已知-π/2＜x＜0,sinx+cosx=1/5.（1）求sinx-cosx的值（2）求sin2x+2sin²x/1-tanx的值

Question

已知-π/2＜x＜0,sinx+cosx=1/5.（1）求sinx-cosx的值
（2）求sin2x+2sin²x/1-tanx的值

Answer 1

(sinx+cosx)²=1+sin2x=1/25
所以：sin2x=-24/25
(sinx-cosx)²=1-sin2x=49/25
因为-π/2＜x＜0
所以sinx<0所以,sinx-cosx<0
因为(sinx-cosx)²=49/25
所以：sinx-cosx=-7/5
(sin2x+2sin²x)/(1-tanx)
=(2sinxcosx+2sin²x)/(1-sinx/cosx)
=2sinx(cosx+sinx) /(1-sinx/cosx)……分子分母同乘以cosx
=2sinx cosx (cosx+sinx) /( cosx -sinx)
=sin2x(cosx+sinx) /( cosx -sinx)
=(-24/25)*(1/5)/(7/5)
=-24/175.