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!

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*}
