If $f\left(x\right)$ is odd polynomial function then $\cos \left(f\left(x\right)\right)$ is 

We know that if $f\left(-x\right)=f\left(x\right)$ then the function $f\left(x\right)$ is even function 

and if $f\left(-x\right)=-f\left(x\right)$ then the function $f\left(x\right)$ is odd function

Given that the function $f\left(x\right)$ is odd function, so that $f\left(-x\right)=-f\left(x\right)$

Consider 

$\cos \left(f\left(-x\right)\right)=\cos \left(-f\left(x\right)\right)$

we know that $\cos \left(-\theta \right)=\cos \theta$

Hence, $\cos \left(f\left(-x\right)\right)=\cos \left(-f\left(x\right)\right)=\cos \left(f\left(x\right)\right)$

Therefore, the function $\cos \left(f\left(x\right)\right)$ is even function