\[ \begin{aligned} \sin(A+B) &= \sin A\cos B + \cos A\sin B \\ \sin(A-B) &= \sin A\cos B - \cos A\sin B \\ \cos(A+B) &= \cos A\cos B - \sin A\sin B \\ \cos(A-B) &= \cos A\cos B + \sin A\sin B \\ \tan(A+B) &= \frac{\tan A + \tan B}{1 - \tan A \tan B} \\ \tan(A-B) &= \frac{\tan A - \tan B}{1 + \tan A \tan B} \\ 1 + \tan^2 x &= \sec^2 x \\ \end{aligned} \]

Note for \(\sin(A+B)\) and \(\cos(A+B)\), Khan Academy has some nice proofs. Notice \(\cos -\theta = \cos \theta \) and \(\sin -\theta = - \sin \theta \). The other identities are not hard to verify.

Euler's Formula

On the other hand, you may never have to memorize these formulas. Here is why. By Euler's formula:

\[ \begin{aligned} e^{i(x+y)} &= \cos(x+y) + i \sin(x+y) \\ &= e^{ix} e^{iy} = (\cos x + i \sin x) \cdot (\cos y + i \sin y) \\ &= (\cos x \cdot \cos y - \sin x \cdot \sin y) + i(\sin x \cdot \cos y + \cos x \cdot \sin y) \\ \end{aligned} \]

and therefore

\[ \begin{aligned} \cos(x+y) &= \cos x \cdot \cos y - \sin x \cdot \sin y \\ \sin(x+y) &= \sin x \cdot \cos y + \cos x \cdot \sin y \\ \end{aligned} \]