解由f(x)=sinxcosx+(cos^2x-1/2)
=1/2*2sinxcosx+(2cos^2x-1)/2
=1/2sin2x+1/2cos2x
=1/2*√2(√2/2sin2x+√2/2cos2x)
=√2/2sin(2x+π/4)
故函数的周期T=2π/2=π
当2kπ-π/2≤2x+π/4≤2kπ+π/2,k属于Z时,y是增函数
即2kπ-3π/8≤x≤kπ+π/8,k属于Z时,y是增函数
故函数的增区间为[2kπ-3π/8,kπ+π/8],k属于Z.
解由f(x)=sinxcosx+(cos^2x-1/2)
=1/2*2sinxcosx+(2cos^2x-1)/2
=1/2sin2x+1/2cos2x
=1/2*√2(√2/2sin2x+√2/2cos2x)
=√2/2sin(2x+π/4)
故函数的周期T=2π/2=π
当2kπ-π/2≤2x+π/4≤2kπ+π/2,k属于Z时,y是增函数
即2kπ-3π/8≤x≤kπ+π/8,k属于Z时,y是增函数
故函数的增区间为[2kπ-3π/8,kπ+π/8],k属于Z.