I am currently using a simulator for simple pendulum, with following update
$$\dot{\theta_t} = \dot{\theta}_{t-1} + (-\frac{3g}{2l}\sin(\theta_{t-1}+\pi) + \frac{3u}{2ml})\cdot \Delta t$$ $$\theta_t = \theta_{t-1} + \dot\theta_t\Delta t$$ where $g = 9.82$ is gravity constant, $m=1$ is mass of bob, $l=1$ is the length, $\theta$ is the angle, and $u$ is the torque.
Is it possible to infer from the equation which phase the angle $\theta$ start from ? e.g. for a small positive $\theta$, should it be at upper left corner ? And the range of $\theta$ lie between $[0, 2\pi]$ or $[-\pi, \pi]$ ?
The $\sin(\theta + \pi)$ part in your first equation gives the clue: when the argument of that is $0$, the pendulum is straight down, $\pi$ is straight up, anything in $(0, \pi)$ is in the right half and $(\pi, 2\pi)$ is on the left. So, small positive $\theta$ means $\theta + \pi$ is just over $\pi$, so indeed the upper left (though I would not say "corner").
The second question is not that simple; if your pendulum is driven by sufficiently strong external torques that make it possible to swing "through" a full circle, you cannot really restrict it to a particular range like that: every time it makes a full circle, $\theta$ has to increase/decrease by $2\pi$ by virtue of your second equation. In topological terms, there is no continuous bijection from a circle to an interval.
If, however, the pendulum cannot pass through the "top" then you can take any interval that "starts" at the upper left "corner" which as we said can be identified with small positive values; so $0$ to $2\pi$ seems most natural, but shifting it by any multiple of $2\pi$ also works.
Finally, is this really a simple pendulum? Looks to me like you are considering a rigid rod, not a bob on a rope.