f（x）=sin（π-wx）coswx+cos²wx
=sinwxcoswx+cos²wx
=(1/2)sin2wx+(1+cos2wx)/2
=(√2/2)sin(2wx+π/4)+1/2
T=2π/2w=π
（1）w=1
（2）f(x)=(√2/2)sin(2x+π/4)+1/2
z(x)=(√2/2)sin(4x+π/4)+1/2
x∈[0, π/16]
4x+π/4∈[π/4,π/2]
所以 当4x+π/4=π/4时,即x=0时,
z(x)有最小值,为(√2/2)*(√2/2)+1/2=1