Question

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

Answer 1

3sin^2x/2-2sinx/2cosx/2+cos^2x/2(分子)
=2sin²x/2-sinx+1
=2-cosx-sinx
=9/5
sinx+cosx=1/5
两边平方,1+2sinxcosx=1/25
sinxcosx=-12/25
tanx+cotx(分母)
=1/sinxcosx=-25/12
∴原式=-108/125

Answer 2

3sin²x/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)²=1-sin2x=49/25
∵-π/2＜x＜0
∴sinx<0<cosx
∴6sinx-cosx<0
∵(sinx-cosx)²=49/25
∴sinx-cosx=-7/5
故3sin²x/2+2sinx/2cosx/2+cos²x/2=2+sinx-cosx=2-7/5=3/5
不懂，请追问，祝愉快O(∩_∩)O~

Answer 3

sinx+cosx=1/5
平方
1+2sinxcosx=1/25
sinxcosx=-12/25

原式=(1+1-cosx-sinx)/(sinx/cosx+cosx/sinx)
=(2-sinx-cosx)/([(sin²x+cos²x)/(sinxcosx)
=(2-1/5)/[1/(-12/25)]
=-108/125