已知函数f(x)=2√3sinXcosX + 2cos²X - 1
求f(π/6)的值 及f(x)的最小正周期
f(x) = 2√3sinXcosX + 2cos²X - 1
=√3sin2x + (1 + cos2x) - 1
= √3sin2x + cos2x
= √(3 + 1) sin(2x + 30°) ------- Acosx + Bsinx = √(A² + B²) sin(2x + φ)
= 2sin(2x + 30°)
f(π/6) = 2 sin60° =√3
最小正周期 2π/2 = π