The quadratic equation whose roots are $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$ is :
The quadratic equation whose roots are $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$ is :
$16 x^2-12 x+1=0$
$16 x^2+12 x+1=0$
$16 x^2-12 x-1=0$
$16 x^2+10 x+1=0$
Solution
Since $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$ are the roots of a quadratic equation.
$\therefore \text { Sum of roots }=\sin ^2 18^{\circ}+\cos ^2 36^{\circ}$
$=\left(\frac{\sqrt{5}-1}{4}\right)^2+\left(\frac{\sqrt{5}+1}{4}\right)^2$
$=\frac{5+1-2 \sqrt{5}}{16}+\frac{5+1+2 \sqrt{5}}{16}$
$=\frac{12}{16}=\frac{3}{4}$
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
$x^2-(\text { sum of roots }) x+(\text { product of roots })=0$
$\Rightarrow \quad x^2-\frac{3}{4} x+\frac{1}{16}=0$
$\Rightarrow \quad 16 x^2-12 x+1=0$