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!
 A: Hope this image helps (pythagorean identity?):

And as to the $\pm$, it simply because $\cos(\arcsin(x))$ is only equal to one value, and this value can be only $+$ or $-$, but not both (so we just choose, and $+$ is just more logical).
A: Let $\arcsin x = \theta$.  Then, by definition of the arcsine function, $\sin\theta = x$, where $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, and 
$\cos(\arcsin x) = \cos\theta$.  Using the Pythagorean Identity $\sin^2\theta + \cos^2\theta = 1$, we obtain 
\begin{align*}
\sin^2\theta + \cos^2\theta & = 1\\
\cos^2\theta & = 1 - \sin^2\theta\\
\cos^2\theta & = 1 - x^2\\
|\cos\theta| & = \sqrt{1 - x^2}
\end{align*}
Since $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, $\cos\theta \geq 0$.  Thus, $|\cos\theta| = \cos\theta$, whence
\begin{align*}
\cos\theta & = \sqrt{1 - x^2}\\
\cos(\arcsin x) & = \sqrt{1 - x^2}
\end{align*}
A: 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.
A: $\cos^2+\sin^2=1\;$ implies $\;\cos(\arcsin(x))=\sqrt{1-\sin^2(\arcsin(x))}=\sqrt{1-x^2}$. 
