Divergence of a specific electrical field I need to show that the divergence of the electrical field given as 
$$\vec{E}=\vec{e_{\theta}}\frac{A\sin\theta}{r}\exp[i\omega(t-r/c)]$$
is zero. As the vector (in sperical coordinates) containes only the $\theta $ component, I looked up the definition of the divergence in Wikipedia for the mentioned term which is:
$$\vec{\nabla}\cdot\vec{E_{(\theta)}}=\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta \, E_{\theta}).$$
Plugging in the above and deriving yields:
$$\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\sin^2\theta\frac{A}{r}\exp[i\omega(t-r/c)] = \frac{2A\cos\theta}{r^2}\exp[i\omega(t-r/c)]\neq 0.$$
Where am I wrong as the term I get seems not to be zero?
 A: You are not wrong; your need (to show that the divergence of this particular vector field is zero) can never be satisfied because it is not zero. However do not lose hope; you might notice for example that you've proven that the divergence goes to zero as $r\to\infty$, and that's a start!
You might be able to determine, however, that this is just the $\theta$-component of some other field which has zero divergence. For example, if instead of having just an $r$ in the denominator, you have an $r^3$, then your field becomes the $\theta$-component for the famous electric dipole field that comes from a voltage $V(r, \theta) \propto \cos\theta / r^2.$ Since this represents two point charges at $r=0$, its fields have no divergence anywhere else. In particular it has an $r$-component which cancels out the divergence due to its $\theta$-component.
Similarly what you're looking at is probably, given its oscillation and $1/r$ dependence, a far-field limit for electric dipole radiation. Let $\psi = k (c t - r)$, this in its fuller form consists of $$
E_r =A~\cos\theta~ \left[\frac{2\cos\psi}{r^3} - \frac{2k\sin\psi}{r^2}\right],\\
E_\theta =A~\sin\theta~ \left[\frac{\cos\psi}{r^3} - \frac{k\sin\psi}{r^2} - \frac{k^2\cos\psi}{r}\right].
$$
The component of divergence due to $E_r$ is $$\frac{1}{r^2} \frac{\partial}{\partial r} 
\left(r^2 E_r\right) = \frac{A\cos\theta}{r^2} \frac{\partial}{\partial r} 
\left(\frac{2\cos\psi}{r} - 2k\sin\psi\right),$$and do not make the mistake of thinking that this latter term vanishes! It gives you a term $(A\cos\theta/r^2)~2k^2\cos\psi,$ which perfectly balances out the corresponding term from the far field, where the $k^2$ are present due to me having a slightly different definition of $A$ above from what you have in your expression.
So: try to adjoin the $1/r^2$ term from $E_r$ and see if that fixes anything?
