Express in terms of : `sin^(-1)(2xsqrt(1-x^(2)))` to `sin^(-1)x` for `1gexgt1/(sqrt(2))`

Options

A. `2sin^(-1)x`
B. `2cos^(-1)x`
C. `-2sin^(-1)x`
D. `-2cos^(-1)x`

Answer

To express \( \sin^{-1}(2x\sqrt{1-x^2}) \) in terms of \( \sin^{-1}x \) for \( \frac{1}{\sqrt{2}} < x \leq 1 \), we can follow these steps:

### Step 1: Substitute \( x \) with \( \sin \theta \)
Let \( x = \sin \theta \). Then we have:
\[ \sqrt{1 - x^2} = \sqrt{1 - \sin^2 \theta} = \cos \theta \]
Thus, we can rewrite \( 2x\sqrt{1-x^2} \) as:
\[ 2x\sqrt{1-x^2} = 2\sin \theta \cos \theta \]

### Step 2: Use the double angle identity
Using the double angle identity for sine, we have:
\[ 2\sin \theta \cos \theta = \sin(2\theta) \]
So, we can express \( \sin^{-1}(2x\sqrt{1-x^2}) \) as:
\[ \sin^{-1}(2x\sqrt{1-x^2}) = \sin^{-1}(\sin(2\theta)) \]

### Step 3: Determine the angle \( 2\theta \)
Since \( x = \sin \theta \), we have:
\[ \theta = \sin^{-1} x \]
Therefore:
\[ 2\theta = 2 \sin^{-1} x \]
Thus, we can write:
\[ \sin^{-1}(2x\sqrt{1-x^2}) = \sin^{-1}(\sin(2\theta)) = 2\theta \]

### Step 4: Adjust for the range
Now, we need to consider the range of \( \theta \). Given that \( \frac{1}{\sqrt{2}} < x \leq 1 \), we find:
- When \( x = \frac{1}{\sqrt{2}} \), \( \theta = \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} \).
- When \( x = 1 \), \( \theta = \sin^{-1}(1) = \frac{\pi}{2} \).

Thus, \( \frac{\pi}{4} < \theta \leq \frac{\pi}{2} \).

### Step 5: Final expression
Since \( 2\theta \) will range from:
\[ \frac{\pi}{2} < 2\theta \leq \pi \]
We can express \( \sin^{-1}(2x\sqrt{1-x^2}) \) as:
\[ \sin^{-1}(2x\sqrt{1-x^2}) = \pi - 2\theta = \pi - 2\sin^{-1}x \]

### Conclusion
Thus, the final result is:
\[ \sin^{-1}(2x\sqrt{1-x^2}) = \pi - 2\sin^{-1}x \]