If $P_n={\cos }^n\theta +{\sin }^n\theta ,$ then 

$\begin{array}{l}2P_6-3P_4+1\\ =2\left({\cos }^6\theta +{\sin }^6\theta \right)-3\left({\cos }^4\theta +{\sin }^4\theta \right)+1\\ =2\left[{\left({\cos }^2\theta +{\sin }^2\theta \right)}^3-3{\sin }^2\theta {\cos }^2\theta \right.\left.\left({\cos }^2\theta +{\sin }^2\theta \right)\right]\end{array}$

$\begin{array}{l}-3\left[{\left({\cos }^2\theta +{\sin }^2\theta \right)}^2-2{\sin }^2\theta {\cos }^2\theta \right]+1\\ =2\left(1-3{\sin }^2\theta {\cos }^2\theta \right)-3\left(1-2{\sin }^2\theta {\cos }^2\theta \right)+1=0\end{array}$

Again for $n \ge 4,$ we have

$P_n-P_{n-2}$

       $\begin{array}{l}={\cos }^n\theta +{\sin }^n\theta -\left({\cos }^{n-2}\theta +{\sin }^{n-2}\theta \right)\\ ={\cos }^{n-2}\theta \left({\cos }^2\theta -1\right)+{\sin }^{n-2}\theta \left({\sin }^2\theta -1\right)\\ \begin{array}{l}=-{\sin }^2\theta {\cos }^{n-2}\theta -{\cos }^2\theta {\sin }^{n-2}\theta \\ =-{\sin }^2\theta {\cos }^2\theta \left({\cos }^{n-4}\theta +{\sin }^{n-4}\theta \right)\\ =-{\sin }^2\theta {\cos }^2\theta P_{n-4}6P_{10}-15P_8+10P_6-1\end{array}\end{array}$

       $$\begin{gathered} =6\left(P_{10}-P_8\right)-9\left(P_8-P_6\right)+\left(P_6-P_4\right)+P_4-P_2 \\ \begin{array}{l}=-{\sin }^2\theta {\cos }^2\theta \left(6P_6-9P_4+P_2+P_0\right)\\ =-3{\sin }^2\theta {\cos }^2\theta \left(2P_6-3P_4\right)-{\sin }^2\theta {\cos }^2\theta (1+2)\\ \left[\therefore P_2=1,P_0=2\right] \end{array} \end{gathered}$$

      $=-3{\sin }^2\theta {\cos }^2\theta (-1)-3{\sin }^2\theta {\cos }^2\theta =0.$

               $\left[\therefore 2P_6-3P_4+1=0\text{ (as proved) }\right]$