已知函数f(x)=sinx·sin(x+π/2)-根号3cos(3π+x)+根号3/2(x属于R)如题

1个回答

  • f(x)=sinxcosx-√3cosx+√3/2=(1/2)sin2x-√3×(1+cos2x)/2+√3/2=(1/2)sin2x-(√3/2)cos2x =sin2xcosπ/3-sinπ/3cos2x=sin(2x-π/3) 1)最小正周期T=2π/2=π 2)单调增区间:2x-π/3∈[2kπ-π/2,2kπ+π/2],解得x∈[kπ-π/12,kπ+5π/12] ∴单调增区间为[kπ-π/12,kπ+5π/12](k∈Z) 3)对称轴:2x-π/3=kπ+π/2,即x=kπ/2+5π/12,∴对称轴为x=kπ/2+5π/12(k∈Z) 对称中心:2x-π/3=kπ,x=kπ/2+π/6,∴对称中心为(kπ/2+π/6,0)(k∈Z)