Trigonometry-circumcircle and sides of triangle. How to prove that 
$$4R\sin A\sin B\sin C=a \cos A+b \cos B+c\cos C$$ where R is the radius of the circumcircle and $a$,$b$ and $c$ the respective sides of the triangle.
I wrote $R=a/2\sin A$ and then got $\sin A$ cancelled then changed $\sin B \sin C$ to the sums of cosine but that didn't work. 

We have $4R=\frac{abc}{\Delta}$ and $\sin A=\frac{2\Delta}{bc}$, hence the LHS equals
$$ \frac{abc}{\Delta}\cdot\frac{8\Delta^3}{a^2 b^2 c^2}= \frac{8\Delta^2}{abc}.$$
Now exploit $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$ and Heron's formula.
