To solve the given problem, we need to find the values of certain trigonometric expressions based on the provided sets. Let's break down the solution step by step.
### Step 1: Define the Sets
We will define the sets based on the problem statement:
1. **Set A**: \( A = \{ \pm \sqrt{n \pi} \mid n \in \mathbb{W} \} \) where \( \mathbb{W} \) is the set of whole numbers (0, 1, 2, 3,...).
2. **Set B**: \( B = \{ \pm \sqrt{n \pi} \mid n \in \mathbb{N} \} \) where \( \mathbb{N} \) is the set of natural numbers (1, 2, 3,...).
3. **Set C**: \( C = \{ \frac{\pi}{2} + 2n\pi \mid n \in \mathbb{Z} \} \) where \( \mathbb{Z} \) is the set of integers (..., -2, -1, 0, 1, 2,...).
4. **Set D**: \( D = \{ 2n\pi \mid n \in \mathbb{Z} \} \).
### Step 2: Calculate Elements of Set A
For Set A, we will calculate the elements for \( n = 0, 1, 2, 3, 4 \):
- For \( n = 0 \): \( \pm \sqrt{0 \cdot \pi} = 0 \)
- For \( n = 1 \): \( \pm \sqrt{1 \cdot \pi} = \pm \sqrt{\pi} \)
- For \( n = 2 \): \( \pm \sqrt{2 \cdot \pi} = \pm \sqrt{2\pi} \)
- For \( n = 3 \): \( \pm \sqrt{3 \cdot \pi} = \pm \sqrt{3\pi} \)
- For \( n = 4 \): \( \pm \sqrt{4 \cdot \pi} = \pm 2\sqrt{\pi} \)
Thus, Set A is:
\[ A = \{ 0, \pm \sqrt{\pi}, \pm \sqrt{2\pi}, \pm \sqrt{3\pi}, \pm 2\sqrt{\pi} \} \]
### Step 3: Calculate Elements of Set B
For Set B, we will calculate the elements for \( n = 1, 2, 3, 4 \):
- For \( n = 1 \): \( \pm \sqrt{1 \cdot \pi} = \pm \sqrt{\pi} \)
- For \( n = 2 \): \( \pm \sqrt{2 \cdot \pi} = \pm \sqrt{2\pi} \)
- For \( n = 3 \): \( \pm \sqrt{3 \cdot \pi} = \pm \sqrt{3\pi} \)
- For \( n = 4 \): \( \pm \sqrt{4 \cdot \pi} = \pm 2\sqrt{\pi} \)
Thus, Set B is:
\[ B = \{ \pm \sqrt{\pi}, \pm \sqrt{2\pi}, \pm \sqrt{3\pi}, \pm 2\sqrt{\pi} \} \]
### Step 4: Calculate Elements of Set C
For Set C, we will calculate the elements for \( n = -2, -1, 0, 1, 2 \):
- For \( n = 0 \): \( \frac{\pi}{2} + 2 \cdot 0 \cdot \pi = \frac{\pi}{2} \)
- For \( n = 1 \): \( \frac{\pi}{2} + 2 \cdot 1 \cdot \pi = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2} \)
- For \( n = -1 \): \( \frac{\pi}{2} + 2 \cdot (-1) \cdot \pi = \frac{\pi}{2} - 2\pi = -\frac{3\pi}{2} \)
- For \( n = 2 \): \( \frac{\pi}{2} + 2 \cdot 2 \cdot \pi = \frac{\pi}{2} + 4\pi = \frac{9\pi}{2} \)
- For \( n = -2 \): \( \frac{\pi}{2} + 2 \cdot (-2) \cdot \pi = \frac{\pi}{2} - 4\pi = -\frac{7\pi}{2} \)
Thus, Set C is:
\[ C = \{ \frac{\pi}{2}, \frac{5\pi}{2}, -\frac{3\pi}{2}, \frac{9\pi}{2}, -\frac{7\pi}{2} \} \]
### Step 5: Calculate Elements of Set D
For Set D, we will calculate the elements for \( n = -2, -1, 0, 1, 2 \):
- For \( n = 0 \): \( 2 \cdot 0 \cdot \pi = 0 \)
- For \( n = 1 \): \( 2 \cdot 1 \cdot \pi = 2\pi \)
- For \( n = -1 \): \( 2 \cdot (-1) \cdot \pi = -2\pi \)
- For \( n = 2 \): \( 2 \cdot 2 \cdot \pi = 4\pi \)
- For \( n = -2 \): \( 2 \cdot (-2) \cdot \pi = -4\pi \)
Thus, Set D is:
\[ D = \{ 0, 2\pi, -2\pi, 4\pi, -4\pi \} \]
### Final Result
The values of the sets are:
- Set A: \( A = \{ 0, \pm \sqrt{\pi}, \pm \sqrt{2\pi}, \pm \sqrt{3\pi}, \pm 2\sqrt{\pi} \} \)
- Set B: \( B = \{ \pm \sqrt{\pi}, \pm \sqrt{2\pi}, \pm \sqrt{3\pi}, \pm 2\sqrt{\pi} \} \)
- Set C: \( C = \{ \frac{\pi}{2}, \frac{5\pi}{2}, -\frac{3\pi}{2}, \frac{9\pi}{2}, -\frac{7\pi}{2} \} \)
- Set D: \( D = \{ 0, 2\pi, -2\pi, 4\pi, -4\pi \} \)
