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3.Trigonometrical Ratios, Functions and Identities
easy
$\cos 20^\circ \cos 40^\circ \cos 80^\circ = $
A
$1/2$
B
$1/4$
C
$1/6$
D
$1/8$
Solution
(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}$.
Standard 11
Mathematics
