sina=x,cosb=y,cos(a+b)=-3/5,求X的取值范围

1个回答

  • sin(a+b)=√(1-9/25)=4/5

    sinacosb+cosasinb=4/5

    cosacosb-sinasinb=-3/5

    (tana+tanb)/(1-tanatanb)=-4/3

    3tana+3tanb=-4+4tanatanb

    tana(4tanb-1)=4+3tanb

    tana=(4+3tanb)/(4tanb-1)

    =(4cosb+3sinb)/4sinb-cosb)

    (1+tan^2a)(4sinb-cosb)^2=16sin^2b-8sinbcosb+cos^2b+

    16cos^2b+24sinbcosb+9sin^2b

    =17+8sin2b+8sin^2b

    =17+8sin2b+4-4cos2b

    =21+4(2sin2b-cos2b)

    sina=x=tana/√(1+tan^2a)

    (4cosb+3sinb)(4sinb-cosb)=16sinbcosb+12sin^2b-4cos^2b-3sinbcosb

    =13sinbcosb-4cos2b+4-4cos2b

    =(13sin2b)/2-8cos2b+4

    x=(6.5sin2b-8cos2b+4)/(21+4(2sin2b-cos2b))

    =[4+√106.25sin(2b-A)]/[21+4√5sin(2b-B)]

    A=arctan(8/6.5)

    B=arctan(1/2)

    很搞怪的题