Is there a formula that generalizes $\sin A+\sin B+\sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$ (where $A+B+C=\pi$) to four angles? If $A+B+C=\pi$ then we have $$\sin A+\sin B+\sin C = 4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)$$
If $A+B+C+D=\pi$ is there a similar formula?

There is this similar result: if $\,A + B + C + D = 2\pi\,$ (not $\pi$), then
$$\sin(A) + \sin(B) + \sin(C) + \sin(D) = 4 \sin\Big(\frac{A+B}{2}\Big) \sin\Big(\frac{B+C}{2}\Big) \sin\Big(\frac{C+A}{2}\Big),$$ or, alternatively, both equal
$$-4 \sin\Big(\frac{A+B}{2}\Big) \sin\Big(\frac{A+C}{2}\Big) \sin\Big(\frac{A+D}{2}\Big).$$
The method I used to discover this was factoring and exponential substitution. That is, let
 $\,A := \log(a)/i,\, B := \log(b)/i,\, C := \log(c)/i,\, D := \log(d)/i.\,$
where $\,d := 1/(a b c).\,$ The left side factors as $\,i(ab-1)(ac-1)(bc-1)/(2abc).\,$ The $\,(ab-1)\,$ factor is $\,\sin((A+B)/2)\,$ up to some simple factors, and similarly for the other two factors.
The original identity has a geometric application. If $\,A,B,C\,$ are the three angles of a triangle, then $\,A+B+C=\pi\,$ and the identity is equating the sum of the sines of the three angles to four times the product of three cosines of half the angles. Similarly, if $\,A,B,C,D\,$ are the four angles of a quadrilateral, then $\,A + B + C + D = 2\pi\,$ and the new identity is equating the sum of the sines of the four angles to four times the product of three sines of half of sum of two angles.
