To simplify the expression $\sqrt{1 - \sin^2(\sin^{-1}(x))}$, let's go step by step:

Step 1: Understanding $\sin^{-1}(x)$

The term $\sin^{-1}(x)$ represents the inverse sine (or arcsine) of $x$, which gives us an angle $\theta$ such that: $\sin(\theta) = x$

Thus, $\sin^{-1}(x)$ is an angle $\theta$, and we can replace $\sin^{-1}(x)$ with $\theta$: $\theta = \sin^{-1}(x)$ This implies $\sin(\theta) = x$.

Step 2: Simplifying the expression

Now substitute this back into the expression: $\sqrt{1 - \sin^2(\theta)}$

We already know that $\sin(\theta) = x$, so: $\sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}$

Final Answer:

The simplified form of the expression is: $\sqrt{1 - x^2}$

Would you like a further explanation, or do you have any other questions?

Related Questions:

1. What are the properties of inverse trigonometric functions?
2. How do you derive the identity $\sin^2(\theta) + \cos^2(\theta) = 1$?
3. How do you find the derivative of $\sin^{-1}(x)$?
4. What is the domain and range of the function $\sin^{-1}(x)$?
5. Can you explain how to simplify expressions involving inverse trigonometric functions?

Tip: Remember that $\sin^{-1}(x)$ gives you an angle whose sine is $x$, and this identity helps simplify many trigonometric expressions.