# Prove: sin A + sin B + sin C - 2sinA/2.sinB/2 = 2 sinC/2 so C= 120o   and cos A/2 cos B/2 cos C/2 - sin A/2 sin B/2 sin C/2 = 1/2 so ABC is a right-angled triangle ? `sin A + sin B = 2 sin (A+B)/2*cos(A-B)/2`

You need to use double angle identity such that:

`sinC = sin 2(C/2)=2sin(C/2)*cos(C/2)`

Since, A,B,C are angles of triangle, then the sum of the measures of angles is of `180^o` .

`A+B+C = 180^o`

`A+B=180^o - C`

`sin (A+B)/2=sin(180^o/2 - C/2)=cos C/2`

Hence `sin A...

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`sin A + sin B = 2 sin (A+B)/2*cos(A-B)/2`

You need to use double angle identity such that:

`sinC = sin 2(C/2)=2sin(C/2)*cos(C/2)`

Since, A,B,C are angles of triangle, then the sum of the measures of angles is of `180^o` .

`A+B+C = 180^o`

`A+B=180^o - C`

`sin (A+B)/2=sin(180^o/2 - C/2)=cos C/2`

Hence `sin A + sin B + sin C = 2 cos (C/2)*cos(A-B)/2 + 2sin(C/2)*cos(C/2) - 2sin(A/2)*sin(B/2) = 2 sin(C/2)` You need to factor out `2 cos (C/2)`  such that:

`2 cos (C/2)*(cos(A-B)/2 + sin(C/2))- 2sin(A/2)*sin(B/2) = 2 sin(C/2)`

Substituting `sin C/2`  by `cos(pi/2 - C/2)` yields:

`2 cos (C/2)*(cos(A-B)/2 + cos(pi/2 - C/2)) - 2sin(A/2)*sin(B/2) = 2 sin(C/2)`

`(cos(A-B)/2 + cos(pi/2 - C/2)) = 2cos(A/2)*cos(B/2)`

`4cos (C/2)*cos(A/2)*cos(B/2)- 2sin(A/2)*sin(B/2) = 2 sin(C/2)`

Notice that, considering the given conditions, the left side is not equal to the right side, hence `sin A + sin B + sin C - 2sinA/2*sinB/2 != 2 sinC/2.`

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