Question

在三角形ABC中,已知sin(B+C/2)=4/5 求cos(A-B)

Answer 1

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