Answer
$$2x\sqrt {1 - {x^2}} $$
Work Step by Step
$$\eqalign{
& \sin \left( {2{{\cos }^{ - 1}}x} \right) \cr
& {\text{Let }}\theta = {\cos ^{ - 1}}x \cr
& {\text{Using the identity sin2}}\theta {\text{ = 2sin}}\theta {\text{cos}}\theta ,{\text{ then }} \cr
& \sin \left( {2{{\cos }^{ - 1}}x} \right) = 2\sin \theta \cos \theta \cr
& {\text{From the triangle we have that }}\sin \theta = \sqrt {1 - {x^2}} ,{\text{ cos}}\theta = x \cr
& {\text{Therefore,}} \cr
& \sin \left( {2{{\cos }^{ - 1}}x} \right) = 2x\sqrt {1 - {x^2}} \cr} $$
