Proof verification - Sum formulas in trig

Question

We have $a: \mathbb{R}^2 \to \mathbb{R}^2$ represent rotation about $(0,0)$ over an angle $\alpha$. We know that $a$ is a linear map and we know that $a$ corresponds to the matrix $A = \begin{pmatrix} \cos(\alpha) & -\sin(\alpha) \\ \sin(\alpha) & \cos(\alpha) \\ \end{pmatrix}$. Rotation over an angle $\alpha + \beta$ would thus be given by the matrix $ \begin{pmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) \\ \sin(\alpha + \beta) & \cos(\alpha + \beta) \\ \end{pmatrix}$. By the additivity property of a linear map, we know that $$ \begin{pmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) \\ \sin(\alpha + \beta) & \cos(\alpha + \beta) \\ \end{pmatrix} = \begin{pmatrix} \cos(\alpha ) & -\sin(\alpha) \\ \sin(\alpha ) & \cos(\alpha ) \\ \end{pmatrix} \begin{pmatrix} \cos(\beta) & -\sin(\beta) \\ \sin(\beta) & \cos(\beta) \\ \end{pmatrix} $$

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

QED.

Is this proof correct? My only gripe with it at the moment is explaining why $$ \begin{pmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) \\ \sin(\alpha + \beta) & \cos(\alpha + \beta) \\ \end{pmatrix} = \begin{pmatrix} \cos(\alpha ) & -\sin(\alpha) \\ \sin(\alpha ) & \cos(\alpha ) \\ \end{pmatrix} \begin{pmatrix} \cos(\beta) & -\sin(\beta) \\ \sin(\beta) & \cos(\beta) \\ \end{pmatrix} $$

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

since additivity is $f(x+y) = f(x) + f(y)$. Intuitively, I completely understand it, but I don't know how to explain it mathematically, or which mathematical property of a linear map explains it.

Answer 1

Rotating by $\alpha + \beta$ degrees can be thought of as first rotating by $\beta$ degrees, then rotating by $\alpha$ degrees. In other words, if $R_\theta:\Bbb R^2 \to \Bbb R^2$ is the rotation by angle $\theta$, then $$ R_\beta (R_\alpha(x)) = R_{\alpha + \beta}(x) $$ That is, the combined rotation is the composition of two linear maps, not their sum. In order to find the linear map of a composition, one "multiplies" the corresponding matrices.