Let |cos θ cos (60-θ) cos (60+θ)| ≤ 1/8, θ ε[0,2 π]. Then, the sum of all θ ∈[0,2 π], where cos 3 θ attains its maximum value, is :

Solution

Maths sir · 4 steps

The product is just 1/4 cos 3θ, so the bound caps cos 3θ at 1/2, not 1.

1. Simplify the product. Here 60 means 60°. Use cos(A-B)cos(A+B) = cos² A - sin² B to get cos(60°-θ)cos(60°+θ) = 1/4 - sin² θ = cos² θ - 3/4. Multiplying by cos θ gives 1/4(4 cos³ θ - 3 cos θ).

   cos θ cos(60°-θ) cos(60°+θ) = 1/4 cos 3θ

2. Use the condition. The condition becomes |1/4 cos 3θ| ≤ 1/8, that is |cos 3θ| ≤ 1/2. So on the allowed values of θ, the maximum of cos 3θ is 1/2.
3. Find every angle. For θ ∈ [0, 2π], 3θ ∈ [0, 6π]. Then cos 3θ = 1/2 gives 3θ = π/3, 5π/3, 7π/3, 11π/3, 13π/3, 17π/3. Dividing by 3 gives θ = π/9, 5π/9, 7π/9, 11π/9, 13π/9, 17π/9.
4. So. The sum is (1+5+7+11+13+17)π/9 = 54π/9 = 6π, which is option (a).

The idea to keep. Whenever you see cos θ cos(60°-θ)cos(60°+θ), replace it with 1/4 cos 3θ at once. Then find the maximum on the restricted set, not the usual maximum of 1.

Answer from the NTA final answer key. Solution by PrepWiser.