Chapter 5 - Section 5.3 - Double-Angle Formulas - 5.3 Problem Set - Page 296: 15

$\dfrac{120}{169}$

$\sec{\theta} = \sqrt{1+\tan^2{\theta}}= \sqrt{1+(\dfrac{5}{12})^2}$ $\sec{\theta} = \dfrac{13}{12}$ $\cos{\theta} = \dfrac{1}{\sec{\theta}} = \dfrac{12}{13}$ $\sin{\theta} = \sqrt{1-(\dfrac{12}{13})^2} = \dfrac{5}{13}$ $\sin{2\theta} = 2 \sin{\theta} \cos{\theta} = 2 \times \dfrac{5}{13} \times \dfrac{12}{13} = \dfrac{120}{169}$