I have been asked to respond to the following:
Define a binary relation R on R as {(x,y)∈R×R∣sin(x)=sin(y)}. Prove that R
is an equivalence relation. What are its equivalence classes?
Given that the relation R is based on equality, the first part of the question is rather simple:
Is R reflexive?
Let a∈R, then (a,a)∈R because sin(a)=sin(a).
Is R symmetric?
Let a,b∈R∣(a,b)∈R. Then (b,a)∈R by the symmetric property of equality.
Is R transitive?
Let (a,b)∈R; thus, sin(a)=sin(b).
Let (b,c)∈R; thus, sin(b)=sin(c).
Thus, (a,c)∈R, as sin(a)=sin(c).
However, I am having trouble following a process to find the equivalence classes for R. As we have demonstrated that R is an equivalence relation, we know that we can decompose R into a series of equivalence classes such that, for any x∈R, then x∈[x], and that [x]={y∈R∣(x,y)∈R} (or, more specifically, [x]={y∈R∣sin(x)=sin(y)}).
Examining the unit circle, we know that, for any value x∈R, the value of sin(x) will be equivalent to that of any value x1∈R such that x1=x∗2πk, where k∈Z (after all, sin is a periodic function).
We can also see that sin(x)=sin(π−x) for any x in the range [0,π/2]; similarly, we know that sin(π+x) and sin(2π−x) are both equal to −sin(x). Along a single period of sin(θ), there are exactly two values for θ (in that domain) for which sin(θ) will be equal.
I don't know which of the above information is relevant to the task at hand, and I'm unsure about how to proceed. Any helpful explanations or clues would be appreciated!
EDIT:
I can begin to define some of the equivalence classes of R:
[0]={kπ∣k∈Z}
[1]={π/2+2kπ∣k∈Z}
[−1]={3π/2+2kπ∣k∈Z}
[π/6]={π/6+2kπ,5π/6+2kπ∣k∈Z}
[−π/6]={−π/6+2kπ,7π/6+2kπ∣k∈Z}
How can I generalize this to include all possible equivalence classes (accounting for all possible values of sin x)?
