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The quadratic equation whose roots are $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$ is
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1685 Upvotes
Verified Answer
The correct answer is:
$16 x^2-12 x+1=0$
$\sin ^2 18^{\circ}=\left(\frac{\sqrt{5}-1}{4}\right)^2=\frac{6-2 \sqrt{5}}{16}=\frac{3-\sqrt{5}}{8}$
$$
\cos ^2 36^{\circ}=\left(\frac{\sqrt{5}+1}{4}\right)^2=\frac{3+\sqrt{5}}{8}
$$
$\therefore$ Required equation is
$$
\begin{aligned}
& x^2-\left(\frac{3-\sqrt{5}}{8}+\frac{3+\sqrt{5}}{8}\right) x+\left(\frac{3-\sqrt{5}}{8} \times \frac{3+\sqrt{5}}{8}\right)=0 \\
& x^2-\frac{3}{4} x+\frac{1}{16}=0 \Rightarrow 16 x^2-12 x+1=0
\end{aligned}
$$
$$
\cos ^2 36^{\circ}=\left(\frac{\sqrt{5}+1}{4}\right)^2=\frac{3+\sqrt{5}}{8}
$$
$\therefore$ Required equation is
$$
\begin{aligned}
& x^2-\left(\frac{3-\sqrt{5}}{8}+\frac{3+\sqrt{5}}{8}\right) x+\left(\frac{3-\sqrt{5}}{8} \times \frac{3+\sqrt{5}}{8}\right)=0 \\
& x^2-\frac{3}{4} x+\frac{1}{16}=0 \Rightarrow 16 x^2-12 x+1=0
\end{aligned}
$$
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