Factorization of trigonometric matrices Consider two real square matrices $A_1$ and $A_2$ and $t_1,t_2\in\mathbb{R}$. $A_1$ and $A_2$ do not commute. Consider the following matrix involving matrix trigonometric functions:
\begin{equation}
M_1(t)=\begin{bmatrix} \cos(tA_1) & t\mathrm{sinc}(t A_1) \\ -A_1\sin(tA_1) & \cos(tA_1) \end{bmatrix} \end{equation}
and $M_2(t)$ defined similarly by changing $A_1$ to $A_2$. Using the double-angle identities, it can be shown that
\begin{align}
\Delta &= M_1(2t_1)-M_2(2t_2) \\
&= 2\begin{bmatrix} t_1\mathrm{sinc}(t_1A_1) & -t_2\mathrm{sinc}(t_2A_2) \\ \cos (t_1 A_1) & -\cos(t_2A_2) \end{bmatrix} \begin{bmatrix}-A_1\sin(t_1A_1) & \cos(t_1 A_1) \\ -A_2\sin(t_2A_2) & \cos(t_2 A_2) \end{bmatrix}
\end{align}
which provides a factorization of the difference $\Delta$.
Is there a similar factorization for $M_1(2t_1)M_2(2t_2) - M_2(2t_4)M_1(2t_3)$ as the product of two matrices (or more)? What about the factorization of the more general case
$$\prod_{i=1}^k M_{\epsilon(i)}(2t_i) - \prod_{i=1}^k M_{\epsilon(i+1)} (2t_{2k-i+1})$$
with $\epsilon(i)=1$ if $i$ is odd and $2$ if $i$ is even?
 A: Even if we replace the $A_i$ by $1\times1$ scalars $a_i$, I don't think that there is such a factorization, at least not for $k=2$ and a forteriori neither for bigger $k$. Putting $u_i:=2t_i$, we have, modulo sign errors,$$  
M:=M_1(2t_1)M_2(2t_2) - M_2(2t_4)M_1(2t_3)=\begin{pmatrix}
a_1^{-1}&0\\
0&1\\
\end{pmatrix}\Bigl( (a_1-a_2) M_-+(a_1+a_2) M_+\Bigr)
\begin{pmatrix}
1&0\\
0&a_2^{-1}\\
\end{pmatrix},$$ where
$ M_-=\begin{pmatrix}
\cos(u_1-u_2) -\cos(u_3-u_4) &   \sin(u_1-u_2) -\sin(u_3-u_4) \\
-  \sin(u_1-u_2) +\sin(u_3-u_4) & -\cos(u_1-u_2) +\cos(u_3-u_4)\\
\end{pmatrix}$  
and $
M_+=\begin{pmatrix}
\cos(u_1+u_2) -\cos(u_3+u_4) &   -\sin(u_1+u_2) +\sin(u_3+u_4) \\
  -\sin(u_1+u_2) +\sin(u_3+u_4) & \cos(u_1+u_2) -\cos(u_3+u_4) \\
\end{pmatrix}.
$ 
Using formulas like 
$\displaystyle \cos \alpha-\cos \beta=-2\sin {\frac {\alpha+\beta}{2}}\sin {\frac {\alpha-\beta}{2}}$, the entries of $M_-$ and $M_+$ can be written as products, which turn out to have common factors:
$$ M_-=2\sin(-t_1+t_2+t_3-t_4)\begin{pmatrix}
\sin(t_1-t_2+t_3-t_4)  &   -\cos(t_1+t_2+t_3+t_4)  \\
\cos(t_1+t_2+t_3+t_4)  &\sin(-t_1+t_2-t_3+t_4) \\
\end{pmatrix}\\ =:2\sin x\begin{pmatrix}
\sin z  &   -\cos w  \\
\cos w  &-\sin z \\
\end{pmatrix}\\
M_+=2\sin(-t_1-t_2+t_3+t_4)\begin{pmatrix}
\sin(t_1+t_2+t_3+t_4)  &   \cos(t_1+t_2+t_3+t_4)  \\
\cos(t_1+t_2+t_3+t_4)  &\sin(t_1+t_2+t_3+t_4) \\
\end{pmatrix}\\ =:2\sin y\begin{pmatrix}
\sin w  &   \cos w  \\
\cos w  &\sin w \\
\end{pmatrix}.$$ The four linear combinations $x,y,z,w$ of $t_1,t_2,t_3,t_4$ are linearly independent, so by virtue of this form there is no chance that $(a_1-a_2) M_-+(a_1+a_2) M_+,$ or equivalently $M,$ can be factorized as a whole.  
