Solve this following
Question:

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ 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 :

 

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

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

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

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


Correct Option: , 3

Solution:

$f(x)=\cos \lambda x$

$\because \mathrm{f}\left(\frac{1}{2}\right)=-1$

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

$\Rightarrow \lambda=2 \pi$

Thus $f(x)=\cos 2 \pi x$

Now $\mathrm{k}$ is natural number

Thus $f(k)=1$

$\sum_{k=1}^{20} \frac{1}{\sin k \sin (k+1)}=\frac{1}{\sin 1} \sum_{k=1}^{20}\left[\frac{\sin ((k+1)-k)}{\sin k \cdot \sin (k+1)}\right]$

$=\frac{1}{\sin 1} \sum_{k=1}^{20}(\cot k-\cot (k+1)$

$=\frac{\cot 1-\cot 21}{\sin 1}=\operatorname{cosec}^{2} 1 \operatorname{cosec}(21) \cdot \sin 20$

 

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