Consider the surjective function cos: R ? [-1, 1] and let ~ be the associated equivalence relation x ~ y ?? cos x = cos y. Find the Equivalence Class

Question

Consider the surjective function cos: R → [−1, 1] and let ∼ be the associated equivalence relation
x ∼ y ⇐⇒ cos x = cos y.

 

Describe the equivalence classes of ∼.

(b) Do the analogous exercise for the surjective function cot: X → R,
where
X = R \{ πk : k ∈ Z }.

Answer 1

cos is an injection on [0,\pi].

 

cos is an even function cos -x = cos x, and is periodic with period 2\pi.

 

Therefore since it is an injection on [0,\pi) it is an injection on [\pi,2\pi), and x = 2\pi k ± y for some integer k iff x ~ y.

 

I.e.

Suppose cos x = cos y.

Then let j,k be integers such that 2\pi k <= x and 2\pi (k+1) > x
and 2\pi j <= y and 2\pi (j+1) > y. Then put a = x - 2\pi k and b = y - 2\pi j.

Then a,b \in [0,2\pi), and cos x = cos (x - 2\pi k) = cos a, and likewise cos y = cos b.

By the initial considerations, a and b are equal and both in interval [0,\pi) or equal and both in interval [\pi,2\pi), or one of a and b is in interval [0,\pi) and one is in interval [\pi,2\pi). In the last case without loss of generality take a in the first inerval and b in the second; then cos b = cos a = cos (-a) = cos(2\pi - a), and we have

 

~ = the set of all pairs (x,y) in \mathbb{R}^2 such that \exists k \in \mathbb{Z} s.t. x = 2\pi k ± y 

 

Verify symmetry, reflexivity, and transitivity, and agreement with the definition given in the problem.

 

So an equivalence class is a value a in [0,\pi) along with 2\pi k ± a for all integers k. Each such a gives a distinct equivalence class.

 

I hope that helps