If A+B+C=180° , then prove that sinA+sinB+sinC = 4 cos(A/2) cos(B/2) cos(C/2)

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Solution

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

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

But a+b=pi-c

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

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

So, sin a+sin b+sin c=2cos c/22cos(a/2)*cos(b/2)

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

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