<s>[INST] Given the matrix $A^k$, how to get $A^{k+1}$? Given: $$A^k = \left(\begin{array}{rr} \cos kx & \sin kx \\ -\sin kx & \cos kx\end{array}\right)$$ 
$$A^{k+1} \overbrace{=}^? \left(\begin{array}{rr} \cos kx & \sin kx \\ -\sin kx & \cos kx\end{array}\right) \left(\begin{array}{rr} \cos x & \sin x \\ -\sin x & \cos x\end{array}\right) \\ = \left(\begin{array}{rr} \cos kx \cos x-\sin kx \sin x & \cos kx\sin x+\sin kx\cos x \\ -\sin kx \cos x+\cos kx (-\sin x) & \cos kx\cos x-\sin kx \sin x\end{array}\right)\\=\left(\begin{array}{rr} \cos (k+1)x & \sin (k+1)x \\ -\sin (k+1)x & \cos (k+1)x\end{array}\right) $$
How? I really don't understand the final part... 
How can for exp.
$$
(-\sin kx)(\cos x) + (\cos kx)(-\sin x)=-\sin (k+1)x
$$
I don't understand much English.
 [/INST] Hint:
$$
\cos a \cos b - \sin a \sin b = \cos (a+b)\\
\sin a \cos b + \sin b \cos a = \sin (a+b)
$$

to remember this:
if the vector $e_a$ of coordinates $(\cos a, \sin a)$ then
$$ \cos a\cos (-b) + \sin a \sin (-b) =
e_a\cdot e_{-b} = \cos(a-(-b))
$$
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