Question

If $\sin x+{\sin }^{2}x=1$, then
${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x+2{\cos }^{4}x+{\cos }^{2}x-2=$

• A. $0$
• B. $1$
• C. ${\sin }^{2}x$
• D. ${\cos }^{2}x$

Answer 1

The correct option is C ${\sin }^{2}x$

$\sin x+{\sin }^{2}x=1\Rightarrow \sin x={\cos }^{2}x\therefore {\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x+2{\cos }^{4}x+{\cos }^{2}x-2$
$={\sin }^{6}x+3{\sin }^{5}x+3{\sin }^{4}x+{\sin }^{3}x+{\sin }^{2}x+\sin x-2$
$=[{\sin }^{6}x+{\sin }^{3}x+3{\sin }^{5}x+3{\sin }^{4}x]+[{\sin }^{2}x+\sin x]-2+{\sin }^{2}x$
$=\left[\right({\sin }^{2}x{)}^3+(\sin x{)}^3+3({\sin }^{2}x\left)\right(\sin x\left)\right({\sin }^{2}x+\sin x\left)\right]+1-2+{\sin }^{2}x=(\sin x+{\sin }^{2}x{)}^3-1+{\sin }^{2}x=1+1-2+{\sin }^{2}x={\sin }^{2}x$