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.
A: Observe that the $y$ coordinate of $C$ is $C_y = \sin(\angle DAC).$ Observe also that $\angle DBC$ is inscribed in the circle shown in the figure and $\angle DAC$ is the corresponding central angle, and therefore 
$\angle DAC = 2 \angle DBC = 2x.$
So $C_y = \sin(2x).$
You now have a triangle with base $BD = 2$ and height $C_y = \sin(2x).$
Its area is $$\frac12 bh = \frac12 \cdot 2 \cdot \sin(2x) = \sin(2x).$$
A: If we first recognise that $BD$ is the diameter of a circle and $C$ is always on the circle, then by Thales's theorem we know that $BCD$ is right-angled. For right-angled triangles we know how to express its sides in terms of a given angle. We use the trigonometric relations in terms of hypothenuse, adjacent and opposite. The hypothenuse is always $2$ since the radius of the circle is $2$, if we let $x$ be our angle $DBC$ we can write:
$$ BC= 2\cos(x)$$
$$ CD= 2\sin(x)$$
Now we calculate the area by taking base times the height, so: 
$$ \frac{1}{2} \cdot BC \cdot CD = 2 \sin(x) \cos(x) = \sin(2x)$$
Here we also used a trigonometric identity in the final step, a so-called double angle formula
