If $f\left(x\right)=\cos \left[{\pi }^2\right]x+\cos \left[-{\pi }^2\right]xwhere\left[x\right]$ is the step function, then

$\begin{array}{l}f\left(x\right)=\cos \left[{\pi }^2\right]x+\cos \left[-{\pi }^2\right]x\\ =\cos \left[9.85\right]x+\cos \left[-9.85\right]x\\ =\cos 9x+\cos \left(-10\right)x\\ =\cos 9x+\cos 10x\\ f\left(10\right)=\cos 0+\cos 0=1+1=2\\ f\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)=\cos \frac{9\pi }{4}+\cos \frac{10\pi }{4}=\frac{1}{\sqrt{2}}+0=\frac{1}{\sqrt{2}}\\ f\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=\cos \frac{9\pi }{2}+\cos \frac{10\pi }{2}=0-1=-1\\ f\left(\pi \right)=\cos 9\pi +\cos 10\pi =-1+1=0\end{array}$