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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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?

Similar Questions

If $\alpha = \frac{180^{\circ}}{7}$,then $3 \sin \alpha - 4 \sin^{3} \alpha$ is equal to

If $\sin x + \cos x = \frac{1}{5},$ then $\tan 2x$ is

$\sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}$ is equal to

The value of $\cot\, x + \cot\, (60^\circ + x) + \cot\, (120^\circ + x)$ is equal to:

$\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module