已知函数y=sin(x/2)+(√3)cos(x/2),求(1)函数y的最大值、最小值及最小正周期;(2)函数y的单调递增区间
解:y=2[(1/2)sin(x/2)+(√3/2)cos(x/2)]=2[sin(x/2)cos(π/3)+cos(x/2)sin(π/3)]=2sin(x/2+π/3)
故最小正周期T=2π/(1/2)=4π,ymax=2,ymin=-2;
由-π/2+2kπ≦x/2+π/3≦π/2+2kπ,即-5π/6+2kπ≦x/2≦π/6+2kπ,
得单调增区间为:-5π/3+4kπ≦x≦π/3+4kπ
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