sin 12^circ sin 24^circ sin 48^circ sin 84^ci | vedclass

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Question

(A) We need to evaluate the product $P = \sin 12^\circ \sin 24^\circ \sin 48^\circ \sin 84^\circ$.Using the formula $2 \sin A \sin B = \cos(A-B) - \co

Vedclass · Accepted Answer

$\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ $

Vedclass · Answer

(A) We need to evaluate the product $P = \sin 12^\circ \sin 24^\circ \sin 48^\circ \sin 84^\circ$.Using the formula $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:$P = \frac{1}{4} (2 \sin 12^\circ \sin 48^\circ) (2 \sin 24^\circ \sin 84^\circ)$$P = \frac{1}{4} (\cos 36^\circ - \cos 60^\circ) (\cos 60^\circ - \cos 108^\circ)$Since $\cos 60^\circ = \frac{1}{2}$,$\cos 36^\circ = \frac{\sqrt{5}+1}{4}$,and $\cos 108^\circ = -\sin 18^\circ = -\frac{\sqrt{5}-1}{4}$:$P = \frac{1}{4} (\frac{\sqrt{5}+1}{4} - \frac{1}{2}) (\frac{1}{2} + \frac{\sqrt{5}-1}{4}) = \frac{1}{4} (\frac{\sqrt{5}-1}{4}) (\frac{\sqrt{5}+1}{4}) = \frac{1}{4} \cdot \frac{5-1}{16} = \frac{1}{16}$.Now,evaluate option $A$: $\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$.Using the identity $\cos 	heta \cos(60^\circ - 	heta) \cos(60^\circ + 	heta) = \frac{1}{4} \cos 3	heta$:$\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{4} \cos(3 	imes 20^\circ) = \frac{1}{4} \cos 60^\circ = \frac{1}{4} 	imes \frac{1}{2} = \frac{1}{8}$.Thus,$\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{8} 	imes \frac{1}{2} = \frac{1}{16}$.Both expressions equal $\frac{1}{16}$,so option $A$ is correct.

$\sin 12^\circ \sin 24^\circ \sin 48^\circ \sin 84^\circ = $

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