Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is C $-1$

$\sin x+{\sin }^{2}x=1$

$\sin x+(1-{\cos }^{2}x)=1$

$\sin x={\cos }^{2}x$

${\sin }^{2}x={\cos }^{4}x$

Now,

$\cos x^{12}+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x-2$

$={({\cos }^{4}x+{\cos }^{2}x)}^3-2$

$={(\sin x+{\sin }^{2}x)}^3-2$

$=(1{)}^3-2$

$=1-2$

$=-1$

Correct option is (C).