Question

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

• 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)

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