Proving that $\cos(\arcsin(x))=\sqrt{1-x^2}$ I am asked to prove that $\cos(\arcsin(x)) = \sqrt{1-x^2}$
I have used the trig identity to show that $\cos^2(x) = 1 - x^2$
Therefore why isn't the answer denoted with the plus-or-minus sign?
as in $\pm \sqrt{1-x^2}$.
Thank you!

Note:  this answer does not explain the formula $\sqrt{1-x^2}$, but does address why you don't need $\pm$.
Let $x$ be any number between $-1$ and $1$.  Then $\arcsin(x)$ is an angle $\theta$ with $\sin(\theta)=x$.  But which angle $\theta$?  There are lots of different angles that all have the same sine.  By definition, $\arcsin(x)$ is an angle between $-90$° and $90$° (or, if you prefer, between $-\pi/2$ and $\pi/2$ radians).
All right, now we want $\cos(\arcsin(x))$, or $\cos(\theta)$.  That means we are taking the cosine of an angle between $-90$° and $90$°.  The cosine of such an angle is never negative (we only get a negative cosine from an obtuse angle).  So $\cos(\arcsin(x))$ is always non-negative.
