Setting: Let's consider a moving mirror at constant speed in z-direction. A light ray of frequency $\omega$ falls onto the mirror. Its wave vector $k$ forms an angle $\theta$ with the vector normal to the mirror.
I would now like to find the angle of reflection with respect to the vector $\vec{e}_z$ normal to the mirror by using the wave four-vector $k^{\mu}=(\omega/c,\vec{k})$. First I wanted to transform the four-vector to a system $K'$ where the mirror is at rest. I know that the transformed position-vector can be written as $$\begin{Bmatrix}x^{{0}^{'}}=\gamma(x^0-\beta x_\parallel)\\x_\parallel'=\gamma(x_\parallel-\beta x^0)\\x_\perp'=x_\perp \end{Bmatrix}$$ The k-vector in the K-system looks like this $$\begin{Bmatrix}k^0=\omega/c=k\\k_\parallel=k_z=k*cos(\theta)\\k_\perp = k_y=k*sin(\theta)\end{Bmatrix}$$
My problem is now how to transform the k vector into the primed system K'.
It should look like this: $$\begin{Bmatrix}k'^0=k \gamma(1-\beta *cos(\theta))\\k'_\parallel=k \gamma(cos\theta -\beta)\\k_\perp'=k_\perp = k*sin(\theta)\end{Bmatrix}$$
The main question is: How do the different cosines come to be?
Note: I know that a similar, but broader question already has been asked here with the title "law of reflection for a moving mirror". But the given references do not provide any insight into this specific problem.