Given trigonometric values as coordinates, Find the value of $t$ for which the Quadrilateral has maximum area 
In a Cartesian Coordinate System we're given four points:

$$A(2\cos(t),2\sin(t))$$
$$ B(-\cos(2-t),-sin(2-t))$$
$$C(-2\cos(t),-2\sin(t))$$
$$ D(\cos(2-t),\sin(2-t))$$

For what value of $t\in(0;1)$ will the Area of quadrilateral $ABCD$ be maximum?

Well, this is an unusual problem. The way I tried to solve it is first by noticing that points $A$-$C$ and $B$-$D$ are $180^\circ$ apart. Also, We can map them on the circle with center at $(0,0)$. I think the points $A$ and $C$ will be on the circle with radius $2$? So $AC=4$ ? I'm really not sure.
Evaluating  $\sin(2-t) = \sin(2)\cos(t) - \cos(2)\sin(t)$. I don't know what to do with this since I've never sin expressed like this before. I mean, I'm guessing $\sin(2)$ means radians, but I don't know how to evaluate that. But perhaps there's no need to...
So I'd really like to hear your thoughts on this, how do I go about solving this problem?
 A: Points $A$ and $C$ are opposite points on a circle of radius 2, moving anti-clockwise as $t$ increases, while points $B$ and $D$ are opposite points on a circle of radius 1, moving clock-wise.
Maximum area is achieved when the lines $AC$ and $BD$ are perpendicular (so the constituent triangles have maximum height). In that case the area is $4\times\frac{1}{2}(2\times1)=4$.
A: It's easy to see that the origin is the midpoint of the diagonals. In other words, the diagonals bisect each other. It's also easy to see that the opposite sides are parallel, hence we conclude that the quadrilateral must be a parallelogram. Now I used the formula for calculating the area of a parallelogram $$\frac{1}{2}D_1D_2\sin\theta$$ where $D_1$ and $D_2$ are the lengths of the two diagonals and $\sin\theta$ is the angle between the diagonals (which one doesn't matter since $\sin\theta=\sin\pi-\theta$). This gives $$\frac{1}{2}(2)(4)(\sin\theta)$$
Now, we use the formula for calculating angle between any two lines with a given slope $\tan\theta=\frac{m_1-m_2}{1+m_1m_2}$ to give us $$\tan\theta=\frac{\tan(t)-\tan(2-t)}{1+\tan(t)\tan(2-t)}$$
We can draw a triangle with hypotenuse $$\sqrt{[\tan(t)-\tan(2-t)]^2+[1+\tan(t)\tan(2-t)]^2}$$ to find $\sin\theta$. Expand the squares inside the root in the expression for the hypotenuse. After cancelling terms and taking terms common, we get $\sqrt{[\tan^2(2-t)+1][\tan^2(t)+1]}$ which after applying a trigonometric identity simplifies to $$\sec(2-t)\sec(t)$$ Final expression for $\sin\theta$ is $$\frac{\tan(t)-\tan(2-t)}{\sec(2-t)\sec(t)}$$ Converting everything to $\sin\theta$ and $\cos\theta$, we can easily simplify it to $\sin(t)\cos(2-t)-\sin(2-t)\cos(t)$ Applying the $\sin(\alpha+\beta)$ formula in reverse on this we get

$$\sin\theta=sin(2-2t)$$

We know maximum value of $\sin\theta$ is at $\theta=\frac{\pi}{2}<2$. For given range of $t$, $π/2$ is included in range of $2-2t$. So as our final answer we get $$t=1-\frac{\pi}{4}$$
A: You can try with $$A=\frac{1}{2}\begin{vmatrix} x_1-x_3 & y_1-y_3 \\ x_2-x_4 & y_2-y_4 \end{vmatrix}.$$
