$\sin (A+B)=\sin A\cos B+\cos A\sin B$

$\sin (A-B)=\sin A\cos B-\cos A\sin B$

$\cos (A+B)=\cos A\cos B-\sin A\sin B$

$\cos (A-B)=\cos A\cos B+\sin A\sin B$

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

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

$\sin 2x=2\sin x\cos x$

$\cos 2x={\cos }^{2}x-{\sin }^{2}x$

$\cos 2x=2{\cos }^{2}x-1$

$\cos 2x=1-2{\sin }^{2}x$

$\tan 2x={\displaystyle \frac{2\tan x}{1-{\tan }^{2}x}}$

$\sin 3x=3\sin x-4{\sin }^{3}x$

$\cos 3x=4{\cos }^{3}x-\cos x$

$\tan 3x={\displaystyle \frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}}$

${\cos }^{2}x=\frac{1}{2}+\frac{1}{2}\cos 2x$

${\sin }^{2}x=\frac{1}{2}-\frac{1}{2}\cos 2x$

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

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

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

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

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

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

$2\cos A\cos B=\cos (A+B)+\cos (A-B)$

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