1.设x为第二象限的角,且cosx/2+sinx/2=-根号5/2,求;(1)sinx/2-cosx/2;(2)sin2

2个回答

  • 1、(1)∵x为第二象限的角,且cosx/2+sinx/2=-根号5/2

    ∴kπ+π/4<x/2<kπ+π/2,且k为奇数.

    (cosx/2+sinx/2)^2=(-根号5/2)^2

    1+sinx=5/4

    sinx=1/4

    (sinx/2-cosx/2)^2=1-sinx=1-1/4=3/4

    ∵sinx/2-cosx/2<0

    ∴sinx/2-cosx/2=-根号3/2

    (2)sinx=1/4,x为第二象限的角

    ∴cosx=-根号15/4

    2、(1)tana/2=2

    ∴tana=(2tana/2)/[1-(tana/2)^2]

    =4/(-3)=-4/3

    (2)(6sina+cosa)/(3sina-2cosa)

    =(6tana+1)/(3tana-2)

    =-7/(-6)=7/6

    3、tan2a=-2根号2,2a属于(圆周率/2,圆周率),

    tan2a=2tana/[1-(tana)^2]

    2tana/[1-(tana)^2]=-2根号2

    ∴tana=-根号2/2 (舍去) tana=根号2

    ∴sina=根号6/3

    cosa=根号3/3

    ∴(cos^2(a/2)-sina-1)/(根号2*sin(a+圆周率/4))

    =[(1/2)cosa-sina-1/2]/(sina+cosa)

    =(-9-3根号2-2根号2-2根号3)/2

    把(cos^2(a/2)-sina-1)/(根号2*sin(a+圆周率/4))改为

    (2cos^2(a/2)-sina-1)/(根号2*sin(a+圆周率/4))

    ∴(2cos^2(a/2)-sina-1)/(根号2*sin(a+圆周率/4))

    =(cosa-sina)/(cosa+sina)

    =(1-tana)/(1+tana)

    =(1-根号2)/(1+根号2)

    =2根号2-3