Chapter 6 - Analytic Trigonometry - Section 6.2 The Inverse Trigonometric Functions (Continued) - 6.2 Assess Your Understanding - Page 480: 28

$\frac{\sqrt 7}{3}$.

1. Let $sin^{-1}\frac{\sqrt 2}{3}=t$, we have $sin(t)=\frac{\sqrt 2}{3}$ and $t$ in quadrant I. 2. Let $r=3, y=\sqrt 2$, we have $x=\sqrt {3^2-(\sqrt 2)^2}=\sqrt 7$. 3. Thus $cos(sin^{-1}\frac{\sqrt 2}{3})=cos(t)=\frac{x}{r}=\frac{\sqrt 7}{3}$.