已知函数f(x)=1/2cos2x+根号3sinxcosx-2cos^2x
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已知函数f(x)=1/2cos2x+根号3sinxcosx-2cos^2x
(1)求f(x)的最大值 (2)若f(a)=-1/5,求cos(2a+π/3) (3)若π/3
(1)求f(x)的最大值 (2)若f(a)=-1/5,求cos(2a+π/3) (3)若π/3
f(x)=1/2cos2x+根号3/2sin2x-1-cos2x
=√3/2 ×sin2x-1/2 ×cos2x -1
=sin(2x-π/6) -1
当sin(2x-π/6)=1时
f(x)有最大值=0
f(a)=-1/5
则,sin(2a-π/6)-1=-1/5
sin(2a-π/6)=4/5
所以,cos(2a+π/3)
=cos(2a-π/6+π/2)
=-sin(2a-π/6)
=-4/5
f(a)=-1/5
则,sin(2a-π/6)-1=-1/5
cos(2a+π/3)=-4/5
π/3
=√3/2 ×sin2x-1/2 ×cos2x -1
=sin(2x-π/6) -1
当sin(2x-π/6)=1时
f(x)有最大值=0
f(a)=-1/5
则,sin(2a-π/6)-1=-1/5
sin(2a-π/6)=4/5
所以,cos(2a+π/3)
=cos(2a-π/6+π/2)
=-sin(2a-π/6)
=-4/5
f(a)=-1/5
则,sin(2a-π/6)-1=-1/5
cos(2a+π/3)=-4/5
π/3
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