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

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Question

(d) $\cos {20^o}\cos {40^o}\cos {80^o} = \frac{{\sin {2^3}{{20}^o}}}{{{2^3}\sin {{20}^o}}}$

$ = \frac{{\sin {{160}^o}}}{{8\sin {{20}^o}}} = \frac{1

Vedclass · Accepted Answer

$1/8$

Vedclass · Answer

(d) $\cos {20^o}\cos {40^o}\cos {80^o} = \frac{{\sin {2^3}{{20}^o}}}{{{2^3}\sin {{20}^o}}}$

$ = \frac{{\sin {{160}^o}}}{{8\sin {{20}^o}}} = \frac{1}{8}$.

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

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