# Prove Sin A + Sin B + Sin C - Sin(A+B+C) = 4 Sin ((A+B)/2) Sin ((B+C)/2) Sin ((C+A)/2)The actual question does not specify the traditional enenunciation of the sum of the angles A+B+C=180...

Prove Sin A + Sin B + Sin C - Sin(A+B+C) =

4 Sin ((A+B)/2) Sin ((B+C)/2) Sin ((C+A)/2)

The actual question does not specify the traditional enenunciation of the sum of the angles A+B+C=180 degrees. A recent post by giorgiana1976 answering a similar problem is very helpful however the formal proof of the non-conditional (angle) form of this identify is being posed. PS. Enotes - a much appreciated resource

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### 1 Answer

Prove Sin A + Sin B + Sin C - Sin(A+B+C) = 4 Sin ((A+B)/2) Sin ((B+C)/2) Sin ((C+A)/2).

Given A+B+C = 180 degree.

We use sin x+sin y = 2sin(x+y)/2 *cos(x-y)/2.

So Sin A+sin B = 2 sin (A+B)/2 * sin(A-B)/2.

Therefore Sin A+Sin B+ sin C = 2sin (A+B)/2*co sA-B)/2+ sin (180 -(A+B)),as A+B+C = 180 degrees by data.

= 2sin (A+B)/2* cos (A-B)/2 + sin (A+B) , as sin (180-x) = sin x.

= 2sin (A+B)/2cos (A-B)/2 + 2sin (A+B)/2*cos A+B)/2, as sin 2x = 2sin x*cos x.

= 2sin (A+B)/2 {cos (A-B)/2 +cos (A+B)/2}

= 2sin (A+B)/2 {2 cos A/2*cos B/2}.

= 4( cos C/2 )(cos A/2)(cos B/2), as sin (A+B)/2 = sin (90-C/2) = cos C/2.

Therefore sin A+sin B+ sin C = sin A+sin B+ sin C - sin (A+B+C) = 4cos A/2*cos B/2*cos C/2 = 4sin (A+B)/2*sin (B+C)/2* sin(C+A)/2.

as sin (A+B+C) = sin 180 deg = 0. And sin (A+B)/2 = cos C/2, sin(B+C)/2 = cos A/2 and sin (C+A)/2 = cos B/2.