Let f : R → R be defined as f x + y + f x - y = 2 f x f y , f 1 2 = - 1 . Then the value of ∑ k = 1 20 1 sin k sin k + f k is equal to :

Step-by-step solution

Let f x = cos λ x ∵   f 1 2 = - 1 So, - 1 = cos λ 2 ⇒ λ 2 = π Thus f x = cos 2 π x Now k is a natural number Thus f k = 1 ∑ k = 1 20 1 sin k sin k + 1 = 1 sin 1 ∑ k = 1 20 sin k + 1 - k sin k · sin k + 1 1 sin 1 ∑ k = 1 20 sin k + 1 cos k - sin k cos k + 1 sin k · sin k + 1 = 1 sin 1 ∑ k = 1 20 cot k - cot k + 1 = cot 1 - cot 21 sin 1 = cos 1 sin 1 - cos 21 sin 21 sin 1 = sin 21 cos 1 - sin 1 cos 21 sin 2 1 sin 21 = sin 21 - 1 cosec 2 1 cos