Question
Fri June 22, 2012 By: Anneshwa

prove that sin a+ sin b+sin c-sin(a+b+c)=4sin a+b/2 sin b+c/2 sin c+a/2

Expert Reply
Sat June 23, 2012

We have to prove that: 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 degrees

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

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

Therefore Sin A+Sin B+ sin C = 2sin (A+B)/2*cos(A-B)/2+ sin (180 -(A+B)), as A+B+C = 180

= 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)

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

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