The formula `sin^(-1){2x(1-x^(2))}=2sin^(-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//sqrt2,1//sqrt2]`

Answer

To solve the equation \( \sin^{-1}(2x(1-x^2)) = 2\sin^{-1}(x) \) and determine the interval for which this holds true, we can follow these steps:

### Step 1: Substitute \( x \) with \( \sin(\theta) \)
Let \( x = \sin(\theta) \). Then, we can rewrite the left-hand side (LHS) of the equation:
\[ \sin^{-1}(2\sin(\theta)(1-\sin^2(\theta))) \]

### Step 2: Simplify the expression
Using the identity \( 1 - \sin^2(\theta) = \cos^2(\theta) \), we can simplify the expression:
\[ \sin^{-1}(2\sin(\theta)\cos^2(\theta)) \]

### Step 3: Recognize the double angle identity
The expression \( 2\sin(\theta)\cos(\theta) \) can be recognized as \( \sin(2\theta) \). Therefore:
\[ \sin^{-1}(2\sin(\theta)\cos(\theta)) = \sin^{-1}(\sin(2\theta)) \]

### Step 4: Apply the inverse sine function
Since \( \sin^{-1}(\sin(2\theta)) = 2\theta \) when \( 2\theta \) is within the range of \( \sin^{-1} \), we have:
\[ 2\theta = 2\sin^{-1}(x) \]

### Step 5: Determine the range for \( 2\theta \)
For the equation \( 2\theta \) to be valid, it must lie within the range of \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \):
\[ -\frac{\pi}{2} < 2\theta < \frac{\pi}{2} \]

### Step 6: Divide the inequality by 2
Dividing the entire inequality by 2 gives:
\[ -\frac{\pi}{4} < \theta < \frac{\pi}{4} \]

### Step 7: Convert back to \( x \)
Since \( \theta = \sin^{-1}(x) \), we can apply the sine function to the inequality:
\[ \sin\left(-\frac{\pi}{4}\right) < x < \sin\left(\frac{\pi}{4}\right) \]

### Step 8: Calculate the sine values
Calculating the sine values gives:
\[ -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \]

### Conclusion
Thus, the interval for which the formula \( \sin^{-1}(2x(1-x^2)) = 2\sin^{-1}(x) \) holds true is:
\[ \boxed{\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)} \]