Question

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

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

Answer 1

Answer: (A)

$4$

$\sin x+{\sin }^{2}x+{\sin }^{3}x=1$
$\Rightarrow \sin x(1+{\sin }^{2}x)=1-{\sin }^{2}x$
$\Rightarrow \sin x(2-{\cos }^{2}x)={\cos }^{2}x$
Squaring on both sides, we get
${\sin }^{2}x(2-{\cos }^{2}x{)}^2={\cos }^{4}x$
$\Rightarrow (1-{\cos }^{2}x)(4-4{\cos }^{2}x+{\cos }^{4}x)={\cos }^{4}x$
$\Rightarrow {\cos }^{6}x-4{\cos }^{4}x+8{\cos }^{2}x=4$.