Q:

Prove that the area of this triangle is given by $f(x)=\sin(2x)$

I've encountered this exercise in multiple textbooks now, and I've always been unable to solve it. 
Let angle $DBC = x$. $C$ moves through the arc $BD$ such that $x\in (0;\frac{\pi }{2})$. Prove that the area of triangle $[BCD]$ can be given by sin(2$x$).

A:

$BC=2\cos{x}$ and $CD=2\sin{x}.$
Now, $$S_{\Delta BCD}=\frac{2\sin{x}\cdot2\cos{x}}{2}.$$
Can you end it now?

A:

Hint:
It is crucial to note that the angle $BCD$ is right. Then the area is
$$\frac{BC\cdot CD}2$$ and the sides are obtained by projecting the diameter.

