题目

已知fx=cos^2x+cosxcos(x+π/3)+sin^2(π/6-x) 1.求fx最小正周期2.

提问时间：2021-11-03

答案

1.f(x) = cos 2 x+ cosxcos(x + π/3) + sin 2 (π/6 –x) = (1/2)(1 + cos2x) + (cosx)[cosxcos(π/3) – sinxsin(π/3)] + (1/2)[1– cos(π/3 – 2x)] = 1/2 + (1/2)cos2x+ (1/2)cos 2 x– (√3/2)cosxsinx+ 1/2 – (1/2)cos(π/3 – 2x) = 1 +(1/2)cos2x + (1/2)*(1/2)(1 + cos2x) – (√3/4)sin2x – (1/2)[cos(π/3)cos2x+ sin(π/3)sin2x] = 5/4 + (3/4)cos2x– (√3/4)sin2x – (1/4)cos2x – (√3/4)sin2x = 5/4 + (1/2)cos2x – (√3/2)sin2x = 5/4 + sin(2x + 5π/6),即f(x) = 5/4 +sin(2x + 5π/6),T = 2π/2 = π,所以函数f(x)的最小正周期为 π ； 2.因为x∈R,所以sin(2x + 5π/6)∈[-1,1],所以[5/4 + sin(2x+ 5π/6)] ∈[1/4,9/4],即f(x)∈[1/4,9/4],所以函数f(x)的最小值为 1/4 ,最大值为 9/4 .