​ Let f : R → R be defined as f(x + y) + f(x – y) = 2 f(x) f(y), f() = - 1. 2 ​ Then, the value of ∑_{k = 1}^20 1/sin(k) sin (k + f(k))

Question

Let f : R → R be defined as 

f(x + y) + f(x – y) = 2 f(x) f(y), f(\(\frac{1}{2}\)) = - 1. 2 

Then, the value of \(\displaystyle \sum_{k=1}^{20}\) \(\frac{1}{sin(k)sin(k + f(k))}\) is equal to:

(1) cosec² (21) cos(20) cos(2) 

(2) sec² (1) sec(21) cos(20) 

(3) cosec² (1) cosec(21) sin(20) 

(4) sec² (21) sin(20) sin(2)

Answer 1

Correct answer (3) cosec² (1) cosec(21) sin(20)

 f(x) = cos \(\lambda\)x

\(\because\) f(\(\frac{1}{2}\)) = -1

So, -1 = cos\(\frac{\lambda }{2}\)

\(\Rightarrow\) \(\lambda\) = 2\(\pi\)

Thus f(x) = cos 2\(\pi\)x 

Now k is natural number 

Thus f(k) = 1