Question

$P_n={\cos }^n\theta +{\sin }^n\theta$, then $2P_6-3P_4+1=$

• A. $2$
• B. $3$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

$P_n=co{s}^n\theta +si{n}^n\theta$
$f\left(\theta \right)=2P_6-3P_4+1=2co{s}^6\theta +2si{n}^6\theta -3co{s}^4\theta -3si{n}^4\theta +1$
$=2(co{s}^6\theta +si{n}^6\theta )-3\left(\right(co{s}^2\theta +si{n}^2\theta {)}^2-2co{s}^2\theta si{n}^2\theta )+1$
$=2(co{s}^6\theta +si{n}^6\theta )+6co{s}^2\theta si{n}^2\theta -2$
$=2\left[\right(co{s}^2\theta +si{n}^2\theta {)}^3-3co{s}^2\theta si{n}^2\theta \left(1\right)]+6co{s}^2\theta si{n}^2\theta -2$
$=2-6co{s}^2\theta si{n}^2\theta +6co{s}^2\theta si{n}^2\theta -2$
$f\left(\theta \right)=0$
$[\because a^3+b^3=(a+b{)}^3-3ab(a+b),a^2+b^2=(a+b{)}^2-2ab,si{n}^2\theta +co{s}^2\theta =1]$