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? 

Taking the approach that $R'' = -R$ we see that 
$$\ddot R_1(t) = \ddot R(t+t_2) = -R(t+t_2) = -R_1(t)$$ $$R_1(0) = R_1(t_2)$$
$$\ddot R_2(t) = R(t_2)\ddot R(t) = R(t_2)(-R(t)) = -R_2(t)$$
$$R_2(0) = R(t_2)$$
By the existence and uniqueness theorem, 
$$R_1(t) = R_2(t)$$
$$R(t+t_2) = R_1(t) = R_2(t) = R(t_2)(R(t))$$
$$R(t+t_2) = R(t_2)(R(t))$$
Setting t = $t_1$:
$$R(t_1+t_2) = R(t_2)(R(t_1))$$
All you have you to do now is prove that $R'' = -R$
