Question

If $\sin x+{\sin }^{2}x=1$, then ${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x+2{\cos }^{4}x+{\cos }^{2}x-2$ is equal to

• A. $0$
• B. ${\cos }^{2}x$
• C. ${\sin }^{2}x$
• D. $-{\sin }^{2}x$

Answer 1

The correct option is C ${\sin }^{2}x$

Given that,

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

To find out,

The value of the expression: ${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x+2{\cos }^{4}x+{\cos }^{2}x-2$

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

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

$\Rightarrow \sin x={\cos }^{2}x[\because {\sin }^{2}x+{\cos }^{2}x=1]$

${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x+2{\cos }^{4}x+{\cos }^{2}x-2$

We can rewrite it as:

$({\cos }^{2}x{)}^6+3({\cos }^{2}x{)}^5+3({\cos }^{2}x{)}^4+({\cos }^{2}x{)}^3+2({\cos }^{2}x{)}^2+{\cos }^{2}x-2$

Now, substituting ${\cos }^{2}x=\sin x$ in the above expression, we get:

${\sin }^{6}x+3{\sin }^{5}x+3{\sin }^{4}x+{\sin }^{3}x+2{\sin }^{2}x+\sin x-2$

$\Rightarrow \left[\right({\sin }^{2}x{)}^3+(\sin x{)}^3+3\sin x{\sin }^{2}x({\sin }^{2}x+\sin x\left)\right]-2(1-{\sin }^{2}x)+\sin x$

We know that, $a^3+b^3+3ab(a+b)=(a+b{)}^3$

On applying the identity, the expression becomes,

$({\sin }^{2}x+\sin x{)}^3-2(1-{\sin }^{2}x)+\sin x$

Now, $\sin x+{\sin }^{2}x=1$ and $1-{\sin }^{2}x=\sin x$

Substituting the above values, the expression becomes,

$(1{)}^3-2(\sin x)+\sin x$

$\Rightarrow 1-2\sin x+\sin x$

$\Rightarrow 1-\sin x$

But, $1-\sin x={\sin }^{2}x[\because \sin x+{\sin }^{2}x=1]$

Hence, the expression reduces to,

${\sin }^{2}x$

$\therefore {\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x+2{\cos }^{4}x+{\cos }^{2}x-2={\sin }^{2}x$

Hence, option C is correct.