sin 36^circ sin 72^circ sin 108^circ sin 144^ | vedclass

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(D) We know that $\sin 108^\circ = \sin(180^\circ - 72^\circ) = \sin 72^\circ$ and $\sin 144^\circ = \sin(180^\circ - 36^\circ) = \sin 36^\circ$.So,th

Vedclass · Accepted Answer

$5/16$

Vedclass · Answer

(D) We know that $\sin 108^\circ = \sin(180^\circ - 72^\circ) = \sin 72^\circ$ and $\sin 144^\circ = \sin(180^\circ - 36^\circ) = \sin 36^\circ$.So,the expression becomes $\sin^2 36^\circ \sin^2 72^\circ$.Using the identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we have:$\sin^2 36^\circ \sin^2 72^\circ = \left(\frac{1 - \cos 72^\circ}{2}ight) \left(\frac{1 - \cos 144^\circ}{2}ight)$$= \frac{1}{4} (1 - \cos 72^\circ)(1 - \cos 144^\circ)$$= \frac{1}{4} (1 - \sin 18^\circ)(1 + \cos 36^\circ)$Substituting the values $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$ and $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$:$= \frac{1}{4} \left(1 - \frac{\sqrt{5}-1}{4}ight) \left(1 + \frac{\sqrt{5}+1}{4}ight)$$= \frac{1}{4} \left(\frac{5-\sqrt{5}}{4}ight) \left(\frac{5+\sqrt{5}}{4}ight)$$= \frac{1}{4} \left(\frac{25-5}{16}ight) = \frac{1}{4} 	imes \frac{20}{16} = \frac{5}{16}$.

$\sin 36^\circ \sin 72^\circ \sin 108^\circ \sin 144^\circ = $

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