Sketching a coupled system without solving it

Question

I need to hand sketch $$\dot{\theta_1}(t) = 2\pi + \sin(\theta_2 - \theta_1)$$ and $$\dot{\theta_2}(t) = 2\pi + \sin(\theta_1 - \theta_2)$$ with $\theta_1(0) = \pi/2$ and $\theta_2(0) = 0$. Is the only way to do this to just use numerical methods with the initial conditions? I've noticed $\dot{\theta}_2 = 4\pi - \dot{\theta}_1$, so I only need to calculate $\dot{\theta_1}$.

I'm not allowed to solve this problem explicitly, so doing this sketch is baffling me. I suppose what I'm asking, is there any qualitative features I have missed that would make a hand sketching of these two functions easier?

Answer 1

A MATHEMATICA script which helps.

gr2 = StreamPlot[{2 Pi + Sin[y - x], 2 Pi + Sin[x - y]}, {x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi}]; gr2a = ContourPlot[2 Pi (x + y) - Cos[x - y], {x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi}, Contours -> 40, ContourStyle -> Black, ContourShading -> Black]; Show[gr2, gr2a]

Answer 2

$\dot{\theta_2}(t) = 2\pi + \sin(\theta_1 - \theta_2)$

$\dot{\theta_1}(t) = 2\pi + \sin(\theta_2 - \theta_1)$

$\Rightarrow \dot{\theta_2}(t) -\dot{\theta_1}(t) = -2 \sin(\theta_2 - \theta_1) $

Use transposition.