cos 36^{circ} - cos 72^{circ} is equal to | vedclass

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Question

(B) Using the formula $\cos C - \cos D = -2 \sin \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$,we have:$\cos 36^{\circ} - \cos 72^{\circ

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) Using the formula $\cos C - \cos D = -2 \sin \left(\frac{C+D}{2}ight) \sin \left(\frac{C-D}{2}ight)$,we have:$\cos 36^{\circ} - \cos 72^{\circ} = -2 \sin \left(\frac{36^{\circ}+72^{\circ}}{2}ight) \sin \left(\frac{36^{\circ}-72^{\circ}}{2}ight)$$= -2 \sin 54^{\circ} \sin (-18^{\circ})$$= 2 \sin 54^{\circ} \sin 18^{\circ}$Substituting the values $\sin 54^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$:$= 2 	imes \left(\frac{\sqrt{5}+1}{4}ight) 	imes \left(\frac{\sqrt{5}-1}{4}ight)$$= 2 	imes \frac{(\sqrt{5})^2 - 1^2}{16}$$= 2 	imes \frac{5-1}{16} = 2 	imes \frac{4}{16} = 2 	imes \frac{1}{4} = \frac{1}{2}$

$\cos 36^{\circ} - \cos 72^{\circ}$ is equal to

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