The quadratic equation whose roots are sin ^2 | vedclass

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(C) We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.$\sin ^2 18^{\circ} = \left(\frac{\sqrt{5}-1}{4

Vedclass · Accepted Answer

$16 x^2-12 x+1=0$

Vedclass · Answer

(C) We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.$\sin ^2 18^{\circ} = \left(\frac{\sqrt{5}-1}{4}ight)^2 = \frac{5+1-2\sqrt{5}}{16} = \frac{6-2\sqrt{5}}{16} = \frac{3-\sqrt{5}}{8}$.$\cos ^2 36^{\circ} = \left(\frac{\sqrt{5}+1}{4}ight)^2 = \frac{5+1+2\sqrt{5}}{16} = \frac{6+2\sqrt{5}}{16} = \frac{3+\sqrt{5}}{8}$.The sum of the roots is $\frac{3-\sqrt{5}}{8} + \frac{3+\sqrt{5}}{8} = \frac{6}{8} = \frac{3}{4}$.The product of the roots is $\left(\frac{3-\sqrt{5}}{8}ight) \left(\frac{3+\sqrt{5}}{8}ight) = \frac{9-5}{64} = \frac{4}{64} = \frac{1}{16}$.The quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^2 - \frac{3}{4}x + \frac{1}{16} = 0$.Multiplying by $16$,we get $16x^2 - 12x + 1 = 0$.

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

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