f(x)=sinxcosθ+cosxsinθ
=sin(x+θ)
f(x)最小正周期T=2π
最大值为1,最小值为-1
若f(π/4)=4/5
即sinπ/4cosθ+cosπ/4sinθ=4/5
那么sinθ+cosθ=4√2/5
cosθ=4√2/5-sinθ
又sin²θ+cos²θ=1
∴sin²θ+(4√2/5-sinθ)²=1
∴2sin²θ-8√2/5sinθ+7/25=0
50sin²θ-40√2sinθ+7=0
解得sinθ=7√2/10或sinθ=√2/10
∴f(0)=sinθ=7√2/10或√2/10