1,已知:sina+sinB=m,cosa+cosB=n,则cos(a-B)=?

2个回答

  • 1、(sinα)^2+(sinβ)^2+2sinαsinβ=m^2,(1)

    (cosα)^2+(cosβ)^2+2cosαcosβ=n^2,(2)

    (1)+(2),

    1+1+2cos(α-β)=m^2+n^2,

    cos(α-β)=(m^2+n^2)/2-1.

    2、tanx+cotx= sinx/cosx+cosx/sinx=1/(sinxcosx),

    sinx+cosx=1/2,

    1+2sinxcosx=1/4,

    sinxcosx=-3/8,

    tanx+cotx=-8/3.

    3、sin(α+β)=1/2,

    sinαcosβ+cosαsinβ=1/2,(1)

    sinαcosβ-cosαsinβ=1/3,(2)

    (1)+(2),

    2sinαcosβ=5/6,

    sinαcosβ=5/12,(3)

    (1)-(2),

    cosαsinβ=1/12,(4)

    (3)/(4)

    tanαcotβ=5.

    4、tanx=1/cotx=1/2,

    tan(y-x)=-tan(x-y)=2/5,

    tany=tan(y-x+x)=[tan(y-x)+tanx]/[1-tan(y-x)tanx]

    =(2/5+1/2)/(1-1/5)

    =9/8.

    5、cosα=1/3,cos(α+β)=-3/5,

    α 、β是锐角,α+β