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
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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
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
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
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/3sin(2a+π/3)=-3/5
cos2a=cos(2a+π/3-π/3)
=cos(2a+π/3)×1/2+sin(2a+π/3)×√3/2
= -4/5 ×1/2-3/5×√3/2
=-(4+3√3)/10
f(x)=1/2cos2x+根号3sinxcosx-2cos^2x=1/2cos2x+2分之根号3sin(2x)-cos(2x)-1=2分之根号3sin(2x)-1/2cos2x-1=sin(2x-30度)-1 所以f(x)的最大值为0
f(x)=1/2cos2x+√3/2 * sin2x-1-cos2x = √3/2 * sin2x - 1/2 * cos2x -1
= sin(2x -π/6)-1 = cos(2x - 2π/3 )-1
∴ (1)f(x)的最大值 = 0
(2)cos(2a+π/3) = cos(2a -2π/3 + π)= - cos(2a - 2π/3)=...
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f(x)=1/2cos2x+√3/2 * sin2x-1-cos2x = √3/2 * sin2x - 1/2 * cos2x -1
= sin(2x -π/6)-1 = cos(2x - 2π/3 )-1
∴ (1)f(x)的最大值 = 0
(2)cos(2a+π/3) = cos(2a -2π/3 + π)= - cos(2a - 2π/3)= -[f(a)+1] = -(-1/5 +1) = -4/5
(3) ∵π/3∴ 2π/3 < 2a < π ; π < 2a+π/3 < 4π/3
∴ cos2a < 0 ; sin (2a+π/3) < 0
根据(2)问知道:当f(a)=-1/5时 cos(2a+π/3) = -4/5
∴ sin(2a+π/3) = -3/5
cos2a = cos[(2a + π/3)-π/3] = 1/2cos(2a+π/3) + √3/2sin(2a+π/3) = -4/5 × 1/2 - 3/5×√3/2
= -(4+3√3)/10
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