已知函数f(x)=4cos(wπ)sin(wx+π/4)(w>0)的最小正周期为π（1）求w的值（2）讨论f(x)在区间[0,π/2]上的单调性
（1）解析：∵函数f(x)=4cos(wπ)sin(wx+π/4)(w>0)的最小正周期为π
∴T=π==>w=2π/π=2
f(x)=4cos(2π)sin(2x+π/4)=4sin(2x+π/4)
（2）解析：2kπ-π/2<=2x+π/4<=2kπ+π/2==>kπ-3π/8<=x<=kπ+π/8,f(x)单调增；
2kπ+π/2<=2x+π/4<=2kπ+3π/2==>kπ+π/8<=x<=kπ+5π/8,f(x)单调减；
∵区间[0,π/2]
∴在[0,π/8]上单调增；在[π/8,π/2]上单调减；