Chapter 7 - Analytic Trigonometry - 7.6 Double-angle and Half-angle Formulas - 7.6 Assess Your Understanding - Page 498: 87

$\frac{24}{25}$

Step 1. Letting $cos^{-1}(\frac{4}{5})=t$, we have $cos(t)=\frac{4}{5}$ and $t\in (0,\frac{\pi}{2})$ Step 2. We have $sin(t)=\sqrt {1-16/25}=\frac{3}{5}$ Step 3. Thus $sin(2cos^{-1}(\frac{4}{5})=sin(2t)=2sin(t)cos(t)=2(\frac{3}{5})(\frac{4}{5})=\frac{24}{25}$