Let's see if the following helps.
First, we don't need to worry about the expansions of $E$ and $H$ themselves; we are told they exist and are of the form
$$
E(z,t)= \sum_{n=0}^\infty A_n(t) \sin(k_nz) \\
H(z,t)= \sum_{n=0}^\infty H_n(t) \cos(k_n z),
$$
whatever the coefficients $A_n(t)$ and $H_n(t)$ are (they have been computed using the formulas you mention).
Now, we want to expand their derivatives. Let's start with the electric field:
$$
\frac{\partial E(t,z)}{\partial z} = \frac{B_0(t)}{2} + \sum_{n=1}^\infty B_n(t)\cos (k_n z),
$$
(the electric field is odd, so its derivative is even, thus the cosines)
where the coefficients $B_n(t)$ are computed as we know (as the theorem you have posted tells us):
$$
B_n(t)=\frac{2}{L} \int_{-L}^0 \frac{\partial E(t,z)}{\partial z} \cos (k_nz)
dz.$$
I mean $B_n(t)$ and not $A_n(t)$ because, for the momment, we are expanding $\partial E/\partial z$ as if we knew nothing about $E(t,z)$. Suppose that $\partial E/\partial z$ is a function $f(t,z)$ and we expand it.
For $n\neq 0$, the integral above is computed by parts:
$$ \int_{-L}^0 \frac{\partial E(t,z)}{\partial z} \cos (k_nz)
dz= E(t,z)\cos(k_nz)\bigg|_{-L}^0 - \int_{-L}^0 E(t,z)\frac{d}{dz}\cos(k_n z)= E(t,z)\cos(k_nz)\bigg|_{-L}^0 + \sum_{m=0}^\infty k_n\int_{-L}^0 A_n(t)\sin(k_m z) (z,t)\sin(k_nz)dz = E(t,z)\cos(k_nz)\bigg|_{-L}^0 - A_n(t)k_n\frac{L}{2}
$$
(you can see the computations here ($m\neq n$) and here ($m=n$), so that it results:
$$B_n(t)= \frac{2}{L}\left(E(t,z)\cos(k_nz)\bigg|_{-L}^0 + k_n\frac{L}{2}\right).
$$
Now let's compute $B_0$:
$$
B_0(t)=\frac{2}{L}\int_{-L}^0 \frac{\partial E(t,z)}{\partial z}dz = \sum_{n=0}^\infty k_n \int_{-L}^0 A_n(t)\cos(n\pi z/L)dz = 0
$$
(answer here.
Hence:
$$
\frac{\partial E(t,z)/\partial z} = \frac{B_0(t)}{2} + \sum_{n=1}^\infty B_n(t)\cos(n\pi z/L) =\sum_{n=1}^\infty \frac{2}{L}\left(E(t,z)\cos(k_nz)\bigg|_{-L}^0 + k_n\frac{L}{2}\right)\cos(n\pi z/L). $$
There is one missing detail, something I do not really understand. For $n>0$, the result for $B_n(t)$ involves $E(0,t)$ and $E(t,-L)\cos(-n\pi)$.
If we use the series to compute $E(t,-L)$, we get a zero, so that term vanishes, so it would not appear in the final expression, that is OK. But the same should happen to $E(t,0)$.
So check the boundary conditions and see what the values of $E(t,0)$ and $E(t,-L)$ are.
The magnetic field seems to follow the same procedure, although you can check that the result is exactly the same as if we computed the derivative $\partial H/\partial z$ directly. Be careful because in doing so we are igonoring the boundary conditions at $z=0,-L$.
Compute $\partial H/\partial z$ using both approaches and check that indeed they agree.
Best Answer
Note: This answer is based on the actual problem as provided in a comment by the OP above.
To simplify notation, let $c_1(z,t) = \cos(kz-wt-\phi_1), c_2(z,t) = \cos(kz-wt-\phi_2)$.
The $E$ field is given by $E_0(c_1(z,t), c_2(z,t))$, or, $E_0c_1(z,t)i + E_0 c_2(z,t)) j$.
The $B$ field is given by $B_0(-c_2(z,t), c_1(z,t))$, or, $-B_0 c_2(z,t)i + B_0 c_1(z,t)) j$.
The dot or inner product is given by $E \cdot B = E_0 B_0 c_1(z,t) c_2(z,t) + E_0 B_0 (-c_2(z,t)) c_1(z,t) = 0$.