How prove this triangulation with indentity 
let $x,y,z\in (0,\pi)$, prove or disprove
$$\sin{(x+y)}\cdot\sin{(y+z)}\cdot\sin{(x+z)}\cdot\sin{(x+y+z)}
=[\sin{(x+y+z)}\cdot\sin{x}+\sin{y}\cdot\sin{z}]\times[\sin{(x+2y+z)}\cdot\sin{z}+\sin{(x+y)}\cdot\sin{y}]$$

if this is ture, we can use this
$$2\sin{x}\sin{y}=\cos{(x-y)}-\cos{(x+y)}$$
But I fell very ugly.

HINT:
From the RHS,
Using Werner Formula
$$2\sin(x+2y+z)\sin z=\cos(x+2y)-\cos(x+2y+2z)$$
and $$2\sin(x+y)\sin y=\cos x-\cos(x+2y)$$
Using Prosthaphaeresis formula,
$$2\sin(x+2y+z)\sin z+2\sin(x+y)\sin y=\cos x-\cos(x+2y+2z)=2\sin(y+z)\sin(x+y+z)$$
Can you apply the same method on $$[\sin(x+y+z)\sin x]+[\sin y\sin z]?$$
