To determine the intervals where the functions \( y_1 = \frac{\sin x}{|\sin x|} \) and \( y_2 = \frac{|\cos x|}{\cos x} \) are identical, we need to analyze the behavior of these functions across the specified intervals.
### Step-by-Step Solution:
1. **Identify the intervals**: We are given four intervals to check:
- I. \( (0, \frac{\pi}{2}) \)
- II. \( (\frac{\pi}{2}, \pi) \)
- III. \( (\pi, \frac{3\pi}{2}) \)
- IV. \( (\frac{3\pi}{2}, 2\pi) \)
2. **Evaluate \( y_1 \) and \( y_2 \) in each interval**:
**Interval I: \( (0, \frac{\pi}{2}) \)**
- In this interval, both \( \sin x \) and \( \cos x \) are positive.
- Therefore, \( |\sin x| = \sin x \) and \( |\cos x| = \cos x \).
- Thus, \( y_1 = \frac{\sin x}{\sin x} = 1 \) and \( y_2 = \frac{\cos x}{\cos x} = 1 \).
- Hence, \( y_1 = y_2 \).
**Interval II: \( (\frac{\pi}{2}, \pi) \)**
- Here, \( \sin x \) is positive and \( \cos x \) is negative.
- Thus, \( |\sin x| = \sin x \) and \( |\cos x| = -\cos x \).
- Therefore, \( y_1 = \frac{\sin x}{\sin x} = 1 \) and \( y_2 = \frac{-\cos x}{\cos x} = -1 \).
- Hence, \( y_1 \neq y_2 \).
**Interval III: \( (\pi, \frac{3\pi}{2}) \)**
- In this interval, both \( \sin x \) and \( \cos x \) are negative.
- Thus, \( |\sin x| = -\sin x \) and \( |\cos x| = -\cos x \).
- Therefore, \( y_1 = \frac{\sin x}{-\sin x} = -1 \) and \( y_2 = \frac{-\cos x}{\cos x} = -1 \).
- Hence, \( y_1 = y_2 \).
**Interval IV: \( (\frac{3\pi}{2}, 2\pi) \)**
- Here, \( \sin x \) is negative and \( \cos x \) is positive.
- Thus, \( |\sin x| = -\sin x \) and \( |\cos x| = \cos x \).
- Therefore, \( y_1 = \frac{\sin x}{-\sin x} = -1 \) and \( y_2 = \frac{\cos x}{\cos x} = 1 \).
- Hence, \( y_1 \neq y_2 \).
3. **Conclusion**: The functions \( y_1 \) and \( y_2 \) are identical in the intervals:
- I. \( (0, \frac{\pi}{2}) \)
- III. \( (\pi, \frac{3\pi}{2}) \)
Thus, the answer is that \( y_1 \) and \( y_2 \) are identical functions for intervals I and III.
