cos 6^{circ} sin 24^{circ} cos 72^{circ} = | vedclass

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Question

(C) We have the expression $E = \cos 6^{\circ} \sin 24^{\circ} \cos 72^{\circ}$.Since $\cos 72^{\circ} = \sin 18^{\circ}$,we can write $E = \cos 6^{\c

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(C) We have the expression $E = \cos 6^{\circ} \sin 24^{\circ} \cos 72^{\circ}$.Since $\cos 72^{\circ} = \sin 18^{\circ}$,we can write $E = \cos 6^{\circ} \sin 24^{\circ} \sin 18^{\circ}$.Multiply and divide by $2$: $E = \frac{1}{2} (2 \sin 24^{\circ} \cos 6^{\circ}) \sin 18^{\circ}$.Using the identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$E = \frac{1}{2} (\sin 30^{\circ} + \sin 18^{\circ}) \sin 18^{\circ}$.$E = \frac{1}{2} (\frac{1}{2} \sin 18^{\circ} + \sin^2 18^{\circ})$.Given $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$,then $\sin^2 18^{\circ} = \frac{5+1-2\sqrt{5}}{16} = \frac{6-2\sqrt{5}}{16} = \frac{3-\sqrt{5}}{8}$.Substituting these values:$E = \frac{1}{2} (\frac{1}{2} \cdot \frac{\sqrt{5}-1}{4} + \frac{3-\sqrt{5}}{8}) = \frac{1}{2} (\frac{\sqrt{5}-1}{8} + \frac{3-\sqrt{5}}{8}) = \frac{1}{2} (\frac{2}{8}) = \frac{1}{8}$.

List-$I$	List-$II$
$(I)$ $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \sin^2 15^{\circ} + \dots + \sin^2 90^{\circ}$	$(A)$ $0$
$(II)$ $\tan^2 5^{\circ} \cdot \tan^2 10^{\circ} \cdot \tan^2 15^{\circ} \dots \tan^2 85^{\circ}$	$(B)$ $\frac{19}{2}$
$(III)$ $\cos^2 5^{\circ} + \cos^2 10^{\circ} + \cos^2 15^{\circ} + \dots + \cos^2 180^{\circ}$	$(C)$ $18$
$(IV)$ $\cot 5^{\circ} + \cot 10^{\circ} + \cot 15^{\circ} + \dots + \cot 175^{\circ}$	$(D)$ $1$
	$(E)$ $-1$

$\cos 6^{\circ} \sin 24^{\circ} \cos 72^{\circ} = $

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