sin 12^circ sin 48^circ sin 54^circ = | vedclass

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Question

(C) We know that $\sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5} + 1}{4}$.Consider the expression $E = \sin 12^\circ \sin 48^\circ \sin 54^\circ$.Usin

Vedclass · Accepted Answer

$1/8$

Vedclass · Answer

(C) We know that $\sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5} + 1}{4}$.Consider the expression $E = \sin 12^\circ \sin 48^\circ \sin 54^\circ$.Using the formula $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$:$E = \frac{1}{2} [\cos(48^\circ - 12^\circ) - \cos(48^\circ + 12^\circ)] \sin 54^\circ$$E = \frac{1}{2} [\cos 36^\circ - \cos 60^\circ] \sin 54^\circ$Since $\cos 60^\circ = 1/2$ and $\sin 54^\circ = \cos 36^\circ$:$E = \frac{1}{2} [\cos 36^\circ - 1/2] \cos 36^\circ$$E = \frac{1}{2} [\cos^2 36^\circ - \frac{1}{2} \cos 36^\circ]$Substitute $\cos 36^\circ = \frac{\sqrt{5} + 1}{4}$:$E = \frac{1}{2} [(\frac{\sqrt{5} + 1}{4})^2 - \frac{1}{2} (\frac{\sqrt{5} + 1}{4})]$$E = \frac{1}{2} [\frac{5 + 1 + 2\sqrt{5}}{16} - \frac{\sqrt{5} + 1}{8}]$$E = \frac{1}{2} [\frac{6 + 2\sqrt{5}}{16} - \frac{2\sqrt{5} + 2}{16}]$$E = \frac{1}{2} [\frac{4}{16}] = \frac{1}{2} 	imes \frac{1}{4} = \frac{1}{8}$.

$\sin 12^\circ \sin 48^\circ \sin 54^\circ = $

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