If alpha = sin 36^{circ},then alpha is a root | vedclass

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(C) We know that $\cos 72^{\circ} = \frac{\sqrt{5}-1}{4}$.Using the identity $\cos 2	heta = 1 - 2\sin^{2}	heta$,we have $\cos 72^{\circ} = 1 - 2\sin

Vedclass · Accepted Answer

$16 x^{4}-20 x^{2}+5=0$

Vedclass · Answer

(C) We know that $\cos 72^{\circ} = \frac{\sqrt{5}-1}{4}$.Using the identity $\cos 2	heta = 1 - 2\sin^{2}	heta$,we have $\cos 72^{\circ} = 1 - 2\sin^{2} 36^{\circ}$.Substituting $\alpha = \sin 36^{\circ}$,we get $1 - 2\alpha^{2} = \frac{\sqrt{5}-1}{4}$.Multiplying by $4$,we get $4 - 8\alpha^{2} = \sqrt{5} - 1$.Rearranging terms,$5 - 8\alpha^{2} = \sqrt{5}$.Squaring both sides,$(5 - 8\alpha^{2})^{2} = 5$.$25 + 64\alpha^{4} - 80\alpha^{2} = 5$.$64\alpha^{4} - 80\alpha^{2} + 20 = 0$.Dividing by $4$,we get $16\alpha^{4} - 20\alpha^{2} + 5 = 0$.Thus,$\alpha$ is a root of $16x^{4} - 20x^{2} + 5 = 0$.

If $\alpha = \sin 36^{\circ}$,then $\alpha$ is a root of which of the following equations?

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