MCQ
$\cos 20^\circ \cos 40^\circ \cos 80^\circ = $
- A$1/2$
- B$1/4$
- C$1/6$
- ✓$1/8$
$ = \frac{{\sin {{160}^o}}}{{8\sin {{20}^o}}} = \frac{1}{8}$.
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