To simplify the expression $\cos(\arcsin(x))$, we can approach it geometrically using a right triangle.

Steps to Simplify:

1. Interpret the Inverse Sine Function: $y = \arcsin(x)$ This means that $\sin(y) = x$. In terms of a right triangle, this implies that the opposite side of the angle $y$ is $x$, and the hypotenuse is 1 (since $\sin(y) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{1}$).

2. Find the Adjacent Side: To find $\cos(y)$, we need the adjacent side. Using the Pythagorean theorem, the adjacent side is: $\text{adjacent} = \sqrt{1^2 - x^2} = \sqrt{1 - x^2}$

3. Find $\cos(y)$: Since $\cos(y)$ is the ratio of the adjacent side to the hypotenuse: $\cos(\arcsin(x)) = \cos(y) = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}$

Final Answer:

$\cos(\arcsin(x)) = \sqrt{1 - x^2}$

Let me know if you'd like further clarification or if you have any questions!

Here are five related questions to deepen your understanding:

1. How do we define the domain of the arcsin function?
2. What is the range of values for $\cos(\arcsin(x))$ when $x$ is in the domain of $\arcsin(x)$?
3. Can the identity $\cos(\arcsin(x)) = \sqrt{1 - x^2}$ be extended to complex numbers?
4. How would you express $\sin(\arccos(x))$?
5. What happens if you apply both $\cos$ and $\arcsin$ to a number outside the typical range of $\arcsin$?

Tip: When working with inverse trigonometric functions, it's helpful to think in terms of right triangles to simplify expressions involving compositions of these functions.