To simplify $\cos(\arcsin(x))$, we proceed as follows:

1. Let $\theta = \arcsin(x)$. This means $\sin(\theta) = x$, and $\theta$ is in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
2. From the Pythagorean identity, $\sin^2(\theta) + \cos^2(\theta) = 1$, we find $\cos^2(\theta) = 1 - \sin^2(\theta)$.
3. Substitute $\sin(\theta) = x$:
   $\cos^2(\theta) = 1 - x^2$
4. Taking the square root, $\cos(\theta) = \sqrt{1 - x^2}$, since $\cos(\theta) \geq 0$ in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

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

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Do you want more details or have any questions?

Here are some related questions for further exploration:

1. What is $\sin(\arccos(x))$?
2. How does the Pythagorean identity work with trigonometric functions?
3. Can $x$ exceed the range $[-1, 1]$ for $\arcsin(x)$?
4. How do we simplify $\tan(\arcsin(x))$?
5. What happens when $x = 1$ or $x = -1$ in this expression?

Tip: Always remember that $\arcsin(x)$ and $\arccos(x)$ involve principal branches, which restrict their output ranges to ensure they are functions.