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. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is A $4$

$\sin x+{\sin }^{2}x+{\sin }^{3}x=1$
$\sin x(1+{\sin }^{2}x)=1-{\sin }^{2}x$
$\sin x(2-{\cos }^{2}x)={\cos }^{2}x$
we square both the sides
${\sin }^{2}x(2-{\cos }^{2}x{)}^2={\cos }^{4}x$
$(1-{\cos }^{2}x)(2-{\cos }^{2}x{)}^2={\cos }^{4}x$
$(1-{\cos }^{2}x)(4-4{\cos }^{2}x+{\cos }^{4}x)={\cos }^{4}x$

$8{\cos }^{2}x+{\cos }^{6}x-4{\cos }^{4}x=4$