Question

Prove that \( \sin^{-1}(-x) = -\sin^{-1}(x) \), where \( x \in [-1, 1] \)

Hint:

The property \( \sin^{-1}(-x) = -\sin^{-1}(x) \) holds due to the odd nature of the sine function.

Answer 1

Step 1: Understanding the inverse sine function.
The inverse sine function, \( \sin^{-1}(x) \), is defined such that if \( y = \sin^{-1}(x) \), then \( \sin(y) = x \) and \( y \in \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).

Step 2: Proving the identity.
Let \( y = \sin^{-1}(x) \). By the definition of inverse sine: \[ \sin(y) = x \] Now, consider \( \sin^{-1}(-x) \). Let \( z = \sin^{-1}(-x) \), then by the definition of inverse sine: \[ \sin(z) = -x \] We know that \( \sin(-y) = -\sin(y) \), so: \[ \sin(-y) = -x \] Thus, \( z = -y \), meaning that: \[ \sin^{-1}(-x) = -\sin^{-1}(x) \]

Step 3: Conclusion.
Thus, we have proved that \( \sin^{-1}(-x) = -\sin^{-1}(x) \) for \( x \in [-1, 1] \).