Преобразование сумм тригонометрических функций в произведение. Задача 21

ОТВЕТ: 4.

Воспользуемся формулой суммы тангенсов:  \({\rm{tg}}\alpha {\rm{ + tg}}\beta {\rm{ = }}\dfrac{{\sin \left( {\alpha  + \beta } \right)}}{{\cos \alpha \cos \beta }}.\)

\({\rm{tg}}{{\rm{9}}^ \circ }-{\rm{tg2}}{{\rm{7}}^ \circ }-{\rm{tg6}}{{\rm{3}}^ \circ } + {\rm{tg8}}{{\rm{1}}^ \circ } = {\rm{tg}}{{\rm{9}}^ \circ } + {\rm{tg8}}{{\rm{1}}^ \circ }-\left( {{\rm{tg2}}{{\rm{7}}^ \circ } + {\rm{tg6}}{{\rm{3}}^ \circ }} \right) = \)

\( = \dfrac{{\sin {{90}^ \circ }}}{{\sin {9^ \circ }\sin {{81}^ \circ }}}-\dfrac{{\sin {{90}^ \circ }}}{{\sin {{27}^ \circ }\sin {{63}^ \circ }}} = \dfrac{1}{{\sin {9^ \circ }\sin \left( {{{90}^ \circ }-{9^ \circ }} \right)}}-\dfrac{1}{{\sin {{27}^ \circ }\sin \left( {{{90}^ \circ }-{{27}^ \circ }} \right)}} = \)

\( = \dfrac{1}{{\sin {9^ \circ }\cos {9^ \circ }}}-\dfrac{1}{{\sin {{27}^ \circ }\cos {{27}^ \circ }}} = \dfrac{2}{{2\sin {9^ \circ }\cos {9^ \circ }}}-\dfrac{2}{{2\sin {{27}^ \circ }\cos {{27}^ \circ }}} = \)

\( = \dfrac{2}{{\sin {{18}^ \circ }}}-\dfrac{2}{{\sin {{54}^ \circ }}} = \dfrac{{2\sin {{54}^ \circ }-2\sin {{18}^ \circ }}}{{\sin {{18}^ \circ }\sin {{54}^ \circ }}} = \dfrac{{2\left( {\sin {{54}^ \circ }-\sin {{18}^ \circ }} \right)}}{{\sin {{18}^ \circ }\sin {{54}^ \circ }}} = \)

\( = \dfrac{{2 \cdot 2 \cdot \sin {{18}^ \circ }\cos {{36}^ \circ }}}{{\sin {{18}^ \circ }\sin {{54}^ \circ }}} = \dfrac{{4\cos \left( {{{90}^ \circ }-{{54}^ \circ }} \right)}}{{\sin {{54}^ \circ }}} = \dfrac{{4\sin {{54}^ \circ }}}{{\sin {{54}^ \circ }}} = 4.\)

Ответ:  4.