The maximum value of the expression sin ⁡ θ c o s 2 θ ∀ θ ∈ 0 , π is

Step-by-step solution

As s i n 2 θ + c o s 2 θ = 1 , ⇒ s i n 2 θ + c o s 2 θ 2 + c o s 2 θ 2 = 1 Applying A M ≥ G M , we have s i n 2 θ + c o s 2 θ 2 + c o s 2 θ 2 3 ≥ s i n 2 θ c o s 4 θ 4 1 3 ⇒ 1 3 3 ≥ s i n 2 θ c o s 4 θ 4 ⇒ sin ⁡ θ ⋅ c o s 2 θ 2 ≤ 4 27 ⇒ sin ⁡ θ ⋅ c o s 2 θ ≤ 2 3 3