Question

已知函数f(x)=2sinxsin(x+π/3) 求函数f(x)的最大值

Answer 1

f(x)=2sinxsin(x+π/3)
=2sinx(sinxcosπ/3+cosxsinπ/3)
=2sinx(1/2*sinx+√3/2*cosx)
=sin²x+√3sinxcosx
=(1-cos2x)/2+√3/2sin2x
=1/2+√3/2*sin2x-1/2*cos2x
=1/2+sin(2x-π/6)
所以函数f(x)的最大值是1/2+1=3/2

Answer 2

f(x)=2sinxsin(x+π/3)
=2sinx(sinxcosπ/3+cosxsinπ/3)
=2sinx(1/2*sinx+√3/2*cosx)
=sin²x+√3sinxcosx
=(1-cos2x)/2 + √3*sin2x/2
=1/2-cos2x / 2 + √3*sin2x/2
=1/2-cosπ/3cos2x+sinπ/3sin2x
=1/2-(cosπ/3cos2x-sinπ/3sin2x)
=1/2-cos(π/3+2x)
显然，cos(π/3+2x) 取值范围是 [-1,1]，
所以，-cos(π/3+2x) 取值范围是 [-1,1]，
所以，1/2-cos(π/3+2x) 取值范围是 [-1/2,3/2]
即 函数的最大值为 3/ 2

Answer 3

f(x)=－[cos(x＋x＋π/3)－cos(x－x－π/3)]
=－cos(2x＋π/3)＋1/2
最大值为3/2。

Answer 4

由公式：sinxsiny=-1/2[cos(x+y)-cos(x-y)]
==>2sinxsiny=-[cos(x+y)-cos(x-y)]=cos(x-y)-cos(x+y)

得 f(x)=2sinxsin(x+π/3)
=cosπ/3-cos(2x+π/3)
=1/2-cos(2x+π/3)
由于cos(2x+π/3)的取值范围为[-1,1]，因此-1/2≤f(x)≤3/2
所以f(x)的最大值为3/2