Chapter 7 - Trigonometric Identities and Equations - 7.5 Inverse Circular Functions - 7.5 Exercises - Page 710: 95

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

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