Problem
If \( \sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{3} \) and \( \cos^{-1}(x) – \cos^{-1}(y) = \frac{\pi}{6} \), find x and y.
Solution
Step 1: Convert cos⁻¹ into sin⁻¹
\[ \cos^{-1}t = \frac{\pi}{2} – \sin^{-1}t \]
So,
\[ \cos^{-1}x – \cos^{-1}y = \left(\frac{\pi}{2} – \sin^{-1}x\right) – \left(\frac{\pi}{2} – \sin^{-1}y\right) \]
\[ = -\sin^{-1}x + \sin^{-1}y \]
Given:
\[ -\sin^{-1}x + \sin^{-1}y = \frac{\pi}{6} \]
Step 2: Form equations
\[ \sin^{-1}x + \sin^{-1}y = \frac{\pi}{3} \quad …(1) \]
\[ -\sin^{-1}x + \sin^{-1}y = \frac{\pi}{6} \quad …(2) \]
Step 3: Solve system
Add (1) and (2):
\[ 2\sin^{-1}y = \frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2} \]
\[ \sin^{-1}y = \frac{\pi}{4} \Rightarrow y = \frac{1}{\sqrt{2}} \]
Substitute in (1):
\[ \sin^{-1}x + \frac{\pi}{4} = \frac{\pi}{3} \]
\[ \sin^{-1}x = \frac{\pi}{3} – \frac{\pi}{4} = \frac{\pi}{12} \]
\[ x = \sin\left(\frac{\pi}{12}\right) = \sin(15^\circ) = \frac{\sqrt{6} – \sqrt{2}}{4} \]
Final Answer
\[ \boxed{ x = \frac{\sqrt{6} – \sqrt{2}}{4}, \quad y = \frac{1}{\sqrt{2}} } \]
Explanation
Convert cos⁻¹ into sin⁻¹, form linear equations, and solve systematically.