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 $-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$.

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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.