the Zassenhaus /Baker–Campbell–Hausdorff formula for cosine. This question concerns the expansion of non-commutative algebra $[X,Y] \neq 0$ for two operators $X,Y$. One can think of $X$ and $Y$ as some matrices.
If $[X,Y] = 0$, we have
$$e^{t(X+Y)}= e^{tX}~  e^{tY}$$
If $[X,Y] \neq 0$, We know the Zassenhaus formula or the Baker–Campbell–Hausdorff formula: $$e^{t(X+Y)}= e^{tX}~  e^{tY} ~e^{-\frac{t^2}{2} [X,Y]} ~
e^{\frac{t^3}{6}(2[Y,[X,Y]]+ [X,[X,Y]] )} ~
e^{\frac{-t^4}{24}([[[X,Y],X],X] + 3[[[X,Y],X],Y] + 3[[[X,Y],Y],Y]) } \cdots$$
This expresses $e^{t(X+Y)}$ in terms of $e^{tX}$ and $e^{tY}$, and their further commutators $[X,Y]$.

Question: Do we have a similar form for $\cos(A+B)$ when $[A,B]=C \neq 0$? (We may take $[C,A]=[C,B]=0$ for the simplest case to extract the first order term.)
If $[A,B]=0$, we have
  $$\cos(A+B)=\cos(A)\cos(B) -\sin(A)\sin(B).$$
If $[A,B]=C \neq 0$, do we have some similar expression like the Zassenhaus formula or the Baker–Campbell–Hausdorff formula:
  $$\cos(A+B)=\cos(A)\cos(B) \dots-\sin(A)\sin(B) \dots+ \dots$$

Can we express $\cos(A+B)$ in terms of $\cos(A)$,$\cos(B)$,$\sin(A)$,$\sin(B)$ and some function of $C$?
 A: It seems to me that if $[A,B]=C$ with $[C,A]=[C,B]=0$, it will be a simple case.
We have
$$e^{A+B}=e^A e^B e^{-\frac{1}{2}[A,B]}$$
and 
$$\cos(A+B)=\frac{e^{i(A+B)}+e^{-i(A+B)}}{2}=\frac{e^{iA}e^{iB}e^{\frac{1}{2}[A,B]}+e^{-iA}e^{-iB}e^{\frac{1}{2}[A,B]}}{2}$$
$$=e^{\frac{1}{2}[A,B]}\frac{e^{iA}e^{iB}+e^{-iA}e^{-iB}}{2}=e^{\frac{1}{2}[A,B]}\big(\cos(A)\cos(B) -\sin(A)\sin(B)\big).$$
Notice that $\frac{e^{iA}e^{iB}+e^{-iA}e^{-iB}}{2}=\cos(A)\cos(B) -\sin(A)\sin(B)$ is true.
and notice that we use the condition $[A,B]=C$ with $[C,A]=[C,B]=0$.

Thus my answer is 
  $$\cos(A+B)=e^{\frac{1}{2}[A,B]}\big(\cos(A)\cos(B) -\sin(A)\sin(B)\big)$$
  up to higher order terms when $[A,B]=C$ with $[C,A]\neq0, [C,B]\neq 0$.

Maybe someone else can fill in the higher order term computation?
A: Let us write the Zassenhaus formula in the form
$$
e^{X+Y}=e^Xe^Ye^{C_2(X,Y)}e^{C_3(X,Y)}\ldots,
$$
where $C_n(X,Y)$ satisfies $C_n(\alpha X,\alpha Y)=\alpha^nC_n(X,Y)$. The first couple of these are
$$
C_2(X,Y)=-\frac12[X,Y],\qquad C_3(X,Y)=\frac13[Y,[X,Y]]+\frac16[X,[X,Y]].
$$
Applying this to the definition of $\cos(A+B)$, we get
$$
2\cos(A+B)=e^{i(A+B)}+e^{-i(A+B)}\\=e^{iA}e^{iB}e^{-C_2}e^{-iC_3}e^{C_4}e^{iC_5}e^{-C_6}\ldots+e^{-iA}e^{-iB}e^{-C_2}e^{iC_3}e^{C_4}e^{-iC_5}e^{-C_6}\ldots,
$$
where we have used the abbreviation $C_n=C_n(X,Y)$.
To get a sense of what is going on, let us assume $C_n=0$ for all $n>3$.
Thus we have
$$
e^{iA}e^{iB}e^{-C_2}e^{-iC_3}+e^{-iA}e^{-iB}e^{-C_2}e^{iC_3}
\\=
(\cos A+i\sin A)(\cos B+i\sin B)e^{-C_2}(\cos C_3-i\sin C_3)\\
+(\cos A-i\sin A)(\cos B-i\sin B)e^{-C_2}(\cos C_3+i\sin C_3)\\
=2(\cos A\cos B-\sin A\sin B)e^{-C_2}\cos C_3+2(\cos A\sin B+\sin A\cos B)e^{-C_2}\sin C_3,
$$
yielding
$$
\cos(A+B)=(\cos A\cos B-\sin A\sin B)e^{-C_2}\cos C_3\\+(\cos A\sin B+\sin A\cos B)e^{-C_2}\sin C_3.
$$
It is now easy to see that the general formula is given by
$$
\cos(A+B)=\big[\cos(x_0+x_1-x_3+x_5-x_7+\ldots)e^{-x_2+x_4-x_6+\ldots}\big]\Big|_{x_0=A,x_1=B,x_2=C_2,x_3=C_3,\ldots},
$$
where it is understood that we expand $\cos(a+b-x_3+x_5-x_7+\ldots)e^{-x_2+x_4-x_6+\ldots}$ as if all the variables were simply numbers, with the terms ordered in such a way that the terms with $x_n$ come after all $x_1,\ldots,x_{n-1}$, and then substitute the matrices for the variables.
