Question

If $p_n={\cos }^{n}x+{\sin }^{n}x$, then $2p_6-3p_4+1$ is equal to

Answer 1

$P_n={\cos }^{n}x+{\sin }^{n}x$

$\therefore P_6={\cos }^{6}x+{\sin }^{6}x$

$\therefore P_y={\cos }^{4}x+{\sin }^{4}x$

$\therefore$ putting $x=0$, we get

$P_6=1+0=1$

$P_4=1+0=1$

$\therefore {}^2{P}_6-{}^3{P}_4+1=2-3+1$

$=3-3$

$=0$