The set of values of 'x' for which the formula `2sin^(-1)x=sin^(-1)2x sqrt(1-x^(2))` is true is

Options

A. `(-1,0)`
B. `[0,1]`
C. `[-sqrt3/2,sqrt3/2]`
D. `[-1/sqrt2,1/sqrt2]`

Answer

To solve the equation \( 2 \sin^{-1} x = \sin^{-1} (2x \sqrt{1 - x^2}) \), we will follow these steps:

### Step 1: Set Up the Equation
Let \( \theta = 2 \sin^{-1} x \). Then, we can express \( \sin^{-1} x \) as \( \sin^{-1} x = \frac{\theta}{2} \).

### Step 2: Express \( x \) in Terms of \( \theta \)
From the definition of the inverse sine function, we have:
\[ x = \sin\left(\frac{\theta}{2}\right) \]

### Step 3: Substitute \( x \) into the Right-Hand Side
Now, substitute \( x = \sin\left(\frac{\theta}{2}\right) \) into the right-hand side of the equation:
\[ \sin^{-1}\left(2x \sqrt{1 - x^2}\right) = \sin^{-1}\left(2 \sin\left(\frac{\theta}{2}\right) \sqrt{1 - \sin^2\left(\frac{\theta}{2}\right)}\right) \]
Using the identity \( \sqrt{1 - \sin^2\left(\frac{\theta}{2}\right)} = \cos\left(\frac{\theta}{2}\right) \), we get:
\[ = \sin^{-1}\left(2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)\right) \]

### Step 4: Apply the Double Angle Formula
Using the double angle formula for sine, we have:
\[ 2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right) = \sin(\theta) \]
Thus, we can rewrite the right-hand side:
\[ \sin^{-1}(\sin(\theta)) \]

### Step 5: Solve for \( \theta \)
Since \( \sin^{-1}(\sin(\theta)) = \theta \) for \( \theta \) in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\), we have:
\[ \theta = 2 \sin^{-1} x \]

### Step 6: Determine the Range of \( \theta \)
For \( \theta \) to be in the range \([- \frac{\pi}{2}, \frac{\pi}{2}]\), we need:
\[ -\frac{\pi}{2} \leq 2 \sin^{-1} x \leq \frac{\pi}{2} \]
Dividing the entire inequality by 2 gives:
\[ -\frac{\pi}{4} \leq \sin^{-1} x \leq \frac{\pi}{4} \]

### Step 7: Apply the Sine Function
Taking the sine of all parts of the inequality (since sine is an increasing function), we have:
\[ \sin\left(-\frac{\pi}{4}\right) \leq x \leq \sin\left(\frac{\pi}{4}\right) \]
This simplifies to:
\[ -\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \]

### Conclusion
The set of values of \( x \) for which the formula \( 2 \sin^{-1} x = \sin^{-1} (2x \sqrt{1 - x^2}) \) is true is:
\[ x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right] \]