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

题目

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

答案

原式可化为y=sin²（π/3-x）+sin(π/3-x)
令t=sin(π/3-x)
则y=t²+t=(t+1/2)²-1/4
∵x∈[-π/3,2π/3]
∴π/3-x∈[-π/3,2π/3]
∴t∈[负二分之根号三,1]
所以y∈[-1/4,2]
即y最大值为2,最小值为-1/4