Question

The quadratic equation whose roots are ${\sin }^{2}18^{\circ }$ and ${\cos }^{2}36^{\circ }$ is :

• A. $16x^2-12x+1=0$
• B. $16x^2+12x+1=0$
• C. $16x^2-12x-1=0$
• D. $16x^2+10x+1=0$

Answer 1

Answer: (A)

Since ${\sin }^{2}18^{\circ }$ and ${\cos }^{2}36^{\circ }$ are the roots of a quadratic equation.
$\therefore$ Sum of roots $={\sin }^{2}18^{\circ }+{\cos }^{2}36^{\circ }$
$={\left(\frac{\sqrt{5}-1}{4}\right)}^2+{\left(\frac{\sqrt{5}+1}{4}\right)}^2$
$=\frac{5+1-2\sqrt{5}}{16}+\frac{5+1+2\sqrt{5}}{16}$
$=\frac{12}{16}=\frac{3}{4}$
and product of roots $={\sin }^{2}18^{\circ }\cdot {\cos }^{2}36^{\circ }$
$={\left(\frac{\sqrt{5}-1}{4}\right)}^2{\left(\frac{\sqrt{5}+1}{4}\right)}^2$
$={\left(\frac{5-1}{4\times 4}\right)}^2={\left(\frac{1}{4}\right)}^2=\frac{1}{16}$
Required equation whose roots are ${\sin }^{2}18^{\circ }$ and ${\cos }^{2}36^{\circ }$, is
$x^2-($ sum of roots $)x+$ (product of roots) $=0$
$\Rightarrow x^2-\frac{3}{4}x+\frac{1}{16}=0$
$\Rightarrow 16x^2-12x+1=0$