在三角形ABC中,已知sin(B+C/2)=4/5 求cos(A-B)

网友回答

sin(B+C/2)
=sin[B+(π-A-B)/2])
=sin[π/2 +(B-A)/2]
=cos{π/2 -[π/2 +(B-A)/2]}
=cos[(A-B)/2)
=4/5 cos(A-B)
=2cos²[(A-B)/2]-1
=7/25