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*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