<s>[INST] Find $\sin(\frac{1}{2}\arccos(\frac{1}{9}))$ 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.
 [/INST] 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*}
</s>