f(x)=sin inverse 2x(root 1-x^2),-1/root2<=x<=1/root2 . find f(x) in sim[ple step put x=sinx.explain each step

To find the function \( f(x) = \sin^{-1}(2x \sqrt{1-x^2}) \) and simplify it by substituting \( x = \sin \theta \), we will go through the following steps:

### Step 1: Substituting \( x = \sin \theta \)
Given \( x = \sin \theta \), where \( -\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \), it implies that \( \theta \) is within the range \( -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \).

### Step 2: Express the function in terms of \( \theta \)
Substitute \( x = \sin \theta \) into the function:
\[
f(x) = \sin^{-1} \left( 2x \sqrt{1-x^2} \right)
\]
\[
f(\theta) = \sin^{-1} \left( 2 \sin \theta \sqrt{1-\sin^2 \theta} \right)
\]

### Step 3: Simplify the expression under the square root
Recall the trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \). Therefore, \( \sqrt{1 - \sin^2 \theta} = \cos \theta \).

Substitute this into the expression:
\[
f(\theta) = \sin^{-1} \left( 2 \sin \theta \cos \theta \right)
\]

### Step 4: Use the double-angle identity
The expression \( 2 \sin \theta \cos \theta \) can be recognized as the sine of a double angle:
\[
2 \sin \theta \cos \theta = \sin(2\theta)
\]

Thus, the function simplifies to:
\[
f(\theta) = \sin^{-1}(\sin(2\theta))
\]

### Step 5: Simplify the inverse sine function
Since \( \theta \) is in the range \( -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \), \( 2\theta \) will be in the range \( -\frac{\pi}{2} \leq 2\theta \leq \frac{\pi}{2} \). Within this range, the function \( \sin^{-1}(\sin(2\theta)) \) simplifies to:
\[
f(\theta) = 2\theta
\]

### Step 6: Substitute back \( \theta = \sin^{-1}(x) \)
Since \( \theta = \sin^{-1}(x) \), the final function is:
\[
f(x) = 2 \sin^{-1}(x)
\]

### Conclusion:
The simplified form of \( f(x) \) is \( f(x) = 2 \sin^{-1}(x) \) for \( -\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \).

---

Would you like to delve into any specific part of the solution? Here are some related questions to consider:

1. What is the significance of the range \( -\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}} \) in this problem?
2. How does the identity \( 2 \sin \theta \cos \theta = \sin(2\theta) \) help simplify trigonometric expressions?
3. What are the properties of the inverse sine function \( \sin^{-1}(x) \)?
4. How does the range of \( \theta \) influence the simplification of \( \sin^{-1}(\sin(2\theta)) \)?
5. What happens if \( \theta \) is outside the range \( -\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4} \)?

**Tip:** Always remember that the range of the inverse trigonometric functions is crucial in simplifying expressions correctly.

f(x)=sin inverse 2x(root 1-x^2), -1/root2<=x<=1/root2. Find f(x) in simple steps by substituting x=sin(theta). Explain each step.

This detailed solution demonstrates how to simplify f(x) = sin^{-1}(2x√(1-x²)) by substituting x = sin(θ). Follow step-by-step explanations of inverse trigonometric functions, double angle identities, and simplifying using sin(2θ). Suitable for Grades 11-12 or early college students.

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