函数y=sin平方x+根号3sinxcosx在区间[π/4,π/2]上的最大值

y=sin²x+√3sinxcosx
=(1-cos2x)/2+√3/2sin2x
=√3/2sin2x-1/2cos2x+1/2
=cosπ/6sin2x-sinπ/6cos2x+1/2
=sin(2x-π/6)+1/2
因为π/4≤x≤π/2
所以π/2≤2x≤π
π/2-π/6≤2x-π/6≤π-π/6
π/3≤2x-π/6≤5π/6
所以当2x-π/6=π/2,时y取得最大值,最大值为y=1+1/2=3/2
答案：3/2