# Prove that (sin a)+(sin b)+(sin c)-sin(a+b+c)=4sin(a+b/2)sin(b+c/2)sin(a+c/2).

Sami Ullah
46 Points
4 years ago
It looks quite difficult but its very easy if you try to solve it.It is pretty straight-forward.

L.H.S:-

$=[\sin (a)+\sin (b)]+[sin(c)+sin(a+b+c)]$

$=[2sin(\frac{a+b}{2})cos(\frac{a-b}{2})]+[2cos(\frac{c+a+b+c}{2})sin(\frac{c-a-b-c}{2})]$

$=[2sin(\frac{a+b}{2})cos(\frac{a-b}{2})]+[2cos(\frac{a+b+2c}{2})sin(\frac{-(a+b)}{2})]$

$=[2sin(\frac{a+b}{2})cos(\frac{a-b}{2})]-[2cos(\frac{a+b+2c}{2})sin(\frac{(a+b)}{2})]$

$=2sin(\frac{a+b}{2})[cos(\frac{a-b}{2})]-cos(\frac{a+b+2c}{2})]$

$=2sin(\frac{a+b}{2})[-2sin(\frac{\frac{a-b+2c+a+b}{2}}{2})sin(\frac{\frac{a-b-2c-a-b}{2}}{2})]$

$=-4sin(\frac{a+b}{2})[sin(\frac{2a+2c}{4})]sin(\frac{-2b-2c}{4})]$

$=-4sin(\frac{a+b}{2})[sin(\frac{2(a+c)}{4})]sin(\frac{-2(b+c)}{4})]$

$=-(-4)sin(\frac{a+b}{2})[sin(\frac{a+c}{2})sin(\frac{b+c}{2})]$

$=4sin(\frac{a+b}{2})sin(\frac{a+c}{2})sin(\frac{b+c}{2})$

I hope that awnsers your question.Actually I saw it today although you posted the question 11 days ago.Sorry for the delay.