f(θ)=(sinθ+cosθ)2+(sinθ−cosθ)4
=sin2θ+cos2θ+2sinθcosθ+((sinθ−cosθ)2)2
=1+sin2θ+(sin2θ+cos2θ−2sinθcosθ)2
=1+sin2θ+(1−sin2θ)2
=1+sin2θ+1+sin22θ−2sin2θ
=sin22θ−sin2θ+2
⇒f(θ)=(sin2θ−12)2+74
Since θ∈(0,π),
∴2θ∈(0,2π)
f(θ) is minimum when sin2θ=12
∴2θ=π6,5π6
⇒θ=π12,5π12
λ1=112, λ2=512
λ1+λ2=112+512=12=0.50