Question

$\text{If}{\text{u}}_n={\cos }^n\theta +{\sin }^n\theta ,\text{then 2}{\text{u}}_6-3{\text{u}}_4+1=0.$

• A. True
• B. False

Answer 1

The correct option is A True

Given

$u_n={\cos }^n\theta +{\sin }^n\theta$

$u_6={\cos }^6\theta +{\sin }^6\theta$

$u_6=({\cos }^2\theta {)}^3+({\sin }^2\theta {)}^3$

$u_6=({\cos }^2\theta +{\sin }^2\theta {)}^3-3({\cos }^2\theta \left)\right({\sin }^2\theta \left)\right({\cos }^2\theta +{\sin }^2\theta )$

$u_6=1-3{\cos }^2\theta {\sin }^2\theta \cdots \left(1\right)$

$u_4={\cos }^4\theta +{\sin }^4\theta$

$u_4=({\cos }^2\theta {)}^2+({\sin }^2\theta {)}^2$

$u_4=({\cos }^2\theta +{\sin }^2\theta {)}^2-2({\cos }^2\theta \left)\right({\sin }^2\theta )$

$u_4=1-2{\cos }^2\theta {\sin }^2\theta \cdots \left(2\right)$

$2u_6-3u_4+1$

From $\left(1\right)\&\left(2\right)$

$⟹2(1-3{\cos }^2\theta {\sin }^2\theta )-3(1-2{\cos }^2\theta {\sin }^2\theta )+1$

$⟹2-6{\cos }^2\theta {\sin }^2\theta )-3+6{\cos }^2\theta {\sin }^2\theta )+1$

$⟹2-3+1$

$⟹0$