For non-negative integers n , let f n = ∑ k = 0 n sin ⁡ k + 1 n + 2 π sin ⁡ k + 2 n + 2 π ∑ k = 0 n sin 2 ⁡ k + 1 n + 2 π Assuming cos - 1 ⁡ x takes values in 0 , π , which of the following options is/are correct?

Step-by-step solution

f n = ∑ k = 0 n sin ⁡ k + 1 n + 2 π sin ⁡ k + 2 n + 2 π   ∑ k = 0 n s i n 2 k + 1 n + 2 π ∵ 2 sin C ⁡ sin ⁡ D = cos ⁡ C - D - cos ⁡ C + D ∵ 2 sin 2 ⁡ θ = 1 - cos ⁡ 2 θ f n = ∑ k = 0 n cos ⁡ π n + 2 - cos ⁡ 2 k + 3 n + 2 π ∑ k = 0 n 1 - cos ⁡ 2 k + 2 n + 2 π f n = n + 1 cos ⁡ π n + 2 - ∑ k = 0 n cos ⁡ 2 k + 3 n + 2 π n + 1 - ∑ k = 0 n cos