16 sin 12^{circ} cos 18^{circ} sin 48^{circ} | vedclass

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Question

(A) We have the expression $E = 16 \sin 12^{\circ} \cos 18^{\circ} \sin 48^{\circ}$.Using the formula $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,we get

Vedclass · Accepted Answer

$\sqrt{10-2 \sqrt{5}}$

Vedclass · Answer

(A) We have the expression $E = 16 \sin 12^{\circ} \cos 18^{\circ} \sin 48^{\circ}$.Using the formula $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$,we get:$E = 8 \cos 18^{\circ} [2 \sin 48^{\circ} \sin 12^{\circ}]$$E = 8 \cos 18^{\circ} [\cos(48^{\circ}-12^{\circ}) - \cos(48^{\circ}+12^{\circ})]$$E = 8 \cos 18^{\circ} [\cos 36^{\circ} - \cos 60^{\circ}]$Since $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\cos 60^{\circ} = \frac{1}{2}$,we have:$E = 8 \cos 18^{\circ} [\frac{\sqrt{5}+1}{4} - \frac{1}{2}] = 8 \cos 18^{\circ} [\frac{\sqrt{5}+1-2}{4}] = 8 \cos 18^{\circ} [\frac{\sqrt{5}-1}{4}]$$E = 2 (\sqrt{5}-1) \cos 18^{\circ}$.Using $\cos 18^{\circ} = \frac{\sqrt{10+2\sqrt{5}}}{4}$,we get:$E = 2 (\sqrt{5}-1) \frac{\sqrt{10+2\sqrt{5}}}{4} = \frac{\sqrt{5}-1}{2} \sqrt{10+2\sqrt{5}}$.Squaring the expression: $E^2 = \frac{(\sqrt{5}-1)^2}{4} (10+2\sqrt{5}) = \frac{5+1-2\sqrt{5}}{4} (10+2\sqrt{5}) = \frac{6-2\sqrt{5}}{4} (10+2\sqrt{5}) = \frac{3-\sqrt{5}}{2} (10+2\sqrt{5}) = \frac{30+6\sqrt{5}-10\sqrt{5}-10}{2} = \frac{20-4\sqrt{5}}{2} = 10-2\sqrt{5}$.Thus,$E = \sqrt{10-2\sqrt{5}}$.

$16 \sin 12^{\circ} \cos 18^{\circ} \sin 48^{\circ} = $

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