若x∈[-π/3，2π/3]，求函数y=cos^2（x+π/6）+sin（x+2π/3）的最大值与最小值

见图

解:
y
=cos^2(x+π/6)+sin(x+2π/3)
=[cos(x+π/6)]^2+sin(x+2π/3)
={cos[π/2-(x+π/3)]}^2+sin[π-(x+π/3)]
=[sin(x+π/3)]^2+sin(x+π/3)

设T=sin(x+π/3)
则:
y=T^2+T
=T^2+T+1/4-1/4
=(T+1/2)^2-1/4

由于x∈[-π/3，2π/3]
则:(x+π/3)∈[0,π]
则:T=sin(x+π/3)∈[0,1]
则:Y=(T+1/2)^2-1/4
属于[0,2]

则:函数y=cos^2(x+π/6)+sin(x+2π/3)
最大值为2
最小值为0