已知-π/2<x<0,sinx+cosx=1/5
(1)求sin2x,cos2x的值
(2)求sinx-cosx的值
(3)求（3sin^2(x/2)-2(sinx/2）(cosx/2)+cos^2(x/2))除以tanx+cotx的值

1)sinx+cosx=1/5两边平方得
1+2sinxcosx=1/25
--->2sinxcosx=-24/25
--->sin2x=-24/25
-pi/2<x<0--->sinx<0,cosx>0
解sinx+cosx=1/5,sinxcosx=-12/25的方程组得sinx=-3/5,cosx=4/5
--->cos2x=(cosx)^2-(sinx)^2=(4/5)^2-(-3/5)=7/25
2)sinx-cosx=(-3/5)-(4/5)=-7/5
3)cosx=4/5,-pi/2<x<0--->-pi/4<x/2<0
sin(x/2)=-√[(1-cosx)/2]=-√(1+4/5)/2]=-3/√10
cos(x/2)=1/√10.
3[sin(x/2)]^2-2sin(x/2)cos(x/2)+[cos(x/2)]^2
=3(-3/√10)^2-sinx+(1/√10)^2
=27/10-(-3/5)+1/10
=17/5.
tanx+cotx=sinx/cosx+cosx/sinx=-3/5-5/3=-34/15
因此，商=(17/5)/(-34/15)=-3/2.