Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 710: 97

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

$$\eqalign{ & \cos \left( {\arcsin u} \right) \cr & {\text{From the triangle we have that}} \cr & \sin \theta = u \cr & \theta = \arcsin u \cr & \cos \theta = \cos \left( {\arcsin u} \right) = \frac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}} \cr & \cos \theta = \frac{{\sqrt {1 - {u^2}} }}{1} \cr & \cos \theta = \sqrt {1 - {u^2}} \cr} $$