求函数y=2sinxcosx+sinx-cosx的值域
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解：
y=2sinxcosx+sinx-cosx
=sin(2x) +√2[(√2/2)sinx-(√2/2)cosx]
=cos(π/2 -2x)+√2[sinxcos(π/4)-cosxsin(π/4)]
=cos(2x- π/2)+√2sin(x-π/4)
=cos[2(x-π/4)]+√2sin(x-π/4)
=1-2sin²(x-π/4)+√2sin(x-π/4)
=-2sin²(x-π/4)+√2sin(x-π/4)+1
=-2[sin(x-π/4) -√2/4]² +5/4
sin(x-π/4)=√2/4时，y取得最大值ymax=5/4
sin(x-π/4)=-1时，y取得最小值ymin=-1-√2
综上，得：函数的值域为[-1-√2，5/4]
y=2sinxcosx+sinx-cosx
=sin(2x) +√2[(√2/2)sinx-(√2/2)cosx]
=cos(π/2 -2x)+√2[sinxcos(π/4)-cosxsin(π/4)]
=cos(2x- π/2)+√2sin(x-π/4)
=cos[2(x-π/4)]+√2sin(x-π/4)
=1-2sin²(x-π/4)+√2sin(x-π/4)
=-2sin²(x-π/4)+√2sin(x-π/4)+1
=-2[sin(x-π/4) -√2/4]² +5/4
sin(x-π/4)=√2/4时，y取得最大值ymax=5/4
sin(x-π/4)=-1时，y取得最小值ymin=-1-√2
综上，得：函数的值域为[-1-√2，5/4]
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