EDIT: 1.The problem statement, all variables and given/known data
The problem asks to simplify this: cos(arcsin x)

2. Relevant equations
let arc sin x = θ

3. The attempt at a solution

EDIT: I have to edit because I received a warning for not posting for #3 and #2. I originally posted "cos(arcsin x)" for #2 because that was what I know that is relevant to the problem. And I apologize for I didn't know it wasn't enough (I'm new here and sorry if I didn't understand the rules well, I thought it was okay to remove the format instructions and just put what is being asked instead. Quoted above is my previous post #3. As I said there, I really don't have any idea and therefore asks for some (can't put any attempt) When I got the hint from the mentors, I was able to work on this:

Let arcsin x = θ

if arcsin x = θ, then x = sinθ

get the cos of θ from given x/1 and should get for cosθ = √1-x^{2} (by pythagorean theorem)

Let [itex]\theta = \arcsin(x)[/itex] so then [itex]x=\sin(\theta)[/itex]. This expression is just simple trig where you first learn that the sine of an angle is OPP/HYP in a right triangle right? So do just that. Create the right-triangle with an angle of [itex]\theta[/itex] such that [itex]\sin(\theta)=x[/itex] (which is equivalent to x/1).

Good! I am glad you were able to get the answer- I hope you were able to work it out for yourself from the hint acen gr gave you.

For others who might wonder: Yes, "the answer is [itex]cos(\theta)[/itex]" but that is not the end of it. As acen gr suggested, "Create the right-triangle with an angle of θ such that sin(θ)=x (which is equivalent to x/1)." Then we have a right triangle with "opposite side" of length x and "hypotenuse" of length 1. By the Pythagorean theorem, the "near side" has length [itex]\sqrt{1- x^2}[/itex]. Since cosine is "near side over hypotenuse", [itex]cos(\theta)= cos(cos^{-1}(x))= \frac{(\sqrt{1- x^2}}{1}= \sqrt{1- x^2}[/itex]

You could also do it this way: for any angle [itex]\theta[/itex], [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] so that [itex]cos(\theta)= \pm\sqrt{1- sin^2(\theta)}[/itex]. That is, [itex]cos(sin^{-1}(x))= \pm\sqrt{1- sin^2(sin^{-1}(x))}= \pm\sqrt{1- x^2}[/itex]. Because the standard domain of [itex]sin^{-1}[/itex] is [itex]-\pi/2[/itex] to [itex]\pi/2[/itex], the sign is the same as the sign of x.

(The sign did not come up in the previous paragraph because being able to draw a triangle with side "x" assumes x is positive.)

For just a single square root symbol, if you click on the "Go Advanced", you will see a table of "Quick Symbols" and there is a "√" there- notice that you had better use ( ) with that!

Just to clarify, we're talking about [tex]\cos(\sin^{-1}\theta)[/tex] and not [tex]\cos(\sin\theta)[/tex]

Also, the range of [itex]\sin^{-1}\theta[/itex] is [itex][-\pi/2, \pi/2][/itex] and thus the range of [itex]\cos(\sin^{-1}\theta)[/itex] is [itex][0,1][/itex] so the answer is only [itex]\sqrt{1-x^2}[/itex]