To solve this question, we need to identify which trigonometric functions have a domain of \([-1, 1]\).
Firstly, let's recall the definitions and domains of some common inverse trigonometric functions:
1. \( y = \arcsin x \):
- Definition: This is the inverse function of \( \sin y \).
- Domain: The domain for \( \arcsin x \) is \([-1, 1]\).
2. \( y = \arccos x \):
- Definition: This is the inverse function of \( \cos y \).
- Domain: The domain for \( \arccos x \) is \([-1, 1]\).
3. \( y = \arctan x \):
- Definition: This is the inverse function of \( \tan y \).
- Domain: The domain for \( \arctan x \) is all real numbers \((-\infty, \infty)\).
4. \( y = \operatorname{arcsec} x \):
- Definition: This is the inverse function of \( \sec y \).
- Domain: The domain for \( \operatorname{arcsec} x \) is \( (-\infty, -1] \cup [1, \infty)\).
5. \( y = \operatorname{arccsc} x \):
- Definition: This is the inverse function of \( \csc y \).
- Domain: The domain for \( \operatorname{arccsc} x \) is \( (-\infty, -1] \cup [1, \infty)\).
Now, we check which of the given pairs have the domain \([-1, 1]\):
1. \( y = \arcsin x \) and \( y = \arccos x \):
- Both \( \arcsin x \) and \( \arccos x \) have the domain \([-1, 1]\).
2. \( y = \arccos x \) and \( y = \arctan x \):
- \( \arccos x \) has the domain \([-1, 1]\), but \( \arctan x \) has the domain \((-\infty, \infty)\).
3. \( y = \arcsin x \) and \( y = \arctan x \):
- \( \arcsin x \) has the domain \([-1, 1]\), but \( \arctan x \) has the domain \((-\infty, \infty)\).
4. \( y = \operatorname{arcsec} x \) and \( y = \operatorname{arccsc} x \):
- Both \( \operatorname{arcsec} x \) and \( \operatorname{arccsc} x \) have the domain \( (-\infty, -1] \cup [1, \infty)\).
The correct functions with the domain \([-1, 1]\) are \( y = \arcsin x \) and \( y = \arccos x \).
Thus, the correct answer is:
[tex]\[ y = \arcsin x \text{ and } y = \arccos x \][/tex].
