Proving using algebraic form of rotations

Question

I am lost as to how to prove, using algebraic form of rotations, that $R_a$ $\circ$ $R_b$ = $R_{a+b}$. To solve this, one needs to create separate matrix such that $R_a$ and $R_b$ are 2x2 matrices with arbitrary values and then add those 2 matrices to get $R_{a+b}$ to find the composition right? I have been trying to mentally come up with a strategy to prove this but have yet to come to anything that seems correct. Any tips are welcome.

Answer 1

The key here is that $R_a \circ R_b \neq R_a + R_b$, so you do not add the matrices in order to compose their corresponding functions, but you multiply them using matrix multiplication. Furthermore you can use that every rotation matrix is given by

$$R_a = \begin{bmatrix} \cos(a) & -\sin(a) \\ \sin(a) & \cos(a) \end{bmatrix}$$