Question

In `DeltaABC`, prove that:
`sinA+sinB-sinC=4sinA/2sinB/2cosC/2`

Answer 1

LHS `=sinA+sinB-sinC`
`=2sin(A+B)/(2).cos(A-B)/2-2sinC/2cosC/2`
`=2sin(180^(@)-C)/(2)cos(A-B)/2(2)` `(therefore A+B+C=180^(@))`
`=2sin(90^(@)-C/2)cos(A-B)/(2)-2sin{90^(@)-(A+B)/(2)}cosC/2`
`=2cosC/2cos(A-B)/(2)-2cos(A+B)/(2)cosC/2`
`=2cosC/2[cos(A-B)/(2)-cos(A+B)/2]`
`=2cosC/2.2sinA/2sinB/2`
`=4sinA/2sinB/2cosC/2`=RHS Hence Proved.