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

$$\begin{gathered} \cos 5\theta =\cos (4\theta +\theta ) \\ \cos 4\theta .\cos \theta -\sin 4\theta .\sin \theta \\ =(2{\cos }^{2}2\theta -1).\cos \theta -2\sin 2\theta .\cos 2\theta .\sin \theta \\ =[2{(2{\cos }^2\theta -1)}^2-1]\cos \theta -4\cos \theta (1-{\cos }^2\theta )(2{\cos }^2\theta -1) \\ =8{\cos }^5\theta -8{\cos }^3\theta +\cos \theta -8{\cos }^3\theta +4\cos \theta +8{\cos }^5\theta -4{\cos }^3\theta \\ =16{\cos }^5\theta -20{\cos }^3\theta +5\cos \theta \\ \therefore a=5,b=-20,c=16,d=0 \end{gathered}$$