Question

函数f（x）=sin x^2+√3 sinx cosx在区间【π/4,π/2】上有最小值是___

Answer 1

f(x)=sin^2x+√3 sinxcosx
f(x)=(1-cos2x)/2+√3 sin2x/2
f(x)=1/2-((1/2)*cos2x-(√3/2)*sin2x)
f(x)=1/2-(sinπ/6*cos2x-cosπ/6*sin2x)
f(x)=1/2-sin(π/6-2x)
x在R上的f(x)最小值f(π/2+2kπ)=1/2-1=-1/2
π/6-2x=π/2+2kπ
x=-π/6-kπ不在区间【,π/2】
讨论区间两端点
f(π/4)=1/2-sin(π/6-2π/4)=1/2-sin(-π/6)=(1+√3)/2
f(π/2)=1/2-sin(π/6-2π/2)=1/2-sin(-5π/6)=1
f(π/4)>f(π/2)
f(x)=sin x^2+√3 sinx cosx在区间【π/4,π/2】上有最小值是1

Answer 2

函数f（x）=sin² x+(√3 )sinx cosx在区间【π/4,π/2】上有最小值是___
解：f(x)=(1-cos2x)/2+(√3/2)sin2x=(1/2)+(√3/2)sin2x-(1/2)cos2x
=(1/2)+sin2xcos(π/6)-cos2xsin(π/6)=(1/2)+sin(2x-π/6)
用五点作图法不难判断在区间[π/4，π/2]上的最小值为f(π/2)=(1/2)+sin(π-π/6)=(1/2)+sin(π/6)
=(1/2)+1/2=1