Answer
$sin(2~tan^{-1}(\frac{12}{5})) = \frac{120}{169}$
Work Step by Step
$\theta = tan^{-1}(\frac{12}{5})$
$tan~\theta = \frac{12}{5} = \frac{opposite}{adjacent}$
Note that $\theta$ is in quadrant I. We can find the magnitude of the hypotenuse:
$hypotenuse = \sqrt{5^2+12^2} = 13$
We need to find $sin~2\theta$ where $2\theta$ is in either quadrant I or quadrant II. In quadrant I and quadrant II, $sine$ is positive. We can find the value of $sin~2\theta$:
$sin~2\theta = 2~sin\theta~cos\theta$
$sin~2\theta = (2)~(\frac{opposite}{hypotenuse})~(\frac{adjacent}{hypotenuse})$
$sin~2\theta = (2)~(\frac{12}{13})~(\frac{5}{13})$
$sin~2\theta = \frac{120}{169}$
Therefore, $sin(2~tan^{-1}(\frac{12}{5})) = \frac{120}{169}$
