Rotate a straight line by angle $\theta$ from the origin with a starting coordinate:

1. $(1,0) \rightarrow (\cos \theta, \sin \theta)$
2. $(0,1) \rightarrow (-\sin \theta, \cos \theta)$
3. $(x,0) \rightarrow (x\cos \theta, x\sin \theta)$
4. $(0,y) \rightarrow (-y\sin \theta, y\cos \theta)$
5. $(x,y) \rightarrow (x\cos \theta - y\sin \theta, x\sin \theta + y\cos \theta)$

Therefore, if we rotate $(1,0)$ first by $\alpha$ and then by $\beta$, we would end up first at $(\cos \alpha, \sin \alpha)$, and then at

$(\cos \alpha \cos \beta - \sin \alpha \sin \beta,   \cos \alpha \sin \beta + \sin \alpha \cos \beta)$

On the other hand, if we rotate $(1,0)$ by $\alpha + \beta$, we would end up at

$(\cos(\alpha + \beta),   \sin(\alpha + \beta))$

This means:

$$\begin{aligned} \cos(\alpha + \beta) &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \sin(\alpha + \beta) &= \cos \alpha \sin \beta + \sin \alpha \cos \beta \end{aligned}$$