Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 53

$-\frac{8\sqrt 3}{49}$

Step 1. Since $\theta$ is in Quadrant II, we have $\frac{\pi}{2}\lt\theta\lt\pi$ which means that $cos\theta$ will be negative. Step 2. Given $sin\theta=\frac{1}{7}$, use the Pythagorean Identity, we have $cos\theta=-\sqrt {1-(\frac{1}{7})^2}=-\frac{4\sqrt 3}{7}$ Step 3. Use the Double-Angle Formula, we have $sin2\theta=2sin\theta cos\theta=2\times\frac{1}{7}\times(-\frac{4\sqrt 3}{7})=-\frac{8\sqrt 3}{49}$