So there is not an easy way - I had to grind out all the algebra. Starting with $$\nabla^2 E +k^2E = 0$$ The Electric field forms are $$E_r = P_n^0(cos\theta)[\frac{A_n j_n(k r)}{k r} + \frac {B_n y_n(k r)}{k r}]$$ $$E_\phi = P_n^1(cos\theta)[C_n j_n(k r) + D_n y_n(k r)]$$ Using $\nabla \cdot E = 0$ and the symmetry over $\phi$ I find $$E_\theta = -\frac {P_n^1(cos\theta)} {n(n+1)} \{ A_n [\frac {j_n(k r)}{k r} + j'_n(k r)] + B_n[ \frac {y_n(k r)}{k r} + y'_n(k r)]\}$$ Clearly $n = 0$ is not allowed.
Using $\nabla \times E = i \omega B$ I find the following for the $B$ field $$B_r = \frac {-i} \omega k \frac {P_n^0(\cos\theta)} {n(n+1)}[C_n \frac {j_n(k r)} {k r} + D_n \frac {y_n(k r)} {k r}]$$ $$B_\theta = i \frac k \omega P_n^1(cos\theta) \{C_n [\frac {j_n(k r)}{k r} + j'_n(k r)] + D_n [\frac {y_n(k r)}{k r} + y'_n(k r)] \}$$ $$B_\phi = \frac {-i} \omega k \frac {P_n^1(cos\theta)} {n(n+1)}[A_n j_n(k r) + B_n y_n(k r)]$$ Note that the imaginary factor simply means that the B field is 90 degrees out of phase with the E field in time.
Now let's look at boundary conditions. Both $E_\phi$ and $E_\theta$ are zero at $r_1$ and $r_2$. We also have $E_\phi$ and $E_r$ are zero at $\theta_w$. The actual field is the sum over all $n$ but each mode is orthogonal to all other modes so each mode must match the boundary conditions independently. For $E_\phi$, given arbitrary $\theta_w$, $P_n^1(cos \theta_w) \neq 0$ so all coefficients $C_n$ and $D_n$ must be $0$. That knocks out $E_\phi$, $B_r$ and $B_\theta$. The same is true for $E_r$ - which says that unless the walls are at a specific angle which is a zero for $P_n^0$ or $P_n^1$, no fields will resonate.
In other words, if the EmDrive guys don't build the cavity to specific angles, it will simply reflect all power and won't have any RF in it at all! So let's suppose they DO build it to a specific angle so it will resonate, then the Poynting vector is $S = E \times B$. Only $E_\theta B_\phi$ is in the $\hat r$ direction, but this is $0$ at the $r_1$ and $r_2$ walls. So even if it resonates, it won't push. QED.
Life gets much more interesting if there is a center pipe so we have $\theta_1$ and $\theta_2$. Then we have both $P_n^m$ and $Q_n^m$ as angular solutions. It will be possible to find a set of coefficients which match all boundary conditions and the system will resonate. (Replace all $P_n^m$ above with ($P_n^m(cos \theta) + K_n Q_n^m(cos \theta))$
So that's a rather terse reply that covers some 20+ pages of algebra (and calculus I suppose). It is obvious on its face that the EmDrive can't work, but it can't work in so many ways it's ridiculous. The resonance of a spherical cone is still an interesting problem and hopefully someone will find this useful.
