Answer
$$\frac{{4\sqrt 6 }}{{25}}$$
Work Step by Step
$$\eqalign{
& \sin \left( {2{{\cos }^{ - 1}}\frac{1}{5}} \right) \cr
& {\text{Let }}\theta = {\cos ^{ - 1}}\frac{1}{5}{\text{, thus}} \cr
& \cos \theta = \frac{1}{5} \cr
& {\text{Recall that }}\cos \theta = \frac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}} \cr
& {\text{adjacent side}} = 1 \cr
& {\text{hypotenuse}} = 5 \cr
& {\text{opposite side }} = 2\sqrt 6 \cr
& \cr
& {\text{We have that }}\sin \left( {2{{\cos }^{ - 1}}\frac{1}{5}} \right) = \sin \left( {2\theta } \right) \cr
& {\text{Use the identity }}\sin 2\theta = 2\sin \theta \cos \theta \cr
& \sin \left( {2{{\cos }^{ - 1}}\frac{1}{5}} \right) = 2\sin \theta \cos \theta \cr
& \sin \left( {2{{\cos }^{ - 1}}\frac{1}{5}} \right) = 2\left( {\frac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}}} \right)\left( {\frac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}}} \right) \cr
& \sin \left( {2{{\cos }^{ - 1}}\frac{1}{5}} \right) = 2\left( {\frac{{2\sqrt 6 }}{5}} \right)\left( {\frac{{\text{1}}}{{\text{5}}}} \right) \cr
& \sin \left( {2{{\cos }^{ - 1}}\frac{1}{5}} \right) = \frac{{4\sqrt 6 }}{{25}} \cr} $$
