question related to something in karlin and taylor stochastic processes one text This question is essentially a question about something in Karlin and Taylor's Stochastic Processes One text in the spectral chapter. Since this is a DSP list, Karlin and Taylor may not be so popular so I will describe the setting which I think I've seen in other texts also.

They define the random variable $X_n =  A  \cos(\omega n) + B \sin(\omega  n)$.

So, $A$ and $B$ are zero mean normal random variables. $\omega$ is fixed at some frequency between $0$ and $\pi$. The process is on  the integers so the process start at some value of $-k$ and continues through to positive $k$. So $k$ can be any integer and $X_n$ represents a discrete stochastic process.

They then say that the $X_n$ process is stationary in the mean. Evidently this is so obvious that there's no need to explain it. I don't get it. Well, I sort of half get it.

A and B are  mean zero so, okay the mean will always be zero no matter what the values of the cos or sin term.  But since $n$ is increasing by 1 as the process proceeds, isn't the trajectory of $X_n$ changing at each $n$ because it multiplies omega ? So, because A and B are zero, stationary in the mean works. But, say A and B were both 1.0 say. Would the $X_n$ process still be stationary in the mean ? Since $X_n$ and $X_{n+1}$ will be at a different frequencies, $\omega  n$ and $\omega (n+1)$ respectively, doesn't that make the  mean different at times $n$ and $n + 1$ ? so it can't be a stationary process. 

So, I guess my question is: For this process,  does stationary in mean only hold because $A$ and $B$ are zero in mean or would it work no matter what their means were ? If the latter is true, then I'm totally not understanding the process. Thanks.
The mean, i.e., the expected value of $X_n$ is

$$E\{X_n\}=E\{A\cos(\omega n)+B\sin(\omega n)\}\tag{1}$$

Since the expectation operator is linear, $(1)$ is equivalent to

$$E\{X_n\}=E\{A\cos(\omega n)\}+E\{B\sin(\omega n)\}\tag{2}$$

and since $\cos(\omega n)$ and $\sin(\omega n)$ are deterministic you finally get

$$E\{X_n\}=E\{A\}\cos(\omega n)+E\{B\}\sin(\omega n)\tag{3}$$

which equals zero if both $E\{A\}$ and $E\{B\}$ are zero. If at least one of the two random variables $A$ and $B$ has a non-zero mean, then $(3)$ will generally depend on $n$ (except for very specific choices of $\omega$), and, consequently, the mean of $X_n$ will also depend on $n$, i.e., $X_n$ cannot be stationary. 

As a side note, a constant mean does of course generally not imply stationarity of $X_n$. For the given process, however, if $A$ and $B$ are identically distributed and $E\{A\}=E\{B\}=0$, and under the additional assumption that $E\{AB\}=0$ holds, $X_n$ is a wide-sense stationary (WSS) process.