Question

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 $\displaystyle\sum_{k=1}^{20} \frac{1}{\sin (k) \sin (k+f(k))}$ is equal to:

• A. $\text{cosec}^{2}(21) \cos (20) \cos$
• B. $\sec ^{2}(1) \sec (21) \cos (20)$
• C. $\text{cosec}^{2}(1) \text{cosec}(21) \sin (20)$
• D. $\sec ^{2}(21) \sin (20) \sin (2)$

Answer 1

Answer: (C)

$f(x)=\cos \lambda x$
$\because f\left(\frac{1}{2}\right)=-1 $
So, $-1=\cos \frac{\lambda}{2}$
$\Rightarrow \lambda=2 \pi $
Thus $ f(x)=\cos 2 \pi x$
Thus $f(x)=\cos 2 \pi x$
Now $k$ is natural number
Thus $f(k)=1$
$\displaystyle\sum_{k=1}^{20} \frac{1}{\sin k \sin (k+1)}=\frac{1}{\sin 1} \displaystyle\sum_{k=1}^{20}\left[\frac{\sin ((k+1)-k)}{\sin k \cdot \sin (k+1)}\right]$
$=\frac{1}{\sin 1} \displaystyle\sum_{k=1}^{20}(\cot k-\cot (k+1)$
$=\frac{\cot 1-\cot 21}{\sin 1}=\text{cosec}^{2} 1 \text{cosec}(21) \cdot \sin 20$