Chapter 7 - Analytic Trigonometry - 7.6 Double-angle and Half-angle Formulas - 7.6 Assess Your Understanding - Page 498: 88

$-\frac{7}{25}$

Step 1. Letting $tan^{-1}(-\frac{4}{3})=t$, we have $tan(t)=-\frac{4}{3}$ and $t\in (\frac{\pi}{2},\pi)$ Step 2. We have $sin(t)=\frac{4}{\sqrt {3^2+4^2}}=\frac{4}{5}$ Step 3. Thus $cos(2tan^{-1}(-\frac{4}{3}))=cos(2t)=1-2sin^2(t)=1-2(\frac{4}{5})^2=-\frac{7}{25}$