Let A = {x ∈ R| cos x ∈ Z}, B = {x ∈ R| sin x ∈ Z}. Is A ⊆ B? Is B ⊆ A? Is either A or B a proper subset of the where other?
I'd say they are the same size.
All multiples of pi/2 are in both sets.
Then, since cosx = sin(pi/2 - x), wherever x is in A, pi/2-x is in B.
Well, A and B are both quite selective clubs. A is the club where the real numbers x make the cosine of x a whole number, and B is the club where the real numbers x make the sine of x a whole number.
Now, is A ⊆ B? Let's analyze this. For a number to be in A, it means that its cosine must be a whole number. And surprise, surprise, any whole number can be written as the sine of another real number, namely arcsin! So, every element in A is an element in B. Therefore, A is ⊆ B, like a small club fitting inside a larger one.
Next up, is B ⊆ A? To be in B, a number must have a sine that is a whole number. But there are plenty of real numbers whose sine is not whole numbers, like 1/2, pi/4, or pi/2. So, not every element in B is an element in A. Hence, B is not ⊆ A.
Now, which one is a proper subset of the other? Well, we know A is a subset of B, and B is not a subset of A. Therefore, A is a proper subset of B, like a small clown nose compared to a big red clown wig.
Hope that clears things up with a touch of clownish clarity!
To determine if A is a subset of B, we need to check if every element in A is also an element of B.
A = {x ∈ R| cos x ∈ Z}
B = {x ∈ R| sin x ∈ Z}
Let's analyze A first:
Since the values of cos x are bounded between -1 and 1, there are only a few possible integers that cos x can take on. These are -1, 0, and 1. Therefore, A consists of the values of x where cos x can take on these three integers.
Now let's analyze B:
The values of sin x are also bounded between -1 and 1. Similarly, there are only a few possible integers that sin x can take on. These are -1, 0, and 1. Therefore, B consists of the values of x where sin x can take on these three integers.
From this analysis, we can conclude that A ⊆ B, since every element in A is also an element of B.
To determine if B is a subset of A, we need to check if every element in B is also an element of A.
Since A and B are sets of real numbers, there is no overlap between the two sets. In other words, there are no real numbers x that satisfy both sin x = 0 or sin x = 1 (from B) and cos x = 0 or cos x = 1 (from A). Therefore, B ⊆ A does not hold.
Neither A nor B is a proper subset of the other, as there is no element in one set that is not present in the other.
To determine if set A is a subset of set B, we need to check if every element in A is also in B. Similarly, to determine if set B is a subset of set A, we need to check if every element in B is also in A.
To understand the sets A and B, let's analyze the values of cos(x) and sin(x):
- The cosine function (cos(x)) returns values between -1 and 1 in the real number system (R). Therefore, cos(x) cannot be an integer for any value of x in the real number system.
- The sine function (sin(x)) also returns values between -1 and 1 in the real number system. Hence, sin(x) cannot be an integer for any x in the real number system.
Since there are no real values of x for which cos(x) or sin(x) can both be integers, set A and B will be empty sets. This means that both A and B are proper subsets of each other.
To summarize:
- A is an empty set because there are no real values for cos(x) to be an integer.
- B is an empty set because there are no real values for sin(x) to be an integer.
Therefore, neither A nor B is a subset of the other, and both are proper subsets of each other.