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