Prove that the rotation of sums is equal to the rotation of products So the question starts off: 
Prove  $$\ e^{t_1+t_2} = e^{t_1}e^{t_2}$$
E(t) is a unique solution to $\dot{E} = E, E(0) = 1$.
Let $E_1(t) = E(t_1 + t)$, and E_2(t) = E(t_1)E(t) 
$$\dot E_1 (t) = \dot E_1 (t_1 + t) = E(t_1+t) = E_1 (t); E_1(0) = E(t_1)$$
$$\dot E_2 (t) = E(t_1)\dot E (t) = E(t_1)E(t) = E_2(t); E_2(0) = E(t_1)$$
Therefore(by uniqueness and existence theorem I think): 
$$E_1(t) = E_2(t)$$
Which implies: 
$$E(t_1+t) = E_1(t) = E_2(t) = E(t_1)E(t)$$
Then by setting $t = t_2$ we get our desired result.
Then the question asks to prove:
$$R(t_1+t_2) = R(t_1)R(t_2)$$
where $$R(t) = \begin{bmatrix}\cos(t)&-\sin(t)\\\sin(t)&\cos(t)\end{bmatrix} $$
Given: $R_1(t) = R(t+t_2)$ and $R_2(t) = R(t)R(t_2)$
Prove by a similar argument to the one above. 
Any suggestions on how to approach this? 

If you want to repeat the argument, you can look at the equation $R''=-R$. 
Another way is to notice that 
$$
R(t)=e^{it}\begin{bmatrix}1/2&i/2\\ -i/2&1/2\end{bmatrix} + e^{-it}\begin{bmatrix} 1/2&-i/2\\ i/2&1/2\end{bmatrix}. 
$$
Then
\begin{align}
R(s)R(t)&=e^{it}e^{is}\begin{bmatrix}1/2&i/2\\ -i/2&1/2\end{bmatrix}^2+e^{-it}e^{-is}\begin{bmatrix}1/2&-i/2\\ i/2&1/2\end{bmatrix}^2+2\operatorname{Re}e^{it}e^{-is}\begin{bmatrix}1/2&i/2\\ -i/2&1/2\end{bmatrix}\begin{bmatrix}1/2&-i/2\\ i/2&1/2\end{bmatrix}\\ \ \\
&=e^{i(t+s)}\begin{bmatrix}1/2&i/2\\ -i/2&1/2\end{bmatrix}+e^{-i(t+s)}\begin{bmatrix}1/2&-i/2\\ i/2&1/2\end{bmatrix}+2\operatorname{Re}e^{i(t-s)}\begin{bmatrix}0&0\\ 0&0\end{bmatrix}\\ \ \\
&=\begin{bmatrix} 
\operatorname{Re} e^{i(t+s)}&-\operatorname{Im} e^{i(t+s)}\\ \operatorname{Im} e^{i(t+s)}&\operatorname{Re} e^{i(t+s)}
\end{bmatrix}\\ \ \\
&=\begin{bmatrix} \cos(t+s) &-\sin(t+s)\\ \sin(t+s)&\cos(t+s)\end{bmatrix}\\ \ \\
&=R(t+s)
\end{align}
