作业帮已知函数f(x)=-根号3sin^2x+sinxcosx (1)求f((23π)/6) (2)设x属于（0,π）,求f(x)的最大值和最小值
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已知函数f(x)=-根号3sin^2x+sinxcosx (1)求f((23π)/6) (2)设x属于（0,π）,求f(x)的最大值和最小值
已知函数f(x)=-根号3sin^2x+sinxcosx (1)求f((23π)/6) (2)设x属于（0,π）,求f(x)的最大值和最小值
已知函数f(x)=-根号3sin^2x+sinxcosx (1)求f((23π)/6) (2)设x属于（0,π）,求f(x)的最大值和最小值
f(x)=-根号3sin^2x+sinxcosx
=-√3/2(1-cos2x)+1/2sin2x
=√3/2cos2x+1/2sin2x-√3/2
=sin(2x+π/3)-√3/2
(1)f((23π)/6)=-√3/2
(2)因为x属于（0,π）所以
π/3
f(x)=-根号3sin^2x+sinxcosx
=√3/2(1-2sin^2x)+1/2 (2sinxcosx)-√3/2
=√3/2cos2x+1/2sin2x-√3/2
=sin(2x+π/3)-√3/2
1、f[(23π)/6]=sin(2X23π/6+π/3)-√3/2
=s...
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f(x)=-根号3sin^2x+sinxcosx
=√3/2(1-2sin^2x)+1/2 (2sinxcosx)-√3/2
=√3/2cos2x+1/2sin2x-√3/2
=sin(2x+π/3)-√3/2
1、f[(23π)/6]=sin(2X23π/6+π/3)-√3/2
=sin8π-√3/2
=-√3/2
2、当0
当sin(2x+π/3)=1时有最大值为：1-√3/2
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23π/6=4π-π/6
sin(23π/6)=-sinπ/6=-1/2 cos(23π/6)=cosπ/6=√3/2
∴f(23π/6)=√3sin²(π/6)-sinπ/6cosπ/6=√3/4-√3/4=0
f(x)=√3sin²x+sinxcosx=√3/2(2sin²x-1)+1/2+1/2 (2sinxcos)
...
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23π/6=4π-π/6
sin(23π/6)=-sinπ/6=-1/2 cos(23π/6)=cosπ/6=√3/2
∴f(23π/6)=√3sin²(π/6)-sinπ/6cosπ/6=√3/4-√3/4=0
f(x)=√3sin²x+sinxcosx=√3/2(2sin²x-1)+1/2+1/2 (2sinxcos)
=√3/2(cos2x)+1/2sin2x+1/2
=sinπ/3cos2x+cosπ/3sin2x+1/2
=sin(2x+π/3)+1/2
∵0＜x＜π
∴0＜2x＜2π
π/3＜2x+π/3＜2π+π/3
∴-1≤sin(2x+π/3)≤1
f(x)的最大值是3/2，最小值是-1/2
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