cos 20^{circ} cos 40^{circ} cos 60^{circ} cos | vedclass

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Question

(D) We have,$\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ} = \cos 20^{\circ} \cos 40^{\circ} \left(\frac{1}{2}\right) \cos 80^{\circ

Vedclass · Accepted Answer

$\frac{1}{16}$

Vedclass · Answer

(D) We have,$\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ} = \cos 20^{\circ} \cos 40^{\circ} \left(\frac{1}{2}ight) \cos 80^{\circ}$ $= \frac{1}{2} (\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ})$ Using the formula $\cos A \cos 2A \cos 4A \dots \cos 2^{n-1}A = \frac{\sin(2^n A)}{2^n \sin A}$,where $A = 20^{\circ}$ and $n = 3$: $= \frac{1}{2} \left[ \frac{\sin(2^3 	imes 20^{\circ})}{2^3 \sin 20^{\circ}} ight]$ $= \frac{1}{2} \cdot \frac{\sin 160^{\circ}}{8 \sin 20^{\circ}}$ $= \frac{1}{16} \cdot \frac{\sin(180^{\circ} - 20^{\circ})}{\sin 20^{\circ}}$ Since $\sin(180^{\circ} - 	heta) = \sin 	heta$,we have $\sin 160^{\circ} = \sin 20^{\circ}$. $= \frac{1}{16} \cdot \frac{\sin 20^{\circ}}{\sin 20^{\circ}} = \frac{1}{16}$

$\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ} = $

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