Let f- be defined as fx-y-fx-y-2fxfy- f 12 -1- Then the value of -k-1201sin ksink-fk is equal to

∵ f(x+y)+f(x y)=2f(x).f(y) ∴ f(x)=cos (λ x) Given that f ( 12 )= 1, cos( λ2 )= 1 ⇒λ =2n π, n ∈𝕀 ∴ f(x)=cos(2π x)⇒ f(k)=1, k ∈𝕀 nbsp; ∑_k=1^201sin k sin (k+f ...

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Standard XII

Mathematics

Sum of Infinite Terms of a GP

Let f:ℝ→ℝ be ...

Question

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 $20 \sum k = 1 \frac{1}{sin (k) sin (k + f (k))}$ is equal to

{cosec}^{2} (21) cos (20) cos (2)

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{sec}^{2} (21) sin (20) sin (2)

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

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{sec}^{2} (1) sec (21) cos (20)

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Solution

The correct option is C ${cosec}^{2} (1) cosec (21) sin (20)$
$∵ f (x + y) + f (x - y) = 2 f (x) . f (y) ∴ f (x) = cos (λ x)$
Given that $f (\frac{1}{2}) = - 1,$
$cos (\frac{λ}{2}) = - 1 \Rightarrow λ = 2 n π, n \in I$
$∴ f (x) = cos (2 π x) \Rightarrow f (k) = 1, k \in I$

$20 \sum k = 1 \frac{1}{sin k sin (k + f (k))} = 20 \sum k = 1 \frac{1}{sin k \cdot sin (k + 1)} = 20 \sum k = 1 \frac{1}{sin 1} \cdot \frac{sin {(k + 1) - k}}{sin k \cdot sin (k + 1)} = \frac{1}{sin 1} 20 \sum k = 1 (cot k - cot (k + 1)) = \frac{1}{sin 1} (cot 1 - cot 21) = \frac{1}{sin 1} \cdot \frac{sin (21 - 1)}{sin 1 \cdot sin 21} = {cosec}^{2} (1) \cdot cosec (21) \cdot sin (20)$

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Let f:R→R be defined as f(x+y)+f(x−y)=2f(x)f(y), f(12)=−1. Then the value of 20∑k=11sin(k)sin(k+f(k)) is equal to

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