Question

If \( -1 \leq x \leq 1 \), then prove that \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \).

Hint:

Use trigonometric identities and angle ranges to prove inverse function relationships.

Answer 1

Let \( \theta = \sin^{-1} x \), so \( \sin \theta = x \), \( \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] \). Then: \[ \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - x^2} \quad (\text{since } \cos \theta \geq 0 \text{ in } [-\frac{\pi}{2}, \frac{\pi}{2}]). \] Thus, \( \theta = \cos^{-1} \sqrt{1 - x^2} \). Consider: \[ \sin^{-1} x + \cos^{-1} x = \theta + \cos^{-1} x. \] Let \( \phi = \cos^{-1} x \), so \( \cos \phi = x \), \( \phi \in [0, \pi] \). Then: \[ \sin \phi = \sqrt{1 - x^2} \quad (\text{since } \sin \phi \geq 0 \text{ in } [0, \pi]). \] Thus, \( \phi = \sin^{-1} \sqrt{1 - x^2} \). But: \[ \theta + \phi = \sin^{-1} x + \cos^{-1} x. \] Since \( \sin \theta = x = \cos \phi \), consider \( \theta + \phi \): \[ \sin (\theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi = x \cdot x + \sqrt{1 - x^2} \cdot \sqrt{1 - x^2} = x^2 + (1 - x^2) = 1. \] \[ \sin (\theta + \phi) = 1 \Rightarrow \theta + \phi = \frac{\pi}{2} + 2k\pi. \] Since \( \theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] \), \( \phi \in [0, \pi] \), \( \theta + \phi \in [-\frac{\pi}{2}, \frac{3\pi}{2}] \), and \( \sin (\theta + \phi) = 1 \Rightarrow \theta + \phi = \frac{\pi}{2} \).
Thus, \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \).