Question

In $A_n={\cos }^n\theta +{\sin }^n\theta ,nϵN$ and $\theta \in R$

The value of $2A_{10}-5A_8+6A_6-4A_4+10A_2-A_0$ is equal to

• A. $10$
• B. $9$
• C. $8$
• D. $6$

Answer 1

Answer: (B)

$9$

As $A_n={\sin }^n\theta +{\cos }^n\theta$

$\therefore A_0=2,A_2=1$

$A_n={\cos }^n\theta -{\sin }^n\theta$

$\Rightarrow A_{n-2}={\cos }^{n-2}\theta -{\sin }^{n-2}\theta$

$\therefore A_n-A_{n-2}={\cos }^n\theta -{\cos }^{n-2}\theta -{\sin }^n\theta +{\sin }^{n-2}\theta$

$={\cos }^{n-2}({\cos }^2\theta -1)-{\sin }^{n-2}\theta ({\sin }^2\theta -1)$

$={\cos }^2\theta {\sin }^2\theta ({\cos }^{n-4}\theta -{\sin }^{n-4}\theta )$

$={\cos }^2\theta {\sin }^2\theta A_{n-4}$

$\therefore A_n-A_{n-2}=-{\sin }^2\theta {\cos }^2\theta A_{n-4}=-\lambda A_{n-4}$

(Where $\lambda ={\sin }^2\theta {\cos }^2\theta$)

Substitute $n=4,6,8,10$

$A_4=A_2-\lambda A_0=1-2\lambda$

$A_6=A_4-\lambda A_2=1-3\lambda$

$A_8=A_6-\lambda A_4=2{\lambda }^2-4\lambda +1$

$A_{10}=A_8-\lambda A_6=5{\lambda }^2-5\lambda +1$

$\therefore 2A_{10}-5A_8+6A_6-4A_4+12A^2-A_0$

$=2(5{\lambda }^2-5\lambda +1)-5(2{\lambda }^2-4\lambda +1)+6(1-3\lambda )-4(1-2\lambda )+12-2$

$=2(5{\lambda }^2-5{\lambda }^2)(28\lambda -28\lambda )+(2-5+6-4+10)=9$