$$\begin{aligned} & \text { – } \sin (A+B)=\sin A \cdot \cos B+\cos A \cdot \sin B \\ & \text { – } \sin (A-B)=\sin A \cdot \cos B-\cos A \cdot \sin B \\ & \text { – } \cos (A+B)=\cos A \cdot \cos B-\sin A \cdot \sin B \\ & \text { – } \cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B \\ & \text { – } \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B} \\ & \text { – } \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \cdot \tan B} \\ & \text { – } \cot (A+B)=\frac{\cot B \cot A-1}{\cot B+\cot A} \\ & \text { – } \cot (A-B)=\frac{\cot B \cdot \cot A+1}{\cot B-\cot A}\end{aligned}$$

$$\begin{aligned} \cdot \sin C+\sin D & =2 \cdot \sin \left(\frac{C+D}{2}\right) \cdot \cos \left(\frac{C-D}{2}\right) \\ \cdot \sin C-\sin D & =2 \cos \left(\frac{C+D}{2}\right) \cdot \sin \left(\frac{C-D}{2}\right) \\ \cdot \cos C+\cos D & =2 \cos \left(\frac{C+D}{2}\right) \cdot \cos \left(\frac{C-D}{2}\right) \\ \cdot \cos C-\cos D & =-2 \sin \left(\frac{C+D}{2}\right) \cdot \sin \left(\frac{C-D}{2}\right) \\ & =2 \sin \left(\frac{C+D}{2}\right) \cdot \sin \left(\frac{D-C}{2}\right)\end{aligned}$$