Question
The value of $$\cos {12^ \circ } \cdot \cos {24^ \circ } \cdot \cos {36^ \circ } \cdot \cos {48^ \circ } \cdot \cos {72^ \circ } \cdot \cos {84^ \circ }$$ is
A.
$$\frac{1}{{64}}$$
B.
$$\frac{1}{{32}}$$
C.
$$\frac{1}{{16}}$$
D.
$$\frac{1}{{128}}$$
Answer :
$$\frac{1}{{64}}$$
Solution :
Value $$ = \cos {12^ \circ } \cdot \cos {24^ \circ } \cdot \cos {48^ \circ } \cdot \cos \left( {{{180}^ \circ } - {{96}^ \circ }} \right) \cdot \cos {36^ \circ } \cdot \cos {72^ \circ }$$
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\, = - \left( {\cos {{12}^ \circ } \cdot \cos {{24}^ \circ } \cdot \cos {{48}^ \circ } \cdot \cos {{96}^ \circ }} \right)\left( {\cos {{36}^ \circ } \cdot \cos {{72}^ \circ }} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = - \left( {\frac{{2\sin{{12}^ \circ } \cdot \cos {{12}^ \circ } \cdot \cos {{24}^ \circ } \cdot \cos {{48}^ \circ } \cdot \cos {{96}^ \circ }}}{{2\sin{{12}^ \circ }}}} \right) \times \left( {\frac{{2\sin{{36}^ \circ } \cdot \cos {{36}^ \circ } \cdot \cos {{72}^ \circ }}}{{2\sin{{36}^ \circ }}}} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\, = - \frac{{\sin {{192}^ \circ }}}{{{2^4} \cdot \sin {{12}^ \circ }}} \cdot \frac{{\sin {{144}^ \circ }}}{{{2^2}\sin {{36}^ \circ }}} = \frac{1}{{{2^4}}} \cdot \frac{1}{{{2^2}}}. \cr} $$