Упрощение тригонометрических выражений. Задача 60math100admin44242025-03-24T16:44:41+03:00
Задача 60. Вычислите \(\sin {20^ \circ }\cos {50^ \circ }\sin {60^ \circ }\cos {10^ \circ }\)
Решение
\(\sin {20^ \circ }\cos {50^ \circ }\sin {60^ \circ }\cos {10^ \circ } = \sin {20^ \circ }\sin {40^ \circ }\sin {60^ \circ }\sin {80^ \circ } = \left( {\sin {{20}^ \circ }\sin {{80}^ \circ }} \right)\left( {\sin {{40}^ \circ }\sin {{60}^ \circ }} \right) = \)
\( = \dfrac{1}{2}\left( {\cos {{60}^ \circ }-\cos {{100}^ \circ }} \right) \cdot \dfrac{1}{2} \cdot \left( {\cos {{20}^ \circ }-\cos {{100}^ \circ }} \right) = \dfrac{1}{4}\left( {\dfrac{1}{2}-\cos {{100}^ \circ }} \right)\left( {\cos {{20}^ \circ }-\cos {{100}^ \circ }} \right) = \)\( = \dfrac{1}{4}\left( {\dfrac{1}{2}\cos {{20}^ \circ }-\dfrac{1}{2}\cos {{100}^ \circ }-\cos {{100}^ \circ }\cos {{20}^ \circ } + {{\cos }^2}{{100}^ \circ }} \right) = \)
\( = \dfrac{1}{4}\left( {\dfrac{1}{2}\cos {{20}^ \circ }-\dfrac{1}{2}\cos {{100}^ \circ }-\dfrac{1}{2}\left( {\cos {{80}^ \circ } + \cos {{120}^ \circ }} \right) + \dfrac{1}{2}\left( {1 + \cos {{200}^ \circ }} \right)} \right) = \)
\( = \dfrac{1}{8}\left( {\cos {{20}^ \circ }-\cos {{100}^ \circ }-\cos {{80}^ \circ }-\cos {{120}^ \circ } + 1 + \cos {{200}^ \circ }} \right) = \)
\( = \dfrac{1}{8}\left( {\cos {{20}^ \circ }-\cos \left( {{{180}^ \circ }-{{80}^ \circ }} \right)-\cos {{80}^ \circ } + \dfrac{1}{2} + 1 + \cos \left( {{{180}^ \circ } + {{20}^ \circ }} \right)} \right) = \)
\( = \dfrac{1}{8}\left( {\cos {{20}^ \circ } + \cos {{80}^ \circ }-\cos {{80}^ \circ } + \dfrac{3}{2}-\cos {{20}^ \circ }} \right) = \dfrac{1}{8} \cdot \dfrac{3}{2} = \dfrac{3}{{16}}.\)
Ответ: \(\dfrac{3}{{16}}.\)