\tan A+\tan B+\tan C}\\ \frac{1}{\tan B\tan C}& =\frac{\tan A}{\tan A+\tan B+\tan C}\end{array}$$

三个式子相加即可

正余割半角

$$\begin{array}{}\text{(7)}& \cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}=\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}\end{array}$$

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：

$$\begin{array}{c}\tan \frac{C}{2}=\frac{\cot \frac{A}{2}\cot \frac{B}{2}-1}{\cot \frac{A}{2}+\cot \frac{B}{2}}\\ \Leftarrow \tan \frac{C}{2}=\frac{1-\tan \frac{A}{2}\tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}}\end{array}$$

推论：

$$\begin{array}{}\text{(8)}& \tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{A}{2}\tan \frac{C}{2}=1\end{array}$$

由 (7) 移项得到

$$\begin{array}{c}\tan \frac{A}{2}\tan \frac{B}{2}=\frac{\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{B}{2}\cot \frac{C}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}}\\ \tan \frac{A}{2}\tan \frac{C}{2}=\frac{\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{B}{2}\cot \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}}\\ \tan \frac{B}{2}\tan \frac{C}{2}=\frac{\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{B}{2}\cot \frac{A}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}}\end{array}$$

三个式子相加即可