y=cosx+√3sinx,x∈[π/6,2π/3]
问题描述:
y=cosx+√3sinx,x∈[π/6,2π/3]
求最值
答
y=cosx+√3sinx=2*((1/2)cosx+(√3/2)sinx )
=2* [ sin(π/6)cosx+ cos(π/6)sinx ]
=2* sin [ x+(π/6) ],
从方括号可以看出,π/3 ≦[ x+(π/6) ]≦5π/6
,所以y最小值为1,最大值为2.