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求y=(sinx)/(2-cosx)的值域
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解:
方法1:
y=(sinx)/(2-cosx)即
sinx=2y-ycosx
sinx+ycosx=2y
左边必然可以化为√(y^2+1)sin(x+A)的形式
则原式化为:
√(y^2+1)sin(x+A)=2y
sin(x+A)=2y/√(y^2+1)
根据正玄函数的有界性可知:-1<=sin(x+A)<=1
则-1<=2y/√(y^2+1)<=1
-1<=2y/√(y^2+1)即:
0<√(y^2+1)<=-2y(必有y<0)
即y^2+1>=4y^2
解得:y∈[-√3/3,√3/3]
2y/√(y^2+1)<=1
即√(y^2+1)>=2y
y^2+1>=4y^2
解得:y∈[-√3/3,√3/3]
综上:值域为y∈[-√3/3,√3/3]
方法2:
解:∵y=sinx/(2-cosx)
=2sin(x/2)*cos(x/2)/[2sin²(x/2)+2cos²(x/2)-cos²(x/2)+sin²(x/2)]
=2sin(x/2)*cos(x/2)/[3sin²(x/2)+cos²(x/2)]
=2tan(x/2)/[3tan²(x/2)+1]
∴1/y=[3tan²(x/2)+1]/2tan(x/2)=3tan(x/2)/2+1/2tan(x/2)
①当tan(x/2)<0时,有
1/y=-[-3tan(x/2)/2+1/-2tan(x/2)]≤-2√[-3tan(x/2)]/2×1/-2tan(x/2)]=-√3
当且仅当-3tan(x/2)/2=1/-2tan(x/2)即tan(x/2)=-√3/3时,取得“=”
则 y≥-√3/3
②当tan(x/2)>0时
1/y=3tan(x/2)2+1/2tan(x/2)≥2√[3tan(x/2)/2×1/2tan(x/2)]=√3
当且仅当3tan(x/2)/2=1/2tan(x/2)即tan(x/2)=√3/3时,取得“=”
则 y≤√3/3
因此,y的值域为[-√3/3,√3/3]
方法1:
y=(sinx)/(2-cosx)即
sinx=2y-ycosx
sinx+ycosx=2y
左边必然可以化为√(y^2+1)sin(x+A)的形式
则原式化为:
√(y^2+1)sin(x+A)=2y
sin(x+A)=2y/√(y^2+1)
根据正玄函数的有界性可知:-1<=sin(x+A)<=1
则-1<=2y/√(y^2+1)<=1
-1<=2y/√(y^2+1)即:
0<√(y^2+1)<=-2y(必有y<0)
即y^2+1>=4y^2
解得:y∈[-√3/3,√3/3]
2y/√(y^2+1)<=1
即√(y^2+1)>=2y
y^2+1>=4y^2
解得:y∈[-√3/3,√3/3]
综上:值域为y∈[-√3/3,√3/3]
方法2:
解:∵y=sinx/(2-cosx)
=2sin(x/2)*cos(x/2)/[2sin²(x/2)+2cos²(x/2)-cos²(x/2)+sin²(x/2)]
=2sin(x/2)*cos(x/2)/[3sin²(x/2)+cos²(x/2)]
=2tan(x/2)/[3tan²(x/2)+1]
∴1/y=[3tan²(x/2)+1]/2tan(x/2)=3tan(x/2)/2+1/2tan(x/2)
①当tan(x/2)<0时,有
1/y=-[-3tan(x/2)/2+1/-2tan(x/2)]≤-2√[-3tan(x/2)]/2×1/-2tan(x/2)]=-√3
当且仅当-3tan(x/2)/2=1/-2tan(x/2)即tan(x/2)=-√3/3时,取得“=”
则 y≥-√3/3
②当tan(x/2)>0时
1/y=3tan(x/2)2+1/2tan(x/2)≥2√[3tan(x/2)/2×1/2tan(x/2)]=√3
当且仅当3tan(x/2)/2=1/2tan(x/2)即tan(x/2)=√3/3时,取得“=”
则 y≤√3/3
因此,y的值域为[-√3/3,√3/3]
展开全部
解:∵y=sinx/(2-cosx)
=2sin(x/2)*cos(x/2)/[2sin²(x/2)+2cos²(x/2)-cos²(x/2)+sin²(x/2)]
=2sin(x/2)*cos(x/2)/[3sin²(x/2)+cos²(x/2)]
=2tan(x/2)/[3tan²(x/2)+1]
∴1/y=[3tan²(x/2)+1]/2tan(x/2)=3tan(x/2)/2+1/2tan(x/2)
①当tan(x/2)<0时,有
1/y=-[-3tan(x/2)/2+1/-2tan(x/2)]≤-2√[-3tan(x/2)]/2×1/-2tan(x/2)]=-√3
当且仅当-3tan(x/2)/2=1/-2tan(x/2)即tan(x/2)=-√3/3时,取得“=”
则 y≥-√3/3
②当tan(x/2)>0时
1/y=3tan(x/2)2+1/2tan(x/2)≥2√[3tan(x/2)/2×1/2tan(x/2)]=√3
当且仅当3tan(x/2)/2=1/2tan(x/2)即tan(x/2)=√3/3时,取得“=”
则 y≤√3/3
因此,y的值域为[-√3/3,√3/3]
=2sin(x/2)*cos(x/2)/[2sin²(x/2)+2cos²(x/2)-cos²(x/2)+sin²(x/2)]
=2sin(x/2)*cos(x/2)/[3sin²(x/2)+cos²(x/2)]
=2tan(x/2)/[3tan²(x/2)+1]
∴1/y=[3tan²(x/2)+1]/2tan(x/2)=3tan(x/2)/2+1/2tan(x/2)
①当tan(x/2)<0时,有
1/y=-[-3tan(x/2)/2+1/-2tan(x/2)]≤-2√[-3tan(x/2)]/2×1/-2tan(x/2)]=-√3
当且仅当-3tan(x/2)/2=1/-2tan(x/2)即tan(x/2)=-√3/3时,取得“=”
则 y≥-√3/3
②当tan(x/2)>0时
1/y=3tan(x/2)2+1/2tan(x/2)≥2√[3tan(x/2)/2×1/2tan(x/2)]=√3
当且仅当3tan(x/2)/2=1/2tan(x/2)即tan(x/2)=√3/3时,取得“=”
则 y≤√3/3
因此,y的值域为[-√3/3,√3/3]
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