If $\cos 5\theta =a\cos \theta +b{\cos }^3\theta +c{\cos }^5\theta +d,$ then

$\cos 5\theta =\cos (4\theta +\theta )=\cos 4\theta \cos \theta -\sin 4\theta \sin \theta$

           $\begin{array}{l}=\left(2{\cos }^{2}2\theta -1\right)\cos \theta -2\sin 2\theta \cos 2\theta \sin \theta \\ =\left[2{\left(2{\cos }^2\theta -1\right)}^2-1\right]\cos \theta -2\cdot 2\cos \theta {\sin }^2\theta \left(2{\cos }^2\theta -1\right)\\ =\left[2\left(4{\cos }^4\theta -4{\cos }^2\theta +1\right)-1\right]\cos \theta -4\cos \theta \left(2{\cos }^2\theta -1\right)\left(1-{\cos }^2\theta \right)\end{array}$

            $$\begin{gathered} =\cos \theta \left(8{\cos }^4\theta -8{\cos }^2\theta +1\right)-4\cos \theta \left(3{\cos }^2\theta -2{\cos }^4\theta -1\right) \\ \begin{array}{l}=\cos \theta \left(16{\cos }^4\theta -20{\cos }^2\theta +5\right)\\ =16{\cos }^5\theta -20{\cos }^3\theta +5\cos \theta \end{array} \end{gathered}$$

clearly, $a=5,b=-20,c=16\text{ and }d=0$ Satisfy the given relation.