The formula `sin^(-1){2xsqrt(1-x^(2))}=2 sin^(-1) x` is true for all values of x lying in the interval

Options

A. [-1, 1]
B. [0, 1]
C. [-1, 0]
D. `[-1// sqrt(2), 1// sqrt(2)]`

Explanation

To solve the problem, we need to verify the formula \( \sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}(x) \) and find the interval of \( x \) for which this holds true.

### Step-by-Step Solution:

1. **Substitution**: Let \( x = \sin(\theta) \). Then, we can rewrite the left-hand side:
\[ \sin^{-1}(2x\sqrt{1-x^2}) = \sin^{-1}(2\sin(\theta)\sqrt{1-\sin^2(\theta)}) \]
Since \( \sqrt{1-\sin^2(\theta)} = \cos(\theta) \), we have:
\[ \sin^{-1}(2\sin(\theta)\cos(\theta)) \]

2. **Using the Double Angle Identity**: Recall that \( 2\sin(\theta)\cos(\theta) = \sin(2\theta) \). Therefore, we can simplify:
\[ \sin^{-1}(\sin(2\theta)) = 2\theta \]
This is valid as long as \( 2\theta \) is within the range of \( \sin^{-1} \), which is \( -\frac{\pi}{2} \leq 2\theta \leq \frac{\pi}{2} \).

3. **Finding the Range of \( \theta \)**: From the condition \( -\frac{\pi}{2} \leq 2\theta \leq \frac{\pi}{2} \), we can divide by 2:
\[ -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \]

4. **Relating \( \theta \) to \( x \)**: Since \( \theta = \sin^{-1}(x) \), we can find the range of \( x \):
- The sine function is increasing in the interval \( -\frac{\pi}{4} \) to \( \frac{\pi}{4} \).
- Therefore, we have:
\[ \sin\left(-\frac{\pi}{4}\right) \leq x \leq \sin\left(\frac{\pi}{4}\right) \]
- This gives us:
\[ -\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \]

5. **Conclusion**: The formula \( \sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}(x) \) is true for all values of \( x \) in the interval:
\[ x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] \]