$\begin{array}{l}Letf:R\to Rbedefinedasf(x+y)+f(x-y)=2f\left(x\right)f\left(y\right),f\left(\frac{1}{2}\right)=-1.Then,thevalueof\sum _{k=1}^{20}\frac{1}{\sin \left(k\right)\sin (k+f(k\left)\right)}\\ isequalto:\end{array}$

$$\begin{gathered} f\left(x\right)=\cos 2\mathrm{\pi x} \\ \mathrm{f}\left(\mathrm{k}\right)=\cos 2\mathrm{k\pi }=1 \end{gathered}$$
$$\begin{gathered} \sum _{k=1}^{20} \frac{1}{\sin \left(k\right) \sin (k+f(k\left)\right))} \\ =\sum _{k=1}^{20}\frac{1}{\sin {\displaystyle }{\displaystyle 1}}\left[\frac{\sin {\displaystyle (}{\displaystyle k}{\displaystyle +}{\displaystyle 1}{\displaystyle -}{\displaystyle k}{\displaystyle )}}{\sin \left(k\right) \sin (k+f(k)}\right] \\ =\sum _{k=1}^{20}\frac{{\displaystyle 1}}{{\displaystyle \sin 1}}\left[\frac{\sin {\displaystyle (}{\displaystyle k}{\displaystyle +}{\displaystyle 1)}{\displaystyle \cos }{\displaystyle k}{\displaystyle -}{\displaystyle \cos }{\displaystyle }{\displaystyle (}{\displaystyle k}{\displaystyle +}{\displaystyle 1}{\displaystyle ) \sin k}}{\sin \left(k\right) \sin (k+1)}\right] \\ =\sum _{k=1}^{20}\frac{{\displaystyle 1}}{{\displaystyle \sin 1}}\left[cot k-cot (k+1)\right] \\ =\frac{{\displaystyle 1}}{{\displaystyle \sin 1}}\left[cot 1-cot 21\right] \\ =\frac{{\displaystyle 1}}{{\displaystyle \sin 1}}\frac{\sin {\displaystyle }{\displaystyle 20}}{\sin {\displaystyle 1}{\displaystyle }{\displaystyle \sin }{\displaystyle }{\displaystyle 21}} \\ \cos e{c}^{2}1 \cos ec 21\sin 20 \end{gathered}$$