Question
If
\[ \sin^{-1}x – \cos^{-1}x = \frac{\pi}{6} \]
Find \( x \).
Solution
We use identity:
\[ \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \]
Let \( \sin^{-1}x = \theta \)
Then:
\[ \theta – \left(\frac{\pi}{2} – \theta\right) = \frac{\pi}{6} \]
\[ 2\theta – \frac{\pi}{2} = \frac{\pi}{6} \]
\[ 2\theta = \frac{\pi}{6} + \frac{\pi}{2} = \frac{4\pi}{6} = \frac{2\pi}{3} \]
\[ \theta = \frac{\pi}{3} \]
Thus,
\[ x = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Final Answer:
\[ \boxed{\frac{\sqrt{3}}{2}} \]
Key Concept
Use identity \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \) to convert difference into a solvable equation.