Prove that `sin^(-1). ((x + sqrt(1 - x^(2))/(sqrt2)) = sin^(-1) x + (pi)/(4)`, where `- (1)/(sqrt2) lt x lt(1)/(sqrt2)`

Explanation

To prove that

\[ \sin^{-1}\left(\frac{x + \sqrt{1 - x^2}}{\sqrt{2}}\right) = \sin^{-1}(x) + \frac{\pi}{4} \]

for \(-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}\), we will follow these steps:

### Step 1: Substitute \(x\) with \(\sin \theta\)

Let \(x = \sin \theta\). Given the range \(-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}\), this implies that \(-\frac{\pi}{4} < \theta < \frac{\pi}{4}\).

### Step 2: Rewrite the left-hand side (LHS)

Now substitute \(x\) in the LHS:

\[ \sin^{-1}\left(\frac{\sin \theta + \sqrt{1 - \sin^2 \theta}}{\sqrt{2}}\right) \]

Using the identity \(\sqrt{1 - \sin^2 \theta} = \cos \theta\), we have:

\[ \sin^{-1}\left(\frac{\sin \theta + \cos \theta}{\sqrt{2}}\right) \]

### Step 3: Simplify the expression inside the inverse sine

The expression \(\sin \theta + \cos \theta\) can be rewritten using the sine addition formula. We know that:

\[ \sin \theta + \cos \theta = \sqrt{2} \left(\sin \theta \cdot \frac{1}{\sqrt{2}} + \cos \theta \cdot \frac{1}{\sqrt{2}}\right) = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) \]

Thus, we can rewrite the LHS as:

\[ \sin^{-1}\left(\frac{\sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right)}{\sqrt{2}}\right) = \sin^{-1}\left(\sin\left(\theta + \frac{\pi}{4}\right)\right) \]

### Step 4: Use the property of inverse sine

Since \(-\frac{\pi}{4} < \theta < \frac{\pi}{4}\), it follows that:

\[ \sin^{-1}\left(\sin\left(\theta + \frac{\pi}{4}\right)\right) = \theta + \frac{\pi}{4} \]

### Step 5: Substitute back for \(\theta\)

Recall that \(\theta = \sin^{-1}(x)\). Therefore, we have:

\[ \theta + \frac{\pi}{4} = \sin^{-1}(x) + \frac{\pi}{4} \]

### Conclusion

Thus, we have shown that:

\[ \sin^{-1}\left(\frac{x + \sqrt{1 - x^2}}{\sqrt{2}}\right) = \sin^{-1}(x) + \frac{\pi}{4} \]

This proves the required identity.

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