Let $f: R \rightarrow R$ be defined as $f(x+y)+f(x-y)=2 f(x) f(y), f\left(\frac{1}{2}\right)=-1$. Then, the value of $\sum_{k=1}^{20} \frac{1}{\sin (k) \sin (k+f(k))}$ is equal to :

(D) $\operatorname{cosec}^{2}(1) \operatorname{cosec}(21) \sin (20)$

(A) $\operatorname{cosec}^{2}(21) \cos (20) \cos (2)$

(B) $\sec ^{2}(1) \sec (21) \cos (20)$

(C) $\sec ^{2}(21) \sin (20) \sin (2)$

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Sets Relations and Functions

Let

f : R \to R

be defined as

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

. Then, the value of

\sum_{k = 1}^{20} \frac{1}{s i n ( k ) s i n ( k + f ( k ))}

is equal to :

Answer: (D)

Putting

x = y = 0

gives

2 f (0) = 2 f (0)^{2}

, so

f (0) = 0

f (0) = 1

; if

f (0) = 0

then

y = 0

forces

f \equiv 0

, contradicting

f (\frac{1}{2}) = - 1

, so

f (0) = 1

. Putting

x = y = \frac{1}{2}

gives

f (1) + f (0) = 2 f (\frac{1}{2})^{2} = 2

, so

f (1) = 1

. Putting

y = 1

gives the recursion

f (x + 1) = 2 f (x) - f (x - 1)

, which with

f (0) = f (1) = 1

gives

f (k) = 1

for every positive integer

k

.
The sum is therefore

S = k = 1 \sum 20 \frac{1}{sin k sin ( k + 1 )} = \frac{1}{sin 1} k = 1 \sum 20 \frac{sin ( ( k + 1 ) - k )}{sin k sin ( k + 1 )} = \frac{1}{sin 1} k = 1 \sum 20 (cot k - cot (k + 1))

The telescoping leaves

S = \frac{cot 1 - cot 21}{sin 1} = \frac{1}{sin 1} \cdot \frac{sin 21 cos 1 - cos 21 sin 1}{sin 1 sin 21} = \frac{sin 20}{sin ^{2} 1 sin 21}

S = cosec^{2} (1) cosec (21) sin (20)

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option