已知函数f(x)等于sinx乘cosx加cosx的平方减二分之一 求函数的周期的单调递增区间!

f(x)=sinxcosx+(cosx)^2-1/2
=(sin2x)/2+(1+cos2x)/2-1/2
=(sin2x+cos2x)/2
=(√2/2)(sin2xcosπ/4+cos2xsinπ/4)
=(√2/2)sin(2x+π/4)
所以f(x)的最小正周期为T= π
当(2x+π/4)∈(-π/2+2kπ,π/2+2kπ),k∈N时
即x∈(-3π/8+kπ,π/8+kπ)时
f(x)单调递增

f(x) = sinxcosx + cos^2x-1/2
= 1/2sin2x + 1/2(cos2x+1) - 1/2
= 1/2sin2x+1/2cos2x
= √2/2(sin2xcosπ/4+cos2xsinπ/4)
= √2/2sin(2x+π/4)
最小正周期 = 2π/2 = π
2x+π/4∈（2kπ-π/2，2kπ-π/2）时单调增
所以单调递增区间（kπ-3π/8，kπ+π/8）

f(x)=sinxcosx+(cosx)^2-1/2=0.5sin(2x)+0.5(1+cos2x)-1/2=0.5(sin2x+cos2x)=0.5√2sin(2x+π/4)
单调增区间：2kπ-π/2