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+√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 ∴ cos2a 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

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+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＝0f(a)=-1/5则,sin(2a-π/6)-1＝-1/5sin(2a-π/6)=4/5所以,cos(2a+π/3)＝cos(2a...