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?



Factorization of trigonometric matrices
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. 


