I have always found it clear that since a spherical potential has all components of angular momentum conserved since the entire system is symmetric under any rotation. However, I was trying to prove this to myself using Noether's theorem. The Lagrangian in spherical coordinates for such a system:
$$L = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta^2} + r^2 \sin^2(\theta) \dot{\phi}^2) - V(r).$$
Since the lagrangian does not have an explicity $\phi$ dependence, the conservation holds. However, when I try this on the $\theta$ axis, I am running into some troubles. Let $\theta \rightarrow \theta + \lambda$. Then by Noether's theorem, I need to show that the lagrangian does not change in the first order of $\lambda$:
$$L' = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta^2} + r^2 \sin^2(\theta + \lambda) \dot{\phi}^2) - V(r).$$
The problem is with the sine term. My first intuition is to taylor expand:
$$\sin(x) = \sin(\theta) + \cos(\theta)(x - \theta) - \frac{1}{2} \sin(\theta) (x - \theta)^2+...$$
$$\sin(\theta + \lambda) = \sin(\theta) + \cos(\theta)\lambda - \frac{1}{2} \sin(\theta) \lambda^2+...$$
$$\sin^2(\theta + \lambda) = (\sin(\theta) + \cos(\theta)\lambda - \frac{1}{2} \sin(\theta) \lambda^2+...)(\sin(\theta) + \cos(\theta)\lambda - \frac{1}{2} \sin(\theta) \lambda^2+...)$$ $$ = \sin^2(\theta) + 2\sin(\theta)\cos(\theta) \lambda + ....$$
Which is not a first order change. In other words:
$$\frac{\partial L}{\partial \lambda} = \frac{1}{2}mr^22\sin(\theta)\cos(\theta)\dot{\phi^2} \neq0.$$
However, I know intuitively that around $\theta$, there is symmetry. What is exactly wrong in my line of logic?