# 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)

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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.**

a+b+c=π

a+b=π-c

∴ a/2 + b/2 = π/2 - c/2

sin(a/2+b/2) = sin(π/2-c/2) = cos c/2

cos(a/2+b/2) = cos(π/2-c/2) = sin c/2

L:H:S = sin a + sin b + sin c

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

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

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

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

= 2cos c/2 * 2cosa/2.cosb/2

= 4cos a/2.cos b/2.cos c/2

= R:H:S

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

= 2sin(90-c/2)*cos(a-b)/2+2sinc/2*cosc/2

=2cosc/2*cos(a-b)/2 + 2sinc/2*cosc/2

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

=2cosc/2 {cos(a-b)/2 + cos (a+b)/2)}

=2cosc/2{ 2 cosa/2 * cos b/2}

=4cos a/2 * cosb/2 * cos c/2