Question

求函数y=√3cos(3/2π+2π)+cos²x-sin²x的周期,当x取何值时,y取最大值、最小值

Answer 1

题目错了一个字母，改正后为：

y=√3cos(3/2π+2x)+cos²x-sin²x
=√3sin2x+cos2x
=2
sin(2x+π/6)
当2x+π/6=π/2+2kπ,即x=π/4-π/12+kπ（k属于Z）,y最大=2
当2x+π/6=-π/2+2kπ,即x=-π/4-π/12+kπ（k属于Z）,y最大=-2