f(x)=sin2x-2(sinx)^2 在[0,π/3]上的最小值

题目

f(x)=sin2x-2(sinx)^2 在[0,π/3]上的最小值

答案

f(x)=sin2x+cos2x-1
=√2sin(2x+π/4)-1
0<=x<=π/3
π/4<=2x+π/4<=11π/12
sin(11π/12)
=sin(2π/3+π/4)
=(√6-√2)/4
所以最小值=√2(√6-√2)/4-1=(√3-3)/2