函数f(x)=sin2x/3+cos(2x/3-π/6)的最大值和最小值

题目

函数f(x)=sin2x/3+cos(2x/3-π/6)的最大值和最小值

答案

f(x)=sin(2x/3)+cos(2x/3)cox(π/6)+sin(2x/3)sin(π/6)
=sin(2x/3)+((根号3)/2)cos(2x/3)+(1/2)sin(2x/3)
=((根号3)/2)cos(2x/3)+(3/2)sin(2x/3)
=(根号3)((1/2)cos(2x/3)+((根号3)/2)sin(2x/3)
=(根号3)(sin(π/6)cos(2x/3)+cos(π/6)sin(2x/3)
=(根号3)sin(2x/3+π/6)
最大值根号3,最小值－根号3