Problem
Simplify: \( \sin^{-1}\left(\frac{x + \sqrt{1 – x^2}}{\sqrt{2}}\right), \quad -\frac{1}{2} < x < \frac{1}{\sqrt{2}} \)
Solution (Substitution Method)
Let:
\[ x = \sin \theta \]
Then,
\[ \sqrt{1 – x^2} = \cos \theta \]
So the expression becomes:
\[ \sin^{-1}\left(\frac{\sin \theta + \cos \theta}{\sqrt{2}}\right) \]
Using identity:
\[ \sin \theta + \cos \theta = \sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right) \]
Thus,
\[ \frac{\sin \theta + \cos \theta}{\sqrt{2}} = \sin\left(\theta + \frac{\pi}{4}\right) \]
Hence,
\[ \sin^{-1}\left(\frac{x + \sqrt{1 – x^2}}{\sqrt{2}}\right) = \sin^{-1}\left(\sin\left(\theta + \frac{\pi}{4}\right)\right) \]
Since the given domain ensures the angle lies in principal range:
\[ = \theta + \frac{\pi}{4} \]
But \( \theta = \sin^{-1} x \)
Final Answer
\[ \boxed{\sin^{-1}x + \frac{\pi}{4}} \]