Упрощение тригонометрических выражений. Задача 37math100admin44242025-03-24T15:57:55+03:00
Задача 37. Упростите выражение \(\dfrac{{{{\left( {1 + {\text{tg}}\,2\alpha } \right)}^2}-2{\text{t}}{{\text{g}}^2}2\alpha }}{{1 + {\text{t}}{{\text{g}}^2}2\alpha }}-\sin 4\alpha -1\)
Ответ
ОТВЕТ: \(-2{\sin ^2}2\alpha \).
Решение
\(\dfrac{{{{\left( {1 + {\rm{tg}}2\alpha } \right)}^2}-2{\rm{t}}{{\rm{g}}^2}2\alpha }}{{1 + {\rm{t}}{{\rm{g}}^2}2\alpha }}-\sin 4\alpha -1 = \dfrac{{1 + 2{\rm{tg}}2\alpha + {\rm{t}}{{\rm{g}}^2}2\alpha -2{\rm{t}}{{\rm{g}}^2}2\alpha }}{{\dfrac{1}{{{{\cos }^2}2\alpha }}}}-\sin 4\alpha -1 = \)
\( = {\cos ^2}2\alpha \left( {1 + 2{\rm{tg}}2\alpha -{\rm{t}}{{\rm{g}}^2}2\alpha } \right)-\sin 4\alpha -1 = \)
\( = {\cos ^2}2\alpha + 2{\cos ^2}2\alpha \cdot \dfrac{{\sin 2\alpha }}{{\cos 2\alpha }}-{\cos ^2}2\alpha \cdot \dfrac{{{{\sin }^2}2\alpha }}{{{{\cos }^2}2\alpha }}-\sin 4\alpha -1 = \)
\( = {\cos ^2}2\alpha + 2\sin 2\alpha \cos 2\alpha -{\sin ^2}2\alpha -\sin 4\alpha -1 = \)
\( = {\cos ^2}2\alpha -{\sin ^2}2\alpha + \sin 4\alpha -\sin 4\alpha -1 = 1-{\sin ^2}2\alpha -{\sin ^2}2\alpha -1 = -2{\sin ^2}2\alpha .\)
Ответ: \(-2{\sin ^2}2\alpha .\)