After an explanation of the restricted domains and ranges of inverse trigonometric functions, I.M. Gelfand's Trigonometry gives the following exercise:
Show that $$\sin(\arccos b) = \pm \sqrt{1-b^2}$$ What determines whether we should choose the positive sign or the negative sign?
It is my understanding that:
$b^2 + \sin^2 x = 1$
So $\sin x = \pm \sqrt{1-b^2}$.
But $0\leq \arccos b \leq \pi$
So $-1\leq b \leq 1$
So $\sin 0 \leq \sin x \leq \sin \pi $
And $0 \leq \sin x \leq 1$.
I must be missing something because it seems to me that we would not need the ambiguous sign and the original expression should be $$\sin(\arccos b) = \sqrt{1-b^2}$$ because $\sin x$ where $x = \arccos b$ will always be positive.
What am I missing?
Your answer is correct. We can verify it in a slightly different way by noting that $$ \sin^2 x + \cos^2x = 1 $$ and so $$ \left|\sin x\right| = \sqrt{1-\cos^2x} \, . $$ Hence, \begin{align} \left|\sin(\arccos(b))\right| &= \sqrt{1-\cos^2(\arccos(b))} \\ &= \sqrt{1-b^2} \, . \end{align} Because $\arccos(b) \in [0,\pi]$, we know that $\sin(\arccos b) \geq 0$. Hence, we are left with $$ \sin(\arccos(b))=\sqrt{1-b^2} \, . $$ The crux of solving this problem is understanding the range of the $\arccos$ function, so that we can draw this conclusion. Strictly speaking, $\arccos$ is not the inverse of $\cos$—rather, it is the inverse of $\operatorname{Cos}$, which is the function $f:[0,\pi] \mapsto \mathbb{R}$ defined by $f(x)=\cos(x)$. On the other hand, $\cos$ is the inverse of $\arccos$, since $\arccos$ is one-to-one. It is because of this that $\cos(\arccos(x))$ is always equal to $x$, but $\arccos(\cos(x))$ might not be equal to $x$.