Упрощение тригонометрических выражений. Задача 21math100admin44242025-03-24T12:16:30+03:00
Задача 21. Докажите тождество \(\dfrac{{{{\sin }^4}\alpha + {{\cos }^4}\alpha -1}}{{{{\sin }^6}\alpha + {{\cos }^6}\alpha -1}} = \dfrac{2}{3}\)
Решение
\(\dfrac{{{{\sin }^4}\alpha + {{\cos }^4}\alpha -1}}{{{{\sin }^6}\alpha + {{\cos }^6}\alpha -1}} = \dfrac{{{{\left( {{{\sin }^2}\alpha } \right)}^2} + 2{{\sin }^2}\alpha {{\cos }^2}\alpha + {{\left( {{{\cos }^2}\alpha } \right)}^2}-2{{\sin }^2}\alpha {{\cos }^2}\alpha -1}}{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)\left( {{{\sin }^4}\alpha -{{\sin }^2}\alpha {{\cos }^2}\alpha + {{\cos }^4}\alpha } \right)-1}} = \)
\( = \dfrac{{{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)}^2}-2{{\sin }^2}\alpha {{\cos }^2}\alpha -1}}{{{{\left( {{{\sin }^2}\alpha } \right)}^2} + 2{{\sin }^2}\alpha {{\cos }^2}\alpha + {{\left( {{{\cos }^2}\alpha } \right)}^2}-3{{\sin }^2}\alpha {{\cos }^2}\alpha -1}} = \)
\( = \dfrac{{{1^2}-2{{\sin }^2}\alpha {{\cos }^2}\alpha -1}}{{{{\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right)}^2}-3{{\sin }^2}\alpha {{\cos }^2}\alpha -1}} = \dfrac{{-2{{\sin }^2}\alpha {{\cos }^2}\alpha }}{{1-3{{\sin }^2}\alpha {{\cos }^2}\alpha -1}} = \dfrac{{-2{{\sin }^2}\alpha {{\cos }^2}\alpha }}{{-3{{\sin }^2}\alpha {{\cos }^2}\alpha }} = \dfrac{2}{3}.\)