Probability of an event in relationship with a certain geometrical representation A recent post (https://math.stackexchange.com/q/1979286) asked the following question [in substance: I have slightly modified it]: "What is the probability of the event
$$\tag{1}\cos(\theta_1)+\cos(\theta_2)+\cos(\theta_2-\theta_1)+1 \leq 0 $$
given that $\theta_1$ and $\theta_2$ are independent, uniformly distributed random variables on $[0,2\pi)$"? 
I gave an answer relying on a geometrical interpretation. While elaborating this answer, at a certain step, I made an error, and replaced, in (1), the constant term $+1$ by $-1$. Surprisingly, the unfavorable regions of the square $[0,2\pi) \times [0,2\pi)$, appeared as the interiors of four truncated elliptical regions as displayed on the bottom figure  (obtained by Geogebra).  
My question is therefore twofold: 
1) What is the probability of the event
$$\tag{2}\cos(\theta_1)+\cos(\theta_2)+\cos(\theta_2-\theta_1)-1 \leq 0 ? $$
with $\theta_1$ and $\theta_2$ independent, uniformly distributed random variables on $[0,2\pi)$? 
2) Why are these frontier curves elliptical, with puzzling equation for the "mother"-ellipse centered in $0$:
$$(\theta_1)^2+(\theta_2)^2-\theta_1\theta_2=\left(\frac{\pi}{2}\right)^2 \ \ ?$$
the other ones being translated from this one.
Remark about question 1:  this probability  can hopefully be obtained using a different (non geometrical) approach.

 A: $$(\theta_1)^2+(\theta_2)^2-\theta_1\theta_2=\left(\frac{\pi}{2}\right)^2$$
can be written as
$$\frac{(\theta_1+\theta_2)^2}{4}+\frac{(\theta_1-\theta_2)^2}{\frac{4}3}=\left(\frac{\pi}{2}\right)^2$$
by changing the coordinate sytem
$$\left\{\begin{matrix}
x=\frac{\sqrt{2}}{2}(\theta_1+\theta_2)
\\ 
y=\frac{\sqrt{2}}{2}(\theta_1-\theta_2)
\end{matrix}\right.$$
it becomes
$$\frac{x^2}{\frac{\pi^2}{2}}+\frac{y^2}{\frac{\pi^2}{6}}=1$$
which is a standard ellipse whose area is
$$A=\pi(\frac{\pi}{\sqrt{2}})(\frac{\pi}{\sqrt{6}})$$
Due to symmetry, the four unfavorable regions constitute a single ellipse. So the undesired region is $A$.
The desired probability is therefore $$\frac{4\pi^2-\sqrt{3}\pi^3/6}{4\pi^2}=1-\frac{\sqrt{3}\pi}{24}\approx 0.7733$$
Using Monte-Carlo method
$$
\bbox[0.5em,#efe,border:0.1em groove navy]{\
\Pr(\cos(\theta_1)+\cos(\theta_2)+\cos(\theta_2-\theta_1)-1 \leq 0) \approx 
 0.7738}
$$
So considering the other answer, there would be an error of the order of $10^{-4}$ by assuming elliptic boundaries.

I used the following Matlab code for the Monte-Carlo simulation
n=1e6;
t1=rand(1,n)*2*pi;
t2=rand(1,n)*2*pi;
f=cos(t1)+cos(t2)+cos(t1-t2)-1;
p=mean(f<0);

A: The shapes in your probability graph are not ellipses. (They just look very close.)
Let us start from
$$\cos(\theta_1)+\cos(\theta_2)+\cos(\theta_2-\theta_1)-1 = 0 $$
Apply the substitution $y=\frac{\theta_1+\theta_2}{\sqrt2}$ and $x=\frac{\theta_1-\theta_2}{\sqrt2}$ which is a rotation by $\frac{\pi}{4}$ to better orientation your pseudo-ellipses:
$$\cos\left(\frac{x+y}{\sqrt2}\right)+\cos\left(\frac{y-x}{\sqrt2}\right)+\cos\left(\sqrt2x\right)-1=0$$
Using sums to products:
$$2\cos\left(\frac{x}{\sqrt{2}}\right)\cos\left(\frac{y}{\sqrt{2}}\right)+\cos(\sqrt{2}x)-1=0$$
This can be solved for $y$:
$$y=\sqrt{2}\arccos\left(\frac{1-\cos(\sqrt{2}x)}{2\cos\left(\frac{x}{\sqrt{2}}\right)}\right)$$
$$y=\sqrt{2}\arccos\left(\frac{2\sin^2\left(\frac{x}{\sqrt{2}}\right)}{2\cos\left(\frac{x}{\sqrt{2}}\right)}\right)$$
$$y=\sqrt{2}\arccos\left(\sin\left(\frac{x}{\sqrt{2}}\right)\tan\left(\frac{x}{\sqrt{2}}\right)\right)$$
Visually this looks similar to a half-ellipse.

Some calculations give a y-intercept of $\frac{\pi}{\sqrt{2}}$ and an x-intercepts of $\pm\sqrt{2}\arctan(\frac{1}{2}(1+\sqrt{5}))$. Plotting an ellipse with these values for the two radii looks basically the same.

However plotting the difference shows that they are not equal.

Using the above formula for the curve rather than the ellipse gives a slightly different area:
$$4\sqrt{2}\int_0^{\sqrt{2}\arctan(\sqrt{\frac{1}{2}(1+\sqrt{5}}))}\ \arccos\left(\sin\left(\dfrac{x}{\sqrt{2}}\right)\tan\left(\dfrac{x}{\sqrt{2}}\right)\right)\ dx$$
The integral isn't exact so using Wolfram and dividing by $4\pi^2$ gives the proportion: $0.2261376180596750632\cdots$
So the probability is: $0.7738623819403249367\cdots$
