sin left( frac{pi}{10} right) sin left( frac{ | vedclass

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(C) We know that $\frac{\pi}{10} = 18^\circ$ and $\frac{3\pi}{10} = 54^\circ$. Thus,$\sin \left( \frac{\pi}{10} \right) \sin \left( \frac{3\pi}{10} \r

Vedclass · Accepted Answer

$1/4$

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(C) We know that $\frac{\pi}{10} = 18^\circ$ and $\frac{3\pi}{10} = 54^\circ$. Thus,$\sin \left( \frac{\pi}{10} ight) \sin \left( \frac{3\pi}{10} ight) = \sin 18^\circ \sin 54^\circ$. Since $\sin 54^\circ = \cos(90^\circ - 54^\circ) = \cos 36^\circ$,the expression becomes $\sin 18^\circ \cos 36^\circ$. Substituting the values $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$ and $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$: $\frac{\sqrt{5}-1}{4} 	imes \frac{\sqrt{5}+1}{4} = \frac{(\sqrt{5})^2 - 1^2}{16} = \frac{5-1}{16} = \frac{4}{16} = \frac{1}{4}$.

$\sin \left( \frac{\pi}{10} \right) \sin \left( \frac{3\pi}{10} \right) = $

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