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I would some help with the steps to decompose the below matrix exponential.

$\exp\left[ \zeta \left ( \begin{matrix} -\cos(x) & i \sin(x) \\ -i \sin(x) & \cos(x) \end{matrix} \right ) \right] = \left ( \begin{matrix} \cosh(\zeta)-\sinh(\zeta)\cos(x) & i \sinh(\zeta)\sin(x) \\ -i \sinh(\zeta)\sin(x) & \cosh(\zeta)+\sinh(\zeta)\cos(x) \end{matrix} \right )$

I think the first matrix could be factored as Pauli matrices but I have not had any luck trying this.

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Let $A = \begin{bmatrix} -\cos(x) & i \sin(x) \\ -i \sin(x) & \cos(x)\end{bmatrix}$. The key is $A^2 = I_2$.

Using this information, we have:

$$\begin{align} \exp(\zeta A) = \sum_{n=0}^\infty \frac{\zeta^n}{n!} A^n &= \left( \sum_{k=0}^\infty \frac{\zeta^{2k}}{2k!} A^{2k} \right) + \left( \sum_{k=0}^\infty \frac{\zeta^{2k+1}}{(2k+1)!} A^{2k} \right) A\\ &= \left( \sum_{k=0}^\infty \frac{\zeta^{2k}}{2k!} \right) I_2 + \left( \sum_{k=0}^\infty \frac{\zeta^{2k+1}}{(2k+1)!} \right) A\\ &= \cosh(\zeta) I_2 + \sinh(\zeta) A\\ &= \begin{bmatrix} \cosh(\zeta)-\sinh(\zeta)\cos(x) & i \sinh(\zeta)\sin(x) \\ -i \sinh(\zeta)\sin(x) & \cosh(\zeta)+\sinh(\zeta)\cos(x) \end{bmatrix} \end{align}$$

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First, check your equality when $\sin(x)=0$.

Second, notice that $\zeta\begin{pmatrix}-\cos(x)& i\sin(x)\\-i\sin(x)& \cos(x)\end{pmatrix}=\zeta\begin{pmatrix}i& 0\\ 0& 1\end{pmatrix}\begin{pmatrix}-\cos(x)& \sin(x)\\\sin(x)& \cos(x)\end{pmatrix}\begin{pmatrix}-i& 0\\ 0& 1\end{pmatrix}$

and $\begin{pmatrix}i& 0\\ 0& 1\end{pmatrix}\begin{pmatrix}-i& 0\\ 0& 1\end{pmatrix}=\begin{pmatrix}1& 0\\ 0& 1\end{pmatrix}$.

Next, $\zeta\begin{pmatrix}-\cos(x)& \sin(x)\\\sin(x)& \cos(x)\end{pmatrix}=\begin{pmatrix}1+\cos(x)& 1-\cos(x)\\-\sin(x)& \sin(x)\end{pmatrix}\begin{pmatrix}-\zeta& 0\\ 0& \zeta\end{pmatrix}\begin{pmatrix}\dfrac{1}{2}& \dfrac{-1+\cos(x)}{2\sin(x)}\\\dfrac{1}{2}& \dfrac{1+\cos(x)}{2\sin(x)}\end{pmatrix}$.

Notice that $\begin{pmatrix}1+\cos(x)& 1-\cos(x)\\-\sin(x)& \sin(x)\end{pmatrix}\begin{pmatrix}\dfrac{1}{2}& \dfrac{-1+\cos(x)}{2\sin(x)}\\\dfrac{1}{2}& \dfrac{1+\cos(x)}{2\sin(x)}\end{pmatrix}=\begin{pmatrix}1& 0\\ 0& 1\end{pmatrix}$.

Thus, you need to compute $\exp(R\begin{pmatrix}-\zeta& 0\\ 0& \zeta\end{pmatrix}R^{-1})=R\begin{pmatrix} e^{-\zeta}& 0\\ 0 & e^{\zeta} \end{pmatrix}R^{-1}$, where $R=\begin{pmatrix}i& 0\\ 0& 1\end{pmatrix}\begin{pmatrix}1+\cos(x)& 1-\cos(x)\\-\sin(x)& \sin(x)\end{pmatrix}$ and $R^{-1}=\begin{pmatrix}\dfrac{1}{2}& \dfrac{-1+\cos(x)}{2\sin(x)}\\\dfrac{1}{2}& \dfrac{1+\cos(x)}{2\sin(x)}\end{pmatrix}\begin{pmatrix}-i& 0\\ 0& 1\end{pmatrix}$.

In order to finish you must use the formulas for the $\cosh(\zeta)$ and $\sinh(\zeta)$.

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