sinA+cosB=(根号下3)/2,cosA+sinB=根号下2,求tanAcotB

2个回答

  • sinA+cosB=√3/2,(平方)

    (sinA+cosB)^2=3/4

    (sinA)^2+2sinAcosB+(cosB)^2=3/4.1

    cosA+sinB=√2,(平方)

    (cosA+sinB)^2=2

    (cosA)^2+2cosAsinB+(sinB)^2=2.2

    1式+2式得

    2+2sinAcosB+2cosAsinB=11/4

    2sinAcosB+2cosAsinB=3/4

    sinAcosB+cosAsinB=3/8.3

    sin(A+B)=3/8

    1式-2式得

    (sinA)^2+2sinAcosB+(cosB)^2-(cosA)^2-2cosAsinB-(sinB)^2=-5/8

    cos2B-cos2A+2sin(A-B)=-5/8

    cos2A-cos2B+2sin(B-A)=5/8

    sin(A+B)sin(B-A)+sin(B-A)=5/8

    [sin(A+B)+1]sin(B-A)=5/8

    [3/8+1]sin(B-A)=5/8

    (11/8)sin(B-A)=5/8

    11sin(B-A)=5

    sin(B-A)=5/11

    sinBcosA-sinAcosB=5/11.4

    3式+4式得

    2cosAsinB=3/8+5/11=33/88+40/88=73/88

    cosAsinB=73/176.5

    3式-4式得

    2sinAcosB=3/8-5/11=33/88-40/88=-7/88

    sinAcosB=-7/176.6

    6式除以5式得

    tanAcotB=(-7/176)/(73/176)=-7/73