How is sinA+sinB+sinC=4 cosA/2 cosB/2 cosC/2

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Solution

If A+B+C = pai, only the theorem can be satisfied.
Proof:
A+B+C = pai

Dividing by 2 on both sides

C/2 = pi/2 - (A/2+B/2)

sinA +sinB +sinC

= 2Sin(A+B)/2 Cos(A-B)/2 + 2SinC/2 Cos C/2

= 2 Sin(pi/2 -C/2)Cos(A-B)/2 + 2SinC/2 Cos C/2

=2 Cos C/2 Cos(A-B)/2 + 2SinC/2 Cos C/2

=2Cos C/2[Cos(A-B)/2 + SinC/2 ]

=2Cos C/2[Cos(A-B)/2 +Sin[pi/2 - (A+B)/2]]

=2Cos C/2[Cos(A-B)/2 +Cos (A+B)/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

Hope it makes you clear.

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