Proof verification - Sum formulas in trig 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. 

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.
