已知函數(shù)f(x)=Asin（ωx+π/4）(其中x∈R,A＞0,ω＞0)的最大值為2,最小周期為8
(1)求函數(shù)f(x)的解析式(2)若函數(shù)f(x)圖像上的兩點(diǎn)P,Q的橫坐標(biāo)依次為2,4,0為坐標(biāo)原點(diǎn),求cos∠P0Q的值
(1)解析：∵函數(shù)f(x)=Asin(ωx+π/4)(其中x∈R,A＞0,ω＞0)的最大值為2,最小周期為8
∴ω=2π/8=π/4
∴f(x)=2sin(π/4x+π/4)
(2)解析：f(2)=2sin(π/2+π/4)=√2==>P(2,√2)
f(4)=2sin(π+π/4)=-√2==>Q(4,-√2)
∴|OP|=√6,|OQ|=3√2,|PQ|=2√3
∴cos∠POQ=(OP^2+OQ^2-PQ^3)/(2OP*OQ)=(6+18-12)/(2*√6*3√2)=√3/3