Chapter 7 - Analytic Trigonometry - 7.6 Double-angle and Half-angle Formulas - 7.6 Assess Your Understanding - Page 498: 83

$\frac{7}{25}$

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 $cos(2sin^{-1}\frac{3}{5})=cos(2t)=1-2sin^2(t)=1-2(\frac{3}{5})^2=\frac{7}{25}$