Prove that sinx + siny +sinz - sin(x+y+z) = 4sin((x+y/2)sin((y+z/2)sin((z+x/2)

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

whats the question if no question pls tell me