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

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

wang le

1. (1+tan75°) / (1-tan75°) = tan(45°+75°) = tan120° = -√3
2. sinx+cosx=3/5 => ( sinx+cosx)^2 = 1 + sin(2x) = 9/25
sin(2x) = -16/25
3. sinx + cosy = 1/3 => (sinx)^2 + 2sinxcosy +(cosy)^...