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?

Note that $\sin 2\beta = 2\sin \beta \cos \beta$ and $1 + \tan^2 \beta = \frac{1}{\cos^2 \beta}$ so $\cos^2 \beta = \frac{4}{5}$. Also $\sin^2 \beta + \cos^2 \beta = 1$ so $\sin^2 \beta = \frac{1}{5}$ (alternatively you know $\sin \beta = \frac{1}{2}\cos \beta$ from what you've shown).
Hence $\sin 2\beta = 2 \sqrt{\frac{1}{5}}\sqrt{\frac{4}{5}} = \frac{4}{5}$ where we justify the positive root by the fact that $\beta$ is acute. 

Alternatively, you can draw a right angles triangle with $\beta$ labelled and opposite and adjacent with length $1$ and $2$ so the hypotenuse is $\sqrt{5}$. Then basic trigonometry gives $\sin \beta = \frac{1}{\sqrt{5}}$ and $\cos \beta = \frac{2}{\sqrt{5}}$
