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