Question

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

Hint:

Use the relationship between inverse sine and cosine functions to simplify such expressions.

Answer 1

We are given \( \sin^{-1}{x} + \cos^{-1}{x} \) and need to prove that it equals \( \frac{\pi}{2} \). Step 1: Let \( y = \sin^{-1}{x} \). Then, by the definition of the inverse sine function: \[ \sin{y} = x, \quad 0 \leq y \leq \frac{\pi}{2} \] Step 2: We know that \( \cos^{-1}{x} \) is the angle whose cosine is \( x \). Therefore, we have: \[ \cos^{-1}{x} = \frac{\pi}{2} - y \] Step 3: Therefore: \[ \sin^{-1}{x} + \cos^{-1}{x} = y + \left( \frac{\pi}{2} - y \right) = \frac{\pi}{2} \] Thus, the equation is proved.