What is the probability of cos(θ1)+cos(θ2)+cos(θ1−θ2)+1≤0\cos(\theta_1) + \cos(\theta_2) + \cos(\theta_1 – \theta_2) + 1 \le 0?

What is the probability of cos(θ1)+cos(θ2)+cos(θ1−θ2)+1≤0\cos(\theta_1) + \cos(\theta_2) + \cos(\theta_1 – \theta_2) + 1 \le 0 given that θ1\theta_1 and θ2\theta_2 are chosen randomly between 00 and 2π2\pi?

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What have you attempted ? You should show that you have already worked on the question you ask…
– JeanMarie
2 days ago

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1 Answer
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Let us denote:

f(θ1,θ2):=cos(θ1)+cos(θ2)+cos(θ1−θ2)+1.\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)=2cos(a+b2)cos(a−b2)\cos(a)+\cos(b)=2 \cos\left(\frac{a+b}{2}\right)\cos\left(\frac{a-b}{2}\right)

1+cos(a)=2cos(a2)2,1+\cos(a)=2\cos\left(\frac{a}{2}\right)^2,

one can write f(θ1,θ2)f(\theta_1,\theta_2) under a product form:

f(θ1,θ2)=4cos(θ12)cos(θ22)cos(θ1−θ22)≤0\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

f(θ1,θ2)≤0\tag{3} f(\theta_1,\theta_2)\le 0

amounts to a regionalization of the domain (θ1,θ2)∈[0,2π)×[0,2π)(\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):

for factors  {cos(θ1/2):line FG with equ. θ1=πcos(θ2/2):line EH with equ. θ2=πcos((θ1−θ2)/2):lines EF and GH with equ. θ1−θ2=±π\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(θ1,θ2)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(θ1,θ2)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(θ1,θ2)>0f(\theta_1,\theta_2)>0 for test point J(θ1,θ2)=(π/2,π/2),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

Probability = 14.\text{Probability} \ = \ \frac{1}{4}.

  

 

Suggestion: Could you change the scale on the graph to be in terms of π\pi?
– Ian Miller
2 days ago

  

 

I could do it, but it is very late (2 am, CET, while, being in China as I have seen in your profile, you are awake) thus pardon me to refer different ameliorations to tomorrow…
– JeanMarie
2 days ago

  

 

No hurry. The graph is readable as it. I just thought that would be a nice improvement. Have a good sleep.
– Ian Miller
2 days ago

  

 

Thanks. Have a good day.
– JeanMarie
2 days ago

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@Ian Miller New graphics available.
– JeanMarie
yesterday