What is the probability of $\cos(\theta_1) + \cos(\theta_2) + \cos(\theta_1 - \theta_2) + 1 \le 0$?

Question

What is the probability of $\cos(\theta_1) + \cos(\theta_2) + \cos(\theta_1 - \theta_2) + 1 \le 0$ given that $\theta_1$ and $\theta_2$ are chosen randomly between $0$ and $2\pi$?

Answer 1

Let us denote:

$$\tag{1}f(\theta_1,\theta_2):=\cos(\theta_1) + \cos(\theta_2) + \cos(\theta_1 - \theta_2) + 1.$$

Using relationships:

$$\cos(a)+\cos(b)=2 \cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)$$

$$1+\cos(a)=2\cos\left(\frac{a}{2}\right)^2,$$

one can write $f(\theta_1,\theta_2)$ under a product form:

$$\tag{2} f(\theta_1,\theta_2)=4 \cos\left(\dfrac{\theta_1}{2}\right) \cos\left(\dfrac{\theta_2}{2}\right)\cos\left(\dfrac{\theta_1-\theta_2}{2}\right) \le 0$$

Thus the discussion on inequation

$$\tag{3} f(\theta_1,\theta_2)\le 0$$

amounts to a regionalization of the domain $(\theta_1,\theta_2) \in [0,2\pi) \times [0,2 \pi)$, materialized by the square ABCD (see graphics below).

More precisely, the regions where inequality (3) is verified are determined by their boundaries (characterized by an occurence of a transition between positive and negative values):

$$\text{for factors } \ \begin{cases}\cos(\theta_1/2):&\text{line FG with equ.} \ \theta_1=\pi \\\cos(\theta_2/2):&\text{line EH with equ.} \ \theta_2=\pi\\ \cos((\theta_1-\theta_2)/2):&\text{lines EF and GH with equ.} \ \theta_1-\theta_2=\pm\pi\end{cases}$$

These boundaries define 6 regions. Which regions are the good ones ?

Determining the sign of $f(\theta_1,\theta_2)$ for each region can be done by obtaining the sign of each factor; the regions to be selected are those where the product of these 3 signs is negative.

This process, rather long, generates a kind of checkerboard pattern (one and only one sign changes when a boundary is crossed). Whence the idea of a simpler method: it suffices to test a single point inside an arbitrary region, color this region in white (resp. red) according to the sign of $f(\theta_1,\theta_2)$ [negative (resp. positive)], then color all the neighbouring regions with the opposite color, etc...

For example, the lower left square AGIE, tested positive because $f(\theta_1,\theta_2)>0$ for test point $J(\theta_1,\theta_2)=(\pi/2,\pi/2),$ will be colored red. Its neighbouring regions, EFI and GHI, will then be colored in white, etc...

In a last step, the areas of the white regions are added, and the result is the ratio between this sum of areas and the area of the square. One finds in this way

$$\text{Probability} \ = \ \frac{1}{4}.$$