Question

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

• 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 $f(k)=1$
$\sum _{k=1}^{20}\frac{1}{\sin k\sin (k+1)}=\frac{1}{\sin 1}\sum _{k=1}^{20}\left[\frac{\sin ((k+1)-k)}{\sin k\cdot \sin (k+1)}\right]$
$=\frac{1}{\sin 1}\sum _{k=1}^{20}(\cot k-\cot (k+1)$
$=\frac{\cot 1-\cot 21}{\sin 1}={\text{cosec}}^{2}1\text{cosec}(21)\cdot \sin 20$