How do you find the exact value of sin(tan^-1(sqrt3/3))?

The exact value of $\sin \left({\tan}^{-} 1 \left(\frac{\sqrt{3}}{3}\right)\right) = \frac{1}{2}$
$\sin \left({\tan}^{-} 1 \left(\frac{\sqrt{3}}{3}\right)\right)$. Let ${\tan}^{-} 1 \left(\frac{\sqrt{3}}{3}\right) = \theta \therefore \tan \theta = \frac{\sqrt{3}}{3}$
We know $\tan \theta$= perpendicular/base=$\frac{\sqrt{3}}{3} \therefore$Hypotenuse = sqrt(3^2+(sqrt3^2))= sqrt12=2sqrt2 :. sin theta=perp./hypotenuse $= \frac{\sqrt{3}}{2 \sqrt{3}} = \frac{1}{2} \therefore \theta = {\sin}^{-} 1 \left(\frac{1}{2}\right)$
Hence $\sin \left({\tan}^{-} 1 \left(\frac{\sqrt{3}}{3}\right)\right) = \sin \left({\sin}^{-} 1 \left(\frac{1}{2}\right)\right) = \frac{1}{2}$ [Ans]