We have to prove that,
sinx+siny+sinz-sin(x+y+z)=4sin((x+y)/2)sin((y+z)/2)sin((z+x)/2)
Keep in mind
LHS:
As, in a triangle , sum of all interior angles is 180°.
x+y+z=180°
x+y=180°-z
Similarly,
and, cos(x-y)+cos(x+y)=2 cos x cos y
=Sinx +Sin y +Sin z-sin (180°)
=Sin x +SIn y +Sin z-0
