Chapter 1 - Functions - 1.4 Trigonometric Functions and Their Inverse - 1.4 Exercises - Page 48: 57

$$\sqrt {1 - {x^2}} $$

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