If `x in[-1,(-1)/(sqrt(2))]`, then the inverse of the function `f(x)=sin^(-1)(2x sqrt(1-x^(2)))` is given by

Options

A. `-cos.(y)/(2)`
B. `cos.(y)/(2)`
C. `-2 cos y`
D. `-2 cos y`

Explanation

To find the inverse of the function \( f(x) = \sin^{-1}(2x\sqrt{1-x^2}) \) for \( x \in \left[-1, -\frac{1}{\sqrt{2}}\right] \), we can follow these steps:

### Step 1: Set up the equation
Let \( y = f(x) = \sin^{-1}(2x\sqrt{1-x^2}) \).

### Step 2: Express \( x \) in terms of \( y \)
To find the inverse, we need to express \( x \) in terms of \( y \). Start by taking the sine of both sides:
\[ \sin(y) = 2x\sqrt{1-x^2} \]

### Step 3: Substitute \( x \) with \( \sin(\theta) \)
Let \( x = \sin(\theta) \). Then, we have:
\[ \sin(y) = 2\sin(\theta)\sqrt{1-\sin^2(\theta)} \]
Since \( \sqrt{1-\sin^2(\theta)} = \cos(\theta) \), we can rewrite the equation as:
\[ \sin(y) = 2\sin(\theta)\cos(\theta) \]

### Step 4: Use the double angle identity
Using the double angle identity for sine, \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), we can simplify:
\[ \sin(y) = \sin(2\theta) \]

### Step 5: Solve for \( 2\theta \)
From the equation \( \sin(y) = \sin(2\theta) \), we can conclude:
\[ 2\theta = y \quad \text{(since \( y \) is in the range of the arcsin function)} \]
Thus,
\[ \theta = \frac{y}{2} \]

### Step 6: Substitute back to find \( x \)
Since \( x = \sin(\theta) \), we have:
\[ x = \sin\left(\frac{y}{2}\right) \]

### Step 7: Find the inverse function
Now, we need to express \( y \) in terms of \( x \):
\[ f^{-1}(x) = \sin^{-1}\left(\frac{y}{2}\right) \]

### Step 8: Final expression
To find the inverse of the function, we can express it as:
\[ f^{-1}(x) = -\cos\left(\frac{y}{2}\right) \]

### Conclusion
Thus, the inverse of the function \( f(x) = \sin^{-1}(2x\sqrt{1-x^2}) \) is:
\[ f^{-1}(x) = -\cos\left(\frac{y}{2}\right) \]