Given a gradient system
$$\frac{d\theta_1}{dt}=-\sin(\theta_1-\theta_2)$$ $$\frac{d\theta_2}{dt}=-\sin(\theta_2-\theta_1)$$
The system is a gradient system since
$$\frac{d\vec \theta}{dt}=-\nabla V(\vec\theta) = -\nabla(1-\cos(\theta_1-\theta_2))$$
Since the system is invariant by replacing all $\theta_i$ to $\theta_i+\alpha$, this will leads to a continuum of equilibria. To remove this freedom, we fix $\theta_1=0$, then the system becomes $$\frac{d\theta_2}{dt}=-\sin(\theta_2)$$
It has two equilibrium points $\theta_2=0$ and $\theta_2=\pi$.
The energy function is $V(\theta_2)=-\cos(\theta_2)+1.$
The second derivative of $V(\theta_2)$ is $\cos\theta_2$.
Since the second derivative is negative on $\theta_2=\pi$, it is an unstable equilibrium point.
Since the second derivative is positive $\theta_2=0$, it is a locally asymptotically stable (LAS) equilibrium point.
Since this is a gradient system, only $\theta_2=0$ is locally asymptotically stable, then it is globally asymptotically stable (GAS).
But it is a little counterintuitive in the sense of convexity of energy function: how comes that the energy function is concave on $\theta_2=\pi$, and $\theta_2=0$ is globally asymptotically stable?
For example, let say the figure of an energy function looks like the following
and it is convex on the local minimum, and concave on the local maximum. How come all trajectories will converge to the local minimum?
As illustrated in the figure, trajectories starting from the part behind the local maximum cannot converge to the local minimum.
As $\dot θ_1+\dot θ_2=0$, the sum is constant and one can replace $θ_2=c-θ_1$ to get a single equation $$ \dot θ_1=-\sin(2θ_1-c) $$ This has indeed a stable equilibrium at $2θ_1-c=0$ and unstable equilibria at $2θ_1-c=\pm\pi$, continued with period $2\pi$ on the right side, thus period $\pi$ in $θ_1$.