几道三角函数问题!1、1+tan75/1-tan75的值2、已知sinx+cosx=3/5,求sin2x的值3、sinx

3个回答

  • 1+tan75/1-tan75

    =(tan75+1)/(1-tan75)

    =(tan75+tan45)/(1-tan75tan45)

    =tan(75+45)

    =tan120

    =tan(180-60)

    =-tan60

    ==√3/3

    sinx+cosx=3/5

    (sinx+cosx)*(sinx+cosx)=3/5*3/5

    1+2sinx*cosx=9/25

    sin2x=-16/25

    sinx+cosy=1/3 ,cosx+siny=1/2

    (sinx+cosy)^2 = 1/9,(cosx+siny)^2 = 1/4

    sin^2x +cos^2y +2sinxcosy +cos^2x+sin^2y +2cosxsiny = 1/9+1/4 = 15/36

    sin^2x +cos^2x+sin^2y +cos^2y +2sinxcosy +2cosxsiny = 15/36

    1 + 1 +2sinxcosy +2cosxsiny = 15/36

    1 +sinxcosy +cosxsiny = 15/72

    sinxcosy +cosxsiny = -57/72

    sin(x+y)= -57/72