If a+b+c=pi, prove that the identity is true. sin a + sin b + sin c=4cos(a/2)cos(b/2)cos(c/2)

We'll choose to work on the left side of the identity and
to transform the addition between the 2 terms into a
product.

sin a + sin b = 2 sin
(a+b)/2*cos(a-b)/2

We'll write sin c= sin (c/2 +
c/2)=2sin(c/2)*cos(c/2)

We'll work on the constraint given
by the
enunciation;

a+b+c=pi

a+b=pi-c

sin
(a+b)/2=sin(pi/2 - c/2)=cos c/2

We'll substitute the
calculated values, to the left side of the identity.

sin
a+sin b+sin c=2sin(a+b)/2*cos(a-b)/2+2sin(c/2)*cos(c/2)

sin
a+sin b+sin c=2cos c/2*cos(a-b)/2+2sin(c/2)*cos(c/2)

After
factorization, we'll have:

sin a+sin b+sin c=2cos
c/2*[cos(a-b)/2+sin(c/2)]

But sin c/2=cos(pi/2 -
c/2)

sin a+sin b+sin c=2cos c/2*[cos(a-b)/2+cos(pi/2 -
c/2)]

[cos(a-b)/2+cos(pi/2 -
c/2)]=

=2cos(a-b+pi-c)/4*cos(a-b-pi+c)/4

But
a+b=pi-c

2cos(a-b+pi-c)/4*cos(a-b-pi+c)/4=

=2cos(a-b+a+b)/4*cos(a-b-a-b)/4=2cos(a/2)*cos(b/2)

So,
sin a+sin b+sin c=2cos
c/2*2*cos(a/2)*cos(b/2)

sin a+sin b+sin
c=4*cos(a/2)*cos(b/2)*cos (c/2) q.e.d.