If sin ⁡ θ = cos 2 ⁡ θ , then the value of cos 12 ⁡ θ + 3 cos 10 ⁡ θ + 3 cos 8 ⁡ θ + cos 6 ⁡ θ - 1 is equal to

Step-by-step solution

sin ⁡ θ = cos 2 ⁡ θ ⇒ sin ⁡ θ + sin 2 ⁡ θ = 1 Now, cos 6 ⁡ θ cos 6 ⁡ θ + 3 cos 4 ⁡ θ + 3 cos 2 ⁡ θ + 1 - 1 = cos 6 ⁡ θ cos 2 ⁡ θ + 1 3 - 1 = cos 2 ⁡ θ 3 sin ⁡ θ + 1 3 - 1 = sin 3 ⁡ θ sin ⁡ θ + 1 3 - 1 = sin 2 ⁡ θ + sin ⁡ θ 3 - 1 = 1 - 1 = 0