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Q.
The quadratic equation whose roots are $\sin ^{2} 18^{\circ}$ and $\cos ^{2} 36^{\circ}$ is :
EAMCETEAMCET 2006
Solution:
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) $=0$
$\Rightarrow x^{2}-\frac{3}{4} x+\frac{1}{16}=0$
$\Rightarrow 16 x^{2}-12 x+1=0$