已知函数f(x)=2根号3sinxsin(兀/2-x)+2cos平方x+2求f(x)的最小正周期与递减区间.

Question

已知函数f(x)=2根号3sinxsin(兀/2-x)+2cos平方x+2
求f(x)的最小正周期与递减区间.

Answer 1

已知函数f(x)=2(√3)sinxsin(π/2-x)+2cos²x+2,求f(x)的最小正周期与递减区间.
f(x)=2(√3)sinxsin(π/2-x)+2cos²x+2=2(√3)sinxcosx+(1+cos2x)+2
=(√3)sin2x+cos2x+3=2[(√3/2)sin2x+(1/2)cos2x]+3=2[sin2xcos(π/6)+cos2xsin(π/6)]+3
=2sin(2x+π/6)+3
最小正周期T=2π/2=π；
单减区间：由π/2+2kπ≦2x+π/6≦3π/2+2kπ,得π/3+2kπ≦2x≦4π/3+2kπ,于是得单减区间为：π/6+kπ≦x≦2π/3+kπ,K∈Z.