Chapter 7 - Analytic Trigonometry - 7.6 Double-angle and Half-angle Formulas - 7.6 Assess Your Understanding - Page 498: 82

$\frac{\sqrt 3}{2}$

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