Answer
$$2x\sqrt {1 - {x^2}} $$
Work Step by Step
$$\eqalign{
& \sin \left( {2{{\cos }^{ - 1}}x} \right) \cr
& {\text{Use the Hint }}\sin 2\theta = 2\sin \theta \cos \theta \cr
& \sin \left( {2{{\cos }^{ - 1}}x} \right) = 2\sin \left( {{{\cos }^{ - 1}}x} \right)\cos \left( {{{\cos }^{ - 1}}x} \right) \cr
& \sin \left( {2{{\cos }^{ - 1}}x} \right) = 2\sin \left( {{{\cos }^{ - 1}}x} \right)x \cr
& {\text{From the triangle shown below}} \cr
& \cos \theta = \frac{x}{1} \cr
& \cos \theta = x \cr
& \theta = {\cos ^{ - 1}}x,{\text{ then}} \cr
& \sin \left( {{{\cos }^{ - 1}}x} \right) = \sin \theta \cr
& {\text{From the triangle we obtain }}\cos \theta \cr
& \sin \left( {{{\cos }^{ - 1}}x} \right) = \frac{{\sqrt {1 - {x^2}} }}{1} \cr
& \sin \left( {{{\cos }^{ - 1}}x} \right) = \sqrt {1 - {x^2}} \cr
& {\text{Therefore,}} \cr
& \sin \left( {2{{\cos }^{ - 1}}x} \right) = 2\sin \left( {{{\cos }^{ - 1}}x} \right)x \cr
& \sin \left( {2{{\cos }^{ - 1}}x} \right) = 2x\sqrt {1 - {x^2}} \cr} $$
