At one vertex of a pentagon inscribed in a circle of unit diameter (unit diameter, not unit radius) let the angles between adjacent diagonals be $\alpha,\beta,\gamma$, at the next, $\beta,\gamma,\delta$, at the next $\gamma,\delta,\varepsilon$, then $\delta,\varepsilon,\alpha$, and finally $\varepsilon,\alpha,\beta$. Note that $ \alpha+\beta+\gamma+\delta+\varepsilon=\pi. \tag{constraint} $ Later note: (Lest anything be misunderstood, notice that what I wrote above is true of all pentagons inscribed in a circle. Angles with vertices on the circle have the same measure if they're subtended by the same arc. Consequently if the three angles between adjacent diagonals at one vertex are $\alpha,\beta,\gamma$, in that order, then two of the angles between adjacent diagonals at one of the neighboring vertices must be $\alpha$ and $\beta$, and two of those at the other neighboring vertex must be $\beta$ and $\gamma$. And regardless of the shape of the pentagon, the sum of the five angles must be a half-circle. That's a general proposition about polygons inscribed in a circle, which, when applied to triangles, says the sum of the three angles is a half-circle.) End of later note
It's not hard to show that the area of the pentagon is $ \frac{\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)+\sin(2\delta)+\sin(2\varepsilon)}{8}.\tag{1} $ It's somewhat more work than that to show that if the "constraint" above holds, then $(1)$ is equal to $ \frac 1 2 \left(\overbrace{\sin\alpha\sin\beta\sin\gamma}^\text{sines}\ \overbrace{\cos\delta\cos\varepsilon}^\text{cosines} +\ \cdots\text{nine more terms }\cdots\ - \overbrace{2\sin\alpha\sin\beta\sin\gamma\sin\delta\sin\varepsilon}^\text{all sines} \right). $ (It should be obvious what the nine more terms are: choose three factors in each term to be sines and then the other two are cosines.)
(As far as I know, this is my own. I've mentioned it here at least once before.)
Can the eleven terms be interpreted as areas?
LATER EDIT: Even for quadrilaterals it seems mysterious. If the angles between adjacent diagonals are $\alpha+\beta+\gamma+\delta=\pi$, and two of them occur at each vertex, and each occurs at two of the four vertices, the the area is $ \frac{\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)+\sin(2\delta)}{8} = \frac 1 2\Big(\overbrace{\sin\alpha\sin\beta\sin\gamma}^\text{sines}\ \overbrace{\cos\delta}^\text{cosine}+\cdots\text{ three more terms }\cdots\Big) $
You might think that the four terms are areas of the four triangles into which the polygon is divided by the diagonals. But guess what?? They're not! Similarly, the pentagram divides the pentagon into $11$ triangles, and there are $11$ terms on the right side, but they don't correspond to the areas.