Question

If $\cos x+{\cos }^{2}x=1$, then the value of ${\sin }^{4}x+{\sin }^{6}x$ is

• A. $-1+\sqrt{5}$
• B. $\frac{-1-\sqrt{5}}{2}$
• C. $\frac{1-\sqrt{5}}{2}$
• D. $\frac{-1+\sqrt{5}}{2}$
• E. $1-\sqrt{5}$

Answer 1

Answer: (D)

$\frac{-1+\sqrt{5}}{2}$

Given, $\cos x+{\cos }^{2}x=1$ .... (i)
$\Rightarrow {\cos }^{2}x+\cos x-1=0$
Therefore, $\cos x=\frac{-1\pm \sqrt{1+4}}{2}(\because$ quadratic in $\cos x)$
$=\frac{-1\pm \sqrt{5}}{2}=\frac{-1+\sqrt{5}}{2}(\because \cos x\ne \frac{-1-\sqrt{5}}{2})$
Also from Eq. (i),
$\cos x=1-{\cos }^{2}x$
$\Rightarrow \cos x={\sin }^{2}x$ .... (ii)
Now, ${\sin }^{4}x+{\sin }^{6}x$
$={\cos }^{2}x+{\cos }^{3}x$
$={\cos }^{2}x(1+\cos x)$
$={\left(\frac{-1+\sqrt{5}}{2}\right)}^2(1+\frac{-1+\sqrt{5}}{2})$
$={\left(\frac{-1+\sqrt{5}}{2}\right)}^2\left(\frac{1+\sqrt{5}}{2}\right)$
$=\left(\frac{-1+\sqrt{5}}{2}\right)\left(\frac{5-1}{4}\right)=\left(\frac{-1+\sqrt{5}}{2}\right)\times 1$
$=\frac{-1+\sqrt{5}}{2}$