Answer
$\frac{4\sqrt2}{9}$
Work Step by Step
Let $\theta = \cos^{-1}{(\frac{1}{3})}$
This means that $\cos{\theta}=\frac{1}{3}$
Since $\cos{\theta} = \frac{x}{r}$ and $\cos{\theta}=\frac{1}{3}$, then we can assume that $x=1$ and $r=3$.
RECALL:
In a unit circle, $r^2=x^2+y^2$.
Use the formula above to solve for $y$ to obtain:
$r^2=x^2+y^2
\\3^2 = 1^2 + y^2
\\9=1+y^2
\\9-1=y^2
\\8=y^2
\\\sqrt8=\sqrt{y^2}
\\\sqrt{4(2)}=y
\\2\sqrt2=y$
Since $\sin{\theta} = \frac{y}{r}$, then
$\sin{\theta} = \frac{2\sqrt2}{3}$
With $\theta=\cos^{-1}{(\frac{1}{3})}$, then:
$\sin{\left(2\cos^{-1}{\frac{1}{3}}\right)}=\sin{2\theta}$
RECALL:
$\sin{2\theta} = 2\sin{\theta}\cos{\theta}$
Thus,
$\sin{\left(2\cos^{-1}{\frac{1}{3}}\right)}
\\=\sin{2\theta}
\\= 2\sin{\theta}\cos{\theta}
\\=2(\frac{2\sqrt2}{3})(\frac{1}{3})
\\=\frac{4\sqrt2}{9}$
