Question

已知函数f（x）=√3sin^2x+sinxcosx-√3/2（x∈R）求最小正周期,单调递减区间,函数的对称轴方程,对

Answer 1

f(x) = -√3sin²x + sinx cosx

= -√3 × (1 - cos(2x))/2 + (sin(2x))/2 (利用二倍角公式)

= -√3/2 + (√3/2)cos(2x) + (1/2)sin(2x)

= -√3/2 + sin(π/3) cos(2x) + cos(π/3) sin(2x)

= -√3/2 + sin(π/3 + 2x) (利用两角和的正弦公式)

(1) 周期 = 2π / 2 = π (T = 2π / w)