Question

If $\cos x+{\cos }^{2}x=1$ then the value of $4{\sin }^{2}x(2-{\cos }^{2}x)$

Answer 1

Given, $\cos x+{\cos }^{2}x=1$

$\Rightarrow {\cos }^{2}x+\cos x-1=0$

$\Rightarrow \cos x=\frac{-1\pm \sqrt{1+4}}{2}$.

But, $\cos x\ne \frac{-1-\sqrt{5}}{2}$ {Because, it is less than$-1$].

$\therefore \cos x=\frac{-1+\sqrt{5}}{2}$

$\Rightarrow {\cos }^{2}x=\frac{(\sqrt{5}-1{)}^2}{4}=\frac{5+1-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}$ and

${\sin }^{2}x=1-{\cos }^{2}x=1-\frac{(3-\sqrt{5})}{2}=\frac{(\sqrt{5}-1)}{2}$.

Now, $4.{\sin }^{2}x(2-{\cos }^{2}x)$

$=4.\left(\frac{\sqrt{5}-1}{2}\right).\left(2-\frac{3-\sqrt{5}}{2}\right)$

$=2.(\sqrt{5}-1).\left(\frac{4-3+\sqrt{5}}{2}\right)$

$=(\sqrt{5}-1).(\sqrt{5}+1)=4$