Find the exact value of $A(\beta)=8\pi-16\sin(2\beta)$ with $\tan(\beta)= \frac{1}{2}$ 
The picture below represents a semi-circumference of diameter [AB] and
  center C. Point D belongs to the semi-circumference and it's one of
  the vertices of the triangle $ABC$. Consider that BÂD = $\beta (\beta
 \in ]0,\frac{\pi}{2}[)$ and AC = 4.

The area of the pink part of the picture is given by $$A(\beta) =
 8\pi-16\sin(2\beta)$$
Find the exact value of the area of the pink part with $\tan(\beta)=
 \frac{1}{2}$

I tried:
$$\tan \beta = \frac{\sin(\beta)}{\cos(\beta)} = \frac{1}{2}\\ \Leftrightarrow \sin(\beta) = \frac{\cos(\beta)}{2} \\ \Leftrightarrow  \sin(\beta) = \frac{\sqrt{1-\sin^2\beta}}{2}\\ \Leftrightarrow ???$$
What do I do next?

Given: $\tan \beta = \frac 12$, so $2\sin\beta = \cos \beta$
$\sin 2\beta = 2\sin\beta\cos\beta = \cos^2\beta = \frac 1{\sec^2 \beta} = \frac 1{1 + \tan^2 \beta} = \frac 1{ 1 + \frac 14} = \frac 45$
