Question

If $\sin x+{\sin }^{2}x=1$, then the value of ${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x-1$ is equal to

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

Answer: (A)

$0$

Using $\sin x+{\sin }^{2}x=1\Rightarrow \sin x=1-{\sin }^{2}x={\cos }^{2}x$

${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x-1={\sin }^{6}x+3{\sin }^{5}x+3{\sin }^{4}x+{\sin }^{3}x-1={\sin }^{3}x({\sin }^{3}x+3{\sin }^{2}x+3\sin x+1)-1={\sin }^{3}x{(\sin x+1)}^3-1={({\sin }^{2}x+\sin x)}^3-1=1^3-1=1-1=0$