(1)
f(χ)=2sinχcosχ+cos2χ
=sin2x+cos2x
=√2sin(2x+π/4)
f(χ)的最小正周期T=2π/2=π
当2x+π/4=2kπ+π/2,k∈Z时,f(x)取得最大值√2
(2)∵f(θ+π/8)=√2/3
又f(θ+π/8)=√2sin[2(θ+π/8)+π/4]
=√2sin(2θ+π/2)=√2cos2θ
∴√2cos2θ=√2/3 ∴cos2θ=1/3
∵ θ为锐角,cos2θ=1/3 ∴2θ为锐角
∴sin2θ=√(1-cos²θ)=2√2/3
∴tan2θ=sin2θ/cos2θ=2√2