Упрощение тригонометрических выражений. Задача 55

Ответ: 1/16.

\(\sin {10^ \circ }\sin {30^ \circ }\sin {50^ \circ }\sin {70^ \circ } = \dfrac{1}{2}\sin \left( {{{90}^ \circ }-{{80}^ \circ }} \right) \cdot \sin \left( {{{90}^ \circ }-{{40}^ \circ }} \right) \cdot \sin \left( {{{90}^ \circ }-{{20}^ \circ }} \right) = \)

\( = \dfrac{1}{2}\cos {80^ \circ } \cdot \cos {40^ \circ } \cdot \cos {20^ \circ } = \dfrac{{2\sin {{20}^ \circ }\cos {{20}^ \circ }\cos {{40}^ \circ }\cos {{80}^ \circ }}}{{4\sin {{20}^ \circ }}} = \dfrac{{2\sin {{40}^ \circ }\cos {{40}^ \circ }\cos {{80}^ \circ }}}{{8\sin {{20}^ \circ }}} = \)

\( = \dfrac{{2\sin {{80}^ \circ }\cos {{80}^ \circ }}}{{16\sin {{20}^ \circ }}} = \dfrac{{\sin {{160}^ \circ }}}{{16\sin {{20}^ \circ }}} = \dfrac{{\sin \left( {{{180}^ \circ }-{{20}^ \circ }} \right)}}{{16\sin {{20}^ \circ }}} = \dfrac{{\sin {{20}^ \circ }}}{{16\sin {{20}^ \circ }}} = \dfrac{1}{{16}}.\)

Ответ:  \(\dfrac{1}{{16}}.\)