Question

已知函数f(x)=sin²x+3cos²x+二倍根号三sinxcosx-2

1.求函数f(X)的最大值及单调递减区间；
2.若f(a)=8\5(π\6＜a＜2π\3)，求cos2a的值。

Answer 1

你好！

f(x) =(sin²x+cos²x)+2cos²x +2√3 sinx cosx -2
= 1 + 1+cos2x +√3 sin2x -2
= √3sin2x + cos2x
= 2 sin (2x + π/6)

最大值2
π/2 + 2kπ < 2x+π/6 < 3π/2+ 2kπ
解得减区间 (π/6+kπ,2π/3+kπ)

2sin(2a+π/6) = 8/5
sin(2a+π/6) = 4/5
π/6＜a＜2π/3
π/3 < 2a < 4π/3
π/2 < 2a+π/6 < 3π/2
∴cos(2a+π/6) = - 3/5
cos(2a) = cos[(2a+π/6) - π/6]
= cos(2a+π/6)cos(π/6) + sin(2a+π/6)sin(π/6)
= - 3/5 × √3 /2 + 4/5 ×1/2
= (4-3√3)/10