Question

在△ABC中,角A,B,C的对边分别为a,b,c,且2cos∧2[（A-B）/2]cosB-sin（A-B）sinB+c

Answer 1

1

2cos^2((A-B)/2)cosB-sin(A-B)sinB+cos(A+C)

=(1+cos(A-B))cosB-sin(A-B)sinB+cos(A+C)

=cosB+cos(A-B)cosB-sin(A-B)sinB+cos(A+C)

=cosB+cosA-cosB=cosA=-3/5

2

cosA=-3/5,即：sinA=4/5

a/sinA=b/sinB,即：sinB=bsinA/a=5*(4/5)/4√2=√2/2

π/2