How do I evaluate $ \sin \left( {\arccos \left( {\dfrac{1}{3}} \right)} \right) $ ?

Hint: In order to determine the simplification of the above question, let the $ \arccos \left( {\dfrac{1}{3}} \right) = \theta $ ,and transposing $ \arccos $ on the right-hand side. Now replacing the $ \cos \theta $ with its value in the identity of trigonometry which state that $ {\sin ^2}\theta + {\cos ^2}\theta = 1 $ and after solving this for $ \sin \theta $ ,put back the $ \theta $ assumed earlier to obtain your required result.
Formula :
 $ \sin \theta = \dfrac{{\tan \theta }}{{\sec \theta }} $
 $ {\sec ^2}\theta = {\tan ^2}\theta + 1 $
 $ {\sin ^2}\theta + {\cos ^2}\theta = 1 $

Complete step-by-step answer:
We are given a trigonometric expression $ \sin \left( {\arccos \left( {\dfrac{1}{3}} \right)} \right) $
Let $ \arccos \left( {\dfrac{1}{3}} \right) = \theta $ ---------(1)
Transposing $ \arccos $ on the right hand side of the equation we get,
 $ \cos \theta = \dfrac{1}{3} $ ---------(2)
Since, we know the identity of trigonometry which states that the sum of square of sine and square of cosine is equal to one. i.e.
 $ {\sin ^2}\theta + {\cos ^2}\theta = 1 $
Taking $ {\cos ^2}\theta $ on the right-hand side
 $ {\sin ^2}\theta = 1 - {\cos ^2}\theta $
Now taking square root on both the sides, we get
 $ \sin \theta = \pm \sqrt {1 - {{\cos }^2}\theta } $
Now putting the value of $ \cos \theta $ from equation (2),
 $
  \sin \theta = \pm \sqrt {1 - {{\left( {\dfrac{1}{3}} \right)}^2}} \\
   = \pm \sqrt {1 - \left( {\dfrac{1}{9}} \right)} \\
   = \pm \sqrt {1 - \dfrac{1}{9}} \\
   = \pm \sqrt {\dfrac{{9 - 1}}{9}} \\
   = \pm \sqrt {\dfrac{8}{9}} \;
  $
Since $ \sqrt 9 = 3 $ and $ \sqrt 8 = \sqrt {4 \times 2} = 2\sqrt 2 $
 $ \sin \theta = \pm \dfrac{{2\sqrt 2 }}{3} $
Putting back the value of $ \theta $ ,which we have assumed in equation 1,
 $ \sin \left( {\arccos \left( {\dfrac{1}{3}} \right)} \right) = \pm \dfrac{{2\sqrt 2 }}{3} $
Therefore, the value of $ \sin \left( {\arccos \left( {\dfrac{1}{3}} \right)} \right) $ is equal to $ \pm \dfrac{{2\sqrt 2 }}{3} $ .
So, the correct answer is “ $ \pm \dfrac{{2\sqrt 2 }}{3} $ ”.

Note: 1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2.Domain and range of inverse of cosine is $ \left[ { - 1,1} \right] $ and $ \left[ {0,\pi } \right] $ respectively.
3. Domain and range of inverse of cosine is $ \left[ { - 1,1} \right] $ and $ \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right] $ respectively.