Show that $z(t) = A \cos(bt) + B \sin(bt)$ is second order weakly stationary process I know that in order for a stochastic process to be a  second-order weakly stationary process. 
Then for every $t$, the following conditions should hold:
$$E(Z(t)) = \mu$$
$$D(Z(t)) = \sigma$$
and $$\operatorname{cov}(Z(t), Z(t+p)) = \gamma(p)$$
but I just cannot show it mathematically like
$$E(z(t)) = E (A \cos(bt) + B \sin(bt) ) = A (E(\cos(bt))) + B (E(\sin(bt)))$$
and I have no clue after that. So, any help will be appreciated on solving this
 A: When all $A$, $B$ and $b$ are constant, the whole expression will be a function of $t$. No uncertainty remains in it, and although it is still valid, there is no reason to call it as stochastic. In this case, the mean is also a function of $t$, so the process is not mean stationary, which means it is also not WSS: $$E[Z(t)]=Z(t)=A\cos(bt)+B\sin(bt)=\mu(t)$$
The covariance will be $0$ because both $Z(t),Z(t+p)$ are constants. Typically, what I encounter is $A,B$ being random. In that case, for the mean being stationary, you should have $E[A]=E[B]=0$ (for this specific case), since:
$$E[A\cos(bt)+B\sin(bt)]=E[A]\cos(bt)+E[B]\sin(bt)$$
and there is no combination of constants to make this expression constant, other than zero-means.
A: Assuming that $A$ and $B$ are independent random variables (and $b$ is a constant), and using standard notation for the moments of these random variables, you have the mean function:
$$\begin{align}
\mu(t) 
&\equiv \mathbb{E}(Z(t)) \\[6pt]
&= \mathbb{E}(A \cos(bt) + B \sin(bt)) \\[6pt]
&= \mathbb{E}(A) \cos(bt) + \mathbb{E}(B) \sin(bt) \\[6pt]
&= \mu_A \cos(bt) + \mu_B \sin(bt), \\[6pt]
\end{align} \quad \quad \quad \quad \quad \quad \quad$$
and autocovariance function:
$$\begin{align}
\quad \quad \quad \quad \gamma(t,k) 
&\equiv \mathbb{C}(Z(t),Z(t+k)) \\[6pt]
&= \mathbb{C}(A \cos(bt) + B \sin(bt), A \cos(bt+bk) + B \sin(bt+bk)) \\[6pt]
&= \mathbb{V}(A) \cos(bt)\cos(bt+bk) + \mathbb{V}(B) \sin(bt)\sin(bt+bk) \\[6pt]
&= \sigma_A^2 \cos(bt)\cos(bt+bk) + \sigma_B^2 \sin(bt)\sin(bt+bk) \\[6pt]
&= \sigma_A^2 \cos(bk) + [\sigma_B^2 - \sigma_A^2] \sin(bt)\sin(bt+bk). \\[6pt]
\end{align}$$
Second-order weak stationarity occurs when neither of these functions depend on $t$.  Assuming that $b \neq 0$ (giving a non-trivial problem), a sufficient condition for this is to have $\mu_A = \mu_B = 0$ and $\sigma_A^2 = \sigma_B^2 \equiv \sigma^2$, which gives $\mu(t) = 0$ and $\gamma(t,k) = \sigma^2 \cos(bk)$.  (Note also that weak-sense stationarity is only achieved when these amplitude values are random; once you condition on the amplitude values the series is no longer stationary.)
