Question

高三数学，三角函数

已知：-π/2<x<0, sinx+cosx=1/5
1：求cos2x的值
2：求sin2x+2sin^2x/1-tanx的值
求解

Answer 1

1.(sinx+cosx)²=1/25
sinxcosx=-12/25
(sinx-cosx)²=1-2sinxcosx
=49/25
-π/2<x<0,
故 sinx-cosx=-7/5
而sinx+cosx=1/5
sinx=-3/5， cosx=4/5

cos2x=cos²x-sin²x=7/25

2.sin2x=2sinxcosx=-24/25
tanx=-3/4
sin2x+2sin²x/1-tanx=2sinxcosx+2sin²x/1-tanx
=-96/175

Answer 2

1.(sinx+cosx)^2=1/25
sinxcosx=-12/25
(sinx-cosx)^2=(sinx+cosx)^2-4sinxcosx
=49/25
sinx-cosx=7/5
cos2x=cosx^2-sinx^2
=(cosx+sinx)(cosx-sinx)
=7/25