Question

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

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

Answer 1

The correct option is D ${\text{cosec}}^2\left(1\right)\text{cosec}\left(21\right)\sin \left(20\right)$

$\because f(x+y)+f(x-y)=2f\left(x\right).f\left(y\right)\therefore f\left(x\right)=\cos \left(\lambda x\right)$
Given that $f\left(\frac{1}{2}\right)=-1,$
$\cos \left(\frac{\lambda }{2}\right)=-1\Rightarrow \lambda =2n\pi ,n\in I$
$\therefore f\left(x\right)=\cos \left(2\pi x\right)\Rightarrow f\left(k\right)=1,k\in I$

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