Chapter 7 - Analytic Trigonometry - 7.6 Double-angle and Half-angle Formulas - 7.6 Assess Your Understanding - Page 498: 90

$\frac{9}{10}$

Step 1. Letting $sin^{-1}(\frac{3}{5})=t$, we have $sin(t)=\frac{3}{5}$ and $t\in (0,\frac{\pi}{2})$ Step 2. We have $cos(t)=\sqrt {1-9/25}=\frac{4}{5}$ Step 3. Thus $cos^2(\frac{1}{2}sin^{-1}(\frac{3}{5}))=cos^2(\frac{1}{2}t)=\frac{1+cos(t)}{2}=\frac{1+\frac{4}{5}}{2}=\frac{9}{10}$