Geometrical Proof of $\sin A+\sin B+\sin C= 4\cos(A/2)\cos(B/2)\cos(C/2)$

Question

I have been seeking for geometrical proof of

$\sin A+\sin B+\sin C= 4\cos(A/2)\cos(B/2)\cos(C/2)$

Where $A, B, C$ are angles of a triangle.

Do you have any ideas about it?

Answer 1

First note $\sin A=\frac{2\Delta}{bc}$, where $\Delta$ is the triangle's area, so $\sin A+\sin B+\sin C=\frac{4s\Delta}{abc}$ with $s$ the semiperimeter. By the cosine rule, $$\cos^2\frac{A}{2}=\frac{1+\cos A}{2}=\frac{(b+c)^2-a^2}{4bc}=\frac{s(s-a)}{bc},$$so$$\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}=\frac{s\sqrt{s(s-a)(s-b)(s-c)}}{abc}.$$Heron's formula then finishes the proof.