Question

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

• A. $4$
• B. $1$
• C. $0$
• D. None of the above

Answer 1

The correct option is A $4$

$\sin x+{\sin }^{2}x+{\sin }^{3}x=1$
$\Rightarrow {\cos }^{2}x=\sin x+{\sin }^{3}x$
Squaring both side
$\Rightarrow {\cos }^{4}x={\sin }^{2}x+{\sin }^{6}x+2{\sin }^{4}x$
$\Rightarrow {\cos }^{4}x=1-{\cos }^{2}x+(1-{\cos }^{2}x{)}^3+2(1-{\cos }^{2}x{)}^2$
$\Rightarrow {\cos }^{6}x-4{\cos }^{4}x+8{\cos }^{2}x=4$