已知-π/2＜x＜0,sinx+cosx=1/5 求sin2x的值 求3sin²x/2+2sinx/2cosx/2+cos²x/2的值

-π/2＜x＜0
cosx>sinx
sinx-cosxsinx+cosx=1/5平方
(sinx+cosx)²=1/25
sin²x+cos²x+2sinxcosx=1/25
1+2sinxcosx=1/25
2sinxcosx=-24/25
-2sinxcosx=24/25
1-2sinxcosx=24/25+1
sin²x+cos²x-2sinxcosx=49/25
(sinx-cosx)²=49/25
sinx-cosx=±7/5
所以sinx-cosx=-7/5
3sin²x/2+2sinx/2cosx/2+cos²x/2
=2sin²x/2+2sinx/2cosx/2+sin²x/2+cos²x/2
=2sin²x/2+sinx+1
=2*(1-cosx)/2+sinx+1
=1-cosx+sinx+1
=sinx-cosx+2
=-7/5+2
=3/5

(sinx+cosx)²=1+sin2x=1/25所以：sin2x=-24/253sin²x/2+2sinx/2cosx/2+cos²x/2=2sin²(x/2)+sinx+sin²(x/2)+cos²(x/2)=2sin²(x/2)+sinx+1=1-cosx+sinx+1=2+sinx-cosx(sinx-cosx)&...