cos 20^circ cos 40^circ cos 80^circ = | vedclass

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(D) We use the trigonometric identity: $\cos 	heta \cos(2	heta) \cos(4	heta) \dots \cos(2^{n-1}	heta) = \frac{\sin(2^n 	heta)}{2^n \sin 	heta}$.

Vedclass · Accepted Answer

$1/8$

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(D) We use the trigonometric identity: $\cos 	heta \cos(2	heta) \cos(4	heta) \dots \cos(2^{n-1}	heta) = \frac{\sin(2^n 	heta)}{2^n \sin 	heta}$.Here,$	heta = 20^\circ$ and $n = 3$.Substituting these values into the formula:$\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{\sin(2^3 	imes 20^\circ)}{2^3 \sin 20^\circ}$.$= \frac{\sin(160^\circ)}{8 \sin 20^\circ}$.Since $\sin(160^\circ) = \sin(180^\circ - 20^\circ) = \sin 20^\circ$,we have:$= \frac{\sin 20^\circ}{8 \sin 20^\circ} = \frac{1}{8}$.

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

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