f(x)=sin2x-2(sinx)^2 在[0,π/3]上的最小值_数学解答

f(x)=sin2x-2(sinx)^2 在[0,π/3]上的最小值

人气：379 ℃　时间：2020-06-18 10:08:49

解答

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