Question

f(x)=sin²x+√3sinxcosx在区间[π/4,π/2]上的最大值是? 求过程

Answer 1

f(x)=sin²x+√3sinxcosx
=-cos²x+1+√3/2*2sinxcosx
=-(2cos²x-1-1)/2+√3/2sin2x
=1/2-1/2cos2x+√3/2sin2x
=√3/2sin2x-1/2cos2x+1/2
=sin(2x-π/6)+1/2
在(-π/2，π/2)的增区间为
-π/2≤2x-π/6≤π/2
-π/6≤x≤π/3
所以[π/4,π/2]上的最大值是x=π/3时，f(x)max=3/2

注：π/3之前是递增的，之后的是递减的，所以π/3最大

Answer 2

f(x)=(1-cos2x)/2+3^0.5/2*sin2x=0.5+3^0.5/2*sin2x-0.5*cos2x
=0.5+sin(2x-pi/6) =>max{f(x)}=0.5+1=1.5 (x=pi/3)