A) 1/2
B) 1/4
C) 1/6
D) 1/8
Correct Answer: D
Solution :
\[\cos {{20}^{o}}\cos {{40}^{o}}\cos {{80}^{o}}=\frac{\sin {{2}^{3}}{{20}^{o}}}{{{2}^{3}}\sin {{20}^{o}}}\]\[=\frac{\sin {{160}^{o}}}{8\sin {{20}^{o}}}=\frac{1}{8}\].
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