(Goldstein 3rd edition pg 72)
After reducing two-body problem to one-body problem
We now restrict ourselves to conservative central forces, where the potential is $V(r)$ function of $r$ only, so that the force is always along $\mathbf{r}$. By the results of the preceding section, we need only consider the problem of a single particle of reduced mass $m$ moving about a fixed center of force, which will be taken as the origin of the coordinate system. Since potential energy involves only the radial distance, the problem has spherical symmetry; i.e., any rotation, ahout any fixed axis, can have no effect on the solution. Hence, an angle coordinate representing rotation about a fixed axis must be cyclic.
But the 3D kinetic energy has a form
$$T = \frac{1}{2}m\left(\dot r^2 + r^2\dot\theta^2 + r^2\sin^2\theta\dot\phi^2\right)$$ and hence the Lagrangian $L$ depends on $\theta$ and hence $\theta$ is not cyclic.
Doubt on Mr Joseph's answer :
Goldstein pg 59
It can be shown that if a cyclic coordinate $q_{j}$ is such that $d q_{j}$ corresponds to a rotation of the system of particles around some axis, then the conservation of its conjugate momentum corresponds to conservation of an angular momentum.
In our case $\phi$ is cyclic as is seen from equation of kinetic energy and change in $\phi$ corresponds to a rotation of our system so one component of angular momentum is conserved.
We can't say the same about $\theta$ since its not cyclic to begin with.
Best Answer
The Lagrangian is $$\mathcal L=T-V(r)$$
This problem has spherical symmetry and the potential energy is a function of $r$ only as stated above and the particle is in a conservative central force. This means that the angular momentum is conserved and motion lies in a plane perpendicular to the angular momentum vector.
In such cases, one can chose the motion of the particle to be in a plane at right angles to the polar axis meaning we can reduce the Lagrangian to $$\mathcal{L}=T -V= \frac{1}{2}m\left(\dot r^2 + r^2\dot\theta^2\right)-V(r)$$ by loosing the $\phi$ degree of freedom.
By definition, a coordinate $q_i$ is cyclic if the Lagrangian $\mathcal L$ doesn't explicitly depend on it. That is $$\frac{\mathcal\partial L}{\partial q_i}=0$$
Since the Lagrangian has that $$\frac{\mathcal\partial L}{\partial \theta}=\frac{\mathcal\partial L}{\partial \phi}=0$$ then the coordinates $\theta$ (and $\phi$) are cyclic but note that $\phi$ does not explicitly appear in the original Lagrangian (the kinetic energy expression you quoted in your question).