Question

If $P_n={\cos }^n\theta +{\sin }^n\theta ,$ then $2p_6-3p_4+5=..........$

• A. 3
• B. 4
• C. 0
• D. 1

Answer 1

The correct option is B 4

Given $P_n={\cos }^n\theta +{\sin }^n\theta$

$\Rightarrow 2P_6-3P_4+5$

$\Rightarrow 2({\cos }^6\theta +{\sin }^6\theta )-3({\cos }^4\theta +{\sin }^4\theta )+5$

$\Rightarrow 2\left[\right({\cos }^2\theta {)}^3+\left({\sin }^2\theta {)}^3\right]-3({\cos }^4\theta +{\sin }^4\theta )+5$

$\Rightarrow 2\left[\right({\sin }^2\theta +{\cos }^2\theta \left)\right({\sin }^4\theta -{\sin }^2{\cos }^2\theta +{\cos }^4\theta \left)\right]-3({\cos }^4\theta +{\sin }^4\theta )+5$

$\Rightarrow 2\left[\right({\sin }^4\theta +{\cos }^4\theta )-{\sin }^2\theta {\cos }^2\theta ]-3({\cos }^4\theta +{\sin }^4\theta )+5$

$\Rightarrow 2({\sin }^4\theta +{\cos }^4\theta )-2{\sin }^2\theta {\cos }^2\theta -3({\cos }^4\theta +{\sin }^4\theta )+5$

`$\Rightarrow -[{\sin }^4\theta +{\cos }^4\theta ]-2{\sin }^2\theta {\cos }^2\theta +5$

$\Rightarrow -\left[\right({\sin }^2\theta {)}^2+\left({\cos }^2\theta {)}^2\right]-2{\sin }^2\theta {\cos }^2\theta +5$

$\Rightarrow -\left[\right({\sin }^2\theta +{\cos }^2\theta )-2{\sin }^2\theta {\cos }^2\theta ]-2{\sin }^2\theta {\cos }^2\theta +5$

$\Rightarrow -(1{)}^2+2{\sin }^2\theta {\cos }^2\theta -2{\sin }^2\theta {\cos }^2\theta +5$

$\Rightarrow 4$