Answer
$sin~(2~sec^{-1}~\frac{u}{2}) = \frac{4\sqrt{u^2-4}}{u^2}$
Work Step by Step
Let $~~\theta = sec^{-1}~\frac{u}{2}$
Then $~~sec~\theta = \frac{u}{2}$
$cos~\theta = \frac{1}{sec~\theta} = \frac{2}{u}$
We can find an expression for $sin~\theta$:
$sin~\theta = \sqrt{1-cos^2~\theta}$
$sin~\theta = \sqrt{1-(\frac{2}{u})^2}$
$sin~\theta = \sqrt{\frac{u^2-4}{u^2}}$
$sin~\theta = \frac{\sqrt{u^2-4}}{u}$
We can find $sin~(2\theta)$:
$sin~(2\theta) = 2~sin~\theta~cos~\theta$
$sin~(2\theta) = (2)(\frac{\sqrt{u^2-4}}{u})(\frac{2}{u})$
$sin~(2\theta) = \frac{4\sqrt{u^2-4}}{u^2}$
Therefore, $~~sin~(2~sec^{-1}~\frac{u}{2}) = \frac{4\sqrt{u^2-4}}{u^2}$
