Question

sinA + sinB + sinC = 4cosA/2.cosB/2.cosC/2

Joshi sir comment

sinA + sinB + sinC = 2sin[(A+B)/2]cos[(A-B)/2] + 2sinC/2cosC/2

                                  = 2sin[(π-C)/2]cos[(A-B)/2] + 2sinC/2cosC/2

                                  = 2cosC/2cos[(A-B)/2] + 2sinC/2cosC/2

                                  = 2cosC/2* [cos[(A-B)/2]+[cos(A+B)/2]                     here sinC/2 = cos(A+B)/2

now use cosC + cosD = 2 cos[(C+D)/2]cos[(C-D)/2]

You should remember that the given formula is based on a triangle. 

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