The quadratic equation whose roots are sin ^2 | vedclass

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Question

(A) Given roots are $\alpha = \sin ^2 18^{\circ}$ and $\beta = \cos ^2 36^{\circ}$.We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\cos 36^

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

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(A) Given roots are $\alpha = \sin ^2 18^{\circ}$ and $\beta = \cos ^2 36^{\circ}$.We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.Sum of roots $= \sin ^2 18^{\circ} + \cos ^2 36^{\circ} = \left(\frac{\sqrt{5}-1}{4}ight)^2 + \left(\frac{\sqrt{5}+1}{4}ight)^2$.$= \frac{5+1-2\sqrt{5}}{16} + \frac{5+1+2\sqrt{5}}{16} = \frac{12}{16} = \frac{3}{4}$.Product of roots $= \sin ^2 18^{\circ} \cdot \cos ^2 36^{\circ} = \left(\frac{\sqrt{5}-1}{4}ight)^2 \cdot \left(\frac{\sqrt{5}+1}{4}ight)^2$.$= \left(\frac{(\sqrt{5}-1)(\sqrt{5}+1)}{16}ight)^2 = \left(\frac{5-1}{16}ight)^2 = \left(\frac{4}{16}ight)^2 = \left(\frac{1}{4}ight)^2 = \frac{1}{16}$.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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