Question

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

• A. $2P_6-3P_4=-1$
• B. $2P_6-3P_4=1$
• C. $6P_{10}-15P_8+10P_6=0$
• D. $3P_6-2P_4+1=0$

Answer 1

The correct option is A $2P_6-3P_4=-1$

$p_m={\cos }^n\Theta +\sin \Theta$

$p_6={\cos }^6\Theta +{\sin }^6\Theta$

$=({\cos }^2\Theta {)}^3+({\sin }^2\Theta {)}^3$

$=({\cos }^2\Theta +{\sin }^2\Theta )({\cos }^4\Theta -{\cos }^2\Theta {\sin }^2\Theta +{\sin }^4\Theta )$

$={\cos }^4\Theta -{\cos }^2\Theta {\sin }^2\Theta +{\sin }^4\Theta ---\left(1\right)$

$P_m={\cos }^4\Theta +{\sin }^4\Theta ---\left(2\right)$

Now, 2p_6 -3p_4$

$=2{\cos }^4\Theta -2\cos \Theta {\sin }^2\Theta +2{\sin }^4\Theta -3{\cos }^4\Theta -3{\sin }^4\Theta$

$=-{\cos }^4\Theta -2{\cos }^2\Theta {\sin }^2\Theta -{\sin }^4\Theta$

$=-({\cos }^4\Theta +2{\cos }^2\Theta {\sin }^2\Theta +{\sin }^4\Theta )$

$=-({\cos }^2\Theta +{\sin }^2\Theta {)}^2$

$=-1-----proved$