Question

已知函数y=sinxcos(x+∏／6)-1/2cos2x，求该函数的最大值并求y取最大值时x的集合

Answer 1

解析：
y=sinxcos(x+∏／6)-1/2cos2x
=sinxcos(x+π／6)-1/2 *(1-2sin²x)
=sinxcos(x+π／6)+sin²x - 1/2
=sinx*[cosxcos(π/6)-sinxsin(π/6) +sinx] -1/2
=sinx*[cosxcos(π/6)+sinxsin(π/6)] -1/2
=sinxcos[x-(π/6)] -1/2
=(1/2){sin[x+(x-π/6)] +sin[x-(x-π/6)]} -1/2
=(1/2)[sin(2x-π/6)+sin(π/6)] -1/2
=(1/2)sin(2x-π/6) -1/4
所以当sin(2x-π/6)=1即2x-π/6=π/2 +2kπ，k属于Z时，函数y取得最大值为1/4
此时x对应的集合为{ x | x=π/3 +kπ,k属于Z }