已知函数f(x)=sinxcosx+(cosx)^2(1)求f(x)的最小正周期（2）求f(x)在区间[-兀/6,兀/3

f(x)=sinxcosx+cos^2 x
=sin2x/2+(1+cos2x)/2
=1/2*(sin2x+cos2x)+1/2
=√2/2*sin(2x+π/4)+1/2
∴f(x)的最小正周期T=2π/2=π
∵f(x)的单调增区间为[-π/4+2kπ,3π/4+2kπ],k∈Z
当k=1时,单调增区间为[-π/4,3π/4],且[-π/6,π/3]在此区间内
∴此时f(x)的最大值为
f(π/3)=√2/2*sin(2*π/3+π/4)+1/2
=√2/2*sin(11π/12)+1/2
=√2/2*(√6-√2)/4+1/2
=(√3-1)/4+1/2
最小值为
f(-π/6)=√2/2*sin(-2*π/6+π/4)+1/2
=√2/2*sin(-π/12)+1/2
=√2/2*[-(√6-√2)/4]+1/2
= -(√3-1)/4+1/2

解:
f(x)=sinxcosx+(cosx)^2
=1/2sin2x+1/2cos2x+1/2
=根号2/2sin(2x+π/4)+1/2
(1)
T=2π/2=π
(2)
-π/6<=x<=π/3
-π/3<=2x<=2π/3
-π/12<=2x+π/4<=11π/12
所以sin(2x+π/4)∈[(√2 - √6)/ 4,1]
所以f(x)=根号2/2sin(2x+π/4)+1/2∈[(3-根号12)/2,(根号2 + 1)/2]