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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?

Blue
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Joshua Farrell
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  • You could actually try that out by picking four values for $A,B,C ,D$. If it works, hmmmm. If it doesn't work...tataaaa! – imranfat Jun 02 '18 at 16:54
  • I was worried there may be too many things to try, a different constant in front, a different ratio of the angle. – Joshua Farrell Jun 02 '18 at 16:55
  • Ok, going off a tangent here, the reason why the above formula exists in a rather nice way is perhaps because $A,B,C$ could be seen as three angles of a triangle (adds up to pi), but with four angles, not sure how that relates to any nice geometric shape. I am not so sure that a similar nice formula exists.... – imranfat Jun 02 '18 at 17:00
  • I did have that realisation as well, I'd also be interested if the four angles added to $2\pi$. – Joshua Farrell Jun 02 '18 at 17:03
  • It is not going to be that easy. If we assume $sinA+sinB+sinC+sinD=xsin(A/2)cos(B/2)cos(C/2)cos(D/2)$ and pick for all angles $\pi/2$ then it follows $x=16$ to make the formula work. But then if I pick for the angles $\pi/3,\pi/2,\pi/6,\pi$, then the formula does not work. It ain't gonna be easy...Good post! I hope somebody else pitches in here... – imranfat Jun 02 '18 at 17:08

1 Answers1

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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.

Somos
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