Question

已知f(x)=5cos²x+sin²x-（4根号3）sinxcosx

(1)化简f（x）的解析式，并求f（x）的最小正周期
（2）当x∈[-π/6，π/4]时，求f（x）的值域

Answer 1

f(x) = 5cos²x + sin²x-4√3sinxcosx
= 4cos²x + cos²x + sin²x - 2√3sin2x
= 2(cos2x+1) + 1 - 2√3sin2x
= -2√3sin2x+2cos2x+3
= -4(sin2xcosπ/6-cos2xsinπ/6) + 3
= -4sin(2x-π/6) + 3

最小正周期：2π/2 = π

x∈[-π/6，π/4]时
2x-π/6∈[-π/2，π/3]
sin(2x-π/6)∈[-1，√3/2]
-4sin(2x-π/6)∈[-2√3，4]
f(x)= -4sin(2x-π/6) + 3的值域[3-2√3，7]

Answer 2

1

Answer 3

解答：
f(x)=5*(1+cos2x)/2+(1-cos2x)/2-2√3sin2x
=2cos2x-2√3sin2x+3
=4[cos(2x)*(1/2)-sin(2x)*(√3/2)]+3
=4[cos(2x)cos(π/3)-sin(2x)sin(π/3)]+3
=4cos(2x+π/3)+3
(1) T=2π/2=π
(2) x∈[-π/6，π/4]
2x+π/3∈[0，5π/6]

cos(2x+π/3)∈[-√3/2，1]
所以 y∈[-2√3+3,7]