Let f(θ)=sinθ(sinθ+sin3θ). Then f(θ) is [2000]
(3)
f(θ)=sinθ(sinθ+sin3θ)
=sinθ(sinθ+3sinθ-4sin3θ)
=sinθ(4sinθ-4sin3θ)=sin2θ(4-4sin2θ)
=4sin2θ(1-sin2θ)
=4sin2θcos2θ=(2sinθcosθ)2=(sin2θ)2≥0, which is true for all θ.