Prove that the rotation of sums is equal to the rotation of products

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}