已知0＜x＜π/2,f(x)=1/2sin²x（cotx/2-tanx/2)+（根号3）/2*cos2x,求f(x)单调区间?

f(x)=1/2sin²x（cotx/2-tanx/2)+（根号3）/2*cos2x
=1/2sin²x{[cos(x/2)]^2-[sin(x/2)]^2}/[cos(x/2)sin(x/2)]+(根号3)/2*cos2x
=sinxcosx+(根号3)/2*cos2x
=sin2x/2+(根号3)/2*cos2x
=sin(2x+π/3)
然后利用sin函数的单调区间即可求所需结果.
(-5π/12+kπ,π/12+kπ)单调递增,k为任意整数
(π/12+kπ,7π/12+kπ)单调递减,k为任意整数（cotx/2-tanx/2)这一步详细点 谢谢cotx/2-tanx/2=sin(x/2)/cos(x/2)-cos(x/2)/sin(x/2)通分即可得上面结果。{[cos(x/2)]^2-[sin(x/2)]^2}/[cos(x/2)sin(x/2)]