Question

已知sinA+cosB=1/3,sinB-cosA=1/2,则sin（A-B)=?(带上过程,

Answer 1

sin(A-B)

=sinAcosB-cosAsinB

sinA+cosB=1/3

(sinA+cosB)^2=1/9

也就是sin^2A+cos^2B+2sinAcosB=1/9①

sinB-cosA=1/2

(sinB-cosA)^2=1/4

也就是sin^2B+cos^2A-2sinBcosA=1/4②

①+②得

sin^2A+cos^2B+2sinAcosB+sin^2B+cos^A-2sinBcosA=13/36

sin^2A+cos^2A+sin^2B+cos^2B+2sinAcosB-2sinBcosA=13/36

1+1+2(sinAcosB-cosAsinB)=13/36

2+2(sinAcosB-cosAsinB)=13/36

2(sinAcosB-cosAsinB)=-59/36

sinAcosB-cosAsinB=-59/72