已知sina/2-cosa/2=√10/5,a属于(π/2,π),tan(π-b)=1/2,求tan(a-2b)的值

2个回答

  • sina/2-cosa/2=√10/5

    sina/2*cosπ/4-cosa/2*sinπ/4=√5/5

    sin(a/2-π/4)=√5/5

    因为a属于(π/2,π),所以a/2-π/4属于(0,π/4)

    所以cos(a/2-π/4)>0

    cos(a/2-π/4)=√(1-sin²(a/2-π/4))=2√5/5

    tan(a/2-π/4)=sin(a/2-π/4)/cos(a/2-π/4)=1/2

    tan(a-π/2)=2*1/2/(1-(1/2)²)=4/3

    tan(a-π/2)=-tan(π/2-a)=-cota=4/3

    cota=-4/3

    tana=-3/4

    tan(π-b)=1/2

    -tanb=1/2

    tanb=-1/2

    tab2b=2*(-1/2)/(1-(-1/2)²)=-4/3

    tan(a-2b)=(-3/4-(-4/3))(1+(-3/4)(-4/3))

    =7/24