Question

If $A(n)={\sin }^n\alpha +{\cos }^n\alpha$, then
$A(1)A(4)+A(2)A(5)=$

• A. $A(1)A(2)+A(4)A(5)$
• B. $A(1)A(6)+A(2)A(3)$
• C. $A(1)A(3)+A(2)A(6)$
• D. $A(1)A(2)+A(3)A(6)$

Answer 1

Answer: (B)

$\because A(n)={\sin }^n\alpha +{\cos }^n\alpha$
Then, $A(1)A(4)+A(2)A(5)$
$=(\sin \alpha +\cos \alpha )\left({\sin }^4\alpha +{\cos }^4\alpha \right)$
$+\left({\sin }^2\alpha +{\cos }^2\alpha \right)\left({\sin }^5\alpha +{\cos }^5\alpha \right)$

$=(\sin \alpha +\cos \alpha )[{\left({\sin }^2\alpha +{\cos }^2\alpha \right)}^2$
$-2{\sin }^2\alpha {\cos }^2\alpha ]+\left({\sin }^5\alpha +{\cos }^5\alpha \right)$
$=(\sin \alpha +\cos \alpha )\left[{\left({\sin }^2\alpha +{\cos }^2\alpha \right)}^3-2{\sin }^2\alpha {\cos }^2\alpha \right]$
$+\left({\sin }^5\alpha +{\cos }^5\alpha \right)$
$=(\sin \alpha +\cos \alpha )\left[{\sin }^6\alpha +{\cos }^6\alpha +{\sin }^2\alpha {\cos }^2\alpha \right]$
$+\left({\sin }^5\alpha +{\cos }^5\alpha \right)$
$=(\sin \alpha +\cos \alpha )\left({\sin }^6\alpha +{\cos }^6\alpha \right)$
$+{\sin }^2\alpha {\cos }^2\alpha (\sin \alpha +\cos \alpha )+\left({\sin }^5\alpha +{\cos }^5\alpha \right)$
$=(\sin \alpha +\cos \alpha )\left({\sin }^6\alpha +{\cos }^6\alpha \right)$
$+\left({\sin }^3\alpha {\cos }^2\alpha +{\sin }^5\alpha \right)+\left({\sin }^2\alpha {\cos }^3\alpha +{\cos }^5\alpha \right)$
$=(\sin \alpha +\cos \alpha )\left({\sin }^6\alpha +{\cos }^6\alpha \right)$
$+\left({\sin }^2\alpha +{\cos }^2\alpha \right)\left({\sin }^3\alpha +{\cos }^3\alpha \right)$
$=A(1)A(6)+A(2)A(3)$