剩下的自己做
将sinxcosx变为1/2sin2x
sinx的平方变为1-cos2x/2
再进行化简
f(x)=(1/2)*sin2x-(√3/2)[1-cos2x]+√3/2
=(1/2)*sin2x+(√3/2)cos2x
=sin(2x+π/3),
∴f(x)的最小正周期为T=2π/2=π,
f(π/12）=sin(2*π/12+π/3)=sin(π/2)=1,
∵-π/4≤x≤π/4,得
-π/6≤2x+π/3≤5π/6,∴-1/2≤sin(2x+π/3)≤1,
即f(x)的值域为【-1/2,1]
f(x)的单调增区间为
2kπ-π/2≤2x+π/3≤2kπ-π/2,解得kπ-5π/12≤x≤kπ+π/12,k∈z
f(x)的单调减区间为
2kπ+π/2≤2x+π/3≤2kπ+3π/2,解得kπ+π/12≤x≤kπ+7π/12,k∈z
f(x)的对称轴为2x+π/3=kπ+π/2,得x=kπ/2+π/12,k∈z
f(x)的对称中心为：2x+π/3=kπ,得x=kπ/2-π/6,即为（kπ/2-π/6，0），k∈z
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