1

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?

Dr. Sonnhard Graubner
  • 95,283
Soling
  • 478

1 Answers1

3

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.

J.G.
  • 115,835
  • this is not a geometrical proof – logo Apr 28 '19 at 10:47
  • 1
    @logo Why not? I think we'd all agree $\Delta=\frac{1}{2}bc\sin A$ is a geometric finding, and I'd argue the same is true of the cosine rule, and Heron's formula follows from the cosine rule. The only trigonometric step is the cosine's double-angle formula. But the compound angle formula has an easy proof from geometry. – J.G. Apr 28 '19 at 10:51
  • it must contain a geometrical Plot – logo Apr 28 '19 at 10:56
  • @logo Well, I'm sure such diagrams would be indispensable in the proofs of the theorems I've used. Now, do you have any criteria in mind the OP made clear? – J.G. Apr 28 '19 at 11:00
  • 1
    fair enough..... – logo May 01 '19 at 14:32