设0

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/07 13:59:30
设0

设0
设0

设0
楼上最小值有问题,以下给出正
p=sin2x+sinx-cosx
=-(sinx-cosx)^2+sinx-cosx+1
=-N^2+N+1 令N=sinx-cosx=√2sina(x-π/4)∈【-1,√2】,-π/4

∵p=sin2x+sinx-cosx ,
=2sinxcosx-1+(sinx-cosx )+1
=-(sin²x+cos²x-2sinxcosx)+(sinx-cosx )+1
=-(sinx-cosx )²+(sinx-cosx)+1
设T= sinx-cos...

全部展开

∵p=sin2x+sinx-cosx ,
=2sinxcosx-1+(sinx-cosx )+1
=-(sin²x+cos²x-2sinxcosx)+(sinx-cosx )+1
=-(sinx-cosx )²+(sinx-cosx)+1
设T= sinx-cosx =√2sin(x-π/4) ∈【-√2,√2】
∴P =-T²+T+1 =-(T-1/2)²+5/4
∴T=1/2时,Pmax=5/4
T=-√2时,Pmin=-(-√2-1/2)²+5/4=-1-√2

收起