Question

In `DeltaABC`, prove that:
`sinA/2+sinB/2+sinC/2=1+4sin(pi-A)/(4)sin(pi-B)/(4)sin(pi-C)/(4)`

Answer 1

LHS`=sinA/2+sinB/2+sinC/2`
`=2sin(A+B)/4cos(A-B)/4+1-2sin^(2)(pi-C)/(4)`
`=1+2sin(pi-C)(4)cos(A-B)/(4)-2sin^(2)(pi-C)/4`
`=1+2sin(pi-C)/(4)[cos(A-B)/(4)-sin(pi-C)/(4)]`
`=1+2sin(pi-C)/(4)[cos(A-B)/(4)-sin(A+B)/(4)]`
`=1+2sin(pi-C)[cos(A-B)/4-cos{pi/2-(A+B)/(4)}]`
`=1+2sin(pi-C)/(4)[cos(A-B)/(4)-cos(2pi-(A+B))/(4)]`
`=1+2sin(pi-C)/(4).2sin(A-B+2pi-(A+B)/(8).sin(2pi-(A+B)-(A-B))/(8)`
`=1+2sin(pi-C)/(4).sin(pi-8)/(4)sin(pi-C)/(4)`
=RHS Hence Proved.