Question

If $P_n={\cos }^n\theta +{\sin }^n\theta ,$ then prove that $2P_6-3P_4+1=0$ and $6P_{10}-15P_8+10P_6-1=0$.

Answer 1

$P_n-P_{n-1}={\cos }^n\theta +{\sin }^n\theta -{\cos }^{n-2}\theta -{\sin }^{n-2}\theta ={\cos }^{n-2}\theta \left({\cos }^2\theta \right)+{\sin }^{n-2}\theta ({\sin }^2\theta -1)=-{\cos }^2\theta {\sin }^2\theta P_{n-4}\therefore P_6-P_4=-{\cos }^2\theta {\sin }^2\theta P_2{P}_4-P_2=-{\cos }^2\theta {\sin }^2\theta P_0=-2{\cos }^2\theta {\sin }^2\theta P_4-({\cos }^2\theta +{\sin }^2\theta )=-2{\sin }^2\theta {\cos }^2\theta P_4=1-2{\sin }^2\theta {\cos }^2\theta \therefore P_6=-{\cos }^2\theta {\sin }^2\theta +1-2{\sin }^2\theta {\cos }^2\theta =-1-3{\cos }^2\theta {\sin }^2\theta 2P_6-3P_4+1=2(1-3{\cos }^2\theta {\sin }^2\theta )-3(1-2{\sin }^2\theta {\cos }^2\theta )=2-3+1=06P_{10}-15P_8+10P_6-1=6(-{\cos }^2\theta {\sin }^2\theta P_6+P_8)-15(-{\cos }^2\theta {\sin }^2\theta P_4+P_6)+10P_6-1=6(-{\cos }^2\theta {\sin }^2\theta (1-3{\sin }^2\theta {\cos }^2\theta ))+(-{\cos }^2\theta {\sin }^2\theta (P_4+P_6))-15(-{\cos }^2\theta {\sin }^2\theta (1-2{\sin }^2\theta {\cos }^2\theta )+1-3{\sin }^2\theta {\cos }^2\theta )+10(1-3{\cos }^2\theta {\sin }^2\theta )-1=0$