This question puts me in mind of the previous alike question beginning with tangent. So, tried to solve it using that method: if $\arccos(\frac{1}{9})=\alpha,$ then $\sin(\frac{\alpha}{2}).$ The formula: $$\sin(\frac{x}{2})=\pm\sqrt{\frac{1-\cos(x)}{2}}$$ doesn't work because there are signs and there wasn't said in which quarter $\frac{1}{9}$ is in. So is representing $\sin(\frac{\alpha}{2})$ via a tangent of a half angle.
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$\begingroup$Hint: You know that $0 \leq \arccos(x) \leq \pi$ for all $-1 \leq x \leq 1$ because of the way $\arccos$ is defined. So $0 \leq \frac{1}{2}\arccos(\frac19) \leq \frac12\pi$. This gives you the sign of $\sin(\frac{1}{2}\arccos(\frac19))$.
Now you can use the formula $\sin(\frac{x}{2})=\pm\sqrt{\frac{1-\cos(x)}{2}}$, as in your post.
$\endgroup$ $\begingroup$$$\sin\left(\frac{1}{2}\arccos\frac{1}{9}\right)=\sqrt{\frac{1-\frac{1}{9}}{2}}=\frac{2}{3}$$ The right formula is $$|\sin\frac{x}{2}|=\sqrt{\frac{1-\cos{x}}{2}}$$ and since it's obvious that $\sin\left(\frac{1}{2}\arccos\frac{1}{9}\right)>0$, we get, which got.
$\endgroup$ $\begingroup$Let $\alpha = \cos^{-1}(\frac{1}{9})$ so $\cos(\alpha)= \frac{1}{9}$, we have \begin{eqnarray*} \sin(\frac{1}{2} \cos^{-1}(\frac{1}{9}))= \sin(\frac{1}{2} \alpha) = \sqrt{\frac{1-\cos(\alpha)}{2}} = \sqrt{\frac{1-\frac{1}{9}}{2}}=\color{red}{\frac{2}{3}}. \end{eqnarray*}
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